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Fundamentals of Differential Equations

EIGHTH EDITION

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Fundamentals of Differential Equations

R. Kent Nagle

Edward B. Saff Vanderbilt University

Arthur David Snider University of South Florida

EIGHTH EDITION

Addison-Wesley

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Library of Congress Cataloging-in-Publication Data Nagle, R. Kent. Fundamentals of differential equations. -- 8th ed. / R. Kent Nagle,

Edward B. Saff, David Snider. p. cm.

Includes index. ISBN-13: 978-0-321-74773-0 ISBN-10: 0-321-74773-9 1. Differential equations--Textbooks. I. Saff, E. B., 1944- II. Snider, Arthur David, 1940- III. Title.

QA371.N24 2012 515'.35--dc22 2011002688

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ISBN-13: 978-0-321-74773-0 www.pearsonhighered.com ISBN-10: 0-321-74773-7

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Dedicated to R. Kent Nagle

He has left his imprint not only on these pages but upon all who knew him. He was that rare mathematician who could effectively communicate at all levels, imparting his love for the subject with the same ease to undergraduates, graduates, precollege stu- dents, public school teachers, and his colleagues at the University of South Florida.

Kent was at peace in life—a peace that emanated from the depth of his under- standing of the human condition and the strength of his beliefs in the institutions of family, religion, and education. He was a research mathematician, an accomplished author, a Sunday school teacher, and a devoted husband and father.

Kent was also my dear friend and my jogging partner who has left me behind still struggling to keep pace with his high ideals.

E. B. Saff

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Preface

OUR GOAL

Fundamentals of Differential Equations is designed to serve the needs of a one-semester course in basic theory as well as applications of differential equations. The flexibility of the text provides the instructor substantial latitude in designing a syllabus to match the emphasis of the course. Sample syllabi are provided in this preface that illustrate the inherent flexibility of this text to balance theory, methodology, applications, and numerical methods, as well as the incorporation of commercially available computer software for this course.

NEW TO THIS EDITION

With this edition we are pleased to feature some new pedagogical and reference tools, as well as some new projects and discussions that bear upon current issues in the news and in acade- mia. In brief:

• In response to our colleagues’ perception that many of today’s students’ skills in integra- tion have gotten rusty by the time they enter the differential equations course, we have included a new appendix offering a quick review of the methods for integrating func- tions analytically. We trust that our light overview will prove refreshing (Appendix A, page A-1).

• A new project models the spread of staph infections in the unlikely setting of a hospi- tal, an ongoing problem in the health community’s battle to contain and confine dan- gerous infectious strains in the population (Chapter 5, Project B, page 310).

• By including five new sketches of the various eigenfunctions arising from separating variables in our chapter on partial differential equations (Chapter 10), we are able to enhance the visualization of their qualitative features.

• We finesse the abstruseness of generalized eigenvector chain theory with a novel tech- nique for computing the matrix exponential for defective matrices (Chapter 9, Section 8, page 554).

• The rectangular window (or “boxcar function”) has become a standard mathematical tool in the communications industry, the backbone of such schemes as pulse code modulation, etc. Our revised chapter on Laplace transforms incorporates it to facilitate the analysis of switching functions for differential equations (Chapter 7, Section 6, page 384).

vii

• We have added a project presenting a cursory analysis of oil dispersion after a spill, based roughly on an incident that occurred in the Mississippi River (Chapter 2, Project A, page 79).

• Even as computer science basks in the limelight as the current “glamour” technology profession, we unearthed an application of the grande dame of differential equations techniques—power series—to predict the performance of Quicksort, a machine algo- rithm that alphabetizes large lists (Chapter 8, Project A, page 493).

• Closely related to the current interest in hydroponics is our project describing the growth of phytoplankton by controlling the supply of the necessary nutrients in a chemostat tank (Chapter 5, Project F, pages 316).

• The basic theorems on linear difference equations closely resemble those for differen- tial equations (but are easier to prove), so we have included a project exploring this kinship (Chapter 6, Project D, page 347).

• We conclude our chapter on power series expansions with a tabulation of the histori- cally significant second-order differential equations, the practical considerations that inspired them, the mathematicians who analyzed them, and the standard notations for their solutions (Chapter 8, pages 485–486).

Additionally, we have added dozens of new problems and have updated the references to relevant literature and Web sites, especially those facilitating the online implementation of numerical methods.

PREREQUISITES

While some universities make linear algebra a prerequisite for differential equations, many schools (especially engineering) only require calculus. With this in mind, we have designed the text so that only Chapter 6 (Theory of Higher-Order Linear Differential Equations) and Chapter 9 (Matrix Methods for Linear Systems) require more than high school level linear algebra. Moreover, Chapter 9 contains review sections on matrices and vectors as well as spe- cific references for the deeper results used from the theory of linear algebra. We have also writ- ten Chapter 5 so as to give an introduction to systems of differential equations—including methods of solving, phase plane analysis, applications, numerical procedures, and Poincaré maps—that does not require a background in linear algebra.

SAMPLE SYLLABI

As a rough guide in designing a one-semester syllabus related to this text, we provide three sam- ples that can be used for a 15-week course that meets three hours per week. The first emphasizes applications and computations including phase plane analysis; the second is designed for courses that place more emphasis on theory; and the third stresses methodology and partial differential equations. Chapters 1, 2, and 4 provide the core for any first course. The rest of the chapters are, for the most part, independent of each other. For students with a background in linear algebra, the instructor may prefer to replace Chapter 7 (Laplace Transforms) or Chapter 8 (Series Solutions of Differential Equations) with sections from Chapter 9 (Matrix Methods for Linear Systems).

viii Preface

RETAINED FEATURES

Most of the material is modular in nature to allow for various course configurations and emphasis (theory, applications and techniques, and concepts).

The availability of computer packages such as Mathcad®, Mathematica®, MATLAB®, and Maple™ provides an opportunity for the student to conduct numerical experiments and tackle realistic applications that give additional insights into the subject. Consequently, we have inserted several exercises and projects throughout the text that are designed for the student to employ available software in phase plane analysis, eigenvalue computations, and the numerical solutions of various equations.

Because of syllabus constraints, some courses will have little or no time for sections (such as those in Chapters 3 and 5) that exclusively deal with applications. Therefore, we have made the sections in these chapters independent of each other. To afford the instructor even greater flexi- bility, we have built in a variety of applications in the exercises for the theoretical sections. In addition, we have included many projects that deal with such applications.

At the end of each chapter are group projects relating to the material covered in the chapter. A project might involve a more challenging application, delve deeper into the theory, or introduce more advanced topics in differential equations. Although these projects can be tackled by an individual student, classroom testing has shown that working in groups lends a valuable added dimension to the learning experience. Indeed, it simulates the interactions that take place in the professional arena.

Preface ix

Methods, Theory and Methods and Computations, Methods (linear Partial Differential and Applications algebra prerequisite) Equations

Week Sections Sections Sections

1 1.1, 1.2, 1.3 1.1, 1.2, 1.3 1.1, 1.2, 1.3 2 1.4, 2.2 1.4, 2.2, 2.3 1.4, 2.2 3 2.3, 2.4, 3.2 2.4, 3.2, 4.1 2.3, 2.4 4 3.4, 3.5, 3.6 4.2, 4.3, 4.4 3.2, 3.4 5 3.7, 4.1 4.5, 4.6 4.2, 4.3 6 4.2, 4.3, 4.4 4.7, 5.2, 5.3 4.4, 4.5, 4.6 7 4.5, 4.6, 4.7 5.4, 6.1 4.7, 5.1, 5.2 8 4.8, 4.9 6.2, 6.3, 7.2 7.1, 7.2, 7.3 9 4.10, 5.1, 5.2 7.3, 7.4, 7.5 7.4, 7.5

10 5.3, 5.4, 5.5 7.6, 7.7 7.6, 7.7 11 5.6, 5.7 8.2, 8.3 7.8, 8.2 12 7.2, 7.3, 7.4 8.4, 8.6, 9.1 8.3, 8.5, 8.6 13 7.5, 7.6, 7.7 9.2, 9.3 10.2, 10.3 14 8.1, 8.2, 8.3 9.4, 9.5, 9.6 10.4, 10.5 15 8.4, 8.6 9.7, 9.8 10.6, 10.7

Flexible Organization

Optional Use of Computer

Software

Choice of Applications

Group Projects

Communication skills are, of course, an essential aspect of professional activities. Yet few texts provide opportunities for the reader to develop these skills. Thus, we have added at the end of most chapters a set of clearly marked technical writing exercises that invite students to make documented responses to questions dealing with the concepts in the chapter. In so doing, stu- dents are encouraged to make comparisons between various methods and to present examples that support their analysis.

Throughout the text historical footnotes are set off by colored daggers (†). These footnotes typ- ically provide the name of the person who developed the technique, the date, and the context of the original research.

Most chapters begin with a discussion of a problem from physics or engineering that motivates the topic presented and illustrates the methodology.

All of the main chapters contain a set of review problems along with a synopsis of the major concepts presented.

Most of the figures in the text were generated via computer. Computer graphics not only ensure greater accuracy in the illustrations, they demonstrate the use of numerical experimentation in studying the behavior of solutions.

While more pragmatic students may balk at proofs, most instructors regard these justifications as an essential ingredient in a textbook on differential equations. As with any text at this level, certain details in the proofs must be omitted. When this occurs, we flag the instance and refer readers either to a problem in the exercises or to another text. For convenience, the end of a proof is marked by the symbol (◆).

We have developed the theory of linear differential equations in a gradual manner. In Chapter 4 (Linear Second-Order Equations) we first present the basic theory for linear second-order equations with constant coefficients and discuss various techniques for solving these equations. Section 4.7 surveys the extension of these ideas to variable-coefficient second-order equations. A more general and detailed discussion of linear differential equations is given in Chapter 6 (Theory of Higher-Order Linear Differential Equations). For a beginning course emphasizing methods of solution, the presentation in Chapter 4 may be sufficient and Chapter 6 can be skipped.

Several numerical methods for approximating solutions to differential equations are presented along with program outlines that are easily implemented on a computer. These methods are introduced early in the text so that teachers and/or students can use them for numerical experi- mentation and for tackling complicated applications. Where appropriate we direct the student to software packages or web-based applets for implementation of these algorithms.

An abundance of exercises is graduated in difficulty from straightforward, routine problems to more challenging ones. Deeper theoretical questions, along with applications, usually occur toward the end of the exercise sets. Throughout the text we have included problems and projects that require the use of a calculator or computer. These exercises are denoted by the symbol ( ).

These sections can be omitted without affecting the logical development of the material. They are marked with an asterisk in the table of contents.

x Preface

Technical Writing Exercises

Historical Footnotes

Motivating Problem

Chapter Summary and Review

Problems

Computer Graphics

Proofs

Linear Theory

Numerical Algorithms

Exercises

Optional Sections

We provide a detailed chapter on Laplace transforms (Chapter 7), since this is a recurring topic for engineers. Our treatment emphasizes discontinuous forcing terms and includes a section on the Dirac delta function.

Power series solutions is a topic that occasionally causes student anxiety. Possibly, this is due to inadequate preparation in calculus where the more subtle subject of convergent series is (fre- quently) covered at a rapid pace. Our solution has been to provide a graceful initiation into the theory of power series solutions with an exposition of Taylor polynomial approximants to solu- tions, deferring the sophisticated issues of convergence to later sections. Unlike many texts, ours provides an extensive section on the method of Frobenius (Section 8.6) as well as a section on finding a second linearly independent solution.

While we have given considerable space to power series solutions, we have also taken great care to accommodate the instructor who only wishes to give a basic introduction to the topic. An introduction to solving differential equations using power series and the method of Frobenius can be accomplished by covering the materials in Sections 8.1, 8.2, 8.3, and 8.6.

An introduction to this subject is provided in Chapter 10, which covers the method of separa- tion of variables, Fourier series, the heat equation, the wave equation, and Laplace’s equation. Examples in two and three dimensions are included.

Chapter 5 describes how qualitative information for two-dimensional systems can be gleaned about the solutions to intractable autonomous equations by observing their direction fields and critical points on the phase plane. With the assistance of suitable software, this approach pro- vides a refreshing, almost recreational alternative to the traditional analytic methodology as we discuss applications in nonlinear mechanics, ecosystems, and epidemiology.

Motivation for Chapter 4 on linear differential equations is provided in an introductory section describing the mass–spring oscillator. We exploit the reader’s familiarity with common vibra- tory motions to anticipate the exposition of the theoretical and analytical aspects of linear equations. Not only does this model provide an anchor for the discourse on constant-coefficient equations, but a liberal interpretation of its features enables us to predict the qualitative behav- ior of variable-coefficient and nonlinear equations as well.

The chapter on matrix methods for linear systems (Chapter 9) begins with two (optional) intro- ductory sections reviewing the theory of linear algebraic systems and matrix algebra.

SUPPLEMENTS

By Viktor Maymeskul. Contains complete, worked-out solutions to odd-numbered exer- cises, providing students with an excellent study tool. ISBN 13: 978-0-321-74834-8; ISBN 10: 0-321-74834-4.

Contains answers to all even-numbered exercises, detailed solutions to the even-numbered prob- lems in several of the main chapters, and additional group projects. ISBN 13: 978-0-321-74835-5; ISBN 10: 0-321-74835-2.

Provides additional resources for both instructors and students, including helpful links keyed to sections of the text, access to Interactive Differential Equations, suggestions for incorporating Interactive Differential Equations modules, suggested syllabi, index of applications, and study tips for students. Access: www.pearsonhighered.com/nagle

Preface xi

Power Series

Partial Differential Equations

Phase Plane

Vibrations

Student’s Solutions Manual

Review of Linear Algebraic

Equations and Matrices

Instructor’s Solutions Manual

Companion Web site

Laplace Transforms

www.pearsonhighered.com/nagle

By Thomas W. Polaski (Winthrop University), Bruno Welfert (Arizona State University), and Maurino Bautista (Rochester Institute of Technology). A collection of worksheets and projects to aid instructors in integrating computer algebra systems into their courses. Available in the Pearson Instructor Resource Center, www.pearsonhighered.com/irc.

ACKNOWLEDGMENTS

The staging of this text involved considerable behind-the-scenes activity. We thank, first of all, Philip Crooke, Joanna Pressley, and Glenn Webb (Vanderbilt University), who have continued to provide novel biomathematical projects.

We also want to thank Frank Glaser (California State Polytechnic University, Pomona) for many of the historical footnotes. We are indebted to Herbert E. Rauch (Lockheed Research Laboratory) for help with Section 3.3 on heating and cooling of buildings, Group Project B in Chapter 3 on aquaculture, and other application problems. Our appreciation goes to Richard H. Elderkin (Pomona College), Jerrold Marsden (California Institute of Technology), T. G. Proc- tor (Clemson University), and Philip W. Schaefer (University of Tennessee), who read and reread the manuscript for the original text, making numerous suggestions that greatly improved our work. Thanks also to the following reviewers of this and previous editions:

Amin Boumenir, University of West Georgia

*Mark Brittenham, University of Nebraska

*Weiming Cao, University of Texas at San Antonio

*Richard Carmichael, Wake Forest University

Karen Clark, The College of New Jersey

Patric Dowling, Miami University

Sanford Geraci, Northern Virginia

*David S. Gilliam, Texas Tech University at Lubbock

Scott Gordon, State University of West Georgia

Richard Rubin, Florida International University

*John Sylvester, University of Washington at Seattle

*Steven Taliaferro, Texas A&M University at College Station

Michael M. Tom, Louisiana State University

Shu-Yi Tu, University of Michigan, Flint

Klaus Volpert, Villanova University

E. B. Saff, A. D. Snider

xii Preface

*Denote reviewers of the current edition.

Instructor’s MATLAB, Maple,

and Mathematica Manuals

www.pearsonhighered.com/irc

1.1 Background 1

1.2 Solutions and Initial Value Problems 6

1.3 Direction Fields 15

1.4 The Approximation Method of Euler 23

Chapter Summary 29

Technical Writing Exercises 29

Group Projects for Chapter 1 30

A. Taylor Series Method 30 B. Picard’s Method 31 C. The Phase Line 32

2.1 Introduction: Motion of a Falling Body 35

2.2 Separable Equations 38

2.3 Linear Equations 46

2.4 Exact Equations 55

2.5 Special Integrating Factors 64

2.6 Substitutions and Transformations 68

Chapter Summary 76

Review Problems 77

Technical Writing Exercises 78

Group Projects for Chapter 2 79

A. Oil Spill in a Canal 79 B. Differential Equations in Clinical Medicine 80 C. Torricelli’s Law of Fluid Flow 82 D. The Snowplow Problem 83 E. Two Snowplows 83

CHAPTER 2 First-Order Differential Equations

Contents

CHAPTER 1 Introduction

xiii

xiv Contents

4.1 Introduction: The Mass-Spring Oscillator 153

4.2 Homogeneous Linear Equations: The General Solution 158

4.3 Auxiliary Equations with Complex Roots 167

4.4 Nonhomogeneous Equations: The Method of Undetermined Coefficients 175

4.5 The Superposition Principle and Undetermined Coefficients Revisited 182

4.6 Variation of Parameters 189

4.7 Variable-Coefficient Equations 193

CHAPTER 4 Linear Second-Order Equations

F. Clairaut Equations and Singular Solutions 84 G. Multiple Solutions of a First-Order Initial Value Problem 85 H. Utility Functions and Risk Aversion 85 I. Designing a Solar Collector 86 J. Asymptotic Behavior of Solutions to Linear Equations 87

3.1 Mathematical Modeling 89

3.2 Compartmental Analysis 91

3.3 Heating and Cooling of Buildings 101

3.4 Newtonian Mechanics 108

3.5 Electrical Circuits 117

3.6 Improved Euler’s Method 121

3.7 Higher-Order Numerical Methods: Taylor and Runge-Kutta 132

Group Projects for Chapter 3 414

A. Dynamics of HIV Infection 141 B. Aquaculture 144 C. Curve of Pursuit 145 D. Aircraft Guidance in a Crosswind 146 E. Feedback and the Op Amp 147 F. Bang-Bang Controls 148 G. Market Equilibrium: Stability and Time Paths 149 H. Stability of Numerical Methods 150 I. Period Doubling and Chaos 151

Mathematical Models and Numerical Methods CHAPTER 3 Involving First-Order Equations

Contents xv

5.1 Interconnected Fluid Tanks 242

5.2 Differential Operators and the Elimination Method for Systems 244

5.3 Solving Systems and Higher-Order Equations Numerically 253

5.4 Introduction to the Phase Plane 263

5.5 Applications to Biomathematics: Epidemic and Tumor Growth Models 276

5.6 Coupled Mass-Spring Systems 285

5.7 Electrical Systems 291

5.8 Dynamical Systems, Poincaré Maps, and Chaos 297

Chapter Summary 307

Review Problems 308

Group Projects for Chapter 5 309

A. Designing a Landing System for Interplanetary Travel 309 B. Spread of Staph Infections in Hospitals—Part 1 310 C. Things That Bob 312 D. Hamiltonian Systems 313 E. Cleaning Up the Great Lakes 315 F. A Growth Model for Phytoplankton—Part I 316

CHAPTER 5 Introduction to Systems and Phase Plane Analysis

4.8 Qualitative Considerations for Variable-Coefficient and Nonlinear Equations 203

4.9 A Closer Look at Free Mechanical Vibrations 214

4.10 A Closer Look at Forced Mechanical Vibrations 223

Chapter Summary 231

Review Problems 233

Technical Writing Exercises 234

Group Projects for Chapter 4 235

A. Nonlinear Equations Solvable by First-Order Techniques 235 B. Apollo Reentry 236 C. Simple Pendulum 237 D. Linearization of Nonlinear Problems 238 E. Convolution Method 239 F. Undetermined Coefficients Using Complex Arithmetic 239 G. Asymptotic Behavior of Solutions 241

xvi

xvi Contents

7.2 Definition of the Laplace Transform 353

7.3 Properties of the Laplace Transform 361

7.4 Inverse Laplace Transform 366

7.5 Solving Initial Value Problems 376

7.6 Transforms of Discontinuous and Periodic Functions 383

7.7 Convolution 397

7.8 Impulses and the Dirac Delta Function 404

7.9 Solving Linear Systems with Laplace Transforms 412

Chapter Summary 414

Review Problems 416

Technical Writing Exercises 417

Group Projects for Chapter 7 418

A. Duhamel’s Formulas 418 B. Frequency Response Modeling 419 C. Determining System Parameters 421

CHAPTER 7 Laplace Transforms

6.1 Basic Theory of Linear Differential Equations 318

6.2 Homogeneous Linear Equations with Constant Coefficients 327

6.3 Undetermined Coefficients and the Annihilator Method 333

6.4 Method of Variation of Parameters 338

Chapter Summary 342

Review Problems 343

Technical Writing Exercises 344

Group Projects for Chapter 6 345

A. Computer Algebra Systems and Exponential Shift 345 B. Justifying the Method of Undetermined Coefficients 346 C. Transverse Vibrations of a Beam 347 D. Higher-Order Difference Equations 347

7.1 Introduction: A Mixing Problem 350

CHAPTER 6 Theory of Higher-Order Linear Differential Equations

xvii

Contents xvii

8.1 Introduction: The Taylor Polynomial Approximation 422

8.2 Power Series and Analytic Functions 427

8.3 Power Series Solutions to Linear Differential Equations 436

8.4 Equations with Analytic Coefficients 446

8.5 Cauchy-Euler (Equidimensional) Equations 452

8.6 Method of Frobenius 455

8.7 Finding a Second Linearly Independent Solution 467

8.8 Special Functions 476

Chapter Summary 489

Review Problems 491

Technical Writing Exercises 492

Group Projects for Chapter 8 493

A. Alphabetization Algorithms 493 B. Spherically Symmetric Solutions to Shrödinger’s Equation

for the Hydrogen Atom 494 C. Airy’s Equation 495 D. Buckling of a Tower 495 E. Aging Spring and Bessel Functions 497

CHAPTER 8 Series Solutions of Differential Equations

9.1 Introduction 498

9.2 Review 1: Linear Algebraic Equations 502

9.3 Review 2: Matrices and Vectors 507

9.4 Linear Systems in Normal Form 517

9.5 Homogeneous Linear Systems with Constant Coefficients 526

9.6 Complex Eigenvalues 538

9.7 Nonhomogeneous Linear Systems 543

9.8 The Matrix Exponential Function 550

Chapter Summary 558

Review Problems 561

CHAPTER 9 Matrix Methods for Linear Systems

xviii

xviii Contents

10.1 Introduction: A Model for Heat Flow 566

10.2 Method of Separation of Variables 569

10.3 Fourier Series 578

10.4 Fourier Cosine and Sine Series 594

10.5 The Heat Equation 599

10.6 The Wave Equation 611

10.7 Laplace’s Equation 623

Chapter Summary 636

Technical Writing Exercises 637

Group Projects for Chapter 10 638

A. Steady-State Temperature Distribution in a Circular Cylinder 638 B. Laplace Transform Solution of the Wave Equation 640 C. Green’s Function 641 D. Numerical Method for u = ƒ on a Rectangle 642

APPENDICES A. Review of Integration Techniques A-1 B. Newton’s Method A-8 C. Simpson’s Rule A-10 D. Cramer’s Rule A-11 E. Method of Least Squares A-13 F. Runge-Kutta Procedure for n Equations A-15

ANSWERS TO ODD-NUMBERED PROBLEMS B-1

INDEX I-1

CHAPTER 10 Partial Differential Equations

Technical Writing Exercises 562

Group Projects for Chapter 9 563

A. Uncoupling Normal Systems 563 B. Matrix Laplace Transform Method 564 C. Undamped Second-Order Systems 565

In the sciences and engineering, mathematical models are developed to aid in the understanding of physical phenomena. These models often yield an equation that contains some derivatives of an unknown function. Such an equation is called a differential equation. Two examples of models developed in calculus are the free fall of a body and the decay of a radioactive substance.

In the case of free fall, an object is released from a certain height above the ground and falls under the force of gravity.† Newton’s second law, which states that an object’s mass times its acceleration equals the total force acting on it, can be applied to the falling object. This leads to the equation (see Figure 1.1)

where m is the mass of the object, h is the height above the ground, is its acceleration, g is the (constant) gravitational acceleration, and !mg is the force due to gravity. This is a differ- ential equation containing the second derivative of the unknown height h as a function of time.

Fortunately, the above equation is easy to solve for h. All we have to do is divide by m and integrate twice with respect to t. That is,

and

h " h AtB " !gt2 2

# c1t # c2 .

dh dt

" !gt # c1

d2h dt2

" !g ,

d2h/dt2 m

d2h dt2

" !mg ,

Introduction

CHAPTER 1

BACKGROUND1.1

†We are assuming here that gravity is the only force acting on the object and that this force is constant. More general models would take into account other forces, such as air resistance.

h−mg

Figure 1.1 Apple in free fall

2 Chapter 1 Introduction

We will see that the constants of integration, c1 and c2, are determined if we know the initial height and the initial velocity of the object. We then have a formula for the height of the object at time t.

In the case of radioactive decay (Figure 1.2), we begin from the premise that the rate of decay is proportional to the amount of radioactive substance present. This leads to the equation

k $ 0

where is the unknown amount of radioactive substance present at time t and k is the pro- portionality constant. To solve this differential equation, we rewrite it in the form

and integrate to obtain

Solving for A yields

A " " eln A " e!kt " Ce!kt

where C is the combination of integration constants . The value of C, as we will see later, is determined if the initial amount of radioactive substance is given. We then have a for- mula for the amount of radioactive substance at any future time t.

Even though the above examples were easily solved by methods learned in calculus, they do give us some insight into the study of differential equations in general. First, notice that the solu- tion of a differential equation is a function, like h or A , not merely a number. Second, inte- gration† is an important tool in solving differential equations (not surprisingly!). Third, we cannot expect to get a unique solution to a differential equation, since there will be arbitrary “constants of integration.” The second derivative in the free-fall equation gave rise to two constants, c1 and c2, and the first derivative in the decay equation gave rise, ultimately, to one constant, C.

Whenever a mathematical model involves the rate of change of one variable with respect to another, a differential equation is apt to appear. Unfortunately, in contrast to the examples for free fall and radioactive decay, the differential equation may be very complicated and diffi- cult to analyze.

d2h/dt2

AtBAtB eC2!C1

,eC2!C1A AtB ln A # C1 " !kt # C2 .

! 1A dA " ! !k dt

1 A

dA " !k dt

A A$0B ,

dA dt

" !kA ,

Figure 1.2 Radioactive decay

†For a review of integration techniques, see Appendix A.

Section 1.1 Background 3

Differential equations arise in a variety of subject areas, including not only the physical sci- ences but also such diverse fields as economics, medicine, psychology, and operations research. We now list a few specific examples.

1. In banking practice, if P is the number of dollars in a savings bank account that pays a yearly interest rate of r% compounded continuously, then P satisfies the dif- ferential equation

(1)

2. A classic application of differential equations is found in the study of an electric cir- cuit consisting of a resistor, an inductor, and a capacitor driven by an electromotive force (see Figure 1.3). Here an application of Kirchhoff’s laws† leads to the equation

(2)

where L is the inductance, R is the resistance, C is the capacitance, E is the electro- motive force, q is the charge on the capacitor, and t is the time.

3. In psychology, one model of the learning of a task involves the equation

(3)

Here the variable y represents the learner’s skill level as a function of time t. The constants p and n depend on the individual learner and the nature of the task.

4. In the study of vibrating strings and the propagation of waves, we find the partial differential equation

(4) ††

where t represents time, x the location along the string, c the wave speed, and u the displacement of the string, which is a function of time and location.

02u 0t2

! c2 02u 0x2

" 0 ,

dy/dt y3/2 A1 ! yB3/2 " 2p1n .

AtB AtB L

d2q

dt2 # R

dt #

C q " E AtB ,

dP dt

" r

100 P , t in years.

AtB

†We will discuss Kirchhoff’s laws in Section 3.5. ††Historical Footnote: This partial differential equation was first discovered by Jean le Rond d’Alembert (1717–1783) in 1747.

Figure 1.3 Schematic for a series RLC circuit

R L

emf

−

4 Chapter 1 Introduction

To begin our study of differential equations, we need some common terminology. If an equation involves the derivative of one variable with respect to another, then the former is called a dependent variable and the latter an independent variable. Thus, in the equation

(5)

t is the independent variable and x is the dependent variable. We refer to a and k as coefficients in equation (5). In the equation

(6)

x and y are independent variables and u is the dependent variable. A differential equation involving only ordinary derivatives with respect to a single indepen-

dent variable is called an ordinary differential equation. A differential equation involving partial derivatives with respect to more than one independent variable is a partial differential equation. Equation (5) is an ordinary differential equation, and equation (6) is a partial differential equation.

The order of a differential equation is the order of the highest-order derivatives present in the equation. Equation (5) is a second-order equation because is the highest-order derivative present. Equation (6) is a first-order equation because only first-order partial deriva- tives occur.

It will be useful to classify ordinary differential equations as being either linear or nonlin- ear. Remember that lines (in two dimensions) and planes (in three dimensions) are especially easy to visualize, when compared to nonlinear objects such as cubic curves or quadric surfaces. For example, all the points on a line can be found if we know just two of them. Correspond- ingly, linear differential equations are more amenable to solution than nonlinear ones. Now the equations for lines ax # by " c and planes ax # by # cz " d have the feature that the variables appear in additive combinations of their first powers only. By analogy a linear differential equation is one in which the dependent variable y and its derivatives appear in additive combi- nations of their first powers.

More precisely, a differential equation is linear if it has the format

(7)

where an , an!1 , . . . , a0 and F depend only on the independent variable x. The addi- tive combinations are permitted to have multipliers (coefficients) that depend on x; no restric- tions are made on the nature of this x-dependence. If an ordinary differential equation is not linear, then we call it nonlinear. For example,

is a nonlinear second-order ordinary differential equation because of the y3 term, whereas

is linear (despite the t3 terms). The equation

" cos x

is nonlinear because of the term.y dy/dx

d2y

dx2 ! y

t3 dx dt

" t3 # x

d2y

dx2 # y3 " 0

AxBAxBAxBAxB an AxB dnydxn ! an!1 AxB dn"1ydxn"1 ! p ! a1 AxB dydx ! a0 AxBy # F AxB ,

d2x/dt2

0u 0x !

0u 0y " x ! 2y ,

d2x dt2

# a dx dt

# kx " 0 ,

Section 1.1 Background 5

In Problems 1–12, a differential equation is given along with the field or problem area in which it arises. Classify each as an ordinary differential equation (ODE) or a partial differential equation (PDE), give the order, and indicate the independent and dependent variables. If the equation is an ordinary differential equation, indicate whether the equation is linear or nonlinear.

(Hermite’s equation, quantum-mechanical harmonic oscillator)

2. # 9x " 2 cos 3t

(mechanical vibrations, electrical circuits, seismology)

(Laplace’s equation, potential theory, electricity, heat, aerodynamics)

(competition between two species, ecology)

5. , where k is a constant

(chemical reaction rates)

6. , where C is a constant

(brachistochrone problem,† calculus of variations)

(Kidder’s equation, flow of gases through a porous medium)

11 ! y d 2 y dx2

# 2x dy

dx " 0

y c1 # ady dx b 2 d " C

dx dt

" k A4 ! xB A1 ! xB dy

dx "

y A2 ! 3xB x A1 ! 3yB

02u 0x2

# 02u 0y2

" 0

5 d 2x dt2

# 4 dx dt

d 2y

dx2 ! 2x

dx # 2y " 0

8. , where k and P are constants

(logistic curve, epidemiology, economics)

(deflection of beams)

10.

(aerodynamics, stress analysis)

11. where k is a constant

(nuclear fission)

12.

(van der Pol’s equation, triode vacuum tube)

In Problems 13–16, write a differential equation that fits the physical description. 13. The rate of change of the population p of bacteria at

time t is proportional to the population at time t. 14. The velocity at time t of a particle moving along a

straight line is proportional to the fourth power of its position x.

15. The rate of change in the temperature T of coffee at time t is proportional to the difference between the temperature M of the air at time t and the tempera- ture of the coffee at time t.

16. The rate of change of the mass A of salt at time t is proportional to the square of the mass of salt present at time t.

17. Drag Race. Two drivers, Alison and Kevin, are par- ticipating in a drag race. Beginning from a standing start, they each proceed with a constant acceleration. Alison covers the last of the distance in 3 sec- onds, whereas Kevin covers the last of the dis- tance in 4 seconds. Who wins and by how much time?

1/3 1/4

d2y

dx2 ! 0.1 A1 ! y2B dy

dx # 9y " 0

0N 0t

" 02N 0r 2

# 1 r 0N 0r

# kN,

x d2y

dx2 #

dx # xy " 0

8 d4y

dx4 " x A1 ! xB

dp dt

" kp AP ! pB

Although the majority of equations one is likely to encounter in practice fall into the nonlin- ear category, knowing how to deal with the simpler linear equations is an important first step (just as tangent lines help our understanding of complicated curves by providing local approximations).

†Historical Footnote: In 1630 Galileo formulated the brachistochrone problem shortest, " time , that is, to determine a path down which a particle will fall from one given point to another in the shortest time. It was reproposed by John Bernoulli in 1696 and solved by him the following year.

Bxróno%Abráxísto% "

1.1 EXERCISES

6 Chapter 1 Introduction

An nth-order ordinary differential equation is an equality relating the independent variable to the nth derivative (and usually lower-order derivatives as well) of the dependent variable. Examples are

(second-order, x independent, y dependent)

(second-order, t independent, y dependent)

(fourth-order, t independent, x dependent).

Thus, a general form for an nth-order equation with x independent, y dependent, can be expressed as

(1)

where F is a function that depends on x, y, and the derivatives of y up to order n; that is, on x, y, . . . , . We assume that the equation holds for all x in an open interval I ( where a or b could be infinite). In many cases we can isolate the highest-order term and write equation (1) as

(2)

which is often preferable to (1) for theoretical and computational purposes.

dny

dxn " f ax, y, dy

dx , . . . ,

dn!1y

dxn!1 b ,

dny/dxn a 6 x 6 b,dny/dxn

Fax, y, dy dx

, . . . , dny dxn b " 0 ,

d4x dt4

" xt

B1 ! ad2ydt2b ! y " 0 x2

d2y

dx2 # x

dx # y " x3

1.2 SOLUTIONS AND INITIAL VALUE PROBLEMS

Explicit Solution

Definition 1. A function that when substituted for y in equation (1) [or (2)] satisfies the equation for all x in the interval I is called an explicit solution to the equation on I.

f AxB

Show that is an explicit solution to the linear equation

(3)

but is not.

The functions and are defined for all Substitution of for y in equation (3) givesA2 ! 2x!3B ! 2

x2 Ax2 ! x!1B " A2 ! 2x!3B ! A2 ! 2x!3B " 0 .f AxBx & 0.

f– AxB " 2 ! 2x!3f AxB " x2 ! x!1, f¿ AxB " 2x # x!2,c AxB " x3

d2y

dx2 !

2 x2

y " 0 ,

f AxB " x2 ! x!1Example 1

Solution

Section 1.2 Solutions and Initial Value Problems 7

Since this is valid for any x 0, the function is an explicit solution to (3) on and also on .

For we have , , and substitution into (3) gives

which is valid only at the point and not on an interval. Hence is not a solution. ◆

Show that for any choice of the constants c1 and c2, the function

is an explicit solution to the linear equation

(4)

We compute and . Substitution of , and for y, y , and y in equation (4) yields

Since equality holds for all x in , then is an explicit solution to (4) on the interval for any choice of the constants c1 and c2. ◆

As we will see in Chapter 2, the methods for solving differential equations do not always yield an explicit solution for the equation. We may have to settle for a solution that is defined implicitly. Consider the following example.

Show that the relation

(5)

implicitly defines a solution to the nonlinear equation

(6)

on the interval .

When we solve (5) for y, we obtain y " ' . Let’s try to see if it is an

explicit solution. Since , both and are defined on . Substituting them into (6) yields

which is indeed valid for all x in . [You can check that is also an explicit solution to (6).] ◆

c AxB " !2x3 ! 8A2, q B 3x2

22x3 ! 8 " 3x22 A2x3 ! 8 B , A2, q Bdf/dxfdf/dx " 3x2/ A22x3 ! 8 B f(x) " 2x3 ! 82x3 ! 8

A2, q B dy dx

" 3x2

y2 ! x3 # 8 " 0

A!q, q B f AxB " c1e!x # c2e2xA!q, q B Ac1 # c1 ! 2c1Be!x # A4c2 ! 2c2 ! 2c2Be2x " 0 .Ac1e!x # 4c2e2xB ! A!c1e!x # 2c2e2xB ! 2 Ac1e!x # c2e2xB

–¿f– f, f¿f– AxB " c1e!x # 4c2e2xf¿ AxB " !c1e!x # 2c2e2x

y– ! y¿ ! 2y " 0 .

f AxB " c1e!x # c2e2x c(x)x " 0

6x ! 2 x2

x3 " 4x " 0 ,

c– AxB " 6xc¿ AxB " 3x2c AxB " x3 A0, q BA!q, 0B f AxB " x2 ! x!1&

Example 2

Solution

Example 3

Solution

8 Chapter 1 Introduction

Show that

(7)

is an implicit solution to the nonlinear equation

(8)

First, we observe that we are unable to solve (7) directly for y in terms of x alone. However, for (7) to hold, we realize that any change in x requires a change in y, so we expect the relation (7) to define implicitly at least one function . This is difficult to show directly but can be rigor- ously verified using the implicit function theorem† of advanced calculus, which guarantees that such a function exists that is also differentiable (see Problem 30).

Once we know that y is a differentiable function of x, we can use the technique of implicit differentiation. Indeed, from (7) we obtain on differentiating with respect to x and applying the product and chain rules,

which is identical to the differential equation (8). Thus, relation (7) is an implicit solution on some interval guaranteed by the implicit function theorem. ◆

Verify that for every constant C the relation is an implicit solution to

(9)

Graph the solution curves for C " 0, '1, '4. (We call the collection of all such solutions a one-parameter family of solutions.)

When we implicitly differentiate the equation with respect to x, we find

8x ! 2y dy dx

" 0 ,

4x2 ! y2 " C

y dy dx

! 4x " 0 .

4x2 ! y2 " C

A1 # xexyB dy dx

# 1 # yexy " 0 ,

1 # dy dx

# exy ay # x dy dx b " 0d

dx Ax # y # exyB "

y AxB y AxB

A1 # xexyB dy dx

# 1 # yexy " 0 .

x # y # exy " 0

Example 4

Solution

†See Vector Calculus, 5th ed, by J. E. Marsden and A. J. Tromba (Freeman, San Francisco, 2004).

Example 5

Implicit Solution

Definition 2. A relation is said to be an implicit solution to equation (1) on the interval I if it defines one or more explicit solutions on I.

G Ax, yB " 0

Solution

Section 1.2 Solutions and Initial Value Problems 9

C = 0

C = 1

C = −4

C = 4 C = 4

C = 0

C = 1

−1 1

−2

C = −1

Figure 1.4 Implicit solutions 4x2 ! y2 " C

which is equivalent to (9). In Figure 1.4 we have sketched the implicit solutions for C " 0, '1, '4. The curves are hyperbolas with common asymptotes y " '2x. Notice that the implicit solution curves (with C arbitrary) fill the entire plane and are nonintersecting for C 0. For C " 0, the implicit solution gives rise to the two explicit solutions y " 2x and y " !2x, both of which pass through the origin. ◆

For brevity we hereafter use the term solution to mean either an explicit or an implicit solution.

In the beginning of Section 1.1, we saw that the solution of the second-order free-fall equation invoked two arbitrary constants of integration c1, c2:

whereas the solution of the first-order radioactive decay equation contained a single constant C:

It is clear that integration of the simple fourth-order equation

brings in four undetermined constants:

It will be shown later in the text that in general the methods for solving nth-order differential equations evoke n arbitrary constants. In most cases, we will be able to evaluate these constants if we know n initial values y , y( , . . . , .Ax0By An!1BAx0BAx0B

y AxB " c1x3 # c2x2 # c3x # c4 . d4y

dx4 " 0

A AtB " Ce!kt . h AtB " !gt2

2 # c1t # c2 ,

10 Chapter 1 Introduction

In the case of a first-order equation, the initial conditions reduce to the single requirement

y " y0 ,

and in the case of a second-order equation, the initial conditions have the form

The terminology initial conditions comes from mechanics, where the independent variable x represents time and is customarily symbolized as t. Then if t0 is the starting time, y " y0 represents the initial location of an object and gives its initial velocity.

Show that " sin x ! cos x is a solution to the initial value problem

(10)

Observe that " sin x ! cos x, cos x # sin x, and " !sin x # cos x are all defined on . Substituting into the differential equation gives

which holds for all Hence, is a solution to the differential equation in (10) on . When we check the initial conditions, we find

" sin 0 ! cos 0 " !1 ,

" cos 0 # sin 0 " 1 ,

which meets the requirements of (10). Therefore, is a solution to the given initial value problem. ◆

f AxB df dx

A0Bf A0B A!q, q B f AxBx $ A!q, q B.

A!sin x # cos xB # Asin x ! cos xB " 0 ,A!q, q B d2f/dx2df/dx "f AxB

d2y

dx2 # y " 0 ; y A0B " !1 , dy

dx A0B " 1 .

AxBf y¿ At0B At0B

y Ax0B " y0 , dydx Ax0B " y1 . Ax0B

Initial Value Problem

Definition 3. By an initial value problem for an nth-order differential equation

we mean: Find a solution to the differential equation on an interval I that satisfies at x0 the n initial conditions

···

where and y0, y1, . . . , yn – 1 are given constants.x0 $ I

dn!1y

dxn!1 Ax0B " yn!1 ,

dy dx Ax0B " y1 ,

y Ax0B " y0 , Fax, y, dy

dx , . . . ,

dny dxn b " 0 ,

Example 6

Solution

Section 1.2 Solutions and Initial Value Problems 11

As shown in Example 2, the function is a solution to

for any choice of the constants c1 and c2. Determine c1 and c2 so that the initial conditions

and

are satisfied.

To determine the constants c1 and c2, we first compute to get Substituting in our initial conditions gives the following system of equations:

Adding the last two equations yields 3c2 " !1, so c2 " . Since c1 # c2 " 2, we find c1 " . Hence, the solution to the initial value problem is ◆

We now state an existence and uniqueness theorem for first-order initial value problems. We presume the differential equation has been cast into the format

Of course, the right-hand side, , must be well defined at the starting value x0 for x and at the stipulated initial value y0 " y for y. The hypotheses of the theorem, moreover, require continuity of both f and for x in some interval a ) x ) b containing x0, and for y in some interval c ) y ) d containing y0. Notice that the set of points in the xy-plane that satisfy a ) x ) b and c ) y ) d constitutes a rectangle. Figure 1.5 on page 12 depicts this “rectangle of continuity” with the initial point in its interior and a sketch of a portion of the solu- tion curve contained therein.

Ax0, y0B 0f/ 0y

Ax0Bf Ax, yB dy dx

" f Ax, yB .

f AxB " A7/3Be!x ! A1/3Be2x.7/3 !1/3 " c1 # c2 " 2 ,!c1 # 2c2 " !3 ."

f A0B " c1e0 # c2e0 " 2 , df dx

A0B " !c1e0 # 2c2e0 " !3 ,

!c1e !x # 2c2e

2x.df/dx "df/dx

dy dx

A0B " !3y A0B " 2

d2y

dx2 !

dx ! 2y " 0

f AxB " c1e!x # c2e2xExample 7

Solution

Existence and Uniqueness of Solution

Theorem 1. Consider the initial value problem

If f and are continuous functions in some rectangle

that contains the point , then the initial value problem has a unique solution in some interval x0 ! d ) x ) x0 # d, where d is a positive number.f AxB Ax0, y0B

R " E Ax, yB: a 6 x 6 b, c 6 y 6 dF0f/ 0y dy dx

" f Ax, yB , y Ax0B " y0 .

12 Chapter 1 Introduction

The preceding theorem tells us two things. First, when an equation satisfies the hypotheses of Theorem 1, we are assured that a solution to the initial value problem exists. Naturally, it is desirable to know whether the equation we are trying to solve actually has a solution before we spend too much time trying to solve it. Second, when the hypotheses are satisfied, there is a unique solution to the initial value problem. This uniqueness tells us that if we can find a solu- tion, then it is the only solution for the initial value problem. Graphically, the theorem says that there is only one solution curve that passes through the point . In other words, for this first-order equation, two solutions cannot cross anywhere in the rectangle. Notice that the exis- tence and uniqueness of the solution holds only in some neighborhood . Unfortunately, the theorem does not tell us the span of this neighborhood (merely that it is not zero). Problem 18 elaborates on this feature.

Problem 19 gives an example of an equation with no solution. Problem 29 displays an ini- tial value problem for which the solution is not unique. Of course, the hypotheses of Theorem 1 are not met for these cases.

When initial value problems are used to model physical phenomena, many practitioners tacitly presume the conclusions of Theorem 1 to be valid. Indeed, for the initial value problem to be a reasonable model, we certainly expect it to have a solution, since physically “something does happen.” Moreover, the solution should be unique in those cases when repetition of the experiment under identical conditions yields the same results.†

The proof of Theorem 1 involves converting the initial value problem into an integral equation and then using Picard’s method to generate a sequence of successive approximations that converge to the solution. The conversion to an integral equation and Picard’s method are discussed in Group Project B at the end of this chapter. A detailed discussion and proof of the theorem are given in Chapter 13.††

A2dB Ax0 ! d, x0 # dB Ax0, y0B

y = (x)

a bx0 − * x0 x0 + *

Figure 1.5 Layout for the existence–uniqueness theorem

†At least this is the case when we are considering a deterministic model, as opposed to a probabilistic model. ††All references to Chapters 11–13 refer to the expanded text Fundamentals of Differential Equations and Boundary Value Problems, 6th ed.

Section 1.2 Solutions and Initial Value Problems 13

For the initial value problem

(11)

does Theorem 1 imply the existence of a unique solution?

Dividing by 3 to conform to the statement of the theorem, we identify as and as . Both of these functions are continuous in any rectangle containing the point (1, 6), so the hypotheses of Theorem 1 are satisfied. It then follows from the theorem that the initial value problem (11) has a unique solution in an interval about of the form

, where is some positive number. ◆

For the initial value problem

(12)

does Theorem 1 imply the existence of a unique solution?

Here and Unfortunately is not continuous or even defined when y " 0. Consequently, there is no rectangle containing in which both f and are continuous. Because the hypotheses of Theorem 1 do not hold, we cannot use Theorem 1 to determine whether the initial value problem does or does not have a unique solution. It turns out that this initial value problem has more than one solution. We refer you to Problem 29 and Group Project G of Chapter 2 for the details. ◆

In Example 9 suppose the initial condition is changed to . Then, since f and are continuous in any rectangle that contains the point but does not intersect the x-axis— say, —it follows from Theorem 1 that this new initial value problem has a unique solution in some interval about x " 2.

R " E Ax, yB: 0 6 x 6 10, 0 6 y 6 5F A2, 1B 0f/ 0yy A2B " 1

0f/ 0yA2, 0B0f/ 0y0f/ 0y " 2y!1/3.f Ax, yB " 3y2/3 dy dx

" 3y2/3 , y A2B " 0 , dA1 ! d,1 # dB x " 1

!xy20f/ 0y Ax2 ! xy3B/3f Ax, yB

3 dy dx

" x2 ! xy3 , y(1) " 6 ,

Example 8

Solution

Example 9

Solution

1. (a) Show that is an implicit solution to on the interval

(b) Show that sin x = 1 is an implicit solu- tion to

on the interval . 2. (a) Show that is an explicit solution to

on the interval (b) Show that is an explicit solution to

on the interval A!q, q B. dy dx

# y2 " e2x # A1 ! 2xBex # x2 ! 1f AxB " e x ! x

A!q, q B.x dy dx

" 2y

f AxB " x2A0, p/2B dy

dx " Ax cos x # sin x ! 1By

3 Ax ! x sin xB xy3 ! xy3

A!q, 3B.dy/dx " !1/ A2yBy2 # x ! 3 " 0 (c) Show that is an explicit solu-tion to on the interval In Problems 3–8, determine whether the given function is a solution to the given differential equation.

3. x " 2 cos t ! 3 sin t , x # x " 0

4. y " sin x # ,

5. x " cos 2t ,

7. y " 3 sin

8. d2y

dx2 !

dx ! 2y " 0y " e2x ! 3e!x ,

y– # 4y " 5e!x2x # e!x ,

d2u dt2

! u du dt

# 3u " !2e2tu " 2e3t ! e2t ,

dx dt

# tx " sin 2t

d2y

dx2 # y " x2 # 2x2

–

A0, q B.x2d2y/dx2 " 2yf AxB " x2 ! x!1 1.2 EXERCISES

14 Chapter 1 Introduction

In Problems 9–13, determine whether the given relation is an implicit solution to the given differential equation. Assume that the relationship does define y implicitly as a function of x and use implicit differentiation.

9. y ! ln y " # 1 ,

10. # " 4 ,

11. # y " x ! 1 ,

12. ! sin " 1 , " 2x sec ! 1

13. sin y # xy ! " 2 ,

14. Show that sin x # c2 cos x is a solution to # y " 0 for any choice of the constants c1

and c2. Thus, c1 sin x # c2 cos x is a two-parameter family of solutions to the differential equation.

15. Verify that where c is an arbi- trary constant, is a one-parameter family of solutions to

Graph the solution curves corresponding to c " 0, '1, '2 using the same coordinate axes.

16. Verify that # " 1, where c is an arbitrary nonzero constant, is a one-parameter family of implicit solutions to

and graph several of the solution curves using the same coordinate axes.

17. Show that is a solution to ! 3y " !3 for any choice of the constant C.

Thus, Ce3x # 1 is a one-parameter family of solu- tions to the differential equation. Graph several of the solution curves using the same coordinate axes.

18. Let Show that the function " is a solution to the initial value prob-

lem , , on the interval Note that this solution becomes

unbounded as approaches Thus, the solution exists on the interval with , but not for larger d. This illustrates that in Theorem 1 the existence

d " c(!d, d) 'c.x

!c 6 x 6 c. y(0) " 1/c2dy/dx " 2xy2

(c2 ! x2)!1 f AxBc 7 0.

dy/dx f AxB " Ce3x # 1

dx "

x2 ! 1

cy2x2

dy dx

" y A y ! 2B

2 .

f AxB " 2/ A1 ! cexB, d2y/dx2

f AxB " c1 y– "

6xy¿ # Ay¿ B3sin y ! 2 Ay¿ B2 3x2 ! y

Ax # yBdy dx

Ax # yBx2 dy dx

" e!xy ! y e!xy # x

exy

dy dx

" x y

y2x2

dy dx

" 2xy

y ! 1 x2

interval can be quite small (if c is small) or quite large (if c is large). Notice also that there is no clue from the equation itself, or from the initial value, that the solution will “blow up” at x " 'c.

19. Show that the equation has no (real-valued) solution.

20. Determine for which values of m the function is a solution to the given equation.

(a)

(b)

21. Determine for which values of m the function is a solution to the given equation.

(a)

(b)

22. Verify that the function is a solution to the linear equation

for any choice of the constants c1 and c2. Determine c1 and c2 so that each of the following initial condi- tions is satisfied. (a) (b)

In Problems 23–28, determine whether Theorem 1 implies that the given initial value problem has a unique solution.

23. " 7

24.

25.

26. # cos x " sin t ,

27.

28. y A2B " 1dy dx

" 3x ! 23 y ! 1 , y A1B " 0y dy dx

" x ,

x ApB " 0dx dt

x A2B " !p3x dx dt

# 4t " 0 ,

y ApB " 5dy dt

! ty " sin2t ,

y A0Bdy dx

" y4 ! x4 ,

y A1B " 1 , y¿ A1B " 0y A0B " 2 , y¿ A0B " 1

d2y

dx2 #

dx ! 2y " 0

f AxB " c1ex # c2e!2x x2

d2y

dx2 ! x

dx ! 5y " 0

3x2 d2y

dx2 # 11x

dx ! 3y " 0

f AxB " xm d3y

dx3 # 3

d2y

dx2 # 2

dx " 0

d2y

dx2 # 6

dx # 5y " 0

f AxB " emx Ady/dxB2 # y2 # 4 " 0

dy/dx " 2xy2

Section 1.3 Direction Fields 15

29. (a) For the initial value problem (12) of Example 9, show that are solutions. Hence, this initial value problem has multiple solutions. (See also Group Project G in Chapter 2.)

(b) Does the initial value problem have a unique solution in a neigh-

borhood of ? 30. Implicit Function Theorem. Let G have

continuous first partial derivatives in the rectangle containing

the point . If G " 0 and the partial derivative Gy 0, then there exists a differ- entiable function , defined in some interval I " , that satisfies for all x $ I.

G Ax, f AxBB " 0Ax0 ! d, x0 # dBy " f AxB&Ax0, y0B Ax0, y0BAx0, y0BR " E Ax, yB: a 6 x 6 b, c 6 y 6 dF

Ax, yBx " 0 y(0) " 10!7,

y¿ " 3y2/ 3,

f1 AxB # 0 and f2 AxB " Ax ! 2B3 The implicit function theorem gives conditionsunder which the relationship G " 0 defines y implicitly as a function of x. Use the implicit function theorem to show that the relationship x # y # " 0, given in Example 4, defines y implicitly as a function of x near the point .

31. Consider the equation of Example 5,

(13)

(a) Does Theorem 1 imply the existence of a unique solution to (13) that satisfies " 0?

(b) Show that when equation (13) can’t possibly have a solution in a neighborhood of x " x0 that satisfies " 0.

y Ax0B x0 & 0,

y Ax0B y

dy dx

! 4x " 0 .

A0, !1Bexy Ax, yB

The existence and uniqueness theorem discussed in Section 1.2 certainly has great value, but it stops short of telling us anything about the nature of the solution to a differential equation. For practical reasons we may need to know the value of the solution at a certain point, or the inter- vals where the solution is increasing, or the points where the solution attains a maximum value. Certainly, knowing an explicit representation (a formula) for the solution would be a consider- able help in answering these questions. However, for many of the differential equations that we are likely to encounter in real-world applications, it will be impossible to find such a formula. Moreover, even if we are lucky enough to obtain an implicit solution, using this relationship to determine an explicit form may be difficult. Thus, we must rely on other methods to analyze or approximate the solution.

One technique that is useful in visualizing (graphing) the solutions to a first-order differen- tial equation is to sketch the direction field for the equation. To describe this method, we need to make a general observation. Namely, a first-order equation

specifies a slope at each point in the xy-plane where f is defined. In other words, it gives the direc- tion that a graph of a solution to the equation must have at each point. Consider, for example, the equation

(1)

The graph of a solution to (1) that passes through the point must have slope 2 ! 1 " 3 at that point, and a solution through has zero slope at that point.

A plot of short line segments drawn at various points in the xy-plane showing the slope of the solution curve there is called a direction field for the differential equation. Because the

A!1, 1B A!2BA!2, 1B dy dx

" x2 ! y .

dy dx

" f Ax, yB

1.3 DIRECTION FIELDS

16 Chapter 1 Introduction

direction field gives the “flow of solutions,” it facilitates the drawing of any particular solution (such as the solution to an initial value problem). In Figure 1.6(a) we have sketched the direc- tion field for equation (1) and in Figure 1.6(b) have drawn several solution curves in color.

Some other interesting direction field patterns are displayed in Figure 1.7. Depicted in Fig- ure 1.7(a) is the pattern for the radioactive decay equation " !2y (recall that in Section 1.1 we analyzed this equation in the form ). From the flow patterns, we can see that all solutions tend asymptotically to the positive x-axis as x gets larger. In other words, any material decaying according to this law eventually dwindles to practically nothing. This is con- sistent with the solution formula we derived earlier,

A " Ce!kt , or y " Ce!2x .

dA/dt " !kA dy/dx

(a) (b)

x x

Figure 1.6 (a) Direction field for (b) Solutions to dy/dx " x2 ! ydy/dx " x2 ! y

dy dx

(a) = −2y dy dx

(b) = − yx

Figure 1.7 (a) Direction field for (b) Direction field for dy/dx " !y/xdy/dx " !2y

Section 1.3 Direction Fields 17

From the direction field in Figure 1.7(b), we can anticipate that all solutions to " also approach the x-axis as x approaches infinity (plus or minus infinity, in fact). But

more interesting is the observation that no solution can make it across the y-axis; “blows up” as x goes to zero from either direction. Exception: On close examination, it appears the function might just make it through this barrier. As a matter of fact, in Problem 19 you are invited to show that the solutions to this differential equation are given by y " , with C an arbitrary constant. So they do diverge at x " 0, unless C " 0.

Let’s interpret the existence–uniqueness theorem of Section 1.2 for these direction fields. For Figure 1.7(a), where " " !2y, we select a starting point x0 and an initial value " y0, as in Figure 1.8(a). Because the right-hand side " !2y is continu- ously differentiable for all x and y, we can enclose any initial point in a “rectangle of continuity.” We conclude that the equation has one and only one solution curve passing through

, as depicted in the figure. For the equation

the right-hand side " does not meet the continuity conditions when x " 0 (i.e., for points on the y-axis). However, for any nonzero starting value x0 and any initial value " y0, we can enclose in a rectangle of continuity that excludes the y-axis, as in Figure 1.8(b). Thus, we can be assured of one and only one solution curve passing through such a point.

The direction field for the equation

is intriguing because Example 9 of Section 1.2 showed that the hypotheses of Theorem 1 do not hold in any rectangle enclosing the initial point x0 " 2, y0 " 0. Indeed, Problem 29 of that section demonstrated the violation of uniqueness by exhibiting two solutions, y AxB # 0

dy dx

" 3y2/3

Ax0, y0B y Ax0B !y/xf Ax, yB

dy dx

" f Ax, yB " ! y x ,

Ax0, y0B Ax0, y0Bf Ax, yBy Ax0B

f Ax, yBdy/dx C/x

y AxB # 0 0 y AxB 0 !y/x

dy/dx

(a)

0 x0

(b)

0 x0

Figure 1.8 (a) A solution for (b) A solution for dy/dx " !y/xdy/dx " !2y

18 Chapter 1 Introduction

and passing through . Figure 1.9(a) displays this direction field, and Figure 1.9(b) demonstrates how both solution curves can successfully “negotiate” this flow pattern.

Clearly, a sketch of the direction field of a first-order differential equation can be helpful in visualizing the solutions. However, such a sketch is not sufficient to enable us to trace, unam- biguously, the solution curve passing through a given initial point . If we tried to trace one of the solution curves in Figure 1.6(b) on page 16, for example, we could easily “slip” over to an adjacent curve. For nonunique situations like that in Figure 1.9(b), as one negotiates the flow along the curve and reaches the inflection point, one cannot decide whether to turn or to (literally) go off on the tangent .

The logistic equation for the population p (in thousands) at time t of a certain species is given by

(2)

(Of course, p is nonnegative. The interpretation of the terms in the logistic equation is discussed in Section 3.2.) From the direction field sketched in Figure 1.10 on page 19, answer the following:

(a) If the initial population is 3000 , what can you say about the limit- ing population ?

(b) Can a population of 1000 ever decline to 500?

(a) The direction field indicates that all solution curves will approach the horizontal line p " 2 as that is, this line is an asymptote for all positive solutions. Thus, limtS#q p AtB " 2.t S #q;

3other than p AtB # 0 4 p AtBlimtS#q 3 that is, p A0B " 3 4

dp dt

" p A2 ! pB .

A y " 0By " Ax ! 2B3 Ax0, y0B

A2, 0By AxB " Ax ! 2B3,

Example 1

Solution

(a)

(b)

12 2

x y(x) = 0

y(x) = (x − 2)3 y

Figure 1.9 (a) Direction field for (b) Solutions for , y A2B " 0dy/dx " 3y2/3dy/dx " 3y2/3

Section 1.3 Direction Fields 19

(b) The direction field further shows that populations greater than 2000 will steadily decrease, whereas those less than 2000 will increase. In particular, a population of 1000 can never decline to 500.

(c) As mentioned in part (b), a population of 1000 will increase with time. But the direc- tion field indicates it can never reach 2000 or any larger value; i.e., the solution curve cannot cross the line p " 2. Indeed, the constant function is a solution to equation (2), and the uniqueness part of Theorem 1, page 11, precludes intersections of solution curves. ◆

Notice that the direction field in Figure 1.10 has the nice feature that the slopes do not depend on t; that is, the slopes are the same along each horizontal line. The same is true for Figures 1.8(a) and 1.9. This is the key property of so-called autonomous equations where the right-hand side is a function of the dependent variable only. Group Project C, page 32, investigates such equations in more detail.

Hand sketching the direction field for a differential equation is often tedious. Fortunately, several software programs have been developed to obviate this task†. When hand sketching is necessary, however, the method of isoclines can be helpful in reducing the work.

The Method of Isoclines An isocline for the differential equation

is a set of points in the xy-plane where all the solutions have the same slope ; thus, it is a level curve for the function . For example, if

(3)

the isoclines are simply the curves (straight lines) x # y " c or y " !x # c. Here c is an arbi- trary constant. But c can be interpreted as the numerical value of the slope of every solu- tion curve as it crosses the isocline. (Note that c is not the slope of the isocline itself; the latter is, obviously, !1.) Figure 1.11(a) on page 20 depicts the isoclines for equation (3).

dy/dx

y¿ " f Ax, yB " x # y ,f Ax, yB dy/dx

y¿ " f Ax, yB

y¿ " f AyB,

p AtB # 2

t 43210

p (in thousands)

Figure 1.10 Direction field for logistic equation

†An applet, maintained on the web at http://alamos.math.arizona.edu/~rychlik/JOde/index.html, sketches direction fields and automates most of the differential equation algorithms discussed in this book.

http://alamos.math.arizona.edu/~rychlik/JOde/index.html

20 Chapter 1 Introduction

(b)

(c)

(a)

c = −5

c = −4 c = −3 c = −2 c = −1 c = 0

c = 1

c = 2 c = 3

c = 4

c = 5

Figure 1.11 (a) Isoclines for (b) Direction field for (c) Solutions to y¿ " x # yy¿ " x # yy¿ " x # y

To implement the method of isoclines for sketching direction fields, we draw hash marks with slope c along the isocline " c for a few selected values of c. If we then erase the underlying isocline curves, the hash marks constitute a part of the direction field for the differ- ential equation. Figure 1.11(b) depicts this process for the isoclines shown in Figure 1.11(a), and Figure 1.11(c) displays some solution curves.

Remark. The isoclines themselves are not always straight lines. For equation (1) at the beginning of this section (page 15), they are parabolas ! y " c. When the isocline curves are complicated, this method is not practical.

f Ax, yB

Section 1.3 Direction Fields 21

1. The direction field for = 2x + y is shown in Figure 1.12. (a) Sketch the solution curve that passes through

. From this sketch, write the equation for the solution.

(b) Sketch the solution curve that passes through .

A!1, 3B A0, !2B

dy/dx

and 15. Why is the value y " 8 called the “terminal velocity”?

4. If the viscous force in Problem 3 is nonlinear, a pos- sible model would be provided by the differential equation

Redraw the direction field in Figure 1.14 to incorpo- rate this dependence. Sketch the solutions with initial conditions " 0, 1, 2, 3. What is the ter- minal velocity in this case?

y A0By3 dy dt

" 1 ! y3

8 .

1.3 EXERCISES

y y = −2x y = 2x

x 1 32 4

Figure 1.13 Direction field for dy/dx " 4x/y

x 1

1 3 52 4 6

Figure 1.12 Direction field for dy/dx " 2x # y

t 0

Figure 1.14 Direction field for dy dt

" 1 ! y 8

2. The direction field for is shown in Figure 1.13. (a) Verify that the straight lines y " '2x are solu-

tion curves, provided (b) Sketch the solution curve with initial condition

" 2. (c) Sketch the solution curve with initial condition

" 1. (d) What can you say about the behavior of the above

solutions as ? How about ? 3. A model for the velocity at time t of a certain

object falling under the influence of gravity in a vis- cous medium is given by the equation

From the direction field shown in Figure 1.14, sketch the solutions with the initial conditions " 5, 8,y A0B

dy dt

" 1 ! y 8

x S !qx S #q

y A2B y A0B

x & 0.

dy/dx " 4x/y

22 Chapter 1 Introduction

5. The logistic equation for the population (in thou- sands) of a certain species is given by

(a) Sketch the direction field by using either a com- puter software package or the method of iso- clines.

(b) If the initial population is 3000 that is, " 3 , what can you say about the limiting popula- tion ?

6. Consider the differential equation

(a) A solution curve passes through the point What is its slope at this point?

(b) Argue that every solution curve is increasing for x $ 1.

(d) A solution curve passes through . Prove that this curve has a relative minimum at .

7. Consider the differential equation

for the population p (in thousands) of a certain species at time t. (a) Sketch the direction field by using either a com-

puter software package or the method of isoclines. (b) If the initial population is 4000 that is, "

4 , what can you say about the limiting popula- tion ?

8. The motion of a set of particles moving along the x-axis is governed by the differential equation

where denotes the position at time t of the particle. (a) If a particle is located at x " 1 when t " 2, what

is its velocity at this time?

x AtB dx dt

" t3 ! x3 ,

lim tS#q p AtBp A0B lim tS#q p AtBp A0B lim tS#q p AtB4 p A0B3

dp dt

" p A p ! 1B A2 ! pB A0, 0BA0, 0B

d2y

dx2 " 1 # x cos y #

2 sin 2y .

A1, p/2B. dy dx

" x # sin y .

lim tS#q p AtBp A0BlimtS#q p AtB 4 p A0B3

dp dt

" 3p ! 2p2 .

(b) Show that the acceleration of a particle is given by

9. Let denote the solution to the initial value problem

(a) Show that (b) Argue that the graph of is decreasing for x

near zero and that as x increases from zero, decreases until it crosses the line y " x, where its derivative is zero.

(c) Let x* be the abscissa of the point where the solution curve crosses the line y " x. Consider the sign of and argue that has a relative minimum at x*.

(d) What can you say about the graph of for x $ x*?

(e) Verify that y " x ! 1 is a solution to " x ! y and explain why the graph of always stays above the line y " x ! 1.

(f) Sketch the direction field for " x ! y by using the method of isoclines or a computer software package.

(g) Sketch the solution using the direction field in part .

10. Use a computer software package to sketch the direction field for the following differential equa- tions. Sketch some of the solution curves. (a) " sin x (b) " sin y (c) " sin x sin y (d) " # (e) " !

In Problems 11–16, draw the isoclines with their direction markers and sketch several solution curves, including the curve satisfying the given initial conditions. 11. " , " 4 12. " y , " 1 13. " 2x , " !1 14. " , " !1 15. " 2 ! y , " 0 16. " x # 2y , " 1y A0Bdy/dx y A0Bx2dy/dx

y A0Bx/ydy/dx y A0Bdy/dx y A0Bdy/dx y A0B!x/ydy/dx

2y2x2dy/dx 2y2x2dy/dx

dy/dx dy/dx dy/dx

Af B y " f AxB dy/dx

f AxBdy/dx y " f AxB

ff– Ax*By " f AxB f AxBf

f–AxB " 1 ! f¿AxB " 1 ! x # f AxB. dy dx

" x ! y , y A0B " 1 . fAxB3Hint: t3 ! x3 " At ! xB At2 # xt # x2B. 4 d2x dt2

" 3t2 ! 3t3x2 # 3x5 .

Section 1.4 The Approximation Method of Euler 23

17. From a sketch of the direction field, what can one say about the behavior as x approaches of a solution to the following?

18. From a sketch of the direction field, what can one say about the behavior as x approaches of a solution to the following?

19. By rewriting the differential equation " in the form

integrate both sides to obtain the solution y " for an arbitrary constant C.

20. A bar magnet is often modeled as a magnetic dipole with one end labeled the north pole N and the oppo- site end labeled the south pole S. The magnetic field for the magnetic dipole is symmetric with respect to rotation about the axis passing lengthwise through the center of the bar. Hence we can study the mag- netic field by restricting ourselves to a plane with the bar magnet centered on the x-axis.

For a point P that is located a distance r from the origin, where r is much greater than the length of the

C/x

1 y

dy " !1 x

!y/xdy/dx

dy dx

" !y

dy dx

" 3 ! y # 1 x

#q magnet, the magnetic field lines satisfy the differen- tial equation

(4)

and the equipotential lines satisfy the equation

(5)

(a) Show that the two families of curves are per- pendicular where they intersect. [Hint: Con- sider the slopes of the tangent lines of the two curves at a point of intersection.]

(b) Sketch the direction field for equation (4) for , You can use a soft-

ware package to generate the direction field or use the method of isoclines. The direction field should remind you of the experiment where iron filings are sprinkled on a sheet of paper that is held above a bar magnet. The iron filings correspond to the hash marks.

(d) Apply the statement of part (a) to the curves in part (c) to sketch the equipotential lines that are solutions to (5). The magnetic field lines and the equipotential lines are examples of orthogo- nal trajectories. (See Problem 32 in Exercises 2.4, pages 62–63.)†

!5 + y + 5.!5 + x + 5

dy dx #

y2 " 2x2

3xy .

dy dx #

3xy

2x2 " y2

†Equations (4) and (5) can be solved using the method for homogeneous equations in Section 2.6 (see Exercises 2.6, Problem 47).

Euler’s method (or the tangent-line method) is a procedure for constructing approximate solu- tions to an initial value problem for a first-order differential equation

(1)

It could be described as a “mechanical” or “computerized” implementation of the informal procedure for hand sketching the solution curve from a picture of the direction field. As such, we will see that it remains subject to the failing that it may skip across solution curves. However, under fairly general conditions, iterations of the procedure do converge to true solutions.

y¿ " f Ax, yB , y Ax0B " y0 . 1.4 THE APPROXIMATION METHOD OF EULER

24 Chapter 1 Introduction

The method is illustrated in Figure 1.15. Starting at the initial point , we follow the straight line with slope , the tangent line, for some distance to the point . Then we reset the slope to the value and follow this line to . In this way we construct polygonal (broken line) approximations to the solution. As we take smaller spacings between points (and thus employ more points), we may expect to converge to the true solution.

To be more precise, assume that the initial value problem (1) has a unique solution in some interval centered at x0. Let h be a fixed positive number (called the step size) and consider the equally spaced points†

The construction of values yn that approximate the solution values proceeds as follows. At the point , the slope of the solution to (1) is given by " . Hence, the tangent line to the solution curve at the initial point is

y " y0 #

Using this tangent line to approximate we find that for the point x1 " x0 # h

Next, starting at the point , we construct the line with slope given by the direction field at the point — that is, with slope equal to . If we follow this line†† namely, y " y1 # in stepping from x1 to x2 " x1 # h, we arrive at the approximation

Repeating the process (as illustrated in Figure 1.15), we get

etc.f Ax4B $ y4 J y3 # h f Ax3, y3B , f Ax3B $ y3 J y2 # h f Ax2, y2B , f Ax2B $ y2 J y1 # h f Ax1, y1B . Ax ! x1B f Ax1, y1B 4 3f Ax1, y1BAx1, y1B Ax1, y1B f Ax1B $ y1 J y0 # h f Ax0, y0B .

f AxB,Ax ! x0B f Ax0, y0B . Ax0, y0B f Ax0, y0Bdy/dxAx0, y0B

f AxnB xn J x0 # nh , n " 0, 1, 2, . . . .

f AxB Ax2, y2Bf Ax1, y1B Ax1, y1Bf Ax0, y0B

Ax0, y0B

†The symbol means “is defined to be.” ††Because y1 is an approximation to we cannot assert that this line is tangent to the solution curve y " f AxB.f Ax1B,J

0 x0 x1 x2 x3

(x0, y 0)

(x1, y 1) (x2, y 2)

(x3, y 3)

Slope f(x0, y 0)

Slope f(x1, y 1)

Slope f(x2, y 2)

Figure 1.15 Polygonal-line approximation given by Euler’s method

Section 1.4 The Approximation Method of Euler 25

This simple procedure is Euler’s method and can be summarized by the recursive formulas

(2)

(3)

Use Euler’s method with step size h " 0.1 to approximate the solution to the initial value problem

(4)

at the points x " 1.1, 1.2, 1.3, 1.4, and 1.5.

Here x0 " 1, y0 " 4, h " 0.1, and Thus, the recursive formula (3) for yn is

Substituting n " 0, we get

Putting n " 1 yields

Continuing in this manner, we obtain the results listed in Table 1.1. For comparison we have included the exact value (to five decimal places) of the solution to (4), which can be obtained using separation of variables (see Section 2.2). As one might expect, the approximation deteriorates as x moves farther away from 1. ◆

f AxB " Ax2 # 7B2/16 y2 " y1 # A0.1B x12y1 " 4.2 # A0.1B A1.1B24.2 $ 4.42543 .x2 " x1 # 0.1 " 1.1 # 0.1 " 1.2 , y1 " y0 # A0.1B x02y0 " 4 # A0.1B A1B24 " 4.2 .x1 " x0 # 0.1 " 1 # 0.1 " 1.1 , yn#1 " yn # h f Axn, ynB " yn # A0.1B xn2yn .f Ax, yB " x2y. y¿ " x2y , y A1B " 4 yn!1 # yn ! h f Axn, ynB , n # 0, 1, 2, . . . .xn!1 # xn ! h ,

Example 1

Solution

TABLE 1.1 Computations for y , y(1) # 4

Euler’s n xn Method Exact Value

0 1 4 4 1 1.1 4.2 4.21276 2 1.2 4.42543 4.45210 3 1.3 4.67787 4.71976 4 1.4 4.95904 5.01760 5 1.5 5.27081 5.34766

% # x2y

Given the initial value problem (1) and a specific point x, how can Euler’s method be used to approximate ? Starting at x0, we can take one giant step that lands on x, or we can take several smaller steps to arrive at x. If we wish to take N steps, then we set h " so that the step size h and the number of steps N are related in a specific way. For example, if x0 " 1.5 and we wish to approximate using 10 steps, then we would take h "

" 0.05. It is expected that the more steps we take, the better will be the approx- imation. (But keep in mind that more steps mean more computations and hence greater accumulated roundoff error.)

A2 ! 1.5B /10 f A2B Ax ! x0B /Nf AxB

26 Chapter 1 Introduction

Use Euler’s method to find approximations to the solution of the initial value problem

(5)

at x " 1, taking 1, 2, 4, 8, and 16 steps.

Remark. Observe that the solution to (5) is just so Euler’s method will generate algebraic approximations to the transcendental number e " 2.71828. . . .

Here " y, x0 " 0, and y0 " 1. The recursive formula for Euler’s method is

To obtain approximations at x " 1 with N steps, we take the step size h " . For N " 1, we have

For In this case we get

For where

(In the above computations, we have rounded to five decimal places.) Similarly, taking N " 8 and 16, we obtain even better estimates for These approximations are shown in Table 1.2. For comparison, Figure 1.16 on page 27 displays the polygonal-line approximations to using Euler’s method with h " and h " . Notice that the smaller step size yields the better approximation. ◆

1/8 AN " 8B1/4 AN " 4B ex f A1B.

f A1B $ y4 " A1 # 0.25B A1.95313B " 2.44141 . y3 " A1 # 0.25B A1.5625B " 1.95313 , y2 " A1 # 0.25B A1.25B " 1.5625 , y1 " A1 # 0.25B A1B " 1.25 ,

N " 4, f Ax4B " f A1B $ y4, f A1B $ y2 " A1 # 0.5B A1.5B " 2.25 . y1 " A1 # 0.5B A1B " 1.5 ,

N " 2, f Ax2B " f A1B $ y2. f A1B $ y1 " A1 # 1B A1B " 2 .

1/N

yn#1 " yn # hyn " A1 # hByn . f Ax, yB

f AxB " ex, y¿ " y , y A0B " 1Example 2

Solution

TABLE 1.2 Euler’s Method for y % = y, y(0) = 1

Approximation N h for # e

1 1.0 2.0 2 0.5 2.25 4 0.25 2.44141 8 0.125 2.56578

16 0.0625 2.63793

F A1B

Section 1.4 The Approximation Method of Euler 27

How good (or bad) is Euler’s method? In judging a numerical scheme, we must begin with two fundamental questions. Does the method converge? And, if so, what is the rate of convergence? These important issues are discussed in Section 3.6, where improvements in Euler’s method are introduced (see also Problems 12 and 13 of this section).

Suppose satisfies the initial value problem

By experimenting with Euler’s method, determine to within one decimal place the value of and the time it will take to reach zero.

Determining rigorous estimates of the accuracy of the answers obtained by Euler’s method can be quite a challenging problem. The common practice is to repeatedly approximate and the zero crossing, using smaller and smaller values of h, until the digits of the computed values stabilize at the required accuracy level. For this example, Euler’s algorithm yields the following values:

v A0.2B v AtBv A0.2B A'0.1B

dv dt

" !3 ! 2v2 , v(0) " 2 .

v AtB

Solution

Example 3

Acknowledging the remote possibility that finer values of h might reveal aberrations, we state with reasonable confidence that . The Intermediate Value Theorem would imply that at some time t0 satisfying if the computations were perfect; they clearly provide evidence that ◆t0 " 0.4 ' 0.1.

0.375 6 t0 6 0.4,v At0B " 0 v A0.2B " 0.7 ' 0.1 v A0.2B $ 0.7071h " 0.00625 v A0.2B $ 0.6938h " 0.0125 v

A0.4B $ !0.0003v A0.375B $ 0.0750v A0.2B $ 0.6659h " 0.025 v A0.4B $ !0.0574v A0.35B $ 0.0935v A0.2B $ 0.6036h " 0.05 v A0.4B $ !0.2024v A0.3B $ 0.0996v A0.2B $ 0.4380h " 0.1

y = e x

1.25

1.5

1.75

2.25

2.5

2.75

0 1/8 1/4 3/8 1/2 5/8 3/4 7/8 1

h = 1/8

h = 1/4

Figure 1.16 Approximations of using Euler’s method with h = and 1/81/4ex

28 Chapter 1 Introduction

In many of the problems below, it will be helpful to have a calculator or computer available.† You may also find it convenient to write a program for solving initial value problems using Euler’s method. (Remember, all trigono- metric calculations are done in radians.)

In Problems 1–4, use Euler’s method to approximate the solution to the given initial value problem at the points x " 0.1, 0.2, 0.3, 0.4, and 0.5, using steps of size 0.1 .

5. Use Euler’s method with step size h " 0.1 to approximate the solution to the initial value problem

at the points x " 1.1, 1.2, 1.3, 1.4, and 1.5.

6. Use Euler’s method with step size h " 0.2 to approximate the solution to the initial value problem

at the points x " 1.2, 1.4, 1.6, and 1.8.

7. Use Euler’s method to find approximations to the solution of the initial value problem

at taking 1, 2, 4, and 8 steps.

8. Use Euler’s method to find approximations to the solution of the initial value problem

at t " 1, taking 1, 2, 4, and 8 steps.

9. Use Euler’s method with h " 0.1 to approximate the solution to the initial value problem

on the interval Compare these approxi- mations with the actual solution y " (verify!) by graphing the polygonal-line approximation and the actual solution on the same coordinate system.

10. Use the strategy of Example 3 to find a value of h for Euler’s method such that is approximated to within if y(x) satisfies the initial value problem

y¿ " x ! y , y A0B " 0 .'0.01, y A1B

!1/x 1 + x + 2.

y¿ " 1

x2 !

x ! y2 , y A1B " !1

dx dt

" 1 # t sin AtxB , x A0B " 0 x " p,

y¿ " 1 ! sin y , y A0B " 0

y¿ " 1 x

A y2 # yB , y A1B " 1

y¿ " x ! y2 , y A1B " 0 dy/dx " x/y , y A0B " !1dy/dx " x # y , y A0B " 1 dy/dx " y A2 ! yB , y A0B " 3dy/dx " !x/y , y A0B " 4 Ah " 0.1B

Also find, to within the value of x0 such that Compare your answers with those given

by the actual solution (verify!). Graph the polygonal-line approximation and the actual solution on the same coordinate system.

11. Use the strategy of Example 3 to find a value of h for Euler’s method such that is approximated to within if satisfies the initial value problem

Also find, to within the value of t0 such that Compare your answers with those given

by the actual solution (verify!). 12. In Example 2 we approximated the transcendental

number e by using Euler’s method to solve the initial value problem

Show that the Euler approximation yn obtained by using the step size is given by the formula

Recall from calculus that

and hence Euler’s method converges (theoretically) to the correct value.

13. Prove that the “rate of convergence” for Euler’s method in Problem 12 is comparable to by showing that

[Hint: Use L’Hôpital’s rule and the Maclaurin expansion for .]

14. Use Euler’s method with the spacings h " 0.5, 0.1, 0.05, 0.01 to approximate the solution to the initial value problem

on the interval (The explanation for the erratic results lies in Problem 18 of Exercises 1.2.)

Heat Exchange. There are basically two mecha- nisms by which a physical body exchanges heat with its environment. The contact heat transfer across the body’s surface is driven by the difference in the body’s

0 + x + 2. y¿ " 2xy2 , y A0B " 1

ln A1 # tB lim

nSq

e ! yn 1/n

" e 2

1/n

lim nSq

a1 # 1 n b n " e ,

yn " a1 # 1nb n , n " 1, 2, . . . 1/n

y¿ " y , y A0B " 1 . x " tan t

x At0B " 1. '0.02, dx dt

" 1 # x2 , x A0B " 0 .x AtB'0.01, x A1B

y " e!x # x ! 1 y Ax0B " 0.2. '0.05,

1.4 EXERCISES

†An applet, maintained on the web at http://alamos.math.arizona.edu/~rychlik/JOde/index.html, automates most of the differential equation algo- rithms discussed in this book.

http://alamos.math.arizona.edu/~rychlik/JOde/index.html

Technical Writing Exercises 29

temperature and that of the environment; this is known as Newton’s law of cooling. However, heat transfer also occurs due to thermal radiation, which according to Stefan’s law of radiation is governed by the difference of the fourth powers of these tempera- tures. In most cases one of these modes dominates the other. Problems 15 and 16 invite you to simulate each mode numerically for a given set of initial conditions.

15. Newton’s Law of Cooling. Newton’s law of cool- ing states that the rate of change in the temperature T of a body is proportional to the difference between the temperature of the medium M and the temperature of the body. That is,

where K is a constant. Let K " 1 (min)!1 and the tem- perature of the medium be constant, IfM AtB # 70º.

dT dt

" K 3M AtB ! T AtB 4 , AtBAtB

the body is initially at 100º, use Euler’s method with h " 0.1 to approximate the temperature of the body after (a) 1 minute. (b) 2 minutes.

16. Stefan’s Law of Radiation. Stefan’s law of radia- tion states that the rate of change in temperature of a body at T degrees in a medium at M degrees is proportional to . That is,

where K is a constant. Let K " and assume that the medium temperature is constant, If " 100º, use Euler’s method with h " 0.1 to approximate and .T A2BT A1BT A0B

M AtB # 70º.A40B!4 dT dt

" K AM(t)4 ! T(t)4B ,M 4 ! T 4

AtBAtB

Chapter Summary

In this chapter we introduced some basic terminology for differential equations. The order of a differential equation is the order of the highest derivative present. The subject of this text is ordinary differential equations, which involve derivatives with respect to a single independent variable. Such equations are classified as linear or nonlinear.

An explicit solution of a differential equation is a function of the independent variable that satisfies the equation on some interval. An implicit solution is a relation between the dependent and independent variables that implicitly defines a function that is an explicit solution. A differ- ential equation typically has infinitely many solutions. In contrast, some theorems ensure that a unique solution exists for certain initial value problems in which one must find a solution to the differential equation that also satisfies given initial conditions. For an nth-order equation, these conditions refer to the values of the solution and its first n – 1 derivatives at some point.

Even if one is not successful in finding explicit solutions to a differential equation, several techniques can be used to help analyze the solutions. One such method for first-order equations views the differential equation as specifying directions (slopes) at points on the plane. The conglomerate of such slopes is the direction field for the equation. Knowing the “flow of solutions” is helpful in sketching the solution to an initial value problem. Further- more, carrying out this method algebraically leads to numerical approximations to the desired solution. This numerical process is called Euler’s method.

dy/dx " f Ax, yB

1. Select four fields (for example, astronomy, geology, biology, and economics) and for each field discuss a situation in which differential equations are used to solve a problem. Select examples that are not covered in Section 1.1.

2. Compare the different types of solutions discussed in this chapter—explicit, implicit, graphical, and numerical. What are advantages and disadvantages of each?

TECHNICAL WRITING EXERCISES

Group Projects for Chapter 1

Euler’s method is based on the fact that the tangent line gives a good local approximation for the function. But why restrict ourselves to linear approximants when higher-degree polynomial approximants are available? For example, we can use the Taylor polynomial of degree n about x " x0, which is defined by

This polynomial is the nth partial sum of the Taylor series representation

To determine the Taylor series for the solution to the initial value problem

we need only determine the values of the derivatives of (assuming they exist) at x0; that is, , . . . . The initial condition gives the first value Using the equation

, we find To determine we differentiate the equation implicitly with respect to x to obtain

and thereby we can compute

(a) Compute the Taylor polynomials of degree 4 for the solutions to the given initial value problems. Use these Taylor polynomials to approximate the solution at x " 1.

(i) (ii)

(b) Compare the use of Euler’s method with that of the Taylor series to approximate the solution to the initial value problem

Do this by completing Table 1.3 on page 31. Give the approximations for and to the nearest thousandth. Verify that and use this formula together with a calculator or tables to find the exact values of to the nearest thousandth. Finally, decide which of the first four methods in Table 1.3 will yield the closest approximation to and give the reasons for your choice. (Remember that the com- putation of trigonometric functions must be done in the radian mode.)

d2y

dx2 " xy

f A10B f AxBf AxB " cos x # e!x

f A3Bf A1B dy dx

# y " cos x ! sin x , y A0B " 2 . f AxB

dy dx

" y A2 ! yB ; y A0B " 4 .dy dx

" x ! 2y ; y A0B " 1 . f– Ax0B.

y– " 0f 0x #

0f 0y

dy dx

" 0f 0x #

0f 0y f

y¿ " f Ax, yB f– Ax0B,f¿ Ax0B " f Ax0, y0B.y¿ " f Ax, yB f Ax0B " y0.f Ax0B, f¿ Ax0B f

dy/dx " f Ax, yB , y Ax0B " y0 , f AxB a q

k"0

y AkB Ax0B k!

Ax ! x0Bk . Pn AxB J y Ax0B # y¿ Ax0B Ax ! x0B # y– Ax0B2! Ax ! x0B2 # p # y AnB Ax0Bn! Ax ! x0Bn .

Taylor Series MethodA

Group Projects for Chapter 1 31

with the initial conditions 1, 0. Do you see how, in general, the Taylor series method for an nth-order differential equation will employ each of the n initial conditions mentioned in Definition 3, Section 1.2?

y¿A0B "y A0B "

TABLE 1.3

Approximation Approximation Method of of

Euler’s method using steps of size 0.1

Euler’s method using steps of size 0.01

Taylor polynomial of degree 2

Taylor polynomial of degree 5

Exact value of to nearest thousandthf AxB

F A3BF A1B

The initial value problem

(1)

can be rewritten as an integral equation. This is obtained by integrating both sides of (1) with respect to x from x " x0 to x " x1:

(2)

Substituting " y0 and solving for gives

(3)

If we use t instead of x as the variable of integration, we can let x " x1 be the upper limit of integration. Equation (3) then becomes

(4)

Equation (4) can be used to generate successive approximations of a solution to (1). Let the function be an initial guess or approximation of a solution to (1). Then a new approxima- tion function is given by

where we have replaced by the approximation in the argument of f. In a similar fashion, we can use to generate a new approximation , and so on. In general, we obtain the

st approximation from the relation

(5) J y0 ! ! x

x0 f At, Fn AtBBdt .Fn!1 AxBAn # 1B

f2 AxBf1 AxB f0 AtBy AtB f1 AxB J y0 # ! x

f At, f0 AtBBdt , f0 AxB y AxB # y0 ! ! x

x0 f At, y AtBBdt .

y Ax1B " y0 # ! x1 x0

f Ax, y AxBBdx .y Ax1By Ax0B !

y¿ AxBdx " y Ax1B ! y Ax0B " ! x1 x0

f Ax, y AxBB dx . y¿ AxB " f Ax, yB , y Ax0B " y0 Picard’s MethodB

32 Chapter 1 Introduction

This procedure is called Picard’s method.† Under certain assumptions on f and , the sequence is known to converge to a solution to (1). These assumptions and the proof of convergence are given in Chapter 13.††

Without further information about the solution to (1), it is common practice to take

(a) Use Picard’s method with to obtain the next four successive approximations of the solution to

(6)

Show that these approximations are just the partial sums of the Maclaurin series for the actual solution .

(b) Use Picard’s method with to obtain the next three successive approximations of the solution to the nonlinear problem

(7)

Graph these approximations for (c) In Problem 29 in Exercises 1.2, we showed that the initial value problem

(8)

does not have a unique solution. Show that Picard’s method beginning with con- verges to the solution whereas Picard’s method beginning with converges to the second solution . [Hint: For the guess , show that has the form , where and as ]n S q.rn S 3cn S 1cn Ax ! 2Brnfn AxB f0 AxB " x ! 2y AxB " Ax ! 2B3

f0 AxB " x ! 2y AxB # 0, f0 AxB # 0 y¿ AxB " 3 3 y AxB 4 2/3 , y A2B " 0

0 + x + 1.

y¿ AxB " 3x ! 3 y AxB 4 2 , y A0B " 0 . f0 AxB # 0ex

y¿ AxB " y AxB , y A0B " 1 . f0 AxB # 1f0 AxB # y0.

Efn AxBF f0 AxB

Sketching the direction field of a differential equation is particularly easy when the equation is autonomous—that is, the independent variable t does not appear explicitly:

(9)

In Figure 1.17(a) the graph exhibits the direction field for and and some solutions are sketched. Note the following properties of the graphs and explain how they follow from the fact that the equation is autonomous:

(a) The slopes in the direction field are all identical along horizontal lines. (b) New solutions can be generated from old ones by time shifting [i.e., replacing with

]

From observation (a) it follows that the entire direction field can be described by a single direction “line,” as in Figure 1.17(b).

y At ! t0B. y AtB A 7 0,

y¿ " !A Ay ! y1B Ay ! y2B Ay ! y3B2 dy dt # f

A yB . dy/dt " f At, yB

The Phase LineC

†Historical Footnote: This approximation method is a by-product of the famous Picard–Lindelöf existence theorem formulated at the end of the 19th century. ††All references to Chapters 11–13 refer to the expanded text Fundamentals of Differential Equations and Boundary Value Problems, 6th ed.

Group Projects for Chapter 1 33

Of particular interest for autonomous equations are the constant, or equilibrium, solutions The equilibrium is called a stable equilibrium, or sink, because the

neighboring solutions are attracted to it as Equilibria that repel neighboring solutions, like , are known as sources; all other equilibria are called nodes, illustrated by Sources and nodes are unstable equilibria.

(c) Describe how equilibria are characterized by the zeros of the function in equation (1) and how the sink–source–node distinction can be decided on the basis of the signs of on either side of its zeros.

Therefore, the simple phase line depicted in Figure 1.17(c), which indicates with dots and arrows only the zeros and signs of is sufficient to describe the nature of the equilibrium solutions for an autonomous equation.

(d) Sketch the phase line for and state the nature of its equi- libria.

(e) Use the phase line for to predict the asymptotic behavior as of the solution satisfying

(f) Sketch the phase line for sin y and state the nature of its equilibria. (g) Sketch the phase lines for and . Discuss the effect

of the small perturbation on the equilibria.

The splitting of the equilibrium at that you observed in part (g) is an illustration of what is known as bifurcation. The following problem provides a dramatic illustration of the effects of bifurcation, in the context of a herd-management situation.

(h) When the logistic model, to be discussed in Section 3.2, is applied to the existing data for the alligator population on the grounds of Kennedy Space Center in Florida, the following differential equation is derived:

y¿ " ! y Ay ! 1500B

3200 .

y " 0

'0.1 y¿ " y sin y ! 0.1y¿ " y sin y # 0.1

y¿ " y y A0B " 2.1.t S q y¿ " ! Ay ! 1B5/3 Ay ! 2B2 Ay ! 3B

y¿ " Ay ! 1B Ay ! 2B Ay ! 3B f AyB,

f AyB f AyB

y " y3.y " y2 t S q.

y " y1y AtB " yi, i " 1, 2, 3.

(a) (b)

(c)

y y(t − 5)y(t)

Figure 1.17 Direction field and solutions for an autonomous equation

Here is the population and time t is measured in years. If hunters were allowed to thin the population at a rate of s alligators per year, the equation would be modified to

Draw the phase lines for s " 0, 50, 100, 125, 150, 175, and 200. Discuss the signifi- cance of the equilibria. Note the bifurcation at s " 175; should a depletion rate near 175 be avoided?

y¿ " ! y Ay ! 1500B

3200 ! s .

y(t)

34 Chapter 1 Introduction

Newton’s second law states that force is equal to mass times acceleration. We can express this by the equation

where F represents the total force on the object, m is the mass of the object, and is the acceleration, expressed as the derivative of velocity with respect to time. It will be convenient in the future to define y as positive when it is directed downward (as opposed to the analysis in Section 1.1).

Near Earth’s surface, the force due to gravity is just the weight of the objects and is also directed downward. This force can be expressed by mg, where g is the acceleration due to grav- ity. No general law precisely models the air resistance acting on the object, since this force seems to depend on the velocity of the object, the density of the air, and the shape of the object, among other things. However, in some instances air resistance can be reasonably represented by !by, where b is a positive constant depending on the density of the air and the shape of the object. We use the negative sign because air resistance is a force that opposes the motion. The forces acting on the object are depicted in Figure 2.1 on page 36. (Note that we have general- ized the free-fall model in Section 1.1 by including air resistance.)

Applying Newton’s law, we obtain the first-order differential equation

(1)

To solve this equation, we exploit a technique called separation of variables, which was used to analyze the radioactive decay model in Section 1.1 and will be developed in full detail in Section 2.2. Treating dy and dt as differentials, we rewrite equation (1) so as to isolate the variables y and t on opposite sides of the equation:

(Hence, the nomenclature “separation of variables.”)

dy mg ! by

" dt m

m dYdt ! mg " bY .

dy/dt

m dy dt

" F ,

An object falls through the air toward Earth. Assuming that the only forces acting on the object are gravity and air resistance, determine the velocity of the object as a function of time.

First-Order Differential Equations

CHAPTER 2

INTRODUCTION: MOTION OF A FALLING BODY2.1

Next we integrate the separated equation

(2)

and derive

(3)

Therefore,

where the new constant A has magnitude e!bc and the same sign (#) as (mg ! by). Solving for y, we obtain

(4)

which is called a general solution to the differential equation because, as we will see in Section 2.3, every solution to (1) can be expressed in the form given in (4).

In a specific case, we would be given the values of m, g, and b. To determine the constant A in the general solution, we can use the initial velocity of the object y0. That is, we solve the initial value problem

Substituting and into the general solution to the differential equation, we can solve for A. With this value for A, the solution to the initial value problem is

(5) Y ! mg b # aY0 " mgb b e "bt/m .

t " 0y " y0

m dy dt

" mg ! by , y(0) " y0 .

y " mg b

! A b

e!bt/m ,

mg ! by " Ae !bt/m ,

0mg ! by 0 " e !bce !bt/m !

1 b

ln 0mg ! by 0 " t m

$ c .

! dymg ! by " ! dtm

36 Chapter 2 First-Order Differential Equations

–b

Gravity

Velocity

Air resistance

Figure 2.1 Forces on falling object

The preceding formula gives the velocity of the object falling through the air as a func- tion of time if the initial velocity of the object is y0. In Figure 2.2 we have sketched the graph of for various values of y0. It appears from Figure 2.2 that the velocity approaches

regardless of the initial velocity y0. This is easy to see from formula (5) by letting The constant is referred to as the limiting or terminal velocity of the

object. From this model for a falling body, we can make certain observations. Because

rapidly tends to zero, the velocity is approximately the weight, mg, divided by the coeffi- cient of air resistance, b. Thus, in the presence of air resistance, the heavier the object, the faster it will fall, assuming shapes and sizes are the same. Also, when air resistance is less- ened (b is made smaller), the object will fall faster. These observations certainly agree with our experience.

Many other physical problems,† when formulated mathematically, lead to first-order dif- ferential equations or initial value problems. Several of these are discussed in Chapter 3. In this chapter we learn how to recognize and obtain solutions for some special types of first-order equations. We begin by studying separable equations, then linear equations, and then exact equations. The methods for solving these are the most basic. In the last two sections, we illus- trate how devices such as integrating factors, substitutions, and transformations can be used to transform certain equations into either separable, exact, or linear equations that we can solve. Through our discussion of these special types of equations, you will gain insight into the behavior of solutions to more general equations and the possible difficulties in finding these solutions.

A word of warning is in order: In solving differential equations, integration plays an essen- tial role. In particular, the separable equations in Section 2.2 always entail integration, as demonstrated in equations (2) and (3) above. For your convenience, Appendix A reviews three standard techniques for integrating the functions encountered in this text.

e!bt/m

mg/bt S $q. 4 3mg/b y AtBy AtB

Section 2.1 Introduction: Motion of a Falling Body 37

m g b

(m/sec)

t (sec)

object slows down

object speeds up

0 >m g /b, so

0 <m g /b, so

Figure 2.2 Graph of for six different initial velocities y0. (g " 9.8 m/sec2, 5 sec)m/b "y AtB

†The physical problem just discussed has other mathematical models. For example, one could take into account the variations in the gravitational field of Earth and the more general equations for air resistance.

38 Chapter 2 First-Order Differential Equations

Separable Equation

Definition 1. If the right-hand side of the equation

can be expressed as a function that depends only on x times a function that depends only on y, then the differential equation is called separable.†

p AyBg AxB dy dx

" f Ax, yB

†Historical Footnote: A procedure for solving separable equations was discovered implicitly by Gottfried Leibniz in 1691. The explicit technique called separation of variables was formalized by John Bernoulli in 1694.

In other words, a first-order equation is separable if it can be written in the form

For example, the equation

is separable, since (if one is sufficiently alert to detect the factorization)

However, the equation

admits no such factorization of the right-hand side and so is not separable. Informally speaking, one solves separable equations by performing the separation and

then integrating each side.

dy dx

" 1 $ xy

2x $ xy

y2 $ 1 " x

2 $ y

y2 $ 1 " g AxB p A yB .

dx "

2x $ xy

y2 $ 1

dy dx ! g

AxB p A yB .

A simple class of first-order differential equations that can be solved using integration is the class of separable equations. These are equations

(1)

that can be rewritten to isolate the variables x and y (together with their differentials dx and dy) on opposite sides of the equation, as in

So the original right-hand side must have the factored form

More formally, we write and present the following definition.p A yB " 1/h A yB f Ax, yB " g AxB # 1

h A yB . f Ax, yBh A yB dy " g AxB dx .

dy dx

" f Ax, yB , 2.2 SEPARABLE EQUATIONS

Caution: Constant functions y c such that p(c) " 0 are also solutions to (2), which may or may not be included in (3) (as we shall see in Example 3).

We will look at the mathematical justification of this “streamlined” procedure shortly, but first we study some examples.

Solve the nonlinear equation

Following the streamlined approach, we separate the variables and rewrite the equation in the form

Integrating, we have

and solving this last equation for y gives

Since C is a constant of integration that can be any real number, 3C can also be any real num- ber. Replacing 3C by the single symbol K, we then have

If we wish to abide by the custom of letting C represent an arbitrary constant, we can go one step further and use C instead of K in the final answer. This solution family is graphed in Figure 2.3 on page 40. ◆

As Example 1 attests, separable equations are among the easiest to solve. However, the procedure does require a facility for computing integrals. Many of the procedures to be dis- cussed in the text also require a familiarity with the techniques of integration. For this reason

y " a3x2 2

! 15x $ Kb 1/3 . y " a3x2

2 ! 15x $ 3Cb 1/3 .

3 "

2 ! 5x $ C ,

! y2 dy " ! Ax ! 5B dx y2 dy " Ax ! 5B dx . dy

dx "

x ! 5

y2 .

Section 2.2 Separable Equations 39

Method for Solving Separable Equations To solve the equation

(2)

multiply by dx and by to obtain

Then integrate both sides:

(3)

where we have merged the two constants of integration into a single symbol C. The last equation gives an implicit solution to the differential equation.

H A yB " G AxB $ C ,!h A yB dy " !g AxB dx , h A yB dy " g AxB dx .h A yB J 1/p A yB

dy dx

" g AxB p A yB

Solution

Example 1

we have provided a review of integration methods in Appendix A and a brief table of integrals on the inside front cover.

Solve the initial value problem

(4)

Separating the variables and integrating gives

(5)

At this point, we can either solve for y explicitly (retaining the constant C ) or use the initial condition to determine C and then solve explicitly for y. Let’s try the first approach.

Exponentiating equation (5), we have

(6)

where †† Now, depending on the values of y, we have ; and similarly, . Thus, (6) can be written as

or y " 1 # C1 Ax $ 3B ,y ! 1 " #C1 Ax $ 3B 0 x $ 3 0 " # Ax $ 3B 0 y ! 1 0 " # A y ! 1BC1 J eC.

0 y ! 1 0 " eC 0 x $ 3 0 " C1 0 x $ 3 0 , e ln 0 y!1 0 " e ln 0 x$3 0$C " eCe ln 0 x$3 0 ,

ln 0 y ! 1 0 " ln 0 x $ 3 0 $ C . ! dy

y ! 1 " ! dxx $ 3 ,

y ! 1 "

dx x $ 3

dy dx

" y ! 1 x $ 3

, y A!1B " 0 .

40 Chapter 2 First-Order Differential Equations

1 1

K = 24

K = 0 K = −12

K = −24 K = 12

Figure 2.3 Family of solutions for Example 1†

Example 2

Solution

†The gaps in the curves reflect the fact that in the original differential equation, y appears in the denominator, so that y " 0 must be excluded. ††Recall that the symbol means “is defined to be.”J

where the choice of sign depends (as we said) on the values of x and y. Because C1 is a positive constant (recall that C1 " e

C % 0), we can replace #C1 by K, where K now represents an arbitrary nonzero constant. We then obtain

(7)

Finally, we determine K such that the initial condition is satisfied. Putting and in equation (7) gives

and so . Thus, the solution to the initial value problem is

(8)

Alternative Approach. The second approach is to first set x " !1 and y " 0 in equation (5) and solve for C. In this case, we obtain

and so C " !ln 2. Thus, from (5), the solution y is given implicitly by

Here we have replaced y ! 1 by 1 ! y and x $ 3 by x $ 3, since we are interested in x and y near the initial values x " !1, y " 0 (for such values, y ! 1 & 0 and x $ 3 % 0). Solving for y, we find

which agrees with the solution (8) found by the first method. ◆

Solve the nonlinear equation

(9)

Separating variables and integrating, we find

At this point, we reach an impasse. We would like to solve for y explicitly, but we cannot. This is often the case in solving nonlinear first-order equations. Consequently, when we say “solve the equa- tion,” we must on occasion be content if only an implicit form of the solution has been found. ◆

The separation of variables technique, as well as several other techniques discussed in this book, entails rewriting a differential equation by performing certain algebraic operations on it.

sin y $ ey " x6 ! x2 $ x $ C .

! Acos y $ eyB dy " ! A6x5 ! 2x $ 1B dx , Acos y $ eyB dy " A6x5 ! 2x $ 1B dx ,

dx "

6x5 ! 2x $ 1

cos y $ ey .

y " 1 ! 1 2

Ax $ 3B " ! 1 2

Ax $ 1B , 1 ! y "

x $ 3 2

ln A1 ! yB " ln Ax $ 3B ! ln 2 " lnax $ 3 2 b ,

0000ln A1 ! yB " ln Ax $ 3B ! ln 2 . 0 " ln 1 " ln 2 $ C ,

ln 00 ! 1 0 " ln 0!1 $ 3 0 $ C , y " 1 !

1 2

Ax $ 3B " ! 1 2

Ax $ 1B .K " !1/2 0 " 1 $ K A!1 $ 3B " 1 $ 2K ,y " 0

x " !1y A!1B " 0y " 1 $ K Ax $ 3B .

Section 2.2 Separable Equations 41

Example 3

Solution

“Rewriting as ” amounts to dividing both sides by . You may recall from your algebra days that doing this can be treacherous. For example, the equation has two solutions: x " 2 and x " 4. But if we “rewrite” the equation as x " 4 by dividing both sides by , we lose track of the root x " 2. Thus, we should record the zeros of itself before dividing by this factor.

By the same token we must take note of the zeros of in the separable equation prior to dividing. After all, if (say) , then

observe that the constant function solves the differential equation :

Indeed, in solving the equation of Example 2,

we obtained as the set of solutions, where K was a nonzero constant (since K replaced #eC). But notice that the constant function (which in this case corresponds to K " 0) is also a solution to the differential equation. The reason we lost this solution can be traced back to a division by y ! 1 in the separation process. (See Problem 30 for an example of where a solution is lost and cannot be retrieved by setting the constant K " 0.)

Formal Justification of Method We close this section by reviewing the separation of variables procedure in a more rigorous framework. The original differential equation (2) is rewritten in the form

(10)

where Letting and denote antiderivatives (indefinite integrals) of and , respectively—that is,

we recast equation (10) as

By the chain rule for differentiation, the left-hand side is the derivative of the composite function :

Thus, if is a solution to equation (2), then and are two functions of x that have the same derivative. Therefore, they differ by a constant:

(11)

Equation (11) agrees with equation (3), which was derived informally, and we have thus verified that the latter can be used to construct implicit solutions.

H Ay AxBB " G AxB $ C . G AxBH Ay AxBBy AxB d dx

H Ay AxBB " H¿ Ay AxBB dydx .H Ay AxBB H¿ A yB dy

dx " G¿ AxB .

H¿ A yB " h A yB , G¿ AxB " g AxB ,g AxBh A yB G AxBH A yBh A yB J 1/p A yB.

h A yB dy dx

" g AxB ,

y " 1 y " 1 $ K Ax $ 3B

dy dx

" y ! 1 x $ 3

g AxBp A yB " Ax ! 2B2 A13 ! 13B " 0 . dy dx

" d A13B

dx " 0 ,

dy/dx " g AxBp A yBy(x) " 13 g AxBp A yB " Ax ! 2B2 A y ! 13Bg AxBp A yBdy/dx " p A yBAx ! 2B

Ax ! 2Bx Ax ! 2B " 4 Ax ! 2B p A yBh A yB dy " g AxB dxdy/dx " g AxBp A yB

42 Chapter 2 First-Order Differential Equations

In Problems 1–6, determine whether the given differen- tial equation is separable.

1. 2.

3. 4.

In Problems 7–16, solve the equation.

7. 8.

9. 10.

11. 12.

13. 14.

15. 16.

In Problems 17–26, solve the initial value problem. 17.

18.

19.

20.

21.

22.

23.

24.

25.

26. 2y dx $ A1 $ xB dy " 0 , y A0B " 1dydx " x2 A1 $ yB , y A0B " 3 dy dx

" 8x3e!2y , y A1B " 0 t!1

dy dt

" 2 cos 2y , y (0) " p/4

x2 dx $ 2y dy " 0 , y A0B " 2 dy du

" y sin u

y2 $ 1 , y(p) " 11

x2 dy dx

" 4x2 ! x ! 2

(x $ 1)(y $ 1) , y(1) " 1

1 2

dy dx

" 2y $ 1 cos x , y(p) " 0 dy dx

" A1 $ y2Btan x , y A0B " 23y¿ " x3 A1 ! yB , y A0B " 3 Ax $ xy2B dx $ ex2y dy " 0y!1 dy $ yecos x sin x dx " 0

dy dx

" 3x2(1 $ y2)3/2 dx dt

! x3 " x

x dy dx

" 1 ! 4y2

3y dy

dx "

sec2y

1 $ x2

dx dt

" t

xet$2x dy dx

" x

y221 $ x x dy dx

" 1 y3

dx dt

" 3xt2

s2 $ ds dt

" s $ 1

Axy2 $ 3y2B dy ! 2x dx " 0 dy

dx "

yex$y

x2 $ 2 ds dt

" t ln As2tB $ 8t2 dy dx

" 4y2 ! 3y $ 1 dy dx

! sin Ax $ yB " 0

Section 2.2 Separable Equations 43

27. Solutions Not Expressible in Terms of Elemen- tary Functions. As discussed in calculus, certain indefinite integrals (antiderivatives) such as cannot be expressed in finite terms using elementary functions. When such an integral is encountered while solving a differential equation, it is often helpful to use definite integration (integrals with variable upper limit). For example, consider the initial value problem

The differential equation separates if we divide by y2

and multiply by dx. We integrate the separated equa- tion from x " 2 to x " x1 and find

If we let t be the variable of integration and replace x1 by x and by 1, then we can express the solu- tion to the initial value problem by

Use definite integration to find an explicit solution to the initial value problems in parts (a)–(c).

(a) (b)

Simpson’s rule, described in Appendix C) to approximate the solution to part (b) at x " 0.5 to three decimal places.

28. Sketch the solution to the initial value problem

and determine its maximum value.

29. Uniqueness Questions. In Chapter 1 we indicated that in applications most initial value problems will have a unique solution. In fact, the existence of unique

dy dt

" 2y ! 2yt , y A0B " 3

dy/dx " 21 $ sin x A1 $ y2B , y A0B " 1dy/dx " ex 2y!2 , y A0B " 1dy/dx " e x 2 , y A0B " 0

y AxB " a1 ! ! x 2

et 2 dtb!1 .

y A2B " !

1 y Ax1B $ 1y A2B .

" ! 1 y

` x"x1 x"2

! x"x1

x"2 ex

2 dx " !

x"x1

x"2 dy

dy dx

" ex 2 y2 , y A2B " 1 .

# ex 2 dx

2.2 EXERCISES

tions depend on the initial conditions. For the ini- tial value problem with a, a % 0, show that as a approaches zero from the right the domain approaches the whole real line

and as a approaches the domain shrinks to a single point.

(e) Sketch the solutions to the initial value problem with a for , #1,

and #2.

32. Analyze the solution to the initial value problem

using approximation methods and then compare with its exact form as follows. (a) Sketch the direction field of the differential

equation and use it to guess the value of lim .

(b) Use Euler’s method with a step size of 0.1 to find an approximation of .

(d) What is the exact value of ? Compare with your approximation in part (b).

(e) Using the exact solution obtained in part (c), determine lim and compare with your guess in part (a).

33. Mixing. Suppose a brine containing 0.3 kilogram (kg) of salt per liter (L) runs into a tank initially filled with 400 L of water containing 2 kg of salt. If the brine enters at 10 L/min, the mixture is kept uni- form by stirring, and the mixture flows out at the same rate. Find the mass of salt in the tank after 10 min (see Figure 2.4). [Hint: Let A denote the number of kilograms of salt in the tank at t min after the process begins and use the fact that

rate of increase in A ! rate of input " rate of exit.

A further discussion of mixing problems is given in Section 3.2.]

xSq f AxB f(1)

f(x)f(x) f(1)

xSq f AxB

dy dx

" y2 ! 3y $ 2 , y A0B " 1.5 y " f(x)

a " #1/2y A0B "dy/dx " xy3 $qA!q, q B

y A0B "dy/dx " xy3 solutions was so important that we stated an exis- tence and uniqueness theorem, Theorem 1, page 11. The method for separable equations can give us a solution, but it may not give us all the solutions (also see Problem 30). To illustrate this, consider the equation . (a) Use the method of separation of variables to

show that

is a solution. (b) Show that the initial value problem

with is satisfied for by

for x ' 0. (c) Now show that the constant function also

satisfies the initial value problem given in part (b). Hence, this initial value problem does not have a unique solution.

(d) Finally, show that the conditions of Theorem 1 on page 11 are not satisfied.

(The solution was lost because of the division by zero in the separation process.)

30. As stated in this section, the separation of equation (2) on page 39 requires division by , and this may disguise the fact that the roots of the equation

"0 are actually constant solutions to the differ- ential equation. (a) To explore this further, separate the equation

to derive the solution,

(b) Show that satisfies the original equation .

(c) Show that there is no choice of the constant C that will make the solution in part (a) yield the solution Thus, we lost the solution

when we divided by .

31. Interval of Definition. By looking at an initial value problem with , it is not always possible to determine the domain of the solution or the interval over which the function

satisfies the differential equation. (a) Solve the equation . (b) Give explicitly the solutions to the initial

value problem with 1; ; 2.y A0B " 1/2y A0B "y A0B "

dy/dx " xy3 y AxB y AxB

y Ax0B " y0dy/dx " f Ax, yB A y $ 1B2/3y " !1 y " !1.

dy/dx " Ax ! 3B A y $ 1B2/3y " !1 y " !1 $ Ax2/6 ! x $ CB3 . dy dx

" Ax ! 3B A y $ 1B2/3 p(y)

p A yB y " 0

y " 0 y " A2x/3B3/2 C " 0y A0B " 0y1/3

dy/dx "

y " a2x 3

$ Cb 3/2 dy/dx " y1/3

44 Chapter 2 First-Order Differential Equations

A(t)

400 L

A( 0) = 2 kg

10 L/min 0.3 kg/L

10 L/min

Figure 2.4 Schematic representation of a mixing problem

Section 2.2 Separable Equations 45

34. Newton’s Law of Cooling. According to New- ton’s law of cooling, if an object at temperature T is immersed in a medium having the constant tempera- ture M, then the rate of change of T is proportional to the difference of temperature M ! T. This gives the differential equation

(a) Solve the differential equation for T. (b) A thermometer reading 100(F is placed in a

medium having a constant temperature of 70(F. After 6 min, the thermometer reads 80(F. What is the reading after 20 min?

(Further applications of Newton’s law of cooling appear in Section 3.3.)

35. Blood plasma is stored at 40(F. Before the plasma can be used, it must be at 90(F. When the plasma is placed in an oven at 120(F, it takes 45 min for the plasma to warm to 90(F. Assume Newton’s law of cooling (Problem 34) applies. How long will it take for the plasma to warm to 90(F if the oven tempera- ture is set at (a) 100(F, (b) 140(F, and (c) 80(F?

36. A pot of boiling water at 100(C is removed from a stove at time t " 0 and left to cool in the kitchen. After 5 min, the water temperature has decreased to 80(C, and another 5 min later it has dropped to 65(C. Assum- ing Newton’s law of cooling (Problem 34) applies, determine the (constant) temperature of the kitchen.

37. Compound Interest. If is the amount of dol- lars in a savings bank account that pays a yearly interest rate of r% compounded continuously, then

t in years.

Assume the interest is 5% annually, $1000, and no monies are withdrawn. (a) How much will be in the account after 2 yr? (b) When will the account reach $4000? (c) If $1000 is added to the account every 12

months, how much will be in the account after yr?

38. Free Fall. In Section 2.1, we discussed a model for an object falling toward Earth. Assuming that only air resistance and gravity are acting on the object, we found that the velocity y must satisfy the equation

where m is the mass, g is the acceleration due to gravity, and b % 0 is a constant (see Figure 2.1).

m dY dt

! mg " bY ,

312

P A0B " dP dt !

r 100 P ,

P AtB

dT/dt ! k AM " TB .

If m " 100 kg, g " 9.8 m sec2, b " 5 kg sec, and 10 m sec, solve for . What is the limit-

ing (i.e., terminal) velocity of the object? 39. Grand Prix Race. Driver A had been leading

archrival B for a while by a steady 3 miles. Only 2 miles from the finish, driver A ran out of gas and decelerated thereafter at a rate proportional to the square of his remaining speed. One mile later, driver A’s speed was exactly halved. If driver B’s speed remained constant, who won the race?

40. The atmospheric pressure (force per unit area) on a surface at an altitude z is due to the weight of the column of air situated above the surface. Therefore, the difference in air pressure p between the top and bottom of a cylindrical volume element of height )z and cross-section area A equals the weight of the air enclosed (density times volume times gravity g), per unit area:

Let to derive the differential equation . To analyze this further we must pos-

tulate a formula that relates pressure and density. The perfect gas law relates pressure, volume, mass m, and absolute temperature T according to , where R is the universal gas constant and M is the molar mass of the air. Therefore, density and pressure are related by .

(a) Derive the equation and solve it

for the “isothermal” case where T is constant to obtain the barometric pressure equation p z " p z0 exp[!Mg z!z0 RT].

(b) If the temperature also varies with altitude , derive the solution

(c) Suppose an engineer measures the barometric pressure at the top of a building to be 99,000 Pa (pascals), and 101,000 Pa at the base z " z0 . If the absolute temperature varies as

, determine the height of the building. Take R " 8.31 N-m mol-K, M " 0.029 kg mol, and g " 9.8 m sec2. (An amusing story concerning this problem can be found at http://www.snopes.com/college/exam/ barometer.asp)

// /

T AzB " 288 ! 0.0065 Az ! z0B BA

p AzB " p Az0B exp e!MgR !z z0

T AzB f T " T AzB /

BABABA dp dz

" ! Mg RT

r :" m/V " Mp/RT

pV " mRT/M

dp/dz " !rg ¢z S 0

p Az $ ¢zB ! p AzB " !r AzB AA¢zBg A

" !r AzBg¢z V " A¢zr

y AtB/y A0B " //

http://www.snopes.com/college/exam/barometer.asp

A type of first-order differential equation that occurs frequently in applications is the linear equation. Recall from Section 1.1 that a linear first-order equation is an equation that can be expressed in the form

(1)

where , , and depend only on the independent variable x, not on y. For example, the equation

is linear, because it can be rewritten in the form

However, the equation

is not linear; it cannot be put in the form of equation (1) due to the presence of the and terms.

There are two situations for which the solution of a linear differential equation is quite immediate. The first arises if the coefficient is identically zero, for then equation (1) reduces to

(2)

which is equivalent to

as long as is not zero . The second is less trivial. Note that if happens to equal the derivative of —that

is, —then the two terms on the left-hand side of equation (1) simply comprise the derivative of the product :

Therefore equation (1) becomes

(3) d dx

3a1 AxBy 4 " b AxB a1 AxBy¿ $ a0 AxBy " a1 AxBy¿ $ a¿1 AxBy " ddx 3a1 AxBy 4 .

a1 AxBya0 AxB " a¿1 AxB a1 AxBa0 AxB4a1 AxB3

y AxB " ! b AxBa1 AxB dx $ C

a1 AxB dydx " b AxB , a0 AxB

y dy/dx y3

y dy dx

$ Asin xBy3 " ex $ 1 Asin xB dy

dx $ Acos xBy " x2sin x .

x2sin x ! Acos xBy " Asin xB dy dx

b AxBa0 AxBa1 AxB a1 AxB dydx # a0 AxBy ! b AxB ,

46 Chapter 2 First-Order Differential Equations

LINEAR EQUATIONS2.3

and the solution is again elementary:

One can seldom rewrite a linear differential equation so that it reduces to a form as simple as (2). However, the form (3) can be achieved through multiplication of the original equation (1) by a well-chosen function . Such a function is then called an “integrating factor” for equation (1). The easiest way to see this is first to divide the original equation (1) by and put it into standard form

(4)

where and Next we wish to determine so that the left-hand side of the multiplied equation

(5)

is just the derivative of the product :

Clearly, this requires that m satisfy

(6)

To find such a function, we recognize that equation (6) is a separable differential equation, which we can write as Integrating both sides gives

(7)

With this choice† for , equation (5) becomes

which has the solution

(8)

Here C is an arbitrary constant, so (8) gives a one-parameter family of solutions to (4). This form is known as the general solution to (4).

y AxB " 1 m AxB c !m AxBQ AxB dx $ C d .

d dx

3m AxBy 4 " m AxBQ AxB ,m AxB M AxB " e# PAxB dx .

A1/mB dm " P AxBdx. m¿ " mP .

m AxB dy dx

$ M AxBP AxBy " d dx

3m AxBy 4 " m AxB dy dx

$ M$ AxBy . m AxBy

m AxB dy dx

$ m AxBP AxBy " m AxBQ AxB m AxBQ AxB " b AxB /a1 AxB.P AxB " a0 AxB /a1 AxB

dy dx # P

AxBy ! Q AxB , a1 AxBm AxBm AxB

y AxB " 1 a1 AxB c !b AxB dx $ C d .

a1 AxBy " !b AxB dx $ C ,

Section 2.3 Linear Equations 47

†Any choice of the integration constant in will produce a suitable .m AxB# P AxB dx

Find the general solution to

(9)

To put this linear equation in standard form, we multiply by x to obtain

(10)

Here , so

Thus, an integrating factor is

Multiplying equation (10) by yields

d dx

Ax!2yB " cos x . x!2

dy dx

! 2x!3y " cos x ,

m AxBm AxB " e !2 ln 0 x 0 " e lnAx!2B " x!2 .

!P AxB dx " !!2x dx " !2 ln 0 x 0 . P AxB " !2/x

dy dx

! 2 x

y " x2cos x .

x dy

dx !

x2 " x cos x , x 7 0 .

48 Chapter 2 First-Order Differential Equations

Example 1

Solution

µ Method for Solving Linear Equations (a) Write the equation in the standard form

(b) Calculate the integrating factor by the formula

side is just obtain

(d) Integrate the last equation and solve for y by dividing by to obtain (8).m AxB d dx

3m AxBy 4 " m AxBQ AxB . m AxB dydx $ P AxBm AxBy " m AxBQ AxB , d dx 3m AxBy 4 , m AxB

m(x) " exp c!P(x)dx d . m AxB dy dx

$ P AxBy " Q AxB .

We can summarize the method for solving linear equations as follows.

We now integrate both sides and solve for y to find

(11)

It is easily checked that this solution is valid for all x % 0. In Figure 2.5 we have sketched solu- tions for various values of the constant C in (11). ◆

y " x2 sin x $ Cx2 .

x!2y " ! cos x dx " sin x $ C

Section 2.3 Linear Equations 49

Example 2

Solution

C = − 1 / 2

100

−100

−200

2 4 6 8 10 12

C = 1 C = 1 / 2 C = 0

C = −1

Figure 2.5 Graph of for five values of the constant Cy " x2 sin x $ Cx2

In the next example, we encounter a linear equation that arises in the study of the radioac- tive decay of an isotope.

A rock contains two radioactive isotopes, RA1 and RA2, that belong to the same radioactive series; that is, RA1 decays into RA2, which then decays into stable atoms. Assume that the rate at which RA1 decays into RA2 is kg/sec. Because the rate of decay of RA2 is propor- tional to the mass of RA2 present, the rate of change in RA2 is

(12)

where k % 0 is the decay constant. If k " and initially 40 kg, find the mass of RA2 for

Equation (12) is linear, so we begin by writing it in standard form

(13)

where we have substituted k " 2 and displayed the initial condition. We now see that , so Thus, an integrating factor is Multiplying equation (13) by yields

d dt

Ae2tyB " 50e!8t . e2t

dy dt

$ 2e2ty " 50e!10t$2t " 50e!8t ,

m AtB m AtB " e2t.# P AtBdt " # 2 dt " 2t. P AtB " 2

dy dt

$ 2y " 50e!10t , y A0B " 40 , t ' 0.

y AtBy A0B "2/sec dy dt

" 50e!10t ! ky ,

dy dt

" rate of creation ! rate of decay ,

y AtB 50e!10t

Integrating both sides and solving for y, we find

Substituting t " 0 and y(0) " 40 gives

so . Thus, the mass of RA2 at time t is given by

(14) ◆

For the initial value problem

find the value of .

The integrating factor for the differential equation is, from equation (7),

The general solution form (8) thus reads

However, this indefinite integral cannot be expressed in finite terms with elementary functions (recall a similar situation in Problem 27 of Exercises 2.2). Because we can use numerical algo- rithms such as Simpson’s rule (Appendix C) to perform definite integration, we revert to the form (5), which in this case reads

and take the definite integral from the initial value x " 1 to the desired value x " 2:

Inserting the given value of and solving, we express

Using Simpson’s rule, we find that the definite integral is approximately 4.841, so

◆y A2B $ 4e!1 $ 4.841e!2 $ 2.127 . y A2B " e!2$1 A4B $ e!2 ! 2

1 ex21 $ cos2x dx .y A1B

exy ` x"2 x"1

" e2y A2B ! e1y A1B " ! x"2 x"1

ex21 $ cos2x dx . d dx

AexyB " ex21 $ cos2x ,

y AxB " e!x a! ex21 $ cos2x dx $ Cb . m AxB " e# 1 dx " ex .

y A2By¿ $ y " 21 $ cos2x , y A1B " 4 , y AtB " a185

4 b e!2t ! a25

4 b e!10t , t ' 0 .y AtBC " 40 $ 25/4 " 185/4

40 " ! 25 4

e0 $ Ce0 " ! 25 4

$ C ,

y " ! 25 4

e!10t $ Ce!2t .

e2ty " ! 25 4

e!8t $ C ,

50 Chapter 2 First-Order Differential Equations

Example 3

Solution

In Example 3 we had no difficulty expressing the integral for the integrating factor Clearly, situations will arise where this integral, too, cannot be expressed

with elementary functions. In such cases we must again resort to a numerical procedure such as Euler’s method (Section 1.4) or to a “nested loop” implementation of Simpson’s rule. You are invited to explore such a possibility in Problem 27.

Because we have established explicit formulas for the solutions to linear first-order differ- ential equations, we get as a dividend a direct proof of the following theorem.

m AxB " e# 1 dx " ex. Section 2.3 Linear Equations 51

Existence and Uniqueness of Solution

Theorem 1. Suppose and are continuous on an interval that contains the point x0. Then for any choice of initial value y0, there exists a unique solution on to the initial value problem

(15)

In fact, the solution is given by (8) for a suitable value of C.

dy dx

$ P AxBy " Q AxB , y Ax0B " y0 . Aa, bB y AxB

Aa, bBQ AxBP AxB

The essentials of the proof of Theorem 1 are contained in the deliberations leading to equa- tion (8); Problem 34 provides the details. This theorem differs from Theorem 1 on page 11 in that for the linear initial value problem (15), we have the existence and uniqueness of the solution on the whole interval , rather than on some smaller unspecified interval about x0.

The theory of linear differential equations is an important branch of mathematics not only because these equations occur in applications but also because of the elegant structure associ- ated with them. For example, first-order linear equations always have a general solution given by equation (8). Some further properties of first-order linear equations are described in Prob- lems 28 and 36. Higher-order linear equations are treated in Chapters 4, 6, and 8.

Aa, bB

In Problems 1–6, determine whether the given equation is separable, linear, neither, or both.

1. 2.

3. 4.

5. 6.

In Problems 7–16, obtain the general solution to the equation.

7. 8. dy dx

! y ! e3x " 0 dy dx

" y x

$ 2x $ 1

x dx dt

$ t2x " sin t3r " dr du

! u3

At2 $ 1B dy dt

" yt ! y3t " et dy dt

$ y ln t

x2 dy dx

$ sin x ! y " 0 dx dt

$ xt " ex

9. 10.

11. 12.

13.

14.

15.

16. A1 ! x2Bdy dx

! x2y " A1 $ xB21 ! x2Ax2 $ 1B dy dx

$ xy ! x " 0

x dy dx

$ 3 Ay $ x2B " sin x x

y dx dy

$ 2x " 5y3

dy dx

" x2e!4x ! 4yAt $ y $ 1B dt ! dy " 0 dr du

$ r tan u " sec ux dy dx

$ 2y " x!3

2.3 EXERCISES

In Problems 17–22, solve the initial value problem.

17.

18.

19.

20.

21.

22.

23. Radioactive Decay. In Example 2 assume that the rate at which RA1 decays into RA2 is kg/sec and the decay constant for RA2 is k " 5/sec. Find the mass of RA2 for if initially 10 kg.

24. In Example 2 the decay constant for isotope RA1 was 10/sec, which expresses itself in the exponent of the rate term kg/sec. When the decay constant for RA2 is k " 2/sec, we see that in formula (14) for y the term eventually dominates (has greater magnitude for t large). (a) Redo Example 2 taking k " 20/sec. Now which

term in the solution eventually dominates? (b) Redo Example 2 taking k " 10/sec.

25. (a) Using definite integration, show that the solution to the initial value problem

can be expressed as

(b) Use numerical integration (such as Simpson’s rule, Appendix C) to approximate the solution at x " 3.

26. Use numerical integration (such as Simpson’s rule, Appendix C) to approximate the solution, at x " 1, to the initial value problem

Ensure your approximation is accurate to three deci- mal places.

dx $

sin 2x

2 A1 $ sin2xB y " 1 , y A0B " 0 .

y AxB " e!x 2 ae4 $ ! x 2

et 2 dtb .

dy dx

$ 2xy " 1 , y A2B " 1 ,

A185/4Be!2t 50e!10t

y A0B "t ' 0y AtB 40e!20t

sin x dy dx

$ y cos x " x sin x , y ap 2 b " 2

y ap 4 b " !1522p2

cos x dy dx

$ y sin x " 2x cos2x ,

dy dx

$ 3y x

$ 2 " 3x , y A1B " 1 t 2

dx dt

$ 3tx " t 4 ln t $ 1, x(1) " 0

dy dx

$ 4y ! e!x " 0 , y A0B " 4 3

dy dx

! y x

" xex , y A1B " e ! 1

52 Chapter 2 First-Order Differential Equations

27. Consider the initial value problem

(a) Using definite integration, show that the inte- grating factor for the differential equation can be written as

and that the solution to the initial value problem is

(b) Obtain an approximation to the solution at x " 1 by using numerical integration (such as Simp- son’s rule, Appendix C) in a nested loop to esti- mate values of and, thereby, the value of

[Hint: First, use Simpson’s rule to approximate at x " 0.1, 0.2, . . . , 1. Then use these

values and apply Simpson’s rule again to approximate .]

[A direct comparison of the merits of the two numer- ical schemes in parts (b) and (c) is very complicated, since it should take into account the number of func- tional evaluations in each algorithm as well as the inherent accuracies.]

28. Constant Multiples of Solutions. (a) Show that is a solution of the linear

equation

(16)

and is a solution of the nonlinear equation

(17)

(b) Show that for any constant the function is a solution of equation (16), while is a solution of equation (17) only when C " 0 or 1.

if is a solution, then for any constant C the function is also a solution.C ŷ AxBŷ AxB dy dx

$ P AxBy " 0 , Cx!1

Ce!xC,

dy dx

$ y2 " 0 .

y " x!1

dy dx

$ y " 0 ,

y " e!x

#10 m AsB s ds m AxB !

0 m AsB s ds .

m AxB y AxB " 1

m AxB ! x0 m AsB s ds $ 2m AxB . m AxB " expa! x

0 21 $ sin2t dtb

dy dx

$ 21 $ sin2x y " x , y A0B " 2 .

29. Use your ingenuity to solve the equation

[Hint: The roles of the independent and dependent variables may be reversed.]

30. Bernoulli Equations. The equation

(18)

is an example of a Bernoulli equation. (Further dis- cussion of Bernoulli equations is in Section 2.6.) (a) Show that the substitution reduces equa-

tion (18) to the equation

(19)

(b) Solve equation (19) for y. Then make the substi- tution to obtain the solution to equation (18).

31. Discontinuous Coefficients. As we will see in Chapter 3, occasions arise when the coefficient in a linear equation fails to be continuous because of jump discontinuities. Fortunately, we may still obtain a “reasonable” solution. For example, consider the initial value problem

where

(a) Find the general solution for . (b) Choose the constant in the solution of part (a) so

that the initial condition is satisfied. (c) Find the general solution for x > 2. (d) Now choose the constant in the general solution

from part (c) so that the solution from part (b) and the solution from part (c) agree at x " 2. By patching the two solutions together, we can obtain a continuous function that satisfies the differential equation except at x " 2, where its derivative is undefined.

(e) Sketch the graph of the solution from x " 0 to x " 5.

32. Discontinuous Forcing Terms. There are occa- sions when the forcing term in a linear equation fails to be continuous because of jump discontinuities. Fortunately, we may still obtain a reasonable solution

Q AxB

0 * x * 2

P AxB J e1 , 0 * x * 2 , 3 , x 7 2 .

dy dx

$ P AxBy " x , y A0B " 1 ,

P AxB y " y3

dy dx

$ 6y " 3x .

y " y3

dy dx

$ 2y " xy!2

dx "

e4y $ 2x .

Section 2.3 Linear Equations 53

imitating the procedure discussed in Problem 31. Use this procedure to find the continuous solution to the initial value problem.

where

Sketch the graph of the solution from x " 0 to x " 7. 33. Singular Points. Those values of x for which

in equation (4) is not defined are called singular points of the equation. For example, x " 0 is a singu- lar point of the equation since when the equation is written in the standard form,

we see that is not defined at x " 0. On an interval containing a singular point, the questions of the existence and uniqueness of a solution are left unanswered, since Theorem 1 does not apply. To show the possible behavior of solutions near a singular point, consider the following equations. (a) Show that has only one solution

defined at x " 0. Then show that the initial value problem for this equation with initial condition

has a unique solution when and no solution when

(b) Show that has an infinite num- ber of solutions defined at x " 0. Then show that the initial value problem for this equation with initial condition has an infinite number of solutions.

34. Existence and Uniqueness. Under the assump- tions of Theorem 1, we will prove that equation (8) gives a solution to equation (4) on . We can then choose the constant C in equation (8) so that the initial value problem (15) is solved. (a) Show that since is continuous on ,

then defined in (7) is a positive, continuous function satisfying on .

(b) Since

verify that y given in equation (8) satisfies equation (4) by differentiating both sides of equation (8).

) and choose C to be the initial condition is satisfied.y Ax0B " y0 y0 m Ax0B,#xx0 m AtBQ AtB dt

# m AxBQ AxB dx d dx

!m AxBQ AxB dx " m AxBQ AxB , Aa, bBdm/dx " P AxBm(x)m AxB Aa, bBP AxB

Aa, bB y A0B " 0

xy¿ ! 2y " 3x y0 + 0.

y0 " 0y A0B " y0 xy¿ $ 2y " 3x

P AxB " 2/xy¿ $ A2/xBy " 3, xy¿ $ 2y " 3x,

P AxB Q AxB J e 2 , 0 * x * 3 ,

!2 , x 7 3 .

dy dx

$ 2y " Q AxB , y A0B " 0 ,

(d) Start with the assumption that is a solution to the initial value problem (15) and argue that the discussion leading to equation (8) implies that must obey equation (8). Then argue that the initial condition in (15) determines the constant C uniquely.

35. Mixing. Suppose a brine containing 0.2 kg of salt per liter runs into a tank initially filled with 500 L of water containing 5 kg of salt. The brine enters the tank at a rate of 5 L/min. The mixture, kept uniform by stirring, is flowing out at the rate of 5 L/min (see Figure 2.6).

y AxB y AxB

54 Chapter 2 First-Order Differential Equations

idea that just by knowing the form of the solution, we can substitute into the given equation and solve for any unknowns. Here we illustrate the method for first-order equations (see Sections 4.6 and 6.4 for the generalization to higher-order equations). (a) Show that the general solution to

(20)

has the form

where yh ( ) is a solution to equation (20) when C is a constant, and "

for a suitable function . [Hint: Show that we can take and then use equation (8).]

We can in fact determine the unknown func- tion yh by solving a separable equation. Then direct substitution of yyh in the original equation will give a simple equation that can be solved for y.

Use this procedure to find the general solution to

(21)

by completing the following steps: (b) Find a nontrivial solution yh to the separable

equation

(22)

(21), and simplify to obtain (d) Now integrate to get . (e) Verify that is a gen-

eral solution to (21). 37. Secretion of Hormones. The secretion of hor-

mones into the blood is often a periodic activity. If a hormone is secreted on a 24-h cycle, then the rate of change of the level of the hormone in the blood may be represented by the initial value problem

where is the amount of the hormone in the blood at time t, is the average secretion rate, is the amount of daily variation in the secretion, and k is a positive constant reflecting the rate at which the body removes the hormone from the blood. If

and x0 " 10, solve for .x AtBa " b " 1, k " 2, ba

x AtB dx dt

" a ! b cos pt 12

! kx , x A0B " x0 ,

y AxB " Cyh AxB $ y AxByh AxBy AxB y¿ AxB " x2/yh AxB.yp AxB " y AxByh AxB,

dy dx

$ 3 x

y " 0 , x 7 0 .

dy dx

$ 3 x

y " x2 , x 7 0 ,

yh " m !1 AxB y AxBy AxByh AxB

yp AxBQ AxB " 0,[ 0 y AxB " Cyh AxB $ yp AxB ,

dy dx

$ P AxBy " Q AxB

A(t)

? L

A (10) = ? kg

5 L/min 0.2 kg/L

5 L/min

1 L/min

Figure 2.7 Mixing problem with unequal flow rates

A(t)

500 L

A (0) = 5 kg

5 L/min 0.2 kg/L

5 L/min

Figure 2.6 Mixing problem with equal flow rates

(a) Find the concentration, in kilograms per liter, of salt in the tank after 10 min. [Hint: Let A denote the number of kilograms of salt in the tank at t minutes after the process begins and use the fact that

rate of increase in A ! rate of input " rate of exit. A further discussion of mixing problems is given in Section 3.2.]

(b) After 10 min, a leak develops in the tank and an additional liter per minute of mixture flows out of the tank (see Figure 2.7). What will be the concentration, in kilograms per liter, of salt in the tank 20 min after the leak develops? [Hint: Use the method discussed in Problems 31 and 32.]

36. Variation of Parameters. Here is another proce- dure for solving linear equations that is particularly useful for higher-order linear equations. This method is called variation of parameters. It is based on the

38. Use the separation of variables technique to derive the solution (7) to the differential equation (6).

39. The temperature T (in units of 100(F) of a university classroom on a cold winter day varies with time t (in hours) as

dT dt

" e1 ! T , if heating unit is ON. !T , if heating unit is OFF.

Section 2.4 Exact Equations 55

Suppose T " 0 at 9:00 A.M., the heating unit is ON from 9–10 A.M., OFF from 10–11 A.M., ON again from 11 A.M.–noon, and so on for the rest of the day. How warm will the classroom be at noon? At 5:00 P.M.?

Suppose the mathematical function F(x,y) represents some physical quantity, such as tempera- ture, in a region of the xy-plane. Then the level curves of F, where F(x,y) " constant, could be interpreted as isotherms on a weather map, as depicted in Figure 2.8.

EXACT EQUATIONS2.4

50° 60°

70° 80°

90°

Figure 2.8 Level curves of F Ax, yB How does one calculate the slope of the tangent to a level curve? It is accomplished by

implicit differentiation: One takes the derivative, with respect to x, of both sides of the equation F(x,y) " C, taking into account that y depends on x along the curve:

(1) ,

and solves for the slope:

(2) .

The expression obtained by formally multiplying the left-hand member of (1) by dx is known as the total differential of F, written dF:

and our procedure for obtaining the equation for the slope f(x,y) of the level curve F(x,y) " C can be expressed as setting the total differential dF " 0 and solving.

Because equation (2) has the form of a differential equation, we should be able to reverse this logic and come up with a very easy technique for solving some differential equations. After all, any first-order differential equation can be rewritten in the (differential) form

(3) M Ax, yB dx $ N Ax, yB dy " 0dy/dx " f Ax, yB

dF :" 0F 0x dx $

0F 0y dy

dy dx

" f Ax,yB " ! 0F/0x0F/0y

0F 0x $

0F 0y

dy dx

" 0

d dx

F Ax,yB " d dx

ACB ˛

(in a variety of ways). Now, if the left-hand side of equation (3) can be identified as a total differential,

then its solutions are given (implicitly) by the level curves

for an arbitrary constant C.

Solve the differential equation

Some of the choices of differential forms corresponding to this equation are

However, the first form is best for our purposes because it is a total differential of the function :

Thus, the solutions are given implicitly by the formula See Figure 2.9. ◆x2y2 $ x " C.

" 0 0x Ax2y2 $ xB dx $ 00y Ax2y2 $ xB dy .

A2xy2 $ 1B dx $ 2x2y dy " d 3 x2y2 $ x 4F Ax, yB " x2y2 $ x dx $

2x2y

2xy2 $ 1 dy " 0 , etc.

2xy2 $ 1

2x2y dx $ dy " 0 ,

A2xy2 $ 1B dx $ 2x2y dy " 0 , dy

dx " !

2xy2 $ 1

2x2y .

F Ax, yB " C M Ax, yB dx $ N Ax, yB dy " 0F0x dx $ 0F0y dy " dF Ax, yB ,

56 Chapter 2 First-Order Differential Equations

Example 1

Solution

C = 4

C = 2 C = 2

C = 0

C = −2

C = 4

Figure 2.9 Solutions of Example 1

As you might suspect, in applications a differential equation is rarely given to us in exact differential form. However, the solution procedure is so quick and simple for such equations that we devote this section to it. From Example 1, we see that what is needed is (i) a test to determine if a differential form is exact and, if so, (ii) a procedure for finding the function itself.

The test for exactness arises from the following observation. If

then the calculus theorem concerning the equality of continuous mixed partial derivatives

would dictate a “compatibility condition” on the functions M and N:

In fact, Theorem 2 states that the compatibility condition is also sufficient for the differential form to be exact.

0 0y M Ax, yB " 00x N Ax, yB . 0 0y

0F 0x "

0 0x

0F 0y

M Ax, yB dx $ N Ax, yB dy " 0F0x dx $ 0F0y dy , F Ax, yB M Ax, yB dx $ N Ax, yB dy

Section 2.4 Exact Equations 57

Next we introduce some terminology.

Exact Differential Form

Definition 2. The differential form is said to be exact in a rectangle R if there is a function such that

(4)

for all in R. That is, the total differential of satisfies

If is an exact differential form, then the equation

is called an exact equation.

M Ax, yB dx $ N Ax, yB dy " 0M Ax, yB dx $ N Ax, yB dy dF Ax, yB " M Ax, yB dx $ N Ax, yB dy . F Ax, yBAx, yB %F %x Ax, yB ! M Ax, yB and %F

%y Ax, yB ! N Ax, yBF

Ax, yBM Ax, yB dx $ N Ax, yB dy

Test for Exactness

Theorem 2. Suppose the first partial derivatives of are continuous in a rectangle R. Then

is an exact equation in R if and only if the compatibility condition

(5)

holds for all in R.†Ax, yB %M %y Ax, yB ! %N

%x Ax, yB

M Ax, yB dx $ N Ax, yB dy " 0 M Ax, yB and N Ax, yB

Before we address the proof of Theorem 2, note that in Example 1 the differential form that led to the total differential wasA2xy2 $ 1B dx $ A2x2yB dy " 0 . †Historical Footnote: This theorem was proven by Leonhard Euler in 1734.

The compatibility conditions are easily confirmed:

Also clear is the fact that the other differential forms considered,

do not meet the compatibility conditions.

Proof of Theorem 2. There are two parts to the theorem: Exactness implies compatibil- ity, and compatibility implies exactness. First, we have seen that if the differential equation is exact, then the two members of equation (5) are simply the mixed second partials of a function

. As such, their equality is ensured by the theorem of calculus that states that mixed sec- ond partials are equal if they are continuous. Because the hypothesis of Theorem 2 guarantees the latter condition, equation (5) is validated.

Rather than proceed directly with the proof of the second part of the theorem, let’s derive a formula for a function that satisfies Integrating the first equation with respect to x yields

(6)

Notice that instead of using C to represent the constant of integration, we have written . This is because y is held fixed while integrating with respect to x, and so our “constant” may well depend on y. To determine , we differentiate both sides of (6) with respect to y to obtain

(7)

As g is a function of y alone, we can write and solving (7) for gives

Since this last equation becomes

(8)

Notice that although the right-hand side of (8) indicates a possible dependence on x, the appearances of this variable must cancel because the left-hand side, depends only on y. By integrating (8), we can determine up to a numerical constant, and therefore we can deter- mine the function up to a numerical constant from the functions and

To finish the proof of Theorem 2, we need to show that the condition (5) implies that is an exact equation. This we do by actually exhibiting a function that

satisfies Fortunately, we needn’t look too far for such a function.0F/ 0x " M and 0F/ 0y " N. F Ax, yBM dx $ N dy " 0 N Ax, yB.M Ax, yBF Ax, yB g A yB g¿

A yB, g¿ A yB " N Ax, yB ! 00y !M Ax, yB dx .

0F/ 0y " N,

g¿ A yB " 0F0y Ax, yB ! 00y !M Ax, yB dx . g¿ A yB0g/ 0y " g¿ A yB,

0F 0y Ax, yB " 00y !M Ax, yB dx $ 00y g A yB .

g A yB g A yBF Ax, yB " !M Ax, yB dx $ g A yB .

0F/ 0x " M and 0F/ 0y " N.F Ax, yB F Ax, yB

2xy2 $ 1

2x2y dx $ dy " 0 , dx $

2x2y

2xy2 $ 1 dy " 0 ,

0N 0x "

0 0x A2x2yB " 4xy .

0M 0y "

0 0y A2xy2 $ 1B " 4xy ,

58 Chapter 2 First-Order Differential Equations

The discussion in the first part of the proof suggests (6) as a candidate, where is given by (8). Namely, we by

(9)

where is a fixed point in the rectangle R and is determined, up to a numerical con- stant, by the equation

(10)

Before proceeding we must address an extremely important question concerning the defin- ition of . That is, how can we be sure (in this portion of the proof) that as given in equation (10), is really a function of just y alone? To show that the right-hand side of (10) is independent of x (that is, that the appearances of the variable x cancel), all we need to do is show that its partial derivative with respect to x is zero. This is where condition (5) is utilized. We leave to the reader this computation and the verification that satisfies conditions (4) (see Problems 35 and 36). ◆

The construction in the proof of Theorem 2 actually provides an explicit procedure for solving exact equations. Let’s recap and look at some examples.

F Ax, yB g¿ A yB,F Ax, yB

g¿ AyB J N Ax, yB ! 00y ! xx0 M At, yB dt . g A yBAx0, y0B

F Ax, yB J ! x x0

M At, yB dt $ g A yB ,define F Ax, yB g¿ AyB

Section 2.4 Exact Equations 59

Method for Solving Exact Equations (a) If is exact, then Integrate this last equation with

respect to x to get

(11)

(b) To determine , take the partial derivative with respect to y of both sides of equa- tion (11) and substitute N for We can now solve for

(d) The solution to is given implicitly by

(Alternatively, starting with the implicit solution can be found by first integrating with respect to y; see Example 3.)

0F/ 0y " N, F Ax, yB " C .M dx $ N dy " 0

F Ax, yB g A yBg A yBg¿ A yB g¿ A yB.0F/ 0y.g A yB

F Ax, yB " !M Ax, yB dx $ g A yB . 0F/ 0x " M.M dx $ N dy " 0

Solve

(12)

Here Because

0M 0y " 2x "

0N 0x ,

M Ax, yB " 2xy ! sec2x and N Ax, yB " x2 $ 2y.A2xy ! sec2xB dx $ Ax2 $ 2yB dy " 0 . Example 2

Solution

equation (12) is exact. To find , we begin by integrating M with respect to x:

(13)

Next we take the partial derivative of (13) with respect to y and substitute for N:

Thus, and since the choice of the constant of integration is not important, we can take Hence, from (13), we have and the solution to equation (12) is given implicitly by ◆

Remark. The procedure for solving exact equations requires several steps. As a check on our work, we observe that when we solve for we must obtain a function that is independent of x. If this is not the case, then we have erred either in our computation of or in computing

In the construction of we can first integrate with respect to y to get

(14)

and then proceed to find . We illustrate this alternative method in the next example.

Solve

(15)

Here and . Because

equation (15) is exact. If we now integrate with respect to y, we obtain

When we take the partial derivative with respect to x and substitute for M, we get

Thus, so we take Hence, and the solution to equation (15) is given implicitly by In this case we can solve explicitly for y to obtain ◆y " AC ! xB / A2 $ xexB.xexy $ 2y $ x " C.

F Ax, yB " xexy $ 2y $ x,h AxB " x.h¿ AxB " 1, exy $ xexy $ h¿ AxB " 1 $ exy $ xexy .

0F 0x Ax, yB " M Ax, yB

F Ax, yB " ! Axex $ 2B dy $ h AxB " xexy $ 2y $ h AxB . N Ax, yB

0M 0y " e

x $ xex " 0N 0x ,

N " xex $ 2M " 1 $ exy $ xexy

A1 $ exy $ xexyB dx $ Axex $ 2B dy " 0 . h AxBF

Ax, yB ! !N Ax, yB dy # h AxB N Ax, yBF Ax, yB,0M/ 0y or 0N/ 0x.

F Ax, yBg¿ A yB, x2y ! tan x $ y2 " C.

F Ax, yB " x2y ! tan x $ y2,g A yB " y2.g¿ A yB " 2y, x2 $ g¿ A yB " x2 $ 2y .

0F 0y Ax, yB " N Ax, yB ,

x2 $ 2y

" x2y ! tan x $ g A yB . F Ax, yB " ! A2xy ! sec2xB dx $ g A yB

F Ax, yB 60 Chapter 2 First-Order Differential Equations

Example 3

Solution

Remark. Since we can use either procedure for finding it may be worthwhile to con- sider each of the integrals and If one is easier to evaluate than the other, this would be sufficient reason for us to use one method over the other. [The skeptical reader should try solving equation (15) by first integrating .]

Show that

(16)

is not exact but that multiplying this equation by the factor yields an exact equation. Use this fact to solve (16).

In equation (16), and Because

equation (16) is not exact. When we multiply (16) by the factor we obtain

(17)

For this new equation, If we test for exactness, we now find that

and hence (17) is exact. Upon solving (17), we find that the solution is given implicitly by Since equations (16) and (17) differ only by a factor of x, then any solution

to one will be a solution for the other whenever Hence the solution to equation (16) is given implicitly by ◆

In Section 2.5 we discuss methods for finding factors that, like in Example 4, change inexact equations into exact equations.

x!1

x $ x3sin y " C. x + 0.

x $ x3sin y " C.

0M 0y " 3x

2cos y " 0N 0x ,

M " 1 $ 3x2sin y and N " x3cos y.

A1 $ 3x2sin yB dx $ Ax3cos yB dy " 0 . x !1, 0M 0y " 3x

3cos y [ 4x3cos y " 0N0x ,

N " x4cos y.M " x $ 3x3sin y

x!1

Ax $ 3x3sin yB dx $ Ax4cos yB dy " 0 M Ax, yB

#N Ax, yB dy.#M Ax, yB dx F Ax, yB, Section 2.4 Exact Equations 61

In Problems 1–8, classify the equation as separable, lin- ear, exact, or none of these. Notice that some equations may have more than one classification.

1. 2.

3. 4. 5. xy dx $ dy " 0 Ayexy $ 2xB dx $ Axexy ! 2yB dy " 02!2y ! y2 dx $ A3 $ 2x ! x2B dy " 0Ax

2y $ x4cos xB dx ! x3 dy " 0Ax10/3 ! 2yB dx $ x dy " 0 6.

In Problems 9–20, determine whether the equation is exact. If it is, then solve it.

10. A2xy $ 3B dx $ Ax2 ! 1B dy " 0A2x $ yB dx $ Ax ! 2yB dy " 0 u dr $ A3r ! u ! 1B du " 03 2x $ y cos AxyB 4 dx $ 3 x cos AxyB ! 2y 4 dy " 0y

2 dx $ A2xy $ cos yB dy " 0

Example 4

Solution

2.4 EXERCISES

11.

12.

13.

14.

15.

16.

17.

18.

19.

20.

In Problems 21–26, solve the initial value problem.

21.

22.

23.

24.

25.

26.

27. For each of the following equations, find the most general function so that the equation is exact. (a) (b)

28. For each of the following equations, find the most general function so that the equation is exact. (a) (b)

29. Consider the equationA y2 $ 2xyB dx ! x2 dy " 0 . Ayexy ! 4x3y $ 2B dx $ N Ax, yB dy " 03 y cos AxyB $ ex 4 dx $ N Ax, yB dy " 0N Ax, yB M Ax, yB dx $ Asin x cos y ! xy ! e!yB dy " 0M Ax, yB dx $ Asec2y ! x/yB dy " 0

M Ax, yB y A0B " 1Atan y ! 2B dx $ Ax sec2y $ 1/yB dy " 0 ,

A y2 sin xB dx $ A1/x ! y/xB dy " 0 , y ApB " 1Aetx $ 1B dt $ Aet ! 1B dx " 0 , x A1B " 1 Aety $ tetyB dt $ Atet $ 2B dy " 0 , y A0B " !1y A1B " 1 A yexy ! 1/yB dx $ Axexy $ x/y2B dy " 0 , y A1B " p A1/x $ 2y2xB dx $ A2yx2 ! cos yB dy " 0 ,

$ 3 x cos AxyB ! y!1/3 4 dy " 0c 221 ! x2 $ y cos AxyB d dx

a2x $ y 1 $ x2y2

b dx $ a x 1 $ x2y2

! 2yb dy " 0$ 3 2xy ! cos Ax $ yB ! e y 4 dy " 032x $ y2 ! cos Ax $ yB 4 dxA1/yB dx ! A3y ! x/y

2B dy " 0A yexy ! 1 /yB dx $ Axexy $ x/y2B dy " 0 cos u dr ! Ar sin u ! euB du " 0et A y ! tB dt $ A1 $ etB dy " 0 At /yB dy $ A1 $ ln yB dt " 0Aexsin y ! 3x2B dx $ Aexcos y $ y!2/3/3B dy " 0 Acos x cos y $ 2xB dx ! Asin x sin y $ 2yB dy " 0

62 Chapter 2 First-Order Differential Equations

(a) Show that this equation is not exact. (b) Show that multiplying both sides of the equation

by yields a new equation that is exact. (c) Use the solution of the resulting exact equation

to solve the original equation. (d) Were any solutions lost in the process?

30. Consider the equation

(a) Show that the equation is not exact. (b) Multiply the equation by and determine

values for n and m that make the resulting equa- tion exact.

31. Argue that in the proof of Theorem 2 the function g can be taken as

which can be expressed as

This leads ultimately to the representation

(18)

Evaluate this formula directly with to rework (a) Example 1. (b) Example 2. (c) Example 3.

32. Orthogonal Trajectories. A geometric problem occurring often in engineering is that of finding a family of curves (orthogonal trajectories) that inter- sects a given family of curves orthogonally at each point. For example, we may be given the lines of force of an electric field and want to find the equation

x0 " 0, y0 " 0

F Ax, yB " ! y y0

N Ax, tB dt $ ! x x0

M As, y0B ds . $ !

M As, y0B ds . g AyB " ! y

N Ax, tB dt ! ! x x0

M As, yB ds g AyB " ! y

N Ax, tB dt ! ! y y0

c 00t ! xx0 M As, tB ds ddt ,

xnym

$ A2x3 $ 3x4y $ 3x2yB dy " 0 .A5x2y $ 6x3y2 $ 4xy2B dx

y !2

for the equipotential curves. Consider the family of curves described by , where k is a param- eter. Recall from the discussion of equation (2) that for each curve in the family, the slope is given by

(a) Recall that the slope of a curve that is orthogo- nal (perpendicular) to a given curve is just the negative reciprocal of the slope of the given curve. Using this fact, show that the curves orthogonal to the family satisfy the differential equation

(b) Using the preceding differential equation, show that the orthogonal trajectories to the family of circles are just straight lines through the origin (see Figure 2.10).

x2 $ y2 " k

%F %y Ax, yB dx " %F

%x Ax, yB dy ! 0 .

F Ax, yB " k

dy dx ! "

%F %x~%F%y .

F Ax, yB " k Section 2.4 Exact Equations 63

33. Use the method in Problem 32 to find the orthogo- nal trajectories for each of the given families of curves, where k is a parameter.

(a) (b) (c) (d) [Hint: First express the family in the form .]

34. Use the method described in Problem 32 to show that the orthogonal trajectories to the family of curves a parameter, satisfy

Find the orthogonal trajectories by solving the above equation. Sketch the family of curves, along with their orthogonal trajectories. [Hint: Try multiplying the equation by as in Problem 30.]

35. Using condition (5), show that the right-hand side of (10) is independent of x by showing that its partial derivative with respect to x is zero. [Hint: Since the partial derivatives of M are continuous, Leibniz’s theorem allows you to interchange the operations of integration and differentiation.]

36. Verify that as defined by (9) and (10) satisfies conditions (4).

F Ax, yB

xmyn

A2yx!1B dx $ A y2x!2 ! 1B dy " 0 .x2 $ y2 " kx, k F(x, y) " k

y2 " kx y " ekx y " kx4 2x2 $ y2 " k

Figure 2.10 Orthogonal trajectories for concentric circles are lines through the center

Figure 2.11 Families of orthogonal hyperbolas

xy " k

If we take the standard form for the linear differential equation of Section 2.3,

and rewrite it in differential form by multiplying through by dx, we obtain

This form is certainly not exact, but it becomes exact upon multiplication by the integrating

factor We have

as the form, and the compatibility condition is precisely the identity (see Problem 20).

This leads us to generalize the notion of an integrating factor.

m AxBP AxB " m¿AxB3m AxBP AxBy ! m AxBQ AxB 4 dx $ m AxB dy " 0 m AxB " e# P AxB dx. 3P AxBy ! Q AxB 4 dx $ dy " 0 . dy dx

$ P AxBy " Q AxB ,

64 Chapter 2 First-Order Differential Equations

†Historical Footnote: A general theory of integrating factors was developed by Alexis Clairaut in 1739. Leonhard Euler also studied classes of equations that could be solved using a specific integrating factor.

2.5 SPECIAL INTEGRATING FACTORS

Integrating Factor

Definition 3. If the equation

(1)

is not exact, but the equation

(2)

which results from multiplying equation (1) by the function exact, then is called an integrating factor† of the equation (1).

m Ax, yBm Ax, yB, ism Ax, yBM Ax, yB dx $ m Ax, yBN Ax, yB dy " 0 , M Ax, yB dx $ N Ax, yB dy " 0

Show that is an integrating factor for

(3)

Use this integrating factor to solve the equation.

We leave it to you to show that (3) is not exact. Multiplying (3) by we obtain

(4)

For this equation we have Because

equation (4) is exact. Hence, is indeed an integrating factor of equation (3).m Ax, yB " xy2 0M 0y Ax, yB " 6xy2 ! 12x2y " 0N0x Ax, yB ,

M " 2xy3 ! 6x2y2 and N " 3x2y2 ! 4x3y.

A2xy3 ! 6x2y2B dx $ A3x2y2 ! 4x3yB dy " 0 . m Ax, yB " xy2, A2y ! 6xB dx $ A3x ! 4x2y!1B dy " 0 .m Ax, yB " xy2Example 1

Solution

Let’s now solve equation (4) using the procedure of Section 2.4. To find we begin by integrating M with respect to x:

When we take the partial derivative with respect to y and substitute for N, we find

Thus, so we can take Hence, and the solution to equation (4) is given implicitly by

Although equations (3) and (4) have essentially the same solutions, it is possible to lose or gain solutions when multiplying by In this case is a solution of equation (4) but not of equation (3). The extraneous solution arises because, when we multiply (3) by to obtain (4), we are actually multiplying both sides of (3) by zero if This gives us

as a solution to (4), but it is not a solution to (3). ◆

Generally speaking, when using integrating factors, you should check whether any solu- tions to are in fact solutions to the original differential equation.

How do we find an integrating factor? If is an integrating factor of (1) with contin- uous first partial derivatives, then testing (2) for exactness, we must have

By use of the product rule, this reduces to the equation

(5)

But solving the partial differential equation (5) for m is usually more difficult than solving the original equation (1). There are, however, two important exceptions.

Let’s assume that equation (1) has an integrating factor that depends only on x; that is, In this case equation (5) reduces to the separable equation

(6)

where is (presumably) just a function of x. In a similar fashion, if equa- tion (1) has an integrating factor that depends only on y, then equation (5) reduces to the separable equation

(7)

where is just a function of y.A0N/ 0x ! 0M/ 0yB /M dm dy

" a0N/ 0x ! 0M/ 0y M

bm , A0M/ 0y ! 0N/ 0xB /N

dm dx

" a0M/ 0y ! 0N/ 0x N

bm ,m " m AxB. M

%y " N %M

%x ! a%N%x " %M%y bM . 0 0y 3m Ax, yBM Ax, yB 4 " 00x 3m Ax, yBN Ax, yB 4 .

m Ax, yBm Ax, yB " 0

y " 0 y " 0.

m " xy2 y " 0m Ax, yB.

x2y3 ! 2x3y2 " C .

F Ax, yB " x2y3 ! 2x3y2,g A yB " 0.g¿ A yB " 0, 3x2y2 ! 4x3y $ g¿A yB " 3x2y2 ! 4x3y .

0F 0y Ax, yB " N Ax, yB

F Ax, yB " ! A2xy3 ! 6x2y2B dx $ g A yB " x2y3 ! 2x3y2 $ g A yB . F Ax, yB,

Section 2.5 Special Integrating Factors 65

We can reverse the above argument. In particular, if is a function that depends only on x, then we can solve the separable equation (6) to obtain the integrating factor

for equation (1). We summarize these observations in the following theorem.

m AxB " exp c! a0M/ 0y ! 0N/ 0xN bdx d A0M/ 0y ! 0N/ 0xB /N

66 Chapter 2 First-Order Differential Equations

Special Integrating Factors

Theorem 3. If is continuous and depends only on x, then

(8)

is an integrating factor for equation (1). If is continuous and depends only on y, then

(9)

is an integrating factor for equation (1).

m A yB " exp c! a0N/ 0x ! 0M/ 0yM bdy d A0N/ 0x ! 0M/ 0yB /M m AxB " exp c!a0M/ 0y ! 0N/ 0xN bdx d

A0M/ 0y ! 0N/ 0xB /N

Theorem 3 suggests the following procedure.

Method for Finding Special Integrating Factors If is neither separable nor linear, compute . If

then the equation is exact. If it is not exact, consider

(10)

If (10) is a function of just x, then an integrating factor is given by formula (8). If not, consider

(11)

If (11) is a function of just y, then an integrating factor is given by formula (9).

0N/ 0x ! 0M/ 0y M

0M/ 0y ! 0N/ 0x N

0M/ 0y " 0N/ 0x, 0M/ 0y and 0N/ 0xM dx $ N dy " 0

Solve

(12)

A quick inspection shows that equation (12) is neither separable nor linear. We also note that

0M 0y " 1 [ A2xy ! 1B " 0N0x . A2x2 $ yB dx $ Ax2y ! xB dy " 0 .Example 2

Solution

Because (12) is not exact, we compute

We obtain a function of only x, so an integrating factor for (12) is given by formula (8). That is,

When we multiply (12) by we get the exact equation

Solving this equation, we ultimately derive the implicit solution

(13)

Notice that the solution was lost in multiplying by Hence, (13) and are solutions to equation (12). ◆

There are many differential equations that are not covered by Theorem 3 but for which an integrating factor nevertheless exists. The major difficulty, however, is in finding an explicit formula for these integrating factors, which in general will depend on both x and y.

x " 0m " x!2.x " 0

2x ! yx!1 $ y2

2 " C .

A2 $ yx!2B dx $ A y ! x!1B dy " 0 .m " x !2,

m AxB " expa!!2x dxb " x!2 . 0M/ 0y ! 0N/ 0x

N "

1 ! A2xy ! 1B x2y ! x

" 2 A1 ! xyB

!x A1 ! xyB " !2x .

Section 2.5 Special Integrating Factors 67

In Problems 1–6, identify the equation as separable, lin- ear, exact, or having an integrating factor that is a func- tion of either x alone or y alone.

1. 2. 3. 4. 5. 6.

In Problems 7–12, solve the equation. 7. 8. 9.

10. 11. 12.

In Problems 13 and 14, find an integrating factor of the form and solve the equation.

13. 14. A12 $ 5xyB dx $ A6xy!1 $ 3x2B dy " 0A2y2 ! 6xyB dx $ A3xy ! 4x2B dy " 0

xnym

A3x2y2 ! y!1B dy " 0A2xy3 $ 1B dx $A y2 $ 2xyB dx ! x2 dy " 0 A2y2 $ 2y $ 4x2B dx $ A2xy $ xB dy " 0Ax4 ! x $ yB dx ! x dy " 0 A3x2 $ yB dx $ Ax2y ! xB dy " 0A2xyB dx $ A y2 ! 3x2B dy " 0 A2y2x ! yB dx $ x dy " 0Ax2sin x $ 4yB dx $ x dy " 0 A y2 $ 2xyB dx ! x2 dy " 0A2x $ yB dx $ Ax ! 2yB dy " 0 A2y3 $ 2y2B dx $ A3y2x $ 2xyB dy " 0A2x $ yx!1B dx $ Axy ! 1B dy " 0

15. (a) Show that if depends only on the product xy, that is,

then the equation has an integrating factor of the form Give the general formula for

(b) Use your answer to part (a) to find an implicit solution to

satisfying the initial condition

16. (a) Prove that has an integrating factor that depends only on the sum if and only if the expression

depends only on . (b) Use part (a) to solve the equation

(3 $ y $ xy)dx $ (3 $ x $ xy)dy " 0.

x $ y

0N/ 0x ! 0M/ 0y M ! N

x $ y Mdx $ Ndy " 0

y(1) " 1.

(3y $ 2xy2)dx $ (x $ 2x2y)dy " 0 ,

m AxyB. m AxyB. M Ax, yB dx $ N Ax, yB dy " 0

0N/ 0x ! 0M/ 0y xM ! yN

" H AxyB , A0N/ 0x ! 0M/ 0yB / AxM ! yNB

2.5 EXERCISES

17. (a) Find a condition on M and N that is necessary and sufficient for to have an integrating factor that depends only on the product .

(b) Use part (a) to solve the equation

18. If find the solution to the equation

19. Fluid Flow. The streamlines associated with a cer- tain fluid flow are represented by the family of curves

The velocity potentials of the flow are just the orthogonal trajectories of this family. y " x ! 1 $ ke!x.

M Ax, yB dx $ N Ax, yB dy " 0.xM Ax, yB $ yN Ax, yB " 0, $ (2x $ x4 $ 2x3y) dy " 0 .

(2x $ 2y $ 2x3y $ 4x2y2) dx

x2y

Mdx $ Ndy " 0

68 Chapter 2 First-Order Differential Equations

(a) Use the method described in Problem 32 of Exercises 2.4 to show that the velocity potentials satisfy

[Hint: First express the family in the form " k.] (b) Find the velocity potentials by solving the equa-

tion obtained in part (a). 20. Verify that when the linear differential equation

is multiplied by the result is exact.e# PAxB dx, m AxB "3P AxBy ! QAxB 4 dx $ dy " 0 F Ax, yB y " x ! 1 $ ke!x

dx $ Ax ! yB dy " 0.

SUBSTITUTIONS AND TRANSFORMATIONS2.6 When the equation

is not a separable, exact, or linear equation, it may still be possible to transform it into one that we know how to solve. This was in fact our approach in Section 2.5, where we used an integrating factor to transform our original equation into an exact equation.

In this section we study four types of equations that can be transformed into either a sepa- rable or linear equation by means of a suitable substitution or transformation.

M Ax, yB dx $ N Ax, yB dy " 0

Substitution Procedure (a) Identify the type of equation and determine the appropriate substitution or

transformation. (b) Rewrite the original equation in terms of new variables. (c) Solve the transformed equation. (d) Express the solution in terms of the original variables.

Homogeneous Equations

Homogeneous Equation

Definition 4. If the right-hand side of the equation

(1)

can be expressed as a function of the ratio alone, then we say the equation is homogeneous.

y/x

dy dx

" f Ax, yB

For example, the equation

(2)

can be written in the form

Since we have expressed as a function of the ratio that is, where then equation (2) is homogeneous.

The equation

(3)

can be written in the form

Here the right-hand side cannot be expressed as a function of alone because of the term in the numerator. Hence, equation (3) is not homogeneous.

One test for the homogeneity of equation (1) is to replace x by tx and y by ty. Then (1) is homogeneous if and only if

for all [see Problem 43(a)]. To solve a homogeneous equation, we make a rather obvious substitution. Let

Our homogeneous equation now has the form

(4)

and all we need is to express in terms of x and y. Since , then . Keeping in mind that both y and y are functions of x, we use the product rule for differentiation to deduce from y " yx that

We then substitute the above expression for into equation (4) to obtain

(5)

The new equation (5) is separable, and we can obtain its implicit solution from

All that remains to do is to express the solution in terms of the original variables x and y.

! 1G AyB ! y dy " ! 1x dx . y $ x

dy dx

" G AyB . dy/dx

dy dx ! y # x

dy dx .

y " yxy " y/xdy/dx

dy dx

" G AyB , Y !

y x .

t + 0

f Atx, tyB " f Ax, yB 1/xy/x

dy dx

" x ! 2y $ 1

y ! x "

1 ! 2 A y/xB $ A1/xBA y/xB ! 1 . Ax ! 2y $ 1B dx $ Ax ! yB dy " 0

G AyB J y ! 1 4 , A y ! xB /x " G A y/xB,3y/xA y ! xB /x dy dx

" y ! x

x "

y x

! 1 .

Ax ! yB dx $ x dy " 0 Section 2.6 Substitutions and Transformations 69

Solve

(6)

A check will show that equation (6) is not separable, exact, or linear. If we express (6) in the derivative form

(7)

then we see that the right-hand side of (7) is a function of just . Thus, equation (6) is homo- geneous.

Now let and recall that With these substitutions, equation (7) becomes

The above equation is separable, and, on separating the variables and integrating, we obtain

Hence,

Finally, we substitute for y and solve for y to get

as an explicit solution to equation (6). Also note that is a solution. ◆

Equations of the Form dy/dx ! G(ax # by) When the right-hand side of the equation can be expressed as a function of the combination ax $ by, where a and b are constants, that is,

then the substitution

transforms the equation into a separable one. The method is illustrated in the next example.

Solve

(8)

The right-hand side can be expressed as a function of that is,

y ! x ! 1 $ Ax ! y $ 2B!1 " ! Ax ! yB ! 1 $ 3 Ax ! yB $ 2 4!1 ,x ! y, dy dx

" y ! x ! 1 $ Ax ! y $ 2B!1 .

z ! ax # by

dy dx

" G Aax $ byB , dy/dx " f Ax, yB

x " 0

y " x tan Aln 0 x 0 $ CBy/x y " tan Aln 0 x 0 $ CB .

arctan y " ln 0 x 0 $ C . ! 1y2 $ 1 dy " !

1 x

dx ,

y $ x dy dx

" y $ y2 $ 1 .

dy/dx " y $ x Ady/dxB.y " y/x y/x

dx "

xy $ y2 $ x2

x2 "

x $ ay

x b 2 $ 1 ,

Axy $ y2 $ x2B dx ! x2 dy " 0 . 70 Chapter 2 First-Order Differential Equations

Example 1

Solution

Example 2

Solution

so let To solve for , we differentiate with respect to x to obtain and so Substituting into (8) yields

Solving this separable equation, we obtain

from which it follows that

Finally, replacing z by yields

as an implicit solution to equation (8). ◆

Bernoulli Equations

Ax ! y $ 2B2 " Ce2x $ 1x ! y Az $ 2B2 " Ce2x $ 1 . 1 2

ln 0 Az $ 2B2 ! 1 0 " x $ C1 , ! z $ 2Az $ 2B2 ! 1 dz " !dx ,

dz dx

" Az $ 2B ! Az $ 2B!1 . 1 !

dz dx

" !z ! 1 $ Az $ 2B!1 , dy/dx " 1 ! dz/dx.dz/dx " 1 ! dy/dx,

z " x ! ydy/dxz " x ! y.

Section 2.6 Substitutions and Transformations 71

Bernoulli Equation

Definition 5. A first-order equation that can be written in the form

(9)

where and are continuous on an interval and n is a real number, is called a Bernoulli equation.†

Aa, bBQ AxBP AxB dy dx # P

AxBy ! Q AxByn ,

†Historical Footnote: This equation was proposed for solution by James Bernoulli in 1695. It was solved by his brother John Bernoulli. (James and John were two of eight mathematicians in the Bernoulli family.) In 1696, Gottfried Leibniz showed that the Bernoulli equation can be reduced to a linear equation by making the substitution y " y1!n.

Notice that when n " 0 or 1, equation (9) is also a linear equation and can be solved by the method discussed in Section 2.3. For other values of n, the substitution

transforms the Bernoulli equation into a linear equation, as we now show.

Y ! y1"n

Dividing equation (9) by yields

(10)

Taking we find via the chain rule that

and so equation (10) becomes

Because is just a constant, the last equation is indeed linear.

Solve

(11)

This is a Bernoulli equation with To transform (11) into a linear equation, we first divide by y3 to obtain

Next we make the substitution Since the transformed equation is

(12)

Equation (12) is linear, so we can solve it for y using the method discussed in Section 2.3. When we do this, it turns out that

Substituting gives the solution

Not included in the last equation is the solution that was lost in the process of dividing (11) by . ◆

Equations with Linear Coefficients We have used various substitutions for y to transform the original equation into a new equation that we could solve. In some cases we must transform both x and y into new variables, say, u and y. This is the situation for equations with linear coefficients—that is, equations of the form

(13) Aa1x # b1 y # c1B dx # Aa2x # b2 y # c2B dy ! 0 ,

y3 y " 0

y!2 " x 2

! 1 20

$ Ce!10x .

y " y!2

y " x 2

! 1 20

$ Ce!10x .

dy dx

$ 10y " 5x .

! 1 2

dy dx

! 5y " ! 5 2

x ,

dy/dx " !2y!3 dy/dx,y " y!2.

y!3 dy dx

! 5y!2 " ! 5 2

x .

n " 3, P AxB " !5, and Q AxB " !5x/2. dy dx

! 5y " ! 5 2

xy3 .

1/ A1 ! nB 1

1 ! n dy dx

$ P AxBy " Q AxB . dy dx

" A1 ! nBy!n dy dx

y " y1!n,

y!n dy dx

$ P AxBy1!n " Q AxB . yn

72 Chapter 2 First-Order Differential Equations

Example 3

Solution

where the ai’s, bi’s, and ci’s are constants. We leave it as an exercise to show that when equation (13) can be put in the form which we solved via

the substitution Before considering the general case when let’s first look at the special situa-

tion when Equation (13) then becomes

which can be rewritten in the form

This equation is homogeneous, so we can solve it using the method discussed earlier in this section.

The above discussion suggests the following procedure for solving (13). If then we seek a translation of axes of the form

where h and k are constants, that will change and change Some elementary algebra shows that such a transformation

exists if the system of equations

(14)

has a solution. This is ensured by the assumption , which is geometrically equiva- lent to assuming that the two lines described by the system (14) intersect. Now if satisfies (14), then the substitutions transform equation (13) into the homo- geneous equation

(15)

which we know how to solve.

Solve

(16)

Since we will use the translation of axes , , where h and k satisfy the system

Solving the above system for h and k gives Hence, we let and . Because dy " dy and dx " du, substituting in equation (16) for x and y yields

dy du

" 3 ! (y/u) 1 $ (y/u) .

A!3u $ yB du $ Au $ yB dy " 0y " y ! 3 x " u $ 1h " 1, k " !3.

h $ k $ 2 " 0 .

!3h $ k $ 6 " 0 ,

y " y $ k x " u $ ha1b2 " A!3B A1B + A1B A1B " a2b1,A!3x $ y $ 6B dx $ Ax $ y $ 2B dy " 0 .

du " !

a1u $ b1y

a2u $ b2y " !

a1 $ b1 Ay/uB a2 $ b2 Ay/uB ,

x " u $ h and y " y $ k Ah, kBa1b2 + a2b1

a1h # b1k # c1 ! 0 , a2h # b2k # c2 ! 0

a2x $ b2y $ c2 into a2u $ b2y. a1x $ b1y $ c1 into a1u $ b1y

x ! u # h and y ! Y # k ,

a1b2 + a2b1,

dx " !

a1x $ b1y

a2x $ b2y " !

a1 $ b1 A y/xB a2 $ b2 A y/xB .

Aa1x $ b1yB dx $ Aa2x $ b2yB dy " 0 ,c1 " c2 " 0. a1b2 + a2b1,

z " ax $ by. dy/dx " G Aax $ byB,a1b2 " a2b1,

Section 2.6 Substitutions and Transformations 73

Example 4

Solution

The last equation is homogeneous, so we let . Then and, sub- stituting for , we obtain

Separating variables gives

from which it follows that

When we substitute back in for z, u, and y, we find

This last equation gives an implicit solution to (16). ◆

A y $ 3B2 $ 2 Ax ! 1B A y $ 3B ! 3 Ax ! 1B2 " C .y2 $ 2uy ! 3u2 " C , Ay/uB2 $ 2 Ay/uB ! 3 " Cu!2 ,

z2 $ 2z ! 3 " Cu!2 .

1 2

ln 0 z2 $ 2z ! 3 0 " !ln 0u 0 $ C1 ,! z $ 1

z2 $ 2z ! 3 dz " !! 1u du ,

z $ u dz du

" 3 ! z 1 $ z

y/u dy/du " z $ u Adz/duB,z " y/u

74 Chapter 2 First-Order Differential Equations

In Problems 1–8, identify (do not solve) the equation as homogeneous, Bernoulli, linear coefficients, or of the form

1. 2. 3. 4.

Use the method discussed under “Homogeneous Equa- tions” to solve Problems 9–16.

9. 10. 11. 12.

13.

14. dy du

" u sec A y/uB $ y

dx dt

" x2 $ t2t2 $ x2

Ax2 $ y2B dx $ 2xy dy " 0A y2 ! xyB dx $ x2 dy " 0 A3x2 ! y2B dx $ Axy ! x3y!1B dy " 0Axy $ y2B dx ! x2 dy " 0 A y3 ! uy2B du $ 2u2y dy " 0cos Ax $ yB dy " sin Ax $ yB dx A ye!2x $ y3B dx ! e!2x dy " 0u dy ! y du " 2uy du At $ x $ 2B dx $ A3t ! x ! 6B dt " 0dy/dx $ y/x " x3y2 A y ! 4x ! 1B2 dx ! dy " 02tx dx $ At2 ! x2B dt " 0 y¿ " G Aax $ byB. 15. 16.

Use the method discussed under “Equations of the Form ” to solve Problems 17–20.

17. 18. 19. 20.

Use the method discussed under “Bernoulli Equations” to solve Problems 21–28.

21.

22.

23.

24.

25. 26.

27. 28. dy dx

$ y3x $ y " 0 dr du

" r2 $ 2ru u2

dy dx

$ y " exy!2 dx dt

$ tx3 $ x t

" 0

dx $

y x ! 2

" 5 Ax ! 2By1/2 dy dx

" 2y x

! x2y2

dy dx

! y " e2xy3

dy dx

$ y x

" x2y2

dy/dx " sinAx ! yBdy/dx " Ax ! y $ 5B2 dy/dx " Ax $y$2B2dy/dx " 2x $ y ! 1dy/dx " G Aax $ byB dy dx

" y Aln y ! ln x $ 1B

x dy dx

" x2 ! y2

3xy

2.6 EXERCISES

Use the method discussed under “Equations with Linear Coefficients” to solve Problems 29–32. 29. 30. 31. 32.

In Problems 33–40, solve the equation given in: 33. Problem 1. 34. Problem 2. 35. Problem 3. 36. Problem 4. 37. Problem 5. 38. Problem 6. 39. Problem 7. 40. Problem 8.

41. Use the substitution to solve equa- tion (8).

42. Use the substitution to solve

43. (a) Show that the equation is homogeneous if and only if [Hint: Let .]

(b) A function is called homogeneous of order n if Show that the equation

is homogeneous if are both homogeneous of the same order.

44. Show that equation (13) reduces to an equation of the form

when [Hint: If , then so that .]

45. Coupled Equations. In analyzing coupled equa- tions of the form

dx dt

" ax $ by ,

dy dt

" ax $ by ,

a2 " ka1 and b2 " kb1a2/a1 " b2/b1 " k, a1b2 " a2b1a1b2 " a2b1.

dy dx

" G Aax $ byB ,

M Ax, yB and N Ax, yBM Ax, yB dx $ N Ax, yB dy " 0 H Atx, tyB " tnH Ax, yB.H Ax, yB

t " 1/x f Atx, tyB " f Ax, yB.dy/dx " f Ax, yB

dy dx

" 2y x

$ cos A y/x2B . y " yx2

y " x ! y $ 2

A2x $ y $ 4B dx $ Ax ! 2y ! 2B dy " 0A2x ! yB dx $ A4x $ y ! 3B dy " 0 Ax $ y ! 1B dx $ A y ! x ! 5B dy " 0A!3x $ y ! 1B dx $ Ax $ y $ 3B dy " 0

Section 2.6 Substitutions and Transformations 75

†Historical Footnote: Count Jacopo Riccati studied a particular case of this equation in 1724 during his investigation of curves whose radii of curvature depend only on the variable y and not the variable x.

where are constants, we may wish to determine the relationship between x and y rather than the individual solutions For this purpose, divide the first equation by the second to obtain

(17)

This new equation is homogeneous, so we can solve it via the substitution We refer to the solu- tions of (17) as integral curves. Determine the inte- gral curves for the system

46. Magnetic Field Lines. As described in Problem 20 of Exercises 1.3, the magnetic field lines of a dipole satisfy

Solve this equation and sketch several of these lines. 47. Riccati Equation. An equation of the form

(18)

is called a generalized Riccati equation.†

(a) If one solution—say, —of (18) is known, show that the substitution reduces (18) to a linear equation in y.

(b) Given that is a solution to

use the result of part (a) to find all the other solu- tions to this equation. (The particular solution

can be found by inspection or by using a Taylor series method; see Section 8.1.) u AxB " x dy dx

" x3 A y ! xB2 $ y x ,

u AxB " x y " u $ 1/y

u AxB dy dx ! P

AxB y2 # Q AxB y # R AxB

dy dx

" 3xy

2x2 ! y2 .

dx dt

" 2x ! y .

dy dt

" !4x ! y ,

y " y/x.

dy dx

" ax $ by ax $ by

x AtB, y AtB. a, b, a, and b

In this chapter we have discussed various types of first-order differential equations. The most important were the separable, linear, and exact equations. Their principal features and method of solution are outlined below.

Separable Equations: Separate the variables and integrate.

Linear Equations: . The integrating factor reduces the equation to

Exact Equations: Solutions are given implicitly by . If , then is exact and F is given by

When an equation is not separable, linear, or exact, it may be possible to find an integrat- ing factor or perform a substitution that will enable us to solve the equation.

Special Integrating Factors: is exact. If depends only on x, then

is an integrating factor. If depends only on y, then

is an integrating factor.

Homogeneous Equations: . Let . Then and the transformed equation in the variables y and x is separable.

Equations of the Form: . Let z " ax $ by. Then and the transformed equation in the variables z and x is separable.

Bernoulli Equations: . For or 1, let . Then and the transformed equation in the variables y and x is linear.

Linear Coefficients: For , let and , where h and k satisfy

Then the transformed equation in the variables u and y is homogeneous.

a2h $ b2k $ c2 " 0 .

a1h $ b1k $ c1 " 0 ,

y " y $ kx " u $ h a1b2 + a2b1Aa1x # b1 y # c1B dx # Aa2x # b2 y # c2B dy ! 0.

dy/dx " A1 ! nBy!n Ady/dxB, y " y1!nn + 0dy/dx # P AxBy ! Q AxB yn dz/dx " a $ b Ady/dxB,dy/dx ! G Aax # byB dy/dx " y $ x Ady/dxB,y " y/xdy/dx ! G A y/xB

m A yB " exp c!a0N/ 0x ! 0M/ 0yM bdy d A0N/ 0x ! 0M/ 0yB /M

m AxB " exp c!a0M/ 0y ! 0N/ 0xN bdx d A0M/ 0y ! 0N/ 0xB /NMM dx # MN dy ! 0

F " !N dy $ h AxB , where h¿ AxB " M ! 00x !N dy . F " !M dx $ g A yB , where g¿ A yB " N ! 00y !M dx

M dx $ N dy " 00N/ 0x 0M/ 0y "F Ax, yB " CdF Ax, yB ! 0.

d AmyB /dx " mQ, so that my " #mQ dx $ C. m " exp 3#P AxB dx 4dy/dx # P AxBy ! Q AxBdy/dx ! g AxBp A yB.

76 Chapter 2 First-Order Differential Equations

Chapter Summary

In Problems 1–30, solve the equation.

1. 2.

5. 6. 7.

9. 10.

11.

12.

13.

14.

15.

16.

17.

18.

19.

20.

21. 22. 23. Ay ! xB dx $ Ax $ yB dy " 0A2x ! 2y ! 8B dx $ Ax ! 3y ! 6B dy " 0

A y ! 2x ! 1B dx $ Ax $ y ! 4B dy " 0 dy du

$ y u

" !4uy!2

Ax2 ! 3y2B dx $ 2xy dy " 0 dy dx

" A2x $ y ! 1B2 dy du

$ 2y " y2

dy dx

$ y tan x $ sin x " 0

dy dx

" 2 ! 22x ! y $ 3 dx dt

! x

t ! 1 " t2 $ 2

dy dx

! y x

" x2sin 2x

A y3 $ 4exyB dx $ A2ex $ 3y2B dy " 0 dx dt

" 1 $ cos2 At ! xB$ 3 y!1/2 $ A1 $ x2 $ 2xy $ y2B!1 4 dy " 0 31 $ A1 $ x2 $ 2xy $ y2B!1 4 dxAx2 $ y2B dx $ 3xy dy " 0 dy dx

$ 2y x

" 2x2y2

t3y2 dt $ t4y!6 dy " 0 2xy3 dx ! A1 ! x2B dy " 03sin AxyB $ xy cos AxyB 4 dx $ 31 $ x2cos AxyB 4 dy " 0 dy dx

$ 3y x

" x2 ! 4x $ 3

Ax2 ! 2y!3B dy $ A2xy ! 3x2B dx " 0 dy dx

! 4y " 32x2 dy dx

" ex$y

y ! 1

Review Problems 77

REVIEW PROBLEMS

24. 25.

26.

27.

28.

29.

30.

In Problems 31–40, solve the initial value problem. 31.

32.

33.

34.

35.

36.

37.

38.

39.

40.

41. Express the solution to the following initial value problem using a definite integral:

Then use your expression and numerical integration to estimate to four decimal places.y(3)

dy dt

" 1

1 $ t2 ! y , y A2B " 3 .

dy dx

! 4y " 2xy2 , y A0B " !4 dy dx

! 2y x

" x!1y!1 , y A1B " 32y dx $ Ax2 $ 4B dy " 0 , y A0B " 4 A2x ! yB dx $ Ax $ y ! 3B dy " 0 , y A0B " 2y A1B " 0$ 3 cos A2x $ yB $ ey 4 dy " 0 , 32 cos A2x $ yB ! x2 4 dx A2y2 $ 4x2B dx ! xy dy " 0 , y A1B " !2 dy dx

! 2y x

" x2cos x , y ApB " 2At $ x $ 3B dt $ dx " 0 , x A0B " 1 dy dx

" ax y

$ y x b , y A1B " !4(x 3 ! y) dx $ x dy " 0 , y(1) " 3

dy dx

" Ax $ y $ 1B2 ! Ax $ y ! 1B2$ A3x 2y2 ! 6xy $ 2x2yB dy " 0A4xy3 ! 9y2 $ 4xy2B dx

dy dx

" x ! y ! 1 x $ y $ 5

A3x ! y ! 5B dx $ Ax ! y $ 1B dy " 0 dy dx

$ xy " 0

y Ax ! y ! 2B dx $ x Ay ! x $ 4B dy " 0A2y/x $ cos xB dx $ A2x/y $ sin yB dy " 0

1. An instructor at Ivey U. asserted: “All you need to know about first-order differential equations is how to solve those that are exact.” Give arguments that support and arguments that refute the instructor’s claim.

2. What properties do solutions to linear equations have that are not shared by solutions to either sepa- rable or exact equations? Give some specific exam- ples to support your conclusions.

78 Chapter 2 First-Order Differential Equations

TECHNICAL WRITING EXERCISES

3. Consider the differential equation

where a, b, and c are constants. Describe what hap- pens to the asymptotic behavior as of the solution when the constants a, b, and c are varied. Illustrate with figures and/or graphs.

x S $q

dy dx

" ay $ be!x , y A0B " c ,

Oil Spill in a Canal

In 1973 an oil barge collided with a bridge in the Mississippi River, leaking oil into the water at a rate estimated at 50 gallons per minute. In 1989 the Exxon Valdez spilled an estimated 11,000,000 gallons of oil into Prudhoe Bay in 6 hours†, and in 2010 the Deepwater Horizon well leaked into the Gulf of Mexico at a rate estimated to be 15,000 barrels per day†† (1 barrel = 42 gallons). In this project you are going to use differential equations to analyze a simplified model of the dissipation of heavy crude oil spilled at a rate of S ft3/sec into a flowing body of water. The flow region is a canal, namely a straight channel of rectangular cross section, w feet wide by d feet deep, having a constant flow rate of v ft/sec; the oil is presumed to float in a thin layer of thickness s (feet) on top of the water, without mixing.

In Figure 2.12, the oil that passes through the cross-section window in a short time )t occu- pies a box of dimensions s by w by v)t. To make the analysis easier, presume that the canal is conceptually partitioned into cells of length L ft. each, and that within each particular cell the oil instantaneously disperses and forms a uniform layer of thickness si(t) in cell i (cell 1 starts at the point of the spill). So, at time t, the ith cell contains si t wL ft

3 of oil. Oil flows out of cell i at a rate equal to si t wv ft

3/sec, and it flows into cell i at the rate si!1 t wv; it flows into the first cell at S ft3/sec.

BABA BA

†Cutler J. Cleveland (Lead Author); C Michael Hogan and Peter Saundry (Topic Editor). 2010. “Deepwater Horizon oil spill.” In: Encyclopedia of Earth, ed. Cutler J. Cleveland (Washington, D.C.: Environmental Information Coalition, National Council for Science and the Environment). ††Cutler J. Cleveland (Contributing Author); National Oceanic and Atmospheric Administration (Content source); Peter Saundry (Topic Editor). 2010. “Exxon Valdez oil spill.” In: Encyclopedia of Earth, ed. Cutler J. Cleveland (Washington, D.C.: Environmental Information Coalition, National Council for Science and the Environment).

Group Projects for Chapter 2

Figure 2.12 Oil leak in a canal.

(a) Formulate a system of differential equations and initial conditions for the oil thickness in the first three cells. Take S " 50 gallons/min, which was roughly the spillage rate for the Mississippi River incident, and take w " 200 ft, d " 25 ft, and v " 1 mi/hr (which are reasonable estimates for the Mississippi River†). Take L " 1000 ft.

(b) Solve for [Caution: Make sure your units are consistent.] (c) If the spillage lasts for T seconds, what is the maximum oil layer thickness in cell 1? (d) Solve for What is the maximum oil layer thickness in cell 2? (e) Probably the least tenable simplification in this analysis lies in regarding the layer

thickness as uniform over distances of length L. Reevaluate your answer to part (c) with L reduced to 500 ft. By what fraction does the answer change?

s2 AtB. s1 AtB.

80 Chapter 2 First-Order Differential Equations

Differential Equations in Clinical MedicineB

Figure 2.13 Lung ventilation pressures

lung elastic pressure, Pe , and residual pressure, Pex

airway-resistance pressure drop, Pr

Papp during inspiration

Courtesy of Philip Crooke, Vanderbilt University

In medicine, mechanical ventilation is a procedure that assists or replaces spontaneous breath- ing for critically ill patients, using a medical device called a ventilator. Some people attribute the first mechanical ventilation to Andreas Vesalius in 1555. Negative pressure ventilators (iron lungs) came into use in the 1940s–1950s in response to poliomyelitis (polio) epidemics. Philip Drinker and Louis Shaw are credited with its invention. Modern ventilators use positive pressure to inflate the lungs of the patient. In the ICU (intensive care unit), common indications for the ini- tiation of mechanical ventilation are acute respiratory failure, acute exacerbation of chronic obstructive pulmonary disease, coma, and neuromuscular disorders. The goals of mechanical ventilation are to provide oxygen to the lungs and to remove carbon dioxide.

In this project, we model the mechanical process performed by the ventilator. We make the following assumptions about this process of filling the lungs with air and then letting them deflate to some rest volume (see Figure 2.13).

(i) The length (in seconds) of each breath is fixed (ttot ) and is set by the clinician, with each breath being identical to the previous breath.

(ii) Each breath is divided into two parts: inspiration (air flowing into the patient) and expiration (air flowing out of the patient). We assume that inspiration takes place over the interval [0, ti] and expiration over the time interval [ti, ttot]. The time ti is called the inspiratory time.

†http://www.nps.gov/miss/riverfacts.htm

http://www.nps.gov/miss/riverfacts.htm

(iii) During inspiration the ventilator applies a constant pressure Papp to the patient’s air- way, and during expiration this pressure is zero, relative to atmospheric pressure. This is called pressure-controlled ventilation.

(iv) We assume that the pulmonary system (lung) is modeled by a single compartment. Hence, the action of the ventilator is similar to inflating a balloon and then releasing the pressure.

(v) At the airway there is a pressure balance:

(1)

where denotes pressure losses due to resistance to flow into and out of the lung, is the elastic pressure due to changes in volume of the lung, is a residual pressure that remains in the lung at the completion of a breath, and denotes the pressure applied to the airway. ( during inspiration and during expiration.) The residual pressure is called the end-expiratory pressure.

(vi) Let denote the volume of the lung at time t, with denoting its volume during inspiration and its volume during expiration. We assume that The number is called the tidal volume of the breath.

(vii) We assume that the resistive pressure is proportional to the flows into and out of the lung such that , and we assume that the proportionality constant R is the same for inspiration and expiration.

(viii) Furthermore, we assume that the elastic pressure is proportional to the instantaneous volume of the lung. That is, where the constant C is called the compliance of the lung.

Using the pressure equation in (1) together with the above assumptions, a mathematical model for the instantaneous volume in the single compartment lung is given by the following pair of first-order linear differential equations:

(2)

(3)

The initial conditions, as indicated in assumption (vi), are and . The constant is not known a priori but is determined from the end condition on the expiratory volume: . This will make each breath identical to the previous breath. To obtain a for- mula for , complete the following steps.

(a) Solve equation (2) for with the initial condition

(b) Solve equation (3) for with the initial condition .

(d) For R " 10 cm (H2O)/L/sec, C " 0.02 L/cm (H2O), Papp " 20 cm (H2O), ti " 1 sec and ttot " 3 sec, plot the graphs of Vi (t) and Ve(t) over the interval [0, ttot]. Compute Pex for

Pex " (eti /RC ! 1) Papp

ettot /RC ! 1 .

Vi (ti) " VT

Ve (ti) " VTVe (t)

Vi (0) " 0.Vi (t)

Pex Ve (ttot) " 0

Pex Ve (ti) " Vi (ti) " VTVi (0) " 0

R adVe dt b $ a1

C bVe $ Pex " 0 , ti * t * ttot .

R adVi dt b $ a1

C bVi $ Pex " Papp , 0 * t * ti ,

Pe " (1/C)V,

Pr " R(dV /dt ) Pr

Vi AtiB " VTVi A0B " Ve AttotB " 0.Ve AtB, ti * t * ttot , Vi AtB, 0 * t * ti ,V AtB Pex

Paw " 0Paw " Papp Paw

Pex PePr

Pr $ Pe $ Pex " Paw ,

Group Projects for Chapter 2 81

these parameters. (e) The mean alveolar pressure is the average pressure in the lung during inspiration and is

given by the formula

Compute this quantity using your expression for Vi(t) in part (a).

Torricelli’s Law of Fluid Flow Courtesy of Randall K. Campbell-Wright

How long does it take for water to drain through a hole in the bottom of a tank? Consider the tank pictured in Figure 2.14, which drains through a small, round hole. Torricelli’s law† states that when the surface of the water is at a height h, the water drains with the velocity it would have if it fell freely from a height h (ignoring various forms of friction).

(a) Show that the standard gravity differential equation

leads to the conclusion that an object that falls from a height will land with a velocity of

(b) Let A(h) be the cross-sectional area of the water in the tank at height h and a the area of the drain hole. The rate at which water is flowing out of the tank at time t can be expressed as the cross-sectional area at height h times the rate at which the height of the water is changing. Alternatively, the rate at which water flows out of the hole can be expressed as the area of the hole times the velocity of the draining water. Set these two equal to each other and insert Torricelli’s law to derive the differential equation

(4)

(c) The conical tank of Figure 2.14 has a radius of 30 cm when it is filled to an initial depth of 50 cm. A small round hole at the bottom has a diameter of 1 cm. Determine and a and then solve the differential equation in (4), thus deriving a formula relating time and the height of the water in this tank.

A AhB A AhB dh

dt " !a22gh .

!22gh A0B. h A0B d2h dt2

" !g

Pm " 1 ti

! ti

0 aVi (t)

C b dt $ Pex .

82 Chapter 2 First-Order Differential Equations

Figure 2.14 Conical tank

A(h)

50 cm

30 cm

†Historical Footnote: Evangelista Torricelli (1608–1647) invented the barometer and worked on computing the value of the acceleration of gravity as well as observing this principle of fluid flow.

(d) Use your solution to (c) to predict how long it will take for the tank to drain entirely. (e) Which would drain faster, the tank pictured or an upside-down conical tank of the same

dimensions draining through a hole of the same size (1-cm diameter)? How long would it take to drain the upside-down tank?

(f) Find a water tank and time how long it takes to drain. (You may be able to borrow a “separatory funnel” from your chemistry department or use a large water cooler.) The tank should be large enough to take several minutes to drain, and the drain hole should be large enough to allow water to flow freely. The top of the tank should be open (so that the water will not “glug”). Repeat steps (c) and (d) for your tank and compare the prediction of Torricelli’s law to your experimental results.

The Snowplow Problem To apply the techniques discussed in this chapter to real-world problems, it is necessary to trans- late these problems into questions that can be answered mathematically. The process of reformu- lating a real-world problem as a mathematical one often requires making certain simplifying assumptions. To illustrate this, consider the following snowplow problem:

One morning it began to snow very hard and continued snowing steadily throughout the day. A snowplow set out at 9:00 A.M. to clear a road, clearing 2 mi by 11:00 A.M. and an additional mile by 1:00 P.M. At what time did it start snowing?

To solve this problem, you can make two physical assumptions concerning the rate at which it is snowing and the rate at which the snowplow can clear the road. Because it is snowing steadily, it is reasonable to assume it is snowing at a constant rate. From the data given (and from our experience), the deeper the snow, the slower the snowplow moves. With this in mind, assume that the rate (in mph) at which a snowplow can clear a road is inversely proportional to the depth of the snow.

Two Snowplows

Courtesy of Alar Toomre, Massachusetts Institute of Technology

One day it began to snow exactly at noon at a heavy and steady rate. A snowplow left its garage at 1:00 P.M., and another one followed in its tracks at 2:00 P.M. (see Figure 2.15 on page 84).

(a) At what time did the second snowplow crash into the first? To answer this question, assume as in Project D that the rate (in mph) at which a snowplow can clear the road is inversely proportional to the depth of the snow (and hence to the time elapsed since the road was clear of snow). [Hint: Begin by writing differential equations for x(t) and y(t), the distances traveled by the first and second snowplows, respectively, at t hours past noon. To solve the differential equation involving y, let t rather than y be the dependent variable!]

(b) Could the crash have been avoided by dispatching the second snowplow at 3:00 P.M. instead?

Group Projects for Chapter 2 83

Clairaut Equations and Singular Solutions

An equation of the form

(5)

where the continuously differentiable function is evaluated at , is called a Clairaut equation.† Interest in these equations is due to the fact that (5) has a one-parameter family of solutions that consist of straight lines. Further, the envelope of this family—that is, the curve whose tangent lines are given by the family—is also a solution to (5) and is called the singular solution.

To solve a Clairaut equation:

(a) Differentiate equation (5) with respect to x and simplify to show that

(6)

(b) From (6), conclude that or Assume that and substitute back into (5) to obtain the family of straight-line solutions

where the parameter . This solution is the singular solution. (d) Use the above method to find the family of straight-line solutions and the singular solu-

tion to the equation

Here Sketch several of the straight-line solutions along with the singular solution on the same coordinate system. Observe that the straight-line solutions are all tangent to the singular solution.

(e) Repeat part (d) for the equation

x(dy/dx)3 ! y(dy/dx)2 $ 2 " 0 .

f AtB " 2t2. y " x ady

dx b $ 2 ady

dx b 2 . p " dy/dx

y " f A pB ! pf ¿A pB , x " !f ¿A pB , y " cx $ f AcB .

dy/dx " cf ¿ Ady/dxB " !x.dy/dx " c 3 x $ f ¿Ady/dxB 4 d2ydx2 " 0 , where f ¿ AtB " ddt f AtB .

t " dy/dxf AtB y ! x

dy dx

# f Ady /dxB ,

84 Chapter 2 First-Order Differential Equations

0 y(t) x(t) Miles from garage

Figure 2.15 Method of successive snowplows

†Historical Footnote: These equations were studied by Alexis Clairaut in 1734.

Multiple Solutions of a First-Order Initial Value Problem

Courtesy of Bruce W. Atkinson, Samford University

The initial value problem (IVP),

(7)

which was discussed in Example 9 and Problem 29 of Section 1.2, is an example of an IVP that has more than one solution. The goal of this project is to find all the solutions to (7) on ( ). It turns out that there are infinitely many! These solutions can be obtained by con- catenating the three functions for the constant 0 for and for

where as can be seen by completing the following steps:

(a) Show that if is a solution to the differential equation that is not zero on an open interval I, then on this interval must be of the form

for some constant c not in I. (b) Prove that if is a solution to the differential equation on

( , ) and where then for [Hint: Consider the sign of .]

(c ) Now let be a solution to the IVP (7) on ( ). Of course If vanishes at some point then let b be the largest of such points; otherwise, set

. Similarly, if g vanishes at some point then let a be the smallest (furthest to the left) of such points; otherwise, set . Here we allow and . (Because g is a continuous function, it can be proved that there always exist such largest and smallest points.) Using the results of parts (a) and (b) prove that if both a and b are finite, then g has the following form:

What is the form of g if ? If ? If both and ?

(d) Verify directly that the above concatenated function g is indeed a solution to the IVP (7) for all choices of a and b with Also sketch the graph of several of the solu- tion function g in part (c) for various values of a and b, including infinite values.

We have analyzed here a first-order IVP that not only fails to have a unique solution but has a solution set consisting of a doubly infinite family of functions (with a and b as the two parameters).

Utility Functions and Risk Aversion

Courtesy of James E. Foster, George Washington University

Would you rather have $5 with certainty or a gamble involving a 50% chance of receiving $1 and a 50% chance of receiving $11? The gamble has a higher expected value ($6); however, it also has a greater level of risk. Economists model the behavior of consumers or other agents facing

a * 2 * b.

a " !qb " $qa " !qb " $q

g(x) " % (x ! a)3 if x * a0 if a 6 x * b .(x ! b)3 if x 7 b

a " !qb " $qa " 2 x 6 2,b " 2

x 7 2, gg (2) " 0.!q, $qy " g(x)

f ¿ a * x * b.f (x) " 0a 6 b,f (a) " f (b) " 0,$q!q

dy /dx " 3y 2/3y " f (x) f (x) " (x ! c)3

f (x) dy /dx " 3y 2/3y " f (x)

a * 2 * b,x 7 b, (x ! b)3a * x * b,x 6 a,(x ! a)3

!q, $q

dy dx

" 3y2/3 , y A2B " 0 ,

Group Projects for Chapter 2 85

risky decisions with the help of a (von Neumann–Morgenstern) utility function u and the criterion of expected utility.

Rather than using expected values of the dollar payoffs, the payoffs are first transformed into utility levels and then weighted by probabilities to obtain expected utility. Following the sugges- tion of Daniel Bernoulli, we might set and then compare to [0.5 ln 1 $ 0.5 ln 11] 0.1969, which would result in the sure thing being chosen in this case rather than the gamble. This utility function is strictly concave, which corresponds to the agent being risk averse, or wanting to avoid gambles (unless of course the extra risk is sufficiently compensated by a high enough increase in the mean or expected payoff).

Alternatively, the utility function might be , which is strictly convex and corre- sponds to the agent being risk loving. This agent would surely select the above gamble. The case of occurs when the agent is risk neutral and would select according to the expected value of the payoff. It is normally assumed that at all payoff levels, x; in other words, higher payoffs are desirable.

In addition to knowing if an agent is risk averse or risk loving, economists are often inter- ested in knowing how risk averse (or risk loving) an agent is. Clearly this has something to do with the second derivative of the utility function. The measure of relative risk aversion of an agent with utility function and payoff x is defined as . Normally, is a function of the payoff level. However, economists often find it convenient to restrict considera- tion to utility functions for which is constant, say, for all x. It is easily shown that each of u(x) " ln x, u(x) " x2, and exhibits constant relative risk aversion (with levels

and respectively). A question naturally arises: What is the set of all utility functions that have constant relative risk aversion?

(a) State the second-order differential equation defined by the above question. (b) Convert this into a separable first-order differential equation for solve, and use the

solution to determine the possible forms that can take. (c) Integrate to obtain the set of all constant relative risk-aversion utility functions. This

class is used extensively throughout economics. (d) An alternative measure of risk aversion is the measure of absolute

risk aversion. Find the set of all utility functions exhibiting constant absolute risk aversion.

(e) Which functions are both constant absolute and constant relative risk-aversion utility functions?

For further reading, see, for example, the economics text Microeconomic Theory, by A. Mas-Colell, M. Whinston, and J. Green (Oxford University Press, Oxford, 1995).

Designing a Solar Collector You want to design a solar collector that will concentrate the sun’s rays at a point. By symmetry this surface will have a shape that is a surface of revolution obtained by revolving a curve about an axis. Without loss of generality, you can assume that this axis is the x-axis and the rays parallel to this axis are focused at the origin (see Figure 2.16). To derive the equation for the curve, proceed as follows:

(a) The law of reflection says that the angles and are equal. Use this and results from geometry to show that b " 2a.

u(x)

a(x) " u–(x)/u¿(x),

u¿(x) u¿(x),

s " 0,s " 1, s " !1, u(x) " x

r (x) " sr (x)

r (x)r (x) " !u–(x)x/u¿(x)u(x)

u¿(x) 7 0 u(x) " x

u(x) " x2

$ ln 5 $ 0.8047u(x) " ln x

86 Chapter 2 First-Order Differential Equations

(b) From calculus recall that . Use this, the fact that , and the double angle formula to show that

(8)

(d) Solve equation (8). [Hint: See Section 2.6.] (e) Describe the solutions and identify the type of collector obtained.

dy dx

" !x $ 2x2 $ y2

y .

y x

" 2 dy/dx

1 ! (dy/dx)2 .

y/x " tan bdy/dx " tan a

Group Projects for Chapter 2 87

Figure 2.16 Curve that generates a solar collector

tangent line

unknown curve

sun rays

α β

δ (x, y)

Asymptotic Behavior of Solutions to Linear Equations

To illustrate how the asymptotic behavior of the forcing term affects the solution to a linear equation, consider the equation

(9)

where the constant a is positive and is continuous on .

(a) Show that the general solution to equation (9) can be written in the form

where x0 is a nonnegative constant.

(b) If for , where k and are nonnegative constants, show that 0 y AxB 0 * 0 y Ax0B 0 e!aAx!x0B $ ka 31 ! e!aAx!x0B 4 for x ' x0 .x0x ' x00Q AxB 0 * k y AxB " y Ax0Be!aAx!x0B $ e!ax ! x

eatQ AtB dt , 3 0, q BQ AxB

dy dx

$ ay " Q AxB , Q AxB

where is continuous on Show that if

then

(d) Now show that if as , then any solution of equation (9) satisfies [Hint: Take in part (c).]

(e) As an application of part (d), suppose a brine solution containing kg of salt per liter at time t runs into a tank of water at a fixed rate and that the mixture, kept uniform by stirring, flows out at the same rate. Given that use the result of part (d) to determine the limiting concentration of the salt in the tank as (see Exer- cises 2.3, Problem 35).

t S q q AtB S b as t S q,

q AtB Q ~ AxB " b and z AxB " b/ay AxB S b/a as x S q. y AxBx S qQ AxB S b

0 z AxB ! y AxB 0 * 0 z Ax0B ! y Ax0B 0 e!aAx!x0B $ Ka 31 ! e!aAx!x0B 4 for x ' x0 . 0Q~ AxB ! Q AxB 0 * K for x ' x0 ,30, q B.Q~ AxB dz dx

$ az " Q ~ AxB ,

Q ~ AxBz AxB

88 Chapter 2 First-Order Differential Equations

Adopting the Babylonian practices of careful measurement and detailed observations, the ancient Greeks sought to comprehend nature by logical analysis. Aristotle’s convincing arguments that the world was not flat, but spherical, led the intellectuals of that day to ponder the question: What is the circumference of Earth? And it was astonishing that Eratosthenes managed to obtain a fairly accurate answer to this problem without having to set foot beyond the ancient city of Alexandria. His method involved certain assumptions and simplifications: Earth is a perfect sphere, the Sun’s rays travel parallel paths, the city of Syene was 5000 stadia due south of Alexandria, and so on. With these idealizations Eratosthenes created a mathemati- cal context in which the principles of geometry could be applied.†

Today, as scientists seek to further our understanding of nature and as engineers seek, on a more pragmatic level, to find answers to technical problems, the technique of repre- senting our “real world” in mathematical terms has become an invaluable tool. This process of mimicking reality by using the language of mathematics is known as mathematical modeling.

Formulating problems in mathematical terms has several benefits. First, it requires that we clearly state our premises. Real-world problems are often complex, involving several different and possibly interrelated processes. Before mathematical treatment can proceed, one must determine which variables are significant and which can be ignored. Often, for the relevant variables, relationships are postulated in the form of laws, formulas, theories, and the like. These assumptions constitute the idealizations of the model.

Mathematics contains a wealth of theorems and techniques for making logical deductions and manipulating equations. Hence, it provides a context in which analysis can take place free of any preconceived notions of the outcome. It is also of great practical importance that mathe- matics provides a format for obtaining numerical answers via a computer.

The process of building an effective mathematical model takes skill, imagination, and objective evaluation. Certainly an exposure to several existing models that illustrate various aspects of modeling can lead to a better feel for the process. Several excellent books and articles are devoted exclusively to the subject.†† In this chapter we concentrate on examples of

Mathematical Models and Numerical Methods Involving First-Order Equations

CHAPTER 3

MATHEMATICAL MODELING3.1

†For further reading, see, for example, The Mapmakers, by John Noble Wilford (Vintage Books, New York, 1982), Chapter 2. ††See, for example, A First Course in Mathematical Modeling, 4th ed., by F. T. Giordano, W. P. Fox, S. B. Horton, and M. D. Weir (Brooks/Cole, Pacific Grove, California, 2009) or Concepts of Mathematical Modeling, by W. J. Meyer (Dover Publications, Mineola, New York, 2004).

models that involve first-order differential equations. In studying these and in building your own models, the following broad outline of the process may be helpful.

Formulate the Problem Here you must pose the problem in such a way that it can be “answered” mathematically. This requires an understanding of the problem area as well as the mathematics. At this stage you may need to confer with experts in that area and read the relevant literature.

Develop the Model There are two things to be done here. First, you must decide which variables are important and which are not. The former are then classified as independent variables or dependent variables. The unimportant variables are those that have very little or no effect on the process. (For exam- ple, in studying the motion of a falling body, its color is usually of little interest.) The indepen- dent variables are those whose effect is significant and that will serve as input for the model.†

For the falling body, its shape, mass, initial position, initial velocity, and time from release are possible independent variables. The dependent variables are those that are affected by the inde- pendent variables and that are important to solving the problem. Again, for a falling body, its velocity, location, and time of impact are all possible dependent variables.

Second, you must determine or specify the relationships (for example, a differential equa- tion) that exist among the relevant variables. This requires a good background in the area and insight into the problem. You may begin with a crude model and then, based upon testing, refine the model as needed. For example, you might begin by ignoring any friction acting on the falling body. Then, if it is necessary to obtain a more acceptable answer, try to take into account any frictional forces that may affect the motion.

Test the Model Before attempting to “verify” a model by comparing its output with experimental data, the following questions should be considered:

Are the assumptions reasonable?

Are the equations dimensionally consistent? (For example, we don’t want to add units of force to units of velocity.)

Is the model internally consistent in the sense that equations do not contradict one another?

Do the relevant equations have solutions?

Are the solutions unique?

How difficult is it to obtain the solutions?

Do the solutions provide an answer for the problem being studied?

When possible, try to validate the model by comparing its predictions with any experimental data. Begin with rather simple predictions that involve little computation or analysis. Then, as the

90 Chapter 3 Mathematical Models and Numerical Methods Involving First-Order Equations

†In the mathematical formulation of the model, certain of the independent variables may be called parameters.

model is refined, check to see that the accuracy of the model’s predictions is acceptable to you. In some cases validation is impossible or socially, politically, economically, or morally unreason- able. For example, how does one validate a model that predicts when our Sun will die out?

Each time the model is used to predict the outcome of a process and hence solve a prob- lem, it provides a test of the model that may lead to further refinements or simplifications. In many cases a model is simplified to give a quicker or less expensive answer—provided, of course, that sufficient accuracy is maintained.

One should always keep in mind that a model is not reality but only a representation of reality. The more refined models may provide an understanding of the underlying processes of nature. For this reason applied mathematicians strive for better, more refined models. Still, the real test of a model is its ability to find an acceptable answer for the posed problem.

In this chapter we discuss various models that involve differential equations. Section 3.2, Compartmental Analysis, studies mixing problems and population models. Sections 3.3 through 3.5 are physics-based and examine heating and cooling, Newtonian mechanics, and electrical circuits. Finally Sections 3.6 and 3.7 introduce some numerical methods for solving first-order initial value problems. This will enable us to consider more realistic models that cannot be solved using the methods of Chapter 2.

Section 3.2 Compartmental Analysis 91

3.2 COMPARTMENTAL ANALYSIS Many complicated processes can be broken down into distinct stages and the entire system modeled by describing the interactions between the various stages. Such systems are called compartmental and are graphically depicted by block diagrams. In this section we study the basic unit of these systems, a single compartment, and analyze some simple processes that can be handled by such a model.

The basic one-compartment system consists of a function that represents the amount of a substance in the compartment at time t, an input rate at which the substance enters the com- partment, and an output rate at which the substance leaves the compartment (see Figure 3.1).

Because the derivative of x with respect to t can be interpreted as the rate of change in the amount of the substance in the compartment with respect to time, the one-compartment system suggests

(1)

as a mathematical model for the process.

dx dt ! input rate " output rate

x AtB

Output rate

Input rate x(t)

Figure 3.1 Schematic representation of a one-compartment system

Mixing Problems A problem for which the one-compartment system provides a useful representation is the mix- ing of fluids in a tank. Let represent the amount of a substance in a tank (compartment) at time t. To use the compartmental analysis model, we must be able to determine the rates at

x AtB

which this substance enters and leaves the tank. In mixing problems one is often given the rate at which a fluid containing the substance flows into the tank, along with the concentration of the substance in that fluid. Hence, multiplying the flow rate (volume/time) by the concentration (amount/volume) yields the input rate (amount/time).

The output rate of the substance is usually more difficult to determine. If we are given the exit rate of the mixture of fluids in the tank, then how do we determine the concentration of the substance in the mixture? One simplifying assumption that we might make is that the concen- tration is kept uniform in the mixture. Then we can compute the concentration of the substance in the mixture by dividing the amount by the volume of the mixture in the tank at time t. Multiplying this concentration by the exit rate of the mixture then gives the desired output rate of the substance. This model is used in Examples 1 and 2.

Consider a large tank holding 1000 L of pure water into which a brine solution of salt begins to flow at a constant rate of 6 L/min. The solution inside the tank is kept well stirred and is flowing out of the tank at a rate of 6 L/min. If the concentration of salt in the brine entering the tank is 0.1 kg/L, determine when the concentration of salt in the tank will reach 0.05 kg/L (see Figure 3.2).

We can view the tank as a compartment containing salt. If we let denote the mass of salt in the tank at time t, we can determine the concentration of salt in the tank by dividing by the volume of fluid in the tank at time t. We use the mathematical model described by equation (1) to solve for .

First we must determine the rate at which salt enters the tank. We are given that brine flows into the tank at a rate of 6 L/min. Since the concentration is 0.1 kg/L, we conclude that the input rate of salt into the tank is

(2)

We must now determine the output rate of salt from the tank. The brine solution in the tank is kept well stirred, so let’s assume that the concentration of salt in the tank is uniform. That is, the concentration of salt in any part of the tank at time t is just divided by the vol- ume of fluid in the tank. Because the tank initially contains 1000 L and the rate of flow into the tank is the same as the rate of flow out, the volume is a constant 1000 L. Hence, the output rate of salt is

(3)

The tank initially contained pure water, so we set . Substituting the rates in (2) and (3) into equation (1) then gives the initial value problem

(4)

as a mathematical model for the mixing problem.

dx dt

! 0.6 " 3x

500 , x A0B ! 0 ,

x A0B ! 0A6 L/minB c x AtB

1000 kg/L d ! 3x AtB

500 kg/min .

x AtB A6 L/minB A0.1 kg/ LB ! 0.6 kg/min .

x AtB x AtBx AtB

x AtB

92 Chapter 3 Mathematical Models and Numerical Methods Involving First-Order Equations

x(t)

1000 L

x (0) = 0 kg

6 L/min 0.1 kg/L

6 L/min

Figure 3.2 Mixing problem with equal flow rates

Example 1

Solution

The equation in (4) is separable (and linear) and easy to solve. Using the initial condition to evaluate the arbitrary constant, we obtain

(5)

Thus, the concentration of salt in the tank at time t is

To determine when the concentration of salt is 0.05 kg/L, we set the right-hand side equal to 0.05 and solve for t. This gives

and hence

min .

Consequently the concentration of salt in the tank will be 0.05 kg/L after 115.52 min. ◆

From equation (5), we observe that the mass of salt in the tank steadily increases and has the limiting value

kg .

Thus, the limiting concentration of salt in the tank is 0.1 kg/L, which is the same as the con- centration of salt in the brine flowing into the tank. This certainly agrees with our expectations!

It might be interesting to see what would happen to the concentration if the flow rate into the tank is greater than the flow rate out.

For the mixing problem described in Example 1, assume now that the brine leaves the tank at a rate of 5 L/min instead of 6 L/min, with all else being the same (see Figure 3.3). Determine the concentration of salt in the tank as a function of time.

The difference between the rate of flow into the tank and the rate of flow out is 1 L/min, so the volume of fluid in the tank after t minutes is L. Hence, the rate at which salt leaves the tank is

A5 L/minB c x AtB 1000 # t

kg/ L d ! 5x AtB 1000 # t

kg/min .

A1000 # tB 6 " 5 !

lim tSq

x AtB ! lim tSq

100 A1 " e"3t/500B ! 100

t ! 500 ln 2

3 ! 115.52

0.1 A1 " e"3t/500B ! 0.05 or e"3t/500 ! 0.5 , x AtB

1000 ! 0.1 A1 " e"3t/500B kg/L .

x AtB ! 100 A1 " e"3t/500B .x A0B ! 0 Section 3.2 Compartmental Analysis 93

x(t)

? L

x (0) = 0 kg

6 L/min 0.1 kg/L

5 L/min

Figure 3.3 Mixing problem with unequal flow rates

Example 2

Solution

Using this in place of (3) for the output rate gives the initial value problem

(6)

as a mathematical model for the mixing problem. The differential equation in (6) is linear, so we can use the procedure outlined on page 48 to

solve for . The integrating factor is Thus,

Using the initial condition , we find and thus the solution to (6) is

Hence, the concentration of salt in the tank at time t is

(7) kg/L . ◆

As in Example 1, the concentration given by (7) approaches 0.1 kg/L as . How- ever, in Example 2 the volume of fluid in the tank becomes unbounded, and when the tank begins to overflow, the model in (6) is no longer appropriate.

Population Models How does one predict the growth of a population? If we are interested in a single population, we can think of the species as being contained in a compartment (a petri dish, an island, a country, etc.) and study the growth process as a one-compartment system.

Let be the population at time t. While the population is always an integer, it is usually large enough so that very little error is introduced in assuming that is a continuous function. We now need to determine the growth (input) rate and the death (output) rate for the population.

Let’s consider a population of bacteria that reproduce by simple cell division. In our model, we assume that the growth rate is proportional to the population present. This assump- tion is consistent with observations of bacterial growth. As long as there are sufficient space and ample food supply for the bacteria, we can also assume that the death rate is zero. (Remember that in cell division, the parent cell does not die, but becomes two new cells.) Hence, a mathematical model for a population of bacteria is

(8)

where k1 $ 0 is the proportionality constant for the growth rate and p0 is the population at time t ! 0. For human populations the assumption that the death rate is zero is certainly wrong! However, if we assume that people die only of natural causes, we might expect the death rate also to be proportional to the size of the population. That is, we revise (8) to read

(9) dp dt

! k1p " k2p ! Ak1 " k2Bp ! kp ,

dp dt

! k1p , p A0B ! p0 ,

p AtBp AtB

t S q

x AtB 1000 # t

! 0.1 31 " A1000B6 A1000 # tB"6 4 x AtB ! 0.1 3A1000 # tB " A1000B6 A1000 # tB"5 4 .c ! "0.1 A1000B6,x A0B ! 0

x AtB ! 0.1 A1000 # tB # c A1000 # tB"5 . A1000 # tB5x ! 0.1 A1000 # tB6 # c d dt

3 A1000 # tB5x 4 ! 0.6 A1000 # tB5m AtB ! A1000 # tB 5.x AtB

dx dt

! 0.6 " 5x

1000 # t , x A0B ! 0 ,

94 Chapter 3 Mathematical Models and Numerical Methods Involving First-Order Equations

where and k2 is the proportionality constant for the death rate. Let’s assume that k1 $ k2 so that k $ 0. This gives the mathematical model

(10)

which is called the Malthusian,† or exponential, law of population growth. This equation is separable, and solving the initial value problem for gives

(11)

To test the Malthusian model, let’s apply it to the demographic history of the United States.

In 1790 the population of the United States was 3.93 million, and in 1890 it was 62.98 million. Using the Malthusian model, estimate the U.S. population as a function of time.

If we set t ! 0 to be the year 1790, then by formula (11) we have

(12)

where is the population in millions. One way to obtain a value for k would be to make the model fit the data for some specific year, such as 1890 (t ! 100 years).†† We have

Solving for k yields

Substituting this value in equation (12), we find

(13) ◆

In Table 3.1 on page 96 we list the U.S. population as given by the U.S. Bureau of the Census and the population predicted by the Malthusian model using equation (13). From Table 3.1 we see that the predictions based on the Malthusian model are in reasonable agreement with the census data until about 1900. After 1900 the predicted population is too large, and the Malthusian model is unacceptable.

We remark that a Malthusian model can be generated using the census data for any two different years. We selected 1790 and 1890 for purposes of comparison with the logistic model that we now describe.

The Malthusian model considered only death by natural causes. What about premature deaths due to malnutrition, inadequate medical supplies, communicable diseases, violent crimes, etc.? These factors involve a competition within the population, so we might assume

p AtB ! A3.93Be A0.027742Bt . k !

ln A62.98B " ln A3.93B 100

! 0.027742 .

p A100B ! 62.98 ! A3.93Be100k . p AtBp AtB ! A3.93Bekt ,

p AtB ! p0ekt . p AtB

dp dt ! kp , p A0B ! p0 ,

k J k1 " k2

Section 3.2 Compartmental Analysis 95

†Historical Footnote: Thomas R. Malthus (1766–1834) was a British economist who studied population models. ††The choice of the year 1890 is purely arbitrary, of course; a more democratic (and better) way of extracting parameters from data is described after Example 4.

Example 3

Solution

that another component of the death rate is proportional to the number of two-party interactions. There are such possible interactions for a population of size p. Thus, if we combine the birth rate (8) with the death rate and rearrange constants, we get the logistic model

(14)

where

Equation (14) has two equilibrium (constant) solutions: and . The non- equilibrium solutions can be found by separating variables and using the integral table on the inside front cover:

or or `1 " p1 p ` ! c2e"Ap1t .1p1 ln ` p " p1p ` ! "At # c1" dpp Ap " p1B ! "A"dt

p AtB # 0p AtB # p1A ! k3/2 and p1 ! A2k1/k3B # 1. dp dt ! "Ap

Ap " p1B , p A0B ! p0 , dp dt

! k1p " k3 p Ap " 1B

p Ap " 1B /2

96 Chapter 3 Mathematical Models and Numerical Methods Involving First-Order Equations

TABLE 3.1 A Comparison of the Malthusian and Logistic Models with U.S. Census Data (Population is given in Millions)

Malthusian Logistic Logistic Year U.S. Census (Example 3) (Example 4) (Least Squares)

1790 3.93 3.93 3.93 4.11 1800 5.31 5.19 5.30 0.0312 5.42 1810 7.24 6.84 7.13 0.0299 7.14 1820 9.64 9.03 9.58 0.0292 9.39 1830 12.87 11.92 12.82 0.0289 12.33 1840 17.07 15.73 17.07 0.0302 16.14 1850 23.19 20.76 22.60 0.0310 21.05 1860 31.44 27.40 29.70 0.0265 27.33 1870 39.82 36.16 38.66 0.0235 35.28 1880 50.19 47.72 49.71 0.0231 45.21 1890 62.98 62.98 62.98 0.0207 57.41 1900 76.21 83.12 78.42 0.0192 72.11 1910 92.23 109.69 95.73 0.0162 89.37 1920 106.02 144.76 114.34 0.0146 109.10 1930 123.20 191.05 133.48 0.0106 130.92 1940 132.16 252.13 152.26 0.0106 154.20 1950 151.33 333.74 169.90 0.0156 178.12 1960 179.32 439.12 185.76 0.0145 201.75 1970 203.30 579.52 199.50 0.0116 224.21 1980 226.54 764.80 211.00 0.0100 244.79 1990 248.71 1009.33 220.38 0.0110 263.01 2000 281.42 1332.03 227.84 0.0107 278.68 2010 308.75 1757.91 233.68 291.80 2020 ? 2319.95 238.17 302.56

1 p

dp dt

If at t ! 0, and then solving for , we find

(15)

The function given in (15) is called the logistic function, and graphs of logistic curves are displayed in Figure 3.4.† Note that for and , the limit population as , is .

Let’s test the logistic model on the population growth of the United States.

Taking the 1790 population of 3.93 million as the initial population and given the 1840 and 1890 populations of 17.07 and 62.98 million, respectively, use the logistic model to estimate the population at time t.

With t ! 0 corresponding to the year 1790, we know that p0 ! 3.93. We must now determine the parameters A, p1 in equation (15). For this purpose, we use the given facts that ! 17.07 and ! 62.98; that is,

(16)

(17)

Equations (16) and (17) are two nonlinear equations in the two unknowns A, p1. To solve such a system, we would usually resort to a numerical approximation scheme such as Newton’s method. However, for the case at hand, it is possible to find the solutions directly because the data are given at times ta and tb with tb ! 2ta (see Problem 12). Carrying out the algebra described in Problem 12, we ultimately find that

(18)

Thus, the logistic model for the given data is

(19) ◆p AtB ! 989.50 3.93 # A247.85Be"A0.030463B t .

p1 ! 251.7812 and A ! 0.0001210 .

62.98 ! 3.93 p1

3.93 # Ap1 " 3.93Be"100Ap1 . 17.07 !

3.93 p1 3.93 # Ap1 " 3.93Be"50Ap1 ,

p A100B p A50B

p1 t S qp0 7 0A 7 0

p AtB p AtB ! p1

1 " c3e"Ap1 t !

p0p1 p0 # A p1 " p0Be"Ap1 t .

p AtBc3 ! 1 " p1/ p0,p AtB ! p0

Section 3.2 Compartmental Analysis 97

p 0 > p1

p 0

0 < p 0 < p1

p 0

(a) (b)

p1 p1

Figure 3.4 The logistic curves

†Historical Footnote: The logistic model for population growth was first developed by P. F. Verhulst around 1840.

Example 4

Solution

The population data predicted by (19) are displayed in column 4 of Table 3.1. As you can see, these predictions are in better agreement with the census data than the Malthusian model is. And, of course, the agreement is perfect in the years 1790, 1840, and 1890. However, the choice of these particular years for estimating the parameters p0, A, and p1 is quite arbitrary, and we would expect that a more robust model would use all of the data, in some way, for the estimation. One way to implement this idea is the following.

Observe from equation (14) that the logistic model predicts a linear relationship between and p:

with Ap1 as the intercept and "A as the slope. In column five of Table 3.1, we list values of , which are estimated from centered differences according to

(20)

(see Problem 16). In Figure 3.5 these estimated values of are plotted against p in what is called a scatter diagram. The linear relationship predicted by the logistic model sug- gests that we approximate the plot by a straight line. A standard technique for doing this is the so-called least-squares linear fit, which is discussed in Appendix E. This yields the straight line

which is also depicted in Figure 3.5. Now with A ! 0.00008231 and ! 341.4, we can solve equation (15) for p0:

(21) p0 ! p AtBp1e"Ap1t

p1 " p AtB 31 " e"Ap1t 4 . p1 ! A0.0280960/AB

1 p

dp dt

! 0.0280960 " 0.00008231 p ,

Adp/dtB /p 1

p AtB dpdt AtB ! 1p AtB p At # 10B " p At " 10B20 Adp/dtB /p

1 p

dp dt

! Ap1 " Ap ,

Adp/dtB /p

98 Chapter 3 Mathematical Models and Numerical Methods Involving First-Order Equations

p (millions)

p 1

0.03

0.02

0.01

250200 300150100500

dt dp

Figure 3.5 Scatter data and straight line fit

By averaging the right-hand side of (21) over all the data, we obtain the estimate Finally, the insertion of these estimates for the parameters in equation (15) leads to the predic- tions listed in column six of Table 3.1.

Note that this model yields p1 ! 341.4 million as the limit on the future population of the United States.

p0 ! 4.107.

Section 3.2 Compartmental Analysis 99

1. A brine solution of salt flows at a constant rate of 8 L/min into a large tank that initially held 100 L of brine solution in which was dissolved 0.5 kg of salt. The solution inside the tank is kept well stirred and flows out of the tank at the same rate. If the con- centration of salt in the brine entering the tank is 0.05 kg/L, determine the mass of salt in the tank after t min. When will the concentration of salt in the tank reach 0.02 kg/L?

2. A brine solution of salt flows at a constant rate of 6 L/min into a large tank that initially held 50 L of brine solution in which was dissolved 0.5 kg of salt. The solution inside the tank is kept well stirred and flows out of the tank at the same rate. If the concen- tration of salt in the brine entering the tank is 0.05 kg/L, determine the mass of salt in the tank after t min. When will the concentration of salt in the tank reach 0.03 kg/L?

3. A nitric acid solution flows at a constant rate of 6 L/min into a large tank that initially held 200 L of a 0.5% nitric acid solution. The solution inside the tank is kept well stirred and flows out of the tank at a rate of 8 L/min. If the solution entering the tank is 20% nitric acid, determine the volume of nitric acid in the tank after t min. When will the percentage of nitric acid in the tank reach 10%?

4. A brine solution of salt flows at a constant rate of 4 L/min into a large tank that initially held 100 L of pure water. The solution inside the tank is kept well stirred and flows out of the tank at a rate of 3 L/min. If the concentration of salt in the brine entering the tank is 0.2 kg/L, determine the mass of salt in the tank after t min. When will the concentration of salt in the tank reach 0.1 kg/L?

5. A swimming pool whose volume is 10,000 gal con- tains water that is 0.01% chlorine. Starting at t ! 0, city water containing 0.001% chlorine is pumped into the pool at a rate of 5 gal/min. The pool water flows out at the same rate. What is the percentage of

chlorine in the pool after 1 h? When will the pool water be 0.002% chlorine?

6. The air in a small room 12 ft by 8 ft by 8 ft is 3% carbon monoxide. Starting at t ! 0, fresh air con- taining no carbon monoxide is blown into the room at a rate of 100 ft3/min. If air in the room flows out through a vent at the same rate, when will the air in the room be 0.01% carbon monoxide?

7. Beginning at time t ! 0, fresh water is pumped at the rate of 3 gal/min into a 60-gal tank initially filled with brine. The resulting less-and-less salty mixture overflows at the same rate into a second 60-gal tank that initially contained only pure water, and from there it eventually spills onto the ground. Assuming perfect mixing in both tanks, when will the water in the second tank taste saltiest? And exactly how salty will it then be, compared with the original brine?

8. A tank initially contains of salt dissolved in 200 gal of water, where is some positive number. Starting at time , water containing 0.5 lb of salt per gallon enters the tank at a rate of 4 gal/min, and the well-stirred solution leaves the tank at the same rate. Letting be the concentration of salt in the tank at time show that the limiting concentration— that is, —is 0.5 lb/gal.

9. In 1990 the Department of Natural Resources released 1000 splake (a crossbreed of fish) into a lake. In 1997 the population of splake in the lake was estimated to be 3000. Using the Malthusian law for population growth, estimate the population of splake in the lake in the year 2020.

10. Use a sketch of the phase line (see Group Project C, Chapter 1) to argue that any solution to the mixing problem model

approaches the equilibrium solution as t approaches that is, is a sink.a/b#q;

x AtB # a/b dx dt

! a " bx ; a, b 7 0 ,

limtSqc(t) t, c (t)

t ! 0 s0

s0 lb

3.2 EXERCISES

11. Use a sketch of the phase line (see Group Project C, Chapter 1) to argue that any solution to the logistic model

where a, b, and p0 are positive constants, approaches the equilibrium solution as t approaches

12. For the logistic curve (15), assume and are given with . Show

that

[Hint: Equate the expressions (21) for p0 at times ta and tb. Set and and solve for . Insert into one of the earlier expres- sions and solve for p1.]

13. In Problem 9, suppose we have the additional infor- mation that the population of splake in 2004 was estimated to be 5000. Use a logistic model to esti- mate the population of splake in the year 2020. What is the predicted limiting population? [Hint: Use the formulas in Problem 12.]

14. In 1980 the population of alligators on the Kennedy Space Center grounds was estimated to be 1500. In 2006 the population had grown to an estimated 6000. Using the Malthusian law for population growth, estimate the alligator population on the Kennedy Space Center grounds in the year 2020.

15. In Problem 14, suppose we have the additional infor- mation that the population of alligators on the grounds of the Kennedy Space Center in 1993 was estimated to be 4100. Use a logistic model to esti- mate the population of alligators in the year 2020. What is the predicted limiting population? [Hint: Use the formulas in Problem 12.]

16. Show that for a differentiable function p(t), we have

which is the basis of the centered difference approx- imation used in (20).

lim hS0

p At # hB " p At " hB

2h ! p¿ AtB ,

x2 ! exp A"Ap1tbBx ! exp A"Ap1taB A !

p1ta ln c pb A pa " p0B

p0 A pb " paB d . p1 ! c papb " 2p0 pb # p0 pap2a " p0 pb d pa ,

tb ! 2ta Ata 7 0Bpb J p AtbB pa J p AtaB #q.

p AtB # a/b dp dt

! Aa " bpBp ; p At0B ! p0 ,

100 Chapter 3 Mathematical Models and Numerical Methods Involving First-Order Equations

17. (a) For the U.S. census data, use the forward differ- ence approximation to the derivative, that is,

to recompute column 5 of Table 3.1. (b) Using the data from part (a), determine the con-

stants A, p1 in the least-squares fit

(see Appendix E). (c) With the values for A and p1 found in part (b),

determine p0 by averaging formula (21) over the data.

(d) Substitute A, p1, and p0 as determined above into the logistic formula (15) and calculate the popu- lations predicted for each of the years listed in Table 3.1.

(e) Compare this model with that of the centered difference-based model in column 6 of Table 3.1.

18. Using the U.S. census data in Table 3.1 for 1900, 1920, and 1940 to determine parameters in the logistic equation model, what populations does the model predict for 2000 and 2010? Compare your answers with the census data for those years.

19. The initial mass of a certain species of fish is 7 million tons. The mass of fish, if left alone, would increase at a rate proportional to the mass, with a proportion- ality constant of 2/yr. However, commercial fishing removes fish mass at a constant rate of 15 million tons per year. When will all the fish be gone? If the fishing rate is changed so that the mass of fish remains constant, what should that rate be?

20. From theoretical considerations, it is known that light from a certain star should reach Earth with intensity I0. However, the path taken by the light from the star to Earth passes through a dust cloud, with absorption coefficient 0.1/light-year. The light reaching Earth has intensity . How thick is the dust cloud? (The rate of change of light intensity with respect to thickness is proportional to the inten- sity. One light-year is the distance traveled by light during 1 yr.)

21. A snowball melts in such a way that the rate of change in its volume is proportional to its surface area. If the snowball was initially 4 in. in diameter and after 30 min its diameter is 3 in., when will its diameter be 2 in.? Mathematically speaking, when will the snowball disappear?

1 /2 I0

1 p

dp dt

! Ap1 " Ap

p AtB dpdt AtB ! 1p AtB p At # 10B " p AtB10 ,

22. Suppose the snowball in Problem 21 melts so that the rate of change in its diameter is proportional to its surface area. Using the same given data, deter- mine when its diameter will be 2 in. Mathematically speaking, when will the snowball disappear?

In Problems 23–27, assume that the rate of decay of a radioactive substance is proportional to the amount of the substance present. The half-life of a radioactive sub- stance is the time it takes for one-half of the substance to disintegrate. 23. If initially there are 50 g of a radioactive substance and

after 3 days there are only 10 g remaining, what per- centage of the original amount remains after 4 days?

24. If initially there are 300 g of a radioactive substance and after 5 yr there are 200 g remaining, how much time must elapse before only 10 g remain?

25. Carbon dating is often used to determine the age of a fossil. For example, a humanoid skull was found in a cave in South Africa along with the remains of a campfire. Archaeologists believe the age of the skull to be the same age as the campfire. It is determined that only 2% of the original amount of carbon-14 remains in the burnt wood of the campfire. Estimate the age of the skull if the half-life of carbon-14 is about 5600 years.

Section 3.3 Heating and Cooling of Buildings 101

26. To see how sensitive the technique of carbon dating of Problem 25 is, (a) Redo Problem 25 assuming the half-life of

carbon-14 is 5550 yr. (b) Redo Problem 25 assuming 3% of the

original mass remains. (c) If each of the figures in parts (a) and (b)

represents a 1% error in measuring the two parameters of half-life and percent of mass remaining, to which parameter is the model more sensitive?

27. The only undiscovered isotopes of the two unknown elements hohum and inertium (sym- bols Hh and It) are radioactive. Hohum decays into inertium with a decay constant of 2/yr, and inertium decays into the nonradioactive isotope of bunkum (symbol Bu) with a decay constant of 1/yr. An initial mass of 1 kg of hohum is put into a nonradiaoctive container, with no other source of hohum, inertium, or bunkum. How much of each of the three elements is in the container after t yr? (The decay constant is the constant of proportionality in the statement that the rate of loss of mass of the element at any time is proportional to the mass of the element at that time.)

3.3 HEATING AND COOLING OF BUILDINGS Our goal is to formulate a mathematical model that describes the 24-hr temperature profile inside a building as a function of the outside temperature, the heat generated inside the build- ing, and the furnace heating or air conditioner cooling. From this model we would like to answer the following three questions:

(a) How long does it take to change the building temperature substantially?

(b) How does the building temperature vary during spring and fall when there is no furnace heating or air conditioning?

(c) How does the building temperature vary in summer when there is air conditioning or in the winter when there is furnace heating?

A natural approach to modeling the temperature inside a building is to use compartmental analysis. Let represent the temperature inside the building at time t and view the building as a single compartment. Then the rate of change in the temperature is determined by all the factors that generate or dissipate heat.

We will consider three main factors that affect the temperature inside the building. First is the heat produced by people, lights, and machines inside the building. This causes a rate of increase in temperature that we will denote by . Second is the heating (or cooling) suppliedH AtB

T AtB

by the furnace (or air conditioner). This rate of increase (or decrease) in temperature will be represented by . In general, the additional heating rate and the furnace (or air condi- tioner) rate are described in terms of energy per unit time (such as British thermal units per hour). However, by multiplying by the heat capacity of the building (in units of degrees temperature change per unit heat energy), we can express the two quantities and in terms of temperature per unit time.

The third factor is the effect of the outside temperature on the temperature inside the building. Experimental evidence has shown that this factor can be modeled using Newton’s law of cooling. This law states that the rate of change in the temperature is proportional to the difference between the outside temperature and the inside temperature . That is, the rate of change in the building temperature due to is

The positive constant K depends on the physical properties of the building, such as the number of doors and windows and the type of insulation, but K does not depend on M, T, or t. Hence, when the outside temperature is greater than the inside temperature, " $ 0 and there is an increase in the building temperature due to . On the other hand, when the outside temperature is less than the inside temperature, then " % 0 and the building temper- ature decreases.

Summarizing, we find

(1)

where the additional heating rate is always nonnegative and is positive for furnace heating and negative for air conditioner cooling. A more detailed model of the temperature dynamics of the building could involve more variables to represent different temperatures in different rooms or zones. Such an approach would use compartmental analysis, with the rooms as different compartments (see Problems 35–37, Exercises 5.2).

Because equation (1) is linear, it can be solved using the method discussed in Section 2.3. Rewriting (1) in the standard form

(2)

where

(3)

we find that the integrating factor is

To solve (2), multiply each side by and integrate:

eKtT AtB ! " eKtQ AtBdt # C . eKt

dT dt

AtB # KeKtT AtB ! eKtQ AtB , eKt

m AtB ! exp a"K dtb ! eKt . Q AtB J KM AtB # H AtB # U AtB , P AtB J K , dT dt

AtB # P AtBT AtB ! Q AtB ,

U AtBH AtB dT dt ! K 3M AtB " T AtB 4 # H AtB # U AtB ,

T AtBM AtBM AtB T AtBM AtB

K 3M AtB " T AtB 4 . M AtB T AtBM AtB T AtB

M AtB U AtBH AtBU

AtB H AtBU AtB 102 Chapter 3 Mathematical Models and Numerical Methods Involving First-Order Equations

Solving for gives

(4)

Suppose at the end of the day (at time t0), when people leave the building, the outside tempera- ture stays constant at M0, the additional heating rate H inside the building is zero, and the furnace/air conditioner rate U is zero. Determine , given the initial condition .

With M ! M0, H ! 0, and U ! 0, equation (4) becomes

Setting t ! t0 and using the initial value T0 of the temperature, we find that the constant C is . Hence,

(5) ◆

When M0 % T0, the solution in (5) decreases exponentially from the initial temperature T0 to the final temperature M0. To determine a measure of the time it takes for the temperature to change “substantially,” consider the simple linear equation whose solutions have the form Now, as the function either decays exponen- tially or grows exponentially . In either case the time it takes for to change from to is just because

The quantity which is independent of , is called the time constant for the equation. For linear equations of the more general form we again refer to as the time constant.

Returning to Example 1, we see that the temperature satisfies the equations

for M0 a constant. In either case, the time constant is just , which represents the time it takes for the temperature difference to change from to . We also call the time constant for the building (without heating or air conditioning). A typical value for the time constant of a building is 2 to 4 hr, but the time constant can be much shorter if windows are open or if there is a fan circulating air. Or it can be much longer if the building is well insulated.

In the context of Example 1, we can use the notion of time constant to answer our initial question (a): The building temperature changes exponentially with a time constant of . An answer to question (b) about the temperature inside the building during spring and fall is given in the next example.

1/K

1/KAT0 " M0B /eT0 " M0T " M0 1/K dT dt

AtB ! " KT AtB # KM0 , d AT " M0Bdt AtB ! "K 3T AtB " M0 4 , T AtB 1/ 0a 0dA/dt ! "aA # g AtB,

A A0B1/ 0a 0 , A a1 a b ! A A0Be"aA1/aB ! A A0B

e .

1/aA A0B /e A ! 0.368 A A0BBA A0B A AtBAa 6 0BAa 7 0B A AtBt S #q,A AtB ! A A0Be "at. dA/dt ! "aA,

T AtB ! M0 # AT0 " M0Be"KAt" t0B . AT0 " M0B exp AKt0B

! M0 # Ce "Kt .

T AtB ! e"Kt c " eKtKM0 dt # C d ! e"Kt 3M0eKt # C 4 T At0B ! T0T AtB

! e"Kt e " eKt 3KM AtB # H AtB # U AtB 4dt # C f . T AtB ! e"Kt" eKtQ AtBdt # Ce"Kt

T AtB Section 3.3 Heating and Cooling of Buildings 103

Example 1

Solution

Find the building temperature if the additional heating rate is equal to the constant H0, there is no heating or cooling and the outside temperature M varies as a sine wave over a 24-hr period, with its minimum at t ! 0 (midnight) and its maximum at t ! 12 (noon). That is,

where B is a positive constant, M0 is the average outside temperature, and radians/hr. (This could be the situation during the spring or fall when there is neither

furnace heating nor air conditioning.)

The function in (3) is now

Setting we can rewrite Q as

(6)

where KB0 represents the daily average value of ; that is,

When the forcing function in (6) is substituted into the expression for the temperature in equation (4), the result (after using integration by parts) is

(7)

where

The constant C is chosen so that at midnight , the value of the temperature T is equal to some initial temperature T0. Thus,

◆

Notice that the third term in solution (7) involving the constant C tends to zero exponentially. The constant term B0 in (7) is equal to and represents the daily average temperature inside the building (neglecting the exponential term). When there is no additional heating rate inside the building , this average temperature is equal to the average outside temperature M0. The term in (7) represents the sinusoidal variation of temperature inside the building responding to the outside temperature variation. Since can be written in the form

(8)

where tan (see Problem 16), the sinusoidal variation inside the building lags behind the outside variation by hours. Further, the magnitude of the variation inside the building is slightly less, by a factor of than the outside variation. The angular frequency31 # Av/KB2 4 "1/2,f/vf ! v/K

F AtB ! 31 # Av/KB2 4"1/ 2 cos Avt " fB , F AtB BF AtBAH0 ! 0B

M0 # H0/K

C ! T0 " B0 # BF A0B ! T0 " B0 # B1 # Av/KB2 . At ! 0B

F AtB J cos vt # Av/KBsin vt 1 # Av/KB2 .

! B0 " BF AtB # Ce"Kt , T AtB ! e "Kt c " eKt AKB0 " KB cos vtB dt # C d

Q AtB KB0 !

1 24 "

0 Q AtBdt .

Q AtBQ AtB ! K AB0 " B cos vtB , B0 J M0 # H0/K,

Q AtB ! K AM0 " B cos vtB # H0 .Q AtB p/12

v ! 2p/24 !

M AtB ! M0 " B cos vt , AU AtB # 0B , H AtBT AtB

104 Chapter 3 Mathematical Models and Numerical Methods Involving First-Order Equations

Example 2

Solution

of variation radians/hr (which is about ). Typical values for the dimensionless ratio lie between and 1. For this range, the lag between inside and outside temperature is approximately 1.8 to 3 hr and the magnitude of the inside variation is between 89% and 71% of the variation outside. Figure 3.6 shows the 24-hr sinusoidal variation of the outside tempera- ture for a typical moderate day as well as the temperature variations inside the building for a dimensionless ratio of unity, which corresponds to a time constant of approximately 4 hr. In sketching the latter curve, we have assumed that the exponential term has died out.

Suppose, in the building of Example 2, a simple thermostat is installed that is used to compare the actual temperature inside the building with a desired temperature TD. If the actual tempera- ture is below the desired temperature, the furnace supplies heating; otherwise, it is turned off. If the actual temperature is above the desired temperature, the air conditioner supplies cooling; otherwise, it is off. (In practice, there is some dead zone around the desired temperature in which the temperature difference is not sufficient to activate the thermostat, but that is to be ignored here.) Assuming that the amount of heating or cooling supplied is proportional to the difference in temperature—that is,

where KU is the (positive) proportionality constant—find .

If the proportional control is substituted directly into the differential equation (1) for the building temperature, we get

(9)

A comparison of equation (9) with the first-order linear differential equation (2) shows that for this example the quantity P is equal to K # KU and the quantity representing the forcing function includes the desired temperature TD. That is,

Q AtB ! KM AtB # H AtB # KUTD . P ! K # KU , Q AtB

dT AtB dt

! K 3M AtB " T AtB 4 # H AtB # KU 3TD " T AtB 4 . U AtB T AtB

U AtB ! KU 3TD " T AtB 4 ,

1/Kv/K

1/2v/K 1/4v is 2p/24

Section 3.3 Heating and Cooling of Buildings 105

Example 3

Solution

90°

80°

70°

60°

50° 0

Midnight 6 12

Noon 18 24

Midnight

Outside

Inside

T , °F

Figure 3.6 Temperature variation inside and outside an unheated building

When the additional heating rate is a constant H0 and the outside temperature M varies as a sine wave over a 24-hr period in the same way as it did in Example 2, the forcing function is

The function has a constant term and a cosine term just as in equation (6). This equiva- lence becomes more apparent after substituting

(10)

where

(11)

The expressions for the constant P and the forcing function of equation (10) are the same as the expressions in Example 2, except that the constants K, B0, and B are replaced, respectively, by the constants K1, B2, and B1. Hence, the solution to the differential equation (9) will be the same as the temperature solution in Example 2, except that the three constant terms are changed. Thus,

(12)

where

The constant C is chosen so that at time t ! 0 the value of the temperature T equals T0. Thus,

◆

In the above example, the time constant for equation (9) is , where K1 ! K # KU. Here is referred to as the time constant for the building with heating and air conditioning. For a typical heating and cooling system, KU is somewhat less than 2; for a typical building, the constant K is between and . Hence, the sum gives a value for K1 of about 2, and the time constant for the building with heating and air conditioning is about hr.

When the heating or cooling is turned on, it takes about 30 min for the exponential term in (12) to die off. If we neglect this exponential term, the average temperature inside the building is B2. Since K1 is much larger than K and H0 is small, it follows from (11) that B2 is roughly TD, the desired temperature. In other words, after a certain period of time, the temperature inside the building is roughly TD, with a small sinusoidal variation. (The outside average M0 and inside heating rate H0 have only a small effect.) Thus, to save energy, the heating or cooling system may be left off during the night. When it is turned on in the morning, it will take roughly 30 min for the inside of the building to attain the desired temperature. These observa- tions provide an answer to question (c), regarding the temperature inside the building during summer and winter, that was posed at the beginning of this section.

1/2

1/41/2

1/K1 1/P ! 1/K1

C ! T0 " B2 # B1F1 A0B . F1 AtB J cos vt # Av/K1Bsin vt1 # Av/K1B2 . T AtB ! B2 " B1F1 AtB # C exp A"K1tB ,

Q AtB B2 J

KUTD # KM0 # H0 K1

, B1 J BK K1

v J 2p 24

! p 12

, K1 J K # KU ,

Q AtB ! K1 AB2 " B1 cos vtB , Q AtBQ AtB ! K AM0 " B cos vtB # H0 # KUTD .

106 Chapter 3 Mathematical Models and Numerical Methods Involving First-Order Equations

The assumption made in Example 3 that the amount of heating or cooling is ! may not always be suitable. We have used it here and in the exercises to illustrate

the use of the time constant. More adventuresome readers may want to experiment with other models for , especially if they have available the numerical techniques discussed in Sections 3.6 and 3.7. Group Project F, page 148, addresses temperature regulation with fixed-rate controllers.

U AtBKU 3TD " T AtB 4 U AtB

Section 3.3 Heating and Cooling of Buildings 107

1. A cup of hot coffee initially at 95ºC cools to 80ºC in 5 min while sitting in a room of temperature 21ºC. Using just Newton’s law of cooling, determine when the temperature of the coffee will be a nice 50ºC.

2. A cold beer initially at 35ºF warms up to 40ºF in 3 min while sitting in a room of temperature 70ºF. How warm will the beer be if left out for 20 min?

3. A white wine at room temperature 70ºF is chilled in ice (32ºF). If it takes 15 min for the wine to chill to 60ºF, how long will it take for the wine to reach 56ºF?

4. A red wine is brought up from the wine cellar, which is a cool 10ºC, and left to breathe in a room of tem- perature 23ºC. If it takes 10 min for the wine to reach 15ºC, when will the temperature of the wine reach 18ºC?

5. It was noon on a cold December day in Tampa: 16ºC. Detective Taylor arrived at the crime scene to find the sergeant leaning over the body. The sergeant said there were several suspects. If they knew the exact time of death, then they could narrow the list. Detec- tive Taylor took out a thermometer and measured the temperature of the body: 34.5ºC. He then left for lunch. Upon returning at 1:00 P.M., he found the body temperature to be 33.7ºC. When did the murder occur? [Hint: Normal body temperature is 37ºC.]

6. On a mild Saturday morning while people are work- ing inside, the furnace keeps the temperature inside the building at 21ºC. At noon the furnace is turned off, and the people go home. The temperature outside is a constant 12ºC for the rest of the afternoon. If the time constant for the building is 3 hr, when will the temperature inside the building reach 16ºC? If some windows are left open and the time constant drops to 2 hr, when will the temperature inside reach 16ºC?

7. On a hot Saturday morning while people are work- ing inside, the air conditioner keeps the temperature inside the building at 24ºC. At noon the air condi- tioner is turned off, and the people go home. The temperature outside is a constant 35ºC for the rest of

the afternoon. If the time constant for the building is 4 hr, what will be the temperature inside the building at 2:00 P.M.? At 6:00 P.M.? When will the tempera- ture inside the building reach 27ºC?

8. A garage with no heating or cooling has a time con- stant of 2 hr. If the outside temperature varies as a sine wave with a minimum of 50ºF at 2:00 A.M. and a maximum of 80ºF at 2:00 P.M., determine the times at which the building reaches its lowest temperature and its highest temperature, assuming the exponen- tial term has died off.

9. A warehouse is being built that will have neither heating nor cooling. Depending on the amount of insulation, the time constant for the building may range from 1 to 5 hr. To illustrate the effect insulation will have on the temperature inside the warehouse, assume the outside temperature varies as a sine wave, with a minimum of 16ºC at 2:00 A.M. and a maxi- mum of 32ºC at 2:00 P.M. Assuming the exponential term (which involves the initial temperature T0) has died off, what is the lowest temperature inside the building if the time constant is 1 hr? If it is 5 hr? What is the highest temperature inside the building if the time constant is 1 hr? If it is 5 hr?

10. Early Monday morning, the temperature in the lec- ture hall has fallen to 40ºF, the same as the tempera- ture outside. At 7:00 A.M., the janitor turns on the furnace with the thermostat set at 70ºF. The time constant for the building is 2 hr and that for the building along with its heating system is !

hr. Assuming that the outside temperature remains constant, what will be the temperature inside the lecture hall at 8:00 A.M.? When will the temperature inside the hall reach 65ºF?

11. During the summer the temperature inside a van reaches 55ºC, while that outside is a constant 35ºC. When the driver gets into the van, she turns on the air conditioner with the thermostat set at 16ºC. If the time constant for the van is 2 hr and that for1 /K !

1 /2 1 /K1

1 /K !

3.3 EXERCISES

the van with its air conditioning system is ! hr, when will the temperature inside the van

reach 27ºC? 12. Two friends sit down to talk and enjoy a cup of coffee.

When the coffee is served, the impatient friend imme- diately adds a teaspoon of cream to his coffee. The relaxed friend waits 5 min before adding a teaspoon of cream (which has been kept at a constant tempera- ture). The two now begin to drink their coffee. Who has the hotter coffee? Assume that the cream is cooler than the air and use Newton’s law of cooling.

13. A solar hot-water-heating system consists of a hot- water tank and a solar panel. The tank is well insu- lated and has a time constant of 64 hr. The solar panel generates 2000 Btu/hr during the day, and the tank has a heat capacity of 2ºF per thousand Btu. If the water in the tank is initially 110ºF and the room temperature outside the tank is 80ºF, what will be the temperature in the tank after 12 hr of sunlight?

14. In Problem 13, if a larger tank with a heat capacity of 1ºF per thousand Btu and a time constant of 72 hr

1 /3 1 /K1

108 Chapter 3 Mathematical Models and Numerical Methods Involving First-Order Equations

is used instead (with all other factors being the same), what will be the temperature in the tank after 12 hr?

15. Stefan’s law of radiation states that the rate of change of temperature of a body at T degrees Kelvin in a medium at M degrees Kelvin is proportional to

. That is,

where k is a positive constant. Solve this equation using separation of variables. Explain why Newton’s law and Stefan’s law are nearly the same when T is close to M and M is constant. Hint: Factor .

16. Show that can be written in

the form , where and [Hint: Use a standard trigono- metric identity with , Use this fact to verify the alternate representation (8) of discussed in Example 2.F AtB

C2 ! A sin f. ]C1 ! A cos f tan f ! C2 /C1.

A ! 2C21 # C22A cos Avt " fBC1 cos vt # C2 sin vt 4M 4 " T 43

dT dt

! k AM4 " T 4B , M4 " T 4

3.4 NEWTONIAN MECHANICS Mechanics is the study of the motion of objects and the effect of forces acting on those objects. It is the foundation of several branches of physics and engineering. Newtonian, or classical, mechanics deals with the motion of ordinary objects—that is, objects that are large compared to an atom and slow moving compared with the speed of light. A model for Newtonian mechanics can be based on Newton’s laws of motion:†

1. When a body is subject to no resultant external force, it moves with a constant velocity.

2. When a body is subject to one or more external forces, the time rate of change of the body’s momentum is equal to the vector sum of the external forces acting on it.

3. When one body interacts with a second body, the force of the first body on the second is equal in magnitude, but opposite in direction, to the force of the second body on the first.

Experimental results for more than two centuries verify that these laws are extremely use- ful for studying the motion of ordinary objects in an inertial reference frame—that is, a refer- ence frame in which an undisturbed body moves with a constant velocity. It is Newton’s second law, which applies only to inertial reference frames, that enables us to formulate the equations of motion for a moving body. We can express Newton’s second law by

dp dt

! F At, x, yB , †For a discussion of Newton’s laws of motion, see Sears and Zemansky’s University Physics, 12th ed., by H. D. Young, R. A. Freedman, J. R. Sandin, and A. L. Ford (Pearson Addison Wesley, San Francisco, 2008).

where is the resultant force on the body at time t, location x, and velocity y, and is the momentum of the body at time t. The momentum is the product of the mass of the body and its velocity—that is,

—so we can express Newton’s second law as

(1)

where is the acceleration of the body at time t. Typically one substitutes for the velocity in (1) and obtains a second-order

differential equation in the dependent variable x. However, in the present section, we will focus on situations where the force F does not depend on x. This enables us to regard (1) as a first-order equation

(2)

in . To apply Newton’s laws of motion to a problem in mechanics, the following general

procedure may be useful.

y AtB m dY dt ! F

At, YB y ! dx/dt

a ! dy/dt

m dy dt

! ma ! F At, x, yB , p AtB ! my AtB

p AtBF At, x, yB Section 3.4 Newtonian Mechanics 109

Procedure for Newtonian Models (a) Determine all relevant forces acting on the object being studied. It is helpful to draw

a simple diagram of the object that depicts these forces. (b) Choose an appropriate axis or coordinate system in which to represent the motion of

the object and the forces acting on it. Keep in mind that this coordinate system must be an inertial reference frame.

In this section we express Newton’s second law in either of two systems of units: the U.S. Customary System or the meter-kilogram-second (MKS) system. The various units in these systems are summarized in Table 3.2, along with approximate values for the gravitational acceleration (on the surface of Earth).

TABLE 3.2 Mechanical Units in the U.S. Customary and MKS Systems

U.S. Customary Unit System MKS System

Distance foot (ft) meter (m) Mass slug kilogram (kg) Time second (sec) second (sec) Force pound (lb) newton (N) g (Earth) 32 ft/sec2 9.81 m/sec2

An object of mass m is given an initial downward velocity y0 and allowed to fall under the influence of gravity. Assuming the gravitational force is constant and the force due to air resistance is proportional to the velocity of the object, determine the equation of motion for this object.

Two forces are acting on the object: a constant force due to the downward pull of gravity and a force due to air resistance that is proportional to the velocity of the object and acts in opposition to the motion of the object. Hence, the motion of the object will take place along a vertical axis. On this axis we choose the origin to be the point where the object was initially dropped and let denote the distance the object has fallen in time t (see Figure 3.7).

The forces acting on the object along this axis can be expressed as follows. The force due to gravity is

where g is the acceleration due to gravity at Earth’s surface (see Table 3.2). The force due to air resistance is

where is the proportionality constant† and the negative sign is present because air resis- tance acts in opposition to the motion of the object. Hence, the net force acting on the object (see Figure 3.7) is

(3)

We now apply Newton’s second law by substituting (3) into (2) to obtain

Since the initial velocity of the object is y0, a model for the velocity of the falling body is expressed by the initial value problem

(4)

where g and b are positive constants. The model (4) is the same as the one we obtained in Section 2.1. Using separation of

variables or the method of Section 2.3 for linear equations, we get

(5) Y AtB ! mgb # aY0 " mgb b e"bt/m . m

dy dt

! mg " by , y A0B ! y0 , m

dy dt

! mg " by .

F ! F1 # F2 ! mg " by AtB . b A$0BF2 ! "by AtB ,

F1 ! mg ,

x AtB

110 Chapter 3 Mathematical Models and Numerical Methods Involving First-Order Equations

x(t)

t = 0

t F 2 = −b (t)

F 1 = mg

Figure 3.7 Forces on a falling object

Solution

†The units of b are lb-sec/ft in the U.S. system, and N-sec/m in the MKS system.

Example 1

Since we have taken x ! 0 when t ! 0, we can determine the equation of motion of the object by integrating with respect to t. Thus, from (5) we obtain

and setting x ! 0 when t ! 0, we find

Hence, the equation of motion is

(6) ◆

In Figure 3.8, we have sketched the graphs of the velocity and the position as functions of t. Observe that the velocity approaches the horizontal asymptote as and that the position asymptotically approaches the line

as The value of the horizontal asymptote for is called the limiting, or terminal, velocity of the object, and, in fact, is a solution of (4). Since the two forces are in balance, this is called an “equilibrium” solution.

Notice that the terminal velocity depends on the mass but not the initial velocity of the object; the velocity of every free-falling body approaches the limiting value . Heavier objects do, in the presence of friction, ultimately fall faster than lighter ones, but for a given object the terminal velocity is the same whether it initially is tossed upward or downward, or simply dropped from rest.

Now that we have obtained the equation of motion for a falling object with air resistance proportional to y, we can answer a variety of questions.

mg/b

y ! mg/b ! constant y AtBmg/bt S #q.

x ! mg

b t "

m2g

b2 #

m b

x AtB t S #qy ! mg/by AtB x AtB ! mgb t # mb aY0 " mgb b A1 " e"bt/mB . c !

m b

ay0 " mgb b . 0 ! "

m b

ay0 " mgb b # c , x AtB ! "y AtB dt ! mgb t " mb ay0 " mgb b e"bt/m # c ,

y ! dx/dt

Section 3.4 Newtonian Mechanics 111

Position Velocity

mg ––– b

0 mg ––– b

x = t − m 2

––– b 2 +

m ––– b 0

0 0

(t)

x(t)

t t

Figure 3.8 Graphs of the position and velocity of a falling object when y0 6 mg/b

An object of mass 3 kg is released from rest 500 m above the ground and allowed to fall under the influence of gravity. Assume the gravitational force is constant, with g ! 9.81 m/sec2, and the force due to air resistance is proportional to the velocity of the object† with proportionality constant b ! 3 N-sec/m. Determine when the object will strike the ground.

We can use the model discussed in Example 1 with y0 ! 0, m ! 3, b ! 3, and g ! 9.81. From (6), the equation of motion in this case is

Because the object is released 500 m above the ground, we can determine when the object strikes the ground by setting and solving for t. Thus, we put

where we have rounded the computations to two decimal places. Unfortunately, this equation cannot be solved explicitly for t. We might try to approximate t using Newton’s approximation method (see Appendix B), but in this case, it is not necessary. Since will be very small for t near , we simply ignore the term and obtain as our approximation

. ◆

A parachutist whose mass is 75 kg drops from a helicopter hovering 4000 m above the ground and falls toward the earth under the influence of gravity. Assume the gravitational force is constant. Assume also that the force due to air resistance is proportional to the velocity of the parachutist, with the proportionality constant b1 ! 15 N-sec/m when the chute is closed and with constant b2 ! 105 N-sec/m when the chute is open. If the chute does not open until 1 min after the parachutist leaves the helicopter, after how many seconds will she hit the ground?

We are interested only in when the parachutist will hit the ground, not where. Thus, we con- sider only the vertical component of her descent. For this, we need to use two equations: one to describe the motion before the chute opens and the other to apply after it opens. Before the chute opens, the model is the same as in Example 1 with y0 ! 0, m ! 75 kg, b ! b1 ! 15 N-sec/m, and g ! 9.81 m/sec2. If we let be the distance the parachutist has fallen in t sec and let , then substituting into equations (5) and (6), we have

! A49.05B A1 " e"0.2tB , y1 AtB ! A75B A9.81B15 A1 " e"A15/75B tB y1 ! dx1/dt

x1 AtB

t ! 51.97 sec e"t51.97 Ae"51.97 ! 10"22B e"t

t # e"t ! 509.81 9.81

! 51.97 ,

500 ! A9.81Bt " 9.81 # A9.81Be"t x AtB ! 500

x AtB ! A3B A9.81B 3

t " A3B2 A9.81BA3B2 A1 " e"3t/3B ! A9.81Bt " A9.81B A1 " e"tB .

112 Chapter 3 Mathematical Models and Numerical Methods Involving First-Order Equations

†The effects of more sophisticated air resistance models (such as a quadratic drag law) are analyzed numerically in Exercises 3.6, Problem 20.

Example 2

Solution

Example 3

Solution

and

Hence, after 1 min, when t ! 60, the parachutist is falling at the rate

and has fallen

(In these and other computations for this problem, we round our answers to two decimal places.) Now when the chute opens, the parachutist is 4000 " 2697.75 or 1302.25 m above the

ground and traveling at a velocity of 49.05 m/sec. To determine the equation of motion after the chute opens, let denote the position of the parachutist T sec after the chute opens (so that T ! t " 60), taking at (see Figure 3.9). Further assume that the initial velocity of the parachutist after the chute opens is the same as the final velocity before it opens—that is, m/sec. Because the forces acting on the parachutist are the same as those acting on the object in Example 1, we can again use equations (5) and (6). With y0 ! 49.05, m ! 75, b ! b2 ! 105, and g ! 9.81, we find from (6) that

To determine when the parachutist will hit the ground, we set 1302.25, the height the parachutist was above the ground when her parachute opened. This gives

(7)

Again we cannot solve (7) explicitly for T. However, observe that is very small for T near 181.49, so we ignore the exponential term and obtain T ! 181.49. Hence, the parachutist will strike the ground 181.49 sec after the parachute opens, or 241.49 sec after dropping from the helicopter. ◆

e"1.4T

T " 4.28e"1.4T " 181.49 ! 0 .

7.01T # 30.03 " 30.03e"1.4T ! 1302.25

x2 AT B ! ! 7.01T # 30.03 A1 " e"1.4T B . x2 ATB ! A75B A9.81B105 T # 75105 c49.05 " A75B A9.81B105 d A1 " e"A105/ 75BT B

x¿2 A0B ! x¿1 A60B ! 49.05 x1 A60Bx2 A0B ! 0x2 ATB

x1 A60B ! A49.05B A60B " A245.25B A1 " e"0.2A60BB ! 2697.75 m . y1 A60B ! A49.05B A1 " e"0.2A60BB ! 49.05 m/sec ,

! 49.05t " 245.25 A1 " e"0.2tB . x1 AtB ! A75B A9.81B15 t " A75B2 A9.81BA15B2 A1 " e"A15/75BtB

Section 3.4 Newtonian Mechanics 113

x 1 (t)

t = 0

T = 0

1302.25

2697.75

Chute opens

x 2 (T)

x 2 (0) = 0 at x 1 (60)

Figure 3.9 The fall of the parachutist

In the computation for T in equation (7), we found that the exponential was negligi- ble. Consequently, ignoring the corresponding exponential term in equation (5), we see that the parachutist’s velocity at impact is

which is the limiting velocity for her fall with the chute open.

In some situations the resistive drag force on an object is proportional to a power of other than 1. Then when the velocity is positive, Newton’s second law for a falling object gen- eralizes to

(8)

where m and g have their usual interpretation and b $0 and the exponent r are experimental constants. [More generally, the drag force would be written as sign y .] Express the solution to (8) for the case r ! 2.

The (positive) equilibrium solution, with the forces in balance, is Other- wise, we write (8) as a separable equation that we can solve using partial fractions or the integral tables on the inside front cover:

and after some algebra,

Here is determined by the initial conditions. Again we see that y approaches the terminal velocity y0 as t → ∞. ◆

c3 ! c2 sign 3 Ay0 # yB / Ay0 " yB 4y ! y0 c3 " e

"2y0bt/m

c3 # e "2y0bt/m .

` y0 # y y0 " y

` ! c2e2y0bt/m , " dyy20 " y2 !

1 2y0

ln ` y0 # y y0 " y

` ! b m

t # c1 ,

y¿ ! b Ay20 " y2B /m, y ! y0 ! 2mg/b. BA"b 0y 0 rBA

m dydt ! mg " by r ,

0y 0 mg b2

! A75B A9.81B

105 ! 7.01 m/sec ,

e"1.4T

114 Chapter 3 Mathematical Models and Numerical Methods Involving First-Order Equations

Unless otherwise stated, in the following problems we assume that the gravitational force is constant with g ! 9.81 m/sec2 in the MKS system and g ! 32 ft/sec2 in the U.S. Customary System.

1. An object of mass 5 kg is released from rest 1000 m above the ground and allowed to fall under the influ- ence of gravity. Assuming the force due to air resis- tance is proportional to the velocity of the object with proportionality constant b ! 50 N-sec/m, deter- mine the equation of motion of the object. When will the object strike the ground?

2. A 400-lb object is released from rest 500 ft above the ground and allowed to fall under the influence of gravity. Assuming that the force in pounds due to air resistance is "10y, where y is the velocity of the object in ft/sec, determine the equation of motion of the object. When will the object hit the ground?

3. If the object in Problem 1 has a mass of 500 kg instead of 5 kg, when will it strike the ground? [Hint: Here the exponential term is too large to ignore. Use Newton’s method to approximate the time t when the object strikes the ground (see Appendix B).]

Example 4

Solution

3.4 EXERCISES

4. If the object in Problem 2 is released from rest 30 ft above the ground instead of 500 ft, when will it strike the ground? [Hint: Use Newton’s method to solve for t.]

5. An object of mass 5 kg is given an initial downward velocity of 50 m/sec and then allowed to fall under the influence of gravity. Assume that the force in newtons due to air resistance is "10y, where y is the velocity of the object in m/sec. Determine the equa- tion of motion of the object. If the object is initially 500 m above the ground, determine when the object will strike the ground.

6. An object of mass 8 kg is given an upward initial velocity of 20 m/sec and then allowed to fall under the influence of gravity. Assume that the force in newtons due to air resistance is "16y, where y is the velocity of the object in m/sec. Determine the equa- tion of motion of the object. If the object is initially 100 m above the ground, determine when the object will strike the ground.

7. A parachutist whose mass is 75 kg drops from a heli- copter hovering 2000 m above the ground and falls toward the ground under the influence of gravity. Assume that the force due to air resistance is propor- tional to the velocity of the parachutist, with the proportionality constant b1 ! 30 N-sec/m when the chute is closed and b2 ! 90 N-sec/m when the chute is open. If the chute does not open until the velocity of the parachutist reaches 20 m/sec, after how many seconds will she reach the ground?

8. A parachutist whose mass is 100 kg drops from a heli- copter hovering 3000 m above the ground and falls under the influence of gravity. Assume that the force due to air resistance is proportional to the velocity of the parachutist, with the proportionality constant b3 ! 20 N-sec/m when the chute is closed and b4 ! 100 N-sec/m when the chute is open. If the chute does not open until 30 sec after the parachutist leaves the helicopter, after how many seconds will he hit the ground? If the chute does not open until 1 min after he leaves the helicopter, after how many seconds will he hit the ground?

9. An object of mass 100 kg is released from rest from a boat into the water and allowed to sink. While gravity is pulling the object down, a buoyancy force of times the weight of the object is pushing the object up (weight ! mg). If we assume that water resistance exerts a force on the object that is proportional to the velocity of the object, with

1 /40

Section 3.4 Newtonian Mechanics 115

proportionality constant 10 N-sec/m, find the equa- tion of motion of the object. After how many seconds will the velocity of the object be 70 m/sec?

10. An object of mass 2 kg is released from rest from a platform 30 m above the water and allowed to fall under the influence of gravity. After the object strikes the water, it begins to sink with gravity pulling down and a buoyancy force pushing up. Assume that the force of gravity is constant, no change in momentum occurs on impact with the water, the buoyancy force is the weight (weight ! mg), and the force due to air resistance or water resistance is proportional to the velocity, with proportionality constant b1 ! 10 N-sec/m in the air and b2 ! 100 N-sec/m in the water. Find the equa- tion of motion of the object. What is the velocity of the object 1 min after it is released?

11. In Example 1, we solved for the velocity of the object as a function of time (equation (5)). In some cases, it is useful to have an expression, independent of t, that relates y and x. Find this relation for the motion in Example 1. [Hint: Letting , then .]

12. A shell of mass 2 kg is shot upward with an initial velocity of 200 m/sec. The magnitude of the force on the shell due to air resistance is When will the shell reach its maximum height above the ground? What is the maximum height?

13. When the velocity y of an object is very large, the mag- nitude of the force due to air resistance is proportional to y2 with the force acting in opposition to the motion of the object. A shell of mass 3 kg is shot upward from the ground with an initial velocity of 500 m/sec. If the magnitude of the force due to air resistance is , when will the shell reach its maximum height above the ground? What is the maximum height?

14. An object of mass m is released from rest and falls under the influence of gravity. If the magnitude of the force due to air resistance is byn, where b and n are positive constants, find the limiting velocity of the object (assuming this limit exists). [Hint: Argue that the existence of a (finite) limiting velocity implies that .]

15. A rotating flywheel is being turned by a motor that exerts a constant torque T (see Figure 3.10 on page 116). A retarding torque due to friction is pro- portional to the angular velocity If the moment of inertia of the flywheel is I and its initial angular veloc- ity is find the equation for the angular velocity vv0,

dy/dt S 0 as t S #q

A0.1By2

0y 0 /20. dy/dt ! AdV/dxBV y AtB ! V Ax AtB B

1 /2

116 Chapter 3 Mathematical Models and Numerical Methods Involving First-Order Equations

19. An object of mass 60 kg starts from rest at the top of a 45º inclined plane. Assume that the coeffi- cient of kinetic friction is 0.05 (see Problem 18). If the force due to air resistance is proportional to the velocity of the object, say, "3y, find the equa- tion of motion of the object. How long will it take the object to reach the bottom of the inclined plane if the incline is 10 m long?

20. An object at rest on an inclined plane will not slide until the component of the gravitational force down the incline is sufficient to overcome the force due to static friction. Static friction is governed by an experimental law somewhat like that of kinetic friction (Problem 18); it has a magnitude of at most mN, where m is the coefficient of static fric- tion and N is, again, the magnitude of the normal force exerted by the surface on the object. If the plane is inclined at an angle determine the criti- cal value for which the object will slide if

but will not move for 21. A sailboat has been running (on a straight course)

under a light wind at 1 m/sec. Suddenly the wind picks up, blowing hard enough to apply a constant force of 600 N to the sailboat. The only other force acting on the boat is water resistance that is proportional to the velocity of the boat. If the pro- portionality constant for water resistance is b ! 100 N-sec/m and the mass of the sailboat is 50 kg, find the equation of motion of the sailboat. What is the limiting velocity of the sailboat under this wind?

22. In Problem 21 it is observed that when the velocity of the sailboat reaches 5 m/sec, the boat begins to rise out of the water and “plane.” When this hap- pens, the proportionality constant for the water resistance drops to b0 ! 60 N-sec/m. Now find the equation of motion of the sailboat. What is the lim- iting velocity of the sailboat under this wind as it is planing?

23. Sailboats A and B each have a mass of 60 kg and cross the starting line at the same time on the first leg of a race. Each has an initial velocity of 2 m/sec. The wind applies a constant force of 650 N to each boat, and the force due to water resistance is proportional to the velocity of the boat. For sail- boat A the proportionality constants are b1 ! 80 N-sec/m before planing when the velocity is less

a 6 a0.a 7 a0 a0

as a function of time. [Hint: Use Newton’s second law for rotational motion, that is, moment of inertia & angular acceleration ! sum of the torques.]

16. Find the equation for the angular velocity in Problem 15, assuming that the retarding torque is proportional to

17. In Problem 16, let I ! 50 kg-m2 and the retarding torque be N-m. If the motor is turned off with the angular velocity at 225 rad/sec, determine how long it will take for the flywheel to come to rest.

18. When an object slides on a surface, it encounters a resistance force called friction. This force has a mag- nitude of mN, where m is the coefficient of kinetic friction and N is the magnitude of the normal force that the surface applies to the object. Suppose an object of mass 30 kg is released from the top of an inclined plane that is inclined 30º to the horizontal (see Figure 3.11). Assume the gravitational force is con- stant, air resistance is negligible, and the coefficient of kinetic friction m ! 0.2. Determine the equation of motion for the object as it slides down the plane. If the top surface of the plane is 5 m long, what is the veloc- ity of the object when it reaches the bottom?

51v1v . v

Motor

T, Torque from motor

Retarding torque = d /dt I

Figure 3.10 Motor-driven flywheel

x (0) = 0 x(t)

− N mg sin 30°

30°

mg cos 30°

Figure 3.11 Forces on an object on an inclined plane

than 5 m/sec and b2 ! 60 N-sec/m when the velocity is above 5 m/sec. For sailboat B the proportionality constants are b3 ! 100 N-sec/m before planing when the velocity is less than 6 m/sec and b4 ! 50 N-sec/m when the velocity is above 6 m/sec. If the first leg of the race is 500 m long, which sailboat will be leading at the end of the first leg?

24. Rocket Flight. A model rocket having initial mass m0 kg is launched vertically from the ground. The rocket expels gas at a constant rate of kg/sec and at a constant velocity of m/sec relative to the rocket. Assume that the magnitude of the gravitational force is proportional to the mass with proportionality con- stant g. Because the mass is not constant, Newton’s second law leads to the equation

where is the velocity of the rocket, x is its height above the ground, and is the mass of the rocket at t sec after launch. If the initial velocity is zero, solve the above equation to determine the velocity of the rocket and its height above ground for

25. Escape Velocity. According to Newton’s law of gravitation, the attractive force between two objects varies inversely as the square of the dis- tances between them. That is, where M1 and M2 are the masses of the objects, r is the distance between them (center to center), Fg is the attractive force, and G is the constant of propor- tionality. Consider a projectile of constant mass m being fired vertically from Earth (see Figure 3.12). Let t represent time and y the velocity of the pro- jectile.

Fg ! GM1M2 /r 2,

0 ' t 6 m0 /a.

m0 " at y ! dx/dt

Am0 " atB dydt " ab ! "g Am0 " atB ,

Section 3.5 Electrical Circuits 117

r R

Earth

Figure 3.12 Projectile escaping from Earth

(a) Show that the motion of the projectile, under Earth’s gravitational force, is governed by the equation

where r is the distance between the projectile and the center of Earth, R is the radius of Earth, M is the mass of Earth, and

(b) Use the fact that to obtain

(d) Use the result of part (c) to show that the veloc- ity of the projectile remains positive if and only if The velocity is called the escape velocity of Earth.

(e) If g ! 9.81 m/sec2 and R ! 6370 km for Earth, what is Earth’s escape velocity?

(f) If the acceleration due to gravity for the Moon is and the radius of the Moon is Rm !

1738 km, what is the escape velocity of the Moon?

gm ! g/6

ye ! 22gRy20 " 2gR 7 0. y2 !

2gR2

r # y20 " 2gR .

y dy

dr ! "

gR2

r 2 .

dr/dt ! y g ! GM/R2.

dt ! "

gR2

r 2 ,

ELECTRICAL CIRCUITS3.5 In this section we consider the application of first-order differential equations to simple electri- cal circuits consisting of a voltage source (e.g., a battery or a generator), a resistor, and either an inductor or a capacitor. These so-called RL and RC circuits are shown in Figure 3.13 on page 118. More general circuits will be discussed in Section 5.7.

118 Chapter 3 Mathematical Models and Numerical Methods Involving First-Order Equations

Voltage source

Inductance

CE E

Voltage source

(a) (b)

Capacitance

Resistance Resistance

Figure 3.13 (a) RL circuit and (b) RC circuit

†Historical Footnote: Gustav Robert Kirchhoff (1824–1887) was a German physicist noted for his research in spectrum analysis, optics, and electricity.

The physical principles governing electrical circuits were formulated by G. R. Kirchhoff†

in 1859. They are the following:

1. Kirchhoff’s current law The algebraic sum of the currents flowing into any junction point must be zero.

2. Kirchhoff’s voltage law The algebraic sum of the instantaneous changes in potential (voltage drops) around any closed loop must be zero.

Kirchhoff’s current law implies that the same current passes through all elements in each circuit of Figure 3.13.

To apply Kirchhoff’s voltage law, we need to know the voltage drop across each element of the circuit. These voltage formulas are stated below (you can consult an introductory physics text for further details).

(a) According to Ohm’s law, the voltage drop ER across a resistor is proportional to the current I passing through the resistor:

The proportionality constant R is called the resistance.

(b) It can be shown using Faraday’s law and Lenz’s law that the voltage drop EL across an inductor is proportional to the instantaneous rate of change of the current I:

The proportionality constant L is called the inductance.

The constant C is called the capacitance.

The common units and symbols used for electrical circuits are listed in Table 3.3.

EC ! 1 C

q .

EL ! L dI dt

ER ! RI .

A voltage source is assumed to add voltage or potential energy to the circuit. If we let E t denote the voltage supplied to the circuit at time t, then applying Kirchhoff’s voltage law to the RL circuit in Figure 3.13(a) gives

(1) EL # ER ! E t .

Substituting into (1) the expressions for EL and ER gives

(2)

Note that this equation is linear (compare Section 2.3), and upon writing it in standard form we obtain the integrating factor

which leads to the following general solution [see equation (8), Section 2.3, page 47]

(3)

For the RL circuit, one is usually given the initial current I 0 as an initial condition.

An RL circuit with a 1-Ω resistor and a 0.01-H inductor is driven by a voltage E t ! sin 100t V. If the initial inductor current is zero, determine the subsequent resistor and inductor voltages and the current.

From equation (3) and the integral tables, we find that the general solution to the linear equa- tion (2) is given by

! sin 100t " cos 100t

2 # Ke"100t .

! e"100t c100 e100t A100 sin 100t " 100 cos 100tB 10,000 # 10,000

# K d I AtB ! e"100t a" e100t sin 100t0.01 dt # Kb

BABA I AtB ! e"Rt/L c " eRt/L E AtBL dt # K d . m(t) ! e "AR/LB dt ! eRt/L ,

L dI dt

# RI ! E AtB . BA

Section 3.5 Electrical Circuits 119

TABLE 3.3 Common Units and Symbols Used With Electrical Circuits

Letter Symbol Quantity Representation Units Representation

Voltage source E volt (V) Generator

Battery

Resistance R ohm (() Inductance L henry (H) Capacitance C farad (F) Charge q coulomb (C) Current I ampere (A)

Example 1

Solution

For I 0 ! 0, we obtain so and the current is

I t ! 0.5 sin 100t " cos 100t # e"100t .

The inductor and resistor voltages are then given by

◆

Now we turn to the RC circuit in Figure 3.13(b). Applying Kirchhoff’s voltage law yields

RI # q C ! E t .

The capacitor current, however, is the rate of change of its charge: I ! dq dt. So

(4)

is the governing differential equation for the RC circuit. The initial condition for a capacitor is its charge q at t ! 0.

Suppose a capacitor of C farads holds an initial charge of Q coulombs. To alter the charge, a constant voltage source of V volts is applied through a resistance of R ohms. Describe the capacitor charge for t $ 0.

Since E t ! V is constant in equation (4), the latter is both separable and linear, and its general solution is easily derived:

The solution meeting the prescribed initial condition is

The capacitor charge changes exponentially from Q to CV as time increases. ◆

If we take V ! 0 in Example 2, we see that the time constant—that is, the time required for the capacitor charge to drop to 1 e times its initial value—is RC. Thus, a capacitor is a leaky energy-storage device; even the very high resistivity of the surrounding air can dissipate its charge, particularly on a humid day. Capacitors are used in cellular phones to store electrical energy from the battery while the phone is in a (more-or-less) idle receiving mode and then assist the battery in delivering energy during the transmitting mode.

The time constant for inductor current in the RL circuit can be gleaned from Example 1 to be L R. An application of the RL circuit is the spark plug of a combustion engine. If a voltage source establishes a nonzero current in an inductor and the source is suddenly disconnected, the rapid change of current produces a high dI dt and, in accordance with the formula

, the inductor generates a voltage surge sufficient to cause a spark across the terminals—thus igniting the gasoline.

If an inductor and a capacitor both appear in a circuit, the governing differential equation will be second order. We’ll return to RLC circuits in Section 5.7.

EL ! LdI/dt /

q AtB ! CV # AQ " CVBe"t/RC . q AtB ! CV # Ke"t/RC . BA

R dq dt

# q C

! E

/ BA/

EL AtB ! L dIdt ! A0.5B Acos 100t # sin 100t " e"100tB . ER AtB ! RI AtB # I AtB ,

BABA K ! 1/2"1/2 # K ! 0,BA 120 Chapter 3 Mathematical Models and Numerical Methods Involving First-Order Equations

Example 2

Solution

Section 3.6 Improved Euler’s Method 121

1. An RL circuit with a 5-Ω resistor and a 0.05-H induc- tor carries a current of 1 A at t ! 0, at which time a voltage source E t ! 5 cos 120t V is added. Deter- mine the subsequent inductor current and voltage.

2. An RC circuit with a 1-Ω resistor and a 0.000001-F capacitor is driven by a voltage E t ! sin 100t V. If the initial capacitor voltage is zero, determine the sub- sequent resistor and capacitor voltages and the current.

3. The pathway for a binary electrical signal between gates in an integrated circuit can be modeled as an RC circuit, as in Figure 3.13(b); the voltage source models the transmitting gate, and the capacitor models the receiving gate. Typically, the resistance is 100 Ω, and the capacitance is very small, say, 10"12 F (1 picofarad, pF). If the capacitor is initially uncharged and the transmitting gate changes instantaneously from 0 to 5 V, how long will it take for the voltage at the receiving gate to reach (say) 3 V? (This is the time it takes to transmit a logical “1.”)

4. If the resistance in the RL circuit of Figure 3.13(a) is zero, show that the current I t is directly propor- tional to the integral of the applied voltage E t . Similarly show that if the resistance in the RC circuit of Figure 3.13(b) is zero, the current is directly pro- portional to the derivative of the applied voltage. (In engineering applications, it is often necessary to generate a voltage, rather than a current, which is the integral or derivative of another voltage. Group

BABA

BA BA Project E shows how this is accomplished using anoperational amplifier.)5. The power generated or dissipated by a circuit ele-

ment equals the voltage across the element times the current through the element. Show that the power dissipated by a resistor equals I2R, the power associ- ated with an inductor equals the derivative of

LI2, and the power associated with a capacitor equals the derivative of

6. Derive a power balance equation for the RL and RC circuits. (See Problem 5.) Discuss the significance of the signs of the three power terms.

7. An industrial electromagnet can be modeled as an RL circuit, while it is being energized with a voltage source. If the inductance is 10 H and the wire wind- ings contain 3 Ω of resistance, how long does it take a constant applied voltage to energize the electro- magnet to within 90% of its final value (that is, the current equals 90% of its asymptotic value)?

8. A 10"8-F capacitor (10 nanofarads) is charged to 50 V and then disconnected. One can model the charge leakage of the capacitor with a RC circuit with no voltage source and the resistance of the air between the capacitor plates. On a cold dry day, the resistance of the air gap is 5 & 1013 Ω; on a humid day, the resistance is 7 & 106 Ω. How long will it take the capacitor voltage to dissipate to half its original value on each day?

CE2C.A1 /2BA1 /2B

IMPROVED EULER’S METHOD3.6 Although the analytical techniques presented in Chapter 2 were useful for the variety of math- ematical models presented earlier in this chapter, the majority of the differential equations encountered in applications cannot be solved either implicitly or explicitly. This is especially true of higher-order equations and systems of equations, which we study in later chapters. In this section and the next, we discuss methods for obtaining a numerical approximation of the solution to an initial value problem for a first-order differential equation. Our goal is to develop algorithms that you can use with a calculator or computer.† These algorithms also extend natu- rally to higher-order equations (see Section 5.3). We describe the rationale behind each method but leave the more detailed discussion to texts on numerical analysis.††

†An applet, maintained on the web at http://alamos.math.arizona.edu/~rychlik/JOde/index.html automates most of the differential equation algorithms discussed in this book. ††See, for example, A First Course in the Numerical Analysis of Differential Equations, 2nd ed., by A. Iserles (Cambridge University Press, 2008), or Numerical Analysis, 8th ed., by R. L. Burden and J. D. Faires ( city of Brooks/Cole, 2005)

3.5 EXERCISES

http://alamos.math.arizona.edu/~rychlik/JOde/index.html

Consider the initial value problem

(1)

To guarantee that (1) has a unique solution, we assume that f and are continuous in a rec- tangle containing . It follows from Theorem 1 in Chapter 1 that the initial value problem (1) has a unique solution in some interval

where is a positive number. Because is not known a priori, there is no assurance that the solution will exist at a particular point even if x is in the interval

. However, if is continuous and bounded† on the vertical strip

then it turns out that (1) has a unique solution on the whole interval . In describing numerical methods, we assume that this last condition is satisfied and that f possesses as many continuous partial derivatives as needed.

In Section 1.4 we used the concept of direction fields to motivate a scheme for approxi- mating the solution to the initial value problem (1). This scheme, called Euler’s method, is one of the most basic, so it is worthwhile to discuss its advantages, disadvantages, and possible improvements. We begin with a derivation of Euler’s method that is somewhat different from that presented in Section 1.4.

Let h $ 0 be fixed (h is called the step size) and consider the equally spaced points

(2)

Our goal is to obtain an approximation to the solution of the initial value problem (1) at those points xn that lie in the interval . Namely, we will describe a method that generates values y0, y1, y2, . . . that approximate at the respective points x0, x1, x2, . . . ; that is,

Of course, the first “approximant” y0 is exact, since is given. Thus, we must describe how to compute y1, y2, . . . .

For Euler’s method we begin by integrating both sides of equation (1) from xn to to obtain

where we have substituted for y. Solving for we have

(3)

Without knowing we cannot integrate Hence, we must approximate the integral in (3). Assuming we have already found the simplest approach is to approximate the

area under the function by the rectangle with base and height (see Figure 3.14). This gives

Substituting h for and the approximation yn for we arrive at the numerical scheme

(4)

which is Euler’s method.

yn#1 ! yn # hf Axn, ynB , n ! 0, 1, 2, . . . , f AxnB,xn#1 " xnf Axn#1B ! f AxnB # Axn#1 " xnB f Axn, f AxnBB .

f Axn, f AxnBB3 xn, xn#1 4f At, f AtBB yn ! f AxnB,f At, f AtBB .f AtB, f Axn#1B ! f AxnB # " xn#1

f At, f AtBBdt . f Axn#1B,f AxB f Axn#1B " f AxnB ! " xn#1

f¿ AtB dt ! " xn#1 xn

f At, f AtBBdt , xn#1

y0 ! f Ax0Byn ! f AxnB , n ! 0, 1, 2, . . . . f AxBAa, bB f

AxBxn J x0 # nh , n ! 0, 1, 2, . . . .

Aa, bBS J E Ax, yB: a 6 x 6 b, "q 6 y 6 qF , 0f/ 0yAa, bB x A)x0B,

ddx0 " d 6 x 6 x0 # d, f AxBAx0, y0BR J E Ax, yB: a 6 x 6 b, c 6 y 6 dF 0f/ 0y

y$ ! f Ax, yB , y Ax0B ! y0 . 122 Chapter 3 Mathematical Models and Numerical Methods Involving First-Order Equations

†A function is bounded on S if there exists a number M such that for all in S.Ax, yB0g Ax, yB 0 ' Mg Ax, yB

Starting with the given value y0, we use (4) to compute and then use y1 to compute and so on. Several examples of Euler’s method can be found in Section 1.4.

As discussed in Section 1.4, if we wish to use Euler’s method to approximate the solution to the initial value problem (1) at a particular value of x, say, x ! c, then we must first determine a suitable step size h so that x0 # Nh ! c for some integer N. For example, we can take N ! 1 and h ! c " x0 in order to arrive at the approximation after just one step:

If, instead, we wish to take 10 steps in Euler’s method, we choose and ultimately obtain

In general, depending on the size of h, we will get different approximations to It is reasonable to expect that as h gets smaller (or, equivalently, as N gets larger), the Euler approximations approach the exact value On the other hand, as h gets smaller, the number (and cost) of computations increases and hence so do machine errors that arise from round-off. Thus, it is important to analyze how the error in the approximation scheme varies with h.

If Euler’s method is used to approximate the solution to the problem

(5)

at x ! 1, then we obtain approximations to the constant It turns out that these approximations take a particularly simple form that enables us to compare the error in the approximation with the step size h. Indeed, setting in (4) yields

Since y0 ! 1, we get

y3 ! A1 # hBy2 ! A1 # hB A1 # hB2 ! A1 # hB3 , y2 ! A1 # hBy1 ! A1 # hB A1 # hB ! A1 # hB2 y1 ! A1 # hBy0 ! 1 # h yn#1 ! yn # hyn ! A1 # hByn , n ! 0, 1, 2, . . . .f Ax, yB ! y

e ! f A1B.y¿ ! y , y A0B ! 1 , f AxB ! ex

f AcB. f AcB.f AcB ! f Ax0 # 10hB ! f Ax10B ! y10 .

h ! Ac " x0B /10f AcB ! f Ax0 # hB ! y1 .

y2 ! y1 # hf Ax1, y1B, y1 ! y0 # hf Ax0, y0B

Section 3.6 Improved Euler’s Method 123

f(t,

(t))

x n + 1x n

Figure 3.14 Approximation by a rectangle

124 Chapter 3 Mathematical Models and Numerical Methods Involving First-Order Equations

TABLE 3.4 Euler’s Approximations to e ! 2.71828. . .

Euler’s Approximation Error h Error h

1 2.00000 0.71828 0.71828 10*1 2.59374 0.12454 1.24539 10*2 2.70481 0.01347 1.34680 10*3 2.71692 0.00136 1.35790 10*4 2.71815 0.00014 1.35902

/e " A1 # hB1/hA1 # hB1/h

and, in general,

(6)

For the problem in (5) we have x0 ! 0, so to obtain approximations at x ! 1, we must set nh ! 1. That is, h must be the reciprocal of an integer . Replacing n by in (6), we see that Euler’s method gives the (familiar) approximation to the con- stant e. In Table 3.4, we list this approximation for along with the corresponding errors

From the second and third columns in Table 3.4, we see that the approximation gains roughly one decimal place in accuracy as h decreases by a factor of 10; that is, the error is roughly proportional to h. This observation is further confirmed by the entries in the last col- umn of Table 3.4. In fact, using methods of calculus (see Exercises 1.4, Problem 13), it can be shown that

(7)

The general situation is similar: When Euler’s method is used to approximate the solu- tion to the initial value problem (1), the error in the approximation is at worst a constant times the step size h. Moreover, in view of (7), this is the best one can say.

Numerical analysts have a convenient notation for describing the convergence behav- ior of a numerical scheme. For fixed x we denote by the approximation to the solu- tion of (1) obtained via the scheme when using a step size of h. We say that the numer- ical scheme converges at x if

In other words, as the step size h decreases to zero, the approximations for a convergent scheme approach the exact value The rate at which tends to is often expressed in terms of a suitable power of h. If the error tends to zero like a constant times , we write

and say that the method is of order p. Of course, the higher the power p, the faster is the rate of convergence as h S 0.

F AxB " y Ax; hB ! O Ah pBh p

f AxB " y Ax; hB f AxBy Ax; hBf AxB. lim hS0

y Ax; hB ! f AxB . f AxB y Ax; hB

lim hS0

error h

! lim hS0

e " A1 # hB1/h h

! e 2

! 1.35914 .

e " A1 # hB1/h . h ! 1, 10"1, 10"2, 10"3, and 10"4,

A1 # hB1/h 1/hAh ! 1/nB yn ! A1 # hBn , n ! 0, 1, 2, . . . .

As seen from our earlier discussion, the rate of convergence of Euler’s method is O(h); that is, Euler’s method is of order p ! 1. In fact, the limit in (7) shows that for equation (5), the error is roughly 1.36h for small h. This means that to have an error less than 0.01 requires

, or computation steps. Thus Euler’s method converges too slowly to be of practical use.

How can we improve Euler’s method? To answer this, let’s return to the derivation expressed in formulas (3) and (4) and analyze the “errors” that were introduced to get the approximation. The crucial step in the process was to approximate the integral

by using a rectangle (recall Figure 3.14). This step gives rise to what is called the local trunca- tion error in the method. From calculus we know that a better (more accurate) approach to approximating the integral is to use a trapezoid—that is, to apply the trapezoidal rule (see Figure 3.15). This gives

which leads to the numerical scheme

(8)

We call equation (8) the trapezoid scheme. It is an example of an implicit method; that is, unlike Euler’s method, equation (8) gives only an implicit formula for , since appears as an argument of f. Assuming we have already computed yn, some root-finding technique such as Newton’s method (see Appendix B) might be needed to compute . Despite the inconve- nience of working with an implicit method, the trapezoid scheme has two advantages over Euler’s method. First, it is a method of order p ! 2; that is, it converges at a rate that is propor- tional to h2 and hence is faster than Euler’s method. Second, as described in Group Project H, the trapezoid scheme has the desirable feature of being stable.

Can we somehow modify the trapezoid scheme in order to obtain an explicit method? One idea is first to get an estimate, say, , of the value using Euler’s method and then use formula (8) with replaced by on the right-hand side. This two-step process is an example of a predictor–corrector method. That is, we predict using Euler’s method andyn#1

y*n#1yn#1 yn#1y*n#1

yn#1

yn#1yn#1

yn#1 ! yn # h 2 3 f Axn, ynB # f Axn#1, yn#1B 4 , n ! 0, 1, 2, . . . .

" xn#1

f At, f AtBB dt ! h2 c f Axn, f AxnBB # f Axn#1, f Axn#1BB d , " xn#1

f At, f AtBBdt n ! 1/h 7 136h 6 0.01/1.36

Section 3.6 Improved Euler’s Method 125

x n + 1x n

f(t, (t))

Figure 3.15 Approximation by a trapezoid

then use that value in (8) to obtain a “more correct” approximation. Setting ! in the right-hand side of (8), we obtain

(9)

where This explicit scheme is known as the improved Euler’s method.

Compute the improved Euler’s method approximation to the solution of

at x ! 1 using step sizes of h ! 1, 10"1, 10"2, 10"3, and 10"4.

The starting values are x0 ! 0 and y0 ! 1. Since , formula (9) becomes

that is,

(10)

Since y0 ! 1, we see inductively that

To obtain approximations at x ! 1, we must have and so Hence, the improved Euler’s approximations to are just

In Table 3.5 we have computed this approximation for the specified values of h, along with the corresponding errors

Comparing the entries of this table with those of Table 3.4, we observe that the improved Euler’s method converges much more rapidly than the original Euler’s method. In fact, from the first few entries in the second and third columns of Table 3.5, it appears that the approxima- tion gains two decimal places in accuracy each time h is decreased by a factor of 10. In other words, the error is roughly proportional to (see the last column of the table and also Problem 4). The entries in the last two rows of the table must be regarded with caution. Indeed, when h ! 10"3 or h ! 10"4, the true error is so small that our calculator rounded it to zero, to five decimal places. The entries in color in the last column may be inaccurate due to the loss of sig- nificant figures in the calculator arithmetic. ◆

e " a1 # h # h2 2 b 1/h .

a1 # h # h2 2 b 1/h .

e ! f A1B n ! 1/h.1 ! x0 # nh ! nh, yn ! a1 # h # h22 b n , n ! 0, 1, 2, . . . . yn#1 ! a1 # h # h22 b yn . yn#1 ! yn #

h 2

3 yn # Ayn # hynB 4 ! yn # hyn # h22 yn ; f Ax, yB ! y

y¿ ! y , y A0B ! 1 f AxB ! ex

xn#1 ! xn # h.

yn#1 ! yn # h 2 3 f Axn, ynB # f Axn # h, yn # hf Axn, ynBB 4 , n ! 0, 1, . . . ,

yn # hf Axn, ynB yn#1 126 Chapter 3 Mathematical Models and Numerical Methods Involving First-Order Equations

Example 1

Solution

As Example 1 suggests, the improved Euler’s method converges at the rate , and indeed it can be proved that in general this method is of order p ! 2.

A step-by-step outline for a subroutine that implements the improved Euler’s method over a given interval is described below. For programming purposes it is usually more con- venient to input the number of steps N in the interval rather than the step size h itself. For an interval starting at x ! x0 and ending at x ! c, the relation between h and N is

(11)

(Note that the subroutine includes an option for printing x and y.) Of course, the implementa- tion of this algorithm with N steps on the interval only produces approximations to the actual solution at N # 1 equally spaced points. If we wish to use these points to help graph an approximate solution over the whole interval , then we must somehow “fill in” the gaps between these points. A crude method is to simply join the points by straight-line segments producing a polygonal line approximation to More sophisticated techniques for prescrib- ing the intermediate points are used in professional codes.

f AxB.3 x0, c 4 3 x0, c 4

Nh ! c " x0 .

3 x0, c 4 O Ah2B

Section 3.6 Improved Euler’s Method 127

TABLE 3.5 Improved Euler’s Approximation to e ! 2.71828. . .

Approximation

h Error Error h2

1 2.50000 0.21828 0.21828 10"1 2.71408 0.00420 0.42010 10"2 2.71824 0.00004 0.44966 10"3 2.71828 0.00000 0.45271

/ a1 # h # h2

2 b 1/h

IMPROVED EULER’S METHOD SUBROUTINE

Purpose To approximate the solution to the initial value problem

for INPUT x0, y0, c, N (number of steps), PRNTR (!1 to print a table) Step 1 Set step size Step 2 For i ! 1 to N, do Steps 3–5 Step 3 Set

Step 4 Set

Step 5 If PRNTR ! 1, print x, y

y ! y # h AF # GB /2x ! x # h G ! f Ax # h, y # hFBF ! f Ax, yB

h ! Ac " x0B /N, x ! x0, y ! y0 x0 ' x ' c.

y¿ ! f Ax, yB , y Ax0B ! y0 ,f AxB

Now we want to devise a program that will compute As we have seen, the accuracy of the approximation depends on the step size h. Our strategy, then, will be to estimate for a given step size and then halve the step size and recompute the estimate, halve again, and so on. When two consecutive estimates of differ by less than some prescribed tolerance , we take the final estimate as our approximation to . Admit- tedly, this does not guarantee that is known to within but it provides a reasonable stop- ping procedure in practice.† The following procedure also contains a safeguard to stop if the desired tolerance is not reached after M halvings of h.

e,f AcB f AcBe f AcBf AcB

f AcB to a desired accuracy. 128 Chapter 3 Mathematical Models and Numerical Methods Involving First-Order Equations

IMPROVED EULER’S METHOD WITH TOLERANCE

Purpose To approximate the solution to the initial value problem

at x ! c, with tolerance INPUT

M (maximum number of halvings of step size) Step 1 Set z ! y0, PRNTR ! 0 Step 2 For m ! 0 to M, do Steps 3–7††

Step 3 Set N ! 2m

Step 4 Call IMPROVED EULER’S METHOD SUBROUTINE Step 5 Print h, y Step 6 If go to Step 10 Step 7 Set z ! y Step 8 Print “ is approximately”; y; “but may not be within the tolerance”; Step 9 Go to Step 11 Step 10 Print “ is approximately”; y; “with tolerance”; Step 11 STOP OUTPUT Approximations of the solution to the initial value problem at x ! c using 2m

steps

ef AcB ef AcB0 y " z 0 6 e,

x0, y0, c, e , e

y¿ ! f Ax, yB , y Ax0B ! y0 ,

If one desires a stopping procedure that simulates the relative error

then replace Step 6 by

go to Step 10 .

Use the improved Euler’s method with tolerance to approximate the solution to the initial value problem

(12)

at x ! 2. For a tolerance of use a stopping procedure based on the absolute error.e ! 0.001,

y¿ ! x # 2y , y A0B ! 0.25 , Step 6¿. If ` z " y

y ` 6 e ,

` approximation " true value true value

` ,

†Professional codes monitor accuracy much more carefully and vary step size in an adaptive fashion for this purpose. ††To save time, one can start with m ! K % M rather than with m ! 0.

Example 2

The starting values are x0 ! 0, y0 ! 0.25. Because we are computing the approximations for c ! 2, the initial value for h is

For equation (12), we have so the numbers F and G in the subroutine are

and we find

Thus, with we get for the first approximation

To describe the further outputs of the algorithm, we use the notation y(2; h) for the approx- imation obtained with step size h. Thus, y(2; 2) ! 5.25, and we find from the algorithm

Since which is less than we stop.

The exact solution of (12) is so we have determined that

◆

In the next section, we discuss methods with higher rates of convergence than either Euler’s or the improved Euler’s methods.

f A2B ! 1 2 ae4 " 5

2 b ! 26.04880 .f AxB !

1 2 Ae2x " x " 12B , e ! 0.001,0 y A2; 2"9B " y A2; 2"8B 0 ! 0.00083,

y A2; 2"4B ! 25.79127 y A2; 2"9B ! 26.04880 . y A2; 2"3B ! 25.12012 y A2; 2"8B ! 26.04797 y A2; 2"2B ! 23.06067 y A2; 2"7B ! 26.04468 y A2; 2"1B ! 18.28125 y A2; 2"6B ! 26.03172

y A2; 1B ! 11.25000 y A2; 2"5B ! 25.98132 y ! 0.25 # A0 # 1B # 2 A1 # 1B ! 5.25 .x0 ! 0, y0 ! 0.25, and h ! 2, y ! y #

h 2

AF # GB ! y # h 2

A2x # 4yB # h2 2

A1 # 2x # 4yB . x ! x # h G ! Ax # hB # 2 A y # hFB ! x # 2y # h A1 # 2x # 4yB , F ! x # 2y

f Ax, yB ! x # 2y,h ! A2 " 0B2"0 ! 2 .

Section 3.6 Improved Euler’s Method 129

†An applet, maintained on the web at http://alamos.math.arizona.edu/~rychlik/JOde/index.html automates most of the differential equation algo- rithms discussed in this book.

Solution

In many of the following problems, it will be essential to have a calculator or computer available. You may use a software package† or write a program for solving initial value problems using the improved Euler’s method algo- rithms on pages 127 and 128. (Remember, all trigono- metric calculations are done in radians.)

1. Show that when Euler’s method is used to approxi- mate the solution of the initial value problem

y¿ ! 5y , y A0B ! 1 ,

at x ! 1, then the approximation with step size h is .

2. Show that when Euler’s method is used to approxi- mate the solution of the initial value problem

at x ! 2, then the approximation with step size h is

3 a1 " h 2 b 2/h .

y¿ ! " 1 2

y , y A0B ! 3 , A1 # 5hB1/h

3.6 EXERCISES

http://alamos.math.arizona.edu/~rychlik/JOde/index.html

3. Show that when the trapezoid scheme given in for- mula (8) is used to approximate the solution

at x ! 1, then we get

which leads to the approximation

for the constant e. Compute this approximation for h ! 1, 10"1, 10"2, 10"3, and 10"4 and compare your results with those in Tables 3.4 and 3.5.

4. In Example 1 the improved Euler’s method approxi- mation to e with step size h was shown to be

First prove that the error approaches zero as Then use L’Hôpital’s rule to show that

Compare this constant with the entries in the last column of Table 3.5.

5. Show that when the improved Euler’s method is used to approximate the solution of the initial value problem

at , then the approximation with step size h is

6. Since the integral with variable upper limit satisfies (for continuous f ) the initial value problem

any numerical scheme that is used to approximate the solution at x ! 1 will give an approximation to the definite integral

Derive a formula for this approximation of the inte- gral using

" 1

0 f AtB dt .

y¿ ! f AxB , y A0B ! 0 , y AxB J $ x0 f AtB dt

1 3

A1 # 4h # 8h2B1/ A2hB . x ! 1 /2

y¿ ! 4y , y A0B ! 1 3

lim hS0

error

h2 !

e 6

! 0.45305 .

h S 0. J e " A1 # h # h2 /2B1/h

a1 # h # h2 2 b 1/h .

a1 # h/2 1 " h/2

b 1/h yn#1 ! a1 # h/21 " h/2b yn , n ! 0, 1, 2, . . . , y¿ ! y , y A0B ! 1 ,f AxB ! ex

130 Chapter 3 Mathematical Models and Numerical Methods Involving First-Order Equations

(a) Euler’s method. (b) the trapezoid scheme. (c) the improved Euler’s method.

7. Use the improved Euler’s method subroutine with step size h ! 0.1 to approximate the solution to the initial value problem

at the points x ! 1.1, 1.2, 1.3, 1.4, and 1.5. (Thus, input N ! 5.) Compare these approximations with those obtained using Euler’s method (see Exercises 1.4, Problem 5).

8. Use the improved Euler’s method subroutine with step size h ! 0.2 to approximate the solution to the initial value problem

at the points x ! 1.2, 1.4, 1.6, and 1.8. (Thus, input N ! 4.) Compare these approximations with those obtained using Euler’s method (see Exercises 1.4, Problem 6).

9. Use the improved Euler’s method subroutine with step size h ! 0.2 to approximate the solution to

at the points x ! 0, 0.2, 0.4, . . . , 2.0. Use your an- swers to make a rough sketch of the solution on [0, 2].

10. Use the improved Euler’s method subroutine with step size h ! 0.1 to approximate the solution to

at the points x ! 0, 0.1, 0.2, . . . , 1.0. Use your an- swers to make a rough sketch of the solution on [0, 1].

11. Use the improved Euler’s method with tolerance to approximate the solution to

at t ! 1. For a tolerance of use a stopping procedure based on the absolute error.

12. Use the improved Euler’s method with tolerance to approximate the solution to

at . For a tolerance of , use a stop- ping procedure based on the absolute error.

13. Use the improved Euler’s method with tolerance to approximate the solution to

at x ! 1. For a tolerance of use a stop- ping procedure based on the absolute error.

e ! 0.003, y¿ ! 1 " y # y3 , y A0B ! 0 ,

e ! 0.01x ! p y¿ ! 1 " sin y , y A0B ! 0 ,

e ! 0.01,

dx dt

! 1 # t sin AtxB , x A0B ! 0 ,

y¿ ! 4 cos Ax # yB , y A0B ! 1 ,

y¿ ! x # 3 cos AxyB , y A0B ! 0 ,

y¿ ! 1 x

A y2 # yB , y A1B ! 1 ,

y¿ ! x " y2 , y A1B ! 0 ,

14. By experimenting with the improved Euler’s method subroutine, find the maximum value over the interval

of the solution to the initial value problem

Where does this maximum value occur? Give answers to two decimal places.

15. The solution to the initial value problem

crosses the x-axis at a point in the interval . By experimenting with the improved Euler’s method subroutine, determine this point to two decimal places.

16. The solution to the initial value problem

has a vertical asymptote (“blows up”) at some point in the interval . By experimenting with the improved Euler’s method subroutine, determine this point to two decimal places.

17. Use Euler’s method (4) with h ! 0.1 to approximate the solution to the initial value problem

on the interval (that is, at x ! 0, 0.1, . . . , 1.0). Compare your answers with the actual solution

What went wrong? Next, try the step size h ! 0.025 and also h ! 0.2. What conclusions can you draw concerning the choice of step size?

18. Local versus Global Error. In deriving formula (4) for Euler’s method, a rectangle was used to approximate the area under a curve (see Figure 3.14). With this approximation can be written as

(a) Show that if g has a continuous derivative that is bounded in absolute value by B, then the rectan- gle approximation has error ; that is, for some constant M,

This is called the local truncation error of the scheme. [Hint: Write

" xn # 1

g AtBdt " hg AxnB !" xn # 1 xn

3 g AtB "g AxnB 4 dt . ` " xn # 1

g AtB dt " hg AxnB ` ' Mh2 . O Ah2B

" xn # 1

g AtB dt ! hg AxnB , where h ! xn#1 " xn . g AtB J f At, f AtBB ,

y ! e "20x.

0 ' x ' 1

y¿ ! "20y , y A0B ! 1 ,

3 1, 2 4 dy dx

# y x

! x3y2 , y A1B ! 3

3 0, 1.4 4dydx ! Ax # y # 2B2 , y A0B ! "2 y¿ ! sin Ax # yB , y A0B ! 2 .3 0, 2 4

Section 3.6 Improved Euler’s Method 131

Next, using the mean value theorem, show that Then integrate to

obtain the error bound .] (b) In applying Euler’s method, local truncation

errors occur in each step of the process and are propagated throughout the further computations. Show that the sum of the local truncation errors in part (a) that arise after n steps is . This is the global error, which is the same as the convergence rate of Euler’s method.

19. Building Temperature. In Section 3.3 we mod- eled the temperature inside a building by the initial value problem

(13)

where M is the temperature outside the building, T is the temperature inside the building, H is the addi- tional heating rate, U is the furnace heating or air conditioner cooling rate, K is a positive constant, and T0 is the initial temperature at time t0. In a typical model, t0 ! 0 (midnight), T0 ! 65ºF, 0.1, , and

75 " 20 cos The constant K is usually between and , depending on such things as insulation. To study the effect of insulating this building, consider the typical building described above and use the improved Euler’s method subroutine with h ! to approxi- mate the solution to (13) on the interval (1 day) for K ! 0.2, 0.4, and 0.6.

20. Falling Body. In Example 1 of Section 3.4, we modeled the velocity of a falling body by the initial value problem

under the assumption that the force due to air resistance is "by. However, in certain cases the force due to air resistance behaves more like where r is some constant. This leads to the model

(14)

To study the effect of changing the parameter r in (14), take m ! 1, g ! 9.81, b ! 2, and y0 ! 0. Then use the improved Euler’s method subroutine with h ! 0.2 to approximate the solution to (14) on the interval for r ! 1.0, 1.5, and 2.0. What is the relationship between these solutions and the constant solution ?y AtB # A9.81 /2B1/r

0 ' t ' 5

m dy dt

! mg " byr , y A0B ! y0 .A$1B "byr,

m dy dt

! mg " by , y A0B ! y0 ,

0 ' t ' 24 2 /3

1 /21 /4 Apt /12B .M AtB !U AtB ! 1.5 3 70 " T AtB 4 H AtB !

T At0B ! T0 , dT dt

! K 3M AtB " T AtB 4 # H AtB # U AtB

O AhB

AB/2Bh20 g AtB " g AxnB 0 ' B 0 t " xn 0 .

In Sections 1.4 and 3.6, we discussed a simple numerical procedure, Euler’s method, for obtaining a numerical approximation of the solution to the initial value problem

(1)

Euler’s method is easy to implement because it involves only linear approximations to the solution But it suffers from slow convergence, being a method of order 1; that is, the error is . Even the improved Euler’s method discussed in Section 3.6 has order of only 2. In this section we present numerical methods that have faster rates of convergence. These include Taylor methods, which are natural extensions of the Euler procedure, and Runge–Kutta methods, which are the more popular schemes for solving initial value problems because they have fast rates of convergence and are easy to program.

As in the previous section, we assume that f and are continuous and bounded on the vertical strip and that f possesses as many continuous partial derivatives as needed.

To derive the Taylor methods, let be the exact solution of the related initial value problem

(2)

The Taylor series for about the point xn is

where h ! x " xn. Since satisfies (2), we can write this series in the form

(3)

Observe that the recursive formula for in Euler’s method is obtained by truncating the Taylor series after the linear term. For a better approximation, we will use more terms in the Taylor series. This requires that we express the higher-order derivatives of the solution in terms of the function .

If y satisfies we can compute by using the chain rule:

(4)

In a similar fashion, define f3, f4, . . . , that correspond to the expressions , etc. If we truncate the expansion in (3) after the term, then, with the above notation, the recursive formulas for the Taylor method of order p are

(5)

(6) yn#1 ! yn # hf Axn, ynB # h22! f2 Axn, ynB # p # hpp! fp Axn, ynB . xn#1 ! xn # h ,

hp y‡ AxB, y A4B AxB !: f2 Ax, yB .

! 0f 0x Ax, yB # 0f0y Ax, yB f Ax, yB

y– ! 0f 0x Ax, yB # 0f0y Ax, yB y¿

y–y¿ ! f Ax, yB,f Ax, yB yn#1

fn AxB ! yn # hf Axn, ynB # h22! f–n AxnB # p . fn

fn AxB ! fn AxnB # hf¿n AxnB # h22! f–n AxnB # p , fn AxB

f¿n ! f Ax, fnB , fn AxnB ! yn . fn AxB

E Ax, yB: a 6 x 6 b, "q 6 y 6 qF 0f/ 0y O AhB f AxB.

y¿ ! f Ax, yB , y Ax0B ! y0 . f(x)

132 Chapter 3 Mathematical Models and Numerical Methods Involving First-Order Equations

3.7 HIGHER-ORDER NUMERICAL METHODS:TAYLOR AND RUNGE–KUTTA

As before, , where is the solution to the initial value problem (1). It can be shown† that the Taylor method of order p has the rate of convergence .

Determine the recursive formulas for the Taylor method of order 2 for the initial value problem

(7)

We must compute as defined in (4). Since ,

Substituting into (4), we have

and the recursive formulas (5) and (6) become

where x0 ! 0, are the starting values. ◆

The convergence rate, , of the pth-order Taylor method raises an interesting question: If we could somehow let p go to infinity, would we obtain exact solutions for the interval ? This possibility is explored in depth in Chapter 8. Of course, a practical difficulty in employing high-order Taylor methods is the tedious computation of the partial derivatives needed to determine fp (typically these computations grow exponen- tially with p). One way to circumvent this difficulty is to use one of the Runge–Kutta methods.††

Observe that the general Taylor method has the form

(8)

where the choice of F depends on p. In particular [compare (6)], for

(9)

The idea behind the Runge–Kutta method of order 2 is to choose F in (8) of the form

(10) F ! K2 Ax, y; hB J f Ax # ah, y # bhf Ax, yBB , p ! 2 , F ! T2 Ax, y; hB J f Ax, yB # h2 c 0f0x Ax, yB # 0f0y Ax, yB f Ax, yB d . p ! 1 , F ! T1 Ax, y; hB J f Ax, yB , yn#1 ! yn # hF Axn, yn; hB ,

3 x0, x0 # h 4 O AhpB y0 ! p

yn#1 ! yn # h sin AxnynB # h22 c yn cos AxnynB # xn2 sin A2xnynB d , xn#1 ! xn # h ,

! y cos AxyB # x 2

sin A2xyB , ! y cos AxyB # x cos AxyB sin AxyB f2 Ax, yB ! 0f0x Ax, yB # 0f0y Ax, yB f Ax, yB

0f 0x Ax, yB ! y cos AxyB , 0f0y Ax, yB ! x cos AxyB .

f Ax, yB ! sin AxyBf2 Ax, yB y¿ ! sin AxyB , y A0B ! p .

O AhpBf AxByn ! f AxnB Section 3.7 Higher-Order Numerical Methods: Taylor and Runge–Kutta 133

†See Introduction to Numerical Analysis by J. Stoer and R. Bulirsch (Springer-Verlag, New York, 2002). ††Historical Footnote: These methods were developed by C. Runge in 1895 and W. Kutta in 1901.

Example 1

Solution

where the constants are to be selected so that (8) has the rate of convergence . The advantage here is that K2 is computed by two evaluations of the original function and does not involve the derivatives of .

To ensure convergence, we compare this new scheme with the Taylor method of order 2 and require

That is, we choose so that the Taylor expansions for T2 and K2 agree through terms of order h. For (x, y) fixed, when we expand as given in (10) about h ! 0, we find

(11)

where the expression in brackets for , evaluated at h ! 0, follows from the chain rule. Comparing (11) with (9), we see that for T2 and K2 to agree through terms of order h, we must have Thus,

The Runge–Kutta method we have derived is called the midpoint method and it has the recursive formulas

(12)

(13)

By construction, the midpoint method has the same rate of convergence as the Taylor method of order 2; that is, . This is the same rate as the improved Euler’s method.

In a similar fashion, one can work with the Taylor method of order 4 and, after some elaborate calculations, obtain the classical fourth-order Runge–Kutta method. The recursive formulas for this method are

(14)

where

k4 ! hf Axn # h, yn # k3B . k3 ! hf axn # h2, yn # k22 b , k2 ! hf axn # h2, yn # k12 b , k1 ! hf Axn, ynB , yn#1 ! yn #

1 6 Ak1 # 2k2 # 2k3 # k4B ,xn#1 ! xn # h ,

O Ah2B yn#1 ! yn # hf axn # h2, yn # h2 f Axn, ynBb . xn#1 ! xn # h ,

K2 Ax, y; hB ! f ax # h2, y # h2 f Ax, yBb . a ! b ! 1/2.

dK2 /dh

! f Ax, yB # ca 0f0x Ax, yB # b 0f0y Ax, yB f Ax, yB dh # O Ah2B , K2 AhB ! K2 A0B # dK2dh A0Bh # O Ah2B

K2 ! K2 AhBa, b T2 Ax, y; hB " K2 Ax, y; hB ! O Ah2B , as h S 0 .

O Ah2B f Ax, yB f Ax, yBO

Ah2Ba, b 134 Chapter 3 Mathematical Models and Numerical Methods Involving First-Order Equations

The classical fourth-order Runge–Kutta method is one of the more popular methods because its rate of convergence is and it is easy to program. Typically, it produces very accurate approximations even when the number of iterations is reasonably small. However, as the num- ber of iterations becomes large, other types of errors may creep in.

Program outlines for the fourth-order Runge–Kutta method are given below. Just as with the algorithms for the improved Euler’s method, the first program (the Runge–Kutta subroutine) is useful for approximating the solution over an interval and takes the number of steps in the interval as input. As in Section 3.6, the number of steps N is related to the step size h and the interval by

The subroutine has the option to print out a table of values of x and y. The second algo- rithm (Runge–Kutta with tolerance) on page 136 is used to approximate, for a given toler- ance, the solution at an inputted value x ! c. This algorithm† automatically halves the step sizes successively until the two approximations and differ by less than the prescribed tolerance For a stopping procedure based on the relative error, Step 6 of the algorithm should be replaced by

go to Step 10 .Step 6¿ If ` y " z y ` 6 e,

e. y Ac; h/2By Ac; hB

Nh ! c " x0 .

3 x0, c 4 3 x0, c 4 O Ah4B

Section 3.7 Higher-Order Numerical Methods: Taylor and Runge–Kutta 135

CLASSICAL FOURTH-ORDER RUNGE–KUTTA SUBROUTINE

Purpose To approximate the solution to the initial value problem

for INPUT x0, y0, c, N (number of steps), PRNTR (!1 to print a table) Step 1 Set step size , x ! x0, y ! y0 Step 2 For i ! 1 to N, do Steps 3–5 Step 3 Set

Step 4 Set

Step 5 If PRNTR ! 1, print x, y

y ! y # 1 6

Ak1 # 2k2 # 2k3 # k4B x ! x # h k4 ! hf Ax # h, y # k3B k3 ! hf ax # h2, y # k22b k2 ! hf ax # h2, y # k12b k1 ! hf Ax, yB

h ! Ac " x0B /N x0 ' x ' c

y¿ ! f Ax, yB , y Ax0B ! y0

†Note that the form of the algorithm on page 136 is the same as that for the improved Euler’s method on page 128 except for Step 4, where the Runge–Kutta subroutine is called. More sophisticated stopping procedures are used in production- grade codes.

Use the classical fourth-order Runge–Kutta algorithm to approximate the solution of the initial value problem

at x ! 1 with a tolerance of 0.001.

The inputs are x0 ! 0, y0 ! 1, c ! 1, and M ! 25 (say). Since , the formulas in Step 3 of the subroutine become

The initial value for N in this algorithm is N ! 1, so

Thus, in Step 3 of the subroutine, we compute

and, in Step 4 of the subroutine, we get for the first approximation

! 2.70833 ,

! 1 # 1 6

31 # 2 A1.5B # 2 A1.75B # 2.75 4 y ! y0 # 1 6

Ak1 # 2k2 # 2k3 # k4B k3 ! A1B A1 # 0.75B ! 1.75 , k4 ! A1B A1 # 1.75B ! 2.75 ,k1 ! A1B A1B ! 1 , k2 ! A1B A1 # 0.5B ! 1.5 , h ! A1 " 0B /1 ! 1 . k ! hy , k2 ! h ay # k12b , k3 ! h ay # k22b , k4 ! h A y # k3B .

f Ax, yB ! ye ! 0.001, y¿ ! y , y A0B ! 1 ,

f(x)

136 Chapter 3 Mathematical Models and Numerical Methods Involving First-Order Equations

CLASSICAL FOURTH-ORDER RUNGE–KUTTA ALGORITHM WITH TOLERANCE

Purpose To approximate the solution to the initial value problem

at x ! c, with tolerance INPUT (maximum number of iterations) Step 1 Set z ! y0, PRNTR ! 0 Step 2 For m ! 0 to M, do Steps 3–7 (or, to save time, start with m $ 0) Step 3 Set N ! 2m

Step 4 Call FOURTH-ORDER RUNGE–KUTTA SUBROUTINE Step 5 Print h, y Step 6 If go to Step 10 Step 7 Set z ! y Step 8 Print “ is approximately”; y; “but may not be within the tolerance”; Step 9 Go to Step 11 Step 10 Print “ is approximately”; y; “with tolerance”; Step 11 STOP OUTPUT Approximations of the solution to the initial value problem at x ! c,

using 2m steps.

ef AcB ef AcB0 z " y 0 6 e,

x0, y0, c, e, M e

y¿ ! f Ax, yB , y Ax0B ! y0

Example 2

Solution

where we have rounded to five decimal places. Because

we start over and reset N ! 2, h ! 0.5. Doing Steps 3 and 4 for i ! 1 and 2, we ultimately obtain (for i ! 2) the approximation

Since we again start over and reset N ! 4, h ! 0.25. This leads to the approximation

so that

which is less than Hence ◆

In Example 2 we were able to obtain a better approximation for with h ! 0.25 than we obtained in Section 3.6 using Euler’s method with h ! 0.001 (see Table 3.4, page 124) and roughly the same accuracy as we obtained in Section 3.6 using the improved Euler’s method with h ! 0.01 (see Table 3.5, page 127).

Use the fourth-order Runge–Kutta subroutine to approximate the solution of the initial value problem

(15)

on the interval using N ! 8 steps (i.e., h ! 0.25).

Here the starting values are x0 ! 0 and y0 ! 1. Since , the formulas in Step 3 of the subroutine are

From the output, we find

What happened? Fortunately, the equation in (15) is separable, and, solving for we obtain It is now obvious where the problem lies: The true solution is not definedf AxBf AxB ! A1 " xB"1. f AxB,

x ! 1.50 y ! overflow . x ! 1.25 y ! 4.09664 & 1011 , x ! 1.00 y ! 32.82820 , x ! 0.75 y ! 3.97238 , x ! 0.50 y ! 1.99884 , x ! 0.25 y ! 1.33322 ,

k3 ! h ay # k22b 2 , k4 ! h A y # k3B2 . k1 ! hy

2 , k2 ! h ay # k12b 2 , f Ax, yB ! y20 ' x ' 2

y¿ ! y2 , y A0B ! 1 , f AxB

f A1B ! e f A1B ! e ! 2.71821.e ! 0.001.0 z " y 0 ! 02.71735 " 2.71821 0 ! 0.00086 ,

y ! 2.71821 ,

0 z " y 0 ! 02.70833 " 2.71735 0 ! 0.00902 7 e,y ! 2.71735 . 0 z " y 0 ! 0 y0 " y 0 ! 01 " 2.70833 0 ! 1.70833 7 e , Section 3.7 Higher-Order Numerical Methods: Taylor and Runge–Kutta 137

Example 3

Solution

at x ! 1. If we had been more cautious, we would have realized that is not bounded for all y. Hence, the existence of a unique solution is not guaranteed for all x between 0 and 2, and in this case, the method does not give meaningful approximations for x near (or greater than) 1. ◆

Use the fourth-order Runge–Kutta algorithm to approximate the solution of the initial value problem

at x ! 2 with a tolerance of 0.0001.

This time we check to see whether is bounded. Here which is certainly unbounded in any vertical strip. However, let’s consider the qualitative behavior of the solution

The solution curve starts at (0, 1), where begins decreasing and continues to decrease until it crosses the curve . After crossing this curve, begins to increase, since increases, it remains below the curve This is because if the solution were to get “close” to the curve then the derivative of would approach zero, so that overtaking the function is impossible.

Therefore, although the existence-uniqueness theorem does not guarantee a solution, we are inclined to try the algorithm anyway. The above argument shows that probably exists for x $ 0, so we feel reasonably sure the fourth-order Runge–Kutta method will give a good approximation of the true solution Proceeding with the algorithm, we use the starting val- ues x0 ! 0 and y0 ! 1. Since the formulas in Step 3 of the subroutine become

In Table 3.6, we give the approximations and 4. The algorithm stops at m ! 4, since

Hence, with a tolerance of 0.0001. ◆f A2B ! 1.251320 y A2; 0.125B " y A2; 0.25B 0 ! 0.00000 . y A2; 2 "m#1B for f A2B for m ! 0, 1, 2, 3,

k3 ! h c ax # h2b " ay # k22b 2 d , k4 ! h 3 Ax # hB " A y # k3B2 4 . k1 ! h Ax " y2B , k2 ! h c ax # h2b " ay # k12b 2 d ,

f Ax, yB ! x " y2,f AxB. f AxB1xf AxB

y ! 1x,y ! 1x. f¿ AxB ! x " f2 AxB 7 0. As f AxB f AxBy ! 1x f¿ A0B ! 0 " 1 6 0, so f AxBf AxB. 0f/ 0y ! "2y,0f/ 0y

y¿ ! x " y2 , y A0B ! 1 , f AxB

0f/ 0y ! 2y

138 Chapter 3 Mathematical Models and Numerical Methods Involving First-Order Equations

Example 4

Solution

TABLE 3.6 Classical Fourth-Order Runge–Kutta Approximation for F(2)

Approximation m h for

0 2.0 "8.33333 1 1.0 1.27504 9.60837 2 0.5 1.25170 0.02334 3 0.25 1.25132 0.00038 4 0.125 1.25132 0.00000

0 y A2; hB " y A2; 2hB 0F A2B

Section 3.7 Higher-Order Numerical Methods: Taylor and Runge–Kutta 139

As in Exercises 3.6, for some problems you will find it essential to have a calculator or computer available.† For Problems 1–17, note whether or not is bounded.

1. Determine the recursive formulas for the Taylor method of order 2 for the initial value problem

2. Determine the recursive formulas for the Taylor method of order 2 for the initial value problem

3. Determine the recursive formulas for the Taylor method of order 4 for the initial value problem

4. Determine the recursive formulas for the Taylor method of order 4 for the initial value problem

5. Use the Taylor methods of orders 2 and 4 with h ! 0.25 to approximate the solution to the initial value problem

at x ! 1. Compare these approximations to the actual solution evaluated at x ! 1.

6. Use the Taylor methods of orders 2 and 4 with h ! 0.25 to approximate the solution to the initial value problem

at x ! 1. Compare these approximations to the actual solution evaluated at x ! 1.

7. Use the fourth-order Runge–Kutta subroutine with h ! 0.25 to approximate the solution to the initial value problem

at x ! 1. (Thus, input N ! 4.) Compare this approx- imation to the actual solution evalu- ated at x ! 1.

8. Use the fourth-order Runge–Kutta subroutine with h ! 0.25 to approximate the solution to the initial value problem

at x ! 1. Compare this approximation with the one obtained in Problem 6 using the Taylor method of order 4.

y¿ ! 1 " y , y A0B ! 0 ,

y ! 3 " 2e2x

y¿ ! 2y " 6 , y A0B ! 1 , y ! 1 " e"x

y¿ ! 1 " y , y A0B ! 0 , y ! x # e"x

y¿ ! x # 1 " y , y A0B ! 1 , y¿ ! x2 # y , y A0B ! 0 . y¿ ! x " y , y A0B ! 0 . y¿ ! xy " y2 , y A0B ! "1 . y¿ ! cos Ax # yB , y A0B ! p .

0f / 0y

9. Use the fourth-order Runge–Kutta subroutine with h ! 0.25 to approximate the solution to the initial value problem

at x ! 1. Compare this approximation with the one ob- tained in Problem 5 using the Taylor method of order 4.

10. Use the fourth-order Runge–Kutta algorithm to approximate the solution to the initial value problem

at x ! 2. For a tolerance of use a stop- ping procedure based on the absolute error.

11. The solution to the initial value problem

crosses the x-axis at a point in the interval . By experimenting with the fourth-order Runge–Kutta subroutine, determine this point to two decimal places.

12. By experimenting with the fourth-order Runge–Kutta subroutine, find the maximum value over the interval

of the solution to the initial value problem

Where does this maximum occur? Give your answers to two decimal places.

13. The solution to the initial value problem

has a vertical asymptote (“blows up”) at some point in the interval . By experimenting with the fourth-order Runge–Kutta subroutine, determine this point to two decimal places.

14. Use the fourth-order Runge–Kutta algorithm to approximate the solution to the initial value problem

at For a tolerance of use a stop- ping procedure based on the absolute error.

e ! 0.01,x ! p.

y¿ ! y cos x , y A0B ! 1 ,

3 0, 2 4 dy dx

! y2 " 2exy # e2x # ex , y A0B ! 3

y¿ ! 1.8 x4

" y2 , y A1B ! "1 . 3 1, 2 4

31, 24y¿ ! 2 x4

" y2 , y A1B ! "0.414 e ! 0.001,

y¿ ! 1 " xy , y A1B ! 1 ,

y¿ ! x # 1 " y , y A0B ! 1 ,

3.7 EXERCISES

†An applet, maintained on the web at http://alamos.math.arizona.edu/~rychlik/JOde/index.html automates most of the differential equation algo- rithms discussed in this book.

http://alamos.math.arizona.edu/~rychlik/JOde/index.html

15. Use the fourth-order Runge–Kutta subroutine with h ! 0.1 to approximate the solution to

at the points x ! 0, 0.1, 0.2, . . . , 3.0. Use your answers to make a rough sketch of the solution on .

16. Use the fourth-order Runge–Kutta subroutine with h ! 0.1 to approximate the solution to

at the points x ! 0, 0.1, 0.2, . . . , 4.0. Use your an- swers to make a rough sketch of the solution on [0, 4].

17. The Taylor method of order 2 can be used to approx- imate the solution to the initial value problem

at x ! 1. Show that the approximation yn obtained by using the Taylor method of order 2 with the step size is given by the formula

The solution to the initial value problem is , so yn is an approximation to the constant e.

18. If the Taylor method of order p is used in Problem 17, show that

19. Fluid Flow. In the study of the nonisothermal flow of a Newtonian fluid between parallel plates, the equation

was encountered. By a series of substitutions, this equation can be transformed into the first-order equation

dy du

! u au 2

# 1b y3 # au # 5 2 b y2 .

d 2y

dx2 # x2ey ! 0 , x 7 0 ,

n ! 1, 2, . . . .

yn ! a1 # 1n # 12n2 # 16n3 # # # # # 1p!npb n , y ! ex

yn ! a1 # 1n # 12n2b n , n ! 1, 2, . . . . 1 /n

y¿ ! y , y A0B ! 1 ,

y¿ ! 3 cos A y " 5xB , y A0B ! 0 , 3 0, 3 4y¿ ! cos A5yB " x , y A0B ! 0 ,

140 Chapter 3 Mathematical Models and Numerical Methods Involving First-Order Equations

Use the fourth-order Runge–Kutta algorithm to approximate if satisfies . For a tolerance of use a stopping procedure based on the relative error.

20. Chemical Reactions. The reaction between nitrous oxide and oxygen to form nitrogen dioxide is given by the balanced chemical equation 2NO # O2 ! 2NO2. At high temperatures the dependence of the rate of this reaction on the concentrations of NO, O2, and NO2 is complicated. However, at 25ºC the rate at which NO2 is formed obeys the law of mass action and is given by the rate equation

where denotes the concentration of NO2 at time t, k is the rate constant, is the initial concentration of NO, and is the initial concentration of O2. At 25ºC, the constant k is 7.13 & 103 (liter)2/(mole)2(sec- ond). Let mole/L, mole/L, and mole/L. Use the fourth-order Runge– Kutta algorithm to approximate . For a toler- ance of use a stopping procedure based on the relative error.

21. Transmission Lines. In the study of the electric field that is induced by two nearby transmission lines, an equation of the form

arises. Let If ! 1, use the fourth-order Runge–Kutta algorithm to approximate . For a tolerance of use a stopping procedure based on the absolute error.

e ! 0.0001,z A1B z A0Bf AxB ! 5x # 2 and g AxB ! x2.

dz dx

# g AxBz2 ! f AxB

e ! 0.000001, x A10Bx A0B ! 0 b ! 0.0041a ! 0.0010

x AtB dx dt

! k Aa " xB2 ab " x 2 b ,

e ! 0.0001, y A2B ! 0.1y AuBy A3B

A Dynamics of HIV Infection Courtesy of Glenn Webb, Vanderbilt University

The dynamics of HIV (human immunodeficiency virus) infection within a human host involve the interaction of the HIV virions and CD4# T lymphocytes. CD4# T lymphocytes are long- lived white blood cells that play a major role in the defense of the human body against microbial invaders. HIV targets these very cells. When HIV first appeared as a new and major health threat, it was recognized that the disease typically exhibited a lengthy gradual progression lasting 10 or more years. It was widely believed that the dynamics of HIV destruction of CD4# T-cell popula- tion involved a very low rate of infection and a very slow turnover of virus and infected cells. In 1995 differential equation models of HIV-CD4# T-cell interaction revealed that the turnover rate for the infected CD4# T cells was very much faster than this (about 2 days)—a scientific break- through reported simultaneously in the papers of D. D. Ho et al., “Rapid Turnover of Plasma Viri- ons and CD4 Lymphocytes in HIV-1 Infection,” Nature 1995; and of G. M. Shaw et al., “Viral Dynamics in Human Immunodeficiency Virus Type I Infection,” Nature 1995.

Underlying the models in these papers is the knowledge that within a person infected with HIV, the virus spends part of its existence free and part inside an infected CD4# T cell. The time spent free was known to be very short—on the order of 30 minutes. The time spent inside an invaded CD4# T cell was believed to be very long—on the order of years. When a cell was invaded, a virion (a complete viral particle, consisting of RNA surrounded by a protein shell) took over the cell’s DNA and used it to replicate its own RNA, thereby creating new virions; then it budded, or burst the cell, to release multiple virus particles.

The differential equations of the model are similar to those for compartmental analysis dis- cussed in Section 3.2. They involve compartments and parameters. The compartments of the model are

the population of uninfected CD4# T cells at time t . the population of infected CD4# T cells at time t .

the population of virus at time t . The parameters (followed by their units) of the model are

constant input source of uninfected cells per day (the human body produces these cells daily in the thymus) .

normal loss rate constant of uninfected cells the average lifespan of an uninfected cell in days .

infection rate constant of uninfected cells per infected cell the rate is of mass action form, i.e.,

loss rate constant of infected cells the average lifespan of an infected cell in days .

loss rate constant of free virus the average lifespan of a free virion in days .

number of virions produced per day per infected cell (the burst number of an infected cell) . N Avirions # cell"1B !B

A1/g !g Aday"1B ! B A1/m !m Aday"1B ! bV AtB T AtB B .

Ab Avirions"1 # day"1B ! B A1/d !d Aday"1B !

l Acells # day"1B ! V AtB !I AtB ! T AtB !

Group Projects for Chapter 3

141

The independent variable of the model is time t in days and the dependent variables of the model are The equations are as follows (see Figure 3.16).

! " "

Time change Source Normal loss Infection rate

! "

Time change Gain from infection Loss rate

! "

Time change Viral production Decay rate

As mentioned above, the average lifespan of a free virion, , is approximately 30 minutes, which means . On the other hand, it was thought that , the average length of time an infected CD4# T cell lasts before bursting to produce new virions, should be several years, implying that must be quite small on the order of However, when drugs to treat HIV infection first became available in the mid-1990s, researchers were able to deduce a surprisingly different value from patient data and the differential equation models. By completing the following steps, you will be able to determine a better approximation to in the manner utilized by Ho et al.

To incorporate the effect of treatment in the differential equations model, set that is, assume the action of the drug completely inhibits the infection process. This is a reasonable approximation and it simplifies the analysis.

(a) With the assumption of treatment, what are the reduced forms of the differential equa- tions for ?

(b) Solve these reduced equations for , with the initial conditions

(c) Argue from your formula for , that the graph of on a log scale (i.e., the graph of log V) over an extended period of time (say, several weeks) will tend toward a graph of a straight line whose slope is either (the negative reciprocal of the average lifespan of a free virus) or (the negative reciprocal of the average lifespan of an infected CD4# T cell), according to whether or is smaller.mg

"m "g

V AtBV AtBT A0B ! T0, I A0B ! I0, and V A0B ! V0. T AtB, I AtB, and V AtBT AtB, I AtB, and V AtB

b ! 0;

1/m

10"3 day"1B.Am 1/mg ! 48 day"1

1/g

gV AtBN mI AtBd dt

V AtB mI AtBbV AtBT AtBd

dt I AtB

bV AtBT AtBdT AtBld dt

T AtB T AtB, I AtB, and V AtB.

142 Chapter 3 Mathematical Models and Numerical Methods Involving First-Order Equations

Figure 3.16 Compartmental views of virus, uninfected T cells, and infected T cells

HIV

V(t) Virus

T(t) Uninfected

cells

I(t) Infected

cells

CD4+ T cells

gV(t)

NmI(t)

b V(t)T(t)

dT(t) mI(t)

}

} } } }

} }

}} }

(d) For patients who undergo therapy, it is possible to measure the viral load V and to deter- mine the rate at which their viral load declines. In Ho et al., the data in Figure 3.17 were presented for the decrease (on a logarithmic scale) in viral loads of three patients. Using these data and part (c), explain why it follows that must be the approximate slope of the nearly linear curve for log V and thereby deduce the revised estimate for the average time an infected cell lasts between being invaded and bursting.

(e) Check out the recent literature of mathematical models of HIV dynamics in infected hosts (try a Google Scholar search) and find out how the estimate of the lifespan of infected CD4# T cells has been improved using more refined ordinary differential equation models. (Other models have been used to estimate the lifespan of the free virus. Also, models have been developed to track the long-term effects of patients undergoing therapy, optimal ways to schedule treatment, and the problems that arise when drug resistance develops.)

The dynamics of HIV-1 replication in patients receiving anti-retroviral therapy is a subject of contin- uing investigation and mathematical modeling. It is well known that therapy does not eliminate the virus in patients, and it is necessary to continue treatment indefinitely. The reasons are complex and connected to the presence of viral reservoirs, which allow the virus population to restore if treatment is discontinued. Further investigations of this subject may be found in the following references:

Quantifying residual HIV-1 replication in patients receiving combination anti-retroviral ther- apy. Zhang LQ, Ramratnam B, Tenner-Racz K, He YX, Vesanen M, Lewin S, Talal A, Racz P, Perelson AS, Korber BT, Markowitz M, and Ho DD. New England Journal of Medicine 340:1605–1613, 1999.

The decay of the latent reservoir of replication-competent HIV-1 is inversely correlated with the extent of residual viral replication during prolonged anti-retroviral therapy. Ramratnam B, Mittler JE, Zhang LQ, Boden D, Hurley A, Fang F, Macken CA, Perelson AS, Markowitz M, and Ho DD. Nature Medicine 6:82–85, 2000.

Group Projects for Chapter 3 143

Figure 3.17 Viral load decrease in three HIV patients

10,000

1,000

100

0.1 –10 –5 5 10 15 20 25 30 0 –10 –5 5 10 15 20 25 30 0 –10 –5 5 10 15 20 25 300

R N

A c

op ie

s pe

r m L

(& 1

03 )

303 403 409 a b c

Slope: –0.21 t : 3.3 days 1

Slope: –0.32 t : 2.2 days 1

Slope: –0.47 t : 1.5 days 1

B Aquaculture Aquaculture is the art of cultivating the plants and animals indigenous to water. In the example considered here, it is assumed that a batch of catfish are raised in a pond. We are interested in determining the best time for harvesting the fish so that the cost per pound for raising the fish is minimized.

A differential equation describing the growth of fish may be expressed as

(1)

where W(t) is the weight of the fish at time t and K and are empirically determined growth con- stants. The functional form of this relationship is similar to that of the growth models for other species. Modeling the growth rate or metabolic rate by a term like is a common assumption. Biologists often refer to equation (1) as the allometric equation. It can be supported by plausi- bility arguments such as growth rate depending on the surface area of the gut (which varies like

) or depending on the volume of the animal (which varies like W).

(a) Solve equation (1) when (b) The solution obtained in part (a) grows large without bound, but in practice there is

some limiting maximum weight Wmax for the fish. This limiting weight may be included in the differential equation describing growth by inserting a dimensionless variable S that can range between 0 and 1 and involves an empirically determined parameter m. Namely, we now assume that

(2)

where When equation (2) has a closed form solu- tion. Solve equation (2) when (ounces), and

1 (ounce). The constants are given for t measured in months. (c) The differential equation describing the total cost in dollars of raising a fish for t

months has one constant term K1 that specifies the cost per month (due to costs such as interest, depreciation, and labor) and a second constant K2 that multiplies the growth rate (because the amount of food consumed by the fish is approximately proportional to the growth rate). That is,

(3)

Solve equation (3) when K1 ! 0.4, K2 ! 0.1, 1.1 (dollars), and is as deter- mined in part (b).

(d) Sketch the curve obtained in part (b) that represents the weight of the fish as a function of time. Next, sketch the curve obtained in part (c) that represents the total cost of rais- ing the fish as a function of time.

(e) To determine the optimal time for harvesting the fish, sketch the ratio . This ratio represents the total cost per ounce as a function of time. When this ratio reaches its minimum—that is, when the total cost per ounce is at its lowest—it is the optimal time to harvest the fish. Determine this optimal time to the nearest month.

C AtB /W AtB

W AtBC A0B ! dC dt

! K1 # K2 dW dt

C AtBW A0B ! a ! 3 /4, m ! 1 /4, Wmax ! 81K ! 10,

m ! 1 " a,S J 1 " AW/WmaxBm. dW dt

! KWaS ,

a ) 1.

W2/3

dW dt

! KWa ,

144 Chapter 3 Mathematical Models and Numerical Methods Involving First-Order Equations

C Curve of Pursuit An interesting geometric model arises when one tries to determine the path of a pursuer chasing its prey. This path is called a curve of pursuit. These problems were analyzed using methods of calculus circa 1730 (more than two centuries after Leonardo da Vinci had considered them). The simplest problem is to find the curve along which a vessel moves in pursuing another vessel that flees along a straight line, assuming the speeds of the two vessels are constant.

Let’s assume that vessel A, traveling at a speed is pursuing vessel B, which is traveling at a speed In addition, assume that vessel A begins (at time t ! 0) at the origin and pursues vessel B, which begins at the point (1, 0) and travels up the line x ! 1. After t hours, vessel A is located at the point , and vessel B is located at the point (see Figure 3.18). The goal is to describe the locus of points P; that is, to find y as a function of x.

(a) Vessel A is pursuing vessel B, so at time t, vessel A must be heading right at vessel B. That is, the tangent line to the curve of pursuit at P must pass through the point Q (see Figure 3.18). For this to be true, show that

(4)

(b) We know the speed at which vessel A is traveling, so we know that the distance it travels in time t is This distance is also the length of the pursuit curve from to

. Using the arc length formula from calculus, show that

(5)

Solving for t in equations (4) and (5), conclude that

(6)

where w J dy/dx.

Ax " 1B dw dx

! " b

a 21 # w2 ,

y " Ax " 1B Ady/dxB b

! 1 a

" x

0 21 # 3 y¿ AuB 4 2 du .

at ! " x

0 21 # 3 y¿ AuB 4 2 du .Ax, yB

A0, 0Bat. dy dx

! y " bt x " 1

Q ! A1, btBP ! Ax, yB b.

Group Projects for Chapter 3 145

0 ( 1 , 0)

Q = ( 1, t )

P = ( x, y )

Figure 3.18 The path of vessel A as it pursues vessel B

(d) Using separation of variables and the initial conditions x ! 0 and when t ! 0, show that

(7)

(e) For — that is, the pursuing vessel A travels faster than the pursued vessel B— use equation (7) and the initial conditions x ! 0 and y ! 0 when t ! 0, to derive the curve of pursuit

(f) Find the location where vessel A intercepts vessel B if (g) Show that if then the curve of pursuit is given by

Will vessel A ever reach vessel B?

D Aircraft Guidance in a Crosswind Courtesy of T. L. Pearson, Professor of Mathematics, Acadia University (Retired), Nova Scotia, Canada

An aircraft flying under the guidance of a nondirectional beacon (a fixed radio transmitter, abbre- viated NDB) moves so that its longitudinal axis always points toward the beacon (see Figure 3.19). A pilot sets out toward an NDB from a point at which the wind is at right angles to the initial direction of the aircraft; the wind maintains this direction. Assume that the wind speed and the speed of the aircraft through the air (its “airspeed”) remain constant. (Keep in mind that the latter is different from the aircraft’s speed with respect to the ground.)

(a) Locate the flight in the xy-plane, placing the start of the trip at and the destination at . Set up the differential equation describing the aircraft’s path over the ground.

Hint: (b) Make an appropriate substitution and solve this equation. [Hint: See Section 2.6.]

dy/dx ! Ady/dtB / Adx/dtB. 43 A0, 0B A2, 0B

y ! 1 2 e 1

2 3 A1 " xB2 " 1 4 " ln A1 " xB f .a ! b,

a 7 b.

y ! 1

2 c A1 " xB1#b/a

1 # b/a " A1 " xB1"b/a

1 " b/a d # ab a2 " b2

a 7 b

dy dx

! w ! 1 2

3 A1 " xB"b/a " A1 " xBb/a 4 . w ! dy/dx ! 0

146 Chapter 3 Mathematical Models and Numerical Methods Involving First-Order Equations

Wind NDB

Figure 3.19 Guided aircraft

(c) Use the fact that x ! 2 and y ! 0 at t ! 0 to determine the appropriate value of the arbitrary constant in the solution set.

(d) Solve to get y explicitly in terms of x. Write your solution in terms of a hyperbolic function. (e) Let be the ratio of windspeed to airspeed. Using a software package, graph the

solutions for the cases 0.1, 0.3, 0.5, and 0.7 all on the same set of axes. Interpret these graphs.

(f) Discuss the (terrifying!) cases

E Feedback and the Op Amp The operational amplifier (op amp) depicted in Figure 3.20(a) is a nonlinear device. Thanks to internal power sources, concatenated transistors, etc., it delivers a huge negative voltage at the output terminal O whenever the voltage at its inverting terminal " exceeds that at its noninvert- ing terminal # , and a huge positive voltage when the situation is reversed. One could express

with a large gain G (sometimes 1000 or more), but the approximation would be too unreliable for many applications. Engineers have come up with a way to tame this unruly device by employing negative feedback, as illustrated in Figure 3.20(b). By connecting the output to the inverting input terminal, the op amp acts like a policeman, preventing any unbalance between the inverting and noninverting input voltages. With such a connection, then, the inverting and noninverting voltages are maintained at the same value: 0 V (electrical ground), for the situa- tion depicted.

Furthermore, the input terminals of the op amp do not draw any current; whatever current is fed to the inverting terminal is immediately redirected to the feedback path. As a result the current drawn from the indicated source E t is governed by the equivalent circuit shown in Figure 3.20(c):

E AtB ! 1 C " I AtBdt or I AtB ! C dEdt ,

Eout ! G AE#in " E"in BBA BA

g ! 1 and g 7 1.

g ! g

Group Projects for Chapter 3 147

(a)

C O

−

(b)

Eout

E−in

E+in E(t)

−

(c)

C I

E(t)

0 V

(d)

R I

Eout

Figure 3.20 Op amp differentiator

and this current I flows through the resistor R in Figure 3.20(d), causing a voltage drop from 0 to "RI. In other words, the output voltage Eout ! "RI ! "RC dE dt is a scaled and inverted replica of the derivative of the source voltage. The circuit is an op amp differentiator.

(a) Mimic this analysis to show that the circuit in Figure 3.21 is an op amp integrator with

up to a constant that depends on the initial charge on the capacitor. (b) Design op amp integrators and differentiators using negative feedback but with induc-

tors instead of capacitors. (In most situations, capacitors are less expensive than induc- tors, so the previous designs are preferred.)

F Bang-Bang Controls In Example 3 of Section 3.3 (page 105), it was assumed that the amount of heating or cooling supplied by a furnace or air conditioner is proportional to the difference between the actual tem- perature and the desired temperature; recall the equation

In many homes the heating/cooling mechanisms deliver a constant rate of heat flow, say,

(with K1 % 0).

(a) Modify the differential equation (9) in Example 3 on page 105 so that it describes the temperature of a home employing this “bang-bang” control law.

(b) Suppose the initial temperature is greater than TD. Modify the constants in the solution (12), page 106, so that the formula is valid as long as .

(c) If the initial temperature is less than TD, what values should the constants in (12) take to make the formula valid for ?

(d) How does one piece the solutions in (b) and (c) to obtain a complete time description of the temperature ?T AtB

T AtB 6 TDT A0B T AtB 7 TDT A0B

U AtB ! eK1 , if T AtB 7 TD , K2 , if T AtB 6 TD

U AtB ! KU 3TD " T AtB 4 .

Eout ! " 1

RC "E AtB dt ,

B/A

148 Chapter 3 Mathematical Models and Numerical Methods Involving First-Order Equations

EoutE(t) −

0 V

Figure 3.21 Op amp integrator

G Market Equilibrium: Stability and Time Paths

Courtesy of James E. Foster, George Washington University

A perfectly competitive market is made up of many buyers and sellers of an economic product, each of whom has no control over the market price. In this model, the overall quantity demanded by the buyers of the product is taken to be a function of the price of the product (among other things) called the demand function. Similarly, the overall quantity supplied by the sellers of the product is a function of the price of the product (among other things) called the supply function. A market is in equilibrium at a price where the quantity demanded is just equal to the quantity supplied.

The linear model assumes that the demand and supply functions have the form and , respectively, where p is the market price of the product, is the associated quantity demanded, is the associated quantity supplied, and and are all positive constants. The functional forms ensure that the “laws” of downward sloping demand and upward sloping supply are being satisfied. It is easy to show that the equilibrium price is

. Economists typically assume that markets are in equilibrium and justify this assumption with

the help of stability arguments. For example, consider the simple price adjustment equation

where is a constant indicating the speed of adjustments. This follows the intuitive require- ment that price rises when demand exceeds supply and falls when supply exceeds demand. The market equilibrium is said to be globally stable if, for every initial price level , the price adjustment path satisfies as

(a) Find the price adjustment path: Substitute the expressions for and into the price adjustment equation and show that the solution to the resulting differential equation is

where (b) Is the market equilibrium globally stable?

Now consider a model that takes into account the expectations of agents. Let the market demand and supply functions over time be given by

respectively, where is the market price of the product, is the associated quantity demanded, is the associated quantity supplied, and are all positive constants. The functional forms ensure that, when faced with an increasing price, demanders will tend to purchase more (before prices rise further) while suppliers will tend to offer less (to take advantage of the higher prices in the future). Now given the above stability argument, we restrict consideration to market clearing time paths satisfying , for all and explore the evolution of price over time. We say that the market is in dynamic equilibrium if

for all t. It is easy to show that the dynamic equilibrium in this model is given byp¿ AtB ! 0 t + 0,qd AtB ! qs AtBp AtB

d0, d1, d2, s0, s1, and s2qs AtB qd AtBp AtB qd AtB ! d0 " d1 p AtB # d2 p¿ AtB and qs AtB ! "s0 # s1 p AtB " s2 p¿ AtB ,

t + 0

c ! "l Ad1 # s1B .p AtB ! 3p A0B " p* 4 ect # p*, qsqd t S q.p AtB S p*p AtB p A0B

l 7 0

dp dt

! l Aqd " qsB , Ad0 # s0B / Ad1 # s1B p* !

s1d0, d1, s0,qs qdqs ! "s0 # s1p

qd ! d0 " d1p

Group Projects for Chapter 3 149

for all t, where is the market equilibrium price defined above. However, many other market clearing time paths are possible.

(c) Find a market clearing time path: Equate and solve the resulting differen- tial equation in terms of its initial value

(d) Is it true that for any market clearing time path we must have (e) If the price of a product is $5 at months and demand and supply functions

are modeled as and what will be the price after 10 months? As t becomes very large? What is happening to and how are the expectations of demanders and suppliers evolving?

For further reading, see, for example, Mathematical Economics, 2nd ed. by Akira Takayama (Cambridge University Press, Cambridge, 1985), Chapter 3; and Fundamental Methods of Mathematical Economics, 4th ed. by Alpha Chiang, and Kevin Wainwright (McGraw-Hill/Irvin, Boston, 2008), Chapter 14.

H Stability of Numerical Methods Numerical methods are often tested on simple initial value problems of the form

(8)

which has the solution Notice that for each the solution tends to zero as Thus, a desirable property for any numerical scheme that generates approximations

at the points 0, h, 2h, 3h, . . . is that, for

(9)

For single-step linear methods, property (9) is called absolute stability.

(a) Show that for xn ! nh, Euler’s method, when applied to the initial value problem (8), yields the approximations

and deduce that this method is absolutely stable only when (This means that for a given we must choose the step size h sufficiently small in order for property (9) to hold.) Further show that for grows large exponentially!

(b) Show that for xn ! nh the trapezoid scheme of Section 3.6, applied to problem (8), yields the approximations

and deduce that this scheme is absolutely stable for all (c) Show that the improved Euler’s method applied to problem (8) is absolutely stable for

Multistep Methods. When multistep numerical methods are used, instability problems may arise that cannot be circumvented by simply choosing a sufficiently small step size h. This is because multistep methods yield “extraneous solutions,” which may dominate the calculations. To see what can happen, consider the two-step method

(10)

for the equation y¿ ! f Ax, yB.yn#1 ! yn"1 # 2hf Axn, ynB , n ! 1, 2, . . . ,

0 6 lh 6 2.

l 7 0, h 7 0.

yn ! a1 " lh/21 # lh/2b n , n ! 0, 1, 2, . . . , h 7 2 /l, the error yn " f AxnBl 7 0,

0 6 lh 6 2. yn ! A1 " lhBn , n ! 0, 1, 2, . . . ,

yn S 0 as n S q . l 7 0,y0, y1, y2, y3, . . . to f AxBx S #q.

f AxBl 7 0f AxB ! e"lx.y¿ # ly ! 0 , y A0B ! 1 , Al ! constantB ,

p¿ AtBqs AtB ! "20 # p AtB " 6p¿ AtB,qd AtB ! 30 " 2p AtB # 4p¿ AtB t ! 0p AtB p AtB S p* as t S q?

p0 ! p A0B.p AtB qd AtB and qs AtB p*p AtB ! p*

150 Chapter 3 Mathematical Models and Numerical Methods Involving First-Order Equations

(d) Show that for the initial value problem (11) the recurrence formula (10), with xn ! nh, becomes (12)

Equation (12), which is called a difference equation, can be solved by using the following approach. We postulate a solution of the form , where r is a constant to be determined.

(e) Show that substituting in (12) leads to the “characteristic equation”

which has roots

By analogy with the theory for second-order differential equations, it can be shown that a general solution of (12) is

where c1 and c2 are arbitrary constants. Thus, the general solution to the difference equation (12) has two independent constants, whereas the differential equation in (11) has only one, namely,

(f) Show that for each h $ 0,

Hence, the term behaves like the solution However, the extraneous solution grows large without bound.

(g) Applying the scheme of (10) to the initial value problem (11) requires two starting values y0, y1. The exact values are However, regardless of the choice of starting values and the size of h, the term will eventually dominate the full solution to the recur- rence equation as xn increases. Illustrate this instability taking and using a calculator or computer to compute y2, y3, . . . , y100 from the recurrence formula (12) for h ! 0.5 and h ! 0.05. (Note: Even if initial conditions are chosen so that c2 ! 0, round- off error will inevitably “excite” the extraneous dominant solution.)

I Period Doubling and Chaos In the study of dynamical systems, the phenomena of period doubling and chaos are observed. These phenomena can be seen when one uses a numerical scheme to approximate the solution to an initial value problem for a nonlinear differential equation such as the following logistic model for population growth:

(13)

(See Section 3.2.)

(a) Solve the initial value problem (13) and show that approaches 1 as (b) Show that using Euler’s method (see Sections 1.4 and 3.6) with step size h to approxi-

mate the solution to (13) gives

(14)

(c) For h ! 0.18, 0.23, 0.25, and 0.3, show that the first 40 iterations of (14) appear to (i) converge to 1 when h ! 0.18, (ii) jump between 1.18 and 0.69 when h ! 0.23, (iii) jump between 1.23, 0.54, 1.16, and 0.70 when h ! 0.25, and (iv) display no discernible pattern when h ! 0.3.

pn#1 ! A1 # 10hBpn " A10hBp2n , p0 ! 0.1 . t S #q.p AtB

dp dt

! 10p A1 " pB , p A0B ! 0.1 .

y1 ! e "2h,y0 ! 1,

c2r n 2

y0 ! 1, y1 ! e "2h.

r n2 f AxnB ! e"2xn as n S q.r n1limnSq r n1 ! 0 but limnSq 0 r n2 0 ! q .

f AxB ! ce"2x. yn ! c1r

n 1 # c2r

n 2 ,

r1 ! "2h # 21 # 4h2 and r2 ! "2h " 21 # 4h2 .r 2 # 4hr " 1 ! 0 ,

yn ! r n

yn#1 # 4hyn " yn"1 ! 0 .

y¿ # 2y ! 0 , y A0B ! 1 , Group Projects for Chapter 3 151

152 Chapter 3 Mathematical Models and Numerical Methods Involving First-Order Equations

0.18 0.20 0.22 0.24 0.28 0.30 h

0.2

0.4

0.6

0.8

1.0

1.2

0.26

Figure 3.22 Period doubling to chaos

The transitions from convergence to jumping between two numbers, then four numbers, and so on, are called period doubling. The phenomenon displayed when h ! 0.3 is referred to as chaos. This transition from period doubling to chaos as h increases is frequently observed in dynamical systems.

The transition to chaos is nicely illustrated in the bifurcation diagram (see Figure 3.22). This dia- gram is generated for equation (14) as follows. Beginning at h ! 0.18, compute the sequence using (14) and, starting at n ! 201, plot the next 30 values—that is, p201, p202, . . . , p230. Next, incre- ment h by 0.001 to 0.181 and repeat. Continue this process until h ! 0.30. Notice how the figure splits from one branch to two, then four, and then finally gives way to chaos.

Our concern is with the instabilities of the numerical procedure when h is not chosen small enough. Fortunately, the instability observed for Euler’s method—the period doubling and chaos—was immediately recognized because we know that this type of behavior is not expected of a solution to the logistic equation. Consequently, if we had tried Euler’s method with h ! 0.23, 0.25, or 0.3 to solve (13) numerically, we would have realized that h was not chosen small enough.

The situation for the classical fourth-order Runge–Kutta method (see Section 3.7) is more troublesome. It may happen that for a certain choice of h period doubling occurs, but it is also possible that for other choices of h the numerical solution actually converges to a limiting value that is not the limiting value for any solution to the logistic equation in (13).

(d) Approximate the solution to (13) by computing the first 60 iterations of the classical fourth-order Runge–Kutta method using the step size h ! 0.3. (Thus, for the subroutine on page 144, input N ! 60 and ) Repeat with h ! 0.325 and h ! 0.35. Which values of h (if any) do you feel are giving the “correct” approximation to the solution? Why?

A further discussion of chaos appears in Section 5.8.

c ! 60 A0.3B ! 18.

EpnF

153

Newton’s second law—force equals mass times acceleration —is without a doubt the most commonly encountered differential equation in practice. It is an ordinary differential equation of the second order since acceleration is the second derivative of position with respect to time

When the second law is applied to a mass–spring oscillator, the resulting motions are com- mon experiences of everyday life, and we can exploit our familiarity with these vibrations to obtain a qualitative description of the solutions of more general second-order equations.

We begin by referring to Figure 4.1, which depicts the mass–spring oscillator. When the spring is unstretched and the inertial mass m is still, the system is at equilibrium; we measure the coordinate y of the mass by its displacement from the equilibrium position. When the mass m is displaced from equilibrium, the spring is stretched or compressed and it exerts a force that resists the displacement. For most springs this force is directly proportional to the displace- ment y and is thus given by

(1)

where the positive constant k is known as the stiffness and the negative sign reflects the oppos- ing nature of the force. Hooke’s law, as equation (1) is commonly known, is only valid for sufficiently small displacements; if the spring is compressed so strongly that the coils press against each other, the opposing force obviously becomes much stronger.

Fspring ! "ky ,

Aa ! d2y/dt2B. AyB AF ! maB

A damped mass–spring oscillator consists of a mass m attached to a spring fixed at one end, as shown in Figure 4.1. Devise a differential equation that governs the motion of this oscillator, taking into account the forces acting on it due to the spring elasticity, damping friction, and possible external influences.

Linear Second-Order Equations

CHAPTER 4

INTRODUCTION: THE MASS–SPRING OSCILLATOR4.1

k m

yEquilibrium point

Figure 4.1 Damped mass–spring oscillator

Practically all mechanical systems also experience friction, and for vibrational motion this force is usually modeled accurately by a term proportional to velocity:

(2)

where is the damping coefficient and the negative sign has the same significance as in equation (1).

The other forces on the oscillator are usually regarded as external to the system. Although they may be gravitational, electrical, or magnetic, commonly the most important external forces are transmitted to the mass by shaking the supports holding the system. For the moment we lump all the external forces into a single, known function Newton’s law then provides the differential equation for the mass–spring oscillator:

(3)

What do mass–spring motions look like? From our everyday experience with weak auto suspensions, musical gongs, and bowls of jelly, we expect that when there is no friction

or external force, the (idealized) motions would be perpetual vibrations like the ones depicted in Figure 4.2. These vibrations resemble sinusoidal functions, with their amplitude depending on the initial displacement and velocity. The frequency of the oscillations increases for stiffer springs but decreases for heavier masses.

In Section 4.3 we will show how to find these solutions. Example 1 demonstrates a quick calculation that confirms our intuitive predictions.

Verify that if b ! 0 and , equation (3) has a solution of the form ! cos #t and that the angular frequency v increases with k and decreases with m.

Under the conditions stated, equation (3) simplifies to

(4)

The second derivative of is and if we insert it into (4), we find

which is indeed zero if This # increases with k and decreases with m, as predicted. ◆

v ! 2k/m.my– $ ky ! "mv2 cos vt $ k cos vt , "v2 cos vt,y AtBmy– $ ky ! 0 .

y AtBFext AtB ! 0

Ab ! 0B my! " by# " ky $ Fext AtB . my– ! "ky " by¿ $ Fext AtB

Fext AtB. b A%0BFfriction ! "b

dy dt

! "by¿ ,

154 Chapter 4 Linear Second-Order Equations

(a) (b) (c)

Figure 4.2 (a) Sinusoidal oscillation, (b) stiffer spring, and (c) heavier mass

Example 1

Solution

When damping is present, the oscillations die out, and the motions resemble Figure 4.3. In Figure 4.3(a) the graph displays a damped oscillation; damping has slowed the frequency, and the amplitude appears to diminish exponentially with time. In Figure 4.3(b) the damping is so dominant that it has prevented the system from oscillating at all. Devices that are supposed to vibrate, like tuning forks or crystal oscillators, behave like Figure 4.3(a), and the damping effect is usually regarded as an undesirable loss mechanism. Good automotive suspension systems, on the other hand, behave like Figure 4.3(b); they exploit damping to suppress the oscillations.

The procedures for solving (unforced) mass–spring systems with damping are also described in Section 4.3, but as Examples 2 and 3 below show, the calculations are more com- plex. Example 2 has a relatively low damping coefficient and illustrates the solutions for the “underdamped” case in Figure 4.3(a). In Example 3 the damping is more severe

, and the solution is “overdamped” as in Figure 4.3(b).

Verify that the exponentially damped sinusoid given by is a solution to equa- tion (3) if , and .

The derivatives of y are

and insertion into (3) gives

◆

Verify that the simple exponential function is a solution to equation (3) if , and .

The derivatives of y are and insertion into (3) produces

◆

Now if a mass–spring system is driven by an external force that is sinusoidal at the angular frequency #, our experiences indicate that although the initial response of the system may be somewhat erratic, eventually it will respond in “sync” with the driver and oscillate at the same frequency, as illustrated in Figure 4.4 on page 156.

my– $ by¿ $ ky ! A1By– $ 10y¿ $ 25y ! 25e"5t $ 10 A"5e"5tB $ 25e"5t ! 0 .y¿ AtB ! "5e"5t, y– AtB ! 25e"5t b ! 10Fext ! 0, m ! 1, k ! 25

y AtB ! e"5t ! 0 . $ 25e"3t cos 4t

! "7e"3t cos 4t $ 24e"3t sin 4t $ 6 A"3e"3t cos 4t " 4e"3t sin tBmy– $ by¿ $ ky ! A1By– $ 6y¿ $ 25y ! "7e"3t cos 4t $ 24e"3t sin 4t ,

y– AtB ! 9e"3t cos 4t $ 12e"3t sin 4t $ 12e"3t sin 4t " 16e"3t cos 4t y¿ AtB ! "3e"3t cos 4t " 4e"3t sin 4t , b ! 6Fext ! 0, m ! 1, k ! 25

y AtB ! e"3t cos 4t Ab ! 10B

Ab ! 6B

Section 4.1 Introduction: The Mass–Spring Oscillator 155

Example 2

Solution

Example 3

Solution

(a) (b)

Figure 4.3 (a) Low damping and (b) high damping

Common examples of systems vibrating in synchronization with their drivers are sound system speakers, cyclists bicycling over railroad tracks, electronic amplifier circuits, and ocean tides (driven by the periodic pull of the moon). However, there is more to the story than is revealed above. Systems can be enormously sensitive to the particular frequency # at which they are driven. Thus, accurately tuned musical notes can shatter fine crystal, wind-induced vibrations at the right (wrong?) frequency can bring down a bridge, and a dripping faucet can cause inordinate headaches. These “resonance” responses (for which the responses have maxi- mum amplitudes) may be quite destructive, and structural engineers have to be very careful to ensure that their products will not resonate with any of the vibrations likely to occur in the operating environment. Radio engineers, on the other hand, do want their receivers to resonate selectively to the desired broadcasting channel.

The calculation of these forced solutions is the subject of Sections 4.4 and 4.5. The next example illustrates some of the features of synchronous response and resonance.

Find the synchronous response of the mass–spring oscillator with m ! 1, b ! 1, k ! 25 to the force sin &t.

We seek solutions of the differential equation

(5)

that are sinusoids in sync with sin &t; so let’s try the form ! Acos &t $ Bsin &t. Since

we can simply insert these forms into equation (5), collect terms, and match coefficients to obtain a solution:

A"&2 $ 25BA $ &B ! 0 ."&A $ A"&2 $ 25BB ! 1 ! 3"&2B " &A $ 25B 4 sin &t $ 3"&2A $ &B $ 25A 4 cos &t , $25[A cos &t $ B sin &t] ! "&

2A cos &t " &2B sin &t$ 3"&A sin &t $ &B cos &t 4 sin &t ! y– $ y¿ $ 25y y– ! "&2A cos &t " &2B sin &t , y¿ ! "&A sin &t $ &B cos &t ,

y AtBy– $ y¿ $ 25y ! sin &t

156 Chapter 4 Linear Second-Order Equations

Fext

(a) (b)

Figure 4.4 (a) Driving force and (b) response

Example 4

Solution

Section 4.1 Introduction: The Mass–Spring Oscillator 157

–0.2

–0.15

–0.1

–0.05

0 5 10 15 20 Ω

0.1

0.15

0.05

Figure 4.5 Vibration amplitudes around resonance

We find

Figure 4.5 displays A and B as functions of the driving frequency &. A resonance clearly occurs around . ◆

In most of this chapter, we are going to restrict our attention to differential equations of the form

(6)

where [or , or , etc.] is the unknown function that we seek; a, b, and c are constants; and [or ] is a known function. The proper nomenclature for (6) is the linear, second- order ordinary differential equation with constant coefficients. In Sections 4.7 and 4.8, we will generalize our focus to equations with nonconstant coefficients, as well as to nonlinear equa- tions. However, (6) is an excellent starting point because we are able to obtain explicit solu- tions and observe, in concrete form, the theoretical properties that are predicted for more gen- eral equations. For motivation of the mathematical procedures and theory for solving (6), we will consistently compare it with the mass–spring paradigm:3 inertia 4 ' y– $ 3damping 4 ' y¿ $ 3 stiffness 4 ' y ! Fext .

f AxBf AtB x AtBy AxBy AtB ay– $ by¿ $ cy ! f AtB ,

& ! 5

A ! "&

&2 $ A&2 " 25B2 , B ! "&2 $ 25&2 $ A&2 " 25B2 .

1. Verify that for and , equation (3) has a solution of the form

where .

2. If , equation (3) becomes

For this equation, verify the following: my– $ by¿ $ ky ! 0 .

Fext AtB ! 0 v ! 2k/my AtB ! cos vt, Fext AtB ! 0b ! 0 (a) If y(t) is a solution, so is cy(t), for any constant c.

(b) If and are solutions, so is their sum .

3. Show that if , and , then equation (3) has the “critically damped” solu- tions and . What is the limit of these solutions as ?t S q

y2 AtB ! te"3ty1 AtB ! e"3t b ! 6Fext AtB ! 0, m ! 1, k ! 9y1 AtB $ y2 AtB

y2 AtBy1 AtB 4.1 EXERCISES

We begin our study of the linear second-order constant-coefficient differential equation

(1)

with the special case where the function f (t) is zero:

(2)

This case arises when we consider mass–spring oscillators vibrating freely—that is, without external forces applied. Equation (2) is called the homogeneous form of equation (1); is the “nonhomogeneity” in (1). (This nomenclature is not related to the way we used the term for first-order equations in Section 2.6.)

A look at equation (2) tells us that a solution of (2) must have the property that its second derivative is expressible as a linear combination of its first and zeroth derivatives.† This sug- gests that we try to find a solution of the form , since derivatives of are just constants times . If we substitute into (2), we obtain

ert Aar2 $ br $ cB ! 0 .ar2 ert $ brert $ cert ! 0 , y ! ertert

erty ! ert

f AtB ay– $ by¿ $ cy ! 0 .

ay– $ by¿ $ cy ! f AtB Aa ( 0B

4. Verify that y ! sin 3t $ 2 cos 3t is a solution to the initial value problem

Find the maximum of for . 5. Verify that the exponentially damped sinusoid

is a solution to equation (3) if , and . What is the

limit of this solution as ? 6. An external force F(t) ! 2 cos 2t is applied to a

mass–spring system with m ! 1, b ! 0, and k ! 4, which is initially at rest; i.e., . Verify that gives the motion of this spring. What will eventually (as t increases) happen to the spring?

In Problems 7–9, find a synchronous solution of the form A cos &t $ B sin &t to the given forced oscillator equa- tion using the method of Example 4 to solve for A and B.

7. 8. 9.

10. Undamped oscillators that are driven at resonance

y– $ 2y¿ $ 4y ! 6 cos 2t $ 8 sin 2t, & ! 2 y– $ 2y¿ $ 5y ! "50 sin 5t, & ! 5 y– $ 2y¿ $ 4y ! 5 sin 3t, & ! 3

y AtB ! 12 t sin 2t y A0B ! 0, y¿ A0B ! 0 t S q

k ! 12Fext AtB ! 0, m ! 1, b ! 6y AtB ! e"3t sin A23 tB "q 6 t 6 qƒ y AtB ƒy A0B ! 2 , y¿ A0B ! 3 .2y– $ 18y ! 0 ;

158 Chapter 4 Linear Second-Order Equations

have unusual (and nonphysical) solutions.

(a) To investigate this, find the synchronous solu- tion A cos &t $ B sin &t to the generic forced oscillator equation

(7)

(b) Sketch graphs of the coefficients A and B, as functions of &, for m ! 1, b ! 0.1, and k ! 25.

(c) Now set b ! 0 in your formulas for A and B and resketch the graphs in part (b), with m ! 1, and k ! 25. What happens at & ! 5? Notice that the amplitudes of the synchronous solutions grow without bound as & approaches 5.

(d) Show directly, by substituting the form A cos &t $ B sin &t into equation (7), that when b ! 0 there are no synchronous solutions if .

(e) Verify that solves equation (7) when b ! 0 and . Notice that this nonsynchronous solution grows in time, without bound.

Clearly one cannot neglect damping in analyz- ing an oscillator forced at resonance, because otherwise the solutions, as shown in part (e), are nonphysical. This behavior will be studied later in this chapter.

& ! 2k/mA2m&B"1t sin &t & ! 2k/m

my– $ by¿ $ ky ! cos &t .

HOMOGENEOUS LINEAR EQUATIONS: THE GENERAL SOLUTION4.2

†The zeroth derivative of a function is the function itself.

Section 4.2 Homogeneous Linear Equations: The General Solution 159

Example 1

Solution

Because is never zero, we can divide by it to obtain

(3)

Consequently, is a solution to (2) if and only if r satisfies equation (3). Equation (3) is called the auxiliary equation (also known as the characteristic equation) associated with the homogeneous equation (2).

Now the auxiliary equation is just a quadratic, and its roots are

When the discriminant, , is positive, the roots and are real and distinct. If , the roots are real and equal. And when , the roots are complex

conjugate numbers. We consider the first two cases in this section; the complex case is deferred to Section 4.3.

Find a pair of solutions to

(4)

The auxiliary equation associated with (4) is

which has the roots . Thus, and are solutions. ◆

Notice that the identically zero function, , is always a solution to (2). Further- more, when we have a pair of solutions and to this equation, as in Example 1, we can construct an infinite number of other solutions by forming linear combinations:

(5)

for any choice of the constants and . The fact that (5) is a solution to (2) can be seen by direct substitution and rearrangement:

The two “degrees of freedom” and in the combination (5) suggest that solutions to the differential equation (2) can be found meeting additional conditions, such as the initial con- ditions for the first-order equations in Chapter 1. But the presence of and leads one to anticipate that two such conditions, rather than just one, can be imposed. This is consistent with the mass–spring interpretation of equation (2), since predicting the motion of a mechani- cal system requires knowledge not only of the forces but also of the initial position y(0) and velocity of the mass. A typical initial value problem for these second-order equations is given in the following example.

y¿(0)

c2c1

! 0 $ 0 . ! c1 Aay–1 $ by¿1 $ cy1B $ c2 Aay–2 $ by¿2 $ cy2B ! a Ac1y–1 $ c2y–2B $ b Ac1y¿1 $ c2y¿2B $ c Ac1y1 $ c2y2B

ay– $ by¿ $ cy ! a Ac1y1 $ c2y2B– $ b Ac1y1 $ c2y2B ¿ $ c Ac1y1 $ c2y2B c2c1

y AtB ! c1 y1 AtB $ c2 y2 AtB y2 AtBy1 AtB y AtB " 0

e"6tetr1 ! 1, r2 ! "6

r 2 $ 5r " 6 ! Ar " 1B Ar $ 6B ! 0 , y– $ 5y¿ " 6y ! 0 .

b2 " 4ac 6 0b2 " 4ac ! 0 r2r1b

2 " 4ac

r1 ! "b $ 2b2 " 4ac

2a and r2 !

"b " 2b2 " 4ac 2a

y ! ert

ar 2 $ br $ c ! 0 .

ert

Solve the initial value problem

(6)

We will first find a pair of solutions as in the previous example. Then we will adjust the con- stants and in (5) to obtain a solution that matches the initial conditions on and . The auxiliary equation is

Using the quadratic formula, we find that the roots of this equation are

Consequently, the given differential equation has solutions of the form

(7)

To find the specific solution that satisfies the initial conditions given in (6), we first differenti- ate y as given in (7), then plug y and into the initial conditions of (6). This gives

Solving this system yields and . Thus,

is the desired solution. ◆

To gain more insight into the significance of the two-parameter solution form (5), we need to look at some of the properties of the second-order equation (2). First of all, there is an existence-and-uniqueness theorem for solutions to (2); it is somewhat like the corresponding Theorem 1 in Section 1.2 for first-order equations but updated to reflect the fact that two initial conditions are appropriate for second-order equations. As motivation for the theorem, suppose the differential equation (2) were really easy, with b ! 0 and c ! 0. Then would merely say that the graph of is simply a straight line, so it is uniquely determined by spec- ifying a point on the line,

(8)

and the slope of the line,

(9)

Theorem 1 states that conditions (8) and (9) suffice to determine the solution uniquely for the more general equation (2).

y¿ At0B ! Y1 . y At0B ! Y0 ,

y AtB y– ! 0

y AtB ! " 22 4

e A"1$22 B t $ 22 4

e A"1"22 B t c2 ! 22/4c1 ! "22/4"1 ! A"1 $ 22 B c1 $ A"1 " 22 B c2 .

0 ! c1 $ c2 ,

y¿ A0B ! A"1 $ 22 B c1e0 $ A"1 " 22 B c2e0 , y A0B ! c1e0 $ c2e0 , y¿

y AtB ! c1e A"1$22 B t $ c2e A"1"22 B t . r1 ! "1 $ 22 and r2 ! "1 " 22 . r2 $ 2r " 1 ! 0 .

y¿ A0By A0Bc2c1 y A0B ! 0 , y¿ A0B ! "1 .y– $ 2y¿ " y ! 0 ;

160 Chapter 4 Linear Second-Order Equations

Solution

Example 2

Note in particular that if a solution and its derivative vanish simultaneously at a point t0 then must be the identically zero solution.

In this section and the next, we will construct explicit solutions to (10), so the question of existence of a solution is not really an issue. It is extremely valuable to know, however, that the solution is unique. The proof of uniqueness is rather different from anything else in this chapter, so we defer it to Chapter 13.†

Now we want to use this theorem to show that, given two solutions and to equa- tion (2), we can always find values of and so that meets specified initial conditions in (10) and therefore is the (unique) solution to the initial value problem. But we need to be a little more precise; if, for example, is simply the identically zero solution, then actually has only one constant and cannot be expected to sat- isfy two conditions. Furthermore, if is simply a constant multiple of —say,

—then again actually has only one constant. The condition we need is linear independence.

c1y1 AtB $ c2y2 AtB ! Ac1 $ kc2By1 AtB ! Cy1 AtBy2 AtB ! ky1 AtB y1 AtBy2 AtB c1y1 AtB $ c2y2 AtB ! c1y1 AtB y2 AtB

c1y1 AtB $ c2y2 AtBc2c1 y2 AtBy1 AtB y AtB(i.e., Y0 ! Y1 ! 0), y AtB

Section 4.2 Homogeneous Linear Equations: The General Solution 161

Existence and Uniqueness: Homogeneous Case

Theorem 1. For any real numbers , and , there exists a unique solu- tion to the initial value problem

(10)

The solution is valid for all t in .A"q, $q By At0B ! Y0 , y¿ At0B ! Y1 .ay– $ by¿ $ cy ! 0 ; Y1a((0), b, c, t0, Y0

Representation of Solutions to Initial Value Problem

Theorem 2. If and are any two solutions to the differential equation (2) that are linearly independent on , then unique constants and can always be found so that satisfies the initial value problem (10) on .("q, q)c1y1 AtB $ c2y2 AtB c2c1("q, q)

y2 AtBy1 AtB

Linear Independence of Two Functions

Definition 1. A pair of functions and is said to be linearly independent on the interval I if and only if neither of them is a constant multiple of the other on all of I.†† We say that and are linearly dependent on I if one of them is a constant multiple of the other on all of I.

y2y1

y2 AtBy1 AtB

†All references to Chapters 11–13 refer to the expanded text Fundamentals of Differential Equations and Boundary Value Problems, 6th ed. ††This definition will be generalized to three or more functions in Problem 35 and Chapter 6.

The proof of Theorem 2 will be easy once we establish the following technical lemma.

Proof of Lemma 1. Case 1. If , then let k equal and consider the solution to (2) given by . It satisfies the same “initial conditions” at t ! t as does

; ,

where the last equality follows from (11). By uniqueness, must be the same function as on I.

Case 2. If but , then (11) implies . Let . Then the solution to (2) given by (again) satisfies the same “initial conditions” at t ! t as does :

By uniqueness, then, on I.

Case 3. If , then is a solution to the differential equation (2) satis- fying the initial conditions ; but is the unique solution to this initial value problem. Thus, and is a constant multiple of . ◆

Proof of Theorem 2. We already know that is a solution to (2); we must show that and can be chosen so that

and

But simple algebra shows these equations have the solution†

as long as the denominator is nonzero, and the technical lemma assures us that this condition is met. ◆

Now we can honestly say that if and are linearly independent solutions to (2) on , then (5) is a general solution, since any solution of (2) can be expressed in

this form; simply pick and so that matches the value and the derivative of at any point. By uniqueness, and have to be the same function. See Figure 4.6 on page 163.

How do we find a general solution for the differential equation (2)? We already know the

ygc1y1 $ c2y2 ygc1y1 $ c2y2c2c1

yg AtBA"q, $q B y2y1 c1 !

Y0y¿2 At0B " Y1y2 At0B y1 At0By¿2 At0B " y¿1 At0By2 At0B and c2 ! Y1y1 At0B " Y0y¿1 At0By1 At0By¿2 At0B " y¿1 At0By2 At0B

y¿ At0B ! c1y¿1 At0B $ c2y¿2 At0B ! Y1 . y At0B ! c1y1 At0B $ c2y2 At0B ! Y0c2c1

y AtB ! c1y1 AtB $ c2y2 AtBy2 AtB 43y1 AtB " 0 y AtB " 0y1 AtB ! y¿1 AtB ! 0y1 AtBy1 AtB ! y¿1 AtB ! 0

y2 AtB ! ky1 AtB y¿ AtB ! y¿2 AtB

y¿1 AtB y¿1 AtB ! y¿2 AtB .y AtB ! y¿2 AtBy¿1 AtB y1 AtB ! 0 ! y2 AtB ; y2 AtB y AtB ! ky1 AtB

k ! y¿2 AtB /y¿1 AtBy2 AtB ! 0y¿1 AtB ( 0y1 AtB ! 0ky1 AtB y2 AtB

y¿ AtB ! y2 AtB y1 AtB y¿1 AtB ! y¿2 AtBy AtB ! y2 AtBy1 AtB y1 AtB ! y2 AtB

y2 AtB y AtB ! ky1 AtB y2 AtB /y1 AtBy1 AtB ( 0

162 Chapter 4 Linear Second-Order Equations

†To solve for , for example, multiply the first equation by and the second by and subtract.y2 At0By¿2 At0Bc1

A Condition for Linear Dependence of Solutions

Lemma 1. For any real numbers , if and are any two solu- tions to the differential equation (2) on and if the equality

(11)

holds at any point t, then and are linearly dependent on . (The expres- sion on the left-hand side of (11) is called the Wronskian of and at the point t; see Problem 34.)

y2y1 ("q, q)y2y1

y1 AtBy2¿ AtB " y¿1 AtBy2 AtB ! 0 ("q, q) y2 AtBy1 AtBa A( 0B, b, and c

Section 4.2 Homogeneous Linear Equations: The General Solution 163

t t0

y(t0)

Slope y)(t0)

(Can’t both be solutions)

Figure 4.6 determine a unique solutiony At0B, y¿ At0B answer if the roots of the auxiliary equation (3) are real and distinct because clearly is not a constant multiple of if .r1 ( r2y2 AtB ! er2t y1 AtB ! er1t

DISTINCT REAL ROOTS

If the auxiliary equation (3) has distinct real roots and , then both and are solutions to (2) and is a general solution.y AtB ! c1er1t $ c2er2ty2 AtB ! er2t y1 AtB ! er1tr2r1

REPEATED ROOT

If the auxiliary equation (3) has a repeated root r, then both and are solutions to (2), and is a general solution.y AtB ! c1ert $ c2tert y2 AtB ! terty1 AtB ! ert

We illustrate this result before giving its proof.

Find a solution to the initial value problem

(12)

The auxiliary equation for (12) is

Because r ! "2 is a double root, the rule says that (12) has solutions and . Let’s confirm that is a solution:

y–2 $ 4y¿2 $ 4y2 ! "4e"2t $ 4te"2t $ 4(e"2t " 2te"2t) $ 4te"2t ! 0 . y–2 AtB ! "2e"2t " 2e"2t $ 4te"2t ! "4e"2t $ 4te"2t , y¿2 AtB ! e"2t " 2te"2t , y2 AtB ! te"2t ,y2 AtB

y2 ! te "2ty1 ! e

"2t

r2 $ 4r $ 4 ! Ar $ 2B2 ! 0 . y A0B ! 1 , y¿ A0B ! 3 .y– $ 4y¿ $ 4y ! 0 ;Example 3

Solution

When the roots of the auxiliary equation are equal, we only get one nontrivial solution, . To satisfy two initial conditions, y(t0) and y)(t0), then, we will need a second, linearly

independent solution. The following rule is the key to finding a second solution. y1 ! e

Further observe that and are linearly independent since neither is a constant multiple of the other on . Finally, we insert the general solution into the initial conditions,

and solve to find . Thus is the desired solution. ◆

Why is it that is a solution to the differential equation (2) when r is a double root (and not otherwise)? In later chapters we will see a theoretical justification of this rule in very general circumstances; for present purposes, though, simply note what happens if we substitute into the differential equation (2):

Now if r is a root of the auxiliary equation (3), the expression in the second brackets is zero. However, if r is a double root, the expression in the first brackets is zero also:

(13)

hence, 2ar $ b ! 0 for a double root. In such a case, then, is a solution. The method we have described for solving homogeneous linear second-order equations

with constant coefficients applies to any order (even first-order) homogeneous linear equations with constant coefficients. We give a detailed treatment of such higher-order equations in Chapter 6. For now, we will be content to illustrate the method by means of an example. We remark briefly that a homogeneous linear nth-order equation has a general solution of the form

where the individual solutions are “linearly independent.” By this we mean that no is expressible as a linear combination of the others; see Problem 35.

Find a general solution to

(14)

If we try to find solutions of the form , then, as with second-order equations, we are led to finding roots of the auxiliary equation

(15) r3 $ 3r2 " r " 3 ! 0 .

y ! ert

y‡ $ 3y– " y¿ " 3y ! 0 .

yiyi AtB y AtB ! c1y1 AtB $ c2y2 AtB $ p $ cnyn AtB

r ! "b * 2b2 " 4ac

2a !

"b * A0B 2a

;

ay–2 $ by¿2 $ cy2 ! 32ar $ b 4 ert $ 3ar2 $ br $ c 4 tert .y–2 AtB ! rert $ rert $ r2tert ! 2rert $ r2tert , y¿2 AtB ! ert $ rtert ,y2 AtB ! tert , y2

y2 AtB ! tert y ! e"2t $ 5te"2tc1 ! 1, c2 ! 5

y¿ A0B ! "2c1e0 $ c2e0 " 2c2 A0Be0 ! 3 ,y A0B ! c1e0 $ c2 A0Be0 ! 1 , y AtB ! c1e"2t $ c2te"2tA"q, q B te"2te"2t

164 Chapter 4 Linear Second-Order Equations

Example 4

Solution

Section 4.2 Homogeneous Linear Equations: The General Solution 165

We observe that r ! 1 is a root of the above equation, and dividing the polynomial on the left-hand side of (15) by r " 1 leads to the factorization

Hence, the roots of the auxiliary equation are 1, "1, and "3, and so three solutions of (14) are , and . The linear independence of these three exponential functions is proved in

Problem 40. A general solution to (14) is then

(16) . ◆

So far we have seen only exponential solutions to the linear second-order constant coeffi- cient equation. You may wonder where the vibratory solutions that govern mass–spring oscilla- tors are. In the next section, it will be seen that they arise when the solutions to the auxiliary equation are complex.

y AtB ! c1et $ c2e"t $ c3e"3t e"3tet, e"t

Ar " 1B Ar2 $ 4r $ 3B ! Ar " 1B Ar $ 1B Ar $ 3B ! 0 .

In Problems 1–12, find a general solution to the given differential equation.

1. 2.

3. 4.

5. 6.

7. 8.

9. 10.

11.

12.

In Problems 13–20, solve the given initial value problem. 13.

14.

15.

16.

17.

18.

19.

20.

21. First-Order Constant-Coefficient Equations. (a) Substituting , find the auxiliary equation

for the first-order linear equation

where a and b are constants with . (b) Use the result of part (a) to find the general

solution.

a ( 0 ay¿ $ by ! 0 ,

y ! ert

y– " 4y¿ $ 4y ! 0 ; y A1B ! 1 , y¿ A1B ! 1y– $ 2y¿ $ y ! 0 ; y A0B ! 1 , y¿ A0B ! "3 y– " 6y¿ $ 9y ! 0 ; y A0B ! 2 , y¿ A0B ! 25/3z– " 2z¿ " 2z ! 0 ; z A0B ! 0 , z¿ A0B ! 3 y– " 4y¿ $ 3y ! 0 ; y A0B ! 1 , y¿ A0B ! 1/3y– " 4y¿ " 5y ! 0 ; y A"1B ! 3 , y¿ A"1B ! 9 y– $ y¿ ! 0 ; y A0B ! 2 , y¿ A0B ! 1y– $ 2y¿ " 8y ! 0 ; y(0) ! 3 , y¿ A0B ! "12 3y– $ 11y¿ " 7y ! 0 4w– $ 20w¿ $ 25w ! 0

y– " y¿ " 11y ! 04y– " 4y¿ $ y ! 0 z– $ z¿ " z ! 06y– $ y¿ " 2y ! 0 y– $ 8y¿ $ 16y ! 0y– " 5y¿ $ 6y ! 0 y– $ 5y¿ $ 6y ! 0y– " y¿ " 2y ! 0 2y– $ 7y¿ " 4y ! 0y– $ 6y¿ $ 9y ! 0

In Problems 22–25, use the method described in Problem 21 to find a general solution to the given equation. 22. 23. 24. 25.

26. Boundary Value Problems. When the values of a solution to a differential equation are specified at two different points, these conditions are called boundary conditions. (In contrast, initial conditions specify the values of a function and its derivative at the same point.) The purpose of this exercise is to show that for boundary value problems there is no existence–uniqueness theorem that is analogous to Theorem 1. Given that every solution to (17) is of the form

where and are arbitrary constants, show that

(a) There is a unique solution to (17) that satisfies the boundary conditions and .

(b) There is no solution to (17) that satisfies y(0) ! 2 and ! 0.

In Problems 27–32, use Definition 1 to determine whether the functions and are linearly dependent on the interval . 27. 28. y1 AtB ! e3t , y2(tB ! e"4ty1 AtB ! cos t sin t , y2 AtB ! sin 2t

A0, 1B y2y1 y ApB ! "2.y A0B ! 2

y ApB y Ap/2B ! 0y A0B ! 2

c2c1

y(t) ! c1 cos t $ c2 sin t ,

y– $ y ! 0

6w¿ " 13w ! 03z¿ $ 11z ! 0 5y¿ $ 4y ! 03y¿ " 7y ! 0

4.2 EXERCISES

29. 30. 31. 32.

33. Explain why two functions are linearly dependent on an interval I if and only if there exist constants and , not both zero, such that

for all t in I . 34. Wronskian. For any two differentiable functions

and , the function

(18)

is called the Wronskian† of and . This function plays a crucial role in proof of Theorem 2. (a) Show that can be conveniently

expressed as the determinant

(b) Let be a pair of solutions to the ho- mogeneous equation (with

) on an open interval I. Prove that and are linearly independent on I if and only if

their Wronskian is never zero on I. [Hint: This is just a reformulation of Lemma 1.]

(c) Show that if and are any two differen- tiable functions that are linearly dependent on I, then their Wronskian is identically zero on I.

35. Linear Dependence of Three Functions. Three functions , and are said to be linearly dependent on an interval I if, on I, at least one of these functions is a linear combination of the remaining two [e.g., if ]. Equivalently (compare Problem 33), and are linearly dependent on I if there exist constants

and , not all zero, such that for all t in I.

Otherwise, we say that these functions are linearly independent on I.

For each of the following, determine whether the given three functions are linearly dependent or lin- early independent on : (a)

(b) (c)

(d) y1 AtB ! et , y2 AtB ! e"t , y3 AtB ! cosh t .y1 AtB ! et , y2 AtB ! tet , y3 AtB ! t2et . y1 AtB ! 1 , y2 AtB ! t , y3 AtB ! t2 .A"q, q B

C1y1 AtB $ C2y2 AtB $ C3y3 AtB ! 0C3C1, C2, y3y1, y2,

y1 AtB ! c1y2 AtB $ c2y3 AtB y3 AtBy1 AtB, y2 AtB y2 AtBy1 AtB

y2 AtB y1 AtBa ( 0 ay– $ by¿ $ cy ! 0

y1(t), y2(t)

W 3 y1, y2 4 AtB ! ` y1 AtB y2 AtBy¿1 AtB y¿2 AtB ` .2 ' 2 W 3 y1, y2 4 y2y1

W 3 y1, y2 4 AtB ! y1 AtBy¿2 AtB " y¿1 AtBy2 AtBy2y1 c1y1 AtB $ c2y2 AtB ! 0c2

y1 AtB " 0 , y2 AtB ! ety1 AtB ! tan 2t " sec2t , y2 AtB " 3 y1 AtB ! t2 cos Aln tB , y2(tB ! t2 sin Aln tBy1 AtB ! te2t , y2 AtB ! e2t

166 Chapter 4 Linear Second-Order Equations

36. Using the definition in Problem 35, prove that if , and are distinct real numbers, then the func-

tions , and are linearly independent on . [Hint: Assume to the contrary that, say,

for all t. Divide by to get and then differentiate to

deduce that and are linearly depen- dent, which is a contradiction. (Why?)]

In Problems 37–41, find three linearly independent solu- tions (see Problem 35) of the given third-order differen- tial equation and write a general solution as an arbitrary linear combination of these. 37. 38. 39. 40. 41.

42. (True or False): If f1, f2, f3 are three functions defined on that are pairwise linearly independent on , then f1, f2, f3 form a linearly indepen- dent set on . Justify your answer.

43. Solve the initial value problem:

44. Solve the initial value problem:

45. By using Newton’s method or some other numerical procedure to approximate the roots of the auxiliary equation, find general solutions to the following equations: (a) (b) (c)

46. One way to define hyperbolic functions is by means of differential equations. Consider the equation

The hyperbolic cosine, cosh t, is defined as the solution of this equation subject to the initial values: and The hyperbolic sine, sinh t, is defined as the solution of this equation subject to the initial values: and

(a) Solve these initial value problems to derive explicit formulas for cosh t, and sinh t. Also

y¿(0) ! 1. y(0) ! 0

y¿(0) ! 0.y (0) ! 1

y– " y ! 0.

$ 4y¿ " 12y ! 0 .yv " 3y iv " 5y‡ $ 15y– yiv " 5y– $ 5y ! 0 . 3y‡ $ 18y– $ 13y¿ " 19y ! 0 .

y A0B ! 2 , y¿ A0B ! 3 , y– A0B ! 5 .y‡ " 2y– " y¿ $ 2y ! 0 ; y– A0B ! "1 .y¿ A0B ! 3 , y A0B ! 2 ,y‡ " y¿ ! 0 ;

A"q, q BA"q, q B A"q, q B

y‡ $ 3y– " 4y¿ " 12y ! 0 y‡ " 7y– $ 7y¿ $ 15y ! 0 z‡ $ 2z– " 4z¿ " 8z ! 0 y‡ " 6y– " y¿ $ 6y ! 0 y‡ $ y– " 6y¿ $ 4y ! 0

e(r3"r2)te(r1"r2)t e(r1"r2)t ! c1 $ c2e

(r3"r2)t er2ter1t ! c1e

r2t $ c2e r3t

A"q, q B er3ter1t, er2t r3r1, r2

y1 AtB ! "3 , y2 AtB ! 5 sin2 t , y3 AtB ! cos2 t . †Historical Footnote: The Wronskian was named after the Polish mathematician H. Wronski (1778–1863).

The simple harmonic equation , so called because of its relation to the fundamental vibration of a musical tone, has as solutions and . Notice, however, that the auxiliary equation associated with the harmonic equation is , which has imaginary roots , where i denotes .† In the previous section, we expressed the solutions to a linear second-order equation with constant coefficients in terms of exponential functions. It would appear, then, that one might be able to attribute a meaning to the forms and and that these “functions” should be related to cos t and sin t. This matchup is accom- plished by Euler’s formula, which is discussed in this section.

When , the roots of the auxiliary equation

(1)

associated with the homogeneous equation

(2)

are the complex conjugate numbers

where a, b are the real numbers

(3)

As in the previous section, we would like to assert that the functions and are solutions to the equation (2). This is in fact the case, but before we can proceed, we need to address some fundamental questions. For example, if is a complex number, what do we mean by the expression ? If we assume that the law of exponents applies to complex numbers, then

(4)

We now need only clarify the meaning of For this purpose, let’s assume that the Maclaurin series for is the same for complex

numbers z as it is for real numbers. Observing that , then for real we have

! a1 " u2 2!

$ u4

4! $ p b $ i au " u3

3! $ u5

5! $ p b . ! 1 $ iu "

2! "

iu3

3! $ u4

4! $

iu5

5! $ p

eiu ! 1 $ AiuB $ AiuB2 2!

$ p $ AiuBn n!

$ p

ui2 ! "1 ez

eibt.

e Aa$ibB t ! eat$ibt ! eateibt . eAa$ibB t r1 ! a $ ib

er2ter1t

a ! " b 2a

and b ! 24ac " b2 2a

r1 ! a $ ib and r2 ! a " ib Ai ! 2"1 B , ay– $ by¿ $ cy ! 0

ar2 $ br $ c ! 0

b2 " 4ac 6 0

e"it eit

2"1r ! *i r2 $ 1 ! 0y2 AtB ! sin ty1 AtB ! cos t y– $ y ! 0

Section 4.3 Auxiliary Equations with Complex Roots 167

AUXILIARY EQUATIONS WITH COMPLEX ROOTS4.3

†Electrical engineers frequently use the symbol j to denote .2"1

show that cosh t ! sinh t and sinh t !

cosh t. (b) Prove that a general solution of the equation

is given by . (c) Suppose a, b, and c are given constants for which

has two distinct real roots. If thear 2 $ br $ c ! 0

y ! c1 cosh t $ c2 sinh ty– " y ! 0

d dt

two roots are expressed in the form and , show that a general solution of the equation

is .

(d) Use the result of part (c) to solve the initial value problem:

. "17/2 y– $ y¿ " 6y ! 0, y(0) ! 2, y¿(0) !

c2e at sinh(bt)

y ! c1e at cosh(bt) $ay– $ by¿ $ cy ! 0

a $ b a " b

Now recall the Maclaurin series for cos

Recognizing these expansions in the proposed series for we make the identification

(5)

which is known as Euler’s formula.†

When Euler’s formula (with ) is used in equation (4), we find

(6)

which expresses the complex function in terms of familiar real functions. Having made sense out of we can now show (see Problem 30) that

(7)

and, with the choices of and as given in (3), the complex function is indeed a solu- tion to equation (2), as is , and a general solution is given by

(8)

Example 1 shows that in general the constants and that go into (8), for a specific ini- tial value problem, are complex.

Use the general solution (8) to solve the initial value problem

The auxiliary equation is , which has roots

Hence, with , a general solution is given by

For initial conditions we have

As a result, , , and , or sim- ply . ◆2e"t sin t

y AtB ! "ie"t A cos t $ i sin tB $ ie"t A cos t " i sin tBc2 ! ic1 ! "i ! 2 . ! A"1 $ iBc1 $ A"1 " iBc2 "c2e0 A cos 0 " i sin 0B $ c2e0 A" sin 0 " i cos 0B

y¿ A0B ! "c1e0 A cos 0 $ i sin 0B $ c1e0 A" sin 0 $ i cos 0B y A0B ! c1e0 A cos 0 $ i sin 0B $ c2e0 A cos 0 " i sin 0B ! c1 $ c2 ! 0 , y AtB ! c1e"t A cos t $ i sin tB $ c2e"t A cos t " i sin tB .a ! "1, b ! 1 r !

"2 * 24 " 8 2

! "1 * i .

r2 $ 2r $ 2 ! 0

y– $ 2y¿ $ 2y ! 0 ; y A0B ! 0, y¿ A0B ! 2 c2c1

! c1e at A cos bt $ i sin btB $ c2eat A cos bt " i sin btBy AtB ! c1e Aa$ ibB t $ c2e Aa" ibB t

e Aa" ibB t eAa$ibB tba d dt

e Aa$ibB t ! Aa $ ibBe Aa$ibB t ,e Aa$ibB t, eAa$ibB t

e Aa$ibB t ! eat Acos bt $ i sin btB ,u ! bt eiU $ cos U " i sin U ,

eiu,

sin u ! u " u3

3! $ u5

5! $ p .

cos u ! 1 " u2

2! $ u4

4! $ p ,

u and sin u:

168 Chapter 4 Linear Second-Order Equations

†Historical Footnote: This formula first appeared in Leonhard Euler’s monumental two-volume Introduction in Analysin Infinitorum (1748).

Example 1

Solution

The final form of the answer to Example 1 suggests that we should seek an alternative pair of solutions to the differential equation (2) that don’t require complex arithmetic, and we now turn to that task.

In general, if is a complex-valued function of the real variable t, we can write ! $ , where and are real-valued functions. The derivatives of are then given by

With the following lemma, we show that the complex-valued solution gives rise to two linearly independent real-valued solutions.

eAa$ibB t dz dt

! du dt

$ i dy dt

, d 2z

dt2 !

d2u dt2

$ i d2y dt2

z AtBy AtBu AtBiy AtBu AtB z AtBz AtB

Section 4.3 Auxiliary Equations with Complex Roots 169

Real Solutions Derived from Complex Solutions

Lemma 2. Let be a solution to equation (2), where a, b, and c are real numbers. Then, the real part and the imaginary part are real-valued solutions of (2).†

y AtBu AtBz AtB ! u AtB $ iy AtB

Proof. By assumption, , and hence

But a complex number is zero if and only if its real and imaginary parts are both zero. Thus, we must have

and

which means that both and are real-valued solutions of (2). ◆

When we apply Lemma 2 to the solution

we obtain the following.

e Aa$ibB t ! eat cos bt $ ieat sin bt ,

y AtBu AtB ay– $ by¿ $ cy ! 0 ,au– $ bu¿ $ cu ! 0 Aau– $ bu¿ $ cuB $ i Aay– $ by¿ $ cyB ! 0 .a Au– $ iy– B $ b Au¿ $ iy¿ B $ c Au $ iyB ! 0 ,

az– $ bz¿ $ cz ! 0

†It will be clear from the proof that this property holds for any linear homogeneous differential equation having real-valued coefficients.

COMPLEX CONJUGATE ROOTS

If the auxiliary equation has complex conjugate roots then two linearly independent solutions to (2) are

and

and a general solution is

(9)

where c1 and c2 are arbitrary constants.

y AtB $ c1eAt cos Bt " c2eAt sin Bt , eatsin bt ,eatcos bt

a * ib,

170 Chapter 4 Linear Second-Order Equations

†For a detailed treatment of these topics see, for example, Fundamentals of Complex Analysis, 3rd ed., by E. B. Saff and A. D. Snider (Prentice Hall, Upper Saddle River, New Jersey, 2003).

In the preceding discussion, we glossed over some important details concerning com- plex numbers and complex-valued functions. In particular, further analysis is required to jus- tify the use of the law of exponents, Euler’s formula, and even the fact that the derivative of

is when r is a complex constant.† If you feel uneasy about our conclusions, we encourage you to substitute the expression in (9) into equation (2) to verify that it is, indeed, a solution.

You may also be wondering what would have happened if we had worked with the function instead of We leave it as an exercise to verify that gives rise to the same general solution (9). Indeed, the sum of these two complex solutions, divided by two, gives the first real-valued solution, while their difference, divided by 2i, gives the second.

Find a general solution to

(10)

The auxiliary equation is

which has roots

Hence, with a general solution for (10) is

◆

When the auxiliary equation has complex conjugate roots, the (real) solutions oscillate between positive and negative values. This type of behavior is observed in vibrating springs.

In Section 4.1 we discussed the mechanics of the mass–spring oscillator (Figure 4.1, page 153), and we saw how Newton’s second law implies that the position of the mass m is governed by the second-order differential equation

(11)

where the terms are physically identified as

Determine the equation of motion for a spring system when m ! 36 kg, b ! 12 kg/sec (which is equivalent to 12 N-sec/m), k ! 37 kg/sec2, m, and m/sec. Also find

, the displacement after 10 sec.

The equation of motion is given by , the solution of the initial value problem for the speci- fied values of m, b, k, , and . That is, we seek the solution to

(12)

The auxiliary equation for (12) is

36r2 $ 12r $ 37 ! 0 ,

36y– $ 12y¿ $ 37y ! 0 ; y A0B ! 0.7 , y¿ A0B ! 0.1 .y¿ A0By A0B y AtBy A10B

y¿ A0B ! 0.1y A0B ! 0.7 3 inertia 4 y– $ 3damping 4 y¿ $ 3 stiffness 4 y ! 0 . my– AtB $ by¿ AtB $ ky AtB ! 0 ,

y AtB

y AtB ! c1e"tcos A23 tB $ c2e"tsin A23 tB .a ! "1, b ! 23, r !

"2 * 24 " 16 2

! "2 * 2"12

2 ! "1 * i23 .

r2 $ 2r $ 4 ! 0 ,

y– $ 2y¿ $ 4y ! 0 .

e Aa"ibBte Aa$ibBt.e Aa"ibBt

rertert

Example 3

Solution

Example 2

Solution

which has roots

Hence, with the displacement can be expressed in the form

(13)

We can find c1 and c2 by substituting and into the initial conditions given in (12). Differentiating (13), we get a formula for :

Substituting into the initial conditions now results in the system

Upon solving, we find c1 ! 0.7 and c2 ! . With these values, the equation of motion becomes

and

◆

From Example 3 we see that any second-order constant-coefficient differential equation can be interpreted as describing a mass–spring system with mass a,

damping coefficient b, spring stiffness c, and displacement y, if these constants make sense physically; that is, if a is positive and b and c are nonnegative. From the discussion in Section 4.1, then, we expect on physical grounds to see damped oscillatory solutions in such a case. This is consistent with the display in equation (9). With a ! m and c ! k, the exponen- tial decay rate equals , and the angular frequency equals by equation (3).

It is a little surprising, then, that the solutions to the equation do not oscillate; the general solution was shown in Example 3 of Section 4.2 (page 163) to be

The physical significance of this is simply that when the damping coefficient b is too high, the resulting friction prevents the mass from oscillating. Rather than overshoot the spring’s equilibrium point, it merely settles in lazily. This could happen if a light mass on a weak spring were submerged in a viscous fluid.

From the above formula for the oscillation frequency we can see that the oscillations will not occur for This overdamping phenomenon is discussed in more detail in Section 4.9.

It is extremely enlightening to contemplate the predictions of the mass–spring analogy when the coefficients b and c in the equation are negative.ay– $ by¿ $ cy ! 0

b 7 24mk. b, c1e

"2t $ c2te "2t.

y– $ 4y¿ $ 4y ! 0

24mk " b2/ A2mB,b"b/ A2mBa ay– $ by¿ $ cy ! 0

y A10B ! 0.7e"5/3cos 10 $ 1.3 6

e"5/3sin 10 ! "0.13 m.

y AtB ! 0.7e"t/6 cos t $ 1.3 6

e"t/6 sin t ,

1.3/6

" c1 6

$ c2 ! 0.1 .

c1 ! 0.7 ,

y¿ AtB ! a" c1 6

$ c2b e"t/6cos t $ a"c1 " c26b e"t/6sin t . y¿ AtB y¿ AtBy AtB

y AtB ! c1e"t/6 cos t $ c2e"t/6 sin t . y AtBa ! "1/6, b ! 1,

r ! "12 * 2144 " 4 A36B A37B

72 !

"12 * 1221 " 37 72

! " 1 6

* i .

Section 4.3 Auxiliary Equations with Complex Roots 171

Interpret the equation

(14)

in terms of the mass–spring system.

Equation (14) is a minor alteration of equation (12) in Example 3; the auxiliary equation has roots Thus, its general solution becomes

(15)

Comparing equation (14) with the mass–spring model

(16)

we have to envision a negative damping coefficient b ! "12, giving rise to a friction force that imparts energy to the system instead of draining it. The increase in

energy over time must then reveal itself in oscillations of ever-greater amplitude–precisely in accordance with formula (15), for which a typical graph is drawn in Figure 4.7. ◆

Interpret the equation

(17)

in terms of the mass–spring system.

Comparing the given equation with (16), we have to envision a spring with a negative stiffness k ! "6. What does this mean? As the mass is moved away from the spring’s equilibrium point, the spring repels the mass farther with a force Fspring ! "ky that intensifies as the displacement increases. Clearly the spring must “exile” the mass to (plus or minus) infinity, and we expect all solutions to approach as t increases (except for the equilibrium solution ).

In fact, in Example 1 of Section 4.2, we showed the general solution to equation (17) to be

(18)

Indeed, if we examine the solutions that start with a unit displacement and velocity , we find

(19) y AtB ! 6 $ y0 7

et $ 1 " y0

7 e"6t ,

y¿ A0B ! y0 y A0B ! 1y AtB c1e

t $ c2e "6t .

y AtB " 0*qy AtB

y– $ 5y¿ " 6y ! 0

Ffriction ! "by¿

3 inertia 4 y! " 3damping 4 y# " 3 stiffness 4 y $ 0 , y AtB ! c1e$t/6 cos t $ c2e$t/6 sin t .r ! A$B

1 6 * i.36r

2 " 12r $ 37

36y– " 12y¿ $ 37y ! 0

172 Chapter 4 Linear Second-Order Equations

Solution

Example 5

Solution

et/6

−et/6

Figure 4.7 Solution graph for Example 4

Example 4

and the plots in Figure 4.8, confirm our prediction that all (nonequilibrium) solutions diverge—except for the one with .

What is the physical significance of this isolated bounded solution? Evidently, if the mass is given an initial inwardly directed velocity of "6, it has barely enough energy to overcome the effect of the spring banishing it to but not enough energy to cross the equilibrium point (and get pushed to ). So it asymptotically approaches the (extremely delicate) equilibrium position y ! 0. ◆

In Section 4.8, we will see that taking further liberties with the mass–spring interpretation enables us to predict qualitative features of more complicated equations.

Throughout this section we have assumed that the coefficients a, b, and c in the differential equation were real numbers. If we now allow them to be complex constants, then the roots r1, r2 of the auxiliary equation (1) are, in general, also complex but not necessarily conjugates of each other. When a general solution to equation (2) still has the form

but c1 and c2 are now arbitrary complex-valued constants, and we have to resort to the clumsy calculations of Example 1.

We also remark that a complex differential equation can be regarded as a system of two real differential equations since we can always work separately with its real and imaginary parts. Systems are discussed in Chapters 5 and 9.

y AtB ! c1er1t $ c2er2t ,r1 ( r2,

"q $q

y0 ! "6

Section 4.3 Auxiliary Equations with Complex Roots 173

υ0 = 6 υ0 = 0

υ0 = −6

υ0 = −12

υ0 = −18

Figure 4.8 Solution graphs for Example 5

In Problems 1–8, the auxiliary equation for the given dif- ferential equation has complex roots. Find a general solution.

1. 2. 3. 4. 5. 6. 7. 8. 4y–" 4y¿ $ 26y ! 04y– $ 4y¿ $ 6y ! 0

w– $ 4w¿ $ 6w ! 0y– " 4y¿ $ 7y ! 0 z– " 6z¿ $ 10z ! 0y– " 10y¿ $ 26y ! 0 y– $ 9y ! 0y– $ y ! 0

In Problems 9–20, find a general solution. 9. 10.

11. 12. 13. 14. 15. 16. 17. 18. 19. 20. y‡ " y– $ 2y ! 0y‡ $ y– $ 3y¿ " 5y ! 0

2y–$13y¿" 7y ! 0y– " y¿ $ 7y ! 0 y–" 3y¿ " 11y ! 0y– $ 10y¿ $ 41y ! 0 y–" 2y¿ $ 26y ! 0y– $ 2y¿ $ 5y ! 0 u– $ 7u ! 0z– $ 10z¿ $ 25z ! 0 y– " 8y¿ $ 7y ! 0y– $ 4y¿ $ 8y ! 0

4.3 EXERCISES

174 Chapter 4 Linear Second-Order Equations

) q ( t C E ( t )

I ( t )

Figure 4.9 RLC series circuit

In Problems 21–27, solve the given initial value problem. 21. 22. 23. 24. 25. 26. 27.

28. To see the effect of changing the parameter b in the initial value problem

solve the problem for b ! 5, 4, and 2 and sketch the solutions.

29. Find a general solution to the following higher-order equations. (a) (b) (c)

30. Using the representation for in (6), verify the differentiation formula (7).

31. Using the mass–spring analogy, predict the behavior as of the solution to the given initial value problem. Then confirm your prediction by actually solving the problem. (a) (b) (c) (d) (e)

32. Vibrating Spring without Damping. A vibrating spring without damping can be modeled by the initial value problem (11) in Example 3 by taking b ! 0. (a) If m ! 10 kg, k ! 250 kg/sec2, m,

and m/sec, find the equation of motion for this undamped vibrating spring.

(b) When the equation of motion is of the form displayed in (9), the motion is said to be oscil- latory with frequency Find the fre- quency of oscillation for the spring system of part (a).

33. Vibrating Spring with Damping. Using the model for a vibrating spring with damping discussed in Example 3: (a) Find the equation of motion for the vibrating spring

with damping if kg, kg/sec,b ! 60m ! 10

b/2p.

y¿ A0B ! "0.1 y A0B ! 0.3 y– " y¿ " 6y ! 0 ; y A0B ! 1 , y¿A0B ! 1y– $ 2y¿ " 3y ! 0 ; y A0B ! "2 , y¿A0B ! 0 y–" 6y¿ $ 8y ! 0 ; y A0B ! 1 , y¿A0B ! 0y– $ 100y¿ $ y ! 0 ; y A0B ! 1 , y¿A0B ! 0 y– $ 16y ! 0 ; y A0B ! 2 , y¿ A0B ! 0

t S $q

e Aa$ibBty iv $ 13y– $ 36y ! 0

y‡ $ 2y– $ 5y¿ " 26y ! 0 y‡ " y– $ y¿ $ 3y ! 0

y– $ by¿ $ 4y ! 0 ; y A0B ! 1 , y¿A0B ! 0 , y– A0B ! 0 y¿A0B ! 0 ,yA0B ! 1 ,y‡" 4y–$ 7y¿ " 6y ! 0 ;

y¿A0B!"2yA0B!1 ,y– " 2y¿ $ y ! 0 ; y¿ApB!0yApB!ep ,y– " 2y¿ $ 2y ! 0 ; y¿A0B ! 1yA0B ! 1 ,y– $ 9y ! 0 ; w¿A0B!1wA0B!0 ,w– " 4w¿ $ 2w ! 0 ; y¿A0B!"1yA0B!1 ,y– $ 2y¿ $ 17y ! 0 ; y¿A0B ! 1 y A0B ! 2 ,y– $ 2y¿ $ 2y ! 0 ;

k ! 250 kg/sec2, m, and ! m/sec.

(b) Find the frequency of oscillation for the spring system of part (a). [Hint: See the definition of frequency given in Problem 32(b).]

(c) Compare the results of Problems 32 and 33 and determine what effect the damping has on the frequency of oscillation. What other effects does it have on the solution?

34. RLC Series Circuit. In the study of an electrical circuit consisting of a resistor, capacitor, inductor, and an electromotive force (see Figure 4.9), we are led to an initial value problem of the form

(20)

where L is the inductance in henrys, R is the resis- tance in ohms, C is the capacitance in farads, is the electromotive force in volts, is the charge in coulombs on the capacitor at time t, and is the current in amperes. Find the current at time t if the charge on the capacitor is initially zero, the initial current is zero, L ! 10 H, R ! 20 &, F, and V. [Hint: Differentiate both sides of the differential equation in (20) to obtain a homo- geneous linear second-order equation for . Then use (20) to determine at t ! 0.]dI/dt

I AtB E AtB ! 100 C ! A6260B"1

I ! dq/dt q AtB E AtB

I A0B $ I0 ,q A0B $ q0 , LdIdt " RI "

q C $ E

AtB ;

"0.1 y¿ A0By A0B ! 0.3

35. Swinging Door. The motion of a swinging door with an adjustment screw that controls the amount of friction on the hinges is governed by the initial value problem

where is the angle that the door is open, I is the moment of inertia of the door about its hinges, b + 0 is a damping constant that varies with the amount of fric- tion on the door, k + 0 is the spring constant associated with the swinging door, is the initial angle that theu0

Iu– $ bu¿ $ ku! 0 ; u A0B ! u0 , u¿A0B ! y0 ,

door is opened, and is the initial angular velocity imparted to the door (see Figure 4.10). If I and k are fixed, determine for which values of b the door will not continually swing back and forth when closing.

Section 4.4 Nonhomogeneous Equations: The Method of Undetermined Coefficients 175

(b) Show that, in general, d1 and d2 in (21) must be complex conjugates in order that the solution be real.

37. The auxiliary equations for the following differential equations have repeated complex roots. Adapt the “repeated root” procedure of Section 4.2 to find their general solutions: (a) (b) [Hint:

The auxiliary equation is .]

38. Prove the sum of angle formula for the sine function by following these steps. Fix x. (a) Let f(t):! sin (x $ t). Show that f ,(t) $ f(t) ! 0,

f(0) ! sin x, and f )(0) ! cos x. (b) Use the auxiliary equation technique to solve the

initial value problem y, $ y ! 0, y(0) ! sin x, and y)(0) ! cos x.

(c) By uniqueness, the solution in part (b) is the same as f(t) from part (a). Write this equality; this should be the standard sum of angle formula for sin (x $ t).

Ar2 $ 2r $ 4B2 ! 0y iv $ 4y‡ $ 12y– $ 16y¿ $ 16y ! 0 . y iv $ 2y– $ y ! 0 .

Figure 4.10 Top view of swinging door

36. Although the real general solution form (9) is conve- nient, it is also possible to use the form

(21) to solve initial value problems, as illustrated in Example 1. The coefficients d1 and d2 are complex constants. (a) Use the form (21) to solve Problem 21. Verify

that your form is equivalent to the one derived using (9).

d1e Aa$ibB t $ d2e Aa"ibB t

In this section we employ “judicious guessing” to derive a simple procedure for finding a solu- tion to a nonhomogeneous linear equation with constant coefficients

(1)

when the nonhomogeneity f(t) is a single term of a special type. Our experience in Section 4.3 indicates that (1) will have an infinite number of solutions. For the moment we are content to find one, particular, solution. To motivate the procedure, let’s first look at a few instructive examples.

Find a particular solution to

(2)

We need to find a function y(t) such that the combination is a linear function of t—namely, 3t. Now what kind of function y “ends up” as a linear function after having its zeroth, first, and second derivatives combined? One immediate answer is: another linear func- tion. So we might try and attempt to match up with 3t.

Perhaps you can see that this won’t work: and gives us

y–1 $ 3y¿1 $ 2y1 ! 3A $ 2At , y–1 ! 0y1 ! At, y¿1 ! A

y–1 $ 3y ¿1 $ 2y1y1 AtB ! At y– $ 3y¿ $ 2y

y– $ 3y¿ $ 2y ! 3t .

ay– $ by¿ $ cy ! f(t) ,

NONHOMOGENEOUS EQUATIONS: THE METHOD OF UNDETERMINED COEFFICIENTS4.4

Example 1

Solution

176 Chapter 4 Linear Second-Order Equations

Example 2

Example 3

Solution

†In this case the coefficient of in will be zero for and C for .k ! mk ( may– $ by¿ $ cytk

and for this to equal 3t, we require both that A ! 0 and A ! . We’ll have better luck if we append a constant term to the trial function: . Then , and

which successfully matches up with 3t if 2A ! 3 and 3A $ 2B ! 0 . Solving this system gives A ! and B ! " . Thus, the function

is a solution to (2). ◆

Example 1 suggests the following method for finding a particular solution to the equation

namely, we guess a solution of the form

with undetermined coefficients and match the corresponding powers of t in $ with .† This procedure involves solving m $ 1 linear equations in the m $ 1 unknowns

, and hopefully they have a solution. The technique is called the method of undetermined coefficients. Note that, as Example 1 demonstrates, we must retain all the pow- ers in the trial solution even though they are not present in the nonhomo- geneity f(t).

Find a particular solution to

(3) .

We guess because then and will retain the same exponential form:

Setting and solving for A gives A ! ; hence,

is a solution to (3). ◆

Find a particular solution to

(4) .

Our initial action might be to guess , but this will fail because the derivatives introduce cosine terms:

,y–1 $ 3y¿1 $ 2y1 ! "A sin t $ 3A cos t $ 2A sin t ! A sin t $ 3A cos t

y1 AtB ! A sin t y– $ 3y¿ $ 2y ! sin t

yp AtB ! e3t2 1/220Ae3t ! 10e3t

y–p $ 3y¿p $ 2yp ! 9Ae3t $ 3 A3Ae3tB $ 2 AAe3tB ! 20Ae3ty–py¿pyp AtB ! Ae3t y– $ 3y¿ $ 2y ! 10e3t

t m, t m"1, p , t1, t 0

A0, A1, p , Am Ct m

by¿ $ cyay–Aj,

yp AtB ! Amtm $ p $ A1t $ A0 ay– $ by¿ $ cy ! Ct m, m ! 0, 1, 2, p ;

y2 AtB ! 32 t " 94 9/43/2

y–2 $ 3y¿2 $ 2y2 ! 3A $ 2 AAt $ BB ! 2At $ A3A $ 2BB ,y¿2 ! A, y–2 ! 0y2 AtB ! At $ B 3/2

and matching this with sin t would require that A equal both 1 and 0. So we include the cosine term in the trial solution:

and (4) becomes

The equations have the solution . Thus, the function

is a particular solution to (4). ◆

More generally, for an equation of the form

(5) ,

the method of undetermined coefficients suggests that we guess

(6)

and solve (5) for the unknowns A and B.

If we compare equation (5) with the mass–spring system equation

(7) ,

we can interpret (5) as describing a damped oscillator, shaken with a sinusoidal force. Accord- ing to our discussion in Section 4.1, then, we would expect the mass ultimately to respond by moving in synchronization with the forcing sinusoid. In other words, the form (6) is suggested by physical, as well as mathematical, experience. A complete description of forced oscillators will be given in Section 4.10.

Find a particular solution to

(8) .

Our experience with Example 1 suggests that we take a trial solution of the form , to match the nonhomogeneity in (8). We find

! 5t2et . y–p $ 4yp ! et A2A $ 2B $ C $ 4CB $ tet A4A $ B $ 4BB $ t2et AA $ 4AB y–p ! 2Aet $ 2 A2At $ BBet $ AAt2 $ Bt $ CBet

y¿p ! A2At $ BBet $ AAt2 $ Bt $ CBet yp ! AAt2 $ Bt $ CBet AAt2 $ Bt $ CBet yp AtB !

y– $ 4y ! 5t2et

3 inertia 4 ' y– $ 3damping 4 ' y¿ $ 3 stiffness 4 ' y ! Fext yp AtB ! A cos bt $ B sin bt ay– $ by¿ $ cy ! C sin bt Aor C cos btB

yp AtB ! 0.1 sin t " 0.3 cos t A ! 0.1, B ! "0.3A " 3B ! 1, B $ 3A ! 0

! sin t . ! AA " 3BB sin t $ AB $ 3AB cos t $ 2A sin t $ 2B cos t

y–p AtB $ 3y¿p AtB $ 2yp AtB ! "A sin t " B cos t $ 3A cos t " 3B sin t y–p AtB ! "A sin t " B cos ty¿p AtB ! A cos t " B sin t yp AtB ! A sin t $ B cos t

Section 4.4 Nonhomogeneous Equations: The Method of Undetermined Coefficients 177

Solution

Example 4

Matching like terms yields . A solution is given by

. ◆

As our examples illustrate, when the nonhomogeneous term is an exponential, a sine, a cosine function, or a nonnegative integer power of t times any of these, the function itself suggests the form of a particular solution. However, certain situations thwart the straightforward application of the method of undetermined coefficients. Consider, for example, the equation

(9) .

Example 1 suggests that we guess a zero-degree polynomial. But substitution into (9) proves futile:

The problem arises because any constant function, such as is a solution to the corre- sponding homogeneous equation , and the undetermined coefficient A gets lost upon substitution into the equation. We would encounter the same situation if we tried to find a solution to

(10)

of the form , because solves the associated homogeneous equation and

The “trick” for refining the method of undetermined coefficients in these situations smacks of the same logic as in Section 4.2, when a method was prescribed for finding second solutions to homogeneous equations with double roots. Basically, we append an extra factor of t to the trial solution suggested by the basic procedure. In other words, to solve (9) we try At instead of A:

(9#) , ,

Similarly, to solve (10) we try instead of . The trick won’t work this time, because the characteristic equation of (10) has a double root and, consequently, also solves the homogeneous equation:

But if we append another factor of t, , we succeed in finding a particular solution:†

so and .yp AtB ! t2e3t /2A ! 1/2 ! 2Ae3t ! e3t , yp– " 6yp¿ $ 9yp ! A2Ae3t $ 12Ate3t $ 9At 2e3tB " 6 A2Ate3t $ 3At 2e3tB $ 9 AAt 2e3tByp ! At2e3t, y ¿p ! 2Ate3t $ 3At2e3t, y–p ! 2Ae3t $ 12Ate3t $ 9At2e3t

yp ! At 2e3t

3Ate3t 4 – " 6 3Ate3t 4 ¿ $ 9 3Ate3t 4 ! 0 ( e3t Ate 3t

Ae3typ ! Ate 3t

A ! 5 , yp AtB ! 5ty–p $ y¿p ! 0 $ A ! 5 yp ! At , y¿p ! A , y–p ! 0

yp AtB ! 3Ae3t 4 – " 6 3Ae3t 4 ¿ $ 9 3Ae3t 4 ! 0 ( e3te3ty1 ! Ae3t y– " 6y¿ $ 9y ! e3t

y– $ y¿ ! 0 y1 AtB ! A,AAB– $ AAB ¿ ! 0 ( 5

y1 AtB ! A,y– $ y¿ ! 5 f AtBf AtB

yp AtB ! at2 " 4t5 " 225b et A ! 1, B ! "4/5, and C ! "2/25

178 Chapter 4 Linear Second-Order Equations

†Indeed, the solution to the equation , computed by simple integration, can also be derived by appending two factors of t to the solution of the associated homogeneous equation.y " 1

y– ! 2t2

To see why this strategy resolves the problem and to generalize it, recall the form of the original differential equation (1), . Its associated auxiliary equation is

(11) ,

and if r is a double root, then

(12)

holds also [equation (13), Section 4.2, page 164]. Now suppose the nonhomogeneity f has the form , and we seek to match this f

by substituting into (1), with the power n to be determined. For simplicity we merely list the leading terms in , and :

(lower-order terms)

Then the left-hand member of (1) becomes

(13)

and we observe the following:

Case 1. If r is not a root of the auxiliary equation, the leading term in (13) is , and to match we must take n ! m:

Case 2. If r is a simple root of the auxiliary equation, (11) holds and the leading term in (13) is , and to match we must take n ! m $ 1:

However, now the final term can be dropped, since it solves the associated homogeneous equation, so we can factor out t and for simplicity renumber the coefficients to write

.yp AtB ! t AAmtm $ p $ A1t $ A0Bert A0e

yp AtB ! AAm$1t m$1 $ Amt m $ p $ A1t $ A0Bertf AtB ! Ct mert$ bBt n"1ertAnn A2ar

yp AtB ! AAmtm $ p $ A1t $ A0Bertf AtB ! Ct mert$ br $ cBt nertAn Aar2

$ Al.o.t.B$ 3Ann An " 1Ba $ An"1 An " 1B A2ar $ bB $ An"2 Aar2 $ br $ cB 4 t n"2ert ! An Aar2 $ br $ cBt nert $ 3Ann A2ar $ bB $ An"1 Aar2 $ br $ cB 4 t n"1ertay–p $ by¿p $ cyp $ An"1r

2t n"1ert $ 2An"1r An " 1Bt n"2ert $ An"2r2t n"2ert $ Al.o.tBy–p ! Anr2t nert $ 2Annrt n"1ert $ Ann An " 1Bt n"2ert $ An"2rt

n"2ert $ Al.o.t.By¿p ! Anrt nert $ Annt n"1ert $ An"1rt n"1ert $ An"1 An " 1Bt n"2ert yp ! Ant

nert $ An"1t n"1ert $ An"2t

n"2ert $

y–pyp , y¿p yp AtB ! AAnt n $ An"1t n"1 $ p $ A1t $ A0Bert AtBCt mertAtB

2ar $ b ! 0

ar2 " br " c $ 0

ay– $ by¿ $ cy ! f AtB Section 4.4 Nonhomogeneous Equations: The Method of Undetermined Coefficients 179

180 Chapter 4 Linear Second-Order Equations

Method of Undetermined Coefficients To find a particular solution to the differential equation

, where m is a nonnegative integer, use the form

(14) ,

with

(i) s ! 0 if r is not a root of the associated auxiliary equation; (ii) s ! 1 if r is a simple root of the associated auxiliary equation; and (iii) s ! 2 if r is a double root of the associated auxiliary equation.

To find a particular solution to the differential equation

for use the form

(15) ,

with

(iv) s ! 0 if is not a root of the associated auxiliary equation; and (v) s ! 1 if is a root of the associated auxiliary equation.a $ ib

a $ ib

$ t s ABmt m $ p $ B1 t $ B0Beat sin b typ AtB ! t s AAmt m $ p $ A1t $ A0Beat cos bt b ( 0,

ay– $ by¿ $ cy ! cCtmeatcos btor Ctmeatsin bt

yp AtB ! t s AAmt m $ p $ A1t $ A0Bert ay– $ by¿ $ cy ! Ctmert

Example 5

[The (cos, sin) formulation (15) is easily derived from the exponential formulation (14) by putting and employing Euler’s formula, as in Section 4.3.]

Remark 1. The nonhomogeneity corresponds to the case when r ! 0.

Remark 2. The rigorous justification of the method of undetermined coefficients [including the analysis of the terms we dropped in (13)] will be presented in a more general context in Chapter 6.

Find the form for a particular solution to

(16) ,

where f equals

(a) (b) (c) (d) (e) (f) t2et3tet5e"3tt2 cos pt2tet sin t7 cos 3t

AtBy– $ 2y¿ " 3y ! f AtB

Ct m r ! a $ ib

Case 3. If r is a double root of the auxiliary equation, (11) and (12) hold and the leading term in (13) is , and to match we must take n ! m $ 2:

but again we drop the solutions to the associated homogeneous equation and renumber to write

We summarize with the following rule.

yp AtB ! t2 AAmtm $ p $ A1t $ A0Bert yp AtB ! AAm$2t m$2 $ Am$1t m$1 $ p $ A2t2 $ A1t $ A0Bertf AtB ! Ct mert" 1Bat n"2ertAnn An

The auxiliary equation for the homogeneous equation corresponding to (16), , has roots and . Notice that the functions in (a), (b), and (c) are associated with complex roots (because of the trigonometric factors). These are clearly different from and , so the solution forms correspond to (15) with

(a)

(b)

(c)

For the nonhomogeneity in (d) we appeal to (ii) and take since "3 is a sim- ple root of the auxiliary equation. Similarly, for (e) we take and for (f ) we take . ◆

Find the form of a particular solution to

, for the same set of nonhomogeneities as in Example 5.

Now the auxiliary equation for the corresponding homogeneous equation is , with the double root r ! 1. This root is not linked with any of the nonhomo-

geneities (a) through (d), so the same trial forms should be used for (a), (b), and (c) as in the previous example, and will work for (d).

Since r ! 1 is a double root, we have s ! 2 in (14) and the trial forms for (e) and (f) have to be changed to

(e)

(f)

respectively, in accordance with (iii). ◆

Find the form of a particular solution to

Now the auxiliary equation for the corresponding homogeneous equation is , and it has complex roots . Since the nonhomogeneity involves with ; that is, , the solution takes the form

. ◆

The nonhomogeneity tan t in an equation like is not one of the forms for which the method of undetermined coefficients can be used; the derivatives of the “trial solution” tan t, for example, get complicated, and it is not clear what additional terms need to be added to obtain a true solution. In Section 4.6 we discuss a different procedure that can handle such nonhomogeneous terms. Keep in mind that the method of undetermined coefficients applies only to nonhomogeneities that are polynomials, exponentials, sines or cosines, or products of these functions. The superposition principle in Section 4.5 shows how the method can be extended to the sums of such nonhomogeneities. Also, it provides the key to assembling a general solution to (1) that can accommodate initial value problems, which we have avoided so far in our examples.

y AtB ! A y– $ y¿ $ y ! tan t yp AtB ! t AA1t $ A0Bet cos t $ t AB1t $ B0Bet sin ta $ ib ! 1 $ i ! r1a ! b ! 1

eat cos btr1 ! 1 $ i, r2 ! 1 " i r2 " 2r $ 2 ! 0

y– " 2y¿ $ 2y ! 5tet cos t

yp AtB ! t 2 AA2t 2 $ A1t $ A0Betyp AtB ! t 2 AA1t $ A0Bet y AtB ! Ae"3t

Ar " 1B2 ! 0 r2 " 2r $ 1 ! f AtBy– " 2y¿ $ y ! f AtB

yp AtB ! t AA2t 2 $ A1t $ A0Bet yp AtB ! t AA1t $ A0Bet yp AtB ! Ate"3typ AtB ! AA2t

2 $ A1t $ A0B cos pt $ AB2t2 $ B1t $ B0B sin ptyp AtB ! AA1t $ A0Bet cos t $ AB1t $ B0Bet sin t yp AtB ! A cos 3t $ B sin 3t s ! 0:

r2r1 r2 ! "3r1 ! 1

r2 $ 2r " 3 ! 0

Section 4.4 Nonhomogeneous Equations: The Method of Undetermined Coefficients 181

Solution

Example 6

Example 7

Solution

The next theorem describes the superposition principle, a very simple observation which nonetheless endows the solution set for our equations with a powerful structure. It extends the applicability of the method of undetermined coefficients and enables us to solve initial value problems for nonhomogeneous differential equations.

182 Chapter 4 Linear Second-Order Equations

THE SUPERPOSITION PRINCIPLE AND UNDETERMINED COEFFICIENTS REVISITED4.5

Superposition Principle

Theorem 3. Let be a solution to the differential equation

and be a solution to

Then for any constants and , the function is a solution to the differential equation

.ay– $ by¿ $ cy ! k1 f1 AtB $ k2 f2 AtB k1y1 $ k2 y2k2k1

ay– $ by¿ $ cy ! f2 AtBy2 ay– $ by¿ $ cy ! f1 AtBy1

In Problems 1–8, decide whether or not the method of undetermined coefficients can be applied to find a partic- ular solution of the given equation.

1. 2. 3. 4. 5. 6. 7. 8.

In Problems 9–26, find a particular solution to the differ- ential equation. 9. 10.

11. 12. 13. 14.

15. 16.

17. 18. 19. 20. 21. x– AtB " 4x¿ AtB $ 4x AtB ! te2ty– $ 4y ! 16t sin 2t

4y– $ 11y¿ " 3y ! "2te"3t y– $ 4y ! 8 sin 2ty– " 2y¿ $ y ! 8et u– AtB " u AtB ! t sin td 2y

dx2 " 5

dy dx

$ 6y ! xex

2z– $ z ! 9e2ty– " y¿ $ 9y ! 3 sin 3t 2x¿ $ x ! 3t2y– AxB $ y AxB ! 2x y– $ 3y ! "9y– $ 2y¿ " y ! 10

8z¿ AxB " 2z AxB ! 3x100e4x cos 25xty– " y¿ $ 2y ! sin 3t y– AuB $ 3y¿ AuB " y AuB ! sec u2v– AxB " 3v AxB ! 4x sin 2x $ 4x cos 2x x– $ 5x¿ " 3x ! 3t 2y– AxB " 6y¿ AxB $ y AxB ! A sinxB /e4x5y– " 3y¿ $ 2y ! t3 cos 4t y– $ 2y¿ " y ! t"1et

22. 23. 24. 25. 26.

In Problems 27–32, determine the form of a particular solution for the differential equation. (Do not evaluate coefficients.) 27. 28. 28. 29. 30. 31. 32.

In Problems 33–36, use the method of undetermined coefficients to find a particular solution to the given higher-order equation. 33. 34. 35. 36. y(4) " 3y– " 8y ! sin t

y‡ $ y– " 2y ! tet 2y‡ $ 3y– $ y¿ " 4y ! e"t y‡ " y– $ y ! sin t

y– " y¿ " 12y ! 2t6e"3t y– $ 2y¿ $ 2y ! 8t3e"t sin t y– " 2y¿ $ y ! 7et cos t y– " 6y¿ $ 9y ! 5t6e3t y– $ 3y¿ " 7y ! t4et

y– $ 3y¿ " 7y ! t4ety– $ 9y ! 4t3 sin 3t

y– $ 2y¿ $ 2y ! 4te"t cos t y– $ 2y¿ $ 4y ! 111e2t cos 3t y– AxB $ y AxB ! 4x cos xy– AuB " 7y¿ AuB ! u2 x– AtB " 2x¿ AtB $ x AtB ! 24t2et

4.4 EXERCISES

Section 4.5 The Superposition Principle and Undetermined Coefficients Revisited 183

Proof. This is straightforward; by substituting and rearranging we find

. ◆

Find a particular solution to

(1) and

(2) .

In Example 1, Section 4.4, we found that was a solution to $ , and in Example 2 we found that solved . By

superposition, then, solves equation (1). The right-hand member of (2) equals minus three times (3t) plus two times (10 ). There-

fore, this same combination of and will solve (2):

. ◆

If we take a particular solution to a nonhomogeneous equation like

(3)

and add it to a general solution of the homogeneous equation associated with (3),

(4) ,

the sum

(5)

is again, according to the superposition principle, a solution to (3):

Since (5) contains two parameters, one would suspect that and can be chosen to make it satisfy arbitrary initial conditions. It is easy to verify that this is indeed the case.

c2c1

! f AtB $ 0 $ 0 ! f AtB .a Ayp $ c1y1 $ c2y2B– $ b Ayp $ c1y1 $ c2y2B ¿ $ c Ayp $ c1y1 $ c2y2B y AtB ! yp AtB $ c1y1 AtB $ c2 y2 AtB ay– $ by¿ $ cy ! 0

c1y1 $ c2y2

ay– $ by¿ $ cy ! f AtB yp y AtB ! "3y1 $ 2y2 ! "3 A3t/2 " 9/4B $ 2 Ae3t /2B ! "9t/2 $ 27/4 $ e3ty2y1

e3t y1 $ y2 ! 3t/2 " 9/4 $ e3t /2

y– $ 3y¿ $ 2y ! 10e3ty2 AtB ! e3t /22y ! 3t y– $ 3y¿y1 AtB ! 3t/2 " 9/4 y– $ 3y¿ $ 2y ! "9t $ 20e3t y– $ 3y¿ $ 2y ! 3t $ 10e3t

! k1 f1 AtB $ k2 f2 AtB! k1 Aay–1 $ by¿1 $ cy1B $ k2 Aay–2 $ by¿2 $ cy2B a Ak1y1 $ k2y2B– $ b Ak1y1 $ k2 y2B ¿ $ c Ak1y1 $ k2y2B

Example 1

Solution

Existence and Uniqueness: Nonhomogeneous Case

Theorem 4. For any real numbers , b, c, , , and , suppose is a partic- ular solution to (3) in an interval I containing and that and are linearly independent solutions to the associated homogeneous equation (4) in I. Then there exists a unique solution in I to the initial value problem

(6) ,

and it is given by (5), for the appropriate choice of the constants .c1, c2

ay– $ by¿ $ cy ! f AtB , y At0B ! Y0, y¿ At0B ! Y1 y2 AtBy1 AtBt0 yp AtBY1Y0t0a A( 0B

Proof. We have already seen that the superposition principle implies that (5) solves the dif- ferential equation. To satisfy the initial conditions in (6) we need to choose the constants so that

(7) b yp At0B $ c1y1 At0B $ c2y2 At0B ! Y0 , y¿p At0B $ c1y¿1 At0B $ c2 y¿2 At0B ! Y1 .

184 Chapter 4 Linear Second-Order Equations

But as in the proof of Theorem 2 in Section 4.2, simple algebra shows that the choice

and

solves (7) unless the denominator is zero; Lemma 1, Section 4.2, assures us that it is not. Why is the solution unique? If were another solution to (6), then the difference

would satisfy

(8)

But the initial value problem (8) admits the identically zero solution, and Theorem 1 in Section 4.2 applies since the differential equation in (8) is homogeneous. Consequently, (8) has only the identically zero solution. Thus, and . ◆

These deliberations entitle us to say that is a general solution to the nonhomogeneous equation (3), since any solution can be expressed in this form. (Proof: As in Section 4.2, we simply pick and so that matches the value and the derivative of at any point; by uniqueness, and have to be the same function.)

Given that is a particular solution to

find a general solution and a solution satisfying .

The corresponding homogeneous equation,

has the associated auxiliary equation . Because are the roots of this equa- tion, a general solution to the homogeneous equation is . Combining this with the particular solution of the nonhomogeneous equation, we find that a general solution is

To meet the initial conditions, set

which yields . The answer is

. ◆y AtB ! t2 $ 1 2 Aet $ e"tB ! t2 $ cosh t

c1 ! c2 ! 1 2

y¿ A0B ! 2 ' 0 $ c1e0 " c2e"0 ! 0 ,y A0B ! 02 $ c1e0 $ c2e"0 ! 1 , y AtB ! t2 $ c1et $ c2e"t .yp AtB ! t

2 c1e

t $ c2e "t

r ! *1r2 " 1 ! 0

y– " y ! 0 ,

y A0B ! 1, y¿ A0B ! 0y– " y ! 2 " t2 , yp AtB ! t2

ygyp $ c1y1 $ c2y2yg yp $ c1y1 $ c2y2c2c1 yg AtBy ! yp $ c1y1 $ c2y2

yI ! yp $ c1y1 $ c2y2yII " 0

e ay–II $ by¿II $ cyII ! f AtB " f AtB ! 0 , yII At0B ! Y0 " Y0 ! 0 , y¿II At0B ! Y1 " Y1 ! 0 .

yII AtB :! yp AtB $ c1y1 AtB $ c2y2 AtB " yI AtByI AtB c2 !

3Y1 " y¿p At0B 4 y1 At0B " 3Y0 " yp At0B 4 y¿1At0B y1 At0By¿2 At0B " y¿1 At0By2 At0B

c1 ! 3Y0 " yp At0B 4 y¿2 At0B " 3Y1 " y¿p At0B 4 y2 At0B

y1 At0By¿2 At0B " y¿1 At0By2 At0B

Example 2

Solution

A mass–spring system is driven by a sinusoidal external force 5 sin t $ 5 cos t . The mass equals 1, the spring constant equals 2, and the damping coefficient equals 2 (in appropriate units), so the deliberations of Section 4.1 imply that the motion is governed by the differential equation

(9)

If the mass is initially located at , with a velocity , find its equation of motion.

The associated homogeneous equation was studied in Example 1, Section 4.3; the roots of the auxiliary equation were found to be , leading to a general solution

. The method of undetermined coefficients dictates that we try to find a particular solution

of the form A sin t $ B cos t for the first nonhomogeneity 5 sin t:

(10)

Matching coefficients requires and so . The second nonhomogeneity 5 cos t calls for the identical form for and leads to

, or . Hence ! .

By the superposition principle, a general solution to (9) is given by the sum

The initial conditions are

requiring , and thus

(11) ◆

The solution (11) exemplifies the features of forced, damped oscillations that we antici- pated in Section 4.1. There is a sinusoidal component 3 sint t " cos t that is synchronous with the driving force 5 sin t $ 5 cos t , and a component that dies out. When the system is “pumped” sinusoidally, the response is a synchronous sinusoidal oscilla- tion, after an initial transient that depends on the initial conditions; the synchronous response is the particular solution supplied by the method of undetermined coefficients, and the transient is the solution to the associated homogeneous equation. This interpretation will be discussed in detail in Sections 4.9 and 4.10.

You may have observed that, since the two undetermined-coefficient forms in the last

A2e"t cos t $ e"t sin tBBA BA y AtB ! 2e"t cos t $ e"t sin t $ 3 sin t " cos t .c1 ! 2, c2 ! 1

! "c1 $ c2 $ 3 , $ 3 cos 0 $ sin 0

y¿ A0B ! 2 ! c1 3"e"t cos t " e"t sin t 4 t!0 $ c2 3"e"t sin t $ e"t cos t 4 t!0 y A0B ! 1 ! c1e"0 cos 0 $ c2e"0 sin 0 $ 3 sin 0 " cos 0 ! c1 " 1 , ! c1e

"t cos t $ c2e "t sin t $ 3 sin t " cos t .

y ! c1e "t cos t $ c2e

"t sin t $ sin t " 2 cos t $ 2 sin t $ cos t

2 sin t $ cos t ypA ! 2, B ! 1A"A " 2B $ 2AB sin t $ A"B $ 2A $ 2BB cos t ! 5 cos t yp

yp ! sin t " 2 cos tA ! 1, B ! "2

y–p $ 2y¿p $ 2yp ! A"A " 2B $ 2AB sin t $ A"B $ 2A $ 2BB cos t ! 5 sin t .yp ! A sin t $ B cos t , y¿p ! A cos t " B sin t , y–p ! "A sin t " B cos t ; c1e

"t cos t $ c2e "t sin t

"1 * i y– $ 2y¿ $ 2y ! 0

y¿ A0B ! 2y A0B ! 1y– $ 2y¿ $ 2y ! 5 sin t $ 5 cos t . BA

Section 4.5 The Superposition Principle and Undetermined Coefficients Revisited 185

Solution

Example 3

example were identical and were destined to be added together, we could have used the form (10) to match both nonhomogeneities at the same time, deriving the condition

with solution . The next example illustrates this “streamlined” procedure.

Find a particular solution to

(12)

A general solution to the associated homogeneous equation is easily seen to be . Thus, a particular solution for matching the nonhomogeneity has the form , whereas matching requires the form . Therefore, we can match both with the first form:

Thus

which yields , and so . ◆

We generalize this procedure by modifying the method of undetermined coefficients as follows.

yp ! A2t2 " tBetA1 ! 2, A0 ! "1 ! 8tet $ 2et , y–p " yp ! 34A1t $ A2A1 $ 2A0B 4 et

! 3A1t2 $ A4A1 $ A0Bt $ A2A1 $ 2A0B 4 et .y–p ! 32A1t $ A2A1 $ A0B 4 et $ 3A1t 2 $ A2A1 $ A0Bt $ A0 4 et y¿p ! AA1t2 $ A0tBet $ A2A1t $ A0Bet ! 3A1t2 $ A2A1 $ A0Bt $ A0 4 et , yp ! t AA1t $ A0Bet ! AA1t 2 $ A0tBet ,

A0te t2et

t AA1t $ A0Bet8tet c1et $ c2e"t y– " y ! 8tet $ 2et .

yp ! 3 sin t " cos t

y–p $ 2y¿p $ 2yp ! A"A " 2B $ 2AB sin t $ A"B $ 2A $ 2BB cos t ! 5 sin t $ 5 cos t ,

186 Chapter 4 Linear Second-Order Equations

Solution

Method of Undetermined Coefficients (Revisited) To find a particular solution to the differential equation

where is a polynomial of degree m, use the form

(13)

if r is not a root of the associated auxiliary equation, take s ! 0; if r is a simple root of the associated auxiliary equation, take s ! 1; and if r is a double root of the associated auxiliary equation, take s ! 2.

To find a particular solution to the differential equation

where is a polynomial of degree m and is a polynomial of degree n, use the form

(14)

where k is the larger of m and n. If a$ ib is not a root of the associated auxiliary equa- tion, take s ! 0; if a$ ib is a root of the associated auxiliary equation, take s ! 1.

$ t s ABkt k $ p $ B1t $ B0Beat sin bt ,yp AtB ! t s AAkt k $ p $ A1t $ A0Beat cos bt Qn AtBPm AtBay– $ by¿ $ cy ! Pm AtBeat cos b t $ Qn AtBeat sin b t , b ( 0,

yp AtB ! t s AAmtm $ p $ A1t $ A0Bert ;Pm AtB ay– $ by¿ $ cy ! Pm AtBert ,

Example 4

The roots of the associated homogeneous equation were identified in Example 3 as "1 * i. Application of (14) dictates the form

◆

The method of undetermined coefficients applies to higher-order linear differential equations with constant coefficients. Details will be provided in Chapter 6, but the following example should be clear.

Write down the form of a particular solution to the equation

The auxiliary equation for the associated homogeneous is , with a double root r ! "1 and a single root r ! 0. Term by term, the nonhomogeneities call for the forms

(If "1 were a triple root, we would need for .) Of course, we have to rename the coefficients, so the general form is

◆yp AtB ! Ae"tcos t $ Be"tsin t $ tC $ t2 ADt $ EBe"t . 7te"tt3 AA1t $ A0Be"tt2 AA1t $ A0Be"t Afor 7te"tB .

t A0 Afor 3B ,A0 e"t cos t $ B0 e"t sin t Afor 5e"t sin tB , r3 $ 2r2 $ r ! r Ar $ 1B2 ! 0y‡ $ 2y– $ y¿ ! 5e"t sin t $ 3 $ 7te"t .

yp AtB ! t AA3t3 $ A2t2 $ A1t $ A0Be"t cos t $ t AB3t3 $ B2t2 $ B1t $ B0Be"t sin t . y– $ 2y¿ $ 2y ! 0

Section 4.5 The Superposition Principle and Undetermined Coefficients Revisited 187

Write down the form of a particular solution to the equation

y– $ 2y¿ $ 2y ! 5e"t sin t $ 5t3e"t cos t .

Solution

Example 5

Example 6

Solution

1. Given that is a solution to and that

is a solution to , use the super- position principle to find solutions to the following: (a) . (b) . (c) .

2. Given that is a solution to

and is a solution to

use the superposition principle to find solutions to the following differential equations: (a) . (b) . (c) .

In Problems 3–8, a nonhomogeneous equation and a particular solution are given. Find a general solution for the equation.

3. y– $ y¿ ! 1 , yp AtB ! t y– " y¿ $ y ! 4 sin t $ 18e2t y– " y¿ $ y ! sin t " 3e2t y– " y¿ $ y ! 5 sin t

y– " y¿ $ y ! e2t , y2 AtB ! e2t /3y– " y¿ $ y ! sin t

y1 AtB ! cos ty– $ 2y¿ $ 4y ! 11t " 12 cos 2t y– $ 2y¿ $ 4y ! 2t " 3 cos 2t y– $ 2y¿ $ 4y ! t $ cos 2t

y– $ 2y¿ $ 4y ! t y2 AtB ! t /4 " 1/8y– $ 2y¿ $ 4y ! cos 2ty1 AtB ! A1/4B sin 2t 4.5.

6. 7. 8.

In Problems 9–16 decide whether the method of undeter- mined coefficients together with superposition can be applied to find a particular solution of the given equa- tion. Do not solve the equation.

9. 10. 11. 12. 13. 14. 15. 16. 2y– " y¿ $ 6y ! t2e"t sin t " 8t cos 3t $ 10t

y– $ ety¿ $ y ! 7 $ 3t y– " 2y¿ $ 3y ! cosh t $ sin3 t 2y– $ 3y¿ " 4y ! 2t $ sin2t $ 3 y– $ y¿ $ ty ! et $ 7 y– " 6y¿ " 4y ! 4 sin 3t " e3tt2 $ 1/ t 3y– $ 2y¿ $ 8y ! t2 $ 4t " t2et sin t y– " y¿ $ y ! Aet $ tB2

y– ! 2y¿ " y $ 2ex , yp AxB ! x2exy– ! 2y $ 2 tan3x , yp AxB ! tan x u– " u¿ " 2u ! 1 " 2t , up AtB ! t " 1yp AxB ! ex $ x2 y– $ 5y¿ $ 6y ! 6x2 $ 10x $ 2 $ 12ex , y– " y ! t , yp AtB ! "t

4.5 EXERCISES

In Problems 17–22, find a general solution to the differ- ential equation. 17. 18. 19. 20. 21. 22.

In Problems 23–30, find the solution to the initial value problem. 23. 24. 25. 26. 27.

28.

29.

30.

In Problems 31–36, determine the form of a particular solution for the differential equation. Do not solve. 31. 32. 33. 34. 35. 36.

In Problems 37–40, find a particular solution to the given higher-order equation. 37. 38. 39. 40.

41. Discontinuous Forcing Term. In certain physical models, the nonhomogeneous term, or forcing term, g(t) in the equation

may not be continuous but may have a jump ay– $ by¿ $ cy ! g AtB

y A4B " 3y‡ $ 3y– " y¿ ! 6t " 20y‡ $ y– " 2y ! te t $ 1

y A4B " 5y– $ 4y ! 10 cos t " 20 sin ty‡ " 2y– " y¿ $ 2y ! 2t 2 $ 4t " 9

y– " 4y¿ $ 4y ! t2e2t " e2t y– " 4y¿ $ 5y ! e5t $ t sin 3t " cos 3t y– $ 5y¿ $ 6y ! sin t " cos 2t x– " x¿ " 2x ! et cos t " t2 $ cos3 t y– " y ! e2t $ te2t $ t2e2t y– $ y ! sin t $ t cos t $ 10t

y A0B ! 0 , y¿ A0B ! 2y– $ 2y¿ $ y ! t2 $ 1 " et ; y A0B ! 1 , y¿ A0B ! "1y– AuB " y AuB ! sin u " e2u ; y A0B ! 1 , y¿ A0B ! 3y– $ y¿ " 12y ! et $ e2t " 1 ; y A0B ! "7/20 , y¿ A0B ! 1/5y– AxB " y¿ AxB " 2y AxB ! cos x " sin 2x ; y– $ 9y ! 27 ; y A0B ! 4 , y¿ A0B ! 6 z– AxB $ z AxB ! 2e"x ; z A0B ! 0 , z¿ A0B ! 0 y– ! 6t ; y A0B ! 3 , y¿ A0B ! "1y¿ " y ! 1 , y A0B ! 0 ! 10x4 $ 24x3 $ 2x2 " 12x $ 18

y– AxB $ 6y¿ AxB $ 10y AxBy– AuB $ 2y¿ AuB $ 2y AuB ! e"u cos u y– AuB $ 4y AuB ! sin u " cos uy– AxB " 3y¿ AxB $ 2y AxB ! ex sin x y– " 2y¿ " 3y ! 3t2 " 5 y– " y ! "11t $ 1

188 Chapter 4 Linear Second-Order Equations

discontinuity. If this occurs, we can still obtain a rea- sonable solution using the following procedure. Con- sider the initial value problem

where

(a) Find a solution to the initial value problem for .

(b) Find a general solution for . (c) Now choose the constants in the general solution

from part (b) so that the solution from part (a) and the solution from part (b) agree, together with their first derivatives, at . This gives us a con- tinuously differentiable function that satisfies the differential equation except at .

42. Forced Vibrations. As discussed in Section 4.1, a vibrating spring with damping that is under external force can be modeled by

(15) ,

where m + 0 is the mass of the spring system, b + 0 is the damping constant, k + 0 is the spring constant, g(t) is the force on the system at time t, and y(t) is the displacement from the equilibrium of the spring system at time t. Assume . (a) Determine the form of the equation of motion for

the spring system when by finding a general solution to equation (15).

(b) Discuss the long-term behavior of this system. [Hint: Consider what happens to the general solution obtained in part (a) as ]

43. A mass–spring system is driven by a sinusoidal exter- nal force . The mass equals 1, the spring constant equals 3, and the damping coefficient equals 4. If the mass is initially located at and at rest, i.e., , find its equation of motion.

44. A mass–spring system is driven by the external force . The mass equals 1, the

spring constant equals 5, and the damping coeffi- cient equals 2. If the mass is initially located at

, with initial velocity , find its equation of motion.

y¿ A0B ! 5y A0B ! "1 g AtB ! 2 sin 3t $ 10 cos 3t

y¿ A0B ! 0 y A0B ! 1/2 g AtB ! 5 sin t

t S $q.

g AtB ! sin bt b2 6 4mk

my– $ by¿ $ ky ! g AtB

t ! 3p/2

t 7 3p/2 0 - t - 3p/2

g AtB ! b10 if 0 - t - 3p /2 0 if t 7 3p /2 .

y– $ 2y¿ $ 5y ! g AtB ; y A0B ! 0 , y¿ A0B ! 0 ,

45. Speed Bumps. Often bumps like the one depicted in Figure 4.11 are built into roads to discourage speeding. The figure suggests that a crude model of the vertical motion y(t) of a car encountering the speed bump with the speed V is given by

(The absence of a damping term indicates that the car’s shock absorbers are not functioning.) (a) Taking , and in appro-

priate units, solve this initial value problem. Thereby show that the formula for the oscillatory motion after the car has traversed the speed bump is where the constant A depends on the speed V.

y AtB ! A sin t, F0 ! 1m ! k ! 1, L ! p

! e F0 cos ApVt/LB for 0 t 0 6 L/ A2VB 0 for t % L/ A2VB. fmy– $ ky

y AtB ! 0 for t - "L/ A2VB

Section 4.6 Variation of Parameters 189

y(t) m

Speed

x = L/2x = −L/2

cos( x/L)

Figure 4.11 Speed bump

(b) Plot the amplitude of the solution found in part (a) versus the car’s speed V. From the graph, estimate the speed that produces the most violent shaking of the vehicle.

46. Show that the boundary value problem ,

has a solution if and only if . 47. Find the solution(s) to

(if it exists) satisfying the boundary conditions (a) (b) (c)

48. All that is known concerning a mysterious sec- ond-order constant-coefficient differential equation y, $ py) $ qy = g(t) is that t2 $ 1 $ et cos t, t2 $ 1 $ et sin t, and t2 $ 1 $ et cos t $ et sin t are solutions. (a) Determine two linearly independent solutions to

the corresponding homogeneous equation. (b) Find a suitable choice of p, q, and g(t) that

enables these solutions.

y A0B ! "1 , y Ap/3B ! "1 .y A0B ! "1 , y Ap/3B ! 5 . y A0B ! "1 , y Ap/6B ! 3 . y– $ 9y ! 27 cos 6t

l ( *1, *2, *3, p y– $ l2y ! sin t ; y A0B ! 0 , y ApB ! 1

y AtBƒ A ƒ

VARIATION OF PARAMETERS4.6 We have seen that the method of undetermined coefficients is a simple procedure for determin- ing a particular solution when the equation has constant coefficients and the nonhomogene- ous term is of a special type. Here we present a more general method, called variation of parameters,† for finding a particular solution.

Consider the nonhomogeneous linear second-order equation

(1)

and let be two linearly independent solutions for the corresponding homogeneous equation

Then we know that a general solution to this homogeneous equation is given by

(2)

where and are constants. To find a particular solution to the nonhomogeneous equation,c2c1

yh AtB ! c1y1 AtB $ c2y2 AtB , ay– $ by¿ $ cy ! 0 .

Ey1 AtB, y2 AtB Fay– $ by¿ $ cy ! ƒ AtB

†Historical Footnote: The method of variation of parameters was invented by Joseph Lagrange in 1774

the strategy of variation of parameters is to replace the constants in (2) by functions of t. That is, we seek a solution of (1) of the form†

(3)

Because we have introduced two unknown functions, and , it is reasonable to expect that we can impose two equations (requirements) on these functions. Naturally, one of these equations should come from (1). Let’s therefore plug given by (3) into (1). To accomplish this, we must first compute and From (3) we obtain

To simplify the computation and to avoid second-order derivatives for the unknowns , in the expression for we impose the requirement

(4)

Thus, the formula for becomes

(5)

and so

(6)

Now, substituting and as given in (3), (5), and (6), into (1), we find

(7)

since and are solutions to the homogeneous equation. Thus, (7) reduces to

(8)

To summarize, if we can find and that satisfy both (4) and (8), that is,

(9)

then given by (3) will be a particular solution to (1). To determine and , we first solve the linear system (9) for and . Algebraic manipulation or Cramer’s rule (see Appendix D) immediately gives

y¿1 AtB ! "f AtBy2 AtBa 3 y1 AtBy¿2 AtB " y¿1AtBy2 AtB 4 and y¿2 AtB ! f AtBy1 AtBa 3 y1 AtBy¿2 AtB " y¿1AtBy2 AtB 4 , y¿2y¿1

y2y1yp

y#1Y#1 " y#2Y#2 $ f a

y1Y#1 " y2Y#2 $ 0 ,

y2y1

y¿1y¿1 $ y¿2y¿2 ! f a

y2y1

! a Ay¿1y¿1 $ y¿2y¿2B $ 0 $ 0 ! a Ay¿1y¿1 $ y¿2y¿2B $ y1 Aay–1 $ by¿1 $ cy1B $ y2 Aay–2 $ by¿2 $ cy2B ! a Ay¿1y¿1 $ y1y–1 $ y¿2y¿2 $ y2y–2B $ b Ay1y¿1 $ y2y¿2B $ c Ay1y1 $ y2y2Bƒ ! ay–p $ by¿p $ cyp

y–p,yp, y¿p,

y–p ! y¿1y¿1 $ y1y–1 $ y¿2y¿2 $ y2y–2 .

y¿p ! y1y¿1 $ y2y¿2 ,

y¿p

y¿1y1 $ y¿2y2 ! 0 .

y–p, y2y1

y¿p ! Ay¿1y1 $ y¿2y2B $ Ay1 y¿1 $ y2y¿2B . y–p AtB.y¿p AtB yp AtB

y2 AtBy1 AtByp AtB $ Y1 AtBy1 AtB " Y2 AtBy2 AtB .

190 Chapter 4 Linear Second-Order Equations

†In Exercises 2.3, Problem 36, we developed this approach for first-order linear equations. Because of the similarity of equations (2) and (3), this technique is sometimes known as “variation of constants.”

Of course, in step (b) one could use the formulas in (10), but and are so easy to derive that you are advised not to memorize them.

Find a general solution on to

(11)

Observe that two independent solutions to the homogeneous equation are and . We now set

(12)

and, referring to (9), solve the system

for and This gives

Integrating, we obtain

(13)

(14) y2 AtB ! # sin t dt ! " cos t $ C2 . ! sin t " ln 0 sec t $ tan t 0 $ C1 , ! "# 1 " cos

2t cos t

dt ! # A cos t " sec tB dt y1 AtB ! "# tan t sin t dt ! "# sin 2t

cos t dt

y¿2 AtB ! tan t cos t ! sin t .y¿1 AtB ! " tan t sin t , y¿2 AtB.y¿1 AtB A" sin tBy¿1 AtB $ A cos tBy¿2 AtB ! tan t ,

A cos tBy¿1 AtB $ A sin tBy¿2 AtB ! 0 , yp AtB ! y1 AtB cos t $ y2 AtB sin tsin t

cos ty– $ y ! 0

d 2y

dt 2 $ y ! tan t .

A"p/2, p/2B y2 AtBy1 AtB

Section 4.6 Variation of Parameters 191

Method of Variation of Parameters To determine a particular solution to

(a) Find two linearly independent solutions to the corresponding homoge- neous equation and take

(b) Determine and by solving the system in (9) for and and integrating.

(c) Substitute and into the expression for to obtain a particular solution.yp AtBy2 AtBy1 AtB y¿2 AtBy¿1 AtBy2 AtBy1 AtByp AtB ! y1 AtBy1 AtB $ y2 AtBy2 AtB .

Ey1 AtB, y2 AtB Fay– $ by¿ $ cy ! f :

Solution

Example 1

where the bracketed expression in the denominator (the Wronskian) is never zero because of Lemma 1, Section 4.2. Upon integrating these equations, we finally obtain

(10)

Let’s review this procedure.

y1 AtB ! # "f AtBy2 AtBa 3 y1 AtBy¿2AtB"y¿1AtBy2 AtB 4 dt and y2 AtB ! # f AtBy1 AtBa 3 y1 AtBy¿2 AtB" y¿1AtBy2 AtB 4 dt .

We need only one particular solution, so we take both and to be zero for simplicity. Then, substituting and in (12), we obtain

which simplifies to

We may drop the absolute value symbols because sec for .

Recall that a general solution to a nonhomogeneous equation is given by the sum of a gen- eral solution to the homogeneous equation and a particular solution. Consequently, a general solution to equation (11) on the interval is

(15) ◆

Note that in the above example the constants C1 and C2 appearing in (13) and (14) were chosen to be zero. If we had retained these arbitrary constants, the ultimate effect would be just to add C1 cos t $ C2 sin t to (15), which is clearly redundant.

Find a particular solution on to

(16)

With tan t $ 3t " 1, the variation of parameters procedure will lead to a solution. But it is simpler in this case to consider separately the equations

(17)

(18)

and then use the superposition principle (Theorem 3, page 182).

In Example 1 we found that

is a particular solution for equation (17). For equation (18) the method of undetermined coeffi- cients can be applied. On seeking a solution to (18) of the form we quickly obtain

Finally, we apply the superposition principle to get

as a particular solution for equation (16). ◆

Note that we could not have solved Example 1 by the method of undetermined coeffi- cients; the nonhomogeneity tan t is unsuitable. In Section 4.7 we’ll see that another important advantage of the method of variation of parameters is its applicability to equations whose coef- ficients are functions of t.

! " A cos tB ln Asec t $ tan tB $ 3t " 1yp AtB ! yq AtB $ yr AtB yr AtB ! 3t " 1 . yr AtB ! At $ B, yq AtB ! " A cos tB ln Asec t $ tan tB

d2y

dt2 $ y ! 3t " 1

d2y

dt2 $ y ! tan t ,

f AtB ! d2y

dt2 $ y ! tan t $ 3t " 1 .

A"p/2, p/2B

y AtB ! c1 cos t $ c2 sin t " A cos tB ln Asec t $ tan tB .A"p/2, p/2B "p/2 6 t 6 p/2

t $ tan t ! A1 $ sin tB / cos t 7 0yp AtB ! " A cos tB ln 0 sec t $ tan t 0 . yp AtB ! A sin t " ln 0 sec t $ tan t 0 B cos t " cos t sin t ,y2 AtBy1 AtB

C2C1

192 Chapter 4 Linear Second-Order Equations

Solution

Example 2

7. 8.

In Problems 9 and 10, find a particular solution first by undetermined coefficients, and then by variation of para- meters. Which method was quicker? 9.

10.

In Problems 11–18, find a general solution to the differ- ential equation. 11. 12. 13. 14. 15. y– $ y ! 3 sec t " t2 $ 1

y– AuB $ y AuB ! sec3 uy– $ 4y ! sec4 A2tB y– $ y ! tan t $ e3t " 1 y– $ y ! tan 2 t

2x– AtB " 2x¿ AtB " 4x AtB ! 2e2ty– " y ! 2t $ 4 y– $ 4y ! csc2 A2tBy– $ 4y¿ $ 4y ! e"2t ln t

Section 4.7 Variable-Coefficient Equations 193

In Problems 1–8, find a general solution to the differential equation using the method of variation of parameters. 1. 2. y– $ 4y ! tan 2t

y– $ y ! sec t

3. 4. 5. 6. y– AuB $ 16y AuB ! sec 4uy– $ 9y ! sec2 A3tB

y– " 2y¿ $ y ! t"1et y– $ 2y¿ $ y ! e"t

16.

17.

18.

19. Express the solution to the initial value problem

using definite integrals. Using numerical integration (Appendix C) to approximate the integrals, find an approximation for y(2) to two decimal places.

20. Use the method of variation of parameters to show that

is a general solution to the differential equation

, where is a continuous function on . [Hint: Use the trigonometric identity

] 21. Suppose y satisfies the equation y, $ 10y) $ 25y !

et 3

subject to y(0) ! 1 and y)(0) ! "5. Estimate y(0.2) to within *0.0001 by numerically approxi- mating the integrals in the variation of parameters formula.

sin t cos s " sin s cos t . sin At " sB !A"q, q Bf AtB

y– $ y ! f AtB y AtB ! c1 cos t $ c2 sin t $ # t

0 f AsB sin At " sB ds

y– " y ! 1 t , y A1B ! 0 , y¿ A1B ! "2 ,

y– " 6y¿ $ 9y ! t"3e3t

1 2

y– $ 2y ! tan 2t " 1 2

y– $ 5y¿ $ 6y ! 18t2

VARIABLE-COEFFICIENT EQUATIONS4.7 The techniques of Sections 4.2 and 4.3 have explicitly demonstrated that solutions to a linear homogeneous constant-coefficient differential equation,

(1) ay, $ by) $ cy ! 0 ,

are defined and satisfy the equation over the whole interval . After all, such solu- tions are combinations of exponentials, sinusoids, and polynomials.

The variation of parameters formula of Section 4.6 extended this to nonhomogeneous constant-coefficient problems,

(2) ay, $ by) $ cy ! ƒ(t) ,

yielding solutions valid over all intervals where ƒ(t) is continuous (ensuring that the integrals in (10) of Section 4.6 containing ƒ(t) exist and are differentiable). We could hardly hope for more; indeed, it is debatable what meaning the differential equation (2) would have at a point where f(t) is undefined, or discontinuous.

("q, $q)

4.6 EXERCISES

Therefore, when we move to the realm of equations with variable coefficients of the form

(3) a2(t)y, $ a1(t)y) $ a0(t)y ! ƒ(t) ,

the most we can expect is that there are solutions that are valid over intervals where all four “governing” functions— a2(t), a1(t), a0(t), and ƒ(t)—are continuous. Fortunately, this expecta- tion is fulfilled except for an important technical requirement—namely, that the coefficient function a2(t) must be nonzero over the interval.

†

Typically, one divides by the nonzero coefficient a2(t) and expresses the theorem for the equation in standard form [see (4), below] as follows.

194 Chapter 4 Linear Second-Order Equations

Existence and Uniqueness of Solutions

Theorem 5. Suppose p(t), q(t), and g(t) are continuous on an interval (a, b) that con- tains the point t0. Then, for any choice of the initial values Y0 and Y1, there exists a unique solution y(t) on the same interval (a, b) to the initial value problem

(4) y–(t) $ p(t)y¿(t) $ q(t)y (t) ! g(t) ; y(t0) ! Y0 , y¿(t0) ! Y1 .

Cauchy–Euler, or Equidimensional, Equations

Definition 2. A linear second-order equation that can be expressed in the form

(6)

where and are constants, is called a Cauchy–Euler, or equidimensional, equation.

ca, b,

at2y–(t) $ bty¿(t) $ cy ! f(t) ,

Determine the largest interval for which Theorem 5 ensures the existence and uniqueness of a solution to the initial value problem

(5)

The data p(t), q(t), and g(t) in the standard form of the equation,

are simultaneously continuous in the intervals and . The former con- tains the point t0 ! 1, where the initial conditions are specified, so Theorem 5 guarantees (5) has a unique solution in . ◆

Theorem 5, embracing existence and uniqueness for the variable-coefficient case, is difficult to prove because we can’t construct explicit solutions in the general case. So the proof is deferred to Chapter 13.†† However, it is instructive to examine a special case that we can solve explicitly.

0 6 t 6 3

3 6 t 6 q0 6 t 6 3

y– $ py¿ $ qy ! d 2y

dt 2 $

1 (t " 3)

dy dt

$ 2t

(t " 3) y !

ln t (t " 3)

! g ,

(t " 3) d2y

dt2 $

dy dt

$ 2t y ! ln t ; y(1) ! 3 , y¿(1) ! "5 . Example 1

Solution

†Indeed, the whole nature of the equation—reduction from second-order to first-order—changes at points where a2(t) is zero. ††All references to Chapters 11–13 refer to the expanded text Fundamentals of Differential Equations and Boundary Value Problems, 6th ed.

The nomenclature equidimensional comes about because if y has the dimensions of, say, meters and t has dimensions of time, then each term and y has the same dimen- sions (meters). The coefficient of in (6) is , and it is zero at ; equivalently, the standard form

has discontinuous coefficients at t ! 0. Therefore, we can expect the solutions to be valid only for t + 0 or t . 0. Discontinuities in ƒ, of course, will impose further restrictions.

To solve a homogeneous Cauchy–Euler equation, for t + 0, we exploit the equidimensional feature by looking for solutions of the form , because then and y each have the form (constant) ' t r:

and substitution into the homogeneous form of (6) (that is, with g ! 0) yields a simple quadratic equation for r:

(7)

which we call the associated characteristic equation.

Find two linearly independent solutions to the equation

Inserting yields, according to (7),

whose roots r ! 1/3 and r ! "3 produce the independent solutions

◆

Clearly, the substitution into a homogeneous equidimensional equation has the same simplifying effect as the insertion of into the homogeneous constant-coefficient equation in Section 4.2. That means we will have to deal with the same encumbrances:

1. What to do when the roots of (7) are complex 2. What to do when the roots of (7) are equal

If r is complex, , we can interpret by using the identity and invoking Euler’s formula [equation (5), Section 4.3]:

t/$i0 ! t/ti0 ! t/ei0 ln t ! t/ [cos(0 ln t) $ i sin(0 ln t)] .

t ! eln tta$ibr ! a $ ib

y ! ert y ! t r

y1(t) ! t 1/3 , y2(t) ! t"3 (for t 7 0) .

3r2 $ (11 " 3)r " 3 ! 3r2 $ 8r " 3 ! 0 ,

y ! t r

3t2y– $ 11ty " 3y ! 0 , t 7 0 .

ar2 $ (b " a)r $ c ! 0 ,

ar(r " 1)t r $ brtr $ ctr ! [ar2 $ (b " a)r $ c] tr ! 0 , or

y ! t r , ty ¿ ! trt r"1 ! rt r , t 2y– ! t 2r (r " 1)t r"2 ! r (r " 1)t r ,

t2y–, ty¿,y ! t r

y– $ bat y¿ $ c

at2 y !

f(t)

at2

t ! 0at2y–(t) t2y–, ty¿,

Section 4.7 Variable-Coefficient Equations 195

For example, the differential equation

is a Cauchy–Euler equation, whereas

is not because the coefficient of is 2, which is not a constant times .t2y– 2y– " 3ty¿ $ 11y ! 3t " 1

3t2y– $ 11ty¿ " 3y ! sin t

Example 2

Solution

Then we simplify as in Section 4.3 by taking the real and imaginary parts to form independent solutions:

(8)

If r is a double root of the characteristic equation (7), then independent solutions of the Cauchy–Euler equation on are given by

(9)

This can be verified by direct substitution into the differential equation. Alternatively, the second, linearly independent, solution can be obtained by reduction of order, a procedure to be discussed shortly in Theorem 8. Furthermore, Problem 23 demonstrates that the substitu- tion changes the homogeneous Cauchy–Euler equation into a homogeneous constant- coefficient equation, and the formats (8) and (9) then follow from our earlier deliberations.

We remark that if a homogeneous Cauchy–Euler equation is to be solved for t . 0, then one simply introduces the change of variable , where + 0. The reader should verify via the chain rule that the identical characteristic equation (7) arises when is substi- tuted in the equation. Thus the solutions take the same form as (8), (9), but with t replaced by "t; for example, if r is a double root of (7), we get and as two linearly inde- pendent solutions on .

Find a pair of linearly independent solutions to the following Cauchy–Euler equations for t + 0.

(a) (b)

For part (a), the characteristic equation becomes , with the roots , and (8) produces the real solutions cos(ln t) and sin(ln t).

For part (b), the characteristic equation becomes simply with the double root , and (9) yields the solutions and ln t. ◆

In Chapter 8 we will see how one can obtain power series expansions for solutions to variable-coefficient equations when the coefficients are analytic functions. But, as we said, there is no procedure for explicitly solving the general case. Nonetheless, thanks to the existence/ uniqueness result of Theorem 5, most of the other theorems and concepts of the preceding sec- tions are easily extended to the variable-coefficient case, with the proviso that they apply only over intervals in which the governing functions p(t), q(t), g(t) are continuous. Thus we have the following analog of Lemma 1, page 162.

t0 ! 1 r ! 0r2 ! 0

t"2t"2 r ! "2 * ir2 $ 4r $ 5 ! 0

t 2y– $ ty¿ ! 0t2 y– $ 5ty¿ $ 5y ! 0

A"q, 0B A"tBr ln A"tBA"tBr tr ! A"tBrtt ! "t

t ! ex

y1 ! t r , y2 ! t r ln t .

(0, q)

y1 ! t% cos(& ln t) , y2 ! t% sin(& ln t) .

196 Chapter 4 Linear Second-Order Equations

Example 3

Solution

†The determinant representation of the Wronskian was introduced in Problem 34, Section 4.2.

A Condition for Linear Dependence of Solutions

Lemma 3. If and are any two solutions to the homogeneous differential equation

(10)

on an interval I where the functions p(t) and q(t) are continuous and if the Wronskian†

is zero at any point t of I, then y1 and y2 are linearly dependent on I.

W[y1, y2](t) J y1(t)y2¿(t) " y1 ¿(t)y2(t) ! ` y1(t)y1 ¿(t) y2(t)y2¿(t) ` y–(t) $ p(t)y¿(t) $ q(t)y(t) ! 0

y2(t)y1(t)

As in the constant-coefficient case, is called a general solution to (10) on I if are linearly independent solutions on I.

For the nonhomogeneous equation

(11)

a general solution on I is given by , where is a general solu- tion to the corresponding homogeneous equation (10) on I and is a particular solution to (11) on I. In other words, the solution to the initial value problem stated in Theorem 5 must be of this form for a suitable choice of the constants , . This follows, just as before, from a straightforward extension of the superposition principle for variable-coefficient equations described in Problem 30.

If linearly independent solutions to the homogeneous equation (10) are known, then can be determined for (11) by the variation of parameters method.

c2c1

yp yh ! c1y1 $ c2y2y ! yp $ yh

y–(t) $ p(t)y¿(t) $ q(t)y(t) ! g(t) ,

y1, y2 yh ! c1y1 $ c2y2

Section 4.7 Variable-Coefficient Equations 197

Representation of Solutions to Initial Value Problems

Theorem 6. If and are any two solutions to the homogeneous differential equation (10) that are linearly independent on an interval I containing t0, then unique constants and can always be found so that satisfies the initial conditions for any and .Y1Y0y(t0) ! Y0, y¿(t0) ! Y1

c1y1(t) $ c2y2(t)c2c1

y2(t)y1(t)

Variation of Parameters

Theorem 7. If and are two linearly independent solutions to the homogeneous equation (10) on an interval I where p(t), q(t), and g(t) are continuous, then a particular solution to (11) is given by , where and are determined up to a constant by the pair of equations

which have the solution

(12)

Note the formulation (12) presumes that the differential equation has been put into standard form [that is, divided by a2(t)].

y1(t) ! # "g(t) y2(t)W[y1, y2](t) dt , y2(t) ! # g(t)y1(t)

W[y1, y2](t) dt .

y1¿Y1¿ " y2¿ Y2¿ $ g , y1Y#1 " y2Y#2 $ 0 ,

y2y1yp ! y1y1 $ y2y2

y2y1

As in the constant-coefficient case, the Wronskian of two solutions is either identically zero or never zero on I, with the latter implying linear independence on I.

Precisely as in the proof for the constant-coefficient case, it can be verified that any linear combination of solutions and to (10) is also a solution.y2y1c1y1 $ c2y2

The proofs of the constant-coefficient versions of these theorems in Sections 4.2 and 4.5 did not make use of the constant-coefficient property, so one can prove them in the general case by literally copying those proofs but interpreting the coefficients as variables. Unfortunately, however, there is no construction analogous to the method of undetermined coefficients for the variable-coefficient case.

What does all this mean? The only stumbling block for our completely solving nonhomo- geneous initial value problems for equations with variable coefficients,

is the lack of an explicit procedure for constructing independent solutions to the associated homogeneous equation (10). If we had and as described in the variation of parameters formula, we could implement (12) to find , formulate the general solution of (11) as

and (with the assurance that the Wronskian is nonzero) fit the constants to the initial conditions. But with the exception of the Cauchy–Euler equation and the ponderous power series machinery of Chapter 8, we are stymied at the outset; there is no general proce- dure for finding and .

Ironically, we only need one nontrivial solution to the associated homogeneous equation, thanks to a procedure known as reduction of order that constructs a second, linearly indepen- dent solution from a known one . So one might well feel that the following theorem rubs salt into the wound.

y1y2

y2y1

yp $ c1y1 $ c2y2, yp

y2y1

y– $ p(t)y¿ $ q(t)y ! g(t) ; y(t0) ! Y0 , y¿(t0) ! Y1 ,

198 Chapter 4 Linear Second-Order Equations

Reduction of Order

Theorem 8. Let be a solution, not identically zero, to the homogeneous differential equation (10) in an interval I (see page 196). Then

(13)

is a second, linearly independent solution.

y2(t) ! y1(t) # e"$p(t)dty1(t)2 dt

y1(t)

This remarkable formula can be confirmed directly, but the following derivation shows how the procedure got its name.

Proof of Theorem 8. Our strategy is similar to that used in the derivation of the variation of parameters formula, Section 4.6. Bearing in mind that is a solution of (10) for any con- stant c, we replace c by a function v(t) and propose the trial solution , spawning the formulas

Substituting these expressions into the differential equation (10) yields

or, on regrouping,

(14)

The group in front of the undifferentiated v(t) is simply a copy of the left-hand member of the original differential equation (10), so it is zero.† Thus (14) reduces to

(15)

which is actually a first-order equation in the variable :

(16) y1w¿ $ A2y1¿ $ py1Bw ! 0 . w " v¿ y1v– $ A2y1¿ $ py1Bv¿ ! 0 , A y1 – $ py1 ¿ $ qy1Bv $ y1v– $ A2y1¿ $ py1Bv¿ ! 0 . Avy1 – $ 2v¿y1 ¿ $ v–y1B $ p Avy1 ¿ $ v¿y1B $ qvy1 ! 0 , y2 ¿ ! vy1 ¿ $ v¿y1 , y2 – ! vy1 – $ 2v¿y1 ¿ $ v–y1 .

y2(t) ! v(t)y1(t) cy1

†This is hardly a surprise; if v were constant, vy would be a solution with ! ! 0 in (14).v–v¿

Indeed, (16) is separable and can be solved immediately using the procedure of Section 2.2. Problem 50 carries out the details of this procedure to complete the derivation of (13). ◆

Given that is a solution to

(17)

use the reduction of order procedure to determine a second linearly independent solution for t + 0.

Rather than implementing the formula (13), let’s apply the strategy used to derive it. We set and substitute , into (17) to find

(18) .

As promised, (18) is a separable first-order equation in , simplifying to with a solution , or (taking integration constants to be zero). Therefore, a sec- ond solution to (17) is .

Of course (17) is a Cauchy–Euler equation for which (7) has equal roots:

and y2 is precisely the form for the independent solution predicted by (9). ◆

The following equation arises in the mathematical modeling of reverse osmosis.†

(19)

Find a general solution.

As we indicated above, the tricky part is to find a single nontrivial solution. Inspection of (19) suggests that y ! sin t or y ! cos t, combined with a little luck with trigonometric identities, might be solutions. In fact, trial and error shows that the cosine function works:

Unfortunately, the sine function fails (try it). So we use reduction of order to construct a second, independent solution. Setting

and computing , , we substitute into (19) to derive

which is equivalent to the separated first-order equation

. Av¿ B ¿Av¿ B ! 2Asin tB Acos tB ! 2sec2 ttan t

! v– A sin tB A cos tB " 2v¿ A sin2 t $ cos2 tB ! 0 , A sin tB 3 v– cos t " 2v¿ sin t " v cos t 4 " 2 A cos tB 3 v¿ cos t " v sin t 4 " A sin tB 3 v cos t 4 v cos t

y2– ! v– cos t " 2v¿ sin t "y2¿ ! v¿ cos t " v sin ty2 AtB ! v AtBy1 At) ! v AtB cos t (sin t)y1– " 2(cos t)y1¿ " (sin t)y1 ! (sin t) ("cos t) " 2(cos t)("sin t) " (sin t)(cos t) ! 0 .

y1 ! cos t , y1¿ ! "sin t , y1– ! "cos t ,

Asin tB y– " 2 Acos tB y¿ " Asin tB y ! 0 , 0 6 t 6 p . ar 2 $ Ab " aBr $ c ! r 2 " 2r $ 1 ! Ar " 1B2 ! 0 ,

y2 ! vt ! t ln t v ! ln tv¿ ! 1/t

Av¿ B ¿ / Av¿ B ! "1/tv¿ v–t $ 2v¿ " 1

t Av¿t $ vB $ 1

t2 vt ! v–t $ A2v¿ " v¿ B ! v–t $ v¿ ! 0

y2– ! v–t $ 2v¿y2¿ ! v¿t $ vy2 AtB ! v AtBy1 AtB ! v AtBt

y– " 1 t y¿ $ 1

t 2 y ! 0 ,

y1 AtB ! t

Section 4.7 Variable-Coefficient Equations 199

Example 4

Solution

Example 5

Solution

†Reverse osmosis is a process used to fortify the alcoholic content of wine, among other applications.

Taking integration constants to be zero yields or , and . Therefore, a second solution to (19) is .

We conclude that a general solution is ◆

In this section we have seen that the theory for variable-coefficient equations differs only slightly from the constant-coefficient case (in that solution domains are restricted to intervals), but explicit solutions can be hard to come by. In the next section, we will supplement our expo- sition by describing some nonrigorous procedures that sometimes can be used to predict quali- tative features of the solutions.

c1cos t $ c2 Asin t " t cos tB.y2 ! Atan t " tB cos t ! sin t " tcos tv ! tan t " t v¿ ! tan2 tln v¿ ! 2 lnˇ A tan tB

200 Chapter 4 Linear Second-Order Equations

In Problems 1 through 4, use Theorem 5 to discuss the existence and uniqueness of a solution to the differential equation that satisfies the initial conditions ,

, where Y0 and Y1 are real constants. 1. 2. 3.

In Problems 5 through 8, determine whether Theorem 5 applies. If it does, then discuss what conclusions can be drawn. If it does not, explain why.

5. 6.

7. 8.

In Problems 9 through 14, find a general solution to the given Cauchy–Euler equation for

10.

11.

12.

13.

14. t 2y– AtB " 3ty¿ AtB $ 4y AtB ! 09t 2y– AtB $ 15ty¿ AtB $ y AtB ! 0 t 2

d 2z

dt 2 $ 5t

dz dt

$ 4z ! 0

d 2w

dt 2 $

6 t

dw dt

$ 4

t 2 w ! 0

t2y– AtB $ 7ty¿ AtB " 7y AtB ! 0 t 2

d 2y

dt 2 $ 2t

dy dt

" 6y ! 0

t 7 0.

y¿ A0B ! 1y A0B ! 1 ,A1 " tBy– $ ty¿ " 2y ! sin t ; y– $ ty¿ " t 2y ! 0 ; y A0B ! 0 , y A1B ! 0y– $ yy¿ ! t 2 " 1 ; y A0B ! 1 , y¿ A0B ! "1 t 2z– $ tz¿ $ z ! cos t ; z A0B ! 1 , z¿ A0B ! 0

ety– " y¿

t " 3 $ y ! ln t

t 2y– $ y ! cos t A1 $ t2By– $ ty¿ " y ! tan tt At " 3By– $ 2ty¿ " y ! t2

y¿ A1B ! Y1 y A1B ! Y0 In Problems 15 through 18, find a general solution for

15.

16.

17.

18.

In Problems 19 and 20, solve the given initial value prob- lem for the Cauchy–Euler equation.

19.

20.

In Problems 21 and 22, devise a modification of the method for Cauchy–Euler equations to find a general solution to the given equation.

21.

22.

23. To justify the solution formulas (8) and (9), perform the following analysis. (a) Show that if the substitution is made in

the function and x is regarded as the new independent variable in , the chain rule implies the following relationships:

t dy dt

! dY dx

, t 2 d 2y

dt 2 !

d 2Y

dx 2 "

dY dx

Y AxB J y AexBy AtB t ! ex

t 7 "1 At $ 1B2y– AtB $ 10 At $ 1By¿ AtB $ 14y AtB ! 0 ,t 7 2 At " 2B2y– AtB " 7 At " 2By¿ AtB $ 7y AtB ! 0 ,

y A1B ! "1 , y¿ A1B ! 13t 2y– AtB $ 7ty¿ AtB $ 5y AtB ! 0 ; y A1B ! "2 , y¿ A1B ! "11t 2y– AtB " 4ty¿ AtB $ 4y AtB ! 0 ; t 2y– AtB $ 3ty¿ AtB $ 5y AtB ! 0t 2y– AtB $ 9ty¿ AtB $ 17y AtB ! 0 t 2y– AtB " 3ty¿ AtB $ 6y AtB ! 0y– AtB "

1 t

y¿ AtB $ 5 t2

y(t) ! 0

t 6 0.

4.7 EXERCISES

Section 4.7 Variable-Coefficient Equations 201

(b) Using part (a), show that if the substitution is made in the Cauchy–Euler differential equation (6), the result is a constant-coefficient equation for , namely,

(20)

(c) Observe that the auxiliary equation (recall Section 4.2) for the homogeneous form of (20) is the same as (7) in this section. If the roots of the former are complex, linearly independent solu- tions of (20) have the form eax cos bx and eax sin bx; if they are equal, linearly independent solu- tions of (20) have the form erx and xerx. Express x in terms of t to derive the corresponding solution forms (8) and (9).

24. Solve the following Cauchy–Euler equations by using the substitution described in Problem 23 to change them to constant coefficient equations, find- ing their general solutions by the methods of the pre- ceding sections, and restoring the original indepen- dent variable t. (a) (b) (c) (d)

25. Let y1 and y2 be two functions defined on . (a) True or False: If y1 and y2 are linearly depen-

dent on the interval , then y1 and y2 are linearly dependent on the smaller interval

. (b) True or False: If y1 and y2 are linearly dependent

on the interval , then y1 and y2 are linearly dependent on the larger interval .

26. Let and . Are y1 and y2 lin- early independent on the following intervals? (a) (b) (c) (d) Compute the Wronskian on the

interval .

27. Consider the linear equation

(21)

(a) Verify that and are two solu- tions to (21) on . Furthermore, show that for .t0 ! 1y1 At0By¿2 At0B " y¿1 At0By2 At0B ( 0A"q, q B

y2 AtB J t3y1 AtB J t t 2y– " 3ty¿ $ 3y ! 0 , for "q 6 t 6 q .

A"q, q B W 3 y1, y2 4 AtBA"q, q BA"q, 0 430, q B y2 AtB ! 0 t 3 0y1 AtB ! t 3 3C, D 4 ) 3a, b 4 3a, b 43 c, d 4 ( 3a, b 4

3a, b 4 A"q, q B t 2y– $ ty¿ $ 9y ! "tan A3 ln tBt 2y– $ 3ty¿ $ y ! t $ t"1 t 2y– $ 3ty¿ $ 10y ! 0 t 2y– $ ty¿ " 9y ! 0

a d 2Y dx 2

$ Ab " aB dY dx

$ cY ! ƒ Ae xB . Y AxB ! y Ae xB

t ! e x (b) Prove that and are linearly indepen- dent on .

(d) Prove that there is no choice of constants c1, c2 such that for all t in . [Hint: Argue that the contrary assumption leads to a contradiction.]

(e) From parts (c) and (d), we see that there is at least one solution to (21) on that is not expressible as a linear combination of the solutions . Does this provide a counterexample to the theory in this section? Explain.

28. Let and . Are y1 and y2 lin- early independent on the interval: (a) ? (b) ? (c) ? (d) Compute the Wronskian on the

interval . 29. Prove that if y1 and y2 are linearly independent solu-

tions of on , then they can- not both be zero at the same point t0 in .

30. Superposition Principle. Let y1 be a solution to

on the interval I and let y2 be a solution to

on the same interval. Show that for any constants k1 and k2, the function is a solution on I to

31. Determine whether the following functions can be Wronskians on for a pair of solutions to some equation (with p and q continuous). (a) (b) (c) (d)

32. By completing the following steps, prove that the Wronskian of any two solutions y1, y2 to the equa- tion on is given by Abel’s formula†

where the constant C depends on y1 and y2. (a) Show that the Wronskian W satisfies the equa-

tion .W¿ $ pW ! 0

t0 and t in (a, b) ,

W 3 y1, y2 4 AtB $ C exp E' # t t 0

p ATB dTF , Aa, bBy– $ py¿ $ qy ! 0

w AtB " 0w AtB ! At $ 1)"1 w AtB ! t 3w AtB ! 6e4t y– $ py¿ $ qy ! 0

"1 6 t 6 1

y– AtB $ p AtBy¿ AtB $ q AtBy AtB ! k1g1 AtB $ k2g2 AtB .k1y1 $ k2y2 y– AtB $ p AtBy¿ AtB $ q AtBy AtB ! g2 AtB y– AtB $ p AtBy¿ AtB $ q AtBy AtB ! g1 AtB

Aa, bBAa, bBy– $ py¿ $ qy ! 0 A"q, q B W 3 y1, y2 4 AtBA"q, q BA"q, 0 430, q B

y2 AtB ! 2t 0 t 0y1 AtB ! t 2 y1 AtB, y2 AtB

A"q, q B A"q, q B y3 AtB ! c1y1 AtB $ c2y2 AtB

A"q, q By3 AtB J 0 t 0 3A"q, q B y2 AtBy1 AtB

†Historical footnote: Niels Abel derived this identity in 1827.

202 Chapter 4 Linear Second-Order Equations

(b) Solve the separable equation in part (a). (c) How does Abel’s formula clarify the fact that the

Wronskian is either identically zero or never zero on ?

33. Use Abel’s formula (Problem 32) to determine (up to a constant multiple) the Wronskian of two solutions on to

34. All that is known concerning a mysterious differen- tial equation is that the functions , and are solutions. (a) Determine two linearly independent solutions

to the corresponding homogeneous differential equation.

(b) Find the solution to the original equation satisfy- ing the initial conditions

35. Given that and are solutions to the differential equation find the solution to this equation that satisfies

36. Verify that the given functions y1 and y2 are linearly independent solutions of the following differential equation and find the solution that satisfies the given initial conditions.

In Problems 37 through 40, use variation of parameters to find a general solution to the differential equation given that the functions y1 and y2 are linearly independent solutions to the corresponding homogeneous equation for

Remember to put the equation in standard form.

37.

38.

39.

40.

In Problems 41 through 43, find general solutions to the nonhomogeneous Cauchy–Euler equations using varia- tion of parameters.

41. t 2z– $ tz¿ $ 9z ! "tan(3 ln t)

y1 ! e t , y2 ! et ln t

ty– $ (1 " 2t)y¿ $ (t " 1)y ! tet ; y1 ! 5t " 1 , y2 ! e"5t ty– $ (5t " 1)y¿ " 5y ! t 2e"5t ; y1 ! t

2 , y2 ! t 3 t 2y– " 4ty¿ $ 6y ! t 3 $ 1 ; y1 ! e

t , y2 ! t $ 1 ty– " (t $ 1)y ¿ $ y ! t 2 ;

t 7 0.

ty– " At $ 2By¿ $ 2y ! 0 ; y1 AtB ! et , y2 AtB ! t2 $ 2t $ 2 ; y A1B ! 0 , y¿ A1B ! 1

y¿ A1B ! 0.y A1B ! 2, y– $ p AtBy¿ $ q AtBy ! g AtB,1 $ 3t21 $ t, 1 $ 2t,

p AtB y A2B ! 2, y¿ A2B ! 5.

t 3t, t 2 y– $ p AtBy¿ $ q AtBy ! g AtB

ty– $ At " 1)y¿ $ 3y ! 0 .A0, q B Aa, bB

42.

43.

44. The Bessel equation of order one-half

has two linearly independent solutions,

Find a general solution to the nonhomogeneous equation

In Problems 45 through 48, a differential equation and a non-trivial solution f are given. Find a second linearly independent solution using reduction of order.

45.

46.

47.

48.

49. In quantum mechanics, the study of the Schrödinger equation for the case of a harmonic oscillator leads to a consideration of Hermite’s equation,

where 1 is a parameter. Use the reduction of order formula to obtain an integral representation of a second linearly independent solution to Hermite’s equation for the given value of 1 and corresponding solution . (a) (b)

50. Complete the proof of Theorem 8 by solving equa- tion (16).

51. The reduction of order procedure can be used more generally to reduce a homogeneous linear nth-order equation to a homogeneous linear th-order equation. For the equation

which has as a solution, use the substitu- tion to reduce this third-order equa- tion to a homogeneous linear second-order equation in the variable w ! v¿.

y (t) ! v (t) f (t) f (t) ! et

ty–¿ " ty– $ y¿ " y ! 0 ,

(n " 1)

l ! 6 , f (t) ! 3t " 2t 3 l ! 4 , f (t) ! 1 " 2t 2

f (t)

y– ' 2 ty¿ " Ly $ 0

f (t) ! et ty– $ (1 " 2t)y¿ $ (t " 1)y ! 0 , t 7 0 ; tx– " (t $ 1)x¿ $ x ! 0 , t 7 0 ; f (t) ! et t 2y– $ 6ty¿ $ 6y ! 0 , t 7 0 ; f (t) ! t"2 t 2y– " 2ty¿ " 4y ! 0 , t 7 0 ; f (t) ! t"1

t 2y– $ ty¿ $ ¢t 2 " 1 4 ≤ y ! t 5/2 , t 7 0 .y1(t) ! t"1/2cos t , y2(t) ! t"1/2sin t .

t 2y– $ ty¿ $ at 2 " 14b y ! 0 , t 7 0 t 2z– " tz¿ $ z ! t ¢1 $ 3

ln t ≤t 2y– $ 3ty¿ $ y ! t"1

Section 4.8 Qualitative Considerations for Variable-Coefficient and Nonlinear Equations 203

52. The equation

has as a solution. Use the substitution to reduce this third-order equation to a

homogeneous linear second-order equation in the variable .

53. Isolated Zeros. Let be a solution to on (a, b), where p, q are continuous

on (a, b). By completing the following steps, prove that if f is not identically zero, then its zeros in (a, b) are isolated, i.e., if then there exists a

such that for (a) Suppose and assume to the contrary

that for each the function f has a zero at where Show that this implies [Hint: Consider the difference quotient for at ]t0.f

f¿(t0) ! 0. 0 6 ƒ t0 " tn ƒ 6 1/n.tn,

n ! 1, 2, p , f(t 0) ! 0

0 6 ƒ t " t0 ƒ 6 d.f(t) ( 0d 7 0 f (t 0) ! 0,

py¿ $ qy ! 0 y– $f(t)

w ! v¿

v (t) f (t) y (t) !f (t) ! t

ty–¿ $ (1 " t)y– $ ty¿ " y ! 0 (b) With the assumptions of part (a), we have

Conclude from this that f must be identically zero, which is a contradic- tion. Hence, there is some integer such that

is not zero for

54. The reduction of order formula (13) can also be derived from Abels’ identity (Problem 32). Let be a nontrivial solution to (10) and a second lin- early independent solution. Show that

and then use Abel’s identity for the Wronskian to obtain the reduction of order formula.W[ f, y]

¢y f ≤¿ ! W[ f, y]

f 2

y(t) f (t)

0 6 ƒ t " t0 ƒ 6 1/n0.f(t) n0

f(t0) ! f¿(t0) ! 0.

4.8 QUALITATIVE CONSIDERATIONS FOR VARIABLE-COEFFICIENT AND NONLINEAR EQUATIONS

There are no techniques for obtaining explicit, closed-form solutions to second-order linear differential equations with variable coefficients (with certain exceptions) or for nonlinear equa- tions. In general, we will have to settle for numerical solutions or power series expansions. So it would be helpful to be able to derive, with simple calculations, some nonrigorous, qualitative conclusions about the behavior of the solutions before we launch the heavy computational machinery. In this section we first display a few examples that illustrate the profound differ- ences that can occur when the equations have variable coefficients or are nonlinear. Then we show how the mass–spring analogy, discussed in Section 4.1, can be exploited to predict some of the attributes of solutions of these more complicated equations.

To begin our discussion we display a linear constant-coefficient, a linear variable-coefficient, and two nonlinear equations.

(a) The equation

(1)

is linear, homogeneous with constant coefficients. We know everything about such equations; the solutions are, at worst, polynomials times exponentials times sinusoids in t, and unique solutions can be found to match any prescribed data at any instant t ! a. It has the superposition property: If and are solutions, so is .y AtB ! c1y1 AtB $ c2y2 AtB y2 AtBy1 AtB

y AaB, y¿ AaB 3y– $ 2y¿ $ 4y ! 0

204 Chapter 4 Linear Second-Order Equations

(b) The equation

(2)

also has the superposition property (Problem 30, Exercises 4.7). It is a linear variable-coefficient equation and is a special case of Legendre’s equation

which arises in the analysis of wave and diffusion phenomena in spherical coordinates.

(3)

(4)

do not share the superposition property because of the square and the cube root of y terms (e.g., the quadratic term does not reduce to . They are nonlinear†

equations.

The Legendre equation (2) has one simple solution, , as can easily be verified by mental calculation. A second, linearly independent, solution for can be derived by the reduction of order procedure of Section 4.7. Traditionally, the second solution is taken to be

(5)

Notice in particular the behavior near t ! *1; none of the solutions of our constant-coefficient equations ever diverged at a finite point!

We would have anticipated troublesome behavior for (2) at t ! *1 if we had rewritten it in standard form as

(6)

since Theorem 5, page 194, only promises existence and uniqueness between these points. As we have noted, there are no general solution procedures for solving nonlinear equa-

tions. However, the following lemma is very useful in some situations such as equations (3), (4). It has an extremely significant physical interpretation, which we will explore in Project D of Chapter 5; for now we will merely tantalize the reader by giving it a suggestive name.

y– " 2t 1 " t2

y¿ $ 2 1 " t2

y ! 0 ,

y2 AtB ! t2 ln a1 $ t1 " tb " 1 . "1 6 t 6 1

y1 AtB ! t y21 $ y

2 2)y

y– " 24y1/3 ! 0

y– " 6y2 ! 0 ,

A1 " t2By– " 2ty¿ $ ly ! 0, A1 ' t2By! ' 2 ty# " 2y $ 0

†Although the quadratic renders equation (3) nonlinear, the occurrence of in (2) does not spoil its linearity (in y).t2y2

The Energy Integral Lemma

Lemma 4. Let y be a solution to the differential equation

(7)

where is a continuous function that does not depend on y) or the independent variable t. Let be an indefinite integral of , that is,

f AyB ! d dy

F AyB . f AyBF AyBf AyB

y– ! f AyB ,AtB

Proof. This is immediate; we insert (8), differentiate, and apply the differential equation (7):

◆

As a result, an equation of the form (7) can be reduced to

(10) ,

for some constant K, which is equivalent to the separable first-order equation

having the implicit two-parameter solution (Section 2.2)

(11)

We will use formula (11) to illustrate some startling features of nonlinear equations.

Apply the energy integral lemma to explore the solutions of the nonlinear equation given in (3).

Since the solution form (11) becomes

For simplicity we take the plus sign and focus attention on solutions with K ! 0. Then we find or

(12)

for any value of the constant c. Clearly, this equation is an enigma; solutions can blow up at or

anywhere—and there is no clue in equation (3) as to why this should happen! Moreover, we have found an infinite family of pairwise linearly independent solutions (rather than the expected two). Yet we still cannot assemble, out of these, a solution matching (say)

all our solutions (12) have and the absence of a super- position principle voids the use of linear combinations . ◆c1y1 AtB $ c2y2 AtBy¿ A0B ! 2y A0B3/2,y A0B ! 1, y¿ A0B ! 3;

t ! 1, t ! 2, t ! p,

y AtB ! Ac " tB"2 ,t ! $ 12 y"3/2 dy ! "y"1/2 $ c, t ! *# dy22 32y3 $ K 4 $ c .

6y2 ! d dy A2y3B,

y– ! 6y2,

t ! *# dy22 3F AyB $ K 4 $ c . dy dt

! *22 3F AyB $ K 4 1 2 Ay¿ B2 " F AyB ! K

! 0 . ! y¿ 3 y– " f AyB 4! 12 2y¿y– " dFdy y¿

d dt

E AtB ! d dt c 1 2

y¿ AtB2 " F Ay AtBB d

Section 4.8 Qualitative Considerations for Variable-Coefficient and Nonlinear Equations 205

Example 1

Solution

Then the quantity

(8)

is constant; i.e.,

(9) d dt

E AtB ! 0 . E AtB J 1

2 y¿ AtB2 " F Ay AtBB

Apply the energy integral lemma to explore the solutions of the nonlinear equation given in (4).

Since formula (11) gives

Again we take the plus sign and focus attention on solutions with K ! 0. Then we find In particular, is a solution, and it satisfies the initial conditions

But note that is also a solution to (4), and it satisfies the same initial conditions! Hence, the uniqueness feature, guaranteed for linear equations by Theorem 5 of Section 4.7, can fail in the nonlinear case. ◆

So these examples have demonstrated violations of the existence and uniqueness proper- ties (as well as “finiteness”) that we have come to expect from the constant-coefficient case. It should not be surprising, then, that the solution techniques for variable-coefficient and nonlin- ear second-order equations are more complicated—when, indeed, they exist. Recall, however, that in Section 4.3 we saw that our familiarity with the mass–spring oscillator equation

(13)

was helpful in picturing the qualitative features of the solutions of other constant-coefficient equations. (See Figure 4.1, page 153.) By pushing these analogies further, we can also anticipate some of the features of the solutions in the variable-coefficient and nonlinear cases. One of the simplest linear second-order differential equations with variable coefficients is

(14)

Using the mass–spring analogy, predict the nature of the solutions to equation (14) for t + 0.

Comparing (13) with (14), we see that the latter equation describes a mass–spring oscillator where the spring stiffness “k” varies in time—in fact, it stiffens as time passes [“k” ! t in equation (14)]. Physically, then, we would expect to see oscillations whose frequency increases with time, while the amplitude of the oscillations diminishes (because the spring gets harder to stretch). The numer- ically computed solution in Figure 4.12 displays precisely this behavior. ◆

y– $ ty ! 0 .

$ my! " by# " ky Fext $ 3 inertia 4 y! " 3damping 4 y# " 3 stiffness 4 y

y2 AtB " 0y A0B ! 0, y¿ A0B ! 0. y1 AtB ! 8t3t ! y1/3/2 $ c. t ! *# dy22 318y4/3 $ K 4 $ c .

24y1/3 ! d dy A18y4/3B,y– ! 24y

1/3,

206 Chapter 4 Linear Second-Order Equations

Figure 4.12 Solution to equation (14)

Example 3

Solution

Example 2

Solution

It is easy to verify (Problem 1) that if is a solution of the Airy equation

(15) ,

then solves so “Airy functions” exhibit the behavior shown in Figure 4.12 for negative time. For positive t the Airy equation has a negative stiffness “k” ! "t, with magni- tude increasing in time. As we observed in Example 5 of Section 4.3, negative stiffness tends to reinforce, rather than oppose, displacements, and the solutions grow rapidly with (positive) time. The solution known as the Airy function of the second kind , depicted in Figure 4.13, behaves exactly as expected.

In Section 4.3 we also pointed out that mass–spring systems with negative spring stiff- ness can have isolated bounded solutions if the initial displacement and velocity are precisely selected to counteract the repulsive spring force. The Airy function of the first kind

, also depicted in Figure 4.13, is such a solution for the “Airy spring”; the initial inwardly directed velocity is just adequate to overcome the outward push of the stiffening spring, and the mass approaches a delicate equilibrium state

Now let’s look at Bessel’s equation. It arises in the analysis of wave or diffusion phenom- ena in cylindrical coordinates. The Bessel equation of order n is written

(16)

Clearly, there are irregularities at t ! 0, analogous to those at t ! *1 for the Legendre equation (2); we will explore these in depth in Chapter 8.

y! " 1t y# " a1 ' n2t2b y $ 0 . y AtB " 0.Ai

AtB y¿ A0By A0B

Bi AtBy AtB y– $ ty ! 0,y A"tBy– " ty ! 0

y AtB

Section 4.8 Qualitative Considerations for Variable-Coefficient and Nonlinear Equations 207

Bi(t)

Ai(t) t

Figure 4.13 Airy functions

Remark. Schemes for numerically computing solutions to second-order equations will be dis- cussed in Section 5.3. If such schemes are available, then why is there a need for the qualitative analysis discussed in this section? The answer is that numerical methods provide only approxi- mations to solutions of initial value problems, and their accuracy is sometimes difficult to predict (especially for nonlinear equations). For example, numerical methods are often ineffective near points of discontinuity or over the long time intervals needed to study the asymptotic behavior. And this is precisely when qualitative arguments can lend insight into the reasonableness of the computed solution.

Apply the mass–spring analogy to predict qualitative features of solutions to Bessel’s equation for t + 0.

Comparing (16) with the paradigm (13), we observe that

• the inertia “m” ! 1 is fixed at unity; • there is positive damping , although it weakens with time; and • the stiffness is positive when t + n and tends to 1 as

Solutions, then, should be expected to oscillate with amplitudes that diminish slowly (due to the damping), and the frequency of the oscillations should settle at a constant value (given, ac- cording to the procedures of Section 4.3, by radian per unit time). The graphs of the Bessel functions and of the first and second kind of order exemplify these qualitative predictions; see Figure 4.14. The effect of the singularities in the coefficients at t ! 0 is manifested in the graph of ◆

Although most Bessel functions have to be computed by power series methods, if the order

n is a half-integer, then and have closed-form expressions. In fact,

and . You can verify directly that these functions solve equation (16).

Give a qualitative analysis of the modified Bessel equation of order n:

(17)

This equation also exhibits unit mass and positive, diminishing, damping. However, the stiffness now converges to negative 1. Accordingly, we expect typical solutions to diverge as The modified Bessel function of the first kind, of order n ! 2 in Figure 4.15 on page 209, follows this prediction, whereas the modified Bessel function of the second kind, of order n ! 2 in Figure 4.15, exhibits the same sort of balance of initial position and velocity as we saw for the Airy function . Again, the effect of the singularity at t ! 0 is evident. ◆

Use the mass–spring model to predict qualitative features of the solutions to the nonlinear Duffing equation

(18)

Although equation (18) is nonlinear, it can be matched with the paradigm (13) if we envision

y! " y " y3 $ y– " A1 " y2By $ 0 . Ai AtB Kn

AtBIn AtB t S $q. y! " 1t y# ' a1 ' n2t2b y $ 0 .

Y1/2 AtB ! "22/(pt) cos t J1/2 AtB ! 22/(pt) sin tYn AtBJn AtB Y1/2 AtB.

n ! 12Yn AtBJn AtB 2k/m ! 1 t S $ q.A“k” ! 1 " n2/t2BA“b” ! 1/tB

208 Chapter 4 Linear Second-Order Equations

Y1/2(t)

J1/2(t)

Figure 4.14 Bessel functions

Example 5

Solution

Example 6

Solution

Example 4

The fascinating van der Pol equation

(19)

originated in the study of the electrical oscillations observed in vacuum tubes.

Predict the behavior of the solutions to equation (19) using the mass–spring model.

By comparison to the paradigm (13), we observe unit mass and stiffness, positive damping when , and negative damping when . Friction thus

dampens large-amplitude motions but energizes small oscillations. The result, then, is that all (nonzero) solutions tend to a limit cycle whose friction penalty incurred while is balanced by the negative-friction boost received while . The computer-generated Figure 4.17 on page 210 illustrates the convergence to the limit cycle for some solutions to the van der Pol equation. ◆

0 y AtB 0 6 1 0 y AtB 0 7 1 0 y AtB 0 6 10 y AtB 0 7 13 “b” ! " A1 " y2B 4

y! ' A1 " y2By¿ " y $ 0

Section 4.8 Qualitative Considerations for Variable-Coefficient and Nonlinear Equations 209

Example 7

Solution

K2(t) I2(t)

Figure 4.15 Modified Bessel functions

Figure 4.16 Solution graphs for the Duffing equation

†Graphic depictions of oscillations on a therapeutic mattress are best left to one’s imagination, the editors say.

unit mass, no damping, and a (positive) stiffness , which increases as the displacement y gets larger. (This increasing-stiffness effect is built into some popular mat- tresses for therapeutic reasons.)† Such a spring grows stiffer as the mass moves farther away, but it restores to its original value when the mass returns. Thus, high-amplitude excursions should oscillate faster than low-amplitude ones, and the sinusoidal shapes in the graphs of should be “pinched in” somewhat at their peaks. These qualitative predictions are demon- strated by the numerically computed solutions plotted in Figure 4.16. ◆

y AtB “k” ! 1 $ y2

Finally, we consider the motion of the pendulum depicted in Figure 4.18. This motion is measured by the angle that the pendulum makes with the vertical line through O at time t. As the diagram shows, the component of gravity, which exerts a torque on the pendulum and thus accelerates the angular velocity , is given by Consequently, the rotational analog of Newton’s second law, torque equals rate of change of angular momentum, dictates (see Problem 7)

(20)

(21)

Give a qualitative analysis of the motion of the pendulum.

If we rewrite (21) as

and compare with the paradigm (13), we see fixed mass m, no damping, and a stiffness given by

This stiffness is plotted in Figure 4.19, where we see that small-amplitude motions are driven by

“k” ! mg /

sin u u

mu– $ mg /

sin u u

u ! 0

mU! " m g O sin U $ 0 .

m/2u– ! "/mg sin u ,

"mg sin u.du/dt

u AtB

210 Chapter 4 Linear Second-Order Equations

mg sin

! 2

Figure 4.18 A pendulum

Figure 4.17 Solutions to the van der Pol equation

Example 8

Solution

π 2π−2π

−π

mg/!

mg( )! sin

Figure 4.19 Pendulum “stiffness”

a nearly constant spring stiffness of value and the considerations of Section 4.3 dictate the familiar formula

for the angular frequency of the nearly sinusoidal oscillations. See Figure 4.20, which com- pares a computer-generated solution to equation (21) to the solution of the constant-stiffness equation with the same initial conditions.†

For larger motions, however, the diminishing stiffness distorts the sinusoidal nature of the graph of and lowers the frequency. This is evident in Figure 4.21.

Finally, if the motion is so energetic that reaches the value the stiffness changes sign and abets the displacement; the pendulum passes the apex and gains speed as it falls, and this spinning motion repeats continuously. See Figure 4.22. ◆

p,u u AtB,

v ! A km ! Ag/ mg//,

Section 4.8 Qualitative Considerations for Variable-Coefficient and Nonlinear Equations 211

True stiffness

6 rad

Constant stiffness

Figure 4.20 Small-amplitude pendulum motion

8 rad 3π

Figure 4.21 Large-amplitude pendulum motion

2π 3π 5π 6ππ 4π t

Figure 4.22 Very-large-amplitude pendulum motion

†The latter is identified as the linearized equation in Group Project D.

212 Chapter 4 Linear Second-Order Equations

1. Show that if y satisfies , then satisfies

2. Using the paradigm (13), what are the inertia, damp- ing, and stiffness for the equation ? If y + 0, what is the sign of the “stiffness constant”? Does your answer help explain the runaway behavior of the solutions ?

3. Try to predict the qualitative features of the solution to that satisfies the initial conditions

Compare with the computer- generated Figure 4.23. [Hint: Consider the sign of the spring stiffness.]

y A0B ! "1, y¿A0B ! "1.y– " 6y2 ! 0 y AtB ! 1 / Ac " tB2

y– " 6y2 ! 0

y– $ ty ! 0. y A"tBy– " ty ! 0AtB 6. Use the energy integral lemma to show that

motions of the free undamped mass–spring oscil- lator obey

constant .

7. Pendulum Equation. To derive the pendulum equation (21), complete the following steps. (a) The angular momentum of the pendulum mass

m measured about the support O in Figure 4.18 on page 210 is given by the product of the “lever arm” length and the component of the vector momentum perpendicular to the lever arm. Show that this gives

(b) The torque produced by gravity equals the prod- uct of the lever arm length and the component of gravitational (vector) force mg perpendicular to the lever arm. Show that this gives

8. Use the energy integral lemma to show that pendu- lum motions obey

9. Use the result of Problem 8 to find the value of , the initial velocity, that must be imparted to a pendu- lum at rest to make it approach (but not cross over) the apex of its motion. Take for simplicity.

10. Use the result of Problem 8 to prove that if the pendu- lum in Figure 4.18 is released from rest at the angle

, then for all t. [Hint: The initial conditions are argue that the constant in Problem 8 equals .]" Ag//Bcos au A0B ! a, u¿ A0B ! 0;

0 u AtB 0 - aa, 0 6 a 6 p / ! g

u¿ A0B Au¿ B2

2 "

g / cos u ! constant .

torque ! "/mg sin u .

angular momentum ! m/2 du dt

my /

m Ay¿ B2 $ ky2 !my– $ ky ! 0

−1

Figure 4.23 Solution for Problem 3

4. Show that the three solutions and to are linearly inde- pendent on . (See Problem 35, Exercises 4.2, page 166.)

5. (a) Use the energy integral lemma to derive the fam- ily of solutions to the equation

(b) For show that these solutions are pairwise linearly independent for different values of c in an appropriate interval around t ! 0.

c ( 0 y– ! 2y3.

y AtB ! 1 / At " cB A"1, 1By– " 6y2 ! 01 / A3 " tB2

1 / A2 " tB2,1 / A1 " tB2,

4.8 EXERCISES

The computation of the solutions of the Legendre, Bessel, and Airy equations and the analysis of the nonlinear equations of Duffing, van der Pol, and the pendulum have challenged many of the great mathematicians of the past. It is gratifying, then, to note that so many of their salient features are susceptible to the qualitative reasoning we have used herein.

11. Use the mass–spring analogy to explain the qualita- tive nature of the solutions to the Rayleigh equation

(22)

depicted in Figures 4.24 and 4.25.

12. Use reduction of order to derive the solution in equation (5) for Legendre’s equation.

13. Figure 4.26 contains graphs of solutions to the Duffing, Airy, and van der Pol equations. Try to match the solution to the equation.

14. Verify that the formulas for the Bessel functions do indeed solve equation (16).

15. Use the mass–spring oscillator analogy to decide whether all solutions to each of the following differ- ential equations are bounded as

(a) (b) (c) (d) (e) (Mathieu’s equation) (f ) (g)

16. Use the energy integral lemma to show that every solution to the Duffing equation (18) is bounded; that is, for some M. [Hint: First argue that for some K.]y2 /2 $ y4 /4 - K

0 y AtB 0 - M y– " ty¿ " y ! 0 y– $ ty¿ $ y ! 0 y– $ A4 $ 2 cos tBy ! 0 y– $ y6 ! 0y– $ y5 ! 0

y– " t 2y ! 0y– $ t 2y ! 0 t S $q.

J1/2 AtB, Y1/2 AtB

y2 AtB y! ' 3 1 ' Ay#B2 4 y# " y $ 0

Section 4.8 Qualitative Considerations for Variable-Coefficient and Nonlinear Equations 213

Figure 4.24 Solution to the Rayleigh equation

Figure 4.25 Solution to the Rayleigh equation

(b)

(c)

(a)

Figure 4.26 Solution graphs for Problem 13

In this section we return to the mass–spring system depicted in Figure 4.1 (page 153) and analyze its motion in more detail. The governing equation is

(1)

Let’s focus on the simple case in which b ! 0 and , the so-called undamped, free case. Then equation (1) reduces to

(2)

and, when divided by m, becomes

(3)

where The auxiliary equation associated with (3) is which has complex conjugate roots Hence, a general solution to (3) is

(4)

We can express y in the more convenient form

(5)

with by letting and That is,

Solving for A and in terms of and , we find

(6) and

where the quadrant in which lies is determined by the signs of and . This is because sin has the same sign as has the same sign as cos For example, if and , then is in Quadrant II. (Note, in particular, that is not simply the arctangent of , which would lie in Quadrant IV.)C1/C2

ffC2 6 0C1 7 0 f ! C2/AB.AC2Asin f ! C1/AB and cos fC1 fC2C1f

tan F $ C1 C2

,A $ 2C21 $ C22 C2C1f ! C1 cos vt $ C2 sin vt .

A sin Avt $ fB ! A cos vt sin f $ A sin vt cos fC2 ! A cos f.C1 ! A sin fA % 0, y AtB $ A sin AVt " FB ,AtB y AtB $ C1 cos Vt " C2 sin Vt .*vi.

r 2 $ v2 ! 0,v ! 2k/m.d 2y

dt2 " V2y $ 0 ,

m d2y

dt2 $ ky ! 0

Fext ! 0

! my– $ by¿ $ ky .

Fext ! 3 inertia 4 d2y dt2

$ 3damping 4 dy dt

$ 3 stiffness 4 y

214 Chapter 4 Linear Second-Order Equations

4.9 A CLOSER LOOK AT FREE MECHANICAL VIBRATIONS

17. Armageddon. Earth revolves around the sun in an approximately circular orbit with radius r ! a, com- pleting a revolution in the time which is one Earth year; here M is the mass of the sun and G is the universal gravitational constant. The gravitational force of the sun on Earth is given by

, where m is the mass of Earth. Therefore, if Earth “stood still,” losing its orbital velocity, it would fall on a straight line into the sun in accor- dance with Newton’s second law:

GMm/r 2

2p Aa3 /GM B1/2,T ! If this calamity occurred, what fraction of the nor- mal year T would it take for Earth to splash into the sun (i.e., achieve r ! 0)? [Hint: Use the energy inte- gral lemma and the initial conditions ,

.]r¿A0B ! 0 r A0B ! a

m d 2r

dt 2 ! "

GMm

r 2 .

It is evident from (5) that, as we predicted in Section 4.1, the motion of a mass in an undamped, free system is simply a sine wave, or what is called simple harmonic motion. (See Figure 4.27.) The constant A is the amplitude of the motion and is the phase angle. The motion is periodic with period and natural frequency , where v ! 2k/m.v/2p2p/v f The period is measured in units of time, and the natural frequency has the dimensions of peri- ods (or cycles) per unit time. The constant is the angular frequency for the sine function in (5) and has dimensions of radians per unit time. To summarize:

.period ! 2p/v (sec) natural frequency ! v/2p (cycles/sec) angular frequency ! v ! 2k/m (rad/sec)

Observe that the amplitude and phase angle depend on the constants and , which, in turn, are determined by the initial position and initial velocity of the mass. However, the period and frequency depend only on k and m and not on the initial conditions.

A -kg mass is attached to a spring with stiffness k ! 16 N/m, as depicted in Figure 4.1. The mass is displaced m to the right of the equilibrium point and given an outward velocity (to

the right) of m/sec. Neglecting any damping or external forces that may be present, deter- mine the equation of motion of the mass along with its amplitude, period, and natural frequency. How long after release does the mass pass through the equilibrium position?

Because we have a case of undamped, free vibration, the equation governing the motion is (3). Thus, we find the angular frequency to be

Substituting this value for into (4) gives

(7)

Now we use the initial conditions, m and m/sec, to solve for and in (7). That is,

22 ! y¿ A0B ! 822C2 , 1/2 ! y A0B ! C1 , C2

C1y¿ A0B ! 22y A0B ! 1/2y AtB ! C1 cos A822 tB $ C2 sin A822 tB . v

v ! A km ! A 161/8 ! 822 rad /sec .

22 1/21/8 C2C1

Section 4.9 A Closer Look at Free Mechanical Vibrations 215

Period Period A − / 2 / 2 /

2 –– – ––

Figure 4.27 Simple harmonic motion of undamped, free vibrations

Example 1

Solution

and so ! and ! . Hence, the equation of motion of the mass is

(8)

To express y in the alternative form (5), we set

Since both and are positive, is in Quadrant I, so Hence,

(9)

Thus, the amplitude A is m, and the phase angle is approximately 1.326 rad. The

period is sec, and the natural frequency is ! cycles per sec.

Finally, to determine when the mass will pass through the equilibrium position, y ! 0, we must solve the trigonometric equation

(10)

for t. Equation (10) will be satisfied whenever

(11) or

n an integer. Putting n ! 1 in (11) determines the first time t when the mass crosses its equilib- rium position:

◆

In most applications of vibrational analysis, of course, there is some type of frictional or damping force affecting the vibrations. This force may be due to a component in the system, such as a shock absorber in a car, or to the medium that surrounds the system, such as air or some liquid. So we turn to a study of the effects of damping on free vibrations, and equation (2) generalizes to

(12)

The auxiliary equation associated with (12) is

(13)

and its roots are

(14)

As we found in Sections 4.2 and 4.3, the form of the solution to (12) depends on the nature of these roots and, in particular, on the discriminant .b2 " 4mk

"b * 2b2 " 4mk 2m

! " b

2m *

1 2m

2b2 " 4mk . mr2 $ br $ k ! 0 ,

m d2y

dt2 " b

dt " ky $ 0 .

t ! p " f

822 ! 0.16 sec . t !

np " f

822 ! np " 1.326822 ,822 t $ f ! np y AtB ! 217

8 sin A822 t $ fB ! 0

8/ A22pB1/PP ! 2p/v ! 2p/ A822 B !22p/8 f217/8 y AtB ! 217

8 sin A822 t $ fB . f ! arctan 4 ! 1.326.fC2C1

tan f ! C1 C2

! 1/2 1/8

! 4 .

A ! 2C21 $ C22 ! 2 A1/2B2 $ A1/8B2 ! 2178 ,AtB y AtB ! 1

2 cos A822tB $ 1

8 sin A822 tB .1/8C21/2C1

216 Chapter 4 Linear Second-Order Equations

Underdamped or Oscillatory Motion When , the discriminant is negative, and there are two complex conjugate roots to the auxiliary equation (13). These roots are where

(15)

Hence, a general solution to (12) is

(16)

As we did with simple harmonic motion, we can express in the alternate form

(17)

where and It is now evident that is the product of an exponential damping factor,

and a sine factor, which accounts for the oscillatory motion. Because the sine factor varies between "1 and 1 with period the solution varies between and with

quasiperiod and quasifrequency . Moreover, since b and m are positive, is negative, and thus the exponential factor tends to zero as A graph of a typical solution is given in Figure 4.28. The system is called underdamped because there is not enough damping present (b is too small) to prevent the system from oscillating.

It is easily seen that as the damping factor approaches the constant A and the quasi- frequency approaches the natural frequency of the corresponding undamped harmonic motion. Figure 4.28 demonstrates that the values of t where the graph of touches the exponential curves are close to (but not exactly) the same values of t at which attains its relative maximum and minimum values (see Problem 13).

y AtB*Aeat y AtB b S 0

y AtB t S $q.a ! "b/2m 1/PP ! 2p/b ! 4mp/24mk " b2 Aeat"Aeaty AtB2p/b,sin Abt $ fB,

Aeat ! Ae"Ab/2mB t , y AtBtan f ! C1/C2.A ! 2C 21 $ C 22y AtB ! Aeat sin Abt $ fB ,

y AtBy AtB ! eat AC1 cos bt $ C2 sin btB . a J " b

2m , b J 1

2m 24mk " b2 . a * ib, b2 " 4mkb2 6 4mk

(b2 6 4mk)

Section 4.9 A Closer Look at Free Mechanical Vibrations 217

Quasiperiod = 2π/β

−A

−Aeαt

Aeαt

Aeαtsin(βt + φ)

Figure 4.28 Damped oscillatory motion

218 Chapter 4 Linear Second-Order Equations

no local max or min

one local max

one local min

(a) (b) (c)

y y y

ttt

Figure 4.29 Overdamped vibrations

Overdamped Motion When , the discriminant is positive, and there are two distinct real roots to the auxiliary equation (13):

(18)

Hence, a general solution to (12) in this case is

(19)

Obviously, is negative. And since (that is, it follows that is also negative. Therefore, as both of the exponentials in (19) decay and

Moreover, since

we see that the derivative is either identically zero (when ) or vanishes for at most one value of t (when the factor in parentheses is zero). If the trivial solution is ignored, it follows that has at most one local maximum or minimum for t + 0. Therefore,

does not oscillate. This leaves, qualitatively, only three possibilities for the motion of , depending on the initial conditions. These are illustrated in Figure 4.29. This case where

+ 4mk is called overdamped motion.

Critically Damped Motion When , the discriminant is zero, and the auxiliary equation has the repeated root . Hence, a general solution to (12) is now

(20)

To understand the motion described by in (20), we first consider the behavior of as By L’Hôpital’s rule,

(21) lim tS$q

y AtB ! lim tS$q

C1 $ C2t

eAb/2mB t ! limtS$q C2Ab/2mBeAb/2mB t ! 0 t S $q.

y AtBy AtBy AtB ! AC1 $ C2tBe "Ab/2mBt .

"b/2m b2 " 4mkb2 ! 4mk

(b2 ! 4mk)

y AtBy AtB y AtB y AtB " 0C1 ! C2 ! 0

y¿ AtB ! C1r1er1t $ C2r2er2t ! er1t AC1r1 $ C2r2eAr2"r1BtB , y AtB S 0.t S $q,r1 b 7 2b2 " 4mkB,b2 7 b2 " 4mkr2y AtB ! C1e

r1t $ C2e r2t .

r1 ! " b

2m $

1 2m

2b2 " 4mk , r2 ! " b2m " 12m 2b2 " 4mk . b2 " 4mkb2 7 4mk

(b2 7 4mk)

(recall that ). Hence, dies off to zero as Next, since

we see again that a nontrivial solution can have at most one local maximum or minimum for t + 0, so motion is nonoscillatory. If b were any smaller, oscillation would occur. Thus, the special case where is called critically damped motion. Qualitatively, critically damped motions are similar to overdamped motions (see Figure 4.29 again).

Assume that the motion of a mass–spring system with damping is governed by

(22)

Find the equation of motion and sketch its graph for the three cases where b ! 6, 10, and 12.

The auxiliary equation for (22) is

(23)

whose roots are

(24)

Case 1. When b ! 6, the roots (24) are "3 * 4i. This is thus a case of underdamping, and the equation of motion has the form

(25)

Setting and gives the system

whose solution is ! . To express as the product of a damping factor and a sine factor [recall equation (17)], we set

where is a Quadrant I angle, since and are both positive. Then

(26)

where arctan The underdamped spring motion is shown in Figure 4.30(a) on page 220.

Case 2. When b ! 10, there is only one (repeated) root to the auxiliary equation (23), namely, r ! "5. This is a case of critical damping, and the equation of motion has the form

(27)

Setting and now gives

and so . Thus,

(28) y AtB ! A1 $ 5tBe"5t .C1 ! 1, C2 ! 5 C1 ! 1 , C2 " 5C1 ! 0 ,

y¿ A0B ! 0y A0B ! 1y AtB ! AC1 $ C2tBe"5t .

A4/3B ! 0.9273.f !y AtB ! 5 4

e"3t sin A4t $ fB , C2C1f A ! 2C 21 $ C 22 ! 54 , tan f ! C1C2 ! 43 ,

y AtB3/4C1 ! 1, C2C1 ! 1 , "3C1 $ 4C2 ! 0 , y¿ A0B ! 0y A0B ! 1y AtB ! C1e"3t cos 4t $ C2e"3t sin 4t .

r ! " b 2

* 1 2

2b2 " 100 . r 2 $ br $ 25 ! 0 ,

d2y

dt2 $ b

dt $ 25y ! 0 ; y A0B ! 1 , y¿ A0B ! 0 .

b2 ! 4mk

y¿ AtB ! aC2 " b2m C1 " b2m C2tb e"Ab/2mB t , t S $q.y AtBb/2m 7 0

Section 4.9 A Closer Look at Free Mechanical Vibrations 219

Example 2

Solution

The graph of given in (28) is represented by the lower curve in Figure 4.30(b). Notice that is zero only for and hence does not cross the t-axis for t + 0.

Case 3. When b ! 12, the roots to the auxiliary equation are This is a case of overdamping, and the equation of motion has the form

(29)

Setting and gives

from which we find and Hence,

(30)

The graph of this overdamped motion is represented by the upper curve in Figure 4.30(b). ◆

It is interesting to observe in Example 2 that when the system is underdamped , the solution goes to zero like when the system is critically damped , the solution tends to zero roughly like and when the system is overdamped , the solution goes to zero like This means that if the system is underdamped, it not only oscillates but also dies off slower than if it were critically damped. Moreover, if the system is overdamped, it again dies off more slowly than if it were critically damped (in agree- ment with our physical intuition that the damping forces hinder the return to equilibrium).

eA"6$211 Bt ! e"2.68t. Ab ! 12Be"5t; Ab ! 10Be"3t; Ab ! 6B

! e A"6$211 B t

22 U11 $ 6211 $ A11 " 6211 B e"2211t V .

y AtB ! 11 $ 6211 22

e A"6$211 B t $ 11 " 6211 22

e A"6"211 B t C2 ! A11 " 6211 B /22.C1 ! A11 $ 6211 B /22C1 $ C2 ! 1 , A"6 $ 211 BC1 $ A"6 " 211 BC2 ! 0 ,

y¿ A0B ! 0y A0B ! 1y AtB ! C1e A"6$211 B t $ C2e A"6"211 B t .

"6 * 211.t ! "1/5y AtB y AtB

220 Chapter 4 Linear Second-Order Equations

e−3t 5 4

b = 6

−

e−3t 5 4

e−3t 5 4 sin(4t + 0.9273)

b = 10

e(−6 + )t11 + 6 11 22

(1 + 5t)e−5t

b = 12

11 − 6 11 22

11 e(−6 − )t11+

Figure 4.30 Solutions for various values of b

(a) (b)

A -kg mass is attached to a spring with a stiffness 4 N/m as shown in Figure 4.31(a). The damping constant b for the system is 1 N-sec/m. If the mass is displaced m to the left and given an initial velocity of 1 m/sec to the left, find the equation of motion. What is the maximum displacement that the mass will attain?

Substituting the values for m, b, and k into equation (12) and enforcing the initial conditions, we obtain the initial value problem

(31)

The negative signs for the initial conditions reflect the facts that the initial displacement and push are to the left.

It can readily be verified that the solution to (31) is

(32)

(33)

where tan and lies in Quadrant III because and are both negative. [See Figure 4.31(b) for a sketch of .]

To determine the maximum displacement from equilibrium, we must determine the maximum value of on the graph in Figure 4.31(b). Because dies off exponentially, this will occur at the first critical point of . Computing from (32), setting it equal to zero, and solving gives

tan A223tB ! 23 5

523 sin A223tB ! cos A223tB ,

y¿ AtB ! e"2t e 523 sin A223tB " cos A223 tB f ! 0 , y¿ AtBy AtB y AtB0 y AtB 0

y AtB C2 ! "1/23C1 ! "1/2ff ! 23/2 y AtB ! A 712 e"2t sin A223t $ fB , y AtB ! " 1

2 e"2t cos A223tB " 123 e"2t sin A223tB ,

4 d2y

dt2 $

dt $ 4y ! 0 ; y A0B ! " 1

2 , y¿ A0B ! "1 .

1/2 1/4

Section 4.9 A Closer Look at Free Mechanical Vibrations 221

Example 3

(b)

k = 4 N/m kg

b = 1 N-sec/m

(a)

Equilibrium −.5

maximum displacement is y(0.096) ≈ −0.55 m

1 4

m12

Figure 4.31 Mass–spring system and graph of motion for Example 3

Solution

222 Chapter 4 Linear Second-Order Equations

All problems refer to the mass–spring configuration depicted in Figure 4.1, page 153.

1. A 2-kg mass is attached to a spring with stiffness k ! 50 N/m. The mass is displaced m to the left of the equilibrium point and given a velocity of 1 m/sec to the left. Neglecting damping, find the equation of motion of the mass along with the amplitude, period, and frequency. How long after release does the mass pass through the equilibrium position?

2. A 3-kg mass is attached to a spring with stiffness k ! 48 N/m. The mass is displaced m to the left of the equilibrium point and given a velocity of 2 m/sec to the right. The damping force is negligible. Find the equation of motion of the mass along with the amplitude, period, and frequency. How long after release does the mass pass through the equilib- rium position?

3. The motion of a mass–spring system with damping is governed by

Find the equation of motion and sketch its graph for b ! 0, 6, 8, and 10.

4. The motion of a mass–spring system with damping is governed by

Find the equation of motion and sketch its graph for b ! 0, 10, 16, and 20.

5. The motion of a mass–spring system with damping is governed by

Find the equation of motion and sketch its graph for k ! 20, 25, and 30.

y A0B ! 1 , y¿ A0B ! 0 .y– AtB $ 10y¿ AtB $ ky AtB ! 0 ;

y A0B ! 1 , y¿A0B ! 0 .y– AtB $ by¿ AtB $ 64y AtB ! 0 ;

y A0B ! 1 , y¿ A0B ! 0 .y– AtB $ by¿ AtB $ 16y AtB ! 0 ;

1/2

1/4

6. The motion of a mass–spring system with damping is governed by

Find the equation of motion and sketch its graph for k ! 2, 4, and 6.

7. A -kg mass is attached to a spring with stiffness 16 N/m. The damping constant for the system is 2 N-sec/m. If the mass is moved m to the left of equilibrium and given an initial leftward velocity of 2 m/sec, determine the equation of motion of the mass and give its damping factor, quasiperiod, and quasifrequency.

8. A 20-kg mass is attached to a spring with stiffness 200 N/m. The damping constant for the system is 140 N-sec/m. If the mass is pulled 25 cm to the right of equilibrium and given an initial leftward velocity of 1 m/sec, when will it first return to its equilibrium position?

9. A 2-kg mass is attached to a spring with stiffness 40 N/m. The damping constant for the system is N-sec/m. If the mass is pulled 10 cm to the right of equilibrium and given an initial rightward velocity of 2 m/sec, what is the maximum displacement from equilibrium that it will attain?

10. A -kg mass is attached to a spring with stiffness 8 N/m. The damping constant for the system is N-sec/m. If the mass is moved 1 m to the left of equilibrium and released, what is the maximum dis- placement to the right that it will attain?

11. A 1-kg mass is attached to a spring with stiffness 100 N/m. The damping constant for the system is 0.2 N-sec/m. If the mass is pushed rightward from the equilibrium position with a velocity of 1 m/sec, when will it attain its maximum displacement to the right?

1/4 1/4

815

3/4

1/8

y A0B ! 1 , y¿ A0B ! 0 .y– AtB $ 4y¿ AtB $ ky AtB ! 0 ; 4.9 EXERCISES

Thus, the first positive root is

Substituting this value for t back into equation (32) or (33) gives Hence, the maximum displacement, which occurs to the left of equilibrium, is approximately 0.55 m. ◆

y A0.096B ! "0.55. t !

223 arctan 235 ! 0.096 .

12. A -kg mass is attached to a spring with stiffness 8 N/m. The damping constant for the system is 2 N-sec/m. If the mass is pushed 50 cm to the left of equilibrium and given a leftward velocity of 2 m/sec, when will the mass attain its maximum displacement to the left?

13. Show that for the underdamped system of Example 3, the times when the solution curve in (33) touches the exponential curves are not the same values of t for which the function attains its rela- tive extrema.

14. For an underdamped system, verify that as the damping factor approaches the constant A and the quasifrequency approaches the natural frequency

15. How can one deduce the value of the damping con- stant b by observing the motion of an underdamped system? Assume that the mass m is known.

16. A mass attached to a spring oscillates with a period of 3 sec. After 2 kg are added, the period becomes

2k/m/ A2pB. b S 0

y AtB*27/12e"2ty AtB

1/4

Section 4.10 A Closer Look at Forced Mechanical Vibrations 223

4 sec. Assuming that we can neglect any damping or external forces, determine how much mass was orig- inally attached to the spring.

17. Consider the equation for free mechanical vibra- tion, , and assume the motion is critically damped. Let and assume (a) Prove that the mass will pass through its equilib-

rium at exactly one positive time if and only if

(b) Use computer software to illustrate part (a) for a specific choice of m, b, k, y0, and y0. Be sure to include an appropriate graph in your illustration.

18. Consider the equation for free mechanical vibration, and assume the motion is

overdamped. Suppose and . Prove that the mass will never pass through its equi- librium at any positive time.

y¿(0) 7 0y(0) 7 0 my– $ by¿ $ ky ! 0,

"2my0 2m30 $ by0

7 0.

y0 ( 0. y (0) ! y0, y¿(0) ! 30

my– $ by¿ $ ky ! 0

We now consider the vibrations of a mass–spring system when an external force is applied. Of particular interest is the response of the system to a sinusoidal forcing term. As a paradigm, let’s investigate the effect of a cosine forcing function on the system governed by the differential equation

(1)

where F0 and are nonnegative constants and 0 . b 2 . 4mk (so the system is underdamped).

A solution to (1) has the form where yp is a particular solution and yh is a general solution to the corresponding homogeneous equation. We found in equation (17) of Section 4.9 that

(2)

where A and are constants. To determine yp, we can use the method of undetermined coefficients (Section 4.4). From

the form of the nonhomogeneous term, we know that

(3)

where A1 and A2 are constants to be determined. Substituting this expression into equation (1) and simplifying gives

(4) 3 Ak " mg 2BA1 $ bgA2 4 cos gt $ 3 Ak " mg 2BA2 " bgA1 4 sin gt ! F0 cos gt . yp AtB ! A1 cos gt $ A2 sin gt , f

yh AtB ! Ae"Ab/2mB t sina24mk " b22m t $ fb , y ! yh $ yp,

m d2y

dt2 " b

dy dt

" ky $ F0 cos Gt ,

4.10 A CLOSER LOOK AT FORCED MECHANICAL VIBRATIONS

Setting the corresponding coefficients on both sides equal, we have

Solving, we obtain

(5)

Hence, a particular solution to (1) is

(6)

The expression in brackets can also be written as

so we can express yp in the alternative form

(7)

where tan and the quadrant in which lies is determined by the signs of A1 and A2.

Combining equations (2) and (7), we have the following representation of a general solu- tion to (1) in the case 0 . b2 . 4mk:

(8)

The solution (8) is the sum of two terms. The first term, yh, represents damped oscillation and depends only on the parameters of the system and the initial conditions. Because of the damp- ing factor A this term tends to zero as Consequently, it is referred to as the transient part of the solution. The second term, yp, in (8) is the offspring of the external forcing function Like the forcing function, yp is a sinusoid with angular frequency It is the synchronous solution that we anticipated in Section 4.1. However, yp is out of phase with (by the angle and its magnitude is different by the factor

(9) .

As the transient term dies off, the motion of the mass–spring system becomes essentially that of the second term yp (see Figure 4.32, page 225). Hence, this term is called the steady-state solution. The factor appearing in (9) is referred to as the frequency gain, or gain factor, since it represents the ratio of the magnitude of the sinusoidal response to that of the input force. Note that this factor depends on the frequency and has units of length/force.

A 10-kg mass is attached to a spring with stiffness k ! 49 N/m. At time t ! 0, an external force cos 4t N is applied to the system. The damping constant for the system is 3 N-sec/m.

Determine the steady-state solution for the system. f AtB ! 20

12 Ak " mg 2B2 $ b 2g 2 u " p/2),f AtB g.f AtB ! F0 cos gt.

t S $q.e"Ab/2mB t,

y AtB ! Ae"Ab/2mB t sin a24mk " b 2 2m

t $ fb $ F02 Ak " mg 2B2 $ b 2g 2 sin Agt $ uB . uu ! A1/A2 ! Ak " mg 2B / AbgB

yp AtB ! F02 Ak " mg 2B2 $ b 2g 2 sin Agt $ uB , 2 Ak " mg 2B2 $ b 2g 2 sin Agt $ uB , yp AtB ! F0Ak " mg 2B2 $ b 2g 2 3 Ak " mg 2Bcos gt $ bg sin gt 4 . A1 !

F0 Ak " mg 2BAk " mg 2B2 $ b 2g 2 , A2 ! F0 bgAk " mg 2B 2 $ b 2g 2 . "bgA1 $ Ak " mg 2BA2 ! 0 . Ak " mg2BA1 $ bgA2 ! F0 ,

224 Chapter 4 Linear Second-Order Equations

Example 1

Substituting the given parameters into equation (1), we obtain

(10)

where is the displacement (from equilibrium) of the mass at time t. To find the steady-state response, we must produce a particular solution to (10) that is a

sinusoid. We can do this using the method of undetermined coefficients, guessing a solution of the form A1 cos 4t $ A2 sin 4t. But this is precisely how we derived equation (7). Thus, we sub- stitute directly into (7) and find

(11)

where tan Since the numerator, , is negative and the denominator, 12, positive, is a Quadrant IV angle. Thus,

and the steady-state solution is given (approximately) by

(12) ◆

The above example illustrates an important point made earlier: The steady-state response (12) to the sinusoidal forcing function 20 cos 4t is a sinusoid of the same frequency but differ- ent amplitude. The gain factor [see (9)] in this case is m/N.

In general, the amplitude of the steady-state solution to equation (1) depends on the angular frequency of the forcing function and is given by where

(13) M AgB J 12 Ak " mg2B2 $ b2g2 A AgB ! F0M AgB,g A0.18B/20 ! 0.009

yp AtB ! A0.18B sin A4t " 1.46B . u ! arctan A"9.25B ! "1.46 ,u

A49 " 160Bu ! A49 " 160B /12 ! "9.25. yp AtB ! 202 A49 " 160B2 $ A9B A16B sin A4t $ uB ! A0.18B sin A4t $ uB ,

y AtB10 d2y

dt2 $ 3

dt $ 49y ! 20 cos 4t ,

Section 4.10 A Closer Look at Forced Mechanical Vibrations 225

y ( t )

y p ( t ) 0.05

0.1

0.15

0.2

2 4 6 8 10

Figure 4.32 Convergence of to the steady-state solution when m ! 4, b ! 6, k ! 3, F0 ! 2, g ! 4yp AtBy AtB Solution

is the frequency gain [see (9)]. This formula is valid even when For a given system (m, b, and k fixed), it is often of interest to know how this system reacts to sinusoidal inputs of various frequencies ( is a variable). For this purpose the graph of the gain called the frequency response curve, or resonance curve, for the system, is enlightening.

To sketch the frequency response curve, we first observe that for we find . Of course, implies the force is static; there is no motion in the

steady state, so this value of is appropriate. Also note that as the gain the inertia of the system limits the extent to which it can respond to extremely rapid vibrations. As a further aid in describing the graph, we compute from (13)

(14)

It follows from (14) that if and only if

(15) or

Now when the system is overdamped or critically damped, so , the term inside the radical in (15) is negative, and hence only when . In this case, as increases from 0 to infinity, decreases from to a limiting value of zero.

When b2 . 2mk (which implies the system is underdamped), then is real and positive, and it is easy to verify that has a maximum at . Substituting into (13) gives

(16)

The value is called the resonance frequency for the system. When the system is stimu- lated by an external force at this frequency, it is said to be at resonance.

To illustrate the effect of the damping constant b on the resonance curve, we consider a system in which m ! k ! 1. In this case the frequency response curves are given by

(17)

and, for b . the resonance frequency is Figure 4.33 displays the graphs of these frequency response curves for , , 1, , and 2. Observe that as the maximum magnitude of the frequency gain increases and the reso- nance frequency for the damped system approaches the natural frequency for the undamped system.

To understand what is occurring, consider the undamped system with forcing term F0 cos This system is governed by

(18) m d2y

dt2 $ ky ! F0 cos gt .

gt. Ab ! 0B2k/m / 2p ! 1/2p,gr/2p

b S 0 3/21/2b ! 1/4

gr/2p ! A1/2pB21 " b2/2.22, M AgB ! 12 A1 " g2B2 $ b2g2 , gr/2p

M AgrB ! 1/bA km " b24m2 . grgrM AgB gr

M A0B ! 1/kM AgB gg ! 0M¿ AgB ! 0 b2 % 4mk 7 2mk

g ! gr J A km " b22m2 .g ! 0 M¿ AgB ! 0

M¿ AgB ! " 2m2g 3g2 " A km " b22m2B 43 Ak " mg2B2 $ b2g2 4 3/2 . M AgB S 0;gS qM A0B F0 cos gtg ! 0M A0B ! 1/k g ! 0

M AgB, g b2 % 4mk.

226 Chapter 4 Linear Second-Order Equations

A general solution to (18) is the sum of a particular solution and a general solution to the homogeneous equation. In Section 4.9 we showed that the latter describes simple har- monic motion:

(19)

The formula for the particular solution given in (7) is valid for b ! 0, provided

However, when b ! 0 and then the form we used with undetermined coefficients to derive (7) does not work because cos are solutions to the corre- sponding homogeneous equation. The correct form is

(20)

which leads to the solution

(21)

[The verification of (21) is straightforward.] Hence, in the undamped resonant case (when v), a general solution to (18) is

(22)

Returning to the question of resonance, observe that the particular solution in (21) oscillates between . Hence, as the maximum magnitude of (21) approaches (see Figure 4.34 on page 228).

qt S $q* AF0tB / A2mvB y AtB ! A sin Avt $ fB $ F0

2mv t sin vt .

g !

yp AtB $ F02mV t sin Vt . yp AtB ! A1t cos vt $ A2t sin vt ,

vt and sin vt g ! v,g ( v ! 2k/m.

yh AtB ! A sin Avt $ fB , v J 2k/m .

Section 4.10 A Closer Look at Forced Mechanical Vibrations 227

M ( )

0 1 2

b =

b = 1

b = – 2 b = 2

– 2 1

– 4 1

Figure 4.33 Frequency response curves for various values of b

It is obvious from the above discussion that if the damping constant b is very small, the system is subject to large oscillations when the forcing function has a frequency near the resonance frequency for the system. It is these large vibrations at resonance that concern engineers. Indeed, resonance vibrations have been known to cause airplane wings to snap, bridges to collapse,† and (less catastrophically) wine glasses to shatter.

When the mass–spring system is hung vertically as in Figure 4.35, the force of gravity must be taken into account. This is accomplished very easily. With y measured downward from the unstretched spring position, the governing equation is

my– $ by¿ $ ky ! mg ,

228 Chapter 4 Linear Second-Order Equations

y p

t F 0 ––––

2 m

t F 0 ––––

2 m −

Figure 4.34 Undamped oscillation of the particular solution in (21)

L L

(a)

(b)

(c)

y yp = mg/k

ynew

mg/k

Figure 4.35 Spring (a) in natural position, (b) in equilibrium, and (c) in motion

†An interesting discussion of one such disaster, involving the Tacoma Narrows bridge in Washington State, can be found in Differential Equations and Their Applications, 4th ed., by M. Braun (Springer-Verlag, New York, 1993). See also the articles “Large-Amplitude Periodic Oscillations in Suspension Bridges: Some New Connections with Nonlinear Analysis,” by A. C. Lazer and P. J. McKenna, SIAM Review, Vol. 32 (1990): 537–578; or “Still Twisting,” by Henry Petroski, American Scientist, Vol. 19 (1991): 398–401.

and if the right-hand side is recognized as a sinusoidal forcing term with frequency zero , then the synchronous steady-state response is a constant, which is easily seen to be

. Now if we redefine to be measured from this (true) equilibrium level, as indicated in Figure 4.35(c),

then the governing equation

simplifies; we find

(23)

Thus, the gravitational force can be ignored if is measured from the equilibrium position. Adopting this convention, we drop the “new” subscript hereafter.

Suppose the mass–spring system in Example 1 is hung vertically. Find the steady-state solution.

This is trivial; the steady-state solution is identical to what we derived before,

[equation (11)], but now yp is measured from the equilibrium position, which is m below the unstretched spring position. ◆

A 64-lb weight is attached to a vertical spring, causing it to stretch 3 in. upon coming to rest at equilibrium. The damping constant for the system is 3 lb-sec/ft. An external force

lb is applied to the weight. Find the steady-state solution for the system.

If a weight of 64 lb stretches a spring by 3 in. (0.25 ft), then the spring stiffness must be lb/ft. Thus, if we measured the displacement from the (true) equilibrium level,

equation (23) becomes

(24)

with b ! 3 and k ! 256. But recall that the unit of mass in the U.S. Customary System is the slug, which equals the weight divided by the gravitational acceleration constant ft/sec2

(Table 3.2, page 109). Therefore, m in equation (24) is slugs, and we have

The steady-state solution is given by equation (6) with F0 ! 3 and

◆ ! 3

580 A"8 cos 12t $ 9 sin 12tB .

yp AtB ! 3A256 " 2 # 122B2 $ 32 # 122 3 A256 " 2 # 122B cos 12t $ 3 # 12 sin 12t 4 g ! 12:

2y– $ 3y¿ $ 256y ! 3 cos 12t .

64/32 ! 2 g ! 32

my– $ by¿ $ ky ! 3 cos 12t ,

64/0.25 ! 256

F AtB ! 3 cos 12t 10 ' 9.8/49 ! 2

mg/k !

yp AtB ! 202 A49 " 160B2 $ A9B A16B sin A4t $ uB y AtBmy–new $ by¿new $ kynew ! Fext AtB .

! mg $ Fext AtB " mg , ! my– $ by¿ $ ky " mg my–new $ by¿new $ kynew ! m Ay " mg/kB– $ b Ay " mg/kB ¿ $ k Ay " mg/kB my– $ by¿ $ ky ! mg $ Fext AtB ynew AtB J y AtB " mg/k ,

y AtByp AtB ! mg/kAmg cos 0tB Section 4.10 A Closer Look at Forced Mechanical Vibrations 229

Example 2

Solution

Example 3

Solution

230 Chapter 4 Linear Second-Order Equations

In the following problems, take g ! 32 ft/sec2 for the U.S. Customary System and g ! 9.8 m/sec2 for the MKS system.

1. Sketch the frequency response curve (13) for the system in which m ! 4, k ! 1, b ! 2.

2. Sketch the frequency response curve (13) for the system in which m ! 2, k ! 3, b ! 3.

3. Determine the equation of motion for an undamped system at resonance governed by

;

Sketch the solution.

4. Determine the equation of motion for an undamped system at resonance governed by

;

Sketch the solution.

5. An undamped system is governed by

;

where (a) Find the equation of motion of the system. (b) Use trigonometric identities to show that the

solution can be written in the form

Hence, can be viewed as the product of a slowly varying sine function, and a rapidly varying sine function, The net effect is a sine function with frequency which serves as the time-varying amplitude of a sine function with frequency This vibration phenom- enon is referred to as beats and is used in tuning stringed instruments. This same phe-

Av $ gB /4p. Av " gB /4p, y AtB

sin 3 Av $ gBt /2 4 .sin 3 Av"gBt /24 ,y AtB v " g.v $ g

v " gvg

y AtB ! 2F0 m Av2 " g2B sin av $ g2 tb sin av " g2 tb .

g ( v J 2k/m.y A0B ! y¿ A0B ! 0 , m

d2y

dt2 $ ky ! F0 cos gt

y A0B ! 0 , y¿ A0B ! 1 . d2y

dt2 $ y ! 5 cos t

y A0B ! 1 , y¿ A0B ! 0 . d2y

dt2 $ 9y ! 2 cos 3t

nomenon in electronics is called amplitude modulation. To illustrate this phenomenon, sketch the curve for F0 ! 32, m ! 2,

and

6. Derive the formula for given in (21).

7. Shock absorbers in automobiles and aircraft can be described as forced overdamped mass–spring sys- tems. Derive an expression analogous to equation (8) for the general solution to the differential equa- tion (1) when b2 + 4mk.

8. The response of an overdamped system to a constant force is governed by equation (1) with m ! 2, b ! 8, k ! 6, F0 ! 18, and If the system starts from rest compute and sketch the displacement . What is the limiting value of as ? Interpret this physically.

9. An 8-kg mass is attached to a spring hanging from the ceiling, thereby causing the spring to stretch 1.96 m upon coming to rest at equilibrium. At time t ! 0, an external force ! cos 2t N is applied to the system. The damping constant for the system is 3 N-sec/m. Determine the steady-state solution for the system.

10. Show that the period of the simple harmonic motion of a mass hanging from a spring is where l denotes the amount (beyond its natural length) that the spring is stretched when the mass is at equilibrium.

11. A mass weighing 8 lb is attached to a spring hanging from the ceiling and comes to rest at its equilibrium position. At t ! 0, an external force ! 2 cos 2t lb is applied to the system. If the spring constant is 10 lb/ft and the damping constant is 1 lb-sec/ft, find the equation of motion of the mass. What is the reso- nance frequency for the system?

12. A 2-kg mass is attached to a spring hanging from the ceiling, thereby causing the spring to stretch 20 cm upon coming to rest at equilibrium. At time t ! 0, the mass is displaced 5 cm below the equilibrium position and released. At this same instant, an exter- nal force ! 0.3 cos t N is applied to the system. If the damping constant for the system is 5 N-sec/m, determine the equation of motion for the mass. What is the resonance frequency for the system?

F AtB

2p2l /g, F AtB

t S $q y AtBy AtB3 y A0B ! y¿ A0B ! 0 4 ,g ! 0.

yp AtBg ! 7.v ! 9, y AtB

4.10 EXERCISES

In this chapter we discussed the theory of second-order linear differential equations and pre- sented explicit solution methods for equations with constant coefficients. Much of the method- ology can also be applied to the more general case of variable coefficients. We also studied the mathematical description of vibrating mechanical systems, and we saw how the mass–spring analogy could be used to predict qualitative features of solutions to some variable-coefficient and nonlinear equations.

The important features and solution techniques for the constant-coefficient case are listed below.

Homogeneous Linear Equations (Constant Coefficients)

constants .

Linearly Independent Solutions: . Two solutions y1 and y2 to the homogeneous equa- tion on the interval I are said to be linearly independent on I if neither function is a constant times the other on I. This will be true provided their Wronskian,

is different from zero for some (and hence all) t in I.

General Solution to Homogeneous Equation: . If and are linearly inde- pendent solutions to the homogeneous equation, then a general solution is

where c1 and c2 are arbitrary constants.

Form of General Solution. The form of a general solution for a homogeneous equation with constant coefficients depends on the roots

r1 ! "b $ 2b2 " 4ac

2a , r2 !

"b " 2b2 " 4ac 2a

y AtB ! c1y1 AtB $ c2y2 AtB , y2y1c1y1 " c2y2

W 3 y1, y2 4 AtB J y1 AtBy¿2 AtB " y¿1 AtBy2 AtB , y1, y2

ay– $ by¿ $ cy ! 0, a A(0B, b, c

Chapter Summary 231

13. A mass weighing 32 lb is attached to a spring hang- ing from the ceiling and comes to rest at its equilib- rium position. At time t ! 0, an external force !

lb is applied to the system. If the spring con- stant is 5 lb/ft and the damping constant is 2 lb-sec/ft, find the steady-state solution for the system.

14. An 8-kg mass is attached to a spring hanging from the ceiling and allowed to come to rest. Assume that the spring constant is 40 N/m and the damping con- stant is 3 N-sec/m. At time t ! 0, an external force of

3 cos 4t F AtB 2 sin 4 is applied to the system. Deter-mine the amplitude and frequency of the steady-statesolution.

15. An 8-kg mass is attached to a spring hanging from the ceiling and allowed to come to rest. Assume that the spring constant is 40 N/m and the damping con- stant is 3 N-sec/m. At time t ! 0, an external force of 2 sin 2t cos 2t N is applied to the system. Deter- mine the amplitude and frequency of the steady- state solution.

(2t $ p/4)

Chapter Summary

of the auxiliary equation

(a) When b2 " 4ac + 0, the auxiliary equation has two distinct real roots and and a general solution is

(b) When b2 " 4ac ! 0, the auxiliary equation has a repeated real root r ! r1 ! r2 and a general solution is

Nonhomogeneous Linear Equations (Constant Coefficients)

General Solution to Nonhomogeneous Equation: . If yp is any particular solution to the nonhomogeneous equation and y1 and y2 are linearly independent solutions to the corresponding homogeneous equation, then a general solution is

where c1 and c2 are arbitrary constants. Two methods for finding a particular solution yp are those of undetermined coefficients

and variation of parameters.

Undetermined Coefficients: . If the right-hand side f of a

nonhomogeneous equation with constant coefficients is a polynomial (t), an exponential of the form a trigonometric function of the form cos or any product of these spe- cial types of functions, then a particular solution of an appropriate form can be found. The form of the particular solution involves unknown coefficients and depends on whether a $ ib is a root of the corresponding auxiliary equation. See the summary box on page 186. The unknown coefficients are found by substituting the form into the differential equation and equating coefficients of like terms.

Variation of Parameters: . If and are two linearly inde- pendent solutions to the corresponding homogeneous equation, then a particular solution to the nonhomogeneous equation is

y AtB ! y1 AtBy1 AtB $ y2 At)y2 AtB , y2y1y AtB $ y1 AtBy1 AtB " y2 AtBy2 AtB

or sin bt,bteat, pn

AtBf AtB $ pn AtBeA t e cos Btsin Bt f

y(tB ! yp AtB $ c1y1 AtB $ c2y2 AtB , yp " c1y1 " c2y2

ay– $ by¿ $ cy ! f AtB

y AtB ! c1eat cos bt $ c2eat sin bt . r ! a * ib

y AtB ! c1ert $ c2tert . y AtB ! c1er1t $ c2er2t .

r2r1

ar2 $ br $ c ! 0 , a ( 0 .

232 Chapter 4 Linear Second-Order Equations

where and are determined by the equations

Superposition Principle. If and are solutions to the equations

and

respectively, then $ is a solution to the equation

The superposition principle facilitates finding a particular solution when the nonhomogeneous term is the sum of nonhomogeneities for which particular solutions can be determined.

Cauchy–Euler (Equidimensional) Equations

Substituting yields the associated characteristic equation

for the corresponding homogeneous Cauchy–Euler equation. A general solution to the homo- geneous equation for t > 0 is given by

(i) , if r1 and r2 are distinct real roots; (ii) ln t, if r is a repeated root;

(iii) if is a complex root.

A general solution to the nonhomogeneous equation is where is a particu- lar solution and is a general solution to the corresponding homogeneous equation. The method of variation of parameters (but not the method of undetermined coefficients) can be used to find a particular solution.

yh ypy ! yp $ yh,

/ $ i0c1t /cos(0 ln t) $ c2t

/sin(0 ln t), c1t

r $ c2t r

c1t r1 $ c2t

ar 2 $ (b " a)r $ c ! 0

y ! t r at2y– " bty¿ " cy $ f(t)

ay– $ by¿ $ cy ! k1 f1 $ k2 f2 .

k2y2k1y1

ay– $ by¿ $ cy ! f2 ,ay– $ by¿ $ cy ! f1

y2y1

y¿1y¿1 $ y¿2 y¿2 ! f AtB /ay¿1 y1 $ y¿2 y2 ! 0 y¿2y¿1

Review Problems 233

REVIEW PROBLEMS

In Problems 1–28, find a general solution to the given differential equation.

1. 2. 3. 4. 5. 6. 7. 8. 9. 10.

11. 12. 2y‡ " 3y– " 12y¿ $ 20y ! 0

t2x– AtB $ 5x AtB ! 0 , t 7 0 u– $ 11u ! 0 16z–" 56z¿$ 49z ! 0 25y–$20y¿$ 4y ! 0 36y– $ 24y¿ $ 5y ! 0 y– $ 8y¿ " 14y ! 0 6y– " 11y¿ $ 3y ! 0 9y–" 30y¿$ 25y ! 0 4y– " 4y¿ $ 10y ! 0 49y– $ 14y¿ $ y ! 0 y– $ 8y¿ " 9y ! 0

13. 14. 15. 16. 17. 18. 19. 20.

21. y– " 3y¿ $ 7y ! 7t2 " et 2y– " y ! t sin t 4y‡ $ 8y– " 11y¿ $ 3y ! 0 yA4B ! 120t y‡ $ 10y¿ " 11y ! 0 y‡ $ 3y– $ 5y¿ $ 3y ! 0 3y‡ $ 10y– $ 9y¿ $ 2y ! 0 y– " 4y¿ $ 7y ! 0 y– $ 16y ! tet

234 Chapter 4 Linear Second-Order Equations

TECHNICAL WRITING EXERCISES

1. Compare the two methods— undetermined coeffi- cients and variation of parameters—for determining a particular solution to a nonhomogeneous equation. What are the advantages and disadvantages of each?

2. Consider the differential equation

where b is a constant. Describe how the behavior of solutions to this equation changes as b varies.

3. Consider the differential equation d2y

dx2 $ cy ! 0 ,

d2y

dx2 $ 2b

dx $ y ! 0 ,

where c is a constant. Describe how the behavior of solutions to this equation changes as c varies.

4. For students with a background in linear algebra: Compare the theory for linear second-order equa- tions with that for systems of n linear equations in n unknowns whose coefficient matrix has rank n " 2. Use the terminology from linear algebra; for exam- ple, subspace, basis, dimension, linear transforma- tion, and kernel. Discuss both homogeneous and nonhomogeneous equations.

22.

23.

24.

25.

26.

27.

28.

In Problems 29–36, find the solution to the given initial value problem.

29. ;

30. ;

31. ;

32. ;

33. ;

34. ;

35. 36. ;

y A0B ! "3 , y¿ A0B ! 39y– $ 12y¿ $ 4y ! 0 y A0B ! 1 , y¿ A0B ! 2y–(u) $ y(u) ! sec u ;y A0B ! 5 , y¿ A0B ! 1

y– $ 5y¿ " 14y ! 0 y– A0B ! "12 y A0B ! "3 , y¿ A0B ! "6 , y‡ " 12y– $ 27y¿ $ 40y ! 0

y A0B ! 1 , y¿ A0B ! "11 /2 4y– " 4y¿ $ 5y ! 0 y A0B ! 2 , y¿ A0B ! "8y– " 2y¿ $ 10y ! 6 cos 3t " sin 3t y A0B ! 3 , y¿ A0B ! 0 y–(u) $ 2y¿(u) $ y(u) ! 2 cos u

y¿ A0B ! "2y A0B ! 1 , y– $ 4y¿ $ 7y ! 0 y– ! 5x"1y¿ " 13x"2y , x 7 0 x2y– $ 2xy¿ " 2y ! 6x"2 $ 3x , x 7 0 y– $ 6y¿ $ 15y ! e2t $ 75 4y– " 12y¿ $ 9y ! e5t $ e3t 10y– $ y¿ " 3y ! t " et/2 y– AuB $ 16y AuB ! tan 4uy– " 8y¿ " 33y ! 546 sin t 37. Use the mass–spring oscillator analogy to decidewhether all solutions to each of the following differ-

ential equations are bounded as (a) (b) (c) (d) (e) (f) (g)

38. A 3-kg mass is attached to a spring with stiffness k ! 75 N/m, as in Figure 4.1, page 153. The mass is displaced m to the left and given a velocity of 1 m/sec to the right. The damping force is negligible. Find the equation of motion of the mass along with the amplitude, period, and frequency. How long after release does the mass pass through the equilibrium position?

39. A 32-lb weight is attached to a vertical spring, caus- ing it to stretch 6 in. upon coming to rest at equilib- rium. The damping constant for the system is 2 lb- sec/ft. An external force ! 4 cos 8t lb is applied to the weight. Find the steady-state solution for the system. What is its resonant frequency?

F AtB

1 /4

y– " t2y¿ " y ! 0 y– $ t2y¿ $ y ! 0 y– $ A3 $ sin tBy ! 0 y– $ y8 ! 0 y– $ y7 ! 0 y– " t4y ! 0 y– $ t4y ! 0

t S $q.

A Nonlinear Equations Solvable by First-Order Techniques

Certain nonlinear second-order equations—namely, those with dependent or independent vari- ables missing—can be solved by reducing them to a pair of first-order equations. This is accom- plished by making the substitution , where x is the independent variable.

(a) To solve an equation of the form in which the dependent variable y is missing, setting (so that yields the pair of equations

Because is a first-order equation, we have available the techniques of Chapter 2 to solve it for w . Once w is determined, we integrate it to obtain .

Using this method, solve

(b) To solve an equation of the form in which the independent variable x is missing, setting yields, via the chain rule,

Thus, is equivalent to the pair of equations

(1)

(2)

In equation (1) notice that y plays the role of the independent variable; hence, solving it yields . Then substituting into (2), we obtain a separable equation that determines .

Using this method, solve the following equations:

i(i) (ii)

(c) Suspended Cable. In the study of a cable suspended between two fixed points (see Figure 4.36 on page 236), one encounters the initial value problem

where a is a constant. Solve this initial value problem for y. The resulting curve is called a catenary.

A( 0B d2y

dx2 !

a B1 $ adydxb 2 ; y A0B ! a , y¿ A0B ! 0 ,

d2y

dx2 $ y

dx ! 0 .2y

d2y

dx2 ! 1 $ ady

dx b 2 .

y AxB w AyBw A yB dy dx

! w .

w dw dy

! F Ay, wB ,y– ! F Ay, y¿ B d2y

dx2 !

dx !

dy dy

dx ! w

dy .

w ! dy/dx y– ! F Ay, y¿ B2xy– " y¿ $

1 y¿ ! 0 , x 7 0 .

y AxBAxBAxBw¿ ! F Ax, wB y¿ ! w . w¿ ! F Ax, wB , w¿ ! y–)w ! y¿

y– ! F Ax, y¿ Bw ! dy/dx

Group Projects for Chapter 4

235

B Apollo Reentry Courtesy of Alar Toomre, Massachusetts Institute of Technology

Each time the Apollo astronauts returned from the moon circa 1970, they took great care to reenter Earth’s atmosphere along a path that was only a small angle from the horizontal. (See Figure 4.37.) This was necessary in order to avoid intolerably large “g” forces during their reentry.

To appreciate their grounds for concern, consider the idealized problem

where K and H are constants and distance s is measured downrange from some reference point on the trajectory, as shown in the figure. This approximate equation pretends that the only force on the capsule during reentry is air drag. For a bluff body such as the Apollo, drag is proportional to the square of the speed and to the local atmospheric density, which falls off exponentially with height. Intuitively, one might expect that the deceleration predicted by this model would depend heavily on the constant K (which takes into account the vehicle’s mass, area, etc.); but, remark- ably, for capsules entering the atmosphere (at “ ”) with a common speed the maxi- mum deceleration turns out to be independent of K.

(a) Verify this last assertion by demonstrating that this maximum deceleration is just [Hint: The independent variable t does not appear in the differential equa-

tion, so it is helpful to make the substitution ; see Project A, part (b).] (b) Also verify that any such spacecraft at the instant when it is decelerating most fiercely

will be traveling exactly with speed having by then lost almost 40% of its original velocity.

(c) Using the plausible data V0 ! 11 km/sec and H ! 10/(sin /) km, estimate how small had to be chosen so as to inconvenience the returning travelers with no more than 10 g’s.

V0/1e,y ! ds/dt V 20 / A2eHB.

V0,s ! "q

d2s dt2

! "Kes /H ads dt b 2 ,

236 Chapter 4 Linear Second-Order Equations

Figure 4.36 Suspended cable

Distance s

s = o

Figure 4.37 Reentry path

C Simple Pendulum In Section 4.8, we discussed the simple pendulum consisting of a mass m suspended by a rod of length " having negligible mass and derived the nonlinear initial value problem

(3)

where g is the acceleration due to gravity and is the angle the rod makes with the vertical at time t (see Figure 4.18, page 210). Here it is assumed that the mass is released with zero velocity at an initial angle We would like to determine the equation of motion for the pen- dulum and its period of oscillation.

(a) Use equation (3) and the energy integral lemma discussed in Section 4.8 to show that

and hence

(b) Use the trigonometric identity to express dt by

(c) Make the change of variables and show that the elapsed time, T, for the pendulum to fall from the angle (corresponding to ) to the angle (corresponding to ), when , is given by

(4) ,

where . (d) The period P of the pendulum is defined to be the time required for it to swing from one

extreme to the other and back—that is, from to and back to . Show that the period is given by

(5) .

The integral in (5) is called an elliptic integral of the first kind and is denoted by . As you might expect, the period of the simple pendulum depends on the

length of the rod and the initial displacement . In fact, a check of an elliptic integral table will show that the period nearly doubles as the initial displacement increases from

to (for fixed ). What happens as approaches ?

(e) From equation (5) show that

(6) , where .F Ak, £ B :! #£ f!0

df21 " k2 sin 2f" T $ P/4 ! A /g F Ak, £ B pa/15p/16p/8

a/ F Ak, p/2B

P ! 4A /g #p/2 0

df21 " k2 sin 2f a"aa

k :! sin Aa/2B T ! #

dt ! " # b

1 2A /g du2 sin 2 Aa/2B " sin 2 Au/2B ! "A /g #£p/2 df21 " k2 sin 2f

a % b % 0f ! £u ! b f ! p/2u ! a

sin Au/2B ! sin Aa/2B sin f dt ! "

1 2A /g du3 sin2 Aa/2B " sin2 Au/2B .

cos x ! 1 " 2 sin2 Ax/2B dt ! "A /2g du3cos u " cos a . adu

dt b 2 ! 2g/ Acos u " cos aB

a, 0 6 a 6 p.

u AtB d2U

dt2 "

g O sin U $ 0 ; U A0B $ A , U# A0B $ 0 ,

Group Projects for Chapter 4 237

For fixed k, has an “inverse,” denoted by sn , that satisfies if and only if sn = sin . The function sn is called a Jacobi elliptic function and has many properties that resemble those of the sine function. Using the Jacobi ellip- tic function sn , express the equation of motion for the pendulum in the form

(7) .

(f) Take = 1m, g = 9.8 m/sec, radians. Use Runge–Kutta algorithms or tabu- lated values of the Jacobi elliptic function to determine the period of the pendulum.

D Linearization of Nonlinear Problems A useful approach to analyzing a nonlinear equation is to study its linearized equation, which is obtained by replacing the nonlinear terms by linear approximations. For example, the nonlinear equation

(8)

which governs the motion of a simple pendulum, has

(9)

as a linearization for small (The nonlinear term sin has been replaced by the linear approxi- mation u.)

A general solution to equation (8) involves Jacobi elliptic functions (see Project C), which are rather complicated, so let’s try to approximate the solutions. For this purpose we consider two methods: Taylor series and linearization.

(a) Derive the first six terms of the Taylor series about t ! 0 of the solution to equation (8) with initial conditions (The Taylor series method is discussed in Project A of Chapter 1 and Section 8.1.)

(b) Solve equation (9) subject to the same initial conditions (c) On the same coordinate axes, graph the two approximations found in parts (a) and (b). (d) Discuss the advantages and disadvantages of the Taylor series method and the

linearization method. (e) Give a linearization for the initial value problem.

, for x small. Solve this linearized problem to obtain an approximation for the nonlinear problem.

x– AtB $ 0.1 3 1 " x2 AtB 4 x¿ AtB $ x AtB ! 0 x A0B ! 0.4 , x¿ A0B ! 0 u A0B ! p/12, u¿ A0B ! 0.

u A0B ! p/12, u¿ A0B ! 0.

uu.

d2u dt2

$ u ! 0

d2u dt2

$ sin u ! 0 ,

a ! p/4/

b ! 2 arcsin Ek sn 3 k,Ag/ A"T $ P/4B 4 F, 0 - T - P/4 Ak, uB

Ak, uB£Ak, uB u ! F Ak, £ BAk, uBF Ak, £ B 238 Chapter 4 Linear Second-Order Equations

E Convolution Method The convolution of two functions g and f is the function defined by

The aim of this project is to show how convolutions can be used to obtain a particular solution to a nonhomogeneous equation of the form

(10) where a, b, and c are constants,

(a) Use Leibniz’s rule,

to show the following:

assuming y and f are sufficiently differentiable. (b) Let be the solution to the homogeneous equation that satis-

fies Show that is the particular solution to equation (10) satisfying

is the unique solution to the initial value problem

(11)

(d) Use the result of part (c) to determine the solution to each of the following initial value prob- lems. Carry out all integrations and express your answers in terms of elementary functions.

ii(i) i(ii) (iii)

F Undetermined Coefficients Using Complex Arithmetic

The technique of undetermined coefficients described in Section 4.5 can be streamlined with the aid of complex arithmetic and the properties of the complex exponential function. The essential formulas are

Re e Aa$ibB t ! eat cos bt , Im e Aa$ibB t ! eat sin bt . d dt

e Aa$ibB t ! Aa $ ibBe Aa$ibB t ,e Aa$ibB t ! eat Acos bt $ i sin btB ,

y– " 2y¿ $ y ! 1tet ; y A0B ! 2 , y¿ A0B ! 0 2y– $ y¿ " y ! e"t sin t ; y A0B ! 1 , y¿ A0B ! 1 y– $ y ! tan t ; y A0B ! 0 , y¿ A0B ! "1 ay– $ by¿ $ cy ! f AtB ; y A0B ! Y0 , y¿ A0B ! Y1 .

Ays * f B AtB $ yk AtB ysy A0B ! Y0, y¿ A0B ! Y1, ay– $ by¿ $ cy ! 0yk AtB y A0B ! y¿ A0B ! 0.

ys * fys A0B ! 0, y¿s A0B ! 1/a. ay– $ by¿ $ cy ! 0ys AtB Ay * f B– AtB ! Ay– * f B AtB $ y¿ A0B f AtB $ y A0B f ¿ AtB ,Ay * f B ¿ AtB ! Ay¿ * f B AtB $ y A0B f AtB d dt

# t

a h At, yB dy ! # t

a 0h 0t At, yB dy $ h At, tB ,

a ( 0.ay– $ by¿ $ cy ! f AtB , Ag * f B AtB J # t

0 g At " YB f AYB dy . g * f

Group Projects for Chapter 4 239

(a) From the preceding formulas derive the equations

(12)

(13)

Now consider a second-order equation of the form

(14)

where a, b, and c are real numbers and g is of the special form

(15)

with the and real numbers. Such a function can always be expressed as the real or imaginary part of a function involving the complex exponential. For example, using equation (12), one can quickly check that

(16)

where

(17)

Now suppose for the moment that we can find a complex-valued solution Y to the equation

(18)

Then, since a, b, and c are real numbers, we get a real-valued solution y to (14) by sim- ply taking the real part of Y; that is, y ! Re Y solves (14). (Recall that in Lemma 2, page 169, we proved this fact for homogeneous equations.) Thus, we need focus only on finding a solu- tion to (18).

The method of undetermined coefficients implies that any differential equation of the form

(19)

has a solution of the form

(20)

where An, . . . , A0 are complex constants and s is the multiplicity of as a root of the auxil- iary equation for the corresponding homogeneous equation . We can solve for the unknown constants Aj by substituting (20) into (19) and equating coefficients of like terms. With these facts in mind, we can (for the small price of using complex arithmetic) dispense with the methods of Section 4.5 and avoid the unpleasant task of computing derivatives of a function like

, which involves both exponential and trigonometric factors. Carry out this procedure to determine particular solutions to the following equations:

(b) (c) (d)

The use of complex arithmetic not only streamlines the computations but also proves very useful in analyzing the response of a linear system to a sinusoidal input. Electrical engineers make good use of this in their study of RLC circuits by introducing the concept of impedance.

y– " 2y¿ $ 10y ! tet sin 3t . y– $ y ! e"t Acos 2t " 3 sin 2tB .y– " y¿ " 2y ! cos t " sin 2t .

e3t A2 $ 3t $ t2Bsin A2tB L 3Y 4 ! 0a $ ibYp AtB ! t

se Aa*ibB t 3Ant n $ p $ A1t $ A0 4 , L 3Y 4 ! e Aa*ibB t 3 Aan $ ibnBt n $ p $ Aa1 $ ib1Bt $ Aa0 $ ib0B 4

L 3Y 4 ! aY– $ bY¿ $ cY ! G . G AtB $ e AA' iBB t 3 Aan " ibnB tn " p " Aa1 " ib1B t " Aa0 " ib0B 4 . g AtB $ Re 3G AtB 4 ,

baj’s, bj’s, a,

g AtB $ eAt 3 Aantn " p " a1t " a0B cos Bt " Abntn " p " b1t " b0B sin Bt 4 , L 3 y 4 J ay– $ by¿ $ cy ! g ,

Im 3 Aa $ ibBe Aa$ibBt 4 ! eat Ab cos bt $ a sin btB . Re 3 Aa $ ibBe Aa$ibBt 4 ! eat Aa cos bt " b sin btB , 240 Chapter 4 Linear Second-Order Equations

G Asymptotic Behavior of Solutions In the application of linear systems theory to mechanical problems, we have encountered the equation

(21)

where p and q are positive constants with p2 . 4q and is a forcing function for the system. In many cases it is important for the design engineer to know that a bounded forcing function gives rise only to bounded solutions. More specifically, how does the behavior of for large values of t affect the asymptotic behavior of the solution? To answer this question, do the following:

(a) Show that the homogeneous equation associated with equation (21) has two linearly independent solutions given by

where and

(b) Let be a continuous function defined on the interval Use the variation of para- meters formula to show that any solution to (21) on can be expressed in the form

(22)

(c) Assuming that f is bounded on (that is, there exists a constant K such that for all ), use the triangle inequality and other properties of the

absolute value to show that given in (22) satisfies

for all t + 0.

(d) In a similar fashion, show that if and are two bounded continuous functions on such that for all t + t0, and if is a solution to (21) with f ! f1 and is a solution to (21) with f ! f2, then

for all t + t0, where M is a constant that depends on and but not on t.

(e) Now assume as where F0 is a constant. Use the result of part (d) to prove that any solution to (21) must satisfy as [Hint: Choose

.]f1 ! f, f2 " F0, f1 ! f, f2 " F0/q t S $q.f AtB S F0/qf t S $q,f AtB S F0 f2f1

0f1 AtB " f2 AtB 0 - Meat $ 2e0a 0b A1 " eaAt"t0B B f2

f10 f1 AtB " f2 AtB 0 - e3 0, q B f2 AtBf1 AtB 0 y AtB 0 - A 0 c1 0 $ 0 c2 0 Beat $ 2K0a 0b A1 " eatB

y AtBv % 00 f Av B 0 - K 3 0, q B $

1 b

eat sin bt # t

0 f AvBe"av cos bv dv .

" 1 b

eat cos bt # t

0 f AvBe"av sin bv dv

y AtB ! c1eat cos bt $ c2eat sin bt 3 0, q B3 0, q B.f AtB b ! 1224q " p2 .a ! "p/2 6 0e

at cos bt , eat sin bt ,

f AtB f AtBy– $ py¿ $ qy ! f AtB ,

Group Projects for Chapter 4 241

242

Note that the volume of liquid in each tank remains constant at 24 L because of the balance between the inflow and outflow volume rates. Hence, we have two unknown functions of t: the mass of salt in tank A and the mass of salt in tank B. By focusing attention on one tank at a time, we can derive two equations relating these unknowns. Since the system is being flushed with fresh water, we expect that the salt content of each tank will diminish to zero as

To formulate the equations for this system, we equate the rate of change of salt in each tank with the net rate at which salt is transferred to that tank. The salt concentration in tank A is kg/L, so the upper interconnecting pipe carries salt out of tank A at a rate of kg/min; similarly, the lower interconnecting pipe brings salt into tank A at the rate kg/min (the concentration of salt in tank B is kg/L). The fresh water inlet, of course, transfers no salt (it simply maintains the volume in tank A at 24 L). From our premise,

dx dt ! input rate " output rate ,

y/24 2y/24 8x/24x AtB /24

t S !q.

y AtBx AtB

Introduction to Systems and Phase Plane Analysis

CHAPTER 5

INTERCONNECTED FLUID TANKS5.1 Two large tanks, each holding 24 liters of a brine solution, are interconnected by pipes as shown in Figure 5.1. Fresh water flows into tank A at a rate of 6 L/min, and fluid is drained out of tank B at the same rate; also 8 L/min of fluid are pumped from tank A to tank B, and 2 L/min from tank B to tank A. The liquids inside each tank are kept well stirred so that each mixture is homogeneous. If, initially, the brine solution in tank A contains x0 kg of salt and that in tank B initially contains y0 kg of salt, determine the mass of salt in each tank at time .†t 7 0

†For this application we simplify the analysis by assuming the lengths and volumes of the pipes are sufficiently small that we can ignore the diffusive and advective dynamics taking place therein.

6 L/min x(t)

24 L 24 L

x(0) = x0 kg

A 8 L/min

y(t)

y(0) = y0 kg

6 L/min

2 L/min

Figure 5.1 Interconnected fluid tanks

so the rate of change of the mass of salt in tank A is

The rate of change of salt in tank B is determined by the same interconnecting pipes and by the drain pipe, carrying away kg/min:

The interconnected tanks are thus governed by a system of differential equations:

(1)

Although both unknowns and appear in each of equations (1) (they are “coupled”), the structure is so transparent that we can obtain an equation for y alone by solving the second equation for x,

(2)

and substituting (2) in the first equation to eliminate x:

This last equation, which is linear with constant coefficients, is readily solved by the methods of Section 4.2. Since the auxiliary equation

has roots , , a general solution is given by

(3)

Having determined y, we use equation (2) to deduce a formula for x:

(4)

Formulas (3) and (4) contain two undetermined parameters, c1 and c2, which can be adjusted to meet the specified initial conditions:

x A0B " # 1 2

c1 ! 1 2

c2 " x0 , y A0B " c1 ! c2 " y0 , x AtB " 3 a# c1

2 e#t/2 #

c2 6

e#t/6b ! c1e#t/2 ! c2e#t/6 " # 12 c1e#t/2 ! 12 c2e#t/6 . y AtB " c1e#t/2 ! c2e#t/6 .

#1/6#1/2

3r 2 ! 2r ! 1 4

" 0

3y– ! 2y¿ ! 1 4

y " 0 .

3y– ! y¿ " #y¿ # 1 3

y ! 1 12

y ,

A3y¿ ! yB ¿ " # 1 3

A3y¿ ! yB ! 1 12

y ,

x " 3y¿ ! y ,

y AtBx AtB y¿ " 1

3 x #

1 3

y .

x¿ " # 1 3

x ! 1 12

y ,

dy dt

" 8 24

x # 2 24

y # 6 24

y " 1 3

x # 1 3

y .

6y/24

dx dt

" 2 24

y # 8 24

x " 1 12

y # 1 3

x .

Section 5.1 Interconnected Fluid Tanks 243

Thus, the mass of salt in tanks A and B at time t are, respectively,

(5)

The ad hoc elimination procedure that we used to solve this example will be generalized and formalized in the next section, to find solutions of all linear systems with constant coeffi- cients. Furthermore, in later sections we will show how to extend our numerical algorithms for first-order equations to general systems and will consider applications to coupled oscillators and electrical systems.

It is interesting to note from (5) that all solutions of the interconnected-tanks problem tend to the constant solution as t . (This is of course consistent with our phys- ical expectations.) This constant solution will be identified as a stable equilibrium solution in Section 5.4, in which we introduce phase plane analysis. It turns out that, for a general class of systems, equilibria can be identified and classified so as to give qualitative information about the other solutions even when we cannot solve the system explicitly.

S !qx AtB ! 0, y AtB ! 0

ay0 ! 2x0 2

b e#t/6 .y AtB " ay0 # 2x0 2

b e#t/2 ! ay0 ! 2x0

4 b e#t/6 , x AtB " # ay0 # 2x0

4 b e#t/2 !

c1 " y0 # 2x0

2 , c2 "

y0 ! 2x0 2

244 Chapter 5 Introduction to Systems and Phase Plane Analysis

5.2 DIFFERENTIAL OPERATORS AND THE ELIMINATIONMETHOD FOR SYSTEMS The notation was devised to suggest that the derivative of a function y is the

result of operating on the function y with the differentiation operator . Indeed, second d dt

y¿(t) " dy dt

" d dt

†Some authors utilize the identity operator I, defined by and write more formally D2 ! 4D ! 3I instead of D2 ! 4D ! 3.

I 3 y 4 " y,

derivatives are formed by iterating the operation: . Commonly, the sym-y–(t) " d2y

dt2 "

d dt

bol D is used instead of , and the second-order differential equation

is represented† by

D2y ! 4Dy ! 3y " D2 ! 4D ! 3 " 0 .

So, we have implicitly adopted the convention that the operator “product,” D times D, is interpreted as the composition of D with itself, when it operates on functions: D2y means D D ; i.e., the second derivative. Similarly, the product D ! 3 D ! 1 operates on a function via

D ! 3 D ! 1 " D ! 3 D ! 1 " D ! 3 y$ ! y

" D y$ ! y ! 3 y$ ! y

" y% ! y$ ! 3y$ ! 3y " y% ! 4y$ ! 3y " D2 ! 4D ! 3 .3 y 4BABABA 4343 43BA43 y 4BA3BA3 y 4BABA BABA B3 y 4A

3 y 4BA y– ! 4y¿ ! 3y " 0

d dt

Thus, D ! 3 D ! 1 is the same operator as D2 ! 4D ! 3; when they are applied to twice-differentiable functions, the results are identical.

Show that the operator D ! 1 D ! 3 is also the same as D2 ! 4D ! 3.

For any twice-differentiable function y(t), we have

D ! 1 D ! 3 " D ! 1 D ! 3 " D ! 1 y$ ! 3y

" D y$ ! 3y ! 1 y$ ! 3y " y% ! 3y$ ! y$ ! 3y

" y% ! 4y$ ! 3y " D2 ! 4D ! 3 .

Hence, D ! 1 D ! 3 " D2 ! 4D ! 3. ◆

Since D ! 1 D ! 3 " D ! 3 D ! 1 " D2 ! 4D ! 3, it is tempting to generalize and propose that one can treat expressions like aD2 ! bD ! c as if they were ordinary polynomials in D. This is true, as long as we restrict the coefficients a, b, c to be constants. The following example, which has variable coefficients, is instructive.

Show that D ! 3t D is not the same as D D ! 3t .

With y t as before,

D ! 3t D " D ! 3t " y% ! 3ty$ ;

D D ! 3t " D y$ ! 3ty " y% ! 3y ! 3ty$ .

They are not the same! ◆

Because the coefficient 3t is not a constant, it “interrupts” the interaction of the differenti- ation operator D with the function y(t). As long as we only deal with expressions like aD2 ! bD ! c with constant coefficients a, b, and c, the “algebra” of differential operators fol- lows the same rules as the algebra of polynomials. (See Problem 39 for elaboration on this point.)

This means that the familiar elimination method, used for solving algebraic systems like

can be adapted to solve any system of linear differential equations with constant coefficients. In fact, we used this approach in solving the system that arose in the interconnected tanks prob- lem of Section 5.1. Our goal in this section is to formalize this elimination method so that we can tackle more general linear constant coefficient systems.

We first demonstrate how the method applies to a linear system of two first-order differen- tial equations of the form

where a1, a2, . . . , a8 are constants and is the function pair to be determined. In opera- tor notation this becomes

Aa5D ! a6B 3 x 4 ! Aa7D ! a8B 3 y 4 " f2 . Aa1D ! a2B 3 x 4 ! Aa3D ! a4B 3 y 4 " f1 , y AtBx AtB, a5x¿ AtB ! a6x AtB ! a7y¿ AtB ! a8y AtB " f2 AtB ,

a1x¿ AtB ! a2x AtB ! a3y¿ AtB ! a4y AtB " f1 AtB ,

2x # y ! 3z " 6 , x ! y # z " 0 ,

3x # 2y ! z " 4 ,

433 y 4BA 3 y¿ 4BA3 y 4BA BA BABA

BABABABA BABA 3 y 4BA BABA4343 43BA43 y 4BA3BA3 y 4BABA

BABA BABA

Section 5.2 Differential Operators and the Elimination Method for Systems 245

Example 1

Example 2

Solution

Solve the system

(1)

The alert reader may observe that since is absent from the first equation, we could use the latter to express y in terms of x and and substitute into the second equation to derive an “uncoupled” equation containing only x and its derivatives. However, this simple trick will not work on more general systems (Problem 18 is an example).

To utilize the elimination method, we first write the system using the operator notation:

(2)

Imitating the elimination procedure for algebraic systems, we can eliminate x from this system by adding 4 times the first equation to applied to the second equation. This gives

which simplifies to

(3)

Now equation (3) is just a second-order linear equation in y with constant coefficients that has the general solution

(4)

which can be found using undetermined coefficients. To find , we have two options.

Method 1. We return to system (2) and eliminate y. This is accomplished by “multiplying” the first equation in (2) by and the second equation by #4 and then adding to obtain

This equation can likewise be solved using undetermined coefficients to yield

(5)

where we have taken K1 and K2 to be the arbitrary constants, which are not necessarily the same as C1 and C2 used in formula (4).

It is reasonable to expect that system (1) will involve only two arbitrary constants, since it consists of two first-order equations. Thus, the four constants C1, C2, K1, and K2 are not inde- pendent. To determine the relationships, we substitute the expressions for and given in (4) and (5) into one of the equations in (1), say, the first one. This yields

which simplifies toA4C1 # 8K1Be#5t ! A4C2 # 2K2Bet " 0 . 3K1e

#5t ! 3K2e t ! 24t ! 15 # 4C1e

#5t # 4C2e t # 24t # 8 ! 1 ,

#5K1e #5t ! K2e

t ! 8 "

y AtBx AtB

x AtB " K1e#5t ! K2et ! 8t ! 5 , AD2 ! 4D # 5B 3 x 4 " 7 # 40t .AD ! 7B

x AtB y AtB " C1e#5t ! C2et ! 6t ! 2 , AD2 ! 4D # 5B 3 y 4 " 14 # 30t . A16 ! AD # 3B AD ! 7BB 3 y 4 " 4 # 1 ! AD # 3B 310t 4 " 4 ! 10 # 30t ,AD # 3B #4x ! AD ! 7B 3 y 4 " 10t . AD # 3B 3 x 4 ! 4y " 1 ,

x¿ y¿

y¿ AtB " 4x AtB # 7y AtB ! 10t . x¿ AtB " 3x AtB # 4y AtB ! 1 , 246 Chapter 5 Introduction to Systems and Phase Plane Analysis

Example 3

Solution

Because are linearly independent functions on any interval, this last equation holds for all t only if

Therefore, and K2 " 2C2. A solution to system (1) is then given by the pair

(6)

As you might expect, this pair is a general solution to (1) in the sense that any solution to (1) can be expressed in this fashion.

Method 2. A simpler method for determining once is known is to use the system to obtain an equation for in terms of and . In this example we can directly solve the second equation in (1) for :

Substituting as given in (4) yields

which agrees with (6). ◆

The above procedure works, more generally, for any linear system of two equations and two unknowns with constant coefficients regardless of the order of the equations. For example, if we let , , , and denote linear differential operators with constant coefficients (i.e., polynomials in D), then the method can be applied to the linear system

Because the system has constant coefficients, the operators commute (e.g., ) and we can eliminate variables in the usual algebraic fashion. Eliminating the variable y gives

(7)

where Similarly, eliminating the variable x yields

(8)

where is a differential operator of order n, then a general solution for (7) contains n arbitrary constants, and a general solution for (8) also contains n arbitrary constants. Thus, a total of 2n constants arise. However, as we saw in Exam- ple 3, there are only n of these that are independent for the system; the remaining constants can be expressed in terms of these.† The pair of general solutions to (7) and (8) written in terms of the n independent constants is called a general solution for the system.

g2 J L1[ f2] # L3[ f1] . Now if L1L4 # L2L3

AL1L4 # L2L3B 3 y 4 " g2 ,g1 J L4[ f1] # L2[ f2] . AL1L4 # L2L3B 3 x 4 " g1 ,

L2L4 " L4L2

L3 3 x 4 ! L4 3 y 4 " f2 . L1 3 x 4 ! L2 3 y 4 " f1 , L4L3L2L1

" 1 2

C1e #5t ! 2C2e

t ! 8t ! 5 ,

x AtB " 1 4

3#5C1e#5t ! C2et ! 6 4 ! 74 3C1e#5t ! C2et ! 6t ! 2 4 # 52 ty AtB x AtB " 1

4 y¿ AtB ! 7

4 y AtB # 5

2 t .

x AtB y¿ AtBy AtBx AtB y AtBx AtB

x AtB " 1 2

C1e #5t ! 2C2e

t ! 8t ! 5 , y AtB " C1e#5t ! C2et ! 6t ! 2 . K1 " C1/2

4C1 # 8K1 " 0 and 4C2 # 2K2 " 0 .

et and e#5t

Section 5.2 Differential Operators and the Elimination Method for Systems 247

†For a proof of this fact, see Ordinary Differential Equations, by M. Tenenbaum and H. Pollard (Dover, New York, 1985), Chapter 7.

If it turns out that is the zero operator, the system is said to be degenerate. As with the anomalous problem of solving for the points of intersection of two parallel or coin- cident lines, a degenerate system may have no solutions, or if it does possess solutions, they may involve any number of arbitrary constants (see Problems 23 and 24).

L1L4 # L2L3

248 Chapter 5 Introduction to Systems and Phase Plane Analysis

Find a general solution for

(9)

We begin by expressing the system in operator notation:

(10)

Here Eliminating y gives [see (7)]:

which reduces to

(11)

Since is third order, we should expect three arbitrary constants in a general solution to system (9).

Although the methods of Chapter 4 focused on solving second-order equations, we have seen several examples of how they extend in a natural way to higher-order

AD # 1B2 AD ! 1BAD # 1B2 AD ! 1B 3 x 4 " #2t # t2 . AD2 # 1B AD # 1B 3 x 4 " #2t # t2 , A AD2 # 1BD # AD ! 1B AD # 1BB 3 x 4 " D[#1] # AD ! 1B 3 t2 4 ,

L1 J D2 # 1, L2 J D ! 1, L3 J D # 1, and L4 J D.

AD # 1B 3 x 4 ! D 3 y 4 " t2 . AD2 # 1B 3 x 4 ! AD ! 1B 3 y 4 " #1 , x¿ AtB ! y¿ AtB # x AtB " t2 . x– AtB ! y¿ AtB # x AtB ! y AtB " #1 ,

Elimination Procedure for 2 # 2 Systems To find a general solution for the system

where , , , and are polynomials in :

(a) Make sure that the system is written in operator form. (b) Eliminate one of the variables, say, y, and solve the resulting equation for . If the

system is degenerate, stop! A separate analysis is required to determine whether or not there are solutions.

(c) (Shortcut) If possible, use the system to derive an equation that involves but not its derivatives. [Otherwise, go to step (d).] Substitute the found expression for into this equation to get a formula for . The expressions for , give the desired general solution.

(d) Eliminate x from the system and solve for . [Solving for gives more constants—in fact, twice as many as needed.]

(e) Remove the extra constants by substituting the expressions for and into one or both of the equations in the system. Write the expressions for and in terms of the remaining constants.

y AtBx AtBy AtBx AtB y AtBy AtB y

AtBx AtBy AtB x AtBy AtB x AtB

D " d/dtL4L3L2L1 L3 3 x 4 ! L4 3 y 4 " f2 ,L1 3 x 4 ! L2 3 y 4 " f1 ,

Example 4

Solution

equations.† Applying this strategy to the third-order equation (11), we observe that the corre- sponding homogeneous equation has the auxiliary equation with roots

Hence, a general solution for the homogeneous equation is

To find a particular solution to (11), we use the method of undetermined coefficients with Substituting into (11) and solving for A, B, and C yields (after a little

algebra)

Thus, a general solution to equation (11) is

(12)

To find , we take the shortcut described in step (c) of the elimination procedure box. Subtracting the second equation in (10) from the first, we find

so that

Inserting the expression for , given in (12), we obtain

(13)

The formulas for in (12) and in (13) give the desired general solution to (9). ◆

The elimination method also applies to linear systems with three or more equations and unknowns; however, the process becomes more cumbersome as the number of equations and unknowns increases. The matrix methods presented in Chapter 9 are better suited for handling larger systems. Here we illustrate the elimination technique for a 3 & 3 system.

Find a general solution to

(14)

We begin by expressing the system in operator notation:

(15)

Eliminating z from the first two equations (by adding them) and then from the last two equations yields (after some algebra, which we omit) we find

(16) # AD # 1B 3 x 4 ! AD # 1B AD # 4B 3 y 4 " 0 . AD # 2B 3 x 4 ! AD # 2B 3 y 4 " 0 , #4x ! 4y ! AD # 5B 3 z 4 " 0 . #x ! D 3 y 4 # z " 0 , AD # 1B 3 x 4 # 2y ! z " 0 , z¿ AtB " 4x AtB # 4y AtB ! 5z AtB . y¿ AtB " x AtB ! z AtB , x¿ AtB " x AtB ! 2y AtB # z AtB ,

y AtBx AtB y AtB " #C2et # 2C3e#t # t2 # 2t # 3 . # 3C1et ! C2 Atet ! 2etB ! C3e#t # 2 4 # 1 # t2 , y AtB " C1et ! C2 Atet ! etB # C3e#t # 2t # 4

x AtBy " AD # D2B 3 x 4 # 1 # t2 . AD2 # DB 3 x 4 ! y " #1 # t2 ,

y AtBx AtB " xh AtB ! xp AtB " C1et ! C2tet ! C3e#t # t2 # 4t # 6 . xp AtB " #t2 # 4t # 6 .

xp AtB " At2 ! Bt ! C. xh AtB " C1et ! C2tet ! C3e#t .r " 1, 1, #1.

Ar # 1B2 Ar ! 1B " 0 Section 5.2 Differential Operators and the Elimination Method for Systems 249

†More detailed treatment of higher-order equations is given in Chapter 6.

Example 5

Solution

On eliminating x from this 2 & 2 system, we eventually obtain

which has the general solution

(17)

Taking the shortcut approach, we add the two equations in (16) to get an expression for x in terms of y and its derivatives, which simplifies to

When we substitute the expression (17) for into this equation, we find

(18)

Finally, using the second equation in (14) to solve for , we get

and substituting in for and yields

(19)

The expressions for in (18), in (17), and in (19) give a general solution with C1, C2, and C3 as arbitrary constants. ◆

z AtBy AtBx AtBz AtB " 2C1et ! 4C2e2t ! 4C3e3t . x AtBy AtBz AtB " y¿ AtB # x AtB ,

z AtBx AtB " #C1et # 2C2e2t # C3e3t . y AtBx " AD2 # 4D ! 2B 3 y 4 " y– # 4y¿ ! 2y .

y AtB " C1et ! C2e2t ! C3e3t . AD # 1B AD # 2B AD # 3B 3 y 4 " 0 ,

250 Chapter 5 Introduction to Systems and Phase Plane Analysis

1. Let where . For , compute

(a) A[y] (b) B[A[y]] (c) B[y] (d) A[B[y]] (e) C[y]

2. Show that the operator (D # 1)(D ! 2) is the same as the operator D2 ! D # 2.

In Problems 3–18, use the elimination method to find a general solution for the given linear system, where dif- ferentiation is with respect to t.

3. 4.

5. 6.

10. x¿ ! y¿ ! 2x ! y " et 2x¿ ! y¿ # x # y " e#t , x¿ ! y¿ # x # y " sin t x¿ ! y¿ ! 2x " 0 , AD ! 1B 3 x 4 ! AD ! 4B 3 y 4 " 1AD # 3B 3 x 4 ! AD # 1B 3 y 4 " t ,AD # 1B 3 u 4 ! A2D ! 1B 3y 4 " 5 AD ! 1B 3 u 4 # AD ! 1B 3y 4 " et ,y¿" 4x # y # cos tx¿ ! y¿ ! y " 1

x¿" 3x # 2y ! sin t ,x¿ ! y¿ # x " 5 , y¿ " y # 4xx¿ # y¿ " 0 x¿ " x # y ,x¿ ! 2y " 0 ,

y " t3 # 8D " d/dt A " D # 1, B " D ! 2, C " D2 ! D # 2, 11. 12.

13. 14.

15. 16.

17.

18. ,

In Problems 19–21, solve the given initial value problem.

19.

20.

dy dt

" x ! 2y ; y A0B " #1 dx dt

" 2x ! y # e2t ; x A0B " 1 , dy dt

" #2x ! y ; y A0B " 0 dx dt

" 4x ! y ; x A0B " 1 , x– ! y¿ # x ! y " #1 x– ! y– # x¿ " 2t #x ! y– ! 2y " 0 x– ! 5x # 4y " 0 ,

2x ! d2y

dt2 " 0

dt " 3w ! 4z ! 17t

dx dt

! x ! dy dt

" e4t , dw dt

" 5w ! 2z ! 5t ,

#x ! dy dt

" 1 dy dt

" x ! y

dx dt

! y " t2 , dx dt

" x # 4y ,

4u ! D 3y 4 " 62u ! AD2 ! 2B 3y 4 " 0 D2 3u 4 ! D 3y 4 " 2 ,AD2 # 1B 3u 4 ! 5y" et , 5.2 EXERCISES

21.

22. Verify that the solution to the initial value problem

satisfies ! as

In Problems 23 and 24, show that the given linear system is degenerate. In attempting to solve the system, determine whether it has no solutions or infinitely many solutions.

23.

24.

In Problems 25–28, use the elimination method to find a general solution for the given system of three equa- tions in the three unknown functions , , . 25. 26.

27. 28.

In Problems 29 and 30, determine the range of values (if any) of the parameter that will ensure all solutions ,

of the given system remain bounded as

29. 30.

31. Two large tanks, each holding 100 L of liquid, are interconnected by pipes, with the liquid flowing from

dy dt

" x # y dy dt

" 3x ! y

dx dt

" #x ! ly , dx dt

" lx # y ,

t S !q.y AtB x AtBl z¿ " #x # 2y # z z¿ " #2x # 4y ! 4z y¿ " 6x # y , y¿ " 4y # 2z , x¿ " x ! 2y ! z , x¿ " 4x # 4z , z¿ " 3x ! 3y # z z¿ " 4x # 4y ! 5z y¿ " x ! 2y # z , y¿ " x ! z , x¿ " 3x ! y # z , x¿ " x ! 2y # z ,

z AtBy AtBx AtB D2 3 x 4 ! AD2 ! DB 3 y 4 " 0 D 3 x 4 ! AD ! 1B 3 y 4 " et , AD ! 2B 3 x 4 ! AD ! 2B 3 y 4 " 3et AD # 1B 3 x 4 ! AD # 1B 3 y 4 " #3e#2t ,

t S !q.0 y AtB 0 S !q0 x AtB 0 y¿ " 4x # 3y # 1 ; y A0B " 0 x¿ " 5x # 3y # 2 ; x A0B " 2 ,

d2y

dt2 " x ; y A0B " 1 , y¿ A0B " #1

d2x dt2

" y ; x A0B " 3 , x¿ A0B " 1 , Section 5.2 Differential Operators and the Elimination Method for Systems 251

tank A into tank B at a rate of 3 L/min and from B into A at a rate of 1 L/min (see Figure 5.2). The liq- uid inside each tank is kept well stirred. A brine solu- tion with a concentration of 0.2 kg/L of salt flows into tank A at a rate of 6 L/min. The (diluted) solu- tion flows out of the system from tank A at 4 L/min and from tank B at 2 L/min. If, initially, tank A con- tains pure water and tank B contains 20 kg of salt, determine the mass of salt in each tank at time

32. In Problem 31, 3 L/min of liquid flowed from tank A into tank B and 1 L/min from B into A. Determine the mass of salt in each tank at time if, instead, 5 L/min flows from A into B and 3 L/min flows from B into A, with all other data the same.

33. In Problem 31, assume that no solution flows out of the system from tank B, only 1 L/min flows from A into B, and only 4 L/min of brine flows into the sys- tem at tank A, other data being the same. Determine the mass of salt in each tank at time

34. Feedback System with Pooling Delay. Many physical and biological systems involve time delays. A pure time delay has its output the same as its input but shifted in time. A more common type of delay is pooling delay. An example of such a feedback system is shown in Figure 5.3 on page 252. Here the level of fluid in tank B determines the rate at which fluid enters tank A. Suppose this rate is given by

and V are positive constants and is the volume of fluid in tank B at time t. (a) If the outflow rate R3 from tank B is constant

and the flow rate R2 from tank A into B is , where K is a positive constant

and is the volume of fluid in tank A at time t, then show that this feedback system is governed by the system

dV2 dt

" KV1 AtB # R3 . dV1 dt

" a AV # V2 AtBB # KV1 AtB , V1 AtBR2 AtB " KV1 AtB

V2 AtBR1 AtB " a 3V # V2 AtB 4 , where a

t ' 0.

t ' 0

t ' 0.

6 L/min

0.2 kg/L

4 L/min

x(t)

100 L

x (0) = 0 kg

A 3 L/min

y(t)

100 L

y (0) = 20 kg

2 L/min

1 L/min

Figure 5.2 Mixing problem for interconnected tanks

(b) Find a general solution for the system in part (a) when , , and R3 " 10 L/min.

(c) Using the general solution obtained in part (b), what can be said about the volume of fluid in each of the tanks as ?

35. A house, for cooling purposes, consists of two zones: the attic area zone A and the living area zone B (see Figure 5.4). The living area is cooled by a 2-ton air conditioning unit that removes 24,000 Btu/hr. The heat capacity of zone B is per thousand Btu. The time constant for heat transfer between zone A and the outside is 2 hr, between zone B and the outside is 4 hr, and between the two zones is 4 hr. If the outside temperature stays at 100ºF, how warm does it eventually get in the attic zone A? (Heating and cooling of buildings was treated in Section 3.3.)

36. A building consists of two zones A and B (see Figure 5.5). Only zone A is heated by a furnace, which generates 80,000 Btu/hr. The heat capacity of zone A is per thousand Btu. The time constant for heat transfer between zone A and the outside is 4 hr,

1 /4ºF

1/ 2ºF

t S !q

K " 2 (min)#1,V " 20 La" 5 (min)#1

252 Chapter 5 Introduction to Systems and Phase Plane Analysis

4 hr 4 hr

24,000 Btu/hr

2 hr

Figure 5.4 Air-conditioned house with attic

R 1

Pump Tank A

Tank B

R 3

R 2

V 1 ( t )

V 2 ( t )

Power

Figure 5.3 Feedback system with pooling delay

between the unheated zone B and the outside is 5 hr, and between the two zones is 2 hr. If the outside temperature stays at 0ºF, how cold does it eventually get in the unheated zone B?

37. In Problem 36, if a small furnace that generates 1000 Btu/hr is placed in zone B, determine the coldest it would eventually get in zone B if zone B has a heat capacity of 2ºF per thousand Btu.

38. Arms Race. A simplified mathematical model for an arms race between two countries whose expendi- tures for defense are expressed by the variables and is given by the linear system

where a and b are constants that measure the trust (or distrust) each country has for the other. Determine whether there is going to be disarmament (x and y approach 0 as t increases), a stabilized arms race (x and y approach a constant as or a run- away arms race (x and y approach as

39. Let A, B, and C represent three linear differential operators with constant coefficients; for example,

where the a’s, b’s, and c’s are constants. Verify the following properties:†

(a) Commutative laws:

(b) Associative laws:

C :" c2D 2 ! c1D ! c0 ,

A:" a2D 2 ! a1D ! a0 , B:" b2D

2 ! b1D ! b0 ̌,

t S !q).!q t S !q),

dy dt

" 4x # 3y ! b ; y A0B " 4 , dx dt

" 2y # x ! a ; x A0B " 1 ,y AtB x AtB

4 hr

x(t) 2 hr y(t) 5 hr

Figure 5.5 Two-zone building with one zone heated

†We say that two operators A and B are equal if A[y] " B[y] for all functions y with the necessary derivatives.

Section 5.3 Solving Systems and Higher-Order Equations Numerically 253

5.3 SOLVING SYSTEMS AND HIGHER-ORDER EQUATIONS NUMERICALLY Although we studied a half-dozen analytic methods for obtaining solutions to first-order ordi- nary differential equations in Chapter 2, the techniques for higher-order equations, or systems of equations, are much more limited. Chapter 4 focused on solving the linear constant- coefficient second-order equation. The elimination method of the previous section is also restricted to constant-coefficient systems. And, indeed, higher-order linear constant-coefficient equations and systems can be solved analytically by extensions of these methods, as we will see in Chapters 6, 7, and 9.

However, if the equations—even a single second-order linear equation—have variable coefficients, the solution process is much less satisfactory. As will be seen in Chapter 8, the solutions are expressed as infinite series, and their computation can be very laborious (with the notable exception of the Cauchy–Euler, or equidimensional, equation). And we know virtually nothing about how to obtain exact solutions to nonlinear second-order equations.

Fortunately, all the cases that arise (constant or variable coefficients, nonlinear, higher- order equations or systems) can be addressed by a single formulation that lends itself to a mul- titude of numerical approaches. In this section we’ll see how to express differential equations as a system in normal form and then show how the basic Euler method for computer solution can be easily “vectorized” to apply to such systems. Although subsequent chapters will return to analytic solution methods, the vectorized version of the Euler technique or the more efficient Runge–Kutta technique will hereafter be available as fallback methods for numerical explo- ration of intractable problems.

Normal Form A system of m differential equations in the m unknown functions x1(t), x2(t), . . . , xm(t) expressed as

x$1(t) " f1 t, x1, x2, . . . , xm ,

(1) x$2(t) " f2 t, x1, x2, . . . , xm ,

x$m(t) " fm t, x1, x2, . . . , xm

is said to be in normal form. Notice that (1) consists of m first-order equations that collec- tively look like a vectorized version of the single generic first-order equation

(2) x$ " f(t, x) ,

and that the system expressed in equation (1) of Section 5.1 takes this form, as do equations (1) and (14) in Section 5.2. An initial value problem for (1) entails finding a solution to this system that satisfies the initial conditions

for prescribed values The importance of the normal form is underscored by the fact that most professional codes

for initial value problems presume that the system is written in this form. Furthermore, for a linear system in normal form, the powerful machinery of linear algebra can be readily applied. [Indeed, in Chapter 9 we will show how the solutions of the simple equation

can be generalized to constant-coefficient systems in normal form.]x¿ " ax x AtB " ceat

t0, a1, a2, . . . , am.

x1 At0B " a1, x2 At0B " a2, . . . , xm At0B " am

BAo BA BA

For these reasons it is gratifying to note that a (single) higher-order equation can always be converted to an equivalent system of first-order equations.

To convert an mth-order differential equation

(3)

into a first-order system, we introduce, as additional unknowns, the sequence of derivatives of y:

With this scheme, we obtain m # 1 first-order equations quite trivially:

(4)

The mth and final equation then constitutes a restatement of the original equation (3) in terms of the new unknowns:

(5)

If equation (3) has initial conditions then the system (4)–(5) has initial conditions

Convert the initial value problem

(6) y% t ! 3ty$ t ! y t 2 " sin t ; y 0 " 1, y$ 0 " 5

into an initial value problem for a system in normal form.

We first express the differential equation in (6) as

y% t " #3ty$ t # y t 2 ! sin t .

Setting x1 t :" y t and x2 t :" y$ t , we obtain

x$1 t " x2 t ,

x$2 t " #3tx2 t # x1 t 2 ! sin t .

The initial conditions transform to x1(0) " 1, x2 0 " 5 . ◆

Euler’s Method for Systems in Normal Form Recall from Section 1.4 that Euler’s method for solving a single first-order equation (2) is based on estimating the solution x at time (t0 ! h) using the approximation

(7) x t0 ! h " x t0 ! hx$ t0 " x t0 ! hf t0, x t0 ,

and that as a consequence the algorithm can be summarized by the recursive formulas

(8) tn!1 " tn ! h ,

(9) xn!1 " xn ! hf tn, xn , n " 0, 1, 2, . . .

[compare equations (2) and (3), Section 1.4]. Now we can apply the approximation (7) to each of the equations in the system (1):

(10) xk t0 ! h " xk t0 ! hxk$ t0 " xk t0 ! hfk t0, x1 t0 , x2 t0 , . . . , xm t0 ,BBABABAABABABABA BA

BBAABABABABA BABABABA

BABA BABABABA BABABA

BABABABABA x1 At0B " a1, x2 At0B " a2, . . . , xm At0B " am.y At0B " a1, y¿ At0B " a2, . . . , y Am#1B At0B " am,

x$m AtB ! yAmB AtB ! f At, x1, x2, . . . , xmB . x$m"1 AtB ! yAm"1B AtB ! xm AtB .o

x$2 AtB ! y– AtB ! x3 AtB , x$1 AtB ! y$ AtB ! x2 AtB , x1 AtB :! y AtB, x2 AtB :! y¿ AtB, . . . , xm AtB :! yAm#1B AtB . y AmB AtB " f At, y, y¿, p , y Am#1BB

254 Chapter 5 Introduction to Systems and Phase Plane Analysis

Example 1

Solution

and for k " 1, 2, . . . m, we are led to the recursive formulas

(11) tn!1 " tn ! h ,

(12)

x1; n!1 " x1; n ! hf1 tn, x1;n, x2;n, . . . , xm;n ,

x2; n!1 " x2; n ! hf2 tn, x1; n, x2; n, . . . , xm;n ,

xm; n!1 " xm;n ! hfm tn, x1; n, x2;n, . . . , xm;n n " 0, 1, 2, . . . .

Here we are burdened with the ungainly notation xp;n for the approximation to the value of the pth-function xp at time t " t0 ! nh; i.e., xp;n " xp t0 ! nh . However, if we treat the unknowns and right-hand members of (1) as components of vectors

x t : " x1 t , x2 t , . . . , xm t , f t, x " f1 t, x1, x2, . . . , xm , f2 t, x1, x2, . . . , xm , . . . , fm t, x1, x2, . . . , xm ,

then (12) can be expressed in the much neater form

(13) xn!1 " xn ! .

Use the vectorized Euler method with step size h " 0.1 to find an approximation for the solu- tion to the initial value problem

(14) y% t ! 4y$ t ! 3y t " 0; y 0 " 1.5 , y$ 0 " #2.5 ,

on the interval [0, 1].

For the given step size, the method will yield approximations for y 0.1 , y 0.2 , . . . , y 1.0 . To apply the vectorized Euler method to (14), we first convert it to normal form. Setting x1 " y and x2 " y$, we obtain the system

(15) x1$ " x2; x1 0 " 1.5 ,

x2$ " #4x2 # 3x1; x2 0 " #2.5 .

Comparing (15) with (1) we see that f1 t, x1, x2 " x2 and f2 t, x1, x2 " #4x2 # 3x1. With the starting values of t0 " 0, x1;0 " 1.5, and x2;0 " #2.5, we compute

BABABA BA

BABABA BABABABABA

h f Atn, xnB 4BABABA3BA 4BABABA3BA

BA BABAo

BA BA

Section 5.3 Solving Systems and Higher-Order Equations Numerically 255

Example 2

Solution

cx1 A0.2B # x1;2 " x1;1 ! hx2;1 " 1.25 ! 0.1 A#1.95B " 1.055 , x2 A0.2) # x2;2 " x2;1 ! h A#4x2;1 # 3x1;1B " #1.95 ! 0.1 3#4 A#1.95B # 3 # 1.25 4 " #1.545 . cx1 A0.1B # x1;1 " x1;0 ! hx2;0 " 1.5 ! 0.1 A#2.5B " 1.25 ,

x2 A0.1) # x2;1 " x2;0 ! h A#4x2;0 # 3x1;0B " #2.5 ! 0.1 3#4 A#2.5B # 3 # 1.5 4 " #1.95 ; Continuing the algorithm we compute the remaining values. These are listed in Table 5.1

on page 256, along with the exact values calculated via the methods of Chapter 4. Note that the x2; n column gives approximations to y$(t), since x2(t) ! y$(t). ◆

Euler’s method is modestly accurate for this problem with a step size of h " 0.1. The next example demonstrates the effects of using a sequence of smaller values of h to improve the accuracy.

For the initial value problem of Example 2, use Euler’s method to estimate y(1) for succes- sively halved step sizes h " 0.1, 0.05, 0.025, 0.0125, 0.00625.

Using the same scheme as in Example 2, we find the following approximations, denoted by y(1;h) (obtained with step size h):

h 0.1 0.05 0.025 0.0125 0.00625 y(1;h) 0.36280 0.37787 0.38535 0.38907 0.39092

[Recall that the exact value, rounded to 5 decimal places, is y(1) " 0.39277.] ◆

The Runge–Kutta scheme described in Section 3.7 is easy to vectorize also; details are given on the following page. As would be expected, its performance is considerably more accurate, yielding five-decimal agreement with the exact solution for a step size of 0.05:

h 0.1 0.05 0.025 0.0125 0.00625 y(1;h) 0.39278 0.39277 0.39277 0.39277 0.39277

As in Section 3.7, both algorithms can be coded so as to repeat the calculation of y(1) with a sequence of smaller step sizes until two consecutive estimates agree to within some prespeci- fied tolerance e. Here one should interpret “two estimates agree to within (” to mean that each component of the successive vector approximants [i.e., approximants to y(1) and y$(1)] should agree to within (.

An Application to Population Dynamics A mathematical model for the population dynamics of competing species, one a predator with population and the other its prey with population , was developed independently in thex1 AtBx2 AtB

256 Chapter 5 Introduction to Systems and Phase Plane Analysis

TABLE 5.1 Approximations of the Solution to (14) in Example 2

t ! n(0.1) x1;n y Exact x2;n y$ Exact

0 1.5 1.5 #2.5 #2.5 0.1 1.25 1.275246528 #1.95 #2.016064749 0.2 1.055 1.093136571 #1.545 #1.641948207 0.3 0.9005 0.944103051 #1.2435 #1.35067271 0.4 0.77615 0.820917152 #1.01625 #1.122111364 0.5 0.674525 0.71809574 #0.842595 #0.9412259 0.6 0.5902655 0.63146108 #0.7079145 #0.796759968 0.7 0.51947405 0.557813518 #0.60182835 #0.680269946 0.8 0.459291215 0.494687941 #0.516939225 #0.585405894 0.9 0.407597293 0.440172416 #0.4479509 #0.507377929 1 0.362802203 0.392772975 #0.391049727 #0.442560044

Example 3

Solution

early 1900s by A. J. Lotka and V. Volterra. It assumes that there is plenty of food available for the prey to eat, so the birthrate of the prey should follow the Malthusian or exponential law (see Sec- tion 3.2); that is, the birthrate of the prey is Ax1, where A is a positive constant. The death rate of the prey depends on the number of interactions between the predators and the prey. This is mod- eled by the expression Bx1x2, where B is a positive constant. Therefore, the rate of change in the population of the prey per unit time is Assuming that the predators depend entirely on the prey for their food, it is argued that the birthrate of the predators depends on the number of interactions with the prey; that is, the birthrate of predators is Dx1x2, where D is a positive constant. The death rate of the predators is assumed to be Cx2 because without food the population would die off at a rate proportional to the population present. Hence, the rate of change in the population of predators per unit time is " Combining these two equations, we obtain the Volterra–Lotka system for the population dynamics of two compet- ing species:

(16)

Such systems are in general not explicitly solvable. In the following example, we obtain an approximate solution for such a system by utilizing the vectorized form of the Runge–Kutta algorithm.

For the system of two equations

with initial conditions the vectorized form of the Runge–Kutta recursive equations (cf. (14), page 134) becomes

(17)

where h is the step size and, for i " 1 and 2,

(18)

It is important to note that both k1,1 and k2,1 must be computed before either k1,2 or k2,2. Similarly, both k1,2 and k2,2 are needed to compute k1,3 and k2,3, etc. In Appendix F, program outlines are given for applying the method to graph approximate solutions over a specified interval or to obtain approximations of the solutions at a specified point to within a desired tolerance.

3 t0, t1 4

fki,1 J hfi Atn, x1;n, x2;nB ,ki,2 J hfi Atn ! h2, x1;n ! 12k1,1, x2;n ! 12k2,1B , ki,3 J hfi Atn ! h2, x1;n ! 12k1,2, x2;n ! 12k2,2B , ki,4 J hfi Atn ! h, x1;n ! k1,3, x2;n ! k2,3B .

e tn!1 J tn ! h An " 0, 1, 2, . . .B ,x1;n!1 J x1;n ! 16 Ak1,1 ! 2k1,2 ! 2k1,3 ! k1,4B , x2;n!1 J x2;n ! 16 Ak2,1 ! 2k2,2 ! 2k2,3 ! k2,4B ,

x1 At0B " x1;0, x2 At0B " x2;0, x2¿ " f2 At, x1, x2B ,x1¿ " f1 At, x1, x2B ,

x$2 ! "Cx2 ! Dx1x2 . x$1 ! Ax1 " Bx1x2 ,

#Cx2 ! Dx1x2.dx2/dt

dx1/dt " Ax1 # Bx1x2.

Section 5.3 Solving Systems and Higher-Order Equations Numerically 257

Use the classical fourth-order Runge–Kutta algorithm for systems to approximate the solution of the initial value problem

(19)

at t " 1. Starting with h " 1, continue halving the step size until two successive approxima- tions of and of differ by at most 0.0001.

Here With the inputs t0 " 0, x1;0 " 1, x2;0 " 3, we proceed with the algorithm to compute and , the approxi- mations to , using h " 1. We find from the formulas in (18) that

and similarly we compute

Inserting these values into formula (17), we get

as the respective approximations to and x2 A1B.x1 A1B " 3 !

1 6 A0 # 12 ! 0 ! 18B " 4 ,

x2;1 " x2;0 ! 1 6 Ak2,1 ! 2k2,2 ! 2k2,3 ! k2,4B ,

" 1 ! 1 6 A#4 ! 8 ! 12 # 28B " #1 ,

x1;1 " x1;0 ! 1 6 Ak1,1 ! 2k1,2 ! 2k1,3 ! k1,4B

k2,4 " h 3 Ax1;0 ! k1,3B Ax2;0 ! k2,3B # Ax2;0 ! k2,3B 4 " #18 .k1,4 " h 32 Ax1;0 ! k1,3B # 2 Ax1;0 ! k1,3B Ax2;0 ! k2,3B 4 " #28 , k2,3 " h 3 Ax1;0 ! 12k1;2B Ax2;0 ! 12k2,2B # Ax2;0 ! 12k2,2B 4 " 0 ,k1,3 " h 32 Ax1;0 ! 12k1,2B # 2 Ax1;0 ! 12k1,2B Ax2;0 ! 12k2,2B 4 " 6 ,

" A#1B A3B # 3 " #6 , " 31 ! 12 A#4B 4 33 ! 12 A0B 4 # 33 ! 12 A0B 4 k2,2 " h 3 Ax1;0 ! 12k1,1B Ax2;0 ! 12k2,1B # Ax2;0 ! 12k2,1B 4 " #2 ! 2 A3B " 4 ,

" 2 31 ! 12 A#4B 4 # 2 31 ! 12 A#4B 4 33 ! 12 A0B 4k1,2 " h 32 Ax1;0 ! 12k1,1B # 2 Ax1;0 ! 12k1,1B Ax2;0 ! 12k2,1B 4 k2,1 " h Ax1;0x2;0 # x2;0B " A1B A3B # 3 " 0 ,k1,1 " h A2x1;0 # 2x1;0x2;0B " 2 A1B # 2 A1B A3B " #4 ,

x2 A1Bx1 A1B x2 A1; 1Bx1 A1; 1B f1 At, x1, x2B " 2x1 # 2x1x2 and f2 At, x1, x2B " x1x2 # x2.

x2 A1Bx1 A1B x¿2 " x1x2 # x2 ; x2 A0B " 3 x¿1 " 2x1 # 2x1x2 ; x1 A0B " 1 ,

258 Chapter 5 Introduction to Systems and Phase Plane Analysis

Example 4

Solution

Repeating the algorithm with we obtain the approximations and for and . In Table 5.2, we list the approximations and

for and using step size for m " 0, 1, 2, 3, and 4. We stopped at m " 4, since both

and

Hence, with tolerance 0.0001. ◆x1 A1B # 0.07735 and x2 A1B # 1.46445, 0 x2 A1; 2#3B # x2 A1; 2#4B 0 " 0.00001 6 0.0001 . 0 x1 A1; 2#3B # x1 A1; 2#4B 0 " 0.00006 6 0.0001

h " 2#mx2 A1Bx1 A1Bx2 A1; 2#mB x1 A1; 2#mBx2 A1Bx1 A1Bx2 A1; 2#1B x1 A1; 2#1Bh " 1/2 AN " 2B

Section 5.3 Solving Systems and Higher-Order Equations Numerically 259

To get a better feel for the solution to system (19), we have graphed in Figure 5.6 an approximation of the solution for using linear interpolation to connect the vector- ized Runge–Kutta approximants for the points t " 0, 0.125, 0.25, . . . , 12.0 (i.e., with h " 0.125). From the graph it appears that the components x1 and x2 are periodic in the variable t. Phase plane analysis is used in Section 5.5 to show that, indeed, Volterra–Lotka equations have periodic solutions.

0 ) t ) 12,

2 3 4 5 6 7

8 9 10 11 12 t

Figure 5.6 Graphs of the components of an approximate solution to the Volterra–Lotka system (17)

TABLE 5.2 Approximations of the Solution to System (19) in Example 4

m h

0 1.0 #1.0 4.0 1 0.5 0.14662 1.47356 2 0.25 0.07885 1.46469 3 0.125 0.07741 1.46446 4 0.0625 0.07735 1.46445

x2 A1; hBx1 A1; hB

In Problems 1–7, convert the given initial value problem into an initial value problem for a system in normal form.

[Hint: Set 6.

8. Sturm–Liouville Form. A second-order equation is said to be in Sturm–Liouville form if it is expressed as

Show that the substitutions result in the normal form

If are the initial values for the Sturm–Liouville problem, what are and ?

9. In Section 3.6, we discussed the improved Euler’s method for approximating the solution to a first- order equation. Extend this method to normal sys- tems and give the recursive formulas for solving the initial value problem.

In Problems 10–13, use the vectorized Euler method with h " 0.25 to find an approximation for the solution to the given initial value problem on the specified interval.

10. on 3 0, 1 4y A0B " 1 , y¿ A0B " 0 y– ! ty¿ ! y " 0 ;

x2 A0Bx1 A0By A0B " a and y¿ A0B " b x ¿2 " #qx1 . x ¿1 " x2 /p ,

x1 " y, x2 " py¿ 3 p AtBy¿ AtB 4 ¿ ! q AtBy AtB " 0 .

2x– ! 5y– # 2y " 1 ; y A0B " y¿ A0B " 1x‡ # y " t ; x A0B " x¿ A0B " x– A0B " 4 , 4y– ! 2y # 6x " 0 ; y A0B " 1 , y¿A0B " 23x–! 5x # 2y "0 ; x A0B"#1 , x¿A0B"0 ,

x4 " y¿. ]x3 " y ,x2 " x¿ ,x1 " x , y– # x ! y " #1 ; y A3B " 1 , y¿A3B " #1x– ! y # x¿ " 2t ; x A3B " 5 , x¿A3B " 2 , y A0B " y¿ A0B " p " y A5B A0B " 0y A6B AtB " 3 y¿ AtB 4 3 # sin Ay AtBB ! e2t ;y A0B " y¿A0B " 1 , y–A0B " 0 , y

A3B A0B " 2y A4B AtB # y A3B AtB ! 7y AtB " cos t ; y A0B " 1 , y¿ A0B " 0y– AtB " cos At # yB ! y2 AtB ; y A0B " 3 , y¿ A0B " #6y– AtB ! ty¿ AtB # 3y AtB " t2 ;

260 Chapter 5 Introduction to Systems and Phase Plane Analysis

11. on

12. on

13. on

(Can you guess the solution?)

In Problems 14–24, you will need a computer and a pro- grammed version of the vectorized classical fourth-order Runge–Kutta algorithm. (At the instructor’s discretion, other algorithms may be used.)†

14. Using the vectorized Runge–Kutta algorithm with h " 0.5, approximate the solution to the initial value problem

at t " 8. Compare this approximation to the actual solution

15. Using the vectorized Runge–Kutta algorithm, ap- proximate the solution to the initial value problem

at t " 1. Starting with continue halving the step size until two successive approximations of both and differ by at most 0.01.

16. Using the vectorized Runge–Kutta algorithm for systems with h " 0.125, approximate the solution to the initial value problem

at t " 1. Compare this approximation to the actual solution

17. Using the vectorized Runge–Kutta algorithm, ap- proximate the solution to the initial value problem

dy dx

" 2u # 3y ; y A0B " 1 du dx

" 3u # 4y ; u A0B " 1 , x AtB " e5t # e3t , y AtB " e3t # 3e5t . y¿ " 3x ! 6y ; y A0B " #2x¿ " 2x # y ; x A0B " 0 ,

y¿ A1B 4y A1B 3h " 1, y– " t2 ! y2 ; y A0B " 1 , y¿ A0B " 0

y AtB " t5/3 # t. y A1B " 0 , y¿ A1B " 2

3t2y– # 5ty¿ ! 5y " 0 ;

3 0, 1 4y A0B " 0 , y¿ A0B " 1 y– " t2 # y2 ; 3 1, 2 4y A1B " 1 , y¿ A1B " #1 t2y– ! y " t ! 2 ;

3 0, 1 4y A0B " 1 , y¿ A0B " #1 A1 ! t2By– ! y¿ # y " 0 ;

†An applet, maintained on the Web at http://alamos.math.arizona.edu/~rychlik/JOde/index.html, automates most of the differential equation algorithms discussed in this book.

5.3 EXERCISES

http://alamos.math.arizona.edu/~rychlik/JOde/index.html

at x " 1. Starting with continue halving the step size until two successive approximations of

and differ by at most 0.001.

18. Combat Model. A simplified mathematical model for conventional versus guerrilla combat is given by the system

where x1 and x2 are the strengths of guerrilla and conventional troops, respectively, and 0.1 and 1 are the combat effectiveness coefficients. Who will win the conflict: the conventional troops or the guerril- las? [Hint: Use the vectorized Runge–Kutta algo- rithm for systems with h " 0.1 to approximate the solutions.]

19. Predator–Prey Model. The Volterra–Lotka predator– prey model predicts some rather interesting behavior that is evident in certain biological systems. For example, suppose you fix the initial population of prey but increase the initial population of predators. Then the population cycle for the prey becomes more severe in the sense that there is a long period of time with a reduced population of prey followed by a short period when the population of prey is very large. To demonstrate this behavior, use the vector- ized Runge–Kutta algorithm for systems with h " 0.5 to approximate the populations of prey x and of predators y over the period that satisfy the Volterra–Lotka system

under each of the following initial conditions: (a) . (b) . (c) .

20. In Group Project C of Chapter 4, it was shown that the simple pendulum equation

has periodic solutions when the initial displacement and velocity are small. Show that the period of the solution may depend on the initial conditions by using the vectorized Runge–Kutta algorithm with

u– AtB ! sin u AtB " 0 x A0B " 2 , y A0B " 7x A0B " 2 , y A0B " 5 x A0B " 2 , y A0B " 4 y¿ " y Ax # 3Bx¿ " x A3 # yB ,

3 0, 5 4

x ¿2 " #x1 ; x2 A0B " 15 ,x ¿1 " # A0.1Bx1x2 ; x1 A0B " 10 ,

y A1Bu A1B h " 1,

Section 5.3 Solving Systems and Higher-Order Equations Numerically 261

h " 0.02 to approximate the solutions to the simple pendulum problem on for the initial conditions: (a) . (b) . (c) [Hint: Approximate the length of time it takes to reach

21. Fluid Ejection. In the design of a sewage treatment plant, the following equation arises:†

where H is the level of the fluid in an ejection chamber and t is the time in seconds. Use the vectorized Runge–Kutta algorithm with h " 0.5 to approximate over the interval

22. Oscillations and Nonlinear Equations. For the initial value problem

use the vectorized Runge–Kutta algorithm with h " 0.02 to illustrate that as t increases from 0 to 20, the solution x exhibits damped oscilla- tions when x0 " 1, whereas x exhibits expand- ing oscillations when x0 " 2.1.

23. Nonlinear Spring. The Duffing equation

where r is a constant, is a model for the vibra- tions of a mass attached to a nonlinear spring. For this model, does the period of vibration vary as the parameter r is varied? Does the period vary as the initial conditions are varied? [Hint: Use the vectorized Runge–Kutta algo- rithm with h " 0.1 to approximate the solutions for r " 1 and 2, with initial conditions

for a " 1, 2, and 3.] 24. Pendulum with Varying Length. A pendu-

lum is formed by a mass m attached to the end of a wire that is attached to the ceiling. Assume that the length of the wire varies with time in some predetermined fashion. If is theu AtBl AtB

y¿ A0B " 0y A0B " a,

y– ! y ! ry3 " 0 ,

x A0B " x0 , x¿ A0B " 0 ,x– ! A0.1B A1 # x2Bx¿ ! x " 0 ; 3 0, 5 4 H AtB

H A0B " H¿ A0B " 0 ,60 # H " A77.7BH– ! A19.42B AH¿ B2 ; #u A0B. ]

u A0B " 1.0 , u¿ A0B " 0 .u A0B " 0.5 , u¿ A0B " 0 u A0B " 0.1 , u¿ A0B " 0

3 0, 4 4

†See Numerical Solution of Differential Equations, by William Milne (Dover, New York, 1970), p. 82.

angle between the pendulum and the vertical, then the motion of the pendulum is governed by the ini- tial value problem

where g is the acceleration due to gravity. Assume that

where l1 is much smaller than l0. (This might be a model for a person on a swing, where the pumping action changes the distance from the center of mass of the swing to the point where the swing is attached.) To simplify the computations, take g " 1. Using the Runge–Kutta algorithm with h " 0.1, study the motion of the pendulum when

and In particular, does the pendulum ever attain an angle greater in absolute value than the initial angle ? Does the total arc traversed during one-half of a swing ever exceed 1?

In Problems 25–30, use a software package or the SUB- ROUTINE in Appendix F.

25. Using the Runge–Kutta algorithm for systems with approximate the solution to the initial

value problem

at t " 1.

26. Use the Runge–Kutta algorithm for systems with h " 0.1 to approximate the solution to the initial value problem

at t " 1.

27. Generalized Blasius Equation. H. Blasius, in his study of laminar flow of a fluid, encountered an equation of the form

Use the Runge–Kutta algorithm for systems with h " 0.1 to approximate the solution that satisfies

y‡ ! yy– " A y¿ B2 # 1 .

z¿ " #xy/2 ; z A0B " 1 ,y¿ " #xz ; y A0B " 1 , x¿ " yz ; x A0B " 0 ,

y A0B " 1 , y¿ A0B " 0 , y– A0B " 1y‡ ! y– ! y2 " t ; h " 0.05,

f " 0.02.v " 1,l1 " 0.1,l0 " 1,u1 " 0, u0 " 0.5,

l AtB " l0 ! l1 cos Avt # fB , u A0B " u0 , u¿ A0B " u1 ,l 2 AtBu– AtB ! 2l AtBl¿ AtBu¿ AtB ! gl AtBu AtB " 0 ;

262 Chapter 5 Introduction to Systems and Phase Plane Analysis

the initial conditions and Sketch this solution on the inter-

val . 28. Lunar Orbit. The motion of a moon moving in a

planar orbit about a planet is governed by the equations

where is the gravitational con- stant, and m is the mass of the moon. Assume Gm " 1. When the motion is a circular orbit of radius 1 and period (a) Setting

express the governing equations as a first-order system in normal form.

(b) Using compute one orbit of this moon (i.e., do N " 100 steps?). Do your approximations agree with the fact that the orbit is a circle of radius 1?

29. Competing Species. Let denote, respectively, the populations of three competing species Si, i " 1, 2, 3. Suppose these species have the same growth rates, and the maximum population that the habitat can support is the same for each species. (We assume it to be one unit.) Also suppose the competi- tive advantage that S1 has over S2 is the same as that of S2 over S3 and S3 over S1. This situation is mod- eled by the system

where a and b are positive constants. To demonstrate the population dynamics of this system when a " b " 0.5, use the Runge–Kutta algorithm for systems with h " 0.1 to approximate the populations pi over the time interval under each of the following initial conditions:

(a) . (b) . (c)

On the basis of the results of parts (a)–(c), decide what you think will happen to these populations as t S ! q.

p1 A0B " 0.1 , p2 A0B " 0.1 , p3 A0B " 1.0 .p1 A0B " 0.1 , p2 A0B " 1.0 , p3 A0B " 0.1 p1 A0B " 1.0 , p2 A0B " 0.1 , p3 A0B " 0.1

3 0, 10 4 p¿3 " p3 A1 # ap1 # bp2 # p3B ,p¿2 " p2 A1 # bp1 # p2 # ap3B , p¿1 " p1 A1 # p1 # ap2 # bp3B ,

pi AtB h " 2p /100 # 0.0628318,

x1 " x, x2 " x¿, x3 " y, x4 " y¿, 2p.

x A0B " 1, x¿ A0B " y A0B " 0, and y¿ A0B " 1, r J Ax2 ! y2B1/2, G

d 2x

dt2 " #G

r 3 ,

d 2y

dt2 " #G

r 3 ,

3 0, 2 4y– A0B " 1.32824. y A0B " 0, y¿ A0B " 0,

30. Spring Pendulum. Let a mass be attached to one end of a spring with spring constant k and the other end attached to the ceiling. Let l0 be the natural length of the spring and let be its length at time t. If is the angle between the pendulum and the vertical, then the motion of the spring pendulum is governed by the system

l 2 AtBu–AtB ! 2l AtBl¿AtBu¿AtB ! gl AtB sin u AtB " 0 . l–AtB # l AtBu¿AtB # g cos u AtB ! k

m Al # l0B " 0 ,

u AtB l AtB

Section 5.4 Introduction to the Phase Plane 263

Assume , , and . When the system is at rest, . (a) Describe the motion of the pendulum when

, , , and (b) When the pendulum is both stretched and given

an angular displacement, the motion of the pendulum is more complicated. Using the Runge–Kutta algorithm for systems with h " 0.1 to approximate the solution, sketch the graphs of the length l and the angular displace- ment on the interval if

u A0B " 0.5, and u¿A0B " 0.l¿A0B " 0, l A0B " 5.5,3 0, 10 4u

u¿ A0B " 0.u A0B " 0l¿ A0B " 0l A0B " 5.5 l " l0 ! mg/k " 5

l0 " 4k " m " 1g " 1

5.4 INTRODUCTION TO THE PHASE PLANE In this section, we study systems of two first-order equations of the form

(1)

Note that the independent variable t does not appear in the right-hand terms and ; such systems are called autonomous. For example, the system that modeled the intercon- nected tanks problem in Section 5.1,

is autonomous. So is the Volterra–Lotka system,

(with A, B, C, D constants), which was discussed in Example 4 of Section 5.3 as a model for population dynamics.

For future reference, we note that the solutions to autonomous systems have a “time-shift immunity,” in the sense that if the pair solves (1), so does the time-shifted pair

for any constant c. Specifically, if we let and then by the chain rule

proving that is also a solution to (1).X AtB, Y AtB dY dt

AtB " dy dt

At ! cB " g Ax At ! cB, y At ! cBB " g AX AtB, Y AtBB , dXdt AtB " dxdt At ! cB " f Ax At ! cB, y At ! cBB " f AX AtB, Y AtBB , Y AtB J y At ! cB,X AtB J x At ! cBx At ! cB, y At ! cB x AtB, y AtB

y¿ " #Cy ! Dxy , x¿ " Ax # Bxy ,

y¿ " 1 3

x # 1 3

y ,

x¿ " # 1 3

x ! 1 12

y ,

g Ax, yBf Ax, yB dy dt ! g

Ax, yB . dx dt ! f

Ax, yB ,

Since t does not appear explicitly in the system (1), it is certainly tempting to divide the two equations, invoke the chain rule

and consider the single first-order differential equation

(2)

We will refer to (2) as the phase plane equation. In Chapters 1 and 2 we mastered several approaches to equations like (2): the use of direction fields to visualize the solution graphs, and the analytic techniques for the cases of separability, linearity, exactness, etc.

So the form (2) certainly has advantages over (1), but it is important to maintain our perspective by noting these distinctions:

(i) A solution to the original problem (1) is a pair of functions of t—namely, and —that satisfies (1) for all t in some interval I. These functions can be visualized

as a pair of graphs, as in Figure 5.7. If, in the xy-plane, we plot the points as t varies over I, the resulting curve is known as the trajectory of the solution pair

and the xy-plane is called the phase plane in this context (see Figure 5.8 on page 265). Note, however, that the trajectory in this plane contains less information than the original graphs, because the t-dependence has been suppressed. (Typically, though, we indicate the direction of time with an arrow on the curve.) In principle we can construct, point by point, the trajectory from the solution graphs, but we cannot reconstruct the solution graphs from the phase plane trajectory alone (because we would not know what value of t to assign to each point).

(ii) Nonetheless, the slope of a trajectory in the phase plane is given by the right- hand side of (2). So, in solving equation (2) we are indeed locating the trajectories of the system (1) in the phase plane. More precisely, we have shown that the trajectories satisfy equation (2), and thus lie on its solution curves.

(iii) In Chapters 1 and 2, we regarded x as the independent variable and y as the depen- dent variable, in equations of the form (2). This is no longer true in the context of the system (1); x and y are both dependent variables on an equal footing, and t is the independent variable.

Thus, it appears that a phase plane portrait may be a useful, albeit incomplete, tool for analyz- ing first-order autonomous systems like (1).

dy/dx

x AtB, y AtB, Ax AtB, y AtBB y AtB x AtB

dx !

g Ax, yB f Ax, yB .

dx "

dy/dt dx/dt

264 Chapter 5 Introduction to Systems and Phase Plane Analysis

t1 t2 t3

x(t)

t t1 t2 t3

y(t)

Figure 5.7 Solution pair for system (1)

Except for the very special case of linear systems with constant coefficients that was dis- cussed in Section 5.2, finding all solutions to the system (1) is generally an impossible task. But it is relatively easy to find constant solutions; if and then the constant functions solve (1). For such solutions the following terminology is used.

x AtB ! x0, y AtB ! y0 g Ax0, y0B " 0,f Ax0, y0B " 0

Section 5.4 Introduction to the Phase Plane 265

(x(t3), y(t3))

(x(t2), y(t2))

(x(t1), y(t1))

Figure 5.8 Phase plane trajectory of the solution pair for system (1)

Critical Points and Equilibrium Solutions

Definition 1. A point where is called a critical point, or equilibrium point, of the system and the corresponding constant solution is called an equilibrium solution. The set of all critical points is called the critical point set.

x AtB ! x0, y AtB ! y0dx/dt " f Ax, yB, dy/dt " g Ax, yB, f Ax0, y0B " 0 and g Ax0, y0B " 0Ax0, y0B

Notice that trajectories of equilibrium solutions consist of just single points (the equilib- rium points). But what can be said about the other trajectories? Can we predict any of their features from closer examination of the equilibrium points? To explore this we focus on the phase plane equation (2) and exploit its direction field (recall Section 1.3, page 15). However, we’ll augment the direction field plot by attaching arrowheads to the line segments, indicating the direction of the “flow” of solutions as t increases. This is easy: When is positive, increases so the trajectory flows to the right. Therefore, according to (1), all direction field seg- ments drawn in a region where is positive should point to the right [and, of course, they point to the left if is negative]. If is zero, we can use to decide if the flow is upward increases] or downward decreases]. [What if both and are zero?]

In the examples that follow, one can use computers or calculators for generating these direction fields.

Sketch the direction field in the phase plane for the system

(3)

and identify its critical point.

dy dt

" #2y

dx dt

" #x ,

g Ax, yBf Ax, yB3 y AtB[ y AtB g Ax, yBf Ax, yBf Ax, yB f Ax, yB x AtBdx/dt

Example 1

Here are both zero when x " y " 0, so is the critical point. The direction field for the phase plane equation

(4)

is given in Figure 5.9. Since in (3), trajectories in the right half-plane (where x * 0) flow to the left, and vice versa. From the figure we can see that all solutions “flow into” the critical point . Such a critical point is called asymptotically stable.† ◆

Remark. For this simple example, we can actually solve the system (3) explicitly; indeed, (3) constitutes an uncoupled pair of linear equations whose solutions are " and

By elimination of t, we obtain the equation " So the trajectories lie along the parabolas . [Alternatively, we could have separated variables in (4) and identified these parabolas as the phase plane solution curves.] Notice that each such parabola is made up of three trajectories: an incoming trajectory approaching the origin in the right half-plane; its mirror-image trajectory approaching the origin in the left half-plane; and the origin itself, an equilibrium point. Sample trajectories are indicated in Figure 5.10.

Sketch the direction field in the phase plane for the system

(5)

and describe the behavior of solutions near the critical point .

This example is almost identical to the previous one; in fact, one could say we have merely “reversed time” in (3). The direction field segments for

(6)

are the same as those of (4), but the direction arrows are reversed. Now all solutions flow away from the critical point ; the equilibrium is unstable. ◆A0, 0B

dy dx

" 2y x

A0, 0B dy dt

" 2y

dx dt

" x ,

y " cx2 c2[x AtB /c1]2 " cx2.y " c2e#2ty AtB " c2e#2t. c1e#tx AtB

A0, 0B dx/dt " #x dy dx

" #2y #x

" 2y x

A0, 0Bf Ax, yB " #x and g Ax, yB " #2y

266 Chapter 5 Introduction to Systems and Phase Plane Analysis

Solution

Figure 5.9 Direction field for Example 1

Figure 5.10 Trajectories for Example 1

†See Section 12.3 for a rigorous exposition of stability and critical points. All references to Chapters 11–13 refer to the expanded text Fundamentals of Differential Equations and Boundary Value Problems, 6th ed.

Example 2

Solution

For the system (7) below, find the critical points, sketch the direction field in the phase plane, and predict the asymptotic nature (i.e., behavior as ) of the solution starting at x " 2, y " 0 when t " 0.

(7)

The only critical point is the solution of the simultaneous equations :

(8)

from which we find The direction field for the phase plane equation

(9)

is shown in Figure 5.11, with some trajectories rough-sketched in by hand.† Note that solutions flow to the right for i.e., for all points below the line

The phase plane solution curve passing through in Figure 5.11 apparently extends to infinity. Does this imply the corresponding system solution , also approaches infinity in the sense that or could its trajectory “stall” at some point along the phase plane solution curve, or possibly even “backtrack”? It cannot backtrack, because the direction arrows along the trajectory point, unambiguously, to the right. And if stalls at some point , then intuitively we would conclude that was an equilibrium point (since the “speeds” and would approach zero there). But we have already found the only critical point. So we conclude, with a high degree of confidence,†† that the system solution does indeed go to infinity.

dy/dtdx/dt Ax1, y1BAx1, y1B Ax AtB, y AtBB

0 x AtB 0 ! 0 y AtB 0 S !q as t S !q, y AtBx AtBA2, 0B 5x # 3y # 2 " 0.5x # 3y # 2 7 0,

dy dx

" 4x # 3y # 1 5x # 3y # 2

x0 " y0 " 1.

4x0 # 3y0 # 1 " 0 ,

5x0 # 3y0 # 2 " 0 ,

f Ax, yB " g Ax, yB " 0 dy dt

" 4x # 3y # 1 .

dx dt

" 5x # 3y # 2 ,

t S !q

Section 5.4 Introduction to the Phase Plane 267

y = 2x − 1

−2

−1

Figure 5.11 Direction field and trajectories for Example 3

Example 3

Solution

†The phase plane solution curves could be obtained analytically by solving equation (9) using the methods of Section 2.6. ††These informal arguments are made more rigorous in Chapter 12. All references to Chapters 11–13 refer to the expanded text Fundamentals of Differential Equations and Boundary Value Problems, 6th edition.

The critical point is unstable because, although many solutions get arbitrarily close to , most of them eventually flow away. Solutions that lie on the line how- ever, do converge to . Such an equilibrium is an example of a saddle point. ◆

In the preceding example, we informally argued that if a trajectory “stalls”—that is, if it has an endpoint—then this endpoint would have to be a critical point. This is more carefully stated in the following theorem, whose proof is outlined in Problem 30.

A1, 1B y " 2x # 1,A1, 1B A1, 1B

268 Chapter 5 Introduction to Systems and Phase Plane Analysis

†An applet, maintained on the Web at http://alamos.math.arizona.edu/~rychlik/JOde/index.html, automates most of the differential equation algorithms discussed in this book.

Endpoints Are Critical Points

Theorem 1. Let the pair , be a solution on to the autonomous system where f and g are continuous in the plane. If the limits

exist and are finite, then the point is a critical point for the system.Ax*, y*Bx* J limtS!q x AtB and y* J limtS!q y AtB dx/dt " f Ax, yB, dy/dt " g Ax, yB, [0, !q By AtBx AtB

Some typical trajectory configurations near critical points are displayed and classified in Figure 5.12. These phase plane portraits arise from the systems listed in Problem 29, and can be generated by software packages having trajectory-sketching options†. A more complete dis- cussion of the nature of various types of equilibrium solutions and their stability is deferred to

Node (asymptotically stable)

Center (stable)

Node (unstable)

Spiral (asymptotically stable)

Spiral (unstable)

Saddle (unstable)

Figure 5.12 Examples of different trajectory behaviors near critical point at origin

http://alamos.math.arizona.edu/~rychlik/JOde/index.html

Chapter 12.† For the moment, however, notice that unstable critical points are distinguished by “runaway” trajectories emanating from arbitrarily nearby points, while stable equilibria “trap” all neighboring trajectories. The asymptotically stable critical points attract their neighboring trajectories as

Historically, the phase plane was introduced to facilitate the analysis of mechanical systems governed by Newton’s second law, force equals mass times acceleration. An autonomous mechanical system arises when this force is independent of time and can be mod- eled by a second-order equation of the form

(10)

As we have seen in Section 5.3, this equation can be converted to a normal first-order system by introducing the velocity and writing

(11)

Thus, we can analyze the behavior of an autonomous mechanical system by studying its phase plane diagram in the yy-plane. Notice that with y as the vertical axis, trajectories flow to the right in the upper half-plane (where * 0), and to the left in the lower half-plane.

Sketch the direction field in the phase plane for the first-order system corresponding to the unforced, undamped mass–spring oscillator described in Section 4.1 (Figure 4.1, page 153). Sketch several trajectories and interpret them physically.

The equation derived in Section 4.1 for this oscillator is or, equivalently, Hence, the system (11) takes the form

(12)

The critical point is at the origin y " " 0. The direction field in Figure 5.13 on page 270 indi- cates that the trajectories appear to be either closed curves (ellipses?) or spirals that encircle the critical point.

We saw in Section 4.9 that the undamped oscillator motions are periodic; they cycle repeatedly through the same sets of points, with the same velocities. Their trajectories in the phase plane, then, must be closed curves.†† Let’s confirm this mathematically by solving the phase plane equation

(13)

Equation (13) is separable, and we find

y dy " # ky m

dy or d ay2 2 b " # k

m d ay2

2 b ,

dy dy

" # ky my

y¿ " # ky m

y¿ " y , y– " #ky/m.

my– ! ky " 0

y Ay AtB, y AtBB

dy dt

" ƒ(y, y) .

dy dt

" y ,

y " dy/dt

y– " ƒ Ay, y¿ B .

t S !q.

Section 5.4 Introduction to the Phase Plane 269

†All references to Chapters 11–13 refer to the expanded text Fundamentals of Differential Equations and Boundary Value Problems, 6th edition. ††By the same reasoning, underdamped oscillations would correspond to spiral trajectories asymptotically approach- ing the origin as t S !q.

Example 4

Solution

so its solutions are the ellipses as shown in Figure 5.14. The solutions of (12) are confined to these ellipses and hence flow neither toward nor away from the equilib- rium solution. The critical point is thus identified as a center in Figure 5.12 on page 268.

Furthermore, the system solutions must continually circulate around the ellipses, since there are no critical points to stop them. This confirms that all solutions are periodic. ◆

Remark. More generally, we argue that if a solution to an autonomous system like (1) passes through a point in the phase plane twice and if it is sufficiently well behaved to satisfy a uniqueness theorem, then the second “tour” satisfies the same initial conditions as the first tour and so must replicate it. In other words, closed trajectories containing no critical points correspond to periodic solutions.

y2/2 ! ky2/2m " C

270 Chapter 5 Introduction to Systems and Phase Plane Analysis

Figure 5.13 Direction field for Example 4

Figure 5.14 Trajectories for Example 4

Through these examples we have seen how, by studying the phase plane, one can often anticipate some of the features (boundedness, periodicity, etc.) of solutions of autonomous systems without solving them explicitly. Much of this information can be predicted simply from the critical points and the direction field (oriented by arrowheads), which are obtainable through standard software packages. The final example ties together several of these ideas.

Find the critical points and solve the phase plane equation (2) for

(14)

What is the asymptotic behavior of the solutions starting from , , and ?

To find the critical points, we solve the system

One family of solutions to this system is given by y " 2 with x arbitrary; that is, the line y " 2. If then the system simplifies to which has the solution x " 2, y " 0. Hence, the critical point set consists of the isolated point and the horizontal line y " 2. The corresponding equilibrium solutions are and the family ,

where c is an arbitrary constant. The trajectories in the phase plane satisfy the equation

(15)

Solving (15) by separating variables,

demonstrates that the trajectories lie on concentric circles centered at . See Figure 5.15.A2, 0By dy " # Ax # 2B dx or y 2 ! Ax # 2B2 " C ,

dx " Ax # 2B A y # 2B

#y A y # 2B " # x # 2y . y AtB ! 2, x AtB ! cx AtB ! 2, y AtB ! 0,

A2, 0B#y " 0, and x # 2 " 0,y + 2, #y A y # 2B " 0, Ax # 2B A y # 2B " 0 .

A2, 3BA5, 0BA3, 0B dy dt

" Ax # 2B A y # 2B . dx dt

" #y A y # 2B ,

Section 5.4 Introduction to the Phase Plane 271

Example 5

Solution

y = 2

0 1 2

(2, 3)

3 4 5 6

(2 − 5 2),

Figure 5.15 Phase plane diagram for Example 5

Next we analyze the flow along each trajectory. From the equation we see that x is decreasing when y * 2. This means the flow is from right to left along the arc of a circle that lies above the line y " 2. For 0 , y , 2, we have , so in this region the flow is from left to right. Furthermore, for y , 0, we have , and again the flow is from right to left.

We now observe in Figure 5.15 that there are four types of trajectories associated with sys- tem (14): (a) those that begin above the line y " 2 and follow the arc of a circle counterclock- wise back to that line; (b) those that begin below the line y " 2 and follow the arc of a circle clockwise back to that line; (c) those that continually move clockwise around a circle centered at with radius less than 2 (i.e., they do not intersect the line y " 2); and finally, (d) the critical points and y " 2, x arbitrary.

The solution starting at lies on a circle with no critical points; therefore, it is a periodic solution, and the critical point is a center. But the circle containing the solutions

starting at and at has critical points at The direction

arrows indicate that both solutions approach asymptotically (as They lie on the same circle (or phase plane solution curve), but they are quite different trajectories. ◆

Note that for the system (14) the critical points on the line y " 2 are not isolated, so they do not fit into any of the categories depicted in Figure 5.12. Observe also that all solutions of this system are bounded, since they are confined to circles.

t S !q B.A2 # 25, 2BA2 # 25, 2B and A2 ! 25, 2B .A2, 3BA5, 0B A2, 0B A3, 0BA2, 0B

A2, 0B dx/dt 6 0

dx/dt 7 0

dx/dt " #y A y # 2B, 272 Chapter 5 Introduction to Systems and Phase Plane Analysis

In Problems 1 and 2, verify that the pair , is a solution to the given system. Sketch the trajectory of the given solution in the phase plane.

In Problems 3–6, find the critical point set for the given system.

3. 4.

5. 6.

dy dt

" Ax # 1B A y # 2Bdy dt

" 3xy # y2

dx dt

" y2 # 3y ! 2 , dx dt

" x2 # 2xy ,

dy dt

" x ! y ! 5 dy dt

" x2 ! y2 # 1

dx dt

" y # 1 , dx dt

" x # y ,

x AtB " t ! 1, y AtB " t3 ! 3t2 ! 3t dx dt

" 1 , dy dt

" 3x2 ;

x AtB " e3t , y AtB " et dx dt

" 3y3 , dy dt

" y ;

y AtBx AtB In Problems 7–9, solve the phase plane equation (2), page 264, for the given system.

7. 8.

10. Find all the critical points of the system

and the xy-phase plane solution curves. Thereby prove that there are two trajectories that are gen- uine semicircles. What are the endpoints of the semicircles?

dy dt

" xy ,

dx dt

" x2 # 1 ,

dy dt

" ex ! y

dx dt

" 2y # x ,

dy dt

" 3x2 # 2xy dy dt

" ex!y

dx dt

" x2 # 2y#3 , dx dt

" y # 1 ,

5.4 EXERCISES

In Problems 11–14, solve the phase plane equation for the given system. Then sketch by hand several represen- tative trajectories (with their flow arrows).

11. 12.

13. 14.

In Problems 15–18, find all critical points for the given system. Then use a software package to sketch the direc- tion field in the phase plane and from this describe the sta- bility of the critical points (i.e., compare with Figure 5.12).

15. 16.

17. 18.

In Problems 19–24, convert the given second-order equation into a first-order system by setting . Then find all the critical points in the -plane. Finally, sketch (by hand or software) the direction fields, and describe the stability of the critical points (i.e., compare with Fig- ure 5.12).

19. 20.

21. 22.

23.

24.

25. Using software, sketch the direction field in the phase plane for the system

From the sketch, conjecture whether the solution passing through each given point is periodic: (a) (b) (c) A1, 0BA2, 2BA0.25, 0.25B

dy/dt " #x ! x3 . dx/dt " y ,

y– AtB ! y AtB # y AtB3 " 0 y– AtB ! y AtB # y AtB4 " 0 d 2y

dt2 ! y3 " 0

d 2y

dt2 ! y ! y5 " 0

d 2y

dt2 ! y " 0

d 2y

dt2 # y " 0

yy y " y¿

dy dt

" y A5 # x # yB dy dt

" #x # 2y

dx dt

" x A7 # x # 2yB ,dx dt

" 2x ! 13y ,

dy dt

" x # 4y dy dt

" #3x # 2y # 4

dx dt

" #5x ! 2y , dx dt

" 2x ! y ! 3 ,

dy dt

" 2 x

dy dt

" Ax # yB Ax # 1B dx dt

" 3 y

, dx dt

" A y # xB A y # 1B , dy dt

" 18x dy dt

" 2x

dx dt

" #8y , dx dt

" 2y ,

Section 5.4 Introduction to the Phase Plane 273

26. Using software, sketch the direction field in the phase plane for the system

From the sketch, conjecture whether all solutions of this system are bounded. Solve the phase plane equa- tion and confirm your conjecture.

27. Using software, sketch the direction field in the phase plane for the system

From the sketch, predict the asymptotic limit (as of the solution starting at .

28. Figure 5.16 displays some trajectories for the system

What types of critical points (compare Figure 5.12) occur at and ?A1, 0BA0, 0B

dy/dt " #x ! x2 . dx/dt " y ,

A1, 1Bt S !q) dy/dt " #5x # 4y . dx/dt " #2x ! y ,

dy/dt " #x # x3 . dx/dt " y ,

x 1

Figure 5.16 Phase plane for Problem 28

29. The phase plane diagrams depicted in Figure 5.12 were derived from the following systems. Use any method (except software) to match the systems to the graphs. (a) (b)

(e) (f) dy/dt " x # y dy/dt " 4x # 3y dx/dt " #y , dx/dt " 5x # 3y , dy/dt " x ! 2y dy/dt " x # 4y dx/dt " 2x # y , dx/dt " #5x ! 2y , dy/dt " #2x dy/dt " 3y dx/dt " y/2 , dx/dt " x ,

30. A proof of Theorem 1, page 268, is outlined below. The goal is to show that . Justify each step. (a) From the given hypotheses, deduce that

and

(b) Suppose Then, by continuity, for all large t (say, for

Deduce from this that for where C is some constant.

(c) Conclude from part (b) that contradicting the fact that this limit is the finite number x*. Thus, cannot be positive.

(d) Argue similarly that the supposition that also leads to a contradiction; hence,

must be zero. (e) In the same manner, argue that must

be zero. Therefore, , and is a critical point.

31. Phase plane analysis provides a quick derivation of the energy integral lemma of Section 4.8 (page 204). By completing the following steps, prove that solutions of equations of the special form satisfy

where is an antiderivative of . (a) Set as an equiva-

lent first-order system. (b) Show that the solutions to the -phase plane

equation for the system in part (a) satisfy . Replacing by then com-

pletes the proof.

32. Use the result of Problem 31 to prove that all solu- tions to the equation

remain bounded. [Hint: Argue that is bounded above by the constant appearing in Problem 31.]

33. A Problem of Current Interest. The motion of an iron bar attracted by the magnetic field produced by a parallel current wire and restrained by springs (see Figure 5.17) is governed by the equation

d2x dt2

" #x ! 1 l # x

, for #x0 6 x 6 l ,

y4/4 y– ! y3 " 0

y¿yy2/2 " F A yB ! K yy

y " y¿ and write y– " f A yBf A yBF A yB 1 2

A y¿ B2 # F A yB " constant ,y– " f A yB

Ax*, y*Bf Ax*, y*B " g Ax*, y*B " 0 g Ax*, y*Bf Ax*, y*B

f Ax*, y*B 6 0 f Ax*, y*B

limtS!q x AtB" !q,t 7 T, x AtB 7 t f Ax*, y*B /2 ! Ct ' T ).x¿ AtB 7 f Ax*, y*B /2

f Ax*, y*B 7 0.g Ax*, y*B. limtS!q y¿ AtB "limtS!q x¿ AtB " f Ax*, y*B

f Ax*, y*B " g Ax*, y*B " 0 274 Chapter 5 Introduction to Systems and Phase Plane Analysis

where the constants x0 and are, respectively, the distances from the bar to the wall and to the wire when the bar is at equilibrium (rest) with the current off. (a) Setting , convert the second-order

equation to an equivalent first-order system. (b) Solve the phase plane equation for the system in

part (a) and thereby show that its solutions are given by

where C is a constant. (c) Show that if there are no critical points in

the -phase plane, whereas if there are two critical points. For the latter case, determine these critical points.

(d) Physically, the case corresponds to a current so high that the magnetic attraction completely overpowers the spring. To gain insight into this, use software to plot the phase plane diagrams for the system when and when .

(e) From your phase plane diagrams in part (d), describe the possible motions of the bar when

and when under various initial conditions.

34. Falling Object. The motion of an object moving vertically through the air is governed by the equation

d2y

dt2 " #g #

V 2 dy

dt ` dy dt ` ,

l " 3,l " 1

l " 3 l " 1

l 6 2

l 7 2xy l 6 2

y " -2C # x2 # 2 ln Al # xB , y " dx/dt

wire

x 0

Figure 5.17 Bar restrained by springs and attracted by a parallel current

where y is the upward vertical displacement and V is a constant called the terminal speed. Take g " 32 ft /sec2

and V " 50 ft/sec. Sketch trajectories in the -phase plane for starting from y " 0 and , #50, #25, 0, 25, 50, and 75 ft/sec. Interpret the trajectories physi- cally; why is V called the terminal speed?

35. Sticky Friction. An alternative for the damping friction model discussed in Section 4.1 is the “sticky friction” model. For a mass sliding on a surface as depicted in Figure 5.18, the contact friction is more complicated than simply We observe, for example, that even if the mass is displaced slightly off the equilibrium location y " 0, it may nonetheless remain stationary due to the fact that the spring force

is insufficient to break the static friction’s grip. If the maximum force that the friction can exert is denoted by then a feasible model is given by

if and if and if y¿ + 0 .#m sign A y¿B , y¿ " 0 ,0 ky 0 ' mFfriction " $m sign A yB ,

y¿ " 0 , 0 ky 0 6 mky ,m,

#ky

#by¿.

F " #by¿

y " #75 #100 ) y ) 100,#100 ) y ) 100,

Section 5.4 Introduction to the Phase Plane 275

(a) Taking convert (16) into the first-order system

if and(17) if and if

(b) Form the phase plane equation for (17) when and solve it to derive the solutions

where the plus sign prevails for and the minus sign for .

(c) Identify the trajectories in the phase plane as two families of concentric semicircles. What is the center of the semicircles in the upper half- plane? The lower half-plane?

(d) What are the critical points for (17)? (e) Sketch the trajectory in the phase plane of the

mass released from rest at y " 7.5. At what value for y does the mass come to rest?

36. Rigid Body Nutation. Euler’s equations describe the motion of the principal-axis components of the angular velocity of a freely rotating rigid body (such as a space station), as seen by an observer rotating with the body (the astronauts, for example). This motion is called nutation. If the angular velocity components are denoted by x, y, and z, then an example of Euler’s equations is the three-dimensional autonomous system

The trajectory of a solution to these equations is the curve generated by the points

in xyz-phase space as t varies over an interval I. (a) Show that each trajectory of this system lies

on the surface of a (possibly degenerate) sphere centered at the origin . [Hint: Compute .] What does this say about the magnitude of the angular velocity vector?

d dt

Ax2 ! y2 ! z2B A0, 0, 0B

Ax AtB, y AtB, z AtBB x AtB, y AtB, z AtB dz/dt " xy . dy/dt " #2xz , dx/dt " yz ,

y 6 0 y 7 0

y2 ! A y - 1B2 " c ,y + 0 y + 0 .#y # sign AyB , y " 0 ,0 y 0 ' 1y¿ " $#y ! sign A yB , y " 0 ,

0 y 0 6 10 ,y¿ " y , m " m " k " 1,

Friction

y = 0

k m

Figure 5.18 Mass–spring system with friction

(The function sign(s) is !1 when s * 0, #1 when s , 0, and 0 when s " 0.) The motion is governed by the equation

(16)

Thus, if the mass is at rest, friction balances the spring force if but simply opposes it with intensity m if . If the mass is moving, fric- tion opposes the velocity with the same intensity m.

0 y 0 ' m/k0 y 0 6 m/k m

d2y

dt2 " #ky ! Ffriction .

(b) Find all the critical points of the system, i.e., all points such that , ,

is a solution. For such solutions, the angular velocity vector remains constant in the body system.

(c) Show that the trajectories of the system lie along the intersection of a sphere and an elliptic cylinder of the form for some constant C. [Hint: Consider the expression for implied by Euler’s equations.]

(d) Using the results of parts (b) and (c), argue that the trajectories of this system are closed curves. What does this say about the corresponding solutions?

(e) Figure 5.19 displays some typical trajectories for this system. Discuss the stability of the three critical points indicated on the positive axes.

dy/dx y2 ! 2x2 " C,

z AtB ! z0 y AtB ! y0x AtB ! x0Ax0, y0, z0B 276 Chapter 5 Introduction to Systems and Phase Plane Analysis

Figure 5.19 Trajectories for Euler’s system

5.5 APPLICATIONS TO BIOMATHEMATICS:EPIDEMIC AND TUMOR GROWTH MODELS In this section we are going to survey some issues in biological systems that have been successfully modeled by differential equations. We begin by reviewing the population models described in Sections 3.2 and 5.3.

In the Malthusian model, the rate of growth of a population p(t) is proportional to the size of the existing population:

(1)

Cells that reproduce by splitting, such as amoebae and bacteria, are obvious biological examples of this type of growth. Equation (1) implies that a Malthusian population grows exponentially; there is no mechanism for constraining the growth. In Section 3.2 we saw that certain popula- tions exhibit Malthusian growth over limited periods of time (as does compound interest).†

Inserting a negative growth rate,

(2)

results in solutions that decay exponentially. Their average lifetime is 1/k, and their half-life is (ln 2)/k (Problems 6 and 8). In animals, certain organs such as the kidney serve to cleanse the bloodstream of unwanted components (creatinine clearance, renal clearance), and their con- centrations diminish exponentially. As a general rule, the body tends to dissipate ingested drugs in such a manner. (Of course, the most familiar physical instance of Malthusian disinte- gration is radioactive decay.) Note that if there are both growth and extinction processes,

and the equation in (1) still holds with .k " k! # k#dp/dt " k!p # k#p

dp dt

" #kp ,

dp dt

" kp (k 7 0) .

†Gordon E. Moore (1929–) has observed that the number of transistors on new integrated circuits produced by the elec- tronics industry doubles every 24 months. “Moore’s law” is commonly cited by industrialists.

When there are two-party interactions occurring in the population that decrease the growth rate, such as competition for resources or violent crime, the logistic model might be applicable; it assumes that the extinction rate is proportional to the number of possible pairs in the popula- tion, p (p # 1)/2:

(3) †

Rodent, bird, and plant populations exhibit logistic growth rates due to social structure, territo- riality, and competition for light and space, respectively. The logistic function

was shown in Section 3.2 to be the solution of (3), and typical graphs of p(t) were displayed there.

In Section 5.3 we observed that the Volterra–Lotka model for two different populations, a predator x2(t) and a prey x1(t), postulates a Malthusian growth rate for the prey and an extinc- tion rate governed by x1x2, the number of possible pairings of one from each population,

(4)

while predators follow a Malthusian extinction rate and pairwise growth rate

(5)

Volterra–Lotka dynamics have been observed in blood vessel growth (predator " new capil- lary tips; prey " chemoattractant), fish populations, and several animal–plant interactions.

Systems like (4)–(5) were studied in Section 5.3 with the aid of the Runge–Kutta algorithm. Now, armed with the insights of Section 5.4, we can further explore this model theoretically.

First, we perform a “reality check” by proving that the populations x1(t), x2(t) in the Volterra–Lotka model never change sign. Separating (4) leads to

while integrating from 0 to t results in

(6)

and the exponential factor is always positive. Thus x1(t) [and similarly x2(t)] retains its initial sign (negative populations never arise).

Find and interpret the critical points for the Volterra–Lotka model (4)–(5).

The system

(7) dx2 dt

" #Cx2 ! Dx1x2 " Dx2 ax1 # CDb " 0 dx1 dt

" Ax1 # Bx1x2 " #Bx1 ax2 # ABb " 0 ,

x1(t) " x1(0)e %

0 EA#Bx2(/)F d/ ,

1 x1

dx1 dt

" d ln x1

dt " A # Bx2 ,

dx2 dt

" #Cx2 ! Dx1x2 .

dx1 dt

" Ax1 # Bx1x2 ,

p (t) " p0p1

p0 ! ( p1 # p0)e #Ap1t

, p0 J p(0)

dp dt

" k1p # k2 p ( p # 1)

2 or, equivalently,

dp dt

" #Ap ( p # p1) .

Section 5.5 Applications to Biomathematics: Epidemic and Tumor Growth Models 277

† One might propose that the growth rate in animal populations is due to two-party interactions as well. (Wink, wink.) However, in monogamous societies, the number of pairs participating in procreation is proportional to p/2, leading to (1). A growth rate determined by all possible pairings p(p # 1)/2 would indicate an extremely utopian social order.

Example 1

Solution

has the trivial solution x1(t) ≡ x2(t) ≡ 0, with an obvious interpretation in terms of populations. If all four coefficients A, B, C, and D are positive, there is also the more interesting solution

(8)

At these population levels, the growth and extinction rates for each species cancel. The direction field diagram in Figure 5.20 for the phase plane equation

(9)

suggests that this equilibrium is a center (compare Figure 5.12) with closed (periodic) neigh- boring trajectories, in accordance with the simulations in Section 5.3. However it is conceiv- able that some spiral trajectories might snake through the field pattern and approach the critical point asymptotically. A rather tricky argument in Problem 4 demonstrates that this is not the case. ◆

The SIR Epidemic Model. The SIR† model for an epidemic addresses the spread of diseases that are only contracted by contact with an infected individual; its victims, once recovered, are immune to further infection and are themselves noninfectious. So the members of a population of size N fall into three classes:

S(t) " the number of susceptible individuals—that is, those who have not been infected; s :" S/N is the fraction of susceptibles.

I(t) " the number of individuals who are currently infected, comprising a fraction i :" I/N of the population.

R(t) " the number of individuals who have recovered from infection, comprising the fraction r :" R/N.

dx2 dx1

" #Cx2 ! Dx1x2 Ax1 # Bx1x2

" x2 x1

#C ! Dx1 A # Bx2

x2(t) ! A B

, x1(t) ! C D

278 Chapter 5 Introduction to Systems and Phase Plane Analysis

Figure 5.20 Typical direction field diagram for the Volterra–Lotka system

(A/B, C/D)

†Introduced by W. O. Kermack and A. G. McKendrick in “A Contribution to the Mathematical Theory of Epidemics,” Proc. Royal Soc. London, Vol. A115 (1927): 700–721.

The classic SIR epidemic model assumes that on the average, an infectious individual encounters a people per unit time (usually per week). Thus, a total of aI people per week are contacted by infectees, but only a fraction s " S/N of them are susceptible. So the susceptible population diminishes at a rate

(10)

(11)

The parameter a is crucial in disease control. Crowded conditions, or high a, make it difficult to combat the spread of infection. Ideally, we would quarantine the infectees (low a) to fight the epidemic.

The infected population is (obviously) increased whenever a susceptible individual is infected. Additionally, infectees recover in a Malthusian-disintegrative manner over an average time of, say, 1/k weeks [recall (1)], so the infected population changes at a rate

(12)

(13)

And, of course, the population of recovered individuals increases whenever an infectee is healed:

(14)

With the SIR model, the total population count remains unchanged:

Thus, any fatalities are tallied in the “recovered/noninfectious” population R. Interestingly, equations (11) and (13) do not contain R or r, so they are suitable for phase

plane analysis. In fact they constitute a Volterra–Lotka system with A " 0, B " D " a, and C " k. Because the coefficient A is zero, the critical point structure is different from that discussed in Example 1. Specifically, if #asi in (11) is zero, then only #ki remains on the right in (13), so I(t) " i(t) ≡ 0 is necessary and sufficient for a critical point, with S unrestricted. (Physi- cally, this means the populations remain stable only if there are no carriers of the infection.)

Our earlier argument has shown that if s(t) and i(t) are initially positive, they remain so. As a result we conclude from (11) immediately that s(t) decreases monotonically; as such, it has a limiting value s(q) as t → q. Does i(t) have a limiting value also? If so, {s(q), i(q)} would be a critical point by Theorem 1 of Section 5.4, and thus i(q) " 0.

To analyze i(t) consider the phase plane equation for (11) and (13):

(15)

which has solutions

(16) i " #s ! k a

ln s ! K .

di ds

" asi # ki

#asi " #1 !

k as

d(S ! I ! R) dt

" #saI ! saI # kI ! kI " 0 .

dR dt

" kI or dr dt

" ki .

di dt " a as # kab i . dI dt

" saI # kI " aas # k a b I or

ds dt " #asi .

dS dt

" #saI or (dividing by N )

Section 5.5 Applications to Biomathematics: Epidemic and Tumor Growth Models 279

From (16) we see that s(q) cannot be zero; otherwise the right-hand side would eventually be negative, contradicting i(t) * 0. Therefore, (16) demonstrates that i(t) does have a limiting value i(q) " #s(q) ! (k/a)ln s(q) ! K. As noted, i(q) must then be zero.

From (13) we further conclude that if s(0) exceeds the “threshold value” k/a, the infected fraction i(t) will initially increase (di/dt * 0 at t " 0) before eventually dying out. The peak value of i(t) occurs when di/dt " 0 " a[s # (k/a)]i, i.e., when s(t) passes through the value k/a. In the jargon of epidemiology, this phenomenon defines an “epidemic.” You will be directed in Problem 10 to show that if s(0) ) k/a, the infected population diminishes monotonically, and no epidemic develops.

According to data issued by the Centers for Disease Control and Prevention (CDC) in Atlanta, Georgia, the Hong Kong flu epidemic during the winter of 1968–1969 was responsible for 1035 deaths in New York City (population 7,900,000), according to the time chart in Figure 5.21. Analyze this data with the SIR model.

Of course, we need to make some assumptions about the parameters. First of all, only a small percentage of people who contract Hong Kong flu perish, so let’s assume that the chart reflects a scaled version of the infected population fraction i(t). It is known that the recovery period for this flu is around 5 days, or 5/7 week, so we try k " 7/5 " 1.4. And since the infectees spend much of their convalescence in bed, the average contact rate a is probably less than 1 person per day or 7 per week. The CDC estimated that the initial infected population I(0) was about 10, so the initial data for (11), (13), and (14) are

s(0) " 7,900,000 # 10

7,900,000 L 0.9999987, i(0) " 10

7,900,000 L 1.2658 & 10#6, r(0) " 0 .

280 Chapter 5 Introduction to Systems and Phase Plane Analysis

Solution

Example 2†

†We borrow liberally from “The SIR Model for Spread of Disease” by Duke University’s David Smith and Lang Moore, Journal of Online Mathematics and Its Applications, The MAA Mathematical Sciences Digital Library, http://mathdl.maa.org/mathDL/4/?pa"content&sa"viewDocument&nodeId"479&bodyId"612, copyright 2000, CCP and the authors, published December, 2001. The article contains much interesting information about this epidemic.

Figure 5.21 Mortality data for Hong Kong flu, New York City

Week

D ea

th s

du e

to fl

1 4 5 9 108 11 13127632

200

100

120

140

160

180

http://mathdl.maa.org/mathDL/4/?pa=content&sa=viewDocument&nodeId=479&bodyId=612

Numerical simulations of this system are displayed in Figure 5.22.† The contact rate a " 3.5 per week generates an infection fraction curve that closely matches the mortality data’s characteristics: time of peak and duration of epidemic. ◆

A Tumor Growth Model.†† The observed growth of certain tumors can be explained by a model that is mathematically similar to the epidemic model. The total number of cells N in the tumor subdivides into a population P that proliferates by splitting (Malthusian growth) and a population Q that remains quiescent. However, the proliferating cells also can make a transi- tion to the quiescent state, and this occurrence is modeled as a Malthusian-like decay with a “rate” r(N) that increases with the overall size of the tumor:

(17)

(18)

Thus the total population N increases only when the proliferating cells split, as can be seen by adding the equations (17) and (18):

(19)

We take (17) and (19) as the system for our analysis. The phase plane equation

(20)

can be integrated, leading to a formula for P in terms of N

(21) P " N # 1 c % r(N) dN ! K .

dP dN

" cP # r(N)P

cP " 1 #

r(N) c

dN dt "

d(P ! Q) dt " cP .

dQ dt " r(N)P .

dP dt " cP # r (N)P ,

Section 5.5 Applications to Biomathematics: Epidemic and Tumor Growth Models 281

†An applet, maintained on the Web at http://alamos.math.arizona.edu/~rychlik/JOde/index.html, automates most of the differential equation algorithms discussed in this book. ††The authors wish to thank Dr. Glenn Webb of Vanderbilt University for this application. See M. Gyllenberg and G. F. Webb, “Quiescence as an Explanation of Gompertz Tumor Growth,” Growth, Development, and Aging, Vol. 53 (1989): 25–55.

(a)

t 30252015105

0.9 1.0

0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1

(b)

t 30252015105

0.9 1.0

0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1

I t

30252015100 5

1.0

0.8

0.6

0.4

0.2

–0.2

S R

(c)

Figure 5.22 SIR simulations (a) k " 1.4, a " 2.0; (b) k " 1.4, a " 3.5; (c) k " 1.4, a " 6.0

http://alamos.math.arizona.edu/~rychlik/JOde/index.html

Suppose the initial conditions are P(0) " 1, Q(0) " 0, and N(0) " 1 (a single proliferating cell). Then we can eliminate the nuisance constant K by taking the indefinite integral in (21) to run from 1 to N and evaluating at t " 0:

Insertion of (21) with K " 0 into (19) produces a differential equation for N alone:

(22)

The Gompertz law

(23)

has been observed experimentally for the growth of some tumors. Show that a transition rate r(N) of the form b(1 ! ln N) predicts Gompertzian growth.

If the indicated integral of the rate r(N) is carried out, (22) becomes

(24)

Dividing by N we obtain a linear differential equation for the function ln N

whose solution, for the initial condition N(0) " 1, is found by the methods of Section 2.3 to be

confirming (23). ◆

Problem 9 invites the reader to show that if the growth rate is modeled as r(N) " s(2N # 1), then the solution of (22) describes logistic growth. Typical curves for the Gompertz and logistic models are displayed in Figure 5.23. See also Figure 3.4 on page 97.

Other applications of differential equations to biomathematics appear in the discussions of arti- ficial respiration (Project B) in Chapter 2, HIV infection (Project A) and aquaculture (Project B) in Chapter 3, and spread of staph infections (Project B) and growth of phytoplankton (Project F) in this chapter.

ln N(t) " c b

(1 # e#bt) ,

d ln N dt

" #b ln N ! c

dN dt

" cN # b(N # 1) # b(N ln N # N ! 1) " (c # b ln N) N .

N(t) " ec(1#e #bt)/b

dN dt

" cN # % N

1 r(u) du .

1 " 1 # 1 c %

1 r(N) dN ! K 1 K " 0 .

282 Chapter 5 Introduction to Systems and Phase Plane Analysis

Figure 5.23 (a) Gompertz and (b) logistic curves

(a)

(b)

Example 3

Solution

1. Logistic Model. In Section 3.2 we discussed the logistic equation

and its use in modeling population growth. A more general model might involve the equation

(25)

where r * 1. To see the effect of changing the param- eter r in (25), take p1 " 3, A " 1, and p0 " 1. Then use a numerical scheme such as Runge–Kutta with h " 0.25 to approximate the solution to (25) on the interval 0 ) t ) 5 for r " 1.5, 2, and 3. What is the limiting population in each case? For r * 1, deter- mine a general formula for the limiting population.

2. Radioisotopes and Cancer Detection. A radioiso- tope commonly used in the detection of breast cancer is technetium-99m. This radionuclide is attached to a chemical that upon injection into a patient accumu- lates at cancer sites. The isotope’s radiation is then detected and the site located, using gamma cameras or other tomographic devices.

Technetium-99m decays radioactively in accor- dance with the equation dy/dt " #ky, with k " 0.1155/h. The short half-life of technetium-99m has the advantage that its radioactivity does not endanger the patient. A disadvantage is that the isotope must be manufactured in a cyclotron. Since hospitals are not equipped with cyclotrons, doses of technetium-99m have to be ordered in advance from medical suppliers.

Suppose a dosage of 5 millicuries (mCi) of technetium-99m is to be administered to a patient. Estimate the delivery time from production at the manufacturer to arrival at the hospital treatment room to be 24 h and calculate the amount of the radionuclide that the hospital must order, to be able to administer the proper dosage.

3. Secretion of Hormones. The secretion of hor- mones into the blood is often a periodic activity. If a hormone is secreted on a 24-h cycle, then the rate of

dp dt

" Ap1p # Ap r , p(0) " p0 ,

dp dt

" Ap1p # Ap 2 , p(0) " p0 ,

Section 5.5 Applications to Biomathematics: Epidemic and Tumor Growth Models 283

change of the level of the hormone in the blood may be represented by the initial value problem

where x(t) is the amount of the hormone in the blood at time t, 0 is the average secretion rate, 1 is the amount of daily variation in the secretion, and k is a positive constant reflecting the rate at which the body removes the hormone from the blood. If 0 " 1 " 1, k " 2, and x0 " 10, solve for x(t).

4. Prove that the critical point (8) of the Volterra–Lotka system is a center; that is, the neighboring trajecto- ries are periodic. Hint: Observe that (9) is separable and show that its solutions can be expressed as

(26)

Prove that the maximum of the function xpe#qx is (p/qe)p, occurring at the unique value x " p/q (see Figure 5.24), so the critical values (8) maximize the factors on the left in (26). Argue that if K takes the corresponding maximum value (A/Be)A(C/De)C, the critical point (8) is the (unique) solution of (26), and it cannot be an endpoint of any trajectory for (26) with a lower value of K.†

[x 2 A e#Bx2 ] # [x 1C e#Dx1 ] " K .

dx dt

" 0 # 1 cos pt 12

# kx , x(0) " x0 ,

xe−x e 1

Figure 5.24 Graph of xe#x

†In fact, the periodic fluctuations predicted by the Volterra–Lotka model were observed in fish populations by Lotka’s son-in-law, Humberto D’Ancona.

5. Suppose for a certain disease described by the SIR model it is determined that a " 0.003 and b " 0.5. (a) In the SI-phase plane, sketch the trajectory cor-

responding to the initial condition that one per- son is infected and 700 persons are susceptible.

5.5 EXERCISES

(b) From your graph in part (a), estimate the peak number of infected persons. Compare this with the theoretical prediction S " k/a # 167 persons when the epidemic is at its peak.

6. Show that the half-life of solutions to (2)—that is, the time required for the solution to decay to one- half of its value—equals (ln 2)/k.

7. Complete the solution of the tumor growth model for Example 3 by finding P(t) and Q(t).

8. If p(t) is a Malthusian population that diminishes according to (2), then p(t2) # p(t1) is the number of individuals in the population whose lifetime lies between t1 and t2. Argue that the average lifetime of the population is given by the formula

and show that this equals 1/k.

% q

0 t ` dp(t)dt ` dt

% q

0 p(t) dt

284 Chapter 5 Introduction to Systems and Phase Plane Analysis

9. Show that with the transition rate formula r(N) " s(2N # 1), equation (22) takes the form of the equa- tion for the logistic model (Section 3.2, equation (14)). Solve (22) for this case.

10. Prove that the infected population I(t) in the SIR model does not increase if S(0) is less than or equal to k/a.

11. An epidemic reported by the British Communicable Disease Surveillance Center in the British Medical Journal (March 4, 1978, p. 587) took place in a boarding school with 763 residents.† The statistics for the infected population are shown in the graph in Figure 5.25.

Assuming that the average duration of the infection is 2 days, use a numerical differential equation solver such as the Web-based one described in Example 2 to try to reproduce the data. Take S(0) " 762, I(0) " 1, R(0) " 0 as initial conditions. Experiment with rea- sonable estimates for the average number of contacts per day by the infected students, who were confined to bed after the infection was detected. What value of this parameter seems to fit the curve best?

†See also the discussion of this epidemic in Mathematical Biology I, An Introduction, by J. D. Murray (Springer- Verlag, New York, 2002), 325–326.

Day

In fe

ct ed

S tu

de nt

1 4 5 9 108 11 151413127632

100

150

200

300

250

Figure 5.25 Flu data for Problem 11

In this section we extend the mass–spring model of Chapter 4 to include situations in which coupled springs connect two masses, both of which are free to move. The resulting motions can be very intriguing. For simplicity we’ll neglect the effects of friction, gravity, and external forces. Let’s analyze the following experiment.

On a smooth horizontal surface, a mass m1 " 2 kg is attached to a fixed wall by a spring with spring constant k1 " 4 N/m. Another mass m2 " 1 kg is attached to the first object by a spring with spring constant k2 " 2 N/m. The objects are aligned horizontally so that the springs are their natural lengths (Figure 5.26). If both objects are displaced 3 m to the right of their equilibrium positions (Figure 5.27) and then released, what are the equations of motion for the two objects?

Section 5.6 Coupled Mass–Spring Systems 285

5.6 COUPLED MASS–SPRING SYSTEMS

x = 0 y = 0

k 1 = 4 k 2 = 2 2 kg 1 kg

x = 3 y = 3

Figure 5.26 Coupled system at equilibrium

x = 0 y = 0

x > 0 y > 0

k 1 = 4 k 2 = 2 2 kg 1 kg

Figure 5.27 Coupled system at initial displacement

Example 1

Solution From our assumptions, the only forces we need to consider are those due to the springs them- selves. Recall that Hooke’s law asserts that the force acting on an object due to a spring has magnitude proportional to the displacement of the spring from its natural length and has direc- tion opposite to its displacement. That is, if the spring is either stretched or compressed, then it tries to return to its natural length.

Because each mass is free to move, we apply Newton’s second law to each object. Let denote the displacement (to the right) of the 2-kg mass from its equilibrium position and similarly, let denote the corresponding displacement for the 1-kg mass. The 2-kg mass has a force F1 acting on its left side due to one spring and a force F2 acting on its right side due to the second spring. Referring to Figure 5.27 and applying Hooke’s law, we see that

since is the net displacement of the second spring from its natural length. There is only one force acting on the 1-kg mass: the force due to the second spring, which is

Applying Newton’s second law to these objects, we obtain the system

(1)

(2)

m2 d2y

dt2 % k2y " k2x ! 0 .

m1 d2x dt2

% Ak1 % k2Bx " k2y ! 0 , m2

d2y

dt2 " F3 " #k2 A y # xB ,

m1 d2x dt2

" F1 ! F2 " #k1x ! k2 A y # xB , F3 " #k2 A y # xB .

A y # xBF1 " #k1x , F2 " !k2 A y # xB , y AtB x

AtB

In this problem, we know that m1 " 2, m2 " 1, k1 " 4, and k2 " 2. Substituting these val- ues into system (2) yields

(3)

(4)

We’ll use the elimination method of Section 5.2 to solve (3)–(4). With we rewrite the system as

(5) (6)

Adding applied to equation (5) to 2 times equation (6) eliminates y:

which simplifies to

(7)

Notice that equation (7) is linear with constant coefficients. To solve it let’s proceed as we did with linear second-order equations and try to find solutions of the form . Substitut- ing in equation (7) gives

Thus, we get a solution to (7) when r satisfies the auxiliary equation

From the factorization we see that the roots of the auxil- iary equation are complex numbers Using Euler’s formula, it follows that

are complex-valued solutions to equation (7). To obtain real-valued solutions, we take the real and imaginary parts of and . Thus, four real-valued solutions are

and a general solution is

(8)

where a1, a2, a3, and a4 are arbitrary constants. †

To obtain a formula for , we use equation (3) to express y in terms of x:

! 3a1 cos t ! 3a2 sin t ! 3a3 cos 2t ! 3a4 sin 2t , " #a1 cos t # a2 sin t # 4a3 cos 2t # 4a4 sin 2t

y AtB " d2x dt2

! 3x

y AtB x AtB ! a1 cos t % a2 sin t % a3 cos 2t % a4 sin 2t , x1 AtB " cos t , x2 AtB " sin t , x3 AtB " cos 2t , x4 AtB " sin 2t ,z2 AtBz1 AtB z1 AtB " eit " cos t ! i sin t and z2 AtB " e2it " cos 2t ! i sin 2ti, #i, 2i, #2i.

r 4 ! 5r 2 ! 4 " Ar 2 ! 1B Ar 2 ! 4B,r 4 ! 5r 2 ! 4 " 0 . 2 Ar 4 ! 5r 2 ! 4Bert " 0 .ert

x " ert

2 d4x dt4

! 10 d2x dt2

! 8x " 0 .

3 AD2 ! 2B A2D2 ! 6B # 4 4 3 x 4 " 0 ,AD2 ! 2B #2x ! AD2 ! 2B 3 y 4 " 0 . A2D2 ! 6B 3 x 4 # 2y " 0 ,

D J d/dt

d2y

dt2 ! 2y # 2x " 0 .

2 d2x dt2

! 6x # 2y " 0 ,

286 Chapter 5 Introduction to Systems and Phase Plane Analysis

†A more detailed discussion of general solutions is given in Chapter 6.

and so

(9)

To determine the constants a1, a2, a3, and a4, let’s return to the original problem. We were told that the objects were originally displaced 3 m to the right and then released. Hence,

(10)

On differentiating equations (8) and (9), we find

Now, if we put t " 0 in the formulas for x, , y, and , the initial conditions (10) give the four equations

From this system, we find a1 " 2, a2 " 0, a3 " 1, and a4 " 0. Hence, the equations of motion for the two objects are

which are depicted in Figure 5.28. ◆

y AtB " 4 cos t # cos 2t ,x AtB " 2 cos t ! cos 2t , y A0B " 2a1 # a3 " 3 , dydt A0B " 2a2 # 2a4 " 0 . x A0B " a1 ! a3 " 3 , dxdt A0B " a2 ! 2a4 " 0 ,

dy/dtdx/dt

dy dt

" #2a1 sin t ! 2a2 cos t ! 2a3 sin 2t # 2a4 cos 2t .

dx dt

" #a1 sin t ! a2 cos t # 2a3 sin 2t ! 2a4 cos 2t ,

x A0B " 3 , dx dt

A0B " 0 ; y A0B " 3 , dy dt

A0B " 0 . y AtB ! 2a1 cos t % 2a2 sin t " a3 cos 2 t " a4 sin 2 t .

Section 5.6 Coupled Mass–Spring Systems 287

2 4

t t

−2−4 2 4

y −2−4

Figure 5.28 Graphs of the motion of the two masses in the coupled mass–spring system

The general solution pair (8), (9) that we have obtained is a combination of sinusoids oscillating at two different angular frequencies: 1 rad/sec and 2 rad/sec. These frequencies extend the notion of the natural frequency of the single (free, undamped) mass–spring oscillator (page 214, Section 4.9) and are called the natural (or normal) angular frequencies† of the system. A com- plex system consisting of more masses and springs would have many normal frequencies.

Notice that if the initial conditions were altered so that the constants a3 and a4 in (8) and (9) were zero, the motion would be a pure sinusoid oscillating at the single frequency 1 rad/sec. Similarly, if a1 and a2 were zero, only the 2 rad/sec oscillation would be “excited.” Such solutions, wherein the entire motion is described by a single sinusoid, are called the normal modes of the system.†† The normal modes in the following example are particularly easy to visualize because we will take all the masses and all the spring constants to be equal.

Three identical springs with spring constant k and two identical masses m are attached in a straight line with the ends of the outside springs fixed (see Figure 5.29). Determine and interpret the normal modes of the system.

We define the displacements from equilibrium, x and y, as in Example 1. The equations expressing Newton’s second law for the masses are quite analogous to (1), except for the effect of the third spring on the second mass:

(11) (12)

Eliminating y in the usual manner results in

(13)

This has the auxiliary equation

with roots . Setting we get the following general solution to (13):

(14) x AtB " C1 cos vt ! C2 sin vt ! C3 cos A23vtB ! C4 sin A23vtB .v J 2k/m,-i2k/m, -i23k/m Amr 2 ! 2kB2 # k2 " Amr 2 ! kB Amr 2 ! 3kB " 0 , 3 AmD2 ! 2kB2 # k2 4 3 x 4 " 0 . #kx ! AmD2 ! 2kB 3 y 4 " 0 .

AmD2 ! 2kB 3 x 4 # ky " 0 , my– " #k A y # xB # ky , mx– " #kx ! k A y # xB ,

288 Chapter 5 Introduction to Systems and Phase Plane Analysis

†The study of the natural frequencies of oscillations of complex systems is known in engineering as modal analysis. ††The normal modes are more naturally characterized in terms of eigenvalues (see Section 9.5).

m k

m kk

x = 0

x * 0 y * 0

y = 0

Figure 5.29 Coupled mass–spring system with fixed ends

Example 2

Solution

To obtain , we solve for in (11) and substitute as given in (14). Upon simplifying, we get

(15)

From the formulas (14) and (15), we see that the normal angular frequencies are

Indeed, if C3 " C4 " 0, we have a solution where oscillating at the

angular frequency rad/sec (equivalent to a frequency periods/sec). Now if in Figure 5.29, the two masses are moving as if they were a single rigid body of mass 2m, forced by a “double spring” with a spring constant given by 2k. And indeed, according to equation (4) of Section 4.9 (page 214), we would expect such a system to oscillate

at the angular frequency (!). This motion is depicted in Figure 5.30(a). Similarly, if C1 " C2 " 0, we find the second normal mode where so that in

Figure 5.29 there are two mirror-image systems, each with mass m and a “spring and a half” with spring constant k ! 2k " 3k. (The half-spring will be twice as stiff.) Section 4.9’s

equation (4) then predicts an angular oscillation frequency for each system of which again is consistent with (14) and (15). This motion is shown in Figure 5.30(b). ◆

23k/m " 23 v, y AtB " #x AtB,22k/2m " 2k/m

x AtB ! y AtB 2k/m /2pv " 2k/m y AtB ! x AtB,v and 23 v. y AtB " C1 cos vt ! C2 sin vt # C3 cos A23vtB # C4 sin A23vtB .x AtBy AtBy AtB

Section 5.6 Coupled Mass–Spring Systems 289

t t

y x

1−1

(a) (b)

Figure 5.30 Normal modes for Example 2

1. Two springs and two masses are attached in a straight line on a horizontal frictionless surface as illustrated in Figure 5.31. The system is set in motion by holding the mass m2 at its equilibrium position and pulling the mass m1 to the left of its equilibrium position a distance 1 m and then releasing both masses. Express Newton’s law for the

k 1 k 2

x * 0 y * 0

x = 0 y = 0

m1 m2

Figure 5.31 Coupled mass–spring system with one end free

5.6 EXERCISES

system and determine the equations of motion for the two masses if m1 " 1 kg, m2 " 2 kg, k1 " 4 N/m, and N/m.

2. Determine the equations of motion for the two masses described in Problem 1 if m1 " 1 kg, m2 " 1 kg, k1 " 3 N/m, and k2 " 2 N/m.

3. Four springs with the same spring constant and three equal masses are attached in a straight line on a hori- zontal frictionless surface as illustrated in Figure 5.32. Determine the normal frequencies for the system and describe the three normal modes of vibration.

k2 " 10 /3

290 Chapter 5 Introduction to Systems and Phase Plane Analysis

6. Referring to the coupled mass–spring system discussed in Example 1, suppose an external force

cos 3t is applied to the second object of mass 1 kg. The displacement functions and now satisfy the system

(16) (17)

(a) Show that satisfies the equation

(18)

(b) Find a general solution to equation (18). [Hint: Use undetermined coefficients with xp " A cos 3t ! B sin 3t.]

(d) If both masses are displaced 2 m to the right of their equilibrium positions and then released, find the displacement functions and .

7. Suppose the displacement functions and for a coupled mass–spring system (similar to the one dis- cussed in Problem 6) satisfy the initial value problem

Solve for and . 8. A double pendulum swinging in a vertical plane

under the influence of gravity (see Figure 5.35 on page 291) satisfies the system

m2l 2 2u–2 ! m2l1l2u–1 ! m2l2gu2 " 0 ,

Am1 ! m2Bl 21u–1 ! m2l1l2u–2 ! Am1 ! m2Bl1gu1 " 0 , y AtBx AtBy A0B ! 1 , y¿ A0B ! 0 .

x A0B ! x$ A0B ! 0 ,y& AtB % 2y AtB " 2x AtB ! 3 sin 2t ; x& AtB % 5x AtB " 2y AtB ! 0 ,

y AtBx AtB y AtBx AtB y AtB x AtB

x AtBx A4B AtB % 5x& AtB % 4x AtB ! 37 cos 3t .x AtB

y& AtB % 2y AtB " 2x AtB ! 37 cos 3t .2x& AtB % 6x AtB " 2y AtB ! 0 , y AtBx AtBE AtB " 37

k k k k m m m

x * 0 y * 0 z * 0

x = 0 y = 0 z = 0

Figure 5.32 Coupled mass–spring system with three degrees of freedom

4. Two springs, two masses, and a dashpot are attached in a straight line on a horizontal frictionless surface as shown in Figure 5.33. The dashpot provides a damping force on mass m2, given by Derive the system of differential equations for the displacements x and y.

F " #by¿.

k 1 k 2 b m 2 m 1

x * 0 y * 0

x y = 0 = 0

Figure 5.33 Coupled mass–spring system with one end damped

5. Two springs, two masses, and a dashpot are attached in a straight line on a horizontal frictionless surface as shown in Figure 5.34. The system is set in motion by holding the mass m2 at its equilibrium position and pushing the mass m1 to the left of its equilibrium position a distance 2 m and then releasing both masses. Determine the equations of motion for the two masses if m1 " m2 " 1 kg, k1 " k2 " 1 N/m, and b " 1 N-sec/m. [Hint: The dashpot damps both m1 and m2 with a force whose magnitude equals b 0 y¿ # x¿ 0 . ]

k 1 k 2 b m 2 m 1

x * 0 y * 0

x y = 0 = 0

Figure 5.34 Coupled mass–spring system with damping between the masses

Section 5.7 Electrical Systems 291

Figure 5.35 Double pendulum

l l

x1 x2

m m

Figure 5.36 Coupled pendulums

10. Suppose the coupled mass–spring system of Problem 1 (Figure 5.31) is hung vertically from a support (with mass m2 above m1), as in Section 4.10, page 228. (a) Argue that at equilibrium, the lower spring is

stretched a distance l1 from its natural length L1, given by .

(b) Argue that at equilibrium, the upper spring is stretched a distance .

(c) Show that if x1 and x2 are redefined to be dis- placements from the equilibrium positions of the masses m1 and m2, then the equations of motion are identical with those derived in Problem 1.

l2 " Am1 ! m2Bg/k2 l1 " m1g/k1

when are small angles. Solve the system when m1 " 3 kg, m2 " 2 kg, l1 " l2 " 5 m,

, 9. The motion of a pair of identical pendulums coupled

by a spring is modeled by the system

for small displacements (see Figure 5.36). Deter- mine the two normal frequencies for the system.

mx–2 " # mg l

x2 ! k Ax1 # x2B mx–1 " # mg l

x1 # k Ax1 # x2B , u2 A0B " u¿1 A0B " u¿2 A0B " 0.u1 A0B " p /6

u1 and u2

5.7 ELECTRICAL SYSTEMS The equations governing the voltage–current relations for the resistor, inductor, and capacitor were given in Section 3.5, together with Kirchhoff’s laws that constrain how these quantities behave when the elements are electrically connected into a circuit. Now that we have the tools for solving linear equations of higher-order systems, we are in a position to analyze more com- plex electrical circuits.

The series RLC circuit in Figure 5.37 on page 292 has a voltage source given by E t " sin 100t volts (V), a resistor of 0.02 ohms (Ω), an inductor of 0.001 henrys (H), and a capacitor of 2 farads (F). (These values are selected for numerical convenience; typical capacitance values are much smaller.) If the initial current and the initial charge on the capacitor are both zero, determine the current in the circuit for t * 0.

Using the notation of Section 3.5, we have L " 0.001 H, R " 0.02 Ω, C " 2 F, and E t " sin 100t. According to Kirchhoff’s current law, the same current I passes through each circuit ele- ment. The current through the capacitor equals the instantaneous rate of change of its charge q:

(1) I ! dq/dt .

BAExample 1

Solution

From the physics equations in Section 3.5, we observe that the voltage drops across the capacitor (EC), the resistor (ER), and the inductor (EL) are expressed as

(2)

Therefore, Kirchhoff’s voltage law, which implies EL ! ER ! EC " E, can be expressed as

(3)

In most applications we will be interested in determining the current . If we differentiate (3) with respect to t and substitute I for , we obtain

(4)

After substitution of the given values this becomes

or, equivalently,

(5)

The homogeneous equation associated with (5) has the auxiliary equation

whose roots are Hence, the solution to the homogeneous equation is

(6)

To find a particular solution for (5), we can use the method of undetermined coefficients. Setting

and carrying out the procedure in Section 4.5, we ultimately find, to three decimals,

Hence, a particular solution to (5) is given by

(7) Ip AtB " #10.080 cos 100t ! 2.122 sin 100t . A " #10.080 , B " 2.122 .

Ip AtB " A cos 100t ! B sin 100t Ih AtB " C1e#10t cos 20t ! C2e#10t sin 20t .#10 - 20i. r 2 ! 20r ! 500 " Ar ! 10B2 ! A20B2 " 0 , d2I dt2

! 20 dI dt

! 500I " 100,000 cos 100t .

A0.001) d2I dt2

! A0.02B dI dt

! (0.5)I " 100 cos 100t ,

L d 2I

dt2 % R dI

dt %

1 C

I ! dE dt

dq/dt I AtBL

dI dt % RI %

C q ! E AtB .

EC ! q C , ER ! RI , EL ! L

dI dt .

292 Chapter 5 Introduction to Systems and Phase Plane Analysis

Resistance R

L Voltage source

Capacitance C

Inductance

Figure 5.37 Schematic representation of an RLC series circuit

Since I " Ih ! Ip, we find from (6) and (7) that

(8)

To determine the constants C1 and C2, we need the values and We were given . To find we substitute the values for L, R, and C into equation (3) and

equate the two sides at t " 0. This gives

Because , we find Finally, using in (8) and the initial conditions we obtain the system

Solving this system yields C1 " 10.080 and C2 " #5.570. Hence, the current in the RLC series circuit is

(9) ◆

Observe that, as was the case with forced mechanical vibrations, the current in (9) is made up of two components. The first, Ih, is a transient current that tends to zero as The second,

is a sinusoidal steady-state current that remains. It is straightforward to verify that the steady-state solution that arises from the more

general voltage source is

(10)

where tan (compare Section 4.10, page 224).

At time t " 0, the charge on the capacitor in the electrical network shown in Figure 5.38 is 2 coulombs (C), while the current through the capacitor is zero. Determine the charge on the capacitor and the currents in the various branches of the network at any time t * 0.

u " A1/C # Lg2B / AgRB Ip AtB " E0 sin Agt ! uB2R2 ! 3gL # 1/ AgCB 4 2 ,

E AtB " E0 sin gt Ip AtB Ip AtB " #10.080 cos 100t ! 2.122 sin 100t ,

t S !q.

# 10.080 cos 100t !2.122 sin 100t .I AtB" e#10t A10.080 cos 20t #5.570 sin 20tB I¿ A0B " #10C1 ! 20C2 ! 212.2 " 0 . I A0B " C1 # 10.080 " 0 ,

I A0B " I¿ A0B " 0, I AtBI¿ A0B " 0.I A0B " q A0B " 0 A0.001BI¿ A0B ! A0.02BI A0B ! A0.5Bq A0B " sin 0 .

I¿ A0B,I A0B " q A0B " 0 I¿ A0B.I A0B I AtB " e#10t AC1 cos 20t ! C2 sin 20tB # 10.080 cos 100t ! 2.122 sin 100t .

Section 5.7 Electrical Systems 293

5 volts (V)

20 ohms (Ω)

loop 1

loop 3 loop 2

1 henry (H) 1160 farads (F)

+q3

−q3

I1 I3

Β I3

Figure 5.38 Schematic of an electrical network

Example 2

To determine the charge and currents in the electrical network, we begin by observing that the network consists of three closed circuits: loop 1 through the battery, resistor, and induc- tor; loop 2 through the battery, resistor, and capacitor; and loop 3 containing the capacitor and inductor. Taking advantage of Kirchhoff’s current law, we denote the current passing through the battery and the resistor by I1, the current through the inductor by I2, and the current through the capacitor by I3. For consistency of notation, we denote the charge on the capacitor by q3. Hence, .

As discussed at the beginning of this section, the voltage drop at a resistor is RI, at an inductor , and at a capacitor . So, applying Kirchhoff’s voltage law to the electrical network in Figure 5.38, we find for loop 1,

(11)

(inductor) (resistor) (battery)

for loop 2,

(12) (resistor) (capacitor) (battery)

and for loop 3,

(13)

(inductor) (capacitor)

[The minus sign in (13) arises from taking a clockwise path around loop 3 so that the current passing through the inductor is #I2.] Notice that these three equations are not independent: We can obtain equation (13) by subtracting (11) from (12). Hence, we have only two equations from which to determine the three unknowns I1, I2, and q3. If we now apply Kirchhoff’s current law to the two junction points in the network, we find at point A that and at point B that In both cases, we get

(14)

since . Assembling equations (11), (12), and (14) into a system, we have (with )

(15) (16) (17)

We solve these by the elimination method of Section 5.2. Using equation (16) to eliminate I1 from the others, we are left with

(18) (19)

Elimination of I2 then leads to

(20) 20D2q3 ! 160Dq3 ! 3200q3 " 0 .

20I2 ! A20D ! 160Bq3 " 5 . DI2 # 160q3 " 0 , #I2 ! I1 # Dq3 " 0 .

20I1 ! 160q3 " 5 ,

DI2 ! 20I1 " 5 ,

D " d/dt I3 " dq3/dt

I1 # I2 # dq3 dt

" 0 ,

I2 ! I3 # I1 " 0. I1 # I2 # I3 " 0

" dI2 dt % 160q3 ! 0 .

20I1 ! 160q3 " 5 ;

dI2 dt

! 20I1 " 5 ;

q/CLdI/dt

I3 " dq3/dt

294 Chapter 5 Introduction to Systems and Phase Plane Analysis

Solution

To obtain the initial conditions for the second-order equation (20), recall that at time t " 0, the charge on the capacitor is 2 coulombs and the current is zero. Hence,

(21)

We can now solve the initial value problem (20)–(21) using the techniques of Chapter 4. Ultimately, we find

Next, to determine I2, we substitute these expressions into (19) and obtain

Finally, from we get

◆

Note that the differential equations that describe mechanical vibrations and RLC series circuits are essentially the same. And, in fact, there is a natural identification of the parameters m, b, and k for a mass–spring system with the parameters L, R, and C that describe circuits. This is illustrated in Table 5.3. Moreover, the terms transient, steady-state, overdamped, criti- cally damped, underdamped, and resonant frequency described in Sections 4.9 and 4.10 apply to electrical circuits as well.

This analogy between a mechanical system and an electrical circuit extends to large-scale systems and circuits. An interesting consequence of this is the use of analog simulation and, in particular, analog computers to analyze mechanical systems. Large-scale mechanical systems are modeled by building a corresponding electrical system and then measuring the charges and currents .I AtB q AtB

I1 AtB " 14 # 16e#4t cos 12t # 163 e#4t sin 12t . I1 " I2 ! I3,

" 1 4

# 16e#4t cos 12t ! 64 3

e#4t sin 12t .

I2 AtB " 14 # dq3dt AtB # 8q3 AtB I3 AtB " dq3dt AtB " # 803 e#4t sin 12t . q3 AtB " 2e#4t cos 12t ! 23 e#4t sin 12t ,

q3 A0B " 2 , dq3dt A0B " 0 .

Section 5.7 Electrical Systems 295

TABLE 5.3 Analogy Between Mechanical and Electrical Systems

Mechanical Mass–Spring Electrical RLC System with Damping Series Circuit

Displacement x Charge q Velocity Current Mass m Inductance L Damping constant b Resistance R Spring constant k (Capacitance)#1

External force Voltage source E AtBf AtB 1/C q¿ " Ix¿

Lq– ! Rq¿ ! A1/CBq " E AtBmx– ! bx¿ ! kx " f AtB

Although such analog simulations are important, both large-scale mechanical and electri- cal systems are currently modeled using digital computer simulation. This involves the numer- ical solution of the initial value problem governing the system. Still, the analogy between mechanical and electrical systems means that basically the same computer software can be used to analyze both systems.

296 Chapter 5 Introduction to Systems and Phase Plane Analysis

1. An RLC series circuit has a voltage source given by V, a resistor of 100 Ω, an inductor of 4 H,

and a capacitor of 0.01 F. If the initial current is zero and the initial charge on the capacitor is 4 C, deter- mine the current in the circuit for t * 0.

2. An RLC series circuit has a voltage source given by cos 2t V, a resistor of 2 Ω, an inductor of

H, and a capacitor of F. If the initial cur- rent is zero and the initial charge on the capacitor is 3.5 C, determine the charge on the capacitor for t * 0.

3. An RLC series circuit has a voltage source given by cos 20t V, a resistor of 120 Ω, an inductor

of 4 H, and a capacitor of F. Find the steady-state current (solution) for this circuit. What is the resonance frequency of the circuit?

4. An LC series circuit has a voltage source given by sin 50t V, an inductor of 2 H, and a capac-

itor of 0.02 F (but no resistor). What is the current in this circuit for t * 0 if at t " 0, ?

5. An RLC series circuit has a voltage source of the form V, a resistor of 10 Ω, an inductor of

4 H, and a capacitor of 0.01 F. Sketch the frequency response curve for this circuit.

6. Show that when the voltage source in (4) is of the form then the steady-state solution Ip is as given in equation (10).

7. A mass–spring system with damping consists of a 7-kg mass, a spring with spring constant 3 N/m, a fric- tional component with damping constant 2 N-sec/m, and an external force given by cos 10t N. Using a 10-2 resistor, construct an RLC series circuit that is the analog of this mechanical system in the sense that the two systems are governed by the same differential equation.

8. A mass–spring system with damping consists of a 16-lb weight, a spring with spring constant 64 lb/ft, a frictional component with damping constant 10 lb-sec/ft, and an external force given by 20 cos 8t lb. Using an inductor of 0.01 H, construct

f AtB "

f AtB " 10

E AtB " E0 sin gt, E AtB " E0 cos gt

I A0B " q A0B " 0 E AtB " 30

A2200B#1E AtB " 10 1 /131 /4

E AtB " 40

E AtB " 20 an RLC series circuit that is the analog of thismechanical system. 9. Because of Euler’s formula, it

is often convenient to treat the voltage sources and simultaneously, using

In this case, equation (3) becomes

(22)

where q is now complex (recall ). (a) Use the method of undetermined coefficients to

show that the steady-state solution to (22) is

The technique is discussed in detail in Project F at the end of Chapter 4.

(b) Now show that the steady-state current is

where tan

(d) The imaginary part of so the imaginary part of the solution to (22) must be the solution to equation (3) for Verify that this is also the case for the current by showing that the imaginary part of Ip in part (c) is the same as that given in equation (10).

In Problems 10–13, find a system of differential equations and initial conditions for the currents in the networks given in the schematic diagrams (Figures 5.39–5.42 on page 297). Assume that all initial currents are zero. Solve for the currents in each branch of the network.

E AtB " E0 sin gt. eigt is sin gt,

u " A1 /C # Lg2B / AgRB. Ip AtB " E02R2 ! 3gL # 1 / AgCB 4 2 eiAgt!uB ,

u " b /a, a ! ib " 2a2 ! b2eiu,Ip AtB " E0R ! i 3gL # 1 / AgCB 4 eigt .

qp AtB " E01 /C # g2L ! igR eigt . I " q¿, I¿ " q–

L d2q

dt2 % R

dt %

C q ! E0eiGt ,

E AtB " E0eigt. E0 sin gtE0 cos gt e iu " cos u ! i sin u,

5.7 EXERCISES

Section 5.8 Dynamical Systems, Poincaré Maps, and Chaos 297

10 H 30 H

40 Ω20 V 10 Ω

I1 I2 I3

Figure 5.39 RL network for Problem 10

10 Ω 5 Ω

F13010 V 20 H

I1 I3 I2

Figure 5.40 RLC network for Problem 11

0.025 H10 V 0.02 H

I1 I3 I2

10 Ω

Figure 5.41 RL network for Problem 12

0.5 F

cos 3t V0.5 H 1 Ω

I1 I3 I2

Figure 5.42 RLC network for Problem 13

10.

11.

12.

13.

5.8 DYNAMICAL SYSTEMS, POINCARÉ MAPS, AND CHAOS In this section we take an excursion through an area of mathematics that has received a lot of attention both for the interesting mathematical phenomena being observed and for its applica- tion to fields such as meteorology, heat conduction, fluid mechanics, lasers, chemical reactions, and nonlinear circuits, among others. The area is that of nonlinear dynamical systems.†

A dynamical system is any system that allows one to determine (at least theoretically) the future states of the system given its present or past state. For example, the recursive formula (difference equation)

is a dynamical system, since we can determine the next state, given the previous state, . If we know x0, then we can compute any future state indeed,

Another example of a dynamical system is provided by the differential equation

where the solution specifies the state of the system at “time” t. If we know , then we can determine the state of the system at any future time t * t0 by solving the initial value problem

Indeed, a simple calculation yields x AtB " x0e#2At#t0B for t ' t0. dx dt

" #2x , x At0B " x0 . x At0B " x0x AtB

dx dt

" #2x ,

xn!1 " x0 A1.05Bn!1 4 .3 xnxn!1, xn!1 " A1.05Bxn , n " 0, 1, 2, . . .

†For a more detailed study of dynamical systems, see An Introduction to Chaotic Dynamical Systems, 2nd ed., by R. L. Devaney (Addison-Wesley, Reading, Mass., 1989) and Nonlinear Oscillations, Dynamical Systems and Bifurcations of Vector Fields, by J. Guckenheimer and P. J. Holmes (Springer-Verlag, New York, 1983).

For a dynamical system defined by a differential equation, it is often helpful to work with a related dynamical system defined by a difference equation. For example, when we cannot express the solution to a differential equation using elementary functions, we can use a numer- ical technique such as the improved Euler’s method or Runge–Kutta to approximate the solution to an initial value problem. This numerical scheme defines a new (but related) dynam- ical system that is often easier to study.

In Section 5.4, we used phase plane diagrams to study autonomous systems in the plane. Many important features of the system can be detected just by looking at these diagrams. For example, a closed trajectory corresponds to a periodic solution. The trajectories for nonautonomous systems in the phase plane are much more complicated to decipher. One tech- nique that is helpful in this regard is the so-called Poincaré map. As we will see, these maps replace the study of a nonautonomous system with the study of a dynamical system defined by the location in the -plane of the solution at regularly spaced moments in time such as where n " 0, 1, 2, . . . . The advantage in using the Poincaré map will become clear when the method is applied to a nonlinear problem for which no explicit solution is known. In such a case, the trajectories are computed using a numerical scheme such as Runge–Kutta. Sev- eral software packages have options that will construct Poincaré maps for a given system.

To illustrate the Poincaré map, consider the equation

(1)

where F and are positive constants. We studied similar equations in Section 4.10 and found that a general solution for is given by

(2)

where the amplitude A and the phase angle are arbitrary constants. Since ,

Because the forcing function F cos t is -periodic, it is natural to seek -periodic solutions to (1). For this purpose, we define the Poincaré map as

(3)

for n " 0, 1, 2, . . . . In Figure 5.43 on page 299, we plotted the first 100 values of in the -plane for different choices of For simplicity, we have taken A " F " 1 and These

graphs are called Poincaré sections. We will interpret them shortly. Now let’s play the following game. We agree to ignore the fact that we already know the

formula for for all We want to see what information about the solution we can glean just from the Poincaré section and the form of the differential equation.

Notice that the first two Poincaré sections in Figure 5.43, corresponding to and 3, consist of a single point. This tells us that, starting with t " 0, every increment of t returns us to the same point in the phase plane. This in turn implies that equation (1) has a -periodic solution, which can be proved as follows: For let be the solution to (1) with

and let Since the Poincaré section is just the point , we have Thus, and have the

same initial values at t " 0. Further, because cos t is -periodic, we also have

X– AtB ! v2X AtB " x– At ! 2pB ! v2x At ! 2pB " cos At ! 2pB " cos t .2p X AtBx AtBX A0B " x A2pB " 1/3 and X¿ A0B " x¿ A2pB " 2.A1/3, 2B X AtB J x At ! 2pB.Ax A0B, y A0BB " A1/3, 2B x AtBv " 2, 2p

2p v " 2

t ' 0.x AtB f " 0.v.xy Axn, ynByn J y A2pnB " vA cos A2pvn ! fB ,

xn J x A2pnB " A sin A2pvn ! fB ! F/ Av2 # 1B , 2p2p

y AtB " vA cos Avt ! fB # F v2 # 1

sin t .

y " x¿f

x AtB " A sin Avt ! fB ! F v2 # 1

cos t ,

v + 1 v

x– AtB ! v2x AtB " F cos t ,

t " 2pn, Ay " dx/dtBxy

298 Chapter 5 Introduction to Systems and Phase Plane Analysis

Consequently, and satisfy the same initial value problem. By the uniqueness theorems of Sections 4.2 and 4.5, these functions must agree on the interval Hence,

for all that is, is -periodic. (The same reasoning works for With a similar argument, it follows from the Poincaré section for that there is

a solution of period that alternates between the two points displayed in Figure 5.43(c) as t is incremented by For the case we deduce that there is a solution of period rotating among three points, and for there is an -periodic solution rotating among four points. We call these last three solutions subharmonics.

The case is different. So far, in Figure 5.43(f), none of the points has repeated.

Did we stop too soon? Will the points ever repeat? Here, the fact that is irrational plays a crucial role. It turns out that every integer n yields a distinct point in the Poincaré section (see Problem 8). However, there is a pattern developing. The points all appear to lie on a simple

curve, possibly an ellipse. To see that this is indeed the case, notice that when and we have

xn " sin A222pnB ! 1 , yn " 22 cos A222pnB , n " 0, 1, 2, . . . .f " 0,A " F " 1, v " 22, 12v " 12

8pv " 1/4, 6pv " 1/3,2p.

4p v " 1/2v " 3.B 2px AtBt ' 0;x At ! 2pBX AtB "

x AtB "[0, q B.X AtBx AtB Section 5.8 Dynamical Systems, Poincaré Maps, and Chaos 299

−1

−1 1 x

−2

(e)

1 –– 4

n = 0

(c)

1 –– 2

−1

−2 −1 1

n = 0

−1

= √2

(f)

n = 0

−1

−1 1 −2

(d)

1 –– 3

n = 0

= 2

(a)

−1

−1 1 2

n = 0, 1, 2, . . . ( , 2)1 ––

−1 1 2

(b)

= 3

n = 0, 1, 2, . . .

Figure 5.43 Poincaré sections for equation (1) for various values of v

It is then an easy computation to show that each lies on the ellipse

In our investigation of equation (1), we concentrated on -periodic solutions because the forcing term F cos t has period [We observed subharmonics when and — that is, solutions with periods When a damping term is introduced into the differential equation, the Poincaré map displays a different behavior. Recall that the solution will now be the sum of a transient and a steady-state term. For example, let’s analyze the equation

(4)

where b, F, and are positive constants. When the solution to (4) can be expressed as

(5)

where tan and A and are arbitrary constants [compare equations (7) and (8) in Section 4.10]. The first term on the right-hand side of (5) is the transient and the second term, the steady-state solution. Let’s construct the Poincaré map using n " 0, 1, 2, . . . . We will take b " 0.22, to simplify the computations. Because tan as well. Then we have

The Poincaré section for n " 0, 1, 2, . . . , 10 is shown in Figure 5.44 (black points). After just a few iterations, we observe that that is, the points of the Poincaré section are approaching a single point in the -plane (colored point). Thus, we might expect that there is a -periodic solution corresponding to a particular choice of A and [In this example, where we can explicitly represent the solution, we see that indeed a -periodic solution arises when we take A " 0 in (5).]

2p f.2p

xy xn # 0 and yn # 1/ A0.22B # 4.545;

! 20.9879 e#0.22pncos A20.9879 2pnB ! 1A0.22B . x¿ A2pnB " yn " #0.11e#0.22pnsin A20.9879 2pnB x A2pnB " xn " e#0.22pnsin A20.9879 2pnB ,u " Av2 # 1B /b " 0, we will take u " 0v " A " F " 1, and f " 0

t " 2pn,

fu " Av2 # 1B /b x AtB " Ae#Ab/2Bt sin a24v2 # b2

2 t ! fb ! F2 Av2 # 1B2 ! b2 sin At ! uB ,

b2 6 4v2, v

x– AtB ! bx¿ AtB ! v2x AtB " F cos t , 2 A2pB, 3 A2pB, and 4 A2pB. ] 1/4v " 1/2, 1/3,2p. 2p

Ax # 1B2 ! y2 2

" 1 .

Axn, ynB 300 Chapter 5 Introduction to Systems and Phase Plane Analysis

5.4

5.2

4.8

4.6

−0.005−0.01−0.015 x

Figure 5.44 Poincaré section for equation (4) with , and v " 1F " 1, b " 0.22

There is an important difference between the Poincaré sections for equation (1) and those for equation (4). In Figure 5.43, the location of all of the points in (a)–(e) depends on the initial value selected (here A " 1 and ). (See Problem 10.) However, in Figure 5.44, the first few points (black points) depend on the initial conditions, while the limit point (colored point) does not (see Problem 6). The latter behavior is typical for equations that have a “damping” term (i.e., b * 0); namely, the Poincaré section has a limit set† that is essentially independent of the initial conditions.

For equations with damping, the limit set may be more complicated than just a point. For example, the Poincaré map for the equation

(6)

has a limit set consisting of an ellipse (see Problem 11). This is illustrated in Figure 5.45 for the initial values x0 " 2, and x0 " 2, .

So far we have seen limit sets for the Poincaré map that were either a single point or an ellipse—independent of the initial values. These particular limit sets are attractors. In general, an attractor is a set A with the property that there exists an open set†† B containing A such that whenever the Poincaré map enters B, its points remain in B and the limit set of the Poincaré map is a subset of A. Further, we assume A has the invariance property: Whenever the Poincaré map starts at a point in A, it remains in A.

In the previous examples, the attractors of the dynamical system (Poincaré map) were easy to describe. In recent years, however, many investigators, working on a variety of applications, have encountered dynamical systems that do not behave orderly—their attractor sets are very complicated (not just isolated points or familiar geometric objects such as ellipses). The behav- ior of such systems is called chaotic, and the corresponding limit sets are referred to as strange attractors.

y0 " 6y0 " 4

x– AtB ! A0.22Bx¿ AtB ! x AtB " cos t ! cos A22 tB f " 0

Section 5.8 Dynamical Systems, Poincaré Maps, and Chaos 301

†The limit set for a map , n " 1, 2, 3, . . . , is the set of points such that lim where is some subsequence of the positive integers.

††A set is an open set if for each point there is an open disk V containing p such that V ( B.p ' BB ( R2 n1 6 n2 6 n3 6 p

kSq Axnk, ynkB " A p, qB,A p, qBAxn, ynB

−1 1 2 x

(a) x 0 = 2, 0 = 4

−2

1 2 x

(b) x 0 = 2, 0 = 6

−1−2

Figure 5.45 Poincaré section for equation (6) with initial values x0, y0

To illustrate chaotic behavior and what is meant by a strange attractor, we discuss two nonlinear differential equations and a simple difference equation. First, let’s consider the forced Duffing equation

(7)

We cannot express the solution to (7) in any explicit form, so we must obtain the Poincaré map by numerically approximating the solution to (7) for fixed initial values and then plot the approximations for (Because the forcing term F sin has period we seek -periodic solutions and subharmonics.) In Figure 5.46, we display the limit sets (attractors) when b " 0.3 and in the cases (a) F " 0.2, (b) F " 0.28, (c) F " 0.29, and (d) F " 0.37.

Notice that as the constant F increases, the Poincaré map changes character. When F " 0.2, the Poincaré section tells us that there is a -periodic solution. For F " 0.28, there is a subharmonic of period and for F " 0.29 and 0.37, there are subharmonics with periods

and respectively. Things are dramatically different when F " 0.5: The solution is neither -periodic nor

subharmonic. The Poincaré section for F " 0.5 is illustrated in Figure 5.47 on page 303. This section was generated by numerically approximating the solution to (7) when b " 0.3, and F " 0.5, for fixed initial values.† Not all of the approximations that were calculated are graphed; because of the presence of a transient solution, the first few points were omitted. It turns out that the plotted set is essentially independent of the initial values and has the property that once a point is in the set, all subsequent points lie in the set. Because of the

x A2pn/gB and y A2pn/gBg " 1.2, 2p/g

10p/g,8p/g 4p/g,

2p/g

g " 1.2 2p/g2p/g,gt

x A2pn/gB and y A2pn/gB " x¿ A2pn/gB. x& AtB % bx$ AtB " x AtB % x3 AtB ! F sin Gt .

302 Chapter 5 Introduction to Systems and Phase Plane Analysis

(d) F = 0.37

1 x

−1

Figure 5.46 Poincaré sections for the Duffing equation (7) with b " 0.3 and g " 1.2

(a) F = 0.2

−1 1 x

−1 1

(b) F = 0.28

1 x

−1

†Historical Footnote: When researchers first encountered these strange-looking Poincaré sections, they would check their computations using different computers and different numerical schemes [see Hénon and Heiles, “The Applica- bility of the Third Integral of Motion: Some Numerical Experiments,” Astronomical Journal, Vol. 69 (1964): 75]. For special types of dynamical systems, such as the Hénon map, it can be shown that there exists a true trajectory that shadows the numerical trajectory [see M. Hammel, J. A. Yorke, and C. Grebogi, “Numerical Orbits of Chaotic Processes Represent True Orbits,” Bulletin American Mathematical Society, Vol. 19 (1988): 466–469].

complicated shape of the set, it is indeed a strange attractor. While the shape of the strange attractor does not depend on the initial values, the picture does change if we consider different sections; for example, n " 0, 1, 2, . . . yields a different configuration.

Another example of a strange attractor occurs when we consider the forced pendulum equation

(8)

where the term in (4) has been replaced by . Here is the angle between the pendulum and the vertical rest position, b is related to damping, and F represents the strength of the forcing function (see Figure 5.48). For F " 2.7 and b " 0.22, we have graphed in Figure 5.49 approximately 90,000 points in the Poincaré map. Since we cannot express the solution to (8) in any explicit form, the Poincaré map was obtained by numerically approximating the solution to (8) for fixed initial values and plotting the approximations for

and The Poincaré maps for the forced Duffing equation and for the forced pendulum equation

not only illustrate the idea of a strange attractor; they also exhibit another peculiar behavior called chaos. Chaos occurs when small changes in the initial conditions lead to major changes in the behavior of the solution. Henri Poincaré described the situation as follows:

It may happen that small differences in the initial conditions will produce very large ones in the final phenomena. A small error in the former produces an enormous error in the latter. Prediction becomes impossible . . . .

y A2pnB " x¿ A2pnB.x A2pnB

x AtBsin Ax AtBBx AtBx& AtB % bx$ AtB % sin Ax AtBB ! F cos t , t " A2pn ! p/2B /g,

Section 5.8 Dynamical Systems, Poincaré Maps, and Chaos 303

x −1

−1

Figure 5.47 Poincaré section for the Duffing equation (7) with b " 0.3, and F " 0.5g " 1.2,

F cos t

−bx ′(t)

x(t)

Figure 5.48 Forced damped pendulum

x −3 −2 −1

−1

3210

Figure 5.49 Poincaré section for the forced pendulum equation (8) with b " 0.22 and F " 2.7

In a physical experiment, we can never exactly (with infinite accuracy) reproduce the same initial conditions. Consequently, if the behavior is chaotic, even a slight difference in the initial conditions may lead to quite different values for the corresponding Poincaré map when n is large. Such behavior does not occur for solutions to either equation (4) or equation (1) (see Problems 6 and 7). However, two solutions to the Duffing equation (7) with F " 0.5 that corre- spond to two different but close initial values have Poincaré maps that do not remain close together. Although they both are attracted to the same set, their locations with respect to this set may be relatively far apart.

The phenomenon of chaos can also be illustrated by the following simple map. Let x0 lie in and define

(9) (mod 1) ,

where by (mod 1) we mean the decimal part of the number if it is greater than or equal to 1; that is

When , we find

etc.

Written as a sequence, we get , where the overbar denotes the repeated pattern.

What happens when we pick a starting value x0 near ? Does the sequence cluster about and as does the mapping when ? For example, when x0 " 0.3, we get the sequence

In Figure 5.50, we have plotted the values of xn for x0 " 0.3, 0.33, and 0.333. We have not plot- ted the first few terms, but only those that repeat. (This omission of the first few terms parallels the situation depicted in Figure 5.47, where transient solutions arise.)

It is clear from Figure 5.50 that while the values for x0 are getting closer to , the corre- sponding maps are spreading out over the whole interval and not clustering near and

. This behavior is chaotic, since the Poincaré maps for initial values near behave quite differently from the map for . If we had selected x0 to be irrational (which we can’t do with a calculator), the sequence would not repeat and would be dense in .30, 1 4x0 " 1/3 1/32/3

1/330, 1 4 1/3 E0.3, 0.6, 0.2, 0.4, 0.8, 0.6, 0.2, 0.4, 0.8, . . .F .x0 " 1/32/3

1/31/3

E1/3, 2/3, 1/3, 2/3, . . .Fx4 " 2 # A2/3B Amod 1B " 1/3 , x3 " 2 # A1/3B Amod 1B " 2/3 ,x2 " 2 # A2/3B Amod 1B " 4/3 Amod 1B " 1/3 , x1 " 2 # A1/3B Amod 1B " 2/3 ,x0 " 1/3 xn!1 " e2xn , for 0 ) xn 6 1/2 ,

2xn # 1 , for 1/2 ) xn 6 1 .

xn!1 " 2xn

30, 1B

304 Chapter 5 Introduction to Systems and Phase Plane Analysis

0 1 3

2 3

(a)

x0 = 0.3

1 0 1 3

2 3

(b)

x0 = 0.33

0 2 3

(c)

x0 = 0.333

11 3

Figure 5.50 Plots of the map for x0 " 0.3, 0.33, and 0.333xn!1 " 2xn Amod 1B

Systems that exhibit chaotic behavior arise in many applications. The challenge to engineers is to design systems that avoid this chaos and, instead, enjoy the property of stability. The topic of stable systems is discussed at length in Chapter 12.†

Section 5.8 Dynamical Systems, Poincaré Maps, and Chaos 305

†All references to Chapters 11–13 refer to the expanded text Fundamentals of Differential Equations and Boundary Value Problems, 6th ed.

A software package that supports the construction of Poincaré maps is required for the problems in this section.

1. Compute and graph the points of the Poincaré map with n " 0, 1, . . . , 20 for equation (1), taking Repeat, taking Do you think the equation has a

-periodic solution for either choice of ? A sub- harmonic solution?

2. Compute and graph the points of the Poincaré map with n " 0, 1, . . . , 20 for equation (1), taking Describe the limit set for this system.

3. Compute and graph the points of the Poincaré map with n " 0, 1, . . . , 20 for equation (4), taking and What is happening to these points as

4. Compute and graph the Poincaré map with n " 0, 1, . . . , 20 for equation (4), taking

and b " 0.1. Describe the attractor for this system.

5. Compute and graph the Poincaré map with n " 0, 1, . . . , 20 for equation (4), taking

and b " 0.22. Describe the attrac- tor for this system.

6. Show that for , the Poincaré map for equation (4) is not chaotic by showing that as t gets large

independent of the initial values and .

7. Show that the Poincaré map for equation (1) is not chaotic by showing that if and are two initial values that define the Poincaré maps

Ax*0, y*0 BAx0, y0B y0 " x¿ A0B x0 " x A0B yn " x¿ A2pnB # F2 Av2 # 1B2 ! b2 cos A2pn ! uB xn " x A2pnB # F2 Av2 # 1B2 ! b2 sin A2pn ! uB ,

b 7 0

v " 1 /3,f " 0, A " F " 1,

t " 2pn,

v " 1,f " 0, A " F " 1,

t " 2pn,

n S q? b " #0.1.v " 1,f " 0,A " F " 1,

t " 2pn,

A " F " 1, f " 0, and v " 1 /13.t " 2pn, v2p

v " 3 /5. A " F " 1, f " 0, and v " 3 /2.

t " 2pn,

and respectively, using the recursive formulas in (3), then one can make the distance between and small by making the distance between and small. [Hint: Let and be the polar coordinates of two points in the plane. From the law of cosines, it follows that the distance d between them is given by !

8. Consider the Poincaré maps defined in (3) with If this map were

ever to repeat, then, for two distinct positive integers n and m, sin Using basic properties of the sine function, show that this would imply that is rational. It follows from this contradiction that the points of the Poincaré map do not repeat.

9. The doubling modulo 1 map defined by equation (9) exhibits some fascinating behavior. Compute the sequence obtained when (a) (b) (c) where j is a positive integer and k "

1, 2, . . . , . Numbers of the form are called dyadic num- bers and are dense in . That is, there is a dyadic number arbitrarily close to any real number (rational or irrational).

10. To show that the limit set of the Poincaré map given in (3) depends on the initial values, do the following: (a) Show that when or 3, the Poincaré map

consists of the single point

Ax, yB " aA sin f ! F v2 # 1

, vA cos fb . v " 2

3 0, 1 4k/2 j2 j # 1 x0 " k/2 j, x0 " k/15 for k " 1, 2, . . . , 14 . x0 " k/7 for k " 1, 2, . . . , 6 .

12A212pnB " sin A212pmB. v " 12, A " F " 1, and f " 0.2AA* 3 1 # cos(f # f*B 4 . ]

d 2 " AA # A*B2 AA*, f*BAA, fB Ax*0, y*0 BAx0, y0B Ax*n , y*n BAxn, ynB

E Ax*n , y*n B F,E Axn, ynB F 5.8 EXERCISES

(b) Show that when the Poincaré map alternates between the two points

(c) Use the results of parts (a) and (b) to show that when " 2, 3, or , the Poincaré map (3) depends on the initial values .

11. To show that the limit set for the Poincaré map where is a solu-

tion to equation (6), is an ellipse and that this ellipse is the same for any initial values x0, , do the following: (a) Argue that since the initial values affect only

the transient solution to (6), the limit set for the Poincaré map is independent of the initial values.

(b) Now show that for n large,

where , and

contains the limit set of the Poincaré map.

12. Using a numerical scheme such as Runge–Kutta or a software package, calculate the Poincaré map for equation (7) when and F " 0.2. (Notice that the closer you start to the limiting point, the sooner the transient part will die out.) Compare your map with Figure 5.46(a) on page 302. Redo for F " 0.28.

13. Redo Problem 12 with F " 0.31. What kind of behavior does the solution exhibit?

14. Redo Problem 12 with F " 0.65. What kind of behavior does the solution exhibit?

b " 0.3, g " 1.2,

x2 ! Ay # cB2

2 " a2

c " arctan E# 3 A0.22B12 4#1F.a " A1 ! 2 A0.22B2B#1/2, c " A0.22B#1yn # c ! 12a cos A212pn ! cB , xn # a sin A212pn ! cB ,

x AtBxn J x A2pnB, yn J x¿ A2pnB, Ax0, y0B1 /2v

a F v2 # 1

- A sin f , - vA cos fb . v " 1 /2,

306 Chapter 5 Introduction to Systems and Phase Plane Analysis

15. Chaos Machine. Chaos can be illustrated using a long ruler, a short ruler, a pin, and a tie tack (pivot). Construct the double pendulum as shown in Figure 5.51(a). The pendulum is set in motion by releasing it from a position such as the one shown in Figure 5.51(b). Repeatedly set the pendulum in motion, each time trying to release it from the same position. Record the number of times the short ruler flips over and the direction in which it was moving. If the pendulum was released in exactly the same position each time, then the motion would be the same. However, from your experiments you will observe that even beginning close to the same posi- tion leads to very different motions. This double pendulum exhibits chaotic behavior.

long ruler

side of table

tie tack (pivot)

short ruler

(a) double pendulum

(b) release position

Figure 5.51 Double pendulum as a chaos machine

Systems of differential equations arise in the study of biological systems, coupled mass–spring oscillators, electrical circuits, ecological models, and many other areas.

Linear systems with constant coefficients can be solved explicitly using a variant of the Gauss elimination process. For this purpose we begin by writing the system with operator notation, using , , and so on. A system of two equations in two unknown functions then takes the form

where L1, L2, L3, and L4 are polynomial expressions in D. Applying L4 to the first equation, L2 to the second, and subtracting, we get a single (typically higher-order) equation in , namely,

We then solve this constant coefficient equation for . Similarly, we can eliminate x from the system to obtain a single equation for , which we can also solve. This procedure introduces some extraneous constants, but by substituting the expressions for x and y back into one of the original equations, we can determine the relationships among these constants.

A preliminary step for the application of numerical algorithms for solving systems or sin- gle equations of higher order is to rewrite them as an equivalent system of first-order equations in normal form:

(1)

For example, by setting , we can rewrite the second-order equation as the normal system

The normal system (1) has the outward appearance of a vectorized version of a single first- order equation, and as such it suggests how to generalize numerical algorithms such as those of Euler and Runge–Kutta.

A technique for studying the qualitative behavior of solutions to the autonomous system

(2)

is phase plane analysis. We begin by finding the critical points of (2)—namely, points where

and .

The corresponding constant solution pairs are called equilibrium solutions to (2). We then sketch the direction field for the phase plane equation

(3) dy dx

" g Ax, yB ƒ Ax, yB

x AtB ! x0, y AtB ! y0g Ax0, y0B " 0ƒ Ax0, y0B " 0 Ax0, y0B

dx dt

" ƒ(x, y) , dy dt

" g(x, y)

y¿ " ƒ At, y, yB .y¿ " y , y– " ƒ At, y, y¿ By " y¿x¿m AtB " fm At, x1, x2, . . . , xmB .

x¿2 AtB " f2 At, x1, x2, . . . , xmB ,x¿1 AtB " f1 At, x1, x2, . . . , xmB ,

y AtB x AtB AL4L1 # L2L3B 3 x 4 " L4 3 f1 4 # L2 3 f2 4 x AtB

L3 3 x 4 ! L4 3 y 4 " f2 ,L1 3 x 4 ! L2 3 y 4 " f1 , D2 J d2/dt2D J d/dt

Chapter Summary 307

Chapter Summary

with appropriate direction arrows (oriented by the sign of or ). From this we can usually conjecture qualitative features of the solutions and of the critical points, such as stabil- ity and asymptotic behavior. Software is typically used to visualize the solution curves to (3), which contain the trajectories of the system (2).

Nonautonomous systems can be studied by considering a Poincaré map for the system. A Poincaré map can be used to detect periodic and subharmonic solutions and to study systems whose solutions exhibit chaotic behavior.

dy/dtdx/dt

308 Chapter 5 Introduction to Systems and Phase Plane Analysis

REVIEW PROBLEMS

In Problems 1–4, find a general solution , for the given system.

1. 2.

In Problems 5 and 6, solve the given initial value problem. 5.

7. For the interconnected tanks problem of Section 5.1, page 242, suppose that instead of pure water being fed into tank A, a brine solution with concentration 0.2 kg/L is used; all other data remain the same. Determine the mass of salt in each tank at time t if the initial masses are x0 " 0.1 kg and y0 " 0.3 kg.

In Problems 8–11, write the given higher-order equation or system in an equivalent normal form (compare Section 5.3).

10.

11.

In Problems 12 and 13, solve the phase plane equation for the given system. Then sketch by hand several

x– # x¿ ! y‡ " 0 x‡ ! y¿ ! y– " t , x¿ # y ! y– " 0 x– # x ! y " 0 , 3y‡ ! 2y¿ # ety " 5 2y– # ty¿ ! 8y " sin t

z¿ " x ! y ; z A0B " #1 y¿ " x ! z ; y A0B " 2 , x¿ " y ! z ; x A0B " 2 , z¿ " z # x ; z A0B " 2 y¿ " z ; y A0B " 0 , x¿ " z # y ; x A0B " 0 , x– # x ! y– " 0 x– ! x # y– " 2e#t , 3y¿ # 4x¿ " y # 15x ! e#t 2x¿ # y¿ " y ! 3x ! et ,

y¿ " #4x # 3yx– ! y¿ " 0 x¿ " x ! 2y , x¿ ! y– ! y " 0 ,

y AtBx AtB representative trajectories (with their flow arrows) and describe the stability of the critical points (i.e., compare with Figure 5.12, page 268).

12. 13.

14. Find all the critical points and determine the phase plane solution curves for the system

In Problems 15 and 16, sketch some typical trajectories for the given system, and by comparing with Figure 5.12, page 268 identify the type of critical point at the origin.

15. 16.

17. In the electrical circuit of Figure 5.52, take R1 " R2 " 1 2, C " 1 F, and L " 1 H. Derive three equations for the unknown currents I1, I2, and I3 by writing Kirchhoff’s voltage law for loops 1 and 2, and Kirchhoff’s current law for the top juncture. Find the general solution.

y¿ " x ! y y¿ " 3x # y x¿ " #x ! 2y , x¿ " #2x # y ,

dy dt

" cos x sin y .

dx dt

" sin x cos y ,

y¿ " #4x y¿ " 2 # x x¿ " 4 # 4y , x¿ " y # 2 ,

18. In the coupled mass–spring system depicted in Figure 5.26, page 285, take each mass to be 1 kg and let k1 " 8 N/m, while k2 " 3 N/m. What are the natural angular frequencies of the system? What is the general solution?

loop 1 loop 2

I2 I3

Figure 5.52 Electrical circuit for Problem 17

A Designing a Landing System for Interplanetary Travel

Courtesy of Alfred Clark, Jr., Professor Emeritus, University of Rochester, Rochester, NY

You are a second-year Starfleet Academy Cadet aboard the U.S.S. Enterprise on a continuing study of the star system Glia. The object of study on the present expedition is the large airless planet Glia-4. A class 1 sensor probe of mass m is to be sent to the planet’s surface to collect data. The probe has a modifiable landing system so that it can be used on planets of different gravity. The system consists of a linear spring (force " #kx, where x is displacement), a nonlinear spring (force ), and a shock-damper (force " ),† all in parallel. Figure 5.53 shows a schematic of the system. During the landing process, the probe’s thrusters are used to create a constant rate of descent. The velocity at impact varies; the symbol VL is used to denote the largest velocity likely to happen in practice. At the instant of impact, (1) the thrust is turned off, and (2) the suspension springs are at their unstretched natural length.

(a) Let the displacement x be measured from the unstretched length of the springs and be taken negative downward (i.e., compression gives a negative x). Show that the equation governing the oscillations after impact is

(b) The probe has a mass m " 1220 kg. The linear spring is permanently installed and has a stiffness k " 35,600 N/m. The gravity on the surface of Glia-4 is g " 17.5 m/sec2. The nonlinear spring is removable; an appropriate spring must be chosen for each mission. These nonlinear springs are made of coralidium, a rare and difficult-to-fabricate alloy. Therefore, the Enterprise stocks only four different strengths: a " 150,000, 300,000, 450,000, and 600,000 N/m3. Determine which springs give a compression as close as

mx $

! bx #

! kx ! ax3 " #mg .

#bx #

" #ax3

309

Group Projects for Chapter 5

†The symbol denotes .dx/dtx#

(b)

Probe body

surface of Glia-4Landing pod

(a)

k b a

Probe body

Landing pod

V = 0

k b a

Figure 5.53 Schematic of the probe landing system. (a) The system at the instant of impact. The springs are not stretched or compressed, the thrusters have been turned off, and the velocity is VL downward. (b) The probe has reached a state of rest on the surface, and the springs are compressed enough to support the weight. Between states (a) and (b), the probe oscillates relative to the landing pod.

possible to 0.3 m without exceeding 0.3 m, when the ship is resting on the surface of Glia-4. (The limit of 0.3 m is imposed by unloading clearance requirements.)

(c) The other adjustable component on the landing system is the linear shock-damper, which may be adjusted in increments of N-sec/m, from a low value of 1000 N-sec/m to a high value of 10,000 N-sec/m. It is desirable to make b as small as possible because a large b produces large forces at impact. However, if b is too small, there is some danger that the probe will rebound after impact. To minimize the chance of this, find the smallest value of b such that the springs are always in compression during the oscillations after impact. Use a minimum impact velocity VL " 5 m/sec downward. To find this value of b, you will need to use a software package to integrate the differential equation.

B Spread of Staph Infections in Hospitals—Part I

Courtesy of Joanna Pressley, Assistant Professor, and Professor Glenn Webb, Vanderbilt University

Methicillin-resistant Staphylococcus aureus (MRSA), commonly referred to as staph, is a bacterium that causes serious infections in humans and is resistant to treatment with the widely used antibiotic methicillin. MRSA has traditionally been a problem inside hospitals, where elderly patients or patients with compromised immune systems could more easily contract the bacteria and develop bloodstream infections. MRSA is implicated in a large percentage of hospital fatalities, causing more deaths per year than AIDS. Recently, a genetically different strain of MRSA has been found in the community at large. The new strain (CA-MRSA) is able to infect healthy and young people, which the traditional strain (HA-MRSA) rarely does. As CA-MRSA appears in the community, it is inevitably being spread into hospitals. Some studies suggest that CA-MRSA will overtake HA-MRSA in the hospital, which would increase the severity of the problem and likely cause more deaths per year.

To predict whether or not CA-MRSA will overtake HA-MRSA, a compartmental model has been developed by mathematicians in collaboration with medical professionals (see references [1], [2] on page 312). This model classifies all patients in the hospital into three groups:

• " patients colonized with the traditional hospital strain, HA-MRSA. • " patients colonized with the community strain, CA-MRSA. • " susceptible patients, those not colonized with either strain.

The parameters of the model are

• " the rate (per day) at which CA-MRSA is transmitted between patients. • " the rate (per day) at which HA-MRSA is transmitted between patients. • " the rate (per day) at which patients who are colonized with CA-MRSA exit the

hospital by death or discharge. • " the rate (per day) at which patients who are colonized with HA-MRSA exit the

hospital by death or discharge. • " the rate (per day) at which susceptible patients exit the hospital by death or discharge. • " the rate (per day) at which patients who are colonized with CA-MRSA successfully

undergo decolonization measures. • " the rate (per day) at which patients who are colonized with HA-MRSA successfully

undergo decolonization measures. • = the total number of patients in the hospital. • = the rate (per day) at which patients enter the hospital. ¶

S AtBC AtB H AtB

¢b " 500

310 Chapter 5 Introduction to Systems and Phase Plane Analysis

Patients move between compartments as they become colonized or decolonized (see Figure 5.54). This type of model is typically known as an SIS (susceptible-infected-susceptible) model, in which patients who become colonized can become susceptible again and colonized again (there is no immunity).

The transition between states is described by the following system of differential equations:

If we assume that the hospital is always full, we can conserve the system by letting . In this case for all (assuming you

start with a population of size ).

(a) Show that this assumption simplifies the above system of equations to

(1)

is then determined by the equation .

Parameter values obtained from the Beth Israel Deaconess Medical Center are given in Table 5.4 on page 312. Plug these values into the model and then complete the following problems.

(b) Find the three equilibria (critical points) of the system (1). (c) Using a computer, sketch the direction field for the system (1). (d) Which trajectory configuration exists near each critical point (node, spiral, saddle, or

center)? What do they represent in terms of how many patients are susceptible, colo- nized with HA-MRSA, and colonized with CA-MRSA over time?

(e) Examining the direction field, do you think CA-MRSA will overtake HA-MRSA in the hospital?

Further discussion of this model appears in Project E of Chapter 12.†

S AtB " N # H AtB # C AtBS dH dt

" AbH/NB AN # C # HBH # AdH ! aHBH. dC dt

" AbC/NB AN # C # HBC # AdC ! aCBC N

tS AtB ! C AtB ! H AtB " N¶ " dSS AtB ! dHH AtB ! dCC AtB dC dt

" bC S AtBC AtB

N # aC C AtB # dC C AtB.

dH dt

" bHS AtBH AtB

N # aHH AtB # dHH AtB

dSS AtB#aCC AtB!aHH AtB! bC S AtBC AtB

N #

bHS AtBH AtB N

#¶dS dt

Group Projects for Chapter 5 311

S(t)

C(t)

aC aH

bC bH

3 dS

H(t) dH

Figure 5.54 A diagram of how patients transit between the compartments

}

}} } }}

} }

} entrance rate

acquire HA-MRSA acquire CA-MRSA

HA-MRSA decolonized CA-MRSA decolonized exit hospital

from S decolonized exit hospital

}}}

from S decolonized exit hospital

†All references to Chapters 11–13 refer to the expanded text Fundamentals of Differential Equations and Boundary Value Problems, 6th ed.

References

1. D’Agata, E. M. C., Webb, G. F., Pressley, J. 2010. Rapid emergence of co-colonization with community- acquired and hospital-acquired methicillin-resistant Staphylococcus aureus strains in the hospital setting. Mathematical Modelling of Natural Phenomena 5(3): 76–93.

2. Pressley, J., D’Agata, E. M. C., Webb, G. F. 2010. The effect of co-colonization with community- acquired and hospital-acquired methicillin-resistant Staphylococcus aureus strains on competitive exclu- sion. Journal of Theoretical Biology 265(3): 645–656.

C Things That Bob Courtesy of Richard Bernatz, Department of Mathematics, Luther College

The motion of various-shaped objects that bob in a pool of water can be modeled by a second- order differential equation derived from Newton’s second law of motion, F " ma. The forces acting on the object include the force due to gravity, a frictional force due to the motion of the object in the water, and a buoyant force based on Archimedes’ principle: An object that is completely or partially submerged in a fluid is acted on by an upward (buoyant) force equal to the weight of the water it displaces.

(a) The first step is to write down the governing differential equation. The dependent vari- able is the depth z of the object’s lowest point in the water. Take z to be negative down- ward so that z " #1 means 1 ft of the object has submerged. Let be the submerged volume of the object, m be the mass of the object, be the density of water (in pounds per cubic foot), g be the acceleration due to gravity, and be the coefficient of friction for water. Assuming that the frictional force is proportional to the vertical velocity of the object, write down the governing second-order ODE.

(b) For the time being, neglect the effect of friction and assume the object is a cube measur- ing L feet on a side. Write down the governing differential equation for this case. Next, designate z " l to be the depth of submersion such that the buoyant force is equal and

V AzB

312 Chapter 5 Introduction to Systems and Phase Plane Analysis

TABLE 5.4 Parameter Values for the Transmission Dynamics of Community-Acquired and Hospital-Acquired Methicillin-Resistant Staphylococcus aureus Colonization (CA-MRSA and HA-MRSA)

Parameter Symbol Baseline Value

Total number of patients N 400

Length of stay Susceptible 5 days Colonized CA-MRSA 7 days Colonized HA-MRSA 5 days

Transmission rate per susceptible patient to Colonized CA-MRSA per colonized CA-MRSA 0.45 per day Colonized HA-MRSA per colonized HA-MRSA 0.4 per day

Decolonization rate per colonized patient per day per length of stay CA-MRSA 0.1 per day HA-MRSA 0.1 per dayaH

1/dH 1/dC 1/dS

opposite the gravitational force. Introduce a new variable, that gives the displacement of the object from its equilibrium position l (that is, ). You can now write the ODE in a more familiar form. [Hint: Recall the mass–spring system and the equilibrium case.] Now you should recognize the type of solution for this problem. What is the natural frequency?

(c) In this task you consider the effect of friction. The bobbing object is a cube, 1 ft on a side, that weighs 32 lb. Let lb-sec/ft, lb/ft3, and suppose the object is initially placed on the surface of the water. Solve the governing ODE by hand to find the general solution. Next, find the particular solution for the case in which the cube is initially placed on the surface of the water and is given no initial velocity. Provide a plot of the position of the object as a function of time t.

(d) In this step of the project, you develop a numerical solution to the same problem pre- sented in part (c). The numerical solution will be useful (indeed necessary) for subse- quent parts of the project. This case provides a trial to verify that your numerical solution is correct. Go back to the initial ODE you developed in part (a). Using parameter values given in part (c), solve the initial value problem for the cube starting on the surface with no initial velocity. To solve this problem numerically, you will have to write the second- order ODE as a system of two first-order ODEs, one for vertical position z and one for vertical velocity w. Plot your results for vertical position as a function of time t for the first 3 or 4 sec and compare with the analytical solution you found in part (c). Are they in close agreement? What might you have to do in order to compare these solutions? Pro- vide a plot of both your analytical and numerical solutions on the same graph.

(e) Suppose a sphere of radius R is allowed to bob in the water. Derive the governing second-order equation for the sphere using Archimedes’ principle and allowing for friction due to its motion in the water. Suppose a sphere weighs 32 lb, has a radius of

ft, and lb-sec/ft. Determine the limiting value of the position of the sphere without solving the ODE. Next, solve the governing ODE for the velocity and position of the sphere as a function of time for a sphere placed on the surface of the water. You will need to write the governing second-order ODE as a system of two ODEs, one for velocity and one for position. What is the limiting position of the sphere for your solution? Does it agree with the equilibrium solution you found above? How does it compare with the equilibrium position of the cube? If it is different, explain why.

(f) Suppose the sphere in part (d) is a volleyball. Calculate the position of the sphere as a func- tion of time t for the first 3 sec if the ball is submerged so that its lowest point is 5 ft under water. Will the ball leave the water? How high will it go? Next, calculate the ball’s trajectory for initial depths lower than 5 ft and higher than 5 ft. Provide plots of velocity and position for each case and comment on what you see. Specifically, comment on the relationship between the initial depth of the ball and the maximum height the ball eventually attains.

You might consider taking a volleyball into a swimming pool to gather real data in order to verify and improve on your model. If you do so, report the data you found and explain how you used it for verification and improvement of your model.

D Hamiltonian Systems The problems in this project explore the Hamiltonian† formulation of the laws of motion of a system and its phase plane implications. This formulation replaces Newton’s second law

and is based on three mathematical manipulations:F " ma " my–

gw " 3.01 /2

r " 62.57gw " 3

z " z ! l z,

Group Projects for Chapter 5 313

†Historical Footnote: Sir William Rowan Hamilton (1805–1865) was an Irish mathematical physicist. Besides his work in mechanics, he invented quaternions and discovered the anticommutative law for vector products.

(i) It is presumed that the force depends only on y and has an antiderivative that is,

(ii) The velocity variable is replaced throughout by the momentum so y$ " p/m). (iii) The Hamiltonian of the system is defined as

(a) Express Newton’s law as an equivalent first-order system in the manner prescribed in Section 5.3.

(b) Show that this system is equivalent to Hamilton’s equations

(2)

(3)

In the formula for the Hamiltonian function , the first term, " is the kinetic energy of the mass. By analogy, then, the second term is known as the potential energy of the mass, and the Hamiltonian is the total energy. The total (mechanical)† energy is constant—hence “conserved”—when the forces do not depend on time t or velocity ; such forces are called conservative. The energy integral lemma of Section 4.8 (page 204) is simply an alternate statement of the conservation of energy.

Hamilton’s formulation for mechanical systems and the conservation of energy principle imply that the phase plane trajectories of conservative systems lie on the curves where the Hamil- tonian is constant, and plotting these curves may be considerably easier than solving for the trajectories directly (which, in turn, is easier than solving the original system!).

(d) For the mass–spring oscillator of Section 4.1, the spring force is given by (where k is the spring constant). Find the Hamiltonian, express Hamilton’s equations, and show that the phase plane trajectories constant for this system are the ellipses given by constant. See Figure 5.14, page 270.

The damping force considered in Section 4.1 is not conservative, of course. Physically speaking, we know that damping drains the energy from a system until it grinds to a halt at an equilibrium point. In the phase plane, we can qualitatively describe the trajectory as continuously migrating to successively lower constant-energy orbits; stable centers become asymptotically stable spiral points when damping is taken into consideration.

(e) The second Hamiltonian equation (3), which effectively states has to be changed to

when damping is present. Show that the Hamiltonian decreases along trajectories in this case (for b * 0):

d dt

H A y, pB " #b ap m b 2 " #b A y¿ B2 .

p¿ " # 0H 0y # by¿ " #

0H 0y #

bp m

p¿ " my– " F,

#by¿

p2 / A2mB ! ky2 /2 " H A y, pB " F " #ky

H A y, pB

y¿F A yB V A yB m A y¿ B2 /2,p2 / A2mBH A y, pB

d dt

H A y, pB " 0 . dp dt

" # 0H 0y a" # dVdyb .

dy dt

" 0H 0p a" pmb ,

F " my–

H " H A y, pB " p2 2m

! V A yB . Ap " my¿y¿F " F A yB " #dV A yB /dy.#V A yB,

F At, y, y¿ B 314 Chapter 5 Introduction to Systems and Phase Plane Analysis

†Physics states that when all forms of energy, such as heat and radiation, are taken into account, energy is conserved even when the forces are not conservative.

(f) The force on a mass–spring system suspended vertically in a gravitational field was shown in Section 4.10 (page 228) to be Derive the Hamiltonian and sketch the phase plane trajectories. Sketch the trajectories when damping is present.

(g) As indicated in Section 4.8 (page 209), the Duffing spring force is modeled by Derive the Hamiltonian and sketch the phase plane trajectories. Sketch

the trajectories when damping is present. (h) For the pendulum system studied in Section 4.8, Example 8, the force is given by

(cf. Figure 4.18, page 210)

(where is the length of the pendulum). For angular variables, the Hamiltonian formu- lation dictates expressing the angular velocity variable in terms of the angular momentum the kinetic energy, , is expressed as

Derive the Hamiltonian for the pendulum and sketch the phase plane trajectories. Sketch the trajectories when damping is present.

(i) The Coulomb force field is a force that varies as the reciprocal square of the distance from the origin: . The force is attractive if k , 0 and repulsive if k * 0. Sketch the phase plane trajectories for this motion. Sketch the trajectories when damping is present.

(j) For an attractive Coulomb force field, what is the escape velocity for a particle situated at a position y? That is, what is the minimal (outward-directed) velocity required for the trajectory to reach ?

E Cleaning Up the Great Lakes A simple mathematical model that can be used to determine the time it would take to clean up the Great Lakes can be developed using a multiple compartmental analysis approach.† In particular, we can view each lake as a tank that contains a liquid in which is dissolved a particular pollutant (DDT, phosphorus, mercury). Schematically, we view the lakes as consisting of five tanks connected as indicated in Figure 5.55.

y " q

F " k/y2

m A/u¿ B2 /2 " p2 / A2m/2B. mass & velocity2 /2p " m/2u¿; u¿

F " #/mg sin u " # 00u A#/mg cos uB " # 00u V AuB F " #y # y3.

F " #ky ! mg.

Group Projects for Chapter 5 315

1515

Superior

2900 mi3

Huron

850 mi3

Erie

Onta rio 3

93 m i3

M ic

hi ga

n 1

18 0

m i3

116 mi3

38 38

Figure 5.55 Compartmental model of the Great Lakes with flow rates (mi3/yr) and volumes (mi3)

†For a detailed discussion of this model, see An Introduction to Mathematical Modeling by Edward A. Bender (Krieger, New York, 1991), Chapter 8.

For our model, we make the following assumptions:

1. The volume of each lake remains constant. 2. The flow rates are constant throughout the year. 3. When a liquid enters the lake, perfect mixing occurs and the pollutants are uniformly

distributed. 4. Pollutants are dissolved in the water and enter or leave by inflow or outflow of solution.

Before using this model to obtain estimates on the cleanup times for the lakes, we consider some simpler models:

(a) Use the outflow rates given in Figure 5.55 to determine the time it would take to “drain” each lake. This gives a lower bound on how long it would take to remove all the pollutants.

(b) A better estimate is obtained by assuming that each lake is a separate tank with only clean water flowing in. Use this approach to determine how long it would take the pol- lution level in each lake to be reduced to 50% of its original level. How long would it take to reduce the pollution to 5% of its original level?

(c) Finally, to take into account the fact that pollution from one lake flows into the next lake in the chain, use the entire multiple compartment model given in Figure 5.55 to determine when the pollution level in each lake has been reduced to 50% of its original level, assuming pollution has ceased (that is, inflows not from a lake are clean water). Assume that all the lakes initially have the same pollution concentration p. How long would it take for the pollution to be reduced to 5% of its original level?

F A Growth Model for Phytoplankton—Part I Courtesy of Dr. Olivier Bernard and Dr. Jean-Luc Gouzé, INRIA

A chemostat is a stirred tank in which phytoplankton grow by consuming a nutrient (e.g., nitrate). The nutrient is supplied to the tank at a given rate, and a solution containing the phytoplankton and remaining nutrient is removed at an equal rate (cf. Figure 5.56). The chemostat reproduces in vitro the conditions of the growth of phytoplankton in the ocean; the phytoplankton is the first element of the marine food chain.

316 Chapter 5 Introduction to Systems and Phase Plane Analysis

: microorganisms : nutrients

Inflow

Outflow

Figure 5.56 Chemostat

Let S denote the concentration (in mol/liter) of the nutrient and X the biovolume (which is to be taken as an estimation of the biomass) of phytoplankton (in mm3 of cells per liter of solu- tion). A classical model (J. Monod, La technique de culture continue: théorie et applications. Annales de I’Institut Pasteur, 79, 1950) of the behavior of the chemostat is the following:

(4)

The unit for time is the day; the dilution rate and the growth rate are in day#1; and the input concentration si and the constant k have the same units as S. Experimental (smoothed) data, obtained from the Station Zoologique of Villefrance-sur-Mer in France, are displayed in Figure 5.57. They can be downloaded from the textbook’s Web site at www.pearsonhighered.com/nagle.

(i) When t , 2.5, the dilution rate is zero (“batch culture”). It is known that k is in the range 0.1 4 k 4 1.

(a) What are the units for the yield factor y? (b) Write a linear approximation of the system (4) for S ** k (i.e., S is much larger than k). (c) Solve the approximated system in part (b) and use the solution for X and the experimen-

tal data on the Web site to obtain a numerical value for [Hint: Plot the logarithm of X against the time, and estimate the slope.] Use the equation for S and the data to obtain an estimation of y.

(d) Take k " 0.5 and the values for and y obtained in (c). Using a computer software package and the initial conditions X(0) " 0.15, S(0) " 45.84, draw the numerical solu- tions X(t), S(t) for the system (4) and for the approximated system of part (b). Is the approximation of part (b) a reasonable one?

(ii) For t * 2.5, the dilution rate is " 1.06 day#1. After a delay, the growth rate of the phytoplankton changes because the cells adapt themselves to their new environment.

(e) Estimate the time T when the growth rate changes and obtain the new value for . (As above, take k " 0.5.)

Further discussion of this model appears in Project F of Chapter 12.†

$ dX

dt " r

S ! k # aX

dt " a Asi # SB # ry SXS ! k .

Group Projects for Chapter 5 317

20S (m

m ol

/L )

X (

m m

3 / L

)

40 45

2 3 4 5 6 Time (Days)

S X

Figure 5.57 Nutrient/biovolume data

†All references to Chapters 11–13 refer to the expanded text Fundamentals of Differential Equations and Boundary Value Problems, 6th ed.

www.pearsonhighered.com/nagle

318

A linear differential equation of order n is an equation that can be written in the form

(1)

where , , . . . , and depend only on x, not y. When a0, a1, . . . , an are all con- stants, we say equation (1) has constant coefficients; otherwise it has variable coefficients. If

equation (1) is called homogeneous; otherwise it is nonhomogeneous. In developing a basic theory, we assume that , , . . . , and are all con-

tinuous on an interval I and on I. Then, on dividing by , we can rewrite (1) in the standard form

(2)

where the functions , . . . , , and are continuous on I. For a linear higher-order differential equation, the initial value problem always has a

unique solution.

g AxBpn AxBp1 AxBy AnB AxB ! p1 AxBy An"1B AxB ! p ! pn AxBy AxB # g AxB ,

an AxBan AxB $ 0 b AxBan AxBa1 AxBa0 AxB b AxB ! 0, b

AxBan AxBa1 AxBa0 AxBan AxBy AnB AxB ! an"1 AxBy An"1B AxB ! p ! a0 AxBy AxB # b AxB ,

Theory of Higher-Order Linear Differential Equations

CHAPTER 6

BASIC THEORY OF LINEAR DIFFERENTIAL EQUATIONS6.1

In this chapter we discuss the basic theory of linear higher-order differential equations. The material is a generalization of the results we obtained in Chapter 4 for second-order constant- coefficient equations. In the statements and proofs of these results, we use concepts usually covered in an elementary linear algebra course—namely, linear dependence, determinants, and methods for solving systems of linear equations. These concepts also arise in the matrix approach for solving systems of differential equations and are discussed in Chapter 9, which includes a brief review of linear algebraic equations and determinants.

Since this chapter is more mathematically oriented—that is, not tied to any particular physical application—we revert to the customary practice of calling the independent variable “x” and the dependent variable “y.”

Existence and Uniqueness

Theorem 1. Suppose , . . . , and are each continuous on an interval that contains the point x0. Then, for any choice of the initial values

there exists a unique solution on the whole interval to the initial value problem

(3)

(4) y Ax0B ! G0, y" Ax0B ! G1, . . . , yAn#1B Ax0B ! Gn#1 .y AnB AxB $ p1 AxByAn#1B AxB $ p $ pn AxBy AxB ! g AxB ,

Aa, bBy AxBg0, g1, . . . , gn"1,Aa, bB g AxBpn AxBp1 AxB

The proof of Theorem 1 can be found in Chapter 13.†

For the initial value problem

(5)

(6)

determine the values of x0 and the intervals containing x0 for which Theorem 1 guaran- tees the existence of a unique solution on .

Putting equation (5) in standard form, we find that ,

, and Now and are continuous on every interval not containing x # 1, while is continuous on every interval not containing x # 0 or x # 1. The function is not defined for x % "5, x # 0, and x # 1, but is continuous on , , and Hence, the functions p1, p2, p3, and g are simultaneously continuous on the intervals , , and From Theorem 1 it fol- lows that if we choose then there exists a unique solution to the initial value problem (5)–(6) on the whole interval . Similarly, for there is a unique solution on and, for a unique solution on ◆

If we let the left-hand side of equation (3) define the differential operator L,

(7)

then we can express equation (3) in the operator form

(8)

It is essential to keep in mind that L is a linear operator—that is, it satisfies

(9)

(10) (c any constant) .

These are familiar properties for the differentiation operator D, from which (9) and (10) follow (see Problem 25).

As a consequence of this linearity, if y1, . . . , ym are solutions to the homogeneous equation

(11)

then any linear combination of these functions, is also a solution, because

Imagine now that we have found n solutions y1, . . . , yn to the nth-order linear equation (11). Is it true that every solution to (11) can be represented by

(12) C1y1 ! p ! Cnyn

L 3C1y1 ! C2y2 ! p ! Cmym 4 # C1 # 0 ! C2 # 0 ! p ! Cm # 0 # 0 .C1y1 ! p ! Cmym, L 3 y 4 AxB # 0 ,

L 3 cy 4 # cL 3 y 4 L 3 y1 ! y2 ! p ! ym 4 # L 3 y1 4 ! L 3 y2 4 ! p ! L 3 ym 4 , L 3 y 4 AxB # g AxB . L 3 y 4 J dny

dxn $ p1

dn#1y

dxn#1 $ p $ pny ! ADn $ p1Dn#1 $ p $ pnB 3 y 4 ,

A1, q B.x0 % A1, q B,A0, 1B x0 % A0, 1B,A"5, 0B x0 % A"5, 0B, A1, q B.A0, 1BA"5, 0B

A1, q B.A0, 1BA"5, 0B g AxB p3 AxB p2 AxBp1 AxBg AxB # 2x ! 5/ 3 x Ax " 1B 4 .p3 AxB # " Acos xB / 3 x Ax " 1B 4 p2 AxB # 6x/ Ax " 1),p1 AxB # "3/ Ax " 1)

Aa, bBAa, bB y Ax0B # 1 , y¿ Ax0B # 0 , y– Ax0B # 7 ,x Ax " 1By‡ " 3xy– ! 6x

2y¿ " Acos xBy # 2x ! 5 ; Section 6.1 Basic Theory of Linear Differential Equations 319

†All references to Chapters 11–13 refer to the expanded text Fundamentals of Differential Equations and Boundary Value Problems, 6th ed.

Example 1

Solution

for appropriate choices of the constants C1, . . . , Cn? The answer is yes, provided the solutions y1, . . . , yn satisfy a certain property that we now derive.

Let be a solution to (11) on the interval and let x0 be a fixed number in . If it is possible to choose the constants C1, . . . , Cn so that

(13) o o o

then, since and are two solutions satisfying the same initial conditions at x0, the uniqueness conclusion of Theorem 1 gives

(14)

for all x in . The system (13) consists of n linear equations in the n unknowns C1, . . . , Cn. It has a

unique solution for all possible values of if and only if the deter- minant† of the coefficients is different from zero; that is, if and only if

(15)

Hence, if y1, . . . , yn are solutions to equation (11) and there is some point x0 in such that (15) holds, then every solution to (11) is a linear combination of y1, . . . , yn. Before for- mulating this fact as a theorem, it is convenient to identify the determinant by name.

f AxB Aa, bB ∞ y1 Ax0B y2 Ax0B p yn Ax0By¿1 Ax0B y¿2 Ax0B p y¿n Ax0B o o o y An"1B1 Ax0B y An"1B2 Ax0B p y An"1Bn Ax0B ∞ $ 0 .

f Ax0B, f¿ Ax0B, . . . , fAn"1B Ax0B Aa, bBf AxB # C1y1 AxB ! p ! Cnyn AxB

C1y1 AxB ! p ! Cnyn AxBf AxB C1y

An"1B 1 Ax0B ! p ! Cny An"1Bn Ax0B # fAn"1B Ax0B ,

C1y¿1 Ax0B ! p ! Cny¿n Ax0B # f¿ Ax0B ,C1y1 Ax0B ! p ! Cnyn Ax0B # f Ax0B , Aa, bBAa, bBf AxB

320 Chapter 6 Theory of Higher-Order Linear Differential Equations

†Determinants are discussed in Section 9.3.

Wronskian

Definition 1. Let f1, . . . , fn be any n functions that are times differentiable. The function

(16)

is called the Wronskian of f1, . . . , fn.

W 3 f1, . . . , fn 4 AxB :! ∞ f1 AxB f2 AxB p fn AxBf"1 AxB f"2 AxB p f"n AxB f An#1B1 AxB f An#1B2 AxB p f An#1Bn AxB ∞

An " 1B

We now state the representation theorem that we proved above for solutions to homoge- neous linear differential equations.

··· ··· ···

The linear combination of y1, . . . , yn in (19), written with arbitrary constants C1, . . . , Cn, is referred to as a general solution to (17).

In linear algebra a set of m column vectors each having m components, is said to be linearly dependent if and only if at least one of them can be expressed as a linear combination of the others.† A basic theorem then states that if a determinant is zero, its column vectors are linearly dependent, and conversely. So if a Wronskian of solutions to (17) is zero at a point x0, one of its columns (the final column, say; we can always renumber!) equals a linear combination of the others:

(20)

Now consider the two functions and They are both solutions to (17), and we can interpret (20) as stating that they satisfy the same ini- tial conditions at By the uniqueness theorem, then, they are one and the same function:

(21)

for all x in the interval I. Consequently, their derivatives are the same also, and so

(22)

for all x in I. Hence, the final column of the Wronskian is always a linear combination of the other columns, and consequently the Wronskian is always zero.

In summary, the Wronskian of n solutions to the homogeneous equation (17) is either identically zero, or never zero, on the interval . We have also shown that, in the formerAa, bB

W 3 y1, y2, p , yn 4 Dyn AxByn¿ AxBo yn An"1B AxBT # d1Dy1 AxBy1¿ AxBoy1An"1B AxBT ! d2Dy2 AxBy2¿ AxBoy2An"1B AxBT ! p ! dn"1Dyn"1 AxBy¿n"1 AxBoyn"1An"1B AxBT

yn AxB # d1y1 AxB ! d2y2 AxB ! p ! dn"1yn"1 AxBx # x0. 3d1y1 AxB ! d2y2 AxB ! p ! dn"1yn"1 AxB 4 .yn AxB

Dyn Ax0Byn¿ Ax0B o

yn An"1B Ax0BT # d1Dy1 Ax0By1¿ Ax0Boy1An"1B Ax0BT ! d2Dy2 Ax0By2¿ Ax0Boy2An"1B Ax0BT ! p ! dn"1 Dyn"1 Ax0By¿n"1 Ax0Boyn"1An"1B Ax0BT .

Ev1, v2, p ,vmF,

Section 6.1 Basic Theory of Linear Differential Equations 321

Representation of Solutions (Homogeneous Case)

Theorem 2. Let y1, . . . , yn be n solutions on of

(17)

where p1, . . . , pn are continuous on . If at some point x0 in these solutions satisfy

(18)

then every solution of (17) on can be expressed in the form

(19)

where C1, . . . , Cn are constants.

y AxB ! C1y1 AxB $ p $ Cnyn AxB ,Aa, bB W 3 y1, . . . , yn 4 Ax0B & 0 ,

Aa, bBAa, bBy AnB AxB ! p1 AxBy An"1B AxB ! p ! pn AxBy AxB # 0 , Aa, bB

†This is equivalent to saying there exist constants not all zero, such that equals the zero vector.

c1v1 ! c2v2 ! p ! cmvmc1, c2, p , cm

case, (21) holds throughout . Such a relationship among functions is an extension of the notion of linear dependence introduced in Section 4.2. We employ the same nomenclature for the general case.

Aa, bB 322 Chapter 6 Theory of Higher-Order Linear Differential Equations

Show that the functions and are linearly depen- dent on

Obviously, f3 is a linear combination of f1 and f2 :

Note further that the corresponding identity matches the pattern (23). Moreover, observe that f1, f2, and f3 are pairwise linearly independent on , but this does not suffice to make the triplet independent. ◆

To prove that functions f1, f2, . . . , fm are linearly independent on the interval , a con- venient approach is the following: Assume that equation (23) holds on and show that this forces

Show that the functions , and are linearly independent on

Assume c1, c2, and c3 are constants for which

(24)

holds at every x. If we can prove that (24) implies c1 # c2 # c3 # 0, then linear independence follows. Let’s set x # 0, 1, and "1 in equation (24). These x values are, essentially, “picked out of a hat” but will get the job done. Substituting in (24) gives

(25)

When we solve this system (or compute the determinant of the coefficients), we find that the only possible solution is c1 # c2 # c3 # 0. Consequently, the functions f1, f2, and f3 are linearly independent on A"q, q).

"c1 ! c2 " c3 # 0 Ax # "1) .c1 ! c2 " c3 # 0 Ax # 1) , c3 # 0 Ax # 0B ,

c1x ! c2x 2 ! c3 A1 " 2x2B # 0

A"q, q B. f3 AxB # 1 " 2x2f1 AxB # x, f2 AxB # x2 c1 # c2 # p # cm # 0.

Aa, bB Aa, bB A"q, q B3f1 AxB " 2f2 AxB " f3 AxB # 0

f3 AxB # 3ex " 2e"2x # 3f1 AxB " 2f2 AxB . A"q, q B. f3 AxB # 3ex " 2e"2xf1 AxB # ex, f2 AxB # e"2x,

Linear Dependence of Functions

Definition 2. The m functions are said to be linearly dependent on an interval I if at least one of them can be expressed as a linear combination of the others on I; equivalently, they are linearly dependent if there exist constants not all zero, such that

(23)

for all x in I. Otherwise, they are said to be linearly independent on I.

c1 f1 AxB ! c2 f2 AxB ! p ! cm fm(xB # 0 c1, c2, p , cm,

f1, f2, p , fm

Example 2

Solution

Example 3

A neater solution is to note that if (24) holds for all x, so do its first and second derivatives. At these conditions are and Obviously, each coeffi- cient must be zero. ◆

Linear dependence of functions is, prima facie, different from linear dependence of vec- tors in the Euclidean space because (23) is a functional equation that imposes a condition at every point of an interval. However, we have seen in (21) that when the functions are all solu- tions to the same homogeneous differential equation, linear dependence of the column vectors of the Wronskian (at any point x0) implies linear dependence of the functions. The converse is also true, as demonstrated by (21) and (22). Theorem 3 summarizes our deliberations.

Rn,

2c2 " 4c3 # 0.c3 # 0, c1 # 0,x # 0

Section 6.1 Basic Theory of Linear Differential Equations 323

Linear Dependence and the Wronskian

Theorem 3. If are n solutions to on the interval with continuous on , then the following statements are equivalent:

(i) are linearly dependent on . (ii) The Wronskian is zero at some point x0 in . (iii) The Wronskian is identically zero on .

The contrapositives of these statements are also equivalent:

(iv) are linearly independent on . (v) The Wronskian is nonzero at some point x0 in . (vi) The Wronskian is never zero on .

Whenever (iv), (v), or (vi) is met, is called a fundamental solution set for (17) on .Aa, bB Ey1, y2, p , ynF

Aa, bBW 3 y1, y2, p , yn 4 AxB Aa, bBW 3 y1, y2, p , yn 4 Ax0B Aa, bBy1, y2, p , yn Aa, bBW 3 y1, y2, p , yn 4 AxB Aa, bBW 3 y1, y2, p , yn 4 Ax0B Aa, bBy1, y2, p , yn

Aa, bBp1, p2, p , pnAa, bB, y AnB ! p1y An"1B ! p ! pny # 0y1, y2, p ,yn

The Wronskian is a curious function. If we take for n arbitrary functions, we simply get a function of x with no particularly interesting properties. But if the n functions are all solutions to the same homogeneous differential equation, then either it is identically zero or never zero. In fact, one can prove Abel’s identity when the functions are all solutions to (17):

(26)

which clearly exhibits this property. Problem 30 outlines a proof of (26) for †

It is useful to keep in mind that the following sets consist of functions that are linearly independent on every open interval :

1, cos x, sin x, cos 2x, sin 2x, . . . , cos nx, sin nx ’s distinct constants

[See Problems 27 and 28, and Section 6.2 (page 328).] If we combine the linearity (superposition) properties (9) and (10) with the representation

theorem for solutions of the homogeneous equation, we obtain the following representation theorem for nonhomogeneous equations.

B .EeA1x, eA2x, . . . , eAnxF AAi F ,E E1, x, x2, . . . , xnF Aa, bB

n # 3.

W 3 y1, y2, p , yn 4 AxB # W 3 y1, y2, p , yn 4 Ax0B exp a" x x0

p1 AtBdtb , W 3 f1, f2, p , fn 4 AxB

†See Problem 32, Exercises 4.7, for the case n # 2.

Proof. Let be any solution to (27). Because both and are solutions to (27), by linearity the difference is a solution to the homogeneous equation (28). It then follows from Theorem 2 that

for suitable constants C1, . . . , Cn. The last equation is equivalent to (29) with in place of , so the theorem is proved. ◆

The linear combination of yp, y1, . . . , yn in (29) written with arbitrary constants C1, . . . , Cn is, for obvious reasons, referred to as a general solution to (27). Theorem 4 can be easily generalized. For example, if L denotes the operator appearing as the left-hand side in equation (27) and if and , then any solution of

can be expressed as

for a suitable choice of the constants C1, C2, . . . , Cn.

Find a general solution on the interval to

(30)

given that is a particular solution to , that is a particular solution to , and that , , and are solutions to the corresponding homogeneous equation.

We previously remarked that the functions are linearly independent because the exponents and 2 are distinct. Since each of these functions is a solution to the corre- sponding homogeneous equation, then is a fundamental solution set. It now follows from the remarks above for nonhomogeneous equations that a general solution to (30) is

(31) ◆# x2 ! 2e"2x ! C1e

"x ! C2e x ! C3e

2x .

y AxB # yp1 ! 2yp2 ! C1y1 ! C2y2 ! C3y3 Ee"x, ex, e2xF"1, 1, e

"x, ex, e2x

y3 AxB # e2xy2 AxB # exy1 AxB # e"xL 3 y 4 # "12e"2x yp2 AxB # e"2xL 3 y 4 # 2x2 " 2x " 4yp1 AxB # x2 L 3 y 4 J y‡ " 2y– " y¿ ! 2y # 2x2 " 2x " 4 " 24e"2x ,A"q, q B y AxB # c1yp1 AxB ! c2yp2 AxB ! C1y1 AxB ! C2y2 AxB ! p ! Cnyn AxB ,

L 3 y 4 # c1g1 ! c2g2 L 3 yp2 4 # g2L 3 yp1 4 # g1 y AxB 4 f AxB3f AxB " yp AxB # C1y1 AxB ! p ! Cnyn AxB

f AxB " yp AxB yp AxBf AxBf AxB

324 Chapter 6 Theory of Higher-Order Linear Differential Equations

Representation of Solutions (Nonhomogeneous Case)

Theorem 4. Let be a particular solution to the nonhomogeneous equation

(27)

on the interval with p1, p2, . . . , pn continuous on , and let be a fundamental solution set for the corresponding homogeneous equation

(28)

Then every solution of (27) on the interval can be expressed in the form

(29) y AxB ! yp AxB $ C1y1 AxB $ p $ Cnyn AxB .Aa, bB y AnB AxB ! p1 AxBy An"1B AxB ! p ! pn AxBy AxB # 0 .

Ey1, . . . , ynFAa, bBAa, bBy AnB AxB ! p1 AxBy An"1B AxB ! p ! pn AxBy AxB # g AxB yp AxB

Solution

Example 4

Section 6.1 Basic Theory of Linear Differential Equations 325

In Problems 1–6, determine the largest interval for which Theorem 1 guarantees the existence of a unique solution on to the given initial value problem.

In Problems 7–14, determine whether the given functions are linearly dependent or linearly independent on the specified interval. Justify your decisions.

7. 8. 9.

10. 11. 12. 13. 14.

Using the Wronskian in Problems 15–18, verify that the given functions form a fundamental solution set for the given differential equation and find a general solution. 15.

16.

17.

18.

In Problems 19–22, a particular solution and a funda- mental solution set are given for a nonhomogeneous

yA4B " y # 0 ; Eex, e"x, cos x, sin xFEx, x2, x3F x3y‡ " 3x2y– ! 6xy¿ " 6y # 0 , x 7 0 ;

U ex, cos 2x, sin 2x Vy‡ " y– ! 4y¿ " 4y # 0 ; U e3x, e"x, e"4x Vy‡ ! 2y– " 11y¿ " 12y # 0 ; Ex, xex, 1F on A"q, q BEx, x2, x3, x4F on A"q, q B Ecos 2x, cos2 x, sin2 xF on A"q, q BEx"1, x1/2, xF on A0, q B Esin x, cos x, tan xF on A"p/2, p/2BEsin2 x, cos2 x, 1F on A"q, q B Ex2, x2 " 1, 5F on A"q, q BEe3x, e5x, e"xF on A"q, q B y A3/4B # 1 , y¿ A3/4B # y– A3/4B # 0Ax2 " 1By‡ ! exy # ln x ; y A1/2B # y¿ A1/2B # "1 , y– A1/2B # 1x1x ! 1y‡ " y¿ ! xy # 0 ;y A"1/2B # 1 , y¿ A"1/2B # y– A"1/2B # 0 x Ax ! 1By‡ " 3xy¿ ! y # 0 ;y A5B # y¿ A5B # y– A5B # 1 y‡ " y– ! 1x " 1y # tan x ;y ApB # 0 , y¿ ApB # 11 , y– ApB # 3y‡ " 1xy # sin x ; y A"2B # 1 , y¿ A"2B # 0 , y– A"2B # 2xy‡ " 3y¿ ! exy # x2 " 1 ;

Aa, bB Aa, bB equation and its corresponding homogeneous equation.

(a) Find a general solution to the nonhomogeneous equation. (b) Find the solution that satisfies the specified initial conditions.

19.

20.

21.

22.

23. Let , and Verify that and

Then use the superposition prin- ciple (linearity) to find a solution to the differential equation: (a) (b)

24. Let , J , and . Verify that and Then use the superposition prin- ciple (linearity) to find a solution to the differential equation: (a) (b)

25. Prove that L defined in (7) is a linear operator by verifying that properties (9) and (10) hold for any n-times differentiable functions y, y1, on .

26. Existence of Fundamental Solution Sets. By Theorem 1, for each j # 1, 2, . . . , n there is a unique solution to equation (17) satisfying the initial conditions

(a) Show that is a fundamental solution set for (17). [Hint: Write out the Wronskian at x0.]

Ey1, y2, . . . , ynF yj AkB Ax0B # e 1 ,0 , for k # j " 1 ,for k $ j " 1, 0 & k & n " 1 .

yj AxB Aa, bBp , ym

L 3 y 4 # "6x cos 2x ! 11x .L 3 y 4 # 7x cos 2x " 3x . L 3 y2 4 AxB # x. L 3 y1 4 AxB # x cos 2xy2 AxB J "1 /3 cos 2xy1 AxBL 3 y 4 J y‡ " xy– ! 4y¿ " 3xy L 3 y 4 # 4x2 ! 4 " 6x sin x .L 3 y 4 # 2x sin x " x2 " 1 .

L 3 y2 4 AxB # x2 ! 1. L 3 y1 4 AxB # x sin xy2 AxB J x. y1 AxB J sin x,L 3 y 4 J y‡ ! y¿ ! xy Eexcos x, exsin x, e"xcos x, e"xsin xFyp # cos x ; y‡ A0B # "2 ;

y– A0B # "1 , y¿ A0B # 1 , y A0B # 2 , y A4B ! 4y # 5 cos x ; Ex, x ln x, x Aln xB2Fyp # ln x ; y– A1B # 0 ; y¿ A1B # 3 , y A1B # 3 ,

x3y‡ ! xy¿ " y # 3 " ln x , x 7 0 ; yp # x

2 ; E1, x, x3Fy– A1B # "4 ; y¿ A1B # "1 , y A1B # 2 , xy‡ " y– # "2 ; yp # x

2 ; Eex, e"xcos 2x, e"xsin 2xFy A0B # "1 , y¿ A0B # 1 , y– A0B # "3 ; y‡ ! y– ! 3y¿ " 5y # 2 ! 6x " 5x2 ;

6.1 EXERCISES

(b) For given initial values express the solution to (17) satisfying

[as in equa- tions (4)] in terms of this fundamental solution set.

27. Show that the set of functions where n is a positive integer, is linearly independent on every open interval . [Hint: Use the fact that a polynomial of degree at most n has no more than n zeros unless it is identically zero.]

28. The set of functions

1, cos x, sin x, . . . , cos nx, sin nx ,

where n is a positive integer, is linearly independent on every interval . Prove this in the special case n # 2 and

29. (a) Show that if f1, . . . , fm are linearly independent on then they are linearly independent on

(b) Give an example to show that if f1, . . . , fm are linearly independent on then they need not be linearly independent on

30. To prove Abel’s identity (26) for n # 3, proceed as fol- lows: (a) Let Use the product

rule for differentiation to show

(b) Show that the above expression reduces to

(32)

(33)

(d) Substituting the expressions in (33) into (32), show that

(34) W" AxB ! #p1 AxBW AxB . Ai ! 1, 2, 3B .yi

A3B AxB ! #a3 k!1

pk AxByiA3#kB AxB

W" AxB ! † y1 y2 y3y"1 y"2 y"3 y'1 y'2 y'3

† . ! † y1 y2 y3y¿1 y¿2 y¿3

y‡1 y‡2 y‡3 † .

W ¿ AxB # † y¿1 y¿2 y¿3y¿1 y¿2 y¿3 y–1 y–2 y–3

† ! † y1 y2 y3y–1 y–2 y–3 y–1 y–2 y–3

† W AxB J W 3 y1, y2, y3 4 AxB.

A"1, 1B.A"q, q B, A"q, q B.A"1, 1B, Aa, bB # A"q, q B.Aa, bB

FE Aa, bB

E1, x, x2, . . . , xnF ,k # 0, . . . , n " 1,y AkB Ax0B # gk, y AxB

g0, g1, . . . , gn"1,

326 Chapter 6 Theory of Higher-Order Linear Differential Equations

(e) Deduce Abel’s identity by solving the first-order differential equation (34).

31. Reduction of Order. If a nontrivial solution is known for the homogeneous equation

the substitution can be used to reduce the order of the equation, as was shown in Section 4.7 for second-order equations. By complet- ing the following steps, demonstrate the method for the third-order equation

(35)

given that is a solution.

(a) Set and compute , , and (b) Substitute your expressions from (a) into (35) to

obtain a second-order equation in (c) Solve the second-order equation in part (b) for w

and integrate to find y. Determine two linearly independent choices for y, say, y1 and y2.

(d) By part (c), the functions and are two solutions to (35). Verify

that the three solutions , and are linearly independent on

32. Given that the function is a solution to " show that the substitution

reduces this equation to where

33. Use the reduction of order method described in Prob- lem 31 to find three linearly independent solutions to

given that is a solution.

34. Constructing Differential Equations. Given three functions , , that are each three times differentiable and whose Wronskian is never zero on

, show that the equation

is a third-order linear differential equation for which is a fundamental solution set. What is the

coefficient of in this equation? 35. Use the result of Problem 34 to construct a third-

order differential equation for which is a fundamental solution set.

Ex, sin x, cos xFy‡ E f1, f2, f3F

∞ f1 AxB f2 AxB f3 AxB yf ¿1 AxB f ¿2 AxB f ¿3 AxB y¿ f –1 AxB f –2 AxB f –3 AxB y– f ‡1 AxB f ‡2 AxB f ‡3 AxB y‡ ∞ # 0

Aa, bB f3 AxBf2 AxBf1 AxB

f AxB # e2xy‡ " 2y– ! y¿ " 2y # 0, w # y¿.xw– ! 3w¿ " x3w # 0,

y AxB # y AxB f AxB # y AxBxx2y¿ ! xy # 0,y‡ f AxB # xA"q, q B.

y2 AxBex, y1 AxBy2 AxB # y2 AxBex y1 AxB # y1 AxBex

w J y¿.

y‡.y–y¿y AxB # y AxBexf AxB # ex y‡ " 2y– " 5y¿ ! 6y # 0 ,

y AxB # y AxB f AxBy AnB ! p1 AxBy An"1B ! p ! pn AxBy # 0 , f AxB

Our goal in this section is to obtain a general solution to an nth-order linear differential equa- tion with constant coefficients. Based on the experience gained with second-order equations in Section 4.2, you should have little trouble guessing the form of such a solution. However, our interest here is to help you understand why these techniques work. This is done using an opera- tor approach—a technique that is useful in tackling many other problems in analysis such as solving partial differential equations.

Let’s consider the homogeneous linear nth-order differential equation

(1)

where are real constants.† Since constant functions are everywhere continuous, equation (1) has solutions defined for all x in (recall Theorem 1 in Section 6.1). If we can find n linearly independent solutions to (1) on , say, then we can express a general solution to (1) in the form

(2)

with C1, . . . , Cn as arbitrary constants. To find these n linearly independent solutions, we capitalize on our previous success with

second-order equations. Namely, experience suggests that we begin by trying a function of the form

If we let L be the differential operator defined by the left-hand side of (1), that is,

(3)

then we can write (1) in the operator form

(4)

For we find

(5)

where is the polynomial . Thus, is a solution to equation (4), provided r is a root of the auxiliary (or characteristic) equation

(6)

According to the fundamental theorem of algebra, the auxiliary equation has n roots (counting multiplicities), which may be either real or complex. However, there are no formulas for determining the zeros of an arbitrary polynomial of degree greater than four, although if we can determine one zero r1, then we can divide out the factor and be left with a poly- nomial of lower degree. (For convenience, we have chosen most of our examples and exercises so that 0, '1, or '2 are zeros of any polynomial of degree greater than two that we must factor.)

r " r1

P ArB ! anrn $ an#1rn#1 $ p $ a0 ! 0 . erxanr

n ! an"1r n"1 ! p ! a0P ArB # e

rx Aanr n ! an"1r n"1 ! p ! a0B # erxP ArB ,L 3 erx 4 AxB # anr nerx ! an"1r n"1erx ! p ! a0erx y # erx,

L 3 y 4 AxB # 0 . L 3 y 4 :! anyAnB $ an#1yAn#1B $ p $ a1y" $ a0y ,

y # erx.

y AxB # C1y1 AxB ! p ! Cnyn AxB , y1, . . . , yn,A"q, q BA"q, q B

an A$0B, an"1, . . . , a0 any

AnB AxB ! an"1y An"1B AxB ! p ! a1y¿ AxB ! a0y AxB # 0 ,

Section 6.2 Homogeneous Linear Equations with Constant Coefficients 327

†Historical Footnote: In a letter to John Bernoulli dated September 15, 1739, Leonhard Euler claimed to have solved the general case of the homogeneous linear nth-order equation with constant coefficients.

6.2 HOMOGENEOUS LINEAR EQUATIONS WITH CONSTANT COEFFICIENTS

When a zero cannot be exactly determined, numerical algorithms such as Newton’s method or the quotient-difference algorithm can be used to compute approximate roots of the polynomial equation.† Some pocket calculators even have these algorithms built in.

We proceed to discuss the various possibilities.

Distinct Real Roots If the roots r1, . . . , rn of the auxiliary equation (6) are real and distinct, then n solutions to equation (1) are

(7)

As stated in the previous section, these functions are linearly independent on a fact that we now officially verify. Let’s assume that c1, . . . , cn are constants such that

(8)

for all x in Our goal is to prove that One way to show this is to construct a linear operator Lk that annihilates (maps to zero)

everything on the left-hand side of (8) except the kth term. For this purpose, we note that since r1, . . . , rn are the zeros of the auxiliary polynomial , then can be factored as

(9)

Consequently, the operator can be expressed in terms of the differentiation operator D as the following composition:†

(10)

We now construct the polynomial by deleting the factor from Then we set that is,

(11)

Applying Lk to both sides of (8), we get, via linearity,

(12)

Also, since we find [just as in equation (5)] that for all r. Thus (12) can be written as

which simplifies to

(13)

because for Since rk is not a root of , then It now follows from (13) that ck # 0. But as k is arbitrary, all the constants c1, . . . , cn must be zero. Thus,

as given in (7) are linearly independent. (See Problem 26 for an alternative proof.)

We have proved that, in the case of n distinct real roots, a general solution to (1) is

(14)

where C1, . . . , Cn are arbitrary constants.

y AxB ! C1er1x $ p $ Cnernx , y1 AxB, . . . , yn AxB

Pk ArkB $ 0.Pk ArBi $ k.Pk AriB # 0ckerkxPk ArkB # 0 , c1e

r1xPk Ar1B ! p ! cnernxPk ArnB # 0 , Lk 3 erx 4 AxB # erxPk ArBLk # Pk ADB,c1Lk 3 er1x 4 ! p ! cnLk 3 ernx 4 # 0 .

Lk J Pk ADB # an AD " r1) p AD " rk"1B AD " rk!1B p AD " rnB .Lk J Pk ADB; P ArB.Ar " rkBPk ArBL # P ADB # an AD " r1B p AD " rnB .

L 3 y 4 # any AnB ! an"1y An"1B ! p ! a0yP ArB # an Ar " r1B p Ar " rnB . P ArBP ArB

c1 # c2 # p # cn # 0.A"q, q B.c1er1x ! p ! cnernx # 0 A"q, q B,y1 AxB ! er1x, y2 AxB ! er2x, . . . , yn AxB ! ernx .

328 Chapter 6 Theory of Higher-Order Linear Differential Equations

†See, for example, Applied and Computational Complex Analysis, by P. Henrici (Wiley-Interscience, New York, 1974), Volume 1, or Numerical Analysis, 9th ed., by R. L. Burden and J. D. Faires (Brooks/Cole Cengage Learning, 2011). †Historical Footnote: The symbolic notation was introduced by Augustin Cauchy in 1827.P ADB

Find a general solution to

(15)

The auxiliary equation is

(16)

By inspection we find that r # 1 is a root. Then, using polynomial division, we get

which further factors into Hence the roots of equation (16) are , , Since these roots are real and distinct, a general solution to (15) is

◆

Complex Roots If is a complex root of the auxiliary equation (6), then so is its complex conjugate since the coefficients of are real-valued (see Problem 24). If we accept complex-valued functions as solutions, then both and are solutions to (1). Moreover, if there are no repeated roots, then a general solution to (1) is again given by (14). To find two real-valued solutions corresponding to the roots , we can just take the real and imaginary parts of That is, since

(17)

then two linearly independent solutions to (1) are

(18)

In fact, using these solutions in place of and in (14) preserves the linear inde- pendence of the set of n solutions. Thus, treating each of the conjugate pairs of roots in this manner, we obtain a real-valued general solution to (1).

Find a general solution to

(19)

The auxiliary equation is

(20)

which has distinct roots Thus, a general solution is

(21) ◆

Repeated Roots If r1 is a root of multiplicity m, then the n solutions given in (7) are not even distinct, let alone lin- early independent. Recall that for a second-order equation, when we had a repeated root r1 to the auxiliary equation, we obtained two linearly independent solutions by taking and So if r1 is a root of (6) of multiplicity m, we might expect that m linearly independent solutions are

(22)

To see that this is the case, observe that if r1 is a root of multiplicity m, then the auxiliary equation can be written in the form

(23) an Ar " r1Bm Ar " rm!1B p Ar " rnB # Ar " r1Bm P~ ArB # 0 , er1x , xer1x , x2er1x , . . . , xm#1er1x .

xer1x.er1x

y AxB # C1ex ! C2e"x cos 2x ! C3e"x sin 2x .r1 # 1, r2 # "1 ! 2i, r3 # "1 " 2i. r 3 ! r 2 ! 3r " 5 # Ar " 1B Ar 2 ! 2r ! 5) # 0 , y‡ ! y– ! 3y¿ " 5y # 0 .

e Aa"ibBxe Aa!ibBx eax cos bx , eax sin bx .

e Aa!ibBx # eax cos bx ! ieax sin bx ,e Aa!ibBx. a ' ib

e Aa"ibBxe Aa!ibBxP ArBa " ib, a ! ib Aa, b realB

y AxB # C1ex ! C2e"2x ! C3e3x .r3 # 3.r2 # "2 r1 # 1Ar " 1B Ar ! 2B Ar " 3B.r 3 " 2r 2 " 5r ! 6 # Ar " 1B Ar 2 " r " 6B ,

r 3 " 2r 2 " 5r ! 6 # 0 .

y‡ " 2y– " 5y¿ ! 6y # 0 .

Section 6.2 Homogeneous Linear Equations with Constant Coefficients 329

Example 1

Solution

Example 2

Solution

where and With this notation, we have the identity

(24)

[see (5)]. Setting r # r1 in (24), we again see that is a solution to To find other solutions, we take the kth partial derivative with respect to r of both sides

of (24):

(25)

Carrying out the differentiation on the right-hand side of (25), we find that the resulting expression will still have as a factor, provided Thus, setting in (25) gives

(26)

Now notice that the function has continuous partial derivatives of all orders with respect to r and x. Hence, for mixed partial derivatives of it makes no difference whether the differentiation is done first with respect to x, then with respect to r, or vice versa. Since L involves derivatives with respect to x, this means we can interchange the order of differentia- tion in (26) to obtain

Thus,

(27)

will be a solution to (1) for So m distinct solutions to (1), due to the root r # r1 of multiplicity m, are indeed given by (22). We leave it as an exercise to show that the m functions in (22) are linearly independent on (see Problem 25).

If is a repeated complex root of multiplicity m, then we can replace the 2m complex-valued functions

by the 2m linearly independent real-valued functions

(28)

Using the results of the three cases discussed above, we can obtain a set of n linearly independent solutions that yield a real-valued general solution for (1).

eAx sin Bx , xeAx sin Bx , . . . , xm#1eAx sin Bx . eAx cos Bx , xeAx cos Bx , . . . , xm#1eAx cos Bx ,

e Aa"ibBx , xe Aa"ibBx , . . . , xm"1e Aa"ibBxe Aa!ibBx , xe Aa!ibBx , . . . , xm"1e Aa!ibBx ,

a ! ib A"q, q)k # 0, 1, . . . , m " 1.

0 k 0r k

AerxB ` r#r1

# xker1x

L c 0 k 0r k

AerxB ` r#r1

d AxB # 0 . erx,

erx

0 k 0rk

L 3 erx 4 AxB ` r#r1

# 0 if k & m " 1 .

r # r1k & m " 1.Ar " r1B 0 k 0r k

L 3 erx 4 AxB # 0 k 0r k

3 erx Ar " r1Bm P~ ArB 4 . L 3 y 4 # 0.er1xL 3 erx 4 AxB # erx Ar " r1Bm P~ ArB

P ~ Ar1B $ 0.P~ ArB J an Ar " rm!1B p Ar " rnB

330 Chapter 6 Theory of Higher-Order Linear Differential Equations

Find a general solution to

(29)

The auxiliary equation is

which has roots Because the root at 1 has multiplicity 3, a general solution is

(30) ◆

Find a general solution to

(31)

whose auxiliary equation can be factored as

(32)

The auxiliary equation (32) has repeated complex roots: , and Hence a general solution is

◆y AxB # C1e2x cos x ! C2xe2x cos x ! C3e2x sin x ! C4xe2x sin x .r4 # 2 " i. r1 # 2 ! i, r2 # 2 ! i, r3 # 2 " i

r 4 " 8r 3 ! 26r 2 " 40r ! 25 # Ar 2 " 4r ! 5B2 # 0 . y A4B " 8y A3B ! 26y– " 40y¿ ! 25y # 0 , y AxB # C1ex ! C2xex ! C3x2ex ! C4e"2x .

r1 # 1, r2 # 1, r3 # 1, r4 # "2.

r 4 " r 3 " 3r 2 ! 5r " 2 # Ar " 1B3 Ar ! 2B # 0 , y A4B " y A3B " 3y– ! 5y¿ " 2y # 0 .

Section 6.2 Homogeneous Linear Equations with Constant Coefficients 331

Example 3

Solution

Example 4

Solution

In Problems 1–14, find a general solution for the differ- ential equation with x as the independent variable.

1. 2. 3. 4. 5. 6. 7. 8. 9.

10. 11. 12. 13. 14.

[Hint: is a solution.]

In Problems 15–18, find a general solution to the given homogeneous equation. 15. 16.

# AD2 ! 4B 3 y 4 # 0AD ! 1B2 AD " 6B3 AD ! 5B AD2 ! 1BAD " 1B 2 AD ! 3B AD2 ! 2D ! 5B2 3 y 4 # 0

y AxB # sin 3xyA4B ! 2y‡ ! 10y– ! 18y¿ ! 9y # 0 y A4B ! 4y– ! 4y # 0y‡ ! 5y– ! 3y¿ " 9y # 0 y A4B ! 4y‡ ! 6y– ! 4y¿ ! y # 0y‡ ! 3y– " 4y¿ " 6y # 0 u‡ " 9u– ! 27u¿ " 27u # 0 y‡ ! 5y– " 13y¿ ! 7y # 0 2y‡ " y– " 10y¿ " 7y # 0 y‡ " y– ! 2y # 0 y‡ ! 3y– ! 28y¿ ! 26y # 0 y‡ ! 2y– " 19y¿ " 20y # 0 6z‡ ! 7z– " z¿ " 2z # 0 y‡ " 3y– " y¿ ! 3y # 0 y‡ ! 2y– " 8y¿ # 0

17.

18.

In Problems 19–21, solve the given initial value problem. 19.

20.

21.

In Problems 22 and 23, find a general solution for the given linear system using the elimination method of Section 5.2.

22.

23.

dx/dt " x ! y # 0 d3x/dt3 " x ! dy/dt ! y # 0 , 2x ! d2y/dt2 ! 2y # 0 d2x/dt2 " x ! 5y # 0 ,

y A0B # 1 , y¿ A0B # 0 , y– A0B # 0y‡ " 4y– ! 7y¿ " 6y # 0 ; y A0B # 1 , y¿ A0B # "3 , y– A0B # 13y‡ ! 7y– ! 14y¿ ! 8y # 0 ; y A0B # "4 , y¿ A0B # "1 , y– A0B # "19y‡ " y– " 4y¿ ! 4y # 0 ;

# AD2 ! 6D ! 10B3 3 y 4 # 0AD " 1B3 AD " 2B AD2 ! D ! 1B # D 5 3 y 4 # 0AD ! 4B AD " 3B AD ! 2B3 AD2 ! 4D ! 5B2

6.2 EXERCISES

24. Let be a polyno- mial with real coefficients an, . . . , a0. Prove that if r1 is a zero of then so is its complex conjugate [Hint: Show that where the bar denotes complex conjugation.]

25. Show that the m functions are linearly independent on [Hint: Show that these functions are linearly independent if and only if are linearly independent.]

26. As an alternative proof that the functions are linearly independent on when

rn are distinct, assume

(33)

holds for all x in and proceed as follows: (a) Because the ri’s are distinct we can (if neces-

sary) relabel them so that

Divide equation (33) by to obtain

Now let on the left-hand side to obtain

(b) Since equation (33) becomes

for all x in Divide this equation by and let to conclude that

(c) Continuing in the manner of (b), argue that all the coefficients, C1, C2, . . . , Cn are zero and hence are linearly indepen- dent on

27. Find a general solution to

by using Newton’s method (Appendix B) or some other numerical procedure to approximate the roots of the auxiliary equation.

28. Find a general solution to by using Newton’s method or some other numerical procedure to approximate the roots of the auxiliary equation.

29. Find a general solution to

y A4B ! 2y A3B ! 4y– ! 3y¿ ! 2y # 0

y‡ " 3y¿ " y # 0

y A4B ! 2y‡ " 3y– " y¿ ! 1 2

y # 0

A"q, q B.er1x, er2x, . . . , ernx C2 # 0.x S ! qer2x

A"q, q B.C2e r2x ! C3e

r3x ! p ! Cnernx # 0

C1 # 0, C1 # 0.

x S ! q

C1 ! C2 er2x

er1x ! p ! Cn

ernx

er1x # 0 .

er1x

r1 7 r2 7 p 7 rn .

A"q, q BC1e r1x ! C2e

r2x ! p ! Cnernx # 0

r1, r2, . . . , A"q, q Bernx er1x, er2x, . . . ,

1, x, . . . , xm"1

A"q, q B.erx, xerx, . . . , xm"1erx P ArB # P ArB, r1.P ArB,

P ArB # anr n ! p ! a1r ! a0 332 Chapter 6 Theory of Higher-Order Linear Differential Equations

by using Newton’s method to approximate numeri- cally the roots of the auxiliary equation. [Hint: To find complex roots, use the Newton recursion formula and start with a complex initial guess z0.]

30. (a) Derive the form

for the general solution to the equation from the observation that the fourth roots of unity are , i, and

(b) Derive the form

for the general solution to the equation from the observation that the cube roots of unity are , and

31. Higher-Order Cauchy–Euler Equations. A dif- ferential equation that can be expressed in the form

where are constants, is called a homogeneous Cauchy–Euler equation. (The second- order case is discussed in Section 4.7.) Use the substi- tution to help determine a fundamental solution set for the following Cauchy–Euler equations: (a) (b)

(c)

[Hint:

32. Let , where and r are real num- bers, be a solution to a differential equation. Sup- pose we cannot determine r exactly but can only approximate it by . Let and consider the error (a) If r and are positive, show that the error

grows exponentially large as x approaches ! (b) If r and are negative, show that the error

goes to zero exponentially as x approaches ! 33. On a smooth horizontal surface, a mass of m1 kg is

attached to a fixed wall by a spring with spring con- stant k1 N/m. Another mass of m2 kg is attached to the first object by a spring with spring constant k2 N/m. The objects are aligned horizontally so that the springs are their natural lengths. As we showed in

q. r $ r~,r~

@ y AxB " y~ AxB @ . y~ AxB J Cer~xr~ C A$ 0By AxB # Cerx# xaEcos Ab ln xB ! i sin Ab ln xB F. 4x

a!ib # e Aa!ibBln xx 7 0 x3y‡ " 2x2y– ! 13xy¿ " 13y # 0 , x 7 0 . x4y A4B ! 6x3y‡ ! 2x2y– " 4xy¿ ! 4y # 0 ,x

3y‡ ! x2y– " 2xy¿ ! 2y # 0 , x 7 0 .

y # xr

an, an"1, . . . , a0

a0y AxB # 0 ,an xny AnB AxB ! an"1xn"1y An"1B AxB ! p e"i2p/3.1, ei2p/3

y A3B # y,A3e "x/2 sin A23x/2By AxB # A1ex ! A2e"x/2 cos A23x/2B

"i.1, "1

y A4B # y, y AxB # A1ex ! A2e"x ! A3 cos x ! A4 sin x

zn!1 # zn " f AznB / f ¿ AznB

Section 5.6, this coupled mass–spring system is governed by the system of differential equations

(34)

(35)

Let’s assume that , and If both objects are displaced 1 m to the right

of their equilibrium positions (compare Figure 5.26, page 285) and then released, determine the equa- tions of motion for the objects as follows: (a) Show that satisfies the equation

(36) (b) Find a general solution to (36). (c) Substitute back into (34) to obtain a general

solution for (d) Use the initial conditions to determine the solu-

tions, and which are the equations of motion.

y AtB,x AtB y AtB.x AtB

x AtBx A4B AtB $ 7x( AtB $ 6 x AtB ! 0 . x AtB

k2 # 2. m1 # m2 # 1, k1 # 3

m2 d2y

dt2 # k2x $ k2 y ! 0 .

m1 d2x dt2

$ Ak1 $ k2Bx # k2 y ! 0 ,

Section 6.3 Undetermined Coefficients and the Annihilator Method 333

34. Suppose the two springs in the coupled mass–spring system discussed in Problem 33 are switched, giving the new data , and If both objects are now displaced 1 m to the right of their equilibrium positions and then released, deter- mine the equations of motion of the two objects.

35. Vibrating Beam. In studying the transverse vibra- tions of a beam, one encounters the homogeneous equation

where is related to the displacement of the beam at position x, the constant E is Young’s modulus, I is the area moment of inertia, and k is a parameter. Assuming E, I, and k are positive constants, find a general solution in terms of sines, cosines, hyper- bolic sines, and hyperbolic cosines.

y AxB EI

d 4y

dx4 " ky # 0 ,

k2 # 3.m1 # m2 # 1, k1 # 2

In Sections 4.4 and 4.5 we mastered an easy method for obtaining a particular solution to a nonho- mogeneous linear second-order constant coefficient equation,

(1)

when the nonhomogeneity had a particular form (namely, a product of a polynomial, an expo- nential, and a sinusoid). Roughly speaking, we were motivated by the observation that if a function ƒ, of this type, resulted from operating on y with an operator L of the form then we must have started with a y of the same type. So we solved (1) by postulating a solution form yp that resembled ƒ, but with undetermined coefficients, and we inserted this form into the equation to fix the values of these coefficients. Eventually, we realized that we had to make certain accommoda- tions when ƒ was a solution to the homogeneous equation

In this section we are going to reexamine the method of undetermined coefficients from another, more rigorous, point of view—partly with the objective of tying up the loose ends in our previous exposition and more importantly with the goal of extending the method to higher- order equations (with constant coefficients). At the outset we’ll describe the new point of view that will be adopted for the analysis. Then we illustrate its implications and ultimately derive a simplified set of rules for its implementation: rules that justify and extend the procedures of Section 4.4. The rigorous approach is known as the annihilator method.

The first premise of the annihilator method is the observation, gleaned from the analysis of the previous section, that all of the “suitable types” of nonhomogeneities (products ofƒ AxB

L 3 y 4 # 0. AaD2 ! bD ! cB,ƒ

AxBL 3 y 4 # AaD2 ! bD ! cB 3 y 4 # ƒ AxB , 6.3 UNDETERMINED COEFFICIENTS AND THE ANNIHILATOR METHOD

polynomials times exponentials times sinusoids) are themselves solutions to homogeneous dif- ferential equations with constant coefficients. Observe the following:

(i) Any nonhomogeneous term of the form satisfies (ii) Any nonhomogeneous term of the form satisfies for

(iii) Any nonhomogeneous term of the form satisfies

(iv) Any nonhomogeneous term of the form satisfies for

In other words, each of these nonhomogeneities is annihilated by a differential operator with constant coefficients.

k # 0, 1, p , m " 1 .3 AD " aB2 ! b2 4m 3 f 4 # 0 ƒ AxB # xkeax cos bx or xkeax sin bxAD2 ! b2B 3 f 4 # 0. ƒ AxB # cos bx or sin bxk # 0, 1, p , m " 1 .

AD " rBm 3 f 4 # 0ƒ AxB # xkerx AD " rB 3 f 4 # 0.ƒ AxB # erx 334 Chapter 6 Theory of Higher-Order Linear Differential Equations

Annihilator

Definition 3. A linear differential operator A is said to annihilate a function f if

(2)

for all x. That is, A annihilates f if f is a solution to the homogeneous linear differential equation (2) on A"q, q B.

A 3 f 4 AxB # 0 , Find a differential operator that annihilates

(3)

Consider the two functions whose sum appears in (3). Observe that annihilates the function . Further, is annihilated by the operator

Hence, the composite operator

which is the same as the operator

annihilates both f1 and f2. But then, by linearity, A also annihilates the sum f1 ! f2. ◆

We now show how annihilators can be used to determine particular solutions to certain nonhomogeneous equations. Consider the nth-order differential equation with constant coeffi- cients

(4)

which can be written in the operator form

(5)

where

Assume that A is a linear differential operator with constant coefficients that annihilates Then

A 3L 3 y 4 4 AxB # A 3 f 4 AxB # 0 , ƒ AxB.L # anD n ! an"1D

n"1 ! p ! a0 .

L 3 y 4 AxB # ƒ AxB , any

AnB AxB ! an"1y An"1B AxB ! p ! a0y AxB # ƒ AxB ,

3 AD " 1B2 ! 4 4 AD ! 4B2 , A J AD ! 4B2 3 AD " 1B2 ! 4 4 ,AD " 1B2 ! 4. f2 AxB J 5e

xsin 2xf1 AxB J 6xe"4x AD ! 4B2 6xe"4x ! 5ex sin 2x .

Example 1

Solution

so any solution to (5) is also a solution to the homogeneous equation

(6)

involving the composition of the operators A and L. But (6) has constant coefficients, and we are experts on differential equations with constant coefficients! In particular, we can use the methods of Section 6.2 to write down a general solution of (6). From this we can deduce the form of a particular solution to (5). Let’s look at some examples and then summarize our findings. The differential equa- tion in the next example is second order, so we will be able to see exactly how the annihilator method is related to the techniques of Sections 4.4 and 4.5.

Find a general solution to

(7)

First let’s solve this by the methods of Sections 4.4 and 4.5, to get a perspective for the annihilator method. The homogeneous equation corresponding to (7) is with the general solution

Since is a solution of the homogeneous equation, the nonhomogeneity demands a solution form To accommodate the nonhomogeneity sin x, we need an undetermined coefficient form Values for C3 through C6 in the particular solution are determined by substitution:

eventually leading to the conclusion and Thus (for future reference), a general solution to (7) is

(8)

For the annihilator method, observe that annihilates sin x and annihi- lates Therefore, any solution to (7), expressed for convenience in operator form as

is annihilated by the composition that is, it satisfies the constant coefficient homogeneous equation

(9)

From Section 6.2 we deduce that the general solution to (9) is given by

(10)

This is precisely the solution form generated by the methods of Chapter 4; the first two terms are the general solution to the associated homogeneous equation, and the remaining four terms express the particular solution to the nonhomogeneous equation with undeter- mined coefficients. Substitution of (10) into (7) will lead to the quoted values for C3 through C6, and indeterminant values for C1 and C2; the latter are available to fit initial conditions.

Note how the annihilator method automatically accounts for the fact that the nonhomo- geneity requires the form in the particular solution, by counting the total number of factors of in the annihilator and the original differential operator. ◆AD " 1B C3xex ! C4x2exxex

y # C1e "x ! C2e

x ! C3xe x ! C4x

2ex ! C5 sin x ! C6 cos x .

AD2 ! 1B AD " 1B2 AD2 " 1B 3 y 4 # AD ! 1B AD " 1B3 AD2 ! 1B 3 y 4 # 0 . AD2 ! 1B AD " 1B2 AD2 " 1B;AD2 " 1B 3 y 4 AxB # xex ! sin x,xex. AD " 1B2AD2 ! 1B

y AxB # C1e"x ! C2ex ! x a" 14 ! 14xb ex " 12 sin x . C6 # 0.C3 # "1/4, C4 # 1/4, C5 # "1/2,

" 3C3xex ! C4x2ex ! C5 sin x ! C6 cos x 4 # sin x ! xex ,yp– " yp # 3C3xex ! C4x2ex ! C5 sin x ! C6 cos x 4 – C5 sin x ! C6 cos x.

x AC3 ! C4xBex. xexexC1e"x ! C2ex. y– " y # 0,

y– " y # xex ! sin x .

AL 3 y 4 AxB # 0 , Section 6.3 Undetermined Coefficients and the Annihilator Method 335

Example 2

Solution

Find a general solution, using the annihilator method, to

(11)

The associated homogeneous equation takes the operator form

(12)

The nonhomogeneity is annihilated by Therefore, every solution of (11) also satisfies

(13)

A general solution to (13) is

(14)

Comparison with (12) shows that the first three terms of (14) give a general solution to the associated homogeneous equation and the last two terms constitute a particular solution form with undetermined coefficients. Direct substitution reveals and and so a general solution to (11) is

◆

The annihilator method, then, rigorously justifies the method of undetermined coefficients of Section 4.4. It also tells us how to upgrade that procedure for higher-order equations with constant coefficients. Note that we don’t have to implement the annihilator method directly; we simply need to introduce the following modifications to the method of undetermined coefficients described in the procedural box on page 180.

y AxB # C1e"x ! C2e2x ! C3xe2x " 118x2e2x ! 118x3e2x . C5 # 1/18C4 # "1/18

y AxB # C1e"x ! C2e2x ! C3xe2x ! C4x2e2x ! C5x3e2x . AD " 2B2 AD3 " 3D2 ! 4B 3 y 4 # AD ! 1B AD " 2B4 3 y 4 # 0 .

AD " 2B2.xe2xAD 3 " 3D2 ! 4B 3 y 4 # AD ! 1B AD " 2B2 3 y 4 # 0 .

y‡ " 3y– ! 4y # xe2x .

336 Chapter 6 Theory of Higher-Order Linear Differential Equations

Example 3

Solution

Method of Undetermined Coefficients To find a particular solution to the constant-coefficient differential equation

where m is a nonnegative integer, use the form

(15)

with if r is not a root of the associated auxiliary equation; otherwise, take s equal to the multiplicity of this root.

To find a particular solution to the constant-coefficient differential equation or where use the form

(16)

with if is not a root of the associated auxiliary equation; otherwise, take s equal to the multiplicity of this root.

a ! ibs # 0

! xs 3Bmxm ! p ! B1x ! B0 4 eax sin bx ,yp AxB # xs 3Amxm ! p ! A1x ! A0 4 eax cos bx b $ 0,L 3 y 4 # Cxmeax sin bx,L 3 y 4 # Cxmeax cos bx

s # 0

yp A xB # xs 3Amxm ! p ! A1x ! A0 4 erx ,L 3 y 4 # Cxmerx,

Section 6.3 Undetermined Coefficients and the Annihilator Method 337

In Problems 1–4, use the method of undetermined coeffi- cients to determine the form of a particular solution for the given equation.

1. 2. 3. 4.

In Problems 5–10, find a general solution to the given equation.

5. 6. 7. 8. 9.

10.

In Problems 11–20, find a differential operator that anni- hilates the given function. 11. 12. 13. 14. 15. 16. 17. 18. 19. 20.

In Problems 21–30, use the annihilator method to deter- mine the form of a particular solution for the given equa- tion. 21. 22. 23. 24. 25. 26. 27. 28. 29. 30.

In Problems 31–33, solve the given initial value problem. 31.

y A0B # "2 , y¿ A0B # "8 , y– A0B # "12y‡ ! 2y– " 9y¿ " 18y # "18x2 " 18x ! 22 ; y‡ ! 2y– " y¿ " 2y # ex " 1 z‡ " 2z– ! z¿ # x " ex y– " 6y¿ ! 10y # e3x " x y– ! 2y¿ ! 2y # e"x cos x ! x2 y– ! 2y¿ ! y # x2 " x ! 1 y– " 6y¿ ! 9y # sin 2x ! x u– " u # xex y– " 5y¿ ! 6y # e3x " x2 y– ! 6y¿ ! 8y # e3x " sin x u– " 5u¿ ! 6u # cos 2x ! 1

x2ex " x sin 4x ! x3xe"2 x ! xe"5x sin 3x xe3x cos 5xx2e"x sin 2x x2 " exe2x " 6ex e5xe"7x 3x2 " 6x ! 1x4 " x2 ! 11

y‡ ! 4y– ! y¿ " 26y # e"3x sin 2x ! x y‡ " 3y– ! 3y¿ " y # ex y‡ ! y– " 2y # xex ! 1 y‡ ! 3y– " 4y # e"2x y‡ ! y– " 5y¿ ! 3y # e"x ! sin x y‡ " 2y– " 5y¿ ! 6y # ex ! x2

y‡ ! y– " 2y # xex ! 1 y‡ ! 3y– " 4y # e"2x y‡ ! y– " 5y¿ ! 3y # e"x ! sin x y‡ " 2y– " 5y¿ ! 6y # ex ! x2

32.

33.

34. Use the annihilator method to show that if in equation (4) and has the form

(17)

then

is the form of a particular solution to equation (4). 35. Use the annihilator method to show that if and

in (4) and has the form given in (17), then equation (4) has a particular solution of the form

36. Use the annihilator method to show that if in (4) has the form then equation (4) has a particular solution of the form where s is chosen to be the smallest nonnegative integer such that is not a solution to the corre- sponding homogeneous equation.

37. Use the annihilator method to show that if in (4) has the form

then equation (4) has a particular solution of the form

(18)

where s is chosen to be the smallest nonnegative integer such that and are not solu- tions to the corresponding homogeneous equation.

In Problems 38 and 39, use the elimination method of Section 5.2 to find a general solution to the given system.

38.

39.

x ! d2y/dt2 " y # e3t d2x/dt2 " x ! y # 0 , dx/dt ! dy/dt " 2y # et x " d2y/dt2 # t ! 1 ,

xs sin bxxs cos bx

yp AxB ! xsEA cos Bx $ B sin BxF , f AxB # a cos bx ! b sin bx ,

f AxB xsea x

yp AxB # xsBea x,f AxB # Bea x, f AxByp AxB # xEBm x m ! Bm"1x m"1 ! p ! B1x ! B0F .

f AxBa1 $ 0 a0 # 0 yp AxB # Bm x m ! Bm"1x m"1 ! p ! B1x ! B0

f AxB # bm x m ! bm"1x m"1 ! p ! b1x ! b0 ,f AxB a0 $ 0

y A0B # 3 , y¿ A0B # 0 , y– A0B # 0# 34xe"2x " 16e"2x " 10x2 ! 6x ! 34 ; y‡ " 2y– " 3y¿ ! 10y y A0B # 4 , y¿ A0B # "1 , y– A0B # "5y‡ " 2y– ! 5y¿ # "24e3x ;

6.3 EXERCISES

40. The currents in the electrical network in Figure 6.1 satisfy the system

where I1, I2, and I3 are the currents through the dif- ferent branches of the network. Using the elimina- tion method of Section 5.2, determine the currents if initially , ,

, and I¿3 A0B # 16 /3.I¿2 A0B # 3 /4 I¿1 A0B # 73 /12I1 A0B # I2 A0B # I3 A0B # 0 I1 # I2 ! I3 ,

64 I3 ! 9I–3 " 64I–2 # 0 ,

1 9

I1 ! 64I–2 # "2 sin t

24 ,

338 Chapter 6 Theory of Higher-Order Linear Differential Equations

48 cos( t /24) volts 9 henrys

9 farads 64 farads

64 henrys

I 1 I 2 I 3

Figure 6.1 An electrical network

In the previous section, we discussed the method of undetermined coefficients and the annihila- tor method. These methods work only for linear equations with constant coefficients and when the nonhomogeneous term is a solution to some homogeneous linear equation with constant coefficients. In this section we show how the method of variation of parameters discussed in Sections 4.6 and 4.7 generalizes to higher-order linear equations with variable coefficients.

Our goal is to find a particular solution to the standard form equation

(1)

where and the coefficient functions p1, . . . , pn, as well as g, are continuous on . The method to be described requires that we already know a funda- mental solution set for the corresponding homogeneous equation

(2)

A general solution to (2) is then

(3)

where C1, . . . , Cn are arbitrary constants. In the method of variation of parameters, we assume there exists a particular solution to (1) of the form

(4)

and try to determine the functions There are n unknown functions, so we will need n conditions (equations) to determine

them. These conditions are obtained as follows. Differentiating yp in (4) gives

(5)

To prevent second derivatives of the unknowns from entering the formula for we impose the condition

y¿1y1 ! p ! y¿nyn # 0 .

y–p,y1, . . . , yn

y¿p # Ay1y¿1 ! p ! yny¿nB ! Ay¿1y1 ! p ! y¿nynB . y1, . . . , yn.

yp AxB ! Y1 AxBy1 AxB $ p $ Yn AxByn AxB yh AxB # C1y1 AxB ! p ! Cnyn AxB , L 3 y 4 AxB # 0 .Ey1, . . . , ynF

Aa, bBL 3 y 4 J y AnB ! p1y An"1B ! p ! pny L 3 y 4 AxB # g AxB ,

6.4 METHOD OF VARIATION OF PARAMETERS

In a like manner, as we compute we impose additional conditions involving namely,

Finally, the nth condition that we impose is that yp satisfy the given equation (1). Using the previous conditions and the fact that y1, . . . , yn are solutions to the homogeneous equation, then reduces to

(6)

(see Problem 12). We therefore seek n functions that satisfy the system

(7)

Caution. This system was derived under the assumption that the coefficient of the highest derivative y(n) in (1) is one. If, instead, the coefficient of this term is the constant a, then in the last equation in (7) the right-hand side becomes g/a.

A sufficient condition for the existence of a solution to system (7) for x in is that the determinant of the matrix made up of the coefficients of be different from zero for all x in . But this determinant is just the Wronskian:

(8)

which is never zero on because is a fundamental solution set. Solving (7) via Cramer’s rule (Appendix D), we find

(9)

where is the determinant of the matrix obtained from the Wronskian by replacing the kth column by . Using a cofactor expansion about this column, we can express in terms of an th-order Wronskian:

(10)

Integrating in (9) gives

(11)

Finally, substituting the ’s back into (4), we obtain a particular solution to equation (1):

(12)

Note that in equation (1) we presumed that the coefficient of the leading term, was unity. If, instead, it is we must replace by in (12).g AxB /p0 AxBg AxBp0 AxB, y AnB,

yp AxB ! an k!1

yk AxB " g AxBWk AxBW 3 y1, . . . , yn 4 AxB dx . yk

yk AxB # " g AxBWk AxBW 3 y1, . . . , yn 4 AxB dx , k # 1, . . . , n . y¿k AxBWk AxB # A"1B

n"kW 3 y1, . . . , yk"1, yk!1, . . . , yn 4 AxB , k # 1, . . . , n .An " 1BWk AxB col 30, . . . , 0, 1 4 W 3 y1, . . . yn 4 AxBWk AxB

y¿k AxB # g AxBWk AxBW 3 y1, . . . , yn 4 AxB , k # 1, . . . , n , Ey1, . . . , ynFAa, bB

∞ y1 . . . yno oy An"2B1 . . . y An"2Bn y An"1B1 . . . y An"1Bn

∞ # W 3 y1, . . . , yn 4 AxB , Aa, bB y¿1, . . . , y¿n

Aa, bB yAn#1B1 Y"1 $ p $ yAn#1Bn Y"n ! g .y An#2B 1 Y"1 $ p $ yAn#2Bn Y"n ! 0 ,oooo

y1Y"1 $ p $ ynY"n ! 0 , y¿1, . . . , y¿n

y¿1y An"1B 1 ! p ! y¿ny

An"1B n # g

L 3 yp 4 # g y¿1y¿1 ! p ! y¿ny¿n # 0, . . . , y¿1y

An"2B 1 ! p ! y¿ny

An"2B n # 0 .

y¿1, . . . , y¿n; An " 2By–p, y‡p , . . . , y An"1Bp ,

Section 6.4 Method of Variation of Parameters 339

Although equation (12) gives a neat formula for a particular solution to (1), its implemen- tation requires one to evaluate determinants and then perform n integrations. This may entail several tedious computations. However, the method does work in cases when the technique of undetermined coefficients does not apply (provided, of course, we know a funda- mental solution set).

Find a general solution to the Cauchy–Euler equation

(13)

given that is a fundamental solution set to the corresponding homogeneous equation.

An important first step is to divide (13) by to obtain the standard form

(14)

from which we see that . Since is a fundamental solution set, we can obtain a particular solution of the form

(15)

To use formula (12), we must first evaluate the four determinants:

Substituting the above expressions into (12), we find

which after some labor simplifies to

(16)

where and denote the constants of integration. Since is a fundamental solution set for the homogeneous equation, we can take and to be arbitrary constants. The right-hand side of (16) then gives the desired general solution. ◆

In the preceding example, the fundamental solution set can be derived by substituting into the homogeneous equation corresponding to (13) (see Problem 31, Exercises 6.2). However, in dealing with other equations that have variable coefficients, the determination of a fundamental set may be extremely difficult. In Chapter 8 we tackle this problem using power series methods.

y # xr Ex, x"1, x2F

C3C1, C2, Ex, x"1, x2FC3C1, C2,yp AxB # cos x " x"1sin x ! C1x ! C2x"1 ! C3x2 ,

# x" a" 12 x sin xb dx ! x"1" 16 x3 sin x dx ! x2" 13 sin x dx , yp AxB # x" Asin xB3"6x"1 dx ! x"1" Asin xB A"x2B"6x"1 dx ! x2" Asin xB A"2x"1B "6x"1 dx

W3 AxB # A"1B A3"3B ` x x"11 "x"2 ` # "2x"1 . W2 AxB # A"1B A3"2B ` x x21 2x2 ` # "x2 , W1 AxB # A"1B A3"1BW 3 x"1, x2 4 AxB # A"1B2 ` x"1 x2"x"2 2x2 ` # 3 ,

W 3 x, x"1, x2 4 AxB # † x x"1 x21 "x"2 2x 0 2x"3 2

† # "6x"1 , yp AxB # y1 AxBx ! y2 AxBx"1 ! y3 AxBx2 .

Ex, x"1, x2Fg AxB # sin xy‡ ! 1 x

y– " 2 x2

y¿ ! 2 x3

y # sin x , x 7 0 ,

Ex, x"1, x2Fx3y‡ ! x2y– " 2xy¿ ! 2y # x3 sin x , x 7 0 ,

n ! 1

340 Chapter 6 Theory of Higher-Order Linear Differential Equations

Example 1

Solution

In Problems 1–6, use the method of variation of parame- ters to determine a particular solution to the given equation.

1. 2. 3.

7. Find a general solution to the Cauchy–Euler equa- tion

given that is a fundamental solution set for the corresponding homogeneous equation.

8. Find a general solution to the Cauchy–Euler equation

given that is a fundamental solution set for the corresponding homogeneous equation.

9. Given that is a fundamental solution set for the homogeneous equation corresponding to the equation

determine a formula involving integrals for a partic- ular solution.

10. Given that is a fundamental solution set for the homogeneous equation corresponding to the equation

determine a formula involving integrals for a partic- ular solution.

11. Find a general solution to the Cauchy–Euler equation

12. Derive the system (7) in the special case when n # 3. [Hint: To determine the last equation, require that and use the fact that , and satisfy the corresponding homogeneous equation.]

13. Show that

Wk AxB # A"1B An"kBW 3 y1, . . . , yk"1, yk!1, . . . , yn 4 AxB . y3y1, y2L 3 yp 4 # g

x3y‡ " 3xy¿ ! 3y # x4 cos x , x 7 0

x 7 0 , x3y‡ " x2y– " 4xy¿ ! 4y # g AxB ,

Ex, x"1, x4F y‡ " 2y– " y¿ ! 2y # g AxB ,

Eex, e"x, e2xFEx, x ln x, x 3Fx 7 0 ,

x3y‡ " 2x2y– ! 3xy¿ " 3y # x2 ,

Ex, x2, x3Fx 7 0 , x3y‡ " 3x2y– ! 6xy¿ " 6y # x"1 ,

y‡ ! y¿ # sec u tan u , 0 6 u 6 p /2 y‡ ! y¿ # tan x , 0 6 x 6 p /2 y‡ " 3y– ! 3y¿ " y # ex z‡ ! 3z– " 4z # e2x y‡ " 2y– ! y¿ # x y‡ " 3y– ! 4y # e2x

Section 6.4 Method of Variation of Parameters 341

14. Deflection of a Beam Under Axial Force. A uniform beam under a load and subject to a constant axial force is governed by the differential equation

where is the deflection of the beam, L is the length of the beam, is proportional to the axial force, and

is proportional to the load (see Figure 6.2). (a) Show that a general solution can be written in

the form

(b) Show that the general solution in part (a) can be rewritten in the form

where

(c) Let First compute the general solution using the formula in part (a) and then using the formula in part (b). Compare these two general solutions with the general solution

which one would obtain using the method of undetermined coefficients.

(d) What are some advantages of the formula in part (b)?

y AxB # B1 ! B2 x ! B3ekx ! B4e"kx " 12k2 x2 , q AxB ! 1.G As, xB J

s " x

k2 "

sin h 3 k As " xB 4 k3

! " x

0 q AsBG As, xB ds ,

y AxB # c1 ! c2x ! c3ekx ! c4e"kx !

ekx

2k3"q AxBe"kx dx " e"kx2k3 "q AxBekx dx . !

1 k2 " q AxB x dx " xk2 " q AxB dx

y AxB # C1 ! C2x ! C3ekx ! C4e"kx q AxB k2

y AxBy A4B AxB " k2y– AxB # q AxB , 0 6 x 6 L,

x L

y ( x )

Load

Axial force

Figure 6.2 Deformation of a beam under axial force and load

6.4 EXERCISES

The theory and techniques for solving an nth-order linear differential equation

(1)

are natural extensions of the development for linear second-order equations given in Chapter 4. Assuming that and g are continuous functions on an open interval I, there is a unique solution to (1) on I that satisfies the n initial conditions: ,

# , where For the corresponding homogeneous equation

(2)

there exists a set of n linearly independent solutions on I. Such functions are said to form a fundamental solution set, and every solution to (2) can be written as a linear combi- nation of these functions:

The linear independence of solutions to (2) is equivalent to the nonvanishing on I of the Wronskian

When equation (2) has (real) constant coefficients so that it is of the form

(3)

then the problem of determining a fundamental solution set can be reduced to the algebraic problem of solving the auxiliary or characteristic equation

(4)

If the n roots of (4) — say, — are all distinct, then

(5)

is a fundamental solution set for (3). If some real root — say, r1 — occurs with multiplicity m (e.g., ), then m of the functions in (5) are replaced by

When a complex root to (4) occurs with multiplicity m, then so does its conjugate and 2m members of the set (5) are replaced by the real-valued functions

A general solution to the nonhomogeneous equation (1) can be written as

y AxB ! yp AxB $ yh AxB , ea x cos bx, xea x cos bx, . . . , xm"1ea x cos bx .

ea x sin bx, xea x sin bx, . . . , xm"1ea x sin bx ,

a ! ib

er1x, xer1x, . . . , xm"1er1x .

r1 # r2 # p # rm

Eer1x, er2x, . . . , ern xFr1, r2, . . . , rn anr

n ! an"1r n"1 ! p ! a0 # 0 .

any AnB ! an"1y An"1B ! p ! a0y # 0 , an $ 0 ,

W 3 y1, . . . , yn 4 AxB J det ≥y1 AxB . . . yn AxBy¿1 AxB . . . y¿n AxBo o y An"1B1 AxB . . . y An"1Bn AxB¥ .

y AxB ! C1y1 AxB $ C2y2 AxB $ p $ Cnyn AxB . Ey1, . . . , ynFyAnB $ p1 AxByAn#1B $ p $ pn AxBy ! 0 ,

x0 % I.g1, . . . , y An"1B Ax0B # gn"1y¿ Ax0B y Ax0B # g0

p1, . . . , pn

yAnB $ p1 AxByAn#1B $ p $ pn AxBy ! g AxB

342 Chapter 6 Theory of Higher-Order Linear Differential Equations

Chapter Summary

where is some particular solution to (1) and is a general solution to the corresponding homogeneous equation. Two useful techniques for finding particular solutions are the annihi- lator method (undetermined coefficients) and the method of variation of parameters.

The annihilator method applies to equations of the form

(6)

where L is a linear differential operator with constant coefficients and the forcing term is a polynomial, exponential, sine, or cosine, or a linear combination of products of these. Such a func- tion is annihilated (mapped to zero) by a linear differential operator A that also has constant coefficients. Every solution to the nonhomogeneous equation (6) is then a solution to the homoge- neous equation and, by comparing the solutions of the latter equation with a general solution to we can obtain the form of a particular solution to (6). These forms have previously been studied in Section 4.4 for the method of undetermined coefficients.

The method of variation of parameters is more general in that it applies to arbitrary equa- tions of the form (1). The idea is, starting with a fundamental solution set for (2), to determine functions such that

(7)

satisfies (1). This method leads to the formula

(8)

where

Wk AxB # A"1Bn"kW 3 y1, . . . , yk"1, yk!1, . . . , yn 4 AxB , k # 1, . . . , n . yp AxB # an

k#1 yk AxB " g AxBWk AxBW 3 y1, . . . , yn 4 AxB dx ,

yp AxB # y1 AxBy1 AxB ! p ! yn AxByn AxBy1, . . . , yn Ey1, . . . , ynF

L 3 y 4 # 0,AL 3 y 4 # 0, g AxB g

AxBL 3 y 4 # g AxB , yhyp

Review Problems 343

REVIEW PROBLEMS

1. Determine the intervals for which Theorem 1 on page 318 guarantees the existence of a solution in that interval. (a) (b)

2. Determine whether the given functions are linearly dependent or linearly independent on the interval

(a) (b) (c)

3. Show that the set of functions , , , is linearly independent on

4. Find a general solution for the given differential equation.

A"q, q B.x3 sin xF x2 sin xx sin xEsin x E2e2x " ex, e2x ! 1, e2x " 3, ex ! 1FEexsin 2x, xexsin 2x, ex, xexF Ee2x, x2e2x, e"xFA0, q B. x2 ! 3

Ax2 " 1By‡ ! Asin xBy– ! 2x ! 4 y¿ ! exy #y A4B " Aln xBy– ! xy¿ ! 2y # cos 3x (a) (b) (c) (d)

5. Find a general solution for the homogeneous linear differential equation with constant coefficients whose auxiliary equation is (a) . (b) .

6. Given that is a particular solution to

on find a general solution.

7. Find a differential operator that annihilates the given function. (a) (b) (c) (d) (e) x2 " 2x ! xe"x ! sin 2x " cos 3x

x2e"2x cos 3xx sin 2x e3x ! x " 1x2 " 2x ! 5

A0, q B,y A4B ! y # A16x4 " 11B sin Ax2B " 48x2cos Ax2B yp # sin Ax2Br4 Ar " 1B2 Ar2 ! 2r ! 4B2 # 0

Ar ! 5B2 Ar " 2B3 Ar2 ! 1B2 # 0 y‡ " 2y– " y¿ ! 2y # ex ! x y A5B " y A4B ! 2y‡ " 2y– ! y¿ " y # 0y‡ ! 3y– " 5y¿ ! y # 0 y A4B ! 2y‡ " 4y– " 2y¿ ! 3y # 0

8. Use the annihilator method to determine the form of a particular solution for the given equation. (a) (b) (c) (d)

9. Find a general solution to the Cauchy–Euler equation

x 7 0 , x3y‡ " 2x2y– " 5xy¿ ! 5y # x"2 ,

y‡ " y– ! 2y # x sin x y A4B ! 6y– ! 9y # x2 " sin 3xy‡ ! 2y– " 19y¿ " 20y # xe"x y– ! 6y¿ ! 5y # e"x ! x2 " 1

344 Chapter 6 Theory of Higher-Order Linear Differential Equations

given that is a fundamental solution set to the corresponding homogeneous equation.

10. Find a general solution to the given Cauchy–Euler equation. (a) (b) x3y‡ ! 2x2y– ! 2xy¿ ! 4y # 0 , x 7 0

4x3y‡ ! 8x2y– " xy¿ ! y # 0 , x 7 0

Ex, x5, x"1F

TECHNICAL WRITING EXERCISES

1. Describe the differences and similarities between second-order and higher-order linear differential equations. Include in your comparisons both theoret- ical results and the methods of solution. For exam- ple, what complications arise in solving higher-order equations that are not present for the second-order case?

2. Explain the relationship between the method of undetermined coefficients and the annihilator method. What difficulties would you encounter in

applying the annihilator method if the linear equa- tion did not have constant coefficients?

3. For students with a background in linear algebra: Compare the theory for kth-order linear differential equations with that for systems of n linear equations in n unknowns whose coefficient matrix has rank

Use the terminology from linear algebra; for example, subspaces, basis, dimension, linear trans- formation, and kernel. Discuss both homogeneous and nonhomogeneous equations.

n " k.

A Computer Algebra Systems and Exponential Shift

Courtesy of Bruce W. Atkinson, Samford University

This project shows that the solution of constant-coefficient equations can be obtained by perform- ing successive integrations, each of which, at worst, involves integration by parts which is nicely automated with a computer algebra system (CAS). This technique could be programmed on a computer and thus sheds some light on the inner workings of the CAS.

(a) Use mathematical induction to prove the following exponential shift property. Given a nonnegative integer n, an n-times differentiable function y(x), and a real number r, then

Dn[e"rxy] # e"rx(D " r)n[y].

(b) Let n be a positive integer and r a real number. Use the exponential shift property to prove that the solution of the homogeneous equation

(D " r)n[y] # 0

is given by

y(x) # (C0 ! C1x ! ((( ! Cn"1 x n"1) erx ,

where C0, C1, . . . , Cn"1 are real constants.

We can apply the exponential shift property to nonhomogeneous equations.

yÔ(x) ! 9yfl(x) ! 27y)(x) ! 27y(x) # (24 ! 47x)ex cos x

! (45 ! 52x)ex sin x " 6xe"3x cos x

! ("6 ! x2)e"3x sin x .

[Hint: Rewrite the equation in operator form; then for the factor of the form (D " r)n, multiply by e"rx, apply the exponential shift property, and perform n integrals using the CAS. These integrals would involve integration by parts if done by hand.]

Note that using the superposition principle together with the method of undetermined coeffi- cients, in part (c) we would have to set up and solve two linear systems, one 4 * 4 and the other 6 * 6. For the equation in part (c), this new method is much faster.

We can use a “peeling off” method to solve equations for which the corresponding auxiliary

345

Group Projects for Chapter 6

equation has more than one root.

(d) Find a general solution to the equation

y(4)(x) " 2yÔ(x) " 3yfl(x) ! 4y)(x) ! 4y(x) # 6e"x ["6 ! 9x ! 2e3x(2 ! 12x ! 9x2)] .

[Hint: Rewrite the equation in operator form; then for the first factor of the form (D " r)n, multiply by e"rx, apply the exponential shift property, and perform n integrals using the CAS. Repeat the procedure for each new factor.]

The exponential shift property also holds with r replaced by a complex number a ! ib. In this case we have to use Euler’s formula on page 168. Now, suppose the factored form of an oper- ator has a power of (D " a)2 ! b2. Then we could write (D " a)2 ! b2 # (D " z)(D " z̄), where z # a ! ib.

(e) Find a general solution to the equation

[(D " 1)2 ! 4]2[y] # sin2 x .

[Hint: So as to only involve integrating polynomials times complex exponentials, write . But make sure your answer is given as a real-valued function.]

B Justifying the Method of Undetermined Coefficients

The annihilator method discussed in Section 6.3 can be used to derive the rules in Sections 4.4 and 4.5 for the method of undetermined coefficients. To show this, it suffices to work with functions of the form

(1)

where pn and qm are polynomials of degrees n and m, respectively—since the other types in Section 4.5 are just special cases of (1).

Consider the nonhomogeneous equation

(2)

where L is the linear operator

(3)

with constants and as given in equation (1). Let

(a) Show that

is an annihilator for g. (b) Show that the auxiliary equation associated with is of the form

(4)

where is the multiplicity of as roots of the auxiliary equation associated with , and are the remaining roots of this equation.

(c) Find a general solution for and compare it with a general solution for to verify that equation (2) has a particular solution of the form

where and are polynomials of degree N.QNPN

yp AxB # xseaxEPN AxB cos bx ! QN AxB sin bxF ,L 3 y 4 # 0 AL 3 y 4 # 0 r2s!1, . . . , rnL 3 y 4 # 0 a ' ibs A+0Ban 3

Ar " aB2 ! b2 4 s!N!1 Ar " r2s!1B p Ar " rnB # 0 ,AL 3 y 4 # 0 A J 3 AD " aB2 ! b2 4 N!1

N J max An, mB.g AxBan, an"1, . . . , a0L 3 y 4 J an y AnB ! an"1y An"1B ! p ! a0 y , L 3 y 4 AxB # g AxB , g AxB # pn AxBeaxcos bx ! qm AxBeaxsin bx ,

sin x # Aeix " e" ix)/(2iB

346 Chapter 6 Theory of Higher-Order Linear Differential Equations

C Transverse Vibrations of a Beam In applying elasticity theory to study the transverse vibrations of a beam, one encounters the equation

where is related to the displacement of the beam at position x; the constant E is Young’s modulus; I is the area moment of inertia, which we assume is constant; is the constant mass per unit length of the beam; and is a positive parameter to be determined. We can simplify the equa- tion by letting that is, we consider

(5)

When the beam is clamped at each end, we seek a solution to (5) that satisfies the boundary conditions

(6)

where L is the length of the beam. The problem is to determine those nonnegative values of r for which equation (5) has a nontrivial solution that satisfies (6). To do this, proceed as follows:

(a) Show that there are no nontrivial solutions to the boundary value problem (5)–(6) when

(b) Represent the general solution to (5) for in terms of sines, cosines, hyperbolic sines, and hyperbolic cosines.

(c) Substitute the general solution obtained in part (b) into the equations (6) to obtain four linear algebraic equations for the four coefficients appearing in the general solution.

(d) Show that the system of equations in part (c) has nontrivial solutions only for those values of r satisfying

(7)

(e) On the same coordinate system, sketch the graphs of cosh(rL) and sec(rL) versus r for L # 1 and argue that equation (7) has an infinite number of positive solutions.

(f ) For L # 1, determine the first two positive solutions to (7) numerically, and plot the corresponding solutions to the boundary value problem (5)–(6). [Hint: You may want to use Newton’s method in Appendix B.]

D Higher-Order Difference Equations Difference equations occur in mathematical models of physical processes and as tools in numerical analysis. The theory of linear difference equations parallels the theory of linear differential equations. Using the results of this chapter as models, both for the statements of theorems and for their proofs, we can develop a theory for linear difference equations.

A Kth order linear difference equation is an equation of the form

(8)

where and gn are defined for all nonnegative integers n. By a solution to (8) we mean a sequence of real numbers that satisfies (8) for all integers n 0.+EynFqn#0aK AnB, . . . , a0 AnB,

aK AnByn!K ! aK"1 AnByn!K"1 ! p ! a1 AnByn!1 ! a0 AnByn # gn, n + 0,

cosh ArLB # sec ArLB .

r 7 0 r # 0.

Ay AxB [ 0B y A0B ! y" A0B ! 0 and y ALB ! y" ALB ! 0 , yA4B AxB # r4y AxB ! 0 .r 4 J gl /EI;

y AxBEIy A4B AxB " gly AxB # 0 ,

Group Projects for Chapter 6 347

(a) Prove the following existence and uniqueness result.

Theorem Let c0, c1, . . . , be given constants and assume that for all integers Then there exists a unique solution to (8) that satisfies the initial conditions [Hint: Begin by solving for yK in terms of ]

(b) The homogeneous equation associated with (8) is

(9)

Prove that any linear combination of solutions to (9) is also a solution to (9). (By a linear combination of sequences and we mean a sequence of the form

where a, b are constants.)

(10)

Prove the following representation theorem.

Theorem Assume for all integers and let be any K solutions to (9) satisfying

(11)

for all integers Then every solution to (9) can be expressed in the form

where are constants.

(d) When the ’s are independent of n—that is, when are constants— then equation (8) (or (9)) is said to have constant coefficients. In this case, equation (9) looks like

(12)

Show that equation (12) has a solution of the form yn = r n, where is a fixed

number (real or complex), if and only if r satisfies the auxiliary equation

(13)

In parts (e) through (h), use the results of parts (c) and (d) and your experience with linear differ- ential equations with constant coefficients to find a “general solution” to the given homogeneous difference equation. When necessary, use Euler’s formula to obtain real-valued solutions.

(e) (f) (g) (h) yn!3 ! 3yn!2 ! 3yn!1 ! yn # 0.

yn!3 " 3yn!2 ! 4yn!1 " 2yn # 0. yn!3 " 4yn!2 ! 5yn!1 " 2yn # 0. yn!3 ! yn!2 " 4yn!1 " 4yn # 0.

aKr K ! aK"1r

K"1 ! . . . ! a1r ! a0 # 0 .

r $ 0

aKyn!K ! aK"1yn!K"1 ! . . . ! a0yn # 0 .

aK, aK"1, . . . , a0aj AnBC1, . . . CK yn # C1yn

A1B ! C2ynA2B ! . . . ! CKynAKB , n + 0.

!n 3 Eyn(1)F, . . . , Eyn(K)F 4 $ 0 EynA1BF, . . . , EynAKBFn + 0,aK(n) $ 0

y A1Bn y A2Bn p y AKBn y A1Bn!1 y A2Bn!1 p y AKBn!1 o o o y A1Bn!K"1 y A2Bn!K"1 p y AKBn!K"1

!n 3 Eyn(1)F, . . . , Eyn(K)F 4 :# Eyn(1)F, Eyn(2)F, . . . , Eyn(K)F: Eayn ! bznFqn#0, EznFqn#0EynFqn#0

aK AnByn!K ! aK"1 AnByn!K"1 ! p ! a0 AnByn # 0, n + 0. yK"1, . . . , y0.

y0 # c0, y1 # c1, . . . , yK"1 # cK"1. EynFn + 0. aK AnB $ 0cK"1

348 Chapter 6 Theory of Higher-Order Linear Differential Equations

In 1202 the mathematician Fibonacci (Leonardo da Pisa) modeled the growth of a rabbit population by assuming that each mating pair produces one new pair (one male, one female) every month. A newly-born pair of rabbits, one male, one female, are put in a field, and they begin mating at the age of one month. (And, in this Utopian setting, they are immortal.)

(i) Show that the numbers of pairs of rabbits yn after n months satisfies the difference equation system

(14)

(j) Equations (14) generate the Fibonacci sequence 0, 1, 1, 2, 3, 5, 8, 13, . . . . Show that

The number is known as the golden ratio. If you rough-sketch a rectangle, without think- ing about it, its sides will probably be in the proportion 1.618; it’s just a aesthetically appealing dimension. (The Parthenon’s dimensions conform to the golden ratio. Next time you visit Athens, you will understand.)

Many biological phenomena can be modeled by Fibonacci’s rule (14), and you will see the sequence appearing in tree branching, arrangement of leaves on a stem, pineapple and artichoke fruitlets, ferns, pine cones, honeybee populations, and nautilus spirals. For further discussion visit the web site http://www.maths.surrey.ac.uk/hosted-ites/R.Knott/Fibonacci/fibInArt.html

yn # fn " A"fB"n25 , where f # 1 ! 252 # 1.618 .

yn!2 # yn!1 ! yn , y0 # 0, y1 # 1 .

Group Projects for Chapter 6 349

http://www.maths.surrey.ac.uk/hosted-ites/R.Knott/Fibonacci/fibInArt.html

350

We analyzed a simpler version of this problem in Example 1 of Section 3.2. Let be the amount of salt (in kilograms) in the tank at time t. Then of course is the concentration, in kilograms per liter. The salt content is depleted at the rate (6 L/min) !

kg/min through the exit valve. Simultaneously, it is enriched through valves A and B at the rate given by

(1)

Thus, changes at a rate

(2) dx dt

" 3

500 x # g AtB ,

d dt

x AtB # g AtB $ 3x AtB 500

x AtBg AtB # e0.4 kg/L ! 6 L/min # 2.4 kg/min , 0 6 t 6 10 (valve A) ,

0.2 kg/L ! 6 L/min # 1.2 kg/min , t 7 10 (valve B) .

g AtB,3x AtB /500 Ax AtB /1000 kg/LB #x AtB /1000

x AtB

Laplace Transforms

CHAPTER 7

INTRODUCTION: A MIXING PROBLEM7.1 Figure 7.1 depicts a mixing problem with valved input feeders. At time t ! 0, valve A is opened, delivering 6 L/min of a brine solution containing 0.4 kg of salt per liter. At t ! 10 min, valve A is closed and valve B is opened, delivering 6 L/min of brine at a concentration of 0.2 kg/L. Initially, 30 kg of salt are dissolved in 1000 L of water in the tank. The exit valve C, which empties the tank at 6 L/min, main- tains the contents of the tank at constant volume. Assuming the solution is kept well stirred, determine the amount of salt in the tank at all times t % 0.

x(t)

x(0) = 30 kg 6 L/min

1000 L C

6 L/min, 0.4 kg/L

6 L/min, 0.2 kg/L

Figure 7.1 Mixing tank with valve A open

with initial condition

(3) To solve the initial value problem (2)–(3) using the techniques of Chapter 4, we would

have to break up the time interval into two subintervals (0, 10) and On these subintervals, the nonhomogeneous term is constant, and the method of undetermined coef- ficients could be applied to equation (2) to determine general solutions for each subinterval, each containing one arbitrary constant (in the associated homogeneous solutions). The initial condition (3) would fix this constant for but then we would need to evaluate

and use it to reset the constant in the general solution for Our purpose here is to illustrate a new approach using Laplace transforms. As we will see,

this method offers several advantages over the previous techniques. For one thing, it is much more convenient in solving initial value problems for linear constant coefficient equations when the forcing term contains jump discontinuities.

The Laplace transform of a function defined on is given by†

(4)

Thus we multiply by and integrate with respect to t from 0 to This takes a function of t and produces a function of s.

In this chapter we’ll scrutinize many of the details on this “exchange of functions,” but for now let’s simply state the main advantage of executing the transform. The Laplace transform replaces linear constant coefficient differential equations in the t-domain by (simpler) alge- braic equations in the s-domain! In particular, if is the Laplace transform of then the transform of is simply Therefore, the information in the differential equa- tion (2) and initial condition (3) is transformed from the t-domain to the s-domain simply as

t-Domain s-Domain

(5)

where is the Laplace transform of (Notice that we have taken certain linearity prop- erties for granted, such as the fact that the transform preserves sums and multiplications by constants.) We can find in the s-domain without solving any differential equations: the solution is simply

(6)

For this procedure to be useful, there has to be an easy way to convert from the t-domain to the s-domain and vice versa. There are, in fact, tables and theorems that facilitate this conver- sion in many useful circumstances. We’ll see, for instance, that the transform of g(t), despite its unpleasant piecewise specification in equation (1), is given by the single formula

and as a consequence the transform of equals

X AsB # 30 s " 3/500

" 2.4

s As " 3/500B $ 1.2e$10ss As " 3/500B . x AtBG AsB #

2.4 s

$ 1.2 s

e$10s ,

X AsB # 30 s " 3/500

" G AsB

s " 3/500 .

X AsB g AtB.G AsBx¿

AtB " 3 500

x AtB # g AtB , x A0B # 30 ; sX AsB $ 30 " 3 500

X AsB # G AsB , sX AsB $ x A0B.x¿ AtB x AtB,X AsB

q.e$stf AtB F AsB J !q

0 e$stf AtB dt . 30, q B,f AtB,

t 7 10.x A10B 0 6 t 6 10, g AtB A10, q B.A0, q B

x A0B # 30 . Section 7.1 Introduction: A Mixing Problem 351

†Historical Footnote: The Laplace transform was first introduced by Pierre Laplace in 1779 in his research on proba- bility. G. Doetsch helped develop the use of Laplace transforms to solve differential equations. His work in the 1930s served to justify the operational calculus procedures earlier used by Oliver Heaviside.

Again by table lookup (and a little theory), we can deduce that

(7)

See Figure 7.2.

x AtB # 400 $ 370e$3t/ 500 $ 200 # e0 , t & 10 ,31 $ e$3At$10B/500 4 , t ' 10 . 352 Chapter 7 Laplace Transforms

x(t)

500 0

100

150

200

100 150 200 250

Figure 7.2 Solution to mixing tank example

Note that to arrive at (7) we did not have to take derivatives of trial solutions, break up intervals, or evaluate constants through initial data. The Laplace transform machinery replaces all of these operations by basic algebra: addition, subtraction, multiplication, division—and, of course, the judicious use of the table. Figure 7.3 depicts the advantages of the transform method.

Unfortunately, the Laplace transform method is less helpful with equations containing variable coefficients or nonlinear equations (and sometimes determining inverse transforms can be a Herculean task!). But it is ideally suited for many problems arising in applications. Thus, we devote the present chapter to this important topic.

Differential equation

Inverse transform

Laplace transform

Break into subintervals

Trial solutions

Solution

Algebra: +, −, × , ÷

, ∫ dtCalculus:

Fit constants to initial data

t-domain

s-domain

d dt

Figure 7.3 Comparison of solution methods

In earlier chapters we studied differential operators. These operators took a function and mapped or transformed it (via differentiation) into another function. The Laplace transform, which is an integral operator, is another such transformation.

Section 7.2 Definition of the Laplace Transform 353

7.2 DEFINITION OF THE LAPLACE TRANSFORM

Laplace Transform

Definition 1. Let be a function on The Laplace transform of f is the function F defined by the integral

(1)

The domain of is all the values of s for which the integral in (1) exists.† The Laplace transform of f is denoted by both F and !E f F.F AsB

F AsB :! !" 0

e$stf AtB dt . 30, q B.f AtB

†We treat s as real-valued, but in certain applications s may be a complex variable. For a detailed treatment of complex- valued Laplace transforms, see Complex Variables and the Laplace Transform for Engineers, by Wilbur R. LePage (Dover Publications, New York, 1980), or Fundamentals of Complex Analysis with Applications to Engineering and Science (3rd ed.), by E. B. Saff and A. D. Snider (Prentice Hall, Englewood Cliffs, N.J., 2003).

Notice that the integral in (1) is an improper integral. More precisely,

whenever the limit exists.

Determine the Laplace transform of the constant function

Using the definition of the transform, we compute

Since when is fixed and we get

for

When the integral diverges. (Why?) Hence with the domain of being all s % 0. ◆

F AsBF AsB # 1/s,"q0 e$st dts & 0, s 7 0 .F AsB # 1

N S q,s 7 0e$sN S 0

# lim NSq

$e$st

s ` t#Nt#0 # limNSq c 1s $ e$sNs d . F AsB # !q

0 e$st # 1 dt # lim

NSq ! N

0 e$st dt

f AtB # 1, t ' 0 . !

0 e$stf AtB dt J lim

NSq ! N

0 e$stf AtB dt

Example 1

Solution

Determine the Laplace transform of where a is a constant.

Using the definition of the transform,

Again, if the integral diverges, and hence the domain of is all ◆

It is comforting to note from Example 2 that the transform of the constant function is which agrees with the solution in Example 1.

Find where b is a nonzero constant.

We need to compute

Referring to the table of integrals on the inside front cover, we see that

(since for such s we have see Problem 32). ◆

Determine the Laplace transform of

f AtB # #2 , 0 6 t 6 5 ,0 , 5 6 t 6 10 , e4t , 10 6 t .

limNSq e $sN As sin bN " b cos bNB # 0;

# b

s2 " b2 for s 7 0

# lim NSq c b s2 " b2

$ e$sN

s2 " b2 As sin bN " b cos bNB d

!Esin btF AsB # lim NSq c e$sts2 " b2 A$s sin bt $ b cos btB ` N0 d

!Esin btF AsB # !q 0

e$st sin bt dt # lim NSq !

0 e$st sin bt dt .

!Esin btF, 1/ As $ 0B # 1/s,f AtB # 1 # e0t

s 7 a.F AsBs & a #

1 s $ a

for s 7 a .

# lim NSq c 1 s $ a

$ e$As$aBN s $ a

d # lim

NSq ! N

0 e$As$aBt dt # lim

NSq

$e$As$aBt s $ a ` N0

F AsB # !q 0

e$steat dt # ! q

0 e$As$aBt dt

f AtB # eat, 354 Chapter 7 Laplace Transforms

Example 2

Solution

Example 3

Example 4

Solution

Since is defined by a different formula on different intervals, we begin by breaking up the integral in (1) into three separate parts.† Thus,

◆

Notice that the function of Example 4 has jump discontinuities at t # 5 and t # 10. These values are reflected in the exponential terms and that appear in the formula for

We’ll make this connection more precise when we discuss the unit step function in Section 7.6.

An important property of the Laplace transform is its linearity. That is, the Laplace transform is a linear operator.!

F AsB. e$10se$5sf AtB#

2 s

$ 2e$5s

s "

e$10As$4B s $ 4

for s 7 4 .

# 2 s

$ 2e$5s

s " lim

NSq c e$10As$4B

s $ 4 $

e$As$4BN s $ 4

d # 2!

0 e$st dt " lim

NSq ! N

10 e$As$4Bt dt

# ! 5

0 e$st # 2 dt " !

5 e$st # 0 dt " !

10 e$ste4t dt

F AsB # !q 0

e$stf AtB dt f AtB

Section 7.2 Definition of the Laplace Transform 355

Solution

†Notice that is not defined at the points t # 0, 5, and 10. Nevertheless, the integral in (1) is still meaningful and unaffected by the function’s values at finitely many points.

f AtB

Linearity of the Transform

Theorem 1. Let f, f1, and f2 be functions whose Laplace transforms exist for and let c be a constant. Then, for

(2)

(3) !Ecf F # c!E f F .!E f1 " f2F # !E f1F " !E f2F , s 7 a,

s 7 a

Proof. Using the linearity properties of integration, we have for

Hence, equation (2) is satisfied. In a similar fashion, we see that

◆# c!E f F AsB .!Ecf F AsB # ! q

0 e$st 3 cf AtB 4dt # c !q

0 e$stf AtB dt

# !E f1F AsB " !E f2F AsB .# ! q

0 e$stf1 AtB dt " !q

0 e$stf2 AtB dt

!E f1 " f2F AsB # !q 0

e$st 3 f1 AtB " f2 AtB 4dt s 7 a

Determine

From the linearity property, we know that the Laplace transform of the sum of any finite num- ber of functions is the sum of their Laplace transforms. Thus,

In Examples 1, 2, and 3, we determined that

Using these results, we find

Since and are all defined for s % 4, so is the transform ◆

Existence of the Transform There are functions for which the improper integral in (1) fails to converge for any value of s. For example, this is the case for the function which grows too fast near zero. Like- wise, no Laplace transform exists for the function which increases too rapidly as

Fortunately, the set of functions for which the Laplace transform is defined includes many of the functions that arise in applications involving linear differential equations. We now discuss some properties that will (collectively) ensure the existence of the Laplace transform.

A function on is said to have a jump discontinuity at if is discontinuous at t0, but the one-sided limits

and exist as finite numbers. If the discontinuity occurs at an endpoint, t0 # a (or b), a jump discon- tinuity occurs if the one-sided limit of as exists as a finite number. We can now define piecewise continuity.

t S a" At S b$Bf AtB lim

tSt"0 f AtBlim

tSt$0 f AtB

f AtBt0 # Aa, bB3a, b 4f AtB t S q.

f AtB # et2,f AtB # 1/t,

!E11 " 5e4t $ 6 sin 2tF. !Esin 2tF!E1F, !Ee4tF, #

11 s

" 5

s $ 4 $

12 s2 " 4

!E11 " 5e4t $ 6 sin 2tF AsB # 11 a1 s b " 5 a 1

s $ 4 b $ 6 a 2

s2 " 4 b

!E1F AsB # 1 s , !Ee4tF AsB # 1

s $ 4 , !Esin 2tF AsB # 2

s2 " 22 .

# 11!E1F " 5!Ee4tF $ 6!Esin 2tF .!E11 " 5e4t $ 6 sin 2tF # !E11F " !E5e4tF " !E$6 sin 2tF !E11 " 5e4t $ 6 sin 2tF .

356 Chapter 7 Laplace Transforms

Example 5

Solution

Piecewise Continuity

Definition 2. A function is said to be piecewise continuous on a finite interval if is continuous at every point in , except possibly for a finite number of

points at which has a jump discontinuity. A function is said to be piecewise continuous on if is piecewise

continuous on for all N % 0.30, N 4 f AtB30, "Bf AtBf AtB 3a, b 4f AtB3a, b 4 f AtB

Show that

whose graph is sketched in Figure 7.4 is piecewise continuous on .

From the graph of we see that is continuous on the intervals (0, 1), (1, 2), and (2, 3]. Moreover, at the points of discontinuity, t # 0, 1, and 2, the function has jump discontinuities, since the one-sided limits exist as finite numbers. In particular, at t # 1, the left-hand limit is 1 and the right-hand limit is 2. Therefore is piecewise continuous on . ◆

Observe that the function of Example 4 is piecewise continuous on because it is piecewise continuous on every finite interval of the form , with N % 0. In contrast, the function is not piecewise continuous on any interval containing the origin, since it has an “infinite jump” at the origin (see Figure 7.5).

f AtB # 1/t 30, N 4 30, q Bf AtB 30, 3 4f AtB

f AtBf AtB 30, 3 4 f AtB # # t , 0 6 t 6 1 ,2 , 1 6 t 6 2 ,At $ 2B2 , 2 & t & 3 ,

Section 7.2 Definition of the Laplace Transform 357

Solution

Example 6

f (t)

0 1 2 3

Figure 7.4 Graph of in Example 6f AtB

−10

−5

−10 −5 5 10 t

f(t) = 1/ t

1–t = + (

1–t = −(

lim

t → 0+

t → 0−

Figure 7.5 Infinite jump at origin

A function that is piecewise continuous on a finite interval is necessarily integrable over that interval. However, piecewise continuity on is not enough to guarantee the existence (as a finite number) of the improper integral over we also need to consider the growth of the integrand for large t. Roughly speaking, we’ll show that the Laplace transform of a piecewise continuous function exists, provided the function does not grow “faster than an exponential.”

30, q B;30, q B 358 Chapter 7 Laplace Transforms

For example, is of exponential order since

and hence (4) holds with M # 1 and T any positive constant. We use the phrase is of exponential order to mean that for some value of the func-

tion satisfies the conditions of Definition 3; that is, grows no faster than a function of the form The function is not of exponential order. To see this, observe that

for any Consequently, grows faster than for every choice of The functions usually encountered in solving linear differential equations with constant

coefficients (e.g., polynomials, exponentials, sines, and cosines) are both piecewise continu- ous and of exponential order. As we now show, the Laplace transforms of such functions exist for large enough values of s.

a.eate t 2

lim tSq

et 2

eat # lim

tSq et At$aB # "q

et 2

Meat. f AtBf AtB a,f AtB

0 e5t sin 2t 0 & e5t , a # 5f AtB # e5t sin 2t

Exponential Order a

Definition 3. A function is said to be of exponential order A if there exist positive constants T and M such that (4) for all t ' T .0 f AtB 0 & Meat , f AtB

Conditions for Existence of the Transform

Theorem 2. If is piecewise continuous on and of exponential order then exists for s 7 a.!E f F AsB a,30, q Bf AtB

Proof. We need to show that the integral

converges for We begin by breaking up this integral into two separate integrals:

(5)

where T is chosen so that inequality (4) holds. The first integral in (5) exists because and hence are piecewise continuous on the interval for any fixed s. To see that the second integral in (5) converges, we use the comparison test for improper integrals.

30, T 4e$stf AtB f AtB ! T

0 e$stf AtB dt " !q

T e$stf AtB dt ,

s 7 a.

! q

0 e$stf AtB dt

Since is of exponential order we have for

and hence

for all Now for

Since for and the improper integral of the larger function converges for then, by the comparison test, the integral

converges for Finally, because the two integrals in (5) exist, the Laplace transform exists for ◆

Table 7.1 lists the Laplace transforms of some of the elementary functions. You should become familiar with these, since they are frequently encountered in solving linear differential equations with constant coefficients. The entries in the table can be derived from the definition of the Laplace transform. A more elaborate table of transforms is given on the inside back cover of this book.

s 7 a.!E f F AsB s 7 a. !

T e$stf AtB dts 7 a,

t ' T0 e$stf AtB 0 & Me$As$aBt! q

T Me$As$aB t dt # M!q

T e$As$aB t dt # Me$As$aBT

s $ a 6 q .

s 7 a.t ' T.

0 e$stf AtB 0 # e$st 0 f AtB 0 & Me$As$aBt , 0 f AtB 0 & Meat , t ' Ta,f AtB

Section 7.2 Definition of the Laplace Transform 359

TABLE 7.1 Brief Table of Laplace Transforms

sin bt

cos bt

s $ aAs $ aB2 " b2 , s 7 aeat cos bt bAs $ aB2 " b2 , s 7 aeat sin bt

n!As $ aBn"1 , s 7 aeattn , n # 1, 2, . . . s

s2 " b2 , s 7 0

b s2 " b2

, s 7 0

n! sn"1

, s 7 0tn , n # 1, 2, . . .

1 s $ a

, s 7 aeat

1 s , s 7 0

F AsB ! !E f F AsBf AtB

In Problems 1–12, use Definition 1 to determine the Laplace transform of the given function.

1. 2. 3. 4. 5. 6. cos bt, b a constant 7. 8.

10.

11.

12.

In Problems 13–20, use the Laplace transform table and the linearity of the Laplace transform to determine the following transforms.

13. 14. 15. 16. 17. 18. 19. 20.

In Problems 21–28, determine whether is continu- ous, piecewise continuous, or neither on and sketch the graph of

21.

22.

23.

24.

25. f AtB # t2 $ t $ 20 t2 " 7t " 10

f AtB # t2 $ 3t " 2 t2 $ 4

f AtB # #1 , 0 & t 6 1 ,t $ 1 , 1 6 t 6 3 , t2 $ 4 , 3 6 t & 10

f AtB # e0 , 0 & t 6 2 , t , 2 & t & 10

f AtB # e 1 , 0 & t & 1 ,At $ 2B2 , 1 6 t & 10 f AtB. 3 0, 10 4f AtB

!Ee$2t cos23t $ t2e$2tF!Et4e5t $ et cos27tF!Et4 $ t2 $ t " sin22 tF !Ee3t sin 6t $ t3 " etF!Et2 $ 3t $ 2e$t sin 3tF !Et3 $ tet " e4t cos tF!E5 $ e2t " 6t2F !E6e$3t $ t2 " 2t $ 8F

f AtB # e e2t , 0 6 t 6 3 , 1 , 3 6 t

f AtB # e sin t , 0 6 t 6 p , 0 , p 6 t

f AtB # e1 $ t , 0 6 t 6 1 , 0 , 1 6 t

f AtB # e0 , 0 6 t 6 2 , t , 2 6 t

e$t sin 2te2t cos 3t cos 2t

te3te6t t2t

360 Chapter 7 Laplace Transforms

26.

27.

28.

29. Which of the following functions are of exponential order? (a) (b) (c) (d)

(e) (f)

(g) (h) (i) ( j)

30. For the transforms in Table 7.1, what can be said about lim ?

31. Thanks to Euler’s formula (page 168) and the alge- braic properties of complex numbers, several of the entries of Table 7.1 can be derived from a single formula; namely,

(6)

(a) By computing the integral in the definition of the Laplace transform on page 353 with #

show that

(b) Deduce (6) from part (a) by showing that

32. Prove that for fixed s % 0, we have

33. Prove that if f is piecewise continuous on and g is continuous on , then the product fg is piecewise continuous on .3 a, b 43 a, b 4 3 a, b 4

lim NSq

e$sN As sin bN " b cos bNB # 0 .

1 s $ Aa " ibB # s $ a " ibAs $ aB2 " b2 . ! U e Aa"ibBt V AsB # 1

s $ Aa " ibB , s 7 a. e Aa"ibBt, f AtB ! U e Aa" ibBt V AsB # s $ a " ibAs $ aB2 " b2 , s 7 a.

sSq F AsBF AsB sin Aet2B " esin texpEt2/ At " 1B F 3 $ et2 " cos 4tsin At2B " t4e6t

1 t2 " 1

cosh At2B t ln tet 3

100e49tt3 sin t

f AtB # # sin tt , t ) 0 ,1 , t # 0

f AtB # #1/t , 0 6 t 6 1 ,1 , 1 & t & 2 , 1 $ t , 2 6 t & 10

f AtB # t t2 $ 1

7.2 EXERCISES

In the previous section, we defined the Laplace transform of a function as

Using this definition to get an explicit expression for requires the evaluation of the improper integral—frequently a tedious task! We have already seen how the linearity property of the transform can help relieve this burden. In this section we discuss some further properties of the Laplace transform that simplify its computation. These new properties will also enable us to use the Laplace transform to solve initial value problems.

!E f F!E f F AsB J ! q

0 e$stf AtB dt .

f AtB

Section 7.3 Properties of the Laplace Transform 361

7.3 PROPERTIES OF THE LAPLACE TRANSFORM

Translation in s

Theorem 3. If the Laplace transform exists for then

(1)

for s 7 a " a .

!Eeatf AtB F AsB ! F As $ aB s 7 a,!E f F AsB # F AsB

Laplace Transform of the Derivative

Theorem 4. Let be continuous on and be piecewise continuous on , with both of exponential order Then, for

(2) !E f$F AsB ! s!E f F AsB % f A0B . s 7 a ,a.30, q B f ¿AtB30, q Bf AtB

Proof. We simply compute

◆

Theorem 3 illustrates the effect on the Laplace transform of multiplication of a function by

Determine the Laplace transform of

In Example 3 in Section 7.2, we found that

Thus, by the translation property of we have

◆!Eeat sin btF AsB # F As $ aB # bAs $ aB2 " b2 . F AsB,

!Esin btF AsB # F AsB # b s2 " b2

eat sin bt.

eat .f AtB # F As $ aB . # !

0 e$As$aBtf AtB dt

!Eeatf AtB F AsB # !q 0

e$steatf AtB dt

Example 1

Solution

Proof. Since exists, we can use integration by parts with and # to obtain

(3)

To evaluate lim we observe that since is of exponential order there exists a constant M such that for N large,

Hence, for

for Equation (3) now reduces to

◆

Using induction, we can extend the last theorem to higher-order derivatives of For example,

which simplifies to

In general, we obtain the following result.

!E f &F AsB ! s2!E f F AsB $ sf A0B $ f$ A0B . # s 3 s!E f F AsB $ f A0B 4 $ f ¿A0B ,!E f –F AsB # s!E f ¿F AsB $ f ¿A0B

f AtB. !E f ¿F AsB # s!E f F AsB $ f A0B .s 7 a. lim

NSq e$sNf ANB # 0

0 & lim NSq 0 e$sNf ANB 0 & lim

NSq Me$As$aBN # 0 ,

s 7 a,

0 e$sNf ANB 0 & e$sNMeaN # Me$As$aBN . a,f AtBNSq e$sNf ANB,

# lim NSq

e$sNf ANB $ f A0B " s!E f F AsB . # lim

NSq e$sNf ANB $ f A0B " s lim

NSq ! N

0 e$stf AtB dt

# lim NSq c e$stf AtB ` N

0 " s !

0 e$stf AtB dt d

!E f ¿F AsB # !q 0

e$stf ¿ AtB dt # lim NSq !

0 e$stf ¿ AtB dt

f ¿AtBdt 4 dyu # e$st3!E f ¿F 362 Chapter 7 Laplace Transforms

Laplace Transform of Higher-Order Derivatives

Theorem 5. Let be continuous on and let be piecewise continuous on with all these functions of exponential order Then, for

(4) ! U f AnB V AsB ! sn!E f F AsB % sn%1f A0B % sn%2f$A0B % p % f An%1B A0B .s 7 a, a.30, q B, f AnB AtB30, q Bf AtB, f ¿AtB, . . . , f An$1B AtB

The last two theorems shed light on the reason why the Laplace transform is such a useful tool in solving initial value problems. Roughly speaking, they tell us that by using the Laplace

transform we can replace “differentiation with respect to t ” with “multiplication by s,” thereby converting a differential equation into an algebraic one. This idea is explored in Section 7.5. For now, we show how Theorem 4 can be helpful in computing a Laplace transform.

Using Theorem 4 and the fact that

determine

Let Then and Substituting into equation (2), we have

Dividing by b gives

◆

Prove the following identity for continuous functions (assuming the transforms exist):

(5)

Use it to verify the solution to Example 2.

Define the function by the integral

Observe that and Thus, if we apply Theorem 4 to instead of equation (2) on page 361 reads

which is equivalent to equation (5). Now since

equation (5) predicts

This identity is indeed valid for the transforms in Example 2. ◆

!Esin btF AsB # 1 s

!Eb cos btF AsB # b s

!Ecos btF AsB . sin bt # !

0 b cos bt dt ,

!E f AtB F AsB # s! e ! t 0 f AtB dt f AsB $ 0 ,

f AtB 4 ,3g AtBg¿ AtB # f AtB.g A0B # 0g AtB J ! t

0 f AtB dt . g AtB

! e ! t 0 f AtBdt f AsB # 1

s !E f AtB F AsB .

f AtB !Ecos btF AsB # s

s2 " b2 .

b!Ecos btF AsB # sb s2 " b2

!Eb cos btF AsB # s!Esin btF AsB $ 0 ,!E f ¿F AsB # s!E f F AsB $ f A0B , f ¿AtB # b cos bt.f A0B # 0f AtB J sin bt.!Ecos btF .

!Esin btF AsB # b s2 " b2

Section 7.3 Properties of the Laplace Transform 363

Example 2

Solution

Example 3

Solution

Proof. Consider the identity

Because of the assumptions on we can apply a theorem from advanced calculus (some- times called Leibniz’s rule) to interchange the order of integration and differentiation:

Thus,

The general result (6) now follows by induction on n. ◆

A consequence of the above theorem is that if is piecewise continuous and of exponential order, then its transform has derivatives of all orders.

Determine

We already know that

Differentiating we obtain

Hence, using formula (6), we have

◆!Et sin btF AsB # $ dF ds

AsB # 2bsAs2 " b2B2 . dF ds

AsB # $2bsAs2 " b2B2 . F AsB,

!Esin btF AsB # F AsB # b s2 " b2

!Et sin btF. F AsB f AtB

!Etf AtB F AsB # A$1B dF ds

AsB . # $!

0 e$sttf AtB dt # $!Etf AtB F AsB .

dF ds

AsB # !q 0

d ds Ae$stB f AtB dt

f AtB, dF ds

AsB # d ds

! q

0 e$stf AtB dt .

364 Chapter 7 Laplace Transforms

Derivatives of the Laplace Transform

Theorem 6. Let and assume is piecewise continuous on and of exponential order Then, for

(6) !Etnf AtB F AsB ! A%1Bn dnF dsn

AsB .s 7 a,a. 30, q Bf AtBF AsB # !E f F AsB

Example 4

Solution

Another question arises concerning the Laplace transform. If is the Laplace transform of is also a Laplace transform of some function of t ? The answer is yes:

In fact, the following more general assertion holds.

F¿ AsB # !E$t f AtB F AsB .F¿ AsBf AtB, F AsB

In Problems 1–20, determine the Laplace transform of the given function using Table 7.1 and the properties of the transform given in Table 7.2. [Hint: In Problems 12–20, use an appropriate trigonometric identity.]

1. 2. 3. 4. 5. 6. 7. 8. 9. 10.

11. cosh bt 12. sin 3t cos 3t 13. 14. 15. 16. 17. 18.

19. 20.

21. Given that use the translation property to compute

22. Starting with the transform use for- mula (6) for the derivatives of the Laplace transform to show that , , and, by using induction, that

.n # 1, 2, . . . !EtnF AsB # n!/sn"1,!Et2F AsB # 2!/s3!EtF AsB # 1/s2

!E1F AsB # 1 /s,!Eeat cos btF. !Ecos btF AsB # s/ As2 " b2B,m ) n

t sin 2t sin 5tcos nt sin mt , m ) n cos nt cos mt ,sin 2t sin 5t t sin2 tcos3 t e7t sin2 tsin2 t

te2t cos 5te$tt sin 2t

A1 " e$tB2At $ 1B4 e$2t sin 2t " e3tt22t2e$t $ t " cos 4t 3t4 $ 2t2 " 1e$t cos 3t " e6t $ 1 3t2 $ e2tt2 " et sin 2t

Section 7.3 Properties of the Laplace Transform 365

TABLE 7.2 Properties of Laplace Transforms

for any constant c .

!Etnf AtB F AsB # A$1Bn dn dsn

A!E f F AsBB .! U f AnBV AsB # sn!E f F AsB $ sn$1f A0B $ sn$2f ¿A0B $ p $ f An$1B A0B . !E f –F AsB # s2!E f F AsB $ sf A0B $ f ¿ A0B .!E f ¿F AsB # s!E f F AsB $ f A0B . !Eeatf AtB F AsB # !E f F As $ aB .!Ecf F # c!E f F !E f " gF # !E f F " !EgF .

For easy reference, Table 7.2 lists some of the basic properties of the Laplace transform derived so far.

23. Use Theorem 4 to show how entry 32 follows from entry 31 in the Laplace transform table on the inside back cover of the text.

24. Show that in two ways: (a) Use the translation property for (b) Use formula (6) for the derivatives of the

Laplace transform. 25. Use formula (6) to help determine

(a) (b)

26. Let be piecewise continuous on and of exponential order. (a) Show that there exist constants K and such that

for all (b) By using the definition of the transform and

estimating the integral with the help of part (a), prove that

27. Let be piecewise continuous on and of exponential order and assume lim exists. Show that

! e f AtB t f AsB # ! q

s F AuB du ,

tS0" 3 f AtB / t 4a 3 0, q Bf AtB lim sSq

!E f F AsB # 0 . t ' 0 .0 f AtB 0 & Keat a

3 0, q Bf AtB !Et2cos btF .!Et cos btF . F AsB.!EeattnF AsB # n! / As $ aBn"1

7.3 EXERCISES

where [Hint: First show that and then use the result of

Problem 26.] 28. Verify the identity in Problem 27 for the following

functions. (Use the table of Laplace transforms on the inside back cover.) (a) (b)

29. The transfer function of a linear system is defined as the ratio of the Laplace transform of the output function to the Laplace transform of the input function when all initial conditions are zero. If a linear system is governed by the differential equation

use the linearity property of the Laplace transform and Theorem 5 on the Laplace transform of higher- order derivatives to determine the transfer function

for this system. 30. Find the transfer function, as defined in Problem 29,

for the linear system governed by

31. Translation in t. Show that for c % 0, the trans- lated function

has Laplace transform !EgF AsB # e$cs!E f F AsB . g AtB # e 0 , 0 6 t 6 c ,

f At $ cB , c 6 t y– AtB " 5y¿ AtB " 6y AtB # g AtB , t 7 0 .

H AsB # Y AsB /G AsB y– AtB " 6y¿ AtB " 10y AtB # g AtB , t 7 0 ,

g AtB,y AtB f AtB # t3/ 2f AtB # t5

d ds !E f AtB / tF AsB # $F AsBF AsB # !E f F AsB.

366 Chapter 7 Laplace Transforms

In Problems 32–35, let be the given function translated to the right by c units. Sketch and and find (See Problem 31.) 32. 33. 34. 35.

36. Use equation (5) to provide another derivation of the formula # [Hint: Start with

and use induction.]

37. Initial Value Theorem. Apply the relation

(7)

to argue that for any function whose derivative is piecewise continuous and of exponential order on

38. Verify the initial value theorem (Problem 37) for the following functions. (Use the table of Laplace trans- forms on the inside back cover.) (a) 1 (b) (c) (d) cos t (e) sin t (f) (g) t cos tt2

e$te t

f A0B # lim sSq

s!E f F AsB .3 0, q B, f AtB

!E f ¿F AsB # ! q 0

e$stf ¿AtB dt # s!E f F AsB $ f A0B !E1F AsB # 1/s n! /sn"1.!EtnF AsB

c # p/2f AtB # sin t , c # pf AtB # sin t , c # 1f AtB # t , c # 2f AtB $ 1 ,

!Eg AtB F AsB. g AtBf AtB f AtBg AtB

In Section 7.2 we defined the Laplace transform as an integral operator that maps a function into a function In this section we consider the problem of finding the function

when we are given the transform That is, we seek an inverse mapping for the Laplace transform.

To see the usefulness of such an inverse, let’s consider the simple initial value problem

(1)

If we take the transform of both sides of equation (1) and use the linearity property of the transform, we find

!Ey–F AsB $ Y AsB # $ 1 s2

y– $ y # $t ; y A0B # 0 , y¿ A0B # 1 . F AsB. f AtBF AsB.f AtB

7.4 INVERSE LAPLACE TRANSFORM

where We know the initial values of the solution so we can use Theorem 5, page 362, on the Laplace transform of higher-order derivatives to express

Substituting for yields

Solving this algebraic equation for gives

We now recall that and since we have

It therefore seems reasonable to conclude that is the solution to the initial value prob- lem (1). A quick check confirms this!

Notice that in the above procedure, a crucial step is to determine from its Laplace transform As we noted, is such a function, but it is not the only function whose Laplace function is For example, the transform of

is also This is because the transform is an integral, and integrals are not affected by chang- ing a function’s values at isolated points. The significant difference between and as far as we are concerned is that is continuous on whereas is not. Naturally, we pre- fer to work with continuous functions, since solutions to differential equations are continuous. Fortunately, it can be shown that if two different functions have the same Laplace transform, at most one of them can be continuous.† With this in mind we give the following definition.

g AtB30, q B,y AtB g AtBy AtB1/s2. g AtB J e t , t ) 6 ,

0 , t # 6

1/s2. y AtB # tY AsB # 1/s2. y AtB

y AtB # t!EyF AsB # 1/s2 # !EtF AsB . Y AsB # !EyF AsB,!EtF AsB # 1/s2,

Y AsB # 1 $ a 1s2b s2 $ 1

# s2 $ 1

s2 As2 $ 1B # 1s2 . Y AsB

s2Y AsB $ 1 $ Y AsB # $ 1 s2

!Ey–F AsB!Ey–F AsB # s2Y AsB $ sy A0B $ y¿ A0B # s2Y AsB $ 1 . y AtB,Y AsB J !EyF AsB.

Section 7.4 Inverse Laplace Transform 367

†For this result and further properties of the Laplace transform and its inverse, we refer you to Operational Mathematics, 3rd ed., by R. V. Churchill (McGraw-Hill, New York, 1972).

Inverse Laplace Transform

Definition 4. Given a function if there is a function that is continuous on and satisfies

(2)

then we say that is the inverse Laplace transform of and employ the notation f # !$1EFF. F AsBf AtB

!E f F # F ,30, q B f AtBF AsB,

In case every function satisfying (2) is discontinuous (and hence not a solution of a differential equation), one could choose any one of them to be the inverse transform; the distinction among them has no physical significance. [Indeed, two piecewise continuous func- tions satisfying (2) can only differ at their points of discontinuity.]

f AtB

Naturally the Laplace transform tables will be a great help in determining the inverse Laplace transform of a given function

Determine where

(a) (b) (c)

To compute we refer to the Laplace transform table on page 359.

(a)

(b)

(c)

In part (c) we used the technique of completing the square to rewrite the denominator in a form that we could find in the table. ◆

In practice, we do not always encounter a transform that exactly corresponds to an entry in the second column of the Laplace transform table. To handle more complicated functions we use properties of just as we used properties of One such tool is the linearity of the inverse Laplace transform, a property that is inherited from the linearity of the operator !.

!.!$1, F AsB,F AsB

!$1e s $ 1 s2 $ 2s " 5

f AtB # !$1e s $ 1As $ 1B2 " 22 f AtB # etcos 2t !$1e 3

s2 " 9 f AtB # !$1e 3

s2 " 32 f AtB # sin 3t

!$1e 2 s3 f AtB # !$1e 2!

s3 f AtB # t2!

$1EFF, F AsB # s $ 1

s2 $ 2s " 5 .F AsB # 3

s2 " 9 .F AsB # 2

s3 .

!$1EFF, F AsB. 368 Chapter 7 Laplace Transforms

Example 1

Solution

Linearity of the Inverse Transform

Theorem 7. Assume that and exist and are continuous on and let c be any constant. Then

(3) (4) !$1EcFF # c!$1EFF .!$1EF1 " F2F # !$1EF1F " !$1EF2F , 30, q B !$1EF2F!$1EFF, !$1EF1F,

The proof of Theorem 7 is outlined in Problem 37. We illustrate the usefulness of this theorem in the next example.

Determine

We begin by using the linearity property. Thus,

Referring to the Laplace transform tables, we see that

and !$1 e s s2 " 32

f AtB # cos 3t .!$1 e 1 s $ 6

f AtB # e6t # 5!$1 e 1

s $ 6 f $ 6!$1 e s

s2 " 9 f " 3

2 !$1 e 1

s2 " 4s " 5 f .

!$1 e 5 s $ 6

$ 6s

s2 " 9 "

3 2 As2 " 4s " 5B f

!$1 e 5 s $ 6

$ 6s

s2 " 9 "

3 2s2 " 8s " 10

f .Example 2 Solution

This gives us the first two terms. To determine we complete the square of the denominator to obtain We now recognize from the tables that

Hence,

◆

Determine

The in the denominator suggests that we work with the formula

Here we have and so Using the linearity prop- erty, we find

◆

Determine

By completing the square, the quadratic in the denominator can be written as

The form of now suggests that we use one or both of the formulas

In this case, and The next step is to express

(5)

where A, B are constants to be determined. Multiplying both sides of (5) by leaves

which is an identity between two polynomials in s. Equating the coefficients of like terms gives

A # 3 , A " 3B # 2 ,

3s " 2 # A As " 1B " 3B # As " AA " 3BB , s2 " 2s " 10

3s " 2 s2 " 2s " 10

# A s " 1As " 1B2 " 32 " B 3As " 1B2 " 32 ,

b # 3.a # $1

!$1 e bAs $ aB2 " b2 f AtB # eat sin bt . !$1 e s $ aAs $ aB2 " b2 f AtB # eat cos bt ,

F AsBs2 " 2s " 10 # s2 " 2s " 1 " 9 # As " 1B2 " 32 . !$1 e 3s " 2

s2 " 2s " 10 f .

!$1 e 5As " 2B4 f AtB # 56 !$1 e 3!As " 2B4 f AtB # 56 e$2tt3 . !$1E6/ As " 2B4F AtB # e$2tt3.n # 3,a # $2!

$1 e n!As $ aBn"1 f AtB # eattn . As " 2B4

!$1 e 5As " 2B4 f . !$1 e 5

s $ 6 $

6s s2 " 9

" 3

2s2 " 8s " 10 f AtB # 5e6t $ 6 cos 3t " 3e$2t

2 sin t .

!$1 e 1As " 2B2 " 12 f AtB # e$2t sin t . s2 " 4s " 5 # As " 2B2 " 1.!$1E1/ As2 " 4s " 5B F,

Section 7.4 Inverse Laplace Transform 369

Example 3

Solution

Example 4

Solution

so and Finally, from (5) and the linearity property, we find

◆

Given the choice of finding the inverse Laplace transform of

or of

which would you select? No doubt is the easier one. Actually, the two functions and are identical. This can be checked by combining the simple fractions that form

Thus, if we are faced with the problem of computing of a rational function such as we will first express it, as we did as a sum of simple rational functions. This is accom- plished by the method of partial fractions.

We briefly review this method. Recall from calculus that a rational function of the form where and are polynomials with the degree of P less than the degree of Q,

has a partial fraction expansion whose form is based on the linear and quadratic factors of (We assume the coefficients of the polynomials to be real numbers.) There are three cases

to consider: 1. Nonrepeated linear factors. 2. Repeated linear factors. 3. Quadratic factors.

1. Nonrepeated Linear Factors If can be factored into a product of distinct linear factors,

where the ’s are all distinct real numbers, then the partial fraction expansion has the form

where the ’s are real numbers. There are various ways of determining the constants In the next example, we demonstrate two such methods.

Determine where

We begin by finding the partial fraction expansion for The denominator consists of three distinct linear factors, so the expansion has the form

(6)

where A, B, and C are real numbers to be determined.

7s $ 1As " 1B As " 2B As $ 3B # As " 1 " Bs " 2 " Cs $ 3 , F AsB.F AsB #

7s $ 1As " 1B As " 2B As $ 3B . !$1EFF,A1, . . . , An. Ai

P AsB Q AsB # A1s % r1 ' A2s % r2 ' p ' Ans % rn ,

Q AsB # As $ r1B As $ r2B p As $ rnB ,Q AsB

Q AsB. Q AsBP AsBP AsB /Q AsB,

F2 AsB, F1 AsB,!$1 F2 AsB.F2 AsB F1 AsBF2 AsB

F2 AsB # 2s $ 1 " 1s " 1 " 4s " 3 , F1 AsB # 7s2 " 10s $ 1s3 " 3s2 $ s $ 3

# 3e$t cos 3t $ 1 3

e$t sin 3t .

!$1 e 3s " 2 s2 " 2s " 10

f AtB # 3!$1 e s " 1As " 1B2 " 32 f AtB $ 13 !$1 e 3As " 1B2 " 32 f AtB B # $1/3.A # 3

370 Chapter 7 Laplace Transforms

Example 5

Solution

One procedure that works for all partial fraction expansions is first to multiply the expansion equation by the denominator of the given rational function. This leaves us with two identical polynomials. Equating the coefficients of leads to a system of linear equations that we can solve to determine the unknown constants. In this example, we multiply (6) by and find

(7) †

which reduces to

Equating the coefficients of , s, and 1 gives the system of linear equations

Solving this system yields , , and . Hence,

(8)

An alternative method for finding the constants A, B, and C from (7) is to choose three values for s and substitute them into (7) to obtain three linear equations in the three unknowns. If we are careful in our choice of the values for s, the system is easy to solve. In this case, equa- tion (7) obviously simplifies if , , or 3. Putting gives

Hence . Next, setting gives

and so . Finally, letting , we similarly find that . In the case of nonrepeated linear factors, the alternative method is easier to use.

Now that we have obtained the partial fraction expansion (8), we use linearity to compute

◆ # 2e$t $ 3e$2t " e3t .

# " !$1 e 1 s $ 3

f AtB # 2!$1 e 1

s " 1 f AtB $ 3!$1 e 1

s " 2 f AtB

!$1 e 7s $ 1As " 1B As " 2B As $ 3B f AtB # !$1 e 2s " 1 $ 3s " 2 " 1s $ 3 f AtB C # 1s # 3B # $3

$15 # 5B ,

$14 $ 1 # A A0B " B A$1B A$5B " C A0B ,s # $2A # 2 $8 # $4A .

$7 $ 1 # A A1B A$4B " B A0B " C A0B , s # $1$2s # $1

7s $ 1As " 1B As " 2B As $ 3B # 2s " 1 $ 3s " 2 " 1s $ 3 . C # 1B # $3A # 2

$6A $ 3B " 2C # $1 .

$A $ 2B " 3C # 7 ,

A " B " C # 0 ,

s2 7s $ 1 # AA " B " CBs2 " A$A $ 2B " 3CBs " A$6A $ 3B " 2CB . 7s $ 1 # A As " 2B As $ 3B " B As " 1B As $ 3B " C As " 1B As " 2B ,As " 1B As " 2B As $ 3B

Section 7.4 Inverse Laplace Transform 371

†Rigorously speaking, equation (7) was derived for s different from and 3, but by continuity it holds for these values as well.

$1, $2,

2. Repeated Linear Factors Let be a factor of and suppose is the highest power of that divides

Then the portion of the partial fraction expansion of that corresponds to the term is

where the ’s are real numbers.

Determine

Since is a repeated linear factor with multiplicity two and is a nonrepeated linear factor, the partial fraction expansion has the form

We begin by multiplying both sides by to obtain

(9)

Now observe that when we set (or ), two terms on the right-hand side of (9) vanish, leaving a linear equation that we can solve for B (or C). Setting in (9) gives

and, hence, Similarly, setting in (9) gives

Thus, Finally, to find A, we pick a different value for s, say Then, since and plugging into (9) yields

so that Hence,

(10)

We could also have determined the constants A, B, and C by first rewriting equation (9) in the form

Then, equating the corresponding coefficients of and 1 and solving the resulting system, we again find and C # $1 .A # 2, B # 3,

s2, s,

s2 " 9s " 2 # AA " CBs2 " A2A " B $ 2CBs " A$3A " 3B " CB .

s2 " 9s " 2As $ 1B2 As " 3B # 2s $ 1 " 3As $ 1B2 $ 1s " 3 . A # 2.

2 # $3A " 3B " C # $3A " 9 $ 1

s # 0C # $1, B # 3s # 0.C # $1.

$16 # 16C .

9 $ 27 " 2 # A A0B " B A0B " 16Cs # $3B # 3. 12 # 4B ,

1 " 9 " 2 # A A0B " 4B " C A0B , s # 1 s # $3s # 1

s2 " 9s " 2 # A As $ 1B As " 3B " B As " 3B " C As $ 1B2 .As $ 1B 2 As " 3B

s2 " 9s " 2As $ 1B2 As " 3B # As $ 1 " BAs $ 1B2 " Cs " 3 . s " 3s $ 1

!$1 e s2 " 9s " 2As $ 1B2 As " 3B f . Ai

A1 s % r

' A2As % rB2 ' p ' AmAs % rBm ,

As $ rBm P AsB /Q AsBQ AsB. s $ r As $ rBmQ AsBs $ r

372 Chapter 7 Laplace Transforms

Example 6

Solution

Now that we have derived the partial fraction expansion (10) for the given rational func- tion, we can determine its inverse Laplace transform:

◆

3. Quadratic Factors Let be a quadratic factor of that cannot be reduced to linear factors with real coefficients. Suppose m is the highest power of that divides Then the portion of the partial fraction expansion that corresponds to is

As we saw in Example 4, it is more convenient to express in the form when we look up the Laplace transforms. So let’s agree to write this portion

of the partial fraction expansion in the equivalent form

Determine

We first observe that the quadratic factor is irreducible (check the sign of the discriminant in the quadratic formula). Next we write the quadratic in the form by completing the square:

Since and are nonrepeated factors, the partial fraction expansion has the form

When we multiply both sides by the common denominator, we obtain

(11)

In equation (11), let’s put and 0. With we find

$8 # 8C ,

2 $ 10 # 3A A$2B " 2B 4 A0B " C A8B , s # $1,s # $1, 1, 2s2 " 10s # 3A As $ 1B " 2B 4 As " 1B " C As2 $ 2s " 5B .

2s2 " 10sAs2 $ 2s " 5B As " 1B # A As $ 1B " 2BAs $ 1B2 " 22 " Cs " 1 . s " 1s2 $ 2s " 5

s2 $ 2s " 5 # As $ 1B2 " 22 . As $ aB2 " b2s2 $ 2s " 5

!$1 e 2s2 " 10sAs2 $ 2s " 5B As " 1B f . A1 As % AB ' BB1As % AB2 ' B2 ' A2 As % AB ' BB23 As % AB2 ' B2 4 2 ' p ' Am As % AB ' BBm3 As % AB2 ' B2 4m .

Ai As $ aB " bBi Ci s " Di C1s " D1As $ aB2 " b2 " C2 s " D23 As $ aB2 " b2 4 2 " p " Cm s " Dm3 As $ aB2 " b2 4m .

As $ aB2 " b2 Q AsB.As $ aB2 " b2Q AsBAs $ aB2 " b2

# 2et " 3tet $ e$3t .

# $ !$1 e 1 s " 3

f AtB # 2!$1 e 1

s $ 1 f AtB " 3!$1 e 1As $ 1B2 f AtB

!$1 e s2 " 9s " 2As $ 1B2 As " 3B f AtB # !$1 e 2s $ 1 " 3As $ 1B2 $ 1s " 3 f AtB

Section 7.4 Inverse Laplace Transform 373

Example 7

Solution

and, hence, With in (11), we obtain

and since the last equation becomes Thus Finally, setting in (11) and using and gives

Hence, and so that

With this partial fraction expansion in hand, we can immediately determine the inverse Laplace transform:

◆

In Section 7.7, we discuss a different method (involving convolutions) for computing inverse transforms that does not require partial fraction decompositions. Moreover, the convo- lution method is convenient in the case of a rational function with a repeated quadratic factor in the denominator. Other helpful tools are described in Problems 33–36 and 38–43.

# 3et cos 2t " 4et sin 2t $ e$t .

# " 4!$1 e 2As $ 1B2 " 22 f AtB $ !$1 e 1s " 1 f AtB # 3!$1 e s $ 1As $ 1B2 " 22 f AtB

!$1 e 2s2 " 10sAs2 $ 2s " 5B As " 1B f AtB # !$1 e 3 As $ 1B " 2 A4BAs $ 1B2 " 22 $ 1s " 1 f AtB

2s2 " 10sAs2 $ 2s " 5B As " 1B # 3 As $ 1B " 2 A4BAs $ 1B2 " 22 $ 1s " 1 . C # $1A # 3, B # 4,

A # 3 .

0 # $A " 8 $ 5 ,

0 # 3A A$1B " 2B 4 A1B " C A5B ,B # 4C # $1s # 0 B # 4.12 # 4B $ 4.C # $1,

2 " 10 # 3A A0B " 2B 4 A2B " C A4B ,s # 1C # $1. 374 Chapter 7 Laplace Transforms

7.4 EXERCISES In Problems 1–10, determine the inverse Laplace trans- form of the given function.

1. 2.

3. 4.

5. 6.

7. 8.

9. 10. s $ 1 2s2 " s " 6

3s $ 15 2s2 $ 4s " 10

1 s5

2s " 16 s2 " 4s " 13

3A2s " 5B31s2 " 4s " 8 4

s2 " 9 s " 1

s2 " 2s " 10

2 s2 " 4

6As $ 1B4 In Problems 11–20, determine the partial fraction expan- sions for the given rational function.

11. 12.

13.

14.

15. 16.

17. 18. 3s 2 " 5s " 3 s4 " s3

3s " 5 s As2 " s $ 6B

$5s $ 36As " 2B As2 " 9B8s $ 2s2 $ 14As " 1B As2 $ 2s " 5B $8s2 $ 5s " 9As " 1B As2 $ 3s " 2B

$2s2 $ 3s $ 2 s As " 1B2

$s $ 7As " 1B As $ 2Bs2 $ 26s $ 47As $ 1B As " 2B As " 5B

19. 20.

In Problems 21–30, determine

21.

22.

23.

24.

25.

26.

27.

28.

29.

30.

31. Determine the Laplace transform of each of the following functions:

(a)

(b)

32. Determine the Laplace transform of each of the following functions:

(a)

(b) f2 AtB # # e t , t ) 5, 8 ,

6 , t # 5 ,

0 , t # 8 .

f1 AtB # e t , t # 1, 2, 3, . . . , et , t ) 1, 2, 3, . . . .

1 /s2

f3 AtB # t . f2 AtB # #5 , t # 1 ,2 , t # 6 ,

t , t ) 1, 6 .

f1 AtB # e0 , t # 2 , t , t ) 2 .

sF AsB $ F AsB # 2s " 5 s2 " 2s " 1

sF AsB " 2F AsB # 10s2 " 12s " 14 s2 $ 2s " 2

s2F AsB " sF AsB $ 6F AsB # s2 " 4 s2 " s

s2F AsB $ 4F AsB # 5 s " 1

F AsB # 7s3 $ 2s2 $ 3s " 6 s3 As $ 2B

F AsB # 7s2 " 23s " 30As $ 2B As2 " 2s " 5B F AsB # 7s2 $ 41s " 84As $ 1B As2 $ 4s " 13B F AsB # 5s2 " 34s " 53As " 3B2 As " 1B F AsB # s " 11As $ 1B As " 3B F AsB # 6s2 $ 13s " 2

s As $ 1B As $ 6B !$1EFF .

sAs $ 1B As2 $ 1B1As $ 3B As2 " 2s " 2B Section 7.4 Inverse Laplace Transform 375

Theorem 6 in Section 7.3 can be expressed in terms of the inverse Laplace transform as

where . Use this equation in Problems 33–36 to compute

33. 34.

35. 36.

37. Prove Theorem 7 on the linearity of the inverse transform. [Hint: Show that the right-hand side of equation (3) is a continuous function on whose Laplace transform is

38. Residue Computation. Let be a rational function with deg P * deg Q and suppose is a nonrepeated linear factor of Prove that the por- tion of the partial fraction expansion of corresponding to is

where A (called the residue) is given by the formula

39. Use the residue computation formula derived in Problem 38 to determine quickly the partial fraction expansion for

40. Heaviside’s Expansion Formula.† Let and be polynomials with the degree of less

than the degree of Let

where the ’s are distinct real numbers. Show that

!$1 e P Q f AtB #an

i#1

P AriB Q¿AriB erit .

Q AsB # As $ r1B As $ r2B p As $ rnB ,Q AsB. P AsBQ AsB P AsB

F AsB # 2s " 1 s As $ 1B As " 2B .

A ! lim sSr

As % rBP AsB Q AsB .

A s % r ,

s $ r P AsB /Q AsBQ AsB. s $ r

P AsB /Q AsBF1 AsB " F2 AsB. 4 3 0, q B F AsB # arctan A1/sBF AsB # ln as2 " 9s2 " 1b F AsB # ln as $ 4

s $ 3 bF AsB # ln as " 2

s $ 5 b

!$1EFF.f # !$1EFF !$1 e dnF

dsn f AtB ! A$tBnf AtB ,

1 / As $ 1B f3 AtB # et .

†Historical Footnote: This formula played an important role in the “operational solution” to ordinary differential equations developed by Oliver Heaviside in the 1890s.

41. Use Heaviside’s expansion formula derived in Prob- lem 40 to determine the inverse Laplace transform of

42. Complex Residues. Let be a rational function with deg P * deg Q and suppose

is a nonrepeated quadratic factor of Q. (That is, are complex conjugate zeros of Q.) Prove that the portion of the partial fraction expansion of corresponding to is

A As $ aB " bBAs $ aB2 " b2 , As $ aB2 " b2P AsB /Q AsB

a + ib As $ aB2 " b2

P AsB /Q AsB F AsB # 3s2 $ 16s " 5As " 1B As $ 3B As $ 2B .

376 Chapter 7 Laplace Transforms

where the complex residue is given by the formula

(Thus we can determine B and A by taking the real and imaginary parts of the limit and dividing them by

43. Use the residue formulas derived in Problems 38 and 42 to determine the partial fraction expansion for

F AsB # 6s2 " 28As2 $ 2s " 5B As " 2B . b.B bB " ibA # lim

sSa"ib

3 As $ aB2 " b2 4P AsB Q AsB .

bB " ibA

Our goal is to show how Laplace transforms can be used to solve initial value problems for linear differential equations. Recall that we have already studied ways of solving such initial value problems in Chapter 4. These previous methods required that we first find a general solution of the differential equation and then use the initial conditions to determine the desired solution. As we will see, the method of Laplace transforms leads to the solution of the initial value problem without first finding a general solution.

Other advantages to the transform method are worth noting. For example, the technique can easily handle equations involving forcing functions having jump discontinuities, as illus- trated in Section 7.1. Further, the method can be used for certain linear differential equations with variable coefficients, a special class of integral equations, systems of differential equa- tions, and partial differential equations.

7.5 SOLVING INITIAL VALUE PROBLEMS

Method of Laplace Transforms To solve an initial value problem:

(a) Take the Laplace transform of both sides of the equation. (b) Use the properties of the Laplace transform and the initial conditions to obtain an

equation for the Laplace transform of the solution and then solve this equation for the transform.

(c) Determine the inverse Laplace transform of the solution by looking it up in a table or by using a suitable method (such as partial fractions) in combination with the table.

In step (a) we are tacitly assuming the solution is piecewise continuous on and of exponential order. Once we have obtained the inverse Laplace transform in step (c), we can verify that these tacit assumptions are satisfied.

Solve the initial value problem

(1) y– $ 2y¿ " 5y # $8e$t ; y A0B # 2 , y¿ A0B # 12 .

30, q B Example 1

The differential equation in (1) is an identity between two functions of t. Hence equality holds for the Laplace transforms of these functions:

Using the linearity property of and the previously computed transform of the exponential function, we can write

(2)

Now let From the formulas for the Laplace transform of higher-order derivatives (see Section 7.3) and the initial conditions in (1), we find

Substituting these expressions into (2) and solving for yields

Our remaining task is to compute the inverse transform of the rational function This was done in Example 7 of Section 7.4, where, using a partial fraction expansion, we found

(3)

which is the solution to the initial value problem (1). ◆

As a quick check on the accuracy of our computations, the reader is advised to verify that the computed solution satisfies the given initial conditions.

The reader is probably questioning the wisdom of using the Laplace transform method to solve an initial value problem that can be easily handled by the methods discussed in Chapter 4. The objective of the first few examples in this section is simply to make the reader familiar with the Laplace transform procedure. We will see in Example 4 and in later sections that the method is applicable to problems that cannot be readily handled by the techniques discussed in the previous chapters.

Solve the initial value problem

(4)

Let Taking the Laplace transform of both sides of the differential equation in (4) gives

(5) !Ey–F AsB " 4!Ey¿F AsB $ 5Y AsB # 1As $ 1B2 . Y AsB J !EyF AsB.y– " 4y¿ $ 5y # tet ; y A0B # 1 , y¿A0B # 0 .

y AtB # 3et cos 2t " 4et sin 2t $ e$t , Y AsB.

Y AsB # 2s2 " 10sAs2 $ 2s " 5B As " 1B . As2 $ 2s " 5BY AsB # 2s2 " 10s

s " 1

As2 $ 2s " 5BY AsB # 2s " 8 $ 8 s " 1

3 s2Y AsB $ 2s $ 12 4 $ 2 3 sY AsB $ 2 4 " 5Y AsB # $8 s " 1

Y AsB!Ey–F AsB # s2Y AsB $ sy A0B $ y¿ A0B # s2Y AsB $ 2s $ 12 . !Ey¿F AsB # sY AsB $ y A0B # sY AsB $ 2 ,

Y AsB J !EyF AsB.!Ey–F AsB $ 2!Ey¿F AsB " 5!EyF AsB # $8

s " 1 .

!Ey– $ 2y¿ " 5yF # !E$8e$tF .

Section 7.5 Solving Initial Value Problems 377

Solution

Example 2

Solution

Using the initial conditions, we can express and in terms of That is,

Substituting back into (5) and solving for gives

The partial fraction expansion for has the form

(6)

Solving for the numerators, we ultimately obtain , , and . Substituting these values into (6) gives

where we have written to facilitate the final step of taking the inverse transform. From the tables, we now obtain

(7)

as the solution to the initial value problem (4). ◆

Solve the initial value problem

(8)

To use the method of Laplace transforms, we first move the initial conditions to This can be done by setting Then

Replacing t by in the differential equation in (8), we have

(9)

Substituting in (9), the initial value problem in (8) becomes

Because the initial conditions are now given at the origin, the Laplace transform method is applicable. In fact, we carried out the procedure in Example 1, where we found

(10) y AtB # 3et cos 2t " 4et sin 2t $ e$t . y– AtB $ 2y¿ AtB " 5y AtB # $8e$t ; y A0B # 2 , y¿ A0B # 12 .y AtB # w At " pB w– At " pB $ 2w¿ At " pB " 5w At " pB # $8ep$At"pB # $8e$t .t " p y¿ AtB # w¿ At " pB , y– AtB # w– At " pB .y AtB J w At " pB.

t # 0.

w– AtB $ 2w¿ AtB " 5w AtB # $8ep$t ; w ApB # 2 , w¿ ApB # 12 .

y AtB # 35 216

e$5t " 181 216

et $ 1 36

tet " 1 12

t2et

D # 1/6 # (1/12)2

Y AsB # 35 216

a 1 s " 5

b " 181 216

a 1 s $ 1

b $ 1 36

a 1As $ 1B2b " 112 a 2As $ 1B3b , D # 1/6

C # $1/36,B # 181/216A # 35/216

s3 " 2s2 $ 7s " 5As " 5B As $ 1B3 # As " 5 " Bs $ 1 " CAs $ 1B2 " DAs $ 1B3 . Y AsB

Y AsB # s3 " 2s2 $ 7s " 5As " 5B As $ 1B3 . As " 5B As $ 1BY AsB # s3 " 2s2 $ 7s " 5As $ 1B2 As2 " 4s $ 5BY AsB # s " 4 " 1As $ 1B2

3 s2Y AsB $ s 4 " 4 3 sY AsB $ 1 4 $ 5Y AsB # 1As $ 1B2 Y AsB!Ey–F AsB # s2Y AsB $ sy A0B $ y¿ A0B # s2Y AsB $ s .

!Ey¿F AsB # sY AsB $ y A0B # sY AsB $ 1 , Y AsB.!Ey–F AsB!Ey¿F AsB 378 Chapter 7 Laplace Transforms

Example 3

Solution

Since then Hence, replacing t by in (10) gives

◆

Thus far we have applied the Laplace transform method only to linear equations with constant coefficients. Yet several important equations in mathematical physics involve linear equations whose coefficients are polynomials in t. To solve such equations using Laplace transforms, we apply Theorem 6, page 364, where we proved that

(11)

If we let and we find

Similarly, with and we obtain from (11)

Thus, we see that for a linear differential equation in whose coefficients are polynomials in t, the method of Laplace transforms will convert the given equation into a linear differential equation in whose coefficients are polynomials in s. Moreover, if the coefficients of the given equation are polynomials of degree 1 in t, then (regardless of the order of the given equa- tion) the differential equation for is just a linear first-order equation. Since we know how to solve this first-order equation, the only serious obstacle we may encounter is obtaining the inverse Laplace transform of [This problem may be insurmountable, since the solution

may not have a Laplace transform.] In illustrating the technique, we make use of the following fact. If is piecewise

continuous on and of exponential order, then

(12)

(You may have already guessed this from the entries in Table 7.1, page 359.) An outline of the proof of (12) is given in Exercises 7.3, Problem 26.

Solve the initial value problem

(13)

Let and take the Laplace transform of both sides of the equation in (13):

(14)

Using the initial conditions, we find

!Ey–F AsB # s2Y AsB $ sy A0B $ y¿ A0B # s2Y AsB !Ey–F AsB " 2!Ety¿ AtB F AsB $ 4Y AsB # 1

s .

Y AsB # !EyF AsBy– " 2ty¿ $ 4y # 1 , y A0B # y¿ A0B # 0 .

lim sS"

!E f F AsB ! 0 .30, q B f AtBy AtB Y

AsB.Y AsBY

AsB y AtB# $s2Y¿ AsB $ 2sY AsB " y A0B .

# $ d ds

3 s2Y AsB $ sy A0B $ y¿A0B 4!Ety– AtB F AsB # $ d ds

!Ey–F AsBf AtB # y–AtB,n # 1 # $

d ds

3 sY AsB $ y A0B 4 # $sY ¿AsB $ Y AsB .!Ety¿ AtB F AsB # $ d ds

!Ey¿F AsBf AtB # y¿ AtB,n # 1 !Et nf AtB F AsB # A$1Bn dnF

dsn AsB .

# 3et$pcos 2t " 4et$psin 2t $ ep$ t .

w AtB # y At $ pB # 3et$pcos 32 At $ pB 4 " 4et$psin 32 At $ pB 4 $ e$At$pBt $ pw AtB # y At $ pB.w At " pB # y AtB, Section 7.5 Solving Initial Value Problems 379

Example 4

Solution

and

Substituting these expressions into (14) gives

(15)

Equation (15) is a linear first-order equation and has the integrating factor

(see Section 2.3). Multiplying (15) by we obtain

Integrating and solving for yields

(16)

Now if is the Laplace transform of a piecewise continuous function of exponential order, then it follows from equation (12) that

For this to occur, the constant C in equation (16) must be zero. Hence, and taking the inverse transform gives We can easily verify that is the solution to the given initial value problem by substituting it into (13). ◆

We end this section with an application from control theory. Let’s consider a servomech- anism that models an automatic pilot. Such a mechanism applies a torque to the steering control shaft so that a plane or boat will follow a prescribed course. If we let be the true direction (angle) of the craft at time t and be the desired direction at time t, then

denotes the error or deviation between the desired direction and the true direction. Let’s assume that the servomechanism can measure the error and feed back to the

steering shaft a component of torque that is proportional to but opposite in sign (see Figure 7.6 on page 381). Newton’s second law, expressed in terms of torques, states thatAmoment of inertiaB ( Aangular accelerationB ! total torque.

e AtB e AtB e AtB J y AtB $ g AtB g AtB

y AtB y AtB # t2/2y AtB # t2/2. Y AsB # 1/s3,

lim sSq

Y AsB # 0 . Y AsB

Y AsB # 1 s3

" C es

2/4 s3

s3e$s 2/4Y AsB # $! s2 e$s2/4 ds # e$s2/4 " C

Y AsB d ds

Em AsBY AsB F # d ds

U s3e$s2/4Y AsB V # $ s2 e$s2/4 . m AsB,m AsB # e"A3/s$s/2Bds # e ln s3$s2/4 # s3e$s2/4

Y ¿AsB " a3 s

$ s 2 bY AsB # $1

2s2 .

$2sY ¿AsB " As2 $ 6BY AsB # 1 s

s2Y AsB " 2 3$sY ¿AsB $ Y AsB 4 $ 4Y AsB # 1 s

# $ d ds

3 sY AsB $ y A0B 4 # $sY ¿AsB $ Y AsB .!Ety¿AtB F AsB # $ d ds

!Ey¿F AsB 380 Chapter 7 Laplace Transforms

For the servomechanism described, this becomes

(17)

where I is the moment of inertia of the steering shaft and k is a positive proportionality constant.

Determine the error for the automatic pilot if the steering shaft is initially at rest in the zero direction and the desired direction is given by where a is a constant.

Based on the discussion leading to equation (17), a model for the mechanism is given by the initial value problem

(18)

where We begin by taking the Laplace transform of both sides of (18):

(19)

where and Since

we find from (19) that

Solving this equation for gives

Hence, on taking the inverse Laplace transform, we obtain the error

(20) ◆

As we can see from equation (20), the automatic pilot will oscillate back and forth about the desired course, always “oversteering” by the factor Clearly, we can make the error a/2k/I.

e AtB # $ a2k/I sin A2k/I tB . E AsB # $ aI

s2I " k #

$a2k/I 2k/Is2 " k/I . E AsBs2IE AsB " aI # $kE AsB .

E AsB # !Ey AtB $ atF AsB # Y AsB $ !EatF AsB # Y AsB $ as$2 ,E AsB # !EeF AsB.Y AsB # !EyF AsB s2IY AsB # $kE AsB ,I 3 s2Y AsB $ sy A0B $ y¿ A0B 4 # $kE AsB

I!Ey–F AsB # $k!EeF AsB e AtB # y AtB $ g AtB # y AtB $ at.Iy– AtB # $ke AtB ; y A0B # 0 , y¿ A0B # 0 ,

g AtB # at,e AtB Iy– AtB # $ke AtB ,

Section 7.5 Solving Initial Value Problems 381

e(t): = y(t) −

Desired direction

Error

Feedback

True direction

y(t)

g(t)

Iy" = − ke g(t)

Figure 7.6 Servomechanism with feedback

Example 5

Solution

small by making k large relative to I, but then the term becomes large, causing the error to oscil- late more rapidly. (See Figure 7.7.) As with vibrations, the oscillations or oversteering can be con- trolled by introducing a damping torque proportional to but opposite in sign (see Problem 40).e¿ AtB2k/I

382 Chapter 7 Laplace Transforms

e(t)

8642

0.5 k = 16 I

−0.5

−1

k = 1

Figure 7.7 Error for automatic pilot when and when k/I # 16k/I # 1

7.5 EXERCISES In Problems 1–14, solve the given initial value problem using the method of Laplace transforms.

10. y A0B # 0 , y¿ A0B # 5y– $ 4y # 4t $ 8e$2t ; z A1B # $1 , z¿ A1B # 9z– " 5z¿ $ 6z # 21et$1 ; y A0B # 0 , y¿ A0B # 3y– " 4y # 4t2 $ 4t " 10 ; y A0B # 5 , y¿ A0B # $4y– $ 7y¿ " 10y # 9 cos t " 7 sin t ; y A0B # 2 , y¿ A0B # 7y– $ 4y¿ " 5y # 4e3t ; w A0B # 1 , w¿ A0B # $1w– " w # t2 " 2 ; y A0B # $1 , y¿ A0B # 7y– " 6y¿ " 5y # 12et ; y A0B # $1 , y¿ A0B # 6y– " 6y¿ " 9y # 0 ; y A0B # $2 , y¿ A0B # 5y– $ y¿ $ 2y # 0 ; y A0B # 2 , y¿A0B # 4y– $ 2y¿ " 5y # 0 ;

11. 12.

13.

14.

In Problems 15–24, solve for the Laplace transform of the solution to the given initial value problem. 15.

16.

17.

18.

19.

20. 21.

y A0B # 1 , y¿ A0B # 3y– $ 2y¿ " y # cos t $ sin t ; y– " 3y # t3 ; y A0B # 0 , y¿ A0B # 0y A0B # 1 , y¿ A0B # 1 y– " 5y¿ $ y # et $ 1 ; y A0B # 1 , y¿ A0B # 3y– $ 2y¿ $ y # e2t $ et ; y A0B # 1 , y¿ A0B # 0y– " y¿ $ y # t3 ; y A0B # 0 , y¿ A0B # $1y– " 6y # t2 $ 1 ; y A0B # 0 , y¿ A0B # $1y– $ 3y¿ " 2y # cos t ;

y AtB Y AsB, y– " y # t ; y ApB # 0 , y¿ ApB # 0y Ap/2B # 1 , y¿ Ap/2B # 0 y– $ y¿ $ 2y # $8 cos t $ 2 sin t ; w A$1B # 3 ; w¿ A$1B # 7w– $ 2w¿ " w # 6t $ 2 ; y– $ y # t $ 2 ; y A2B # 3 , y¿ A2B # 0

22.

23. where

24. where

In Problems 25–28, solve the given third-order initial value problem for using the method of Laplace transforms. 25.

26.

27.

28.

In Problems 29–32, use the method of Laplace trans- forms to find a general solution to the given differential equation by assuming and where a and b are arbitrary constants. 29. 30. 31. 32.

33. Use Theorem 6 in Section 7.3 to show that

where Y AsB # !EyF AsB .!Et2y¿ AtB F AsB # sY–AsB " 2Y¿AsB , y– $ 5y¿ " 6y # $6te2t y– " 2y¿ " 2y # 5

y– " 6y¿ " 5y # ty– $ 4y¿ " 3y # 0

y¿ A0B # b,y A0B # a y A0B # 0 , y¿ A0B # 2 , y– A0B # $4y‡ " y– " 3y¿ $ 5y # 16e$t ; y A0B # $4 , y¿ A0B # 4 , y– A0B # $2y‡ " 3y– " 3y¿ " y # 0 ; y A0B # 1 , y¿ A0B # 4 , y– A0B # $2y‡ " 4y– " y¿ $ 6y # $12 ; y A0B # 1 , y¿ A0B # 1 , y– A0B # 3y‡ $ y– " y¿ $ y # 0 ;

y AtB g AtB # e 1 , t 6 3 ,

t , t 7 3

y– $ y # g AtB ; y A0B # 1 , y¿ A0B # 2 ,g AtB # e t , t 6 2 , 5 , t 7 2

y– " 4y # g AtB ; y A0B # $1 ; y¿ A0B # 0 ,y A0B # 2 , y¿ A0B # $1 y– $ 6y¿ " 5y # te t ;

Section 7.6 Transforms of Discontinuous and Periodic Functions 383

34. Use Theorem 6 in Section 7.3 to show that

where

In Problems 35–38, find solutions to the given initial value problem. 35.

36.

37.

[Hint: 38.

39. Determine the error for the automatic pilot in Example 5 if the shaft is initially at rest in the zero direction and the desired direction is where a is a constant.

40. In Example 5 assume that in order to control oscillations a component of torque proportional to

but opposite in sign, is also fed back to the steering shaft. Show that equation (17) is now replaced by

where is a positive constant. Determine the error for the automatic pilot with mild damping (i.e.,

if the steering shaft is initially at rest in the zero direction and the desired direction is given by where a is a constant.

41. In Problem 40 determine the error when the desired direction is given by where a is a constant.

g AtB # at,e AtB g AtB # a,

m 6 22IkBe AtB m Iy– AtB # $ke AtB $ me¿ AtB ,

e¿ AtB, g AtB # a,

e AtBy A0B # 0 , y¿ A0B # 3 y– " ty¿ $ y # 0 ;

!$1E1 / As2 " 1B2F AtB # Asin t $ t cos tB /2. 4y A0B # 1 , y¿ A0B # 0ty– $ 2y¿ " ty # 0 ; y A0B # 2 , y¿ A0B # $1ty– $ ty¿ " y # 2 ; y A0B # 0 , y¿ A0B # 0y– " 3ty¿ $ 6y # 1 ;

Y AsB # !EyF AsB .!Et2y– AtB F AsB # s2Y– AsB " 4sY¿ AsB " 2Y AsB ,

7.6 TRANSFORMS OF DISCONTINUOUS AND PERIODIC FUNCTIONS In this section we study special functions that often arise when the method of Laplace trans- forms is applied to physical problems. Of particular interest are methods for handling functions with jump discontinuities. Jump discontinuities occur naturally in physical problems such as electric circuits with on/off switches. To handle such behavior, Oliver Heaviside introduced the following step function.

By shifting the argument of the jump can be moved to a different location. That is,

(2)

has its jump at t # a. By multiplying by a constant M, the height of the jump can also be modified:

Mu At $ aB # e0 , t 6 a , M , a 6 t .

u At $ aB # e0 , t $ a 6 0 , 1 , 0 6 t $ a

# e0 , t 6 a 1 , a 6 t

u AtB,

384 Chapter 7 Laplace Transforms

Unit Step Function

Definition 5. The unit step function is defined by

(1) u AtB :! e0 , t 6 0 ,1 , 0 6 t . u AtB

Rectangular Window Function

Definition 6. The rectangular window function is defined by†

(3) ßa,b AtB :# u At $ aB $ u At $ bB # #0, t 6 a ,1, a 6 t 6 b ,0, b 6 t . ßa,b AtB

1 2 3 4 0 t

2u(t – 1)

u(t – 2)

Figure 7.8 Two-step functions expressed using the unit step function

†Also known as the square pulse, or the boxcar function.

To express piecewise continuous functions, we employ the rectangular window, which turns the step function on and then turns it back off.

Section 7.6 Transforms of Discontinuous and Periodic Functions 385

Figure 7.9 The rectangular window

Figure 7.10 The windowing effect of ßa,b AtB

The function is displayed in Figure 7.9, and Figure 7.10, illustrating multiplication of a function by , justifies its name.ßa,b AtBßa,b AtB

Any piecewise continuous function can be expressed in terms of window and step functions.

Write the function

(4)

(see Figure 7.11 on page 386) in terms of window and step functions.

Clearly, from the figure we want to window the function in the intervals (0, 2), (2, 5), and (5, 8), and to introduce a step for t % 8. From (5) we read off the desired representation as

(5) . ◆

The Laplace transform of with is

(6)

since, for

# lim NSq

$e$st

s ` N

a #

e$as

s .

!Eu At $ aB F AsB # !q 0

e$stu At $ aB dt # !q a

e$st dt

s 7 0,

!Eu At % aB F AsB ! e%ass , a ' 0u At $ aB

f AtB # 3ß0,2 AtB " 1ß2,5 AtB " tß5,8 AtB " At2/10)u At $ 8B

f AtB # # 3 , t 6 2 , 1 , 2 6 t 6 5 , t , 5 6 t 6 8 , t2/10 , 8 6 t

Example 1

Solution

a b

,a,b(t)

a b

f (t)

a b

f (t),a,b(t)

Conversely, for a % 0, we say that the piecewise continuous function is an inverse Laplace transform for and we write†

For the rectangular window function, we deduce from (6) that

(7) .

The translation property of discussed in Section 7.3 described the effect on the Laplace transform of multiplying a function by The next theorem illustrates an analogous effect of multiplying the Laplace transform of a function by e$as.

eat. F AsB! Eu At $ aB $ u At $ bBF AsB # 3 e$sa $ e$sb 4 /s , 0 6 a 6 b!Eßa,b AtB F AsB #

!$1 e e$as s f AtB # u At $ aB .e$as/s

u At $ aB

386 Chapter 7 Laplace Transforms

†The absence of a specific value for u(0) in Definition 5 is a reflection of the ambiguity of the inverse Laplace transform, when no continuous inverse transform exists. ††This inverse transform is in fact a continuous function of t if and is continuous for the values of

for are of no consequence, since the factor is zero there.u At $ aBt 6 0f AtB t ' 0;f AtBf A0B # 0

Translation in t

Theorem 8. Let exist for If a is a positive constant, then (8) and, conversely, an inverse Laplace transform†† of is given by

(9) !%1Ee%asF AsB F AtB ! f At % aBu At % aB . e$asF AsB !E f At % aBu At % aB F AsB ! e%asF AsB ,s 7 a ' 0.F AsB # !E f F AsB

Proof. By the definition of the Laplace transform, we have

(10)

# ! q

a e$stf At $ aB dt ,

!E f At $ aBu At $ aB F AsB # !q 0

e$stf At $ aBu At $ aB dt

f(t)

0 4 6 8 1210

Figure 7.11 Graph of in equation (4)f AtB

where, in the last equation, we used the fact that is zero for and equals 1 for Now let Then we have and equation (10) becomes

◆

Notice that formula (8) includes as a special case the formula for indeed, if we take then and (8) becomes

In practice it is more common to be faced with the problem of computing the transform of a function expressed as rather than To compute

we simply identify with so that Equation (8) then gives

(11)

Determine the Laplace transform of

To apply equation (11), we take and Then

Now the Laplace transform of is

So, by formula (11), we have

◆

Determine

Here and Hence,

and so the Laplace transform of is

Thus, from formula (11), we get

◆!E Acos tBu At $ pB F AsB # $e$ps s s2 " 1

!Eg At " aB F AsB # $!Ecos tF AsB # $ s s2 " 1

g At " aBg At " aB # g At " pB # cos At " pB # $cos t , a # p.g AtB # cos t!E Acos tBu At $ pB F .

!Et2u At $ 1B F AsB # e$s e 2 s3

" 2 s2

" 1 s f .

!Eg At " aB F AsB # !Et2 " 2t " 1F AsB # 2 s3

" 2 s2

" 1 s .

g At " aBg At " aB # g At " 1B # At " 1B 2 # t2 " 2t " 1 .

a # 1.g AtB # t2t 2u At $ 1B .

!Eg AtBu At % aB F AsB ! e%as!Eg At ' aB F AsB . f AtB # g At " aB.f At $ aBg AtB!Eg AtBu At $ aB F, f At $ aBu At $ aB.g AtBu At $ aB

!Eu At $ aB F AsB # e$as/s.F AsB # 1/sf AtB $ 1, !Eu At $ aB F; # e$as !

0 e$syf AyB dy # e$asF AsB .

!E f At $ aBu At $ aB F AsB # !q 0

e$ase$syf AyB dy dy # dt,y # t $ a.t 7 a.

t 6 au(t $ aB Section 7.6 Transforms of Discontinuous and Periodic Functions 387

Example 2

Solution

Example 3

Solution

In Examples 2 and 3, we could also have computed the Laplace transform directly from the definition. In dealing with inverse transforms, however, we do not have a simple alternative formula† upon which to rely, and so formula (9) is especially useful whenever the transform has as a factor.

Determine and sketch its graph.

To use the translation property (9), we first express as the product For this purpose, we put and Thus, and

It now follows from the translation property that

See Figure 7.12. ◆

As illustrated by the next example, step functions arise in the modeling of on/off switches, changes in polarity, etc.

The current I in an LC series circuit is governed by the initial value problem

(12) where

Determine the current as a function of time t.

Let Then we have !EI–F AsB # s2J AsB.J AsB J !EIF AsB. g AtB J # 1 , 0 6 t 6 1 ,$1 , 1 6 t 6 2 ,

0 , 2 6 t .

I– AtB " 4I AtB # g AtB ; I A0B # 0 , I¿ A0B # 0 ,

!$1 e e$2s s2 f AtB # f At $ 2Bu At $ 2B # At $ 2Bu At $ 2B .

f AtB # !$1 e 1 s2 f AtB # t . a # 2F AsB # 1/s

2.e$as # e$2s e$asF AsB.e$2s/s2

!$1 e e$2s s2 fe

$as

388 Chapter 7 Laplace Transforms

†Under certain conditions, the inverse transform is given by the contour integral

See, for example, Complex Variables and the Laplace Transform for Engineers, by Wilbur R. LePage (Dover Publica- tions, New York, 1980), or Fundamentals of Complex Analysis with Applications to Engineering and Science, 3rd ed., by E. B. Saff and A. D. Snider (Prentice Hall, Englewood Cliffs, N.J., 2003).

!$1EFF AtB # 1 2pi !

a"i q

a$ i q estF AsB ds .

Example 4

Solution

(t − 2) u(t − 2)

0 2 t

Figure 7.12 Graph of solution to Example 4

Example 5

Solution

Writing g in terms of the rectangular window function we get

and so

Thus, when we take the Laplace transform of both sides of (12), we obtain

To find we first observe that

where

Computing the inverse transform of gives

Hence, via the translation property (9), we find

The current is graphed in Figure 7.13. Note that is smoother than the former has discontinuities in its second derivative at the points where the latter has jumps. ◆

g AtB;I AtB # " c 14 $ 14 cos 2 At $ 2B du At $ 2B . # a1

4 $

1 4

cos 2tb $ c 1 2

$ 1 2

cos 2 At $ 1B du At $ 1B # f AtB $ 2 f At $ 1Bu At $ 1B " f At $ 2Bu At $ 2B I AtB # !$1EF AsB $ 2e$sF AsB " e$2sF AsB F AtB f AtB J !$1EFF AtB # 1

4 $

1 4

cos 2t .

F AsBF AsB J 1

s As2 " 4B # 14 a1sb $ 14 a ss2 " 4b . J AsB # F AsB $ 2e$sF AsB " e$2sF AsB ,I # !$1EJF,

J AsB # 1 s As2 " 4B $ 2e$ss As2 " 4B " e$2ss As2 " 4B .

s2J AsB " 4J AsB # 1 s

$ 2e$s

s "

e$2s

!EI–F AsB " 4!EIF AsB # !EgF AsB !EgF AsB # 1

s $

2e$s

s "

e$2s

s .

# 1 $ 2u At $ 1B " u At $ 2Bg AtB # ß0,1 AtB " A$1Bß1,2 AtB # u AtB $ u At $ 1B $ 3u At $ 1B $ u At $ 2B 4 ßa,b AtB # u At $ aB $ u At $ bB,AtB

Section 7.6 Transforms of Discontinuous and Periodic Functions 389

I(t)

2 3 4

Figure 7.13 Solution to Example 5

Periodic functions are another class of functions that occur frequently in applications.

390 Chapter 7 Laplace Transforms

Periodic Function

Definition 7. A function is said to be periodic of period T ()0) if

for all t in the domain of f.

f At " TB # f AtB f AtB

As we know, the sine and cosine functions are periodic with period and the tangent function is periodic with period † To specify a periodic function, it is sufficient to give its val- ues over one period. For example, the square wave function in Figure 7.14 can be expressed as

(13) and has period 2.f AtBf AtB J e1 , 0 6 t 6 1 , $1 , 1 6 t 6 2 ,

p. 2p

†A function that has period T will also have period 2T, 3T, etc. For example, the sine function has periods 2p, 4p, 6p, etc. Some authors refer to the smallest period as the fundamental period or just the period of the function.

Figure 7.14 Graph of square wave function f AtB

t 0 T

fT (t) f (t)

Figure 7.15 Windowed version of periodic function

It is convenient to introduce a notation for the “windowed” version of a periodic function (using a rectangular window whose width is the period):

(14)

(See Figure 7.15.) The Laplace transform of is given by

FT AsB # !q 0

e$st fT AtB dt # ! T 0

e$st f AtB dt .fT AtBfT

AtB J f AtBß0,T AtB # f AtB 3u AtB $ u At $ TB 4 # e f AtB , 0 6 t 6 T ,0 , otherwise .f AtB

t –2 –1

–1

0 1

2 3

It is related to the Laplace transform of as follows.f AtB Section 7.6 Transforms of Discontinuous and Periodic Functions 391

Transform of Periodic Function

Theorem 9. If f has period T and is piecewise continuous on then the Laplace

transforms and are related by

(15) or .F AsB ! FT AsB 1%e%sT

FT AsB # F AsB 31 $ e$sT 4 FT AsB # ! T

0 e$stf AtBdtF AsB # !q

0 e$st f AtBdt 30, T 4 ,

Proof. From (14) and the periodicity of ƒ, we have

(16) ,

so taking transforms and applying (8) yields , which is equivalent to (15). ◆

Determine , where f is the square wave function in Figure 7.14.

Here T = 2. Windowing the function results in , so by (7) we get Therefore (15) implies

. ◆

For functions with power series expansions we can find their transforms by using the formula .

Determine where

We begin by expressing in a Taylor series† about Since

then dividing by t, we obtain

for t % 0. This representation also holds at t # 0 since

lim tS0

f AtB # lim tS0

sin t

t # 1 .

f AtB # sin t t

# 1 $ t2

3! "

5! $

7! " p

sin t # t $ t3

3! "

5! $

7! " p ,

t # 0.f AtB f AtB J #

sin t

t , t ) 0 ,

1 , t # 0 .

!E f F,!Et nF AsB # n!/sn"1, n # 0, 1, 2, . . .

A1 $ e$sB2/s 1 $ e$2s

# 1 $ e$sA1 " e$sBs!E f F #

FT AsB # A1 $ e$sB /s $ Ae$s $ e$2sB /s # A1 $ e$sB2/s.fT AtB # ß0,1 AtB $ ß1,2 AtB !E f F

FT AsB # F AsB $ e$sTF AsBfT AtB # f AtBu AtB $ f AtBu At $ TB # f AtBu AtB $ f At $ TBu At $ TB

Example 6

Solution

Example 7

Solution

†For a discussion of Taylor series, see Sections 8.1 and 8.2.

Observe that is continuous on and of exponential order. Hence, its Laplace trans- form exists for all s sufficiently large. Because of the linearity of the Laplace transform, we would expect that

Indeed, using tools from analysis, it can be verified that this series representation is valid for all s % 1. Moreover, one can show that the series converges to the function (see Prob- lem 54). Thus,

(17) ◆

A similar procedure involving the series expansion for in powers of can be used to compute (see Problems 55–57).

We have previously shown, for every nonnegative integer n, that But what if the power of t is not an integer? Is this formula still valid? To answer this question, we need to extend the idea of “factorial.” This is accomplished by the gamma function.†

!EtnF AsB # n!/sn"1.f AtB # !$1EFF AtB 1/sF AsB ! e sin t

t f AsB # arctan 1

s .

arctan A1/sB #

1 s

$ 1

3s3 "

1 5s5

$ 1

7s7 " p .

# 1 s

$ 2!

3!s3 "

4! 5!s5

$ 6!

7!s7 " p

!E f F AsB # !E1F AsB $ 1 3!

!Et2F AsB " 1 5!

!Et4F AsB " p 30, q Bf AtB

392 Chapter 7 Laplace Transforms

Gamma Function

Definition 8. The gamma function is defined by

(18) ) AtB :! !" 0

e%uut%1 du , t 7 0 .

- AtB

It can be shown that the integral in (18) converges for t % 0. A useful property of the gamma function is the recursive relation

(19)

This identity follows from the definition (18) after performing an integration by parts:

# 0 " t- AtB # t- AtB . # lim

NSq A$e$NNtB " t lim

NSq ! N

0 e$uut$1 du

# lim NSq

e$e$uut ` N 0

" ! N

0 te$uut$1 du f

- At " 1B # !q 0

e$uut du # lim NSq !

0 e$uut du

) At ' 1B ! t) AtB .

†Historical Footnote: The gamma function was introduced by Leonhard Euler.

When t is a positive integer, say then the recursive relation (19) can be repeatedly applied to obtain

It follows from the definition (18) that so we find

Thus, the gamma function extends the notion of factorial! As an application of the gamma function, let’s return to the problem of determining the

Laplace transform of an arbitrary power of t. We will verify that the formula

(20)

holds for every constant By definition,

Let’s make the substitution Then and we find

Notice that when is a nonnegative integer, then and so formula (20) reduces to the familiar formula for !EtnF. - An " 1B # n!,r # n

# 1

sr"1 ! q

0 e$uur du #

- Ar " 1B sr"1

!EtrF AsB # !q 0

e$u au s b r a1

s bdu du # s dt,u # st.

!EtrF AsB # !q 0

e$sttr dt .

r 7 $1 .

!EtrF AsB ! ) Ar ' 1B sr'1

) An ' 1B ! n! . - A1B # 1, # n An $ 1B An $ 2B p 2- A1B .- An " 1B # n- AnB # n An $ 1B- An $ 1B # p

t # n,

Section 7.6 Transforms of Discontinuous and Periodic Functions 393

7.6 EXERCISES In Problems 1–4, sketch the graph of the given function and determine its Laplace transform.

1. 2. 3. 4.

In Problems 5–10, express the given function using window and step functions and compute its Laplace transform.

5. g AtB # # 0 , 0 6 t 6 1 , 2 , 1 6 t 6 2 , 1 , 2 6 t 6 3 , 3 , 3 6 t

tu At $ 1Bt2u At $ 2B u At $ 1B $ u At $ 4BAt $ 1B2u At $ 1B

6. g AtB # e 0 , 0 6 t 6 2 , t " 1 , 2 6 t

g(t)

0 1 2

Figure 7.16 Function in Problem 7

394 Chapter 7 Laplace Transforms

g(t)

−1

sin t

Figure 7.17 Function in Problem 8

g (t)

10 2 3 4

Figure 7.18 Function in Problem 9

t 43

(t − 1)2

g(t)

Figure 7.19 Function in Problem 10

10.

In Problems 11–18, determine an inverse Laplace trans- form of the given function.

11. 12.

13. 14.

15. 16.

17. 18.

19. The current in an RLC series circuit is governed by the initial value problem

I A0B # 10 , I¿ A0B # 0 ,I– AtB " 2I¿ AtB " 2I AtB # g AtB ; I AtB

e$s A3s2 $ s " 2BAs $ 1B As2 " 1Be$3s As $ 5BAs " 1B As " 2B e$s

s2 " 4 se$3s

s2 " 4s " 5

e$3s

s2 " 9 e$2s $ 3e$4s

s " 2

e$3s

s2 e$2s

s $ 1

where

Determine the current as a function of time t. Sketch for

20. The current in an LC series circuit is governed by the initial value problem

where

Determine the current as a function of time t.

In Problems 21–24, determine , where is peri- odic with the given period. Also graph 21. and has period 2. 22. and has period 1.

23.

and has period 2.

24.

and has period 2.

In Problems 25–28, determine where the periodic function is described by its graph.

!E f F,f AtB f AtB # e t , 0 6 t 6 1 ,

1 $ t , 1 6 t 6 2 ,

f AtBf AtB # e e$t , 0 6 t 6 1 , 1 , 1 6 t 6 2 ,

f AtBf AtB # et , 0 6 t 6 1 , f AtBf AtB # t , 0 6 t 6 2 , f AtB. f AtB!E f F

g AtB J e 3 sin t , 0 & t & 2p , 0 , 2p 6 t .

I¿ A0B # 3 ,I A0B # 1 , I– AtB " 4I AtB # g AtB ; I AtB0 6 t 6 8p.I AtB

g AtB J # 20 , 0 6 t 6 3p ,0 , 3p 6 t 6 4p , 20 , 4p 6 t .

f (t)

t a 3a 4a0

Figure 7.20 Square wave

5a2a

f (t)

t a 3a 4a0

Figure 7.21 Sawtooth wave

25.

26.

Section 7.6 Transforms of Discontinuous and Periodic Functions 395

f (t)

t a 3a 4a0

Figure 7.22 Triangular wave

f (t)

t 0

Figure 7.23 Half-rectified sine wave

27.

28.

In Problems 29–32, solve the given initial value problem using the method of Laplace transforms. Sketch the graph of the solution. 29.

30.

31.

32.

In Problems 33–40, solve the given initial value problem using the method of Laplace transforms. 33.

34.

35.

36.

37.

where

38.

where g AtB # #10 , 0 & t & 10 ,20 , 10 6 t 6 20 , 0 , 20 6 t

y¿ A0B # 0 ,y A0B # $1 , y– " 2y¿ " 10y # g AtB ; g AtB # e sin t , 0 & t & 2p ,

0 , 2p 6 t

y¿ A0B # 3 ,y– " 4y # g AtB ; y A0B # 1 , y A0B # 0 , y¿ A0B # 1 y– " 5y¿ " 6y # tu At $ 2B ;z A0B # 2 , z¿ A0B # $3 z– " 3z¿ " 2z # e$3t u At $ 2B ;y A0B # 0 , y¿ A0B # 0 y– " 4y¿ " 4y # u At $ pB $ u At $ 2pB ;y A0B # 1 , y¿ A0B # 1 y– " 2y¿ " 2y # u At $ 2pB $ u At $ 4pB ; y A0B # 1 , y¿ A0B # $2y– " y # 3 sin 2t $ 3 Asin 2tBu At $ 2pB ; y A0B # 0 , y¿ A0B # 1y– " y # t $ At $ 4Bu At $ 2B ; w A0B # 1 , w¿ A0B # 0w– " w # u At $ 2B $ u At $ 4B ;

y¿ A0B # 1y A0B # 0 , y– " y # u At $ 3B ;

39.

where

40.

where

41. Show that if where is fixed, then

(21)

[Hint: Use the fact that

42. The function in (21) can be expressed in a more convenient form as follows: (a) Show that for each n # 0, 1, 2, . . . ,

for [Hint: Use the fact that #

(b) Let Show that when * then and

(22)

(c) Use the facts that the first term in (22) is periodic with period T and the second term is independent of n to sketch the graph of in (22) for and

43. Show that if # then

44. Use the result of Problem 43 to show that

where is periodic with period and

g AtB J e sin t , 0 & t & p , 0 , p & t & 2p .

2pg AtB! $1 e 1As2 " 1B A1 $ e$psB f AtB # g AtB ,

' 3 sin B At % 3TB 4u At % 3TB ' p .' 3 sin B At % 2TB 4u At % 2TBg AtB ! sin Bt ' 3 sin B At % TB 4u At % TB b 3 As2 " b2B A1 $ e$TsB 4$1,!EgF AsB T # 2.a # 1 g AtB

g AtB ! e%AY eAT % 1

$ e%At

eAT % 1 .

$T 6 y 6 0t 6 An " 1BT,nT y # t $ An " 1BT. Axn"1 $ 1B / Ax $ 1B. 4 1 " x " x2 " p " xn nT 6 t 6 An " 1BT. g AtB # e$at c e An"1BaT $ 1

eaT $ 1 d

g AtB1 " x " x2 " p # 1 / A1 $ xB. 4 ' e%AAt%3TBu At % 3TB ' p .' e%AAt%2TBu At % 2TB

g AtB ! e%at ' e%AAt%TBu At % TBT 7 0 !EgF AsB # 3 As " aB A1 $ e$TsB 4$1,g AtB # e e$t , 0 & t 6 3 , 1 , 3 6 t

y¿ A0B # $1 ,y A0B # 2 , y– " 3y¿ " 2y # g AtB ; g AtB # #0 , 0 & t 6 1 ,t , 1 6 t 6 5 ,

1 , 5 6 t

y¿ A0B # 2 ,y A0B # 0 , y– " 5y¿ " 6y # g AtB ;

396 Chapter 7 Laplace Transforms

In Problems 45 and 46, use the method of Laplace trans- forms and the results of Problems 41 and 42 to solve the initial value problem.

where is the periodic function defined in the stated problem. 45. Problem 22 46. Problem 25 with

In Problems 47–50, find a Taylor series for about Assuming the Laplace transform of can be

computed term by term, find an expansion for in powers of . If possible, sum the series. 47. 48.

49. 50.

51. Using the recursive relation (19) and the fact that determine

(a) (b) 52. Use the recursive relation (19) and the fact that

to show that

where n is a positive integer.

53. Verify (15) in Theorem 9 for the function taking the period as Repeat, taking

the period as

54. By replacing s by in the Maclaurin series expan- sion for arctan s, show that

55. Find an expansion for in powers of . Use the expansion for to obtain an expansion for

in terms of Assuming the inverse Laplace transform can be computed term by term, show that

[Hint: Use the result of Problem 52.]

56. Use the procedure discussed in Problem 55 to show that

!$1Es$3/2e$1/sF AtB # 11p sin 21t . !$1Es$1/2e$1/sF AtB # 11pt cos 21t .

1 /sn"1/2.s$1/2e$1/s e$1/s

1 /se$1/s arctan

1 s

# 1 s

$ 1

3s3 "

1 5s5

$ 1

7s7 " p .

1 /s 4p.

2p.ƒ AtB # sin t, !$1 U s$An"1/2B V AtB # 2ntn$1/2

1 # 3 # 5 p A2n $ 1B1p ,- A1 /2B # 1p !Et7/2F .!Et$1/2F .- A1 /2B # 1p,

f AtB # e$t2f AtB # 1 $ cos t t

f AtB # sin tf AtB # et1 /s !E f F AsBf AtBt # 0. f AtB

a # 1

f AtBy A0B # 0 , y¿ A0B # 0 , y– " 3y¿ " 2y # f AtB ;

57. Find an expansion for ln in powers of . Assuming the inverse Laplace transform can be

computed term by term, show that

58. The unit triangular pulse is defined by

(a) Sketch the graph of . Why is it so named? Why is its symbol appropriate?

(b) Show that .

59. The mixing tank in Figure 7.24 initially holds 500 L of a brine solution with a salt concentration of 0.2 kg/L. For the first 10 min of operation, valve A is open, adding 12 L/min of brine containing a 0.4 kg/L salt concentration. After 10 min, valve B is switched in, adding a 0.6 kg/L concentration at 12 L/min. The exit valve C removes 12 L/min, thereby keeping the volume constant. Find the con- centration of salt in the tank as a function of time.

¶ AtB¶ AtB #! t

$q 2Eß0,1/2 AtB $ ß1/2,1 AtB Fdt ¶ AtB

¶ AtB J # 0 , t 6 0 ,2t , 0 6 t 6 1 /2 ,2 $ 2t , 1 /2 6 t 6 1 ,0 , t 7 1 .

¶ AtB !$1 e ln a1 " 1

s2 b f AtB # 2

t A1 $ cos tB .

1 /s 3 1 " A1 /s2B 4

12 L/min 0.4 kg/L

12 L/min 0.6 kg/L

Figure 7.24 Mixing tank

60. Suppose in Problem 59 valve B is initially opened for 10 min and then valve A is switched in for 10 min. Finally, valve B is switched back in. Find the concentration of salt in the tank as a function of time.

61. Suppose valve C removes only 6 L/min in Prob- lem 59. Can Laplace transforms be used to solve the problem? Discuss.

Consider the initial value problem

(1)

If we let and then taking the Laplace transform of both sides of (1) yields

and hence

(2)

That is, the Laplace transform of the solution to (1) is the product of the Laplace transform of sin t and the Laplace transform of the forcing term What we would now like to have is a simple formula for in terms of sin t and Just as the integral of a product is not the product of the integrals, is not the product of sin t and However, we can express as the “convolution” of sin t and g AtB. y AtBg AtB.y AtB g

AtB.y AtB g AtB. Y AsB # a 1

s2 " 1 b G AsB .

s2Y AsB " Y AsB # G AsB , G AsB # !EgF AsB,Y AsB # !EyF AsBy– " y # g AtB ; y A0B # 0 , y¿ A0B # 0 .

Section 7.7 Convolution 397

7.7 CONVOLUTION

Convolution

Definition 9. Let and be piecewise continuous on The convolution of and denoted is defined by

(3) A f * gB AtB :! ! t 0

f At % YBg AYB dY . f * g,g AtB,f AtB 30, q B.g AtBf AtB

Properties of Convolution

Theorem 10. Let , and be piecewise continuous on Then

(4) (5) (6) (7) f * 0 # 0 .

A f * gB * h # f * Ag * hB ,f * Ag " hB # A f * gB " A f * hB , f * g # g * f ,

30, q B.h AtBf AtB, g AtB

For example, the convolution of t and is

Convolution is certainly different from ordinary multiplication. For example, and in general However, convolution does satisfy some of the same

properties as multiplication. 1 * f ) f.1 * 1 # t ) 1

# aty3 3

$ y4

4 b ` t

0 #

3 $

4 #

12 .

t * t2 # ! t

0 At $ yBy2 dy # ! t

0 Aty2 $ y3B dyt 2

Proof. To prove equation (4), we begin with the definition

Using the change of variables we have

which proves (4). The proofs of equations (5) and (6) are left to the exercises (see Problems 33 and 34). Equation (7) is obvious, since ◆

Returning to our original goal, we now prove that if is the product of the Laplace transforms and then is the convolution A f * gB AtB.y AtBG AsB,F AsB Y AsB

f At $ yB # 0 $ 0. A f * gB AtB # ! 0

t f AwBg At $ wB A$dwB # ! t

0 g At $ wB f AwBdw # Ag * f B AtB ,

w # t $ y,

A f * gB AtB J ! t 0

f At $ yBg AyB dy .

398 Chapter 7 Laplace Transforms

Convolution Theorem

Theorem 11. Let and be piecewise continuous on and of exponential order and set and . Then

(8)

or, equivalently,

(9) !%1EF AsBG AsB F AtB ! A f * gB AtB . !E f * gF AsB ! F AsBG AsB , G AsB # !EgF AsBF AsB # !E f F AsBa

30, q Bg AtBf AtB

Proof. Starting with the left-hand side of (8), we use the definition of convolution to write for

To simplify the evaluation of this iterated integral, we introduce the unit step function and write

where we have used the fact that if Reversing the order of integration†

gives

(10)

Recall from the translation property in Section 7.6 that the integral in brackets in equation (10) equals Hence,

This proves formula (8). ◆

!E f * gF AsB # !q 0

g AyBe$syF AsB dy # F AsB !q 0

e$syg AyB dy # F AsBG AsB .e $syF AsB. !E f * gF AsB # !q

0 g AyB c !q

0 e$stu At $ yB f At $ yB dt d dy .

y 7 t.u At $ yB # 0 !E f * gF AsB # !q

0 e$st c !q

0 u At $ yB f At $ yBg AyB dy d dt ,

u At $ yB !E f * gF AsB # !q

0 e$st c ! t

0 f At $ yBg AyB dy d dt .s 7 a

†This is permitted since, for each the absolute value of the integrand is integrable on A0, q B ! A0, q B.s 7 a,

For the initial value problem (1), recall that we found

It now follows from the convolution theorem that

Thus we have obtained an integral representation for the solution to the initial value problem (1) for any forcing function that is piecewise continuous on and of exponential order.

Use the convolution theorem to solve the initial value problem

(11)

where is piecewise continuous on and of exponential order.

Let and Taking the Laplace transform of both sides of the differential equation in (11) and using the initial conditions gives

Solving for we have

Hence,

Referring to the table of Laplace transforms on the inside back cover, we find

so we can now express

Thus

is the solution to the initial value problem (11). ◆

y AtB # et " ! t 0 sinh At $ yBg AyB dy

!$1 e 1 s2 $ 1

G AsB f AtB # sinh t * g AtB . !Esinh tF AsB # 1

s2 $ 1 ,

# et " !$1 e 1 s2 $ 1

G AsB f AtB . y AtB # !$1 e 1

s $ 1 f AtB " !$1 e 1

s2 $ 1 G AsB f AtB

Y AsB # s " 1 s2 $ 1

" a 1 s2 $ 1

bG AsB # 1 s $ 1

" a 1 s2 $ 1

bG AsB . Y AsB,s2Y AsB $ s $ 1 $ Y AsB # G AsB .

G AsB # !EgF AsB.Y AsB # !EyF AsB 30, q Bg AtB y– $ y # g AtB ; y A0B # 1 , y¿ A0B # 1 ,

30, q Bg AtB y AtB # sin t * g AtB # ! t

0 sin At $ yBg AyB dy .

Y AsB # a 1 s2 " 1

bG AsB # !Esin tF AsB !EgF AsB . Section 7.7 Convolution 399

Example 1

Solution

Use the convolution theorem to find

Write

Since it follows from the convolution theorem that

†

◆

As the preceding example attests, the convolution theorem is useful in determining the inverse transforms of rational functions of s. In fact, it provides an alternative to the method of partial fractions. For example,

and all that remains in finding the inverse is to compute the convolution

In the early 1900s, V. Volterra introduced integro-differential equations in his study of population growth. These equations enabled him to take into account “hereditary influences.” In certain cases, these equations involved a convolution. As the next example shows, the convolution theorem helps to solve such integro-differential equations.

Solve the integro-differential equation

(12)

Equation (12) can be written as

(13) y¿ AtB # 1 $ y AtB * e$2t .

y¿ AtB # 1 $ ! t 0

y At $ yBe$2y dy , y A0B # 1 .

eat * ebt.

!$1 e 1As $ aB As $ bB f AtB # !$1 e a 1s $ ab a 1s $ bb f AtB # eat * ebt ,

# sin t $ t cos t

2 .

# 1 2 c sin t

2 $

sin A$tB 2 d $ 1

2 t cos t

# 1 2 c sin A2y $ tB

2 d t

0 $

1 2

t cos t

# 1 2 !

0 3 cos A2y $ tB $ cos t 4dy

!$1 e 1As2 " 1B2 f AtB # sin t * sin t # ! t0 sin At $ yB sin y dy !Esin tF AsB # 1/ As2 " 1B,

1As2 " 1B2 # a 1s2 " 1b a 1s2 " 1b . !$1E1/ As2 " 1B2F.

400 Chapter 7 Laplace Transforms

Example 2

Solution

Example 3

Solution

†Here we used the identity sin a sin b # 12 3 cos Ab $ aB $ cos Ab " aB 4 .

Let Taking the Laplace transform of (13) (with the help of the convolution theorem) and solving for we obtain

Hence, ◆

The transfer function of a linear system is defined as the ratio of the Laplace transform of the output function to the Laplace transform of the input function under the assumption that all initial conditions are zero. That is, If the linear system is governed by the differential equation

(14)

where a, b, and c are constants, we can compute the transfer function as follows. Take the Laplace transform of both sides of (14) to get

Because the initial conditions are assumed to be zero, the equation reduces to

Thus the transfer function for equation (14) is

(15)

You may note the similarity of these calculations to those for finding the auxiliary equation for the homogeneous equation associated with (14) (recall Section 4.2, page 158). Indeed, the first step in inverting would be to find the roots of the denominator

which is identical to solving the characteristic equation for (14). The function is called the impulse response function for the system

because, physically speaking, it describes the solution when a mass–spring system is struck by a hammer (see Section 7.8). We can also characterize as the unique solution to the homo- geneous problem

(16)

Indeed, observe that taking the Laplace transform of the equation in (16) gives

(17)

Substituting in and and solving for yields

which is the same as the formula for the transfer function given in equation (15).

H AsB # 1 as2 " bs " c

H AsBh¿ A0B # 1/ah A0B # 0a 3 s2H AsB $ sh A0B $ h¿ A0B 4 " b 3 sH AsB $ h A0B 4 " cH AsB # 0 . ah– " bh¿ " ch # 0 ; h A0B # 0 , h¿ A0B # 1/a .

h AtBh AtB J !$1EHF AtBas2 " bs " c, Y AsB # G AsB / Aas

2 " bs " cB H AsB ! Y AsB

G AsB ! 1as2 ' bs ' c . Aas2 " bs " cBY AsB # G AsB . as2Y AsB $ asy A0B $ ay¿ A0B " bsY AsB $ by A0B " cY AsB # G AsB . ay– " by¿ " cy # g AtB , t 7 0 ,

H AsB # Y AsB /G AsB. g AtB,y AtB H AsBy

AtB # 2 $ e$t . Y AsB # 2 s

$ 1

s " 1 .

Y AsB # As " 1B As " 2B s As " 1B2 # s " 2s As " 1B

as2 " 2s " 1 s " 2

bY AsB # s " 1 s

sY AsB " a 1 s " 2

bY AsB # 1 " 1 s

sY AsB $ 1 # 1 s

$ Y AsB a 1 s " 2

bY AsB, Y AsB # !EyF AsB.

Section 7.7 Convolution 401

One nice feature of the impulse response function h is that it can help us describe the solution to the general initial value problem

(18)

From the discussion of equation (14), we can see that the convolution is the solution to (18) in the special case when the initial conditions are zero (i.e., To deal with nonzero initial conditions, let denote the solution to the corresponding homogeneous initial value problem; that is, solves

(19)

Then, the desired solution to the general initial value problem (18) must be Indeed, it follows from the superposition principle (see Theorem 3 in Section 4.5) that since is a solution to equation (14) and is a solution to the corresponding homogeneous equation, then

is a solution to equation (14). Moreover, since has initial conditions zero,

We summarize these observations in the following theorem.

Ah * gB ¿ A0B " y¿k A0B # 0 " y1 # y1 .Ah * gB A0B " yk A0B # 0 " y0 # y0 , h * gh * g " yk

yk h * g

h * g " yk.

ay– " by¿ " cy # 0 ; y A0B # y0 , y¿ A0B # y1 .yk yk

y0 # y1 # 0B.h * g ay– " by¿ " cy # g AtB ; y A0B # y0 , y¿ A0B # y1 .

402 Chapter 7 Laplace Transforms

Solution Using Impulse Response Function

Theorem 12. Let I be an interval containing the origin. The unique solution to the initial value problem

where a, b, and c are constants and g is continuous on I, is given by

(20)

where h is the impulse response function for the system and is the unique solution to (19).yk

y AtB # Ah * gB AtB " yk AtB # ! t 0

h At $ yBg AyB dy " yk AtB , ay– " by¿ " cy # g ; y A0B # y0 , y¿ A0B # y1 ,

Equation (20) is instructive in that it highlights how the value of y at time t depends on the initial conditions (through and on the nonhomogeneity (through the convolution integral). It even displays the causal nature of the dependence, in that the value of cannot influence until

A proof of Theorem 12 that does not involve Laplace transforms is outlined in Group Project E in Chapter 4.

In the next example, we use Theorem 12 to find a formula for the solution to an initial value problem.

A linear system is governed by the differential equation

(21)

Find the transfer function for the system, the impulse response function, and a formula for the solution.

According to formula (15), the transfer function for (21) is

H AsB # 1 as2 " bs " c

# 1

s2 " 2s " 5 #

1As " 1B2 " 22 .

y– " 2y¿ " 5y # g AtB ; y A0B # 2 , y¿ A0B # $2 .

t ' y.y AtB g AyBg AtByk AtB B

Example 4

Solution

The inverse Laplace transform of is the impulse response function

To solve the initial value problem, we need the solution to the corresponding homoge- neous problem. The auxiliary equation for the homogeneous equation is which has roots Thus a general solution is Choos- ing and so that the initial conditions in (21) are satisfied, we obtain

Hence, a formula for the solution to the initial value problem (21) is

◆Ah * gB AtB " yk AtB # 12 ! t0e$At$yB sin 32 At $ yB 4g AyBdy " 2e$t cos 2t . yk AtB # 2e$t cos 2t.C2C1 C1e$t cos 2t " C2e$t sin 2t.r # $1 + 2i.

r 2 " 2r " 5 # 0,

# 1 2

e$t sin 2t .

h AtB # !$1EHF AtB # 1 2

!$1 e 2As " 1B2 " 22 f AtB H AsB

Section 7.7 Convolution 403

7.7 EXERCISES In Problems 1–4, use the convolution theorem to obtain a formula for the solution to the given initial value prob- lem, where is piecewise continuous on and of exponential order.

2. 3.

In Problems 5–12, use the convolution theorem to find the inverse Laplace transform of the given function.

5. 6.

7. 8.

9. 10.

11.

12.

13. Find the Laplace transform of

f AtB J ! t 0 At $ yBe3ydy .

s " 1As2 " 1B2 cHint: s

s $ 1 # 1 "

1 s $ 1

. dsAs $ 1B As " 2B 1

s3 As2 " 1BsAs2 " 1B2 1As2 " 4B214As " 2B As $ 5B

1As " 1B As " 2B1s As2 " 1B

y¿ A0B # 1y– " y # g AtB ; y A0B # 0 , y¿ A0B # 1y A0B # 1 , y– " 4y¿ " 5y # g AtB ; y¿ A0B # 0y– " 9y # g AtB ; y A0B # 1 ,

y¿ A0B # 1y A0B # $1 , y– $ 2y¿ " y # g AtB ; 3 0, q Bg AtB 14. Find the Laplace transform of

In Problems 15–22, solve the given integral equation or integro-differential equation for

15.

16.

17.

18.

19.

20.

21.

22.

In Problems 23–28, a linear system is governed by the given initial value problem. Find the transfer function

y¿ AtB $ 2 ! t 0

et$yy AyB dy # t , y A0B # 2y A0B # 1 y¿ AtB " y AtB $ ! t

0 y AyBsin At $ yB dy # $sin t ,

y¿ AtB " ! t 0 At $ yBy AyB dy # t , y A0B # 0

y AtB " ! t 0 At $ yB2y AyB dy # t3 " 3

y AtB " ! t 0 At $ yBy AyB dy # t2

y AtB " ! t 0 At $ yBy AyB dy # 1

y AtB " ! t 0

et$yy AyB dy # sin t y AtB " 3 ! t

0 y AyBsin At $ yB dy # ty AtB.

f AtB J ! t 0

ey sin At $ yB dy .

for the system and the impulse response function and give a formula for the solution to the initial

value problem. 23.

24. 25.

26.

27.

28.

In Problems 29 and 30, the current in an RLC circuit with voltage source is governed by the initial value problem

where (see Figure 7.25). For the given constants R, L, C, a, and b, find a formula for the solu- tion in terms of e AtB .I AtB

e AtB # E¿ AtB I¿ A0B # b ,I A0B # a , LI– AtB " RI¿ AtB " 1

C I AtB # e AtB ,

E AtB I AtB y¿ A0B # 1y A0B # 0 , y– $ 4y¿ " 5y # g AtB ; y¿ A0B # 2y A0B # 0 , y– $ 2y¿ " 5y # g AtB ; y¿ A0B # 8y A0B # 0 , y– " 2y¿ $ 15y # g AtB ; y¿ A0B # 8y A0B # 1 , y– $ y¿ $ 6y # g AtB ;

y¿ A0B # 0y A0B # 2 , y– $ 9y # g AtB ; y¿ A0B # $3y A0B # 2 , y– " 9y # g AtB ;

h AtBH AsB 404 Chapter 7 Laplace Transforms

29. Ω, H, F, A, A/sec.

30. Ω, H, F, A, A/sec.

31. Use the convolution theorem and Laplace transforms to compute

32. Use the convolution theorem and Laplace transforms to compute

33. Prove property (5) in Theorem 10.

34. Prove property (6) in Theorem 10.

35. Use the convolution theorem to show that

where

36. Using Theorem 5 in Section 7.3 and the convolution theorem, show that

where

37. Prove directly that if is the impulse response func- tion characterized by equation (16), then for any con- tinuous we have # [Hint: Use Leibniz’s rule, described in Group Project E of Chapter 4.]

Ah * gB ¿ A0B # 0.Ah * gB A0Bg AtB, h AtBF AsB # !E f F AsB. # t !

0 f AyB dy $ ! t

0 y f AyB dy ,

! t

0 ! y

0 f AzB dz dy # !$1 e F AsB

s2 f AtB

F AsB # !E f F AsB. !$1 e F AsB

s f AtB # ! t

0 f AyB dy ,

1 * t * t2.

1 * 1 * 1.

b # $8 a # 2C # 1 /410L # 10R # 80

b # 8 a # $1C # 0.005L # 5R # 20

Resistance R

Voltage source

Capacitance C

Inductance L

Figure 7.25 Schematic representation of an RLC series circuit

7.8 IMPULSES AND THE DIRAC DELTA FUNCTION In mechanical systems, electrical circuits, bending of beams, and other applications, one encounters functions that have a very large value over a very short interval. For example, the strike of a hammer exerts a relatively large force over a relatively short time, and a heavy weight concentrated at a spot on a suspended beam exerts a large force over a very small section of the beam. To deal with violent forces of short duration, physicists and engineers use the delta function introduced by Paul A. M. Dirac. Relaxing our standards of rigor for the moment, we present the following somewhat informal definition.

By shifting the argument of we have and

(3)

for any function that is continuous on an interval containing It is obvious that is not a function in the usual sense; instead it is an example of

what is called a generalized function or a distribution. Despite this shortcoming, the Dirac delta function was successfully used for several years to solve various physics and engineering problems before Laurent Schwartz mathematically justified its use!

A heuristic argument for the existence of the Dirac delta function can be made by consid- ering the impulse of a force over a short interval. If a force is applied from time to time

then the impulse due to " is the integral

Impulse

By Newton’s second law, we see that

(4)

where m denotes mass and denotes velocity. Since represents the momentum, we can interpret equation (4) as saying: The impulse equals the change in momentum.

When a hammer strikes an object, it transfers momentum to the object. This change in momentum takes place over a very short period of time, say, If we let represent the force due to the hammer, then the area under the curve is the impulse or change in momentum (see Figure 7.26 on page 406). If, as is illustrated in Figure 7.27 on page 406, the same change in momentum takes place over shorter and shorter time intervals—say, or —then the average force must get greater and greater in order for the impulses (the areas under the curves to remain the same. In fact, if the forces having the same impulse act, respectively, over the inter- vals where as then approaches a function that is zero for but has an infinite value for Moreover, the areas under the ’s have a common value. Normalizing this value to be 1 gives

for all n .! q

$q "n AtB dt # 1

"nt # t0. t ) t0"nn S q,tn S t03 t0, tn 4 , "n "n)

3 t0, t3 43 t0, t2 4"1AtB "1 AtB3 t0, t1 4 .

myy

! t1

" AtB dt # ! t1 t0

m dy dt

dt # my At1B $ my At0B , J !

" AtB dt .t1, t0" AtB

d At $ aB t # a.f AtB !

$q f AtBd At $ aB dt # f AaB

d At $ aB # 0, t ) a,d AtB,

Section 7.8 Impulses and the Dirac Delta Function 405

Dirac Delta Function

Definition 10. The Dirac delta function is characterized by the following two properties:

(1)

and

(2)

for any function that is continuous on an open interval containing t # 0.f AtB! "

%" f AtBD AtB dt ! f A0B

D AtB ! e0 , t * 0 , “infinite,” t ! 0 ,

d AtB

When we derive from the limiting properties of the ’s a “function” that satisfies property (1) and the integral condition

(5)

Notice that (5) is a special case of property (2) that is obtained by taking It is inter- esting to note that (1) and (5) actually imply the general property (2) (see Problem 33).

The Laplace transform of the Dirac delta function can be quickly derived from property (3). Since for then setting in (3), we find for

Thus, for

(6)

An interesting connection exists between the unit step function and the Dirac delta func- tion. Observe that as a consequence of equation (5) and the fact that is zero for and for we have

(7)

If we formally differentiate both sides of (7) with respect to t (in the spirit of the fundamental theorem of calculus), we find

d At $ aB # u¿ At $ aB . # u At $ aB .!

$q d Ax $ aBdx # e0 , t 6 a ,

1 , t 7 a

x 7 a, x 6 ad Ax $ aB

!ED At % aB F AsB ! e$as .a ' 0, !

0 e$std At $ aBdt # !q

$q e$std At $ aBdt # e$as . a ' 0f

AtB # e$stt ) a,d At $ aB # 0 f AtB $ 1.!

$q d AtB dt # 1 .

d"nt0 # 0,

406 Chapter 7 Laplace Transforms

t00 t

" 1

Impulse

Figure 7.26 Force due to a blow from a hammer

t00 t

t3 t2 t1

" 3

" 2

" 1

Figure 7.27 Forces with the same impulse

Thus it appears that the Dirac delta function is the derivative of the unit step function. That is, in fact, the case if we consider “differentiation” in a more general sense.†

The Dirac delta function is used in modeling mechanical vibration problems involving an impulse. For example, a mass–spring system at rest that is struck by a hammer exerting an impulse on the mass might be governed by the symbolic initial value problem

(8)

where denotes the displacement from equilibrium at time t. We refer to this as a symbolic problem because while the left-hand side of equation (8) represents an ordinary function, the right-hand side does not. Consequently, it is not clear what we mean by a solution to problem (8). Because is zero everywhere except at t # 0, one might be tempted to treat (8) as a homogeneous equation with zero initial conditions. But the solution to the latter is zero every- where, which certainly does not describe the motion of the spring after the mass is struck by the hammer.

To define what is meant by a solution to (8), recall that is depicted as the limit of forces having unit impulse and acting over shorter and shorter intervals. If we let be the solution to the initial value problem

(9)

where is replaced by then we can think of the solution to (8) as the limit (as of the solutions

For example, let

Taking the Laplace transform of equation (9), we find

and so

Now

Hence,

(10)

# en $ n cos t , 0 6 t 6 1/n , n cos At $ 1/nB $ n cos t , 1/n 6 t .

yn AtB # n $ n cos t $ 3n $ n cos At $ 1/nB 4 u At $ 1/nB n

s As2 " 1B # ns $ nss2 " 1 # !En $ n cos tF AsB . Yn AsB # ns As2 " 1B $ e$s/n ns As2 " 1B . As2 " 1BYn AsB # ns A1 $ e$s/nB , "n AtB J n $ nu At $ 1/nB # en , 0 6 t 6 1/n ,

0 , otherwise .

yn AtB. n S q Bx AtB"n,d y–n " yn # "n AtB ; yn A0B # 0 , y¿n A0B # 0 ,

yn AtB"n AtB d AtB d AtB

x AtBx– " x # d AtB ; x A0B # 0 , x¿ A0B # 0 ,

Section 7.8 Impulses and the Dirac Delta Function 407

†See Distributions, Complex Variables, and Fourier Transforms, by H. J. Bremermann (Addison-Wesley, Reading, Mass., 1965).

Fix Then for n large enough, we have Thus,

Also, for Therefore,

Hence, the solution to the symbolic initial value problem (8) is Fortunately, we do not have to go through the tedious process of solving for each in

order to find the solution x of the symbolic problem. It turns out that the Laplace transform method when applied directly to (8) yields the derived solution Indeed, simply taking the Laplace transform of both sides of (8), we obtain from (6) (with a # 0)

which gives

A peculiarity of using the Dirac delta function is that the solution of the sym- bolic initial value problem (8) does not satisfy both initial conditions; that is, This reflects the fact that the impulse is applied at t # 0. Thus, the momentum (observe that in equation (8) the mass equals 1) jumps abruptly from to

In the next example, the Dirac delta function is used in modeling a mechanical vibration problem.

A mass attached to a spring is released from rest 1 m below the equilibrium position for the mass–spring system and begins to vibrate. After seconds, the mass is struck by a hammer exerting an impulse on the mass. The system is governed by the symbolic initial value problem

(11)

where denotes the displacement from equilibrium at time t. Determine

Let Since

and !Ed At $ pB F AsB # e$ps ,!Ex–F AsB # s2X AsB $ s X AsB # !ExF AsB. x AtB.x AtB

d2x dt2

" 9x # 3d At $ pB ; x A0B # 1 , dx dt A0B # 0 ,

x¿ A0"B # 1.x¿ A0B # 0 x¿d AtB x¿ A0B # 1 ) 0.x AtB # sin t

x AtB # !$1 e 1 s2 " 1

f AtB # sin t . X AsB # 1

s2 " 1 ,

As2 " 1BX AsB # 1 , x AtB. yn

x AtB # sin t.limnSq yn AtB # sin t .

t # 0, we have limnSq yn A0B # 0 # sin 0. # $

d dt Acos tB # sin t .

# $lim hS0

cos At " hB $ cos t h

(where h # $1/n) ,

# $ lim nSq

cos At $ 1/nB $ cos t $1/n

lim nSq

yn AtB # limnSq 3n cos At $ 1/nB $ n cos t 41/n 6 t.t 7 0. 408 Chapter 7 Laplace Transforms

Example 1

Solution

taking the Laplace transform of both sides of (11) and solving for yields

Using the translation property (cf. page 386) to determine the inverse Laplace transform of we find

The graph of is given in color in Figure 7.28. For comparison, the dashed curve depicts the displacement of an undisturbed vibrating spring. Note that the impulse effectively adds 3 units to the momentum at time ◆

In Section 7.7 we defined the impulse response function for

(12)

as the function where is the transfer function. Recall that is the ratio

H AsB J Y AsB G AsB ,

H AsBH AsBh AtB J !$1EHF AtB,ay– " by¿ " cy # g AtB t # p.

x AtB # #

cos 3t , t 6 p ,22 cos a3t " p 4 b , p 6 t .

# e cos 3t , t 6 p , cos 3t $ sin 3t , p 6 t

x AtB # cos 3t " 3 sin 3 At $ pB 4u At $ pBX AsB, # !Ecos 3tF AsB " e$ps!Esin 3tF AsB .X AsB #

s s2 " 9

" e$ps 3

s2 " 9

s2X AsB $ s " 9X AsB # 3e$ps X AsB

Section 7.8 Impulses and the Dirac Delta Function 409

x(t)

0 3

3 4 12

−1

−

Figure 7.28 Displacement of a vibrating spring that is struck by a hammer at t # p

where is the Laplace transform of the solution to (12) with zero initial conditions and is the Laplace transform of It is important to note that and hence does not depend on the choice of the function in (12) [see equation (15) in Section 7.7]. However, it is useful to think of the impulse response function as the solution of the symbolic initial value problem

(13)

Indeed, with we have and hence Consequently So we see that the function is the response to the impulse for a mechanical system governed by the symbolic initial value problem (13).

d AtBh AtB h AtB # y AtB.H AsB # Y AsB.G AsB # 1,g AtB # d AtB, ay& ' by$ ' cy ! D AtB ; y A0B ! 0 , y$ A0B ! 0 .

g AtB h AtB,H AsB,g AtB. G AsBY AsB

410 Chapter 7 Laplace Transforms

7.8 EXERCISES In Problems 1–6, evaluate the given integral.

In Problems 7–12, determine the Laplace transform of the given generalized function.

7. 8. 9. 10.

11. 12.

In Problems 13–20, solve the given symbolic initial value problem. 13.

14.

15. y A0B # 2 , y¿ A0B # $2y– " 2y¿ $ 3y # d At $ 1B $ d At $ 2B ; y A0B # 1 , y¿ A0B # 1y– " 2y¿ " 2y # d At $ pB ; w A0B # 0 , w¿ A0B # 0w– " w # d At $ pB ;

etd At $ 3Bd At $ pBsin t t3d At $ 3Btd At $ 1B 3d At $ 1Bd At $ 1B $ d At $ 3B

! 1

$1 Acos 2tBd AtB dt

! q

0 e$2td At $ 1B dt

! q

$q e$2td At " 1B dt

! q

$q Asin 3tBd at $ p

2 bdt!

$q e3td AtB dt

! q

$q At2 $ 1Bd AtB dt 16.

17.

18.

19.

20.

In Problems 21–24, solve the given symbolic initial value problem and sketch a graph of the solution. 21.

22.

23.

24.

In Problems 25–28, find the impulse response function by using the fact that is the solution to the sym-

bolic initial value problem with and zero ini- tial conditions. 25. 26. 27. 28. y– $ y # g AtBy– $ 2y¿ " 5y # g AtBy– $ 6y¿ " 13y # g AtB

y– " 4y¿ " 8y # g AtB g AtB # d AtBh AtBh AtB

y A0B # 0 , y¿ A0B # 1y– " y # d At $ pB $ d At $ 2pB ; y A0B # 0 , y¿ A0B # 1y– " y # $d At $ pB " d At $ 2pB ;

y¿ A0B # 1y A0B # 0 , y– " y # d At $ p/2B ; y¿ A0B # 1y A0B # 0 , y– " y # d At $ 2pB ;

y A0B # 2 , y¿ A0B # $5y– " 5y¿ " 6y # e$td At $ 2B ; w A0B # 0 , w¿ A0B # 4w– " 6w¿ " 5w # etd At $ 1B ;

y¿ A0B # 3y A0B # 0 , y– $ y¿ $ 2y # 3d At $ 1B " et ; y¿ A0B # 2y A0B # 0 , y– $ y # 4d At $ 2B " t2 ;

y A0B # 2 , y¿ A0B # 2y– $ 2y¿ $ 3y # 2d At $ 1B $ d At $ 3B ;

29. A mass attached to a spring is released from rest 1 m below the equilibrium position for the mass–spring system and begins to vibrate. After sec, the mass is struck by a hammer exerting an impulse on the mass. The system is governed by the symbolic initial value problem

where denotes the displacement from equilib- rium at time t. What happens to the mass after it is struck?

30. You have probably heard that soldiers are told not to march in cadence when crossing a bridge. By solv- ing the symbolic initial value problem

explain why soldiers are so instructed. [Hint: See Section 4.10.]

31. A linear system is said to be stable if its impulse response function remains bounded as If the linear system is governed by

where b and c are not both zero, show that the system is stable if and only if the real parts of the roots to

are less than or equal to zero. 32. A linear system is said to be asymptotically stable

if its impulse response function satisfies as If the linear system is governed by

show that the system is asymptotically stable if and only if the real parts of the roots to

are strictly less than zero. 33. The Dirac delta function may also be characterized

by the properties

and ! q

$q d AtB dt # 1 .

d AtB # e 0 , t ) 0 , “infinite,” t # 0 ,

ar 2 " br " c # 0

ay– " by¿ " cy # g AtB ,t S "q. h AtB S 0

ar 2 " br " c # 0

ay– " by¿ " cy # g AtB , t S "q.h AtB

y¿ A0B # 0 ,y A0B # 0 , y– " y # a

k#1 d At $ 2kpB ;

x AtB dx dt A0B # 0 ,x A0B # 1 ,

d2x dt2

" 9x # $3d at $ p 2 b

p /2

Section 7.8 Impulses and the Dirac Delta Function 411

Formally using the mean value theorem for definite integrals, verify that if is continuous, then the above properties imply

34. Formally using integration by parts, show that

Also show that, in general,

35. Figure 7.29 shows a beam of length that is imbedded in a support on the left side and free on the right. The vertical deflection of the beam a distance x from the support is denoted by If the beam has a concentrated load L acting on it in the center of the beam, then the deflection must satisfy the symbolic boundary value problem

where E, the modulus of elasticity, and I, the moment of inertia, are constants. Find a formula for the displacement in terms of the constants

and I. [Hint: Let and First solve the fourth-order symbolic initial value problem and then use the conditions

to determine A and B.]y‡ A2lB # 0 y– A2lB # y‡ A0B # B.y– A0B # Al, L, E, y AxB

y A0B # y¿ A0B # y– A2lB # y‡ A2lB # 0 ,EIy A4B AxB # Ld Ax $ lB ; y AxB.

! q

$q f AtBdAnB AtB dt # A$1Bnf AnB A0B .

! q

$q f AtBd¿ AtB dt # $f ¿ A0B .

! q

$q f AtBd AtB dt # f A0B .

f AtB

Imbedded in support

y(x)

x Free

Figure 7.29 Beam imbedded in a support under a concentrated load at x # l

412 Chapter 7 Laplace Transforms

7.9 SOLVING LINEAR SYSTEMS WITH LAPLACETRANSFORMS We can use the Laplace transform to reduce certain systems of linear differential equations with initial conditions to a system of linear algebraic equations, where again the unknowns are the transforms of the functions that make up the solution. Solving for these unknowns and taking their inverse Laplace transforms, we can then obtain the solution to the initial value problem for the system.

Solve the initial value problem

(1)

Taking the Laplace transform of both sides of the differential equations gives

(2)

Let and Then, we have, by Theorem 4 on page 361,

Substituting these expressions into system (2) and simplifying, we find

(3)

To eliminate from the system, we multiply the first equation by and the second by 2 and then add to obtain

This simplifies to

X AsB # 4s $ 2As " 4B As $ 2B .

3 s As " 2B $ 8 4X AsB # As " 2B A4s2 " 4B s2

$ 10s2 " 4s " 8

s2 .

As " 2BY AsB $4X AsB " As " 2BY AsB # $ 5s2 " 2s " 4

s2 .

sX AsB $ 2Y AsB # 4s2 " 4 s2

!Ey¿F AsB # sY AsB $ y A0B # sY AsB " 5 .!Ex¿F AsB # sX AsB $ x A0B # sX AsB $ 4 , Y AsB J !EyF AsB.X AsB J !ExF AsB

!Ey¿F AsB " 2!EyF AsB $ 4!ExF AsB # $ 4 s2

$ 2 s .

!Ex¿F AsB $ 2!EyF AsB # 4 s2

y¿ AtB " 2y AtB $ 4x AtB # $4t $ 2 ; y A0B # $5 .x¿ AtB $ 2y AtB # 4t ; x A0B # 4 , Example 1

Solution

To compute the inverse transform, we first write in the partial fraction form

Hence, from the Laplace transform table on the inside back cover, we find that

(4)

To determine we could solve system (3) for and then compute its inverse Laplace transform. However, it is easier just to solve the first equation in system (1) for in terms of

Thus,

Substituting from equation (4), we find that

(5)

The solution to the initial value problem (1) consists of the pair of functions given by equations (4) and (5). ◆

x AtB, y AtBy AtB # $6e$4t " e2t $ 2t . x AtB

y AtB # 1 2

x¿ AtB $ 2t . x AtB. y AtBY

AsBy AtB,x AtB # 3e $4t " e2t .

X AsB # 3 s " 4

" 1

s $ 2 .

X AsB Section 7.9 Solving Linear Systems with Laplace Transforms 413

7.9 EXERCISES In Problems 1–19, use the method of Laplace transforms to solve the given initial value problem. Here etc., denotes differentiation with respect to t; so does the sym- bol D.

8. 4x " D 3 y 4 # 3 ; y A0B # 4D 3 x 4 " y # 0 ; x A0B # 7 /4 ,x $ AD $ 1B 3 y 4 # 5e

$3t ; y A0B # 4AD $ 4B 3 x 4 " 6y # 9e$3t ; x A0B # $9 ,$x " y¿ $ y # 0 ; y A0B # $5 /2 x¿ $ x $ y # 1 ; x A0B # 0 ,y¿ # x " 2 cos t ; y A0B # 0 x¿ # y " sin t ; x A0B # 2 ,4x $ y¿ $ y # cos t ; y A0B # 0 x¿ $ 3x " 2y # sin t ; x A0B # 0 ,z¿ $ w¿ # z $ w ; w A0B # 0 z¿ " w¿ # z $ w ; z A0B # 1 ,y¿ # 2x " 4y ; y A0B # 0 x¿ # x $ y ; x A0B # $1 ,y¿ # 3y $ 2x ; y A0B # 1 x¿ # 3x $ 2y ; x A0B # 1 ,

x¿, y¿, 9.

10.

11.

12.

13.

14.

15.

16.

17. x– $ x¿ $ 2y # $et$2 ; x¿ A2B # 1 , y A2B # 1x¿ " x $ y¿ # 2 At $ 2Bet$2 ; x A2B # 0 , 2x " y¿ " y # sin t " 3 cos t ; y ApB # 3x¿ $ 2x " y¿ # $ Acos t " 4 sin tB ; x ApB # 0 , x¿ " x $ y¿ # t2 " 2t $ 1 ; y A1B # 0x¿ $ 2y # 2 ; x A1B # 1 ,

y¿ A0B # 0y– # x " 1 $ u At $ 1B ; y A0B # 0 , x¿ A0B # 0 ,x– # y " u At $ 1B ; x A0B # 1 , x " y¿ # 0 ; y A0B # 1

x¿ $ y¿ # Asin tBu At $ pB ; x A0B # 1 , 2x¿ " y– # u At $ 3B ; y¿ A0B # $1 x¿ " y # x ; x A0B # 0 , y A0B # 1 ,x " y¿ # 0 ; y A0B # 0 x¿ " y # 1 $ u At $ 2B ; x A0B # 0 ,y¿ A0B # $1x " y– # $1 ; y A0B # 1 ,

x¿ A0B # 1 ,x– " y # 1 ; x A0B # 1 , $3x– " 2y– # 3x $ 4y ; y A0B # 4 , y¿ A0B # $9 x– " 2y¿ # $x ; x A0B # 2 , x¿ A0B # $7 ,

The use of the Laplace transform helps to simplify the process of solving initial value problems for certain differential and integral equations, especially when a forcing function with jump discontinuities is involved. The Laplace transform of a function is defined by

for all values of s for which the improper integral exists. If is piecewise continuous on and of exponential order [that is, grows no faster than a constant times as

], then exists for all The Laplace transform can be interpreted as an integral operator that maps a function

to a function The transforms of commonly occurring functions appear in Table 7.1, page 359,F AsB. f AtBs 7 a.!E f F AsBt S q eat0 f AtB 0a30, q B f AtB

!E f F AsB J !q 0

e%stf AtB dt f AtB!E f F

18.

19.

20. Use the method of Laplace transforms to solve

[Hint: Let and then solve for c.] 21. For the interconnected tanks problem of Section 5.1,

page 242, suppose that the input to tank A is now controlled by a valve which for the first 5 min deliv- ers 6 L/min of pure water, but thereafter delivers 6 L/min of brine at a concentration of 2 kg/L. Assuming that all other data remain the same (see Figure 5.1, page 242), determine the mass of salt in each tank for t % 0 if and .

22. Recompute the coupled mass–spring oscillator motion in Problem 1, Exercises 5.6 (page 289), using Laplace transforms.

In Problems 23 and 24, find a system of differential equa- tions and initial conditions for the currents in the

y0 # 4x0 # 0

y A0B # c 4x " y¿ # 6 ; y A1B # 4 . x– " y¿ # 2 ; x A0B # 3 , x¿ A0B # 0 , z¿ # 4x " y $ 3z ; z A0B # $12y¿ # $x " 2y " z ; y A0B # 2 , x¿ # 3x " y $ 2z ; x A0B # $6 ,x " y¿ $ z # 3 ; z A0B # $2 x¿ $ z¿ # 0 ; y A0B # 0 ,x¿ $ 2y # 0 ; x A0B # 0 ,

414 Chapter 7 Laplace Transforms

networks given by the schematic diagrams; the initial currents are all assumed to be zero. Solve for the currents in each branch of the network. (See Section 5.7 for a discussion of electrical networks.)

6 V

2 Ω

I2 I3

1 Ω

0.2 H

0.1 H

Figure 7.30 RL network for Problem 23

23.

50 V

I1 I3 I2

20 Ω10 Ω

0.005 H 0.01 H

Figure 7.31 RL network for Problem 24

24.

Chapter Summary

Chapter Summary 415

and on the inside back cover of this book. The use of these tables is enhanced by several impor- tant properties of the operator

Linearity:

Translation in s: where

Translation in t: where is the step function that equals 1 for t % a and 0 for t * a. If is continuous and , then

where

Convolution Property: where denotes the convolu- tion function

One reason for the usefulness of the Laplace transform lies in the simple formula for the transform of the derivative :

(1) where

This formula shows that by using the Laplace transform, “differentiation with respect to t” can essentially be replaced by the simple operation of “multiplication by s.” The extension of (1) to higher-order derivatives is

(2)

To solve an initial value problem of the form

(3)

via the Laplace transform method, one takes the transform of both sides of the differential equa- tion in (3). Using the linearity of and formula (2) leads to an equation involving the Laplace transform of the (unknown) solution The next step is to solve this simpler equation for

Finally, one computes the inverse Laplace transform of to obtain the desired solution. This last step of finding is often the most difficult. Sometimes it requires a partial frac- tions decomposition, a judicious use of the properties of the transform, an excursion through the Laplace transform tables, or the evaluation of a contour integral in the complex plane.

For the special problem in (3), where a, b, and c are constants, the differential equation is transformed to a simple algebraic equation for Another nice feature of this latter equation is that it incorporates the initial conditions. When the coefficients of the equation in (3) depend on t, the following formula may be helpful in taking the transform:

where

If the forcing function in equation (3) has jump discontinuities, it is often convenient to write in terms of unit step functions before proceeding with the Laplace transformu At $ aBf AtB f AtB

F # !E f F .!Etnf AtB F AsB # A$1Bn dnF dsn

AsB , Y AsB.

!$1EYF Y AsBY AsB. y AtB.Y AsB !

ay– " by¿ " cy # f AtB ; y A0B # a , y¿ A0B # b !E f AnBF AsB # snF AsB $ sn$1f A0B $ sn$2f ¿ A0B $ p $ f An$1B A0B .

F # !E f F .!E f ¿F AsB # sF AsB $ f A0B , f ¿ A f * gB AtB J ! t

0 f At $ yBg AyB dy .

f * g!E f * gF # !E f F!EgF,f # !$1EFF. !$1Ee$asF AsB F AtB # f At $ aBu At $ aB , f A0B # 0f AtB

u At $ aB!Eg AtBu At $ aBF AsB # e$as!Eg At " aBF AsB,F # !E f F.!Eeatf AtB F AsB # F As $ aB, !Eaf " bgF # a!E f F " b!EgF.!.

method. For this purpose, the rectangular window function is useful. The transform of a periodic forcing function with period T is given by

The Dirac delta function is useful in modeling a system that is excited by a large force applied over a short time interval. It is not a function in the usual sense but can be roughly interpreted as the derivative of a unit step function. The transform of is

!Ed At $ aB F AsB # e$as , a ' 0 . d At $ aBd

AtB !E f F AsB # ! T0 e$stf AtB dt

1 $ e$sT .

f AtB ßa,b AtB # u At $ aB $ u At $ bB 416 Chapter 7 Laplace Transforms

REVIEW PROBLEMS

In Problems 1 and 2, use the definition of the Laplace transform to determine .

In Problems 3–10, determine the Laplace transform of the given function.

3. 4. 5. 6. 7. 8. 9.

10. and has period p.

In Problems 11–17, determine the inverse Laplace trans- form of the given function.

11. 12.

13.

14.

15. 16.

17. e$2s A4s " 2BAs $ 1B As " 2B

1As2 " 9B22s2 " 3s $ 1As " 1B2 As " 2B s2 " 16s " 9As " 1B As " 3B As $ 2B

4s2 " 13s " 19As $ 1B As2 " 4s " 13B 2s $ 1

s2 $ 4s " 6 7As " 3B3

f AtBf AtB # cos t, $p/2 & t & p/2 t2u At $ 4BAt " 3B2 $ Aet " 3B2 t cos 6t 7e2t cos 3t $ 2e7t sin 5t e2t $ t3 " t2 $ sin 5t

e3t sin 4tt2e$9t

f AtB # e e$t , 0 & t & 5 , $1 , 5 6 t

f AtB # e 3 , 0 & t & 2 , 6 $ t , 2 6 t

!E f F 18. Find the Taylor series for about t # 0.Then, assuming that the Laplace transform of can be computed term by term, find an expansion for

in powers of .

In Problems 19–24, solve the given initial value problem for using the method of Laplace transforms.

19.

20.

21.

22.

23.

24.

In Problems 25 and 26, find solutions to the given initial value problem.

25.

26.

y A0B # 1 , y¿ A0B # $1ty– " 2 At $ 1By¿ " At $ 2By # 0 ; y¿ A0B # 0y A0B # 0 , ty– " 2 At $ 1By¿ $ 2y # 0 ; y¿ A0B # 0y A0B # 0 , y– $ 4y¿ " 4y # t2et ; y¿ A0B # 1y A0B # 0 , y– " 3y¿ " 4y # u At $ 1B ;

y¿ A0B # 5y A0B # $1 , y– " 9y # 10e2t ; y A0B # 0 , y¿ A0B # $1y– " 2y¿ " 2y # t2 " 4t ;

y¿ A0B # 10y A0B # $3 , y– " 6y¿ " 9y # 0 ; y¿ A0B # $3y A0B # 0 , y– $ 7y¿ " 10y # 0 ;

y AtB 1 /s!E f F AsB f

AtBf AtB # e$t2

In Problems 27 and 28, solve the given equation for

27.

28.

29. A linear system is governed by

Find the transfer function and the impulse response function.

y– $ 5y¿ " 6y # g AtB . y A0B # $1y¿ AtB $ 2 !

0 y AyBsin At $ yB dy # 1 ;

y AtB " ! t 0 At $ yBy AyB dy # e$3t y AtB.

Technical Writing Exercises 417

TECHNICAL WRITING EXERCISES

30. Solve the symbolic initial value problem

In Problems 31 and 32, use Laplace transforms to solve the given system. 31.

32. y A0B # 0x " y¿ # y ; x– " 2y¿ # u At $ 3B ; x A0B # 1 , x ¿A0B # $1 ,

x " y¿ # 1 $ u At $ 2B ; y A0B # 0x¿ " y # 0 ; x A0B # 0 , y A0B # 0 , y¿ A0B # 1 .y– " 4y # d at $ p2b ;

1. Compare the use of Laplace transforms in solving linear differential equations with constant coeffi- cients with the use of logarithms in solving algebraic equations of the form

2. Explain why the method of Laplace transforms works so well for linear differential equations with constant coefficients and integro-differential equa- tions involving a convolution.

3. Discuss several examples of initial value problems in which the method of Laplace transforms cannot be applied.

xr # a.

4. A linear system is said to be asymptotically stable if its impulse response function as Assume the Laplace transform of is a rational function in reduced form with the degree of its numerator less than the degree of its denominator. Explain in detail how the asymptotic stability of the linear system can be characterized in terms of the zeros of the denominator of Give examples.

5. Compare and contrast the solution of initial value problems by Laplace transforms versus the methods of Chapter 4.

H AsB.

h AtB,H AsB, t S "q.h AtB S 0

A Duhamel’s Formulas For a linear system governed by the equation

(1)

where a, b, c are real constants, the function

(2)

where all initial conditions are taken to be zero, is called the transfer function for the system. (As mentioned in Section 7.7, the transfer function depends only on the constants a, b, c of the system; it is not affected by the choice of g.) If the output function is the unit step func- tion then equation (2) yields

The solution (output function) in this special case is called the indicial admittance and is denoted by Hence,

It is possible to express the response of the system to a general input function in terms of and the indicial admittance To derive these relations, proceed as follows:

(a) Show that

(3)

(b) Now apply the convolution theorem to (3) and show that

(4)

where f and are assumed continuous in and t, and are differentiable functions of t. Applying this rule to (4), derive the formulas

(5)

(6)

[Hint: Recall that the initial conditions on A are A A0B # A¿ A0B # 0. 4y AtB # ! t

0 A AyBg¿ At $ yBdy " A AtBg A0B .

y AtB # ! t 0

A¿ At $ yBg AyB dy , a AtB, b AtBy0f / 0t

d dt c ! bAtB

aAtB f Ay, tB dy d # ! bAtBaAtB 0f0t Ay, tB dy " f Ab AtB, tB dbdt AtB $ f Aa AtB, tB dadt AtB , y AtB # d

dt c ! t

0 A At $ yBg AyB dy d # d

dt c ! t

0 A AyBg At $ yB dy d .

!EyF AsB # s!EAF AsB!EgF AsB . A AtB.g AtB g AtBy AtB

!EAF AsB # H AsB /s.A AtB. !EyF AsB # !EuF AsBH AsB # H AsB

s .

u AtB, g AtB H AsB

H AsB J !EyF AsB !EgF AsB # !EoutputF!EinputF ,

ay– " by¿ " cy # g AtB ,

Group Projects for Chapter 7

418

(d) In equations (5) and (6), make the change of variables and show that

(7)

(8)

Equations (5)–(8) are referred to as Duhamel’s formulas, in honor of the French mathe- matician J. M. C. Duhamel. These formulas are helpful in determining the response of the system to a general input since the indicial admittance of the system can be determined experimen- tally by measuring the response of the system to a unit step function.

(e) The impulse response function is defined as where is the transfer function. Show that so that equations (5) and (7) can be written in the form

(9)

We remark that the indicial admittance is the response of the system to a unit step function, and the impulse response function is the response to the unit impulse or delta function (see Sec- tion 7.8). But the delta function is the derivative (in a generalized sense) of the unit step function. Therefore, the fact that is not really surprising.

B Frequency Response Modeling Frequency response modeling of a linear system is based on the premise that the dynamics of a linear system can be recovered from a knowledge of how the system responds to sinusoidal inputs. (This will be made mathematically precise in Theorem 13.) In other words, to determine (or identify) a linear system, all one has to do is observe how the system reacts to sinusoidal inputs.

Let’s assume that we have a linear system governed by

(10)

where p and q are real constants. The function is called the forcing function or input func- tion. When is a sinusoid, the particular solution to (10) obtained by the method of undeter- mined coefficients is the steady-state solution or output function corresponding to We can think of a linear system as a compartment or block into which goes an input function g and out of which comes the output function (see Figure 7.32). To identify a linear system means to determine the coefficients p and q in equation (10).

It will be convenient for us to work with complex variables. A complex number z is usually expressed in the form with real numbers and i denoting We can also express z in polar form, where and Here is called the magnitude and the phase angle of z.u

r A'0Btan u # b /a.r 2 # a2 " b2z # reiu, 1$1.a, bz # a " ib, yss

g AtB.yss AtBg AtB g AtBy– " py¿ " qy # g AtB ,

h AtB # A¿ AtB y AtB # ! t

0 h At $ yBg AyB dy # ! t

0 h AyBg At $ yB dy .

h AtB # A¿ AtB, H AsBh AtB J !$1EHF AtB,h AtB g AtB, y AtB # ! t

0 A At $ wBg¿ AwB dw " A AtBg A0B .

y AtB # ! t 0

A¿ AwBg At $ wB dw , w # t $ y Group Projects for Chapter 7 419

g(t) Input Output

yss(t) Linear system y" + py' + qy = g(t)

Figure 7.32 Block diagram depicting a linear system

The following theorem gives the relationship between the linear system and its response to sinusoidal inputs in terms of the transfer function [see Project A, equation (2)].H AsB

420 Chapter 7 Laplace Transforms

1.0

0.5

−0.5

−1.0

−1.5

−2.0 0.1 0.2 107420.4

Ph as

e an

gl e

(d eg

)

0.7 1.0 0.1

0 −20 −40 −60 −80

−100 −120 −140 −160 −180

0.2 107420.4 0.7 1.0

Figure 7.33 Bode plots for H AivB # 31 " 0.2 AivB " AivB2 4$1

†Frequency response curves are also discussed in Section 4.10.

Steady-State Solutions to Sinusoidal Inputs

Theorem 13. Let be the transfer function for equation (10). If is finite at with real, then the steady-state solution to (10) for is

(11) yss AtB ! H AiVBeiVt ! H AiVBEcos Vt ' i sin VtF .g AtB # eivtv s # iv,H AsBH AsB

(a) Prove Theorem 13. Hint: Guess and show that (b) Use Theorem 13 to show that if then the steady-state solution to (10) is

where is the polar form for (c) Solve for and in terms of p and q. (d) Experimental results for modeling done by frequency response methods are usually pre-

sented in frequency response† or Bode plots. There are two types of Bode plots. The first is the log of the magnitude of versus the angular frequency using a log scale for The second is a plot of the phase angle or argument of ver- sus the angular frequency using a log scale for The Bode plots for the transfer func- tion are given in Figure 7.33.H AsB # A1 " 0.2s " s2B$1 v.

H AivBN AvBv. vH AivBM AvB N AvBM AvB H AivB.H AivB # M AvBeiNAvBM AvBsin 3vt " N AvB 4 , g AtB # sin vt, A # H AivB. 4yss AtB # Ae

ivt3

Sketch the Bode plots of the linear system governed by equation (10) with p # 0.4 and q # 1.0. Use # 0.3, 0.6, 0.9, 1.2, and 1.5 for the plot of and # 0.5, 0.8, 1, 2, and 5 for the plot of

(e) Assume we know that q # 1. When we input a sine wave with the system settles into a steady-state sinusoidal output with magnitude Find p and thus identify the linear system.

(f ) Suppose a sine wave input with produces a steady-state sinusoidal output with magnitude and that when then Find p and q and thus identify the system.

M A4B # 0.1.v # 4,M A2B # 0.5 v # 2 M A2B # 0.333.v # 2,

N AvB. vM AvBv

In most applications there are some inaccuracies in the measurement of the magnitudes and frequencies. To compensate for these errors, sinusoids with several different frequencies are used as input. A least-squares approximation for p and q is then found. For a discussion of frequency response modeling as a mathematical modeling tool, see the chapter by W. F. Powers, “Modeling Linear Systems by Frequency Response Methods,” in Differential Equations Models, by M. Braun, C. Coleman, and D. Drew (eds.) (Springer-Verlag, New York, 1983), Chapter 9. Additional examples may be found in Schaum’s Outline on Feedback and Control Systems, by J. J. DiStefano, A. R. Stubberud, and I. J. Williams (McGraw-Hill, New York, 1995, 3rd edition), Chapter 15.

C Determining System Parameters In mechanical design one sometimes must determine system parameters even though information on the system forces is incomplete. For example, the differential equation governing the motion of an externally forced damped mass–spring oscillator was shown in Section 4.1 (page 153) to be

where is the external force; the other parameters are defined in Section 4.1. Assume the sys- tem is underdamped see Section 4.9) and starts from rest and that the force is bounded: for all t.

(a) Show that the transforms and of and are related by

(b) Use the convolution theorem to derive the formula

where

and also by

(d) Suppose the mass m # 5 kg, the spring constant k # 3000 N/m, and the force is bounded by A # 10 N. What range of values for the damping constant b will ensure a displacement of 1 cm (0.01 m) or less?

0 x AtB 0 & 2A/bb . 0 x AtB 0 & At/bm x AtB b J 1

2m 24mk $ b2 .

x AtB # 1 bm

! t

0 f At $ yBe$by/2m sin by dy ,

X AsB # 1 ms2 " bs " k

F AsB . f AtBx AtBF AsBX AsB 0 f AtB 0 & A 3 x A0B # 0, x¿ A0B # 0 4Ab2 6 4mk;f AtB

m d 2x

dt2 ' b dx

dt ' kx ! f AtB ,

x AtB

Group Projects for Chapter 7 421

422

Probably the best tool for numerically approximating a function near a particular point is the Taylor polynomial. The formula for the Taylor polynomial of degree n centered at approximating a function possessing n derivatives at is given by

(1)

This polynomial matches the value of f and the values of its derivatives, up to the order of the polynomial, at the point :

For example, the first four Taylor polynomials for expanded around are

(2)

Their efficacy in approximating the exponential function is demonstrated in Figure 8.1. The Taylor polynomial of degree n differs from the polynomial of the next lower degree

only in the addition of a single term:

pn AxB ! pn"1 AxB # f AnB Ax0Bn! Ax " x0Bn ,

p3 AxB ! 1 # x # x22 # x36 . p2 AxB ! 1 # x # x22 , p1 AxB ! 1 # x , p0 AxB ! 1 ,

x0 ! 0,e x,

pAnBn Ax0B ! f AnB Ax0B . p–n Ax0B ! f –Ax0B , p¿n Ax0B ! f ¿Ax0B , pn Ax0B ! f Ax0B ,x0

! a n

j!0

f A jB Ax0B j!

Ax " x0B j . ! #

f ‡Ax0B 3!

Ax " x0B3 # p # f AnB Ax0Bn! Ax " x0Bn pn AxB ! f Ax0B # f ¿Ax0B Ax " x0B # f –Ax0B2! Ax " x0B2

x0,f AxB x0, x0f AxB

Series Solutions of Differential Equations

CHAPTER 8

INTRODUCTION: THE TAYLOR POLYNOMIAL APPROXIMATION8.1

so a listing like (2) is clearly redundant—one can read off and from the formula for In fact, if f is infinitely differentiable, is just the st partial sum of the Taylor series†

(3)

Determine the fourth-degree Taylor polynomials matching the functions and at

For we have for each j ! 0, 1, . . . , so from (1) we obtain

For we have so that

In a similar fashion we find

◆# sin 2

4! Ax " 2B4 .

sin x ! sin 2 # Acos 2B Ax " 2B " sin 2 2!

Ax " 2B2 " cos 2 3!

Ax " 2B3 cos x ! cos 2 " Asin 2B Ax " 2B " cos 2

2! Ax " 2B2 # sin 2

3! Ax " 2B3 # cos 2

4! Ax " 2B4 .

f ¿AxB ! "sin x, f –AxB ! "cos x, f ‡AxB ! sin x, f A4B AxB ! cos x,f AxB ! cos x, ex ! e2 # e2 Ax " 2B # e2

2! Ax " 2B2 # e2

3! Ax " 2B3 # e2

4! Ax " 2B4 .

f A jB A2B ! e2f AxB ! exx0 ! 2. sin xex, cos x,

a! j"0

f A jB Ax0B j!

Ax # x0B j . An # 1Bpn AxBp3 AxB. p2 AxBp0 AxB, p1 AxB,

Section 8.1 Introduction: The Taylor Polynomial Approximation 423

1 + x +

1 + x

2 3

−1

−1−2−3

1 + x + + x 3

6 x2

2 x2

Figure 8.1 Graphs of Taylor polynomials for ex

†Truncated Taylor series were introduced in Section 3.7 (page 132) as a tool for constructing recursive formulas for approximate solutions of differential equations.

Example 1

Solution

To relate this approximation scheme to our theme (the solution of differential equations), we alter our point of view; we regard a differential equation not as a “condition to be satisfied,” but as a prescription for constructing the Taylor polynomials for its solutions. Besides provid- ing a very general method for computing accurate approximate solutions to the equation near any particular “starting” point, this interpretation also provides insight into the role of the initial conditions. The following example illustrates the method.

Find the first few Taylor polynomials approximating the solution around of the initial value problem

To construct

we need the values of etc. The first two are provided by the given initial conditions. The value of can be deduced from the differential equation itself and the values of the lower derivatives:

Now since holds for some interval around we can differentiate both sides to derive

Thus on substituting x ! 0 we deduce, in turn, that

does not exist!

Consequently, we can only construct the Taylor polynomials of degree 0 through 4 for the solution, and is given by

◆

The next example demonstrates the application of the Taylor polynomial method to a nonlinear equation.

! 10 # 5x # 15 2

x2 # 15 2

x3 # 45 8

x4 .

p4 AxB ! 10 # 5x # 152 x2 # 456 x3 # 13524 x4 p4 AxB

y A5B A0B ! 3 # 135 # 28 27

# 0"2/3 # 10 # A p B y A4B A0B ! 3 # 45 # 28

9 # 01/3 # 10 # 14

3 # 04/3 # 5 # 07/3 # 15 ! 135 ,

y‡ A0B ! 3 # 15 # 7 3

# 04/3 # 10 # 07/3 # 5 ! 45 ,

y A5B ! 3y A4B # 28 27

x"2/3y # A p B . y A4B ! 3y‡ # 28

9 x1/3y # 14

3 x4/3y¿ # x7/3y– ,

y‡ ! 3y– # 7 3

x4/3y # x7/3y¿ ,

x0 ! 0,y– ! 3y¿ # x7/3y

y– A0B ! 3y¿ A0B # 07/3y A0B ! 3 # 5 # 0 # 10 ! 15 . y– A0By A0B, y¿ A0B, y– A0B, y‡ A0B,

pn AxB ! y A0B # y¿ A0Bx # y– A0B2! x2 # y‡ A0B3! x3 # p # y AnB A0Bn! xn , y– ! 3y¿ # x7/3y ; y A0B ! 10 , y¿ A0B ! 5 .

x0 ! 0

424 Chapter 8 Series Solutions of Differential Equations

Example 2

Solution

Determine the Taylor polynomial of degree 3 for the solution to the initial value problem

(4)

Using we substitute and into equation (4) and find that To determine we differentiate both sides of the equation in (4) with respect to x, thereby getting an expression for in terms of and That is,

(5)

Substituting and in (5), we obtain

Similarly, differentiating (5) and substituting, we obtain

Thus, the Taylor polynomial of degree 3 is

◆

In a theoretical sense we can estimate the accuracy to which a Taylor polynomial approximates its target function for near Indeed, if we let measure the accuracy of the approximation,

then calculus provides us with several formulas for estimating The most transparent is due to Lagrange: if the st derivative of f exists and is continuous on an interval containing x0 and x, then

(6)

where , although unknown, is guaranteed to lie between x0 and x.†

Figure 8.1 and equation (6) suggest that one might control the error in the Taylor polyno- mial approximation by increasing the degree n of the polynomial (i.e., taking more terms), thereby increasing the factor in the denominator. This possibility is limited, of course, by the number of times f can be differentiated. In Example 2, for instance, the solution did not have a fifth derivative at is “infinite”). Thus, we could not construct nor could we conclude anything about the accuracy of from the Lagrange formula.

However, for Example 3 we could, in theory, compute every derivative of the solution at , and speculate on the convergence of the Taylor series

a q

j!0

y A jB Ax0B j!

Ax " x0B j ! limnSq an j!0

y A jB Ax0B j!

Ax " x0B j x0 ! 0

y AxBp4 AxB p5 AxB,( f A5B A0Bx0 ! 0

An # 1B! j

En AxB " f An$1B AJBAn $ 1B! Ax # x0Bn$1 , An # 1B en.

en AxB J f AxB " pn AxB , en AxBx0.xf AxB pn AxB

p3 AxB ! 0 # x " x2 # 106 x3 ! x " x2 # 53 x3 . y‡ A0B ! 2 A1B"3 A1 # 1B2 " A1B"2 A"2B ! 10 .y‡ AxB ! 2 3 x # y AxB # 1 4"3 31 # y¿ AxB 4 2 " 3 x # y AxB # 1 4"2y– AxB , y– A0B ! A"1B A1B"2 A1 # 1B ! "2 .y¿ A0B ! 1x ! 0, y A0B ! 0, y– AxB ! A"1B 3 x # y AxB # 1 4"2 31 # y¿ AxB 4 .y¿ AxB.x, y AxB,y– AxBy– A0B,

y¿ A0B ! 1.y ! 0x ! 0y A0B ! 0,y¿ ! 1

x # y # 1 , y A0B ! 0 .

Section 8.1 Introduction: The Taylor Polynomial Approximation 425

Example 3

Solution

†Equation (6) is proved by invoking the mean value theorem; see, e.g., Principles of Mathematical Analysis, 3rd ed., by Walter Rudin (McGraw-Hill, New York, 1976).

to the solution . Now for nonlinear equations such as (4), the factor in the Lagrange error formula may grow too rapidly with n, and the convergence can be thwarted. But if the differential equation is linear and its coefficients and nonhomogeneous term enjoy a feature known as analyticity, our wish is granted; the error does indeed diminish to zero as the degree n goes to infinity, and the sequence of Taylor polynomials can be guaranteed to con- verge to the actual solution on a certain (known) interval. For instance, the exponential, sine, and cosine functions in Example 1 all satisfy linear differential equations with constant coeffi- cients, and their Taylor series converge to the corresponding function values. (Indeed, all their derivatives are bounded on any interval of finite length, so the Lagrange formula for their approximation errors approaches zero as n increases, for each value of x.) This topic is the theme for the early sections of this chapter.

In Sections 8.5 and 8.6 we’ll see that solutions to second-order linear equations can exhibit very wild behavior near points x0 where the coefficient of is zero; so wild, in fact, that no Euler or Runge–Kutta algorithm could hope to keep up with them. But a clever modifi- cation of the Taylor polynomial method, due to Frobenius, provides very accurate approxima- tions to the solutions in such regions. It is this latter feature, perhaps, that underscores the value of the Taylor methodology in the current practice of applied mathematics.

y–

f (n#1) AjBy AxB 426 Chapter 8 Series Solutions of Differential Equations

8.1 EXERCISES In Problems 1–8, determine the first three nonzero terms in the Taylor polynomial approximations for the given initial value problem.

9. (a) Construct the Taylor polynomial of degree 3 for the function around

(b) Using the error formula (6), show that

(d) Sketch the graphs of ln x and (on the same axes) for 0 < x < 2.

10. (a) Construct the Taylor polynomial of degree 3 for the function around x ! 0.

f AxB ! 1/ A2 " xBp3 AxB p3 AxB@ ln A1.5B " p3 A1.5B @ .

@ ln A1.5B " p3 A1.5B @ $ A0.5B44 ! 0.015625 . x ! 1.f AxB ! ln x p3 AxB

y– # sin y ! 0 ; y A0B ! 1 , y¿ A0B ! 0y A0B ! 0 , y¿ A0B ! 0 y– AuB # y AuB3 ! sin u ;y– # y ! 0 ; y A0B ! 0 , y¿ A0B ! 1 x– # tx ! 0 ; x A0B ! 1 , x¿ A0B ! 0y¿ ! sin Ax # yB ; y A0B ! 0 y¿ ! sin y # ex ; y A0B ! 0y¿ ! y2 ; y A0B ! 2 y¿ ! x2 # y2 ; y A0B ! 1

(b) Using the error formula (6), show that

(d) Sketch the graphs of and (on the same axes) for

11. Argue that if is a solution to the differential equation on the interval

, where p, q, and g are each twice-differentiable, then the fourth derivative of exists on .

12. Argue that if is a solution to the differential equation on the interval

, where p, q, and g possess derivatives of all orders, then has derivatives of all orders on .

13. Duffing’s Equation. In the study of a nonlinear spring with periodic forcing, the following equation arises:

Let and Find the first three nonzero terms in the Taylor polynomial approxima- tions to the solution with initial values , y¿ A0B ! 1. y A0B ! 0

v ! 10.k ! r ! A ! 1

y– # ky # ry3 ! A cos vt .

Aa, bBfAa, bB y– # p AxBy¿ # q AxBy ! g AxBy ! f AxB

Aa, bBf AxBAa, bB y– # p AxBy¿ # q AxBy ! g AxBy ! f AxB

"2 6 x 6 2. p3 AxB1/ A2 " xB

` 2 3

" p3 a12b ` . ` f a1

2 b " p3 a12b ` ! ` 23 " p3 a12b ` $ 235 .

14. Soft versus Hard Springs. For Duffing’s equation given in Problem 13, the behavior of the solutions changes as r changes sign. When r > 0, the restoring force becomes stronger than for the linear spring Such a spring is called hard. When r < 0, the restoring force becomes weaker than the linear spring and the spring is called soft. Pendu- lums act like soft springs. (a) Redo Problem 13 with Notice that for

the initial conditions the soft and hard springs appear to respond in the same way for t small.

(b) Keeping and change the initial conditions to and Now redo Problem 13 with r ! %1.

(c) Based on the results of part (b), is there a differ- ence between the behavior of soft and hard springs for t small? Describe.

y¿ A0B ! 0.y A0B ! 1v ! 10,k ! A ! 1 y A0B ! 0, y¿ A0B ! 1,r ! "1.

Ar ! 0B.ky # ry3

Section 8.2 Power Series and Analytic Functions 427

15. The solution to the initial value problem

has derivatives of all orders at x ! 0 (although this is far from obvious). Use L’Hôpital’s rule to compute the Taylor polynomial of degree 2 approximating this solution.

16. van der Pol Equation. In the study of the vacuum tube, the following equation is encountered:

Find the Taylor polynomial of degree 4 approximat- ing the solution with the initial values , y¿ A0B ! 0. y A0B ! 1

y– # A0.1B Ay2 " 1By¿ # y ! 0 .

y¿ A0B ! 0y A0B ! 1 , xy– AxB # 2y¿ AxB # xy AxB ! 0 ;

8.2 POWER SERIES AND ANALYTIC FUNCTIONS The differential equations studied in earlier sections often possessed solutions that could be written in terms of elementary functions such as polynomials, exponentials, sines, and cosines. However, many important equations arise whose solutions cannot be so expressed. In the previous chapters, when we encountered such an equation we either settled for expressing the solution as an integral (see Exercises 2.2, Problem 27, page 43) or as a numerical approxi- mation (Sections 3.6, 3.7, and 5.3). However, the Taylor polynomial approximation scheme of the preceding section suggests another possibility. Suppose the differential equation (and initial conditions) permit the computation of every derivative at the expansion point Are there any conditions that would guarantee that the sequence of Taylor polynomials would converge to the solution as the degree of the polynomials tends to infinity:

In other words, when can we be sure that a solution to a differential equation is represented by its Taylor series? As we’ll see, the answer is quite favorable, and it enables a powerful new technique for solving equations.

The most efficient way to begin an exploration of this issue is by investigating the algebraic and convergence properties of generic expressions that include Taylor series—“long polynomials” so to speak, or more conventionally, power series.

Power Series A power series about the point is an expression of the form

(1) a q

n!0 an Ax " x0Bn ! a0 # a1 Ax " x0B # a2 Ax " x0B2 # p ,

lim nSq

a n

j!0

y A jB Ax0B j!

Ax " x0B j ! aq j!0

y A jB Ax0B j!

Ax " x0B j ! y AxB ? y AxB x0.y

AnB

y AxB

where x is a variable and the an’s are constants. We say that (1) converges at the point x ! c if the infinite series (of real numbers) converges; that is, the limit of the partial sums,

exists (as a finite number). If this limit does not exist, the power series is said to diverge at x ! c. Observe that (1) converges at since

But what about convergence for other values of x? As stated in the following Theorem 1, a power series of the form (1) converges for all values of x in some “interval” centered at and diverges for x outside this interval. Moreover, at the interior points of this interval, the power series converges absolutely in the sense that converges. [Recall that absolute convergence of a series implies (ordinary) convergence of the series.]

©qn!0 0an Ax " x0Bn 0 x0 a q

n!0 an Ax0 " x0Bn ! a0 # 0 # 0 # p ! a0 .

x ! x0,

lim NSq

a N

n!0 an Ac " x0Bn ,

©qn!0 an Ac " x0Bn 428 Chapter 8 Series Solutions of Differential Equations

" # " #

Absolute convergence DivergenceDivergence ? ?

x0x0 − x0 +

Figure 8.2 Interval of convergence

Radius of Convergence

Theorem 1. For each power series of the form (1), there is a number called the radius of convergence of the power series, such that (1) converges absolutely for and diverges for (See Figure 8.2.)

If the series (1) converges for all values of x, then When the series (1) con- verges only at then r ! 0.x0,

r ! q. 0 x " x0 0 7 r.0 x " x0 0 6 r r A0 $ r $ q B,

Ratio Test for Power Series

Theorem 2. If, for n large, the coefficients are nonzero and satisfy

then the radius of convergence of the power series is r ! L.©qn!0 an Ax " x0Bn lim nSq ` an an#1

` ! L A0 $ L $ q B , an

Notice that Theorem 1 settles the question of convergence except at the endpoints Thus, these two points require separate analysis. To determine the radius of convergence one method that is often easy to apply is the ratio test.

r, x0 % r.

Remark. We caution that if the ratio does not have a limit, then methods other than the ratio test (e.g., root test) must be used to determine In particular, if infinitely many of the an’s are zero, then the ratio test cannot be directly applied. (However, Problem 7 demon- strates how to apply the result for series containing only “even-order” or “odd-order” terms.)

r. 0an /an#1 0

Determine the convergence set of

(2)

Since we have

By the ratio test, the radius of convergence is Hence, the series (2) converges abso- lutely for and diverges when It remains only to determine what happens when that is, when and .

Set , and the series (2) becomes the harmonic series which is known to diverge. When , the series (2) becomes an alternating harmonic series, which is known to converge. Thus, the power series converges for each x in the half-open interval ; outside this interval it diverges. ◆

For each value of x for which the power series converges, we get a number that is the sum of the series. It is appropriate to denote this sum by since its value depends on the choice of x. Thus, we write

for all numbers x in the convergence interval. For example, the geometric series has the radius of convergence and the sum function that is,

(3)

In this chapter we’ll frequently appeal to the following basic property of power series.

1 1 # x " 1 $ x $ x

2 $ p " a! n"0

xn for #1 6 x 6 1 .

f AxB ! 1/ A1 " xB;r ! 1 ©qn!0 xn f AxB ! aq

n!0 an Ax " x0Bn

f AxB,©qn!0 an Ax " x0Bn A5/2, 7/2 4 x ! 7/2

©qn!0 An # 1B"1,x ! 5/2 x ! 7/2x ! 5/20 x " 3 0 ! 1/2, 0 x " 3 0 7 1/2.0 x " 3 0 6 1/2 r ! 1/2.

! lim nSq

n # 2 2 An # 1B ! 12 ! L .

lim nSq ` an an#1

` ! lim nSq ` A"2Bn An # 2BA"2Bn#1 An # 1B `

an ! A"2Bn/ An # 1B, a q

n!0

A"2Bn n # 1

Ax " 3Bn .

Section 8.2 Power Series and Analytic Functions 429

Power Series Vanishing on an Interval

Theorem 3. If for all x in some open interval, then each coefficient equals zero.an

©qn!0 an Ax " x0Bn ! 0

Example 1

Solution

Given two power series

(4)

with nonzero radii of convergence, we want to find power series representations for the sum, product, and quotient of the functions and The sum is simply obtained by termwise addition:

f AxB # g AxB ! aq n!0 Aan # bnB Ax " x0Bn

g AxB.f AxB f AxB ! aq

n!0 an Ax " x0Bn , g AxB ! aq

n!0 bn Ax " x0Bn ,

for all x in the common interval of convergence of the power series in (4). The power series representation for the product is a bit more complicated. To provide motivation for the formula, we treat the power series for and as “long polynomials,” apply the distribu- tive law, and group the terms in powers of

The general formula for the product is

(5)

where

(6)

The power series in (5) is called the Cauchy product, and it will converge for all x in the common open interval of convergence for the power series of f and g.†

The quotient will also have a power series expansion about provided However, the radius of convergence for this quotient series may be smaller than that

for or Unfortunately, there is no nice formula for obtaining the coefficients in the power series for However, we can use the Cauchy product to divide power series indirectly (see Problem 15). The quotient series can also be obtained by formally carrying out polynomial long division (see Problem 16).

The next theorem explains, in part, why power series are so useful.

f AxB /g AxB.g AxB.f AxB g Ax0B & 0. x0,f AxB /g AxB

cn J a n

k!0 akbn"k .

f AxBg AxB ! aq n!0

cn Ax " x0Bn , ! a0b0 # Aa0b1 # a1b0B Ax " x0B # Aa0b2 # a1b1 # a2b0B Ax " x0B2 # p .3a0 # a1 Ax " x0B # a2 Ax " x0B2 # p 4 # 3b0 # b1 Ax " x0B # b2 Ax " x0B2 # p 4

Ax " x0B:g AxBf AxB f AxBg AxB

430 Chapter 8 Series Solutions of Differential Equations

†Actually, it may happen that the radius of convergence of the power series for or is larger than that for the power series of f or g.

f AxB # g AxBf AxBg AxB

Differentiation and Integration of Power Series

Theorem 4. If the series has a positive radius of convergence then f is differentiable in the interval and termwise differentiation gives the power series for the derivative:

for

Furthermore, termwise integration gives the power series for the integral of f :

for 0 x " x0 0 6 r .$ f AxB dx ! aqn!0 ann # 1 Ax " x0Bn#1 # C 0 x " x0 0 6 r .f ¿AxB ! aq

n!1 nan Ax " x0Bn"1

0 x " x0 0 6 rr, f AxB ! ©qn!0 an Ax " x0Bn

Starting with the geometric series (3) for find a power series for each of the following functions:

(a) (b) (c) arctan x .

(a) Replacing x by in (3) immediately gives

(7) 1 1 # x2

! 1 " x2 # x4 " x6 # p # A"1Bnx2n # p ."x 2

1A1 " xB2 .11 # x2 . 1/ A1 " xB,Example 2

Solution

(b) Notice that is the derivative of the function Hence, on differentiating (3) term by term, we obtain

(8)

we can integrate the series in (7) termwise to obtain the series for arctan x. Thus,

(9) ◆

It is important to keep in mind that since the geometric series (3) has the (open) interval of convergence the representations (7), (8), and (9) are at least valid in this interval. [Actually, the series (9) for arctan x converges for all ]

Shifting the Summation Index The index of summation in a power series is a dummy index just like the variable of integration in a definite integral. Hence,

Just as there are times when we want to change the variable of integration, there are situa- tions (and we will encounter many in this chapter) when it is desirable to change or shift the index of summation. This is particularly important when one has to combine two different power series.

Express the series

as a series where the generic term is instead of

Setting we have and Note that when n ! 2, then k ! 0. Hence, substituting into the given series, we find

◆

Show that

x3a q

n!0 n2 An " 2Banxn ! aq

n!3 An " 3B2 An " 5Ban"3xn .

a q

n!2 n An " 1Banxn"2 ! aq

k!0 Ak # 2B Ak # 1Bak#2xk .

n " 1 ! k # 1.n ! k # 2k ! n " 2,

xn"2.xk

a q

n!2 n An " 1Banxn"2

a q

n!0 an Ax " x0Bn ! aq

k!0 ak Ax " x0Bk ! aq

i!0 ai Ax " x0Bi .

0 x 0 $ 1.A"1, 1B, arctan x ! x "

1 3

x3 # 1 5

x5 " 1 7

x7 # p # A"1Bnx2n#1

2n # 1 # p .

$ x

1 1 # t2

dt ! $ x

0 E1 " t2 # t4 " t6 # p # A"1Bnt2n # p F dt

arctan x ! $ x

1 1 # t2

dt ,

f ¿AxB ! 1A1 " xB2 ! 1 # 2x # 3x2 # 4x3 # p # nxn"1 # p . f AxB ! 1/ A1 " xB.1/ A1 " xB2

Section 8.2 Power Series and Analytic Functions 431

Example 3

Solution

Example 4

We start by taking the inside the summation on the left-hand side:

To rewrite this with generic term we set Thus and corre- sponds to Straightforward substitution thus yields

By replacing k by n, we obtain the desired form. ◆

Show that the identity

implies that and for

First, we rewrite both series in terms of For the first series, we set and hence to write

Then with the second series becomes

The identity thus states

and so we have a power series that sums to zero; consequently, by Theorem 3, each of its coefficients equals zero. For k ! 3, 4, . . . , both series contribute to the coefficient of and thus we confirm that

or for For k ! 0, 1, or 2, only the first series contributes, and we find, in turn,

Hence, ◆a0 ! a1 ! a2 ! 0.

A2 # 1Ba2 ! 0 .A1 # 1Ba1 ! 0 , A0 # 1Ba0 ! 0 ,

k ' 3.ak ! "bk"1/ Ak # 1B Ak # 1Bak # bk"1 ! 0

xk,

a q

k!0 Ak # 1Bakxk # aq

k!3 bk"1x

k ! 0 ,

a q

n!2 bnx

n#1 ! a q

k!3 bk"1x

k .

k ! n # 1, n ! k " 1,

a q

n!1 nan"1x

n"1 ! a q

k!0 Ak # 1Bakxk .

n ! k # 1, k ! n " 1,xk.

n ' 3.an ! "bn"1/ An # 1Ba0 ! a1 ! a2 ! 0 a q

n!1 nan"1x

n"1 # a q

n!2 bnx

n#1 ! 0

a q

n!0 n2 An " 2Banxn#3 ! aq

k!3 Ak " 3B2 Ak " 5Bak"3xk .

k ! 3. n ! 0n ! k " 3,k ! n # 3.xk,

x3a q

n!0 n2 An " 2Banxn ! aq

n!0 n2 An " 2Banxn#3 .

432 Chapter 8 Series Solutions of Differential Equations

Example 5

Solution

Analytic Functions Not all functions are expressible as power series. Those distinguished functions that can be so represented are called analytic.

Section 8.2 Power Series and Analytic Functions 433

Analytic Function

Definition 1. A function f is said to be analytic at if, in an open interval about this function is the sum of a power series that has a positive radius of convergence.

©qn!0 an Ax " x0Bn x0,x0

For example, a polynomial function is analytic at every since we can always rewrite it in the form A rational func- tion where and are polynomials without a common factor, is an analytic function except at those for which As you may recall from calculus, the elementary functions and cos x are analytic for all x, while ln x is analytic for x > 0. Some familiar representations are

(10)

(11)

(12)

(13)

where (10), (11), and (12) are valid for all x, whereas (13) is valid for x in the half-open interval .

From Theorem 4 on the differentiation of power series, we see that a function f analytic at is differentiable in a neighborhood of Moreover, because has a power series represen-

tation in this neighborhood, it too is analytic at Repeating this argument, we see that etc., exist and are analytic at Consequently, if a function does not have derivatives of all orders at , then it cannot be analytic at The function is not analytic at

because does not exist; and is not analytic at because does not exist.

Now we can deduce something very specific about the possible power series that can represent an analytic function. If is analytic at then (by definition) it is the sum of some power series that converges in a neighborhood of

f AxB ! aq n!0

an Ax " x0Bn . x0:

x0,f AxB f ‡A0Bx0 ! 0f AxB ! x7/3f ¿A1Bx0 ! 1 f AxB ! 0 x " 1 0x0.x0 x0.

f –, f A3B,x0. f ¿x0.x0 A0, 2 4

ln x ! Ax " 1B " 1 2

Ax " 1B2 # 1 3

Ax " 1B3 " p ! aq n!1

A"1Bn"1 n

Ax " 1Bn , cos x ! 1 "

2! #

4! " p ! a

n!0

A"1BnA2nB! x2n , sin x ! x "

3! #

5! " p ! a

n!0

A"1BnA2n # 1B! x2n#1 , ex ! 1 # x #

2! #

3! # p ! a

n!0 xn

n! ,

ex, sin x, Q Ax0B ! 0.x0 Q AxBP AxBP AxB /Q AxB,

a0 # a1 Ax " x0B # p # an Ax " x0Bn. x0,b0 # b1x # p # bnxn

By the reasoning in the previous paragraph, the derivatives of f have convergent power series representations

o But if we evaluate these series at we learn that

o that is, and the power series must coincide with the Taylor series†

about Any power series—regardless of how it is derived—that converges in some neighbor- hood of to a function has to be the Taylor series of that function. For example, the expan- sion for arctan x given in (9) of Example 2 must be its Taylor expansion.

With these facts in mind we are ready to turn to the study of the effectiveness of power series techniques for solving differential equations. In the next sections, you will find it helpful to keep in mind that if f and g are analytic at , then so are and if These facts follow from the algebraic properties of power series discussed earlier.

g Ax0B & 0.f/gf # g, cf, fg,x0 x0

x0.

a q

j!0

f A jB Ax0B j!

Ax " x0B j aj ! f

A jB Ax0B /j! f A jB Ax0B ! j! # aj , f –Ax0B ! 2 # 1 # a2 ,f ¿Ax0B ! 1 # a1 , f Ax0B ! a0 , x ! x0,

! a q

n! j n An " 1B p An " 3 j " 1 4 B an Ax " x0Bn" j ,

f A jB AxB ! aq n!0

n An " 1B p An " 3 j " 1 4 B an Ax " x0Bn" j , f –AxB ! aq

n!0 n An " 1Ban Ax " x0Bn"2 ! 0 # 0 # aq

n!2 n An " 1Ban Ax " x0Bn"2 ,

f ¿AxB ! aq n!0

nan Ax " x0Bn"1 ! 0 # aq n!1

nan Ax " x0Bn"1 ,

434 Chapter 8 Series Solutions of Differential Equations

†When the expansion point is zero, the Taylor series is also known as the Maclaurin series.x0

8.2 EXERCISES In Problems 1–6, determine the convergence set of the given power series.

1. 2. a q

n!0 3n

n! xna

n!0

2"n

n # 1 Ax " 1Bn

3. 4.

5. 6. a q

n!0

An # 2B! n!

Ax # 2Bnaq n!1

n3 Ax " 2Bn

a q

n!1

4 n2 # 2n

Ax " 3Bnaq n!0

2n Ax # 2Bn

7. Sometimes the ratio test (Theorem 2) can be applied to a power series containing an infinite number of zero coefficients, provided the zero pattern is regular. Use Theorem 2 to show, for example, that the series

has a radius of convergence if

and that

has a radius of convergence if

[Hint: Let 8. Determine the convergence set of the given power series.

(a) (b)

(e)

(f)

In Problems 9 and 10, find the power series expansion for , given the expansions for

10.

In Problems 11–14, find the first three nonzero terms in the power series expansion for the product

11.

12.

g AxB ! cos x ! aq k!0

A"1BkA2kB! x2k f AxB ! sin x ! aq

k!0

A"1BkA2k # 1B! x2k#1 , g AxB ! sin x ! aq

k!0

A"1BkA2k # 1B! x2k#1 f AxB ! ex ! aq

n!0 1 n!

xn ,

f AxBg AxB. g AxB ! aq

n!1 n2

2n Ax " 1Bn"1

f AxB ! aq n!3

n! Ax " 1Bn"2 ,

f AxB ! aq n!0

1 n # 1

xn , g AxB ! aq n!1

2"nxn"1

f AxB and g AxB. f AxB # g AxB©qn!0 anxn a q

k!0 22kx4k

Asin xB /x ! aq n!0 A"1Bnx2n/ A2n # 1B!

a q

k!0 22k#1x2k#1a

k!0 22kx2k

z ! x2. 4limkSq ` a2k#1a2k#3 ` ! M . r ! 2M,! aqk!0 a2k#1x2k#1

a1x # a3x 3 # a5x

5 # a7x 7 # p

lim kSq ` a2k a2k#2

` ! L , r ! 2L, a0 # a2x

2 # a4x 4 # a6x

6 # p ! a q

k!0 a2kx

Section 8.2 Power Series and Analytic Functions 435

13.

14.

15. Find the first few terms of the power series for the quotient

by completing the following: (a) Let where the coefficients

are to be determined. Argue that is the Cauchy product of and

(b) Use formula (6) of the Cauchy product to deduce the equations

16. To find the first few terms in the power series for the quotient in Problem 15, treat the power series in the numerator and denominator as “long polynomi- als” and carry out long division. That is, perform

In Problems 17–20, find a power series expansion for given the expansion for

17.

18.

19.

20. f AxB ! aq n!1

nanx n"1

f AxB ! aq k!0

akx 2k

f AxB ! sin x ! aq k!0

A"1BkA2k # 1B! x2k#1 f AxB ! A1 # xB"1 ! aq

n!0 A"1Bnxnf AxB.f ¿AxB,

1 # x # 1 2

x2 # p ` 1 # 1 2

x # 1 4

x2 # p .

q AxB a0, a1, a2, a3.

1 23

! a0 6

# a1 2

# a2 # a3 , . . . .

1 22

! a0 2

# a1 # a2 ,

1 20

! a0 , 1 2

! a0 # a1 ,

©qn!0 xn/n!.q AxB ©qn!0 xn/2n anq AxB ! ©qn!0 anxn,

q AxB ! aaq n!0

1 2n

xnb~aaq n!0

1 n!

xnb g AxB ! e"x ! aq

n!0

A"1Bn n!

f AxB ! ex ! aq n!0

1 n!

xn ,

g AxB ! A1 # xB"1 ! aq n!0 A"1Bnxn

f AxB ! e"x ! aq n!0

A"1Bn n!

xn ,

In Problems 21 and 22, find a power series expansion for given the expansion for

21.

22.

In Problems 23–26, express the given power series as a series with generic term x k.

23. 24.

25. 26.

27. Show that

28. Show that

In Problems 29–34, determine the Taylor series about the point for the given functions and values of 29.

30.

31.

32.

33.

34. f AxB ! 1x , x0 ! 1f AxB ! x3 # 3x " 4 , x0 ! 1 f AxB ! ln A1 # xB , x0 ! 0f AxB !

1 # x 1 " x

, x0 ! 0

f AxB ! x"1 , x0 ! 1f AxB ! cos x , x0 ! p x0.x0

! b1 # a q

n!1 3 2an"1 # An # 1Bbn#1 4 xn .2a

n!0 anx

n#1 # a q

n!1 nbnx

n"1

! a q

n!2 An " 2B An " 1Ban"2 xn .x2aq

n!0 n An # 1Banxn

a q

n!1

an n # 3

xn#3a q

n!0 anx

n#1

a q

n!2 n An " 1Banxn#2aq

n!1 nanx

n"1

f AxB ! sin x x

! a q

k!0

A"1BkA2k # 1B! x2k f AxB ! A1 # xB"1 ! aq

n!0 A"1Bnxn f AxB.g AxB J %

x 0 f AtB dt,

436 Chapter 8 Series Solutions of Differential Equations

35. The Taylor series for about given in equation (13) can also be obtained as follows: (a) Starting with the expansion !

and observing that

obtain the Taylor series for about (b) Since ln use the result of part (a)

and termwise integration to obtain the Taylor series for about

36. Let and be analytic at Determine whether the following statements are always true or sometimes false: (a) is analytic at

(b) is analytic at

(d) is analytic at

37. Let

Show that for n ! 0, 1, 2, . . . and hence that the Maclaurin series for is

which converges for all x but is equal to only when x ! 0. This is an example of a function possessing derivatives of all orders (at

), whose Taylor series converges, but the Taylor series (about ) does not converge to the original function! Consequently, this function is not analytic at x ! 0.

38. Compute the Taylor series for about [Hint: Multiply the series for

by 2x and integrate.]A1 # x2B"1x0 ! 0. f AxB ! ln A1 # x2B

x0 ! 0 x0 ! 0

f AxB0 # 0 # 0 # p , f AxBf AnB A0B ! 0

f AxB ! e e"1/x2 , x & 0 , 0 , x ! 0 .

x03 f AxB 4 3 " %xx0 g AtB dtx0f ¿AxB x0f AxB /g AxB x03f AxB # g AxB

x0.g AxBf AxB x0 ! 1.f AxB ! ln x x ! %x1 1 / t dt,

x0 ! 1.1 /x

1 x

! 1

1 # Ax " 1B , ©qn!0 sn

1 / A1 " sB x0 ! 1f AxB ! ln x

8.3 POWER SERIES SOLUTIONS TO LINEAR DIFFERENTIAL EQUATIONS In this section we demonstrate a method for obtaining a power series solution to a linear differ- ential equation with polynomial coefficients. This method is easier to use than the Taylor series method discussed in Section 8.1 and sometimes gives a nice expression for the general term in the power series expansion. Knowing the form of the general term also allows us to test for the radius of convergence of the power series.

We begin by writing the linear differential equation

(1)

in the standard form

(2)

where and q AxB J a0 AxB /a2 AxB.p AxB J a1 AxB /a2 AxBy– # p AxBy¿ # q AxBy ! 0 , a2 AxBy– # a1 AxBy¿ # a0 AxBy ! 0

Section 8.3 Power Series Solutions to Linear Differential Equations 437

Determine all the singular points of

Dividing the equation by x, we find that

The singular points are those points where fails to be analytic. Observe that and are the ratios of functions that are everywhere analytic. Hence, and are analytic except, perhaps, when their denominators are zero. For this occurs at x ! 0 and x ! 1. But since we can cancel an x in the numerator and denominator of that is,

we see that is actually analytic at x ! 0.† Therefore, is analytic except at x ! 1. For the denominator is zero at x ! 0. Just as with this zero is removable since has

the power series expansion

Thus, is everywhere analytic. Consequently, the only singular point of the given equation is x ! 1. ◆

At an ordinary point of equation (1) (or (2)), the coefficient functions and are analytic. Hence, we might expect that the solutions to these equations inherit this property. From the discussion in Section 6.1 on linear equations, the continuity of p and q in a neighbor- hood of is sufficient to imply that equation (2) has two linearly independent solutions defined in that neighborhood. But analytic functions are not merely continuous—they possess

q AxBp AxBx0 q AxBq AxB ! sin x

x !

x " x3

3! #

5! " p

x ! 1 "

3! #

5! " p .

q AxBp AxB,q AxB, p AxBp AxB p AxB ! x

x A1 " xB ! 11 " x , p AxB,p AxB q

AxBp AxBq AxB p AxBp AxB or q AxB p AxB ! x

x A1 " xB , q AxB ! sin xx . xy– # x A1 " xB"1y¿ # Asin xBy ! 0 .

†Such points are called removable singularities. In this chapter we assume in such cases that the function has been defined (or redefined) so that it is analytic at the point.

Ordinary and Singular Points

Definition 2. A point is called an ordinary point of equation (1) if both and are analytic at If is not an ordinary point, it is called a singular point of the equation.

x0x0.q ! a0/a2 p ! a1/a2x0

Example 1

Solution

derivatives of all orders in a neighborhood of . Thus we can differentiate equation (2) to show that exists and, by a “bootstrap” argument, prove that solutions to (2) must likewise possess derivatives of all orders. Although we cannot conclude by this reasoning that the solu- tions enjoy the stronger property of analyticity, this is nonetheless the case (see Theorem 5 in Section 8.4). Hence, in a neighborhood of an ordinary point the solutions to (1) (or (2)) can be expressed as a power series about

To illustrate the power series method about an ordinary point, let’s look at a simple first- order linear differential equation.

Find a power series solution about x ! 0 to

(3)

The coefficient of y is the polynomial 2x, which is analytic everywhere, so x ! 0 is an ordinary point† of equation (3). Thus, we expect to find a power series solution of the form

(4)

Our task is to determine the coefficients For this purpose we need the expansion for that is given by termwise differentiation

of (4):

We now substitute the series expansions for and into (3) and obtain

which simplifies to

(5)

To add the two power series in (5), we add the coefficients of like powers of x. If we write out the first few terms of these summations and add, we get

(6)

For the power series on the left-hand side of equation (6) to be identically zero, we must have all the coefficients equal to zero. Thus,

etc.

Solving the preceding system, we find

a4 ! " 1 2

a2 ! " 1 2

A"a0B ! 12 a0 . a1 ! 0 , a2 ! "a0 , a3 ! "

2 3

a1 ! 0 ,

3a3 # 2a1 ! 0 , 4a4 # 2a2 ! 0 , a1 ! 0 , 2a2 # 2a0 ! 0 ,

a1 # A2a2 # 2a0Bx # A3a3 # 2a1Bx2 # A4a4 # 2a2Bx3 # p ! 0 .Aa1 # 2a2x # 3a3x2 # 4a4x3 # p B # A2a0x # 2a1x2 # 2a2x3 # p B ! 0 , a q

n!1 nanx

n"1 # a q

n!0 2anx

n#1 ! 0 .

a q

n!1 nanx

n"1 # 2xa q

n!0 anx

n ! 0 ,

y¿y

y¿ AxB ! 0 # a1 # 2a2x # 3a3x2 # p ! aq n!1

nanx n"1 .

y¿ AxBan. y AxB ! a0 # a1x # a2x2 # p ! aq

n!0 anx

n .

y¿ # 2xy ! 0 .

x0. x0,

y A3B x0 438 Chapter 8 Series Solutions of Differential Equations

Example 2

Solution

†By an ordinary point of a first-order equation we mean a point where is analytic.q AxBy¿ # q AxBy ! 0,

Hence, the power series for the solution takes the form

(7)

Although the first few terms displayed in (7) are useful, it is sometimes advantageous to have a formula for the general term in the power series expansion for the solution. To achieve this goal, let’s return to equation (5). This time, instead of just writing out a few terms, let’s shift the indices in the two power series so that they sum over the same powers of x, say, To do this we shift the index in the first summation in (5) by letting Then and when Hence, the first summation in (5) becomes

(8)

In the second summation of (5), we put so that and when This gives

(9)

Substituting (8) and (9) into (5) yields

(10)

Since the first summation in (10) begins at k ! 0 and the second at k ! 1, we break up the first into

Then (10) becomes

(11)

When we set all the coefficients in (11) equal to zero, we find

and, for all

(12)

Equation (12) is a recurrence relation that we can use to determine the coefficient in terms of that is,

Setting k ! 1, 2, . . . , 8 and using the fact that we find

a8 ! " 2 8

a6 ! 1 4! a0

Ak ! 7B , a9 ! " 29 a7 ! 0 Ak ! 8B . a6 ! "

2 6

a4 ! # 1 3! a0 Ak ! 5B , a7 ! " 27 a5 ! 0 Ak ! 6B ,

a4 ! " 2 4

a2 ! 1 2 a0

Ak ! 3B , a5 ! " 25 a3 ! 0 Ak ! 4B , a2 ! "

2 2

a0 ! #a0 Ak ! 1B , a3 ! " 23 a1 ! 0 Ak ! 2B , a1 ! 0,

ak#1 ! " 2

k # 1 ak"1 .

ak"1; ak#1

Ak # 1Bak#1 # 2ak"1 ! 0 .k ' 1, a1 ! 0 ,

a1 # a q

k!1 3 Ak # 1Bak#1 # 2ak"1 4 xk ! 0 .

a q

k!0 Ak # 1Bak#1xk ! a1 # aq

k!1 Ak # 1Bak#1xk .

a q

k!0 Ak # 1Bak#1xk # aq

k!1 2ak"1x

k ! 0 .

a q

n!0 2anx

n#1 ! a q

k!1 2ak"1x

k .

n ! 0.k ! 1n ! k " 1k ! n # 1

a q

n!1 nanx

n"1 ! a q

k!0 Ak # 1Bak#1xk .

n ! 1.k ! 0 n ! k # 1k ! n " 1.

xk.

y AxB ! a0 " a0x2 # 12 a0x4 # p . Section 8.3 Power Series Solutions to Linear Differential Equations 439

After a moment’s reflection, we realize that

Substituting back into the expression (4), we obtain the power series solution

(13)

Since is left undetermined, it serves as an arbitrary constant, and hence (13) gives a general solution to equation (3). ◆

Applying the ratio test as described in Problem 7, Exercises 8.2 (page 435), we can verify that the power series in (13) has radius of convergence Moreover, (13) is reminiscent of the expansion for the exponential function; you can check that it converges to

This general solution to the simple equation (3) can also be obtained by the method of separa- tion of variables.

The convergence of the partial sums of (13), with to the actual solution is depicted in Figure 8.3. Notice that taking more terms results in better approximations around x ! 0. Note also, however, that every partial sum is a polynomial and hence must diverge at

while converges to zero, of course. Thus, each partial sum approximation even- tually deteriorates for large This is a typical feature of power series approximations.

In the next example we use the power series method to obtain a general solution to a linear second-order differential equation.

0 x 0 .e"x2x ! %q, e"x

2 a0 ! 1,

y AxB ! a0 e"x2 . r ! q.

y AxB ! a0 " a0x2 # 12! a0x4 # p ! a0 aqn!0 A"1Bnn! x2n . a2n#1 ! 0 , n ! 0, 1, 2, . . . .

a2n ! A"1Bn

n! a0 , n ! 1, 2, . . . ,

440 Chapter 8 Series Solutions of Differential Equations

−1

−2

−3 −2 −1 1 320

degree 4

degree 20

degree 0

degree 2

e−x 2

Figure 8.3 Partial sum approximations to e"x 2

Find a general solution to

(14)

in the form of a power series about the ordinary point x ! 0.

Writing

(15)

we differentiate termwise to obtain

Substituting these power series into equation (14), we find

(16)

To simplify the addition of the three summations in (16), let’s shift the indices so that the general term in each is a constant times For the first summation, we substitute and get

In the second and third summations, we simply substitute k for n. With these changes of indices, equation (16) becomes

Next, we separate the terms from the others and then combine the like powers of x in the three summations to get

Setting the coefficients of this power series equal to zero yields

(17)

and the recurrence relation

(18)

We can now use (17) and (18) to determine all the coefficients of the solution in terms of and Solving (18) for gives

(19) ak#2 ! "1

2 Ak # 2B ak , k ' 1 . ak#2a1.

a0ak

2 Ak # 2B Ak # 1B ak#2 # Ak # 1Bak ! 0 , k ' 1 . 4a2 # a0 ! 0

4a2 # a0 # a q

k!1 32 Ak # 2B Ak # 1Bak#2 # kak # ak 4 xk ! 0 .

a q

k!0 2 Ak # 2B Ak # 1Bak#2xk # aq

k!1 kakx

k # a q

k!0 akx

k ! 0 .

a q

n!2 2n An " 1B anxn"2 ! aq

k!0 2 Ak # 2B Ak # 1Bak#2xk .

k ! n " 2xk.

a q

n!2 2n An " 1B anxn"2 # aq

n!1 nanx

n # a q

n!0 anx

n ! 0 .

y– AxB ! 2a2 # 6a3x # 12a4x2 # p ! aq n!2

n An " 1Banxn"2 . y¿ AxB ! a1 # 2a2x # 3a3x2 # p ! aq

n!1 nanx

n"1 ,

y AxB ! a0 # a1x # a2x2 # p ! aq n!0

anx n ,

2y– # xy¿ # y ! 0

Section 8.3 Power Series Solutions to Linear Differential Equations 441

Example 3

Solution

Thus,

The pattern for the coefficients is now apparent. Since and are not restricted, we find

and

From this, two linearly independent solutions emerge; namely,

(20)

(21)

Hence, a general solution to (14) is Approximations to the solutions are depicted in Figure 8.4 on page 443. ◆

The method illustrated in Example 3 can also be used to solve initial value problems. Sup- pose we are given the values of and then, from equation (15), we see that and Knowing these two coefficients leads to a unique power series solution for the initial value problem.

The recurrence relation (18) in Example 3 involved just two of the coefficients, and and we were fortunate in being able to deduce from this relation the general form for the

coefficient However, many cases arise that lead to more complicated two-term or even to many-term recurrence relations. When this occurs, it may be impossible to determine the general form for the coefficients In the next example, we consider an equation that gives rise to a three-term recurrence relation.

an.

an. ak,

ak#2

a1 ! y¿ A0B. a0 ! y A0By¿ A0B;y A0B y1 AxB, y2 AxB a0y1 AxB # a1y2 AxB.

y2 AxB ! aq n!0

A"1Bn 2n 31 # 3 # 5 p A2n # 1B 4 x2n#1 Atake a0 ! 0, a1 ! 1B .

y1 AxB ! aq n!0

A"1Bn 22nn!

x2n Atake a0 ! 1, a1 ! 0B , a2n#1 !

A"1Bn 2n 31 # 3 # 5 p A2n # 1B 4 a1 , n ' 1

a2n ! A"1Bn 22nn!

a0 , n ' 1 ,

a1a0

a8 ! "1 2 # 8 a6 !

1 24 # 2 # 4 # 6 # 8 a0 !

1 28 # 4! a0 Ak ! 6B .

a7 ! "1 2 # 7 a5 !

"1 23 # 3 # 5 # 7 a1 Ak ! 5B ,

a6 ! "1 2 # 6 a4 !

"1 23 # 2 # 4 # 6 a0 !

"1 26 # 3! a0 Ak ! 4B ,

a5 ! "1 2 # 5 a3 !

1 22 # 3 # 5 a1 Ak ! 3B ,

a4 ! "1 2 # 4 a2 !

1 22 # 2 # 4 a0 Ak ! 2B ,

a3 ! "1 2 # 3 a1 Ak ! 1B ,

a2 ! "1 22

a0 ,

442 Chapter 8 Series Solutions of Differential Equations

Find the first few terms in a power series expansion about x ! 0 for a general solution to

(22)

Since and are analytic at then is an ordinary point for equation (22). Hence, we can express its general solution in the form

Substituting this expansion into (22) yields

(23)

To sum over like powers we put in the first summation of (23), in the third, and in the second and fourth. This gives

Separating the terms corresponding to k ! 0 and k ! 1 and combining the rest under one sum- mation, we have

Setting the coefficients equal to zero gives

(24)

(25) , 6a3 " 2a2 # a1 ! 0

2a2 " a1 # a0 ! 0 ,

# a q

k!2 3 Ak # 2B Ak # 1Bak#2 " Ak # 1Bak#1 # Ak Ak " 1B # 1B ak 4 xk ! 0 .A2a2 " a1 # a0B # A6a3 " 2a2 # a1Bx

a q

k!0 Ak # 2B Ak # 1Bak#2xk #aq

k!2 k Ak " 1Bakxk "aq

k!0 Ak # 1Bak#1xk # aq

k!0 akx

k ! 0 .

k ! n k ! n " 1k ! n " 2xk,

a q

n!2 n An " 1B anxn"2 # aq

n!2 n An " 1Banxn " aq

n!1 nanx

n"1 # a q

n!0 anx

n ! 0 .

A1 # x2B aq n!2

n An " 1Banxn"2 " aq n!1

nanx n"1 # a

n!0 anx

n ! 0 ,

y AxB ! aq n!0

anx n .

x ! 0x ! 0,q AxB ! A1 # x2B"1p AxB ! " A1 # x2B"1A1 # x2By– " y¿ # y ! 0 .

Section 8.3 Power Series Solutions to Linear Differential Equations 443

Degree 5

Degree 9

Degree 21

Degree 20

Degree 4

Degree 8

−2

−4

2 4−4 x

partial sums for y2(x)

partial sums for y1(x)

Figure 8.4 Partial sum approximations to solutions for Example 3

Example 4

Solution

and the recurrence relation

(26)

We can solve (24) for in terms of and

Now that we have we can use (25) to express in terms of and

Solving the recurrence relation (26) for we obtain

(27)

For k ! 2, 3, and 4, this gives

We can now express a general solution in terms up to order 6, using and as the arbitrary constants. Thus,

(28)

◆

Given specific values for and will the partial sums of the power series representation (28) yield useful approximations to the solution when x ! 0.5? What about when x ! 2.3 or x ! 7.8? The answers to these questions certainly depend on the radius of convergence of the power series in (28). But since we were not able to determine a general form for the coefficients in this example, we cannot use the ratio test (or other methods such as the root test, integral test, or com- parison test) to compute the radius In the next section we remedy this situation by giving a simple procedure that determines a lower bound for the radius of convergence of power series solutions.

a1,a0

# a1 ax # 12 x2 " 18 x4 " 140 x5 # 120 x6 # p b . ! a0 a1 " 12 x2 " 16 x3 # 112 x4 # 340 x5 " 17720 x6 # p b

# a2a0 " 3a1 24

b x4 # a3a0 " a1 40

b x5 # a36a1 " 17a0 720

b x6 # p y AxB ! a0 # a1x # aa1 " a02 b x2 " a06 x3

a1a0

a6 ! 5a5 " 13a4

6 # 5 ! 36a1 " 17a0

720 Ak ! 4B .

a5 ! 4a4 " 7a3

5 # 4 ! 3a0 " a1

40 Ak ! 3B ,

"a0 6

" aa1 " a0 2 b

4 !

2a0 " 3a1 24

Ak ! 2B , a4 !

3a3 " 3a2 4 # 3 !

a3 " a2 4

ak#2 ! Ak # 1Bak#1 " Ak2 " k # 1BakAk # 2B Ak # 1B , k ' 2 .

ak#2,

a3 ! 2a2 " a1

6 ! Aa1 " a0B " a1

6 !

"a0 6

a1:a0a3a2,

a2 ! a1 " a0

2 .

a1:a0a2

Ak # 2B Ak # 1Bak#2 " Ak # 1Bak#1 # Ak2 " k # 1Bak ! 0 , k ' 2 . 444 Chapter 8 Series Solutions of Differential Equations

In Problems 1–10, determine all the singular points of the given differential equation.

1. 2. 3. 4. 5. 6. 7. 8. 9.

10.

In Problems 11–18, find at least the first four nonzero terms in a power series expansion about for a general solution to the given differential equation. 11. 12. 13. 14. 15. 16. 17. 18.

In Problems 19–24, find a power series expansion about for a general solution to the given differential

equation. Your answer should include a general formula for the coefficients. 19. 20. 21. 22. 23. 24.

In Problems 25–28, find at least the first four nonzero terms in a power series expansion about for the solution to the given initial value problem. 25.

26.

27.

28. y A0B ! "1 , y¿ A0B ! 0y– # Ax " 2By¿ " y ! 0 ; y A0B ! 0 , y¿ A0B ! 1Ax # 1By– " y ! 0 ; y A0B ! 0 , y¿ A0B ! 1Ax2 " x # 1By– " y¿ " y ! 0 ; w A0B ! 2 , w¿ A0B ! 0w– # 3xw¿ " w ! 0 ;

x ! 0

Ax2 # 1By– " xy¿ # y ! 0z– " x2z¿ " xz ! 0 y– " xy ! 0y– " xy¿ # 4y ! 0 y– # y ! 0y¿ " 2xy ! 0

x ! 0

A2x " 3By– " xy¿ # y ! 0w– " x2w¿ # w ! 0 y– " 2y¿ # y ! 0 y– # Ax " 1By¿ # y ! 0 Ax2 # 1By–# y ! 0z– " x2z ! 0

y¿ " y ! 0y¿ # Ax # 2By ! 0 x ! 0

3 ln Ax " 1B 4 y– # Asin 2xBy¿ " exy ! 0Asin uBy– " Aln uBy ! 0e xy– " Ax2 " 1By¿ # 2xy ! 0Asin xBy– # Acos xBy ! 0 Ax2 " 1By– # A1 " xBy¿ # Ax2 " 2x # 1By ! 0At2 " t " 2Bx– # At # 1Bx¿ " At " 2Bx ! 0 Ax2 # xBy– # 3y¿ " 6xy ! 0Au2 " 2By– # 2y¿ # Asin uBy ! 0 x2y– # 3y¿ " xy ! 0 Ax # 1By– " x2y¿ # 3y ! 0

Section 8.3 Power Series Solutions to Linear Differential Equations 445

In Problems 29–31, use the first few terms of the power series expansion to find a cubic polynomial approxima- tion for the solution to the given initial value problem. Graph the linear, quadratic, and cubic polynomial approximations for 29.

30.

31.

32. Consider the initial value problem

where and are constants. (a) Show that if then the solution will be an

odd function [that is, for all x]. What happens when ?

(b) Show that if and are positive, then the solution is increasing on

(d) What conditions on and would guarantee that the solution is increasing on

33. Use the ratio test to show that the radius of conver- gence of the series in equation (13) is infinite. [Hint: See Problem 7, Exercises 8.2, page 435.]

34. Emden’s Equation. A classical nonlinear equa- tion that occurs in the study of the thermal behavior of a spherical cloud is Emden’s equation

with initial conditions Even though x ! 0 is not an ordinary point for this equa- tion (which is nonlinear for ), it turns out that there does exist a solution analytic at x ! 0. Assum- ing that n is a positive integer, show that the first few terms in a power series solution are

[Hint: Substitute # into the equation and carefully compute

the first few terms in the expansion for yn. 4c5x5 # p y ! 1 # c2x2 # c3x3 # c4x4 y ! 1 "

3! # n

5! # p .

n & 1

y A0B ! 1, y¿ A0B ! 0.y– # 2 x

y¿ # yn ! 0 ,

A"q, q B?a1a0 A"q, 0B.a1a0

A0, q B.a1a0 a1 ! 0 y A"xB ! "y AxBa0 ! 0,

a1a0

y A0B ! a0 , y¿ A0B ! a1 ,y– " 2xy¿ " 2y ! 0 ; y¿ A0B ! 2y A0B ! 1 , Ax2 # 2By– # 2xy¿ # 3y ! 0 ;

y A0B ! "1 , y¿ A0B ! 1y– " 4xy¿ # 5y ! 0 ; y A0B ! 1 , y¿ A0B ! "2y– # y¿ " xy ! 0 ;

"5 $ x $ 5.

8.3 EXERCISES

35. Variable Resistor. In Section 5.7, we showed that the charge q on the capacitor in a simple RLC circuit is governed by the equation

where L is the inductance, R the resistance, C the capacitance, and E the voltage source. Since the resistance of a resistor increases with temperature, let’s assume that the resistor is heated so that the resis- tance at time t is Ω (see Figure 8.5). If L ! 0.1 H, C ! 2 F, C, and

A, find at least the first four nonzero termsq¿ A0B ! 0 E AtB & 0, q A0B ! 10 R AtB ! 1 # t /10

Lq– AtB # Rq¿ AtB # 1 C

q AtB ! E AtB ,

446 Chapter 8 Series Solutions of Differential Equations

in a power series expansion about t ! 0 for the charge on the capacitor.

36. Variable Spring Constant. As a spring is heated, its spring “constant” decreases. Suppose the spring is heated so that the spring “constant” at time t is

N/m (see Figure 8.6). If the unforced mass–spring system has mass m ! 2 kg and a damp- ing constant b ! 1 N-sec/m with initial conditions

m and m/sec, then the displace- ment is governed by the initial value problem

Find at least the first four nonzero terms in a power series expansion about t ! 0 for the displacement.

x¿ A0B ! 0 .x A0B ! 3 , 2x– AtB # x¿ AtB # A6 " tBx(t) ! 0 ; x AtB x¿ A0B ! 0x A0B ! 3

k(t) ! 6 " t

0.1 henrys

R(t) = 1 + t/10 ohms

2 farads q(0) = 10 coulombs q′(0) = 0 amps

Figure 8.5 An RLC circuit whose resistor is being heated

2 kg

k(t) = 6 – t N/m 1 N-sec/m

x(t) x(0) = 3 m x'(0) = 0 m/sec

Figure 8.6 A mass–spring system whose spring is being heated

8.4 EQUATIONS WITH ANALYTIC COEFFICIENTS In Section 8.3 we introduced a method for obtaining a power series solution about an ordinary point. In this section we continue the discussion of this procedure. We begin by stating a basic existence theorem for the equation

(1)

which justifies the power series method.

y– AxB # p AxBy¿ AxB # q AxBy AxB ! 0 , Existence of Analytic Solutions

Theorem 5. Suppose is an ordinary point for equation (1). Then (1) has two linearly independent analytic solutions of the form

(2)

Moreover, the radius of convergence of any power series solution of the form given by (2) is at least as large as the distance from to the nearest singular point (real or complex- valued) of equation (1).

y AxB ! aq n!0

an Ax " x0Bn . x0

The key element in the proof of Theorem 5 is the construction of a convergent geometric series that dominates the series expansion (2) of a solution to equation (1). The convergence of the series in (2) then follows by the comparison test. The details of the proof can be found in more advanced books on differential equations.†

As we saw in Section 8.3, the power series method gives us a general solution in the same form as (2), with and as arbitrary constants. The two linearly independent solutions referred to in Theorem 5 can be obtained by taking for the first and ,

for the second. Thus, we can extend Theorem 5 by saying that equation (1) has a general solution of the form (2) with as the arbitrary constants.

The second part of Theorem 5 gives a simple way of determining a minimum value for the radius of convergence of the power series. We need only find the singular points of equation (1) and then determine the distance between the ordinary point and the nearest singular point.

Find a minimum value for the radius of convergence of a power series solution about x ! 0 to

(3)

For this equation, and Both of these functions are analytic for all real or complex values of x. Since equation (3) has no singular points, the distance between the ordinary point x ! 0 and the nearest singular point is infinite. Hence, the radius of convergence is infinite. ◆

The next example helps to answer the questions posed at the end of the last section.

Find a minimum value for the radius of convergence of a power series solution about x ! 0 to

(4)

Here , and so the singular points of equation (4)

occur when that is, when Since the only singular points of equation (4) are the complex numbers we see that x ! 0 is an ordinary point. Moreover, the distance†† from 0 to either is 1. Thus, the radius of convergence of a power series solution about x ! 0 is at least 1. ◆

In equation (28) of Section 8.3, we found the first few terms of a power series solution to equation (4). Because we now know that this series has radius of convergence at least 1, the partial sums of this series will converge to the solution for However, when

we have no basis on which to decide whether we can use the series to approximate the solution.

Power series expansions about are somewhat easier to manipulate than expansions about nonzero points. As the next example shows, a simple shift in variable enables us always to expand about the origin.

Find the first few terms in a power series expansion about x ! 1 for a general solution to

(5)

Also determine the radius of convergence of the series.

2y– # xy¿ # y ! 0 .

x0 ! 0

0 x 0 ' 1, 0x 0 6 1. %i

%i, x ! %2"1 ! %i.1 # x2 ! 0; q AxB ! 1/ A1 # x2B,p AxB ! "1/ A1 # x2B

A1 # x2By– " y¿ # y ! 0 .

q AxB ! 1/2.p AxB ! x/22y– # xy¿ # y ! 0 . x0

a0 and a1 a1 ! 1

a0 ! 0a0 ! 1, a1 ! 0 a1a0

Section 8.4 Equations with Analytic Coefficients 447

†See, for example, Ordinary Differential Equations, 4th ed., by G. Birkhoff and G.-C. Rota (Wiley, New York, 1989). ††Recall that the distance between the two complex numbers and is given by 2 Aa " cB2 # Ab " dB2. w ! c # diz ! a # bi

Example 1

Solution

Example 2

Solution

Example 3

As seen in Example 1, there are no singular points for equation (5). Thus, x ! 1 is an ordinary point, and, as a consequence of Theorem 5, equation (5) has a general solution of the form

(6)

Moreover, the radius of convergence of the series in (6) must be infinite. We can simplify the computation of the coefficients by shifting the center of the expan-

sion (6) from to This is accomplished by the substitution Setting we find via the chain rule

and, hence, equation (5) is changed into

(7)

We now seek a general solution of the form

(8)

where the ’s in equations (6) and (8) are the same. Proceeding as usual, we substitute the power series for into (7), derive a recurrence relation for the coefficients, and ultimately find that

(the details are left as an exercise). Thus, restoring we have

(9)

◆

When the coefficients of a linear equation are not polynomials in x, but are analytic func- tions, we can still find analytic solutions by essentially the same method.

Find a power series expansion for the solution to

(10)

Here and and both are analytic for all x. Thus, by Theorem 5, the initial value problem (10) has a power series solution

(11)

that converges for all To find the first few terms of this series, we first expand in its Maclaurin series:

ex ! 1 # x # x2

2! #

3! # p .

p AxB ! ex x Ar ! q B. y AxB ! aq

n!0 anx

q AxB ! 1 # x2,p AxB ! exy– AxB # exy¿ AxB # A1 # x2By AxB ! 0 ; y A0B ! 1 , y¿ A0B ! 0 .

# a1 e Ax " 1B " 14 Ax " 1B2 " 18 Ax " 1B3 # p f . y AxB ! a0 e1 " 14 Ax " 1B2 # 124 Ax " 1B3 # p f

t ! x " 1

Y AtB ! a0 e1 " 14 t2 # 124 t3 # p f # a1 e t " 14 t2 " 18 t3 # p f Y AtBan

Y AtB ! aq n!0

ant n ,

2 d2Y dt2

# At # 1B dY dt

# Y ! 0 .

dx !

dt ,

d2y

dx2 !

d2Y

dt2 ,

Y AtB ! y At # 1B, x ! t # 1.t0 ! 0.x0 ! 1 an y AxB ! aq

n!0 an Ax " 1Bn .

448 Chapter 8 Series Solutions of Differential Equations

Solution

Example 4

Solution

Substituting the expansions for and into (10) gives

(12)

Because of the computational difficulties due to the appearance of the product of the power series for and we concern ourselves with just those terms up to order 4. Writing out (12) and keeping track of all such terms, we find

(13)

Grouping the like powers of x in equation (13) (for example, the terms are shown in color) and then setting the coefficients equal to zero yields the system of equations

The initial conditions in (10) imply that and Using these values for and we can solve the above system first for then and so on:

30a6 " 1 24

# 0 # 1 4

" 2 3

# 0 ! 0 1 a6 ! 11 720

20a5 # 0 # 2 3

" 1 2

# 0 ! 0 1 a5 ! " 1

120 ,

12a4 # 1 2

" 3 2

# 0 # 1 ! 0 1 a4 ! 0 ,

6a3 " 1 # 0 ! 0 1 a3 ! 1 6

2a2 # 0 # 1 ! 0 1 a2 ! " 1 2

a3,a2,a1,a0 y¿ A0B ! a1 ! 0.y A0B ! a0 ! 1

30a6 # 5a5 # 5a4 # 3 2

a3 # 4 3

a2 # 1 24

a1 ! 0 Ax4 termB . 20a5 # 4a4 # 4a3 # a2 #

7 6

a1 ! 0 Ax3 termB , 12a4 $ 3a3 $ 3a2 $

1 2 a1 $ a0 " 0 Ax2 termB ,

6a3 # 2a2 # 2a1 ! 0 Ax1 termB , 2a2 # a1 # a0 ! 0 Ax0 termB , x2

x2 # © anxn# Aa0x2 # a1x3 # a2x4 # p B ! 0 . 1 # © anxn# Aa0 # a1x # a2x2 # a3x3 # a4x4 # p B # a 1

24 a1x

4 # p b # a1

6 a1x

3 # 1 3

a2x 4 # p b

2 # © nanxn"1# a12 a1x2 # a2x3 # 32 a3x4 # p b

x # © nanxn"1# Aa1x # 2a2x2 # 3a3x3 # 4a4x4 # p B 1 # © nanxn"1# Aa1 # 2a2x # 3a3x2 # 4a4x3 # 5a5x4 # p B A2a2 # 6a3x # 12a4x2 # 20a5x3 # 30a6x4 # p B

y¿ AxB,ex # A1 # x2B aq

n!0 anx

n ! 0 .

a q

n!2 n An " 1B anxn"2 # a1 # x # x22 # x36 # x424 # p b aqn!1 nanxn"1

exy AxB, y¿ AxB, y– AxB, Section 8.4 Equations with Analytic Coefficients 449

Thus, the solution to the initial value problem in (10) is

(14) ◆

Thus far we have used the power series method only for homogeneous equations. But the same method applies, with obvious modifications, to nonhomogeneous equations of the form

(15)

provided the forcing term and the coefficient functions are analytic at For example, to find a power series about x ! 0 for a general solution to

(16)

we use the substitution to obtain a power series expansion for the left-hand side of (16). We then equate the coefficients of this series with the corresponding coefficients of the Maclaurin expansion for sin x:

Carrying out the details (see Problem 20), we ultimately find that an expansion for a general solution to (16) is

(17)

where

(18)

(19)

are the solutions to the homogeneous equation associated with equation (16) and

(20)

is a particular solution to equation (16).

yp AxB ! 16 x3 # 140 x5 # 195040 x7 # p y2 AxB ! x # 13 x3 # 115 x5 # 1105 x7 # p y1 AxB ! 1 # 12 x2 # 18 x4 # 148 x6 # p , y AxB ! a0y1 AxB # a1y2 AxB # yp AxB , sin x ! a

n!0

A"1BnA2n # 1B! x2n#1 . y AxB ! © anxny– AxB " xy¿ AxB " y AxB ! sin x ,

x0.g AxBy– AxB # p AxBy¿ AxB # q AxBy AxB ! g AxB , y AxB ! 1 " 1

2 x2 #

1 6

x3 " 1

120 x5 #

11 720

x6 # p .

450 Chapter 8 Series Solutions of Differential Equations

In Problems 1–6, find a minimum value for the radius of convergence of a power series solution about

1. 2. 3. 4. 5. 6.

In Problems 7–12, find at least the first four nonzero terms in a power series expansion about for a general solution to the given differential equation with the given value for x0.

A1 # x3By– " xy¿ # 3x2y ! 0 ; x0 ! 1y– " Atan xBy¿ # y ! 0 ; x0 ! 0 Ax2 " 5x # 6By– " 3xy¿ " y ! 0 ; x0 ! 0A1 # x # x2By– " 3y ! 0 ; x0 ! 1 y– " xy¿ " 3y ! 0 ; x0 ! 2 Ax # 1By– " 3xy¿ # 2y ! 0 ; x0 ! 1x0.

7. 8. 9.

10. 11. 12.

In Problems 13–19, find at least the first four nonzero terms in a power series expansion of the solution to the given initial value problem. 13. 14. y¿ " exy ! 0 ; y A0B ! 1x¿ # Asin tBx ! 0 ; x A0B ! 1

y– # A3x " 1By¿ " y ! 0 ; x0 ! "1x2y– " y¿ # y ! 0 ; x0 ! 2 x2y– " xy¿ # 2y ! 0 ; x0 ! 2 Ax2 " 2xBy– # 2y ! 0 ; x0 ! 1y¿ " 2xy ! 0 ; x0 ! "1 y¿ # 2 Ax " 1By ! 0 ; x0 ! 1

8.4 EXERCISES

15.

16.

17.

18.

19.

20. To derive the general solution given by equations (17)–(20) for the nonhomogeneous equation (16), complete the following steps: (a) Substitute and the Maclaurin

series for sin x into equation (16) to obtain

(b) Equate the coefficients of like powers of x on both sides of the equation in part (a) and thereby deduce the equations

In Problems 21–28, use the procedure illustrated in Problem 20 to find at least the first four nonzero terms in a power series expansion about of a general solution to the given differential equation. 21. 22. 23. 24. 25. 26. y– " xy¿ # 2y ! cos x

A1 # x2By– " xy¿ # y ! e"xy– " 2xy¿ # 3y ! x2 z– # xz¿ # z ! x2 # 2x # 1 w¿ # xw ! ex y¿ " xy ! sin x

x ! 0

a7 ! 19

5040 #

a1 105

a5 ! 1 40

# a1 15

, a6 ! a0 48

a2 ! a0 2

, a3 ! 1 6

# a1 3

, a4 ! a0 8

! a q

n!0

A"1BnA2n # 1B! x2n#1 . A2a2 " a0B #aq

k!1 3 Ak # 2B Ak # 1B ak#2 " Ak # 1B ak 4 xk y AxB ! ©qn!0 anxn

y A0B ! "1 , y¿ A0B ! 1y– " e2xy¿ # Acos xBy ! 0 ; y Ap/2B ! 1 , y¿ Ap/2B ! 1y– " Acos xBy¿ " y ! 0 ; y ApB ! 1 , y¿ ApB ! 0y– " Asin xBy ! 0 ; y A0B ! 0 , y¿ A0B ! "1y– # ty¿ # ety ! 0 ; y A0B ! 1 , y¿ A0B ! 1Ax2 # 1By– " exy¿ # y ! 0 ;

Section 8.4 Equations with Analytic Coefficients 451

27. 28.

29. The equation

where n is an unspecified parameter, is called Legendre’s equation. This equation occurs in appli- cations of differential equations to engineering sys- tems in spherical coordinates. (a) Find a power series expansion about x ! 0 for a

solution to Legendre’s equation. (b) Show that for n a nonnegative integer, there

exists an nth-degree polynomial that is a solution to Legendre’s equation. These polyno- mials, up to a constant multiple, are called Legendre polynomials.

30. Aging Spring. As a spring ages, its spring “con- stant” decreases in value. One such model for a mass–spring system with an aging spring is

where m is the mass, b the damping constant, k and positive constants, and the displacement of the

spring from its equilibrium position. Let m ! 1 kg, b ! 2 N-sec/m, k ! 1 N/m, and The system is set in motion by displacing the mass 1 m from its equilibrium position and then releasing it

Find at least the first four nonzero terms in a power series expansion about t ! 0 for the displacement.

31. Aging Spring without Damping. In the mass– spring system for an aging spring discussed in Prob- lem 30, assume that there is no damping (i.e., b ! 0), m ! 1, and k ! 1. To see the effect of aging, consider as a positive parameter. (a) Redo Problem 30 with b ! 0 and arbitrary but

fixed. (b) Set in the expansion obtained in part (a).

Does this expansion agree with the expansion for the solution to the problem with ? [Hint: When the solution is cos t. 4x AtB !h ! 0, h ! 0 h ! 0

Ax A0B ! 1, x¿ A0B ! 0B. h ! 1 AsecB"1.

x AtBh mx– AtB # bx¿ AtB # ke"htx AtB ! 0 ,

A1 " x2By– " 2xy¿ # n An # 1By ! 0 , y– " Asin xBy ! cos xA1 " x2By– " y¿ # y ! tan x

In the previous sections we considered methods for obtaining power series solutions about an ordinary point for a linear second-order equation. However, in certain cases we may want a series expansion about a singular point of the equation. To motivate a procedure for finding such expansions, we consider the class of Cauchy–Euler equations. In Section 4.7 we studied this topic briefly. Here we will use the operator approach to rederive and extend our conclusions.

A second-order homogeneous Cauchy–Euler equation has the form

(1)

where b, and c are (real) constants. Since here and it follows that x ! 0 is a singular point for (1).

Equation (1) has solutions of the form To determine the values for r, we can proceed as follows. Let L be the differential operator defined by the left-hand side of equation (1), that is,

(2)

and set

(3)

When we substitute for in (2), we find

From this, we see that is a solution to (1) if and only if r satisfies

(4)

This equation is referred to as the characteristic, or indicial, equation for (1). When the indicial equation has two distinct roots, we have

from which it follows that equation (1) has the two solutions

(5) (6)

which are linearly independent. When and are complex conjugates, we can use Euler’s formula to express

Since the real and imaginary parts of must also be solutions to (1), we can replace (5) and (6) by the two linearly independent real-valued solutions

If the indicial equation (4) has a repeated real root then it turns out that and ln x are two linearly independent solutions. This can be deduced using the reduction of order approach of Section 4.7. However, it is more instructive for later applications to see how these

xr0xr0r0,

y1 AxB ! xa cos Ab ln xB , y2 AxB ! xa sin Ab ln xB . xa# ib

! xa cos Ab ln xB # ixa sin Ab ln xB .xa# ib ! e Aa# ibB ln x ! ea ln x cos Ab ln xB # iea ln x sin Ab ln xB a % ib,r2r1

y2 AxB ! w Ar2, xB ! xr2 , x 7 0 ,y1 AxB ! w Ar1, xB ! xr1 , x 7 0 , L 3w 4 AxB ! a Ar " r1B Ar " r2Bxr , ar2 $ Ab # aBr $ c " 0 .w ! xr

! Ear 2 # Ab " aBr # cFxr .L 3w 4 AxB ! ax2r Ar " 1Bxr"2 # bxrxr"1 # cxr y AxBw Ar, xBw Ar, xB J xr .

L 3 y 4 AxB J ax2y– AxB # bxy¿ AxB # cy AxB ,y ! x r.

q AxB ! c/ Aax2B,p AxB ! b/ AaxBa A&0B,ax 2y– AxB $ bxy¿ AxB $ cy AxB " 0 , x 7 0 ,

452 Chapter 8 Series Solutions of Differential Equations

8.5 CAUCHY–EULER (EQUIDIMENSIONAL) EQUATIONS

two linearly independent solutions can be obtained via the operator approach. If is a repeated root, then

(7)

Setting gives the solution

(8)

To find a second linearly independent solution, we make the following observation. The right- hand side of (7) has the factor so taking the partial derivative of (7) with respect to r and setting we get zero. That is,

(9)

While it may not appear that we have made any progress toward finding a second solution, a closer look at the expression on the left in (9) will soon vindicate our efforts.

First note that has continuous partial derivatives of all orders with respect to both r and x. Hence, the mixed partial derivatives are equal:

Consequently, for the differential operator L, we have

that is, the operators and L commute. With this fact, (9) can be written as

Thus for the case of a repeated root a second linearly independent solution to (1) is

(10)

Find a general solution to

(11)

Let and let L denote the left-hand side of (11). A short calculation gives

Solving the indicial equation

4r 2 " 4r # 1 ! A2r " 1B2 ! 4 ar " 1 2 b 2 ! 0

L 3w 4 AxB ! A4r 2 " 4r # 1Bxr .w Ar, xB ! xr 4x2y– AxB # y AxB ! 0 , x 7 0 . y2 AxB ! 0w0r Ar0, xB ! 00r AxrB ` r!r0 ! xr0 ln x , x 7 0 .

r0,

L c 0w0r d ` r"r0 ! 0 . 0 / 0r

! L c 0w0r d ; ! ax2

03w 0x20r

# bx 02w 0x0r

# c 0w 0r

! ax2 03w

0r0x2 # bx

02w 0r0x

# c 0w 0r

0 0r

L 3w 4 ! 0 0r

eax2 02w 0x2

# bx 0w 0x

# cw f 03w

0r0x2 !

03w 0x20r

, 0 2w

0r0x !

02w 0x0r

w Ar, xB ! xr 0 0r EL 3w 4 AxB F ` r!r0 ! Ea Ar " r0B2xr ln x # 2a Ar " r0BxrF ` r!r0 ! 0 .

r ! r0, Ar " r0B2,

y1 AxB ! w Ar0, xB ! xr0 , x 7 0 .r ! r0 L 3w 4 AxB ! a Ar " r0B2xr .

Section 8.5 Cauchy–Euler (Equidimensional) Equations 453

Example 1

Solution

yields the repeated root Thus, a general solution to (11) is obtained from equations (8) and (10) by setting That is,

◆

In closing we note that the solutions to Cauchy–Euler equations can exhibit some extraor- dinary features near their singular points—features unlike anything we have encountered in the case of constant coefficient equations. Observe that:

1. and for 0 < r < 1, have infinite slope at the origin; 2. for r < 0, grows without bound near the origin; and 3. and oscillate “infinitely rapidly” near the origin. (See

Figure 8.7.)

Numerical solvers based on Euler or Runge–Kutta procedures would be completely frustrated in trying to chart such behavior; analytic methods are essential for characterizing solutions near singular points.

In Section 8.7 we discuss the problem of finding a second linearly independent series solution to certain differential equations. As we will see, operator methods similar to those described in this section will lead to the desired second solution.

xa sin Ab ln xBxa cos Ab ln xBxr, x r ln x,xr

y AxB ! C11x # C21x ln x , x 7 0 .r0 ! 1/2. r0 ! 1/2.

454 Chapter 8 Series Solutions of Differential Equations

−1

−0.5

0.5

0.2 0.4−0.4 −0.2 0.6 0.8 1 1.2 1.4 1.6 1.8 2

Figure 8.7 Graph of x0.3cos A12 ln xB

8.5 EXERCISES In Problems 1–10, use the substitution to find a general solution to the given equation for

1. 2. 3. 4.

6. d 2y

dx2 !

dx "

x2 y

d 2y

dx2 !

dx "

x2 y

x2y– AxB # 2xy¿ AxB " 3y AxB ! 0x2y– AxB " xy¿ AxB # 17y AxB ! 0 2x2y– AxB # 13xy¿ AxB # 15y AxB ! 0x2y– AxB # 6xy¿ AxB # 6y AxB ! 0

x 7 0. y ! xr 7.

8. 9.

10.

In Problems 11 and 12, use a substitution of the form to find a general solution to the given

equation for 11. 12. 4 Ax # 2B2y– AxB # 5y AxB ! 02 Ax " 3B2y– AxB # 5 Ax " 3By¿ AxB " 2y AxB ! 0

x 7 c. y ! Ax " cBr

x3y‡ AxB # 9x2y– AxB # 19xy¿ AxB # 8y AxB ! 0x3y‡ AxB # 3x2y– AxB # 5xy¿ AxB " 5y AxB ! 0 x3y‡ AxB # 4x2y– AxB # xy¿ AxB ! 0x3y‡ AxB # 4x2y– AxB # 10xy¿ AxB " 10y AxB ! 0

In Problems 13 and 14, use variation of parameters to find a general solution to the given equation for 13. 14.

In Problems 15–17, solve the given initial value problem. 15.

16.

17.

18. Suppose is a repeated root of the auxiliary equation Then, as we well ar2 # br # c ! 0.

y– A1B ! 19y A1B ! 2 , y¿ A1B ! "3 , x3y‡ AxB # 6x2y– AxB # 29xy¿ AxB " 29y AxB ! 0 ; y A1B ! 3 , y¿ A1B ! 7x2y– AxB # 5xy¿ AxB # 4y AxB ! 0 ; x A1B ! 3 , x¿ A1B ! 5t2x– AtB " 12x AtB ! 0 ; x2y– AxB # 2xy¿ AxB " 2y AxB ! 6x"2 # 3xx2y– AxB " 2xy¿ AxB # 2y AxB ! x"1/ 2

x 7 0.

Section 8.6 Method of Frobenius 455

know, is a solution to the equation where a, b, and c are con-

stants. Use a derivation similar to the one given in this section for the case when the indicial equation has a repeated root to show that a second linearly independent solution is

19. Let (a) Show that (b) Using an extension of the argument given in this

section for the case when the indicial equation has a double root, show that has the general solution

y AxB ! C1x # C2x ln x # C3x Aln xB2 . L 3 y 4 ! 0

L 3 xr 4 AxB ! Ar " 1B3xr .L 3 y 4 AxB J x3y‡ AxB # xy¿ AxB " y AxB .y2 AtB ! te r0t.

by¿ # cy ! 0,ay– # y1 AtB ! er0t

8.6 METHOD OF FROBENIUS In the previous section we showed that a homogeneous Cauchy–Euler equation has a solution of the form where r is a certain constant. Cauchy–Euler equations have, of course, a very special form with only one singular point (at x ! 0). In this section we show how the theory for Cauchy–Euler equations generalizes to other equations that have a special type of singularity.

To motivate the procedure, let’s rewrite the Cauchy–Euler equation,

(1)

in the standard form

(2)

where

and are the constants and , respectively. When we substitute for y into equation (2), we get

which yields the indicial equation

(3)

Thus, if is a root of (3), then is a solution to equations (1) and (2). Let’s now assume, more generally, that (2) is an equation for which and

instead of being constants, are analytic functions. That is, in some open interval about x ! 0,

(4)

(5) x2q AxB ! q0 # q1x # q2x2 # p ! aq n!0

qnx n .

xp AxB ! p0 # p1x # p2x2 # p ! aq n!0

pnx n ,

x2q AxB,xp AxBw Ar1, xB ! xr1r1 r Ar " 1B # p0r # q0 ! 0 . 3 r Ar " 1B # p0r # q0 4 xr"2 ! 0 , w

Ar, xB ! xrc/ab/ap0, q0 p AxB ! p0

x , q AxB ! q0

x2 ,

y– AxB # p AxBy¿ AxB # q AxBy AxB ! 0 , x 7 0 , ax2y– AxB # bxy¿ AxB # cy AxB ! 0 , x 7 0 ,

y AxB ! xr, x 7 0,

It follows from (4) and (5) that

(6) and

and hence, for x near 0 we have and Therefore, it is reasonable to expect that the solutions to (2) will behave (for x near 0) like the solutions to the Cauchy–Euler equation

When and satisfy (4) and (5), we say that the singular point at x ! 0 is regular. More generally, we state the following.

q AxBp AxBx 2y– # p0xy¿ # q0y ! 0 .

x2q AxB ! q0.xp AxB ! p0 limxS0 x2q AxB ! q0 ,limxS0 xp AxB ! p0 456 Chapter 8 Series Solutions of Differential Equations

Classify the singular points of the equation

(8)

Here

from which we see that %1 are the singular points of (8). For the singularity at 1, we have

which is not analytic at x ! 1. Therefore, x ! 1 is an irregular singular point. For the singularity at we have

both of which are analytic at Hence, is a regular singular point. ◆

Let’s assume that x ! 0 is a regular singular point for equation (7) so that and satisfy (4) and (5); that is,

(9) p AxB ! aq n!0

pnx n"1 , q AxB ! aq

n!0 qnx

n"2 .

q AxBp AxBx ! "1x ! "1. Ax # 1Bp AxB ! 1Ax " 1B2 , Ax # 1B2q AxB ! "1Ax " 1B2 ,

"1,

Ax " 1Bp AxB ! 1Ax # 1B Ax " 1B , q AxB ! "1Ax2 " 1B2 ! "1Ax # 1B2 Ax " 1B2 , p AxB ! x # 1Ax2 " 1B2 ! 1Ax # 1B Ax " 1B2 , Ax2 " 1B2y– AxB # Ax # 1By¿ AxB " y AxB ! 0 .

Regular Singular Point

Definition 3. A singular point of

(7)

is said to be a regular singular point if both and are analytic at † Otherwise is called an irregular singular point.x0x0.

Ax " x0B2q AxBAx " x0Bp AxBy– AxB # p AxBy¿ AxB # q AxBy AxB ! 0 x0

†In the terminology of complex variables, p has a pole of order at most 1, and q has a pole of order at most 2, at x0.

Example 1

Solution

The idea of the mathematician Frobenius was that since Cauchy–Euler equations have solu- tions of the form then for the regular singular point x ! 0, there should be solutions to (7) of the form times an analytic function.† Hence we seek solutions to (7) of the form

(10)

In writing (10), we have assumed is the first nonzero coefficient, so we are left with determining r and the coefficients Differentiating with respect to x, we have

(11)

(12)

If we substitute the above expansions for and into (7), we obtain

(13)

Now we use the Cauchy product to perform the series multiplications and then group like powers of x, starting with the lowest power, This gives

(14)

For the expansion on the left-hand side of equation (14) to sum to zero, each coefficient must be zero. Considering the first term, we find

(15)

We have assumed that so the quantity in brackets must be zero. This gives the indicial equation; it is the same as the one we derived for Cauchy–Euler equations.

a0 & 0,

3 r Ar " 1B # p0r # q0 4a0 ! 0 .xr"2, # 3 Ar # 1Bra1 # Ar # 1Bp0a1 # p1ra0 # q0a1 # q1a0 4 xr"1 # p ! 0 .3 r Ar " 1B # p0r # q0 4a0xr"2

xr"2.

# aaq n!0

qnx n"2b aaq

n!0 anx

n#rb ! 0 . a q

n!0 An # rB An # r " 1Banxn#r"2 # aaq

n!0 pnx

n"1b aaq n!0 An # rBanxn#r"1bq AxBw Ar, xB, w¿Ar, xB, w–Ar, xB, p AxB,

w– Ar, xB ! aq n!0 An # rB An # r " 1Banxn#r"2 .

w¿ Ar, xB ! aq n!0 An # rBanxn#r"1 ,

w Ar, xBan, n ' 1.a0 w Ar, xB " xr a!

n"0 anxn " a

n"0 anxn$r , x 7 0 .

xr xr,

Section 8.6 Method of Frobenius 457

Indicial Equation

Definition 4. If is a regular singular point of then the indicial equation for this point is

(16)

where

The roots of the indicial equation are called the exponents (indices) of the singularity x0.

p0 J limxSx0 Ax " x0Bp AxB , q0 J limxSx0 Ax " x0B2q AxB .

r Ar # 1B $ p0r $ q0 " 0 , y– # py¿ # qy ! 0,x0

†Historical Footnote: George Frobenius (1848–1917) developed this method in 1873. He is also known for his research on group theory.

Find the indicial equation and the exponents at the singularity of

(17)

In Example 1 we showed that is a regular singular point. Since ! and we find

Substituting these values for and into (16), we obtain the indicial equation

(18)

Multiplying by 4 and factoring gives Hence, are the exponents. ◆

As we have seen, we can use the indicial equation to determine those values of r for which the coefficient of in (14) is zero. If we set the coefficient of in (14) equal to zero, we have

(19)

Since is arbitrary and we know the ’s, ’s, and r, we can solve equation (19) for provided the coefficient of in (19) is not zero. This will be the case if we take r to be the larger of the two roots of the indicial equation (see Problem 43).† Similarly, when we set the coefficient of equal to zero, we can solve for in terms of the ’s, ’s, r, and Con- tinuing in this manner, we can recursively solve for the ’s. The procedure is illustrated in the following example.

Find a series expansion about the regular singular point x ! 0 for a solution to

(20)

Here and so

Since x ! 0 is a regular singular point, we seek a solution to (20) of the form

(21) w Ar, xB ! xr aq n!0

anx n ! a

n!0 anx

n#r .

q0 ! limxS0 x 2q AxB ! lim

xS0 Ax # 2B"1 A1 # xB ! 1

2 .

p0 ! limxS0 xp AxB ! lim

xS0 3" Ax # 2B"1 4 ! " 1

2 ,

q AxB ! x"2 Ax # 2B"1 A1 # xB,p AxB ! "x"1 Ax # 2B"1 Ax # 2Bx2y– AxB " xy¿ AxB # A1 # xBy AxB ! 0 , x 7 0 .

an a1.a0,qipia2x

a1 a1,qipia0

3 Ar # 1Br # Ar # 1Bp0 # q0 4a1 # Ap1r # q1Ba0 ! 0 . xr"1xr"2

r ! 1, "1/4A4r # 1B Ar " 1B ! 0. r Ar " 1B # 1

4 r "

1 4

! 0 .

q0p0

q0 ! limxS"1 Ax # 1B2q AxB ! lim

xS"1 3" Ax " 1B"2 4 ! " 1

4 .

p0 ! limxS"1 Ax # 1Bp AxB ! lim

xS"1 Ax " 1B"2 ! 1

4 ,

q AxB ! " Ax # 1B"2 Ax " 1B"2,Ax # 1B"1 Ax " 1B"2 p AxBx ! "1 Ax2 " 1B2y– AxB # Ax # 1By¿ AxB " y AxB ! 0 . x ! "1

458 Chapter 8 Series Solutions of Differential Equations

Example 3

Solution

Example 2

Solution

†“Larger” in the sense of Problem 43.

By the previous discussion, r must satisfy the indicial equation (16). Substituting for and in (16), we obtain

which simplifies to Thus, r ! 1 and are the roots of the indicial equation associated with x ! 0.

Let’s use the larger root r ! 1 and solve for etc., to obtain the solution We can simplify the computations by substituting directly into equation (20), where the coefficients are polynomials in x, rather than dividing by and having to work with the rational functions and Inserting in (20) and recalling the formulas for

and in (11) and (12) gives (with r ! 1)

(22)

which we can write as

(23)

Next we shift the indices so that each summation in (23) is in powers With in the first and last summations and in the rest, (23) becomes

(24)

Separating off the k ! 1 term and combining the rest under one summation yields

(25)

Notice that the coefficient of x in (25) is zero. This is because r ! 1 is a root of the indicial equation, which is the equation we obtained by setting the coefficient of the lowest power of x equal to zero.

We can now determine the ’s in terms of by setting the coefficients of in equation (25) equal to zero for k ! 2, 3, etc. This gives the recurrence relation

(26)

or, equivalently,

(27) ak"1 ! " k2 " 3k # 3A2k " 1B Ak " 1B ak"2 , k ' 2 .

Ak2 " 3k # 3Bak"2 # A2k " 1B Ak " 1Bak"1 ! 0 , xka0ak

32 A1B A0B " 1 # 1 4a0x # aq k!2 3 Ak2 " 3k # 3Bak"2 # A2k " 1B Ak " 1Bak"1 4 xk ! 0 .

a q

k!2 3 Ak " 1B Ak " 2B # 1 4ak"2 xk # aq

k!1 32k Ak " 1B " k # 1 4ak"1xk ! 0 .

k ! n # 1 k ! n # 2xk.

# a q

n!0 anx

n#1 # a q

n!0 anx

n#2 ! 0 .

a q

n!0 An # 1Bnanxn#2 # aq

n!0 2 An # 1Bnanxn#1 " aq

n!0 An # 1Banxn#1

# A1 # xB aq n!0

anx n#1 ! 0 ,

Ax # 2Bx2 aq n!0 An # 1Bnanxn"1 " x aq

n!0 An # 1Banxn

w– Ar, xBw¿ Ar, xB w Ar, xBq AxB.p AxB Ax # 2Bx2w Ar, xB w

A1, xB.a1, a2, r ! 1/22r 2 " 3r # 1 ! A2r " 1B Ar " 1B ! 0.

r Ar " 1B " 1 2

r # 1 2

! 0 ,

q0p0

Section 8.6 Method of Frobenius 459

Setting k ! 2, 3, and 4 in (27), we find

Substituting these values for and into (21) gives

(28)

where is arbitrary. In particular, for we get the solution

See Figure 8.8. ◆

To find a second linearly independent solution to equation (20), we could try setting and solving for . . . to obtain a solution (see Problem 44). In this par-

ticular case, the approach would work. However, if we encounter an indicial equation that has a repeated root, then the method of Frobenius would yield just one solution (apart from constant multiples). To find the desired second solution, we must use another technique, such as the reduction of order procedure discussed in Section 4.7 or Exercises 6.1, Problem 31. We tackle the problem of finding a second linearly independent solution in the next section.

w A1/2, xBa1, a2,r ! 1/2 y1 AxB ! x " 13 x2 # 110 x3 " 130 x4 # p Ax 7 0B .

a0 ! 1,a0

w A1, xB ! a0x1 a1 " 13 x # 110 x2 " 130 x3 # p b , a3r, a1, a2,

a3 ! " 1 3

a2 ! " 1 30

a0 Ak ! 4B . a2 ! "

3 10

a1 ! 1 10

a0 Ak ! 3B , a1 ! "

1 3

a0 Ak ! 2B ,

460 Chapter 8 Series Solutions of Differential Equations

degree 4

degree 20

2 3 4

−1

−1−2−3

−2

−3

−4

−5

Figure 8.8 Partial sums approximating the solution of Example 3y1 AxB

One important question that remains concerns the radius of convergence of the power series that appears in (31). The following theorem contains an answer.†

Section 8.6 Method of Frobenius 461

Method of Frobenius To derive a series solution about the singular point of

(29)

(a) Set If both and are analytic at then is a regular singular point and the remaining steps apply.

(b) Let

(30)

and, using termwise differentiation, substitute into equation (29) to obtain an equation of the form

(c) Set the coefficients . . . equal to zero. [Notice that the equation is just a constant multiple of the indicial equation

(d) Use the system of equations

to find a recurrence relation involving and (e) Take the larger root of the indicial equation, and use the relation obtained in

step (d) to determine . . . recursively in terms of and (f ) A series expansion of a solution to (29) is

(31)

where is arbitrary and the ’s are defined in terms of and r1.a0ana0

w Ar1, xB ! Ax " x0Br1 aq n!0

an Ax " x0Bn , x 7 x0 , r1.a0a1, a2,

r ! r1, a0, a1, . . . , ak"1.ak

A0 ! 0 , A1 ! 0 , . . . , Ak ! 0

r Ar " 1B # p0r # q0 ! 0.]A0 ! 0A0, A1, A2, A0 Ax " x0Br#J # A1 Ax " x0Br#J#1 # p ! 0 .

w Ar, xB w Ar, xB ! Ax " x0Br aq

n!0 an Ax " x0Bn ! aq

n!0 an Ax " x0Bn#r ,

x0x0, Ax " x0B2q AxBAx " x0Bp AxBp AxB J a1 AxB /a2 AxB, q AxB J a0 AxB /a2 AxB.a2 AxBy– AxB # a1 AxBy¿ AxB # a0 AxBy AxB ! 0 , x 7 x0 :

†For a proof of this theorem, see Ordinary Differential Equations, by E. L. Ince (Dover Publications, New York, 1956), Chapter 16.

Frobenius’s Theorem

Theorem 6. If is a regular singular point of equation (29), then there exists at least one series solution of the form (30), where is the larger root of the associated indicial equation. Moreover, this series converges for all x such that

where R is the distance from to the nearest other singular point (real or complex) of (29).

x0x " x0 6 R,0 6

r ! r1 x0

For simplicity, in the examples that follow we consider only series expansions about the regular singular point x ! 0 and only those equations for which the associated indicial equa- tion has real roots.

The method of Frobenius can be summarized as follows.

The following three examples not only illustrate the method of Frobenius but also are important models to which we refer in later sections.

Find a series solution about the regular singular point x ! 0 of

(32)

Here and It is easy to check that x ! 0 is a regular singular point of (32), so we compute

Then the indicial equation is

which has the roots Next we substitute

(33)

into (32) and obtain

(34)

which we write as

(35)

Shifting the indices so that each summation in (35) is in powers we take in the last summation and in the rest. This gives

(36)

Singling out the term corresponding to k ! 0 and combining the rest under one summation yields

(37)

When we set the coefficients equal to zero, we recover the indicial equation

(38) 3 r Ar " 1B " r # 1 4a0 ! 0 , # a

k!1 E 3 Ak # rB Ak # r " 1B " Ak # rB # 1 4ak " ak"1Fxk#r ! 0 .3 r Ar " 1B " r # 1 4a0xr

a q

k!0 3 Ak # rB Ak # r " 1B " Ak # rB # 1 4akxk#r " aq

k!1 ak"1x

k#r ! 0 .

k ! n k ! n # 1xk#r,

# a q

n!0 anx

n#r " a q

n!0 anx

n#r#1 ! 0 .

a q

n!0 An # rB An # r " 1Banxn#r " aq

n!0 An # rBanxn#r

# A1 " xB aq n!0

anx n#r ! 0 ,

x2 a q

n!0 An # rB An # r " 1Banxn#r"2 " x aq

n!0 An # rBanxn#r"1

w Ar, xB ! xr aq n!0

anx n ! a

n!0 anx

n#r

r1 ! r2 ! 1.

r Ar " 1B " r # 1 ! r 2 " 2r # 1 ! Ar " 1B2 ! 0 , q0 ! limxS0 x

2q AxB ! lim xS0 A1 " xB ! 1 .p0 ! limxS0 xp AxB ! limxS0 A"1B ! "1 ,

q AxB ! A1 " xBx"2.p AxB ! "x"1x 2y– AxB " xy¿ AxB # A1 " xBy AxB ! 0 , x 7 0 .

462 Chapter 8 Series Solutions of Differential Equations

Example 4

Solution

and obtain for the recurrence relation

(39)

which reduces to

(40)

Relation (40) can be used to solve for in terms of

(41)

Setting in (38) gives (as expected) and in (41) gives

(42)

For k ! 1, 2, and 3, we now find

In general, we have

(43)

Hence, equation (32) has a series solution given by

(44)

◆

Since x ! 0 is the only singular point for equation (32), it follows from Frobenius’s theo- rem or directly by the ratio test that the series solution (44) converges for all x > 0.

In the next two examples, we only outline the method; we leave it to you to furnish the intermediate steps.

Find a series solution about the regular singular point x ! 0 of

(45)

Since and we see that x ! 0 is indeed a regular singular point and

The indicial equation is

with roots and r2 ! "3.r1 ! 0

r Ar " 1B # 4r ! r2 # 3r ! r Ar # 3B ! 0 , p0 ! limxS0 xp

AxB ! 4 , q0 ! limxS0 x2q AxB ! 0 . q AxB ! "1,p AxB ! 4/xxy–

AxB # 4y¿ AxB " xy AxB ! 0 , x 7 0 .

! a0x a q

k!0

1Ak!B2 xk , x 7 0 . w A1, xB ! a0x e1 # x # 14 x2 # 136 x3 # p f ak !

1Ak!B2 a0 . a3 !

1 32

a2 ! 1A3 # 2 # 1B2 a0 ! 136 a0 Ak ! 3B .

a2 ! 1 22

a1 ! 1A2 # 1B2 a0 ! 14 a0 Ak ! 2B ,

a1 ! 1 12

a0 ! a0 Ak ! 1B , ak "

1 k2

ak#1 , k ' 1 .

0 # a0 ! 0,r ! r1 ! 1 ak !

1Ak # r " 1B2 ak"1 , k ' 1 . ak"1:ak

Ak # r " 1B2ak " ak"1 ! 0 . 3 Ak # rB2 " 2 Ak # rB # 1 4ak " ak"1 ! 0 ,k ' 1,

Section 8.6 Method of Frobenius 463

Example 5

Solution

Now we substitute

(46)

into (45). After a little algebra and a shift in indices, we get

(47)

Next we set the coefficients equal to zero and find

(48) (49)

and, for the recurrence relation

(50)

For equation (48) becomes and (49) becomes Hence, although is arbitrary, must be zero. Setting in (50), we find

(51)

from which it follows (after a few experimental computations) that for k ! 0, 1, . . . , and

(52)

Hence equation (45) has the power series solution

(53) ◆

If in Example 5 we had worked with the root then we would actually have obtained two linearly independent solutions (see Problem 45).

Find a series solution about the regular singular point of x ! 0 of

(54)

Since and we see that x ! 0 is a regular singular point. Moreover,

So the indicial equation is

(55)

with roots and Substituting

(56) w Ar, xB ! xr aq n!0

anx n ! a

n!0 anx

n#r

r2 ! "2.r1 ! 0

r Ar " 1B # 3r ! r2 # 2r ! r Ar # 2B ! 0 , p0 ! limxS0 xp

AxB ! 3 , q0 ! limxS0 x2q AxB ! 0 .q AxB ! "1,p AxB ! 3/x xy– AxB # 3y¿ AxB " xy AxB ! 0 , x 7 0 .

r ! r2 ! "3,

wA0, xB ! a0 e1 #aq k!1

1 2kk! 35 # 7 p A2k # 3B 4 x2kf , x 7 0 .

! 1

2kk! 35 # 7 p A2k # 3B 4 a0 , k ' 1 . a2k !

132 # 4 p A2kB 4 35 # 7 p A2k # 3B 4 a0 a2k#1 ! 0

ak$1 " 1Ak $ 1B Ak $ 4B ak#1 , k ' 1 ,

r ! r1 ! 0a1a0 4 # a1 ! 0.0 # a0 ! 0r ! r1 ! 0,

Ak # r # 1B Ak # r # 4Bak#1 " ak"1 ! 0 .k ' 1, 3 Ar # 1Br # 4 Ar # 1B 4a1 ! 0 ,3 r Ar " 1B # 4r 4a0 ! 0 ,

# a q

k!1 3 Ak # r # 1B Ak # r # 4Bak#1 " ak"1 4 xk#r ! 0 .3 r Ar " 1B # 4r 4a0xr"1 # 3 Ar # 1Br # 4 Ar # 1B 4a1xr

w Ar, xB ! xr aq n!0

anx n ! a

n!0 anx

n#r

464 Chapter 8 Series Solutions of Differential Equations

Example 6

Solution

into (54) ultimately gives

(57)

Setting the coefficients equal to zero, we have

(58) (59)

and, for the recurrence relation

(60)

With these equations lead to the following formulas: k ! 0, 1, . . . , and

(61)

Hence equation (54) has the power series solution

(62) ◆

Unlike in Example 5, if we work with the second root in Example 6, then we do not obtain a second linearly independent solution (see Problem 46).

In the preceding examples we were able to use the method of Frobenius to find a series solution valid to the right (x > 0) of the regular singular point x ! 0. For x < 0, we can use the change of variables and then solve the resulting equation for t > 0.

The method of Frobenius also applies to higher-order linear equations (see Problems 35–38).

x ! "t

r ! r2 ! "2

w A0, xB ! a0 aq k!0

22kk! Ak # 1B! x2k , x 7 0 . a2k !

132 # 4 p A2kB 4 34 # 6 p A2k # 2B 4 a0 ! 122kk! Ak # 1B! a0 , k ' 0 . a2k#1 ! 0,r ! r1 ! 0,

Ak $ r $ 1B Ak $ r $ 3Bak$1 # ak#1 " 0 .k ' 1, 3 Ar # 1Br # 3 Ar # 1B 4a1 ! 0 ,3 r Ar " 1B # 3r 4a0 ! 0 ,

# a q

k!1 3 Ak # r # 1B Ak # r # 3Bak#1 " ak"1 4 xk#r ! 0 .3 r Ar " 1B # 3r 4a0xr"1 # 3 Ar # 1Br # 3 Ar # 1B 4a1xr

Section 8.6 Method of Frobenius 465

8.6 EXERCISES In Problems 1–10, classify each singular point (real or complex) of the given equation as regular or irregular.

1. 2. 3. 4. 5. 6. 7. 8. 9.

10.

In Problems 11–18, find the indicial equation and the exponents for the specified singularity of the given differ- ential equations.

x3 Ax " 1By– # Ax2 " 3xB Asin xBy¿ " xy ! 0Ax2 # 2x " 8B2y– # A3x # 12By¿ " x2y ! 0 Ax2 " xBy– # xy¿ # 7y ! 0At2 " t " 2B2x– # At2 " 4Bx¿ " tx ! 0 Ax2 " 4By– # Ax # 2By¿ # 3y ! 0Ax2 " 1B2y– " Ax " 1By¿ # 3y ! 0 x2y– " 5xy¿ # 7y ! 0 Ax2 # 1Bz– # 7x2z¿ " 3xz ! 0x2y– # 8xy¿ " 3xy ! 0 Ax2 " 1By– # xy¿ # 3y ! 0

11. at 12. at 13.

at 14.

at 15. at 16. at 17.

at 18. at

In Problems 19–24, use the method of Frobenius to find at least the first four nonzero terms in the series expan- sion about for a solution to the given equation for

19. 9x2y– # 9x2y¿ # 2y ! 0 x 7 0.

x ! 0

x ! 04x Asin xBy– " 3y ! 0 , x ! 1 Ax " 1B2y– # Ax2 " 1By¿ " 12y ! 0 , x ! 1Ax2 " 1By– " Ax " 1By¿ " 3y ! 0 ,

u ! 0u3y– # u Asin uBy¿ " Atan uBy ! 0 , x ! "2 Ax2 " 4By– # Ax # 2By¿ # 3y ! 0 , x ! 2 Ax2 " x " 2B2z– # Ax2 " 4Bz¿ " 6xz ! 0 , x ! 0x2y– # 4xy¿ # 2y ! 0 ,

x ! 0x2y– " 2xy¿ " 10y ! 0 ,

20. 21. 22. 23. 24.

In Problems 25–30, use the method of Frobenius to find a general formula for the coefficient in a series expan- sion about for a solution to the given equation for

25. 26. 27. 28. 29. 30.

In Problems 31–34, first determine a recurrence formula for the coefficients in the (Frobenius) series expansion of the solution about Use this recurrence formula to determine if there exists a solution to the differential equation that is decreasing for 31. 32. 33. 34.

In Problems 35–38, use the method of Frobenius to find at least the first four nonzero terms in the series expan- sion about for a solution to the given linear third- order equation for 35. 36. 37. 38.

In Problems 39 and 40, try to use the method of Frobenius to find a series expansion about the irregular singular point for a solution to the given differen- tial equation. If the method works, give at least the first four nonzero terms in the expansion. If the method does not work, explain what went wrong. 39. 40.

In certain applications, it is desirable to have an expan- sion about the point at infinity. To obtain such an expan- sion, we use the change of variables and expandz ! 1/x

x2y– # y¿ " 2y ! 0 x2y– # A3x " 1By¿ # y ! 0

x ! 0

6x3y‡ # A13x2 " x3By– # xy¿ " xy ! 06x3y‡ # 13x2y– " Ax2 # 3xBy¿ " xy ! 0 6x3y‡ # 11x2y– " 2xy¿ " Ax " 2By ! 06x3y‡ # 13x2y– # Ax # x2By¿ # xy ! 0

x 7 0. x ! 0

xy– # Ax # 2By¿ " y ! 03xy– # 2 A1 " xBy¿ " 4y ! 0 x2y– " x A1 # xBy¿ # y ! 0xy– # A1 " xBy¿ " y ! 0

x 7 0.

x ! 0.

x Ax # 1By– # Ax # 5By¿ " 4y ! 0xy– # Ax " 1By¿ " 2y ! 0 3x2y– # 8xy¿ # Ax " 2By ! 0xw– " w¿ " xw ! 0 x2y– # Ax2 " xBy¿ # y ! 04x2y– # 2x2y¿ " Ax # 3By ! 0

x 7 0. x ! 0

3xy– # A2 " xBy¿ " y ! 0x2z– # Ax2 # xBz¿ " z ! 0 xy– # y¿ " 4y ! 0 x2y– # xy¿ # x2y ! 0 2x Ax " 1By– # 3 Ax " 1By¿ " y ! 0

466 Chapter 8 Series Solutions of Differential Equations

about In Problems 41 and 42, show that infinity is a regular singular point of the given differential equation by showing that is a regular singular point for the transformed equation in z. Also find at least the first four nonzero terms in the series expansion about infinity of a solution to the original equation in x. 41. 42.

43. Show that if and are roots of the indicial equa- tion (16), with the larger root (Re then the coefficient of in equation (19) is not zero when

44. To obtain a second linearly independent solution to equation (20): (a) Substitute given in (21) into (20) and

conclude that the coefficients must satisfy the recurrence relation

(b) Use the recurrence relation with to derive the second series solution

45. In Example 5, show that if we choose then we obtain two linearly independent solutions to equation (45). [Hint: and are arbitrary constants.]

46. In Example 6, show that if we choose then we obtain a solution that is a constant multiple of the solution given in (62). [Hint: Show that and

must be zero while is arbitrary.] 47. In applying the method of Frobenius, the following

recurrence relation arose: k ! 0, 1, 2, . . . . (a) Show that the coefficients are given by the

formula k ! 0, 1, 2, . . . . (b) Use the formula obtained in part (a) with

to compute and on your computer or calculator. What goes wrong?

(d) What advantage does the recurrence relation have over the formula?

a0 ! 1. ak

a25a5, a10, a15, a20, a0 ! 1

ak ! 15 7ka0/ Ak!B9,

ak#1 ! 15 7ak/ Ak # 1B9,

a2a1 a0

r ! r2 ! "2, a3a0

r ! r2 ! "3, w A1, xB

" 133

1920 x7/2 # p b .a0 ax1/2 " 34 x3/2 # 732 x5/2

w a1 2

, xb ! r ! 1/2

# 3 Ak # r " 1B Ak # r " 2B # 1 4 ak"1 ! 0 .Ak # r " 1B A2k # 2r " 1Bak ak, k ' 1,

w Ar, xB r ! r1 .

a1 r1 ' Re r2B,r1 r2r1

18 Ax " 4B2 Ax " 6By– # 9x Ax " 4By¿ " 32y ! 0x3y– " x2y¿ " y ! 0 z ! 0

z ! 0.

In the previous section we showed that if x ! 0 is a regular singular point of

(1)

then the method of Frobenius can be used to find a series solution valid for x near zero. The first step in the method is to find the roots and of the associated indicial equation

(2)

Then, utilizing the larger root equation (1) has a series solution of the form

(3)

where To find a second linearly independent solution, our first inclination is to set and seek a solution of the form

(4)

We’ll see that this procedure works, provided is not an integer. However, when is an integer, the Frobenius method with may just lead to the same solution that we obtained using the root (This is obviously true when

Find the first few terms in the series expansion about the regular singular point x ! 0 for a general solution to

(5)

In Example 3 of Section 8.6, we used the method of Frobenius to find a series solution for (5). In the process we determined that and that the indicial equation has roots Since these roots do not differ by an integer the method of Frobenius will give two linearly independent solutions of the form

(6)

In Problem 44 of Exercises 8.6, you were requested to show that substituting into (5) leads to the recurrence relation

(7)

With and we find

(8)

as obtained in the previous section. Moreover, taking and in (7) leads to the second solution,

(9)

Consequently, a general solution to equation (5) is

(10)

where and are the series solutions given in equations (8) and (9). See Figure 8.9 on page 468. ◆

y2 AxBy1 AxBy AxB ! c1y1 AxB # c2y2 AxB , x 7 0 , y2 AxB ! x1/2 " 34 x3/2 # 732 x5/2 " 1331920 x7/2 # p .

a0 ! 1r ! r2 ! 1/2 y1 AxB ! x " 13 x2 # 110 x3 " 130 x4 # p

a0 ! 1,r ! r1 ! 1

Ak # r " 1B A2k # 2r " 1Bak # 3 Ak # r " 1B Ak # r " 2B # 1 4ak"1 ! 0 , k ' 1 . w Ar, xBw Ar, xB J xr a

n!0 anx

n ! a q

n!0 anx

n#r .

Ar1 " r2 ! 1/2B,r1 ! 1, r2 ! 1/2. p0 ! "1/2, q0 ! 1/2 Ax # 2Bx2y– AxB " xy¿ AxB # A1 # xBy AxB ! 0 , x 7 0 .

r1 ! r2.Br1. r ! r2 r1 " r2r1 " r2

w Ar2, xB ! xr2 aq n!0

anx n ! a

n!0 anx

n#r2 .

r ! r2 a0 & 0.

w Ar1, xB ! xr1 aq n!0

anx n ! a

n!0 anx

n#r1 ,

r1,

r Ar " 1B # p0r # q0 ! 0 . r2 ARe r1 ' Re r2Br1 y– AxB # p AxBy¿ AxB # q AxBy AxB ! 0 , x 7 0 ,

Section 8.7 Finding a Second Linearly Independent Solution 467

8.7 FINDING A SECOND LINEARLY INDEPENDENT SOLUTION

Example 1

Solution

When the indicial equation has repeated roots, obviously substituting just gives us back the first solution and gets us nowhere. One possibility is to use the reduction of order method described in Section 4.7 or Exercises 6.1, Problem 31. However, this method has drawbacks in that it requires manipulations of series that often make it difficult to determine the general term in the series expansion for the second linearly independent solution. A more direct approach is to use the next theorem, which provides the form of this second solution. You will not be surprised to see that, analogous to the situation for a Cauchy–Euler equation whose indicial equation has repeated roots, a second linearly independent solution will involve the first solution multiplied by a logarithmic function.

In the following theorem, we give the general form of two linearly independent solutions for the three cases where the roots of the indicial equation (a) do not differ by an integer, (b) are equal, or (c) differ by a nonzero integer.

r ! r2r1 ! r2,

468 Chapter 8 Series Solutions of Differential Equations

0.8 10.60.40.20

0.2

0.4

0.6

0.8

Equation (8)

Equation (9)

Figure 8.9 Partial sums approximating solutions to Example 1

†For a proof of Theorem 7, see the text Ordinary Differential Equations, 4th ed., by G. Birkhoff and G.-C. Rota (Wiley, New York, 1989).

Form of Second Linearly Independent Solution

Theorem 7.† Let be a regular singular point for and let and be the roots of the associated indicial equation, where Re

(a) If is not an integer, then there exist two linearly independent solutions of the form

(11)

(12)

(b) If then there exist two linearly independent solutions of the form

(13)

(14)

r1 " r2

y2 AxB ! y1 AxB ln Ax " x0B # aq n!1

bn Ax " x0Bn#r1 . y1 AxB ! aq

n!0 an Ax " x0Bn#r1 , a0 & 0 ,

r1 ! r2,

y2 AxB ! aq n!0

bn Ax " x0Bn#r2 , b0 & 0 . y1 AxB ! aq

n!0 an Ax " x0Bn#r1 , a0 & 0 ,

r1 " r2 r1 ' Re r2.r2

r1y– # py¿ # qy ! 0x0

In each case of the theorem, is just the series solution obtained by the method of Frobe- nius, with When is not an integer, the method of Frobenius yields a second linearly independent solution by taking Let’s see how knowing the form of the second solution enables us to obtain it. Again, for simplicity, we consider only indicial equations having real roots.

Find the first few terms in the series expansion about the regular singular point for two linearly independent solutions to

(17)

In Example 4 of Section 8.6, we used the method of Frobenius to obtain a series solution to equation (17). In the process, we found the indicial equation to be which has roots Working with we derived the series solution

(18)

[see equation (44) in Section 8.6 with To find a second linearly independent solution we appeal to Theorem 7, part (b),

which asserts that with has the form

(19)

Our goal is to determine the coefficients by substituting directly into equation (17). We begin by differentiating in (19) to obtain

Substituting into (17) yields

(20)

which simplifies to

(21)

# a q

n!1 n An # 1Bbnxn#1 " aq

n!1 An # 1Bbnxn#1 # aq

n!1 bnx

n#1 " a q

n!1 bnx

n#2 ! 0 .

Ex2y%1 AxB # xy&1 AxB $ A1 # xBy1 AxB Fln x " 2y1 AxB # 2xy¿1 AxB # A1 " xB e y1 AxB ln x # aq

n!1 bnx

n#1f ! 0 ," xe y¿1 AxB ln x # x"1y1 AxB # aq

n!1 An # 1Bbnxnf

x2e y–1 AxB ln x " x"2y1 AxB # 2x"1y¿1 AxB # aq n!1

n An # 1Bbnxn"1fy2 AxB y–2 AxB ! y–1 AxB ln x " x"2y1 AxB # 2x"1y¿1 AxB # aq

n!1 n An # 1Bbnxn"1 .

y¿2 AxB ! y¿1 AxB ln x # x"1y1 AxB # aq n!1 An # 1Bbnxn ,

y2 AxB y2 AxBbn y2 AxB " y1 AxB ln x $ a!

n"1 bnxn$1 .

y2 AxBx0 ! 0BA y2 AxB, a0 ! 1 4 .y1 AxB ! x # x

2 # 1 4

x3 # 1 36

x4 # 1

576 x5 # p ! a

k!0

1Ak!B2 xk#1 r1 ! 1,r1 ! r2 ! 1.

r2 " 2r # 1 ! 0,

x2y– AxB " xy¿ AxB # A1 " xBy AxB ! 0 , x 7 0 . x ! 0

r ! r2. r1 " r2r ! r1.

y1 AxB

Section 8.7 Finding a Second Linearly Independent Solution 469

(15)

(16)

where C is a constant that could be zero.

y2 AxB ! Cy1 AxB ln Ax " x0B # aq n!0

bn Ax " x0Bn#r2 , b0 & 0 , y1 AxB ! aq

n!0 an Ax " x0Bn#r1 , a0 & 0 ,

Example 2

Solution

Notice that the factor in front of ln x is just the left-hand side of equation (17) with Since is a solution to (17), this factor is zero. With this observation and a shift in the indices of summation, equation (21) can be rewritten as

(22)

To identify the coefficients in (22) so that we can set them equal to zero, we must substi- tute back in the series expansions for and . From (18), we get that !

Inserting this series together with (18) into (22), we find

(23)

Listing separately the k ! 0 and k ! 1 terms and combining the remaining terms gives

(24)

Next, we set the coefficients in (24) equal to zero. From the term we have so From the term, we obtain

(25)

Taking k ! 2 and 3, we compute

(26)

Hence, a second linearly independent solution to (17) is

(27)

See Figure 8.10. ◆

y2 AxB ! y1 AxB ln x " 2x2 " 34 x3 " 11108 x4 # p . b2 !

1 22 3b1 " 1 4 ! "34 , b3 ! 19 c" 34 " 636 d ! "11108 .

bk ! 1 k2 cbk"1 " 2kAk!B2 d Ak ' 2B .

2kAk!B2 # k2bk " bk"1 ! 0 xk#1b1 ! "2.

2 # b1 ! 0,x 2

A2 # b1Bx2 #aq k!2 c 2kAk!B2 # k2bk " bk"1 d xk#1 ! 0 .

a q

k!0

2 Ak # 1B " 2Ak!B2 xk#1 # b1x2 # aqk!2 Ak2bk " bk"1Bxk#1 ! 0 . ©qk!0 Ak # 1Bxk/ Ak!B2. y¿1 AxBy¿1 AxBy1 AxB

2xy¿1 AxB " 2y1 AxB # b1x2 # aq k!2 Ak2bk " bk"1Bxk#1 ! 0 .

y1 y ! y1.

470 Chapter 8 Series Solutions of Differential Equations

10.80.6

Equation (18)

Equation (27)

0.40.20

−1

−2

Figure 8.10 Partial sum approximations to solutions of Example 2

In the next two examples we consider the case when the difference between the roots of the indicial equation is a positive integer. In Example 3, it will turn out that the constant C in formula (16) must be taken to be zero (i.e., no ln x term is present), while in Example 4, this constant is nonzero (i.e., a ln x term is present). Since the solutions to these examples require several intermediate computations, we do not display all the details. Rather, we encourage you to take an active part by bridging the gaps.

Find the first few terms in the series expansion about the regular singular point x ! 0 for a general solution to

(28)

In Example 5 of Section 8.6, we used the method of Frobenius to find a series expansion about x ! 0 for a solution to equation (28). There we found the indicial equation to be

which has roots and Working with we obtained the series solution

(29)

[see equation (53) in Section 8.6, with Since is a positive integer, it follows from Theorem 7 that equation (28) has a

second linearly independent solution of the form

(30)

When we substitute this expression for into equation (28), we obtain

(31)

which simplifies to

(32)

The factor in braces is zero because satisfies equation (28). After we combine the summa- tions and simplify, equation (32) becomes

(33)

Substituting in the series for and writing out the first few terms of the summation in (33), we obtain

# p ! 0 .

# a 7 10

C # 10b5 " b3b x # A18b6 " b4Bx2 # a 11280 C # 28b7 " b5b x3 "2b1x

"3 # A"2b2 " b0Bx"2 # A3C " b1Bx"1 # A4b4 " b2B y1 AxB3Cx

"1y1 AxB # 2Cy¿1 AxB " 2b1x"3 # aq k!2 3 k Ak " 3Bbk " bk"2 4 xk"4 ! 0 .

y1 AxB# a q

n!0 An " 3B An " 4Bbnxn"4 # aq

n!0 4 An " 3Bbnxn"4 " aq

n!0 bnx

n"2 ! 0 .

Exy–1 AxB # 4y¿1 AxB " xy1 AxB FC ln x # 3Cx"1y1 AxB # 2Cy¿1 AxB " xeCy1 AxB ln x # aq

n!0 bnx

n"3f ! 0 ,# 4eCy¿1 AxB ln x # Cx "1y1 AxB # aq

n!0 An " 3Bbnxn"4f

xeCy–1 AxB ln x # 2Cx"1y¿1 AxB " Cx"2y1 AxB # aq n!0 An " 3B An " 4Bbnxn"5fy2

y2 AxB ! Cy1 AxB ln x # aq n!0

bnx n"3 .

r1 " r2 ! 3 a0 ! 1 4 .y1 AxB ! 1 #

1 10

x2 # 1

280 x4 # p

r1 ! 0,r2 ! "3.r1 ! 0r 2 # 3r ! 0,

xy– AxB # 4y¿ AxB " xy AxB ! 0 , x 7 0 .

Section 8.7 Finding a Second Linearly Independent Solution 471

Example 3

Solution

Next, we set the coefficients equal to zero:

Note, in particular, that C must equal zero. Substituting the above values for the ’s and C ! 0 back into equation (30) gives

(34)

where and are arbitrary constants. Observe that the expression in braces following is just the series expansion for given in equation (29). Hence, in order to obtain a second linearly independent solution, we must choose to be nonzero. Taking and gives

(35)

Therefore, a general solution to equation (28) is

(36)

where and are given in (29) and (35). [Notice that the right-hand side of (34) coincides with (36) if we identify as and as ] See Figure 8.11. ◆ c1.b3c2b0

y2 AxBy1 AxB y AxB ! c1y1 AxB # c2y2 AxB , x 7 0 , y2 AxB ! x"3 " 12 x"1 " 18 x " 1144 x3 # p .

b3 ! 0b0 ! 1b0 y1 AxB b3b3b0

# b3 e1 # 110 x2 # 1280 x4 # p f , y2 AxB ! b0 e x"3 " 12 x"1 " 18 x " 1144 x3 # p f

11 280

C # 28b7 " b5 ! 0 1 b7 ! b5 "

11 280

28 !

1 280

b3 .

18b6 " b4 ! 0 1 b6 ! 1 18

b4 ! " 1

144 b0 ,

7 10

C # 10b5 " b3 ! 0 1 b5 ! b3 "

7 10

10 !

1 10

b3 ,

3C " b1 ! 0 1 C " 1 3 b1 " 0 , 4b4 " b2 ! 0 1 b4 !

1 4

b2 ! " 1 8

b0 ,

"2b1 ! 0 1 b1 ! 0 , "2b2 " b0 ! 0 1 b2 ! " 1 2

b0 ,

472 Chapter 8 Series Solutions of Differential Equations

Equation (35)Equation (29)

1.0

1.5

0.5

0.5 1.0 x

Figure 8.11 Partial sum approximations to solutions of Example 3

Find the first few terms in the series expansion about the regular singular point x ! 0 for two linearly independent solutions to

(37)

In Example 6 of Section 8.6, we used the method of Frobenius to find a series expansion about x ! 0 for a solution to equation (37). The indicial equation turned out to be which has roots and Using we obtained the series solution

(38)

[see equation (62) in Section 8.6 and put Because is a positive integer, it follows from Theorem 7 that equation (37)

has a second linearly independent solution of the form

(39)

Plugging the expansion for into equation (37) and simplifying yields

(40)

Again the factor in braces is zero because is a solution to equation (37). If we combine the summations and simplify, equation (40) becomes

(41)

Substituting in the series expansions for and and writing out the first few terms of the summation in (41) leads to

(42)

When we set the coefficients in (42) equal to zero, it turns out that we are free to choose C and as arbitrary constants:

24b6 " b4 # 5 96

C ! 0 1 b6 ! b4 "

5 96

24 !

1 192

b2 " 7

1152 C .

15b5 " b3 ! 0 1 b5 ! 1 15

b3 ! 0 ,

8b4 " b2 # 3 4

C ! 0 1 b4 ! b2 "

3 4

8 !

1 8

b2 " 3 32

C Ab2 arbitraryB, 3b3 " b1 ! 0 1 b3 !

1 3

b1 ! 0 ,

2C " b0 ! 0 1 b0 ! 2C AC arbitraryB,"b1 ! 0 1 b1 ! 0 , b2

# A15b5 " b3Bx2 # a 596 C # 24b6 " b4b x3 # p ! 0 . "b1x

"2 # A2C " b0Bx"1 # A3b3 " b1B # a34 C # 8b4 " b2b x y¿1 AxBy1 AxB

2Cx"1y1 AxB # 2Cy¿1 AxB " b1x"2 # aq k!2 3 k Ak " 2Bbk " bk"2 4 xk"3 ! 0 .

y1 AxB # a

n!0 An " 2B An " 3Bbnxn"3 # aq

n!0 3 An " 2Bbnxn"3 " aq

n!0 bnx

n"1 ! 0 .

Exy–1 AxB # 3y¿1 AxB " xy1 AxB FC ln x # 2Cx"1y1 AxB # 2Cy¿1 AxBy2 AxB y2 AxB ! Cy1 AxB ln x # aq

n!0 bnx

n"2 .

r1 " r2 ! 2 a0 ! 1 4 .y1 AxB ! 1 #

1 8

x2 # 1

192 x4 #

1 9216

x6 # p

r1 ! 0,r2 ! "2.r1 ! 0 r2 # 2r ! 0,

xy– AxB # 3y¿ AxB " xy AxB ! 0 , x 7 0 .

Section 8.7 Finding a Second Linearly Independent Solution 473

Example 4

Solution

Substituting these values for the ’s back into (39), we obtain the solution

(43)

where C and are arbitrary constants. Since the factor multiplying is the first solution we can obtain a second linearly independent solution by choosing and

(44)

See Figure 8.12. ◆

In closing we note that if the roots and of the indicial equation associated with a dif- ferential equation are complex, then they are complex conjugates. Thus, the difference is imaginary and, hence, not an integer, and we are in case (a) of Theorem 7. However, rather than employing the display (11)–(12) for the linearly independent solutions, one usually takes the real and imaginary parts of (11); Problem 26 provides an elaboration of this situation.

r1 " r2 r2r1

y2 AxB ! y1 AxB ln x # 2x"2 " 332 x2 " 71152 x4 # p . b2 ! 0:C ! 1y1 AxB, b2b2

# b2 e1 # 18 x2 # 1192 x4 # p f , y2 AxB ! C e y1 AxB ln x # 2x"2 " 332 x2 " 71152 x4 # p f

474 Chapter 8 Series Solutions of Differential Equations

21.510.5

Equation (44)

Equation (38)

Figure 8.12 Partial sum approximations to the solutions of Example 4

8.7 EXERCISES In Problems 1–14, find at least the first three nonzero terms in the series expansion about for a general solution to the given equation for (These are the same equations as in Problems 19–32 of Exercises 8.6.)

1. 2. 3. x2y– # xy¿ # x2y ! 0

2x Ax " 1By– # 3 Ax " 1By¿ " y ! 09x2y– # 9x2y¿ # 2y ! 0 x 7 0.

x ! 0 4. 5. 6. 7. 8. 9.

10. 3x2y– # 8xy¿ # Ax " 2By ! 0xw– " w¿ " xw ! 0 x2y– # Ax2 " xBy¿ # y ! 04x2y– # 2x2y¿ " Ax # 3By ! 0 3xy– # A2 " xBy¿ " y ! 0x2z– # Ax2 # xBz¿ " z ! 0 xy– # y¿ " 4y ! 0

11. 12. 13. 14.

In Problems 15 and 16, determine whether the given equation has a solution that is bounded near the origin, all solutions are bounded near the origin, or none of the solutions are bounded near the origin. (These are the same equations as in Problems 33 and 34 of Exercises 8.6.) Note that you need to analyze only the indicial equation in order to answer the question. 15. 16.

In Problems 17–20, find at least the first three nonzero terms in the series expansion about for a general solution to the given linear third-order equation for

(These are the same equations as in Problems 35–38 in Exercises 8.6.) 17. 18. 19. 20.

21. Buckling Columns. In the study of the buckling of a column whose cross section varies, one encoun- ters the equation (45) where x is related to the height above the ground and y is the deflection away from the vertical. The positive constant depends on the rigidity of the column, its moment of inertia at the top, and the load. The positive integer n depends on the type of column. For example, when the column is a truncated cone [see Figure 8.13(a)], we have n ! 4. (a) Use the substitution to reduce (45) with

n ! 4 to the form

(b) Find at least the first six nonzero terms in the series expansion about t ! 0 for a general solu- tion to the equation obtained in part (a).

22. In Problem 21 consider a column with a rectangular cross section with two sides constant and the other

x ! q

d2y

dt2 #

dt # a2y ! 0 , t 7 0 .

x ! t"1

xny% AxB $ A2y AxB " 0 , x 7 0 , 6x3y‡ # A13x2 " x3By– # xy¿ " xy ! 06x3y‡ # 13x2y– " Ax2 # 3xBy¿ " xy ! 0 6x3y‡ # 11x2y– " 2xy¿ " Ax " 2By ! 06x3y‡ # 13x2y– # Ax # x2By¿ # xy ! 0

x 7 0.

x ! 0

xy– # Ax # 2By¿ " y ! 03xy– # 2 A1 " xBy¿ " 4y ! 0

x2y– " x A1 # xBy¿ # y ! 0xy– # A1 " xBy¿ " y ! 0 x Ax # 1By– # Ax # 5By¿ " 4y ! 0xy– # Ax " 1By¿ " 2y ! 0

Section 8.7 Finding a Second Linearly Independent Solution 475

two changing linearly [see Figure 8.13(b)]. In this case, n ! 1. Find at least the first four nonzero terms in the series expansion about x ! 0 for a general solution to equation (45) when n ! 1.

23. Use the method of Frobenius and a reduction of order procedure (see page 198) to find at least the first three nonzero terms in the series expansion about the irregular singular point x ! 0 for a general solution to the differential equation

24. The equation

where n is a nonnegative integer, is called Laguerre’s differential equation. Show that for each n, this equation has a polynomial solution of degree n. These polynomials are denoted by and are called Laguerre polynomials. The first few Laguerre poly- nomials are

25. Use the results of Problem 24 to obtain the first few terms in a series expansion about x ! 0 for a general solution for x > 0 to Laguerre’s differential equation for n ! 0 and 1.

26. To obtain two linearly independent solutions to

(46)

complete the following steps. (a) Verify that (46) has a regular singular point at

x ! 0 and that the associated indicial equation has complex roots %i.

x2y% $ Ax $ x2By&$ y " 0 , x 7 0 ,

L2 AxB ! x2 " 4x # 2 .L0 AxB ! 1 , L1 AxB ! "x # 1 . Ln AxB

xy% AxB $ A1 # xBy& AxB $ ny AxB " 0 , x2y– # y¿ " 2y ! 0 .

(a) Truncated cone

Truncated pyramid with fixed thickness T and varying width

(b)

Figure 8.13 Buckling columns

(b) As discussed in Section 8.5, we can express

Deduce from this formula that !

(c) Set where the coeffi- cients now are complex constants, and substi- tute this series into equation (46) using the result of part (b).

(d) Setting the coefficients of like powers equal to zero, derive the recurrence relation

for n ' 1 . an ! "

n " 1 # iAn # iB2 # 1 an"1 ,

y AxB ! ©qn!0 anxn# i,Aa # ibBxa"1# ib. d dx

xa# ib

ixasin Ab ln xB .! xacos Ab ln xBxa# ib ! xaxib 476 Chapter 8 Series Solutions of Differential Equations

(e) Taking compute the coefficients and and thereby obtain the first few terms of a

complex solution to (46). (f) By computing the real and imaginary parts of the

solution obtained in part (e), derive the following linearly independent real solutions to (46):

# 3 sin Aln xB 4 e1 " 2 5

x # 1 10

x2 # p f .y2 AxB ! 3 cos Aln xB 4 e" 1 5

x # 1 20

x2 # p f 3 sin Aln xB 4 e 1

5 x "

1 20

x2 # p f ,y1 AxB ! 3 cos Aln xB 4 e1 " 25 x # 110 x2 # p f a2

a1a0 ! 1,

In advanced work in applied mathematics, engineering, and physics, a few special second-order equations arise very frequently. These equations have been extensively studied, and volumes have been written on the properties of their solutions, which are often referred to as special functions.

For reference purposes we include a brief survey of three of these equations: the hypergeo- metric equation, Bessel’s equation, and Legendre’s equation. Bessel’s equation governs the radial dependence of the solutions to the classical partial differential equations of physics in spherical coordinate systems (see Section 10.7, page 625); Legendre’s equation governs their latitudinal dependence. C. F. Gauss formulated the hypergeometric equation as a generic equation whose solutions include the special functions of Legendre, Chebyshev, Gegenbauer, and Jacobi. For a more detailed study of special functions, we refer you to Basic Hypergeometric Series, 2nd ed., by G. Gasper and M. Rahman (Cambridge University Press, Cambridge, 2004); Special Func- tions, by E. D. Rainville (Chelsea, New York, 1971); and Higher Transcendental Functions, by A. Erdelyi (ed.) (McGraw-Hill, New York, 1953), 3 volumes.

Hypergeometric Equation The linear second-order differential equation

(1)

where and are fixed parameters, is called the hypergeometric equation. This equation has singular points at x ! 0 and 1, both of which are regular. Thus a series expansion about x ! 0 for a solution to (1) obtained by the method of Frobenius will converge at least for 0 < x < 1 (see Theo- rem 6, page 461). To find this expansion, observe that the indicial equation associated with x ! 0 is

which has roots 0 and Let us assume that is not an integer and use the root r ! 0 to obtain a solution to (1) of the form

(2) y1 AxB ! aq n!0

anx n .

g1 " g.

r Ar " 1B # gr ! r 3 r " A1 " gB 4 ! 0 , ga, b,

x A1 # xBy% $ 3G # AA $ B $ 1Bx 4 y& # ABy " 0 ,

8.8 SPECIAL FUNCTIONS

Substituting given in (2) into (1), shifting indices, and simplifying ultimately leads to the equation

(3)

Setting the series coefficients equal to zero yields the recurrence relation

(4)

Since and is not an integer, there is no fear of dividing by zero when we rewrite (4) as

(5)

Solving recursively for we obtain

(6)

If we employ the factorial function which is defined for nonnegative integers n by

(7)

then we can express more compactly as

(8)

[In fact, we can write n! as If we take and then substitute the expression for in (8) into (2), we obtain the following solution to the hypergeometric equation:

(9)

The solution given in (9) is called a Gaussian hypergeometric function and is denoted by the symbol † That is,

(10)

Hypergeometric functions are generalizations of the geometric series. To see this, observe that for any constant that is not zero or a negative integer,

It is interesting to note that many other familiar functions can be expressed in terms of the hypergeometric function. For example,

(11)

(12) F A1, 1; 2; xB ! "x"1 ln A1 " xB ,F Aa, b; b; xB ! A1 " xB"a , F A1, b; b; xB ! 1 # x # x2 # x3 # p .b F AA, B; G; xB :" 1 $ a!

n"1

AABn ABBn n! AGBn xn .

F Aa, b; g; xB. y1 AxB ! 1 # aq

n!1

AaBn AbBn n! AgBn xn .

ana0 ! 1A1Bn. ] an !

AaBn AbBn n! AgBn a0 , n ' 1 .

AtB0 J 1 , t & 0 ,AtBn J t At # 1B At # 2B p At # n " 1B , n ' 1 , AtBn,

an ! a Aa # 1B p Aa # n " 1Bb Ab # 1B p Ab # n " 1B

n!g Ag # 1B p Ag # n " 1B a0 , n ' 1 . an,

an ! An # a " 1B An # b " 1B

n An # g " 1B an"1 . gn ' 1

n An # g " 1Ban " An # a " 1B An # b " 1Ban"1 ! 0 , n ' 1 . a q

n!1 3n An # g " 1Ban " An # a " 1B An # b " 1Ban"1 4 xn"1 ! 0 . y1 AxB

Section 8.8 Special Functions 477

†Historical Footnote: A detailed study of this function was done by Carl Friedrich Gauss in 1813. In Men of Mathematics (Simon & Schuster, London, 1986), the mathematical historian E. T. Bell refers to Gauss as the Prince of Mathematicians.

(13)

(14)

We omit the verification of these formulas. To obtain a second linearly independent solution to (1) when is not an integer, we use the

other root, of the indicial equation and seek a solution of the form

(15)

Substituting into equation (1) and solving for we eventually arrive at

(16)

Factoring out we see that the second solution can be expressed in terms of a hyper- geometric function. That is,

(17)

When is an integer, one of the formulas given in (9) or (16) (corresponding to the larger root, 0 or still gives a solution. We then use the techniques of this chapter to obtain a second linearly independent solution, which may or may not involve a logarithmic term. We omit a discussion of these solutions.

In many books, the hypergeometric function is expressed in terms of the gamma function instead of the factorial function. Recall that in Section 7.6, we defined

(18)

and showed that

(19)

It follows from repeated use of relation (19) that

(20) ,

for t > 0 and n any nonnegative integer. Using relation (20), we can express the hypergeometric function as

(21)

Bessel’s Equation The linear second-order differential equation

(22) x2y% $ xy& $ Ax2 # N2By " 0 ,

F Aa, b; g; xB " ( AgB ( AaB( AbB a!n"0 ( Aa $ nB( Ab $ nBn!( Ag $ nB x n .

AtBn ! ( At # nB ( AtB

( Ax # 1B ! x ( AxB , x 7 0 . ( AxB J $q

0 e"uux"1 du , x 7 0

( AxB 1 " gBg

y2 AxB ! x1"gF Aa # 1 " g, b # 1 " g; 2 " g; xB . y2 AxBx1"g,

y2 AxB ! x1"g # aq n!1

Aa # 1 " gBn Ab # 1 " gBn n! A2 " gBn xn#1"g .

bn,y2 AxB y2 AxB ! aq

n!0 bnx

n#1"g .

1 " g, g

F a1 2

, 1; 3 2

; "x2b ! x"1 arctan x . Fa1

2 , 1;

3 2

; x2b ! 1 2

x"1 lna1 # x 1 " x

b , 478 Chapter 8 Series Solutions of Differential Equations

where is a fixed parameter, is called Bessel’s equation of order This equation has a regular singular point at x ! 0 and no other singular points in the complex plane. Hence, a series solution for (22) obtained by the method of Frobenius will converge for The indicial equation for (22) is

which has roots and If is not an integer, then the method of Frobenius yields two linearly independent solutions, given by

(23)

(24)

If in (23) we take

then it follows from relation (20) that the function

(25)

is a solution to (22). We call the Bessel function of the first kind of order † Similarly, taking

in equation (24) gives the solution

(26)

which is the Bessel function of the first kind of order When is not an integer, we know by Theorem 7 of Section 8.7 that and are linearly independent. More- over, it can be shown that if is not an integer, even though is, then and are still linearly independent.

What happens in the remaining case when is a nonnegative integer, say ? While is still a solution, the function is not even properly defined because formula (26)

will involve the gamma function evaluated at a nonpositive integer. From a more in-depth study of the gamma function, it turns out that for . . . . Hence, (26) becomes

(27) J"m AxB ! aq n!0

A"1Bn n!( A1 " m # nB ax2b 2n"m ! aqn!m A"1Bnn!( A1 " m # nB ax2b 2n"m .

k ! 0, "1, "2,1/( AkB ! 0, J"m AxBJm AxB n ! mn

J"n AxBJn AxB2nn J"n AxBJn AxB r1 " r2 ! 2n"n.

J#N AxB :" a! n"0

A#1Bn n!' A1 # N $ nB ax2b 2n#N ,

b0 ! 1

2"n( A1 " nB n.Jn AxB

JN AxB :" a! n"0

A#1Bn n!' A1 $ N $ nB ax2b 2n$N

a0 ! 1

2n( A1 # nB , y2 AxB ! b0 aq

n!0

A"1Bn 22nn! A1 " nBn x2n"n .

y1 AxB ! a0 aq n!0

A"1Bn 22nn! A1 # nBn x2n#n ,

2nr2 ! "n.r1 ! n

r Ar " 1B # r " n2 ! Ar " nB Ar # nB ! 0 , 0 6 x 6 q.

n.n ' 0

Section 8.8 Special Functions 479

†Historical Footnote: Frederic Wilhelm Bessel (1784–1846) started his career in commercial navigation and later became an astronomer. In 1817 Bessel introduced the functions in his study of planetary orbits.Jn AxB

Comparing (27) with the formula (25) for we see (after a shift in index) that

(28)

which means that and are linearly dependent. To resolve the problem of finding a second linearly independent solution in the case when is a nonnegative integer, we can use Theorem 7 in Section 8.7. There is, however, another approach to this problem, which we now describe.

For not an integer, we can take linear combinations of and to obtain other solutions to (22). In particular, let

(29)

for x > 0. The function is called the Bessel function of the second kind of order , and, as can be verified, and are linearly independent. Notice that when is an integer, the denominator in (29) is zero; but, by formula (28), so is the numerator! Hence, it is reasonable to hope that, in a limiting sense, formula (29) is still meaningful. In fact, using L’Hôpital’s rule, it is possible to show that for m a nonnegative integer, the function defined by

(30)

for x > 0 is a solution to (22) with Furthermore, and are linearly indepen- dent. We again call the Bessel function of the second kind of order m. In the literature the function is sometimes denoted by variations of it have been called the Neumann function and the Weber function.

Figure 8.14 shows the graphs of and and Figure 8.15 (page 481) the graphs of and Notice that the curves for and behave like damped sine waves and

have zeros that interlace (see Problem 28). The fact that the Bessel functions have an infinite number of zeros is helpful in certain applications (see Group Project A in Chapter 10).

Jn J1 AxBJ0 AxBY1 AxB.Y0 AxB J1 AxBJ0 AxB

Nm AxB;Ym AxB Ym AxB Ym AxBJm AxBn ! m.

Ym AxB :" lim NSm

cos ANPBJN AxB # J#N AxB sin ANPB

nYn AxBJn AxB nYn AxB YN AxB :" cos ANPBJN AxB # J#N AxBsin ANPB , N not an integer ,

J"n AxBJn AxBn n

Jm AxBJ"m AxB J"m AxB ! A"1BmJm AxB ,

Jm AxB, 480 Chapter 8 Series Solutions of Differential Equations

1.0

5 10

J0(x)

J1(x)0.5

−0.4

Figure 8.14 Graphs of the Bessel functions and J1 AxBJ0 AxB

There are several useful recurrence relations involving Bessel functions. For example,

(31) (32)

(33) (34)

Furthermore, analogous equations hold for Bessel functions of the second kind. To illustrate the techniques involved in proving the recurrence relations, let’s verify rela-

tion (31). We begin by substituting series (25) for into the left-hand side of (31). Differen- tiating, we get

(35)

Since we have from (35)

Factoring out an gives

as claimed in equation (31). We leave the verifications of the remaining relations as exercises (see Problems 22–24).

dx 3 xnJn AxB 4 ! xn aq

n!0

A"1Bn n!( A1 # An " 1B # nB ax2b 2n#n"1 ! xnJn"1 AxB ,

dx 3 xnJn AxB 4 ! aq

n!0

A"1Bn2x2n#2n"1 n!( An # nB22n#n .

( A1 # n # nB ! An # nB( An # nB, ! a

n!0

A"1Bn A2n # 2nBx2n#2n"1 n!( A1 # n # nB22n#n .

! d

dx e aq

n!0

A"1Bnx2n#2n n!( A1 # n # nB22n#n f

dx 3 xnJn AxB 4 ! ddx e xn aqn!0 A"1Bnn!( A1 # n # nB ax2b 2n#n f

Jn AxB Jn#1 AxB ! Jn"1 AxB " 2J ¿n AxB .Jn#1 AxB ! 2nx Jn AxB " Jn"1 AxB , d dx

3 x"nJn AxB 4 ! "x"nJn#1 AxB ,ddx 3 xnJn AxB 4 ! xnJn"1 AxB ,

Section 8.8 Special Functions 481

Y0(x)

Y1(x)

2.5 5 7.5 10 12.5 15 17.5

0.6

0.4

0.2

−0.2

−0.4

−0.6

−0.8

Figure 8.15 Graphs of the Bessel functions and Y1 AxBY0 AxB

The following approximations are useful when analyzing the behavior of Bessel functions for large arguments:

(36)

This oscillatory behavior is demonstrated in Figures 8.14 and 8.15. Problem 41 provides some justification of formula (36). Of course, for small x the leading terms of the power series expansion dominate and we have

(37)

Legendre’s Equation The linear second-order differential equation

(38)

where n is a fixed parameter, is called Legendre’s equation.† This equation has a regular singular point at 1, and hence a series solution for (38) about x ! 1 may be obtained by the method of Frobenius. By setting equation (38) is transformed into

(39)

The indicial equation for (39) at z ! 0 is

which has roots Upon substituting

into (39) and proceeding as usual, we arrive at the solution

(40)

where we have expressed in terms of the original variable x and have taken We have written the solution in the above form because it is now obvious from (40) that

(41)

where F is the Gaussian hypergeometric function defined in (10). For n a nonnegative integer, the factorA"nBk ! A"nB A"n # 1B A"n # 2B p A"n # k " 1B

y1 AxB ! Fa"n, n # 1; 1; 1 " x2 b , a0 ! 1.y1

y1 AxB ! 1 # aq k!1

A"nBk An # 1Bk k! A1Bk a1 " x2 b k ,

y AzB ! aq k!0

akz k

r1 ! r2 ! 0.

r Ar " 1B # r ! r2 ! 0 , z Az $ 2B d2y

dz2 $ 2 Az $ 1B dydz # n An $ 1By " 0 .

z ! x " 1,

A1 # x2By% # 2 xy& $ n An $ 1By " 0 ,

0 6 x V 1 . Yn70 AxB ! "( AnB2n/ ApxnB ,Jn AxB ! xn/ 32n( A1 # nB 4 , Y0 AxB ! A2 ln xB /p ,

Jn AxB ! A 2px cosax " np2 " p4b , Yn AxB ! A 2px sinax " np2 " p4b , x W 1 . 482 Chapter 8 Series Solutions of Differential Equations

†Historical Footnote: Solutions to this equation were obtained by Adrien Marie Legendre (1752–1833) in 1785 and are referred to as Legendre functions.

Hence, the solution given in (40) and (41) is a polynomial of degree n. Moreover, These polynomial solutions of equation (38) are called the Legendre polynomials or spherical polynomials and are traditionally denoted by That is,

(42)

If we expand about x ! 0, then takes the form

(43)

where is the greatest integer less than or equal to (see Problem 34). The first three Legendre polynomials are

The Legendre polynomials satisfy the orthogonality condition

(44) for

To see this, we first rewrite equation (38) in what is called the selfadjoint form:

(45)

Since and satisfy (45) with parameters n and m, respectively, we have

(46)

(47)

Multiplying (46) by and (47) by and then subtracting, we find

which can be rewritten in the form

(48)

It is a straightforward calculation to show that the right-hand side of (48) is just

Using this fact and the identity we rewrite equa- tion (48) as

(49) An " mB An # m # 1BPm AxBPn AxB ! E A1 " x2B 3Pn AxBP¿m AxB " P¿n AxBPm AxB 4 F ¿ . n2 " m2 # n " m ! An " mB An # m # 1B,E A1 " x

2B 3Pn AxBP¿m AxB " P¿n AxBPm AxB 4 F ¿ . An2 " m2 # n " mBPm AxBPn AxB ! Pn AxB 3 A1 " x2BP¿m AxB 4 ¿ " Pm AxB 3 A1 " x2BP¿n AxB 4 ¿ .

# 3n An # 1B " m Am # 1B 4Pm AxBPn AxB ! 0 ,Pm AxB 3 A1 " x2BP¿n AxB 4 ¿ " Pn AxB 3 A1 " x2BP¿m AxB 4 ¿ Pn AxBPm AxB

3 A1 " x2BP¿m AxB 4 ¿ # m Am # 1BPm AxB ! 0 .3 A1 " x 2BP¿n AxB 4 ¿ # n An # 1BPn AxB ! 0 ,Pm AxBPn AxB

3 A1 " x2By¿ 4 ¿ # n An # 1By ! 0 . n & m .$ 1

#1 Pm AxBPn AxBdx " 0

P0 AxB ! 1 , P1 AxB ! x , P2 AxB ! 32 x2 " 12 . n/23n/2 4

Pn AxB " 2#n a3n/24 m"0

A#1Bm A2n # 2mB!An # mB!m! An # 2mB! xn#2m , Pn AxB

Pn AxB J 1 # aq k!1

A"nBk An # 1Bk k! A1Bk a1 " x2 b k .

Pn AxB.y1 A1B ! 1. will be zero for k ' n # 1.

Section 8.8 Special Functions 483

Integrating both sides of (49) from to x ! 1 yields

(50)

because for Since n and m are nonnegative integers with then and so equation (44) follows from (50).

Legendre polynomials also satisfy the recurrence formula

(51)

and Rodrigues’s formula

(52)

(see Problems 32 and 33). The Legendre polynomials are generated by the function in the sense

that

(53)

That is, if we expand in a Taylor series about z ! 0, treating x as a fixed parameter, then the coefficients of are the Legendre polynomials The function

is called a generating function for and can be derived from the recurrence formula (51) (see Problem 35).

The Legendre polynomials are an example of a class of special functions called classical orthogonal polynomials.† The latter includes, for example, the Jacobi polynomials, the Gegenbauer or ultraspherical polynomials, the Chebyshev (Tchebichef) polyno- mials, and ; the Laguerre polynomials, and the Hermite polynomials,

The properties of the classical orthogonal polynomials can be found in the books mentioned earlier in this section or in the Handbook of Mathematical Functions with Formu- las, Graphs, and Mathematical Tables, by M. Abramowitz and I. A. Stegun (eds.) (Dover Publications, New York, 1972), Chapter 22. The latter has been updated by the National Institute of Standards and Technology and is freely available from their website http://dlmf.nist.gov/.

We close this chapter with a table listing the forms of various well-known differential equations, the names of the researchers associated with their solutions, and the areas of application in which they arise. Many of them have been discussed in the body or exercises of our text.

Hn AxB. Lan AxB;Un AxBTn AxB Cln AxB; P Aa,bBn AxB;

Pn AxBA1 " 2xz # z2B"1/2 Pn AxB.zn A1 " 2xz # z2B"1/2

A1 " 2xz # z2B"1/2 ! aq n!0

Pn AxBzn , 0 z 0 6 1 , 0 x 0 6 1 . A1 " 2xz # z2B"1/2

Pn AxB ! 12nn! d ndxn E Ax2 " 1BnF An $ 1BPn$1 A xB " A2n $ 1BxPn AxB # nPn#1 AxB

An " mB An # m # 1B & 0, n & m,x ! %1.1 " x2 ! 0 ! 0

! E A1 " x2B 3Pn AxBP¿m AxB " P¿n AxBPm AxB 4 F 0 1"1! $ 1"1E A1 " x2B 3Pn AxBP¿m AxB " P¿n AxBPm AxB 4 F ¿dx An " mB An # m # 1B $ 1

"1 Pm AxBPn AxB dx x ! "1

484 Chapter 8 Series Solutions of Differential Equations

†Here orthogonality is used in the more general sense that for where is a weight function on the interval .Aa, bB w AxBn & m,%ba pn AxBpm AxBw AxB dx ! 0

http://dlmf.nist.gov/

Differential Equation Researchers Areas of Application Solutions

(Harmonic oscillator) Vibrations, waves in Cartesian coordinates

(Damped oscillator) Vibrations

Cauchy, Euler, Mellin Electrostatics in polar coordinates

Airy Caustics

Bessel, Weber, Waves in cylindrical coordinates Neumann, Hankel

(Modified Bessel) Electrostatics in cylindrical coordinates

(Generalized Bessel)

Legendre Spherical coordinates ( )

Laguerre Hydrogen atom

Hermite Quantum mechanical harmonic oscillator

Weber Quantum mechanical harmonic oscillator

Chebyshev Approximation theory, filters

Mathieu Waves in elliptic coordinates

Gauss Hypergeometric equation

Kummer, Whittaker Confluent hypergeometric equation (Continued )

1F1 Aa; c; xBxy– # Ac " xBy¿ " ay ! 0 F Aa, b; g; xBx A1 " xBy– # 3g " Aa # b # 1B x 4 y¿

" aby ! 0

F Aa, q, xBy– # Aa " 2qcos2xBy ! 0 Tn AxBA1 " x2By– " xy¿ # n2y ! 0 e"x

2/ 2Hn AxBy– # A2n # 1 " x2By ! 0 Hn AxBy– " 2xy¿ # 2ny ! 0 Lkn AxBxy– # Ak # 1 " xBy¿ # ny ! 0 Pm/ AxB, Qm/ AxB, l ! "/ A/ # 1Bx ! cosfA1 " x2By– " 2xy¿

" 3l # m2/ A1 " x2B 4 y ! 0 x (1"a)/2e"bx

r/rJp A2dxs/s B, p ! A2 A1 " aB2/4 " c B /sx2y– # Aa # 2bxrBxy¿# 3 c # dx2s " b A1 " a " rBxr # b2x2r 4 y ! 0

In AxB, Kn AxBx2y– # xy¿ " Ax2 # n2By ! 0 Jn AxB, Yn AxB, H (1)n AxB, H(2)n AxBx2y– # xy¿ # Ax2 " n2By ! 0 Ai AxB , Bi AxBy– " xy ! 0 xacos Ab lnxB, xasin Ab lnxBax2y– # bxy¿ # cy ! 0 eaxcosbx, eaxsinbxmy– # by¿ # ky ! 0 cos2lx, sin2lx, e%2"lx, cosh2"lx, sinh2"lxy– # ly ! 0

485

486

Differential Equation Researchers Areas of Application Solutions

Jacobi Orthogonal Polynomials

Gegenbauer Ultraspherical electrostatics (2a # 1 dimensions)

Riemann

Rayleigh Limit cycle

van der Pol Vacuum tubes

Duffing Nonlinear spring

Emden, Fowler Reaction/diffusion

Ricatti Circuit synthesis, geometry

Bernoulliy¿ # p AxBy " q AxByn ! 0y¿ " p AxB " q AxBy " r AxBy2 ! 0 Axry¿ B ¿ " xsyn ! 0y– # y # ey3 ! 0 y– " m A1 " y2By¿ # y ! 0y– " m A1 " y¿2By¿ # y ! 0 y– # a1 " a1 " a2

x #

1 " b1 " b2 x " 1

b y¿ 3 A1 " x2Ba#1y¿ 4 ¿ # n An # 2a # 1B A1 " x2Ba y ! 0 A1 " x2By– # 3b " a " Aa # b # 2Bx 4 y¿ # n An # a # b # 1By ! 0

# c a1a2 x2

# b1b2Ax " 1B2 # c1c2 " a1a2 " b1b2x Ax " 1B d y ! 0

Section 8.8 Special Functions 487

8.8 EXERCISES In Problems 1–4, express a general solution to the given equation using Gaussian hypergeometric functions.

2. 3. 4.

In Problems 5–8, verify the following formulas by expanding each function in a power series about

5. 6.

In Problems 9 and 10, use the method in Section 8.7 to obtain two linearly independent solutions to the given hypergeometric equation.

10.

11. Show that the confluent hypergeometric equation

where and are fixed parameters and is not an integer, has two linearly independent solutions

and

12. Use the property of the gamma function given in (19) to derive relation (20).

In Problems 13–18, express a general solution to the given equation using Bessel functions of either the first or second kind.

13. 14. 15. 16. 17. 18. x2z– # xz¿ # Ax2 " 16Bz ! 09t2x– # 9tx¿ # A9t2 " 4Bx ! 0

x2y– # xy¿ # x2y ! 0 x2y– # xy¿ # Ax2 " 1By ! 09x2y– # 9xy¿ # A9x2 " 16By ! 0 4x2y– # 4xy¿ # A4x2 " 1By ! 0

y2 AxB ! x1"g1F1 Aa # 1 " g; 2 " g; xB . y1 AxB ! 1F1 Aa; g; xB J 1 # aq

n!1

AaBn n! AgBn xn gga

xy– # Ag " xBy¿ " ay ! 0 , x A1 " xBy– # A2 " 2xBy¿ " 1

4 y ! 0

x A1 " xBy– # A1 " 3xBy¿ " y ! 0

F a1 2

, 1; 3 2

; "x2b ! x"1 arctan x F a1

2 , 1;

3 2

; x2b ! 1 2

x"1 ln a1 # x 1 " x

bF Aa, b; b; xB ! A1 " xB"a F A1, 1; 2; xB ! "x"1 ln A1 " xB x ! 0. 2x A1 " xBy– # A3 " 10xBy¿ " 6y ! 02x A1 " xBy– # A1 " 6xBy¿ " 2y ! 0 3x A1 " xBy– # A1 " 27xBy¿ " 45y ! 0x A1 " xBy– # a12 " 4xb y¿ " 2y ! 0

In Problems 19 and 20, a Bessel equation is given. For the appropriate choice of the Bessel function is one solution. Use the method in Section 8.7 to obtain a second linearly independent solution.

19. 20.

21. Show that satisfies the equation

and use this result to find a solution for the equation

In Problems 22 through 24, derive the indicated recur- rence formulas. 22. Formula (32) 23. Formula (33) 24. Formula (34)

25. Show that and

26. The Bessel functions of order any integer, are related to the spherical Bessel func- tions. Use relation (33) and the results of Problem 25 to show that such Bessel functions can be repre- sented in terms of sin x, cos x, and powers of x. Demonstrate this by determining a closed form for

and 27. Use Theorem 7 in Section 8.7 to determine a second

linearly independent solution to Bessel’s equation of order 0 in terms of the Bessel function

28. Show that between two consecutive positive roots (zeros) of there is a root of This interlac- ing property of the roots of Bessel functions is illus- trated in Figure 8.14. [Hint: Use relation (31) and Rolle’s theorem from calculus.]

29. Use formula (43) to determine the first five Legendre polynomials.

30. Show that the Legendre polynomials of even degree are even functions of x, while those of odd degree are odd functions.

31. (a) Show that the orthogonality condition (44) for Legendre polynomials implies that

for any polynomial of degree at most n " 1. [Hint: The polynomials areP0, P1, . . . , Pn"1

q AxB$ 1

"1 Pn AxBq AxBdx ! 0

J0 AxB.J1 AxB, J0 AxB.

J5/2 AxB.J"3/2 AxB

n ! n # 1/2, n J"1/2 AxB ! A2/pxB1/2 cos x .J1/2 AxB ! A2/pxB

1/2 sin x

xy– " 2y¿ # xy ! 0 , x 7 0 .

xy– # A1 " 2nBy¿ # xy ! 0 , x 7 0 ,xnJn AxB x2y– # xy¿ # Ax2 " 4By ! 0x2y– # xy¿ # Ax2 " 1By ! 0

Jn AxBn,

linearly independent and hence span the space of all polynomials of degree at most Thus, for suitable constants

(b) Prove that if is a polynomial of degree n such that

for then for some constant c. [Hint: Select c so that the coefficient of for is zero. Then, since

is a basis, #

Multiply the last equation by and integrate from to

to show that each is zero.]

32. Deduce the recurrence formula (51) for Legendre polynomials by completing the following steps. (a) Show that the function

is a polynomial of degree [Hint: Compute the coefficient of the term using the representation (42). The coefficient of is also zero because

and are both odd or both even func- tions, a consequence of Problem 30.] (b) Using the result of Problem 31(a), show that

for

[Hint: Recall that for all m.] From the definition of in part (a), the recurrence formula now follows.

33. To prove Rodrigues’s formula (52) for Legendre polynomials, complete the following steps. (a) Let and show that

is a polynomial of degree n with the coeffi- cient of equal to

(b) Use integration by parts n times to show that, for any polynomial of degree less than n,

$ 1

"1 yn AxBq AxBdx ! 0 .

q AxB A2nB!/n!.xnyn AxB

yn J Adn/dxnBE Ax2 " 1BnF Qn"1 AxBPm A1B ! 1c ! "n.

cPn"1 AxB Qn"1 AxB ! k ! 0, 1, . . . , n " 2 .$

"1 Qn"1 AxBPk AxBdx ! 0

xPn AxBPn#1 AxB xn xn#1

n " 1.

Qn"1 AxB J An # 1BPn#1 AxB " A2n # 1BxPn AxB

akx ! 1 x ! "1A0 $ k $ n " 1B Pk AxB

an"1Pn"1 AxB .Qn AxB " cPn AxB ! a0P0 AxB # pP0, . . . , Pn"1 Qn AxB " cPn AxB xn

Qn AxB ! cPn AxBk ! 0, 1, . . . , n " 1 , $

"1 Qn AxBPk AxBdx ! 0

Qn AxBak. 4 an"1Pn"1 AxBq AxB ! a0P0 AxB # p # n " 1.

488 Chapter 8 Series Solutions of Differential Equations

Hint: For example, when n ! 2,

Since n ! 2, the degree of is at most 1, and so Thus

comparing the coefficients of in and

34. Use Rodrigues’s formula (52) to obtain the represen- tation (43) for the Legendre polynomials Hint: From the binomial formula,

35. The generating function in (53) for Legendre poly- nomials can be derived from the recurrence formula (51) as follows. Let x be fixed and set

The goal is to determine an explicit formula for (a) Show that multiplying each term in the recur-

rence formula (51) by and summing the terms from n ! 1 to leads to the differential equation

Hint:

(b) Solve the differential equation derived in part (a) and use the initial conditions to obtain f AzB ! A1 " 2xz # z2B"1/2.f A0B ! P0 AxB & 1

! df dz

" x .

! a q

n!0 An # 1BPn#1 AxBzn " P1 AxB

a q

n!1 An # 1BPn#1 AxBzn

dz !

x " z

1 " 2xz # z2 f .

q zn

f AzB.f AzB J ©qn!0 Pn AxBzn. !

2nn!

dxn e an

m!0

n! A"1BmAn " mB!m! x2n"2m f . Pn AxB ! 12nn! dndxn E Ax2 " 1BnF

Pn AxB. yn AxB . Pn AxBxn

c ! 1/2nn!Pn AxB ! cyn AxB $

"1 d2

dx2 E Ax2 " 1B2Fq AxBdx ! 0 .q– AxB & 0.

q AxB# $ 1

"1 q– AxB Ax2 " 1B2dx .

! q AxB d dx

E Ax2 " 1B2F 0 1 "1

" Eq¿ AxB Ax2 " 1B2F 0 1 "1

$ 1

dx2 E Ax2 " 1B2Fq AxBdx

36. Find a general solution about x ! 0 for the equation

by first finding a polynomial solution and then using the reduction of order formula given in Exercises 6.1, Problem 31 to find a second (series) solution.

37. The Hermite polynomials are polynomial solutions to Hermite’s equation

The Hermite polynomials are generated by

Use this equation to determine the first four Hermite polynomials.

38. The Chebyshev (Tchebichef) polynomials are polynomial solutions to Chebyshev’s equation

The Chebyshev polynomials satisfy the recurrence relation

with and Use this recurrence relation to determine the next three Chebyshev polynomials.

39. The Laguerre polynomials are polynomial solutions to Laguerre’s equation

The Laguerre polynomials satisfy Rodrigues’s formula,

Use this formula to determine the first four Laguerre polynomials.

Ln AxB ! exn! dndxn Axne"xB . xy– # A1 " xBy¿ # ny ! 0 .

Ln AxB T1 AxB ! x.T0 AxB ! 1Tn#1 AxB ! 2xTn AxB " Tn"1 AxB ,

A1 " x2By– " xy¿ # n2y ! 0 . Tn AxB

e2tx" t 2

! a q

n!0

Hn AxB n!

tn .

y– " 2xy¿ # 2ny ! 0 .

Hn AxB A1 " x2By– " 2xy¿ # 2y ! 0

Section 8.8 Special Functions 489

40. Reduction to Bessel’s Equation. The class of equations of the form (54) where c and n are positive constants, can be solved by transforming the equation into Bessel’s equation. (a) First, use the substitution to transform

(54) into an equation involving x and z. (b) Second, use the substitution

to transform the equation obtained in part (a) into the Bessel equation

(c) A general solution to the equation in part (b) can be given in terms of Bessel functions of the first and second kind. Substituting back in for s and z, obtain a general solution for equation (54).

41. (a) Show that the substitution ren- ders Bessel’s equation (22) in the form

(55)

(b) For equation (55) would apparently be approximated by the equation (56) Write down the general solution to (56), reset

and argue the plausibility of formula (36).

n ! %1/2,

y AxB ! z AxB /1x,z– # z ! 0 . x W 1,

z– # a1 # 1 " 4n2 4x2

b z ! 0 . z AxB ! 1x y AxB

s2 d2z ds2

# s dz ds

#as2 " 1An # 2B2b z ! 0 , s 7 0 . s !

21c n # 2

xn/2#1

y ! x1/2z

y% AxB $ cxny AxB " 0 , x 7 0 ,

Chapter Summary

Initial value problems that do not fall into the “solvable” categories that have been studied (such as constant-coefficient or equidimensional equations) can often be analyzed by interpret- ing the differential equation as a prescription for computing the higher derivatives of the unknown function. The Taylor polynomial method uses the equation to construct a polyno- mial approximation matching the initial values of a finite number of derivatives of the unknown. If the equation permits the extrapolation of this procedure to polynomials of arbitrar- ily high degree, power series representations of the solution can be constructed.

Power Series Every power series has a radius of convergence such that the series converges absolutely for and diverges when By the ratio test,

provided that this limit exists as an extended real number. A function that is the sum of a power series in some open interval about is said to be analytic at x0 . If f is analytic at its power series representation about is the Taylor series

Power Series Method for an Ordinary Point In the case of a linear equation of the form

(1)

where p and q are analytic at the point is called an ordinary point, and the equation has a pair of linearly independent solutions expressible as power series about The radii of con- vergence of these series solutions are at least as large as the distance from to the nearest sin- gularity (real or complex) of the equation. To find power series solutions to (1), we substitute

into (1), group like terms, and set the coefficients of the resulting power series equal to zero. This leads to a recurrence relation for the coefficients which, in some cases, may even yield a general formula for the The same method applies to the non- homogeneous version of (1), provided the forcing function is also analytic at

Regular Singular Points If, in equation (1), either p or q fails to be analytic at then is a singular point of (1). If is a singular point for which and are both analytic at then is a regular singular point. The Cauchy–Euler equation,

(2)

has a regular singular point at x ! 0, and a general solution to (2) can be obtained by substitut- ing and examining the roots of the resulting indicial equation

Method of Frobenius For an equation of the form (1) with a regular singular point at a series solution can be found by the method of Frobenius. This is obtained by substituting

into (1), finding a recurrence relation for the coefficients, and choosing the larger root of the indicial equation

(3)

where p0 J limxSx0 Ax " x0Bp AxB, q0 J limxSx0 Ax " x0B2q AxB.r Ar " 1B # p0r # q0 ! 0 , r ! r1,

w Ar, xB ! Ax " x0Br aq n!0

an Ax " x0Bn x0,

ar2 # Ab " aBr # c ! 0.y ! xr ax2

d2y

dx2 # bx

dx # cy ! 0 , x 7 0 ,

x0x0,Ax " x0B2q AxBAx " x0Bp AxB x0x0x0, x0.

an. an,

y AxB ! ©qn!0 an Ax " x0Bn x0 x0.

x0x0,

y– # p AxBy¿ # q AxBy ! 0 ,

f AxB ! aq n!0

f AnB Ax0B n!

Ax " x0Bn . x0

x0,x0 f AxB

r ! lim nSq

0an 00an#1 0 , 0 x " x0 0 7 r.0 x " x0 0 6 r r, 0 $ r $ q,©qn!0 an Ax " x0Bn

490 Chapter 8 Series Solutions of Differential Equations

Finding a Second Linearly Independent Solution If the two roots of the indicial equation (3) do not differ by an integer, then a second lin- early independent solution to (1) can be found by taking in the method of Frobenius. However, if or is a positive integer, then discovering a second solution requires a different approach. This may be a reduction of order procedure or the utilization of Theorem 7, page 468, which gives the forms of the solutions.

Special Functions Some special functions in physics and engineering that arise from series solutions to linear second-order equations with polynomial coefficients are Gaussian hypergeometric functions,

Bessel functions and orthogonal polynomials such as those of Legendre, Chebyshev, Laguerre, and Hermite.

Jn AxB;F Aa, b; g; xB;

r1 " r2r1 ! r2 r ! r2

r1, r2

Review Problems 491

1. Find the first four nonzero terms in the Taylor polyno- mial approximation for the given initial value problem. (a) (b)

2. Determine all the singular points of the given equa- tion and classify them as regular or irregular. (a) (b)

3. Find at least the first four nonzero terms in a power series expansion about x ! 0 for a general solution to the given equation. (a) (b)

4. Find a general formula for the coefficient in a power series expansion about x ! 0 for a general solution to the given equation. (a) (b)

5. Find at least the first four nonzero terms in a power series expansion about x ! 2 for a general solution to

6. Use the substitution to find a general solution to the given equation for x > 0. (a) (b) x3y‡ AxB # 3x2y– AxB " 2xy¿ AxB " 2y AxB ! 02x2y– AxB # 5xy¿ AxB " 12y AxB ! 0

y ! xr w– # Ax " 2Bw¿ " w ! 0 . Ax2 " 2By– # 3y ! 0A1 " x2By– # xy¿ # 3y ! 0

y– # e"xy¿ " y ! 0 y– # x2y¿ " 2y ! 0

Asin xBy– # y ! 0Ax2 " 4B2y– # Ax " 4By¿ # xy ! 0 z¿ A0B ! 1z– " x3z¿ # xz2 ! 0 ; z A0B ! "1 , y¿ ! xy " y2 ; y A0B ! 1

7. Use the method of Frobenius to find at least the first four nonzero terms in the series expansion about x ! 0 for a solution to the given equation for x > 0. (a) (b)

8. Find the indicial equation and its roots and state (but do not compute) the form of the series expansion about x ! 0 (as in Theorem 7 on page 468) for two linearly independent solutions of the given equation for x > 0. (a) (b) (c)

9. Find at least the first three nonzero terms in the series expansion about x ! 0 for a general solution to the given equation for . (a) (b) (c) (d)

10. Express a general solution to the given equation using Gaussian hypergeometric functions or Bessel functions.

(a)

(b) 9u2y– # 9uy¿ # A9u2 " 1By ! 0x A1 " xBy– # a12 " 6xb y¿ " 6y ! 0 x2y– # Ax " 2By ! 02xy– # 6y¿ # y ! 0 xy– # y¿ " 2y ! 0 x2y– " x A1 # xBy¿ # y ! 0x 7 0 Ax sin xBy– # xy¿ # Atan xBy ! 02xy– # 5y¿ # xy ! 0 x2y– # Asin xBy¿ " 4y ! 0

x2y– # Ax2 # 2xBy¿ " 2y ! 0x2y– " 5xy¿ # A9 " xBy ! 0

REVIEW PROBLEMS

1. Knowing that a general solution to a nonhomoge- neous linear second-order equation can be expressed as a particular solution plus a general solution to the corresponding homogeneous equation, what can you say about the form of a general power series solution to the nonhomogeneous equation about an ordinary point?

2. Discuss advantages and disadvantages of power series solutions over numerical solutions generated by the Euler or Runge–Kutta methods.

492 Chapter 8 Series Solutions of Differential Equations

3. The factors and all can arise in solutions generated using the method of Frobenius to solve a differential equation with 0 as a regular singular point. Discuss the problems encoun- tered in attempting to approximate such solutions using the methods of Euler and of Runge–Kutta.

4. Discuss the issues involved when attempting to use the Taylor polynomials (expanded around to study the asymptotic behavior (as of solutions to the differential equation y– # y ! 0.

t S % q Bx0 ! 0B

x0.3#12ix0.3, x0.3 ln x, x"0.3,

TECHNICAL WRITING EXERCISES

A Alphabetization Algorithms In 1961, C. A. R. Hoare published the algorithm Quicksort, for alphabetizing a large list of words (“Algorithm 64: Quicksort,” by C. A. R. Hoare, Comm. ACM 4 (1961): 324). First, one of the n words is selected at random, and the rest of the list is separated into two sublists—those that pre- cede the selected word and those that follow it. This entails comparisons. Then the same procedure is applied to each sublist, and so on.

If the selected words happen to belong in the middle of the collections—say, 50th out of 100— then the sort is very efficient. One sort involving 99 comparisons, then 2 sorts of 49 comparisons each, and so forth. But in the worst case, if the randomly selected words happen to belong at the beginning or end of the list, the number of comparisons is .

(a) Show that, for a list of length n, if all the selected words happen to belong at the mid- point of their sublists, then the total number of comparisons is roughly n ln2n .

(b) Show that if all the selected words happen to belong at the beginning of their sublists, the number of comparisons is .

(c) Show that if en is the expected number of comparisons (in the statistical sense) to alpha- betize n words, then , , and for all n ) 0,

where e0 ! 0 by definition. To express en explicitly, we are going to construct a differential equation whose solutions

have Taylor coefficients satisfying the same recursion as in part (c), solve the equation explicitly, and perform the Taylor expansion to extract en.

Consider the nonhomogeneous linear equation . It is singular at x ! 1, of course, but it is regular at x ! 0 if (the known function) is analytic

there. Assume and .

(d) Substitute these series directly into the differential equation and, recalling the geometric

series , derive the recursion relation .

(e) Identify the value of bn that renders the recursion relations in (c) and (d) identical, and

explicitly construct . [Hint: Use the derivatives of the geometric series.]

(f) Solve the above differential equation, with initial condition , explicitly to obtain . [You may find it convenient to refer to Section 2.3.]

(g) Multiply out the Taylor series for ln and in and deduce the Quicksort comparisons formula

.en ! 2 an " 12 # n " 23 # n " 34 # p # 1nb ! 2anr!2 n " r # 1r , n ' 2 y AxBA1 " xB"2A1 " xB

y AxB ! " 3 2x # 2 ln A1 " xB 4 / A1 " xB2 y A0B ! e0 ! 0 f AxB ! aq

m!0 bm x

an ! 2 n a

n"1

r!0 ar # bn"1 /n1/ A1 " xB ! aqm!0 xm

f AxB ! aq m!0

bmx my AxB ! aq

m!0 amx

f AxBy¿ ! 2y/ A1 " xB # f AxB

en ! n " 1 # 1 n a

n"1

r!0 3 er # en"1"r 4 ! 2n an"1r!0er # n " 1

e2 ! 1e1 ! 0

an"1k!1k ! n An " 1B /2 99 # 98 # 97 # p # 1

An " 1B

Group Projects for Chapter 8

493

Several authors have devised improvements to Quicksort to reduce the probability of worst-case scenarios. (For example, instead of choosing one word at random, one chooses three words and uses the median for comparison.) The article “An Asymptotic Theory for Cauchy-Euler Differential Equations with Applications to the Analysis of Algorithms,” by H.-H. Chern, H.-K. Hwang, and T.-H. Tsai (Journal of Algorithms, 44 (2002); 177–225) shows how the methodology in this project can be extended to ana- lyze many of these strategies.

B Spherically Symmetric Solutions to Schrödinger’s Equation for the Hydrogen Atom

In quantum mechanics one is interested in determining the wave function and energy states of an atom. These are determined from Schrödinger’s equation. In the case of the hydrogen atom, it is possible to find wave functions that are functions only of r, the distance from the proton to the electron. Such functions are called spherically symmetric and satisfy the simpler equation

(1)

where , m, and h are constants and E, also a constant, represents the energy of the atom, which we assume here to be negative.

(a) Show that with the substitutions

where is a negative constant, equation (1) reduces to

(b) If then the preceding equation becomes

(2)

Show that the substitution where is a positive constant, transforms (2) into

(3)

(4)

Show that a power series solution (starting with k ! 1) for (4) must have coefficients that satisfy the recurrence relation

(5) ak#1 ! 2 Aak " 1B k Ak # 1B ak , k ' 1 .

ak g ArB ! gqk!1 akrk

d2g

dr2 " 2a

dg dr

# 2 r

g ! 0 .

a2 ! "e Ae d2g

dr2 " 2a

dr # a2 r

# e # a2bg ! 0 . af ArB ! e"argArB,

d2f

dr2 ! " ae # 2

r b f .f J rc,

d2 ArcB dr2

! " ae # 2 r brc .e

r ! h2

4p2me20 r , E !

2p2me40 h2

e ,

e20

dr 2 ArcB ! "8mp2

h2 aE # e20

r bc ,

494 Chapter 8 Series Solutions of Differential Equations

(d) Now for and very large, and so which are the coefficients for Hence, g acts like so Going back further, we then see that Therefore, when is large, so is . Roughly speaking, is proportional to the probability of finding an electron a distance r from the proton. Thus, the above argument would imply that the electron in a hydrogen atom is more likely to be found at a very large distance from the proton! Since this makes no sense physically, we ask: Do there exist positive values for

for which remains bounded as r becomes large? Show that when then is a polynomial of degree n and

argue that is therefore bounded. (e) Let and denote, respectively, the energy state and wave function correspond-

ing to Find (in terms of the constants and h) and for n ! 1, 2, and 3.

C Airy’s Equation In aerodynamics one encounters the following initial value problem for Airy’s equation:

(a) Find the first five nonzero terms in a power series expansion about x ! 0 for the solu- tion and graph this polynomial for

(b) Using the Runge–Kutta subroutine (see Section 5.3) with h ! 0.05, approximate the solution on the interval [0, 10], i.e., at the points 0.05, 0.1, 0.15, etc.

(c) Using the Runge–Kutta subroutine with h ! 0.05, approximate the solution on the interval [Hint: With the change of variables it suffices to approxi- mate the solution to on the interval .]

(d) Using your knowledge of constant-coefficient equations as a basis for guessing the behavior of the solutions to Airy’s equation, decide whether the power series approxi- mation obtained in part (a) or the numerical approximation obtained in parts (b) and (c) better describes the true behavior of the solution on the interval

D Buckling of a Tower A tower is constructed of four angle beams connected by diagonals (see Figure 8.16 on page 496). The deflection curve for the tower is governed by the equation

(6)

where x is the vertical coordinate measured down from the extended top of the tower, y is the deflection from the vertical passing through the center of the unbuckled tower, L is the tower height, a is the length of the truncation, P is the load, E is the modulus of elasticity, and I is the moment of inertia. The appropriate boundary conditions for this design are

(7)

(8) y¿ Aa # LB ! 0 .y AaB ! 0 ,

x2 d2y

dx2 #

Pa2

EI y ! 0 , a 6 x 6 a # L ,

y AxB

3"10, 10 4 . 30, 10 4y– " zy ! 0; y A0B ! 1, y¿ A0B ! 0,z ! "x,3"10, 0 4 .

"10 $ x $ 10.

y– # xy ! 0 , y A0B ! 1 , y¿ A0B ! 0 .

cn ArBe20, m,Ena ! 1/n.cn ArBEn c

g ArBa ! 1/n, n ! 1, 2, 3, . . . ,ca c2 ArBc r ! h2r/4p2me20c ! ear.

f ArB ! e"arg ArB is like rear.re2ar,re2ar. ak#1 ! A2aBk/k!,ak#1 ! A2a/kBakka1 ! 1 Group Projects for Chapter 8 495

Clearly, the solution is always at hand. However, when the load P is heavy enough, the tower can buckle and a nontrivial solution is also possible. We want to predict this phenomenon.

(a) Solve equation (6). [Hint: Equation (6) is a Cauchy–Euler equation.] (b) Show that the first boundary condition (7) implies

where A is an arbitrary constant and (c) Show that the second boundary condition (8) gives

(d) Use the result of part (c) to argue that no buckling takes place (i.e., the only possibility is if where is the smallest positive real number that makes the expression in braces zero.

(e) The value of the load corresponding to is called the critical load Solve for in terms of and I.

(f ) Find the critical load if a ! 10, L ! 40, and EI ! 1000.Pc bc, a, E,

PcPc.bc

bc0 6 b 6 bc,A ! 0B A e tan cb ln aa # L

a b d # 2b f ! 0 .

b J 2Pa2/EI " 1/4.y ! Ax1/2 sin 3b ln Ax/aB 4 , y AxBy AxB & 0

496 Chapter 8 Series Solutions of Differential Equations

V er

tic al

V er

tic al

y(x)

(a) Unbuckled (b) Buckled

Figure 8.16 Buckling tower

E Aging Spring and Bessel Functions In Problems 30 and 31 in Exercises 8.4, page 451, we discussed a model for a mass–spring sys- tem with an aging spring. Without damping, the displacement at time t is governed by the equation

(9)

where m, k, and are positive constants. The general solution to this equation can be expressed using Bessel functions.

(a) The coefficient of x suggests a change of variables of the form Show that (9) transforms into

(10)

(b) Show that choosing and to satisfy

and

transforms (10) into the Bessel equation of order 0 in s and x.

where and are the Bessel functions of order 0 of the first and second kind, respectively.

(d) Discuss the behavior of the displacement for positive, negative, and zero.

(e) Compare the behavior of the displacement for a small positive number and a large positive number.

hhx AtB c2x AtB Y0J0

x AtB ! c1J0 a2h2k/me"ht/2b # c2Y0 a2h2k/me"ht/2b ,

mb2a"h/b ! 1

b ! 2

b2s2 d2x ds2

# b2s dx ds

# k m as a b"h/bx ! 0 .

s ! aebt.

mx– AtB # ke"htx AtB ! 0 , x AtB

Group Projects for Chapter 8 497

498

In this chapter we return to the analysis of systems of differential equations. When the equa- tions in the system are linear, matrix algebra provides a compact notation for expressing the system. In fact, the notation itself suggests new and elegant ways of characterizing the solution properties, as well as novel, efficient techniques for explicitly obtaining solutions.

Matrix Methods for Linear Systems

CHAPTER 9

INTRODUCTION 9.1

The right-hand side of the first member of (1) possesses a mathematical structure that is famil- iar from vector calculus; namely, it is the dot product† of two vectors:

(2)

Similarly, the second right-hand side in (1) is the dot product

The frequent occurrence in mathematics of arrays of dot products, such as evidenced in the system (1), led to the development of matrix algebra, a mathematical discipline whose basic operation—the matrix product—is the arrangement of a set of dot products according to the following plan:

In general, the product of a matrix—i.e., an m by n rectangular array of numbers—and a column vector is defined to be the collection of dot products of the rows of the matrix with the

c!4 2 4 !4

d c x y d " c 3!4 2 4 # 3 x y 434 !4 4 # 3 x y 4 d " c!4x # 2y4x ! 4y d .

4x ! 4y " 34 !4 4 # 3 x y 4 . !4x # 2y " 3!4 2 4 # 3 x y 4 .

†Recall that the dot product of two vectors u and v equals the length of u times the length of v times the cosine of the angle between u and v. However, it is more conveniently computed from the components of u and v by the “inner product” indicated in equation (2).

In Chapter 5 we analyzed physical situations wherein two fluid tanks containing brine solutions were interconnected and pumped so as ultimately to deplete the salt content in each tank. By accounting for the various influxes and outfluxes of brine, a system of differential equations for the salt contents and of each tank was derived; a typical model is

(1)

Express this system in matrix notation as a single equation.

dy/dt ! 4x " 4y . dx/dt ! "4x # 2y ,

y AtB BAx AtB

vector, arranged as a column vector:

where the vector v has n components; the dot product of two n-dimensional vectors is com- puted in the obvious way:

Using the notation for the matrix product, we can write the system (1) for the intercon- nected tanks as

The following example demonstrates a four-dimensional implementation of this notation. Note that the coefficients in the linear system need not be constants.

Express the system

(3)

as a matrix equation.

We express the right-hand side of the first member of (3) as the dot product

The other dot products are similarly identified, and the matrix form is given by

◆

In general, if a system of differential equations is expressed as

o x¿n " an1 AtBx1 # an2 AtBx2 # p # ann AtBxn , x¿2 " a21 AtBx1 # a22 AtBx2 # p # a2n AtBxn x¿1 " a11 AtBx1 # a12 AtBx2 # p # a1n AtBxn

Dx¿1x¿2 x¿3 x¿4

T " D2 t2 0 A4t # etB0 sin t cos t 0 1 1 1 1 0 0 0 0

T Dx1x2 x3 x4

T . 2x1 # t

2x2 # A4t # etBx4 " 32 t2 0 A4t # etB 4 # 3 x1 x2 x3 x4 4 . x¿4 " 0 x¿3 " x1 # x2 # x3 # x4 , x¿2 " Asin tBx2 # Acos tBx3 , x¿1 " 2x1 # t2x2 # A4t # etBx4 ,

c x¿ y¿ d " c!4 24 !4 d c xy d . 3a1 a2 p an 4 # 3 x1 x2 p xn 4 " a1x1 # a2x2 # p # anxn .

D row # 1row # 2o row # m

T DvT " D 3 row # 1 4 # v3 row # 2 4 # vo3 row # m 4 # vT , Section 9.1 Introduction 499

Example 1

Solution

it is said to be a linear homogeneous system in normal form.† The matrix formulation of such a system is then

where A is the coefficient matrix

and x is the solution vector

Note that we have used to denote the vector of derivatives

Express the differential equation for the undamped, unforced mass–spring oscillator (recall Section 4.1, page 154)

(4)

as an equivalent system of first-order equations in normal form, expressed in matrix notation.

We have to express the second derivative, as a first derivative in order to formulate (4) as a first-order system. This is easy; the acceleration is the derivative of the velocity so (4) becomes

(5)

The first-order system is then assembled by identifying with and appending it to (5):

To put this system in normal form and express it as a matrix equation, we need to divide the second equation by the mass m:

◆c y y d ¿" c 0 1

!k/m 0 d c yy d . my¿ " !ky .

y¿ " y

y¿,y

my¿ # ky " 0 .

y " y¿,y– y–,

my– # ky " 0

x¿ " Dx1x2 o

T ¿" Dx¿1x¿2 o

x¿n

T .x¿ x " Dx1x2

o xn

T . A " A AtB " Da11 AtB a12 AtB p a1n AtBa21 AtB a22 AtB p a2n AtB

o o o an1 AtB an2 AtB p ann AtBT

x$ ! Ax ,

500 Chapter 9 Matrix Methods for Linear Systems

†The normal form was defined for general systems in Section 5.3, page 253.

Example 2

Solution

In general, the customary way to write an nth-order linear homogeneous differential equation

as an equivalent system in normal form is to define the first derivatives of y (including y, the zeroth derivative, itself) to be new unknowns:

Then the system consists of the identification of as the derivative of together with the original differential equation expressed in these variables and divided by :

For systems of two or more higher-order differential equations, the same procedure is applied to each unknown function in turn; an example will make this clear.

The coupled mass–spring oscillator depicted in Figure 5.26 on page 285 was shown to be governed by the system

(6)

Write (6) in matrix notation.

We introduce notation for the lower-order derivatives:

(7)

In these variables, the system (6) states

(8)

The normal form is then

x¿4 " 2x1 ! 2x3 x¿3 " x4 , x¿2 " !3x1 # x3 , x¿1 " x2 ,

x¿4 # 2x3 ! 2x1 " 0 . 2x¿2 # 6x1 ! 2x3 " 0 ,

x1 " x , x2 " x¿ , x3 " y , x4 " y¿ .

d2y

dt2 # 2y ! 2x " 0 .

2 d2x dt2

# 6x ! 2y " 0 ,

x¿n " ! a0 AtB an AtB x1 ! a1 AtBan AtB x2 ! p ! an!1 AtBan AtB xn .

x¿n!1 " xn ,

x¿2 " x3 , x¿1 " x2 ,

Ban AtBA xj!1 AtB,xj AtB xn AtB " y An!1B AtB . x2 AtB " y¿ AtB , x1 AtB " y AtB ,

An ! 1Ban AtBy AnB # an!1 AtBy An!1B # p # a1 AtBy¿ # a0 AtBy " 0

Section 9.1 Introduction 501

Example 3

Solution

or in matrix notation

◆Dx1x2 x3 x4

T œ" D 0 1 0 0!3 0 1 0 0 0 0 1 2 0 !2 0

T Dx1x2 x3 x4

T . 502 Chapter 9 Matrix Methods for Linear Systems

9.1 EXERCISES In Problems 1–6, express the given system of differential equations in matrix notation.

x¿3 " x1 ! x2 x ¿2 " Asin 2tBx2 , x ¿1 " Acos 2tBx1 , y¿ " Acos tBx # Aa # bt3By x¿ " Asin tBx # ety , x¿4 " 0 x ¿3 " 1px1 ! x3 , x ¿2 " x1 # x4 , x ¿1 " x1 ! x2 # x3 ! x4 , z¿ " 4y y¿ " 2z ! x , x¿ " x # y # z , y¿ " !x x¿ " y , y¿ " 3x ! 2y x¿ " 7x # 2y ,

In Problems 7–10, express the given higher-order differ- ential equation as a matrix system in normal form.

7. The damped mass–spring oscillator equation

8. Legendre’s equation 9. The Airy equation

10. Bessel’s equation

In Problems 11–13, express the given system of higher- order differential equations as a matrix system in normal form. 11.

12.

13. y‡ # y– ! tx¿ # y¿ # etx " 0 x– ! 3x¿ # t2y ! Acos tBx " 0 ,y– # x¿ # 3y¿ # y " 0 x– # 3x¿ ! y¿ # 2y " 0 , y– ! 2x " 0 x– # 3x # 2y " 0 ,

y– # 1 t

y¿ # a1 ! n2 t2 b y " 0y– ! ty " 0

A1 ! t2By– ! 2ty¿ # 2y " 0my– # by¿ # ky " 0

9.2 REVIEW 1: LINEAR ALGEBRAIC EQUATIONS Here and in the next section we review some basic facts concerning linear algebraic systems and matrix algebra that will be useful in solving linear systems of differential equations in normal form. Readers competent in these areas may proceed to Section 9.4.

A set of equations of the form

(where the ’s and ’s are given constants) is called a linear system of n algebraic equations in the n unknowns The procedure for solving the system using elimination methods is well known. Herein we describe a particularly convenient implementation of the

x1, x2, . . . , xn. biaij

an1x1 # an2x2 # p # annxn " bn

a21x1 # a22x2 # p # a2nxn " b2 , a11x1 # a12x2 # p # a1nxn " b1 ,

method called the Gauss–Jordan elimination algorithm.† The basic idea of this formulation is to use the first equation to eliminate in all the other equations; then use the second equation to eliminate in all the others; and so on. If all goes well, the resulting system will be “uncoupled,” and the values of the unknowns will be apparent. A short example will make this clear.

Solve the system

By subtracting 2 times the first equation from the second, we eliminate from the latter. Sim- ilarly, is eliminated from the third equation by subtracting 1 times the first equation from it:

Next we subtract multiples of the second equation from the first and third to eliminate in them; the appropriate multiples are 2 and respectively:

Finally, we eliminate from the first two equations by subtracting multiples (2 and 3, respec- tively) of the third equation:

The system is now uncoupled; i.e., we can solve each equation separately:

◆

Two complications can disrupt the straightforward execution of the Gauss–Jordan algo- rithm. The first occurs when the impending variable to be eliminated (say, ) does not occur in the jth equation. The solution is usually obvious; we employ one of the subsequent equations to eliminate Example 2 illustrates this maneuver.

Solve the system

x1 # 4x2 # 2x3 # " !3 .

!2x1 ! 4x2 ! 8x3 # 2x4 " 4 ,

!x1 ! 2x2 ! 2x3 # " 1 ,

x1 # 2x2 # 4x3 # x4 " 0 ,

xj.

x1 " 0 , x2 " 0 , x3 " 2 .

2x13x2x3 " 2 .

2x13x2x3 " 0 ,

x3 " 2 .

2x13x2 # 3x3 " 6 ,

2x13x2 # 2x3 " 4 ,

!2, x2

!6x2 ! 5x3 " !10 .

3x2 # 3x3 " 6 ,

2x1 # 6x2 # 8x3 " 16 ,

x1 x1

2x1 # 3x3 " 6 .

4x1 # 15x2 # 19x3 " 38 ,

2x1 # 6x2 # 8x3 " 16 ,

x1, x2, . . . , xn x2

Section 9.2 Review 1: Linear Algebraic Equations 503

†The Gauss–Jordan algorithm is neither the fastest nor the most accurate computer algorithm for solving a linear system of algebraic equations, but for solutions executed by hand it has many pedagogical advantages. Usually it is much faster than Cramer’s rule, described in Appendix D.

Example 1

Solution

Example 2

504 Chapter 9 Matrix Methods for Linear Systems

Solution The first unknown is eliminated from the last three equations by subtracting multiples of the first equation:

Now, we cannot use the second equation to eliminate the second unknown because is not present. The next equation that does contain is the fourth, so we switch the second and fourth equation:

and proceed to eliminate :

To eliminate we have to switch again,

and eliminate, in turn, and This gives

and

The solution to the uncoupled equations is

◆

The other complication that can disrupt the Gauss–Jordan algorithm is much more pro- found. What if, when we are “scheduled” to eliminate the unknown it is absent from all of the subsequent equations? The first thing to do is to move on to the elimination of the next unknown as demonstrated in Example 3.

Apply the Gauss–Jordan algorithm to the system

(1)

!4x1 ! 8x2 # x3 " !10 .

2x1 # 4x2 # " 6 ,

2x1 # 4x2 # x3 " 8 ,

xj#1,

xj,

x1 " 1 , x2 " !1 , x3 " 0 , x4 " 1 .

x12x22x34x4 " 4 . 4x4 " 4 ,

x12x22x34x4 " 0 , x12x22x3 # x4 " 1 ,

x12x22x34x4 " !2 , x12x22x3 # " !2 ,

x12x22x34x4 " 1 , x12x22x3 ! x4 " 0 ,

x4.x3

4x4 " 4 ,

2x3 # x4 " 1 ,

x12x2 ! 2x3 ! x4 " !3 ,

x12x2 # 6x3 # 2x4 " 3 ,

x3,

2x3 # x4 " 1 .

4x4 " 4 ,

x12x2 ! 2x3 ! x4 " !3 ,

x12x2 # 6x3 # 2x4 " 3 ,

2x3 # x4 " 1 ,

4x4 " 4 ,

2x2 ! 2x3 ! x4 " !3 ,

x1 # 2x2 # 4x3 # x4 " 0 ,

x2 x2

2x2 ! 2x3 ! x4 " !3 .

4x4 " 4 ,

2x3 # x4 " 1 ,

x1 # 2x2 # 4x3 # x4 " 0 ,

Example 3

Elimination of proceeds as usual:

Now since is absent from the second and third equations, we use the second equation to eliminate :

(2)

How do we interpret the system (2)? The final equation contains no information, of course, and we ignore it.† The second equation implies that

The first equation implies that but there is no equation for Evidently, is a “free” variable, and we can assign any value to it—as long as we take to be Thus (1) has an infinite number of solutions, and a convenient way of characterizing them is

We remark that an equivalent solution can be obtained by treating x1 as the free variable, say and taking ◆

The final example is contrived to demonstrate all the features that we have encountered.

Find all solutions to the system

We use the first equation to eliminate :

Now, both are absent from all subsequent equations, so we use the second equation to eliminate

0 " 0 .

0 " 0 ,

!x4 " 1 ,

x1 ! x2 # 2x3 " 2 ,

x4. x2 and x3

0 " 0 .

3x4 " !3 ,

! x4 " 1 ,

x1 ! x2 # 2x3 # 2x4 " 0 ,

4x1 ! 4x2 # 8x3 # 8x4 " 0 .

3x1 ! 3x2 # 6x3 # 9x4 " !3 ,

2x1 ! 2x2 # 4x3 # 3x4 " 1 ,

x1 ! x2 # 2x3 # 2x4 " 0 ,

x2 " A3 ! sB /2, x3 " 2.x1 " s, x1 " 3 ! 2s , x2 " s , x3 " 2 ; !q 6 s 6 q .

3 ! 2x2.x1 x2x2.x1 " 3 ! 2x2,

x3 " 2.

0 " 0 .

! x3 " !2 ,

2x1 # 4x2 # " 6 ,

x3 x2

3x3 " 6 .

! x3 " !2 ,

2x1 # 4x2 # x3 " 8 ,

Section 9.2 Review 1: Linear Algebraic Equations 505

Solution

†The occurrence of the identity 0 " 0 in the Gauss–Jordan algorithm implies that one of the original equations was redundant. In this case you may observe that the final equation in (1) can be derived by subtracting 3 times the second equation from the first.

Example 4

Solution

There are no constraints on either or ; thus we take them to be free variables and charac- terize the solutions by

◆

In closing, we note that if the execution of the Gauss-Jordan algorithm results in a display of the form 0 " 1 (or where ), the original system has no solutions; it is inconsistent. This is explored in Problem 12.

k $ 00 " k,

x1 " 2 # s ! 2t , x2 " s , x3 " t , x4 " !1, !q 6 s, t 6 q .

x3x2

506 Chapter 9 Matrix Methods for Linear Systems

In Problems 1–11, find all solutions to the system using the Gauss–Jordan elimination algorithm.

!x1! A1# iB x2 " 0A1! iBx1 # 2x2 " 0 , x1 # 3x2 ! 2x3 " 3

2x1 # 4x2 ! x3 " 0 , x1 # 2x2 # x3 " !3 ,

!3x1 # 9x2 " 0 !x1 # 3x2 " 0 ,

4x1 ! 4x2 # 2x3 " 0 x1 ! 3x2 # x3 " 0 ,

!2x1 # 2x2 ! x3 " 0 ,

2x1 # 3x2 " 0 !x1 # 2x2 " 0 ,

2x1 ! x2 # x3 # x4 " 0 2x1 ! x2 # x3 # 2x4 " 0 ,

x1 # x2 # x3 # x4 " 1 , x3 # x4 " 0 ,

x1 # x2 ! x3 " 0 !x1 ! x2 # x3 " 0 ,

x1 # x2 ! x3 " 0 ,

x1 # 2x2 ! x3 # x4 " 0 2x1 # 2x2 ! x3 # x4 " 0 ,

x1 # # x4 " 0 , x1 # x2 # x3 # x4 " 1 ,

x1 # x2 # 3x3 " 6 2x1 # x2 # x3 " 6 ,

x1 # 2x2 # 2x3 " 6 ,

10.

11.

12. Use the Gauss–Jordan elimination algorithm to show that the following systems of equations are inconsistent. That is, demonstrate that the exis- tence of a solution would imply a mathematical contradiction. (a)

(b)

13. Use the Gauss–Jordan elimination algorithm to show that the following system of equations has a unique solution for r " 2, but an infinite number of solutions for r " 1.

14. Use the Gauss–Jordan elimination algorithm to show that the following system of equations has a unique solution for but an infinite number of solutions for r " 2.

4x1 ! 4x2 # 5x3 " rx3

x1 # x3 " rx2 , x1 # 2x2 ! x3 " rx1 ,

r " !1,

x1 ! 2x2 " rx2

2x1 ! 3x2 " rx1 ,

!x1 # x2 # 5x3 " 1 !3x1 # x2 # 4x3 " 1 ,

2x1 # x3 " !1 ,

!6x1 # 3x2 " 4 2x1 ! x2 " 2 ,

!x1 # x2 # 5x3 " 0 !3x1 # x2 # 4x3 " 1 ,

2x1 # x3 " !1 , x1 # 2x2 # x3 " i

2x1 # 3x2 ! ix3 " 0 , x1 # x2 # x3 " i ,

9.2 EXERCISES

A matrix is a rectangular array of numbers arranged in rows and columns. An matrix—that is, a matrix with m rows and n columns—is usually denoted by

where the element in the ith row and jth column is The notation is also used to desig- nate A. The matrices we will work with usually consist of real numbers, but in certain instances we allow complex-number entries.

Some matrices of special interest are square matrices, which have the same number of rows and columns; diagonal matrices, which are square matrices with only zero entries off the main diagonal (that is, if and (column) vectors, which are matrices. For example, if

then A is a square matrix, B is a diagonal matrix, and x is a vector. An matrix whose entries are all zero is called a zero matrix and is denoted by 0. For consistency, we denote matrices by boldfaced capitals, such as A, B, C, I, X, and Y, and reserve boldfaced lowercase letters, such as c, x, y, and z, for vectors.

Algebra of Matrices Matrix Addition and Scalar Multiplication. The operations of matrix addition and scalar mul- tiplication are very straightforward. Addition is performed by adding corresponding elements:

Formally, the sum of two matrices is given by

(The sole novelty here is that addition is not defined for two matrices whose dimensions m, n differ.)

To multiply a matrix by a scalar (number), we simply multiply each element in the matrix by the number:

In other words, The notation stands for A.A!1B!ArA " r 3aij 4 " 3 raij 4 .3 c 1 2 3 4 5 6

d " c 3 6 9 12 15 18

d .

A # B " 3aij 4 # 3bij 4 " 3aij # bij 4 .m % n c1 2 3 4 5 6

d # c1 1 1 1 1 1

d " c2 3 4 5 6 7

d .

m % n

A " £3 4 !12 6 5 0 1 4

§ , B " £3 0 00 0 0 0 0 7

§ , x " £42 1 § ,

n % 1i $ j);aij " 0

3aij 4aij. A J Da11 a12 a13 p a1na21 a22 a23 p a2n

o o o p o am1 am2 am3 p amn

T , m % n Section 9.3 Review 2: Matrices and Vectors 507

9.3 REVIEW 2: MATRICES AND VECTORS

Properties of Matrix Addition and Scalar Multiplication. Matrix addition and scalar multiplication are nothing more than mere bookkeeping, and the usual algebraic properties hold. If A, B, and C are matrices and r, s are scalars, then

Matrix Multiplication. The matrix product is what makes matrix algebra interesting and useful. We indicated in Section 9.1 that the product of a matrix A and a column vector x is the column vector composed of dot products of the rows of A with x:

More generally, the product of two matrices A and B is formed by taking the array of dot products of the rows of the first “factor” A with the columns of the second factor B; the dot product of the ith row of A with the jth column of B is written as the ijth entry of the product AB:

Note that AB is only defined when the number of columns of A matches the number of rows of B. A useful formula for the product of an matrix A and an matrix B is

where

The dot product of the ith row of A and the jth column of B is seen in the “sum of products” expression for

Since AB is computed in terms of the rows of the first factor and the columns of the second factor, it should not be surprising that, in general, AB does not equal BA (matrix multiplication does not commute):

but

In fact, the dimensions of A and B may render one or the other of these products undefined:

not defined .c1 2 3 4

d c0 1 d " c2

4 d ; c0

1 d c1 2

3 4 d c0 1 1 0

d c1 2 3 4

d " c3 4 1 2

d .c1 2 3 4

d c0 1 1 0

d " c2 1 4 3

d , cij.

cij J a n

k"1 aikbkj .AB J 3 cij 4 ,

n % pm % n

" c 5 3 x # z 12 9 3x ! y # 2z

d . c1 0 1

3 !1 2 d £ 1 2 x!1 "1 y

4 1 z § " c1 # 0 # 4 2 # 0 # 1 x # 0 # z

3 # 1 # 8 6 # 1 # 2 3x ! y # 2z d

c1 2 3 4 5 6

d £10 2 § " c1 # 1 # 2 # 0 # 3 # 2

4 # 1 # 5 # 0 # 6 # 2 d " c 716 d .

r AsAB " ArsBA " s ArAB . Ar # sBA " rA # sA ,r AA # BB " rA # rB , A # A!AB " 0 ,A # 0 " A , A # B " B # A ,A # AB # CB " AA # BB # C ,

m % n

508 Chapter 9 Matrix Methods for Linear Systems

By the same token, one might not expect (AB)C to equal A(BC), since in (AB)C we take dot products with the columns of B, whereas in A(BC) we employ the rows of B. So it is a pleasant surprise that this complication does not arise, and the “parenthesis grouping” rules are the customary ones:

Section 9.3 Review 2: Matrices and Vectors 509

Properties of Matrix Multiplication

(AB)C " A(BC) (Associativity) (A # B)C " AC # BC (Distributivity) A(B # C) " AB # AC (Distributivity) (rA)B " r(AB) " A(rB) (Associativity)

To summarize, the algebra of matrices proceeds much like the standard algebra of num- bers, except that we should never presume that we can switch the order of matrix factors.

Matrices as Linear Operators. Let A be an matrix and let x and y be vectors. Then Ax is an vector, and so we can think of multiplication by A as defining an operator that maps vectors into vectors. A consequence of the distributivity and associa- tivity properties is that multiplication by A defines a linear operator, since A(x # y) " Ax # Ay and A(rx) " rAx. Moreover, if A is an matrix and B is an matrix, then the

matrix AB defines a linear operator that is the composition of the linear operator defined by B with the linear operator defined by A. That is, (AB)x " A(Bx), where x is a vector.

Examples of linear operations are (i) stretching or contracting the components of a vector by constant factors;

(ii) rotating a vector through some angle about a fixed axis; (iii) reflecting a vector in a plane mirror.

The Matrix Formulation of Linear Algebraic Systems. Matrix algebra was developed to provide a convenient tool for expressing and analyzing linear algebraic systems. Note that the set of equations

can be written using the matrix product

(1)

In general, we express the linear system

o an1x1 # an2x2 # p # annxn " bn

a21x1 # a22x2 # p # a2nxn " b2 , a11x1 # a12x2 # p # a1nxn " b1 ,

£1 2 11 3 2 1 0 1

§ £ x1x2 x3 § " £ 1!1

0 § .

x1 # x3 " 0

x1 # 3x2 # 2x3 " !1 ,

x1 # 2x2 # x3 " 1 ,

p % 1 m % p

n % pm % n

m % 1n % 1 m % 1

n % 1m % n

in matrix notation as Ax " b, where A is the coefficient matrix, x is the vector of unknowns, and b is the vector of constants occurring on the right-hand side:

If b " 0, the system Ax " b is said to be homogeneous (analogous to the nomenclature of Section 4.2).

Matrix Transpose. The matrix obtained from A by interchanging its rows and columns is called the transpose of A and is denoted by AT. For example, if

In general, we have Properties of the transpose are explored in Problem 7.

Matrix Identity. There is a “multiplicative identity” in matrix algebra, namely, a square diagonal matrix I with ones down the main diagonal. Multiplying I on the right or left by any other matrix (with compatible dimensions) reproduces the latter matrix:

(The notation In is used if it is convenient to specify the dimensions, of the identity matrix.)

Matrix Inverse. Some square matrices A can be paired with other (square) matrices B having the property that BA " I:

(2)

When this happens, it can be shown that

(i) B is the unique matrix satisfying BA " I, and (ii) B also satisfies AB " I.

In such a case, we say that B is the inverse of A and write B " A Not every matrix possesses an inverse; the zero matrix 0, for example, can never satisfy

the equation 0B " I. A matrix that has no inverse is said to be singular.

!1.

≥ 32 !1 1212 0 ! 12 ! 32 1

1 2

¥ £1 2 11 3 2 1 0 1

§ " £1 0 00 1 0 0 0 1

§ .

n % n,

£1 0 00 1 0 0 0 1

§ £1 2 11 3 2 1 0 1

§ " £1 2 11 3 2 1 0 1

§ £1 0 00 1 0 0 0 1

§ .

3aij 4 T " 3bij 4 , where bij " aji. AT " £1 !12 2

6 !1 § .

A " c 1 2 6 !1 2 !1

d , then

A " Da11 a12 p a1na21 a22 p a2n o o o

an1 an2 p ann

T , x " Dx1x2 o

T , b " Db1b2 o

T . 510 Chapter 9 Matrix Methods for Linear Systems

If we know an inverse for the coefficient matrix A in a system of linear equations Ax " b, the solution can be calculated directly by computing A b, as the following derivation shows:

Ax " b implies A Ax " A b implies x " A b .

Using (2), for example, we can solve equation (1) quite efficiently:

When is known, this calculation is certainly easier than applying the Gauss–Jordan algorithm of the previous section. So it would appear advantageous to be able to find matrix inverses. Some inverses can be obtained directly from the interpretation of the matrix as a linear operator. For example, the inverse of a matrix that rotates a vector is the matrix that rotates it in the opposite direction. A matrix that performs a mirror reflection is its own inverse (what do you get if you reflect twice?). But in general one must employ an algorithm to com- pute a matrix inverse. The underlying strategy for this algorithm is based on the observation that if X denotes the inverse of A, then X must satisfy the equation AX " I; finding X amounts to solving n linear systems of equations for the columns {x1, x2, . . . , xn} of X:

The implementation of this operation is efficiently executed by the following variation of the Gauss–Jordan algorithm.

Finding the Inverse of a Matrix. By a row operation, we mean any one of the following: (a) Interchanging two rows of the matrix (b) Multiplying a row of the matrix by a nonzero scalar (c) Adding a scalar multiple of one row of the matrix to another row.

If the matrix A has an inverse, then A can be determined by performing row opera- tions on the matrix obtained by writing A and I side by side. In particular, we perform row operations on the matrix until the first n rows and columns form the iden- tity matrix; that is, the new matrix is Then . We remark that if this procedure fails to produce a matrix of the form then A has no inverse.

Find the inverse of A " £1 2 11 3 2 1 0 1

§ . 3 I!B 4 , A!1 " B3 I!B 4 .3A!I 4

3A!I 4n % 2n !1n % n

Ax1 " F100o 0 0

V , Ax2 " F010o 0 0

V , p , Axn " F000o 0 1

V .

A!1

£ x1x2 x3 § " ≥ 32 !1 1212 0 ! 12

! 32 1 1 2

¥ £ 1!1 0 § " ≥ 5212

! 52

¥ . !1!1!1

Section 9.3 Review 2: Matrices and Vectors 511

Example 1

We first form the matrix and row-reduce the matrix to Computing, we find the following:

The matrix

The matrix shown in color is [Compare equation (2).] ◆

Determinants. For a matrix A, the determinant of A, denoted det A or is defined by

We can define the determinant of a matrix A in terms of its cofactor expansion about the first row; that is,

For example,

" 1 A!3 ! 5B ! 2 A0 ! 10B # 1 A0 ! 6B " 6 . † 1 2 10 3 5 2 1 !1

† " 1 ` 3 5 1 !1

` ! 2 ` 0 5 2 !1

` # 1 ` 0 3 2 1

` det A J † a11 a12 a13a21 a22 a23

a31 a32 a33 † " a11 ` a22 a23a32 a33 ` ! a12 ` a21 a23a31 a33 ` # a13 ` a21 a22a31 a32 ` .

3 % 3

det A J ` a11 a12 a21 a22

` " a11a22 ! a12a21 . 0A 0 ,2 % 2A

!1.

≥ 1 0 0 ! 32 "1 120 1 0 ! 12 0 " 12 0 0 1 ! " 32 1

1 2

¥ .Add the third row to the first andthen subtract the third row from the second to obtain

≥ 1 0 !1 ! 3 !2 00 1 1 ! !1 1 0 0 0 1 ! ! 32 1

1 2

¥ .Multiply the third row by1/2 to obtain £1 0 !1 ! 3 !2 00 1 1 ! !1 1 0

0 0 2 ! !3 2 1 § .Subtract 2 times the second

row from the first to obtain

£1 2 1 ! 1 0 00 1 1 ! !1 1 0 0 0 2 ! !3 2 1

§ .Add 2 times the second row to the third row to obtain

£1 2 1 ! 1 0 00 1 1 ! !1 1 0 0 !2 0 ! !1 0 1

§ .Subtract the first row from the second and third to obtain

£1 2 1 ! 1 0 01 3 2 ! 0 1 0 1 0 1 ! 0 0 1

§ .3A!I 4 3 I!A!1 4 .3A!I 4

512 Chapter 9 Matrix Methods for Linear Systems

Solution

The determinant of an matrix can be similarly defined by a cofactor expansion involving st-order determinants. However, a more practical way to evaluate the deter- minant when n is large involves the row-reduction of the matrix to upper triangular form. Here we will deal mainly with and matrices, and leave further discussion of evaluating determinants to an introductory text on linear algebra.†

Determinants—particularly the higher-order ones—are laborious to compute directly. They have a geometric interpretation: det A is the volume (in n-dimensional space) of the parallelepiped whose edges are given by the column vectors of A. But their chief value lies in the role they play in the following theorem, which summarizes many of the results from linear algebra that we shall need, and in Cramer’s rule, described in Appendix D.

3 % 32 % 2

An ! 1B n % n Section 9.3 Review 2: Matrices and Vectors 513

†See Linear Algebra and Its Applications, 3rd updated ed., by David C. Lay (Addison-Wesley, Reading, Mass., 2006).

Matrices and Systems of Equations

Theorem 1. Let A be an matrix. The following statements are equivalent: (a) A is singular (does not have an inverse). (b) The determinant of A is zero. (c) Ax " 0 has nontrivial solutions (x 0). (d) The columns (rows) of A form a linearly dependent set.

n % n

In part (d), the statement that the n columns of A are linearly dependent means that there exist scalars c1, . . . , cn, not all zero, such that

where aj is the vector forming the jth column of A. If A is a singular square matrix (so det A " 0), then Ax " 0 has infinitely many solutions.

Indeed, Theorem 1 asserts that there is a vector such that Ax0 " 0, and we can get infinitely many other solutions by multiplying x0 by any scalar, i.e., taking x " cx0. Further- more, Ax " b either has no solutions or it has infinitely many of them of the form

x " xp # xh ,

where xp is a particular solution to Ax " b and xh is any of the infinity of solutions to Ax " 0 (see Problem 15). The resemblance of this situation to that of solving nonhomogeneous linear differential equations should be quite apparent.

To illustrate, in Example 3 of Section 9.2 (page 504) we saw that the system

has solutions

x1 " 3 ! 2s , x2 " s , x3 " 2 ; !q 6 s 6 q .

£ 2 4 12 4 0 !4 !8 1

§ £ x1x2 x3 § " £ 86

!10 §

x0 $ 0

c1a1 # c2a2 # p # cnan " 0 ,

Writing these in matrix notation, we can identify the vectors xp and xh mentioned above:

Note further that the determinant of A is indeed zero,

and that the linear dependence of the columns of A is exhibited by the identity

If A is a nonsingular square matrix (i.e., A has an inverse and det then the homo- geneous system Ax " 0 has x " 0 as its only solution. More generally, when det the system Ax " b has a unique solution, (namely,

Calculus of Matrices If we allow the entries in a matrix to be functions of the variable t, then is a matrix function of t. Similarly, if the entries of a vector are functions of t, then is a vector function of t.

These matrix and vector functions have a calculus much like that of real-valued functions. A matrix is said to be continuous at if each entry is continuous at Moreover,

is differentiable at if each entry is differentiable at and we write

(3)

Similarly, we define

(4)

Let

Find: (a) (b)

Using formulas (3) and (4), we compute

(a) (b) ◆! 1

A AtB dt " s 43 sin 1 e ! 1 1

t .A¿ AtB " c2t !sin t et 0

d . !

0 A AtB dt .A¿ AtB .

A AtB " c t2 # 1 cos t et 1

d . !

a A AtB dt J c ! b

a aij AtB dt d .

dA dt

At0B " A¿ At0B J 3a¿ij At0B 4 . t0,aij AtBt0A AtB t0.aij AtBt0A AtB

x AtBx AtBxi AtB A AtBA AtBaij AtB x " A!1bB. A $ 0,A $ 0

B, !2 £ 22

!4 § # 1 £ 44

!8 § # 0 £10

1 § " £00

0 § .

det A " 2 ` 4 0 !8 1

` ! 4 ` 2 0 !4 1

` # 1 ` 2 4 !4 !8

` " 2 # 4 ! 4 # 2 # 1 # 0 " 0 , x " £3 ! 2ss

2 § " £30

2 § # s £!21

0 § " xp # xh .

514 Chapter 9 Matrix Methods for Linear Systems

Example 2

Solution

Show that is a solution of the matrix differential equation , where

We simply verify that and are the same vector function:

◆

The basic properties of differentiation are valid for matrix functions.

x¿ AtB " c!v sin vt v cos vt

d ; Ax " c 0 !v v 0

d c cos vt sin vt

d " c!v sin vt v cos vt

d . Ax AtBx¿ AtB

A " c 0 !v v 0

d . x¿ " Axx AtB " c cos vt

sin vt d

Section 9.3 Review 2: Matrices and Vectors 515

Example 3

Solution

Differentiation Formulas for Matrix Functions

d dt

AABB " A dB dt

# dA dt

B .

d dt

AA # BB " dA dt

# dB dt

d dt

ACAB " C dA dt

(C a constant matrix) .

In the last formula, the order in which the matrices are written is very important because, as we have emphasized, matrix multiplication does not always commute.

9.3 EXERCISES

1. Let and

Find: (a) (b)

2. Let and

Find: (a) (b)

3. Let and

Find: (a) (b) (c)

4. Let and

Find: (a) (b) BA .AB .

B J c1 1 !1 0 3 1

d .A J £ 2 10 4 !1 3

§ B2 " BB .A2 " AA .AB .

B J c!1 3 5 2

d .A J c 2 4 1 1

d 7A ! 4B .A # B . B J c1 !1 2

0 3 !2 d .A J c 2 0 5

2 1 1 d 3A ! B .A # B .

B J c!1 0 2 !3

d .A J c 2 1 3 5

d 5. Let and .

Find: (a) (b) (c)

6. Let and

Find: (a) (b) (c) 7. (a) Show that if u and v are each column vec-

tors, then the matrix product is the same as the dot product u # v.

uTv n % 1

AA # BBC .AABBC .AB . C J c1 !4

1 1 d .

A J c 1 2 1 1

d , B J c 0 3 1 2

d , A AB # CB .AC .AB . C J c!1 1

2 1 d

A J c 1 !2 2 !3

d , B J c 1 0 1 1

d ,

(b) Let v be a column vector with Show that, for A as given in

Example 1, (c) Does hold for every

matrix A and vector (d) Does hold for every pair of

matrices A, B such that both matrix products are defined? Justify your answer.

8. Let and

Verify that

In Problems 9–14, compute the inverse of the given matrix, if it exists.

9. 10.

11. 12.

13. 14.

15. Prove that if satisfies then every solu- tion to the nonhomogeneous system Ax " b is of the form , where xh is a solution to the corresponding homogeneous system Ax ! 0.

16. Let

(a) Show that A is singular.

(b) Show that has no solutions.

solutions.

In Problems 17–20, find the matrix function whose value at t is the inverse of the given matrix

17.

18. X AtB " c sin 2t cos 2t 2 cos 2t !2 sin 2t

d X AtB " c et e4t

et 4e4t d

X AtB.X!1 AtB Ax " £ 30

3 §

Ax " £ 31 3 §

A " £ 2 !1 1!1 2 1 1 1 2

§ . x " xp # xh

Axp " b,xp

£ 1 1 11 !1 2 1 1 4

§£!2 !1 12 1 0 3 1 !1

§ £ 1 1 11 2 3

0 1 1 §£ 1 1 11 2 1

2 3 2 §

c 4 1 5 9

dc 2 1 !1 4

d AB $ BA.

B J c1 2 3 2

d .A J c 2 !1 !3 4

d AABBT " BTAT v?n % 1

m % nAAvBT " vTATAAvBT " vTAT. vT " 3 2 3 5 4 .3 % 1

516 Chapter 9 Matrix Methods for Linear Systems

19.

20.

In Problems 21–26, evaluate the given determinant.

21. 22.

23. 24.

25. 26.

In Problems 27–29, determine the values of r for which det

27. 28.

29.

30. Illustrate the equivalence of the assertions (a)–(d) in Theorem 1 (page 513) for the matrix

as follows. (a) Show that the row-reduction procedure applied

to fails to produce the inverse of A. (b) Calculate det A. (c) Determine a nontrivial solution x to Ax ! 0. (d) Find scalars and not all zero, so that

where and are the columns of A.

In Problems 31 and 32, find for the given vector functions.

31. 32. x AtB " £ e!t sin 3t0 !e!t sin 3t

§x AtB " £ e3t2e3t !e3t

§ dx /dt

a3a1, a2,c1a1 # c2a2 # c3a3 " 0, c3,c1, c2,

3A!I 4 A " £ 4 !2 2!2 4 2

2 2 4 §

A " £ 0 0 00 1 0 1 0 1

§ A " c 3 3

2 4 dA " c 1 1

!2 4 dAA ! rIB " 0.

† 1 4 43 0 !3 1 6 2

†† 1 4 3!1 !1 2 4 5 2

† † 1 0 20 3 !1 !1 2 1

†† 1 0 03 1 2 1 5 !2

† ` 12 8

3 2 `` 4 3

!1 2 `

X AtB " £ e3t 1 t3e3t 0 1 9e3t 0 0

§ X AtB " £ et e!t e2tet !e!t 2e2t

et e!t 4e2t §

In Problems 33 and 34, find for the given matrix functions.

33.

34.

In Problems 35 and 36, verify that the given vector function satisfies the given system.

35.

36.

In Problems 37 and 38, verify that the given matrix function satisfies the given matrix differential equation.

37.

38.

X AtB " £ et 0 00 et e5t 0 et !e5t

§ X¿ " £ 1 0 00 3 !2

0 !2 3 § X ,

X¿ " c 1 !1 2 4

d X , X AtB " c e2t e3t !e2t !2e3t

x¿ " £ 0 0 00 1 0 1 0 1

§ x , x AtB " £ 0et !3et

§ x¿ " c 1 1

!2 4 d x , x AtB " c e3t

2e3t d

X AtB " £ sin 2t cos 2t e!2t!sin 2t 2 cos 2t 3e!2t 3 sin 2t cos 2t e!2t

§ X AtB " c e5t 3e2t

!2e5t !e2t d

dX /dt

Section 9.4 Linear Systems in Normal Form 517

In Problems 39 and 40, the matrices are given. Find

(a) (b) (c)

39.

40.

41. An matrix A is called symmetric if that is, if for all Show that if A is an matrix, then is a symmetric matrix.

42. Let A be an matrix. Show that is a sym- metric matrix and is a symmetric

matrix (see Problem 41). 43. The inner product of two vectors is a generalization

of the dot product, for vectors with complex entries. It is defined by

where

, y " col Ay1, y2, . . . , ynBx " col Ax1, x2, . . . , xnBAx, yB J a n

i"1 xiyi ,

m % m AATn % n

ATAm % n

A # ATn % n i, j " 1, . . . , n.aij " aji,

AT " A;n % n

A AtB " c 1 e!2t 3 e!2t

d , B AtB " c e!t e!t !e!t 3e!t

d B AtB " c cos t !sin t

sin t cos t dA AtB " c t et

1 et d ,

d dt

3A AtBB AtB 4 .! 1 0

B AtB dt .! A AtB dt . A AtB and B AtB

9.4 LINEAR SYSTEMS IN NORMAL FORM In keeping with the introduction presented in Section 9.1, we say that a system of n linear differential equations is in normal form if it is expressed as

(1)

where col , col and is an matrix. As with a scalar linear differential equation, a system is called homogeneous

when otherwise, it is called nonhomogeneous. When the elements of A are allf AtB " 0;n % n A AtB " 3aij AtB 4A f1 AtB, . . . , fn AtBB ,f AtB "Ax1 AtB, . . . , xn AtBBx AtB "x¿ AtB " A AtBx AtB # f AtB ,

are complex vectors and the overbar denotes complex conjugation.

(a) Show that , where col

(b) Prove that for any vectors x, y, z and any complex number we have

Ax, lyB " l Ax, yB .Alx, yB " l Ax, y) , Ax, y # zB " Ax, yB # Ax, zB , Ax, y) " Ay, xB , l,

n % 1 Ay1, y2, . . . , ynB .y " Ax, yB " xTy

constants, the system is said to have constant coefficients. Recall that an nth-order linear dif- ferential equation

(2)

can be rewritten as a first-order system in normal form using the substitution indeed, equation (2) is equivalent to

where col col and

The theory for systems in normal form parallels very closely the theory of linear differen- tial equations presented in Chapters 4 and 6. In many cases the proofs for scalar linear differ- ential equations carry over to normal systems with appropriate modifications. Conversely, results for normal systems apply to scalar linear equations since, as we showed, any scalar lin- ear equation can be expressed as a normal system. This is the case with the existence and uniqueness theorems for linear differential equations.

The initial value problem for the normal system (1) is the problem of finding a differen- tiable vector function that satisfies the system on an interval I and also satisfies the initial condition where is a given point of I and col is a given vector.

Ax1,0, . . . , xn,0Bx0 "t0x At0B " x0,x AtB

A AtB J E 0 1 0 p 0 00 0 1 0 0o o o o o 0 0 0 p 0 1

!p0 AtB !p1 AtB !p2 AtB p !pn!2 AtB !pn!1 AtB U .A0, . . . , 0, g AtBB ,Ax1 AtB, . . . , xn AtBB , f AtB Jx AtB "f AtB,

A AtBx AtB #x¿ AtB "x2 AtB J y¿ AtB, . . . , xn AtB J y An!1B AtB; x1 AtB J y AtB, y AnB AtB # pn!1 AtBy An!1B AtB # p # p0 AtBy AtB " g AtB

518 Chapter 9 Matrix Methods for Linear Systems

Existence and Uniqueness

Theorem 2. Suppose and are continuous on an open interval I that contains the point Then, for any choice of the initial vector there exists a unique solution

on the whole interval I to the initial value problem

x¿ AtB " A AtBx AtB # f AtB , x At0B " x0 .x AtB x0,t0.

f AtBA AtB

We give a proof of this result in Chapter 13† and obtain as corollaries the existence and uniqueness theorems for second-order equations (Theorem 4, Section 4.5) and higher-order linear equations (Theorem 1, Section 6.1).

If we rewrite system (1) as and define the operator then we can express system (1) in the operator form Here the operator L maps vector functions into vector functions. Moreover, L is a linear operator in the sense that for any scalars a, b and differentiable vector functions x, y, we have

The proof of this linearity follows from the properties of matrix multiplication (see Problem 27).

L 3ax # by 4 " aL 3x 4 # bL 3y 4 . L 3x 4 " f. L 3x 4 J x¿ ! Ax,x¿ ! Ax " f

†All references to Chapters 11–13 refer to the expanded text Fundamentals of Differential Equations and Boundary Value Problems, 6th ed.

As a consequence of the linearity of L, if are solutions to the homogeneous sys- tem or in operator notation, then any linear combination of these vectors,

is also a solution. Moreover, we will see that if the solutions are linearly independent, then every solution to can be expressed as for an appropriate choice of the constants c1, . . . , cn.

c1x1 # p # cnxnL 3x 4 " 0 x1, . . . , xnc1x1 # p # cnxn,L 3x 4 " 0x¿ " Ax, x1, . . . , xn

Section 9.4 Linear Systems in Normal Form 519

Linear Dependence of Vector Functions

Definition 1. The m vector functions are said to be linearly dependent on an interval I if there exist constants not all zero, such that

(3)

for all t in I. If the vectors are not linearly dependent, they are said to be linearly independent on I.

c1x1 AtB # p # cmxm AtB " 0c1, . . . , cm, x1, . . . , xm

Show that the vector functions col col and col are linearly dependent on

Notice that is just 3 times and therefore for all t. Hence, and are linearly dependent on ◆

Show that

are linearly independent on .

Note that at every instant t0, the column vector is a multiple of ; indeed, for , and for . Nonetheless, the vector functions are not depen- dent, because the c’s in condition (3) are not allowed to change with t; for t & 0, the equation

implies , but for t ' 0 it implies . Thus and the functions are independent. ◆

Show that the vector functions col col and col are linearly independent on

To prove independence, we assume and are constants for which

holds at every t in and show that this forces In particular, when we obtain

which is equivalent to the system of linear equations

(4) c1 ! c2 # c3 " 0 .

c2 # 2c3 " 0 , c1 # c2 # c3 " 0 ,

c1 £10 1 § # c2 £ 11

!1 § # c3 £12

1 § " 0 ,

t " 0 c1 " c2 " c3 " 0.A!q, q Bc1x1 AtB # c2x2 AtB # c3x3 AtB " 0

c3c1, c2,

A!q, q B.Aet, 2et, etBx3 AtB " Ae2t, e2t, !e2tB,Ae2t, 0, e2tB, x2 AtB "x1 AtB " c1 " c2 " 0

c1 # c2 " 0c1 ! c2 " 0c1x1 AtB # c2x2 AtB " 0 t0 ( 0x1 At0B " !x2 At0Bt0 ) 0 x1 At0B " x2 At0Bx2 At0Bx1 At0B

A!q, q Bx1 AtB " c t0 t 0 d , x2 AtB " c 0 t 0t d A!q, q B.x3x1, x2, 3x1 AtB ! x2 AtB # 0 # x3 AtB " 0x1x2

A!q, q B.At, 1, 0Bx3 AtB " A3et, 0, 3etB,Aet, 0, etB, x2 AtB "x1 AtB "Example 1

Example 2

Solution

Example 3

Solution

Either by solving (4) or by checking that the determinant of its coefficients is nonzero (recall Theorem 1 on page 513), we can verify that (4) has only the trivial solution Therefore the vector functions and are linearly independent on (in fact, on any interval containing t " 0). ◆

As Example 3 illustrates, if are n vector functions, each having n components, we can establish their linear independence on an interval I if we can find one point t0 in I where the determinant

is not zero. Because of the analogy with scalar equations, we call this determinant the Wronskian.

det 3x1 At0B . . . xn At0B 4 x1 AtB, x2 AtB, . . . , xn AtB

A!q, q Bx3x1, x2, c1 " c2 " c3 " 0. 520 Chapter 9 Matrix Methods for Linear Systems

Wronskian

Definition 2. The Wronskian of n vector functions col col is defined to be the real-valued functionAx1,n, . . . , xn,nBxn AtB " Ax1,1, . . . , xn,1B, . . . ,x1 AtB "

W 3x1, . . . , xn 4 AtB J ∞ x1,1 AtB x1,2 AtB p x1,n AtBx2,1 AtB x2,2 AtB p x2,n AtBo o o xn,1 AtB xn,2 AtB p xn,n AtB ∞ .

We now show that if are linearly independent solutions on I to the homoge- neous system where is an matrix of continuous functions, then the Wronskian is never zero on I. For suppose to the contrary that

at some point Then by Theorem 1, page 513, the vanishing of the determinant implies that the column vectors , are linearly dependent. Thus there exist scalars not all zero, such that

However, and the vector function are both solutions to on I, and they agree at the point . So these solutions must be identical on I according

to the existence-uniqueness theorem (Theorem 2). That is,

for all t in I. But this contradicts the given information that are linearly independent on I. We have shown that and since is an arbitrary point, it follows that for all

The preceding argument has two important implications that parallel the scalar case. First, the Wronskian of solutions to is either identically zero or never zero on I (see also Problem 33). Second, a set of n solutions on I is linearly independent on I if and only if their Wronskian is never zero on I. With these facts in hand, we can imitate the proof given for the scalar case in Section 6.1 (Theorem 2) to obtain the following representa- tion theorem for the solutions to x¿ " Ax.

x1, . . . , xn to x¿ " Ax x¿ " Ax

t % I. W AtB $ 0t0W At0B $ 0, x1, . . . , xn

c1x1 AtB # p # cnxn AtB " 0 t0x¿ " Ax

z AtB " 0c1x1 AtB # p # cnxn AtBc1x1 At0B # p # cnxn At0B " 0 .c1, . . . , cn

x2 At0B, . . . , xn At0Bx1 At0Bt0.W At0B " 0 W AtB J det 3x1, x2, . . . , xn 4 n % nAx¿ " Ax, x1, x2, . . . , xn

A set of solutions that are linearly independent on I or, equivalently, whose Wronskian does not vanish on I, is called a fundamental solution set for (5) on I. The linear combination in (6), written with arbitrary constants, is referred to as a general solution to (5).

If we take the vectors in a fundamental solution set and let them form the columns of a matrix that is,

then the matrix is called a fundamental matrix for (5). We can use it to express the general solution (6) as

where col is an arbitrary constant vector. Since det is never zero on I, it follows from Theorem 1 in Section 9.3 that is invertible for every t in I.

Verify that the set

is a fundamental solution set for the system

(7)

on the interval and find a fundamental matrix for (7). Also determine a general solution for (7).

Substituting the first vector in the set S into the right-hand side of (7) gives

Ax " £0 1 11 0 1 1 1 0

§ £ e2te2t e2t § " £2e2t2e2t

2e2t § " x¿ AtB .

A!q, q B x¿ AtB " £0 1 11 0 1

1 1 0 § x AtB

S " # £ e2te2te2t § , £!e !t

0 e!t § , £ 0e!t

!e!t § $

X AtB X " W 3x1, . . . , xn 4Ac1, . . . , cnBc " x AtB " X AtBc ,

X AtB X AtB " 3x1 AtB x2 AtB . . . xn AtB 4 " Dx1,1 AtB x1,2 AtB p x1,n AtBx2,1 AtB x2,2 AtB p x2,n AtBo o o

xn,1 AtB xn,2 AtB p xn,n AtBT , X AtB,

Ex1, . . . , xnF

Section 9.4 Linear Systems in Normal Form 521

Representation of Solutions (Homogeneous Case)

Theorem 3. Let be n linearly independent solutions to the homogeneous system

(5)

on the interval I, where is an matrix function continuous on I. Then every solution to (5) on I can be expressed in the form

(6)

where are constants.c1, . . . , cn

x AtB " c1x1 AtB # p # cnxn AtB , n % nA AtBx¿ AtB " A AtBx AtB

x1, . . . , xn

Solution

Example 4

Hence this vector satisfies system (7) for all t. Similar computations verify that the remaining vectors in S are also solutions to (7) on For us to show that S is a fundamental solu- tion set, it is enough to observe that the Wronskian

is never zero. A fundamental matrix for (7) is just the matrix we used to compute the Wronskian;

that is,

(8)

A general solution to (7) can now be expressed as

◆

It is easy to check that the fundamental matrix in (8) satisfies the equation

indeed, this is equivalent to showing that for each column x in S. In general, a funda- mental matrix for a system satisfies the corresponding matrix differential equation

Another consequence of the linearity of the operator L defined by is the superposition principle for linear systems. It states that if and are solutions, respectively, to the nonhomogeneous systems

and

then is a solution to

Using the superposition principle and the representation theorem for homogeneous systems, we can prove the following theorem.

L 3x 4 " c1g1 # c2g2 .c1x1 # c2x2 L 3x 4 " g2 ,L 3x 4 " g1 x2x1

L 3x 4 J x¿ ! AxX¿ " AX. x¿ " Ax x¿ " Ax

X¿ AtB " £0 1 11 0 1 1 1 0

§ X AtB ; x AtB " X AtBc " c1 £ e2te2t

e2t § # c2 £!e!t0

e!t § # c3 £ 0e!t

!e!t § .

X AtB J £ e2t !e!t 0e2t 0 e!t e2t e!t !e!t

§ . X AtB

W AtB " † e2t !e!t 0e2t 0 e!t e2t e!t !e!t

† " e2t ` 0 e!t e!t !e!t

` # e!t ` e2t e!t e2t !e!t

` " !3 A!q, q B.

522 Chapter 9 Matrix Methods for Linear Systems

Representation of Solutions (Nonhomogeneous Case)

Theorem 4. Let be a particular solution to the nonhomogeneous system

(9)

on the interval I, and let be a fundamental solution set on I for the corre- sponding homogeneous system Then every solution to (9) on I can be expressed in the form

(10)

where are constants.c1, . . . , cn

x AtB " xp AtB # c1x1 AtB # p # cnxn AtB , x AtB " A AtBx AtB.Ex1, . . . , xnF

x¿ AtB " A AtBx AtB # f AtBxp

The proof of this theorem is almost identical to the proofs of Theorem 4 in Section 4.5 and Theorem 4 in Section 6.1. We leave the proof as an exercise.

The linear combination of in (10) written with arbitrary constants is called a general solution of (9). This general solution can also be expressed as

where X is a fundamental matrix for the homogeneous system and c is an arbitrary constant vector.

We now summarize the results of this section as they apply to the problem of finding a general solution to a system of n linear first-order differential equations in normal form.

x " xp # Xc,

c1, . . . , cnxp, x1, . . . , xn

Section 9.4 Linear Systems in Normal Form 523

We devote the rest of this chapter to methods for finding fundamental solution sets for homo- geneous systems and particular solutions for nonhomogeneous systems.

In Problems 1–4, write the given system in the matrix form

3. 4.

dz dt

" x # 5z dz dt

" tx ! y # 3z ! et

dy dt

" 2x ! y # 3z , dy dt

" etz # 5 ,

dx dt

" x # y # z , dx dt

" t2x ! y ! z # t ,

u¿ AtB " r AtB ! u AtB # 1r¿ AtB " 2r AtB # sin t , y¿ AtB " !x AtB # 2y AtB # etx¿ AtB " 3x AtB ! y AtB # t2 , x¿ " Ax # f.

Approach to Solving Normal Systems 1. To determine a general solution to the homogeneous system :

(a) Find a fundamental solution set that consists of n linearly independent solutions to the homogeneous system.

(b) Form the linear combination

where col is any constant vector and is the fundamental matrix, to obtain a general solution.

2. To determine a general solution to the nonhomogeneous system :

(a) Find a particular solution to the nonhomogeneous system. (b) Form the sum of the particular solution and the general solution

to the corresponding homogeneous system in part 1,

to obtain a general solution to the given system.

x " xp # Xc " xp # c1x1 # . . . # cnxn ,

c1x1 # . . . # cnxn Xc "

x¿ " Ax # f

X " 3x1 . . . xn 4Ac1, . . . , cnBc "x " Xc " c1x1 # . . . # cnxn , Ex1, . . . , xnF x¿ " Axn % n

9.4 EXERCISES In Problems 5–8, rewrite the given scalar equation as a first-order system in normal form. Express the system in the matrix form

6. 7.

In Problems 9–12, write the given system as a set of scalar equations.

9. x¿ " c 5 0 !2 4

d x # e!2t c 2 !3 d

d3y

dt3 !

dy dt

# y " cos t

d4w dt4

# w " t2x– AtB # x AtB " t2y– AtB ! 3y¿ AtB ! 10y AtB " sin t x¿ " Ax # f.

10.

11.

12.

In Problems 13–19, determine whether the given vector functions are linearly dependent or linearly inde- pendent on the interval

13. 14.

15. 16.

17.

18.

19.

20. Let

(a) Compute the Wronskian. (b) Are these vector functions linearly independent

on ? (c) Is there a homogeneous linear system for which

these functions are solutions?

In Problems 21–24, the given vector functions are solu- tions to the system Determine whether they form a fundamental solution set. If they do, find a fundamental matrix for the system and give a general solution.

21. x1 " e2t c 1!2 d , x2 " e2t c!24 d x¿ AtB " Ax AtB.

(!q, q)

x1 " £ cos t0 0

§ , x2 " £sin tcos t cos t

§ , x3 " £ cos tsin t cos t

£ 10 1 § , £ t0

t § , £ t20

t2 §

c sin t cos t

d , c sin t sin t d , c cos t

cos t d

e2t £ 10 5 § , e2t £ 11

!1 § , e3t £ 01

0 §

c sin t cos t

d , c sin 2t cos 2t

det c 1 5 d , et c !3

!15 d

c te!t e!t d , c e!t

e!t dc t

3 d , c 4

1 d A!q, q B.ALIB

ALDB x¿ " £ 0 1 00 0 1

!1 1 2 § x # t £ 1!1

2 § # £ 31

0 §

x¿ " £ 1 0 1!1 2 5 0 5 1

§ x # et £ 10 0 § # t £ 01

0 §

x¿ " c 2 1 !1 3

d x # et c t 1 d

524 Chapter 9 Matrix Methods for Linear Systems

22.

23.

24.

25. Verify that the vector functions

and

are solutions to the homogeneous system

on and that

is a particular solution to the nonhomogeneous system where col Find a general solution to

26. Verify that the vector functions

are solutions to the homogeneous system

on and that

is a particular solution to where col Find a general solution to

x¿ " Ax # f AtB.A!9t, 0, !18tB.f AtB " x¿ " Ax # f AtB,

xp " £ 5t # 12t 4t # 2

§ A!q, q B,

x¿ " Ax " £ 1 !2 2!2 1 2 2 2 1

§ x x3 " £!e!3t!e!3t

e!3t §x1 " £ e3t0

e3t § , x2 " £!e3te3t

0 § ,

x¿ " Ax # f AtB. Aet, tB.f AtB "x¿ " Ax # f AtB, xp "

3 2

c tet tet d ! 1

4 c et

3et d # c t

2t d ! c 0

1 dA!q, q B,

x¿ " Ax " c 2 !1 3 !2

d x , x2 " c e!t3e!t dx1 " c etet d

x3 " £!cos tsin t cos t

§ x1 " £ etet

et § , x2 " £ sin tcos t

!sin t § ,

x3 " £ e3t!e3t 2e3t §x1 " £ e!t2e!t

e!t § , x2 " £ et0

et § ,

x1 " e!t c 32 d , x2 " e4t c 1!1 d

27. Prove that the operator defined by J where is an matrix function and x is an

differentiable vector function, is a linear oper- ator.

28. Let be a fundamental matrix for the system Show that is the

solution to the initial value problem ,

In Problems 29–30, verify that is a fundamental matrix for the given system and compute Use the result of Problem 28 to find the solution to the given initial value problem.

29.

30.

31. Show that

on but that the two vector functions

are linearly independent on 32. Abel’s Formula. If are any n solutions

to the system then Abel’s formula gives a representation for the Wronskian

Namely,

where are the main diagonal ele- ments of Prove this formula in the special case when n " 3. [Hint: Follow the outline in Problem 30 of Exercises 6.1.]

33. Using Abel’s formula (Problem 32), prove that the Wronskian of n solutions to on the interval I is either identically zero on I or never zero on I.

34. Prove that a fundamental solution set for the homo- geneous system always exists on anx¿ AtB " A AtBx AtB

x¿ " Ax

A AsB.a11 AsB, . . . , ann AsB W AtB ! W At0Bexp a! t

t0 Ea11 AsB # p # ann AsB F dsb ,W AtB J W 3 x1, . . . , xn 4 AtB. x¿ AtB " A AtBx AtB,n % n x1, . . . , xn

A!q, q B.c t 2

2t d , c t 0 t 0

2 0 t 0 d A!q, q B,`

t2 t 0 t 0 2t 2 0 t 0 ` " 0

X AtB " c e!t e5t !e!t e5t

dx¿ " c 2 3 3 2

d x , x A0B " c 3 !1 d ;

X AtB " £ 6e!t !3e!2t 2e3t!e!t e!2t e3t !5e!t e!2t e3t

§ x¿ " £ 0 6 01 0 1

1 1 0 § x , x A0B " £!10

1 § ;

X!1 AtB.X AtB x At0B " x0. x¿ " Ax

x AtB " X AtBX!1 At0Bx0x¿ " Ax.X AtB n % 1

n % nA x¿ ! Ax,L 3 x 4

Section 9.4 Linear Systems in Normal Form 525

interval I, provided is continuous on I. Hint: Use the existence and uniqueness theorem (Theorem 2) and make judicious choices for

35. Prove Theorem 3 on the representation of solutions of the homogeneous system.

36. Prove Theorem 4 on the representation of solutions of the nonhomogeneous system.

37. To illustrate the connection between a higher-order equation and the equivalent first-order system, con- sider the equation

(11)

(a) Show that is a fundamental solution set for (11).

(b) Using the definition of Section 6.1, compute the Wronskian of

(12)

where

(d) The substitution used in part (c) suggests that

is a fundamental solution set for system (12). Verify that this is the case.

(e) Compute the Wronskian of S. How does it compare with the Wronskian computed in part (b)?

38. Define x1 , x2 , and x3 , for , by

(a) Show that for the three scalar functions in each individual row there are nontrivial linear combi- nations that sum to zero for all t.

(b) Show that, nonetheless, the three vector functions are linearly independent. (No single nontrivial combination works for each row, for all t.)

tial equation system having as solutions?

x1 AtB, x2 AtB, and x3 AtB W 3x1, x2, x3 4 AtB

x1 AtB " £sin tsin t 0 § , x2 AtB " £sin t0

sin t § , x3 AtB " £0sin t

sin t § .

!q 6 t 6 qAtBAtBAtB

S J # £ etetet § , £ e 2t

2e2t

4e2t § , £ e3t3e3t

9e3t § $

A J £ 0 1 00 0 1 6 !11 6

§ . x$ ! Ax ,

x1 " y, x2 " y¿, x3 " y–, Eet, e2t, e3tF.

Eet, e2t, e3tFy& AtB " 6y' AtB # 11y$ AtB " 6y AtB ! 0 .

x0. 4 3A AtB

526 Chapter 9 Matrix Methods for Linear Systems

In this section we discuss a procedure for obtaining a general solution for the homogeneous system

(1)

where A is a (real) constant matrix. The general solution we seek will be defined for all t because the elements of A are just constant functions, which are continuous on (recall Theorem 2, page 518). In Section 9.4 we showed that a general solution to (1) can be constructed from a fundamental solution set consisting of n linearly independent solutions to (1). Thus our goal is to find n such vector solutions.

In Chapter 4 we were successful in solving homogeneous linear equations with constant coefficients by guessing that the equation had a solution of the form Because any scalar linear equation can be expressed as a system, it is reasonable to expect system (1) to have solutions of the form

where r is a constant and u is a constant vector, both of which must be determined. Substituting for in (1) gives

Canceling the factor and rearranging terms, we find that

(2)

where denotes the diagonal matrix with r’s along its main diagonal. The preceding calculation shows that is a solution to (1) if and only if r and u

satisfy equation (2). Since the trivial case, u " 0, is of no help in finding linearly independent solutions to (1), we require that Such vectors are given a special name, as follows.u $ 0.

x AtB " erturI AA ! rIBu " 0 ,ert rertu " Aertu " ertAu .

x AtBertu x AtB " ertu ,

ert.

A!q, q Bn % n x¿ AtB " Ax AtB ,

9.5 HOMOGENEOUS LINEAR SYSTEMS WITH CONSTANT COEFFICIENTS

†We will allow u to have complex-number entries.

As stated in Theorem 1 of Section 9.3, a linear homogeneous system of n algebraic equa- tions in n unknowns has a nontrivial solution if and only if the determinant of its coefficients is zero. Hence, a necessary and sufficient condition for (2) to have a nontrivial solution is that

(3)

Expanding the determinant of in terms of its cofactors, we find that it is an nth- degree polynomial in r; that is,

(4) 0A " rI 0 ! p ArB . A ! rI

0A " rI 0 ! 0 .

Eigenvalues and Eigenvectors

Definition 3. Let be an constant matrix. The eigenvalues of A are those (real or complex) numbers r for which has at least one nontrivial solution† u. The corresponding nontrivial solutions u are called the eigenvectors of A associated with r.

AA ! rIBu " 0n % nA " 3aij 4

Therefore, finding the eigenvalues of a matrix A is equivalent to finding the zeros of the poly- nomial Equation (3) is called the characteristic equation of A, and in (4) is the characteristic polynomial of A. The characteristic equation plays a role for systems similar to the role played by the auxiliary equation for scalar equations.

Many commercially available software packages can be used to compute the eigenvalues and eigenvectors for a given matrix. Three such packages are MATLAB, available from The MathWorks, Inc.; MATHEMATICA, available from Wolfram Research; and MAPLE, avail- able from Waterloo Maple Inc. Although you are encouraged to make use of such packages, the examples and most exercises in this text can be easily carried out without them. Those exer- cises for which a computer package is desirable are flagged with the icon .

Find the eigenvalues and eigenvectors of the matrix

The characteristic equation for A is

Hence the eigenvalues of A are To find the eigenvectors corresponding to we must solve Substituting for A and gives

(5)

Notice that this matrix equation is equivalent to the single scalar equation Therefore, the solutions to (5) are obtained by assigning an arbitrary value for (say, ) and setting Consequently, the eigenvectors associated with can be expressed as

(6)

For the equation becomes

Solving, we obtain and with s arbitrary. Therefore, the eigenvectors associated with the eigenvalue are

(7) ◆

We remark that in the above example the collection (6) of all eigenvectors associated with forms a one-dimensional subspace when the zero vector is adjoined. The same is true

for These subspaces are called eigenspaces.r2 " !1. r1 " 1

u2 " s c11 d . r2 " !1

u2 " s,u1 " s

c3 !3 1 !1

d cu1 u2 d " c0

0 d . AA ! r2IBu " 0r2 " !1,

u1 " s c31 d . r1 " 1u1 " 3u2 " 3s.

u2 " su2 u1 ! 3u2 " 0.

c1 !3 1 !3

d cu1 u2 d " c0

0 d .

r1AA ! r1IBu " 0.r1 " 1, r1 " 1, r2 " !1. 0A ! rI 0 " ` 2 ! r !3

1 !2 ! r ` " A2 ! rB A!2 ! rB # 3 " r2 ! 1 " 0 .

A J c2 !3 1 !2

d .

p ArBp ArB. Section 9.5 Homogeneous Linear Systems with Constant Coefficients 527

Example 1

Solution

Find the eigenvalues and eigenvectors of the matrix

The characteristic equation for A is

which simplifies to Hence, the eigenvalues of A are and To find the eigenvectors corresponding to we set in This gives

(8)

Using elementary row operations (Gaussian elimination), we see that (8) is equivalent to the two equations

Thus, we can obtain the solutions to (8) by assigning an arbitrary value to (say, solving for to get and then solving for to get

Hence, the eigenvectors associated with are

(9)

For we solve

in a similar fashion to obtain the eigenvectors

(10)

Finally, for we solve

and get the eigenvectors

(11) ◆ u3 " s £!11 4 § .

£!2 2 !11 !3 1 4 !4 2

§ £u1u2 u3 § " £00

0 §

r3 " 3,

u2 " s £!21 4 § .

£!1 2 !11 !2 1 4 !4 3

§ £u1u2 u3 § " £00

0 §

r2 " 2,

u1 " s £!11 2 § .

r1 " 1u1 " !s. u1u1 ! u2 # u3 " 0u3 " 2s,u32u2 ! u3 " 0 u2 " s),u2

2u2 ! u3 " 0 .

u1 ! u2 # u3 " 0 ,

£0 2 !11 !1 1 4 !4 4

§ £u1u2 u3 § " £00

0 § .

AA ! rIBu " 0. r " 1r1 " 1,r3 " 3.r1 " 1, r2 " 2, Ar ! 1B Ar ! 2B Ar ! 3B " 0.

0A ! rI 0 " † 1 ! r 2 !11 !r 1 4 !4 5 ! r

† " 0 , A J £1 2 !11 0 1

4 !4 5 § .

528 Chapter 9 Matrix Methods for Linear Systems

Example 2

Solution

Let’s return to the problem of finding a general solution to a homogeneous system of differential equations. We have already shown that is a solution to (1) if r is an eigenvalue and u a corresponding eigenvector. The question is: Can we obtain n linearly independent solu- tions to the homogeneous system by finding all the eigenvalues and eigenvectors of A? The answer is yes, if A has n linearly independent eigenvectors.

ertu

Section 9.5 Homogeneous Linear Systems with Constant Coefficients 529

n Linearly Independent Eigenvectors

Theorem 5. Suppose the constant matrix A has n linearly independent eigenvectors Let be the eigenvalue† corresponding to Then

(12)

is a fundamental solution set (and X(t) = is a fundamental matrix) on for the homogeneous system Consequently, a general solution of is

(13)

where are arbitrary constants.c1, . . . , cn

x AtB " c1er1tu1 # c2er2tu2 # p # cnerntun ,x¿ " Ax x¿ " Ax.A!q, q B 3 er1tu1 er2tu2 p erntun 4Ee

r1tu1, er2tu2, . . . , erntunF ui.riu1, u2, . . . , un. n % n

†The eigenvalues may be real or complex and need not be distinct. In this section the cases we discuss have real eigenvalues. We consider complex eigenvalues in Section 9.6.

r1, . . . , rn

Proof. As we have seen, the vector functions listed in (12) are solutions to the homo- geneous system. Moreover, their Wronskian is

Since the eigenvectors are assumed to be linearly independent, it follows from Theorem 1 in Section 9.3 that is not zero. Hence the Wronskian is never zero. This shows that (12) is a fundamental solution set, and consequently a general solution is given by (13). ◆

An application of Theorem 5 is given in the next example.

Find a general solution of

(14) where

In Example 1 we showed that the matrix A has eigenvalues and Taking, say, in equations (6) and (7), we get the corresponding eigenvectors

and

Because and are linearly independent, it follows from Theorem 5 that a general solu- tion to (14) is

(15) ◆x AtB " c1et c31 d # c2e!t c11 d . u2u1

u2 " c11 d .u1 " c31 d s " 1

r2 " !1.r1 " 1

A " c2 !3 1 !2

d .x¿ AtB " Ax AtB ,

W AtBdet 3u1, . . . , un 4 W AtB " det 3 er1tu1, . . . , erntun 4 " e Ar1#. . . #rn B t det 3u1, . . . , un 4 .

Example 3

Solution

If we sum the vectors on the right-hand side of equation (15) and then write out the expressions for the components of col we get

This is the familiar form of a general solution for a system, as discussed in Section 5.2. Example 3 nicely illustrates the geometric role played by the eigenvectors and If the

initial vector is a scalar multiple of then the vector solution to the system, will always have the same or opposite direction as That is, it will lie along the straight line determined by (see Figure 9.1). Furthermore, the trajectory of this solution, as t increases, will tend to infinity, since the corresponding eigenvalue is posi- tive (observe the term). A similar assertion holds if the initial vector is a scalar multiple of

except that since is negative, the trajectory will approach the origin as t increases (because of ). For an initial vector that involves both and such as

the resulting trajectory is a blend of the above motions, with the contri- bution due to the larger eigenvalue dominating as t increases; see Figure 9.1.

The straight-line trajectories in the -plane (the phase plane), then, point along the directions of the eigenvectors of the matrix A. (See Section 5.4, Figure 5.10, for example.)

A useful property of eigenvectors that concerns their linear independence is stated in the next theorem.

x1x2 r1 " 1

x A0B " 12 Au1 # u2B, u2,u1e!t x AtB " c2e!tu2r2 " !1u2, et

r1 " 1 u1

u1.x AtB " c1etu1, u1 Ai.e., x A0B " c1u1B,x A0B u2.u1

x2 AtB " c1et # c2e!t .x1 AtB " 3c1et # c2e!t , Ax1 AtB, x2 AtBB ,x AtB "

530 Chapter 9 Matrix Methods for Linear Systems

u1 u2

1.5

0.5

0 1 2 3 4 5

x(0) = (u1 + u2) 1 2

Figure 9.1 Trajectories of solutions for Example 3

Linear Independence of Eigenvectors

Theorem 6. If are distinct eigenvalues for the matrix A and is an eigen- vector associated with then are linearly independent.u1, . . . , umri,

uir1, . . . , rm

Proof. Let’s first treat the case m " 2. Suppose, to the contrary, that and are linearly dependent so that

(16) u1 " cu2

u2u1

for some constant c. Multiplying both sides of (16) by A and using the fact that and are eigenvectors with corresponding eigenvalues and we obtain

(17)

Next we multiply (16) by and then subtract from (17) to get

Since is not the zero vector, we must have But this violates the assumption that the eigenvalues are distinct! Hence and are linearly independent.

The cases follow by induction. The details of the proof are left as Problem 48. ◆

Combining Theorems 5 and 6, we get the following corollary.

m 7 2 u2u1

r1 " r2.u1

Ar1 ! r2Bu1 " 0 .r2 r1u1 " cr2u2 .

r2,r1 u2u1

Section 9.5 Homogeneous Linear Systems with Constant Coefficients 531

n Distinct Eigenvalues

Corollary 1. If the constant matrix A has n distinct eigenvalues and is an eigenvector associated with then

is a fundamental solution set for the homogeneous system x¿ " Ax.

Eer1tu1, . . . , erntunF ri,ui r1, . . . , rnn % n

Solve the initial value problem

(18)

In Example 2 we showed that the coefficient matrix A has the three distinct eigenvalues and If we set in equations (9), (10), and (11), we obtain the cor-

responding eigenvectors

whose linear independence is guaranteed by Theorem 6. Hence, a general solution to (18) is

(19)

" £!et !2e2t !e3tet e2t e3t 2et 4e2t 4e3t

§ £ c1c2 c3 § .

x AtB " c1et £!11 2 § # c2e2t £!21

4 § # c3e3t £!11

4 §

u1 " £!11 2 § , u2 " £!21

4 § , u3 " £!11

4 § ,

s " 1r3 " 3.r1 " 1, r2 " 2, 3 % 3

x¿ AtB " £1 2 !11 0 1 4 !4 5

§ x AtB , x A0B " £!10 0 § .

Example 4

Solution

To satisfy the initial condition in (18), we solve

and find that and Inserting these values into (19) gives the desired solution. ◆

There is a special class of matrices that always have real eigenvalues and always have n linearly independent eigenvectors. These are the real symmetric matrices.

n % n

c3 " !1.c1 " 0, c2 " 1,

x A0B " £!1 !2 !11 1 1 2 4 4

§ £ c1c2 c3 § " £!10

0 §

532 Chapter 9 Matrix Methods for Linear Systems

Real Symmetric Matrices

Definition 4. A real symmetric matrix A is a matrix with real entries that satisfies AT " A.

Taking the transpose of a matrix interchanges its rows and columns. Doing this is equiva- lent to “flipping” the matrix about its main diagonal. Consequently, if and only if A is symmetric about its main diagonal.

If A is an real symmetric matrix, it is known† that there always exist n linearly inde- pendent eigenvectors. Thus, Theorem 5 applies and a general solution to is given by (13).

Find a general solution of

(20) where

A is symmetric, so we are assured that A has three linearly independent eigenvectors. To find them, we first compute the characteristic equation for A:

Thus the eigenvalues of A are and Notice that the eigenvalue r " 3 has multiplicity 2 when considered as a root of the

characteristic equation. Therefore, we must find two linearly independent eigenvectors associated with Substituting in gives

This system is equivalent to the single equation so we can obtain its solu- tions by assigning an arbitrary value to say and an arbitrary value to say u3 " s.u3,u2 " y,u2,

!u1 ! u2 # u3 " 0,

£!2 !2 2!2 !2 2 2 2 !2

§ £u1u2 u3 § " £00

0 § .

AA ! rIBu " 0r " 3r " 3. r3 " !3.r1 " r2 " 3

0A ! rI 0 " † 1 ! r !2 2!2 1 ! r 2 2 2 1 ! r

† " ! Ar ! 3B2 Ar # 3B " 0 .

A " £ 1 !2 2!2 1 2 2 2 1

§ .x¿ AtB " Ax AtB , x¿ " Ax

n % n

AT " A

†See Linear Algebra and Its Applications, 3rd updated ed., by David C. Lay (Addison-Wesley, Reading, Mass., 2006).

Solution

Example 5

Solving for we find Therefore, the eigenvectors associated with can be expressed as

By first taking and then taking we get the two linearly independent eigenvectors

(21)

For we solve

to obtain the eigenvectors col Taking gives

Since the eigenvectors and are linearly independent, a general solution to (20) is

◆

If a matrix A is not symmetric, it is possible for A to have a repeated eigenvalue but not to have two linearly independent corresponding eigenvectors. In particular, the matrix

(22)

has the repeated eigenvalue but Problem 35 shows that all the eigenvectors associated with are of the form col Consequently, no two eigenvectors are linearly independent.

A procedure for finding a general solution in such a case is illustrated in Problems 35–40, but the underlying theory is deferred to Section 9.8, where we discuss the matrix exponential.

A final note. If an matrix A has n linearly independent eigenvectors ui with eigen- values ri, a little inspection reveals that property (2) is expressed columnwise by the equation

(23) £ A

§ £ o o p ou1 u2 p un∞ o o p o

§ " £ o o p ou1 u2 p un∞ o o p o

§ £ r1 0 p 00 r2 p 0∞ 0 0 p rn

§ n % n

A1, 2B.u " sr " !1 r1 " r2 " !1, A " c1 !1

4 !3 d

x AtB " c1e3t £10 1 § # c2e3t £!11

0 § # c3e!3t £!1!1

1 § .

u3u1, u2,

u3 " £!1!1 1 § .

s " 1A!s, !s, sB. AA # 3IBu " £ 4 !2 2!2 4 2

2 2 4 § £u1u2

u3 § " £00

0 §

r3 " !3,

u1 " £10 1 § , u2 " £!11

0 § .

s " 0, y " 1,s " 1, y " 0

u " £ s ! yy s § " s £10

1 § # y £!11

0 § .

r1 " r2 " 3 u1 " u3 ! u2 " s ! y.u1,

Section 9.5 Homogeneous Linear Systems with Constant Coefficients 533

or AU " UD, where U is the matrix whose column vectors are eigenvectors and D is a diago- nal matrix whose diagonal entries are the eigenvalues. Since U’s columns are independent, U is invertible and we can write

(24) A " UDU!1 or D " U!1AU ,

and we say that A is diagonalizable. [In this context equation (24) expresses a similarity trans- formation.] Because the argument that leads from (2) to (23) to (24) can be reversed, we have a new characterization: An matrix has n linearly independent eigenvectors if, and only if, it is diagonalizable.

n % n

534 Chapter 9 Matrix Methods for Linear Systems

9.5 EXERCISES In Problems 1–8, find the eigenvalues and eigenvectors of the given matrix.

1. 2.

3. 4.

5. 6.

7. 8.

In Problems 9 and 10, some of the eigenvalues of the given matrix are complex. Find all the eigenvalues and eigenvectors.

9. 10.

In Problems 11–16, find a general solution of the system for the given matrix A.

11. 12.

13. A " £ 1 2 22 0 3 2 3 0

§ A " c 1 3

12 1 dA " £!1 34

!5 3 §

x¿ AtB " Ax AtB £ 1 2 !10 1 1

0 !1 1 §c 0 !1

1 0 d

£!3 1 00 !3 1 4 !8 2

§£ 1 0 02 3 1 0 2 4

§ £ 0 1 11 0 1

1 1 0 §£ 1 0 00 0 2

0 2 0 §

c 1 5 1 !3

dc 1 !1 2 4

d c 6 !3 2 1

dc!4 2 2 !1

d 14. 15. 16.

17. Consider the system with

(a) Show that the matrix A has eigenvalues and with corresponding eigenvectors

col and col (b) Sketch the trajectory of the solution having

initial vector (c) Sketch the trajectory of the solution having

initial vector (d) Sketch the trajectory of the solution having

initial vector

18. Consider the system with

(a) Show that the matrix A has eigenvalues and with corresponding eigenvectors

col and col (b) Sketch the trajectory of the solution having

initial vector x A0B " u1. A1, !1B.u2 "A1, 1Bu1 "r2 " !3

r1 " !1

A " c!2 1 1 !2

d .x¿ AtB " Ax AtB, t ) 0, x A0B " u2 ! u1. x A0B " u2. x A0B " !u1.

A1, !23 B .u2 "A23, 1Bu1 "r2 " !2 r1 " 2 A " C 1 2323 !1S .x¿ AtB " Ax AtB, t ) 0,

A " £!7 0 60 5 0 6 0 2

§A " £ 1 2 30 1 0 2 1 2

§ A " £!1 1 01 2 1

0 3 !1 §

(d) Sketch the trajectory of the solution having initial vector

In Problems 19–24, find a fundamental matrix for the system for the given matrix A.

19. 20.

21.

22.

23.

24.

25. Using matrix algebra techniques, find a general solu- tion of the system

26. Using matrix algebra techniques, find a general solu- tion of the system

In Problems 27–30, use a linear algebra software pack- age such as MATLAB, MAPLE, or MATHEMATICA to compute the required eigenvalues and eigenvectors and then give a fundamental matrix for the system

for the given matrix A.

27. 28.

29. A " D0 1 0 00 0 1 0 0 0 0 1 2 !6 3 3

T A " £ 2 1 1

!1 1 0 3 3 3

§A " £ 0 1.1 00 0 1.3 0.9 1.1 !6.9

§ x¿ AtB " Ax AtB

y¿ " 4x ! 7y . x¿ " 3x ! 4y ,

z¿ " 4x ! 4y # 5z . y¿ " x # z , x¿ " x # 2y ! z ,

A " D4 !1 0 00 0 0 0 0 0 2 !3 0 0 1 !2

T A " D2 1 1 !10 !1 0 1

0 0 3 1 0 0 0 7

TA " £ 3 1 !1 1 3 !1 3 3 !1

§ A " £ 0 1 00 0 1

8 !14 7 §

A " c 5 4 !1 0

dA " c!1 1 8 1

dx¿ AtB " Ax AtB x A0B " u1 ! u2. x A0B " !u2.

Section 9.5 Homogeneous Linear Systems with Constant Coefficients 535

30.

In Problems 31–34, solve the given initial value problem.

31.

32.

33.

34.

35. (a) Show that the matrix

has the repeated eigenvalue and that all the eigenvectors are of the form s col

(b) Use the result of part (a) to obtain a nontrivial solution to the system

Since must be an eigenvector, set col and solve for

(d) What is ? (In Section 9.8, will be identified as a generalized eigenvector.)

36. Use the method discussed in Problem 35 to find a general solution to the system

37. (a) Show that the matrix

A " £ 2 1 60 2 5 0 0 2

§ x¿ AtB " c 5 !3

3 !1 d x AtB .

u2AA # IB2u2 u2. 4A1, 2B u1 "u1 AA # IBu1 " 0 , AA # IBu2 " u1 .

x¿ " Axx2 3x2 AtB " te!tu1 # e!tu2.x¿ " Ax, x¿ " Ax.x1 AtB

A1, 2B. u " r " !1

A " c 1 !1 4 !3

d x¿ AtB " £ 0 1 11 0 1

1 1 0 § x AtB , x A0B " £!14

0 §

x A0B " £!2!3 2 §

x¿ AtB " £ 1 !2 2!2 1 !2 2 !2 1

§ x AtB , x¿ AtB " c 6 !3

2 1 d x AtB , x A0B " c!10

!6 d

x¿ AtB " c 1 3 3 1

d x AtB , x A0B " c 3 1 d

A " D0 1 0 01 !1 0 0 0 0 0 1 0 0 !2 4

has the repeated eigenvalue r " 2 with multi- plicity 3 and that all the eigenvectors of A are of the form u " s col(1, 0, 0).

(b) Use the result of part (a) to obtain a solution to the system of the form

(d) To obtain a third linearly independent solution to try

Hint: Show that and must satisfy

(e) Show that

38. Use the method discussed in Problem 37 to find a general solution to the system

39. (a) Show that the matrix

has the repeated eigenvalue r " 1 of multiplicity 3 and that all the eigenvectors of A are of the form col col

(b) Use the result of part (a) to obtain two linearly independent solutions to the system of the form

and (c) To obtain a third linearly independent solution

to try Hint: Show that and must satisfy

Choose an eigenvector of A, so that you can solve for

(d) What is ?

40. Use the method discussed in Problem 39 to find a general solution to the system

x¿ AtB " £ 1 3 !20 7 !4 0 9 !5

§ x AtB . AA ! IB2u4u4. 4u3,

AA ! IBu3 " 0 , AA ! IBu4 " u3 .u4u3 3x3 AtB " tetu3 # etu4.x¿ " Ax, x2 AtB " e

tu2.x1 AtB " etu1 x¿ " Ax

A!1, 0, 1B.A!1, 1, 0B # yu " s A " £ 2 1 11 2 1

!2 !2 !1 §

x¿ AtB " £ 3 !2 12 !1 1 !4 4 1

§ x AtB. AA ! 2IB2u2 " AA ! 2IB3u3 " 0.AA ! 2IBu3 " u2. 4

AA ! 2IBu1 " 0 , AA ! 2IBu2 " u1 ,u3u1, u2,3 x3 AtB " t22 e2tu1 # te2tu2 # e2tu3 .

x¿ " Ax,

AA ! 2IBu1 " 0 , AA ! 2IBu2 " u1. 4u2u1 3x2 AtB " te 2tu1 # e2tu2.x¿ " Ax,

x1 AtB " e2tu1.x¿ " Ax

536 Chapter 9 Matrix Methods for Linear Systems

41. Use the substitution to convert the linear equation where and c are constants, into a normal system. Show that the characteristic equation for this system is the same as the auxiliary equation for the original equation.

42. (a) Show that the Cauchy–Euler equation can be written as a

Cauchy–Euler system

(25)

with a constant coefficient matrix A, by setting x1 " y/t and x2 " .

(b) Show that for t ' 0 any system of the form (25) with A an n % n constant matrix has nontrivial solutions of the form if and only if r is an eigenvalue of A and u is a corresponding eigenvector.

In Problems 43 and 44, use the result of Problem 42 to find a general solution of the given system.

43.

44.

45. Mixing Between Interconnected Tanks. Two tanks, each holding 50 L of liquid, are intercon- nected by pipes with liquid flowing from tank A into tank B at a rate of 4 L/min and from tank B into tank A at 1 L/min (see Figure 9.2). The liquid inside each tank is kept well stirred. Pure water flows into tank A at a rate of 3 L/min, and the solution flows out of tank B at 3 L/min. If, initially, tank A contains 2.5 kg of salt and tank B contains no salt (only water), determine the mass of salt in each tank at time Graph on the same axes the two quanti- ties and where is the mass of salt in tank A and is the mass in tank B.x2 AtB x1 AtBx2 AtB,x1 AtB

t ) 0.

tx¿ AtB " c!4 2 2 !1

d x AtB , t 7 0 tx¿ AtB " c 1 3

!1 5 d x AtB , t 7 0

x AtB " tru y¿

t x¿ " Ax

at2y– # bty¿ # cy " 0

a, b,ay– # by¿ # cy " 0, x1 " y, x2 " y¿

3 L/min

Pure water x 1 (t)

50 L

x 1 (0) = 2.5 kg

A 4 L/min

x 2 (t)

50 L

x 2 (0) = 0 kg

3 L/min

1 L/min

Figure 9.2 Mixing problem for interconnected tanks

46. Mixing with a Common Drain. Two tanks, each holding 1 L of liquid, are connected by a pipe through which liquid flows from tank A into tank B at a rate of The liquid inside each tank is kept well stirred. Pure water flows into tank A at a rate of 3 L/min. Solution flows out of tank A at L/min and out of tank B at L/min. If, initially, tank B contains no salt (only water) and tank A contains 0.1 kg of salt, determine the mass of salt in each tank at time How does the mass of salt in tank A depend on the choice of ? What is the maximum mass of salt in tank B? (See Figure 9.3.)

t ) 0.

3 ! aa

3 ! a L/min A0 6 a 6 3B.

Section 9.5 Homogeneous Linear Systems with Constant Coefficients 537

3 L/min

Pure water x 1 (t)

1 L

x 1 (0) = 0.1 kg

A L/min

x 2 (t)

1 L

x 2 (0) = 0 kg

3 L/min

L/min ) L/min

Figure 9.3 Mixing problem for a common drain, 0 6 a 6 3

48. To complete the proof of Theorem 6, assume the induction hypothesis that are linearly independent. (a) Show that if

then

(b) Use the result of part (a) and the induction hypoth- esis to conclude that are linearly independent. The theorem follows by induction.

49. Stability. A homogeneous system with constant coefficients is stable if it has a fundamental matrix whose entries all remain bounded as

(It will follow from Lemma 1 in Section 9.8 that if one fundamental matrix of the system has this property, then all fundamental matrices for the system do.) Otherwise, the system is unstable. A stable system is asymptotically stable if all solu- tions approach the zero solution as Stabil- ity is discussed in more detail in Chapter 12.†

(a) Show that if A has all distinct real eigenvalues, then is stable if and only if all eigenvalues are nonpositive.

(b) Show that if A has all distinct real eigenvalues, then is asymptotically stable if and only if all eigenvalues are negative.

(c) Argue that in parts (a) and (b), we can replace “has distinct real eigenvalues” by “is symmet- ric” and the statements are still true.

50. In an ice tray, the water level in any particular ice cube cell will change at a rate proportional to the dif- ference between that cell’s water level and the level in the adjacent cells. (a) Argue that a reasonable differentiable equation

model for the water levels x, y, and z in the sim- plified three-cell tray depicted in Figure 9.4 is given by

(b) Use eigenvectors to solve this system for the ini- tial conditions , y A0B " z A0B " 0.x A0B " 3

z¿ " y ! z .x¿ " y ! x , y¿ " x # z ! 2y ,

x¿ AtB " Ax AtB x¿ AtB " Ax AtB

t S #q.

x¿ " Ax

u1, . . . , uk#1

c1 Ar1 ! rk#1Bu1 # p # ck Ark ! rk#1Buk " 0 . c1u1 # p # ckuk # ck#1uk#1 " 0 ,

u1, . . . , uk, 2 ( k,

47. To find a general solution to the system

proceed as follows: (a) Use a numerical root-finding procedure to

approximate the eigenvalues. (b) If r is an eigenvalue, then let col

be an eigenvector associated with r. To solve for u, assume (If not then either or may be chosen to be 1. Why?) Now solve the system

for and Use this procedure to find approx- imations for three linearly independent eigen- vectors for A.

u3.u2

AA ! rIB £ 1u2 u3 § " £ 00

0 §

u3u2u1,u1 " 1.

Au1, u2, u3Bu "

x¿ " Ax " £ 1 3 !13 0 1 !1 1 2

§ x ,

x y z

Figure 9.4 Ice tray

†All references to Chapters 11–13 refer to the expanded text Fundamentals of Differential Equations and Boundary Value Problems, 6th ed.

In the previous section, we showed that the homogeneous system

(1)

where A is a constant matrix, has a solution of the form if and only if r is an eigenvalue of A and u is a corresponding eigenvector. In this section we show how to obtain two real vector solutions to system (1) when A is real and has a pair† of complex conjugate eigenvalues and

Suppose and real numbers) is an eigenvalue of A with corresponding eigenvector where a and b are real constant vectors. We first observe that the com- plex conjugate of z, namely is an eigenvector associated with the eigenvalue

To see this, note that taking the complex conjugate of yields because the conjugate of the product is the product of the conjugates and A

and I have real entries Since we see that is an eigenvector associ- ated with Therefore, two linearly independent complex vector solutions to (1) are

(2) (3)

As in Section 4.3, where we handled complex roots to the auxiliary equation, let’s use one of these complex solutions and Euler’s formula to obtain two real vector solutions. With the aid of Euler’s formula, we rewrite as

We have thereby expressed in the form where and are the two real vector functions

(4) (5)

Since is a solution to , then

Equating the real and imaginary parts yields

and

Hence, and are real vector solutions to (1) associated with the complex conjugate eigenvalues Because a and b are not both the zero vector, it can be shown that and are linearly independent vector functions on (see Problem 15).A!q, q Bx2 AtB x1 AtBa * ib.

x2 AtBx1 AtB x¿2 AtB " Ax2 AtB .x¿1 AtB " Ax1 AtB

x¿1 # ix¿2 " Ax1 # iAx2 . w¿1 AtB " Aw1 AtB , A1Bw1 AtB

x2 AtB J eat sin bt a # eat cos bt b .x1 AtB J eat cos bt a ! eat sin bt b , x2 AtBx1 AtBw1 AtB " x1 AtB # ix2 AtB,w1 AtB " e

atE Acos bt a ! sin bt bB # i Asin bt a # cos bt bB F . w1 AtB " eat Acos bt # i sin btB Aa # ibB w1 AtB

w2 AtB " er2tz " e Aa!ibBt Aa ! ibB .w1 AtB " er1tz " e Aa#ibBt Aa # ibB , r2.

zr2 " r1,AA " A, I " IB.AA ! r1IBz " 0 AA ! r1IBz " 0r2 " a ! ib. z J a ! ib,

z " a # ib, br1 " a # ib Aaa ! ib.a # ib

x AtB " ertun % nx¿ AtB " Ax AtB ,

538 Chapter 9 Matrix Methods for Linear Systems

9.6 COMPLEX EIGENVALUES

†Recall that the complex roots of a polynomial equation with real coefficients must occur in complex conjugate pairs.

Let’s summarize our findings.

Section 9.6 Complex Eigenvalues 539

Complex Eigenvalues

If the real matrix A has complex conjugate eigenvalues with corresponding eigenvectors then two linearly independent real vector solutions to

are

(6) (7) eat sin bt a # eat cos bt b .

eat cos bt a ! eat sin bt b ,

x¿ AtB " Ax AtBa * ib, a * ib

Find a general solution of

(8)

The characteristic equation for A is

Hence, A has eigenvalues To find a general solution, we need only find an eigenvector associated with the eigenvalue

Substituting into gives

The solutions can be expressed as and with s arbitrary. Hence, the eigen- vectors associated with are col Taking gives the eigenvector

We have found that and with col and b " col so a general solution to (8) is

(9) ◆x AtB " c1 c 2e!2t cos t!e!2t Acos t # sin tB d # c2 c 2e!2t sin te!2t Acos t ! sin tB d . " # c2 ce!2t sin t c 2!1 d # e!2t cos t c01 ds x AtB " c1 ce!2t cos t c 2!1 d ! e!2t sin t c01 ds

A0, 1B, A2, !1B,a "z " a # iba " !2, b " 1, z " c 2

!1 # i d " c 2

!1 d # i c0

1 d .

s " 1A2, !1 # iB.z " sr " !2 # i z2 " A!1 # iBs,z1 " 2s c1 ! i 2

!1 !1 ! i d c z1

z2 d " c0

0 d . AA ! rIBz " 0r " !2 # ir " !2 # i.

r " !2 * i.

0A ! rI 0 " `!1 ! r 2 !1 !3 ! r

` " r2 # 4r # 5 " 0 . x¿ AtB " c!1 2

!1 !3 d x AtB .Example 1

Solution

Complex eigenvalues occur in modeling coupled mass–spring systems. For example, the motion of the mass–spring system illustrated in Figure 9.5 is governed by the second-order system

(10)

where and represent the displacements of the masses and to the right of their equi- librium positions and are the spring constants of the three springs (see the discussion in Section 5.6). If we introduce the new variables then we can rewrite the system in the normal form

(11)

For such a system, it turns out that A has only imaginary eigenvalues and they occur in complex conjugate pairs: Hence, any solution will consist of sums of sine and cosine functions. The frequencies of these functions

and

are called the normal or natural frequencies of the system ( and are the angular frequencies of the system).

In some engineering applications, the only information that is required about a particular device is a knowledge of its normal frequencies; one must ensure that they are far from the fre- quencies that occur naturally in the device’s operating environment (so that no resonances will be excited).

Determine the normal frequencies for the coupled mass–spring system governed by system (11) when kg, N/m, N/m, and N/m.

To find the eigenvalues of A, we must solve the characteristic equation

0A ! rI 0 " ∞ !r 1 0 0!3 !r 2 0 0 0 !r 1 2 0 !5 !r

∞ " r4 # 8r2 # 11 " 0 . k3 " 3k2 " 2k1 " 1m1 " m2 " 1

b2b1

f2 J b2 2p

f1 J b1 2p

*ib1, *ib2.

y¿ AtB " Ay AtB " D 0 1 0 0! Ak1 # k2B /m1 0 k2/m1 0 0 0 0 1

k2/m2 0 ! Ak2 # k3B /m2 0T y AtB . y1 J x1, y2 J x¿1, y3 J x2, y4 J x¿2,

k1, k2, k3 m2m1x2x1

m2x–2 " !k2 Ax2 ! x1B ! k3x2 ,m1x–1 " !k1x1 # k2 Ax2 ! x1B ,

540 Chapter 9 Matrix Methods for Linear Systems

k 1

x 1 > 0

x 1 = 0

k 2

x 2 > 0

x 2 = 0

k 3 m 1 m 2

Figure 9.5 Coupled mass–spring system with fixed ends

Example 2

Solution

From the quadratic formula we find so the four eigenvalues of A are

and Hence, the two normal frequencies for this system are

and cycles per second. ◆ 34 # 25

2p % 0.39734 ! 25

2p % 0.211

*i34 # 25.*i34 ! 25 r2 " !4 * 25, Section 9.6 Complex Eigenvalues 541

9.6 EXERCISES In Problems 1–4, find a general solution of the system

for the given matrix A.

1. 2.

In Problems 5–8, find a fundamental matrix for the sys- tem for the given matrix A.

5. 6.

In Problems 9–12, use a linear algebra software package to compute the required eigenvalues and eigenvectors for the given matrix A and then give a fundamental matrix for the system

9. A " £ 0 1 1!1 0 1 !1 !1 0

§ x¿ AtB " Ax AtB.

A " D0 1 0 01 0 0 0 0 0 0 1 0 0 !13 4

TA " £ 0 0 1 0 0 !1 0 1 0

§ A " c!2 !2

4 2 dA " c!1 !2

8 !1 dx¿ AtB " Ax AtB

A " £ 5 !5 !5!1 4 2 3 !5 !3

§ A " £ 1 2 !10 1 1

0 !1 1 §

A " c!2 !5 1 2

dA " c 2 !4 2 !2

dx¿ AtB " Ax AtB 10. 11.

12.

In Problems 13 and 14, find the solution to the given sys- tem that satisfies the given initial condition.

13.

(a) (b)

14.

(a) (b) x A!pB " £ 01 1 §x A0B " £!22

!1 §

x¿ AtB " £ 1 0 !10 2 0 1 0 1

§ x AtB , x Ap/2B " c 01 dx A!2pB " c 21 d x ApB " c 1

!1 dx A0B " c!1

0 d

x¿ AtB " c!3 !1 2 !1

d x AtB ,

A " E1 0 0 0 00 0 1 0 00 1 0 0 0 0 0 0 0 1 0 0 0 !29 !4

U A " D 0 1 0 00 0 1 0

0 0 0 1 !2 2 !3 2

T A " D 0 1 0 00 0 1 0

0 0 0 1 13 !4 !12 4

15. Show that and given by equations (4) and (5) are linearly independent on provided

and a and b are not both the zero vector. 16. Show that and given by equations (4) and

(5) can be obtained as linear combinations of and given by equations (2) and (3). [Hint: Show that

In Problems 17 and 18, use the results of Problem 42 in Exercises 9.5 to find a general solution to the given Cauchy–Euler system for

17.

18.

19. For the coupled mass–spring system governed by sys- tem (10), assume kg, N/m, and N/m. Determine the normal frequencies for this coupled mass–spring system.

20. For the coupled mass–spring system governed by system (10), assume kg, "

N/m, and assume initially that m, m/sec, m, and m/sec.

Using matrix algebra techniques, solve this initial value problem.

21. RLC Network. The currents in the RLC network given by the schematic diagram in Figure 9.6 are governed by the following equations:

I1 AtB " I2 AtB # I3 AtB ,13I3 AtB # 52q1 AtB " 10 , 4I¿2 AtB # 52q1 AtB " 10 ,

x¿2 A0B " 0x2 A0B " 2x¿1 A0B " 0 x1 A0B " 0k3 " 1 k1 " k2m1 " m2 " 1

k3 " 3 k1 " k2 " 2m1 " m2 " 1

tx¿ AtB " c!1 !1 9 !1

d x AtB tx¿ AtB " £!1 !1 02 !1 1

0 1 !1 § x AtB

t 7 0.

x2 AtB " w1 AtB ! w2 AtB2i . dx1 AtB " w1 AtB # w2 AtB2 , w2 AtB w1 AtB

x2 AtBx1 AtBb $ 0 A!q, q B,x2 AtBx1 AtB

542 Chapter 9 Matrix Methods for Linear Systems

where is the charge on the capacitor, and initially coulombs and

amps. Solve for the currents and Hint: Differentiate the first two equations, elimi-

nate and form a normal system with and

22. RLC Network. The currents in the RLC network given by the schematic diagram in Figure 9.7 are governed by the following equations:

where is the charge on the capacitor, and initially coulombs

and amps. Solve for the currents and Hint: Differentiate the first two equations, use

the third equation to eliminate and form a normal system with and x3 " I2. 4x1 " I1, x2 " I¿1, I3,3I3.

I1, I2,I3 A0B " 0 q3 A0B " 0.5I3 AtB " q¿3 AtB, q3 AtBI1 AtB " I2 AtB # I3 AtB ,

50I¿1 AtB # 800q3 AtB " 160 ,50I¿1 AtB # 80I2 AtB " 160 , x3 " I3. 4x2 " I¿2, x1 " I2,I1,3

I3.I1, I2,I1 A0B " 0 q1 A0B " 0I1 AtB " q¿1 AtB, q1 AtB

10 volts

I 1

I 2

farads

4 henries 13 ohms

I 3

1 —– 52

Figure 9.6 RLC network for Problem 21

800

160 volts

I 1

I 2

I 3

50 henries

80 ohms farads

Figure 9.7 RLC network for Problem 22

23. Stability. In Problem 49 of Exercises 9.5, we discussed the notion of stability and asymptotic sta- bility for a linear system of the form " Assume that A has all distinct eigenvalues (real or complex). (a) Show that the system is stable if and only if all

the eigenvalues of A have nonpositive real part. (b) Show that the system is asymptotically stable if

and only if all the eigenvalues of A have nega- tive real part.

24. (a) For Example 1, page 539, verify that

is another general solution to equation (8). (b) How can the general solution of part (a) be directly

obtained from the general solution derived in (9) on page 539?

# c2 c!e!2t sin t ! e!2t cos te!2t sin t d x AtB " c1 c!e!2t cos t # e!2t sin te!2t cos t d

Ax AtB.x¿ AtB

The techniques discussed in Chapters 4 and 6 for finding a particular solution to the nonhomo- geneous equation have natural extensions to nonhomogeneous linear systems.

Undetermined Coefficients The method of undetermined coefficients can be used to find a particular solution to the nonhomogeneous linear system

when A is an constant matrix and the entries of are polynomials, exponential func- tions, sines and cosines, or finite sums and products of these functions. We can use the proce- dure box in Section 4.5 and reproduced on the inside back cover as a guide in choosing the form of a particular solution Some exceptions are discussed in the exercises (see Problems 25–28).

Find a general solution of

(1) where and

In Example 5 in Section 9.5, we found that a general solution to the corresponding homoge- neous system is

(2)

Since the entries in are just linear functions of t, we are inclined to seek a partic- ular solution of the form

where the constant vectors a and b are to be determined. Substituting this expression for into system (1) yields

which can be written as

Setting the “coefficients” of this vector polynomial equal to zero yields the two systems

(3) (4) Ab " a .

Aa " !g ,

t AAa # gB # AAb ! aB " 0 . a " A Ata # bB # tg ,

xp AtB xp AtB " ta # b " t £a1a2

a3 § # £b1b2

b3 § ,

f AtB J tg xh AtB " c1e3t £10

1 § # c2e3t £!11

0 § # c3e!3t £!1!1

1 § .

x¿ " Ax

g " £ !90 !18 § .A " £ 1 !2 2!2 1 2

2 2 1 §x$ AtB " Ax AtB # tg ,

xp AtB. f AtBn % nx¿ AtB " Ax AtB # f AtB

y– # p AxBy¿ # q AxBy " g AxB

Section 9.7 Nonhomogeneous Linear Systems 543

9.7 NONHOMOGENEOUS LINEAR SYSTEMS

Example 1

Solution

By Gaussian elimination or by using a linear algebra software package, we can solve (3) for a and we find a " col(5, 2, 4). Next we substitute for a in (4) and solve for b to obtain b " col(1, 0, 2). Hence a particular solution for (1) is

(5)

A general solution for (1) is where is given in (2) and in (5). ◆

In the preceding example, the nonhomogeneous term was a vector polynomial. If, instead, has the form

col

then, using the superposition principle, we would seek a particular solution of the form

Similarly, if

col

we would take

Of course, we must modify our guess, should one of the terms be a solution to the corre- sponding homogeneous system. If this is the case, the annihilator method [equations (15) and (16) of Section 6.3, page 336] would appear to suggest that for a nonhomogeneity f(t) of the form , where r is an eigenvalue of A, m is a nonnegative integer, and g is a constant vec- tor, a particular solution of x Ax + f can be found in the form

for a suitable choice of s. We omit the details.

Variation of Parameters In Section 4.6 we discussed the method of variation of parameters for a general constant- coefficient second-order linear equation. Simply put, the idea is that if a general solution to the homogeneous equation has the form where and are lin- early independent solutions to the homogeneous equation, then a particular solution to the non- homogeneous equation would have the form where and

are certain functions of t. A similar idea can be used for systems. Let be a fundamental matrix for the homogeneous system

(6)

where now the entries of A may be any continuous functions of t. Because a general solution to (6) is given by where c is a constant vector, we seek a particular solution to the nonhomogeneous system

(7) x¿ AtB " A AtBx AtB # f AtB n % 1X AtBc,

x¿ AtB " A AtBx AtB ,X AtB y2 AtB y1 AtBxp AtB " y1 AtBx1 AtB # y2 AtBx2 AtB,

x2 AtBx1 AtBxh AtB " c1x1 AtB # c2x2 AtB,

xp AtB " ertEtm#sam#s # tm#s!1am#s!1 # . . . # ta1 # a0F¿ " erttmg

xp AtB " t2a # tb # c # etd . At, et, t2B ,f AtB "

xp AtB " ta # b # Asin tBc # Acos tBd . A1, t, sin tB ,f AtB "f AtB

f AtB xp AtBxh AtBx AtB " xh AtB # xp AtB,

xp AtB " ta # b " t £52 4 § # £10

2 § " £5t # 12t

4t # 2 § .

544 Chapter 9 Matrix Methods for Linear Systems

of the form

(8)

where col is a vector function of t to be determined. To derive a formula for we first differentiate (8) using the matrix version of the

product rule to obtain

Substituting the expressions for and into (7) yields

(9)

Since satisfies the matrix equation equation (9) becomes

Multiplying both sides of the last equation by [which exists since the columns of are linearly independent] gives

Integrating, we obtain

Hence, a particular solution to (7) is

(10)

Combining (10) with the solution to the homogeneous system yields the following general solution to (7):

(11)

The elegance of the derivation of the variation of parameters formula (10) for systems becomes evident when one compares it with the more lengthy derivations for the scalar case in Sections 4.6 and 6.4.

Given an initial value problem of the form

(12)

we can use the initial condition to solve for c in (11). Expressing using a defi- nite integral, we have

x AtB " X AtBc # X AtB ! t t0

X!1 AsBf AsB ds . x AtBx At0B " x0

x¿ AtB " A AtBx AtB # f AtB , x At0B " x0 ,

x AtB ! X AtBc # X AtB ! X"1 AtBf AtB dt . X AtBc

xp AtB ! X AtBY AtB ! X AtB ! X"1 AtBf AtB dt . Y AtB " ! X!1 AtBf AtB dt . Y¿ AtB " X!1 AtBf AtB .

X AtBX!1 AtBXY¿ " f . XY¿ # AXY " AXY # f ,

X¿ AtB " A AtBX AtB,X AtBX AtBY¿ AtB # X¿ AtBY AtB " A AtBX AtBY AtB # f AtB .x¿p

AtBxp AtB x¿p AtB " X AtBY¿ AtB # X¿ AtBY AtB .

Y AtB,Ay1 AtB, . . . , yn AtBBY AtB " xp AtB " X AtBY AtB ,

Section 9.7 Nonhomogeneous Linear Systems 545

Using the initial condition we find

Solving for c, we have Thus, the solution to (12) is given by the formula

(13)

To apply the variation of parameters formulas, we first must determine a fundamental matrix for the homogeneous system. In the case when the coefficient matrix A is constant, we have discussed methods for finding However, if the entries of A depend on t, the determination of may be extremely difficult (entailing, perhaps, a matrix power series!).

Find the solution to the initial value problem

(14)

In Example 3 in Section 9.5, we found two linearly independent solutions to the corresponding homogeneous system; namely,

and

Hence a fundamental matrix for the homogeneous system is

Although the solution to (14) can be found via the method of undetermined coefficients, we shall find it directly from formula (13). For this purpose, we need One way† to obtain is to form the augmented matrix

and row-reduce this matrix to the matrix This gives

X!1 AtB " D! 12 e!t ! 12 e!t ! 12 e

t ! 32 e t T . 3 I!X!1 AtB 4 .

c3et e!t ! 1 0 et e!t ! 0 1

d X!1 AtB X!1 AtB.

X AtB " c3et e!t et e!t

d . x2 AtB " c e!te!t d .x1 AtB " c3etet d

x¿ AtB " c2 !3 1 !2

d x AtB # c e2t 1 d , x A0B " c!1

0 d .

X AtB X AtB.X AtB x AtB ! X AtBX"1 At0Bx0 # X AtB ! t

t0 X"1 AsBf AsB ds .

c " X!1 At0Bx0. x0 " x At0B " X At0Bc # X At0B ! t0

X!1 AsBf AsB ds " X At0Bc . x At0B " x0,

546 Chapter 9 Matrix Methods for Linear Systems

Example 2

Solution

†This procedure works for an invertible matrix of any dimension. For an arbitrary invertible matrix , a formula for is derived in Problem 32.U!1 AtB U AtB2 % 2

Substituting into formula (13), we obtain the solution

◆ " £!92 et ! 56 e!t # 43 e2t # 3 ! 32 e

t ! 5 6

e!t # 13 e 2t # 2

§ . " £! 32 et # 12 e!t

! 12 e t # 12 e

!t § # c3et e!t

et e!t d £ 12 et # 12 e!t ! 1

3 2 e

t ! 1 6

e3t ! 43

§ " £! 32 et # 12 e!t

! 12 e t # 12 e

!t § # c3et e!t

et e!t d ! t

£ 12 es ! 12 e!s ! 12 e

3s # 32 e s § ds

" # c3et e!t et e!t

d ! t 0

£! 12 e!s ! 12 e!s ! 12 e

s ! 32

es § c e2s

1 d ds

x AtB " c3et e!t et e!t

d £ 12 ! 12 ! 12

3 2

§ c!1 0 d

Section 9.7 Nonhomogeneous Linear Systems 547

9.7 EXERCISES In Problems 1–6, use the method of undetermined coeffi- cients to find a general solution to the system

are given.

6. A " c 1 1 0 2

d , f AtB " e!2t c t 3 d

A " £ 0 !1 0!1 0 0 0 0 1

§ , f AtB " £ e2 tsin t t §

A " c 2 2 2 2

d , f AtB " c!4 cos t !sin t

d A " £ 1 !2 2!2 1 2

2 2 1 § , f AtB " £ 2et4et

!2et §

A " c 1 1 4 1

d , f AtB " c !t ! 1 !4t ! 2

d A " c 6 1

4 3 d , f AtB " c!11

!5 dAx AtB # f AtB, where A and f AtB

x¿ AtB " In Problems 7–10, use the method of undetermined coeffi-cients to determine only the form of a particular solution for the system , where and are given.

10.

In Problems 11–16, use the variation of parameters for- mula (11) to find a general solution of the system

are given.

11. A " c 0 1 !1 0

d , f AtB " c 1 0 dx¿ AtB " Ax AtB # f AtB, where A and f AtB

A " c 2 !1 1 5

d , f AtB " c te!t 3e!t d

A " £ 0 !1 0!1 0 1 0 0 1

§ , f AtB " £ e2tsin t t §

A " c!1 0 2 2

d , f AtB " c t2 t # 1

d A " c 0 1

2 0 d , f AtB " c sin 3t

t d

f AtBAx¿ AtB " Ax AtB # f AtB

12.

13.

14.

15.

16.

In Problems 17–20, use the variation of parameters for- mulas (11) and possibly a linear algebra software pack- age to find a general solution of the system

are given.

17.

18.

19.

20.

In Problems 21 and 22, find the solution to the given sys- tem that satisfies the given initial condition.

21.

(a) (b)

0 d

x A1B " c 0 1 dx A0B " c 5

4 d

x¿ AtB " c 0 2 !1 3

d x AtB # c et !et d ,

A " D0 1 0 00 0 1 0 0 0 0 1 8 !4 !2 !1

T , f AtB " Det0 1 0

T A " D 0 1 0 0!1 0 0 0

0 0 0 1 0 0 1 0

T , f AtB " D t0 e!t

TA " £ 1 !1 1 0 0 1 0 !1 2

§ , f AtB " £ 0et et §

A " £ 0 1 11 0 1 1 1 0

§ , f AtB " £ 3et!et !et §

Ax AtB # f AtB, where A and f AtB x¿ AtB " A " c 0 1

!1 0 d , f AtB " c 8 sin t

0 d

A " c!4 2 2 !1

d , f AtB " c t!1 4 # 2t!1

d A " c 0 !1

1 0 d , f AtB " c t2

1 d

A " c 2 1 !3 !2

d , f AtB " c 2et 4et d

A " c 1 2 3 2

d , f AtB " c 1 !1 d

548 Chapter 9 Matrix Methods for Linear Systems

22.

(a) (b)

23. Using matrix algebra techniques and the method of undetermined coefficients, find a general solution for

Compare your solution with the solution in Example 4 in Section 5.2.

24. Using matrix algebra techniques and the method of undetermined coefficients, solve the initial value problem

Compare your solution with the solution in Example 1 in Section 7.9.

25. To find a general solution to the system

proceed as follows: (a) Find a fundamental solution set for the corre-

sponding homogeneous system. (b) The obvious choice for a particular solution

would be a vector function of the form however, the homogeneous system

has a solution of this form. The next choice would be Show that this choice does not work.

Use this guess to find a particular solution of the given system.

(d) Use the results of parts (a) and (c) to find a general solution of the given system.

26. For the system of Problem 25, we found that a proper guess for a particular solution is "

In some cases a or b may be zero. (a) Find a particular solution for the system of

Problem 25 if col (b) Find a particular solution for the system of

Problem 25 if col Aet, etB.f AtB " A3et, 6etB.f AtB "

teta # etb. xp AtB

xp AtB " teta # etb . xp AtB " teta.

xp AtB " eta;

x$AtB ! c 0 1 "2 3 d x AtB # f AtB , where f AtB ! c et0 d ,

y A0B " !5 .y¿AtB # 2y AtB ! 4x AtB " !4t ! 2 , x¿ AtB ! 2y AtB " 4t , x A0B " 4 ;

x¿ AtB # y¿ AtB ! x AtB " t2 .x– AtB # y¿ AtB ! x AtB # y AtB " !1 ,

x A2B " c 1 1 dx A0B " c 4

!5 d

x¿ AtB " c 0 2 4 !2

d x AtB # c 4t !4t ! 2

d ,

27. Find a general solution of the system

Hint: Try

28. Find a particular solution for the system

Hint: Try

In Problems 29 and 30, find a general solution to the given Cauchy–Euler system for (See Problem 42 in Exercises 9.5.) Remember to express the system in the form before using the variation of parameters formula.

29.

30.

31. Use the variation of parameters formula (10) to derive a formula for a particular solution to the scalar equation in terms of two linearly independent solutions of the corresponding homogeneous equation. Show that your answer agrees with the formulas derived in Section 4.6. [Hint: First write the scalar equation in system form.]

32. Let be the invertible matrix

Show that

33. RL Network. The currents in the RL network given by the schematic diagram in Figure 9.8 are governed by the following equations:

I3 AtB " I4 AtB # I5 AtB .I1 AtB " I2 AtB # I3 AtB , 60I5 AtB ! 30I4 AtB " 0 ,I¿3 AtB # 30I4 AtB ! 90I2 AtB " 0 , 2I¿1 AtB # 90I2 AtB " 9 ,

U!1 AtB " 13 a AtBd AtB ! b AtBc AtB 4 c d AtB !b AtB!c AtB a AtB d . U AtB J c a AtB b AtB

c AtB d AtB d . 2 % 2U AtB

y1 AtB, y2 AtBy– # p AtBy¿ # q AtBy " g AtB yp

tx¿ AtB " c 4 !3 8 !6

d x AtB # c t 2t d

tx¿ AtB " c 2 !1 3 !2

d x AtB # c t!1 1 d

x¿ AtB " A AtBx AtB # f AtB t 7 0.

xp AtB " ta # b. 43 x¿ AtB " c 1 !1

!1 1 d x AtB # c!3

1 d .

xp AtB " e!ta # te!tb # c. 43 x¿ AtB " £ 0 1 11 0 1

1 1 0 § x AtB # £ !1!1 ! e!t

!2e!t § .

Section 9.7 Nonhomogeneous Linear Systems 549

Assume the currents are initially zero. Solve for the five currents Hint: Eliminate all unknowns except and and form a normal sys- tem with and

34. Conventional Combat Model. A simplistic model of a pair of conventional forces in combat yields the following system:

where col The variables and represent the strengths of opposing forces at time t. The terms and represent the operational loss rates, and the terms and represent the combat loss rates for the troops and respectively. The constants p and q represent the respective rates of reinforcement. Let , ,

, and By solving the appropriate initial value problem, determine which forces will win if (a) (b) (c)

35. Mixing Problem. Two tanks A and B, each hold- ing 50 L of liquid, are interconnected by pipes. The liquid flows from tank A into tank B at a rate of 4 L/min and from B into A at a rate of 1 L/min (see Figure 9.9). The liquid inside each tank is kept well

x1 A0B " 20 , x2 A0B " 21 .x1 A0B " 21 , x2 A0B " 20 . x1 A0B " 20 , x2 A0B " 20 .

p " q " 5.d " 2,c " 3 b " 4a " 1

x2,x1 !cx1!bx2

!dx2!ax1

x2 AtBx1 AtBAx1, x2B.x " x¿ " c!a !b

!c !d d x # c p

q d ,

x2 " I5. 4x1 " I2 I5,I2 3I1, . . . , I5.

9 volts

I 1

I 4 I 5

2 henries

90 ohms

I 2

I 3 1 henry I 3

I 3

30 ohms 60 ohms

Figure 9.8 RL network for Problem 33

4 L/min 0.2 kg/L

1 L/min 0.1 kg/L x 1 (t)

50 L

x 1 (0) = 0 kg

A 4 L/min

x 2 (t)

50 L

x 2 (0) = 0.5 kg

4 L/min

1 L/min

Figure 9.9 Mixing problem for interconnected tanks

stirred. A brine solution that has a concentration of 0.2 kg/L of salt flows into tank A at a rate of 4 L/min. A brine solution that has a concentration of 0.1 kg/L of salt flows into tank B at a rate of 1 L/min. The solutions flow out of the system from both tanks— from tank A at 1 L/min and from tank B at 4 L/min.

550 Chapter 9 Matrix Methods for Linear Systems

If, initially, tank A contains pure water and tank B contains 0.5 kg of salt, determine the mass of salt in each tank at time After several minutes have elapsed, which tank has the higher concentration of salt? What is its limiting concentration?

t ) 0.

In this chapter we have developed various ways to extend techniques for scalar differential equations to systems. In this section we take a substantial step further by showing that with the right notation, the formulas for solving normal systems with constant coefficients are identical to the formulas for solving first-order equations with constant coefficients. For example, we know that a general solution to the equation where a is a constant, is Analogously, we show that a general solution to the normal system

(1)

where A is a constant matrix, is Our first task is to define the matrix exponential

If A is a constant matrix, we define by taking the series expansion for and replacing a by A; that is,

(2)

(Note that we also replace 1 by I.) By the right-hand side of (2), we mean the matrix whose elements are power series with coefficients given by the corresponding entries in the matrices

If A is a diagonal matrix, then the computation of is straightforward. For example, if

then

and so

More generally, if A is an matrix with down its main diagonal, then is the diagonal matrix with down its main diagonal (see Problem 26). If A is not a diagonal matrix, the computation of is more involved. We deal with this impor- tant problem later in this section.

eAt er1t, er2t, . . . , ernteAt

r1, r2, . . . , rnn % n diagonal

eAt " a q

n"0 An t

n! " Daqn"0 A!1Bn t nn! 0

0 a q

n"0 2n

t n

T " c e!t 0 0 e2t

d . A2 " AA " c1 0

0 4 d , A3 " c!1 0

0 8 d , . . . , An " c A!1Bn 0

0 2n d ,

A " c!1 0 0 2

d , eAt

I, A, A2/2!, . . . .

n % n

eAt :! I # At # A2 t 2

2! # p # An tnn! # p .

eateAtn % n eAt.

x AtB " eAtc.n % nx¿ AtB " Ax AtB , x AtB " ceat.x¿ AtB " ax AtB,

9.8 THE MATRIX EXPONENTIAL FUNCTION

It can be shown that the series (2) converges for all t and has many of the same properties†

as the scalar exponential eat.

Section 9.8 The Matrix Exponential Function 551

Properties of the Matrix Exponential Function

Theorem 7. Let A and B be constant matrices and r, s, and t be real (or complex) numbers. Then,

(a)

(b)

(c)

(d) provided that AB " BA .

(e) erIt " ertI .

e AA#BBt " eAteBt, AeAtB!1 " e!At .eAAt#sB " eAteAs . eA0 " e0 " I .

n % n

Property (c) has profound implications. First, it asserts that for any matrix A, the matrix has an inverse for all t. Moreover, this inverse is obtained by simply replacing t by

(Note that (c) follows from (a) and (b) with s " !t.) In applying property (d) (the law of expo- nents), one must exercise care because of the stipulation that the matrices A and B commute (see Problem 25).

Another important property of the matrix exponential arises from the fact that we can dif- ferentiate the series in (2) term by term. This gives

Hence,

and so is a solution to the matrix differential equation Since is invertible [property (c)], it follows that the columns of are linearly independent solutions to system (1). Combining these facts we have the following.

eAt eAtX¿ " AX.eAt

d dt AeAtB ! AeAt ,

" A c I # At # A2 t2 2

# p # An!1 t n!1An ! 1B! # p d .

" A # A2t # A3 t 2

2 # p # An t

n!1An ! 1B! # p d dt

AeAtB " d dt

aI # At # A2 t2 2

# p # An t n

n! # p b

!t.eAt

†For proofs of these and other properties of the matrix exponential function, see Matrix Computations, 3rd ed., by Gene H. Golub and Charles F. van Loan (Johns Hopkins University Press, Baltimore, 1996), Chapter 11. See also the amusing articles “Nineteen Dubious Ways to Compute the Exponential of a Matrix,” by Cleve Moler and Charles van Loan, SIAM Review, Vol. 20, No. 4 (Oct. 1978), “Nineteen . . . Matrix, Twenty-Five Years Later,” ibid., Vol. 45, No. 1 (Jan. 2003).

eAt Is a Fundamental Matrix

Theorem 8. If A is an constant matrix, then the columns of the matrix exponen- tial form a fundamental solution set for the system Therefore, is a fundamental matrix for the system, and a general solution is x AtB " eAtc. eAtx¿ AtB " Ax AtB.eAt n % n

If a fundamental matrix X(t) for the system " Ax has somehow been determined, it is easy to compute eAt, as the next theorem describes.

x¿

552 Chapter 9 Matrix Methods for Linear Systems

Relationship Between Fundamental Matrices

Theorem 9. Let X(t) and Y(t) be two fundamental matrices for the same system " Ax . Then there exists a constant matrix C such that Y(t) = X(t)C for all t. In particular,

(3) eAt " X AtBX A0B!1 . x¿

Proof. Since is a fundamental matrix, every column of can be expressed as c for a suitable constant vector c, so column-by-column we have

If we choose , then (3) follows by setting t " 0. ◆

If the matrix A has n linearly independent eigenvectors ui, then Theorem 5 of Sec- tion 9.5 provides us with X(t) and (3) gives us

(4)

Are there any other ways that we can compute eAt? As we observed, if A is a diagonal matrix, then we simply exponentiate the diagonal elements of At to obtain eAt. Also, if A is a nilpotent matrix, that is, Ak " 0 for some positive integer k, then the series for eAt has only a finite number of terms—it “truncates”—since . In such cases, eAt

reduces to

Thus eAt can be calculated in finite terms if A is diagonal or nilpotent. Can we take this any further? Yes; a consequence of the Cayley–Hamilton theorem† is that when the characteris- tic polynomial for A has the form , then (r1I ! A)n " 0 = (!1)n(A ! r1I)n. So if A has only one (multiple) eigenvalue r1, then A ! r1I is nilpotent, and we exploit that by writing A " r1I # A ! r1I:

.eAt " er1Ite (A!r1I)t " er1t 3 I # AA ! r1IBt # p # AA ! r1IBn!1 tn!1An ! 1B! 4 p(r) " (r1 ! r)

eAt " I # At # p # Ak!1 t k!1Ak ! 1B! # 0 # 0 # p " I # At # p # Ak!1 tk!1Ak ! 1B! .

Ak " Ak#1 " p " 0

eAt " 3 er1tu1 er2tu2 p erntun 4 3u1 u2 p un 4!1 . n % n

Y AtB " eAt " X AtBC £ Y AtB

§ " £ X AtB

§ £ o o p oc1 c2 p cn

o o p o § " X AtBC .

X AtB Y AtBX AtB

†The Cayley–Hamilton theorem states that a matrix satisfies its own characteristic equation, that is, For a discussion of this theorem, see Matrices and Linear Transformation, 2nd ed., by Charles G. Cullen (Dover Publica- tions, New York, 1990), Chapter 5.

p AAB " 0.

Find the fundamental matrix for the system

where

We begin by computing the characteristic polynomial for A:

Thus, r " 1 is an eigenvalue of A with multiplicity 3. By the Cayley–Hamilton theorem, , and so

(5)

Computing, we find

and

Substituting into (5) yields

(6) ◆

We are not through with nilpotency yet. What if we have a nonzero vector u, an exponent m, and a scalar r satisfying so that is “nilpotent when restricted to u”? Such vectors are given a (predictable) name.

A ! r IAA ! r IBm u " 0, eAt " et £1 0 00 1 0

0 0 1 § # tet £ 1 1 11 1 1

!2 !2 !2 § " £ et # tet tet tettet et # tet tet

!2tet !2tet et ! 2tet § .

AA ! IB2 " £0 0 00 0 0 0 0 0

§ .A ! I " £ 1 1 11 1 1 !2 !2 !2

§ eAt " ete AA!IBt " et e I # AA ! IBt # AA ! IB2 t2

2 f .AA ! IB3 " 0

p ArB " 0A ! rI 0 " † 2 ! r 1 11 2 ! r 1 !2 !2 !1 ! r

† " !r3 # 3r2 ! 3r # 1 " ! Ar ! 1B3 . A " £ 2 1 11 2 1

!2 !2 !1 § .x¿ " Ax ,

eAt

Section 9.8 The Matrix Exponential Function 553

Example 1

Solution

Generalized Eigenvectors

Definition 5. Let A be a square matrix. A nonzero vector u satisfying

(7)

for some scalar r and some positive integer m is called a generalized eigenvector associated with r.

[Note that r must be an eigenvalue of A, since the final nonzero vector in the list u, is a “regular” eigenvector.]AA ! rIBu, AA ! rIB2u, p , AA ! rIBm!1u

AA ! rIBmu " 0

A valuable feature of generalized eigenvectors u is that we can compute in finite terms without knowing because

(8)

Moreover, is a solution to the system (recall Theorem 8). Hence if we can find n generalized eigenvectors for the matrix A that are linearly independent, then % nui

x¿ " AxeAtu

" ert cu # t AA ! rIBu # p # tm!1Am ! 1B! AA ! rIBm!1u # 0 # p d . " ert c Iu # t AA ! rIBu # p # tm!1Am ! 1B! AA ! rIBm!1u # tmm! AA ! rIBmu # p d

eAtu " erIte AA!rIBt u eAt,

eAtu

corresponding solutions will form a fundamental solution set and can be assem- bled into a fundamental matrix (Since the are solutions that reduce to the linearly independent at they are always linearly independent.) Finally, we get the matrix exponential by applying (3) from Theorem 9:

(9) ,

computed in a finite number of steps.

Of course, any (regular) eigenvector is a generalized eigenvector (corresponding to ), and if A has a full set of n linearly independent eigenvectors, then (9) is simply the representa- tion (4). But what if A is defective, that is, possesses fewer than n linearly independent eigen- vectors? Luckily, the primary decomposition theorem in advanced linear algebra† guarantees that when the characteristic polynomial of A is

(10)

where the are the distinct eigenvalues of A and is the multiplicity of the eigenvalue then for each i there exist linearly independent generalized eigenvectors satisfying

(11) .

Furthermore, the conglomeration of these generalized eigenvectors is linearly independent.††

We’re home! The following scheme will always yield a fundamental solution set, for any square matrix A.

n " m1 # m2 # p # mk

AA ! riIBmi u " 0mi ri,miri ’s

p ArB " Ar1 ! rBm1 Ar2 ! rBm2 p Ark ! rBmk ,

m " 1

eAt " X AtBX A0B!1 " 3 eAtu1 eAtu2 p eAtun 4 3u1 u2 p un 4!1t " 0,ui’s xi’sX AtB.xi AtB " eAtui

554 Chapter 9 Matrix Methods for Linear Systems

†Matrices and Linear Transformations, 2nd ed., by Charles G. Cullen, ibid. ††Some of the generalized eigenvectors satisfy (11) with a lower exponent, but such details do not concern us here. †††With defective matrices this algorithm is not recommended for computer implementation; rounding effects inevitably foil the machine’s ability to detect multiple eigenvalues.

Solving x = Ax To obtain a fundamental solution set for for any constant square matrix A:

(a) Compute the characteristic polynomial (b) Find the zeros of p(r) and express it as

where are the distinct zeros (i.e., eigenvalues) and are their multiplicities.

(c) For each eigenvalue find linearly independent generalized eigenvectors by applying the Gauss–Jordan algorithm to the system

(d) Form linearly independent solutions to by computing

for each generalized eigenvector u found in part (c) and corresponding eigenvalue r. If r has multiplicity m, this series terminates after m or fewer terms.

We can then, if desired, assemble the fundamental matrix X(t) from the n solutions and obtain the matrix exponential using (9).†††eAt

x AtB:" eAtu " ert 3u # t AA ! rIBu # t2 2! AA ! rI)2u # p 4x¿ " Axn " m1 # m2 # p # mk AA ! riIBmiu " 0.miri

m1, m2, p , mkr1, r2, p , rk p ArB " Ar1 ! rBm1 Ar2 ! rBm2 p Ark ! rBmk,p ArB " 0A ! rI 0 .

x¿ " Ax ¿

Find the fundamental matrix for the system

(12) where A " £1 0 01 3 0 0 1 1

§ .x¿ " Ax , eAtExample 2

We begin by finding the characteristic polynomial for A:

Hence, the eigenvalues of A are r " 1 with multiplicity 2 and r " 3 with multiplicity 1. Since r " 1 has multiplicity 2, we must determine two linearly independent associated

generalized eigenvectors satisfying From

we find and where s and are arbitrary. Taking s " 0 and we obtain the generalized eigenvector u1 " col(0, 0, 1). The cor-

responding solution to (12) is

(13)

(u1 is, in fact, a regular eigenvector). Next we take s " 1 and to derive the second linearly independent generalized

eigenvector u2 " col(!2, 1, 0) and (linearly independent) solution

(14)

†

For the eigenvalue r " 3, we solve that is,

to obtain the eigenvector col Hence, a third linearly independent solution to (12) is

(15) x3 AtB " e3tu3 " e3t £02 1 § " £ 02e3t

e3t § .

A0, 2, 1B.u3 " £!2 0 01 0 0

0 1 !2 § £u1u2

u3 § " £00

0 § , AA ! 3IBu " 0,

" et £!21 0 § # tet £00

1 § " £!2etet

tet § .

" et £!21 0 § # tet £0 0 01 2 0

0 1 0 § £!21

0 §

x2 AtB " eAtu2 " etEu2 # t AA ! IBu2F y " 0

x1 AtB " etEu1 # t AA ! lBu1F " et £00 1 § # tet £0 0 01 2 0

0 1 0 § £00

1 § " £00

et §

y " 1, yu3 " y,u2 " s, u1 " !2u2 " !2s,

AA ! IB2 u " £0 0 02 4 0 1 2 0

§ £u1u2 u3 § " £00

0 § ,

AA ! IB2 u " 0. p ArB " 0A ! rI 0 " †1 ! r 0 01 3 ! r 0

0 1 1 ! r † " ! Ar ! 1B2 Ar ! 3B .

Section 9.8 The Matrix Exponential Function 555

Solution

†Note that has been expressed in the format discussed in Problem 35 of Exercises 9.5, page 535.x2 AtB

The matrix whose columns are the vectors and ,

is a fundamental matrix for (12). Setting t " 0 and then computing , we find

and

It now follows from formula (3) that

◆

Use of the fundamental matrix simplifies many computations. For example, the prop- erties and enable us to rewrite the variation of parameters formula (13) in Section 9.7 in a simpler form. Namely, the solution to the initial value problem

is given by

(16)

which is a system version of the formula for the solution to the scalar initial value problem

In closing, we remark that the software packages for eigenvalue computation listed in Sec- tion 9.5 (page 527) also contain subroutines for computing the matrix exponential.

x¿ " ax # f AtB, x At0B " x0. x AtB " eAAt!t0Bx0 # ! t

eAAt!sBf AsB ds ,x¿ " Ax # f AtB, x At0B " x0

AeAt0B!1 " e!At0eAte!As " eAAt!sB eAt " E et 0 0! 12 et # 12 e3t e3t 0

! 14 e t ! 12 te

t # 14 e 3t ! 12 e

t # 12 e 3t et

U . eAt " X AtBX!1 A0B " £ 0 !2et 00 et 2e3t

et tet e3t § E! 14 ! 12 1! 12 0 0

1 4

1 2 0

U X!1 A0B " E! 14 ! 12 1! 12 0 0

1 4

1 2 0

U .X A0B " £0 !2 00 1 2 1 0 1

§ X!1 A0B

X AtB " £ 0 !2et 00 et 2e3t et tet e3t

§ , x3 AtBx1 AtB, x2 AtB,X AtB

556 Chapter 9 Matrix Methods for Linear Systems

In Problems 1–6, (a) show that the given matrix A satis- fies for some number r and some positive integer k and (b) use this fact to determine the matrix [Hint: Compute the characteristic polynomial and use the Cayley–Hamilton theorem.]

1. 2.

In Problems 7–10, determine by first finding a fundamental matrix and then using formula (3).

7. 8.

9. 10.

In Problems 11 and 12, determine by using general- ized eigenvectors to find a fundamental matrix and then using formula (3).

11.

12. A " £ 1 1 12 1 !1 0 !1 1

§ A " £ 5 !4 01 0 2

0 2 5 §

eAt

A " £ 0 2 22 0 2 2 2 0

§A " £ 0 1 00 0 1 1 !1 1

§ A " c 1 1

4 1 dA " c 0 1

!1 0 d

X AtB for x¿ " AxeAt A " £ 0 1 00 0 1

!1 !3 !3 §

A " £!2 0 04 !2 0 1 0 !2

§ A " £ 2 1 30 2 !1

0 0 2 §

A " £ 2 1 !1!3 !1 1 9 3 !4

§ A " c 1 !1

1 3 dA " c 3 !2

0 3 d

eAt. AA ! rIBk " 0

Section 9.8 The Matrix Exponential Function 557

9.8 EXERCISES In Problems 13–16, use a linear algebra software pack- age for help in determining

13.

14.

15.

16.

In Problems 17–20, use the generalized eigenvectors of A to find a general solution to the system where A is given.

17. 18.

19.

20.

21. Use the results of Problem 5 to find the solution to the initial value problem

x A0B " £ 11 !1 § .x¿ AtB " £!2 0 04 !2 0

1 0 !2 § x AtB ,

A " £!1 !8 1!1 !3 2 !4 !16 7

§ A " D1 0 1 21 1 2 1

0 0 2 0 0 0 1 1

T A " £ 0 0 1 0 1 2 0 0 1

§A " £ 0 1 00 0 1 !2 !5 !4

§ x¿ AtB " Ax AtB,

A " E!1 0 0 0 00 0 1 0 00 !1 !2 0 0 0 0 0 0 1 0 0 0 !4 !4

U A " E 0 1 0 0 00 0 1 0 0!1 !3 !3 0 0

0 0 0 0 1 0 0 0 !4 !4

U A " E1 0 0 0 00 0 1 0 00 !1 !2 0 0

0 0 0 0 1 0 0 0 !1 0

U A " E0 1 0 0 00 0 1 0 01 !3 3 0 0

0 0 0 0 1 0 0 0 !1 0

UeAt.

22. Use your answer to Problem 12 to find the solution to the initial value problem

23. Use the results of Problem 3 and the variation of parameters formula (16) to find the solution to the initial value problem

24. Use your answer to Problem 9 and the variation of parameters formula (16) to find the solution to the initial value problem

25. Let

and

(a) Show that (b) Show that property (d) in Theorem 7 does not

hold for these matrices. That is, show that

26. Let A be a diagonal matrix with entries down its main diagonal. To compute

proceed as follows: eAt,r1, . . . , rn

n % n e AA#BBt $ eAteBt.

AB $ BA.

B " c 2 1 0 1

d .A " c 1 2 !1 3

d x A0B " £ 1!1

0 § .

x¿ AtB " £ 0 1 00 0 1 1 !1 1

§ x AtB # £ 00 t § ,

x A0B " £ 03 0 § .

x¿ AtB " £ 2 1 !1!3 !1 1 9 3 !4

§ x AtB # £ 0t 0 § ,

x A0B " £!10 3 § .x¿ AtB " £ 1 1 12 1 !1

0 !1 1 § x AtB ,

558 Chapter 9 Matrix Methods for Linear Systems

(a) Show that is the diagonal matrix with entries down its main diagonal.

(b) Use the result of part (a) and definition (2) to show that is the diagonal matrix with entries

down its main diagonal. 27. In Problems 35–40 of Exercises 9.5, page 535, some

ad hoc formulas were invoked to find general solu- tions to the system when A had repeated eigenvalues. Using the generalized eigenvector procedure outlined on page 554, justify the ad hoc formulas proposed in (a) Problem 35, Exercises 9.5. (b) Problem 37, Exercises 9.5. (c) Problem 39, Exercises 9.5.

28. Let

(a) Find a general solution to x Ax. (b) Determine which initial conditions

yield a solution that remains bounded for all that is, satisfies

for some constant M and all 29. For the matrix A in Problem 28, solve the initial

value problem

x A0B " £ 01 !1 § .

x¿ AtB " Ax AtB # sin A2tB £ 20 4 § ,

t ) 0.

0 0x AtB 0 0 :" 2x21 AtB # x22 AtB # x23 AtB ( Mt ) 0; x AtB " col Ax1 AtB, x2 AtB, x3 AtB Bx A0B " x0

¿ "

A " s 5 2 !40 3 0 4 !5 !5

x¿ " Ax

er1t, . . . , ern t eAt

rk1, . . . , r k n

Chapter Summary

In this chapter we discussed the theory of linear systems in normal form and presented methods for solving such systems. The theory and methods are natural extensions of the devel- opment for second-order and higher-order linear equations. The important properties and techniques are as follows.

Homogeneous Normal Systems

The matrix function is assumed to be continuous on an interval I.A AtBn % nx¿ AtB " A AtBx AtB

Fundamental Solution Set: The n vector solutions of the homogeneous system on the interval I form a fundamental solution set, provided they are lin- early independent on I or, equivalently, their Wronskian

is never zero on I.

Fundamental Matrix: An matrix function whose column vectors form a fundamental solution set for the homogeneous system is called a fundamental matrix. The determinant of is the Wronskian of the fundamental solution set. Since the Wronskian is never zero on the interval I, then exists for t in I.

General Solution to Homogeneous System: If is a funda- mental matrix whose column vectors are then a general solution to the homoge- neous system is

where col is an arbitrary constant vector.

Homogeneous Systems with Constant Coefficients. The form of a general solution for a homogeneous system with constant coefficients depends on the eigenvalues and eigenvectors of the constant matrix A. An eigenvalue of A is a number r such that the system

has a nontrivial solution u, called an eigenvector of A associated with the eigenvalue r. Finding the eigenvalues of A is equivalent to finding the roots of the characteristic equation

The corresponding eigenvectors are found by solving the system If the matrix A has n linearly independent eigenvectors and is the eigenvalue

corresponding to the eigenvector then

is a fundamental solution set for the homogeneous system. A class of matrices that always has n linearly independent eigenvectors is the set of symmetric matrices—that is, matrices that satisfy

If A has complex conjugate eigenvalues and associated eigenvectors where a and b are real vectors, then two linearly independent real vector solutions to the homo- geneous system are

When A has a repeated eigenvalue r of multiplicity m, then it is possible that A does not have n linearly independent eigenvectors. However, associated with r are m linearly indepen- dent generalized eigenvectors that can be used to generate m linearly independent solutions to the homogeneous system (see page 561 under “Generalized Eigenvectors”).

eat cos bt a ! eat sin bt b , eat sin bt a # eat cos bt b .

z " a * ib,a * ib A " AT.

Eer1tu1, er2tu2, . . . , erntunFui, riu1, . . . , un

AA ! rIBu " 0.0A ! rI 0 " 0 . Au " ru

n % n

Ac1, . . . , cnBc " x AtB " X AtBc " c1x1 AtB # c2x2 AtB # p # cnxn AtB ,

x1, . . . , xn, X AtBXc " c1x1 # p # cnxn.X!1 AtB

X AtB X AtBn % nX AtB.

W 3x1, . . . , xn 4 AtB J det 3x1, . . . , xn 4 " ∞ x1,1 AtB x1,2 AtB p x1,n AtBx2,1 AtB x2,2 AtB p x2,n AtBo o o xn,1 AtB xn,2 AtB p xn,n AtB ∞

x1 AtB, . . . , xn AtBEx1, . . . , xnF. Chapter Summary 559

Nonhomogeneous Normal Systems

The matrix function and the vector function are assumed continuous on an interval I.

General Solution to Nonhomogeneous System: If is any particular solution for the nonhomogeneous system and is a fundamental matrix for the associated homoge- neous system, then a general solution for the nonhomogeneous system is

where are the column vectors of and col is an arbitrary constant vector.

Undetermined Coefficients. If the nonhomogeneous term is a vector whose components are polynomials, exponential or sinusoidal functions, and A is a constant matrix, then one can use an extension of the method of undetermined coefficients to decide the form of a particular solution to the nonhomogeneous system.

Variation of Parameters: Let be a fundamental matrix for the homogeneous system. A particular solution to the nonhomogeneous system is given by the variation of parameters formula

Matrix Exponential Function If A is a constant matrix, then the matrix exponential function

is a fundamental matrix for the homogeneous system The matrix exponential has some of the same properties as the scalar exponential In particular,

If for some r and k, then the series for has only a finite number of terms:

The matrix exponential function can also be computed from any fundamental matrix via the formula

eAt " X AtBX!1 A0B . X AtBeAt

eAt " ert e I # AA ! rIBt # p # AA ! rIBk!1 tk!1Ak ! 1B! f . eAtAA ! rIBk " 0e

0 " I , eAAt#sB " eAteAs , AeAtB!1 " e!At . eat.

x¿ AtB " Ax AtB. eAt J I # At # A2 t

2! # p # An t

n! # p

n % n

xp AtB " X AtBY AtB " X AtB ! X!1 AtBf AtB dt . X AtBX AtBY AtB.

f AtB Ac1, . . . , cnBc "X AtBx1 AtB, . . . , xn AtB

x AtB " xp AtB # X AtBc " xp AtB # c1x1 AtB # p # cnxn AtB , X AtB xp AtBxp # Xc.

f AtBA AtBn % nx¿ AtB " A AtBx AtB # f AtB

560 Chapter 9 Matrix Methods for Linear Systems

In Problems 1–4, find a general solution for the system where A is given.

1. 2.

3. 4.

In Problems 5 and 6, find a fundamental matrix for the system where A is given.

5. 6.

In Problems 7–10, find a general solution for the sys- tem where A and are given.

9. A " £ 2 1 !1!3 !1 1 9 3 !4

§ , f AtB " £ t0 1 §

A " c!4 2 2 !1

d , f AtB " c e4t 3e4t d

A " c 1 1 4 1

d , f AtB " c 5 6 d f AtBx¿ AtB " Ax AtB # f AtB, A " £ 5 0 00 !4 3

0 3 4 §A " c 1 !1

2 4 dx¿

AtB " Ax AtB, A " £ 1 1 00 1 0

0 0 2 §A " D1 2 0 02 1 0 0

0 0 1 2 0 0 2 1

T A " c 3 2!5 1 dA " c 6 !32 1 d x¿ AtB " Ax AtB,

Review Problems 561

10.

In Problems 11 and 12, solve the given initial value problem.

11.

12.

In Problems 13 and 14, find a general solution for the Cauchy–Euler system where A is given.

13.

14.

In Problems 15 and 16, find the fundamental matrix where A is given.

15. 16. A " £ 0 1 40 0 2 0 0 0

§A " £ 4 2 32 1 2 !1 2 0

§ eAt for the system x¿ AtB " Ax AtB,

A " £ 1 2 !12 1 1 !1 1 0

§ A " £ 0 3 11 2 1

1 3 0 §

tx¿ AtB " Ax AtB, x¿ AtB " c 2 1

!4 2 d x AtB # c te2t

e2t d , x A0B " c 2

2 d

x¿ AtB " c 0 1 !2 3

d x AtB , x A0B " c 1 !1 d

A " £ 2 !2 !30 3 2 0 !1 2

§ , f AtB " £ e!t2 1 §

REVIEW PROBLEMS

Generalized Eigenvectors If an eigenvalue ri of a constant matrix A has multiplicity mi there exist mi linearly independent generalized eigenvectors u satisfying Each such u determines a solution to the system of the form

and the totality of all such solutions is linearly independent on and forms a funda- mental solution set for the system.

A!q, q B x AtB " eri t c I # AA ! ri IBt # p # AA ! riIBmi!1 tmi!1Ami ! 1B! du

x¿ " Ax AA ! riIBmi u " 0.n % n

562 Chapter 9 Matrix Methods for Linear Systems

TECHNICAL WRITING EXERCISES

1. Explain how the theory of homogeneous linear differential equations (as described in Section 6.1) follows from the theory of linear systems in normal form (as described in Section 9.4).

2. Discuss the similarities and differences between the method for finding solutions to a linear constant coefficient differential equation (see Chapters 4 and 6) and the method for finding solutions to a linear sys- tem in normal form that has constant coefficients (see Sections 9.5 and 9.6).

3. Explain how the variation of parameters formulas for linear second-order equations derived in Section

4.6 follow from the formulas derived in Section 9.7 for linear systems in normal form.

4. Explain how you would define the matrix functions sin At and cos At, where A is a constant matrix. How are these functions related to the matrix exponential and how are they connected to the solu- tions of the system Your discussion should include the cases when A is

and c 1 !11 !1 d .c 0 00 0 d , c 0 10 0 d , c 0 01 0 d , x– # A2x " 0?

n % n

Group Projects for Chapter 9

A Uncoupling Normal Systems The easiest normal systems to solve are systems of the form

(1)

where D is an diagonal matrix. Such a system actually consists of n uncoupled equations

(2)

whose solution is

where the ’s are arbitrary constants. This raises the following question: When can we uncouple a normal system?

To answer this question, we need the following result from linear algebra. An matrix A is diagonalizable if and only if A has n linearly independent eigenvectors More- over, if P is the matrix whose columns are then

(3)

where D is the diagonal matrix whose entry is the eigenvalue associated with the vector

(a) Use the above result to show that the system

(4)

where A is an diagonalizable matrix, is equivalent to an uncoupled system

(5)

where and (b) Solve system (5). (c) Use the results of parts (a) and (b) to show that a general solution to (4) is given by

(d) Use the procedure discussed in parts (a)–(c) to obtain a general solution for the system

Specify and y.P, D, P!1,

x¿ AtB " £ 5 !4 41 0 1 !1 2 1

§ x AtB . x AtB " c1ed11tp1 # c2ed22tp2 # p # cnednntpn .

D " P!1AP.y " P!1x

y¿ AtB " Dy AtB ,n % n x¿ AtB " Ax AtB ,

pi.dii

P!1AP " D ,

p1, . . . , pn, p1, . . . , pn.

n % n

xi AtB " ciediit , x¿i AtB " diixi AtB , i " 1, . . . , n ,n % n x¿ AtB " Dx AtB ,

563

B Matrix Laplace Transform Method The Laplace transform method for solving systems of linear differential equations with constant coefficients was discussed in Chapter 7. To apply the procedure for equations given in matrix form, we first extend the definition of the Laplace operator to a column vector of functions

col by taking the transform of each component:

col

With this notation, the vector analogue of the important property relating the Laplace transform of the derivative of a function (see Theorem 4, Chapter 7, page 361) becomes

(6)

Now suppose we are given the initial value problem

(7)

where A is a constant matrix. Let denote the Laplace transform of and denote the transform of Then, taking the transform of the system and using the relation (6), we get

Next we collect the terms and solve for by premultiplying by

Finally, we obtain the solution by taking the inverse Laplace transform:

(8)

In applying the matrix Laplace transform method, it is straightforward (but possibly tedious) to compute but the computation of the inverse transform may require some of the special techniques (such as partial fractions) discussed in Chapter 7.

(a) In the above procedure, we used the property that for any constant matrix A. Show that this property follows from the linearity of the transform in

the scalar case. (b) Use the matrix Laplace transform method to solve the following initial value problems:

(i)

(ii)

(c) By comparing the Laplace transform solution formula (8) with the matrix exponential solution formula given in Section 9.8 [relation (16), page 556] for the homogeneous case

and derive the Laplace transform formula for the matrix exponential

(9)

(d) What is for the coefficient matrices in part (b) above? (e) Use (9) to rework Problems 1, 2, 7, and 8 in Exercises 9.8, page 557.

eAt

eAt " "!1E AsI ! AB!1F AtB .t0 " 0,f AtB " 0 x¿ AtB " c 3 !2

4 !1 d x AtB # c sin t

!cos t d , x A0B " c 0

0 d .

x¿ AtB " c 0 2 !1 3

d x AtB , x A0B " c!1 3 d .

n % n "EAxF " A"ExF

AsI ! AB!1, x " "!1Ex̂F " "!1E AsI ! AB!1 A f̂ # x0B F .x AtB

x̂ " AsI ! AB!1 A f̂ # x0B . AsI ! ABx̂ " f̂ # x0 , AsI ! AB!1:x̂x̂ sx̂ ! x0 " Ax̂ # f̂ .

"Ex¿F " "EAx # fF ,f AtB. f̂ AsBx AtBx̂ AsBn % nx¿ " Ax # f AtB , x A0B " x0 ,

"Ex¿F AsB " s"ExF AsB ! x A0B . A"Ex1F AsB, . . . , "ExnF AsBB ."ExF AsB JAx1 AtB, . . . , xn AtBBx "

564 Chapter 9 Matrix Methods for Linear Systems

C Undamped Second-Order Systems We have seen that the coupled mass–spring system depicted in Figure 9.5, page 540, is governed by equations (10) of Section 9.6, which we reproduce here:

This system was rewritten in normal form as equation (11) in Section 9.6; however, there is some advantage in expressing it as a second-order system in the form

This structure,

(10)

with a diagonal mass matrix M and a symmetric stiffness matrix K, is typical for most undamped vibrating systems. Our experience with other mass–spring systems (Section 5.6) suggests that we seek solutions to (10) of the form

(11) or

where v is a constant vector and is a positive constant.

(a) Show that the system (10) has a nontrivial solution of the form (11) if and only if and v satisfy the “generalized eigenvalue problem”

(b) By employing the inverse of the mass matrix, one can rewrite (10) as

Show that must be an eigenvalue of B if (11) is a nontrivial solution. (c) If B is an constant matrix, then can be written as a system of 2n first-

order equations in normal form. Thus, a general solution can be formed from 2n lin- early independent solutions. Use the observation in part (b) to find a general solution to the following second-order systems:

(i)

(ii)

(iii)

(iv) x– " £ 2 !1 !1!1 2 0 !1 0 2

§ x . x– " £!5 4 !4!1 0 !1

1 !2 !1 § x .

x– " c!2 2 2 !5

d x . x– " c 0 1

!4 !5 d x .

x– " Bxn % n !v2

x– " !M!1Kx ": Bx .

Kv " v2 Mv. v

x " Asin vtB v ,x " Acos vtB v

Mx– " !Kx ,

cm1 0 0 m2

d c x1 x2 d – " ! c k1 # k2 !k2

!k2 k2 # k3 d c x1

x2 d .

m2x–2 " !k2 Ax2 ! x1B ! k3x2 .m1x–1 " !k1x1 # k2 Ax2 ! x1B ,

Group Projects for Chapter 9 565

566

Suppose the wire is placed along the x-axis with x ! 0 at the left end of the wire and x ! L at the right end (see Figure 10.1). If we let u denote the temperature of the wire, then u depends on the time t and on the position x within the wire. (We will assume the wire is thin and hence u is constant throughout a cross section of the wire corresponding to a fixed value of x.) Because the wire is insulated, we assume no heat enters or leaves through the sides of the wire.

To develop a model for heat flow through the thin wire, let’s consider the small volume element V of wire between the two cross-sectional planes A and B that are perpendicular to the x-axis, with plane A located at x and plane B located at (see Figure 10.1).

The temperature on plane A at time t is and on plane is We will need the following principles of physics that describe heat flow.†

1. Heat Conduction: The rate of heat flow (the amount of heat per unit time flowing through a unit of cross-sectional area at A) is proportional to the temperature gradient at A (Fick’s law). The proportionality constant k is called the thermal con- ductivity of the material. In general, the thermal conductivity can vary from point to point:

2. Direction of Heat Flow: The direction of heat flow is always from points of higher temperature to points of lower temperature.

k ! k AxB. 0u/ 0x,

u Ax " ¢x, tB.Bu Ax, tB x " ¢x

Develop a model for the flow of heat through a thin, insulated wire whose ends are kept at a constant temperature of 0°C and whose initial temperature distribution is to be specified.

Partial Differential Equations

CHAPTER 10

INTRODUCTION: A MODEL FOR HEAT FLOW 10.1

0 x

A B

x + ∆ x

Figure 10.1 Heat flow through a thin piece of wire

†For a discussion of heat transfer, see University Physics with Modern Physics with MasteringPhysics™, 12th ed., by H. D. Young, R. A. Freedman, and L. Ford (Addison-Wesley, Reading, Mass., 2008).

Section 10.1 Introduction: A Model for Heat Flow 567

3. Specific Heat Capacity: The amount of heat necessary to raise the temperature of an object of mass m by an amount is where the constant c is the specific heat capacity of the material. The specific heat capacity, like the thermal conductivity, can vary with position:

If we let H represent the amount of heat flowing from left to right through the surface A during an interval of time then the formula for heat conduction becomes

where a is the cross-sectional area of the wire. The negative sign follows from the second principle—if is positive, then heat flows from right to left (hotter to colder).

Similarly, the amount of heat flowing from left to right across plane B during an interval of time is

The net change in the heat in volume V is the amount entering at end A minus the amount leaving at end B, plus any heat generated by sources (such as electric currents, chemi- cal reactions, heaters, etc.). The latter is modeled by a term where Q is the energy rate (power) density. Therefore,

(1)

Now by the third principle, the net change is given by where is the change in temperature and c is the specific heat capacity. If we assume that the change in tem- perature in the volume V is essentially equal to the change in temperature at x—that is,

, and that the mass of the volume V of wire is where is the density of the wire, then

(2)

Equating the two expressions for given in equations (1) and (2) yields

Now dividing both sides by a and then taking the limits as and approach zero, we obtain

(3)

If the physical parameters and are uniform along the length of the wire, then (3) reduces to the one-dimensional heat flow equation

(4)

where the positive constant is the diffusivity of the material and J Q Ax, tB /cr. P Ax, tBb J k/cr

0u 0t

Ax, tB ! b 02u 0x2

Ax, tB " P Ax, tB , rk, c,

0 0x c k AxB 0u0x Ax, tB d " Q Ax, tB ! c AxBr AxB 0u0t Ax, tB .

¢t¢x¢t¢x

! c AxBar AxB ¢x 3u Ax, t " ¢tB # u Ax, tB 4 .a ¢t c k Ax " ¢xB 0u0x Ax " ¢x, tB # k AxB 0u0x Ax, tB d " Q Ax, tB ¢x a ¢t ¢E

¢E ! c AxBar AxB ¢x 3u Ax, t " ¢tB # u Ax, tB 4 .r ! r AxB ar ¢x,¢u ! u Ax, t " ¢tB # u Ax, tB

¢u¢E ! cm ¢u,

! a ¢t c k Ax " ¢xB 0u0x Ax " ¢x, tB # k AxB 0u0x Ax, tB d " Q Ax, tB¢x a ¢t . ¢E ! H AxB # H Ax " ¢xB " Q Ax, tB¢x a ¢t

Q Ax, tB¢x a ¢t,¢E H Ax " ¢xB ! #k Ax " ¢xB a ¢t 0u0x Ax " ¢x, tB .

¢t

0u/ 0x

H AxB ! #k AxB a ¢t 0u0x Ax, tB , ¢t,

c ! c AxB. cm ¢u,¢u

568 Chapter 10 Partial Differential Equations

Equation (4) governs the flow of heat in the wire. We have two other constraints in our original problem. First, we are keeping the ends of the wire at 0°C. Thus, we require that

(5)

for all t. These are called boundary conditions. Second, we must be given the initial tempera- ture distribution That is, we require

(6)

Equation (6) is referred to as the initial condition on u. Combining equations (4), (5), and (6), we have the following mathematical model for the

heat flow in a uniform wire without internal sources whose ends are kept at the constant temperature 0°C:

(7)

(8) (9)

This model is an example of an initial-boundary value problem. Intuitively we expect that equations (7)–(9) completely and unambiguously specify the temperature in the wire. Once we have found a function that meets all three of these conditions, we can be assured that u is the temperature. (Theorem 7 in Section 10.5 will bear this out.)

In higher dimensions, the heat flow equation (or just heat equation) is adjusted to account for the additional heat flow contribution along the other axes by the simple modification

where known as the Laplacian in this context,† is defined in two and three dimensions, respectively, as

and

When the temperature reaches a steady state—that is, when u does not depend on time, and there are no sources—then and the temperature satisfies Laplace’s equation

One classical technique for solving the initial-boundary value problem for the heat equation (7)–(9) is the method of separation of variables, which effectively allows us to replace the partial differential equation by ordinary differential equations. This technique is discussed in the next section. In using separation of variables, one is often required to express a given function as a trigonometric series. Such series are called Fourier series; their properties are discussed in Sections 10.3 and 10.4. We devote the remaining three sections to the three basic partial differential equations that arise most commonly in applications: the heat equation, the wave equation, and Laplace’s equation.

Many computer-based algorithms for solving partial differential equations have been devel- oped. These are based on finite differences, finite elements, variational principles, and projection methods that include the method of moments and, more recently, its wavelet-based implementa- tions. Like the Runge–Kutta or Euler methods for ordinary differential equations, such tech- niques are more universally applicable than the analytic procedures of separation of variables, but their accuracy is often difficult to assess. Indeed, it is customary practice to assess any newly proposed numerical procedure by comparing its predictions with those of separation of variables.

¢u ! 0 .

0u/ 0t ! 0

¢u J 0 2u

0x2 "

02u 0y2

" 02u 0z2

.¢u J 0 2u

0x2 "

02u 0y2

¢u,

0u 0t ! b¢u " P Ax, y, z, tB ,

u Ax, tB 0 6 x 6 L .u Ax, 0B ! f AxB , t 7 0 ,u A0, tB ! u AL, tB ! 0 , 0 6 x 6 L , t 7 0 ,"u

"t Ax, tB ! B"2u

"x2 Ax, tB ,

AP ! 0B u Ax, 0B ! f AxB , 0 6 x 6 L .f AxB. u A0, tB ! u AL, tB ! 0

†Regrettably, it is the same notation as we just used for Some authors prefer the symbol for the Laplacian.

§2u Ax, t " ¢tB # u Ax, tB.

Section 10.2 Method of Separation of Variables 569

The method of separation of variables is a classical technique that is effective in solving several types of partial differential equations. The idea is roughly the following. We think of a solution, say to a partial differential equation as being a linear combination of simple component functions which also satisfy the equation and certain boundary condi- tions. (This is a reasonable assumption provided the partial differential equation and the boundary conditions are linear.) To determine a component solution, we assume it can be written with its variables separated; that is, as

Substituting this form for a solution into the partial differential equation and using the bound- ary conditions leads, in many circumstances, to two ordinary differential equations for the unknown functions and In this way we have reduced the problem of solving a par- tial differential equation to the more familiar problem of solving a differential equation that involves only one variable. In this section we will illustrate this technique for the heat equation and the wave equation.

In the previous section we derived the following initial-boundary value problem as a math- ematical model for the sourceless heat flow in a uniform wire whose ends are kept at the con- stant temperature zero:

(1)

(2) (3)

To solve this problem by the method of separation of variables, we begin by addressing equation (1). We propose that it has solutions of the form

where X is a function of x alone and T is a function of t alone. To determine X and T, we first compute the partial derivatives of u to obtain

and

Substituting these expressions into (1) gives

and separating variables yields

(4)

We now observe that the functions on the left-hand side of (4) depend only on t, while those on the right-hand side depend only on x. If we fix t and vary x, the ratio on the right cannot change; it must be constant. Most authors define this constant with a minus sign, so we adhere to the convention:

and , T¿ AtB bT AtB ! #lX– AxBX AxB ! #l

T ¿ AtB bT AtB ! X– AxBX AxB . X AxBT ¿ AtB ! bX– AxBT AtB ,

02u 0x2

! X– AxBT AtB .0u 0t

! X AxBT ¿ AtB u Ax, tB ! X AxBT AtB ,

0 6 x 6 L .u Ax, 0B ! f AxB , t 7 0 ,u A0, tB ! u AL, tB ! 0 , 0 6 x 6 L , t 7 0 ,"u

"t Ax, tB ! B"2u

"x2 Ax, tB ,

Tn AtB.Xn AxB un Ax, tB ! Xn AxBTn AtB .

un Ax, tB, un Ax, tB, n ! 0, 1, 2, . . . ,u Ax, tB,

10.2 METHOD OF SEPARATION OF VARIABLES

570 Chapter 10 Partial Differential Equations

(5) and

Consequently, for separable solutions, we have reduced the problem of solving the partial dif- ferential equation (1) to solving the two ordinary differential equations in (5).

Next we consider the boundary conditions in (2). Since these condi- tions are

and

Hence, either for all which implies that or

(6)

Ignoring the trivial solution we combine the boundary conditions in (6) with the differential equation for X in (5) and obtain the boundary value problem

(7)

where can be any constant. Notice that the function is a solution of (7) for every . Depending on the choice

of , this may be the only solution to the boundary value problem (7). Thus, if we seek a non- trivial solution to (1)–(2), we must first determine those values of for which the boundary value problem (7) has nontrivial solutions. These solutions are called the eigenfunctions of the problem; the eigenvalues are the special values of . If we write the first equation in (7) as and compare with the matrix eigenvector equation Au = ru (Section 9.5, page 526), the designation of as an eigenvalue (of ) becomes clear.

To solve the (constant coefficient) equation in (7), we try derive the auxiliary equation and consider three cases.

Case 1. In this case, the roots of the auxiliary equation are so a general solution to the differential equation in (7) is

To determine and we appeal to the boundary conditions:

From the first equation, we see that The second equation can then be written

as or Since it follows that

Therefore and hence is zero. Consequently, there is no nontrivial solution to (7) for

Case 2. Here is a repeated root to the auxiliary equation, and a general solution to the differential equation is

The boundary conditions in (7) yield and which imply that Thus, for there is no nontrivial solution to (7).

Case 3. In this case the roots of the auxiliary equation are Thus a general solution to is

(8) X AxB ! C1 cos 1lx " C2 sin 1lx .X– " lX ! 0 $i1l.l 7 0. l ! 0,C1 ! C2 ! 0.

C1 " C2L ! 0,C1 ! 0

X AxB ! C1 " C2x . r ! 0l ! 0.

l 6 0. C2,C1,Ae21#lL # 1B 7 0. #l 7 0,C1 Ae21#lL # 1B ! 0.C1 Ae1#lL # e#1#lLB ! 0 C2 ! #C1.

X ALB ! C1e1#lL " C2e#1#lL ! 0 .X A0B ! C1 " C2 ! 0 , C2,C1

X AxB ! C1e1#lx " C2e#1#lx . $1#l,l 6 0.r 2 " l ! 0, X AxB ! erx,

#D2l #D2 3X 4 ! lX l

lu Ax, tB ! X AxBT AtBl lX AxB ! 0l

X# AxB $ lX AxB ! 0 , X A0B ! X ALB ! 0 , u Ax, tB ! 0,X A0B ! X ALB ! 0 .

u Ax, tB ! 0,t 7 0,T AtB ! 0 X ALBT AtB ! 0 , t 7 0 .X A0BT AtB ! 0 u Ax, tB ! X AxBT AtB,

T¿ AtB ! %LBT AtB .X– AxB ! %LX AxB

Section 10.2 Method of Separation of Variables 571

This time the boundary conditions give the system

Because the system reduces to solving Hence, either or Now only when where n is an

integer. Therefore, (7) has a nontrivial solution when or n ! 1, 2, 3, . . . (we exclude n ! 0, since it makes l! 0). Furthermore, the nontrivial solutions (eigenfunctions) corresponding to the eigenvalue are given by [cf. (8)]

(9)

where the ’s are arbitrary nonzero constants. The eigenvalues and eigenfunctions for this example have many features that are common

to all of the separation of variables solutions that we will study. Take note of these features of Figure 10.2.

SEPARATION OF VARIABLES: EIGENFUNCTION PROPERTIES

(i) The eigenfunctions are solutions to a second-order ordinary differential equation (7) containing a parameter called the eigenvalue.

(ii) Each eigenfunction satisfies a homogeneous boundary con- dition at each end (7).

(iii) The only values of that admit nontrivial solutions, i.e., the eigenvalues, form an infinite set accumulating at .

(iv) The eigenfunctions oscillate faster as increases; the first contains no interior zeros, and each subsequent eigenfunc- tion contains one more zero than its predecessor.

(v) If any two distinct eigenfunctions are multiplied together, the resulting function is zero on the average; specifically,

and both cosines integrate to 0 over

Having determined that for any positive integer n, let’s consider the second equation in (5):

For each n ! 1, 2, 3, . . . , a general solution to this linear first-order equation is

Tn AtB ! bne#bAnp/LB2t . T ¿AtB " b anp

L b 2T AtB ! 0 .

l ! Anp/LB2 30, L 4 .sin

mpx L

sin npx

L !

1 2 c cos Am # nBpx

L # cos

Am " nBpx L

q l

Xn AxB ! an sin anpxL b , l ! Anp/LB2Xn

l ! Anp/LB2,1lL ! npAC2 % 0B 1lL ! np,sin 1lL ! 0C2 ! 0.sin 1lL ! 0 C2 sin 1lL ! 0.C1 ! 0, C1 cos 1lL " C2 sin 1lL ! 0 .C1 ! 0 ,

X A0B ! X ALB ! 0

Eigenvalue Eigenfunction

L 2

216 x

Figure 10.2 Eigenfunctions

572 Chapter 10 Partial Differential Equations

(Note that the time factor has none of the eigenfunction properties.) Combining this with equa- tion (9) we obtain, for each n ! 1, 2, 3, . . . , the functions

(10)

where is also an arbitrary constant. It is easy to see that each is a solution to (1)–(2). A simple computation also shows

that if and are solutions to (1)–(2), then so is any linear combination (This is a consequence of the facts that the operator is a linear operator and the boundary conditions in (2) are homogeneous.)

This enables us to solve the following example.

Find the solution to the heat flow problem

(11)

(12)

(13)

Comparing equation (11) with (1), we see that and Hence, we need only find a combination of terms like (10) that satisfies the initial condition (13):

For these data the task is simple; and . The solution to the heat flow problem (11)–(13) is

◆

What would we do if the initial condition (13) had been an arbitrary function, rather than a simple combination of a few of the eigenfunctions we found? Possibly the most beautiful property of the eigenfunctions—one that we did not list on page 571—is completeness; virtually any function ƒ likely to arise in applications can be expressed as a convergent series of eigenfunctions! For the sines we have been working with, the Fourier sine series looks like

(14)

This enables the complete solution to the generic problem given by (1)–(3):

(15)

provided this expansion and its first two derivatives converge. The convergence of the sine series (among others), as well as the procedure for finding the

coefficients cn, will be discussed in the next two sections (and generalized in Chapter 11). For now, let us turn to the mathematical description of a vibrating string, another situation in which the separation of variables approach applies. This concerns the transverse vibrations of a string stretched between two points, such as a guitar string or piano wire. The goal is to find a function

that gives the displacement (deflection) of the string at any point x and anyA0 ' x ' LBu Ax, tB

u Ax, tB ! a& n!1

cn e#B Anp/LB2 tsin anpxL b ,

ƒ AxB ! aq n!1

cnsin anpxL b for 0 6 x 6 L .

! 3e#28t sin 2x # 6e#175t sin 5x .

u Ax, tB ! c2e#bA2p/LB2t sin A2px/LB " c5e#bA5p/LB2t sin A5px/LB c5 ! #6c2 ! 3

u Ax, 0B ! a cne0 sin nx ! 3 sin 2x # 6 sin 5x. L ! p.b ! 7

0 6 x 6 p .u Ax, 0B ! 3 sin 2x # 6 sin 5x , t 7 0 ,u A0, tB ! u Ap, tB ! 0 , 0 6 x 6 p , t 7 0 ,0u

0t ! 7

02u 0x2

L J 0 / 0t # b02/ 0x2 aun " bum.umun

un Ax, tBcn ! cne

#bAnp/LB2t sin Anpx/LB ,un Ax, tB J Xn AxBTn AtB ! an sin Anpx/LBbne#bAnp/LB 2t

Example 1

Solution

Section 10.2 Method of Separation of Variables 573

time (see Figure 10.3). In developing the mathematical model, we assume that the string is perfectly flexible and has constant linear density, the tension on the string is constant, gravity is negligible, and no other forces are acting on the string. Under these conditions and the addi- tional assumption that the displacements are small in comparison to the length of the string, it turns out that the motion of the string is governed by the following initial-boundary value problem.†

(16)

(17)

(18)

(19)

Equation (16) essentially states Newton’s third law, F ! ma, for the string. The second (time) derivative on the left is the acceleration, and the second (spatial) derivative on the right arises because the restoring force is produced by the string’s curvature. The constant is strictly positive and equals the ratio of the tension to the (linear) density of the string. The physical significance of which has units of velocity, will be revealed in Section 10.6. The boundary conditions in (17) reflect the fact that the string is held fixed at the two endpoints x ! 0 and x ! L. Equations (18) and (19), respectively, specify the initial displacement of the string and the initial velocity of each point on the string. Recall that these are the typical initial data for all mechanical systems. For the initial and boundary conditions to be consistent, we assume and .

Let’s apply the method of separation of variables to the initial-boundary value problem for the vibrating string (16)–(19). Thus, we begin by assuming equation (16) has a solution of the form

where X is a function of x alone and T is a function of t alone. Differentiating u, we obtain

Substituting these expressions into (16), we have

X AxBT – AtB ! a2X– AxBT AtB , 02u 0t2

! X AxBT –AtB , 02u 0x2

! X– AxBT AtB . u Ax, tB ! X AxBT AtB ,

g A0B ! g ALB ! 0f A0B ! f ALB ! 0 a,

0 ' x ' L . "u "t Ax, 0B ! g AxB , 0 ' x ' L , u Ax, 0B ! f AxB ,

t ( 0 , u A0, tB ! u AL, tB ! 0 , 0 6 x 6 L , t 7 0 , "2u "t2

! A2 "2u "x2

u Ax, tB t ( 0

L x

u(x, t)

Figure 10.3 Displacement of string at time t

†For a derivation of this mathematical model, see Partial Differential Equations: Sources and Solutions, by Arthur D. Snider (Dover Publications, N.Y., 2006).

574 Chapter 10 Partial Differential Equations

and separating variables gives

Just as before, these ratios must equal some constant

(20) and

Furthermore, with the boundary conditions in (17) give

In order for these equations to hold for all either which implies that or

Ignoring the trivial solution, we combine these boundary conditions with the differential equation for X in (20) and obtain the boundary value problem

(21)

where l can be any constant. This is the same boundary value problem that we encountered earlier while solving the

heat equation. There we found that the suitable values for are

with corresponding eigenfunctions (nontrivial solutions)

(22)

where the ’s are arbitrary nonzero constants. Recall Figure 10.2, page 571. Having determined that for some positive integer n, let’s consider the second

equation in (20) for such :

For each n ! 1, 2, 3, . . . , a general solution is

Combining this with equation (22), we obtain, for each n ! 1, 2, 3, . . . , the function

or, reassembling the constants,

(23) un Ax, tB ! aan cos npaL t " bn sin npaL tb sin npxL . un Ax, tB ! Xn AxBTn AtB ! acn sin npxL b acn,1 cos npaL t " cn,2 sin npaL tb , Tn AtB ! cn,1 cos npaL t " cn,2 sin npaL t . T – AtB " a2n2p2

L2 T AtB ! 0 . l l ! Anp/LB2cn

Xn AxB ! cn sin anpxL b , l ! anp

L b 2 , n ! 1, 2, 3, . . . ,

X– AxB " lX AxB ! 0 , X A0B ! X ALB ! 0 , X A0B ! X ALB ! 0 .u Ax, tB ! 0,

T AtB ! 0,t ( 0,X A0BT AtB ! 0 , X ALBT AtB ! 0 , t ( 0 . u Ax, tB ! X AxBT AtB,

T – AtB A2T AtB ! #L .X# AxBX AxB ! #L

#l:

T – AtB a2T AtB ! X– AxBX AxB .

Section 10.2 Method of Separation of Variables 575

Using the fact that linear combinations of solutions to (16)–(17) are again solutions, we consider a superposition of the functions in (23):

(24)

For a solution of the form (24), substitution into the initial conditions (18)–(19) gives

(25)

(26)

We have now reduced the vibrating string problem (16)–(19) to the problem of determin- ing the Fourier sine series expansions for and

(27)

where If we choose the ’s and ’s so that the equations in (25) and (26) hold, then the expansion for in (24) is a formal solution to the vibrating string problem (16)–(19). If this expansion is finite, or converges to a function with continuous second partial derivatives, then the formal solution is an actual (genuine) solution.

Find the solution to the vibrating string problem

(28)

(29) (30)

(31)

Comparing equation (28) with equation (16), we see that and Hence, we need only determine the values of the coefficients and in formula (24). The ’s are chosen so that equation (25) holds; that is,

Equating coefficients of like terms, we see that

and the remaining ’s are zero. Similarly, referring to equation (26), we must choose the ’s so that

Comparing coefficients, we find

or b6 ! 1 12

,1 ! A6B A2Bb6 b4 !

1 4

,2 ! A4B A2Bb4 0u 0t Ax, 0B ! 2 sin 4x " sin 6x ! aqn!1 n2bn sin nx .

bnan

a3 ! 1 , a10 ! #4 ,

u Ax, 0B ! sin 3x # 4 sin 10x ! aq n!1

an sin nx .

anbnan L ! p.a ! 2

0u 0t Ax, 0B ! 2 sin 4x " sin 6x , 0 ' x ' p . u Ax, 0B ! sin 3x # 4 sin 10x , 0 ' x ' p ,u A0, tB ! u Ap, tB ! 0 , t ( 0 ,

02u 0t2

! 4 02u 0x2

, 0 6 x 6 p , t 7 0 ,

u Ax, tB bnanBn ! Anpa/LBbn. f AxB ! aq

n!1 an sin

npx L

, g AxB ! aq n!1

Bn sin npx

L ,

g AxB:f AxB 0u 0t Ax, 0B ! aqn!1 npaL bn sin npxL ! g AxB , 0 ' x ' L . u Ax, 0B ! aq

n!1 an sin

npx L

! f AxB , 0 ' x ' L , u Ax, tB ! a&

n!1 can cos nPAL t $ bn sin nPAL t d sin nPxL .

Example 2

Solution

576 Chapter 10 Partial Differential Equations

and the remaining ’s are zero. Hence, from formula (24), the solution to the vibrating string problem (28)–(31) is

(32) ◆

In later sections the method of separation of variables is used to study a wide variety of problems for the heat, wave, and Laplace’s equations. However, to use the method effectively, one must be able to compute trigonometric series (or, more generally, eigenfunction expan- sions) such as the Fourier sine series that we encountered here. These expansions are discussed in the next two sections.

u Ax, tB ! cos 6t sin 3x " 1 4

sin 8t sin 4x " 1 12

sin 12t sin 6x # 4 cos 20t sin 10x .

In Problems 1– 8, determine all the solutions, if any, to the given boundary value problem by first finding a general solution to the differential equation.

In Problems 9–14, find the values of (eigenvalues) for which the given problem has a nontrivial solution. Also determine the corresponding nontrivial solutions (eigenfunctions).

10.

11.

12. y¿ A0B ! 0 , y¿ Ap /2B ! 0y– " ly ! 0 ; 0 6 x 6 p/2 , y A0B ! y A2pB , y¿ A0B ! y¿ A2pBy– " ly ! 0 ; 0 6 x 6 2p ,

y ApB ! 0y¿ A0B ! 0 , y– " ly ! 0 ; 0 6 x 6 p , y A0B ! 0 , y¿ ApB ! 0y– " ly ! 0 ; 0 6 x 6 p ,

y A#1B ! 0 , y A1B ! 2y– # 2y¿ " y ! 0 ; #1 6 x 6 1 , y A2pB ! 1y A0B ! 1 , y– " y ! 0 ; 0 6 x 6 2p ,

y A0B ! 0 , y A2pB ! 1y– " y ! 0 ; 0 6 x 6 2p , y A0B ! 0 , y A1B ! 1 " ey– # y ! 1 # 2x ; 0 6 x 6 1 , y A0B ! 0 , y¿ ApB ! #6y– " 9y ! 0 ; 0 6 x 6 p ,

y¿ ApB ! 0y A0B ! 0 , y– " 4y ! 0 ; 0 6 x 6 p , y A0B ! 1 , y A2B ! 1y– # 6y¿ " 5y ! 0 ; 0 6 x 6 2 ,

y A1B ! #4y A0B ! 0 , y– # y ! 0 ; 0 6 x 6 1 , 13.

14.

In Problems 15–18, solve the heat flow problem (1)–(3) with and the given function 15. 16. 17. 18.

In Problems 19–22, solve the vibrating string problem (16)–(19) with and the given initial functions 19. 20.

21.

22.

23. Find the formal solution to the heat flow problem (1)–(3) with and if

24. Find the formal solution to the vibrating string prob- lem (16)–(19) with and

g AxB ! aq n!1

A#1Bn"1

n sin nx .

f AxB ! aq n!1

1 n2

sin nx ,

a ! 4, L ! p,

f AxB ! aq n!1

1 n2

sin npx .

L ! 1b ! 2

g AxB ! 6 sin 3x # 7 sin 5x f AxB ! sin x # sin 2x " sin 3x , g AxB ! 11 sin 9x # 14 sin 15x f AxB ! 6 sin 2x " 2 sin 6x , g AxB ! #2 sin 3x " 9 sin 7x # sin 10x f AxB ! 0 , f AxB ! 3 sin 2x " 12 sin 13x , g AxB ! 0 f AxB and gAxB.

a ! 3, L ! p,

f AxB ! sin 4x " 3 sin 6x # sin 10x f AxB ! sin x # 7 sin 3x " sin 5x f AxB ! sin 3x " 5 sin 7x # 2 sin 13x f AxB ! sin x # 6 sin 4x

f AxB.b ! 3, L ! p, y A0B ! 0 , y ApB ! 0y– # 2y¿ " ly ! 0 ; 0 6 x 6 p , y A0B # y¿ A0B ! 0 , y ApB ! 0y– " ly ! 0 ; 0 6 x 6 p ,

10.2 EXERCISES

Section 10.2 Method of Separation of Variables 577

25. By considering the behavior of the solutions of the equation

give an argument that is based on physical grounds to rule out the case where in equation (5).

26. Verify that given in equation (10) satisfies equation (1) and the boundary conditions in (2) by substituting directly into the equations involved.

In Problems 27–30, a partial differential equation (PDE) is given along with the form of a solution having sepa- rated variables. Show that such a solution must satisfy the indicated set of ordinary differential equations.

27.

with yields

where is a constant.

28.

with yields

where is a constant.

29.

with yields

where l, m are constants. Y– AyB " Al # mBY AyB ! 0 ,X– AxB " mX AxB ! 0 , T ¿ AtB " blT AtB ! 0 ,u Ax, y, tB ! X AxBY AyBT AtB

0u 0t

! b e 02u 0x2

" 02u 0y2 fl

T – AtB " T ¿ AtB " A1 " la2BT AtB ! 0 ,X– AxB " lX AxB ! 0 , u Ax, tB ! X AxBT AtB

02u 0t2

" 0u 0t

" u ! a2 02u 0x2

T– AuB " lT AuB ! 0 ,r 2R– ArB " rR¿ ArB # lR ArB ! 0 , u Ar, uB ! R ArBT AuB

02u 0r 2

" 1 r

0u 0r

" 1 r 2

02u 0u2

! 0

un Ax, tB un Ax, tB l 6 0

T ¿ AtB ! #lbT AtB , t 7 0 , 30.

with yields

where are constants.

31. For the PDE in Problem 27, assume that the follow- ing boundary conditions are imposed:

remains bounded as

Show that a nontrivial solution of the form must satisfy the boundary conditions

remains bounded as

32. For the PDE in Problem 29, assume that the follow- ing boundary conditions are imposed:

Show that a nontrivial solution of the form must satisfy the boundary conditions

33. When the temperature in a wire reaches a steady state, that is, when u depends only on x, then satisfies Laplace’s equation (a) Find the steady-state solution when the ends of

the wire are kept at a constant temperature of 50°C, that is, when

(b) Find the steady-state solution when one end of the wire is kept at 10°C, while the other is kept at 40°C, that is, when and u ALB ! 40.u A0B ! 10

u A0B ! u ALB ! 50. 02u/ 0x2 ! 0.

u AxB Y¿ A0B ! Y¿ AbB ! 0 .X A0B ! X AaB ! 0 ,

X AxBY AyBT AtB u Ax, y, tB ! 0u 0y Ax, 0, tB ! 0u0y Ax, b, tB ! 0 ; 0 ' x ' a , t ( 0 . u A0, y, tB ! u Aa, y, tB ! 0 ; 0 ' y ' b , t ( 0 ,

r S 0" .R ArBT A0B ! T ApB ! 0 , R ArBT AuB u Ar, uB !

r S 0" .u Ar, uBu Ar, 0B ! u Ar, pB ! 0 , m, l

r 2R– ArB " rR¿ ArB # Ar 2l " mBR ArB ! 0 ,Z– AzB " lZ AzB ! 0 , T– AuB " mT AuB ! 0 ,u Ar, u, zB ! R ArBT AuBZ AzB

02u 0r 2

" 1 r

0u 0r

" 1 r 2

02u 0u2

" 02u 0z2

! 0

578 Chapter 10 Partial Differential Equations

While analyzing heat flow and vibrating strings in the previous section, we encountered the problem of expressing a function in a trigonometric series [compare equations (12) and (25) in Section 10.2]. In the next two sections, we discuss the theory of Fourier series, which deals with trigonometric series expansions. First, however, we review some function properties that are particularly relevant to this study: piecewise continuity, periodicity, and even and odd symmetry.

In Section 7.2 we defined a piecewise continuous function on as a function f that is continuous at every point in , except possibly for a finite number of points at which f has a jump discontinuity. Such functions are necessarily integrable over any finite interval on which they are piecewise continuous.

Recall also that a function is periodic of period T if for all x in the domain of f. The smallest positive value of T is called the fundamental period. The trigono- metric functions sin x and cos x are examples of periodic functions with fundamental period

and tan x is periodic with fundamental period A constant function is a periodic function with arbitrary period T.

Two symmetry properties of functions will be useful in the study of Fourier series. A function f that satisfies for all x in the domain of f has a graph that is symmetric with respect to the y-axis [see Figure 10.4(a)]. We say that such a function is even. A function f that satisfies for all x in the domain of f has a graph that is symmetric with respect to the origin [see Figure 10.4(b)]. It is said to be an odd function. The functions . . . are examples of even functions, while the functions . . . are odd. The trigonometric functions sin x and tan x are odd functions and cos x is an even function.

x, x3, x5, 1, x2, x4,

f A#xB ! #f AxBf A#xB ! f AxB

p.2p

f Ax " TB ! f AxB 3a, b 4 3a, b 4

10.3 FOURIER SERIES

a−a x x A A

f(x)

(a)

a −a

−A

f(x)

(b)

Figure 10.4 (a) Even function (b) Odd function "a #a

f ! A # A ! 0"a #a

f ! A " A ! 2"a 0

Determine whether the given function is even, odd, or neither.

(a) (b) (c)

(a) Since then is an even function. (b) Because

it follows that is an odd function. (c) Here Since only when and is never equal to

then is neither an even nor an odd function. ◆h AxB #ex,e#xx ! 0e#x ! exh A#xB ! e#x.g AxBg A#xB ! A#xB1/3 # sin A#xB ! #x1/3 " sin x ! # Ax1/3 # sin xB ! #g AxB,f

AxBf A#xB ! 21 " A#xB2 ! 21 " x2 ! f AxB, h AxB ! ex g AxB ! x1/3 # sin x f AxB ! 21 " x2 Example 1

Solution

Section 10.3 Fourier Series 579

Example 2

Properties of Symmetric Functions

Theorem 1. If f is an even piecewise continuous function on then

(1)

If f is an odd piecewise continuous function on then

(2) # a

#a f AxBdx ! 0 . 3#a, a 4 ,

# a

#a f AxBdx ! 2 # a

0 f AxBdx . 3#a, a 4 ,

Proof. If f is an even function, then Hence,

where we used the change of variables This is illustrated in Figure 10.4(a). Formula (2) can be proved by a similar argument and is illustrated in Figure 10.4(b). ◆

The next example deals with certain integrals that are crucial in Fourier series.

Evaluate the following integrals when m and n are nonnegative integers:

(a) (b)

(c)

The even and odd functions occurring in the integrands are sketched in Figure 10.5 on page 580. The given integrals are easily evaluated by invoking the trigonometric formula for prod-

ucts of sines and cosines:

Thus if , each of the integrals calls for the (signed) area under an oscillating sinusoid over a whole number of periods, and is therefore zero. In fact, the only way to avoid these “oscillators” is to take ; whence and subtends an area of 2L, while subtending zero area (again). But note that if m and n are both zero, then also subtends area 2L. Inserting the factors of 1/2, we summarize with

cos 3 Am " nBpx/L 4 ! cos 0sin 3 Am # nBpx/L 4 ! sin 0 ! 0cos 3 Am # nBpx/L 4 ! cos 0 ! 1m ! n m % n

cos mpx

L cos

npx L

! 1 2

cos Am # nBpx

L "

1 2

cos Am " nBpx

L .

sin mpx

L sin

npx L

! 1 2

cos Am # nBpx

L #

1 2

cos Am " nBpx

L ,

sin mpx

L cos

npx L

! 1 2

sin Am # nBpx

L "

1 2

sin Am " nBpx

L ,

# L

#L cos

mpx L

cos npx

L dx .

# L

#L sin

mpx L

sin npx

L dx .#

#L sin

mpx L

cos npx

L dx .

u ! #x.

! # # 0

a f AuB du " # a

0 f AxB dx ! 2 # a

0 f AxB dx ,

# a

#a f AxB dx ! # 0

#a f AxB dx " # a

0 f AxB dx f A#xB ! f AxB.

Solution

Knowing that a function is even or odd can be useful in evaluating definite integrals.

580 Chapter 10 Partial Differential Equations

(3)

(4)

(5) ◆

Equations (3)–(5) express an orthogonality condition† satisfied by the set of trigonometric functions where We will say more about this later in this section.

L ! p.Ecos x, sin x, cos 2x, sin 2x, . . .F, #

#L cos

mpx L

cos npx

L dx ! $0 , m % n ,L , m ! n % 0 ,2L , m ! n ! 0 .

# L

#L sin

mpx L

sin npx

L dx ! e0 , m % n ,

L , m ! n ,

# L

#L sin

mpx L

cos npx

L dx ! 0 ,

†This nomenclature is suggested by the fact that the formulas for the Riemann sums approximating the integrals (3)–(5) look like dot products (Section 9.1, page 498) of higher-dimensional vectors. In most calculus texts, it is shown that the dot product of orthogonal vectors in two dimensions is zero.

−1

L 0cos x

L 0sin xx

−L L

−1

L 1cos x

L 1sin xx

−1

L 2cos x

L 2sin xx

−1

L 3cos x

L 3sin xx

−1

Figure 10.5 The sinusoids

Section 10.3 Fourier Series 581

It is easy to verify that if each of the functions is periodic of period T, then so is any linear combination

For example, the sum has period 2, since each term has period 2. Furthermore, if the infinite series

consisting of 2L-periodic functions converges for all x, then the function to which it converges will be periodic of period 2L.

Just as we can associate a Taylor series with a function that has derivatives of all orders at a fixed point, we can identify a particular trigonometric series with a piecewise continuous function. To illustrate this, let’s assume that has the series expansion†

(6)

where the ’s and ’s are constants. (Necessarily, f has period 2L.) To determine the coefficients , , , , . . . , we proceed as follows. Let’s integrate from to L, assuming that we can integrate term by term:

The (signed) area under the “oscillators” is zero. Hence,

and so

(Notice that is the average value of f over one period 2L.) Next, to find the coefficient when we multiply (6) by and integrate:

(7)

The orthogonality conditions (3)–(5) render the integrals on the right-hand side quite immedi- ately. We have already observed that

# L

#L cos

mpx L

dx ! 0 , m ( 1 ,

" a q

n!1 bn #

#L sin

npx L

cos mpx

L dx .

# L #L

f AxB cos mpx L

dx ! a0 2 #

#L cos

mpx L

dx " a q

n!1 an #

#L cos

npx L

cos mpx

L dx

cos Ampx/LBm ( 1, ama0/2 a0 !

1 L #

#L f AxBdx .

# L

#L f AxBdx ! # L

a0 2

dx ! a0L ,

# L

#L f AxBdx ! # L

a0 2

dx " a q

n!1 an #

#L cos

npx L

dx " a q

n!1 bn #

#L sin

npx L

dx .

#Lf AxB b2,a2b1a1a0bnan f AxB ! a0

2 " a

n!1 ean cos npxL " bn sin npxL f ,

f AxB

a0 2

" a q

n!1 aan cos npxL " bn sin npxL b

7 " 3 cos px # 8 sin px " 4 cos 2px # 6 sin 2px

c1 f1 AxB " . . . " cn fn AxB . f1, . . . , fn

†The choice of a constant instead of just will make the formulas easier to remember.a0a0/2

582 Chapter 10 Partial Differential Equations

and, by formulas (3) and (5), we have

Hence, in (7) we see that only one term on the right-hand side survives the integration:

Thus, we have a formula for the coefficient :

Similarly, multiplying (6) by and integrating yields

so that the formula for is

Motivated by the above computations, we now make the following definition.

bm ! 1 L #

#L f AxB sin mpx

L dx .

# L

#L f AxB sin mpx

L dx ! bmL

sin Ampx/LB am !

1 L #

#L f AxB cos mpx

L dx .

# L

#L f AxB cos mpx

L dx ! amL .

# L

#L cos

npx L

cos mpx

L dx ! e0 , n % m ,

L , n ! m .

# L

#L sin

npx L

cos mpx

L dx ! 0 ,

Fourier Series

Definition 1. Let f be a piecewise continuous function on the interval The Fourier series† of f is the trigonometric series

(8)

where the ’s and ’s are given by the formulas††

(9)

(10) bn ! 1 L #

#L f AxB sin npx

L dx , n ! 1, 2, 3, . . . .

an ! 1 L #

#L f AxB cos npx

L dx , n ! 0, 1, 2, . . . ,

bnan

f AxB % a0 2

" a q

n!1 ean cos npxL " bn sin npxL f ,

3#L, L 4 .

†Historical Footnote: Joseph B. J. Fourier (1768–1830) developed his series for solving heat flow problems. Lagrange expressed doubts about the validity of the representation, but Dirichlet devised conditions that ensured its convergence. ††Notice that need not be defined for every x in we need only that the integrals in (9) and (10) exist.3#L, L 4 ;f AxB

Section 10.3 Fourier Series 583

Compute the Fourier series for

Here Using formulas (9) and (10), we have

Therefore,

(11)

Some partial sums of this series are displayed in Figure 10.6. ◆

" e sin x # 1 2

sin 2x " 1 3

sin 3x " p f . ! p 4

# 2 p

e cos x " 1 9

cos 3x " 1 25

cos 5x " p f f AxB % p

4 " a

n!1 e 1 pn2

3 A#1Bn # 1 4 cos nx " A#1Bn"1 n

sin nx f !

#cos np n

! A#1Bn"1

n , n ! 1, 2, 3, . . . .

! 1 pn2

# pn

0 u sin u du !

1 pn2

3 sin u # u cos u 4 ` pn 0

bn ! 1 p

# p

#p f AxB sin nx dx ! 1

p # p

0 x sin nx dx

a0 ! 1 p

# p

#p f AxB dx ! 1

p # p

0 x dx !

2p ` p0 ! p2 , !

pn2 Acos np # 1B ! 1

pn2 3 A#1Bn # 1 4 , n ! 1, 2, 3, . . . , !

1 pn2

# pn

0 u cos u du !

1 pn2

3 cos u " u sin u 4 ` pn 0

an ! 1 p

# p

#p f AxB cos nx dx ! 1

p # p

0 x cos nx dx

L ! p.

f AxB ! e0 , #p 6 x 6 0 , x , 0 6 x 6 p .

Example 3

Solution

Formulas (9) and (10) are called the Euler formulas. We use the symbol in (8) to remind us that this series is associated with but may not converge to We will return to the ques- tion of convergence later in this section. Let’s first consider a few examples of Fourier series.

f AxB.f AxB %

4 terms 9 terms f(x)3

20−2 4−4 6−6 8 x

−8 −1

Figure 10.6 Partial sums of Fourier series in Example 3

584 Chapter 10 Partial Differential Equations

Compute the Fourier series for

Again, Notice that f is an odd function. Since the product of an odd function and an even function is odd (see Problem 7), is also an odd function. Thus,

Furthermore, is the product of two odd functions and therefore is an even func- tion, so

Thus,

(12)

Some partial sums of (12) are sketched in Figure 10.7. ◆

f AxB % 2 p

a q

n!1 31 # A#1Bn 4

n sin nx !

4 p

c sin x " 1 3

sin 3x " 1 5

sin 5x " p d . ! $0 , n even ,4

pn , n odd .

! 2 p

c#cos nx n

d ` p 0

! 2 p

c 1 n

# A#1Bn

n d , n ! 1, 2, 3, . . . ,

bn ! 1 p

# p

#p f AxB sin nx dx ! 2

p # p

0 sin nx dx

f AxB sin nx an !

1 p

# p

#p f AxB cos nx dx ! 0 , n ! 0, 1, 2, . . . .

f AxB cos nxL ! p. f AxB ! e#1 , #p 6 x 6 0 ,

1 , 0 6 x 6 p .

In Example 4 the odd function f has a Fourier series consisting only of sine functions. It is easy to see that, in general, if f is any odd function, then its Fourier series consists only of sine terms.

0 2−2 4−4 6−6 8 x

−8

−2

2 terms 9 terms

f(x)

Figure 10.7 Partial sums of Fourier series in Example 4

Example 4

Solution

Section 10.3 Fourier Series 585

Compute the Fourier series for

Here L ! 1. Since f is an even function, is an odd function. Therefore,

Since is an even function, we have

Therefore,

(13)

Partial sums for (13) are displayed in Figure 10.8. ◆

! 1 2

# 4 p2

e cos ApxB " 1 9

cos A3pxB " 1 25

cos A5pxB " p f . f AxB % 1

2 " a

n!1

2 p2n2

3 A#1Bn # 1 4 cos AnpxB !

p2n2 3 A#1Bn # 1 4 , n ! 1, 2, 3, . . . .

! 2 p2n2

# pn

0 u cos u du !

2 p2n2

3 cos u " u sin u 4 ` pn 0

! 2 p2n2

Acos np # 1B an ! #

#1 f AxB cos AnpxB dx ! 2 # 1

0 x cos AnpxB dx

a0 ! # 1

#1 f AxB dx ! 2 # 1

0 x dx ! x2 ` 1

0 ! 1 ,

f AxB cos AnpxB bn ! #

#1 f AxB sin AnpxB dx ! 0 , n ! 1, 2, 3, . . . .

f AxB sin AnpxB f AxB ! 0 x 0 , #1 6 x 6 1.Example 5

Solution

1 9 terms

2 terms

0 1−1 2 x

−2

Figure 10.8 Partial sums of Fourier series in Example 5

Notice that the even function f of Example 5 has a Fourier series consisting only of cosine functions and the constant function In general, if f is an even function, then its Fourier series consists only of cosine functions [including ].cos A0pxB1 ! cos A0pxB.

586 Chapter 10 Partial Differential Equations

Orthogonal Expansions Fourier series are examples of orthogonal expansions.† A set of functions is said to be an orthogonal system or just orthogonal with respect to the nonnegative weight function

on the interval if

(14) whenever

As we have seen, the set of trigonometric functions

(15)

is orthogonal on with respect to the weight function If we define the norm of f as

(16)

then we say that a set of functions is an orthonormal system with respect to if (14) holds and also for each n. Equivalently, we say the set is an orthonormal system if

(17)

We can always obtain an orthonormal system from an orthogonal system just by dividing each function by its norm. In particular, since

and

then the orthogonal system (15) gives rise on to the orthonormal system

If is an orthogonal system with respect to on we might ask if we can expand a function in terms of these functions; that is, can we express f in the form

(18)

for a suitable choice of constants . . . ? Such an expansion is called an orthogonal expansion, or a generalized Fourier series.

c1, c2,

f AxB ! c1 f1 AxB " c2 f2 AxB " c3 f3 AxB " p f AxB 3a, b 4 ,w AxBE fn AxB Fqn!1

E A2pB#1/2, p#1/2 cos x, p#1/2 sin x, p#1/2 cos 2x, p#1/2 sin 2x, . . .F .3#p, p 4 # p

#p 1dx ! 2p ,

# p

#p cos2nxdx ! #

#p sin2nxdx ! p , n ! 1, 2, 3, . . .

# b

a fm AxB fn AxBw AxB dx ! e0 , m % n ,

1 , m ! n .

& fn & ! 1w AxB E fn AxB Fqn!1 Aor E fn AxB FNn!1B & f & J c # b

a f 2 AxBw AxBdx d 1/2 ,

w AxB ! 1.3#p, p 4E1, cos x, sin x, cos 2x, sin 2x, . . .F m % n .#

a fm AxB fn AxBw AxBdx ! 0 ,

3a, b 4w AxB E fn AxB Fqn!1

†Orthogonality is also discussed in Section 8.8 on page 483.

Section 10.3 Fourier Series 587

To determine the constants in (18), we can proceed as we did in deriving Euler’s formu- las for the coefficients of a Fourier series, this time using the orthogonality of the system. Presuming the representation (18) is valid, we multiply by and integrate to obtain

(19)

(Here we have also assumed that we can integrate term by term.) Because the system is orthog- onal with respect to every integral on the right-hand side of (19) is zero except when

Solving for gives

(20)

The derivation of the formula for was only formal, since the question of the convergence of the expansion in (18) was not answered. If the series converges uniformly to on then each step can be justified, and indeed, the coefficients are given by formula (20). The notion of uniform convergence is discussed in the next subsection and in Section 13.2.†

Convergence of Fourier Series Let’s turn to the question of the convergence of a Fourier series. Figures 10.6–10.8 provide some clues. In Example 5 it is possible to use a comparison or limit comparison test to show that the series is absolutely dominated by a p-series of the form which converges. How- ever, this is much harder to do in Example 4, since the terms go to zero like . Matters can be even worse, since there exist Fourier series that diverge.†† We state two theorems that deal with the convergence of a Fourier series and two dealing with the properties of termwise differenti- ation and integration. For proofs of these results, see Partial Differential Equations of Mathe- matical Physics, 2nd ed., by Tyn Myint-U (Elsevier North Holland, Inc., New York, 1983), Chapter 5; Advanced Calculus with Applications, by N. J. DeLillo (Macmillan, New York, 1982), Chapter 9; or an advanced text on the theory of Fourier series.

Before proceeding, we need a notation for the left- and right-hand limits of a function. Let

and

We now present the fundamental pointwise convergence theorem for Fourier series.

f Ax#B J lim hS0"

f Ax # hB .f Ax"B J lim hS0"

f Ax " hB

1/n gqn!1 1/n2,

3a, b 4 ,f AxB gqn!1 cn fn AxBcm cm !

# b

a f AxB fm AxBw AxBdx

# b

a f 2m AxBw AxBdx !

# b

c f AxB fm AxBw AxBdx

& fm &2 , n ! 1, 2, 3, . . . .

cmn ! m. w AxB,

! a q

n!1 cn #

a fn AxB fm AxBw AxBdx .

# b

a f AxB fm AxBw AxBdx ! c1 # b

a f1 AxB fm AxBw AxBdx " c2 # b

a f2 AxB fm AxBw AxBdx " p

fm AxBw AxB

†All references to Chapters 11–13 refer to the expanded text Fundamentals of Differential Equations and Boundary Value Problems, 6th ed. ††In fact, there are trigonometric series that converge but are not Fourier series; an example is a

n!1

sin nx ln An " 1B .

588 Chapter 10 Partial Differential Equations

In other words, when and are piecewise continuous on the Fourier series converges to whenever f is continuous at x and converges to the average of the left- and right-hand limits at points where f is discontinuous.

Observe that the left-hand side of (21) is periodic of period 2L. This means that if we extend from the interval to the entire real line using 2L-periodicity, then equation (21) holds for all x for the 2L-periodic extension of

To which function does the Fourier series for

converge?

In Example 4 we found that the Fourier series for is given by (12), and in Figure 10.7, we sketched the graphs of two of its partial sums. Now and are piecewise continuous in

Moreover, f is continuous except at x ! 0. Thus, by Theorem 2, the Fourier series of f in (12) converges to the -periodic function where for

for and at

we have The graph of is given in Figure 10.9. ◆

g AxBg A$pB ! 3 f A#p"B " f Ap#B 4 /2 ! A#1 " 1B /2 ! 0. $p0 6 x 6 p, g A0B ! 3 f A0"B " f A0#B 4 /2 ! 0,#p 6 x 6 0, g AxB ! f AxB ! 1 g AxB ! f AxB ! #1g AxB,2p 3#p, p 4 . f ¿ AxBf AxBf AxB

f AxB ! e #1 , #p 6 x 6 0 , 1 , 0 6 x 6 p , f AxB.A#L, LBf AxB

f AxB 3#L, L 4 ,f ¿f

Pointwise Convergence of Fourier Series

Theorem 2. If and are piecewise continuous on then for any x in

(21)

where the ’s and ’s are given by the Euler formulas (9) and (10). For the

series converges to † 1 2 3 f A#L"B " f AL#B 4 . x ! $L,bnan

a0 2

" a q

n!1 ean cos npxL " bn sin npxL f ! 12 3 f Ax"B " f Ax#B 4 ,A#L, LB

3#L, L 4 ,f ¿f

Example 6

Solution

x 20

−

−1

Figure 10.9 The limit function of the Fourier series for f AxB ! e#1 , #p 6 x 6 0 , 1 , 0 6 x 6 p

†From formula (21), we see that it doesn’t matter how we define at its points of discontinuity, since only the left- and right-hand limits are involved. The derivative of course, is undefined at such points.f ¿AxB, f AxB

Section 10.3 Fourier Series 589

In Example 5 we obtained the Fourier series expansion given in (13) for , Since the periodic extension of (see Figure 10.11 on page 590) is con-

tinuous on and

is piecewise continuous on the Fourier series expansion (13) converges uniformly to on Compare Figure 10.8. The term-by-term differentiation of a Fourier series is not always permissible. For

example, the Fourier series for (see Problem 9), is

(22) f AxB % 2 aq n!1

A#1Bn"1 sin nx n

f AxB ! x, #p 6 x 6 p3#1, 1 4 .0 x 0 3#1, 1 4 ,f ¿ AxB ! e

#1 , #1 6 x 6 0 , 1 , 0 6 x 6 1 ,

A#q, q B f AxBg AxB,#1 6 x 6 1. f AxB ! 0 x 0

f ( x ) +

f ( x ) f ( x ) –

s N ( x )

Figure 10.10 An -corridor about fe

Uniform Convergence of Fourier Series

Theorem 3. Let f be a continuous function on and periodic of period 2L. If is piecewise continuous on then the Fourier series for f converges uniformly to f on and hence on any interval. That is, for each there exists an integer (that depends on ) such that

for all and all x ' A#q, q B.N ( N0, p f AxB # e a02 " aNn!1 ean cos npxL " bn sin npxL f f p 6 e ,

eN0 e 7 0,3#L, L 4 3#L, L 4 ,f ¿ A#q, q B

When f is a 2L-periodic function that is continuous on and has a piecewise continuous derivative, its Fourier series not only converges at each point—it converges uniformly on This means that for any prescribed tolerance the graph of the partial sum

will, for all N sufficiently large, lie in an -corridor about the graph of f on (see Figure 10.10). The property of uniform convergence of Fourier series is particularly helpful when one needs to verify that a formal solution to a partial differential equation is an actual (genuine) solution.

A#q, q Be sN AxB J a02 " aNn!1 ean cos npxL " bn sin npxL f

e 7 0,A#q, q B. A#q, q B

590 Chapter 10 Partial Differential Equations

Notice that Theorem 4 does not apply to the example function and its Fourier series expansion shown in (22), since the -periodic extension of this fails to be continuous on

Termwise integration of a Fourier series is permissible under much weaker conditions. A#q, q B. f AxB2p f

AxB

x 0

2−2−4 4

g(x)

Figure 10.11 Periodic extension of f AxB ! 0 x 0 , #1 6 x 6 1 Differentiation of Fourier Series

Theorem 4. Let be continuous on and 2L-periodic. Let and be piecewise continuous on Then, the Fourier series of can be obtained from the Fourier series for by termwise differentiation. In particular, if

then

f ¿AxB % aq n!1

pn L

e#an sin npxL " bn cos npxL f . f AxB ! a0

2 " a

n!1 ean cos npxL " bn sin npxL f , f AxB f ¿AxB3#L, L 4 . f –AxBf ¿AxBA#q, q Bf AxB

which converges for all x, whereas its derived series

diverges for every x. The following theorem gives sufficient conditions for using termwise differentiation.

2 a q

n!1 A#1Bn"1 cos nx

Integration of Fourier Series

Theorem 5. Let be piecewise continuous on with Fourier series

Then, for any x in we have

# x

#L f AtB dt ! # x

#L a0 2

dt " a q

n!1 #

#L ean cos nptL " bn sin nptL f dt .

3#L, L 4 ,f AxB % a02 " a q

n!1 ean cos npxL " bn sin npxL f .

3#L, L 4f AxB

Section 10.3 Fourier Series 591

A final note on the convergence issue: The question as to under what conditions the Fourier series converges has led to a tremendous amount of beautiful mathematics. As a practi- cal matter, however, time and economics will often dictate that you can afford to compute only a few terms of an eigenfunction series for your particular partial differential equation. Conver- gence fades into the background, and you need to know how to “do the best with what you’ve got.” So it is extremely gratifying to know that even if you’re going to use only part of an eigenfunction expansion, the Fourier coefficients (9)–(10) are still the best choice; Problem 37 demonstrates that every partial sum of a Fourier series outperforms any comparable superposi- tion of trigometric functions, in terms of mean-square approximation.

In Problems 1–6, determine whether the given function is even, odd, or neither.

1. 2. 3. 4. 5. 6.

7. Prove the following properties: (a) If f and g are even functions, then so is the

product fg. (b) If f and g are odd functions, then fg is an even

function. (c) If f is an even function and g is an odd function,

then fg is an odd function. 8. Verify formula (5). Hint: Use the identity

In Problems 9–16, compute the Fourier series for the given function f on the specified interval. Use a computer or graphing calculator to plot a few partial sums of the Fourier series.

10.

11.

12.

13.

14.

15.

16.

#p 6 x 6 #p /2 , #p /2 6 x 6 0 ,

0 6 x 6 p /2 , p /2 6 x 6 p

f AxB ! $ 0 ,#1 ,1 ,0 , f AxB ! ex , #p 6 x 6 pf AxB ! e

x , 0 6 x 6 p , x " p , #p 6 x 6 0

f AxB ! x2 , #1 6 x 6 1f AxB ! e 0 , #p 6 x 6 0 , x2 , 0 6 x 6 p

f AxB ! e 1 , #2 6 x 6 0 , x , 0 6 x 6 2

f AxB ! 0 x 0 , #p 6 x 6 pf AxB ! x , #p 6 x 6 p

cos B ! cos AA " BB " cos AA # BB. 4 2 cos A3

f AxB ! x1/5 cos x2f AxB ! e#x cos 3x f AxB ! sin Ax " 1Bf AxB ! A1 # x2B#1/2 f AxB ! sin2xf AxB ! x3 " sin 2x

In Problems 17–24, determine the function to which the Fourier series for given in the indicated problem, converges. 17. Problem 9 18. Problem 10 19. Problem 11 20. Problem 12 21. Problem 13 22. Problem 14 23. Problem 15 24. Problem 16

25. Find the functions represented by the series obtained by the termwise integration of the given series from

(a)

(b)

26. Show that the set of functions

is an orthonormal system on with respect to the weight function

27. Find the orthogonal expansion (generalized Fourier series) for

in terms of the orthonormal system of Problem 26.

28. (a) Show that the function has the Fourier series, on ,

f AxB % p2 3

" 4 a q

n!1 A#1Bn

n2 cos nx .

#p 6 x 6 p f AxB ! x2

f AxB ! e 0 , #1 6 x 6 0 , 1 , 0 6 x 6 1 ,

w AxB ! 1.3#1, 1 4 cos A2n # 1Bp

2 x, sin

A2n # 1Bp 2

x, . . . f e cos p

2 x, sin

p 2

x, cos 3p 2

x, sin 3p 2

x, . . . ,

f AxB ! e#1 , #p 6 x 6 0 , 1 , 0 6 x 6 p

4 p

a q

n!0 sin A2n " 1BxA2n " 1B % f AxB ,

2 a q

n!1

A#1Bn"1 n

sin nx % x , #p 6 x 6 p

x.#p

f AxB, 10.3 EXERCISES

592 Chapter 10 Partial Differential Equations

(b) Use the result of part (a) and Theorem 2 to show that

29. In Section 8.8, it was shown that the Legendre poly- nomials are orthogonal on the interval

with respect to the weight function Using the fact that the first three

Legendre polynomials are

find the first three coefficients in the expansion

where is the function

30. As in Problem 29, find the first three coefficients in the expansion

when 31. The Hermite polynomials are orthogonal on

the interval with respect to the weight function Verify this fact for the first three Hermite polynomials:

32. The Chebyshev (Tchebichef) polynomials are orthogonal on the interval with respect to the weight function Verify this fact for the first three Chebyshev polynomials:

33. Let be an orthogonal set of functions on the interval with respect to the weight function

Show that they satisfy the Pythagorean property

if m % n. & fm " fn& 2 ! & fm& 2 " & fn& 2

w AxB. 3 a, b 4E fn AxB F T2 AxB ! 2x2 # 1 .T0 AxB ! 1 , T1 AxB ! x ,

w AxB ! A1 # x2B#1/2.3#1, 1 4 Tn AxB H2 AxB ! 4x2 # 2 .H0 AxB ! 1 , H1 AxB ! 2x ,

W AxB ! e#x2.A#q, q B Hn AxBf AxB ! 0 x 0 , #1 6 x 6 1.

f AxB ! a0P0 AxB " a1P1 AxB " a2P2 AxB " p , f AxB J e#1 , #1 6 x 6 0 ,

1 , 0 6 x 6 1 .

f AxBf AxB ! a0P0 AxB " a1P1 AxB " a2P2 AxB " p , P2 AxB ! A3 /2Bx2 # A1 /2B ,P1 AxB ! x , P0 AxB ! 1 ,

w AxB ! 1.3#1, 1 4 Pn AxB a q

n!1

1 n2

! p2

6 .

a q

n!1

A#1Bn"1 n2

! p2

12 .

34. Norm. The norm of a function is like the length of a vector in In particular, show that the norm defined in (16) satisfies the following proper- ties associated with length assume f and g are continuous and on (a) and if and only if (b) where c is any real number. (c)

35. Inner Product. The integral in the orthogonality condition (14) is like the dot product of two vectors in In particular, show that the inner product of two functions defined by

(23)

where is a positive weight function, satisfies the following properties associated with the dot product assume f, g, and h are continuous on

(a) (b) , where c is any real number. (c)

36. Complex Form of the Fourier Series. (a) Using the Euler formula ,

prove that

and

(b) Show that the Fourier series

where

cn ! 1

2p # p

#p f AxB e#inx dx .

f AxB % aq n!#q

cne inx , #p 6 x 6 p ,

c0 ! a0 2

, cn ! an # ibn

2 , c#n !

an " ibn 2

! c0 " a q

n!1 Ecneinx " c#ne#inxF ,

f AxB % a0 2

" a q

n!1 Ean cos nx " bn sin nxF

sin nx ! einx # e#inx

2i .cos nx !

einx " e#inx

i ! 1#1, eiu ! cos u " i sin u ' f, g ( ! 'g, f ( . 'cf, h ( ! c' f, h ( ' f " g, h ( ! ' f, h ( " 'g, h ( .

3 a, b 4 B :A w AxB ' f, g ( J #

a f AxBg AxBw AxB dx ,

Rn.

& f " g& ' & f & " &g& . &cf & ! 0 c 0 & f & , f ! 0.& f & ! 0& f & ( 0,

3 a, b 4 B :w AxB 7 0 A Rn.

& f &

Section 10.3 Fourier Series 593

37. Least-Squares Approximation Property. The Nth partial sum of the Fourier series

gives the best mean-square approximation of f by a trigonometric polynomial. To prove this, let denote an arbitrary trigonometric polynomial of degree N:

and define

which is the total square error. Expanding the inte- grand, we get

(a) Use the orthogonality of the functions to show that

and

(b) Let be the error when we approximate f by the Nth partial sum of its Fourier series, that is, when we choose and Show that

" b21 " p " b2Nb . E* ! #

#p f 2 AxBdx # p aa20

2 " a21 " p " a2N

bn ! bn.an ! an

" b1b1 " p " bN bNb . # p

#p f AxBFN AxB dx ! p aa0a02 " a1a1 " p " aNaN

" b21 " p " b2Nb # p

#p F2N AxB dx ! p aa202 " a21 " p " a2N

sin x, cos 2x, . . .F E1, cos x, " #

#p F2N AxB dx .

E ! # p

#p f 2 AxB dx # 2 # p

#p f AxBFN AxB dx

E J # p

#p 3 f AxB # FN AxB 4 2 dx ,

FN AxB ! a02 " aNn!1 Ean cos nx " bn sin nxF , FN AxB

f AxB % a0 2

" a q

n!1 Ean cos nx " bn sin nxF

Hence, the Nth partial sum of the Fourier series gives the least total square error, since

38. Bessel’s Inequality. Use the fact that defined in part (b) of Problem 37, is nonnegative to prove Bessel’s inequality

(24)

(If f is piecewise continuous on then we have equality in (24). This result is called Parseval’s identity.)

39. Gibbs Phenomenon.† The American mathemati- cian Josiah Willard Gibbs (1839–1903) observed that near points of discontinuity of f, the partial sums of the Fourier series for f may overshoot by approximately 9% of the jump, regardless of the number of terms. This is illustrated in Figure 10.12, on page 594, for the function

whose Fourier series has the partial sums

To verify this for proceed as follows:

(a) Show that

(b) Infer from part (a) and the figure that the maxi- mum occurs at and has the value

" p " 1

2n # 1 sin A2n # 1Bp

2n d .f2n#1 a

p 2n b ! 4

p c sin p

2n "

1 3

sin 3p 2n

x ! p / A2nB ! 2 sin 2nx . ! " p " cos A2n # 1Bx 4 p Asin xB f ¿2n#1 AxB ! 4 sin x 3 cos x " cos 3x

f AxB," p "

sin A2n # 1BxA2n # 1B d . f2n#1 AxB ! 4p c sin x " 13 sin 3x f AxB ! e#1 , #p 6 x 6 0 ,

1 , 0 6 x 6 p ,

3#p, p 4 ,a 2 0

2 " a q

n!1 Ea2n " b2nF ' 1p # p#p f 2 AxB dx .

E*, E ( E*.

" p " AbN # bNB2 f ." p " AaN # aNB 2 " Ab1 # b1B2E # E* ! p e

Aa0 # a0B2 2

" Aa1 # a1B2 E ( E*,E # E* ( 0,

†Historical Footnote: Actually, H. Wilbraham discovered this phenomenon some 50 years earlier than Gibbs did. It is more appropriately called the Gibbs–Wilbraham phenomenon.

594 Chapter 10 Partial Differential Equations

using the partition , n, and choosing the mid-

point of each interval as the place to evaluate the integrand, then

! p 2

f2n#1 ap2nb . ! "

sin 3 A2n # 1Bp /2n 4A2n # 1Bp /2n pn # p

0 sin x

x dx )

sin Ap /2nB p /2n

n " p

¢xk ! p /n2, . . . xk J A2k # 1B Ap /2nB, k ! 1,#

0 sin x

x dx

(d) Use the result of part (c) to show that the over- shoot satisfies

(e) Using the result of part (d) and a numerical inte- gration algorithm (such as Simpson’s rule, Appendix C) for the sine integral function

show that Thus, the approximations overshoot the true value of

by 0.18, or 9% of the jump from to f A0" B.f A0#Bf A0" B ! 1

limnSq f2n#1 Ap / A2nBB ) 1.18.Si AzB J # z

0 sin x

x dx ,

lim nSq

f2n#1 ap2nb ! 2p # p0 sin xx dx .

x 1 0 2 3 −3 −2 −1

−1

x 1 0 2 3 −3 −2 −1

−1

Overshoot Overshoot

(a) Graph of f 11

( x ) (b) Graph of f 51

( x )

Figure 10.12 Gibbs phenomenon for partial sums of Fourier series

10.4 FOURIER COSINE AND SINE SERIES A typical problem encountered in using separation of variables to solve a partial differential equation is the problem of representing a function defined on some finite interval by a trigono- metric series consisting of only sine functions or only cosine functions. For example, in Sec- tion 10.2, equation (25), page 575, we needed to express the initial values ,

of the solution to the initial-boundary value problem associated with the problem of a vibrating string as a trigonometric series of the form

(1)

Recalling that the Fourier series for an odd function defined on consists entirely of sine terms, we might try to achieve (1) by artificially extending the function f AxB, 0 6 x 6 L,3#L, L 4

f AxB ! aq n!1

an sin anpxL b . 0 6 x 6 L,

u Ax, 0B ! f AxB

Section 10.4 Fourier Cosine and Sine Series 595

to the interval in such a way that the extended function is odd. This is accomplished by defining the function

and extending to all x using 2L-periodicity.† Since is an odd function, it has a Fourier series consisting entirely of sine terms. Moreover, is an extension of since

on This extension is called the odd 2L-periodic extension of The resulting Fourier series expansion is called a half-range expansion for since it represents the function on which is half of the interval where it represents

In a similar fashion, we can define the even 2L-periodic extension of as the function

with To illustrate the various extensions, let’s consider the function

If we extend to the interval using is given by

with In Problem 14 of Exercises 10.3, the Fourier series for was found to be

which consists of both odd functions (the sine terms) and even functions (the constant term), since the -periodic extension is neither an even nor an odd function. The odd 2 -periodic extension of is just which has the Fourier series expansion

(2)

(see Problem 9 in Exercises 10.3). Because on the interval the expansion in (2) is a half-range expansion for The even -periodic extension of is the function

which has the Fourier series expansion

(3)

(see Problem 10 in Exercises 10.3).

fe AxB ! p2 # 4p aqn!0 1A2n " 1B2 cos A2n " 1Bx fe AxB ! 0 x 0 , #p 6 x 6 p, f AxB2pf AxB. A0, pB,fo AxB ! f AxB

fo AxB % 2 aq n!1

A#1Bn"1 n

sin nx

fo AxB ! x, #p 6 x 6 p,f AxBp f~ AxBp f ~ AxB % p

2 # a

n!1

1 n

sin 2nx ,

f ~ AxBf~ Ax " 2pB ! f~ AxB.

f ~ AxB ! e x , 0 6 x 6 p ,

x " p , #p 6 x 6 0 ,

p-periodicity, then the extension f ~A#p, pBf AxB f AxB ! x, 0 6 x 6 p.fe

Ax " 2LB ! fe AxB. fe AxB J e f AxB , 0 6 x 6 L ,

f A#xB , #L 6 x 6 0 , f AxBfo AxB.A#L, LBA0, LB,f AxB f

AxB, f AxB.A0, LB.fo AxB ! f AxB f AxB,fo AxBfo AxBfo AxB

fo AxB J e f AxB , 0 6 x 6 L , #f A#xB , #L 6 x 6 0 ,

A#L, LB

†Strictly speaking, we have extended to all x other than the integer multiples of L. Continuity considerations often suggest appropriate values for the extended functions at some of these points, as well. Figure 10.13 on page 596 illus- trates this.

fo AxB

596 Chapter 10 Partial Differential Equations

x 2

g ( x )

3 − −2−3

x 2 3 − −2−3

f (x) ~

x 2 3 − −2−3

fe(x)

x 2 3 − −2−3

fo(x)

Figure 10.13 Extensions of f AxB ! x, 0 6 x 6 p

The preceding three extensions, the -periodic function the odd -periodic func- tion and the even -periodic function are natural extensions of There are many other ways of extending For example, the function

which we studied in Example 3 of Section 10.3, is also an extension of However, its Fourier series contains both sine and cosine terms and hence is not as useful as previous exten- sions. The graphs of these extensions of are given in Figure 10.13.f AxB f

AxB.g AxB ! e x , 0 6 x 6 p ,

0 , #p 6 x 6 0 , g Ax " 2pB ! g AxB ,

f AxB. f AxB.fe AxB,2pfo AxB, 2pf ~ AxB,p

The Fourier series expansions for and given in (2) and (3) represent on the interval actually, they equal on This motivates the following definitions.A0, pBB .f AxBAA0, pB f AxBfe AxBfo AxB Fourier Cosine and Sine Series

Definition 2. Let be piecewise continuous on the interval The Fourier cosine series of on is

(4)

where

(5)

The Fourier sine series of on is

(6)

where

(7) bn ! 2 L

# L

0 f AxB sin npx

L dx , n ! 1, 2, . . . .

a q

n!1 bn sin

npx L

30, L 4f AxBan ! 2 L

# L

0 f AxB cos npx

L dx , n ! 0, 1, . . . .

a0 2

" a q

n!1 an cos

npx L

30, L 4f AxB 30, L 4 .f AxB

Section 10.4 Fourier Cosine and Sine Series 597

The trigonometric series in (4) is just the Fourier series for the even 2L-periodic extension of and that in (6) is the Fourier series for the odd 2L-periodic extension of

These are called half-range expansions for

Compute the Fourier sine series for

Using formula (7) with we find

So on letting we find the Fourier sine series for to be

(8) ◆

Since, in Example 1, the function is continuous and is piecewise continuous on it follows from Theorem 2 on pointwise convergence of Fourier series that

for all x in Let’s return to the problem of heat flow in one dimension.

Find the solution to the heat flow problem

(9)

(10)

(11) u Ax, 0B ! e x , 0 6 x ' p/2 , p # x , p/2 ' x 6 p .

u A0, tB ! u Ap, tB ! 0 , t 7 0 , 0u 0t

! 2 02u 0x2

, 0 6 x 6 p , t 7 0 ,

30, p 4 .f AxB ! 4 p e sin x # 1

9 sin 3x "

1 25

sin 5x # 1 49

sin 7x " p f A0, pB, f ¿ AxBf AxB

p a q

k!0

A#1BkA2k " 1B2 sin A2k " 1Bx ! 4p e sin x # 19 sin 3x " 125 sin 5x " p f . f AxBn ! 2k " 1,

! 4

pn2 sin

np 2

! $0 , n even ,4 A#1B An#1B/2 pn2

, n odd .

# 2 pn2 3 sin u # u cos u 4 ` pn

pn/ 2

! 2 pn2 3 sin u # u cos u 4 ` pn/2

0 #

2 n c cos pn # cos np

2 d

! 2 pn2 #

pn/2

0 u sin udu " 2 #

p/2 sin nx dx #

2 pn2 #

pn/2 u sin udu

bn ! 2 p #

0 f AxB sin nxdx ! 2

p # p/2

0 x sin nx dx "

2 p #

p/2 Ap # xB sin nx dx

L ! p,

f AxB ! e x , 0 ' x ' p/2 , p # x , p/2 ' x ' p .

f AxB.f AxB. fo AxB,f AxB, fe AxB,

Example 1

Solution

Example 2

10.4 EXERCISES

598 Chapter 10 Partial Differential Equations

Solution Comparing equation (9) with equation (1) in Section 10.2 (page 569), we see that and Hence, we need only represent in a Fourier sine series of the form

In Example 1 we obtained this expansion and showed that

Hence, from equation (15) on page 572, the solution to the heat flow problem (9)–(11) is

A sketch of a partial sum for is displayed in Figure 10.14. ◆u Ax, tB !

4 p

e e#2t sin x # 1 9

e#18t sin 3x " 1 25

e#50t sin 5x " p f . !

p a

k!0 A#1BkA2k " 1B2 e#2 A2k"1B 2t sin A2k " 1Bx

u Ax, tB ! aq n!1

cne #2n2t sin nx

cn ! 4

pn2 sin

np 2

! $0 , n even ,4 A#1B An#1B/2 pn2

, n odd .

a q

n!1 cn sin nx .

u Ax, 0B ! f AxBL ! p. b ! 2

)

Figure 10.14 Partial sum for in Example 2u Ax, tB

In Problems 1–4, determine (a) the -periodic extension , (b) the odd 2 -periodic extension , and (c) the even

2 -periodic extension for the given function f and sketch their graphs.

1. 2.

3. f AxB ! e 0 , 0 6 x 6 p /2 , 1 , p /2 6 x 6 p

f AxB ! sin 2x , 0 6 x 6 p f AxB ! x2 , 0 6 x 6 p fep

fopf ~

p 4.

In Problems 5–10, compute the Fourier sine series for the given function.

5. 6. 7. 8. f AxB ! p # x , 0 6 x 6 p f AxB ! x2 , 0 6 x 6 p

f AxB ! cos x , 0 6 x 6 p f AxB ! #1 , 0 6 x 6 1 f AxB ! p # x , 0 6 x 6 p

Section 10.5 The Heat Equation 599

In Problems 17–19, find the solution to the heat flow problem.

where is given.

17. 18.

19. f AxB ! e#x , 0 6 x ' p /2 , x # p , p /2 ' x 6 p

f AxB ! x Ap # xBf AxB ! 1 # cos 2xf AxB u Ax, 0B ! f AxB , 0 6 x 6 p ,u A0, tB ! u Ap, tB ! 0 , t 7 0 ,

0u 0t

! 5 02u 0x2

, 0 6 x 6 p , t 7 0 ,

9. 10.

In Problems 11–16, compute the Fourier cosine series for the given function. 11. 12. 13. 14. 15. 16. f AxB ! x # x2 , 0 6 x 6 1f AxB ! sin x , 0 6 x 6 p

f AxB ! e#x , 0 6 x 6 1f AxB ! ex , 0 6 x 6 1 f AxB ! 1 " x , 0 6 x 6 pf AxB ! p # x , 0 6 x 6 p f AxB ! ex , 0 6 x 6 1f AxB ! x # x2 , 0 6 x 6 1

10.5 THE HEAT EQUATION In Section 10.1 we developed a model for heat flow in an insulated uniform wire whose ends are kept at the constant temperature 0°C. In particular, we found that the temperature in the wire is governed by the initial-boundary value problem

(1)

(2) (3)

[see equations (7)–(9) in Section 10.1, page 568]. Here equation (2) specifies that the temperature at the ends of the wire is zero, whereas equation (3) specifies the initial temperature distribution.

In Section 10.2 we also derived a formal solution to (1)–(3) using separation of variables. There we found the solution to (1)–(3) to have the form

(4)

where the ’s are the coefficients in the Fourier sine series for

(5)

In other words, solving (1)–(3) reduces to computing the Fourier sine series for the initial value function

In this section we discuss heat flow problems where the ends of the wire are insulated or kept at a constant, but nonzero, temperature. (The latter involves nonhomogeneous boundary condi- tions.) We will also discuss the problem in which a heat source is adding heat to the wire. (This results in a nonhomogeneous partial differential equation.) The problem of heat flow in a rectan- gular plate is also discussed and leads to the topic of double Fourier series. We conclude this sec- tion with a discussion of the existence and uniqueness of solutions to the heat flow problem.

In the model of heat flow in a uniform wire, let’s replace the assumption that the ends of the wire are kept at a constant temperature zero and instead assume that the ends of the wire are insulated—that is, no heat flows out (or in) at the ends of the wire. It follows from the principle of heat conduction (see Section 10.1) that the temperature gradient must be zero at these end- points, that is,

0u 0x A0, tB ! 0u0x AL, tB ! 0 , t 7 0 .

f AxB. f AxB ! aq

n!1 cn sin

npx L

f AxB:cn u Ax, tB ! aq

n!1 cne

#b Anp/LB 2t sin npx L

u Ax, 0B ! f AxB , 0 6 x 6 Lu A0, tB ! u AL, tB ! 0 , t 7 0 , 0u 0t

! b 02u 0x2

, 0 6 x 6 L , t 7 0 ,

u Ax, tB

Solution

600 Chapter 10 Partial Differential Equations

Example 1

In the next example we obtain the formal solution to the heat flow problem with these bound- ary conditions.

Find a formal solution to the heat flow problem governed by the initial-boundary value problem

(6)

(7)

(8)

Using the method of separation of variables, we first assume that

Substituting into equation (6) and separating variables, as was done in Section 10.2 [compare equation (5) on page 570], we get the two equations

(9)

(10)

where l is some constant. The boundary conditions in (7) become

and

For these equations to hold for all either which implies that or

(11)

Combining the boundary conditions in (11) with equation (9) gives the boundary value problem

(12)

where l can be any constant. To solve for the nontrivial solutions to (12), we begin with the auxiliary equation

When arguments similar to those used in Section 10.2 show that there are no nontrivial solutions to (12).

When the auxiliary equation has the repeated root 0 and a general solution to the differential equation is

The boundary conditions in (12) reduce to B ! 0 with A arbitrary. Thus, for the nontrivial solutions to (12) are of the form

where is an arbitrary nonzero constant. When the auxiliary equation has the roots Thus, a general solution to

the differential equation in (12) is

X AxB ! C1 cos1lx " C2 sin1lx . r ! $i1l.l 7 0,c0 ! AX AxB ! c0 ,

l ! 0,

X AxB ! A " Bx . l ! 0,

l 6 0,r 2 " l ! 0.

X– AxB " lX AxB ! 0 ; X¿ A0B ! X¿ ALB ! 0 , X¿ A0B ! X¿ ALB ! 0 . u Ax, tB ! 0,T AtB ! 0,t 7 0,

X¿ ALBT AtB ! 0 .X¿ A0BT AtB ! 0 T ¿ AtB " blT AtB ! 0 ,X– AxB " lX AxB ! 0 , u Ax, tB ! X AxBT AtB .

u Ax, 0B ! f AxB , 0 6 x 6 L . 0u 0x A0, tB ! 0u0x AL, tB ! 0 , t 7 0 ,

0u 0t

! b 02u 0x2

, 0 6 x 6 L , t 7 0 ,

Section 10.5 The Heat Equation 601

The boundary conditions in (12) lead to the system

Hence, and the system reduces to solving Since only when where n is an integer, we obtain a nontrivial solution only when

or , Furthermore, the nontrivial solutions (eigen- functions) corresponding to the eigenvalue are given by

(13)

where the ’s are arbitrary nonzero constants. In fact, formula (13) also holds for n ! 0, since l ! 0 has the eigenfunctions

Note that these eigenvalues and eigenfunctions (see Figure 10.15) have the same properties as those on page 571 (which we now recognize as the components of the sine series): homogeneous boundary conditions, eigenvalues clustering at infinity, increasingly oscillatory, orthogonal.

Having determined that let’s consider equation (10) for such

T ¿ AtB " b Anp/LB2T AtB ! 0 .l: l ! Anp/LB2, n ! 0, 1, 2, . . . , X0 AxB ! c0.cn

Xn AxB ! cn cos npxL , l ! Anp/LB2Xn n ! 1, 2, 3, . . . .l ! Anp/LB21l ! np/L1lL ! np,

sin1lL ! 0C1 sin1lL ! 0.C2 ! 0#1lC1 sin1lL " 1lC2 cos1lL ! 0 . 1lC2 ! 0 ,

Figure 10.15 Cosine eigenfunctions

Eigenvalue Eigenfunction

L 2

216

For n ! 0, 1, 2, . . . , the general solution is

where the ’s are arbitrary constants. Combining this with equation (13), we obtain the functions

where is, again, an arbitrary constant. If we take an infinite series of these functions, we obtain

(14)

which will be a solution to (6)–(7) provided the series has the proper convergence behavior. Notice that in (14) we have altered the constant term and written it as thus producing the stan- dard form for cosine expansions.

a0/2,

u Ax, tB ! a0 2

" a q

n!1 ane

#b Anp/LB2t cos npx L

an ! bncn

un Ax, tB ! ane#b Anp/LB2t cos npxL , un Ax, tB ! Xn AxBTn AtB ! c cn cos npxL d 3bne#b Anp/LB 2t 4 ,

Tn AtB ! bne#b Anp/LB2t ,

Assuming a solution to (6)–(7) is given by the series in (14) and substituting into the initial condition (8), we get

(15) u Ax, 0B ! a0 2

" a q

n!1 an cos

npx L

! f AxB , 0 6 x 6 L .

(a) n = 0

u u

(b) n = 1

Figure 10.16 Modes for equation (14)

602 Chapter 10 Partial Differential Equations

This means that if we choose the ’s as the coefficients in the Fourier cosine series for f,

then given in (14) will be a formal solution to the heat flow problem (6)–(8). Again, if this expansion converges to a continuous function with continuous second partial derivatives, then the formal solution is an actual solution. ◆

Note that the individual terms in the series (4) and (14) are genuine solutions to the heat equation and the associated boundary conditions (but not the initial condition, of course). These solutions are called the modes of the system. Observe that the time factor decays faster for the more oscillatory (higher n) modes. This makes sense physically; a temper- ature pattern of many alternating hot and cold strips arranged closely together will equilibrate faster than a more uniform pattern. See Figure 10.16.

e#b Anp/LB2t

u Ax, tBan ! 2 L

# L

0 f AxB cos npx

L dx , n ! 0, 1, 2, . . . ,

A good way of expressing how separation of variables “works” is as follows: The initial temperature profile is decomposed into modes through Fourier analysis—equations (5) or (15). Then each mode is matched with a time decay factor—equations (4) or (14)—and it evolves in time accordingly.

When the ends of the wire are kept at 0°C or when the ends are insulated, the boundary conditions are said to be homogeneous. But, when the ends of the wire are kept at constant temperatures different from zero, that is,

f AxB

Section 10.5 The Heat Equation 603

(16) and

then the boundary conditions are called nonhomogeneous. From our experience with vibration problems in Section 4.9 we expect that the solution to

the heat flow problem with nonhomogeneous boundary conditions will consist of a steady- state solution that satisfies the nonhomogeneous boundary conditions in (16) plus a transient solution That is,

where and its partial derivatives tend to zero as The function will then satisfy homogeneous boundary conditions, as illustrated in the next example.

Find a formal solution to the heat flow problem governed by the initial-boundary value problem

(17)

(18) (19)

Let’s assume that the solution consists of a steady-state solution and a transient solution —that is,

(20)

Substituting for in equations (17)–(19) leads to

(21)

(22)

(23)

If we allow in (21)–(22), assuming that is a transient solution, we obtain the steady-state boundary value problem

Solving for we obtain and choosing A and B so that the boundary condi- tions are satisfied yields

(24)

as the steady-state solution.

With this choice for the initial-boundary value problem (21)–(23) reduces to the fol- lowing initial-boundary value problem for

(25)

(26)

(27) w Ax, 0B ! f AxB # U1 # AU2 # U1BxL , 0 6 x 6 L . w A0, tB ! w AL, tB ! 0 , t 7 0 ,

0w 0t

! b 02w 0x2

, 0 6 x 6 L , t 7 0 ,

w Ax, tB:y AxB, y AxB ! U1 " AU2 # U1BxL

y AxB ! Ax " B,y,y A0B ! U1 , y ALB ! U2 . y– AxB ! 0 , 0 6 x 6 L ,

w Ax, tBt S qy AxB " w Ax, 0B ! f AxB , 0 6 x 6 L . y A0B " w A0, tB ! U1 , y ALB " w AL, tB ! U2 , t 7 0 ,

0u 0t

! 0w 0t

! by– AxB " b 02w 0x2

, 0 6 x 6 L , t 7 0 ,

u Ax, tBu Ax, tB ! y AxB " w Ax, tB . w Ax, tB y AxBu Ax, tB u Ax, 0B ! f AxB , 0 6 x 6 L .u A0, tB ! U1 , u AL, tB ! U2 , t 7 0 ,

0u 0t

! b 02u 0x2

, 0 6 x 6 L , t 7 0 ,

w Ax, tBt S q.w Ax, tBu Ax, tB ! Y AxB $ w Ax, tB , w Ax, tB.y AxB

u AL, tB ! U2 , t 7 0 ,u A0, tB ! U1

Solution

Example 2

604 Chapter 10 Partial Differential Equations

Recall that a formal solution to (25)–(27) is given by equation (4). Hence,

where the ’s are the coefficients of the Fourier sine series expansion

Therefore, the formal solution to (17)–(19) is

(28)

with

◆

In the next example, we consider the heat flow problem when a heat source is present but is independent of time. (Recall the derivation in Section 10.1.)

Find a formal solution to the heat flow problem governed by the initial-boundary value problem

(29)

(30)

(31)

We begin by assuming that the solution consists of a steady-state solution and a transient solution , namely,

where and its partial derivatives tend to zero as Substituting for in (29)–(31) yields

(32)

(33)

(34)

Letting in (32)–(33), we obtain the steady-state boundary value problem

y A0B ! U1 , y ALB ! U2 .y– AxB ! # 1 b

P AxB , 0 6 x 6 L ,t S q y AxB " w Ax, 0B ! f AxB , 0 6 x 6 L .y A0B " w A0, tB ! U1 , y ALB " w AL, tB ! U2 , t 7 0 ,

0u 0t

! 0w 0t

! by– AxB " b 02w 0x2

" P AxB , 0 6 x 6 L , t 7 0 , u Ax, tBt S q.w Ax, tBu Ax, tB ! y AxB " w Ax, tB ,

w Ax, tB y AxB u Ax, 0B ! f AxB , 0 6 x 6 L .u A0, tB ! U1 , u AL, tB ! U2 , t 7 0 ,

0u 0t

! b 02u 0x2

" P AxB , 0 6 x 6 L , t 7 0 ,

cn ! 2 L #

0 c f AxB # U1 # AU2 # U1BxL d sin npxL dx .

u Ax, tB ! U1 " AU2 # U1BxL " aqn!1 cne#b Anp/LB2t sin npxL , f AxB # U1 # AU2 # U1BxL ! aqn!1 cn sin npxL . cn

w Ax, tB ! aq n!1

cne #b Anp/LB2t sin npx

L ,

Example 3

Solution

Section 10.5 The Heat Equation 605

The solution to this boundary value problem can be obtained by two integrations using the boundary conditions to determine the constants of integration. The reader can verify that the solution is given by the formula

(35)

With this choice for we find that the initial-boundary value problem (32)–(34) reduces to the following initial-boundary value problem for

(36)

(37) (38)

where is given by formula (35). As before, the solution to this initial-boundary value prob- lem is

(39)

where the ’s are determined from the Fourier sine series expansion of

(40)

Thus the formal solution to (29)–(31) is given by

where is given in (35) and is prescribed by (39)–(40). ◆

The method of separation of variables is also applicable to problems in higher dimensions. For example, consider the problem of heat flow in a rectangular plate with sides , ,

, and If the two sides are kept at a constant temperature of 0°C and the two sides x ! 0, x ! L are perfectly insulated, then heat flow is governed by the initial- boundary value problem in the following example (see Figure 10.17).

y ! 0, y ! Wy ! W.y ! 0 x ! Lx ! 0

w Ax, tBy AxBu Ax, tB ! y AxB " w Ax, tB , f AxB # y AxB ! aq

n!1 cn sin

npx L

f AxB # y AxB:cnw Ax, tB ! aq

n!1 cne

#b Anp/LB2t sin npx L

y AxB w Ax, 0B ! f AxB # y AxB , 0 6 x 6 L , w A0, tB ! w AL, tB ! 0 , t 7 0 ,

0w 0t

! b 02w 0x2

, 0 6 x 6 L , t 7 0 ,

w Ax, tB:y AxB, y AxB ! cU2 # U1 " # L

0 a# z

0 1 b

P AsB dsb dz d x L

" U1 # # x

0 a# z

0 1 b

P AsB dsbdz .

x L

0°

= 0 = 0

Figure 10.17 Plate with insulated sides

Solution

606 Chapter 10 Partial Differential Equations

Example 4 Find a formal solution to the initial-boundary value problem

(41)

(42)

(43)

(44)

If we assume a solution of the form then equation (41) separates into the two equations

(45)

(46)

where l can be any constant. To solve equation (46), we again use separation of variables. Here we assume This allows us to separate equation (46) into the two equations

(47) (48)

where m can be any constant (see Problem 29 in Exercises 10.2). To solve for we observe that the boundary conditions in (42), in terms of the sepa-

rated variables, become

Hence, in order to get a nontrivial solution, we must have

(49)

The boundary value problem for X given in equations (47) and (49) was solved in Example 1 [compare equations (12) and (13)]. Here and

where the ’s are arbitrary. To solve for we first observe that the boundary conditions in (43) become

(50)

Next, substituting into equation (48) yields

which we can rewrite as

(51)

where The boundary value problem for Y consisting of (50)–(51) has also been solved before. In Section 10.2 [compare equations (7) and (9), pages 570 and 571] we

E ! l # Amp/LB2.Y– AyB " EY AyB ! 0 , Y– AyB " Al # Amp/LB2BY AyB ! 0 ,m ! Amp/LB2 Y A0B ! Y AWB ! 0 .Y AyB,

Xm AxB ! cm cos mpxL , m ! Amp/LB2, m ! 0, 1, 2, . . . ,

X¿ A0B ! X¿ ALB ! 0 . X¿ A0BY AyBT AtB ! X¿ ALBY AyBT AtB ! 0 ; 0 6 y 6 W , t 7 0 .

X AxB, Y– AyB " Al # mBY AyB ! 0 ,X– AxB " mX AxB ! 0 ,

V Ax, yB ! X AxBY AyB. 02V 0x2 Ax, yB " 02V

0y2 Ax, yB " lV Ax, yB ! 0 ,T ¿ AtB " blT AtB ! 0 ,

u Ax, y, tB ! V Ax, yBT AtB,u Ax, y, 0B ! f Ax, yB , 0 6 x 6 L , 0 6 y 6 W . u Ax, 0, tB ! u Ax, W, tB ! 0 , 0 6 x 6 L , t 7 0 ,

0u 0x A0, y, tB ! 0u0x AL, y, tB ! 0 , 0 6 y 6 W , t 7 0 ,

0u 0t

! b e 02u 0x2

" 02u 0y2 f , 0 6 x 6 L , 0 6 y 6 W , t 7 0 ,u Ax, y, tB

Section 10.5 The Heat Equation 607

showed that and the nontrivial solutions are given by

where the ’s are arbitrary. Since we have

Substituting l into equation (45), we can solve for and obtain

Substituting in for and we get

where are arbitrary constants. If we now take a doubly infinite series of such functions, then we obtain the formal series

(52)

We are now ready to apply the initial conditions (44). Setting t ! 0, we obtain

(53)

This is a double Fourier series.† The formulas for the coefficients are obtained by exploit- ing the orthogonality conditions twice. Presuming (53) is valid and permits term-by-term integration, we multiply each side by cos and integrate over x and y:

According to the orthogonality conditions, each integral on the right is zero, except when m ! p and n ! q; thus,

Hence,

(54) a0q ! 2

LW # L

0 #

0 f Ax, yB sin qpy

W dy dx , q ! 1, 2, 3, . . . ,

! apq # L

0 cos2

ppx L

dx # W

0 sin2

qpy W

dy ! $ LW

4 apq , p % 0 ,

2 apq , p ! 0 .

# L

0 #

0 f Ax, yB cos ppx

L sin

qpy W

dy dx

! a q

m!0 a q

n!1 amn #

0 #

0 cos

mpx L

sin npy W

cos ppx

L sin

qpy W

dy dx .

# L

0 #

0 f Ax, yB cos ppx

L sin

qpy W

dydx

Appx/LB sin Aqpy/ WB amn

u Ax, y, 0B ! f Ax, yB ! aq m!0

a q

n!1 amn cos

mpx L

sin npy W

u Ax, y, tB ! aq m!0

a q

n!1 amne

#Am2/L2"n2/W2Bbp2t cos mpx L

sin npy W

amn J anbmncm Am ! 0, 1, 2, . . . , n ! 1, 2, 3, . . .B umn Ax, y, tB ! amne#Am2/L2"n2/W 2Bbp2t cos mpxL sin npyW , umn Ax, y, tB ! acm cos mpxL b aan sin npyW b Abmne#Am2/L2"n2 /W 2Bbp2tB ,

Tmn,Xm, Yn,

Tmn AtB ! bmne#Am2/L2"n2/W 2Bbp2t . T AtB l ! Anp/WB2 " Amp/LB2 ; m ! 0, 1, 2, . . . , n ! 1, 2, 3, . . . .l ! E " Amp/LB

2, an

Yn AyB ! an sin npyW , E ! Anp/WB2, n ! 1, 2, 3, . . . ,

†For a discussion of double Fourier series, see Partial Differential Equations of Mathematical Physics, 2nd ed. by Tyn Myint-U (Elsevier North Holland, New York, 1983), Sec. 5.14.

608 Chapter 10 Partial Differential Equations

and for

(55)

Finally, the solution to the initial-boundary value problem (41)–(44) is given by equation (52), where the coefficients are prescribed by equations (54) and (55). ◆

We derived formula (55) under the assumption that the double series (53) truly con- verged (uniformly). How can we justify the assumption that a double Fourier series converges? A rough argument goes as follows. If we fix y in , then presumably has a convergent cosine series in x: . But the coefficients dm depend on the value of the y that we fixed: So presumably each function has a convergent sine series in y: . Assembling all this, we get

Existence and Uniqueness of Solutions In the examples that we have studied in this section and in Section 10.2, we were able to obtain formal solutions in the sense that we could express the solution in terms of a series expansion consisting of exponentials, sines, and cosines. To prove that these series converge to actual solutions requires results on the convergence of Fourier series and results from real analysis on uniform convergence. We will not go into these details here, but refer the reader to Section 6.5 of the text by Tyn Myint-U (see footnote on page 607) for a proof of the existence of a solution to the heat flow problem discussed in Sections 10.1 and 10.2. (A proof of uniqueness is also given there.)

As might be expected, by using Fourier series and the method of separation of variables one can also obtain “solutions” when the initial data are discontinuous, since the formal solu- tions require only the existence of a convergent Fourier series. This allows one to study ideal- ized problems in which the initial conditions do not agree with the boundary conditions or the initial conditions involve a jump discontinuity. For example, we may assume that initially one half of the wire is at one temperature, whereas the other half is at a different temperature, that is,

The formal solution that we obtain will make sense for but near the points of discontinuity and L, we have to expect behavior like that exhibited in Figure 10.7 of Section 10.3, page 584.

The question of the uniqueness of the solution to the heat flow problem can be answered in various ways. One is tempted to argue that the method of separation of variables yields for- mulas for the solutions and therefore a unique solution. However, this does not exclude the possibility of solutions existing that cannot be obtained by the method of separation of variables.

From physical considerations, we know that the peak temperature along a wire does not increase spontaneously if there are no heat sources to drive it up. Indeed, if hot objects could draw heat from colder objects without external intervention, we would have violations of the second law of thermodynamics. Thus no point on an unheated wire will ever reach a higher

x ! 0, L/2, 0 6 x 6 L, t 7 0,

f AxB ! eU1 , 0 6 x 6 L/2 , U2 , L/2 6 x 6 L .

cos Ampx/LB sin Anpy/WBƒ Ax, yB ! aqm!0aqn!1amn dm AyB ! aqn!1amnsin Anpy/WB dm AyBdm ! dm AyB.ƒ Ax, yB ! aqm!0dmcos Ampx/LB

ƒ Ax, yBƒ Ax, yB

apq ! 4

LW # L

0 #

0 f Ax, yB cos ppx

L sin

qpy W

dy dx .

p ( 1, q ( 1,

Section 10.5 The Heat Equation 609

†For a discussion of maximum principles and their applications, see Maximum Principles in Differential Equations, by M. H. Protter and H. F. Weinberger (Springer-Verlag, New York, 1984).

We can use the maximum principle to show that the heat flow problem has a unique solution.

Maximum Principle for the Heat Equation

Theorem 6. Let be a continuously differentiable function that satisfies the heat equation

(56)

and the boundary conditions

(57)

Then attains its maximum value at t ! 0, for some x in —that is,

max 0'x'L

t(0 u Ax, tB ! max

0'x'L u Ax, 0B . 30, L 4u Ax, tB

u A0, tB ! u AL, tB ! 0 , t 7 0 . 0u 0t

! b 02u 0x2

, 0 6 x 6 L , t 7 0 ,

u Ax, tB

Proof. Assume and are continuously differentiable functions that satisfy the initial-boundary value problem (58)–(60). Let Now w is a continuously differ- entiable solution to the boundary value problem (56)–(57). By the maximum principle, w must attain its maximum at t ! 0, and since

we have Hence, for all A similar argument using yields Therefore we have for all

Thus, there is at most one continuously differentiable solution to the problem (58)–(60). ◆

t ( 0.0 ' x ' L, u Ax, tB ! y Ax, tBy Ax, tB ' u Ax, tB.ŵ ! y # u 0 ' x ' L, t ( 0.u Ax, tB ' y Ax, tBw Ax, tB ' 0.

w Ax, 0B ! u Ax, 0B # y Ax, 0B ! f AxB # f AxB ! 0 , w ! u # y.

y Ax, tBu Ax, tB

Uniqueness of Solution

Theorem 7. The initial-boundary value problem

(58)

(59) (60)

has at most one continuously differentiable solution.

u Ax, 0B ! f AxB , 0 6 x 6 L ,u A0, tB ! u AL, tB ! 0 , t 7 0 , 0u 0t

! b 02u 0x2

, 0 6 x 6 L , t 7 0 ,

temperature than the initial peak temperature. Such statements are called maximum princi- ples and can be proved mathematically for the heat equation and Laplace’s equation. One such result is the following.†

610 Chapter 10 Partial Differential Equations

In Problems 1–10, find a formal solution to the given ini- tial-boundary value problem.

u Ax, 0B ! sin 2x , 0 6 x 6 pu A0, tB ! u Ap, tB ! 0 , t 7 0 , 0u 0t

! 02u 0x2

" e#x , 0 6 x 6 p , t 7 0 ,

u Ax, 0B ! 0 , 0 6 x 6 pu A0, tB ! 0 , u Ap, tB ! 3p , t 7 0 , 0u 0t

! 02u 0x2

, 0 6 x 6 p , t 7 0 ,

u Ax, 0B ! sin 3x # sin 5x , 0 6 x 6 pu A0, tB ! 5 , u Ap, tB ! 10 , t 7 0 , 0u 0t

! 2 02u 0x2

, 0 6 x 6 p , t 7 0 ,

u Ax, 0B ! 1 # sin x , 0 6 x 6 p 0u 0x A0, tB ! 0u0x Ap, tB ! 0 , t 7 0 , 0u 0t

! 7 02u 0x2

, 0 6 x 6 p , t 7 0 ,

u Ax, 0B ! ex , 0 6 x 6 p 0u 0x A0, tB ! 0u0x Ap, tB ! 0 , t 7 0 , 0u 0t

! 02u 0x2

, 0 6 x 6 p , t 7 0 ,

u Ax, 0B ! x A1 # xB , 0 6 x 6 1 0u 0x A0, tB ! 0u0x A1, tB ! 0 , t 7 0 , 0u 0t

! 2 02u 0x2

, 0 6 x 6 1 , t 7 0 ,

u Ax, 0B ! x , 0 6 x 6 p 0u 0x A0, tB ! 0u0x Ap, tB ! 0 , t 7 0 , 0u 0t

! 3 02u 0x2

, 0 6 x 6 p , t 7 0 ,

u Ax, 0B ! x2 , 0 6 x 6 pu A0, tB ! u Ap, tB ! 0 , t 7 0 , 0u 0t

! 02u 0x2

, 0 6 x 6 p , t 7 0 ,

u Ax, 0B ! A1 # xBx2 , 0 6 x 6 1u A0, tB ! u A1, tB ! 0 , t 7 0 , 0u 0t

! 5 02u 0x2

, 0 6 x 6 1 , t 7 0 ,

10.

11. Find a formal solution to the initial-boundary value problem

12. Find a formal solution to the initial-boundary value problem

13. Find a formal solution to the initial-boundary value problem

14. Find a formal solution to the initial-boundary value problem

In Problems 15–18, find a formal solution to the initial- boundary value problem.

for the given function

15. f Ax, yB ! cos 6x sin 4y # 3 cos x sin 11y f Ax, yB. u Ax, y, 0B ! f Ax, yB , 0 6 x 6 p , 0 6 y 6 p ,

u Ax, 0, tB ! u Ax, p, tB ! 0 , 0 6 x 6 p , t 7 0 , 0u 0x A0, y, tB ! 0u0x Ap, y, tB ! 0 , 0 6 y 6 p , t 7 0 ,

t 7 0 ,0u 0t

! 02u 0x2

" 02u 0y2

, 0 6 x 6 p , 0 6 y 6 p ,

u Ax, 0B ! 1 , 0 6 x 6 p .u A0, tB ! u Ap, tB ! 1 , t 7 0 , 0u 0t

! 3 02u 0x2

" 5 , 0 6 x 6 p , t 7 0 ,

u Ax, 0B ! sin x , 0 6 x 6 p .u A0, tB ! u Ap, tB ! 0 , t 7 0 , 0u 0t

! 2 02u 0x2

" 4x , 0 6 x 6 p , t 7 0 ,

u Ax, 0B ! f AxB , 0 6 x 6 p . t 7 0 ,u A0, tB ! 0 , u Ap, tB " 0u 0x Ap, tB ! 0 ,

0u 0t

! 02u 0x2

, 0 6 x 6 p , t 7 0 ,

u Ax, 0B ! f AxB , 0 6 x 6 p . 0u 0x A0, tB ! 0 , u Ap, tB ! 0 , t 7 0 , 0u 0t

! 4 02u 0x2

, 0 6 x 6 p , t 7 0 ,

u Ax, 0B ! sin x , 0 6 x 6 pu A0, tB ! u Ap, tB ! 0 , t 7 0 , 0u 0t

! 3 02u 0x2

" x , 0 6 x 6 p , t 7 0 ,

10.5 EXERCISES

Section 10.6 The Wave Equation 611

16. #

17. 18.

19. Chemical Diffusion. Chemical diffusion through a thin layer is governed by the equation

where is the concentration in moles/cm3, the diffusivity k is a positive constant with units cm2/sec,

C Ax, tB 0C 0t

! k 02C 0x2

# LC ,

f Ax, yB ! x sin y f Ax, yB ! y 3 cos 3x sin 4y

f Ax, yB ! cos x sin y " 4 cos 2x sin y and is a consumption rate with units sec Assume the boundary conditions are

and the initial concentration is given by

Use the method of separation of variables to solve for- mally for the concentration What happens to the concentration as ?t S "q

C Ax, tB. C Ax, 0B ! f AxB , 0 6 x 6 a . C A0, tB ! C Aa, tB ! 0 , t 7 0 ,

#1.L 7 0

In Section 10.2 we presented a model for the motion of a vibrating string. If represents the displacement (deflection) of the string and the ends of the string are held fixed, then the motion of the string is governed by the initial-boundary value problem

(1)

(2) (3)

(4)

Equation (1) is called the wave equation. The constant appearing in (1) is strictly positive and depends on the linear density and

tension of the string. The boundary conditions in (2) reflect the fact that the string is held fixed at the two endpoints x ! 0 and x ! L.

Equations (3) and (4) specify, respectively, the initial displacement and the initial velocity of each point on the string. For the initial and boundary conditions to be consistent, we assume

and Using the method of separation of variables, we found in Section 10.2 that a formal solu-

tion to (1)–(4) is given by [compare equations (24)–(26) on page 575]

(5)

where the ’s and ’s are determined from the Fourier sine series

(6)

(7)

See Figure 10.18 on page 612 for a sketch of partial sums for (5).

g AxB ! aq n!1

bn anpaL b sin npxL . f AxB ! aq

n!1 an sin

npx L

bnan

u Ax, tB ! aq n!1

can cos npaL t " bn sin npaL t d sin npxL , g A0B ! g ALB ! 0.f A0B ! f ALB ! 0

0u 0t Ax, 0B ! g AxB , 0 6 x 6 L . u Ax, 0B ! f AxB , 0 6 x 6 L , u A0, tB ! u AL, tB ! 0 , t 7 0 ,

02u 0t2

! a2 02u 0x2

, 0 6 x 6 L , t 7 0 ,

u Ax, tBTHE WAVE EQUATION 10.6

0 L

n = 1

Node

n = 2

0 L 0 L

Nodes

n = 3

Figure 10.19 Standing waves. Time-varying amplitudes are shown by dashed curves

612 Chapter 10 Partial Differential Equations

Each term (or mode) in expansion (5) can be viewed as a standing wave (a wave that vibrates in place without lateral motion along the string). For example, the first term,

consists of a sinusoidal shape function multiplied by a time-varying amplitude. The second term is also a sinusoid with a time-varying amplitude. In the latter case, there is a node in the middle at that never moves. For the nth term, we have a sinusoid

with a time-varying amplitude and nodes. This is illustrated in Figure 10.19. Thus, separation of variables decomposes the initial data into sinusoids or modes [equations (6)–(7)], assigns each mode a frequency at which to vibrate [equation (5)], and represents the solution as the superposition of infinitely many standing waves. Modes with more nodes vibrate at higher frequencies. By choosing the initial configuration of the string to have the same shape as one of the individual terms in the solution, we can activate only that mode.

An # 1Bsin Anpx/LB x ! L/2 sin A2px/LBsin Apx/LB

aa1 cos paL t " b1 sin paL tb sin pxL ,

The different modes of vibration of a guitar string are distinguishable to the human ear by the frequency of the sound they generate; different frequencies are discerned as different “pitches.” The fundamental is the frequency of the lowest mode, and integer multiples of the fundamental frequency are called “harmonics.” A cello and a trombone sound different even if they are playing the same fundamental, say F#, because of the difference in the intensities of the harmonics they generate.

tx u =

Figure 10.18 Partial sums for equation (5) with f AxB ! e x, 0 6 x ' p/2, p # x, p/2 ' x 6 p, and g AxB ! 0

Section 10.6 The Wave Equation 613

Many engineering devices operate better in some modes than in others. For example, cer- tain modes propagate down a waveguide or an optical fiber with less attenuation than the others. In such cases, engineers design the systems to suppress the undesirable modes, either by shap- ing the initial excitation or by introducing devices (like resistor cards in a waveguide) which damp out certain modes preferentially.

The fundamental mode on a guitar string can be suppressed by holding the finger lightly in contact with the midpoint, thereby creating a stationary point or node there. According to equa- tion (5), the pitch of the note then doubles, thus producing an “octave.” This style of fingering, called “playing harmonics,” is used frequently in musical performance.

An important characteristic of an engineering device is the set of angular frequencies supported by its eigenmodes—its “eigenfrequencies.” According to equation (5), the eigenfre- quencies of the vibrating string are the harmonics These blend together pleasantly to the human ear, and we enjoy melodies played on string instruments. The eigenfrequencies of a drum are not harmonics (see Problem 21). Therefore, drums are used for rhythm, not for melody.

As we have seen in the preceding section, the method of separation of variables can be used to solve problems with nonhomogeneous boundary conditions and nonhomogeneous equations where the forcing term is time independent. In the next example, we will consider a problem with a time-dependent forcing term

For given functions f, g, and h, find a formal solution to the initial-boundary value problem

(8)

(9) (10)

(11)

The boundary conditions in (9) certainly require that the solution be zero for x ! 0 and x ! L. Motivated by the fact that the solution to the corresponding homogeneous system (1)–(4) consists of a superposition of standing waves, let’s try to find a solution to (8)–(11) of the form

(12)

where the ’s are functions of t to be determined. For each fixed t, we can compute a Fourier sine series for If we assume that the

series is convergent to then

(13)

where the coefficient is given by [recall equation (7) on page 596]

If the series in (13) has the proper convergence properties, then we can substitute (12) and (13) into equation (8) and obtain

a q

n!1 cu–n AtB " anpaL b 2un AtB d sin npxL ! aqn!1 hn AtB sin npxL .

hn AtB ! 2L # L0 h Ax, tB sin npxL dx , n ! 1, 2, . . . . hn AtB

h Ax, tB ! aq n!1

hn AtB sin npxL , h Ax, tB, h Ax, tB.un

AtBu Ax, tB ! a q

n!1 un AtB sin npxL ,

0u 0t Ax, 0B ! g AxB , 0 6 x 6 L .u Ax, 0B ! f AxB , 0 6 x 6 L ,u A0, tB ! u AL, tB ! 0 , t 7 0 ,

02u 0t2

! a2 02u 0x2

$ h Ax, tB , 0 6 x 6 L , t 7 0 , h Ax, tB.

vn ! npa/L, n ! 1, 2, 3, . . . .

Example 1

Solution

614 Chapter 10 Partial Differential Equations

Equating the coefficients in each series (why?), we have

This is a nonhomogeneous, constant-coefficient equation that can be solved using variation of parameters. You should verify that

[compare Problem 20 of Exercises 4.6]. Hence, with this choice of the series in (12) is a formal solution to the partial differential equation (8).

Since†

and

substituting (12) into the initial conditions (10)–(11) yields

(14)

(15)

Thus, if we choose the ’s and ’s so that equations (14) and (15) are satisfied, a formal solu- tion to (8)–(11) is given by

(16)

◆

The method of separation of variables can also be used to solve initial-boundary value problems for the wave equation in higher dimensions. For example, a vibrating rectangular membrane of length L and width W (see Figure 10.20 on page 615) is governed by the following initial-boundary value problem for

(17)

(18) (19) (20)

(21) 0u0t Ax, y, 0B ! g Ax, yB , 0 6 x 6 L , 0 6 y 6 W . u Ax, y, 0B ! f Ax, yB , 0 6 x 6 L , 0 6 y 6 W ,u Ax, 0, tB ! u Ax, W, tB ! 0 , 0 6 x 6 L , t 7 0 , u A0, y, tB ! u AL, y, tB ! 0 , 0 6 y 6 W , t 7 0 ,

02u 0t2

! a2 a02u 0x2

" 02u 0y2 b , 0 6 x 6 L , 0 6 y 6 W , t 7 0 ,u Ax, y, tB:

" L

npa # t

0 hn AsB sin c npaL At # sB dds f sin npxL .

u Ax, tB ! aq n!1 ean cos npaL t " bn sin npaL t

bnan

0u 0t Ax, 0B ! g AxB ! aqn!1 bn anpaL b sin npxL . u Ax, 0B ! f AxB ! aq

n!1 an sin

npx L

u¿n A0B ! bn anpaL b ,un A0B ! an un AtB,

un AtB ! an cos npaL t " bn sin npaL t " Lnpa # t0 hn AsB sin c npaL At # sB d ds u–n AtB " anpaL b 2un AtB ! hn AtB .

†To compute we use the fact that d dt #

0 G As, tBds ! G At, tB " # t

0G 0t As, tB ds.u¿n A0B,

Section 10.6 The Wave Equation 615

Using an argument similar to the one given for the problem of heat flow in a rectangular plate (Example 4 in Section 10.5), we find that the initial-boundary value problem (17)–(21) has a formal solution

(22)

where the constants and are determined from the double Fourier series

In particular,

(23)

(24)

We leave the derivation of this solution as an exercise (see Problem 19).

We mentioned earlier that the solution to the vibrating string problem (1)–(4) consisted of a superposition of standing waves. There are also “traveling waves” associated with the wave equation. Traveling waves arise naturally out of d’Alembert’s solution to the wave equation for an “infinite” string.

To obtain d’Alembert’s solution to the wave equation

02u 0t2

! a2 02u 0x2

bmn ! 4

LWpaBm2L2 " n2W2 # L0 #W0 g Ax, yB sin mpxL sin npyW dy dx . amn !

4 LW #

0 #

0 f Ax, yB sin mpx

L sin

npy W

dydx ,

g Ax, yB ! aq m!1

a q

n!1 apBm2L2 " n2W2 bmn sin mpxL sin npyW .

f Ax, yB ! aq m!1

a q

n!1 amn sin

mpx L

sin npy W

bmnamn

" bmn sin °Bm2L2 " n2W2 apt¢ * sin mpxL sin npyW , u Ax, y, tB ! aq

m!1 a q

n!1 $amn cos °Bm2L2 " n2W2 apt¢

u = 0

u = 0 u = 0

= + ((

Figure 10.20 Vibrating membrane

Solution

we use the change of variables

If u has continuous second partial derivatives, then and from which we obtain

Substituting these expressions into the wave equation and simplifying yields

We can solve this equation directly by first integrating with respect to to obtain

where is an arbitrary function of and then integrating with respect to to find

where and are arbitrary functions. Substituting the original variables x and t gives d’Alembert’s solution

(25)

It is easy to check by direct substitution that defined by formula (25), is indeed a solu- tion to the wave equation, provided A and B are twice-differentiable functions.

Using d’Alembert’s formula (25), find a solution to the initial value problem

(26)

(27)

(28)

A solution to (26) is given by formula (25), so we need only choose the functions A and B so that the initial conditions (27)–(28) are satisfied. For this we need

(29)

(30)

Integrating equation (30) from to x and dividing by gives

(31) A AxB # B AxB ! 1 a #

g AsB ds " C , aAx0 arbitraryBx0 0u 0t Ax, 0B ! aA¿ AxB # aB¿ AxB ! g AxB . u Ax, 0B ! A AxB " B AxB ! f AxB ,

0u 0t Ax, 0B ! gAxB #q 6 x 6 q . u Ax, 0B ! f AxB #q 6 x 6 q ,

02u 0t2

! a2 02u 0x2

#q 6 x 6 q , t 7 0 ,

u Ax, tB,u Ax, tB ! A Ax $ AtB $ B Ax % AtB .

B AhB ! " b AhB dhA AcBu Ac, hB ! A AcB " B AhB , hh,b

AhB 0u 0h ! b AhB ,

02u 0c0h ! 0 .

02u 0t2

! a2 e 02u 0c2

# 2 02u

0c0h "

02u 0h2 f .

02u 0x2

! 02u 0c2

" 2 02u

0c0h "

02u 0h2

a A0u/ 0c # 0u/ 0hB, 0u/ 0t !0u/ 0x ! 0u/ 0c " 0u/ 0h c ! x " at , h ! x # at .

616 Chapter 10 Partial Differential Equations

Example 2

Section 10.6 The Wave Equation 617

where C is also arbitrary. Solving the system (29) and (31), we obtain

Using these functions in formula (25) gives

which simplifies to

(32) ◆

Find the solution to the initial value problem

(33)

(34)

(35)

This is just a special case of the preceding example where and Substituting into (32), we obtain the solution

(36)

◆

We now use d’Alembert’s formula to show that the solution to the “infinite” string prob- lem consists of traveling waves.

Let be a function defined on The function where is a trans- lation of the function in the sense that its “shape” is the same as but its position has been shifted to the left by an amount a. This is illustrated in Figure 10.21 for a function whose graph consists of a triangular “bump.” If we let be a parameter (say, time), then the functions

represent a family of functions with the same shape but shifted farther and farther to the left as (Figure 10.22 on page 618). We say that is a traveling wave moving to the left with speed In a similar fashion, is a traveling wave moving to the right with speed a.

h Ax # atBa. h Ax " atBt S qh Ax " atB t ( 0 h

AxBh AxB,h AxB a 7 0,h Ax " aB,A#q, q B.h AxB

! sin x cos 2t " t .

u Ax, tB ! 1 2

3 sin Ax " 2tB " sin Ax # 2tB 4 " 1 4

# x"2t

x#2t ds

g AxB ! 1.a ! 2, f AxB ! sin x, 0u 0t Ax, 0B ! 1 , #q 6 x 6 q . u Ax, 0B ! sin x , #q 6 x 6 q ,

02u 0t2

! 4 02u 0x2

, #q 6 x 6 q , t 7 0 ,

u Ax, tB ! 1 2

3 f Ax " atB " f Ax # atB 4 " 1 2a

# x"at

x#at g AsBds .

u Ax, tB ! 1 2

3 f Ax " atB " f Ax # atB 4 " 1 2a

c # x"at x0

g AsBds # # x#at x0

g AsB ds d , B AxB ! 1

2 f AxB # 1

2a #

g AsBds # C 2

A AxB ! 1 2

f AxB " 1 2a

# x

g AsBds " C 2

Example 3

Solution

x0 − a x1 − a

h(x + a) h(x)

x0 x1 x

Figure 10.21 Graphs of and h Ax " aBh AxB

Example 4

Solution

618 Chapter 10 Partial Differential Equations

If we refer to formula (25), we find that the solution to consists of traveling waves moving to the left with speed and moving to the right at the same speed.

In the special case when the initial velocity we have

Hence, is the sum of the traveling waves

and

These waves are initially superimposed, since

As t increases, the two waves move away from each other with speed This is illustrated in Figure 10.23 for a triangular wave.

2a.

u Ax, 0B ! 1 2

f AxB " 1 2

f AxB ! f AxB . 1 2

f Ax # atB .1 2

f Ax " atBu Ax, tBu Ax, tB !

1 2

3 f Ax " atB " f Ax # atB 4 . g AxB ! 0, B Ax # atBaA Ax " atB 02u/ 0t2 ! a202u/ 0x2

x 0

h(x)

Figure 10.22 Traveling wave

t = 0 0

1t =

0 1t =

0 2t =

Figure 10.23 Decomposition of initial displacement into traveling waves

Express the standing wave as a superposition of traveling waves.

By a familiar trigonometric identity,

where

This standing wave results from adding a wave traveling to the right to the same waveform traveling to the left. ◆

h AxB ! 1 2

sin px L

! h Ax # atB " h Ax " atB , cos a paL tb sin apxL b ! 12 sin pL Ax # atB " 12 sin pL Ax " atB cos apa

L tb sin apx

L b

Section 10.6 The Wave Equation 619

Existence and Uniqueness of Solutions In Example 1 the method of separation of variables was used to derive a formal solution to the given initial-boundary value problem. To show that these series converge to an actual solution requires results from real analysis, just as was the case for the formal solutions to the heat equa- tion in Section 10.5. In Examples 2 and 3, we can establish the validity of d’Alembert’s solu- tion by direct substitution into the initial value problem, assuming sufficient differentiability of the initial functions. We leave it as an exercise for you to show that if f has a continuous second derivative and g has a continuous first derivative, then d’Alembert’s solution is a true solution (see Problem 12).

The question of the uniqueness of the solution to the initial-boundary value problem (1)–(4) can be answered using an energy argument.

Uniqueness of the Solution to the Vibrating String Problem

Theorem 8. The initial-boundary value problem

(37)

(38) (39)

(40)

has at most one twice continuously differentiable solution.

0u 0t Ax, 0B ! g AxB , 0 6 x 6 L , u Ax, 0B ! f AxB , 0 6 x 6 L ,u A0, tB ! u AL, tB ! 0 , t 7 0 ,

02u 0t2

! a2 02u 0x2

, 0 6 x 6 L , t 7 0 ,

Proof. Assume both and are twice continuously differentiable solutions to (37)–(40) and let It is easy to check that satisfies the initial- boundary value problem (37)–(40) with zero initial data; that is, for

(41) and

We now show that for If is the displacement of the vibrating string at location x for time t, then with the

appropriate units, the total energy of the vibrating string at time t is defined by the integral

(42)

(The first term in the integrand relates to the stretching of the string at location x and represents the potential energy. The second term is the square of the velocity of the vibrating string at x and represents the kinetic energy.)

We now consider the derivative of

dE dt

! d dt e 1

2 # L

0 ca2 a0w0xb 2 " a0w0t b 2 d dx f .

E AtB:

E AtB J 1 2 #

0 ca2 a0w0xb 2 " a0w0t b 2 d dx .

E AtBw Ax, tB 0 ' x ' L, t ( 0.w Ax, tB ! 0

0w 0t Ax, 0B ! 0 .w Ax, 0B ! 0

0 ' x ' L, w Ax, tBw Ax, tB J u Ax, tB # y Ax, tB.y Ax, tBu Ax, tB

620 Chapter 10 Partial Differential Equations

Since w has continuous second partial derivatives (because u and do), we can interchange the order of integration and differentiation. This gives

(43)

Again the continuity of the second partials of w guarantees that the mixed partials are equal; that is,

Combining this fact with integration by parts, we obtain

(44)

The boundary conditions , imply that ! ! 0, This reduces equation (44) to

Substituting this in for the first integrand in (43), we find

Since w satisfies equation (37), the integrand is zero for all x. Thus, and so where C is a constant. This means that the total energy is conserved within the

vibrating string. The first boundary condition in (41) states that for Hence,

for Combining this with the second boundary condition in (41), we find that when t ! 0, the integrand in (42) is zero for Therefore,

Since we must have C ! 0. Hence,

(45)

That is, the total energy of this system is zero. Because the integrand in (45) is nonnegative and continuous and the integral is zero, the

integrand must be zero for Moreover, the integrand is the sum of two squares and so each term must be zero. Hence,

and 0w 0t Ax, tB ! 00w0x Ax, tB ! 0

0 ' x ' L.

E AtB ! 1 2

# L

0 ca2 a0w0xb 2 " a0w0t b 2 d dx ! 0 .

E AtB ! C,E A0B ! 0. 0 6 x 6 L.0 6 x 6 L. A0w/ 0xB Ax, 0B ! 0 0 ' x ' L.w Ax, 0B ! 0 E AtB ! C, dE/dt ! 0,

dE dt

! # L

0 0w 0t

c 02w 0t2

# a2 02w 0x2 d dx .

# L

0 a2

0w 0x

02w 0t0x

dx ! ## L

0 a2

0w 0t

02w 0x2

dx .

t ( 0.A0w/ 0tB AL, tB A0w/ 0tB A0, tBt ( 0,w A0, tB ! w AL, tB ! 0 ! # #

0 a2

02w 0x2

0w 0t

dx .

! a2 0w 0x AL, tB 0w0t AL, tB # a2 0w0x A0, tB 0w0t A0, tB

# L

0 a2

0w 0x

02w 0t0x dx ! #

0 a2

0w 0x

02w 0x0t dx

02w 0t0x !

02w 0x0t .

dE dt

! # L

0 ca2 0w

02w 0t0x

" 0w 0t

02w 0t2 d dx .

Section 10.6 The Wave Equation 621

for all Thus where K is a constant. Physically, this says that there is no motion in the string.

Finally, since w is constant and w is zero when t ! 0, then Consequently, and the initial-boundary value problem has at most one solution. ◆u Ax, tB ! y Ax, tB w Ax, tB ! 0.

w Ax, tB ! K,0 ' x ' L, t ( 0.

In Problems 1–4, find a formal solution to the vibrating string problem governed by the given initial-boundary value problem.

5. The Plucked String. A vibrating string is gov- erned by the initial-boundary value problem (1)–(4). If the string is lifted to a height at and released, then the initial conditions are

and Find a formal solution.g AxB ! 0. f AxB ! e h0x/a , 0 6 x ' a ,

h0 AL # xB / AL # aB , a 6 x 6 L , x ! ah0

0u 0t Ax, 0B ! e x , 0 6 x 6 p /2 ,p # x , p /2 6 x 6 p u Ax, 0B ! sin 4x " 7 sin 5x , 0 6 x 6 p ,u A0, tB ! u Ap, tB ! 0 , t 7 0 , 02u 0t2

! 9 02u 0x2

, 0 6 x 6 p , t 7 0 ,

0u 0t Ax, 0B ! 0 , 0 6 x 6 p u Ax, 0B ! x2 Ap # xB , 0 6 x 6 p ,u A0, tB ! u Ap, tB ! 0 , t 7 0 , 02u 0t2

! 4 02u 0x2

, 0 6 x 6 p , t 7 0 ,

0u 0t Ax, 0B ! 1 # cos x , 0 6 x 6 p u Ax, 0B ! sin2x , 0 6 x 6 p ,u A0, tB ! u Ap, tB ! 0 , t 7 0 , 02u 0t2

! 16 02u 0x2

, 0 6 x 6 p , t 7 0 ,

0u 0t Ax, 0B ! sin 7px , 0 6 x 6 1 u Ax, 0B ! x A1 # xB , 0 6 x 6 1 ,u A0, tB ! u A1, tB ! 0 , t 7 0 , 02u 0t2

! 02u 0x2

, 0 6 x 6 1 , t 7 0 ,

6. The Struck String. A vibrating string is governed by the initial-boundary value problem (1)–(4). If the string is struck at x ! a, then the initial conditions may be approximated by and

where is a constant. Find a formal solution.

In Problems 7 and 8, find a formal solution to the vibrat- ing string problem governed by the given nonhomoge- neous initial-boundary value problem.

9. If one end of a string is held fixed while the other is free, then the motion of the string is governed by the initial-boundary value problem

Derive a formula for a formal solution.

0u 0t Ax, 0B ! g AxB , 0 6 x 6 L . u Ax, 0B ! f AxB , 0 6 x 6 L , u A0, tB ! 0 and

0u 0x AL, tB ! 0 t 7 0 ,

02u 0t2

! a2 02u 0x2

, 0 6 x 6 L , t 7 0 ,

0u 0t Ax, 0B ! 0 , 0 6 x 6 p u Ax, 0B ! 0 , 0 6 x 6 p ,u A0, tB ! u Ap, tB ! 0 , t 7 0 , 02u 0t2

! 02u 0x2

" x sin t , 0 6 x 6 p , t 7 0 ,

0u 0t Ax, 0B ! 5 sin 2x # 3 sin 5x , 0 6 x 6 p u Ax, 0B ! sin x , 0 6 x 6 p ,u A0, tB ! u Ap, tB ! 0 , t 7 0 , 02u 0t2

! 02u 0x2

" tx , 0 6 x 6 p , t 7 0 ,

g AxB ! ey0x/a , 0 6 x ' a , y0 AL # xB / AL # aB a 6 x 6 L ,

f AxB ! 0 10.6 EXERCISES

622 Chapter 10 Partial Differential Equations

10. Derive a formula for the solution to the following initial-boundary value problem involving nonhomo- geneous boundary conditions

where and are constants. 11. The Telegraph Problem.† Use the method of sep-

aration of variables to derive a formal solution to the telegraph problem

12. Verify d’Alembert’s solution (32) to the initial value problem (26)–(28) when has a continuous sec- ond derivative and has a continuous first deriva- tive by substituting it directly into the equations.

In Problems 13–18, find the solution to the initial value problem.

for the given functions 13. 14. 15. 16. 17. 18.

19. Derive the formal solution given in equations (22)–(24) to the vibrating membrane problem governed by the initial-boundary value problem (17)–(21).

f AxB ! cos 2x , g AxB ! 1 # xf AxB ! e#x2 , g AxB ! sin x f AxB ! sin 3x , g AxB ! 1f AxB ! x , g AxB ! x f AxB ! x2 , g AxB ! 0f AxB ! 0 , g AxB ! cos x

f AxB and g AxB. 0u 0t Ax, 0B ! g AxB , #q 6 x 6 q , u Ax, 0B ! f AxB , #q 6 x 6 q ,

02u 0t2

! a2 02u 0x2

, #q 6 x 6 q , t 7 0 ,

g AxB f AxB 0u 0t Ax, 0B ! 0 , 0 6 x 6 L . u Ax, 0B ! f AxB , 0 6 x 6 L ,u A0, tB ! u AL, tB ! 0 , t 7 0 ,

02u 0t2

" 0u 0t

" u ! a2 02u 0x2

, 0 6 x 6 L , t 7 0 ,

U2U1

0u 0t Ax, 0B ! g AxB , 0 6 x 6 L , u Ax, 0B ! f AxB , 0 6 x 6 L ,u A0, tB ! U1 , u AL, tB ! U2 , t 7 0 ,

02u 0t2

! a2 02u 0x2

, 0 6 x 6 L , t 7 0 ,

20. Long Water Waves. The motion of long water waves in a channel of constant depth is governed by the linearized Korteweg and de Vries (KdV) equation

(46)

where is the displacement of the water from its equilibrium depth at location x and at time t, and

and are positive constants. (a) Show that equation (46) has a solution of the

form

(47)

where k is a fixed constant and is a function of k, provided V satisfies

(48)

These solutions, defined by (47), are called uni- form waves.

(b) Physically, we are interested only in solutions that are bounded and nonconstant on the

infinite interval Show that such solu- tions exist only if

, where A and B are arbitrary constants. [Hint: Solve for w in terms of and k and use (47).]

(d) Since both and k can be chosen arbitrarily and they always appear together as the product we can set without loss of generality. Hence, we have

as a uniform wave solution to (46). The defin- ing relation is called the dispersion relation, the ratio is called the phase velocity, and the derivative

is called the group velocity. When the group velocity is not constant, the waves are called dispersive. Show that the stan- dard wave equation has only nondis- persive waves.

21. Vibrating Drum. A vibrating circular membrane of unit radius whose edges are held fixed in a plane

utt ! a 2uxx

dw/dk ! a # 3bk2 w AkB /k ! a # bk2w AkB ! ak # bk3

V Ax, tB ! A sin 3 kx # Aak # bk3Bt " B 4l ! 1 lk,

V Ax, tB ! A sin 3lkx # Aalk # bl3k3Bt " B 4l 2 ! Aak # wB /(bk3B.ak # w 7 0.

A#q, q B.V AzB

#w dV dz

" ak dV dz

" bk3 d3V dz3

! 0 .

w AkBz ! kx # w AkBt , u Ax, tB ! V AzB ,

u Ax, tBut $ Aux $ Buxxx ! 0 ,

†For a discussion of the telegraph problem, see Methods of Mathematical Physics, by R. Courant and D. Hilbert, Vol. 2 (Wiley-VCH, Weinheim, Germany, 2004).

Section 10.7 Laplace’s Equation 623

and whose displacement depends only on the radial distance r from the center and on the time t is governed by the initial-boundary value problem.

0u 0t Ar, 0B ! g ArB , 0 6 r 6 1 , u Ar, 0B ! f ArB , 0 6 r 6 1 ,u Ar, tB remains finite as r S 0" , u A1, tB ! 0 , t 7 0 , 0 6 r 6 1 , t 7 0 ,

02u 0t2

! a2 a02u 0r2

" 1 r

0u 0r b ,

u Ar, tB where f and g are the initial displacements and velocities, respectively. Use the method of separa- tion of variables to derive a formal solution to the vibrating drum problem. [Hint: Show that there is a family of solutions of the form

where is the Bessel function of the first kind of order zero and are the positive zeros of Now use superposition.] See Figure 10.24.

J0. 0 6 k1 6 k2 6 p 6 kn 6 p

un Ar, tB ! 3an cos AknatB " bn sin AknatB 4 J0 AknrB ,

In Section 10.1 we showed how Laplace’s equation,

arises in the study of steady-state or time-independent solutions to the heat equation. Because these solutions do not depend on time, initial conditions are irrelevant and only boundary con- ditions are specified. Other applications include the static displacement of a stretched membrane fastened in space along the boundary of a region (here u must satisfy Laplace’s equation inside the region); the electrostatic and gravitational potentials in certain force fields (here u must satisfy Laplace’s equation in any region that is free of electrical charges or mass); and, in fluid mechanics for an idealized fluid, the stream function whose level curves (stream lines), represent the path of particles in the fluid (again u satisfies Laplace’s equation in the flow region).

There are two basic types of boundary conditions that are usually associated with Laplace’s equation: Dirichlet boundary conditions, where the solution to Laplace’s equation in a domain D is required to satisfy

with a specified function defined on the boundary of D; and Neumann boundary conditions, where the directional derivative along the outward normal to the boundary is required to satisfy

on 0D ,0u0n Ax, yB ! g Ax, yB 0u/ 0n

0Df Ax, yB 0D ,u Ax, yB ! f Ax, yB u Ax, yB

u Ax, yB ! constant, u Ax, yB u Ax, yB

¢u ! 0 2u

0x2 "

02u 0y2

! 0 ,

10.7 LAPLACE’S EQUATION

(a) (b) (c)

Figure 10.24 Mode shapes for vibrating drum. (a) (b) (c) J0 A8.654rBJ0 A5.520rB,J0 A2.405rB,

624 Chapter 10 Partial Differential Equations

with a specified function defined on † We say that the boundary conditions are mixed if the solution is required to satisfy on part of the boundary and

on the remaining portion of the boundary. In this section we use the method of separation of variables to find solutions to Laplace’s

equation with various boundary conditions for rectangular, circular, and cylindrical domains. We also discuss the existence and uniqueness of such solutions.

Find a solution to the following mixed boundary value problem for a rectangle (see Figure 10.25):

(1)

(2)

(3)

(4) u Ax, 0B ! f AxB , 0 ' x ' a .u Ax, bB ! 0 , 0 ' x ' a , 0u 0x A0, yB ! 0u0x Aa, yB ! 0 , 0 ' y ' b , 02u 0x2

" 02u 0y2

! 0 , 0 6 x 6 a , 0 6 y 6 b ,

A0u/ 0nB Ax, yB ! g Ax, yB u Ax, yB ! f Ax, yB 0D.g Ax, yB

Example 1

Solution

u(x, b ) = 0

au(x, 0) = f(x)

(0, y) = 0 *u *x

*u *x

(a, y) = 0

Figure 10.25 Mixed boundary value problem

†These generalize the boundary conditions (5) in Section 10.1 and (42) in Section 10.5.

Separating variables, we first let Substituting into equation (1), we have

which separates into

where l is some constant. This leads to the two ordinary differential equations

(5)

(6)

From the boundary condition (2), we observe that

(7) X¿ A0B ! X¿ AaB ! 0 . Y– AyB # lY AyB ! 0 .X– AxB " lX AxB ! 0 , X– AxB X AxB ! # Y– AyBY AyB ! #l ,

X– AxBY AyB " X AxBY– AyB ! 0 ,u Ax, yB ! X AxBY AyB.

Section 10.7 Laplace’s Equation 625

We have encountered the eigenvalue problem in (5) and (7) before (see Example 1 in Section 10.5). The eigenvalues are with corresponding solutions

(8)

where the ’s are arbitrary constants. Setting in equation (6) and solving for Y gives†

(9)

Now the boundary condition in (3) will be satisfied if Setting y ! b in (9), we see that we want and

or equivalently,

(see Problem 18). Thus we find that there are solutions to (1)–(3) of the form

where the ’s are constants. In fact, by the superposition principle,

(10)

is a formal solution to (1)–(3). Applying the remaining nonhomogeneous boundary condition in (4), we have

u Ax, 0B ! f AxB ! #E0b " aq n!1

En sinh a# npba b cos anpxa b . u Ax, yB ! E0 Ay # bB " aq

n!1 En cos anpxa b sinh c npa Ay # bB d

! En cos anpxa b sinh c npa A y # bB d , n ! 1, 2, . . . , un Ax, yB ! Xn AxBYn AyB ! an cos anpxa b Cn sinh c npa Ay # bB d u0 Ax, yB ! X0 AxBY0 AyB ! a0B0 Ay # bB ! E0 Ay # bB , Yn AyB ! Cn sinh c npa Ay # bB d , n ! 1, 2, . . .

An ! #Bn tanh anpba b , A0 ! #bB0

Y AbB ! 0.u Ax, bB ! 0 Yn AyB ! An cosh anpya b " Bn sinh anpya b , n ! 1, 2, . . . Y0 Ay B ! A0 " B0y , l ! ln ! Anp/aB2an

Xn AxB ! an cos anpxa b , l ! ln ! Anp/aB2, n ! 0, 1, 2, . . . ,

†We usually express However, computation is simplified in this case by using the hyper- bolic functions cosh and sinh z ! Aez # e#zB /2 .z ! Aez " e#zB /2Yn AyB ! anenpy/a " bne#npy/a.

626 Chapter 10 Partial Differential Equations

This is a Fourier cosine series for and hence the coefficients are given by the formulas

(11)

Thus a formal solution is given by (10) with the constants given by (11). See Figure 10.26 for a sketch of partial sums for (10). ◆

En ! 2

a sinh a# npb a b # a0 f AxB cos anpxa b dx , n ! 1, 2, . . . .

E0 ! 1A#baB # a0 f AxB dx ,

f AxB

Figure 10.26 Partial sums for Example 1 with f AxB ! e x, 0 6 x ' p/2, p # x, p/2 ' x 6 p

In Example 1 the boundary conditions were homogeneous on three sides of the rectangle and nonhomogeneous on the fourth side, It is important to note that the method used in Example 1 can also be used to solve problems for which the boundary conditions are nonhomogeneous on all sides. This is accomplished by solving four separate boundary value problems in which three sides have homogeneous boundary conditions and only one side is nonhomogeneous. The solution is then obtained by summing these four solu- tions (see Problem 5).

For problems involving circular domains, it is usually more convenient to use polar coor- dinates. In rectangular coordinates the Laplacian has the form

In polar coordinates we let

so that

r ! 2x2 " y2 , tan u ! y/x . x ! r cos u , y ! r sin u

Ar, uB, ¢u ! 0

2u 0x2

" 02u 0y2

E Ax, yB: y ! 0, 0 ' x ' aF.

With patience and a little care in applying the chain rule, one can show that the Laplacian in polar coordinates is

(12)

(see Problem 6). In the next example we obtain a solution to the Dirichlet problem in a disk of radius a.

A circular metal disk of radius a has its top and bottom insulated. The edge of the disk is kept at a specified temperature that depends on its location (varies with ). The steady-state temperature inside the disk satisfies Laplace’s equation. Determine the temperature distribution

inside the disk by finding the solution to the following Dirichlet boundary value prob- lem, depicted in Figure 10.27:

(13)

(14) u Aa, uB ! f AuB , #p ' u ' p . 02u 0r2

" 1 r

0u 0r

" 1

r2 02u 0u2

! 0 , 0 ' r 6 a , #p ' u ' p ,

u Ar, uB u Ar ! aB

(u ! " 2u

"r2 $

1 r

"u "r

$ 1 r2

"2u "U2

Section 10.7 Laplace’s Equation 627

Example 2

(r, )

u(a, ) = f( )

Figure 10.27 Steady-state temperature distribution in a disk

Solution To use the method of separation of variables, we first set

where and Substituting into (13) and separating variables give

where is any constant. This leads to the two ordinary differential equations

(15) (16) T – AuB " lT AuB ! 0 .r 2R– ArB " rR¿ ArB # lR ArB ! 0 ,

r 2R– ArB " rR¿ ArB R ArB ! # T – AuBT AuB ! l ,

#p ' u ' p.0 ' r 6 a

u Ar, uB ! R ArBT AuB ,

628 Chapter 10 Partial Differential Equations

For to be continuous in the disk we need to be -periodic; in par- ticular, we require

(17) and

Therefore, we seek nontrivial solutions to the eigenvalue problem (16)–(17). When the general solution to (16) is the sum of two exponentials. Hence we have

only trivial -periodic solutions. When we find to be the solution to (16). This linear function is

periodic only when that is, is the only -periodic solution corresponding to

When the general solution to (16) is

Here we get a nontrivial -periodic solution only when [You can check this using (17).] Hence, we obtain the nontrivial -periodic solutions

(18)

corresponding to Notice that the eigenfunctions are the terms of the Fourier series discussed in Section 10.3 (recall Figure 10.5, page 580).

Now for equation (15) is the Cauchy–Euler equation

(19)

(see Section 4.7, page 194). When n ! 0, the general solution is

Since ln as this solution is unbounded near r ! 0 when Therefore, we must choose D ! 0 if is to be continuous at r ! 0. We now have and so

which for convenience we write in the form

(20)

where is an arbitrary constant. When you should verify that equation (19) has the general solution

Since as we must set in order for to be bounded at r ! 0. Thus,

Now for each n ! 1, 2, . . . , we have the solutions

(21)

and by forming an infinite series from the solutions in (20) and (21), we get the following for- mal solution to (13):

It is more convenient to write this series in the equivalent form

(22) u Ar, uB ! a0 2

" a q

n!1 ar a b n Aan cos nu " bn sin nuB ,

u Ar, uB ! A0 2

" a q

n!1 Cnr

n AAn cos nu " Bn sin nuB . un Ar, uB ! Rn ArBTn AuB ! Cnr n AAn cos nu " Bn sin nuB , Rn ArB ! Cnr n . u Ar, uBDn ! 0r S 0",r #n S q Rn ArB ! Cnr n " Dnr #n .l ! n

2, n ! 1, 2, . . . , A0

u0 Ar, uB ! A02 , u0 Ar, uB ! R0 ArBT AuB ! CB, R0 ArB ! Cu Ar, uB

D % 0.r S 0",r S #q

R0 ArB ! C " D ln r . r 2R– ArB " rR¿ ArB # n2R ArB ! 0l ! n2, n ! 0, 1, 2, . . . ,

1l ! n, n ! 1, 2, . . . .Tn AuB ! An cos nu " Bn sin nu 2p

1l ! n, n ! 1, 2, . . . .2pT AuB ! A cos1lu " B sin1lu . l 7 0,

l ! 0. 2pT0 AuB ! BA ! 0, T AuB ! Au " Bl ! 0,

2p l 6 0,

T ¿ A#pB ! T ¿ ApB .T A#pB ! T ApB 2pT AuB0 ' r 6 a,u Ar, uB

Section 10.7 Laplace’s Equation 629

where the ’s and ’s are constants. These constants can be determined from the boundary condition, indeed, with r ! a in (22), condition (14) becomes

Hence, we recognize that are Fourier coefficients for . Thus,

(23)

(24)

To summarize, if and are defined by formulas (23) and (24), then given in (22) is a formal solution to the Dirichlet problem (13)–(14). ◆

The procedure in Example 2 can also be used to study the Neumann problem in a disk:

(25)

(26)

For this problem there is no unique solution, since if u is a solution, then the function u plus a constant is also a solution. Moreover, f must satisfy the consistency condition

(27)

If we interpret the solution of equation (25) as the steady-state temperature distribution inside a circular disk that does not contain either a heat source or heat sink, then equation (26) specifies the flow of heat across the boundary of the disk. Here the consistency condition (27) is simply the requirement that the net flow of heat across the boundary is zero. (If we keep pumping heat in, the temperature won’t reach equilibrium!) We leave the solution of the Neumann problem and the derivation of the consistency condition for the exercises.

The technique used in Example 2 also applies to annular domains, and to exterior domains, We also leave these applications as exercises.

Laplace’s equation in cylindrical coordinates arises in the study of steady-state tempera- ture distributions in a solid cylinder and in determining the electric potential inside a cylinder. In cylindrical coordinates,

Laplace’s equation becomes

(28)

The Dirichlet problem for the cylinder has the boundary conditions

(29) (30) (31) #p ' u ' p .0 ' r ' a ,u Ar, u, bB ! h Ar, uB , #p ' u ' p ,0 ' r ' a ,u Ar, u, 0B ! g Ar, uB ,

0 ' z ' b ,#p ' u ' p ,u Aa, u, zB ! f Au, zB , E Ar, u, zB: 0 ' r ' a, 0 ' z ' bF

(u ! " 2u

"r2 $

1 r

"u "r

$ 1 r2

"2u "U2

$ "2u "z2

! 0 .

x ! r cosu , y ! r sinu , z ! z ,

E Ar, uB: a 6 rF. E Ar, uB: 0 6 a 6 r 6 bF,

u Ar, uB# p

#p f AuB du ! 0 .

0u 0r Aa, uB ! f AuB , #p ' u ' p . ¢u ! 0 , 0 ' r 6 a , #p ' u ' p ,

u Ar, uBbnan bn !

1 p #

#p f AuB sin nu du , n ! 1, 2, . . . .

an ! 1 p #

#p f AuB cos nu du , n ! 0, 1, . . . ,

f AuBan, bn f AuB ! a0

2 " a

n!1 Aan cos nu " bn sin nuB .

bnan

630 Chapter 10 Partial Differential Equations

To solve the Dirichlet boundary value problem (28)–(31), we first solve the three bound- ary value problems corresponding to: (i) and (ii) and and (iii) and Then by the superposition principle, the solution to (28)–(31) will be the sum of these three solutions. This is the same method that was discussed in dealing with Dirichlet problems on rectangular domains. (See the remarks following Example 1.) In the next example we solve the Dirichlet problem when and

The base and the top of a charge-free cylinder are grounded and therefore are at zero potential. The potential on the lateral surface of the cylinder is given by

where Inside the cylinder, the potential satisfies Laplace’s equation. Determine the potential u inside the cylinder by finding a solution to the Dirichlet boundary value problem

(32)

(33) (34)

Using the method of separation of variables, we first assume that

Substituting into equation (32) and separating out the Z’s, we find

where can be any constant. Separating further the R’s and T’s gives

where can also be any constant. We now have the three ordinary differential equations:

(35) (36) (37)

For u to be continuous in the cylinder, must be periodic. Thus, let’s begin with the eigenvalue problem

and

In Example 2 we showed that this problem has nontrivial solutions for that are given by the Fourier series components

(38)

where the ’s and ’s are arbitrary constants. The boundary conditions in (34) imply that Therefore, Z must satisfy

the eigenvalue problem

Z A0B ! Z AbB ! 0 .Z– AzB " lZ AzB ! 0 , 0 6 z 6 b , Z A0B ! Z AbB ! 0.BnAn

Tn AuB ! An cos nu " Bn sin nu ,n ! 0, 1, 2, . . . , m ! n2,

T ¿ A#pB ! T ¿ ApB .T A#pB ! T ApBT –AuB " mT AuB ! 0 , #p 6 u 6 p , 2p-T AuBZ –AzB " lZ AzB ! 0 .

T –AuB " mT AuB ! 0 ,r 2R– ArB " rR¿ ArB # Ar 2l " mBR ArB ! 0 , m

r 2R– ArB " rR¿ ArB R ArB # r 2l ! # T– AuBT AuB ! m ,

R –ArB " A1/rBR ¿ArB R ArB " 1r 2 T –AuBT AuB ! # Z –AzBZ AzB ! l ,

u Ar, u, zB ! R ArBT AuBZ AzB . u Ar, u, 0B ! u Ar, u, bB ! 0 , 0 ' r 6 a , #p ' u ' p .u Aa, u, zB ! f Au, zB , #p ' u ' p , 0 ' z ' b , 02u 0r 2

" 1 r

0u 0r

" 1 r 2

02u 0u2

" 02u 0z2

! 0 , 0 ' r 6 a , #p ' u ' p , 0 6 z 6 b ,

u Ar, u, zBf Au, 0B ! f Au, bB ! 0.u Aa, u, zB ! f Au, zB Ar ! aB Az ! bBAz ! 0B h ! 0.g ! 0

g ! 0. f ! 0h ! 0;f ! 0h ! 0;g ! 0

Solution

Example 3

Section 10.7 Laplace’s Equation 631

We have seen this eigenvalue problem several times before. Nontrivial solutions exist for and are given by the sine series components

(39)

where the ’s are arbitrary constants. Substituting for and in equation (35) gives

You should verify that the change of variables transforms this equation into the modified Bessel’s equation of order n†

(40)

The modified Bessel’s equation of order n has two linearly independent solutions—the modi- fied Bessel function of the first kind:

which remains bounded near zero, and the modified Bessel function of the second kind:

which becomes unbounded as (Recall that is the gamma function discussed in Section 7.6.) A general solution to (40) has the form where C and D are constants. Since u must remain bounded near s ! 0, we must take C ! 0. Thus, the desired solutions to (40) have the form

(41)

where the ’s are arbitrary constants. If we multiply the functions in (38), (39), and (41) and then sum over m and n, we obtain

the following series solution to (32) and (34):

(42)

where the ’s and ’s are arbitrary constants.bmnamn

" a q

n!1 a q

m!1 Aamn cos nu " bmn sin nuBIn amprb b sin ampzb b ,

u Ar, u, zB ! aq m!1

am0I0 amprb b sin ampzb b Dmn

Rmn ArB ! DmnIn amprb b ; n ! 0, 1, . . . , m ! 1, 2, . . . , CKn " DIn,

+s S 0.

Kn AsB ! lim ySn

2 I#y AsB # Iy AsB

sinyp ,

In AsB ! aq k!0

As/2B2k"n k!+ Ak " n " 1B ,

s2R– AsB " sR¿ AsB # As2 " n2BR AsB ! 0 , 0 ' s 6 mpa b

s ! Ampr/bB r 2R– ArB " rR¿ ArB # ar 2 amp

b b 2 " n2b R ArB ! 0 , 0 ' r 6 a .lm

Zm AzB ! Cm sin ampzb b , l ! Amp/bB2, m ! 1, 2, 3, . . .

†The modified Bessel’s equation of order n arises in many applications and has been studied extensively. We refer the reader to the text Special Functions by E. D. Rainville (Chelsea Publishing, New York, 1972), for details about its solution.

632 Chapter 10 Partial Differential Equations

The constants in (42) can be obtained by imposing the boundary condition (33). Setting r ! a and rearranging terms, we have

(43)

Treating this double Fourier series like the one in Section 10.5 (page 607), we find

(44)

(45)

(46)

Consequently, a formal solution to (32)–(34) is given by equation (42) with the constants and determined by equations (44)–(46). ◆

Existence and Uniqueness of Solutions The existence of solutions to the boundary value problems for Laplace’s equation can be estab- lished by studying the convergence of the formal solutions that we obtained using the method of separation of variables.

To answer the question of the uniqueness of the solution to a Dirichlet boundary value problem for Laplace’s equation, recall that Laplace’s equation arises in the search for steady- state solutions to the heat equation. Just as there are maximum principles for the heat equation, there are also maximum principles for Laplace’s equation. We state one such result here.†

bmnamn

bmn ! 2 # b

0 # p

#p f Au, zB sin ampz

b b sinnududznpbIn ampab b , n ( 1 .

amn ! 2 # b

0 # p

#p f Au, zB sin ampz

b b cosnududznpbIn ampab b , n ( 1 ,

am0 ! # b

0 # p

#p f Au, zB sin ampz

b b dudznpbI0 ampab b ,

" a q

n!1 c aq

m!1 bmnIn ampab b sin ampzb b d sin nu .

" a q

n!1 c aq

m!1 amnIn ampab b sin ampzb b d cosnu

f Au, zB ! aq m!1

am0I0 ampab b sin ampzb b

†A proof can be found in Section 8.2 of the text Partial Differential Equations of Mathematical Physics, 2nd ed., by Tyn Myint-U (Elsevier North Holland, New York, 1983).

Maximum Principle for Laplace’s Equation

Theorem 9. Let be a solution to Laplace’s equation in a bounded domain D with continuous in the closure of where denotes the boundary of D.) Then attains its maximum value on 0D.u Ax, yB 0DD. A ˛˛D ! D ´ 0D,D,u Ax, yB u

Ax, yB

The uniqueness of the solution to the Dirichlet boundary value problem follows from the maximum principle. We state this result in the next theorem but leave its proof as an exercise (see Problem 19).

Section 10.7 Laplace’s Equation 633

Solutions to Laplace’s equation in two variables are called harmonic functions. These functions arise naturally in the study of analytic functions of a single complex variable. More- over, complex analysis provides many useful results about harmonic functions. For a discus- sion of this interaction, we refer the reader to an introductory text on complex analysis such as Fundamentals of Complex Analysis, 3rd ed., by E. B. Saff and A. D. Snider (Prentice Hall, Englewood Cliffs, N.J., 2003).

Uniqueness of Solution

Theorem 10. Let D be a bounded domain. If there is a continuous solution to the Dirichlet boundary value problem

then the solution is unique.

0D ,u Ax, yB ! f Ax, yBD ,¢u Ax, yB ! 0

In Problems 1–5, find a formal solution to the given boundary value problem.

u Ax, pB ! 0 , 0 ' x ' pu Ax, 0B ! sin x " sin 4x , 0 ' x ' p , u A0, yB ! u Ap, yB ! 0 , 0 ' y ' p ,0 6 y 6 p , 02u 0x2

" 02u 0y2

! 0 , 0 6 x 6 p ,

u Ax, pB ! 0 , 0 ' x ' pu Ax, 0B ! f AxB , 0 ' x ' p , u A0, yB ! u Ap, yB ! 0 , 0 ' y ' p ,0 6 y 6 p , 02u 0x2

" 02u 0y2

! 0 , 0 6 x 6 p ,

u Ax, pB ! 0 , 0 ' x ' pu Ax, 0B ! cos x # 2 cos 4x , 0 ' x ' p , 0u 0x A0, yB ! 0u0x Ap, yB ! 0 , 0 ' y ' p ,

0 6 y 6 p ,0 2u

0x2 "

02u 0y2

! 0 , 0 6 x 6 p ,

u Ax, 1B ! 0 , 0 ' x ' pu Ax, 0B ! 4 cos 6x " cos 7x , 0 ' x ' p , 0u 0x A0, yB ! 0u0x Ap, yB ! 0 , 0 ' y ' 1 ,

0 6 y 6 1 ,0 2u

0x2 "

02u 0y2

! 0 , 0 6 x 6 p ,

6. Derive the polar coordinate form of the Laplacian given in equation (12).

In Problems 7 and 8, find a solution to the Dirichlet boundary value problem for a disk:

for the given function 7. 8.

9. Find a solution to the Neumann boundary value problem for a disk:

#p ' u ' p . 0u 0r Aa, uB ! f AuB ,

#p ' u ' p , 0 ' r 6 a ,

02u 0r2

" 1 r

0u 0r

" 1

r2 02u 0u2

! 0 ,

#p ' u ' pf AuB ! cos2 u , #p ' u ' pf AuB ! 0 u 0 , f AuB.u A2, uB ! f AuB , #p ' u ' p

#p ' u ' p ,0 ' r 6 2 ,

02u 0r2

" 1 r

0u 0r

" 1

r2 02u 0u2

! 0 ,

u Ax, 1B ! cos 2x , 0 ' x ' pu Ax, 0B ! cos x # cos 3x , 0 ' x ' p , 0u 0x A0, yB ! 0u0x Ap, yB ! 0 , 0 ' y ' 1 ,

0 6 y 6 1 ,0 2u

0x2 "

02u 0y2

! 0 , 0 6 x 6 p ,

10.7 EXERCISES

634 Chapter 10 Partial Differential Equations

10. A solution to the Neumann problem (25)–(26) must also satisfy the consistency condition in (27). To show this, use Green’s second formula

where is the outward normal derivative and ds is the differential of arc length. Hint: Take and observe that

11. Find a solution to the following Dirichlet problem for an annulus:

12. Find a solution to the following Dirichlet problem for an annulus:

13. Find a solution to the following Dirichlet problem for an exterior domain:

remains bounded as

14. Find a solution to the following Neumann problem for an exterior domain:

remains bounded as

15. Find a solution to the following Dirichlet problem for a half disk:

0 6 r 6 1 , 0 6 u 6 p ,

02u 0r 2

" 1 r

0u 0r

" 1 r 2

02u 0u2

! 0 ,

r S q.u Ar, uB 0u 0r A1, uB ! f AuB , #p ' u ' p ,

1 6 r , #p ' u ' p ,

02u 0r2

" 1 r

0u 0r

" 1

r2 02u 0u2

! 0 ,

r S q.u Ar, uBu A1, uB ! f AuB #p ' u ' p , 1 6 r , #p ' u ' p ,

02u 0r 2

" 1 r

0u 0r

" 1 r 2

02u 0u2

! 0 ,

u A3, uB ! cos 3u " sin 5u , #p ' u ' p .u A1, uB ! 0 , #p ' u ' p , 1 6 r 6 3 , #p ' u ' p ,

02u 0r 2

" 1 r

0u 0r

" 1 r 2

02u 0u2

! 0 ,

u A2, uB ! sin u , #p ' u ' p .u A1, uB ! sin 4u # cos u , #p ' u ' p , #p ' u ' p ,1 6 r 6 2 ,

02u 0r2

" 1 r

0u 0r "

r2 02u 0u2

! 0 ,

0u/ 0n ! 0u/ 0r. 4 y ! 130 / 0n # #

D AY¢u # u¢YB dxdy ! #

"D aY"u

"n # u "Y "nb ds , 16. Find a solution to the following Dirichlet problem

for a half annulus:

17. Find a solution to the mixed boundary value problem

18. Show that

where

19. Prove Theorem 10 on the uniqueness of the solution to the Dirichlet problem.

20. Stability. Use the maximum principle to prove the following theorem on the continuous dependence of the solution on the boundary conditions: Theorem. Let and be continuous functions on , where D is a bounded domain. For i ! 1 and 2, let be the solution to the Dirichlet problem

in on

If the boundary values satisfy for all on

where is some constant, then for all in D .Ax, yB0 u1Ax, yB # u2 Ax, yB 0 ' ee 7 0

0D ,Ax, yB0 f1 Ax, yB # f2 Ax, yB 0 ' e 0D .u ! fi , D ,¢u ! 0 ,

ui 0D

f2f1

Cn ! Bn / cosh anpba b . ! Cn sinh c npa Ay # bB d ,

#Bn tanh anpba b cosh anpya b " Bn sinh anpya b 0u 0r A3, uB ! g AuB , #p ' u ' p . u A1, uB ! f AuB , #p ' u ' p ,1 6 r 6 3 , #p ' u ' p , 02u 0r2

" 1 r

0u 0r

" 1

r2 02u 0u2

! 0 ,

u Ap, uB ! u A2p, uB ! 0 , 0 ' u ' p .u Ar, pB ! 0 , p ' r ' 2p , u Ar, 0B ! sin r , p ' r ' 2p ,p 6 r 6 2p , 0 6 u 6 p , 02u 0r2

" 1 r

0u 0r

" 1

r2 02u 0u2

! 0 ,

u A0, uB bounded.u A1, uB ! sin 3u , 0 ' u ' p , u Ar, pB ! 0 , 0 ' r ' 1 ,u Ar, 0B ! 0 , 0 ' r ' 1 ,

Section 10.7 Laplace’s Equation 635

21. For the Dirichlet problem described in Example 3, let and assume the potential on the lat- eral side of the cylinder is Use equations (44)–(46) to compute the solution given by equation (42).

22. Invariance of Laplace’s Equation. A complex- valued function of the complex variable z ! x " iy can be written in the form

where and are real-valued functions. If is analytic in a planar region D, then its real and imaginary parts satisfy the Cauchy–Riemann equations in D; that is, in D

Let f be analytic and one-to-one in D and assume its inverse is analytic in where is the image of D under f. Then

are just the Cauchy–Riemann equations for Show that if satisfies Laplace’s equation for

in then satis- fies Laplace’s equation for in

23. Fluid Flow Around a Corner. The stream lines that describe the fluid flow around a corner (see Figure 10.28) are given by where k is a constant and the stream function, satisfies the boundary value problem

f A0, yB ! 0 , 0 ' y .f Ax, 0B ! 0 , 0 ' x , 02f 0x2

" 02f 0y2

! 0 , x 7 0 , y 7 0 ,

f, f Ax, yB ! k,

D¿.Au, yBc Au, yB J f Ax Au, yB, y Au, yBBD,Ax, yB f Ax, yB f #1.

0x 0u !

0y 0y ,

0x 0y ! #

0y 0u

D¿D¿,f #1

0u 0x !

0y 0y ,

0u 0y ! #

0y 0x .

f AzB yuu Ax, yB " iy Ax, yB, f AzB !f AzB

f Au, zB ! sin z.Ar ! pBa ! b ! p

(a) Using the results of Problem 22, show that this problem can be reduced to finding the flow above a flat plate (see Figure 10.29). That is, show that the problem reduces to finding the solution to

where and are related as follows: with the mapping

between and given by the analytic function

(b) Verify that a nonconstant solution to the prob- lem in part (a) is given by

24. Unbounded Domain. Using separation of vari- ables, find a solution of Laplace’s equation in the infinite rectangle that is zero on the sides and approaches zero as and equals for y ! 0.f AxBy S q, x ! p,x ! 0

0 6 x 6 p, 0 6 y 6 q

f Ax, yB c Au, yB ! y.

f AzB ! z2. Ax, yBAu, yB c Au, yB ! f Ax Au, yB, y Au, yBBfcc Au, 0B ! 0 , #q 6 u 6 q ,

y 7 0 , #q 6 u 6 q ,

02c 0u2

" 02c 0y2

! 0 ,

( x, y) = constant = 0

= 0

Figure 10.28 Flow around a corner

( u, ) = constant

= 0

Figure 10.29 Flow above a flat surface

636 Chapter 10 Partial Differential Equations

Chapter Summary

Separation of Variables. A classical technique that is effective in solving boundary value problems for partial differential equations is the method of separation of variables. Briefly, the idea is to assume first that there exists a solution that can be written with the variables sepa- rated: e.g., Substituting into the partial differential equation and then imposing the boundary conditions leads to the problem of finding the eigenvalues and eigenfunctions for a boundary value problem for an ordinary differential equation. Solving for the eigenvalues and eigenfunctions, one eventually obtains solutions ,

that solve the partial differential equation and the boundary conditions. Taking infinite linear combinations of the yields solutions to the partial differential equation and boundary conditions of the form

The coefficients are then obtained using the initial conditions or some other boundary conditions.

Fourier Series: Let be a piecewise continuous function on the interval The Fourier series of f is the trigonometric series

where the ’s and ’s are given by Euler’s formulas:

Let f be piecewise continuous on the interval The Fourier cosine series of on is

where

The Fourier sine series of on is

where

bn ! 2 L #

0 f AxB sin anpx

L b dx .

f AxB % aq n!1

bn sin anpxL b , 30, L 4f AxBan !

2 L #

0 f AxB cos anpx

L bdx .

f AxB % a0 2

" a q

n!1 an cos anpxL b ,

30, L 4 f AxB30, L 4 . bn !

1 L #

#L f AxB sin anpx

L bdx , n ! 1, 2, 3, . . . .

an ! 1 L #

#L f AxB cos anpx

L bdx , n ! 0, 1, 2, . . . ,

bnan

f AxB % a0 2

" a q

n!1 ean cos anpxL b " bn sin anpxL b f ,

3#L, L 4 .f AxBcn u Ax, tB ! aq

n!1 cnun Ax, tB .

un Ax, tBn ! 1, 2, 3, . . . un Ax, tB ! Xn AxBTn AtB

u ! XTu Ax, tB ! X AxBT AtB.

Technical Writing Exercises 637

Fourier series and the method of separation of variables are used to solve boundary value problems and initial-boundary value problems for the three classical equations:

Heat equation

Wave equation

Laplace’s equation

For the heat equation, there is a unique solution when the boundary values at the ends of a conducting wire are specified along with the initial temperature distribution. The wave equation yields the displacement of a vibrating string when the ends are held fixed and the initial displacement and initial velocity at each point of the string are given. In such a case, separation of variables yields a solution that is the sum of standing waves. For an infinite string with specified initial displacement and velocity, d’Alembert’s solution yields traveling waves. Laplace’s equation arises in the study of steady-state solutions to the heat equation where either Dirichlet or Neumann boundary conditions are specified.

u Ax, tB u Ax, tB

02u 0x2

" 02u 0y2

! 0 .

02u 0t2

! a2 02u 0x2

0u 0t

! b 02u 0x2

1. The method of separation of variables is an important technique in solving initial-boundary value problems and boundary value problems for linear partial differ- ential equations. Explain where the linearity of the differential equation plays a crucial role in the method of separation of variables.

2. In applying the method of separation of variables, we have encountered a variety of special functions, such as sines, cosines, Bessel functions, and modified Bessel functions. Describe three or four examples of partial differential equations that involve other special func- tions, such as Legendre polynomials, Hermite polyno- mials, and Laguerre polynomials. (Some exploring in the library may be needed.)

3. A constant-coefficient second-order partial differen- tial equation of the form

can be classified using the discriminant In particular, the equation is called

hyperbolic if and elliptic if

Verify that the wave equation is hyperbolic and Laplace’s equation is elliptic. It can be shown that such hyperbolic (elliptic) equations can be transformed by a linear change of variables into the wave (Laplace’s) equation. Based on your knowledge of the latter equa- tions, describe which types of problems (initial value, boundary value, etc.) are appropriate for hyperbolic equations and elliptic equations.

D 6 0. D 7 0,

b2 # 4ac. D J

a 02u 0x2

" b 02u

0x0y " c

02u 0y2

! 0

TECHNICAL WRITING EXERCISES

638

Steady-State Temperature Distribution in a Circular Cylinder

When the temperature u inside a circular cylinder reaches a steady state, it satisfies Laplace’s equation If the temperature on the lateral surface is kept at zero, the temperature on the top is kept at zero, and the temperature on the bottom is given by

then the steady-state temperature satisfies the boundary value problem

where for (see Figure 10.30). To find a solution to this boundary value problem, proceed as follows:

(a) Let Show that R, T, and Z must satisfy the three ordinary dif- ferential equations

(b) Show that has the form

for (c) Show that has the form

for where b 7 0.l ! #b2, Z AzB ! C sinh 3b Ab # zB 4Z AzBm ! n

2, n ! 0, 1, 2, . . . .

T AuB ! A cos nu " B sin nuT AuB Z – " lZ ! 0 . T – " mT ! 0 ,

r 2R– " rR¿ # Ar 2l " mBR ! 0 , u Ar, u, zB ! R ArBT AuBZ AzB.

#p ' u ' pf Aa, uB ! 0u Ar, u, 0B ! f Ar, uB , 0 ' r 6 a , #p ' u ' p , u Ar, u, bB ! 0 , 0 ' r 6 a , #p ' u ' p ,u Aa, u, zB ! 0 , #p ' u ' p , 0 ' z ' b , 02u 0r 2

" 1 r

0u 0r

" 1 r 2

02u 0u2

" 02u 0z2

! 0 , 0 ' r 6 a , #p ' u ' p , 0 6 z 6 b ,

u Ar, u, 0B ! f Ar, uB, Az ! 0BAz ! bB Ar ! aB¢u ! 0.

z u ( r, , b) = 0

u ( r, , 0 ) = f ( r, )

u ! 0

u ( a, , z) = 0

Figure 10.30 Dirichlet problem for cylinder

Group Projects for Chapter 10

Group Projects for Chapter 10 639

(d) Show that has the form, for each n,

where is the Bessel function of the first kind.

(e) Show that the boundary conditions require and so

Hence, for each n, if are the zeros of then

Moreover,

Some typical eigenfunctions are displayed in Figure 10.31.

(f) Use the preceding results to show that has the form

where and are constants.

(g) Use the final boundary condition and a double orthogonal expansion involving Bessel

bnmanm

u Ar, u, zB ! aq n!0

a q

m!1 Jn aanmra b Aanm cos nu " bnm sin nuB sinh aanm Ab # zBa b ,

u Ar, u, zB Rn ArB ! DJn Aanmr/aB . bnm ! anm/a .

Jn,0 6 an1 6 an2 6 p 6 anm 6 p

Jn AbaB ! 0 . R AaB ! 0, Jn

Rn ArB ! DJn AbrB ,R ArB

Eigenvalue Eigenfunction

a r

J0(2.405r /a)

J0(5.520r /a)

J0(8.654r /a)

J0(11.792r /a)

a r

Figure 10.31 Bessel eigenfunctions

640 Chapter 10 Partial Differential Equations

functions and trigonometric functions to derive the formulas

for m ! 1, 2, 3, . . . , and for n, m ! 1, 2, 3, . . . ,

Laplace Transform Solution of the Wave Equation

Laplace transforms can be used to solve certain partial differential equations. To illustrate this technique, consider the initial-boundary value problem

(1)

(2)

(3)

(4)

(5)

This problem arises in studying a semi-infinite string that is initially horizontal and at rest and where one end is being moved vertically. Let be the solution to (1)–(5). For each x, let

(a) Using the fact that

show that satisfies the equation

(6)

(b) Show that the general solution to (6) is

where and are arbitrary functions of s.B AsBA AsBU Ax, sB ! A AsBe#sx/a " B AsBesx/a , s2U Ax, sB ! a2 02U

0x2 , 0 6 x 6 q .

U Ax, sB ! e 02u

0x2 f ! 02

0x2 ! EuF ,

U Ax, sB J ! Eu Ax, tB F Ax, sB ! # q 0

e#stu Ax, tB dt . u Ax, tB

lim xSq

u Ax, tB ! 0 , t ( 0 . 0u 0t Ax, 0B ! 0 , 0 6 x 6 q , u Ax, 0B ! 0 , 0 6 x 6 q ,u A0, tB ! h AtB , t 7 0 ,

02u 0t2

! a2 02u 0x2

, 0 6 x 6 q , t 7 0 ,

bnm ! 2

pa2 sinh Aanmb/aB 3 Jn"1 AanmB 4 2 # a0 # 2p0 f Ar, uB Jn aanmra b sin AnuB rdrdu . anm !

pa2 sinh Aanmb/aB 3 Jn"1 AanmB 4 2 # a0 # 2p0 f Ar, uB Jn aanmra b cos AnuB rdrdu , a0m !

pa2 sinh Aa0mb/aB 3 J1 Aa0mB 4 2 # a0 # 2p0 f Ar, uB J0 aa0mra b rdrdu ,

Group Projects for Chapter 10 641

(d) Using equation (2), show that

where is given in part (b). (e) Use the results of parts (b), (c), and (d) to obtain a formal solution to (1)–(5).

Green’s Function

Let be a region in the xy-plane having a smooth boundary Associated with is a Green’s function defined for pairs of distinct points in The function

has the following property. Let denote the Laplacian operator on let be a given

continuous function on and let be a given continuous function on Then, a continuous solution to the Dirichlet boundary value problem

is given by

(7)

where n is the outward normal to the boundary of and the second integral is the line integral around the boundary of with the interior of on the left as the boundary is traversed. In (7) we assume that is sufficiently smooth so that the integrands and integrals exist.

When is the upper half-plane, the Green’s function is†

(a) Using (7), show that a solution to

(8)

(9)

is given by

" 1

4p # q

0 #

#q ln c Ax # jB2 " Ay # hB2Ax # jB2 " Ay " hB2 d h Aj, hB dj dh .

u Ax, yB ! y p #

f AjBAx # jB2 " y2 dj u Ax, 0B ! f AxB , #q 6 x 6 q ,¢u Ax, yB ! h Ax, yB , #q 6 x 6 q , 0 6 y ,

G Ax, y; j, hB ! 1 4p

ln c Ax # jB2 " Ay # hB2Ax # jB2 " Ay " hB2 d . ,

0, ,,

,0,

" # 0,

f Aj, hB 0G0n Ax, y; j, hB ds Aj, hB , u Ax, yB ! ##

G Ax, y; j, hBh Aj, hB dj dh 0, ,u Ax, yB ! f Ax, yB , , ,¢u Ax, yB ! h Ax, yB ,

0,.f Ax, yB,; h Ax, yB,;¢u J 02u/ 0x2 " 02u/ 0y2 G Ax, y; j, hB ,.Ax, yB, Aj, hBG Ax, y; j, hB

,0,.,

A AsBA AsB ! H AsB ! l EhF AsB , B AsB x S q.U Ax, sB S 00 ' t 6 q,u Ax, tB S 0 as x S q

†Techniques for determining Green’s functions can be found in texts on partial differential equations such as Partial Differential Equations of Mathematical Physics, 2nd ed., by Tyn Myint-U (Elsevier North Holland, New York, 1983), Chapter 10.

642 Chapter 10 Partial Differential Equations

(b) Use the result of part (a) to determine a solution to (8)–(9) when and (c) When is the interior of the unit circle , the Green’s function, in polar

coordinates and is given by

Using (7), show that the solution to

is given by

This is known as Poisson’s integral formula. (d) Use Poisson’s integral formula to derive the following mean value property for solu-

tions to Laplace’s equation.

u Ar, uB ! 1 2p #

1 # r 2

1 " r 2 # 2r cos Af # uB f AfB df . u A1, uB ! f AuB , 0 ' u ' 2p , 02u 0r 2

" 1 r

0u 0r

" 1 r 2

02u 0u2

! 0 , 0 ' r 6 1 , 0 ' u ' 2p ,

# 1

4p ln 3 r 2 " r#2 # 2rr#1 cos Af # uB 4 # 1

4p lnr .

G Ar, u; r, fB ! 1 4p

ln 3 r 2 " r2 # 2rr cos Af # uB 4Ar, fB,Ar, uB x2 " y2 ! 1,

f ! 1.h ! 0

Mean Value Property

Theorem 11. Let satisfy in a bounded domain in and let lie in Then

(10)

for all for which the disk lies entirely in ,.

B Ax0, y0; rB ! E Ax, yB: Ax # x0B2 " Ay # y0B2 ' r 2Fr 7 0u Ax0, y0B ! 1

2p # 2p

0 u Ax0 " r cos u, y0 " r sin uB du

,. Ax0, y0BR2,¢u ! 0u Ax, yB

Hint: Use a change of variables that maps the disk to the unit disk B A0, 0; 1B. 4B Ax0, y0; rB3 Numerical Method for ¢u ! ƒ on a Rectangle

Let R denote the open rectangle

and be its boundary. Here we describe a numerical technique for solving the generalized Dirichlet problem

(11) for in

for on 0R .Ax, yBu Ax, yB ! g Ax, yB , R ,Ax, yB 02u 0x2

" 02u 0y2

! f Ax, yB , 0R

R ! E Ax, yB: a 6 x 6 b, c 6 y 6 dF D

Group Projects for Chapter 10 643

The method is similar to the finite-difference technique discussed in Chapter 4. We begin by selecting positive integers m and n and step sizes h and k so that and The interval is now partitioned into m equal subintervals and into n equal subinter- vals. The partition points are

(see Figure 10.32). The (dashed) lines and are called grid lines and their intersec- tions are the mesh points of the partition. Our goal is to obtain approximations to the solution of problem (11) at each interior mesh point, i.e., at where ,

Of course, from (11), we are given the values of at the boundary mesh points, e.g.,

The next step is to approximate the partial derivatives and using the centered-difference approximations

(12)

and

(13)

These approximations are based on the generic formula

which follows from the Taylor series expansion for

(a) Show that substituting the approximations (12) and (13) into the Dirichlet problem (11) yields the following system:

(14)

for and

ui,0 ! g Axi, cB , ui,n ! g Axi, dB , i ! 1, 2, . . . , m # 1 ,u0, j ! g Aa, yjB , um, j ! g Ab, yjB , j ! 0, 1, . . . , n , j ! 1, 2, . . . , n # 1;i ! 1, 2, . . . , m # 1,

2 c ah k b 2 " 1 d ui, j # Aui"1, j " ui#1, jB # ahkb 2 Aui, j"1 " ui, j#1B ! #h2f Axi, yjB

y AxB.y– AxB ! 3 y Ax " hB # 2y AxB " y Ax # hB 4 /h2 " 3 terms involving h2, h3, . . . 4 , Afor 1 ' i ' n # 1B .02u

0y2 Axi, yjB ) 1k2 3 u Axi, yj"1B # 2u Axi, yjB " u Axi, yj#1B 4

Afor 1 ' i ' m # 1B ,02u 0x2 Axi, yjB ) 1h2 3 u Axi"1, yjB # 2u Axi, yjB " u Axi#1, yjB 4

02u/ 0y202u/ 0x2 u Ax0, yjB ! g Ax0, yjB, 0 ' j ' n. 4 u Ax, yB31 ' j ' n # 1. 1 ' i ' m # 1Axi, yjBu Ax, yB

Axi, yjB y ! yjx ! xi yj ! c " jk , 0 ' j ' n xi ! a " ih , 0 ' i ' m ,

3 c, d 43 a, b 4 d # c ! kn.b # a ! hm

yn = d

y0 = c

h x1x0 = a xm − 1

xm = b

Figure 10.32 Rectangular grid

644 Chapter 10 Partial Differential Equations

where approximates Notice that each equation in (14) involves approxima- tions to the solution that appear in a cross centered at a mesh point (see Figure 10.33).

(b) Show that the system in part (a) is a linear system of unknowns in equations.

Compare this averaging formula with the mean value property of Theorem 11 (Project C).

(d) For the square plate

the boundary is maintained at the following temperatures:

Using the system in part (a) with and find approximations to the steady-state temperatures at the mesh points of the plate. [Hint: It is helpful to label these grid points with a single index, say choosing the ordering in a book-reading sequence.]

p1, p2, . . . , pq,

h ! k ! 0.1,f Ax, yB ! 0, m ! n ! 4,u Ax, 0B ! 0 , u Ax, 0.4B ! 10x , 0 ' x ' 0.4 . u A0, yB ! 0 , u A0.4, yB ! 10y , 0 ' y ' 0.4 . R ! E Ax, yB: 0 ' x ' 0.4, 0 ' y ' 0.4F ,

ui, j ! 1 4 Aui#1, j " ui"1, j " ui, j#1 " ui, j"1B .

f Ax, yB ! 0,Am # 1B An # 1B Am # 1B An # 1B

u Axi, yjB.ui, j

(xi , y j + 1 )

(xi − 1 , y j ) ( xi + 1 , y j )

(xi , y j −1)

(xi , y j)

Figure 10.33 Approximate solution at is obtained from values at the ends of the crossAxi, yjBui, j

Appendices

A-1

In solving differential equations, we cannot overemphasize the importance of integration tech- niques. Indeed, we sometimes refer to the problem of solving a differential equation by asking, “how can we integrate this equation?”

This section is devoted to reviewing three of the standard techniques that will be useful for integrating many of the functions encountered in this text. Of course, one can always search in a table of integrals (a brief table appears on the inside front cover of this text), but familiarity with tricks of integration is a worthwhile investment of time for several reasons: (i) we may not find the particular integral we need in the table; (ii) using one of the techniques may be quicker; and (iii) we might be curious to know how the various integrals in the table were obtained.

So what techniques can we use for the following integrals?

(a) (b) (c)

(d) (e)

For (a) we can use algebraic substitution, while (b) can be handled with trigonometric substitution. For (c) we have available integration by parts, and the integral in (d) will yield to partial fractions. But try as we may, no trick is sufficient to give (e) in a finite number of terms. By this we mean that if we have available the elementary functions (rational functions, trigono- metric functions, inverse trigonometric functions, exponential functions, and logarithmic func- tions) and permit use of the standard operations—addition, subtraction, multiplication, divi- sion, taking roots, and composite functions—a finite number of times, we cannot find a formula for the antiderivative of A similar conundrum arises with integration of

So we can add a fourth reason for honing our skills with integration techniques: They will help us to recognize those integrals that can be expressed in a finite number of terms.

We begin our review of integration with the method of substitution (also called change of variable), which is derived from the chain rule for differentiation.

Method of Substitution

The chain rule asserts that and so

!ƒ¿ Ag AxB Bg¿ AxB dx ! ƒ Ag AxB B " C . d dx

ƒ Ag AxB B ! ƒ¿ Ag AxB Bg¿(x)

21 " cos 2x. ex2.

! ex2dx! 7x2 " 10x # 1x3 " 3x2 # x # 3 dx ! x ln x dx! x229 # x2 dx! x22x " 1 dx

A REVIEW OF INTEGRATION TECHNIQUES

Thus if we can recognize an integrand as being of the form then integration is immediate via the substitution Here are some simple illustrations:

•

Since the substitution reduces the integration to that of

•

Observe that is essentially (except for a constant factor) the derivative of so we set and integrate

•

Recognizing that is the derivative of we put and integrate

•

Here it is convenient to use the identity and write the integrand as Recalling that the substitution

reduces the problem to integrating

When a portion of the integrand involves some fractional power of an expression, an algebraic substitution can sometimes simplify the integration process. The technique is amply illustrated in the following example.

Find

(1)

The difficulty here arises from the square root factor, so let’s try to rid ourselves of this problem term by making the substitution Here are the details:

(2)

If we use these expressions in (1), we find that in terms of the new variable u,

(3)

Returning to the original variable x, via we deduce that

◆

We remark that the substitution is not the only substitution that will allow us to compute the integral in (1). Another substitution that works just as well is

In a similar manner we can compute the integrals

by making the substitution in the first and in the second.u ! x1/3 " 1u ! 1 " x2

! x3A1 " x2B3/2 dx and ! ex1/3"1x2/3 dx u ! 2x " 1.

u ! 22x " 1I1 ! A2x " 1B 3/2

30 33 A2x " 1B # 5 4 " C ! A2x " 1B3/2

15 A3x # 1B " C .

u ! 22x " 1, ! u 3

30 a3u2 # 5b " C . !

1 2 au5

5 #

3 b " C! x22x " 1 dx ! ! au

2 # 1 2 b AuB Au duB ! 1

2 ! Au4 # u2B du u2 ! 2x " 1, x ! u

2 # 1 2

, dx ! u du .

u ! 22x " 1.I1 :!! x22x " 1 dx . Au2 # 1B du.u ! sec x Asec xB ¿ ! sec x tan x,Asec2 x # 1B sec x tan x. tan 2 x ! sec2 x # 1

! tan3 x sec x dx 6u du.

u ! ln Ax " 5Bln Ax " 5B,1/ Ax " 5B! 6 ln Ax " 5B

x " 5 dx

eu/20 du.u ! 5x4 5x4,x3

! x3e5x4 dx u3/2 du.

u ! sin 2xd Asin 2xB ! 2 cos 2x dx,! sin3 2x cos 2x dx u ! g AxB. ƒ¿ Ag AxB Bg¿ AxB,

A-2 Appendices

Example 1

Solution

The integration of certain algebraic expressions can sometimes be simplified by utilizing trigonometric substitutions. The effectiveness of these substitutions in handling quadratic expressions under a radical derives from the familiar identities

Accordingly, if the integrand involves:

(I) set

(II) set

(III) set

These substitutions are easy to remember if we associate with each of the cases a right tri- angle whose sides are a, x, and a suitable radical†. The labeling of the triangle depends on the particular case at hand. For example, in (II) it is clear that the hypotenuse must be while in (I) the hypotenuse must be a, and in (III) the hypotenuse is x.

Find the indefinite integral

This integral is of type (I) with We first construct a triangle as shown in Figure A.1.a ! 3.

I2 :!! x2dx29 # x2. 2a2 " x2,

x ! a sec u.2x2 # a2, x ! a tan u,2a2 " x2, x ! a sin u,2a2 # x2,

1 # sin2 u ! cos2 u, 1 " tan2 u ! sec2 u, sec2 u # 1 ! tan2 u .

Section A Review of Integration Techniques A-3

†The radical sign may not be visible. For example, if the integrand is this comes under case (II), because 1/ Aa2 " x2B5 ! 1/ A2a2 " x2B10. 1/ Aa2 " x2B5,

Figure A.1 Triangle for Example 2

√9 − x2

√9 − x2 x = 3 sin !,

= 3 cos !, dx = 3 cos ! d!.

Example 2

Solution

From this triangle we see immediately that

To integrate we use the trigonometric identity

and obtain

! 9 2 u #

9 2

sin u cos u " C .

I2 ! 9 2 ! A1 # cos 2uB du ! 92 u # 94 sin 2u " C

sin2 u ! 1 # cos 2u

sin2 u

I2 ! ! x2dx29 # x2 ! ! A3sin uB2 3 cos u du3cosu ! 9! sin2 u du .

Returning to the original variable x, we can derive the expressions for and directly from the triangle in Figure A.1. We find that

Hence

◆

Integration by Parts The method of integration by parts is of both practical and theoretical importance. It is based on the product rule for differentiation:

Integrating both sides of this equation and transposing, we obtain

(4)

Since and equation (4) can be put in the more compact form

(5)

Equation (5) is the basic equation for integration by parts. The idea is that the integrand on the left side may be very complicated but if we select the part for u and the part for dy quite carefully, the integral on the right side may be much easier. (Some experimentation may be necessary to find the right selection.)

Find the indefinite integral

There are two natural ways of selecting u and dy so that the product is x ln x dx:

First possibility: Second possibility:

With the first selection we cannot easily find y, the integral of ln x, so we come to a dead end. In the second case we can find y. We arrange the work as follows. Let

Then

Making these substitutions in (5), we have for the left side

and for the right side

(6) uy # !y du ! x22 ln x # ! x 2

2 1 x

dx ! x2

2 ln x #

1 2 ! x dx .

!u dy ! ! Aln xB Ax dxB ! ! x ln x dx , du !

1 x

dx, y ! x 2

2 .

u ! ln x, dy ! x dx .

dy ! x dx.let u ! ln x, dy ! ln x dx.let u ! x,

I3:!! x lnx dx.

!udy ! uy # !y du . du ! u¿dx,dy ! y¿dx

!uy¿dx ! uy # !yu¿dx .

d dx AuyB ! u dy

dx " y

du dx

! uy¿ " yu¿ .

I2 ! 9 2

arcsin x 3

# 9 2

x 3 29 # x2

3 " C !

9 2

arcsin x 3

# 1 2

x29 # x2 " C . u ! arcsin

x 3

, sinu ! x 3

, cosu ! 29 # x2 3

cosuu, sinu,

A-4 Appendices

Example 3

Solution

Therefore, by (5),

(7) ◆

Note that we used rather than It is clear that equation (5) is cor- rect for every value of C, and in most cases it is simpler to set C = 0. The arbitrary constant that we need can be added in the last step, as we have done in going from equation (6) to (7).

It is sometimes necessary to apply integration by parts more than once in order to evaluate a particular integral, such as

In the first application, we set to deduce that

(8)

and to evaluate the last integral in (8) we perform a second application of integration by parts with

There are some integrals that may appear at first not to be split into two parts, but we need to keep in mind that “1” can be considered as a factor of any expression. For example, to com- pute the integrals

an effective splitting for evaluating is while for an appropriate choice is The reader should verify that integration by parts then reduces these integrals to routine problems.

Partial Fractions Any rational function, that is, any function of the form where and are polynomials, can be integrated in finite terms. The first step in the process is to compare the degree of with that of If deg then division gives

(9)

where and are polynomials and Since the integration of the polyno- mial is a trivial matter, we need only focus on the integration of the remainder term

The idea of the method of partial fractions is to express as the sum of simpler fractions that are easy to integrate. For example, knowing that

(10)

can be written as

2 x # 1

" 1

x " 1 "

4 x " 3

R AxB ! 7x2 " 10x # 1 x3 " 3x2 # x # 3

p AxB /Q AxBp AxB /Q AxB. s AxB deg p 6 deg Q.p AxBs AxB

P AxB Q AxB ! s AxB " p AxBQ AxB ,

P $ deg Q,Q AxB.P AxB Q AxBP AxBR AxB ! P AxB /Q AxB,

u ! ln x, dy ! 1dx. I6u ! arctan A3xB, dy ! 1dx,I5

I5 :!! arctan A3xB dx and I6 :!! ln x dx ,

u ! x, dy ! exdx.

I4 ! x 2ex # 2! xexdx

u ! x2, dy ! exdx

I4 :!! x2exdx .

y ! x2/2 " C.y ! x2/2

I3 ! ! x ln x dx ! x22 ln x # 14x2 " C .

Section A Review of Integration Techniques A-5

leads immediately to the evaluation

The key to finding such a simple fraction representation is to first factor the denominator poly- nomial Indeed, in (10) the denominator is just the product In general, we recall that any polynomial with real coefficients can be factored into a prod- uct of linear and irreducible quadratic factors with real coefficients. There are four cases that may arise in the factorization: (i) nonrepeated linear factors; (ii) repeated linear factors; (iii) nonrepeated quadratic factors; and (iv) repeated quadratic factors.

In the case when a nonrepeated linear factor occurs in the factorization of we associate the partial fraction where A is a constant to be determined. If a linear factor

is repeated m times in the factorization of we associate the sum of m fractions

where the m constants are to be determined. In the case of a nonrepeated quadratic factor (where we associate the single fraction

where we now have two constants in the numerator to be determined, while if this quadratic factor is repeated m times, then we associate the sum

which involves the determination of 2m constants and We illustrate the approach in the following example. Further discussion of partial fractions

appears in Section 7.4, page 370.

Find the indefinite integral

Since the factor x is repeated three times in the denominator, while is a nonrepeated factor, our partial fraction decomposition must be

(11)

where A, B, C, and D are unknowns to be determined. Multiplying both sides of (11) by we have

(12)

In order for (12) to be true for all x, the corresponding coefficients must be the same:

The coefficients of yield A B

The coefficients of yield B C

The coefficients of yield C D

The coefficients of yield D ! 2.x0 ! 3."x1 ! 3."x2 ! 3."x3

3x3 " 3x2 " 3x " 2 ! Ax3 " Bx2 Ax " 1B " Cx Ax " 1B " D Ax " 1B .x 3 Ax " 1B,

3x3 " 3x2 " 3x " 2 x3 Ax " 1B ! Ax " 1 " Bx " Cx2 " Dx3 ,

Ax " 1B I7 ! ! 3x3 " 3x2 " 3x " 2x3 Ax " 1B dx.

Bk.Ak

a m

k!1

Akx " BkAax2 " bx " cBk ,

Ax " B ax2 " bx " c

b2 # 4ac 6 0)ax2 " bx " c Ak

a m

k!1

AkAx # aBk , Q AxB,Ax # aB A/ Ax # aB, Q

AxB,Ax # aB Q AxB Ax # 1B Ax " 1B Ax " 3B.Q AxB.

!R AxB dx ! 2 ln 0 x # 1 0 " ln 0 x " 1 0 " 4 ln 0 x " 3 0 " C . A-6 Appendices

Example 4

Solution

Solving this system of four linear equations in four unknowns gives and Hence

◆

We remark that the coefficients A and D in the representation (11) can be determined directly. Indeed, if we set in (12), we obtain and if we set in (12), we obtain or But B and C cannot be obtained so readily. If we are fortunate to have a denominator with only nonrepeated linear factors, then this simple sub- stitution method will work to quickly determine all the unknown coefficients.

In dealing with the integration of rational functions that have a quadratic factor in the denominator, the familiar formula

plays a central role. For example, by completing the square and making a simple change of variables, we find

! 3x2 " 4x " 5 dx ! ! 3Ax " 2B2 " 1 dx ! 3 arctan Ax " 2B " C .

! 11 " x2 dx ! arctan x " C

A ! 1.#3 " 3 # 3 " 2 ! A A#1B x ! #12 ! D,x ! 0 I7 ! ! a 1x " 1 " 2x " 1x2 " 2x3bdx ! ln @ Ax " 1Bx2 @ # 1x # 1x2 " C .

D ! 2. C ! 1,B ! 2,A ! 1,

Section A Review of Integration Techniques A-7

In Problems 1–26, find the indicated indefinite integral.

1. 2.

3. 4.

5. 6.

7. 8.

9. 10.

11.

12. 13.

14. 15.

16. 17. ! 1x ln x dx! x 3 " 4x " 1

x3 " 3x2 " 4x " 2 dx

! x3 " 2x2 " 8x2 Ax2 " 4B dx! 2x " 41x2 " 5x # 14 dx !29x2 # 1x dx! sec A2xB tan A2xB dx!

4Ax # 1B Ax # 2B Ax # 3B dx ! 12x2 " 4x " 4 dx!

121 # 2x2 dx ! 10 sec2 A5xB dx! t " 1t2 " 4 dt ! x1/3x1/3 " 1 dx! cos 2 ApuB du ! x3e3xdx! 4t4e6t5dt ! x cos A2xB dx! x3 A1 " x2B3/2dx 18. 19.

20. 21.

22. 23.

24. 25.

26.

In Problems 27–34, use an appropriate trigonometric identity to help find the indicated integral.

27. 28.

29. 30.

31. 32.

33. 34. ! cot2x csc4 x dx! sin2 A3xB cos2 A3xB dx ! cos1/3x sin 3 x dx! tan4u sec6u du ! cot3u du! cos A3xB cos A7xB dx ! tan3x2sec x dx! sin 2/5x cos3x dx

! e3y1 " e6y dy ! sin t1 " cos 2t dt! sin 2x dx !

t ln At " 3B dt! 36x 5exp A2x 3B dx ! 1Ax2 " 4B3/2 dx! sec A3u " 1B du ! y sinh y dy! x2x2 # 1 dx

A EXERCISES

A-8 Appendices

To solve an equation we must find the point or points where the graph of meets the x-axis. One procedure for approximating a solution is Newton’s method.

To motivate Newton’s method geometrically, we let be a root of and let be our guess at the value of If we are done. If then we are off by some amount that we call dy (see Figure B.1). Then

and so

(1) dx ! dy

g¿ Ax1B .

dy dx

! g¿ Ax1B , g Ax1B % 0,g Ax1B ! 0,x~. x1g AxB ! 0x~

y ! g AxBg AxB ! 0,B NEWTON’S METHOD

y = g(x)

xx2 x1

dy = g(x1)

(x1, g(x1))

Figure B.1 Tangent line approximation of root

Now or where is at the point where the tangent line through intersects the x-axis (Figure B.1). Using equation (1) and the fact that

we obtain

which we use as the next approximation to the root Repeating this process with in place of we obtain the next approximation to the

root In general, we find the next approximation by the formula

(2)

The process is illustrated in Figure B.2. If the initial guess is sufficiently close to a root then the sequence of iterations

usually converges to the root However, if we make a bad guess for then the process may lead away from x~.

x1,x~.ExnFqn!1 x~,x1 xn!1 " xn #

g AxnB g$ AxnB , n " 1, 2, . . . .

xn"1x~. x3x1,x2

x~.

x2 ! x1 # dy

g¿ Ax1B ! x1 # g Ax1Bg¿ Ax1B , dy ! g Ax1B,Ax1, g Ax1B B x2x2 ! x1 # dx,dx ! x1 # x2

Find a root to four decimal places of the equation

(3)

Setting we find that is positive for all x. Hence, g is increasing and has at most one zero. Furthermore, since and this zero must lie between 1 and 2. Thus, we begin the procedure with the initial guess For

equation (2) becomes

(4)

With equation (4) gives

Using to compute and so on, we find

where we have rounded off the computations to five decimal places. Since and agree to four decimal places and we are uncertain of the fifth decimal place because of roundoff, we surmise that the root of (3) agrees with 1.1795 to four decimal places. Indeed,

and

so Consequently, ◆

Observe that Newton’s method transforms the problem of finding a root to the equation into the problem of finding a fixed point for the function that

is, finding a number x such that [See equation (2).] Several theorems give conditions that guarantee that the sequence of iterations

defined by (2) will converge to a zero of We mention one such result.g AxB. ExnFqn!1x ! h AxB. h AxB ! x # g AxB /g¿ AxB;g AxB ! 0

x~ ! 1.1795 . . . .1.1795 6 x~ 6 1.1796.

g A1.1796B ! 0.00056 . . . ,g A1.1795B ! #0.00005 . . .x ~

x5x4

x5 ! 1.17951 ,

x4 ! 1.17951 ,

x3 ! 1.18085 ,

x3x2

x2 ! 1.5 # A1.5B3 " 2 A1.5B # 4

3 A1.5B2 " 2 ! 1.5 # 2.3758.75 " 1.22857 . x1 ! 1.5,

xn"1 ! xn # x3n " 2xn # 4

3x2n " 2 , n ! 1, 2, . . . .

g AxB ! x3 " 2x # 4, x1 ! 1.5.g A2B ! 8,g A1B ! #1g¿ AxB ! 3x2 " 2g AxB ! x3 " 2x # 4,

x3 " 2x # 4 ! 0 .

Section B Newton’s Method A-9

y = g(x)

x x4 x1

x2x3

Figure B.2 Sequence of iterations converging to root

Example 1

Solution

We do not give a proof of this theorem, but refer the reader to an introductory numerical analysis text such as Numerical Analysis, 9th ed., by R. Burden and J. Faires (Brooks/Cole, Pacific Grove, Calif., 2010).

A-10 Appendices

Convergence of Newton’s Method

Theorem 1. Suppose a zero of lies in the interval and that in this interval

and

If we select so that then the sequence of iterations defined by (2) will decrease to x~.

x~ 6 x1 6 b,x1

g– AxB 7 0 .g¿ AxB 7 0 Aa, bBg AxBx~

A useful procedure for approximating the value of a definite integral is Simpson’s rule. Let the interval [a, b] be divided into 2n equal parts and let be the points of

the partition, that is,

where If

then the Simpson’s rule approximation for the value of the definite integral

is given by

(1)

is the error that results from using Simpson’s rule to approximate the value of the definite integral, then

(2)

where for all in 3a, b 4 .xM J max 0 f A4B AxB 00E 0 & Ab # aB

180 h4M ,

E J ! b

a f AxB dx # IS

! h 3

a n

k!1 Ay2k#2 " 4y2k#1 " y2kB .

IS " h 3 3 y0 ! 4y1 ! 2y2 ! 4y3 ! p ! 2y2n#2 ! 4y2n#1 ! y2n 4

! b

a f AxB dx

yk J f AxkB , k ! 0, 1, . . . , 2n ,h J Ab # aB / A2nB. xk J a " kh , k ! 0, 1, . . . , 2n ,

x0, x1, . . . , x2n

C SIMPSON’S RULE

Use Simpson’s rule with n = 4 to approximate the value of the definite integral

(3)

Here and

By Simpson’s rule (1), we find

Hence, the value of the definite integral in (3) is approximately ◆

For a more detailed discussion of Simpson’s rule, we refer the reader to a numerical analy- sis book such as Numerical Analysis, 9th ed., by R. Burden and J. Faires (Brooks/Cole, Pacific Grove, Calif., 2010).

IS ! 0.7854.

" 4 a 64 64 " 49

b " a 64 64 " 64

b d ! 0.7854 . " 2 a 64

64 " 16 b " 4 a 64

64 " 25 b " 2 a 64

64 " 36 b

IS ! A18B 3

c1 " 4 a 64 64 " 1

b " 2 a 64 64 " 4

b " 4 a 64 64 " 9

b yk ! A1 " x2kB#1 ! 1

1 " k2

! 64

64 " k2 .

h ! 1/8, xk ! k/8, k ! 0, 1, . . . , 8,

! 1

1 1 " x2

dx .

Section D Cramer’s Rule A-11

Example 1

Solution

When a system of n linear equations in n unknowns has a unique solution, determinants can be used to obtain a formula for the unknowns. This procedure is called Cramer’s rule. When n is small, these formulas provide a simple procedure for solving the system.

Suppose that for a system of n linear equations in n unknowns,

(1) o o o o

the coefficient matrix

(2) A J Da11 a12 p a1na21 a22 p a2n o o o

an1 an2 p ann

T an1x1 " an2x2 " p " annxn ! bn ,

a21x1 " a22x2 " p " a2nxn ! b2 , a11x1 " a12x2 " p " a1nxn ! b1 ,

D CRAMER’S RULE

has a nonzero determinant. Then Cramer’s rule gives the solutions

(3)

where is the matrix obtained from A by replacing the ith column of A by the column vector

consisting of the constants on the right-hand side of system (1).

Use Cramer’s rule to solve the system

We first compute the determinant of the coefficient matrix:

Using formula (3), we find

◆

For a more detailed discussion of Cramer’s rule, we refer the reader to an introductory linear algebra text such as Linear Algebra and Its Applications, 3rd ed., by David C. Lay (Pearson Addison Wesley, 2006).

x3 ! 1 12

det £1 2 02 1 9 1 #1 1

§ ! 24 12

! 2 .

x2 ! 1 12

det £1 0 #12 9 1 1 1 #2

§ ! #12 12

! #1 ,

x1 ! 1 12

det £0 2 #19 1 1 1 #1 #2

§ ! 48 12

! 4 ,

det £1 2 #12 1 1 1 #1 #2

§ ! 12 . x1 # x2 # 2x3 ! 1 . 2x1 " x2 " x3 ! 9 , x1 " 2x2 # x3 ! 0 ,

Db1b2 o

TAi xi !

det Ai det A , i ! 1, 2, . . . , n ,

A-12 Appendices

Example 1

Solution

If the y-axis intercept and the slope of the line are known, then the y-value on the line corresponding to the measured x-value is given by The corresponding measured y-value, thus deviates from the line by and the total sum of squares of deviations is

The interesting feature of the function S is that the symbols and are constants, while and are the (unknown) variables. The values of and that minimize S will force its partial derivatives to be zero:

Displaying these conditions in a form that emphasizes the roles of and we rewrite them after a little algebra as

a a N

i!1 xi " b a

i!1 x2i ! a

i!1 xiyi

aN " b a N

i!1 xi ! a

i!1 yi ,

b,a

0 ! 0S 0b ! a

i!1 2 3 yi # Aa " bxiB 4 A#xiB .

0 ! 0S 0a ! a

i!1 2 3 yi # Aa " bxiB 4 A#1B ,

bab ayixi

S J a N

i!1 3 yi # Aa " bxiB 4 2 .

3 yi # Aa " bxiB 4 ,yi, a " bxi.xiba

Section E Method of Least Squares A-13

The method of least squares is a procedure for fitting a straight line to a set of measured data. Consider the scatter diagram in Figure E.1, consisting of an x, y-plot of some data points

It is desired to construct the straight line that best fits the data points, in the sense that the sum of the squares of the vertical deviations from the points to the line is minimized.

y ! a " bxE Axi, yiB; i ! 1, 2, . . . , N F. E METHOD OF LEAST SQUARES

x 0

1 2 3 4 5

P1 P2

P4 P5

Figure E.1 Scatter diagram and least-squares linear fit

and obtain the formulas for the optimal values of intercept and slope:

Find the least-squares linear fit to the data points and which are plotted in Figure E.1.

Arranging the data as in Table E.1 yields

b ! 5 A34.5B # 15 A9.6B

5 A55B # A15B2 ! 28.550 ! 0.57 . a !

55 A9.6B # 15 A34.5B 5 A55B # A15B2 ! 10.550 ! 0.21 ,

P5 A5, 3.1B, P1 A1, 1B, P2 A2, 1B, P3 A3, 2B, P4 A4, 2.5B, b !

N a N

i!1 xiyi # aaN

i!1 xib aaN

i!1 yib

N a N

i!1 x2i # aaN

i!1 xib 2 .

a !

aaN i!1

x2ib aaN i!1

yib # aaN i!1

xib aaN i!1

xiyib N a

i!1 x2i # aaN

i!1 xib 2 ,

A-14 Appendices

Example 1

Solution

Thus, is the equation for the best-fitting line, which is graphed in Figure E.1. ◆

y ! 0.21 " 0.57x

TABLE E.1

1 1 1 1 1 2 2 1 2 4 3 3 2 6 9 4 4 2.5 10 16 5 5 3.1 15.5 25

Sums 15 9.6 34.5 55

x2ixi yiyixi

Section F Runge–Kutta Procedure For n Equations A-15

Below are program outlines for the classical fourth-order Runge–Kutta algorithm for systems dis- cussed in Section 5.3. The first of these, called the “subroutine,” provides approximations to the solution functions of a system of n first-order ordinary differential equations over an interval. The second outline on page A-16 utilizes the method of halving the step size in order to obtain approximations (to within a prescribed tolerance) to the solution functions at a single given point.

F RUNGE–KUTTA PROCEDURE FOR n EQUATIONS

Classical Fourth-Order Runge–Kutta Subroutine (n Equations) Purpose To approximate the solution to the initial value problem

for INPUT (number of steps), PRNTR (!1 to print table).

Step 1 Set step size

Step 2 For j ! 1 to N do Steps 3–5. Step 3 Set

Step 4 Set

Step 5 If PRNTR ! 1, print t, x1, x2, . . . , xn.

xi ! xi " 1 6

Aki,1 " 2ki,2 " 2ki,3 " ki,4B , i ! 1, . . . , n . t ! t " h ; i ! 1, . . . , n .

ki,4 ! hfi At " h, x1 " k1,3, . . . , xn " kn,3B ,i ! 1, . . . , n ; ki,3 ! hfi at " h2, x1 " 12 k1,2, . . . , xn " 12 kn,2b ,i ! 1, . . . , n ; ki,2 ! hfi at " h2, x1 " 12 k1,1, . . . , xn " 12 kn,1b ,i ! 1, . . . , n ; ki,1 ! hfi At, x1, . . . , xnB , h ! Ac # t0B /N , t ! t0 , x1 ! a1 , . . . , xn ! an

n, t0, a1, . . . , an, c, N t0 & t & c.

xi At0B ! ai , i ! 1, 2, . . . , n x¿i ! fi At, x1, . . . , xnB ;

A-16 Appendices

Classical Fourth-Order Runge–Kutta Algorithm with Tolerance (n Equations) Purpose To approximate the solution to the initial value problem

at t ! c, with tolerance . INPUT

(tolerance) M (maximum number of iterations)

Step 1 Set set PRNTR ! 0. Step 2 For m ! 0 to M do Steps 3–7 (or, to save time, start with m ' 0). Step 3 Set Step 4 Call FOURTH-ORDER RUNGE–KUTTA SUBROUTINE

(n EQUATIONS). Step 5 Print h, x1, x2, . . . , xn. Step 6 If for go to Step 10. Step 7 Set Step 8 Print “ is approximately”; ; “but may not be within

the tolerance”; Step 9 Go to Step 11. Step 10 Print “ is approximately”; ; “with tolerance”; Step 11 STOP. OUTPUT Approximations of the solution to the initial value problem at t ! c, using

steps. 2m

e.xi Afor i ! 1, . . . , nBxi AcB e.

xi Afor i ! 1, . . . , nBxi AcBzi ! xi , i ! 1, . . . , n . i ! 1, . . . , n,0 zi # xi 0 6 e

N ! 2m.

zi ! ai, i ! 1, 2, . . . , n;

e n, t0, a1, . . . , an, c

xi At0B ! ai , i ! 1, 2, . . . , nx¿i ! fi At, x1, . . . , xnB ;

B-1

CHAPTER 1

Exercises 1.1, page 5 1. ODE, 2nd-order, ind. var. x, dep. var. y, linear 3. PDE, 2nd-order, ind. var. x, y, dep. var. u 5. ODE, 1st-order, ind. var. t, dep. var. x, nonlinear 7. ODE, 2nd-order, ind. var. x, dep. var. y, nonlinear 9. ODE, 4th-order, ind. var. x, dep. var. y, linear

11. PDE, 2nd-order, ind. var. t, r, dep. var. N 13. where k is the proportionality constant 15. where k is the proportionality

constant 17. Kevin wins by sec.

Exercises 1.2, page 13 3. Yes 5. No 7. Yes 9. Yes 11. Yes

13. Yes 19. The left-hand side is always 21. (a) , (b) 23. Yes 25. Yes 27. No 29. (b) Yes 31. (a) No

(c)

Exercises 1.3, page 21 1. (a) (b)

y ! 2x and y ! "2x

1 # 26"31/3 $ 4. 623 " 426 ! 0.594dT/dt ! k AM " TB,

dp/dt ! kp,

3. All solutions have limiting value 8 as t S %q.

5. (a)

Answers to Odd-Numbered Problems

x–1–2–3

x1–1

–2

–4

Figure B.1 The solution curve in Problem 1(a).

Figure B.2 The solution curve in Problem 1(b).

t 1

Figure B.3 Solutions to Problem 3 satisfying y A0B ! 5, y A0B ! 8, and y A0B ! 15

t 1

Figure B.4 Direction field for Problem 5(a)

(a)

(b) lim (c) lim (d) NotSq p AtB ! 3/2tSq p AtB ! 3/2 7. (a)

(b) 2 (c) 2 (d) 0 (e) No

t 1

Figure B.5 Direction field for Problem 7(a)

15.

11. 13.

y = x – 1

(x)

Figure B.6 Direction field and sketch of for Problems 9 ( f, g)

f(x)

y y = (x)

' y

(0, 4)

y = (x)'

Figure B.7 Solution to Problem 11

Figure B.8 Solution to Problem 13

c = 0c = –1 c = 2

0 1

Figure B.9 Solution to Problem 15

3. (rounded to three decimal places)

0.1 0.2 0.3 0.4 0.5

1.100 1.220 1.362 1.528 1.721

5. (rounded to three decimal places)

1.1 1.2 1.3 1.4 1.5

0.100 0.209 0.325 0.444 0.564

7. (rounded to three decimal places)

N h 9.

1 3.142 1.1 "0.90000 2 1.571 1.2 "0.81654 4 1.207 1.3 "0.74572 8 1.148 1.4 "0.68480

1.5 "0.63176 1.6 "0.58511 1.7 "0.54371 1.8 "0.50669 1.9 "0.47335 2.0 "0.44314

11. stabilizes at for h below for some

15. (a) (b)

CHAPTER 2

Exercises 2.2, page 43 1. No 3. Yes 5. Yes 7. 9.

11. 13. 15.

17. 19. 21. 23.

25.

27. (a)

(b)

(c)

(d) 29. (d) is not continuous at (0, 0) . 31. (a)

(b) 121 " x2 ; 124 " x2 ; 121/4 " x2x2 % y"2 ! C 0f/ 0y y A0.5B ! 1.381y AxB ! tan a"

0 21 % sin t dt % p/4by AxB ! a1 % 3 "

0 et

2 dtb 1/3y AxB ! "

0 et

2 dt

y ! 4ex 3/3 " 1

y ! arctan At 2 % 1By2/2 % ln y ! sin u " u cos u % 1/2 " p y ! Asin x % 1B2 " 1y ! 2e"x4/4 % 1y ! 1/ AC " ecos xB

x ! # 2Ce2t / A1 " Ce2tB, C $ 02y % sin A2yB! 4 arctan x % Cy ! 32 A1 % xB3/2 " 6 A1 % xB1/2 % C 4 1/3 x ! Cet

T A2B ! 73.647ºT A1B ! 80.460º0.785 p 4 .3 arctan A1B ! t0 in A0.78, 0.80B;x At0B ! 11.5574 p . 4 2"10; 3 tan A1B !1.56 # 0.01x A1B

p/8 p/4 p/2 p

ynxnyN

B-2 Answers to Odd-Numbered Problems

9. (d) It increases and asymptotically approaches the line

(f ) and (g) y ! x " 1.

17. It approaches 3.

Exercises 1.4, page 28 1. (rounded to three decimal places)

0.1 0.2 0.3 0.4 0.5

4.000 3.998 3.992 3.985 3.975yn

(c)

(d) is domain .

33. 28.1 kg 35. (a) 82.2 min (b) 31.8 min

Exercises 2.3, page 51 1. Neither 3. Linear 5. Linear 7. 9.

11. 13. 15. 17.

19.

21. 23. 25. (b) 27. (b) 0.9960 (c) 0.9486 , 0.9729 29. 31. (a) (b)

(c)

(d)

(e)

y ! e x " 1 % 2e"x , if 0 ( x ( 2 x/3 " 1/9 % A4e6/9 % 2e4Be"3x , if 2 6 x

y ! x/3 " 1/9 % Ce"3x y ! x " 1 % 2e"xy ! x " 1 % Ce"x

x ! e4y/2 % Ce2y y A3B ! 0.183y AtB ! A38/3Be"5t " A8/3Be"20t

y ! x 2cos x " p2cos x

x ! t 3

6 ln t "

t 3

36 %

1 2t

" 17

36t 3

y ! xex " xy ! 1 % C Ax2 % 1B"1/2 x ! y3 % Cy"2y ! "t " 2 % Cet y ! C/x2 " 1/x3y ! x A2x % ln 0 x 0 % CB

0 x 0 6 1 a

"1 6 x 6 1 ; "2 6 x 6 2 ; "1 2

6 x 6 1 2

15. 17. Not exact 19. 21. 23. 25. (equation is sep-

arable, not exact) 27. (a)

(b) where f is a function of x only

29. (c) (d) Yes, 33. (a)

(b) (c) (d)

Exercises 2.5, page 67 1. Integrating factor depending on x alone 3. Exact 5. Linear with x as dep. var., has integrating factor depend-

ing on x alone 7. 9.

11. 13. 15. (a)

(b)

17. (a) depends only on

(b) 19. (b)

Exercises 2.6, page 74 1. Homogeneous and Bernoulli 3. Bernoulli 5. Homogeneous and Bernoulli 7. 9.

11.

13. 15. 17. 19. 21. 23. 25. 27. 29. 31. 33. 35.

37. 39. 41. 45. Ay " 4xB2 Ay % xB3 ! CAx " y % 2B2 ! Ce2x % 1 uy2 ! C Au % yB2 and y ! "uy"2 ! "e2x/2 % Ce"2x and y # 0

% A2/23Barctan 3 Ax % 3B /23 At " 1B 4 ! Cln 3 Ax % 3B2 % 3 At " 1B2 4y ! 4x % A3 % Ce4xB / A1 " Ce4xB and y ! 4x " 1 A2x % 2y " 3B3 ! C A2x % y " 2B2Ay % 2B2 % 2 Ax % 1B Ay % 2B " 3 Ax % 1B2 ! C r ! u2/ AC " uB and r # 0x"2 ! 2t 2ln 0 t 0 % Ct 2 and x # 0 y ! 5x2/ Ax5 % CB and y # 0y ! 2/ ACx " x3B and y # 0 y ! x % A6 % 4Ce2xB / A1 % Ce2xB and y ! x % 4y ! Ax % CB2/4 " x and y ! "x Ax2 " 4y2B3x2 ! C21 % x2/t2 ! ln 0 t 0 % Cy ! x/ Aln 0 x 0 % CB and y # 0, x # 0 y ! "x/ Aln 0 x 0 % CB and y # 0 y¿ ! G Aax % byB m ! ey ; x ! y " 1 % Ce"y exp(x2y) [x2 % 2xy ] ! C

x2y . 0N/0x " 0M/0y

x2M " 2xyN

ln (x3y) % 2xy ! 2

m AzB ! exp 3$H AzB dz 4 ; z ! xym ! xy ; x2y3 " 2x3y2 ! Cm ! y"2 ; x2y"1 % x ! C and y # 0m ! x "2 ; y ! x4/3 " x ln 0 x 0 % Cx and x # 0m ! y"4 ; x2 y"3 " y"1 ! C and y # 0

2x2 % y2 ! c 2y2ln y " y2 % 2x2 ! cx2 % 4y2 ! c

x ! cy2 ; x # 0 ; y # 0 y # 0y ! x2/ AC " xB

cos x sin y " y2/2 % f AxB,"ln 0 y 0 % f AxB sin x " x cos x ! ln y % 1/y % p " 1

y ! "2/ Atet % 2Bln x % x2y2 " sin y ! p2x2 " y2 % arctan AxyB ! C r ! AC " euBsec u

Answers to Odd-Numbered Problems B-3

1 2 3 4 5

Figure B.18 Solution to Problem 31(e)

33. (a) is only solution in a neighborhood of x ! 0. (b) for any C.

35. (a) 0.0281 kg L (b) 0.0598 kg L

37.

39. 5 P.M.

Exercises 2.4, page 61 1. Linear 3. Separable 5. Separable, also linear with x as dep. var. 7. Exact 9.

11. 13. t ln y % t ! Csin x cos y % x2 " y2 ! C x2 % xy " y2 ! C

71.8ºF at noon; 26.9ºF at

% a19 2

% 2

4 % Ap/12B2b e"2t x AtB ! 1

2 "

2 cos Apt/12B 4 % Ap/12B2 " Ap/12Bsin Apt/12B4 % Ap/12B2

// y ! "3x % Cx2 satisfies y A0B ! 0y ! x

47. (a) (b)

Review Problems, page 77 1. 3. 5. 7. 9. 11.

13. 15. 17. 19. 21. 23. 25. 27.

29. 31. 33. 35. 37.

39.

41.

CHAPTER 3

Exercises 3.2, page 99 1. 5 " 4.5e"2t/25 kg; 5.07 min 3. 5. 0.0097%; 7. times as salty 9. 110,868 13. 5970; 6000

15. 6572; 6693

17. (a) Logistic Year (Least Squares)

1790 0.0351 3.29 1800 0.0363 4.51 1810 0.0331 6.16 1820 0.0335 8.41 1830 0.0326 11.45 1840 0.0359 15.53 1850 0.0356 20.97 1860 0.0267 28.14 1870 0.0260 37.45 1880 0.0255 49.31 1890 0.0210 64.07

1 p

dp dt

1/e20 min later; 73.24 h

19.96 minA0.4B A100 " tB " A3.9 ) 10"7B A100 " tB4 L;

y (t) ! e"t" t

1 % r 2 dr % 3e"(t"2), y(3) ! 1.1883...

y ! 2 A19x4 " 1B /2% 22 arctan c y " 222 Ax " 1B d ! ln 2 ln 3 Ay " 2B2 % 2 Ax " 1B2 4 y ! "2x22x2 " 1x ! "t " 2 % 3e"ty ! "x3/2 % 7x/2 x4y3 " 3x3y2 % x4y2 ! C

/ 323 Ax " 3B " Ay " 4B 4 0 1/23 ! C3 Ay " 4B2 " 3 Ax " 3B2 4 0 323 Ax " 3B % Ay " 4B 4x2y"2 " 2xy"1 " 4xy"2 ! C and y # 0 y2 % 2xy " x2 ! C xy " x2 " x % y2/2 " 4y ! C y2 ! x2 % Cx3 and x # 0 y ! 2/ A1 % Ce2uB and y # 0y ! 2x % 3 " Ax % CB2/4 y ! " Ax2/2Bcos A2xB % Ax/4Bsin A2xB % Cxtan At " xB % t ! CAx2 % 4y2B3x2 ! C

t ! Cexp A1/ A7y7B By % x sin AxyB ! C x2y " x3 % y"2 ! Cex % ye"y ! C y ! x % 5x/ AC " x5Bv¿ % [2 Pu % Q]v ! "P LogisticYear (Least Squares)

1900 0.0210 81.89 1910 0.0150 102.65 1920 0.0162 125.87 1930 0.0073 150.64 1940 0.0145 175.80 1950 0.0185 200.10 1960 0.0134 222.47 1970 0.0114 242.16 1980 0.0098 258.81 1990 0.0132 272.44 2000 0.0097 283.28 2010 291.73

(b) (c) (using all data, including 2010) (d) See table in part (a).

19. 14 million tons per yr 21. 1 hr; 2 h 23. 11.7% 25. 31,606 yr 27. kg of Hh, kg of It, and

kg of Bu

Exercises 3.3, page 107 1. 20.7 min 3. 22.6 min 5. 9:08 A.M. 7. 1:16 P.M. 9. 11. 39.5 min

13. 15.

and so

Exercises 3.4, page 114 1. 3. 18.6 sec 5. 7. 241 sec 9.

11. 13. 2.69 sec; 101.19 m 15. 17. 300 sec 19. 21. 23. Sailboat B 25. (e) 11.18 km/sec (f) 2.38 km/sec

Exercises 3.5, page 121 1.

7. "(10 ln .1)/3 ! 7.68 sec VI ! EC

dCEC dt

! d dt

CE 2C

VI ! L dI dt

I ! d dt

LI 2

2 ;VI ! IRI ! I 2R;

"(ln .4) ) 10"10 ! 9.2 ) 10"11 sec EL ! A"7.2e"100t " 6 sin 120t % 7.2 cos 120tB /2.44I ! E1.44e"100t % cos 120t % 1.2 sin 120tF /2.44; 5e"2t /2 % 6t " 5/2; 6 m/sec 2636e"t/20 % 131.8t " 2636 m; 1.768 sec

Av0 " T/kBe"kt/I % T/keby 0by " mg 0mg ! ey0b 0by0 " mg 0mge"b2x/m 95.65t % 956.5e"t/10 " 956.5 m; 13.2 sec

4.91t % 22.55 " 22.55e"2t m; 97.3 sec A0.981Bt % A0.0981Be"10t " 0.0981 m; 1019 sec dT/dt ! k1 AM " TB, where k1 ! 4M3kfor T near M, M4 " T 4 ! 4M3 AM " TB, T " M ! C AT % MBexp 32 arctan AT/MB " 4M3kt 4 ;148.6ºF16.3ºC; 19.1ºC; 31.7ºC; 28.9ºC 28.3ºC; 32.5ºC;

1 " 2e"t % e"2t 2e"t " 2e"2te"2t

A1/2Bln 15 ! 1.354 yr; p0 ! 3.28780 p1 ! 316.920, A ! 0.00010050

1 p

dp dt

B-4 Answers to Odd-Numbered Problems

Answers to Odd-Numbered Problems B-5

Exercises 3.6, page 129 3. h “e” 7.

1 3 1.1 0.10450 0.1 2.72055 1.2 0.21668

0.01 2.71830 1.3 0.33382 0.001 2.71828 1.4 0.45300 0.0001 2.71828 1.5 0.57135

0.2 0.61784 0.4 1.23864 0.6 1.73653 0.8 1.98111 1.0 1.99705 1.2 1.88461 1.4 1.72447 1.6 1.56184 1.8 1.41732 2.0 1.29779

11. 13. 15.

17.

0.1 "1 0.06250 0.2 "3 1 0.00391 0.3 "1 0.00024 0.4 9 1 0.00002 0.5 "1 0.00000 0.6 "27 1 0.00000 0.7 "1 0.00000 0.8 81 1 0.00000 0.9 "1 0.00000 1.0 "243 1 0.00000

We conclude that step size can dramatically affect convergence.

19.

Time

Midnight 65.0000 65.0000 65.0000 4 A.M. 69.1639 68.5644 68.1300 8 A.M. 71.4836 72.6669 73.6678 Noon 72.9089 75.1605 76.9783 4 P.M. 72.0714 73.5977 74.7853 8 P.M. 69.8095 69.5425 69.2831 Midnight 68.3852 67.0500 65.9740

Exercises 3.7, page 139 1.

" h2

2 sin Axn % ynB 31 % cos Axn % ynB 4yn%1 ! yn % h cos Axn % ynB

K ! 0.6K ! 0.4K ! 0.2

yn Ah ! 0.025Byn Ah ! 0.1Byn Ah ! 0.2Bxnx ! 1.27 f A1B ! y A1; 2"3B ! 0.71698f A1B ! x A1; 2"3B ! 1.25494

ynxn

ynxn 3.

5. Order 2, 7. 9. 1.36789 11.

13.

15.

0.5 0.21462 1.0 0.13890 1.5 "0.02668 2.0 "0.81879 2.5 "1.69491 3.0 "2.99510

19. with 21. with

CHAPTER 4

Exercises 4.1, page 157 3. Both approach zero. 5. 0 7. 9.

Exercises 4.2, page 165 1. 3. 5. 7. 9.

11. 13. 15. 17. 19. 21. (a) (b) 23. 25. 27. Lin. dep. 29. Lin. indep. 31. Lin. dep. 33. If then 35. (a) Lin. indep. (b) Lin. dep. (c) Lin. indep.

(d) Lin. dep.

37. 39. 41. 43. 45. (a) (where ,

and ) (b)

(where ) (c) c1e"t % c2et % c3e"2t % c4e2t % c5e3t

r1 ! 1.176, r2 ! 1.902 c1e

r1t % c2e "r1t % c3e

r2t % c4e "r2t

r3 ! 0.701r2 ! "1.869, r1 ! "4.832c1e

r1t % c2e r2t % c3e

r3t 3 % et " 2e"t c1e

"3t % c2e "2t % c3e

2t c1e

"2t % c2te "2t % c3e

2t c1e

t % c2e A"1"25Bt % c3eA"1%25Bt

y1 ! "(c2/c1)y2 .c1 * 0,

ce13t/6 ce"4t/5

ce"bt/aar % b ! 0 e"t " 2te"t A23/2B 3 eA1%23Bt " eA1"23Bt 42e5At%1B % e"At%1B3e"4tc1e

"5t/ 2 % c2te"5t/ 2 c1e

t/ 2 % c2tet/ 2 c1e

t/ 2 % c2e"2t/ 3 c1e

2t % c2e 3t

c1e 2t % c2e

"t c1e

"3t % c2te "3t

y AtB ! "2 cos 2t % A3/2B sin 2ty AtB ! " A30/61B cos 3t " A25/61B sin 3t

h ! 0.03125z A1B ! 2.87083 h ! 0.0625y A3B ! 0.24193

ynxn

x ! 0.50 x ! 1.41"11.7679

f A1B ! 1.3725; order 4, f A1B ! 1.3679" h3

6 A1 " xn % ynB % h4

24 A1 " xn % ynB

yn%1 ! yn % h Axn " ynB % h22 A1 " xn % ynB Answers to Odd-Numbered Problems B-5

Exercises 4.3, page 173 1. 3.

5. 7. 9.

11. 13. 15. 17. 19. 21. 23. 25. 27. 29. (a)

(b) (c)

31. (a) Oscillatory (b) Tends to zero (c) Tends to "+ (d) Tends to "+ (e) Tends to %+

33. (a) (b) (c) Decreases the frequency of oscillation, introduces

the factor , causing the solution to decay to zero 35. 37. (a)

(b)

Exercises 4.4, page 181 1. No 3. Yes 5. Yes 7. No, not a constant coefficient equation 9. 11.

13.

15. 17.

19. 21.

23.

25.

27.

29.

31.

33.

35. ¢ 1 10

t2 " 4

25 t≤etA1/5B cos t % A2/5B sin t% AB3t4 % B2t 3 % B1t2 % B0tBe"t sin t

AA3t4 % A2t3 % A1t 2 % A0tBe"t cos te3t AA6t8 % A5t7 % A4t6 % A3t 5 % A2t4 % A1t 3 % A0t 2B % AB3t4 % B2t3 % B1t2 % B0tB sin 3tAA3t4 % A2t3 % A1t2 % A0tB cos 3t

e2t Acos 3t % 6 sin 3tB" 121 u3 " 149 u2 " 2343 u t3e2t /6¢ t13 % 8169≤te"3t 4t2etxex/2 % 3ex/4 cos 3t yp AxB ! 3 Aln2B

2 % 1 4"12xyp AtB # "10 Ac1 % c2tBe"t cos A23tB % Ac3 % c4tBe"t sin A23tBc1 cos t % c2 sin t % c3t cos t % c4t sin tb $ 22Ik

e"3t

2/p y AtB ! 0.3e"3tcos 4t % 0.2e"3tsin 4t c1 cos 2t % c2 sin 2t % c3 cos 3t % c4 sin 3t c1e

2t % c2e "2t

cos 3t % c3e "2t

sin 3t c1e

"t % c2e t cos 22t % c3et sin 22te2t " 22 et sin 22tet sin t " et cos t

A22/4B 3 eA2%22Bt " eA2"22Bt 42e"t cos t % 3e"t sin tc1et % c2e"t cos 2t % c3e"t sin 2t c1e

t/2 cos A323t/2B % c2et/2 sin A323t/2Bc1e"5t cos 4t % c2e"5t sin 4tc1e"t cos 2t % c2e"t sin 2t c1e

"5t % c2te "5t

c1e "2tcos2t % c2e

"2tsin2t c1e

"t/2cos A25t/2B % c2e"t/2sin A25t/2Bc1e2tcos A23tB % c2e2tsin A23tBc1e 5tcos t % c2e

5tsin t c1cos t % c2sin t

Exercises 4.5, page 187 1. (a)

(b) (c)

3. 5.

7. 9. Yes 11. No 13. Yes 15. No

17. 19. 21. 23. 25. 27.

29. 31. 33.

35. 37. 39. 41. (a)

for (b)

for (c)

43.

45. (a)

(b) 47. (a) (b) No solution

Exercises 4.6, page 192 1. 3. 5.

7. 9.

11. 13.

15.

17. " A1/2B A cos 2tB ln ƒ sec 2t % tan 2t ƒc1 cos 2t % c2 sin 2t " et/5 %3 A cos tB ln ƒ cos t ƒc1 cos t % c2 sin t " t2 % 3 % 3t sin t % A1/8B A sin 2tB ln ƒ sec 2t % tan 2t ƒc1cos 2t % c2sin 2t % A1/24Bsec2 2t " 1/8

c1cos t % c2sin t % A sin tB ln ƒ sec t % tan t ƒ " 2yp ! "2t " 4 A2ln t " 3Bt2e"2t /4 % c1e"2t % c2te"2t% c1cos 3t % c2sin3 t " A1/9B % A1/9B A sin 3tB ln ƒ sec3t % tan 3t ƒt2e"t /2 % c1e"t % c2te"t A cos tB ˛ ln ƒ cos t ƒ % tsin t % c1cos t % c2sin t

c sin 3t " cos 6t , 2 sin 3t " cos 6t V ! 0.73

y AtB ! p 2

sin t for V ! 1

y AtB ! 2Vcos Ap/2VB V 2 " 1

sin t for V * 1;

y ! " cos t % A1/2B sin t " A1/2Be"3t % 2e"tc1 ! "2 Ae3p/ 2 % 1B , c2 ! " Ae3p/ 2 % 1B t 7 3p/2

y2 ! yh ! Ac1 cos2t % c2 sin2tBe"t0 ( t ( 3p/2 y1 ! " A2 cos 2t % sin 2tBe"t % 2yp ! At/10 " 4/25Btet " 1/2

yp ! t 2 % 3t " 1

yp ! AA1t % A0B cos 3t % AB1t % B0B sin 3t % Ce5t% D1cos3t % D2sin3t % E1cos t % E2sin t xp AtB ! AAcos t % Bsin tBet % C2t2 % C1t % C0yp ! AA1t % A0Bt cos t % AB1t % B0Bt sin t % C # 10t

y ! " A1/2B sin u " A1/3Be2u % A3/4Beu % A7/12Be"u% A3/20B sin 2x y ! " A3/10B cos x " A1/10B sin x " A1/20B cos 2xz ! e"x " cos x % sin x y ! et " 1 y ! A1/2B u e"usin u % Ac1 cos u % c2 sin uBe"uy ! A cos x " sin xBex/2 % c1ex % c2e2x y ! 11t " 1 % c1e

t % c2e "t

y ! tanx % c1e 22x % c2e"22xy ! ex % x2 % c1e"2x % c2e"3x

y ! t % c1 % c2e "t

11t/4 " 11/8 " 3sin(2t) t/2 " 1/4 " A3/4B sin A2tBt/4 " 1/8 % 3 sin A2tB 4 /4

B-6 Answers to Odd-Numbered Problems

19.

21. 0.3785

Exercises 4.7, page 200 1. Unique solution on (0, 3) 3. Unique solution on 5. Does not apply; t ! 0 is a point of discontinuity 7. Does not apply; not an initial value problem 9.

11. 13. 15. 17. 19. 21. 23. (c) 25. (a) True (b) False 27. (e) No, because the coefficient of vanishes at

and the equation cannot be written in standard form. 29. Otherwise their Wronskian would be zero at t0, contra-

dicting linear independence. 31. (a) Yes (b) No (c) Yes (d) Yes 33. Cte"t

35. 37. 39. 41.

43.

45. 47.

49. (a)

(b)

51.

53. (a)

Exercises 4.8, page 212 1. Let . Then

. But , so or .

3. The spring stiffness is , so it opposes negative dis- placements and reinforces positive displacements

. Initially y , 0 and , so the (positive) stiffness reverses the negative velocity and restores y to 0.

y¿ 6 0Ay 7 0B Ay 6 0B A"6yBY– AtB % tY AtB ! 0 y– A"tB % ty A"tB ! 0y– AsB " sy AsB ! 0Y¿ AtB ! "y¿ A"tB, Y– AtB ! y– A"tB

Y AtB ! y A"tB f¿(t0) ! limnSq

f(tn) " f(t0) tn "t0

! lim nSq

0 " 0 tn"t0

! 0

tw– % 2tw¿ % (t % 1)w ! 0

(3t " 2t3)" (3t " 2t3)"2 et 2 dt (1 " 2t2)" (1 " 2t2)"2 et 2 dt

t % 1 t4 c1t % c2t ln t % (1/2) t (ln t)

2 % 3t(ln t) [ ln 0 ln t 0 ]% (1/9) cos (3 ln t)ln 0 sec (3 ln t) % tan (3 ln t) 0c1 cos (3 ln t) % c2 sin (3 ln t) c1(5t " 1) % c2e

"5t " t 2e"5t /10 c1e

t % c2(t % 1) " t 2

1 % 2t " t2

t ! 0y–

t a cos(b ln 0 t 0 ), t a sin (b ln 0 t 0 ); t r, t r ln 0 t 0c1(t " 2) % c2(t " 2)7t " 3t 4

t"4 Ec1 cos 3 ln A"tB 4 % c2 sin [ ln A"tB 4 Fc1 t cos 32 ln A"tB 4 % c2t sin 32 ln A"tB 4c1t "1/3 % c2t"1/ 3ln t

c1t "1 % c2t

"4 c1t

"3 % c2t 2

A0, q B

3 y A2B ! "1.93 4y ! e1" t " et"1 % e t

2 " t

e"u

u du "

e"t

2 " t

u du

Thereafter, y - 0 and the negative stiffness drives y to %+.

5. (a) . Thus, by setting K ! 0 and

choosing the (") sign in equation (11), we get

, or .

(b) Linear dependence would imply

constant

in a neighborhood of 0, which is false if . (c) If , then ,

, which is false for the given data.

7. (a) The velocity, which is always perpendicular to the lever arm, is . Thus, (lever arm) times (per- pendicular momentum) ! .

(b) The component of the gravitational force perpendic- ular to lever arm is mg sin u, and is directed toward decreasing u. Thus, torque ! .

. 9. 2 or "2

11. The sign of the damping coefficient indicates that low velocities are boosted by negative damping but that high velocities are slowed. Hence, one expects a limit cycle.

13. (a) Airy (b) Duffing (c) van der Pol 15. (a) Yes ( positive stiffness)

(b) No ( negative stiffness) (c) Yes ( positive stiffness) (d) No ( negative stiffness for y , 0) (e) Yes (4 % 2 cos t ! positive stiffness) (f) Yes (positive stiffness and damping) (g) No (negative stiffness and damping)

17.

Exercises 4.9, page 222 1.

amplitude ! ; period ! ; frequency ! ; sec

3. b ! 0: y(t) ! cos 4t 3p " arctan A5/4B 4 /55/2p 2p/5241/20y AtB ! " A1/4B cos 5t " A1/5B sin 5t

1/422 y5 ! y4 ! "t 2 ! t 2 !

Ay¿ B2 " 1 "/mg sin u ! Am/2u¿ B ¿ ! m/2u–ddt

"/mg sin u

/ m/ du/dt ! m/2du/dt / du/dt

y¿ A0B ! "1/c2 ! "y A0B2y A0B ! "1/cy AtB ! 1/ At " cB c1 * c2

y1 AtB y2 AtB ! 1/ At " c1B1/ At " c2B ! t " c2t " c1 #

y ! 1/ At " cBt ! "" dy22y4/2 % c ! 1y % c y– ! 2y3 ! d

dy Ay4/2B

π 2π

1 b = 0

−1

Figure B.19 b ! 0

Answers to Odd-Numbered Problems B-7

% 3 A1 " 25B /2 4 eA"5"25Btk ! 20: y AtB ! 3 A1 % 25B /2 4 eA"5%25Bt

, where f ! arctan 27/3 ! 0.723! A4/27Be"3t sin A27t % fBb ! 6: y AtB ! e

"3t cos 27t % A3/27Be"3t sin 27t

b ! 8: y AtB ! A1 % 4tBe"4t

b ! 10: y AtB ! A4/3Be"2t " A1/3Be"8t

b = 6

–1

1 e –3t4

√7

e–3t4 √7

Figure B.20 b ! 6

y b = 8

1 4

Figure B.21 b ! 8

y b = 10

1 6 ln 4

Figure B.22 b ! 10

y k = 20

Figure B.23 k ! 20

y k = 25

1 5

Figure B.24 k ! 25

where f ! arctan A1/25B ! 0.421! 26e"5t sin A25t % fB ,k ! 30: y AtB ! e

"5t cos 25t % 25e"5t sin 25t

k ! 25: y AtB ! A1 % 5tBe"5t

7. , where

arctan ; damp. factor ! ; quasiperiod ; quasifreq.

9. 0.242 m

11. arctan sec

13. Relative extrema at

for ; but touches curves at for

15. First measure half the quasiperiod P as the time between two successive zero crossings. Then compute the ratio

Exercises 4.10, page 229

1. M AgB ! 1/2 A1 " 4g2B2 % 4g2 y At % PB /y AtB ! e"Ab/2mBP

m ! 0, 1, 2, p . t ! 3p/2 % mp " arctan A23/2B 4 / A223B#27/12e"2tn ! 0, 1, 2, pt ! 3p/3 % np " arctan A23/2B 4 / A223B

A29999B ! 0.156A10/29999B ! 4/p! p/4 A5/4Be"8t! 3.785A3/4Bf ! p %

! A5/4Be"8t sin A8t % fBy AtB ! A"3/4Be"8t cos 8t " e"8t sin 8t

k = 30

–1

1 √6e–5t

–√6e–5t

√6

–√6 Figure B.25 k ! 30

3. y AtB ! cos 3t % A1/3Bt sin 3t

5. (a)

! AF0/ 3m Av2 " g2B 4 B A cos gt " cos vtB% 3F0/ Ak " mg2B 4 cos gt y AtB ! " 3F0/ Ak " mg2B 4 cos A2k/m tB

1 2 3 4

Figure B.26

t 1

3 2π

3 4π

3 5π

3 π

2π

Figure B.27

B-8 Answers to Odd-Numbered Problems

CHAPTER 5

Exercises 5.2, page 250 1. (a) (b)

3. 5. 7.

11.

13.

15.

17.

19.

21.

23. Infinitely many solutions satisfying

25.

27.

29.

31.

r2 ! "5 % 27

100 ! "0.0235

r1 ! "5 " 27

100 ! "0.0765 ,

y AtB ! 3027 er1t " 3027 er2t % 20 kg, where% 20 kg , x AtB ! " a10 % 2027b er1t " a10 " 2027b er2t"3 ( l ( "1 z AtB ! "c1e8t % c3y AtB !

1 2

Ac1e8t " c2e4t % c3B ,x AtB ! c1e 8t % c2e

4t % c3 ,

z AtB ! 2c1et % 4c2e2t % 4c3e3ty AtB ! c1et % c2e2t % c3e3t , x AtB ! "c1et " 2c2e2t " c3e3t ,x % y ! et % e"2t y ! et % e"t " cos t " sin t x ! et % e"t % cos t % sin t ; y ! "2e3t % 2e2t x ! 2e3t " e2t ; y ! c1 cos t % c2 sin t % c3 cos A26 tB % c4 sin A26 tB" 4c4 sin A26 tB ; x ! c1 cos t % c2 sin t " 4c3 cos A26 tBz ! c1e2t % c2e7t " 5t " 2w ! " A2/3Bc1e

2t % c2e 7t % t % 1 ;

y ! c1e t cos 2t % c2e

t sin 2t x ! 2c2e

t cos 2t " 2c1e t sin 2t ;

% A1/5Bet" A2/5Bc4e"23ty ! c1 cos 2t % c2 sin 2t " A2/5Bc3e23t " A3/10Bet ;u ! c1 cos 2t % c2 sin 2t % c3e23t % c4e"23t

y ! "3c1e t " A3/4Bcos t " A1/4Bsin tx ! c1et % A1/4Bcos t " A1/4Bsin t ;

y ! c1 % c2e "t % A5/3Btu ! c1 " A1/2Bc2e"t % A1/2Bet % A5/3Bt ;

x # "5 ; y # 1 x ! c1 % c2e

"2t ; y ! c2e"2t " 2t3 % 3t2 % 6t % 16

" 2t3 % 3t2 % 6t % 162t3 % 3t2 " 16 " 2t3 % 3t2 % 6t % 16" t3 % 3t2 % 8

7. ,

where and tan as in equation (7)

9. , where

11. , where

; res. freq.

13.

15.

freq. ! 2//

Review Problems, page 233 1. 3. 5. 7. 9.

11. 13. 15. 17. 19. 21.

23.

25. 27. 29. 31. 33. 35. 37. (a), (c), (e), and (f) have all solutions bounded as

39. yp AtB ! A1/4B sin 8t ; 262/2pt S%q cos u % 2 sin u % u sin u % Acos uB ln ƒ cos u ƒ"e"t " 3e5t % e8t 2et cos 3t " A7/3Bet sin 3t " sin 3te"2t cos A23tBc1x % c2x

"2 " 2x"2 ln x % x ln x c1e

3t/2 % c2te3t/2 % e3t /9 % e5t /49 " A1/16B A cos 4uB ln ƒ sec 4u % tan 4u ƒc1 cos 4u % c2 sin 4u %t2 % 6t/7 % 4/49

c1e 3t/2 cos A219t/2B % c2e3t/2 sin A219t/2B " et/5c1e"3t % c2et/2 % c3tet/2c1et % c2e"t/2 cos A243t/2B % c3e"t/2 sin A243t/2B

c1e "2t % c2e

"t % c3e "t/ 3

c1 cos 4t % c2 sin 4t % A1/17Btet " A2/289Bett1/ 2Ec1 cos 3 A219/2B ln t 4 % c2 sin 3 A219/2B ln t 4 Fc1e7t/4 % c2te7t/4 c1e

"t/3 cos At/6B % c2e"t/3 sin At/6Bc1e3t/2 % c2et/3 c1e

t/2 cos A3t/2B % c2et/2 sin A3t/2Bc1e"9t % c2et

Amp ! B a" 11986b 2 % a" 31972b 2 ! 0.01(m) ,yp AtB ! A3/185B A8 sin 4t " 11 cos 4tB ! 222/p cycles/secu ! arctan A9/2B ! 1.352% A2/285B sin A2t % uB

y AtB ! " A18/85Be"2t cos 6t " A22/255Be"2t sin 6tu ! arctan A4/3B ! 0.927 ! A0.1B sin A2t % uByp AtB ! A0.08B cos 2t % A0.06B sin 2t u ! Ak " mg2B / AbgBr1, r2 ! " Ab/2mB # A1/2mB2b2 " 4mk

y AtB ! c1er1t % c2er2t % F0 sin Agt % uB2 Ak " mg2B2 % b2g2 t

sin t

–sin t

–1

/ 2/

Figure B.28

33.

35. 37.

Exercises 5.3, page 260 1. , ; ,

11. i 13. i

1 0.250 0.750000 1 0.250 0.25000 2 0.500 0.625000 2 0.500 0.50000 3 0.750 0.573529 3 0.750 0.75000 4 1.000 0.563603 4 1.000 1.00000

15. 17.

19. Part (a) Part (b) Part (c)

11 0.5 1.95247 2.25065 1.48118 2.42311 0.91390 2.79704 12 1.0 3.34588 1.83601 2.66294 1.45358 1.63657 1.13415 13 1.5 4.53662 3.36527 5.19629 2.40348 4.49334 1.07811 14 2.0 2.47788 4.32906 3.10706 4.64923 5.96115 5.47788 15 2.5 1.96093 2.71900 1.92574 3.32426 1.51830 5.93110 16 3.0 2.86412 1.96166 2.34143 2.05910 0.95601 2.18079 17 3.5 4.28449 2.77457 3.90106 2.18977 2.06006 0.98131 18 4.0 3.00965 4.11886 3.83241 3.89043 5.62642 1.38072 19 4.5 2.18643 3.14344 2.32171 3.79362 5.10594 5.10462 10 5.0 2.63187 2.25824 2.21926 2.49307 1.74187 5.02491

y AtiBx AtiBy AtiBx AtiBy AtiBx AtiBti u A1; 2"2B ! y A1; 2"2B ! 0.36789y A1B ! x1 A1, 2"2B ! 1.69, y¿ A1B ! 1.82

y AtiBtiy AtiBtii ! 1, 2, . . . , m xm,n % hƒm Atn, x1,n, . . . , xm,nB 4 . . . , fi Atn % h, x1,n % hƒ1 Atn, x1,n, . . . , xm,nB,

x i, n%1 ! x i,n % h 2

3 fi Atn, x1,n, . . . , xm,nB %tn%1 ! tn % h, n ! 0, 1, 2, . . . ; x4 A0B ! x5 A0B ! 1x¿5 ! A2x4 " 2x3 % 1B /5 ; x1 A0B ! x2 A0B ! x3 A0B ! 4 , x¿1 ! x2 , x¿2 ! x3 , x¿3 ! x4 % t , x¿4 ! x5 , x1 A3B ! 5 , x2 A3B ! 2 , x3 A3B ! 1 , x4 A3B ! "1x¿3 ! x4 , x¿4 ! x1 " x3 " 1 ; x¿1 ! x2 , x¿2 ! x2 " x3 % 2t , x3 A0B ! 0 , x4 A0B ! 2x¿4 ! x4 " 7x1 % cos t ; x1 A0B ! x2 A0B ! 1 , x¿1 ! x2 , x¿2 ! x3 , x¿3 ! x4 , x2 A0B ! "6 x1 A0B ! 3x¿2 ! 3x1 " tx2 % t2x¿1 ! x2

460/11 ! 41.8ºF90.4ºF % 20

y ! " a 1025b e A"3%25 Bt/100 % a 1025b e A"3"25 B t/100 " c 20 % 102525 d e A"3"25 Bt/100 % 20 ;

x ! c 20 " 102525 d e A"3%25 Bt/100 21. i11 0.5 0.09573 12 1.0 0.37389 13 1.5 0.81045 14 2.0 1.37361 15 2.5 2.03111 16 3.0 2.75497 17 3.5 3.52322 18 4.0 4.31970 19 4.5 5.13307 10 5.0 5.95554

23. Yes, yes 25. 27. 29. (a)

(b) (c)

All populations approach 0.5.

Exercises 5.4, page 272 1. x ! y3 , y 7 0

P1 A10B ! 0.463, P2 A10B ! 0.463, P3 A10B ! 0.567P1 A10B ! 0.463, P2 A10B ! 0.567, P3 A10B ! 0.463 P1 A10B ! 0.567, P2 A10B ! 0.463, P3 A10B ! 0.463y A0.1B ! 0.00647, . . . , y A2.0B ! 1.60009

y A1B ! x1 A1; 2"3B ! 1.25958

x1 AtiB ! H AtiBti B-10 Answers to Odd-Numbered Problems

Figure B.29

3. 5. 7. 9.

11. 13. Ax " 1B2 % Ay " 1B2 ! cy2 " x2 ! c ex % xy " y2 ! cex % ye"y ! c A0, 0BA1/22, 1/22 B , A"1/22, "1/22 B

Figure B.30

y = x

–1 –2 –3

Figure B.31

Answers to Odd-Numbered Problems B-11

21. is a center (stable).A0, 0B

Figure B.35

23. is a center (stable); is a saddle point (unstable).A1, 0BA0, 0B

27. approaches .A0, 0BAx AtB, y AtBB

−3

−2

−1 −1−2

Figure B.36

−1

−2

−1−2

Figure B.38

15. is a saddle point (unstable).A"2, 1B y

x −1

−1

−2−3−4

Figure B.32

17. is a center (stable).A0, 0B

19. is a saddle point (unstable).A0, 0B

Figure B.33

Figure B.34

25. (a) Periodic (b) Not periodic (c) Critical point (periodic)

2 31

−1

Figure B.37

29. (a) Unstable node (b) Center (stable) (c) Stable node (d) Unstable spiral (e) Saddle (unstable) (f) Asymptotically stable spiral

35. (c) For upper half-plane, center is at for lower half-plane, center is at ,

(d) All points on the segment (e) see Figure B.41.y ! "0.5;

y ! 0, "1 ( y ( 1 y ! 1.y ! 0

y ! 0, y ! "1;

Exercises 5.5, page 283 1. Runge–Kutta approximations

tn pn (r ! 1.5) pn (r ! 2) pn (r ! 3)

0.25 1.59600 1.54249 1.43911 0.5 2.37945 2.07410 1.64557 0.75 3.30824 2.47727 1.70801 1.0 4.30243 2.72769 1.72545 1.25 5.27054 2.86458 1.73024 1.5 6.13869 2.93427 1.73156 1.75 6.86600 2.96848 1.73192 2.0 7.44350 2.98498 1.73201 2.25 7.88372 2.99286 1.73204 2.5 8.20933 2.99661 1.73205 2.75 8.44497 2.99839 1.73205 3.0 8.61286 2.99924 1.73205 3.25 8.73117 2.99964 1.73205 3.5 8.81392 2.99983 1.73205 3.75 8.87147 2.99992 1.73205 4.0 8.91136 2.99996 1.73205 4.25 8.93893 2.99998 1.73205 4.5 8.95796 2.99999 1.73205 4.75 8.97107 3.00000 1.73205 5.0 8.98010 3.00000 1.73205

Limiting population is 31/(r"1).

% a19 2

% 2

4 % (p/12)2 b e"2t

x(t) ! 1 2

" 2 cos(pt /12)

4 % (p/12)2 "

(p/12)sin (pt /12)

4 % (p/12)2

B-12 Answers to Odd-Numbered Problems

2 31

−0.5

−1

0.5

Figure B.40 l ! 3

7.5

0 y

−1 1

3.5 −0.5

−1.5

−5.5

Figure B.41

(e) See Figures B.39 and B.40. When one critical point is a center and the other is a saddle. For the bar is attracted to the magnet. For the bar may oscillate periodically, or (rarely) come to rest at the saddle critical point.

l ! 3, l ! 1,

l ! 3,

10.5

−1

−0.5−1

Figure B.39 l ! 1

31. (a)

(b) implies constant, or

33. (a)

(b) implies

constant, or

constant. The solu- tion equation follows with the choice for the constant.

Solutions for the latter are and

are real only for (d)

l $ 2.

x ! l # 2l2 " 4

2 ,

y ! 0 and "x % 1/ Al " xB ! 0. C/2

y2/2 ! "x2/2 " ln Al " xB %"y dy ! " a"x % 1l " xbdx % dy dx

! "x % 1/ Al " xB

x¿ ! y, y¿ ! "x % 1/ Al " xBy2/2 ! F AyB % K "y dy ! " f AyB dy %dydy ! f AyBy

y¿ ! y , y¿ ! f AyB

Answers to Odd-Numbered Problems B-13

0.01

0.02

√175/8 Figure B.43

7. H, Ω, F, and

11.

13.

Exercises 5.8, page 305 1. For The Poincaré map alternates between the

points and There is a subharmonic solution of period For The Poincaré map cycles through the points

, and There is a

subharmonic solution of period 3. The points become unbounded. 5. The attractor is the point . 9. (a)

(b)

(c) xn ! 0 for n $ j E7/15, 14/15, 13/15, 11/15, 7/15, . . . FE1/3, 2/3, 1/3, . . . F , E1/5, 2/5, 4/5, 3/5, 1/5, . . . F , E1/15, 2/15, 4/15, 8/15, 1/15, . . . F , E3/7, 6/7, 5/7, 3/7, . . . FE1/7, 2/7, 4/7, 1/7, . . .F ,

A"1.0601, 0.2624B 10p.

A"0.9747, "0.4854).A"2.5136, 0.1854B, A"0.6114, 0.1854BA"2.1503, "0.4854B, A"1.5625, 0.6B, v ! 3/5:4p.

A0.8, "1.5B.A0.8, 1.5Bv ! 3/2: " A24/61Bsin 3t" A323/61B e"t sin23t % A81/61Bcos 3tI3 ! " A81/61Be

"t cos23t" A45/61Bcos 3t % A54/61Bsin 3t ,I2 ! A45/61Be "t cos23t " A3923/61B e"t sin23t% A30/61Bsin 3t ," A4223/61B e

"t sin23t % A36/61Bcos 3tI1 ! " A36/61Be"t cos23tI1 A0B ! I2 A0B ! I3 A0B ! 0 ; A1/2BI ¿1 % I2 ! 0, I1 ! I2 % I3 :A1/2BI ¿1 % 2q3 ! cos 3t Awhere I3 ! q¿3B , I3 ! A2/5Be"3t/2 " A2/5Be"2t/3I2 ! A1/5Be"3t/2 " A6/5Be"2t/3 % 1 , I1 ! A3/5Be"3t/2 " A8/5Be"2t/3 % 1 ,E AtB ! 50 cos 10t V

C ! 1/15R ! 10L ! 35

(b) 295 persons

11. Roughly, 2

Exercises 5.6, page 289 1.

The normal frequency has the

mode the normal frequency

has the mode

and the normal frequency

has the mode 5. 7.

Exercises 5.7, page 296 1. 3. resonance

frequency is 5/p. Ip AtB ! A4/51Bcos 20t " A1/51Bsin 20t ;I AtB ! A19/221 B 3 eA"25"5221 Bt/2 " eA"25%5221 Bt/2 4 A1/2pB2g/l ; A1/2pB2 Ag/lB % A2k/mB" A26/10B sin26 t " A1/2Bsin 2t y AtB ! A4/5Bcos t % A8/5Bsin t % A1/5Bcos 26 t% A26/5B sin 26 t " sin 2t ;x AtB ! A2/5Bcos t % A4/5Bsin t " A2/5Bcos 26 t x AtB! "e"t " te"t " cos t ; y AtB! e"t % te"t " cos tx AtB ! "z AtB, y AtB # 0 .A1/2pB22k/mx AtB ! z AtB ! A1/22 B y AtB; A1/2pB3 A2 "22 B Ak/mBx AtB ! z AtB ! " A1/22 B y AtB;

A1/2pB3 A2 %22 B Ak/mBmz– ! "k Az " yB " kz ;my– ! "k Ay " xB % k Az " yB , mx– ! "kx % k Ay " xB ,y AtB ! " A6/17Bcos t % A6/17Bcos A220/3 tBx AtB ! " A8/17Bcos t " A9/17Bcos A220/3 tB , x A0B ! "1 , x¿ A0B ! 0 , y A0B ! 0 , y¿ A0B ! 0m2y– ! "k1 Ay " xB " k2y ; m1x– ! k1 Ay " xB ,

N(t) ! N0 (c % s)

N0s % (c % s " N0s) e "(c%s)t

, N0 J N(0)

Q(t) ! exp e c[1 " exp("bt) ] b

" bt f [exp(bt) " 1]P(t) ! exp e c [1 " exp("bt) ]

b " bt f ,

150

300

167 700

Figure B.42

5. (a) 5. M AgB ! 1/2 A100 " 4g2B2 % 100g2

Review Problems, page 308 1.

5. , ,

9. 11.

13. critical point is saddle (unstable).

15. Asymptotically stable spiral point 19. , ,

where q is the charge on the capacitor

CHAPTER 6

Exercises 6.1, page 325 1. 3. 5. 7. Lin. indep. 9. Lin. dep.

11. Lin. indep. 13. Lin. indep. 15. 17. 19. (a)

(b) 21. (a)

(b) 23. (a) (b) 29. (b) Let and 33. 35.

Exercises 6.2, page 331 1. 3. 5. c1e"x % c2e"x cos 5x % c3e"x sin 5x

c1e "x % c2e

"2x/3 % c3ex/2c1 % c2e2x % c3e"4x

xy‡ " y– % xy¿ " y ! 0 e2x , Asin x " 2 cos xB /5 , " A2 sin x % cos xB /5f2 AxB ! x " 1f1 AxB ! 0 x " 1 0 4x " 6 sin x2 sin x " x

3x " x ln x % x Aln xB2 % ln xc1x % c2x ln x % c3x Aln xB2 % ln x "ex % e"x sin 2x % x2 c1e

x % c2e "x cos 2x % c3e

"x sin 2x % x2 c1x % c2x

2 % c3x 3c1e

3x % c2e "x % c3e

"4x

A0, q BA3p/2, 5p/2BA"q, 0B

I1 ! e "t 3 AA " BBsin t " AA % BBcos t 4 .I2 ! e"t AB cos t " A sin tB ,I3 ! e"t AA cos t % B sin tB , AI1 ! dq/dtB;R2I2 ! R1I3 % L dI3/dt ,

q/C ! R2I2I1 % I2 % I3 ! 0

A0, 1Bx2 " Ay " 1B2 ! c ;x¿4 ! x5 , x¿5 ! x6 , x¿6 ! x2 " x3 x¿1 ! x2 , x¿2 ! x3 , x¿3 ! t " x5 " x6 , x¿1 ! x2 , x¿2 ! x3 , x¿3 ! A1/3B A5 % etx1 " 2x2By AtB ! " A13.9/2Be"t/6 % A4.9/2Be"t/2 % 4.8 x AtB ! " A13.9/4Be"t/6 " A4.9/4Be"t/2 % 4.8 ,z AtB ! et % cos t % sin t

y AtB ! et " cos t % sin tx AtB ! 2 sin t" A11/20Bet " A1/4Be"t y AtB ! A3/2B Ac1 % c2Bcos 3t " A3/2B Ac1 " c2Bsin 3tx AtB ! c1 cos 3t % c2 sin 3t % et/10 , y AtB ! c1t2 % c2t % c3x AtB ! " Ac1/3Bt3 " Ac2/2Bt2 " Ac3 % 2c1Bt % c4 ,

7. 9.

11.

13. 15.

17.

19. 21. 23. 27. 29.

31. (a) (b) (c)

33. (b)

(c)

(d)

35. , where

Exercises 6.3, page 337 1. 3. 5.

7. 9.

11. 13. 15. 17. 19. 21. 23. 25. 27. 29. 31. 33.

39.

Exercises 6.4, page 341 1. 3. 5. 7.

11. c1x % c2x"1 % c3x3 " x sin x " 3 cos x % 3x"1 sin x

% A1/3Be2x " e"2xg AxBdx " A1/2Bex " e"xg AxBdx % A1/6Be"x " exg AxBdxc1x % c2x

2 % c3x 3 " A1/24Bx"1ln Asec xB " Asin xBln Asec x % tan xB e2x/16A1/6Bx2e2x

y AtB ! A8/63Be3t % c1 % c2t % c3e22t % c4e"22tx AtB ! " A1/63Be3t % c1 % c2t " c3e22t " c4e"22t , x2e"2x " x2 % 3"2e3x % e"2x % x2 " 1

c2x % c3x 2 % c6 x

2ex c3xe

"x cos x % c4xe "x sin x % c5x

2 % c6 x % c7 c3 % c4x % c5 cos 2x % c6 sin 2x c3xe

3x % c4x 2 % c5x % c6

c3 cos 2x % c4 sin 2x % c5 AD % 2B2 3 AD % 5B2 % 9 4 23 AD % 1B2 % 4 4 3 AD " 2B AD " 1BD % 7D5c1ex % c2xex % c3x2ex % A1/6Bx3ex

c1e x % c2e

"2x % c3xe "2x " A1/6Bx2e"2x% A5/18Bx % 37/108

c1e x % c2e

3x % c3e "2x " A1/6Bxex % A1/6Bx2c1x2e"2x

c1xe x % c2 % c3x % c4x

r4 ! k/ AEIBc1 cosh rx % c2 sinh rx % c3 cos rx % c4 sin rx y AtB ! A6/5Bcos t " A1/5Bcos26 tx AtB ! A3/5Bcos t % A2/5Bcos 26 t ," Ac4/2Bsin26 t y AtB ! 2c1 cos t % 2c2 sin t " Ac3/2B cos 26 t% c4 sin26 tx AtB ! c1 cos t % c2 sin t % c3 cos 26 t Ex, x2 cos A3 ln xB, x2 sin A3 ln xBFEx, x2, x"1, x"2FEx, x"1, x2F % c3e

"0.5x cos A1.323xB % c4e"0.5x sin A1.323xBc1e"0.5x cos A0.866xB % c2 e"0.5x sin A0.866xB c1e

1.120x % c2e 0.296x % c3e

"0.520x % c4e "2.896x

x AtB ! c1 % c2t % c3et, y AtB ! c1 " c2 % c2te2x " 22ex sin22xex " 2e"2x " 3e2x% c11x % c12x2 % c13x3 % c14x4 % Ac6 % c7xBe"2x cos x % Ac8 % c9xBe"2x sin x % c10c1e"4x % c2e3x % Ac3 % c4x % c5x2Be"2x

% Ac6 % c7xBe"x sin 2xc1ex % c2xex % c3e"3x % Ac4 % c5xBe"x cos 2x c1 cos22x % c2x cos22x % c3 sin22x % c4x sin22xc1e"x % c2xe"x % c3x2e"x % c4x3e"xc1e3x % c2xe3x % c3x2e3x c1e

"x % c2eA3%265 Bx/4 % c3eA3"265 Bx/4 B-14 Answers to Odd-Numbered Problems

x 2.20–2.2

–2

Figure B.44

13.

Review Problems, page 343 1. (a) (b) 5. (a)

(b)

7. (a) (b) (c) (d) (e)

CHAPTER 7

Exercises 7.2, page 360

1. 3.

9. 11.

13.

15.

17.

19.

21. Continuous (hence piecewise continuous) 23. Piecewise continuous 25. Continuous (hence piecewise continuous) 27. Neither 29. All but functions (c), (e), and (h)

Exercises 7.3, page 365

9. 11.

13. 15. 3s

4 As2 % 1B % s4 As2 % 9B12s " s2 As2 % 4B s

s2 " b2 4 As % 1B3 As % 1B2 % 4 4 2

s5 "

s4 %

s3 "

s2 %

1 s

4As % 1B3 " 1s2 % ss2 % 16 s % 1As % 1B2 % 9 % 1s " 6 " 1s

s3 %

2As " 1B2 % 4

24As " 5B5 " s " 1As " 1B2 % 7 , s 7 5 6As " 3B2 % 36 " 6s4 % 1s " 1 , s 7 3

s4 "

1As " 1B2 % s " 4As " 4B2 % 1 , s 7 4 6

s % 3 "

s3 %

s2 "

8 s , s 7 0

e"ps % 1

s2 % 1 , all s e"2s a2s % 1

s2 b , s 7 0

s " 2As " 2B2 % 9 , s 7 2 s

s2 % 4 , s 7 0

1 s " 6

, s 7 6 1 s2

, s 7 0

c1x % c2x 5 % c3x

"1 " A1/21Bx"2D3 AD % 1B2 AD2 % 4B AD2 % 9B 3 AD % 2B2 % 9 4 3 3D2 % 4 4 2D2 AD " 3BD3% Ae"x sin23xB Ac9 % c10 xB % Ae"x cos23xB Ac7 % c8xBc1 % c2x % c3x2 % c4x3 % ex Ac5 % c6xB% Acos xB Ac6 % c7xB % Asin xB Ac8 % c9 xB

e"5x Ac1 % c2xB % e2x Ac3 % c4x % c5 x2BA"4, "1B , A"1, 1B , A1, q BA0, q B 17.

19.

21.

25. (a) (b)

29. 33.

35.

Exercises 7.4, page 374 1. 3. 5. 7. 9.

11. 13.

15.

17.

19.

21.

23.

25.

27.

29. 31.

33. 35.

39.

41.

43.

Exercises 7.5, page 382 1. 3.

5. 7.

9. 3 At " 1Bet"1 " e"6 At"1B cos t " 4e5t % 8e2tt2 % cos t " sin t "e"3t % 3te"3t2etcos 2t % etsin 2t

s % 2 %

2 As " 1B % 2 A3BAs " 1B2 % 22 2e"t " 4e3t % 5e2t

C ! 2s % 1

s As " 1B ` s!"2 ! "12 B !

2s % 1 s As % 2B ` s!1 ! 1 ,

A ! 2s % 1As " 1B As % 2B ` s!0 ! "12 ,

A s

% B

s " 1 %

C s % 2

, where

2 Acos t " cos 3tB /te5t / t " e"2t / t!"1 E1/ s2F AtB ! f3 AtB ! t F1 AsB ! F2 AsB ! F3 AsB ! 1/s2 ,3e"2t % 7etcos t % 11etsin t " A5/3Be"t % A5/12Be2t % A5/4Be"2t8e2t " e"tcos 2t % 3e"tsin 2t "e"3t % 2te"3t % 6e"t A1/3B % et % A14/3Be6t 1 17 c 1 s " 3

" s % 1As % 1B2 % 1 " 4As % 1B2 % 1 d

" 5

6s %

11 10 As " 2B " 415 As % 3B

" 3

s % 1 % As " 1B % 2As " 1B2 % 4

1As % 1B2 " 2s6s % 5 " 1s % 2 " 4s " 1 A3/2Betcos 2t " 3etsin 2t2e"2tcos 3t % 4e"2tsin 3t

A1/2Be"2tsin 2te"tcos 3tett3 e"Ap/2Bs/ As2 % 1B

e"s/s21 s2 % 6s % 10

2s3 " 6sb2As2 % b2B3s2 " b2As2 % b2B2 s " aAs " aB2 % b2

n % m

2 3 s2 % An % mB2 4 % m " n2 3 s2 % Am " nB2 4 s

2 As2 % 9B " s2 As2 % 49B Answers to Odd-Numbered Problems B-15

11. 13.

15.

17. 19.

21. 23.

25.

27. 29.

31.

35.

37.

39.

41.

Exercises 7.6, page 393 1. 3.

9. 11. 13. 15. 17. 19.

21. 1 " 2se "2s " e"2s

s2 A1 " e"2sB % 10u At " 4pB 31 " e"At"4pB Acos t % sin tB 410 " 10u At " 3pB 31 % e"At"3pB Acos t % sin tB 4A7e

6"2t " 6e3" tBu At " 3Be"2At"3B 3 cos At " 3B " 2 sin At " 3B 4u At " 3Be"2At"2Bu At " 2B " 3e"2At"4Bu At " 4B et"2u At " 2BAe"s " 2e"2s % e"3sB /s23 Ae"s " e"2sB As % 1B 4 /s2

A2e"s " e"2s % 2e"3sB /se"2s A4s2 % 4s % 2B /s32e"s/s3 e AtB ! A"2aI/24Ik "m2 B e"mt/ A2IBsin A24Ik "m2 t/ A2IB Be AtB ! "a cos A2k/I tBcos t % t sin t % c Asin t " t cos tB, Ac arbitraryB t2/2 5/2 % Aa " 5/2Be"tcos t % Aa % b " 5/2Be"tsin tA3a " bBet/2 % Ab " aBe3t /2At2 " 4Be"t 2et " cos t " sin t

"s3 % 1 % 3se"2s " e"2s

s2 As2 % 4Bs3 % s2 % 2sAs2 % 1B As " 1B2 s3 % 5s2 " 6s % 1

s As " 1B As2 % 5s " 1Bs5 % s4 % 6s4 As2 % s " 1B "s2 % s " 1As2 % 1B As " 1B As " 2B

A7/5Bsin t % A11/5Bcos t % A3/5Be2t"p " e"At"p/2B2 " t % e2" t % 2et"2 29. sin t % 3 1 " cos At " 3B 4u At " 3B B-16 Answers to Odd-Numbered Problems

f(t)

10 2 3 4 5 6 Figure B.45

23. 1

1 " e"2s c 1 " e"s"1

s % 1 %

e"s " e"2s

s d

f(t)

1 2 3 4 5 Figure B.46

25. 27. 1 " e "as

as2 A1 % e"asB1s A1 % e"asB

–1 321 4 5 6

Figure B.47

642 8 10 Figure B.48

31. t % 34 " t % sin At " 2B " 2 cos At " 2B 4 u At " 2B

33.

35.

37.

39.

45.

for

47.

49.

51. (a) (b)

59. 61. The resulting differential equation has polynomial

coefficients, so the Laplace transform method will result in a differential equation for the transform.

Exercises 7.7, page 403

3. " t

0 g AyBe2y"2tsin At " yBdy % e"2tcos t % 3e"2tsin t

2tet " et % " t

0 et"y At " yBg AyBdy

0.4 " 0.2e"3t/125 % 0.2u At " 10B 31 " e"3At"10B/125 41052p/ A16s9/2B2p/s a q

n!1

A"1Bn%1 2ns2n

! 1

2 ln a1 % 1

s2 b

a q

n!1

1 sn

! 1

s " 1

n 6 t 6 n % 1

% e"2t

3 c 1 % e % e2 " e2n%2

e % 1 d ,

et"n

6 "

e"t

2 A1 % e " en%1B" A7/4Be

"2At"5B % A11/9Be"3At"5B 4u At " 5B" 319/36 % A1/6B At " 5B % A2/9Be"3At"1B 4u At " 1B% 31/36 % A1/6B At " 1B " A1/4Be"2At"1B

2e"2t " 2e"3t % A1/6B 38 % u At " 2pB 4 sin 2tcos 2t % A1/3B 31 " u At " 2pB 4 sin t

e"t % e"2t % A1/2B 3 e"3t " 2e"2At%1B% e"At%4B 4u At " 2B" A1/2B 31 " e4p"t Acos t % sin tB 4u At " 4pB % A1/2B 31 " e2p"t Acos t % sin tB 4u At " 2pBe"tcos t % 2e"tsin t

Exercises 7.9, page 413 1. 3. 5.

9. 11.

13.

15. 17. 19. 21.

23.

Review Problems, page 416

1. 3.

5. 7.

9. 11.

13. 15. 17. 19. 21.

23.

25.

27. 29. 31.

y ! Aet " e"tB /2 " A1/2B 3 et"2 " e"At"2B 4u At " 2B " 31 " Aet"2 % e"At"2B B /2 4u At " 2B ,x ! 1 " Aet % e"tB /2 As2 " 5s % 6B"1 ; e3t " e2t A9/10Be"3t % A1/10Bcos t " A3/10Bsin t c 3 t % te"2t % e"2t " 1 4 " A1/4Be"3At"1B/2cos 327 At " 1B /2 4 Vu At " 1B % U A1/4B " 33/ A427 B 4 e"3At"1B/2sin 327 At " 1B /2 4 A2/27 B e"3t/2sin A27 t/2B" A3/2B % t % t2/2 % A3/2Be"tcos t " A1/2Be"tsin t e2t " e5t 32et"2 % 2e4"2t 4u At " 2B e"2t % e"t " 2te"t 2et % 2e"2tcos 3t % e"2tsin 3t

A7/2Bt2e"3t 2e"4s c 1 s3

% 4

s2 %

8 s d

s2 " 36As2 % 36B2 1s " 2 " 6s4 % 2s3 " 5s2 % 25 2As % 9B3 3s % e"2s c 1s " 1s2 d

I3 ! e "20t " 4e"5t % 3

I2 ! "2e "20t % 2e"5t ,

I1 ! "e "20t " 2e"5t % 3 ,

I1 ! I2 % I3 ; I1 A0B ! I2 A0B ! I3 A0B ! 0 ;2I1 % A0.1BI¿3 % A0.2BI¿1 ! 6 , A0.1BI¿3 " I2 ! 0 , % 348 " 72e"At"5B/6 % 24e"At"5B/2 4u At " 5B y ! 2e"t/2 % 2e"t/6% 348 " 36e"At"5B/6 " 12e"At"5B/2 4u At " 5B ;

x ! "e"t/2 % e"t/6 x ! "7e"t % et ; y ! 2e"t ; z ! "13e"t % et x ! At " 2Bet"2 ; y ! et"2 x ! t2 ; y ! t " 1

% A1/2Bsin t 4u At " pB y ! e"t % 31 " A1/2Be"At"pB % A1/ 2Bcos tx ! e "t " A1/2B 3 e"At"pB % cos t " sin t 4 u At " pB ;" 31 " Aet"2 % e"At"2B B /2 4 u At " 2B

y ! 1 " Aet % e"tB /2x ! Aet " e"tB /2 " A1/2B 3 et"2 " e " At"2B 4u At " 2B ;x ! 4e"2t " e"t " cos t ; y ! 5e"2t " e"t " A146215/85B e5t/2sin A215 t/2B % A22/17Be"3t y ! A46/17Be5t/2cos A215 t/2B" A334215/85B e

5t/2sin A215 t/2B " A3/17Be"3t ;x ! " A150/17Be5t/2cos A215 t/2By ! A7/4Be t " A7/4Be"t % A1/2Bsin t x ! A7/4Bet % A7/4Be"t " A3/2Bcos t ;

z ! et ; w # 0 x ! et ; y ! et

Answers to Odd-Numbered Problems B-17

5. 7. 9. 11. 13. 15. 17. 19. 3

21.

23.

25.

27.

29.

31.

Exercises 7.8, page 410 1. 3. 5. 7. 9. 11. 0 13.

15.

17.

19. 21. ; see Fig. B.49.sin t % Asin tBu At " 2pB e"t " e"5t % Ae/4B Ae1" t " e5"5tBu At " 1B

2 Aet"2 " e"At"2B Bu At " 2B % 2et " t2 " 2 " A1/4B Aet"2 " e6"3tBu At " 2B et % e"3t % A1/4B Aet"1 " e3"3tBu At " 1B" Asin tBu At " pB e"s

e"s " e"3s e"2 "1 "1

t2/2 % e"2tsin 6t

A1/30B " t 0 e"2At"yB 3sin 6 At " yB 4 e AyBdy " e"2tcos 6ty AtB ! A1/2B " t0 eAt"yB 3sin 2 At " yB 4g AyBdy % etsin 2t

yk AtB ! etsin 2t ;H AsB ! As2 " 2s % 5B"1 ; h AtB ! A1/2Betsin 2t ; y AtB! A1/5B " t

0 3 e3At"yB" e"2At"yB 4g AyBdy% 2e3t " e"2t yk AtB ! 2e3t " e"2t ;H

AsB ! As2 " s " 6B"1 ; h AtB ! Ae3t " e"2tB /5 ;y AtB! A1/3B " t

0 3sin 3 At " yB 4g AyBdy% 2 cos 3t " sin 3t yk AtB ! 2 cos 3t " sin 3t ;H AsB ! As2 % 9B"1 ; h AtB ! A1/3Bsin 3t ;

e"t/2cos A23 t/2B " A1/23 B e"t/2sin A23 t/2B cos t t/4 % A3/8Bsin 2t s"2 As " 3B"1 A2/3Be"2t % A1/3Bet At/2B sin t 2e5t " 2e"2t 1 " cos t

23. sin t % Asin tBu At " pB % Asin tBu At " 2pB

25. 27. 29. The mass remains stopped at

35. L 6EI

33lx2 " x3 % Ax " lB3u Ax " lB 4 x AtB # 0 , t 7 p/2 A1/2Betsin 2t A1/2Be"2tsin 2t

–1 –2

2 1

–3

2π 4π

3π

Figure B.49

–1 –2

2 1

–3

2π 4π

3π

Figure B.50

CHAPTER 8

Exercises 8.1, page 426 1. 3. 5. 7.

9. (a)

(b) (c) (d) 0 ln A1.5B " p3 A1.5B 0 ! 0.011202. . .0 e3 A1.5B 0 ! 0 f A4B AjB A1.5 " 1B4/4! 0 ( 6 A0.5B4/4! p3 AxB ! Ax " 1B " 1

2 Ax " 1B2 % 1

3 Ax " 1B3 A1/6Bu3 " A1/120Bu5 % A1/5040Bu7 % p

1 " A1/6Bt3 % A1/180Bt6 % px % x2 % A1/2Bx3 % p1 % x % x2 % p 11. 13. 15.

17.

19.

21.

23.

25. 27. 29. 31. 35.

Exercises 8.4, page 450 1. 2 3. 5. 7.

11.

13. 15. 17.

19. 21.

23.

25.

27. % a1 3 x % A1/2Bx2 % p 4 % 3 A1/6Bx3 % p 4a0 31 " A1/2Bx2 " A1/6Bx3 % p 4% 3 A1/2Bx

2 " A1/6Bx3 % p 4a0 31 " A1/2Bx2 % p 4 % a1 3 x % p 4% 3 A1/2Bx2 % A1/3Bx3 % p 4 a0 31 " A1/2Bx2 % p 4 % a1 3 x " A1/3Bx3 % p 4% 3 A1/2Bx2 % A1/12Bx4 % A11/720Bx6 % p 4a0 31 % A1/2Bx

2 % A1/8Bx4 % A1/48Bx6 % p 4"1 % x % x2 % A1/2Bx3 % p% A1/180B Ax " pB6 % p 1 " A1/6B Ax " pB3 % A1/120B Ax " pB51 % x % A1/24Bx4 % A1/60Bx5 % p 1 " A1/2Bt2 % A1/6Bt4 " A31/720Bt6 % p" A7/96B Ax " 2B3 % p 4 % a1 3 Ax " 2B % A1/8B Ax " 2B

2 a0 31 " A1/8B Ax " 2B2 % A1/32B Ax " 2B3 % p 4% a1 3 Ax " 1B % A1/3B Ax " 1B3 % p 4a0 31 % Ax " 1B2 % p 4

" A1/6B Ax " 1B6 % p 4a0 31 " Ax " 1B2 % A1/2B Ax " 1B4p/223 10 " 25t2 % A250/3Bt3 " A775/4Bt4 % p1 % 2x " A3/4Bx2 " A5/6Bx3 1 " 2x % x2 " A1/6Bx3 x % A1/6Bx3 " A1/12Bx4 % A7/120Bx5 % p 2 % x2 " A5/12Bx4 % A11/72Bx6 % p% a1 ax % a

k!1 32 # 5 # 8 p A3k " 1B 4 2A3k % 1B! x3k%1b

a0 a1 % aq k!1

31 # 4 # 7 p A3k " 2B 4 2A3kB! x3kb

a3k%2 ! 0 , k ! 0, 1, . . .

% a1 c x % aq k!1

A"3B A"1B p A2k " 5BA2k % 1B! x2k%1 da0 A1 " 2x

2 % x4/3Ba0 a q

n!0 1 n!

x2n

a0 A1 " x2/2 % p B % a1 Ax " x3/6 % p B% a1 Ax % x2/2 " x3/6 % p B a0 A1 " x2/2 " x3/6 % p Ba0 A1 % x4/12 % p B % a1 Ax % x5/20 % p B y ! a0 A1 " 2x % A3/2Bx2 " x3/3 % p B

B-18 Answers to Odd-Numbered Problems

x 2

"2 "4

1 ln x

p3(x)

Figure B.51

13. 15.

Exercises 8.2, page 434 1. 3. 5.

11. 13. 15. (c)

17. 19.

21.

23.

25. 29.

31.

33.

35. (a) (b)

Exercises 8.3, page 445 1. 3. 5. 7. 9. u ( 0 and u ! np , n ! 1, 2, 3, . . .

x ! np , n an integer "1, 2# 22 "1

a q

n!1 A"1Bn"1

n Ax " 1Bnaq

n!0 A"1Bn Ax " 1Bn6 Ax " 1B % 3 Ax " 1B2 % Ax " 1B3

1 % a q

n!1 2xn

a q

n!0 A"1Bn%1 Ax " pB2nA2nB! aqk!1 ak"1xk

a q

k!0 Ak % 1Bak%1xk

ln Ax % 1B ! aq n!0

A"1Bn n % 1

xn%1

a q

k!1 ak 2kx

2k"1a q

n!1 A"1Bnnxn"11 " A1/2Bx % A1/4Bx

2 " A1/24Bx3 % p1 " 2x % A5/2Bx2 % p x % x2 % A1/3Bx3 % pa

n!0 c 1

n % 1 % 2"n"1 d xn 31, 3 4A"4, 0B3"1, 3B

1 " x2/6 t % A1/2Bt2 " A1/6Bt3 % p

Exercises 8.5, page 454

1. 3. 5. 7.

11. 13. 15. 17.

Exercises 8.6, page 465 1. are regular. 3. are regular. 5. 1 is regular, is irregular. 7. 2 is regular, is irregular. 9. is regular and 2 is irregular.

11.

13.

15. 17. 19. 21. 23.

25. 27.

29. 31.

33.

35.

37.

39. The expansion

41. The transformed equation is

are analytic at ; hence, is a regular singular point.

% A1/360Bx"3 % p 4y1 AxB ! a0 31 % A1/3Bx"1 % A1/24Bx"2 z ! 0z ! 0z2q AzB ! "zz d2y/dz2 % 3 dy/dz " y ! 0. Also zp AzB ! 3 and

a q

n!0 n!xn diverges for x * 0.

% A1/68,034Bx13/3 % p 4a0 3 x4/3 % A1/17Bx7/3 % A1/782Bx10/3 " A1/25,806Bx23/6 % p 4a0 3 x5/6 " A1/11Bx11/6 % A1/374Bx17/6

a0 c x1/3 % aq n!1

2n"2 A3n % 4Bxn% A1/3B 3nn!

d ; yes, a0 6 0 a0a

n!0

n! ! a0e

x ; yes, a0 6 0a0 x2

a0 a q

n!0

x2n%2

22n An % 1B!n!a0 aqn!0 A"1Bnxn% A3/2B2n"1 An % 2B! a0 3 x " A1/3Bx2 % A1/12Bx3 " A1/60Bx4 % p 4a0 31 " A1/4Bx2 % A1/64Bx4 " A1/2304Bx6 % p 4a0 3 x

2/3 " A1/2Bx5/3 % A5/28Bx8/3 " A1/21Bx11/3 % p 4r2 % r " 12 ! 0 ; r1 ! 3 , r2 ! "4r 2 " 1 ! 0 ; r1 ! 1 , r2 ! "1

r2 ! A5 " 2457 B /18r2 " 5r/9 " 4/3 ! 0 ; r1 ! A5 % 2457 B /18 ,r 2 " 3r " 10 ! 0 ; r1 ! 5 , r2 ! "2

"4 "1 "1

#i#1

" A76/85Bx"2sin A5 ln xBA31/17Bx % A3/17Bx"2cos A5 ln xB 2t4 % t"3c1x % c2x

2 % A4/15Bx"1/2c1 Ax " 3B"2 % c2 Ax " 3B1/2 % c3x

"1/2sin 3 A219 /2B ln x 4c1x % c2x"1/2cos 3 A219 /2B ln x 4c1x % c2x"1cos A3 ln xB % c3x"1sin A3 ln xB c1x

3cos A2 ln xB % c2x3sin A2 ln xBc1x cos A4 ln xB % c2x sin A4 ln xBc1x"2 % c2x"3 45.

Exercises 8.7, page 474 1. , where

and

3. where and

5. 7. where

and

9. where and

11. where and

13. where and

15. where and

all solutions are bounded near the origin.

17. where

and

19. where

and

21. (b) where

and

and y2 AxB ! x 31 " A1/2B Aa/xB2 % A1/24B Aa/xB4 % p 4y1 AxB ! 1 " A1/6B Aa/xB2 % A1/120B Aa/xB4 % p , c1y1 AxB % c2y2 AxB ,y2 AtB ! t"1 31 " A1/2B AatB2 % A1/24B AatB4 % p 4 y1 AtB ! 1 " A1/6B AatB2 % A1/120B AatB4 % p ,c1y1 AtB % c2y2 AtB ,

y3 AxB ! x"1/2 " A1/5Bx1/2 " A1/10Bx3/2 % py2 AxB ! 1 " A1/3Bx " A1/30Bx2 % p , y1 AxB ! x4/3 % A1/17Bx7/3 % A1/782Bx10/3 % p ,c1y1 AxB % c2y2 AxB % c3y3 AxB ,

% A4997/298116Bx3 % py3 AxB ! y2 AxB ln x % 7x " A117/196Bx2 y2 AxB ! 1 " x % A1/14Bx2 % p ,y1 AxB ! x5/6 " A1/11Bx11/6 % A1/374Bx17/6 % p , c1y1 AxB % c2y2 AxB % c3y3 AxB , y2 AxB ! 1 % 2x % A6/5Bx2 % p ;y1 AxB ! x1/3 % A7/6Bx4/3 % A5/9Bx7/3 % p c1y1 AxB % c2y2 AxB ,y2 AxB ! y1 AxBln x " 3 x % A3/4Bx2 % A11/36Bx3 % p 4y1 AxB ! 1 % x % A1/2Bx2 % p c1y1 AxB % c2y2 AxB ,y2 AxB ! x2 ln x " 1 % 2x " A1/3Bx3 % A1/24Bx4 % p

y1 AxB ! x2c1y1 AxB % c2y2 AxB ,w2 AxB ! w1 AxBln x % 2 " A3/32Bx4 " A7/1152Bx6 % p w1 AxB ! x2 % A1/8Bx4 % A1/192Bx6 % pc1w1 % c2w2 , y2 AxB ! x"1/2 " A1/2Bx1/2y1 AxB ! x3/2 " A1/6Bx5/2 % A1/48Bx7/2 % p c1y1 AxB % c2y2 AxB ,c1 3 x " A1/3Bx2 % A1/12Bx3 % p 4 % c2 3 x"1 " 1 4% A11/13,824Bx6 % p y2 AxB ! y1 AxB ln x % A1/4Bx2 " A3/128Bx4y1 AxB ! 1 " A1/4Bx2 % A1/64Bx4 % p c1y1 AxB % c2y2 AxB ,y2 AxB ! x1/3 " A1/2Bx4/3 % A1/5Bx7/3 % p y1 AxB ! x2/3 " A1/2Bx5/3 % A5/28Bx8/3 % pc1y1 AxB % c2y2 AxB

% A1/15,120Bx6 % p 4% a3 31 % A1/10Bx2 % A1/280Bx4a0 3 x"3 " A1/2Bx"1 " A1/8Bx " A1/144Bx3 % p 4

Answers to Odd-Numbered Problems B-19

29. (a)

% a1 c x % aq k!1 A"1Bk An " 1B An " 3B p An " 2k % 1B An % 2B An % 4B p An % 2kBA2k % 1B! x2k%1 d

a0 c1 % aq k!1 A"1Bk n An " 2B An " 4B p An " 2k % 2B An % 1B An % 3B p An % 2k " 1BA2kB! x2k d

23. where and

25. For , ;

and for

Exercises 8.8, page 487

9. 13. 15.

17.

19.

21.

27.

29.

37. 39.

41. (b) so for large,

Review Problems, page 491 1. (a)

(b) 3. (a)

(b)

7. (a) (b)

9. (a) where and

(b) where and

y2 AxB! y1 AxBln x " 34x % 3x2 % A22/27Bx3 % p 4y1 AxB ! 1 % 2x % x2 % pc1y1 AxB % c2y2 AxB , y2 AxB! y1 AxBln x " x2 " A3/4Bx3 " A11/36Bx4 % py1 AxB ! x % x2 % A1/2Bx3 % p ! xex c1y1 AxB % c2y2 AxB ,a0 3 x " A1/4Bx2 % A1/20Bx3 " A1/120Bx4 % p 4a0 3 x3 % x4 % A1/4Bx5 % A1/36Bx6 % p 4 % a1 Ax " 2Ba0 31 % A1/2B Ax " 2B2 " A1/24B Ax " 2B4 % p 4% a1 3 x " A1/2Bx

2 % A1/2Bx3 % p 4a0 31 % A1/2Bx2 " A1/6Bx3 % p 4a0 31 % x2 % p 4 % a1 3 x % A1/3Bx3 % p 4 "1 % x " A1/6Bx3 % A1/6Bx4 % p1 " x % A3/2Bx2 " A5/3Bx3 % p y AxB ! c1x"1/2 cos x % c2x"1/2 sin xxz ! c1 cos x % c2 sin x ,

A6 " 18x % 9x2 " x3B /61 , 1 " x , A2 " 4x % x2B /2 , 1 , 2x , 4x2 " 2 , 8x3 " 12x A5x3 " 3xB /2 , A35x4 " 30x2 % 3B /81 , x , A3x2 " 1B /2 , J0 AxB ln x % aq

n!1

A"1Bn%1 22n An!B2 c1 % 12 % p % 1n d x2nx

3/2J3/2 AxBJ1 AxB ln x " x"1 % A3/64Bx3 " A7/2304Bx5 % p c1J2/3 AxB % c2J"2/3 AxB c1J1 AxB % c2Y1 AxBc1J1/2 AxB % c2J"1/2 AxB A1 " xB"1 , A1 " xB"1 ln x!

1 2

x"1 ln a1 % x 1 " x

b! a q

n!0 x2n/ A2n % 1B

F a1 2

, 1; 3 2

; x2b ! aq n!0 A1/2Bn x2n/ A3/2Bn

F A1, 1; 2; xB ! aq n!0

1An % 1B xn ! "x"1 ln A1 " xB c1F a1, 1; 12; xb % c2x1/2F a32, 32; 32; xb c1F a1, 2; 12; xb % c2x1/2F a32, 52; 32; xb

" A1/36Bx3 % p 4c1 A1 " xB % c2 3 A1 " xB ln x % 3x " A1/4Bx2n ! 1, c1 % c2 3 ln x % x % A1/4Bx2 % A1/18Bx3 % p 4n ! 0y2 AxB ! " A1/6Bx"1 " A1/24Bx"2 " A1/120Bx"3 % p y1 AxB ! 1 % 2x % 2x2c1y1 AxB % c2y2 AxB , (c) where and

(d) where and

CHAPTER 9

Exercises 9.1, page 502

11.

13.

Exercises 9.2, page 506 1. 3. 5. 7. 9. with s any complex number

11.

13. For r ! 2, unique solution is For r ! 1, solutions are , ,

Exercises 9.3, page 515

1. (a) (b)

3. (a) (b) (c) c16 3 5 19

dc8 12 3 5

dc18 14 4 5

d c 7 3 7 18

dc1 1 5 2

d "q 6 s 6 q .

x2 ! sx1 ! 3s x1 ! x2 ! 0.

A"q 6 s 6 q B x3 ! sx2 ! " A11s % 1B /2 ,x1 ! " As % 1B /2 , x1 ! 2s , x2 ! A"1 % iBs ,x1 ! 3s , x2 ! s A"q 6 s 6 q B x1 ! 0 , x2 ! 0 x1 ! "s % t , x2 ! s , x3 ! t x1 ! 2 , x2 ! 1 , x3 ! 1

x1 ! x , x2 ! x¿ , x3 ! y , x4 ! y¿ , x5 ! y–

Ex1x2x3 x4 x5

U ¿ ! E 0 1 0 0 0cos t 3 "t2 0 00 0 0 1 0 0 0 0 0 1

"et t 0 "1 "1

U Ex1x2x3 x4 x5

U ;x1 ! x , x2 ! x¿ , x3 ! y , x4 ! y¿ Dx1x2 x3 x4

T ¿ ! D 0 1 0 0"3 0 "2 0 0 0 0 1 2 0 0 0

T Dx1x2 x3 x4

T ;c x1x2 d ¿ ! c0 1t 0 d c x1x2 d ; x1 ! y , x2 ! y¿ c y y d ¿ ! c 0 1

"k/m "b/m d c yy d c x y d ¿ ! c sin t et

cos t a % bt3 d c x

y d

£ xy z § ¿ ! £ 1 1 1"1 0 2

0 4 0 § £ xy

z §

c x y d ¿ ! c7 2

3 "2 d c x

y d

y2 AxB ! x"1 % A1/2B % A1/4Bx % py1 AxB ! x2 " A1/4Bx3 % A1/40Bx4 % p c1y1 AxB % c2y2 AxB ,y2 AxB ! y1 AxBln x " 8x"2 " 4x"1 % A29/36B % p

y1 AxB ! 1 " A1/6Bx % A1/96Bx2 % pc1y1 AxB % c2y2 AxB , B-20 Answers to Odd-Numbered Problems

25.

29.

37. (b) (e)

Exercises 9.5, page 534 1. Eigenvalues are and with associated

eigenvectors and .

3. Eigenvalues are and with associated

eigenvectors and .

5. Eigenvalues are and with

associated eigenvectors

and

7. Eigenvalues are and with

associated eigenvectors

and

9. Eigenvalues are and with associated

eigenvectors and

11.

13.

15.

17. (b)

c1e "t £ 30

"2 § % c2et £ 1"6

4 § % c3e4t £10

1 §

c1e "t £"21

1 § % c2e"3t £ 0"1

1 § % c3e5t £11

1 §

c1e 3t/2 c 3

10 d % c2et/2 c12 d

u2 ! s c"i1 d .u1 ! s c i1 d r2 ! "i,r1 ! i u3 ! s £01

2 § .

u1 ! s £ 2"3 2 § , u2 ! s £ 0"1

1 § ,r3 ! 5r1 ! 1, r2 ! 2,

u3 ! s £ 0"1 1 § .

u1 ! s £10 0 § , u2 ! s £01

1 § ,r3 ! "2r1 ! 1, r2 ! 2,

u2 ! s c 1"2 du1 ! s c 1"1 d r2 ! 3r1 ! 2 u2 ! s c 2s"s du1 ! c s2s d r2 ! "5r1 ! 0

2e6t ; same2e6t

x(t) ! £"(3/2)e"t % (3/5)e"2t " (1/10)e3t(1/4)e"t " (1/5)e"2t " (1/20)e3t (5/4)e"t " (1/5)e"2t " (1/20)e3t

§ X"1 AtB ! £ 0 A1/4Bet " A1/4Bet" A1/5Be2t A4/5Be2t " A2/5Be2tA1/5Be"3t A9/20Be"3t A3/20Be"3t § ;

% c2 c e"t3e"t d32 c tettet d " 14 c et3et d % c t2t d " c01 d % c1 c etet d5. (a) (b) (c) 7. (c) Yes (d) Yes 9. 11. Doesn’t exist

13. 17.

19.

21. 11 23. 25. 25 27. 2, 3

29. 0, 1, 1 31. 33.

39. (a)

(b)

(c)

Exercises 9.4, page 523

11.

13. LI 15. LD 17. LI 19. LI

21. Not a fundamental solution set

23. Yes;

c1 £ e"t2e"t e"t § % c2 £ et0

et § % c3 £ e3t"e3t

2e3t §

£ e"t e t e3t2e"t 0 "e3t e"t e t 2e3t

§ ; x¿3 AtB ! 5x2 AtB % x3 AtBx¿2 AtB ! "x1 AtB % 2x2 AtB % 5x3 AtB % t ; x¿1 AtB ! x1 AtB % x3 AtB % et ;x¿2 AtB ! "2x1 AtB % 4x2 AtB " 3e"2t x¿1 AtB ! 5x1 AtB % 2e"2t ;x1 ! w , x2 ! w¿ , x3 ! w– , x4 ! w‡ Dx¿1 AtBx¿2 AtB x¿3 AtB x¿4 AtBT ! D 0 1 0 00 0 1 00 0 0 1"1 0 0 0T Dx1 AtBx2 AtBx3 AtBx4 AtBT % D000t2T ;x1 ! y , x2 ! y¿ c x¿1 AtB x¿2 AtB d ! c 0 110 3 d c x1 AtBx2 AtB d % c 0sin t d ; £ dx/dtdy/dt

dz/dt § ! £ t2 "1 "10 0 et

t "1 3 § £ xy

z § % £ t5

"et §

c x¿(t) y¿(t) d ! c 3 "1"1 2 d c x(t)y(t) d % c t2et d BA1 % etB cos t % Aet " tB sin t Aet " tB cos t " Aet % 1B sin tAet " 1B sin t % et cos t Aet " 1B cos t " et sin t Rc sin 1 "1 % cos 11 " cos 1 sin 1 d

c A1/2Bt2 % c1 et % c2 t % c3 e

t % c4 d

c 5e5t 6e2t "10e5t "2e2t

d£ 3e3t6e3t "3e3t

§ "12

£ e"t A1/2Be"t " A1/2Be"tA1/3Bet " A1/2Bet A1/6Bet " A1/3Be"2t 0 A1/3Be"2t §

c A4/3Be"t " A1/3Be"t " A1/3Be"4t A1/3Be"4t d£ 1 0 1"2 1 "21 1 0 §

c4/9 "1/9 1/9 2/9 d

c"6 "3 "9 "4

dc"5 "1 "8 "1

dc"1 "2 "1 "3

d Answers to Odd-Numbered Problems B-21

(–√3, –1)

Figure B.52

B-22 Answers to Odd-Numbered Problems

(c)

(d)

19. 21.

23.

25.

27.

29. D0.0251e3.4142t "0.2361e"1.6180t e0.6180t e0.5858t0.0858e3.4142t 0.3820e"1.6180t 0.6180e0.6180t 0.5858e0.5858t 0.2929e3.4142t "0.6180e"1.6180t 0.3820e0.6180t 0.3431e0.5858t

e3.4142t e"1.6180t 0.2361e0.6180t 0.2010e0.5858t T

£ e"0.3473t e0.5237t 0.0286e"7.0764t"0.3157e"0.3473t 0.4761e0.5237t "0.1837e"7.0764t 0.0844e"0.3473t 0.1918e0.5237t e"7.0764t

§ z ! 2c1e

t % 4c2e 2t % 4c3e

3t y ! c1e

t % c2e 2t % c3e

3t , x ! "c1e

t " 2c2e 2t " c3e

3t ,

De2t0 0 0

e"t

"3e"t

0 0

e3t

0 e3t

"e7t

e7t

2e7t

8e7t T £

et e2t e4t

et 2e2t 4e4t

et 4e2t 16e4t §c e3t e"3t

4e3t "2e"3t d

31. 33.

35. (b)

(c)

37. (b)

(c)

(d)

39. (b)

43.

45.

47. (a)

(b)

c1e r1tu1 % c2er2tu2 % c3er3tu3,

u3 ! £ 10.81004 "0.12236

§ u1 ! £"1"2.64178

"9.31625 § ; u2 ! £"1"1.16825

"0.43862 § ;

r1 ! 2.39091 , r2 ! "2.94338 , r3 ! 3.55247 x2 AtB ! A2.5B Ae"t/25 " e"3t/25B x1 AtB ! A2.5/2B Ae"3t/25 % e"t/25B , c1 c3t2t2 d % c2 c t4t4 d

x3 AtB ! tet £ 11 "2 § % et £10

0 §

x1 AtB ! et £"11 0 § ; x2 AtB ! et £"10

1 §

x3 AtB ! t2 2

e2t £10 0 § % te2t £01

0 § % e2t £ 0"6/5

1/5 §

x2 AtB ! te2t £10 0 § % e2t £01

0 §

x1 AtB ! e2t £10 0 §

x2 AtB ! te"t c12 d % e"t c11 d x1 AtB ! e"t c12 d

£"3e"t % e5t"2e"t " e5t e"t % e5t

§c2e4t % e"2t 2e4t " e"2t

dx2 x1

(1, –√3)

Figure B.53

(1 – √3, –√3 – 1)

Figure B.54

Exercises 9.6, page 541

Note: Equivalent answer is obtained if and are replaced by and , respectively.col A0, "2, 1Bcol A"5, 1, 2B col A"2, 0, 1Bcol A"1, 1, 0B c1e

t cos t £"11 0 § " c1e t sin t £"20

1 § % c2e t sin t £"11

0 § % c2e t cos t £"20

1 § % c3e t £10

0 §

c1 c 2 cos 2tcos 2t % sin 2t d % c2 c 2 sin 2tsin 2t " cos 2t d

Exercises 9.7, page 547

11.

13.

15. c1 c12 d % c2e"5t c"21 d % c ln 0 t 0 % A8/5Bt " 8/252 ln 0 t 0 % A16/5Bt % 4/25 d c1e

t c 1 "1 d % c2e"t c 1"3 d % c 5tet % A3/4Bet"5tet % A9/4Bet d

c1 c sin tcos t d % c2 c cos t"sin t d % c 0"1 d x AtB ! ta % b % e2tc % A sin tBd % A cos tB exp ! ta % b % sin 3t c % cos 3t d

% c1e " t £ 11

0 § % c2et £"11

0 § % c3et £00

1 §

t s 00 "1 § % £ 00

"1 § % e2t

3 £ 2"1

0 § % sin t

2 £10

0 § % cos t

2 £ 0"1

0 §

c1e "3t £ 11

"1 § % c2e3t £10

1 § % c3e3t £"11

0 § % et £ 10

"1 §

c1e 7t c1

1 d % c2e2t c 1"4 d % c 2"1 d 17.

19.

21. (a)

(b)

(c)

(d)

23. y ! "c2e

t " 2c3e "t " t2 " 2t " 3

x ! Ac1 % c2tBet % c3e"t " t2 " 4t " 6 ; c"18et%1 % 14e2t%2 " 3e2t%1 % A4t % 7Bet

"9et%1 % 14e2t%2 " 3e2t%1 % A2t % 5Bet d c ("20 % 2e"5)et % ("3e"5 " e"10)e2t % (4t % 3)et

("10 % e"5)et % ("3e"5 " e"10)e2t % (2t % 3)et d

c"2et"1 % 2e2At"1B " 3e2t"1 % A4t " 1Bet "et"1 % 2e2At"1B " 3e2t"1 % A2t % 1Bet d c4tet % 5et 2tet % 4et

d Dc1 cos t % c2 sin t % 1"c1 sin t % c2 cos t " t c3e

t % c4e "t " A1/4Be"t " t % A1/2Bte"t

c3e t " c4e

"t " A1/4Be"t " 1 " A1/2Bte"tT c1e

2t £11 1 § % c2e"t £"10

1 § % c3e"t £"11

0 § % et £ 1"1

"1 §

Answers to Odd-Numbered Problems B-23

5. 7.

11.

13. (a) (b)

17.

19.

21.

I3 ! A10/13Be"2t cos 3t % A20/39Be"2t sin 3tI2 ! " A10/13Be"2t cos 3t % A25/78Be"2t sin 3t , I1 ! A5/6Be"2t sin 3t , 39 " 217

222p ! 0.249 ; 39 % 217222p ! 0.408 c1 £ t"10

"2t"1 § % c2 £ t"1cos Aln tBt"1sin Aln tB

"t"1cos Aln tB § % c3 £ t "1sin Aln tB

"t"1cos Aln tB "t"1sin Aln tB § c ep"2t cos t ep"2t Asin t " cos tB dc e"2At%2pB A2 cos t " 3 sin tBe"2At%2pB Acos t % 5 sin tB d

c"e"2At"pB cos t e"2At"pB Acos t " sin tB dc e"2t Asin t " cos tB"2e"2t sin t d

D et sin t " et cos t "et sin t " et cos t "cos t "sin t2et sin t "2et cos t sin t "cos t 2et sin t % 2et cos t 2et sin t " 2et cos t cos t sin t

4et cos t 4et sin t "sin t cos t

T D 1"1

23 sin(23 t) " cos(23 t)23 sin(23 t) % cos(23 t)

2 cos(23 t) "sin(23 t) " 23 cos(23 t)sin(23 t) " 23 cos(23 t)2 sin(23 t) T £1 cos t sin t0 "cos t "sin t

0 "sin t cos t §c e"t cos 4t e"t sin 4t

2e"t sin 4t "2e"t cos 4t d

Exercises 9.8, page 557

1. (a) (b)

3. (a) (b)

5. (a) (b) 7.

11.

13.

15.

17. 19. c1e2tD361 1

T % c2etD010 0

T % c3etD1t0 0

T % c4etD 2tt % t20 1

Tc1e"t £ 1"1 1 § % c2e"t £ "t"1 % t

2 " t § % c3e"2t £ 1"2

4 §

E(1 % t % t2/2)e"t (t % t2)e"t (t2/2)e"t 0 0("t2/2)e"t (1 % t " t2)e"t (t " t2/2)e"t 0 0("t % t2/2)e"t ("3t % t2)e"t (1 " 2t % t2/2)e"t 0 0 0 0 0 (1 % 2t)e"2t te"2t

0 0 0 "4te"2t (1 " 2t)e"2t

U E(1 " t % t2/2)et (t " t2)et (t2/2)et 0 0(t2/2)et (1 " t % t2)et (t % t2/2)et 0 0(t % t2/2)et ("3t " t2)et (1 % 2t % t2/2)et 0 0

0 0 0 cos t sin t 0 0 0 "sin t cos t

U 1

25 £"4 % 29e5t " 20te5t 20 " 20e5t "8 % 8e5t " 40te5t"5 % 5e5t 25 "10 % 10e5t

2 " 2e5t % 10te5t "10 % 10e5t 4 % 21e5t % 20te5t §

1 2 £ et % cos t " sin t 2 sin t et " cos t " sin tet " sin t " cos t 2 cos t et % sin t " cos t

et " cos t % sin t "2 sin t et % cos t % sin t § c cos t sin t "sin t cos t

de"2t £ 1 0 04t 1 0 t 0 1

§r ! "2 ; k ! 2 e"t £1 % 3t " (3/2)t2 t "t % (1/2)t2"3t 1 t

9t " (9/2)t2 3t 1 " 3t % (3/2)t2 §r ! "1 ; k ! 3

e3t c1 "2t 0 1

dr ! 3 ; k ! 2

33. where

35.

Tank A will have the higher concentration. The limiting concentration in tank A is kg/L and in tank B is

kg/L.3.7/21 4.1/21

x2 ! A185 " A584.5/2Be"3t/50 % A235.5/2Be"7t/50B /21 .x1 ! A205 " A584.5/4Be"3t/50 " A235.5/4Be"7t/50B /21 , " A31 % 2817 B eA"31%2817 B5t/2 4 % 3/20I5 ! A3/ A402817 B B 3 A31 " 2817 B e"A31%2817 B5t/2" A13 % 2817 B e

A"31%2817 B5t/2 4 % 1/10 ,I2 ! A1/ A202817 B B 3 A13 " 2817 B e"A31%2817 B5t/2I1 ! I2 % 3I5, I3 ! 3I5, I4 ! 2I5,25. (a) (c)(d) 27.

29.

% c" A3/4Bt"1 " A1/2Bt"1 ln t % 1 " A3/4Bt"1 " A3/2Bt"1 ln t % 2 d

c1t c11 d % c2t"1 c13 d % te"t £ 10

"1 § % e"t £10

0 § % £00

1 §

c1e 2t £11

1 § % c2e"t £"10

1 § % c3e"t £"11

0 §

c1e t c1

1 d % c2e2t c12 d % tet c22 d % et c01 d

tet c2 2 d % et c0

1 de c et

et d , c e2t

2e2t d f

B-24 Answers to Odd-Numbered Problems

21. 23. where is the matrix in the answer to Problem 3.

29.

Review Problems, page 561

1. 3. 5.

7. 9.

11. 13.

15. £ (1/2)et % (2/3)e5t " (1/6)e"t "et % (2/3)e5t % (1/3)e"t "(1/2)et % (2/3)e5t " (1/6)e"t(1/3)e5t " (1/3)e"t (1/3)e5t % (2/3)e"t (1/3)e5t " (1/3)e"t "(1/2)et % (1/2)e"t et " e"t (1/2)et % (1/2)e"t

§ c1t

"1 £"31 0 § % c2t"1 £"10

1 § % c3t4 £11

1 §c3et " 2e2t

3et " 4e2t d

c1e "t £10

3 § % c2e"t £ t1

3t § % c3e"t £ "t % (1/2)t2t

1 " 3t % (3/2)t2 § % £ 2 % t7 " 3t

10 §c1e"t c 1"2 d % c2e3t c12 d % c "1/3"14/3 d

c e2t e3t "e2t "2e3t

dc1e"tD 1"10 0

T % c2e"tD 001 "1

T % c3e3tD110 0

T % c4e3tD001 1

Tc1e3t c11 d % c2e4t c32 d x AtB ! £ A4/13Be"3t % 6te3t % A2/13B A3sin 2t " 2cos 2tBe3tA8/13Be"3t % A3t " 1Be3t % A4/13B A3sin 2t " 2cos 2tB § .

eAte"t £3t3 9t § % eAt £ 2 " et At2 " 2t % 2B1 % et At " 1B

6 " 3et At2 " 2t % 2B § ,£ e "2t

4te"2t % e"2t

te"2t " e"2t § .

Answers to Odd-Numbered Problems B-25

CHAPTER 10

Exercises 10.2, page 576 1. 3.

5. 7. arbitrary

9. and where and the ’s are arbitrary.

11. and

where

and are arbitrary.

13. The eigenvalues are the roots of

where For n large, is a positive integer. The eigenfunctions are

where the ’s are arbitrary.

15.

17.

19.

21.

" A14/45B sin 45t sin 15x% A11/27B sin 27t sin 9x u Ax, tB ! 6 cos 6t sin 2x % 2 cos 18t sin 6xu Ax, tB ! 3 cos 6t sin 2x % 12 cos 39t sin 13x u Ax, tB ! e"3t sin x " 7e"27t sin 3x % e"75t sin 5xu Ax, tB ! e"3t sin x " 6e"48t sin 4x

cn yn ! cn 3 sin A1lnxB % 1ln cos A1lnxB 4 ,ln ! A2n " 1B2/4, n

ln 7 0.tan A1ln pB % 1ln ! 0,bna0, an, yn ! an cos nx % bn sin nx, n ! 1, 2, 3, . . . ,

ln ! n 2, n ! 0, 1, 2, . . . ; y0 ! a0

cnn ! 1, 2, 3, . . . yn ! cn sin 3 A2n " 1Bx/2 4 ,ln ! A2n " 1B2/4 y ! cos x % c sin x; cy ! e

x % 2x " 1

y # 0y ! 34/ Ae " e"1B 4 Ae"x " exB 23.

25. If then becomes unbounded as and so the temperature becomes unbounded at each position x. Since the temperature must remain bounded for all time, K 0 0.

33. (a) (b)

Exercises 10.3, page 591 1. Odd 3. Even 5. Neither

11.

13.

15.

% A"1Bn%1n 1 % n2

sin nx db f AxB % 3 Asinh pB /p 4 a1 % 2aq

n!1 c A"1Bn 1 % n2

cos nx

f AxB % 1 3

% a q

n!1

4 A"1Bn n2p2

cos npx

% 1 pn A A"1Bn%1 " 1B sin npx2 df

AxB % 1 % aq n!1 c 2 p2n2

A"1 % A"1BnB cos npx 2

f AxB % aq n!1

2 A"1Bn%1 n

sin nx

u AxB ! 30x/L % 10u AxB # 50 u Ax, tB ! X AxBT AtB t S q,T AtBK 7 0,

u Ax, tB ! aq n!1

n"2e"2p 2n2t sin npx

17. The -periodic function where

19. The 4-periodic function where

21. The 2-periodic function where

23. The -periodic function where

25. (a) (b)

27.

29.

Exercises 10.4, page 598

1. (a) The -periodic function where

(b) The -periodic function where

3. (a) The -periodic function where

(b) The -periodic function where

fe AxB ! e1 A"p 6 x 6 "p/2B ,0 A"p/2 6 x 6 0B , 0 A0 6 x 6 p/2B , 1 Ap/2 6 x 6 pB

fe AxB,2p fo AxB ! e"1 A"p 6 x 6 "p/2B , 0 A"p/2 6 x 6 0B ,

0 A0 6 x 6 p/2B , 1 Ap/2 6 x 6 pB

fo AxB,2pf ~ AxB ! e0 A0 6 x 6 p/2B ,

1 Ap/2 6 x 6 pB f ~ AxB,pfe AxB ! e

x2 A0 6 x 6 pB , x2 A"p 6 x 6 0B

fe AxB,2pfo AxB ! e x2 A0 6 x 6 pB , "x2 A"p 6 x 6 0B

fo AxB,2pf~ AxB ! x2 A0 6 x 6 pB f ~ AxB,p

a0 ! 0 ; a1 ! 3/2 ; a2 ! 0 % sin aA2n " 1Bpx

2 b d

f AxB % aq n!1

2A2n " 1Bp c A"1Bn%1 cos aA2n " 1Bpx2 b F AxB ! 0 x 0 " pF AxB ! Ax2 " p2B /2g AxB ! e

ex A"p 6 x 6 pB ,Aep % e"pB /2 Ax ! #pB g AxB,2pg AxB ! x2 A"1 ( x ( 1B

g AxB, g AxB ! e1 A"2 6 x 6 0B ,x A0 6 x 6 2B ,

1/2 Ax ! 0B , 3/2 Ax ! #2B

g AxB,g AxB ! e x A"p 6 x 6 pB ,0 Ax ! #pB g AxB,2p

11.

13.

15.

17.

19.

Exercises 10.5, page 610

where

cn ! g2e"p " 2pn A"1Bn % 2np A1 % n2B 3 A"1Bn%1e"p % 1 4 % 2pn 3 A"1Bn " 1 4 An * 2B , e"p " 1

p %

5p A1 " e"pB % 1 An ! 2B

u Ax, tB ! e"p " 1 p

x " e"x % 1 % a q

n!1 cne

"n2t sin nx ,

% a q

n!6

10 pn 32 A"1Bn " 1 4 e"2n2t sin nx%

5 2p

e"32t sin 4x " a1 % 6 p b e"50t sin 5x

% a1 " 10 p b e"18t sin 3xu Ax, tB ! 5 %

5 p

x " 30 p

e"2t sin x % 5 p

e"8t sin 2x

u Ax, tB ! 2 Aep " 1B p

% a q

n!1

2ep A"1Bn " 2 p A1 % n2B e"n2t cos nx

u Ax, tB ! p 2

" a q

k!0

4 p A2k % 1B2 e"3A2k%1B2t cos A2k % 1Bx

u Ax, tB ! aq n!1

8 A"1Bn%1 " 4 p3n3

e"5p 2n2t sin npx

u Ax, tB ! 4 p a q

k!0

A"1Bk%1A2k % 1B2 e"5A2k%1B2t sin A2k % 1Bx 2 p a q

k!1 c 2 2k " 1

" 1

2k % 1 "

1 2k " 3

de"5A2k"1B2tsin A2k " 1Bxu Ax, tB ! f AxB % 2

p %

2 p a q

k!1 a 1 2k % 1

" 1

2k " 1 b cos 2kx

f AxB % e " 1 % 2 aq n!1

A"1Bne " 1 1 % p2n2

cos npx

f AxB % p 2

% 4 p a q

k!1

1A2k " 1B2 cos A2k " 1Bx f AxB % aq

k!0

8A2k % 1B3p3 sin A2k % 1Bpx f AxB % aq

n!1 c 2p A"1Bn%1

n %

pn3 A A"1Bn " 1B d sin nx

f AxB % " 4 p a q

k!1

1 2k " 1

sin A2k " 1Bpx B-26 Answers to Odd-Numbered Problems

11. where

13.

15.

17.

19. where

Concentration goes to zero as

Exercises 10.6, page 621

where

and

11. where

and

where

bn ! 1

2L 23L2 % 4a2p2n2Tn AtB ! e"t/2 acos bnt % 12bn sin bntb ,

an ! 2 L "

0 f AxB sin npx

L dx ,

u Ax, tB ! aq n!1

anTn AtB sin npxL , g AxB ! aq

n!0 bn A2n % 1Bpa

2L sin

A2n % 1Bpx 2L

f AxB ! aq n!0

an sin A2n % 1Bpx

% bn sin A2n % 1Bpat

2L d sin A2n % 1Bpx

2L ,

u Ax, tB ! aq n!0 can cos A2n % 1Bpat2L

% 2 a q

n!1

A"1Bn%1 n3

c t " sin nt n d sin nx

u Ax, tB ! cos t sin x % 5 2

sin 2t sin 2x " 3 5

sin 5t sin 5x

u Ax, tB ! 2h0L2 p2a AL " aB aqn!1 1n2 sin npaL sinnpxL cos npatL

u Ax, tB ! aq n!1

4 n3 3 2 A"1Bn%1 " 1 4 cos 2nt sin nx

% a q

k!0

8A A2k % 1BpB 3 cos A2k % 1Bpt sin A2k % 1Bpx u Ax, tB ! 1

7p sin 7pt sin 7px

t S % q.

cn ! 2 a "

0 f AxB sin anpx

a b dxC Ax, tB ! a

n!1 cne

"3L%kn2p2/a24 t sin anpx a b ,u Ax, y, tB ! 2 a

n!1

A"1Bn%1 n

e"n 2t sin ny

u Ax, y, tB ! e"52t cos 6x sin 4y " 3e"122t cos x sin 11y% a q

n!2

4 A"1Bn n3

e"2n 2t sin nx

u Ax, tB ! p2 3

x " 1 3

x3 " 3e"2t sin x

f AxB ! aq n!0

an cos An % 1/2Bx u Ax, tB ! aq

n!0 ane

"4An%1/2B2t cos An % 1/2Bx, 13. 15.

17.

21.

where

and

with

Exercises 10.7, page 633

3. where

9. where

is arbitrary, and for n ! 1, 2, 3, . . .

11.

% a" 1 255

r4 % 256 255

r"4b sin 4uu Ar, uB ! a 1 3

r " 4 3

r "1b cos u % a2 3

r " 2 3

r "1b sin u bn !

a np "

"p f AuB sin nudu

an ! a pn "

"p f AuB cos nudu ,a0

u Ar, uB ! a0 2

% a q

n!1 ar a b n Aan cos nu % bn sin nuB,

u Ar, uB ! p 2

" a q

k!0

r 2k%1A2k % 1B2p22k"1 cos A2k % 1Bu %

cos 2x sinh 2y

sinh A2B u Ax, yB ! cos x sinh A y " 1B

sinh A"1B " cos 3x sinh A3y " 3Bsinh A"3B An !

2 p sinh A"npB " p0 f AxB sin nx dx

u Ax, yB ! aq n!1

An sin nx sinh Any " npB, %

cos 7x sinh 37 A y " 1B 4 sinh A"7B

u Ax, yB ! 4 cos 6x sinh 36 A y " 1B 4 sinh A"6B

cn ! " 1

0 J20 AknrBr dr

bn ! 1 akncn "

0 g ArBJ0 AknrBr dr ,

an ! 1 cn "

0 f ArBJ0 AknrBr dr ,

u Ar, tB ! aq n!1 3an cos AknatB % bn sin AknatB 4 J0 AknrB ,%

cos Ax " atB " cos Ax % atB a

du Ax, tB ! 1 2 c e"Ax%atB2 % e"Ax"atB2u Ax, tB ! x % tx

! 1 a

sin at cos x

u Ax, tB ! 1 2a 3 sin Ax % atB " sin Ax " atB 4

Answers to Odd-Numbered Problems B-27

APPENDIX A

Exercises A, page A-7

1. 3.

11.

13.

15. 17.

19. 21.

23.

25.

27.

29.

31.

33. 1 8

x " 1

96 sin A12xB % C

1 9

tan9 x % 2 7

tan7 x % 1 5

tan5 x % C

" 1

20 sin A10xB " 1

8 sin A4xB % C

5 7

sin7/5 x " 5

17 sin17/5 x % C

" arctan Acos tB % C t2

2 ln 0 t % 3 0 " 1

2 c t " 3 % 9

t % 3 d % C

42x2 % 4 % Cy cosh y " sinh y % C ln 0 ln x 0 % C 1 2

ln Ax2 % 4B " 2 x

% C

29x2 " 1 " arccos a 1 3x b % C2 ln 0 x " 1 0 " 4 ln 0 x " 2 0 % 2 ln 0 x " 3 0 % C

arcsin A12xB % C12 ln At 2 % 4B % 12 arctan At/2B % C 1 2 u %

1 4p

sin A2puB % C 215e6t 5

% CAx2 % 1B5/2 A5x2 " 2B /35 % C

B-28 Answers to Odd-Numbered Problems

13.

where

and

15.

17.

where

and for n ! 1, 2, 3, . . .

and

21.

23. (c) f Ax, yB ! 2xyu Ar, u, zB ! 3 I0 ArB /I0 ApB 4 sin z n3n"1cn " n3

"n"1dn ! 1 p "

"p g AuB sin nudu

cn % dn ! 1 p "

"p f AuB sin nudu ,

n3n"1an " n3 "n"1bn !

1 p "

"p g AuB cos nudu ,

an % bn ! 1 p "

"p f AuB cos nudu ,

b0 ! 3

2p " p

"p g AuBdu ,a0 ! 12p " p"p f AuBdu ,

% Acnr n % dnr "nBsin nu 4 ,u Ar, uB ! a0 % b0 ln r % a q

n!1 3 Aanrn % bnr"nBcos nuu Ar, uB ! r3 sin 3u

bn ! 1 p "

"p f AuB sin nudu (n ! 1, 2, 3, . . .)

an ! 1 p "

"p f AuB cos nudu (n ! 0, 1, 2, . . .) ,

u Ar, uB ! a0 2

% a q

n!1 r"n Aan cos nu % bn sin nuB,

Note: n indicates footnote.

Abel’s formula, 201, 323, 326, 525 Absolute stability, 150 Addition, matrix, 507–508 Aging spring, 451, 496–497 Air resistance models, 111 Aircraft guidance in crosswind, 146–147 Airy equation

explanation of, 207 initial value problem for, 495

Airy functions, 207 Algebra, of matrices, 498, 507–514 Algebraic equations, linear, 502–506 Algorithms

alphabetization, 493–494 classical 4th order Runge-Kutta, 134–138,

253–259 Gauss-Jordan elimination, 503–506 Improved Euler, 121–129

Alligator problem, 100 Allometric equation, 144 Alphabetization algorithms, 493–494 Amplitude modulation, 230 Analog simulation, 295 Analytic coefficients, equations with, 446 Analytic functions, 433–434 Angular frequency, 215, 288, 540 Angular momentum, 209–210 Annihilator method

examples using, 334–336 explanation of, 333–334 undetermined coefficients and, 333–336

Apollo reentry, 236 Aquaculture, 144 Archimedes principle, 312 Armageddon, 214 Arms race, 252 Asymptotic behavior of solutions, 87–88,

241 Asymptotically stable

critical points, 266 explanation of, 266 homogeneous linear systems, 537 linear systems, 411

Atmospheric pressure, 45 Attractors, 301

I-1

PrefaceIndex

Building temperature, 101–102 Buoyancy, 312

Calculus of matrices, 514–515 Cancer detection, 283 Capacitance, 118 Carbon dating problem, 101 Casoratian, 347 Cauchy, Augustin, 328n Cauchy-Euler equations, 194, 485

higher-order, 332 second-order, 201 theory for, 455–457

Cauchy-Euler system, 536 Cauchy product, 430 Cayley-Hamilton theorem, 552 Center (critical point), 268, 270 Centered-difference approximations,

643 Centered-difference formula, 100 Chaos, 151–152, 301, 303 Chaos machine, 306 Characteristic equations, 159 Characteristic polynomials, 527 Chebyshev polynomials, 484, 489 Chemical diffusion, 611 Chemical reactions, 140 Circular cylinders, 638–640 Clairaut, Alexis, 64n, 84n Clairaut equations, 84 Classical differential equations, 485–486 Classical mechanics. See Newtonian mechanics Classical orthogonal polynomials, 484 Classical 4th order Runge-Kutta

algorithm with tolerance, 136, A-16 method, 134–138, 258–259 subroutine, 135, A-15

Coefficient of kinetic friction, 116 Coefficients

constant, 157, 327–331 discontinuous, 53 equations with analytic, 446–450 linear, 72–74 undetermined, 175–181, 186–187,

239–240, 333–336, 346, 543–544 variable, 193–200

Coffee problems, 107, 108

Automatic pilot, 381 Autonomous equations, 32 Autonomous systems

critical points of, 264–265 explanation of, 263

Auxiliary equations complex roots and, 167–173, 329 explanation of, 159, 327 repeated roots and, 163, 329–331

Bang-bang controls, 148–149 Barometric equation, 45 Beams

under axial force, 341 vibrating, 347

Beats, 230 Bernoulli, James, 71n Bernoulli, John, 5n, 71n, 327n Bernoulli equation, 53, 71–72 Bessel, Frederic Wilhelm, 479n Bessel equation

explanation of, 202, 207, 478–482 modified, 208 reduction to, 207–208, 489

Bessel functions aging spring and, 496–497 explanation of, 478–482 spherical, 487

Bessel inequality, 593 Bifurcation, 33 Blasius equation, 262 Block diagrams, 91 Bobbing objects, 312–313 Bode plots, 420 Boxcar function, 384 Boundary conditions

Dirichlet, 623, 627, 630 explanation of, 165, 568 homogeneous and nonhomogeneous, 602 Neumann, 623, 629

Boundary value problems explanation of, 165 initial, 568, 573–574, 603–608

Brachistochrone problem, 5n Buckling

of columns, 475 of towers, 495–496

Columns, buckling, 475 Combat models, 261, 549 Comparison test for improper integrals, 358 Compartmental analysis

explanation of, 91 mixing problems and, 91–94, 536 population models and, 94–99

Competing species, Runge-Kutta subroutine and, 262

Complex roots of auxiliary equations, 167–173, 329 of polynomial equations, 538n

Compound interest, 3, 37, 45, 276 Computer algebra system, 345–346 Confluent hypergeometric equation, 487 Conservative system, 314 Consistency condition, 629 Constant coefficients

differential operators with, 157, 245, 327 explanation of, 157, 327 homogeneous equations with, 158–165,

327–331 homogeneous linear systems with,

526–534 linear systems with, 518

Control theory, 380 Conventional combat model, 549 Convergence

Euler’s method and, 124 of Fourier series, 587–590 power series and, 427–431 radius of, 428, 446–447 uniform, 589

Convolution explanation of, 239 Laplace transforms and, 397–403 properties of, 397

Cooling of buildings, 101–107 Coulomb force field, 315 Coupled mass-spring systems (see also Mass-

spring systems), 285–289 Cramer’s rule, 503n, A-11–A-12 Critical load, 496 Critical points

asymptotically stable, 266 examples of, 268 explanation of, 265 limit points as, 268 phase plane analysis and, 265–272

Critically damped motion, 218–221 Crosswinds, 146–147 Curve of pursuit, 145–146

d’Alembert, Jean le Rond, 3n d’Alembert’s solution, 615–618 Damped oscillations, 155

I-2 Index

Damping Hamilton’s equations and, 314 mass-spring oscillators and, 153–155 vibrating spring with, 174 vibrating spring without, 174

Decay constant, 101 Deflection of beams, 341 Determinants, 512–513 Diagonalizable matrices, 534 Diagonal matrices, 507 Difference equation, 151, 347 Differential equations (see also Equations;

specific types of equations) autonomous, 32, 263 construction of, 326 exact, 55–61 first-order. See First-order equations higher-order linear. See Higher-order

differential equations linear. See Linear second-order

equations nonlinear, 4, 203–211 ordinary, 4 overview of, 1–5 partial. See Partial differential equations second-order. See Linear second-order

equations separable, 38–42 series solutions of. See Series solutions table listing, 484–486

Differential operators, 318 Diffusivity, 567 Dirac delta function

explanation of, 404–406 Laplace transform and, 406–410 properties of, 411

Direction fields for autonomous equation, 33 explanation of, 15–20 method of isoclines and, 19–20

Dirichlet boundary conditions. See Boundary conditions

Dirichlet problem, 627–631, 630, 642–644 Discontinuous coefficients, 53 Discontinuous forcing term, 53, 188 Discontinuous functions, 356, 383–389,

|578 Dispersion relation, 622 Distinct real roots, 163, 328–329 Distribution, 405 Doetsch, G., 351n Dot products, 498–500 Double Fourier series, 607 Drag race, 5 Duffing equation, 208–209, 302, 426 Duhamel, J.M.C., 419

Duhamel’s formulas, 418–419 Dynamical systems, 297–305

Eigenfunctions, 570 Eigenspaces, 527 Eigenvalues

complex, 538–541 distinct, 531–532 explanation of, 526, 570 real-valued, 532

Eigenvectors explanation of, 526 generalized, 553–554 linear independence of, 529–531 matrix of, 527–530

Electrical circuits analogy between mechanical and,

295 explanation of, 117–120

Electrical systems, 291–296 Electrical units, 119 Elimination method, 244–250 Elliptical integral of first kind, 237 Emden’s equation, 445 Emphatic remark, 393 Energy integral lemma, 204–205, 274 Epidemic models

boarding school, 284 SIR model, 278–281

Equations (see also Differential equations; specific types of equations)

with analytic coefficients, 447–450 auxiliary, 159, 327 characteristic, 159 coupled, 75 of form dy/dx = G(ax + by), 70–71 indicial, 452, 457 integral, 31 linear algebraic, 412–413 with linear coefficients, 72–74

Equidimensional equations. See Cauchy-Euler equations

Equilibrium points, 265 Equilibrium solutions, 265 Escape velocity, 117 Euler, Leonhard, 57n, 64n, 168n, 327n Euler’s formulas, 168, 583 Euler’s method

convergence and, 27 explanation of, 23–25 improved, 121–129 systems, 254–256

Exact differential equations explanation of, 55–57 method for solving, 59 test for, 57–59

Index I-3

Existence and uniqueness of solutions direction fields and, 15 explanation of, 11, 13, 318 first-order equations and, 11, 51, 53–54 initial value problems and, 43–44 Laplace’s equation and, 632–633 limitations of, 13 linear second-order equations and, 161,

184, 194 linear systems and, 518–519 wave equation and, 619–621

Explicit solutions, 6 Exponential law, 95 Exponential of a matrix, 550–556 Exponential order (alpha??), 358 Exponential shift property, 345–346 Exponents of a regular singular point, 457

Factorial function, 477 Factors

integrating, 64–67 quadratic, 373–374 repeated linear, 372–373

Falling bodies motion of, 1, 35–37, 274–275 velocity of, 131

Feedback system, 147–148, 251–252 Fibonacci, 349 Fibonacci sequence, 349 First-order equations

compartmental analysis and, 91–99 exact equations and, 55–59 heating and cooling of buildings and,

101–107 higher-order numerical methods and,

132–138 improved Euler’s method and, 121–129 linear equations and, 46–51 mathematical modeling and, 89–91 motion of falling body and, 35–37 multiple solutions of, 85 Newtonian mechanics and, 108–114 Runge-Kutta method and, 133–138 separable equations and, 38–42 special integrating factors and, 64–67 substitutions and transformations and,

68–74 Taylor method and, 132–134

Fluid ejection, 261 Fluid flow

around a corner, 635 nonisothermal, 140 Torricelli’s law of, 82–83 velocity potentials, 68

Fluid tanks, interconnected, 242–244, 536, 549–550

Forced Duffing equation, 302 Forced mechanical vibrations, 223–229 Forced pendulum equation, 303 Forced vibrations, 188–189 Forcing function, 419 Forcing term, 53, 188 Formal solutions, 575 Fourier, Joseph B.J., 582n Fourier cosine series, 594–598 Fourier series

complex forms of, 592 convergence of, 587–590 differentiation of, 590 double, 607 examples of, 586 explanation of, 582 generalized, 586 integrals crucial in, 579–580 integration of, 590 orthogonal expansions and, 586–587

Fourier sine series, 572, 594–598 Free fall

examples of, 1, 35–37 problem in, 45

Frequency gain, 224 Frequency response curve, 225–226 Frequency response modeling, 419–421 Friction

coefficient of kinetic, 116 explanation of, 116 static, 116 sticky, 275

Frobenius, George, 457n Frobenius’s method

explanation of, 455–465 linearly independent solutions and, 468–469

Frobenius’s theorem, 461 Functions

analytic, 433–434 discontinuous, 356, 383–389, 578 even, 578 gamma, 392–393 generalized, 405 harmonic, 633 linear dependence of, 161, 322 matrix, 515 odd, 578 periodic, 390–392 piecewise continuous (see also Piecewise

continuity), 356, 578 special, 476–486 spherically symmetric, 494 symmetric, 579

Fundamental matrices, 521, 551–553 Fundamental period, 578 Fundamental solution sets, 323, 521

Gain factor, 224 Galileo, 5n Gamma function, 392–393 Gauss, Carl Friedrich, 477n Gauss-Jordan elimination algorithm, 503–506,

511 Gaussian hypergeometric equation, 477 Gegenbauer polynomials, 484 General solutions

explanation of, 36, 47, 163, 321 to homogeneous equations, 158–165 to linear differential equations, 163, 321 to linear systems in normal form, 523 matrix systems and, 521 to nonhomogeneous equations, 184 for systems, 247

Generalized Blasius equation, 262 Generalized eigenvectors, 553–554 Generalized Fourier series, 586 Generalized function, 405 Generating functions for Legendre

polynomials, 484, 488 Geometric series, 429 Gibbs phenomenon, 593–594 Gibbs-Wilbraham phenomenon, 593n Global error, 131 Golden ratio, 349 Gompertz equation, 282 Grand Prix race problem, 45 Gravitational force, 35, 213 Great Lakes clean up, 315–316 Green’s formula, 634 Green’s functions, 641–642 Group velocity, 622 Guitar strings, 612–614

Half life, 284 Half-range expansions, 595, 597 Hamilton, Sir William Rowan, 313n Hamiltonian systems, 313–315 Hamilton’s equations, 314 Harmonic functions, 633 Harmonic motion, 215 Harmonic oscillators, 215 Harmonics, 612 Heat capacity, specific, 567 Heat conduction, 566 Heat equation, 568, 599–609 Heat flow

direction of, 566 heat equation and, 599–609 model for, 566–568

Heating and cooling of buildings, 101–107

Heaviside, Oliver, 351n, 375n Heaviside’s Expansion formula, 375

Henon map, 302n Hermite polynomials, 484, 489 Hermites equation

Hermite polynomials and, 484, 489 in quantum mechanics, 202

Higher-order differential equations basic theory of, 318–324 constant coefficients with homogeneous,

327–331 conversion to normal form, 253–254,

517–523 numerical methods for, 253–259 undetermined coefficients and annihilator

method, 333–336 variation of parameters method and,

338–340 HIV infection, 141–143 Hoare, C.A.R., 493 Homogeneous boundary conditions, 602 Homogeneous equations

with constant coefficients, 158–165, 327–331

explanation of, 68–70, 158–159, 318 fundamental solutions of, 323, 521 general solutions to, 158–165 substitutions and transformations and,

68–70 Homogeneous linear systems, 526–534 Homogeneous of order n, 75 Hong Kong flu, 280 Hooke’s law, 153 Hormone secretion problem, 54 Hormones, 283 Hypergeometric equation

confluent, 487 explanation of, 476–478

Identity, matrix, 510–511 Impedance, 240 Implicit function theorem, 8, 15 Implicit numerical method, 125 Implicit solutions, 8–9 Improved Euler’s method

explanation of, 121–129 subroutine, 127 with tolerance, 128

Impulse response function, 401–402, 409 Impulses, Dirac delta function and, 404–410 Indicial admittance, 418 Indicial equations, 452, 457–460 Inductance, 118 Inertial reference frame, 108 Initial-boundary value problems

example of, 568 heat flow and, 603–608 separation of variables and, 573–574

Initial conditions, 10, 159, 568 Initial value problems

convolution theorem to solve, 399 example of, 36

I-4 Index

explanation of, 10–13 Laplace transforms and, 376–382 Taylor polynomial and, 424–425 unique solution and, 11, 43–44, 197

Inner products, 517, 592 Integral equations, 31 Integrating factors

explanation of, 48, 65–66 finding, 65–66 special, 66

Integration techniques integration by parts, A-4–A-5 method of substitution, A-1–A-4 partial fractions, A-5–A-7

Integro-differential equations convolution theorem and, 400–402 explanation of, 400

Interconnected fluid tanks, 242–244, 536, 549–550

Interplanetary travel, 309–310 Interval of definition problem, 44 Inverse, matrix, 510–512 Inverse Laplace transforms

explanation of, 366–367 linearity of, 368–370 nonrepeating linear factors and, 370–371 quadratic factors and, 373–374 repeated linear factors and, 372–373

Irregular singular points, 456 Isoclines, 19–20 Isolated zeros, 203

Jacobi polynomials, 484 Jump discontinuity, 356, 578

Kinetic energy, 314 Kirchhoff, Gustav Robert, 118n Kirchhoff’s current law, 3, 118 Kirchhoff’s voltage law, 3, 118 Korteweg-de Vries equation, 622 Kutta, W., 133n

Laguerre equation, 475, 489 Laguerre polynomials, 475, 484, 489 Lane-Emden equation, 445 Laplace, Pierre, 351n Laplace transforms

convolution and, 397–403 definition of, 351, 353–359 of derivative, 361–362 discontinuous and periodic functions and,

383–393 existence of, 356–359 impulses and Dirac delta function and,

404–410 initial value problems and, 376–382 inverse, 366–374 linearity and, 355 matrix, 564 mixing problem and, 350–352

properties of, 361–365 table of, 359 wave equation and, 640–641

Laplace’s equation boundary conditions associated with,

623–627 Dirichlet problem and, 627–631 existence and uniqueness of solutions and,

632–633 explanation of, 623–633 heat equation and, 568 invariance of, 635

Laplacian, 568 Least-squares approximation property, 593 Least-squares linear fit, 98 Least-squares method, A-13–A-14 Legendre, Adrien Marie, 482n Legendre polynomials, 483, 487 Legendre’s equation

explanation of, 204, 451, 482 special functions and, 482–484

Leibniz, Gottfried, 38n, 71n Leibniz’s rule, 364, 418 Leonardo da Pisa, 349 Lifetime, average, 284 Limiting (terminal) velocity, 37, 111 Linear algebraic equations, 502–506 Linear coefficients, 72–74 Linear dependence

of functions, 162, 196, 322 of three functions, 166 of vector functions, 519–520

Linear differential equations (see also Higher- order differential equations)

asymptotic behavior of solutions to, 87–88

basic theory of, 318–324 definition of, 4 existence and uniqueness of solution and,

51, 161, 183–184, 318 power series solutions to, 436–444

Linear differential operators, 245, 247, 318 Linear independence

explanation of, 161, 322 of functions, 161, 196, 322 fundamental solution sets and, 323, 521 of vector functions, 519–520

Linear operator explanation of, 509 Laplace transform as, 355, 368 matrices as, 509

Linear second-order equations auxiliary equations for, 159 Cauchy-Euler, 194–200, 452–454 forced mechanical vibrations and, 223–229 free mechanical vibrations and, 214–222 homogeneous linear equations with con-

stant coefficients and, 169–186 mass-spring oscillator and, 153–157,

170–173

Index I-5

method of undetermined coefficients and, 175–181, 186–187, 239–240

power series method and, 436–444 qualitative considerations for variable coef-

ficient and, 203–212 reduction of order and, 198, 326 superposition and nonhomogeneous equa-

tions and, 182–184, 201 undetermined coefficients and, 175–181,

186, 239–240 variation of parameters and, 189–192, 197

Linear systems Cauchy-Euler, 536 frequency response modeling of, 419–421 Laplace transforms to solve, 412–413 matrix methods for (see also Matrix meth-

ods), 498–550 nonhomogeneous, 543–547 in normal form, 517–523 stability of, 411

Linearized equation, 622 Local truncation error, 131 Logistic function, 97, 277 Logistic model

biomathematics applications, 277 Malthusian vs., 95–96, 277 phase line for, 33 for population growth, 18, 96–99, 283

Long water waves, 622 Lunar orbit, 262 Lungs, 80–82 Lymphocytes, 143

Maclaurin series, 434n Magnetic field, 25, 73, 274 Malthus, Thomas R., 95n Malthusian law, 95 Malthusian model, 95–96 Market equilibrium, 149–150 Mass-spring systems

analogies, 206–212 coupled, 285–289 examples with, 170–173, 269–270,

285–289 explanation of, 153–157 forced mechanical vibrations and, 223–229 free mechanical vibrations and, 214–222 matrix methods and, 501–502 spring stiffness and, 153, 206

Mathematical models applications for, 89–91 compartmental analysis and, 91–99 development of, 90 heating and cooling of buildings and,

101–107 historical background of, 89 Newtonian mechanics and, 108–114 population dynamics and, 256–259 tests of, 90–91

Mathieu’s equation, 213

Matrices algebra of, 498, 507–514 calculus of, 514–515 coefficient, 500 diagonal, 507 explanation of, 507 fundamental, 521, 551–553 nilpotent, 552–553 product of, 498–500 real symmetric, 532–534 square, 507 transpose, 510 zero, 507

Matrix exponential function, 550–556 Matrix methods

complex eigenvalues and, 538–541 homogeneous linear systems with constant

coefficients and, 526–534 introduction to, 498–502 linear algebraic equations and, 502–506 linear systems in normal form and,

517–523 matrices and vectors and, 507–515 matrix exponential function and, 550–556 nonhomogeneous linear systems and,

543–547 Maximum principle, 608–609 Mean value property, 642 Mechanical units, 109 Mechanical vibrations

forced, 223–229 free, 214–222

Mechanics explanation of, 108 Newtonian, 108–114

Method of Frobenius, 455–465 Method of isoclines, 19–20 Method of least squares, A-13–A-14 Midpoint method, 134 Mixing problems

common drains and, 537 compartmental analysis and, 91–94 first-order equations and, 44, 54 interconnected tanks and, 242–244, 536,

549–550 Modal analysis, 288n Models. See Mathematical models Modes, 602 Modified Bessel equation, 208, 631 Modified Bessel function

of the first kind, 631 of the second kind, 631

Moore, Gordon E., 276n Moore’s law, 276n Motion

critically damped, 218–222 of falling bodies, 1, 35–37 of objects that bob, 312–313 overdamped, 217–218 underdamped, 217

MRSA, 310–312 Multiplication

matrix, 508–509 scalar, 507–508

Multistep methods, 150–151

Neumann boundary conditions. See Boundary conditions

Neumann function, 480 Newtonian mechanics

explanation of, 108 laws of motion and, 108–109 procedure for Newtonian models and,

109 Newton’s approximation method, A-8–A-10 Newton’s law of cooling, 29, 45, 102 Newton’s second law

explanation of, 1, 35, 269 mass-spring systems and, 153, 170, 285,

288 Newtonian mechanics and, 108–114

Nilpotent matrix, 552–553 Node, 268 Nonhomogeneous boundary conditions, 602 Nonhomogeneous equations

explanation of, 158, 318 power series and, 450 superposition and, 182–184 undetermined coefficients and, 175–181,

186, 239–240, 333–336 variation of parameters and, 189–192, 197,

338–340 Nonhomogeneous linear systems

explanation of, 517 representation of solutions for, 522 undetermined coefficients and, 543–544 variation of parameters and, 544–547

Nonlinear equations explanation of, 4 linearization of, 238 oscillations and, 261 qualitative considerations for, 203–212 solvable by first-order techniques,

235–236 Nonlinear springs, 261 Norm, 592 Normal form

explanation of, 253–254, 500 linear systems in, 517–523 matrix algebra and, 500

Normal frequencies, angular, 288, 540 Normal modes, 288 Normal systems, uncoupling, 563 Numerical methods

Euler’s method, 23–27, 254–256 global error, 131 higher-order, 132–138 for higher-order equations and systems,

254–259 improved Euler’s method and, 126–129

Numerical methods (continued) local truncation error, 131 Runge-Kutta, 133–138, 258–259,

A-15–A-16 for solving generalized Dirichlet problem,

642–644 stability of, 150–151 Taylor, 132–134

Nutation, 275–276

Oil spill, 79–80 One-dimensional heat flow equation, 567 Operators

differential, 245, 247, 253, 318 integral, 353 linear, 355, 368, 509

Ordinary differential equations, 4 Ordinary points, 437 Orthogonal expansions, 586–587 Orthogonal trajectories, 62–63 Orthogonality

Fourier series and, 580–581 Legendre polynomials and, 483

Oscillation explanation of, 217 nonlinear equations and, 261

Oscillators harmonic, 215 mass-spring (see also Mass-spring sys-

tems), 153–157, 170–173, 214–229, 285–289

Overdamped motion, 218

Parachutist example, 112–113 Parameters (see also Variation of parameters),

90n Parseval’s identity, 593 Partial differential equations

explanation of, 4 Fourier cosine and sine series and, 594–598 Fourier series and, 578–591 heat equation and, 599–609 Laplace’s equation and, 623–633 model for heat flow and, 566–568 separation of variables method and,

569–576 wave equation and, 611–621

Partial fractions method, 370, A-5–A-7 Pendulum

motion of, 210–211, 237–238, 290–291 spring, 263 with varying length, 261–262

Period doubling, 151–152 Periodic functions, 390–392 Periodic solutions, 270 Phase angle, 419 Phase line, 32–34 Phase plane, introduction to, 263–272 Phase plane equation, 264

I-6 Index

Phase velocity, 622 Phytoplankton growth model, 316–317 Picard-Lindelöf existence theorem, 32n Picard’s method, 31–32 Piecewise continuity

explanation of, 356, 578 Laplace transform and, 356, 358, 379

Plucked string, 621 Poincaré maps, 298–305 Poincaré sections, 298–303 Poisson’s integral formula, 642 Polynomials

characteristic, 527 Chebyshev, 484, 489 classical orthogonal, 484 Gegenbauer, 484 Hermite, 484, 489 Jacobi, 484 Laguerre, 484, 489 Legendre, 483, 487 spherical, 483 Taylor, 422–426 ultraspherical, 484

Pooling delay, 251–252 Population dynamics, 256–259 Population models, 94–99 Potential energy, 314 Power series

analytic functions and, 433 differentiation and integration of, 430 explanation of, 427–434 linear differential equations and, 436–444

Predator-prey models, 256–259, 261 Predictor-corrector method, 125 Pythagorean property, 592

Quadratic factors, 373–374 Quasifrequency, 217 Quasiperiod, 217 Quicksort algorithm, 493–494

Radioactive decay, 1, 35, 49–50, 52, 276 Radioisotopes, 283 Radius of convergence

explanation of, 428 minimum value for, 446–447

Rate of change, 2 Ratio test, 428, 435 Rayleigh equation, 213 Rectangular window function, 384 Recurrence relation, 439, 484 Reduction of order, 198–200, 326 Regular singular points, 456 Regular window function, 384 Relative error, 128 Removable singularities, 437n Repeated roots, 163, 329–331 Residue computation, 375 Resistance, 118

Resonance curve, 225–226 Resonance frequencies, 226 Riccati, Jacopo, 75n Riccati equation, 75 Rigid body nutation, 275–276 Risk aversion, 85–86 RL network, 549 RLC network, 542 RLC series circuit, 291 Rocket flight, 117 Rodrigues’s formula, 484, 488 Roots

complex, 167–173, 329 complex conjugate, 169, 538n distinct real, 163, 328–329 repeated, 163, 329–331

Row operation, 511–512 Runge, C., 133n Runge-Kutta method

classical 4th order, 134–138, 258–259, A- 15–A-16

for systems, 258–259

Saddle point, 268 Sawtooth wave, 394 Scalar multiplication, 507–508 Scatter diagrams, 98 Schrödinger’s equation, 494–495 Second-order linear equations. See Linear

second-order equations Secretion of hormones problem, 54 Selfadjoint, 483 Separable equations

explanation of, 38 method of solving, 39–42

Separation of variables eigenfunction properties, 571 explanation of, 35–36, 569 heat flow and, 600–603 Laplace’s equation and, 623–633 partial differential equations and, 569–576 to solve initial-boundary problems, 615

Series solutions Cauchy-Euler equations and, 452–454 equations with analytic coefficients and,

446–450 linear differential equations and power,

436–444 method of Frobenius and, 455–465 power series and analytical functions and,

427–434 second linearly independent, 467–474 special functions and, 476–486 Taylor polynomial approximation and,

422–426 Similarity transformation, 534 Simple harmonic equation, 167 Simple harmonic motion, 215 Simpson’s rule, A-10–A-11

Index I-7

Singular points explanation of, 437 first-order, 53 irregular, 456 regular, 456

Singular solutions, 84 Sink, 33 Sinusoidal inputs, 419–420 SIR epidemic model, 278–281 Snowplow problems, 83–84 Solar collector, 86–87 Solar water heater, 108 Solution sets, fundamental, 323, 521 Solutions

asymptotic behavior of, 241 equilibrium, 265 existence and uniqueness, 11–13, 161,

183–184, 194, 318–320, 518–519, 608–609, 619, 632–633

explicit, 7 formal, 575 general. See General solutions implicit, 8–9 linear combinations of, 159 linear dependence of, 162, 196, 323, 521 linear independence of, 161, 196, 323, 521 singular, 84 spherically symmetric, 494 unique. see Unique solutions

Source, 33 Special functions

Bessel’s equation, 478–482 explanation of, 476 hypergeometric equation, 476–478 Legendre’s equation, 482–484

Special integrating factors, 64–67 Specific heat capacity, 567 Speed bumps, 189 Spherical Bessel functions, 487 Spherical polynomials, 483 Spherically symmetric solutions, 494 Spiral point, 268 Springs (see also Mass-spring systems)

aging, 451 Bessel functions and aging, 496–497 nonlinear, 261 soft vs. hard, 427 vibrating, 174

Square pulse, 384 Square wave, 394 Stability

of homogeneous linear systems, 537 of linear systems, 411 of numerical methods, 150–151 of trapezoid scheme, 125

Stable systems, 266, 411, 537 Standard form, linear differential equations in,

47, 318 Standing wave, 612

Staph infections, 310–312 Static friction, 116 Steady-state current, 293 Steady-state solutions

explanation of, 224, 419 to sinusoidal inputs, 420

Steady-state temperature distribution in circular cylinder, 638–640

Stefan’s law of radiation, 29, 108 Step size, 24, 122 Sticky friction, 275 Stopping procedures, 128 Strange attractors

explanation of, 301 forced pendulum equation and, 303

Struck string, 621 Sturm-Liouville equations, 260 Subharmonics, 299 Substitutions

Bernoulli equations and, 71–72 equations of form dy/dx = G(ax + by),

70–71 equations with linear coefficients and,

72–74 homogeneous equations and, 68–70 procedure for, 68

Summation index, 431–432 Superposition principle

explanation of, 182, 522 undetermined coefficients and, 182–187 variable coefficients and, 201

Suspended cable, 235–236 Swinging door, 175 Symmetric functions, 579 Symmetric matrices, 517, 532–534 System parameters, 421

Tacoma Narrows bridge, 228n Taylor method of order p, 132–134 Taylor polynomial approximation

explanation of, 30–31, 422–426 series solutions and, 423–426

Taylor series (see also Power series), 30–31, 423

Tchebichef (Chebyshev) polynomials, 484, 489

Technetium, 283 Telegraph problem, 622 Temperature distribution in a circular cylinder,

638 Terminal velocity, 37, 111, 275 Thermal radiation, 29 Threshold value, 280 Time constant, 103–106 Time-shifted pair, 263–264 Tolerance

classical 4th order Runge-Kutta algorithm with, 136, A-16

improved Euler’s method with, 128

Torricelli’s law of fluid flow, 82–83 Total differential, 55 Towers, buckling of, 495–496 Trajectories

explanation of, 264 orthogonal, 62–63 in phase plane equations, 264–272

Transfer function, 401, 409, 418–419 Transformations. See Substitutions

Transient current, 293 Transient solution

circuit, 293 heat flow, 603 vibrations, 224

Transmission lines, 140 Transpose, 510 Transverse vibrations of a beam, 347 Trapezoid scheme, 125 Traveling wave solutions, 615–618 Triangular wave, 395 Truncation error, 131 Tumor growth models, 276, 281–282

Ultraspherical polynomials, 484 Uncoupling systems, 563 Undamped second-order systems, 565 Underdamped motion, 217 Undetermined coefficients method

annihilator method and, 333–336 explanation of, 175–181, 185–187 nonhomogeneous linear systems and,

543–544 using complex arithmetic, 239–240

Uniform convergence, 589 Uniform waves, 622 Unique solutions, 11–12, 51, 160, 183–184,

318–320, 518–519, 608–609, 619, 632–633

Unit step function, 384 Unit triangular pulse, 396 Utility functions, risk aversion and, 85–86

van der Pol equation, 209, 427 Variable-coefficient equations

of higher order, 318–324 qualitative considerations for, 203–212 of second order, 193–200

Variable resistor, 446 Variable spring constant, 446 Variables, separation of. See Separation of

variables Variation of parameters

explanation of, 54, 189–192, 197 higher-order equations and, 338–340 nonhomogeneous linear systems and,

544–547 Vector functions, 519–520 Vectors, 321, 507

Velocity group, 622 limiting or terminal, 37, 111, 275 phase, 622

Ventilation, 80 Verhulst, P.F., 97n Vibrating beams, 333 Vibrating drums, 622–623 Vibrating membrane, 614 Vibrating springs, 174 Vibrating strings, 3, 612–614, 619

I-8 Index

Vibrations forced, 188, 223–229 transverse, 347

Virions, 143 Virus, 143 Volterra-Lotka system, 256–259, 261, 263,

277

Wave equation, 611–621, 640–641 Weber function, 480 Window function, 384

Wire, heat flow in insulated, 566–568 Wronskian

definition of, 162, 164, 196, 320, 520 fundamental sets, linear independence and,

323, 520 vector functions and, 520

Young’s modulus, 333, 347

Zeros, isolated, 203

A BRIEF TABLE OF INTEGRALS*

*Note: An arbitrary constant is to be added to each indefinite integral.

A BRIEF TABLE OF INTEGRALS* (continued)

SOME POWER SERIES EXPANSIONS

*Note: An arbitrary constant is to be added to each indefinite integral.

LINEAR FIRST-ORDER EQUATIONS A general solution to the first-order linear equation is

METHOD OF UNDETERMINED COEFFICIENTS To find a particular solution to the constant-coefficient differential equation

where is a polynomial of degree m, use the form

if r is not a root of the associated auxiliary equation, take s " 0; if r is a simple root of the associated auxiliary equation, take s " 1; and if r is a double root of the associated auxiliary equation, take s " 2.

To find a particular solution to the differential equation

where is a polynomial of degree m and is a polynomial of degree n, use the form

where k is the larger of m and n. If a# ib is not a root of the associated auxiliary equa- tion, take s " 0; if a# ib is a root of the associated auxiliary equation, take s " 1.

VARIATION OF PARAMETERS FORMULA If and are two linearly independent solutions to then a particu- lar solution to is where

and W 3 y1, y2 4 AtB " y1 AtBy¿2 AtB ! y¿1 AtBy2 AtB. y1 AtB " ! !ƒ AtBy2 AtBaW 3 y1, y2 4 AtB dt, y2 AtB " ! ƒ AtBy1 AtBaW 3 y1, y2 4 AtB dt,

y " y1y1 # y2y2,ay– # by¿ # cy " ƒ ay– # by¿ # cy " 0,y2y1

# t s ABkt k # p # B1t # B0Beat sin bt ,yp AtB " t s AAkt k # p # A1t # A0Beat cos bt

Qn AtBPm AtB b & 0 ,ay– # by¿ # cy " Pm AtBeat cos b t # Qn AtBeat sin b t ,

yp AtB " t s AAmt m # p # A1t # A0Bert ;Pm AtB ay– # by¿ # cy " Pm AtBert ,

y AxB " 3m AxB 4!1 a!m AxBQ AxBdx # Cb , where m AxB " exp a!P AxBdxb .dy/dx # P AxBy " Q AxB

A TABLE OF LAPLACE TRANSFORMS

,a,b(t),

Cover
Title Page
Copyright Page
Preface
ACKNOWLEDGMENTS
Contents
CHAPTER 1 Introduction

1.1 Background
1.2 Solutions and Initial Value Problems
1.3 Direction Fields
1.4 The Approximation Method of Euler
Chapter Summary
Technical Writing Exercises
Group Projects for Chapter 1

A. Taylor Series Method
B. Picard’s Method
C. The Phase Line

CHAPTER 2 First-Order Differential Equations

2.1 Introduction: Motion of a Falling Body
2.2 Separable Equations
2.3 Linear Equations
2.4 Exact Equations
2.5 Special Integrating Factors
2.6 Substitutions and Transformations
Chapter Summary
Review Problems
Technical Writing Exercises
Group Projects for Chapter 2

A. Oil Spill in a Canal
B. Differential Equations in Clinical Medicine
C. Torricelli’s Law of Fluid Flow
D. The Snowplow Problem
E. Two Snowplows
F. Clairaut Equations and Singular Solutions
G. Multiple Solutions of a First-Order Initial Value Problem
H. Utility Functions and Risk Aversion
I. Designing a Solar Collector
J. Asymptotic Behavior of Solutions to Linear Equations

CHAPTER 3 Mathematical Models and Numerical Methods Involving First-Order Equations

3.1 Mathematical Modeling
3.2 Compartmental Analysis
3.3 Heating and Cooling of Buildings
3.4 Newtonian Mechanics
3.5 Electrical Circuits
3.6 Improved Euler’s Method
3.7 Higher-Order Numerical Methods: Taylor and Runge-Kutta
Group Projects for Chapter 3

A. Dynamics of HIV Infection
B. Aquaculture
C. Curve of Pursuit
D. Aircraft Guidance in a Crosswind
E. Feedback and the Op Amp
F. Bang-Bang Controls
G. Market Equilibrium: Stability and Time Paths
H. Stability of Numerical Methods
I. Period Doubling and Chaos

CHAPTER 4 Linear Second-Order Equations

4.1 Introduction: The Mass-Spring Oscillator
4.2 Homogeneous Linear Equations: The General Solution
4.3 Auxiliary Equations with Complex Roots
4.4 Nonhomogeneous Equations: The Method of Undetermined Coefficients
4.5 The Superposition Principle and Undetermined Coefficients Revisited
4.6 Variation of Parameters
4.7 Variable-Coefficient Equations
4.8 Qualitative Considerations for Variable-Coefficient and Nonlinear Equations
4.9 A Closer Look at Free Mechanical Vibrations
4.10 A Closer Look at Forced Mechanical Vibrations
Chapter Summary
Review Problems
Technical Writing Exercises
Group Projects for Chapter 4

A. Nonlinear Equations Solvable by First-Order Techniques
B. Apollo Reentry
C. Simple Pendulum
D. Linearization of Nonlinear Problems
E. Convolution Method
F. Undetermined Coefficients Using Complex Arithmetic
G. Asymptotic Behavior of Solutions

CHAPTER 5 Introduction to Systems and Phase Plane Analysis

5.1 Interconnected Fluid Tanks
5.2 Differential Operators and the Elimination Method for Systems
5.3 Solving Systems and Higher-Order Equations Numerically
5.4 Introduction to the Phase Plane
5.5 Applications to Biomathematics: Epidemic and Tumor Growth Models
5.6 Coupled Mass-Spring Systems
5.7 Electrical Systems
5.8 Dynamical Systems, Poincaré Maps, and Chaos
Chapter Summary
Review Problems
Group Projects for Chapter 5

A. Designing a Landing System for Interplanetary Travel
B. Spread of Staph Infections in Hospitals—Part 1
C. Things That Bob
D. Hamiltonian Systems
E. Cleaning Up the Great Lakes
F. A Growth Model for Phytoplankton—Part I

CHAPTER 6 Theory of Higher-Order Linear Differential Equations

6.1 Basic Theory of Linear Differential Equations
6.2 Homogeneous Linear Equations with Constant Coefficients
6.3 Undetermined Coefficients and the Annihilator Method
6.4 Method of Variation of Parameters
Chapter Summary
Review Problems
Technical Writing Exercises
Group Projects for Chapter 6

A. Computer Algebra Systems and Exponential Shift
B. Justifying the Method of Undetermined Coefficients
C. Transverse Vibrations of a Beam
D. Higher-Order Difference Equations

CHAPTER 7 Laplace Transforms

7.1 Introduction: A Mixing Problem
7.2 Definition of the Laplace Transform
7.3 Properties of the Laplace Transform
7.4 Inverse Laplace Transform
7.5 Solving Initial Value Problems
7.6 Transforms of Discontinuous and Periodic Functions
7.7 Convolution
7.8 Impulses and the Dirac Delta Function
7.9 Solving Linear Systems with Laplace Transforms
Chapter Summary
Review Problems
Technical Writing Exercises
Group Projects for Chapter 7

A. Duhamel’s Formulas
B. Frequency Response Modeling
C. Determining System Parameters

CHAPTER 8 Series Solutions of Differential Equations

8.1 Introduction: The Taylor Polynomial Approximation
8.2 Power Series and Analytic Functions
8.3 Power Series Solutions to Linear Differential Equations
8.4 Equations with Analytic Coefficients
8.5 Cauchy-Euler (Equidimensional) Equations
8.6 Method of Frobenius
8.7 Finding a Second Linearly Independent Solution
8.8 Special Functions
Chapter Summary
Review Problems
Technical Writing Exercises
Group Projects for Chapter 8

A. Alphabetization Algorithms
B. Spherically Symmetric Solutions to Shrödinger’s Equation for the Hydrogen Atom
C. Airy’s Equation
D. Buckling of a Tower
E. Aging Spring and Bessel Functions

CHAPTER 9 Matrix Methods for Linear Systems

9.1 Introduction
9.2 Review 1: Linear Algebraic Equations
9.3 Review 2: Matrices and Vectors
9.4 Linear Systems in Normal Form
9.5 Homogeneous Linear Systems with Constant Coefficients
9.6 Complex Eigenvalues
9.7 Nonhomogeneous Linear Systems
9.8 The Matrix Exponential Function
Chapter Summary
Review Problems
Technical Writing Exercises
Group Projects for Chapter 9

A. Uncoupling Normal Systems
B. Matrix Laplace Transform Method
C. Undamped Second-Order Systems

CHAPTER 10 Partial Differential Equations

10.1 Introduction: A Model for Heat Flow
10.2 Method of Separation of Variables
10.3 Fourier Series
10.4 Fourier Cosine and Sine Series
10.5 The Heat Equation
10.6 The Wave Equation
10.7 Laplace’s Equation
Chapter Summary
Technical Writing Exercises
Group Projects for Chapter 10

A. Steady-State Temperature Distribution in a Circular Cylinder
B. Laplace Transform Solution of the Wave Equation
C. Green’s Function
D. Numerical Method for Δu = ƒ on a Rectangle

APPENDICES

A. Review of Integration Techniques
B. Newton’s Method
C. Simpson’s Rule
D. Cramer’s Rule
E. Method of Least Squares
F. Runge-Kutta Procedure for n Equations

ANSWERS TO ODD-NUMBERED PROBLEMS
INDEX