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Applied Numerical Methods with MATLAB® for Engineers and Scientists

Fourth Edition

Steven C. Chapra Berger Chair in Computing and Engineering

Tufts University

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APPLIED NUMERICAL METHODS WITH MATLAB® FOR ENGINEERS AND SCIENTISTS, FOURTH EDITION

Published by McGraw-Hill Education, 2 Penn Plaza, New York, NY 10121. Copyright © 2018 by McGraw-Hill Education. All rights reserved. Printed in the United States of America. Previous editions © 2012, 2008, and 2005. No part of this publication may be reproduced or distributed in any form or by any means, or stored in a database or retrieval system, without the prior written consent of McGraw-Hill Education, including, but not limited to, in any network or other electronic storage or transmission, or broadcast for distance learning.

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Library of Congress Cataloging-in-Publication Data Chapra, Steven C., author. Applied numerical methods with MATLAB for engineers and scientists / Steven C. Chapra, Berger Chair in Computing and Engineering, Tufts University. Fourth edition. | New York, NY : McGraw-Hill Education, [2018] | Includes bibliographical references and index. LCCN 2016038044 | ISBN 9780073397962 (alk. paper) | ISBN 0073397962 (alk. paper) LCSH: Numerical analysis—Data processing—Textbooks. | Engineering mathematics—Textbooks. | MATLAB—Textbooks. LCC QA297 .C4185 2018 | DDC 518—dc23 LC record available at https://lccn.loc.gov/2016038044 The Internet addresses listed in the text were accurate at the time of publication. The inclusion of a website does not indicate an endorsement by the authors or McGraw-Hill Education, and McGraw-Hill Education does not guarantee the accuracy of the information presented at these sites.

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My brothers,

John and Bob Chapra

and

Fred Berger (1947–2015)

who I miss as a good friend, a good man.

and a comrade in bringing the light of engineering

to some of world’s darker corners.

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ABOUT THE AUTHOR

Steve Chapra teaches in the Civil and Environmental Engineering Department at Tufts University, where he holds the Louis Berger Chair in Computing and Engineering. His other books include Numerical Methods for Engineers and Surface Water-Quality Modeling.

Steve received engineering degrees from Manhattan College and the University of Michigan. Before joining the faculty at Tufts, he worked for the Environmental Protection Agency and the National Oceanic and Atmospheric Administration, and taught at Texas A&M University and the University of Colorado. His general research interests focus on surface water-quality modeling and advanced computer applications in environmental engineering.

He has received a number of awards for his scholarly contributions, including the Rudolph Hering Medal, the Meriam/Wiley Distinguished Author Award, and the Chandler- Misener Award. He has also been recognized as the outstanding teacher at Texas A&M University (1986 Tenneco Award), the University of Colorado (1992 Hutchinson Award), and Tufts University (2011 Professor of the Year Award).

Steve was originally drawn to environmental engineering and science because of his love of the outdoors. He is an avid fly fisherman and hiker. An unapologetic nerd, his love affair with computing began when he was first introduced to Fortran programming as an undergraduate in 1966. Today, he feels truly blessed to be able to meld his love of math- ematics, science, and computing with his passion for the natural environment. In addition, he gets the bonus of sharing it with others through his teaching and writing!

Beyond his professional interests, he enjoys art, music (especially classical music, jazz, and bluegrass), and reading history. Despite unfounded rumors to the contrary, he never has, and never will, voluntarily bungee jump or sky dive.

If you would like to contact Steve, or learn more about him, visit his home page at http://engineering.tufts.edu/cee/people/chapra/ or e-mail him at [email protected].

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CONTENTS

About the Author iv

Preface xiv

Part One Modeling, Computers, and Error Analysis 1

1.1 Motivation 1 1.2 Part Organization 2

CHAPTER 1

Mathematical Modeling, Numerical Methods, and Problem Solving 4

1.1 A Simple Mathematical Model 5 1.2 Conservation Laws in Engineering and Science 12 1.3 Numerical Methods Covered in This Book 13 1.4 Case Study: It’s a Real Drag 17 Problems 20

CHAPTER 2

MATLAB Fundamentals 27

2.1 The MATLAB Environment 28 2.2 Assignment 29 2.3 Mathematical Operations 36 2.4 Use of Built-In Functions 39 2.5 Graphics 42 2.6 Other Resources 46 2.7 Case Study: Exploratory Data Analysis 46 Problems 49

CHAPTER 3

Programming with MATLAB 53

3.1 M-Files 54 3.2 Input-Output 61

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vi CONTENTS

3.3 Structured Programming 65 3.4 Nesting and Indentation 79 3.5 Passing Functions to M-Files 81 3.6 Case Study: Bungee Jumper Velocity 87 Problems 91

CHAPTER 4

Roundoff and Truncation Errors 99

4.1 Errors 100 4.2 Roundoff Errors 106 4.3 Truncation Errors 114 4.4 Total Numerical Error 125 4.5 Blunders, Model Errors, and Data Uncertainty 130 Problems 131

Part Two Roots and Optimization 135

2.1 Overview 135 2.2 Part Organization 136

CHAPTER 5

Roots: Bracketing Methods 138

5.1 Roots in Engineering and Science 139 5.2 Graphical Methods 140 5.3 Bracketing Methods and Initial Guesses 141 5.4 Bisection 146 5.5 False Position 152 5.6 Case Study: Greenhouse Gases and Rainwater 156 Problems 159

CHAPTER 6

Roots: Open Methods 164

6.1 Simple Fixed-Point Iteration 165 6.2 Newton-Raphson 169 6.3 Secant Methods 174 6.4 Brent’s Method 176 6.5 MATLAB Function: fzero 181 6.6 Polynomials 183 6.7 Case Study: Pipe Friction 186 Problems 191

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CONTENTS vii

CHAPTER 7

Optimization 198

7.1 Introduction and Background 199 7.2 One-Dimensional Optimization 202 7.3 Multidimensional Optimization 211 7.4 Case Study: Equilibrium and Minimum Potential Energy 213 Problems 215

Part Three Linear Systems 223

3.1 Overview 223 3.2 Part Organization 225

CHAPTER 8

Linear Algebraic Equations and Matrices 227

8.1 Matrix Algebra Overview 229 8.2 Solving Linear Algebraic Equations with MATLAB 238 8.3 Case Study: Currents and Voltages in Circuits 240 Problems 244

CHAPTER 9

Gauss Elimination 248

9.1 Solving Small Numbers of Equations 249 9.2 Naive Gauss Elimination 254 9.3 Pivoting 261 9.4 Tridiagonal Systems 264 9.5 Case Study: Model of a Heated Rod 266 Problems 270

CHAPTER 10

LU Factorization 274

10.1 Overview of LU Factorization 275 10.2 Gauss Elimination as LU Factorization 276 10.3 Cholesky Factorization 283 10.4 MATLAB Left Division 286 Problems 287

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viii CONTENTS

CHAPTER 11

Matrix Inverse and Condition 288

11.1 The Matrix Inverse 288 11.2 Error Analysis and System Condition 292 11.3 Case Study: Indoor Air Pollution 297 Problems 300

CHAPTER 12

Iterative Methods 305

12.1 Linear Systems: Gauss-Seidel 305 12.2 Nonlinear Systems 312 12.3 Case Study: Chemical Reactions 320 Problems 323

CHAPTER 13

Eigenvalues 326

13.1 Mathematical Background 328 13.2 Physical Background 331 13.3 The Power Method 333 13.4 MATLAB Function: eig 336 13.5 Case Study: Eigenvalues and Earthquakes 337 Problems 340

Part Four Curve Fitting 343

4.1 Overview 343 4.2 Part Organization 345

CHAPTER 14

Linear Regression 346

14.1 Statistics Review 348 14.2 Random Numbers and Simulation 353 14.3 Linear Least-Squares Regression 358 14.4 Linearization of Nonlinear Relationships 366 14.5 Computer Applications 370 14.6 Case Study: Enzyme Kinetics 373 Problems 378

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CONTENTS ix

CHAPTER 15

General Linear Least-Squares and Nonlinear Regression 385

15.1 Polynomial Regression 385 15.2 Multiple Linear Regression 389 15.3 General Linear Least Squares 391 15.4 QR Factorization and the Backslash Operator 394 15.5 Nonlinear Regression 395 15.6 Case Study: Fitting Experimental Data 397 Problems 399

CHAPTER 16

Fourier Analysis 404

16.1 Curve Fitting with Sinusoidal Functions 405 16.2 Continuous Fourier Series 411 16.3 Frequency and Time Domains 414 16.4 Fourier Integral and Transform 415 16.5 Discrete Fourier Transform (DFT) 418 16.6 The Power Spectrum 423 16.7 Case Study: Sunspots 425 Problems 426

CHAPTER 17

Polynomial Interpolation 429

17.1 Introduction to Interpolation 430 17.2 Newton Interpolating Polynomial 433 17.3 Lagrange Interpolating Polynomial 441 17.4 Inverse Interpolation 444 17.5 Extrapolation and Oscillations 445 Problems 449

CHAPTER 18

Splines and Piecewise Interpolation 453

18.1 Introduction to Splines 453 18.2 Linear Splines 455 18.3 Quadratic Splines 459 18.4 Cubic Splines 462 18.5 Piecewise Interpolation in MATLAB 468 18.6 Multidimensional Interpolation 473 18.7 Case Study: Heat Transfer 476 Problems 480

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x CONTENTS

Part Five Integration and Differentiation 485

5.1 Overview 485 5.2 Part Organization 486

CHAPTER 19

Numerical Integration Formulas 488

19.1 Introduction and Background 489 19.2 Newton-Cotes Formulas 492 19.3 The Trapezoidal Rule 494 19.4 Simpson’s Rules 501 19.5 Higher-Order Newton-Cotes Formulas 507 19.6 Integration with Unequal Segments 508 19.7 Open Methods 512 19.8 Multiple Integrals 512 19.9 Case Study: Computing Work with Numerical Integration 515 Problems 518

CHAPTER 20

Numerical Integration of Functions 524

20.1 Introduction 524 20.2 Romberg Integration 525 20.3 Gauss Quadrature 530 20.4 Adaptive Quadrature 537 20.5 Case Study: Root-Mean-Square Current 540 Problems 544

CHAPTER 21

Numerical Differentiation 548

21.1 Introduction and Background 549 21.2 High-Accuracy Differentiation Formulas 552 21.3 Richardson Extrapolation 555 21.4 Derivatives of Unequally Spaced Data 557 21.5 Derivatives and Integrals for Data with Errors 558 21.6 Partial Derivatives 559 21.7 Numerical Differentiation with MATLAB 560 21.8 Case Study: Visualizing Fields 565 Problems 567

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CONTENTS xi

Part six Ordinary Differential Equations 573

6.1 Overview 573 6.2 Part Organization 577

CHAPTER 22

Initial-Value Problems 579

22.1 Overview 581 22.2 Euler’s Method 581 22.3 Improvements of Euler’s Method 587 22.4 Runge-Kutta Methods 593 22.5 Systems of Equations 598 22.6 Case Study: Predator-Prey Models and Chaos 604 Problems 609

CHAPTER 23

Adaptive Methods and Stiff Systems 615

23.1 Adaptive Runge-Kutta Methods 615 23.2 Multistep Methods 624 23.3 Stiffness 628 23.4 MATLAB Application: Bungee Jumper with Cord 634 23.5 Case Study: Pliny’s Intermittent Fountain 635 Problems 640

CHAPTER 24

Boundary-Value Problems 646

24.1 Introduction and Background 647 24.2 The Shooting Method 651 24.3 Finite-Difference Methods 658 24.4 MATLAB Function: bvp4c 665 Problems 668

APPENDIX A: MATLAB BUILT-IN FUNCTIONS 674

APPENDIX B: MATLAB M-FILE FUNCTIONS 676

APPENDIX C: INTRODUCTION TO SIMULINK 677

BIBLIOGRAPHY 685

INDEX 687

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xiv

PREFACE

This book is designed to support a one-semester course in numerical methods. It has been written for students who want to learn and apply numerical methods in order to solve prob- lems in engineering and science. As such, the methods are motivated by problems rather than by mathematics. That said, sufficient theory is provided so that students come away with insight into the techniques and their shortcomings.

MATLAB® provides a great environment for such a course. Although other en- vironments (e.g., Excel/VBA, Mathcad) or languages (e.g., Fortran 90, C++) could have been chosen, MATLAB presently offers a nice combination of handy program- ming features with powerful built-in numerical capabilities. On the one hand, its M-file programming environment allows students to implement moderately compli- cated algorithms in a structured and coherent fashion. On the other hand, its built-in, numerical capabilities empower students to solve more difficult problems without try- ing to “reinvent the wheel.”

The basic content, organization, and pedagogy of the third edition are essentially pre- served in the fourth edition. In particular, the conversational writing style is intentionally maintained in order to make the book easier to read. This book tries to speak directly to the reader and is designed in part to be a tool for self-teaching.

That said, this edition differs from the past edition in three major ways: (1) new material, (2) new and revised homework problems, and (3) an appendix introducing Simulink.

1. New Content. I have included new and enhanced sections on a number of topics. The primary additions include material on some MATLAB functions not included in previ- ous editions (e.g., fsolve, integrate, bvp4c), some new applications of Monte Carlo for problems such as integration and optimization, and MATLAB’s new way to pass parameters to function functions.

2. New Homework Problems. Most of the end-of-chapter problems have been modified, and a variety of new problems have been added. In particular, an effort has been made to include several new problems for each chapter that are more challenging and dif- ficult than the problems in the previous edition.

3. I have developed a short primer on Simulink which I have my students read prior to covering that topic. Although I recognize that some professors may not choose to cover Simulink, I included it as a teaching aid for those that do.

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Aside from the new material and problems, the fourth edition is very similar to the third. In particular, I have endeavored to maintain most of the features contributing to its pedagogical effectiveness including extensive use of worked examples and engineering and scien tific applications. As with the previous edition, I have made a concerted effort to make this book as “student-friendly” as possible. Thus, I’ve tried to keep my explanations straight- forward and practical.

Although my primary intent is to empower students by providing them with a sound introduction to numerical problem solving, I have the ancillary objective of making this introduction exciting and pleasurable. I believe that motivated students who enjoy engi- neering and science, problem solving, mathematics—and yes—programming, will ulti- mately make better professionals. If my book fosters enthusiasm and appreciation for these subjects, I will consider the effort a success.

Acknowledgments. Several members of the McGraw-Hill team have contributed to this project. Special thanks are due to Jolynn Kilburg, Thomas Scaife, Ph.D., Chelsea Haupt, Ph.D., and Jeni McAtee for their encouragement, support, and direction.

During the course of this project, the folks at The MathWorks, Inc., have truly dem- onstrated their overall excellence as well as their strong commitment to engineering and science education. In particular, Naomi Fernandes of The MathWorks, Inc., Book Program has been especially helpful and Jared Wasserman of the MathWorks Technical Support Department was of great help with technical questions.

The generosity of the Berger family has provided me with the opportunity to work on creative projects such as this book dealing with computing and engineering. In addition, my colleagues in the School of Engineering at Tufts, notably Masoud Sanayei, Babak Moaveni, Luis Dorfmann, Rob White, Linda Abriola, and Laurie Baise, have been very supportive and helpful.

Significant suggestions were also given by a number of colleagues. In particular, Dave Clough (University of Colorado–Boulder), and Mike Gustafson (Duke University) pro- vided valuable ideas and suggestions. In addition, a number of reviewers provided use- ful feedback and advice including Karen Dow Ambtman (University of Alberta), Jalal Behzadi (Shahid Chamran University), Eric Cochran (Iowa State University), Frederic Gibou (University of California at Santa Barbara), Jane Grande-Allen (Rice University), Raphael Haftka (University of Florida), Scott Hendricks (Virginia Tech University), Ming Huang (University of San Diego), Oleg Igoshin (Rice University), David Jack (Baylor Uni- versity), Se Won Lee (Sungkyunkwan University), Clare McCabe (Vanderbilt University), Eckart Meiburg (University of California at Santa Barbara), Luis Ricardez (University of Waterloo), James Rottman (University of California, San Diego), Bingjing Su (University of Cincinnati), Chin-An Tan (Wayne State University), Joseph Tipton (The University of Evansville), Marion W. Vance (Arizona State University), Jonathan Vande Geest (University of Arizona), Leah J. Walker (Arkansas State University), Qiang Hu (University of Alabama, Huntsville), Yukinobu Tanimoto (Tufts University), Henning T. Søgaard (Aarhus University), and Jimmy Feng (University of British Columbia).

It should be stressed that although I received useful advice from the aforementioned individuals, I am responsible for any inaccuracies or mistakes you may find in this book. Please contact me via e-mail if you should detect any errors.

PREFACE xv

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xvi PREFACE

Finally, I want to thank my family, and in particular my wife, Cynthia, for the love, patience, and support they have provided through the time I’ve spent on this project.

Steven C. Chapra Tufts University

Medford, Massachusetts [email protected]

PEDAGOGICAL TOOLS

Theory Presented as It Informs Key Concepts. The text is intended for Numerical Methods users, not developers. Therefore, theory is not included for “theory’s sake,” for ex- ample no proofs. Theory is included as it informs key concepts such as the Taylor series, con- vergence, condition, etc. Hence, the student is shown how the theory connects with practical issues in problem solving.

Introductory MATLAB Material. The text in cludes two introductory chapters on how to use MATLAB. Chapter 2 shows students how to per form computations and create graphs in MATLAB’s standard command mode. Chapter 3 provides a primer on developing numerical programs via MATLAB M-file functions. Thus, the text provides students with the means to develop their own nu merical algorithms as well as to tap into MATLAB’s powerful built-in routines.

Algorithms Presented Using MATLAB M-files. Instead of using pseudocode, this book presents algorithms as well-structured MATLAB M-files. Aside from being useful com- puter programs, these provide students with models for their own M-files that they will develop as homework exercises.

Worked Examples and Case Studies. Extensive worked examples are laid out in detail so that students can clearly follow the steps in each numerical computation. The case stud- ies consist of engineering and science applications which are more complex and richer than the worked examples. They are placed at the ends of selected chapters with the intention of (1) illustrating the nuances of the methods and (2) showing more realistically how the methods along with MATLAB are applied for problem solving.

Problem Sets. The text includes a wide variety of problems. Many are drawn from en- gineering and scientific disciplines. Others are used to illustrate numerical techniques and theoretical concepts. Problems include those that can be solved with a pocket calculator as well as others that require computer solution with MATLAB.

Useful Appendices and Indexes. Appendix A contains MATLAB commands, Appendix B contains M-file functions, and new Appendix C contains a brief Simulink primer.

Instructor Resources. Solutions Manual, Lecture PowerPoints, Text images in Power- Point, M-files and additional MATLAB resources are available through Connect®.

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Part One

Modeling, Computers, and Error Analysis

1.1 MOTIVATION

What are numerical methods and why should you study them? Numerical methods are techniques by which mathematical problems are formulated

so that they can be solved with arithmetic and logical operations. Because digital comput- ers excel at performing such operations, numerical methods are sometimes referred to as computer mathematics.

In the pre–computer era, the time and drudgery of implementing such calculations seriously limited their practical use. However, with the advent of fast, inexpensive digital computers, the role of numerical methods in engineering and scientific problem solving has exploded. Because they figure so prominently in much of our work, I believe that numerical methods should be a part of every engineer’s and scientist’s basic education. Just as we all must have solid foundations in the other areas of mathematics and science, we should also have a fundamental understanding of numerical methods. In particular, we

should have a solid appreciation of both their capabilities and their limitations.

Beyond contributing to your overall education, there are several additional reasons why you should study numerical methods:

1. Numerical methods greatly expand the types of problems you can address. They are capable of handling large sys- tems of equations, nonlinearities, and complicated geometries that are not uncommon in engineering and science and that are often impossible to solve analytically with standard calculus. As such, they greatly enhance your prob- lem-solving skills.

2. Numerical methods allow you to use “canned” software with insight. During

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2 PART 1 ModEling, CoMPuTERs, And ERRoR AnAlysis

your career, you will invariably have occasion to use commercially available prepack- aged computer programs that involve numerical methods. The intelligent use of these programs is greatly enhanced by an understanding of the basic theory underlying the methods. In the absence of such understanding, you will be left to treat such packages as “black boxes” with little critical insight into their inner workings or the validity of the results they produce.

3. Many problems cannot be approached using canned programs. If you are conversant with numerical methods, and are adept at computer programming, you can design your own programs to solve problems without having to buy or commission expensive software.

4. Numerical methods are an efficient vehicle for learning to use computers. Because nu- merical methods are expressly designed for computer implementation, they are ideal for illustrating the computer’s powers and limitations. When you successfully implement numerical methods on a computer, and then apply them to solve otherwise intractable problems, you will be provided with a dramatic demonstration of how computers can serve your professional development. At the same time, you will also learn to acknowl- edge and control the errors of approximation that are part and parcel of large-scale numerical calculations.

5. Numerical methods provide a vehicle for you to reinforce your understanding of math- ematics. Because one function of numerical methods is to reduce higher mathematics to basic arithmetic operations, they get at the “nuts and bolts” of some otherwise obscure topics. Enhanced understanding and insight can result from this alternative perspective.

With these reasons as motivation, we can now set out to understand how numerical methods and digital computers work in tandem to generate reliable solutions to mathemati- cal problems. The remainder of this book is devoted to this task.

1.2 PART ORGANIZATION

This book is divided into six parts. The latter five parts focus on the major areas of nu- merical methods. Although it might be tempting to jump right into this material, Part One consists of four chapters dealing with essential background material.

Chapter 1 provides a concrete example of how a numerical method can be employed to solve a real problem. To do this, we develop a mathematical model of a free-falling bungee jumper. The model, which is based on Newton’s second law, results in an ordinary differential equation. After first using calculus to develop a closed-form solution, we then show how a comparable solution can be generated with a simple numerical method. We end the chapter with an overview of the major areas of numerical methods that we cover in Parts Two through Six.

Chapters 2 and 3 provide an introduction to the MATLAB® software environment. Chapter 2 deals with the standard way of operating MATLAB by entering commands one at a time in the so-called calculator, or command, mode. This interactive mode provides a straightforward means to orient you to the environment and illustrates how it is used for common operations such as performing calculations and creating plots.

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1.2 PART oRgAniZATion 3

Chapter 3 shows how MATLAB’s programming mode provides a vehicle for assem- bling individual commands into algorithms. Thus, our intent is to illustrate how MATLAB serves as a convenient programming environment to develop your own software.

Chapter 4 deals with the important topic of error analysis, which must be understood for the effective use of numerical methods. The first part of the chapter focuses on the roundoff errors that result because digital computers cannot represent some quantities exactly. The latter part addresses truncation errors that arise from using an approximation in place of an exact mathematical procedure.

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Mathematical Modeling, numerical Methods, and Problem solving

CHAPTER OBJECTIVES

The primary objective of this chapter is to provide you with a concrete idea of what numerical methods are and how they relate to engineering and scientific problem solving. Specific objectives and topics covered are

• Learning how mathematical models can be formulated on the basis of scientific principles to simulate the behavior of a simple physical system.

• Understanding how numerical methods afford a means to generate solutions in a manner that can be implemented on a digital computer.

• Understanding the different types of conservation laws that lie beneath the models used in the various engineering disciplines and appreciating the difference between steady-state and dynamic solutions of these models.

• Learning about the different types of numerical methods we will cover in this book.

YOU’VE GOT A PROBLEM

suppose that a bungee-jumping company hires you. You’re given the task of predicting the velocity of a jumper (Fig. 1.1) as a function of time during the free-fall part of the jump. This information will be used as part of a larger analysis to determine the length and required strength of the bungee cord for jumpers of different mass.

You know from your studies of physics that the acceleration should be equal to the ratio of the force to the mass (Newton’s second law). Based on this insight and your knowledge

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1.1 A siMPlE MAThEMATiCAl ModEl 5

of physics and fluid mechanics, you develop the following mathematical model for the rate of change of velocity with respect to time,

dυ ___ dt = g − cd ___ m υ

where υ = downward vertical velocity (m/s), t = time (s), g = the acceleration due to gravity (≅ 9.81 m/s2), cd = a lumped drag coefficient (kg/m), and m = the jumper’s mass (kg). The drag coefficient is called “lumped” because its magnitude depends on fac- tors such as the jumper’s area and the fluid density (see Sec. 1.4).

Because this is a differential equation, you know that calculus might be used to obtain an analytical or exact solution for υ as a function of t. However, in the following pages, we will illustrate an alternative solution approach. This will involve developing a computer- oriented numerical or approximate solution.

Aside from showing you how the computer can be used to solve this particular prob- lem, our more general objective will be to illustrate (a) what numerical methods are and (b) how they figure in engineering and scientific problem solving. In so doing, we will also show how mathematical models figure prominently in the way engineers and scientists use numerical methods in their work.

1.1 A SIMPLE MATHEMATICAL MODEL

A mathematical model can be broadly defined as a formulation or equation that expresses the essential features of a physical system or process in mathematical terms. In a very gen- eral sense, it can be represented as a functional relationship of the form

Dependent variable = f ( independent variables , parameters, forcing functions ) (1.1)

where the dependent variable is a characteristic that typically reflects the behavior or state of the system; the independent variables are usually dimensions, such as time and space, along which the system’s behavior is being determined; the parameters are reflective of the system’s properties or composition; and the forcing functions are external influences acting upon it.

The actual mathematical expression of Eq. (1.1) can range from a simple algebraic relationship to large complicated sets of differential equations. For example, on the basis of his observations, Newton formulated his second law of motion, which states that the time rate of change of momentum of a body is equal to the resultant force acting on it. The mathematical expression, or model, of the second law is the well-known equation

F = ma (1.2)

where F is the net force acting on the body (N, or kg m/s2), m is the mass of the object (kg), and a is its acceleration (m/s2).

Upward force due to air resistance

Downward force due to gravity

FIGURE 1.1 Forces acting on a free-falling bungee jumper.

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6 MAThEMATiCAl ModEling, nuMERiCAl METhods, And PRoblEM solving

The second law can be recast in the format of Eq. (1.1) by merely dividing both sides by m to give

a = F __ m (1.3)

where a is the dependent variable reflecting the system’s behavior, F is the forcing func- tion, and m is a parameter. Note that for this simple case there is no independent variable because we are not yet predicting how acceleration varies in time or space.

Equation (1.3) has a number of characteristics that are typical of mathematical models of the physical world.

• It describes a natural process or system in mathematical terms. • It represents an idealization and simplification of reality. That is, the model ignores

negligible details of the natural process and focuses on its essential manifestations. Thus, the second law does not include the effects of relativity that are of minimal importance when applied to objects and forces that interact on or about the earth’s surface at velocities and on scales visible to humans.

• Finally, it yields reproducible results and, consequently, can be used for predictive purposes. For example, if the force on an object and its mass are known, Eq. (1.3) can be used to compute acceleration.

Because of its simple algebraic form, the solution of Eq. (1.2) was obtained easily. However, other mathematical models of physical phenomena may be much more complex, and either cannot be solved exactly or require more sophisticated mathematical techniques than simple algebra for their solution. To illustrate a more complex model of this kind, Newton’s second law can be used to determine the terminal velocity of a free-falling body near the earth’s surface. Our falling body will be a bungee jumper (Fig. 1.1). For this case, a model can be derived by expressing the acceleration as the time rate of change of the velocity (dυ/dt) and substituting it into Eq. (1.3) to yield

dυ ___ dt = F __ m (1.4)

where υ is velocity (in meters per second). Thus, the rate of change of the velocity is equal to the net force acting on the body normalized to its mass. If the net force is positive, the object will accelerate. If it is negative, the object will decelerate. If the net force is zero, the object’s velocity will remain at a constant level.

Next, we will express the net force in terms of measurable variables and parameters. For a body falling within the vicinity of the earth, the net force is composed of two opposing forces: the downward pull of gravity FD and the upward force of air resistance FU (Fig. 1.1):

F = FD + FU (1.5)

If force in the downward direction is assigned a positive sign, the second law can be used to formulate the force due to gravity as

FD = mg (1.6)

where g is the acceleration due to gravity (9.81 m/s2).

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1.1 A siMPlE MAThEMATiCAl ModEl 7

Air resistance can be formulated in a variety of ways. Knowledge from the science of fluid mechanics suggests that a good first approximation would be to assume that it is proportional to the square of the velocity,

FU = −cd υ2 (1.7)

where cd is a proportionality constant called the lumped drag coefficient (kg/m). Thus, the greater the fall velocity, the greater the upward force due to air resistance. The parameter cd accounts for properties of the falling object, such as shape or surface roughness, that af- fect air resistance. For the present case, cd might be a function of the type of clothing or the orientation used by the jumper during free fall.

The net force is the difference between the downward and upward force. Therefore, Eqs. (1.4) through (1.7) can be combined to yield

dυ ___ dt = g − cd ___ m υ

2 (1.8)

Equation (1.8) is a model that relates the acceleration of a falling object to the forces acting on it. It is a differential equation because it is written in terms of the differential rate of change (dυ/dt) of the variable that we are interested in predicting. However, in contrast to the solution of Newton’s second law in Eq. (1.3), the exact solution of Eq. (1.8) for the velocity of the jumper cannot be obtained using simple algebraic manipulation. Rather, more advanced techniques such as those of calculus must be applied to obtain an exact or analytical solution. For example, if the jumper is initially at rest (υ = 0 at t = 0), calculus can be used to solve Eq. (1.8) for

υ(t) = √ ___

gm ___ cd tanh ( √ ___

gcd ___ m t ) (1.9)

where tanh is the hyperbolic tangent that can be either computed directly1 or via the more elementary exponential function as in

tanh x = e x − e−x _______ ex + e−x (1.10)

Note that Eq. (1.9) is cast in the general form of Eq. (1.1) where υ(t) is the dependent variable, t is the independent variable, cd and m are parameters, and g is the forcing function.

EXAMPlE 1.1 Analytical solution to the bungee Jumper Problem

Problem statement. A bungee jumper with a mass of 68.1 kg leaps from a stationary hot air balloon. Use Eq. (1.9) to compute velocity for the first 12 s of free fall. Also deter- mine the terminal velocity that will be attained for an infinitely long cord (or alternatively, the jumpmaster is having a particularly bad day!). Use a drag coefficient of 0.25 kg/m.

1 MATLAB allows direct calculation of the hyperbolic tangent via the built-in function tanh(x).

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8 MAThEMATiCAl ModEling, nuMERiCAl METhods, And PRoblEM solving

solution. Inserting the parameters into Eq. (1.9) yields

υ(t) = √ _________

9.81(68.1) _________ 0.25 tanh ( √ _________

9.81(0.25) _________ 68.1 t ) = 51.6938 tanh(0.18977t) which can be used to compute

t, s υ, m/s

0 0 2 18.7292 4 33.1118 6 42.0762 8 46.9575 10 49.4214 12 50.6175 ∞ 51.6938

According to the model, the jumper accelerates rapidly (Fig. 1.2). A velocity of 49.4214 m/s (about 110 mi/hr) is attained after 10 s. Note also that after a sufficiently

0 4 8 12 t, s

Terminal velocity

υ, m

FIGURE 1.2 The analytical solution for the bungee jumper problem as computed in Example 1.1. Velocity increases with time and asymptotically approaches a terminal velocity.

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1.1 A siMPlE MAThEMATiCAl ModEl 9

long time, a constant velocity, called the terminal velocity, of 51.6983 m/s (115.6 mi/hr) is reached. This velocity is constant because, eventually, the force of gravity will be in balance with the air resistance. Thus, the net force is zero and acceleration has ceased.

Equation (1.9) is called an analytical or closed-form solution because it exactly satis- fies the original differential equation. Unfortunately, there are many mathematical models that cannot be solved exactly. In many of these cases, the only alternative is to develop a numerical solution that approximates the exact solution.

Numerical methods are those in which the mathematical problem is reformulated so it can be solved by arithmetic operations. This can be illustrated for Eq. (1.8) by realizing that the time rate of change of velocity can be approximated by (Fig. 1.3):

dυ ___ dt ≅ Δυ ___ Δt =

υ(ti+1) − υ(ti ) ___________ ti+1 − ti (1.11)

where Δυ and Δt are differences in velocity and time computed over finite intervals, υ(ti) is velocity at an initial time ti, and υ(ti+1) is velocity at some later time ti+1. Note that dυ/dt ≅ Δυ/Δt is approximate because Δt is finite. Remember from calculus that

dυ ___ dt = lim Δt→0

Δυ ___ Δt

Equation (1.11) represents the reverse process.

υ(ti+1)

ti+1 t

∆t

υ(ti)

∆υ

True slope dυ/dt

Approximate slope ∆υ υ(ti+1) − υ(ti)

ti+1 − ti∆t =

FIGURE 1.3 The use of a finite difference to approximate the first derivative of υ with respect to t.

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10 MAThEMATiCAl ModEling, nuMERiCAl METhods, And PRoblEM solving

Equation (1.11) is called a finite-difference approximation of the derivative at time ti. It can be substituted into Eq. (1.8) to give

υ(ti+1) − υ(ti ) ___________ ti+1 − ti

= g − cd ___ m υ(ti )

This equation can then be rearranged to yield

υ(ti+1) = υ(ti ) + [ g − cd ___ m υ(ti )2 ] (ti+1 − ti ) (1.12) Notice that the term in brackets is the right-hand side of the differential equation itself

[Eq. (1.8)]. That is, it provides a means to compute the rate of change or slope of υ. Thus, the equation can be rewritten more concisely as

υi+1 = υi + dυi ___ dt Δt (1.13)

where the nomenclature υi designates velocity at time ti, and Δt = ti+1 − ti. We can now see that the differential equation has been transformed into an equation that

can be used to determine the velocity algebraically at ti+1 using the slope and previous values of υ and t. If you are given an initial value for velocity at some time ti, you can easily compute velocity at a later time ti+1. This new value of velocity at ti+1 can in turn be employed to extend the computation to velocity at ti+2 and so on. Thus at any time along the way,

New value = old value + slope × step size

This approach is formally called Euler’s method. We’ll discuss it in more detail when we turn to differential equations later in this book.

EXAMPlE 1.2 numerical solution to the bungee Jumper Problem

Problem statement. Perform the same computation as in Example 1.1 but use Eq. (1.12) to compute velocity with Euler’s method. Employ a step size of 2 s for the calculation.

solution. At the start of the computation (t0 = 0), the velocity of the jumper is zero. Using this information and the parameter values from Example 1.1, Eq. (1.12) can be used to compute velocity at t1 = 2 s:

υ = 0 + [ 9.81 − 0.25 ____ 68.1 (0)2 ] × 2 = 19.62 m/s For the next interval (from t = 2 to 4 s), the computation is repeated, with the result

υ = 19.62 + [ 9.81 − 0.25 ____ 68.1 (19.62)2 ] × 2 = 36.4137 m/s

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1.1 A siMPlE MAThEMATiCAl ModEl 11

The calculation is continued in a similar fashion to obtain additional values:

t, s υ, m/s

0 0 2 19.6200 4 36.4137 6 46.2983 8 50.1802 10 51.3123 12 51.6008 ∞ 51.6938

The results are plotted in Fig. 1.4 along with the exact solution. We can see that the numeri- cal method captures the essential features of the exact solution. However, because we have employed straight-line segments to approximate a continuously curving function, there is some discrepancy between the two results. One way to minimize such discrepancies is to use a smaller step size. For example, applying Eq. (1.12) at 1-s intervals results in a smaller error, as the straight-line segments track closer to the true solution. Using hand calcula- tions, the effort associated with using smaller and smaller step sizes would make such numerical solutions impractical. However, with the aid of the computer, large numbers of calculations can be performed easily. Thus, you can accurately model the velocity of the jumper without having to solve the differential equation exactly.

0 4 8 12 t, s

Terminal velocity

υ, m

Approximate, numerical solution

Exact, analytical solution

FIGURE 1.4 Comparison of the numerical and analytical solutions for the bungee jumper problem.

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12 MAThEMATiCAl ModEling, nuMERiCAl METhods, And PRoblEM solving

As in Example 1.2, a computational price must be paid for a more accurate numeri- cal result. Each halving of the step size to attain more accuracy leads to a doubling of the number of computations. Thus, we see that there is a trade-off between accuracy and com- putational effort. Such trade-offs figure prominently in numerical methods and constitute an important theme of this book.

1.2 CONSERVATION LAWS IN ENGINEERING AND SCIENCE

Aside from Newton’s second law, there are other major organizing principles in science and engineering. Among the most important of these are the conservation laws. Although they form the basis for a variety of complicated and powerful mathematical models, the great conservation laws of science and engineering are conceptually easy to understand. They all boil down to

Change = increases − decreases (1.14)

This is precisely the format that we employed when using Newton’s law to develop a force balance for the bungee jumper [Eq. (1.8)].

Although simple, Eq. (1.14) embodies one of the most fundamental ways in which conservation laws are used in engineering and science—that is, to predict changes with respect to time. We will give it a special name—the time-variable (or transient) computation.

Aside from predicting changes, another way in which conservation laws are applied is for cases where change is nonexistent. If change is zero, Eq. (1.14) becomes

Change = 0 = increases − decreases or

Increases = decreases (1.15) Thus, if no change occurs, the increases and decreases must be in balance. This case, which is also given a special name—the steady-state calculation—has many applications in engi- neering and science. For example, for steady-state incompressible fluid flow in pipes, the flow into a junction must be balanced by flow going out, as in

Flow in = flow out For the junction in Fig. 1.5, the balance that can be used to compute that the flow out of the fourth pipe must be 60.

For the bungee jumper, the steady-state condition would correspond to the case where the net force was zero or [Eq. (1.8) with dυ/dt = 0]

mg = cd υ2 (1.16) Thus, at steady state, the downward and upward forces are in balance and Eq. (1.16) can be solved for the terminal velocity

υ = √ ___

gm ___ cd

Although Eqs. (1.14) and (1.15) might appear trivially simple, they embody the two funda- mental ways that conservation laws are employed in engineering and science. As such, they will form an important part of our efforts in subsequent chapters to illustrate the connection between numerical methods and engineering and science.

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1.3 nuMERiCAl METhods CovEREd in This book 13

Table 1.1 summarizes some models and associated conservation laws that figure prominently in engineering. Many chemical engineering problems involve mass balances for reactors. The mass balance is derived from the conservation of mass. It specifies that the change of mass of a chemical in the reactor depends on the amount of mass flowing in minus the mass flowing out.

Civil and mechanical engineers often focus on models developed from the conserva- tion of momentum. For civil engineering, force balances are utilized to analyze structures such as the simple truss in Table 1.1. The same principles are employed for the mechanical engineering case studies to analyze the transient up-and-down motion or vibrations of an automobile.

Finally, electrical engineering studies employ both current and energy balances to model electric circuits. The current balance, which results from the conservation of charge, is simi- lar in spirit to the flow balance depicted in Fig. 1.5. Just as flow must balance at the junction of pipes, electric current must balance at the junction of electric wires. The energy balance specifies that the changes of voltage around any loop of the circuit must add up to zero.

It should be noted that there are many other branches of engineering beyond chemical, civil, electrical, and mechanical. Many of these are related to the Big Four. For example, chem- ical engineering skills are used extensively in areas such as environmental, petroleum, and bio- medical engineering. Similarly, aerospace engineering has much in common with mechanical engineering. I will endeavor to include examples from these areas in the coming pages.

1.3 NUMERICAL METHODS COVERED IN THIS BOOK

Euler’s method was chosen for this introductory chapter because it is typical of many other classes of numerical methods. In essence, most consist of recasting mathematical operations into the simple kind of algebraic and logical operations compatible with digital computers. Figure 1.6 summarizes the major areas covered in this text.

Pipe 2 Flow in = 80

Pipe 3 Flow out = 120

Pipe 4 Flow out = ?

Pipe 1 Flow in = 100

FIGURE 1.5 A flow balance for steady incompressible fluid flow at the junction of pipes.

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14 MAThEMATiCAl ModEling, nuMERiCAl METhods, And PRoblEM solving

TABLE 1.1 devices and types of balances that are commonly used in the four major areas of engineering. For each case, the conservation law on which the balance is based is specified.

Field Organizing Principle Mathematical ExpressionDevice

Force balance:Mechanical engineering

Conservation of momentum

Upward force

Downward force

x = 0

= downward force − upward forcem d 2x

dt2

Machine

Current balance:

Voltage balance:

Around each loop Σ emf’s − Σ voltage drops for resistors = 0 Σ ξ − Σ iR = 0

For each node Σ current (i) = 0

Electrical engineering

Conservation of energy

Conservation of charge

+i2

−i3+i1

i1R1

i3R3

i2R2 ξ

−

Circuit

Chemical engineering

Conservation of mass

Over a unit of time period ∆mass = inputs − outputs

Mass balance:

Reactors

Force balance:

At each node Σ horizontal forces (FH) = 0 Σ vertical forces (FV) = 0

Civil engineering

Conservation of momentum

Structure

+FV

−FV

+FH−FH

Input Output

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1.3 nuMERiCAl METhods CovEREd in This book 15

∆t

Slope = f(ti, yi)

y(e) Part 6 : Di�erential equations

Given

solve for y as a function of t

dy dt

∆y =∆t≈ f (t, y)

(d) Part 5 : Integration and di�erentiation

Integration: Find the area under the curve

Di�erentiation: Find the slope of the curve

Regression

Interpolation

(a) Part 2 : Roots and optimization

Roots: Solve for x so that f (x) = 0

Optimization: Solve for x so that f ' (x) = 0

(b) Part 3 : Linear algebraic equations

Given the a’s and the b’s, solve for the x’s a11x1 + a12x2 = b1 a21x1 + a22x2 = b2

Solution

Roots

Optima

f (x)

f (x)f (x)

y dy/dx

yi+1 = yi + f (ti, yi)∆t

FIGURE 1.6 Summary of the numerical methods covered in this book.

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16 MAThEMATiCAl ModEling, nuMERiCAl METhods, And PRoblEM solving

Part Two deals with two related topics: root finding and optimization. As depicted in Fig. 1.6a, root location involves searching for the zeros of a function. In contrast, optimiza- tion involves determining a value or values of an independent variable that correspond to a “best” or optimal value of a function. Thus, as in Fig. 1.6a, optimization involves identify- ing maxima and minima. Although somewhat different approaches are used, root location and optimization both typically arise in design contexts.

Part Three is devoted to solving systems of simultaneous linear algebraic equations (Fig. 1.6b). Such systems are similar in spirit to roots of equations in the sense that they are concerned with values that satisfy equations. However, in contrast to satisfying a single equation, a set of values is sought that simultaneously satisfies a set of linear algebraic equations. Such equations arise in a variety of problem contexts and in all disciplines of engineering and science. In particular, they originate in the mathematical modeling of large systems of interconnected elements such as structures, electric circuits, and fluid networks. However, they are also encountered in other areas of numerical methods such as curve fitting and differential equations.

As an engineer or scientist, you will often have occasion to fit curves to data points. The techniques developed for this purpose can be divided into two general categories: re- gression and interpolation. As described in Part Four (Fig. 1.6c), regression is employed where there is a significant degree of error associated with the data. Experimental results are often of this kind. For these situations, the strategy is to derive a single curve that represents the general trend of the data without necessarily matching any individual points.

In contrast, interpolation is used where the objective is to determine intermediate val- ues between relatively error-free data points. Such is usually the case for tabulated informa- tion. The strategy in such cases is to fit a curve directly through the data points and use the curve to predict the intermediate values.

As depicted in Fig. 1.6d, Part Five is devoted to integration and differentiation. A physical interpretation of numerical integration is the determination of the area under a curve. Integration has many applications in engineering and science, ranging from the determination of the centroids of oddly shaped objects to the calculation of total quantities based on sets of discrete measurements. In addition, numerical integration formulas play an important role in the solution of differential equations. Part Five also covers methods for numerical differentiation. As you know from your study of calculus, this involves the determination of a function’s slope or its rate of change.

Finally, Part Six focuses on the solution of ordinary differential equations (Fig. 1.6e). Such equations are of great significance in all areas of engineering and science. This is because many physical laws are couched in terms of the rate of change of a quantity rather than the magnitude of the quantity itself. Examples range from population-forecasting models (rate of change of population) to the acceleration of a falling body (rate of change of velocity). Two types of problems are addressed: initial-value and boundary-value problems.

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1.4 CASE STUDY iT’s A REAl dRAg

Background. In our model of the free-falling bungee jumper, we assumed that drag depends on the square of velocity (Eq. 1.7). A more detailed representation, which was originally formulated by Lord Rayleigh, can be written as

Fd = − 1 __ 2 ρυ

2ACd _ › υ (1.17)

where Fd = the drag force (N), ρ = fluid density (kg/m3), A = the frontal area of the object on a plane perpendicular to the direction of motion (m2), Cd = a dimensionless drag coef- ficient, and

_ › υ = a unit vector indicating the direction of velocity. This relationship, which assumes turbulent conditions (i.e., a high Reynolds number),

allows us to express the lumped drag coefficient from Eq. (1.7) in more fundamental terms as

cd = 1 __ 2 ρACd (1.18)

Thus, the lumped drag coefficient depends on the object’s area, the fluid’s density, and a dimensionless drag coefficient. The latter accounts for all the other factors that contribute to air resistance such as the object’s “roughness.” For example, a jumper wearing a baggy outfit will have a higher Cd than one wearing a sleek jumpsuit.

Note that for cases where velocity is very low, the flow regime around the object will be laminar and the relationship between the drag force and velocity becomes linear. This is referred to as Stokes drag.

In developing our bungee jumper model, we assumed that the downward direction was positive. Thus, Eq. (1.7) is an accurate representation of Eq. (1.17), because

_ › υ = +1 and the drag force is negative. Hence, drag reduces velocity.

But what happens if the jumper has an upward (i.e., negative) velocity? In this case, _ › υ = −1 and Eq. (1.17) yields a positive drag force. Again, this is physically correct as the positive drag force acts downward against the upward negative velocity.

Unfortunately, for this case, Eq. (1.7) yields a negative drag force because it does not include the unit directional vector. In other words, by squaring the velocity, its sign and hence its direction is lost. Consequently, the model yields the physically unrealistic result that air resistance acts to accelerate an upward velocity!

In this case study, we will modify our model so that it works properly for both downward and upward velocities. We will test the modified model for the same case as Example 1.2, but with an initial value of υ(0) = −40 m/s. In addition, we will also illustrate how we can extend the numerical analysis to determine the jumper’s position.

Solution. The following simple modification allows the sign to be incorporated into the drag force:

Fd = − 1 __ 2 ρυ∣υ∣ ACd (1.19)

1.4 CAsE sTudy 17

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18 MAThEMATiCAl ModEling, nuMERiCAl METhods, And PRoblEM solving

1.4 CASE STUDY continued

or in terms of the lumped drag:

Fd = −cd υ∣υ∣ (1.20)

Thus, the differential equation to be solved is

dυ ___ dt = g − cd __ m υ∣υ∣ (1.21)

In order to determine the jumper’s position, we recognize that distance traveled, x (m), is related to velocity by

dx ___ dt = −υ (1.22)

In contrast to velocity, this formulation assumes that upward distance is positive. In the same fashion as Eq. (1.12), this equation can be integrated numerically with Euler’s method:

xi +1 = xi − υ(ti )Δt (1.23)

Assuming that the jumper’s initial position is defined as x(0) = 0, and using the parameter values from Examples 1.1 and 1.2, the velocity and distance at t = 2 s can be computed as

υ(2) = −40 + [ 9.81 − 0.25 ____ 68.1 (−40)(40) ] 2 = −8.6326 m/s x(2) = 0 − (−40)2 = 80 m

Note that if we had used the incorrect drag formulation, the results would be −32.1274 m/s and 80 m.

The computation can be repeated for the next interval (t = 2 to 4 s):

υ(4) = −8.6326 + [ 9.81 − 0.25 ____ 68.1 (−8.6326)(8.6326) ] 2 = 11.5346 m/s x(4) = 80 − (−8.6326)2 = 97.2651 m

The incorrect drag formulation gives –20.0858 m/s and 144.2549 m. The calculation is continued and the results shown in Fig. 1.7 along with those obtained

with the incorrect drag model. Notice that the correct formulation decelerates more rapidly because drag always diminishes the velocity.

With time, both velocity solutions converge on the same terminal velocity because eventually both are directed downward in which case, Eq. (1.7) is correct. However, the impact on the height prediction is quite dramatic with the incorrect drag case resulting in a much higher trajectory.

This case study demonstrates how important it is to have the correct physical model. In some cases, the solution will yield results that are clearly unrealistic. The current example is more insidious as there is no visual evidence that the incorrect solution is wrong. That is, the incorrect solution “looks” reasonable.

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1.4 CASE STUDY continued

1.4 CAsE sTudy 19

FIGURE 1.7 Plots of (a) velocity and (b) height for the free-falling bungee jumper with an upward (negative) initial velocity generated with Euler’s method. Results for both the correct (Eq. 1.20) and incorrect (Eq. 1.7) drag formulations are displayed.

υ, m

−40

−20

4 8 12 t, s

Correct drag

Incorrect drag

(a) Velocity, m/s

(b) Height, m

100

200

−200

−100

4 8 12 t, s

x, m

Incorrect drag

Correct drag

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20 MAThEMATiCAl ModEling, nuMERiCAl METhods, And PRoblEM solving

PROBLEMS

1.1 Use calculus to verify that Eq. (1.9) is a solution of Eq. (1.8) for the initial condition υ(0) = 0. 1.2 Use calculus to solve Eq. (1.21) for the case where the ini- tial velocity is (a) positive and (b) negative. (c) Based on your results for (a) and (b), perform the same computation as in Example 1.1 but with an initial velocity of −40 m/s. Compute values of the velocity from t = 0 to 12 s at intervals of 2 s. Note that for this case, the zero velocity occurs at t = 3.470239 s. 1.3 The following information is available for a bank account:

Date Deposits Withdrawals Balance

5/1 1512.33 220.13 327.26 6/1 216.80 378.61 7/1 450.25 106.80 8/1 127.31 350.61 9/1

Note that the money earns interest which is computed as

Interest = iBi

where i = the interest rate expressed as a fraction per month, and Bi the initial balance at the beginning of the month. (a) Use the conservation of cash to compute the balance on

6∕1, 7∕1, 8∕1, and 9∕1 if the interest rate is 1% per month (i = 0.01∕month). Show each step in the computation.

(b) Write a differential equation for the cash balance in the form

dB ___ dt = f [D(t), W(t), i]

where t = time (months), D(t) = deposits as a function of time ($/month), W(t) = withdrawals as a function of time ($/month). For this case, assume that interest is compounded continuously; that is, interest = iB.

(c) Use Euler’s method with a time step of 0.5 month to simulate the balance. Assume that the deposits and with- drawals are applied uniformly over the month.

(d) Develop a plot of balance versus time for (a) and (c). 1.4 Repeat Example 1.2. Compute the velocity to t = 12 s, with a step size of (a) 1 and (b) 0.5 s. Can you make any statement regarding the errors of the calculation based on the results?

1.5 Rather than the nonlinear relationship of Eq. (1.7), you might choose to model the upward force on the bungee jumper as a linear relationship:

FU = −c′υ

where c′ = a first-order drag coefficient (kg/s). (a) Using calculus, obtain the closed-form solution for the

case where the jumper is initially at rest (υ = 0 at t = 0). (b) Repeat the numerical calculation in Example 1.2 with

the same initial condition and parameter values. Use a value of 11.5 kg/s for c′.

1.6 For the free-falling bungee jumper with linear drag (Prob. 1.5), assume a first jumper is 70 kg and has a drag co- efficient of 12 kg/s. If a second jumper has a drag coefficient of 15 kg/s and a mass of 80 kg, how long will it take her to reach the same velocity jumper 1 reached in 9 s? 1.7 For the second-order drag model (Eq. 1.8), compute the velocity of a free-falling parachutist using Euler’s method for the case where m = 80 kg and cd = 0.25 kg/m. Perform the calculation from t = 0 to 20 s with a step size of 1 s. Use an initial condition that the parachutist has an upward veloc- ity of 20 m/s at t = 0. At t = 10 s, assume that the chute is instantaneously deployed so that the drag coefficient jumps to 1.5 kg/m. 1.8 The amount of a uniformly distributed radioactive con- taminant contained in a closed reactor is measured by its concentration c (becquerel/liter or Bq/L). The contaminant decreases at a decay rate proportional to its concentration; that is

Decay rate = −kc

where k is a constant with units of day−1. Therefore, ac- cording to Eq. (1.14), a mass balance for the reactor can be written as

dc ___ dt = −kc

( change in mass ) = ( decrease by decay ) (a) Use Euler’s method to solve this equation from t = 0 to

1 d with k = 0.175 d–1. Employ a step size of Δt = 0.1 d. The concentration at t = 0 is 100 Bq/L.

(b) Plot the solution on a semilog graph (i.e., ln c versus t) and determine the slope. Interpret your results.

1.9 A storage tank (Fig. P1.9) contains a liquid at depth y where y = 0 when the tank is half full. Liquid is withdrawn at a constant flow rate Q to meet demands. The contents are

cha97962_ch01_001-026.indd 20 08/11/16 12:51 pm

resupplied at a sinusoidal rate 3Q sin2(t). Equation (1.14) can be written for this system as

d(Ay) _____ dt = 3 Q sin 2(t) − Q

( change in volume ) = (inflow) − (outflow) or, since the surface area A is constant

dy ___ dt = 3 Q __ A sin

2(t) − Q __ A

Use Euler’s method to solve for the depth y from t = 0 to 10 d with a step size of 0.5 d. The parameter values are A = 1250 m2 and Q = 450 m3/d. Assume that the initial condition is y = 0. 1.10 For the same storage tank described in Prob. 1.9, sup- pose that the outflow is not constant but rather depends on the depth. For this case, the differential equation for depth can be written as

dy ___ dt = 3 Q __ A sin

2(t) − α(1 + y) 1.5 _________ A

Use Euler’s method to solve for the depth y from t = 0 to 10 d with a step size of 0.5 d. The parameter values are A = 1250 m2, Q = 450 m3/d, and α = 150. Assume that the initial condition is y = 0. 1.11 Apply the conservation of volume (see Prob. 1.9) to sim- ulate the level of liquid in a conical storage tank (Fig. P1.11).

FIGURE P1.9

ytop

yout

rtop Qin

Qout

FIGURE P1.11

PRoblEMs 21

The liquid flows in at a sinusoidal rate of Qin = 3 sin2(t) and flows out according to

Qout = 3(y − yout)1.5 y > yout Qout = 0 y ≤ yout

where flow has units of m3/d and y = the elevation of the water surface above the bottom of the tank (m). Use Euler’s method to solve for the depth y from t = 0 to 10 d with a step size of 0.5 d. The parameter values are rtop = 2.5 m, ytop = 4 m, and yout = 1 m. Assume that the level is initially below the outlet pipe with y(0) = 0.8 m. 1.12 A group of 35 students attend a class in an insu- lated room which measures 11 by 8 by 3 m. Each student takes up about 0.075 m3 and gives out about 80 W of heat (1 W = 1 J/s). Calculate the air temperature rise during the first 20 minutes of the class if the room is completely sealed and insulated. Assume the heat capacity Cυ for air is 0.718 kJ/(kg K). Assume air is an ideal gas at 20 °C and 101.325 kPa. Note that the heat absorbed by the air Q is related to the mass of the air m the heat capacity, and the change in temperature by the following relationship:

Q = m ∫ T1

T2 Cυ dT = mCυ (T2 − T1)

The mass of air can be obtained from the ideal gas law:

PV = m ____ Mwt RT

where P is the gas pressure, V is the volume of the gas, Mwt is the molecular weight of the gas (for air, 28.97 kg/kmol), and R is the ideal gas constant [8.314 kPa m3/(kmol K)].

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22 MAThEMATiCAl ModEling, nuMERiCAl METhods, And PRoblEM solving

1.13 Figure P1.13 depicts the various ways in which an average man gains and loses water in one day. One liter is ingested as food, and the body metabolically produces 0.3 liters. In breathing air, the exchange is 0.05 liters while inhaling, and 0.4 liters while exhaling over a one-day period. The body will also lose 0.3, 1.4, 0.2, and 0.35 liters through sweat, urine, feces, and through the skin, respectively. To maintain steady state, how much water must be drunk per day? 1.14 In our example of the free-falling bungee jumper, we assumed that the acceleration due to gravity was a constant value of 9.81 m/s2. Although this is a decent approxima- tion when we are examining falling objects near the surface of the earth, the gravitational force decreases as we move above sea level. A more general representation based on Newton’s inverse square law of gravitational attraction can be written as

g(x) = g(0) R 2 _______

(R + x)2

where g(x) = gravitational acceleration at altitude x (in m) measured upward from the earth’s surface (m/s2), g(0) = gravitational acceleration at the earth’s surface (≅ 9.81 m/s2), and R = the earth’s radius (≅ 6.37 × 106 m). (a) In a fashion similar to the derivation of Eq. (1.8), use a

force balance to derive a differential equation for veloc- ity as a function of time that utilizes this more complete representation of gravitation. However, for this deriva- tion, assume that upward velocity is positive.

(b) For the case where drag is negligible, use the chain rule to express the differential equation as a function of alti- tude rather than time. Recall that the chain rule is

dυ ___ dt = dυ ___ dx

dx ___ dt

(d) Use Euler’s method to obtain a numerical solution from x = 0 to 100,000 m using a step of 10,000 m where the initial velocity is 1500 m/s upward. Compare your result with the analytical solution.

1.15 Suppose that a spherical droplet of liquid evaporates at a rate that is proportional to its surface area.

dV ___ dt = −kA

where V = volume (mm3), t = time (min), k = the evapo- ration rate (mm/min), and A = surface area (mm2). Use Euler’s method to compute the volume of the droplet from t = 0 to 10 min using a step size of 0.25 min. Assume that k = 0.08 mm/min and that the droplet initially has a radius of 2.5 mm. Assess the validity of your results by determining the radius of your final computed volume and verifying that it is consistent with the evaporation rate. 1.16 A fluid is pumped into the network shown in Fig. P1.16. If Q2 = 0.7, Q3 = 0.5, Q7 = 0.1, and Q8 = 0.3 m3/s, determine the other flows. 1.17 Newton’s law of cooling says that the temperature of a body changes at a rate proportional to the difference be- tween its temperature and that of the surrounding medium (the ambient temperature),

dT ___ dt = −k(T − Ta)

where T = the temperature of the body (°C), t = time (min), k = the proportionality constant (per minute), and Ta = the ambient temperature (°C). Suppose that a cup of coffee originally has a temperature of 70 °C. Use Euler’s method to compute the temperature from t = 0 to 20 min using a step size of 2 min if Ta = 20 °C and k = 0.019/min.

Urine Feces

Skin

Food

Drink

Air

Sweat

Metabolism

BODY

FIGURE P1.13

FIGURE P1.16

Q10

Q6 Q7

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1.18 You are working as a crime scene investigator and must predict the temperature of a homicide victim over a 5-hr period. You know that the room where the victim was found was at 10 °C when the body was discovered. (a) Use Newton’s law of cooling (Prob. 1.17) and Euler’s

method to compute the victim’s body temperature for the 5-hr period using values of k = 0.12/hr and Δt = 0.5 hr. Assume that the victim’s body temperature at the time of death was 37 °C, and that the room temperature was at a constant value of 10 °C over the 5-hr period.

(b) Further investigation reveals that the room temperature had actually dropped linearly from 20 to 10 °C over the 5-hr period. Repeat the same calculation as in (a) but incorporate this new information.

1.19 The velocity is equal to the rate of change of distance, x (m):

dx ___ dt = υ(t) (P1.19)

Use Euler’s method to numerically integrate Eqs. (P1.19) and (1.8) in order to determine both the velocity and distance fallen as a function of time for the first 10 seconds of freefall using the same parameters and conditions as in Example 1.2. Develop a plot of your results. 1.20 In addition to the downward force of gravity (weight) and drag, an object falling through a fluid is also subject to a buoyancy force which is proportional to the displaced volume (Archimedes’ principle). For example, for a sphere with diameter d (m), the sphere’s volume is V = π d3/6, and its projected area is A = πd2/4. The buoyancy force can then be computed as Fb = −ρVg. We neglected buoyancy in our derivation of Eq. (1.8) because it is relatively small for an object like a bungee jumper moving through air. However, for a more dense fluid like water, it becomes more prominent. (a) Derive a differential equation in the same fashion as

Eq. (1.8), but include the buoyancy force and represent the drag force as described in Sec. 1.4.

(b) Rewrite the differential equation from (a) for the special case of a sphere.

(c) Use the equation developed in (b) to compute the terminal velocity (i.e., for the steady-state case). Use the follow- ing parameter values for a sphere falling through water: sphere diameter = 1 cm, sphere density = 2700 kg/m3, water density = 1000 kg/m3, and Cd = 0.47.

(d) Use Euler’s method with a step size of Δt = 0.03125 s to numerically solve for the velocity from t = 0 to 0.25 s with an initial velocity of zero.

1.21 As noted in Sec. 1.4, a fundamental representation of the drag force, which assumes turbulent conditions (i.e., a high Reynolds number), can be formulated as

Fd = − 1 __ 2 ρACd υ∣υ∣

where Fd = the drag force (N), ρ = fluid density (kg/m3), A = the frontal area of the object on a plane perpendicular to the direction of motion (m2), υ = velocity (m/s), and Cd = a dimen- sionless drag coefficient. (a) Write the pair of differential equations for velocity and

position (see Prob. 1.19) to describe the vertical motion of a sphere with diameter, d (m), and a density of ρs (kg/m3). The differential equation for velocity should be written as a function of the sphere’s diameter.

(b) Use Euler’s method with a step size of Δt = 2 s to com- pute the position and velocity of a sphere over the first 14 seconds. Employ the following parameters in your calculation: d = 120 cm, ρ = 1.3 kg/m3, ρs = 2700 kg/m3, and Cd = 0.47. Assume that the sphere has the initial conditions: x(0) = 100 m and υ(0) = −40 m/s.

(c) Develop a plot of your results (i.e., y and υ versus t) and use it to graphically estimate when the sphere would hit the ground.

(d) Compute the value for the bulk second-order drag coef- ficient, cd′ (kg/m). Note that the bulk second-order drag coefficient is the term in the final differential equation for velocity that multiplies the term υ |υ|.

1.22 As depicted in Fig. P1.22, a spherical particle set- tling through a quiescent fluid is subject to three forces: the downward force of gravity (FG), and the upward forces of buoyancy (FB) and drag (FD). Both the gravity and buoyancy forces can be computed with Newton’s second law with the latter equal to the weight of the displaced fluid. For laminar flow, the drag force can be computed with Stoke’s law,

FD = 3πμdυ

where μ = the dynamic viscosity of the fluid (N s/m2), d = the particle diameter (m), and υ = the particle’s settling velocity (m/s). The mass of the particle can be expressed as the product of the particle’s volume and density, ρs (kg/m3), and the mass of the displaced fluid can be computed as the product of the particle’s volume and the fluid’s density, ρ (kg/m3). The volume of a sphere is πd3/6. In addition, laminar flow corresponds to the case where the dimension- less Reynolds number, Re, is less than 1, where Re = ρdυ/μ. (a) Use a force balance for the particle to develop the dif-

ferential equation for dυ/dt as a function of d, ρ, ρs, and μ.

PRoblEMs 23

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24 MAThEMATiCAl ModEling, nuMERiCAl METhods, And PRoblEM solving

1.24 Use Archimedes’ principle to develop a steady-state force balance for a spherical ball of ice floating in seawater. The force balance should be expressed as a third-order poly- nomial (cubic) in terms of height of the cap above the water line (h), and the seawater’s density (ρf), the ball’s density (ρs) and radius (r). 1.25 Beyond fluids, Archimedes’ principle has proven useful in geology when applied to solids on the earth’s crust. Figure P1.25 depicts one such case where a lighter conical granite mountain “floats on” a denser basalt layer at the earth’s surface. Note that the part of the cone below the surface is formally referred to as a frustum. Develop a steady-state force balance for this case in terms of the fol- lowing parameters: basalt’s density (ρb), granite’s density (ρg), the cone’s bottom radius (r), and the height above (h1) and below (h2) the earth’s surface.

FD FB

FIGURE P1.22

FIGURE P1.24

(b) At steady-state, use this equation to solve for the par- ticle’s terminal velocity.

(c) Employ the result of (b) to compute the particle’s ter- minal velocity in m/s for a spherical silt particle settling in water: d = 10 μm, ρ = 1 g/cm3, ρs = 2.65 g/cm3, and μ = 0.014 g/(cm·s).

(d) Check whether flow is laminar. (e) Use Euler’s method to compute the velocity from t = 0 to

2−15 s with Δt = 2−18 s given the initial condition: υ(0) = 0. 1.23 As depicted in Fig. P1.23, the downward deflection, y (m), of a cantilever beam with a uniform load, w = 10,000 kg/m, can be computed as

y = w _____ 24EI (x 4 − 4L x3 + 6L2x2)

where x = distance (m), E = the modulus of elasticity = 2 × 1011 Pa, I = moment of inertia = 3.25 × 10–4 m4, and L = length = 4 m. This equation can be differentiated to yield the slope of the downward deflection as a function of x

dy ___ dx = w _____ 24EI (4x

3 − 12 L x2 + 12L2x)

If y = 0 at x = 0, use this equation with Euler’s method (Δx = 0.125 m) to compute the deflection from x = 0 to L. Develop a plot of your results along with the analytical solution com- puted with the first equation.

x = L

x = 0

FIGURE P1.23 A cantilever beam.

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1.26 As depicted in Fig. P1.26, an RLC circuit consists of three elements: a resistor (R), an inductor (L), and a capaci- tor (C). The flow of current across each element induces a voltage drop. Kirchhoff’s second voltage law states that the algebraic sum of these voltage drops around a closed circuit is zero,

iR + L di __ dt + q __ C = 0

where i = current, R = resistance, L = inductance, t = time, q = charge, and C = capacitance. In addition, the current is related to charge as in

dq ___ dt = i

(a) If the initial values are i(0) = 0 and q(0) = 1 C, use Euler’s method to solve this pair of differential equa- tions from t = 0 to 0.1 s using a step size of Δt = 0.01 s. Employ the following parameters for your calculation: R = 200 Ω, L = 5 H, and C = 10−4 F.

(b) Develop a plot of i and q versus t. 1.27 Suppose that a parachutist with linear drag (m = 70 kg, c = 12.5 kg/s) jumps from an airplane flying at an altitude of 200 m with a horizontal velocity of 180 m/s relative to the ground. (a) Write a system of four differential equations for x, y, υx =

dx/dt and υy = dy/dt. (b) If the initial horizontal position is defined as x = 0, use

Euler’s methods with Δt = 1 s to compute the jumper’s position over the first 10 seconds.

FIGURE P1.25 FIGURE P1.26

FIGURE P1.28 Forces on a hot air balloon: FB = buoyancy, FG = weight of gas, FP = weight of payload (including the balloon envelope), and FD = drag. Note that the direction of the drag is downward when the balloon is rising.

Granite

Basalt

Inductor CapacitorResistor

iR L di dt

q C

(c) Develop plots of y versus t and y versus x. Use the plot to graphically estimate when and where the jumper would hit the ground if the chute failed to open.

1.28 Figure P1.28 shows the forces exerted on a hot air bal- loon system.

PRoblEMs 25

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26 MAThEMATiCAl ModEling, nuMERiCAl METhods, And PRoblEM solving

Formulate the drag force as

FD = 1 __ 2 ρa υ

2ACd

where ρa = air density (kg/m3), υ = velocity (m/s), A = pro- jected frontal area (m2), and Cd = the dimensionless drag co- efficient (≅ 0.47 for a sphere). Note also that the total mass of the balloon consists of two components:

m = mG + mP

inside the envelope is at roughly the same pressure as the outside air.

Other necessary parameters are:

Normal atmospheric pressure, P = 101,300 Pa The gas constant for dry air, R = 287 Joules/(kg K) The air inside the balloon is heated to an average tempera-

ture, T = 100 ºC The normal (ambient) air density, ρ = 1.2 kg/m3.

(a) Use a force balance to develop the differential equa- tion for dυ/dt as a function of the model’s fundamental parameters.

(b) At steady-state, calculate the particle’s terminal velocity. (c) Use Euler’s method and Excel to compute the velocity

from t = 0 to 60 s with Δt = 2 s given the previous parameters along with the initial condition: υ(0) = 0. Develop a plot of your results.

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CHAPTER OBJECTIVES

Knowledge and understanding are prerequisites for the effective implementation of any tool.

No matter how impressive your tool chest, you will be hard-pressed to repair a car if you do not understand how it works.

• This is the first chapter objectives entry. • Second chapter objective entry, the entries use ic/lc per manuscript, the first and

last entry have space above or below, middle entries do not. • Third chapter entry copy goes here.

MATLAB Fundamentals

2 MATLAB Fundamentals

CHAPTER OBJECTIVES

The primary objective of this chapter is to provide an introduction and overview of how MATLAB’s calculator mode is used to implement interactive computations. Specific objectives and topics covered are

• Learning how real and complex numbers are assigned to variables. • Learning how vectors and matrices are assigned values using simple assignment,

the colon operator, and the linspace and logspace functions. • Understanding the priority rules for constructing mathematical expressions. • Gaining a general understanding of built-in functions and how you can learn more

about them with MATLAB’s Help facilities. • Learning how to use vectors to create a simple line plot based on an equation.

YOU’VE GOT A PROBLEM

In Chap. 1, we used a force balance to determine the terminal velocity of a free-falling object like a bungee jumper: υt = √

___ gm ___ cd

where υt = terminal velocity (m/s), g = gravitational acceleration (m/s2), m = mass (kg), and cd = a drag coefficient (kg/m). Aside from predicting the terminal velocity, this equa- tion can also be rearranged to compute the drag coefficient

cd = mg ___ υ t 2

(2.1)

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28 MATLAB FundAMenTALs

Thus, if we measure the terminal velocity of a number of jumpers of known mass, this equation provides a means to estimate the drag coefficient. The data in Table 2.1 were col- lected for this purpose.

In this chapter, we will learn how MATLAB can be used to analyze such data. Beyond showing how MATLAB can be employed to compute quantities like drag coefficients, we will also illustrate how its graphical capabilities provide additional insight into such analyses.

2.1 THE MATLAB ENVIRONMENT

MATLAB is a computer program that provides the user with a convenient environment for performing many types of calculations. In particular, it provides a very nice tool to imple- ment numerical methods.

The most common way to operate MATLAB is by entering commands one at a time in the command window. In this chapter, we use this interactive or calculator mode to in- troduce you to common operations such as performing calculations and creating plots. In Chap. 3, we show how such commands can be used to create MATLAB programs.

One further note. This chapter has been written as a hands-on exercise. That is, you should read it while sitting in front of your computer. The most efficient way to become proficient is to actually implement the commands on MATLAB as you proceed through the following material.

MATLAB uses three primary windows: • Command window. Used to enter commands and data. • Graphics window. Used to display plots and graphs. • Edit window. Used to create and edit M-files.

In this chapter, we will make use of the command and graphics windows. In Chap. 3 we will use the edit window to create M-files.

After starting MATLAB, the command window will open with the command prompt being displayed

The calculator mode of MATLAB operates in a sequential fashion as you type in com- mands line by line. For each command, you get a result. Thus, you can think of it as operat- ing like a very fancy calculator. For example, if you type in

>> 55 − 16

MATLAB will display the result1

ans = 39

TABLE 2.1 data for the mass and associated terminal velocities of a number of jumpers.

m, kg 83.6 60.2 72.1 91.1 92.9 65.3 80.9 υt , m/s 53.4 48.5 50.9 55.7 54 47.7 51.1

1 MATLAB skips a line between the label (ans =) and the number (39). Here, we omit such blank lines for conciseness. You can control whether blank lines are included with the format compact and format loose commands.

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2.2 AssIGnMenT 29

Notice that MATLAB has automatically assigned the answer to a variable, ans. Thus, you could now use ans in a subsequent calculation:

>> ans + 11

with the result

ans = 50

MATLAB assigns the result to ans whenever you do not explicitly assign the calculation to a variable of your own choosing.

2.2 ASSIGNMENT

Assignment refers to assigning values to variable names. This results in the storage of the values in the memory location corresponding to the variable name.

2.2.1 Scalars

The assignment of values to scalar variables is similar to other computer languages. Try typing

>> a = 4

Note how the assignment echo prints to confirm what you have done:

a = 4

Echo printing is a characteristic of MATLAB. It can be suppressed by terminating the com- mand line with the semicolon (;) character. Try typing

>> A = 6;

You can type several commands on the same line by separating them with commas or semicolons. If you separate them with commas, they will be displayed, and if you use the semicolon, they will not. For example,

>> a = 4,A = 6;x = 1;

a = 4

MATLAB treats names in a case-sensitive manner—that is, the variable a is not the same as A. To illustrate this, enter

>> a

and then enter

>> A

See how their values are distinct. They are distinct names.

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30 MATLAB FundAMenTALs

We can assign complex values to variables, since MATLAB handles complex arith- metic automatically. The unit imaginary number √

___ −1 is preassigned to the variable i.

Consequently, a complex value can be assigned simply as in

>> x = 2 +i*4

x = 2.0000 + 4.0000i

It should be noted that MATLAB allows the symbol j to be used to represent the unit imaginary number for input. However, it always uses an i for display. For example,

>> x = 2 +j*4

x = 2.0000 + 4.0000i

There are several predefined variables, for example, pi.

>> pi

ans = 3.1416

Notice how MATLAB displays four decimal places. If you desire additional precision, enter the following:

>> format long

Now when pi is entered the result is displayed to 15 significant figures:

>> pi

ans = 3.14159265358979

To return to the four decimal version, type

>> format short

The following is a summary of the format commands you will employ routinely in engi- neering and scientific calculations. They all have the syntax: format type.

type Result Example

short Scaled fixed-point format with 5 digits 3.1416 long Scaled fixed-point format with 15 digits for double and 7 digits for single 3.14159265358979 short e Floating-point format with 5 digits 3.1416e+000 long e Floating-point format with 15 digits for double and 7 digits for single 3.141592653589793e+000 short g Best of fixed- or floating-point format with 5 digits 3.1416 long g Best of fixed- or floating-point format with 15 digits for double 3.14159265358979 and 7 digits for single short eng Engineering format with at least 5 digits and a power that is a multiple of 3 3.1416e+000 long eng Engineering format with exactly 16 significant digits and a power 3.14159265358979e+000 that is a multiple of 3 bank Fixed dollars and cents 3.14

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2.2 AssIGnMenT 31

2.2.2 Arrays, Vectors, and Matrices

An array is a collection of values that are represented by a single variable name. One- dimensional arrays are called vectors and two-dimensional arrays are called matrices. The scalars used in Sec. 2.2.1 are actually matrices with one row and one column.

Brackets are used to enter arrays in the command mode. For example, a row vector can be assigned as follows:

>> a = [1 2 3 4 5]

a = 1 2 3 4 5

Note that this assignment overrides the previous assignment of a = 4. In practice, row vectors are rarely used to solve mathematical problems. When we

speak of vectors, we usually refer to column vectors, which are more commonly used. A column vector can be entered in several ways. Try them.

>> b = [2;4;6;8;10]

or >> b = [2 4 6 8 10]

or, by transposing a row vector with the ' operator, >> b = [2 4 6 8 10]'

The result in all three cases will be b = 2 4 6 8 10

A matrix of values can be assigned as follows:

>> A = [1 2 3; 4 5 6; 7 8 9]

A = 1 2 3 4 5 6 7 8 9

In addition, the Enter key (carriage return) can be used to separate the rows. For example, in the following case, the Enter key would be struck after the 3, the 6, and the ] to assign the matrix:

>> A = [1 2 3 4 5 6 7 8 9]

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32 MATLAB FundAMenTALs

Finally, we could construct the same matrix by concatenating (i.e., joining) the vectors representing each column:

>> A = [[1 4 7]' [2 5 8]' [3 6 9]']

At any point in a session, a list of all current variables can be obtained by entering the who command:

>> who

Your variables are: A a ans b x

or, with more detail, enter the whos command: >> whos

Name Size Bytes Class

A 3x3 72 double array a 1x5 40 double array ans 1x1 8 double array b 5x1 40 double array x 1x1 16 double array (complex)

Grand total is 21 elements using 176 bytes

Note that subscript notation can be used to access an individual element of an array. For example, the fourth element of the column vector b can be displayed as

>> b(4)

ans = 8

For an array, A(m,n) selects the element in mth row and the nth column. For example, >> A(2,3)

ans = 6

There are several built-in functions that can be used to create matrices. For exam- ple, the ones and zeros functions create vectors or matrices filled with ones and zeros, respectively. Both have two arguments, the first for the number of rows and the second for the number of columns. For example, to create a 2 × 3 matrix of zeros:

>> E = zeros(2,3)

E = 0 0 0 0 0 0

Similarly, the ones function can be used to create a row vector of ones: >> u = ones(1,3)

u = 1 1 1

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2.2 AssIGnMenT 33

2.2.3 The Colon Operator

The colon operator is a powerful tool for creating and manipulating arrays. If a colon is used to separate two numbers, MATLAB generates the numbers between them using an increment of one:

>> t = 1:5

t = 1 2 3 4 5

If colons are used to separate three numbers, MATLAB generates the numbers between the first and third numbers using an increment equal to the second number:

>> t = 1:0.5:3

t = 1.0000 1.5000 2.0000 2.5000 3.0000

Note that negative increments can also be used

>> t = 10:−1:5

t = 10 9 8 7 6 5

Aside from creating series of numbers, the colon can also be used as a wildcard to select the individual rows and columns of a matrix. When a colon is used in place of a specific subscript, the colon represents the entire row or column. For example, the second row of the matrix A can be selected as in

>> A(2,:)

ans = 4 5 6

We can also use the colon notation to selectively extract a series of elements from within an array. For example, based on the previous definition of the vector t:

>> t(2:4)

ans = 9 8 7

Thus, the second through the fourth elements are returned.

2.2.4 The linspace and logspace Functions

The linspace and logspace functions provide other handy tools to generate vectors of spaced points. The linspace function generates a row vector of equally spaced points. It has the form

linspace(x1, x2, n)

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34 MATLAB FundAMenTALs

which generates n points between x1 and x2. For example >> linspace(0,1,6)

ans = 0 0.2000 0.4000 0.6000 0.8000 1.0000

If the n is omitted, the function automatically generates 100 points. The logspace function generates a row vector that is logarithmically equally spaced. It

has the form

logspace(x1, x2, n)

which generates n logarithmically equally spaced points between decades 10x1 and 10x2. For example,

>> logspace(-1,2,4)

ans = 0.1000 1.0000 10.0000 100.0000

If n is omitted, it automatically generates 50 points.

2.2.5 Character Strings

Aside from numbers, alphanumeric information or character strings can be represented by enclosing the strings within single quotation marks. For example,

>> f = 'Miles '; >> s = 'Davis';

Each character in a string is one element in an array. Thus, we can concatenate (i.e., paste together) strings as in

>> x = [f s]

x = Miles Davis

Note that very long lines can be continued by placing an ellipsis (three consecutive periods) at the end of the line to be continued. For example, a row vector could be entered as

>> a = [1 2 3 4 5 ... 6 7 8]

a = 1 2 3 4 5 6 7 8

However, you cannot use an ellipsis within single quotes to continue a string. To enter a string that extends beyond a single line, piece together shorter strings as in

>> quote = ['Any fool can make a rule,' ... ' and any fool will mind it']

quote = Any fool can make a rule, and any fool will mind it

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2.2 AssIGnMenT 35

TABLE 2.2 some useful string functions.

Function Description

n=length(s) Number of characters, n, in a string, s.

b=strcmp(s1,s2) Compares two strings, s1 and s2; if equal returns true (b = 1). If not equal, returns false (b = 0).

n=str2num(s) Converts a string, s, to a number, n.

s=num2str(n) Converts a number, n, to a string, s.

s2=strrep(s1,c1,c2) Replaces characters in a string with different characters.

i=strfind(s1,s2) Returns the starting indices of any occurrences of the string s2 in the string s1.

S=upper(s) Converts a string to upper case.

s=lower(S) Converts a string to lower case.

A number of built-in MATLAB functions are available to operate on strings. Table 2.2 lists a few of the more commonly used ones. For example,

>> x1 = ‘Canada’; x2 = ‘Mexico’; x3 = ‘USA’; x4 = ‘2010’; x5 = 810;

>> strcmp(a1,a2)

ans =

>> strcmp(x2,’Mexico’)

ans =

>> str2num(x4)

ans =

2010

>> num2str(x5)

ans =

810

>> strrep

>> lower

>> upper

Note, if you want to display strings in multiple lines, use the sprint function and insert the two-character sequence \n between the strings. For example,

>> disp(sprintf('Yo\nAdrian!'))

yields

Adrian!

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36 MATLAB FundAMenTALs

2.3 MATHEMATICAL OPERATIONS

Operations with scalar quantities are handled in a straightforward manner, similar to other computer languages. The common operators, in order of priority, are

^ Exponentiation − Negation * / Multiplication and division \ Left division2

+ − Addition and subtraction

These operators will work in calculator fashion. Try

>> 2*pi

ans = 6.2832

Also, scalar real variables can be included:

>> y = pi/4; >> y ^ 2.45

ans = 0.5533

Results of calculations can be assigned to a variable, as in the next-to-last example, or simply displayed, as in the last example.

As with other computer calculation, the priority order can be overridden with paren- theses. For example, because exponentiation has higher priority than negation, the follow- ing result would be obtained:

>> y = −4 ^ 2

y = −16

Thus, 4 is first squared and then negated. Parentheses can be used to override the priorities as in

>> y = (−4) ^ 2

y = 16

Within each precedence level, operators have equal precedence and are evaluated from left to right. As an example,

>> 4^2^3 >> 4^(2^3) >> (4^2)^3

2 Left division applies to matrix algebra. It will be discussed in detail later in this book.

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2.3 MATheMATIcAL OperATIOns 37

In the first case 42 = 16 is evaluated first, which is then cubed to give 4096. In the second case 23 = 8 is evaluated first and then 48 = 65,536. The third case is the same as the first, but uses parentheses to be clearer.

One potentially confusing operation is negation; that is, when a minus sign is em- ployed with a single argument to indicate a sign change. For example,

>> 2*−4

The −4 is treated as a number, so you get −8. As this might be unclear, you can use paren- theses to clarify the operation

>> 2*(−4)

Here is a final example where the minus is used for negation

>> 2^−4

Again −4 is treated as a number, so 2^−4 = 2−4 = 1/24 = 1/16 = 0.0625. Parentheses can make the operation clearer

>> 2^(−4)

Calculations can also involve complex quantities. Here are some examples that use the values of x (2 + 4i) and y (16) defined previously:

>> 3 * x

ans = 6.0000 + 12.0000i

>> 1 / x

ans = 0.1000 − 0.2000i

>> x ^ 2

ans = −12.0000 + 16.0000i

>> x + y

ans = 18.0000 + 4.0000i

The real power of MATLAB is illustrated in its ability to carry out vector-matrix calculations. Although we will describe such calculations in detail in Chap. 8, it is worth introducing some examples here.

The inner product of two vectors (dot product) can be calculated using the * operator,

>> a * b

ans = 110

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38 MATLAB FundAMenTALs

and likewise, the outer product

>> b * a

ans = 2 4 6 8 10 4 8 12 16 20 6 12 18 24 30 8 16 24 32 40 10 20 30 40 50

To further illustrate vector-matrix multiplication, first redefine a and b: >> a = [1 2 3];

and >> b = [4 5 6]';

Now, try >> a * A

ans = 30 36 42

or >> A * b

ans = 32 77 122

Matrices cannot be multiplied if the inner dimensions are unequal. Here is what happens when the dimensions are not those required by the operations. Try

>> A * a

MATLAB automatically displays the error message:

??? Error using ==> mtimes Inner matrix dimensions must agree.

Matrix-matrix multiplication is carried out in likewise fashion:

>> A * A

ans = 30 36 42 66 81 96 102 126 150

Mixed operations with scalars are also possible:

>> A/pi

ans = 0.3183 0.6366 0.9549 1.2732 1.5915 1.9099 2.2282 2.5465 2.8648

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2.4 use OF BuILT-In FuncTIOns 39

We must always remember that MATLAB will apply the simple arithmetic operators in vector-matrix fashion if possible. At times, you will want to carry out calculations item by item in a matrix or vector. MATLAB provides for that too. For example,

>> A^2

ans = 30 36 42 66 81 96 102 126 150

results in matrix multiplication of A with itself. What if you want to square each element of A? That can be done with >> A.^2

ans = 1 4 9 16 25 36 49 64 81

The . preceding the ^ operator signifies that the operation is to be carried out element by element. The MATLAB manual calls these array operations. They are also often referred to as element-by-element operations.

MATLAB contains a helpful shortcut for performing calculations that you’ve already done. Press the up-arrow key. You should get back the last line you typed in.

>> A.^2

Pressing Enter will perform the calculation again. But you can also edit this line. For example, change it to the line below and then press Enter.

>> A.^3

ans = 1 8 27 64 125 216 343 512 729

Using the up-arrow key, you can go back to any command that you entered. Press the up- arrow until you get back the line

>> b * a

Alternatively, you can type b and press the up-arrow once and it will automatically bring up the last command beginning with the letter b. The up-arrow shortcut is a quick way to fix errors without having to retype the entire line.

2.4 USE OF BUILT-IN FUNCTIONS

MATLAB and its Toolboxes have a rich collection of built-in functions. You can use online help to find out more about them. For example, if you want to learn about the log function, type in

>> help log

LOG Natural logarithm.

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40 MATLAB FundAMenTALs

LOG(X) is the natural logarithm of the elements of X. Complex results are produced if X is not positive.

See also LOG2, LOG10, EXP, LOGM.

For a list of all the elementary functions, type >> help elfun

One of their important properties of MATLAB’s built-in functions is that they will operate directly on vector and matrix quantities. For example, try

>> log(A)

ans = 0 0.6931 1.0986 1.3863 1.6094 1.7918 1.9459 2.0794 2.1972

and you will see that the natural logarithm function is applied in array style, element by element, to the matrix A. Most functions, such as sqrt, abs, sin, acos, tanh, and exp, operate in array fashion. Certain functions, such as exponential and square root, have matrix defini- tions also. MATLAB will evaluate the matrix version when the letter m is appended to the function name. Try

>> sqrtm(A)

ans = 0.4498 + 0.7623i 0.5526 + 0.2068i 0.6555 − 0.3487i 1.0185 + 0.0842i 1.2515 + 0.0228i 1.4844 − 0.0385i 1.5873 − 0.5940i 1.9503 − 0.1611i 2.3134 + 0.2717i

There are several functions for rounding. For example, suppose that we enter a vector: >> E = [−1.6 −1.5 −1.4 1.4 1.5 1.6];

The round function rounds the elements of E to the nearest integers: >> round(E)

ans = −2 −2 −1 1 2 2

The ceil (short for ceiling) function rounds to the nearest integers toward infinity: >> ceil(E)

ans = −1 −1 −1 2 2 2

The floor function rounds down to the nearest integers toward minus infinity: >> floor(E)

ans = −2 −2 −2 1 1 1

There are also functions that perform special actions on the elements of matrices and arrays. For example, the sum function returns the sum of the elements:

>> F = [3 5 4 6 1]; >> sum(F)

ans = 19

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2.4 use OF BuILT-In FuncTIOns 41

In a similar way, it should be pretty obvious what’s happening with the following commands:

>> min(F),max(F),mean(F),prod(F),sort(F)

ans = 1

ans = 6

ans = 3.8000

ans = 360

ans = 1 3 4 5 6

A common use of functions is to evaluate a formula for a series of arguments. Recall that the velocity of a free-falling bungee jumper can be computed with [Eq. (1.9)]:

υ = √ ___

gm ___ cd tanh ( √ ____

gcd ____ m t )

where υ is velocity (m/s), g is the acceleration due to gravity (9.81 m/s2), m is mass (kg), cd is the drag coefficient (kg/m), and t is time (s).

Create a column vector t that contains values from 0 to 20 in steps of 2: >> t = [0:2:20]'

t = 0 2 4 6 8 10 12 14 16 18 20

Check the number of items in the t array with the length function: >> length(t)

ans = 11

Assign values to the parameters: >> g = 9.81; m = 68.1; cd = 0.25;

MATLAB allows you to evaluate a formula such as υ = f (t), where the formula is com- puted for each value of the t array, and the result is assigned to a corresponding position in the υ array. For our case,

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42 MATLAB FundAMenTALs

>> v = sqrt(g*m/cd)*tanh(sqrt(g*cd/m)*t)

v = 0 18.7292 33.1118 42.0762 46.9575 49.4214 50.6175 51.1871 51.4560 51.5823 51.6416

2.5 GRAPHICS

MATLAB allows graphs to be created quickly and conveniently. For example, to create a graph of the t and v arrays from the data above, enter

>> plot(t, v)

The graph appears in the graphics window and can be printed or transferred via the clip- board to other programs.

0 0 2 4 6 8 10 12 16 1814 20

You can customize the graph a bit with commands such as the following: >> title('Plot of v versus t') >> xlabel('Values of t') >> ylabel('Values of v') >> grid

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2.5 GrAphIcs 43

The plot command displays a solid thin blue line by default. If you want to plot each point with a symbol, you can include a specifier enclosed in single quotes in the plot function. Table 2.3 lists the available specifiers. For example, if you want to use open circles enter

>> plot(t, v, 'o')

0 0 2 4 6 8 10

Values of t

Plot of υ versus t

V al

ue s

of υ

16 181412 20

TABLE 2.3 specifiers for colors, symbols, and line types.

Colors Symbols Line Types

Blue b Point . Solid − Green g Circle o Dotted : Red r X-mark x Dashdot -. Cyan c Plus + Dashed -- Magenta m Star * Yellow y Square s Black k Diamond d White w Triangle(down)

Triangle(up) ^ Triangle(left) < Triangle(right) > Pentagram p Hexagram h

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44 MATLAB FundAMenTALs

You can also combine several specifiers. For example, if you want to use square green markers connected by green dashed lines, you could enter

>> plot(t, v, 's−−g')

You can also control the line width as well as the marker’s size and its edge and face (i.e., interior) colors. For example, the following command uses a heavier (2-point), dashed, cyan line to connect larger (10-point) diamond-shaped markers with black edges and magenta faces:

>> plot(x,y,'−−dc','LineWidth', 2,... 'MarkerSize',10,... 'MarkerEdgeColor','k',... 'MarkerFaceColor','m')

Note that the default line width is 1 point. For the markers, the default size is 6 point with blue edge color and no face color.

MATLAB allows you to display more than one data set on the same plot. For example, an alternative way to connect each data marker with a straight line would be to type

>> plot(t, v, t, v, 'o')

It should be mentioned that, by default, previous plots are erased every time the plot command is implemented. The hold on command holds the current plot and all axis proper- ties so that additional graphing commands can be added to the existing plot. The hold off command returns to the default mode. For example, if we had typed the following com- mands, the final plot would only display symbols:

>> plot(t, v) >> plot(t, v, 'o')

In contrast, the following commands would result in both lines and symbols being displayed: >> plot(t, v) >> hold on >> plot(t, v, 'o') >> hold off

In addition to hold, another handy function is subplot, which allows you to split the graph window into subwindows or panes. It has the syntax

subplot(m, n, p)

This command breaks the graph window into an m-by-n matrix of small axes, and selects the p-th axes for the current plot.

We can demonstrate subplot by examining MATLAB’s capability to generate three- dimensional plots. The simplest manifestation of this capability is the plot3 command which has the syntax

plot3(x, y, z)

where x, y, and z are three vectors of the same length. The result is a line in three-dimen- sional space through the points whose coordinates are the elements of x, y, and z.

Plotting a helix provides a nice example to illustrate its utility. First, let’s graph a circle with the two-dimensional plot function using the parametric representation: x = sin(t) and y = cos(t). We employ the subplot command so we can subsequently add the three- dimensional plot.

>> t = 0:pi/50:10*pi; >> subplot(1,2,1);plot(sin(t),cos(t))

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2.5 GrAphIcs 45

>> axis square >> title('(a)')

As in Fig. 2.1a, the result is a circle. Note that the circle would have been distorted if we had not used the axis square command.

Now, let’s add the helix to the graph’s right pane. To do this, we again employ a para- metric representation: x = sin(t), y = cos(t), and z = t

>> subplot(1,2,2);plot3(sin(t),cos(t),t); >> title('(b)')

The result is shown in Fig. 2.1b. Can you visualize what’s going on? As time evolves, the x and y coordinates sketch out the circumference of the circle in the x–y plane in the same fashion as the two-dimensional plot. However, simultaneously, the curve rises verti- cally as the z coordinate increases linearly with time. The net result is the characteristic spring or spiral staircase shape of the helix.

There are other features of graphics that are useful—for example, plotting objects instead of lines, families of curves plots, plotting on the complex plane, log-log or semilog plots, three-dimensional mesh plots, and contour plots. As described next, a variety of re- sources are available to learn about these as well as other MATLAB capabilities.

0 1

−1

(a)

(b)

0.5

−0.5

−1 −1 −0.5 0 0.5 1

FIGURE 2.1 A two-pane plot of (a) a two-dimensional circle and (b) a three-dimensional helix.

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46 MATLAB FundAMenTALs

2.6 OTHER RESOURCES

The foregoing was designed to focus on those features of MATLAB that we will be using in the remainder of this book. As such, it is obviously not a comprehensive overview of all of MATLAB’s capabilities. If you are interested in learning more, you should consult one of the excellent books devoted to MATLAB (e.g., Attaway, 2009; Palm, 2007; Hanselman and Littlefield, 2005; and Moore, 2008).

Further, the package itself includes an extensive Help facility that can be accessed by clicking on the Help menu in the command window. This will provide you with a number of different options for exploring and searching through MATLAB’s Help material. In ad- dition, it provides access to a number of instructive demos.

As described in this chapter, help is also available in interactive mode by typing the help command followed by the name of a command or function.

If you do not know the name, you can use the lookfor command to search the MATLAB Help files for occurrences of text. For example, suppose that you want to find all the com- mands and functions that relate to logarithms, you could enter

>> lookfor logarithm

and MATLAB will display all references that include the word logarithm. Finally, you can obtain help from The MathWorks, Inc., website at www.mathworks

.com. There you will find links to product information, newsgroups, books, and technical support as well as a variety of other useful resources.

2.7 CASE STUDY eXpLOrATOrY dATA AnALYsIs

Background. Your textbooks are filled with formulas developed in the past by re- nowned scientists and engineers. Although these are of great utility, engineers and sci- entists often must supplement these relationships by collecting and analyzing their own data. Sometimes this leads to a new formula. However, prior to arriving at a final predic- tive equation, we usually “play” with the data by performing calculations and developing plots. In most cases, our intent is to gain insight into the patterns and mechanisms hidden in the data.

In this case study, we will illustrate how MATLAB facilitates such exploratory data analysis. We will do this by estimating the drag coefficient of a free-falling human based on Eq. (2.1) and the data from Table 2.1. However, beyond merely computing the drag coefficient, we will use MATLAB’s graphical capabilities to discern patterns in the data.

Solution. The data from Table 2.1 along with gravitational acceleration can be entered as >> m = [83.6 60.2 72.1 91.1 92.9 65.3 80.9]; >> vt = [53.4 48.5 50.9 55.7 54 47.7 51.1]; >> g = 9.81;

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2.7 cAse sTudY 47

2.7 CASE STUDY continued

The drag coefficients can then be computed with Eq. (2.1). Because we are performing element-by-element operations on vectors, we must include periods prior to the operators:

>> cd = g*m./vt.^2

cd = 0.2876 0.2511 0.2730 0.2881 0.3125 0.2815 0.3039

We can now use some of MATLAB’s built-in functions to generate some statistics for the results:

>> cdavg = mean(cd),cdmin = min(cd),cdmax = max(cd) cdavg = 0.2854 cdmin = 0.2511 cdmax = 0.3125

Thus, the average value is 0.2854 with a range from 0.2511 to 0.3125 kg/m. Now, let’s start to play with these data by using Eq. (2.1) to make a prediction of the

terminal velocity based on the average drag: >> vpred=sqrt(g*m/cdavg)

vpred = 53.6065 45.4897 49.7831 55.9595 56.5096 47.3774 52.7338

Notice that we do not have to use periods prior to the operators in this formula? Do you understand why?

We can plot these values versus the actual measured terminal velocities. We will also superimpose a line indicating exact predictions (the 1:1 line) to help assess the results. Because we are going to eventually generate a second plot, we employ the subplot command:

>> subplot(2,1,1);plot(vt,vpred,'o',vt,vt) >> xlabel('measured') >> ylabel('predicted') >> title('Plot of predicted versus measured velocities')

As in the top plot of Fig. 2.2, because the predictions generally follow the 1:1 line, you might initially conclude that the average drag coefficient yields decent results. However, notice how the model tends to underpredict the low velocities and overpredict the high. This suggests that rather than being constant, there might be a trend in the drag coeffi- cients. This can be seen by plotting the estimated drag coefficients versus mass:

>> subplot(2,1,2);plot(m,cd,'o') >> xlabel('mass (kg)') >> ylabel('estimated drag coefficient (kg/m)') >> title('Plot of drag coefficient versus mass')

The resulting plot, which is the bottom graph in Fig. 2.2, suggests that rather than being constant, the drag coefficient seems to be increasing as the mass of the jumper increases. Based on this result, you might conclude that your model needs to be improved. At the

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48 MATLAB FundAMenTALs

2.7 CASE STUDY continued

least, it might motivate you to conduct further experiments with a larger number of jumpers to confirm your preliminary finding.

In addition, the result might also stimulate you to go to the fluid mechanics literature and learn more about the science of drag. As described previously in Sec. 1.4, you would discover that the parameter cd is actually a lumped drag coefficient that along with the true drag includes other factors such as the jumper’s frontal area and air density:

cd = CD ρ A _____ 2 (2.2)

where CD = a dimensionless drag coefficient, ρ = air density (kg/m3), and A = frontal area (m2), which is the area projected on a plane normal to the direction of the velocity.

Assuming that the densities were relatively constant during data collection (a pretty good assumption if the jumpers all took off from the same height on the same day), Eq. (2.2) suggests that heavier jumpers might have larger areas. This hypothesis could be substanti- ated by measuring the frontal areas of individuals of varying masses.

FIGURE 2.2 Two plots created with MATLAB.

60 Plot of predicted versus measured velocities

Measured

Pr ed

ic te

d 55

45 47 48 49 50 51 52 53 54 55 56

0.35 Plot of drag coe�cient versus mass

Mass (kg)

Es tim

at ed

d ra

g co

e� ci

en t (

kg /m

)

0.3

0.25

0.2 60 65 70 75 80 85 90 95

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prOBLeMs 49

PROBLEMS

2.1 What is the output when the following commands are implemented? A = [1:3;2:2:6;3:−1:1] A = A' A(:,3) = [] A = [A(:,1) [4 5 7]' A(:,2)] A = sum(diag(A)) 2.2 You want to write MATLAB equations to compute a vector of y values using the following equations

(a) y = 6t 3 − 3t − 4 _______ 8 sin(5t)

(b) y = 6t − 4 ______ 8t − π __ 2 t

where t is a vector. Make sure that you use periods only where necessary so the equation handles vector operations properly. Extra periods will be considered incorrect. 2.3 Write a MATLAB expression to compute and dis- play the values of a vector of x values using the following equation

x = y (a + bz) 1.8 __________ z(1 − y)

Assume that y and z are vector quantities of equal length and a and b are scalars. 2.4 What is displayed when the following MATLAB state- ments are executed? (a) A = [1 2; 3 4; 5 6]; A(2,:)' (b) y = [0:1.5:7]' (c) a = 2; b = 8; c = 4; a + b / c 2.5 The MATLAB humps function defines a curve that has 2 maxima (peaks) of unequal height over the interval 0 ≤ x ≤ 2,

f (x) = 1 ______________ (x − 0.3)2 + 0.01

+ 1 ______________ (x − 0.9)2 + 0.04

− 6

Use MATLAB to generate a plot of f (x) versus x with x = [0:1/256:2];

Do not use MATLAB’s built-in humps function to generate the values of f (x). Also, employ the minimum number of periods to perform the vector operations needed to generate f (x) values for the plot. 2.6 Use the linspace function to create vectors identical to the following created with colon notation: (a) t = 4:6:35 (b) x = −4:2 2.7 Use colon notation to create vectors identical to the following created with the linspace function: (a) v = linspace(−2,1.5,8) (b) r = linspace(8,4.5,8)

2.8 The command linspace(a, b, n) generates a row vec- tor of n equally spaced points between a and b. Use colon notation to write an alternative one-line command to gener- ate the same vector. Test your formulation for a = −3, b = 5, n = 6. 2.9 The following matrix is entered in MATLAB:

>> A = [3 2 1;0:0.5:1;linspace(6, 8, 3)]

(a) Write out the resulting matrix. (b) Use colon notation to write a single-line MATLAB

command to multiply the second row by the third col- umn and assign the result to the variable c.

2.10 The following equation can be used to compute values of y as a function of x:

y = be−ax sin(bx) (0.012x4 − 0.15x3 + 0.075x2 + 2.5x)

where a and b are parameters. Write the equation for imple- mentation with MATLAB, where a = 2, b = 5, and x is a vector holding values from 0 to π∕2 in increments of Δx = π∕40. Employ the minimum number of periods (i.e., dot no- tation) so that your formulation yields a vector for y. In ad- dition, compute the vector z = y2 where each element holds the square of each element of y. Combine x, y, and z into a matrix w, where each column holds one of the variables, and display w using the short g format. In addition, gener- ate a labeled plot of y and z versus x. Include a legend on the plot (use help to understand how to do this). For y, use a 1.5-point, dashdotted red line with 14-point, red-edged, white-faced pentagram-shaped markers. For z, use a stan- dard-sized (i.e., default) solid blue line with standard-sized, blue-edged, green-faced square markers. 2.11 A simple electric circuit consisting of a resistor, a ca- pacitor, and an inductor is depicted in Fig. P2.11. The charge on the capacitor q(t) as a function of time can be computed as

q(t) = q0e−Rt∕(2L) cos [ √__________ 1 ___ LC − ( R ___ 2L ) 2 t ] where t = time, q0 the initial charge, R = the resistance, L = inductance, and C = capacitance. Use MATLAB to generate a plot of this function from t = 0 to 0.8, given that q0 = 10, R = 60, L = 9, and C = 0.00005. 2.12 The standard normal probability density function is a bell-shaped curve that can be represented as

f (z) = 1 _____ √

___ 2π e−z2∕2

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50 MATLAB FundAMenTALs

Use MATLAB to generate a plot of this function from z = −5 to 5. Label the ordinate as frequency and the abscissa as z. 2.13 If a force F (N) is applied to compress a spring, its displacement x (m) can often be modeled by Hooke’s law:

F = kx

where k = the spring constant (N/m). The potential energy stored in the spring U (J) can then be computed as

U = 1 __ 2 kx 2

Five springs are tested and the following data compiled:

F, N 14 18 8 9 13 x, m 0.013 0.020 0.009 0.010 0.012

Use MATLAB to store F and x as vectors and then compute vectors of the spring constants and the potential energies. Use the max function to determine the maximum potential energy. 2.14 The density of freshwater can be computed as a func- tion of temperature with the following cubic equation:

ρ = 5.5289 × 10−8 T C 3 − 8.5016 × 10−6 T C

+ 6.5622 × 10−5TC + 0.99987

where ρ = density (g/cm3) and TC = temperature (°C). Use MATLAB to generate a vector of temperatures ranging from 32 °F to 93.2 °F using increments of 3.6 °F. Convert this vector to degrees Celsius and then compute a vector of den- sities based on the cubic formula. Create a plot of ρ versus TC. Recall that TC = 5/9(TF − 32). 2.15 Manning’s equation can be used to compute the veloc- ity of water in a rectangular open channel:

U = √ __

S ____ n ( BH _______ B + 2H ) 2∕3

where U = velocity (m/s), S = channel slope, n = roughness coefficient, B = width (m), and H = depth (m). The follow- ing data are available for five channels:

n S B H

0.035 0.0001 10 2 0.020 0.0002 8 1 0.015 0.0010 20 1.5 0.030 0.0007 24 3 0.022 0.0003 15 2.5

Store these values in a matrix where each row represents one of the channels and each column represents one of the param- eters. Write a single-line MATLAB statement to compute a column vector containing the velocities based on the values in the parameter matrix. 2.16 It is general practice in engineering and science that equations be plotted as lines and discrete data as symbols. Here are some data for concentration (c) versus time (t) for the photodegradation of aqueous bromine:

t, min 10 20 30 40 50 60 c, ppm 3.4 2.6 1.6 1.3 1.0 0.5

These data can be described by the following function:

c = 4.84e−0.034t

Use MATLAB to create a plot displaying both the data (using diamond-shaped, filled-red symbols) and the func- tion (using a green, dashed line). Plot the function for t = 0 to 70 min. 2.17 The semilogy function operates in an identical fashion to the plot function except that a logarithmic (base-10) scale is used for the y axis. Use this function to plot the data and function as described in Prob. 2.16. Explain the results. 2.18 Here are some wind tunnel data for force (F) versus velocity (υ):

υ, m/s 10 20 30 40 50 60 70 80 F, N 25 70 380 550 610 1220 830 1450

These data can be described by the following function:

F = 0.2741υ1.9842

Switch

Resistor

Capacitor −

+ V0

i −

+ Battery Inductor

FIGURE P2.11

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prOBLeMs 51

(b) >> q = 4:2:12; >> r = [7 8 4; 3 6 –5]; >> sum(q) * r(2, 3) 2.24 The trajectory of an object can be modeled as

y = (tan θ0)x − g ________

2 υ 0 2 cos2θ0

x2 + y0

where y = height (m), θ0 = initial angle (radians), x = horizontal distance (m), g = gravitational acceleration (= 9.81 m/s2), υ0 = initial velocity (m/s), and y0 = initial height. Use MATLAB to find the trajectories for y0 = 0 and υ0 = 28 m/s for initial angles ranging from 15 to 75° in incre- ments of 15°. Employ a range of horizontal distances from x = 0 to 80 m in increments of 5 m. The results should be as- sembled in an array where the first dimension (rows) corre- sponds to the distances, and the second dimension (columns) corresponds to the different initial angles. Use this matrix to generate a single plot of the heights versus horizontal distances for each of the initial angles. Employ a legend to distinguish among the different cases, and scale the plot so that the minimum height is zero using the axis command. 2.25 The temperature dependence of chemical reactions can be computed with the Arrhenius equation:

k = Ae−E∕(RTa)

where k = reaction rate (s−1), A = the preexponential (or fre- quency) factor, E = activation energy (J/mol), R = gas con- stant [8.314 J/(mole · K)], and Ta = absolute temperature (K). A compound has E = 1 × 105 J/mol and A = 7 × 1016. Use MATLAB to generate values of reaction rates for temperatures ranging from 253 to 325 K. Use subplot to gen- erate a side-by-side graph of (a) k versus Ta (green line) and (b) log10 k (red line) versus 1∕Ta. Employ the semilogy func- tion to create (b). Include axis labels and titles for both sub- plots. Interpret your results. 2.26 Figure P2.26a shows a uniform beam subject to a lin- early increasing distributed load. As depicted in Fig. P2.26b, deflection y (m) can be computed with

y = w0 _______ 120EIL (−x

5 + 2L2x3 − L4x)

where E = the modulus of elasticity and I = the moment of inertia (m4). Employ this equation and calculus to generate MATLAB plots of the following quantities versus distance along the beam: (a) displacement (y), (b) slope [θ (x) = dy/dx], (c) moment [M(x) = EId2 y/dx2], (d) shear [V(x) = EId3 y/dx3], and (e) loading [w(x) = −EId 4 y/dx4].

Use MATLAB to create a plot displaying both the data (using circular magenta symbols) and the function (using a black dash-dotted line). Plot the function for υ = 0 to 100 m/s and label the plot’s axes. 2.19 The loglog function operates in an identical fash- ion to the plot function except that logarithmic scales are used for both the x and y axes. Use this function to plot the data and function as described in Prob. 2.18. Explain the results. 2.20 The Maclaurin series expansion for the cosine is

cos x = 1 − x 2 __ 2! +

x4 __ 4! − x6 __ 6! +

x8 __ 8! − · · ·

Use MATLAB to create a plot of the cosine (solid line) along with a plot of the series expansion (black dashed line) up to and including the term x8∕8!. Use the built-in func- tion factorial in computing the series expansion. Make the range of the abscissa from x = 0 to 3π∕2. 2.21 You contact the jumpers used to generate the data in Table 2.1 and measure their frontal areas. The result- ing values, which are ordered in the same sequence as the corresponding values in Table 2.1, are

A, m2 0.455 0.402 0.452 0.486 0.531 0.475 0.487

(a) If the air density is ρ = 1.223 kg/m3, use MATLAB to compute values of the dimensionless drag coefficient CD.

(b) Determine the average, minimum, and maximum of the resulting values.

(c) Develop a stacked plot of A versus m (upper) and CD versus m (lower). Include descriptive axis labels and titles on the plots.

2.22 The following parametric equations generate a conical helix.

x = t cos(6t) y = t sin(6t) z = t

Compute values of x, y, and z for t = 0 to 6π with Δt = π/64. Use subplot to generate a two-dimensional line plot (red solid line) of (x, y) in the top pane and a three-dimensional line plot (cyan solid line) of (x, y, z) in the bottom pane. Label the axes for both plots. 2.23 Exactly what will be displayed after the following MATLAB commands are typed? (a) >> x = 5; >> x ^ 3; >> y = 8 – x

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52 MATLAB FundAMenTALs

Use the following parameters for your computation: L = 600 cm, E = 50,000 kN/cm2, I = 30,000 cm4, w0 = 2.5 kN/cm, and Δx = 10 cm. Employ the subplot function to display all the plots vertically on the same page in the order (a) to (e).

Include labels and use consistent MKS units when developing the plots. 2.27 The butterfly curve is given by the following paramet- ric equations:

x = sin(t) ( ecos t − 2 cos 4t − sin5 t ___ 12 )

y = cos(t) ( ecos t − 2 cos 4t − sin5 t ___ 12 ) Generate values of x and y for values of t from 0 to 100 with Δt = 1/16. Construct plots of (a) x and y versus t and (b) y versus x. Use subplot to stack these plots vertically and make the plot in (b) square. Include titles and axis labels on both plots and a legend for (a). For (a), employ a dotted line for y in order to distinguish it from x. 2.28 The butterfly curve from Prob. 2.27 can also be repre- sented in polar coordinates as

r = esin θ − 2 cos(4θ ) − sin5 ( 2θ − π ______ 24 ) Generate values of r for values of θ from 0 to 8π with Δθ = π /32. Use the MATLAB function polar to generate the polar plot of the butterfly curve with a dashed red line. Employ the MATLAB Help to understand how to generate the plot.

L (a)

(x = 0, y = 0) (x = L, y = 0)

(b)

FIGURE P2.26

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CHAPTER OBJECTIVES

The primary objective of this chapter is to learn how to write M-file programs to implement numerical methods. Specific objectives and topics covered are

• Learning how to create well-documented M-files in the edit window and invoke them from the command window.

• Understanding how script and function files differ. • Understanding how to incorporate help comments in functions. • Knowing how to set up M-files so that they interactively prompt users for

information and display results in the command window. • Understanding the role of subfunctions and how they are accessed. • Knowing how to create and retrieve data files. • Learning how to write clear and well-documented M-files by employing

structured programming constructs to implement logic and repetition. • Recognizing the difference between if...elseif and switch constructs. • Recognizing the difference between for...end and while structures. • Knowing how to animate MATLAB plots. • Understanding what is meant by vectorization and why it is beneficial. • Understanding how anonymous functions can be employed to pass functions to

function function M-files.

3 Programming with MATLAB

YOU’VE GOT A PROBLEM

In Chap. 1, we used a force balance to develop a mathematical model to predict the fall velocity of a bungee jumper. This model took the form of the following differential equation: dυ ___ dt = g −

cd __ m υ∣υ∣

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54 ProgrAMMIng wITh MATLAB

We also learned that a numerical solution of this equation could be obtained with Euler’s method:

υi+1 = υi + dυi ___ dt Δt

This equation can be implemented repeatedly to compute velocity as a function of time. However, to obtain good accuracy, many small steps must be taken. This would be extremely laborious and time consuming to implement by hand. However, with the aid of MATLAB, such calculations can be performed easily.

So our problem now is to figure out how to do this. This chapter will introduce you to how MATLAB M-files can be used to obtain such solutions.

3.1 M-FILES

The most common way to operate MATLAB is by entering commands one at a time in the command window. M-files provide an alternative way of performing operations that greatly expand MATLAB’s problem-solving capabilities. An M-file consists of a series of statements that can be run all at once. Note that the nomenclature “M-file” comes from the fact that such files are stored with a .m extension. M-files come in two flavors: script files and function files.

3.1.1 Script Files

A script file is merely a series of MATLAB commands that are saved on a file. They are useful for retaining a series of commands that you want to execute on more than one oc- casion. The script can be executed by typing the file name in the command window or by pressing the Run button.

EXAMPLE 3.1 Script File

Problem Statement. Develop a script file to compute the velocity of the free-falling bun- gee jumper for the case where the initial velocity is zero.

Solution. Open the editor with the selection: New, Script. Type in the following state- ments to compute the velocity of the free-falling bungee jumper at a specific time [recall Eq. (1.9)]:

g = 9.81; m = 68.1; t = 12; cd = 0.25; v = sqrt(g * m / cd) * tanh(sqrt(g * cd / m) * t)

Save the file as scriptdemo.m. Return to the command window and type

>>scriptdemo

The result will be displayed as

v = 50.6175

Thus, the script executes just as if you had typed each of its lines in the command window.

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3.1 M-FILES 55

As a final step, determine the value of g by typing >> g

g = 9.8100

So you can see that even though g was defined within the script, it retains its value back in the command workspace. As we will see in the following section, this is an important distinction between scripts and functions.

3.1.2 Function Files

Function files are M-files that start with the word function. In contrast to script files, they can accept input arguments and return outputs. Hence they are analogous to user- defined functions in programming languages such as Fortran, Visual Basic or C.

The syntax for the function file can be represented generally as

function outvar = funcname(arglist) % helpcomments statements outvar = value;

where outvar = the name of the output variable, funcname = the function’s name, arglist = the function’s argument list (i.e., comma-delimited values that are passed into the function), helpcomments = text that provides the user with information regarding the function (these can be invoked by typing Help funcname in the command window), and statements = MATLAB statements that compute the value that is assigned to outvar.

Beyond its role in describing the function, the first line of the helpcomments, called the H1 line, is the line that is searched by the lookfor command (recall Sec. 2.6). Thus, you should include key descriptive words related to the file on this line.

The M-file should be saved as funcname.m. The function can then be run by typing funcname in the command window as illustrated in the following example. Note that even though MATLAB is case-sensitive, your computer’s operating system may not be. Whereas MATLAB would treat function names like freefall and FreeFall as two different variables, your operating system might not.

EXAMPLE 3.2 Function File

Problem Statement. As in Example 3.1, compute the velocity of the free-falling bungee jumper but now use a function file for the task.

Solution. Type the following statements in the file editor:

function v = freefall(t, m, cd) % freefall: bungee velocity with second-order drag % v=freefall(t,m,cd) computes the free-fall velocity % of an object with second-order drag % input:

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56 ProgrAMMIng wITh MATLAB

% t = time (s) % m = mass (kg) % cd = second-order drag coefficient (kg/m) % output: % v = downward velocity (m/s)

g = 9.81; % acceleration of gravity v = sqrt(g * m / cd)*tanh(sqrt(g * cd / m) * t);

Save the file as freefall.m. To invoke the function, return to the command window and type in

>> freefall(12,68.1,0.25)

The result will be displayed as

ans = 50.6175

One advantage of a function M-file is that it can be invoked repeatedly for different argument values. Suppose that you wanted to compute the velocity of a 100-kg jumper after 8 s:

>> freefall(8,100,0.25)

ans = 53.1878

To invoke the help comments type

>> help freefall

which results in the comments being displayed

freefall: bungee velocity with second-order drag v = freefall(t,m,cd) computes the free-fall velocity of an object with second-order drag input: t = time (s) m = mass (kg) cd = second-order drag coefficient (kg/m) output: v = downward velocity (m/s)

If at a later date, you forgot the name of this function, but remembered that it involved bungee jumping, you could enter

>> lookfor bungee

and the following information would be displayed

freefall.m − bungee velocity with second-order drag

Note that, at the end of the previous example, if we had typed >> g

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3.1 M-FILES 57

the following message would have been displayed

??? Undefined function or variable 'g'.

So even though g had a value of 9.81 within the M-file, it would not have a value in the command workspace. As noted previously at the end of Example 3.1, this is an important distinction between functions and scripts. The variables within a function are said to be local and are erased after the function is executed. In contrast, the variables in a script retain their existence after the script is executed.

Function M-files can return more than one result. In such cases, the variables containing the results are comma-delimited and enclosed in brackets. For example, the following function, stats.m, computes the mean and the standard deviation of a vector:

function [mean, stdev] = stats(x) n = length(x); mean = sum(x)/n; stdev = sqrt(sum((x-mean).^2/(n − 1)));

Here is an example of how it can be applied:

>> y = [8 5 10 12 6 7.5 4]; >> [m,s] = stats(y)

m = 7.5000

s = 2.8137

Although we will also make use of script M-files, function M-files will be our primary programming tool for the remainder of this book. Hence, we will often refer to function M-files as simply M-files.

3.1.3 Variable Scope

MATLAB variables have a property known as scope that refers to the context of the com- puting environment in which the variable has a unique identity and value. Typically, a variable’s scope is limited either to the MATLAB workspace or within a function. This principle prevents errors when a programmer unintentionally gives the same name to vari- ables in different contexts.

Any variables defined through the command line are within the MATLAB workspace and you can readily inspect a workspace variable’s value by entering its name at the com- mand line. However, workspace variables are not directly accessible to functions but rather are passed to functions via their arguments. For example, here is a function that adds two numbers

function c = adder(a,b) x = 88 a c = a + b

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58 ProgrAMMIng wITh MATLAB

Suppose in the command window we type

>> x = 1; y = 4; c = 8 c = 8

So as expected, the value of c in the workspace is 8. If you type >> d = adder(x,y)

the result will be

x = 88 a = 1 c = 5 d = 5

But, if you then type >> c, x, a

The result is

c = 8 x = 1

Undefined function or variable 'a'. Error in ScopeScript (line 6) c, x, a

The point here is that even though x was assigned a new value inside the function, the variable of the same name in the MATLAB workspace is unchanged. Even though they have the same name, the scope of each is limited to their context and does not overlap. In the function, the variables a and b are limited in scope to that function, and only exist while that function is being executed. Such variables are formally called local variables. Thus, when we try to display the value of a in the workspace, an error message is generated be- cause the workspace has no access to the a in the function.

Another obvious consequence of limited-scope variables is that any parameter needed by a function must be passed as an input argument or by some other explicit means. A func- tion cannot otherwise access variables in the workspace or in other functions.

3.1.4 Global Variables

As we have just illustrated, the function’s argument list is like a window through which information is selectively passed between the workspace and a function, or between two functions. Sometimes, however, it might be convenient to have access to a vari- able in several contexts without passing it as an argument. In such cases, this can be

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3.1 M-FILES 59

accomplished by defining the variable as global. This is done with the global command, which is defined as

global X Y Z

where X, Y, and Z are global in scope. If several functions (and possibly the workspace), all declare a particular name as global, then they all share a single value of that variable. Any change to that variable, in any function, is then made to all the other functions that declare it global. Stylistically, MATLAB recommends that global variables use all capital letters, but this is not required.

EXAMPLE 3.3 Use of global Variables

Problem Statement. The Stefan-Boltzmann law is used to compute the radiation flux from a black body1 as in

J = σ T a 4

where J = radiation flux [W/(m2 s)], σ = the Stefan-Boltzmann constant (5.670367 × 10−8 W m−2 K−4), and Ta = absolute temperature (K). In assessing the impact of climate change on water temperature, it is used to compute the radiation terms in a waterbody’s heat balance. For example, the long-wave radiation from the atmosphere to the waterbody, Jan [W/(m2 s)], can be calculated as

Jan = 0.97σ(Tair + 273.15)4 ( 0.6 + 0.031 √ ___

eair ) where Tair = the temperature of the air above the waterbody (°C), and eair = the vapor pres- sure of the air above the water body (mmHg),

eair = 4.596e 17.27 T d _ 237.3 + T d (E3.3.1)

where Td = the dew-point temperature (°C). The radiation from the water surface back into the atmosphere, Jbr [W/(m2 s)], is calculated as

Jbr = 0.97σ(Tw + 273.15)4 (E3.3.2)

where Tw = the water temperature (°C). Write a script that utilizes two functions to com- pute the net long-wave radiation (i.e., the difference between the atmospheric radiation in and the water back radiation out) for a cold lake with a surface temperature of Tw = 15 °C on a hot (Tair = 30 °C), humid (Td = 27.7 °C) summer day. Use global to share the Stefan- Boltzmann constant between the script and functions.

Solution. Here is the script

clc, format compact global SIGMA SIGMA = 5.670367e−8; Tair = 30; Tw = 15; Td = 27.7;

1A black body is an object that absorbs all incident electromagnetic radiation, regardless of frequency or angle of incidence. A white body is one that reflects all incident rays completely and uniformly in all directions.

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60 ProgrAMMIng wITh MATLAB

Jan = AtmLongWaveRad(Tair, Td) Jbr = WaterBackRad(Tw) JnetLongWave = Jan −Jbr

Here is a function to compute the incoming long wave radiation from the atmosphere into the lake

function Ja = AtmLongWaveRad(Tair, Td) global SIGMA eair = 4.596*exp(17.27*Td/(237.3+Td)); Ja = 0.97*SIGMA*(Tair+273.15)^4*(0.6+0.031*sqrt(eair)); end

and here is a function to compute the long wave radiation from the lake back into the atmosphere

function Jb = WaterBackRad(Twater) global SIGMA Jb = 0.97*SIGMA*(Twater+273.15)^4; end

When the script is run, the output is

Jan = 354.8483 Jbr = 379.1905 JnetLongWave = −24.3421

Thus, for this case, because the back radiation is larger than the incoming radiation, so the lake loses heat at the rate of 24.3421 W/(m2 s) due to the two long-wave radiation fluxes.

If you require additional information about global variables, you can always type help global at the command prompt. The help facility can also be invoked to learn about other MATLAB commands dealing with scope such as persistent.

3.1.5 Subfunctions

Functions can call other functions. Although such functions can exist as separate M-files, they may also be contained in a single M-file. For example, the M-file in Example 3.2 (without comments) could have been split into two functions and saved as a single M-file2:

function v = freefallsubfunc(t, m, cd) v = vel(t, m, cd); end

function v = vel(t, m, cd) g = 9.81; v = sqrt(g * m / cd)*tanh(sqrt(g * cd / m) * t); end

2Note that although end statements are not used to terminate single-function M-files, they are included when subfunctions are involved to demarcate the boundaries between the main function and the subfunctions.

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3.2 InPUT-oUTPUT 61

This M-file would be saved as freefallsubfunc.m. In such cases, the first function is called the main or primary function. It is the only function that is accessible to the command win- dow and other functions and scripts. All the other functions (in this case, vel) are referred to as subfunctions.

A subfunction is only accessible to the main function and other subfunctions within the M-file in which it resides. If we run freefallsubfunc from the command window, the result is identical to Example 3.2:

>> freefallsubfunc(12,68.1,0.25)

ans = 50.6175

However, if we attempt to run the subfunction vel, an error message occurs:

>> vel(12,68.1,.25) ??? Undefined function or method 'vel' for input arguments of type 'double'.

3.2 INPUT-OUTPUT

As in Sec.3.1, information is passed into the function via the argument list and is output via the function’s name. Two other functions provide ways to enter and display information directly using the command window.

The input Function. This function allows you to prompt the user for values directly from the command window. Its syntax is

n = input('promptstring')

The function displays the promptstring, waits for keyboard input, and then returns the value from the keyboard. For example,

m = input('Mass (kg): ')

When this line is executed, the user is prompted with the message

Mass (kg):

If the user enters a value, it would then be assigned to the variable m. The input function can also return user input as a string. To do this, an 's' is appended

to the function’s argument list. For example,

name = input('Enter your name: ','s')

The disp Function. This function provides a handy way to display a value. Its syntax is

disp(value)

where value = the value you would like to display. It can be a numeric constant or vari- able, or a string message enclosed in hyphens. Its application is illustrated in the following example.

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62 ProgrAMMIng wITh MATLAB

EXAMPLE 3.4 An Interactive M-File Function

Problem Statement. As in Example 3.2, compute the velocity of the free-falling bungee jumper, but now use the input and disp functions for input/output.

Solution. Type the following statements in the file editor:

function freefalli % freefalli: interactive bungee velocity % freefalli interactive computation of the % free-fall velocity of an object % with second-order drag. g = 9.81; % acceleration of gravity m = input('Mass (kg): '); cd = input('Drag coefficient (kg/m): '); t = input('Time (s): '); disp(' ') disp(‘Velocity (m/s):') disp(sqrt(g * m / cd)*tanh(sqrt(g * cd / m) * t))

Save the file as freefalli.m. To invoke the function, return to the command window and type >> freefalli

Mass (kg): 68.1 Drag coefficient (kg/m): 0.25 Time (s): 12

Velocity (m/s): 50.6175

The fprintf Function. This function provides additional control over the display of information. A simple representation of its syntax is

fprintf('format', x, ...)

where format is a string specifying how you want the value of the variable x to be displayed. The operation of this function is best illustrated by examples.

A simple example would be to display a value along with a message. For instance, suppose that the variable velocity has a value of 50.6175. To display the value using eight digits with four digits to the right of the decimal point along with a message, the statement along with the resulting output would be

>> fprintf('The velocity is %8.4f m/s\n', velocity)

The velocity is 50.6175 m/s

This example should make it clear how the format string works. MATLAB starts at the left end of the string and displays the labels until it detects one of the symbols: % or \. In our example, it first encounters a % and recognizes that the following text is a format code. As in Table 3.1, the format codes allow you to specify whether numeric values are displayed in integer, decimal, or scientific format. After displaying the value of velocity, MATLAB continues displaying the character information (in our case the units: m/s) until it detects the symbol \. This tells MATLAB that the following text is a control code. As in Table 3.1, the

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3.2 InPUT-oUTPUT 63

control codes provide a means to perform actions such as skipping to the next line. If we had omitted the code \n in the previous example, the command prompt would appear at the end of the label m/s rather than on the next line as would typically be desired.

The fprintf function can also be used to display several values per line with different formats. For example,

>> fprintf('%5d %10.3f %8.5e\n',100,2*pi,pi);

100 6.283 3.14159e + 000

It can also be used to display vectors and matrices. Here is an M-file that enters two sets of values as vectors. These vectors are then combined into a matrix, which is then displayed as a table with headings:

function fprintfdemo x = [1 2 3 4 5]; y = [20.4 12.6 17.8 88.7 120.4]; z = [x;y]; fprintf(' x y\n'); fprintf('%5d %10.3f\n',z);

The result of running this M-file is

>> fprintfdemo

x y 1 20.400 2 12.600 3 17.800 4 88.700 5 120.400

3.2.1 Creating and Accessing Files

MATLAB has the capability to both read and write data files. The simplest approach in- volves a special type of binary file, called a MAT-file, which is expressly designed for implementation within MATLAB. Such files are created and accessed with the save and load commands.

TABLE 3.1 Commonly used format and control codes employed with the fprintf function.

Format Code Description

%d Integer format %e Scientific format with lowercase e %E Scientific format with uppercase E %f Decimal format %g The more compact of %e or %f

Control Code Description

\n Start new line \t Tab

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64 ProgrAMMIng wITh MATLAB

The save command can be used to generate a MAT-file holding either the entire work- space or a few selected variables. A simple representation of its syntax is

save filename var1 var2 ... varn

This command creates a MAT-file named filename.mat that holds the variables var1 through varn. If the variables are omitted, all the workspace variables are saved. The load command can subsequently be used to retrieve the file:

load filename var1 var2 ... varn

which retrieves the variables var1 through varn from filename.mat. As was the case with save, if the variables are omitted, all the variables are retrieved.

For example, suppose that you use Eq. (1.9) to generate velocities for a set of drag coefficients:

>> g = 9.81;m = 80;t = 5; >> cd = [.25 .267 .245 .28 .273]'; >> v = sqrt(g*m ./cd).*tanh(sqrt(g*cd/m)*t);

You can then create a file holding the values of the drag coefficients and the velocities with

>> save veldrag v cd

To illustrate how the values can be retrieved at a later time, remove all variables from the workspace with the clear command,

>> clear

At this point, if you tried to display the velocities you would get the result:

>> v ??? Undefined function or variable 'v'.

However, you can recover them by entering

>> load veldrag

Now, the velocities are available as can be verified by typing

>> who

Your variables are: cd v

Although MAT-files are quite useful when working exclusively within the MATLAB environment, a somewhat different approach is required when interfacing MATLAB with other programs. In such cases, a simple approach is to create text files written in ASCII format.

ASCII files can be generated in MATLAB by appending –ascii to the save command. In contrast to MAT-files where you might want to save the entire workspace, you would typically save a single rectangular matrix of values. For example,

>> A = [5 7 9 2;3 6 3 9]; >> save simpmatrix.txt –ascii

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In this case, the save command stores the values in A in 8-digit ASCII form. If you want to store the numbers in double precision, just append –ascii –double. In either case, the file can be accessed by other programs such as spreadsheets or word processors. For example, if you open this file with a text editor, you will see 5.0000000e + 000 7.0000000e + 000 9.0000000e + 000 2.0000000e + 000 3.0000000e + 000 6.0000000e + 000 3.0000000e + 000 9.0000000e + 000

Alternatively, you can read the values back into MATLAB with the load command, >> load simpmatrix.txt

Because simpmatrix.txt is not a MAT-file, MATLAB creates a double precision array named after the filename:

>> simpmatrix

simpmatrix = 5 7 9 2 3 6 3 9

Alternatively, you could use the load command as a function and assign its values to a variable as in

>> A = load('simpmatrix.txt')

The foregoing material covers but a small portion of MATLAB’s file management capabilities. For example, a handy import wizard can be invoked with the menu selections: File, Import Data. As an exercise, you can demonstrate the import wizards convenience by using it to open simpmatrix.txt. In addition, you can always consult help to learn more about this and other features.

3.3 STRUCTURED PROGRAMMING

The simplest of all M-files perform instructions sequentially. That is, the program state- ments are executed line by line starting at the top of the function and moving down to the end. Because a strict sequence is highly limiting, all computer languages include state- ments allowing programs to take nonsequential paths. These can be classified as

• Decisions (or Selection). The branching of flow based on a decision. • Loops (or Repetition). The looping of flow to allow statements to be repeated.

3.3.1 Decisions

The if Structure. This structure allows you to execute a set of statements if a logical condition is true. Its general syntax is

if condition statements end

where condition is a logical expression that is either true or false. For example, here is a simple M-file to evaluate whether a grade is passing:

function grader(grade) % grader(grade):

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% determines whether grade is passing % input: % grade = numerical value of grade (0−100) % output: % displayed message if grade >= 60 disp('passing grade') end

The following illustrates the result >> grader(95.6)

passing grade

For cases where only one statement is executed, it is often convenient to implement the if structure as a single line,

if grade > 60, disp('passing grade'), end

This structure is called a single-line if. For cases where more than one statement is imple- mented, the multiline if structure is usually preferable because it is easier to read.

Error Function. A nice example of the utility of a single-line if is to employ it for rudi- mentary error trapping. This involves using the error function which has the syntax,

error(msg)

When this function is encountered, it displays the text message msg, indicates where the error occurred, and causes the M-file to terminate and return to the command window.

An example of its use would be where we might want to terminate an M-file to avoid a division by zero. The following M-file illustrates how this could be done:

function f = errortest(x) if x == 0, error('zero value encountered'), end f = 1/x;

If a nonzero argument is used, the division would be implemented successfully as in >> errortest(10)

ans = 0.1000

However, for a zero argument, the function would terminate prior to the division and the error message would be displayed in red typeface:

>> errortest(0)

??? Error using ==> errortest at 2 zero value encountered

Logical Conditions. The simplest form of the condition is a single relational expression that compares two values as in

value1 relation value2

where the values can be constants, variables, or expressions and the relation is one of the relational operators listed in Table 3.2.

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MATLAB also allows testing of more than one logical condition by employing logical operators. We will emphasize the following:

• ~(Not). Used to perform logical negation on an expression. ~ expression

If the expression is true, the result is false. Conversely, if the expression is false, the result is true.

• & (And). Used to perform a logical conjunction on two expressions. expression1 & expression2

If both expressions evaluate to true, the result is true. If either or both expressions evaluates to false, the result is false.

• || (Or). Used to perform a logical disjunction on two expressions. expression1 || expression2

If either or both expressions evaluate to true, the result is true. Table 3.3 summarizes all possible outcomes for each of these operators. Just as for

arithmetic operations, there is a priority order for evaluating logical operations. These are from highest to lowest: ~, &, and ||. In choosing between operators of equal priority, MATLAB evaluates them from left to right. Finally, as with arithmetic operators, paren- theses can be used to override the priority order.

Let’s investigate how the computer employs the priorities to evaluate a logical expres- sion. If a = −1, b = 2, x = 1, and y = 'b', evaluate whether the following is true or false:

a * b > 0 & b = = 2 & x > 7 || ~(y > 'd')

TABLE 3.2 Summary of relational operators in MATLAB.

Example Operator Relationship

x = = 0 = = Equal unit ~ = 'm' ~ = Not equal a < 0 < Less than s > t > Greater than 3.9 < = a/3 <= Less than or equal to r >= 0 >= Greater than or equal to

TABLE 3.3 A truth table summarizing the possible outcomes for logical operators employed in MATLAB. The order of priority of the operators is shown at the top of the table.

Highest Lowest x y ~x x & y x || y

T T F T T T F F F T F T T F T F F T F F

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To make it easier to evaluate, substitute the values for the variables: −1 * 2 > 0 & 2 = = 2 & 1 > 7 || ~('b' > 'd')

The first thing that MATLAB does is to evaluate any mathematical expressions. In this example, there is only one: −1 * 2,

−2 > 0 & 2 == 2 & 1 > 7 || ~('b' > 'd')

Next, evaluate all the relational expressions −2 > 0 & 2 == 2 & 1 > 7 || ~('b' > 'd') F & T & F || ~ F

At this point, the logical operators are evaluated in priority order. Since the ~ has highest priority, the last expression (~F) is evaluated first to give

F & T & F || T

The & operator is evaluated next. Since there are two, the left-to-right rule is applied and the first expression (F & T) is evaluated:

F & F || T

The & again has highest priority F || T

Finally, the || is evaluated as true. The entire process is depicted in Fig. 3.1.

a * b > 0 & b = = 2 & x > 7 || ~ ( y > ' d ' )

−1 * 2 > 0 & 2 = = 2 & 1 > 7 || ~ ( ' b ' > ' d ' )

F & T & F || ~F

F & F ||

Substitute constants

Evaluate mathematical expressions

Evaluate relational expressions

Evaluate compound expressions

T||

−2 > 0 & 2 = = 2 & 1 > 7 || ~ ( ' b ' > ' d ' )

FIGURE 3.1 A step-by-step evaluation of a complex decision.

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The if...else Structure. This structure allows you to execute a set of statements if a logical condition is true and to execute a second set if the condition is false. Its general syntax is

if condition statements1 else statements2 end

The if...elseif Structure. It often happens that the false option of an if...else structure is another decision. This type of structure often occurs when we have more than two op- tions for a particular problem setting. For such cases, a special form of decision structure, the if...elseif has been developed. It has the general syntax

if condition1 statements1 elseif condition2 statements2 elseif condition3 statements3 · · · else statementselse end

EXAMPLE 3.5 if Structures

Problem Statement. For a scalar, the built-in MATLAB sign function returns the sign of its argument (−1, 0, 1). Here’s a MATLAB session that illustrates how it works:

>> sign(25.6)

ans = 1

>> sign(−0.776)

ans = −1

>> sign(0)

ans = 0

Develop an M-file to perform the same function.

Solution. First, an if structure can be used to return 1 if the argument is positive: function sgn = mysign(x) % mysign(x) returns 1 if x is greater than zero. if x > 0 sgn = 1; end

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This function can be run as >> mysign(25.6)

ans = 1

Although the function handles positive numbers correctly, if it is run with a negative or zero argument, nothing is displayed. To partially remedy this shortcoming, an if...else structure can be used to display –1 if the condition is false: function sgn = mysign(x) % mysign(x) returns 1 if x is greater than zero. % −1 if x is less than or equal to zero. if x > 0 sgn = 1; else sgn = −1; end

This function can be run as >> mysign(−0.776)

ans = −1

Although the positive and negative cases are now handled properly, −1 is erroneously returned if a zero argument is used. An if...elseif structure can be used to incorporate this final case: function sgn = mysign(x) % mysign(x) returns 1 if x is greater than zero. % −1 if x is less than zero. % 0 if x is equal to zero. if x > 0 sgn = 1; elseif x < 0 sgn = −1; else sgn = 0; end

The function now handles all possible cases. For example, >> mysign(0)

ans = 0

The switch Structure. The switch structure is similar in spirit to the if...elseif structure. However, rather than testing individual conditions, the branching is based on the value of a single test expression. Depending on its value, different blocks of code are implemented. In addition, an optional block is implemented if the expression takes on none of the prescribed values. It has the general syntax

switch testexpression case value1

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statements1 case value2 statements2 . . . otherwise statementsotherwise end

As an example, here is function that displays a message depending on the value of the string variable, grade.

grade = 'B'; switch grade case 'A' disp('Excellent') case 'B' disp('Good') case 'C' disp('Mediocre') case 'D' disp('Whoops') case 'F' disp('Would like fries with your order?') otherwise disp('Huh!') end

When this code was executed, the message “Good” would be displayed.

Variable Argument List. MATLAB allows a variable number of arguments to be passed to a function. This feature can come in handy for incorporating default values into your functions. A default value is a number that is automatically assigned in the event that the user does not pass it to a function.

As an example, recall that earlier in this chapter, we developed a function free fall, which had three arguments:

v = freefall(t,m,cd)

Although a user would obviously need to specify the time and mass, they might not have a good idea of an appropriate drag coefficient. Therefore, it would be nice to have the pro- gram supply a value if they omitted it from the argument list.

MATLAB has a function called nargin that provides the number of input arguments supplied to a function by a user. It can be used in conjunction with decision structures like the if or switch constructs to incorporate default values as well as error messages into your functions. The following code illustrates how this can be done for freefall:

function v = freefall2(t, m, cd) % freefall2: bungee velocity with second-order drag % v = freefall2(t,m,cd) computes the free-fall velocity % of an object with second-order drag. % input: % t = time (s) % m = mass (kg)

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% cd = drag coefficient (default = 0.27 kg/m) % output: % v = downward velocity (m/s) switch nargin case 0 error('Must enter time and mass') case 1 error('Must enter mass') case 2 cd = 0.27; end g = 9.81; % acceleration of gravity v = sqrt(g * m / cd)*tanh(sqrt(g * cd / m) * t);

Notice how we have used a switch structure to either display error messages or set the default, depending on the number of arguments passed by the user. Here is a command window session showing the results:

>> freefall2(12,68.1,0.25)

ans = 50.6175

>> freefall2(12,68.1)

ans = 48.8747

>> freefall2(12)

??? Error using ==> freefall2 at 15 Must enter mass

>> freefall2()

??? Error using ==> freefall2 at 13 Must enter time and mass

Note that nargin behaves a little differently when it is invoked in the command window. In the command window, it must include a string argument specifying the function and it returns the number of arguments in the function. For example,

>> nargin('freefall2')

ans = 3

3.3.2 Loops

As the name implies, loops perform operations repetitively. There are two types of loops, depending on how the repetitions are terminated. A for loop ends after a specified number of repetitions. A while loop ends on the basis of a logical condition.

The for...end Structure. A for loop repeats statements a specific number of times. Its general syntax is

for index = start:step:finish statements end

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The for loop operates as follows. The index is a variable that is set at an initial value, start. The program then compares the index with a desired final value, finish. If the index is less than or equal to the finish, the program executes the statements. When it reaches the end line that marks the end of the loop, the index variable is increased by the step and the program loops back up to the for statement. The process continues until the index becomes greater than the finish value. At this point, the loop terminates as the program skips down to the line immediately following the end statement.

Note that if an increment of 1 is desired (as is often the case), the step can be dropped. For example,

for i = 1:5 disp(i) end

When this executes, MATLAB would display in succession, 1, 2, 3, 4, 5. In other words, the default step is 1.

The size of the step can be changed from the default of 1 to any other numeric value. It does not have to be an integer, nor does it have to be positive. For example, step sizes of 0.2, −1, or −5, are all acceptable.

If a negative step is used, the loop will “countdown” in reverse. For such cases, the loop’s logic is reversed. Thus, the finish is less than the start and the loop terminates when the index is less than the finish. For example,

for j = 10:−1:1 disp(j) end

When this executes, MATLAB would display the classic “countdown” sequence: 10, 9, 8, 7, 6, 5, 4, 3, 2, 1.

EXAMPLE 3.6 Using a for Loop to Compute the Factorial

Problem Statement. Develop an M-file to compute the factorial.3

0! = 1 1! = 1 2! = 1 × 2 = 2 3! = 1 × 2 × 3 = 6 4! = 1 × 2 × 3 × 4 = 24 5! = 1 × 2 × 3 × 4 × 5 = 120

· · ·

Solution. A simple function to implement this calculation can be developed as function fout = factor(n) % factor(n): % Computes the product of all the integers from 1 to n. x = 1; for i = 1:n x = x * i; end fout = x; end 3 Note that MATLAB has a built-in function factorial that performs this computation.

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which can be run as >> factor(5)

ans = 120

This loop will execute 5 times (from 1 to 5). At the end of the process, x will hold a value of 5! (meaning 5 factorial or 1 × 2 × 3 × 4 × 5 = 120). Notice what happens if n = 0. For this case, the for loop would not execute, and we would get the desired result, 0! = 1.

Vectorization. The for loop is easy to implement and understand. However, for MATLAB, it is not necessarily the most efficient means to repeat statements a specific number of times. Because of MATLAB’s ability to operate directly on arrays, vectorization provides a much more efficient option. For example, the following for loop structure:

i = 0; for t = 0:0.02:50 i = i + 1; y(i) = cos(t); end

can be represented in vectorized form as t = 0:0.02:50; y = cos(t);

It should be noted that for more complex code, it may not be obvious how to vectorize the code. That said, wherever possible, vectorization is recommended.

Preallocation of Memory. MATLAB automatically increases the size of arrays every time you add a new element. This can become time consuming when you perform actions such as adding new values one at a time within a loop. For example, here is some code that sets value of elements of y depending on whether or not values of t are greater than one:

t = 0:.01:5; for i = 1:length(t) if t(i)>1 y(i) = 1/t(i); else y(i) = 1; end end

For this case, MATLAB must resize y every time a new value is determined. The following code preallocates the proper amount of memory by using a vectorized statement to assign ones to y prior to entering the loop.

t = 0:.01:5; y = ones(size(t)); for i = 1:length(t) if t(i)>1 y(i) = 1/t(i); end end

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Thus, the array is only sized once. In addition, preallocation helps reduce memory frag- mentation, which also enhances efficiency.

The while Structure. A while loop repeats as long as a logical condition is true. Its general syntax is

while condition statements end

The statements between the while and the end are repeated as long as the condition is true. A simple example is

x = 8 while x > 0 x = x − 3; disp(x) end

When this code is run, the result is x = 8 5 2 −1

The while...break Structure. Although the while structure is extremely useful, the fact that it always exits at the beginning of the structure on a false result is somewhat constraining. For this reason, languages such as Fortran 90 and Visual Basic have special structures that allow loop termination on a true condition anywhere in the loop. Although such structures are currently not available in MATLAB, their functionality can be mim- icked by a special version of the while loop. The syntax of this version, called a while... break structure, can be written as

while (1) statements if condition, break, end statements end

where break terminates execution of the loop. Thus, a single line if is used to exit the loop if the condition tests true. Note that as shown, the break can be placed in the middle of the loop (i.e., with statements before and after it). Such a structure is called a midtest loop.

If the problem required it, we could place the break at the very beginning to create a pretest loop. An example is

while (1) If x < 0, break, end x = x − 5; end

Notice how 5 is subtracted from x on each iteration. This represents a mechanism so that the loop eventually terminates. Every decision loop must have such a mechanism. Otherwise it would become a so-called infinite loop that would never stop.

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Alternatively, we could also place the if...break statement at the very end and create a posttest loop,

while (1) x = x − 5; if x < 0, break, end end

It should be clear that, in fact, all three structures are really the same. That is, de- pending on where we put the exit (beginning, middle, or end) dictates whether we have a pre-, mid- or posttest. It is this simplicity that led the computer scientists who developed Fortran 90 and Visual Basic to favor this structure over other forms of the decision loop such as the conventional while structure.

The pause Command. There are often times when you might want a program to temporar- ily halt. The command pause causes a procedure to stop and wait until any key is hit. A nice example involves creating a sequence of plots that a user might want to leisurely peruse before moving on to the next. The following code employs a for loop to create a sequence of interesting plots that can be viewed in this manner:

for n = 3:10 mesh (magic (n)) pause end

The pause can also be formulated as pause (n), in which case the procedure will halt for n seconds. This feature can be demonstrated by implementing it in conjunction with several other useful MATLAB functions. The beep command causes the computer to emit a beep sound. Two other functions, tic and toc, work together to measure elapsed time. The tic command saves the current time that toc later employs to display the elapsed time. The following code then confirms that pause (n) works as advertised complete with sound effects:

tic beep pause (5) beep toc

When this code is run, the computer will beep. Five seconds later it will beep again and display the following message:

Elapsed time is 5.006306 seconds.

By the way, if you ever have the urge to use the command pause (inf), MATLAB will go into an infinite loop. In such cases, you can return to the command prompt by typing Ctrl+c or Ctrl+Break.

Although the foregoing examples might seem a tad frivolous, the commands can be quite useful. For instance, tic and toc can be employed to identify the parts of an algorithm that consume the most execution time. Further, the Ctrl+c or Ctrl+Break key combina- tions come in real handy in the event that you inadvertently create an infinite loop in one of your M-files.

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3.3.3 Animation

There are two simple ways to animate a plot in MATLAB. First, if the computations are sufficiently quick, the standard plot function can be employed in a way that the animation can appear smooth. Here is a code fragment that indicates how a for loop and standard plotting functions can be employed to animate a plot,

% create animation with standard plot functions for j=1:n plot commands end

Thus, because we do not include hold on, the plot will refresh on each loop iteration. Through judicious use of axis commands, this can result in a smoothly changing image.

Second, there are special functions, getframe and movie, that allow you to capture a sequence of plots and then play them back. As the name implies, the getframe function captures a snapshot (pixmap) of the current axes or figure. It is usually used in a for loop to assemble an array of movie frames for later playback with the movie function, which has the following syntax:

movie (m,n,fps)

where m = the vector or matrix holding the sequence of frames constituting the movie, n = an optional variable specifying how many times the movie is to be repeated (if it is omitted, the movie plays once), and fps = an optional variable that specifies the movie’s frame rate (if it is omitted, the default is 12 frames per second). Here is a code fragment that indicates how a for loop along with the two functions can be employed to create a movie,

% create animation with getframe and movie for j = 1:n plot commands M(j) = getframe; end movie(M)

Each time the loop executes, the plot commands create an updated version of a plot, which is then stored in the vector M. After the loop terminates, the n images are then played back by movie.

EXAMPLE 3.7 Animation of Projectile Motion

Problem Statement. In the absence of air resistance, the Cartesian coordinates of a pro- jectile launched with an initial velocity (υ0) and angle (θ0) can be computed with

x = υ0 cos(θ0)t y = υ0 sin(θ0)t − 0.5gt2

where g = 9.81 m/s2. Develop a script to generate an animated plot of the projectile’s trajectory given that υ0 = 5 m/s and θ0 = 45°.

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Solution. A script to generate the animation can be written as clc,clf,clear g = 9.81; theta0 = 45*pi/180; v0 = 5; t(1) = 0;x = 0;y = 0; plot(x,y,'o','MarkerFaceColor','b','MarkerSize',8) axis([0 3 0 0.8]) M(1) = getframe; dt = 1/128; for j = 2:1000 t(j) = t(j − 1) + dt; x = v0*cos(theta0)*t(j); y = v0*sin(theta0)*t(j) − 0.5*g*t(j)^2; plot(x,y,'o','MarkerFaceColor','b','MarkerSize',8) axis([0 3 0 0.8]) M(j) = getframe; if y< = 0, break, end end pause movie(M,1)

Several features of this script bear mention. First, notice that we have fixed the ranges for the x and y axes. If this is not done, the axes will rescale and cause the animation to jump around. Second, we terminate the for loop when the projectile’s height y falls below zero. When the script is executed, two animations will be displayed (we’ve placed a pause between them). The first corresponds to the sequential generation of the frames within the loop, and the second corresponds to the actual movie. Although we cannot show the results here, the trajectory for both cases will look like Fig. 3.2. You should enter and run the foregoing script in MATLAB to see the actual animation.

FIGURE 3.2 Plot of a projectile’s trajectory.

0.8

0.7

0.6

0.5

0.4

0.3

0.2

0.1

0 0 0.5 1 1.5 2 2.5 3

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3.4 nESTIng AnD InDEnTATIon 79

3.4 NESTING AND INDENTATION

We need to understand that structures can be “nested” within each other. Nesting refers to placing structures within other structures. The following example illustrates the concept.

EXAMPLE 3.8 nesting Structures

Problem Statement. The roots of a quadratic equation f (x) = ax2 + bx + c

can be determined with the quadratic formula

x = −b ± √

_______ b2 − 4ac ______________ 2a

Develop a function to implement this formula given values of the coefficients.

Solution. Top-down design provides a nice approach for designing an algorithm to compute the roots. This involves developing the general structure without details and then refining the algorithm. To start, we first recognize that depending on whether the parameter a is zero, we will either have “special” cases (e.g., single roots or trivial values) or conventional cases using the quadratic formula. This “big-picture” version can be programmed as function quadroots(a, b, c) % quadroots: roots of quadratic equation % quadroots(a,b,c): real and complex roots % of quadratic equation % input: % a = second-order coefficient % b = first-order coefficient % c = zero-order coefficient % output: % r1 = real part of first root % i1 = imaginary part of first root % r2 = real part of second root % i2 = imaginary part of second root if a == 0 %special cases else %quadratic formula end

Next, we develop refined code to handle the “special” cases: %special cases if b ~= 0 %single root r1 = −c / b else %trivial solution disp('Trivial solution. Try again') end

And we can develop refined code to handle the quadratic formula cases: %quadratic formula d = b ^ 2 − 4 * a * c;

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if d >= 0 %real roots r1 = (−b + sqrt(d)) / (2 * a) r2 = (−b − sqrt(d)) / (2 * a) else %complex roots r1 = −b / (2 * a) i1 = sqrt(abs(d)) / (2 * a) r2 = r1 i2 = −i1 end

We can then merely substitute these blocks back into the simple “big-picture” frame- work to give the final result: function quadroots(a, b, c) % quadroots: roots of quadratic equation % quadroots(a,b,c): real and complex roots % of quadratic equation % input: % a = second-order coefficient % b = first-order coefficient % c = zero-order coefficient % output: % r1 = real part of first root % i1 = imaginary part of first root % r2 = real part of second root % i2 = imaginary part of second root if a = = 0

%special cases if b ~ = 0 %single root r1 = −c / b else %trivial solution disp('Trivial solution. Try again') end

else

%quadratic formula d = b ^ 2 − 4 * a * c; %discriminant if d >= 0 %real roots r1 = (−b + sqrt(d)) / (2 * a) r2 = (−b − sqrt(d)) / (2 * a) else %complex roots r1 = −b / (2 * a) i1 = sqrt(abs(d)) / (2 * a) r2 = r1 i2 = −i1 end

end

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3.5 PASSIng FUnCTIonS To M-FILES 81

As highlighted by the shading, notice how indentation helps to make the underlying logical structure clear. Also notice how “modular” the structures are. Here is a command window session illustrating how the function performs: >> quadroots(1,1,1)

r1 = −0.5000 i1 = 0.8660 r2 = −0.5000 i2 = −0.8660

>> quadroots(1,5,1)

r1 = −0.2087 r2 = −4.7913

>> quadroots(0,5,1)

r1 = −0.2000

>> quadroots(0,0,0)

Trivial solution. Try again

3.5 PASSING FUNCTIONS TO M-FILES

Much of the remainder of the book involves developing functions to numerically eval- uate other functions. Although a customized function could be developed for every new equation we analyzed, a better alternative is to design a generic function and pass the particular equation we wish to analyze as an argument. In the parlance of MATLAB, these functions are given a special name: function functions. Before de- scribing how they work, we will first introduce anonymous functions, which provide a handy means to define simple user-defined functions without developing a full-blown M-file.

3.5.1 Anonymous Functions

Anonymous functions allow you to create a simple function without creating an M-file. They can be defined within the command window with the following syntax:

fhandle = @(arglist) expression

where fhandle = the function handle you can use to invoke the function, arglist = a comma separated list of input arguments to be passed to the function, and expression = any single valid MATLAB expression. For example,

>> f1= @(x,y) x^2 + y^2;

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82 ProgrAMMIng wITh MATLAB

Once these functions are defined in the command window, they can be used just as other functions:

>> f1(3,4)

ans = 25

Aside from the variables in its argument list, an anonymous function can include vari- ables that exist in the workspace where it is created. For example, we could create an anonymous function f(x) = 4x2 as

>> a = 4; >> b = 2; >> f2 = @(x) a*x^b; >> f2(3)

ans = 36

Note that if subsequently we enter new values for a and b, the anonymous function does not change:

>> a = 3; >> f2(3)

ans = 36

Thus, the function handle holds a snapshot of the function at the time it was created. If we want the variables to take on values, we must recreate the function. For example, having changed a to 3,

>> f2 = @(x) a*x^b;

with the result

>> f2(3)

ans = 27

It should be noted that prior to MATLAB 7, inline functions performed the same role as anonymous functions. For example, the anonymous function developed above, f1, could be written as

>> f1 = inline('x^2 + y^2','x','y');

Although they are being phased out in favor of anonymous function, some readers might be using earlier versions, and so we thought it would be helpful to mention them. MATLAB help can be consulted to learn more about their use and limitations.

3.5.2 Function Functions

Function functions are functions that operate on other functions which are passed to it as input arguments. The function that is passed to the function function is referred to as the passed function. A simple example is the built-in function fplot, which plots the graphs of functions. A simple representation of its syntax is

fplot(func,lims)

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3.5 PASSIng FUnCTIonS To M-FILES 83

where func is the function being plotted between the x-axis limits specified by lims = [xmin xmax]. For this case, func is the passed function. This function is “smart” in that it automati- cally analyzes the function and decides how many values to use so that the plot will exhibit all the function’s features.

Here is an example of how fplot can be used to plot the velocity of the free-falling bungee jumper. The function can be created with an anonymous function:

>> vel = @(t) ... sqrt(9.81*68.1/0.25)*tanh(sqrt(9.81*0.25/68.1)*t);

We can then generate a plot from t = 0 to 12 as

>> fplot(vel,[0 12])

The result is displayed in Fig. 3.3. Note that in the remainder of this book, we will have many occasions to use MATLAB’s

built-in function functions. As in the following example, we will also be developing our own.

FIGURE 3.3 A plot of velocity versus time generated with the fplot function.

0 0 2 4 6 8 10 12

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84 ProgrAMMIng wITh MATLAB

EXAMPLE 3.9 Building and Implementing a Function Function

Problem Statement. Develop an M-file function function to determine the average value of a function over a range. Illustrate its use for the bungee jumper velocity over the range from t = 0 to 12 s:

υ (t) = √ ___

gm ___ cd tanh ( √ ____

gcd ___ m t )

where g = 9.81, m = 68.1, and cd = 0.25.

Solution. The average value of the function can be computed with standard MATLAB commands as

>> t =linspace(0,12); >> v = sqrt(9.81*68.1/0.25)*tanh(sqrt(9.81*0.25/68.1)*t); >> mean(v)

ans = 36.0870

Inspection of a plot of the function (Fig. 3.3) shows that this result is a reasonable estimate of the curve’s average height. We can write an M-file to perform the same computation: function favg = funcavg(a,b,n) % funcavg: average function height % favg = funcavg(a,b,n): computes average value % of function over a range % input: % a = lower bound of range % b = upper bound of range % n = number of intervals % output: % favg = average value of function x = linspace(a,b,n); y = func(x); favg = mean(y); end

function f = func(t) f = sqrt(9.81*68.1/0.25)*tanh(sqrt(9.81*0.25/68.1)*t); end

The main function first uses linspace to generate equally spaced x values across the range. These values are then passed to a subfunction func in order to generate the cor- responding y values. Finally, the average value is computed. The function can be run from the command window as >> funcavg (0,12,60)

ans = 36.0127

Now let’s rewrite the M-file so that rather than being specific to func, it evaluates a nonspecific function name f that is passed in as an argument: function favg = funcavg (f,a,b,n) % funcavg: average function height

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3.5 PASSIng FUnCTIonS To M-FILES 85

% favg = funcavg(f,a,b,n): computes average value % of function over a range % input: % f = function to be evaluated % a = lower bound of range % b = upper bound of range % n = number of intervals % output: % favg = average value of function x = linspace(a,b,n); y = f(x); favg = mean(y);

Because we have removed the subfunction func, this version is truly generic. It can be run from the command window as >> vel = @(t) ... sqrt(9.81*68.1/0.25)*tanh(sqrt(9.81*0.25/68.1)*t); >> funcavg(vel,0,12,60)

ans = 36.0127

To demonstrate its generic nature, funcavg can easily be applied to another case by merely passing it a different function. For example, it could be used to determine the aver- age value of the built-in sin function between 0 and 2π as

>> funcavg(@sin,0,2*pi,180)

ans = −6.3001e − 017

Does this result make sense? We can see that funcavg is now designed to evaluate any valid MATLAB expres- sion. We will do this on numerous occasions throughout the remainder of this text in a number of contexts ranging from nonlinear equation solving to the solution of differential equations.

3.5.3 Passing Parameters

Recall from Chap. 1 that the terms in mathematical models can be divided into depen- dent and independent variables, parameters, and forcing functions. For the bungee jumper model, the velocity (υ) is the dependent variable, time (t) is the independent variable, the mass (m) and drag coefficient (cd) are parameters, and the gravitational constant (g) is the forcing function. It is commonplace to investigate the behavior of such models by perform- ing a sensitivity analysis. This involves observing how the dependent variable changes as the parameters and forcing functions are varied.

In Example 3.9, we developed a function function, funcavg, and used it to determine the average value of the bungee jumper velocity for the case where the parameters were set at m = 68.1 and cd = 0.25. Suppose that we wanted to analyze the same function, but with different parameters. Of course, we could retype the function with new values for each case, but it would be preferable to just change the parameters.

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86 ProgrAMMIng wITh MATLAB

As we learned in Sec. 3.5.1, it is possible to incorporate parameters into anonymous functions. For example, rather than “wiring” the numeric values, we could have done the following:

>> m = 68.1;cd = 0.25; >> vel = @(t) sqrt(9.81*m/cd)*tanh(sqrt(9.81*cd/m)*t); >> funcavg(vel,0,12,60)

ans = 36.0127

However, if we want the parameters to take on new values, we must recreate the anony- mous function.

MATLAB offers a better alternative by adding the term varargin as the function func- tion’s last input argument. In addition, every time the passed function is invoked within the function function, the term varargin{:} should be added to the end of its argument list (note the curly brackets). Here is how both modifications can be implemented for funcavg (omit- ting comments for conciseness):

function favg = funcavg(f,a,b,n,varargin) x = linspace(a,b,n); y = f(x,varargin{:}); favg = mean(y);

When the passed function is defined, the actual parameters should be added at the end of the argument list. If we used an anonymous function, this can be done as in

>> vel = @(t,m,cd) sqrt(9.81*m/cd)*tanh(sqrt(9.81*cd/m)*t);

When all these changes have been made, analyzing different parameters becomes easy. To implement the case where m = 68.1 and cd = 0.25, we could enter

>> funcavg(vel,0,12,60,68.1,0.25)

ans = 36.0127

An alternative case, say m = 100 and cd = 0.28, could be rapidly generated by merely changing the arguments:

>> funcavg(vel,0,12,60,100,0.28)

ans = 38.9345

A new Approach for Passing Parameters. At the time of this edition’s develop- ment, MATLAB is going through a transition to a new and better way of passing parameters to function functions. As in the previous example, if the function being passed is

>> vel = @(t,m,cd) sqrt(9.81*m/cd)*tanh(sqrt(9.81*cd/m)*t);

then you invoke the function as

>> funcavg(vel,0,12,60,68.1,0.25)

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3.6 CASE STUDY 87

The Mathworks developers thought this approach was cumbersome, so they devised the following alternative

>> funcavg(@(t) vel(t,68.1,0.25),0,12,60)

Thus, the extra parameters are not strung out at the end making it clear that the parameter list is in the function.

I’ve described both the “old” and the new way of passing parameters because MATLAB will maintain the old way in functions that support in order to minimize backwards incom- patibilities. So if you have old code that has worked in the past, there is no need to go back and convert old code to the new way. For new code, however, I strongly recommend using the new way, because it is easier to read and more versatile.

3.6 CASE STUDY BUngEE JUMPEr VELoCITY

Background. In this section, we will use MATLAB to solve the free-falling bungee jumper problem we posed at the beginning of this chapter. This involves obtaining a solution of

dυ ___ dt = g − cd __ m υ∣υ∣

Recall that, given an initial condition for time and velocity, the problem involved itera- tively solving the formula,

υi+1 = υi + dυi ___ dt Δt

Now also remember that to attain good accuracy, we would employ small steps. Therefore, we would probably want to apply the formula repeatedly to step out from our initial time to attain the value at the final time. Consequently, an algorithm to solve the problem would be based on a loop.

Solution. Suppose that we started the computation at t = 0 and wanted to predict velocity at t = 12 s using a time step of Δt = 0.5 s. We would therefore need to apply the iterative equation 24 times—that is,

n = 12 ___ 0.5 = 24

where n = the number of iterations of the loop. Because this result is exact (i.e., the ratio is an integer), we can use a for loop as the basis for the algorithm. Here’s an M-file to do this including a subfunction defining the differential equation:

function vend = velocity1(dt, ti, tf, vi) % velocity1: Euler solution for bungee velocity % vend = velocity1(dt, ti, tf, vi) % Euler method solution of bungee % jumper velocity % input: % dt = time step (s) % ti = initial time (s) % tf = final time (s) % vi = initial value of dependent variable (m/s)

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3.6 CASE STUDY continued

% output: % vend = velocity at tf (m/s) t = ti; v = vi; n = (tf − ti) / dt; for i = 1:n dvdt = deriv(v); v = v + dvdt * dt; t = t + dt; end vend = v; end

function dv = deriv(v) dv = 9.81 − (0.25 / 68.1) * v*abs(v); end

This function can be invoked from the command window with the result:

>> velocity1(0.5,0,12,0)

ans = 50.9259

Note that the true value obtained from the analytical solution is 50.6175 (Exam ple 3.1). We can then try a much smaller value of dt to obtain a more accurate numerical result:

>> velocity1(0.001,0,12,0)

ans = 50.6181

Although this function is certainly simple to program, it is not foolproof. In particu- lar, it will not work if the computation interval is not evenly divisible by the time step. To cover such cases, a while . . . break loop can be substituted in place of the shaded area (note that we have omitted the comments for conciseness):

function vend = velocity2(dt, ti, tf, vi) t = ti; v = vi; h = dt; while(1) if t + dt > tf, h = tf − t; end dvdt = deriv(v); v = v + dvdt * h; t = t + h; if t >= tf, break, end end vend = v; end

function dv = deriv(v) dv = 9.81 − (0.25 / 68.1) * v*abs(v); end

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3.6 CASE STUDY 89

-3.6 CASE STUDY continued

As soon as we enter the while loop, we use a single line if structure to test whether adding t + dt will take us beyond the end of the interval. If not (which would usually be the case at first), we do nothing. If so, we would shorten up the interval—that is, we set the variable step h to the interval remaining: tf – t. By doing this, we guarantee that the last step falls exactly on tf. After we implement this final step, the loop will terminate because the condition t >= tf will test true.

Notice that before entering the loop, we assign the value of the time step dt to another variable h. We create this dummy variable so that our routine does not change the given value of dt if and when we shorten the time step. We do this in anticipation that we might need to use the original value of dt somewhere else in the event that this code were inte- grated within a larger program.

If we run this new version, the result will be the same as for the version based on the for loop structure:

>> velocity2(0.5,0,12,0)

ans = 50.9259

Further, we can use a dt that is not evenly divisible into tf – ti:

>> velocity2(0.35,0,12,0)

ans = 50.8348

We should note that the algorithm is still not foolproof. For example, the user could have mistakenly entered a step size greater than the calculation interval (e.g., tf − ti = 5 and dt = 20). Thus, you might want to include error traps in your code to catch such errors and then allow the user to correct the mistake.

As a final note, we should recognize that the foregoing code is not generic. That is, we have designed it to solve the specific problem of the velocity of the bungee jumper. A more generic version can be developed as

function yend = odesimp(dydt, dt, ti, tf, yi) t = ti; y = yi; h = dt; while (1) if t + dt > tf, h = tf − t; end y = y + dydt(y) * h; t = t + h; if t >= tf, break, end end yend = y;

Notice how we have stripped out the parts of the algorithm that were specific to the bungee example (including the subfunction defining the differential equation) while keep- ing the essential features of the solution technique. We can then use this routine to solve

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90 ProgrAMMIng wITh MATLAB

3.6 CASE STUDY continued

the bungee jumper example, by specifying the differential equation with an anonymous function and passing its function handle to odesimp to generate the solution

>> dvdt=@(v) 9.81−(0.25/68.1)*v*abs(v); >> odesimp(dvdt,0.5,0,12,0)

ans = 50.9259

We could then analyze a different function without having to go in and modify the M-file. For example, if y = 10 at t = 0, the differential equation dy∕dt = −0.1y has the analytical solution y = 10e−0.1t. Therefore, the solution at t = 5 would be y(5) = 10e−0.1(5) = 6.0653. We can use odesimp to obtain the same result numerically as in

>> odesimp(@(y) −0.1*y,0.005,0,5,10)

ans = 6.0645

Finally, we can use varargin and the new way of passing parameters to develop a final and superior version. To do this, the odesimp function is first modified by adding the high- lighted code

function yend = odesimp2(dydt, dt, ti, tf, yi, varargin) t = ti; y = yi; h = dt; while (1) if t + dt > tf, h = tf − t; end y = y + dydt(y, varargin{:}) * h; t = t + h; if t >= tf, break, end end yend = y;

Then, we can develop a script to perform the computation,

clc format compact dvdt=@(v,cd,m) 9.81−(cd/m)*v*abs(v); odesimp2(@(v) dvdt(v,0.25,68.1),0.5,0,12,0)

which yields the correct result

ans = 50.9259

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ProBLEMS 91

PROBLEMS

3.1 Figure P3.1 shows a cylindrical tank with a conical base. If the liquid level is quite low, in the conical part, the volume is simply the conical volume of liquid. If the liquid level is midrange in the cylindrical part, the total volume of liquid includes the filled conical part and the partially filled cylindrical part.

Use decisional structures to write an M-file to compute the tank’s volume as a function of given values of R and d. Design the function so that it returns the volume for all cases

FIGURE P3.1

where the depth is less than 3R. Return an error message (“Overtop”) if you overtop the tank—that is, d > 3R. Test it with the following data:

R 0.9 1.5 1.3 1.3 d 1 1.25 3.8 4.0

Note that the tank’s radius is R. 3.2 An amount of money P is invested in an account where interest is compounded at the end of the period. The future worth F yielded at an interest rate i after n periods may be determined from the following formula:

F = P(1 + i)n

Write an M-file that will calculate the future worth of an investment for each year from 1 through n. The input to the function should include the initial investment P, the interest rate i (as a decimal), and the number of years n for which the future worth is to be calculated. The output should consist of a table with headings and columns for n and F. Run the program for P = $100,000, i = 0.05, and n = 10 years. 3.3 Economic formulas are available to compute annual payments for loans. Suppose that you borrow an amount of money P and agree to repay it in n annual payments at

an interest rate of i. The formula to compute the annual payment A is

A = P i (1 + i ) n ___________ (1 + i )n − 1

Write an M-file to compute A. Test it with P = $100,000 and an interest rate of 3.3% (i = 0.033). Compute results for n = 1, 2, 3, 4, and 5 and display the results as a table with head- ings and columns for n and A. 3.4 The average daily temperature for an area can be ap- proximated by the following function:

T = Tmean + (Tpeak − Tmean) cos(ω(t − tpeak))

where Tmean = the average annual temperature, Tpeak = the peak temperature, ω = the frequency of the annual varia- tion (= 2π/365), and tpeak = day of the peak temperature (≅205 d). Parameters for some U.S. towns are listed here:

City Tmean (°C) Tpeak (°C)

Miami, FL 22.1 28.3 Yuma, AZ 23.1 33.6 Bismarck, ND 5.2 22.1 Seattle, WA 10.6 17.6 Boston, MA 10.7 22.9

Develop an M-file that computes the average temperature between two days of the year for a particular city. Test it for (a) January–February in Yuma, AZ (t = 0 to 59) and (b) July–August temperature in Seattle, WA (t = 180 to 242). 3.5 The sine function can be evaluated by the following infinite series:

sin x = x − x 3 __ 3! +

x5 __ 5! − · · ·

Create an M-file to implement this formula so that it com- putes and displays the values of sin x as each term in the series is added. In other words, compute and display in sequence the values for

sin x = x

sin x = x − x 3 __ 3!

sin x = x − x 3 __ 3! +

x5 __ 5!

...

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92 ProgrAMMIng wITh MATLAB

up to the order term of your choosing. For each of the pre- ceding, compute and display the percent relative error as

%error = true − series approximation _______________________ true × 100%

As a test case, employ the program to compute sin(0.9) for up to and including eight terms—that is, up to the term x15∕15!. 3.6 Two distances are required to specify the location of a point relative to an origin in two-dimensional space (Fig. P3.6):

• The horizontal and vertical distances (x, y) in Cartesian coordinates.

• The radius and angle (r, θ ) in polar coordinates.

It is relatively straightforward to compute Cartesian coor- dinates (x, y) on the basis of polar coordinates (r, θ ). The reverse process is not so simple. The radius can be computed by the following formula:

r = √ ______

x2 + y2

If the coordinates lie within the first and fourth coordi- nates (i.e., x > 0), then a simple formula can be used to compute θ:

θ = tan−1 ( y __ x )

The difficulty arises for the other cases. The following table summarizes the possibilities:

x y θ

<0 >0 tan−1( y∕x) + π <0 <0 tan−1(y∕x) − π <0 =0 π =0 >0 π∕2 =0 <0 −π2 =0 =0 0

Write a well-structured M-file using if...elseif structures to calculate r and θ as a function of x and y. Express the final results for θ in degrees. Test your program by evaluating the following cases:

x y r θ

2 0 2 1 0 3

−3 1 −2 0 −1 −2 0 0 0 −2 2 2

3.7 Develop an M-file to determine polar coordinates as described in Prob. 3.6. However, rather than designing the function to evaluate a single case, pass vectors of x and y. Have the function display the results as a table with columns for x, y, r, and θ. Test the program for the cases outlined in Prob. 3.6. 3.8 Develop an M-file function that is passed a numeric grade from 0 to 100 and returns a letter grade according to the scheme:

Letter Criteria

A 90 ≤ numeric grade ≤ 100 B 80 ≤ numeric grade < 90 C 70 ≤ numeric grade < 80 D 60 ≤ numeric grade < 70 F numeric grade < 60

The first line of the function should be

function grade = lettergrade(score) FIGURE P3.6

x θ

III

III IV

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ProBLEMS 93

Design the function so that it displays an error message and terminates in the event that the user enters a value of score that is less than zero or greater than 100. Test your function with 89.9999, 90, 45, and 120. 3.9 Manning’s equation can be used to compute the velocity of water in a rectangular open channel:

U = √ __

S ____ n ( BH _______ B + 2H ) 2∕3

where U = velocity (m/s), S = channel slope, n = roughness coefficient, B = width (m), and H = depth (m). The follow- ing data are available for five channels:

n S B H

0.036 0.0001 10 2 0.020 0.0002 8 1 0.015 0.0012 20 1.5 0.030 0.0007 25 3 0.022 0.0003 15 2.6

Write an M-file that computes the velocity for each of these channels. Enter these values into a matrix where each column represents a parameter and each row represents a channel. Have the M-file display the input data along with the computed velocity in tabular form where velocity is the fifth column. Include headings on the table to label the columns. 3.10 A simply supported beam is loaded as shown in Fig. P3.10. Using singularity functions, the displacement along the beam can be expressed by the equation:

uy(x) = −5 ___ 6 [⟨x − 0⟩

4 − ⟨x − 5⟩4] + 15 ___ 6 ⟨x − 8⟩ 3

+ 75 ⟨x − 7⟩2 + 57 ___ 6 x 3 − 238.25x

By definition, the singularity function can be expressed as follows:

⟨x − a⟩n = { (x − a)n when x > a } 0 when x ≤ a Develop an M-file that creates a plot of displacement (dashed line) versus distance along the beam, x. Note that x = 0 at the left end of the beam. 3.11 The volume V of liquid in a hollow horizontal cylinder of radius r and length L is related to the depth of the liquid h by

V = [ r2 cos–1 ( r – h ____ r ) − (r − h) √ _______

2rh − h2 ] L Develop an M-file to create a plot of volume versus depth. Here are the first few lines:

function cylinder(r, L, plot_title) % volume of horizontal cylinder % inputs: % r = radius % L = length % plot_title = string holding plot title

Test your program with

>> cylinder(3,5,... 'Volume versus depth for horizontal... cylindrical tank')

3.12 Develop a vectorized version of the following code:

tstart = 0; tend = 20; ni = 8; t(1) = tstart; y(1) = 12 + 6*cos(2*pi*t(1)/(tend−tstart)); for i = 2:ni+1 t(i) = t(i–1)+(tend-tstart)/ni; y(i) = 12 + 6*cos(2*pi*t(i)/ ...

(tend–tstart)); end

20 kips/ft

150 kip-ft 15 kips

5ʹ 2ʹ 1ʹ 2ʹ

FIGURE P3.10

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94 ProgrAMMIng wITh MATLAB

3.13 The “divide and average” method, an old-time method for approximating the square root of any positive number a, can be formulated as

x = x + a∕x _______ 2

Write a well-structured M-file function based on the while...break loop structure to implement this algorithm. Use proper indentation so that the structure is clear. At each step estimate the error in your approximation as

ε = | xnew − xold _________ xnew | Repeat the loop until ε is less than or equal to a specified value. Design your program so that it returns both the re- sult and the error. Make sure that it can evaluate the square root of numbers that are equal to and less than zero. For the latter case, display the result as an imaginary number. For example, the square root of −4 would return 2i. Test your program by evaluating a = 0, 2, 10 and −4 for ε = 1 × 10−4. 3.14 Piecewise functions are sometimes useful when the re- lationship between a dependent and an independent variable cannot be adequately represented by a single equation. For example, the velocity of a rocket might be described by

10t2 − 5t 0 ≤ t ≤ 8 624 − 3t 8 ≤ t ≤ 16

υ(t) = { 36t + 12(t − 16)2 16 ≤ t ≤ 26 2136e−0.1(t−26) t > 26 0 otherwise

Develop an M-file function to compute υ as a function of t. Then, develop a script that uses this function to generate a plot of υ versus t for t = −5 to 50. 3.15 Develop an M-file function called rounder to round a number x to a specified number of decimal digits, n. The first line of the function should be set up as

function xr = rounder(x, n)

Test the program by rounding each of the following to 2 dec- imal digits: x = 477.9587, −477.9587, 0.125, 0.135, −0.125, and −0.135. 3.16 Develop an M-file function to determine the elapsed days in a year. The first line of the function should be set up as

function nd = days(mo, da, leap)

where mo = the month (1–12), da = the day (1–31), and leap = (0 for non–leap year and 1 for leap year). Test it for January 1, 1997, February 29, 2004, March 1, 2001, June 21, 2004,

and December 31, 2008. Hint: A nice way to do this com- bines the for and the switch structures. 3.17 Develop an M-file function to determine the elapsed days in a year. The first line of the function should be set up as

function nd = days(mo, da, year)

where mo = the month (1–12), da = the day (1–31), and year = the year. Test it for January 1, 1997, February 29, 2004, March 1, 2001, June 21, 2004, and December 31, 2008. 3.18 Develop a function function M-file that returns the difference between the passed function’s maximum and minimum value given a range of the independent variable. In addition, have the function generate a plot of the function for the range. Test it for the following cases: (a) f(t) = 8e−0.25tsin(t − 2) from t = 0 to 6π. (b) f(x) = e4xsin(1∕x) from x = 0.01 to 0.2. (c) The built-in humps function from x = 0 to 2. 3.19 Modify the function function odesimp developed at the end of Sec. 3.6 so that it can be passed the arguments of the passed function. Test it for the following case:

>> dvdt = @(v,m,cd) 9.81−(cd/m)*v^2; >> odesimp(dvdt,0.5,0,12,−10,70,0.23)

3.20 A Cartesian vector can be thought of as representing magnitudes along the x-, y-, and z-axes multiplied by a unit vector (i, j, k). For such cases, the dot product of two of these vectors {a} and {b} corresponds to the product of their mag- nitudes and the cosine of the angle between their tails as in

{a}·{b} = ab cosθ

The cross product yields another vector, {c} = {a} × {b}, which is perpendicular to the plane defined by {a} and {b} such that its direction is specified by the right-hand rule. Develop an M-file function that is passed two such vectors and returns θ, {c} and the magnitude of {c}, and generates a three-dimensional plot of the three vectors {a}, {b}, and {c) with their origins at zero. Use dashed lines for {a} and {b} and a solid line for {c}. Test your function for the fol- lowing cases:

(a) a = [6 4 2]; b = [2 6 4]; (b) a = [3 2 −6]; b = [4 −3 1]; (c) a = [2 −2 1]; b = [4 2 −4]; (d) a = [−1 0 0]; b = [0 −1 0];

3.21 Based on Example 3.7, develop a script to produce an animation of a bouncing ball where υ0 = 5 m/s and θ0 = 50°. To do this, you must be able to predict exactly when the ball hits the ground. At this point, the direction changes (the new angle will equal the negative of the angle at impact), and the

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ProBLEMS 95

% precomputations d = 2 * r; Ta = TC + 273.15; Ab = pi / 4 * d ^ 2; Vb = pi / 6 * d ^ 3; rhog = P / Rgas / Ta; mG = Vb * rhog; FB = Vb * rhoa * g; FG = mG * g; cdp = rhoa * Ab * CD / 2;

% compute times, velocities and elevations [t,y] = Balloon(FB, FG, mG, cdp, mP, md, zd, ti,vi,zi,tf,dt);

% Display results Your code to display a nice labeled table of times, velocities, and elevations

% Plot results Your code to create a nice labeled plot of velocity and elevation versus time.

Your function should be structured like:

function [tout,yout]=Balloon(FB, FG, mG, cdp, mP, md, zd, ti,vi,zi,tf,dt) global g

% balloon % function [tout,yout]=Balloon(FB, FG, mG, cdp, mP1, md, zd, ti,vi,zi,tf,dt) % Function to generate solutions of vertical velocity and elevation % versus time with Euler’s method for a hot air balloon % Input: % FB = buoyancy force (N) % FG = gravity force (N) % mG = mass (kg) % cdp=dimensional drag coefficient % mP= mass of payload (kg) % md=mass jettisoned (kg) % zd=elevation at which mass is jettisoned (m) % ti = initial time (s) % vi=initial velocity (m/s) % zi=initial elevation (m) % tf = final time (s) % dt=integration time step (s) % Output: % tout = vector of times (s) % yout[:,1] = velocities (m/s) % yout[:,2] = elevations (m) % Code to implement Euler’s method to compute output and plot results

3.25 A general equation for a sinusoid can be written as

y(t) = _ y + Δy sin (2πft − ϕ)

velocity will decrease in magnitude to reflect energy loss due to the collision of the ball with the ground. The change in velocity can be quantified by the coefficient of restitution CR which is equal to the ratio of the velocity after to the velocity before impact. For the present case, use a value of CR = 0.8. 3.22 Develop a function to produce an animation of a par- ticle moving in a circle in Cartesian coordinates based on radial coordinates. Assume a constant radius, r, and allow the angle, θ, to increase from zero to 2π in equal increments. The function’s first lines should be

function phasor(r, nt, nm) % function to show the orbit of a phasor % r = radius % nt = number of increments for theta % nm = number of movies

Test your function with

phasor(1, 256, 10)

3.23 Develop a script to produce a movie for the butterfly plot from Prob. 2.22. Use a particle located at the x-y coordi- nates to visualize how the plot evolves in time. 3.24 Develop a MATLAB script to compute the velocity, υ, and position, z, of a hot air balloon as described in Prob. 1.28. Perform the calculation from t = 0 to 60 s with a step size of 1.6 s. At z = 200 m, assume that part of the payload (100 kg) is dropped out of the balloon. Your script should be struc- tured like:

% YourFullName % Hot Air Balloon Script

clear,clc,clf

g = 9.81; global g

% set parameters r = 8.65; % balloon radius CD = 0.47; % dimensionless drag coefficient mP = 265; % mass of payload P = 101300; % atmospheric pressure Rgas = 287; % Universal gas constant for dry air TC = 100; % air temperature rhoa = 1.2; % air density zd = 200; % elevation at which mass is jettisoned md = 100; % mass jettisoned ti = 0; % initial time (s) tf = 60; % final time (s) vi = 0; % initial velocity zi = 0; % initial elevation dt = 1.6; % integration time step

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96 ProgrAMMIng wITh MATLAB

Step 7: Compute new values for m = m/2 and n = (300 + n)/2, and then go to Step 9.

Step 8: Compute new values for m = (300 + m)/2 and n = (300 + n)/2.

Step 9: If i is less than 1000 then go to Step 10. Otherwise, go to Step 11.

Step 10: Plot a point at the coordinate,(m,n). Step 11: Terminate ii loop. Step 12: Set hold off.

Develop a MATLAB script for this algorithm using for and if structures. Run it for the following two cases (a) m = 2 and n = 1 and (b) m = 100 and n = 200. 3.27 Write a well-structured MATLAB function procedure named Fnorm to calculate the Frobenius norm of an m×n matrix,

∥A∥f = √ _________

∑ i=1

∑ j=1

a i,j 2

Here is a script that uses the function A = [5 7 9; 1 8 4; 7 6 2]; Fn = Fnorm(A)

Here is the first line of the function function Norm = Fnorm(x)

Develop two versions of the function: (a) using nested for loops and (b) using sum functions. 3.28 The pressure and temperature of the atmosphere are constantly changing depending on a number of factors in- cluding altitude, latitude/longitude, time of day, and season. To take all these variations into account when considering the design and performance of flight vehicles is impracti- cal. Therefore, a standard atmosphere is frequently used to provide engineers and scientists with a common reference for their research and development. The International Stan- dard Atmosphere is one such model of how conditions of the earth’s atmosphere change over a wide range of altitudes or elevations. The following table shows values of temperature and pressure at selected altitudes.

where y = the dependent variable, _ y = the mean value, Δy =

the amplitude, f = the ordinary frequency (i.e., the number of oscillations that occur each unit of time), t = the independent variable (in this case time), and ϕ = phase shift. Develop a MATLAB script to generate a 5 panel vertical plot to il- lustrate how the function changes as the parameters change. On each plot display the simple sine wave, y(t) = sin(2πt), as a red line. Then, add the following functions to each of the 5 panels as black lines:

subplot Function Title

5,1,1 y(t) = 1 + sin(2πt) (a) Effect of mean 5,1,2 y(t) = 2sin(2πt) (b) Effect of amplitude 5,1,3 y(t) = sin(4πt) (c) Effect of frequency 5,1,4 y(t) = sin(2πt − π/4) (d) Effect of phase shift 5,1,5 y(t) = cos(2πt −π/2) (e) Relationship of sine to cosine

Employ a range of t = 0 to 2π and scale each subplot so that the abscissa goes from 0 to 2π and the ordinate goes from −2 to 2. Include the titles with each subplot, label each subplot’s ordinate as 'f(t)', and the bottom plot’s abscissa as 't'. 3.26 A fractal is a curve or geometric figure, each part of which has the same statistical character as the whole. Fractals are useful in modeling structures (such as eroded coastlines or snowflakes) in which similar patterns recur at progressively smaller scales, and in describing partly random or chaotic phenomena such as crystal growth, fluid turbulence, and gal- axy formation. Devaney (1990) has written a nice little book that includes a simple algorithm to create an interesting fractal pattern. Here is a step-by-step description of this algorithm:

Step 1: Assign value to m and n and set hold on. Step 2: Start a for loop to iterate over ii = 1:100000 Step 3: Compute a random number, q = 3*rand(1) Step 4: If the value of q is less than 1 go to Step 5. Otherwise

go to Step 6. Step 5: Compute new values for m = m/2 and n = n/2 and

then go to Step 9. Step 6: If the value of q is less than 2 go to Step 7. Otherwise

go to Step 8.

Layer Index, i

Layer Name

Base Geopotential Altitude Above MSL, h (km)

Lapse Rate (ºC/km)

Base Temperature T (ºC)

Base Pressure, p (Pa)

1 Troposphere 0 −6.5 15 101,325 2 Tropopause 11 0 −56.5 22,632 3 Stratosphere 20 1 −56.5 5474.9 4 Stratosphere 32 2.8 −44.5 868.02 5 Stratopause 47 0 −2.5 110.91 6 Mesosphere 51 −2.8 −2.5 66.939 7 Mesosphere 71 −2.0 −58.5 3.9564 8 Mesopause 84.852 — −86.28 0.3734

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The temperature at each altitude can then be computed as

T(h) = Ti + γi (h − hi ) hi < h ≤ hi+1 where T(h) = temperature at altitude h (°C), Ti = the base temperature for layer i (°C), γi = lapse rate or the rate at which atmospheric temperature decreases linearly with increase in altitude for layer i (°C/km), and hi = base geo- potential altitude above mean sea level (MSL) for layer i. The pressure at each altitude can then be computed as

p(h) = pi + pi+1 − pi ________ hi+1 − hi

(h − hi)

where p(h) = pressure at altitude h(Pa ≡ N/m2), pi = the base pressure for layer i (Pa). The density, ρ(kg/m3), can then be calculated according to a molar form of the ideal gas law:

ρ = pM ____ RTa

where M = molar mass (≅ 0.0289644 kg/mol), R = the uni- versal gas constant (8.3144621 J/(mol · K), and Ta = absolute temperature (K) = T + 273.15.

Develop a MATLAB function, StdAtm, to deter- mine values of the three properties for a given altitude. If the user requests a value outside the range of altitudes, have the function display an error message and terminate the application. Use the following script as the starting point to create a 3-panel plot of altitude versus the properties.

% Script to generate a plot of temperature, pressure and density % for the International Standard Atmosphere clc,clf h = [0 11 20 32 47 51 71 84.852]; gamma = [−6.5 0 1 2.8 0 −2.8 −2]; T = [15 −56.5 −56.5 -44.5 −2.5 −2.5 −58.5 −86.28]; p = [101325 22632 5474.9 868.02 110.91 66.939 3.9564 0.3734]; hint = [0:0.1:84.852]; for i = 1:length(hint) [Tint(i),pint(i),rint(i)] = StdAtm(h,T,p,gamma,hint (i)); end

% Create plot % Function call to test error trap [Tint(i),pint(i),rint(i)]=StdAtm(h,T,p,gamma,85);

3.29 Develop a MATLAB function to convert a vec- tor of temperatures from Celsius to Fahrenheit and vice versa. Test it with the following data for the average monthly temperatures at Death Volley, CA and at the South Pole.

Use the following script as the starting point to gener- ate a 2-panel stacked plot of the temperatures versus day for both of the sites with the Celsius time series at the top and the Fahrenheit at the bottom. If the user requests a unit other than 'C' or 'F', have the function display an error message and terminate the application.

% Script to generate stacked plots of temperatures versus time % for Death Valley and the South Pole with Celsius time series % on the top plot and Fahrenheit on the bottom.

clc,clf t=[15 45 75 105 135 165 195 225 255 285 315 345]; TFDV = [54 60 69 77 87 96 102 101 92 78 63 52]; TCSP = [−27 −40 −53 −56 −57 −57 −59 −59 −59 −50 −38 −27]; TCDV=TempConv(TFDV,'C'); TFSP=TempConv(TCSP,'F');

% Create plot

% Test of error trap TKSP=TempConv(TCSP,'K');

3.30 Because there are only two possibilities, as in Prob. 3.29, it’s relatively easy to convert between Cel- sius and Fahrenheit temperature units. Because there are many more units in common use, converting between pressure units is more challenging. Here are some of the possibilities along with the number of Pascals represented by each:

Day

Death Valley °F

South Pole °C

15 54 −27 45 60 −40 75 69 −53 105 77 −56 135 87 −57 165 96 −57 195 102 −59 225 101 −59 255 92 −59 285 78 −50 315 63 −38 345 52 −27

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98 ProgrAMMIng wITh MATLAB

3. Have the user enter the given value of pressure, Pi. 4. Have the user select the desired units by entering the

value of j. If the user enters a correct value within the range, 1–12, continue to step 5.

If the user enters a value outside the range, display an error message and repeat step 4.

5. Use the formula to convert the quantity in input units to its value in the desired output units.

6. Display the original quantity and units and the output quantity and units.

7. Ask if another output result, for the same input, is desired.

If yes, go back to step 4 and continue from there. If no, go on to step 8.

8. Ask if another conversion is desired.

If yes, go back to step 2 and continue from there. If no, end of algorithm.

Develop a well-structured MATLAB script using loop and if structures to implement this algorithm. Test it with the following:

(a) Duplicate the hand calculation example to ensure that you get about 420.304 atm for an input of 28.6 psi.

(b) Try to enter a choice code of i = 13. Does the program trap that error and allow you to correct it? If not, it should. Now, try a choice code of the letter Q. What happens?

(c) For the yes/no questions in the program, try a letter other than y or n. Does the program scold you and allow you to correct your response? What happens if you enter a number like 1.3 as a response? What happens if you just press the Enter key with no response?

The information in the table can be used to implement a conversion calculation. One way to do this is to store both the units and the corresponding number of the Pascals in individual arrays with the subscripts corresponding to each entry. For example,

U(1) = 'psi' C(1)=6894.76 U(2) = 'atm' C(2)=101325. · · · · · ·

The conversion from one unit to another could then be com- puted with the following general formula:

Pd = Cj __ Ci

where Pg = given pressure, Pd = desired pressure, j = the index of the desired unit, and i = the index of the given unit. As an example, to convert a tire pressure of say 28.6 psi to atm, we would use

Pd = C2 ___ C1

Pg = 101325. atm /Pa _____________ 6894.76 psi /Pa 28.6 psi = 420.304 atm

So we see that the conversion from one unit to another in- volves first determining the indices corresponding to the given and the desired units, and then implementing the conversion equation. Here is a step-by-step algorithm to do this: 1. Assign the values of the unit, U, and conversion, C, arrays. 2. Have the user select the input units by entering the

value of i.

If the user enters a correct value within the range, 1–12, continue to step 3. If the user enters a value outside the range, display an error message and repeat step 2.

Index, i Unit, Ui Description or Usage # of Pa, Ci

1 psi Tire pressure, above ambient pressure 6894.76 2 atm Used for high pressure experiments 101,325 3 inHg Atmospheric pressure given by weatherperson 3376.85 4 kg/cm2 Europe metric unit in situations where psi used in U.S. 98,066.5 5 inH2O Used in heating/ventilating systems in buildings 248.843 6 Pa Standard SI (metric) unit of pressure, 1 N/m2 1 7 bar Frequented used by meteorologists 100,000 8 dyne/cm2 Older scientific pressure unit from the CGS system 0.1 9 ftH2O American and English low value pressure unit. 2988.98 10 mmHg Used in laboratory pressure measurements 133.322 11 torr Same as 1 mmHg but used in vacuum measurement 133.322 12 ksi Used in structural engineering 6,894,760

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CHAPTER OBJECTIVES

Knowledge and understanding are prerequisites for the effective implementation of any tool.

No matter how impressive your tool chest, you will be hard-pressed to repair a car if you do not understand how it works.

• This is the first chapter objectives entry. • Second chapter objective entry, the entries use ic/lc per manuscript, the first and

last entry have space above or below, middle entries do not. • Third chapter entry copy goes here.

4 Roundoff and Truncation Errors

CHAPTER OBJECTIVES

The primary objective of this chapter is to acquaint you with the major sources of errors involved in numerical methods. Specific objectives and topics covered are

• Understanding the distinction between accuracy and precision. • Learning how to quantify error. • Learning how error estimates can be used to decide when to terminate an iterative

calculation. • Understanding how roundoff errors occur because digital computers have a

limited ability to represent numbers. • Understanding why floating-point numbers have limits on their range and

precision. • Recognizing that truncation errors occur when exact mathematical formulations

are represented by approximations. • Knowing how to use the Taylor series to estimate truncation errors. • Understanding how to write forward, backward, and centered finite-difference

approximations of first and second derivatives. • Recognizing that efforts to minimize truncation errors can sometimes increase

roundoff errors.

YOU’VE GOT A PROBLEM

In Chap. 1, you developed a numerical model for the velocity of a bungee jumper. To solve the problem with a computer, you had to approximate the derivative of velocity with a finite difference: dυ ___ dt ≅

Δυ ___ Δt = υ(ti+1) − υ(ti ) ___________ ti+1 − ti

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100 Roundoff and TRuncaTIon ERRoRs

Thus, the resulting solution is not exact—that is, it has error. In addition, the computer you use to obtain the solution is also an imperfect tool. Be-

cause it is a digital device, the computer is limited in its ability to represent the magnitudes and precision of numbers. Consequently, the machine itself yields results that contain error.

So both your mathematical approximation and your digital computer cause your re- sulting model prediction to be uncertain. Your problem is: How do you deal with such uncertainty? In particular, is it possible to understand, quantify, and control such errors in order to obtain acceptable results? This chapter introduces you to some approaches and concepts that engineers and scientists use to deal with this dilemma.

4.1 ERRORS

Engineers and scientists constantly find themselves having to accomplish objectives based on uncertain information. Although perfection is a laudable goal, it is rarely if ever at- tained. For example, despite the fact that the model developed from Newton’s second law is an excellent approximation, it would never in practice exactly predict the jumper’s fall. A variety of factors such as winds and slight variations in air resistance would result in deviations from the prediction. If these deviations are systematically high or low, then we might need to develop a new model. However, if they are randomly distributed and tightly grouped around the prediction, then the deviations might be considered negligible and the model deemed adequate. Numerical approximations also introduce similar discrepancies into the analysis.

This chapter covers basic topics related to the identification, quantification, and mini- mization of these errors. General information concerned with the quantification of error is reviewed in this section. This is followed by Sections 4.2 and 4.3, dealing with the two major forms of numerical error: roundoff error (due to computer approximations) and truncation error (due to mathematical approximations). We also describe how strategies to reduce truncation error sometimes increase roundoff. Finally, we briefly discuss errors not directly connected with the numerical methods themselves. These include blunders, model errors, and data uncertainty.

4.1.1 Accuracy and Precision

The errors associated with both calculations and measurements can be characterized with regard to their accuracy and precision. Accuracy refers to how closely a computed or mea- sured value agrees with the true value. Precision refers to how closely individual computed or measured values agree with each other.

These concepts can be illustrated graphically using an analogy from target practice. The bullet holes on each target in Fig. 4.1 can be thought of as the predictions of a numeri- cal technique, whereas the bull’s-eye represents the truth. Inaccuracy (also called bias) is defined as systematic deviation from the truth. Thus, although the shots in Fig. 4.1c are more tightly grouped than in Fig. 4.1a, the two cases are equally biased because they are both centered on the upper left quadrant of the target. Imprecision (also called uncertainty), on the other hand, refers to the magnitude of the scatter. Therefore, although Fig. 4.1b and d are equally accurate (i.e., centered on the bull’s-eye), the latter is more precise because the shots are tightly grouped.

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4.1 ERRoRs 101

Numerical methods should be sufficiently accurate or unbiased to meet the require- ments of a particular problem. They also should be precise enough for adequate design. In this book, we will use the collective term error to represent both the inaccuracy and imprecision of our predictions.

4.1.2 Error Definitions

Numerical errors arise from the use of approximations to represent exact mathematical operations and quantities. For such errors, the relationship between the exact, or true, result and the approximation can be formulated as

True value = approximation + error (4.1)

By rearranging Eq. (4.1), we find that the numerical error is equal to the discrepancy between the truth and the approximation, as in

Et = true value − approximation (4.2)

where Et is used to designate the exact value of the error. The subscript t is included to designate that this is the “true” error. This is in contrast to other cases, as described shortly, where an “approximate” estimate of the error must be employed. Note that the true error is commonly expressed as an absolute value and referred to as the absolute error.

A shortcoming of this definition is that it takes no account of the order of magnitude of the value under examination. For example, an error of a centimeter is much more significant

(c)

(a)

(d )

(b)

Increasing accuracy

In cr

ea si

ng p

re ci

si on

FIGURE 4.1 An example from marksmanship illustrating the concepts of accuracy and precision: (a) inaccurate and imprecise, (b) accurate and imprecise, (c) inaccurate and precise, and (d ) accurate and precise.

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102 Roundoff and TRuncaTIon ERRoRs

if we are measuring a rivet than a bridge. One way to account for the magnitudes of the quantities being evaluated is to normalize the error to the true value, as in

True fractional relative error = true value − approximation _______________________ true value

The relative error can also be multiplied by 100% to express it as

εt = true value − approximation _______________________ true value 100% (4.3)

where εt designates the true percent relative error. For example, suppose that you have the task of measuring the lengths of a bridge and

a rivet and come up with 9999 and 9 cm, respectively. If the true values are 10,000 and 10 cm, respectively, the error in both cases is 1 cm. However, their percent relative errors can be computed using Eq. (4.3) as 0.01% and 10%, respectively. Thus, although both mea- surements have an absolute error of 1 cm, the relative error for the rivet is much greater. We would probably conclude that we have done an adequate job of measuring the bridge, whereas our estimate for the rivet leaves something to be desired.

Notice that for Eqs. (4.2) and (4.3), E and ε are subscripted with a t to signify that the error is based on the true value. For the example of the rivet and the bridge, we were pro- vided with this value. However, in actual situations such information is rarely available. For numerical methods, the true value will only be known when we deal with functions that can be solved analytically. Such will typically be the case when we investigate the theo retical behavior of a particular technique for simple systems. However, in real-world applications, we will obviously not know the true answer a priori. For these situations, an alternative is to normalize the error using the best available estimate of the true value—that is, to the approximation itself, as in

εa = approximate error _______________ approximation 100% (4.4)

where the subscript a signifies that the error is normalized to an approximate value. Note also that for real-world applications, Eq. (4.2) cannot be used to calculate the error term in the numerator of Eq. (4.4). One of the challenges of numerical methods is to determine error estimates in the absence of knowledge regarding the true value. For example, certain numerical methods use iteration to compute answers. In such cases, a present approxima- tion is made on the basis of a previous approximation. This process is performed repeat- edly, or iteratively, to successively compute (hopefully) better and better approximations. For such cases, the error is often estimated as the difference between the previous and present approximations. Thus, percent relative error is determined according to

εa = present approximation − previous approximation ________________________________________ present approximation 100% (4.5)

This and other approaches for expressing errors is elaborated on in subsequent chapters. The signs of Eqs. (4.2) through (4.5) may be either positive or negative. If the approxi-

mation is greater than the true value (or the previous approximation is greater than the current approximation), the error is negative; if the approximation is less than the true value, the error is positive. Also, for Eqs. (4.3) to (4.5), the denominator may be less than zero, which can

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4.1 ERRoRs 103

also lead to a negative error. Often, when performing computations, we may not be concerned with the sign of the error but are interested in whether the absolute value of the percent rela- tive error is lower than a prespecified tolerance εs.Therefore, it is often useful to employ the absolute value of Eq. (4.5). For such cases, the computation is repeated until

∣εa∣ < εs (4.6)

This relationship is referred to as a stopping criterion. If it is satisfied, our result is assumed to be within the prespecified acceptable level εs. Note that for the remainder of this text, we almost always employ absolute values when using relative errors.

It is also convenient to relate these errors to the number of significant figures in the approximation. It can be shown (Scarborough, 1966) that if the following criterion is met, we can be assured that the result is correct to at least n significant figures.

εs = (0.5 × 102−n)% (4.7)

EXaMPLE 4.1 Error Estimates for Iterative Methods

Problem statement. In mathematics, functions can often be represented by infinite se- ries. For example, the exponential function can be computed using

ex = 1 + x + x 2 __ 2 +

x3 __ 3! + · · · + xn __ n! (E4.1.1)

Thus, as more terms are added in sequence, the approximation becomes a better and better estimate of the true value of ex. Equation (E4.1.1) is called a Maclaurin series expansion.

Starting with the simplest version, ex = 1, add terms one at a time in order to estimate e0.5. After each new term is added, compute the true and approximate percent relative errors with Eqs. (4.3) and (4.5), respectively. Note that the true value is e0.5 = 1.648721 . . . . Add terms until the absolute value of the approximate error estimate εa falls below a prespeci- fied error criterion εs conforming to three significant figures.

solution. First, Eq. (4.7) can be employed to determine the error criterion that ensures a result that is correct to at least three significant figures:

εs = (0.5 × 102−3)% = 0.05%

Thus, we will add terms to the series until εa falls below this level. The first estimate is simply equal to Eq. (E4.1.1) with a single term. Thus, the first

estimate is equal to 1. The second estimate is then generated by adding the second term as in

ex = 1 + x or for x = 0.5

e0.5 = 1 + 0.5 = 1.5

This represents a true percent relative error of [Eq. (4.3)]

εt = ∣ 1.648721 − 1.5 _____________ 1.648721 ∣ × 100% = 9.02%

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104 Roundoff and TRuncaTIon ERRoRs

4.1.3 Computer Algorithm for Iterative Calculations

Many of the numerical methods described in the remainder of this text involve iterative calculations of the sort illustrated in Example 4.1. These all entail solving a mathematical problem by computing successive approximations to the solution starting from an initial guess.

The computer implementation of such iterative solutions involves loops. As we saw in Sec. 3.3.2, these come in two basic flavors: count-controlled and decision loops. Most it- erative solutions use decision loops. Thus, rather than employing a prespecified number of iterations, the process typically is repeated until an approximate error estimate falls below a stopping criterion as in Example 4.1.

To do this for the same problem as Example 4.1, the series expansion can be expressed as

ex ≅ ∑ i=0

x n __ n!

An M-file to implement this formula is shown in Fig. 4.2. The function is passed the value to be evaluated (x) along with a stopping error criterion (es) and a maximum allowable number of iterations (maxit). If the user omits either of the latter two parameters, the func- tion assigns default values.

Equation (4.5) can be used to determine an approximate estimate of the error, as in

εa = ∣ 1.5 − 1 _______ 1.5 ∣ × 100% = 33.3% Because εa is not less than the required value of εs, we would continue the computation by adding another term, x2∕2!, and repeating the error calculations. The process is continued until ∣εa∣ < εs. The entire computation can be summarized as

Terms Result εt, % εa, %

1 1 39.3 2 1.5 9.02 33.3 3 1.625 1.44 7.69 4 1.645833333 0.175 1.27 5 1.648437500 0.0172 0.158 6 1.648697917 0.00142 0.0158

Thus, after six terms are included, the approximate error falls below εs = 0.05%, and the computation is terminated. However, notice that, rather than three significant figures, the result is accurate to five! This is because, for this case, both Eqs. (4.5) and (4.7) are conser- vative. That is, they ensure that the result is at least as good as they specify. Although, this is not always the case for Eq. (4.5), it is true most of the time.

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4.1 ERRoRs 105

The function then initializes three variables: (a) iter, which keeps track of the number of iterations, (b) sol, which holds the current estimate of the solution, and (c) a variable, ea,which holds the approximate percent relative error. Note that ea is initially set to a value of 100 to ensure that the loop executes at least once.

These initializations are followed by a decision loop that actually implements the iterative calculation. Prior to generating a new solution, the previous value, sol, is first assigned to solold. Then a new value of sol is computed and the iteration counter is incre- mented. If the new value of sol is nonzero, the percent relative error, ea, is determined. The stopping criteria are then tested. If both are false, the loop repeats. If either is true, the loop terminates and the final solution is sent back to the function call.

When the M-file is implemented, it generates an estimate for the exponential func- tion which is returned along with the approximate error and the number of iterations. For example, e1 can be evaluated as

>> format long >> [approxval, ea, iter] = IterMeth(1,1e − 6,100)

FIGURE 4.2 An M-file to solve an iterative calculation. This example is set up to evaluate the Maclaurin series expansion for ex as described in Example 4.1.

function [fx,ea,iter] = IterMeth(x,es,maxit) % Maclaurin series of exponential function % [fx,ea,iter] = IterMeth(x,es,maxit) % input: % x = value at which series evaluated % es = stopping criterion (default = 0.0001) % maxit = maximum iterations (default = 50) % output: % fx = estimated value % ea = approximate relative error (%) % iter = number of iterations

% defaults: if nargin < 2|isempty(es),es = 0.0001;end if nargin < 3|isempty(maxit),maxit = 50;end % initialization iter = 1; sol = 1; ea = 100; % iterative calculation while (1) solold = sol; sol = sol + x ^ iter / factorial(iter); iter = iter + 1; if sol~ = 0 ea = abs((sol − solold)/sol)*100; end if ea< = es | iter> = maxit,break,end end fx = sol; end

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approxval = 2.718281826198493 ea = 9.216155641522974e − 007 iter = 12

We can see that after 12 iterations, we obtain a result of 2.7182818 with an approx- imate error estimate of = 9.2162 × 10−7%. The result can be verified by using the built- in exp function to directly calculate the exact value and the true percent relative error,

>> trueval = exp(1)

trueval = 2.718281828459046

>> et = abs((trueval− approxval)/trueval)*100

et = 8.316108397236229e − 008

As was the case with Example 4.1, we obtain the desirable outcome that the true error is less than the approximate error.

4.2 ROUNDOFF ERRORS

Roundoff errors arise because digital computers cannot represent some quantities ex- actly. They are important to engineering and scientific problem solving because they can lead to erroneous results. In certain cases, they can actually lead to a calculation going unstable and yielding obviously erroneous results. Such calculations are said to be ill-conditioned. Worse still, they can lead to subtler discrepancies that are difficult to detect.

There are two major facets of roundoff errors involved in numerical calculations:

1. Digital computers have magnitude and precision limits on their ability to represent numbers.

2. Certain numerical manipulations are highly sensitive to roundoff errors. This can re- sult from both mathematical considerations as well as from the way in which comput- ers perform arithmetic operations.

4.2.1 Computer Number Representation

Numerical roundoff errors are directly related to the manner in which numbers are stored in a computer. The fundamental unit whereby information is represented is called a word. This is an entity that consists of a string of binary digits, or bits. Numbers are typically stored in one or more words. To understand how this is accomplished, we must first review some material related to number systems.

A number system is merely a convention for representing quantities. Because we have 10 fingers and 10 toes, the number system that we are most familiar with is the decimal, or base-10, number system. A base is the number used as the reference for constructing the system. The base-10 system uses the 10 digits—0, 1, 2, 3, 4, 5, 6, 7, 8,

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and 9—to represent numbers. By themselves, these digits are satisfactory for counting from 0 to 9.

For larger quantities, combinations of these basic digits are used, with the position or place value specifying the magnitude. The rightmost digit in a whole number represents a number from 0 to 9. The second digit from the right represents a multiple of 10. The third digit from the right represents a multiple of 100 and so on. For example, if we have the number 8642.9, then we have eight groups of 1000, six groups of 100, four groups of 10, two groups of 1, and nine groups of 0.1, or

(8 × 103) + (6 × 102) + (4 × 101) + (2 × 100) + (9 × 10−1) = 8642.9

This type of representation is called positional notation. Now, because the decimal system is so familiar, it is not commonly realized that

there are alternatives. For example, if human beings happened to have eight fingers and toes we would undoubtedly have developed an octal, or base-8, representation. In the same sense, our friend the computer is like a two-fingered animal who is limited to two states—either 0 or 1. This relates to the fact that the primary logic units of digital computers are on/off electronic components. Hence, numbers on the computer are rep- resented with a binary, or base-2, system. Just as with the decimal system, quantities can be represented using positional notation. For example, the binary number 101.1 is equivalent to (1 × 22) + (0 × 21) + (1 × 20) + (1 × 2−1) = 4 + 0 + 1 + 0.5 = 5.5 in the decimal system.

Integer Representation. Now that we have reviewed how base-10 numbers can be rep- resented in binary form, it is simple to conceive of how integers are represented on a com- puter. The most straightforward approach, called the signed magnitude method, employs the first bit of a word to indicate the sign, with a 0 for positive and a 1 for negative. The remaining bits are used to store the number. For example, the integer value of 173 is rep- resented in binary as 10101101:

(10101101)2 = 27 + 25 + 23 + 22 + 20 = 128 + 32 + 8 + 4 + 1 = (173)10 Therefore, the binary equivalent of −173 would be stored on a 16-bit computer, as depicted in Fig. 4.3.

If such a scheme is employed, there clearly is a limited range of integers that can be represented. Again assuming a 16-bit word size, if one bit is used for the sign, the 15 remain- ing bits can represent binary integers from 0 to 111111111111111. The upper limit can be converted to a decimal integer, as in (1 × 214) + (1 × 213) + . . . + (1 × 21) + (1 × 20) = 32,767.

1 0 0 0 0 0 0 0 1 0 1 0 1 1 0 1

Sign Magnitude

FIGURE 4.3 The binary representation of the decimal integer –173 on a 16-bit computer using the signed magnitude method.

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108 Roundoff and TRuncaTIon ERRoRs

Note that this value can be simply evaluated as 215 − 1. Thus, a 16-bit computer word can store decimal integers ranging from −32,767 to 32,767.

In addition, because zero is already defined as 0000000000000000, it is redundant to use the number 1000000000000000 to define a “minus zero.” Therefore, it is conven- tionally employed to represent an additional negative number: −32,768, and the range is from −32,768 to 32,767. For an n-bit word, the range would be from −2n−1 to 2n−1 − 1. Thus, 32-bit integers would range from −2,147,483,648 to +2,147,483,647.

Note that, although it provides a nice way to illustrate our point, the signed mag- nitude method is not actually used to represent integers for conventional computers. A preferred approach called the 2s complement technique directly incorporates the sign into the number’s magnitude rather than providing a separate bit to represent plus or minus. Regardless, the range of numbers is still the same as for the signed magnitude method described above.

The foregoing serves to illustrate how all digital computers are limited in their capability to represent integers. That is, numbers above or below the range cannot be represented. A more serious limitation is encountered in the storage and manipulation of fractional quanti- ties as described next.

floating-Point Representation. Fractional quantities are typically represented in com- puters using floating-point format. In this approach, which is very much like scientific notation, the number is expressed as

±s × be

where s = the significand (or mantissa), b = the base of the number system being used, and e = the exponent.

Prior to being expressed in this form, the number is normalized by moving the decimal place over so that only one significant digit is to the left of the decimal point. This is done so computer memory is not wasted on storing useless nonsignificant zeros. For example, a value like 0.005678 could be represented in a wasteful manner as 0.005678 × 100. However, nor- malization would yield 5.678 × 10−3 which eliminates the useless zeroes.

Before describing the base-2 implementation used on computers, we will first ex- plore the fundamental implications of such floating-point representation. In particular, what are the ramifications of the fact that in order to be stored in the computer, both the mantissa and the exponent must be limited to a finite number of bits? As in the next example, a nice way to do this is within the context of our more familiar base-10 decimal world.

EXaMPLE 4.2 Implications of floating-Point Representation

Problem statement. Suppose that we had a hypothetical base-10 computer with a 5-digit word size. Assume that one digit is used for the sign, two for the exponent, and two for the mantissa. For simplicity, assume that one of the exponent digits is used for its sign, leaving a single digit for its magnitude.

solution. A general representation of the number following normalization would be

s1d1.d2 × 10 s 0 d 0

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where s0 and s1 = the signs, d0 = the magnitude of the exponent, and d1 and d2 = the mag- nitude of the significand digits.

Now, let’s play with this system. First, what is the largest possible positive quantity that can be represented? Clearly, it would correspond to both signs being positive and all magnitude digits set to the largest possible value in base-10, that is, 9:

Largest value = +9.9 × 10+9

So the largest possible number would be a little less than 10 billion. Although this might seem like a big number, it’s really not that big. For example, this computer would be incapable of representing a commonly used constant like Avogadro’s number (6.022 × 1023). In the same sense, the smallest possible positive number would be

Smallest value = +1.0 × 10−9

Again, although this value might seem pretty small, you could not use it to represent a quantity like Planck’s constant (6.626 × 10−34 J · s). Similar negative values could also be developed. The resulting ranges are displayed in Fig. 4.4. Large positive and negative numbers that fall outside the range would cause an overflow error. In a similar sense, for very small quantities there is a “hole” at zero, and very small quantities would usually be converted to zero. Recognize that the exponent overwhelmingly determines these range limitations. For example, if we increase the mantissa by one digit, the maximum value increases slightly to 9.99 × 109. In contrast, a one-digit increase in the exponent raises the maximum by 90 orders of magnitude to 9.9 × 1099! When it comes to precision, however, the situation is reversed. Whereas the significand plays a minor role in defining the range, it has a profound effect on specifying the precision. This is dramatically illustrated for this example where we have limited the significand to only 2 digits. As in Fig. 4.5, just as there is a “hole” at zero, there are also “holes” between values. For example, a simple rational number with a finite number of digits like 2−5 = 0.03125 would have to be stored as 3.1 × 10−2 or 0.031. Thus, a roundoff error is introduced. For this case, it represents a relative error of

0.03125 − 0.031 ______________ 0.03125 = 0.008

−9.9 × 109 9.9 × 109−1.0 × 10−9 1.0 × 10−9

“Hole” at zero

SmallestMinimum Maximum

Underflow OverflowOverflow

FIGURE 4.4 The number line showing the possible ranges corresponding to the hypothetical base-10 floating-point scheme described in Example 4.2.

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While we could store a number like 0.03125 exactly by expanding the digits of the significand, quantities with infinite digits must always be approximated. For example, a commonly used constant such as π (= 3.14159…) would have to be represented as 3.1 × 100 or 3.1. For this case, the relative error is

3.14159 − 3.1 ____________ 3.14159 = 0.0132

Although adding significand digits can improve the approximation, such quantities will always have some roundoff error when stored in a computer. Another more subtle effect of floating-point representation is illustrated by Fig. 4.5. Notice how the interval between numbers increases as we move between orders of mag- nitude. For numbers with an exponent of −1 (i.e., between 0.1 and 1), the spacing is 0.01. Once we cross over into the range from 1 to 10, the spacing increases to 0.1. This means that the roundoff error of a number will be proportional to its magnitude. In addition, it means that the relative error will have an upper bound. For this example, the maxi- mum relative error would be 0.05. This value is called the machine epsilon (or machine precision).

As illustrated in Example 4.2, the fact that both the exponent and significand are finite means that there are both range and precision limits on floating-point representation. Now, let us examine how floating-point quantities are actually represented in a real computer using base-2 or binary numbers.

First, let’s look at normalization. Since binary numbers consist exclusively of 0s and 1s, a bonus occurs when they are normalized. That is, the bit to the left of the binary point will always be one! This means that this leading bit does not have to be stored. Hence, nonzero binary floating-point numbers can be expressed as

±(1 + f ) × 2e

where f = the mantissa (i.e., the fractional part of the significand). For example, if we nor- malized the binary number 1101.1, the result would be 1.1011 × (2)−3 or (1 + 0.1011) × 2−3.

0.98 0.99 1 1.1 1.2

0.01 0.1

FIGURE 4.5 A small portion of the number line corresponding to the hypothetical base-10 floating-point scheme described in Example 4.2. The numbers indicate values that can be represented exactly. All other quantities falling in the “holes” between these values would exhibit some roundoff error.

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Thus, although the original number has five significant bits, we only have to store the four fractional bits: 0.1011.

By default, MATLAB has adopted the IEEE double-precision format in which eight bytes (64 bits) are used to represent floating-point numbers. As in Fig. 4.6, one bit is re- served for the number’s sign. In a similar spirit to the way in which integers are stored, the exponent and its sign are stored in 11 bits. Finally, 52 bits are set aside for the mantissa. However, because of normalization, 53 bits can be stored.

Now, just as in Example 4.2, this means that the numbers will have a limited range and precision. However, because the IEEE format uses many more bits, the resulting number system can be used for practical purposes.

Range. In a fashion similar to the way in which integers are stored, the 11 bits used for the exponent translates into a range from −1022 to 1023. The largest positive number can be represented in binary as

Largest value = +1.1111 . . . 1111 × 2+1023

where the 52 bits in the mantissa are all 1. Since the significand is approximately 2 (it is actually 2 − 2−52), the largest value is therefore 21024 = 1.7977 × 10308. In a similar fashion, the smallest positive number can be represented as

Smallest value = +1.0000 . . . 0000 × 2−1022

This value can be translated into a base-10 value of 2–1022 = 2.2251 × 10–308.

Precision. The 52 bits used for the mantissa correspond to about 15 to 16 base-10 digits. Thus, π would be expressed as

>> format long >> pi

ans = 3.14159265358979

Note that the machine epsilon is 2–52 = 2.2204 × 10–16.

FIGURE 4.6 The manner in which a floating-point number is stored in an 8-byte word in IEEE double- precision format.

Signed exponent

Sign (1 bit)

Mantissa

52 bits11 bits

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MATLAB has a number of built-in functions related to its internal number representa- tion. For example, the realmax function displays the largest positive real number:

>> format long >> realmax

ans = 1.797693134862316e + 308

Numbers occurring in computations that exceed this value create an overflow. In MATLAB they are set to infinity, inf. The realmin function displays the smallest positive real number:

>> realmin

ans = 2.225073858507201e − 308

Numbers that are smaller than this value create an underflow and, in MATLAB, are set to zero. Finally, the eps function displays the machine epsilon:

>> eps

ans = 2.220446049250313e − 016

4.2.2 Arithmetic Manipulations of Computer Numbers

Aside from the limitations of a computer’s number system, the actual arithmetic manipu- lations involving these numbers can also result in roundoff error. To understand how this occurs, let’s look at how the computer performs simple addition and subtraction.

Because of their familiarity, normalized base-10 numbers will be employed to illus- trate the effect of roundoff errors on simple addition and subtraction. Other number bases would behave in a similar fashion. To simplify the discussion, we will employ a hypotheti- cal decimal computer with a 4-digit mantissa and a 1-digit exponent.

When two floating-point numbers are added, the numbers are first expressed so that they have the same exponents. For example, if we want to add 1.557 + 0.04341, the com- puter would express the numbers as 0.1557 × 101 + 0.004341 × 101. Then the mantissas are added to give 0.160041 × 101. Now, because this hypothetical computer only carries a 4-digit mantissa, the excess number of digits get chopped off and the result is 0.1600 × 101. Notice how the last two digits of the second number (41) that were shifted to the right have essentially been lost from the computation.

Subtraction is performed identically to addition except that the sign of the subtrahend is reversed. For example, suppose that we are subtracting 26.86 from 36.41. That is,

0.3641 × 102

−0.2686 × 10 2 _____________

0.0955 × 102

For this case the result must be normalized because the leading zero is unneces- sary. So we must shift the decimal one place to the right to give 0.9550 × 101 = 9.550.

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Notice that the zero added to the end of the mantissa is not significant but is merely ap- pended to fill the empty space created by the shift. Even more dramatic results would be obtained when the numbers are very close as in

0.7642 × 103

−0.7641 × 10 3 _____________

0.0001 × 103

which would be converted to 0.1000 × 100 = 0.1000. Thus, for this case, three nonsig- nificant zeros are appended.

The subtracting of two nearly equal numbers is called subtractive cancellation. It is the classic example of how the manner in which computers handle mathematics can lead to numerical problems. Other calculations that can cause problems include:

Large computations. Certain methods require extremely large numbers of arithmetic manipulations to arrive at their final results. In addition, these computations are often inter- dependent. That is, the later calculations are dependent on the results of earlier ones. Con- sequently, even though an individual roundoff error could be small, the cumulative effect over the course of a large computation can be significant. A very simple case involves summing a round base-10 number that is not round in base-2. Suppose that the following M-file is constructed:

function sout = sumdemo() s = 0; for i = 1:10000 s = s + 0.0001; end sout = s;

When this function is executed, the result is >> format long >> sumdemo

ans = 0.99999999999991

The format long command lets us see the 15 significant-digit representation used by MATLAB. You would expect that sum would be equal to 1. However, although 0.0001 is a nice round number in base-10, it cannot be expressed exactly in base-2. Thus, the sum comes out to be slightly different than 1. We should note that MATLAB has features that are designed to minimize such errors. For example, suppose that you form a vector as in

>> format long >> s = [0:0.0001:1];

For this case, rather than being equal to 0.99999999999991, the last entry will be exactly one as verified by

>> s(10001)

ans = 1

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adding a Large and a small number. Suppose we add a small number, 0.0010, to a large number, 4000, using a hypothetical computer with the 4-digit mantissa and the 1-digit exponent. After modifying the smaller number so that its exponent matches the larger,

0.4000 × 104

0.0000001 × 10 4 ______________

0.4000001 × 104

which is chopped to 0.4000 × 104. Thus, we might as well have not performed the addition! This type of error can occur in the computation of an infinite series. The initial terms in such series are often relatively large in comparison with the later terms. Thus, after a few terms have been added, we are in the situation of adding a small quantity to a large quan- tity. One way to mitigate this type of error is to sum the series in reverse order. In this way, each new term will be of comparable magnitude to the accumulated sum.

smearing. Smearing occurs whenever the individual terms in a summation are larger than the summation itself. One case where this occurs is in a series of mixed signs.

Inner Products. As should be clear from the last sections, some infinite series are par- ticularly prone to roundoff error. Fortunately, the calculation of series is not one of the more common operations in numerical methods. A far more ubiquitous manipulation is the calculation of inner products as in

∑ i=1

xi yi = x1 y1 + x2 y2 + · · · + xn yn

This operation is very common, particularly in the solution of simultaneous linear algebraic equations. Such summations are prone to roundoff error. Consequently, it is often desirable to compute such summations in double precision as is done automatically in MATLAB.

4.3 TRUNCATION ERRORS

Truncation errors are those that result from using an approximation in place of an exact mathematical procedure. For example, in Chap. 1 we approximated the derivative of veloc- ity of a bungee jumper by a finite-difference equation of the form [Eq. (1.11)]

dυ ___ dt ≅ Δυ ___ Δt =

υ(ti+1) − υ(ti ) ___________ ti+1 − ti (4.8)

A truncation error was introduced into the numerical solution because the difference equa- tion only approximates the true value of the derivative (recall Fig. 1.3). To gain insight into the properties of such errors, we now turn to a mathematical formulation that is used widely in numerical methods to express functions in an approximate fashion—the Taylor series.

4.3.1 The Taylor Series

Taylor’s theorem and its associated formula, the Taylor series, is of great value in the study of numerical methods. In essence, the Taylor theorem states that any smooth function can be approximated as a polynomial. The Taylor series then provides a means to express this idea mathematically in a form that can be used to generate practical results.

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A useful way to gain insight into the Taylor series is to build it term by term. A good problem context for this exercise is to predict a function value at one point in terms of the function value and its derivatives at another point.

Suppose that you are blindfolded and taken to a location on the side of a hill facing downslope (Fig. 4.7). We’ll call your horizontal location xi and your vertical distance with respect to the base of the hill f (xi). You are given the task of predicting the height at a position xi+1, which is a distance h away from you.

At first, you are placed on a platform that is completely horizontal so that you have no idea that the hill is sloping down away from you. At this point, what would be your best guess at the height at xi+1? If you think about it (remember you have no idea whatsoever what’s in front of you), the best guess would be the same height as where you’re standing now! You could express this prediction mathematically as

f (xi+1) ≅ f (xi ) (4.9)

This relationship, which is called the zero-order approximation, indicates that the value of f at the new point is the same as the value at the old point. This result makes intuitive sense because if xi and xi+1 are close to each other, it is likely that the new value is probably similar to the old value.

Equation (4.9) provides a perfect estimate if the function being approximated is, in fact, a constant. For our problem, you would be right only if you happened to be standing on a perfectly flat plateau. However, if the function changes at all over the interval, addi- tional terms of the Taylor series are required to provide a better estimate.

So now you are allowed to get off the platform and stand on the hill surface with one leg positioned in front of you and the other behind. You immediately sense that the front

1.0

0.5

Zero order f (x) f (xi)

xi = 0 xi+1 = 1 x

f (xi+1)

f (xi+1) ≃ f (xi) + fʹ(xi)h + f ʺ(xi)

2! h 2

f (xi+1) ≃ f (xi) + fʹ(xi)h

f (xi+1) ≃ f (xi)

Second order

True

First order

FIGURE 4.7 The approximation of f (x) = −0.1 x4 − 0.15x3 − 0.5x2 − 0.25x + 1.2 at x = 1 by zero-order, first-order, and second-order Taylor series expansions.

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foot is lower than the back foot. In fact, you’re allowed to obtain a quantitative estimate of the slope by measuring the difference in elevation and dividing it by the distance between your feet.

With this additional information, you’re clearly in a better position to predict the height at f (xi+1). In essence, you use the slope estimate to project a straight line out to xi+1. You can express this prediction mathematically by

f (xi+1) ≅ f (xi) + f ́ (xi)h (4.10)

This is called a first-order approximation because the additional first-order term consists of a slope f ́ (xi) multiplied by h, the distance between xi and xi+1. Thus, the expression is now in the form of a straight line that is capable of predicting an increase or decrease of the function between xi and xi+1.

Although Eq. (4.10) can predict a change, it is only exact for a straight-line, or linear, trend. To get a better prediction, we need to add more terms to our equation. So now you are allowed to stand on the hill surface and take two measurements. First, you measure the slope behind you by keeping one foot planted at xi and moving the other one back a distance Δx. Let’s call this slope f ́b (xi). Then you measure the slope in front of you by keeping one foot planted at xi and moving the other one forward Δx. Let’s call this slope f ́f (xi). You im- mediately recognize that the slope behind is milder than the one in front. Clearly the drop in height is “accelerating” downward in front of you. Thus, the odds are that f (xi) is even lower than your previous linear prediction.

As you might expect, you’re now going to add a second-order term to your equation and make it into a parabola. The Taylor series provides the correct way to do this as in

f (xi+1) ≅ f (xi ) + f ́ (xi )h + f ̋ (xi ) _____ 2! h

2 (4.11)

To make use of this formula, you need an estimate of the second derivative. You can use the last two slopes you determined to estimate it as

f ̋ (xi +1) ≅ f ́f (xi ) − f ́b (xi ) ___________ Δx (4.12)

Thus, the second derivative is merely a derivative of a derivative; in this case, the rate of change of the slope.

Before proceeding, let’s look carefully at Eq. (4.11). Recognize that all the values subscripted i represent values that you have estimated. That is, they are numbers. Con- sequently, the only unknowns are the values at the prediction position xi+1. Thus, it is a quadratic equation of the form

f (h) ≅ a2h2 + a1h + a0 Thus, we can see that the second-order Taylor series approximates the function with a second-order polynomial.

Clearly, we could keep adding more derivatives to capture more of the function’s cur- vature. Thus, we arrive at the complete Taylor series expansion

f (xi+1) ≅ f (xi ) + f ́ (xi )h + f ̋ (xi ) _____ 2! h

2 + f (3) (xi ) _____ 3! h

3 + · · · + f (n) (xi ) _____ n! h

n + Rn (4.13)

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Note that because Eq. (4.13) is an infinite series, an equal sign replaces the approxi- mate sign that was used in Eqs. (4.9) through (4.11). A remainder term is also included to account for all terms from n + 1 to infinity:

Rn = f (n+1)(ξ) _______ (n + 1)! h

n+1 (4.14)

where the subscript n connotes that this is the remainder for the nth-order approximation and ξ is a value of x that lies somewhere between xi and xi+1.

We can now see why the Taylor theorem states that any smooth function can be ap- proximated as a polynomial and that the Taylor series provides a means to express this idea mathematically.

In general, the nth-order Taylor series expansion will be exact for an nth-order poly- nomial. For other differentiable and continuous functions, such as exponentials and sinu- soids, a finite number of terms will not yield an exact estimate. Each additional term will contribute some improvement, however slight, to the approximation. This behavior will be demonstrated in Example 4.3. Only if an infinite number of terms are added will the series yield an exact result.

Although the foregoing is true, the practical value of Taylor series expansions is that, in most cases, the inclusion of only a few terms will result in an approximation that is close enough to the true value for practical purposes. The assessment of how many terms are required to get “close enough” is based on the remainder term of the expansion (Eq. 4.14). This relationship has two major drawbacks. First, ξ is not known exactly but merely lies somewhere between xi and xi+1. Second, to evaluate Eq. (4.14), we need to determine the (n + 1)th derivative of f (x). To do this, we need to know f (x). However, if we knew f (x), there would be no need to perform the Taylor series expansion in the present context!

Despite this dilemma, Eq. (4.14) is still useful for gaining insight into truncation errors. This is because we do have control over the term h in the equation. In other words, we can choose how far away from x we want to evaluate f (x), and we can con- trol the number of terms we include in the expansion. Consequently, Eq. (4.14) is often expressed as

Rn = O (hn+1)

where the nomenclature O(hn+1) means that the truncation error is of the order of hn+1. That is, the error is proportional to the step size h raised to the (n + 1)th power. Al- though this approximation implies nothing regarding the magnitude of the derivatives that multiply hn+1, it is extremely useful in judging the comparative error of numerical methods based on Taylor series expansions. For example, if the error is O(h), halving the step size will halve the error. On the other hand, if the error is O(h2), halving the step size will quarter the error.

In general, we can usually assume that the truncation error is decreased by the addi- tion of terms to the Taylor series. In many cases, if h is sufficiently small, the first- and other lower-order terms usually account for a disproportionately high percent of the error. Thus, only a few terms are required to obtain an adequate approximation. This property is illustrated by the following example.

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118 Roundoff and TRuncaTIon ERRoRs

EXaMPLE 4.3 approximation of a function with a Taylor series Expansion

Problem statement. Use Taylor series expansions with n = 0 to 6 to approximate f (x) = cos x at xi+1 = π/3 on the basis of the value of f (x) and its derivatives at xi = π/4. Note that this means that h = π/3 − π/4 = π/12.

solution. Our knowledge of the true function allows us to determine the correct value f (π/3) = 0.5. The zero-order approximation is [Eq. (4.9)]

f ( π __ 3 ) ≅ cos ( π __ 4 ) = 0.707106781 which represents a percent relative error of

εt = ∣ 0.5 − 0.707106781 ________________ 0.5 ∣ 100% = 41.4% For the first-order approximation, we add the first derivative term where f ́ (x) = −sin x:

f ( π __ 3 ) ≅ cos ( π __ 4 ) − sin ( π __ 4 ) ( π __ 12 ) = 0.521986659 which has ∣εt∣ = 0.40%. For the second-order approximation, we add the second derivative term where f ̋ (x) = −cos x:

f ( π __ 3 ) ≅ cos ( π __ 4 ) − sin ( π __ 4 ) ( π __ 12 ) − cos (π/4) ________ 2 ( π __ 12 )

2 = 0.497754491

with ∣εt∣ = 0.449%. Thus, the inclusion of additional terms results in an improved estimate. The process can be continued and the results listed as in

Order n f (n)(x) f (π∕3) ∣εt∣

0 cos x 0.707106781 41.4 1 −sin x 0.521986659 4.40 2 −cos x 0.497754491 0.449 3 sin x 0.499869147 2.62 × 10−2 4 cos x 0.500007551 1.51 × 10−3 5 −sin x 0.500000304 6.08 × 10−5 6 −cos x 0.499999988 2.44 × 10−6

Notice that the derivatives never go to zero as would be the case for a polynomial. Therefore, each additional term results in some improvement in the estimate. However, also notice how most of the improvement comes with the initial terms. For this case, by the time we have added the third-order term, the error is reduced to 0.026%, which means that we have attained 99.974% of the true value. Consequently, although the addition of more terms will reduce the error further, the improvement becomes negligible.

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4.3.2 The Remainder for the Taylor Series Expansion

Before demonstrating how the Taylor series is actually used to estimate numerical errors, we must explain why we included the argument ξ in Eq. (4.14). To do this, we will use a simple, visually based explanation.

Suppose that we truncated the Taylor series expansion [Eq. (4.13)] after the zero-order term to yield

f (xi+1) ≅ f (xi )

A visual depiction of this zero-order prediction is shown in Fig. 4.8. The remainder, or error, of this prediction, which is also shown in the illustration, consists of the infinite series of terms that were truncated

R0 = f ́ (xi )h + f ̋ (xi ) _____ 2! h

2 + f (3) (xi ) _____ 3! h

3 + · · ·

It is obviously inconvenient to deal with the remainder in this infinite series format. One simplification might be to truncate the remainder itself, as in

R0 ≅ f ́ (xi )h (4.15) Although, as stated in the previous section, lower-order derivatives usually account for a greater share of the remainder than the higher-order terms, this result is still inexact be- cause of the neglected second- and higher-order terms. This “inexactness” is implied by the approximate equality symbol (≅) employed in Eq. (4.15).

An alternative simplification that transforms the approximation into an equivalence is based on a graphical insight. As in Fig. 4.9, the derivative mean-value theorem states that if a function f (x) and its first derivative are continuous over an interval from xi to xi+1, then

FIGURE 4.8 Graphical depiction of a zero-order Taylor series prediction and remainder.

Zero-order prediction

Exa ct p

red ictio

f (x)

xi xi +1 x

f (xi)

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120 Roundoff and TRuncaTIon ERRoRs

there exists at least one point on the function that has a slope, designated by f ́ (ξ), that is parallel to the line joining f (xi) and f (xi+1). The parameter ξ marks the x value where this slope occurs (Fig. 4.9). A physical illustration of this theorem is that, if you travel between two points with an average velocity, there will be at least one moment during the course of the trip when you will be moving at that average velocity.

By invoking this theorem, it is simple to realize that, as illustrated in Fig. 4.9, the slope f ́ (ξ) is equal to the rise R0 divided by the run h, or

f ́ (ξ) = R0 ___ h

which can be rearranged to give

R0 = f ́ (ξ)h (4.16)

Thus, we have derived the zero-order version of Eq. (4.14). The higher-order versions are merely a logical extension of the reasoning used to derive Eq. (4.16). The first-order version is

R1 = f ̋ (ξ) _____ 2! h

2 (4.17)

For this case, the value of ξ conforms to the x value corresponding to the second derivative that makes Eq. (4.17) exact. Similar higher-order versions can be developed from Eq. (4.14).

4.3.3 Using the Taylor Series to Estimate Truncation Errors

Although the Taylor series will be extremely useful in estimating truncation errors through- out this book, it may not be clear to you how the expansion can actually be applied to

FIGURE 4.9 Graphical depiction of the derivative mean-value theorem.

f (x)

xi xi + 1ξ x

Slope = fʹ(ξ)

Slope = R0h

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numerical methods. In fact, we have already done so in our example of the bungee jumper. Recall that the objective of both Examples 1.1 and 1.2 was to predict velocity as a function of time. That is, we were interested in determining υ(t). As specified by Eq. (4.13), υ(t) can be expanded in a Taylor series:

υ(ti+1) = υ(ti ) + υ ́ (ti )(ti+1 − ti ) + υ ̋ (ti ) _____ 2! (ti+1 − ti )

2 + · · · + Rn

Now let us truncate the series after the first derivative term:

υ(ti+1) = υ(ti ) + υ ́ (ti )(ti+1 − ti ) + R1 (4.18)

Equation (4.18) can be solved for

υ ́ (ti) = υ(ti+1) − υ(ti ) ___________ ti+1 − ti

− R1 _____ ti+1 − ti

(4.19)

The first part of Eq. (4.19) is exactly the same relationship that was used to approximate the derivative in Example 1.2 [Eq. (1.11)]. However, because of the Taylor series approach, we have now obtained an estimate of the truncation error associated with this approxima- tion of the derivative. Using Eqs. (4.14) and (4.19) yields

R1 _____ ti+1 − ti

= υ ̋ (ξ) _____ 2! (ti+1 − ti )

R1 _____ ti+1 − ti

= O (ti+1 − ti )

Thus, the estimate of the derivative [Eq. (1.11) or the first part of Eq. (4.19)] has a trunca- tion error of order ti+1 − ti. In other words, the error of our derivative approximation should be proportional to the step size. Consequently, if we halve the step size, we would expect to halve the error of the derivative.

4.3.4 Numerical Differentiation

Equation (4.19) is given a formal label in numerical methods—it is called a finite differ- ence. It can be represented generally as

f ́ (xi) = f (xi+1) − f (xi) ____________ xi+1 − xi + O (xi+1 − xi) (4.20)

f ́ (xi) = f (xi+1) − f (xi) ____________ h + O (h) (4.21)

where h is called the step size—that is, the length of the interval over which the approxima- tion is made, xi+1 − xi. It is termed a “forward” difference because it utilizes data at i and i + 1 to estimate the derivative (Fig. 4.10a).

⏟First-order approximation

⏟Truncation error

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122 Roundoff and TRuncaTIon ERRoRs

Tru e d

eriv ativ

Tru e d

eriv ativ

Ap pro

xim atio

(c)

Ap pr

ox im

ati on

(b)

Approxi mation

(a)

f (x)

xi xxi+1

xi xxi−1

xxi−1 xi+1

Tru e d

eriv ativ

Ap pro

xim atio

FIGURE 4.10 Graphical depiction of (a) forward, (b) backward, and (c) centered finite-difference approximations of the first derivative.

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4.3 TRuncaTIon ERRoRs 123

This forward difference is but one of many that can be developed from the Taylor series to approximate derivatives numerically. For example, backward and centered dif- ference approximations of the first derivative can be developed in a fashion similar to the derivation of Eq. (4.19). The former utilizes values at xi−1 and xi (Fig. 4.10b), whereas the latter uses values that are equally spaced around the point at which the derivative is estimated (Fig. 4.10c). More accurate approximations of the first derivative can be devel- oped by including higher-order terms of the Taylor series. Finally, all the foregoing versions can also be developed for second, third, and higher derivatives. The following sections pro- vide brief summaries illustrating how some of these cases are derived.

Backward difference approximation of the first derivative. The Taylor series can be expanded backward to calculate a previous value on the basis of a present value, as in

f (xi−1) = f (xi ) − f ́ (xi )h + f ̋ (xi ) _____ 2! h

2 − · · · (4.22)

Truncating this equation after the first derivative and rearranging yields

f ́ (xi ) ≅ f (xi ) − f (xi−1) ____________ h (4.23)

where the error is O(h).

centered difference approximation of the first derivative. A third way to approxi- mate the first derivative is to subtract Eq. (4.22) from the forward Taylor series expansion:

f (xi+1) = f (xi ) + f ́ (xi )h + f ̋ (xi ) _____ 2! h

2 + · · · (4.24)

to yield

f (xi+1) = f (xi−1) + 2 f ́ (xi )h + 2 f (3) (xi ) _____ 3! h

3 + · · ·

which can be solved for

f ́ (xi ) = f (xi+1) − f (xi−1) _____________ 2h −

f (3) (xi ) _____ 6 h 2 + · · ·

f ́ (xi ) = f (xi+1) − f (xi−1) _____________ 2h − O (h

2) (4.25)

Equation (4.25) is a centered finite difference representation of the first derivative. Notice that the truncation error is of the order of h2 in contrast to the forward and backward approximations that were of the order of h. Consequently, the Taylor series analysis yields the practical information that the centered difference is a more accurate representation of the derivative (Fig. 4.10c). For example, if we halve the step size using a forward or back- ward difference, we would approximately halve the truncation error, whereas for the central difference, the error would be quartered.

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124 Roundoff and TRuncaTIon ERRoRs

EXaMPLE 4.4 finite-difference approximations of derivatives

Problem statement. Use forward and backward difference approximations of O(h) and a centered difference approximation of O(h2) to estimate the first derivative of

f (x) = −0.1x4 − 0.15x3 − 0.5x2 − 0.25x + 1.2

at x = 0.5 using a step size h = 0.5. Repeat the computation using h = 0.25. Note that the derivative can be calculated directly as

f ́ (x) = −0.4x3 − 0.45x2 − 1.0x − 0.25 and can be used to compute the true value as f ́ (0.5) = −0.9125.

solution. For h = 0.5, the function can be employed to determine

xi−1 = 0 f (xi−1) = 1.2

xi = 0.5 f (xi ) = 0.925

xi+1 = 1.0 f (xi+1) = 0.2

These values can be used to compute the forward difference [Eq. (4.21)],

f ́ (0.5) ≅ 0.2 − 0.925 __________ 0.5 = −1.45 ∣εt∣ = 58.9%

the backward difference [Eq. (4.23)],

f ́ (0.5) ≅ 0.925 − 1.2 __________ 0.5 = −0.55 ∣εt∣ = 39.7%

and the centered difference [Eq. (4.25)],

f ́ (0.5) ≅ 0.2 − 1.2 __________ 1.0 = −1.0 ∣εt∣ = 9.6%

For h = 0.25,

xi−1 = 0.25 f (xi−1) = 1.10351563

xi = 0.5 f (xi ) = 0.925

xi+1 = 0.75 f (xi+1) = 0.63632813

which can be used to compute the forward difference,

f ́ (0.5) ≅ 0.63632813 − 0.925 __________ 0.25 = −1.155 ∣εt∣ = 26.5%

the backward difference,

f ́ (0.5) ≅ 0.925 − 1.10351563 __________ 0.25 = −0.714 ∣εt∣ = 21.7%

and the centered difference,

f ́ (0.5) ≅ 0.63632813 − 1.10351563 __________ 0.5 = −0.934 ∣εt∣ = 2.4%

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4.4 ToTaL nuMERIcaL ERRoR 125

For both step sizes, the centered difference approximation is more accurate than for- ward or backward differences. Also, as predicted by the Taylor series analysis, halving the step size approximately halves the error of the backward and forward differences and quarters the error of the centered difference.

finite-difference approximations of Higher derivatives. Besides first derivatives, the Taylor series expansion can be used to derive numerical estimates of higher deriva- tives. To do this, we write a forward Taylor series expansion for f (xi+2) in terms of f (xi):

f (xi+2) = f (xi ) + f ́ (xi )(2h) + f ̋ (xi ) _____ 2! (2h)

2 + · · · (4.26)

Equation (4.24) can be multiplied by 2 and subtracted from Eq. (4.26) to give

f (xi+2) − 2 f (xi+1) = − f (xi ) + f ̋ (xi )h2 + · · ·

which can be solved for

f ̋ (xi) = f (xi+2) − 2 f (xi+1) + f (xi ) _____________________

h2 + O(h) (4.27)

This relationship is called the second forward finite difference. Similar manipulations can be employed to derive a backward version

f ̋ (xi) = f (xi ) − 2 f (xi−1) + f (xi−2) _____________________

h2 + O(h)

A centered difference approximation for the second derivative can be derived by add- ing Eqs. (4.22) and (4.24) and rearranging the result to give

f ̋ (xi) = f (xi+1) − 2 f (xi ) + f (xi−1) _____________________

h2 + O(h2)

As was the case with the first-derivative approximations, the centered case is more accurate. Notice also that the centered version can be alternatively expressed as

f ̋ (xi) ≅ f (xi+1) − f (xi ) ____________ h −

f (xi ) − f (xi−1) ___________ h _________________________ h

Thus, just as the second derivative is a derivative of a derivative, the second finite differ- ence approximation is a difference of two first finite differences [recall Eq. (4.12)].

4.4 TOTAL NUMERICAL ERROR

The total numerical error is the summation of the truncation and roundoff errors. In general, the only way to minimize roundoff errors is to increase the number of significant figures of the computer. Further, we have noted that roundoff error may increase due to subtractive cancellation or due to an increase in the number of computations in an analysis. In contrast, Example 4.4 demonstrated that the truncation error can be reduced by decreasing the step size. Because a decrease in step size can lead to subtractive cancellation or to an increase in computations, the truncation errors are decreased as the roundoff errors are increased.

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126 Roundoff and TRuncaTIon ERRoRs

Therefore, we are faced by the following dilemma: The strategy for decreasing one component of the total error leads to an increase of the other component. In a computa- tion, we could conceivably decrease the step size to minimize truncation errors only to discover that in doing so, the roundoff error begins to dominate the solution and the total error grows! Thus, our remedy becomes our problem (Fig. 4.11). One challenge that we face is to determine an appropriate step size for a particular computation. We would like to choose a large step size to decrease the amount of calculations and roundoff errors without incurring the penalty of a large truncation error. If the total error is as shown in Fig. 4.11, the challenge is to identify the point of diminishing returns where roundoff error begins to negate the benefits of step-size reduction.

When using MATLAB, such situations are relatively uncommon because of its 15- to 16- digit precision. Nevertheless, they sometimes do occur and suggest a sort of “numerical un- certainty principle” that places an absolute limit on the accuracy that may be obtained using certain computerized numerical methods. We explore such a case in the following section.

4.4.1 Error Analysis of Numerical Differentiation

As described in Sec. 4.3.4, a centered difference approximation of the first derivative can be written as [Eq. (4.25)]

f ́ (xi ) = f (xi+1) − f (xi−1) _____________________ 2h −

f (3) (ξ) _____ 6 h 2

(4.28) True Finite-difference Truncation value approximation error

Thus, if the two function values in the numerator of the finite-difference approximation have no roundoff error, the only error is due to truncation.

Log step size

Lo g

er ro

Tru nca

tion err

Total error

Point of diminishing

returns

Round-o� error

FIGURE 4.11 A graphical depiction of the trade-off between roundoff and truncation error that sometimes comes into play in the course of a numerical method. The point of diminishing returns is shown, where roundoff error begins to negate the benefits of step-size reduction.

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4.4 ToTaL nuMERIcaL ERRoR 127

However, because we are using digital computers, the function values do include roundoff error as in

f (xi−1) = ̃ f (xi−1) + ei−1

f (xi+1) = ̃ f (xi+1) + ei+1

where the ̃ f ’s are the rounded function values and the e’s are the associated roundoff errors.

Substituting these values into Eq. (4.28) gives

f ́ (xi) = ̃ f (xi+1) − ̃

f (xi−1) ______________ 2h +

ei+1 − ei−1 _________ 2h − f (3) (ξ) _____ 6 h

True Finite-difference Roundoff Truncation value approximation error error

We can see that the total error of the finite-difference approximation consists of a roundoff error that decreases with step size and a truncation error that increases with step size.

Assuming that the absolute value of each component of the roundoff error has an upper bound of ε, the maximum possible value of the difference ei+1 – ei−1 will be 2ε. Fur- ther, assume that the third derivative has a maximum absolute value of M. An upper bound on the absolute value of the total error can therefore be represented as

Total error = ∣ f ́ (xi ) − ̃ f (xi+1) − ̃

f (xi−1) ______________ 2h ∣ ≤ ε __ h + h2M ____ 6 (4.29)

An optimal step size can be determined by differentiating Eq. (4.29), setting the result equal to zero and solving for

hopt = 3 √ ___

3ε ___ M (4.30)

EXaMPLE 4.5 Roundoff and Truncation Errors in numerical differentiation

Problem statement. In Example 4.4, we used a centered difference approximation of O(h2) to estimate the first derivative of the following function at x = 0.5,

f (x) = −0.1x4 − 0.15x3 − 0.5x2 − 0.25x + 1.2

Perform the same computation starting with h = 1. Then progressively divide the step size by a factor of 10 to demonstrate how roundoff becomes dominant as the step size is reduced. Relate your results to Eq. (4.30). Recall that the true value of the derivative is −0.9125.

solution. We can develop the following M-file to perform the computations and plot the results. Notice that we pass both the function and its analytical derivative as arguments: function diffex(func,dfunc,x,n) format long dftrue = dfunc(x); h = 1; H(1) = h; D(1) = (func(x+h) − func(x − h))/(2*h); E(1) = abs(dftrue − D(1));

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128 Roundoff and TRuncaTIon ERRoRs

for i = 2:n h = h/10; H(i) = h; D(i) = (func(x+h)−func(x − h))/(2*h); E(i) = abs(dftrue − D(i)); end L = [H' D' E']'; fprintf(' step size finite difference true error\n'); fprintf('%14.10f %16.14f %16.13f\n',L); loglog(H,E),xlabel('Step Size'),ylabel('Error') title('Plot of Error Versus Step Size') format short

The M-file can then be run using the following commands: >> ff = @(x) −0.1*x^4 − 0.15*x^3 − 0.5*x^2 − 0.25*x+1.2; >> df = @(x) −0.4*x^3 − 0.45*x^2 − x − 0.25; >> diffex(ff,df,0.5,11)

step size finite difference true error 1.0000000000 −1.26250000000000 0.3500000000000 0.1000000000 −0.91600000000000 0.0035000000000 0.0100000000 −0.91253500000000 0.0000350000000 0.0010000000 −0.91250035000001 0.0000003500000 0.0001000000 −0.91250000349985 0.0000000034998 0.0000100000 −0.91250000003318 0.0000000000332 0.0000010000 −0.91250000000542 0.0000000000054 0.0000001000 −0.91249999945031 0.0000000005497 0.0000000100 -0.91250000333609 0.0000000033361 0.0000000010 −0.91250001998944 0.0000000199894 0.0000000001 −0.91250007550059 0.0000000755006

As depicted in Fig. 4.12, the results are as expected. At first, roundoff is minimal and the estimate is dominated by truncation error. Hence, as in Eq. (4.29), the total error drops by a factor of 100 each time we divide the step by 10. However, starting at about h = 0.0001, we see roundoff error begin to creep in and erode the rate at which the error diminishes. A minimum error is reached at h = 10−6. Beyond this point, the error increases as roundoff dominates. Because we are dealing with an easily differentiable function, we can also investigate whether these results are consistent with Eq. (4.30). First, we can estimate M by evaluating the function’s third derivative as

M = ∣ f (3)(0.5)∣ = ∣−2.4(0.5) − 0.9∣ = 2.1

Because MATLAB has a precision of about 15 to 16 base-10 digits, a rough estimate of the upper bound on roundoff would be about ε = 0.5 × 10−16. Substituting these values into Eq. (4.30) gives

hopt = 3 √ ___________

3(0.5 × 10−16)

___________ 2.1 = 4.3 × 10 −6

which is on the same order as the result of 1 × 10–6 obtained with MATLAB.

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4.4.2 Control of Numerical Errors

For most practical cases, we do not know the exact error associated with numerical methods. The exception, of course, is when we know the exact solution, which makes our numerical approximations unnecessary. Therefore, for most engineering and scientific applications we must settle for some estimate of the error in our calculations.

There are no systematic and general approaches to evaluating numerical errors for all problems. In many cases error estimates are based on the experience and judgment of the engineer or scientist.

Although error analysis is to a certain extent an art, there are several practical programming guidelines we can suggest. First and foremost, avoid subtracting two nearly equal numbers. Loss of significance almost always occurs when this is done. Sometimes you can rearrange or refor- mulate the problem to avoid subtractive cancellation. If this is not possible, you may want to use extended-precision arithmetic. Furthermore, when adding and subtracting numbers, it is best to sort the numbers and work with the smallest numbers first. This avoids loss of significance.

Beyond these computational hints, one can attempt to predict total numerical errors using theoretical formulations. The Taylor series is our primary tool for analysis of such errors. Pre- diction of total numerical error is very complicated for even moderately sized problems and tends to be pessimistic. Therefore, it is usually attempted for only small-scale tasks.

Plot of error versus step size 100

10−2

10−4

10−6

10−8

10−10

10−12 10−10 10−8 10−6 10−4 10−2 100

Step size

Er ro

FIGURE 4.12

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130 Roundoff and TRuncaTIon ERRoRs

The tendency is to push forward with the numerical computations and try to estimate the accuracy of your results. This can sometimes be done by seeing if the results satisfy some condition or equation as a check. Or it may be possible to substitute the results back into the original equation to check that it is actually satisfied.

Finally you should be prepared to perform numerical experiments to increase your awareness of computational errors and possible ill-conditioned problems. Such experi- ments may involve repeating the computations with a different step size or method and comparing the results. We may employ sensitivity analysis to see how our solution changes when we change model parameters or input values. We may want to try different numerical algorithms that have different theoretical foundations, are based on different computational strategies, or have different convergence properties and stability characteristics.

When the results of numerical computations are extremely critical and may involve loss of human life or have severe economic ramifications, it is appropriate to take special precautions. This may involve the use of two or more independent groups to solve the same problem so that their results can be compared.

The roles of errors will be a topic of concern and analysis in all sections of this book. We will leave these investigations to specific sections.

4.5 BLUNDERS, MODEL ERRORS, AND DATA UNCERTAINTY

Although the following sources of error are not directly connected with most of the nu- merical methods in this book, they can sometimes have great impact on the success of a modeling effort. Thus, they must always be kept in mind when applying numerical tech- niques in the context of real-world problems.

4.5.1 Blunders

We are all familiar with gross errors, or blunders. In the early years of computers, erroneous nu- merical results could sometimes be attributed to malfunctions of the computer itself. Today, this source of error is highly unlikely, and most blunders must be attributed to human imperfection.

Blunders can occur at any stage of the mathematical modeling process and can con- tribute to all the other components of error. They can be avoided only by sound knowledge of fundamental principles and by the care with which you approach and design your solu- tion to a problem.

Blunders are usually disregarded in discussions of numerical methods. This is no doubt due to the fact that, try as we may, mistakes are to a certain extent unavoidable. However, we believe that there are a number of ways in which their occurrence can be minimized. In particular, the good programming habits that were outlined in Chap. 3 are extremely useful for mitigating programming blunders. In addition, there are usually simple ways to check whether a particular numerical method is working properly. Throughout this book, we dis- cuss ways to check the results of numerical calculations.

4.5.2 Model Errors

Model errors relate to bias that can be ascribed to incomplete mathematical models. An ex- ample of a negligible model error is the fact that Newton’s second law does not account for relativistic effects. This does not detract from the adequacy of the solution in Example 1.1

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PRoBLEMs 131

because these errors are minimal on the time and space scales associated with the bungee jumper problem.

However, suppose that air resistance is not proportional to the square of the fall velocity, as in Eq. (1.7), but is related to velocity and other factors in a different way. If such were the case, both the analytical and numerical solutions obtained in Chap. 1 would be erroneous be- cause of model error. You should be cognizant of this type of error and realize that, if you are working with a poorly conceived model, no numerical method will provide adequate results.

4.5.3 Data Uncertainty

Errors sometimes enter into an analysis because of uncertainty in the physical data on which a model is based. For instance, suppose we wanted to test the bungee jumper model by having an individual make repeated jumps and then measuring his or her velocity after a specified time interval. Uncertainty would undoubtedly be associated with these measure- ments, as the parachutist would fall faster during some jumps than during others. These errors can exhibit both inaccuracy and imprecision. If our instruments consistently under- estimate or overestimate the velocity, we are dealing with an inaccurate, or biased, device. On the other hand, if the measurements are randomly high and low, we are dealing with a question of precision.

Measurement errors can be quantified by summarizing the data with one or more well- chosen statistics that convey as much information as possible regarding specific character- istics of the data. These descriptive statistics are most often selected to represent (1) the location of the center of the distribution of the data and (2) the degree of spread of the data. As such, they provide a measure of the bias and imprecision, respectively. We will return to the topic of characterizing data uncertainty when we discuss regression in Part Four.

Although you must be cognizant of blunders, model errors, and uncertain data, the nu- merical methods used for building models can be studied, for the most part, independently of these errors. Therefore, for most of this book, we will assume that we have not made gross errors, we have a sound model, and we are dealing with error-free measurements. Under these conditions, we can study numerical errors without complicating factors.

PROBLEMS

4.1 The “divide and average” method, an old-time method for approximating the square root of any positive number a, can be formulated as

x = x + a∕x _______ 2

Write a well-structured function to implement this algorithm based on the algorithm outlined in Fig. 4.2. 4.2 Convert the following base-2 numbers to base 10: (a) 1011001, (b) 0.01011, and (c) 110.01001. 4.3 Convert the following base-8 numbers to base 10: 61,565 and 2.71. 4.4 For computers, the machine epsilon ε can also be thought of as the smallest number that when added to one

gives a number greater than 1. An algorithm based on this idea can be developed as

Step 1: Set ε = 1. Step 2: If 1+ ε is less than or equal to 1, then go to Step 5.

Otherwise go to Step 3. Step 3: ε = ε∕2 Step 4: Return to Step 2 Step 5: ε = 2 × ε

Write your own M-file based on this algorithm to determine the machine epsilon. Validate the result by comparing it with the value computed with the built-in function eps. 4.5 In a fashion similar to Prob. 4.4, develop your own M-file to determine the smallest positive real number used

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132 Roundoff and TRuncaTIon ERRoRs

in MATLAB. Base your algorithm on the notion that your computer will be unable to reliably distinguish between zero and a quantity that is smaller than this number. Note that the result you obtain will differ from the value computed with realmin. Challenge question: Investigate the results by taking the base-2 logarithm of the number generated by your code and those obtained with realmin. 4.6 Although it is not commonly used, MATLAB allows numbers to be expressed in single precision. Each value is stored in 4 bytes with 1 bit for the sign, 23 bits for the mantissa, and 8 bits for the signed exponent. Determine the smallest and largest positive floating-point numbers as well as the machine epsilon for single precision representation. Note that the exponents range from −126 to 127. 4.7 For the hypothetical base-10 computer in Example 4.2, prove that the machine epsilon is 0.05. 4.8 The derivative of f (x) = 1∕(1 − 3x2) is given by

6x _________ (1 − 3x2)2

Do you expect to have difficulties evaluating this function at x = 0.577? Try it using 3- and 4-digit arithmetic with chopping. 4.9 (a) Evaluate the polynomial

y = x3 − 7x2 + 8x − 0.35

at x = 1.37. Use 3-digit arithmetic with chopping. Evaluate the percent relative error. (b) Repeat (a) but express y as

y = ((x − 7)x + 8)x − 0.35

Evaluate the error and compare with part (a). 4.10 The following infinite series can be used to approxi- mate ex:

ex = 1 + x + x 2 __ 2 +

x3 __ 3! + · · · + xn __ n!

(a) Prove that this Maclaurin series expansion is a special case of the Taylor series expansion (Eq. 4.13) with xi = 0 and h = x.

(b) Use the Taylor series to estimate f (x) = e−x at xi+1 = 1 for xi = 0.25. Employ the zero-, first-, second-, and third- order versions and compute the ∣εt∣ for each case.

4.11 The Maclaurin series expansion for cos x is

cos x = 1 − x 2 __ 2 +

x4 __ 4! + x6 __ 6! +

x8 __ 8! − · · ·

Starting with the simplest version, cos x = 1, add terms one at a time to estimate cos(π/3). After each new term is

added, compute the true and approximate percent relative er- rors. Use your calculator or MATLAB to determine the true value. Add terms until the absolute value of the approximate error estimate falls below an error criterion conforming to two significant figures. 4.12 Perform the same computation as in Prob. 4.11, but use the Maclaurin series expansion for the sin x to estimate sin(π/3).

sin x = x − x 3 __ 3! +

x5 __ 5! − x7 __ 7! + · · ·

4.13 Use zero- through third-order Taylor series expansions to predict f (3) for

f (x) = 25x3 − 6x2 + 7x − 88

using a base point at x = 1. Compute the true percent relative error for each approximation. 4.14 Prove that Eq. (4.11) is exact for all values of x if f(x) = ax2 + bx + c. 4.15 Use zero- through fourth-order Taylor series expan- sions to predict f(2) for f(x) = ln x using a base point at x = 1. Compute the true percent relative error εt for each approxima- tion. Discuss the meaning of the results. 4.16 Use forward and backward difference approximations of O(h) and a centered difference approximation of O(h2) to estimate the first derivative of the function examined in Prob. 4.13. Evaluate the derivative at x = 2 using a step size of h = 0.25. Compare your results with the true value of the derivative. Interpret your results on the basis of the remain- der term of the Taylor series expansion. 4.17 Use a centered difference approximation of O(h2) to estimate the second derivative of the function examined in Prob. 4.13. Perform the evaluation at x = 2 using step sizes of h = 0.2 and 0.1. Compare your estimates with the true value of the second derivative. Interpret your results on the basis of the remainder term of the Taylor series expansion. 4.18 If ∣x∣ < 1 it is known that

1 _____ 1 − x = 1 + x + x 2 + x3 + · · ·

Repeat Prob. 4.11 for this series for x = 0.1. 4.19 To calculate a planet’s space coordinates, we have to solve the function

f (x) = x − 1 − 0.5 sin x

Let the base point be a = xi = π∕2 on the interval [0, π]. De- termine the highest-order Taylor series expansion resulting in a maximum error of 0.015 on the specified interval. The error is equal to the absolute value of the difference between the given function and the specific Taylor series expansion. (Hint: Solve graphically.)

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PRoBLEMs 133

4.20 Consider the function f (x) = x3 − 2x + 4 on the inter- val [−2, 2] with h = 0.25. Use the forward, backward, and centered finite difference approximations for the first and second derivatives so as to graphically illustrate which ap- proximation is most accurate. Graph all three first-derivative finite difference approximations along with the theoretical, and do the same for the second derivative as well. 4.21 Derive Eq. (4.30). 4.22 Repeat Example 4.5, but for f (x) = cos(x) at x = π∕6. 4.23 Repeat Example 4.5, but for the forward divided dif- ference (Eq. 4.21). 4.24 One common instance where subtractive cancellation occurs involves finding the roots of a parabola, ax2 + bx + c, with the quadratic formula:

x = −b ± √

________ b2 − 4ac ______________ 2a

For cases where b2 >> 4ac, the difference in the numerator can be very small and roundoff errors can occur. In such cases, an alternative formulation can be used to minimize subtractive cancellation:

x = −2c _____________ b ± √

________ b2 − 4ac

Use 5-digit arithmetic with chopping to determine the roots of the following equation with both versions of the quadratic formula.

x2 − 5000.002x + 10

4.25 Develop a well-structured MATLAB function to com- pute the Maclaurin series expansion for the cosine function

as described in Prob. 4.11. Pattern your function after the one for the exponential function in Fig. 4.2. Test your pro- gram for θ = π/3 (60°) and θ = 2π + π/3 = 7π/3 (420°). Explain the difference in the number of iterations required to obtain the correct result with the desired approximate ab- solute error (εa). 4.26 Develop a well-structured MATLAB function to com- pute the Maclaurin series expansion for the sine function as described in Prob. 4.12. Pattern your function after the one for the exponential function in Fig. 4.2. Test your program for θ = π/3 (60°) and θ = 2π + π/3 = 7π/3 (420°). Explain the difference in the number of iterations required to ob- tain the correct result with the desired approximate absolute error (εa). 4.27 Recall from your calculus class that the Maclaurin series, named after the Scottish mathematician Colin Ma- claurin (1698–1746), is a Taylor series expansion of a func- tion about 0. Use the Taylor series to derive the first four terms of the Maclaurin series expansion for the cosine em- ployed in Probs. 4.11 and 4.25. 4.28 The Maclaurin series expansion for the arctangent of x is defined for |x| ≤ 1 as

arctan x = ∑ n=0

∞

(−1) n ______ 2n + 1 x

2n+1

(a) Write out the first 4 terms (n = 0,...,3). (b) Starting with the simplest version, arctan x = x, add

terms one at a time to estimate arctan(π/6). After each new term is added, compute the true and approximate percent relative errors. Use your calculator to determine the true value. Add terms until the absolute value of the approximate error estimate falls below an error criterion conforming to two significant figures.

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135

Part Two

Roots and Optimization

2.1 OVERVIEW

Years ago, you learned to use the quadratic formula

x = −b ± √

________ b2 − 4ac ______________ 2a (PT2.1)

to solve

f (x) = ax2 + bx + c = 0 (PT2.2)

The values calculated with Eq. (PT2.1) are called the “roots” of Eq. (PT2.2). They repre- sent the values of x that make Eq. (PT2.2) equal to zero. For this reason, roots are some- times called the zeros of the equation.

Although the quadratic formula is handy for solving Eq. (PT2.2), there are many other functions for which the root cannot be determined so easily. Before the advent of digi- tal computers, there were a number of ways to solve for the roots of such equations. For some cases, the roots could be obtained by direct methods, as with Eq. (PT2.1). Although

there were equations like this that could be solved directly, there were many more that could not. In such instances, the only alternative is an approximate solution technique.

One method to obtain an approxi- mate solution is to plot the function and determine where it crosses the x axis. This point, which represents the x value for which f (x) = 0, is the root. Although graphical methods are useful for obtain- ing rough estimates of roots, they are lim- ited because of their lack of precision. An alternative approach is to use trial and error. This “technique” consists of guess- ing a value of x and evaluating whether f (x) is zero. If not (as is almost always the case), another guess is made, and f (x) is

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136 PART 2 ROOTs And OPTimizATiOn

again evaluated to determine whether the new value provides a better estimate of the root. The process is repeated until a guess results in an f(x) that is close to zero.

Such haphazard methods are obviously inefficient and inadequate for the requirements of engi neering and science practice. Numerical methods represent alternatives that are also approximate but employ systematic strategies to home in on the true root. As elaborated in the following pages, the combination of these systematic methods and computers makes the solution of most applied roots-of-equations problems a simple and efficient task.

Besides roots, another feature of interest to engineers and scientists are a function’s minimum and maximum values. The determination of such optimal values is referred to as optimization. As you learned in calculus, such solutions can be obtained analytically by determining the value at which the function is flat; that is, where its derivative is zero. Although such analytical solutions are sometimes feasible, most practical optimization prob- lems require numerical, computer solutions. From a numerical standpoint, such optimization methods are similar in spirit to the root-location methods we just discussed. That is, both involve guessing and searching for a location on a function. The fundamental difference be- tween the two types of problems is illustrated in Fig. PT2.1. Root location involves searching for the location where the function equals zero. In contrast, optimization involves searching for the function’s extreme points.

2.2 PART ORGANIZATION

The first two chapters in this part are devoted to root location. Chapter 5 focuses on brack- eting methods for finding roots. These methods start with guesses that bracket, or contain, the root and then systematically reduce the width of the bracket. Two specific methods are covered: bisection and false position. Graphical methods are used to provide visual insight into the techniques. Error formulations are developed to help you determine how much computational effort is required to estimate the root to a prespecified level of precision.

Chapter 6 covers open methods. These methods also involve systematic trial-and-error iterations but do not require that the initial guesses bracket the root. We will discover that these methods are usually more computationally efficient than bracketing methods but

Maximum

Minimum

0 Root

Root

f (x)

f ʹ(x) = 0 f ʺ(x) > 0

f ʹ(x) = 0 f ʺ(x) < 0

f (x) = 0

FIGURE PT2.1 A function of a single variable illustrating the difference between roots and optima.

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2.2 PART ORGAnizATiOn 137

that they do not always work. We illustrate several open methods including the fixed-point iteration, Newton-Raphson, and secant methods.

Following the description of these individual open methods, we then discuss a hybrid approach called Brent’s root-finding method that exhibits the reliability of the bracketing methods while exploiting the speed of the open methods. As such, it forms the basis for MATLAB’s root-finding function, fzero. After illustrating how fzero can be used for en- gineering and scientific problems solving, Chap. 6 ends with a brief discussion of special methods devoted to finding the roots of polynomials. In particular, we describe MATLAB’s excellent built-in capabilities for this task.

Chapter 7 deals with optimization. First, we describe two bracketing methods, golden- section search and parabolic interpolation, for finding the optima of a function of a single variable. Then, we discuss a robust, hybrid approach that combines golden-section search and quadratic interpolation. This approach, which again is attributed to Brent, forms the basis for MATLAB’s one-dimensional root-finding function:fminbnd. After describing and illustrating fminbnd, the last part of the chapter provides a brief description of optimiza- tion of multidimensional functions. The emphasis is on describing and illustrating the use of MATLAB’s capability in this area: the fminsearch function. Finally, the chapter ends with an example of how MATLAB can be employed to solve optimization problems in engineering and science.

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138

Roots: Bracketing methods 5

YOU’VE GOT A PROBLEM

medical studies have established that a bungee jumper’s chances of sustaining a significant vertebrae injury increase significantly if the free-fall velocity exceeds 36 m/s after 4 s of free fall. Your boss at the bungee-jumping company wants you to determine the mass at which this criterion is exceeded given a drag coefficient of 0.25 kg/m.

You know from your previous studies that the following analytical solution can be used to predict fall velocity as a function of time:

υ(t) = √ ___

gm ___ cd tanh ( √ ___

gcd ___ m t ) (5.1)

Try as you might, you cannot manipulate this equation to explicitly solve for m—that is, you cannot isolate the mass on the left side of the equation.

CHAPTER OBJECTIVES

The primary objective of this chapter is to acquaint you with bracketing methods for finding the root of a single nonlinear equation. Specific objectives and topics covered are

• Understanding what roots problems are and where they occur in engineering and science.

• Knowing how to determine a root graphically. • Understanding the incremental search method and its shortcomings. • Knowing how to solve a roots problem with the bisection method. • Knowing how to estimate the error of bisection and why it differs from error

estimates for other types of root-location algorithms. • Understanding false position and how it differs from bisection.

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5.1 ROOTs in enGineeRinG And science 139

An alternative way of looking at the problem involves subtracting υ(t) from both sides to give a new function:

f(m) = √ ___

gm ___ cd tanh ( √ ___

gcd ___ m t ) − υ(t) (5.2)

Now we can see that the answer to the problem is the value of m that makes the function equal to zero. Hence, we call this a “roots” problem. This chapter will introduce you to how the computer is used as a tool to obtain such solutions.

5.1 ROOTS IN ENGINEERING AND SCIENCE

Although they arise in other problem contexts, roots of equations frequently occur in the area of design. Table 5.1 lists a number of fundamental principles that are routinely used in design work. As introduced in Chap. 1, mathematical equations or models derived from these principles are employed to predict dependent variables as a function of independent variables, forcing functions, and parameters. Note that in each case, the dependent vari- ables reflect the state or performance of the system, whereas the parameters represent its properties or composition.

An example of such a model is the equation for the bungee jumper’s velocity. If the parameters are known, Eq. (5.1) can be used to predict the jumper’s velocity. Such com- putations can be performed directly because υ is expressed explicitly as a function of the model parameters. That is, it is isolated on one side of the equal sign.

However, as posed at the start of the chapter, suppose that we had to determine the mass for a jumper with a given drag coefficient to attain a prescribed velocity in a set time period. Although Eq. (5.1) provides a mathematical representation of the interrelationship among the model variables and parameters, it cannot be solved explicitly for mass. In such cases, m is said to be implicit.

TABLE 5.1 Fundamental principles used in design problems.

Fundamental Dependent Independent Principle Variable Variable Parameters

Heat balance Temperature Time and position Thermal properties of material, system geometry Mass balance Concentration or quantity Time and position Chemical behavior of material, mass transfer, of mass system geometry Force balance Magnitude and direction Time and position Strength of material, structural properties, system of forces geometry Energy balance Changes in kinetic and Time and position Thermal properties, mass of material, system potential energy geometry Newton’s laws of Acceleration, velocity, Time and position Mass of material, system geometry, dissipative motion or location parameters Kirchhoff’s laws Currents and voltages Time Electrical properties (resistance, capacitance,

inductance)

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140 ROOTs: BRAckeTinG meThOds

This represents a real dilemma, because many design problems involve specifying the properties or composition of a system (as represented by its parameters) to ensure that it performs in a desired manner (as represented by its variables). Thus, these problems often require the determination of implicit parameters.

The solution to the dilemma is provided by numerical methods for roots of equations. To solve the problem using numerical methods, it is conventional to reexpress Eq. (5.1) by subtracting the dependent variable υ from both sides of the equation to give Eq. (5.2). The value of m that makes f (m) = 0 is, therefore, the root of the equation. This value also represents the mass that solves the design problem.

The following pages deal with a variety of numerical and graphical methods for deter- mining roots of relationships such as Eq. (5.2). These techniques can be applied to many other problems confronted routinely in engineering and science.

5.2 GRAPHICAL METHODS

A simple method for obtaining an estimate of the root of the equation f (x) = 0 is to make a plot of the function and observe where it crosses the x axis. This point, which represents the x value for which f (x) = 0, provides a rough approximation of the root.

eXAmPLe 5.1 The Graphical Approach

Problem statement. Use the graphical approach to determine the mass of the bungee jumper with a drag coefficient of 0.25 kg/m to have a velocity of 36 m/s after 4 s of free fall. Note: The acceleration of gravity is 9.81 m/s2.

solution. The following MATLAB session sets up a plot of Eq. (5.2) versus mass:

>> cd = 0.25; g = 9.81; v = 36; t = 4; >> mp = linspace(50,200); >> fp = sqrt(g*mp/cd).*tanh(sqrt(g*cd./mp)*t)− v; >> plot(mp,fp),grid

Root

−5 50 100 150 200

−4

−3

−2

−1

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5.3 BRAckeTinG meThOds And iniTiAL GUesses 141

The function crosses the m axis between 140 and 150 kg. Visual inspection of the plot provides a rough estimate of the root of 145 kg (about 320 lb). The validity of the graphical estimate can be checked by substituting it into Eq. (5.2) to yield

>> sqrt(g*145/cd)*tanh(sqrt(g*cd/145)*t)− v

ans = 0.0456

which is close to zero. It can also be checked by substituting it into Eq. (5.1) along with the parameter values from this example to give

>> sqrt(g*145/cd)*tanh(sqrt(g*cd/145)*t)

ans = 36.0456

which is close to the desired fall velocity of 36 m/s.

Graphical techniques are of limited practical value because they are not very precise. However, graphical methods can be utilized to obtain rough estimates of roots. These esti- mates can be employed as starting guesses for numerical methods discussed in this chapter.

Aside from providing rough estimates of the root, graphical interpretations are useful for understanding the properties of the functions and anticipating the pitfalls of the numeri- cal methods. For example, Fig. 5.1 shows a number of ways in which roots can occur (or be absent) in an interval prescribed by a lower bound xl and an upper bound xu. Figure 5.1b depicts the case where a single root is bracketed by negative and positive values of f (x). However, Fig. 5.1d, where f(xl ) and f(xu ) are also on opposite sides of the x axis, shows three roots occurring within the interval. In general, if f(xl ) and f(xu ) have opposite signs, there are an odd number of roots in the interval. As indicated by Fig. 5.1a and c, if f(xl ) and f(xu ) have the same sign, there are either no roots or an even number of roots between the values.

Although these generalizations are usually true, there are cases where they do not hold. For example, functions that are tangential to the x axis (Fig. 5.2a) and discontinuous func- tions (Fig. 5.2b) can violate these principles. An example of a function that is tangential to the axis is the cubic equation f (x) = (x − 2)(x − 2)(x − 4). Notice that x = 2 makes two terms in this polynomial equal to zero. Mathematically, x = 2 is called a multiple root. Although they are beyond the scope of this book, there are special techniques that are expressly designed to locate multiple roots (Chapra and Canale, 2010).

The existence of cases of the type depicted in Fig. 5.2 makes it difficult to develop foolproof computer algorithms guaranteed to locate all the roots in an interval. However, when used in conjunction with graphical approaches, the methods described in the fol- lowing sections are extremely useful for solving many problems confronted routinely by engineers, scientists, and applied mathematicians.

5.3 BRACKETING METHODS AND INITIAL GUESSES

If you had a roots problem in the days before computing, you’d often be told to use “trial and error” to come up with the root. That is, you’d repeatedly make guesses until the function was sufficiently close to zero. The process was greatly facilitated by the advent of software

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142 ROOTs: BRAckeTinG meThOds

(a)

(c)

(d)

f (x)

xl xu

(b)

f (x)

FIGURE 5.1 Illustration of a number of general ways that a root may occur in an interval prescribed by a lower bound xl and an upper bound xu. Parts (a) and (c) indicate that if both f(xl) and f(xu) have the same sign, either there will be no roots or there will be an even number of roots within the interval. Parts (b) and (d) indicate that if the function has different signs at the end points, there will be an odd number of roots in the interval.

(a)

(b)

f (x)

xl xu

FIGURE 5.2 Illustration of some exceptions to the general cases depicted in Fig. 5.1. (a) Multiple roots that occur when the function is tangential to the x axis. For this case, although the end points are of opposite signs, there are an even number of axis interceptions for the interval. (b) Discontinuous functions where end points of opposite sign bracket an even number of roots. Special strategies are required for determining the roots for these cases.

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5.3 BRAckeTinG meThOds And iniTiAL GUesses 143

tools such as spreadsheets. By allowing you to make many guesses rapidly, such tools can actually make the trial-and-error approach attractive for some problems.

But, for many other problems, it is preferable to have methods that come up with the correct answer automatically. Interestingly, as with trial and error, these approaches require an initial “guess” to get started. Then they systematically home in on the root in an iterative fashion.

The two major classes of methods available are distinguished by the type of initial guess. They are

• Bracketing methods. As the name implies, these are based on two initial guesses that “bracket” the root—that is, are on either side of the root.

• Open methods. These methods can involve one or more initial guesses, but there is no need for them to bracket the root.

For well-posed problems, the bracketing methods always work but converge slowly (i.e., they typically take more iterations to home in on the answer). In contrast, the open methods do not always work (i.e., they can diverge), but when they do they usually con- verge quicker.

In both cases, initial guesses are required. These may naturally arise from the physical context you are analyzing. However, in other cases, good initial guesses may not be obvi- ous. In such cases, automated approaches to obtain guesses would be useful. The following section describes one such approach, the incremental search.

5.3.1 Incremental Search

When applying the graphical technique in Example 5.1, you observed that f (x) changed sign on opposite sides of the root. In general, if f (x) is real and continuous in the interval from xl to xu and f (xl) and f (xu) have opposite signs, that is,

f (xl ) f (xu ) < 0 (5.3)

then there is at least one real root between xl and xu. Incremental search methods capitalize on this observation by locating an interval

where the function changes sign. A potential problem with an incremental search is the choice of the increment length. If the length is too small, the search can be very time consuming. On the other hand, if the length is too great, there is a possibility that closely spaced roots might be missed (Fig. 5.3). The problem is compounded by the possible exis- tence of multiple roots.

An M-file can be developed1 that implements an incremental search to locate the roots of a function func within the range from xmin to xmax (Fig. 5.4). An optional argument ns allows the user to specify the number of intervals within the range. If ns is omitted, it is automatically set to 50. A for loop is used to step through each interval. In the event that a sign change occurs, the upper and lower bounds are stored in an array xb.

1 This function is a modified version of an M-file originally presented by Recktenwald (2000).

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function xb = incsearch(func,xmin,xmax,ns) % incsearch: incremental search root locator % xb = incsearch(func,xmin,xmax,ns): % finds brackets of x that contain sign changes % of a function on an interval % input: % func = name of function % xmin, xmax = endpoints of interval % ns = number of subintervals (default = 50) % output: % xb(k,1) is the lower bound of the kth sign change % xb(k,2) is the upper bound of the kth sign change % If no brackets found, xb = [].

if nargin < 3, error('at least 3 arguments required'), end if nargin < 4, ns = 50; end %if ns blank set to 50

% Incremental search x = linspace(xmin,xmax,ns); f = func(x); nb = 0; xb = []; %xb is null unless sign change detected for k = 1:length(x)−1 if sign(f(k)) ~ = sign(f(k+1)) %check for sign change nb = nb + 1; xb(nb,1) = x(k); xb(nb,2) = x(k+1); end end if isempty(xb) %display that no brackets were found disp('no brackets found') disp('check interval or increase ns') else disp('number of brackets:') %display number of brackets disp(nb) end

FIGURE 5.4 An M-file to implement an incremental search.

x6x0 x1 x2 x3 x4 x5

f (x)

FIGURE 5.3 Cases where roots could be missed because the incremental length of the search procedure is too large. Note that the last root on the right is multiple and would be missed regardless of the increment length.

144

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5.3 BRAckeTinG meThOds And iniTiAL GUesses 145

eXAmPLe 5.2 incremental search

Problem statement. Use the M-file incsearch (Fig. 5.4) to identify brackets within the interval [3, 6] for the function:

f (x) = sin(10x) + cos(3x) (5.4)

solution. The MATLAB session using the default number of intervals (50) is >> incsearch(@(x) sin(10*x)+cos(3*x),3,6) number of brackets: 5

ans =

3.2449 3.3061 3.3061 3.3673 3.7347 3.7959 4.6531 4.7143 5.6327 5.6939

A plot of Eq. (5.4) along with the root locations is shown here.

3 −2

−1

3.5 4 4.5 5 5.5 6

Although five sign changes are detected, because the subintervals are too wide, the func- tion misses possible roots at x ≅ 4.25 and 5.2. These possible roots look like they might be double roots. However, by using the zoom in tool, it is clear that each represents two real roots that are very close together. The function can be run again with more subintervals with the result that all nine sign changes are located >> incsearch(@(x) sin(10*x) + cos(3*x),3,6,100)

number of brackets: 9

ans = 3.2424 3.2727 3.3636 3.3939 3.7273 3.7576

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4.2121 4.2424 4.2424 4.2727 4.6970 4.7273 5.1515 5.1818 5.1818 5.2121 5.6667 5.6970

3 −2

−1

3.5 4 4.5 5 5.5 6

The foregoing example illustrates that brute-force methods such as incremental search are not foolproof. You would be wise to supplement such automatic techniques with any other information that provides insight into the location of the roots. Such information can be found by plotting the function and through understanding the physical problem from which the equation originated.

5.4 BISECTION

The bisection method is a variation of the incremental search method in which the interval is always divided in half. If a function changes sign over an interval, the function value at the midpoint is evaluated. The location of the root is then determined as lying within the subinterval where the sign change occurs. The subinterval then becomes the interval for the next iteration. The process is repeated until the root is known to the required precision. A graphical depiction of the method is provided in Fig. 5.5. The following example goes through the actual computations involved in the method.

eXAmPLe 5.3 The Bisection method

Problem statement. Use bisection to solve the same problem approached graphically in Example 5.1.

solution. The first step in bisection is to guess two values of the unknown (in the present problem, m) that give values for f(m) with different signs. From the graphical solution in

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5.4 BisecTiOn 147

Example 5.1, we can see that the function changes sign between values of 50 and 200. The plot obviously suggests better initial guesses, say 140 and 150, but for illustrative purposes let’s assume we don’t have the benefit of the plot and have made conservative guesses. Therefore, the initial estimate of the root xr lies at the midpoint of the interval

xr = 50 + 200 ________ 2 = 125

Note that the exact value of the root is 142.7376. This means that the value of 125 calcu- lated here has a true percent relative error of

|εt| = | 142.7376 − 125 ______________ 142.7376 | × 100% = 12.43% Next we compute the product of the function value at the lower bound and at the midpoint:

f (50) f (125) = −4.579(−0.409) = 1.871

which is greater than zero, and hence no sign change occurs between the lower bound and the midpoint. Consequently, the root must be located in the upper interval between 125 and 200. Therefore, we create a new interval by redefining the lower bound as 125.

−6

−4

−2

50 100 150

First iteration

Second iteration

Third iteration

Fourth iteration

Root

f (m)

xl xr xu

FIGURE 5.5 A graphical depiction of the bisection method. This plot corresponds to the first four iterations from Example 5.3.

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148 ROOTs: BRAckeTinG meThOds

At this point, the new interval extends from xl = 125 to xu = 200. A revised root esti- mate can then be calculated as

xr = 125 + 200 ________ 2 = 162.5

which represents a true percent error of |εt| = 13.85%. The process can be repeated to obtain refined estimates. For example,

f (125) f (162.5) = −0.409(0.359) = −0.147

Therefore, the root is now in the lower interval between 125 and 162.5. The upper bound is redefined as 162.5, and the root estimate for the third iteration is calculated as

xr = 125 + 162.5 ________ 2 = 143.75

which represents a percent relative error of εt = 0.709%. The method can be repeated until the result is accurate enough to satisfy your needs.

We ended Example 5.3 with the statement that the method could be continued to ob- tain a refined estimate of the root. We must now develop an objective criterion for deciding when to terminate the method.

An initial suggestion might be to end the calculation when the error falls below some prespecified level. For instance, in Example 5.3, the true relative error dropped from 12.43 to 0.709% during the course of the computation. We might decide that we should terminate when the error drops below, say, 0.5%. This strategy is flawed because the error estimates in the example were based on knowledge of the true root of the function. This would not be the case in an actual situation because there would be no point in using the method if we already knew the root.

Therefore, we require an error estimate that is not contingent on foreknowledge of the root. One way to do this is by estimating an approximate percent relative error as in [recall Eq. (4.5)]

|εa| = | x r new − x r old _________ x r new |100% (5.5) where x r new is the root for the present iteration and x r old is the root from the previous itera- tion. When εa becomes less than a prespecified stopping criterion εs, the computation is terminated.

eXAmPLe 5.4 error estimates for Bisection

Problem statement. Continue Example 5.3 until the approximate error falls below a stopping criterion of εs = 0.5%. Use Eq. (5.5) to compute the errors.

solution. The results of the first two iterations for Example 5.3 were 125 and 162.5. Sub- stituting these values into Eq. (5.5) yields

|εa| = | 162.5 − 125 ___________ 162.5 | 100% = 23.08%

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5.4 BisecTiOn 149

Recall that the true percent relative error for the root estimate of 162.5 was 13.85%. There- fore, |εa| is greater than |εt| . This behavior is manifested for the other iterations:

Iteration xl xu xr | εa| (%) | εt | (%) 1 50 200 125 12.43 2 125 200 162.5 23.08 13.85 3 125 162.5 143.75 13.04 0.71 4 125 143.75 134.375 6.98 5.86 5 134.375 143.75 139.0625 3.37 2.58 6 139.0625 143.75 141.4063 1.66 0.93 7 141.4063 143.75 142.5781 0.82 0.11 8 142.5781 143.75 143.1641 0.41 0.30

Thus after eight iterations |εa| finally falls below εs = 0.5%, and the computation can be terminated. These results are summarized in Fig. 5.6. The “ragged” nature of the true error is due to the fact that, for bisection, the true root can lie anywhere within the bracketing interval. The true and approximate errors are far apart when the interval happens to be centered on the true root. They are close when the true root falls at either end of the interval.

0.1

100

Iterations

Pe rc

en t r

el at

iv e

er ro

Approximate error, ∣εa∣

True error, ∣εt∣

0 2 4 6 8

FIGURE 5.6 Errors for the bisection method. True and approximate errors are plotted versus the number of iterations.

Although the approximate error does not provide an exact estimate of the true error, Fig. 5.6 suggests that |εa| captures the general downward trend of |εt| . In addition, the plot exhibits the extremely attractive characteristic that |εa| is always greater than |εt| . Thus,

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when |εa| falls below εs, the computation could be terminated with confidence that the root is known to be at least as accurate as the prespecified acceptable level.

While it is dangerous to draw general conclusions from a single example, it can be demonstrated that |εa| will always be greater than |εt| for bisection. This is due to the fact that each time an approximate root is located using bisection as xr = (xl + xu )∕2, we know that the true root lies somewhere within an interval of Δx = xu − xl. Therefore, the root must lie within ±Δx∕2 of our estimate. For instance, when Example 5.4 was terminated, we could make the definitive statement that

xr = 143.1641 ± 143.7500 − 142.5781 __________________ 2 = 143.1641 ± 0.5859

In essence, Eq. (5.5) provides an upper bound on the true error. For this bound to be exceeded, the true root would have to fall outside the bracketing interval, which by definition could never occur for bisection. Other root-locating techniques do not always behave as nicely. Although bisection is generally slower than other methods, the neatness of its error analysis is a positive feature that makes it attractive for certain engineering and scientific applications.

Another benefit of the bisection method is that the number of iterations required to attain an absolute error can be computed a priori—that is, before starting the computation. This can be seen by recognizing that before starting the technique, the absolute error is

E a 0 = x u 0 − x l 0 = Δx0

where the superscript designates the iteration. Hence, before starting the method we are at the “zero iteration.” After the first iteration, the error becomes

E a 1 = Δx0 ____ 2

Because each succeeding iteration halves the error, a general formula relating the error and the number of iterations n is

E a n = Δx0 ____ 2n

If Ea,d is the desired error, this equation can be solved for2

n = log(Δx0∕Ea,d ) ____________ log 2 = log2 ( Δx0 ____ Ea,d ) (5.6)

Let’s test the formula. For Example 5.4, the initial interval was Δx0 = 200 − 50 = 150. After eight iterations, the absolute error was

Ea = |143.7500 − 142.5781| ___________________ 2 = 0.5859

We can substitute these values into Eq. (5.6) to give

n = log2(150∕0.5859) = 8

2 MATLAB provides the log2 function to evaluate the base-2 logarithm directly. If the pocket calculator or computer language you are using does not include the base-2 logarithm as an intrinsic function, this equation shows a handy way to compute it. In general, logb(x) = log(x)∕log(b).

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5.4 BisecTiOn 151

Thus, if we knew beforehand that an error of less than 0.5859 was acceptable, the formula tells us that eight iterations would yield the desired result.

Although we have emphasized the use of relative errors for obvious reasons, there will be cases where (usually through knowledge of the problem context) you will be able to specify an absolute error. For these cases, bisection along with Eq. (5.6) can provide a useful root-location algorithm.

5.4.1 MATLAB M-file: bisect

An M-file to implement bisection is displayed in Fig. 5.7. It is passed the function (func) along with lower (xl) and upper (xu) guesses. In addition, an optional stopping criterion (es)

function [root,fx,ea,iter]=bisect(func,xl,xu,es,maxit,varargin) % bisect: root location zeroes % [root,fx,ea,iter]=bisect(func,xl,xu,es,maxit,p1,p2,...): % uses bisection method to find the root of func % input: % func = name of function % xl, xu = lower and upper guesses % es = desired relative error (default = 0.0001%) % maxit = maximum allowable iterations (default = 50) % p1,p2,... = additional parameters used by func % output: % root = real root % fx = function value at root % ea = approximate relative error (%) % iter = number of iterations

if nargin<3,error('at least 3 input arguments required'),end test = func(xl,varargin{:})*func(xu,varargin{:}); if test>0,error('no sign change'),end if nargin<4|isempty(es), es=0.0001;end if nargin<5|isempty(maxit), maxit=50;end iter = 0; xr = xl; ea = 100; while (1) xrold = xr; xr = (xl + xu)/2; iter = iter + 1; if xr ~ = 0,ea = abs((xr − xrold)/xr) * 100;end test = func(xl,varargin{:})*func(xr,varargin{:}); if test < 0 xu = xr; elseif test > 0 xl = xr; else ea = 0; end if ea <= es | iter >= maxit,break,end end root = xr; fx = func(xr, varargin{:});

FIGURE 5.7 An M-file to implement the bisection method.

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152 ROOTs: BRAckeTinG meThOds

and maximum iterations (maxit) can be entered. The function first checks whether there are sufficient arguments and if the initial guesses bracket a sign change. If not, an error message is displayed and the function is terminated. It also assigns default values if maxit and es are not supplied. Then a while...break loop is employed to implement the bisection algorithm until the approximate error falls below es or the iterations exceed maxit.

We can employ this function to solve the problem posed at the beginning of the chapter. Recall that you need to determine the mass at which a bungee jumper’s free-fall velocity exceeds 36 m/s after 4 s of free fall given a drag coefficient of 0.25 kg/m. Thus, you have to find the root of

f(m) = √ ______

9.81m ______ 0.25 tanh ( √ _________

9.81(0.25) _________ m 4 ) − 36 In Example 5.1 we generated a plot of this function versus mass and estimated that the root fell between 140 and 150 kg. The bisect function from Fig. 5.7 can be used to determine the root with the following script

fm = @(m,cd,t,v) sqrt(9.81*m/cd)*tanh(sqrt(9.81*cd/m)*t) − v; [mass fx ea iter] = bisect(@(m) fm(m,0.25,4,36),40,200)

mass = 142.7377 fx = 4.6089e-007 ea = 5.345e-005 iter = 21

Thus, a result of m = 142.74 kg is obtained after 21 iterations with an approximate relative error of εa = 0.00005345%, and a function value close to zero.

5.5 FALSE POSITION

False position (also called the linear interpolation method) is another well-known bracket- ing method. It is very similar to bisection with the exception that it uses a different strategy to come up with its new root estimate. Rather than bisecting the interval, it locates the root by joining f (xl ) and f (xu ) with a straight line (Fig. 5.8). The intersection of this line with the x axis represents an improved estimate of the root. Thus, the shape of the function in- fluences the new root estimate. Using similar triangles, the intersection of the straight line with the x axis can be estimated as (see Chapra and Canale, 2010, for details),

xr = xu − f (xu )(xl − xu ) ___________ f (xl ) − f (xu )

(5.7)

This is the false-position formula. The value of xr computed with Eq. (5.7) then re- places whichever of the two initial guesses, xl or xu, yields a function value with the same sign as f (xr). In this way the values of xl and xu always bracket the true root. The process

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5.5 FALse POsiTiOn 153

is repeated until the root is estimated adequately. The algorithm is identical to the one for bisection (Fig. 5.7) with the exception that Eq. (5.7) is used.

eXAmPLe 5.5 The False-Position method

Problem statement. Use false position to solve the same problem approached graphi- cally and with bisection in Examples 5.1 and 5.3.

solution. As in Example 5.3, initiate the computation with guesses of xl = 50 and xu = 200.

First iteration:

xl = 50 f (xl ) = −4.579387

xu = 200 f (xu) = 0.860291

xr = 200 − 0.860291(50 − 200) ___________________ −4.579387 − 0.860291 = 176.2773

which has a true relative error of 23.5%.

Second iteration:

f (xl) f (xr) = −2.592732

f (x)

f (xl)

f (xu)

FIGURE 5.8 False position.

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154 ROOTs: BRAckeTinG meThOds

Therefore, the root lies in the first subinterval, and xr becomes the upper limit for the next iteration, xu = 176.2773.

xl = 50 f (xl ) = −4.579387

xu = 176.2773 f (xu) = 0.566174

xr = 176.2773 − 0.566174(50 − 176.2773) ___________________ −4.579387 − 0.566174 = 162.3828

which has true and approximate relative errors of 13.76% and 8.56%, respectively. Addi- tional iterations can be performed to refine the estimates of the root.

Although false position often performs better than bisection, there are other cases where it does not. As in the following example, there are certain cases where bisection yields superior results.

eXAmPLe 5.6 A case Where Bisection is Preferable to False Position

Problem statement. Use bisection and false position to locate the root of

f (x) = x10 − 1

between x = 0 and 1.3.

solution. Using bisection, the results can be summarized as

Iteration xl xu xr εa (%) εt (%)

1 0 1.3 0.65 100.0 35 2 0.65 1.3 0.975 33.3 2.5 3 0.975 1.3 1.1375 14.3 13.8 4 0.975 1.1375 1.05625 7.7 5.6 5 0.975 1.05625 1.015625 4.0 1.6

Thus, after five iterations, the true error is reduced to less than 2%. For false position, a very different outcome is obtained:

Iteration xl xu xr εa (%) εt (%)

1 0 1.3 0.09430 90.6 2 0.09430 1.3 0.18176 48.1 81.8 3 0.18176 1.3 0.26287 30.9 73.7 4 0.26287 1.3 0.33811 22.3 66.2 5 0.33811 1.3 0.40788 17.1 59.2

After five iterations, the true error has only been reduced to about 59%. Insight into these results can be gained by examining a plot of the function. As in Fig. 5.9, the curve violates the premise on which false position was based—that is, if f (xl) is much closer to

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5.5 FALse POsiTiOn 155

zero than f (xu ), then the root should be much closer to xl than to xu (recall Fig. 5.8). Because of the shape of the present function, the opposite is true.

The foregoing example illustrates that blanket generalizations regarding root- location methods are usually not possible. Although a method such as false position is often superior to bisection, there are invariably cases that violate this general conclusion. Therefore, in addition to using Eq. (5.5), the results should always be checked by substi- tuting the root estimate into the original equation and determining whether the result is close to zero.

The example also illustrates a major weakness of the false-position method: its one- sidedness. That is, as iterations are proceeding, one of the bracketing points will tend to stay fixed. This can lead to poor convergence, particularly for functions with significant curvature. Possible remedies for this shortcoming are available elsewhere (Chapra and Canale, 2010).

1.0

f (x)

FIGURE 5.9 Plot of f (x) = x10 − 1, illustrating slow convergence of the false-position method.

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5.6 CASE STUDY GReenhOUse GAses And RAinWATeR

Background. It is well documented that the atmospheric levels of several so-called “greenhouse” gases have been increasing over the past 50 years. For example, Fig. 5.10 shows data for the partial pressure of carbon dioxide (CO2 ) collected at Mauna Loa, Hawaii from 1958 through 2008. The trend in these data can be nicely fit with a quadratic polynomial,3

pCO2 = 0.012226(t − 1983) 2 + 1.418542(t − 1983) + 342.38309

where pCO2 = CO2 partial pressure (ppm). These data indicate that levels have increased a little over 22% over the period from 315 to 386 ppm.

One question that we can address is how this trend is affecting the pH of rainwater. Outside of urban and industrial areas, it is well documented that carbon dioxide is the pri- mary determinant of the pH of the rain. pH is the measure of the activity of hydrogen ions and, therefore, its acidity or alkalinity. For dilute aqueous solutions, it can be computed as

pH = −log10[H+] (5.8)

where [H+] is the molar concentration of hydrogen ions. The following five equations govern the chemistry of rainwater:

K1 = 106 [H+][ HCO 3

− ] ___________ KH pCO2

(5.9)

3 In Part Four, we will learn how to determine such polynomials.

310

330

350

370

1950 1960 1970 1980 1990 2000 2010

pCO2 (ppm)

FIGURE 5.10 Average annual partial pressures of atmospheric carbon dioxide (ppm) measured at Mauna Loa, Hawaii.

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5.6 CASE STUDY continued

K2 = [H+][ CO 3

−2 ] ___________ [H CO 3

− ] (5.10)

K𝑤 = [H+][OH−] (5.11)

cT = KH pCO2 ______

106 + [ HCO 3

− ] + [ CO 3 −2 ] (5.12)

0 = [ HCO 3 − ] + 2[ CO 3

−2 ] + [OH−] − [H+] (5.13)

where KH = Henry’s constant, and K1, K2, and K𝑤 are equilibrium coefficients. The five unknowns are cT = total inorganic carbon, [ HCO 3

− ] = bicarbonate, [ CO 3 −2 ] = carbonate,

[H+] = hydrogen ion, and [OH−] = hydroxyl ion. Notice how the partial pressure of CO2 shows up in Eqs. (5.9) and (5.12).

Use these equations to compute the pH of rainwater given that KH = 10−1.46, K1 = 10−6.3, K2 = 10−10.3, and K𝑤 = 10−14. Compare the results in 1958 when the pCO2 was 315 and in 2008 when it was 386 ppm. When selecting a numerical method for your com- putation, consider the following:

• You know with certainty that the pH of rain in pristine areas always falls between 2 and 12.

• You also know that pH can only be measured to two places of decimal precision.

Solution. There are a variety of ways to solve this system of five equations. One way is to eliminate unknowns by combining them to produce a single function that only depends on [H+]. To do this, first solve Eqs. (5.9) and (5.10) for

[ HCO 3 − ] =

K1 _______ 106[H+]

KH pCO2 (5.14)

[ CO 3 −2 ] =

K2 [ HCO 3 − ] _________

[H+] (5.15)

Substitute Eq. (5.14) into (5.15)

[ CO 3 −2 ] =

K2 K1 _______ 106[H+]2

KH pCO2 (5.16)

Equations (5.14) and (5.16) can be substituted along with Eq. (5.11) into Eq. (5.13) to give

0 = K1 _______

106[H+] KH pCO2 + 2

K2 K1 _______ 106[H+]2

KH pCO2 + K𝑤 _____ [H+] − [H

+] (5.17)

Although it might not be immediately apparent, this result is a third-order polynomial in [H+]. Thus, its root can be used to compute the pH of the rainwater.

Now we must decide which numerical method to employ to obtain the solution. There are two reasons why bisection would be a good choice. First, the fact that the pH always falls within the range from 2 to 12, provides us with two good initial guesses. Second, be- cause the pH can only be measured to two decimal places of precision, we will be satisfied

5.6 cAse sTUdY 157

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with an absolute error of Ea,d = ±0.005. Remember that given an initial bracket and the desired error, we can compute the number of iteration a priori. Substituting the present values into Eq. (5.6) gives

>> dx = 12 − 2; >> Ead = 0.005; >> n = log2(dx/Ead)

n = 10.9658

Eleven iterations of bisection will produce the desired precision. Before implementing bisection, we must first express Eq. (5.17) as a function. Because

it is relatively complicated, we will store it as an M-file: function f = fpH(pH,pCO2) K1 = 10^− 6.3;K2=10^− 10.3;Kw = 10^−14; KH = 10^− 1.46; H = 10^− pH; f = K1/(1e6*H)*KH*pCO2+2*K2*K1/(1e6*H)*KH*pCO2+Kw/H − H;

We can then use the M-file from Fig. 5.7 to obtain the solution. Notice how we have set the value of the desired relative error (εa = 1 × 10−8) at a very low level so that the iteration limit (maxit) is reached first so that exactly 11 iterations are implemented

>> [pH1958 fx ea iter] = bisect(@(pH) fpH(pH,315),2,12,1e − 8,11)

pH1958 = 5.6279 fx = −2.7163e − 008 ea = 0.0868 iter = 11

Thus, the pH is computed as 5.6279 with a relative error of 0.0868%. We can be con- fident that the rounded result of 5.63 is correct to two decimal places. This can be verified by performing another run with more iterations. For example, setting maxit to 50 yields

>> [pH1958 fx ea iter] = bisect(@(pH) fpH(pH,315),2,12,1e − 8,50)

pH1958 = 5.6304 fx = 1.615e − 015 ea = 5.1690e − 009 iter = 35

For 2008, the result is >> [pH2008 fx ea iter] = bisect(@(pH) fpH(pH,386),2,12,1e − 8,50)

5.6 CASE STUDY continued

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5.6 CASE STUDY continued

pH2008 = 5.5864 fx = 3.2926e − 015 ea = 5.2098e − 009 iter = 35

Interestingly, the results indicate that the 22.5% rise in atmospheric CO2 levels has produced only a 0.78% drop in pH. Although this is certainly true, remember that the pH represents a logarithmic scale as defined by Eq. (5.8). Consequently, a unit drop in pH rep- resents an order-of-magnitude (i.e., a 10-fold) increase in the hydrogen ion. The concentra- tion can be computed as [H+] = 10−pH and its percent change can be calculated as

>> ((10^− pH2008 − 10^− pH1958)/10^− pH1958)*100

ans = 10.6791

Therefore, the hydrogen ion concentration has increased about 10.7%. There is quite a lot of controversy related to the meaning of the greenhouse gas trends.

Most of this debate focuses on whether the increases are contributing to global warming. However, regardless of the ultimate implications, it is sobering to realize that something as large as our atmosphere has changed so much over a relatively short time period. This case study illustrates how numerical methods and MATLAB can be employed to analyze and interpret such trends. Over the coming years, engineers and scientists can hopefully use such tools to gain increased understanding of such phenomena and help rationalize the debate over their ramifications.

PROBLEMS

5.1 Use bisection to determine the drag coefficient needed so that an 95-kg bungee jumper has a velocity of 46 m/s after 9 s of free fall. Note: The acceleration of gravity is 9.81 m/s2. Start with initial guesses of xl = 0.2 and xu = 0.5 and iterate until the approximate relative error falls below 5%. 5.2 Develop your own M-file for bisection in a similar fash- ion to Fig. 5.7. However, rather than using the maximum iterations and Eq. (5.5), employ Eq. (5.6) as your stopping criterion. Make sure to round the result of Eq. (5.6) up to the next highest integer (Hint: the ceil function provides a handy way to do this). The first line of your function should be

function [root,Ea,ea,n] = bisectnew(func,xl,xu,Ead, varargin)

Note that for the output, Ea = the approximate absolute error and ea = the approximate percent relative error. Then de- velop your own script, called LastNameHmwk04Script to solve Prob. 5.1. Note that you MUST pass the parameters via the argument. In addition, set up the function so that it uses a default value for Ead = 0.000001. 5.3 Figure P5.3 shows a pinned-fixed beam subject to a uni- form load. The equation for the resulting deflections is

y = − w _____ 48EI (2x 4 − 3Lx3 + L3x)

Develop a MATLAB script that (a) plots the function, dy/dx versus x (with appropriate

labels) and

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160 ROOTs: BRAckeTinG meThOds

(b) uses LastNameBisect to determine the point of maxi- mum deflection (i.e., the value of x where dy/dx = 0). Then substitute this value into the deflection equation to determine the value of the maximum deflection. Employ initial guesses of xl = 0 and xu = 0.9L. Use the follow- ing parameter values in your computation (making sure that you use consistent units): L = 400 cm, E = 52,000 kN/cm2, I = 32,000 cm4, and w = 4 kN/cm. In addi- tion, use Ead = 0.0000001 m. Also, set format long in your script so you display 15 significant digits for your results.

5.4 As shown in Fig. P5.4, the velocity of water, υ (m/s), discharged from a cylindrical tank through a long pipe can be computed as

υ = √ ____

2gH tanh ( √____ 2gH ____ 2L t ) where g = 9.81 m/s2, H = initial head (m), L = pipe length (m), and t = elapsed time (s). Develop a MATLAB script that (a) plots the function f(H) versus H for H = 0 to 4 m (make

sure to label the plot) and (b) uses LastNameBisect with initial guesses of xl = 0 and

xu = 4 m to determine the initial head needed to achieve υ = 5 m/s in 2.5 s for a 4-m long pipe. In addition, use Ead = 0.0000001. Also, set format long in your script so you display 15 significant digits for your results.

5.5 Repeat Prob. 5.1, but use the false-position method to obtain your solution. 5.6 Develop an M-file for the false-position method. Test it by solving Prob. 5.1.

5.7 (a) Determine the roots of f (x) = −12 − 21x + 18x2 − 2.75x3 graphically. In addition, determine the first root of the function with

(b) bisection and (c) false position. For (b) and (c) use initial guesses of xl = −1 and xu = 0 and a stopping

criterion of 1%. 5.8 Locate the first nontrivial root of sin(x) = x2 where x is in radians. Use a graphical technique and bisection with the initial interval from 0.5 to 1. Perform the computation until εa is less than εs = 2%. 5.9 Determine the positive real root of ln(x2) = 0.7 (a) graphi cally, (b) using three iterations of the bisection method, with initial guesses of xl = 0.5 and xu = 2, and (c) using three iterations of the false-position method, with the same initial guesses as in (b). 5.10 The saturation concentration of dissolved oxygen in freshwater can be calculated with the equation

ln osf = −139.34411 + 1.575701 × 105 _____________ Ta

− 6.642308 × 10 7 _____________

T a 2 + 1.243800 × 10

10 ______________

T a 3

− 8.621949 × 10 11 ______________

T a 4

where osf = the saturation concentration of dissolved oxygen in freshwater at 1 atm (mg L−1); and Ta = absolute temperature (K). Remember that Ta = T + 273.15, where T = temperature (°C). According to this equation, saturation decreases with increasing temperature. For typical natural waters in temperate climates, the equation can be used to de- termine that oxygen concentration ranges from 14.621 mg/L at 0 °C to 6.949 mg/L at 35 °C. Given a value of oxygen concentration, this formula and the bisection method can be used to solve for temperature in °C. (a) If the initial guesses are set as 0 and 35 °C, how many

bisection iterations would be required to determine tem- perature to an absolute error of 0.05 °C?

(b) Based on (a), develop and test a bisection M-file function to determine T as a function of a given oxygen concen- tration. Test your function for osf = 8, 10, and 14 mg/L. Check your results.

5.11 A beam is loaded as shown in Fig. P5.11. Use the bisection method to solve for the position inside the beam where there is no moment. 5.12 Water is flowing in a trapezoidal channel at a rate of Q = 20 m3/s. The critical depth y for such a channel must satisfy the equation

0 = 1 − Q 2 ____

g A c 3 B

FIGURE P5.3

FIGURE P5.4

H L

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where g = 9.81 m/s2, Ac = the cross-sectional area (m2), and B = the width of the channel at the surface (m). For this case, the width and the cross-sectional area can be related to depth y by

B = 3 + y and

Ac = 3y + y2 __ 2

Solve for the critical depth using (a) the graphical method, (b) bisection, and (c) false position. For (b) and (c) use initial guesses of xl = 0.5 and xu = 2.5, and iterate until the approximate error falls below 1% or the number of iterations exceeds 10. Discuss your results. 5.13 The Michaelis-Menten model describes the kinetics of enzyme mediated reactions:

dS ___ dt = −υm S _____ ks + S

where S = substrate concentration (moles/L), υm = maxi- mum uptake rate (moles/L/d), and ks = the half-saturation constant, which is the substrate level at which uptake is half of the maximum [moles/L]. If the initial substrate level at t = 0 is S0, this differential equation can be solved for

S = S0 − υmt + ks ln(S0/S) Develop an M-file to generate a plot of S versus t for the case where S0 = 8 moles/L, υm = 0.7 moles/L/d, and ks = 2.5 moles/L. 5.14 A reversible chemical reaction

2A + B →← C

can be characterized by the equilibrium relationship

K = cc ____

c a 2 cb

where the nomenclature ci represents the concentration of constituent i. Suppose that we define a variable x as repre- senting the number of moles of C that are produced. Conser- vation of mass can be used to reformulate the equilibrium relationship as

K = (cc,0 + x) _________________

(ca,0 − 2x)2(cb,0 − x)

where the subscript 0 designates the initial concentration of each constituent. If K = 0.016, ca,0 = 42, cb,0 = 28, and cc,0 = 4, determine the value of x. (a) Obtain the solution graphically. (b) On the basis of (a), solve for the root with initial guesses of

xl = 0 and xu = 20 to εs = 0.5%. Choose either bisection or false position to obtain your solution. Justify your choice.

5.15 Figure P5.15a shows a uniform beam subject to a lin- early increasing distributed load. The equation for the result- ing elastic curve is (see Fig. P5.15b)

y = 𝑤0 ________ 120E I L (−x

5 + 2L2x3 − L4x) (P5.15)

Use bisection to determine the point of maximum deflec- tion (i.e., the value of x where dy/dx = 0). Then substitute this value into Eq. (P5.15) to determine the value of the maximum deflection. Use the following parameter val- ues in your computation: L = 600 cm, E = 50,000 kN/cm2, I = 30,000 cm4, and 𝑤0 = 2.5 kN/cm.

FIGURE P5.11

3’ 3’ 2’4’

100 lb100 lb/ft

FIGURE P5.15

(x = 0, y = 0) (x = L, y = 0)

(a)

(b)

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162 ROOTs: BRAckeTinG meThOds

5.16 You buy a $35,000 vehicle for nothing down at $8500 per year for 7 years. Use the bisect function from Fig. 5.7 to determine the interest rate that you are paying. Employ initial guesses for the interest rate of 0.01 and 0.3 and a stopping criterion of 0.00005. The formula relating present worth P, annual payments A, number of years n, and interest rate i is

A = P i(1 + i ) n __________ (1 + i)n − 1

5.17 Many fields of engineering require accurate population estimates. For example, transportation engineers might find it necessary to determine separately the population growth trends of a city and adjacent suburb. The population of the urban area is declining with time according to

Pu(t) = Pu,maxe−ku t + Pu,min

while the suburban population is growing, as in

Ps (t) = Ps,max ___________________

1 + [Ps,max∕P0 − 1]e−ks t

where Pu,max, ku, Ps,max, P0, and ks = empirically derived pa- rameters. Determine the time and corresponding values of Pu(t) and Ps(t) when the suburbs are 20% larger than the city. The parameter values are Pu,max = 80,000, ku = 0.05/yr, Pu,min = 110,000 people, Ps,max = 320,000 people, P0 = 10,000 people, and ks = 0.09/yr. To obtain your solutions, use (a) graphical and (b) false-position methods. 5.18 The resistivity ρ of doped silicon is based on the charge q on an electron, the electron density n, and the electron mo- bility μ. The electron density is given in terms of the doping density N and the intrinsic carrier density ni. The electron mobility is described by the temperature T, the reference temperature T0, and the reference mobility μ0. The equations required to compute the resistivity are

ρ = 1 ____ qnμ

where

n = 1 __ 2 ( N + √ ________

N2 + 4 n i 2 ) and μ = μ0 ( T __ T0 ) −2.42

Determine N, given T0 = 300 K, T = 1000 K, μ0 = 1360 cm2 (V s)−1, q = 1.7 × 10−19 C, ni = 6.21 × 109 cm−3, and a desired ρ = 6.5 × 106 V s cm/C. Employ initial guesses of N = 0 and 2.5 × 1010. Use (a) bisection and (b) the false position method.

5.19 A total charge Q is uniformly distributed around a ring-shaped conductor with radius a. A charge q is located at a distance x from the center of the ring (Fig. P5.19). The force exerted on the charge by the ring is given by

F = 1 ____ 4πe0 qQx _________ (x2 + a2)3/2

where e0 = 8.9 × 10−12 C2/(N m2). Find the distance x where the force is 1.25 N if q and Q are 2 × 10−5 C for a ring with a radius of 0.85 m. 5.20 For fluid flow in pipes, friction is described by a di- mensionless number, the Fanning friction factor f. The Fanning friction factor is dependent on a number of param- eters related to the size of the pipe and the fluid, which can all be represented by another dimensionless quantity, the Reynolds number Re. A formula that predicts f given Re is the von Karman equation:

1 ___ √

_ f = 4 log10 ( Re √

_ f ) − 0.4

Typical values for the Reynolds number for turbulent flow are 10,000 to 500,000 and for the Fanning friction factor are 0.001 to 0.01. Develop a function that uses bisection to solve for f given a user-supplied value of Re between 2500 and 1,000,000. Design the function so that it ensures that the ab- solute error in the result is Ea,d < 0.000005. 5.21 Mechanical engineers, as well as most other engineers, use thermodynamics extensively in their work. The follow- ing polynomial can be used to relate the zero-pressure spe- cific heat of dry air cp kJ/(kg K) to temperature (K):

cp = 0.99403 + 1.671 × 10−4T + 9.7215 × 10−8T2 −9.5838 × 10−11 T 3 + 1.9520 × 10−14 T 4

Develop a plot of cp versus a range of T = 0 to 1200 K, and then use bisection to determine the temperature that corre- sponds to a specific heat of 1.1 kJ/(kg K).

FIGURE P5.19

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5.22 The upward velocity of a rocket can be computed by the following formula:

υ = u ln m0 _______ m0 − qt

− gt

where υ = upward velocity, u = the velocity at which fuel is expelled relative to the rocket, m0 = the initial mass of the rocket at time t = 0, q = the fuel consumption rate, and g = the downward acceleration of gravity (assumed constant = 9.81 m/s2). If u = 1800 m/s, m0 = 160,000 kg, and q = 2600 kg/s, compute the time at which υ = 750 m/s. (Hint: t is somewhere between 10 and 50 s.) Determine your result so that it is within 1% of the true value. Check your answer. 5.23 Although we did not mention it in Sec. 5.6, Eq. (5.13) is an expression of electroneutrality—that is, that positive and negative charges must balance. This can be seen more clearly by expressing it as

[H+] = [ HCO 3 − ] + 2[ CO 3

2− ] + [OH−]

In other words, the positive charges must equal the negative charges. Thus, when you compute the pH of a natural water body such as a lake, you must also account for other ions that may be present. For the case where these ions originate from nonreactive salts, the net negative minus positive charges due to these ions are lumped together in a quantity called alkalinity, and the equation is reformulated as

Alk + [H+] = [ HCO 3 − ] + 2[ CO 3

2− ] + [OH−] (P5.23)

where Alk = alkalinity (eq/L). For example, the alkalinity of Lake Superior is approximately 0.4 × 10–3 eq/L. Perform the same calculations as in Sec. 5.6 to compute the pH of Lake Superior in 2008. Assume that just like the raindrops, the lake is in equilibrium with atmospheric CO2 but account for the alkalinity as in Eq. (P5.23). 5.24 According to Archimedes’ principle, the buoyancy force is equal to the weight of fluid displaced by the submerged portion of the object. For the sphere depicted in Fig. P5.24, use bisection to determine the height, h, of the portion that is above water. Employ the following values for your computa- tion: r = 1 m, ρs = density of sphere = 200 kg/m3, and ρw = density of water = 1000 kg/m3. Note that the volume of the above-water portion of the sphere can be computed with

V = πh 2 ___ 3 (3r − h)

5.25 Perform the same computation as in Prob. 5.24, but for the frustum of a cone as depicted in Fig. P5.25. Employ the following values for your computation: r1 = 0.5 m, r2 = 1 m, h = 1 m, ρf = frustum density = 200 kg/m3, and ρw = water density = 1000 kg/m3. Note that the volume of a frustum is given by

V = πh ___ 3 ( r 1 2 + r 2

2 + r1r2)

PROBLems 163

FIGURE P5.25

FIGURE P5.24

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164

CHAPTER OBJECTIVES

The primary objective of this chapter is to acquaint you with open methods for finding the root of a single nonlinear equation. Specific objectives and topics covered are

• Recognizing the difference between bracketing and open methods for root location.

• Understanding the fixed-point iteration method and how you can evaluate its convergence characteristics.

• Knowing how to solve a roots problem with the Newton-Raphson method and appreciating the concept of quadratic convergence.

• Knowing how to implement both the secant and the modified secant methods. • Understanding how Brent’s method combines reliable bracketing methods with

fast open methods to locate roots in a robust and efficient manner. • Knowing how to use MATLAB’s fzero function to estimate roots. • Learning how to manipulate and determine the roots of polynomials with

MATLAB.

For the bracketing methods in Chap. 5, the root is located within an interval prescribed by a lower and an upper bound. Repeated application of these methods always results in closer estimates of the true value of the root. Such methods are said to be conver- gent because they move closer to the truth as the computation progresses (Fig. 6.1a).

In contrast, the open methods described in this chapter require only a single starting value or two starting values that do not necessarily bracket the root. As such, they some- times diverge or move away from the true root as the computation progresses (Fig. 6.1b). However, when the open methods converge (Fig. 6.1c) they usually do so much more quickly than the bracketing methods. We will begin our discussion of open techniques with a simple approach that is useful for illustrating their general form and also for demonstrat- ing the concept of convergence.

Roots: Open Methods

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6.1 SIMPLE FIXED-POINT ITERATION 165

6.1 SIMPLE FIXED-POINT ITERATION

As just mentioned, open methods employ a formula to predict the root. Such a formula can be developed for simple fixed-point iteration (or, as it is also called, one-point iteration or successive substitution) by rearranging the function f (x) = 0 so that x is on the left-hand side of the equation:

x = g(x) (6.1)

This transformation can be accomplished either by algebraic manipulation or by simply adding x to both sides of the original equation.

The utility of Eq. (6.1) is that it provides a formula to predict a new value of x as a function of an old value of x. Thus, given an initial guess at the root xi, Eq. (6.1) can be used to compute a new estimate xi+1 as expressed by the iterative formula

xi+1 = g(xi) (6.2)

f (x)

(a)

xl xu

f (x)

(b)

xi xi+1

f (x)

(c)

xi xi+1

xl xu

FIGURE 6.1 Graphical depiction of the fundamental difference between the (a) bracketing and (b) and (c) open methods for root location. In (a), which is bisection, the root is constrained within the interval prescribed by xl and xu. In contrast, for the open method depicted in (b) and (c), which is Newton-Raphson, a formula is used to project from xi to xi+1 in an iterative fashion. Thus the method can either (b) diverge or (c) converge rapidly, depending on the shape of the function and the value of the initial guess.

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166 ROOTS: OPEN METhODS

As with many other iterative formulas in this book, the approximate error for this equation can be determined using the error estimator:

εa = | xi+1 − xi _______ xi+1 |100% (6.3) EXAMPLE 6.1 Simple Fixed-Point Iteration

Problem Statement. Use simple fixed-point iteration to locate the root of f (x) = e−x − x.

Solution. The function can be separated directly and expressed in the form of Eq. (6.2) as

xi+1 = e−xi Starting with an initial guess of x0 = 0, this iterative equation can be applied to compute:

i xi |εa|, % |εt|, % |εt|i∕|εt|i−1

0 0.0000 100.000 1 1.0000 100.000 76.322 0.763 2 0.3679 171.828 35.135 0.460 3 0.6922 46.854 22.050 0.628 4 0.5005 38.309 11.755 0.533 5 0.6062 17.447 6.894 0.586 6 0.5454 11.157 3.835 0.556 7 0.5796 5.903 2.199 0.573 8 0.5601 3.481 1.239 0.564 9 0.5711 1.931 0.705 0.569

10 0.5649 1.109 0.399 0.566

Thus, each iteration brings the estimate closer to the true value of the root: 0.56714329.

Notice that the true percent relative error for each iteration of Example 6.1 is roughly proportional (for this case, by a factor of about 0.5 to 0.6) to the error from the previ- ous iteration. This property, called linear convergence, is characteristic of fixed-point iteration.

Aside from the “rate” of convergence, we must comment at this point about the “pos- sibility” of convergence. The concepts of convergence and divergence can be depicted graphically. Recall that in Sec. 5.2, we graphed a function to visualize its structure and behavior. Such an approach is employed in Fig. 6.2a for the function f (x) = e−x − x. An alternative graphical approach is to separate the equation into two component parts, as in

f1(x) = f2(x)

Then the two equations

y1 = f1(x) (6.4)

and

y2 = f2(x) (6.5)

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6.1 SIMPLE FIXED-POINT ITERATION 167

can be plotted separately (Fig. 6.2b). The x values corresponding to the intersections of these functions represent the roots of f (x) = 0.

The two-curve method can now be used to illustrate the convergence and divergence of fixed-point iteration. First, Eq. (6.1) can be reexpressed as a pair of equations y1 = x and y2 = g(x). These two equations can then be plotted separately. As was the case with Eqs. (6.4) and (6.5), the roots of f (x) = 0 correspond to the abscissa value at the intersection of the two curves. The function y1 = x and four different shapes for y2 = g(x) are plotted in Fig. 6.3.

For the first case (Fig. 6.3a), the initial guess of x0 is used to determine the cor- responding point on the y2 curve [x0, g(x0)]. The point [x1, x1] is located by moving left horizontally to the y1 curve. These movements are equivalent to the first iteration of the fixed-point method:

x1 = g(x0)

f (x)

Root

f (x) = e−x − x

f 1(x) = x

f 2(x) = e−x

(a)

(b)

FIGURE 6.2 Two alternative graphical methods for determining the root of f (x) = e−x − x. (a) Root at the point where it crosses the x axis; (b) root at the intersection of the component functions.

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168 ROOTS: OPEN METhODS

Thus, in both the equation and in the plot, a starting value of x0 is used to obtain an estimate of x1. The next iteration consists of moving to [x1, g(x1)] and then to [x2, x2]. This iteration is equivalent to the equation

x2 = g(x1) The solution in Fig. 6.3a is convergent because the estimates of x move closer to the

root with each iteration. The same is true for Fig. 6.3b. However, this is not the case for Fig. 6.3c and d, where the iterations diverge from the root.

A theoretical derivation can be used to gain insight into the process. As described in Chapra and Canale (2010), it can be shown that the error for any iteration is linearly propor- tional to the error from the previous iteration multiplied by the absolute value of the slope of g:

Ei+1 = g′(ξ )Ei

xx1

y1 = x

y2 = g(x)

x2 x0

(a)

y1 = x

y2 = g(x)

(b)

y1= x

y2 = g(x)

(c)

y1 = x

y2 = g(x)

(d )

FIGURE 6.3 “Cobweb plots” depicting convergence (a and b) and divergence (c and d ). Graphs (a) and (c) are called monotone patterns whereas (b) and (c) are called oscillating or spiral patterns. Note that convergence occurs when |g′ (x)| < 1.

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6.2 NEWTON-RAPhSON 169

Consequently, if |g′| < 1, the errors decrease with each iteration. For |g′| > 1 the errors grow. Notice also that if the derivative is positive, the errors will be positive, and hence the errors will have the same sign (Fig. 6.3a and c). If the derivative is negative, the errors will change sign on each iteration (Fig. 6.3b and d ).

6.2 NEWTON-RAPHSON

Perhaps the most widely used of all root-locating formulas is the Newton-Raphson method (Fig. 6.4). If the initial guess at the root is xi, a tangent can be extended from the point [xi, f (xi )]. The point where this tangent crosses the x axis usually represents an improved estimate of the root.

The Newton-Raphson method can be derived on the basis of this geometrical interpre- tation. As in Fig. 6.4, the first derivative at x is equivalent to the slope:

f ′(xi ) = f (xi ) − 0 ________ xi − xi+1

which can be rearranged to yield

xi+1 = xi − f (xi ) _____ f ′(xi )

(6.6)

which is called the Newton-Raphson formula.

EXAMPLE 6.2 Newton-Raphson Method

Problem Statement. Use the Newton-Raphson method to estimate the root of f (x) = e−x −x employing an initial guess of x0 = 0.

Solution. The first derivative of the function can be evaluated as

f ′(x ) = −e−x − 1

which can be substituted along with the original function into Eq. (6.6) to give

xi+1 = xi − e− x i − xi ________

−e− x i − 1

Starting with an initial guess of x0 = 0, this iterative equation can be applied to compute

i xi |εt|, % 0 0 100 1 0.500000000 11.8 2 0.566311003 0.147 3 0.567143165 0.0000220 4 0.567143290 <10−8

Thus, the approach rapidly converges on the true root. Notice that the true percent relative error at each iteration decreases much faster than it does in simple fixed-point iteration (compare with Example 6.1).

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170 ROOTS: OPEN METhODS

As with other root-location methods, Eq. (6.3) can be used as a termination criterion. In addition, a theoretical analysis (Chapra and Canale, 2010) provides insight regarding the rate of convergence as expressed by

Et,i+1 = −f ″(xr ) _______ 2 f ′ (xr)

E t,i 2 (6.7)

Thus, the error should be roughly proportional to the square of the previous error. In other words, the number of significant figures of accuracy approximately doubles with each iteration. This behavior is called quadratic convergence and is one of the major reasons for the popularity of the method.

Although the Newton-Raphson method is often very efficient, there are situations where it performs poorly. A special case—multiple roots—is discussed elsewhere (Chapra and Canale, 2010). However, even when dealing with simple roots, difficulties can also arise, as in the following example.

EXAMPLE 6.3 A Slowly Converging Function with Newton-Raphson

Problem Statement. Determine the positive root of f (x) = x10 − 1 using the Newton- Raphson method and an initial guess of x = 0.5.

Solution. The Newton-Raphson formula for this case is

xi+1 = xi − x i

10 − 1 ______

10x i 9

which can be used to compute

f (x)

f (xi)

f(xi) − 0

Slope = f'(xi)

0 xxi+1 xi xi − xi+1

FIGURE 6.4 Graphical depiction of the Newton-Raphson method. A tangent to the function of xi [i.e., f ′(x)] is extrapolated down to the x axis to provide an estimate of the root at xi+1.

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6.2 NEWTON-RAPhSON 171

i xi | εa|, % 0 0.5 1 51.65 99.032 2 46.485 11.111 3 41.8365 11.111 4 37.65285 11.111 • • •

40 1.002316 2.130 41 1.000024 0.229 42 1 0.002

Thus, after the first poor prediction, the technique is converging on the true root of 1, but at a very slow rate.

Why does this happen? As shown in Fig. 6.5, a simple plot of the first few iterations is helpful in providing insight. Notice how the first guess is in a region where the slope is near zero. Thus, the first iteration flings the solution far away from the initial guess to a new value (x = 51.65) where f (x) has an extremely high value. The solution then plods along for over 40 iterations until converging on the root with adequate accuracy.

2E + 17

1E + 17

0 10 20 30 40 50

0 0.5 1

f (x)

FIGURE 6.5 Graphical depiction of the Newton-Raphson method for a case with slow convergence. The inset shows how a near-zero slope initially shoots the solution far from the root. Thereafter, the solution very slowly converges on the root.

Aside from slow convergence due to the nature of the function, other difficulties can arise, as illustrated in Fig. 6.6. For example, Fig. 6.6a depicts the case where an inflection

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172 ROOTS: OPEN METhODS

f (x)

x2 x1

x0 x1 x

x0 x

x x1 x0

x0 x2x4 x3 x1 x

(a)

(b)

(c)

(d )

FIGURE 6.6 Four cases where the Newton-Raphson method exhibits poor convergence.

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6.2 NEWTON-RAPhSON 173

point (i.e., f ″ (x) = 0) occurs in the vicinity of a root. Notice that iterations beginning at x0 progressively diverge from the root. Fig. 6.6b illustrates the tendency of the Newton- Raphson technique to oscillate around a local maximum or minimum. Such oscillations may persist, or, as in Fig. 6.6b, a near-zero slope is reached whereupon the solution is sent far from the area of interest. Figure 6.6c shows how an initial guess that is close to one root can jump to a location several roots away. This tendency to move away from the area of interest is due to the fact that near-zero slopes are encountered. Obviously, a zero slope [f ′(x) = 0] is a real disaster because it causes division by zero in the Newton-Raphson formula [Eq. (6.6)]. As in Fig. 6.6d, it means that the solution shoots off horizontally and never hits the x axis.

Thus, there is no general convergence criterion for Newton-Raphson. Its convergence depends on the nature of the function and on the accuracy of the initial guess. The only remedy is to have an initial guess that is “sufficiently” close to the root. And for some func- tions, no guess will work! Good guesses are usually predicated on knowledge of the physi- cal problem setting or on devices such as graphs that provide insight into the behavior of the solution. It also suggests that good computer software should be designed to recognize slow convergence or divergence.

6.2.1 MATLAB M-file: newtraph

An algorithm for the Newton-Raphson method can be easily developed (Fig. 6.7). Note that the program must have access to the function (func) and its first derivative (dfunc). These can be simply accomplished by the inclusion of user-defined functions to compute these quantities. Alternatively, as in the algorithm in Fig. 6.7, they can be passed to the function as arguments.

After the M-file is entered and saved, it can be invoked to solve for root. For example, for the simple function x2 − 9, the root can be determined as in

>> newtraph(@(x) x^2 − 9,@(x) 2*x,5)

ans = 3

EXAMPLE 6.4 Newton-Raphson Bungee Jumper Problem

Problem Statement. Use the M-file function from Fig. 6.7 to determine the mass of the bungee jumper with a drag coefficient of 0.25 kg/m to have a velocity of 36 m/s after 4 s of free fall. The acceleration of gravity is 9.81 m/s2.

Solution. The function to be evaluated is

f (m) = √ ___

gm ___ cd tanh ( √ ___

gcd ___ m t ) − υ(t) (E6.4.1)

To apply the Newton-Raphson method, the derivative of this function must be evalu- ated with respect to the unknown, m:

df (m) _____ dm = 1 __ 2 √

____ g ____ mcd

tanh ( √___ gcd ___ m t ) − g ___ 2m t sech2 ( √ ___

gcd ___ m t ) (E6.4.2)

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174 ROOTS: OPEN METhODS

We should mention that although this derivative is not difficult to evaluate in principle, it involves a bit of concentration and effort to arrive at the final result.

The two formulas can now be used in conjunction with the function newtraph to evalu- ate the root: >> y = @(m) sqrt(9.81*m/0.25)*tanh(sqrt(9.81*0.25/m)*4)−36; >> dy = @(m) 1/2*sqrt(9.81/(m*0.25))*tanh((9.81*0.25/m) ...

^(1/2)*4)−9.81/(2*m)*sech(sqrt(9.81*0.25/m)*4)^2;

>> newtraph(y,dy,140,0.00001)

ans = 142.7376

6.3 SECANT METHODS

As in Example 6.4, a potential problem in implementing the Newton-Raphson method is the evaluation of the derivative. Although this is not inconvenient for polynomials and many other functions, there are certain functions whose derivatives may be difficult or

FIGURE 6.7 An M-file to implement the Newton-Raphson method.

function [root,ea,iter] = newtraph(func,dfunc,xr,es,maxit,varargin) % newtraph: Newton − Raphson root location zeroes % [root,ea,iter] = newtraph(func,dfunc,xr,es,maxit,p1,p2,...): % uses Newton − Raphson method to find the root of func % input: % func = name of function % dfunc = name of derivative of function % xr = initial guess % es = desired relative error (default = 0.0001%) % maxit = maximum allowable iterations (default = 50) % p1,p2,... = additional parameters used by function % output: % root = real root % ea = approximate relative error (%) % iter = number of iterations

if nargin<3,error('at least 3 input arguments required'),end if nargin<4|isempty(es),es=0.0001;end if nargin<5|isempty(maxit),maxit = 50;end iter = 0; while (1) xrold = xr; xr = xr − func(xr)/dfunc(xr); iter = iter + 1; if xr ~ = 0, ea = abs((xr − xrold)/xr) * 100; end if ea <= es | iter >= maxit, break, end end root = xr;

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6.3 SECANT METhODS 175

inconvenient to evaluate. For these cases, the derivative can be approximated by a back- ward finite divided difference:

f ′(xi ) ≅ f (xi−1) − f (xi) ____________ xi−1 − xi

This approximation can be substituted into Eq. (6.6) to yield the following iterative equation:

xi+1 = xi − f (xi)(xi−1 − xi) ____________ f (xi−1) − f(xi)

(6.8)

Equation (6.8) is the formula for the secant method. Notice that the approach requires two initial estimates of x. However, because f (x) is not required to change signs between the estimates, it is not classified as a bracketing method.

Rather than using two arbitrary values to estimate the derivative, an alternative ap- proach involves a fractional perturbation of the independent variable to estimate f ′(x),

f ′(xi ) ≅ f (xi + δ xi) − f (xi) _______________

δ xi

where δ = a small perturbation fraction. This approximation can be substituted into Eq. (6.6) to yield the following iterative equation:

xi+1 = xi − δxi f (xi) ______________ f (xi + δxi) − f (xi)

(6.9)

We call this the modified secant method. As in the following example, it provides a nice means to attain the efficiency of Newton-Raphson without having to compute derivatives.

EXAMPLE 6.5 Modified Secant Method

Problem Statement. Use the modified secant method to determine the mass of the bun- gee jumper with a drag coefficient of 0.25 kg/m to have a velocity of 36 m/s after 4 s of free fall. Note: The acceleration of gravity is 9.81 m/s2. Use an initial guess of 50 kg and a value of 10−6 for the perturbation fraction.

Solution. Inserting the parameters into Eq. (6.9) yields

First iteration:

x0 = 50 f (x0) = −4.57938708

x0 + δx0 = 50.00005 f (x0 + δx0) = −4.579381118

x1 = 50 − 10−6(50)(−4.57938708) ___________________________ −4.579381118 − (−4.57938708)

= 88.39931(|εt| = 38.1%; |εa| = 43.4%)

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176 ROOTS: OPEN METhODS

Second iteration:

x1 = 88.39931 f (x1) = −1.69220771

x1 + δx1 = 88.39940 f (x1 + δx1) = −1.692203516

x2 = 88.39931 − 10−6(88.39931)(−1.69220771) ___________________________ −1.692203516 − (−1.69220771)

= 124.08970(|εt| = 13.1%; |εa| = 28.76%)

The calculation can be continued to yield

i xi | εt|, % | εa|, %

0 50.0000 64.971 1 88.3993 38.069 43.438 2 124.0897 13.064 28.762 3 140.5417 1.538 11.706 4 142.7072 0.021 1.517 5 142.7376 4.1 × 10−6 0.021 6 142.7376 3.4 × 10−12 4.1 × 10−6

The choice of a proper value for δ is not automatic. If δ is too small, the method can be swamped by roundoff error caused by subtractive cancellation in the denomina- tor of Eq. (6.9). If it is too big, the technique can become inefficient and even divergent. However, if chosen correctly, it provides a nice alternative for cases where evaluating the derivative is difficult and developing two initial guesses is inconvenient.

Further, in its most general sense, a univariate function is merely an entity that re- turns a single value in return for values sent to it. Perceived in this sense, functions are not always simple formulas like the one-line equations solved in the preceding examples in this chapter. For example, a function might consist of many lines of code that could take a significant amount of execution time to evaluate. In some cases, the function might even represent an independent computer program. For such cases, the secant and modified se- cant methods are valuable.

6.4 BRENT’S METHOD

Wouldn’t it be nice to have a hybrid approach that combined the reliability of bracketing with the speed of the open methods? Brent’s root-location method is a clever algorithm that does just that by applying a speedy open method wherever possible, but reverting to a reliable bracketing method if necessary. The approach was developed by Richard Brent (1973) based on an earlier algorithm of Theodorus Dekker (1969).

The bracketing technique is the trusty bisection method (Sec. 5.4), whereas two dif- ferent open methods are employed. The first is the secant method described in Sec. 6.3. As explained next, the second is inverse quadratic interpolation.

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6.4 BRENT’S METhOD 177

6.4.1 Inverse Quadratic Interpolation

Inverse quadratic interpolation is similar in spirit to the secant method. As in Fig. 6.8a, the secant method is based on computing a straight line that goes through two guesses. The intersection of this straight line with the x axis represents the new root estimate. For this reason, it is sometimes referred to as a linear interpolation method.

Now suppose that we had three points. In that case, we could determine a quadratic function of x that goes through the three points (Fig. 6.8b). Just as with the linear secant method, the intersection of this parabola with the x axis would represent the new root es- timate. And as illustrated in Fig. 6.8b, using a curve rather than a straight line often yields a better estimate.

Although this would seem to represent a great improvement, the approach has a fun- damental flaw: it is possible that the parabola might not intersect the x axis! Such would be the case when the resulting parabola had complex roots. This is illustrated by the parabola, y = f(x), in Fig. 6.9.

The difficulty can be rectified by employing inverse quadratic interpolation. That is, rather than using a parabola in x, we can fit the points with a parabola in y. This amounts to reversing the axes and creating a “sideways” parabola [the curve, x = f (y), in Fig. 6.9].

If the three points are designated as (xi–2, yi–2), (xi–1, yi–1), and (xi, yi ), a quadratic func- tion of y that passes through the points can be generated as

g(y) = (y − yi−1)(y − yi ) __________________ (yi−2 − yi−1)(yi−2 − yi )

xi−2 + (y − yi−2)(y − yi ) _____________________ (yi−1 − yi−2)(yi−1 − yi )

xi−1

+ (y − yi−2)(y − yi−1) __________________ (yi − yi−2)(yi − yi −1)

xi (6.10)

f (x)

(a) (b)

f (x)

FIGURE 6.8 Comparison of (a) the secant method and (b) inverse quadratic interpolation. Note that the approach in (b) is called “inverse” because the quadratic function is written in y rather than in x.

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178 ROOTS: OPEN METhODS

As we will learn in Sec. 18.2, this form is called a Lagrange polynomial. The root, xi+1, corresponds to y = 0, which when substituted into Eq. (6.10) yields

xi+1 = yi−1 yi __________________ ( yi−2 − yi−1)(yi−2 − yi )

xi−2 + yi−2yi _____________________ (yi−1 − yi−2)(yi−1 − yi )

xi−1

+ yi−2 yi−1 __________________ (yi − yi−2)(yi − yi −1)

xi (6.11)

As shown in Fig. 6.9, such a “sideways” parabola always intersects the x axis.

EXAMPLE 6.6 Inverse Quadratic Interpolation

Problem Statement. Develop quadratic equations in both x and y for the data points depicted in Fig. 6.9: (1, 2), (2, 1), and (4, 5). For the first, y = f (x), employ the quadratic formula to illustrate that the roots are complex. For the latter, x = g(y), use inverse quadratic interpolation (Eq. 6.11) to determine the root estimate.

Solution. By reversing the x’s and y’s, Eq. (6.10) can be used to generate a quadratic in x as

f (x) = (x − 2)(x − 4) ____________ (1 − 2)(1 − 4) 2 + (x − 1)(x − 4) ____________ (2 − 1)(2 − 4) 1 +

(x − 1)(x − 2) ____________ (4 − 1)(4 − 2) 5

or collecting terms

f (x) = x2 − 4x + 5

x = f(y)

y = f(x)

Root

FIGURE 6.9 Two parabolas fit to three points. The parabola written as a function of x, y = f ( x ), has com- plex roots and hence does not intersect the x axis. In contrast, if the variables are reversed, and the parabola developed as x = f ( y ), the function does intersect the x axis.

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6.4 BRENT’S METhOD 179

This equation was used to generate the parabola, y = f (x), in Fig. 6.9. The quadratic for- mula can be used to determine that the roots for this case are complex,

x = 4 ± √

_____________ (−4)2 − 4(1)(5) ___________________ 2 = 2

± i

Equation (6.10) can be used to generate the quadratic in y as

g (y) = ( y − 1)( y − 5) _____________ (2 − 1)(2 − 5) 1 + ( y − 2)( y − 5) ____________ (1 − 2)(1 − 5) 2 +

(y − 2)(y − 1) ____________ (5 − 2)(5 − 1) 4

or collecting terms:

g(y) = 0.5x2 − 2.5x + 4

Finally, Eq. (6.11) can be used to determine the root as

xi+1 = −1(−5) ____________ (2 − 1)(2 − 5) 1 +

−2(−5) ____________ (1 − 2)(1 − 5) 2 + −2(−1) ____________ (5 − 2)(5 − 1) 4 = 4

Before proceeding to Brent’s algorithm, we need to mention one more case where inverse quadratic interpolation does not work. If the three y values are not distinct (i.e., yi−2 = yi−1 or yi−1 = yi ), an inverse quadratic function does not exist. So this is where the secant method comes into play. If we arrive at a situation where the y values are not distinct, we can always revert to the less efficient secant method to generate a root using two of the points. If yi−2 = yi−1, we use the secant method with xi−1 and xi. If yi−1 = yi, we use xi−2 and xi−1.

6.4.2 Brent’s Method Algorithm

The general idea behind the Brent’s root-finding method is whenever possible to use one of the quick open methods. In the event that these generate an unacceptable result (i.e., a root estimate that falls outside the bracket), the algorithm reverts to the more conservative bisection method. Although bisection may be slower, it generates an es- timate guaranteed to fall within the bracket. This process is then repeated until the root is located to within an acceptable tolerance. As might be expected, bisection typi- cally dominates at first but as the root is approached, the technique shifts to the faster open methods.

Figure 6.10 presents a function based on a MATLAB M-file developed by Cleve Moler (2004). It represents a stripped down version of the fzero function which is the professional root-location function employed in MATLAB. For that reason, we call the simplified ver- sion: fzerosimp. Note that it requires another function f that holds the equation for which the root is being evaluated.

The fzerosimp function is passed two initial guesses that must bracket the root. Then, the three variables defining the search interval (a, b, c) are initialized, and f is evaluated at the endpoints.

A main loop is then implemented. If necessary, the three points are rearranged to satisfy the conditions required for the algorithm to work effectively. At this point, if the

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180 ROOTS: OPEN METhODS

function b = fzerosimp(xl,xu) a = xl; b = xu; fa = f(a); fb = f(b); c = a; fc = fa; d = b − c; e = d; while (1) if fb == 0, break, end if sign(fa) == sign(fb) %If needed, rearrange points a = c; fa = fc; d = b − c; e = d; end if abs(fa) < abs(fb) c = b; b = a; a = c; fc = fb; fb = fa; fa = fc; end m = 0.5*(a − b); %Termination test and possible exit tol = 2 * eps * max(abs(b), 1); if abs(m) <= tol | fb == 0. break end %Choose open methods or bisection if abs(e) >= tol & abs(fc) > abs(fb) s = fb/fc; if a == c %Secant method p = 2*m*s; q = 1 − s; else %Inverse quadratic interpolation q = fc/fa; r = fb/fa; p = s * (2*m*q * (q − r) − (b − c)*(r − 1)); q = (q − 1)*(r − 1)*(s − 1); end if p > 0, q = −q; else p = −p; end; if 2*p < 3*m*q − abs(tol*q) & p < abs(0.5*e*q) e = d; d = p/q; else d = m; e = m; end else %Bisection d = m; e = m; end c = b; fc = fb; if abs(d) > tol, b = b + d; else b = b−sign(b − a)*tol; end fb = f(b); end

FIGURE 6.10 Function for Brent’s root-finding algorithm based on a MATLAB M-file developed by Cleve Moler (2004).

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6.5 MATLAB FUNCTION: FZERO 181

stopping criteria are met, the loop is terminated. Otherwise, a decision structure chooses among the three methods and checks whether the outcome is acceptable. A final section then evaluates f at the new point and the loop is repeated. Once the stopping criteria are met, the loop terminates and the final root estimate is returned.

6.5 MATLAB FUNCTION: fzero

The fzero function is designed to find the real root of a single equation. A simple represen- tation of its syntax is

fzero(function,x0)

where function is the name of the function being evaluated, and x0 is the initial guess. Note that two guesses that bracket the root can be passed as a vector:

fzero(function,[x0 x1])

where x0 and x1 are guesses that bracket a sign change. Here is a MATLAB session that solves for the root of a simple quadratic: x2 − 9.

Clearly two roots exist at −3 and 3. To find the negative root:

>> x = fzero(@(x) x^2−9,−4)

x = −3

If we want to find the positive root, use a guess that is near it:

>> x = fzero(@(x) x^2 − 9,4)

x = 3

If we put in an initial guess of zero, it finds the negative root:

>> x = fzero(@(x) x^2 − 9,0)

x = −3

If we wanted to ensure that we found the positive root, we could enter two guesses as in

>> x = fzero(@(x) x^2−9,[0 4])

x = 3

Also, if a sign change does not occur between the two guesses, an error message is displayed

>> x = fzero(@(x) x^2−9,[−4 4])

??? Error using ==> fzero The function values at the interval endpoints must ... differ in sign.

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182 ROOTS: OPEN METhODS

The fzero function works as follows. If a single initial guess is passed, it first performs a search to identify a sign change. This search differs from the incremental search described in Sec. 5.3.1, in that the search starts at the single initial guess and then takes increasingly bigger steps in both the positive and negative directions until a sign change is detected.

Thereafter, the fast methods (secant and inverse quadratic interpolation) are used un- less an unacceptable result occurs (e.g., the root estimate falls outside the bracket). If an unacceptable result happens, bisection is implemented until an acceptable root is obtained with one of the fast methods. As might be expected, bisection typically dominates at first but as the root is approached, the technique shifts to the faster methods.

A more complete representation of the fzero syntax can be written as

[x,fx] = fzero(function,x0,options,p1,p2,...)

where [x,fx] = a vector containing the root x and the function evaluated at the root fx, op- tions is a data structure created by the optimset function, and p1, p2... are any parameters that the function requires. Note that if you desire to pass in parameters but not use the options, pass an empty vector [] in its place.

The optimset function has the syntax

options = optimset('par1',val1,'par2',val2,...)

where the parameter pari has the value vali. A complete listing of all the possible param- eters can be obtained by merely entering optimset at the command prompt. The parameters that are commonly used with the fzero function are

display: When set to 'iter' displays a detailed record of all the iterations. tolx: A positive scalar that sets a termination tolerance on x.

EXAMPLE 6.7 The fzero and optimset Functions

Problem Statement. Recall that in Example 6.3, we found the positive root of f (x) = x10 − 1 using the Newton-Raphson method with an initial guess of 0.5. Solve the same problem with optimset and fzero.

Solution. An interactive MATLAB session can be implemented as follows: >> options = optimset('display','iter'); >> [x,fx] = fzero(@(x) x^10 − 1,0.5,options) Func−count x f(x) Procedure 1 0.5 −0.999023 initial 2 0.485858 −0.999267 search 3 0.514142 −0.998709 search 4 0.48 −0.999351 search 5 0.52 −0.998554 search 6 0.471716 −0.999454 search • • • 23 0.952548 −0.385007 search 24 −0.14 −1 search 25 1.14 2.70722 search

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6.6 POLyNOMIALS 183

Looking for a zero in the interval [−0.14, 1.14]

26 0.205272 −1 interpolation 27 0.672636 −0.981042 bisection 28 0.906318 −0.626056 bisection 29 1.02316 0.257278 bisection 30 0.989128 −0.103551 interpolation 31 0.998894 −0.0110017 interpolation 32 1.00001 7.68385e−005 interpolation 33 1 −3.83061e−007 interpolation 34 1 −1.3245e−011 interpolation 35 1 0 interpolation Zero found in the interval: [−0.14, 1.14].

x = 1

fx = 0

Thus, after 25 iterations of searching, fzero finds a sign change. It then uses interpola- tion and bisection until it gets close enough to the root so that interpolation takes over and rapidly converges on the root.

Suppose that we would like to use a less stringent tolerance. We can use the optim- set function to set a low maximum tolerance and a less accurate estimate of the root results:

>> options = optimset ('tolx', 1e − 3); >> [x,fx] = fzero(@(x) x^10 − 1,0.5,options)

x = 1.0009

fx = 0.0090

6.6 POLYNOMIALS

Polynomials are a special type of nonlinear algebraic equation of the general form

fn(x) = a1xn + a2 xn−1 + · · · + an−1x2 + an x + an+1 (6.12)

where n is the order of the polynomial, and the a’s are constant coefficients. In many (but not all) cases, the coefficients will be real. For such cases, the roots can be real and/or com- plex. In general, an nth order polynomial will have n roots.

Polynomials have many applications in engineering and science. For example, they are used extensively in curve fitting. However, one of their most interesting and powerful applications is in characterizing dynamic systems—and, in particular, lin- ear systems. Examples include reactors, mechanical devices, structures, and electrical circuits.

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184 ROOTS: OPEN METhODS

6.6.1 MATLAB Function: roots

If you are dealing with a problem where you must determine a single real root of a polyno- mial, the techniques such as bisection and the Newton-Raphson method can have utility. However, in many cases, engineers desire to determine all the roots, both real and complex. Unfortunately, simple techniques like bisection and Newton-Raphson are not available for determining all the roots of higher-order polynomials. However, MATLAB has an excel- lent built-in capability, the roots function, for this task.

The roots function has the syntax,

x = roots(c)

where x is a column vector containing the roots and c is a row vector containing the poly- nomial’s coefficients.

So how does the roots function work? MATLAB is very good at finding the eigen- values of a matrix. Consequently, the approach is to recast the root evaluation task as an eigenvalue problem. Because we will be describing eigenvalue problems later in the book, we will merely provide an overview here.

Suppose we have a polynomial

a1 x5 + a2 x4 + a3 x3 + a4 x2 + a5 x + a6 = 0 (6.13)

Dividing by a1 and rearranging yields

x5 = − a2 __ a1 x

4 − a3 __ a1 x

3 − a4 __ a1 x

2 − a5 __ a1 x −

a6 __ a1

A special matrix can be constructed by using the coefficients from the right-hand side as the first row and with 1’s and 0’s written for the other rows as shown:

−a2∕a1 −a3∕a1 −a4∕a1 −a5∕a1 −a6∕a1 1 0 0 0 0 0 1 0 0 0 (6.14) 0 0 1 0 0 0 0 0 1 0

Equation (6.14) is called the polynomial’s companion matrix. It has the useful prop- erty that its eigenvalues are the roots of the polynomial. Thus, the algorithm underlying the roots function consists of merely setting up the companion matrix and then using MATLAB’s powerful eigenvalue evaluation function to determine the roots. Its applica- tion, along with some other related polynomial manipulation functions, are described in the following example.

We should note that roots has an inverse function called poly, which when passed the values of the roots, will return the polynomial’s coefficients. Its syntax is

c = poly(r)

where r is a column vector containing the roots and c is a row vector containing the poly- nomial’s coefficients.

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6.6 POLyNOMIALS 185

EXAMPLE 6.8 Using MATLAB to Manipulate Polynomials and Determine Their Roots

Problem Statement. Use the following equation to explore how MATLAB can be em- ployed to manipulate polynomials:

f 5(x) = x5 − 3.5x4 + 2.75x3 + 2.125x2 − 3.875x + 1.25 (E6.8.1)

Note that this polynomial has three real roots: 0.5, −1.0, and 2; and one pair of complex roots: 1 ± 0.5i.

Solution. Polynomials are entered into MATLAB by storing the coefficients as a row vec- tor. For example, entering the following line stores the coefficients in the vector a:

>> a = [1 −3.5 2.75 2.125 −3.875 1.25];

We can then proceed to manipulate the polynomial. For example we can evaluate it at x = 1, by typing

>> polyval(a,1)

with the result, 1(1)5 − 3.5(1)4 + 2.75(1)3 + 2.125(1)2 − 3.875(1) + 1.25 = −0.25: ans =

−0.2500

We can create a quadratic polynomial that has roots corresponding to two of the origi- nal roots of Eq. (E6.8.1): 0.5 and −1. This quadratic is (x − 0.5)(x + 1) = x2 + 0.5x − 0.5. It can be entered into MATLAB as the vector b: >> b = [1 .5 −.5]

b = 1.0000 0.5000 −0.5000

Note that the poly function can be used to perform the same task as in

>> b = poly([0.5 −1])

b = 1.0000 0.5000 −0.5000

We can divide this polynomial into the original polynomial by

>> [q,r] = deconv(a,b)

with the result being a quotient (a third-order polynomial, q) and a remainder (r)

q = 1.0000 −4.0000 5.2500 −2.5000 r = 0 0 0 0 0 0

Because the polynomial is a perfect divisor, the remainder polynomial has zero coeffi- cients. Now, the roots of the quotient polynomial can be determined as

>> x = roots(q)

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186 ROOTS: OPEN METhODS

with the expected result that the remaining roots of the original polynomial Eq. (E6.8.1) are found:

x = 2.0000 1.0000 + 0.5000i 1.0000 − 0.5000i

We can now multiply q by b to come up with the original polynomial:

>> a = conv(q,b)

a = 1.0000 −3.5000 2.7500 2.1250 −3.8750 1.2500

We can then determine all the roots of the original polynomial by

>> x = roots(a)

x = 2.0000 −1.0000 1.0000 + 0.5000i 1.0000 − 0.5000i 0.5000

Finally, we can return to the original polynomial again by using the poly function:

>> a = poly(x)

a = 1.0000 −3.5000 2.7500 2.1250 −3.8750 1.2500

6.7 CASE STUDY PIPE FRICTION

Background. Determining fluid flow through pipes and tubes has great relevance in many areas of engineering and science. In engineering, typical applications include the flow of liquids and gases through pipelines and cooling systems. Scientists are interested in topics ranging from flow in blood vessels to nutrient transmission through a plant’s vascular system.

The resistance to flow in such conduits is parameterized by a dimensionless number called the friction factor. For turbulent flow, the Colebrook equation provides a means to calculate the friction factor:

0 = 1 ____ √

__ f + 2.0 log ( ε _____ 3.7D + 2.51 ______ Re √__ f ) (6.15)

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6.7 CASE STUDy 187

6.7 CASE STUDY continued

where ε = the roughness (m), D = diameter (m), and Re = the Reynolds number:

Re = ρVD ____ μ

where ρ = the fluid’s density (kg/m3), V = its velocity (m/s), and μ = dynamic viscosity (N · s/m2). In addition to appearing in Eq. (6.15), the Reynolds number also serves as the criterion for whether flow is turbulent (Re > 4000).

In this case study, we will illustrate how the numerical methods covered in this part of the book can be employed to determine f for air flow through a smooth, thin tube. For this case, the parameters are ρ = 1.23 kg/m3, μ = 1.79 × 10–5 N · s/m2, D = 0.005 m, V = 40 m/s, and ε = 0.0015 mm. Note that friction factors range from about 0.008 to 0.08. In addition, an explicit formulation called the Swamee-Jain equation provides an approximate estimate:

f = 1.325 ________________ [ ln ( ε _____ 3.7 D + 5.74 ____ Re0.9 ) ] 2 (6.16)

Solution. The Reynolds number can be computed as

Re = ρVD ____ μ = 1.23(40)0.005 ____________

1.79 × 10−5 = 13,743

This value along with the other parameters can be substituted into Eq. (6.15) to give

g ( f ) = 1 ____ √

__ f + 2.0 log ( 0.0000015 _________ 3.7(0.005) + 2.51 _________ 13,743 √__ f )

Before determining the root, it is advisable to plot the function to estimate initial guesses and to anticipate possible difficulties. This can be done easily with MATLAB:

>> rho = 1.23;mu = 1.79e − 5;D = 0.005;V = 40;e = 0.0015/1000; >> Re = rho*V*D/mu; >> g = @(f) 1/sqrt(f)+2*log10(e/(3.7*D)+2.51/(Re*sqrt(f))); >> fplot(g,[0.008 0.08]),grid,xlabel('f'),ylabel('g(f)')

As in Fig. 6.11, the root is located at about 0.03. Because we are supplied initial guesses (xl = 0.008 and xu = 0.08), either of the brack-

eting methods from Chap. 5 could be used. For example, the bisect function developed in Fig. 5.7 gives a value of f = 0.0289678 with a percent relative error of error of 5.926 × 10–5 in 22 iterations. False position yields a result of similar precision in 26 iterations. Thus, although they produce the correct result, they are somewhat inefficient. This would not be important for a single application, but could become prohibitive if many evaluations were made.

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188 ROOTS: OPEN METhODS

6.7 CASE STUDY continued

We could try to attain improved performance by turning to an open method. Because Eq. (6.15) is relatively straightforward to differentiate, the Newton-Raphson method is a good candidate. For example, using an initial guess at the lower end of the range (x0 = 0.008), the newtraph function developed in Fig. 6.7 converges quickly:

>> dg = @(f) −2/log(10)*1.255/Re*f^(−3/2)/(e/D/3.7 ... +2.51/Re/sqrt(f))−0.5/f^(3/2); >> [f ea iter] = newtraph(g,dg,0.008)

f = 0.02896781017144 ea = 6.870124190058040e−006 iter = 6

However, when the initial guess is set at the upper end of the range (x0 = 0.08), the routine diverges,

>> [f ea iter] = newtraph(g,dg,0.08)

f = NaN + NaNi

As can be seen by inspecting Fig. 6.11, this occurs because the function’s slope at the initial guess causes the first iteration to jump to a negative value. Further runs demonstrate that for this case, convergence only occurs when the initial guess is below about 0.066.

−3 0.020.01 0.03 0.04 0.05 0.06 0.07 0.08

−2

−1

g( f)

FIGURE 6.11

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6.7 CASE STUDy 189

6.7 CASE STUDY continued

So we can see that although the Newton-Raphson is very efficient, it requires good ini- tial guesses. For the Colebrook equation, a good strategy might be to employ the Swamee- Jain equation (Eq. 6.16) to provide the initial guess as in

>> fSJ=1.325/log(e/(3.7*D)+5.74/Re^0.9)^2

fSJ = 0.02903099711265

>> [f ea iter] = newtraph(g,dg,fSJ)

f = 0.02896781017144 ea = 8.510189472800060e−010 iter = 3

Aside from our homemade functions, we can also use MATLAB’s built-in fzero func- tion. However, just as with the Newton-Raphson method, divergence also occurs when fzero function is used with a single guess. However, in this case, guesses at the lower end of the range cause problems. For example,

>> fzero(g,0.008)

Exiting fzero: aborting search for an interval containing a sign change because complex function value encountered ... during search. (Function value at −0.0028 is −4.92028−20.2423i.) Check function or try again with a different starting value. ans = NaN

If the iterations are displayed using optimset (recall Example 6.7), it is revealed that a nega- tive value occurs during the search phase before a sign change is detected and the routine aborts. However, for single initial guesses above about 0.016, the routine works nicely. For example, for the guess of 0.08 that caused problems for Newton-Raphson, fzero does just fine:

>> fzero(g,0.08)

ans = 0.02896781017144

As a final note, let’s see whether convergence is possible for simple fixed-point it- eration. The easiest and most straightforward version involves solving for the first f in Eq. (6.15):

fi+1 = 0.25 _____________________

( log ( ε _____ 3.7 D + 2.51 _______ Re √__ fi ) ) 2 (6.17)

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190 ROOTS: OPEN METhODS

6.7 CASE STUDY continued

0.01

0.02

0.03

0.04

0.05

0 0.02 0.04 0.06 0.08

y2 = g(x)

y1 = x

FIGURE 6.12

The two-curve display of this function depicted indicates a surprising result (Fig. 6.12). Recall that fixed-point iteration converges when the y2 curve has a relatively flat slope (i.e., |g′(ξ)| < 1). As indicated by Fig. 6.12, the fact that the y2 curve is quite flat in the range from f = 0.008 to 0.08 means that not only does fixed-point iteration converge, but it con- verges fairly rapidly! In fact, for initial guesses anywhere between 0.008 and 0.08, fixed- point iteration yields predictions with percent relative errors less than 0.008% in six or fewer iterations! Thus, this simple approach that requires only one guess and no derivative estimates performs really well for this particular case.

The take-home message from this case study is that even great, professionally devel- oped software like MATLAB is not always foolproof. Further, there is usually no single method that works best for all problems. Sophisticated users understand the strengths and weaknesses of the available numerical techniques. In addition, they understand enough of the underlying theory so that they can effectively deal with situations where a method breaks down.

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PROBLEMS 191

PROBLEMS

6.1 Employ fixed-point iteration to locate the root of

f (x) = sin ( √ __

x ) − x

Use an initial guess of x0 = 0.5 and iterate until εa ≤ 0.01%. Verify that the process is linearly convergent as described at the end of Sec. 6.1. 6.2 Use (a) fixed-point iteration and (b) the Newton- Raphson method to determine a root of f (x) = −0.9x2 + 1.7x + 2.5 using x0 = 5. Perform the computation until εa is less than εs = 0.01%. Also check your final answer. 6.3 Determine the highest real root of f (x) = x3 − 6x2 + 11x − 6.1: (a) Graphically. (b) Using the Newton-Raphson method (three iterations,

x0 = 3.5 (c) Using the secant method (three iterations, x−1 = 2.5 and

x0 = 3.5). (d) Using the modified secant method (three iterations,

x0 = 3.5, δ = 0.01). (e) Determine all the roots with MATLAB. 6.4 Determine the lowest positive root of f (x) = 7 sin(x)e−x − 1: (a) Graphically. (b) Using the Newton-Raphson method (three iterations,

x0 = 0.3). (c) Using the secant method (three iterations, x−1 = 0.5 and

x0 = 0.4. (d) Using the modified secant method (five iterations,

x0 = 0.3, δ = 0.01). 6.5 Use (a) the Newton-Raphson method and (b) the modi- fied secant method (δ = 0.05) to determine a root of f (x) = x5 − 16.05x4 + 88.75x3 − 192.0375x2 + 116.35x + 31.6875 using an initial guess of x = 0.5825 and εs = 0.01%. Explain your results. 6.6 Develop an M-file for the secant method. Along with the two initial guesses, pass the function as an argument. Test it by solving Prob. 6.3. 6.7 Develop an M-file for the modified secant method. Along with the initial guess and the perturbation frac- tion, pass the function as an argument. Test it by solving Prob. 6.3. 6.8 Differentiate Eq. (E6.4.1) to get Eq. (E6.4.2). 6.9 Employ the Newton-Raphson method to determine a real root for f (x) = −2 + 6x − 4x2 + 0.5x3, using an ini- tial guess of (a) 4.5 and (b) 4.43. Discuss and use graphical and analytical methods to explain any peculiarities in your results.

6.10 The “divide and average” method, an old-time method for approximating the square root of any positive number a, can be formulated as

xi+1 = xi + a∕xi ________ 2

Prove that this formula is based on the Newton-Raphson algorithm. 6.11 (a) Apply the Newton-Raphson method to the function f (x) = tanh(x2 − 9) to evaluate its known real root at x = 3. Use an initial guess of x0 = 3.2 and take a minimum of three iterations. (b) Did the method exhibit convergence onto its real root? Sketch the plot with the results for each iteration labeled. 6.12 The polynomial f (x) = 0.0074x4 − 0.284x3 + 3.355x2− 12.183x + 5 has a real root between 15 and 20. Apply the Newton-Raphson method to this function using an initial guess of x0 =16.15. Explain your results. 6.13 Mechanical engineers, as well as most other engineers, use thermodynamics extensively in their work. The follow- ing polynomial can be used to relate the zero-pressure spe- cific heat of dry air cp in kJ/(kg K) to temperature in K:

cp = 0.99403 + 1.671 × 10−4T + 9.7215 × 10−8T 2

−9.5838 × 10−11T 3 + 1.9520 × 10−14T 4

Write a MATLAB script (a) to plot cp versus a range of T = 0 to 1200 K and (b) to determine the temperature that corre- sponds to a specific heat of 1.1 kJ/(kg K) with MATLAB polynomial functions. 6.14 In a chemical engineering process, water vapor (H2O) is heated to sufficiently high temperatures that a significant portion of the water dissociates, or splits apart, to form oxy- gen (O2) and hydrogen (H2):

H2O⇄H2 + 1 _ 2 O2

If it is assumed that this is the only reaction involved, the mole fraction x of H2O that dissociates can be represented by

K = x _____ 1 − x √ _____

2pt _____ 2 + x (P6.14.1)

where K is the reaction’s equilibrium constant and pt is the total pressure of the mixture. If pt = 3 atm and K = 0.05, determine the value of x that satisfies Eq. (P6.14.1). 6.15 The Redlich-Kwong equation of state is given by

p = RT _____ υ − b −

a ___________ υ(υ + b) √

__ T

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192 ROOTS: OPEN METhODS

where R = the universal gas constant [= 0.518 kJ/(kg K)], T = absolute temperature (K), p = absolute pressure (kPa), and υ = the volume of a kg of gas (m3/kg). The parameters a and b are calculated by

a = 0.427 R2 T c

2.5 ______ pc b = 0.0866R

Tc __ pc

where pc = 4600 kPa and Tc = 191 K. As a chemical engi- neer, you are asked to determine the amount of methane fuel that can be held in a 3-m3 tank at a temperature of−40 °C with a pressure of 65,000 kPa. Use a root-locating method of your choice to calculate υ and then determine the mass of methane contained in the tank. 6.16 The volume of liquid V in a hollow horizontal cylinder of radius r and length L is related to the depth of the liquid h by

V = [ r 2cos−1 ( r − h _____ r ) − (r − h) √ _______

2rh − h2 ] L Determine h given r = 2 m, L = 5 m, and V = 8 m3. 6.17 A catenary cable is one which is hung between two points not in the same vertical line. As depicted in Fig. P6.17a, it is subject to no loads other than its own weight. Thus, its weight acts as a uniform load per unit length along the cable 𝑤 (N/m). A free-body diagram of a section AB is depicted in Fig. P6.17b, where TA and TB are the tension forces at the end. Based on horizontal and verti- cal force balances, the following differential equation model of the cable can be derived:

d 2 y ____

dx 2 = 𝑤 ___ TA

√ ________

1 + ( dy ___ dx ) 2

Calculus can be employed to solve this equation for the height of the cable y as a function of distance x:

y = TA ___ 𝑤 cosh ( 𝑤 ___ TA x ) + y0 − TA ___ 𝑤

(a) Use a numerical method to calculate a value for the parameter TA given values for the parameters 𝑤 = 10 and y0 = 5, such that the cable has a height of y = 15 at x = 50.

(b) Develop a plot of y versus x for x = −50 to 100. 6.18 An oscillating current in an electric circuit is described by I = 9e−t sin(2πt), where t is in seconds. Determine all values of t such that I = 3.5 6.19 Figure P6.19 shows a circuit with a resistor, an induc- tor, and a capacitor in parallel. Kirchhoff’s rules can be used to express the impedance of the system as

1 __ Z = √ _______________

1 ___ R2

+ ( ωC − 1 ___ ωL ) 2

where Z = impedance (Ω), and ω is the angular frequency. Find the ω that results in an impedance of 100 Ω using the fzero function with initial guesses of 1 and 1000 for the following parameters: R = 225 Ω, C = 0.6 × 10−6 F, and L = 0.5 H.

(a) (b)

W = �s

�

y0 TA

FIGURE P6.17

R L C

FIGURE P6.19

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PROBLEMS 193

Find the appropriate initial angle θ0, if υ0 = 30 m/s, and the distance to the catcher is 90 m. Note that the throw leaves the right fielder’s hand at an elevation of 1.8 m and the catcher receives it at 1 m. 6.22 You are designing a spherical tank (Fig. P6.22) to hold water for a small village in a developing country. The vol- ume of liquid it can hold can be computed as

V = πh2 [3R − h] ________ 3

where V = volume [m3], h = depth of water in tank [m], and R = the tank radius [m].

6.20 Real mechanical systems may involve the deflection of nonlinear springs. In Fig. P6.20, a block of mass m is released a distance h above a nonlinear spring. The resis- tance force F of the spring is given by

F = −(k1d + k2d 3∕2)

Conservation of energy can be used to show that

0 = 2k2d 5∕2 ______ 5 +

1 __ 2 k1d 2 − mgd − mgh

Solve for d, given the following parameter values: k1 = 40,000 g/s2, k2 = 40 g/(s2 m0.5), m = 95 g, g = 9.81 m/s2, and h = 0.43 m. 6.21 Aerospace engineers sometimes compute the trajec- tories of projectiles such as rockets. A related problem deals with the trajectory of a thrown ball. The trajectory of a ball thrown by a right fielder is defined by the (x, y) coordinates as displayed in Fig. P6.21. The trajectory can be modeled as

y = (tan θ0)x − g __________

2υ 0 2 cos2 θ0

x2 + y0

υ0

0θ

FIGURE P6.21

FIGURE P6.22

(a) (b)

d h + d

FIGURE P6.20

If R = 3 m, what depth must the tank be filled to so that it holds 30 m3? Use three iterations of the most efficient numerical method possible to determine your answer. De- termine the approximate relative error after each iteration. Also, provide justification for your choice of method. Extra information: (a) For bracketing methods, initial guesses of 0 and R will bracket a single root for this example. (b) For open methods, an initial guess of R will always converge. 6.23 Perform the identical MATLAB operations as those in Example 6.8 to manipulate and find all the roots of the polynomial

f 5(x) = (x + 2)(x + 5)(x − 6)(x − 4)(x − 8)

6.24 In control systems analysis, transfer functions are developed that mathematically relate the dynamics of a sys- tem’s input to its output. A transfer function for a robotic positioning system is given by

G(s) = C(s) ____ N(s) = s3 + 9s2 + 26s + 24 ________________________

s4 + 15s3 + 77s2 + 153s + 90

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194 ROOTS: OPEN METhODS

If you use a bracketing method, use initial guesses of xl = 0. and xu = 2. If you use the Newton-Raphson or the modified secant method, use an initial guess of xi = 0.7. If you use the secant method, use initial guesses of xi−1 = 0 and xi = 2. 6.30 (a) Develop an M-file function to implement Brent’s root-location method. Base your function on Fig. 6.10, but with the beginning of the function changed to

function [b,fb] = fzeronew(f, xl, xu,varargin) % fzeronew: Brent root location zeroes % [b,fb] = fzeronew(f,xl,xu,p1,p2,...): % uses Brent’s method to find the root of f % input: % f = name of function % xl, xu = lower and upper guesses % p1,p2,... = additional parameters used by f % output: % b = real root % fb = function value at root

Make the appropriate modifications so that the function per- forms as outlined in the documentation statements. In addi- tion, include error traps to ensure that the function’s three required arguments (f,xl,xu) are prescribed, and that the initial guesses bracket a root. (b) Test your function by using it to solve for the root of the

function from Example 5.6 using

>> [x,fx] = fzeronew(@(x,n) x^n − 1,0,1.3,10)

6.31 Figure P6.31 shows a side view of a broad crested weir. The symbols shown in Fig. P6.31 are defined as: Hw = the height of the weir (m), Hh = the head above the weir (m), and H = Hw + Hh = the depth of the river upstream of the weir (m).

where G(s) = system gain, C(s) = system output, N(s) = system input, and s = Laplace transform complex frequency. Use MATLAB to find the roots of the numerator and de- nominator and factor these into the form

G(s) = (s + a1)(s + a2)(s + a3) _________________________ (s + b1)(s + b2)(s + b3)(s + b4)

where ai and bi = the roots of the numerator and denomina- tor, respectively. 6.25 The Manning equation can be written for a rectangular open channel as

Q = √

__ S (BH)5∕3 ___________

n(B + 2H)2∕3

where Q = flow (m3/s), S = slope (m/m), H = depth (m), and n = the Manning roughness coefficient. Develop a fixed- point iteration scheme to solve this equation for H given Q = 5, S = 0.0002, B = 20, and n = 0.03. Perform the compu- tation until εa is less than εs = 0.05%. Prove that your scheme converges for all initial guesses greater than or equal to zero. 6.26 See if you can develop a foolproof function to com- pute the friction factor based on the Colebrook equation as described in Sec. 6.7. Your function should return a precise result for Reynolds number ranging from 4000 to 107 and for ε/D ranging from 0.00001 to 0.05. 6.27 Use the Newton-Raphson method to find the root of

f (x) = e−0.5x (4 − x) − 2

Employ initial guesses of (a) 2, (b) 6, and (c) 8. Explain your results. 6.28 Given

f (x) = −2x6 − 1.5x4 + 10x + 2

Use a root-location technique to determine the maximum of this function. Perform iterations until the approximate relative error falls below 5%. If you use a bracketing method, use initial guesses of xl = 0 and xu = 1. If you use the Newton-Raphson or the modified secant method, use an initial guess of xi = 1. If you use the secant method, use initial guesses of xi − 1 = 0 and xi = 1. Assuming that convergence is not an issue, choose the tech- nique that is best suited to this problem. Justify your choice. 6.29 You must determine the root of the following easily differentiable function:

e0.5x = 5 − 5x Pick the best numerical technique, justify your choice, and then use that technique to determine the root. Note that it is known that for positive initial guesses, all techniques ex- cept fixed-point iteration will eventually converge. Perform iterations until the approximate relative error falls below 2%.

FIGURE P6.31 A broad-crested weir used to control depth and velocity of rivers and streams.

Hw Weir

QwH

The flow across the weir, Qw (m3/s), can be computed as (Munson et al., 2009)

Qw = Cw Bw √ __

g ( 2 __ 3 ) 3∕2

H h 3∕2 (P6.31.1)

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PROBLEMS 195

Determine the solution approach that converges for initial guesses in the range of 0 < x < 7. Use either a graphical or analytical approach to prove that your formulation always converges in the given range. 6.35 A circular pipe made out of new cast iron is used to convey water at a volume flow rate of Q = 0.3 m3/s. Assume that the flow is steady and fully developed and the water is incompressible. The head loss, friction, and diameter are related by the Darcy-Weisbach equation,

hL = f LV 2 ____ D 2 g

(P6.35.1)

where f = the friction factor (dimensionless), L = length (m), D = pipe inner diameter (m), V = velocity (m/s), and g = the gravitational constant (= 9.81 m/s2). The velocity can be related to flow by

Q = A c V (P6.35.2)

where Ac = the pipe’s cross-sectional area (m2) = πD2/4 and the friction factor can be determined by the Colebrook equa- tion. If you want the head loss to be less than 0.006 m per meter of pipe, develop a MATLAB function to determine the smallest diameter pipe to achieve this objective. Use the following parameter values: υ = 1.16 × 10–6 m2/s and ε = 0.4 mm. 6.36 Figure P6.36 shows an asymmetric diamond-shaped supersonic airfoil. The orientation of the airfoil relative to the airflow is represented by a number of angles: α = the angle of attack, β = the shock angle, θ = the deflection angle, with the subscripts “l” and “u” designating the lower and upper surfaces of the airfoil. The following formula relates the deflection angle to the oblique shock angle and speed,

tan θ = 2cot β(M 2 sin2 β − 1) __________________

M 2(k + cos2 β + 2)

where Cw = a weir coefficient (dimensionless), Bw = the weir width (m), and g = the gravitational constant (m/s2). Cw can be determined using the weir height (Hw) as in

Cw = 1.125 √ __________

1 + Hh ∕Hw __________ 2 + Hh ∕Hw

(P6.31.2)

Given g = 9.81 m/s2, Hw = 0.8 m, Bw = 8 m, and Qw = 1.3 m3/s, determine the upstream depth, H, using (a) the modi- fied secant method with δ = 10–5, (b) fixed-point iteration, and (c) the MATLAB function fzero. For all cases employ an initial guess of 0.5Hw, which for this case is 0.4. For (b) also prove that your result will be convergent for positive initial guesses. 6.32 The following reversible chemical reaction describes how gaseous phases of methane and water react to form car- bon dioxide and hydrogen in a closed reactor,

CH4 + 2H2 O ⇔ CO2 + 4H2

with the equilibrium relationship

K = [CO2 ] [ H2 ]4 ____________

[CH4 ] [ H2 O ]2

where K = the equilibrium coefficient and the brackets [] designate molar concentrations (mole/L). Conserva- tion of mass can be used to reformulate the equilibrium relationship as

K = ( x __ V ) ( 4x ___ V )

______________________ ( M CH 4 − x _________ V ) ( M H 2 O − 2x __________ V ) 2

where x = the number of moles created in the forward direc- tion (mole), V = the volume of the reactor (L), and Mi = the initial number of moles of constituent i (mole). Given that K = 7 × 10−3, V = 20 L, and M CH 4 = M H 2 O = 1 moles, deter- mine x using (a) fixed-point iteration and (b) fzero. 6.33 The concentration of pollutant bacteria c in a lake de- creases according to

c = 77e−1.5t + 20e−0.08t

Determine the time required for the bacteria concentration to be reduced to 15 using the Newton-Raphson method with an initial guess of t = 6 and a stopping criterion of 1%. Check your result with fzero. 6.34 You are asked to solve for the root of the following equation with fixed-point iteration:

x4 = 5x + 10

α θuβu

βl θl

FIGURE P6.36 A diamond-shaped airfoil.

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196 ROOTS: OPEN METhODS

[Hint: The problem can be solved by integrating Eq. (P6.37) two times to yield an equation for x as a function of t.]

where M = the Mach number which is the ratio of the jet’s speed, υ (m/s), to the speed of sound, c (m/s), where

c = √ _____

kRTa

where k = the ratio of specific heats which for air is cp/cυ (= 1.4), R = the air gas constant (= 287 N m/(kg K)), and Ta = the air’s absolute temperature (K). Given estimates of M, k, and θ, the shock angle can be determined as the root of

f ( β) = 2cot β(M 2 sin2 β − 1) __________________

M2(k + cos2 β + 2) − tan θ

The pressure on the airfoil surface, pa (kPa), can then be computed as

pa = p ( 2k _____ k + 1 (M sin β)2 − k − 1 _____ k + 1 ) Suppose that the airfoil is attached to a jet traveling at a speed υ = 625 m/s through air with a temperature T = 4 °C, pressure p = 110 kPa, and θu = 4°. Develop a MATLAB script to (a) generate a plot of f (βu) versus βu = 2° to 88°, and (b) compute the pressure on the upper surface of the airfoil. 6.37 As described in Sec. 1.4, for objects falling through fluids at very low speeds, the flow regime around the object will be laminar and the relationship between the drag force and velocity is linear. In addition, in such cases, the buoy- ancy force must also be included. For such cases, a force balance can be written as

dv ___ dt = g − ρf V ___ m g −

cd __ m υ (P6.37)

(gravity) (buoyancy) (drag)

where υ = velocity (m/s), t = time (s), m = the mass of the particle (kg), g = the gravitational constant (= 9.81 m/s2), ρf = fluid density (kg/m3), V = particle volume (m3), and cd = the linear drag coefficient (kg/m). Note that the mass of the particle can be computed as Vρs, where ρs = the density of the particle (kg/m3). For a small sphere, Stokes developed the following formula for the drag coefficient, cd = 6πμr, where μ = the fluid’s dynamic viscosity (N s/m2), and r = the sphere’s radius (m). You release an iron sphere at the surface of a container (x = 0) filled with honey (Fig. P6.37) and then measure how long it takes to settle to the bottom (x = L). Use this infor- mation to estimate the honey’s viscosity based on the fol- lowing parameter values: ρf = 1420 kg/m3, ρs = 7850 kg/ m3, r = 0.02 m, L = 0.5 m, and t(x = 0.5) = 3.6 s. Check the Reynolds number (Re = ρf υd/μ, where d = diameter) to con- firm that laminar conditions occurred during the experiment.

x = 0

x = L

FIGURE P6.37 A sphere settling in a cylinder filled with viscous honey.

6.38 As depicted in Fig. P6.38a, a scoreboard is suspended above a sports arena by two cables, pinned at A, B, and C. The cables are initially horizontal and of length L. After the scoreboard is hung, a free-body diagram at node B can be de- veloped as shown in Fig. P6.38b. Assuming that the weight of each cable is negligible, determine the deflection, d (m), that results if the scoreboard weighs W = 9000 N. Also, compute how much each cable is elongated. Note that each cable obeys Hooke’s law such that the axial elongation is represented by L′ – L = FL /(AcE) where F = the axial force (N), Ac = the ca- ble’s cross-sectional area (m2), and E = the modulus of elastic- ity (N/m2). Use the following parameters for your calculations: L = 45 m, Ac = 6.362 × 10–4 m2, and E = 1.5 × 1011 N/m2.

B L′

θ F

Home Visitor Period

The Rabid Hippos

(a)

(b)FIGURE P6.38 (a) Two thin cables pinned at A, B, and C with a score- board suspended from B. (b) Free-body diagram of the pin at B after the scoreboard is hung.

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PROBLEMS 197

d = pipe diameter (m), Le,e = equivalent length for the elbow (m), Le,v = equivalent length for the valve (m), and K = loss coefficient for the contraction at the bottom of the tank. Write a MATLAB script to determine the flow exiting the valve, Q (m3/s), using the following parameter values: h = 24 m, L = 65 m, d = 100 mm, Le,e∕d = 30, Le,υ∕d = 8, and K = 0.5. In addition, the kinematic viscosity of water is υ = μ/ρ = 1.2 × 10–6 m2/s. 6.40 Modify the fzerosimp function (Fig. 6.10) so that it can be passed any function with a single unknown and uses varargin to pass the function’s parameters. Then test it with the following script to obtain a solution for pipe friction based on Case Study 6.3,

clc format long, format compact rho = 1.23;mu = 1.79e − 5;D=0.005;V = 40;e = 0.0015/1000; Re = rho*V*D/mu; g = @(f,e,D) 1/sqrt(f)+2*log10(e/(3.7*D)+2.51/ (Re*sqrt(f))); f = fzerosimp(@(x) g(x,e,D),0.008,0.08)

6.39 A water tower is connected to a pipe with a valve at its end as depicted in Fig. P6.39. Under a number of simplify- ing assumptions (e.g., minor friction losses neglected), the following energy balance can be written

gh − υ 2 __ 2 = f ( L + h _____ d + Le,e ___ d + Le,υ ___ d ) υ2 __ 2 + K υ2 __ 2

where g = gravitational acceleration (= 9.81 m/s2), h = tower height (m), υ = mean water velocity in pipe (m/s), f = the pipe’s friction factor, L = horizontal pipe length (m),

h d

FIGURE P6.39 A water tower connected to a pipe with a valve at its end.

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198

7 Optimization

YOU’VE GOT A PROBLEM

An object like a bungee jumper can be projected upward at a specified velocity. If it is subject to linear drag, its altitude as a function of time can be computed as z = z 0 +

m __ c ( υ0 + mg ___ c ) (1 − e−(c∕m)t ) − mg ___ c t (7.1)

CHAPTER OBJECTIVES

The primary objective of this chapter is to introduce you to how optimization can be used to determine minima and maxima of both one-dimensional and multidimensional functions. Specific objectives and topics covered are

• Understanding why and where optimization occurs in engineering and scientific problem solving.

• Recognizing the difference between one-dimensional and multidimensional optimization.

• Distinguishing between global and local optima. • Knowing how to recast a maximization problem so that it can be solved with a

minimizing algorithm. • Being able to define the golden ratio and understand why it makes one-

dimensional optimization efficient. • Locating the optimum of a single-variable function with the golden-section search. • Locating the optimum of a single-variable function with parabolic interpolation. • Knowing how to apply the fminbnd function to determine the minimum of a

one-dimensional function. • Being able to develop MATLAB contour and surface plots to visualize two-

dimensional functions. • Knowing how to apply the fminsearch function to determine the minimum of a

multidimensional function.

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7.1 INTRODUCTION AND BACKGROUND 199

where z = altitude (m) above the earth’s surface (defined as z = 0), z0 = the initial altitude (m), m = mass (kg), c = a linear drag coefficient (kg/s), υ0 = initial velocity (m/s), and t = time (s). Note that for this formulation, positive velocity is considered to be in the upward direction. Given the following parameter values: g = 9.81 m/s2, z0 = 100 m, υ0 = 55 m/s, m = 80 kg, and c = 15 kg/s, Eq. (7.1) can be used to calculate the jumper’s altitude. As displayed in Fig. 7.1, the jumper rises to a peak elevation of about 190 m at about t = 4 s.

Suppose that you are given the job of determining the exact time of the peak elevation. The determination of such extreme values is referred to as optimization. This chapter will introduce you to how the computer is used to make such determinations.

7.1 INTRODUCTION AND BACKGROUND

In the most general sense, optimization is the process of creating something that is as effective as possible. As engineers, we must continuously design devices and products that perform tasks in an efficient fashion for the least cost. Thus, engineers are always confronting optimization problems that attempt to balance performance and limitations. In addition, scientists have interest in optimal phenomena ranging from the peak elevation of projectiles to the minimum free energy.

From a mathematical perspective, optimization deals with finding the maxima and minima of a function that depends on one or more variables. The goal is to determine the values of the variables that yield maxima or minima for the function. These can then be substituted back into the function to compute its optimal values.

Although these solutions can sometimes be obtained analytically, most practical optimization problems require numerical, computer solutions. From a numerical stand- point, optimization is similar in spirit to the root-location methods we just covered in Chaps. 5 and 6. That is, both involve guessing and searching for a point on a function. The fundamental difference between the two types of problems is illustrated in Fig. 7.2. Root location involves searching for the location where the function equals zero. In contrast, optimization involves searching for the function’s extreme points.

FIGURE 7.1 Elevation as a function of time for an object initially projected upward with an initial velocity.

t, s

z, m

−100

100

200

2 4 6 8 10 12

Maximum elevation

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200 OpTImIzATION

As can be seen in Fig. 7.2, the optimums are the points where the curve is flat. In math- ematical terms, this corresponds to the x value where the derivative f ′(x) is equal to zero. Additionally, the second derivative, f ″(x), indicates whether the optimum is a minimum or a maximum: if f ″(x) < 0, the point is a maximum; if f ″(x) > 0, the point is a minimum.

Now, understanding the relationship between roots and optima would suggest a pos- sible strategy for finding the latter. That is, you can differentiate the function and locate the root (i.e., the zero) of the new function. In fact, some optimization methods do just this by solving the root problem: f ′(x) = 0.

EXAmpLE 7.1 Determining the Optimum Analytically by Root Location

problem Statement. Determine the time and magnitude of the peak elevation based on Eq. (7.1). Use the following parameter values for your calculation: g = 9.81 m/s2, z0 = 100 m, υ0 = 55 m/s, m = 80 kg, and c = 15 kg/s.

Solution. Equation (7.1) can be differentiated to give

dz __ dt = υ0 e −(c∕m)t − mg ___ c (1 − e

−(c∕m)t ) (E7.1.1)

Note that because υ = dz/dt, this is actually the equation for the velocity. The maximum el- evation occurs at the value of t that drives this equation to zero. Thus, the problem amounts to determining the root. For this case, this can be accomplished by setting the derivative to zero and solving Eq. (E7.1.1) analytically for

t = m __ c ln ( 1 + c υ0 ___ mg ) Substituting the parameters gives

t = 80 ___ 15 ln ( 1 + 15(55) ________ 80(9.81) ) = 3.83166 s

Maximum

Minimum

0 Root

Root

f (x)

f ʹ(x) = 0 f ʺ(x) > 0

f ʹ(x) = 0 f ʺ(x) < 0

f (x) = 0

FIGURE 7.2 A function of a single variable illustrating the difference between roots and optima.

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7.1 INTRODUCTION AND BACKGROUND 201

This value along with the parameters can then be substituted into Eq. (7.1) to compute the maximum elevation as

z = 100 + 80 ___ 15 ( 50 + 80(9.81) ________ 15 ) (1 − e−(15/80)3.83166) − 80(9.81) ________ 15 (3.83166) = 192.8609 m We can verify that the result is a maximum by differentiating Eq. (E7.1.1) to obtain

the second derivative

d 2z ___

dt2 = − c __ m υ0e

−(c∕m)t − ge−(c∕m)t = −9.81 m __ s2

The fact that the second derivative is negative tells us that we have a maximum. Further, the result makes physical sense since the acceleration should be solely equal to the force of gravity at the maximum when the vertical velocity (and hence drag) is zero.

Although an analytical solution was possible for this case, we could have obtained the same result using the root-location methods described in Chaps. 5 and 6. This will be left as a homework exercise.

Although it is certainly possible to approach optimization as a roots problem, a variety of direct numerical optimization methods are available. These methods are available for both one-dimensional and multidimensional problems. As the name implies, one-dimensional problems involve functions that depend on a single dependent variable. As in Fig. 7.3a, the search then consists of climbing or descending one-dimensional peaks and valleys. Multi- dimensional problems involve functions that depend on two or more dependent variables.

(b)(a)

Optimum f (x*, y*)

Minimum f (x)

f (x)

−f (x)

Maximum −f (x)

f (x, y)

f (x)

FIGURE 7.3 (a) One-dimensional optimization. This figure also illustrates how minimization of f (x) is equivalent to the maximization of −f (x). (b) Two-dimensional optimization. Note that this figure can be taken to represent either a maximization (contours increase in elevation up to the maximum like a mountain) or a minimization (contours decrease in elevation down to the minimum like a valley).

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202 OpTImIzATION

In the same spirit, a two-dimensional optimization can again be visualized as searching out peaks and valleys (Fig. 7.3b). However, just as in real hiking, we are not constrained to walk a single direction; instead the topography is examined to efficiently reach the goal.

Finally, the process of finding a maximum versus finding a minimum is essentially identical because the same value x* both minimizes f (x) and maximizes −f (x). This equiva- lence is illustrated graphically for a one-dimensional function in Fig. 7.3a.

In the next section, we will describe some of the more common approaches for one- dimensional optimization. Then we will provide a brief description of how MATLAB can be employed to determine optima for multidimensional functions.

7.2 ONE-DIMENSIONAL OPTIMIZATION

This section will describe techniques to find the minimum or maximum of a function of a single variable f (x). A useful image in this regard is the one-dimensional “roller coaster”– like function depicted in Fig. 7.4. Recall from Chaps. 5 and 6 that root location was com- plicated by the fact that several roots can occur for a single function. Similarly, both local and global optima can occur in optimization.

A global optimum represents the very best solution. A local optimum, though not the very best, is better than its immediate neighbors. Cases that include local optima are called multimodal. In such cases, we will almost always be interested in finding the global optimum. In addition, we must be concerned about mistaking a local result for the global optimum.

Just as in root location, optimization in one dimension can be divided into bracketing and open methods. As described in the next section, the golden-section search is an example of a bracketing method that is very similar in spirit to the bisection method for root location. This is followed by a somewhat more sophisticated bracketing approach—parabolic inter- polation. We will then show how these two methods are combined and implemented with MATLAB’s fminbnd function.

Global maximum

Local minimumGlobal

minimum

Local maximum

f (x)

FIGURE 7.4 A function that asymptotically approaches zero at plus and minus ∞ and has two maximum and two minimum points in the vicinity of the origin. The two points to the right are local op- tima, whereas the two to the left are global.

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7.2 ONE-DImENSIONAL OpTImIzATION 203

7.2.1 Golden-Section Search

In many cultures, certain numbers are ascribed magical qualities. For example, we in the West are all familiar with “lucky 7” and “Friday the 13th.” Beyond such superstitious quan- tities, there are several well-known numbers that have such interesting and powerful math- ematical properties that they could truly be called “magical.” The most common of these are the ratio of a circle’s circumference to its diameter π and the base of the natural logarithm e.

Although not as widely known, the golden ratio should surely be included in the pan- theon of remarkable numbers. This quantity, which is typically represented by the Greek letter ϕ (pronounced: fee), was originally defined by Euclid (ca. 300 BCE) because of its role in the construction of the pentagram or five-pointed star. As depicted in Fig. 7.5, Euclid’s definition reads: “A straight line is said to have been cut in extreme and mean ratio when, as the whole line is to the greater segment, so is the greater to the lesser.”

The actual value of the golden ratio can be derived by expressing Euclid’s definition as

ℓ1 + ℓ2 _____ ℓ1

= ℓ1 __ ℓ2

(7.2)

Multiplying by ℓ 1∕ ℓ 2 and collecting terms yields

ϕ2 − ϕ − 1 = 0 (7.3)

where ϕ = ℓ 1∕ ℓ 2. The positive root of this equation is the golden ratio:

ϕ = 1 + √ __

5 _______ 2 = 1.61803398874989 . . . (7.4)

The golden ratio has long been considered aesthetically pleasing in Western cultures. In addition, it arises in a variety of other contexts including biology. For our purposes, it provides the basis for the golden-section search, a simple, general-purpose method for de- termining the optimum of a single-variable function.

The golden-section search is similar in spirit to the bisection approach for locating roots in Chap. 5. Recall that bisection hinged on defining an interval, specified by a lower guess (xl) and an upper guess (xu) that bracketed a single root. The presence of a root be- tween these bounds was verified by determining that f (xl) and f (xu) had different signs. The root was then estimated as the midpoint of this interval:

xr = xl + xu ______ 2 (7.5)

FIGURE 7.5 Euclid’s definition of the golden ratio is based on dividing a line into two segments so that the ratio of the whole line to the larger segment is equal to the ratio of the larger segment to the smaller segment. This ratio is called the golden ratio.

�1 �2

�1 + �2

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204 OpTImIzATION

The final step in a bisection iteration involved determining a new smaller bracket. This was done by replacing whichever of the bounds xl or xu had a function value with the same sign as f (xr). A key advantage of this approach was that the new value xr replaced one of the old bounds.

Now suppose that instead of a root, we were interested in determining the minimum of a one-dimensional function. As with bisection, we can start by defining an interval that contains a single answer. That is, the interval should contain a single minimum, and hence is called unimodal. We can adopt the same nomenclature as for bisection, where xl and xu defined the lower and upper bounds, respectively, of such an interval. However, in contrast to bisection, we need a new strategy for finding a minimum within the interval. Rather than using a single intermediate value (which is sufficient to detect a sign change, and hence a zero), we would need two intermediate function values to detect whether a minimum occurred.

The key to making this approach efficient is the wise choice of the intermediate points. As in bisection, the goal is to minimize function evaluations by replacing old values with new values. For bisection, this was accomplished by choosing the midpoint. For the golden- section search, the two intermediate points are chosen according to the golden ratio:

x1 = xl + d (7.6) x2 = xu − d (7.7)

where

d = (ϕ − 1)(xu − xl) (7.8)

The function is evaluated at these two interior points. Two results can occur:

1. If, as in Fig. 7.6a, f (x1) < f (x2), then f (x1) is the minimum, and the domain of x to the left of x2, from xl to x2, can be eliminated because it does not contain the minimum. For this case, x2 becomes the new xl for the next round.

2. If f (x2) < f (x1), then f (x2) is the minimum and the domain of x to the right of x1, from x1 to xu would be eliminated. For this case, x1 becomes the new xu for the next round.

Now, here is the real benefit from the use of the golden ratio. Because the original x1 and x2 were chosen using the golden ratio, we do not have to recalculate all the function values for the next iteration. For example, for the case illustrated in Fig. 7.6, the old x1 be- comes the new x2. This means that we already have the value for the new f (x2), since it is the same as the function value at the old x1.

To complete the algorithm, we need only determine the new x1. This is done with Eq. (7.6) with d computed with Eq. (7.8) based on the new values of xl and xu. A similar approach would be used for the alternate case where the optimum fell in the left subinterval. For this case, the new x2 would be computed with Eq. (7.7).

As the iterations are repeated, the interval containing the extremum is reduced rapidly. In fact, each round the interval is reduced by a factor of ϕ − 1 (about 61.8%). That means that after 10 rounds, the interval is shrunk to about 0.61810 or 0.008 or 0.8% of its initial length. After 20 rounds, it is about 0.0066%. This is not quite as good as the reduction achieved with bisection (50%), but optimization is a harder problem than root location.

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7.2 ONE-DImENSIONAL OpTImIzATION 205

EXAmpLE 7.2 Golden-Section Search

problem Statement. Use the golden-section search to find the minimum of

f (x) = x 2 ___ 10 − 2 sin x

within the interval from xl = 0 to xu = 4.

Solution. First, the golden ratio is used to create the two interior points:

d = 0.61803(4 − 0) = 2.4721 x1 = 0 + 2.4721 = 2.4721 x2 = 4 − 2.4721 = 1.5279

xl d x1 x2 d xu

Eliminate Minimum

x x2 xux1

Old x2

Old x1

New

f (x)

(b)

(a)

FIGURE 7.6 (a) The initial step of the golden-section search algorithm involves choosing two interior points according to the golden ratio. (b) The second step involves defining a new interval that encompasses the optimum.

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206 OpTImIzATION

The function can be evaluated at the interior points:

f (x2) = 1.52792 _______ 10 − 2 sin(1.5279) = −1.7647

f (x1) = 2.47212 _______ 10 − 2 sin(2.4721) = −0.6300

Because f (x2) < f (x1), our best estimate of the minimum at this point is that it is located at x = 1.5279 with a value of f (x) = –1.7647. In addition, we also know that the minimum is in the interval defined by xl, x2, and x1. Thus, for the next iteration, the lower bound remains xl = 0, and x1 becomes the upper bound, that is, xu = 2.4721. In addition, the former x2 value becomes the new x1, that is, x1 = 1.5279. In addition, we do not have to recalculate f (x1), it was determined on the previous iteration as f (1.5279) = –1.7647. All that remains is to use Eqs. (7.8) and (7.7) to compute the new value of d and x2:

d = 0.61803(2.4721 − 0) = 1.5279

x2 = 2.4721 − 1.5279 = 0.9443

The function evaluation at x2 is f (0.9943) = −1.5310. Since this value is less than the function value at x1, the minimum is f (1.5279) = −1.7647, and it is in the interval pre- scribed by x2, x1, and xu. The process can be repeated, with the results tabulated here:

i xl f (xl) x2 f (x2) x1 f (x1) xu f (xu) d

1 0 0 1.5279 −1.7647 2.4721 −0.6300 4.0000 3.1136 2.4721 2 0 0 0.9443 −1.5310 1.5279 −1.7647 2.4721 −0.6300 1.5279 3 0.9443 −1.5310 1.5279 −1.7647 1.8885 −1.5432 2.4721 −0.6300 0.9443 4 0.9443 −1.5310 1.3050 −1.7595 1.5279 −1.7647 1.8885 −1.5432 0.5836 5 1.3050 −1.7595 1.5279 −1.7647 1.6656 −1.7136 1.8885 −1.5432 0.3607 6 1.3050 −1.7595 1.4427 −1.7755 1.5279 −1.7647 1.6656 −1.7136 0.2229 7 1.3050 −1.7595 1.3901 −1.7742 1.4427 −1.7755 1.5279 −1.7647 0.1378 8 1.3901 −1.7742 1.4427 −1.7755 1.4752 −1.7732 1.5279 −1.7647 0.0851

Note that the current minimum is highlighted for every iteration. After the eighth iteration, the minimum occurs at x = 1.4427 with a function value of −1.7755. Thus, the result is converging on the true value of −1.7757 at x = 1.4276.

Recall that for bisection (Sec. 5.4), an exact upper bound for the error can be calculated at each iteration. Using similar reasoning, an upper bound for golden-section search can be derived as follows: Once an iteration is complete, the optimum will either fall in one of two intervals. If the optimum function value is at x2, it will be in the lower interval (xl, x2, x1). If the optimum function value is at x1, it will be in the upper interval (x2, x1, xu). Because the interior points are symmetrical, either case can be used to define the error.

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7.2 ONE-DImENSIONAL OpTImIzATION 207

Looking at the upper interval (x2, x1, xu), if the true value were at the far left, the maxi- mum distance from the estimate would be

Δxa = x1 − x2 = xl + (ϕ − 1)(xu − xl) − xu + (ϕ − 1)(xu − xl) = (xl − xu) + 2(ϕ − 1)(xu − xl) = (2ϕ − 3)(xu − xl)

or 0.2361 (xu − xl). If the true value were at the far right, the maximum distance from the estimate would be

Δxb = xu − x1 = xu − xl − (ϕ − 1)(xu − xl) = (xu − xl) − (ϕ − 1)(xu − xl) = (2 − ϕ)(xu − xl)

or 0.3820 (xu − xl). Therefore, this case would represent the maximum error. This result can then be normalized to the optimal value for that iteration xopt to yield

εa = (2 − ϕ) | xu − xl ______ xopt | × 100% (7.9) This estimate provides a basis for terminating the iterations.

An M-file function for the golden-section search for minimization is presented in Fig. 7.7. The function returns the location of the minimum, the value of the function, the approximate error, and the number of iterations.

The M-file can be used to solve the problem from Example 7.1.

>> g = 9.81;v0 = 55;m = 80;c = 15;z0 = 100; >> z = @(t) −(z0 + m/c*(v0 + m*g/c)*(1 − exp(−c/m*t)) −m*g/c*t); >> [xmin,fmin,ea] = goldmin(z,0,8)

xmin = 3.8317 fmin = −192.8609 ea = 6.9356e−005

Notice how because this is a maximization, we have entered the negative of Eq. (7.1). Consequently, fmin corresponds to a maximum height of 192.8609.

You may be wondering why we have stressed the reduced function evaluations of the golden-section search. Of course, for solving a single optimization, the speed savings would be negligible. However, there are two important contexts where minimizing the number of function evaluations can be important. These are

1. Many evaluations. There are cases where the golden-section search algorithm may be a part of a much larger calculation. In such cases, it may be called many times. Therefore, keeping function evaluations to a minimum could pay great dividends for such cases.

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208 OpTImIzATION

2. Time-consuming evaluation. For pedagogical reasons, we use simple functions in most of our examples. You should understand that a function can be very complex and time-consuming to evaluate. For example, optimization can be used to estimate the parameters of a model consisting of a system of differential equations. For such cases, the “function” involves time-consuming model integration. Any method that minimizes such evaluations would be advantageous.

FIGURE 7.7 An M-file to determine the minimum of a function with the golden-section search.

function [x,fx,ea,iter] = goldmin(f,xl,xu,es,maxit,varargin) % goldmin: minimization golden section search % [x,fx,ea,iter] = goldmin(f,xl,xu,es,maxit,p1,p2,...): % uses golden section search to find the minimum of f % input: % f = name of function % xl, xu = lower and upper guesses % es = desired relative error (default = 0.0001%) % maxit = maximum allowable iterations (default = 50) % p1,p2,... = additional parameters used by f % output: % x = location of minimum % fx = minimum function value % ea = approximate relative error (%) % iter = number of iterations

if nargin < 3,error('at least 3 input arguments required'),end if nargin < 4|isempty(es), es = 0.0001;end if nargin < 5|isempty(maxit), maxit = 50;end phi = (1+sqrt(5))/2; iter = 0; d = (phi−1)*(xu − xl); x1 = xl + d; x2 = xu − d; f1 = f(x1,varargin{:}); f2 = f(x2,varargin{:}); while(1) xint = xu − xl; if f1 < f2 xopt = x1; xl = x2; x2 = x1; f2 = f1; x1 = xl + (phi − 1)*(xu − xl); f1 = f(x1,varargin{:}); else xopt = x2; xu = x1; x1 = x2; f1 = f2; x2 = xu − (phi−1)*(xu−xl); f2 = f(x2,varargin{:}); end iter = iter + 1; if xopt~ = 0, ea = (2 − phi) * abs(xint / xopt) * 100;end if ea < = es | iter > = maxit,break,end end x = xopt; fx = f(xopt,varargin{:});

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7.2 ONE-DImENSIONAL OpTImIzATION 209

7.2.2 Parabolic Interpolation

Parabolic interpolation takes advantage of the fact that a second-order polynomial often provides a good approximation to the shape of f (x) near an optimum (Fig. 7.8).

Just as there is only one straight line connecting two points, there is only one parabola connecting three points. Thus, if we have three points that jointly bracket an optimum, we can fit a parabola to the points. Then we can differentiate it, set the result equal to zero, and solve for an estimate of the optimal x. It can be shown through some algebraic manipula- tions that the result is

x4 = x2 − 1 __ 2

(x2 − x1)2 [ f (x2) − f (x3)] − (x2 − x3)2 [ f (x2) − f (x1)] ___________________________________________ (x2 − x1) [ f (x2) − f (x3)] − (x2 − x3) [ f (x2) − f (x1)] (7.10)

where x1, x2, and x3 are the initial guesses, and x4 is the value of x that corresponds to the optimum value of the parabolic fit to the guesses.

EXAmpLE 7.3 parabolic Interpolation

problem Statement. Use parabolic interpolation to approximate the minimum of

f (x) = x 2 ___ 10 − 2 sin x

with initial guesses of x1 = 0, x2 = 1, and x3 = 4.

Solution. The function values at the three guesses can be evaluated:

x1 = 0 f (x1) = 0 x2 = 1 f (x2) = −1.5829 x3 = 4 f (x3) = 3.1136

FIGURE 7.8 Graphical depiction of parabolic interpolation.

xx1 x2 x4 x3

Parabolic approximation of maximum

Parabolic function

True maximum

True functionf (x)

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210 OpTImIzATION

and substituted into Eq. (7.10) to give

x4 = 1 − 1 __ 2

(1 − 0)2 [−1.5829 − 3.1136] − (1 − 4)2 [−1.5829 − 0] ____________________________________________ (1 − 0) [−1.5829 − 3.1136] − (1 − 4) [−1.5829 − 0] = 1.5055

which has a function value of f (1.5055) = −1.7691. Next, a strategy similar to the golden-section search can be employed to determine which point should be discarded. Because the function value for the new point is lower than for the intermediate point (x2) and the new x value is to the right of the intermediate point, the lower guess (x1) is discarded. Therefore, for the next iteration:

x1 = 1 f (x1) = −1.5829 x2 = 1.5055 f (x2) = −1.7691 x3 = 4 f (x3) = 3.1136

which can be substituted into Eq. (7.10) to give

x4 = 1.5055 − 1 __ 2

(1.5055 − 1)2 [−1.7691 − 3.1136] − (1.5055 − 4)2 [−1.7691 − (−1.5829)] ____________________________________________________________ (1.5055 − 1) [−1.7691 − 3.1136] − (1.5055 − 4) [−1.7691 − (−1.5829)]

= 1.4903

which has a function value of f (1.4903) = −1.7714. The process can be repeated, with the results tabulated here:

i x1 f (x1) x2 f (x2) x3 f (x3) x4 f (x4)

1 0.0000 0.0000 1.0000 −1.5829 4.0000 3.1136 1.5055 −1.7691 2 1.0000 −1.5829 1.5055 −1.7691 4.0000 3.1136 1.4903 −1.7714 3 1.0000 −1.5829 1.4903 −1.7714 1.5055 −1.7691 1.4256 −1.7757 4 1.0000 −1.5829 1.4256 −1.7757 1.4903 −1.7714 1.4266 −1.7757 5 1.4256 −1.7757 1.4266 −1.7757 1.4903 −1.7714 1.4275 −1.7757

Thus, within five iterations, the result is converging rapidly on the true value of −1.7757 at x = 1.4276.

7.2.3 MATLAB Function: fminbnd

Recall that in Sec. 6.4 we described Brent’s method for root location, which combined sev- eral root-finding methods into a single algorithm that balanced reliability with efficiency. Because of these qualities, it forms the basis for the built-in MATLAB function fzero.

Brent also developed a similar approach for one-dimensional minimization which forms the basis for the MATLAB fminbnd function. It combines the slow, dependable golden-section search with the faster, but possibly unreliable, parabolic interpolation. It first attempts parabolic interpolation and keeps applying it as long as acceptable results are obtained. If not, it uses the golden-section search to get matters in hand.

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7.3 mULTIDImENSIONAL OpTImIzATION 211

A simple expression of its syntax is

[xmin, fval] = fminbnd(function,x1,x2)

where x and fval are the location and value of the minimum, function is the name of the function being evaluated, and x1 and x2 are the bounds of the interval being searched.

Here is a simple MATLAB session that uses fminbnd to solve the problem from Example 7.1.

>> g = 9.81;v0 = 55;m = 80;c = 15;z0 = 100; >> z = @(t) −(z0 + m/c*(v0 + m*g/c)*(1 − exp(−c/m*t)) − m*g/c*t); >> [x,f] = fminbnd(z,0,8)

x = 3.8317 f = −192.8609

As with fzero, optional parameters can be specified using optimset. For example, we can display calculation details:

>> options = optimset('display','iter'); >> fminbnd (z,0,8,options)

Func-count x f(x) Procedure 1 3.05573 −189.759 initial 2 4.94427 −187.19 golden 3 1.88854 −171.871 golden 4 3.87544 −192.851 parabolic 5 3.85836 −192.857 parabolic 6 3.83332 −192.861 parabolic 7 3.83162 −192.861 parabolic 8 3.83166 −192.861 parabolic 9 3.83169 −192.861 parabolic

Optimization terminated: the current x satisfies the termination criteria using OPTIONS.TolX of 1.000000e − 004

ans = 3.8317

Thus, after three iterations, the method switches from golden to parabolic, and after eight iterations, the minimum is determined to a tolerance of 0.0001.

7.3 MULTIDIMENSIONAL OPTIMIZATION

Aside from one-dimensional functions, optimization also deals with multidimensional functions. Recall from Fig. 7.3a that our visual image of a one-dimensional search was like a roller coaster. For two-dimensional cases, the image becomes that of mountains and valleys (Fig. 7.3b). As in the following example, MATLAB’s graphic capabilities provide a handy means to visualize such functions.

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212 OpTImIzATION

EXAmpLE 7.4 Visualizing a Two-Dimensional Function

problem Statement. Use MATLAB’s graphical capabilities to display the following function and visually estimate its minimum in the range –2 ≤ x1 ≤ 0 and 0 ≤ x2 ≤ 3:

f (x1, x2) = 2 + x1 − x2 + 2 x 1 2 + 2x1x2 + x 2

Solution. The following script generates contour and mesh plots of the function: x = linspace(−2,0,40);y = linspace(0,3,40); [X,Y] = meshgrid(x,y); Z = 2 + X − Y + 2 * X.^2 + 2 * X.* Y + Y.^2; subplot(1,2,1); cs = contour(X,Y,Z);clabel(cs); xlabel('x_1');ylabel('x_2'); title('(a) Contour plot');grid; subplot(1,2,2); cs = surfc(X,Y,Z); zmin = floor(min(Z)); zmax = ceil(max(Z)); xlabel('x_1');ylabel('x_2');zlabel('f(x_1,x_2)'); title ('(b) Mesh plot');

As displayed in Fig. 7.9, both plots indicate that function has a minimum value of about f (x1, x2) = 0 to 1 located at about x1 = −1 and x2 = 1.5.

x2 x1

0 −2

−1

+ + +

+2 +4

+6+78 +5

(a) Contour plot (b) Mesh plot

2.5

1.5x 2

f ( x 1

, x 2)

0.5

0 −2 −1.5 −1 −0.5 0

FIGURE 7.9 (a) Contour and (b) mesh plots of a two-dimensional function.

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Techniques for multidimensional unconstrained optimization can be classified in a number of ways. For purposes of the present discussion, we will divide them depending on whether they require derivative evaluation. Those that require derivatives are called gra- dient, or descent (or ascent), methods. The approaches that do not require derivative evalu- ation are called nongradient, or direct, methods. As described next, the built-in MATLAB function fminsearch is a direct method.

7.3.1 MATLAB Function: fminsearch

Standard MATLAB has a function fminsearch that can be used to determine the mini- mum of a multidimensional function. It is based on the Nelder-Mead method, which is a direct-search method that uses only function values (does not require derivatives) and handles non-smooth objective functions. A simple expression of its syntax is

[xmin, fval] = fminsearch (function,x0)

where xmin and fval are the location and value of the minimum, function is the name of the func- tion being evaluated, and x0 is the initial guess. Note that x0 can be a scalar, vector, or a matrix.

Here is a simple MATLAB session that uses fminsearch to determine minimum for the function we just graphed in Example 7.4:

>> f = @ (x) 2 + x (1) − x (2) + 2*x (1)^2 + 2*x (1)*x (2) + x (2)^2; >> [x,fval] = fminsearch (f,[−0.5,0.5])

x = −1.0000 1.5000 fval = 0.7500

7.4 CASE STUDY EQUILIBRIUm AND mINImUm pOTENTIAL ENERGY

Background. As in Fig. 7.10a, an unloaded spring can be attached to a wall mount. When a horizontal force is applied, the spring stretches. The displacement is related to the force by Hookes law, F = kx. The potential energy of the deformed state consists of the differ- ence between the strain energy of the spring and the work done by the force:

PE(x) = 0.5kx 2 − Fx (7.11)

7.4 CASE STUDY 213

(a)

(b) F

FIGURE 7.10 (a) An unloaded spring attached to a wall mount. (b) Application of a horizontal force stretches the spring where the relationship between force and displacement is described by Hooke’s law.

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214 OpTImIzATION

Equation (7.11) defines a parabola. Since the potential energy will be at a minimum at equilibrium, the solution for displacement can be viewed as a one-dimensional optimi- zation problem. Because this equation is so easy to differentiate, we can solve for the dis- placement as x = F∕k. For example, if k = 2 N∕cm and F = 5 N, x = 5N∕(2 N/cm) = 2.5 cm.

A more interesting two-dimensional case is shown in Fig. 7.11. In this system, there are two degrees of freedom in that the system can move both horizontally and vertically. In the same way that we approached the one-dimensional system, the equilibrium deforma- tions are the values of x1 and x2 that minimize the potential energy:

PE(x1, x2) = 0.5ka ( √____________ x 1 2 + (La − x2)2 − La ) 2 + 0.5kb ( √____________ x 1 2 + (Lb + x2)2 − Lb ) 2 − F1x1 − F2 x2 (7.12)

If the parameters are ka = 9 N/cm, kb = 2 N/cm, La = 10 cm, Lb = 10 cm, F1 = 2 N, and F2 = 4 N, use MATLAB to solve for the displacements and the potential energy.

7.4 CASE STUDY continued

(b)(a)

Lb kb

ka F2

F1 x2

FIGURE 7.11 A two-spring system: (a) unloaded and (b) loaded.

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7.4 CASE STUDY continued

Solution. An M-file can be developed to hold the potential energy function: function p = PE(x, ka, kb, La, Lb, F1, F2) PEa = 0.5*ka*(sqrt(x(1)^2 + (La − x(2))^2) − La)^2; PEb = 0.5*kb*(sqrt(x(1)^2 + (Lb + x(2))^2) − Lb)^2; W = F1*x(1) + F2*x(2); p = PEa + PEb − W;

The solution can be obtained with the fminsearch function: >> ka = 9;kb = 2;La = 10;Lb = 10;F1 = 2;F2 = 4; >> [x,f] = fminsearch(@(x) PE(x,ka,kb,La,Lb,F1,F2),[−0.5,0.5])

x = 4.9523 1.2769 f = −9.6422

Thus, at equilibrium, the potential energy is −9.6422 N · cm. The connecting point is located 4.9523 cm to the right and 1.2759 cm above its original position.

PROBLEMS

7.1 Perform three iterations of the Newton-Raphson method to determine the root of Eq. (E7.1.1). Use the parameter val- ues from Example 7.1 along with an initial guess of t = 3 s. 7.2 Given the formula

f (x) = −x2 + 8x − 12

(a) Determine the maximum and the corresponding value of x for this function analytically (i.e., using differentiation).

(b) Verify that Eq. (7.10) yields the same results based on initial guesses of x1 = 0, x2 = 2, and x3 = 6.

7.3 Consider the following function:

f (x) = 3 + 6x + 5x2 + 3x3 + 4x4

Locate the minimum by finding the root of the derivative of this function. Use bisection with initial guesses of xl = −2 and xu = 1. 7.4 Given

f (x) = −1.5x6 − 2x4 + 12x

(a) Plot the function. (b) Use analytical methods to prove that the function is con-

cave for all values of x.

(c) Differentiate the function and then use a root-location method to solve for the maximum f (x) and the corre- sponding value of x.

7.5 Solve for the value of x that maximizes f (x) in Prob. 7.4 using the golden-section search. Employ initial guesses of xl = 0 and xu = 2, and perform three iterations. 7.6 Repeat Prob. 7.5, except use parabolic interpolation. Employ initial guesses of x1 = 0, x2 = 1, and x3 = 2, and perform three iterations. 7.7 Employ the following methods to find the maximum of

f (x) = 4x − 1.8x2 + 1.2x3 − 0.3x4

(a) Golden-section search (xl = – 2, xu = 4, εs = 1%). (b) Parabolic interpolation (x1 = 1.75, x2 = 2, x3 = 2.5,

iterations = 5). 7.8 Consider the following function:

f (x) = x4 + 2x3 + 8x2 + 5x

Use analytical and graphical methods to show the func- tion has a minimum for some value of x in the range −2 ≤ x ≤ 1.

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216 OpTImIzATION

7.9 Employ the following methods to find the minimum of the function from Prob. 7.8: (a) Golden-section search (xl = −2, xu = 1, εs = 1%). (b) Parabolic interpolation (x1 = −2, x2 = −1, x3 = 1,

iterations = 5). 7.10 Consider the following function:

f (x) = 2x + 3 __ x

Perform 10 iterations of parabolic interpolation to locate the minimum. Comment on the convergence of your results (x1 = 0.1, x2 = 0.5, x3 = 5) 7.11 The following function defines a curve with several unequal minima over the interval: 2 ≤ x ≤ 20,

f (x) = sin(x) + sin ( 2 __ 3 x ) Develop a MATLAB script to (a) plot the function over the interval. Determine the minimum (b) with fminbnd and (c) by hand with golden-section search with a stopping cri- terion corresponding to three significant figures. For (b) and (c), use initial guesses of [4, 8]. 7.12 Use the golden-section search to determine the loca- tion, xmax, and maximum, f(xmax), of the following function by hand,

f (x) = −0.8x4 + 2.2x2 + 0.6

Use initial guesses of xl = 0.7 and xu = 1.4 and perform suffi- cient iterations so that εs = 10%. Determine the approximate relative error of your final result. 7.13 Develop a single script to (a) generate contour and mesh subplots of the following temperature field in a similar fashion to Example 7.4:

T (x, y) = 2x2 + 3y2 − 4xy − y − 3x

and (b) determine the minimum with fminsearch. 7.14 The head of a groundwater aquifer is described in Car- tesian coordinates by

h(x, y) = 1 _________________ 1 + x2 + y2 + x + xy

Develop a single script to (a) generate contour and mesh subplots of the function in a similar fashion to Example 7.4, and (b) determine the maximum with fminsearch. 7.15 Recent interest in competitive and recreational cycling has meant that engineers have directed their skills toward the design and testing of mountain bikes (Fig. P7.15a). Sup- pose that you are given the task of predicting the horizontal

and vertical displacement of a bike bracketing system in response to a force. Assume the forces you must analyze can be simplified as depicted in Fig. P7.15b. You are inter- ested in testing the response of the truss to a force exerted in any number of directions designated by the angle θ. The parameters for the problem are E = Young’s modulus = 2 × 1011 Pa, A = cross-sectional area = 0.0001 m2, 𝑤 = width = 0.44 m, ℓ = length = 0.56 m, and h = height = 0.5 m. The displacements x and y can be solved by determining the val- ues that yield a minimum potential energy. Determine the displacements for a force of 10,000 N and a range of θ’s from 0° (horizontal) to 90° (vertical). 7.16 As electric current moves through a wire (Fig. P7.16), heat generated by resistance is conducted through a layer of insulation and then convected to the surrounding air. The steady-state temperature of the wire can be computed as

T = Tair + q ___ 2π [ 1 __ k ln ( r𝑤 + ri ______ r𝑤 ) + 1 __ h 1 ______ r𝑤 + ri ]

h ℓ

�

(b)(a)

FIGURE P7.15 (a) A mountain bike along with (b) a free-body diagram for a part of the frame.

Tair

FIGURE P7.16 Cross-section of an insulated wire.

cha97962_ch07_198-222.indd 216 08/11/16 3:25 pm

Determine the thickness of insulation ri (m) that minimizes the wire’s temperature given the following parameters: q = heat generation rate = 75 W/m, r𝑤 = wire radius = 6 mm, k = thermal conductivity of insulation = 0.17 W/(m K), h = convective heat transfer coefficient = 12 W/(m2 K), and Tair = air temperature = 293 K. 7.17 Develop an M-file that is expressly designed to locate a maximum with the golden-section search. In other words, set it up so that it directly finds the maximum rather than finding the minimum of −f (x). The function should have the following features: • Iterate until the relative error falls below a stopping cri-

terion or exceeds a maximum number of iterations. • Return both the optimal x and f (x).

Test your program with the same problem as Example 7.1. 7.18 Develop an M-file to locate a minimum with the golden-section search. Rather than using the maximum it- erations and Eq. (7.9) as the stopping criteria, determine the number of iterations needed to attain a desired tolerance. Test your function by solving Example 7.2 using Ea,d = 0.0001. 7.19 Develop an M-file to implement parabolic interpo- lation to locate a minimum. The function should have the following features: • Base it on two initial guesses, and have the program

generate the third initial value at the midpoint of the interval.

• Check whether the guesses bracket a maximum. If not, the function should not implement the algorithm, but should return an error message.

• Iterate until the relative error falls below a stopping cri- terion or exceeds a maximum number of iterations.

• Return both the optimal x and f (x).

Test your program with the same problem as Example 7.3. 7.20 Pressure measurements are taken at certain points behind an airfoil over time. These data best fit the curve y = 6 cos x − 1.5 sin x from x = 0 to 6 s. Use four iterations of the golden-search method to find the minimum pressure. Set xl = 2 and xu = 4. 7.21 The trajectory of a ball can be computed with

y = (tan θ0)x − g _________

2υ 0 2 cos2 θ0

x2 + y0

where y = the height (m), θ0 = the initial angle (radians), υ0 = the initial velocity (m/s), g = the gravitational constant = 9.81 m/s2, and y0 = the initial height (m). Use the golden-section search to determine the maximum height

given y0 = 2 m, υ0 = 20 m/s, and θ0 = 45°. Iterate until the approximate error falls below εs = 10% using initial guesses of xl = 10 and xu = 30 m. 7.22 The deflection of a uniform beam subject to a linearly increasing distributed load can be computed as

y = 𝑤0 ________ 120E I L (−x

5 + 2L2x3 − L4x)

Given that L = 600 cm, E = 50,000 kN/cm2, I = 30,000 cm4, and 𝑤0 = 2.5 kN/cm, determine the point of maximum de- flection (a) graphically, (b) using the golden-section search until the approximate error falls below εs = 1% with initial guesses of xl = 0 and xu = L. 7.23 A object with a mass of 90 kg is projected upward from the surface of the earth at a velocity of 60 m/s. If the object is subject to linear drag (c = 15 kg/s), use the golden- section search to determine the maximum height the object attains. 7.24 The normal distribution is a bell-shaped curve defined by

y = e−x2

Use the golden-section search to determine the location of the inflection point of this curve for positive x. 7.25 Use the fminsearch function to determine the minimum of

f (x, y) = 2y2 − 2.25xy − 1.75y + 1.5x2

7.26 Use the fminsearch function to determine the maximum of

f (x, y) = 4x + 2y + x2 − 2x4 + 2xy − 3y2

7.27 Given the following function:

f (x, y) = −8x + x2 + 12y + 4y2 − 2xy

Determine the minimum (a) graphically, (b) numerically with the fminsearch function, and (c) substitute the result of (b) back into the function to determine the minimum f (x, y). 7.28 The specific growth rate of a yeast that produces an antibiotic is a function of the food concentration c:

g = 2c __________________ 4 + 0.8c + c2 + 0.2c3

As depicted in Fig. P7.28, growth goes to zero at very low concentrations due to food limitation. It also goes to zero at high concentrations due to toxicity effects. Find the value of c at which growth is a maximum.

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218 OpTImIzATION

7.29 A compound A will be converted into B in a stirred tank reactor. The product B and unreacted A are purified in a sepa- ration unit. Unreacted A is recycled to the reactor. A process engineer has found that the initial cost of the system is a func- tion of the conversion xA. Find the conversion that will result in the lowest cost system. C is a proportionality constant.

Cost = C [ ( 1 ________ (1 − xA)2 ) 0.6 + 6 ( 1 __ xA ) 0.6 ] 7.30 A finite-element model of a cantilever beam subject to loading and moments (Fig. P7.30) is given by optimizing

f (x, y) = 5x2 − 5xy + 2.5y2 − x − 1.5y

where x = end displacement and y = end moment. Find the values of x and y that minimize f (x, y). 7.31 The Streeter-Phelps model can be used to compute the dissolved oxygen concentration in a river below a point dis- charge of sewage (Fig. P7.31),

o = os − kd Lo __________ kd + ks − ka

(e−ka t − e−(kd + ks)t)

− Sb __ ka

(1 − e−ka t ) (P7.31)

where o = dissolved oxygen concentration (mg/L), os = oxygen saturation concentration (mg/L), t = travel time (d), Lo = biochemical oxygen demand (BOD) concentration at the mixing point (mg/L), kd = rate of decomposition of BOD (d−1), ks = rate of settling of BOD (d−1), ka = reaera- tion rate (d−1), and Sb = sediment oxygen demand (mg/L/d).

As indicated in Fig. P7.31, Eq. (P7.31) produces an oxygen “sag” that reaches a critical minimum level oc, some travel time tc below the point discharge. This point is called “critical” because it represents the location where biota that depend on oxygen (like fish) would be the most stressed. Develop a MATLAB script that (a) generates a plots of the function versus travel time and (b) uses fminbnd to deter- mine the critical travel time and concentration, given the fol- lowing values:

os = 10 mg/L kd = 0.1 d−1 ka = 0.6 d−1

ks = 0.05 d−1 Lo = 50 mg/L Sb = 1 mg/L/d

7.32 The two-dimensional distribution of pollutant concen- tration in a channel can be described by

c(x, y) = 7.9 + 0.13x + 0.21y − 0.05x2

−0.016y2 − 0.007xy

Determine the exact location of the peak concentration given the function and the knowledge that the peak lies within the bounds −10 ≤ x ≤ 10 and 0 ≤ y ≤ 20. 7.33 A total charge Q is uniformly distributed around a ring- shaped conductor with radius a. A charge q is located at a distance x from the center of the ring (Fig. P7.33). The force exerted on the charge by the ring is given by

F = 1 ____ 4π e0 q Qx _________ (x2 + a2)3/2

FIGURE P7.30 A cantilever beam.

15 20

0 0

t (d) 5

o (mg/L) o

FIGURE P7.31 A dissolved oxygen “sag” below a point discharge of sewage into a river.

5 10

0.4

0 0

c (mg/L)

g (d−1) 0.2

FIGURE P7.28 The specific growth rate of a yeast that produces an antibiotic versus the food concentration.

cha97962_ch07_198-222.indd 218 08/11/16 3:25 pm

differently as velocity increases. Whereas friction drag in- creases with velocity, the drag due to lift decreases. The com- bination of the two factors leads to a minimum drag. (a) If σ = 0.6 and W = 16,000, determine the minimum drag

and the velocity at which it occurs. (b) In addition, develop a sensitivity analysis to determine

how this optimum varies in response to a range of W = 12,000 to 20,000 with σ = 0.6.

7.36 Roller bearings are subject to fatigue failure caused by large contact loads F (Fig. P7.36). The problem of finding the location of the maximum stress along the x axis can be shown to be equivalent to maximizing the function:

f (x) = 0.4 ________ √

______ 1 + x2 − √

______ 1 + x2 ( 1 − 0.4 ______ 1 + x2 ) + x

Find the x that maximizes f (x).

where e0 = 8.85 × 10−12 C2/(N m2), q = Q = 2 × 10−5 C, and a = 0.9 m. Determine the distance x where the force is a maximum. 7.34 The torque transmitted to an induction motor is a func- tion of the slip between the rotation of the stator field and the rotor speed s, where slip is defined as

s = n − nR ______ n

where n = revolutions per second of rotating stator speed and nR = rotor speed. Kirchhoff’s laws can be used to show that the torque (expressed in dimensionless form) and slip are related by

T = 15s (1 − s) _________________ (1 − s)(4s2 − 3s + 4)

Figure P7.34 shows this function. Use a numerical method to determine the slip at which the maximum torque occurs. 7.35 The total drag on an airfoil can be estimated by

D = 0.01σ V 2 + 0.95 ____ σ ( W __ V ) 2 Friction Lift

where D = drag, σ = ratio of air density between the flight altitude and sea level, W = weight, and V = velocity. As seen in Fig. P7.35, the two factors contributing to drag are affected

FIGURE P7.33

4 8 10

0 0

FIGURE P7.34 Torque transmitted to an inductor as a function of slip.

400 800 1,200

10,000

20,000

Total

Minimum

Fric Lift

0 0

FIGURE P7.35 Plot of drag versus velocity for an airfoil.

FIGURE P7.36 Roller bearings.

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220 OpTImIzATION

7.37 In a similar fashion to the case study described in Sec. 7.4, develop the potential energy function for the sys- tem depicted in Fig. P7.37. Develop contour and surface plots in MATLAB. Minimize the potential energy function to determine the equilibrium displacements x1 and x2 given the forcing function F = 100 N and the parameters ka = 20 and kb = 15 N/m. 7.38 As an agricultural engineer, you must design a trap- ezoidal open channel to carry irrigation water (Fig. P7.38). Determine the optimal dimensions to minimize the wetted perimeter for a cross-sectional area of 50 m2. Are the relative dimensions universal? 7.39 Use the function fminsearch to determine the length of the shortest ladder that reaches from the ground over the fence to the building’s wall (Fig. P7.39). Test it for the case where h = d = 4 m. 7.40 The length of the longest ladder that can negoti- ate the corner depicted in Fig. P7.40 can be determined by computing the value of θ that minimizes the following function:

L(θ) = 𝑤1 ____ sin θ +

𝑤2 ____________ sin(π − α − θ)

FIGURE P7.37 Two frictionless masses connected to a wall by a pair of linear elastic springs.

1 2 F

kbka

x1 x2

�

��

FIGURE P7.38

FIGURE P7.39 A ladder leaning against a fence and just touching a wall.

FIGURE P7.40 A ladder negotiating a corner formed by two hallways.

�

�2

�1

For the case where 𝑤1 = 𝑤2 = 2 m, use a numerical method described in this chapter (including MATLAB’s built-in capabilities) to develop a plot of L versus a range of α’s from 45 to 135°. 7.41 Figure P7.41 shows a pinned-fixed beam subject to a uniform load. The equation for the resulting deflections is

y = – 𝑤 _____ 48EI (2x 4 – 3Lx3 + L3x)

cha97962_ch07_198-222.indd 220 08/11/16 3:25 pm

Develop a MATLAB script that uses fminbnd to (a) generate a labeled plot of deflection versus distance and (b) deter- mine the location and magnitude of the maximum deflec- tion. Employ an initial guesses of 0 and L and use optimset to display the iterations. Use the following parameter values in your computation (making sure that you use consistent units): L = 400 cm, E = 52,000 kN/cm2, I = 32,000 cm4, and 𝑤 = 4 kN/cm. 7.42 For a jet in steady, level flight, thrust balances drag and lift balances weight (Fig. P7.42). Under these condi- tions, the optimal cruise speed occurs when the ratio of drag force to velocity is minimized. The drag, CD, can be com- puted as

CD = CD0 + C L

2 _____

π · AR

where CD0 = drag coefficient at zero lift, CL = the lift coef- ficient, and AR = the aspect ratio. For steady level flight, the lift coefficient can be computed as

CL = 2W _____

ρυ2 A

where W = the jet’s weight (N), ρ = air density (kg/m3), υ = velocity (m/s), and A = wing planform area (m2). The drag force can then be computed as

FD = W CD ___ CL

FIGURE P7.41

FIGURE P7.44

Use these formulas to determine the optimal steady cruise velocity for a 670 kN jet flying at 10 km above sea level. Employ the following parameters in your computation: A = 150 m2, AR = 6.5, CD0 = 0.018, and ρ = 0.413 kg/m3. 7.43 Develop a MATLAB script to generate a plot of the optimal velocity of the jet from Prob. 7.42 versus elevation above sea level. Employ a mass of 68,300 kg for the jet. Note that the gravitational acceleration at 45° latitude can be computed as a function of elevation with

g(h) = 9.8066 ( re _____ re + h ) 2 where g(h) = gravitational acceleration (m∕s2) at elevation h (m) above sea level, and re = Earth’s mean radius (= 6.371 × 106 m). In addition, air density as a function of elevation can be calculated with

ρ(h) = −9.57926 × 10−14 h3 + 4.71260 × 10−9 h2 − 1.18951 × 10−4 h + 1.22534

Employ the other parameters from Prob. 7.42 and design the plot for elevations ranging from h = 0 to 12 km above sea level. 7.44 As depicted in Fig. P7.44, a mobile fire hose projects a stream of water onto the roof of a building. At what angle, θ, and how far from the building, x1, should the hose be placed in order to maximize the coverage of the roof; that is, to maximize: x2 – x1? Note that the water velocity leaving the nozzle has a constant value of 3 m/s regardless of the angle, and the other parameter values are h1 = 0.06 m, h2 = 0.2 m, and L = 0.12 m. [Hint: The coverage is maximized for the trajectory that just clears the top front corner. That is, we want to choose an x1 and θ that just clear the top corner while maximizing x2 – x1.]

FIGURE P7.42 The four major forces on a jet in steady, level flight.

Lift

Thrust

Weight (gravity)

Drag

x1 x2

L θ

pROBLEmS 221

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222 OpTImIzATION

0 = −Ux ∂c ___ ∂x + E ( ∂2c ___ ∂x2 + ∂2c ___ ∂y2 ) − kc

where the x and y axes are defined to be parallel and per- pendicular to the shoreline, respectively (Fig. P7.45). The parameters and variables are: Ux = the water velocity along the shoreline (m/d), c = concentration, E = the turbulent diffusion coefficient, and k = the first-order decay rate. For the case where a constant loading, W, enters at (0, 0), the solution for the concentration at any coordinate is given by

c = 2 {c(x, y) + ∑ n=1 ∞

[ c(x, y + 2nY ) – c(x, y + 2nY ) ] } where

c(x, y) = W _____ πHE e

Ux x ____ 2E K0 ( √__________________ (x2 + y2) [ k __ E + ( Ux ___ 2E ) 2 ] ) where Y = the width, H = depth, and K0 = the modified Bes- sel function of the second kind. Develop a MATLAB script to generate a contour plot of concentration for a section of a lake with Y = 4.8 km and a length from X = –2.4 to 2.4 km using Δx = Δy = 0.32 km. Employ the following parameters for your calculation: W = 1.2 × 1012, H = 20, E = 5 × 106, Ux = 5 × 103, k = 1, and n = 3.

x−x

E UxY

Open boundary

Solid boundary

FIGURE P7.45 Plan view of a section of a lake with a point source of pollutant entering at the middle of the lower boundary.

7.45 Since many pollutants enter lakes (and other water- bodies for that matter) at their peripheries, an important water-quality problem involves modeling the distribution of contaminants in the vicinity of a waste discharge or a river. For a vertically well-mixed, constant-depth layer, the steady-state distribution of a pollutant reacting with first- order decay is represented by

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223

Part Three

Linear Systems

3.1 OVERVIEW

What Are Linear Algebraic Equations?

In Part Two, we determined the value x that satisfied a single equation, f (x) = 0. Now, we deal with the case of determining the values x1, x2, . . . , xn that simultaneously satisfy a set of equations:

f1(x1, x2, . . . , xn) = 0 f2(x1, x2, . . . , xn) = 0 . . . . . . fn(x1, x2, . . . , xn) = 0

Such systems are either linear or nonlinear. In Part Three, we deal with linear algebraic equations that are of the general form

a11 x1 + a12 x2 + · · · + a1n xn = b1 a21 x1 + a22 x2 + · · · + a2n xn = b2 . . (PT3.1) . . . . an1 x1 + an2 x2 + · · · + ann xn) = bn

where the a’s are constant coefficients, the b’s are constants, the x’s are unknowns, and n is the number of equations. All other algebraic equa- tions are nonlinear.

Linear Algebraic Equations in Engineering and Science

Many of the fundamental equations of engineering and science are based on conservation laws. Some familiar quantities that conform to such laws are mass, energy, and momentum. In mathematical terms, these principles lead to balance or continuity equations that relate system behavior as represented

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224 PART 3 LineAR SySTemS

by the levels or response of the quantity being modeled to the properties or characteristics of the system and the external stimuli or forcing functions acting on the system.

As an example, the principle of mass conservation can be used to formulate a model for a series of chemical reactors (Fig. PT3.1a). For this case, the quantity being modeled is the mass of the chemical in each reactor. The system properties are the reaction character- istics of the chemical and the reactors’ sizes and flow rates. The forcing functions are the feed rates of the chemical into the system.

When we studied roots of equations, you saw how single-component systems result in a single equation that can be solved using root-location techniques. Multicomponent systems result in a coupled set of mathematical equations that must be solved simultaneously. The equations are coupled because the individual parts of the system are influenced by other parts. For example, in Fig. PT3.1a, reactor 4 receives chemical inputs from reactors 2 and 3. Consequently, its response is dependent on the quantity of chemical in these other reactors.

When these dependencies are expressed mathematically, the resulting equations are often of the linear algebraic form of Eq. (PT3.1). The x’s are usually measures of the mag- nitudes of the responses of the individual components. Using Fig. PT3.1a as an example, x1 might quantify the amount of chemical mass in the first reactor, x2 might quantify the amount in the second, and so forth. The a’s typically represent the properties and character- istics that bear on the interactions between components. For instance, the a’s for Fig. PT3.1a might be reflective of the flow rates of mass between the reactors. Finally, the b’s usually represent the forcing functions acting on the system, such as the feed rate.

Multicomponent problems of these types arise from both lumped (macro-) or distrib- uted (micro-) variable mathematical models. Lumped variable problems involve coupled

Feed x1 ... ...xi−1 xi xi+1 xn

(a)

(b)

x2 x4

Feed

x1 x5

FIGURE PT3.1 Two types of systems that can be modeled using linear algebraic equations: (a) lumped variable system that involves coupled finite components and (b) distributed variable system that involves a continuum.

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3.2 PART ORGAniZATiOn 225

finite components. The three interconnected bungee jumpers described at the beginning of Chap. 8 are a lumped system. Other examples include trusses, reactors, and electric circuits.

Conversely, distributed variable problems attempt to describe the spatial detail on a continuous or semicontinuous basis. The distribution of chemicals along the length of an elongated, rectangular reactor (Fig. PT3.1b) is an example of a continuous variable model. Differential equations derived from conservation laws specify the distribution of the depen- dent variable for such systems. These differential equations can be solved numerically by converting them to an equivalent system of simultaneous algebraic equations.

The solution of such sets of equations represents a major application area for the meth- ods in the following chapters. These equations are coupled because the variables at one lo- cation are dependent on the variables in adjoining regions. For example, the concentration at the middle of the reactor in Fig. PT3.1b is a function of the concentration in adjoining regions. Similar examples could be developed for the spatial distribution of temperature, momentum, or electricity.

Aside from physical systems, simultaneous linear algebraic equations also arise in a variety of mathematical problem contexts. These result when mathematical functions are required to satisfy several conditions simultaneously. Each condition results in an equation that contains known coefficients and unknown variables. The techniques discussed in this part can be used to solve for the unknowns when the equations are linear and algebraic. Some widely used numerical techniques that employ simultaneous equations are regres- sion analysis and spline interpolation.

3.2 PART ORGANIZATION

Due to its importance in formulating and solving linear algebraic equations, Chap. 8 provides a brief overview of matrix algebra. Aside from covering the rudiments of matrix representation and manipulation, the chapter also describes how matrices are handled in MATLAB.

Chapter 9 is devoted to the most fundamental technique for solving linear algebraic systems: Gauss elimination. Before launching into a detailed discussion of this technique, a preliminary section deals with simple methods for solving small systems. These approaches are presented to provide you with visual insight and because one of the methods—the elim- ination of unknowns—represents the basis for Gauss elimination.

After this preliminary material, “naive” Gauss elimination is discussed. We start with this “stripped-down” version because it allows the fundamental technique to be elabo- rated on without complicating details. Then, in subsequent sections, we discuss potential problems of the naive approach and present a number of modifications to minimize and circumvent these problems. The focus of this discussion will be the process of switching rows, or partial pivoting. The chapter ends with a brief description of efficient methods for solving tridiagonal matrices.

Chapter 10 illustrates how Gauss elimination can be formulated as an LU factoriza- tion. Such solution techniques are valuable for cases where many right-hand-side vectors need to be evaluated. The chapter ends with a brief outline of how MATLAB solves linear systems.

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226 PART 3 LineAR SySTemS

Chapter 11 starts with a description of how LU factorization can be employed to ef- ficiently calculate the matrix inverse, which has tremendous utility in analyzing stimulus- response relationships of physical systems. The remainder of the chapter is devoted to the important concept of matrix condition. The condition number is introduced as a measure of the roundoff errors that can result when solving ill-conditioned matrices.

Chapter 12 deals with iterative solution techniques, which are similar in spirit to the approximate methods for roots of equations discussed in Chap. 6. That is, they involve guess- ing a solution and then iterating to obtain a refined estimate. The emphasis is on the Gauss- Seidel method, although a description is provided of an alternative approach, the Jacobi method. The chapter ends with a brief description of how nonlinear simultaneous equations can be solved.

Finally, Chap. 13 is devoted to eigenvalue problems. These have general mathematical relevance as well as many applications in engineering and science. We describe two simple methods as well as MATLAB’s capabilities for determining eigenvalues and eigenvectors. In terms of applications, we focus on their use to study the vibrations and oscillations of mechanical systems and structures.

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227

CHAPTER OBJECTIVES

The primary objective of this chapter is to acquaint you with linear algebraic equations and their relationship to matrices and matrix algebra. Specific objectives and topics covered are

• Understanding matrix notation. • Being able to identify the following types of matrices: identity, diagonal,

symmetric, triangular, and tridiagonal. • Knowing how to perform matrix multiplication and being able to assess when it is

feasible. • Knowing how to represent a system of linear algebraic equations in matrix

form. • Knowing how to solve linear algebraic equations with left division and matrix

inversion in MATLAB.

8 Linear Algebraic equations and matrices

YOU’VE GOT A PROBLEM

Suppose that three jumpers are connected by bungee cords. Figure 8.1a shows them being held in place vertically so that each cord is fully extended but unstretched. We can define three distances, x1, x2, and x3, as measured downward from each of their unstretched positions. After they are released, gravity takes hold and the jumpers will eventually come to the equilibrium positions shown in Fig. 8.1b. Suppose that you are asked to compute the displacement of each of the jumpers. If we assume that each cord behaves as a linear spring and follows Hooke’s law, free-body diagrams can be developed for each jumper as depicted in Fig. 8.2.

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228 LineAR ALGebRAic equATiOnS And mATRiceS

Using Newton’s second law, force balances can be written for each jumper:

m1 d2x1 ____ dt2

= m1 g + k2(x2 − x1) − k1x1

m2 d2x2 ____ dt2

= m2 g + k3(x3 − x2) + k2(x1 − x2) (8.1)

m3 d2x3 ____ dt2

= m3 g + k3(x2 − x3)

where mi = the mass of jumper i (kg), t = time (s), kj = the spring constant for cord j (N/m), xi = the displacement of jumper i measured downward from the equilibrium position (m), and g = gravitational acceleration (9.81 m/s2). Because we are interested in the steady-state solution, the second derivatives can be set to zero. Collecting terms gives

(k1 + k2) x1 − k2x2 = m1g

−k2 x1 + (k2 + k3) x2 − k3 x3 = m2g (8.2)

−k3 x2 + k3 x3 = m3 g

Thus, the problem reduces to solving a system of three simultaneous equations for the three unknown displacements. Because we have used a linear law for the cords, these equations are linear algebraic equations. Chapters 8 through 12 will introduce you to how MATLAB is used to solve such systems of equations.

(a ) Unstretched (b ) Stretched

x1 = 0

x2 = 0

x3 = 0

FIGURE 8.1 Three individuals connected by bungee cords.

k3(x3 − x2)m2gk2(x2 − x1)m1g m3g

k3(x3 − x2)k2(x2 − x1)k1x1

FIGURE 8.2 Free-body diagrams.

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8.1 mATRiX ALGebRA OVeRVieW 229

8.1 MATRIX ALGEBRA OVERVIEW

Knowledge of matrices is essential for understanding the solution of linear algebraic equa- tions. The following sections outline how matrices provide a concise way to represent and manipulate linear algebraic equations.

8.1.1 Matrix Notation

A matrix consists of a rectangular array of elements represented by a single symbol. As depicted in Fig. 8.3, [A] is the shorthand notation for the matrix and ai j designates an indi- vidual element of the matrix.

A horizontal set of elements is called a row and a vertical set is called a column. The first subscript i always designates the number of the row in which the element lies. The sec- ond subscript j designates the column. For example, element a23 is in row 2 and column 3.

The matrix in Fig. 8.3 has m rows and n columns and is said to have a dimension of m by n (or m × n). It is referred to as an m by n matrix.

Matrices with row dimension m = 1, such as

[b] = [b1 b2 · · · bn]

are called row vectors. Note that for simplicity, the first subscript of each element is dropped. Also, it should be mentioned that there are times when it is desirable to employ a special shorthand notation to distinguish a row matrix from other types of matrices. One way to accomplish this is to employ special open-topped brackets, as in ⌊b⌋.1

Matrices with column dimension n = 1, such as

c1 c2 [c] = ⋮ (8.3) cm

1 In addition to special brackets, we will use case to distinguish between vectors (lowercase) and matrices (uppercase).

. . .

....

amnam3am2am1

a2na22a21

a11 a1na13a12a11

[A] =

a23

Column 3

Row 2

FIGURE 8.3 A matrix.

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230 LineAR ALGebRAic equATiOnS And mATRiceS

are referred to as column vectors. For simplicity, the second subscript is dropped. As with the row vector, there are occasions when it is desirable to employ a special shorthand notation to distinguish a column matrix from other types of matrices. One way to accom- plish this is to employ special brackets, as in {c}.

Matrices where m = n are called square matrices. For example, a 3 × 3 matrix is a11 a12 a13

[A] = a21 a22 a23 a31 a32 a33

The diagonal consisting of the elements a11, a22, and a33 is termed the principal or main diagonal of the matrix.

Square matrices are particularly important when solving sets of simultaneous linear equa- tions. For such systems, the number of equations (corresponding to rows) and the number of unknowns (corresponding to columns) must be equal for a unique solution to be possible. Con- sequently, square matrices of coefficients are encountered when dealing with such systems.

There are a number of special forms of square matrices that are important and should be noted:

A symmetric matrix is one where the rows equal the columns—that is, ai j = aj i for all i’s and j’s. For example,

5 1 2 [A] = 1 3 7

2 7 8 is a 3 × 3 symmetric matrix.

A diagonal matrix is a square matrix where all elements off the main diagonal are equal to zero, as in

a11 [A] = a22

a33 Note that where large blocks of elements are zero, they are left blank.

An identity matrix is a diagonal matrix where all elements on the main diagonal are equal to 1, as in

1 [I ] = 1

1 The identity matrix has properties similar to unity. That is,

[A][I ] = [I ][A] = [A] An upper triangular matrix is one where all the elements below the main diagonal are

zero, as in a11 a12 a13

[A] = a22 a23 a33

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8.1 mATRiX ALGebRA OVeRVieW 231

A lower triangular matrix is one where all elements above the main diagonal are zero, as in

a11 [A] = a21 a22

a31 a32 a33 A banded matrix has all elements equal to zero, with the exception of a band centered

on the main diagonal:

a11 a12 [A] = a21 a22 a23

a32 a33 a34 a43 a44

The preceding matrix has a bandwidth of 3 and is given a special name—the tridiagonal matrix.

8.1.2 Matrix Operating Rules

Now that we have specified what we mean by a matrix, we can define some operating rules that govern its use. Two m by n matrices are equal if, and only if, every element in the first is equal to every element in the second—that is, [A] = [B] if ai j = bi j for all i and j.

Addition of two matrices, say, [A] and [B], is accomplished by adding corresponding terms in each matrix. The elements of the resulting matrix [C ] are computed as

ci j = ai j + bi j for i = 1, 2, . . . , m and j = 1, 2, . . . , n. Similarly, the subtraction of two matrices, say, [E ] minus [F ], is obtained by subtracting corresponding terms, as in

di j = ei j − fi j for i = 1, 2, . . . , m and j = 1, 2, . . . , n. It follows directly from the preceding definitions that addition and subtraction can be performed only between matrices having the same dimensions.

Both addition and subtraction are commutative:

[A] + [B] = [B] + [A]

and associative:

([A] + [B ]) + [C ] = [A] + ([B ] + [C ])

The multiplication of a matrix [A] by a scalar g is obtained by multiplying every element of [A] by g. For example, for a 3 × 3 matrix:

ga11 ga12 ga13 [D] = g [A] = ga21 ga22 ga23

ga31 ga32 ga33

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232 LineAR ALGebRAic equATiOnS And mATRiceS

The product of two matrices is represented as [C ] = [A][B], where the elements of [C ] are defined as

ci j = ∑ k=1

n aik bk j (8.4)

where n = the column dimension of [A] and the row dimension of [B]. That is, the ci j ele- ment is obtained by adding the product of individual elements from the ith row of the first matrix, in this case [A], by the jth column of the second matrix [B]. Figure 8.4 depicts how the rows and columns line up in matrix multiplication.

According to this definition, matrix multiplication can be performed only if the first matrix has as many columns as the number of rows in the second matrix. Thus, if [A] is an m by n matrix, [B] could be an n by l matrix. For this case, the resulting [C ] matrix would have the dimension of m by l. However, if [B] were an m by l matrix, the multiplication could not be performed. Figure 8.5 provides an easy way to check whether two matrices can be multiplied.

If the dimensions of the matrices are suitable, matrix multiplication is associative:

([A][B]) [C] = [A]([B][C])

and distributive:

[A]([B] + [C ]) = [A][B] + [A][C ]

([A] + [B])[C ] = [A][C ] + [B][C ]

However, multiplication is not generally commutative:

[A][B] ≠ [B][A]

That is, the order of matrix multiplication is important.

0 4

3 × 5 + 1 × 7 = 22

FIGURE 8.4 Visual depiction of how the rows and columns line up in matrix multiplication.

[A]m×n [B]n×l [C]m×l=

Interior dimensions are equal,

multiplication is possible

Exterior dimensions define the dimensions of the result

FIGURE 8.5 Matrix multiplication can be performed only if the inner dimensions are equal.

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8.1 mATRiX ALGebRA OVeRVieW 233

Although multiplication is possible, matrix division is not a defined operation. How- ever, if a matrix [A] is square and nonsingular, there is another matrix [A]−1, called the inverse of [A], for which

[A][A]−1 = [A]−1[A] = [I ]

Thus, the multiplication of a matrix by the inverse is analogous to division, in the sense that a number divided by itself is equal to 1. That is, multiplication of a matrix by its inverse leads to the identity matrix.

The inverse of a 2 × 2 matrix can be represented simply by

[A]−1 = 1 ____________ a11a22 − a12a21 a22 −a12

−a21 a11

Similar formulas for higher-dimensional matrices are much more involved. Chapter 11 will deal with techniques for using numerical methods and the computer to calculate the inverse for such systems.

The transpose of a matrix involves transforming its rows into columns and its columns into rows. For example, for the 3 × 3 matrix:

a11 a12 a13 [A] = a21 a22 a23

a31 a32 a33 the transpose, designated [A]T, is defined as

a11 a21 a31 [A]T = a12 a22 a32

a13 a23 a33 In other words, the element ai j of the transpose is equal to the aj i element of the original matrix.

The transpose has a variety of functions in matrix algebra. One simple advantage is that it allows a column vector to be written as a row, and vice versa. For example, if

c1 {c} = {c1 } c1

then

{c}T = ⌊c1 c2 c3⌋

In addition, the transpose has numerous mathematical applications. A permutation matrix (also called a transposition matrix) is an identity matrix with

rows and columns interchanged. For example, here is a permutation matrix that is con- structed by switching the first and third rows and columns of a 3 × 3 identity matrix:

[P] = [

1 0 1 0

0 ]

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234 LineAR ALGebRAic equATiOnS And mATRiceS

Left multiplying a matrix [A] by this matrix, as in [P][A], will switch the corresponding rows of [A]. Right multiplying, as in [A][P], will switch the corresponding columns. Here is an example of left multiplication:

0 0 1 2 −7 4 5 1 9 [P] [A] = 0 1 0 8 3 −6 = 8 3 −6

1 0 0 5 1 9 2 −7 4

The final matrix manipulation that will have utility in our discussion is augmentation. A matrix is augmented by the addition of a column (or columns) to the original matrix. For example, suppose we have a 3 × 3 matrix of coefficients. We might wish to augment this matrix [A] with a 3 × 3 identity matrix to yield a 3 × 6 dimensional matrix:

a11 a11 a11 1 0 0 a21 a21 a21 0 1 0 a31 a31 a31 0 0 1

Such an expression has utility when we must perform a set of identical operations on the rows of two matrices. Thus, we can perform the operations on the single augmented matrix rather than on the two individual matrices.

eXAmPLe 8.1 mATLAb matrix manipulations

Problem Statement. The following example illustrates how a variety of matrix manipula- tions are implemented with MATLAB. It is best approached as a hands-on exercise on the computer.

Solution. Create a 3 × 3 matrix: >> A = [1 5 6;7 4 2;−3 6 7]

A = 1 5 6 7 4 2 −3 6 7

The transpose of [A] can be obtained using the ' operator: >> A'

ans = 1 7 −3 5 4 6 6 2 7

Next we will create another 3 × 3 matrix on a row basis. First create three row vectors:

>> x = [8 6 9]; >> y = [−5 8 1]; >> z = [4 8 2];

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8.1 mATRiX ALGebRA OVeRVieW 235

Then we can combine these to form the matrix:

>> B = [x; y; z]

B = 8 6 9 −5 8 1 4 8 2

We can add [A] and [B] together:

>> C = A + B

C = 9 11 15 2 12 3 1 14 9

Further, we can subtract [B] from [C ] to arrive back at [A]:

>> A = C − B

A = 1 5 6 7 4 2 −3 6 7

Because their inner dimensions are equal, [A] and [B] can be multiplied

>> A * B

ans = 7 94 26 44 90 71 −26 86 −7

Note that [A] and [B] can also be multiplied on an element-by-element basis by including a period with the multiplication operator as in

>> A.* B

ans = 8 30 54 −35 32 2 −12 48 14

A 2 × 3 matrix can be set up

>> D = [1 4 3;5 8 1];

If [A] is multiplied times [D], an error message will occur

>> A*D

??? Error using ==> mtimes Inner matrix dimensions must agree.

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236 LineAR ALGebRAic equATiOnS And mATRiceS

However, if we reverse the order of multiplication so that the inner dimensions match, matrix multiplication works

>> D * A

ans =

20 39 35 58 63 53

The matrix inverse can be computed with the inv function:

>> AI = inv(A)

AI =

0.2462 0.0154 −0.2154 −0.8462 0.3846 0.6154 0.8308 −0.3231 −0.4769

To test that this is the correct result, the inverse can be multiplied by the original matrix to give the identity matrix:

>> A * AI

ans =

1.0000 −0.0000 −0.0000 0.0000 1.0000 −0.0000 0.0000 −0.0000 1.0000

The eye function can be used to generate an identity matrix:

>> I = eye(3)

I =

1 0 0 0 1 0 0 0 1

We can set up a permutation matrix to switch the first and third rows and columns of a 3 × 3 matrix as

>> P = [0 0 1;0 1 0;1 0 0]

P =

0 0 1 0 1 0 1 0 0

We can then either switch the rows:

>> PA = P*A

PA =

−3 6 7 7 4 2 1 5 6

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8.1 mATRiX ALGebRA OVeRVieW 237

or the columns:

>> AP = A * P

AP =

6 5 1 2 4 7 7 6 −3

Finally, matrices can be augmented simply as in

>> Aug = [A I]

Aug =

1 5 6 1 0 0 7 4 2 0 1 0 −3 6 7 0 0 1

Note that the dimensions of a matrix can be determined by the size function:

>> [n,m] = size(Aug)

n = 3

m = 6

8.1.3 Representing Linear Algebraic Equations in Matrix Form

It should be clear that matrices provide a concise notation for representing simultaneous linear equations. For example, a 3 × 3 set of linear equations,

a11x1 + a12x2 + a13x3 = b1 a21x1 + a22x2 + a23x3 = b2 (8.5)

a31x1 + a32x2 + a33x3 = b3

can be expressed as

[A]{x} = {b} (8.6)

where [A] is the matrix of coefficients:

a11 a12 a13 [A] = a21 a22 a23

a31 a32 a33

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238 LineAR ALGebRAic equATiOnS And mATRiceS

{b} is the column vector of constants:

{b}T = ⌊b1 b2 b3⌋

and {x} is the column vector of unknowns:

{x}T = ⌊x1 x2 x3⌋

Recall the definition of matrix multiplication [Eq. (8.4)] to convince yourself that Eqs. (8.5) and (8.6) are equivalent. Also, realize that Eq. (8.6) is a valid matrix multiplica- tion because the number of columns n of the first matrix [A] is equal to the number of rows n of the second matrix {x}.

This part of the book is devoted to solving Eq. (8.6) for {x}. A formal way to obtain a solution using matrix algebra is to multiply each side of the equation by the inverse of [A] to yield

[A]−1[A]{x} = [A]−1{b}

Because [A]−1[A] equals the identity matrix, the equation becomes

{x} = [A]−1{b}

(8.7)

Therefore, the equation has been solved for {x}. This is another example of how the in- verse plays a role in matrix algebra that is similar to division. It should be noted that this is not a very efficient way to solve a system of equations. Thus, other approaches are em- ployed in numerical algorithms. However, as discussed in Sec. 11.1.2, the matrix inverse itself has great value in the engineering analyses of such systems.

It should be noted that systems with more equations (rows) than unknowns (columns), m > n, are said to be overdetermined. A typical example is least-squares regression where an equation with n coefficients is fit to m data points (x, y). Conversely, systems with less equations than unknowns, m < n, are said to be underdetermined. A typical example of underdetermined systems is numerical optimization.

8.2 SOLVING LINEAR ALGEBRAIC EQUATIONS WITH MATLAB

MATLAB provides two direct ways to solve systems of linear algebraic equations. The most efficient way is to employ the backslash, or “left-division,” operator as in

>> x = A\b

The second is to use matrix inversion:

>> x = inv(A)*b

As stated at the end of Sec. 8.1.3, the matrix inverse solution is less efficient than using the backslash. Both options are illustrated in the following example.

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8.2 SOLVinG LineAR ALGebRAic equATiOnS WiTH mATLAb 239

eXAmPLe 8.2 Solving the bungee Jumper Problem with mATLAb

Problem Statement. Use MATLAB to solve the bungee jumper problem described at the beginning of this chapter. The parameters for the problem are

Spring Constant Unstretched Cord Jumper Mass (kg) (N/m) Length (m)

Top (1) 60 50 20 Middle (2) 70 100 20 Bottom (3) 80 50 20

Solution. Substituting these parameter values into Eq. (8.2) gives

150 −100 0 x1 588.6 −100 150 −50 { x2 } = {686.7 } 0 −50 50 x3 784.8

Start up MATLAB and enter the coefficient matrix and the right-hand-side vector:

>> K = [150 −100 0;−100 150 −50;0 −50 50]

K = 150 −100 0 −100 150 −50 0 −50 50

>> mg = [588.6; 686.7; 784.8]

mg = 588.6000 686.7000 784.8000

Employing left division yields

>> x = K\mg

x = 41.2020 55.9170 71.6130

Alternatively, multiplying the inverse of the coefficient matrix by the right-hand-side vec- tor gives the same result: >> x = inv(K)*mg

x = 41.2020 55.9170 71.6130

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240 LineAR ALGebRAic equATiOnS And mATRiceS

Because the jumpers were connected by 20-m cords, their initial positions relative to the platform is

>> xi = [20;40;60];

Thus, their final positions can be calculated as

>> xf = x+xi

xf = 61.2020 95.9170 131.6130

The results, which are displayed in Fig. 8.6, make sense. The first cord is extended the longest because it has a lower spring constant and is subject to the most weight (all three jumpers). Notice that the second and third cords are extended about the same amount. Because it is subject to the weight of two jumpers, one might expect the second cord to be extended longer than the third. However, because it is stiffer (i.e., it has a higher spring constant), it stretches less than expected based on the weight it carries.

(a) (b)

120

FIGURE 8.6 Positions of three individuals connected by bungee cords. (a) Unstretched and (b) stretched.

8.3 CASE STUDY cuRRenTS And VOLTAGeS in ciRcuiTS

Background. Recall that in Chap. 1 (Table 1.1), we summarized some models and asso- ciated conservation laws that figure prominently in engineering. As in Fig. 8.7, each model represents a system of interacting elements. Consequently, steady-state balances derived from the conservation laws yield systems of simultaneous equations. In many cases, such systems are linear and hence can be expressed in matrix form. The present case study focuses on one such application: circuit analysis.

A common problem in electrical engineering involves determining the currents and voltages at various locations in resistor circuits. These problems are solved using Kirch- hoff’s current and voltage rules. The current (or point) rule states that the algebraic sum of all currents entering a node must be zero (Fig. 8.8a), or

Σ i = 0 (8.8) where all current entering the node is considered positive in sign. The current rule is an application of the principle of conservation of charge (recall Table 1.1).

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8.3 cASe STudy 241

8.3 CASE STUDY continued

The voltage (or loop) rule specifies that the algebraic sum of the potential differences (i.e., voltage changes) in any loop must equal zero. For a resistor circuit, this is expressed as

Σ ξ − Σ iR = 0 (8.9) where ξ is the emf (electromotive force) of the voltage sources, and R is the resistance of any resistors on the loop. Note that the second term derives from Ohm’s law (Fig. 8.8b), which states that the voltage drop across an ideal resistor is equal to the product of the current and the resistance. Kirchhoff’s voltage rule is an expression of the conservation of energy.

Solution. Application of these rules results in systems of simultaneous linear algebraic equations because the various loops within a circuit are interconnected. For example, con- sider the circuit shown in Fig. 8.9. The currents associated with this circuit are unknown both in magnitude and direction. This presents no great difficulty because one simply assumes a direction for each current. If the resultant solution from Kirchhoff’s laws is negative, then the assumed direction was incorrect. For example, Fig. 8.10 shows some assumed currents.

i1 i3

Vi Vj Rij

iij

(a)

(b)

FIGURE 8.8 Schematic representations of (a) Kirchhoff’s current rule and (b) Ohm’s law.

Structure

Machine

Reactors

Circuit

−

(a) Chemical engineering (b) Civil engineering

(d) Mechanical engineering(c) Electrical engineering

FIGURE 8.7 Engineering systems which, at steady state, can be modeled with linear algebraic equations.

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242 LineAR ALGebRAic equATiOnS And mATRiceS

8.3 CASE STUDY continued

Given these assumptions, Kirchhoff’s current rule is applied at each node to yield

i12 + i52 + i32 = 0 i65 − i52 − i54 = 0 i43 − i32 = 0 i54 − i43 = 0

Application of the voltage rule to each of the two loops gives

−i54R54 − i43R43 − i32R32 + i52R52 = 0 −i65R65 − i52R52 + i12R12 − 200 = 0

or, substituting the resistances from Fig. 8.9 and bringing constants to the right-hand side,

−15i54 − 5i43 − 10i32 + 10i52 = 0 −20i65 − 10i52 + 5i12 = 200

Therefore, the problem amounts to solving six equations with six unknown currents. These equations can be expressed in matrix form as

1 1 1 0 0 0 i12 0 0 −1 0 1 −1 0 i52 0 0 0 −1 0 0 1 { i32 } = { 0 }0 0 0 0 1 −1 i65 00 10 −10 0 −15 −5 i54 0 5 −10 0 −20 0 0 i43 200

R = 10 Ω V1 = 200 V

V6 = 0 V

3 2 1

4 5 6

R = 5 Ω

R = 5 Ω R = 10 Ω

R = 15 Ω R = 20 Ω

FIGURE 8.9 A resistor circuit to be solved using simultaneous linear algebraic equations.

3 2 1

4 5 6

i12

i65 i52

i32

i54 i43

FIGURE 8.10 Assumed current directions.

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8.3 cASe STudy 243

8.3 CASE STUDY continued

Although impractical to solve by hand, this system is easily handled by MATLAB. The solution is

>> A=[1 1 1 0 0 0 0 −1 0 1 −1 0 0 0 −1 0 0 1 0 0 0 0 1 −1 0 10 −10 0 −15 −5 5 −10 0 −20 0 0]; >> b = [0 0 0 0 0 200]'; >> current = A\b

current = 6.1538 −4.6154 −1.5385 −6.1538 −1.5385 −1.5385

Thus, with proper interpretation of the signs of the result, the circuit currents and volt- ages are as shown in Fig. 8.11. The advantages of using MATLAB for problems of this type should be evident.

V = 153.85 V = 169.23

i = 1.5385

V = 146.15 V = 123.08 V = 0

V = 200

i = 6.1538

FIGURE 8.11 The solution for currents and voltages obtained using MATLAB.

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244 LineAR ALGebRAic equATiOnS And mATRiceS

PROBLEMS

8.1 Given a square matrix [A], write a single line MATLAB command that will create a new matrix [Aug] that consists of the original matrix [A] augmented by an identity matrix [I ]. 8.2 A number of matrices are defined as

[A] = [

7 1 2

6 ]

[B] = [

7 1 2 7

4 ]

{C} = {

3 6

1 }

[D] = 9 4 3 −6 2 −1 7 5

[E ] = [

8 7 2 3

6 ]

[F ] = [ 3 0 1 1 7 3 ] ⌊G⌋ = ⌊7 6 4⌋

Answer the following questions regarding these matrices: (a) What are the dimensions of the matrices? (b) Identify the square, column, and row matrices. (c) What are the values of the elements: a12, b23, d32, e22,

f12, g12? (d) Perform the following operations:

(1) [E] + [B] (2) [A] + [F] (3) [B] − [E]

(4) 7 × [B] (5) {C}T (6) [E] × [B]

(7) [B] × [A] (8) [D]T (9) [A] × {C}

(10) [I ] × [B] (11) [E]T × [E] (12) {C}T × {C}

8.3 Write the following set of equations in matrix form:

50 = 5x3 − 7x2 4x2 + 7x3 + 30 = 0 x1 − 7x3 = 40 − 3x2 + 5x1

Use MATLAB to solve for the unknowns. In addition, use it to compute the transpose and the inverse of the coefficient matrix. 8.4 Three matrices are defined as

[A] = [

−1 12 8

−5

4 ]

[B ] = [ 4 0 0.5 2 ] [C ] = [

2 −2 3 1 ]

(a) Perform all possible multiplications that can be com- puted between pairs of these matrices.

(b) Justify why the remaining pairs cannot be multiplied.

8.5 Solve the following system with MATLAB:

[ 3 + 2i 4

−i

1 ] {

z1 z2 } = { 2 + i 3 }

8.6 Develop, debug, and test your own M-file to multiply two matrices—that is, [X ] = [Y ][Z ], where [Y ] is m by n and [Z ] is n by p. Employ for...end loops to implement the multiplication and include error traps to flag bad cases. Test the program using the matrices from Prob. 8.4. 8.7 Develop, debug, and test your own M-file to gener- ate the transpose of a matrix. Employ for...end loops to implement the transpose. Test it on the matrices from Prob. 8.4. 8.8 Develop, debug, and test your own M-file function to switch the rows of a matrix using a permutation matrix. The first lines of the function should be as follows:

function B = permut(A,r1,r2) % Permut: Switch rows of matrix A % with a permutation matrix % B = permut(A,r1,r2) % input: % A = original matrix % r1, r2 = rows to be switched % output: % B = matrix with rows switched

Include error traps for erroneous inputs (e.g., user specifies rows that exceed the dimensions of the original matrix). 8.9 Five reactors linked by pipes are shown in Fig. P8.9. The rate of mass flow through each pipe is computed as the product of flow (Q) and concentration (c). At steady state, the mass flow into and out of each reactor must be equal. For example, for the first reactor, a mass balance can be written as

Q01c01 + Q31c3 = Q15c1 + Q12c1

Write mass balances for the remaining reactors in Fig. P8.9 and express the equations in matrix form. Then use MATLAB to solve for the concentrations in each reactor. 8.10 An important problem in structural engineering is that of finding the forces in a statically determinate truss (Fig. P8.10). This type of structure can be described as a system of coupled linear algebraic equations derived from force balances. The sum of the forces in both horizontal and

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PRObLemS 245

vertical directions must be zero at each node, because the sys- tem is at rest. Therefore, for node 1:

ΣFH = 0 = −F1 cos 30° + F3 cos 60° + F1,h ΣFV = 0 = −F1 sin 30° − F3 sin 60° + F1,υ

for node 2:

ΣFH = 0 = F2 + F1 cos 30° + F2,h + H2 ΣFV = 0 = F1 sin 30° + F2, υ + V2

for node 3:

ΣFH = 0 = −F2 − F3 cos 60° + F3,h ΣFV = 0 = F3 sin 60° + F3, υ + V3

where Fi,h is the external horizontal force applied to node i (where a positive force is from left to right) and Fi,υ is the ex- ternal vertical force applied to node i (where a positive force is upward). Thus, in this problem, the 2000-N downward force on node 1 corresponds to Fi,υ = −2000. For this case, all other Fi,υ’s and Fi,h’s are zero. Express this set of linear algebraic equations in matrix form and then use MATLAB to solve for the unknowns. 8.11 Consider the three mass-four spring system in Fig. P8.11. Determining the equations of motion from ΣFx = max for each mass using its free-body diagram results in the

Q15 = 3

Q25 = 1

Q23 = 1

Q55 = 2

Q54 = 2

Q12 = 3

Q31 = 1

Q34 = 8

Q24 = 1 Q44 = 11Q01 = 5

c01 = 10

Q03 = 8

c03 = 20

FIGURE P8.9

60°

90°

30°

1000 N

F1 F3

FIGURE P8.10

k2 k3 k4k1

x1 x2 x3

m1 m2 m3

FIGURE P8.11

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246 LineAR ALGebRAic equATiOnS And mATRiceS

following differential equations:

x ¨ 1 + ( k1 + k2 ______ m1 ) x1 − ( k2 ___ m1 ) x2 = 0 x ¨ 2 − ( k2 ___ m2 ) x1 + ( k2 + k3 ______ m2 ) x2 − ( k3 ___ m2 ) x3 = 0 x ¨ 3 − ( k3 ___ m3 ) x2 + ( k3 + k4 ______ m3 ) x3 = 0

where k1 = k4 = 10 N/m, k2 = k3 = 30 N/m, and m1 = m2 = m3 = 1 kg. The three equations can be written in matrix form:

0 = {Acceleration vector} + [k∕m matrix]{displacement vector x}

At a specific time where x1 = 0.05 m, x2 = 0.04 m, and x3 = 0.03 m, this forms a tridiagonal matrix. Use MATLAB to solve for the acceleration of each mass. 8.12 Perform the same computation as in Example 8.2, but use five jumpers with the following characteristics:

Spring Unstretched Mass Constant Cord Jumper (kg) (N/m) Length (m)

1 55 80 10 2 75 50 10 3 60 70 10 4 75 100 10 5 90 20 10

8.13 Three masses are suspended vertically by a series of identical springs where mass 1 is at the top and mass 3 is at the bottom. If g = 9.81 m/s2, m1 = 2 kg, m2 = 3 kg, m3 = 2.5 kg, and the k’s = 10 kg/s2, use MATLAB to solve for the displacements x. 8.14 Perform the same computation as in Sec. 8.3, but for the circuit in Fig. P8.14. 8.15 Perform the same computation as in Sec. 8.3, but for the circuit in Fig. P8.15. 8.16 Besides solving simultaneous equations, linear algebra has lots of other applications in engineering and science. An example from computer graphics involves rotating an ob- ject in Euclidean space. The following rotation matrix can be employed to rotate a group of points counter-clockwise through an angle θ about the origin of a Cartesian coordinate system,

R = [ cos θ − sin θ sin θ cos θ ]

R = 7 ΩR = 8 Ω R = 10 Ω

R = 30 Ω3 2 1

4 5 6

R = 15 Ω

R = 35 Ω V1 = 20 volts

V6 = 140 volts R = 5 Ω

FIGURE P8.15

R = 20 Ω V1 = 150 volts

V6 = 0 volts

3 2 1

4 5 6

R = 10 Ω

R = 2 Ω R = 5 Ω

R = 5 Ω R = 25 Ω

FIGURE P8.14

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PRObLemS 247

8.3 CASE STUDYTo do this, each point’s position must be represented by a column vector υ, containing the coordinates of the point. For example, here are vectors for the x and y coordinates of the rectangle in Fig. P8.16

x = [1 4 4 1]; y = [1 1 4 4];

The rotated vector is then generated with matrix multipli- cation: [R]{υ}. Develop a MATLAB function to perform this operation and display the initial and the rotated points as filled shapes on the same graph. Here is a script to test your function:

clc;clf;format compact x = [1 4 4 1]; y = [1 1 4 4]; [xt, yt] = Rotate2D(45, x, y);

and here is a skeleton of the function

function [xr, yr] = Rotate2D(thetad, x, y) % two dimensional rotation 2D rotate Cartesian % [xr, yr] = rot2d(thetad, x, y) % Rotation of a two-dimensional object the Cartesian coordinates % of which are contained in the vectors x and y. % input: % thetad = angle of rotation (degrees) % x = vector containing objects x coordinates % y = vector containing objects y coordinates % output: % xr = vector containing objects rotated x coordinates % yr = vector containing objects rotated y coordinates % convert angle to radians and set up rotation matrix

∙ ∙ · ∙

% close shape ∙ ∙ · ∙

% plot original object hold on, grid on

∙ ∙ · ∙

% rotate shape ∙ ∙ · ∙

% plot rotated object ∙ ∙ · ∙

hold off

(1, 4)

(1, 1)

(4, 4)

(4,1)

FIGURE P8.16

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248

9 Gauss Elimination

At the end of Chap. 8, we stated that MATLAB provides two simple and direct methods for solving systems of linear algebraic equations: left division, >> x = A\b

and matrix inversion,

>> x = inv(A)*b

Chapters 9 and 10 provide background on how such solutions are obtained. This mate- rial is included to provide insight into how MATLAB operates. In addition, it is intended to show how you can build your own solution algorithms in computational environments that do not have MATLAB’s built-in capabilities.

CHAPTER OBJECTIVES

The primary objective of this chapter is to describe the Gauss elimination algorithm for solving linear algebraic equations. Specific objectives and topics covered are

• Knowing how to solve small sets of linear equations with the graphical method and Cramer’s rule.

• Understanding how to implement forward elimination and back substitution as in Gauss elimination.

• Understanding how to count flops to evaluate the efficiency of an algorithm. • Understanding the concepts of singularity and ill-condition. • Understanding how partial pivoting is implemented and how it differs from

complete pivoting. • Knowing how to compute the determinant as part of the Gauss elimination

algorithm with partial pivoting. • Recognizing how the banded structure of a tridiagonal system can be exploited

to obtain extremely efficient solutions.

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9.1 SOLVING SMALL NUMBERS OF EQUATIONS 249

The technique described in this chapter is called Gauss elimination because it involves combining equations to eliminate unknowns. Although it is one of the earliest methods for solving simultaneous equations, it remains among the most important algorithms in use today and is the basis for linear equation solving on many popular software packages including MATLAB.

9.1 SOLVING SMALL NUMBERS OF EQUATIONS

Before proceeding to Gauss elimination, we will describe several methods that are ap- propriate for solving small (n ≤ 3) sets of simultaneous equations and that do not require a computer. These are the graphical method, Cramer’s rule, and the elimination of unknowns.

9.1.1 The Graphical Method

A graphical solution is obtainable for two linear equations by plotting them on Cartesian coordinates with one axis corresponding to x1 and the other to x2. Because the equations are linear, each equation will plot as a straight line. For example, suppose that we have the following equations:

3x1 + 2x2 = 18 −x1 + 2x2 = 2

If we assume that x1 is the abscissa, we can solve each of these equations for x2:

x2 = − 3 __ 2 x1 + 9

x2 = 1 __ 2 x1 + 1

The equations are now in the form of straight lines—that is, x2 = (slope) x1 + intercept. When these equations are graphed, the values of x1 and x2 at the intersection of the lines represent the solution (Fig. 9.1). For this case, the solution is x1 = 4 and x2 = 3.

For three simultaneous equations, each equation would be represented by a plane in a three-dimensional coordinate system. The point where the three planes intersect would represent the solution. Beyond three equations, graphical methods break down and, conse- quently, have little practical value for solving simultaneous equations. However, they are useful in visualizing properties of the solutions.

For example, Fig. 9.2 depicts three cases that can pose problems when solving sets of linear equations. Fig. 9.2a shows the case where the two equations represent parallel lines. For such situations, there is no solution because the lines never cross. Figure 9.2b depicts the case where the two lines are coincident. For such situations there is an infinite number of solutions. Both types of systems are said to be singular.

In addition, systems that are very close to being singular (Fig. 9.2c) can also cause problems. These systems are said to be ill-conditioned. Graphically, this corresponds to the fact that it is difficult to identify the exact point at which the lines intersect. Ill-conditioned systems will also pose problems when they are encountered during the numerical solution of linear equations. This is because they will be extremely sensitive to roundoff error.

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250 GAUSS ELIMINATION

9.1.2 Determinants and Cramer’s Rule

Cramer’s rule is another solution technique that is best suited to small numbers of equa- tions. Before describing this method, we will briefly review the concept of the determinant, which is used to implement Cramer’s rule. In addition, the determinant has relevance to the evaluation of the ill-conditioning of a matrix.

0 62 4 0

Solution: x1 = 4; x2 = 3

−x 1 + 2

x 2 = 2

3x 1 + 2x

2 = 18

FIGURE 9.1 Graphical solution of a set of two simultaneous linear algebraic equations. The intersection of the lines represents the solution.

x 1 + x 2 =

x2 x2 x2

x1 x1 x1

x 1 + x 2 =

−x 1 + 2

x 2 = 2

(a) (b) (c)

1 2− x 1

+ x 2 = 1.

1 x 1 +

x 2 = 1

1 2−

x 1 + x 2 =

1 2−

2.3 5−

1 2−

1 2

FIGURE 9.2 Graphical depiction of singular and ill-conditioned systems: (a) no solution, (b) infinite solutions, and (c) ill-conditioned system where the slopes are so close that the point of intersection is difficult to detect visually.

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9.1 SOLVING SMALL NUMBERS OF EQUATIONS 251

Determinants. The determinant can be illustrated for a set of three equations:

[A]{x} = {b}

where [A] is the coefficient matrix

[A] = a11 a12 a13 a21 a22 a23 a31 a32 a33

The determinant of this system is formed from the coefficients of [A] and is represented as

D = a11 a12 a13 a21 a22 a23 a31 a32 a33

Although the determinant D and the coefficient matrix [A] are composed of the same elements, they are completely different mathematical concepts. That is why they are dis- tinguished visually by using brackets to enclose the matrix and straight lines to enclose the determinant. In contrast to a matrix, the determinant is a single number. For example, the value of the determinant for two simultaneous equations

D = a11 a12 a21 a22

is calculated by

D = a11a22 − a12a21 For the third-order case, the determinant can be computed as

D = a11 a22 a23 a32 a33 − + (9.1)

where the 2 × 2 determinants are called minors.

EXAMPLE 9.1 Determinants

Problem Statement. Compute values for the determinants of the systems represented in Figs. 9.1 and 9.2.

Solution. For Fig. 9.1:

D = 3 2 −1 2 = 3(2) − 2(−1) = 8

For Fig. 9.2a:

D = − 1 _ 2 1 − 1 _ 2

1 = −

__ 2 (1) − 1 ( −1 ___ 2 ) = 0

For Fig. 9.2b:

D = − 1 _ 2 1 −1 2

= − 1

__ 2 (2) − 1(−1) = 0

a12 a21 a23 a31 a33

a13 a21 a22 a31 a32

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252 GAUSS ELIMINATION

For Fig. 9.2c:

D = − 1 _ 2 1 − 2.3 _ 5

1 = −

__ 2 (1) − 1

( −2.3 ___ 5 )

= −0.04

In the foregoing example, the singular systems had zero determinants. Additionally, the results suggest that the system that is almost singular (Fig. 9.2c) has a determinant that is close to zero. These ideas will be pursued further in our subsequent discussion of ill- conditioning in Chap. 11.

Cramer’s Rule. This rule states that each unknown in a system of linear algebraic equations may be expressed as a fraction of two determinants with denominator D and with the nu- merator obtained from D by replacing the column of coefficients of the unknown in question by the constants b1, b2, . . . , bn. For example, for three equations, x1 would be computed as

x1 =

b1 a12 a13 b2 a22 a23 b3 a32 a33

EXAMPLE 9.2 Cramer’s Rule

Problem Statement. Use Cramer’s rule to solve

0.3x1 + 0.52x2 + x3 = −0.01 0.5x1 + x2 + 1.9x3 = 0.67 0.1x1 + 0.3 x2 + 0.5x3 = −0.44

Solution. The determinant D can be evaluated as [Eq. (9.1)]:

D = 0.3 1 1.9

0.3 0.5 = −0.0022

The solution can be calculated as

x1 =

−0.01 0.52 1 0.67 1 1.9 −0.44 0.3 0.5

−0.0022

x2 =

0.3 −0.01 1 0.5 0.67 1.9 0.1 −0.44 0.5

−0.0022

x3 =

0.3 0.52 −0.01 0.5 1 0.67 0.1 0.3 −0.44

−0.0022

= 0.03278 ________ −0.0022 = −14.9

= 0.0649 ________ −0.0022 = −29.5

= −0.04356 ________ −0.0022 = 19.8

− 0.52 0.5 1.9

0.1 0.5 + 1

0.5 1 0.1 0.3

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9.1 SOLVING SMALL NUMBERS OF EQUATIONS 253

The det Function. The determinant can be computed directly in MATLAB with the det function. For example, using the system from the previous example:

>> A = [0.3 0.52 1;0.5 1 1.9;0.1 0.3 0.5]; >> D = det(A)

D = −0.0022

Cramer’s rule can be applied to compute x1 as in

>> A(:,1) = [−0.01;0.67;−0.44]

A = −0.0100 0.5200 1.0000 0.6700 1.0000 1.9000 −0.4400 0.3000 0.5000

>> x1 = det(A)/D

x1 = −14.9000

For more than three equations, Cramer’s rule becomes impractical because, as the number of equations increases, the determinants are time consuming to evaluate by hand (or by computer). Consequently, more efficient alternatives are used. Some of these alter- natives are based on the last noncomputer solution technique covered in Sec. 9.1.3—the elimination of unknowns.

9.1.3 Elimination of Unknowns

The elimination of unknowns by combining equations is an algebraic approach that can be illustrated for a set of two equations:

a11x1 + a12x2 = b1 (9.2) a21x1 + a22x2 = b2 (9.3)

The basic strategy is to multiply the equations by constants so that one of the unknowns will be eliminated when the two equations are combined. The result is a single equation that can be solved for the remaining unknown. This value can then be substituted into either of the original equations to compute the other variable.

For example, Eq. (9.2) might be multiplied by a21 and Eq. (9.3) by a11 to give a21a11x1 + a21a12x2 = a21b1 (9.4) a11a21x1 + a11a22x2 = a11b2 (9.5)

Subtracting Eq. (9.4) from Eq. (9.5) will, therefore, eliminate the x1 term from the equations to yield

a11a22x2 − a21a12x2 = a11b2 − a21b1

which can be solved for

x2 = a11b2 − a21b1 ____________ a11a22 − a21a12 (9.6)

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254 GAUSS ELIMINATION

Equation (9.6) can then be substituted into Eq. (9.2), which can be solved for

x1 = a22b1 − a12b2 ____________ a11a22 − a21a12 (9.7)

Notice that Eqs. (9.6) and (9.7) follow directly from Cramer’s rule:

x1 = b1 a12 b2 a22

x2 = a11 b1 a21 b2

The elimination of unknowns can be extended to systems with more than two or three equations. However, the numerous calculations that are required for larger systems make the method extremely tedious to implement by hand. However, as described in Sec. 9.2, the technique can be formalized and readily programmed for the computer.

9.2 NAIVE GAUSS ELIMINATION

In Sec. 9.1.3, the elimination of unknowns was used to solve a pair of simultaneous equa- tions. The procedure consisted of two steps (Fig. 9.3):

1. The equations were manipulated to eliminate one of the unknowns from the equations. The result of this elimination step was that we had one equation with one unknown.

2. Consequently, this equation could be solved directly and the result back-substituted into one of the original equations to solve for the remaining unknown.

This basic approach can be extended to large sets of equations by developing a system- atic scheme or algorithm to eliminate unknowns and to back-substitute. Gauss elimination is the most basic of these schemes.

This section includes the systematic techniques for forward elimination and back substitution that comprise Gauss elimination. Although these techniques are ideally suited for implementation on computers, some modifications will be required to obtain a reliable algorithm. In particular, the computer program must avoid division by zero. The following method is called “naive” Gauss elimination because it does not avoid this problem. Section 9.3 will deal with the additional features required for an effective com- puter program.

The approach is designed to solve a general set of n equations:

a11x1 + a12x2 + a13x3 + · · · + a1nxn = b1 (9.8a)

a21x1 + a22x2 + a23x3 + · · · + a2nxn = b2 (9.8b) ···

··· an1x1 + an2x2 + an3x3 + · · · + annxn = bn (9.8c)

a11 a12 b21 a22

= a22b1 − a12b2 ____________ a11a22 − a21a12

a11 a12 a21 a22

= a11b2 − a21b1 ____________ a11a22 − a21a12

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9.2 NAIVE GAUSS ELIMINATION 255

As was the case with the solution of two equations, the technique for n equations consists of two phases: elimination of unknowns and solution through back substitution.

Forward Elimination of Unknowns. The first phase is designed to reduce the set of equa- tions to an upper triangular system (Fig. 9.3a). The initial step will be to eliminate the first unknown x1 from the second through the nth equations. To do this, multiply Eq. (9.8a) by a21∕a11 to give

a21x1 + a21 ___ a11 a12 x2 +

a21 ___ a11 a13 x3 + · · · + a21 ___ a11 a1n xn =

a21 ___ a11 b1 (9.9)

This equation can be subtracted from Eq. (9.8b) to give

( a22 − a21 ___ a11 a12 ) x2 + · · · + ( a2n − a21 ___ a11 a1n ) xn = b2 − a21 ___ a11 b1 or

a 22 ′ x2 + · · · + a 2n ′ xn = b 2 ′

where the prime indicates that the elements have been changed from their original values. The procedure is then repeated for the remaining equations. For instance, Eq. (9.8a)

can be multiplied by a31∕a11 and the result subtracted from the third equation. Repeating the procedure for the remaining equations results in the following modified system:

a11 x1 + a12 x2 + a13 x3 + · · · + a1n xn = b1 (9.10a) a 22 ′ x2 + a 23 ′ x3 + · · · + a 2n ′ xn = b 2 ′ (9.10b)

(a) Forward elimination

(b) Back substitution

a11

a21

a31

a13a12

a23a22

a33a32

a11 a13a12

aʹ23aʹ22

aʺ33

x3 = bʺ3⧸aʺ33

x2 = (bʹ2 − aʹ23x3)⧸aʹ22

x1 = (b1 − a13x3 − a12x2)⧸a11

bʹ2

bʺ3

FIGURE 9.3 The two phases of Gauss elimination: (a) forward elimination and (b) back substitution.

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256 GAUSS ELIMINATION

a 32 ′ x2 + a 33 ′ x3 + · · · + a 3n ′ xn = b 3 ′ (9.10c) ···

··· a n2 ′ x2 + a n3 ′ x3 + · · · + a nn ′ xn = b n ′ (9.10d)

For the foregoing steps, Eq. (9.8a) is called the pivot equation and a11 is called the pivot element. Note that the process of multiplying the first row by a21∕a11 is equivalent to dividing it by a11 and multiplying it by a21. Sometimes the division operation is referred to as normalization. We make this distinction because a zero pivot element can interfere with normalization by causing a division by zero. We will return to this important issue after we complete our description of naive Gauss elimination.

The next step is to eliminate x2 from Eq. (9.10c) through (9.10d). To do this, mul- tiply Eq. (9.10b) by a 32 ′ ∕ a 22 ′ and subtract the result from Eq. (9.10c). Perform a similar elimination for the remaining equations to yield

a11x1 + a12x2 + a13x3 + · · · + a1n xn = b1 a 22 ′ x2 + a 23 ′ x3 + · · · + a 2n ′ xn = b 2 ′

a 33 ″ x3 + · · · + a 3n ″ xn = b 3 ″ ···

··· a n3 ″ x3 + · · · + a nn ″ xn = b n ″

where the double prime indicates that the elements have been modified twice. The procedure can be continued using the remaining pivot equations. The final

manipulation in the sequence is to use the (n − 1)th equation to eliminate the xn−1 term from the nth equation. At this point, the system will have been transformed to an upper triangular system:

a11x1 + a12x2 + a13x3 + · · · + a1n xn = b1 (9.11a)

a 22 ′ x2 + a 23 ′ x3 + · · · + a 2n ′ xn = b 2 ′ (9.11b)

a 33 ″ x3 + · · · + a 3n ″ xn = b 3 ″ (9.11c) ··· ···

a nn (n−1) xn = b n (n−1) (9.11d)

Back Substitution. Equation (9.11d) can now be solved for xn:

xn = b n (n−1) _____ a nn (n−1)

(9.12)

This result can be back-substituted into the (n − 1)th equation to solve for xn−1. The pro cedure, which is repeated to evaluate the remaining x’s, can be represented by the following formula:

xi = b i

(i−1) − ∑ j=i+1

a ij (i−1) xj ______________

a ii (i−1)

for i = n − 1, n − 2, . . . , 1 (9.13)

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9.2 NAIVE GAUSS ELIMINATION 257

EXAMPLE 9.3 Naive Gauss Elimination

Problem Statement. Use Gauss elimination to solve

3x1 − 0.1x2 − 0.2x3 = 7.85 (E9.3.1) 0.1x1 + 7x2 − 0.3x3 = −19.3 (E9.3.2) 0.3x1 − 0.2x2 + 10x3 = 71.4 (E9.3.3)

Solution. The first part of the procedure is forward elimination. Multiply Eq. (E9.3.1) by 0.1∕3 and subtract the result from Eq. (E9.3.2) to give

7.00333x2 − 0.293333x3 = −19.5617

Then multiply Eq. (E9.3.1) by 0.3∕3 and subtract it from Eq. (E9.3.3). After these opera- tions, the set of equations is

3x1 − 0.1x2 − 0.2x3 = 7.85 (E9.3.4) 7.00333x2 − 0.293333x3 = −19.5617 (E9.3.5)

− 0.190000x2 + 10.0200x3 = 70.6150 (E9.3.6) To complete the forward elimination, x2 must be removed from Eq. (E9.3.6). To ac- complish this, multiply Eq. (E9.3.5) by −0.190000∕7.00333 and subtract the result from Eq. (E9.3.6). This eliminates x2 from the third equation and reduces the system to an upper triangular form, as in

3x1 − 0.1x2 − 0.2x3 = 7.85 (E9.3.7) 7.00333x2 − 0.293333x3 = −19.5617 (E9.3.8)

10.0120x3 = 70.0843 (E9.3.9) We can now solve these equations by back substitution. First, Eq. (E9.3.9) can be solved for

x3 = 70.0843 _______ 10.0120 = 7.00003

This result can be back-substituted into Eq. (E9.3.8), which can then be solved for

x2 = −19.5617 + 0.293333(7.00003) __________________________ 7.00333 = −2.50000

Finally, x3 = 7.00003 and x2 = −2.50000 can be substituted back into Eq. (E9.3.7), which can be solved for

x1 = 7.85 + 0.1(−2.50000) + 0.2(7.00003) _______________________________ 3 = 3.00000

Although there is a slight round off error, the results are very close to the exact solution of x1 = 3, x2 = −2.5, and x3 = 7. This can be verified by substituting the results into the original equation set:

3(3) − 0.1(−2.5) − 0.2(7.00003) = 7.84999 ≅ 7.85 0.1(3) + 7(−2.5) − 0.3(7.00003) = −19.30000 = −19.3 0.3(3) − 0.2(−2.5) + 10(7.00003) = 71.4003 ≅ 71.4

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258 GAUSS ELIMINATION

9.2.1 MATLAB M-file: GaussNaive

An M-file that implements naive Gauss elimination is listed in Fig. 9.4. Notice that the coefficient matrix A and the right-hand-side vector b are combined in the augmented matrix Aug. Thus, the operations are performed on Aug rather than separately on A and b.

Two nested loops provide a concise representation of the forward elimination step. An outer loop moves down the matrix from one pivot row to the next. The inner loop moves below the pivot row to each of the subsequent rows where elimination is to take place. Finally, the actual elimination is represented by a single line that takes advantage of MATLAB’s ability to perform matrix operations.

The back-substitution step follows directly from Eqs. (9.12) and (9.13). Again, MATLAB’s ability to perform matrix operations allows Eq. (9.13) to be programmed as a single line.

9.2.2 Operation Counting

The execution time of Gauss elimination depends on the amount of floating-point operations (or flops) involved in the algorithm. On modern computers using math coprocessors, the time consumed to perform addition/subtraction and multiplication/division is about the same.

function x = GaussNaive(A,b) % GaussNaive: naive Gauss elimination % x = GaussNaive(A,b): Gauss elimination without pivoting. % input: % A = coefficient matrix % b = right hand side vector % output: % x = solution vector

[m,n] = size(A); if m~=n, error('Matrix A must be square'); end nb = n+1; Aug = [A b]; % forward elimination for k = 1:n−1 for i = k+1:n factor = Aug(i,k)/Aug(k,k); Aug(i,k:nb) = Aug(i,k:nb)−factor*Aug(k,k:nb); end end % back substitution x = zeros(n,1); x(n) = Aug(n,nb)/Aug(n,n); for i = n−1:−1:1 x(i) = (Aug(i,nb)−Aug(i,i+1:n)*x(i+1:n))/Aug(i,i); end

FIGURE 9.4 An M-file to implement naive Gauss elimination.

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9.2 NAIVE GAUSS ELIMINATION 259

Therefore, totaling up these operations provides insight into which parts of the algorithm are most time consuming and how computation time increases as the system gets larger.

Before analyzing naive Gauss elimination, we will first define some quantities that facilitate operation counting:

∑ i=1

c f (i ) = c ∑ i=1

f (i) ∑ i=1

f (i) + g(i ) = ∑ i=1

f (i) + ∑ i=1

g(i ) (9.14a,b)

∑ i=1

1 = 1 + 1 + 1 + · · · + 1 = m ∑ i=k

1 = m − k + 1 (9.14c,d)

∑ i=1

i = 1 + 2 + 3 + · · · + m = m(m + 1) ________ 2 = m2 ___ 2 + O(m) (9.14e)

∑ i=1

i2 = 12 + 22 + 32 + · · · + m2 = m(m + 1)(2m + 1) _______________ 6 = m3 ___ 3 + O(m

2) (9.14f )

where O(mn) means “terms of order m n and lower.” Now let us examine the naive Gauss elimination algorithm (Fig. 9.4) in detail. We will

first count the flops in the elimination stage. On the first pass through the outer loop, k = 1. Therefore, the limits on the inner loop are from i = 2 to n. According to Eq. (9.14d ), this means that the number of iterations of the inner loop will be

∑ i=2

1 = n − 2 + 1 = n − 1 (9.15)

For every one of these iterations, there is one division to calculate the factor. The next line then performs a multiplication and a subtraction for each column element from 2 to nb. Because nb = n + 1, going from 2 to nb results in n multiplications and n subtrac- tions. Together with the single division, this amounts to n + 1 multiplications/divisions and n addition/subtractions for every iteration of the inner loop. The total for the first pass through the outer loop is therefore (n − 1)(n + 1) multiplication/divisions and (n − 1)(n) addition/subtractions.

Similar reasoning can be used to estimate the flops for the subsequent iterations of the outer loop. These can be summarized as

Outer Loop Inner Loop Addition/Subtraction Multiplication/Division k i Flops Flops

1 2, n (n − 1)(n) (n − 1)(n + 1) 2 3, n (n − 2)(n − 1) (n − 2)(n) ···

··· k k + 1, n (n − k)(n + 1 − k) (n − k)(n + 2 − k) ···

··· n − 1 n, n (1)(2) (1)(3)

Therefore, the total addition/subtraction flops for elimination can be computed as

∑ k=1

n −1

(n − k)(n + 1 − k) = ∑ k=1

n −1

[n(n + 1) − k(2n + 1) + k2] (9.16)

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260 GAUSS ELIMINATION

n(n + 1) ∑ k=1

n −1

1 − (2n + 1) ∑ k=1

n −1

k + ∑ k=1

n −1

k2 (9.17)

Applying some of the relationships from Eq. (9.14) yields

[n3 + O(n)] − [n3 + O(n2)] + [ 1 __ 3 n3 + O(n2) ] = n 3 __ 3 + O(n) (9.18)

A similar analysis for the multiplication/division flops yields

[n3 + O(n2)] − [n3 + O(n)] + [ 1 __ 3 n3 + O(n2) ] = n 3 __ 3 + O(n

2) (9.19)

Summing these results gives

2n 3 ___ 3 + O(n

2) (9.20)

Thus, the total number of flops is equal to 2n3∕3 plus an additional component pro- portional to terms of order n2 and lower. The result is written in this way because as n gets large, the O(n2) and lower terms become negligible. We are therefore justified in conclud- ing that for large n, the effort involved in forward elimination converges on 2n3∕3.

Because only a single loop is used, back substitution is much simpler to evaluate. The number of addition/subtraction flops is equal to n(n − 1)/2. Because of the extra division prior to the loop, the number of multiplication/division flops is n(n + 1)/2. These can be added to arrive at a total of

n2 + O(n) (9.21)

Thus, the total effort in naive Gauss elimination can be represented as

2n 3 ___ 3 + O(n

2 ) + n2 + O(n) 2n 3 ___ 3 + O(n

2) (9.22)

Two useful general conclusions can be drawn from this analysis:

1. As the system gets larger, the computation time increases greatly. As in Table 9.1, the amount of flops increases nearly three orders of magnitude for every order of magni- tude increase in the number of equations.

Forward elimination ⏟ Back

substitution ⏟

as n increases

TABLE 9.1 Number of flops for naive Gauss elimination.

Back Total Percent Due n Elimination Substitution Flops 2n3∕3 to Elimination

10 705 100 805 667 87.58 100 671550 10000 681550 666667 98.53 1000 6.67 × 108 1 × 106 6.68 × 108 6.67 × 108 99.85

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9.3 PIVOTING 261

2. Most of the effort is incurred in the elimination step. Thus, efforts to make the method more efficient should probably focus on this step.

9.3 PIVOTING

The primary reason that the foregoing technique is called “naive” is that during both the elimination and the back-substitution phases, it is possible that a division by zero can occur. For example, if we use naive Gauss elimination to solve

2x2 + 3x3 = 8 4x1 + 6x2 + 7x3 = −3 2x1 − 3x2 + 6x3 = 5

the normalization of the first row would involve division by a11 = 0. Problems may also arise when the pivot element is close, rather than exactly equal, to zero because if the mag- nitude of the pivot element is small compared to the other elements, then round off errors can be introduced.

Therefore, before each row is normalized, it is advantageous to determine the coefficient with the largest absolute value in the column below the pivot element. The rows can then be switched so that the largest element is the pivot element. This is called partial pivoting.

If columns as well as rows are searched for the largest element and then switched, the procedure is called complete pivoting. Complete pivoting is rarely used because most of the improvement comes from partial pivoting. In addition, switching columns changes the order of the x’s and, consequently, adds significant and usually unjustified complexity to the computer program.

The following example illustrates the advantages of partial pivoting. Aside from avoid- ing division by zero, pivoting also minimizes round off error. As such, it also serves as a partial remedy for ill-conditioning.

EXAMPLE 9.4 Partial Pivoting

Problem Statement. Use Gauss elimination to solve 0.0003x1 + 3.0000x2 = 2.0001 1.0000x1 + 1.0000x2 = 1.0000

Note that in this form the first pivot element, a11 = 0.0003, is very close to zero. Then repeat the computation, but partial pivot by reversing the order of the equations. The exact solution is x1 = 1∕3 and x2 = 2∕3.

Solution. Multiplying the first equation by 1∕(0.0003) yields x1 + 10,000x2 = 6667

which can be used to eliminate x1 from the second equation: −9999x2 = −6666

which can be solved for x2 = 2∕3. This result can be substituted back into the first equation to evaluate x1:

x1 = 2.0001 − 3(2∕3) _____________ 0.0003 (E9.4.1)

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262 GAUSS ELIMINATION

Due to subtractive cancellation, the result is very sensitive to the number of significant figures carried in the computation:

Absolute Value of Significant Percent Relative

Figures x2 x1 Error for x1

3 0.667 −3.33 1099 4 0.6667 0.0000 100 5 0.66667 0.30000 10 6 0.666667 0.330000 1 7 0.6666667 0.3330000 0.1

Note how the solution for x1 is highly dependent on the number of significant figures. This is because in Eq. (E9.4.1), we are subtracting two almost-equal numbers.

On the other hand, if the equations are solved in reverse order, the row with the larger pivot element is normalized. The equations are

1.0000x1 + 1.0000x2 = 1.0000 0.0003x1 + 3.0000x2 = 2.0001

Elimination and substitution again yields x2 = 2∕3. For different numbers of significant figures, x1 can be computed from the first equation, as in

x1 = 1 − (2∕3) _____________ 1

This case is much less sensitive to the number of significant figures in the computation:

Absolute Value of Significant Percent Relative

Figures x2 x1 Error for x1

3 0.667 0.333 0.1 4 0.6667 0.3333 0.01 5 0.66667 0.33333 0.001 6 0.666667 0.333333 0.0001 7 0.6666667 0.3333333 0.0000

Thus, a pivot strategy is much more satisfactory.

9.3.1 MATLAB M-file: GaussPivot

An M-file that implements Gauss elimination with partial pivoting is listed in Fig. 9.5. It is identical to the M-file for naive Gauss elimination presented previously in Sec. 9.2.1 with the exception of the bold portion that implements partial pivoting.

Notice how the built-in MATLAB function max is used to determine the largest avail- able coefficient in the column below the pivot element. The max function has the syntax

[y,i] = max (x)

where y is the largest element in the vector x, and i is the index corresponding to that element.

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9.3 PIVOTING 263

9.3.2 Determinant Evaluation with Gauss Elimination

At the end of Sec. 9.1.2, we suggested that determinant evaluation by expansion of minors was impractical for large sets of equations. However, because the determinant has value in assessing system condition, it would be useful to have a practical method for computing this quantity.

Fortunately, Gauss elimination provides a simple way to do this. The method is based on the fact that the determinant of a triangular matrix can be simply computed as the product of its diagonal elements:

D = a11a22a33 · · · ann

function x = GaussPivot(A,b) % GaussPivot: Gauss elimination pivoting % x = GaussPivot(A,b): Gauss elimination with pivoting. % input: % A = coefficient matrix % b = right hand side vector % output: % x = solution vector

[m,n] = size(A); if m ~ = n, error('Matrix A must be square'); end nb = n + 1; Aug = [A b]; % forward elimination for k = 1:n − 1 % partial pivoting [big,i] = max(abs(Aug(k:n,k))); ipr = i + k−1; if ipr~ = k Aug([k,ipr],:) = Aug([ipr,k],:); end for i = k+1:n factor = Aug(i,k)/Aug(k,k); Aug(i,k:nb) = Aug(i,k:nb)−factor*Aug(k,k:nb); end end % back substitution x = zeros(n,1); x(n) = Aug(n,nb)/Aug(n,n); for i = n − 1:−1:1 x(i) = (Aug(i,nb) − Aug(i,i+1:n)*x(i+1:n))/Aug(i,i); end

FIGURE 9.5 An M-file to implement Gauss elimination with partial pivoting.

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264 GAUSS ELIMINATION

The validity of this formulation can be illustrated for a 3 × 3 system:

D = a11 a12 a13 0 a22 a23 0 0 a33

where the determinant can be evaluated as [recall Eq. (9.1)]:

D = a11 a22 a23 0 a33 − +

or, by evaluating the minors:

D = a11a22a33 − a12(0) + a13(0) = a11a22a33 Recall that the forward-elimination step of Gauss elimination results in an upper

triangular system. Because the value of the determinant is not changed by the forward- elimination process, the determinant can be simply evaluated at the end of this step via

D = a11 a 22 ′ a 33 ″ · · · a nn (n−1)

where the superscripts signify the number of times that the elements have been modified by the elimination process. Thus, we can capitalize on the effort that has already been expended in reducing the system to triangular form and, in the bargain, come up with a simple estimate of the determinant.

There is a slight modification to the above approach when the program employs partial pivoting. For such cases, the determinant changes sign every time a row is switched. One way to represent this is by modifying the determinant calculation as in

D = a11 a 22 ′ a 33 ″ · · · a nn (n−1) (−1) p

where p represents the number of times that rows are pivoted. This modification can be incorporated simply into a program by merely keeping track of the number of pivots that take place during the course of the computation.

9.4 TRIDIAGONAL SYSTEMS

Certain matrices have a particular structure that can be exploited to develop efficient solu- tion schemes. For example, a banded matrix is a square matrix that has all elements equal to zero, with the exception of a band centered on the main diagonal.

A tridiagonal system has a bandwidth of 3 and can be expressed generally as

(9.23)

a12 0 a23

0 a33 a13

0 a22

0 0

f1 g1 e2 f2 g2 e3 f3 g3 · · · · · · · · · en−1 fn−1 gn−1 en fn

x1 x2 x3 . . . xn−1 xn

r1 r2 r3 . . . rn−1 rn

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9.4 TRIDIAGONAL SYSTEMS 265

Notice that we have changed our notation for the coefficients from a’s and b’s to e’s, f ’s, g’s, and r’s. This was done to avoid storing large numbers of useless zeros in the square matrix of a’s. This space-saving modification is advantageous because the resulting algo- rithm requires less computer memory.

An algorithm to solve such systems can be directly patterned after Gauss elimination— that is, using forward elimination and back substitution. However, because most of the matrix elements are already zero, much less effort is expended than for a full matrix. This efficiency is illustrated in the following example.

EXAMPLE 9.5 Solution of a Tridiagonal System

Problem Statement. Solve the following tridiagonal system:

2.04 −1 −1 2.04 −1 −1 2.04 −1 −1 2.04

x1 x2 x3 x4

40.8 0.8 0.8 200.8

Solution. As with Gauss elimination, the first step involves transforming the matrix to upper triangular form. This is done by multiplying the first equation by the factor e2∕f1 and subtracting the result from the second equation. This creates a zero in place of e2 and transforms the other coefficients to new values,

f2 = f2 − e2 __ f1

g1 = 2.04 − −1 ____ 2.04 (−1) = 1.550

r2 = r2 − e2 __ f1

r1 = 0.8 − −1 ____ 2.04 (40.8) = 20.8

Notice that g2 is unmodified because the element above it in the first row is zero. After performing a similar calculation for the third and fourth rows, the system is trans-

formed to the upper triangular form

2.04 −1 1.550 −1 1.395 −1 1.323

x1 x2 x3 x4

40.8 20.8 14.221 210.996

Now back substitution can be applied to generate the final solution:

x4 = r4 __ f4

= 210.996 _______ 1.323 = 159.480

x3 = r3 − g3 x4 ________ f3

= 14.221 − (−1)159.480 ___________________ 1.395 = 124.538

x2 = r2 − g2 x3 ________ f2

= 20.800 − (−1)124.538 ___________________ 1.550 = 93.778

x1 = r1 − g1 x2 ________ f1

= 40.800 − (−1)93.778 ___________________ 2.040 = 65.970

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266 GAUSS ELIMINATION

9.4.1 MATLAB M-file: Tridiag

An M-file that solves a tridiagonal system of equations is listed in Fig. 9.6. Note that the algorithm does not include partial pivoting. Although pivoting is sometimes required, most tridiagonal systems routinely solved in engineering and science do not require pivoting.

Recall that the computational effort for Gauss elimination was proportional to n3. Because of its sparseness, the effort involved in solving tridiagonal systems is proportional to n. Consequently, the algorithm in Fig. 9.6 executes much, much faster than Gauss elimi- nation, particularly for large systems.

function x = Tridiag (e,f,g,r) % Tridiag: Tridiagonal equation solver banded system % x = Tridiag (e,f,g,r): Tridiagonal system solver. % input: % e = subdiagonal vector % f = diagonal vector % g = superdiagonal vector % r = right hand side vector % output: % x = solution vector n = length(f); % forward elimination for k = 2:n factor = e (k)/f(k−1); f(k) = f(k) − factor*g(k − 1); r(k) = r(k) − factor*r(k − 1); end % back substitution x(n) = r(n)/f(n); for k = n − 1: − 1:1 x(k) = (r(k) − g(k)*x(k+1))/f(k); end

FIGURE 9.6 An M-file to solve a tridiagonal system.

9.5 CASE STUDY MODEL OF A HEATED ROD

Background. Linear algebraic equations can arise when modeling distributed systems. For example, Fig. 9.7 shows a long, thin rod positioned between two walls that are held at constant temperatures. Heat flows through the rod as well as between the rod and the sur- rounding air. For the steady-state case, a differential equation based on heat conservation can be written for such a system as

d 2T ____

dx 2 + h′(Ta − T ) = 0 (9.24)

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9.5 CASE STUDY continued

where T = temperature (°C), x = distance along the rod (m), h′ = a heat transfer coefficient between the rod and the surrounding air (m−2), and Ta = the air temperature (°C).

Given values for the parameters, forcing functions, and boundary conditions, calculus can be used to develop an analytical solution. For example, if h′ = 0.01, Ta = 20, T(0) = 40, and T(10) = 200, the solution is

T = 73.4523e0.1x − 53.4523e−0.1x + 20 (9.25)

Although it provided a solution here, calculus does not work for all such problems. In such instances, numerical methods provide a valuable alternative. In this case study, we will use finite differences to transform this differential equation into a tridiagonal system of linear algebraic equations which can be readily solved using the numerical methods described in this chapter.

Solution. Equation (9.24) can be transformed into a set of linear algebraic equations by conceptualizing the rod as consisting of a series of nodes. For example, the rod in Fig. 9.7 is divided into six equispaced nodes. Since the rod has a length of 10, the spacing between nodes is Δx = 2.

Calculus was necessary to solve Eq. (9.24) because it includes a second derivative. As we learned in Sec. 4.3.4, finite-difference approximations provide a means to transform derivatives into algebraic form. For example, the second derivative at each node can be approximated as

d 2T ____

dx 2 =

Ti+1 − 2Ti + Ti−1 ______________ Δx2

where Ti designates the temperature at node i. This approximation can be substituted into Eq. (9.24) to give

Ti+1 − 2Ti + Ti−1 ______________

Δx2 + h′(Ta − Ti) = 0

x = 10

T0 = 40

Ta = 20

T5 = 200

∆x

x = 0

54310 2

FIGURE 9.7 A noninsulated uniform rod positioned between two walls of constant but different temperature. The finite-difference representation employs four interior nodes.

9.5 CASE STUDY 267

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268 GAUSS ELIMINATION

Collecting terms and substituting the parameters gives

−Ti−1 + 2.04Ti − Ti+1 = 0.8 (9.26)

Thus, Eq. (9.24) has been transformed from a differential equation into an algebraic equa- tion. Equation (9.26) can now be applied to each of the interior nodes:

−T0 + 2.04T1 − T2 = 0.8 −T1 + 2.04T2 − T3 = 0.8 −T2 + 2.04T3 − T4 = 0.8 −T3 + 2.04T4 − T5 = 0.8

(9.27)

The values of the fixed end temperatures, T0 = 40 and T5 = 200, can be substituted and moved to the right-hand side. The results are four equations with four unknowns expressed in matrix form as

2.04 −1 0 0 −1 2.04 −1 0 0 −1 2.04 −1 0 0 −1 2.04

T1 T2 T3 T4

40.8 0.8 0.8 200.8

(9.28)

So our original differential equation has been converted into an equivalent system of linear algebraic equations. Consequently, we can use the techniques described in this chap- ter to solve for the temperatures. For example, using MATLAB

>> A = [2.04 −1 0 0 −1 2.04 −1 0 0 −1 2.04 −1 0 0 −1 2.04]; >> b = [40.8 0.8 0.8 200.8]'; >> T = (A\b)'

T = 65.9698 93.7785 124.5382 159.4795

A plot can also be developed comparing these results with the analytical solution obtained with Eq. (9.25),

>> T = [40 T 200]; >> x = [0:2:10]; >> xanal = [0:10]; >> TT = @(x) 73.4523*exp(0.1*x)−53.4523* ...

exp(−0.1*x)+20; >> Tanal = TT(xanal); >> plot(x,T,'o',xanal,Tanal)

As in Fig. 9.8, the numerical results are quite close to those obtained with calculus.

9.5 CASE STUDY continued

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9.5 CASE STUDY continued

In addition to being a linear system, notice that Eq. (9.28) is also tridiagonal. We can use an efficient solution scheme like the M-file in Fig. 9.6 to obtain the solution:

>> e = [0 −1 −1 −1]; >> f = [2.04 2.04 2.04 2.04]; >> g = [−1 −1 −1 0]; >> r = [40.8 0.8 0.8 200.8]; >> Tridiag(e,f,g,r)

ans = 65.9698 93.7785 124.5382 159.4795

The system is tridiagonal because each node depends only on its adjacent nodes. Because we numbered the nodes sequentially, the resulting equations are tridiagonal. Such cases often occur when solving differential equations based on conservation laws.

220 Analytical (line) and numerical (points) solutions

200

180

160

140

120

100

20 0 1 2 3 4 5 6 7 8 9 10

FIGURE 9.8 A plot of temperature versus distance along a heated rod. Both analytical (line) and numerical (points) solutions are displayed.

9.5 CASE STUDY 269

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270 GAUSS ELIMINATION

PROBLEMS

9.1 Determine the number of total flops as a function of the number of equations n for the tridiagonal algorithm (Fig. 9.6). 9.2 Use the graphical method to solve

4x1 − 8x2 = −24 x1 + 6x2 = 34

Check your results by substituting them back into the equations. 9.3 Given the system of equations

−1.1x1 + 10x2 = 120 −2x1 + 17.4x2 = 174

(a) Solve graphically and check your results by substituting them back into the equations.

(b) On the basis of the graphical solution, what do you expect regarding the condition of the system?

−3x2 + 7x3 = 4 x1 + 2x2 − x3 = 0 5x1 − 2x2 = 3

(a) Compute the determinant. (b) Use Cramer’s rule to solve for the x’s. (c) Use Gauss elimination with partial pivoting to solve for

the x’s. As part of the computation, calculate the deter- minant in order to verify the value computed in (a).

(d) Substitute your results back into the original equations to check your solution.

9.5 Given the equations 0.5x1 − x2 = − 9.5 1.02x1 − 2x2 = −18.8

(a) Solve graphically. (b) Compute the determinant. (c) On the basis of (a) and (b), what would you expect

regarding the system’s condition? (d) Solve by the elimination of unknowns. (e) Solve again, but with a11 modified slightly to 0.52.

Interpret your results. 9.6 Given the equations

10x1 + 2x2 − x3 = 27 −3x1 − 5x2 + 2x3 = −61.5 x1 + x2 + 6x3 = −21.5

(a) Solve by naive Gauss elimination. Show all steps of the computation.

(b) Substitute your results into the original equations to check your answers.

9.7 Given the equations 2x1 − 6x2 − x3 = −38 −3x1 − x2 + 7x3 = −34 −8x1 + x2 − 2x3 = −20

(a) Solve by Gauss elimination with partial pivoting. As part of the computation, use the diagonal elements to calcu- late the determinant. Show all steps of the computation.

(b) Substitute your results into the original equations to check your answers.

9.8 Perform the same calculations as in Example 9.5, but for the tridiagonal system:

0.8 −0.4 −0.4 0.8 −0.4 −0.4 0.8

x1 x2 x3

41 25 105

9.9 Figure P9.9 shows three reactors linked by pipes. As indicated, the rate of transfer of chemicals through each pipe is equal to a flow rate (Q, with units of cubic meters per second) multiplied by the concentration of the reactor from which the flow originates (c, with units of milligrams per cubic meter). If the system is at a steady state, the transfer into each reactor will balance the transfer out. Develop mass- balance equations for the reactors and solve the three simul- taneous linear algebraic equations for their concentrations. 9.10 A civil engineer involved in construction requires 4800, 5800, and 5700 m3 of sand, fine gravel, and coarse gravel, respectively, for a building project. There are three

Q33 = 120 Q13 = 40 Q12 = 90 Q23 = 60 Q21 = 30

Q12c1Q21c2 Q23c2

Q33c3Q13c1 500 mg/s

500 mg/s

FIGURE P9.9 Three reactors linked by pipes. The rate of mass trans- fer through each pipe is equal to the product of flow Q and concentration c of the reactor from which the flow originates.

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pits from which these materials can be obtained. The com- position of these pits is

Sand Fine Gravel Coarse Gravel % % %

Pit1 55 30 15 Pit2 25 45 30 Pit3 25 20 55

How many cubic meters must be hauled from each pit in order to meet the engineer’s needs? 9.11 An electrical engineer supervises the production of three types of electrical components. Three kinds of mate- rial—metal, plastic, and rubber—are required for produc- tion. The amounts needed to produce each component are

Metal (g/ Plastic (g/ Rubber (g/ Component component) component) component)

1 15 0.30 1.0 2 17 0.40 1.2 3 19 0.55 1.5

If totals of 3.89, 0.095, and 0.282 kg of metal, plastic, and rubber, respectively, are available each day, how many com- ponents can be produced per day? 9.12 As described in Sec. 9.4, linear algebraic equations can arise in the solution of differential equations. For example, the following differential equation results from a steady-state mass balance for a chemical in a one-dimensional canal:

0 = D d 2c ___

d x 2 − U dc ___ d x − kc

where c = concentration, t = time, x = distance, D = dif- fusion coefficient, U = fluid velocity, and k = a first-order decay rate. Convert this differential equation to an equiva- lent system of simultaneous algebraic equations. Given D = 2, U = 1, k = 0.2, c(0) = 80 and c(10) = 20, solve these equations from x = 0 to 10 and develop a plot of concentra- tion versus distance. 9.13 A stage extraction process is depicted in Fig. P9.13. In such systems, a stream containing a weight fraction yin of a chemical enters from the left at a mass flow rate of F1. Simultaneously, a solvent carrying a weight fraction xin of the same chemical enters from the right at a flow rate of F2. Thus, for stage i, a mass balance can be represented as

F1 yi−1 + F2 xi+1 = F1 yi + F2 xi (P9.13a)

At each stage, an equilibrium is assumed to be established between yi and xi as in

K = xi __ yi (P9.13b)

where K is called a distribution coefficient. Equation (P9.13b) can be solved for xi and substituted into Eq. (P9.13a) to yield

yi−1 − ( 1 + F2 ___ F1 K ) yi + ( F2 ___ F1 K ) yi +1 = 0 (P9.13c) If F1 = 500 kg/h, yin = 0.1, F2 = 1000 kg/h, xin = 0, and K = 4, determine the values of yout and xout if a five-stage reactor is used. Note that Eq. (P9.13c) must be modified to account for the inflow weight fractions when applied to the first and last stages. 9.14 A peristaltic pump delivers a unit flow (Q1) of a highly viscous fluid. The network is depicted in Fig. P9.14. Every pipe section has the same length and diameter. The mass and mechanical energy balance can be simplified to obtain the

Flow = F1

Flow = F2

x2xout x3 xi xi +1 xn −1 xn xin

y1yin y2 yi −1 yi yn −2 yn −1 yout

1 02 0n0i n −1

FIGURE P9.13 A stage extraction process.

PROBLEMS 271

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272 GAUSS ELIMINATION

flows in every pipe. Solve the following system of equations to obtain the flow in every stream.

Q3 + 2Q4 − 2Q2 = 0 Q1 = Q2 + Q3 Q5 + 2Q6 − 2Q4 = 0 Q3 = Q4 + Q5 3Q7 − 2Q6 = 0 Q5 = Q6 + Q7

9.15 A truss is loaded as shown in Fig. P9.15. Using the following set of equations, solve for the 10 unknowns, AB, BC, AD, BD, CD, DE, CE, Ax, Ay, and Ey.

Ax + AD = 0 −24 − CD − (4/5)CE = 0 Ay + AB = 0 −AD + DE − (3/5)BD = 0 74 + BC + (3/5)BD = 0 CD + (4/5)BD = 0 −AB − (4/5)BD = 0 −DE − (3/5)CE = 0 −BC + (3/5)CE = 0 Ey + (4/5)CE = 0

9.16 A pentadiagonal system with a bandwidth of five can be expressed generally asQ1 Q3 Q5

Q2 Q4 Q6 Q7

FIGURE P9.14

3 m 3 m

4 m

D A E

CB74 kN

24 kN

FIGURE P9.15

f1 g1 h1 e2 f2 g2 h2 d3 e3 f3 g3 h3 · · · · · · · · · dn−1 en−1 fn−1 gn−1 dn en fn

x1 x2 x3 .

xn−1 xn

r1 r2 r3 .

rn−1 rn

Develop an M-file to efficiently solve such systems without pivoting in a similar fashion to the algorithm used for tridi- agonal matrices in Sec. 9.4.1. Test it for the following case:

8 −2 −1 0 0 −2 9 −4 −1 0 −1 −3 7 −1 −2 0 −4 −2 12 −5 0 0 −7 −3 −15

x1 x2 x3 x4 x5

5 2 1 1 5

9.17 Develop an M-file function based on Fig. 9.5 to im plement Gauss elimination with partial pivoting. Modify the function so that it computes and returns the determinant (with the correct sign), and detects whether the system is singular based on a near-zero determinant. For the latter, define “near-zero” as being when the absolute value of the determinant is below a tolerance. When this occurs, design the function so that an error message is displayed and the function terminates. Here is the functions first line:

function [x, D] = GaussPivotNew (A, b, tol)

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where D = the determinant and tol = the tolerance. Test your program for Prob. 9.5 with tol = 1 × 10−5. 9.18 As described in Sec. 9.5, linear algebraic equations can arise in the solution of differential equations. The following differential equation results from a steady-state mass bal- ance for a chemical in a one-dimensional canal,

0 = D d 2c ___

dx 2 − U dc ___ dx − kc

where x = distance along the canal (m), c = concentration, t = time, x = distance, D = diffusion coefficient, U = fluid velocity, and k = a first-order decay rate. (a) Convert this differential equation to an equivalent sys-

tem of simultaneous algebraic equations using centered difference approximations for the derivatives.

(b) Develop a function to solve these equations from x = 0 to L and return the resulting distances and concentra- tions. The first line of your function should be

function [x,c] = YourLastName_reactor(D, U, k, c0, cL, L, dx)

(d) Test your script for the following parameters: L = 10 m, Δx = 0.5 m, D = 2 m2/d, U = 1 m/d, k = 0.2/d, c(0) = 80 mg/L, and c(10) = 20 mg/L.

9.19 The following differential equation results from a force balance for a beam with a uniform loading,

0 = EI d 2y ___

dx 2 − wL x ____ 2 +

wx 2 ____ 2

where x = distance along the beam (m), y = deflection (m), L = length (m), E = modulus of elasticity (N/m2), I = moment of inertia (m4), and w = uniform load (N/m). (a) Convert this differential equation to an equivalent system

of simultaneous algebraic equations using a centered- difference approximation for the second derivative.

(b) Develop a function to solve these equations from x = 0 to L and return the resulting distances and deflections. The first line of your function should be

function [x, y] = YourLastName_beam (E, I, w, y0, yL, L, dx)

(d) Test your script for the following parameters: L = 3 m, Δx = 0.2 m, E = 250 × 109 N/m2, I = 3 × 10− 4 m4, w = 22,500 N/m, y(0) = 0, and y(3) = 0.

9.20 Heat is conducted along a metal rod positioned be- tween two fixed temperature walls. Aside from conduction, heat is transferred between the rod and the surrounding air by convection. Based on a heat balance, the distribution of temperature along the rod is described by the following sec- ond-order differential equation

0 = d 2T ___

dx 2 + h′(T∞ − T)

where T = temperature (K), h′ = a bulk heat transfer co- efficient reflecting the relative importance of convection to conduction (m−2), x = distance along the rod (m), and T∞ = temperature of the surrounding fluid (K). (a) Convert this differential equation to an equivalent sys-

tem of simultaneous algebraic equations using a centered difference approximation for the second derivative.

(b) Develop a function to solve these equations from x = 0 to L and return the resulting distances and temperatures. The first line of your function should be

function [x, y] = YourLastName_rod(hp, Tinf, T0, TL, L, dx)

(d) Test your script for the following parameters: h′ = 0.0425 m−2, L = 12 m, T∞ = 220 K, T(0) = 320 K, T(L) = 450 K, and Δx = 0.5 m.

PROBLEMS 273

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274

CHAPTER OBJECTIVES

The primary objective of this chapter is to acquaint you with LU factorization.1 Specific objectives and topics covered are

• Understanding that LU factorization involves decomposing the coefficient matrix into two triangular matrices that can then be used to efficiently evaluate different right-hand-side vectors.

• Knowing how to express Gauss elimination as an LU factorization. • Given an LU factorization, knowing how to evaluate multiple right-hand-side

vectors. • Recognizing that Cholesky’s method provides an efficient way to decompose a

symmetric matrix and that the resulting triangular matrix and its transpose can be used to evaluate right-hand-side vectors efficiently.

• Understanding in general terms what happens when MATLAB’s backslash operator is used to solve linear systems.

As described in Chap. 9, Gauss elimination is designed to solve systems of linear algebraic equations: [A]{x} = {b} (10.1)

Although it certainly represents a sound way to solve such systems, it becomes inefficient when solving equations with the same coefficients [A], but with different right-hand-side constants {b}.

LU Factorization

1 In the parlance of numerical methods, the terms “factorization” and “decomposition” are synonymous. To be consistent with the MATLAB documentation, we have chosen to employ the terminology LU factorization for the subject of this chapter. Note that LU decomposition is very commonly used to describe the same approach.

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10.1 OVERVIEW OF LU FACTORIZATION 275

Recall that Gauss elimination involves two steps: forward elimination and back substi- tution (Fig. 9.3). As we learned in Sec. 9.2.2, the forward-elimination step comprises the bulk of the computational effort. This is particularly true for large systems of equations.

LU factorization methods separate the time-consuming elimination of the matrix [A] from the manipulations of the right-hand side {b}. Thus, once [A] has been “factored” or “decomposed,” multiple right-hand-side vectors can be evaluated in an efficient manner.

Interestingly, Gauss elimination itself can be expressed as an LU factorization. Before showing how this can be done, let us first provide a mathematical overview of the factorization strategy.

10.1 OVERVIEW OF LU FACTORIZATION

Just as was the case with Gauss elimination, LU factorization requires pivoting to avoid division by zero. However, to simplify the following description, we will omit pivoting. In addition, the following explanation is limited to a set of three simultaneous equations. The results can be directly extended to n-dimensional systems.

Equation (10.1) can be rearranged to give

[A]{x} − {b} = 0 (10.2)

Suppose that Eq. (10.2) could be expressed as an upper triangular system. For example, for a 3 × 3 system:

u11 u12 u13 x1 d1 0 u22 u23 x2 = d2 (10.3) 0 0 u33 x3 d3

Recognize that this is similar to the manipulation that occurs in the first step of Gauss elimination. That is, elimination is used to reduce the system to upper triangular form. Equation (10.3) can also be expressed in matrix notation and rearranged to give

[U]{x} − {d} = 0 (10.4)

Now assume that there is a lower diagonal matrix with 1’s on the diagonal,

1 0 0 [L] = l21 1 0 (10.5) l31 l32 1

that has the property that when Eq. (10.4) is premultiplied by it, Eq. (10.2) is the result. That is,

[L]{[U]{x} − {d}} = [A]{x} − {b} (10.6)

If this equation holds, it follows from the rules for matrix multiplication that

[L][U] = [A] (10.7)

and

[L]{d} = {b} (10.8)

{ } { }

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276 LU FACTORIZATION

A two-step strategy (see Fig. 10.1) for obtaining solutions can be based on Eqs. (10.3), (10.7), and (10.8):

1. LU factorization step. [A] is factored or “decomposed” into lower [L] and upper [U] triangular matrices.

2. Substitution step. [L] and [U] are used to determine a solution {x} for a right-hand side {b}. This step itself consists of two steps. First, Eq. (10.8) is used to generate an intermediate vector {d} by forward substitution. Then, the result is substituted into Eq. (10.3) which can be solved by back substitution for {x}.

Now let us show how Gauss elimination can be implemented in this way.

10.2 GAUSS ELIMINATION AS LU FACTORIZATION

Although it might appear at face value to be unrelated to LU factorization, Gauss elimina- tion can be used to decompose [A] into [L] and [U]. This can be easily seen for [U], which is a direct product of the forward elimination. Recall that the forward-elimination step is intended to reduce the original coefficient matrix [A] to the form

a11 a12 a13 [U ] = 0 a′ 22 a′ 23 (10.9) 0 0 a″ 33

which is in the desired upper triangular format.

[A] {x} = {b}

[L] {d } = {b}

{d }

[U] [L]

{x}

(a) Factorization

(b) Forward

Substitution

[U] {x} = {d }

FIGURE 10.1 The steps in LU factorization.

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10.2 GAUSS ELIMINATION AS LU FACTORIZATION 277

Though it might not be as apparent, the matrix [L] is also produced during the step. This can be readily illustrated for a three-equation system,

a11 a12 a13 x1 b1 a21 a22 a23 x2 = b2 a31 a32 a33 x3 b3

The first step in Gauss elimination is to multiply row 1 by the factor [recall Eq. (9.9)]

f21 = a21 ___ a11

and subtract the result from the second row to eliminate a21. Similarly, row 1 is multiplied by

f31 = a31 ___ a11

and the result subtracted from the third row to eliminate a31. The final step is to multiply the modified second row by

f32 = a′ 32 ___ a′ 22

and subtract the result from the third row to eliminate aʹ 32. Now suppose that we merely perform all these manipulations on the matrix [A].

Clearly, if we do not want to change the equations, we also have to do the same to the right- hand side {b}. But there is absolutely no reason that we have to perform the manipulations simultaneously. Thus, we could save the f ’s and manipulate {b} later.

Where do we store the factors f21, f31, and f32? Recall that the whole idea behind the elimination was to create zeros in a21, a31, and a32. Thus, we can store f21 in a21, f31 in a31, and f32 in a32. After elimination, the [A] matrix can therefore be written as

a11 a12 a13 f21 a′ 22 a′ 23 (10.10) f31 f32 a″ 33

This matrix, in fact, represents an efficient storage of the LU factorization of [A],

[A] → [L][U ] (10.11)

where a11 a12 a13 [U ] = 0 a′ 22 a′ 22 (10.12) 0 0 a″ 33

and 1 0 0 [L] = f21 1 0 (10.13) f31 f32 1

The following example confirms that [A] = [L][U].

{ } { }

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278 LU FACTORIZATION

EXAMPLE 10.1 LU Factorization with Gauss Elimination

Problem Statement. Derive an LU factorization based on the Gauss elimination per- formed previously in Example 9.3.

Solution. In Example 9.3, we used Gauss elimination to solve a set of linear algebraic equations that had the following coefficient matrix:

3 −0.1 −0.2 [A] = 0.1 7 −0.3 0.3 −0.2 10

After forward elimination, the following upper triangular matrix was obtained:

3 −0.1 −0.2 [U ] = 0 7.00333 −0.293333 0 0 10.0120

The factors employed to obtain the upper triangular matrix can be assembled into a lower triangular matrix. The elements a21 and a31 were eliminated by using the factors

f21 = 0.1 ___ 3 = 0.0333333 f31 =

0.3 ___ 3 = 0.1000000

and the element a32 was eliminated by using the factor

f32 = −0.19 _______ 7.00333 = −0.0271300

Thus, the lower triangular matrix is

1 0 0 [L ] = 0.0333333 1 0 0.100000 −0.0271300 1

Consequently, the LU factorization is

1 0 0 [A ] = [L ] [U ] = 0.0333333 1 0 0.100000 −0.0271300 1

This result can be verified by performing the multiplication of [L][U] to give

3 −0.1 −0.2 [L ][U ] = 0.0999999 7 −0.3

0.3 −0.2 9.99996

where the minor discrepancies are due to roundoff.

After the matrix is decomposed, a solution can be generated for a particular right-hand- side vector {b}. This is done in two steps. First, a forward-substitution step is executed by solving Eq. (10.8) for {d}. It is important to recognize that this merely amounts to perform- ing the elimination manipulations on {b}. Thus, at the end of this step, the right-hand side

3 −0.1 −0.2 0 7.00333 −0.293333 0 0 10.0120

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10.2 GAUSS ELIMINATION AS LU FACTORIZATION 279

will be in the same state that it would have been had we performed forward manipulation on [A] and {b} simultaneously.

The forward-substitution step can be represented concisely as

di = bi − ∑ j=1

i−1

lij dj for i = 1, 2, . . . , n

The second step then merely amounts to implementing back substitution to solve Eq. (10.3). Again, it is important to recognize that this is identical to the back-substitution phase of conventional Gauss elimination [compare with Eqs. (9.12) and (9.13)]:

xn = dn∕unn

xi = di − ∑

j=i+1

n ui j xj

________ uii for i = n − 1, n − 2, . . . , 1

EXAMPLE 10.2 The Substitution Steps

Problem Statement. Complete the problem initiated in Example 10.1 by generating the final solution with forward and back substitution.

Solution. As just stated, the intent of forward substitution is to impose the elimination manipulations that we had formerly applied to [A] on the right-hand-side vector {b}. Recall that the system being solved is

[

−0.1

−0.2 0.1 7 −0.3

0.3

−0.2

10 ]

{

x2 x3

}

= {

7.85

−19.3 71.4

}

and that the forward-elimination phase of conventional Gauss elimination resulted in

[

−0.1

−0.2 0 7.00333 −0.293333

10.0120

] {

x2 x3

}

= {

7.85

−19.5617 70.0843

}

The forward-substitution phase is implemented by applying Eq. (10.8):

[

0 0.0333333 1 0

0.100000

−0.0271300

1 ]

⎧

⎪

⎨ ⎪

⎩ d1

d2 d3

⎫

⎪

⎬ ⎪

⎭ =

{ 7.85

−19.3 71.4

}

or multiplying out the left-hand side:

d1 = 7.85 0.0333333d1 + d2 = −19.3 0.100000d1 − 0.0271300d2 + d3 = 71.4

We can solve the first equation for d1 = 7.85, which can be substituted into the second equation to solve for

d2 = −19.3 − 0.0333333(7.85) = −19.5617

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280 LU FACTORIZATION

Both d1 and d2 can be substituted into the third equation to give

d3 = 71.4 − 0.1(7.85) + 0.02713(−19.5617) = 70.0843

Thus,

{d} = {

7.85

−19.5617 70.0843

}

This result can then be substituted into Eq. (10.3), [U ]{x} = {d}:

[

−0.1

−0.2 0 7.00333 −0.293333

10.0120

] {

x2 x3

}

= {

7.85

−19.5617 70.0843

}

which can be solved by back substitution (see Example 9.3 for details) for the final solution:

{x} = {

3 −2.5

7.00003 }

10.2.1 LU Factorization with Pivoting

Just as for standard Gauss elimination, partial pivoting is necessary to obtain reliable solu- tions with LU factorization. One way to do this involves using a permutation matrix (recall Sec. 8.1.2). The approach consists of the following steps:

1. Elimination. The LU factorization with pivoting of a matrix [A] can be represented in matrix form as

[P][A] = [L][U]

The upper triangular matrix, [U], is generated by elimination with partial pivoting, while storing the multiplier factors in [L] and employing the permutation matrix, [P], to keep track of the row switches.

2. Forward substitution. The matrices [L] and [P] are used to perform the elimination step with pivoting on {b} in order to generate the intermediate right-hand-side vector, {d}. This step can be represented concisely as the solution of the following matrix formulation:

[L]{d} = [P]{b}

3. Back substitution. The final solution is generated in the same fashion as done previ- ously for Gauss elimination. This step can also be represented concisely as the solution of the matrix formulation:

[U]{x} = {d}

The approach is illustrated in the following example.

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10.2 GAUSS ELIMINATION AS LU FACTORIZATION 281

EXAMPLE 10.3 LU Factorization with Pivoting

Problem Statement. Compute the LU factorization and find the solution for the same system analyzed in Example 9.4

[ 0.0003 3.0000 1.0000 1.0000 ] {

x1 x2 } = { 2.0001 1.0000 }

Solution. Before elimination, we set up the initial permutation matrix:

[P] = [ 1.0000 0.0000 0.0000 1.0000 ]

We immediately see that pivoting is necessary, so prior to elimination we switch the rows:

[A] = [ 1.0000 1.0000 0.0003 3.0000 ]

At the same time, we keep track of the pivot by switching the rows of the permutation matrix:

[P] = [ 0.0000 1.0000 1.0000 0.0000 ]

We then eliminate a21 by subtracting the factor l21 = a21∕a11 = 0.0003∕1 = 0.0003 from the second row of A. In so doing, we compute that the new value of a′ 22 = 3 − 0.0003(1) = 2.9997. Thus, the elimination step is complete with the result:

[U ] = [ 1 1 0 2.9997 ] [L] = [

1 0 0.0003 1 ]

Before implementing forward substitution, the permutation matrix is used to reorder the right-hand-side vector to reflect the pivots as in

[P]{b} = [ 0.0000 1.0000 1.0000 0.0000 ] {

2.0001 1 } = { 1 2.0001 }

Then, forward substitution is applied as in

[ 1 0 0.0003 1 ] {

d1 d2 } = {

1 2.0001 }

which can be solved for d1 = 1 and d2 = 2.0001 − 0.0003(1) = 1.9998. At this point, the system is

[ 1 1 0 2.9997 ] {

x1 x2 } = {

1 1.9998 }

Applying back substitution gives the final result:

x2 = 1.9998 ______ 2.9997 = 0.66667

x1 = 1 − 1(0.66667) _____________ 1 = 0.33333

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282 LU FACTORIZATION

The LU factorization algorithm requires the same total flops as for Gauss elimination. The only difference is that a little less effort is expended in the factorization phase since the operations are not applied to the right-hand side. Conversely, the substitution phase takes a little more effort.

10.2.2 MATLAB Function: lu

MATLAB has a built-in function lu that generates the LU factorization. It has the general syntax:

[L,U] = lu (X)

where L and U are the lower triangular and upper triangular matrices, respectively, derived from the LU factorization of the matrix X. Note that this function uses partial pivoting to avoid division by zero. The following example shows how it can be employed to generate both the factorization and a solution for the same problem that was solved in Examples 10.1 and 10.2.

EXAMPLE 10.4 LU Factorization with MATLAB

Problem Statement. Use MATLAB to compute the LU factorization and find the solution for the same linear system analyzed in Examples 10.1 and 10.2:

[

−0.1

−0.2 0.1 7 −0.3

0.3

−0.2

10 ]

{

x2 x3

}

= {

7.85

−19.3 71.4

}

Solution. The coefficient matrix and the right-hand-side vector can be entered in standard fashion as

>> A = [3 −.1 −.2;.1 7 −.3;.3 −.2 10]; >> b = [7.85; −19.3; 71.4];

Next, the LU factorization can be computed with

>> [L,U] = lu(A)

L = 1.0000 0 0 0.0333 1.0000 0 0.1000 −0.0271 1.0000

U = 3.0000 −0.1000 −0.2000 0 7.0033 −0.2933 0 0 10.0120

This is the same result that we obtained by hand in Example 10.1. We can test that it is correct by computing the original matrix as

>> L*U

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10.3 CHOLESKY FACTORIZATION 283

ans =

3.0000 −0.1000 −0.2000 0.1000 7.0000 −0.3000 0.3000 −0.2000 10.0000

To generate the solution, we first compute >> d = L\b

d = 7.8500 −19.5617 70.0843

And then use this result to compute the solution >> x = U\d

x =

3.0000 −2.5000 7.0000

These results conform to those obtained by hand in Example 10.2.

10.3 CHOLESKY FACTORIZATION

Recall from Chap. 8 that a symmetric matrix is one where ai j = aji for all i and j. In other words, [A] = [A]T. Such systems occur commonly in both mathematical and engineering/ science problem contexts.

Special solution techniques are available for such systems. They offer computational advantages because only half the storage is needed and only half the computation time is required for their solution.

One of the most popular approaches involves Cholesky factorization (also called Cho- lesky decomposition). This algorithm is based on the fact that a symmetric matrix can be decomposed, as in

[A] = [U ]T [U] (10.14)

That is, the resulting triangular factors are the transpose of each other. The terms of Eq. (10.14) can be multiplied out and set equal to each other. The factor-

ization can be generated efficiently by recurrence relations. For the ith row:

uii = √ _________

aii − ∑ k=1

i−1

u ki 2 (10.15)

ui j = ai j −

∑

k=1 i−1 ukiukj

____________

uii for j = i + 1, . . . , n (10.16)

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284 LU FACTORIZATION

EXAMPLE 10.5 Cholesky Factorization

Problem Statement. Compute the Cholesky factorization for the symmetric matrix

[A] = [

55 15 55 225

225

979 ]

Solution. For the first row (i = 1), Eq. (10.15) is employed to compute

u11 = √ ___

a11 = √ __

6 = 2.44949

Then, Eq. (10.16) can be used to determine

u12 = a12 ___ u11 =

15 _______ 2.44949 = 6.123724

u13 = a13 ___ u11 =

55 _______ 2.44949 = 22.45366

For the second row (i = 2):

u22 = √ _______

a22 − u 12 2

= √ _______________

55 − (6.123724)2 = 4.1833

u23 = a23 − u12u13 __________ u22 =

225 − 6.123724(22.45366) ______________________ 4.1833 = 20.9165

For the third row (i = 3):

u33 = √ ____________

a33 − u 13 2 − u 23

2 = √

__________________________ 979 − (22.45366)2 − (20.9165)2 = 6.110101

Thus, the Cholesky factorization yields

2.44949 6.123724 22.45366 [U] = 4.1833 20.9165 6.110101

The validity of this factorization can be verified by substituting it and its transpose into Eq. (10.14) to see if their product yields the original matrix [A]. This is left for an exercise.

After obtaining the factorization, it can be used to determine a solution for a right- hand-side vector {b} in a manner similar to LU factorization. First, an intermediate vector {d} is created by solving

[U ]T {d} = {b} (10.17)

Then, the final solution can be obtained by solving

[U ]{x} = {d} (10.18)

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10.3 CHOLESKY FACTORIZATION 285

10.3.1 MATLAB Function: chol

MATLAB has a built-in function chol that generates the Cholesky factorization. It has the general syntax,

U = chol(X)

where U is an upper triangular matrix so that U'*U = X. The following example shows how it can be employed to generate both the factorization and a solution for the same matrix that we looked at in the previous example.

EXAMPLE 10.6 Cholesky Factorization with MATLAB

Problem Statement. Use MATLAB to compute the Cholesky factorization for the same matrix we analyzed in Example 10.5.

[A] = [

55 15 55 225

225

979 ]

Also obtain a solution for a right-hand-side vector that is the sum of the rows of [A]. Note that for this case, the answer will be a vector of ones.

Solution. The matrix is entered in standard fashion as

>> A = [6 15 55; 15 55 225; 55 225 979];

A right-hand-side vector that is the sum of the rows of [A] can be generated as

>> b = [sum (A (1,:)); sum (A (2,:)); sum (A (3,:))]

b = 76 295 1259

Next, the Cholesky factorization can be computed with

>> U = chol(A)

U = 2.4495 6.1237 22.4537 0 4.1833 20.9165 0 0 6.1101

We can test that this is correct by computing the original matrix as

>> U'*U

ans = 6.0000 15.0000 55.0000 15.0000 55.0000 225.0000 55.0000 225.0000 979.0000

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286 LU FACTORIZATION

To generate the solution, we first compute

>> d = U'\b

d = 31.0269 25.0998 6.1101

And then use this result to compute the solution

>> x = U\d

x = 1.0000 1.0000 1.0000

10.4 MATLAB LEFT DIVISION

We previously introduced left division without any explanation of how it works. Now that we have some background on matrix solution techniques, we can provide a simplified description of its operation.

When we implement left division with the backslash operator, MATLAB invokes a highly sophisticated algorithm to obtain a solution. In essence, MATLAB examines the structure of the coefficient matrix and then implements an optimal method to obtain the solution. Although the details of the algorithm are beyond our scope, a simplified overview can be outlined.

First, MATLAB checks to see whether [A] is in a format where a solution can be obtained without full Gauss elimination. These include systems that are (a) sparse and banded, (b) triangular (or easily transformed into triangular form), or (c) symmetric. If any of these cases are detected, the solution is obtained with the efficient techniques that are available for such systems. Some of the techniques include banded solvers, back and forward substitution, and Cholesky factorization.

If none of these simplified solutions are possible and the matrix is square,2 a general triangular factorization is computed by Gauss elimination with partial pivoting and the solution obtained with substitution.

2 It should be noted that in the event that [A] is not square, a least-squares solution is obtained with an approach called QR factorization.

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PROBLEMS 287

is, develop a function that is passed the symmetric matrix and returns the matrix [U]. Test your function by using it to solve the system in Prob. 10.8 and use the built-in function chol to confirm that your function is working properly. 10.10 Solve the following set of equations with LU factor- ization with pivoting:

3x1 − 2x2 + x3 = −10 2x1 + 6x2 − 4x3 = 44 −x1 − 2x2 + 5x3 = −26

10.11 (a) Determine the LU factorization without pivoting by hand for the following matrix and check your results by validating that [L][U] = [A].

[

1 3 7 2

9 ]

(b) Employ the result of (a) to compute the determinant. (c) Repeat (a) and (b) using MATLAB. 10.12 Use the following LU factorization to (a) compute the determinant and (b) solve [A]{x} = {b} with {b}T = ⌊ −10 44 − 26 ⌋

[A] = [L][U ] = [

0.6667 1

−0.3333

−0.3636

1 ]

× [

−2

1 7.3333 −4.6667

3.6364 ]

10.13 Use Cholesky factorization to determine [U] so that

[A] = [U]T[U] = [

−1

0 −1 2 −1

−1

2 ]

10.14 Compute the Cholesky factorization of

[A] = [

0 0 25 0

4 ]

Do your results make sense in terms of Eqs. (10.15) and (10.16)?

10.1 Determine the total flops as a function of the number of equations n for the (a) factorization, (b) forward substitu- tion, and (c) back-substitution phases of the LU factorization version of Gauss elimination. 10.2 Use the rules of matrix multiplication to prove that Eqs. (10.7) and (10.8) follow from Eq. (10.6). 10.3 Use naive Gauss elimination to factor the following system according to the description in Sec. 10.2:

10x1 + 2x2 − x3 = 27 −3x1 − 6x2 + 2x3 = −61.5 x1 + x2 + 5x3 = −21.5

Then, multiply the resulting [L] and [U] matrices to deter- mine that [A] is produced. 10.4 (a) Use LU factorization to solve the system of equa- tions in Prob. 10.3. Show all the steps in the computation. (b) Also solve the system for an alternative right-hand-side vector

{b}T = ⌊ 12 18 − 6 ⌋

10.5 Solve the following system of equations using LU factorization with partial pivoting:

2x1 − 6x2 − x3 = −38 −3x1 − x2 + 7x3 = −34 −8x1 + x2 − 2x3 = −40

10.6 Develop your own M-file to determine the LU factor- ization of a square matrix without partial pivoting. That is, develop a function that is passed the square matrix and re- turns the triangular matrices [L] and [U]. Test your function by using it to solve the system in Prob. 10.3. Confirm that your function is working properly by verifying that [L][U] = [A] and by using the built-in function lu. 10.7 Confirm the validity of the Cholesky factorization of Example 10.5 by substituting the results into Eq. (10.14) to verify that the product of [U]T and [U] yields [A]. 10.8 (a) Perform a Cholesky factorization of the following symmetric system by hand:

[

15 20 80 50

60 ]

{

x2 x3

}

= {

250 100

}

(b) Verify your hand calculation with the built-in chol func- tion. (c) Employ the results of the factorization [U] to deter- mine the solution for the right-hand-side vector. 10.9 Develop your own M-file to determine the Cholesky factorization of a symmetric matrix without pivoting. That

PROBLEMS

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288

11 Matrix Inverse and Condition

CHAPTER OBJECTIVES

The primary objective of this chapter is to show how to compute the matrix inverse and to illustrate how it can be used to analyze complex linear systems that occur in engineering and science. In addition, a method to assess a matrix solution’s sensitivity to roundoff error is described. Specific objectives and topics covered are

• Knowing how to determine the matrix inverse in an efficient manner based on LU factorization.

• Understanding how the matrix inverse can be used to assess stimulus-response characteristics of engineering systems.

• Understanding the meaning of matrix and vector norms and how they are computed. • Knowing how to use norms to compute the matrix condition number. • Understanding how the magnitude of the condition number can be used to esti-

mate the precision of solutions of linear algebraic equations.

11.1 THE MATRIX INVERSE

In our discussion of matrix operations (Sec. 8.1.2), we introduced the notion that if a matrix [A] is square, there is another matrix [A]−1, called the inverse of [A], for which

[A][A]−1 = [A]−1[A] = [I ] (11.1)

Now we will focus on how the inverse can be computed numerically. Then we will explore how it can be used for engineering analysis.

11.1.1 Calculating the Inverse

The inverse can be computed in a column-by-column fashion by generating solutions with unit vectors as the right-hand-side constants. For example, if the right-hand-side constant

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11.1 THE MATRIX INVERSE 289

has a 1 in the first position and zeros elsewhere,

{b} = {

1 0

0 }

(11.2)

the resulting solution will be the first column of the matrix inverse. Similarly, if a unit vec- tor with a 1 at the second row is used

{b} = {

0 1

0 }

(11.3)

the result will be the second column of the matrix inverse. The best way to implement such a calculation is with LU factorization. Recall that one

of the great strengths of LU factorization is that it provides a very efficient means to evalu- ate multiple right-hand-side vectors. Thus, it is ideal for evaluating the multiple unit vectors needed to compute the inverse.

EXAMPLE 11.1 Matrix Inversion

Problem Statement. Employ LU factorization to determine the matrix inverse for the system from Example 10.1:

[A] = [

−0.1

−0.2 0.1 7 −0.3

0.3

−0.2

10 ]

Recall that the factorization resulted in the following lower and upper triangular matrices:

[U ] = [

−0.1

−0.2 0 7.00333 −0.293333

10.0120

] [L] =

[

0 0.0333333 1 0

0.100000

−0.0271300

1 ]

Solution. The first column of the matrix inverse can be determined by performing the forward-substitution solution procedure with a unit vector (with 1 in the first row) as the right-hand-side vector. Thus, the lower triangular system can be set up as [recall Eq. (10.8)]

[

0 0.0333333 1 0

0.100000

−0.0271300

1 ]

⎧

⎪

⎨ ⎪

⎩

d2 d3

⎫

⎪

⎬ ⎪

⎭

= {

1 0

0 }

and solved with forward substitution for {d}T = ⌊1 −0.03333 −0.1009⌋. This vector can then be used as the right-hand side of the upper triangular system [recall Eq. (10.3)]

[

−0.1

−0.2 0 7.00333 −0.293333

10.0120

] {

x2 x3

}

= {

1 −0.03333

−0.1009 }

which can be solved by back substitution for {x}T = ⌊0.33249 −0.00518 −0.01008⌋, which is the first column of the matrix inverse:

[A]−1 = [

0.33249

0 −0.00518 0 0

−0.01008

0 ]

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290 MATRIX INVERSE ANd CoNdITIoN

To determine the second column, Eq. (10.8) is formulated as

[

0 0.0333333 1 0

0.100000

−0.0271300

1 ]

⎧

⎪

⎨ ⎪

⎩

d2 d3

⎫

⎪

⎬ ⎪

⎭

= {

0 1

0 }

This can be solved for {d }, and the results are used with Eq. (10.3) to determine {x}T = ⌊0.004944 0.142903 0.00271⌋, which is the second column of the matrix inverse:

[A]−1 = [

0.33249

0.004944

0 −0.00518 0.142903 0

−0.01008

0.002710

0 ]

Finally, the same procedures can be implemented with {b}T = ⌊0 0 1⌋ to solve for {x}T = ⌊0.006798 0.004183 0.09988⌋, which is the final column of the matrix inverse:

[A]−1 = [

0.33249

0.004944

0.006798

−0.00518 0.142903 0.004183 −0.01008

0.002710

0.099880

]

The validity of this result can be checked by verifying that [A][A]−1 = [I ].

11.1.2 Stimulus-Response Computations

As discussed in PT 3.1, many of the linear systems of equations arising in engineering and science are derived from conservation laws. The mathematical expression of these laws is some form of balance equation to ensure that a particular property—mass, force, heat, momentum, electrostatic potential—is conserved. For a force balance on a structure, the properties might be horizontal or vertical components of the forces acting on each node of the structure. For a mass balance, the properties might be the mass in each reactor of a chemical process. Other fields of engineering and science would yield similar examples.

A single balance equation can be written for each part of the system, resulting in a set of equations defining the behavior of the property for the entire system. These equations are interrelated, or coupled, in that each equation may include one or more of the variables from the other equations. For many cases, these systems are linear and, therefore, of the exact form dealt with in this chapter:

[A]{x} = {b} (11.4)

Now, for balance equations, the terms of Eq. (11.4) have a definite physical interpre- tation. For example, the elements of {x} are the levels of the property being balanced for each part of the system. In a force balance of a structure, they represent the horizontal and vertical forces in each member. For the mass balance, they are the mass of chemical in each reactor. In either case, they represent the system’s state or response, which we are trying to determine.

The right-hand-side vector {b} contains those elements of the balance that are inde- pendent of behavior of the system—that is, they are constants. In many problems, they represent the forcing functions or external stimuli that drive the system.

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11.1 THE MATRIX INVERSE 291

Finally, the matrix of coefficients [A] usually contains the parameters that express how the parts of the system interact or are coupled. Consequently, Eq. (11.4) might be reexpressed as

[Interactions]{response} = {stimuli}

As we know from previous chapters, there are a variety of ways to solve Eq. (11.4). However, using the matrix inverse yields a particularly interesting result. The formal solu- tion can be expressed as

{x} = [A]−1{b}

or (recalling our definition of matrix multiplication from Sec. 8.1.2)

x1 = a 11 −1 b1 + a 12

−1 b2 + a 13 −1 b3

x2 = a 21 −1 b1 + a 22

−1 b2 + a 23 −1 b3

x3 = a 31 −1 b1 + a 32

−1 b2 + a 33 −1 b3

Thus, we find that the inverted matrix itself, aside from providing a solution, has ex- tremely useful properties. That is, each of its elements represents the response of a single part of the system to a unit stimulus of any other part of the system.

Notice that these formulations are linear and, therefore, superposition and proportion- ality hold. Superposition means that if a system is subject to several different stimuli (the b’s), the responses can be computed individually and the results summed to obtain a total response. Proportionality means that multiplying the stimuli by a quantity results in the response to those stimuli being multiplied by the same quantity. Thus, the coefficient a 11

−1 is

a proportionality constant that gives the value of x1 due to a unit level of b1. This result is independent of the effects of b2 and b3 on x1, which are reflected in the coefficients a 12

−1 and a 13

−1 , respectively. Therefore, we can draw the general conclusion that the element a i j −1 of the

inverted matrix represents the value of xi due to a unit quantity of bj. Using the example of the structure, element a i j

−1 of the matrix inverse would represent the force in member i due to a unit external force at node j. Even for small systems, such behavior of individual stimulus-response interactions would not be intuitively obvious. As such, the matrix inverse provides a powerful technique for understanding the interrelation- ships of component parts of complicated systems.

EXAMPLE 11.2 Analyzing the Bungee Jumper Problem

Problem Statement. At the beginning of Chap. 8, we set up a problem involving three individuals suspended vertically connected by bungee cords. We derived a system of linear algebraic equations based on force balances for each jumper,

[

150

−100

0 −100 150 −50

−50

50 ]

{

x2 x3

}

= {

588.6

686.7 784.8

}

In Example 8.2, we used MATLAB to solve this system for the vertical positions of the jumpers (the x’s). In the present example, use MATLAB to compute the matrix inverse and interpret what it means.

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292 MATRIX INVERSE ANd CoNdITIoN

Solution. Start up MATLAB and enter the coefficient matrix:

>> K = [150 −100 0;−100 150 −50;0 −50 50];

The inverse can then be computed as >> KI = inv(K)

KI = 0.0200 0.0200 0.0200 0.0200 0.0300 0.0300 0.0200 0.0300 0.0500

Each element of the inverse, k i j −1 of the inverted matrix represents the vertical change

in position (in meters) of jumper i due to a unit change in force (in Newtons) applied to jumper j. First, observe that the numbers in the first column ( j = 1) indicate that the position of all three jumpers would increase by 0.02 m if the force on the first jumper was increased by 1 N. This makes sense, because the additional force would only elongate the first cord by that amount. In contrast, the numbers in the second column ( j = 2) indicate that applying a force of 1 N to the second jumper would move the first jumper down by 0.02 m, but the second and third by 0.03 m. The 0.02-m elongation of the first jumper makes sense because the first cord is subject to an extra 1 N regardless of whether the force is applied to the first or second jumper. However, for the second jumper the elongation is now 0.03 m because along with the first cord, the second cord also elongates due to the additional force. And of course, the third jumper shows the identical translation as the second jumper as there is no additional force on the third cord that connects them. As expected, the third column ( j = 3) indicates that applying a force of 1 N to the third jumper results in the first and second jumpers moving the same distances as occurred when the force was applied to the second jumper. However, now because of the additional elongation of the third cord, the third jumper is moved farther downward. Superposition and proportionality can be demonstrated by using the inverse to deter- mine how much farther the third jumper would move downward if additional forces of 10, 50, and 20 N were applied to the first, second, and third jumpers, respectively. This can be done simply by using the appropriate elements of the third row of the inverse to compute,

Δ x3 = k 31 −1 Δ F1 + k 32

−1 Δ F2 + k 33 −1 Δ F3 = 0.02(10) + 0.03(50) + 0.05(20) = 2.7 m

11.2 ERROR ANALYSIS AND SYSTEM CONDITION

Aside from its engineering and scientific applications, the inverse also provides a means to discern whether systems are ill-conditioned. Three direct methods can be devised for this purpose:

1. Scale the matrix of coefficients [A] so that the largest element in each row is 1. Invert the scaled matrix and if there are elements of [A]−1 that are several orders of magnitude greater than one, it is likely that the system is ill-conditioned.

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11.2 ERRoR ANALYSIS ANd SYSTEM CoNdITIoN 293

2. Multiply the inverse by the original coefficient matrix and assess whether the result is close to the identity matrix. If not, it indicates ill-conditioning.

3. Invert the inverted matrix and assess whether the result is sufficiently close to the original coefficient matrix. If not, it again indicates that the system is ill-conditioned.

Although these methods can indicate ill-conditioning, it would be preferable to obtain a single number that could serve as an indicator of the problem. Attempts to formulate such a matrix condition number are based on the mathematical concept of the norm.

11.2.1 Vector and Matrix Norms

A norm is a real-valued function that provides a measure of the size or “length” of multi- component mathematical entities such as vectors and matrices.

A simple example is a vector in three-dimensional Euclidean space (Fig. 11.1) that can be represented as

⌊F⌋ = ⌊a b c⌋

where a, b, and c are the distances along the x, y, and z axes, respectively. The length of this vector—that is, the distance from the coordinate (0, 0, 0) to (a, b, c)—can be simply computed as

∥ F ∥ e = √ __________

a2 + b2 + c2

where the nomenclature ∥ F ∥ e indicates that this length is referred to as the Euclidean norm of [F ].

Similarly, for an n-dimensional vector ⌊X⌋ = ⌊x1 x2 · · · xn ⌋, a Euclidean norm would be computed as

∥ X ∥ e = √ _____

∑ i=1

n x i

‖F ‖ e

= a2

+ b2

+ c2

FIGURE 11.1 Graphical depiction of a vector in Euclidean space.

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294 MATRIX INVERSE ANd CoNdITIoN

The concept can be extended further to a matrix [A], as in

∥ A ∥ f = √ ________

∑ i=1

n ∑

j=1

n a i, j

2 (11.5)

which is given a special name—the Frobenius norm. As with the other vector norms, it provides a single value to quantify the “size” of [A].

It should be noted that there are alternatives to the Euclidean and Frobenius norms. For vectors, there are alternatives called p norms that can be represented generally by

∥ X ∥ p = ( ∑ i=1

n | xi | p ) 1/p

We can see that the Euclidean norm and the 2 norm, ∥ X ∥ 2, are identical for vectors. Other important examples are ( p = 1)

∥ X ∥ 1 = ∑ i=1

n | xi |

which represents the norm as the sum of the absolute values of the elements. Another is the maximum-magnitude or uniform-vector norm ( p = ∞),

∥ X ∥ ∞ = max | xi | which defines the norm as the element with the largest absolute value.

Using a similar approach, norms can be developed for matrices. For example,

∥ A ∥ 1 = max ∑ i=1

n | ai j |

That is, a summation of the absolute values of the coefficients is performed for each col- umn, and the largest of these summations is taken as the norm. This is called the column- sum norm.

A similar determination can be made for the rows, resulting in a uniform-matrix or row-sum norm:

∥ A ∥ ∞ = max ∑ j=1

n | ai j |

It should be noted that, in contrast to vectors, the 2 norm and the Frobenius norm for a matrix are not the same. Whereas the Frobenius norm ∥ A ∥ f can be easily determined by Eq. (11.5), the matrix 2 norm ∥ A ∥ 2 is calculated as

∥ A ∥ 2 = (µmax)1/2

where µmax is the largest eigenvalue of [A]T[A]. In Chap. 13, we will learn more about ei- genvalues. For the time being, the important point is that the ∥ A ∥ 2, or spectral norm, is the minimum norm and, therefore, provides the tightest measure of size (Ortega, 1972).

11.2.2 Matrix Condition Number

Now that we have introduced the concept of the norm, we can use it to define

Cond[A] = ∥ A ∥ · ∥ A−1 ∥

1≤ i≤n

1≤ j≤n

1≤i≤n

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11.2 ERRoR ANALYSIS ANd SYSTEM CoNdITIoN 295

where Cond[A] is called the matrix condition number. Note that for a matrix [A], this number will be greater than or equal to 1. It can be shown (Ralston and Rabinowitz, 1978; Gerald and Wheatley, 1989) that

∥ Δ X ∥ ______ ∥ X ∥ ≤ Cond [A]

∥ Δ A ∥ _____ ∥ A ∥

That is, the relative error of the norm of the computed solution can be as large as the rela- tive error of the norm of the coefficients of [A] multiplied by the condition number. For example, if the coefficients of [A] are known to t-digit precision (i.e., rounding errors are on the order of 10−t ) and Cond[A] = 10c, the solution [X ] may be valid to only t − c digits (rounding errors ≈10c−t ).

EXAMPLE 11.3 Matrix Condition Evaluation

Problem Statement. The Hilbert matrix, which is notoriously ill-conditioned, can be repre sented generally as

⎡

⎢ ⎣

1 _ 2

1 _ 3

· · ·

1 _ n

1 _ 2

1 _ 3 1 _ 4

· · ·

1 ___ n+1 · · ·

· · ·

1 _ n

1 ___ n+1

1 ___ n+2

· · ·

1 ___ 2n−1

⎤

⎥ ⎦

Use the row-sum norm to estimate the matrix condition number for the 3 × 3 Hilbert matrix:

[A] =

⎡

⎢ ⎣ 1

1 _ 2

1 _ 3 1 _ 2

1 _ 3

1 _ 4

1 _ 3

1 _ 4

1 _ 5

⎤

⎥ ⎦

Solution. First, the matrix can be normalized so that the maximum element in each row is 1:

[A] =

⎡

⎢ ⎣ 1

1 _ 2

1 _ 3 1 2 _ 3

1 _ 2 1

3 _ 4

3 _ 5

⎤

⎥ ⎦

Summing each of the rows gives 1.833, 2.1667, and 2.35. Thus, the third row has the larg- est sum and the row-sum norm is

∥ A ∥ ∞ = 1 + 3 __ 4 + 3 __ 5 = 2.35

The inverse of the scaled matrix can be computed as

[A]−1 = [

−18

10 −36 96 −60

−90

60 ]

Note that the elements of this matrix are larger than the original matrix. This is also re- flected in its row-sum norm, which is computed as

∥ A−1 ∥ ∞ = | −36 | + | 96 | + | −60 | = 192

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Thus, the condition number can be calculated as

Cond[A] = 2.35(192) = 451.2

The fact that the condition number is much greater than unity suggests that the sys- tem is ill-conditioned. The extent of the ill-conditioning can be quantified by calculating c = log 451.2 = 2.65. Hence, the last three significant digits of the solution could ex- hibit rounding errors. Note that such estimates almost always overpredict the actual error. However, they are useful in alerting you to the possibility that roundoff errors may be significant.

11.2.3 Norms and Condition Number in MATLAB

MATLAB has built-in functions to compute both norms and condition numbers:

>> norm (X,p) and

>> cond (X,p)

where X is the vector or matrix and p designates the type of norm or condition number (1, 2, inf, or 'fro'). Note that the cond function is equivalent to

>> norm (X,p) * norm (inv(X),p)

Also, note that if p is omitted, it is automatically set to 2.

EXAMPLE 11.4 Matrix Condition Evaluation with MATLAB

Problem Statement. Use MATLAB to evaluate both the norms and condition numbers for the scaled Hilbert matrix previously analyzed in Example 11.3:

[A] =

⎡

⎢ ⎣ 1

1 _ 2

1 _ 3 1 2 _ 3

1 _ 2 1

3 _ 4

3 _ 5

⎤

⎥ ⎦

(a) As in Example 11.3, first compute the row-sum versions (p = inf). (b) Also compute the Frobenius (p = 'fro') and the spectral (p = 2) condition numbers.

Solution: (a) First, enter the matrix: >> A = [1 1/2 1/3;1 2/3 1/2;1 3/4 3/5];

Then, the row-sum norm and condition number can be computed as >> norm (A,inf)

ans = 2.3500

>> cond (A,inf)

ans = 451.2000

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11.3 CASE STUdY 297

These results correspond to those that were calculated by hand in Example 11.3.

(b) The condition numbers based on the Frobenius and spectral norms are

>> cond(A,'fro')

ans = 368.0866

>> cond(A)

ans = 366.3503

11.3 CASE STUDY INdooR AIR PoLLUTIoN

Background. As the name implies, indoor air pollution deals with air contamination in enclosed spaces such as homes, offices, and work areas. Suppose that you are studying the ventilation system for Bubba’s Gas ’N Guzzle, a truck-stop restaurant located adjacent to an eight-lane freeway.

As depicted in Fig. 11.2, the restaurant serving area consists of two rooms for smokers and kids and one elongated room. Room 1 and section 3 have sources of carbon monoxide from smokers and a faulty grill, respectively. In addition, rooms 1 and 2 gain carbon mon- oxide from air intakes that unfortunately are positioned alongside the freeway.

Qc = 150 m3/hr

2 (Kids’ section)

1 (Smoking section)

Grill load (2000 mg/hr)

Smoker load (1000 mg/hr)

25 m3/hr

Qb = 50 m3/hr

cb = 2 mg/m3

Qa = 200 m3/hr

ca = 2 mg/m3

Qd = 100 m3/hr

5 0

m 3 /

FIGURE 11.2 Overhead view of rooms in a restaurant. The one-way arrows represent volumetric airflows, whereas the two-way arrows represent diffusive mixing. The smoker and grill loads add car- bon monoxide mass to the system but negligible airflow.

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298 MATRIX INVERSE ANd CoNdITIoN

11.3 CASE STUDY continued

Write steady-state mass balances for each room and solve the resulting linear al- gebraic equations for the concentration of carbon monoxide in each room. In addition, generate the matrix inverse and use it to analyze how the various sources affect the kids’ room. For example, determine what percent of the carbon monoxide in the kids’ section is due to (1) the smokers, (2) the grill, and (3) the intake vents. In addition, compute the improvement in the kids’ section concentration if the carbon monoxide load is decreased by banning smoking and fixing the grill. Finally, analyze how the concentration in the kids’ area would change if a screen is constructed so that the mixing between areas 2 and 4 is decreased to 5 m3/hr.

Solution. Steady-state mass balances can be written for each room. For example, the bal- ance for the smoking section (room 1) is

0 = Wsmoker + Qaca − Qac1 + E13(c3 − c1) (Load) + (Inflow) − (Outflow) + (Mixing)

Similar balances can be written for the other rooms:

0 = Qbcb + (Qa − Qd )c4 − Qcc2 + E24(c4 − c2 ) 0 = Wgrill + Qac1 + E13(c1 − c3) + E34(c4 − c3) − Qac3 0 = Qac3 + E34(c3 − c4) + E24(c2 − c4) − Qac4

Substituting the parameters yields the final system of equation:

⎡ ⎢

⎣ 225

−25

0 175 0 −125 −225 0 275 −50

−25

−250

275

⎤ ⎥

⎦ ⎧

⎪

⎨ ⎪

⎩

c2 c3 c4

⎫

⎪

⎬ ⎪

⎭

⎧

⎪ ⎨

⎪

⎩ 1400

100 2000

⎫

⎪ ⎬

⎪

⎭

MATLAB can be used to generate the solution. First, we can compute the inverse. Note that we use the “short g” format in order to obtain five significant digits of precision:

>> format short g >> A = [225 0 −25 0 0 175 0 −125 −225 0 275 −50 0 −25 −250 275]; >> AI = inv(A)

AI = 0.0049962 1.5326e − 005 0.00055172 0.00010728 0.0034483 0.0062069 0.0034483 0.0034483 0.0049655 0.00013793 0.0049655 0.00096552 0.0048276 0.00068966 0.0048276 0.0048276

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11.3 CASE STUdY 299

11.3 CASE STUDY continued

The solution can then be generated as >> b = [1400 100 2000 0]'; >> c = AI*b

c = 8.0996 12.345 16.897 16.483

Thus, we get the surprising result that the smoking section has the lowest carbon mon- oxide levels! The highest concentrations occur in rooms 3 and 4 with section 2 having an intermediate level. These results take place because (a) carbon monoxide is conservative and (b) the only air exhausts are out of sections 2 and 4 (Qc and Qd). Room 3 is so bad because not only does it get the load from the faulty grill, but it also receives the effluent from room 1.

Although the foregoing is interesting, the real power of linear systems comes from using the elements of the matrix inverse to understand how the parts of the system interact. For example, the elements of the matrix inverse can be used to determine the percent of the carbon monoxide in the kids’ section due to each source:

The smokers:

c2,smokers = a 21 −1 Wsmokers = 0.0034483(1000) = 3.4483

%smokers = 3.4483 ______ 12.345 × 100% = 27.93%

The grill:

c2,grill = a 23 −1 Wgrill = 0.0034483(2000) = 6.897

%grill = 6.897 ______ 12.345 × 100% = 55.87%

The intakes:

c2,intakes = a 21 −1 Qaca + a 22

−1 Qbcb = 0.0034483(200)2 + 0.0062069(50)2

= 1.37931 + 0.62069 = 2

%grill = 2 ______ 12.345 × 100% = 16.20%

The faulty grill is clearly the most significant source. The inverse can also be employed to determine the impact of proposed remedies such

as banning smoking and fixing the grill. Because the model is linear, superposition holds and the results can be determined individually and summed:

Δc2 = a 21 −1 ΔWsmokers + a 23

−1 ΔWgrill = 0.0034483(−1000) + 0.0034483(−2000)

= −3.4483 − 6.8966 = −10.345

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300 MATRIX INVERSE ANd CoNdITIoN

11.3 CASE STUDY continued

Note that the same computation would be made in MATLAB as

>> AI(2,1)*(−1000) + AI(2,3)*(−2000)

ans = −10.345

Implementing both remedies would reduce the concentration by 10.345 mg/m3. The result would bring the kids’ room concentration to 12.345 − 10.345 = 2 mg/m3. This makes sense, because in the absence of the smoker and grill loads, the only sources are the air intakes which are at 2 mg/m3.

Because all the foregoing calculations involved changing the forcing functions, it was not necessary to recompute the solution. However, if the mixing between the kids’ area and zone 4 is decreased, the matrix is changed

⎡ ⎢

⎣ 225

−25

0 155 0 −105 −225 0 275 −50

−5

−250

255

⎤ ⎥

⎦

⎧

⎪ ⎨

⎪

⎩

c2 c3 c4

⎫

⎪

⎬ ⎪

⎭

⎧

⎪ ⎨

⎪

⎩ 1400

100 2000

⎫

⎪ ⎬

⎪

⎭

The results for this case involve a new solution. Using MATLAB, the result is

⎧

⎪ ⎨

⎪

⎩

c2 c3 c4

⎫

⎪

⎬ ⎪

⎭

⎧

⎪ ⎨

⎪

⎩ 8.1084

12.0800 16.9760

16.8800

⎫

⎪ ⎬

⎪

⎭

Therefore, this remedy would only improve the kids’ area concentration by a paltry 0.265 mg/m3.

PROBLEMS

11.1 Determine the matrix inverse for the following system:

10x1 + 2x2 − x3 = 27 −3x1 − 6x2 + 2x3 = −61.5

x1 + x2 + 5x3 = −21.5

Check your results by verifying that [A][A]−1 = [I ]. Do not use a pivoting strategy.

11.2 Determine the matrix inverse for the following system:

−8x1 + x2 − 2x3 = −20 2x1 − 6x2 − x3 = −38

−3x1 − x2 + 7x3 = −34

11.3 The following system of equations is designed to determine concentrations (the c’s in g/m3) in a series of

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PRoBLEMS 301

coupled reactors as a function of the amount of mass input to each reactor (the right-hand sides in g/day):

15c1 − 3c2 − c3 = 4000 −3c1 + 18c2 − 6c3 = 1200 −4c1 − c2 + 12c3 = 2350

(a) Determine the matrix inverse. (b) Use the inverse to determine the solution. (c) Determine how much the rate of mass input to reactor 3

must be increased to induce a 10 g/m3 rise in the concen- tration of reactor 1.

(d) How much will the concentration in reactor 3 be reduced if the rate of mass input to reactors 1 and 2 is reduced by 500 and 250 g/day, respectively?

11.4 Determine the matrix inverse for the system de- scribed in Prob. 8.9. Use the matrix inverse to determine the concentration in reactor 5 if the inflow concentrations are changed to c01 = 20 and c03 = 50. 11.5 Determine the matrix inverse for the system described in Prob. 8.10. Use the matrix inverse to determine the force in the three members (F1, F2, and F3) if the vertical load at node 1 is doubled to F1,υ = −2000 N and a horizontal load of F3,h = −500 N is applied to node 3. 11.6 Determine ∥ A ∥ f , ∥ A ∥ 1 , and ∥ A ∥ ∞ for

[A] = [

−10 −9 1 3

−1

]

Before determining the norms, scale the matrix by making the maximum element in each row equal to one. 11.7 Determine the Frobenius and row-sum norms for the systems in Probs. 11.2 and 11.3. 11.8 Use MATLAB to determine the spectral condition num- ber for the following system. Do not normalize the system:

⎡

⎢ ⎣

36 9 16 25 36 49

⎤

⎥ ⎦

Compute the condition number based on the row-sum norm. 11.9 Besides the Hilbert matrix, there are other matrices that are inherently ill-conditioned. One such case is the Vandermonde matrix, which has the following form:

⎡

⎢ ⎣

x 1 2

x 2 2 x2 1

x 3 2

⎤

⎥ ⎦

(a) Determine the condition number based on the row-sum norm for the case where x1 = 4, x2 = 2, and x3 = 7.

(b) Use MATLAB to compute the spectral and Frobenius condition numbers.

11.10 Use MATLAB to determine the spectral condition number for a 10-dimensional Hilbert matrix. How many dig- its of precision are expected to be lost due to ill-conditioning? Determine the solution for this system for the case where each element of the right-hand-side vector {b} consists of the summation of the coefficients in its row. In other words, solve for the case where all the unknowns should be exactly one. Compare the resulting errors with those expected based on the condition number. 11.11 Repeat Prob. 11.10, but for the case of a six- dimensional Vandermonde matrix (see Prob. 11.9) where x1 = 4, x2 = 2, x3 = 7, x4 = 10, x5 = 3, and x6 = 5. 11.12 The Lower Colorado River consists of a series of four reservoirs as shown in Fig. P11.12.

Mass balances can be written for each reservoir, and the following set of simultaneous linear algebraic equations results:

⎡ ⎢

⎣ 13.422

0 −13.422 12.252 0 0 0 −12.252 12.377 0

−12.377

11.797

⎤ ⎥

⎦

× ⎧

⎪

⎨ ⎪

⎩

c2 c3 c4

⎫

⎪ ⎬

⎪

⎭

⎧

⎪ ⎨

⎪

⎩ 750.5

300 102

⎫

⎪ ⎬

⎪

⎭

where the right-hand-side vector consists of the loadings of chloride to each of the four lakes and c1, c2, c3, and c4 = the resulting chloride concentrations for Lakes Powell, Mead, Mohave, and Havasu, respectively. (a) Use the matrix inverse to solve for the concentrations in

each of the four lakes. (b) How much must the loading to Lake Powell be reduced

for the chloride concentration of Lake Havasu to be 75? (c) Using the column-sum norm, compute the condition

number and how many suspect digits would be gener- ated by solving this system.

11.13 (a) Determine the matrix inverse and condition num- ber for the following matrix:

[

3 4 5 6

9 ]

(b) Repeat (a) but change a33 slightly to 9.1.

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302 MATRIX INVERSE ANd CoNdITIoN

11.14 Polynomial interpolation consists of determining the unique (n – 1)th-order polynomial that fits n data points. Such polynomials have the general form,

f (x) = p1x n−1 + p2x n−2 + · · · + pn−1x + pn (P11.14)

where the p’s are constant coefficients. A straightforward way for computing the coefficients is to generate n linear algebraic equations that we can solve simultaneously for the coefficients. Suppose that we want to determine the coefficients of the fourth-order polynomial f (x) = p1x4 + p2 x3 + p3 x2 + p4 x + p5 that passes through the following five points: (200, 0.746), (250, 0.675), (300, 0.616), (400, 0.525), and (500, 0.457). Each of these pairs can be substituted into Eq. (P11.14) to yield a system of five equations with five unknowns (the p’s). Use this approach to solve for the coef- ficients. In addition, determine and interpret the condition number. 11.15 A chemical constituent flows between three reactors as depicted in Fig. P11.15. Steady-state mass balances can be written for a substance that reacts with first-order kinet- ics. For example, the mass balance for reactor 1 is

Q1,inc1,in − Q1,2c1 − Q1,3c1 + Q2,1c2 − kV1c1 = 0 (P11.15)

where Q1,in = the volumetric inflow to reactor 1 (m3/min), c1,in = the inflow concentration to reactor 1 (g/m3), Qi, j = the flow from reactor i to reactor j (m3/min), ci = the concentra- tion of reactor i (g/m3), k = a first-order decay rate (/min), and Vi = the volume of reactor i (m3). (a) Write the mass balances for reactors 2 and 3.

(b) If k = 0.1/min, write the mass balances for all three reactors as a system of linear algebraic equations.

(c) Compute the LU decomposition for this system. (d) Use the LU decomposition to compute the matrix inverse. (e) Use the matrix inverse to answer the following questions:

(i) What are the steady-state concentrations for the three reactors? (ii) If the inflow concentration to the second re- actor is set to zero, what is the resulting reduction in con- centration of reactor 1? (iii) If the inflow concentration to reactor 1 is doubled, and the inflow concentration to reactor 2 is halved, what is the concentration of reactor 3?

Upper Colorado

River

Lake Mead

Lake Mohave

Lake Havasu

Lake Powell

FIGURE P11.12 The Lower Colorado River.

FIGURE P11.15

V2 = 50

Q2,in = 10 c2,in = 200

Q1,in = 100 Q1,3 = 117

Q3,2 = 7 Q1,2 = 5

Q2,1 = 22

Q3,out = 110 c1,in = 10

V1 = 100 V3 = 150

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PRoBLEMS 303

11.16 As described in Examples 8.2 and 11.2, use the ma- trix inverse to answer the following: (a) Determine the change in position of the first jumper, if

the mass of the third jumper is increased to 100 kg. (b) What force must be applied to the third jumper so that

the final position of the third jumper is 140 m? 11.17 Determine the matrix inverse for the electric circuit formulated in Sec. 8.3. Use the inverse to determine the new current between nodes 2 and 5 (i52), if a voltage of 200 V is applied at node 6 and the voltage at node 1 is halved. 11.18 (a) Using the same approach as described in Sec. 11.3, develop steady-state mass balances for the room configura- tion depicted in Fig. P11.18. (b) Determine the matrix inverse and use it to calculate the

resulting concentrations in the rooms. (c) Use the matrix inverse to determine how much the room

4 load must be reduced to maintain a concentration of 20 mg/m3 in room 2.

11.19 Write your own well-structured MATLAB Function procedure named Fnorm to calculate the Frobenius norm of an m×n matrix with for…end loops,

∥ A ∥ f = √ ________

∑ i=1

m ∑

j=1

n a i, j 2

Have the function scale the matrix before computing the norm. Test your function with the following script:

A = [5 7 −9; 1 8 4; 7 6 2]; Fn = Fnorm (A)

Here is the first line of your function

function Norm = Fnorm (x)

11.20 Figure P11.20 shows a statically determinate truss.

This type of structure can be described as a system of cou- pled linear algebraic equations by developing free-body force diagrams for the forces at each node in Fig. P11.20. The sum of the forces in both horizontal and vertical direc- tions must be zero at each node, because the system is at rest. Therefore, for node 1,

FH = 0 = −F1 cos 30° + F3 cos 60° + F1,h FV = 0 = −F1 sin 30° − F3 sin 60° + F1,υ

for node 2,

FH = 0 = F2 + F1 cos 30° + F2,h + H2 FV = 0 = F1 sin 30° + F2,υ + V2

FIGURE P11.18

Q = 50 m3/hr

50 m3/hr

90 m3/hr

Load = 5000 mg/hr

Room 4

Room 3

Room 2

Room 1

Q = 50 m3/hr c = 40 mg/m3

Q = 150 m3/hr c = 1 mg/m3

Q = 50 m3/hr

V2 V3

H2 2

1000 lb

3 60°

90°

30°

FIGURE P11.20 Forces on a statically determinate truss.

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304 MATRIX INVERSE ANd CoNdITIoN

for node 3,

FH = 0 = − F2 − F3 cos 60° + F3,h FV = 0 = F3 sin 60° + F3,υ + V3

where Fi,h is the external horizontal force applied to node i (where a positive force is from left to right) and F1,υ is the ex- ternal vertical force applied to node i (where a positive force is upward). Thus, in this problem, the 1000-N downward force on node 1 corresponds to F1,υ =−1000. For this case all other Fi,υ ’s and Fi,h’s are zero. Note that the directions of the internal forces and reactions are unknown. Proper applica- tion of Newton’s laws requires only consistent assumptions regarding direction. Solutions are negative if the directions are assumed incorrectly. Also note that in this problem, the forces in all members are assumed to be in tension and act to pull adjoining nodes together. A negative solution therefore corresponds to compression. When the external forces are substituted and the trigonometric functions evaluated, this problem reduces to a set of six linear algebraic equations with six unknowns. (a) Solve for the forces and reactions for the case displayed

in Fig. P11.20. (b) Determine the system’s matrix inverse. What is your

interpretation of the zeros in the second row of the inverse?

(i) If the force at node 1 was reversed (i.e., directed up- ward), compute the impact on H2 and V2.

(ii) If the force at node 1 was set to zero and horizon- tal forces of 1500 N were applied at nodes 1 and 2 (Fi,h = F2,h = 1500), what would be the vertical reac- tion at node 3 (V3).

11.21 Employing the same approach as in Prob. 11.20, (a) Compute the forces and reactions for the members and

supports for the truss depicted in Fig. P11.21. (b) Compute the matrix inverse. (c) Determine the change in the reactions at the two sup-

ports if the force at the peak is directed upward.

1200

600

500

45°45°30°

FIGURE P11.21

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305

CHAPTER OBJECTIVES

The primary objective of this chapter is to acquaint you with iterative methods for solving simultaneous equations. Specific objectives and topics covered are

• Understanding the difference between the Gauss-Seidel and Jacobi methods. • Knowing how to assess diagonal dominance and knowing what it means. • Recognizing how relaxation can be used to improve the convergence of iterative

methods. • Understanding how to solve systems of nonlinear equations with successive

substitution, Newton-Raphson, and the MATLAB fsolve function.

Iterative or approximate methods provide an alternative to the elimination methods described to this point. Such approaches are similar to the techniques we developed to obtain the roots of a single equation in Chaps. 5 and 6. Those approaches consisted of guessing a value and then using a systematic method to obtain a refined estimate of the root. Because the present part of the book deals with a similar problem—obtaining the values that simultaneously satisfy a set of equations—we might suspect that such approxi- mate methods could be useful in this context. In this chapter, we will present approaches for solving both linear and nonlinear simultaneous equations.

12.1 LINEAR SYSTEMS: GAUSS-SEIDEL

The Gauss-Seidel method is the most commonly used iterative method for solving linear algebraic equations. Assume that we are given a set of n equations:

[A]{x} = {b}

Suppose that for conciseness we limit ourselves to a 3 × 3 set of equations. If the diagonal elements are all nonzero, the first equation can be solved for x1, the second for x2, and the

Iterative Methods

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306 IteratIve Methods

third for x3 to yield

x 1 j =

b1 − a12 x 2 j−1 − a13 x 3

j−1 ___________________ a11 (12.1a)

x 2 j =

b2 − a21 x 1

j − a23 x 3 j−1 __________________

a22 (12.1b)

x 3 j =

b3 − a31 x 1 j −

a32 x 2

j _________________

a33 (12.1c)

where j and j − 1 are the present and previous iterations. To start the solution process, initial guesses must be made for the x’s. A simple ap-

proach is to assume that they are all zero. These zeros can be substituted into Eq. (12.1a), which can be used to calculate a new value for x1 = b1∕a11. Then we substitute this new value of x1 along with the previous guess of zero for x3 into Eq. (12.1b) to compute a new value for x2. The process is repeated for Eq. (12.1c) to calculate a new estimate for x3. Then we return to the first equation and repeat the entire procedure until our solution converges closely enough to the true values. Convergence can be checked using the criterion that for all i,

εa,i = | x i j − x i j−1 _________ x i j | × 100% ≤ εs (12.2) eXaMPLe 12.1 Gauss-seidel Method

Problem statement. Use the Gauss-Seidel method to obtain the solution for

3 x1 − 0.1 x2 − 0.2 x3 = 7.85 0.1 x1 + 7x2 − 0.3 x3 = −19.3 0.3 x1 − 0.2 x2 + 10 x3 = 71.4

Note that the solution is x1 = 3, x2 = −2.5, and x3 = 7.

solution. First, solve each of the equations for its unknown on the diagonal:

x1 = 7.85 + 0.1x2 + 0.2x3 _____________________ 3 (E12.1.1)

x2 = −19.3 − 0.1x1 + 0.3x3 ______________________ 7 (E12.1.2)

x3 = 71.4 − 0.3x1 + 0.2x2 _____________________ 10 (E12.1.3)

By assuming that x2 and x3 are zero, Eq. (E12.1.1) can be used to compute

x1 = 7.85 + 0.1(0) + 0.2(0) ___________________ 3 = 2.616667

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12.1 LINear sYsteMs: GaUss-seIdeL 307

This value, along with the assumed value of x3 = 0, can be substituted into Eq. (E12.1.2) to calculate

x2 = −19.3 − 0.1(2.616667) + 0.3(0) __________________________ 7 = −2.794524

The first iteration is completed by substituting the calculated values for x1 and x2 into Eq. (E12.1.3) to yield

x3 = 71.4 − 0.3(2.616667) + 0.2(−2.794524) _________________________________ 10 = 7.005610

For the second iteration, the same process is repeated to compute

x1 = 7.85 + 0.1(−2.794524) + 0.2(7.005610) _________________________________ 3 = 2.990557

x2 = −19.3 − 0.1(2.990557) + 0.3(7.005610) _________________________________ 7 = −2.499625

x3 = 71.4 − 0.3(2.990557) + 0.2(−2.499625) _________________________________ 10 = 7.000291

The method is, therefore, converging on the true solution. Additional iterations could be applied to improve the answers. However, in an actual problem, we would not know the true answer a priori. Consequently, Eq. (12.2) provides a means to estimate the error. For example, for x1:

εa,1 = | 2.990557 − 2.616667 __________________ 2.990557 | × 100% = 12.5% For x2 and x3, the error estimates are εa,2 = 11.8% and εa,3 = 0.076%. Note that, as was the case when determining roots of a single equation, formulations such as Eq. (12.2) usually provide a conservative appraisal of convergence. Thus, when they are met, they ensure that the result is known to at least the tolerance specified by εs.

As each new x value is computed for the Gauss-Seidel method, it is immediately used in the next equation to determine another x value. Thus, if the solution is converging, the best available estimates will be employed. An alternative approach, called Jacobi itera- tion, utilizes a somewhat different tactic. Rather than using the latest available x’s, this technique uses Eq. (12.1) to compute a set of new x’s on the basis of a set of old x’s. Thus, as new values are generated, they are not immediately used but rather are retained for the next iteration.

The difference between the Gauss-Seidel method and Jacobi iteration is depicted in Fig. 12.1. Although there are certain cases where the Jacobi method is useful, Gauss- Seidel’s utilization of the best available estimates usually makes it the method of preference.

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308 IteratIve Methods

12.1.1 Convergence and Diagonal Dominance

Note that the Gauss-Seidel method is similar in spirit to the technique of simple fixed-point iteration that was used in Sec. 6.1 to solve for the roots of a single equation. Recall that simple fixed-point iteration was sometimes nonconvergent. That is, as the iterations pro- gressed, the answer moved farther and farther from the correct result.

Although the Gauss-Seidel method can also diverge, because it is designed for linear systems, its ability to converge is much more predictable than for fixed-point iteration of nonlinear equations. It can be shown that if the following condition holds, Gauss-Seidel will converge:

|aii| > ∑ j=1

n |ai j| (12.3)

That is, the absolute value of the diagonal coefficient in each of the equations must be larger than the sum of the absolute values of the other coefficients in the equation. Such systems are said to be diagonally dominant. This criterion is sufficient but not necessary for convergence. That is, although the method may sometimes work if Eq. (12.3) is not met, convergence is guaranteed if the condition is satisfied. Fortunately, many engineer- ing and scientific problems of practical importance fulfill this requirement. Therefore, Gauss-Seidel represents a feasible approach to solve many problems in engineering and science.

j≠i

x1 = (b1 − a12x2 − a13x3)∕a11

x2 = (b2 − a21x1 − a23x3)∕a22

x3 = (b3 − a31x1 − a32x2)∕a33

x1 = (b1 − a12x2 − a13x3)∕a11

x2 = (b2 − a21x1 − a23x3)∕a22

x3 = (b3 − a31x1 − a32x2)∕a33

x2 = (b2 − a21x1 − a23x3)∕a22

First iteration

Second iteration

(a) (b)

FIGURE 12.1 Graphical depiction of the difference between (a) the Gauss-Seidel and (b) the Jacobi iterative methods for solving simultaneous linear algebraic equations.

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12.1 LINear sYsteMs: GaUss-seIdeL 309

12.1.2 MATLAB M-file: GaussSeidel

Before developing an algorithm, let us first recast Gauss-Seidel in a form that is com- patible with MATLAB’s ability to perform matrix operations. This is done by expressing Eq. (12.1) as

x 1 new =

b1 ___ a11 − a12 ___ a11 x 2

old − a13 ___ a11 x 3

old

x 2 new =

b2 ___ a22 − a21 ___ a22 x 1

new − a23 ___ a22 x 3

old

x 3 new =

b3 ___ a33 − a31 ___ a33 x 1

new − a32 ___ a33 x 2

new

Notice that the solution can be expressed concisely in matrix form as

{x} = {d} − [C ]{x} (12.4)

where

{d} = ⎧

⎪

⎨ ⎪

⎩ b1∕a11

b2∕a22 b3∕a33

⎫

⎪

⎬ ⎪

⎭

and

[C ] = ⎡ ⎢

⎣

a12∕a11

a13∕a11 a21∕a22 0 a23∕a22

a31∕a33

a32∕a33

⎤ ⎥

⎦

An M-file to implement Eq. (12.4) is listed in Fig. 12.2.

12.1.3 Relaxation

Relaxation represents a slight modification of the Gauss-Seidel method that is designed to enhance convergence. After each new value of x is computed using Eq. (12.1), that value is modified by a weighted average of the results of the previous and the present iterations:

x i new = λx i

new + (1 − λ) x i old (12.5)

where λ is a weighting factor that is assigned a value between 0 and 2. If λ = 1, (1 − λ) is equal to 0 and the result is unmodified. However, if λ is set at a

value between 0 and 1, the result is a weighted average of the present and the previous re- sults. This type of modification is called underrelaxation. It is typically employed to make a nonconvergent system converge or to hasten convergence by dampening out oscillations.

For values of λ from 1 to 2, extra weight is placed on the present value. In this instance, there is an implicit assumption that the new value is moving in the correct direction toward the true solution but at too slow a rate. Thus, the added weight of λ is intended to improve the estimate by pushing it closer to the truth. Hence, this type of modification, which is called overrelaxation, is designed to accelerate the convergence of an already convergent system. The approach is also called successive overrelaxation, or SOR.

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310 IteratIve Methods

The choice of a proper value for λ is highly problem-specific and is often determined empirically. For a single solution of a set of equations it is often unnecessary. However, if the system under study is to be solved repeatedly, the efficiency introduced by a wise choice of λ can be extremely important. Good examples are the very large systems of linear algebraic equations that can occur when solving partial differential equations in a variety of engineering and scientific problem contexts.

function x = GaussSeidel(A,b,es,maxit) % GaussSeidel: Gauss Seidel method % x = GaussSeidel(A,b): Gauss Seidel without relaxation % input: % A = coefficient matrix % b = right hand side vector % es = stop criterion (default = 0.00001%) % maxit = max iterations (default = 50) % output: % x = solution vector

if nargin<2,error('at least 2 input arguments required'),end if nargin<4|isempty(maxit),maxit = 50;end if nargin<3|isempty(es),es = 0.00001;end [m,n] = size(A); if m~ = n, error('Matrix A must be square'); end C = A; for i = 1:n C(i,i) = 0; x(i) = 0; end x = x'; for i = 1:n C(i,1:n) = C(i,1:n)/A(i,i); end for i = 1:n d(i) = b(i)/A(i,i); end iter = 0; while (1) xold = x; for i = 1:n x(i) = d(i) − C(i,:)*x; if x(i) ~ = 0 ea(i) = abs((x(i) − xold(i))/x(i)) * 100; end end iter = iter+1; if max(ea)<=es | iter >= maxit, break, end end

FIGURE 12.2 MATLAB M-file to implement Gauss-Seidel.

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12.1 LINear sYsteMs: GaUss-seIdeL 311

eXaMPLe 12.2 Gauss-seidel Method with relaxation

Problem statement. Solve the following system with Gauss-Seidel using overrelaxation (λ = 1.2) and a stopping criterion of εs = 10%:

−3x1 + 12x2 = 9

10x1 − 2x2 = 8

solution. First rearrange the equations so that they are diagonally dominant and solve the first equation for x1 and the second for x2:

x1 = 8 + 2x2 _______ 10 = 0.8 + 0.2x2

x2 = 9 + 3x1 _______ 12 = 0.75 + 0.25x1

First iteration: Using initial guesses of x1 = x2 = 0, we can solve for x1:

x1 = 0.8 + 0.2(0) = 0.8

Before solving for x2, we first apply relaxation to our result for x1:

x1, r = 1.2(0.8) − 0.2(0) = 0.96

We use the subscript r to indicate that this is the “relaxed” value. This result is then used to compute x2:

x2 = 0.75 + 0.25(0.96) = 0.99

We then apply relaxation to this result to give

x2, r = 1.2(0.99) − 0.2(0) = 1.188

At this point, we could compute estimated errors with Eq. (12.2). However, since we started with assumed values of zero, the errors for both variables will be 100%.

Second iteration: Using the same procedure as for the first iteration, the second iteration yields

x1 = 0.8 + 0.2(1.188) = 1.0376 x1,r = 1.2(1.0376) − 0.2(0.96) = 1.05312

εa,1 = | 1.05312 − 0.96 __________________ 1.05312 | × 100% = 8.84% x2 = 0.75 + 0.25(1.05312) = 1.01328

x2,r = 1.2(1.01328) − 0.2(1.188) = 0.978336

εa,2 = | 0.978336 − 1.188 __________________ 0.978336 | × 100% = 21.43% Because we have now have nonzero values from the first iteration, we can compute ap- proximate error estimates as each new value is computed. At this point, although the error estimate for the first unknown has fallen below the 10% stopping criterion, the second has not. Hence, we must implement another iteration.

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312 IteratIve Methods

Third iteration:

x1 = 0.8 + 0.2(0.978336) = 0.995667 x1,r = 1.2(0.995667) − 0.2(1.05312) = 0.984177

εa,1 = | 0.984177 − 1.05312 __________________ 0.984177 | × 100% = 7.01% x2 = 0.75 + 0.25(0.984177) = 0.996044

x2,r = 1.2(0.996044) − 0.2(0.978336) = 0.999586

εa,2 = | 0.999586 − 0.978336 __________________ 0.999586 | × 100% = 2.13% At this point, we can terminate the computation because both error estimates have fallen below the 10% stopping criterion. The results at this juncture, x1 = 0.984177 and x2 = 0.999586, are converging on the exact solution of x1 = x2 = 1.

12.2 NONLINEAR SYSTEMS

The following is a set of two simultaneous nonlinear equations with two unknowns:

x 1 2 + x1x2 = 10 (12.6a)

x2 + 3x1 x 2 2 = 57 (12.6b)

In contrast to linear systems which plot as straight lines (recall Fig. 9.1), these equations plot as curves on an x2 versus x1 graph. As in Fig. 12.3, the solution is the intersection of the curves.

0 0

1 2 3 x1

x2 + 3x1x22 = 57

x12 + x1x2 = 10

Solution x1 = 2, x2 = 3

FIGURE 12.3 Graphical depiction of the solution of two simultaneous nonlinear equations.

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12.2 NoNLINear sYsteMs 313

Just as we did when we determined roots for single nonlinear equations, such systems of equations can be expressed generally as

f1 (x1, x2, . . . , xn ) = 0 f2 (x1, x2, . . . , xn ) = 0 ...

(12.7)

fn (x1, x2, . . . , xn ) = 0

Therefore, the solution are the values of the x’s that make the equations equal to zero.

12.2.1 Successive Substitution

A simple approach for solving Eq. (12.7) is to use the same strategy that was employed for fixed-point iteration and the Gauss-Seidel method. That is, each one of the nonlinear equa- tions can be solved for one of the unknowns. These equations can then be implemented iteratively to compute new values which (hopefully) will converge on the solutions. This approach, which is called successive substitution, is illustrated in the following example.

eXaMPLe 12.3 successive substitution for a Nonlinear system

Problem statement. Use successive substitution to determine the roots of Eq. (12.6). Note that a correct pair of roots is x1 = 2 and x2 = 3. Initiate the computation with guesses of x1 = 1.5 and x2 = 3.5.

solution. Equation (12.6a) can be solved for

x1 = 10 − x 1

2 _______ x2 (E12.3.1)

and Eq. (12.6b) can be solved for

x2 = 57 − 3x1 x 2 2 (E12.3.2)

On the basis of the initial guesses, Eq. (E12.3.1) can be used to determine a new value of x1:

x1 = 10 − (1.5)2 __________ 3.5 = 2.21429

This result and the initial value of x2 = 3.5 can be substituted into Eq. (E12.3.2) to deter- mine a new value of x2:

x2 = 57 − 3(2.21429)(3.5)2 = −24.37516

Thus, the approach seems to be diverging. This behavior is even more pronounced on the second iteration:

x1 = 10 − (2.21429)2 ______________ −24.37516 = −0.20910

x2 = 57 − 3(−0.20910)(−24.37516)2 = 429.709

Obviously, the approach is deteriorating.

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314 IteratIve Methods

Now we will repeat the computation but with the original equations set up in a differ- ent format. For example, an alternative solution of Eq. (12.6a) is

x1 = √ ________

10 − x1x2

and of Eq. (12.6b) is

x2 = √ _______

57 − x2 _______ 3x1

Now the results are more satisfactory:

x1 = √ ________

10 − 1.5(3.5) = 2.17945

x2 = √ _______

57 − 3.5 _______ 3(2.17945) = 2.86051

x1 = √ ________

10 − 2.17945(2.86051) = 1.94053

x2 = √ _______

57 − 2.86051 _______ 3(1.94053) = 3.04955

Thus, the approach is converging on the true values of x1 = 2 and x2 = 3.

The previous example illustrates the most serious shortcoming of successive substitution—that is, convergence often depends on the manner in which the equations are formulated. Additionally, even in those instances where convergence is possible, divergence can occur if the initial guesses are insufficiently close to the true solution. These criteria are so restrictive that fixed-point iteration has limited utility for solving nonlinear systems.

12.2.2 Newton-Raphson

Just as fixed-point iteration can be used to solve systems of nonlinear equations, other open root location methods such as the Newton-Raphson method can be used for the same purpose. Recall that the Newton-Raphson method was predicated on employing the deriva- tive (i.e., the slope) of a function to estimate its intercept with the axis of the independent variable—that is, the root. In Chap. 6, we used a graphical derivation to compute this esti- mate. An alternative is to derive it from a first-order Taylor series expansion:

f (xi+1) = f (xi ) + (xi+1 − xi ) f ́(xi ) (12.8)

where xi is the initial guess at the root and xi+1 is the point at which the slope intercepts the x axis. At this intercept, f (xi+1) by definition equals zero and Eq. (12.8) can be rearranged to yield

xi+1 = xi − f (xi ) _____ f ́(xi )

(12.9)

which is the single-equation form of the Newton-Raphson method. The multiequation form is derived in an identical fashion. However, a multivariable

Taylor series must be used to account for the fact that more than one independent variable

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12.2 NoNLINear sYsteMs 315

contributes to the determination of the root. For the two-variable case, a first-order Taylor series can be written for each nonlinear equation as

f1,i+1 = f1,i + (x1,i+1 − x1,i ) ∂ f1,i ____ ∂x1

+ (x2,i+1 − x2,i ) ∂ f1,i ____ ∂x2

(12.10a)

f2,i+1 = f2,i + (x1,i+1 − x1,i ) ∂ f2,i ____ ∂x1

+ (x2,i+1 − x2,i ) ∂ f2,i ____ ∂x2

(12.10b)

Just as for the single-equation version, the root estimate corresponds to the values of x1 and x2, where f1,i+1 and f2,i+1 equal zero. For this situation, Eq. (12.10) can be rearranged to give

∂ f1,i ____ ∂x1

x1,i+1 + ∂ f1,i ____ ∂x2

x2,i+1 = −f1,i + x1,i ∂ f1,i ____ ∂x1

+ x2,i ∂ f1,i ____ ∂x2

(12.11a)

∂ f2,i ____ ∂x1

x1,i+1 + ∂ f2,i ____ ∂x2

x2,i+1 = −f2,i + x1,i ∂ f2,i ____ ∂x1

+ x2,i ∂ f2,i ____ ∂x2

(12.11b)

Because all values subscripted with i’s are known (they correspond to the latest guess or approximation), the only unknowns are x1,i+1 and x2,i+1. Thus, Eq. (12.11) is a set of two lin- ear equations with two unknowns. Consequently, algebraic manipulations (e.g., Cramer’s rule) can be employed to solve for

x1,i+1 = x1,i − f1,i

∂ f2,i ____ ∂x2

− f2,i ∂ f1,i ____ ∂x2

__________________

∂ f1,i ____ ∂x1

∂ f2,i ____ ∂x2

− ∂ f1,i ____ ∂x2

∂ f2,i ____ ∂x1

(12.12a)

x2,i+1 = x2,i − f2,i

∂ f1,i ____ ∂x1

− f1,i ∂ f2,i ____ ∂x1

__________________

∂ f1,i ____ ∂x1

∂ f2,i ____ ∂x2

− ∂ f1,i ____ ∂x2

∂ f2,i ____ ∂x1

(12.12b)

The denominator of each of these equations is formally referred to as the determinant of the Jacobian of the system.

Equation (12.12) is the two-equation version of the Newton-Raphson method. As in the following example, it can be employed iteratively to home in on the roots of two simul taneous equations.

eXaMPLe 12.4 Newton-raphson for a Nonlinear system

Problem statement. Use the multiple-equation Newton-Raphson method to determine roots of Eq. (12.6). Initiate the computation with guesses of x1 = 1.5 and x2 = 3.5.

solution. First compute the partial derivatives and evaluate them at the initial guesses of x and y:

∂ f1,0 ____ ∂x1

= 2x1 + x2 = 2(1.5) + 3.5 = 6.5 ∂ f1,0 ____ ∂x2

= x1 = 1.5

∂ f2,0 ____ ∂x1

= 3 x 2 2 = 3(3.5)2 = 36.75

∂ f2,0 ____ ∂x2

= 1 + 6x1x2 = 1 + 6(1.5)(3.5) = 32.5

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316 IteratIve Methods

Thus, the determinant of the Jacobian for the first iteration is

6.5(32.5) − 1.5(36.75) = 156.125

The values of the functions can be evaluated at the initial guesses as

f1,0 = (1.5)2 + 1.5(3.5) − 10 = −2.5

f2,0 = 3.5 + 3(1.5)(3.5)2 − 57 = 1.625

These values can be substituted into Eq. (12.12) to give

x1 = 1.5 − −2.5(32.5) − 1.625(1.5) ____________________ 156.125 = 2.03603

x2 = 3.5 − 1.625(6.5) − (−2.5)(36.75) ______________________ 156.125 = 2.84388

Thus, the results are converging to the true values of x1 = 2 and x2 = 3. The computation can be repeated until an acceptable accuracy is obtained.

When the multiequation Newton-Raphson works, it exhibits the same speedy quadratic convergence as the single-equation version. However, just as with successive substitution, it can diverge if the initial guesses are not sufficiently close to the true roots. Whereas graphical methods could be employed to derive good guesses for the single-equation case, no such simple procedure is available for the multiequation version. Although there are some advanced approaches for obtaining acceptable first estimates, often the initial guesses must be obtained on the basis of trial and error and knowledge of the physical system being modeled.

The two-equation Newton-Raphson approach can be generalized to solve n simultane- ous equations. To do this, Eq. (12.11) can be written for the kth equation as

∂ fk,i ____ ∂x1

x1,i+1 + ∂ fk,i ____ ∂x2

x2,i+1 + · · · + ∂ fk,i ____ ∂xn

xn,i+1 = −f k,i + x1,i ∂ fk,i ____ ∂x1

+ x2,i ∂ fk,i ____ ∂x2

+ · · · + xn,i ∂ fk,i ____ ∂xn

(12.13)

where the first subscript k represents the equation or unknown and the second subscript denotes whether the value or function in question is at the present value (i) or at the next value (i + 1). Notice that the only unknowns in Eq. (12.13) are the xk,i+1 terms on the left-hand side. All other quantities are located at the present value (i) and, thus, are known at any iteration. Consequently, the set of equations generally represented by Eq. (12.13) (i.e., with k = 1, 2, . . . , n) constitutes a set of linear simultaneous equa- tions that can be solved numerically by the elimination methods elaborated in previous chapters.

Matrix notation can be employed to express Eq. (12.13) concisely as

[J ]{xi+1} = −{ f } + [J ]{xi} (12.14)

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12.2 NoNLINear sYsteMs 317

where the partial derivatives evaluated at i are written as the Jacobian matrix consisting of the partial derivatives:

[J ] =

∂fn,i ____ ∂x1

∂fn,i ____ ∂x2

· · · ∂fn,i ____ ∂xn

∂f1,i ____ ∂x1

∂ f1,i ____ ∂x2

· · · ∂ f1,i ____ ∂xn

∂f2,i ____ ∂x1

∂f2,i ____ ∂x2

· · · ∂ f2,i ____ ∂xn

···

··· ···

···

(12.15)

The initial and final values are expressed in vector form as

{xi}T = ⌊ x1,i x2,i · · · xn,i ⌋

and

{xi+1}T = ⌊ x1,i+1 x2,i+1 · · · xn,i+1 ⌋

Finally, the function values at i can be expressed as

{ f }T = ⌊ f1,i f2,i · · · fn,i ⌋

Equation (12.14) can be solved using a technique such as Gauss elimination. This process can be repeated iteratively to obtain refined estimates in a fashion similar to the two-equation case in Example 12.4.

Insight into the solution can be obtained by solving Eq. (12.14) with matrix inversion. Recall that the single-equation version of the Newton-Raphson method is

xi+1 = xi − f (xi ) _____ f ́(xi )

(12.16)

If Eq. (12.14) is solved by multiplying it by the inverse of the Jacobian, the result is

{xi+1} = {xi} − [J ]−1{f } (12.17)

Comparison of Eqs. (12.16) and (12.17) clearly illustrates the parallels between the two equations. In essence, the Jacobian is analogous to the derivative of a multivariate function.

Such matrix calculations can be implemented very efficiently in MATLAB. We can illustrate this by using MATLAB to duplicate the calculations from Example 12.4. After defining the initial guesses, we can compute the Jacobian and the function values as

>> x = [1.5;3.5]; >> J = [2*x(1)+x(2) x(1);3*x(2)^2 1+6*x(1)*x(2)]

J = 6.5000 1.5000 36.7500 32.5000

>> f = [x(1)^2 + x(1)*x(2) − 10;x(2)+3*x(1)*x(2)^2 − 57]

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318 IteratIve Methods

f = −2.5000 1.6250

Then, we can implement Eq. (12.17) to yield the improved estimates

>> x = x−J\f

x = 2.0360 2.8439

Although we could continue the iterations in the command mode, a nicer alternative is to express the algorithm as an M-file. As in Fig. 12.4, this routine is passed an M-file that computes the function values and the Jacobian at a given value of x. It then calls this func- tion and implements Eq. (12.17) in an iterative fashion. The routine iterates until an upper limit of iterations (maxit) or a specified percent relative error (es) is reached.

function [x,f,ea,iter]=newtmult(func,x0,es,maxit,varargin) % newtmult: Newton-Raphson root zeroes nonlinear systems % [x,f,ea,iter] = newtmult(func,x0,es,maxit,p1,p2,...): % uses the Newton-Raphson method to find the roots of % a system of nonlinear equations % input: % func = name of function that returns f and J % x0 = initial guess % es = desired percent relative error (default = 0.0001%) % maxit = maximum allowable iterations (default = 50) % p1,p2,... = additional parameters used by function % output: % x = vector of roots % f = vector of functions evaluated at roots % ea = approximate percent relative error (%) % iter = number of iterations

if nargin<2,error('at least 2 input arguments required'),end if nargin<3|isempty(es),es = 0.0001;end if nargin<4|isempty(maxit),maxit = 50;end iter = 0; x = x0; while (1) [J,f] = func(x,varargin{:}); dx = J\f; x = x − dx; iter = iter + 1; ea=100*max(abs(dx./x)); if iter> = maxit|ea< = es, break, end end

FIGURE 12.4 MATLAB M-file to implement Newton-Raphson method for nonlinear systems of equations.

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12.2 NoNLINear sYsteMs 319

We should note that there are two shortcomings to the foregoing approach. First, Eq. (12.15) is sometimes inconvenient to evaluate. Therefore, variations of the Newton- Raphson approach have been developed to circumvent this dilemma. As might be expected, most are based on using finite-difference approximations for the partial derivatives that comprise [J ]. The second shortcoming of the multiequation Newton-Raphson method is that excellent initial guesses are usually required to ensure convergence. Because these are sometimes difficult or inconvenient to obtain, alternative approaches that are slower than Newton-Raphson but which have better convergence behavior have been developed. One approach is to reformulate the nonlinear system as a single function:

F(x) = ∑ i=1

[ fi (x1, x2, . . . , xn)]2

where fi (x1, x2, . . . , xn) is the ith member of the original system of Eq. (12.7). The values of x that minimize this function also represent the solution of the nonlinear system. Therefore, nonlinear optimization techniques can be employed to obtain solutions.

12.2.3 MATLAB Function: fsolve

The fsolve function solves systems of nonlinear equations with several variables. A general representation of its syntax is

[x, fx] = fsolve(function, x0, options)

where [x, fx] = a vector containing the roots x and a vector containing the values of the functions evaluated at the roots, function = the name of the function containing a vec- tor holding the equations being solved, x0 is a vector holding the initial guesses for the unknowns, and options is a data structure created by the optimset function. Note that if you desire to pass function parameters but not use the options, pass an empty vector [] in its place.

The optimset function has the syntax

options = optimset('par1',val1,'par2',val2,...)

where the parameter pari has the value vali. A complete listing of all the possible param- eters can be obtained by merely entering optimset at the command prompt. The parameters commonly used with the fsolve function are

display: When set to 'iter' displays a detailed record of all the iterations. tolx: A positive scalar that sets a termination tolerance on x. tolfun: A positive scalar that sets a termination tolerance on fx.

As an example, we can solve the system from Eq. (12.6)

f (x1, x2 ) = 2x1 + x1x2 − 10

f (x1, x2 ) = x2 + 3x1 x 2 2 − 57

First, set up a function to hold the equations

function f = fun(x) f = [x(1)^2+x(1)*x(2) − 10;x(2)+3*x(1)*x(2)^2 − 57];

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320 IteratIve Methods

A script can then be used to generate the solution,

clc, format compact [x,fx] = fsolve(@fun,[1.5;3.5])

with the result

x = 2.0000 3.0000 fx = 1.0e−13 * 0 0.1421

12.3 CASE STUDY CheMICaL reaCtIoNs

Background. Nonlinear systems of equations occur frequently in the characterization of chemical reactions. For example, the following chemical reactions take place in a closed system:

2A + B → ← C (12.18)

A + D → ← C (12.19)

At equilibrium, they can be characterized by

K1 = cc ____

c a 2 cb

(12.20)

K2 = cc ____ cacd (12.21)

where the nomenclature ci represents the concentration of constituent i. If x1 and x2 are the number of moles of C that are produced due to the first and second reactions, respectively, formulate the equilibrium relationships as a pair of two simultaneous nonlinear equations. If K1 = 4 × 10−4, K2 = 3.7 × 10−2, ca,0 = 50, cb,0 = 20, cc,0 = 5, and cd,0 = 10, employ the Newton-Raphson method to solve these equations.

Solution. Using the stoichiometry of Eqs. (12.18) and (12.19), the concentrations of each constituent can be represented in terms of x1 and x2 as

ca = ca,0 − 2x1 − x2 (12.22)

cb = cb,0 − x1 (12.23)

cc = cc,0 + x1 + x2 (12.24)

cd = cd,0 − x2 (12.25)

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12.3 Case stUdY 321

12.3 CASE STUDY continued

where the subscript 0 designates the initial concentration of each constituent. These values can be substituted into Eqs. (12.20) and (12.21) to give

K1 = (cc,0 + x1 + x2) ______________________

(ca,0 − 2x1 − x2)2(cb,0 − x1)

K2 = (cc,0 + x1 + x2) ______________________ (ca,0 − 2x1 − x2)(cd,0 − x2)

Given the parameter values, these are two nonlinear equations with two unknowns. Thus, the solution to this problem involves determining the roots of

f1(x1, x2) = 5 + x1 + x2 _____________________

(50 − 2x1 − x2)2(20 − x1) − 4 × 10−4 (12.26)

f2(x1, x2) = 5 + x1 + x2 _____________________ (50 − 2x1 − x2)(10 − x2)

− 3.7 × 10−2 (12.27)

In order to use Newton-Raphson, we must determine the Jacobian by taking the partial derivatives of Eqs. (12.26) and (12.27). Although this is certainly possible, evaluating the derivatives is time consuming. An alternative is to represent them by finite differences in a fashion similar to the approach used for the modified secant method in Sec. 6.3. For example, the partial derivatives comprising the Jacobian can be evaluated as

∂ f1 ___ ∂x1

= f1(x1 + δx1, x2) − f1(x1, x2) _____________________

δx1

∂ f1 ___ ∂x2

= f1(x1, x2 + δx2) − f1(x1, x2) _____________________

δx2

∂ f2 ___ ∂x1

= f2(x1 + δx1, x2) − f2(x1, x2) _____________________

δx1

∂ f2 ___ ∂x2

= f2(x1, x2 + δx2) − f2(x1, x2) _____________________

δx2

These relationships can then be expressed as an M-file to compute both the function values and the Jacobian as

function [J,f]=jfreact(x,varargin) del = 0.000001; df1dx1 = (u(x(1)+del*x(1),x(2))−u(x(1),x(2)))/(del*x(1)); df1dx2 = (u(x(1),x(2)+del*x(2))−u(x(1),x(2)))/(del*x(2)); df2dx1 = (v(x(1)+del*x(1),x(2))−v(x(1),x(2)))/(del*x(1)); df2dx2 = (v(x(1),x(2)+del*x(2))−v(x(1),x(2)))/(del*x(2)); J=[df1dx1 df1dx2;df2dx1 df2dx2]; f1=u(x(1),x(2)); f2=v(x(1),x(2)); f=[f1;f2];

function f = u(x,y) f = (5 + x + y) / (50 − 2 * x − y) ^ 2 / (20 − x) − 0.0004;

function f = v(x,y) f = (5 + x + y) / (50 − 2 * x − y) / (10 − y) − 0.037;

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322 IteratIve Methods

12.3 CASE STUDY continued

The function newtmult (Fig. 12.4) can then be employed to determine the roots given initial guesses of x1 = x2 = 3:

>>> format short e, x0 =[3; 3]; >> [x,f,ea,iter] = newtmult(@jfreact,x0)

x = 3.3366e+000 2.6772e+000

f = −7.1286e−017 8.5973e−014

ea = 5.2237e-010

iter = 4

After four iterations, a solution of x1 = 3.3366 and x2 = 2.6772 is obtained. These values can then be substituted into Eq. (12.22) through (12.25) to compute the equilibrium con- centrations of the four constituents:

ca = 50 − 2(3.3366) − 2.6772 = 40.6496 cb = 20 − 3.3366 = 16.6634 cc = 5 + 3.3366 + 2.6772 = 11.0138 cd = 10 − 2.6772 = 7.3228

Finally, the fsolve function can also be used to obtain the solution by first writing a MATLAB file function to hold the system of nonlinear equations as a vector

function F = myfun(x) F = [(5+x(1)+x(2))/(50−2*x(1)−x(2))^2/(20−x(1))−0.0004;... (5+x(1)+x(2))/(50−2*x(1)−x(2))/(10−x(2))−0.037];

The solution can then be generated by

[x,fx] = fsolve(@myfun,[3;3])

with the result

x = 3.3372 2.6834 fx = 1.0e−04 * 0.0041 0.6087

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ProbLeMs 323

PROBLEMS

12.1 Solve the following system using three iterations with Gauss-Seidel using overrelaxation (λ = 1.25). If necessary, rearrange the equations and show all the steps in your solu- tion including your error estimates. At the end of the compu- tation, compute the true error of your final results.

3x1 + 8x2 = 11 7x1 − x2 = 5

12.2 (a) Use the Gauss-Seidel method to solve the follow- ing system until the percent relative error falls below εs = 5%:

[

0.8

−0.4

−0.4 0.8 −0.4

−0.4

0.8 ]

{

x2 x3

}

= {

25 105

}

(b) Repeat (a) but use overrelaxation with λ = 1.2. 12.3 Use the Gauss-Seidel method to solve the following system until the percent relative error falls below εs = 5%:

10x1 + 2x2 − x3 = 27

−3x1 − 6x2 + 2x3 = −61.5

x1 + x2 + 5x3 = −21.5

12.4 Repeat Prob. 12.3 but use Jacobi iteration. 12.5 The following system of equations is designed to deter- mine concentrations (the c’s in g/m3) in a series of coupled reactors as a function of the amount of mass input to each reactor (the right-hand sides in g/day):

15c1 − 3c2 − c3 = 3800

−3c1 + 18c2 − 6c3 = 1200

−4c1 − c2 + 12c3 = 2350

Solve this problem with the Gauss-Seidel method to εs = 5%. 12.6 Use the Gauss-Seidel method (a) without relaxation and (b) with relaxation (λ = 1.2) to solve the following sys- tem to a tolerance of εs = 5%. If necessary, rearrange the equations to achieve convergence.

2x1 − 6x2 − x3 = −38

−3x1 − x2 + 7x3 = −34

−8x1 + x2 − 2x3 = −20

12.7 Of the following three sets of linear equations, identify the set(s) that you could not solve using an iterative method such as Gauss-Seidel. Show using any number of itera- tions that is necessary that your solution does not converge. Clearly state your convergence criteria (how you know it is not converging).

Set One Set Two Set Three

8x + 3y + z = 13 x + y + 6z = 8 −3x + 4y + 5z = 6 −6x + 8z = 2 x + 5y − z = 5 −2x + 2y − 3z = −3 2x + 5y − z = 6 4x + 2y − 2z = 4 2y − z = 1

12.8 Determine the solution of the simultaneous nonlinear equations

y = −x2 + x + 0.75 y + 5xy = x2

Use the Newton-Raphson method and employ initial guesses of x = y = 1.2. 12.9 Determine the solution of the simultaneous nonlinear equations:

x2 = 5 − y2

y + 1 = x2

(a) Graphically. (b) Successive substitution using initial guesses of x =

y = 1.5. (c) Newton-Raphson using initial guesses of x = y = 1.5. 12.10 Figure P12.10 depicts a chemical exchange process consisting of a series of reactors in which a gas flowing from left to right is passed over a liquid flowing from right to left. The transfer of a chemical from the gas into the liquid occurs at a rate that is proportional to the difference between the gas and liquid concentrations in each reactor. At steady state, a mass balance for the first reactor can be written for the gas as

QGcG0 − QGcG1 + D(cL1 − cG1) = 0

and for the liquid as

QLcL2 − QLcL1 + D(cG1 − cL1) = 0

where QG and QL are the gas and liquid flow rates, respec- tively, and D = the gas-liquid exchange rate. Similar bal- ances can be written for the other reactors. Use Gauss-Seidel without relaxation to solve for the concentrations given the following values: QG = 2, QL = 1, D = 0.8, cG0 = 100, cL6 = 10.

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324 IteratIve Methods

12.11 The steady-state distribution of temperature on a heated plate can be modeled by the Laplace equation:

0 = ∂ 2T ___

∂x2 + ∂

2T ___ ∂y2

If the plate is represented by a series of nodes (Fig. P12.11), centered finite differences can be substituted for the second derivatives, which result in a system of linear algebraic equations. Use the Gauss-Seidel method to solve for the temperatures of the nodes in Fig. P12.11. 12.12 Develop your own M-file function for the Gauss- Seidel method without relaxation based on Fig. 12.2, but

T12

T11

T22

T21

100°C

0°C

75°C 75°C

25°C 25°C

FIGURE P12.11

cG1cG0 cG2 cG3 cG4 QGQG

cG5

cL1 cL2 cL3 cL4 cL5 cL6

FIGURE P12.10

change the first line so that it returns the approximate error and the number of iterations:

function [x,ea,iter] = ... GaussSeidel(A,b,es, maxit)

Test it by duplicating Example 12.1 and then use it to solve Prob. 12.2a. 12.13 Develop your own M-file function for Gauss-Seidel with relaxation. Here is the function’s first line:

function [x,ea,iter] = ... GaussSeidelR(A,b,lambda,es,maxit)

In the event that the user does not enter a value for λ, set the default value as λ = 1.Test it by duplicating Example 12.2 and then use it to solve Prob. 12.2b. 12.14 Develop your own M-file function for the Newton- Raphson method for nonlinear systems of equations based on Fig. 12.4. Test it by solving Example 12.4 and then use it to solve Prob. 12.8. 12.15 Determine the roots of the following simultaneous nonlinear equations using (a) fixed-point iteration, (b) the Newton-Raphson method, and (c) the fsolve function:

y = −x2 + x + 0.75 y + 5xy = x2

Employ initial guesses of x = y = 1.2 and discuss the results. 12.16 Determine the roots of the simultaneous nonlinear equations

(x − 4)2 + (y − 4)2 = 5 x2 + y2 = 16

Use a graphical approach to obtain your initial guesses. De- termine refined estimates with (a) the two-equation Newton- Raphson method and (b) the fsolve function.

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ProbLeMs 325

12.17 Repeat Prob. 12.16 except determine the positive root of

y = x2 + 1 y = 2 cos x

12.18 The following chemical reactions take place in a closed system

2A + B ⇌ C A + D ⇌ C

At equilibrium, they can be characterized by

K1 = cc ____

c a 2 cb

K2 = cc ____ cacd

where the nomenclature ci represents the concentration of constituent i. If x1 and x2 are the number of moles of C that are produced due to the first and second reactions, re- spectively, use an approach to reformulate the equilibrium relationships in terms of the initial concentrations of the constituents. Then, solve the pair of simultaneous nonlinear equations for x1 and x2 if K1 = 4 × 10−4, K2 = 3.7 × 10−2, ca,0 = 50, cb,0 = 20, cc,0 = 5, and cd,0 = 10. (a) Use a graphical approach to develop your initial guesses.

Then use these guesses as the starting point to determine refined estimates with

(b) the Newton-Raphson method, and (c) the fsolve function. 12.19 As previously described in Sec. 5.6, the following system of five nonlinear equations govern the chemistry of rainwater,

K1 = 106 [H+][ HCO 3

− ] ___________ KH p co 2

K2 = [H+][ CO 3

−2 ] ___________ [ HCO 3

− ] Kw = [H +][OH−]

cT = KH p co 2 ______

106 + [ HCO 3

− ] + [ CO 3 −2 ]

0 = [ HCO 3 − ] + 2 [ CO 3

−2 ] + [OH−] + [H+]

where KH = Henry’s constant, K1, K2, and Kw = equilib- rium coefficients, cT = total inorganic carbon, [ HCO 3

− ] = bicarbonate, [ CO 3

−2 ] = carbonate, [H+] = hydrogen ion, and [OH−] = hydroxyl ion. Notice how the partial pressure of CO2 shows up in the equations indicating the impact of this greenhouse gas on the acidity of rain. Use these equations and the fsolve function to compute the pH of rainwater given that KH = 10−1.46, K1 = 10−6.3, K2 = 10−10.3, and Kw = 10−14. Compare the results in 1958 when the p CO 2 was 315 and in 2015 when it was about 400 ppm. Note that this is a difficult problem to solve because the concentrations tend to be very small and vary over many orders of magnitude. Therefore, it is useful to use the trick based on express- ing the unknowns in a negative log scale, pK = −log10(K). That is, the five unknowns, cT = total inorganic carbon, = bicarbonate, carbonate, [H+], [OH−], [ HCO 3

− ], [ CO 3 −2 ] and

cT can be reexpressed as the unknowns pH, pOH, pHCO3, pCO3, and pcT as in

[H+] = 10−pH [OH−] = 10−pOH

[ HCO 3 − ] = 1 0− pHCO 3

[ CO 3 −2 ] = 1 0− pCO 3 cT = 10−pcT

In addition, it is helpful to use optimset to set a stringent criterion for the function tolerance as in the following script which you can use to generate your solutions

clc, format compact xguess = [7;7;3;7;3]; options = optimset('tolfun',1e−12) [x1,fx1] = fsolve(@funpH,xguess,options,315); [x2,fx2] = fsolve(@funpH,xguess,options,400); x1',fx1',x2',fx2'

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326

13 Eigenvalues

YOU’VE GOT A PROBLEM

At the beginning of Chap. 8, we used Newton’s second law and force balances to pre-dict the equilibrium positions of three bungee jumpers connected by cords. Because we assumed that the cords behaved like ideal springs (i.e., followed Hooke’s law), the steady-state solution reduced to solving a system of linear algebraic equations [recall Eq. (8.1) and Example 8.2]. In mechanics, this is referred to as a statics problem.

Now let’s look at a dynamics problem involving the same system. That is, we’ll study the jumpers’ motion as a function of time. To do this, their initial conditions (i.e., their initial po sitions and velocities) must be prescribed. For example, we can set the jumpers’ initial positions at the equilibrium values computed in Example 8.2. If we then set their ini- tial velocities to zero, nothing would happen because the system would be at equilibrium.

Because we are now interested in examining the system’s dynamics, we must set the initial conditions to values that induce motion. Although we set the jumpers’ initial posi- tions to the equilibrium values and the middle jumper’s initial velocity to zero, we set the upper and bottom jumper’s initial velocities to some admittedly extreme values. That is, we

CHAPTER OBJECTIVES

The primary objective of this chapter is to introduce you to eigenvalues. Specific objectives and topics covered are

• Understanding the mathematical definition of eigenvalues and eigenvectors. • Understanding the physical interpretation of eigenvalues and eigenvectors within

the context of engineering systems that vibrate or oscillate. • Knowing how to implement the polynomial method. • Knowing how to implement the power method to evaluate the largest and smallest

eigenvalues and their respective eigenvectors. • Knowing how to use and interpret MATLAB’s eig function.

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327

impose a downward velocity of 200 m/s on jumper 1 and an upward velocity of 100 m/s on jumper 3. (Safety tip: Don’t try this at home!) We then used MATLAB to solve the dif- ferential equations [Eq. (8.1)] to generate the resulting positions and velocities as a function of time.1

As displayed in Fig. 13.1, the outcome is that the jumpers oscillate wildly. Because there are no friction forces (e.g., no air drag or spring dampening), they lurch up and down around their equilibrium positions in a persistent manner that at least visually borders on the chaotic. Closer inspection of the individual trajectories suggests that there may be some pattern to the oscillations. For example, the distances between peaks and troughs might be constant. But when viewed as a time series, it is difficult to perceive whether there is any- thing systematic and predictable going on.

In this chapter, we deal with one approach for extracting something fundamental out of such seemingly chaotic behavior. This entails determining the eigenvalues, or characteristic values, for such systems. As we will see, this involves formulating and solving systems of linear algebraic equations in a fashion that differs from what we’ve done to this point. To do this, let’s first describe exactly what is meant by eigenvalues from a mathematical standpoint.

FIGURE 13.1 The (a) positions and (b) velocities versus time for the system of three interconnected bungee jumpers from Example 8.2.

1 We will show how this is done when we cover ordinary differential equations in Part Six.

EIGENVALUES 327

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328 EIGENVALUES

13.1 MATHEMATICAL BACKGROUND

Chapters 8 through 12 have dealt with methods for solving sets of linear algebraic equa- tions of the general form

[A]{x} = {b} (13.1)

Such systems are called nonhomogeneous because of the presence of the vector {b} on the right-hand side of the equality. If the equations comprising such a system are linearly independent (i.e., have a nonzero determinant), they will have a unique solution. In other words, there is one set of x values that will make the equations balance. As we’ve already seen in Sec. 9.1.1, for two equations with two unknowns, the solution can be visualized as the intersection of two straight lines represented by the equations (recall Fig. 9.1).

In contrast, a homogeneous linear algebraic system has a right-hand side equal to zero:

[A]{x} = 0 (13.2)

At face value, this equation suggests that the only possible solution would be the trivial case for which all x’s = 0. Graphically this would correspond to two straight lines that intersected at zero.

Although this is certainly true, eigenvalue problems associated with engineering are typically of the general form

[[A] − λ[ I ]] {x} = 0 (13.3)

where the parameter λ is the eigenvalue. Thus, rather than setting the x’s to zero, we can determine the value of λ that drives the left-hand side to zero! One way to accomplish this is based on the fact that, for nontrivial solutions to be possible, the determinant of the matrix must equal zero:

|[A] − λ[ I ]| = 0 (13.4) Expanding the determinant yields a polynomial in λ, which is called the characteristic polynomial. The roots of this polynomial are the solutions for the eigenvalues.

In order to better understand these concepts, it is useful to examine the two-equation case,

(a11 − λ)x1 + a12 x2 = 0 a21x1 + (a22 − λ) x2 = 0

(13.5)

Expanding the determinant of the coefficient matrix gives

| a11 − λ a12 a21 a22 − λ | = λ2 − (a11 + a22)λ − a12a21 (13.6) which is the characteristic polynomial. The quadratic formula can then be used to solve for the two eigenvalues:

λ1 λ2

= (a11 − a22) ± √

__________________ (a11 − a22)2 − 4a12 a21 _______________________________ 2 (13.7)

These are the values that solve Eq. (13.5). Before proceeding, let’s convince ourselves that this approach (which, by the way, is called the polynomial method) is correct.

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13.1 MATHEMATICAL BACKGROUND 329

EXAMPLE 13.1 The Polynomial Method

Problem Statement. Use the polynomial method to solve for the eigenvalues of the fol- lowing homogeneous system:

(10 − λ)x1 − 5x2 = 0 −5x1 + (10 − λ)x2 = 0

Solution. Before determining the correct solution, let’s first investigate the case where we have an incorrect eigenvalue. For example, if λ = 3, the equations become

7x1 − 5x2 = 0 −5x1 + 7x2 = 0

Plotting these equations yields two straight lines that intersect at the origin (Fig. 13.2a). Thus, the only solution is the trivial case where x1 = x2 = 0.

To determine the correct eigenvalues, we can expand the determinant to give the char- acteristic polynomial:

| 10 − λ −5 −5 10 − λ | = λ2 − 20λ + 75 which can be solved for

λ1 λ2

= 20 ± √

__________________ 20 2 − 4(1)75 _______________________________ 2 = 15, 5

Therefore, the eigenvalues for this system are 15 and 5. We can now substitute either of these values back into the system and examine the

result. For λ1 = 15, we obtain

−5x1 − 5x2 = 0 −5x1 − 5x2 = 0

Thus, a correct eigenvalue makes the two equations identical (Fig. 13.2b). In essence as we move toward a correct eigenvalue the two lines rotate until they lie on top of each other. Mathematically, this means that there are an infinite number of solutions. But solving either of the equations yields the interesting result that all the solutions have the property that x1 = −x2. Although at first glance this might appear trivial, it’s actually quite interest- ing as it tells us that the ratio of the unknowns is a constant. This result can be expressed in vector form as

{x} = { −1 1 }

which is referred to as the eigenvector corresponding to the eigenvalue λ = 15. In a similar fashion, substituting the second eigenvalue, λ2 = 5, gives

5x1 − 5x2 = 0 −5x1 + 5x2 = 0

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330 EIGENVALUES

Again, the eigenvalue makes the two equations identical (Fig. 13.2b) and we can see that the solution for this case corresponds to x1 = x2, and the eigenvector is

{x} = { 1 1 }

We should recognize that MATLAB has built-in functions to facilitate the polynomial method. For Example 13.1, the poly function can be used to generate the characteristic polynomial as in

>> A = [10 −5;−5 10]; >> p = poly(A)

p = 1 −20 75

Then, the roots function can be employed to compute the eigenvalues: >> d = roots(p)

d = 15 5

The previous example yields the useful mathematical insight that the solution of n homogeneous equations of the form of Eq. (13.3) consists of a set of n eigenvalues and their associated eigenvectors. Further, it showed that the eigenvectors provide the ratios of the unknowns representing the solution.

(a) Incorrect eigenvalue (b) Correct eigenvalues

1 2

Eqs. 1 & 2 (λ = 5)

Eqs. 1 & 2 (λ = 15)

−2 −1

−1

−2

1 2

Eq. 2 (λ = 3)

Eq. 1 (λ = 3)

−2

−1

−2

−1

FIGURE 13.2 Plots of a system of two homogeneous linear equations from Example 13.1. (a) An incorrect ei- genvalue (λ = 3) means that the two equations, which are labeled as Eqs. 1 and 2 in the figure, plot as separate lines and the only solution is the trivial case (x1 = x2 = 0). (b) In contrast, the cases with correct eigenvalues (λ = 5 and 15), the equations fall on top of each other.

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13.2 PHYSICAL BACKGROUND 331

In the next section, we will show how such information has utility in engineering and science by turning back to our physical problem setting of oscillating objects. However, before doing so, we’d like to make two more mathematical points.

First, inspection of Fig. 13.2b indicates that the straight lines representing each eigen- value solution are at right angles to each other. That is, they are orthogonal. This property is true for symmetric matrices with distinct eigenvalues.

Second, multiplying out Eq. (13.3) and separating terms gives

[A]{x} = λ{x}

When viewed in this way, we can see that solving for the eigenvalues and eigenvectors amounts to translating the information content of a matrix [A] into a scalar λ. This might not seem significant for the 2 × 2 system we have been examining, but it is pretty remark- able when we consider that the size of [A] can potentially be much larger.

13.2 PHYSICAL BACKGROUND

The mass-spring system in Fig. 13.3a is a simple context to illustrate how eigenvalues occur in physical problem settings. It will also help to demonstrate some of the mathematical concepts introduced in the previous section.

To simplify the analysis, assume that each mass has no external or damping forces acting on it. In addition, assume that each spring has the same natural length l and the same spring constant k. Finally, assume that the displacement of each spring is measured rela- tive to its own local coordinate system with an origin at the spring’s equilibrium position (Fig. 13.3a). Under these assumptions, Newton’s second law can be employed to develop a force balance for each mass:

m1 d 2x1 ____ dt 2

= −kx1 + k(x2 − x1) (13.8a)

(a) 0

0 x1 0 x2

(b)

m1 m2

FIGURE 13.3 A two mass–three spring system with frictionless rollers vibrating between two fixed walls. The position of the masses can be referenced to local coordinates with origins at their respective equilibrium positions (a). As in (b), positioning the masses away from equilibrium creates forces in the springs that on release lead to oscillations of the masses.

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332 EIGENVALUES

m2 d 2x2 ____ dt 2

= −k(x2 − x1) − kx2 (13.8b)

where xi is the displacement of mass i away from its equilibrium position (Fig. 13.3b). From vibration theory, it is known that solutions to Eq. (13.8) can take the form

xi = Xi sin(ωt) (13.9)

where Xi = the amplitude of the oscillation of mass i (m) and ω = the angular frequency of the oscillation (radians/time), which is equal to

ω = 2π ___ Tp (13.10)

where Tp = the period (time/cycle). Note that the inverse of the period is called the ordi- nary frequency f (cycles/time). If time is measured in seconds, the unit for f is the cycles/s, which is referred to as a Hertz (Hz).

Equation (13.9) can be differentiated twice and substituted into Eq. (13.8). After col- lection of terms, the result is

( 2k ___ m1 − ω2 ) X1 − k ___ m1 X2 = 0 (13.11a) − k ___ m2 X1 + ( 2k ___ m2 − ω2 ) X2 = 0 (13.11b) Comparison of Eq. (13.11) with Eq. (13.3) indicates that at this point, the solution has

been reduced to an eigenvalue problem—where, for this case, the eigenvalue is the square of the frequency. For a two-degree-of-freedom system such as Fig. 13.3, there will be two such values along with their eigenvectors. As shown in the following example, the latter establish the unique relationship between the unknowns.

EXAMPLE 13.2 Physical Interpretation of Eigenvalues and Eigenvectors

Problem Statement. If m1 = m2 = 40 kg and k = 200 N/m, Eq. (13.11) is

(10 − λ)x1 − 5x2 = 0 −5x1 + (10 − λ)x2 = 0

Mathematically, this is the same system we already solved with the polynomial methods in Example 13.2. Thus, the two eigenvalues are ω2 = 15 and 5 s−2 and the corresponding eigenvectors are X1 = X2 and X1 = −X2. Interpret these results as they relate to the mass- spring system of Fig. 13.3.

Solution. This example provides valuable information regarding the behavior of the sys- tem in Fig. 13.3. First, it tells us that the system has two primary modes of oscillation with angular frequencies of ω = 3.873 and 2.36 radians s−1, respectively. These values can be also expressed as periods (1.62 and 2.81 s, respectively) or ordinary frequencies (0.6164 and 0.3559 Hz, respectively). As stated in Sec. 13.1, a unique set of values cannot be obtained for the unknown amplitudes X. However, their ratios are specified by the eigenvectors. Thus, if the system

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13.3 THE POWER METHOD 333

is vibrating in the first mode, the first eigenvector tells us that the amplitude of the second mass will be equal but of opposite sign to the amplitude of the first. As in Fig. 13.4a, the masses vibrate apart and then together indefinitely (like two hands clapping every 1.62 s). In the second mode, the eigenvector specifies that the two masses have equal amplitudes at all times. Thus, as in Fig. 13.4b, they vibrate back and forth in unison every 2.81 s. We should note that the configuration of the amplitudes provides guidance on how to set their initial values to attain pure motion in either of the two modes. Any other configuration will lead to superposition of the modes. It is this superposition that leads to the apparently cha- otic behavior of systems like the bungee jumpers in Fig. 13.1. But as this example should make clear, there is an underlying systematic behavior that is embodied by the eigenvalues.

13.3 THE POWER METHOD

The power method is an iterative approach that can be employed to determine the largest or dominant eigenvalue. With slight modification, it can also be employed to determine the smallest value. It has the additional benefit that the corresponding eigenvector is obtained as a by-product of the method. To implement the power method, the system being analyzed is expressed in the form

[A]{x} = λ{x} (13.12)

TP = 1.62

TP = 2.81

(a) First mode (b) Second mode

FIGURE 13.4 The principal modes of vibration of two equal masses connected by three identical springs between fixed walls.

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334 EIGENVALUES

As illustrated by the following example, Eq. (13.12) forms the basis for an itera- tive solution technique that eventually yields the highest eigenvalue and its associated eigenvector.

EXAMPLE 13.3 Power Method for Highest Eigenvalue

Problem Statement. Using the same approach as in Sec. 13.2, we can derive the follow- ing homogeneous set of equations for a three mass–four spring system between two fixed walls:

( 2k ___ m1 − ω2 ) X1 − k ___ m1 X2 = 0 − k ___ m2 X1 + ( 2k ___ m2 − ω2 ) X2 − k ___ m2 X3 = 0

− k ___ m3 X2 + ( k ___ m3 − ω2 ) X3 = 0 If all the masses m = 1 kg and all the spring constants k = 20 N/m, the system can be expressed in the matrix format of Eq. (13.4) as

[

−20

0 −20 40 −20

−20

40 ]

− λ[ I ] = 0

where the eigenvalue λ is the square of the angular frequency ω2. Employ the power method to determine the highest eigenvalue and its associated eigenvector.

Solution. The system is first written in the form of Eq. (13.12):

40X1 − 20X2 = λX1 −20X1 + 40X2 − 20X3 = λX2

−20X2 + 40X3 = λX3

At this point, we can specify initial values of the X’s and use the left-hand side to compute an eigenvalue and eigenvector. A good first choice is to assume that all the X’s on the left- hand side of the equation are equal to one:

40(1) − 20(1) = 20 −20(1) + 40(1) − 20(1) = 0

− 20(1) + 40(1) = 20

Next, the right-hand side is normalized by 20 to make the largest element equal to one:

{

0 20

}

= 20 {

1 0

1 }

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13.3 THE POWER METHOD 335

Thus, the normalization factor is our first estimate of the eigenvalue (20) and the correspond- ing eigenvector is {1 0 1}T. This iteration can be expressed concisely in matrix form as

[

−20

0 −20 40 −20

−20

40 ]

{

1 1

1 }

= {

0 20

}

= 20 {

1 0

1 }

The next iteration consists of multiplying the matrix by the eigenvector from the last itera- tion, {1 0 1}T to give

[

−20

0 −20 40 −20

−20

40 ]

{

1 0

1 }

= {

−40 40

}

= 40 {

1 −1

1 }

Therefore, the eigenvalue estimate for the second iteration is 40, which can be employed to determine an error estimate:

|εa| = | 40 − 20 _______ 40 | × 100% = 50% The process can then be repeated.

Third iteration:

[

−20

0 −20 40 −20

−20

40 ]

{

−1 1

}

= {

−80 60

}

= −80 {

−0.75

1 −0.75

}

where |εa| = 150% (which is high because of the sign change).

Fourth iteration:

[

−20

0 −20 40 −20

−20

40 ]

{

−0.75

1 −0.75

}

= {

−50

70 −50

}

= 70 {

−0.71429

1 −0.71429

}

where |εa| = 214% (another sign change).

Fifth iteration:

[

−20

0 −20 40 −20

−20

40 ]

{

−0.71429

1 −0.71429

}

= {

−48.51714

68.51714 −48.51714

}

= 68.51714 {

−0.70833

1 −0.70833

}

where |εa| = 2.08%. Thus, the eigenvalue is converging. After several more iterations, it stabilizes on a value of 68.28427 with a corresponding eigenvector of {−0.707107 1 −0.707107}T.

Note that there are some instances where the power method will converge to the second- largest eigenvalue instead of to the largest. James, Smith, and Wolford (1985) provide an illustration of such a case. Other special cases are discussed in Fadeev and Fadeeva (1963).

In addition, there are sometimes cases where we are interested in determining the smallest eigenvalue. This can be done by applying the power method to the matrix inverse of [A]. For this case, the power method will converge on the largest value of 1/ λ—in other

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336 EIGENVALUES

words, the smallest value of λ. An application to find the smallest eigenvalue will be left as a problem exercise.

Finally, after finding the largest eigenvalue, it is possible to determine the next highest by replacing the original matrix by one that includes only the remaining eigenvalues. The process of removing the largest known eigenvalue is called deflation.

We should mention that although the power method can be used to locate intermediate values, better methods are available for cases where we need to determine all the eigen- values as described in the next section. Thus, the power method is primarily used when we want to locate the largest or the smallest eigenvalue.

13.4 MATLAB FUNCTION: eig

As might be expected, MATLAB has powerful and robust capabilities for evaluating ei- genvalues and eigenvectors. The function eig, which is used for this purpose, can be em- ployed to generate a vector of the eigenvalues as in

>> e = eig (A)

where e is a vector containing the eigenvalues of a square matrix A. Alternatively, it can be invoked as

>> [V,D] = eig (A)

where D is a diagonal matrix of the eigenvalues and V is a full matrix whose columns are the corresponding eigenvectors.

It should be noted that MATLAB scales the eigenvectors by dividing them by their Euclidean distance. Thus, as shown in the following example, although their magnitude may be different from values computed with say the polynomial method, the ratio of their elements will be identical.

EXAMPLE 13.4 Eigenvalues and Eigenvectors with MATLAB

Problem Statement. Use MATLAB to determine all the eigenvalues and eigenvectors for the system described in Example 13.3.

Solution. Recall that the matrix to be analyzed is

[

−20

0 −20 40 −20

−20

40 ]

The matrix can be entered as >> A = [40 −20 0;−20 40 −20;0 −20 40];

If we just desire the eigenvalues, we can enter >> e = eig (A)

e = 11.7157 40.0000 68.2843

Notice that the highest eigenvalue (68.2843) is consistent with the value previously deter- mined with the power method in Example 13.3.

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If we want both the eigenvalues and eigenvectors, we can enter >> [v,d] = eig (A)

v = 0.5000 −0.7071 −0.5000 0.7071 −0.0000 0.7071 0.5000 0.7071 −0.5000

d = 11.7157 0 0 0 40.0000 0 0 0 8.2843

Although the results are scaled differently, the eigenvector corresponding to the high- est eigenvalue {−0.5 0.7071 −0.5}T is consistent with the value previously determined with the power method in Example 13.3: {−0.707107 1 −0.707107}T. The can be demon- strated by dividing the eigenvector from the power method by its Euclidean norm: >> vpower = [−0.7071 1 −0.7071]'; >> vMATLAB = vpower/norm (vpower)

vMATLAB = −0.5000 0.7071 −0.5000

Thus, although the magnitudes of the elements differ, their ratios are identical.

13.5 CASE STUDY EIGENVALUES AND EARTHQUAKES

Background. Engineers and scientists use mass-spring models to gain insight into the dy- namics of structures under the influence of disturbances such as earthquakes. Figure 13.5 shows such a model for a three-story building. Each floor mass is represented by mi, and each floor stiffness is represented by ki for i = 1 to 3.

m3 = 8000 kg

k3 = 1800 kN/m

k2 = 2400 kN/m

k1 = 3000 kN/m

m2 = 10000 kg

m1 = 12000 kg

FIGURE 13.5 A three-story building modeled as a mass-spring system.

13.5 CASE STUDY 337

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338 EIGENVALUES

For this case, the analysis is limited to horizontal motion of the structure as it is subjected to horizontal base motion due to earthquakes. Using the same approach as developed in Sec. 13.2, dynamic force balances can be developed for this system as

( k1 + k2 ______ m1 − ω n 2 ) X1 − k2 ___ m1 X2 = 0 −

k2 ___ m2 X1 + ( k2 + k3 ______ m2 − ω n 2 ) X2 − k3 ___ m2 X3 = 0 −

k3 ___ m3 X2 + ( k3 ______ m3 − ω n 2 ) X3 = 0 where Xi represent horizontal floor translations (m), and ωn is the natural, or resonant, fre- quency (radians/s). The resonant frequency can be expressed in Hertz (cycles/s) by dividing it by 2π radians/cycle.

Use MATLAB to determine the eigenvalues and eigenvectors for this system. Graphi- cally represent the modes of vibration for the structure by displaying the amplitudes versus height for each of the eigenvectors. Normalize the amplitudes so that the translation of the third floor is one.

Solution. The parameters can be substituted into the force balances to give

(450 − ω n 2 ) X1 − 200 X2 = 0 −240 X1 + (420 − ω n 2 ) X2 − 180 X3 = 0

−225 X2 + (225 − ω n 2 ) X3 = 0

A MATLAB session can be conducted to evaluate the eigenvalues and eigenvectors as

>> A = [450 −200 0;−240 420 −180;0 −225 225]; >> [v,d] = eig(A)

v = −0.5879 −0.6344 0.2913 0.7307 −0.3506 0.5725 −0.3471 0.6890 0.7664

d = 698.5982 0 0 0 339.4779 0 0 0 56.9239

Therefore, the eigenvalues are 698.6, 339.5, and 56.92 and the resonant frequencies in Hz are

>> wn = sqrt(diag(d))'/2/pi wn = 4.2066 2.9324 1.2008

13.5 CASE STUDY continued

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The corresponding eigenvectors are (normalizing so that the amplitude for the third floor is one)

{

1.6934

−2.1049 1

} {

−0.9207

−0.5088 1

} {

0.3801

0.7470 1

}

A graph can be made showing the three modes (Fig. 13.6). Note that we have ordered them from the lowest to the highest natural frequency as is customary in structural engineering.

Natural frequencies and mode shapes are characteristics of structures in terms of their tendencies to resonate at these frequencies. The frequency content of an earthquake typically has the most energy between 0 and 20 Hz and is influenced by the earthquake magnitude, the epicentral distance, and other factors. Rather than a single frequency, they contain a spectrum of all frequencies with varying amplitudes. Buildings are more receptive to vibration at their lower modes of vibrations due to their simpler deformed shapes and requiring less strain energy to deform in the lower modes. When these amplitudes coincide with the natural frequencies of buildings, large dynamic responses are induced, creating large stresses and strains in the structure’s beams, columns, and foundations. Based on anal- yses like the one in this case study, structural engineers can more wisely design buildings to withstand earthquakes with a good factor of safety.

Mode 1 (ωn = 1.2008 Hz)

Mode 2 (ωn = 2.9324 Hz)

Mode 3 (ωn = 4.2066 Hz)

1 0 01 1 2−1 −2 −1

FIGURE 13.6 The three primary modes of oscillation of the three-story building.

Mode 1 (ωn = 1.2008 Hz)

Mode 2 (ωn = 2.9324 Hz)

Mode 3 (ωn = 4.2066 Hz)

1 0 01 1 2−1 −2 −1

FIGURE 13.6 The three primary modes of oscillation of the three-story building.

Mode 1 (ωn = 1.2008 Hz)

Mode 2 (ωn = 2.9324 Hz)

Mode 3 (ωn = 4.2066 Hz)

1 0 01 1 2−1 −2 −1

FIGURE 13.6 The three primary modes of oscillation of the three-story building.

13.5 CASE STUDY 339

13.5 CASE STUDY continued

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340 EIGENVALUES

PROBLEMS

13.1 Repeat Example 13.1 but for three masses with the m’s = 40 kg and the k’s = 240 N/m. Produce a plot like Fig. 13.4 to identify the principle modes of vibration. 13.2 Use the power method to determine the highest eigen- value and corresponding eigenvector for

[

2 − λ

10 8 4 − λ 5

7 − λ ]

13.3 Use the power method to determine the lowest eigen- value and corresponding eigenvector for the system from Prob. 13.2. 13.4 Derive the set of differential equations for a three mass– four spring system (Fig. P13.4) that describes their time mo- tion. Write the three differential equations in matrix form

{Acceleration vector} + [k/m matrix] {displacement vector x} = 0

Note each equation has been divided by the mass. Solve for the eigenvalues and natural frequencies for the follow- ing values of mass and spring constants: k1 = k4 = 15 N/m, k2 = k3 = 35 N/m, and m1 = m2 = m3 = 1.5 kg. 13.5 Consider the mass-spring system in Fig. P13.5. The fre- quencies for the mass vibrations can be determined by solving for the eigenvalues and by applying M ẍ + kx = 0, which yields

⎡ ⎢

⎣ m1

0 0 m2 0

m3 ⎤ ⎥

⎦ ⎧

⎪

⎨ ⎪

⎩ ẍ 1

ẍ 2 ẍ 3

⎫

⎪

⎬ ⎪

⎭ +

{ 2k

−k

−k 2k −k −k

−k

}

{

x2 x3

}

= {

0 0

0 }

Applying the guess x = x0eiωt as a solution, we get the fol- lowing matrix:

⎡ ⎢

⎣ 2k − m1ω2

−k

−k 2k − m2ω2 −k −k

−k

2k − m3ω2

⎤ ⎥

⎦ {

x01

x02 x03

}

eiωt = {

0 0

0 }

Use MATLAB’s eig command to solve for the eigenvalues of the k − mω2 matrix above. Then use these eigenvalues to solve for the frequencies (ω). Let m1 = m2 = m3 = 1 kg, and k = 2 N/m. 13.6 As displayed in Fig. P13.6, an LC circuit can be mod- eled by the following system of differential equations:

L1 d 2i1 ____ dt 2

+ 1 ___ C1

(i1 − i2) = 0

L2 d 2i2 ____ dt 2

+ 1 ___ C2

(i2 − i3) − 1 ___ C1

(i1 − i2) = 0

L3 d 2i3 ____ dt 2

+ 1 ___ C3

i3 − 1 ___ C2

(i2 − i3) = 0

where L = inductance (H), t = time (s), i = current (A), and C = capacitance (F). Assuming that a solution is of the form ij = Ij sin (ωt), determine the eigenvalues and eigenvectors for this system with L = 1 H and C = 0.25C. Draw the network, illustrating how the currents oscillate in their primary modes. 13.7 Repeat Prob. 13.6 but with only two loops. That is, omit the i3 loop. Draw the network, illustrating how the cur- rents oscillate in their primary modes. 13.8 Repeat the problem in Sec. 13.5 but leave off the third floor. 13.9 Repeat the problem in Sec. 13.5 but add a fourth floor with m4 = 6000 and k4 = 1200 kN/m.

k1 k2 k3 k4

x2 x3

m1 m2 m3

FIGURE P13.4

x1 x2 x3

m1 m2 m3

FIGURE P13.5

C1 C2 C3

L2 L3

−

+ i1 i2 i3

FIGURE P13.6

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13.10 The curvature of a slender column subject to an axial load P (Fig. P13.10) can be modeled by

d 2y ___

dx 2 + p2y = 0

where

p2 = P ___ E I

where E = the modulus of elasticity, and I = the moment of inertia of the cross section about its neutral axis. This model can be converted into an eigenvalue problem by substituting a centered finite-difference approximation for the second derivative to give

yi+1 − 2yi + yi−1 _______________

Δx 2 + p2 yi = 0

where i = a node located at a position along the rod’s inte- rior, and Δx = the spacing between nodes. This equation can be expressed as

yi−1 − (2 − Δx2 p2)yi + yi+1 = 0

Writing this equation for a series of interior nodes along the axis of the column yields a homogeneous system of equa- tions. For example, if the column is divided into five seg- ments (i.e., four interior nodes), the result is

An axially loaded wooden column has the following charac- teristics: E = 10 × 109 Pa, I = 1.25 × 10−5 m4, and L = 3 m. For the five-segment, four-node representation: (a) Implement the polynomial method with MATLAB to

determine the eigenvalues for this system. (b) Use the MATLAB eig function to determine the eigen-

values and eigenvectors. (c) Use the power method to determine the largest eigen-

value and its corresponding eigenvector. 13.11 A system of two homogeneous linear ordinary differ- ential equations with constant coefficients can be written as

dy1 ___ dt

= −5y1 + 3y2, y1(0) = 50

dy2 ___ dt = 100y1 − 301y2, y2(0) = 100

If you have taken a course in differential equations, you know that the solutions for such equations have the form

yi = ci eλt

where c and λ are constants to be determined. Substitut- ing this solution and its derivative into the original equa- tions converts the system into an eigenvalue problem. The resulting eigenvalues and eigenvectors can then be used to derive the general solution to the differential equations. For example, for the two-equation case, the general solution can be written in terms of vectors as

{y} = c1{υ1}e λ 1 t + c2{υ2}e λ 2 t

where {υi} = the eigenvector corresponding to the ith eigen- value (λi) and the c’s are unknown coefficients that can be determined with the initial conditions. (a) Convert the system into an eigenvalue problem. (b) Use MATLAB to solve for the eigenvalues and

eigenvectors. (c) Employ the results of (b) and the initial conditions to

determine the general solution. (d) Develop a MATLAB plot of the solution for t = 0

to 1. 13.12 Water flows between the North American Great Lakes as depicted in Fig. P13.12. Based on mass balances, the following differential equations can be written for the concentrations in each of the lakes for a pollutant that decays with first-order kinetics:

(a) (b)

y y

Pʹ

(0, 0)

(L, 0)

Pʹ

FIGURE P13.10 (a) A slender rod. (b) A freebody diagram of a rod.

⎡

⎢ ⎣

(2 − Δx2 p2)

−1

(2 − Δx2 p2)

−1

(2 − Δx2 p2)

−1

(2 − Δx2 p2)

⎤

⎥ ⎦

y1 y2 y3 y4

= 0

PROBLEMS 341

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342 EIGENVALUES

dc1 ___ dt = −(0.0056 + k)c1

dc2 ___ dt = −(0.01 + k)c2

dc3 ___ dt = 0.01902c1 + 0.01387c2 − (0.047 + k)c3

dc4 ___ dt = 0.33597c3 − (0.376 + k)c4

dc5 ___ dt = 0.11364c4 − (0.133 + k)c5

where k = the first-order decay rate (/yr), which is equal to 0.69315/(half-life). Note that the constants in each of the equations account for the flow between the lakes. Due to the testing of nuclear weapons in the atmosphere, the con- centrations of strontium-90 (90Sr) in the five lakes in 1963 were approximately {c} = {17.7 30.5 43.9 136.3 30.1}T in units of Bq/m3. Assuming that no additional 90Sr entered the system thereafter, use MATLAB and the approach outlined in Prob. 13.11 to compute the concentrations in each of the lakes from 1963 through 2010. Note that 90Sr has a half-life of 28.8 years.

Superior (1)

Michigan (2)

Huron (3)

Erie (4)

Ontario (5)

FIGURE P13.12 The North American Great Lakes. The arrows indicate how water flows between the lakes.

13.13 Develop an M-file function to determine the largest eigenvalue and its associated eigenvector with the power method. Test the program by duplicating Example 13.3 and then use it to solve Prob. 13.2. 13.14 Repeat the computations in Sec. 13.5 but remove the third floor. 13.15 Repeat the computations in Sec. 13.5 but add a fourth floor with a mass of m4 = 6000 kg connected with the third floor by a spring with k4 = 1200 kN/m. 13.16 Recall that at the start of Chap. 8, we suspended three bungee jumpers with zero drag from frictionless cords that followed Hooke’s law. Determine the resulting eigenval- ues and eigenvectors that would eventually characterize the jumpers’ oscillating motions and relative positions if they were instantaneously released from their starting positions as depicted in Fig. 8.1a (i.e., with each cord fully extended, but not stretched). Although bungee cords do not actually behave like true springs, assume that they stretch and com- press in linear proportion to the applied force. Use the pa- rameters from Example 8.2.

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343

Part Four

Curve Fitting

4.1 OVERVIEW

What Is Curve Fitting?

Data are often given for discrete values along a continuum. However, you may require es- timates at points between the discrete values. Chapters 14 through 18 describe techniques to fit curves to such data to obtain intermediate estimates. In addition, you may require a simplified version of a complicated function. One way to do this is to compute values of the function at a number of discrete values along the range of interest. Then, a simpler function may be derived to fit these values. Both of these applications are known as curve fitting.

There are two general approaches for curve fitting that are distinguished from each other on the basis of the amount of error associated with the data. First, where the data exhibit a significant degree of error or “scatter,” the strategy is to derive a single curve that represents the general trend of the data. Because any individual data point may be incor- rect, we make no effort to intersect every point. Rather, the curve is designed to follow the pattern of the points taken as a group. One approach of this nature is called least-squares regression (Fig. PT4.1a).

Second, where the data are known to be very precise, the basic approach is to fit a curve or a series of curves that pass directly through each of the points. Such data usually originate from

tables. Examples are values for the density of water or for the heat capacity of gases as a function of temperature. The estimation of values between well-known discrete points is called interpolation (Fig. PT4.1b and c).

Curve Fitting and Engineering and Science. Your first exposure to curve fitting may have been to determine intermediate values from tabulated data—for instance, from interest ta- bles for engineering eco nomics or from steam tables for thermodynamics. Throughout the re- mainder of your career, you will have frequent occasion to estimate intermediate values from such tables.

Although many of the widely used engineering and scientific properties have been

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344 PART 4 CuRvE FiTTing

tabulated, there are a great many more that are not available in this convenient form. Special cases and new problem contexts often require that you measure your own data and develop your own predictive relationships. Two types of applications are generally encountered when fitting experimental data: trend analysis and hypothesis testing.

Trend analysis represents the process of using the pattern of the data to make predic- tions. For cases where the data are measured with high precision, you might utilize interpo- lating polynomials. Imprecise data are often analyzed with least-squares regression.

Trend analysis may be used to predict or forecast values of the dependent variable. This can involve extrapolation beyond the limits of the observed data or interpolation within the range of the data. All fields of engineering and science involve problems of this type.

A second application of experimental curve fitting is hypothesis testing. Here, an existing mathematical model is compared with measured data. If the model coefficients are unknown, it may be necessary to determine values that best fit the observed data. On the other hand, if estimates of the model coefficients are already available, it may be

FIGURE PT4.1 Three attempts to fit a “best” curve through five data points: (a) least-squares regression, (b) linear interpolation, and (c) curvilinear interpolation.

f (x)

x (a)

(b)

(c)

f (x)

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4.2 PART ORgAniZATiOn 345

appropriate to compare predicted values of the model with observed values to test the ad- equacy of the model. Often, alternative models are compared and the “best” one is selected on the basis of empirical observations.

In addition to the foregoing engineering and scientific applications, curve fitting is important in other numerical methods such as integration and the approximate solution of differential equations. Finally, curve-fitting techniques can be used to derive simple func- tions to approximate complicated functions.

4.2 PART ORGANIZATION

After a brief review of statistics, Chap. 14 focuses on linear regression; that is, how to deter- mine the “best” straight line through a set of uncertain data points. Besides discussing how to calculate the slope and intercept of this straight line, we also present quantitative and visual methods for evaluating the validity of the results. In addition, we describe random number generation as well as several approaches for the linearization of nonlinear equations.

Chapter 15 begins with brief discussions of polynomial and multiple linear regres- sion. Polynomial regression deals with developing a best fit of parabolas, cubics, or higher-order polynomials. This is followed by a description of multiple linear regres- sion, which is designed for the case where the dependent variable y is a linear function of two or more independent variables x1, x2, . . . , xm. This approach has special utility for evaluating experimental data where the variable of interest is dependent on a number of different factors.

After multiple regression, we illustrate how polynomial and multiple regression are both subsets of a general linear least-squares model. Among other things, this will allow us to introduce a concise matrix representation of regression and discuss its general statistical properties. Finally, the last sections of Chap. 15 are devoted to nonlinear regression. This approach is designed to compute a least-squares fit of a nonlinear equation to data.

Chapter 16 deals with Fourier analysis which involves fitting periodic functions to data. Our emphasis will be on the fast Fourier transform or FFT. This method, which is readily implemented with MATLAB, has many engineering applications, ranging from vibration analysis of structures to signal processing.

In Chap. 17, the alternative curve-fitting technique called interpolation is described. As discussed previously, interpolation is used for estimating intermediate values between precise data points. In Chap. 17, polynomials are derived for this purpose. We introduce the basic concept of polynomial interpolation by using straight lines and parabolas to con- nect points. Then, we develop a generalized procedure for fitting an nth-order polynomial. Two formats are presented for expressing these polynomials in equation form. The first, called Newton’s interpolating polynomial, is preferable when the appropriate order of the polynomial is unknown. The second, called the Lagrange interpolating polynomial, has advantages when the proper order is known beforehand.

Finally, Chap. 18 presents an alternative technique for fitting precise data points. This technique, called spline interpolation, fits polynomials to data but in a piecewise fashion. As such, it is particularly well suited for fitting data that are generally smooth but exhibit abrupt local changes. The chapter ends with an overview of how piecewise interpolation is implemented in MATLAB.

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CHAPTER OBJECTIVES

Knowledge and understanding are prerequisites for the effective implementation of any tool.

No matter how impressive your tool chest, you will be hard-pressed to repair a car if you do not understand how it works.

• This is the first chapter objectives entry. • Second chapter objective entry, the entries use ic/lc per manuscript, the first and

last entry have space above or below, middle entries do not. • Third chapter entry copy goes here.

346

Linear Regression

CHAPTER OBJECTIVES

The primary objective of this chapter is to introduce you to how least-squares regression can be used to fit a straight line to measured data. Specific objectives and topics covered are

• Familiarizing yourself with some basic descriptive statistics and the normal distribution.

• Knowing how to compute the slope and intercept of a best-fit straight line with linear regression.

• Knowing how to generate random numbers with MATLAB and how they can be employed for Monte Carlo simulations.

• Knowing how to compute and understand the meaning of the coefficient of determination and the standard error of the estimate.

• Understanding how to use transformations to linearize nonlinear equations so that they can be fit with linear regression.

• Knowing how to implement linear regression with MATLAB.

YOU’VE GOT A PROBLEM

in Chap. 1, we noted that a free-falling object such as a bungee jumper is subject to the upward force of air resistance. As a first approximation, we assumed that this force was proportional to the square of velocity as in FU = cd υ2 (14.1)

where FU = the upward force of air resistance [N = kg m/s2], cd = a drag coefficient (kg/m), and υ = velocity [m/s].

Expressions such as Eq. (14.1) come from the field of fluid mechanics. Although such relationships derive in part from theory, experiments play a critical role in their formula- tion. One such experiment is depicted in Fig. 14.1. An individual is suspended in a wind

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tunnel (any volunteers?) and the force measured for various levels of wind velocity. The result might be as listed in Table 14.1.

The relationship can be visualized by plotting force versus velocity. As in Fig. 14.2, several features of the relationship bear mention. First, the points indicate that the force increases as velocity increases. Second, the points do not increase smoothly, but exhibit rather significant scatter, particularly at the higher velocities. Finally, although it may not be obvious, the relationship between force and velocity may not be linear. This conclusion becomes more apparent if we assume that force is zero for zero velocity.

FIGURE 14.1 Wind tunnel experiment to measure how the force of air resistance depends on velocity.

FIGURE 14.2 Plot of force versus wind velocity for an object suspended in a wind tunnel.

1600

1200

800

400

0 0 20 40 60 80

υ, m/s

F, N

TABLE 14.1 Experimental data for force (n) and velocity (m/s) from a wind tunnel experiment.

υ, m/s 10 20 30 40 50 60 70 80 F, N 25 70 380 550 610 1220 830 1450

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348 LinEAR REgRESSiOn

In Chaps. 14 and 15, we will explore how to fit a “best” line or curve to such data. In so doing, we will illustrate how relationships like Eq. (14.1) arise from experimental data.

14.1 STATISTICS REVIEW

Before describing least-squares regression, we will first review some basic concepts from the field of statistics. These include the mean, standard deviation, residual sum of the squares, and the normal distribution. In addition, we describe how simple descriptive statistics and distributions can be generated in MATLAB. If you are familiar with these subjects, feel free to skip the following pages and proceed directly to Sec. 14.2. If you are unfamiliar with these concepts or are in need of a review, the following material is designed as a brief introduction.

14.1.1 Descriptive Statistics

Suppose that in the course of an engineering study, several measurements were made of a particular quantity. For example, Table 14.2 contains 24 readings of the coefficient of ther- mal expansion of a structural steel. Taken at face value, the data provide a limited amount of information—that is, that the values range from a minimum of 6.395 to a maximum of 6.775. Additional insight can be gained by summarizing the data in one or more well- chosen statistics that convey as much information as possible about specific characteristics of the data set. These descriptive statistics are most often selected to represent (1) the location of the center of the distribution of the data and (2) the degree of spread of the data set.

Measure of Location. The most common measure of central tendency is the arithmetic mean. The arithmetic mean (

_ y ) of a sample is defined as the sum of the individual data

points ( yi ) divided by the number of points (n), or

_ y = Σ yi ____ n (14.2)

where the summation (and all the succeeding summations in this section) is from i = 1 through n.

There are several alternatives to the arithmetic mean. The median is the midpoint of a group of data. It is calculated by first putting the data in ascending order. If the number of measurements is odd, the median is the middle value. If the number is even, it is the arith- metic mean of the two middle values. The median is sometimes called the 50th percentile.

The mode is the value that occurs most frequently. The concept usually has direct util- ity only when dealing with discrete or coarsely rounded data. For continuous variables such as the data in Table 14.2, the concept is not very practical. For example, there are actually

TABLE 14.2 Measurements of the coefficient of thermal expansion of structural steel.

6.495 6.595 6.615 6.635 6.485 6.555 6.665 6.505 6.435 6.625 6.715 6.655 6.755 6.625 6.715 6.575 6.655 6.605 6.565 6.515 6.555 6.395 6.775 6.685

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14.1 STATiSTiCS REviEW 349

four modes for these data: 6.555, 6.625, 6.655, and 6.715, which all occur twice. If the num- bers had not been rounded to 3 decimal digits, it would be unlikely that any of the values would even have repeated twice. However, if continuous data are grouped into equispaced intervals, it can be an informative statistic. We will return to the mode when we describe histograms later in this section.

Measures of Spread. The simplest measure of spread is the range, the difference be- tween the largest and the smallest value. Although it is certainly easy to determine, it is not considered a very reliable measure because it is highly sensitive to the sample size and is very sensitive to extreme values.

The most common measure of spread for a sample is the standard deviation (sy) about the mean:

sy = √ _____

St _____ n − 1 (14.3)

where St is the total sum of the squares of the residuals between the data points and the mean, or

St = Σ ( yi − _ y )2 (14.4)

Thus, if the individual measurements are spread out widely around the mean, St (and, consequently, sy ) will be large. If they are grouped tightly, the standard deviation will be small. The spread can also be represented by the square of the standard deviation, which is called the variance:

s y 2 = Σ ( yi − _ y )2 _________

n − 1 (14.5)

Note that the denominator in both Eqs. (14.3) and (14.5) is n − 1. The quantity n − 1 is referred to as the degrees of freedom. Hence St and sy are said to be based on n − 1 degrees of freedom. This nomenclature derives from the fact that the sum of the quantities upon which St is based (i.e.,

_ y − y1,

_ y − y2, . . . ,

_ y − yn ) is zero. Consequently, if

_ y is known and

n − 1 of the values are specified, the remaining value is fixed. Thus, only n − 1 of the val- ues are said to be freely determined. Another justification for dividing by n − 1 is the fact that there is no such thing as the spread of a single data point. For the case where n = 1, Eqs. (14.3) and (14.5) yield a meaningless result of infinity.

We should note that an alternative, more convenient formula is available to compute the variance:

s y 2 = Σ y i 2 − ( Σ yi ) 2∕n _____________

n − 1 (14.6)

This version does not require precomputation of _ y and yields an identical result as

Eq. (14.5). A final statistic that has utility in quantifying the spread of data is the coefficient of

variation (c.v.). This statistic is the ratio of the standard deviation to the mean. As such, it provides a normalized measure of the spread. It is often multiplied by 100 so that it can be expressed in the form of a percent:

c.v. = sy __ _ y × 100% (14.7)

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350 LinEAR REgRESSiOn

EXAMPLE 14.1 Simple Statistics of a Sample

Problem Statement. Compute the mean, median, variance, standard deviation, and coeffi cient of variation for the data in Table 14.2.

Solution. The data can be assembled in tabular form and the necessary sums computed as in Table 14.3. The mean can be computed as [Eq. (14.2)],

_ y = 158.4 _____ 24 = 6.6

Because there are an even number of values, the median is computed as the arithmetic mean of the middle two values: (6.605 + 6.615)∕2 = 6.61. As in Table 14.3, the sum of the squares of the residuals is 0.217000, which can be used to compute the standard deviation [Eq. (14.3)]:

sy = √ ________

0.217000 ________ 24 − 1 = 0.097133

TABLE 14.3 Data and summations for computing simple descriptive statistics for the coefficients of thermal expansion from Table 14.2.

i yi ( yi − y–)2 y2i

1 6.395 0.04203 40.896 2 6.435 0.02723 41.409 3 6.485 0.01323 42.055 4 6.495 0.01103 42.185 5 6.505 0.00903 42.315 6 6.515 0.00723 42.445 7 6.555 0.00203 42.968 8 6.555 0.00203 42.968 9 6.565 0.00123 43.099

10 6.575 0.00063 43.231 11 6.595 0.00003 43.494 12 6.605 0.00002 43.626 13 6.615 0.00022 43.758 14 6.625 0.00062 43.891 15 6.625 0.00062 43.891 16 6.635 0.00122 44.023 17 6.655 0.00302 44.289 18 6.655 0.00302 44.289 19 6.665 0.00422 44.422 20 6.685 0.00722 44.689 21 6.715 0.01322 45.091 22 6.715 0.01322 45.091 23 6.755 0.02402 45.630 24 6.775 0.03062 45.901

Σ 158.400 0.21700 1045.657

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14.1 STATiSTiCS REviEW 351

the variance [Eq. (14.5)]:

s y 2 = (0.097133)2 = 0.009435

and the coefficient of variation [Eq. (14.7)]:

c.v. = 0.097133 ________ 6.6 × 100% = 1.47%

The validity of Eq. (14.6) can also be verified by computing

s y 2 = 1045.657 − (158.400)2∕24 ______________________ 24 − 1 = 0.009435

14.1.2 The Normal Distribution

Another characteristic that bears on the present discussion is the data distribution—that is, the shape with which the data are spread around the mean. A histogram provides a simple visual representation of the distribution. A histogram is constructed by sorting the mea- surements into intervals, or bins. The units of measurement are plotted on the abscissa and the frequency of occurrence of each interval is plotted on the ordinate.

As an example, a histogram can be created for the data from Table 14.2. The result (Fig. 14.3) suggests that most of the data are grouped close to the mean value of 6.6. Notice also, that now that we have grouped the data, we can see that the bin with the most values is from 6.6 to 6.64. Although we could say that the mode is the midpoint of this bin, 6.62, it is more common to report the most frequent range as the modal class interval.

If we have a very large set of data, the histogram often can be approximated by a smooth curve. The symmetric, bell-shaped curve superimposed on Fig. 14.3 is one such characteristic shape—the normal distribution. Given enough additional measurements, the histogram for this particular case could eventually approach the normal distribution.

Fr eq

ue nc

6.4 6.6 6.8 0

FIGURE 14.3 A histogram used to depict the distribution of data. As the number of data points increases, the histogram often approaches the smooth, bell-shaped curve called the normal distribution.

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352 LinEAR REgRESSiOn

The concepts of the mean, standard deviation, residual sum of the squares, and nor- mal distribution all have great relevance to engineering and science. A very simple exam- ple is their use to quantify the confidence that can be ascribed to a particular measurement. If a quantity is normally distributed, the range defined by

_ y − sy to

_ y + sy will encompass

approximately 68% of the total measurements. Similarly, the range defined by _ y − 2sy

to _ y + 2sy will encompass approximately 95%. For example, for the data in Table 14.2, we calculated in Example 14.1 that

_ y = 6.6

and sy = 0.097133. Based on our analysis, we can tentatively make the statement that approximately 95% of the readings should fall between 6.405734 and 6.794266. Because it is so far outside these bounds, if someone told us that they had measured a value of 7.35, we would suspect that the measurement might be erroneous.

14.1.3 Descriptive Statistics in MATLAB

Standard MATLAB has several functions to compute descriptive statistics.1 For example, the arithmetic mean is computed as mean (x). If x is a vector, the function returns the mean of the vector’s values. If it is a matrix, it returns a row vector containing the arithmetic mean of each column of x. The following is the result of using mean and the other statistical func- tions to analyze a column vector s that holds the data from Table 14.2:

>> format short g >> mean(s),median(s),mode(s)

ans = 6.6 ans = 6.61 ans = 6.555

>> min(s),max(s)

ans = 6.395 ans = 6.775

>> range = max(s) − min(s)

range = 0.38

>> var(s),std(s)

ans = 0.0094348

ans = 0.097133

These results are consistent with those obtained previously in Example 14.1. Note that although there are four values that occur twice, the mode function only returns the first of the values: 6.555.

1 MATLAB also offers a Statistics Toolbox that provides a wide range of common statistical tasks, from random number generation, to curve fitting, to design of experiments and statistical process control.

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14.2 RAnDOM nuMBERS AnD SiMuLATiOn 353

MATLAB can also be used to generate a histogram based on the hist function. The hist function has the syntax

[n, x] = hist (y, x)

where n = the number of elements in each bin, x = a vector specifying the midpoint of each bin, and y is the vector being analyzed. For the data from Table 14.2, the result is

>> [n,x] = hist(s)

n = 1 1 3 1 4 3 5 2 2 2 x = 6.414 6.452 6.49 6.528 6.566 6.604 6.642 6.68 6.718 6.756

The resulting histogram depicted in Fig. 14.4 is similar to the one we generated by hand in Fig. 14.3. Note that all the arguments and outputs with the exception of y are optional. For example, hist (y) without output arguments just produces a histogram bar plot with 10 bins determined automatically based on the range of values in y.

14.2 RANDOM NUMBERS AND SIMULATION

In this section, we will describe two MATLAB functions that can be used to produce a sequence of random numbers. The first (rand) generates numbers that are uniformly distributed, and the second (randn) generates numbers that have a normal distribution.

4.5

3.5

2.5

1.5

0.5

0 6.35 6.4 6.45 6.5 6.55 6.656.6 6.7 6.86.75 6.85

FIGURE 14.4 Histogram generated with the MATLAB hist function.

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354 LinEAR REgRESSiOn

14.2.1 MATLAB Function: rand

This function generates a sequence of numbers that are uniformly distributed between 0 and 1. A simple representation of its syntax is

r = rand(m, n)

where r = an m-by-n matrix of random numbers. The following formula can then be used to generate a uniform distribution on another interval:

runiform = low + (up – low) * rand(m, n)

where low = the lower bound and up = the upper bound.

EXAMPLE 14.2 generating uniform Random values of Drag

Problem Statement. If the initial velocity is zero, the downward velocity of the free- falling bungee jumper can be predicted with the following analytical solution [Eq. (1.9)]:

υ = √ ___

gm ___ cd tanh ( √ ___

gcd ___ m t )

Suppose that g = 9.81m/s2, and m = 68.1 kg, but cd is not known precisely. For example, you might know that it varies uniformly between 0.225 and 0.275 (i.e., ± 10% around a mean value of 0.25 kg/m). Use the rand function to generate 1000 random uniformly distributed values of cd and then employ these values along with the analytical solution to compute the resulting distribution of velocities at t = 4 s.

Solution. Before generating the random numbers, we can first compute the mean velocity:

υmean = √ _________

9.81(68.1) _________ 0.25 tanh ( √ _________

9.81(0.25) _________ 68.1 4 ) = 33.1118 m __ s We can also generate the range:

υlow = √ _________

9.81(68.1) _________ 0.275 tanh ( √ _________

9.81(0.275) _________ 68.1 4 ) = 32.6223 m __ s

υhigh = √ _________

9.81(68.1) _________ 0.225 tanh ( √ _________

9.81(0.225) _________ 68.1 4 ) = 33.6198 m __ s Thus, we can see that the velocity varies by

Δυ = 33.6198 − 32.6223 ________________ 2(33.1118) × 100% = 1.5063%

The following script generates the random values for cd, along with their mean, standard deviation, percent variation, and a histogram:

clc,format short g n = 1000;t = 4;m = 68.1;g = 9.81; cd = 0.25;cdmin = cd−0.025,cdmax = cd + 0.025 r = rand(n,1);

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14.2 RAnDOM nuMBERS AnD SiMuLATiOn 355

cdrand = cdmin+(cdmax − cdmin)*r; meancd = mean(cdrand),stdcd = std(cdrand) Deltacd = (max(cdrand)−min(cdrand))/meancd/2*100. subplot(2,1,1) hist(cdrand),title('(a) Distribution of drag') xlabel('cd (kg/m)')

The results are meancd = 0.25018 stdcd = 0.014528 Deltacd = 9.9762

These results, as well as the histogram (Fig. 14.5a) indicate that rand has yielded 1000 uniformly distributed values with the desired mean value and range. The values can then be employed along with the analytical solution to compute the resulting distribution of velocities at t = 4 s. vrand = sqrt(g*m./cdrand).*tanh(sqrt(g*cdrand/m)*t); meanv = mean(vrand) Deltav = (max(vrand)−min(vrand))/meanv/2*100. subplot(2,1,2) hist(vrand),title('(b) Distribution of velocity') xlabel('v (m/s)')

(b) Distribution of velocity

(a) Distribution of drag

cd (kg/m)

v (m/s)

100

150

0 0.23 0.24 0.25 0.26 0.27 0.280.22

100

150

0 32.6 32.8 33 33.2 33.4 33.6 33.832.4

FIGURE 14.5 Histograms of (a) uniformly distributed drag coefficients and (b) the resulting distribution of velocity.

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356 LinEAR REgRESSiOn

The results are meanv = 33.1151 Deltav = 1.5048

These results, as well as the histogram (Fig. 14.5b), closely conform to our hand calculations.

The foregoing example is formally referred to as a Monte Carlo simulation. The term, which is a reference to Monaco’s Monte Carlo casino, was first used by physicists working on nuclear weapons projects in the 1940s. Although it yields intuitive results for this simple example, there are instances where such computer simulations yield surprising outcomes and provide insights that would otherwise be impossible to determine. The approach is fea- sible only because of the computer’s ability to implement tedious, repetitive computations in an efficient manner.

14.2.2 MATLAB Function: randn

This function generates a sequence of numbers that are normally distributed with a mean of 0 and a standard deviation of 1. A simple representation of its syntax is r = randn(m, n)

where r = an m-by-n matrix of random numbers. The following formula can then be used to generate a normal distribution with a different mean (mn) and standard deviation (s), rnormal = mn + s * randn(m, n)

EXAMPLE 14.3 generating normally Distributed Random values of Drag

Problem Statement. Analyze the same case as in Example 14.2, but rather than employ- ing a uniform distribution, generate normally distributed drag coefficients with a mean of 0.25 and a standard deviation of 0.01443.

Solution. The following script generates the random values for cd, along with their mean, standard deviation, coefficient of variation (expressed as a %), and a histogram: clc,format short g n = 1000;t = 4;m = 68.1;g = 9.81; cd = 0.25; stdev = 0.01443; r = randn(n,1); cdrand = cd+stdev*r; meancd = mean(cdrand),stdevcd = std(cdrand) cvcd = stdevcd/meancd*100. subplot(2,1,1) hist(cdrand),title('(a) Distribution of drag') xlabel('cd (kg/m)')

The results are meancd = 0.24988

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14.2 RAnDOM nuMBERS AnD SiMuLATiOn 357

stdevcd = 0.014465 cvcd = 5.7887

These results, as well as the histogram (Fig. 14.6a) indicate that randn has yielded 1000 uniformly distributed values with the desired mean, standard deviation, and coefficient of variation. The values can then be employed along with the analytical solution to compute the resulting distribution of velocities at t = 4 s. vrand = sqrt(g*m./cdrand).*tanh(sqrt(g*cdrand/m)*t); meanv = mean(vrand),stdevv = std(vrand) cvv = stdevv/meanv*100. subplot(2,1,2) hist(vrand),title('(b) Distribution of velocity') xlabel('v (m/s)')

The results are meanv = 33.117 stdevv = 0.28839 cvv = 0.8708

These results, as well as the histogram (Fig. 14.6b), indicate that the velocities are also normally distributed with a mean that is close to the value that would be computed using the mean and the analytical solution. In addition, we compute the associated standard de- viation which corresponds to a coefficient of variation of ±0.8708%.

(a) Distribution of drag

cd (kg/m)

100

200

300

0 0.2 0.22 0.24 0.26 0.28 0.30.18

(b) Distribution of velocity

v (m/s)

100

200

300

0 32.5 33 33.5 34 34.532

FIGURE 14.6 Histograms of (a) normally distributed drag coefficients and (b) the resulting distribution of velocity.

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358 LinEAR REgRESSiOn

Although simple, the foregoing examples illustrate how random numbers can be easily gen- erated within MATLAB. We will explore additional applications in the end-of-chapter problems.

14.3 LINEAR LEAST-SQUARES REGRESSION

Where substantial error is associated with data, the best curve-fitting strategy is to derive an approximating function that fits the shape or general trend of the data without neces sarily matching the individual points. One approach to do this is to visually inspect the plotted data and then sketch a “best” line through the points. Although such “eyeball” approaches have commonsense appeal and are valid for “back-of-the-envelope” calculations, they are deficient because they are arbitrary. That is, unless the points define a perfect straight line (in which case, interpolation would be appropriate), different analysts would draw differ- ent lines.

To remove this subjectivity, some criterion must be devised to establish a basis for the fit. One way to do this is to derive a curve that minimizes the discrepancy between the data points and the curve. To do this, we must first quantify the discrepancy. The simplest example is fitting a straight line to a set of paired observations: (x1, y1), (x2, y2), . . . , (xn, yn). The mathematical expression for the straight line is

y = a0 + a1x + e (14.8)

where a0 and a1 are coefficients representing the intercept and the slope, respectively, and e is the error, or residual, between the model and the observations, which can be represented by rearranging Eq. (14.8) as

e = y − a0 − a1x (14.9)

Thus, the residual is the discrepancy between the true value of y and the approximate value, a0 + a1x, predicted by the linear equation.

14.3.1 Criteria for a “Best” Fit

One strategy for fitting a “best” line through the data would be to minimize the sum of the residual errors for all the available data, as in

∑ i=1

n ei = ∑

i=1

n (yi − a0 − a1xi ) (14.10)

where n = total number of points. However, this is an inadequate criterion, as illustrated by Fig. 14.7a, which depicts the fit of a straight line to two points. Obviously, the best fit is the line connecting the points. However, any straight line passing through the midpoint of the connecting line (except a perfectly vertical line) results in a minimum value of Eq. (14.10) equal to zero because positive and negative errors cancel.

One way to remove the effect of the signs might be to minimize the sum of the absolute values of the discrepancies, as in

∑ i=1

n |ei | = ∑

i=1

n | yi − a0 − a1xi | (14.11)

Figure 14.7b demonstrates why this criterion is also inadequate. For the four points shown, any straight line falling within the dashed lines will minimize the sum of the absolute val- ues of the residuals. Thus, this criterion also does not yield a unique best fit.

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14.3 LinEAR LEAST-SQuARES REgRESSiOn 359

A third strategy for fitting a best line is the minimax criterion. In this technique, the line is chosen that minimizes the maximum distance that an individual point falls from the line. As depicted in Fig. 14.7c, this strategy is ill-suited for regression because it gives undue influence to an outlier—that is, a single point with a large error. It should be noted that the minimax principle is sometimes well-suited for fitting a simple function to a com- plicated function (Carnahan, Luther, and Wilkes, 1969).

A strategy that overcomes the shortcomings of the aforementioned approaches is to minimize the sum of the squares of the residuals:

Sr = ∑ i=1

n e i

2 = ∑ i=1

n ( yi − a0 − a1xi )2 (14.12)

Midpoint

Outlier

(a)

(b)

(c)

FIGURE 14.7 Examples of some criteria for “best fit” that are inadequate for regression: (a) minimizes the sum of the residuals, (b) minimizes the sum of the absolute values of the residuals, and (c) minimizes the maximum error of any individual point.

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360 LinEAR REgRESSiOn

This criterion, which is called least squares, has a number of advantages, including that it yields a unique line for a given set of data. Before discussing these properties, we will pres- ent a technique for determining the values of a0 and a1 that minimize Eq. (14.12).

14.3.2 Least-Squares Fit of a Straight Line

To determine values for a0 and a1, Eq. (14.12) is differentiated with respect to each unknown coefficient:

∂ Sr ___ ∂ a0

= −2 ∑ ( yi − a0 − a1xi )

∂ Sr ___ ∂ a1

= −2 ∑ [( yi − a0 − a1xi ) xi ]

Note that we have simplified the summation symbols; unless otherwise indicated, all sum- mations are from i = 1 to n. Setting these derivatives equal to zero will result in a minimum Sr. If this is done, the equations can be expressed as

0 = ∑ yi − ∑ a0 − ∑ a1xi

0 = ∑ xi yi − ∑ a0 xi − ∑ a1 x i 2

Now, realizing that ∑ a0 = na0 we can express the equations as a set of two simultaneous linear equations with two unknowns (a0 and a1):

n a0 + ( ∑ xi ) a1 = ∑ yi (14.13)

( ∑ xi ) a0 + ( ∑ x i 2 ) a1 = ∑ xi yi (14.14) These are called the normal equations. They can be solved simultaneously for

a1 = n ∑ xi yi − ∑ xi ∑ yi _________________

n∑ x i 2 − ( ∑ xi ) 2

(14.15)

This result can then be used in conjunction with Eq. (14.13) to solve for

a0 = _ y − a1

_ x (14.16)

where _ y and _ x are the means of y and x, respectively.

EXAMPLE 14.4 Linear Regression

Problem Statement. Fit a straight line to the values in Table 14.1.

Solution. In this application, force is the dependent variable ( y) and velocity is the independent variable (x). The data can be set up in tabular form and the necessary sums computed as in Table 14.4.

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14.3 LinEAR LEAST-SQuARES REgRESSiOn 361

The means can be computed as

_ x = 360 ____ 8 = 45

_ y = 5,135 _____ 8 = 641.875

The slope and the intercept can then be calculated with Eqs. (14.15) and (14.16) as

a1 = 8 (312,850) − 360 (5,135) _____________________

8(20,400) − (360)2 = 19.47024

a0 = 641.875 − 19.47024(45) = −234.2857

Using force and velocity in place of y and x, the least-squares fit is

F = −234.2857 + 19.47024υ

The line, along with the data, is shown in Fig. 14.8.

1600

1200

800

400

−400

0 20 40 60 80

F, N

υ, m/s

FIGURE 14.8 Least-squares fit of a straight line to the data from Table 14.1

TABLE 14.4 Data and summations needed to compute the best-fit line for the data from Table 14.1.

i xi yi x i 2 xiyi

1 10 25 100 250 2 20 70 400 1,400 3 30 380 900 11,400 4 40 550 1,600 22,000 5 50 610 2,500 30,500 6 60 1,220 3,600 73,200 7 70 830 4,900 58,100 8 80 1,450 6,400 116,000

∑ 360 5,135 20,400 312,850

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362 LinEAR REgRESSiOn

Notice that although the line fits the data well, the zero intercept means that the equation predicts physically unrealistic negative forces at low velocities. In Sec. 14.4, we will show how transformations can be employed to derive an alternative best-fit line that is more physically realistic.

14.3.3 Quantification of Error of Linear Regression

Any line other than the one computed in Example 14.4 results in a larger sum of the squares of the residuals. Thus, the line is unique and in terms of our chosen criterion is a “best” line through the points. A number of additional properties of this fit can be elucidated by examining more closely the way in which residuals were computed. Recall that the sum of the squares is defined as [Eq. (14.12)]

Sr = ∑ i=1

n ( yi − a0 − a1xi )2 (14.17)

Notice the similarity between this equation and Eq. (14.4)

St = ∑ ( yi − _ y )2 (14.18)

In Eq. (14.18), the square of the residual represented the square of the discrepancy between the data and a single estimate of the measure of central tendency—the mean. In Eq. (14.17), the square of the residual represents the square of the vertical distance between the data and another measure of central tendency—the straight line (Fig. 14.9).

The analogy can be extended further for cases where (1) the spread of the points around the line is of similar magnitude along the entire range of the data and (2) the distri- bution of these points about the line is normal. It can be demonstrated that if these criteria

xi x

a0 + a1xi

Measurement

yi − a0 − a1xi

Re gre

ssi on

lin e

FIGURE 14.9 The residual in linear regression represents the vertical distance between a data point and the straight line.

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14.3 LinEAR LEAST-SQuARES REgRESSiOn 363

are met, least-squares regression will provide the best (i.e., the most likely) estimates of a0 and a1 (Draper and Smith, 1981). This is called the maximum likelihood principle in statis- tics. In addition, if these criteria are met, a “standard deviation” for the regression line can be determined as [compare with Eq. (14.3)]

sy∕x = √ ______

Sr _____ n − 2 (14.19)

where sy∕x is called the standard error of the estimate. The subscript notation “y∕x” des- ignates that the error is for a predicted value of y corresponding to a particular value of x. Also, notice that we now divide by n − 2 because two data-derived estimates—a0 and a1—were used to compute Sr; thus, we have lost two degrees of freedom. As with our dis- cussion of the standard deviation, another justification for dividing by n − 2 is that there is no such thing as the “spread of data” around a straight line connecting two points. Thus, for the case where n = 2, Eq. (14.19) yields a meaningless result of infinity.

Just as was the case with the standard deviation, the standard error of the estimate quantifies the spread of the data. However, sy∕x quantifies the spread around the regres- sion line as shown in Fig. 14.10b in contrast to the standard deviation sy that quantified the spread around the mean (Fig. 14.10a).

These concepts can be used to quantify the “goodness” of our fit. This is particularly useful for comparison of several regressions (Fig. 14.11). To do this, we return to the original data and determine the total sum of the squares around the mean for the dependent variable (in our case, y). As was the case for Eq. (14.18), this quantity is designated St. This is the magnitude of the residual error associated with the dependent variable prior to regression. After performing the regression, we can compute Sr , the sum of the squares of the residuals around the regression line with Eq. (14.17). This characterizes the residual

(a) (b)

FIGURE 14.10 Regression data showing (a) the spread of the data around the mean of the dependent variable and (b) the spread of the data around the best-fit line. The reduction in the spread in going from (a) to (b), as indicated by the bell-shaped curves at the right, represents the improvement due to linear regression.

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364 LinEAR REgRESSiOn

error that remains after the regression. It is, therefore, sometimes called the unexplained sum of the squares. The difference between the two quantities, St − Sr , quantifies the im- provement or error reduction due to describing the data in terms of a straight line rather than as an average value. Because the magnitude of this quantity is scale-dependent, the difference is normalized to St to yield

r 2 = St − Sr ______ St

(14.20)

where r2 is called the coefficient of determination and r is the correlation coefficient (= √

__ r 2 ). For a perfect fit, Sr = 0 and r 2 = 1, signifying that the line explains 100% of

the variability of the data. For r 2 = 0 Sr = St and the fit represents no improvement. An alternative formulation for r that is more convenient for computer implementation is

r = n∑ (xi yi ) − ( ∑ xi ) ( ∑ yi )

_____________________________ √

_____________ n∑ x i

2 − ( ∑ xi ) 2 √ _____________

n∑ y i 2 − ( ∑ yi ) 2

(14.21)

EXAMPLE 14.5 Estimation of Errors for the Linear Least-Squares Fit

Problem Statement. Compute the total standard deviation, the standard error of the esti- mate, and the correlation coefficient for the fit in Example 14.4.

Solution. The data can be set up in tabular form and the necessary sums computed as in Table 14.5.

x (a)

(b)

FIGURE 14.11 Examples of linear regression with (a) small and (b) large residual errors.

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14.3 LinEAR LEAST-SQuARES REgRESSiOn 365

The standard deviation is [Eq. (14.3)]

sy = √ _________

1,808,297 _________ 8 − 1 = 508.26

and the standard error of the estimate is [Eq. (14.19)]

sy∕x = √ _______

216,118 _______ 8 − 2 = 189.79

Thus, because sy∕x < sy, the linear regression model has merit. The extent of the improve- ment is quantified by [Eq. (14.20)]

r 2 = 1,808,297 − 216,118 __________________ 1,808,297 = 0.8805

or r = √ _______

0.8805 = 0.9383 These results indicate that 88.05% of the original uncertainty has been explained by the linear model.

Before proceeding, a word of caution is in order. Although the coefficient of determi- nation provides a handy measure of goodness-of-fit, you should be careful not to ascribe more meaning to it than is warranted. Just because r 2 is “close” to 1 does not mean that the fit is necessarily “good.” For example, it is possible to obtain a relatively high value of r 2 when the underlying relationship between y and x is not even linear. Draper and Smith (1981) provide guidance and additional material regarding assessment of results for linear regression. In addition, at the minimum, you should always inspect a plot of the data along with your regression curve.

A nice example was developed by Anscombe (1973). As in Fig. 14.12, he came up with four data sets consisting of 11 data points each. Although their graphs are very different, all have the same best-fit equation, y = 3 + 0.5x, and the same coefficient of determination, r 2 = 0.67! This example dramatically illustrates why developing plots is so valuable.

TABLE 14.5 Data and summations needed to compute the goodness-of-fit statistics for the data from Table 14.1.

i xi yi a0 + a1 xi (yi − _ y )2 (yi − a0 − a1 xi)2

1 10 25 −39.58 380,535 4,171 2 20 70 155.12 327,041 7,245 3 30 380 349.82 68,579 911 4 40 550 544.52 8,441 30 5 50 610 739.23 1,016 16,699 6 60 1,220 933.93 334,229 81,837 7 70 830 1,128.63 35,391 89,180 8 80 1,450 1,323.33 653,066 16,044

Σ 360 5,135 1,808,297 216,118

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14.4 LINEARIZATION OF NONLINEAR RELATIONSHIPS

Linear regression provides a powerful technique for fitting a best line to data. However, it is predicated on the fact that the relationship between the dependent and independent variables is linear. This is not always the case, and the first step in any regression analysis should be to plot and visually inspect the data to ascertain whether a linear model applies. In some cases, techniques such as polynomial regression, which is described in Chap. 15, are appropriate. For others, transformations can be used to express the data in a form that is compatible with linear regression.

One example is the exponential model:

y = α1e β1x (14.22)

where α1 and β1 are constants. This model is used in many fields of engineering and sci- ence to characterize quantities that increase (positive β1 ) or decrease (negative β1 ) at a rate that is directly proportional to their own magnitude. For example, population growth or radioactive decay can exhibit such behavior. As depicted in Fig. 14.13a, the equation represents a nonlinear relationship (for β1 ≠ 0) between y and x.

Another example of a nonlinear model is the simple power equation:

y = α2 x β2 (14.23)

where α2 and β2 are constant coefficients. This model has wide applicability in all fields of engineering and science. It is very frequently used to fit experimental data when the underlying model is not known. As depicted in Fig. 14.13b, the equation (for β2 ≠ 0) is nonlinear.

0 0 5 10 15 20

FIGURE 14.12 Anscombe’s four data sets along with the best-fit line, y = 3 + 0.5 x.

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14.4 LinEARiZATiOn OF nOnLinEAR RELATiOnShiPS 367

A third example of a nonlinear model is the saturation-growth-rate equation:

y = α3 x ______

β3 + x (14.24)

where α3 and β3 are constant coefficients. This model, which is particularly well-suited for characterizing population growth rate under limiting conditions, also represents a nonlinear relationship between y and x (Fig. 14.13c) that levels off, or “saturates,” as x increases. It has many applications, particularly in biologically related areas of both engineering and science.

Nonlinear regression techniques are available to fit these equations to experimental data directly. However, a simpler alternative is to use mathematical manipulations to trans- form the equations into a linear form. Then linear regression can be employed to fit the equations to data.

y = α1eβ1x

Li ne

ar iz

at io

y = α2xβ2

Li ne

ar iz

at io

xxx

Li ne

ar iz

at io

ln y

Slope = β1

Intercept = In α1

log y

log x

1∕y

1∕x

Intercept = log α2

Intercept = 1∕α3

Slope = β2 Slope = β3∕α3

(d ) (e) (f )

(a) (b) (c)

y = α3 β3 + x x

FIGURE 14.13 (a) The exponential equation, (b) the power equation, and (c) the saturation-growth-rate equation. Parts (d ), (e), and (f ) are linearized versions of these equations that result from simple transformations.

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368 LinEAR REgRESSiOn

For example, Eq. (14.22) can be linearized by taking its natural logarithm to yield

ln y = ln α1 + β1x (14.25)

Thus, a plot of ln y versus x will yield a straight line with a slope of β1 and an intercept of ln α1 (Fig. 14.13d ).

Equation (14.23) is linearized by taking its base-10 logarithm to give

log y = log α2 + β2 log x (14.26)

Thus, a plot of log y versus log x will yield a straight line with a slope of β2 and an intercept of log α2 (Fig. 14.13e). Note that any base logarithm can be used to linearize this model. However, as done here, the base-10 logarithm is most commonly employed.

Equation (14.24) is linearized by inverting it to give

1 __ y = 1 __ α3 +

β3 __ α3 1 __ x (14.27)

Thus, a plot of 1∕y versus 1∕x will be linear, with a slope of β3∕α3 and an intercept of 1∕α3 (Fig. 14.13f ).

In their transformed forms, these models can be fit with linear regression to evaluate the constant coefficients. They can then be transformed back to their original state and used for predictive purposes. The following illustrates this procedure for the power model.

EXAMPLE 14.6 Fitting Data with the Power Equation

Problem Statement. Fit Eq. (14.23) to the data in Table 14.1 using a logarithmic transformation.

Solution. The data can be set up in tabular form and the necessary sums computed as in Table 14.6.

The means can be computed as

_ x = 12.606 ______ 8 = 1.5757

_ y = 20.515 ______ 8 = 2.5644

TABLE 14.6 Data and summations needed to fit the power model to the data from Table 14.1

i xi yi log xi log yi (log xi)2 log xi log yi

1 10 25 1.000 1.398 1.000 1.398 2 20 70 1.301 1.845 1.693 2.401 3 30 380 1.477 2.580 2.182 3.811 4 40 550 1.602 2.740 2.567 4.390 5 50 610 1.699 2.785 2.886 4.732 6 60 1220 1.778 3.086 3.162 5.488 7 70 830 1.845 2.919 3.404 5.386 8 80 1450 1.903 3.161 3.622 6.016

Σ 12.606 20.515 20.516 33.622

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14.4 LinEARiZATiOn OF nOnLinEAR RELATiOnShiPS 369

The slope and the intercept can then be calculated with Eqs. (14.15) and (14.16) as

a1 = 8 (33.622) − 12.606 (20.515) _______________________

8 (20.516) − (12.606)2 = 1.9842

a0 = 2.5644 − 1.9842(1.5757) = −0.5620

The least-squares fit is

log y = −0.5620 + 1.9842 log x

The fit, along with the data, is shown in Fig. 14.14a.

1000

log x

log y

100

10 10 50 100

(a)

(b)

1600

1200

800

400

0 0 20 40 60 80 x

FIGURE 14.14 Least-squares fit of a power model to the data from Table 14.1. (a) The fit of the transformed data. (b) The power equation fit along with the data.

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370 LinEAR REgRESSiOn

We can also display the fit using the untransformed coordinates. To do this, the coef- ficients of the power model are determined as α2 = 10−0.5620 = 0.2741 and β2 = 1.9842. Using force and velocity in place of y and x, the least-squares fit is

F = 0.2741υ1.9842

This equation, along with the data, is shown in Fig. 14.14b.

The fits in Example 14.6 (Fig. 14.14) should be compared with the one obtained previously in Example 14.4 (Fig. 14.8) using linear regression on the untransformed data. Although both results would appear to be acceptable, the transformed result has the advan- tage that it does not yield negative force predictions at low velocities. Further, it is known from the discipline of fluid mechanics that the drag force on an object moving through a fluid is often well described by a model with velocity squared. Thus, knowledge from the field you are studying often has a large bearing on the choice of the appropriate model equation you use for curve fitting.

14.4.1 General Comments on Linear Regression

Before proceeding to curvilinear and multiple linear regression, we must emphasize the introductory nature of the foregoing material on linear regression. We have focused on the simple derivation and practical use of equations to fit data. You should be cognizant of the fact that there are theoretical aspects of regression that are of practical importance but are beyond the scope of this book. For example, some statistical assumptions that are inherent in the linear least-squares procedures are

1. Each x has a fixed value; it is not random and is known without error. 2. The y values are independent random variables and all have the same variance. 3. The y values for a given x must be normally distributed.

Such assumptions are relevant to the proper derivation and use of regression. For example, the first assumption means that (1) the x values must be error-free and (2) the regression of y versus x is not the same as x versus y. You are urged to consult other refer- ences such as Draper and Smith (1981) to appreciate aspects and nuances of regression that are beyond the scope of this book.

14.5 COMPUTER APPLICATIONS

Linear regression is so commonplace that it can be implemented on most pocket calcula- tors. In this section, we will show how a simple M-file can be developed to determine the slope and intercept as well as to create a plot of the data and the best-fit line. We will also show how linear regression can be implemented with the built-in polyfit function.

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14.5 COMPuTER APPLiCATiOnS 371

14.5.1 MATLAB M-file: linregr

An algorithm for linear regression can be easily developed (Fig. 14.15). The required summations are readily computed with MATLAB’s sum function. These are then used to compute the slope and the intercept with Eqs. (14.15) and (14.16). The routine displays the intercept and slope, the coefficient of determination, and a plot of the best-fit line along with the measurements.

A simple example of the use of this M-file would be to fit the force-velocity data analyzed in Example 14.4:

>> x = [10 20 30 40 50 60 70 80]; >> y = [25 70 380 550 610 1220 830 1450]; >> [a, r2] = linregr (x,y)

a = 19.4702 − 234.2857

r2 = 0.8805

10 −200

200

400

600

800

1000

1200

1400

1600

20 30 40 50 60 70 80

It can just as easily be used to fit the power model (Example 14.6) by applying the log10 function to the data as in

>> [a, r2] = linregr(log10(x),log10(y))

a = 1.9842 − 0.5620

r2 = 0.9481

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372 LinEAR REgRESSiOn

1 1

1.5

2.5

3.5

1.2 1.4 1.6 1.8 2

FIGURE 14.15 An M-file to implement linear regression.

function [a, r2] = linregr(x,y) % linregr: linear regression curve fitting % [a, r2] = linregr(x,y): Least squares fit of straight % line to data by solving the normal equations

% input: % x = independent variable % y = dependent variable % output: % a = vector of slope, a(1), and intercept, a(2) % r2 = coefficient of determination

n = length(x); if length(y)~ = n, error('x and y must be same length'); end x = x(:); y = y(:); % convert to column vectors sx = sum(x); sy = sum(y); sx2 = sum(x.*x); sxy = sum(x.*y); sy2 = sum(y.*y); a(1) = (n*sxy — sx*sy)/(n*sx2—sx^2); a(2) = sy/n — a(1)*sx/n; r2 = ((n*sxy — sx*sy)/sqrt(n*sx2 — sx^2)/sqrt(n*sy2 — sy^2))^2; % create plot of data and best fit line xp = linspace(min(x),max(x),2); yp = a(1)*xp + a(2); plot(x,y,'o',xp,yp) grid on

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14.6 CASE STuDY 373

14.5.2 MATLAB Functions: polyfit and polyval

MATLAB has a built-in function polyfit that fits a least-squares nth-order polynomial to data. It can be applied as in

>> p = polyfit(x, y, n)

where x and y are the vectors of the independent and the dependent variables, respectively, and n = the order of the polynomial. The function returns a vector p containing the poly- nomial’s coefficients. We should note that it represents the polynomial using decreasing powers of x as in the following representation:

f (x) = p1x n + p2 x n−1 + · · · + pn x + pn+1 Because a straight line is a first-order polynomial, polyfit(x,y,1) will return the slope

and the intercept of the best-fit straight line.

>> x = [10 20 30 40 50 60 70 80]; >> y = [25 70 380 550 610 1220 830 1450]; >> a = polyfit(x,y,1)

a = 19.4702 −234.2857

Thus, the slope is 19.4702 and the intercept is −234.2857. Another function, polyval, can then be used to compute a value using the coefficients.

It has the general format:

>> y = polyval(p, x)

where p = the polynomial coefficients, and y = the best-fit value at x. For example,

>> y = polyval(a,45)

y = 641.8750

14.6 CASE STUDY EnZYME KinETiCS

Background. Enzymes act as catalysts to speed up the rate of chemical reactions in living cells. In most cases, they convert one chemical, the substrate, into another, the product. The Michaelis-Menten equation is commonly used to describe such reactions:

υ = υm [S] _______ ks + [S]

(14.28)

where υ = the initial reaction velocity, υm = the maximum initial reaction velocity, [S] = substrate concentration, and ks = a half-saturation constant. As in Fig. 14.16, the equation describes a saturating relationship which levels off with increasing [S]. The graph also illustrates that the half-saturation constant corresponds to the substrate concentration at which the velocity is half the maximum.

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14.6 CASE STUDY continued

Although the Michaelis-Menten model provides a nice starting point, it has been refined and extended to incorporate additional features of enzyme kinetics. One simple extension involves so-called allosteric enzymes, where the binding of a substrate molecule at one site leads to enhanced binding of subsequent molecules at other sites. For cases with two interacting bonding sites, the following second-order version often results in a better fit:

υ = υm [S]2 ________

k s 2 + [S]2 (14.29)

This model also describes a saturating curve but, as depicted in Fig. 14.16, the squared concentrations tend to make the shape more sigmoid, or S-shaped.

Suppose that you are provided with the following data:

[S] 1.3 1.8 3 4.5 6 8 9 υ 0.07 0.13 0.22 0.275 0.335 0.35 0.36

Employ linear regression to fit these data with linearized versions of Eqs. (14.28) and (14.29). Aside from estimating the model parameters, assess the validity of the fits with both statistical measures and graphs.

Solution. Equation (14.28), which is in the format of the saturation-growth-rate model (Eq. 14.24), can be linearized by inverting it to give [recall Eq. (14.27)]

1 __ υ = 1 ___ υm +

ks ___ υm 1 ___ [S]

FIGURE 14.16 Two versions of the Michaelis-Menten model of enzyme kinetics.

υm

0.5υm

ks [S]

Second-order Michaelis-Menten

model

Michaelis-Menten model

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14.6 CASE STUDY continued

The linregr function from Fig. 14.15 can then be used to determine the least-squares fit: >> S = [1.3 1.8 3 4.5 6 8 9]; >> v = [0.07 0.13 0.22 0.275 0.335 0.35 0.36]; >> [a,r2] = linregr(1./S,1./v)

a = 16.4022 0.1902 r2 = 0.9344

The model coefficients can then be calculated as >> vm = 1/a(2)

vm = 5.2570

>> ks = vm*a(1)

ks = 86.2260

Thus, the best-fit model is

υ = 5.2570[S] ____________ 86.2260 + [S]

Although the high value of r2 might lead you to believe that this result is accept- able, inspection of the coefficients might raise doubts. For example, the maximum velocity (5.2570) is much greater than the highest observed velocity (0.36). In addition, the half- saturation rate (86.2260) is much bigger than the maximum substrate concentration (9).

The problem is underscored when the fit is plotted along with the data. Figure 14.17a shows the transformed version. Although the straight line follows the upward trend, the data clearly appear to be curved. When the original equation is plotted along with the data in the untransformed version (Fig. 14.17b), the fit is obviously unacceptable. The data are clearly leveling off at about 0.36 or 0.37. If this is correct, an eyeball estimate would suggest that υm should be about 0.36, and ks should be in the range of 2 to 3.

Beyond the visual evidence, the poorness of the fit is also reflected by statistics like the coefficient of determination. For the untransformed case, a much less acceptable result of r2 = 0.6406 is obtained.

The foregoing analysis can be repeated for the second-order model. Equation (14.28) can also be linearized by inverting it to give

1 __ υ = 1 ___ υm +

k s 2 ___ υm 1 ___ [S]2

The linregr function from Fig. 14.15 can again be used to determine the least-squares fit: >> [a,r2]=linregr(1./S.^2,1./v)

a = 19.3760 2.4492 r2 = 0.9929

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14.6 CASE STUDY continued

The model coefficients can then be calculated as >> vm = 1/a(2)

vm = 0.4083

>> ks = sqrt(vm*a(1))

ks = 2.8127

Substituting these values into Eq. (14.29) gives

υ = 0.4083[S] 2 ___________

7.911 + [S]2

Although we know that a high r2 does not guarantee of a good fit, the fact that it is very high (0.9929) is promising. In addition, the parameters values also seem consistent with the trends in the data; that is, the km is slightly greater than the highest observed velocity and the half-saturation rate is lower than the maximum substrate concentration (9).

0.8

0.6

0.4

0.2

0 0 1 2 3 4

(a) Transformed model

(b) Original model

5 6 7 8 9

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8

1/ υ

[S]

1/[S]

FIGURE 14.17 Plots of least-squares fit (line) of the Michaelis-Menten model along with data (points). The plot in (a) shows the transformed fit, and (b) shows how the fit looks when viewed in the untransformed, original form.

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14.6 CASE STUDY continued

The adequacy of the fit can be assessed graphically. As in Fig. 14.18a, the transformed results appear linear. When the original equation is plotted along with the data in the untransformed version (Fig. 14.18b), the fit nicely follows the trend in the measurements. Beyond the graphs, the goodness of the fit is also reflected by the fact that the coefficient of determination for the untransformed case can be computed as r2 = 0.9896.

Based on our analysis, we can conclude that the second-order model provides a good fit of this data set. This might suggest that we are dealing with an allosteric enzyme.

Beyond this specific result, there are a few other general conclusions that can be drawn from this case study. First, we should never solely rely on statistics such as r2 as the sole basis of assessing goodness of fit. Second, regression equations should always be assessed graphically. And for cases where transformations are employed, a graph of the untrans- formed model and data should always be inspected.

Finally, although transformations may yield a decent fit of the transformed data, this does not always translate into an acceptable fit in the original format. The reason that this might occur is that minimizing squared residuals of transformed data is not the same as for the untransformed data. Linear regression assumes that the scatter of points around the

FIGURE 14.18 Plots of least-squares fit (line) of the second-order Michaelis-Menten model along with data (points). The plot in (a) shows the transformed fit, and (b) shows the untransformed, original form.

0.4

0.3

0.2

0.1

0 0 1 2 3 4

(a) Transformed model

(b) Original model

5 6 7 8 9

0 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7

1/ υ

[S]

1/[S]2

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14.6 CASE STUDY continued

best-fit line follows a Gaussian distribution, and that the standard deviation is the same at every value of the dependent variable. These assumptions are rarely true after transforming data.

As a consequence of the last conclusion, some analysts suggest that rather than using linear transformations, nonlinear regression should be employed to fit curvilinear data. In this approach, a best-fit curve is developed that directly minimizes the untransformed residuals. We will describe how this is done in Chap. 15.

PROBLEMS

14.1 Given the data

0.90 1.42 1.30 1.55 1.63 1.32 1.35 1.47 1.95 1.66 1.96 1.47 1.92 1.35 1.05 1.85 1.74 1.65 1.78 1.71 2.29 1.82 2.06 2.14 1.27

Determine (a) the mean, (b) median, (c) mode, (d) range, (e) standard deviation, (f) variance, and (g) coefficient of variation. 14.2 Construct a histogram from the data from Prob. 14.1. Use a range from 0.8 to 2.4 with intervals of 0.2. 14.3 Given the data

29.65 28.55 28.65 30.15 29.35 29.75 29.25 30.65 28.15 29.85 29.05 30.25 30.85 28.75 29.65 30.45 29.15 30.45 33.65 29.35 29.75 31.25 29.45 30.15 29.65 30.55 29.65 29.25

Determine (a) the mean, (b) median, (c) mode, (d) range, (e) standard deviation, (f) variance, and (g) coefficient of variation. (h) Construct a histogram. Use a range from 28 to 34 with

increments of 0.4. (i) Assuming that the distribution is normal, and that your

estimate of the standard deviation is valid, compute the range (i.e., the lower and the upper values) that encom- passes 68% of the readings. Determine whether this is a valid estimate for the data in this problem.

14.4 Using the same approach as was employed to derive Eqs. (14.15) and (14.16), derive the least-squares fit of the following model:

y = a1x + e

That is, determine the slope that results in the least-squares fit for a straight line with a zero intercept. Fit the following data with this model and display the result graphically.

x 2 4 6 7 10 11 14 17 20 y 4 5 6 5 8 8 6 9 12

14.5 Use least-squares regression to fit a straight line to

x 0 2 4 6 9 11 12 15 17 19 y 5 6 7 6 9 8 8 10 12 12

Along with the slope and intercept, compute the standard error of the estimate and the correlation coefficient. Plot the data and the regression line. Then repeat the problem, but regress x versus y—that is, switch the variables. Interpret your results. 14.6 Fit a power model to the data from Table 14.1, but use natural logarithms to perform the transformations. 14.7 The following data were gathered to determine the relationship between pressure and temperature of a fixed volume of 1 kg of nitrogen. The volume is 10 m3.

T, °C −40 0 40 80 120 160 p, N/m2 6900 8100 9350 10,500 11,700 12,800

Employ the ideal gas law pV = nRT to determine R on the basis of these data. Note that for the law, T must be expressed in kelvins.

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14.8 Beyond the examples in Fig. 14.13, there are other models that can be linearized using transformations. For example,

y = α4xeβ4x

Linearize this model and use it to estimate α4 and β4 based on the following data. Develop a plot of your fit along with the data.

x 0.1 0.2 0.4 0.6 0.9 1.3 1.5 1.7 1.8 y 0.75 1.25 1.45 1.25 0.85 0.55 0.35 0.28 0.18

14.9 The concentration of E. coli bacteria in a swimming area is monitored after a storm:

t (hr) 4 8 12 16 20 24 c (CFU/100 mL) 1600 1320 1000 890 650 560

The time is measured in hours following the end of the storm and the unit CFU is a “colony forming unit.” Use this data to estimate (a) the concentration at the end of the storm (t = 0) and (b) the time at which the concentration will reach 200 CFU/100 mL. Note that your choice of model should be consistent with the fact that negative concentrations are impossible and that the bacteria concentration always de- creases with time. 14.10 Rather than using the base-e exponential model [Eq. (14.22)], a common alternative is to employ a base-10 model:

y = α510β5x

When used for curve fitting, this equation yields identi- cal results to the base-e version, but the value of the ex- ponent parameter ( β5) will differ from that estimated with Eq. (14.22) ( β1). Use the base-10 version to solve Prob. 14.9. In addition, develop a formulation to relate β1 to β5. 14.11 Determine an equation to predict metabolism rate as a function of mass based on the following data. Use it to predict the metabolism rate of a 200-kg tiger.

Animal Mass (kg) Metabolism (watts)

Cow 400 270 Human 70 82 Sheep 45 50 Hen 2 4.8 Rat 0.3 1.45 Dove 0.16 0.97

14.12 On average, the surface area A of human beings is related to weight W and height H. Measurements on a num- ber of individuals of height 180 cm and different weights (kg) give values of A (m2) in the following table:

W (kg) 70 75 77 80 82 84 87 90 A (m2) 2.10 2.12 2.15 2.20 2.22 2.23 2.26 2.30

Show that a power law A = aWb fits these data reasonably well. Evaluate the constants a and b, and predict what the surface area is for a 95-kg person. 14.13 Fit an exponential model to

x 0.4 0.8 1.2 1.6 2 2.3 y 800 985 1490 1950 2850 3600

Plot the data and the equation on both standard and semi- logarithmic graphs with the MATLAB subplot function. 14.14 An investigator has reported the data tabulated below for an experiment to determine the growth rate of bacteria k (per d) as a function of oxygen concentration c (mg/L). It is known that such data can be modeled by the following equation:

k = kmaxc2 ______ cs + c2

where cs and kmax are parameters. Use a transformation to linearize this equation. Then use linear regression to esti- mate cs and kmax and predict the growth rate at c = 2 mg/L.

c 0.5 0.8 1.5 2.5 4 k 1.1 2.5 5.3 7.6 8.9

14.15 Develop an M-file function to compute descriptive statistics for a vector of values. Have the function determine and display number of values, mean, median, mode, range, standard deviation, variance, and coefficient of variation. In addition, have it generate a histogram. Test it with the data from Prob. 14.3. 14.16 Modify the linregr function in Fig. 14.15 so that it (a) computes and returns the standard error of the estimate and (b) uses the subplot function to also display a plot of the residuals (the predicted minus the measured y) versus x. Test it for the data from Examples 14.2 and 14.3. 14.17 Develop an M-file function to fit a power model. Have the function return the best-fit coefficient α2 and power β2 along with the r 2 for the untransformed model. In addition, use the subplot function to display graphs of both the transformed and untransformed equations along with the data. Test it with the data from Prob. 14.11.

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14.18 The following data show the relationship between the viscosity of SAE 70 oil and temperature. After taking the log of the data, use linear regression to find the equation of the line that best fits the data and the r 2 value.

Temperature, °C 26.67 93.33 148.89 315.56 Viscosity, µ, N·s/m2 1.35 0.085 0.012 0.00075

14.19 You perform experiments and determine the follow- ing values of heat capacity c at various temperatures T for a gas:

T −50 −30 0 60 90 110 c 1250 1280 1350 1480 1580 1700

Use regression to determine a model to predict c as a func- tion of T. 14.20 It is known that the tensile strength of a plastic in- creases as a function of the time it is heat treated. The fol- lowing data are collected:

Time 10 15 20 25 40 50 55 60 75 Tensile Strength 5 20 18 40 33 54 70 60 78

(a) Fit a straight line to these data and use the equation to determine the tensile strength at a time of 32 min.

(b) Repeat the analysis but for a straight line with a zero intercept.

14.21 The following data were taken from a stirred tank reactor for the reaction A → B. Use the data to determine the best possible estimates for k01 and E1 for the following kinetic model:

− dA ___ dt = k01e −E1∕RT A

where R is the gas constant and equals 0.00198 kcal/mol/K.

−dA∕dt (moles/L/s) 460 960 2485 1600 1245 A (moles/L) 200 150 50 20 10 T (K) 280 320 450 500 550

14.22 Concentration data were collected at 15 time points for the polymerization reaction:

x A + yB → Ax By We assume the reaction occurs via a complex mechanism consisting of many steps. Several models have been hypoth- esized, and the sum of the squares of the residuals had been calculated for the fits of the models of the data. The results

are shown below. Which model best describes the data (sta- tistically)? Explain your choice.

Model A Model B Model C

Sr 135 105 100 Number of Model Parameters Fit 2 3 5

14.23 Below are data taken from a batch reactor of bacterial growth (after lag phase was over). The bacteria are allowed to grow as fast as possible for the first 2.5 hours, and then they are induced to produce a recombinant protein, the pro- duction of which slows the bacterial growth significantly. The theoretical growth of bacteria can be described by

dX ___ dt = µX

where X is the number of bacteria, and µ is the specific growth rate of the bacteria during exponential growth. Based on the data, estimate the specific growth rate of the bacteria during the first 2 hours of growth and during the next 4 hours of growth.

Time, h 0 1 2 3 4 5 6 [Cells], g/L 0.100 0.335 1.102 1.655 2.453 3.702 5.460

14.24 A transportation engineering study was conducted to determine the proper design of bike lanes. Data were gath- ered on bike-lane widths and average distance between bikes and passing cars. The data from 9 streets are

Distance, m 2.4 1.5 2.4 1.8 1.8 2.9 1.2 3 1.2 Lane Width, m 2.9 2.1 2.3 2.1 1.8 2.7 1.5 2.9 1.5

(a) Plot the data. (b) Fit a straight line to the data with linear regression. Add

this line to the plot. (c) If the minimum safe average distance between bikes and

passing cars is considered to be 1.8 m, determine the corresponding minimum lane width.

14.25 In water-resources engineering, the sizing of reser- voirs depends on accurate estimates of water flow in the river that is being impounded. For some rivers, long-term historical records of such flow data are difficult to obtain. In contrast, meteorological data on precipitation are often available for many years past. Therefore, it is often useful

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PROBLEMS 381

to determine a relationship between flow and precipitation. This relationship can then be used to estimate flows for years when only precipitation measurements were made. The following data are available for a river that is to be dammed:

Precip., cm/yr 88.9 108.5 104.1 139.7 127 94 116.8 99.1 Flow, m3/s 14.6 16.7 15.3 23.2 19.5 16.1 18.1 16.6

(a) Plot the data. (b) Fit a straight line to the data with linear regression.

Superimpose this line on your plot. (c) Use the best-fit line to predict the annual water flow if

the precipitation is 120 cm. (d) If the drainage area is 1100 km2, estimate what fraction

of the precipitation is lost via processes such as evapo- ration, deep groundwater infiltration, and consumptive use.

14.26 The mast of a sailboat has a cross-sectional area of 10.65 cm2 and is constructed of an experimental alumi- num alloy. Tests were performed to define the relationship between stress and strain. The test results are

Strain, cm/cm 0.0032 0.0045 0.0055 0.0016 0.0085 0.0005 Stress, N/cm2 4970 5170 5500 3590 6900 1240

The stress caused by wind can be computed as F/Ac where F = force in the mast and Ac = mast’s cross-sectional area. This value can then be substituted into Hooke’s law to determine the mast’s deflection, Δ L strain × L, where L = the mast’s length. If the wind force is 25,000 N, use the data to estimate the deflection of a 9-m mast. 14.27 The following data were taken from an experiment that measured the current in a wire for various imposed voltages:

V, V 2 3 4 5 7 10 i, A 5.2 7.8 10.7 13 19.3 27.5

(a) On the basis of a linear regression of this data, determine current for a voltage of 3.5 V. Plot the line and the data and evaluate the fit.

(b) Redo the regression and force the intercept to be zero.

14.28 An experiment is performed to determine the % elon- gation of electrical conducting material as a function of tem- perature. The resulting data are listed below. Predict the % elongation for a temperature of 400 °C.

Temperature, °C 200 250 300 375 425 475 600 % Elongation 7.5 8.6 8.7 10 11.3 12.7 15.3

14.29 The population p of a small community on the out- skirts of a city grows rapidly over a 20-year period:

t 0 5 10 15 20 p 100 200 450 950 2000

As an engineer working for a utility company, you must forecast the population 5 years into the future in order to anticipate the demand for power. Employ an exponential model and linear regression to make this prediction. 14.30 The velocity u of air flowing past a flat surface is measured at several distances y away from the surface. Fit a curve to this data assuming that the velocity is zero at the surface (y = 0). Use your result to determine the shear stress ( µ du∕dy) at the surface where µ =1.8 × 10−5 N · s/m2.

y, m 0.002 0.006 0.012 0.018 0.024 u, m/s 0.287 0.899 1.915 3.048 4.299

14.31 Andrade’s equation has been proposed as a model of the effect of temperature on viscosity:

µ = DeB∕Ta

where µ = dynamic viscosity of water (10–3 N ·s/m2), Ta = absolute temperature (K), and D and B are parameters. Fit this model to the following data for water:

T 0 5 10 20 30 40 µ 1.787 1.519 1.307 1.002 0.7975 0.6529

14.32 Perform the same computation as in Example 14.2, but in addition to the drag coefficient, also vary the mass uniformly by ±10%. 14.33 Perform the same computation as in Example 14.3, but in addition to the drag coefficient, also vary the mass normally around its mean value with a coefficient of varia- tion of 5.7887%.

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14.34 Manning’s formula for a rectangular channel can be written as

Q = 1 ___ nm (BH)5∕3 __________

(B + 2H)2∕3 √

__ S

where Q = flow (m3∕s), nm = a roughness coefficient, B = width (m), H = depth (m), and S = slope. You are applying this formula to a stream where you know that the width = 20 m and the depth = 0.3 m. Unfortunately, you know the rough- ness and the slope to only a ±10% precision. That is, you know that the roughness is about 0.03 with a range from 0.027 to 0.033 and the slope is 0.0003 with a range from 0.00027 to 0.00033. Assuming uniform distributions, use a Monte Carlo analysis with n = 10,000 to estimate the distribution of flow. 14.35 A Monte Carlo analysis can be used for optimization. For example, the trajectory of a ball can be computed with

y = (tanθ0)x − g _________

2 υ 0 2 cos2θ0

x2 + y0 (P14.35)

where y = the height (m), θ0 = the initial angle (radians), υ0 = the initial velocity (m/s), g = the gravitational constant = 9.81 m/s2, and y0 = the initial height (m). Given y0 = 1 m, υ0 = 25 m/s, and θ0 = 50o, determine the maximum height and the corresponding x distance (a) analytically with cal- culus and (b) numerically with Monte Carlo simulation. For the latter, develop a script that generates a vector of 10,000 uniformly distributed values of x between 0 and 60 m. Use this vector and Eq. (P14.35) to generate a vector of heights. Then, employ the max function to determine the maximum height and the associated x distance. 14.36 Stokes Settling Law provides a means to compute the settling velocity of spherical particles under laminar conditions

υs = g ___ 18

ρs − ρ ______ µ d 2

where υs = the terminal settling velocity (m/s), g = gravitational acceleration (= 9.81 m/s2), ρ = the fluid density (kg/m3), ρs = the particle density (kg/m3), µ = the dynamic viscosity of the fluid (N s/m2), and d = the particle diameter (m). Suppose that you conduct an ex- periment in which you measure the terminal settling ve- locities of a number of 10-μm spheres having different densities,

ρs , kg/m 3 1500 1600 1700 1800 1900 2000 2100 2200 2300

υs , 10 −3 m/s 1.03 1.12 1.59 1.76 2.42 2.51 3.06 3 3.5

(a) Generate a labeled plot of the data. (b) Fit a straight line to the data with linear regression (polyfit) and superimpose this line on your plot. (c) Use the model to predict the set- tling velocity of a 2500 kg/m3 density sphere. (d) Use the slope and the intercept to estimate the fluid’s viscosity and density. 14.37 Beyond the examples in Fig. 14.13, there are other models that can be linearized using transformations. For ex- ample, the following model applies to third-order chemical reactions in batch reactors

c = c0 1 _________

√ _______

1 + 2 kc 0 2 t

where c = concentration (mg/L), c0 = initial concentration (mg/L), k = reaction rate (L2/(mg2 d)), and t = time (d). Lin- earize this model and use it to estimate k and c0 based on the following data. Develop plots of your fit along with the data for both the transformed, linearized form and the untrans- formed form.

t 0 0.5 1 1.5 2 3 4 5 c 3.26 2.09 1.62 1.48 1.17 1.06 0.9 0.85

14.38 In Chap. 7 we presented optimization techniques to find the optimal values of one- and multi-dimensional func- tions. Random numbers provide an alternative means to solve the same sorts of problems (recall Prob. 14.35). This is done by repeatedly evaluating the function at randomly se- lected values of the independent variable and keeping track of the one that yields the best value of the function being optimized. If a sufficient number of samples are conducted, the optimum will eventually be located. In their simplest manifestations, such approaches are not very efficient. How- ever, they do have the advantage that they can detect global optimums for functions with lots of local optima. Develop a function that uses random numbers to locate the maximum of the humps function

f (x) = 1 _____________ (x − 0.3)2 + 0.01

+ 1 ______________ (x − 0.9)2 + 0.04

− 6

in the domain bounded by x = 0 to 2. Here is a script that you can use to test your function

clear,clc,clf,format compact xmin = 0;xmax = 2;n = 1000 xp = linspace(xmin,xmax,200); yp = f(xp); plot(xp,yp) [xopt,fopt] = RandOpt(@f,n,xmin,xmax)

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PROBLEMS 383

14.39 Using the same approach as described in Prob. 14.38, develop a function that uses random numbers to determine the maximum and corresponding x and y values of the fol- lowing two-dimensional function

f (x, y) = y − x − 2x2 − 2xy − y2

in the domain bounded by x = −2 to 2 and y = 1 to 3. The domain is depicted in Fig. P14.39. Notice that a single maxi- mum of 1.25 occurs at x = −1 and y = 1.5. Here is a script that you can use to test your function

clear,clc,format compact xint = [−2;2];yint = [1;3];n = 10000; [xopt,yopt,fopt] = RandOpt2D(@fxy,n,xint,yint)

14.40 Suppose that a population of particles is confined to motion along a one-dimensional line (Fig. P14.40). Assume that each particle has an equal likelihood of moving a dis- tance, Δx, to either the left or right over a time step, Δt. At t = 0 all particles are grouped at x = 0 and are allowed to take one step in either direction. After Δt, approximately 50% will step to the right and 50% to the left. After 2Δt, 25% would be two steps to the left, 25% would be two steps to the right, and 50% would have stepped back to the origin. With additional time, the particles would spread out with the population greater near the origin and diminishing at the ends. The net result is that the distribution of the particles approaches a spreading bell-shaped distribution. This pro- cess, formally called a random walk (or drunkard’s walk), describes many phenomena in engineering and science with a common example being Brownian motion. Develop a

MATLAB function that is given a stepsize (Δx), and a total number of particles (n) and steps (m). At each step, deter- mine the location along the x axis of each particle and use these results to generate an animated histogram that displays how the distribution’s shape evolves as the computation progresses. 14.41 Repeat Prob. 14.40, but for a two-dimensional ran- dom walk. As depicted in Fig. P14.41, have each particle take a random step of length Δ at a random angle θ ranging from 0 to 2π. Generate an animated two panel stacked plot with the location of all the particles displayed on the top plot (subplot (2,1,1)), and the histogram of the particles’ x coor- dinates on the bottom (subplot (2,1,2)).

−20 −10

−2 −1 0 1 2

Maximum

FIGURE P14.39 A two-dimensional function with a maximum of 1.25 at x = −1 and y = 1.5.

t = ∆t

t = 2∆t

t = 3∆t

t = 0

∆x∆x

FIGURE P14.40 The one-dimensional random or “drunkard’s” walk.

∆ sin θ

∆ cos θ

∆

FIGURE P14.41 Depiction of the steps of a two-dimensional random or random walk.

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14.42 The table below shows the 2015 world record times and holders for outdoor running. Note that all but the 100 m and the marathon (42,195 m) are run on oval tracks.

Fit a power model for each gender and use it to predict the record time for a half marathon (21,097.5 m). Note that the actual records for the half marathon are 3503 s (Tadese) and 3909 s (Kiplagat) for men and women, respectively.

Event (m) Time (s) Men Holder Time (s) Women Holder

100 200 400 800

1000 1500 2000 5000

10,000 20,000 42,195

9.58 19.19 43.18

100.90 131.96 206.00 284.79 757.40

1577.53 3386.00 7377.00

Bolt Bolt Johnson Rudisha Ngeny El Guerrouj El Guerrouj Bekele Bekele Gebrselassie Kimetto

10.49 21.34 47.60 113.28 148.98 230.07 325.35 851.15 1771.78 3926.60 8125.00

Griffith-Joyner Griffith-Joyner Koch Kratochvilova Masterkova Dibaba O’Sullivan Dibaba Wang Loroupe Radcliffe

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385

CHAPTER OBJECTIVES

This chapter takes the concept of fitting a straight line and extends it to (a) fitting a polynomial and (b) fitting a variable that is a linear function of two or more independent variables. We will then show how such applications can be generalized and applied to a broader group of problems. Finally, we will illustrate how optimization techniques can be used to implement nonlinear regression. Specific objectives and topics covered are

• Knowing how to implement polynomial regression. • Knowing how to implement multiple linear regression. • Understanding the formulation of the general linear least-squares model. • Understanding how the general linear least-squares model can be solved with

MATLAB using either the normal equations or left division. • Understanding how to implement nonlinear regression with optimization

techniques.

15.1 POLYNOMIAL REGRESSION

In Chap.14, a procedure was developed to derive the equation of a straight line using the least-squares criterion. Some data, although exhibiting a marked pattern such as seen in Fig. 15.1, are poorly represented by a straight line. For these cases, a curve would be bet- ter suited to fit the data. As discussed in Chap. 14, one method to accomplish this objec- tive is to use transformations. Another alternative is to fit polynomials to the data using polynomial regression.

The least-squares procedure can be readily extended to fit the data to a higher-order polynomial. For example, suppose that we fit a second-order polynomial or quadratic:

y = a0 + a1x + a2x2 + e (15.1)

General Linear Least-Squares and Nonlinear Regression

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386 GeNeRaL LiNeaR LeaSt-SquaReS aNd NoNLiNeaR ReGReSSioN

For this case the sum of the squares of the residuals is

Sr = ∑ i=1

n ( yi − a0 − a1xi − a2 x i 2 )

2 (15.2)

To generate the least-squares fit, we take the derivative of Eq. (15.2) with respect to each of the unknown coefficients of the polynomial, as in

∂ Sr ___ ∂a0

= −2 ∑ ( yi − a0 − a1xi − a2 x i 2 )

∂ Sr ___ ∂a1

= −2 ∑ xi ( yi − a0 − a1xi − a2 x i 2 )

∂ Sr ___ ∂a2

= −2 ∑ x i 2 ( yi − a0 − a1xi − a2 x i 2 )

These equations can be set equal to zero and rearranged to develop the following set of normal equations:

(n)a0 + (∑ xi) a1 + (∑ x i 2 )a2 = ∑yi (∑ xi)a0 + (∑ x i 2 )a1 + (∑ x i 3 )a2 = ∑xi yi

(∑ x i 2 )a0 + (∑ x i 3 )a1 + (∑ x i 4 )a2 = ∑ x i 2 yi

x (a)

x (b)

FIGURE 15.1 (a) Data that are ill-suited for linear least-squares regression. (b) Indication that a parabola is preferable.

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15.1 PoLYNoMiaL ReGReSSioN 387

where all summations are from i = 1 through n. Note that the preceding three equations are linear and have three unknowns: a0, a1, and a2. The coefficients of the unknowns can be calculated directly from the observed data.

For this case, we see that the problem of determining a least-squares second-order polynomial is equivalent to solving a system of three simultaneous linear equations. The two-dimensional case can be easily extended to an mth-order polynomial as in

y = a0 + a1x + a2 x2 + · · · + am xm + e

The foregoing analysis can be easily extended to this more general case. Thus, we can recognize that determining the coefficients of an mth-order polynomial is equivalent to solving a system of m + 1 simultaneous linear equations. For this case, the standard error is formulated as

sy∕x = √ __________

Sr __________ n − (m + 1) (15.3)

This quantity is divided by n − (m + 1) because (m + 1) data-derived coefficients— a0, a1, . . . , am—were used to compute Sr; thus, we have lost m + 1 degrees of freedom. In addition to the standard error, a coefficient of determination can also be computed for polynomial regression with Eq. (14.20).

eXaMPLe 15.1 Polynomial Regression

Problem Statement. Fit a second-order polynomial to the data in the first two columns of Table 15.1.

TABLE 15.1 Computations for an error analysis of the quadratic least-squares fit.

xi yi (yi − y–)2 ( yi − a0 − a1xi − a2 x2i )2

0 2.1 544.44 0.14332 1 7.7 314.47 1.00286 2 13.6 140.03 1.08160 3 27.2 3.12 0.80487 4 40.9 239.22 0.61959 5 61.1 1272.11 0.09434

∑ 152.6 2513.39 3.74657

Solution. The following can be computed from the data:

m = 2 ∑ xi = 15 ∑ x i 4 = 979

n = 6 ∑ yi = 152.6 ∑xi yi = 585.6

_ x = 2.5 ∑ x i 2 = 55 ∑ x i 2 yi = 2488.8

_ y = 25.433 ∑ x i

3 = 225

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Therefore, the simultaneous linear equations are

[

55 15 55 225

225

979 ]

{

a1 a2

}

= {

152.6

585.6 2488.8

}

These equations can be solved to evaluate the coefficients. For example, using MATLAB: >> N = [6 15 55;15 55 225;55 225 979]; >> r = [152.6 585.6 2488.8]; >> a = N\r

a = 2.4786 2.3593 1.8607

Therefore, the least-squares quadratic equation for this case is

y = 2.4786 + 2.3593x + 1.8607x2

The standard error of the estimate based on the regression polynomial is [Eq. (15.3)]

sy∕x = √ __________

3.74657 __________ 6 − (2 + 1) = 1.1175

The coefficient of determination is

r2 = 2513.39 − 3.74657 ________________ 2513.39 = 0.99851

and the correlation coefficient is r = 0.99925

x50

50 Least-squares

parabola

FIGURE 15.2 Fit of a second-order polynomial.

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15.2 MuLtiPLe LiNeaR ReGReSSioN 389

These results indicate that 99.851 percent of the original uncertainty has been ex- plained by the model. This result supports the conclusion that the quadratic equation represents an excellent fit, as is also evident from Fig. 15.2.

15.2 MULTIPLE LINEAR REGRESSION

Another useful extension of linear regression is the case where y is a linear function of two or more independent variables. For example, y might be a linear function of x1 and x2, as in

y = a0 + a1x1 + a2 x2 + e

Such an equation is particularly useful when fitting experimental data where the variable being studied is often a function of two other variables. For this two-dimensional case, the regression “line” becomes a “plane” (Fig. 15.3).

As with the previous cases, the “best” values of the coefficients are determined by formulating the sum of the squares of the residuals:

Sr = ∑ i=1

n ( yi − a0 − a1x1,i − a2 x2,i ) 2 (15.4)

and differentiating with respect to each of the unknown coefficients:

∂ Sr ___ ∂a0

= −2 ∑ ( yi − a0 − a1x1,i − a2 x2,i )

∂ Sr ___ ∂a1

= −2 ∑ x1,i ( yi − a0 − a1x1,i − a2 x2,i )

∂ Sr ___ ∂a2

= −2 ∑ x2,i ( yi − a0 − a1x1,i − a2 x2,i )

FIGURE 15.3 Graphical depiction of multiple linear regression where y is a linear function of x1 and x2.

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390 GeNeRaL LiNeaR LeaSt-SquaReS aNd NoNLiNeaR ReGReSSioN

The coefficients yielding the minimum sum of the squares of the residuals are obtained by setting the partial derivatives equal to zero and expressing the result in matrix form as

⎡ ⎢

⎣

∑ x1,i

∑ x2,i ∑ x1,i ∑ x 1 2 ,i ∑ x1,i x2,i

∑ x2,i

∑ x1,i x2,i

∑ x 2 2 ,i

⎤ ⎥

⎦

{

a0 a1

}

⎧

⎪ ⎨

⎪

⎩

∑ yi

∑ x1,i yi ∑ x2,i yi

⎫

⎪ ⎬

⎪

⎭

(15.5)

eXaMPLe 15.2 Multiple Linear Regression

Problem Statement. The following data were created from the equation y = 5 + 4x1 − 3x2:

x1 x2 y

0 0 5 2 1 10 2.5 2 9 1 3 0 4 6 3 7 2 27

Use multiple linear regression to fit this data.

Solution. The summations required to develop Eq. (15.5) are computed in Table 15.2. Substituting them into Eq. (15.5) gives

[

16.5

14 16.5 76.25 48

54 ]

{

a1 a2

}

= {

243.5 100

}

(15.6)

which can be solved for a0 = 5 a1 = 4 a2 = −3

which is consistent with the original equation from which the data were derived.

The foregoing two-dimensional case can be easily extended to m dimensions, as in

y = a0 + a1x1 + a2 x2 + · · · + amxm + e

TABLE 15.2 Computations required to develop the normal equations for example 15.2.

y x1 x2 x 1 2 x 2

2 x1x2 x1 y x2 y

5 0 0 0 0 0 0 0 10 2 1 4 1 2 20 10 9 2.5 2 6.25 4 5 22.5 18 0 1 3 1 9 3 0 0 3 4 6 16 36 24 12 18 27 7 2 49 4 14 189 54

54 16.5 14 76.25 54 48 243.5 100

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15.3 GeNeRaL LiNeaR LeaSt SquaReS 391

where the standard error is formulated as

sy∕x = √ __________

Sr __________ n − (m + 1)

and the coefficient of determination is computed with Eq. (14.20). Although there may be certain cases where a variable is linearly related to two or more

other variables, multiple linear regression has additional utility in the derivation of power equations of the general form

y = a0 x 1 a 1 x 2

a 2 · · · x m a m

Such equations are extremely useful when fitting experimental data. To use multiple linear regression, the equation is transformed by taking its logarithm to yield

log y = log a0 + a1 log x1 + a2 log x2 + · · · + am log xm

15.3 GENERAL LINEAR LEAST SQUARES

In the preceding pages, we have introduced three types of regression: simple linear, polynomial, and multiple linear. In fact, all three belong to the following general linear least-squares model:

y = a0z0 + a1z1 + a2z2 + · · · + am zm + e (15.7)

where z 0, z1, . . . , z m are m + 1 basis functions. It can easily be seen how simple linear and multiple linear regression fall within this model—that is, z0 = 1, z1 = x1, z2 = x2, . . . , zm = xm. Further, polynomial regression is also included if the basis functions are simple monomials as in z0 = 1, z1 = x, z2 = x2, . . . , zm = x m.

Note that the terminology “linear” refers only to the model’s dependence on its parameters—that is, the a’s. As in the case of polynomial regression, the functions them- selves can be highly nonlinear. For example, the z’s can be sinusoids, as in

y = a0 + a1 cos(ωx) + a2 sin(ωx)

Such a format is the basis of Fourier analysis. On the other hand, a simple-looking model such as

y = a0(1 − e− a 1 x )

is truly nonlinear because it cannot be manipulated into the format of Eq. (15.7). Equation (15.7) can be expressed in matrix notation as

{y} = [Z ]{a} + {e} (15.8)

where [Z ] is a matrix of the calculated values of the basis functions at the measured values of the independent variables:

[Z ] =

⎡

⎢ ⎣

z01

z11

· · ·

zm1

z02

z12

· · ·

zm2

··· ···

···

z0n

z1n

· · ·

zmn

⎤

⎥ ⎦

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where m is the number of variables in the model and n is the number of data points. Because n ≥ m + 1, you should recognize that most of the time, [Z] is not a square matrix.

The column vector {y} contains the observed values of the dependent variable:

{y}T = ⌊ y1 y2 · · · yn ⌋

The column vector {a} contains the unknown coefficients:

{a}T = ⌊ a0 a1 · · · am ⌋

and the column vector {e} contains the residuals:

{e}T = ⌊ e1 e2 · · · en ⌋

The sum of the squares of the residuals for this model can be defined as

Sr = ∑ i=1

n ( yi − ∑ j=0

n aj zji ) 2 (15.9)

This quantity can be minimized by taking its partial derivative with respect to each of the coefficients and setting the resulting equation equal to zero. The outcome of this process is the normal equations that can be expressed concisely in matrix form as

[[Z ]T [Z ]]{a} = {[Z ]T {y}} (15.10)

It can be shown that Eq. (15.10) is, in fact, equivalent to the normal equations developed previously for simple linear, polynomial, and multiple linear regression.

The coefficient of determination and the standard error can also be formulated in terms of matrix algebra. Recall that r2 is defined as

r 2 = St − Sr ______ St

= 1 − Sr __ St

Substituting the definitions of Sr and St gives

r 2 = 1 − ∑ ( yi − ̂ y i )2 __________ ∑ ( yi −

_ y i )2

where ̂ y = the prediction of the least-squares fit. The residuals between the best-fit curve and the data, yi − ̂ y , can be expressed in vector form as

{y} − [Z ]{a} Matrix algebra can then be used to manipulate this vector to compute both the coefficient of determination and the standard error of the estimate as illustrated in the following example.

eXaMPLe 15.3 Polynomial Regression with MatLaB

Problem Statement. Repeat Example 15.1, but use matrix operations as described in this section.

Solution. First, enter the data to be fit

>> x = [0 1 2 3 4 5]'; >> y = [2.1 7.7 13.6 27.2 40.9 61.1]';

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15.3 GeNeRaL LiNeaR LeaSt SquaReS 393

Next, create the [Z ] matrix:

>> Z = [ones(size(x)) x x.^2]

Z = 1 0 0 1 1 1 1 2 4 1 3 9 1 4 16 1 5 25

We can verify that [Z ]T [Z ] results in the coefficient matrix for the normal equations:

>> Z'*Z

ans = 6 15 55 15 55 225 55 225 979

This is the same result we obtained with summations in Example 15.1. We can solve for the coefficients of the least-squares quadratic by implementing Eq. (15.10):

>> a = (Z'*Z)\(Z'*y)

ans = 2.4786 2.3593 1.8607

In order to compute r2 and sy∕x, first compute the sum of the squares of the residuals:

>> Sr = sum((y-Z*a).^2)

Sr = 3.7466

Then r2 can be computed as

>> r2 = 1−Sr/sum((y − mean(y)).^2)

r2 = 0.9985

and sy∕x can be computed as

>> syx = sqrt(Sr/(length(x) − length(a)))

syx = 1.1175

Our primary motivation for the foregoing has been to illustrate the unity among the three approaches and to show how they can all be expressed simply in the same matrix notation. It also sets the stage for the next section where we will gain some insights into the preferred strategies for solving Eq. (15.10). The matrix notation will also have relevance when we turn to nonlinear regression in Sec. 15.5.

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15.4 QR FACTORIZATION AND THE BACKSLASH OPERATOR

Generating a best fit by solving the normal equations is widely used and certainly adequate for many curve-fitting applications in engineering and science. It must be mentioned, how- ever, that the normal equations can be ill-conditioned and hence sensitive to roundoff errors.

Two more advanced methods, QR factorization and singular value decomposition, are more robust in this regard. Although the description of these methods is beyond the scope of this text, we mention them here because they can be implemented with MATLAB.

Further, QR factorization is automatically used in two simple ways within MATLAB. First, for cases where you want to fit a polynomial, the built-in polyfit function automati- cally uses QR factorization to obtain its results.

Second, the general linear least-squares problem can be directly solved with the back- slash operator. Recall that the general model is formulated as Eq. (15.8)

{y} = [Z ]{a} (15.11)

In Sec. 10.4, we used left division with the backslash operator to solve systems of linear alge- braic equations where the number of equations equals the number of unknowns (n = m). For Eq. (15.8) as derived from general least squares, the number of equations is greater than the number of unknowns (n > m). Such systems are said to be overdetermined. When MATLAB senses that you want to solve such systems with left division, it automatically uses QR fac- torization to obtain the solution. The following example illustrates how this is done.

eXaMPLe 15.4 implementing Polynomial Regression with polyfit and Left division

Problem Statement. Repeat Example 15.3, but use the built-in polyfit function and left division to calculate the coefficients.

Solution. As in Example 15.3, the data can be entered and used to create the [Z ] matrix as in >> x = [0 1 2 3 4 5]'; >> y = [2.1 7.7 13.6 27.2 40.9 61.1]'; >> Z = [ones(size(x)) x x.^2];

The polyfit function can be used to compute the coefficients: >> a = polyfit(x,y,2)

a = 1.8607 2.3593 2.4786

The same result can also be calculated using the backslash:

>> a = Z\y

a = 2.4786 2.3593 1.8607

As just stated, both these results are obtained automatically with QR factorization.

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15.5 NoNLiNeaR ReGReSSioN 395

15.5 NONLINEAR REGRESSION

There are many cases in engineering and science where nonlinear models must be fit to data. In the present context, these models are defined as those that have a nonlinear depen- dence on their parameters. For example,

y = a0(1 − e− a 1 x ) + e (15.12)

This equation cannot be manipulated so that it conforms to the general form of Eq. (15.7). As with linear least squares, nonlinear regression is based on determining the values

of the parameters that minimize the sum of the squares of the residuals. However, for the nonlinear case, the solution must proceed in an iterative fashion.

There are techniques expressly designed for nonlinear regression. For example, the Gauss-Newton method uses a Taylor series expansion to express the original nonlinear equation in an approximate, linear form. Then least-squares theory can be used to obtain new estimates of the parameters that move in the direction of minimizing the residual. Details on this approach are provided elsewhere (Chapra and Canale, 2010).

An alternative is to use optimization techniques to directly determine the least-squares fit. For example, Eq. (15.12) can be expressed as an objective function to compute the sum of the squares:

f (a0, a1) = ∑ i=1

n [yi − a0(1 − e− a 1 xi )]2 (15.13)

An optimization routine can then be used to determine the values of a0 and a1 that minimize the function.

As described previously in Sec. 7.3.1, MATLAB’s fminsearch function can be used for this purpose. It has the general syntax

[x, fval] = fminsearch(fun,x0,options,p1,p2,...)

where x = a vector of the values of the parameters that minimize the function fun, fval = the value of the function at the minimum, x0 = a vector of the initial guesses for the param- eters, options = a structure containing values of the optimization parameters as created with the optimset function (recall Sec. 6.5), and p1, p2, etc. = additional arguments that are passed to the objective function. Note that if options is omitted, MATLAB uses default values that are reasonable for most problems. If you would like to pass addi- tional arguments (p1, p2, . . .), but do not want to set the options, use empty brackets [] as a place holder.

eXaMPLe 15.5 Nonlinear Regression with MatLaB

Problem Statement. Recall that in Example 14.6, we fit the power model to data from Table 14.1 by linearization using logarithms. This yielded the model:

F = 0.2741υ1.9842

Repeat this exercise, but use nonlinear regression. Employ initial guesses of 1 for the coefficients.

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Solution. First, an M-file function must be created to compute the sum of the squares. The following file, called fSSR.m, is set up for the power equation: function f = fSSR(a,xm,ym) yp = a(1)*xm.^a(2); f = sum((ym − yp).^2);

In command mode, the data can be entered as >> x = [10 20 30 40 50 60 70 80]; >> y = [25 70 380 550 610 1220 830 1450];

The minimization of the function is then implemented by >> fminsearch(@fSSR, [1, 1], [], x, y)

ans = 2.5384 1.4359

The best-fit model is therefore

F = 2.5384υ1.4359

Both the original transformed fit and the present version are displayed in Fig. 15.4. Note that although the model coefficients are very different, it is difficult to judge which fit is superior based on inspection of the plot.

This example illustrates how different best-fit equations result when fitting the same model using nonlinear regression versus linear regression employing transformations. This is because the former minimizes the residuals of the original data whereas the latter mini- mizes the residuals of the transformed data.

1600

F, N

1200

800

400

0 0 20 40 60 80

υ, m/s

Transformed Untransformed

FIGURE 15.4 Comparison of transformed and untransformed model fits for force versus velocity data from Table 14.1.

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15.6 CaSe StudY 397

15.6 CASE STUDY FittiNG eXPeRiMeNtaL data

Background. As mentioned at the end of Sec. 15.2, although there are many cases where a variable is linearly related to two or more other variables, multiple linear regression has additional utility in the derivation of multivariable power equations of the general form

y = a0 x 1 a 1 x 2

a 2 · · · x m a m (15.14)

Such equations are extremely useful when fitting experimental data. To do this, the equa- tion is transformed by taking its logarithm to yield

log y = log a0 + a1 log x1 + a2 log x2 · · · + am log xm (15.15)

Thus, the logarithm of the dependent variable is linearly dependent on the logarithms of the independent variables.

A simple example relates to gas transfer in natural waters such as rivers, lakes, and estuaries. In particular, it has been found that the mass-transfer coefficient of dis- solved oxygen KL (m/d) is related to a river’s mean water velocity U (m/s) and depth H (m) by

KL = a0 U a 1 H a 2 (15.16)

Taking the common logarithm yields

log KL = log a0 + a1 logU + a2 log H (15.17)

The following data were collected in a laboratory flume at a constant temperature of 20 °C:

U 0.5 2 10 0.5 2 10 0.5 2 10 H 0.15 0.15 0.15 0.3 0.3 0.3 0.5 0.5 0.5 KL 0.48 3.9 57 0.85 5 77 0.8 9 92

Use these data and general linear least squares to evaluate the constants in Eq. (15.16).

Solution. In a similar fashion to Example 15.3, we can develop a script to assign the data, create the [Z] matrix, and compute the coefficients for the least-squares fit:

% Compute best fit of transformed values clc; format short g U = [0.5 2 10 0.5 2 10 0.5 2 10]'; H = [0.15 0.15 0.15 0.3 0.3 0.3 0.5 0.5 0.5]'; KL = [0.48 3.9 57 0.85 5 77 0.8 9 92]'; logU = log10(U);logH =log10(H);logKL=log10(KL); Z = [ones(size(logKL)) logU logH]; a = (Z'*Z)\(Z'*logKL)

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15.6 CASE STUDY continued

with the result: a = 0.57627 1.562 0.50742

Therefore, the best-fit model is

log KL = 0.57627 + 1.562 log U + 0.50742 log H

or in the untransformed form (note, a0 = 100.57627 = 3.7694),

KL = 3.7694U1.5620 H 0.5074

The statistics can also be determined by adding the following lines to the script:

% Compute fit statistics Sr =sum((logKL−Z*a).^2) r2 =1−Sr/sum((logKL−mean(logKL)).^2) syx =sqrt(Sr/(length(logKL)−length(a)))

Sr = 0.024171 r2 = 0.99619 syx = 0.063471

Finally, plots of the fit can be developed. The following statements display the model predictions versus the measured values for KL. Subplots are employed to do this for both the transformed and untransformed versions.

%Generate plots clf KLpred = 10^a(1)*U.^a(2).*H.^a(3); KLmin = min(KL);KLmax = max(KL); dKL=(KLmax-KLmin)/100; KLmod=[KLmin:dKL:KLmax]; subplot(1,2,1) loglog(KLpred,KL,'ko',KLmod,KLmod,'k-') axis square,title('(a) log-log plot’) legend('model prediction','1:1 line','Location','NorthWest') xlabel('log(K_L) measured'),ylabel('log(K_L) predicted') subplot(1,2,2) plot(KLpred,KL,'ko',KLmod,KLmod,'k−') axis square,title('(b) untransformed plot') legend('model prediction','1:1 line','Location','NorthWest') xlabel('K_L measured'),ylabel('K_L predicted')

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PRoBLeMS 399

15.6 CASE STUDY continued

The result is shown in Fig. 15.5.

lo g

(K L)

p re

di ct

log (KL) measured

10−1 10−1

100

100 101 102

101

102

model prediction 1:1 line

(a) log-log plot (b) untransformed plot

K L

pr ed

ic te

KL measured

0 20

40 60 80 100

100

FIGURE 15.5 Plots of predicted versus measured values of the oxygen mass-transfer coefficient as computed with multiple regression. Results are shown for (a ) log transformed and (b) untransformed cases. The 1:1 line, which indicates a perfect correlation, is superimposed on both plots.

PROBLEMS

15.1 Fit a parabola to the data from Table 14.1. Determine the r 2 for the fit and comment on the efficacy of the result. 15.2 Using the same approach as was employed to derive Eqs. (14.15) and (14.16), derive the least-squares fit of the following model:

y = a1 x + a2 x2 + e

That is, determine the coefficients that result in the least- squares fit for a second-order polynomial with a zero in- tercept. Test the approach by using it to fit the data from Table 14.1. 15.3 Fit a cubic polynomial to the following data:

x 3 4 5 7 8 9 11 12 y 1.6 3.6 4.4 3.4 2.2 2.8 3.8 4.6

Along with the coefficients, determine r2 and sy∕x.

15.4 Develop an M-file to implement polynomial re- gression. Pass the M-file two vectors holding the x and y values along with the desired order m. Test it by solving Prob. 15.3. 15.5 For the data from Table P15.5, use polynomial regression to derive a predictive equation for dissolved oxy- gen concentration as a function of temperature for the case where the chloride concentration is equal to zero. Employ a polynomial that is of sufficiently high order that the predic- tions match the number of significant digits displayed in the table. 15.6 Use multiple linear regression to derive a predictive equation for dissolved oxygen concentration as a func- tion of temperature and chloride based on the data from Table P15.5. Use the equation to estimate the concentration of dissolved oxygen for a chloride concentration of 15 g/L at T = 12 °C. Note that the true value is 9.09 mg/L. Compute

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the percent relative error for your prediction. Explain pos- sible causes for the discrepancy. 15.7 As compared with the models from Probs. 15.5 and 15.6, a somewhat more sophisticated model that accounts for the effect of both temperature and chloride on dis- solved oxygen saturation can be hypothesized as being of the form

o = f 3(T ) + f 1(c)

That is, a third-order polynomial in temperature and a lin- ear relationship in chloride is assumed to yield superior re- sults. Use the general linear least-squares approach to fit this model to the data in Table P15.5. Use the resulting equation to estimate the dissolved oxygen concentration for a chloride concentration of 15 g/L at T = 12 °C. Note that the true value is 9.09 mg/L. Compute the percent relative error for your prediction. 15.8 Use multiple linear regression to fit

x1 0 1 1 2 2 3 3 4 4 x2 0 1 2 1 2 1 2 1 2 y 15.1 17.9 12.7 25.6 20.5 35.1 29.7 45.4 40.2

Compute the coefficients, the standard error of the estimate, and the correlation coefficient. 15.9 The following data were collected for the steady flow of water in a concrete circular pipe:

Experiment Diameter, m Slope, m/m Flow, m3/s

1 0.3 0.001 0.04 2 0.6 0.001 0.24 3 0.9 0.001 0.69 4 0.3 0.01 0.13 5 0.6 0.01 0.82 6 0.9 0.01 2.38 7 0.3 0.05 0.31 8 0.6 0.05 1.95 9 0.9 0.05 5.66

Use multiple linear regression to fit the following model to this data:

Q = α0 D α 1 S α 2

where Q = flow, D = diameter, and S = slope. 15.10 Three disease-carrying organisms decay exponen- tially in seawater according to the following model:

p(t) = Ae−1.5t + Be−0.3t + Ce−0.05t

Use general linear least-squares to estimate the initial con- centration of each organism (A, B, and C ) given the follow- ing measurements:

t 0.5 1 2 3 4 5 6 7 9 p(t) 6 4.4 3.2 2.7 2 1.9 1.7 1.4 1.1

15.11 The following model is used to represent the effect of solar radiation on the photosynthesis rate of aquatic plants:

P = Pm I ___ Isat

e − I _ Isat + 1

where P = the photosynthesis rate (mg m−3d−1), Pm = the maximum photosynthesis rate (mg m−3d−1), I = solar radiation ( μE m−2s−1), and Isat = optimal solar radiation ( μE m−2s−1). Use nonlinear regression to evaluate Pm and Isat based on the following data:

I 50 80 130 200 250 350 450 550 700 P 99 177 202 248 229 219 173 142 72

15.12 The following data are provided

x 1 2 3 4 5 y 2.2 2.8 3.6 4.5 5.5

TABLE P15.5 dissolved oxygen concentration in water as a function of temperature (°C) and chloride concentration (g/L).

Dissolved Oxygen (mg/L) for Temperature (°C) and Concentration of Chloride (g/L)

T, °C c = 0 g/L c = 10 g/L c = 20 g/L

0 14.6 12.9 11.4 5 12.8 11.3 10.3 10 11.3 10.1 8.96 15 10.1 9.03 8.08 20 9.09 8.17 7.35 25 8.26 7.46 6.73 30 7.56 6.85 6.20

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use least-squares regression to fit (a) a straight line, (b) a power equation, (c) a saturation-growth-rate equation, and (d) a parabola. For (b) and (c), employ transformations to linearize the data. Plot the data along with all the curves. Is any one of the curves superior? If so, justify. 15.16 The following data represent the bacterial growth in a liquid culture over of number of days:

Day 0 4 8 12 16 20 Amount × 106 67.38 74.67 82.74 91.69 101.60 112.58

Find a best-fit equation to the data trend. Try several possibilities—linear, quadratic, and exponential. Determine the best equation to predict the amount of bacteria after 35 days. 15.17 Dynamic viscosity of water μ(10–3 N · s/m2) is related to temperature T(°C) in the following manner:

T 0 5 10 20 30 40 μ 1.787 1.519 1.307 1.002 0.7975 0.6529

(a) Plot this data. (b) Use linear interpolation to predict μ at T = 7.5 °C. (c) Use polynomial regression to fit a parabola to the data in

order to make the same prediction. 15.18 Use general linear least squares to find the best pos- sible virial constants (A1 and A2) for the following equation of state. R = 82.05 mL atm/gmol K, and T = 303 K.

PV ___ RT = 1 + A1 ___ V +

A2 ___ V 2

P (atm) 0.985 1.108 1.363 1.631 V (mL) 25,000 22,200 18,000 15,000

15.19 Environmental scientists and engineers dealing with the impacts of acid rain must determine the value of the ion product of water K𝑤 as a function of temperature. Sci- entists have suggested the following equation to model this relationship:

−log10 K𝑤 = a __ Ta

+ b log10 Ta + cTa + d

where Ta = absolute temperature (K), and a, b, c, and d are parameters. Employ the following data and regression to

Fit the following model to this data using MATLAB and the general linear least-squares model

y = a + bx + c __ x

15.13 In Prob. 14.8 we used transformations to linearize and fit the following model:

y = α4x e β 4 x

Use nonlinear regression to estimate α4 and β4 based on the following data. Develop a plot of your fit along with the data.

x 0.1 0.2 0.4 0.6 0.9 1.3 1.5 1.7 1.8 y 0.75 1.25 1.45 1.25 0.85 0.55 0.35 0.28 0.18

15.14 Enzymatic reactions are used extensively to charac- terize biologically mediated reactions. The following is an example of a model that is used to fit such reactions:

υ0 = km[S]3 ________

K + [S ]3

where υ0 = the initial rate of the reaction (M/s), [S] = the substrate concentration (M), and km and K are parameters. The following data can be fit with this model:

[S], M υ0, M/s

0.01 6.078 × 10−11 0.05 7.595 × 10−9 0.1 6.063 × 10−8 0.5 5.788 × 10−6 1 1.737 × 10−5 5 2.423 × 10−5 10 2.430 × 10−5 50 2.431 × 10−5 100 2.431 × 10−5

(a) Use a transformation to linearize the model and evaluate the parameters. Display the data and the model fit on a graph.

(b) Perform the same evaluation as in (a) but use nonlinear regression.

15.15 Given the data

x 5 10 15 20 25 30 35 40 45 50 y 17 24 31 33 37 37 40 40 42 41

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15.23 An investigator has reported the data tabulated below for an experiment to determine the growth rate of bacteria k (per d), as a function of oxygen concentration c (mg/L). It is known that such data can be modeled by the following equation:

k = kmaxc2 ______ cs + c2

Use nonlinear regression to estimate cs and kmax and predict the growth rate at c = 2 mg/L.

c 0.5 0.8 1.5 2.5 4 k 1.1 2.4 5.3 7.6 8.9

15.24 A material is tested for cyclic fatigue failure whereby a stress, in MPa, is applied to the material and the number of cycles needed to cause failure is measured. The results are in the table below. Use nonlinear regression to fit a power model to this data.

N, cycles 1 10 100 1000 10,000 100,000 1,000,000 Stress, MPa 1100 1000 925 800 625 550 420

15.25 The following data shows the relationship between the viscosity of SAE 70 oil and temperature. Use nonlinear regression to fit a power equation to this data.

Temperature, T, °C 26.67 93.33 148.89 315.56 Viscosity, μ, N · s/m2 1.35 0.085 0.012 0.00075

15.26 The concentration of E. coli bacteria in a swimming area is monitored after a storm:

t (hr) 4 8 12 16 20 24 c(CFU/100 mL) 1590 1320 1000 900 650 560

The time is measured in hours following the end of the storm and the unit CFU is a “colony forming unit.” Employ non- linear regression to fit an exponential model [Eq. (14.22)] to this data. Use the model to estimate (a) the concentration at the end of the storm (t = 0) and (b) the time at which the concentration will reach 200 CFU/100 mL.

estimate the parameters with MATLAB. Also, generate a plot of predicted Kw versus the data.

T (°C) K𝒘

0 1.164 × 10−15 10 2.950 × 10−15 20 6.846 × 10−15 30 1.467 × 10−14 40 2.929 × 10−14

15.20 The distance required to stop an automobile consists of both thinking and braking components, each of which is a function of its speed. The following experimental data were collected to quantify this relationship. Develop best- fit equations for both the thinking and braking components. Use these equations to estimate the total stopping distance for a car traveling at 110 km/hr.

Speed, km/hr 30 45 60 75 90 120 Thinking, m 5.6 8.5 11.1 14.5 16.7 22.4 Braking, m 5.0 12.3 21.0 32.9 47.6 84.7

15.21 An investigator has reported the data tabulated below. It is known that such data can be modeled by the following equation

x = e( y−b)∕a

where a and b are parameters. Use nonlinear regression to determine a and b. Based on your analysis predict y at x = 2.6.

x 1 2 3 4 5 y 0.5 2 2.9 3.5 4

15.22 It is known that the data tabulated below can be mod- eled by the following equation

y = ( a + √ __

x _______ b √ __

x ) 2

Use nonlinear regression to determine the parameters a and b. Based on your analysis predict y at x = 1.6.

x 0.5 1 2 3 4 y 10.4 5.8 3.3 2.4 2

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where p is the vapor pressure, T is temperature (K), and A, B, and C are component-specific constants. Use MATLAB to determine the best values for the constants for carbon mon- oxide based on the following measurements

T (K ) 50 60 70 80 90 100 110 120 130 p (Pa) 82 2300 18,500 80,500 2.3×105 5×105 9.6×105 1.5×106 2.4×106

In addition to the constants determine the r 2 and sy∕x for your fit. 15.30 The following model, based on a simplification of the Arrhenius equation, is frequently used in environmental en- gineering to parameterize the effect of temperature, T (°C), on pollutant decay rates, k (per day),

k = k20 θ T−20

where the parameters k20 = the decay rate at 20 °C, and θ = the dimensionless temperature dependence coefficient. The following data are collected in the laboratory

T (°C) 6 12 18 24 30 k (per d) 0.15 0.20 0.32 0.45 0.70

(a) Use a transformation to linearize this equation and then employ linear regression to estimate k20 and θ. (b) Employ nonlinear regression to estimate the same parameters. For both (a) and (b) employ the equation to predict the reaction rate at T = 17 °C.

15.27 Employ nonlinear regression and the following set of pressure-volume data to find the best possible virial con- stants (A1 and A2) for the equation of state shown below. R = 82.05 mL atm/gmol K and T = 303 K.

PV ___ RT = 1 + A1 ___ V +

A2 ___ V 2

P (atm) 0.985 1.108 1.363 1.631 V (mL) 25,000 22,200 18,000 15,000

15.28 Three disease-carrying organisms decay exponentially in lake water according to the following model:

p(t) = Ae−1.5t + Be−0.3t + Ce−0.05t

Use nonlinear regression to estimate the initial popula- tion of each organism (A, B, and C) given the following measurements:

t, hr 0.5 1 2 3 4 5 6 7 9 p(t) 6.0 4.4 3.2 2.7 2.2 1.9 1.7 1.4 1.1

15.29 The Antoine equation describes the relation between vapor pressure and temperature for pure components as

ln ( p) = A − B ______ C + T

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404

16 Fourier Analysis

YOU’VE GOT A PROBLEM

At the beginning of Chap. 8, we used Newton’s second law and force balances to predict the equilibrium positions of three bungee jumpers connected by cords. Then, in Chap. 13, we determined the same system’s eigenvalues and eigenvectors in order to identify its resonant frequencies and principal modes of vibration. Although this

CHAPTER OBJECTIVES

The primary objective of this chapter is to introduce you to Fourier analysis. The subject, which is named after Joseph Fourier, involves identifying cycles or patterns within a time series of data. Specific objectives and topics covered in this chapter are

• Understanding sinusoids and how they can be used for curve fitting. • Knowing how to use least-squares regression to fit a sinusoid to data. • Knowing how to fit a Fourier series to a periodic function. • Understanding the relationship between sinusoids and complex exponentials

based on Euler’s formula. • Recognizing the benefits of analyzing mathematical function or signals in the

frequency domain (i.e., as a function of frequency). • Understanding how the Fourier integral and transform extend Fourier analysis to

aperiodic functions. • Understanding how the discrete Fourier transform (DFT) extends Fourier analysis

to discrete signals. • Recognizing how discrete sampling affects the ability of the DFT to distinguish

frequencies. In particular, know how to compute and interpret the Nyquist frequency. • Recognizing how the fast Fourier transform (FFT) provides a highly efficient

means to compute the DFT for cases where the data record length is a power of 2. • Knowing how to use the MATLAB function fft to compute a DFT and understand

how to interpret the results. • Knowing how to compute and interpret a power spectrum.

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16.1 CURVE FITTING WITH SINUSOIDAL FUNCTIONS 405

analysis certainly provided useful results, it required detailed system information includ- ing knowledge of the underlying model and parameters (i.e., the jumpers’ masses and the cords’ spring constants).

So suppose that you have measurements of the jumpers’ positions or velocities at dis- crete, equally spaced times (recall Fig. 13.1). Such information is referred to as a time series. However, suppose further that you do not know the underlying model or the param- eters needed to compute the eigenvalues. For such cases, is there any way to use the time series to learn something fundamental about the system’s dynamics?

In this chapter, we describe such an approach, Fourier analysis, which provides a way to accomplish this objective. The approach is based on the premise that more complicated functions (e.g., a time series) can be represented by the sum of simpler trigonometric func- tions. As a prelude to outlining how this is done, it is useful to explore how data can be fit with sinusoidal functions.

16.1 CURVE FITTING WITH SINUSOIDAL FUNCTIONS

A periodic function f (t) is one for which

f (t) = f (t + T ) (16.1)

where T is a constant called the period that is the smallest value of time for which Eq. (16.1) holds. Common examples include both artificial and natural signals (Fig. 16.1a).

The most fundamental are sinusoidal functions. In this discussion, we will use the term sinusoid to represent any waveform that can be described as a sine or cosine. There is no

(a)

(b)

FIGURE 16.1 Aside from trigonometric functions such as sines and cosines, periodic functions include ide- alized waveforms like the square wave depicted in (a). Beyond such artificial forms, periodic signals in nature can be contaminated by noise like the air temperatures shown in (b).

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406 FOURIER ANALySIS

clear-cut convention for choosing either function, and in any case, the results will be identi- cal because the two functions are simply offset in time by π∕2 radians. For this chapter, we will use the cosine, which can be expressed generally as

f (t) = A0 + C1cos(ω0t + θ) (16.2)

Inspection of Eq. (16.2) indicates that four parameters serve to uniquely characterize the sinusoid (Fig. 16.2a):

• The mean value A0 sets the average height above the abscissa. • The amplitude C1 specifies the height of the oscillation. • The angular frequency ω0 characterizes how often the cycles occur. • The phase angle (or phase shift) θ parameterizes the extent to which the sinusoid is

shifted horizontally.

y(t)

t, s

ωt, rad

0 π 2π 3π

B1 sin (ω0t)

A1 cos (ω0t)

(a)

−1

(b)

FIGURE 16.2 (a) A plot of the sinusoidal function y(t) = A0 + C1 cos(ω0t + θ). For this case, A0 = 1.7, C1 = 1, ω0 = 2π/T = 2π/(1.5 s), and θ = π/3 radians = 1.0472 (= 0.25 s). Other parameters used to describe the curve are the frequency f = ω0/(2π), which for this case is 1 cycle/(1.5 s) = 0.6667 Hz and the period T = 1.5 s. (b) An alternative expression of the same curve is y(t) = A0 + A1 cos(ω0t) + B1 sin(ω0t). The three components of this function are depicted in (b), where A1 = 0.5 and B1 = –0.866. The summation of the three curves in (b) yields the single curve in (a).

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16.1 CURVE FITTING WITH SINUSOIDAL FUNCTIONS 407

Note that the angular frequency (in radians/time) is related to the ordinary frequency f (in cycles/time)1 by

ω0 = 2π f (16.3)

and the ordinary frequency in turn is related to the period T by

f = 1 __ T (16.4)

In addition, the phase angle represents the distance in radians from t = 0 to the point at which the cosine function begins a new cycle. As depicted in Fig. 16.3a, a negative value is referred to as a lagging phase angle because the curve cos(ω0 t − θ) begins a new cycle θ radians after cos(ω0 t). Thus, cos(ω0 t − θ) is said to lag cos(ω0 t). Conversely, as in Fig. 16.3b, a positive value is referred to as a leading phase angle.

Although Eq. (16.2) is an adequate mathematical characterization of a sinusoid, it is awkward to work with from the standpoint of curve fitting because the phase shift is included in the argument of the cosine function. This deficiency can be overcome by invoking the trigonometric identity:

C1cos(ω0 t + θ ) = C1[cos(ω0 t )cos(θ ) − sin(ω0 t )sin(θ )] (16.5)

cos (ω0t)

cos ω0t π2−

cos ω0t π2+

(a)

(b)

FIGURE 16.3 Graphical depictions of (a) a lagging phase angle and (b) a leading phase angle. Note that the lagging curve in (a) can be alternatively described as cos(ω0t + 3π/2). In other words, if a curve lags by an angle of α, it can also be represented as leading by 2π − α.

1 When the time unit is seconds, the unit for the ordinary frequency is a cycle/s or Hertz (Hz).

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408 FOURIER ANALySIS

Substituting Eq. (16.5) into Eq. (16.2) and collecting terms gives (Fig. 16.2b)

f (t) = A0 + A1cos(ω0 t) + B1sin(ω0 t) (16.6)

where

A1 = C1cos(θ ) B1 = −C1sin(θ ) (16.7)

Dividing the two parts of Eq. (16.7) gives

θ = arctan ( − B1 ___ A1 ) (16.8) where, if A1 < 0, add π to θ. Squaring and summing Eq. (16.7) leads to

C1 = √ _______

A 1 2 + B 1

2 (16.9)

Thus, Eq. (16.6) represents an alternative formulation of Eq. (16.2) that still requires four parameters but that is cast in the format of a general linear model [recall Eq. (15.7)]. As we will discuss in the next section, it can be simply applied as the basis for a least-squares fit.

Before proceeding to the next section, however, we should stress that we could have employed a sine rather than a cosine as our fundamental model of Eq. (16.2). For example,

f (t ) = A0 + C1sin(ω0 t + δ )

could have been used. Simple relationships can be applied to convert between the two forms:

sin(ω0 t + δ ) = cos ( ω0 t + δ − π __ 2 ) and

cos(ω0 t + δ ) = sin ( ω0 t + δ + π __ 2 ) (16.10) In other words, θ = δ − π∕2. The only important consideration is that one or the other format should be used consistently. Thus, we will use the cosine version throughout our discussion.

16.1.1 Least-Squares Fit of a Sinusoid

Equation (16.6) can be thought of as a linear least-squares model:

y = A0 + A1cos(ω0 t ) + B1sin(ω0 t ) + e (16.11)

which is just another example of the general model [recall Eq. (15.7)]

y = a0z0 + a1z1 + a2z2 + · · · + amzm + e

where z 0 = 1, z 1 = cos(ω0 t ), z 2 = sin(ω0 t ), and all other z’s = 0. Thus, our goal is to determine coefficient values that minimize

Sr = ∑ i=1

{yi − [A0 + A1cos(ω0 t ) + B1sin(ω0 t )]}2

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16.1 CURVE FITTING WITH SINUSOIDAL FUNCTIONS 409

The normal equations to accomplish this minimization can be expressed in matrix form as [recall Eq. (15.10)]

⎡ ⎢

⎣

∑ cos(ω0 t)

∑ sin(ω0 t) ∑ cos(ω0 t) ∑ cos2(ω0 t) ∑ cos(ω0 t)sin(ω0 t)

∑ sin(ω0 t)

∑ cos(ω0 t)sin(ω0 t)

∑ sin2(ω0 t)

⎤ ⎥

⎦ ⎧

⎪

⎨ ⎪

⎩ A0

B1 B1

⎫

⎪

⎬ ⎪

⎭ =

⎧

⎪

⎨ ⎪

⎩

∑ y

∑ y cos(ω0t ) ∑ y sin(ω0 t )

⎫

⎪

⎬ ⎪

⎭

(16.12)

These equations can be employed to solve for the unknown coefficients. However, rather than do this, we can examine the special case where there are N observations equi- spaced at intervals of Δt and with a total record length of T = (N − 1)Δt. For this situation, the following average values can be determined (see Prob. 16.5):

∑ sin(ω0 t) _________ N = 0

∑ cos(ω0 t) _________ N = 0

(16.13) ∑ sin2(ω0 t) _________ N =

1 __ 2 ∑ cos2(ω0 t) _________ N =

1 __ 2

∑cos(ω0 t ) sin(ω0 t ) ______________ N = 0

Thus, for equispaced points the normal equations become

[

0 0 N∕2 0

N∕2 ]

⎧

⎪

⎨ ⎪

⎩

B1 B2

⎫

⎪

⎬ ⎪

⎭

= ⎧

⎪

⎨ ⎪

⎩

∑ y

∑ y cos(ω0t ) ∑ y sin(ω0 t )

⎫

⎪

⎬ ⎪

⎭

The inverse of a diagonal matrix is merely another diagonal matrix whose elements are the reciprocals of the original. Thus, the coefficients can be determined as

⎧

⎪

⎨ ⎪

⎩ A0

B1 B2

⎫

⎪

⎬ ⎪

⎭ =

⎡ ⎢

⎣ 1∕N

0 0 2∕N 0

2∕N ⎤ ⎥

⎦ ⎧

⎪

⎨ ⎪

⎩

∑ y

∑ y cos(ω0t ) ∑ y sin(ω0 t )

⎫

⎪

⎬ ⎪

⎭

A0 = ∑ y

____ N (16.14)

A1 = 2 __ N ∑ y cos(ω0 t ) (16.15)

B1 = 2 __ N ∑ y sin(ω0 t ) (16.16)

Notice that the first coefficient represents the function’s average value.

EXAMPLE 16.1 Least-Squares Fit of a Sinusoid

Problem Statement. The curve in Fig. 16.2a is described by y = 1.7 + cos(4.189t + 1.0472). Generate 10 discrete values for this curve at intervals of Δt = 0.15 for the range t = 0 to 1.35. Use this information to evaluate the coefficients of Eq. (16.11) by a least- squares fit.

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410 FOURIER ANALySIS

Solution. The data required to evaluate the coefficients with ω = 4.189 are

t y y cos(ω0t) y sin(ω0t)

0 2.200 2.200 0.000 0.15 1.595 1.291 0.938 0.30 1.031 0.319 0.980 0.45 0.722 −0.223 0.687 0.60 0.786 −0.636 0.462 0.75 1.200 −1.200 0.000 0.90 1.805 −1.460 −1.061 1.05 2.369 −0.732 −2.253 1.20 2.678 0.829 −2.547 1.35 2.614 2.114 −1.536

∑ = 17.000 2.502 −4.330

These results can be used to determine [Eqs. (16.14) through (16.16)]

A0 = 17.000 ______ 10 = 1.7 A1 =

2 ___ 10 2.502 = 0.500 B1 = 2 ___ 10 (−4.330) = −0.866

Thus, the least-squares fit is

y = 1.7 + 0.500 cos(ω0 t ) − 0.866 sin(ω0 t )

The model can also be expressed in the format of Eq. (16.2) by calculating [Eq. (16.8)]

θ = arctan ( −0.866 _______ 0.500 ) = 1.0472 and [Eq. (16.9)]

C1 = √ ______________

0.52 + (−0.866)2 = 1.00

to give y = 1.7 + cos(ω0 t + 1.0472)

or alternatively, as a sine by using [Eq. (16.10)] y = 1.7 + sin(ω0 t + 2.618)

The foregoing analysis can be extended to the general model

f (t) = A0 + A1cos(ω0 t) + B1sin(ω0 t) + A2cos(2ω0 t) + B2sin(2ω0 t ) + · · · + Amcos(mω0 t) + Bmsin(mω0 t)

where, for equally spaced data, the coefficients can be evaluated by

A0 = ∑ y

____ N

Aj =

2 __ N ∑ y cos( jω0) t Bj =

2 __ N ∑ y sin( jω0 t) ⎫

⎪

⎬ ⎪

⎭

j = 1, 2, . . . , m

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16.2 CONTINUOUS FOURIER SERIES 411

Although these relationships can be used to fit data in the regression sense (i.e., N > 2m + 1), an alternative application is to employ them for interpolation or collocation—that is, to use them for the case where the number of unknowns 2m + 1 is equal to the number of data points N. This is the approach used in the continuous Fourier series, as described next.

16.2 CONTINUOUS FOURIER SERIES

In the course of studying heat-flow problems, Fourier showed that an arbitrary periodic function can be represented by an infinite series of sinusoids of harmonically related frequencies. For a function with period T, a continuous Fourier series can be written

f (t) = a0 + a1cos(ω0 t ) + b1sin(ω0 t ) + a2cos(2ω0 t ) + b2sin(2ω0 t ) + · · ·

or more concisely,

f (t) = a0 + ∑ k=1

∞

[akcos(kω0 t ) + bksin(kω0 t )] (16.17)

where the angular frequency of the first mode (ω0 = 2π / T ) is called the fundamental frequency and its constant multiples 2ω0, 3ω0, etc., are called harmonics. Thus, Eq. (16.17) expresses f (t) as a linear combination of the basis functions: 1, cos(ω0 t ), sin(ω0 t ), cos(2ω0 t ), sin(2ω0 t), . . . .

The coefficients of Eq. (16.17) can be computed via

ak = 2 __ T ∫ 0

f (t)cos(kω0 t ) dt (16.18)

and

bk = 2 __ T ∫ 0

f (t)sin(kω0 t ) dt (16.19)

for k = 1, 2, . . . and

a0 = 1 __ T ∫ 0

f (t) dt (16.20)

EXAMPLE 16.2 Continuous Fourier Series Approximation

Problem Statement. Use the continuous Fourier series to approximate the square or rect- angular wave function (Fig. 16.1a) with a height of 2 and a period T = 2π∕ω0:

f (t) = ⎧

⎪

⎨ ⎪

⎩

−1

−T∕2 < t <

−T∕4

1 −T∕4 < t < T∕4 −1

T∕4 < t <

T∕2

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412 FOURIER ANALySIS

Solution. Because the average height of the wave is zero, a value of a0 = 0 can be obtained directly. The remaining coefficients can be evaluated as [Eq. (16.18)]

ak = 2 __ T ∫ −T/2

T/2

f (t) cos(kω0 t ) dt

= 2 __ T [− ∫ −T/2 −T/4

cos(kω0 t ) dt + ∫ −T/4

T/4

cos(kω0 t ) dt − ∫ T/4

T/2

cos(kω0 t ) dt]

The integrals can be evaluated to give

ak = { 4∕(kπ)

for k = 1, 5, 9, . . .

−4∕(kπ) for k = 3, 7, 11, . . . 0

for k = even integers

cos (ω0t) 4π

cos (3ω0t) 43π

cos (5ω0t) 45π

(a)

(b)

(c)

FIGURE 16.4 The Fourier series approximation of a square wave. The series of plots shows the summation up to and including the (a) first, (b) second, and (c) third terms. The individual terms that were added or subtracted at each stage are also shown.

cos (ω0t) 4π

cos (3ω0t) 43π

cos (5ω0t) 45π

(a)

(b)

(c)

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16.2 CONTINUOUS FOURIER SERIES 413

Similarly, it can be determined that all the b’s = 0. Therefore, the Fourier series approximation is

f (t) = 4 __ π cos(ω0 t ) − 4 ___ 3π cos(3ω0 t ) +

4 ___ 5π cos(5ω0 t ) − 4 ___ 7π cos(7ω0 t ) + · · ·

The results up to the first three terms are shown in Fig. 16.4.

Before proceeding, the Fourier series can also be expressed in a more compact form using complex notation. This is based on Euler’s formula (Fig. 16.5):

e±i x = cos x ± i sin x (16.21)

where i = √ ___

−1 , and x is in radians. Equation (16.21) can be used to express the Fourier series concisely as (Chapra and Canale, 2010)

f (t) = ∑ k=−∞

∞

~ c k eikω0 t (16.22)

where the coefficients are

~ c k = 1 __ T ∫ −T/2

T/2

f (t) e−ikω0 t dt (16.23)

Note that the tildes ~ are included to stress that the coefficients are complex numbers. Because it is more concise, we will primarily use the complex form in the rest of the chapter. Just remember, that it is identical to the sinusoidal representation.

FIGURE 16.5 Graphical depiction of Euler’s formula. The rotating vector is called a phasor.

Real cos θ

eiθ = cos θ + i sin θ

0 1

sin θ

Imaginary

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414 FOURIER ANALySIS

16.3 FREQUENCY AND TIME DOMAINS

To this point, our discussion of Fourier analysis has been limited to the time domain. We have done this because most of us are fairly comfortable conceptualizing a function’s behavior in this dimension. Although it is not as familiar, the frequency domain provides an alternative perspective for characterizing the behavior of oscillating functions.

Just as amplitude can be plotted versus time, it can also be plotted versus frequency. Both types of expression are depicted in Fig. 16.6a, where we have drawn a three- dimensional graph of a sinusoidal function:

f (t) = C1cos ( t + π __ 2 )

f(t)

f t

f1/T

A m

pl itu

0 0

−π

f1/T

1/T

Ph as

Frequency

Ti m e

(a)

(b)

FIGURE 16.6 (a) A depiction of how a sinusoid can be portrayed in the time and the frequency domains. The time projection is reproduced in (b), whereas the amplitude-frequency projection is reproduced in (c). The phase-frequency projection is shown in (d).

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16.4 FOURIER INTEGRAL AND TRANSFORM 415

In this plot, the magnitude or amplitude of the curve f (t) is the dependent variable, and time t and frequency f = ω0∕2π are the independent variables. Thus, the amplitude and the time axes form a time plane, and the amplitude and the frequency axes form a frequency plane. The sinusoid can, therefore, be conceived of as existing a distance 1∕T out along the frequency axis and running parallel to the time axes. Consequently, when we speak about the behavior of the sinusoid in the time domain, we mean the projection of the curve onto the time plane (Fig. 16.6b). Similarly, the behavior in the frequency domain is merely its projection onto the frequency plane.

As in Fig. 16.6c, this projection is a measure of the sinusoid’s maximum positive amplitude C1. The full peak-to-peak swing is unnecessary because of the symmetry. Together with the location 1∕T along the frequency axis, Fig. 16.6c now defines the amplitude and frequency of the sinusoid. This is enough information to reproduce the shape and size of the curve in the time domain. However, one more parameter—namely, the phase angle—is required to position the curve relative to t = 0. Consequently, a phase diagram, as shown in Fig. 16.6d, must also be included. The phase angle is de- termined as the distance (in radians) from zero to the point at which the positive peak occurs. If the peak occurs after zero, it is said to be delayed (recall our discussion of lags and leads in Sec. 16.1), and by convention, the phase angle is given a negative sign. Conversely, a peak before zero is said to be advanced and the phase angle is posi- tive. Thus, for Fig. 16.6, the peak leads zero and the phase angle is plotted as +π∕2. Figure 16.7 depicts some other possibilities.

We can now see that Fig. 16.6c and d provide an alternative way to present or summa- rize the pertinent features of the sinusoid in Fig. 16.6a. They are referred to as line spectra. Admittedly, for a single sinusoid they are not very interesting. However, when applied to a more complicated situation—say, a Fourier series—their true power and value is revealed. For example, Fig. 16.8 shows the amplitude and phase line spectra for the square-wave function from Example 16.2.

Such spectra provide information that would not be apparent from the time domain. This can be seen by contrasting Fig. 16.4 and Fig. 16.8. Figure 16.4 presents two alter- native time domain perspectives. The first, the original square wave, tells us nothing about the sinusoids that comprise it. The alternative is to display these sinusoids— that is, (4∕π) cos(ω0 t ), −(4∕3π) cos(3ω0 t ), (4∕5π) cos(5ω0 t ), etc. This alternative does not provide an adequate visualization of the structure of these harmonics. In contrast, Fig. 16.8a and b provide a graphic display of this structure. As such, the line spectra represent “fingerprints” that can help us to characterize and understand a complicated waveform. They are particularly valuable for nonidealized cases where they sometimes allow us to discern structure in otherwise obscure signals. In the next section, we will describe the Fourier transform that will allow us to extend such analyses to nonperiodic waveforms.

16.4 FOURIER INTEGRAL AND TRANSFORM

Although the Fourier series is a useful tool for investigating periodic functions, there are many waveforms that do not repeat themselves regularly. For example, a lightning bolt occurs only once (or at least it will be a long time until it occurs again), but it will cause

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416 FOURIER ANALySIS

interference with receivers operating on a broad range of frequencies—for example, TVs, radios, and shortwave receivers. Such evidence suggests that a nonrecurring signal such as that produced by lightning exhibits a continuous frequency spectrum. Because such phenomena are of great interest to engineers, an alternative to the Fourier series would be valuable for analyzing these aperiodic waveforms.

The Fourier integral is the primary tool available for this purpose. It can be derived from the exponential form of the Fourier series [Eqs. (16.22) and (16.23)]. The transi- tion from a periodic to a nonperiodic function can be effected by allowing the period to approach infinity. In other words, as T becomes infinite, the function never repeats itself and thus becomes aperiodic. If this is allowed to occur, it can be demonstrated (e.g., Van Valkenburg, 1974; Hayt and Kemmerly, 1986) that the Fourier series reduces to

f (t) = 1 __ 2π ∫ −∞ ∞

F (ω)eiωt dω (16.24)

FIGURE 16.7 Various phases of a sinusoid showing the associated phase line spectra.

−π

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16.4 FOURIER INTEGRAL AND TRANSFORM 417

and the coefficients become a continuous function of the frequency variable ω, as in

F (ω) = ∫ −∞

∞

f (t)e−iωt dt (16.25)

The function F(ω), as defined by Eq. (16.25), is called the Fourier integral of f (t). In addition, Eqs. (16.24) and (16.25) are collectively referred to as the Fourier transform pair. Thus, along with being called the Fourier integral, F(ω) is also called the Fourier trans- form of f (t). In the same spirit, f (t), as defined by Eq. (16.24), is referred to as the inverse Fourier transform of F(ω). Thus, the pair allows us to transform back and forth between the time and the frequency domains for an aperiodic signal.

The distinction between the Fourier series and transform should now be quite clear. The major difference is that each applies to a different class of functions—the series to periodic and the transform to nonperiodic waveforms. Beyond this major distinction, the two approaches differ in how they move between the time and the frequency domains. The Fourier series converts a continuous, periodic time-domain function to frequency-domain magnitudes at discrete frequencies. In contrast, the Fourier transform converts a continu- ous time-domain function to a continuous frequency-domain function. Thus, the discrete frequency spectrum generated by the Fourier series is analogous to a continuous frequency spectrum generated by the Fourier transform.

FIGURE 16.8 (a) Amplitude and (b) phase line spectra for the square wave from Fig. 16.4.

4/π

f0 3f0 5f0 7f0

2/π

f0 3f0 5f0 7f0

π/2

−π/2

−π

(a)

(b)

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418 FOURIER ANALySIS

Now that we have introduced a way to analyze an aperiodic signal, we will take the final step in our development. In the next section, we will acknowledge the fact that a signal is rarely characterized as a continuous function of the sort needed to implement Eq. (16.25). Rather, the data are invariably in a discrete form. Thus, we will now show how to compute a Fourier transform for such discrete measurements.

16.5 DISCRETE FOURIER TRANSFORM (DFT)

In engineering, functions are often represented by a finite set of discrete values. Addi- tionally, data are often collected in or converted to such a discrete format. As depicted in Fig. 16.9, an interval from 0 to T can be divided into n equispaced subintervals with widths of Δt = T∕n. The subscript j is employed to designate the discrete times at which samples are taken. Thus, fj designates a value of the continuous function f (t) taken at tj. Note that the data points are specified at j = 0, 1, 2, . . . , n − 1. A value is not included at j = n. (See Ramirez, 1985, for the rationale for excluding fn.)

For the system in Fig. 16.9, a discrete Fourier transform can be written as

Fk = ∑ j=0

n−1

fj e−ik ω 0 j for k = 0 to n − 1 (16.26)

and the inverse Fourier transform as

fj = 1 __ n ∑ k=0

n−1

Fk eik ω 0 j for j = 0 to n − 1 (16.27)

where ω0 = 2π∕n.

FIGURE 16.9 The sampling points of the discrete Fourier series.

∆t

f (t)

t0 t 1 t 2

f 1

f 2 f 3

f n −1

t n −1 t n = T

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16.5 DISCRETE FOURIER TRANSFORM (DFT) 419

Equations (16.26) and (16.27) represent the discrete analogs of Eqs. (16.25) and (16.24), respectively. As such, they can be employed to compute both a direct and an in- verse Fourier transform for discrete data. Note that the factor 1∕n in Eq. (16.27) is merely a scale factor that can be included in either Eq. (16.26) or (16.27), but not both. For example, if it is shifted to Eq. (16.26), the first coefficient F0 (which is the analog of the constant a0) is equal to the arithmetic mean of the samples.

Before proceeding, several other aspects of the DFT bear mentioning. The highest frequency that can be measured in a signal, called the Nyquist frequency, is half the sam- pling frequency. Periodic variations that occur more rapidly than the shortest sampled time interval cannot be detected. The lowest frequency you can detect is the inverse of the total sample length.

As an example, suppose that you take 100 samples of data (n = 100 samples) at a sample frequency of fs = 1000 Hz (i.e., 1000 samples per second). This means that the sample interval is

Δt = 1 __ fs = 1 _____________ 1000 samples/s = 0.001 s/sample

The total sample length is

tn = n __ fs

= 100 samples _____________ 1000 samples/s = 0.1 s

and the frequency increment is

Δ f = fs __ n =

1000 samples/s _____________ 100 samples = 10 Hz

The Nyquist frequency is

fmax = 0.5 fs = 0.5(1000 Hz) = 500 Hz

and the lowest detectable frequency is

fmin = 1 ____ 0.1 s = 10 Hz

Thus, for this example, the DFT could detect signals with periods from 1∕500 = 0.002 s up to 1∕10 = 0.1 s.

16.5.1 Fast Fourier Transform (FFT)

Although an algorithm can be developed to compute the DFT based on Eq. (16.26), it is computationally burdensome because n2 operations are required. Consequently, for data samples of even moderate size, the direct determination of the DFT can be extremely time consuming.

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420 FOURIER ANALySIS

The fast Fourier transform, or FFT, is an algorithm that has been developed to com- pute the DFT in an extremely economical fashion. Its speed stems from the fact that it utilizes the results of previous computations to reduce the number of operations. In particular, it exploits the periodicity and symmetry of trigonometric functions to com- pute the transform with approximately n log2n operations (Fig. 16.10). Thus, for n = 50 samples, the FFT is about 10 times faster than the standard DFT. For n = 1000, it is about 100 times faster.

The first FFT algorithm was developed by Gauss in the early nineteenth century (Heideman et al., 1984). Other major contributions were made by Runge, Danielson, Lanczos, and others in the early twentieth century. However, because discrete transforms often took days to weeks to calculate by hand, they did not attract broad interest prior to the development of the modern digital computer.

In 1965, J. W. Cooley and J. W. Tukey published a key paper in which they outlined an algorithm for calculating the FFT. This scheme, which is similar to those of Gauss and other earlier investigators, is called the Cooley-Tukey algorithm. Today, there are a host of other approaches that are offshoots of this method. As described next, MATLAB offers a function called fft that employs such efficient algorithms to compute the DFT.

16.5.2 MATLAB Function: fft

MATLAB’s fft function provides an efficient way to compute the DFT. A simple repre- sentation of its syntax is

F = fft(f, n)

FIGURE 16.10 Plot of number of operations versus sample size for the standard DFT and the FFT.

FFT(∼n lo g2n)

DFT(∼n2)

Samples

2000

1000

O pe

ra tio

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16.5 DISCRETE FOURIER TRANSFORM (DFT) 421

where F = a vector containing the DFT, and f = a vector containing the signal. The parameter n, which is optional, indicates that the user wants to implement an n-point FFT. If f has less than n points, it is padded with zeros and truncated if it has more.

Note that the elements in F are sequenced in what is called reverse-wrap-around order. The first half of the values are the positive frequencies (starting with the constant) and the second half are the negative frequencies. Thus, if n = 8, the order is 0, 1, 2, 3, 4, −3, −2, −1. The following example illustrates the function’s use to calculate the DFT of a simple sinusoid.

EXAMPLE 16.3 Computing the DFT of a Simple Sinusoid with MATLAB

Problem Statement. Apply the MATLAB fft function to determine the discrete Fourier transform for a simple sinusoid:

f (t) = 5 + cos(2π(12.5)t ) + sin(2π(18.75)t )

Generate 8 equispaced points with Δt = 0.02 s. Plot the result versus frequency.

Solution. Before generating the DFT, we can compute a number of quantities. The sam- pling frequency is

fs = 1 ___ Δt =

1 _____ 0.02 s = 50 Hz

The total sample length is

tn = n __ fs

= 8 samples ___________ 50 samples/s = 0.16 s

The Nyquist frequency is

fmax = 0.5 fs = 0.5(50 Hz) = 25 Hz

and the lowest detectable frequency is

fmin = 1 _____ 0.16 s = 6.25 Hz

Thus, the analysis can detect signals with periods from 1∕25 = 0.04 s up to 1∕6.25 = 0.16 s. So we should be able to detect both the 12.5 and 18.75 Hz signals. The following MATLAB statements can be used to generate and plot the sample (Fig. 16.11a):

>> clc >> n = 8; dt = 0.02; fs = 1/dt; T = 0.16; >> tspan = (0:n − 1)/fs; >> y = 5+cos(2*pi*12.5*tspan)+sin(2*pi*18.75*tspan);

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422 FOURIER ANALySIS

>> subplot(3,1,1); >> plot(tspan,y,'−ok','linewidth',2,'MarkerFaceColor','black'); >> title('(a) f(t) versus time (s)');

As was mentioned at the beginning of Sec. 16.5, notice that tspan omits the last point. The fft function can be used to compute the DFT and display the results

>> Y = fft(y)/n; >> Y'

We have divided the transform by n in order that the first coefficient is equal to the arithme- tic mean of the samples. When this code is executed, the results are displayed as

ans = 5.0000 0.0000 − 0.0000i 0.5000 −0.0000 + 0.5000i 0 −0.0000 − 0.5000i 0.5000 0.0000 + 0.0000i

0 0.02 0.04 0.06 0.08 0.1 0.12 0.14 2

(a) f (t) versus time (s)

6 8 10 12 14 16 18 20 22 24 26 −0.5

0.5

(b) Real component versus frequency

6 8 10 12 14 16 18 20 22 24 26 −0.5

0.5

Frequency (Hz)

FIGURE 16.11 Results of computing a DFT with MATLAB’s fft function: (a) the sample; and plots of the (b) real and (c) imaginary parts of the DFT versus frequency.

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16.6 THE POWER SPECTRUM 423

Notice that the first coefficient corresponds to the signal’s mean value. In addition, be- cause of the reverse-wrap-around order, the results can be interpreted as in the following table:

Index k Frequency Period Real Imaginary

1 0 constant 5 0 2 1 6.25 0.16 0 0 3 2 12.5 0.08 0.5 0 4 3 18.75 0.053333 0 0.5 5 4 25 0.04 0 0 6 −3 31.25 0.032 0 −0.5 7 −2 37.5 0.026667 0.5 0 8 −1 43.75 0.022857 0 0

Notice that the fft has detected the 12.5- and 18.75-Hz signals. In addition, we have high- lighted the Nyquist frequency to indicate that the values below it in the table are redundant. That is, they are merely reflections of the results below the Nyquist frequency. If we remove the constant value, we can plot both the real and imaginary parts of the DFT versus frequency

>> nyquist = fs/2;fmin = 1/T; >> f = linspace(fmin,nyquist,n/2); >> Y(1)=[];YP = Y(1:n/2); >> subplot(3,1,2) >> stem(f,real(YP),'linewidth',2,'MarkerFaceColor','blue') >> grid;title('(b) Real component versus frequency') >> subplot(3,1,3) >> stem(f,imag(YP),'linewidth',2,'MarkerFaceColor','blue') >> grid;title('(b) Imaginary component versus frequency') >> xlabel('frequency (Hz)')

As expected (recall Fig. 16.7), a positive peak occurs for the cosine at 12.5 Hz (Fig. 16.11b), and a negative peak occurs for the sine at 18.75 Hz (Fig. 16.11c).

16.6 THE POWER SPECTRUM

Beyond amplitude and phase spectra, power spectra provide another useful way to discern the underlying harmonics of seemingly random signals. As the name implies, it derives from the analysis of the power output of electrical systems. In terms of the DFT, a power spectrum consists of a plot of the power associated with each frequency component versus frequency. The power can be computed by summing the squares of the Fourier coefficients:

Pk = | ~ c k |2

where Pk is the power associated with each frequency kω0.

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424 FOURIER ANALySIS

EXAMPLE 16.4 Computing the Power Spectrum with MATLAB

Problem Statement. Compute the power spectrum for the simple sinusoid for which the DFT was computed in Example 16.3.

Solution. The following script can be developed to compute the power spectrum:

% compute the DFT clc;clf n = 8; dt = 0.02; fs = 1/dt;tspan = (0:n−1)/fs; y = 5+cos(2*pi*12.5*tspan)+sin(2*pi*18.75*tspan); Y = fft(y)/n; f = (0:n−1)*fs/n; Y(1) = [];f(1) = []; % compute and display the power spectrum nyquist=fs/2; f = (1:n/2)/(n/2)*nyquist; Pyy = abs(Y(1:n/2)).^2; stem(f,Pyy,'linewidth',2,'MarkerFaceColor','blue') title('Power spectrum') xlabel('Frequency (Hz)');ylim([0 0.3])

As indicated, the first section merely computes the DFT with the pertinent statements from Example 16.3. The second section then computes and displays the power spectrum. As in Fig. 16.12, the resulting graph indicates that peaks occur at both 12.5 and 18.75 Hz as expected.

6 8 2610 12 14 16 18 20 22 24

0.05

0.1

0.15

0.2

0.25

Power spectrum

Frequency (Hz)

FIGURE 16.12 Power spectrum for a simple sinusoidal function with frequencies of 12.5 and 18.75 Hz.

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16.7 CASE STUDY SUNSPOTS

Background. In 1848, Johann Rudolph Wolf devised a method for quantifying solar ac- tivity by counting the number of individual spots and groups of spots on the sun’s surface. He computed a quantity, now called a Wolf sunspot number, by adding 10 times the num- ber of groups plus the total count of individual spots. As in Fig. 16.13, the data set for the sunspot number extends back to 1700. On the basis of the early historical records, Wolf determined the cycle’s length to be 11.1 years. Use a Fourier analysis to confirm this result by applying an FFT to the data.

Solution. The data for year and sunspot number are contained in a MATLAB file, sunspot.dat. The following statements load the file and assign the year and number infor- mation to vectors of the same name:

>> load sunspot.dat >> year = sunspot(:,1);number = sunspot(:,2);

Before applying the Fourier analysis, it is noted that the data seem to exhibit an upward linear trend (Fig. 16.13). MATLAB can be used to remove this trend:

>> n = length(number); >> a = polyfit(year,number,1); >> lineartrend = polyval(a,year); >> ft = number−lineartrend;

Next, the fft function is employed to generate the DFT F = fft(ft);

The power spectrum can then be computed and plotted fs = 1; f = (0:n/2)*fs/n; pow = abs(F(1:n/2+1)).^2;

FIGURE 16.13 Plot of Wolf sunspot number versus year. The dashed line indicates a mild, upward linear trend.

1700 20001750 1800 1850 1900 1950 0

100

150

200

16.7 CASE STUDy 425

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426 FOURIER ANALySIS

PROBLEMS

16.1 The following equation describes the variations of temperature of a tropical lake:

T(t ) = 12.8 + 4 cos ( 2π ___ 365 t) + 3 sin (

2π ___ 365 t)

What is (a) the mean temperature, (b) the amplitude, and (c) the period? 16.2 The temperature in a pond varies sinusoidally over the course of a year. Use linear least-squares regression to fit Eq. (16.11) to the following data. Use your fit to determine the mean, amplitude, and date of maximum temperature. Note that the period is 365 d.

plot(f,pow) xlabel('Frequency (cycles/year)'); ylabel('Power') title('Power versus frequency')

The result, as shown in Fig. 16.14, indicates a peak at a frequency of about 0.0915 cycles/yr. This corresponds to a period of 1∕0.0915 = 10.93 years. Thus, the Fourier analysis is consis- tent with Wolf’s estimate of 11 years.

16.7 CASE STUDY continued

t, d 15 45 75 105 135 165 225 255 285 315 345

T, °C 3.4 4.78.5 11.7 16 18.7 19.7 17.1 12.7 7.7 5.1

16.3 The pH in a reactor varies sinusoidally over the course of a day. Use least-squares regression to fit Eq. (16.11) to the following data. Use your fit to determine the mean, ampli- tude, and time of maximum pH. Note that the period is 24 hr

Time, hr 0 2 4 5 7 9 pH 7.6 7.2 7 6.5 7.5 7.2

Time, hr 12 15 20 22 24 pH 8.9 9.1 8.9 7.9 7

FIGURE 16.14 Power spectrum for Wolf sunspot number versus year.

0 0.05 0.50.1 0.15 0.2 0.25 0.3 0.35 0.4 0.45 0

2000

3000

4000

5000 Power spectrum

Po w

Cycles/year

1000

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16.4 The solar radiation for Tucson, Arizona, has been tab- ulated as

Time, mo J F M A M J Radiation, W/m2 144 188 245 311 351 359

Time, mo J A S O N D Radiation, W/m2 308 287 260 211 159 131

Assuming each month is 30 days long, fit a sinusoid to these data. Use the resulting equation to predict the radiation in mid-August. 16.5 The average values of a function can be determined by

f̄ = ʃ t

0 f (t) dt ______ t

Use this relationship to verify the results of Eq. (16.13). 16.6 In electric circuits, it is common to see current behavior in the form of a square wave as shown in Fig. P16.6 (notice that square wave differs from the one described in Example 16.2). Solving for the Fourier series from

f (t) = { A0 0 ≤ t ≤ T∕2 −A0 T∕2 ≤ t ≤ T

the Fourier series can be represented as

f (t) = ∑ n=1

∞

( 4A0 ________ (2n − 1)π ) sin ( 2π (2n − 1) t __________ T ) Develop a MATLAB function to generate a plot of the first n terms of the Fourier series individually, as well as the sum of these six terms. Design your function so that it plots the curves from t = 0 to 4T. Use thin dotted red lines for the individual terms and a bold black solid line for the

summation (i.e., 'k−','linewidth',2). The function’s first line should be function [t,f] = FourierSquare(A0,T,n)

Let A0 = 1 and T = 0.25 s. 16.7 Use a continuous Fourier series to approximate the sawtooth wave in Fig. P16.7. Plot the first four terms along with the summation. In addition, construct amplitude and phase line spectra for the first four terms. 16.8 Use a continuous Fourier series to approximate the tri- angular wave form in Fig. P16.8. Plot the first four terms along with the summation. In addition, construct amplitude and phase line spectra for the first four terms. 16.9 Use the Maclaurin series expansions for ex, cos x, and sin x to prove Euler’s formula [Eq. (16.21)]. 16.10 A half-wave rectifier can be characterized by

C1 = [ 1 __ π +

1 __ 2 sin t − 2 ___

3π cos 2t − 2 ____

15π cos 4t

− 2 ____ 35π

cos 6t − . . . ]

0 0.25 0.5 0.75 1

f (t)

−1

FIGURE P16.6

T−1

−1

FIGURE P16.7 A sawtooth wave.

2−2

FIGURE P16.8 A triangular wave.

PROBLEMS 427

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428 FOURIER ANALySIS

where C1 is the amplitude of the wave. (a) Plot the first four terms along with the summation. (b) Construct amplitude and phase line spectra for the first

four terms. 16.11 Duplicate Example 16.3, but for 64 points sampled at a rate of Δt = 0.01 s from the function

f (t) = cos[2π (12.5)t] + cos[2π (25)t]

Use fft to generate a DFT of these values and plot the results. 16.12 Use MATLAB to generate 64 points from the function

f (t) = cos(10t) + sin(3t)

from t = 0 to 2π. Add a random component to the signal with the function randn. Use fft to generate a DFT of these values and plot the results. 16.13 Use MATLAB to generate 32 points for the sinusoid depicted in Fig. 16.2 from t = 0 to 6 s. Compute the DFT and create subplots of (a) the original signal, (b) the real part, and (c) the imaginary part of the DFT versus frequency.

16.14 Use the fft function to compute a DFT for the trian- gular wave from Prob. 16.8. Sample the wave from t = 0 to 4T using 128 sample points. 16.15 Develop an M-file function that uses the fft func- tion to generate a power spectrum plot. Use it to solve Prob. 16.11. 16.16 Use the fft function to compute the DFT for the fol- lowing function:

f (t) = 1.5 + 1.8cos(2π (12)t) + 0.8sin(2π (20)t) − 1.25cos(2π (28)t)

Take n = 64 samples with a sampling frequency of fs = 128 samples/s. Have your script compute values of Δt, tn, Δf, fmin, and fmax. As illustrated in Examples 16.3 and 16.4, have your script generate plots as in Fig. 16.11 and Fig. 16.12. 16.17 If you take 128 samples of data (n = 128 samples) with a total sample length of tn = 0.4 s, compute the follow- ing: (a) the sample frequency, fs (sample/s); (b) the sample interval, Δt (s/sample); (c) the Nyquist frequency, fmax (Hz); (d) the minimum frequency, fmin (Hz).

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429

Polynomial Interpolation

CHAPTER OBJECTIVES

The primary objective of this chapter is to introduce you to polynomial interpolation. Specific objectives and topics covered are

• Recognizing that evaluating polynomial coefficients with simultaneous equations is an ill-conditioned problem.

• Knowing how to evaluate polynomial coefficients and interpolate with MATLAB’s polyfit and polyval functions.

• Knowing how to perform an interpolation with Newton’s polynomial. • Knowing how to perform an interpolation with a Lagrange polynomial. • Knowing how to solve an inverse interpolation problem by recasting it as a roots

problem. • Appreciating the dangers of extrapolation. • Recognizing that higher-order polynomials can manifest large oscillations.

YOU’VE GOT A PROBLEM

If we want to improve the velocity prediction for the free-falling bungee jumper, we might expand our model to account for other factors beyond mass and the drag coefficient. As was previously mentioned in Sec. 1.4, the drag coefficient can itself be formulated as a function of other factors such as the area of the jumper and characteristics such as the air’s density and viscosity.

Air density and viscosity are commonly presented in tabular form as a function of temperature. For example, Table 17.1 is reprinted from a popular fluid mechanics textbook (White, 1999).

Suppose that you desired the density at a temperature not included in the table. In such a case, you would have to interpolate. That is, you would have to estimate the value

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430 PolynomIal InterPolatIon

at the desired temperature based on the densities that bracket it. The simplest approach is to determine the equation for the straight line connecting the two adjacent values and use this equation to estimate the density at the desired intermediate temperature. Although such linear interpolation is perfectly adequate in many cases, error can be introduced when the data exhibit significant curvature. In this chapter, we will explore a number of different approaches for obtaining adequate estimates for such situations.

17.1 INTRODUCTION TO INTERPOLATION

You will frequently have occasion to estimate intermediate values between precise data points. The most common method used for this purpose is polynomial interpolation. The general formula for an (n − 1)th-order polynomial can be written as

f (x) = a1 + a2 x + a3 x 2 + · · · + an x n−1 (17.1)

For n data points, there is one and only one polynomial of order (n − 1) that passes through all the points. For example, there is only one straight line (i.e., a first-order polynomial) that connects two points (Fig. 17.1a). Similarly, only one parabola connects a set of three points (Fig. 17.1b). Polynomial interpolation consists of determining the unique (n − 1) th-order polynomial that fits n data points. This polynomial then provides a formula to compute intermediate values.

Before proceeding, we should note that MATLAB represents polynomial coefficients in a different manner than Eq. (17.1). Rather than using increasing powers of x, it uses decreasing powers as in

f (x) = p1 x n−1 + p2 x n−2 + · · · + pn−1 x + pn (17.2)

To be consistent with MATLAB, we will adopt this scheme in the following section.

TABLE 17.1 Density ( ρ), dynamic viscosity ( µ), and kinematic viscosity (υ) as a function of temperature (T ) at 1 atm as reported by White (1999).

T, °C ρ, kg/m3 µ, N · s/m2 υ, m2/s

−40 1.52 1.51 × 10−5 0.99 × 10−5 0 1.29 1.71 × 10−5 1.33 × 10−5 20 1.20 1.80 × 10−5 1.50 × 10−5 50 1.09 1.95 × 10−5 1.79 × 10−5 100 0.946 2.17 × 10−5 2.30 × 10−5 150 0.835 2.38 × 10−5 2.85 × 10−5 200 0.746 2.57 × 10−5 3.45 × 10−5 250 0.675 2.75 × 10−5 4.08 × 10−5 300 0.616 2.93 × 10−5 4.75 × 10−5 400 0.525 3.25 × 10−5 6.20 × 10−5 500 0.457 3.55 × 10−5 7.77 × 10−5

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17.1 IntroDUCtIon to InterPolatIon 431

17.1.1 Determining Polynomial Coefficients

A straightforward way for computing the coefficients of Eq. (17.2) is based on the fact that n data points are required to determine the n coefficients. As in the following example, this allows us to generate n linear algebraic equations that we can solve simultaneously for the coefficients.

eXamPle 17.1 Determining Polynomial Coefficients with Simultaneous equations

Problem Statement. Suppose that we want to determine the coefficients of the parabola, f (x) = p1x2 + p2 x + p3, that passes through the last three density values from Table 17.1:

x1 = 300 f (x1) = 0.616

x2 = 400 f (x2) = 0.525

x3 = 500 f (x3) = 0.457

Each of these pairs can be substituted into Eq. (17.2) to yield a system of three equations:

0.616 = p1(300)2 + p2(300) + p3 0.525 = p1(400)2 + p2(400) + p3 0.457 = p1(500)2 + p2(500) + p3

or in matrix form:

[

90,000

300

1 160,000 400 1

250,000

500

1 ]

{

p2 p3

}

= {

0.616

0.525 0.457

}

Thus, the problem reduces to solving three simultaneous linear algebraic equations for the three unknown coefficients. A simple MATLAB session can be used to obtain the

(a) (b) (c)

FIGURE 17.1 Examples of interpolating polynomials: (a) first-order (linear) connecting two points, (b) second-order (quadratic or parabolic) connecting three points, and (c) third-order (cubic) connecting four points.

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432 PolynomIal InterPolatIon

solution: >> format long >> A = [90000 300 1;160000 400 1;250000 500 1]; >> b = [0.616 0.525 0.457]'; >> p = A\b

p = 0.00000115000000 −0.00171500000000 1.02700000000000

Thus, the parabola that passes exactly through the three points is

f (x) = 0.00000115x2 − 0.001715x + 1.027

This polynomial then provides a means to determine intermediate points. For example, the value of density at a temperature of 350 °C can be calculated as

f (350) = 0.00000115(350)2 − 0.001715(350) + 1.027 = 0.567625

Although the approach in Example 17.1 provides an easy way to perform interpola- tion, it has a serious deficiency. To understand this flaw, notice that the coefficient matrix in Example 17.1 has a decided structure. This can be seen clearly by expressing it in general terms:

⎡ ⎢

⎣ x 1 2

1 x 2 2 x2 1

x 3 2

⎤ ⎥

⎦ {

p2 p3

}

= ⎧

⎪

⎨ ⎪

⎩ f (x1)

f (x2) f (x3)

⎫

⎪

⎬ ⎪

⎭ (17.3)

Coefficient matrices of this form are referred to as Vandermonde matrices. Such ma- trices are very ill-conditioned. That is, their solutions are very sensitive to round off errors. This can be illustrated by using MATLAB to compute the condition number for the coef- ficient matrix from Example 17.1 as

>> cond(A)

ans =

5.8932e + 006

This condition number, which is quite large for a 3 × 3 matrix, implies that about six digits of the solution would be questionable. The ill-conditioning becomes even worse as the number of simultaneous equations becomes larger.

As a consequence, there are alternative approaches that do not manifest this short- coming. In this chapter, we will also describe two alternatives that are well-suited for computer implementation: the Newton and the Lagrange polynomials. Before doing this, however, we will first briefly review how the coefficients of the interpolating polynomial can be estimated directly with MATLAB’s built-in functions.

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17.2 neWton InterPolatInG PolynomIal 433

17.1.2 MATLAB Functions: polyfit and polyval

Recall from Sec. 14.5.2, that the polyfit function can be used to perform polynomial regression. In such applications, the number of data points is greater than the number of coefficients being estimated. Consequently, the least-squares fit line does not necessarily pass through any of the points, but rather follows the general trend of the data.

For the case where the number of data points equals the number of coefficients, poly- fit performs interpolation. That is, it returns the coefficients of the polynomial that pass directly through the data points. For example, it can be used to determine the coefficients of the parabola that passes through the last three density values from Table 17.1:

>> format long >> T = [300 400 500]; >> density = [0.616 0.525 0.457]; >> p = polyfit(T,density,2)

p = 0.00000115000000 −0.00171500000000 1.02700000000000

We can then use the polyval function to perform an interpolation as in >> d = polyval(p,350)

d = 0.56762500000000

These results agree with those obtained previously in Example 17.1 with simultaneous equations.

17.2 NEWTON INTERPOLATING POLYNOMIAL

There are a variety of alternative forms for expressing an interpolating polynomial beyond the familiar format of Eq. (17.2). Newton’s interpolating polynomial is among the most popular and useful forms. Before presenting the general equation, we will introduce the first- and second-order versions because of their simple visual interpretation.

17.2.1 Linear Interpolation

The simplest form of interpolation is to connect two data points with a straight line. This technique, called linear interpolation, is depicted graphically in Fig. 17.2. Using similar triangles,

f1(x) − f (x1) __________ x − x1 =

f (x2 ) − f (x1) ___________ x2 − x1 (17.4)

which can be rearranged to yield

f1(x) = f (x1) + f (x2 ) − f (x1) ___________ x2 − x1 (x − x1) (17.5)

which is the Newton linear-interpolation formula. The notation f1(x) designates that this is a first-order interpolating polynomial. Notice that besides representing the slope of the line connecting the points, the term [ f (x2) − f (x1)]∕(x2 − x1) is a finite-difference approximation

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434 PolynomIal InterPolatIon

of the first derivative [recall Eq. (4.20)]. In general, the smaller the interval between the data points, the better the approximation. This is due to the fact that, as the interval de- creases, a continuous function will be better approximated by a straight line. This charac- teristic is demonstrated in the following example.

eXamPle 17.2 linear Interpolation

Problem Statement. Estimate the natural logarithm of 2 using linear interpolation. First, perform the computation by interpolating between ln 1 = 0 and ln 6 = 1.791759. Then, repeat the procedure, but use a smaller interval from ln 1 to ln 4 (1.386294). Note that the true value of ln 2 is 0.6931472.

Solution. We use Eq. (17.5) from x1 = 1 to x2 = 6 to give

f1(2) = 0 + 1.791759 − 0 ____________ 6 − 1 (2 − 1) = 0.3583519

which represents an error of εt = 48.3%. Using the smaller interval from x1 = 1 to x2 = 4 yields

f1(2) = 0 + 1.386294 − 0 ____________ 4 − 1 (2 − 1) = 0.4620981

f (x2)

f1(x)

f (x)

f (x1)

x1 x2x x

FIGURE 17.2 Graphical depiction of linear interpolation. The shaded areas indicate the similar triangles used to derive the Newton linear-interpolation formula [Eq. (17.5)].

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17.2 neWton InterPolatInG PolynomIal 435

Thus, using the shorter interval reduces the percent relative error to εt = 33.3%. Both interpolations are shown in Fig. 17.3, along with the true function.

17.2.2 Quadratic Interpolation

The error in Example 17.2 resulted from approximating a curve with a straight line. Con- sequently, a strategy for improving the estimate is to introduce some curvature into the line connecting the points. If three data points are available, this can be accomplished with a second-order polynomial (also called a quadratic polynomial or a parabola). A particularly convenient form for this purpose is

f2(x) = b1 + b2(x − x1) + b3(x − x1)(x − x2) (17.6)

A simple procedure can be used to determine the values of the coefficients. For b1, Eq. (17.6) with x = x1 can be used to compute

b1 = f (x1) (17.7)

Equation (17.7) can be substituted into Eq. (17.6), which can be evaluated at x = x2 for

b2 = f (x2) − f (x1) ___________ x2 − x1 (17.8)

Finally, Eqs. (17.7) and (17.8) can be substituted into Eq. (17.6), which can be evaluated at x = x3 and solved (after some algebraic manipulations) for

b3 = f (x3) − f (x2) ___________ x3 − x2 −

f (x2) − f (x1) ___________ x2 − x1 ________________________ x3 − x1 (17.9)

f (x)

f (x) = ln x

f1(x) True value

Linear estimates

x50

FIGURE 17.3 Two linear interpolations to estimate ln 2. Note how the smaller interval provides a better estimate.

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436 PolynomIal InterPolatIon

Notice that, as was the case with linear interpolation, b2 still represents the slope of the line connecting points x1 and x2. Thus, the first two terms of Eq. (17.6) are equivalent to linear interpolation between x1 and x2, as specified previously in Eq. (17.5). The last term, b3(x − x1)(x − x2), introduces the second-order curvature into the formula.

Before illustrating how to use Eq. (17.6), we should examine the form of the coef- ficient b3. It is very similar to the finite-difference approximation of the second derivative introduced previously in Eq. (4.27). Thus, Eq. (17.6) is beginning to manifest a structure that is very similar to the Taylor series expansion. That is, terms are added sequentially to capture increasingly higher-order curvature.

eXamPle 17.3 Quadratic Interpolation

Problem Statement. Employ a second-order Newton polynomial to estimate ln 2 with the same three points used in Example 17.2:

x1 = 1 f (x1) = 0

x2 = 4 f (x2) = 1.386294

x3 = 6 f (x3) = 1.791759

Solution. Applying Eq. (17.7) yields

b1 = 0

Equation (17.8) gives

b2 = 1.386294 − 0 ____________ 4 − 1 = 0.4620981

f (x)

f(x) = ln x

f2(x)

True value

Linear estimate

Quadratic estimate

x50

FIGURE 17.4 The use of quadratic interpolation to estimate ln 2. The linear interpolation from x = 1 to 4 is also included for comparison.

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17.2 neWton InterPolatInG PolynomIal 437

and Eq. (17.9) yields

b3 = 1.791759 − 1.386294 __________________ 6 − 4 − 0.4620981 _____________________________ 6 − 1 = −0.0518731

Substituting these values into Eq. (17.6) yields the quadratic formula

f2 (x) = 0 + 0.4620981(x − 1) − 0.0518731(x − 1)(x − 4)

which can be evaluated at x = 2 for f2(2) = 0.5658444, which represents a relative error of εt = 18.4%. Thus, the curvature introduced by the quadratic formula (Fig. 17.4) improves the interpolation compared with the result obtained using straight lines in Example 17.2 and Fig. 17.3.

17.2.3 General Form of Newton’s Interpolating Polynomials

The preceding analysis can be generalized to fit an (n − 1)th-order polynomial to n data points. The (n − 1)th-order polynomial is

fn−1(x) = b1 + b2(x − x1) + · · · + bn(x − x1)(x − x2 ) · · · (x − xn−1) (17.10)

As was done previously with linear and quadratic interpolation, data points can be used to evaluate the coefficients b1, b2, . . . , bn. For an (n − 1)th-order polynomial, n data points are required: [x1, f (x1)], [x2, f (x2)], . . . , [xn, f (xn )]. We use these data points and the following equations to evaluate the coefficients:

b1 = f (x1) (17.11)

b2 = f [x2, x1] (17.12)

b3 = f [x3, x2, x1] (17.13) · · ·

bn = f [xn, xn−1, . . . , x2, x1] (17.14)

where the bracketed function evaluations are finite divided differences. For example, the first finite divided difference is represented generally as

f [xi, xj ] = f (xi ) − f (xj) ___________ xi − xj (17.15)

The second finite divided difference, which represents the difference of two first divided differences, is expressed generally as

f [xi, xj , xk ] = f [xi, xj ] − f [xj, xk ] ________________ xi − xk (17.16)

Similarly, the nth finite divided difference is

f [xn, xn−1, . . . , x2, x1] = f [xn, xn−1, . . . , x2] − f [xn−1, xn−2, . . . , x1] __________________________________ xn − x1 (17.17)

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438 PolynomIal InterPolatIon

These differences can be used to evaluate the coefficients in Eqs. (17.11) through (17.14), which can then be substituted into Eq. (17.10) to yield the general form of Newton’s inter- polating polynomial:

fn−1(x) = f (x1) + (x − x1) f [x2, x1] + (x − x1)(x − x2) f [x3, x2, x1] + · · · + (x − x1)(x − x2) · · · (x − xn−1) f [xn, xn−1, . . . , x2, x1] (17.18)

We should note that it is not necessary that the data points used in Eq. (17.18) be equally spaced or that the abscissa values necessarily be in ascending order, as illustrated in the following example. However, the points should be ordered so that they are centered around and as close as possible to the unknown. Also, notice how Eqs. (17.15) through (17.17) are recursive—that is, higher-order differences are computed by taking differences of lower-order differences (Fig. 17.5). This property will be exploited when we develop an efficient M-file to implement the method.

eXamPle 17.4 newton Interpolating Polynomial

Problem Statement. In Example 17.3, data points at x1 = 1, x2 = 4, and x3 = 6 were used to estimate ln 2 with a parabola. Now, adding a fourth point [x4 = 5; f (x4) = 1.609438], estimate ln 2 with a third-order Newton’s interpolating polynomial.

Solution. The third-order polynomial, Eq. (17.10) with n = 4, is

f3(x) = b1 + b2(x − x1) + b3(x − x1)(x − x2) + b4(x − x1)(x − x2)(x − x3 )

The first divided differences for the problem are [Eq. (17.15)]

f [x2, x1] = 1.386294 − 0 ____________ 4 − 1 = 0.4620981

f [x3, x2 ] = 1.791759 − 1.386294 __________________ 6 − 4 = 0.2027326

f [x4, x3] = 1.609438 − 1.791759 __________________ 5 − 6 = 0.1823216

x1 x2 x3 x4

f (x1)

f (x2)

f (x3)

f (x4)

f(xi)

f [x2, x1]

f [x3, x2]

f [x4, x3]

First

f [x3, x2, x1]

f [x4, x3, x2]

Second

f [x4, x3, x2, x1]

Third

FIGURE 17.5 Graphical depiction of the recursive nature of finite divided differences. This representation is referred to as a divided difference table.

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17.2 neWton InterPolatInG PolynomIal 439

The second divided differences are [Eq. (17.16)]

f [x3, x2, x1] = 0.2027326 − 0.4620981 __________________ 6 − 1 = −0.05187311

f [x4, x3, x2 ] = 0.1823216 − 0.2027326 __________________ 5 − 4 = −0.02041100

The third divided difference is [Eq. (17.17) with n = 4]

f [x4, x3, x2, x1] = −0.02041100 − (−0.05187311) __________________ 5 − 1 = 0.007865529

Thus, the divided difference table is

xi f (xi) First Second Third

1 0 0.4620981 −0.05187311 0.007865529 4 1.386294 0.2027326 −0.02041100 6 1.791759 0.1823216 5 1.609438

The results for f (x1), f [x2, x1], f [x3, x2, x1], and f [x4, x3, x2, x1] represent the coefficients b1, b2, b3, and b4, respectively, of Eq. (17.10). Thus, the interpolating cubic is

f3(x) = 0 + 0.4620981(x − 1) − 0.05187311(x − 1)(x − 4) + 0.007865529(x − 1)(x − 4)(x − 6)

which can be used to evaluate f3(2) = 0.6287686, which represents a relative error of εt = 9.3%. The complete cubic polynomial is shown in Fig. 17.6.

f (x)

f (x) = ln x

f3(x)

True value

Cubic estimate

x50

FIGURE 17.6 The use of cubic interpolation to estimate ln 2.

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440 PolynomIal InterPolatIon

17.2.4 MATLAB M-file: Newtint

It is straightforward to develop an M-file to implement Newton interpolation. As in Fig. 17.7, the first step is to compute the finite divided differences and store them in an array. The dif- ferences are then used in conjunction with Eq. (17.18) to perform the interpolation.

An example of a session using the function would be to duplicate the calculation we just performed in Example 17.3:

>> format long >> x = [1 4 6 5]';

function yint = Newtint(x,y,xx) % Newtint: Newton interpolating polynomial % yint = Newtint(x,y,xx): Uses an (n − 1)−order Newton % interpolating polynomial based on n data points (x, y) % to determine a value of the dependent variable (yint) % at a given value of the independent variable, xx. % input: % x = independent variable % y = dependent variable % xx = value of independent variable at which % interpolation is calculated % output: % yint = interpolated value of dependent variable

% compute the finite divided differences in the form of a % difference table n = length(x); if length(y)~=n, error('x and y must be same length'); end b = zeros(n,n); % assign dependent variables to the first column of b. b(:,1) = y(:); % the (:) ensures that y is a column vector. for j = 2:n for i = 1:n−j+1 b(i,j) = (b(i+1,j−1)−b(i,j−1))/(x(i+j−1)−x(i)); end end % use the finite divided differences to interpolate xt = 1; yint = b(1,1); for j = 1:n−1 xt = xt*(xx−x(j)); yint = yint+b(1,j+1)*xt; end

FIGURE 17.7 An M-file to implement Newton interpolation.

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17.3 laGranGe InterPolatInG PolynomIal 441

>> y = log(x); >> Newtint(x,y,2)

ans = 0.62876857890841

17.3 LAGRANGE INTERPOLATING POLYNOMIAL

Suppose we formulate a linear interpolating polynomial as the weighted average of the two values that we are connecting by a straight line:

f (x) = L1 f (x1) + L2 f (x2) (17.19)

where the L’s are the weighting coefficients. It is logical that the first weighting coefficient is the straight line that is equal to 1 at x1 and 0 at x2:

L1 = x − x2 ______ x1 − x2

Similarly, the second coefficient is the straight line that is equal to 1 at x2 and 0 at x1:

L2 = x − x1 ______ x2 − x1

Substituting these coefficients into Eq. (17.19) yields the straight line that connects the points (Fig. 17.8):

f1(x) = x − x2 ______ x1 − x2 f (x1) +

x − x1 ______ x2 − x1 f (x2) (17.20)

where the nomenclature f1(x) designates that this is a first-order polynomial. Equation (17.20) is referred to as the linear Lagrange interpolating polynomial.

The same strategy can be employed to fit a parabola through three points. For this case three parabolas would be used with each one passing through one of the points and equal- ing zero at the other two. Their sum would then represent the unique parabola that connects the three points. Such a second-order Lagrange interpolating polynomial can be written as

f2(x) = (x − x2)(x − x3) ______________ (x1 − x2)(x1 − x3)

f (x1) + (x − x1)(x − x3) ______________ (x2 − x1)(x2 − x3)

f (x2)

+ (x − x1)(x − x2) ______________ (x3 − x1)(x3 − x2)

f (x3) (17.21)

Notice how the first term is equal to f (x1) at x1 and is equal to zero at x2 and x3. The other terms work in a similar fashion.

Both the first- and second-order versions as well as higher-order Lagrange polynomi- als can be represented concisely as

fn−1(x) = ∑ i=1

n Li (x) f (xi) (17.22)

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442 PolynomIal InterPolatIon

where

Li (x) = ∏ j=1

x − xj ______ xi − xj (17.23)

where n = the number of data points and Π designates the “product of.”

eXamPle 17.5 lagrange Interpolating Polynomial

Problem Statement. Use a Lagrange interpolating polynomial of the first and second order to evaluate the density of unused motor oil at T = 15 °C based on the following data:

x1 = 0 f (x1) = 3.85

x2 = 20 f (x2) = 0.800

x3 = 40 f (x3) = 0.212

Solution. The first-order polynomial [Eq. (17.20)] can be used to obtain the estimate at x = 15:

f1(x) = 15 − 20 _______ 0 − 20 3.85 +

15 − 0 ______ 20 − 0 0.800 = 1.5625

j ≠i

f (x2)

L2 f (x2)

L1 f (x1)

f(x)

f (x1)

x1 x2 x

FIGURE 17.8 A visual depiction of the rationale behind Lagrange interpolating polynomials. The figure shows the first-order case. Each of the two terms of Eq. (17.20) passes through one of the points and is zero at the other. The summation of the two terms must, therefore, be the unique straight line that connects the two points.

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17.3 laGranGe InterPolatInG PolynomIal 443

In a similar fashion, the second-order polynomial is developed as [Eq. (17.21)]

f2(x) = (15 − 20)(15 − 40) ________________ (0 − 20)(0 − 40) 3.85 +

(15 − 0)(15 − 40) _______________ (20 − 0)(20 − 40) 0.800

+ (15 − 0)(15 − 20) _______________ (40 − 0)(40 − 20) 0.212 = 1.3316875

17.3.1 MATLAB M-file: Lagrange

It is straightforward to develop an M-file based on Eqs. (17.22) and (17.23). As in Fig. 17.9, the function is passed two vectors containing the independent (x) and the dependent (y) variables. It is also passed the value of the independent variable where you want to interpo- late (xx). The order of the polynomial is based on the length of the x vector that is passed. If n values are passed, an (n − 1)th order polynomial is fit.

function yint = Lagrange(x,y,xx) % Lagrange: Lagrange interpolating polynomial % yint = Lagrange(x,y,xx): Uses an (n − 1) −order % Lagrange interpolating polynomial based on n data points % to determine a value of the dependent variable (yint) at % a given value of the independent variable, xx. % input: % x = independent variable % y = dependent variable % xx = value of independent variable at which the % interpolation is calculated % output: % yint = interpolated value of dependent variable

n = length(x); if length(y) ~ = n, error('x and y must be same length'); end s = 0; for i = 1:n product = y(i); for j = 1:n if i ~ = j product = product*(xx − x(j))/(x(i) −x(j)); end end s = s + product; end yint = s;

FIGURE 17.9 An M-file to implement Lagrange interpolation.

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444 PolynomIal InterPolatIon

An example of a session using the function would be to predict the density of air at 1 atm pressure at a temperature of 15 °C based on the first four values from Table 17.1. Because four values are passed to the function, a third-order polynomial would be imple- mented by the Lagrange function to give:

>> format long >> T = [−40 0 20 50]; >> d = [1.52 1.29 1.2 1.09]; >> density = Lagrange (T,d,15)

density = 1.22112847222222

17.4 INVERSE INTERPOLATION

As the nomenclature implies, the f (x) and x values in most interpolation contexts are the dependent and independent variables, respectively. As a consequence, the values of the x’s are typically uniformly spaced. A simple example is a table of values derived for the function f (x) = 1∕x:

x 1 2 3 4 5 6 7 f (x) 1 0.5 0.3333 0.25 0.2 0.1667 0.1429

Now suppose that you must use the same data, but you are given a value for f (x) and must determine the corresponding value of x. For instance, for the data above, suppose that you were asked to determine the value of x that corresponded to f (x) = 0.3. For this case, because the function is available and easy to manipulate, the correct answer can be deter- mined directly as x = 1∕0.3 = 3.3333.

Such a problem is called inverse interpolation. For a more complicated case, you might be tempted to switch the f (x) and x values [i.e., merely plot x versus f (x)] and use an approach like Newton or Lagrange interpolation to determine the result. Unfor- tunately, when you reverse the variables, there is no guarantee that the values along the new abscissa [the f (x)’s] will be evenly spaced. In fact, in many cases, the values will be “telescoped.” That is, they will have the appearance of a logarithmic scale with some adjacent points bunched together and others spread out widely. For example, for f (x) = 1∕x the result is

f (x) 0.1429 0.1667 0.2 0.25 0.3333 0.5 1 x 7 6 5 4 3 2 1

Such nonuniform spacing on the abscissa often leads to oscillations in the resulting inter- polating polynomial. This can occur even for lower-order polynomials. An alternative strategy is to fit an nth-order interpolating polynomial, fn (x), to the original data [i.e., with f (x) versus x]. In most cases, because the x’s are evenly spaced, this polynomial will not be ill-conditioned. The answer to your problem then amounts to finding the value of x that makes this polynomial equal to the given f (x). Thus, the interpolation problem reduces to a roots problem!

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17.5 eXtraPolatIon anD oSCIllatIonS 445

For example, for the problem just outlined, a simple approach would be to fit a qua- dratic polynomial to the three points: (2, 0.5), (3, 0.3333), and (4, 0.25). The result would be

f2 (x) = 0.041667x2 − 0.375x + 1.08333

The answer to the inverse interpolation problem of finding the x corresponding to f (x) = 0.3 would therefore involve determining the root of

0.3 = 0.041667x2 − 0.375x + 1.08333

For this simple case, the quadratic formula can be used to calculate

x = 0.375 ± √

___________________________ (−0.375)2 − 4(0.041667)0.78333 ____________________________________ 2(0.041667) =

5.704158 3.295842

Thus, the second root, 3.296, is a good approximation of the true value of 3.333. If ad- ditional accuracy were desired, a third- or fourth-order polynomial along with one of the root-location methods from Chaps. 5 or 6 could be employed.

17.5 EXTRAPOLATION AND OSCILLATIONS

Before leaving this chapter, there are two issues related to polynomial interpolation that must be addressed. These are extrapolation and oscillations.

17.5.1 Extrapolation

Extrapolation is the process of estimating a value of f (x) that lies outside the range of the known base points, x1, x2, . . . , xn. As depicted in Fig. 17.10, the open-ended nature of

f (x)

True curve

Extrapolation of interpolating polynomial

Interpolation Extrapolation

x3x2x1

FIGURE 17.10 Illustration of the possible divergence of an extrapolated prediction. The extrapolation is based on fitting a parabola through the first three known points.

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446 PolynomIal InterPolatIon

extrapolation represents a step into the unknown because the process extends the curve beyond the known region. As such, the true curve could easily diverge from the predic- tion. Extreme care should, therefore, be exercised whenever a case arises where one must extrapolate.

eXamPle 17.6 Dangers of extrapolation

Problem Statement. This example is patterned after one originally developed by Forsythe, Malcolm, and Moler.1 The population in millions of the United States from 1920 to 2000 can be tabulated as

Date 1920 1930 1940 1950 1960 1970 1980 1990 2000 Population 106.46 123.08 132.12 152.27 180.67 205.05 227.23 249.46 281.42

Fit a seventh-order polynomial to the first 8 points (1920 to 1990). Use it to compute the population in 2000 by extrapolation and compare your prediction with the actual result.

Solution. First, the data can be entered as >> t = [1920:10:1990]; >> pop = [106.46 123.08 132.12 152.27 180.67 205.05 227.23

249.46];

The polyfit function can be used to compute the coefficients >> p = polyfit(t,pop,7)

However, when this is implemented, the following message is displayed:

Warning: Polynomial is badly conditioned. Remove repeated data points or try centering and scaling as described in HELP POLYFIT.

We can follow MATLAB’s suggestion by scaling and centering the data values as in >> ts = (t − 1955)/35;

Now polyfit works without an error message: >> p = polyfit(ts,pop,7);

We can then use the polynomial coefficients along with the polyval function to predict the population in 2000 as >> polyval(p,(2000 − 1955)/35)

ans = 175.0800

which is much lower that the true value of 281.42. Insight into the problem can be gained by generating a plot of the data and the polynomial, >> tt = linspace (1920,2000); >> pp = polyval(p,(tt − 1955)/35); >> plot(t,pop,'o',tt,pp)

1 Cleve Moler is one of the founders of The MathWorks, Inc., the makers of MATLAB.

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17.5 eXtraPolatIon anD oSCIllatIonS 447

As in Fig. 17.11, the result indicates that the polynomial seems to fit the data nicely from 1920 to 1990. However, once we move beyond the range of the data into the realm of extrapolation, the seventh-order polynomial plunges to the erroneous prediction in 2000.

17.5.2 Oscillations

Although “more is better” in many contexts, it is absolutely not true for polynomial inter- polation. Higher-order polynomials tend to be very ill-conditioned—that is, they tend to be highly sensitive to round off error. The following example illustrates this point nicely.

eXamPle 17.7 Dangers of Higher-order Polynomial Interpolation

Problem Statement. In 1901, Carl Runge published a study on the dangers of higher- order polynomial interpolation. He looked at the following simple-looking function:

f (x) = 1 ________ 1 + 25x2

(17.24)

which is now called Runge’s function. He took equidistantly spaced data points from this function over the interval [–1, 1]. He then used interpolating polynomials of increasing order and found that as he took more points, the polynomials and the original curve dif- fered considerably. Further, the situation deteriorated greatly as the order was increased. Duplicate Runge’s result by using the polyfit and polyval functions to fit fourth- and tenth- order polynomials to 5 and 11 equally spaced points generated with Eq. (17.24). Create plots of your results along with the sampled values and the complete Runge’s function.

200019901980197019601950194019301920

250

200

150

100

FIGURE 17.11 Use of a seventh-order polynomial to make a prediction of U.S. population in 2000 based on data from 1920 through 1990.

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448 PolynomIal InterPolatIon

Solution. The five equally spaced data points can be generated as in

>> x = linspace (−1,1,5); >> y = 1./(1 + 25*x.^2);

Next, a more finally spaced vector of xx values can be computed so that we can create a smooth plot of the results: >> xx = linspace(−1,1);

Recall that linspace automatically creates 100 points if the desired number of points is not specified. The polyfit function can be used to generate the coefficients of the fourth-order polynomial, and the polval function can be used to generate the polynomial interpolation at the finely spaced values of xx: >> p = polyfit(x,y,4); >> y4 = polyval(p,xx);

Finally, we can generate values for Runge’s function itself and plot them along with the polynomial fit and the sampled data: >> yr = 1./(1 + 25*xx.^2); >> plot(x,y,'o',xx,y4,xx,yr,'− −')

As in Fig. 17.12, the polynomial does a poor job of following Runge’s function. Continuing with the analysis, the tenth-order polynomial can be generated and plotted with >> x = linspace(−1,1,11); >> y = 1./(1 + 25*x.^2);

10.50−0.5−1

−0.2

0.2

0.4

0.6

0.8

−0.4

FIGURE 17.12 Comparison of Runge’s function (dashed line) with a fourth-order polynomial fit to 5 points sampled from the function.

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ProBlemS 449

>> p = polyfit (x,y,10); >> y10 = polyval (p,xx); >> plot(x,y,'o',xx,y10,xx,yr,'− −')

As in Fig. 17.13, the fit has gotten even worse, particularly at the ends of the interval! Although there may be certain contexts where higher-order polynomials are neces- sary, they are usually to be avoided. In most engineering and scientific contexts, lower- order polynomials of the type described in this chapter can be used effectively to capture the curving trends of data without suffering from oscillations.

10.50−0.5−1

0.5

1.5

−0.5

FIGURE 17.13 Comparison of Runge’s function (dashed line) with a tenth-order polynomial fit to 11 points sampled from the function.

PROBLEMS

17.1 The following data come from a table that was mea- sured with high precision. Use the best numerical method (for this type of problem) to determine y at x = 3.5. Note that a polynomial will yield an exact value. Your solution should prove that your result is exact.

x 0 1.8 5 6 8.2 9.2 12 y 26 16.415 5.375 3.5 2.015 2.54 8

17.2 Use Newton’s interpolating polynomial to determine y at x = 3.5 to the best possible accuracy. Compute the finite divided differences as in Fig. 17.5, and order your points to attain op- timal accuracy and convergence. That is, the points should be centered around and as close as possible to the unknown.

x 0 1 2.5 3 4.5 5 6 y 2 5.4375 7.3516 7.5625 8.4453 9.1875 12

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450 Polynomial interPolation

17.3 Use Newton’s interpolating polynomial to determine y at x = 8 to the best possible accuracy. Compute the finite divided differences as in Fig. 17.5, and order your points to attain op- timal accuracy and convergence. That is, the points should be centered around and as close as possible to the unknown.

x 0 1 2 5.5 11 13 16 18 y 0.5 3.134 5.3 9.9 10.2 9.35 7.2 6.2

17.4 Given the data

x 1 2 2.5 3 4 5 f (x) 0 5 7 6.5 2 0

(a) Calculate f (3.4) using Newton’s interpolating polynomi- als of order 1 through 3. Choose the sequence of the points for your estimates to attain the best possible accuracy. That is, the points should be centered around and as close as possible to the unknown.

(b) Repeat (a) but use the Lagrange polynomial. 17.5 Given the data

x 1 2 3 5 6 f (x) 4.75 4 5.25 19.75 36

Calculate f (4) using Newton’s interpolating polynomials of order 1 through 4. Choose your base points to attain good accuracy. That is, the points should be centered around and as close as possible to the unknown. What do your results indicate regarding the order of the polynomial used to gener- ate the data in the table? 17.6 Repeat Prob. 17.5 using the Lagrange polynomial of order 1 through 3. 17.7 Table P15.5 lists values for dissolved oxygen concen- tration in water as a function of temperature and chloride concentration. (a) Use quadratic and cubic interpolation to determine the

oxygen concentration for T = 12 °C and c = 10 g/L. (b) Use linear interpolation to determine the oxygen con-

centration for T = 12 °C and c = 15 g/L. (c) Repeat (b) but use quadratic interpolation. 17.8 Employ inverse interpolation using a cubic interpolat- ing polynomial and bisection to determine the value of x that corresponds to f (x) = 1.7 for the following tabulated data:

x 1 2 3 4 5 6 7 f (x) 3.6 1.8 1.2 0.9 0.72 1.5 0.51429

17.9 Employ inverse interpolation to determine the value of x that corresponds to f (x) = 0.93 for the following tabulated data:

x 0 1 2 3 4 5 f (x) 0 0.5 0.8 0.9 0.941176 0.961538

Note that the values in the table were generated with the function f (x) = x2∕(1 + x2). (a) Determine the correct value analytically. (b) Use quadratic interpolation and the quadratic formula to

determine the value numerically. (c) Use cubic interpolation and bisection to determine the

value numerically. 17.10 Use the portion of the given steam table for su- perheated water at 200 MPa to find (a) the correspond- ing entropy s for a specific volume υ of 0.118 with linear interpolation, (b) the same corresponding entropy using qua- dratic interpolation, and (c) the volume corresponding to an entropy of 6.45 using inverse interpolation.

υ, m3/kg 0.10377 0.11144 0.12547 s, kJ/(kg K) 6.4147 6.5453 6.7664

17.11 The following data for the density of nitrogen gas versus temperature come from a table that was measured with high precision. Use first- through fifth-order polynomi- als to estimate the density at a temperature of 330 K. What is your best estimate? Employ this best estimate and inverse interpolation to determine the corresponding temperature.

T, K 200 250 300 350 400 450 Density, 1.708 1.367 1.139 0.967 0.854 0.759 kg/m3

17.12 Ohm’s law states that the voltage drop V across an ideal resistor is linearly proportional to the current i flowing through the resister as in V = i R, where R is the resistance. However, real resistors may not always obey Ohm’s law. Suppose that you performed some very precise experiments to measure the voltage drop and corresponding current for a resistor. The following results suggest a curvilinear relation- ship rather than the straight line represented by Ohm’s law:

i −2 −1 −0.5 0.5 1 2 V −637 −96.5 −20.5 20.5 96.5 637

To quantify this relationship, a curve must be fit to the data. Because of measurement error, regression would typically be the preferred method of curve fitting for analyzing such ex- perimental data. However, the smoothness of the relationship, as well as the precision of the experimental methods, suggests that interpolation might be appropriate. Use a fifth-order inter- polating polynomial to fit the data and compute V for i = 0.10.

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ProBlemS 451

17.13 Bessel functions often arise in advanced engineering analyses such as the study of electric fields. Here are some selected values for the zero-order Bessel function of the first kind

x 1.8 2.0 2.2 2.4 2.6 J1(x) 0.5815 0.5767 0.5560 0.5202 0.4708

Estimate J1 (2.1) using third- and fourth-order interpolating polynomials. Determine the percent relative error for each case based on the true value, which can be determined with MATLAB’s built-in function besselj. 17.14 Repeat Example 17.6 but using first-, second-, third-, and fourth-order interpolating polynomials to pre- dict the population in 2000 based on the most recent data. That is, for the linear prediction use the data from 1980 and 1990, for the quadratic prediction use the data from 1970, 1980, and 1990, and so on. Which approach yields the best result? 17.15 The specific volume of a superheated steam is listed in steam tables for various temperatures.

T, °C 370 382 394 406 418 υ, L/kg 5.9313 7.5838 8.8428 9.796 10.5311

Determine υ at T = 400 ºC. 17.16 The vertical stress σz under the corner of a rectangular area subjected to a uniform load of intensity q is given by the solution of Boussinesq’s equation:

σ = q ___ 4π [ 2mn √ __________

m2 + n2 + 1 ________________ m2 + n2 + 1 + m2n2

m 2 + n2 + 2 __________

m2 + n2+ 1

+sin−1 ( 2mn √__________ m2 + n2 + 1 ________________ m2 + n2 + 1 + m2n2 ) ] Because this equation is inconvenient to solve manually, it has been reformulated as

σz = q fz(m, n)

where fz(m, n) is called the influence value, and m and n are dimensionless ratios, with m = a∕z and n = b∕z and a and b are defined in Fig. P17.16. The influence value is then tabulated, a portion of which is given in Table P17.16. If a = 4.6 and b = 14, use a third-order interpolating poly- nomial to compute σz at a depth 10 m below the cor- ner of a rectangular footing that is subject to a total load

of 100 t (metric tons). Express your answer in tonnes per square meter. Note that q is equal to the load per area.

TABLE P17.16

m n = 1.2 n = 1.4 n = 1.6

0.1 0.02926 0.03007 0.03058 0.2 0.05733 0.05894 0.05994 0.3 0.08323 0.08561 0.08709 0.4 0.10631 0.10941 0.11135 0.5 0.12626 0.13003 0.13241 0.6 0.14309 0.14749 0.15027 0.7 0.15703 0.16199 0.16515 0.8 0.16843 0.17389 0.17739

17.17 You measure the voltage drop V across a resistor for a number of different values of current i. The results are

i 0.5 1.5 2.5 3.0 4.0 V −0.45 −0.6 0.70 1.88 6.0

Use first- through fourth-order polynomial interpolation to estimate the voltage drop for i = 2.3. Interpret your results. 17.18 The current in a wire is measured with great precision as a function of time:

t 0 0.250 0.500 0.750 1.000 i 0 6.24 7.75 4.85 0.0000

Determine i at t = 0.23. 17.19 The acceleration due to gravity at an altitude y above the surface of the earth is given by

y, m 0 30,000 60,000 90,000 120,000 g, m/s2 9.8100 9.7487 9.6879 9.6278 9.5682

Compute g at y = 55,000 m.

FIGURE P17.16

z a

σz

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452 PolynomIal InterPolatIon

17.20 Temperatures are measured at various points on a heated plate (Table P17.20). Estimate the temperature at (a) x = 4, y = 3.2 and (b) x = 4.3, y = 2.7.

17.23 The following data comes from a table that was mea- sured with high precision. Use the Newton interpolating polynomial to determine y at x = 3.5. Properly order all the points and then develop a divided difference table to com- pute the derivatives. Note that a polynomial will yield an exact value. Your solution should prove that your result is exact.

x 0 1 2.5 3 4.5 5 6 y 26 15.5 5.375 3.5 2.375 3.5 8

17.24 The following data are measured precisely:

T 2 2.1 2.2 2.7 3 3.4 Z 6 7.752 10.256 36.576 66 125.168

(a) Use Newton interpolating polynomials to determine z at t = 2.5. Make sure that you order your points to attain the most accurate results. What do your results tell you regard- ing the order of the polynomial used to generate the data?

(b) Use a third-order Lagrange interpolating polynomial to determine y at t = 2.5.

17.25 The following data for the density of water versus temperature come from a table that was measured with high precision. Use inverse interpolation to determine the temperature corresponding to a density of 0.999245 g/cm3. Base your estimate on a third-order interpolating polyno- mial (Even though you’re doing this problem by hand, feel free to use the MATLAB polyfit function to determine the polynomial.) Determine the root with the Newton-Raphson method (by hand) with an initial guess of T = 14 °C. Be care- ful regarding roundoff errors.

T, °C 0 4 8 12 16 Density, 0.99987 1 0.99988 0.99952 0.99897 g/cm3

TABLE P17.20 temperatures (°C) at various points on a square heated plate.

x = 0 x = 2 x = 4 x = 6 x = 8

y = 0 100.00 90.00 80.00 70.00 60.00 y = 2 85.00 64.49 53.50 48.15 50.00 y = 4 70.00 48.90 38.43 35.03 40.00 y = 6 55.00 38.78 30.39 27.07 30.00 y = 8 40.00 35.00 30.00 25.00 20.00

17.21 Use the portion of the given steam table for su- perheated H2O at 200 MPa to (a) find the corresponding entropy s for a specific volume υ of 0.108 m3/kg with linear interpolation, (b) find the same corresponding entropy using quadratic interpolation, and (c) find the volume correspond- ing to an entropy of 6.6 using inverse interpolation.

υ (m3/kg) 0.10377 0.11144 0.12540 s (kJ/kg · K) 6.4147 6.5453 6.7664

17.22 Develop an M-file function, that uses polyfit and polyval for polynomial interpolation. Here is the script you can use to test your function

clear,clc,clf,format compact x = [1 2 4 8]; fx = @(x) 10*exp(−0.2*x); y = fx(x); yint = polyint(x,y,3) ytrue = fx(3) et = abs((ytrue − yint) / ytrue)*100.

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CHAPTER OBJECTIVES

Knowledge and understanding are prerequisites for the effective implementation of any tool.

No matter how impressive your tool chest, you will be hard-pressed to repair a car if you do not understand how it works.

• This is the first chapter objectives entry. • Second chapter objective entry, the entries use ic/lc per manuscript, the first and

last entry have space above or below, middle entries do not. • Third chapter entry copy goes here.

453

18 Splines and Piecewise Interpolation

18.1 INTRODUCTION TO SPLINES

In Chap. 17 (n − 1)th-order polynomials were used to interpolate between n data points. For example, for eight points, we can derive a perfect seventh-order polynomial. This curve would capture all the meanderings (at least up to and including seventh derivatives) suggested by the points. However, there are cases where these functions can lead to errone- ous results because of round off error and oscillations. An alternative approach is to apply lower-order polynomials in a piecewise fashion to subsets of data points. Such connecting polynomials are called spline functions.

For example, third-order curves employed to connect each pair of data points are called cubic splines. These functions can be constructed so that the connections between adjacent cubic equations are visually smooth. On the surface, it would seem that the third-order

CHAPTER OBJECTIVES

The primary objective of this chapter is to introduce you to splines. Specific objectives and topics covered are

• Understanding that splines minimize oscillations by fitting lower-order polynomials to data in a piecewise fashion.

• Knowing how to develop code to perform a table lookup. • Recognizing why cubic polynomials are preferable to quadratic and higher-order

splines. • Understanding the conditions that underlie a cubic spline fit. • Understanding the differences between natural, clamped, and not-a-knot end

conditions. • Knowing how to fit a spline to data with MATLAB’s built-in functions. • Understanding how multidimensional interpolation is implemented with MATLAB.

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454 SPlIneS and PIecewISe InterPolatIon

approximation of the splines would be inferior to the seventh-order expression. You might wonder why a spline would ever be preferable.

Figure 18.1 illustrates a situation where a spline performs better than a higher-order polynomial. This is the case where a function is generally smooth but undergoes an abrupt change somewhere along the region of interest. The step increase depicted in Fig. 18.1 is an extreme example of such a change and serves to illustrate the point.

Figure 18.1a through c illustrates how higher-order polynomials tend to swing through wild oscillations in the vicinity of an abrupt change. In contrast, the spline also connects the points, but because it is limited to lower-order changes, the oscillations are kept to a

(a)

f (x)

(b)

f (x)

(c)

f (x)

(d )

f (x)

FIGURE 18.1 A visual representation of a situation where splines are superior to higher-order interpolating polynomials. The function to be fit undergoes an abrupt increase at x = 0. Parts (a) through (c) indicate that the abrupt change induces oscillations in interpolating polynomials. In contrast, because it is limited to straight-line connections, a linear spline (d) provides a much more acceptable approximation.

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18.2 lInear SPlIneS 455

minimum. As such, the spline usually provides a superior approximation of the behavior of functions that have local, abrupt changes.

The concept of the spline originated from the drafting technique of using a thin, flexible strip (called a spline) to draw smooth curves through a set of points. The process is depicted in Fig. 18.2 for a series of five pins (data points). In this technique, the drafter places paper over a wooden board and hammers nails or pins into the paper (and board) at the location of the data points. A smooth cubic curve results from interweaving the strip between the pins. Hence, the name “cubic spline” has been adopted for polynomials of this type.

In this chapter, simple linear functions will first be used to introduce some basic con- cepts and issues associated with spline interpolation. Then we derive an algorithm for fitting quadratic splines to data. This is followed by material on the cubic spline, which is the most common and useful version in engineering and science. Finally, we describe MATLAB’s capabilities for piecewise interpolation including its ability to generate splines.

18.2 LINEAR SPLINES

The notation used for splines is displayed in Fig. 18.3. For n data points (i = 1, 2, . . . , n), there are n − 1 intervals. Each interval i has its own spline function, si (x). For linear splines, each function is merely the straight line connecting the two points at each end of the interval, which is formulated as

si (x) = ai + bi (x − xi ) (18.1) where ai is the intercept, which is defined as

ai = fi (18.2) and bi is the slope of the straight line connecting the points:

bi = fi+1 − fi ________ xi+1 − xi (18.3)

FIGURE 18.2 The drafting technique of using a spline to draw smooth curves through a series of points. Notice how, at the end points, the spline straightens out. This is called a “natural” spline.

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456 SPlIneS and PIecewISe InterPolatIon

where fi is shorthand for f (xi ). Substituting Eqs. (18.1) and (18.2) into Eq. (18.3) gives

si (x) = fi + fi+1 − fi _______ xi+1 − xi (x − xi) (18.4)

These equations can be used to evaluate the function at any point between x1 and xn by first locating the interval within which the point lies. Then the appropriate equation is used to determine the function value within the interval. Inspection of Eq. (18.4) indicates that the linear spline amounts to using Newton’s first-order polynomial [Eq. (17.5)] to interpolate within each interval.

eXaMPle 18.1 First-order Splines

Problem Statement. Fit the data in Table 18.1 with first-order splines. Evaluate the function at x = 5.

TABLE 18.1 data to be fit with spline functions.

i xi fi

1 3.0 2.5 2 4.5 1.0 3 7.0 2.5 4 9.0 0.5

Solution. The data can be substituted into Eq. (18.4) to generate the linear spline functions. For example, for the second interval from x = 4.5 to x = 7, the function is

s2(x) = 1.0 + 2.5 − 1.0 ________ 7.0 − 4.5 (x − 4.5)

sn−1(x)

Interval 1

f (x) s1(x)

x1 x2 xi xi+1 xn−1 xn

f1 fi fi+1 f2

Interval i

Interval n − 1

si(x)

fn−1 fn

FIGURE 18.3 Notation used to derive splines. Notice that there are n − 1 intervals and n data points.

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18.2 lInear SPlIneS 457

The equations for the other intervals can be computed, and the resulting first-order splines are plotted in Fig. 18.4a. The value at x = 5 is 1.3.

s2(x) = 1.0 + 2.5 − 1.0 ________ 7.0 − 4.5 (5 − 4.5) = 1.3

Visual inspection of Fig. 18.4a indicates that the primary disadvantage of first-order splines is that they are not smooth. In essence, at the data points where two splines meet (called a knot), the slope changes abruptly. In formal terms, the first derivative of the func- tion is discontinuous at these points. This deficiency is overcome by using higher-order polynomial splines that ensure smoothness at the knots by equating derivatives at these points, as will be discussed subsequently. Before doing that, the following section provides an application where linear splines are useful.

(a)

(b)

f (x)

0 2 4 6 8 10

f (x)

(c)

f (x)

First-order spline

Second-order spline

Interpolating cubic

Cubic spline

FIGURE 18.4 Spline fits of a set of four points. (a) Linear spline, (b) quadratic spline, and (c) cubic spline, with a cubic interpolating polynomial also plotted.

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458 SPlIneS and PIecewISe InterPolatIon

18.2.1 Table Lookup

A table lookup is a common task that is frequently encountered in engineering and science computer applications. It is useful for performing repeated interpolations from a table of independent and dependent variables. For example, suppose that you would like to set up an M-file that would use linear interpolation to determine air density at a particular temperature based on the data from Table 17.1. One way to do this would be to pass the M-file the temperature at which you want the interpolation to be performed along with the two adjoining values. A more general approach would be to pass in vec- tors containing all the data and have the M-file determine the bracket. This is called a table lookup.

Thus, the M-file would perform two tasks. First, it would search the independent vari- able vector to find the interval containing the unknown. Then it would perform the linear interpolation using one of the techniques described in this chapter or in Chap. 17.

For ordered data, there are two simple ways to find the interval. The first is called a sequential search. As the name implies, this method involves comparing the desired value with each element of the vector in sequence until the interval is located. For data in ascend- ing order, this can be done by testing whether the unknown is less than the value being as- sessed. If so, we know that the unknown falls between this value and the previous one that we examined. If not, we move to the next value and repeat the comparison. Here is a simple M-file that accomplishes this objective:

function yi = TableLook(x, y, xx)

n = length(x); if xx < x(1) | xx > x(n) error('Interpolation outside range') end % sequential search i = 1; while(1) if xx < = x(i + 1), break, end i = i + 1; end % linear interpolation yi = y(i) + (y(i + 1) − y(i))/(x(i + 1) − x(i))*(xx − x(i));

The table’s independent variables are stored in ascending order in the array x and the dependent variables stored in the array y. Before searching, an error trap is included to ensure that the desired value xx falls within the range of the x’s. A while . . . break loop compares the value at which the interpolation is desired, xx, to determine whether it is less than the value at the top of the interval, x(i + 1). For cases where xx is in the second interval or higher, this will not test true at first. In this case the counter i is incremented by one so that on the next iteration, xx is compared with the value at the top of the sec- ond interval. The loop is repeated until the xx is less than or equal to the interval’s upper bound, in which case the loop is exited. At this point, the interpolation can be performed simply as shown.

For situations for which there are lots of data, the sequential sort is inefficient because it must search through all the preceding points to find values. In these cases, a simple

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18.3 QUadratIc SPlIneS 459

alternative is the binary search. Here is an M-file that performs a binary search followed by linear interpolation:

function yi = TableLookBin(x, y, xx)

n = length(x); if xx < x(1) | xx > x(n) error('Interpolation outside range') end % binary search iL = 1; iU = n; while (1) if iU − iL < = 1, break, end iM = fix((iL + iU) / 2); if x(iM) < xx iL = iM; else iU = iM; end end % linear interpolation yi = y(iL) + (y(iL + 1) − y(iL))/(x(iL + 1) − x(iL))*(xx − x(iL));

The approach is akin to the bisection method for root location. Just as in bisection, the index at the midpoint iM is computed as the average of the first or “lower” index iL = 1 and the last or “upper” index iU = n. The unknown xx is then compared with the value of x at the midpoint x(iM) to assess whether it is in the lower half of the array or in the upper half. Depending on where it lies, either the lower or upper index is redefined as being the middle index. The process is repeated until the difference between the upper and the lower index is less than or equal to zero. At this point, the lower index lies at the lower bound of the interval containing xx, the loop terminates, and the linear interpolation is performed.

Here is a MATLAB session illustrating how the binary search function can be applied to calculate the air density at 350 °C based on the data from Table 17.1. The sequential search would be similar.

>> T = [−40 0 20 50 100 150 200 250 300 400 500]; >> density = [1.52 1.29 1.2 1.09 .946 .935 .746 .675 .616... .525 .457]; >> TableLookBin(T,density,350)

ans = 0.5705

This result can be verified by the hand calculation:

f (350) = 0.616 + 0.525 − 0.616 ____________ 400 − 300 (350 − 300) = 0.5705

18.3 QUADRATIC SPLINES

To ensure that the nth derivatives are continuous at the knots, a spline of at least n + 1 order must be used. Third-order polynomials or cubic splines that ensure continuous first and second derivatives are most frequently used in practice. Although third and higher

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460 SPlIneS and PIecewISe InterPolatIon

derivatives can be discontinuous when using cubic splines, they usually cannot be detected visually and consequently are ignored.

Because the derivation of cubic splines is somewhat involved, we have decided to first illustrate the concept of spline interpolation using second-order polynomials. These “qua- dratic splines” have continuous first derivatives at the knots. Although quadratic splines are not of practical importance, they serve nicely to demonstrate the general approach for developing higher-order splines.

The objective in quadratic splines is to derive a second-order polynomial for each in- terval between data points. The polynomial for each interval can be represented generally as

si (x) = ai + bi (x − xi ) + ci (x − xi )2 (18.5)

where the notation is as in Fig. 18.3. For n data points (i = 1, 2, . . . , n), there are n − 1 intervals and, consequently, 3(n − 1) unknown constants (the a’s, b’s, and c’s) to evaluate. Therefore, 3(n − 1) equations or conditions are required to evaluate the unknowns. These can be developed as follows:

1. The function must pass through all the points. This is called a continuity condition. It can be expressed mathematically as

fi = ai + bi (xi − xi ) + ci (xi − xi )2

which simplifies to

ai = fi (18.6)

Therefore, the constant in each quadratic must be equal to the value of the dependent variable at the beginning of the interval. This result can be incorporated into Eq. (18.5):

si (x) = fi + bi (x − xi ) + ci (x − xi )2

Note that because we have determined one of the coefficients, the number of condi- tions to be evaluated has now been reduced to 2(n − 1).

2. The function values of adjacent polynomials must be equal at the knots. This condition can be written for knot i + 1 as

fi + bi (xi+1 − xi ) + ci (xi+1 − xi )2 = fi+1 + bi+1(xi+1 − xi+1) + ci+1(xi+1 − xi+1)2 (18.7)

This equation can be simplified mathematically by defining the width of the ith interval as

hi = xi+1 − xi

Thus, Eq. (18.7) simplifies to

fi + bi hi + ci h i 2 = fi+1 (18.8)

This equation can be written for the nodes, i = 1, . . . , n − 1. Since this amounts to n − 1 conditions, it means that there are 2(n − 1) − (n − 1) = n − 1 remaining conditions.

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18.3 QUadratIc SPlIneS 461

3. The first derivatives at the interior nodes must be equal. This is an important condition, because it means that adjacent splines will be joined smoothly, rather than in the jagged fashion that we saw for the linear splines. Equation (18.5) can be differentiated to yield

s i ′ (x) = bi + 2ci (x − xi )

The equivalence of the derivatives at an interior node, i + 1 can therefore be written as

bi + 2ci hi = bi+1 (18.9)

Writing this equation for all the interior nodes amounts to n − 2 conditions. This means that there is n − 1 − (n − 2) = 1 remaining condition. Unless we have some additional information regarding the functions or their derivatives, we must make an arbitrary choice to successfully compute the constants. Although there are a number of different choices that can be made, we select the following condition.

4. Assume that the second derivative is zero at the first point. Because the second deriva- tive of Eq. (18.5) is 2ci , this condition can be expressed mathematically as

c1 = 0

The visual interpretation of this condition is that the first two points will be connected by a straight line.

eXaMPle 18.2 Quadratic Splines

Problem Statement. Fit quadratic splines to the same data employed in Example 18.1 (Table 18.1). Use the results to estimate the value at x = 5.

Solution. For the present problem, we have four data points and n = 3 intervals. There- fore, after applying the continuity condition and the zero second-derivative condition, this means that 2(4 − 1) − 1 = 5 conditions are required. Equation (18.8) is written for i = 1 through 3 (with c1 = 0) to give

f1 + b1 h1 = f2 f2 + b2 h2 + c2 h 2

2 = f3 f3 + b3 h3 + c3 h 3

2 = f4 Continuity of derivatives, Eq. (18.9), creates an additional 3 − 1 = 2 conditions (again, recall that c1 = 0):

b1 = b2 b2 + 2c2h2 = b3

The necessary function and interval width values are

f1 = 2.5 h1 = 4.5 − 3.0 = 1.5 f2 = 1.0 h2 = 7.0 − 4.5 = 2.5 f3 = 2.5 h3 = 9.0 − 7.0 = 2.0 f4 = 0.5

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462 SPlIneS and PIecewISe InterPolatIon

These values can be substituted into the conditions which can be expressed in matrix form as

⎡

⎢ ⎣

1.5

2.5

6.25

0 0 0 0 2 4

−1

⎤

⎥ ⎦

⎧

⎪

⎨ ⎪

⎩

c2 b3

⎫

⎪

⎬ ⎪

⎭

⎧

⎪

⎨ ⎪

⎩

−1.5

1.5

−2 0

⎫

⎪

⎬ ⎪

⎭

These equations can be solved using MATLAB with the results:

b1 = −1 b2 = −1 c2 = 0.64 b3 = 2.2 c3 = −1.6

These results, along with the values for the a’s [Eq. (18.6)], can be substituted into the original quadratic equations to develop the following quadratic splines for each interval:

s1(x) = 2.5 − (x − 3)

s2(x) = 1.0 − (x − 4.5) + 0.64(x − 4.5)2

s3(x) = 2.5 + 2.2(x − 7.0) − 1.6(x − 7.0)2

Because x = 5 lies in the second interval, we use s2 to make the prediction,

s2(5) = 1.0 − (5 − 4.5) + 0.64(5 − 4.5)2 = 0.66

The total quadratic spline fit is depicted in Fig. 18.4b. Notice that there are two shortcom- ings that detract from the fit: (1) the straight line connecting the first two points and (2) the spline for the last interval seems to swing too high. The cubic splines in the next section do not exhibit these shortcomings and, as a consequence, are better methods for spline interpolation.

18.4 CUBIC SPLINES

As stated at the beginning of the previous section, cubic splines are most frequently used in practice. The shortcomings of linear and quadratic splines have already been discussed. Quartic or higher-order splines are not used because they tend to exhibit the instabilities inherent in higher-order polynomials. Cubic splines are preferred because they provide the simplest representation that exhibits the desired appearance of smoothness.

The objective in cubic splines is to derive a third-order polynomial for each interval between knots as represented generally by

si (x) = ai + bi (x − xi ) + ci (x − xi )2 + di (x − xi )3 (18.10)

Thus, for n data points (i = 1, 2, . . . , n), there are n − 1 intervals and 4 (n − 1) un- known coefficients to evaluate. Consequently, 4 (n − 1) conditions are required for their evaluation.

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18.4 cUBIc SPlIneS 463

The first conditions are identical to those used for the quadratic case. That is, they are set up so that the functions pass through the points and that the first derivatives at the knots are equal. In addition to these, conditions are developed to ensure that the second deriva- tives at the knots are also equal. This greatly enhances the fit’s smoothness.

After these conditions are developed, two additional conditions are required to obtain the solution. This is a much nicer outcome than occurred for quadratic splines where we needed to specify a single condition. In that case, we had to arbitrarily specify a zero sec- ond derivative for the first interval, hence making the result asymmetric. For cubic splines, we are in the advantageous position of needing two additional conditions and can, there- fore, apply them evenhandedly at both ends.

For cubic splines, these last two conditions can be formulated in several different ways. A very common approach is to assume that the second derivatives at the first and last knots are equal to zero. The visual interpretation of these conditions is that the function becomes a straight line at the end nodes. Specification of such an end condition leads to what is termed a “natural” spline. It is given this name because the drafting spline naturally behaves in this fashion (Fig. 18.2).

There are a variety of other end conditions that can be specified. Two of the more popu- lar are the clamped condition and the not-a-knot conditions. We will describe these options in Sec. 18.4.2. For the following derivation, we will limit ourselves to natural splines.

Once the additional end conditions are specified, we would have the 4(n − 1) condi- tions needed to evaluate the 4(n − 1) unknown coefficients. Whereas it is certainly pos- sible to develop cubic splines in this fashion, we will present an alternative approach that requires the solution of only n − 1 equations. Further, the simultaneous equations will be tridiagonal and hence can be solved very efficiently. Although the derivation of this approach is less straightforward than for quadratic splines, the gain in efficiency is well worth the effort.

18.4.1 Derivation of Cubic Splines

As was the case with quadratic splines, the first condition is that the spline must pass through all the data points:

fi = ai + bi (xi − xi ) + ci (xi − xi )2 + di (xi − xi )3

which simplifies to

ai = fi (18.11)

Therefore, the constant in each cubic must be equal to the value of the dependent variable at the beginning of the interval. This result can be incorporated into Eq. (18.10):

si (x) = fi + bi (x − xi ) + ci (x − xi )2 + di (x − xi )3 (18.12)

Next, we will apply the condition that each of the cubics must join at the knots. For knot i + 1, this can be represented as

fi + bi hi + ci h i 2 + di h i

3 = fi+1 (18.13)

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464 SPlIneS and PIecewISe InterPolatIon

where

hi = xi+1 − xi The first derivatives at the interior nodes must be equal. Equation (18.12) is differenti-

ated to yield

s i ′ (x) = bi + 2ci (x − xi ) + 3di (x − xi ) 2 (18.14)

The equivalence of the derivatives at an interior node, i + 1 can therefore be written as

bi + 2ci hi + 3di h i 2 = bi+1 (18.15)

The second derivatives at the interior nodes must also be equal. Equation (18.14) can be differentiated to yield

s i ″ (x) = 2ci + 6di (x − xi ) (18.16)

The equivalence of the second derivatives at an interior node, i + 1 can therefore be written as

ci + 3di hi = ci+1 (18.17)

Next, we can solve Eq. (18.17) for di:

di = ci+1 − ci _______ 3hi

(18.18)

This can be substituted into Eq. (18.13) to give

fi + bi hi + h i

2 __ 3 (2ci + ci+1) = fi+1 (18.19)

Equation (18.18) can also be substituted into Eq. (18.15) to give

bi+1 = bi + hi (ci + ci+1) (18.20)

Equation (18.19) can be solved for

bi = fi+1 − fi ______ hi

− hi __ 3 (2ci + ci+1) (18.21)

The index of this equation can be reduced by 1:

bi−1 = fi − fi−1 ______ hi−1

− hi−1 ____ 3 (2ci−1 + ci ) (18.22)

The index of Eq. (18.20) can also be reduced by 1:

bi = bi−1 + hi−1(ci−1 + ci ) (18.23)

Equations (18.21) and (18.22) can be substituted into Eq. (18.23) and the result simplified to yield

hi−1ci−1 + 2(hi−1 + hi ) ci + hi ci+1 = 3 fi+1 − fi ______ hi

− 3 fi − fi−1 ______ hi−1

(18.24)

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18.4 cUBIc SPlIneS 465

This equation can be made a little more concise by recognizing that the terms on the right-hand side are finite differences [recall Eq. (17.15)]:

f [xi , xj] = fi − fj ______ xi − xj

Therefore, Eq. (18.24) can be written as

hi−1ci−1 + 2(hi−1 + hi )ci + hi ci+1 = 3 ( f [xi+1, xi ] − f [xi , xi−1]) (18.25)

Equation (18.25) can be written for the interior knots, i = 2, 3, . . . , n − 2, which results in n − 3 simultaneous tridiagonal equations with n − 1 unknown coefficients, c1, c2, . . . , cn–1. Therefore, if we have two additional conditions, we can solve for the c’s. Once this is done, Eqs. (18.21) and (18.18) can be used to determine the remaining coef- ficients, b and d.

As stated previously, the two additional end conditions can be formulated in a number of ways. One common approach, the natural spline, assumes that the second derivatives at the end knots are equal to zero. To see how these can be integrated into the solution scheme, the second derivative at the first node [Eq. (18.16)] can be set to zero as in

s 1 ″ (x1) = 0 = 2c1 + 6d1(x1 − x1)

Thus, this condition amounts to setting c1 equal to zero. The same evaluation can be made at the last node:

s n−1 ″ (xn) = 0 = 2cn−1 + 6dn−1hn−1 (18.26)

Recalling Eq. (18.17), we can conveniently define an extraneous parameter cn, in which case Eq. (18.26) becomes

cn−1 + 3dn−1hn−1 = cn = 0

Thus, to impose a zero second derivative at the last node, we set cn = 0. The final equations can now be written in matrix form as

⎡

⎢ ⎣

2(h1 + h2)

hn−2

2(hn−2 + hn−1)

hn−1

⎤

⎥ ⎦

⎧

⎪ ⎨ ⎪

⎩

cn−1

⎫

⎪ ⎬ ⎪

⎭

⎧

⎪ ⎨ ⎪

⎩

3( f [x3, x2] − f [x2, x1])

3( f [xn, xn−1] − f [xn−1, xn−2])

⎫

⎪ ⎬ ⎪

⎭

(18.27)

As shown, the system is tridiagonal and hence efficient to solve.

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466 SPlIneS and PIecewISe InterPolatIon

eXaMPle 18.3 natural cubic Splines

Problem Statement. Fit cubic splines to the same data used in Examples 18.1 and 18.2 (Table 18.1). Utilize the results to estimate the value at x = 5.

Solution. The first step is to employ Eq. (18.27) to generate the set of simultaneous equa- tions that will be utilized to determine the c coefficients:

⎡

⎢ ⎣

2(h1 + h2) h2

2(h2 + h3) h3

⎤

⎥ ⎦

⎧

⎪

⎨ ⎪

⎩

c2 c3 c4

⎫

⎪

⎬ ⎪

⎭ =

⎧

⎪

⎨ ⎪

⎩

3( f [x3, x2] − f [x2, x1]) 3( f [x4, x3] − f [x3, x2])

⎫

⎪

⎬ ⎪

⎭

The necessary function and interval width values are f1 = 2.5 h1 = 4.5 − 3.0 = 1.5 f2 = 1.0 h2 = 7.0 − 4.5 = 2.5 f3 = 2.5 h3 = 9.0 − 7.0 = 2.0 f4 = 0.5

These can be substituted to yield

⎡ ⎢

⎣ 1

1.5 8 2.5 2.5 9 2

⎤ ⎥

⎦ ⎧

⎪

⎨ ⎪

⎩

c2 c3 c4

⎫

⎪

⎬ ⎪

⎭

⎧

⎪ ⎨

⎪

⎩

0 4.8 −4.8

⎫

⎪ ⎬

⎪

⎭

These equations can be solved using MATLAB with the results: c1 = 0 c2 = 0.839543726 c3 = −0.766539924 c4 = 0

Equations (18.21) and (18.18) can be used to compute the b’s and d’s b1 = −1.419771863 d1 = 0.186565272 b2 = −0.160456274 d2 = −0.214144487 b3 = 0.022053232 d3 = 0.127756654

These results, along with the values for the a’s [Eq. (18.11)], can be substituted into Eq. (18.10) to develop the following cubic splines for each interval:

s1(x) = 2.5 − 1.419771863(x − 3) + 0.186565272(x − 3)3

s2(x) = 1.0 − 0.160456274(x − 4.5) + 0.839543726(x − 4.5)2

− 0.214144487(x − 4.5)3

s3(x) = 2.5 + 0.022053232(x − 7.0) − 0.766539924(x − 7.0)2

+ 0.127756654(x − 7.0)3

The three equations can then be employed to compute values within each interval. For example, the value at x = 5, which falls within the second interval, is calculated as

s2(5) = 1.0 − 0.160456274(5 − 4.5) + 0.839543726(5 − 4.5)2 − 0.214144487(5 − 4.5)3

= 1.102889734. The total cubic spline fit is depicted in Fig. 18.4c.

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18.4 cUBIc SPlIneS 467

The results of Examples 18.1 through 18.3 are summarized in Fig. 18.4. Notice the progressive improvement of the fit as we move from linear to quadratic to cubic splines. We have also superimposed a cubic interpolating polynomial on Fig. 18.4c. Although the cubic spline consists of a series of third-order curves, the resulting fit differs from that obtained using the third-order polynomial. This is due to the fact that the natural spline requires zero second derivatives at the end knots, whereas the cubic polynomial has no such constraint.

18.4.2 End Conditions

Although its graphical basis is appealing, the natural spline is only one of several end con- ditions that can be specified for splines. Two of the most popular are

• Clamped End Condition. This option involves specifying the first derivatives at the first and last nodes. This is sometimes called a “clamped” spline because it is what occurs when you clamp the end of a drafting spline so that it has a desired slope. For example, if zero first derivatives are specified, the spline will level off or become horizontal at the ends.

• “Not-a-Knot” End Condition. A third alternative is to force continuity of the third de- rivative at the second and the next-to-last knots. Since the spline already specifies that the function value and its first and second derivatives are equal at these knots, speci- fying continuous third derivatives means that the same cubic functions will apply to each of the first and last two adjacent segments. Since the first internal knots no longer represent the junction of two different cubic functions, they are no longer true knots. Hence, this case is referred to as the “not-a-knot” condition. It has the additional prop- erty that for four points, it yields the same result as is obtained using an ordinary cubic interpolating polynomial of the sort described in Chap. 17.

These conditions can be readily applied by using Eq. (18.25) for the interior knots, i = 2, 3, . . . , n − 2, and using first (1) and last equations (n − 1) as written in Table 18.2.

Figure 18.5 shows a comparison of the three end conditions as applied to fit the data from Table 18.1. The clamped case is set up so that the derivatives at the ends are equal to zero.

As expected, the spline fit for the clamped case levels off at the ends. In contrast, the natural and not-a-knot cases follow the trend of the data points more closely. Notice how the natural spline tends to straighten out as would be expected because the second deriva- tives go to zero at the ends. Because it has nonzero second derivatives at the ends, the not-a-knot exhibits more curvature.

TABLE 18.2 the first and last equations needed to specify some commonly used end conditions for cubic splines.

Condition First and Last Equations

Natural c1 = 0, cn = 0

Clamped (where f 1 ′ and f n ′ are the specified first 2h1c1 + h1c2 = 3 f [x2, x1] − 3 f 1 ′ derivatives at the first and last nodes, respectively) hn−1 cn−1 + 2hn−1 cn = 3 f n ′ − 3 f [xn, xn−1]

Not-a-knot h2 c1 − (h1 + h2) c2 + h1 c3 = 0 hn−1 cn−2 − (hn−2 + hn−1) cn−1 + hn−2 cn = 0

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468 SPlIneS and PIecewISe InterPolatIon

18.5 PIECEWISE INTERPOLATION IN MATLAB

MATLAB has several built-in functions to implement piecewise interpolation. The spline function performs cubic spline interpolation as described in this chapter. The pchip function implements piecewise cubic Hermite interpolation. The interp1 function can also imple- ment spline and Hermite interpolation, but can also perform a number of other types of piecewise interpolation.

18.5.1 MATLAB Function: spline

Cubic splines can be easily computed with the built-in MATLAB function, spline. It has the general syntax

yy = spline (x, y, xx) (18.28)

where x and y = vectors containing the values that are to be interpolated, and yy = a vector containing the results of the spline interpolation as evaluated at the points in the vector xx.

By default, spline uses the not-a-knot condition. However, if y contains two more values than x has entries, then the first and last value in y are used as the derivatives at the end points. Consequently, this option provides the means to implement the clamped-end condition.

eXaMPle 18.4 Splines in MatlaB

Problem Statement. Runge’s function is a notorious example of a function that cannot be fit well with polynomials (recall Example 17.7):

f (x) = 1 ________ 1 + 25x2

f (x)

2 4 6 8

Natural

Not-a-knot

Clamped

FIGURE 18.5 Comparison of the clamped (with zero first derivatives), not-a-knot, and natural splines for the data from Table 18.1.

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18.5 PIecewISe InterPolatIon In MatlaB 469

Use MATLAB to fit nine equally spaced data points sampled from this function in the interval [−1, 1]. Employ (a) a not-a-knot spline and (b) a clamped spline with end slopes of f 1 ′ = 1 and f n−1 ′ = −4.

Solution. (a) The nine equally spaced data points can be generated as in

>> x = linspace(−1,1,9); >> y = 1./(1 + 25*x.^2);

Next, a more finely spaced vector of values can be generated so that we can create a smooth plot of the results as generated with the spline function:

>> xx = linspace(−1,1); >> yy = spline(x,y,xx);

Recall that linspace automatically creates 100 points if the desired number of points are not specified. Finally, we can generate values for Runge’s function itself and display them along with the spline fit and the original data:

>> yr = 1./(1 + 25*xx.^2); >> plot(x,y,'o',xx,yy,xx,yr,'−−')

As in Fig. 18.6, the not-a-knot spline does a nice job of following Runge’s function without exhibiting wild oscillations between the points.

(b) The clamped condition can be implemented by creating a new vector yc that has the desired first derivatives as its first and last elements. The new vector can then be used to

0.8

0.6

0.2

0.4

0 −1 −0.5 0 0.5 1

FIGURE 18.6 Comparison of Runge’s function (dashed line) with a 9-point not-a-knot spline fit generated with MATLAB (solid line).

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470 SPlIneS and PIecewISe InterPolatIon

generate and plot the spline fit:

>> yc = [1 y − 4]; >> yyc = spline(x,yc,xx); >> plot(x,y,'o',xx,yyc,xx,yr,'−−')

As in Fig. 18.7, the clamped spline now exhibits some oscillations because of the artificial slopes that we have imposed at the boundaries. In other examples, where we have knowl- edge of the true first derivatives, the clamped spline tends to improve the fit.

18.5.2 MATLAB Function: interp1

The built-in function interp1 provides a handy means to implement a number of different types of piecewise one-dimensional interpolation. It has the general syntax

yi = interp1(x, y, xi, 'method')

where x and y = vectors containing values that are to be interpolated, yi = a vector contain- ing the results of the interpolation as evaluated at the points in the vector xi, and 'method' = the desired method. The various methods are

• 'nearest'—nearest neighbor interpolation. This method sets the value of an interpo- lated point to the value of the nearest existing data point. Thus, the interpolation looks like a series of plateaus, which can be thought of as zero-order polynomials.

• 'linear'—linear interpolation. This method uses straight lines to connect the points.

0.8

0.6

0.2

0.4

0 −1 −0.5 0 0.5 1

FIGURE 18.7 Comparison of Runge’s function (dashed line) with a 9-point clamped end spline fit generated with MATLAB (solid line). Note that first derivatives of 1 and −4 are specified at the left and right boundaries, respectively.

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18.5 PIecewISe InterPolatIon In MatlaB 471

• 'spline'—piecewise cubic spline interpolation. This is identical to the spline function. • 'pchip' and 'cubic'—piecewise cubic Hermite interpolation.

If the 'method' argument is omitted, the default is linear interpolation. The pchip option (short for “piecewise cubic Hermite interpolation”) merits more dis-

cussion. As with cubic splines, pchip uses cubic polynomials to connect data points with continuous first derivatives. However, it differs from cubic splines in that the second de- rivatives are not necessarily continuous. Further, the first derivatives at the knots will not be the same as for cubic splines. Rather, they are expressly chosen so that the interpolation is “shape preserving.” That is, the interpolated values do not tend to overshoot the data points as can sometimes happen with cubic splines.

Therefore, there are trade-offs between the spline and the pchip options. The results of using spline will generally appear smoother because the human eye can detect discontinui- ties in the second derivative. In addition, it will be more accurate if the data are values of a smooth function. On the other hand, pchip has no overshoots and less oscillation if the data are not smooth. These trade-offs, as well as those involving the other options, are explored in the following example.

eXaMPle 18.5 trade-offs Using interp1

Problem Statement. You perform a test drive on an automobile where you alternately accelerate the automobile and then hold it at a steady velocity. Note that you never de- celerate during the experiment. The time series of spot measurements of velocity can be tabulated as

t 0 20 40 56 68 80 84 96 104 110 υ 0 20 20 38 80 80 100 100 125 125

Use MATLAB’s interp1 function to fit these data with (a) linear interpolation, (b) nearest neighbor, (c) cubic spline with not-a-knot end conditions, and (d) piecewise cubic Hermite interpolation.

Solution. (a) The data can be entered, fit with linear interpolation, and plotted with the following commands:

>> t = [0 20 40 56 68 80 84 96 104 110]; >> v = [0 20 20 38 80 80 100 100 125 125]; >> tt = linspace(0,110); >> vl = interp1(t,v,tt); >> plot(t,v,'o',tt,vl)

The results (Fig. 18.8a) are not smooth, but do not exhibit any overshoot.

(b) The commands to implement and plot the nearest neighbor interpolation are >> vn = interp1(t,v,tt,'nearest'); >> plot(t,v,'o',tt,vn)

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472 SPlIneS and PIecewISe InterPolatIon

As in Fig. 18.8b, the results look like a series of plateaus. This option is neither a smooth nor an accurate depiction of the underlying process.

>> vs = interp1(t,v,tt,'spline'); >> plot(t,v,'o',tt,vs)

These results (Fig. 18.8c) are quite smooth. However, severe overshoot occurs at several locations. This makes it appear that the automobile decelerated several times during the experiment.

140

120

100

0 0 20 40 60 80 100 120

(a) linear

(b) nearest neighbor

140

120

100

0 0 20 40 60 80 100 120

140

120

100

0 0 20 40 60 80 100 120

140

120

100

0 0 20 40 60 80 100 120

FIGURE 18.8 Use of several options of the interp1 function to perform piecewise polynomial interpolation on a velocity time series for an automobile.

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18.6 MUltIdIMenSIonal InterPolatIon 473

(d) The commands to implement the piecewise cubic Hermite interpolation are

>> vh = interp1 (t,v,tt,'pchip'); >> plot (t,v,'o',tt,vh)

For this case, the results (Fig. 18.8d) are physically realistic. Because of its shape-preserving nature, the velocities increase monotonically and never exhibit deceleration. Although the result is not as smooth as for the cubic splines, continuity of the first derivatives at the knots makes the transitions between points more gradual and hence more realistic.

18.6 MULTIDIMENSIONAL INTERPOLATION

The interpolation methods for one-dimensional problems can be extended to multidimen- sional interpolation. In this section, we will describe the simplest case of two-dimensional interpolation in Cartesian coordinates. In addition, we will describe MATLAB’s capabili- ties for multidimensional interpolation.

18.6.1 Bilinear Interpolation

Two-dimensional interpolation deals with determining intermediate values for functions of two variables z = f (xi , yi ). As depicted in Fig. 18.9, we have values at four points: f (x1, y1), f (x2, y1), f (x1, y2), and f (x2, y2). We want to interpolate between these points

f (x, y)

f (xi, yi)

(xi, yi)

f (x1, y2)

y1 x1

xi x2

x y

f (x2, y2)

f (x2, y1)

f (x1, y1)

FIGURE 18.9 Graphical depiction of two-dimensional bilinear interpolation where an intermediate value (filled circle) is estimated based on four given values (open circles).

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474 SPlIneS and PIecewISe InterPolatIon

to estimate the value at an intermediate point f (xi , yi ). If we use a linear function, the result is a plane connecting the points as in Fig. 18.9. Such functions are called bilinear.

A simple approach for developing the bilinear function is depicted in Fig. 18.10. First, we can hold the y value fixed and apply one-dimensional linear interpolation in the x direction. Using the Lagrange form, the result at (xi , y1 ) is

f (xi , y1 ) = xi − x2 ______ x1 − x2 f (x1, y1 ) +

xi − x1 ______ x2 − x1 f (x2, y1) (18.29)

and at (xi , y2 ) is

f (xi , y2 ) = xi − x2 ______ x1 − x2 f (x1, y2 ) +

xi − x1 ______ x2 − x1 f (x2, y2) (18.30)

These points can then be used to linearly interpolate along the y dimension to yield the final result:

f (xi , yi ) = yi − y2 ______ y1 − y2 f (xi , y1 ) +

yi − y1 ______ y2 − y1 f (xi, y2) (18.31)

A single equation can be developed by substituting Eqs. (18.29) and (18.30) into Eq. (18.31) to give

f (xi , yi ) = xi − x2 ______ x1 − x2

yi − y2 ______ y1 − y2 f (x1, y1) + xi − x1 ______ x2 − x1

yi − y2 ______ y1 − y2 f (x2, y1)

(18.32) +

xi − x2 ______ x1 − x2 yi − y1 ______ y2 − y1 f (x1, y2) +

xi − x1 ______ x2 − x1 yi − y1 ______ y2 − y1 f (x2, y2)

FIGURE 18.10 Two-dimensional bilinear interpolation can be implemented by first applying one-dimensional linear interpolation along the x dimension to determine values at xi. These values can then be used to linearly interpolate along the y dimension to yield the final result at xi, yi.

x1 x2xi

f (x1, y1) f (x2, y1)

f (xi, y1)

f(xi, yi)

f (xi, y2)f (x1, y2) f (x2, y2)

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18.6 MUltIdIMenSIonal InterPolatIon 475

eXaMPle 18.6 Bilinear Interpolation

Problem Statement. Suppose you have measured temperatures at a number of coordi- nates on the surface of a rectangular heated plate:

T(2, 1) = 60 T(9, 1) = 57.5 T(2, 6) = 55 T(9, 6) = 70

Use bilinear interpolation to estimate the temperature at xi = 5.25 and yi = 4.8.

Solution. Substituting these values into Eq. (18.32) gives

f (5.25, 4.8) = 5.25 − 9 ________ 2 − 9 4.8 − 6 _______ 1 − 6 60 +

5.25 − 2 ________ 9 − 2 4.8 − 6 _______ 1 − 6 57.5

+ 5.25 − 9 ________ 2 − 9 4.8 − 1 _______ 6 − 1 55 +

5.25 − 2 ________ 9 − 2 4.8 − 1 _______ 6 − 1 70 = 61.2143

18.6.2 Multidimensional Interpolation in MATLAB

MATLAB has two built-in functions for two- and three-dimensional piecewise interpola- tion: interp2 and interp3. As you might expect from their names, these functions operate in a similar fashion to interp1 (Sec. 18.5.2). For example, a simple representation of the syntax of interp2 is

zi = interp2(x, y, z, xi, yi, 'method')

where x and y = matrices containing the coordinates of the points at which the values in the matrix z are given, zi = a matrix containing the results of the interpolation as evaluated at the points in the matrices xi and yi, and method = the desired method. Note that the methods are identical to those used by interp1; that is, linear, nearest, spline, and cubic.

As with interp1, if the method argument is omitted, the default is linear interpolation. For example, interp2 can be used to make the same evaluation as in Example 18.6 as

>> x = [2 9]; >> y = [1 6]; >> z = [60 57.5;55 70]; >> interp2(x,y,z,5.25,4.8)

ans = 61.2143

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476 SPlIneS and PIecewISe InterPolatIon

18.7 CASE STUDY Heat tranSFer

Background. Lakes in the temperate zone can become thermally stratified during the summer. As depicted in Fig. 18.11, warm, buoyant water near the surface overlies colder, denser bottom water. Such stratification effectively divides the lake vertically into two layers: the epilimnion and the hypolimnion, separated by a plane called the thermocline.

Thermal stratification has great significance for environmental engineers and scientists studying such systems. In particular, the thermocline greatly diminishes mixing between the two layers. As a result, decomposition of organic matter can lead to severe depletion of oxygen in the isolated bottom waters.

The location of the thermocline can be defined as the inflection point of the temperature- depth curve—that is, the point at which d2T∕dz2 = 0. It is also the point at which the absolute value of the first derivative or gradient is a maximum.

The temperature gradient is important in its own right because it can be used in con- junction with Fourier’s law to determine the heat flux across the thermocline:

J = −DρC dT ___ dz (18.33)

where J = heat flux [cal/(cm2 · s)], α = an eddy diffusion coefficient (cm2/s), ρ = density (≅1 g/cm3), and C = specific heat [≅ 1 cal/(g · C)].

In this case study, natural cubic splines are employed to determine the thermocline depth and temperature gradient for Platte Lake, Michigan (Table 18.3). The latter is also used to determine the heat flux for the case where α = 0.01 cm2/s.

TABLE 18.3 temperature versus depth during summer for Platte lake, Michigan.

z, m 0 2.3 4.9 9.1 13.7 18.3 22.9 27.2 T, °C 22.8 22.8 22.8 20.6 13.9 11.7 11.1 11.1

z ( m

)

0 10

Epilimnion

Thermocline

Hypolimnion

T (°C) 20 30

FIGURE 18.11 Temperature versus depth during summer for Platte Lake, Michigan.

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18.7 CASE STUDY continued

Solution. As just described, we want to use natural spline end conditions to perform this analysis. Unfortunately, because it uses not-a-knot end conditions, the built-in MATLAB spline function does not meet our needs. Further, the spline function does not return the first and second derivatives we require for our analysis.

However, it is not difficult to develop our own M-file to implement a natural spline and return the derivatives. Such a code is shown in Fig. 18.12. After some preliminary error trapping, we set up and solve Eq. (18.27) for the second-order coefficients (c). Notice how

function [yy,dy,d2] = natspline(x,y,xx) % natspline: natural spline with differentiation % [yy,dy,d2] = natspline(x,y,xx): uses a natural cubic spline % interpolation to find yy, the values of the underlying function % y at the points in the vector xx. The vector x specifies the % points at which the data y is given. % input: % x = vector of independent variables % y = vector of dependent variables % xx = vector of desired values of dependent variables % output: % yy = interpolated values at xx % dy = first derivatives at xx % d2 = second derivatives at xx

n = length(x); if length(y)~ = n, error('x and y must be same length'); end if any(diff(x)< = 0),error('x not strictly ascending'),end m = length(xx); b = zeros(n,n); aa(1,1) = 1; aa(n,n) = 1; %set up Eq. 18.27 bb(1) = 0; bb(n) = 0; for i = 2:n − 1 aa(i,i−1) = h(x, i − 1); aa(i,i) = 2 * (h(x, i − 1) + h(x, i)); aa(i,i+1) = h(x, i); bb(i) = 3 * (fd(i + 1, i, x, y) − fd(i, i − 1, x, y)); end c = aa\bb'; %solve for c coefficients for i = 1:n − 1 %solve for a, b and d coefficients a(i) = y(i); b(i) = fd(i + 1, i, x, y) − h(x, i) / 3 * (2 * c(i) + c(i + 1)); d(i) = (c(i + 1) − c(i)) / 3 / h(x, i); end

FIGURE 18.12 M-file to determine intermediate values and derivatives with a natural spline. Note that the diff function employed for error trapping is described in Sec. 21.7.1.

(continued)

18.7 caSe StUdY 477

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478 SPlIneS and PIecewISe InterPolatIon

we use two subfunctions, h and fd, to compute the required finite differences. Once Eq. (18.27) is set up, we solve for the c’s with back division. A loop is then employed to generate the other coefficients (a, b, and d).

At this point, we have all we need to generate intermediate values with the cubic equation:

f (x) = ai + bi (x − xi ) + ci (x − xi )2 + di (x − xi )3

We can also determine the first and second derivatives by differentiating this equation twice to give

f ′ (x) = bi + 2ci (x − xi ) + 3di (x − xi )2

f ″ (x) = 2ci + 6di (x − xi )

As in Fig. 18.12, these equations can then be implemented in another subfunction, SplineInterp, to determine the values and the derivatives at the desired intermediate values.

Here is a script file that uses the natspline function to generate the spline and create plots of the results:

z = [0 2.3 4.9 9.1 13.7 18.3 22.9 27.2]; T=[22.8 22.8 22.8 20.6 13.9 11.7 11.1 11.1]; zz = linspace(z(1),z(length(z)));

18.7 CASE STUDY continued

for i = 1:m %perform interpolations at desired values [yy(i),dy(i),d2(i)] = SplineInterp(x, n, a, b, c, d, xx(i)); end end function hh = h(x, i) hh = x(i + 1) − x(i); end function fdd = fd(i, j, x, y) fdd = (y(i) − y(j)) / (x(i) − x(j)); end function [yyy,dyy,d2y] = SplineInterp(x, n, a, b, c, d, xi) for ii = 1:n − 1 if xi > = x(ii) − 0.000001 & xi < = x(ii + 1) + 0.000001 yyy = a(ii) + b(ii)*(xi − x(ii))+c(ii)*(xi − x(ii))^2 + d(ii)... *(xi − x(ii))^3; dyy = b(ii)+2*c(ii)*(xi − x(ii))+3*d(ii)*(xi − x(ii))^2; d2y = 2*c(ii)+6*d(ii)*(xi − x(ii)); break end end end

FIGURE 18.12 (Continued )

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18.7 CASE STUDY continued

[TT,dT,dT2] = natspline(z,T,zz); subplot(1,3,1),plot(T,z,'o',TT,zz) title('(a) T'),legend('data','T') set(gca,'YDir','reverse'),grid subplot(1,3,2),plot(dT,zz) title('(b) dT/dz') set(gca,'YDir','reverse'),grid subplot(1,3,3),plot(dT2,zz) title('(c) d2T/dz2') set(gca,'YDir','reverse'),grid

As in Fig. 18.13, the thermocline appears to be located at a depth of about 11.5 m. We can use root location (zero second derivative) or optimization methods (minimum first derivative) to refine this estimate. The result is that the thermocline is located at 11.35 m where the gradient is −1.61 °C/m.

18.7 caSe StUdY 479

0 (a) T

30 10 20 30

data

0 (b) dT/dz

30 −2 0 2

0 (c) d2T/dz2

30 −0.5 0 0.5

FIGURE 18.13 Plots of (a) temperature, (b) gradient, and (c) second derivative versus depth (m) generated with the cubic spline program. The thermocline is located at the inflection point of the temperature-depth curve.

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480 SPlIneS and PIecewISe InterPolatIon

The gradient can be used to compute the heat flux across the thermocline with Eq. (18.33):

J = −0.01 cm 2 ____ s × 1

g ____ cm3

× 1 cal _____ g · °C × ( −1.61 °C ___ m ) × 1 m _______ 100 cm × 86,400 s _______ d = 13.9

cal ______ cm2 · d

The foregoing analysis demonstrates how spline interpolation can be used for engi- neering and scientific problem solving. However, it also is an example of numerical differ- entiation. As such, it illustrates how numerical approaches from different areas can be used in tandem for problem solving. We will be describing the topic of numerical differentiation in detail in Chap. 21.

18.7 CASE STUDY continued

PROBLEMS

18.1 Given the data

x 1 2 2.5 3 4 5 f (x) 1 5 7 8 2 1

Fit these data with (a) a cubic spline with natural end condi- tions, (b) a cubic spline with not-a-knot end conditions, and (c) piecewise cubic Hermite interpolation. 18.2 A reactor is thermally stratified as in the following table:

Depth, m 0 0.5 1 1.5 2 2.5 3

Temperature, °C 70 70 55 22 13 10 10

Based on these temperatures, the tank can be idealized as two zones separated by a strong temperature gradient or thermocline. The depth of the thermocline can be defined as the inflection point of the temperature-depth curve—that is, the point at which d2T∕dz2 = 0. At this depth, the heat flux from the surface to the bottom layer can be computed with Fourier’s law:

J = −k dT ___ dz

Use a clamped cubic spline fit with zero end derivatives to determine the thermocline depth. If k = 0.01 cal/ (s · cm · °C) compute the flux across this interface.

18.3 The following is the built-in humps function that MATLAB uses to demonstrate some of its numerical capabilities:

f (x) = 1 ______________ (x − 0.3)2 + 0.01

+ 1 __________________ (x − 0.9)2 + 0.04

− 6

The humps function exhibits both flat and steep regions over a relatively short x range. Here are some values that have been generated at intervals of 0.1 over the range from x = 0 to 1:

x 0 0.1 0.2 0.3 0.4 0.5 f (x) 5.176 15.471 45.887 96.500 47.448 19.000

x 0.6 0.7 0.8 0.9 1 f (x) 11.692 12.382 17.846 21.703 16.000

Fit these data with a (a) cubic spline with not-a-knot end conditions and (b) piecewise cubic Hermite interpolation. In both cases, create a plot comparing the fit with the exact humps function. 18.4 Develop a plot of a cubic spline fit of the following data with (a) natural end conditions and (b) not-a-knot end conditions. In addition, develop a plot using (c) piecewise cubic Hermite interpolation (pchip).

x 0 100 200 400 f (x) 0 0.82436 1.00000 0.73576

x 600 800 1000 f (x) 0.40601 0.19915 0.09158

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In each case, compare your plot with the following equation which was used to generate the data:

f (x) = x ____ 200

e−x∕200+1

18.5 The following data are sampled from the step function depicted in Fig. 18.1:

x −1 −0.6 −0.2 0.2 0.6 1 f (x) 0 0 0 1 1 1

Fit these data with a (a) cubic spline with not-a-knot end con- ditions, (b) cubic spline with zero-slope clamped end condi- tions, and (c) piecewise cubic Hermite interpolation. In each case, create a plot comparing the fit with the step function. 18.6 Develop an M-file to compute a cubic spline fit with natural end conditions. Test your code by using it to dupli- cate Example 18.3. 18.7 The following data were generated with the fifth- order polynomial: f (x) = 0.0185x5 − 0.444x4 + 3.9125x3 − 15.456x2 + 27.069x − 14.1:

x 1 3 5 6 7 9 f (x) 1.000 2.172 4.220 5.430 4.912 9.120

(a) Fit these data with a cubic spline with not-a-knot end conditions. Create a plot comparing the fit with the function. (b) Repeat (a) but use clamped end conditions where the end slopes are set at the exact values as determined by differen- tiating the function. 18.8 Bessel functions often arise in advanced engineering and scientific analyses such as the study of electric fields. These functions are usually not amenable to straightforward evaluation and, therefore, are often compiled in standard mathematical tables. For example,

x 1.8 2 2.2 2.4 2.6 J1(x) 0.5815 0.5767 0.556 0.5202 0.4708

Estimate J1(2.1), (a) using an interpolating polynomial and (b) using cubic splines. Note that the true value is 0.5683. 18.9 The following data define the sea-level concentra- tion of dissolved oxygen for fresh water as a function of temperature:

T, °C 0 8 16 24 32 40 o, mg/L 14.621 11.843 9.870 8.418 7.305 6.413

Use MATLAB to fit the data with (a) piecewise linear inter- polation, (b) a fifth-order polynomial, and (c) a spline. Dis- play the results graphically and use each approach to estimate o(27). Note that the exact result is 7.986 mg/L. 18.10 (a) Use MATLAB to fit a cubic spline to the follow- ing data to determine y at x = 1.5:

x 0 2 4 7 10 12 y 20 20 12 7 6 6

(b) Repeat (a), but with zero first derivatives at the end knots. 18.11 Runge’s function is written as

f (x) = 1 ________ 1 + 25x2

Generate five equidistantly spaced values of this function over the interval: [−1, 1]. Fit these data with (a) a fourth- order polynomial, (b) a linear spline, and (c) a cubic spline. Present your results graphically. 18.12 Use MATLAB to generate eight points from the function

f (t) = sin2 t

from t = 0 to 2π. Fit these data using (a) cubic spline with not-a-knot end conditions, (b) cubic spline with derivative end conditions equal to the exact values calculated with dif- ferentiation, and (c) piecewise cubic Hermite interpolation. Develop plots of each fit as well as plots of the absolute error (Et = approximation − true) for each. 18.13 The drag coefficient for spheres such as sporting balls is known to vary as a function of the Reynolds number Re, a dimensionless number that gives a measure of the ratio of inertial forces to viscous forces:

Re = ρVD ____ μ

where ρ = the fluid’s density (kg/m3), V = its velocity (m/s), D = diameter (m), and μ = dynamic viscosity (N · s/m2). Although the relationship of drag to the Reynolds number is sometimes available in equation form, it is frequently tabu- lated. For example, the following table provides values for a smooth spherical ball:

Re ( × 10−4) 2 5.8 16.8 27.2 29.9 33.9 CD 0.52 0.52 0.52 0.5 0.49 0.44

Re ( × 10−4) 36.3 40 46 60 100 200 400 CD 0.18 0.074 0.067 0.08 0.12 0.16 0.19

ProBleMS 481

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482 SPlIneS and PIecewISe InterPolatIon

(a) Develop a MATLAB function that employs an appro- priate interpolation function to return a value of CD as a function of the Reynolds number. The first line of the func- tion should be

function CDout = Drag(ReCD,ReIn)

where ReCD = a 2-row matrix containing the table, ReIn = the Reynolds number at which you want to estimate the drag, and CDout = the corresponding drag coefficient. (b) Write a script that uses the function developed in part (a) to generate a labeled plot of the drag force ver- sus velocity (recall Sec. 1.4). Use the following param- eter values for the script: D = 22 cm, ρ = 1.3 kg/m3, and μ = 1.78 × 10−5 Pa · s. Employ a range of velocities from 4 to 40 m/s for your plot. 18.14 The following function describes the temperature dis- tribution on a rectangular plate for the range −2 ≤ x ≤ 0 and 0 ≤ y ≤ 3

T = 2 + x − y + 2x2 + 2xy + y2

Develop a script to: (a) Generate a meshplot of this function using the MATLAB function surfc. Employ the linspace function with default spacing (i.e., 100 interior points) to generate the x and y values. (b) Use the MATLAB function interp2 with the default interpolation option ('linear') to compute the temperature at x = –1.63 and y = 1.627. Deter- mine the percent relative error of your result. (c) Repeat (b), but with 'spline'. Note: for parts (b) and (c), employ the linspace function with 9 interior points. 18.15 The U.S. Standard Atmosphere specifies atmospheric properties as a function of altitude above sea level. The fol- lowing table shows selected values of temperature, pressure, and density

Altitude (km) T (°C) p (atm) ρ (kg/m3)

−0.5 18.4 1.0607 1.2850

2.5 −1.1 0.73702 0.95697

6 −23.8 0.46589 0.66015

11 −56.2 0.22394 0.36481

20 −56.3 0.054557 0.088911

28 −48.5 0.015946 0.025076 50 −2.3 7.8721 × 10−4 1.0269 × 10−3

60 −17.2 2.2165 × 10−4 3.0588 × 10−4

80 −92.3 1.0227 × 10−5 1.9992 × 10−5

90 −92.3 1.6216 × 10−6 3.1703 × 10−6

Develop a MATLAB function, StdAtm, to determine values of the three properties for a given altitude. Base the func- tion on the pchip option for interp1. If the user requests a value outside the range of altitudes, have the function display an error message and terminate the application. Use the following script as the starting point to create a 3-panel plot of altitude versus the properties as depicted in Fig. P18.15.

% Script to generate a plot of temperature, pressure and density % for the U.S. Standard Atmosphere clc, clf z = [−0.5 2.5 6 11 20 28 50 60 80 90]; T = [18.4 −1.1 −23.8 −56.2 −56.3 −48.5 −2.3 −17.2 −92.3 −92.3]; p = [1.0607 0.73702 0.46589 0.22394 0.054557 0.015946 ... 7.8721e − 4 2.2165e − 4 1.02275e − 05 1.6216e − 06]; rho = [1.285025 0.95697 0.6601525 0.364805 0.0889105 ... 0.02507575 0.001026918 0.000305883 0.000019992 3.1703e − 06]; zint = [− 0.5:0.1:90]; for i = 1:length(zint) [Tint(i),pint(i),rint(i)] = StdAtm(z,T,p,rho,zint(i)); end

% Create plot

Te = StdAtm(z,T,p,rho,−1000); 18.16 Felix Baumgartner ascended to 39 km in a strato- spheric balloon and made a free-fall jump rushing toward earth at supersonic speeds before parachuting to the ground. As he fell, his drag coefficient changed primarily because the air density changed. Recall from Chap. 1 that the ter- minal velocity, υterminal (m/s), of a free-falling object can be computed as

υ terminal = √ ___

gm ___ c d

where g = gravitational acceleration (m/s2), m = mass (kg), and cd = drag coefficient (kg/m). The drag coefficient can be computed as

cd = 0.5ρACd

where ρ = the fluid density (kg/m3), A = the projected area (m2), and Cd = a dimensionless drag coefficient. Note that the gravitational acceleration, g (m/s2), can be related to elevation by

g = 9.806412 − 0.003039734z

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ProBleMS 483

where z = elevation above the earth’s surface (km) and the den- sity of air, ρ (kg/m3), at various elevations can be tabulated as

z (km) ρ (kg/m3) z (km) ρ (kg/m3) z (km) ρ (kg/m3)

−1 1.347 6 0.6601 25 0.04008

0 1.225 7 0.5900 30 0.01841

1 1.112 8 0.5258 40 0.003996

2 1.007 9 0.4671 50 0.001027

3 0.9093 10 0.4135 60 0.0003097

4 0.8194 15 0.1948 70 8.283 × 10−5

5 0.7364 20 0.08891 80 1.846 × 10−5

Assume that m = 80 kg, A = 0.55 m2, and Cd = 1.1. Develop a MATLAB script to create a labeled plot of terminal veloc- ity versus elevation for z = [0:0.5:40]. Use a spline to gener- ate the required densities needed to construct the plot.

(a) Temperature (°C) A

lti tu

de (k

m )

−10

90 (c) Density (kg/m3)

−100 −50 0 50

(b) Pressure (atm)

0.0000 0.5000 1.0000 −10

1.5000 0.0000 0.5000 1.0000 1.5000 −10

FIGURE P18.15

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485

Part Five

Integration and Differentiation

5.1 OVERVIEW

In high school or during your first year of college, you were introduced to differential and integral calculus. There you learned techniques to obtain analytical or exact derivatives and integrals.

Mathematically, the derivative represents the rate of change of a dependent variable with respect to an independent variable. For example, if we are given a function y(t) that specifies an object’s position as a function of time, differentiation provides a means to determine its velocity, as in:

υ(t) = d __ dt y (t)

As in Fig. PT5.1a, the derivative can be visualized as the slope of a function. Integration is the inverse of differentiation. Just as dif-

ferentiation uses differences to quantify an instantaneous process, integration involves summing instantaneous infor- mation to give a total result over an interval. Thus, if we are provided with velocity as a function of time, integration can be used to determine the distance traveled:

y (t) = ∫ 0 t

υ (t) dt

As in Fig. PT5.1b, for functions lying above the abscissa, the integral can be visualized as the area under the curve of υ(t) from 0 to t. Consequently, just as a derivative can be thought of as a slope, an integral can be envisaged as a summation.

Because of the close relationship between differen- tiation and integration, we have opted to devote this part of the book to both processes. Among other things, this will provide the opportunity to highlight their similarities

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486 PART 5 InTegRATIon AnD DIffeRenTIATIon

and differences from a numerical perspective. In addition, the material will have relevance to the next part of the book where we will cover differential equations.

Although differentiation is taught before integration in calculus, we reverse their order in the following chapters. We do this for several reasons. First, we have already introduced you to the basics of numerical differentiation in Chap. 4. Second, in part because it is much less sensitive to roundoff errors, integration represents a more highly developed area of numerical methods. Finally, although numerical differentiation is not as widely employed, it does have great significance for the solution of differential equations. Hence, it makes sense to cover it as the last topic prior to describing differential equations in Part Six.

5.2 PART ORGANIZATION

Chapter 19 is devoted to the most common approaches for numerical integration—the Newton-Cotes formulas. These relationships are based on replacing a complicated func- tion or tabulated data with a simple polynomial that is easy to integrate. Three of the most widely used Newton-Cotes formulas are discussed in detail: the trapezoidal rule, Simpson’s 1∕3 rule, and Simpson’s 3∕8 rule. All these formulas are designed for cases where the data to be integrated are evenly spaced. In addition, we also include a discussion of numerical integration of unequally spaced data. This is a very important topic because many real-world applications deal with data that are in this form.

0 0

200

400

8 124 t

0 0

8 124 t

0 0

8 124

(a)

0 0

200

400

8 124

(b)

FIGURE PT5.1 The contrast between (a) differentiation and (b) integration.

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5.2 PART oRgAnIZATIon 487

All the above material relates to closed integration, where the function values at the ends of the limits of integration are known. At the end of Chap. 19, we present open inte- gration formulas, where the integration limits extend beyond the range of the known data. Although they are not commonly used for definite integration, open integration formulas are presented here because they are utilized in the solution of ordinary differential equa- tions in Part Six.

The formulations covered in Chap. 19 can be employed to analyze both tabulated data and equations. Chapter 20 deals with two techniques that are expressly designed to integrate equations and functions: Romberg integration and Gauss quadrature. Computer algorithms are provided for both of these methods. In addition, adaptive integration is discussed.

In Chap. 21, we present additional information on numerical differentiation to supplement the introductory material from Chap. 4. Topics include high-accuracy finite- difference formulas, Richardson extrapolation, and the differentiation of unequally spaced data. The effect of errors on both numerical differentiation and integration is also discussed.

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488

19 numerical Integration formulas

CHAPTER OBJECTIVES

The primary objective of this chapter is to introduce you to numerical integration. Specific objectives and topics covered are

• Recognizing that Newton-Cotes integration formulas are based on the strategy of replacing a complicated function or tabulated data with a polynomial that is easy to integrate.

• Knowing how to implement the following single application Newton-Cotes formulas:

Trapezoidal rule Simpson’s 1∕3 rule Simpson’s 3∕8 rule

• Knowing how to implement the following composite Newton-Cotes formulas: Trapezoidal rule Simpson’s 1∕3 rule

• Recognizing that even-segment–odd-point formulas like Simpson’s 1∕3 rule achieve higher than expected accuracy.

• Knowing how to use the trapezoidal rule to integrate unequally spaced data. • Understanding the difference between open and closed integration formulas.

YOU’VE GOT A PROBLEM

Recall that the velocity of a free-falling bungee jumper as a function of time can be computed as υ(t) = √

___ gm ___ cd tanh ( √

___

gcd ___ m t ) (19.1)

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19.1 InTRoDUCTIon AnD BACKgRoUnD 489

Suppose that we would like to know the vertical distance z the jumper has fallen after a certain time t. This distance can be evaluated by integration:

z (t) = ∫ 0 t

υ (t) dt (19.2)

Substituting Eq. (19.1) into Eq. (19.2) gives

z (t) = ∫ 0 t

√ ___

gm ___ cd tanh ( √ ___

gcd ___ m t ) dt (19.3)

Thus, integration provides the means to determine the distance from the velocity. Calculus can be used to solve Eq. (19.3) for

z (t) = m __ cd ln [cosh ( √ ___

gcd ___ m t ) ] (19.4)

Although a closed form solution can be developed for this case, there are other func- tions that cannot be integrated analytically. Further, suppose that there was some way to measure the jumper’s velocity at various times during the fall. These velocities along with their associated times could be assembled as a table of discrete values. In this situation, it would also be possible to integrate the discrete data to determine the distance. In both these instances, numerical integration methods are available to obtain solutions. Chapters 19 and 20 will introduce you to some of these methods.

19.1 INTRODUCTION AND BACKGROUND

19.1.1 What Is Integration?

According to the dictionary definition, to integrate means “to bring together, as parts, into a whole; to unite; to indicate the total amount. . . .” Mathematically, definite integration is represented by

I = ∫ a

f (x) dx (19.5)

which stands for the integral of the function f (x) with respect to the independent variable x, evaluated between the limits x = a to x = b.

As suggested by the dictionary definition, the “meaning” of Eq. (19.5) is the total value, or summation, of f (x) dx over the range x = a to b. In fact, the symbol ∫ is actually a stylized capital S that is intended to signify the close connection between integration and summation.

Figure 19.1 represents a graphical manifestation of the concept. For functions lying above the x axis, the integral expressed by Eq. (19.5) corresponds to the area under the curve of f (x) between x = a and b.

Numerical integration is sometimes referred to as quadrature. This is an archaic term that originally meant the construction of a square having the same area as some curvilinear figure. Today, the term quadrature is generally taken to be synonymous with numerical definite integration.

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19.1.2 Integration in Engineering and Science

Integration has so many engineering and scientific applications that you were required to take integral calculus in your first year at college. Many specific examples of such applica- tions could be given in all fields of engineering and science. A number of examples relate directly to the idea of the integral as the area under a curve. Figure 19.2 depicts a few cases where integration is used for this purpose.

Other common applications relate to the analogy between integration and summation. For example, a common application is to determine the mean of a continuous function. Recall that the mean of n discrete data points can be calculated by [Eq. (14.2)].

Mean = ∑ i=1

n yi _____ n (19.6)

where yi are individual measurements. The determination of the mean of discrete points is depicted in Fig. 19.3a.

In contrast, suppose that y is a continuous function of an independent variable x, as depicted in Fig. 19.3b. For this case, there are an infinite number of values between a and b. Just as Eq. (19.6) can be applied to determine the mean of the discrete readings, you might also be interested in computing the mean or average of the continuous function y = f (x) for the interval from a to b. Integration is used for this purpose, as specified by

Mean = ∫ a

b f (x)dx ________ b − a (19.7)

This formula has hundreds of engineering and scientific applications. For example, it is used to calculate the center of gravity of irregular objects in mechanical and civil engineer- ing and to determine the root-mean-square current in electrical engineering.

f (x)

a b x

FIGURE 19.1 Graphical representation of the integral of f (x) between the limits x = a to b. The integral is equivalent to the area under the curve.

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19.1 InTRoDUCTIon AnD BACKgRoUnD 491

(a) (b) (c)

FIGURE 19.2 Examples of how integration is used to evaluate areas in engineering and scientific applications. (a) A surveyor might need to know the area of a field bounded by a meandering stream and two roads. (b) A hydrologist might need to know the cross-sectional area of a river. (c) A structural engineer might need to determine the net force due to a nonuniform wind blowing against the side of a skyscraper.

0 4 62

Mean

3 51

y = f (x)

Mean

(a)

(b)

FIGURE 19.3 An illustration of the mean for (a) discrete and (b) continuous data.

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Integrals are also employed by engineers and scientists to evaluate the total amount or quantity of a given physical variable. The integral may be evaluated over a line, an area, or a volume. For example, the total mass of chemical contained in a reactor is given as the product of the concentration of chemical and the reactor volume, or

Mass = concentration × volume

where concentration has units of mass per volume. However, suppose that concentration varies from location to location within the reactor. In this case, it is necessary to sum the products of local concentrations ci and corresponding elemental volumes ΔVi:

Mass = ∑ i=1

n ci ΔVi

where n is the number of discrete volumes. For the continuous case, where c(x, y, z) is a known function and x, y, and z are independent variables designating position in Cartesian coordinates, integration can be used for the same purpose:

Mass = ∫∫∫ c(x, y, z) dx dy dz or

Mass = ∫ ∫ V

∫ c(V ) d V which is referred to as a volume integral. Notice the strong analogy between summation and integration.

Similar examples could be given in other fields of engineering and science. For ex- ample, the total rate of energy transfer across a plane where the flux (in calories per square centimeter per second) is a function of position is given by

Flux = ∫ ∫ A

flux d A

which is referred to as an areal integral, where A = area. These are just a few of the applications of integration that you might face regularly

in the pursuit of your profession. When the functions to be analyzed are simple, you will normally choose to evaluate them analytically. However, it is often difficult or impossible when the function is complicated, as is typically the case in more realistic examples. In addition, the underlying function is often unknown and defined only by measurement at discrete points. For both these cases, you must have the ability to obtain approximate val- ues for integrals using numerical techniques as described next.

19.2 NEWTON-COTES FORMULAS

The Newton-Cotes formulas are the most common numerical integration schemes. They are based on the strategy of replacing a complicated function or tabulated data with a poly- nomial that is easy to integrate:

I = ∫ a

f (x) dx ≅ ∫ a

fn (x) dx (19.8)

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19.2 neWTon-CoTes foRmUlAs 493

where fn(x) = a polynomial of the form fn(x) = a0 + a1x + · · · + an−1xn−1 + an xn (19.9)

where n is the order of the polynomial. For example, in Fig. 19.4a, a first-order polynomial (a straight line) is used as an approximation. In Fig. 19.4b, a parabola is employed for the same purpose.

The integral can also be approximated using a series of polynomials applied piecewise to the function or data over segments of constant length. For example, in Fig. 19.5, three

f (x)

a b (a) (b)

f (x)

a b x

FIGURE 19.4 The approximation of an integral by the area under (a) a straight line and (b) a parabola.

f (x)

a b x

FIGURE 19.5 The approximation of an integral by the area under three straight-line segments.

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straight-line segments are used to approximate the integral. Higher-order polynomials can be utilized for the same purpose.

Closed and open forms of the Newton-Cotes formulas are available. The closed forms are those where the data points at the beginning and end of the limits of integration are known (Fig. 19.6a). The open forms have integration limits that extend beyond the range of the data (Fig. 19.6b). This chapter emphasizes the closed forms. However, material on open Newton-Cotes formulas is briefly introduced in Sec. 19.7.

19.3 THE TRAPEZOIDAL RULE

The trapezoidal rule is the first of the Newton-Cotes closed integration formulas. It cor- responds to the case where the polynomial in Eq. (19.8) is first-order:

I = ∫ a

[ f (a) + f (b) − f (a) _________ b − a (x − a)] dx (19.10)

The result of the integration is

I = (b − a) f (a) + f (b) _________ 2 (19.11)

which is called the trapezoidal rule. Geometrically, the trapezoidal rule is equivalent to approximating the area of the trap-

ezoid under the straight line connecting f (a) and f (b) in Fig. 19.7. Recall from geometry that the formula for computing the area of a trapezoid is the height times the average of the bases. In our case, the concept is the same but the trapezoid is on its side. Therefore, the integral estimate can be represented as

I = width × average height (19.12)

f (x)

a b (a)

f (x)

a b (b)

FIGURE 19.6 The difference between (a) closed and (b) open integration formulas.

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19.3 THe TRAPeZoIDAl RUle 495

I = (b − a) × average height (19.13)

where, for the trapezoidal rule, the average height is the average of the function values at the end points, or [ f (a) + f (b)]∕2.

All the Newton-Cotes closed formulas can be expressed in the general format of Eq. (19.13). That is, they differ only with respect to the formulation of the average height.

19.3.1 Error of the Trapezoidal Rule

When we employ the integral under a straight-line segment to approximate the integral under a curve, we obviously can incur an error that may be substantial (Fig. 19.8). An esti- mate for the local truncation error of a single application of the trapezoidal rule is

Et = − 1 ___ 12 f ″ (ξ )(b − a)

3 (19.14)

where ξ lies somewhere in the interval from a to b. Equation (19.14) indicates that if the function being integrated is linear, the trapezoidal rule will be exact because the second derivative of a straight line is zero. Otherwise, for functions with second- and higher-order derivatives (i.e., with curvature), some error can occur.

eXAmPle 19.1 single Application of the Trapezoidal Rule

Problem statement. Use Eq. (19.11) to numerically integrate

f (x) = 0.2 + 25x − 200x2 + 675x3 − 900x4 + 400x5

from a = 0 to b = 0.8. Note that the exact value of the integral can be determined analytically to be 1.640533.

f (x)

f (a)

f (b)

a b x

FIGURE 19.7 Graphical depiction of the trapezoidal rule.

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solution. The function values f (0) = 0.2 and f (0.8) = 0.232 can be substituted into Eq. (19.11) to yield

I = (0.8 − 0) 0.2 + 0.232 __________ 2 = 0.1728

which represents an error of Et = 1.640533 − 0.1728 = 1.467733, which corresponds to a percent relative error of εt = 89.5%. The reason for this large error is evident from the graphical depiction in Fig. 19.8. Notice that the area under the straight line neglects a sig- nificant portion of the integral lying above the line.

In actual situations, we would have no foreknowledge of the true value. Therefore, an approximate error estimate is required. To obtain this estimate, the function’s second derivative over the interval can be computed by differentiating the original function twice to give

f ″(x) = −400 + 4,050x − 10,800x 2 + 8,000x 3

The average value of the second derivative can be computed as [Eq. (19.7)]

_ f ″(x) =

∫ 0 0.8 (−400 + 4,050x − 10,800x2 + 8,000x3) dx

______________________________________ 0.8 − 0 = −60

which can be substituted into Eq. (19.14) to yield

Ea = − 1 ___ 12 (−60)(0.8)

3 = 2.56

f (x)

0.80

2.0

Error

Integral estimate

FIGURE 19.8 Graphical depiction of the use of a single application of the trapezoidal rule to approximate the integral of f (x) = 0.2 + 25x − 200x 2 + 675x 3 − 900x 4 + 400x5 from x = 0 to 0.8.

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19.3 THe TRAPeZoIDAl RUle 497

which is of the same order of magnitude and sign as the true error. A discrepancy does exist, however, because of the fact that for an interval of this size, the average second derivative is not necessarily an accurate approximation of f ″(ξ). Thus, we denote that the error is approximate by using the notation Ea, rather than exact by using Et.

19.3.2 The Composite Trapezoidal Rule

One way to improve the accuracy of the trapezoidal rule is to divide the integration interval from a to b into a number of segments and apply the method to each segment (Fig. 19.9). The areas of individual segments can then be added to yield the integral for the entire interval. The resulting equations are called composite, or multiple-segment, integration formulas.

Figure 19.9 shows the general format and nomenclature we will use to characterize composite integrals. There are n + 1 equally spaced base points (x0, x1, x2, . . . , xn ). Conse- quently, there are n segments of equal width:

h = b − a _____ n (19.15)

If a and b are designated as x0, and xn, respectively, the total integral can be repre- sented as

I = ∫ x0 x1

f (x) dx + ∫ x1 x2

f (x) dx + · · · + ∫ xn−1

f (x) dx

f (x)

x0 = a xn = b

x1 x2 x3 x4 x5

b − a nh =

FIGURE 19.9 Composite trapezoidal rule.

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Substituting the trapezoidal rule for each integral yields

I = h f (x0) + f (x1) ___________ 2 + h

f (x1) + f (x2) __________ 2 + · · · + h f (xn−1) + f (xn) ____________ 2 (19.16)

or, grouping terms:

I = h __ 2 [ f (x0) + 2 ∑ i=1

n−1

f (xi ) + f (xn)] (19.17)

or, using Eq. (19.15) to express Eq. (19.17) in the general form of Eq. (19.13):

I = (b − a) ⏟

Width

f (x0) + 2 ∑

i=1

n−1 f (xi ) + f (xn )

_____________________ 2n  Average height (19.18)

Because the summation of the coefficients of f (x) in the numerator divided by 2n is equal to 1, the average height represents a weighted average of the function values. According to Eq. (19.18), the interior points are given twice the weight of the two end points f (x 0) and f (xn ).

An error for the composite trapezoidal rule can be obtained by summing the individual errors for each segment to give

Et = − (b − a)3 _______

12n3 ∑

i=1

n f ″(ξi ) (19.19)

where f ″(ξi ) is the second derivative at a point ξi located in segment i. This result can be simplified by estimating the mean or average value of the second derivative for the entire interval as

_ f ″ ≅

∑ i=1

n f ″(ξi )

________ n (19.20)

Therefore ∑ f ″(ξi ) ≅ n _ f ″ and Eq. (19.19) can be rewritten as

Ea = − (b − a)3 _______

12n2

_ f ″ (19.21)

Thus, if the number of segments is doubled, the truncation error will be quartered. Note that Eq. (19.21) is an approximate error because of the approximate nature of Eq. (19.20).

eXAmPle 19.2 Composite Application of the Trapezoidal Rule

Problem statement. Use the two-segment trapezoidal rule to estimate the integral of

f (x) = 0.2 + 25x − 200x2 + 675x3 − 900x4 + 400x5

from a = 0 to b = 0.8. Employ Eq. (19.21) to estimate the error. Recall that the exact value of the integral is 1.640533.

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19.3 THe TRAPeZoIDAl RUle 499

solution. For n = 2 (h = 0.4):

f (0) = 0.2 f (0.4) = 2.456 f (0.8) = 0.232

I = 0.8 0.2 + 2(2.456) + 0.232 ______________________ 4 = 1.0688

Et = 1.640533 − 1.0688 = 0.57173 εt = 34.9%

Ea = − 0.83 ______

12(2)2 (−60) = 0.64

where −60 is the average second derivative determined previously in Example 19.1.

The results of the previous example, along with three- through ten-segment applications of the trapezoidal rule, are summarized in Table 19.1. Notice how the error decreases as the number of segments increases. However, also notice that the rate of decrease is gradual. This is because the error is inversely related to the square of n [Eq. (19.21)]. Therefore, doubling the number of segments quarters the error. In subse- quent sections we develop higher-order formulas that are more accurate and that con- verge more quickly on the true integral as the segments are increased. However, before investigating these formulas, we will first discuss how MATLAB can be used to imple- ment the trapezoidal rule.

19.3.3 MATLAB M-file: trap

A simple algorithm to implement the composite trapezoidal rule can be written as in Fig. 19.10. The function to be integrated is passed into the M-file along with the limits of integration and the number of segments. A loop is then employed to generate the integral following Eq. (19.18).

TABLE 19.1 Results for the composite trapezoidal rule to estimate the integral of f (x) = 0.2 + 25x − 200x2 + 675x3 − 900x4 + 400x5 from x = 0 to 0.8. The exact value is 1.640533.

n h I εt (%)

2 0.4 1.0688 34.9 3 0.2667 1.3695 16.5 4 0.2 1.4848 9.5 5 0.16 1.5399 6.1 6 0.1333 1.5703 4.3 7 0.1143 1.5887 3.2 8 0.1 1.6008 2.4 9 0.0889 1.6091 1.9 10 0.08 1.6150 1.6

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An application of the M-file can be developed to determine the distance fallen by the free-falling bungee jumper in the first 3 s by evaluating the integral of Eq. (19.3). For this example, assume the following parameter values: g = 9.81 m/s2, m = 68.1 kg, and cd = 0.25 kg/m. Note that the exact value of the integral can be computed with Eq. (19.4) as 41.94805.

The function to be integrated can be developed as an M-file or with an anonymous function,

>> v=@(t) sqrt(9.81*68.1/0.25)*tanh(sqrt(9.81*0.25/68.1)*t)

v = @(t) sqrt(9.81*68.1/0.25)*tanh(sqrt(9.81*0.25/68.1)*t)

First, let’s evaluate the integral with a crude five-segment approximation:

>> format long >> trap(v,0,3,5)

ans = 41.86992959072735

function I = trap(func,a,b,n,varargin) % trap: composite trapezoidal rule quadrature % I = trap(func,a,b,n,pl,p2,...): % composite trapezoidal rule % input: % func = name of function to be integrated % a, b = integration limits % n = number of segments (default = 100) % pl,p2,... = additional parameters used by func % output: % I = integral estimate

if nargin<3,error('at least 3 input arguments required'),end if ~(b>a),error('upper bound must be greater than lower'),end if nargin<4|isempty(n),n = 100;end x = a; h = (b − a)/n; s=func(a,varargin{:}); for i = 1 : n − 1 x = x + h; s = s + 2*func(x,varargin{:}); end s = s + func(b,varargin{:}); I = (b − a) * s/(2*n);

FIGURE 19.10 M-file to implement the composite trapezoidal rule.

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19.4 sImPson’s RUles 501

As would be expected, this result has a relatively high true error of 18.6%. To obtain a more accurate result, we can use a very fine approximation based on 10,000 segments:

>> trap(v,0,3,10000)

x = 41.94804999917528

which is very close to the true value.

19.4 SIMPSON’S RULES

Aside from applying the trapezoidal rule with finer segmentation, another way to obtain a more accurate estimate of an integral is to use higher-order polynomials to connect the points. For example, if there is an extra point midway between f (a) and f (b), the three points can be connected with a parabola (Fig. 19.11a). If there are two points equally spaced between f (a) and f (b), the four points can be connected with a third-order poly- nomial (Fig. 19.11b). The formulas that result from taking the integrals under these polynomials are called Simpson’s rules.

19.4.1 Simpson’s 1∕3 Rule

Simpson’s 1∕3 rule corresponds to the case where the polynomial in Eq. (19.8) is second- order:

I = ∫ x0 x2

[ (x − x1)(x − x2) ______________ (x0 − x1)(x0 − x2)

f (x0) + (x − x0)(x − x2) ______________ (x1 − x0)(x1 − x2)

f (x1)

+ (x − x0)(x − x1) ______________ (x2 − x0)(x2 − x1)

f (x2) ] dx

f (x)

(a) x

f (x)

(b) x

FIGURE 19.11 (a) Graphical depiction of Simpson’s 1∕3 rule: It consists of taking the area under a parabola connecting three points. (b) Graphical depiction of Simpson’s 3∕8 rule: It consists of taking the area under a cubic equation connecting four points.

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where a and b are designated as x0 and x2, respectively. The result of the integration is

I = h __ 3 [ f (x0) + 4 f (x1) + f (x2)] (19.22)

where, for this case, h = (b − a)∕2. This equation is known as Simpson’s 1∕3 rule. The label “1∕3” stems from the fact that h is divided by 3 in Eq. (19.22). Simpson’s 1∕3 rule can also be expressed using the format of Eq. (19.13):

I = (b − a) f (x0) + 4 f (x1) + f (x2) __________________ 6 (19.23)

where a = x0, b = x2, and x1 = the point midway between a and b, which is given by (a + b)∕2. Notice that, according to Eq. (19.23), the middle point is weighted by two- thirds and the two end points by one-sixth.

It can be shown that a single-segment application of Simpson’s 1∕3 rule has a trunca- tion error of

Et = − 1 ___ 90 h

5 f (4)(ξ )

or, because h = (b − a)∕2:

Et = − (b − a)5 _______ 2880 f

(4)(ξ ) (19.24)

where ξ lies somewhere in the interval from a to b. Thus, Simpson’s 1∕3 rule is more ac- curate than the trapezoidal rule. However, comparison with Eq. (19.14) indicates that it is more accurate than expected. Rather than being proportional to the third derivative, the error is proportional to the fourth derivative. Consequently, Simpson’s 1∕3 rule is third- order accurate even though it is based on only three points. In other words, it yields exact results for cubic polynomials even though it is derived from a parabola!

eXAmPle 19.3 single Application of simpson’s 1∕3 Rule

Problem statement. Use Eq. (19.23) to integrate

f (x) = 0.2 + 25x − 200x2 + 675x3 − 900x4 + 400x5

from a = 0 to b = 0.8. Employ Eq. (19.24) to estimate the error. Recall that the exact inte- gral is 1.640533.

solution. n = 2(h = 0.4):

f (0) = 0.2 f (0.4) = 2.456 f (0.8) = 0.232

I = 0.8 0.2 + 4(2.456) + 0.232 ___________________ 6 = 1.367467

Et = 1.640533 − 1.367467 = 0.2730667 εt = 16.6%

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19.4 sImPson’s RUles 503

which is approximately five times more accurate than for a single application of the trap- ezoidal rule (Example 19.1). The approximate error can be estimated as

Ea = − 0.85 _____ 2880 (−2400) = 0.2730667

where −2400 is the average fourth derivative for the interval. As was the case in Example 19.1, the error is approximate (Ea) because the average fourth derivative is generally not an exact estimate of f (4)(ξ ). However, because this case deals with a fifth- order polynomial, the result matches exactly.

19.4.2 The Composite Simpson’s 1∕3 Rule

Just as with the trapezoidal rule, Simpson’s rule can be improved by dividing the integra- tion interval into a number of segments of equal width (Fig. 19.12). The total integral can be represented as

I = ∫ x0 x2

f (x) dx + ∫ x2 x4

f (x) dx + · · · + ∫ xn−2

f (x) dx (19.25)

f (x)

xa b

1 4 1

1 4 11 4 1

1 44

1 4 1

2 42

1 4 1

FIGURE 19.12 Composite Simpson’s 1∕3 rule. The relative weights are depicted above the function values. Note that the method can be employed only if the number of segments is even.

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Substituting Simpson’s 1∕3 rule for each integral yields

I = 2h f (x0) + 4 f (x1) + f (x2) __________________ 6 + 2h

f (x2) + 4 f (x3) + f (x4) __________________ 6

+ · · · + 2h f (xn−2) + 4 f (xn−1) + f (xn) _____________________ 6

or, grouping terms and using Eq. (19.15):

I = (b − a) f (x0) + 4 ∑

i=1,3,5

n−1 f (xi ) + 2 ∑

j=2,4,6

n−2 f (xj ) + f (xn )

____________________________________ 3n (19.26)

Notice that, as illustrated in Fig. 19.12, an even number of segments must be utilized to implement the method. In addition, the coefficients “4” and “2” in Eq. (19.26) might seem peculiar at first glance. However, they follow naturally from Simpson’s 1∕3 rule. As illustrated in Fig. 19.12, the odd points represent the middle term for each application and hence carry the weight of four from Eq. (19.23). The even points are common to adjacent applications and hence are counted twice.

An error estimate for the composite Simpson’s rule is obtained in the same fashion as for the trapezoidal rule by summing the individual errors for the segments and averaging the derivative to yield

Ea = − (b − a)5 _______ 180n4

_ f (4) (19.27)

where f (4) is the average fourth derivative for the interval.

eXAmPle 19.4 Composite simpson’s 1∕3 Rule

Problem statement. Use Eq. (19.26) with n = 4 to estimate the integral of

f (x) = 0.2 + 25x − 200x2 + 675x3 − 900x4 + 400x5

from a = 0 to b = 0.8. Employ Eq. (19.27) to estimate the error. Recall that the exact integral is 1.640533.

solution. n = 4(h = 0.2):

f (0) = 0.2 f (0.2) = 1.288

f (0.4) = 2.456 f (0.6) = 3.464

f (0.8) = 0.232

From Eq. (19.26):

I = 0.8 0.2 + 4(1.288 + 3.464) + 2(2.456) + 0.232 ___________________________________ 12 = 1.623467

Et = 1.640533 − 1.623467 = 0.017067 εt = 1.04%

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19.4 sImPson’s RUles 505

The estimated error [Eq. (19.27)] is

Ea = − (0.8)5 _______

180(4)4 (−2400) = 0.017067

which is exact (as was also the case for Example 19.3).

As in Example 19.4, the composite version of Simpson’s 1∕3 rule is considered supe- rior to the trapezoidal rule for most applications. However, as mentioned previously, it is limited to cases where the values are equispaced. Further, it is limited to situations where there are an even number of segments and an odd number of points. Consequently, as dis- cussed in Sec. 19.4.3, an odd-segment–even-point formula known as Simpson’s 3∕8 rule can be used in conjunction with the 1∕3 rule to permit evaluation of both even and odd numbers of equispaced segments.

19.4.3 Simpson’s 3∕8 Rule

In a similar manner to the derivation of the trapezoidal and Simpson’s 1∕3 rule, a third- order Lagrange polynomial can be fit to four points and integrated to yield

I = 3h ___ 8 [ f (x0) + 3f (x1) + 3 f (x2) + f (x3)]

where h = (b − a)∕3. This equation is known as Simpsons 3∕8 rule because h is multiplied by 3∕8. It is the third Newton-Cotes closed integration formula. The 3∕8 rule can also be expressed in the form of Eq. (19.13):

I = (b − a) f (x0) + 3 f (x1) + 3 f (x2) + f (x3) _________________________ 8 (19.28)

Thus, the two interior points are given weights of three-eighths, whereas the end points are weighted with one-eighth. Simpson’s 3∕8 rule has an error of

Et = − 3 ___ 80 h

5 f (4)(ξ )

or, because h = (b − a)∕3:

Et = − (b − a)5 _______ 6480 f

(4)(ξ ) (19.29)

Because the denominator of Eq. (19.29) is larger than for Eq. (19.24), the 3∕8 rule is some- what more accurate than the 1∕3 rule.

Simpson’s 1∕3 rule is usually the method of preference because it attains third-order accuracy with three points rather than the four points required for the 3∕8 version. How- ever, the 3∕8 rule has utility when the number of segments is odd. For instance, in Example 19.4 we used Simpson’s rule to integrate the function for four segments. Suppose that you desired an estimate for five segments. One option would be to use a composite version of the trapezoidal rule as was done in Example 19.2. This may not be advisable, however, because of the large truncation error associated with this method. An alternative would be to apply Simpson’s 1∕3 rule to the first two segments and Simpson’s 3∕8 rule to the last

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three (Fig. 19.13). In this way, we could obtain an estimate with third-order accuracy across the entire interval.

eXAmPle 19.5 simpson’s 3∕8 Rule

Problem statement. (a) Use Simpson’s 3∕8 rule to integrate

f (x) = 0.2 + 25x − 200x2 + 675x3 − 900x4 + 400x5

from a = 0 to b = 0.8. (b) Use it in conjunction with Simpson’s 1∕3 rule to integrate the same function for five segments.

solution. (a) A single application of Simpson’s 3∕8 rule requires four equally spaced points:

f (0) = 0.2 f (0.2667) = 1.432724

f (0.5333) = 3.487177 f (0.8) = 0.232

Using Eq. (19.28):

I = 0.8 0.2 + 3(1.432724 + 3.487177) + 0.232 ___________________________________ 8 = 1.51917

f (x)

x0.80.60.40.2

3/8 rule1/3 rule

FIGURE 19.13 Illustration of how Simpson’s 1∕3 and 3∕8 rules can be applied in tandem to handle multiple applications with odd numbers of intervals.

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19.5 HIgHeR-oRDeR neWTon-CoTes foRmUlAs 507

(b) The data needed for a five-segment application (h = 0.16) are

f (0) = 0.2 f (0.16) = 1.296919

f (0.32) = 1.743393 f (0.48) = 3.186015

f (0.64) = 3.181929 f (0.80) = 0.232

The integral for the first two segments is obtained using Simpson’s 1∕3 rule:

I = 0.32 0.2 + 4(1.296919) + 1.743393 _________________________ 6 = 0.3803237

For the last three segments, the 3∕8 rule can be used to obtain

I = 0.48 1.743393 + 3(3.186015 + 3.181929) + 0.232 _____________________________________ 8 = 1.264754

The total integral is computed by summing the two results:

I = 0.3803237 + 1.264754 = 1.645077

19.5 HIGHER-ORDER NEWTON-COTES FORMULAS

As noted previously, the trapezoidal rule and both of Simpson’s rules are members of a family of integrating equations known as the Newton-Cotes closed integration formu- las. Some of the formulas are summarized in Table 19.2 along with their truncation-error estimates.

Notice that, as was the case with Simpson’s 1∕3 and 3∕8 rules, the five- and six-point formulas have the same order error. This general characteristic holds for the higher-point formulas and leads to the result that the even-segment–odd-point formulas (e.g., 1∕3 rule and Boole’s rule) are usually the methods of preference.

TABLE 19.2 newton-Cotes closed integration formulas. The formulas are presented in the format of eq. (19.13) so that the weighting of the data points to estimate the average height is apparent. The step size is given by h = (b − a)/n.

Segments (n) Points Name Formula Truncation Error

1 2 Trapezoidal rule (b − a) f (x0) + f (x1) _________________ 2 −(1∕12) h

3 f ″(ξ)

2 3 Simpson’s 1/3 rule (b − a) f (x0) + 4 f (x1) + f (x2) ________________________ 6 − (1∕90) h

5 f (4)(ξ)

3 4 Simpson’s 3/8 rule (b − a) f (x0) + 3 f (x1) + 3 f (x2) + f (x3) _________________________ 8 −(3∕80) h

5 f (4)(ξ)

4 5 Boole’s rule (b − a) 7 f (x0) + 32 f (x1) + 12 f (x2) + 32 f (x3) + 7 f (x4) _____________________________________________ 90 −(8∕945) h

7 f (6)(ξ)

5 6 (b − a) 19 f (x0) + 75 f (x1) + 50 f (x2) + 50 f (x3) + 75 f (x4) + 19 f (x5) __________________________________________________ 288 −(275∕12,096) h

7 f (6)(ξ)

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However, it must also be stressed that, in engineering and science practice, the higher- order (i.e., greater than four-point) formulas are not commonly used. Simpson’s rules are sufficient for most applications. Accuracy can be improved by using the composite ver- sion. Furthermore, when the function is known and high accuracy is required, methods such as Romberg integration or Gauss quadrature, described in Chap. 20, offer viable and attractive alternatives.

19.6 INTEGRATION WITH UNEQUAL SEGMENTS

To this point, all formulas for numerical integration have been based on equispaced data points. In practice, there are many situations where this assumption does not hold and we must deal with unequal-sized segments. For example, experimentally derived data are often of this type. For these cases, one method is to apply the trapezoidal rule to each seg- ment and sum the results:

I = h1 f (x0) + f (x1) ___________ 2 + h2

f (x1) + f (x2) __________ 2 + · · · + hn f (xn−1) + f (xn) ____________ 2 (19.30)

where hi = the width of segment i. Note that this was the same approach used for the com- posite trapezoidal rule. The only difference between Eqs. (19.16) and (19.30) is that the h’s in the former are constant.

eXAmPle 19.6 Trapezoidal Rule with Unequal segments

Problem statement. The information in Table 19.3 was generated using the same poly- nomial employed in Example 19.1. Use Eq. (19.30) to determine the integral for these data. Recall that the correct answer is 1.640533.

solution. Applying Eq. (19.30) yields

I = 0.12 0.2 + 1.309729 _____________ 2 + 0.10 1.309729 + 1.305241 __________________ 2

+ · · · + 0.10 2.363 + 0.232 ____________ 2 = 1.594801

which represents an absolute percent relative error of εt = 2.8%.

TABLE 19.3 Data for f (x) = 0.2 + 25x − 200x2 + 675x3 − 900x4 + 400x5, with unequally spaced values of x.

x f (x) x f (x)

0.00 0.200000 0.44 2.842985 0.12 1.309729 0.54 3.507297 0.22 1.305241 0.64 3.181929 0.32 1.743393 0.70 2.363000 0.36 2.074903 0.80 0.232000 0.40 2.456000

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19.6 InTegRATIon WITH UneQUAl segmenTs 509

19.6.1 MATLAB M-file: trapuneq

A simple algorithm to implement the trapezoidal rule for unequally spaced data can be written as in Fig. 19.14. Two vectors, x and y, holding the independent and dependent variables are passed into the M-file. Two error traps are included to ensure that (a) the two vectors are of the same length and (b) the x’s are in ascending order.1 A loop is employed to generate the integral. Notice that we have modified the subscripts from those of Eq. (19.30) to account for the fact that MATLAB does not allow zero subscripts in arrays.

An application of the M-file can be developed for the same problem that was solved in Example 19.6:

>> x = [0 .12 .22 .32 .36 .4 .44 .54 .64 .7 .8]; >> y = 0.2 + 25*x − 200*x.^2 + 675*x.^3 − 900*x.^4 + 400*x.^5; >> trapuneq(x,y)

ans =

1.5948

which is identical to the result obtained in Example 19.6.

function I = trapuneq(x,y) % trapuneq: unequal spaced trapezoidal rule quadrature % I = trapuneq(x,y): % Applies the trapezoidal rule to determine the integral % for n data points (x, y) where x and y must be of the % same length and x must be monotonically ascending % input: % x = vector of independent variables % y = vector of dependent variables % output: % I = integral estimate

if nargin<2,error('at least 2 input arguments required'),end if any(diff(x)<0),error('x not monotonically ascending'),end n = length(x); if length(y)~ = n,error('x and y must be same length'); end s = 0; for k = 1:n − 1 s = s + (x(k+l) − x(k))*(y(k)+ y(k +l))/2; end I = s;

FIGURE 19.14 M-file to implement the trapezoidal rule for unequally spaced data.

1 The diff function is described in Sec. 21.7.1.

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19.6.2 MATLAB Functions: trapz and cumtrapz

MATLAB has a built-in function that evaluates integrals for data in the same fashion as the M-file we just presented in Fig. 19.14. It has the general syntax

z = trapz(x, y)

where the two vectors, x and y, hold the independent and dependent variables, respec- tively. Here is a simple MATLAB session that uses this function to integrate the data from Table 19.3:

>> x = [0 .12 .22 .32 .36 .4 .44 .54 .64 .7 .8]; >> y = 0.2 + 25*x − 200*x.^2 + 675*x.^3 − 900*x.^4 + 400*x.^5; >> trapz(x,y)

ans =

1.5948

In addition, MATLAB has another function, cumtrapz, that computes the cumulative integral. A simple representation of its syntax is

z = cumtrapz(x, y)

where the two vectors, x and y, hold the independent and dependent variables, respectively, and z = a vector whose elements z(k) hold the integral from x(1) to x(k).

eXAmPle 19.7 Using numerical Integration to Compute Distance from Velocity

Problem statement. As described at the beginning of this chapter, a nice application of integration is to compute the distance z (t) of an object based on its velocity υ (t) as in [recall Eq. (19.2)]:

z(t) = ∫ 0 t

υ(t) dt

Suppose that we had measurements of velocity at a series of discrete unequally spaced times during free fall. Use Eq. (19.2) to synthetically generate such information for a 70-kg jumper with a drag coefficient of 0.275 kg/m. Incorporate some random error by rounding the velocities to the nearest integer. Then use cumtrapz to determine the distance fallen and compare the results to the analytical solution [Eq. (19.4)]. In addi- tion, develop a plot of the analytical and computed distances along with velocity on the same graph.

solution. Some unequally spaced times and rounded velocities can be generated as >> format short g >> t = [0 1 1.4 2 3 4.3 6 6.7 8]; >> g = 9.81;m=70;cd = 0.275; >> v = round(sqrt(g*m/cd)*tanh(sqrt(g*cd/m)*t));

The distances can then be computed as >> z = cumtrapz(t,v)

z = 0 5 9.6 19.2 41.7 80.7 144.45 173.85 231.7

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19.6 InTegRATIon WITH UneQUAl segmenTs 511

Thus, after 8 seconds, the jumper has fallen 231.7 m. This result is reasonably close to the analytical solution [Eq. (19.4)]:

z (t) = 70 _____ 0.275 ln [cosh ( √ __________

9.81(0.275) __________ 70 8 ) ] = 234.1 A graph of the numerical and analytical solutions along with both the exact and rounded velocities can be generated with the following commands:

>> ta = linspace(t(1),t(length(t))); >> za = m/cd*log(cosh(sqrt(g*cd/m)*ta)); >> plot(ta,za,t,z,'o') >> title('Distance versus time') >> xlabel('t (s)'),ylabel('x (m)') >> legend('analytical','numerical')

As in Fig. 19.15, the numerical and analytical results match fairly well.

250 Distance versus time

numerical analytical

200

150

100

0 0 1 2 3 4

t (s)

x (m

)

5 6 7 8

FIGURE 19.15 Plot of distance versus time. The line was computed with the analytical solution, whereas the points were determined numerically with the cumtrapz function.

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512 nUmeRICAl InTegRATIon foRmUlAs

19.7 OPEN METHODS

Recall from Fig. 19.6b that open integration formulas have limits that extend beyond the range of the data. Table 19.4 summarizes the Newton-Cotes open integration formulas. The formulas are expressed in the form of Eq. (19.13) so that the weighting factors are evident. As with the closed versions, successive pairs of the formulas have the same-order error. The even-segment–odd-point formulas are usually the methods of preference because they require fewer points to attain the same accuracy as the odd-segment–even-point formulas.

The open formulas are not often used for definite integration. However, they have util- ity for analyzing improper integrals. In addition, they will have relevance to our discussion of methods for solving ordinary differential equations in Chaps. 22 and 23.

19.8 MULTIPLE INTEGRALS

Multiple integrals are widely used in engineering and science. For example, a general equation to compute the average of a two-dimensional function can be written as [recall Eq. (19.7)]

_ f =

∫ c d ( ∫ a b f (x, y) dx ) _______________ (d − c)(b − a) (19.31)

The numerator is called a double integral. The techniques discussed in this chapter (and Chap. 20) can be readily employed to

evaluate multiple integrals. A simple example would be to take the double integral of a function over a rectangular area (Fig. 19.16).

Recall from calculus that such integrals can be computed as iterated integrals:

∫ c

( ∫ a b

f (x, y) dx ) dy = ∫ a b

( ∫ c

f (x, y) dy ) dx (19.32) Thus, the integral in one of the dimensions is evaluated first. The result of this first integra- tion is integrated in the second dimension. Equation (19.32) states that the order of integra- tion is not important.

TABLE 19.4 newton-Cotes open integration formulas. The formulas are presented in the format of eq. (19.13) so that the weighting of the data points to estimate the average height is apparent. The step size is given by h = (b − a)∕n.

Segments (n) Points Name Formula Truncation Error

2 1 Midpoint method (b − a) f (x1) (1∕3) h3 f ″(ξ)

3 2 (b − a) f (x1) + f (x2) _________________ 2 (3∕4) h

3 f ″(ξ)

4 3 (b − a) 2 f (x1) − f (x2) + 2 f (x3) _________________________ 3 (14∕45) h

5 f (4)(ξ)

5 4 (b − a) 11 f (x1) + f (x2) + f (x3) + 11 f (x4) _____________________________________________ 24 (95∕144) h

5 f (4)(ξ)

6 5 (b − a) 11 f (x1) − 14 f (x2) + 26 f (x3) − 14 f (x4) + 11 f (x5) __________________________________________________ 20 (41∕140) h

7 f (6)(ξ)

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19.8 mUlTIPle InTegRAls 513

A numerical double integral would be based on the same idea. First, methods such as the composite trapezoidal or Simpson’s rule would be applied in the first dimension with each value of the second dimension held constant. Then the method would be applied to integrate the second dimension. The approach is illustrated in the following example.

eXAmPle 19.8 Using Double Integral to Determine Average Temperature

Problem statement. Suppose that the temperature of a rectangular heated plate is de- scribed by the following function:

T (x, y) = 2xy + 2x − x2 − 2y2 + 72

If the plate is 8 m long (x dimension) and 6 m wide (y dimension), compute the average temperature.

solution. First, let us merely use two-segment applications of the trapezoidal rule in each dimension. The temperatures at the necessary x and y values are depicted in Fig. 19.17. Note that a simple average of these values is 47.33. The function can also be evaluated analytically to yield a result of 58.66667. To make the same evaluation numerically, the trapezoidal rule is first implemented along the x dimension for each y value. These values are then integrated along the y dimen- sion to give the final result of 2544. Dividing this by the area yields the average tempera- ture as 2544∕(6 × 8) = 53. Now we can apply a single-segment Simpson’s 1∕3 rule in the same fashion. This results in an integral of 2816 and an average of 58.66667, which is exact. Why does this occur? Recall that Simpson’s 1∕3 rule yielded perfect results for cubic polynomials. Since the highest-order term in the function is second order, the same exact result occurs for the present case.

f (x, y)

FIGURE 19.16 Double integral as the area under the function surface.

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For higher-order algebraic functions as well as transcendental functions, it would be necessary to use composite applications to attain accurate integral estimates. In addition, Chap. 20 introduces techniques that are more efficient than the Newton-Cotes formulas for evaluating integrals of given functions. These often provide a superior means to implement the numerical integrations for multiple integrals.

19.8.1 MATLAB Functions: integral2 and integral3

MATLAB has functions to implement both double (integral2) and triple (integral3) inte- gration. A simple representation of the syntax for integral2 is

q = integral2(fun, xmin, xmax, ymin, ymax)

where q is the double integral of the function fun over the ranges from xmin to xmax and ymin to ymax.

Here is an example of how this function can be used to compute the double integral evaluated in Example 19.7:

>> q = integral2(@(x,y) 2*x*y + 2*x − x.^2 − 2*y.^2 + 72,0,8,0,6)

q = 2816

(8 − 0) 0 + 2(40) + 48 4

(8 − 0) 54 + 2(70) + 54 4

(8 − 0) 72 + 2(64) + 24 4

(6 − 0) = 2544256 + 2(496) + 448 4

256

448

496

FIGURE 19.17 Numerical evaluation of a double integral using the two-segment trapezoidal rule.

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19.9 CAse sTUDY 515

19.9 CASE STUDY ComPUTIng WoRK WITH nUmeRICAl InTegRATIon

Background. The calculation of work is an important component of many areas of engi- neering and science. The general formula is

Work = force × distance

When you were introduced to this concept in high school physics, simple applications were presented using forces that remained constant throughout the displacement. For example, if a force of 10 N was used to pull a block a distance of 5 m, the work would be calculated as 50 J (1 joule = 1 N · m).

Although such a simple computation is useful for introducing the concept, realistic problem settings are usually more complex. For example, suppose that the force varies dur- ing the course of the calculation. In such cases, the work equation is reexpressed as

W = ∫ x0 xn

F (x) dx (19.33)

where W = work (J), x0 and xn = the initial and final positions (m), respectively, and F(x) = a force that varies as a function of position (N). If F(x) is easy to integrate, Eq. (19.33) can be evaluated analytically. However, in a realistic problem setting, the force might not be expressed in such a manner. In fact, when analyzing measured data, the force might be available only in tabular form. For such cases, numerical integration is the only viable op- tion for the evaluation.

Further complexity is introduced if the angle between the force and the direction of movement also varies as a function of position (Fig. 19.18). The work equation can be modified further to account for this effect, as in

W = ∫ x0 xn

F (x) cos[θ(x)] dx (19.34)

Again, if F(x) and θ(x) are simple functions, Eq. (19.34) might be solved analytically. How- ever, as in Fig. 19.18, it is more likely that the functional relationship is complicated. For this situation, numerical methods provide the only alternative for determining the integral.

Suppose that you have to perform the computation for the situation depicted in Fig. 19.18. Although the figure shows the continuous values for F(x) and θ(x), assume that, because of experimental constraints, you are provided with only discrete measurements at x = 5-m intervals (Table 19.5). Use single- and composite versions of the trapezoidal rule and Simpson’s 1∕3 and 3∕8 rules to compute work for these data.

Solution. The results of the analysis are summarized in Table 19.6. A percent relative error εt was computed in reference to a true value of the integral of 129.52 that was esti- mated on the basis of values taken from Fig. 19.18 at 1-m intervals.

The results are interesting because the most accurate outcome occurs for the simple two-segment trapezoidal rule. More refined estimates using more segments, as well as Simpson’s rules, yield less accurate results.

The reason for this apparently counterintuitive result is that the coarse spacing of the points is not adequate to capture the variations of the forces and angles. This is particularly

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19.9 CASE STUDY continued

0 10 20 30

x, m

θ( x)

, r ad

F(x) F(x)

xnx0

F( x)

, N

FIGURE 19.18 The case of a variable force acting on a block. For this case the angle, as well as the magnitude, of the force varies.

TABLE 19.5 Data for force F(x) and angle θ (x) as a function of position x.

x, m F(x), N ϴ, rad F(x) cos ϴ

0 0.0 0.50 0.0000 5 9.0 1.40 1.5297 10 13.0 0.75 9.5120 15 14.0 0.90 8.7025 20 10.5 1.30 2.8087 25 12.0 1.48 1.0881 30 5.0 1.50 0.3537

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19.9 CAse sTUDY 517

19.9 CASE STUDY continued

evident in Fig. 19.19, where we have plotted the continuous curve for the product of F(x) and cos [θ(x)]. Notice how the use of seven points to characterize the continuously varying function misses the two peaks at x = 2.5 and 12.5 m. The omission of these two points ef- fectively limits the accuracy of the numerical integration estimates in Table 19.6. The fact that the two-segment trapezoidal rule yields the most accurate result is due to the chance positioning of the points for this particular problem (Fig. 19.20).

The conclusion to be drawn from Fig. 19.20 is that an adequate number of measure- ments must be made to accurately compute integrals. For the present case, if data were available at F(2.5) cos [θ (2.5)] = 3.9007 and F(12.5) cos [θ (12.5)] = 11.3940, we could

TABLE 19.6 estimates of work calculated using the trapezoidal rule and simpson’s rules. The percent relative error εt as computed in reference to a true value of the integral (129.52 Pa) that was estimated on the basis of values at 1-m intervals.

Technique Segments Work εt, %

Trapezoidal rule 1 5.31 95.9 2 133.19 2.84 3 124.98 3.51 6 119.09 8.05 Simpson’s 1∕3 rule 2 175.82 35.75 6 117.13 9.57 Simpson’s 3∕8 rule 3 139.93 8.04

x, ft 0 30

F (x

) c os

[θ (x

)]

Work

FIGURE 19.19 A continuous plot of F(x) cos [θ (x)] versus position with the seven discrete points used to de- velop the numerical integration estimates in Table 19.6. Notice how the use of seven points to characterize this continuously varying function misses two peaks at x = 2.5 and 12.5 m.

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19.9 CASE STUDY continued

determine an improved integral estimate. For example, using the MATLAB trapz function, we could compute

>> x = [0 2.5 5 10 12.5 15 20 25 30]; >> y = [0 3.9007 1.5297 9.5120 11.3940 8.7025 2.8087 ... 1.0881 0.3537]; >> trapz(x,y)

ans = 132.6458

Including the two additional points yields an improved integral estimate of 132.6458 (εt = 2.16%). Thus, the inclusion of the additional data incorporates the peaks that were missed previously and, as a consequence, lead to better results.

x, ft

0 0

F (x

) c os

[θ (x

)]

Overestimates

Underestimates

FIGURE 19.20 Graphical depiction of why the two-segment trapezoidal rule yields a good estimate of the integral for this particular case. By chance, the use of two trapezoids happens to lead to an even balance between positive and negative errors.

PROBLEMS

19.1 Derive Eq. (19.4) by integrating Eq. (19.3). 19.2 Evaluate the following integral:

∫ 0 4

(1 − e−x ) dx

(a) analytically, (b) single application of the trapezoidal rule, (c) composite trapezoidal rule with n = 2 and 4, (d) single application of Simpson’s 1∕3 rule, (e) composite Simpson’s

1∕3 rule with n = 4, (f) Simpson’s 3∕8 rule, and (g) com- posite Simpson’s rule, with n = 5. For each of the numerical estimates (b) through (g), determine the true percent relative error based on (a). 19.3 Evaluate the following integral:

∫ 0 π∕2

(8 + 4 cos x) dx

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PRoBlems 519

(a) analytically, (b) single application of the trapezoidal rule, (c) composite trapezoidal rule with n = 2 and 4, (d) single application of Simpson’s 1∕3 rule, (e) composite Simpson’s 1∕3 rule with n = 4, (f ) Simpson’s 3∕8 rule, and (g) com- posite Simpson’s rule, with n = 5. For each of the numerical estimates (b) through (g), determine the true percent relative error based on (a). 19.4 Evaluate the following integral:

∫ −2

(1 − x − 4x3 + 2x5 ) dx

(a) analytically, (b) single application of the trapezoidal rule, (c) composite trapezoidal rule with n = 2 and 4, (d) single ap- plication of Simpson’s 1∕3 rule, (e) Simpson’s 3∕8 rule, and (f) Boole’s rule. For each of the numerical estimates (b) through (f), determine the true percent relative error based on (a). 19.5 The function

f (x) = e−x

can be used to generate the following table of unequally spaced data:

x 0 0.1 0.3 0.5 0.7 0.95 1.2 f (x) 1 0.9048 0.7408 0.6065 0.4966 0.3867 0.3012

Evaluate the integral from a = 0 to b = 1.2 using (a) analyti- cal means, (b) the trapezoidal rule, and (c) a combination of the trapezoidal and Simpson’s rules wherever possible to at- tain the highest accuracy. For (b) and (c), compute the true percent relative error. 19.6 Evaluate the double integral

∫ −2

∫ 0 4

(x2 − 3y2 + xy3 ) dx dy

(a) analytically, (b) using the composite trapezoidal rule with n = 2, (c) using single applications of Simpson’s 1∕3 rule, and (d) using the integral2 function. For (b) and (c), compute the percent relative error.

19.7 Evaluate the triple integral

∫ −4

∫ 0 6

∫ −1

(x3 − 2yz ) dx dy dz

(a) analytically, (b) using single applications of Simpson’s 1∕3 rule, and (c) the integral3 function. For (b), compute the true percent relative error. 19.8 Determine the distance traveled from the following velocity data:

t 1 2 3.25 4.5 6 7 8 8.5 9 10 υ 5 6 5.5 7 8.5 8 6 7 7 5

(a) Use the trapezoidal rule. In addition, determine the aver- age velocity.

(b) Fit the data with a cubic equation using polynomial regression. Integrate the cubic equation to determine the distance.

19.9 Water exerts pressure on the upstream face of a dam as shown in Fig. P19.9. The pressure can be characterized by

p(z) = ρg(D − z) where p(z) = pressure in pascals (or N/m2) exerted at an elevation z meters above the reservoir bottom; ρ = density of water, which for this problem is assumed to be a constant 103 kg/m3; g = acceleration due to gravity (9.81 m/s2); and D = elevation (in m) of the water surface above the reser- voir bottom. According to Eq. (P19.9), pressure increases linearly with depth, as depicted in Fig. P19.9a. Omitting at- mospheric pressure (because it works against both sides of the dam face and essentially cancels out), the total force ft can be determined by multiplying pressure times the area of the dam face (as shown in Fig. P19.9b). Because both pres- sure and area vary with elevation, the total force is obtained by evaluating

ft = ∫ 0 D

ρg𝑤(z)(D − z) dz

(a) (b)

122 130 135 160 175 190

200

FIGURE P19.9 Water exerting pressure on the upstream face of a dam: (a) side view showing force increasing linearly with depth; (b) front view showing width of dam in meters.

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520 nUmeRICAl InTegRATIon foRmUlAs

where 𝑤(z) = width of the dam face (m) at elevation z (Fig. P19.9b). The line of action can also be obtained by evaluating

d = ∫ 0

D ρgz𝑤(z)(D − z) dz __________________

∫ 0 D ρg𝑤(z)(D − z) dz

Use Simpson’s rule to compute ft and d. 19.10 The force on a sailboat mast can be represented by the following function:

f (z) = 200 ( z _____ 5 + z ) e−2z∕H where z = the elevation above the deck and H = the height of the mast. The total force F exerted on the mast can be deter- mined by integrating this function over the height of the mast:

F = ∫ 0 H

f (z) dz

The line of action can also be determined by integration:

d = ∫ 0

H z f (z) dz __________________

∫ 0 H f (z) dz

(a) Use the composite trapezoidal rule to compute F and d for the case where H = 30 (n = 6).

(b) Repeat (a), but use the composite Simpson’s 1∕3 rule. 19.11 A wind force distributed against the side of a sky- scraper is measured as

Height l, m 0 30 60 90 120 Force, F(l ), N/m 0 340 1200 1550 2700

Height l, m 150 180 210 240 Force, F(l ), N/m 3100 3200 3500 3750

Compute the net force and the line of action due to this dis- tributed wind. 19.12 An 11-m beam is subjected to a load, and the shear force follows the equation

V(x) = 5 + 0.25x2

where V is the shear force, and x is length in distance along the beam. We know that V = dM∕dx, and M is the bending moment. Integration yields the relationship

M = Mo + ∫ 0 x

V d x

If Mo is zero and x = 11, calculate M using (a) analytical in- tegration, (b) composite trapezoidal rule, and (c) composite Simpson’s rules. For (b) and (c) use 1-m increments.

19.13 The total mass of a variable density rod is given by

m = ∫ 0 L

ρ(x) Ac (x) dx

where m = mass, ρ(x) = density, Ac(x) = cross-sectional area, x = distance along the rod, and L = the total length of the rod. The following data have been measured for a 20-m length rod. Determine the mass in grams to the best possible accuracy.

x, m 0 4 6 8 12 16 20 ρ, g/cm3 4.00 3.95 3.89 3.80 3.60 3.41 3.30 Ac, cm2 100 103 106 110 120 133 150

19.14 A transportation engineering study requires that you determine the number of cars that pass through an intersec- tion traveling during morning rush hour. You stand at the side of the road and count the number of cars that pass every 4 minutes at several times as tabulated below. Use the best numerical method to determine (a) the total number of cars that pass between 7:30 and 9:15 and (b) the rate of cars going through the intersection per minute. (Hint: Be careful with units.)

Time (hr) 7:30 7:45 8:00 8:15 8:45 9:15 Rate (cars per 4 min) 18 23 14 24 20 9

19.15 Determine the average value for the data in Fig. P19.15. Perform the integral needed for the average in the order shown by the following equation:

I = ∫ x0

[ ∫ y0 ym

f (x, y) dy] dx

19.16 Integration provides a means to compute how much mass enters or leaves a reactor over a specified time period, as in

M = ∫ t1

Qc dt

where t1 and t2 = the initial and final times, respectively. This formula makes intuitive sense if you recall the anal- ogy between integration and summation. Thus, the inte- gral represents the summation of the product of flow times concentration to give the total mass entering or leaving from t1 to t2. Use numerical integration to evaluate this equation for the data listed below:

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PRoBlems 521

t, min 0 10 20 30 35 40 45 50 Q, m3/min 4 4.8 5.2 5.0 4.6 4.3 4.3 5.0 c, mg/m3 10 35 55 52 40 37 32 34

19.17 The cross-sectional area of a channel can be com- puted as

Ac = ∫ 0 B

H (y) dy

where B = the total channel width (m), H = the depth (m), and y = distance from the bank (m). In a similar fashion, the average flow Q (m3/s) can be computed as

Q = ∫ 0 B

U (y) H (y) dy

where U = water velocity (m/s). Use these relationships and a numerical method to determine Ac and Q for the following data:

y, m 0 2 4 5 6 9 H, m 0.5 1.3 1.25 1.8 1 0.25 U, m/s 0.03 0.06 0.05 0.13 0.11 0.02

19.18 The average concentration of a substance _ c (g/m3) in a lake where the area As(m2) varies with depth z(m) can be computed by integration:

_ c =

∫ 0 Z c(z)As(z) dz __________________ ∫ 0

Z As(z) dz

where Z = the total depth (m). Determine the average con- centration based on the following data:

z, m 0 4 8 12 16 A, 106 m2 9.8175 5.1051 1.9635 0.3927 0.0000 c, g/m3 10.2 8.5 7.4 5.2 4.1

19.19 As was done in Sec. 19.9, determine the work per- formed if a constant force of 1 N applied at an angle θ results in the following displacements. Use the cumtrapz function to determine the cumulative work and plot the result versus θ.

x, m 0 1 2.8 3.9 3.8 3.2 1.3 ϴ, deg 0 30 60 90 120 150 180

19.20 Compute work as described in Sec. 19.9, but use the following equations for F(x) and θ(x):

F(x) = 1.6x − 0.045x2

θ(x) = −0.00055x3 + 0.0123x2 + 0.13x

The force is in Newtons and the angle is in radians. Perform the integration from x = 0 to 30 m. 19.21 As specified in the following table, a manufactured spherical particle has a density that varies as a function of the distance from its center (r = 0):

r, mm 0 0.12 0.24 0.36 0.49 ρ(g/cm3) 6 5.81 5.14 4.29 3.39

r, mm 0.62 0.79 0.86 0.93 1 ρ(g/cm3) 2.7 2.19 2.1 2.04 2

FIGURE P19.15

−8

−3

−1

6.5

−8 4

0 0 4 8 12

−4

−2

−6

1.5

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522 Numerical iNtegratioN Formulas

Use numerical integration to estimate the particle’s mass (in g) and average density (in g/cm3). 19.22 As specified in the following table, the earth’s density varies as a function of the distance from its center (r = 0):

r, km 0 1100 1500 2450 3400 3630 ρ(g/cm3) 13 12.4 12 11.2 9.7 5.7

r, km 4500 5380 6060 6280 6380 ρ(g/cm3) 5.2 4.7 3.6 3.4 3

Use numerical integration to estimate the earth’s mass (in metric tonnes) and average density (in g/cm3). Develop ver- tically stacked subplots of (top) density versus radius, and (bottom) mass versus radius. Assume that the earth is a per- fect sphere. 19.23 A spherical tank has a circular orifice in its bottom through which the liquid flows out (Fig. P19.23). The fol- lowing data is collected for the flow rate through the orifice as a function of time:

t, s 0 500 1000 1500 2200 2900 Q, m3/hr 10.55 9.576 9.072 8.640 8.100 7.560

t, s 3600 4300 5200 6500 7000 7500 Q, m3/hr 7.020 6.480 5.688 4.752 3.348 1.404

Write a script with supporting functions (a) to estimate the volume of fluid (in liters) drained over the entire measure- ment period and (b) to estimate the liquid level in the tank at t = 0 s. Note that r = 1.5 m.

FIGURE P19.23

0 50 100

F, kN/m

z, m

100

200

300

400

500

600

FIGURE P19.25

19.24 Develop an M-file function to implement the com- posite Simpson’s 1∕3 rule for equispaced data. Have the function print error messages and terminate (1) if the data is not equispaced or (2) if the input vectors hold- ing the data are not of equal length. If there are only 2 data points, implement the trapezoidal rule. If there are an even number of data points n (i.e., an odd number of segments, n – 1), use Simpson’s 3/8 rule for the final 3 segments. 19.25 During a storm a high wind blows along one side of a rectangular skyscraper as depicted in Fig. P19.25. As described in Prob. 19.9, use the best lower-order Newton- Cotes formulas (trapezoidal, Simpsons 1∕3 and 3∕8 rules) to determine (a) the force on the building in Newtons and (b) the line of force in meters. 19.26 The following data is provided for the velocity of an object as a function of time:

t, s 0 4 8 12 16 20 24 28 30 υ, m/s 0 18 31 42 50 56 61 65 70

(a) Limiting yourself to trapezoidal rule and Simpson’s 1∕3 and 3∕8 rules, make the best estimate of how far the object travels from t = 0 to 30 s?

(b) Employ the results of (a) to compute the average velocity.

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Problems 523

19.27 The total mass of a variable density rod is given by

m = ∫ 0 L

ρ(x) Ac (x) dx

where m = mass, ρ(x) = density, Ac(x) = cross-sectional area, and x = distance along the rod. The following data has been measured for a 10-m length rod:

x (m) 0 2 3 4 6 8 10

ρ (g/cm3) 4.00 3.95 3.89 3.80 3.60 3.41 3.30

Ac (cm2) 100 103 106 110 120 133 150

Determine the mass in grams to the best possible accuracy limiting yourself to trapezoidal rule and Simpson’s 1∕3 and 3∕8 rules. 19.28 A gas is expanded in an engine cylinder, following the law

PV1.3 = c

The initial pressure is 2550 kPa and the final pressure is 210 kPa. If the volume at the end of expansion is 0.75 m3, compute the work done by the gas. 19.29 The pressure p and volume υ of a given mass of gas are connected by the relation

( p + a/υ2)(υ – b)

where a, b, and k are constants. Express p in terms of υ, and write a script to compute the work done by the gas in expanding from an initial volume to a final volume. Test your solution with a = 0.01, b = 0.001, initial pressure and volume = 100 kPa and 1 m3, respectively, and final volume = 2 m3.

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524

20 Numerical Integration of Functions

20.1 INTRODUCTION

In Chap. 19, we noted that functions to be integrated numerically will typically be of two forms: a table of values or a function. The form of the data has an important influence on the approaches that can be used to evaluate the integral. For tabulated information, you are limited by the number of points that are given. In contrast, if the function is available, you can generate as many values of f (x) as are required to attain acceptable accuracy.

At face value, the composite Simpson’s 1∕3 rule might seem to be a reasonable tool for such problems. Although it is certainly adequate for many problems, there are more efficient methods that are available. This chapter is devoted to three such techniques, which capitalize on the ability to generate function values to develop efficient schemes for numerical integration.

The first technique is based on Richardson extrapolation, which is a method for combining two numerical integral estimates to obtain a third, more accurate value. The computational algorithm for implementing Richardson extrapolation in a highly efficient manner is called Romberg integration. This technique can be used to generate an integral estimate within a prespecified error tolerance.

CHAPTER OBJECTIVES

The primary objective of this chapter is to introduce you to numerical methods for integrating given functions. Specific objectives and topics covered are

• Understanding how Richardson extrapolation provides a means to create a more accurate integral estimate by combining two less accurate estimates.

• Understanding how Gauss quadrature provides superior integral estimates by picking optimal abscissas at which to evaluate the function.

• Knowing how to use MATLAB’s built-in function integral to integrate functions.

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20.2 ROMBERG INTEGRATION 525

The second method is called Gauss quadrature. Recall that, in Chap. 19, values of f (x) for the Newton-Cotes formulas were determined at specified values of x. For example, if we used the trapezoidal rule to determine an integral, we were constrained to take the weighted average of f (x) at the ends of the interval. Gauss-quadrature formulas employ x values that are positioned between the integration limits in such a manner that a much more accurate integral estimate results.

The third approach is called adaptive quadrature. This techniques applies composite Simpson’s 1∕3 rule to subintervals of the integration range in a way that allows error es- timates to be computed. These error estimates are then used to determine whether more refined estimates are required for a subinterval. In this way, more refined segmentation is only used where it is necessary. A built-in MATLAB function that uses adaptive quadra- ture is illustrated.

20.2 ROMBERG INTEGRATION

Romberg integration is one technique that is designed to attain efficient numerical integrals of functions. It is quite similar to the techniques discussed in Chap. 19 in the sense that it is based on successive application of the trapezoidal rule. However, through mathematical manipulations, superior results are attained for less effort.

20.2.1 Richardson Extrapolation

Techniques are available to improve the results of numerical integration on the basis of the integral estimates themselves. Generally called Richardson extrapolation, these methods use two estimates of an integral to compute a third, more accurate approximation.

The estimate and the error associated with the composite trapezoidal rule can be rep- resented generally as

I = I (h) + E(h)

where I = the exact value of the integral, I (h) = the approximation from an n-segment application of the trapezoidal rule with step size h = (b − a)∕n, and E (h) = the truncation error. If we make two separate estimates using step sizes of h1 and h2 and have exact values for the error:

I (h1) + E (h1) = I (h2) + E (h2) (20.1)

Now recall that the error of the composite trapezoidal rule can be represented approxi- mately by Eq. (19.21) [with n = (b − a)∕h]:

E ≅ − b − a _____ 12 h 2 f̄ ″ (20.2)

If it is assumed that f̄ ″ is constant regardless of step size, Eq. (20.2) can be used to deter- mine that the ratio of the two errors will be

E(h1) _____ E(h2)

≅ h 1

2 __

h 2 2 (20.3)

This calculation has the important effect of removing the term f̄ ″ from the computation. In so doing, we have made it possible to utilize the information embodied by Eq. (20.2)

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526 NuMERIcAl INTEGRATION OF FuNcTIONs

without prior knowledge of the function’s second derivative. To do this, we rearrange Eq. (20.3) to give

E (h1) ≅ E (h2) ( h1 __ h2 ) 2 which can be substituted into Eq. (20.1):

I (h1) + E (h2) ( h1 __ h2 ) 2 = I (h2) + E (h2) which can be solved for

E (h2) = I (h1) − I (h2) ___________ 1 − (h1∕h2)2

Thus, we have developed an estimate of the truncation error in terms of the integral esti- mates and their step sizes. This estimate can then be substituted into

I = I (h2) + E (h2)

to yield an improved estimate of the integral:

I = I (h2) + 1 __________

(h1∕h2)2 − 1 [ I (h2) − I (h1)] (20.4)

It can be shown (Ralston and Rabinowitz, 1978) that the error of this estimate is O(h4). Thus, we have combined two trapezoidal rule estimates of O(h2) to yield a new estimate of O(h4). For the special case where the interval is halved (h2 = h1∕2), this equation becomes

I = 4 __ 3 I (h2) − 1 __ 3 I (h1) (20.5)

EXAMPlE 20.1 Richardson Extrapolation

Problem statement. Use Richardson extrapolation to evaluate the integral of f (x) = 0.2 + 25x − 200x2 + 675x3 − 900x4 + 400x5 from a = 0 to b = 0.8.

solution. Single and composite applications of the trapezoidal rule can be used to evaluate the integral:

Segments h Integral εt

1 0.8 0.1728 89.5% 2 0.4 1.0688 34.9% 4 0.2 1.4848 9.5%

Richardson extrapolation can be used to combine these results to obtain improved estimates of the integral. For example, the estimates for one and two segments can be combined

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20.2 ROMBERG INTEGRATION 527

to yield

I = 4 __ 3 (1.0688) − 1 __ 3 (0.1728) = 1.367467

The error of the improved integral is Et = 1.640533 − 1.367467 = 0.273067(εt = 16.6%), which is superior to the estimates upon which it was based. In the same manner, the estimates for two and four segments can be combined to give

I = 4 __ 3 (1.4848) − 1 __ 3 (1.0688) = 1.623467

which represents an error of Et = 1.640533 − 1.623467 = 0.017067 (εt = 1.0%).

Equation (20.4) provides a way to combine two applications of the trapezoidal rule with error O(h2) to compute a third estimate with error O(h4). This approach is a subset of a more general method for combining integrals to obtain improved estimates. For instance, in Example 20.1, we computed two improved integrals of O(h4) on the basis of three trap- ezoidal rule estimates. These two improved integrals can, in turn, be combined to yield an even better value with O(h6). For the special case where the original trapezoidal estimates are based on successive halving of the step size, the equation used for O(h6) accuracy is

I = 16 ___ 15 Im − 1 ___ 15 Il (20.6)

where Im and Il are the more and less accurate estimates, respectively. Similarly, two O(h6) results can be combined to compute an integral that is O(h8) using

I = 64 ___ 63 Im − 1 ___ 63 Il (20.7)

EXAMPlE 20.2 Higher-Order corrections

Problem statement. In Example 20.1, we used Richardson extrapolation to compute two integral estimates of O(h4). Utilize Eq. (20.6) to combine these estimates to compute an integral with O(h6).

solution. The two integral estimates of O(h4) obtained in Example 20.1 were 1.367467 and 1.623467. These values can be substituted into Eq. (20.6) to yield

I = 16 ___ 15 (1.623467) − 1 ___ 15 (1.367467) = 1.640533

which is the exact value of the integral.

20.2.2 The Romberg Integration Algorithm

Notice that the coefficients in each of the extrapolation equations [Eqs. (20.5), (20.6), and (20.7)] add up to 1. Thus, they represent weighting factors that, as accuracy increases,

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528 NuMERIcAl INTEGRATION OF FuNcTIONs

place relatively greater weight on the superior integral estimate. These formulations can be expressed in a general form that is well suited for computer implementation:

Ij,k = 4k−1 Ij+1, k−1 − Ij, k−1 _______________

4k−1 − 1 (20.8)

where Ij+1, k−1 and Ij, k−1 = the more and less accurate integrals, respectively, and Ij,k = the improved integral. The index k signifies the level of the integration, where k = 1 corresponds to the original trapezoidal rule estimates, k = 2 corresponds to the O(h4) estimates, k = 3 to the O(h6), and so forth. The index j is used to distinguish between the more ( j + 1) and the less ( j) accurate estimates. For example, for k = 2 and j = 1, Eq. (20.8) becomes

I1,2 = 4I2,1 − I1,1 _________ 3

which is equivalent to Eq. (20.5). The general form represented by Eq. (20.8) is attributed to Romberg, and its system-

atic application to evaluate integrals is known as Romberg integration. Figure 20.1 is a graphical depiction of the sequence of integral estimates generated using this approach. Each matrix corresponds to a single iteration. The first column contains the trapezoidal rule evaluations that are designated Ij,1, where j = 1 is for a single-segment application (step size is b − a), j = 2 is for a two-segment application [step size is (b − a)∕2], j = 3 is for a four-segment application [step size is (b − a)∕4], and so forth. The other columns of the matrix are generated by systematically applying Eq. (20.8) to obtain successively better estimates of the integral.

For example, the first iteration (Fig. 20.1a) involves computing the one- and two- segment trapezoidal rule estimates (I1,1 and I2,1). Equation (20.8) is then used to compute the element I1,2 = 1.367467, which has an error of O(h4).

(a) 0.172800 1.068800

O(h2) O(h4) O(h6) O(h8)

(b)

(c)

0.172800 1.068800 1.484800

1.367467

1.367467 1.623467

1.640533

0.172800 1.068800 1.484800 1.600800

1.367467 1.623467 1.639467

1.640533 1.640533

1.640533

FIGURE 20.1 Graphical depiction of the sequence of integral estimates generated using Romberg integra- tion. (a) First iteration. (b) Second iteration. (c) Third iteration.

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20.2 ROMBERG INTEGRATION 529

Now, we must check to determine whether this result is adequate for our needs. As in other approximate methods in this book, a termination, or stopping, criterion is required to assess the accuracy of the results. One method that can be employed for the present purposes is

|εa| = | I1, k − I2, k−1 _________ I1, k | × 100% (20.9) where εa = an estimate of the percent relative error. Thus, as was done previously in other iterative processes, we compare the new estimate with a previous value. For Eq. (20.9), the previous value is the most accurate estimate from the previous level of integration (i.e., the k − 1 level of integration with j = 2). When the change between the old and new values as represented by εa is below a prespecified error criterion εs, the computation is terminated. For Fig. 20.1a, this evaluation indicates the following percent change over the course of the first iteration:

|εa| = | 1.367467 − 1.068800 _________ 1.367467 | × 100% = 21.8% The object of the second iteration (Fig. 20.1b) is to obtain the O(h6) estimate—I1,3.

To do this, a four-segment trapezoidal rule estimate, I3,1 = 1.4848, is determined. Then it is combined with I2,1 using Eq. (20.8) to generate I2,2 = 1.623467. The result is, in turn, combined with I1,2 to yield I1,3 = 1.640533. Equation (20.9) can be applied to determine that this result represents a change of 1.0% when compared with the previous result I2,2.

The third iteration (Fig. 20.1c) continues the process in the same fashion. In this case, an eight-segment trapezoidal estimate is added to the first column, and then Eq. (20.8) is applied to compute successively more accurate integrals along the lower diagonal. After only three iterations, because we are evaluating a fifth-order polynomial, the result (I1,4 = 1.640533) is exact.

Romberg integration is more efficient than the trapezoidal rule and Simpson’s rules. For example, for determination of the integral as shown in Fig. 20.1, Simpson’s 1∕3 rule would require about a 48-segment application in double precision to yield an estimate of the integral to seven significant digits: 1.640533. In contrast, Romberg integration pro- duces the same result based on combining one-, two-, four-, and eight-segment trapezoidal rules—that is, with only 15 function evaluations!

Figure 20.2 presents an M-file for Romberg integration. By using loops, this algorithm implements the method in an efficient manner. Note that the function uses another function trap to implement the composite trapezoidal rule evaluations (recall Fig. 19.10). Here is a MATLAB session showing how it can be used to determine the integral of the polynomial from Example 20.1:

>> f = @(x) 0.2 + 25*x − 200*x^2 + 675*x^3 − 900*x^4 + 400*x^5; >> romberg(f,0,0.8)

ans = 1.6405

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530 NuMERIcAl INTEGRATION OF FuNcTIONs

20.3 GAUSS QUADRATURE

In Chap. 19, we employed the Newton-Cotes equations. A characteristic of these formulas (with the exception of the special case of unequally spaced data) was that the integral esti- mate was based on evenly spaced function values. Consequently, the location of the base points used in these equations was predetermined or fixed.

For example, as depicted in Fig. 20.3a, the trapezoidal rule is based on taking the area under the straight line connecting the function values at the ends of the integration interval. The formula that is used to compute this area is

I ≅ (b − a) f (a) + f (b) _________ 2 (20.10)

function [q,ea,iter] = romberg(func,a,b,es,maxit,varargin) % romberg: Romberg integration quadrature % q = romberg(func,a,b,es,maxit,p1,p2,...): % Romberg integration. % input: % func = name of function to be integrated % a, b = integration limits % es = desired relative error (default = 0.000001%) % maxit = maximum allowable iterations (default = 30) % pl,p2,... = additional parameters used by func % output: % q = integral estimate % ea = approximate relative error (%) % iter = number of iterations

if nargin<3,error('at least 3 input arguments required'),end if nargin<4|isempty(es), es = 0.00000l;end if nargin<5|isempty(maxit), maxit = 50;end n = 1; I(1,1) = trap(func,a,b,n,varargin{:}); iter = 0; while iter<maxit iter = iter+l; n = 2^iter; I(iter+l,l) = trap (func,a,b,n,varargin{:}); for k = 2:iter+l j = 2+iter−k; I(j,k) = (4^(k−1)*I(j + 1,k − 1) − I(j,k − 1))/(4^(k − 1)−1); end ea = abs((I(1,iter+l)−I(2,iter))/I(1,iter+l))*100; if ea<=es, break; end end q = I(1,iter+l);

FIGURE 20.2 M-file to implement Romberg integration.

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20.3 GAuss QuADRATuRE 531

where a and b = the limits of integration and b − a = the width of the integration interval. Because the trapezoidal rule must pass through the end points, there are cases such as Fig. 20.3a where the formula results in a large error.

Now, suppose that the constraint of fixed base points was removed and we were free to evaluate the area under a straight line joining any two points on the curve. By positioning these points wisely, we could define a straight line that would balance the positive and nega- tive errors. Hence, as in Fig. 20.3b, we would arrive at an improved estimate of the integral.

Gauss quadrature is the name for a class of techniques to implement such a strategy. The particular Gauss quadrature formulas described in this section are called Gauss- Legendre formulas. Before describing the approach, we will show how numerical integration formulas such as the trapezoidal rule can be derived using the method of undetermined coefficients. This method will then be employed to develop the Gauss-Legendre formulas.

20.3.1 Method of Undetermined Coefficients

In Chap. 19, we derived the trapezoidal rule by integrating a linear interpolating polynomial and by geometrical reasoning. The method of undetermined coefficients offers a third ap- proach that also has utility in deriving other integration techniques such as Gauss quadrature.

f (x)

(a)

(b)

f (x)

FIGURE 20.3 (a) Graphical depiction of the trapezoidal rule as the area under the straight line joining fixed end points. (b) An improved integral estimate obtained by taking the area under the straight line passing through two intermediate points. By positioning these points wisely, the positive and negative errors are better balanced, and an improved integral estimate results.

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532 NuMERIcAl INTEGRATION OF FuNcTIONs

To illustrate the approach, Eq. (20.10) is expressed as

I ≅ c0 f (a) + c1 f (b) (20.11)

where the c’s = constants. Now realize that the trapezoidal rule should yield exact results when the function being integrated is a constant or a straight line. Two simple equations that represent these cases are y = 1 and y = x (Fig. 20.4). Thus, the following equalities should hold:

c0 + c1 = ∫ −(b−a)∕2 (b−a)∕2

1 dx

and

−c0 b − a _____ 2 + c1

b − a _____ 2 = ∫ −(b−a)∕2 (b−a)∕2

x dx

y = 1

(a)

−(b − a) 2

y = x y

(b)

b − a 2

FIGURE 20.4 Two integrals that should be evaluated exactly by the trapezoidal rule: (a) a constant and (b) a straight line.

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20.3 GAuss QuADRATuRE 533

or, evaluating the integrals,

c0 + c1 = b − a and

−c0 b − a _____ 2 + c1

b − a _____ 2 = 0

These are two equations with two unknowns that can be solved for

c0 = c1 = b − a _____ 2

which, when substituted back into Eq. (20.11), gives

I = b − a _____ 2 f (a) + b − a _____ 2 f (b)

which is equivalent to the trapezoidal rule.

20.3.2 Derivation of the Two-Point Gauss-Legendre Formula

Just as was the case for the previous derivation of the trapezoidal rule, the object of Gauss quadrature is to determine the coefficients of an equation of the form

I ≅ c0 f (x0) + c1 f (x1 ) (20.12)

where the c’s = the unknown coefficients. However, in contrast to the trapezoidal rule that used fixed end points a and b, the function arguments x0 and x1 are not fixed at the end points, but are unknowns (Fig. 20.5). Thus, we now have a total of four unknowns that must be evaluated, and consequently, we require four conditions to determine them exactly.

Just as for the trapezoidal rule, we can obtain two of these conditions by assuming that Eq. (20.12) fits the integral of a constant and a linear function exactly. Then, to arrive at the other two conditions, we merely extend this reasoning by assuming that it also fits the integral of a parabolic (y = x2) and a cubic (y = x3) function. By doing this, we determine

f (x)

f (x0)

f (x1)

−1 1x1 xx0

FIGURE 20.5 Graphical depiction of the unknown variables x0 and x1 for integration by Gauss quadrature.

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534 NuMERIcAl INTEGRATION OF FuNcTIONs

all four unknowns and in the bargain derive a linear two-point integration formula that is exact for cubics. The four equations to be solved are

c0 + c1 = ∫ −1

1 dx = 2 (20.13)

c0x0 + c1x1 = ∫ −1

x dx = 0 (20.14)

c0 x 0 2 + c1 x 1

2 = ∫ −1

x2 dx = 2 __ 3 (20.15)

c0 x 0 3 + c1 x 1

3 = ∫ −1

x3 dx = 0 (20.16)

Equations (20.13) through (20.16) can be solved simultaneously for the four unknowns. First, solve Eq. (20.14) for c1 and substitute the result into Eq. (20.16), which can be solved for

x 0 2 = x 1

Since x0 and x1 cannot be equal, this means that x0 = −x1. Substituting this result into Eq. (20.14) yields c0 = c1. Consequently from Eq. (20.13) it follows that

c0 = c1 = 1

Substituting these results into Eq. (20.15) gives

x0 = − 1 ___

√ _

3 = −0.5773503 . . .

x1 = 1 ___

√ _

3 = 0.5773503 . . .

Therefore, the two-point Gauss-Legendre formula is

I = f ( −1 ___ √ _ 3 ) + f ( 1 ___ √ _ 3 ) (20.17) Thus, we arrive at the interesting result that the simple addition of the function values at x = −1∕ √

_ 3 and 1∕ √

_ 3 yields an integral estimate that is third-order accurate.

Notice that the integration limits in Eqs. (20.13) through (20.16) are from −1 to 1. This was done to simplify the mathematics and to make the formulation as general as possible. A simple change of variable can be used to translate other limits of integration into this form. This is accomplished by assuming that a new variable xd is related to the original variable x in a linear fashion, as in

x = a1 + a2 xd (20.18)

If the lower limit, x = a, corresponds to xd = −1, these values can be substituted into Eq. (20.18) to yield

a = a1 + a2(−1) (20.19)

Similarly, the upper limit, x = b, corresponds to xd = 1, to give

b = a1 + a2(1) (20.20)

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20.3 GAuss QuADRATuRE 535

Equations (20.19) and (20.20) can be solved simultaneously for

a1 = b + a _____ 2 and a2 =

b − a _____ 2 (20.21)

which can be substituted into Eq. (20.18) to yield

x = (b + a) + (b − a)xd ________________ 2 (20.22)

This equation can be differentiated to give

dx = b − a _____ 2 dxd (20.23)

Equations (20.22) and (20.23) can be substituted for x and dx, respectively, in the equation to be integrated. These substitutions effectively transform the integration interval without changing the value of the integral. The following example illustrates how this is done in practice.

EXAMPlE 20.3 Two-Point Gauss-legendre Formula

Problem statement. Use Eq. (20.17) to evaluate the integral of

f (x) = 0.2 + 25x − 200x2 + 675x3 − 900x4 + 400x5

between the limits x = 0 to 0.8. The exact value of the integral is 1.640533.

solution. Before integrating the function, we must perform a change of variable so that the limits are from −1 to +1. To do this, we substitute a = 0 and b = 0.8 into Eqs. (20.22) and (20.23) to yield

x = 0.4 + 0.4xd and dx = 0.4dxd Both of these can be substituted into the original equation to yield

∫ 0

0.8

(0.2 + 25x − 200x2 + 675x3 − 900x4 + 400x5) dx

= ∫ −1

[0.2 + 25(0.4 + 0.4xd ) − 200(0.4 + 0.4xd )2 + 675(0.4 + 0.4xd )3

−900(0.4 + 0.4xd )4 + 400(0.4 + 0.4xd )5]0.4dxd Therefore, the right-hand side is in the form that is suitable for evaluation using Gauss quadrature. The transformed function can be evaluated at xd = −1∕ √

_ 3 as 0.516741 and

at xd = 1∕ √ _

3 as 1.305837. Therefore, the integral according to Eq. (20.17) is 0.516741 + 1.305837 = 1.822578, which represents a percent relative error of −11.1%. This result is comparable in magnitude to a four-segment application of the trapezoidal rule or a single application of Simpson’s 1∕3 and 3∕8 rules. This latter result is to be expected because Simpson’s rules are also third-order accurate. However, because of the clever choice of base points, Gauss quadrature attains this accuracy on the basis of only two function evaluations.

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536 NuMERIcAl INTEGRATION OF FuNcTIONs

20.3.3 Higher-Point Formulas

Beyond the two-point formula described in the previous section, higher-point versions can be developed in the general form

I ≅ c0 f (x0) + c1 f (x1) + · · · + cn−1 f (xn−1) (20.24) where n = the number of points. Values for c’s and x’s for up to and including the six-point formula are summarized in Table 20.1.

EXAMPlE 20.4 Three-Point Gauss-legendre Formula

Problem statement. Use the three-point formula from Table 20.1 to estimate the integral for the same function as in Example 20.3.

solution. According to Table 20.1, the three-point formula is I = 0.5555556 f (−0.7745967) + 0.8888889 f (0) + 0.5555556 f (0.7745967)

which is equal to I = 0.2813013 + 0.8732444 + 0.4859876 = 1.640533

which is exact.

TABLE 20.1 Weighting factors and function arguments used in Gauss-legendre formulas.

Weighting Function Truncation Points Factors Arguments Error

1 c0 = 2 x0 = 0.0 ≅ f (2) (ξ) 2 c0 = 1 x0 = −1∕ √

_ 3 ≅ f (4) (ξ)

c1 = 1 x1 = 1∕ √ _

3 3 c0 = 5∕9 x0 = − √

_ 3∕5 ≅ f (6) (ξ)

c1 = 8∕9 x1 = 0.0 c2 = 5∕9 x2 = √

_ 3∕5

4 c0 = (18 − √ __

30 )∕36 x0 = − √ ______

525 + 70 √ __

30 ∕35 ≅ f (8) (ξ) c1 = (18 + √

__ 30 )∕36 x1 = − √

______ 525 − 70 √

__ 30 ∕35

c2 = (18 + √ __

30 )∕36 x2 = √ ______

525 − 70 √ __

30 ∕35 c3 = (18 − √

__ 30 )∕36 x3 = √

______ 525 + 70 √

__ 30 ∕35

5 c0 = (322 − 13 √ __

70 )∕900 x0 = − √ ______

245 + 14 √ __

70 ∕21 ≅ f (10) (ξ) c1 = (322 + 13 √

__ 70 )∕900 x1 = − √

______ 245 − 14 √

__ 70 ∕21

c2 = 128∕225 x2 = 0.0 c3 = (322 + 13 √

__ 70 )∕900 x3 = √

______ 245 − 14 √

__ 70 ∕21

c4 = (322 − 13 √ __

70 )∕900 x4 = √ ______

245 + 14 √ __

70 ∕21 6 c0 = 0.171324492379170 x0 = −0.932469514203152 ≅ f (12) (ξ) c1 = 0.360761573048139 x1 = −0.661209386466265 c2 = 0.467913934572691 x2 = −0.238619186083197 c3 = 0.467913934572691 x3 = 0.238619186083197 c4 = 0.360761573048131 x4 = 0.661209386466265 c5 = 0.171324492379170 x5 = 0.932469514203152

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20.4 ADAPTIvE QuADRATuRE 537

Because Gauss quadrature requires function evaluations at nonuniformly spaced points within the integration interval, it is not appropriate for cases where the function is unknown. Thus, it is not suited for engineering problems that deal with tabulated data. However, where the function is known, its efficiency can be a decided advantage. This is particularly true when numerous integral evaluations must be performed.

20.4 ADAPTIVE QUADRATURE

Although Romberg integration is more efficient than the composite Simpson’s 1/3 rule, both use equally spaced points. This constraint does not take into account that some func- tions have regions of relatively abrupt changes where more refined spacing might be required. Hence, to achieve a desired accuracy, fine spacing must be applied everywhere even though it is only needed for the regions of sharp change. Adaptive quadrature meth- ods remedy this situation by automatically adjusting the step size so that small steps are taken in regions of sharp variations and larger steps are taken where the function changes gradually.

20.4.1 MATLAB M-file: quadadapt

Adaptive quadrature methods accommodate the fact that many functions have regions of high variability along with other sections where change is gradual. They accomplish this by adjusting the step size so that small intervals are used in regions of rapid variations and larger intervals are used where the function changes gradually. Many of these techniques are based on applying the composite Simpson’s 1∕3 rule to subintervals in a fashion that is very similar to the way in which the composite trapezoidal rule was used in Richardson extrapolation. That is, the 1∕3 rule is applied at two levels of refinement, and the difference between these two levels is used to estimate the truncation error. If the truncation error is acceptable, no further refinement is required, and the integral estimate for the subinterval is deemed acceptable. If the error estimate is too large, the step size is refined and the process repeated until the error falls to acceptable levels. The total integral is then computed as the summation of the integral estimates for the subintervals.

The theoretical basis of the approach can be illustrated for an interval x = a to x = b with a width of h1 = b − a. A first estimate of the integral can be estimated with Simpson’s 1/3 rule:

I (h1) = h1 __ 6 [ f (a) + 4 f (c) + f (b)] (20.25)

where c = (a + b)∕2. As in Richardson extrapolation, a more refined estimate can be obtained by halving the

step size. That is, by applying the composite Simpson’s 1/3 rule with n = 4:

I (h2) = h2 __ 6 [ f (a) + 4 f (d) + 2 f (c) + 4 f (e) + f (b)] (20.26)

where d = (a + c)∕2, e = (c + b)∕2, and h2 = h1∕2. Because both I(h1) and I(h2) are estimates of the same integral, their difference pro-

vides a measure of the error. That is,

E ≅ I (h2) − I (h1) (20.27)

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538 NuMERIcAl INTEGRATION OF FuNcTIONs

In addition, the estimate and error associated with either application can be represented generally as

I = I (h) + E (h) (20.28)

where I = the exact value of the integral, I(h) = the approximation from an n-segment application of the Simpson’s 1/3 rule with step size h = (b − a)∕n, and E(h) = the corre- sponding truncation error.

Using an approach similar to Richardson extrapolation, we can derive an estimate in the error of the more refined estimate I(h2) as a function of the difference between the two integral estimates:

E(h2) = 1 ___ 15 [ I (h2) − I (h1)] (20.29)

The error can then be added to I (h2) to generate an even better estimate:

I = I (h2) + 1 ___ 15 [ I (h2) − I (h1)] (20.30)

This result is equivalent to Boole’s rule (Table 19.2). The equations just developed can now be combined into an efficient algorithm.

Figure 20.6 presents an M-file function that is based on an algorithm originally developed by Cleve Moler (2004).

The function consists of a main calling function quadadapt along with a recursive func- tion qstep that actually performs the integration. The main calling function quadadapt is passed the function f and the integration limits a and b. After setting the tolerance, the func- tion evaluations required for the initial application of Simpson’s 1/3 rule [Eq. (20.25)] are computed. These values along with the integration limits are then passed to qstep. Within qstep, the remaining step sizes and function values are determined, and the two integral estimates [Eqs. (20.25) and (20.26)] are computed.

At this point, the error is estimated as the absolute difference between the integral estimates. Depending on the value of the error, two things can then happen:

1. If the error is less or equal to the tolerance (tol), Boole’s rule is generated; the function terminates and passes back the result.

2. If the error is larger than the tolerance, qstep is invoked twice to evaluate each of the two subintervals of the current call.

The two recursive calls in the second step represent the real beauty of this algorithm. They just keep subdividing until the tolerance is met. Once this occurs, their results are passed back up the recursive path, combining with the other integral estimates along the way. The process ends when the final call is satisfied and the total integral is evaluated and returned to the main calling function.

It should be stressed that the algorithm in Fig. 20.6 is a stripped-down version of the integral function, which is the professional root-location function employed in MATLAB. Thus, it does not guard against failure such as cases where integrals do not exist. Nevertheless,

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20.4 ADAPTIvE QuADRATuRE 539

function q = quadadapt(f,a,b,tol,varargin) % Evaluates definite integral of f(x) from a to b if nargin < 4 | isempty(tol),tol = 1.e−6;end c = (a + b)/2; fa = feval(f,a,varargin{:}); fc = feval(f,c,varargin{:}); fb = feval(f,b,varargin{:}); q = quadstep(f, a, b, tol, fa, fc, fb, varargin{:}); end

function q = quadstep(f,a,b,tol,fa,fc,fb,varargin) % Recursive subfunction used by quadadapt. h = b − a; c = (a + b)/2; fd = feval(f,(a + c)/2,varargin{:}); fe = feval(f,(c + b)/2,varargin{:}); q1 = h/6 * (fa + 4*fc + fb); q2 = h/12 * (fa + 4*fd + 2*fc + 4*fe + fb); if abs(q2 − q1) <= tol q = q2 + (q2 − q1)/15; else qa = quadstep(f, a, c, tol, fa, fd, fc, varargin{:}); qb = quadstep(f, c, b, tol, fc, fe, fb, varargin{:}); q = qa + qb; end end

FIGURE 20.6 An M-file to implement an adaptive quadrature algorithm based on an algorithm originally developed by Cleve Moler (2004).

it works just fine for many applications, and certainly serves to illustrate how adaptive quadra- ture works. Here is a MATLAB session showing how quadadapt can be used to determine the integral of the polynomial from Example 20.1:

>> f = @(x) 0.2 + 25*x − 200*x^2 + 675*x^3 − 900*x^4 + 400*x^5; >> q = quadadapt(f,0,0.8)

q = 1.640533333333336

20.4.2 MATLAB Function: integral

MATLAB has a function for implementing adaptive quadrature:

q = integral(fun, a, b)

where fun is the function to be integrated, and a and b = the integration bounds. It should be noted that array operators .*, ./ and .^ should be used in the definition of fun.

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540 NuMERIcAl INTEGRATION OF FuNcTIONs

EXAMPlE 20.5 Adaptive Quadrature

Problem statement. Use integral to integrate the following function:

f (x) = 1 _____________ (x − q)2 + 0.01

+ 1 _____________ (x − r )2 + 0.04

− s

between the limits x = 0 to 1. Note that for q = 0.3, r = 0.9, and s = 6, this is the built-in humps function that MATLAB uses to demonstrate some of its numerical capabilities. The humps func- tion exhibits both flat and steep regions over a relatively short x range. Hence, it is useful for demonstrating and testing a function like integral. Note that the humps function can be inte- grated analytically between the given limits to yield an exact integral of 29.85832539549867.

solution. First, let’s evaluate the integral using the built-in version of humps

>> format long >> Q = integral(@(x) humps(x),0,1)

ans = 29.85832612842764

Thus, the solution is correct to seven significant digits.

20.5 CASE STUDY ROOT-MEAN-sQuARE cuRRENT

Background. Because it results in efficient energy transmission, the current in an AC circuit is often in the form of a sine wave:

i = ipeak sin(ωt)

where i = the current (A = C/s), ipeak = the peak current (A), ω = the angular frequency (radians/s), and t = time (s). The angular frequency is related to the period T(s) by ω = 2π∕T. The power generated is related to the magnitude of the current. Integration can be used to determine the average current over one cycle:

_ i = 1 __ T ∫ 0

ipeak sin(ωt) dt = ipeak ____ T (−cos(2π) + cos(0)) = 0

Despite the fact that the average is zero, such a current is capable of generating power. Therefore, an alternative to the average current must be derived. To do this, electrical engineers and scientists determine the root mean square current irms (A), which is calculated as

irms = √ ________

1 __ T ∫ 0 T

i peak 2 sin2(ωt) dt =

ipeak ____ √

_ 2 (20.31)

Thus, as the name implies, the rms current is the square root of the mean of the squared current. Because 1∕ √

__ 2 = 0.70707, irms is equal to about 70% of the peak current for our

assumed sinusoidal wave form.

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20.5 CASE STUDY continued

This quantity has meaning because it is directly related to the average power absorbed by an element in an AC circuit. To understand this, recall that Joule’s law states that the instantaneous power absorbed by a circuit element is equal to product of the voltage across it and the current through it:

P = iV (20.32)

where P = the power (W = J/s), and V = voltage (V = J/C). For a resistor, Ohm’s law states that the voltage is directly proportional to the current:

V = iR (20.33)

where R = the resistance (Ω = V/A = J · s/C2). Substituting Eq. (20.33) into (20.32) gives

P = i 2R (20.34)

The average power can be determined by integrating Eq. (20.34) over a period with the result:

P̄ = i rms 2 R

Thus, the AC circuit generates the equivalent power as a DC circuit with a constant current of irms. Now, although the simple sinusoid is widely employed, it is by no means the only waveform that is used. For some of these forms, such as triangular or square waves, the irms can be evaluated analytically with closed-form integration. However, some waveforms must be analyzed with numerical integration methods. In this case study, we will calculate the root-mean-square current of a non sinusoidal wave form. We will use both the Newton-Cotes formulas from Chap. 19 as well as the approaches described in this chapter.

Solution. The integral that must be evaluated is

i rms 2 = ∫ 0

1∕2

(10e−t sin 2π t)2 dt (20.35)

For comparative purposes, the exact value of this integral to fifteen significant digits is 15.41260804810169. Integral estimates for various applications of the trapezoidal rule and Simpson’s 1∕3 rule are listed in Table 20.2. Notice that Simpson’s rule is more accurate than the trapezoidal rule. The value for the integral to seven significant digits is obtained using a 128-segment trapezoidal rule or a 32-segment Simpson’s rule. The M-file we developed in Fig. 20.2 can be used to evaluate the integral with Romberg integration:

>> format long >> i2=@(t) (10*exp(−t).*sin(2*pi*t)).^2; >> [q,ea,iter]=romberg(i2,0,.5)

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542 NuMERIcAl INTEGRATION OF FuNcTIONs

q = 15.41260804288977 ea = 1.480058787326946e−008 iter = 5

Thus, with the default stopping criterion of es = 1 × 10−6, we obtain a result that is correct to over nine significant figures in five iterations. We can obtain an even better result if we impose a more stringent stopping criterion:

>> [q,ea,iter]=romberg(i2,0,.5,1e−15)

q = 15.41260804810169 ea = 0 iter = 7

Gauss quadrature can also be used to make the same estimate. First, a change in variable is performed by applying Eqs. (20.22) and (20.23) to yield

t = 1 __ 4 + 1 __ 4 td dt =

1 __ 4 dtd

These relationships can be substituted into Eq. (20.35) to yield

i rms 2 = ∫ −1

[10e−(0.25+0.25 t d ) sin 2π(0.25 + 0.25td)]2 0.25 dt (20.36)

20.5 CASE STUDY continued

TABLE 20.2 values for the integral calculated using Newton-cotes formulas.

Technique Segments Integral εt (%)

Trapezoidal rule 1 0.0 100.0000 2 15.163266493 1.6178 4 15.401429095 0.0725 8 15.411958360 4.22 × 10−3 16 15.412568151 2.59 × 10−4 32 15.412605565 1.61 × 10−5 64 15.412607893 1.01 × 10−6 128 15.412608038 6.28 × 10−8

Simpson’s 1∕3 rule 2 20.217688657 31.1763 4 15.480816629 0.4426 8 15.415468115 0.0186 16 15.412771415 1.06 × 10−3 32 15.412618037 6.48 × 10−5

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20.5 CASE STUDY continued

For the two-point Gauss-Legendre formula, this function is evaluated at td = −1∕ √ _

3 and 1∕ √

_ 3 , with the results being 7.684096 and 4.313728, respectively. These values can be

substituted into Eq. (20.17) to yield an integral estimate of 11.99782, which represents an error of εt = 22.1%. The three-point formula is (Table 20.1)

I = 0.5555556(1.237449) + 0.8888889(15.16327) + 0.5555556(2.684915) = 15.65755

which has εt = 1.6%. The results of using the higher-point formulas are summarized in Table 20.3. Finally, the integral can be evaluated with the built-in MATLAB function integral:

>> irms2=integral(i2,0,.5)

irms2 = 15.412608049345090

We can now compute the irms by merely taking the square root of the integral. For example, using the result computed with integral, we get

>> irms=sqrt(irms2)

irms = 3.925889459485796

This result could then be employed to guide other aspects of the design and operation of the circuit such as power dissipation computations. As we did for the simple sinusoid in Eq. (20.31), an interesting calculation involves comparing this result with the peak current. Recognizing that this is an optimization prob- lem, we can readily employ the fminbnd function to determine this value. Because we are looking for a maximum, we evaluate the negative of the function:

>> [tmax,imax]=fminbnd(@(t) −10*exp(−t).*sin(2*pi*t),0,.5)

tmax = 0.22487940319321 imax = −7.886853873932577

A maximum current of 7.88685 A occurs at t = 0.2249 s. Hence, for this particular wave form, the root-mean-square value is about 49.8% of the maximum.

TABLE 20.3 Results of using various-point Gauss quadrature formulas to approximate the integral.

Points Estimate εt (%)

2 11.9978243 22.1 3 15.6575502 1.59 4 15.4058023 4.42 × 10−2 5 15.4126391 2.01 × 10−4 6 15.4126109 1.82 × 10−5

20.5 cAsE sTuDY 543

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544 NuMERIcAl INTEGRATION OF FuNcTIONs

PROBLEMS

20.1 Use Romberg integration to evaluate

I = ∫ 1

( x + 1 __ x ) 2 dx

to an accuracy of εs = 0.5%. Your results should be presented in the format of Fig. 20.1. Use the analytical solution of the integral to determine the percent relative error of the result ob- tained with Romberg integration. Check that εt is less than εs. 20.2 Evaluate the following integral (a) analytically, (b) Romberg integration (εs = 0.5%), (c) the three-point Gauss quadrature formula, and (d) MATLAB integral function:

I = ∫ 0 8

− 0.055x4 + 0.86x3 − 4.2x2 + 6.3x + 2 dx

20.3 Evaluate the following integral with (a) Romberg in- tegration (εs = 0.5%), (b) the two-point Gauss quadrature formula, and (c) MATLAB integral function:

I = ∫ 0

x e2 x dx

20.4 There is no closed form solution for the error function

erf(a) = 2 ____ √ __

π ∫ 0 a

e− x 2 dx

Use the (a) two-point and (b) three-point Gauss-Legendre formulas to estimate erf(1.5). Determine the percent relative error for each case based on the true value, which can be determined with MATLAB’s built-in function erf. 20.5 The force on a sailboat mast can be represented by the following function:

F = ∫ 0

200 ( z _____ 5 + z ) e−2z∕H dz where z = the elevation above the deck and H = the height of the mast. Compute F for the case where H = 30 using (a) Romberg integration to a tolerance of εs = 0.5%, (b) the two-point Gauss- Legendre formula, and (c) the MATLAB integral function. 20.6 The root-mean-square current can be computed as

IRMS = √ ________

1 __ T ∫ 0 T

i 2(t) dt

For T = 1, suppose that i(t) is defined as

i (t) = 8e−t∕T sin ( 2π t __ T ) for 0 ≤ t ≤ T∕2 i (t) = 0 for T∕2 ≤ t ≤ T

Evaluate the IRMS using (a) Romberg integration to a toler- ance of 0.1%, (b) the two- and three-point Gauss-Legendre formulas, and (c) the MATLAB integral function.

20.7 The heat required, ΔH (cal), to induce a temperature change, ΔT (°C), of a material can be computed as

ΔH = mCp(T )ΔT

where m = mass (g), and Cp(T) = heat capacity [cal/(g .°C)]. The heat capacity increases with temperature, T (°C), according to

Cp(T ) = 0.132 + 1.56 × 10−4 T + 2.64 × 10−7T 2

Write a script that uses the integral function to generate a plot of ΔH versus ΔT for cases where m = 1 kg, the starting temperature is −100 °C, and ΔT ranges from 0 to 300 °C. 20.8 The amount of mass transported via a pipe over a pe- riod of time can be computed as

M = ∫ t1 t2 Q(t) c(t) dt

where M = mass (mg), t1 = the initial time (min), t2 = the final time (min), Q(t) = flow rate (m3/min), and c (t) = concentration (mg/m3). The following functional repre sentations define the temporal variations in flow and concentration:

Q(t) = 9 + 5 cos2(0.4 t) c(t) = 5e−0.5t + 2 e0.15t

Determine the mass transported between t1 = 2 and t2 = 8 min with (a) Romberg integration to a tolerance of 0.1% and (b) the MATLAB integral function. 20.9 Evaluate the double integral

∫ − 2

∫ 0

(x2 − 3y2 + xy3) dx dy

(a) analytically, with the 2-point Gauss quadrature, and (b) with the integral2 function. 20.10 Compute work as described in Sec. 19.9, but use the following equations for F(x) and θ (x):

F(x) = 1.6x − 0.045x2

θ(x) = −0.00055x3 + 0.0123x2 + 0.13x

The force is in newtons and the angle is in radians. Perform the integration from x = 0 to 30 m. 20.11 Perform the same computation as in Sec. 20.5, but for the current as specified by

i (t) = 6e−1.25t sin 2π t for 0 ≤ t ≤ T∕2 i (t) = 0 for T∕2 < t ≤ T

where T = 1 s.

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20.12 Compute the power absorbed by an element in a cir- cuit as described in Sec. 20.5, but for a simple sinusoidal current i = sin(2π t∕T ) where T = 1 s. (a) Assume that Ohm’s law holds and R = 5 Ω. (b) Assume that Ohm’s law does not hold and that voltage

and current are related by the following nonlinear rela- tionship: V = (5i − 1.25i3).

20.13 Suppose that the current through a resistor is de- scribed by the function

i (t) = (60 − t)2 + (60 − t) sin( √ _ t )

and the resistance is a function of the current:

R = 10i + 2i 2∕3

Compute the average voltage over t = 0 to 60 using the composite Simpson’s 1∕3 rule. 20.14 If a capacitor initially holds no charge, the voltage across it as a function of time can be computed as

V(t) = 1 __ C ∫ 0 t

i (t) dt

Use MATLAB to fit these data with a fifth-order polynomial. Then, use a numerical integration function along with a value of C = 10−5 farad to generate a plot of voltage versus time.

t, s 0 0.2 0.4 0.6 i, 10−3 A 0.2 0.3683 0.3819 0.2282

t, s 0.8 1 1.2 i, 10−3 A 0.0486 0.0082 0.1441

20.15 The work done on an object is equal to the force times the distance moved in the direction of the force. The velocity of an object in the direction of a force is given by

υ = 4t 0 ≤ t ≤ 5 υ = 20 + (5 − t)2 5 ≤ t ≤ 15

where υ is in m/s. Determine the work if a constant force of 200 N is applied for all t. 20.16 A rod subject to an axial load (Fig. P20.16a) will be deformed, as shown in the stress-strain curve in Fig. P20.16b. The area under the curve from zero stress out to the point of rupture is called the modulus of toughness of the material. It provides a measure of the energy per unit volume required to cause the material to rupture. As such, it is representative of the material’s ability to withstand an impact load. Use numerical integration to compute the modulus of toughness for the stress-strain curve seen in Fig. P20.16b.

0.1

s, ks

(b)(a)

e 0.02 0.05 0.10 0.15 0.20 0.25

s 40.0 37.5 43.0 52.0 60.0 55.0

Rupture

0.2 e

Modulus of toughness

FIGURE P20.16 (a) A rod under axial loading and (b) the resulting stress-strain curve, where stress is in kips per square inch (103 lb/in2), and strain is dimensionless.

PROBlEMs 545

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546 NuMERIcAl INTEGRATION OF FuNcTIONs

20.17 If the velocity distribution of a fluid flowing through a pipe is known (Fig. P20.17), the flow rate Q (i.e., the vol- ume of water passing through the pipe per unit time) can be computed by Q = ʃ υ dA, where υ is the velocity, and A is the pipe’s cross-sectional area. (To grasp the meaning of this relationship physically, recall the close connection between summation and integration.) For a circular pipe, A = π r 2 and dA = 2π r dr. Therefore,

Q = ∫ 0 r

υ(2π r) dr

where r is the radial distance measured outward from the center of the pipe. If the velocity distribution is given by

υ = 2 ( 1 − r __ r0 ) 1∕6 where r0 is the total radius (in this case, 3 cm), compute Q using the composite trapezoidal rule. Discuss the results. 20.18 Using the following data, calculate the work done by stretching a spring that has a spring constant of k = 300 N/m to x = 0.35 m. To do this, first fit the data with a polynomial and then integrate the polynomial numerically to compute the work:

F, 103 · N 0 0.01 0.028 0.046 x, m 0 0.05 0.10 0.15

F, 103 · N 0.063 0.082 0.11 0.13 x, m 0.20 0.25 0.30 0.35

20.19 Evaluate the vertical distance traveled by a rocket if the vertical velocity is given by

υ = 11t2 − 5t 0 ≤ t ≤ 10 υ = 1100 − 5t 10 ≤ t ≤ 20 υ = 50t + 2(t − 20)2 20 ≤ t ≤ 30

20.20 The upward velocity of a rocket can be computed by the following formula:

υ = u ln ( m0 _______ m0 − qt ) − gt

where υ = upward velocity, u = velocity at which fuel is expelled relative to the rocket, m0 = initial mass of the rocket at time t = 0, q = fuel consumption rate, and g = downward acceleration of gravity (assumed constant = 9.81 m/s2). If u = 1850 m/s, m0 = 160,000 kg, and q = 2500 kg/s, deter- mine how high the rocket will fly in 30 s. 20.21 The normal distribution is defined as

f (x) = 1 _____ √

___ 2π e− x

2 /2

(a) Use MATLAB to integrate this function from x = −1 to 1 and from −2 to 2.

(b) Use MATLAB to determine the inflection points of this function.

20.22 Use Romberg integration to evaluate

∫ 0

e x sin x ______ 1 + x2

to an accuracy of εs = 0.5%. Your results should be pre- sented in the form of Fig. 20.1. 20.23 Recall that the velocity of the free-falling bungee jumper can be computed analytically as [Eq. (1.9)]:

υ (t) = √ __

gm ___ cd tanh ( √ __

gcd ___ m t )

where v(t) = velocity (m/s), t = time (s), g = 9.81 m/s2, m = mass (kg), cd = drag coefficient (kg/m). (a) Use Romberg integration to compute how far the jumper

travels during the first 8 seconds of free fall given m = 80 kg and cd = 0.2 kg/m. Compute the answer to εs = 1%.

(b) Perform the same computation with integral. 20.24 Prove that Eq. (20.30) is equivalent to Boole’s rule. 20.25 As specified in the following table, the earth’s density varies as a function of the distance from its center (r = 0):

r, km 0 1100 1500 2450 3400 3630 4500 ρ, g/cm3 13 12.4 12 11.2 9.7 5.7 5.2

r, km 5380 6060 6280 6380 ρ, g/cm3 4.7 3.6 3.4 3

Develop a script to fit these data with interp1 using the pchip option. Generate a plot showing the resulting fit along with the data points. Then use one of MATLAB’s integra- tion functions to estimate the earth’s mass (in metric tonnes) by integrating the output of the interp1 function. 20.26 Develop an M-file function to implement Romberg integration based on Fig. 20.2. Test the function by using it

r A

FIGURE P20.17

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to determine the integral of the polynomial from Example 20.1. Then use it to solve Prob. 20.1. 20.27 Develop an M-file function to implement adaptive quadrature based on Fig. 20.6. Test the function by using it to determine the integral of the polynomial from Example 20.1. Then use it to solve Prob. 20.20. 24.28 The average flow in a river channel, Q (m3/s), with an irregular cross-section can be computed as the integral of the product of velocity and depth

Q = ∫ 0

U(y) H (y) dy

where U(y) = water velocity (m/s) at distance y (m) from the bank, and H(y) = water depth (m/s) at distance y from the bank. Use integral along with spline fits of U and H to the following data collected at different distances across the channel to estimate the flow.

y (m) H (m) Y (m) U (m/s)

0 0 0 0

1.1 0.21 1.6 0.08

2.8 0.78 4.1 0.61

4.6 1.87 4.8 0.68

6 1.44 6.1 0.55

8.1 1.28 6.8 0.42

9 0.2 9 0

20.29 Use the two-point Gauss quadrature approach to es- timate the average value of the following function between a = 1 and b = 5

f (x) = 2 ______ 1 + x2

20.30 Evaluate the following integral

I = ∫ 0

x3 dx

(a) Analytically. (b) Using the MATLAB integral function. (c) Using Monte Carlo integration. 20.31 The MATLAB humps function defines a curve that has 2 maxima (peaks) of unequal height over the interval 0 ≤ x ≤ 2. Develop a MATLAB script to determine the integral over the interval with (a) the MATLAB integral function and (b) Monte Carlo integration. 20.32 Evaluate the following double integral

I = ∫ 0

∫ −3

y4 (x2 + xy) dx dy

(a) Using a single application of Simpson’s 1/3 rule across each dimension.

(b) Check your results with the integral2 function.

PROBLEMS 547

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548

21 Numerical Differentiation

CHAPTER OBJECTIVES

The primary objective of this chapter is to introduce you to numerical differentiation. Specific objectives and topics covered are

• Understanding the application of high-accuracy numerical differentiation formulas for equispaced data.

• Knowing how to evaluate derivatives for unequally spaced data. • Understanding how Richardson extrapolation is applied for numerical differentiation. • Recognizing the sensitivity of numerical differentiation to data error. • Knowing how to evaluate derivatives in MATLAB with the diff and gradient

functions. • Knowing how to generate contour plots and vector fields with MATLAB.

YOU’VE GOT A PROBLEM

Recall that the velocity of a free-falling bungee jumper as a function of time can be computed as υ(t) = √

___ gm ___ cd tanh ( √

___

gcd ___ m t ) (21.1)

At the beginning of Chap. 19, we used calculus to integrate this equation to determine the vertical distance z the jumper has fallen after a time t.

z(t) = m __ cd ln [cosh ( √ ____

gcd ____ m t ) ] (21.2)

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21.1 INtRoDuctIoN aND BackgRouND 549

Now suppose that you were given the reverse problem. That is, you were asked to determine velocity based on the jumper’s position as a function of time. Because it is the inverse of integration, differentiation could be used to make the determination:

υ(t) = dz(t) ____ dt (21.3)

Substituting Eq. (21.2) into Eq. (21.3) and differentiating would bring us back to Eq. (21.1). Beyond velocity, you might also be asked to compute the jumper’s acceleration. To

do this, we could either take the first derivative of velocity, or the second derivative of displacement:

a(t) = dυ(t) _____ dt = d 2z(t) _____

dt 2 (21.4)

In either case, the result would be

a(t) = g sech2 ( √___ gcd ___ m t ) (21.5) Although a closed-form solution can be developed for this case, there are other func-

tions that may be difficult or impossible to differentiate analytically. Further, suppose that there was some way to measure the jumper’s position at various times during the fall. These distances along with their associated times could be assembled as a table of discrete val- ues. In this situation, it would be useful to differentiate the discrete data to determine the velocity and the acceleration. In both these instances, numerical differentiation methods are available to obtain solutions. This chapter will introduce you to some of these methods.

21.1 INTRODUCTION AND BACKGROUND

21.1.1 What Is Differentiation?

Calculus is the mathematics of change. Because engineers and scientists must continu- ously deal with systems and processes that change, calculus is an essential tool of our profession. Standing at the heart of calculus is the mathematical concept of differentiation.

According to the dictionary definition, to differentiate means “to mark off by differ- ences; distinguish; . . . to perceive the difference in or between.” Mathematically, the deriva- tive, which serves as the fundamental vehicle for differentiation, represents the rate of change of a dependent variable with respect to an independent variable. As depicted in Fig. 21.1, the mathematical definition of the derivative begins with a difference approximation:

Δy ___ Δx = f (xi + Δx) − f (xi ) ______________ Δx (21.6)

where y and f (x) are alternative representatives for the dependent variable and x is the independent variable. If Δx is allowed to approach zero, as occurs in moving from Fig. 21.1a to c, the difference becomes a derivative:

dy ___ dx = lim Δx→0

f (xi + Δx) − f (xi ) ______________ Δx (21.7)

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where dy∕dx [which can also be designated as y′ or f ′(xi )]1 is the first derivative of y with respect to x evaluated at xi. As seen in the visual depiction of Fig. 21.1c, the derivative is the slope of the tangent to the curve at xi.

The second derivative represents the derivative of the first derivative,

d 2y ___ dx 2 =

d ___ d x ( dy ___ dx ) (21.8) Thus, the second derivative tells us how fast the slope is changing. It is commonly referred to as the curvature, because a high value for the second derivative means high curvature.

Finally, partial derivatives are used for functions that depend on more than one vari- able. Partial derivatives can be thought of as taking the derivative of the function at a point with all but one variable held constant. For example, given a function f that depends on both x and y, the partial derivative of f with respect to x at an arbitrary point (x, y) is defined as

∂ f ___ ∂x =

lim Δx→0 f (x + Δx, y) − f (x, y) ______________ Δx (21.9)

Similarly, the partial derivative of f with respect to y is defined as

∂ f ___ ∂y =

lim Δy→0 f (x, y + Δ y) − f (x, y) ______________ Δy (21.10)

To get an intuitive grasp of partial derivatives, recognize that a function that depends on two variables is a surface rather than a curve. Suppose you are mountain climbing and have access to a function f that yields elevation as a function of longitude (the east-west oriented

f (xi)

xi xi + ∆x x

∆x

f (xi + ∆x)

y y y

∆y

f (xi) f ' (xi)

f (xi + ∆x)

xi xi + ∆x x

∆x (a) (b)

∆y

xi x

(c)

FIGURE 21.1 The graphical definition of a derivative: as Δx approaches zero in going from (a) to (c), the difference approximation becomes a derivative.

1 The form dy∕dx was devised by Leibnitz, whereas y′ is attributed to Lagrange. Note that Newton used the so-called dot notation: ˙ y. Today, the dot notation is usually used for time derivatives.

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21.1 INtRoDuctIoN aND BackgRouND 551

x axis) and latitude (the north-south oriented y axis). If you stop at a particular point (x0, y0), the slope to the east would be ∂f (x0, y0)∕∂x, and the slope to the north would be ∂f (x0, y0)∕∂y.

21.1.2 Differentiation in Engineering and Science

The differentiation of a function has so many engineering and scientific applications that you were required to take differential calculus in your first year at college. Many specific examples of such applications could be given in all fields of engineering and science. Differentiation is commonplace in engineering and science because so much of our work involves characterizing the changes of variables in both time and space. In fact, many of the laws and other generalizations that figure so prominently in our work are based on the predictable ways in which change manifests itself in the physical world. A prime example is Newton’s second law, which is not couched in terms of the position of an object but rather in its change with respect to time.

Aside from such temporal examples, numerous laws involving the spatial behavior of variables are expressed in terms of derivatives. Among the most common of these are the constitutive laws that define how potentials or gradients influence physical processes. For example, Fourier’s law of heat conduction quantifies the observation that heat flows from regions of high to low temperature. For the one-dimensional case, this can be expressed mathematically as

q = −k dT ___ dx (21.11)

where q(x) = heat flux (W/m2), k = coefficient of thermal conductivity [W/(m · K)], T = temperature (K), and x = distance (m). Thus, the derivative, or gradient, provides a mea- sure of the intensity of the spatial temperature change, which drives the transfer of heat (Fig. 21.2).

(b)(a)

Direction of heat flow

< 0 > 0

i T i

FIGURE 21.2 Graphical depiction of a temperature gradient. Because heat moves “downhill” from high to low temperature, the flow in (a) is from left to right. However, due to the orientation of Cartesian coordinates, the slope is negative for this case. Thus, a negative gradient leads to a positive flow. This is the origin of the minus sign in Fourier’s law of heat conduction. The reverse case is depicted in (b), where the positive gradient leads to a negative heat flow from right to left.

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Similar laws provide workable models in many other areas of engineering and science, including the modeling of fluid dynamics, mass transfer, chemical reaction kinetics, elec- tricity, and solid mechanics (Table 21.1). The ability to accurately estimate derivatives is an important facet of our capability to work effectively in these areas.

Beyond direct engineering and scientific applications, numerical differentiation is also important in a variety of general mathematical contexts including other areas of numerical methods. For example, recall that in Chap. 6 the secant method was based on a finite- difference approximation of the derivative. In addition, probably the most important application of numerical differentiation involves the solution of differential equations. We have already seen an example in the form of Euler’s method in Chap. 1. In Chap. 24, we will investigate how numerical differentiation provides the basis for solving boundary- value problems of ordinary differential equations.

These are just a few of the applications of differentiation that you might face regularly in the pursuit of your profession. When the functions to be analyzed are simple, you will nor- mally choose to evaluate them analytically. However, it is often difficult or impossible when the function is complicated. In addition, the underlying function is often unknown and de- fined only by measurement at discrete points. For both these cases, you must have the ability to obtain approximate values for derivatives, using numerical techniques as described next.

21.2 HIGH-ACCURACY DIFFERENTIATION FORMULAS

We have already introduced the notion of numerical differentiation in Chap. 4. Recall that we employed Taylor series expansions to derive finite-difference approximations of deriv- atives. In Chap. 4, we developed forward, backward, and centered difference approxima- tions of first and higher derivatives. Remember that, at best, these estimates had errors that were O(h2)—that is, their errors were proportional to the square of the step size. This level of accuracy is due to the number of terms of the Taylor series that were retained during the

TABLE 21.1 the one-dimensional forms of some constitutive laws commonly used in engineering and science.

Law Equation Physical Area Gradient Flux Proportionality

Fourier’s law q = −k dT ___ dx Heat conduction Temperature Heat flux Thermal Conductivity

Fick’s law J = −D dc ___ dx Mass diffusion Concentration Mass flux Diffusivity

Darcy’s law q = −k dh ___ dx Flow through Head Flow flux Hydraulic porous media Conductivity

Ohm’s law J = −σ dV ___ dx Current flow Voltage Current flux Electrical Conductivity

Newton’s τ = μ du ___ dx Fluids Velocity Shear Dynamic viscosity law Stress Viscosity

Hooke’s law σ = E ΔL ___ L Elasticity Deformation Stress Young’s Modulus

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21.2 HIgH-accuRacy DIffeReNtIatIoN foRmulas 553

derivation of these formulas. We will now illustrate how high-accuracy finite-difference formulas can be generated by including additional terms from the Taylor series expansion.

For example, the forward Taylor series expansion can be written as [recall Eq. (4.13)]

f (xi+1) = f (xi ) + f ′ (xi ) h + f ″(xi ) _____ 2! h

2 + · · · (21.12)

which can be solved for

f ′(xi ) = f (xi+1) − f (xi ) _____________ h −

f ″(xi) _____ 2! h + O(h 2) (21.13)

In Chap. 4, we truncated this result by excluding the second- and higher-derivative terms and were thus left with a forward-difference formula:

f ′(xi ) = f (xi+1) − f (xi ) _____________ h + O(h) (21.14)

In contrast to this approach, we now retain the second-derivative term by substituting the following forward-difference approximation of the second derivative [recall Eq. (4.27)]:

f ″(xi ) = f (xi+2) − 2 f (xi+1) + f (xi) _____________

h2 + O(h) (21.15)

into Eq. (21.13) to yield

f ′(xi ) = f (xi+1) − f (xi ) _____________ h −

f (xi+2) − 2 f (xi+1) + f (xi ) ____________________ 2h2

h + O(h2 ) (21.16)

or, by collecting terms:

f ′(xi ) = −f (xi+2) + 4 f (xi+1) − 3 f (xi) _____________ 2h + O(h

2 ) (21.17)

Notice that inclusion of the second-derivative term has improved the accuracy to O(h2). Similar improved versions can be developed for the backward and centered formulas as well as for the approximations of higher-order derivatives. The formulas are summarized in Fig. 21.3 through Fig. 21.5 along with the lower-order versions from Chap. 4. The fol- lowing example illustrates the utility of these formulas for estimating derivatives.

eXamPle 21.1 High-accuracy Differentiation formulas

Problem statement. Recall that in Example 4.4 we estimated the derivative of

f (x) = −0.1x4 − 0.15x3 − 0.5x2 − 0.25x + 1.2

at x = 0.5 using finite-differences and a step size of h = 0.25. The results are summarized in the following table. Note that the errors are based on the true value of f ′(0.5) = −0.9125.

Backward Centered Forward O(h) O(h2) O(h)

Estimate −0.714 −0.934 −1.155 εt 21.7% −2.4% −26.5%

Repeat this computation, but employ the high-accuracy formulas from Fig. 21.3 through Fig. 21.5.

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solution. The data needed for this example are

xi−2 = 0 f (xi−2) = 1.2 xi−1 = 0.25 f (xi−1) = 1.1035156 xi = 0.5 f (xi ) = 0.925 xi+1 = 0.75 f (xi+1) = 0.6363281 xi+2 = 1 f (xi+2) = 0.2

The forward difference of accuracy O(h2) is computed as (Fig. 21.3)

f ′(0.5) = −0.2 + 4(0.6363281) − 3(0.925) ___________________________ 2(0.25) = −0.859375 εt = 5.82%

The backward difference of accuracy O(h2) is computed as (Fig. 21.4)

f ′(0.5) = 3(0.925) − 4(1.1035156) + 1.2 ___________________________ 2(0.25) = −0.878125 εt = 3.77%

The centered difference of accuracy O(h4) is computed as (Fig. 21.5)

f ′(0.5) = −0.2 + 8(0.6363281) − 8(1.1035156) + 1.2 ___________________________ 12(0.25) = −0.9125 εt = 0%

FIGURE 21.3 Forward finite-difference formulas: two versions are presented for each derivative. The latter version incorporates more terms of the Taylor series expansion and is, consequently, more accurate.

First Derivative Error

f ′(xi) = f (xi+1) − f (xi) ___________ h O(h)

f ′(xi) = − f (xi+2) + 4 f (xi+1) − 3 f (xi) _______________________ 2h O(h

Second Derivative

f ″(xi) = f (xi+2) − 2 f (xi+1) + f (xi) _____________________

h2 O(h)

f ″(xi) = − f (xi+3) + 4 f (xi+2) − 5 f (xi+1) + 2 f (xi) __________________________________

h2 O(h2)

Third Derivative

f ‴(xi) = f (xi+3) − 3 f (xi+2) + 3 f (xi+1) − f (xi) ___________________________

h3 O(h)

f ‴(xi) = −3 f (xi+4) + 14 f (xi+3) − 24 f (xi+2) + 18 f (xi+1) − 5 f (xi) _________________________________________

2h3 O(h2)

Fourth Derivative

f ⁗(xi) = f (xi+4) − 4 f (xi+3) + 6 f (xi+2) − 4 f (xi+1) + f (xi) _________________________________________

h4 O(h)

f ⁗(xi) = −2 f (xi+5) + 11 f (xi+4) − 24 f (xi+3) + 26 f (xi+2) − 14 f (xi+1) + 3 f (xi) _________________________________________

h4 O(h2)

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21.3 RIcHaRDsoN eXtRaPolatIoN 555

As expected, the errors for the forward and backward differences are considerably more accurate than the results from Example 4.4. However, surprisingly, the centered dif- ference yields the exact derivative at x = 0.5. This is because the formula based on the Taylor series is equivalent to passing a fourth-order polynomial through the data points.

21.3 RICHARDSON EXTRAPOLATION

To this point, we have seen that there are two ways to improve derivative estimates when employing finite differences: (1) decrease the step size or (2) use a higher-order formula that employs more points. A third approach, based on Richardson extrapolation, uses two derivative estimates to compute a third, more accurate, approximation.

Recall from Sec. 20.2.1 that Richardson extrapolation provided a means to obtain an improved integral estimate by the formula [Eq. (20.4)]

I = I (h2) + 1 __________

(h1∕h2)2 − 1 [I (h2) − I (h1)] (21.18)

where I (h1) and I (h2) are integral estimates using two step sizes: h1 and h2. Because of its convenience when expressed as a computer algorithm, this formula is usually written for the case where h2 = h1∕2, as in

I = 4 __ 3 I (h2) − 1 __ 3 I (h1) (21.19)

FIGURE 21.4 Backward finite-difference formulas: two versions are presented for each derivative. The latter version incorporates more terms of the Taylor series expansion and is, consequently, more accurate.

First Derivative Error

f ′(xi) = f (xi) − f (xi−1) ___________ h O(h)

f ′(xi) = 3 f (xi) − 4 f (xi−1) + f (xi−2) _______________________ 2h O(h

Second Derivative

f ″(xi) = f (xi) − 2 f (xi−1) + f (xi−2) _____________________

h2 O(h)

f ″(xi) = 2 f (xi) − 5 f (xi−1) + 4 f (xi−2) − f (xi−3) __________________________________

h2 O(h2)

Third Derivative

f ‴(xi) = f (xi) − 3 f (xi−1) + 3 f (xi−2) − f (xi−3) ___________________________

h3 O(h)

f ‴(xi) = 5 f (xi) − 18 f (xi−1) + 24 f (xi−2) − 14 f (xi−3) + 3 f (xi−4) _________________________________________

2h3 O(h2)

Fourth Derivative

f ⁗(xi) = f (xi) − 4 f (xi−1) + 6 f (xi−2) − 4 f (xi−3) + f (xi−4) _________________________________________

h4 O(h)

f ⁗(xi) = 3 f (xi) − 14 f (xi−1) + 26 f (xi−2) − 24 f (xi−3) + 11 f (xi−4) − 2 f (xi−5) _________________________________________

h4 O(h2)

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In a similar fashion, Eq. (21.19) can be written for derivatives as

D = 4 __ 3 D(h2) − 1 __ 3 D(h1) (21.20)

For centered difference approximations with O(h2), the application of this formula will yield a new derivative estimate of O(h4).

eXamPle 21.2 Richardson extrapolation

Problem statement. Using the same function as in Example 21.1, estimate the first de- rivative at x = 0.5 employing step sizes of h1 = 0.5 and h2 = 0.25. Then use Eq. (21.20) to compute an improved estimate with Richardson extrapolation. Recall that the true value is −0.9125.

solution. The first-derivative estimates can be computed with centered differences as

D(0.5) = 0.2 − 1.2 ________ 1 = −1.0 εt = −9.6%

and

D(0.25) = 0.6363281 − 1.103516 ____________________ 0.5 = −0.934375 εt = −2.4%

FIGURE 21.5 Centered finite-difference formulas: two versions are presented for each derivative. The latter version incorporates more terms of the Taylor series expansion and is, consequently, more accurate.

First Derivative Error

f ′(xi) = f (xi+1) − f (xi−1) ___________ 2h O(h

f ′(xi) = − f (xi+2) + 8 f (xi+1) − 8 f (xi−1) + f (xi−2) _______________________ 12h O(h

Second Derivative

f ″(xi) = f (xi+1) − 2 f (xi ) + f (xi−1) _____________________

h2 O(h2)

f ″(xi) = − f (xi+2) + 16 f (xi+1) − 30 f (xi) + 16 f (xi−1) − f (xi−2) __________________________________

12h2 O(h4)

Third Derivative

f ‴(xi) = f (xi+2) − 2 f (xi+1) + 2 f (xi−1) − f (xi−2) ___________________________

2h3 O(h2)

f ‴(xi) = − f (xi+3) + 8 f (xi+2) − 13 f (xi+1) + 13 f (xi−1) − 8 f (xi−2) + f (xi−3) _________________________________________

8h3 O(h4)

Fourth Derivative

f ⁗(xi) = f (xi+2) − 4 f (xi+1) + 6 f (xi) − 4 f (xi−1) + f (xi−2) _________________________________________

h4 O(h2)

f ⁗(xi) = − f (xi+3) + 12 f (xi+2) − 39 f (xi+1) + 56 f (xi) − 39 f (xi−1) + 12 f (xi−2) − f (xi−3) _________________________________________

6h4 O(h4)

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21.4 DeRIVatIVes of uNeQually sPaceD Data 557

The improved estimate can be determined by applying Eq. (21.20) to give

D = 4 __ 3 (−0.934375) − 1 __ 3 (−1) = −0.9125

which for the present case is exact.

The previous example yielded an exact result because the function being analyzed was a fourth-order polynomial. The exact outcome was due to the fact that Richardson extrapolation is actually equivalent to fitting a higher-order polynomial through the data and then evaluating the derivatives by centered divided differences. Thus, the present case matched the derivative of the fourth-order polynomial precisely. For most other functions, of course, this would not occur, and our derivative estimate would be improved but not exact. Consequently, as was the case for the application of Richardson extrapolation, the approach can be applied iteratively using a Romberg algorithm until the result falls below an acceptable error criterion.

21.4 DERIVATIVES OF UNEQUALLY SPACED DATA

The approaches discussed to this point are primarily designed to determine the derivative of a given function. For the finite-difference approximations of Sec. 21.2, the data had to be evenly spaced. For the Richardson extrapolation technique of Sec. 21.3, the data also had to be evenly spaced and generated for successively halved intervals. Such control of data spac- ing is usually available only in cases where we can use a function to generate a table of values.

In contrast, empirically derived information—that is, data from experiments or field studies—are often collected at unequal intervals. Such information cannot be analyzed with the techniques discussed to this point.

One way to handle nonequispaced data is to fit a Lagrange interpolating polynomial [recall Eq. (17.21)] to a set of adjacent points that bracket the location value at which you want to evaluate the derivative. Remember that this polynomial does not require that the points be equispaced. The polynomial can then be differentiated analytically to yield a formula that can be used to estimate the derivative.

For example, you can fit a second-order Lagrange polynomial to three adjacent points (x0, y0), (x1, y1), and (x2, y2). Differentiating the polynomial yields:

f ′(x) = f (x0) 2x − x1 − x2 _______________ (x0 − x1)(x0 − x2)

+ f (x1) 2x − x0 − x2 ______________ (x1 − x0)(x1 − x2)

+ f (x2) 2x − x0 − x1 _______________ (x2 − x0)(x2 − x1)

(21.21)

where x is the value at which you want to estimate the derivative. Although this equation is certainly more complicated than the first-derivative approximation from Fig. 21.3 through Fig. 21.5, it has some important advantages. First, it can provide estimates anywhere within the range prescribed by the three points. Second, the points themselves do not have to be equally spaced. Third, the derivative estimate is of the same accuracy as the centered dif- ference [Eq. (4.25)]. In fact, for equispaced points, Eq. (21.21) evaluated at x = x1 reduces to Eq. (4.25).

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eXamPle 21.3 Differentiating unequally spaced Data

Problem statement. As in Fig. 21.6, a temperature gradient can be measured down into the soil. The heat flux at the soil-air interface can be computed with Fourier’s law (Table 21.1):

q(z = 0) = −k dT ___ dz | z=0 where q (z) = heat flux (W/m2), k = coefficient of thermal conductivity for soil [= 0.5 W/ (m · K)], T = temperature (K), and z = distance measured down from the surface into the soil (m). Note that a positive value for flux means that heat is transferred from the air to the soil. Use numerical differentiation to evaluate the gradient at the soil-air interface and employ this estimate to determine the heat flux into the ground.

solution. Equation (21.21) can be used to calculate the derivative at the air-soil interface as

f ′(0) = 13.5 2(0) − 0.0125 − 0.0375 _____________________ (0 − 0.0125)(0 − 0.0375) + 12 2(0) − 0 − 0.0375 __________________________ (0.0125 − 0)(0.0125 − 0.0375)

+ 10 2(0) − 0 − 0.0125 _________________________ (0.0375 − 0)(0.0375 − 0.0125)

= −1440 + 1440 − 133.333 = −133.333 K/m

which can be used to compute

q(z = 0) = −0.5 W ____ m K ( −133.333 K __ m ) = 66.667 W ___ m2

z, cm

T(°C)10Air

Soil

3.75

13.512

1.25

FIGURE 21.6 Temperature versus depth into the soil.

21.5 DERIVATIVES AND INTEGRALS FOR DATA WITH ERRORS

Aside from unequal spacing, another problem related to differentiating empirical data is that these data usually include measurement error. A shortcoming of numerical differentia- tion is that it tends to amplify errors in the data.

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21.6 PaRtIal DeRIVatIVes 559

Fig. 21.7a shows smooth, error-free data that when numerically differentiated yield a smooth result (Fig. 21.7b). In contrast, Fig. 21.7c uses the same data, but with alternat- ing points raised and lowered slightly. This minor modification is barely apparent from Fig. 21.7c. However, the resulting effect in Fig. 21.7d is significant.

The error amplification occurs because differentiation is subtractive. Hence, random positive and negative errors tend to add. In contrast, the fact that integration is a summing process makes it very forgiving with regard to uncertain data. In essence, as points are summed to form an integral, random positive and negative errors cancel out.

As might be expected, the primary approach for determining derivatives for imprecise data is to use least-squares regression to fit a smooth, differentiable function to the data. In the absence of any other information, a lower-order polynomial regression might be a good first choice. Obviously, if the true functional relationship between the dependent and inde- pendent variable is known, this relationship should form the basis for the least-squares fit.

21.6 PARTIAL DERIVATIVES

Partial derivatives along a single dimension are computed in the same fashion as ordinary derivatives. For example, suppose that we want to determine to partial derivatives for a two-dimensional function f (x, y). For equally spaced data, the partial first derivatives can

t (a)

D i�

er en

tia te

t (c)

D i�

er en

tia te

t (b)

t (d)

dy dt

FIGURE 21.7 Illustration of how small data errors are amplified by numerical differentiation: (a) data with no error, (b) the resulting numerical differentiation of curve (a), (c) data modified slightly, and (d ) the resulting differentiation of curve (c) manifesting increased variability. In contrast, the reverse operation of integration [moving from (d ) to (c ) by taking the area under (d )] tends to attenuate or smooth data errors.

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be approximated with centered differences:

∂ f ___ ∂x =

f (x + Δ x, y) − f (x − Δ x, y) ______________ 2Δx (21.22)

∂ f ___ ∂y =

f (x, y + Δ y) − f (x, y − Δy) ______________ 2Δy (21.23)

All the other formulas and approaches discussed to this point can be applied to evaluate partial derivatives in a similar fashion.

For higher-order derivatives, we might want to differentiate a function with respect to two or more different variables. The result is called a mixed partial derivative. For exam- ple, we might want to take the partial derivative of f (x, y) with respect to both independent variables

∂ 2f ___

∂x∂y = ∂ ___ ∂x ( ∂ f ___ ∂y ) (21.24)

To develop a finite-difference approximation, we can first form a difference in x of the partial derivatives in y:

∂ 2f ___

∂x∂y = ∂ f ___ ∂y (x + Δx, y) −

∂ f ___ ∂y (x − Δx, y) __________________________ 2Δx (21.25)

Then, we can use finite differences to evaluate each of the partials in y:

∂ 2f ___

∂x∂y = f (x + Δx, y + Δy) − f (x + Δx, y − Δy) ________________________________ 2Δy −

f (x − Δx, y + Δy) − f (x − Δx, y − Δy) ________________________________ 2Δy __________________________________________________________________ 2Δx (21.26)

Collecting terms yields the final result

∂ 2f ___

∂x∂y = f (x + Δx, y + Δy) − f (x + Δx, y − Δy) − f (x − Δx, y + Δy) + f (x − Δx, y − Δy) __________________________________________________________________ 4ΔxΔy (21.27)

21.7 NUMERICAL DIFFERENTIATION WITH MATLAB

MATLAB software has the ability to determine the derivatives of data based on two built- in functions: diff and gradient.

21.7.1 MATLAB Function: diff

When it is passed a one-dimensional vector of length n, the diff function returns a vector of length n − 1 containing the differences between adjacent elements. As described in the following example, these can then be employed to determine finite-difference approxima- tions of first derivatives.

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21.7 NumeRIcal DIffeReNtIatIoN wItH matlaB 561

eXamPle 21.4 using diff for Differentiation

Problem statement. Explore how the MATLAB diff function can be employed to dif- ferentiate the function

f (x) = 0.2 + 25x − 200x2 + 675x3 − 900x4 + 400x5

from x = 0 to 0.8. Compare your results with the exact solution:

f ′ (x) = 25 − 400x2 + 2025x2 − 3600x3 + 2000x4

solution. We can first express f (x) as an anonymous function:

>> f = @(x) 0.2 + 25*x − 200*x.^2 + 675*x.^3 − 900*x.^4 + 400*x.^5;

We can then generate a series of equally spaced values of the independent and dependent variables:

>> x = 0:0.1:0.8; >> y = f(x);

The diff function can be used to determine the differences between adjacent elements of each vector. For example,

>> diff(x)

ans = Columns 1 through 5 0.1000 0.1000 0.1000 0.1000 0.1000 Columns 6 through 8 0.1000 0.1000 0.1000

As expected, the result represents the differences between each pair of elements of x. To compute divided-difference approximations of the derivative, we merely perform a vector division of the y differences by the x differences by entering

>> d = diff(y)./diff(x)

d = Columns 1 through 5 10.8900 −0.0100 3.1900 8.4900 8.6900 Columns 6 through 8 1.3900 −11.0100 −21.3100

Note that because we are using equally spaced values, after generating the x values, we could have simply performed the above computation concisely as

>> d = diff(f(x))/0.1;

The vector d now contains derivative estimates corresponding to the midpoint between adjacent elements. Therefore, in order to develop a plot of our results, we must first gener- ate a vector holding the x values for the midpoint of each interval:

>> n = length(x); >> xm = (x(1:n −1)+x(2:n))./2;

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As a final step, we can compute values for the analytical derivative at a finer level of reso- lution to include on the plot for comparison. >> xa = 0:.01:.8; >> ya = 25 − 400*xa + 3*675*xa.^2 − 4*900*xa.^3 + 5*400*xa.^4;

A plot of the numerical and analytical estimates can be generated with >> plot(xm,d,'o',xa,ya)

As displayed in Fig. 21.8, the results compare favorably for this case.

Note that aside from evaluating derivatives, the diff function comes in handy as a programming tool for testing certain characteristics of vectors. For example, the following statement displays an error message and terminates an M-file if it determines that a vector x has unequal spacing:

if any(diff(diff(x)) ~ = 0), error('unequal spacing'), end

Another common use is to detect whether a vector is in ascending or descending order. For example, the following code rejects a vector that is not in ascending order (i.e., mono tonically increasing):

if any(diff(x)< = 0), error('not in ascending order'), end

−10

−20

−5

−15

−25 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8

FIGURE 21.8 Comparison of the exact derivative (line) with numerical estimates (circles) computed with MATLAB’s diff function.

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21.7.2 MATLAB Function: gradient

The gradient function also returns differences. However, it does so in a manner that is more compatible with evaluating derivatives at the values themselves rather than in the intervals between values. A simple representation of its syntax is

fx = gradient(f)

where f = a one-dimensional vector of length n, and fx is a vector of length n containing differences based on f. Just as with the diff function, the first value returned is the differ- ence between the first and second value. However, for the intermediate values, a centered difference based on the adjacent values is returned

diffi = fi+1 − fi−1 ________ 2 (21.28)

The last value is then computed as the difference between the final two values. Hence, the results are akin to using centered differences for all the intermediate values, with forward and backward differences at the ends.

Note that the spacing between points is assumed to be one. If the vector represents equally spaced data, the following version divides all the results by the interval and hence returns the actual values of the derivatives,

fx = gradient(f, h)

where h = the spacing between points.

eXamPle 21.5 using gradient for Differentiation

Problem statement. Use the gradient function to differentiate the same function that we analyzed in Example 21.4 with the diff function.

solution. In the same fashion as Example 21.4, we can generate a series of equally spaced values of the independent and dependent variables: >> f = @(x) 0.2 + 25*x − 200*x.^2 + 675*x.^3 − 900*x.^4 + 400*x.^5; >> x = 0:0.1:0.8; >> y = f(x);

We can then use the gradient function to determine the derivatives as >> dy = gradient(y,0.1)

dy = Columns 1 through 5 10.8900 5.4400 1.5900 5.8400 8.5900 Columns 6 through 9 5.0400 −4.8100 −16.1600 −21.3100

As in Example 21.4, we can generate values for the analytical derivative and display both the numerical and analytical estimates on a plot: >> xa = 0:.01:.8; >> ya = 25 − 400*xa + 3*675*xa.^2 − 4*900*xa.^3 + 5*400*xa.^4; >> plot(x,dy,'o', xa,ya)

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As displayed in Fig. 21.9, the results are not as accurate as those obtained with the diff function in Example 21.4. This is due to the fact that gradient employs intervals that are two times (0.2) as wide as for those used for diff (0.1).

Beyond one-dimensional vectors, the gradient function is particularly well suited for determining the partial derivatives of matrices. For example, for a two-dimensional matrix, f, the function can be invoked as

[fx,fy] = gradient(f, h)

where fx corresponds to the differences in the x (column) direction, and fy corresponds to the differences in the y (row) direction, and h = the spacing between points. If h is omitted, the spacing between points in both dimensions is assumed to be one. In the next section, we will illustrate how gradient can be used to visualize vector fields.

−10

−20

−5

−15

−25 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8

FIGURE 21.9 Comparison of the exact derivative (line) with numerical estimates (circles) computed with MATLAB’s gradient function.

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21.8 case stuDy 565

21.8 CASE STUDY VIsualIZINg fIelDs

Background. Beyond the determination of derivatives in one dimension, the gradient function is also quite useful for determining partial derivatives in two or more dimensions. In particular, it can be used in conjunction with other MATLAB functions to produce vi- sualizations of vector fields.

To understand how this is done, we can return to our discussion of partial derivatives at the end of Sec. 21.1.1. Recall that we used mountain elevation as an example of a two- dimensional function. We can represent such a function mathematically as

z = f (x, y)

where z = elevation, x = distance measured along the east-west axis, and y = distance measured along the north-south axis.

For this example, the partial derivatives provide the slopes in the directions of the axes. However, if you were mountain climbing, you would probably be much more interested in determining the direction of the maximum slope. If we think of the two partial derivatives as component vectors, the answer is provided very neatly by

∇f = ∂ f ___ ∂ x i +

∂ f ___ ∂ y j

where ∇f is referred to as the gradient of f. This vector, which represents the steepest slope, has a magnitude

√ ____________

( ∂ f ___ ∂ x ) 2

+ ( ∂ f ___ ∂ y ) 2

and a direction

θ = tan−1 = ( ∂ f∕∂ y ___ ∂ f∕∂ x ) where θ = the angle measured counterclockwise from the x axis.

Now suppose that we generate a grid of points in the x-y plane and used the foregoing equations to draw the gradient vector at each point. The result would be a field of arrows indicating the steepest route to the peak from any point. Conversely, if we plotted the nega- tive of the gradient, it would indicate how a ball would travel as it rolled downhill from any point.

Such graphical representations are so useful that MATLAB has a special function, called quiver, to create such plots. A simple representation of its syntax is

quiver(x,y,u,v)

where x and y are matrices containing the position coordinates and u and v are matrices containing the partial derivatives. The following example demonstrates the use of quiver to visualize a field.

Employ the gradient function to determine to partial derivatives for the following two- dimensional function:

f (x, y) = y − x − 2x2 − 2xy − y2

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21.8 CASE STUDY continued

from x = −2 to 2 and y = 1 to 3. Then use quiver to superimpose a vector field on a contour plot of the function.

Solution. We can first express f (x, y) as an anonymous function >> f = @(x,y) y − x − 2*x.^2 − 2.*x.*y − y.^2;

A series of equally spaced values of the independent and dependent variables can be gener- ated as

>> [x,y] = meshgrid(−2:.25:0, 1:.25:3); >> z = f(x,y);

The gradient function can be employed to determine the partial derivatives: >> [fx,fy] = gradient(z,0.25);

We can then develop a contour plot of the results: >> cs = contour(x,y,z);clabel(cs);hold on

As a final step, the resultant of the partial derivatives can be superimposed as vectors on the contour plot:

>> quiver(x,y,−fx,−fy);hold off

−1

1 +

0 +

−1 +

−2 + −3+

−4 +

−5 +

2.8

2.4

1.6

1.2

2.6

2.2

1.8

1.4

1 −2 −1.8 −1.6 −1.4 −1.2 −0.8 −0.6 −0.4 −0.2 0−1

FIGURE 21.10 MATLAB generated contour plot of a two-dimensional function with the resultant of the partial derivatives displayed as arrows.

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21.8 CASE STUDY continued

Note that we have displayed the negative of the resultants, in order that they point “downhill.”

The result is shown in Fig. 21.10. The function’s peak occurs at x = −1 and y = 1.5 and then drops away in all directions. As indicated by the lengthening arrows, the gradient drops off more steeply to the northeast and the southwest.

PROBLEMS

21.1 Compute forward and backward difference approxi- mations of O(h) and O(h2), and central difference approxi- mations of O(h2) and O(h4) for the first derivative of y = sin x at x = π∕4 using a value of h = π∕12. Estimate the true percent relative error εt for each approximation. 21.2 Use centered difference approximations to estimate the first and second derivatives of y = ex at x = 2 for h = 0.1. Employ both O(h2) and O(h4) formulas for your estimates. 21.3 Use a Taylor series expansion to derive a centered finite-difference approximation to the third derivative that is second-order accurate. To do this, you will have to use four different expansions for the points xi−2, xi−1, xi+1, and xi+2. In each case, the expansion will be around the point xi. The interval Δx will be used in each case of i − 1 and i + 1, and 2Δx will be used in each case of i − 2 and i + 2. The four equations must then be combined in a way to eliminate the first and second derivatives. Carry enough terms along in each expansion to evaluate the first term that will be trun- cated to determine the order of the approximation. 21.4 Use Richardson extrapolation to estimate the first de- rivative of y = cos x at x = π∕4 using step sizes of h1= π∕3 and h2 = π∕6. Employ centered differences of O(h2) for the initial estimates. 21.5 Repeat Prob. 21.4, but for the first derivative of ln x at x = 5 using h1 = 2 and h2 = 1. 21.6 Employ Eq. (21.21) to determine the first derivative of y = 2x4 − 6x3 − 12x − 8 at x = 0 based on values at x0 = −0.5, x1 = 1, and x2 = 2. Compare this result with the true value and with an estimate obtained using a centered difference approximation based on h = 1. 21.7 Prove that for equispaced data points, Eq. (21.21) reduces to Eq. (4.25) at x = x1. 21.8 Develop an M-file to apply a Romberg algorithm to estimate the derivative of a given function.

21.9 Develop an M-file to obtain first-derivative estimates for unequally spaced data. Test it with the following data:

x 0.6 1.5 1.6 2.5 3.5 f (x) 0.9036 0.3734 0.3261 0.08422 0.01596

where f (x) = 5e−2x x. Compare your results with the true derivatives. 21.10 Develop an M-file function that computes first and second derivative estimates of order O(h2) based on the for- mulas in Fig. 21.3 through Fig. 21.5. The function’s first line should be set up as

function [dydx, d2ydx2] = diffeq(x,y)

where x and y are input vectors of length n containing the values of the independent and dependent variables, respec- tively, and dydx and dy2dx2 are output vectors of length n containing the first- and second-derivative estimates at each value of the independent variable. The function should generate a plot of dydx and dy2dx2 versus x. Have your M-file return an error message if (a) the input vectors are not the same length or (b) the values for the independent vari- able are not equally spaced. Test your program with the data from Prob. 21.11. 21.11 The following data were collected for the distance traveled versus time for a rocket:

t, s 0 25 50 75 100 125 y, km 0 32 58 78 92 100

Use numerical differentiation to estimate the rocket’s veloc- ity and acceleration at each time.

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21.17 The following data were generated from the normal distribution:

x −2 −1.5 −1 −0.5 0 f (x) 0.05399 0.12952 0.24197 0.35207 0.39894

x 0.5 1 1.5 2 f (x) 0.35207 0.24197 0.12952 0.05399

Use MATLAB to estimate the inflection points of these data. 21.18 Use the diff(y) command to develop a MATLAB M-file function to compute finite-difference approximations to the first and second derivative at each x value in the table below. Use finite-difference approximations that are second- order correct, O(x2):

x 0 1 2 3 4 5 6 7 8 9 10 y 1.4 2.1 3.3 4.8 6.8 6.6 8.6 7.5 8.9 10.9 10

21.19 The objective of this problem is to compare second- order accurate forward, backward, and centered finite- difference approximations of the first derivative of a function to the actual value of the derivative. This will be done for

f (x) = e−2x − x

(a) Use calculus to determine the correct value of the de- rivative at x = 2.

(b) Develop an M-file function to evaluate the centered finite-difference approximations, starting with x = 0.5. Thus, for the first evaluation, the x values for the cen- tered difference approximation will be x = 2 ± 0.5 or x = 1.5 and 2.5. Then, decrease in increments of 0.1 down to a minimum value of Δx = 0.01.

(c) Repeat part (b) for the second-order forward and back- ward differences. (Note that these can be done at the same time that the centered difference is computed in the loop.)

(d) Plot the results of (b) and (c) versus x. Include the exact result on the plot for comparison.

21.20 You have to measure the flow rate of water through a small pipe. In order to do it, you place a bucket at the pipe’s outlet and measure the volume in the bucket as a function of time as tabulated below. Estimate the flow rate at t = 7 s.

Time, s 0 1 5 8 Volume, cm3 0 1 8 16.4

21.21 The velocity υ (m/s) of air flowing past a flat surface is measured at several distances y (m) away from the surface.

21.12 A jet fighter’s position on an aircraft carrier’s runway was timed during landing:

t, s 0 0.52 1.04 1.75 2.37 3.25 3.83 x, m 153 185 208 249 261 271 273

where x is the distance from the end of the carrier. Estimate (a) velocity (dx∕dt) and (b) acceleration (dυ∕dt) using nu- merical differentiation. 21.13 Use the following data to find the velocity and accel- eration at t = 10 seconds:

Time, t, s 0 2 4 6 8 10 12 14 16 Position, x, m 0 0.7 1.8 3.4 5.1 6.3 7.3 8.0 8.4

Use second-order correct (a) centered finite-difference, (b) forward finite-difference, and (c) backward finite- difference methods. 21.14 A plane is being tracked by radar, and data are taken every second in polar coordinates θ and r.

t, s 200 202 204 206 208 210 θ, (rad) 0.75 0.72 0.70 0.68 0.67 0.66 r, m 5120 5370 5560 5800 6030 6240

At 206 seconds, use the centered finite-difference (second- order correct) to find the vector expressions for velocity → υ and acceleration → a . The velocity and acceleration given in polar coordinates are

→ υ = r ̇ → e r + r θ ˙ → e θ and

→ a = ( r ̈ − r θ ˙ 2) → e r + (r θ ¨ + 2 r ̇ θ ˙ ) → e θ

21.15 Use regression to estimate the acceleration at each time for the following data with second-, third-, and fourth- order polynomials. Plot the results:

t 1 2 3.25 4.5 6 7 8 8.5 9.3 10 υ 10 12 11 14 17 16 12 14 14 10

21.16 The normal distribution is defined as

f (x) = 1 _____ √

___ 2π e−x2∕2

Use MATLAB to determine the inflection points of this function.

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PRoBlems 569

21.25 The horizontal surface area As (m2) of a lake at a partic- ular depth can be computed from volume by differentiation:

As (z) = dV ___ dz (z)

where V = volume (m3) and z = depth (m) as measured from the surface down to the bottom. The average concentration of a substance that varies with depth,

_ c (g/m3), can be com-

puted by integration:

_ c =

∫ 0 Z c(z)As (z) dz _____________ ∫ 0

Z As (z) dz

where Z = the total depth (m). Determine the average con- centration based on the following data:

z, m 0 4 8 12 16 V, 106 m3 9.8175 5.1051 1.9635 0.3927 0.0000 c, g/m3 10.2 8.5 7.4 5.2 4.1

21.26 Faraday’s law characterizes the voltage drop across an inductor as

VL = L di __ dt

where VL = voltage drop (V), L = inductance (in henrys; 1 H = 1 V · s/A), i = current (A), and t = time (s). Determine the voltage drop as a function of time from the following data for an inductance of 4 H.

t 0 0.1 0.2 0.3 0.5 0.7 i 0 0.16 0.32 0.56 0.84 2.0

21.27 Based on Faraday’s law (Prob. 21.26), use the follow- ing voltage data to estimate the inductance if a current of 2 A is passed through the inductor over 400 milliseconds.

t, ms 0 10 20 40 60 80 120 180 280 400 V, volts 0 18 29 44 49 46 35 26 15 7

21.28 The rate of cooling of a body (Fig. P21.28) can be expressed as

dT ___ dt = −k(T − Ta)

where T = temperature of the body (°C), Ta = temperature of the surrounding medium (°C), and k = a proportionality con- stant (per minute). Thus, this equation (called Newton’s law of cooling) specifies that the rate of cooling is proportional to

Use Newton’s viscosity law to determine the shear stress τ (N/m2) at the surface (y = 0),

τ = μ du ___ dy

Assume a value of dynamic viscosity μ = 1.8 × 10−5 N · s/m2.

y, m 0 0.002 0.006 0.012 0.018 0.024 u, m/s 0 0.287 0.899 1.915 3.048 4.299

21.22 Fick’s first diffusion law states that

Mass flux = −D dc ___ dx (P21.22)

where mass flux = the quantity of mass that passes across a unit area per unit time (g/cm2/s), D = a diffusion coef- ficient (cm2/s), c = concentration (g/cm3), and x = distance (cm). An environmental engineer measures the following con centration of a pollutant in the pore waters of sediments underlying a lake (x = 0 at the sediment-water interface and increases downward):

x, cm 0 1 3 c, 10−6 g/cm3 0.06 0.32 0.6

Use the best numerical differentiation technique available to estimate the derivative at x = 0. Employ this estimate in conjunction with Eq. (P21.22) to compute the mass flux of pollutant out of the sediments and into the overlying waters (D = 1.52 × 10−6 cm2/s). For a lake with 3.6 × 106 m2 of sediments, how much pollutant would be transported into the lake over a year’s time? 21.23 The following data were collected when a large oil tanker was loading:

t, min 0 10 20 30 45 60 75 V, 106 barrels 0.4 0.7 0.77 0.88 1.05 1.17 1.35

Calculate the flow rate Q (i.e., dV∕dt) for each time to the order of h2. 21.24 Fourier’s law is used routinely by architectural engi- neers to determine heat flow through walls. The following temperatures are measured from the surface (x = 0) into a stone wall:

x, m 0 0.08 0.16 T, °C 20.2 17 15

If the flux at x = 0 is 60 W/m2, compute k.

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made for air flowing over a flat plate where y = distance normal to the surface:

y, cm 0 1 3 5 T, K 900 480 270 210

If the plate’s dimensions are 200 cm long and 50 cm wide, and k = 0.028 J/(s · m · K), (a) determine the flux at the sur- face and (b) the heat transfer in watts. Note that 1 J = 1 W · s. 21.31 The pressure gradient for laminar flow through a con- stant radius tube is given by

dp ___ dx

= − 8μQ ____ πr4

where p = pressure (N/m2), x = distance along the tube’s centerline (m), μ = dynamic viscosity (N · s/m2), Q = flow (m3/s), and r = radius (m). (a) Determine the pressure drop for a 10-cm length tube

for a viscous liquid ( μ = 0.005 N · s/m2, density = ρ = 1 × 103 kg/m3) with a flow of 10 × 10−6 m3/s and the following varying radii along its length:

x, cm 0 2 4 5 6 7 10 r, mm 2 1.35 1.34 1.6 1.58 1.42 2

(b) Compare your result with the pressure drop that would have occurred if the tube had a constant radius equal to the average radius.

(c) Determine the average Reynolds number for the tube to verify that flow is truly laminar (Re = ρυD∕μ < 2100 where υ = velocity).

21.32 The following data for the specific heat of benzene were generated with an nth-order polynomial. Use numeri- cal differentiation to determine n.

T, K 300 400 500 600 Cp , kJ/(kmol · K) 82.888 112.136 136.933 157.744

T, K 700 800 900 1000 Cp, kJ/(kmol · K) 175.036 189.273 200.923 210.450

21.33 The specific heat at constant pressure cp [J/(kg · K)] of an ideal gas is related to enthalpy by

cp = dh ___ dT

where h = enthalpy (kJ/kg) and T = absolute tempera- ture (K). The following enthalpies are provided for carbon

the difference in the temperatures of the body and of the sur- rounding medium. If a metal ball heated to 80 °C is dropped into water that is held constant at Ta = 20 °C, the temperature of the ball changes, as in

Time, min 0 5 10 15 20 25 T, °C 80 44.5 30.0 24.1 21.7 20.7

Utilize numerical differentiation to determine dT∕dt at each value of time. Plot dT∕dt versus T − Ta and employ linear regression to evaluate k. 21.29 The enthalpy of a real gas is a function of pressure as described below. The data were taken for a real fluid. Esti- mate the enthalpy of the fluid at 400 K and 50 atm (evaluate the integral from 0.1 to 50 atm).

H = ∫ 0 P

( V − T ( ∂V ___ ∂T ) P ) dP

FIGURE P21.28

V, L

P, atm T = 350 K T = 400 K T = 450 K

0.1 220 250 282.5 5 4.1 4.7 5.23

10 2.2 2.5 2.7 20 1.35 1.49 1.55 25 1.1 1.2 1.24 30 0.90 0.99 1.03 40 0.68 0.75 0.78 45 0.61 0.675 0.7 50 0.54 0.6 0.62

21.30 For fluid flow over a surface, the heat flux to the surface can be computed with Fourier’s law: y = distance normal to the surface (m). The following measurements are

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PRoBlems 571

Based on the following data and H = 0.1 m, use nu- merical differentiation to generate plots of (a) SOD versus time and (b) SOD versus oxygen concentration:

t, d 0 0.125 0.25 0.375 0.5 0.625 0.75 o, mg/L 10 7.11 4.59 2.57 1.15 0.33 0.03

21.36 The following relationships can be used to analyze uniform beams subject to distributed loads:

dy ___ dx \\\ = θ(x) dθ ___ dx =

M(x) _____ E I dM ___ dx = V(x)

dV ___ dx = −𝑤(x)

where x = distance along beam (m), y = deflection (m), θ(x) = slope (m/m), E = modulus of elasticity (Pa = N/m2), I = moment of inertia (m4), M(x) = moment (N m), V(x) = shear (N), and w(x) = distributed load (N/m). For the case of a linearly increasing load (recall Fig. P5.13), the slope can be computed analytically as

θ(x) = 𝑤0 ________ 120E I L (−5x

4 + 6L2x2 − L4) (P21.36)

Employ (a) numerical integration to compute the deflec- tion (in m) and (b) numerical differentiation to compute the moment (in N m) and shear (in N). Base your nu- merical calculations on values of the slope computed with Eq. (P21.36) at equally spaced intervals of Δx = 0.125 m along a 3-m beam. Use the following parameter values in your computation: E = 200 GPa, I = 0.0003 m4, and w0 = 2.5 kN/cm. In addition, the deflections at the ends of the beam are set at y(0) = y(L) = 0. Be careful of units. 21.37 You measure the following deflections along the length of a simply supported uniform beam (see Prob. 21.36)

dioxide (CO2) at several temperatures. Use these values to determine the specific heat in J/(kg · K) for each of the tabu- lated temperatures. Note that the atomic weights of carbon and oxygen are 12.011 and 15.9994 g/mol, respectively

T, K 750 800 900 1000 h, kJ/kmol 29,629 32,179 37,405 42,769

21.34 An nth-order rate law is often used to model chemical reactions that solely depend on the concentration of a single reactant:

dc ___ dt = −kc n

where c = concentration (mole), t = time (min), n = reaction order (dimensionless), and k = reaction rate (min−1 mole1−n). The differential method can be used to evaluate the parameters k and n. This involves applying a logarithmic transform to the rate law to yield,

log ( − dc ___ dt ) = log k + n log c Therefore, if the nth-order rate law holds, a plot of the log(−dc∕dt) versus log c should yield a straight line with a slope of n and an intercept of log k. Use the differential method and linear regression to determine k and n for the following data for the conversion of ammonium cyanate to urea:

t, min 0 5 15 30 45 c, mole 0.750 0.594 0.420 0.291 0.223

21.35 The sediment oxygen demand [SOD in units of g/(m2 · d)] is an important parameter in determining the dissolved oxygen content of a natural water. It is mea- sured by placing a sediment core in a cylindrical container (Fig. P21.35). After carefully introducing a layer of distilled, oxygenated water above the sediments, the container is cov- ered to prevent gas transfer. A stirrer is used to mix the water gently, and an oxygen probe tracks how the water’s oxygen concentration decreases over time. The SOD can then be computed as

SOD = −H do ___ dt

where H = the depth of water (m), o = oxygen concentration (g/m3), and t = time (d).

Lid

Water

Probe

Sediments

SOD H

FIGURE P21.35

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21.39 Develop a script to generate the same computations and plots as in Sec. 21.8, but for the following functions (for x = –3 to 3 and y = –3 to 3): (a) f (x, y) = e−( x 2 + y 2 ) and (b) f (x, y) = xe−( x 2 + y 2 ). 21.40 Develop a script to generate the same computa- tions and plots as in Sec. 21.8, but for the MATLAB peaks function over ranges of both x and y from –3 to 3. 21.41 The velocity (m/s) of an object at time t seconds is given by

υ = 2t _______ √

_____ 1 + t2

Using Richardson’s extrapolation, find the acceleration of the particle at time t = 5 s using h = 0.5 and 0.25. Employ the exact solution to compute the true percent relative error of each estimate.

x, m 0 0.375 0.75 1.125 1.5 y, cm 0 −0.2571 −0.9484 −1.9689 −3.2262

x, m 1.875 2.25 2.625 3 y, cm −4.6414 −6.1503 −7.7051 −9.275

Employ numerical differentiation to compute the slope, the moment (in N m), the shear (in N), and the distributed load (in N/m). Use the following parameter values in your com- putation: E = 200 GPa and I = 0.0003 m4. 21.38 Evaluate ∂f∕∂x, ∂f∕∂y and ∂2f∕(∂x∂y) for the follow- ing function at x = y = 1 (a) analytically and (b) numerically Δx = Δy = 0.0001:

f (x, y) = 3xy + 3x − x3 − 3y3

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573

Part Six

Ordinary Differential Equations

6.1 OVERVIEW

The fundamental laws of physics, mechanics, electricity, and thermodynamics are usually based on empirical observations that explain variations in physical properties and states of systems. Rather than describing the state of physical systems directly, the laws are usu- ally couched in terms of spatial and temporal changes. These laws define mechanisms of change. When combined with continuity laws for energy, mass, or momentum, dif- ferential equations result. Subsequent integration of these differential equations results in mathematical functions that describe the spatial and temporal state of a system in terms of energy, mass, or velocity variations. As in Fig. PT6.1, the integration can be implemented analytically with calculus or numerically with the computer.

The free-falling bungee jumper problem introduced in Chap. 1 is an example of the derivation of a differential equation from a fundamental law. Recall that Newton’s second

law was used to develop an ODE describ- ing the rate of change of velocity of a fall- ing bungee jumper:

dυ ___ dt = g − cd __ m υ

2 (PT6.1)

where g is the gravitational constant, m is the mass, and cd is a drag coefficient. Such equations, which are composed of an unknown function and its derivatives, are called differential equations. They are sometimes referred to as rate equations because they express the rate of change of a variable as a function of variables and parameters.

In Eq. (PT6.1), the quantity being differentiated υ is called the dependent variable. The quantity with respect to which υ is differentiated t is called the independent variable. When the function

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574 PART 6 ORDinARy DiffEREnTiAl EquATiOns

involves one independent variable, the equation is called an ordinary differential equation (or ODE). This is in contrast to a partial differential equation (or PDE) that involves two or more independent variables.

Differential equations are also classified as to their order. For example, Eq. (PT6.1) is called a first-order equation because the highest derivative is a first derivative. A second- order equation would include a second derivative. For example, the equation describing the position x of an unforced mass-spring system with damping is the second-order equation:

m d 2x ___

dt 2 + c dx ___ dt + kx = 0 (PT6.2)

where m is mass, c is a damping coefficient, and k is a spring constant. Similarly, an nth- order equation would include an nth derivative.

Higher-order differential equations can be reduced to a system of first-order equations. This is accomplished by defining the first derivative of the dependent variable as a new variable. For Eq. (PT6.2), this is done by creating a new variable υ as the first derivative of displacement

υ = dx ___ dt (PT6.3)

where υ is velocity. This equation can itself be differentiated to yield

dυ ___ dt = d 2x ___ dt 2

(PT6.4)

Equations (PT6.3) and (PT6.4) can be substituted into Eq. (PT6.2) to convert it into a first- order equation:

m dυ ___ dt + cυ + kx = 0 (PT6.5)

Physical law F = ma

ODE dυ dt

= g − cd m

υ2

Analytical Numerical (calculus) (computer)

Solution υ = gm cd

tanh gcd m

t υi+1 = υi + g − cd m

υ2 ∆t

FIGURE PT6.1 The sequence of events in the development and solution of ODEs for engineering and science. The example shown is for the velocity of the free-falling bungee jumper.

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6.1 OVERViEW 575

As a final step, we can express Eqs. (PT6.3) and (PT6.5) as rate equations:

dx ___ dt = υ (PT6.6)

dυ ___ dt = − c __ m υ −

k __ m x (PT6.7)

Thus, Eqs. (PT6.6) and (PT6.7) are a pair of first-order equations that are equivalent to the original second-order equation [Eq. (PT6.2)]. Because other nth-order differential equations can be similarly reduced, this part of our book focuses on the solution of first- order equations.

A solution of an ordinary differential equation is a specific function of the independent variable and parameters that satisfies the original differential equation. To illustrate this concept, let us start with a simple fourth-order polynomial,

y = − 0.5x4 + 4x3 − 10x2 + 8.5x + 1 (PT6.8)

Now, if we differentiate Eq. (PT6.8), we obtain an ODE:

dy ___ dx

= −2x3 + 12x2 − 20x + 8.5 (PT6.9)

This equation also describes the behavior of the polynomial, but in a manner different from Eq. (PT6.8). Rather than explicitly representing the values of y for each value of x, Eq. (PT6.9) gives the rate of change of y with respect to x (i.e., the slope) at every value of x. Figure PT6.2 shows both the function and the derivative plotted versus x. Notice how the zero values of the derivatives correspond to the point at which the original function is flat—that is, where it has a zero slope. Also, the maximum absolute values of the deriva- tives are at the ends of the interval where the slopes of the function are greatest.

Although, as just demonstrated, we can determine a differential equation given the original function, the object here is to determine the original function given the differential equation. The original function then represents the solution.

Without computers, ODEs are usually solved analytically with calculus. For example, Eq. (PT6.9) could be multiplied by dx and integrated to yield

y = ʃ(−2x 3 + 12x2 − 20x + 8.5) dx (PT6.10) The right-hand side of this equation is called an indefinite integral because the limits of integration are unspecified. This is in contrast to the definite integrals discussed previously in Part Five [compare Eq. (PT6.10) with Eq. (19.5)].

An analytical solution for Eq. (PT6.10) is obtained if the indefinite integral can be eval- uated exactly in equation form. For this simple case, it is possible to do this with the result:

y = − 0.5x4 + 4x3 − 10x2 + 8.5x + C (PT6.11)

which is identical to the original function with one notable exception. In the course of dif- ferentiating and then integrating, we lost the constant value of 1 in the original equation and gained the value C. This C is called a constant of integration. The fact that such an arbitrary constant appears indicates that the solution is not unique. In fact, it is but one of an infinite

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576 PART 6 ORDinARy DiffEREnTiAl EquATiOns

number of possible functions (corresponding to an infinite number of possible values of C) that satisfy the differential equation. For example, Fig. PT6.3 shows six possible functions that satisfy Eq. (PT6.11).

Therefore, to specify the solution completely, a differential equation is usually ac- companied by auxiliary conditions. For first-order ODEs, a type of auxiliary condition called an initial value is required to determine the constant and obtain a unique solution. For example, the original differential equation could be accompanied by the initial condi- tion that at x = 0, y = 1. These values could be substituted into Eq. (PT6.11) to determine C = 1. Therefore, the unique solution that satisfies both the differential equation and the specified initial condition is

y = − 0.5x4 + 4x3 − 10x2 + 8.5x + 1

Thus, we have “pinned down” Eq. (PT6.11) by forcing it to pass through the initial condi- tion, and in so doing, we have developed a unique solution to the ODE and have come full circle to the original function [Eq. (PT6.8)].

Initial conditions usually have very tangible interpretations for differential equations derived from physical problem settings. For example, in the bungee jumper problem, the

(a)

dy/dx 8

(b)

−8

FIGURE PT6.2 Plots of (a) y versus x and (b) dy∕dx versus x for the function y = −0.5x4 + 4x3 − 10x2 + 8.5x + 1.

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6.2 PART ORGAniZATiOn 577

initial condition was reflective of the physical fact that at time zero the vertical velocity was zero. If the bungee jumper had already been in vertical motion at time zero, the solution would have been modified to account for this initial velocity.

When dealing with an nth-order differential equation, n conditions are required to ob- tain a unique solution. If all conditions are specified at the same value of the independent variable (e.g., at x or t = 0), then the problem is called an initial-value problem. This is in contrast to boundary-value problems where specification of conditions occurs at different values of the independent variable. Chapters 22 and 23 will focus on initial-value prob- lems. Boundary-value problems are covered in Chap. 24.

6.2 PART ORGANIZATION

Chapter 22 is devoted to one-step methods for solving initial-value ODEs. As the name suggests, one-step methods compute a future prediction yi+1, based only on information at a single point yi and no other previous information. This is in contrast to multistep ap- proaches that use information from several previous points as the basis for extrapolating to a new value.

With all but a minor exception, the one-step methods presented in Chap. 22 belong to what are called Runge-Kutta techniques. Although the chapter might have been organized around this theoretical notion, we have opted for a more graphical, intuitive approach to introduce the methods. Thus, we begin the chapter with Euler’s method, which has a very straightforward graphical interpretation. In addition, because we have already introduced Euler’s method in Chap. 1, our emphasis here is on quantifying its truncation error and describing its stability.

FIGURE PT6.3 Six possible solutions for the integral of −2x3 + 12x2 − 20x + 8.5. Each conforms to a different value of the constant of integration C.

x C = 0

C = −1

C = −2

C = 3

C = 2

C = 1

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578 PART 6 ORDinARy DiffEREnTiAl EquATiOns

Next, we use visually oriented arguments to develop two improved versions of Euler’s method—the Heun and the midpoint techniques. After this introduction, we formally de- velop the concept of Runge-Kutta (or RK) approaches and demonstrate how the forego- ing techniques are actually first- and second-order RK methods. This is followed by a discussion of the higher-order RK formulations that are frequently used for engineering and scientific problem solving. In addition, we cover the application of one-step methods to systems of ODEs. Note that all the applications in Chap. 22 are limited to cases with a fixed step size.

In Chap. 23, we cover more advanced approaches for solving initial-value problems. First, we describe adaptive RK methods that automatically adjust the step size in response to the truncation error of the computation. These methods are especially pertinent as they are employed by MATLAB to solve ODEs.

Next, we discuss multistep methods. As mentioned above, these algorithms retain in- formation of previous steps to more effectively capture the trajectory of the solution. They also yield the truncation error estimates that can be used to implement step-size control. We describe a simple method—the non-self-starting Heun method—to introduce the essential features of the multistep approaches.

Finally, the chapter ends with a description of stiff ODEs. These are both individual and systems of ODEs that have both fast and slow components to their solution. As a con- sequence, they require special solution approaches. We introduce the idea of an implicit solution technique as one commonly used remedy. We also describe MATLAB’s built-in functions for solving stiff ODEs.

In Chap. 24, we focus on two approaches for obtaining solutions to boundary-value problems: the shooting and finite-difference methods. Aside from demonstrating how these techniques are implemented, we illustrate how they handle derivative boundary conditions and nonlinear ODEs.

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579

initial-Value Problems

CHAPTER OBJECTIVES

The primary objective of this chapter is to introduce you to solving initial-value problems for ODEs (ordinary differential equations). Specific objectives and topics covered are

• Understanding the meaning of local and global truncation errors and their relationship to step size for one-step methods for solving ODEs.

• Knowing how to implement the following Runge-Kutta (RK) methods for a single ODE:

Euler Heun Midpoint Fourth-order RK

• Knowing how to iterate the corrector of Heun’s method. • Knowing how to implement the following Runge-Kutta methods for systems

of ODEs: Euler Fourth-order RK

YOU’VE GOT A PROBLEM

We started this book with the problem of simulating the velocity of a free-falling bungee jumper. This problem amounted to formulating and solving an ordinary differential equation, the topic of this chapter. Now let’s return to this problem and make it more interesting by computing what happens when the jumper reaches the end of the bungee cord.

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580 iniTiAl-VAluE PROblEms

To do this, we should recognize that the jumper will experience different forces de- pending on whether the cord is slack or stretched. If it is slack, the situation is that of free fall where the only forces are gravity and drag. However, because the jumper can now move up as well as down, the sign of the drag force must be modified so that it always tends to retard velocity,

dυ ___ dt = g − sign (υ) cd __ m υ

2 (22.1a)

where υ is velocity (m/s), t is time (s), g is the acceleration due to gravity (9.81 m/s2), cd is the drag coefficient (kg/m), and m is mass (kg). The signum function,1 sign, returns a −1 or a 1 depending on whether its argument is negative or positive, respectively. Thus, when the jumper is falling downward (positive velocity, sign = 1), the drag force will be nega- tive and hence will act to reduce velocity. In contrast, when the jumper is moving upward (negative velocity, sign = −1), the drag force will be positive so that it again reduces the velocity. Once the cord begins to stretch, it obviously exerts an upward force on the jumper. As done previously in Chap. 8, Hooke’s law can be used as a first approximation of this force. In addition, a dampening force should also be included to account for frictional effects as the cord stretches and contracts. These factors can be incorporated along with gravity and drag into a second force balance that applies when the cord is stretched. The result is the following differential equation:

dυ ___ dt = g − sign(υ) cd __ m υ

2 − k __ m (x − L) − γ __ m υ (22.1b)

where k is the cord’s spring constant (N/m), x is vertical distance measured downward from the bungee jump platform (m), L is the length of the unstretched cord (m), and γ is a dampening coefficient (N · s/m). Because Eq. (22.1b) only holds when the cord is stretched (x > L), the spring force will always be negative. That is, it will always act to pull the jumper back up. The dampening force increases in magnitude as the jumper’s velocity increases and always acts to slow the jumper down. If we want to simulate the jumper’s velocity, we would initially solve Eq. (22.1a) until the cord was fully extended. Then, we could switch to Eq. (22.1b) for periods that the cord is stretched. Although this is fairly straightforward, it means that knowledge of the jumper’s position is required. This can be done by formulating another differential equa- tion for distance:

dx ___ dt = υ (22.2)

Thus, solving for the bungee jumper’s velocity amounts to solving two ordinary dif- ferential equations where one of the equations takes different forms depending on the value

1 Some computer languages represent the signum function as sgn(x) . As represented here, MATLAB uses the nomenclature sign(x).

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22.2 EulER’s mETHOD 581

of one of the dependent variables. Chapters 22 and 23 explore methods for solving this and similar problems involving ODEs.

22.1 OVERVIEW

This chapter is devoted to solving ordinary differential equations of the form

dy ___ dt = f (t, y) (22.3)

In Chap. 1, we developed a numerical method to solve such an equation for the velocity of the free-falling bungee jumper. Recall that the method was of the general form

New value = old value + slope × step size

or, in mathematical terms,

yi+1 = yi + ϕh (22.4)

where the slope ϕ is called an increment function. According to this equation, the slope estimate of ϕ is used to extrapolate from an old value yi to a new value yi+1 over a distance h. This formula can be applied step by step to trace out the trajectory of the solution into the future. Such approaches are called one-step methods because the value of the increment function is based on information at a single point i. They are also referred to as Runge- Kutta methods after the two applied mathematicians who first discussed them in the early 1900s. Another class of methods called multistep methods use information from several previous points as the basis for extrapolating to a new value. We will describe multistep methods briefly in Chap. 23.

All one-step methods can be expressed in the general form of Eq. (22.4), with the only difference being the manner in which the slope is estimated. The simplest approach is to use the differential equation to estimate the slope in the form of the first derivative at ti. In other words, the slope at the beginning of the interval is taken as an approximation of the average slope over the whole interval. This approach, called Euler’s method, is discussed next. This is followed by other one-step methods that employ alternative slope estimates that result in more accurate predictions.

22.2 EULER’S METHOD

The first derivative provides a direct estimate of the slope at ti (Fig. 22.1):

ϕ = f (ti, yi )

where f (ti , yi ) is the differential equation evaluated at ti and yi. This estimate can be sub- stituted into Eq. (22.1):

yi+1 = yi + f (ti , yi ) h (22.5)

This formula is referred to as Euler’s method (or the Euler-Cauchy or point-slope method). A new value of y is predicted using the slope (equal to the first derivative at the original value of t) to extrapolate linearly over the step size h (Fig. 22.1).

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EXAmPlE 22.1 Euler’s method

Problem statement. Use Euler’s method to integrate y′ = 4e0.8t − 0.5y from t = 0 to 4 with a step size of 1. The initial condition at t = 0 is y = 2. Note that the exact solution can be determined analytically as

y = 4 ___ 1.3 (e 0.8t − e−0.5t ) + 2e−0.5t

solution. Equation (22.5) can be used to implement Euler’s method:

y(1) = y(0) + f (0, 2)(1)

where y(0) = 2 and the slope estimate at t = 0 is

f (0, 2) = 4e0 − 0.5(2) = 3

Therefore,

y(1) = 2 + 3(1) = 5

The true solution at t = 1 is

y = 4 ___ 1.3 (e 0.8(1) − e−0.5(1)) + 2e−0.5(1) = 6.19463

Thus, the percent relative error is

εt = | 6.19463 − 5 ___________ 6.19463 | × 100% = 19.28% For the second step:

y(2) = y(1) + f (1, 5)(1)

= 5 + [4e0.8(1) − 0.5(5)] (1) = 11.40216

tti ti+1

Predicted

error

True

FIGURE 22.1 Euler’s method.

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22.2 EulER’s mETHOD 583

The true solution at t = 2.0 is 14.84392 and, therefore, the true percent relative error is 23.19%. The computation is repeated, and the results compiled in Table 22.1 and Fig. 22.2. Note that although the computation captures the general trend of the true solution, the error is considerable. As discussed in the next section, this error can be reduced by using a smaller step size.

22.2.1 Error Analysis for Euler’s Method

The numerical solution of ODEs involves two types of error (recall Chap. 4): 1. Truncation, or discretization, errors caused by the nature of the techniques employed

to approximate values of y. 2. Roundoff errors caused by the limited numbers of significant digits that can be retained

by a computer. The truncation errors are composed of two parts. The first is a local truncation error

that results from an application of the method in question over a single step. The second is a propagated truncation error that results from the approximations produced during the

TABLE 22.1 Comparison of true and numerical values of the integral of y′ = 4e0.8t − 0.5y, with the initial condition that y = 2 at t = 0. The numerical values were computed using Euler’s method with a step size of 1.

t ytrue yEuler |εt| (%)

0 2.00000 2.00000 1 6.19463 5.00000 19.28 2 14.84392 11.40216 23.19 3 33.67717 25.51321 24.24 4 75.33896 56.84931 24.54

0 0 1 2 3 4

Euler solution

True solution

FIGURE 22.2 Comparison of the true solution with a numerical solution using Euler’s method for the integral of y′ = 4e0.8t − 0.5y from t = 0 to 4 with a step size of 1.0. The initial condition at t = 0 is y = 2.

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584 iniTiAl-VAluE PROblEms

previous steps. The sum of the two is the total error. It is referred to as the global trunca- tion error.

Insight into the magnitude and properties of the truncation error can be gained by deriving Euler’s method directly from the Taylor series expansion. To do this, realize that the differential equation being integrated will be of the general form of Eq. (22.3), where dy∕dt = y′, and t and y are the independent and the dependent variables, respectively. If the solution—that is, the function describing the behavior of y—has continuous derivatives, it can be represented by a Taylor series expansion about a starting value (ti , yi ), as in [recall Eq. (4.13)]:

yi+1 = yi + y i ′ h + y i ″ __ 2! h

2 + · · · + y i

(n) ___ n! h

n + Rn (22.6)

where h = ti+1 − ti and Rn = the remainder term, defined as

Rn = y(n+1)(ξ ) _________ (n + 1)! h

n+1 (22.7)

where ξ lies somewhere in the interval from ti to ti+1. An alternative form can be developed by substituting Eq. (22.3) into Eqs. (22.6) and (22.7) to yield

yi+1 = yi + f (ti , yi )h + f ′(ti , yi ) _______ 2! h

2 + · · · + f (n−1)(ti , yi ) __________ n! h

n + O(hn+1) (22.8)

where O(hn+1) specifies that the local truncation error is proportional to the step size raised to the (n + 1)th power.

By comparing Eqs. (22.5) and (22.8), it can be seen that Euler’s method corresponds to the Taylor series up to and including the term f (ti , yi )h. Additionally, the comparison indi- cates that a truncation error occurs because we approximate the true solution using a finite number of terms from the Taylor series. We thus truncate, or leave out, a part of the true solution. For example, the truncation error in Euler’s method is attributable to the remain- ing terms in the Taylor series expansion that were not included in Eq. (22.5). Subtracting Eq. (22.5) from Eq. (22.8) yields

Et = f ′(ti , yi ) _______ 2! h

2 + · · · + O(hn+1) (22.9)

where Et = the true local truncation error. For sufficiently small h, the higher-order terms in Eq. (22.9) are usually negligible, and the result is often represented as

Ea = f ′(ti , yi ) _______ 2! h

2 (22.10)

or Ea = O(h2) (22.11)

where Ea = the approximate local truncation error. According to Eq. (22.11), we see that the local error is proportional to the square of

the step size and the first derivative of the differential equation. It can also be demon- strated that the global truncation error is O(h)—that is, it is proportional to the step size

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22.2 EulER’s mETHOD 585

(Carnahan et al., 1969). These observations lead to some useful conclusions:

1. The global error can be reduced by decreasing the step size. 2. The method will provide error-free predictions if the underlying function (i.e., the

solution of the differential equation) is linear, because for a straight line the second derivative would be zero.

This latter conclusion makes intuitive sense because Euler’s method uses straight-line seg- ments to approximate the solution. Hence, Euler’s method is referred to as a first-order method.

It should also be noted that this general pattern holds for the higher-order one-step methods described in the following pages. That is, an nth-order method will yield perfect results if the underlying solution is an nth-order polynomial. Further, the local truncation error will be O(hn+1) and the global error O(hn).

22.2.2 Stability of Euler’s Method

In the preceding section, we learned that the truncation error of Euler’s method depends on the step size in a predictable way based on the Taylor series. This is an accuracy issue.

The stability of a solution method is another important consideration that must be considered when solving ODEs. A numerical solution is said to be unstable if errors grow exponentially for a problem for which there is a bounded solution. The stability of a par- ticular application can depend on three factors: the differential equation, the numerical method, and the step size.

Insight into the step size required for stability can be examined by studying a very simple ODE:

dy ___ dt = −ay (22.12)

If y(0) = y0, calculus can be used to determine the solution as

y = y0e−at

Thus, the solution starts at y0 and asymptotically approaches zero. Now suppose that we use Euler’s method to solve the same problem numerically:

yi+1 = yi + dyi ___ dt h

Substituting Eq. (22.12) gives

yi+1 = yi − ayi h or

yi+1 = yi (1 − ah) (22.13)

The parenthetical quantity 1 − ah is called an amplification factor. If its absolute value is greater than unity, the solution will grow in an unbounded fashion. So clearly, the stability depends on the step size h. That is, if h > 2∕a, |yi| → ∞ as i → ∞. Based on this analysis, Euler’s method is said to be conditionally stable.

Note that there are certain ODEs where errors always grow regardless of the method. Such ODEs are called ill-conditioned.

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Inaccuracy and instability are often confused. This is probably because (a) both repre- sent situations where the numerical solution breaks down and (b) both are affected by step size. However, they are distinct problems. For example, an inaccurate method can be very stable. We will return to the topic when we discuss stiff systems in Chap. 23.

22.2.3 MATLAB M-file Function: eulode

We have already developed a simple M-file to implement Euler’s method for the falling bungee jumper problem in Chap. 3. Recall from Sec. 3.6, that this function used Euler’s method to compute the velocity after a given time of free fall. Now, let’s develop a more general, all-purpose algorithm.

Figure 22.3 shows an M-file that uses Euler’s method to compute values of the depen- dent variable y over a range of values of the independent variable t. The name of the function holding the right-hand side of the differential equation is passed into the function as the

FIGURE 22.3 An M-file to implement Euler’s method.

function [t,y] = eulode (dydt,tspan,y0,h,varargin) % eulode: Euler ODE solver % [t,y] = eulode (dydt,tspan,y0,h,p1,p2,...): % uses Euler's method to integrate an ODE % input: % dydt = name of the M−file that evaluates the ODE % tspan = [ti, tf] where ti and tf = initial and % final values of independent variable % y0 = initial value of dependent variable % h = step size % p1,p2,... = additional parameters used by dydt % output: % t = vector of independent variable % y = vector of solution for dependent variable

if nargin < 4,error('at least 4 input arguments required'),end ti = tspan(1);tf = tspan(2); if ~(tf > ti),error('upper limit must be greater than lower'),end t = (ti:h:tf)'; n = length(t); % if necessary, add an additional value of t % so that range goes from t = ti to tf if t(n)<tf t(n + 1) = tf; n = n + 1; end y = y0*ones(n,1); %preallocate y to improve efficiency for i = 1:n − 1 %implement Euler's method y(i + 1) = y(i) + dydt(t(i),y(i),varargin{:})*(t(i + 1) − t(i)); end

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22.3 imPROVEmEnTs Of EulER’s mETHOD 587

variable dydt. The initial and final values of the desired range of the independent variable is passed as a vector tspan. The initial value and the desired step size are passed as y0 and h, respectively.

The function first generates a vector t over the desired range of the dependent variable using an increment of h. In the event that the step size is not evenly divisible into the range, the last value will fall short of the final value of the range. If this occurs, the final value is added to t so that the series spans the complete range. The length of the t vector is deter- mined as n. In addition, a vector of the dependent variable y is preallocated with n values of the initial condition to improve efficiency.

At this point, Euler’s method [Eq. (22.5)] is implemented by a simple loop:

for i = 1:n − 1 y(i + 1) = y(i) + dydt(t(i),y(i),varargin {:})*(t(i + 1) − t(i)); end

Notice how a function is used to generate a value for the derivative at the appropriate val- ues of the independent and dependent variables. Also notice how the time step is automati- cally calculated based on the difference between adjacent values in the vector t.

The ODE being solved can be set up in several ways. First, the differential equation can be defined as an anonymous function object. For example, for the ODE from Example 22.1:

>> dydt = @ (t,y) 4*exp(0.8*t) − 0.5*y;

The solution can then be generated as

>> [t,y] = eulode(dydt,[0 4],2,1); >> disp([t,y])

with the result (compare with Table 22.1):

0 2.0000 1.0000 5.0000 2.0000 11.4022 3.0000 25.5132 4.0000 56.8493

Although using an anonymous function is feasible for the present case, there will be more complex problems where the definition of the ODE requires several lines of code. In such instances, creating a separate M-file is the only option.

22.3 IMPROVEMENTS OF EULER’S METHOD

A fundamental source of error in Euler’s method is that the derivative at the beginning of the interval is assumed to apply across the entire interval. Two simple modifications are available to help circumvent this shortcoming. As will be demonstrated in Sec. 22.4, both modifications (as well as Euler’s method itself ) actually belong to a larger class of solu- tion techniques called Runge-Kutta methods. However, because they have very straight- forward graphical interpretations, we will present them prior to their formal derivation as Runge-Kutta methods.

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22.3.1 Heun’s Method

One method to improve the estimate of the slope involves the determination of two deriva- tives for the interval—one at the beginning and another at the end. The two derivatives are then averaged to obtain an improved estimate of the slope for the entire interval. This approach, called Heun’s method, is depicted graphically in Fig. 22.4.

Recall that in Euler’s method, the slope at the beginning of an interval

y i ′ = f (ti , yi ) (22.14)

is used to extrapolate linearly to yi+1:

y i+1 0 = yi + f (ti , yi )h (22.15)

For the standard Euler method we would stop at this point. However, in Heun’s method the y i+1

0 calculated in Eq. (22.15) is not the final answer, but an intermediate prediction. This is why we have distinguished it with a superscript 0. Equation (22.15) is called a predic- tor equation. It provides an estimate that allows the calculation of a slope at the end of the interval:

y i+1 ′ = f (ti+1 , y i+1 0 ) (22.16) Thus, the two slopes [Eqs. (22.14) and (22.16)] can be combined to obtain an average slope for the interval:

ȳ ′ = f (ti , yi ) + f (ti+1 , y i+1 0 ) __________________ 2

This average slope is then used to extrapolate linearly from yi to yi+1 using Euler’s method:

yi+1 = yi + f (ti , yi ) + f (ti+1 , y i+1 0 ) __________________ 2 h (22.17)

which is called a corrector equation.

tti ti+1

Slope = f (ti+1, y i0+1)

Slope = f (ti, y i)

tti ti+1

Slope = f (ti, yi) + f (ti+1, y i0+1)

(a) (b)

FIGURE 22.4 Graphical depiction of Heun’s method. (a) Predictor and (b) corrector.

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22.3 imPROVEmEnTs Of EulER’s mETHOD 589

The Heun method is a predictor-corrector approach. As just derived, it can be ex- pressed concisely as

Predictor (Fig. 22.4a): y i+1 0 = y i

m + f (ti , yi )h (22.18)

Corrector (Fig. 22.4b): y i+1 j = y i

m + f (ti , y i

m ) + f (ti+1 , y i+1 j−1 ) __________________ 2 h (22.19) (for j = 1, 2, . . . , m)

Note that because Eq. (22.19) has yi+1 on both sides of the equal sign, it can be applied in an iterative fashion as indicated. That is, an old estimate can be used repeatedly to provide an improved estimate of yi+1. The process is depicted in Fig. 22.5.

As with similar iterative methods discussed in previous sections of the book, a termi- nation criterion for convergence of the corrector is provided by

|εa| = | y i+1 j − y i+1 j−1 _________ y i+1 j | × 100% where y i+1

j−1 and y i+1 j are the result from the prior and the present iteration of the corrector,

respectively. It should be understood that the iterative process does not necessarily con- verge on the true answer but will converge on an estimate with a finite truncation error, as demonstrated in the following example.

EXAmPlE 22.2 Heun’s method

Problem statement. Use Heun’s method with iteration to integrate y′ = 4e0.8t − 0.5y from t = 0 to 4 with a step size of 1. The initial condition at t = 0 is y = 2. Employ a stopping criterion of 0.00001% to terminate the corrector iterations.

solution. First, the slope at (t0, y0 ) is calculated as y 0 ′ = 4e

0 − 0.5(2) = 3 Then, the predictor is used to compute a value at 1.0:

y 1 0 = 2 + 3(1) = 5

f(ti, y im) + f (ti+1, y i+1)+y ij+1 yi m

2 h

FIGURE 22.5 Graphical representation of iterating the corrector of Heun’s method to obtain an improved estimate.

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Note that this is the result that would be obtained by the standard Euler method. The true value in Table 22.2 shows that it corresponds to a percent relative error of 19.28%. Now, to improve the estimate for yi+1 we use the value y 1

0 to predict the slope at the end of the interval

y 1 ′ = f (x1, y 1 0 ) = 4e0.8(1) − 0.5(5) = 6.402164 which can be combined with the initial slope to yield an average slope over the interval from t = 0 to 1:

ȳ ′ = 3 + 6.402164 ____________ 2 = 4.701082

This result can then be substituted into the corrector [Eq. (22.19)] to give the prediction at t = 1:

y 1 1 = 2 + 4.701082(1) = 6.701082

which represents a true percent relative error of −8.18%. Thus, the Heun method without iteration of the corrector reduces the absolute value of the error by a factor of about 2.4 as compared with Euler’s method. At this point, we can also compute an approximate error as

|εa| = | 6.701082 − 5 ____________ 6.701082 | × 100% = 25.39% Now the estimate of y1 can be refined by substituting the new result back into the right- hand side of Eq. (22.19) to give

y 1 2 = 2 + 3 + 4e

0.8(1) − 0.5(6.701082) _______________________ 2 1 = 6.275811

which represents a true percent relative error of 1.31 percent and an approximate error of

|εa| = | 6.275811 − 6.701082 __________________ 6.275811 | × 100% = 6.776%

TABLE 22.2 Comparison of true and numerical values of the integral of y′ = 4e0.8t − 0.5y, with the initial condition that y = 2 at t = 0. The numerical values were computed using the Euler and Heun methods with a step size of 1. The Heun method was implemented both without and with iteration of the corrector.

Without Iteration With Iteration

t ytrue yEuler |εt| (%) yHeun |εt| (%) yHeun |εt| (%)

0 2.00000 2.00000 2.00000 2.00000 1 6.19463 5.00000 19.28 6.70108 8.18 6.36087 2.68 2 14.84392 11.40216 23.19 16.31978 9.94 15.30224 3.09 3 33.67717 25.51321 24.24 37.19925 10.46 34.74328 3.17 4 75.33896 56.84931 24.54 83.33777 10.62 77.73510 3.18

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The next iteration gives

y 1 2 = 2 + 3 + 4e

0.8(1) − 0.5(6.275811) _______________________ 2 1 = 6.382129

which represents a true error of 3.03% and an approximate error of 1.666%. The approximate error will keep dropping as the iterative process converges on a sta- ble final result. In this example, after 12 iterations the approximate error falls below the stopping criterion. At this point, the result at t = 1 is 6.36087, which represents a true rela- tive error of 2.68%. Table 22.2 shows results for the remainder of the computation along with results for Euler’s method and for the Heun method without iteration of the corrector.

Insight into the local error of the Heun method can be gained by recognizing that it is related to the trapezoidal rule. In the previous example, the derivative is a function of both the dependent variable y and the independent variable t. For cases such as polynomi- als, where the ODE is solely a function of the independent variable, the predictor step [Eq. (22.18)] is not required and the corrector is applied only once for each iteration. For such cases, the technique is expressed concisely as

yi+1 = yi + f (ti ) + f (ti+1) ___________ 2 h (22.20)

Notice the similarity between the second term on the right-hand side of Eq. (22.20) and the trapezoidal rule [Eq. (19.11)]. The connection between the two methods can be formally demonstrated by starting with the ordinary differential equation

dy ___ dt = f (t) (22.21)

This equation can be solved for y by integration:

∫ yi yi+1

dy = ∫ ti ti+1

f (t) dt (22.22)

which yields

yi+1 − yi = ∫ ti ti+1

f (t) dt (22.23)

yi+1 = yi + ∫ ti ti+1

f (t) dt (22.24)

Now, recall that the trapezoidal rule [Eq. (19.11)] is defined as

∫ ti ti+1

f (t) dt = f (ti ) + f (ti+1) _____________ 2 h (22.25)

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where h = ti+1 − ti. Substituting Eq. (22.25) into Eq. (22.24) yields

yi+1 = yi + f (ti ) + f (ti+1) ___________ 2 h (22.26)

which is equivalent to Eq. (22.20). For this reason, Heun’s method is sometimes referred to as the trapezoidal rule.

Because Eq. (22.26) is a direct expression of the trapezoidal rule, the local truncation error is given by [recall Eq. (19.14)]

Et = − f ″(ξ ) _____ 12 h

3 (22.27)

where ξ is between ti and ti+1. Thus, the method is second order because the second de- rivative of the ODE is zero when the true solution is a quadratic. In addition, the local and global errors are O(h3) and O(h2), respectively. Therefore, decreasing the step size decreases the error at a faster rate than for Euler’s method.

22.3.2 The Midpoint Method

Figure 22.6 illustrates another simple modification of Euler’s method. Called the midpoint method, this technique uses Euler’s method to predict a value of y at the midpoint of the interval (Fig. 22.6a):

yi+1∕2 = yi + f (ti , yi ) h __ 2 (22.28)

Then, this predicted value is used to calculate a slope at the midpoint:

y i+1∕2 ′ = f (ti+1∕2, yi+1∕2) (22.29)

tti ti+1ti+1∕2

Slope = f (ti+1∕2, y i+1∕2)

tti ti+1 (a) (b)

FIGURE 22.6 Graphical depiction of midpoint method. (a) Predictor and (b) corrector.

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22.4 RunGE-KuTTA mETHODs 593

which is assumed to represent a valid approximation of the average slope for the entire interval. This slope is then used to extrapolate linearly from ti to ti+1 (Fig. 22.6b):

yi+1 = yi + f (ti+1∕2, yi+1∕2)h (22.30) Observe that because yi+1 is not on both sides, the corrector [Eq. (22.30)] cannot be applied iteratively to improve the solution as was done with Heun’s method.

As in our discussion of Heun’s method, the midpoint method can also be linked to Newton-Cotes integration formulas. Recall from Table 19.4 that the simplest Newton-Cotes open integration formula, which is called the midpoint method, can be represented as

∫ a

f (x) dx ≅ (b − a) f (x1) (22.31)

where x1 is the midpoint of the interval (a, b). Using the nomenclature for the present case, it can be expressed as

∫ ti

ti+1 f (t) dt ≅ h f (ti+1∕2) (22.32)

Substitution of this formula into Eq. (22.24) yields Eq. (22.30). Thus, just as the Heun method can be called the trapezoidal rule, the midpoint method gets its name from the underlying integration formula on which it is based.

The midpoint method is superior to Euler’s method because it utilizes a slope estimate at the midpoint of the prediction interval. Recall from our discussion of numerical differ- entiation in Sec. 4.3.4 that centered finite differences are better approximations of deriva- tives than either forward or backward versions. In the same sense, a centered approximation such as Eq. (22.29) has a local truncation error of O(h2) in comparison with the forward approximation of Euler’s method, which has an error of O(h). Consequently, the local and global errors of the midpoint method are O(h3) and O(h2), respectively.

22.4 RUNGE-KUTTA METHODS

Runge-Kutta (RK) methods achieve the accuracy of a Taylor series approach without requiring the calculation of higher derivatives. Many variations exist but all can be cast in the generalized form of Eq. (22.4):

yi+1 = yi + ϕh (22.33)

where ϕ is called an increment function, which can be interpreted as a representative slope over the interval. The increment function can be written in general form as

ϕ = a1k1 + a2k2 + · · · + ankn (22.34)

where the a’s are constants and the k’s are

k1 = f (ti , yi ) (22.34a) k2 = f (ti + p1h, yi + q11k1h) (22.34b) k3 = f (ti + p2h, yi + q21k1h + q22 k2h) (22.34c)

kn = f (ti + pn−1h, yi + qn−1,1k1h + qn−1,2k2h + · · · + qn−1,n−1kn−1h) (22.34d )

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where the p’s and q’s are constants. Notice that the k’s are recurrence relationships. That is, k1 appears in the equation for k 2 which appears in the equation for k3 and so forth. Because each k is a functional evaluation, this recurrence makes RK methods efficient for computer calculations.

Various types of RK methods can be devised by employing different numbers of terms in the increment function as specified by n. Note that the first-order RK method with n = 1 is, in fact, Euler’s method. Once n is chosen, values for the a’s, p’s, and q’s are evalu- ated by setting Eq. (22.33) equal to terms in a Taylor series expansion. Thus, at least for the lower-order versions, the number of terms n usually represents the order of the approach. For example, in Sec. 22.4.1, second-order RK methods use an increment function with two terms (n = 2). These second-order methods will be exact if the solution to the differential equation is quadratic. In addition, because terms with h3 and higher are dropped during the derivation, the local truncation error is O(h3) and the global error is O(h2). In Sec. 22.4.2, the fourth-order RK method (n = 4) is presented for which the global truncation error is O(h4).

22.4.1 Second-Order Runge-Kutta Methods

The second-order version of Eq. (22.33) is

yi+1 = yi + (a1k1 + a2k2)h (22.35)

where

k1 = f (ti , yi ) (22.35a) k2 = f (ti + p1h, yi + q11k1h) (22.35b)

The values for a1, a2, p1, and q11 are evaluated by setting Eq. (22.35) equal to a second-order Taylor series. By doing this, three equations can be derived to evaluate the four unknown constants (see Chapra and Canale, 2010, for details). The three equations are

a1 + a2 = 1 (22.36) a2 p1 = 1∕2 (22.37) a2q11 = 1∕2 (22.38)

Because we have three equations with four unknowns, these equations are said to be underdetermined. We, therefore, must assume a value of one of the unknowns to determine the other three. Suppose that we specify a value for a2. Then Eqs. (22.36) through (22.38) can be solved simultaneously for

a1 = 1 − a2 (22.39)

p1 = q11 = 1 ___ 2a2

(22.40)

Because we can choose an infinite number of values for a2, there are an infinite number of second-order RK methods. Every version would yield exactly the same results if the solution to the ODE were quadratic, linear, or a constant. However, they yield different results when (as is typically the case) the solution is more complicated. Three of the most commonly used and preferred versions are presented next.

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Heun method without iteration (a2 = 1∕2). If a2 is assumed to be 1∕2, Eqs. (22.39) and (22.40) can be solved for a1 = 1∕2 and p1 = q11 = 1. These parameters, when substituted into Eq. (22.35), yield

yi+1 = yi + ( 1 __ 2 k1 + 1 __ 2 k2 ) h (22.41) where

k1 = f (ti , yi ) (22.41a) k2 = f (ti + h, yi + k1h) (22.41b)

Note that k1 is the slope at the beginning of the interval and k2 is the slope at the end of the interval. Consequently, this second-order RK method is actually Heun’s technique without iteration of the corrector.

The midpoint method (a2 = 1). If a2 is assumed to be 1, then a1 = 0, p1 = q11 = 1∕2, and Eq. (22.35) becomes

yi+1 = yi + k2h (22.42)

where

k1 = f (ti , yi ) (22.42a) k2 = f (ti + h∕2, yi + k1h∕2) (22.42b)

This is the midpoint method.

Ralston’s method (a2 = 3∕4). Ralston (1962) and Ralston and Rabinowitz (1978) determined that choosing a2 = 3∕4 provides a minimum bound on the truncation error for the second-order RK algorithms. For this version, a1 = 1∕4 and p1 = q11 = 2∕3, and Eq. (22.35) becomes

yi+1 = yi + ( 1 __ 4 k1 + 3 __ 4 k2 ) h (22.43) where

k1 = f (ti , yi ) (22.43a)

k2 = f ( ti + 2 __ 3 h, yi + 2 __ 3 k1h ) (22.43b)

22.4.2 Classical Fourth-Order Runge-Kutta Method

The most popular RK methods are fourth order. As with the second-order approaches, there are an infinite number of versions. The following is the most commonly used form, and we therefore call it the classical fourth-order RK method:

yi+1 = yi + 1 __ 6 (k1 + 2k2 + 2k3 + k4)h (22.44)

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where

k1 = f (ti , yi ) (22.44a)

k2 = f ( ti + 1 __ 2 h, yi + 1 __ 2 k1h ) (22.44b) k3 = f ( ti + 1 __ 2 h, yi + 1 __ 2 k2h ) (22.44c) k4 = f (ti + h, yi + k3h) (22.44d )

Notice that for ODEs that are a function of t alone, the classical fourth-order RK method is similar to Simpson’s 1∕3 rule. In addition, the fourth-order RK method is simi- lar to the Heun approach in that multiple estimates of the slope are developed to come up with an improved average slope for the interval. As depicted in Fig. 22.7, each of the k’s represents a slope. Equation (22.44) then represents a weighted average of these to arrive at the improved slope.

EXAmPlE 22.3 Classical fourth-Order RK method

Problem statement. Employ the classical fourth-order RK method to integrate y′ = 4e0.8t − 0.5y from t = 0 to 1 using a step size of 1 with y(0) = 2.

solution. For this case, the slope at the beginning of the interval is computed as

k1 = f (0, 2) = 4e0.8(0) − 0.5(2) = 3

tti ti+1ti+1∕2

y k2

k1 k2

FIGURE 22.7 Graphical depiction of the slope estimates comprising the fourth-order RK method.

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22.4 RunGE-KuTTA mETHODs 597

This value is used to compute a value of y and a slope at the midpoint:

y(0.5) = 2 + 3(0.5) = 3.5 k2 = f (0.5, 3.5) = 4e0.8(0.5) − 0.5(3.5) = 4.217299

This slope in turn is used to compute another value of y and another slope at the midpoint:

y(0.5) = 2 + 4.217299(0.5) = 4.108649 k3 = f (0.5, 4.108649) = 4e0.8(0.5) − 0.5(4.108649) = 3.912974

Next, this slope is used to compute a value of y and a slope at the end of the interval:

y(1.0) = 2 + 3.912974(1.0) = 5.912974 k4 = f (1.0, 5.912974) = 4e0.8(1.0) − 0.5(5.912974) = 5.945677

Finally, the four slope estimates are combined to yield an average slope. This average slope is then used to make the final prediction at the end of the interval.

ϕ = 1 __ 6 [3 + 2(4.217299) + 2(3.912974) + 5.945677] = 4.201037

y(1.0) = 2 + 4.201037(1.0) = 6.201037

which compares favorably with the true solution of 6.194631 (εt = 0.103%).

It is certainly possible to develop fifth- and higher-order RK methods. For example, Butcher’s (1964) fifth-order RK method is written as

yi+1 = yi + 1 ___ 90 (7k1 + 32k3 + 12k4 + 32k5 + 7k6)h (22.45)

where

k1 = f (ti , yi ) (22.45a)

k2 = f ( ti + 1 __ 4 h, yi + 1 __ 4 k1h ) (22.45b) k3 = f ( ti + 1 __ 4 h, yi + 1 __ 8 k1h + 1 __ 8 k2h ) (22.45c) k4 = f ( ti + 1 __ 2 h, yi − 1 __ 2 k2h + k3 h ) (22.45d ) k5 = f ( ti + 3 __ 4 h, yi + 3 __ 16 k1h + 9 __ 16 k4h ) (22.45e) k6 = f ( ti + h, yi − 3 __ 7 k1h + 2 __ 7 k2h + 12 ___ 7 k3h − 12 ___ 7 k4h + 8 __ 7 k5 h ) (22.45f )

Note the similarity between Butcher’s method and Boole’s rule in Table 19.2. As expected, this method has a global truncation error of O(h5).

Although the fifth-order version provides more accuracy, notice that six function evalu- ations are required. Recall that up through the fourth-order versions, n function evaluations

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are required for an nth-order RK method. Interestingly, for orders higher than four, one or two additional function evaluations are necessary. Because the function evaluations account for the most computation time, methods of order five and higher are usually considered relatively less efficient than the fourth-order versions. This is one of the main reasons for the popularity of the fourth-order RK method.

22.5 SYSTEMS OF EQUATIONS

Many practical problems in engineering and science require the solution of a system of simultaneous ordinary differential equations rather than a single equation. Such systems may be represented generally as

dy1 ___ dt = f1(t, y1, y2, . . . , yn)

dy2 ___ dt = f2(t, y1, y2, . . . , yn)

dyn ___ dt = fn(t, y1, y2, . . . , yn)

(22.46)

The solution of such a system requires that n initial conditions be known at the starting value of t.

An example is the calculation of the bungee jumper’s velocity and position that we set up at the beginning of this chapter. For the free-fall portion of the jump, this problem amounts to solving the following system of ODEs:

dx ___ dt = υ (22.47)

dυ ___ dt = g − cd __ m υ

2 (22.48)

If the stationary platform from which the jumper launches is defined as x = 0, the initial conditions would be x(0) = υ(0) = 0.

22.5.1 Euler’s Method

All the methods discussed in this chapter for single equations can be extended to systems of ODEs. Engineering applications can involve thousands of simultaneous equations. In each case, the procedure for solving a system of equations simply involves applying the one-step technique for every equation at each step before proceeding to the next step. This is best illustrated by the following example for Euler’s method.

EXAmPlE 22.4 solving systems of ODEs with Euler’s method

Problem statement. Solve for the velocity and position of the free-falling bungee jumper using Euler’s method. Assuming that at t = 0, x = υ = 0, and integrate to t = 10 s with a step size of 2 s. As was done previously in Examples 1.1 and 1.2, the gravitational acceleration is 9.81 m/s2, and the jumper has a mass of 68.1 kg with a drag coefficient of 0.25 kg/m.

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Recall that the analytical solution for velocity is [Eq. (1.9)]:

υ(t) = √ ___

gm ___ cd tanh ( √ ___

gcd ___ m t )

This result can be substituted into Eq. (22.47) which can be integrated to determine an analytical solution for distance as

x(t) = m __ cd ln [ cosh ( √ ___

gcd ___ m t ) ]

Use these analytical solutions to compute the true relative errors of the results.

solution. The ODEs can be used to compute the slopes at t = 0 as

dx ___ dt = 0

dυ ___ dt = 9.81 − 0.25 ____ 68.1 (0)

2 = 9.81

Euler’s method is then used to compute the values at t = 2 s,

x = 0 + 0(2) = 0

υ = 0 + 9.81(2) = 19.62

The analytical solutions can be computed as x (2) = 19.16629 and υ (2) = 18.72919. Thus, the percent relative errors are 100% and 4.756%, respectively. The process can be repeated to compute the results at t = 4 as

x = 0 + 19.62(2) = 39.24

υ = 19.62 + ( 9.81 − 0.25 ____ 68.1 (19.62)2 ) 2 = 36.41368 Proceeding in a like manner gives the results displayed in Table 22.3.

TABLE 22.3 Distance and velocity of a free-falling bungee jumper as computed numerically with Euler’s method.

t xtrue υtrue xEuler υEuler εt (x) εt (υ)

0 0 0 0 0 2 19.1663 18.7292 0 19.6200 100.00% 4.76% 4 71.9304 33.1118 39.2400 36.4137 45.45% 9.97% 6 147.9462 42.0762 112.0674 46.2983 24.25% 10.03% 8 237.5104 46.9575 204.6640 50.1802 13.83% 6.86%

10 334.1782 49.4214 305.0244 51.3123 8.72% 3.83%

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Although the foregoing example illustrates how Euler’s method can be implemented for systems of ODEs, the results are not very accurate because of the large step size. In addition, the results for distance are a bit unsatisfying because x does not change until the second iteration. Using a much smaller step greatly mitigates these deficiencies. As described next, using a higher-order solver provides decent results even with a relatively large step size.

22.5.2 Runge-Kutta Methods

Note that any of the higher-order RK methods in this chapter can be applied to systems of equations. However, care must be taken in determining the slopes. Figure 22.7 is helpful in visualizing the proper way to do this for the fourth-order method. That is, we first develop slopes for all variables at the initial value. These slopes (a set of k1’s) are then used to make predictions of the dependent variable at the midpoint of the interval. These midpoint values are in turn used to compute a set of slopes at the midpoint (the k2’s). These new slopes are then taken back to the starting point to make another set of midpoint predictions that lead to new slope predictions at the midpoint (the k3’s). These are then employed to make predictions at the end of the interval that are used to develop slopes at the end of the interval (the k4’s). Finally, the k’s are combined into a set of increment functions [as in Eq. (22.44)] that are brought back to the beginning to make the final predictions. The fol- lowing example illustrates the approach.

EXAmPlE 22.5 solving systems of ODEs with the fourth-Order RK method

Problem statement. Use the fourth-order RK method to solve for the same problem we addressed in Example 22.4.

solution. First, it is convenient to express the ODEs in the functional format of Eq. (22.46) as

dx ___ dt = f1(t, x, υ) = υ

dυ ___ dt = f2(t, x, υ) = g − cd __ m υ

The first step in obtaining the solution is to solve for all the slopes at the beginning of the interval:

k1,1 = f1(0, 0, 0) = 0

k1,2 = f2(0, 0, 0) = 9.81 − 0.25 ____ 68.1 (0)

2 = 9.81

where ki, j is the ith value of k for the jth dependent variable. Next, we must calculate the first values of x and υ at the midpoint of the first step:

x (1) = x (0) + k1,1 h __ 2 = 0 + 0

2 __ 2 = 0

υ (1) = υ (0) + k1,2 h __ 2 = 0 + 9.81

2 __ 2 = 9.81

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which can be used to compute the first set of midpoint slopes:

k2,1 = f1(1, 0, 9.81) = 9.8100 k2,2 = f2(1, 0, 9.81) = 9.4567

These are used to determine the second set of midpoint predictions:

x (1) = x (0) + k2,1 h __ 2 = 0 + 9.8100

2 __ 2 = 9.8100

υ (1) = υ (0) + k2,2 h __ 2 = 0 + 9.4567

2 __ 2 = 9.4567

which can be used to compute the second set of midpoint slopes:

k3,1 = f1(1, 9.8100, 9.4567) = 9.4567 k3,2 = f2(1, 9.8100, 9.4567) = 9.4817

These are used to determine the predictions at the end of the interval:

x (2) = x (0) + k3,1h = 0 + 9.4567(2) = 18.9134 υ (2) = υ (0) + k3,2h = 0 + 9.4817(2) = 18.9634

which can be used to compute the endpoint slopes:

k4,1 = f1(2, 18.9134, 18.9634) = 18.9634 k4,2 = f2(2, 18.9134, 18.9634) = 8.4898

The values of k can then be used to compute [Eq. (22.44)]:

x (2) = 0 + 1 __ 6 [0 + 2(9.8100 + 9.4567) + 18.9634] 2 = 19.1656

υ (2) = 0 + 1 __ 6 [9.8100 + 2(9.4567 + 9.4817) + 8.4898] 2 = 18.7256

Proceeding in a like manner for the remaining steps yields the values displayed in Table 22.4. In contrast to the results obtained with Euler’s method, the fourth-order RK predictions are much closer to the true values. Further, a highly accurate, nonzero value is computed for distance on the first step.

TABLE 22.4 Distance and velocity of a free-falling bungee jumper as computed numerically with the fourth-order RK method.

t xtrue υtrue xRK4 υRK4 εt (x) εt (υ)

0 0 0 0 0 2 19.1663 18.7292 19.1656 18.7256 0.004% 0.019% 4 71.9304 33.1118 71.9311 33.0995 0.001% 0.037% 6 147.9462 42.0762 147.9521 42.0547 0.004% 0.051% 8 237.5104 46.9575 237.5104 46.9345 0.000% 0.049%

10 334.1782 49.4214 334.1626 49.4027 0.005% 0.038%

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22.5.3 MATLAB M-file Function: rk4sys

Figure 22.8 shows an M-file called rk4sys that uses the fourth-order RK method to solve a sys- tem of ODEs. This code is similar in many ways to the function developed earlier (Fig. 22.3) to solve a single ODE with Euler’s method. For example, it is passed the function name defining the ODEs through its argument.

FIGURE 22.8 An M-file to implement the RK4 method for a system of ODEs.

function [tp,yp] = rk4sys(dydt,tspan,y0,h,varargin) % rk4sys: fourth-order Runge-Kutta for a system of ODEs % [t,y] = rk4sys(dydt,tspan,y0,h,p1,p2,...): integrates % a system of ODEs with fourth-order RK method % input: % dydt = name of the M-file that evaluates the ODEs % tspan = [ti, tf]; initial and final times with output % generated at interval of h, or % = [t0 t1 ... tf]; specific times where solution output % y0 = initial values of dependent variables % h = step size % p1,p2,... = additional parameters used by dydt % output: % tp = vector of independent variable % yp = vector of solution for dependent variables

if nargin < 4,error('at least 4 input arguments required'), end if any(diff(tspan)< = 0),error('tspan not ascending order'), end n = length(tspan); ti = tspan(1);tf = tspan(n); if n = = 2 t = (ti:h:tf)'; n = length(t); if t(n)<tf t(n + 1) = tf; n = n + 1; end else t = tspan; end tt = ti; y(1,:) = y0; np = 1; tp(np) = tt; yp(np,:) = y(1,:); i = 1; while(1) tend = t(np + 1); hh = t(np + 1) − t(np);

(Continued)

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However, it has an additional feature that allows you to generate output in two ways, depending on how the input variable tspan is specified. As was the case for Fig. 22.3, you can set tspan = [ti tf], where ti and tf are the initial and final times, respectively. If done in this way, the routine automatically generates output values between these limits at equal spaced intervals h. Alternatively, if you want to obtain results at specific times, you can define tspan = [t0,t1,...,tf]. Note that in both cases, the tspan values must be in ascend- ing order.

We can employ rk4sys to solve the same problem as in Example 22.5. First, we can develop an M-file to hold the ODEs:

function dy = dydtsys(t, y) dy = [y(2);9.81 − 0.25/68.1*y(2)^2];

where y(1) = distance (x) and y(2) = velocity (υ). The solution can then be generated as

>> [t y] = rk4sys(@dydtsys,[0 10],[0 0],2); >> disp([t' y(:,1) y(:,2)])

0 0 0 2.0000 19.1656 18.7256 4.0000 71.9311 33.0995 6.0000 147.9521 42.0547 8.0000 237.5104 46.9345 10.0000 334.1626 49.4027

FIGURE 22.8 (Continued )

if hh > h,hh = h;end while(1) if tt+hh > tend,hh = tend-tt;end k1 = dydt(tt,y(i,:),varargin{:})'; ymid = y(i,:) + k1.*hh./2; k2 = dydt(tt + hh/2,ymid,varargin{:})'; ymid = y(i,:) + k2*hh/2; k3 = dydt(tt + hh/2,ymid,varargin{:})'; yend = y(i,:) + k3*hh; k4 = dydt(tt + hh,yend,varargin{:})'; phi = (k1 + 2*(k2 + k3) + k4)/6; y(i + 1,:) = y(i,:) + phi*hh; tt = tt + hh; i = i + 1; if tt > = tend,break,end end np = np + 1; tp(np) = tt; yp(np,:) = y(i,:); if tt > = tf,break,end end

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We can also use tspan to generate results at specific values of the independent variable. For example,

>> tspan = [0 6 10]; >> [t y] = rk4sys(@dydtsys,tspan,[0 0],2); >> disp([t' y(:,1) y(:,2)])

0 0 0 6.0000 147.9521 42.0547 10.0000 334.1626 49.4027

22.6 CASE STUDY PREDATOR-PREy mODEls AnD CHAOs

Background. Engineers and scientists deal with a variety of problems involving systems of nonlinear ordinary differential equations. This case study focuses on two of these ap- plications. The first relates to predator-prey models that are used to study species interac- tions. The second are equations derived from fluid dynamics that are used to simulate the atmosphere.

Predator-prey models were developed independently in the early part of the twenti- eth century by the Italian mathematician Vito Volterra and the American biologist Alfred Lotka. These equations are commonly called Lotka-Volterra equations. The simplest ver- sion is the following pairs of ODEs:

dx ___ dt = ax − bxy (22.49)

dy ___ dt = −cy + dxy (22.50)

where x and y = the number of prey and predators, respectively, a = the prey growth rate, c = the predator death rate, and b and d = the rates characterizing the effect of the predator- prey interactions on the prey death and the predator growth, respectively. The multiplica- tive terms (i.e., those involving xy) are what make such equations nonlinear.

An example of a simple nonlinear model based on atmospheric fluid dynamics is the Lorenz equations created by the American meteorologist Edward Lorenz:

dx ___ dt = −σ x + σ y

dy ___ dt = r x − y − xz

dz __ dt = −bz + xy

Lorenz developed these equations to relate the intensity of atmospheric fluid motion x to temperature variations y and z in the horizontal and vertical directions, respectively.

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22.6 CASE STUDY continued

As with the predator-prey model, the nonlinearities stem from the simple multiplicative terms: xz and xy.

Use numerical methods to obtain solutions for these equations. Plot the results to visualize how the dependent variables change temporally. In addition, graph the dependent variables versus each other to see whether any interesting patterns emerge.

Solution. The following parameter values can be used for the predator-prey simulation: a = 1.2, b = 0.6, c = 0.8, and d = 0.3. Employ initial conditions of x = 2 and y = 1 and integrate from t = 0 to 30, using a step size of h = 0.0625.

First, we can develop a function to hold the differential equations:

function yp = predprey(t,y,a,b,c,d) yp = [a*y(1) − b*y(1)*y(2); − c*y(2)+d*y(1)*y(2)];

The following script employs this function to generate solutions with both the Euler and the fourth-order RK methods. Note that the function eulersys was based on modify- ing the rk4sys function (Fig. 22.8). We will leave the development of such an M-file as a homework problem. In addition to displaying the solution as a time-series plot (x and y versus t), the script also generates a plot of y versus x. Such phase-plane plots are often useful in elucidating features of the model’s underlying structure that may not be evident from the time series.

h = 0.0625;tspan = [0 40];y0 = [2 1]; a = 1.2;b = 0.6;c = 0.8;d = 0.3; [t y] = eulersys(@predprey,tspan,y0,h,a,b,c,d); subplot(2,2,1);plot(t,y(:,1),t,y(:,2),'− −') legend('prey','predator');title('(a) Euler time plot') subplot(2,2,2);plot(y(:,1),y(:,2)) title('(b) Euler phase plane plot') [t y] = rk4sys(@predprey,tspan,y0,h,a,b,c,d); subplot(2,2,3);plot(t,y(:,1),t,y(:,2),'− −') title('(c) RK4 time plot') subplot(2,2,4);plot(y(:,1),y(:,2)) title('(d) RK4 phase plane plot')

The solution obtained with Euler’s method is shown at the top of Fig. 22.9. The time series (Fig. 22.9a) indicates that the amplitudes of the oscillations are expanding. This is reinforced by the phase-plane plot (Fig. 22.9b). Hence, these results indicate that the crude Euler method would require a much smaller time step to obtain accurate results.

In contrast, because of its much smaller truncation error, the RK4 method yields good results with the same time step. As in Fig. 22.9c, a cyclical pattern emerges in time. Because the predator population is initially small, the prey grows exponentially. At a certain point, the prey become so numerous that the predator population begins to grow. Eventually, the increased predators cause the prey to decline. This decrease, in turn, leads to a decrease of the predators. Eventually, the process repeats. Notice that, as expected, the predator peak lags the prey. Also, observe that the process has a fixed period—that is, it repeats in a set time.

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The phase-plane representation for the accurate RK4 solution (Fig. 22.9d) indi- cates that the interaction between the predator and the prey amounts to a closed coun- terclockwise orbit. Interestingly, there is a resting or critical point at the center of the orbit. The exact location of this point can be determined by setting Eqs. (22.49) and (22.50) to steady state (dy∕dt = dx∕dt = 0) and solving for (x, y) = (0, 0) and (c∕d, a∕b). The former is the trivial result that if we start with neither predators nor prey, nothing will happen. The latter is the more interesting outcome that if the initial conditions are set at x = c∕d and y = a∕b, the derivatives will be zero, and the populations will remain constant.

Now, let’s use the same approach to investigate the trajectories of the Lorenz equations with the following parameter values: a = 10, b = 8∕3, and r = 28. Employ initial conditions of x = y = z = 5 and integrate from t = 0 to 20. For this case, we will use the fourth-order RK method to obtain solutions with a constant time step of h = 0.03125.

22.6 CASE STUDY continued

(a) Euler time plot

prey

predator 10

0 0 10 20 30 40

(b) Euler phase plane plot

0 0 5 10 15

(d ) RK4 phase plane plot

0 0 2 4 6

FIGURE 22.9 Solution for the Lotka-Volterra model. Euler’s method (a) time-series and (b) phase-plane plots, and RK4 method (c) time-series and (d) phase-plane plots.

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22.6 CASE STUDY continued

22.6 CAsE sTuDy 607

The results are quite different from the behavior of the Lotka-Volterra equations. As in Fig. 22.10, the variable x seems to be undergoing an almost random pattern of oscilla- tions, bouncing around from negative values to positive values. The other variables exhibit similar behavior. However, even though the patterns seem random, the frequency of the oscillation and the amplitudes seem fairly consistent.

An interesting feature of such solutions can be illustrated by changing the initial condition for x slightly (from 5 to 5.001). The results are superimposed as the dashed line in Fig. 22.10. Although the solutions track on each other for a time, after about t = 15 they diverge significantly. Thus, we can see that the Lorenz equations are quite sensitive to their initial conditions. The term chaotic is used to describe such solutions. In his original study, this led Lorenz to the conclusion that long-range weather forecasts might be impossible!

The sensitivity of a dynamical system to small perturbations of its initial conditions is sometimes called the butterfly effect. The idea is that the flapping of a butterfly’s wings might induce tiny changes in the atmosphere that ultimately leads to a large-scale weather phenomenon like a tornado.

20 Lorenz model x versus t

x = 5.001, y = z = 5 x = y = z = 5

−5

−10

−15

−20 0 2 4 6 8 10 12 14 16 18 20

FIGURE 22.10 Time-domain representation of x versus t for the Lorenz equations. The solid time series is for the initial conditions (5, 5, 5). The dashed line is where the initial condition for x is perturbed slightly (5.001, 5, 5).

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22.6 CASE STUDY continued

Although the time-series plots are chaotic, phase-plane plots reveal an underly- ing structure. Because we are dealing with three independent variables, we can generate projections. Figure 22.11 shows projections in the xy, xz, and the yz planes. Notice how a structure is manifest when perceived from the phase-plane perspective. The solution forms orbits around what appear to be critical points. These points are called strange attractors in the jargon of mathematicians who study such nonlinear systems.

Beyond the two-variable projections, MATLAB’s plot3 function provides a vehicle to directly generate a three-dimensional phase-plane plot:

>> plot3(y(:,1),y(:,2),y(:,3)) >> xlabel('x');ylabel('y');zlabel('z');grid

As was the case for Fig. 22.11, the three-dimensional plot (Fig 22.12) depicts trajectories cycling in a definite pattern around a pair of critical points.

As a final note, the sensitivity of chaotic systems to initial conditions has implications for numerical computations. Beyond the initial conditions themselves, different step sizes or different algorithms (and in some cases, even different computers) can introduce small differences in the solutions. In a similar fashion to Fig. 22.10, these discrepancies will eventually lead to large deviations. Some of the problems in this chapter and in Chap. 23 are designed to demonstrate this issue.

(a) y versus x

0 10 20

−10

−20

−30 −20 −10

(b) z versus x

0 10 20

5 −20 −10

0 20 40

5 −40 −20

FIGURE 22.11 Phase-plane representation for the Lorenz equations. (a) xy, (b) xz, and (c) yz projections.

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22.6 CASE STUDY continued

−10

0 50

−50 −20 −10

y x

0 10

FIGURE 22.12 Three-dimensional phase-plane representation for the Lorenz equations generated with MATLAB’s plot3 function.

PROBLEMS

22.1 Solve the following initial value problem over the interval from t = 0 to where y(0) = 1. Display all your results on the same graph.

dy ___ dt = yt 2 − 1.1y

(a) Analytically. (b) Using Euler’s method with h = 0.5 and 0.25. (c) Using the midpoint method with h = 0.5. (d) Using the fourth-order RK method with h = 0.5.

22.2 Solve the following problem over the interval from x = 0 to 1 using a step size of 0.25 where y(0) = 1. Display all your results on the same graph.

dy ___ dx = (1 + 2x) √ __

(a) Analytically. (b) Using Euler’s method. (c) Using Heun’s method without iteration. (d) Using Ralston’s method. (e) Using the fourth-order RK method.

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22.3 Solve the following problem over the interval from t = 0 to 3 using a step size of 0.5 where y(0) = 1. Display all your results on the same graph.

dy ___ dt = −y + t 2

Obtain your solutions with (a) Heun’s method without iterating the corrector, (b) Heun’s method with iterating the corrector until εs < 0.1%, (c) the midpoint method, and (d) Ralston’s method. 22.4 The growth of populations of organisms has many engineering and scientific applications. One of the simplest models assumes that the rate of change of the population p is proportional to the existing population at any time t:

dp ___ dt = kg p (P22.4.1)

where kg = the growth rate. The world population in millions from 1950 through 2000 was

t 1950 1955 1960 1965 1970 1975 p 2555 2780 3040 3346 3708 4087

t 1980 1985 1990 1995 2000 p 4454 4850 5276 5686 6079

(a) Assuming that Eq. (P22.4.1) holds, use the data from 1950 through 1970 to estimate kg.

(b) Use the fourth-order RK method along with the results of (a) to stimulate the world population from 1950 to 2050 with a step size of 5 years. Display your simulation results along with the data on a plot.

22.5 Although the model in Prob. 22.4 works adequately when population growth is unlimited, it breaks down when factors such as food shortages, pollution, and lack of space inhibit growth. In such cases, the growth rate is not a con- stant, but can be formulated as

kg = kgm(1 − p∕pmax) where kgm = the maximum growth rate under unlimited con- ditions, p = population, and pmax = the maximum population. Note that pmax is sometimes called the carrying capacity. Thus, at low population density p ≪ pmax, kg → kgm. As p approaches pmax, the growth rate approaches zero. Using this growth rate formulation, the rate of change of population can be modeled as

dp ___ dt = kgm(1 − p∕pmax)p

This is referred to as the logistic model. The analytical solu- tion to this model is

p = p0 pmax _________________

p0 + ( pmax − p0 )e−kgmt

Simulate the world’s population from 1950 to 2050 using (a) the analytical solution and (b) the fourth-order RK method with a step size of 5 years. Employ the following initial conditions and parameter values for your simulation: p0 (in 1950) = 2555 million people, kgm = 0.026/yr, and pmax = 12,000 million people. Display your results as a plot along with the data from Prob. 22.4. 22.6 Suppose that a projectile is launched upward from the earth’s surface. Assume that the only force acting on the ob- ject is the downward force of gravity. Under these condi- tions, a force balance can be used to derive

dυ ___ dt = −g(0) R2 _______

(R + x)2

where υ = upward velocity (m/s), t = time (s), x = al- titude (m) measured upward from the earth’s surface, g(0) = the gravitational acceleration at the earth’s sur- face (≅ 9.81 m/s2), and R = the earth’s radius (≅ 6.37 × 106 m). Recognizing that dx∕dt = υ, use Euler’s method to determine the maximum height that would be obtained if υ(t = 0) = 1500 m/s. 22.7 Solve the following pair of ODEs over the interval from t = 0 to 0.4 using a step size of 0.1. The initial condi- tions are y(0) = 2 and z(0) = 4. Obtain your solution with (a) Euler’s method and (b) the fourth-order RK method. Dis- play your results as a plot.

dy ___ dt = −2y + 4e −t

dz __ dt = − yz2 ___ 3

22.8 The van der Pol equation is a model of an electronic circuit that arose back in the days of vacuum tubes:

d 2y ___

dt 2 − (1 − y 2) dy ___ dt + y = 0

Given the initial conditions, y(0) = y′(0) = 1, solve this equa- tion from t = 0 to 10 using Euler’s method with a step size of (a) 0.2 and (b) 0.1. Plot both solutions on the same graph. 22.9 Given the initial conditions, y(0) = 1 and y′(0) = 0, solve the following initial-value problem from t = 0 to 4:

d 2y ___

dt 2 + 9y = 0

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Obtain your solutions with (a) Euler’s method and (b) the fourth-order RK method. In both cases, use a step size of 0.1. Plot both solutions on the same graph along with the exact solution y = cos 3t. 22.10 Develop an M-file to solve a single ODE with Heun’s method with iteration. Design the M-file so that it creates a plot of the results. Test your program by using it to solve for population as described in Prob. 22.5. Employ a step size of 5 years and iterate the corrector until εs < 0.1%. 22.11 Develop an M-file to solve a single ODE with the midpoint method. Design the M-file so that it creates a plot of the results. Test your program by using it to solve for population as described in Prob. 22.5. Employ a step size of 5 years. 22.12 Develop an M-file to solve a single ODE with the fourth-order RK method. Design the M-file so that it creates a plot of the results. Test your program by using it to solve Prob. 22.2. Employ a step size of 0.1. 22.13 Develop an M-file to solve a system of ODEs with Euler’s method. Design the M-file so that it creates a plot of the results. Test your program by using it to solve Prob. 22.7 with a step size of 0.25. 22.14 Isle Royale National Park is a 210-square-mile archi- pelago composed of a single large island and many small islands in Lake Superior. Moose arrived around 1900, and by 1930, their population approached 3000, ravaging veg- etation. In 1949, wolves crossed an ice bridge from Ontario. Since the late 1950s, the numbers of the moose and wolves have been tracked.

Year Moose Wolves Year Moose Wolves Year Moose Wolves

1959 563 20 1975 1355 41 1991 1313 12 1960 610 22 1976 1282 44 1992 1590 12 1961 628 22 1977 1143 34 1993 1879 13 1962 639 23 1978 1001 40 1994 1770 17 1963 663 20 1979 1028 43 1995 2422 16 1964 707 26 1980 910 50 1996 1163 22 1965 733 28 1981 863 30 1997 500 24 1966 765 26 1982 872 14 1998 699 14 1967 912 22 1983 932 23 1999 750 25 1968 1042 22 1984 1038 24 2000 850 29 1969 1268 17 1985 1115 22 2001 900 19 1970 1295 18 1986 1192 20 2002 1100 17 1971 1439 20 1987 1268 16 2003 900 19 1972 1493 23 1988 1335 12 2004 750 29 1973 1435 24 1989 1397 12 2005 540 30 1974 1467 31 1990 1216 15 2006 450 30

PROblEms 611

(a) Integrate the Lotka-Volterra equations (Sec. 22.6) from 1960 through 2020 using the following coefficient val- ues: a = 0.23, b = 0.0133, c = 0.4, and d = 0.0004. Com- pare your simulation with the data using a time-series plot and determine the sum of the squares of the residu- als between your model and the data for both the moose and the wolves.

(b) Develop a phase-plane plot of your solution. 22.15 The motion of a damped spring-mass system (Fig. P22.15) is described by the following ordinary differ- ential equation:

m d 2x ___

dt 2 + c dx ___ dt + kx = 0

where x = displacement from equilibrium position (m), t = time (s), m = 20-kg mass, and c = the damping coef- ficient (N · s/m). The damping coefficient c takes on three

FIGURE P22.15

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values of 5 (underdamped), 40 (critically damped), and 200 (overdamped). The spring constant k = 20 N/m. The initial velocity is zero, and the initial displacement x = 1 m. Solve this equation using a numerical method over the time period 0 ≤ t ≤ 15 s. Plot the displacement versus time for each of the three values of the damping coefficient on the same plot. 22.16 A spherical tank has a circular orifice in its bottom through which the liquid flows out (Fig. P22.16). The flow rate through the hole can be estimated as

Qout = CA √ ____

2gh

where Qout = outflow (m3/s), C = an empirically derived coefficient, A = the area of the orifice (m2), g = the gravi- tational constant (= 9.81 m/s2), and h = the depth of liquid in the tank. Use one of the numerical methods described in this chapter to determine how long it will take for the water to flow out of a 3-m diameter tank with an initial height of 2.75 m. Note that the orifice has a diameter of 3 cm and C = 0.55. 22.17 In the investigation of a homicide or accidental death, it is often important to estimate the time of death. From the experimental observations, it is known that the surface tem- perature of an object changes at a rate proportional to the difference between the temperature of the object and that of the surrounding environment or ambient temperature. This is known as Newton’s law of cooling. Thus, if T(t) is the temperature of the object at time t, and Ta is the constant ambient temperature:

dT ___ dt = −K(T − Ta)

where K > 0 is a constant of proportionality. Suppose that at time t = 0 a corpse is discovered and its temperature is measured to be To. We assume that at the time of death, the body temperature Td was at the normal value of 37 °C. Suppose that the temperature of the corpse when it was dis- covered was 29.5 °C, and that two hours later, it is 23.5 °C. The ambient temperature is 20 °C. (a) Determine K and the time of death. (b) Solve the ODE numerically and plot the results. 22.18 The reaction A → B takes place in two reactors in se- ries. The reactors are well mixed but are not at steady state. The unsteady-state mass balance for each stirred tank reactor is shown below:

dCA1 _____ dt =

1 __ τ (CA0 − CA1) − kCA1

dCB1 _____ dt = −

1 __ τ CB1 + kCA1

dCA2 _____ dt =

1 __ τ (CA1 − CA2) − kCA2

dCB2 _____ dt =

1 __ τ (CB1 − CB2) + kCA2

where CA0 = concentration of A at the inlet of the first reactor, CA1 = concentration of A at the outlet of the first reactor (and inlet of the second), CA2 = concentration of A at the outlet of the second reactor, CB1 = concentration of B at the outlet of the first reactor (and inlet of the second), CB2 = concentration of B in the second reactor, τ = residence time for each reactor, and k = the rate constant for reaction of A to produce B. If CA0 is equal to 20, find the concentrations of A and B in both reactors during their first 10 minutes of operation. Use k = 0.12/min and τ = 5 min and assume that the initial conditions of all the dependent variables are zero. 22.19 A nonisothermal batch reactor can be described by the following equations:

dC ___ dt = −e (−10∕(T+273))C

dT ___ dt = 1000e (−10∕(T+273))C − 10( T − 20)

where C is the concentration of the reactant and T is the tem- perature of the reactor. Initially, the reactor is at 15 °C and has a concentration of reactant C of 1.0 gmol/L. Find the concentration and temperature of the reactor as a function of time.

FIGURE P22.16 A spherical tank.

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PROblEms 613

22.20 The following equation can be used to model the de- flection of a sailboat mast subject to a wind force:

d 2y ___

dz 2 = f (z) ____ 2E I (L − z)

where f (z) = wind force, E = modulus of elasticity, L = mast length, and I = moment of inertia. Note that the force varies with height according to

f (z) = 200z _____ 5 + z e −2z∕30

Calculate the deflection if y = 0 and dy∕dz = 0 at z = 0. Use parameter values of L = 30, E = 1.25 × 108, and I = 0.05 for your computation. 22.21 A pond drains through a pipe as shown in Fig. P22.21. Under a number of simplifying assumptions, the following differential equation describes how depth changes with time:

dh ___ dt = − πd 2 _____ 4A(h) √

________ 2g(h + e)

where h = depth (m), t = time (s), d = pipe diameter (m), A(h) = pond surface area as a function of depth (m2), g = gravitational constant (= 9.81 m/s2), and e = depth of pipe outlet below the pond bottom (m). Based on the following area-depth table, solve this differential equation to deter- mine how long it takes for the pond to empty, given that h(0) = 6 m, d = 0.25 m, e = 1 m.

h, m 6 5 4 3 2 1 0

A(h), 104 m2 1.17 0.97 0.67 0.45 0.32 0.18 0

22.22 Engineers and scientists use mass-spring models to gain insight into the dynamics of structures under the influ- ence of disturbances such as earthquakes. Figure P22.22 shows such a representation for a three-story building. For this case, the analysis is limited to horizontal motion of the

structure. Using Newton’s second law, force balances can be developed for this system as

d 2x1 ____ dt 2

= − k1 ___ m1 x1 +

k2 ___ m1 (x2 − x1)

d 2x2 ____ dt 2

= k2 ___ m2 (x1 − x2) +

k3 ___ m2 (x3 − x2)

d 2x3 ____ dt 2

= k3 ___ m3 (x2 − x3)

Simulate the dynamics of this structure from t = 0 to 20 s, given the initial condition that the velocity of the ground floor is dx1∕dt = 1 m/s, and all other initial values of dis- placements and velocities are zero. Present your results as two time-series plots of (a) displacements and (b) velocities. In addition, develop a three-dimensional phase-plane plot of the displacements. 22.23 Repeat the same simulations as in Sec. 22.6 for the Lorenz equations but generate the solutions with the mid- point method.

FIGURE P22.21

A(h)

FIGURE P22.22

m3 = 8000 kg

k3 = 1800 kN/m

k2 = 2400 kN/m

k1 = 3000 kN/m

m2 = 10,000 kg

m1 = 12,000 kg

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614 iniTiAl-VAluE PROblEms

22.24 Perform the same simulations as in Sec. 22.6 for the Lorenz equations but use a value of r = 99.96. Compare your results with those obtained in Sec. 22.6. 22.25 Figure P22.25 shows the kinetic interactions govern- ing the concentrations of a bacteria culture and their nutri- tion source (substrate) in a continuously stirred flow-through bioreactor.

The mass balances for the bacteria biomass, X (gC∕m3), and the substrate concentration, S (gC/m3), can be written as

dX ___ dt = ( kg, max S ______ Ks + S − kd − kr − 1 __ τw ) X dS ___ dt = −

1 __ Y kg, max S ______ Ks + S

X + kd X − 1 __ τw (Sin − S)

where t = time (h), kg,max = maximum bacterial growth rate (/d), Ks = half-saturation constant (gC/m3), kd = death rate (/d), kr = respiration rate (h), Q = flow rate (m3/h), V = reactor volume (m3), Y = yield coefficient (gC-cell/gC- substrate), and Sin = inflow substrate concentration (mgC/m3). Simulate how the substrate, bacteria, and total organic carbon (X + S) change over time in this reactor for three residence times: (a) τw = 20 h, (b) τw = 10 h, and (c) τw = 5 h. Employ the following parameters for the simulation: X(0) = 100 gC/m3, S(0) = 0, kg,max = 0.2/hr, Ks = 150 gC/m3, kd = kr = 0.01/hr, Y = 0.5 gC-cell/gC-substrate, V = 0.01 m3, and Sin = 1000 gC/m3, and display your results graphically.

Bacteria X

Substrate SY

1 Y kgX

kgX

kdX krX

Y−1 Y

Sin

FIGURE P22.25 Continuously stirred flow-through bioreactor to grow a bacterial culture.

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615

23 Adaptive Methods and Stiff Systems

23.1 ADAPTIVE RUNGE-KUTTA METHODS

To this point, we have presented methods for solving ODEs that employ a constant step size. For a significant number of problems, this can represent a serious limitation. For example, suppose that we are integrating an ODE with a solution of the type depicted in Fig. 23.1. For most of the range, the solution changes gradually. Such behavior suggests that a fairly large step size could be employed to obtain adequate results. However, for a localized region from t = 1.75 to 2.25, the solution undergoes an abrupt change. The practical con- sequence of dealing with such functions is that a very small step size would be required to accurately capture the impulsive behavior. If a constant step-size algorithm were employed, the smaller step size required for the region of abrupt change would have to be applied to the entire computation. As a consequence, a much smaller step size than necessary—and, therefore, many more calculations—would be wasted on the regions of gradual change.

Algorithms that automatically adjust the step size can avoid such overkill and hence be of great advantage. Because they “adapt” to the solution’s trajectory, they are said to have

CHAPTER OBJECTIVES

The primary objective of this chapter is to introduce you to more advanced methods for solving initial-value problems for ordinary differential equations. Specific objectives and topics covered are

• Understanding how the Runge-Kutta Fehlberg methods use RK methods of different orders to provide error estimates that are used to adjust the step size.

• Familiarizing yourself with the built-in MATLAB functions for solving ODEs. • Learning how to adjust the options for MATLAB’s ODE solvers. • Learning how to pass parameters to MATLAB’s ODE solvers. • Understanding the difference between one-step and multistep methods for solving

ODEs. • Understanding what is meant by stiffness and its implications for solving ODEs.

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616 AdAptive MethodS And Stiff SySteMS

adaptive step-size control. Implementation of such approaches requires that an estimate of the local truncation error be obtained at each step. This error estimate can then serve as a basis for either shortening or lengthening the step size.

Before proceeding, we should mention that aside from solving ODEs, the methods described in this chapter can also be used to evaluate definite integrals. The evaluation of the definite integral

I = ∫ a b

f (x) d x

is equivalent to solving the differential equation

d y ___ d x = f (x)

for y(b) given the initial condition y(a) = 0. Thus, the following techniques can be em- ployed to efficiently evaluate definite integrals involving functions that are generally smooth but exhibit regions of abrupt change.

There are two primary approaches to incorporate adaptive step-size control into one- step methods. Step halving involves taking each step twice, once as a full step and then as two half steps. The difference in the two results represents an estimate of the local trunca- tion error. The step size can then be adjusted based on this error estimate.

In the second approach, called embedded RK methods, the local truncation error is es- timated as the difference between two predictions using different-order RK methods. These are currently the methods of choice because they are more efficient than step halving.

0 1 2 3 t

FIGURE 23.1 An example of a solution of an ODE that exhibits an abrupt change. Automatic step-size adjustment has great advantages for such cases.

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23.1 AdAptive RUnGe-KUttA MethodS 617

The embedded methods were first developed by Fehlberg. Hence, they are sometimes referred to as RK-Fehlberg methods. At face value, the idea of using two predictions of dif- ferent order might seem too computationally expensive. For example, a fourth- and fifth- order prediction amounts to a total of 10 function evaluations per step [recall Eqs. (22.44) and (22.45)]. Fehlberg cleverly circumvented this problem by deriving a fifth-order RK method that employs most of the same function evaluations required for an accompanying fourth-order RK method. Thus, the approach yielded the error estimate on the basis of only six function evaluations!

23.1.1 MATLAB Functions for Nonstiff Systems

Since Fehlberg originally developed his approach, other even better approaches have been developed. Several of these are available as built-in functions in MATLAB.

ode23. The ode23 function uses the BS23 algorithm (Bogacki and Shampine, 1989; Shampine, 1994), which simultaneously uses second- and third-order RK formulas to solve the ODE and make error estimates for step-size adjustment. The formulas to advance the solution are

yi+1 = yi + 1 __ 9 (2k1 + 3k2 + 4k3)h (23.1)

where

k1 = f (ti, yi ) (23.1a)

k2 = f ( ti + 1 __ 2 h, yi + 1 __ 2 k1h ) (23.1b) k3 = f ( ti + 3 __ 4 h, yi + 3 __ 4 k2 h ) (23.1c)

The error is estimated as

Ei+1 = 1 ___ 72 (−5k1 + 6k2 + 8k3 − 9k4)h (23.2)

where

k4 = f (ti+1, yi+1) (23.2a)

Note that although there appear to be four function evaluations, there are really only three because after the first step, the k1 for the present step will be the k4 from the previous step. Thus, the approach yields a prediction and error estimate based on three evaluations rather than the five that would ordinarily result from using second- (two evaluations) and third- order (three evaluations) RK formulas in tandem.

After each step, the error is checked to determine whether it is within a desired toler- ance. If it is, the value of yi+1 is accepted, and k4 becomes k1 for the next step. If the error is too large, the step is repeated with reduced step sizes until the estimated error satisfies

E ≤ max(RelTol × |y|, AbsTol) (23.3)

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618 AdAptive MethodS And Stiff SySteMS

where RelTol is the relative tolerance (default = 10–3) and AbsTol is the absolute tolerance (default = 10–6). Observe that the criteria for the relative error uses a fraction rather than a percent relative error as we have done on many occasions prior to this point. ode45. The ode45 function uses an algorithm developed by Dormand and Prince (1980), which simultaneously uses fourth- and fifth-order RK formulas to solve the ODE and make error estimates for step-size adjustment. MATLAB recommends that ode45 is the best func- tion to apply as a “first try” for most problems. ode113. The ode113 function uses a variable-order Adams-Bashforth-Moulton solver. It is useful for stringent error tolerances or computationally intensive ODE functions. Note that this is a multistep method as we will describe subsequently in Sec. 23.2.

These functions can be called in a number of different ways. The simplest approach is

[t, y] = ode45(odefun, tspan, y0)

where y is the solution array where each column is one of the dependent variables and each row corresponds to a time in the column vector t, odefun is the name of the function returning a column vector of the right-hand-sides of the differential equations, tspan speci- fies the integration interval, and y0 = a vector containing the initial values.

Note that tspan can be formulated in two ways. First, if it is entered as a vector of two numbers,

tspan = [ti tf];

the integration is performed from ti to tf. Second, to obtain solutions at specific times t0, t1, ... , tn (all increasing or all decreasing), use

tspan = [t0 t1 ... tn];

Here is an example of how ode45 can be used to solve a single ODE, y′ = 4e0.8t − 0.5y from t = 0 to 4 with an initial condition of y(0) = 2. Recall from Example 22.1 that the analytical solution at t = 4 is 75.33896. Representing the ODE as an anonymous function, ode45 can be used to generate the same result numerically as

>> dydt = @(t,y) 4*exp(0.8*t) − 0.5*y; >> [t,y] = ode45(dydt,[0 4],2); >> y(length(t))

ans = 75.3390

As described in the following example, the ODE is typically stored in its own M-file when dealing with systems of equations.

eXAMpLe 23.1 Using MAtLAB to Solve a System of ODEs

problem Statement. Employ ode45 to solve the following set of nonlinear ODEs from t = 0 to 20:

dy1 ___ dt = 1.2y1 − 0.6y1 y2

dy2 ___ dt = −0.8y2 + 0.3y1 y2

where y1 = 2 and y2 = 1 at t = 0. Such equations are referred to as predator-prey equations.

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23.1 AdAptive RUnGe-KUttA MethodS 619

Solution. Before obtaining a solution with MATLAB, you must create a function to com- pute the right-hand side of the ODEs. One way to do this is to create an M-file as in

function yp = predprey(t,y) yp = [1.2*y(1) −0.6*y(1)*y(2);−0.8*y(2)+0.3*y(1)*y(2)];

We stored this M-file under the name: predprey.m. Next, enter the following commands to specify the integration range and the initial

conditions:

>> tspan = [0 20]; >> y0 = [2, 1];

The solver can then be invoked by

>> [t,y] = ode45(@predprey, tspan, y0);

This command will then solve the differential equations in predprey.m over the range defined by tspan using the initial conditions found in y0. The results can be displayed by simply typing

>> plot(t,y)

which yields Fig. 23.2. In addition to a time series plot, it is also instructive to generate a phase-plane plot—

that is, a plot of the dependent variables versus each other by

>> plot(y(:,1),y(:,2))

which yields Fig. 23.3.

0 0

5 10 15 20

y1 y2

FIGURE 23.2 Solution of predator-prey model with MATLAB.

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620 AdAptive MethodS And Stiff SySteMS

As in the previous example, the MATLAB solver uses default parameters to control vari- ous aspects of the integration. In addition, there is also no control over the differential equa- tions’ parameters. To have control over these features, additional arguments are included as in

[t, y] = ode45(odefun, tspan, y0, options, p1, p2,...)

where options is a data structure that is created with the odeset function to control features of the solution, and p1, p2,... are parameters that you want to pass into odefun.

The odeset function has the general syntax options = odeset('par1',val1,'par2',val2,...)

where the parameter pari has the value vali. A complete listing of all the possible param- eters can be obtained by merely entering odeset at the command prompt. Some commonly used parameters are 'RelTol' Allows you to adjust the relative tolerance. 'AbsTol' Allows you to adjust the absolute tolerance. 'InitialStep' The solver automatically determines the initial step. This option allows

you to set your own. 'MaxStep' The maximum step defaults to one-tenth of the tspan interval. This option

allows you to override this default.

eXAMpLe 23.2 Using odeset to Control integration options

problem Statement. Use ode23 to solve the following ODE from t = 0 to 4:

dy ___ dt = 10e −(t−2)2∕[2(0.075)2] − 0.6y

where y(0) = 0.5. Obtain solutions for the default (10–3) and for a more stringent (10–4) relative error tolerance.

0 0

2y2

1 2 3 y1

4 5 6

FIGURE 23.3 Phase-plane plot of predator-prey model with MATLAB.

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23.1 AdAptive RUnGe-KUttA MethodS 621

Solution. First, we will create an M-file to compute the right-hand side of the ODE: function yp = dydt(t, y) yp = 10*exp(−(t − 2)*(t − 2)/(2*.075^2)) − 0.6*y;

Then, we can implement the solver without setting the options. Hence the default value for the relative error (10–3) is automatically used: >> ode23(@dydt, [0 4], 0.5);

Note that we have not set the function equal to output variables [t, y]. When we imple- ment one of the ODE solvers in this way, MATLAB automatically creates a plot of the results displaying circles at the values it has computed. As in Fig. 23.4a, notice how ode23 takes relatively large steps in the smooth regions of the solution whereas it takes smaller steps in the region of rapid change around t = 2.

We can obtain a more accurate solution by using the odeset function to set the relative error tolerance to 10–4: >> options = odeset('RelTol',1e − 4); >> ode23(@ dydt, [0, 4], 0.5, options);

As in Fig. 23.4b, the solver takes more small steps to attain the increased accuracy.

(a) RelTol = 10−3

1.5

0.5

0 0 1 2 3 4

(b) RelTol = 10−4

1.5

0.5

0 0 1 2 3 4

FIGURE 23.4 Solution of ODE with MATLAB. For (b), a smaller relative error tolerance is used and hence many more steps are taken.

23.1.2 Events

MATLAB’s ODE solvers are commonly implemented for a prespecified integration inter- val. That is, they are often used to obtain a solution from an initial to a final value of the de- pendent variable. However, there are many problems where we do not know the final time.

A nice example relates to the free-falling bungee jumper that we have been using throughout this book. Suppose that the jump master inadvertently neglects to attach the cord to the jumper. The final time for this case, which corresponds to the jumper hitting

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622 AdAptive MethodS And Stiff SySteMS

the ground, is not a given. In fact, the objective of solving the ODEs would be to determine when the jumper hit the ground.

MATLAB’s events option provides a means to solve such problems. It works by solv- ing differential equations until one of the dependent variables reaches zero. Of course, there may be cases where we would like to terminate the computation at a value other than zero. As described in the following paragraphs, such cases can be readily accommodated.

We will use our bungee jumper problem to illustrate the approach. The system of ODEs can be formulated as

d x ___ d t = υ

d υ ___ d t = g − cd __ m υ| υ |

where x = distance (m), t = time (s), υ = velocity (m/s) where positive velocity is in the downward direction, g = the acceleration of gravity ( = 9.81 m/s2), cd = a second-order drag coefficient (kg/m), and m = mass (kg). Note that in this formulation, distance and velocity are both positive in the downward direction, and the ground level is defined as zero distance. For the present example, we will assume that the jumper is initially located 200 m above the ground and the initial velocity is 20 m/s in the upward direction—that is, x(0) = –200 and v(0) = 20.

The first step is to express the system of ODEs as an M-file function: function dydt = freefall(t,y,cd,m) % y(1) = x and y(2) = v grav = 9.81; dydt = [y(2);grav − cd/m*y(2)*abs(y(2))];

In order to implement the event, two other M-files need to be developed. These are (1) a function that defines the event and (2) a script that generates the solution.

For our bungee jumper problem, the event function (which we have named endevent) can be written as

function [detect,stopint,direction] = endevent(t,y,varargin) % Locate the time when height passes through zero % and stop integration. detect = y(1); % Detect height = 0 stopint = 1; % Stop the integration direction = 0; % Direction does not matter

This function is passed the values of the independent (t) and dependent variables (y) along with the model parameters (varargin). It then computes and returns three variables. The first, detect, specifies that MATLAB should detect the event when the dependent vari- able y(1) equals zero—that is, when the height x = 0. The second, stopint, is set to 1. This instructs MATLAB to stop when the event occurs. The final variable, direction, is set to 0 if all zeros are to be detected (this is the default), +1 if only the zeros where the event function increases are to be detected, and −1 if only the zeros where the event function decreases are to be detected. In our case, because the direction of the approach to zero is unimportant, we set direction to zero.1

1 Note that, as mentioned previously, we might want to detect a nonzero event. For example, we might want to detect when the jumper reached x = 5. To do this, we would merely set detect = y(1) − 5.

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23.1 AdAptive RUnGe-KUttA MethodS 623

Finally, a script can be developed to generate the solution:

opts = odeset('events',@ endevent); y0 = [−200 −20]; [t,y,te,ye] = ode45(@freefall,[0 inf],y0,opts,0.25,68.1); te,ye plot(t,−y(:,1),'−',t,y(:,2),'−−','LineWidth',2) legend('Height (m)','Velocity (m/s)') xlabel('time (s)'); ylabel('x (m) and v (m/s)')

In the first line, the odeset function is used to invoke the events option and specify that the event we are seeking is defined in the endevent function. Next, we set the initial condi- tions (y0) and the integration interval (tspan). Observe that because we do not know when the jumper will hit the ground, we set the upper limit of the integration interval to infinity. The third line then employs the ode45 function to generate the actual solution. As in all of MATLAB’s ODE solvers, the function returns the answers in the vectors t and y. In addition, when the events option is invoked, ode45 can also return the time at which the event occurs (te), and the corresponding values of the dependent variables (ye). The remaining lines of the script merely display and plot the results. When the script is run, the output is displayed as

te = 9.5475 ye = 0.0000 46.2454

The plot is shown in Fig. 23.5. Thus, the jumper hits the ground in 9.5475 s with a velocity of 46.2454 m/s.

0 1 2 3 4 5 6 7 8 9 10 −50

100

150

200

250

time (s)

x (m

) a nd

v (m

/s )

height (m) velocity (m/s)

FIGURE 23.5 MATLAB-generated plot of the height above the ground and velocity of the free-falling bungee jumper without the cord.

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624 AdAptive MethodS And Stiff SySteMS

23.2 MULTISTEP METHODS

The one-step methods described in the previous sections utilize information at a single point ti to predict a value of the dependent variable yi+1 at a future point ti+1 (Fig. 23.6a). Alternative approaches, called multistep methods (Fig. 23.6b), are based on the insight that, once the computation has begun, valuable information from previous points is at our command. The curvature of the lines connecting these previous values provides informa- tion regarding the trajectory of the solution. Multistep methods exploit this information to solve ODEs. In this section, we will present a simple second-order method that serves to demonstrate the general characteristics of multistep approaches.

23.2.1 The Non-Self-Starting Heun Method

Recall that the Heun approach uses Euler’s method as a predictor [Eq. (22.15)]:

y i+1 0 = yi + f (ti, yi )h (23.4)

and the trapezoidal rule as a corrector [Eq. (22.17)]:

yi+1 = yi + f (ti, yi ) + f ( ti+1, y i+1 0

) __________________ 2 h (23.5)

Thus, the predictor and the corrector have local truncation errors of O(h2) and O(h3), respectively. This suggests that the predictor is the weak link in the method because it has the greatest error. This weakness is significant because the efficiency of the iterative corrector step depends on the accuracy of the initial prediction. Consequently, one way to improve Heun’s method is to develop a predictor that has a local error of O(h3). This

xi (a)

xxi+1

(b)

xxi+1xi−1xi−2

FIGURE 23.6 Graphical depiction of the fundamental difference between (a) one-step and (b) multistep methods for solving ODEs.

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23.2 MULtiStep MethodS 625

can be accomplished by using Euler’s method and the slope at yi, and extra information from a previous point yi–1, as in

y i+1 0 = yi−1 + f (ti, yi )2 h (23.6)

This formula attains O(h3) at the expense of employing a larger step size 2h. In addi- tion, note that the equation is not self-starting because it involves a previous value of the dependent variable yi–1. Such a value would not be available in a typical initial-value problem. Because of this fact, Eqs. (23.5) and (23.6) are called the non-self-starting Heun method. As depicted in Fig. 23.7, the derivative estimate in Eq. (23.6) is now located at the midpoint rather than at the beginning of the interval over which the prediction is made. This centering improves the local error of the predictor to O(h3).

ti−1 ti+1 tti

y Slope = f (ti+1, y i0+1)

Slope = f (ti, yi) + f (ti+1, y i0+1)

ti+1 tti

(a)

(b)

FIGURE 23.7 A graphical depiction of the non-self-starting Heun method. (a) The midpoint method that is used as a predictor. (b) The trapezoidal rule that is employed as a corrector.

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626 AdAptive MethodS And Stiff SySteMS

The non-self-starting Heun method can be summarized as

Predictor (Fig. 23.7a): y i+1 0 = y i−1

m + f (ti, y i m ) 2h (23.7)

Corrector (Fig. 23.7b): y i+1 j = y i

m + f ( ti, y i m ) + f ( ti+1, y i+1 j−1 )

____________________ 2 h (23.8)

(for j = 1, 2, . . . , m)

where the superscripts denote that the corrector is applied iteratively from j = 1 to m to obtain refined solutions. Note that y i

m and y i−1 m are the final results of the corrector iterations

at the previous time steps. The iterations are terminated based on an estimate of the ap- proximate error,

|εa| = | y i+1 j − y i+1 j−1 _________ y i+1 j | × 100% (23.9) When |εa | is less than a prespecified error tolerance εs, the iterations are terminated. At this point, j = m. The use of Eqs. (23.7) through (23.9) to solve an ODE is demonstrated in the following example.

eXAMpLe 23.3 non-Self-Starting heun’s Method

problem Statement. Use the non-self-starting Heun method to perform the same com- putations as were performed previously in Example 22.2 using Heun’s method. That is, integrate y′ = 4e0.8t − 0.5y from t = 0 to 4 with a step size of 1. As with Example 22.2, the initial condition at t = 0 is y = 2. However, because we are now dealing with a multistep method, we require the additional information that y is equal to –0.3929953 at t = –1.

Solution. The predictor [Eq. (23.7)] is used to extrapolate linearly from t = –1 to 1:

y 1 0 = −0.3929953 + [4e0.8(0) − 0.5(2)] 2 = 5.607005

The corrector [Eq. (23.8)] is then used to compute the value:

y 1 1 = 2 + 4e

0.8(0) − 0.5(2) + 4e0.8(1) − 0.5(5.607005) __________________________________ 2 1 = 6.549331

which represents a true percent relative error of –5.73% (true value = 6.194631). This error is somewhat smaller than the value of –8.18% incurred in the self-starting Heun.

Now, Eq. (23.8) can be applied iteratively to improve the solution:

y 1 2 = 2 + 3 + 4e

0.8(1) − 0.5(6.549331) _______________________ 2 1 = 6.313749

which represents an error of –1.92%. An approximate estimate of the error can be deter- mined using Eq. (23.9):

|εa| = | 6.313749 − 6.549331 __________________ 6.313749 | × 100% = 3.7%

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23.2 MULtiStep MethodS 627

Equation (23.8) can be applied iteratively until εa falls below a prespecified value of εs. As was the case with the Heun method (recall Example 22.2), the iterations converge on a value of 6.36087 (εt = –2.68%). However, because the initial predictor value is more accurate, the multistep method converges at a somewhat faster rate.

For the second step, the predictor is

y 2 0 = 2 + [4e0.8(1) − 0.5(6.36087)] 2 = 13.44346 εt = 9.43%

which is superior to the prediction of 12.0826 (εt = 18%) that was computed with the original Heun method. The first corrector yields 15.76693 (εt = 6.8%), and subsequent iterations converge on the same result as was obtained with the self-starting Heun method: 15.30224 (εt = –3.09%). As with the previous step, the rate of convergence of the corrector is somewhat improved because of the better initial prediction.

23.2.2 Error Estimates

Aside from providing increased efficiency, the non-self-starting Heun can also be used to estimate the local truncation error. As with the adaptive RK methods in Sec. 23.1, the error estimate then provides a criterion for changing the step size.

The error estimate can be derived by recognizing that the predictor is equivalent to the midpoint rule. Hence, its local truncation error is (Table 19.4)

Ep = 1 __ 3 h

3 y(3)(ξp) = 1 __ 3 h

3 f ″(ξp) (23.10)

where the subscript p designates that this is the error of the predictor. This error estimate can be combined with the estimate of yi+1 from the predictor step to yield

True value = y i+1 0 + 1 __ 3 h

3 y(3)(ξp) (23.11)

By recognizing that the corrector is equivalent to the trapezoidal rule, a similar esti- mate of the local truncation error for the corrector is (Table 19.2)

Ec = − 1 ___ 12 h

3 y(3)(ξc) = − 1 ___ 12 h

3 f ″(ξc) (23.12)

This error estimate can be combined with the corrector result yi+1 to give

True value = y i+1 m − 1 __ 12 h

3 y(3)(ξc) (23.13)

Equation (23.11) can be subtracted from Eq. (23.13) to yield

0 = y i+1 m

− y i+1 0 − 5 ___ 12 h

3 y(3)(ξ ) (23.14)

where ξ is now between ti–1 and ti. Now, dividing Eq. (23.14) by 5 and rearranging the result gives

y i+1

0 − y i+1

m _________ 5 = −

1 ___ 12 h 3 y(3)(ξ ) (23.15)

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628 AdAptive MethodS And Stiff SySteMS

Notice that the right-hand sides of Eqs. (23.12) and (23.15) are identical, with the excep- tion of the argument of the third derivative. If the third derivative does not vary appreciably over the interval in question, we can assume that the right-hand sides are equal, and there- fore, the left-hand sides should also be equivalent, as in

Ec = − y i+1

0 − y i+1

m _________ 5 (23.16)

Thus, we have arrived at a relationship that can be used to estimate the per-step truncation error on the basis of two quantities that are routine by-products of the computation: the predictor ( y i+1

0 ) and the corrector ( y i+1 m ).

eXAMpLe 23.4 estimate of per-Step truncation error

problem Statement. Use Eq. (23.16) to estimate the per-step truncation error of Example 23.3. Note that the true values at t = 1 and 2 are 6.194631 and 14.84392, respectively.

Solution. At ti+1 = 1, the predictor gives 5.607005 and the corrector yields 6.360865. These values can be substituted into Eq. (23.16) to give

Ec = − 6.360865 − 5.607005 __________________ 5 = −0.150722

which compares well with the exact error,

Et = 6.194631 − 6.360865 = −0.1662341

At ti+1 = 2, the predictor gives 13.44346 and the corrector yields 15.30224, which can be used to compute

Ec = − 15.30224 − 13.44346 __________________ 5 = −0.37176

which also compares favorably with the exact error, Et = 14.84392 − 15.30224 = −0.45831.

The foregoing has been a brief introduction to multistep methods. Additional information can be found elsewhere (e.g., Chapra and Canale, 2010). Although they still have their place for solving certain types of problems, multistep methods are usually not the method of choice for most problems routinely confronted in engineering and science. That said, they are still used. For example, the MATLAB function ode113 is a multistep method. We have therefore included this section to introduce you to their basic principles.

23.3 STIFFNESS

Stiffness is a special problem that can arise in the solution of ordinary differential equa- tions. A stiff system is one involving rapidly changing components together with slowly changing ones. In some cases, the rapidly varying components are ephemeral transients that die away quickly, after which the solution becomes dominated by the slowly varying

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23.3 StiffneSS 629

components. Although the transient phenomena exist for only a short part of the integration interval, they can dictate the time step for the entire solution.

Both individual and systems of ODEs can be stiff. An example of a single stiff ODE is

d y ___ d t = −1000y + 3000 − 2000e −t (23.17)

If y(0) = 0, the analytical solution can be developed as

y = 3 − 0.998e−1000t − 2.002e−t (23.18)

As in Fig. 23.8, the solution is initially dominated by the fast exponential term (e–1000t ). After a short period (t < 0.005), this transient dies out and the solution becomes governed by the slow exponential (e–t ).

Insight into the step size required for stability of such a solution can be gained by ex- amining the homogeneous part of Eq. (23.17):

d y ___ d t = −ay (23.19)

If y(0) = y0, calculus can be used to determine the solution as

y = y0e−at

Thus, the solution starts at y0 and asymptotically approaches zero. Euler’s method can be used to solve the same problem numerically:

yi+1 = yi + d yi ___ d t h

Substituting Eq. (23.19) gives

yi+1 = yi − ayih

0 42 t0

0 0.020.010

FIGURE 23.8 Plot of a stiff solution of a single ODE. Although the solution appears to start at 1, there is actually a fast transient from y = 0 to 1 that occurs in less than the 0.005 time unit. This transient is perceptible only when the response is viewed on the finer timescale in the inset.

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630 AdAptive MethodS And Stiff SySteMS

yi+1 = yi (1 − ah) (23.20)

The stability of this formula clearly depends on the step size h. That is, |1 − ah| must be less than 1. Thus, if h > 2∕a, |yi| → ∞ as i → ∞.

For the fast transient part of Eq. (23.18), this criterion can be used to show that the step size to maintain stability must be < 2∕1000 = 0.002. In addition, we should note that, whereas this criterion maintains stability (i.e., a bounded solution), an even smaller step size would be required to obtain an accurate solution. Thus, although the transient occurs for only a small fraction of the integration interval, it controls the maximum allowable step size.

Rather than using explicit approaches, implicit methods offer an alternative remedy. Such representations are called implicit because the unknown appears on both sides of the equation. An implicit form of Euler’s method can be developed by evaluating the derivative at the future time:

yi+1 = yi + d yi+1 ____ d t h

This is called the backward, or implicit, Euler’s method. Substituting Eq. (23.19) yields

yi+1 = yi − a yi+1 h

which can be solved for

yi+1 = yi ______ 1 + a h (23.21)

For this case, regardless of the size of the step, |yi| → 0 as i → ∞. Hence, the approach is called unconditionally stable.

eXAMpLe 23.5 explicit and implicit euler

problem Statement. Use both the explicit and implicit Euler methods to solve Eq. (23.17), where y(0) = 0. (a) Use the explicit Euler with step sizes of 0.0005 and 0.0015 to solve for y between t = 0 and 0.006. (b) Use the implicit Euler with a step size of 0.05 to solve for y between 0 and 0.4.

Solution. (a) For this problem, the explicit Euler’s method is

yi+1 = yi + (−1000yi + 3000 − 2000 e − t i )h

The result for h = 0.0005 is displayed in Fig. 23.9a along with the analytical solution. Although it exhibits some truncation error, the result captures the general shape of the analytical solution. In contrast, when the step size is increased to a value just below the stability limit (h = 0.0015), the solution manifests oscillations. Using h > 0.002 would result in a totally unstable solution—that is, it would go infinite as the solution progressed.

(b) The implicit Euler’s method is

yi+1 = yi + (−1000yi+1 + 3000 − 2000 e − t i+1 )h

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23.3 StiffneSS 631

Now because the ODE is linear, we can rearrange this equation so that yi+1 is isolated on the left-hand side:

yi+1 = yi + 3000h − 2000h e

− t i+1 _____________________ 1 + 1000h

The result for h = 0.05 is displayed in Fig. 23.9b along with the analytical solution. Notice that even though we have used a much bigger step size than the one that induced instability for the explicit Euler, the numerical result tracks nicely on the analytical solution.

Systems of ODEs can also be stiff. An example is

d y1 ___ d t = −5y1 + 3y2 (23.22a)

d y2 ___ d t = 100y1 − 301y2 (23.22b)

For the initial conditions y1(0) = 52.29 and y2(0) = 83.82, the exact solution is

y1 = 52.96e−3.9899t − 0.67e−302.0101t (23.23a) y2 = 17.83e−3.9899t + 65.99e−302.0101t (23.23b)

1.5

0.5

0 0.0060.004

h = 0.0015

h = 0.0005 Exact

(a)

t0 0.002

0 0.40.3

Exact

h = 0.05

(b)

t0 0.20.1

FIGURE 23.9 Solution of a stiff ODE with (a) the explicit and (b) implicit Euler methods.

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632 AdAptive MethodS And Stiff SySteMS

Note that the exponents are negative and differ by about two orders of magnitude. As with the single equation, it is the large exponents that respond rapidly and are at the heart of the system’s stiffness.

An implicit Euler’s method for systems can be formulated for the present example as

y1,i+1 = y1,i + (−5y1,i+1 + 3y2,i+1)h (23.24a)

y2,i+1 = y2,i + (100y1,i+1 − 301y2,i+1)h (23.24b)

Collecting terms gives

(1 + 5h)y1,i+1 − 3y2,i+1 = y1,i (23.25a)

−100y1,i+1 + (1 + 301h)y2,i+1 = y2,i (23.25b)

Thus, we can see that the problem consists of solving a set of simultaneous equations for each time step.

For nonlinear ODEs, the solution becomes even more difficult since it involves solving a system of nonlinear simultaneous equations (recall Sec. 12.2). Thus, although stability is gained through implicit approaches, a price is paid in the form of added solution complexity.

23.3.1 MATLAB Functions for Stiff Systems

MATLAB has a number of built-in functions for solving stiff systems of ODEs. These are

ode15s. This function is a variable-order solver based on numerical differentiation formulas. It is a multistep solver that optionally uses the Gear backward differentiation formulas. This is used for stiff problems of low to medium accuracy. ode23s. This function is based on a modified Rosenbrock formula of order 2. Because it is a one-step solver, it may be more efficient than ode15s at crude tolerances. It can solve some kinds of stiff problems better than ode15s. ode23t. This function is an implementation of the trapezoidal rule with a “free” inter- polant. This is used for moderately stiff problems with low accuracy where you need a solution without numerical damping. ode23tb. This is an implementation of an implicit Runge-Kutta formula with a first stage that is a trapezoidal rule and a second stage that is a backward differentiation formula of order 2. This solver may also be more efficient than ode15s at crude tolerances.

eXAMpLe 23.6 MAtLAB for Stiff odes

problem Statement. The van der Pol equation is a model of an electronic circuit that arose back in the days of vacuum tubes,

d 2y1 ____ d t 2

− μ ( 1 − y 1 2 ) d y1 ___ d t + y1 = 0 (E23.6.1) The solution to this equation becomes progressively stiffer as μ gets large. Given the initial conditions, y1(0) = dy1∕dt = 1, use MATLAB to solve the following two cases: (a) for μ = 1, use ode45 to solve from t = 0 to 20; and (b) for μ = 1000, use ode23s to solve from t = 0 to 6000.

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23.3 StiffneSS 633

Solution. (a) The first step is to convert the second-order ODE into a pair of first-order ODEs by defining

d y1 ___ d t = y2

Using this equation, Eq. (E23.6.1) can be written as

d y2 ___ d t = μ ( 1 − y 1

2 ) y2 − y1 = 0 An M-file can now be created to hold this pair of differential equations:

function yp = vanderpol(t,y,mu) yp = [y(2);mu*(1−y(1)^2)*y(2)−y(1)];

Notice how the value of μ is passed as a parameter. As in Example 23.1, ode45 can be in- voked and the results plotted:

>> [t,y] = ode45(@ vanderpol,[0 20],[1 1],[],1); >> plot(t,y(:,1),'-',t,y(:,2),'--') >> legend('y1','y2');

Observe that because we are not specifying any options, we must use open brackets [] as a place holder. The smooth nature of the plot (Fig. 23.10a) suggests that the van der Pol equation with μ = 1 is not a stiff system.

(b) If a standard solver like ode45 is used for the stiff case ( μ = 1000), it will fail miserably (try it, if you like). However, ode23s does an efficient job:

>> [t,y] = ode23s(@vanderpol,[0 6000],[1 1],[],1000); >> plot(t,y(:,1))

−1

−2

−3

−1

−2

−3 0 5 10 15 20 0 2000 4000 6000

y1 y2

(b) m = 1000(a) m = 1

FIGURE 23.10 Solutions for van der Pol’s equation. (a) Nonstiff form solved with ode45 and (b) stiff form solved with ode23s.

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634 AdAptive MethodS And Stiff SySteMS

We have only displayed the y1 component because the result for y2 has a much larger scale. Notice how this solution (Fig. 23.10b) has much sharper edges than is the case in Fig. 23.10a. This is a visual manifestation of the “stiffness” of the solution.

23.4 MATLAB APPLICATION: BUNGEE JUMPER WITH CORD

In this section, we will use MATLAB to solve for the vertical dynamics of a jumper con- nected to a stationary platform with a bungee cord. As developed at the beginning of Chap. 22, the problem consisted of solving two coupled ODEs for vertical position and velocity. The differential equation for position is

d x ___ d t = υ (23.26)

The differential equation for velocity is different depending on whether the jumper has fallen to a distance where the cord is fully extended and begins to stretch. Thus, if the distance fallen is less than the cord length, the jumper is only subject to gravitational and drag forces,

d υ ___ d t = g − sign(υ) cd ___ m υ

2 (23.27a)

Once the cord begins to stretch, the spring and dampening forces of the cord must also be included:

d υ ___ d t = g − sign(υ) cd ___ m υ

2 − k __ m (x − L) γ __ m υ (23.27b)

The following example shows how MATLAB can be used to solve this problem.

eXAMpLe 23.7 Bungee Jumper with Cord

problem Statement. Determine the position and velocity of a bungee jumper with the following parameters: L = 30 m, g = 9.81 m/s2, m = 68.1 kg, cd = 0.25 kg/m, k = 40 N/m, and γ = 8 N · s/m. Perform the computation from t = 0 to 50 s and assume that the initial conditions are x(0) = υ(0) = 0.

Solution. The following M-file can be set up to compute the right-hand sides of the ODEs:

function dydt = bungee(t,y,L,cd,m,k,gamma) g = 9.81; cord = 0; if y(1) > L %determine if the cord exerts a force cord = k/m*(y(1)−L) + gamma/m*y(2); end dydt = [y(2); g − sign(y(2))*cd/m*y(2)^2 − cord];

Notice that the derivatives are returned as a column vector because this is the format required by the MATLAB solvers.

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23.5 CASe StUdy 635

Because these equations are not stiff, we can use ode45 to obtain the solutions and display them on a plot:

>> [t,y] = ode45(@bungee,[0 50],[0 0],[],30,0.25,68.1,40,8); >> plot(t,−y(:,1),'−',t,y(:,2),':') >> legend('x (m)','v (m/s)')

As in Fig. 23.11, we have reversed the sign of distance for the plot so that negative distance is in the downward direction. Notice how the simulation captures the jumper’s bouncing motion.

−20

−40

−60

−80 0 10 20 30 40 50

x (m) v (m/s)

FIGURE 23.11 Plot of distance and velocity of a bungee jumper.

23.5 CASE STUDY pLiny’S inteRMittent foUntAin

Background. The Roman natural philosopher, Pliny the Elder, purportedly had an inter- mittent fountain in his garden. As in Fig. 23.12, water enters a cylindrical tank at a constant flow rate Qin and fills until the water reaches yhigh. At this point, water siphons out of the tank through a circular discharge pipe, producing a fountain at the pipe’s exit. The fountain runs until the water level decreases to ylow, whereupon the siphon fills with air and the fountain stops. The cycle then repeats as the tank fills until the water reaches yhigh, and the fountain flows again.

When the siphon is running, the outflow Qout can be computed with the following formula based on Torricelli’s law:

Qout = C √ ____

2gy πr 2 (23.28)

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636 AdAptive MethodS And Stiff SySteMS

23.5 CASE STUDY continued

Qin

Qout

y = yhigh

y = ylow y = 0

FIGURE 23.12 An intermittent fountain.

Neglecting the volume of water in the pipe, compute and plot the level of the water in the tank as a function of time over 100 seconds. Assume an initial condition of an empty tank y(0) = 0, and employ the following parameters for your computation:

RT = 0.05 m r = 0.007 m ylow = 0.025 m

yhigh = 0.1 m C = 0.6 g = 9.81 m/s2

Qin = 50 × 10−6 m3/s

Solution. When the fountain is running, the rate of change in the tank’s volume V (m3) is determined by a simple balance of inflow minus the outflow:

dV ___ dt = Qin − Qout (23.29)

where V = volume (m3). Because the tank is cylindrical, V = π R t 2 y. Substituting this rela-

tionship along with Eq. (23.28) into Eq. (23.29) gives

dy ___ dt = Qin − C √

____ 2gy πr 2 _______________

π R t 2 (23.30)

When the fountain is not running, the second term in the numerator goes to zero. We can incorporate this mechanism in the model by introducing a new dimensionless variable siphon that equals zero when the fountain is off and equals one when it is flowing:

dy ___ dt = Qin − siphon × C √

____ 2gy πr 2 _______________

π R t 2 (23.31)

In the present context, siphon can be thought of as a switch that turns the fountain off and on. Such two-state variables are called Boolean or logical variables, where zero is equiva- lent to false and one is equivalent to true.

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23.5 CASe StUdy 637

23.5 CASE STUDY continued

Next we must relate siphon to the dependent variable y. First, siphon is set to zero whenever the level falls below ylow. Conversely, siphon is set to one whenever the level rises above yhigh. The following M-file function follows this logic in computing the derivative:

function dy = Plinyode(t,y) global siphon Rt = 0.05; r = 0.007; yhi = 0.1; ylo = 0.025; C = 0.6; g = 9.81; Qin = 0.00005; if y(1) <= ylo siphon = 0; elseif y(1) >= yhi siphon = 1; end Qout = siphon * C * sqrt(2 * g * y(1)) * pi * r ^ 2; dy = (Qin − Qout) / (pi * Rt ^ 2);

Notice that because its value must be maintained between function calls, siphon is declared as a global variable. Although the use of global variables is not encouraged (particularly in larger programs), it is useful in the present context.

The following script employs the built-in ode45 function to integrate Plinyode and gen- erate a plot of the solution:

global siphon siphon = 0; tspan = [0 100]; y0 = 0; [tp,yp] = ode45(@ Plinyode,tspan,y0); plot(tp,yp) xlabel('time, (s)') ylabel('water level in tank, (m)')

As shown in Fig. 23.13, the result is clearly incorrect. Except for the original filling period, the level seems to start emptying prior to reaching yhigh. Similarly, when it is drain- ing, the siphon shuts off well before the level drops to ylow.

At this point, suspecting that the problem demands more firepower than the trusty ode45 routine, you might be tempted to use one of the other MATLAB ODE solvers such as ode23s or ode23tb. But if you did, you would discover that although these routines yield somewhat different results, they would still generate incorrect solutions.

The difficulty arises because the ODE is discontinuous at the point that the siphon switches on or off. For example, as the tank is filling, the derivative is dependent only on the constant inflow and for the present parameters has a constant value of 6.366 × 10−3 m/s. However, as soon as the level reaches yhigh, the outflow kicks in and the derivative abruptly drops to −1.013 × 10−2 m/s. Although the adaptive step-size routines used by MATLAB work marvelously for many problems, they often get heartburn when dealing with such discontinuities. Because they infer the behavior of the solution by comparing the results of different steps, a discontinuity represents something akin to stepping into a deep pothole on a dark street.

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638 AdAptive MethodS And Stiff SySteMS

23.5 CASE STUDY continued

At this point, your first inclination might be to just give up. After all, if it’s too hard for MATLAB, no reasonable person could expect you to come up with a solution. Because professional engineers and scientists rarely get away with such excuses, your only recourse is to develop a remedy based on your knowledge of numerical methods.

Because the problem results from adaptively stepping across a discontinuity, you might revert to a simpler approach and use a constant, small step size. If you think about it, that’s precisely the approach you would take if you were traversing a dark, pothole-filled street. We can implement this solution strategy by merely replacing ode45 with the constant-step rk4sys function from Chap. 22 (Fig. 22.8). For the script outlined above, the fourth line would be formulated as

[tp,yp] = rk4sys(@ Plinyode,tspan,y0,0.0625);

As in Fig. 23.14, the solution now evolves as expected. The tank fills to yhigh and then emp- ties until it reaches ylow, when the cycle repeats.

There are a two take-home messages that can be gleaned from this case study. First, although it’s human nature to think the opposite, simpler is sometimes better. After all, to paraphrase Einstein, “Everything should be as simple as possible, but no simpler.” Second, you should never blindly believe every result generated by the computer. You’ve probably heard the old chestnut, “garbage in, garbage out” in reference to the impact of data quality

FIGURE 23.13 The level in Pliny’s fountain versus time as simulated with ode45.

0.12

0.08

0.06

0.04

0.1

0.02

0 0 10 20 30 40 50

Time (s)

W at

er le

ve l i

n ta

nk (m

)

60 70 80 90 100

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23.5 CASe StUdy 639

23.5 CASE STUDY continued

on the validity of computer output. Unfortunately, some individuals think that regardless of what went in (the data) and what’s going on inside (the algorithm), it’s always “gospel out.” Situations like the one depicted in Fig. 23.13 are particularly dangerous—that is, al- though the output is incorrect, it’s not obviously wrong. That is, the simulation does not go unstable or yield negative levels. In fact, the solution moves up and down in the manner of an intermittent fountain, albeit incorrectly.

Hopefully, this case study illustrates that even a great piece of software such as MATLAB is not foolproof. Hence, sophisticated engineers and scientists always examine numerical output with a healthy skepticism based on their considerable experience and knowledge of the problems they are solving.

FIGURE 23.14 The level in Pliny’s fountain versus time as simulated with a small, constant step size using the rk4sys function (Fig. 22.8).

0.1

0.07

0.06

0.05

0.04

0.03

0.08

0.09

0.02

0.01

0 0 10 20 30 40 50

Time (s)

W at

er le

ve l i

n ta

nk (m

)

60 70 80 90 100

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640 AdAptive MethodS And Stiff SySteMS

PROBLEMS

23.1 Repeat the same simulations as in Sec. 23.5 for Pliny’s fountain, but generate the solutions with ode23, ode23s, and ode113. Use subplot to develop a vertical three-pane plot of the time series. 23.2 The following ODEs have been proposed as a model of an epidemic:

d S ___ d t = −a S I

d I __ d t = a S I − r I

d R ___ d t = r I

where S = the susceptible individuals, I = the infected, R = the recovered, a = the infection rate, and r = the recovery rate. A city has 10,000 people, all of whom are susceptible. (a) If a single infectious individual enters the city at t = 0,

compute the progression of the epidemic until the number of infected individuals falls below 10. Use the following parameters: a = 0.002/(person · week) and r = 0.15/d. Develop time-series plots of all the state variables. Also generate a phase-plane plot of S versus I versus R.

(b) Suppose that after recovery, there is a loss of immunity that causes recovered individuals to become suscep- tible. This reinfection mechanism can be computed as ρR, where ρ = the reinfection rate. Modify the model to include this mechanism and repeat the computations in (a) using ρ = 0.03/d.

23.3 Solve the following initial-value problem over the interval from t = 2 to 3:

d y ___ d t = −0.5y + e −t

Use the non-self-starting Heun method with a step size of 0.5 and initial conditions of y(1.5) = 5.222138 and y(2.0) = 4.143883. Iterate the corrector to εs = 0.1%. Com- pute the percent relative errors for your results based on the exact solutions obtained analytically: y(2.5) = 3.273888 and y(3.0) = 2.577988. 23.4 Solve the following initial-value problem over the interval from t = 0 to 0.5:

d y ___ d t = y t 2 − y

Use the fourth-order RK method to predict the first value at t = 0.25. Then use the non-self-starting Heun method to make the prediction at t = 0.5. Note: y(0) = 1.

23.5 Given

d y ___ d t = −100,000y + 99,999e −t

(a) Estimate the step size required to maintain stability using the explicit Euler method.

(b) If y(0) = 0, use the implicit Euler to obtain a solution from t = 0 to 2 using a step size of 0.1.

23.6 Given

d y ___ d t = 30(sin t − y) + 3 cos t

If y(0) = 0, use the implicit Euler to obtain a solution from t = 0 to 4 using a step size of 0.4. 23.7 Given

d x1 ___ d t = 999x1 + 1999x2

d x2 ___ d t = −1000x1 − 2000x2

If x1(0) = x2(0) = 1, obtain a solution from t = 0 to 0.2 using a step size of 0.05 with the (a) explicit and (b) implicit Euler methods. 23.8 The following nonlinear, parasitic ODE was suggested by Hornbeck (1975):

d y ___ d t = 5( y − t 2)

If the initial condition is y(0) = 0.08, obtain a solution from t = 0 to 5: (a) Analytically. (b) Using the fourth-order RK method with a constant step

size of 0.03125. (c) Using the MATLAB function ode45. (d) Using the MATLAB function ode23s. (e) Using the MATLAB function ode23tb. Present your results in graphical form. 23.9 Recall from Example 20.5 that the humps function ex- hibits both flat and steep regions over a relatively short x range,

f (x) = 1 ______________ (x − 0.3)2 + 0.01

+ 1 _______________ (x − 0.9)2 + 0.04

− 6

Determine the value of the definite integral of this function between x = 0 and 1 using (a) the quad and (b) the ode45 functions.

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pRoBLeMS 641

Solve this differential equation (a) analytically and (b) numerically for x = 0 to 5. For (b) use an implicit approach with h = 0.5. Note that the initial conditions are y(0) = 1 and y′ (0) = 0. Display both results graphically. 23.15 Consider the thin rod of length l moving in the x-y plane as shown in Fig. P23.15. The rod is fixed with a pin on one end and a mass at the other. Note that g = 9.81 m/s2 and l = 0.5 m. This system can be solved using

θ ¨ − g __ l θ = 0

Let θ(0) = 0 and θ ˙ (0) = 0.25 rad/s. Solve using any method studied in this chapter. Plot the angle versus time and the angular velocity versus time. (Hint: Decompose the second- order ODE.)

23.16 Given the first-order ODE:

d x ___ d t = −700x − 1000e −t

x(t = 0) = 4

Solve this stiff differential equation using a numerical method over the time period 0 ≤ t ≤ 5. Also solve analyti- cally and plot the analytic and numerical solution for both the fast and slow transient phases of the time scale. 23.17 Solve the following differential equation from t = 0 to 2

d y ___ d t = −10y

with the initial condition y(0) = 1. Use the following tech- niques to obtain your solutions: (a) analytically, (b) the ex- plicit Euler method, and (c) the implicit Euler method. For (b) and (c) use h = 0.1 and 0.2. Plot your results. 23.18 The Lotka-Volterra equations described in Sec. 22.6 have been refined to include additional factors that impact predator-prey dynamics. For example, over and above predation, prey population can be limited by other factors

23.10 The oscillations of a swinging pendulum can be sim- ulated with the following nonlinear model:

d 2θ ____

d t 2 + g __ l sin θ = 0

where θ = the angle of displacement, g = the gravitational constant, and l = the pendulum length. For small angular displacements, the sin θ is approximately equal to θ and the model can be linearized as

d 2θ ____

d t 2 + g __ l θ = 0

Use ode45 to solve for θ as a function of time for both the lin- ear and nonlinear models where l = 0.6 m and g = 9.81 m/s2. First, solve for the case where the initial condition is for a small displacement (θ = π∕8 and dθ∕dt = 0). Then repeat the calculation for a large displacement (θ = π∕2). For each case, plot the linear and nonlinear simulations on the same plot. 23.11 Employ the events option described in Sec. 23.1.2 to determine the period of a 1-m long, linear pendulum (see description in Prob. 23.10). Compute the period for the following initial conditions: (a) θ = π∕8, (b) θ = π∕4, and (c) θ = π∕2. For all three cases, set the initial angular velocity at zero. (Hint: A good way to compute the period is to determine how long it takes for the pendulum to reach θ = 0 [i.e., the bottom of its arc]). The period is equal to four times this value. 23.12 Repeat Prob. 23.11, but for the nonlinear pendulum described in Prob. 23.10. 23.13 The following system is a classic example of stiff ODEs that can occur in the solution of chemical reaction kinetics:

d c1 ___ d t = −0.013c1 − 1000c1c3

d c2 ___ d t = −2500c2c3

d c3 ___ d t = −0.013c1 − 1000c1c3 − 2500c2c3

Solve these equations from t = 0 to 50 with initial condi- tions c1(0) = c2(0) = 1 and c3(0) = 0. If you have access to MATLAB software, use both standard (e.g., ode45) and stiff (e.g., ode23s) functions to obtain your solutions. 23.14 The following second-order ODE is considered to be stiff:

d 2 y ____

d x 2 = −1001 d y ___ d x − 1000y

FIGURE P23.15

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642 AdAptive MethodS And Stiff SySteMS

such as space. Space limitation can be incorporated into the model as a carrying capacity (recall the logistic model de- scribed in Prob. 22.5) as in

d x ___ d t = a ( 1 − x __ K ) x − bx y d y ___ d t = −cy + dx y

where K = the carrying capacity. Use the same parameter values and initial conditions as in Sec. 22.6 to integrate these equations from t = 0 to 100 using ode45, and develop both time series and phase-plane plots of the results. (a) Employ a very large value of K = 108 to validate that

you obtain the same results as in Sec. 22.6. (b) Compare (a) with the more realistic carrying capacity of

K = 200. Discuss your results. 23.19 Two masses are attached to a wall by linear springs (Fig. P23.19). Force balances based on Newton’s second law can be written as

d 2x1 ____ d t 2

= − k1 ___ m1 (x1 − L1) +

k2 ___ m1 (x2 − x1 − 𝑤1 − L2)

d 2x 2 ____ d t2

= − k2 ___ m2 (x2 − x1 − 𝑤1 − L2)

k2k1

L1 �1 L2 �2

x1 0

x m1 m2

FIGURE P23.19

d 2 θ1 _____ dt2

= − g(2m1 + m2)sin θ1 − m2 g sin(θ1 − 2θ2) − 2 sin (θ1 − θ2) m2 (dθ2∕dt)2 L2 + (d θ1∕dt)2 L1 cos( θ1 − θ2)) ___________________________________________________________________________________ L1(2m1 + m2 − m2 cos (2θ1 − θ2)

d 2 θ2 _____ dt2

= 2sin (θ1 − θ2)((dθ1∕dt)2 L1 (m1 + m2) + g (m1 + m2) cos (θ1) + (dθ2∕dt)2 L2 m2 cos(θ1 − θ2)) ___________________________________________________________________________________ L2(2m1 + m2 − m2 cos (2θ1 − θ2)

L 1

L 2

m 1

m 2

θ1

θ2

FIGURE P23.23 A double pendulum.

where k = the spring constants, m = mass, L = the length of the unstretched spring, and w = the width of the mass. Com- pute the positions of the masses as a function of time using the following parameter values: k1 = k2 = 5, m1 = m2 = 2, 𝑤1 = 𝑤2 = 5, and L1 = L2 = 2. Set the initial conditions as x1 = L1 and x2 = L1 + 𝑤1 + L2 + 6. Perform the simulation from t = 0 to 20. Construct time-series plots of both the displacements and the velocities. In addition, pro- duce a phase-plane plot of x1 versus x2. 23.20 Use ode45 to integrate the differential equations for the system described in Prob. 23.19. Generate verti- cally stacked subplots of displacements (top) and velocities (bottom). Employ the fft function to compute the discrete Fourier transform (DFT) of the first mass’s displacement. Generate and plot a power spectrum in order to identify the system’s resonant frequencies. 23.21 Perform the same computations as in Prob. 23.20 but based on the first floor of the structure in Prob. 22.22.

23.22 Use the approach and example outlined in Sec. 23.1.2, but determine the time, height, and velocity when the bun- gee jumper is the farthest above the ground, and generate a plot of the solution. 23.23 As depicted in Fig. P23.23, a double pendulum consists of a pendulum attached to another pendulum. We indicate the upper and lower pendulums by subscripts 1 and 2, respectively, and we place the origin at the pivot point of the upper pendulum with y increasing upward. We further assume that the system oscillates in a vertical plane subject to gravity, that the pendulum rods are massless and rigid, and the pendulum masses are considered to be point masses. Under these assumptions, force balances can be used to de- rive the following equations of motion

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Problems 643

balloon, T = 100 °C; and the normal (ambient) air density, ρ = 1.2 kg/m3. (a) Use a force balance to develop the differential equa-

tion for dv∕dt as a function of the model’s fundamental parameters.

(b) At steady-state, calculate the particle’s terminal velocity. (c) Use ode45 to compute the velocity and position of the

balloon from t = 0 to 60 s given the previous parameters along with the initial condition: υ(0) = 0. Develop a plot of your results.

23.25 Develop a MATLAB script using ode45 to compute the velocity, υ, and position, z, of a hot air balloon as de- scribed in Prob. 23.24. Perform the calculation from t = 0 to 60 s with a step size of 1.6 s. At z = 200 m, assume that part of the payload (100 kg) is dropped out of the balloon. Develop a plot of your results. 23.26 You go on a two-week vacation and place your pet goldfish “Freddie” into your bathtub. Note that you dechlo- rinate the water first! You then place an air tight plexiglass cover over the top of the tub in order to protect Freddie from your cat, Beelzebub. You mistakenly mix one table- spoon of sugar into the tub (you thought it was fish food!).

where the subscripts 1 and 2 designate the top and bot- tom pendulum, respectively, θ = angle (radians) with 0 = vertical downward and counter-clockwise positive, t = time (s), g = gravitational acceleration (= 9.81 m/s2), m = mass (kg), and L = length (m). Note that the x and y coordinates of the masses are functions of the angles as in

x1 = L1 sinθ1 y1 = −L1 cos θ1

x2 = x1 + L2 sin θ2 y2 = y1 − L2 cosθ2

(a) Use ode45 to solve for the angles and angular velocities of the masses as a function of time from t = 0 to 40 s. Employ subplot to create a stacked plot with a time se- ries of the angles in the top panel and a state space plot of θ2 versus θ1 in the bottom panel. (b) Create an ani- mated plot depicting the motion of the pendulum. Test your code for the following:

Case 1 (small displacement): L1 = L2 = 1 m, m1 = m2 = 0.25 kg, with initial conditions: θ1 = 0.5 m and θ2 = dθ1∕dt = dθ2∕dt = 0. Case 2 (large displacement): L1 = L2 = 1 m, m1 = 0.5 kg, m2 = 0.25 kg, with initial conditions: θ1 = 1 m and θ2 = dθ1∕dt = dθ2∕dt = 0.

23.24 Figure P23.24 shows the forces exerted on a hot air balloon system.

Formulate the drag force as

FD = 1 __ 2 ρa υ

2 A Cd

m = mG + mP

where mG = the mass of the gas inside the expanded balloon (kg), and mP = the mass of the payload (basket, passengers, and the unexpanded balloon = 265 kg). Assume that the ideal gas law holds (P = ρRT), that the balloon is a perfect sphere with a diameter of 17.3 m, and that the heated air inside the envelope is at roughly the same pressure as the outside air. Other necessary parameters are normal atmo- spheric pressure, P = 101,300 Pa; gas constant for dry air, R = 287 Joules/kg · K; average temperature of air inside the

FIGURE P23.24 Forces on a hot air balloon: FB = buoyancy, FG = weight of gas, FP = weight of payload (including the balloon envelope), and FD = drag. Note that the direction of the drag is downward when the balloon is rising.

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644 AdAptive MethodS And Stiff SySteMS

(b) Use ode45 to solve for θ and dθ∕dt as a function of time for the case where l = 0.65 m and the initial conditions are θ = π∕8 and dθ∕dt = 0. (c) Generate a plot of your results, and (d) use the diff function to generate a plot of the angu- lar accelerations (d 2θ∕dt 2) versus time based on the vector of angular velocities (dθ∕dt) generated in (b). Use subplot to display all graphs as a single vertical three-panel plot with the top, middle, and bottom plots corresponding to θ, dθ∕dt, and d 2θ∕dt 2, respectively. 23.29 A number of individuals have made skydives from very high altitudes. Suppose that an 80-kg skydiver launches from an elevation of 36.500 km above the earth’s surface. The skydiver has a projected area, A = 0.55 m2; and a dimen- sionless drag coefficient, Cd = 1. Note that the gravitational acceleration, g (m/s2), can be related to elevation by

g = 9.806412 − 0.003039734z

where z = elevation above the earth’s surface (km) and the density of air, ρ (kg/m3), at various elevations can be tabulated as

z (km) ρ (kg/m3) z (km) ρ (kg/m3) z (km) ρ (kg/m3)

−1 1.3470 6 0.6601 25 0.04008 0 1.2250 7 0.5900 30 0.01841 1 1.1120 8 0.5258 40 0.003996 2 1.0070 9 0.4671 50 0.001027 3 0.9093 10 0.4135 60 0.0003097 4 0.8194 15 0.1948 70 8.283 × 10−5

5 0.7364 20 0.08891 80 1.846 × 10−5

(a) Based on a force balance between gravity and drag, derive differential equations for velocity and distance based on a force balance for the skydiver.

(b) Use a numerical method to solve for velocity and dis- tance that terminates when the jumper reaches an eleva- tion that is a kilometer above the earth’s surface.

(c) Plot your results. 23.30 As depicted in Fig. P23.30, a parachutist jumps from an aircraft that is flying in a straight line parallel with the ground. (a) Using force balances derive four differential equations for the rates of change of the x and y components of distances and velocities. [Hint: Recognize that sin θ = υy∕υ and cos θ = υx ∕x]. (b) Employ Ralston’s 2nd-order method with Δt = 0.25 s to generate a solution from t = 0 until the parachutist hits the ground assuming that the chute never opens. The drag coefficient is 0.25 kg/m, the mass is 80 kg, and the ground is 2000 km below the initial vertical position

Unfortunately, there are bacteria in the water (remember you got rid of the chlorine!), which break down the sugar consuming dissolved oxygen in the process. The oxidation reaction follows first-order kinetics with a reaction rate of kd = 0.15∕d. The tub initially has a sugar concentration of 20 mgO2/L and an oxygen concentration of 8.4 mgO2/L. Note that the mass balances for the sugar (expressed in ox- ygen equivalents) and dissolved oxygen can be written as

d L ___ d t = − k d L

d o ___ d t = − k d L

where L = sugar concentration expressed as oxygen equiva- lents (mg/L), t = time (d), and o = dissolved oxygen con- centration (mg/L). Thus, as the sugar gets oxidized, an equivalent amount of oxygen is lost from the tub. Develop a MATLAB script using ode45 to numerically compute the concentrations of sugar and oxygen as a function of time and develop plots of each versus time. Use event to auto- matically stop when the oxygen concentration falls below a critical oxygen level of 2 mgO2/L. 23.27 The growth of bacteria from substrate can be repre- sented by the following pair of differential equations

d X ___ d t = Y kmax S _____ ks + S

dS ___ d t = − k max S ______ ks + S

where X = bacterial biomass, t = time (d), Y = a yield coef- ficient, kmax = maximum bacterial growth rate, S = substrate concentration, and ks = half saturation constant. The param- eter values are Y = 0.75, kmax = 0.3, and ks = 1 × 10–4 and the initial conditions at t = 0 are S(0) = 5 and X(0) = 0.05. Note that neither X nor S can fall below zero as negative val- ues are impossible. (a) Use ode23 to solve for X and S from t = 0 to 20. (b) Repeat the solution, but set the relative toler- ance to 1 × 10–6. (c) Keep repeating the solution with the relative tolerance set to 1 × 10–6, but determine which of the MATLAB ode solvers (including the stiff solvers) obtains correct (i.e., positive) results. Use the tic and toc functions to determine the execution time for each option. 23.28 The oscillations of a swinging pendulum can be sim- ulated with the following nonlinear model:

d 2 θ ____

d t 2 = − g __ l sin θ

where θ = the angle of displacement (radians), g = the gravi- tational constant (= 9.81 m/s2), and l = the pendulum length. (a) Express this equation as a pair of first-order ODEs.

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pRoBLeMS 645

23.32 The following differential equations define the con- centrations of three reactants in a closed system (Fig. P23.32),

d c1 ___ d t = − k 12 c 1 + k 21 c 2 + k 31 c 3

d c2 ___ d t = k12 c1 − k21 c2 − k32 c2

d c3 ___ d t = k32 c2 − k31 c3

An experiments with initial conditions of c1(0) = 100, and c2(0) = c3(0) = 0 yields the following data:

t 1 2 3 4 5 6 8 9 10 12 15

c1 85.366.660.6 56.1 49.1 45.3 41.9 37.833.7 34.4 35.1 c2 16.918.724.1 20.9 18.9 19.9 20.6 13.919.1 14.5 15.4 c3 4.7 7.9 20.1 22.8 32.5 37.7 42.4 47 50.5 52.3 51.3

Use ode45 to integrate the equations and an optimization function to estimate the values of the k’s that minimize the sum of the squares of the discrepancies between the model predictions and the data. Employ initial guesses of 0.15 for all the k’s.

of the aircraft. The initial conditions are υx = 135 m/s, and υy = x = y = 0. (c) Develop a plot of position on Cartesian (x − y) coordinates. 23.31 The basic differential equation of the elastic curve for a cantilever beam (Fig. P23.31) is given as

EI d 2 y ____

d x 2 = − P(L − x)

where E = the modulus of elasticity and I the moment of inertia. Solve for the deflection of the beam using ode45. The following parameter values apply: E = 2 × 1011 Pa, I = 0.00033 m4, P = 4.5 kN, and L = 3 m. Develop a plot of your results along with the analytical solution,

y = − PL x 2 _____ 2EI +

P x 3 ____ 6EI

FIGURE P23.30

FIGURE P23.31

k 21

k 12 k 31

c 1 c 2 c 3 k 23

FIGURE P23.32

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646

24 Boundary-Value Problems

YOU’VE GOT A PROBLEM

To this point, we have been computing the velocity of a free-falling bungee jumper by integrating a single ODE: d υ ___ d t = g −

cd __ m υ 2 (24.1)

CHAPTER OBJECTIVES

The primary objective of this chapter is to introduce you to solving boundary-value problems for ODEs. Specific objectives and topics covered are

• Understanding the difference between initial-value and boundary-value problems. • Knowing how to express an nth-order ODE as a system of n first-order ODEs. • Knowing how to implement the shooting method for linear ODEs by using linear

interpolation to generate accurate “shots.” • Understanding how derivative boundary conditions are incorporated into the

shooting method. • Knowing how to solve nonlinear ODEs with the shooting method by using root

location to generate accurate “shots.” • Knowing how to implement the finite-difference method. • Understanding how derivative boundary conditions are incorporated into the

finite-difference method. • Knowing how to solve nonlinear ODEs with the finite-difference method by using

root-location methods for systems of nonlinear algebraic equations. • Familiarizing yourself with the built-in MATLAB function bvp4c for solving

boundary-value ODEs.

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24.1 INTRODUCTION AND BACKGROUND 647

Suppose that rather than velocity, you are asked to determine the position of the jumper as a function of time. One way to do this is to recognize that velocity is the first derivative of distance:

d x ___ d t = υ (24.2)

Thus, by solving the system of two ODEs represented by Eqs. (24.1) and (24.2), we can simultaneously determine both the velocity and the position.

However, because we are now integrating two ODEs, we require two conditions to obtain the solution. We are already familiar with one way to do this for the case where we have values for both position and velocity at the initial time:

x (t = 0) = xi υ (t = 0) = υi

Given such conditions, we can easily integrate the ODEs using the numerical techniques described in Chaps. 22 and 23. This is referred to as an initial-value problem.

But what if we do not know values for both position and velocity at t = 0? Let’s say that we know the initial position but rather than having the initial velocity, we want the jumper to be at a specified position at a later time. In other words:

x (t = 0) = xi x (t = tf ) = xf

Because the two conditions are given at different values of the independent variable, this is called a boundary-value problem.

Such problems require special solution techniques. Some of these are related to the methods for initial value problems that were described in the previous two chapters. How- ever, others employ entirely different strategies to obtain solutions. This chapter is de- signed to introduce you to the more common of these methods.

24.1 INTRODUCTION AND BACKGROUND

24.1.1 What Are Boundary-Value Problems?

An ordinary differential equation is accompanied by auxiliary conditions, which are used to evaluate the constants of integration that result during the solution of the equation. For an nth-order equation, n conditions are required. If all the conditions are specified at the same value of the independent variable, then we are dealing with an initial-value problem (Fig. 24.1a). To this point, the material in Part Six (Chaps. 22 and 23) has been devoted to this type of problem.

In contrast, there are often cases when the conditions are not known at a single point but rather are given at different values of the independent variable. Because these values are often specified at the extreme points or boundaries of a system, they are customarily referred to as boundary-value problems (Fig. 24.1b). A variety of significant engineering applications fall within this class. In this chapter, we discuss some of the basic approaches for solving such problems.

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648 BOUNDARy-VAlUe PROBlems

24.1.2 Boundary-Value Problems in Engineering and Science

At the beginning of this chapter, we showed how the determination of the position and velocity of a falling object could be formulated as a boundary-value problem. For that example, a pair of ODEs was integrated in time. Although other time-variable examples can be developed, boundary-value problems arise more naturally when integrating in space. This occurs because auxiliary conditions are often specified at different positions in space.

A case in point is the simulation of the steady-state temperature distribution for a long, thin rod positioned between two constant-temperature walls (Fig. 24.2). The rod’s cross- sectional dimensions are small enough so that radial temperature gradients are minimal and, consequently, temperature is a function exclusively of the axial coordinate x. Heat is transferred along the rod’s longitudinal axis by conduction and between the rod and the surrounding gas by convection. For this example, radiation is assumed to be negligible.1

t y2, 0

y1, 0

(a)

(b)

Initial conditions

Boundary condition

Boundary conditiony

0 L

= f1(t, y1, y2) dy1 dt

= f2(t, y1, y2) dy2 dt

where at t = 0, y1 = y1, 0 and y2 = y2, 0

= f (x, y) d2y dx2

where at x = 0, y = y0 x = L, y = yL

FIGURE 24.1 Initial-value versus boundary-value problems. (a) An initial-value problem where all the conditions are specified at the same value of the independent variable. (b) A boundary-value problem where the conditions are specified at different values of the independent variable.

1 We incorporate radiation into this problem later in this chapter in Example 24.4.

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24.1 INTRODUCTION AND BACKGROUND 649

As depicted in Fig. 24.2, a heat balance can be taken around a differential element of thickness Δ x as

0 = q (x) Ac − q (x + Δ x) Ac + h As(T∞ − T ) (24.3) where q(x) = flux into the element due to conduction [J/(m2 · s)]; q(x + Δ x) = flux out of the element due to conduction [J/(m2 · s)]; Ac = cross-sectional area [m2] = π r 2, r = the radius [m]; h = the convection heat transfer coefficient [J/(m2 · K · s)]; As = the element’s surface area [m2] = 2π r Δx; T∞ = the temperature of the surrounding gas [K]; and T = the rod’s temperature [K].

Equation (24.3) can be divided by the element’s volume (π r 2Δ x) to yield

0 = q (x) − q (x + Δ x) ______________ Δ x + 2h ___ r (T∞ − T )

Taking the limit Δx → 0 gives

0 = − d q ___ d x + 2h ___ r (T∞ − T ) (24.4)

The flux can be related to the temperature gradient by Fourier’s law:

q = − k d T ___ d x (24.5)

where k = the coefficient of thermal conductivity [J/(s · m · K)]. Equation (24.5) can be dif- ferentiated with respect to x, substituted into Eq. (24.4), and the result divided by k to yield,

0 = d 2T ____

d x 2 + h′ (T∞ − T ) (24.6)

where h′ = a bulk heat-transfer parameter reflecting the relative impacts of convection and conduction [m−2] = 2h∕(rk).

Equation (24.6) represents a mathematical model that can be used to compute the tem- perature along the rod’s axial dimension. Because it is a second-order ODE, two conditions

T∞

Ta TbConduction

Convection

x + ∆x

∆x

FIGURE 24.2 A heat balance for a differential element of a heated rod subject to conduction and convection.

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650 BOUNDARy-VAlUe PROBlems

are required to obtain a solution. As depicted in Fig. 24.2, a common case is where the temperatures at the ends of the rod are held at fixed values. These can be expressed math- ematically as

T (0) = Ta T (L) = Tb

The fact that they physically represent the conditions at the rod’s “boundaries” is the origin of the terminology: boundary conditions.

Given these conditions, the model represented by Eq. (24.6) can be solved. Because this particular ODE is linear, an analytical solution is possible as illustrated in the follow- ing example.

eXAmPle 24.1 Analytical solution for a Heated Rod

Problem statement. Use calculus to solve Eq. (24.6) for a 10-m rod with h′ = 0.05 m−2[h = 1 J/(m2 · K · s), r = 0.2 m, k = 200 J/(s · m · K)], T∞ = 200 K, and the bound ary conditions:

T (0) = 300 K T (10) = 400 K

solution. This ODE can be solved in a number of ways. A straightforward approach is to first express the equation as

d 2T ____

d x 2 − h′T = −h′T∞

Because this is a linear ODE with constant coefficients, the general solution can be readily obtained by setting the right-hand side to zero and assuming a solution of the form T = eλx. Substituting this solution along with its second derivative into the homogeneous form of the ODE yields

λ2eλx − h′eλx = 0

which can be solved for λ = ± √ __

h′ . Thus, the general solution is

T = Aeλx + Be−λx

where A and B are constants of integration. Using the method of undetermined coefficients we can derive the particular solution T = T∞. Therefore, the total solution is

T = T∞ + Aeλx + Be−λx

The constants can be evaluated by applying the boundary conditions

Ta = T∞ + A + B Tb = T∞ + Ae λ L + Be−λL

These two equations can be solved simultaneously for

A = (Ta − T∞)e−λ L − (Tb − T∞) ______________________

e−λ L − e λ L

B = (Tb − T∞) − (Ta − T∞)e λL ______________________

e−λ L − e λ L

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24.2 THe sHOOTING meTHOD 651

Substituting the parameter values from this problem gives A = 20.4671 and B = 79.5329. Therefore, the final solution is

T = 200 + 20.4671e √ _____

0.05x + 79.5329e− √ _____

0.05x (24.7) As can be seen in Fig. 24.3, the solution is a smooth curve connecting the two bound-

ary temperatures. The temperature in the middle is depressed due to the convective heat loss to the cooler surrounding gas.

In the following sections, we will illustrate numerical approaches for solving the same problem we just solved analytically in Example 24.1. The exact analytical solution will be useful in assessing the accuracy of the solutions obtained with the approximate, numerical methods.

24.2 THE SHOOTING METHOD

The shooting method is based on converting the boundary-value problem into an equiva- lent initial-value problem. A trial-and-error approach is then implemented to develop a solution for the initial-value version that satisfies the given boundary conditions.

Although the method can be employed for higher-order and nonlinear equations, it is nicely illustrated for a second-order, linear ODE such as the heated rod described in the previous section:

0 = d 2T ____

d x 2 + h′(T∞ − T ) (24.8)

subject to the boundary conditions

T (0) = Ta T (L) = Tb We convert this boundary-value problem into an initial-value problem by defining the

rate of change of temperature, or gradient, as

d T ___ d x = z (24.9)

x, m

T, K

400

300

200 0 2 4 6 8 10

FIGURE 24.3 Analytical solution for the heated rod.

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652 BOUNDARy-VAlUe PROBlems

and reexpressing Eq. (24.8) as

d z ___ d x = −h′(T∞ − T ) (24.10)

Thus, we have converted the single second-order equation (Eq. 24.8) into a pair of first- order ODEs [Eqs. (24.9) and (24.10)].

If we had initial conditions for both T and z, we could solve these equations as an initial- value problem with the methods described in Chaps. 22 and 23. However, because we only have an initial value for one of the variables T (0) = Ta we simply make a guess for the other z(0) = za1 and then perform the integration.

After performing the integration, we will have generated a value of T at the end of the interval, which we will call Tb1. Unless we are incredibly lucky, this result will differ from the desired result Tb.

Now, let’s say that the value of Tb1 is too high (Tb1 > Tb ), it would make sense that a lower value of the initial slope z(0) = za2 might result in a better prediction. Using this new guess, we can integrate again to generate a second result at the end of the interval Tb2. We could then continue guessing in a trial-and-error fashion until we arrived at a guess for z(0) that resulted in the correct value of T (L) = Tb.

At this point, the origin of the name shooting method should be pretty clear. Just as you would adjust the angle of a cannon in order to hit a target, we are adjusting the trajectory of our solution by guessing values of z (0) until we hit our target T (L) = Tb.

Although we could certainly keep guessing, a more efficient strategy is possible for linear ODEs. In such cases, the trajectory of the perfect shot z a is linearly related to the results of our two erroneous shots (za1, Tb1) and (za2, Tb2). Consequently, linear interpolation can be employed to arrive at the required trajectory:

za = za1 + za2 − za1 ________ Tb2 − Tb1

(Tb − Tb1) (24.11)

The approach can be illustrated by an example.

eXAmPle 24.2 The shooting method for a linear ODe

Problem statement. Use the shooting method to solve Eq. (24.6) for the same condi- tions as Example 24.1: L = 10 m, h′ = 0.05 m−2, T∞ = 200 K, T (0) = 300 K, and T (10) = 400 K.

solution. Equation (24.6) is first expressed as a pair of first-order ODEs:

d T ___ d x = z

d z ___ d x = − 0.05(200 − T )

Along with the initial value for temperature T (0) = 300 K, we arbitrarily guess a value of za1 = −5 K/m for the initial value for z (0). The solution is then obtained by integrating the pair of ODEs from x = 0 to 10. We can do this with MATLAB’s ode45 function by first setting up an M-file to hold the differential equations:

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24.2 THe sHOOTING meTHOD 653

function dy = Ex2402(x,y) dy = [y(2);−0.05*(200 − y(l))];

We can then generate the solution as >> [t,y] = ode45(@ Ex2402,[0 10],[300,− 5]); >> Tb1 = y(length(y))

Tb1 = 569.7539

Thus, we obtain a value at the end of the interval of Tb1 = 569.7539 (Fig. 24.4a), which differs from the desired boundary condition of Tb = 400. Therefore, we make another guess za2 = −20 and perform the computation again. This time, the result of Tb2 = 259.5131 is obtained (Fig. 24.4b).

600

400

200 (a)

600

400

200 (b)

600

400

200 0 2 4 6 8 10

(c)

FIGURE 24.4 Temperature (K) versus distance (m) computed with the shooting method: (a) the first “shot,” (b) the second “shot,” and (c) the final exact “hit.”

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654 BOUNDARy-VAlUe PROBlems

Now, because the original ODE is linear, we can use Eq. (24.11) to determine the correct trajectory to yield the perfect shot:

za = −5 + −20 − (−5) __________________ 259.5131 − 569.7539 (400 − 569.7539) = −13.2075

This value can then be used in conjunction with ode45 to generate the correct solution, as depicted in Fig. 24.4c.

Although it is not obvious from the graph, the analytical solution is also plotted on Fig. 24.4c. Thus, the shooting method yields a solution that is virtually indistinguishable from the exact result.

24.2.1 Derivative Boundary Conditions

The fixed or Dirichlet boundary condition discussed to this point is but one of several types that are used in engineering and science. A common alternative is the case where the derivative is given. This is commonly referred to as a Neumann boundary condition.

Because it is already set up to compute both the dependent variable and its deriva- tive, incorporating derivative boundary conditions into the shooting method is relatively straightforward.

Just as with the fixed-boundary condition case, we first express the second-order ODE as a pair of first-order ODEs. At this point, one of the required initial conditions, whether the dependent variable or its derivative, will be unknown. Based on guesses for the missing initial condition, we generate solutions to compute the given end condition. As with the initial condition, this end condition can either be for the dependent variable or its deriva- tive. For linear ODEs, interpolation can then be used to determine the value of the missing initial condition required to generate the final, perfect “shot” that hits the end condition.

eXAmPle 24.3 The shooting method with Derivative Boundary Conditions

Problem statement. Use the shooting method to solve Eq. (24.6) for the rod in Exa mple 24.1: L = 10 m, h′ = 0.05 m−2 [ h = 1 J/(m2 · K · s), r = 0.2 m, k = 200 J/ (s · m · K)], T∞ = 200 K, and T(10) = 400 K. However, for this case, rather than hav- ing a fixed temperature of 300 K, the left end is subject to convection as in Fig. 24.5.

T∞

T∞ TbConduction

ConvectionConvection

FIGURE 24.5 A rod with a convective boundary condition at one end and a fixed temperature at the other.

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24.2 THe sHOOTING meTHOD 655

For simplicity, we will assume that the convection heat transfer coefficient for the end area is the same as for the rod’s surface.

solution. As in Example 24.2, Eq. (24.6) is first expressed as

d T ___ d x = z

d z ___ d x = −0.05(200 − T )

Although it might not be obvious, convection through the end is equivalent to speci- fying a gradient boundary condition. In order to see this, we must recognize that because the system is at steady state, convection must equal conduction at the rod’s left boundary (x = 0). Using Fourier’s law [Eq. (24.5)] to represent conduction, the heat balance at the end can be formulated as

h Ac(T∞ − T (0)) = −k Ac d T ___ d x (0) (24.12)

This equation can be solved for the gradient

d T ___ d x (0) = h __ k (T(0) − T∞) (24.13)

If we guess a value for temperature, we can see that this equation specifies the gradient. The shooting method is implemented by arbitrarily guessing a value for T (0). If we

choose a value of T (0) = Ta1 = 300 K, Eq. (24.13) then yields the initial value for the gradient

za1 = d T ___ d x (0) =

1 ____ 200 (300 − 200) = 0.5

The solution is obtained by integrating the pair of ODEs from x = 0 to 10. We can do this with MATLAB’s ode45 function by first setting up an M-file to hold the differential equa- tions in the same fashion as in Example 24.2. We can then generate the solution as

>> [t,y] = ode45(@ Ex2402,[0 10],[300,0.5]); >> Tb1 = y(length(y))

Tb1 = 683.5088

As expected, the value at the end of the interval of Tb1 = 683.5088 K differs from the desired boundary condition of Tb = 400. Therefore, we make another guess Ta2 = 150 K, which corresponds to za2 = −0.25, and perform the computation again.

>> [t,y] = ode45(@ Ex2402,[0 10],[150,−0.25]); >> Tb2 = y(length(y))

Tb2 = −41.7544

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656 BOUNDARy-VAlUe PROBlems

Linear interpolation can then be employed to compute the correct initial temperature:

Ta = 300 + 150 − 300 ___________________ − 41.7544 − 683.5088 (400 − 683.5088) = 241.3643 K

which corresponds to a gradient of z a = 0.2068. Using these initial conditions, ode45 can be employed to generate the correct solution, as depicted in Fig. 24.6.

Note that we can verify that our boundary condition has been satisfied by substituting the initial conditions into Eq. (24.12) to give

1 J ______ m2 K s

π × (0.2 m)2 × (200 K − 241.3643 K ) = − 200 J _____ m K s π × (0.2 m) 2 × 0.2068 K __ m

which can be evaluated to yield −5.1980 J/s = −5.1980 J/s. Thus, conduction and convec- tion are equal and transfer heat out of the left end of the rod at a rate of 5.1980 W.

24.2.2 The Shooting Method for Nonlinear ODEs

For nonlinear boundary-value problems, linear interpolation or extrapolation through two solution points will not necessarily result in an accurate estimate of the required bound- ary condition to attain an exact solution. An alternative is to perform three applications of the shooting method and use a quadratic interpolating polynomial to estimate the proper boundary condition. However, it is unlikely that such an approach would yield the exact answer, and additional iterations would be necessary to home in on the solution.

Another approach for a nonlinear problem involves recasting it as a roots problem. Re- call that the general goal of a roots problem is to find the value of x that makes the function f (x) = 0. Now, let us use the heated rod problem to understand how the shooting method can be recast in this form.

First, recognize that the solution of the pair of differential equations is also a “func- tion” in the sense that we guess a condition at the left-hand end of the rod z a, and the inte- gration yields a prediction of the temperature at the right-hand end Tb. Thus, we can think of the integration as

Tb = f (za)

x, m

T, K

400

300

200 0 2 4 6 8 10

FIGURE 24.6 The solution of a second-order ODE with a convective boundary condition at one end and a fixed temperature at the other.

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24.2 THe sHOOTING meTHOD 657

That is, it represents a process whereby a guess of za yields a prediction of Tb. Viewed in this way, we can see that what we desire is the value of z a that yields a specific value of Tb. If, as in the example, we desire Tb = 400, the problem can be posed as

400 = f (za )

By bringing the goal of 400 over to the right-hand side of the equation, we generate a new function res(za ) that represents the difference, or residual, between what we have, f (za ), and what we want, 400.

res(za ) = f (za ) − 400

If we drive this new function to zero, we will obtain the solution. The next example illus- trates the approach.

eXAmPle 24.4 The shooting method for Nonlinear ODes

Problem statement. Although it served our purposes for illustrating the shooting method, Eq. (24.6) was not a completely realistic model for a heated rod. For one thing, such a rod would lose heat by mechanisms such as radiation that are nonlinear.

Suppose that the following nonlinear ODE is used to simulate the temperature of the heated rod:

0 = d 2T ____

d x 2 + h′ (T∞ − T ) + σ ′′ ( T ∞ 4 − T 4 )

where σ ′ = a bulk heat-transfer parameter reflecting the relative impacts of radiation and conduction = 2.7 × 10−9 K−3 m−2. This equation can serve to illustrate how the shooting method is used to solve a two-point nonlinear boundary-value problem. The remaining problem conditions are as specified in Example 24.2: L = 10 m, h′ = 0.05 m−2, T∞ = 200 K, T (0) = 300 K, and T (10) = 400 K.

solution. Just as with the linear ODE, the nonlinear second-order equation is first ex- pressed as two first-order ODEs:

d T ___ d x = z

d z ___ d x = − 0.05(200 − T ) − 2.7 × 10 −9(1.6 × 109 − T 4 )

An M-file can be developed to compute the right-hand sides of these equations:

function dy = dydxn(x,y) dy = [y(2);−0.05*(200 − y(1)) − 2.7e − 9*(1.6e9 − y(1)^4)];

Next, we can build a function to hold the residual that we will try to drive to zero as

function r = res(za) [x,y] = ode45(@dydxn,[0 10],[300 za]); r = y(length(x),1) − 400;

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658 BOUNDARy-VAlUe PROBlems

Notice how we use the ode45 function to solve the two ODEs to generate the temperature at the rod’s end: y(length(x),1). We can then find the root with the fzero function:

>> fzero(@res,− 50)

ans = − 41.7434

Thus, we see that if we set the initial trajectory z(0) = − 41.7434, the residual function will be driven to zero and the temperature boundary condition T (10) = 400 at the end of the rod should be satisfied. This can be verified by generating the entire solution and plotting the temperatures versus x:

>> [x,y] = ode45(@ dydxn,[0 10],[300 fzero(@ res,− 50)]); >> plot(x,y(:,1))

The result is shown in Fig. 24.7 along with the original linear case from Example 24.2. As expected, the nonlinear case is depressed lower than the linear model due to the addi- tional heat lost to the surrounding gas by radiation.

24.3 FINITE-DIFFERENCE METHODS

The most common alternatives to the shooting method are finite-difference approaches. In these techniques, finite differences (Chap. 21) are substituted for the derivatives in the original equation. Thus, a linear differential equation is transformed into a set of simultane- ous algebraic equations that can be solved using the methods from Part Three.

We can illustrate the approach for the heated rod model [Eq. (24.6)]:

0 = d 2 T ____

d x 2 + h′(T∞ − T ) (24.14)

The solution domain is first divided into a series of nodes (Fig. 24.8). At each node, finite- difference approximations can be written for the derivatives in the equation. For example,

200

300

400

T, K

0 2 4 6 8

Linear

Nonlinear

10 x, m

FIGURE 24.7 The result of using the shooting method to solve a nonlinear problem.

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24.3 FINITe-DIFFeReNCe meTHODs 659

at node i, the second derivative can be represented by (Fig. 21.5):

d 2T ____

dx 2 =

Ti−1 − 2Ti + Ti+1 ______________ Δx 2

(24.15)

This approximation can be substituted into Eq. (24.14) to give

Ti−1 − 2Ti + Ti+1 ______________

Δx 2 + h′(T∞ − Ti ) = 0

Thus, the differential equation has been converted into an algebraic equation. Collecting terms gives

−Ti−1 + (2 + h′Δ x 2) Ti − Ti+1 = h′ x 2T∞ (24.16)

This equation can be written for each of the n − 1 interior nodes of the rod. The first and last nodes T0 and Tn, respectively, are specified by the boundary conditions. Therefore, the problem reduces to solving n − 1 simultaneous linear algebraic equations for the n − 1 unknowns.

Before providing an example, we should mention two nice features of Eq. (24.16). First, observe that since the nodes are numbered consecutively, and since each equation consists of a node (i) and its adjoining neighbors (i − 1 and i + 1), the resulting set of lin- ear algebraic equations will be tridiagonal. As such, they can be solved with the efficient algorithms that are available for such systems (recall Sec. 9.4).

Further, inspection of the coefficients on the left-hand side of Eq. (24.16) indicates that the system of linear equations will also be diagonally dominant. Hence, convergent solutions can also be generated with iterative techniques like the Gauss-Seidel method (Sec. 12.1).

eXAmPle 24.5 Finite-Difference Approximation of Boundary-Value Problems

Problem statement. Use the finite-difference approach to solve the same problem as in Examples 24.1 and 24.2. Use four interior nodes with a segment length of Δ x = 2 m.

solution. Employing the parameters in Example 24.1 and Δ x = 2 m, we can write Eq. (24.16) for each of the rod’s interior nodes. For example, for node 1:

−T0 + 2.2T1 − T2 = 40

Substituting the boundary condition T0 = 300 gives

2.2T1 − T2 = 340

T0 T1 Ti−1

∆x

Ti Ti+1 Tn−1 Tn

FIGURE 24.8 In order to implement the finite-difference approach, the heated rod is divided into a series of nodes.

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660 BOUNDARy-VAlUe PROBlems

After writing Eq. (24.16) for the other interior nodes, the equations can be assembled in matrix form as

⎡

⎢ ⎣

2.2

−1

2.2

−1

0 0 −1 2.2 −1

−1

2.2

⎤

⎥ ⎦

⎧

⎪ ⎨ ⎪

⎩

⎫

⎪ ⎬ ⎪

⎭ =

⎧

⎪ ⎨ ⎪

⎩

340

440

⎫

⎪ ⎬ ⎪

⎭

Notice that the matrix is both tridiagonal and diagonally dominant. MATLAB can be used to generate the solution:

>> A = [2.2 −1 0 0; −1 2.2 −1 0; 0 −1 2.2 −1; 0 0 −1 2.2]; >> b = [340 40 40 440]'; >> T = A\b

T = 283.2660 283.1853 299.7416 336.2462

Table 24.1 provides a comparison between the analytical solution (Eq. 24.7) and the numerical solutions obtained with the shooting method (Example 24.2) and the finite- difference method (Example 24.5). Note that although there are some discrepancies, the numerical approaches agree reasonably well with the analytical solution. Further, the biggest discrepancy occurs for the finite-difference method due to the coarse node spacing we used in Example 24.5. Better agreement would occur if a finer nodal spacing had been used.

TABLE 24.1 Comparison of the exact analytical solution for temperature with the results obtained with the shooting and finite-difference methods.

Analytical Shooting Finite x Solution Method Difference

0 300 300 300 2 282.8634 282.8889 283.2660 4 282.5775 282.6158 283.1853 6 299.0843 299.1254 299.7416 8 335.7404 335.7718 336.2462 10 400 400 400

24.3.1 Derivative Boundary Conditions

As mentioned in our discussion of the shooting method, the fixed or Dirichlet boundary condition is but one of several types that are used in engineering and science. A common

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24.3 FINITe-DIFFeReNCe meTHODs 661

alternative, called the Neumann boundary condition, is the case where the derivative is given.

We can use the heated rod introduced earlier in this chapter to demonstrate how a derivative boundary condition can be incorporated into the finite-difference approach:

0 = d 2 T ____

d x 2 + h′(T∞ − T )

However, in contrast to our previous discussions, we will prescribe a derivative boundary condition at one end of the rod:

d T ___ d x (0) = T a ′

T(L) = Tb Thus, we have a derivative boundary condition at one end of the solution domain and a fixed boundary condition at the other.

Just as in the previous section, the rod is divided into a series of nodes and a finite- difference version of the differential equation (Eq. 24.16) is applied to each interior node. However, because its temperature is not specified, the node at the left end must also be included. Fig. 24.9 depicts the node (0) at the left edge of a heated plate for which the derivative boundary condition applies. Writing Eq. (24.16) for this node gives

−T−1 + (2 + h′ Δ x 2)T0 − T1 = h′ Δ x2 T∞ (24.17)

Notice that an imaginary node (−1) lying to the left of the rod’s end is required for this equation. Although this exterior point might seem to represent a difficulty, it actually serves as the vehicle for incorporating the derivative boundary condition into the problem. This is done by representing the first derivative in the x dimension at (0) by the centered difference [Eq. (4.25)]:

d T ___ d x = T1 − T−1 _______ 2Δ x

which can be solved for

T−1 = T1 − 2Δ x d T ___ d x

T−1 T0 T1

∆x ∆x

FIGURE 24.9 A boundary node at the left end of a heated rod. To approximate the derivative at the boundary, an imaginary node is located a distance Δ x to the left of the rod’s end.

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662 BOUNDARy-VAlUe PROBlems

Now we have a formula for T−1 that actually reflects the impact of the derivative. It can be substituted into Eq. (24.17) to give

(2 + h′Δ x 2)T0 − 2T1 = h′Δ x 2T∞ − 2Δ x d T ___ d x (24.18)

Consequently, we have incorporated the derivative into the balance. A common example of a derivative boundary condition is the situation where the end

of the rod is insulated. In this case, the derivative is set to zero. This conclusion follows directly from Fourier’s law [Eq. (24.5)], because insulating a boundary means that the heat flux (and consequently the gradient) must be zero. The following example illustrates how the solution is affected by such boundary conditions.

eXAmPle 24.6 Incorporating Derivative Boundary Conditions

Problem statement. Generate the finite-difference solution for a 10-m rod with Δ x = 2 m, h′ = 0.05 m−2, T∞ = 200 K, and the boundary conditions: T a ′ = 0 and Tb = 400 K. Note that the first condition means that the slope of the solution should approach zero at the rod’s left end. Aside from this case, also generate the solution for dT∕dx = −20 at x = 0.

solution. Equation (24.18) can be used to represent node 0 as

2.2T0 − 2T1 = 40

We can write Eq. (24.16) for the interior nodes. For example, for node 1,

−T0 + 2.2T1 − T2 = 40

A similar approach can be used for the remaining interior nodes. The final system of equa- tions can be assembled in matrix form as

⎡

⎢ ⎣

2.2

−2

−1

2.2

−1

−1 2.2 −1

−1

2.2

−1

2.2

⎤

⎥ ⎦

⎧

⎪

⎨ ⎪

⎩

T2 T3

⎫

⎪

⎬ ⎪

⎭

⎧

⎪

⎨ ⎪

⎩

40 40

440

⎫

⎪

⎬ ⎪

⎭

These equations can be solved for

T0 = 243.0278 T1 = 247.3306 T2 = 261.0994 T3 = 287.0882 T4 = 330.4946

As displayed in Fig. 24.10, the solution is flat at x = 0 due to the zero derivative condition and then curves upward to the fixed condition of T = 400 at x = 10.

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24.3 FINITe-DIFFeReNCe meTHODs 663

For the case where the derivative at x = 0 is set to −20, the simultaneous equations are

⎡

⎢ ⎣

2.2

−2

−1

2.2

−1

−1 2.2 −1

−1

2.2

−1

2.2

⎤

⎥ ⎦

⎧

⎪

⎨ ⎪

⎩

T2 T3

⎫

⎪

⎬ ⎪

⎭

⎧

⎪

⎨ ⎪

⎩

120

40 40

440

⎫

⎪

⎬ ⎪

⎭

which can be solved for

T0 = 328.2710 T1 = 301.0981 T2 = 294.1448 T3 = 306.0204 T4 = 339.1002

As in Fig. 24.10, the solution at x = 0 now curves downward due to the negative derivative we imposed at the boundary.

24.3.2 Finite-Difference Approaches for Nonlinear ODEs

For nonlinear ODEs, the substitution of finite differences yields a system of nonlinear simultaneous equations. Thus, the most general approach to solving such problems is to use root-location methods for systems of equations such as the Newton-Raphson method described in Sec. 12.2.2. Although this approach is certainly feasible, an adaptation of suc- cessive substitution can sometimes provide a simpler alternative.

The heated rod with convection and radiation introduced in Example 24.4 provides a nice vehicle for demonstrating this approach,

0 = d 2 T ____

d x 2 + h′(T∞ − T ) + σ′′ ( T ∞ 4 − T 4)

200

300

400

T, K

T'(0) = 0

T'(0) = −20

0 2 4 6 8 10 x, m

FIGURE 24.10 The solution of a second-order ODE with a derivative boundary condition at one end and a fixed boundary condition at the other. Two cases are shown reflecting different derivative values at x = 0.

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664 BOUNDARy-VAlUe PROBlems

We can convert this differential equation into algebraic form by writing it for a node i and substituting Eq. (24.15) for the second derivative:

0 = Ti−1 − 2Ti + Ti+1 ______________

Δ x 2 + h′ (T∞ − Ti ) + σ ′′( T ∞ 4 − T i 4 )

Collecting terms gives

−Ti−1 + (2 + h′Δ x2 ) Ti − Ti+1 = h′Δ x2 T∞ + σ ′′Δ x2 ( T ∞ 4 − T i 4 )

Notice that although there is a nonlinear term on the right-hand side, the left-hand side is expressed in the form of a linear algebraic system that is diagonally dominant. If we as- sume that the unknown nonlinear term on the right is equal to its value from the previous iteration, the equation can be solved for

Ti = h′Δ x2 T∞ + σ ′′Δ x2 ( T ∞ 4 − T i

4 ) + Ti−1 + Ti+1 ___________________________________ 2 + h′Δ x2

(24.19)

As in the Gauss-Seidel method, we can use Eq. (24.19) to successively calculate the tem- perature of each node and iterate until the process converges to an acceptable tolerance. Although this approach will not work for all cases, it converges for many ODEs derived from physically based systems. Hence, it can sometimes prove useful for solving problems routinely encountered in engineering and science.

eXAmPle 24.7 The Finite-Difference method for Nonlinear ODes

Problem statement. Use the finite-difference approach to simulate the temperature of a heated rod subject to both convection and radiation:

0 = d 2T ____

d x 2 + h′(T∞ − T ) + σ ′′( T ∞ 4 − T 4)

where σ ′ = 2.7 × 10−9 K−3 m−2, L = 10 m, h′ = 0.05 m−2, T∞ = 200K, T (0) = 300 K, and T (10) = 400 K. Use four interior nodes with a segment length of Δ x = 2 m. Recall that we solved the same problem with the shooting method in Example 24.4.

solution. Using Eq. (24.19) we can successively solve for the temperatures of the rod’s interior nodes. As with the standard Gauss-Seidel technique, the initial values of the interior nodes are zero with the boundary nodes set at the fixed conditions of T0 = 300 and T5 = 400. The results for the first iteration are

T1 = 0.05(2)2 200 + 2.7 × 10−9′(2)2(2004 − 04) + 300 + 0 __________________________________________

2 + 0.05(2)2 = 159.2432

T2 = 0.05(2)2 200 + 2.7 × 10−9′(2)2(2004 − 04) + 159.2432 + 0 __________________________________________

2 + 0.05(2)2 = 97.9674

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24.4 mATlAB FUNCTION: BVP4C 665

T3 = 0.05(2)2 200 + 2.7 × 10−9′(2)2(2004 − 04) + 97.9674 + 0 __________________________________________

2 + 0.05(2)2 = 70.4461

T4 = 0.05(2)2 200 + 2.7 × 10−9′(2)2(2004 − 04) + 70.4461 + 400 __________________________________________

2 + 0.05(2)2 = 226.8704

The process can be continued until we converge on the final result:

T0 = 300 T1 = 250.4827 T2 = 236.2962 T3 = 245.7596 T4 = 286.4921 T5 = 400

These results are displayed in Fig. 24.11 along with the result generated in Example 24.4 with the shooting method.

24.4 MATLAB FUNCTION: bvp4c

The bvp4c function solves ODE boundary-value problems by integrating a system of ordi- nary differential equations of the form y′ = f (x, y) on the interval [a, b], subject to general two-point boundary conditions. A simple representation of its syntax is

sol = bvp4c(odefun,bcfun,solinit)

where sol = a structure containing the solution, odefun = the function that sets up the ODEs to be solved, bcfun = the function that computes the residuals in the boundary conditions, and solinit = a structure with fields holding an initial mesh and initial guesses for the solution.

200

400

T, K

0 2 4 6

Finite di�erence

Shooting

8 10 x, m

FIGURE 24.11 The filled circles are the result of using the finite-difference method to solve a nonlinear problem. The line generated with the shooting method in Example 24.4 is shown for comparison.

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666 BOUNDARy-VAlUe PROBlems

The general format of odefun is

dy = odefun(x,y)

where x = a scalar, y = a column vector holding the dependent variables [y1; y2], and dy = a column vector holding the derivatives [dy1; dy2].

The general format of bcfun is

res = bcfun(ya,yb)

where ya and yb = column vectors holding the values of the dependent variables at the boundary values x = a and x = b, and res = a column vector holding the residuals between the computed and the specified boundary values.

The general format of solinit is

solinit = bvpinit(xmesh, yinit);

where bvpinit = a built-in MATLAB function that creates the guess structure holding the initial mesh and solution guesses, xmesh = a vector holding the ordered nodes of the initial mesh, and yinit = a vector holding the initial guesses. Note that whereas your choices for the initial mesh and guesses will not be of great importance for linear ODEs, they can often be critical for efficiently solving nonlinear equations.

eXAmPle 24.8 solving a Boundary-Value Problem with bvp4c

Problem statement. Use bvp4c to solve the following second-order ODE

d 2 y ___

d x 2 + y = 1

subject to the boundary conditions

y(0) = 1 y(π ∕ 2) = 0

solution. First, express the second-order equation as a pair of first-order ODEs

d y ___ d x = z

d z ___ d x = 1 − y

Next, set up a function to hold the first-order ODEs

function dy = odes(x,y)

dy = [y(2); 1 − y(1)];

We can now develop the function to hold the boundary conditions. This is done just like a roots problem in that we set up two functions that should be zero when the boundary conditions are satisfied. To do this, the vectors of unknowns at the left and right boundaries

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24.4 mATlAB FUNCTION: BVP4C 667

are defined as ya and yb. Hence, the first condition, y(0) = 1, can be formulated as ya(1) − 1; whereas the second condition, y(π∕2) = 0, corresponds to yb(1).

function r = bcs(ya,yb)

r = [ya(1) − 1; yb(1)];

Finally, we can set up solinit to hold the initial mesh and solution guesses with the bvpinit function. We will arbitrarily select 10 equally spaced mesh points, and initial guesses of y = 1 and z = dy∕dx = –1.

solinit = bvpinit (linspace(0,pi/2,10),[1,−1]);

The entire script to generate the solution is

clc solinit = bvpinit(linspace(0,pi/2,10),[1,−1]); sol = bvp4c(@odes,@bcs,solinit); x = linspace(0,pi/2); y = deval(sol,x); plot(x,y(1,:))

where deval is a built-in MATLAB function which evaluates the solution of a differential equation problem with the general syntax

yxint = deval(sol,xint)

where deval evaluates the solution at all the values of the vector xint, and sol is the struc- ture returned by the ODE problem solver (in this case, bvp4c). When the script is run the plot below is generated. Note that the script and functions developed in this example can be applied to other boundary value problems with minor modi- fications. Several end-of-chapter problems are included to test your ability to do just that.

0.9

0.8

0.7

0.6

0.5

0.4

0.3

0.2

0.1

0.20 0.4 0.6 0.8 1 1.2 1.4 1.6 0

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668 BOUNDARy-VAlUe PROBlems

PROBLEMS

24.1 A steady-state heat balance for a rod can be repre- sented as

d 2T ____

d x 2 − 0.15T = 0

Obtain a solution for a 10-m rod with T (0) = 240 and T (10) = 150 (a) analytically, (b) with the shooting method, and (c) using the finite-difference approach with Δ x = 1. 24.2 Repeat Prob. 24.1 but with the right end insulated and the left end temperature fixed at 240. 24.3 Use the shooting method to solve

7 d 2y ___

d x 2 − 2 d y ___ d x − y + x = 0

with the boundary conditions y(0) = 5 and y(20) = 8. 24.4 Solve Prob. 24.3 with the finite-difference approach using Δ x = 2. 24.5 The following nonlinear differential equation was solved in Examples 24.4 and 24.7.

0 = d 2T ____

dx 2 + h′(T∞ − T ) + σ ″ ( T ∞

4 − T 4) (P24.5)

Such equations are sometimes linearized to obtain an ap- proximate solution. This is done by employing a first-order Taylor series expansion to linearize the quartic term in the equation as

σ ′T 4 = σ ′ __

T 4 + 4σ ′ __

T 3(T − __

T )

where __

T is a base temperature about which the term is linear- ized. Substitute this relationship into Eq. (P24.5), and then solve the resulting linear equation with the finite- difference approach. Employ

__ T = 300, Δ x = 1 m, and the parameters

from Example 24.4 to obtain your solution. Plot your re- sults along with those obtained for the nonlinear versions in Examples 24.4 and 24.7. 24.6 Develop an M-file to implement the shooting method for a linear second-order ODE. Test the program by dupli- cating Example 24.2. 24.7 Develop an M-file to implement the finite-differ- ence approach for solving a linear second-order ODE with Dirichlet boundary conditions. Test it by duplicating Example 24.5. 24.8 An insulated heated rod with a uniform heat source can be modeled with the Poisson equation:

d 2T ____

d x 2 = − f (x)

Given a heat source f (x) = 25 °C/m2 and the boundary con- ditions T (x = 0) = 40 °C and T (x = 10) = 200 °C, solve for the temperature distribution with (a) the shooting method and (b) the finite-difference method (Δ x = 2). 24.9 Repeat Prob. 24.8, but for the following spatially vary- ing heat source: f (x) = 0.12 x 3 − 2.4 x 2 + 12 x. 24.10 The temperature distribution in a tapered conical cooling fin (Fig. P24.10) is described by the following dif- ferential equation, which has been nondimensionalized:

d 2u ____

d x 2 + ( 2 __ x ) ( d u ___ d x − pu ) = 0

where u = temperature (0 ≤ u ≤ 1), x = axial distance (0 ≤ x ≤ 1), and p is a nondimensional parameter that de- scribes the heat transfer and geometry:

p = hL ___ k √ _______

1 + 4 ____ 2m 2

where h = a heat transfer coefficient, k = thermal conductiv- ity, L = the length or height of the cone, and m = the slope of the cone wall. The equation has the boundary conditions:

u(x = 0) = 0 u(x = 1) = 1

Solve this equation for the temperature distribution using finite-difference methods. Use second-order accurate finite- difference formulas for the derivatives. Write a computer

u(x = 1) = 1

u(x = 0) = 0

x = 1

FIGURE P24.10

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program to obtain the solution and plot temperature versus axial distance for various values of p = 10, 20, 50, and 100. 24.11 Compound A diffuses through a 4-cm-long tube and reacts as it diffuses. The equation governing diffusion with reaction is

D d 2A ____

d x 2 − kA = 0

At one end of the tube (x = 0), there is a large source of A that results in a fixed concentration of 0.1 M. At the other end of the tube there is a material that quickly absorbs any A, making the concentration 0 M. If D = 1.5 × 10−6 cm2/s and k = 5 × 10−6 s−1, what is the concentration of A as a function of distance in the tube? 24.12 The following differential equation describes the steady-state concentration of a substance that reacts with first-order kinetics in an axially dispersed plug-flow reactor (Fig. P24.12):

D d 2c ____

d x 2 − U dc ___ d x − kc = 0

where D = the dispersion coefficient (m2/hr), c = concentra- tion (mol/L), x = distance (m), U = the velocity (m/hr), and k = the reaction rate (/hr). The boundary conditions can be formulated as

Ucin = Uc (x = 0) − D d c ___ d x (x = 0)

d c ___ d x (x = L) = 0

where cin = the concentration in the inflow (mol/L), L = the length of the reactor (m). These are called Danckwerts boundary conditions.

Use the finite-difference approach to solve for con- centration as a function of distance given the following parameters: D = 5000 m2/hr, U = 100 m/hr, k = 2 /hr, L = 100 m, and cin = 100 mol/L. Employ centered finite- difference approximations with Δ x = 10 m to obtain your

solutions. Compare your numerical results with the analytical solution:

c = Ucin _____________________________

(U − Dλ1)λ 2 e λ 2 L − (U − Dλ2)λ1e λ 1 L

× (λ 2e λ 2 Le λ 1 x − λ 1e λ 1 Le λ 2 x)

where

λ1 __ λ2

= U ___ 2D ( 1 ± √ ________

1 + 4kD ____ U2

) 24.13 A series of first-order, liquid-phase reactions create a desirable product (B) and an undesirable byproduct (C):

A → B → C

If the reactions take place in an axially dispersed plug-flow reactor (Fig. P24.12), steady-state mass balances can be used to develop the following second-order ODEs:

D d 2ca ____ d x2

− U d ca ___ d x − k1ca = 0

D d 2cb ____ d x2

− U d cb ___ d x + k1ca − k2cb = 0

D d 2cc ____ d x2

− U d cc ___ d x + k2cb = 0

Use the finite-difference approach to solve for the concen- tration of each reactant as a function of distance given: D = 0.1 m2/min, U = 1 m/min, k1 = 3/min, k2 = 1/min, L = 0.5 m, ca,in = 10 mol/L. Employ centered finite-difference approximations with Δ x = 0.05 m to obtain your solutions and assume Danckwerts boundary conditions as described in Prob. 24.12. Also, compute the sum of the reactants as a function of distance. Do your results make sense? 24.14 A biofilm with a thickness Lf (cm), grows on the sur- face of a solid (Fig. P24.14). After traversing a diffusion layer of thickness L (cm), a chemical compound A diffuses into the biofilm where it is subject to an irreversible first- order reaction that converts it to a product B.

k 1 k 2

PROBlems 669

x = 0 x = L

FIGURE P24.12 An axially dispersed plug-flow reactor.

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670 BOUNDARy-VAlUe PROBlems

Steady-state mass balances can be used to derive the following ordinary differential equations for compound A:

D d 2ca ____ d x 2

= 0 0 ≤ x < L

Df d 2ca ____ d x 2

− kca = 0 L ≤ x < L + L f

where D = the diffusion coefficient in the diffusion layer = 0.8 cm2/d, Df = the diffusion coefficient in the biofilm = 0.64 cm2/d, and k = the first-order rate for the conversion of A to B = 0.1/d. The following boundary conditions hold:

ca = ca 0 at x = 0

dca ___ d x = 0 at x = L + Lf

where ca 0 = the concentration of A in the bulk liquid = 100 mol/L. Use the finite-difference method to compute the steady-state distribution of A from x = 0 to L + L f, where L = 0.008 cm and Lf = 0.004 cm. Employ centered finite differences with Δ x = 0.001 cm. 24.15 A cable is hanging from two supports at A and B (Fig. P24.15). The cable is loaded with a distributed load whose magnitude varies with x as

𝑤 = 𝑤o [ 1 + sin ( π x ___ 2 lA ) ]

where 𝑤o = 450 N/m. The slope of the cable (dy∕dx) = 0 at x = 0, which is the lowest point for the cable. It is also the point where the tension in the cable is a minimum of To. The

L Lf

Bulk liquid

Di�usion layer Biofilm

Solid surface

FIGURE P24.14 A biofilm growing on a solid surface.

� = �o[1 + sin (πx/2la)]

lA = 60 m

x B

hA = 15 m

FIGURE P24.15

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differential equation which governs the cable is

d 2 y ___

d x 2 =

𝑤o ____ To [

1 + sin ( π x ___ 2 lA ) ] Solve this equation using a numerical method and plot the shape of the cable (y versus x). For the numerical solution, the value of To is unknown, so the solution must use an itera- tive technique, similar to the shooting method, to converge on a correct value of hA for various values of To. 24.16 The basic differential equation of the elastic curve for a simply supported, uniformly loaded beam (Fig. P24.16) is given as

E I d 2 y ___

d x 2 = 𝑤L x ____ 2 −

𝑤x2 ____ 2

where E = the modulus of elasticity and I = the moment of inertia. The boundary conditions are y(0) = y(L) = 0. Solve for the deflection of the beam using (a) the finite-differ-

ence approach (Δ x = 0.6 m) and (b) the shooting method. The following parameter values apply: E = 200 GPa, I = 30,000 cm4, 𝑤 = 15 kN/m, and L = 3 m. Compare your numerical results to the analytical solution:

y = 𝑤L x 3 _____ 12 E I −

𝑤x 4 _____ 24 E I − 𝑤L3x _____ 24 E I

24.17 In Prob. 24.16, the basic differential equation of the elastic curve for a uniformly loaded beam was formulated as

E I d 2 y ___

d x 2 = 𝑤L x ____ 2 −

𝑤x2 ____ 2

Note that the right-hand side represents the moment as a function of x. An equivalent approach can be formulated in terms of the fourth derivative of deflection as

E I d 4y ___

d x 4 = −𝑤

For this formulation, four boundary conditions are required. For the supports shown in Fig. P24.16, the conditions are that the end displacements are zero, y(0) = y(L) = 0, and that the end moments are zero, y″(0) = y″(L) = 0. Solve for the deflection of the beam using the finite-difference ap- proach (Δx = 0.6 m). The following parameter values apply: E = 200 GPa, I = 30,000 cm4, 𝑤 = 15 kN/m, and L = 3 m. Compare your numerical results with the analytical solution given in Prob. 24.16. 24.18 Under a number of simplifying assumptions, the steady-state height of the water table in a one-dimensional, unconfined groundwater aquifer (Fig. P24.18) can be

�

x 0

FIGURE P24.16

Groundwater flow Aquifer

Ground surface

Infiltration

Water table

Confining bed

FIGURE P24.18 An unconfined or “phreatic” aquifer.

PROBlems 671

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672 BOUNDARy-VAlUe PROBlems

modeled with the following second-order ODE:

K h̄ d 2h ___

d x 2 + N = 0

where x = distance (m), K = hydraulic conductivity (m/d), h = height of the water table (m), h̄ = the average height of the water table (m), and N = infiltration rate (m/d).

Solve for the height of the water table for x = 0 to 1000 m where h (0) = 10 m and h(1000) = 5 m. Use the following parameters for the calculation: K = 1 m/d and N = 0.0001 m/d. Set the average height of the water table as the average of the boundary conditions. Obtain your solution with (a) the shooting method and (b) the finite-difference method (Δ x = 100 m). 24.19 In Prob. 24.18, a linearized groundwater model was used to simulate the height of the water table for an uncon- fined aquifer. A more realistic result can be obtained by using the following nonlinear ODE:

d ___ d x ( K h d h ___ d x ) + N = 0 where x = distance (m), K = hydraulic conductivity (m/d), h = height of the water table (m), and N = infiltration rate (m/d). Solve for the height of the water table for the same case as in Prob. 24.18. That is, solve from x = 0 to 1000 m with h(0) = 10 m, h(1000) = 5 m, K = 1 m/d, and N = 0.0001 m /d. Obtain your solution with (a) the shooting method and (b) the finite-difference method (Δ x = 100 m). 24.20 Just as Fourier’s law and the heat balance can be employed to characterize temperature distribution, analo- gous relationships are available to model field problems in other areas of engineering. For example, electrical engi- neers use a similar approach when modeling electrostatic fields. Under a number of simplifying assumptions, an ana- log of Fourier’s law can be represented in one-dimensional form as

D = −ε dV ___ d x

where D is called the electric flux density vector, ε = permit- tivity of the material, and V = electrostatic potential. Simi- larly, a Poisson equation (see Prob. 24.8) for electrostatic fields can be represented in one dimension as

d 2V ____

d x 2 = −

ρυ __ ε

where ρυ = charge density. Use the finite-difference tech- nique with Δ x = 2 to determine V for a wire where V(0) = 1000, V(20) = 0, ε = 2, L = 20, and ρυ = 30.

24.21 Suppose that the position of a falling object is gov- erned by the following differential equation:

d 2x ___

d t 2 + c __ m

d x ___ d t − g = 0

where c = a first-order drag coefficient = 12.5 kg/s, m = mass = 70 kg, and g = gravitational acceleration = 9.81 m/s2. Use the shooting method to solve this equation for the boundary conditions:

x (0) = 0 x (12) = 500

24.22 As in Fig. P24.22, an insulated metal rod has a fixed temperature (T0) boundary condition at its left end. On it right end, it is joined to a thin-walled tube filled with water through which heat is conducted. The tube is insu- lated at its right end and convects heat with the surrounding fixed-temperature air (T∞). The convective heat flux at a lo- cation x along the tube (W/m2) is represented by

Jconv = h(T∞ − T2(x))

where h = the convection heat transfer coefficient [W/(m2 . K)]. Employ the finite-difference method with Δ x = 0.1 m to compute the temperature distribution for the case where both the rod and tube are cylindrical with the same radius r (m). Use the following parameters for your analysis: Lrod = 0.6 m, Ltube = 0.8 m, T0 = 400 K, T∞ = 300 K, r = 3 cm, ρ1 = 7870 kg/m3, Cp1 = 447 J/(kg . K), k1 = 80.2 W/(m . K), ρ2 = 1000 kg/m3, Cp2 = 4.18 kJ/(kg . K), k2 = 0.615 W/(m . K), and h = 3000 W/(m2 . K). The subscripts designate the rod (1) and the tube (2). 24.23 Perform the same calculation as in Prob. 24.22, but for the case where the tube is also insulated (i.e., no con- vection) and the right-hand wall is held at a fixed boundary temperature of 200 K. 24.24 Solve the following problem with the bvp4c,

d 2 y ____

d x 2 + y = 0

Lrod Ltube

T∞

FIGURE P24.22

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subject to the following boundary conditions

y(0) = 1

dy ___ dx (1) = 0

24.25 Figure P24.25a shows a uniform beam subject to a linearly increasing distributed load. The equation for the re- sulting elastic curve is (see Fig. P24.25b)

E I d 2 y ____

d x 2 −

𝑤 0 ___ 6 (Lx − x 3 __ L ) = 0

Note that the analytical solution for the resulting elastic curve is (see Fig. P24.25b)

y = 𝑤 0 _______ 120 EIL

(− x 5 + 2 L 2 x 3 − L 4 x)

Use bvp4c to solve for the differential equation for the elastic curve for L = 600 cm, E = 50,000 kN/cm2, I = 30,000 cm4, and 𝑤0 = 2.5 kN/cm. Then, plot both the numerical (points) and the analytical (lines) solutions on the same graph. 24.26 Use bvp4c to solve the boundary-value ordinary dif- ferential equation

d 2 u ____

d x 2 + 6 du ___ dx − u = 2

with boundary conditions u(0) = 10 and u(2) = 1. Plot the results of u versus x. 24.27 Use bvp4c to solve the following nondimensionalized ODE that describes the temperature distribution in a circular rod with internal heat source S

d 2 T ____

d r 2 + 1 __ r

dT ___ dr + S = 0

over the range 0 ≤ r ≤ 1, with the boundary conditions

T(1) = 1

d T ___ d r (0) = 0

for S = 1, 10, and 20 K/m2. Plot the temperature versus ra- dius for all three cases on the same graph. 24.28 A heated rod with a uniform heat source can be mod- eled with the Poisson equation,

d 2 T ____

d x 2 = − f (x)

Given a heat source f (x) = 25 and the boundary conditions, T (0) = 40 and T (10) = 200, solve for the temperature distri- bution with (a) the shooting method, (b) the finite-difference method, and (c) bvp4c. 24.29 Repeat Prob. 24.28, but for the following heat source: f (x) = 0.12 x3 – 2.4x 2 + 12x.

L (a)

(b)

(x = 0, y = 0) (x = L, y = 0)

FIGURE P24.25

PROBlems 673

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674

APPENDIX A MATLAB BUILT-IN FUNCTIONS

abs, 40 acos, 40 ascii, 64 axis, 51 axis square, 45 beep, 76 besselj, 451 ceil, 40 chol, 285 clabel, 212 clear, 64 cond, 296, 297 contour, 212, 566 conv, 186 cumtrapz, 510, 511 deconv, 185 det, 253 diag, 338 diff, 560, 561, 562 disp, 61 double, 61 eig, 336 elfun, 40 eps, 112 erf, 544 error, 66 event, 622 exp, 40 eye, 336 factorial, 51, 73n3 fft, 420 fix, 459

floor, 40, 212 fminbnd, 137, 210, 211 fminsearch, 213, 215, 199 format bank, 30 format compact, 28n1 format long, 30, 102 format long e, 30, 300 format long eng, 30 format long g, 30 format loose, 28n1 format short, 30 format short e, 30, 300 format short eng, 30 format short g, 30 fplot, 82 fprintf, 62, 63 fzero, 137, 181–183 getframe, 77, 78 gradient, 563, 564 grid, 42 help, 39, 46 help elfun, 40 hist, 353 hold off, 44 hold on, 44 humps, 94, 540 inline, 82 input, 61 interp1, 470 interp2, 475 interp3, 475 inv, 236, 238

isempty, 105 legend, 398, 511 length, 37 LineWidth, 44 linspace, 84 load, 64, 65 log, 39 log10, 371 log2, 150n2 loglog, 51 logspace, 34 lookfor, 46, 55 lu, 282 MarkerEdgeColor, 44 MarkerFaceColor, 44 MarkerSize, 44 max, 41, 262 mean, 57, 352 median, 352 mesh, 76 meshgrid, 212, 566 min, 41, 352 mode, 352 movie, 77–78 nargin, 71–72 norm, 296 ode113, 618 ode15s, 632 ode23, 617 ode23s, 632 ode23t, 632 ode23tb, 632

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APPENDIX A MATLAB BUILT-IN FUNCTIONS 675

ode45, 618, 633 odeset, 620 ones, 32 optimset, 182, 183, 189 pause, 76 pchip, 468, 471 peaks, 572 pi, 30 plot, 42 plot3, 44, 608 poly, 184, 185 polyfit, 373, 433, 446 polyval, 373, 433, 446 prod, 41 quiver, 565 rand, 353–356 randn, 353, 356

realmax, 112 realmin, 112 roots, 184–187 round, 40 save, 63 semilogy, 50, 51 set, 479 sign, 69 sin, 40 size, 237 sort, 41 spline, 468 sqrt, 40 sqrtm, 40 std, 352 stem, 423 subplot, 44

sum, 40, 285 surfc, 212 tanh, 7, 35 tic, 76 title, 42 toc, 76 trapz, 510, 518 var, 352 varargin, 86 who, 32 whos, 32 xlabel, 42 ylabel, 42 ylim, 424 zeros, 32 zlabel, 212

cha97962_appA_674-675.indd 675 06/11/16 3:39 pm

676

APPENDIX B MATLAB M-FILE FUNCTIONS

M-file Name Description Page

bisect Root location with bisection 151 eulode Integration of a single ordinary differential equation with Euler’s method 586 fzerosimp Brent’s method for root location 180 GaussNaive Solving linear systems with Gauss elimination without pivoting 258 GaussPivot Solving linear systems with Gauss elimination with partial pivoting 263 GaussSeidel Solving linear systems with the Gauss-Seidel method 310 goldmin Minimum of one-dimensional function with golden-section search 208 incsearch Root location with an incremental search 144 IterMeth General algorithm for iterative calculation 105 Lagrange Interpolation with the Lagrange polynomial 443 linregr Fitting a straight line with linear regression 372 natspline Cubic spline with natural end conditions 477 Newtint Interpolation with the Newton polynomial 440 newtmult Root location for nonlinear systems of equations 318 newtraph Root location with the Newton-Raphson method 174 quadadapt Adaptive quadrature 539 rk4sys Integration of system of ODEs with 4th-order RK method 602 romberg Integration of a function with Romberg integration 530 TableLook Table lookup with linear interpolation 458 trap Integration of a function with the composite trapezoidal rule 500 trapuneq Integration of unequispaced data with the trapezoidal rule 509 Tridiag Solving tridiagonal linear systems 266

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677

APPENDIX C INTRODUCTION TO SIMULINK

Simulink® is a graphical programming environment for modeling, simulating, and analyz- ing dynamic systems. In short, it allows engineers and scientists to build process models by interconnecting blocks with communication lines. Thus, it provides an easy-to-use com- puting framework to quickly develop dynamic process models of physical systems. Along with offering a variety of numerical integration options for solving differential equations, Simulink includes built-in features for graphical output which significantly enhance visu- alization of a system’s behavior.

As a historical footnote, back in the Pleistocene days of analog computers (aka the 1950s), you had to design information flow diagrams that showed graphically how multiple ODEs in models were interconnected with themselves and with algebraic relationships. The diagrams also showed flaws in modeling where there was information lacking or structural defects. One of the nice features of Simulink is that it also does that. It’s often beneficial to see that aspect separate from numerical methods and then merge the two in MATLAB.

As was done with Chap. 2, most of this appendix has been written as a hands-on exer- cise. That is, you should read it while sitting in front of your computer. The most efficient way to start learning Simulink is to actually implement it on MATLAB as you proceed through the following material.

So let’s get started by setting up a simple Simulink application to solve an initial value problem for a single ODE. A nice candidate is the differential equation we developed for the velocity of the free-falling bungee jumper in Chap. 1,

dυ __ dt = g − cd __ m υ

2 (C.1)

where υ = velocity (m/s), t = time (s), g = 9.81 m/s2, cd = drag coefficient (kg/m), and m = mass (kg). As in Chap. 1 use cd = 0.25 kg/m, m = 68.1 kg, and integrate from 0 to 12 s with an initial condition of υ = 0.

To generate the solution with Simulink, first launch MATLAB. You should eventu- ally see the MATLAB window with an entry prompt, ≫, in the Command Window. After changing MATLAB’s default directory, open the Simulink Library Browser using one of the following approaches:

• On the MATLAB toolbar, click the Simulink button ( ). • At the MATLAB prompt, enter the simulink command.

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678 APPENDIX C INTRODUCTION TO SIMULINK

The Simulink Library Browser window should appear displaying the Simulink block li- braries installed on your system. Note, to keep the Library Browser above all other win- dows on your desktop, in the Library Browser select View, Stay on Top.

Click on the New model command button on the left of the toolbar, and an untitled Simulink model window should appear.

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APPENDIX C INTRODUCTION TO SIMULINK 679

You will build your simulation model in the untitled window by choosing items from the Library Browser and then dragging and dropping them onto the untitled window. First, select the untitled window to activate it (clicking on the title bar is a sure way to do this) and select Save As from the File menu. Save the window as Freefall in the default direc- tory. This file is saved automatically with an .slx extension. As you build your model in this window, it is a good idea to save it frequently. You can do this in three ways: the Save button, the Ctrl+s keyboard shortcut, and the menu selections, File, Save.

We will start by placing an integrator element (for the model’s differential equation) in the Freefall window. To do this, you need to activate the Library Browser and double click on the Commonly Used Blocks item.

The Browser window should show something like

Examine the icon window until you see the Integrator icon.

Input port Output port

Integrator

1 s

The block symbol is what will appear in your model window. Notice how the icon has both input and output ports which are used to feed values in and out of the block. The 1/s symbol represents integration in the Laplace domain. Use the mouse to drag an Integrator icon onto your Freefall window. This icon will be used to integrate the differential

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680 APPENDIX C INTRODUCTION TO SIMULINK

equations. Its input will be the differential equation [the right hand side of Eq. (C.1)] and its output will be the solution (in our example, velocity).

Next you have to build an information flow diagram that describes the differential equation and “feeds” it into the integrator. The first order of business will be to set up constant blocks to assign values to the model parameters. Drag a Constant icon from the Commonly Used Blocks1 branch in the Browser window to the Freefall model window and place it above and to the left of the integrator.

Next, click on the Constant Label and change it to g. Then, double click the Constant block and the Constant Block Dialogue box will be opened. Change the default value in the Constant field to 9.81 and click OK.

Now set up Constant blocks for both the drag coefficient (cd = 0.25) and mass (m = 68.1) positioned below the g block. The result should look like

1 s

9.81

0.25

68.1

cd Integrator

From the Math Operations Library Browser, select a Sum icon and drag it just to the left of the Integrator block. Place the mouse pointer on the output port of the sum block. Notice that the mouse pointer will change to a crosshair shape when it’s on the output port. Then drag a connecting line from the output port to the Integrator block’s input port. As you drag, the mouse pointer retains its crosshair shape until it is on the input port where- upon it changes to a double-lined crosshair. We have now “wired” the two blocks together with the output of the sum block feeding into the Integrator.

Notice that the sum block has two input ports into which can be fed two quantities that will be added as specified by the two positive signs inside the circular block. Recall that our differential equation consists of the difference between two quantities: g − (cd∕m)υ2. We therefore, have to change one of the input ports to a negative. To do this, double click on the Sum Block in order to open the Sum Block Dialogue box. Notice that the List of Signs has two plus signs (+ +). By changing the second to a minus sign (+ −), the value entering the second input port will be subtracted from the first. After closing the Sum Block Dialogue box, the result should look like

1 s

Integrator

+ −

1 All the icons in the Commonly Used Blocks group are available in other groups. For example, the Constant icon is located in the Sources group.

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APPENDIX C INTRODUCTION TO SIMULINK 681

Because the first term in the differential equation is g, wire the output of the g block to the positive input port of the sum block. The system should now look like

68.1

0.25

9.81

1 s

Integrator

+ −

In order to construct the second term that will be subtracted from g, we must first square the velocity. This can be done by dragging a Math Functions block from the Math Operations Library Browser and positioning it to the right and below the integrator block. Double click the Math Functions icon to open the Math Functions Dialogue box and use the pull down menu to change the Math Function to square.

Before connecting the Integrator and Square blocks, it would be nice to rotate the latter so its input port is on top. To do this, select the Square block and then hit ctrl-r once. Then, we can wire the output port of the Integrator block to the input port of the square block. Because the output of the Integrator block is the solution of the differential equation, the output of the square block will be υ 2. To make this clearer, double click the arrow connect- ing the Integrator and Square blocks and a text box will appear. Add the label, v(t), in this text box to indicate that the output of the Integrator block is the velocity. Note that you can label all the connecting wires in this way in order to better document the system diagram. At this point, it should look like

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682 APPENDIX C INTRODUCTION TO SIMULINK

68.1 Math

Function

v(t)

0.25

9.81

1 s

Integrator

+ −

Next, drag a Divide block from the Math Operations Library Browser and position it to the right of the cd and m blocks. Notice that the Divide block has two input ports: one for the dividend (×) and one for the divisor (÷). Note that these can be switched by double clicking the Divide block to open the Divide Block Dialogue box and switching the order in the “number of inputs:” field. Wire the cd block output port to the × input port and the m block output port to the ÷ input port of the Divide block. The output port of the divide block will now carry the ratio, cd∕m.

68.1 Divide Math

Function

v(t)

0.25

9.81

1 s

Integrator

+ −

÷ ×

Drag a Product block from the Math Operations Library Browser and position it to the right of the divide block and just below the sum block. Use ctrl-r to rotate the product block until its output port points upward toward the sum block. Wire the divide block output port to the nearest product input port and the square Math Function block output port to the other product input port. Finally, wire the product output port to the remaining sum input port.

68.1 Divide Math

Function

v(t)

0.25

9.81

1 s

Integrator

Product

+ −

÷ × ×

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APPENDIX C INTRODUCTION TO SIMULINK 683

As displayed above, we have now successfully developed a Simulink program to gen- erate the solution to this problem. At this point, we could run the program but we have not yet set up a way to display the output. For the present case, a simple way to do this employs a Scope Block

Scope

The Scope block displays signals with respect to simulation time. If the input signal is continuous, the Scope draws a point-to-point plot between major time step values. Drag a Scope block from the Commonly Used Blocks browser and position it to the right of the Integrator block. Position the mouse pointer on the Integrator’s output wire (for the present case a nice position would be at the corner). Simultaneously hold down the control key and another line across to the Scope block’s input port.

68.1 Divide Math

Function

v(t)

0.25

9.81

1 s

Integrator Scope

Product

+ −

÷ × ×

We are now ready to generate results. Before doing that, it’s a good idea to save the model. Double click on the Scope block. Then click the run button, . If there are any mistakes, you will have to correct them. Once your have successfully made corrections, the program should execute and the scope display should look something like

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684 APPENDIX C INTRODUCTION TO SIMULINK

Click on the Autoscale button, , and the plot will resize to fit the entire range of results

Note that a Scope window can display multiple y-axes (graphs) with one graph per input port. All of the y-axes have a common time range on the x-axis. By selecting the parameter button on the graph window ( ), you can use scope parameters to change graph features such as figure color and style and axis settings.

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685

BIBLIOGRAPHY

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Attaway, S., MATLAB: A Practical Introduction to Pro- gramming and Problem Solving, Elsevier Science, Burlington, MA, 2009.

Bogacki, P. and L. F. Shampine, “A 3(2) Pair of Runge-Kutta Formulas,” Appl. Math. Letters, 2(1989):1–9, 1989.

Brent, R. P., Algorithms for Minimization Without Deriva- tives, Prentice Hall, Englewood Cliffs, NJ, 1973.

Butcher, J. C., “On Runge-Kutta Processes of Higher Order,” J. Austral. Math. Soc., 4:179, 1964.

Carnahan, B., H. A. Luther, and J. O. Wilkes, Applied Numerical Methods, Wiley, New York, 1969.

Chapra, S. C. and R. P. Canale, Numerical Methods for Engineers, 6th ed., McGraw-Hill, New York, 2010.

Cooley, J. W. and J. W. Tukey, “An Algorithm for the Machine Calculation of Complex Fourier Series,” Math. Comput., 19:297–301, 1965.

Dekker, T. J., “Finding a Zero by Means of Successive Linear Interpolation.” In B. Dejon and P. Henrici (editors), Constructive Aspects of the Fundamental Theorem of Algebra, Wiley-Interscience, New York, 1969, pp. 37–48.

Devaney, R. L. Chaos, Fractals, and Dynamics: Computer Experiments in Mathematics, Menlo Park, CA: Addison- Wesley, 1990.

Dormand, J. R. and P. J. Prince, “A Family of Embed- ded Runge-Kutta Formulae,” J. Comp. Appl. Math., 6:19–26, 1980.

Draper, N. R. and H. Smith, Applied Regression Analysis, 2nd ed., Wiley, New York, 1981.

Fadeev, D. K. and V. N. Fadeeva, Computational Methods of Linear Algebra, Freeman, San Francisco, CA, 1963.

Forsythe, G. E., M. A. Malcolm, and C. B. Moler, Com- puter Methods for Mathematical Computation, Prentice Hall, Englewood Cliffs, NJ, 1977.

Gabel, R. A. and R. A. Roberts, Signals and Linear Systems, Wiley, New York, 1987.

Gander, W. and W. Gautschi, Adaptive Quadrature– Revisited, BIT Num. Math., 40:84–101, 2000.

Gerald, C. F. and P. O. Wheatley, Applied Numerical Analysis, 3rd ed., Addison-Wesley, Reading, MA, 1989.

Hanselman, D. and B. Littlefield, Mastering MATLAB 7, Prentice Hall, Upper Saddle River, NJ, 2005.

Hayt, W. H. and J. E. Kemmerly, Engineering Circuit Analysis, McGraw-Hill, New York, 1986.

Heideman, M. T., D. H. Johnson, and C. S. Burrus, “Gauss and the History of the Fast Fourier Transform,” IEEE ASSP Mag., 1(4):14–21, 1984.

Hornbeck, R. W., Numerical Methods, Quantum, New York, 1975.

James, M. L., G. M. Smith, and J. C. Wolford, Applied Numerical Methods for Digital Computations with FORTRAN and CSMP, 3rd ed., Harper & Row, New York, 1985.

Moler, C. B., Numerical Computing with MATLAB, SIAM, Philadelphia, PA, 2004.

Moore, H., MATLAB for Engineers, 2nd ed., Prentice Hall, Upper Saddle River, NJ, 2008.

Munson, B. R., D. F. Young, T. H. Okiishi, and W. D. Huebsch, Fundamentals of Fluid Mechanics, 6th ed., Wiley, Hoboken, NJ, 2009.

Ortega, J. M., Numerical Analysis–A Second Course, Academic Press, New York, 1972.

Palm, W. J. III, A Concise Introduction to MATLAB, McGraw-Hill, New York, 2007.

Ralston, A., “Runge-Kutta Methods with Minimum Error Bounds,” Match. Comp., 16:431, 1962.

Ralston, A. and P. Rabinowitz, A First Course in Numerical Analysis, 2nd ed., McGraw-Hill, New York, 1978.

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Ramirez, R. W., The FFT, Fundamentals and Concepts, Prentice Hall, Englewood Cliffs, NJ, 1985.

Recktenwald, G., Numerical Methods with MATLAB, Prentice Hall, Englewood Cliffs, NJ, 2000.

Scarborough, I. B., Numerical Mathematical Analysis, 6th ed., Johns Hopkins Press, Baltimore, MD, 1966.

Shampine, L. F., Numerical Solution of Ordinary Differen- tial Equations, Chapman & Hall, New York, 1994.

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687

INDEX

A Absolute error, 101 Absolute tolerance (AbsTol), 618 Accuracy, 100–101 Adaptive integration, 487 Adaptive methods

adaptive Runge-Kutta methods, 615–623

MATLAB application, 634–635 multistep methods, 624–628 Pliny’s intermittent fountain,

635–639 Adaptive quadrature, 525, 537. See also

Gauss quadrature; Romberg integration example, 540 integral function, 539 quadadapt function, 537–539

Adaptive Runge-Kutta methods, 615. See also Runge-Kutta methods (RK methods) adaptive step-size control, 615–616 events, 621–623 MATLAB functions for nonstiff

systems, 617–621 solution of ODE, 616

Adaptive step-size control, 615–616 Allosteric enzymes, 374 Alphanumeric information, 34 Amplification factor, 585 Amplitude, 406 Analytical solution, 9

for heated rod, 650–651 Angular frequency, 332, 406, 407 Animation, 77

of projectile motion, 77–78 Anonymous functions, 81–82

Areal integral, 492 Arithmetic manipulations of computer

numbers, 112 adding large and small number, 114 inner products, 114 large computations, 113 smearing, 114

Arithmetic mean, 348 Arrays, 31–32

operations, 39 Ascent methods, 213 ASCII files in MATLAB, 64–65 Assignment function, 29

arrays, 31–32 character strings, 34–35 colon operator, 33 linspace functions, 33–34 logspace functions, 33–34 matrices, 31–32 scalars, 29–30 vectors, 31–32

Associative equation, 232 Aug function, 258 Augmentation, 234

B Backslash operator, 394 Backward method, 630 Banded matrix, 231, 264 Base-2. See Binary Base-8. See Octal Bias. See Inaccuracy Bilinear function, 474 Binary, 107

digits, 106 search, 458–459

bisect function, 151–152

Bisection method, 146, 147. See also Numerical methods bisect function, 151–152 error estimation for, 148–151 preferable to false position, 154–155

Bits, 106 Blunders, 130 Boole’s rule, 538 Boolean variables, 636 Boundary-value problems, 577, 647

in engineering and science, 648–651 finite-difference methods, 658–665 initial-value problem vs., 647, 648 MATLAB function, 665–667 ordinary differential equation, 647 shooting method, 651–658 single ODE, 646

Bracketing methods. See also Numerical methods; Roots and initial guesses, 141

incremental search, 143–146 Brent’s method, 176. See also Secant

methods algorithm, 179–181 inverse quadratic interpolation,

177–179 Brent’s root-finding method, 179 Brent’s root-location method, 176 Built-in functions, 39–42 Bungee jumper

with cord, 634–635 problem

analysis, 291–292 analytical solution to, 7–10 free-falling, 573 numerical solution to, 10–12 velocity, 87–90

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688 INDEX

Butterfly effect, 607 bvp4c function, 665–667

C Calculator mode, 2, 28 Calculus, 549 Cartesian coordinates, 492 ceil function, 40 Centered finite difference, 123 Chaos, 604–609 Character Strings, 34–35 Characteristic polynomial, 328 Chemical reactions, 320–322 chol function, 285–286 Cholesky decomposition. See also

Cholesky factorization Cholesky factorization, 283

example, 284 MATLAB function, 285–286

Circuits, currents and voltages in, 240–243

Clamped end condition, 467 Classical fourth-order RK method,

595–598 Closed integration formulas, 487, 494 Closed-form solution, 9 Coefficient matrix, 251 Colebrook equation, 186 Colon operator, 33 Column, 229

column-sum norm, 294 vectors, 230

Command mode, 2 Commutative equation, 231, 232 Companion matrix, 184 Complete pivoting, 261 Composite Simpson’s 1∕3 rule,

503–505, 524–525 Composite trapezoidal rule, 497–499 Computer algorithm for iterative calcu-

lations, 104–106 Computer applications, 370

MATLAB functions, polyfit and polyval, 373

MATLAB M-file, linregr, 371–372 Computer mathematics, 1 Computer number representation, 106

floating-point representation, 108–111 integer representation, 107–108 precision, 111–112

range, 111 Computing work with numerical

integration, 515–518 Concatenation, 32, 34 Conditionally stable method, 585 Conservation

of charge, 240 of energy, 241 laws, 223

in engineering and science, 12–13 Constant of integration, 575–576 Constitutive laws, 551, 552 Continuity condition, 460 Continuous Fourier series, 411

approximation, 411–412 Euler’s formula, 413 square wave, 412–413

Control codes, 62–63 Convergence, 164, 308 Cooley-Tukey algorithm, 420 Corrector equation, 588 Correlation coefficient, 364 Cosine, 406

function, 407 Cramer’s rule, 249, 250–253, 315 Creating and accessing files, 63–65 Cubic interpolation, 439 Cubic splines, 453, 462. See also

Linear splines derivation, 463–465 end conditions, 467–468 material on, 455 natural, 466–467

cumtrapz functions, 510 Currents in circuits, 240–243 Curvature, 550 Curve fitting

with sinusoidal functions, 405, 407 alternative formulation, 408 least-squares fit of sinusoid,

408–411 plot of sinusoidal function, 406

techniques, 343 through data points, 344 engineering and science, 343–345 linear regression, 345

D Darcy’s law, 552 Data uncertainty, 131

Decisions, 65 error function, 66 if structure, 65–66, 69–70 if . . . else structure, 69 if . . . elseif structure, 69 logical conditions, 66–68 switch structure, 70–71 variable argument list, 71–72

Default value, 71 Definite integrals equation, 575 Deflation, 336 Degrees of freedom, 349 Dependent variable, 5, 573–574 Derivative, 485, 549, 550

boundary conditions finite-difference methods,

660–662 shooting method with, 654–656

for data with errors, 558–559 mean-value theorem, 119 of unequally spaced data, 557–558

Descent methods, 213 Descriptive statistics, 348

location measurement, 348–349 in MATLAB, 352–353 spread measurement, 349 statistics of sample, 350–351

det function, 253 Determinants, 250–253

evaluation with gauss elimination, 263–264

DFT. See Discrete Fourier transform (DFT)

Diagonal dominance, 308 Diagonal matrix, 230 diff function, 560–562 Differential equations, 7, 573, 574 Differentiation, 485–487, 549–551

in engineering and science, 551–552 one-dimensional forms of

constitutive laws, 552 Direct methods, 213 Dirichlet boundary condition, 654 Discrete Fourier transform (DFT), 418

FFT, 419–420 MATLAB function, 420–423 of simple sinusoid with MATLAB,

421–422 Discretization errors, 583 disp function, 61

cha97962_Index_687-704.indd 688 06/11/16 3:39 pm

INDEX 689

Distributed variable problems, 225 Distributive, 232 Dot product. See Inner product of two

vectors Double integral, 512–513

to determine average temperature, 513–514

Dummy variable, 89

E Earthquakes, eigenvalues and, 337–339 Echo printing, 29 Eigenvalues, 327, 328

and earthquakes, 337–339 eig function, 336 mass–three spring system, 331 with MATLAB, 336–337 physical background, 331 physical interpretation of, 332–333 polynomial method, 329–331 positions and velocities vs.

time, 327 power method, 333–336

Eigenvectors, 226, 329 with MATLAB, 336 physical interpretation of, 332–333

Electromotive force (emf), 241 Element-by-element operations, 39 Elimination of unknowns, 253–254 Ellipsis, 34 Embedded RK methods, 616, 617 emf. See Electromotive force (emf) End conditions, 467–468 Enzyme kinetics, 373–378 Equilibrium, 213–215 error function, 66 Error(s), 100

absolute, 101 accuracy, 100–101 amplification, 559 analysis for Euler’s method, 583–585 blunders, 130 computer algorithm for iterative

calculations, 104–106 data uncertainty, 131 estimation, 627–628

for iterative methods, 103–104 model errors, 130–131 numerical, 101 percent relative, 102

precision, 100–101 roundoff, 106–114 total numerical error, 125–130 truncation, 114–125

Euclid’s definition, 203 Euclidean norm, 293 Euler-Cauchy method. See Euler’s

method Euler’s formula, 413 Euler’s method, 10, 13, 577, 581, 582,

598, 629–630. See also Runge-Kutta methods (RK methods) error analysis for, 583–585 example, 582–583 fundamental source of error in, 587 Heun’s method, 588–592 MATLAB M-file function,

586–587 midpoint method, 592–593 solving systems of ODEs with,

598–600 stability of, 585–586

eulode function, 586–587 Events, 621–623 Explicit Euler method, 630–632 Exploratory data analysis, 46–48 Exponential model, 366 Extrapolation, 445–447. See also

Interpolation “Eyeball” approaches, 358

F Factorial, loops to computing, 73–74 False position, 152, 177, 433–435, 470

bisection method preferable to, 154–155

method, 153–154 False-position formula, 152 Fast Fourier transform (FFT), 345,

419–420 fft function, 420–423 Fick’s law, 552 50th percentile. See Median Finite difference, 121

approximations of derivatives, 124–125 of higher derivatives, 125

methods, 658 of boundary-value problems,

659–660

derivative boundary conditions, 660–662

implementation, 659 incorporating derivative boundary

conditions, 662–663 for nonlinear ODEs, 663–665

First derivative backward difference approximation,

123 centered difference approximation,

123 First-order

approximation, 116 equation, 574 method, 585 splines, 456–457

Fitting experimentatal data, 397–399 Floating-point operations (flops), 258 Floating-point representation, 108–111 floor function, 40 flops. See Floating-point operations

(flops) fmin corresponds, 207 fminbnd function, 210–211 fminsearch function, 213 fminsearch MATLAB function, 395 for... end structure, 72–73 Forcing functions, 5 Format codes, 62 Forward elimination of unknowns,

255–256 Fourier analysis, 345, 405. See also

Gauss elimination continuous Fourier series, 411–413 curve fitting with sinusoidal

functions, 405–411 DFT, 418–423 Fourier integral and transform,

415–418 frequency and time domains,

414–415 power spectrum, 423–424 sunspots, 425–426

Fourier coefficients, 423 Fourier integral, 415

amplitude and phase line spectra, 417 aperiodic signal, 418 various phases of sinusoid, 416

Fourier series, 415 continuous, 411–413

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690 INDEX

Fourier transform, 415–418 pair, 417

Fourier’s law, 552, 655 of heat conduction, 551

Fourth-order RK method, 600–601 fprintf function, 62 Free-falling bungee jumper problem,

573 Frequency domain, 414–415 Frequency plane, 415 Friction factor, 186 Frobenius norm, 294 fsolve function, 319–320 Function files, 55–57 Function functions, 81, 82–83

building and implementing, 84–85 Fundamental frequency, 411 fval function, 213 fzero function, 181–183 fzerosimp function, 179

G Gauss elimination, 225, 249. See also

Fourier analysis determinants and Cramer’s rule,

250–253 elimination of unknowns, 253–254 graphical method, 249–250 as LU factorization, 276

example, 278 MATLAB function, 282–283 matrix, 277 with pivoting, 280–282 substitution steps, 279–280

model of heated rod, 266–269 naive Gauss elimination,

254–261 pivoting, 261–264 solving small numbers of equations,

249 tridiagonal systems, 264–266

Gauss quadrature, 487, 508, 525, 530–531. See also Adaptive quadrature; Romberg integration higher-point formulas, 536–537 three-point Gauss-Legendre formula,

536–537 two-point Gauss-Legendre formula,

533–536, 543

undetermined coefficients method, 531–533

Gauss-Newton method, 395 Gauss-Seidel method, 226, 305–312,

664 with relaxation, 311–312

GaussNaive function, 258 GaussPivot function, 262–263 Gear backward differentiation formulas,

632 General linear least-squares regression,

391–392. See also Nonlinear regression model, 345 multiple linear regression,

389–391 polynomial regression, 385–389

with MATLAB, 392–393 QR factorization and backslash

operator, 394 Global optimum, 202 Global truncation error, 583–584 Global variables, 58–60 Golden ratio, 203 Golden-section search, 203–208 Gradien methods, 213 gradient function, 563–564 Graphical methods, 140–141, 249–250 Graphics, 42–45

H Harmonics, 411 Heated rod model, 266–269 Hertz (Hz), 332 Heun’s method, 588

example, 589–592 without iteration, 595 predictor and corrector, 588 predictor equation, 588 predictor-corrector approach, 589

High-accuracy differentiation formulas, 552 example, 553–555 Taylor series, 552–553

Higher-order corrections, 527 Higher-order differential equations,

574 Higher-order polynomial interpolation,

dangers of, 447–449 Higher-point formulas, 536–537

Hilbert matrix, 295 Histogram, 351–352 Homogeneous linear algebraic system,

328 Hooke’s law, 213, 552, 580 humps function, 540 Hypothesis testing, 344

I Identity matrix, 230 IEEE double-precision format, 111 if structure, 65–66, 69–70 if... else structure, 69 if... elseif structure, 69 Ill-conditioned ODEs, 585 Ill-conditioned systems, 249 Implicit Euler method, 630–632 Implicit method, 630 Implicit Runge-Kutta formula, 632 Imprecision, 100 Inaccuracy, 100 Increment function, 581, 593 Incremental search method, 143–146 Indefinite integral equation, 575 Indentation, 79–81 Independent variables, 5, 573–574 Indoor air pollution, 297–300 Infinite loop, 75 Initial-value problems, 577, 579, 647

Euler’s method, 581–593 MATLAB M-file function, rk4sys,

602–603 predator-prey models and chaos,

604–609 RK methods, 593–598 signum function, 580 solving ODE, 581 systems of equations, 598–604

Inner product of two vectors, 37 input function, 61 Input-output, 61

creating and accessing files, 63–65 interactive M-file function, 62–63

Integer representation, 107–108 integral function, 539, 543

integral2 functions, 514 integral3 functions, 514

Integrals, 492 for data with errors, 558–559

cha97962_Index_687-704.indd 690 06/11/16 3:39 pm

INDEX 691

Integration, 485–487, 489–490 in engineering and science, 490–492 Newton-Cotes open integration

formulas, 512 numerical integration to computing

distance, 510–511 trapezoidal rule with unequal

segments, 508 trapuneq function, 509 trapz and cumtrapz, 510 with unequal segments, 508

Interactive M-file function, 62–63 interp1 function, 468, 470

options, 472 trade-offs using, 471–473

Interpolation, 16, 343, 345, 430. See also Polynomial interpolation determining polynomial coefficients,

431–432 polyfit functions, 433 polyval functions, 433

Inverse Fourier transform, 417, 418 Inverse interpolation, 444–445. See also

Interpolation Inverse quadratic interpolation,

177–179 Iterative methods

linear systems, 305–312 nonlinear systems, 312–322

J Jacobi iteration, 307 Joule’s law, 541 Jumpers motion, 326

K Kirchhoff’s current, 240 Kirchhoff’s voltage rule, 241 Knots, 457

L Lagging phase angle, 407 Lagrange interpolating polynomial, 345,

441. See also Newton interpolating polynomial Lagrange function, 443–444 rationale, 442

Lagrange polynomial, 178 Leading phase angle, 407

Least squares criterion, 360 fit of sinusoid, 408–411 regression, 343

Left division, polynomial regression implemention with, 394

Line spectra, 415 Linear algebraic equations, 223, 227,

248 bungee cords, 228 currents and voltages in circuits,

240–243 in engineering and science,

223–225 free-body diagrams, 228 linear algebraic equations with

MATLAB, 238–240 matrix algebra, 229–238

Linear convergence, 166 Linear interpolation method. See False

position Linear Lagrange interpolating

polynomial, 441 Linear least-squares regression, 358

“criteria for best” fit, 358–360 error quantification of linear

regression, 362 errors estimation for linear

least-squares fit, 364–366 least-squares fit of straight line, 360 linear regression, 360–362 linear regression with residual errors,

364 regression data, 363 residual in linear regression,

362–363 Linear ODE, shooting method for,

652–654 Linear regression, 345, 360–362, 370

computer applications, 370–373 descriptive statistics, 348–351

in MATLAB, 352–353 enzyme kinetics, 373–378 experimental data for force and

velocity, 347–348 fluid mechanics, 346–347 linear least-squares regression,

358–366 linearization of nonlinear

relationships, 366–370

normal distribution, 351–352 random numbers and simulation,

353–358 statistics review, 348 wind tunnel experiment, 347

Linear splines, 455. See also Cubic splines first-order splines, 456–457 notation for, 456 table lookup, 458–459

Linear systems, 305. See also Nonlinear systems chemical reactions, 320–322 convergence and diagonal

dominance, 308 fsolve function, 319–320 Gauss-Seidel method, 306–308

GaussSeidel function, 309 linear algebraic equations, 305–306 relaxation, 309–312

Linearization of nonlinear relationships, 366 comments on linear regression, 370 fitting data with power equation,

368–370 nonlinear regression techniques,

367–368 power equation, 366–367

linregr MATLAB M-file function, 371

linspace functions, 33–34 Local optimum, 202 Local truncation error, 583–584 Local variables, 57, 58 Logical conditions, 66–68 Logical variables, 636 logspace functions, 33–34 Loop(s), 65, 72

to computing factorial, 73–74 for... end structure, 72–73 pause command, 76 preallocation of memory, 74–75 rule, 241 vectorization, 74 while structure, 75 while... break structure, 75–76

Lorenz equations, 604, 606 Lotka-Volterra equations, 604, 607 Lotka-Volterra model, 606 Lower triangular matrix, 231

cha97962_Index_687-704.indd 691 06/11/16 3:39 pm

692 INDEX

LU factorization, 225, 275 Cholesky factorization, 283–286 with Gauss elimination, 278 Gauss elimination as, 276–283 LU function, 282–283 MATLAB left division, 286 n-dimensional systems, 275 two-step strategy, 276

Lumped drag coefficient, 7 Lumped variable problems, 224

M M-files, 54, 143, 637

bisect function, 151–152 eulode function, 586–587 function files, 55–57 GaussNaive, 258 GaussPivot, 262–263 GaussSeidel, 309 global variables, 58–60 to implementing Lagrange

interpolation, 443 to implementing Newton

interpolation, 440 to implementing Romberg

integration, 530 Lagrange function, 443–444 linregr M-file function, 371–372 Newtint function, 440 newtraph function, 173–174 passing functions to, 81

anonymous functions, 81–82 building and implementing

function function, 84–85 function functions, 82–83 passing parameters, 85–87

quadadapt function, 537–539 rk4sys, 602–603 script files, 54–55 subfunctions, 60–61 trap function, 499–501 trapuneq function, 509 Tridiag function, 266 variable scope, 57–58

Machine epsilon, 110 Machine precision. See Machine epsilon Maclaurin series expansion, 103 Main function, 61 Mantissa, 110

Mathematical model, 2, 5 analytical solution to bungee jumper

problem, 7–10 numerical solution to bungee jumper

problem, 10–12 real drag, 17–19

Mathematical operations, 36–39 MATLAB

assignment function, 29–35 bisect function, 151–152 built-in functions, 674–675 bvp4c function, 665–667 chol function, 285–286 Cholesky factorization with,

285–286 cumtrapz functions, 510 descriptive statistics in, 352–353 DFT of simple sinusoid with,

421–422 eig function, 336 eigenvalues, 336–337 eigenvectors, 336–337 environment, 28–29 eulode, 586–587 exploratory data analysis, 46–48 fft function, 420–423 fminsearch, 395 fsolve function, 319–320 fzero function, 181–183 GaussNaive, 258 GaussPivot, 262–263 GaussSeidel, 309 graphics, 42–45 to implement Gauss-Seidel method,

310 integral function, 539 integral2 and integral3 functions, 514 Lagrange function, 443–444 left division, 286 linear algebraic equations with,

238–240 linregr function, 371–372 LU factorization, 282–283 M-file functions, 676 mathematical operations, 36–39 matrix manipulations, 234–237 multidimensional interpolation in,

475 Newtint function, 440 newtraph function, 173–174

numerical differentiation with, 560–564

piecewise interpolation in, 468 interp1 function, 470–473 spline function, 468–470

polyfit functions, 373, 433 polyval functions, 373, 433 power spectrum with, 424 programming mode, 3 quadadapt function, 537–539 rand function, 354–356 randn function, 356–358 resources, 46 rk4sys, 602–603 roots function, 184–186 software environment, 2 for stiffness, 632–634 trap, 499–501 trap function, 499–501 trapuneq function, 509 trapz functions, 510 Tridiag function, 266 use of built-in functions, 39–42 variables, 57 workspace, 57

Matrix, 31–32, 229 algebra, 225, 229 condition number, 294–296

in MATLAB, 296–297 form, 409 linear algebraic equations in matrix

form, 237–238 matrix algebra, 229–238 multiplication, 231–232 norms, 293–294 notation, 229–231 operating rules, 231

higher-dimensional matrices, 233

MATLAB matrix manipulations, 234–237

visual depiction, 232 Matrix inverse, 226, 288

bungee jumper problem analysis, 291–292

calculating inverse, 288–289 error analysis and system condition,

292 matrix condition number,

294–296

cha97962_Index_687-704.indd 692 06/11/16 3:39 pm

INDEX 693

methods, 292–293 vector and matrix norms,

293–294 example, 289–290 indoor air pollution, 297–300 norms and condition number in

MATLAB, 296–297 stimulus-response computations,

290–291 max function, 262 Maximum likelihood principle, 363 Mean value, 406 Measurement errors, 131 Median, 348 Michaelis-Menten equation, 373–374 Midpoint method, 592–593, 595 Midtest loop, 75 Minimax criterion, 359 Minimum potential energy, 213–215 Minors, 251 Mixed partial derivative, 560 Mode, 348 Model errors, 130–131 Modified secant method, 175–176 Multidimensional interpolation, 473.

See also Piecewise interpolation; Polynomial interpolation bilinear interpolation, 473–475 in MATLAB, 475

Multidimensional optimization, 211–213

Multimodal cases, 202 Multiple integrals, 512

double integral, 512–514 MATLAB functions: integral2 and

integral3 functions, 514 Multiple linear regression, 345,

389–391 Multistep methods, 581, 624

error estimation, 627–628 non-self-starting Heun method,

624–627

N Naive Gauss elimination, 254, 261

back substitution, 256 example, 257 forward elimination of unknowns,

255–256

MATLAB M-file, 258 operation counting, 258–261

Natural cubic splines, 466–467 Nearest neighbor interpolation, 470 Nesting, 79–81 Neumann boundary condition, 654 Newtint function, 440–441 Newton interpolating polynomial, 433.

See also Lagrange interpolating polynomial form of, 437–439 linear interpolation, 433–435 Newtint function, 440–441 quadratic interpolation, 435–437

Newton linear-interpolation formula, 433

Newton-Cotes closed integration formulas, 507 formulas, 486, 492–494, 512, 525,

541–542 integration formulas, 593 open integration formulas, 512

Newton-Raphson bungee jumper problem, 173–174

Newton-Raphson method, 169, 170, 314 MATLAB M-file, 173–174, 318–319 multivariable Taylor series, 314–315 near-zero slope, 171 for nonlinear system, 315–318 poor convergence, 172 slowly converging function, 170–171

Newton’s interpolating polynomial, 345 Newton’s second law, 12, 228, 326, 405,

551, 573 Newton’s viscosity law, 552 newtraph function, 173–174 Non-self-starting Heun method, 624

example, 626–627 fundamental difference, 624, 625

Nongradient methods, 213 Nonhomogeneous system, 328 Nonlinear algebraic equation, 183 Nonlinear ODEs

finite-difference methods for, 663–665

shooting method for, 656–658 Nonlinear regression, 345, 395. See also

General linear least-squares regression fitting experimentatal data, 397–399

with MATLAB, 395–396 techniques, 367

Nonlinear relationships, linearization of, 366 comments on linear regression, 370 fitting data with power equation,

368–370 nonlinear regression techniques,

367–368 power equation, 366–367

Nonlinear simultaneous equations, 226 Nonlinear systems, 312. See also Linear

systems Newton-Raphson method, 314–319 simultaneous nonlinear equations,

312–313 successive substitution, 313–314

Nonperiodic function, 416 Nonstiff systems, MATLAB functions

for, 617 MATLAB to solving system of

ODEs, 618–620 ode23 function, 617 odeset to control integration options,

620–621 predator-prey model, 619, 620

Norm, 293 in MATLAB, 296–297

Normal distribution, 351–352 Normalization, 256 “Not-a-knot” end condition, 467 Number system, 106 Numerical differentiation, 16, 121, 487,

549–551 backward difference approximation

of first derivative, 123 centered difference approximation of

first derivative, 123 derivatives

and integrals for data with errors, 558–559

partial, 559–560 of unequally spaced data,

557–558 diff function, 560–562 in engineering and science, 551–552 error analysis, 126–129 finite-difference approximations

of derivatives, 124–125 of higher derivatives, 125

cha97962_Index_687-704.indd 693 06/11/16 3:39 pm

694 INDEX

Numerical differentiation (Continued) free-falling bungee jumper, 548 gradient function, 563–564 high-accuracy differentiation

formulas, 552–555 with MATLAB, 560 Richardson extrapolation, 555–557 roundoff errors in, 127–129 truncation errors in, 127–129 visualizations of vector fields,

565–567 Numerical errors, 101

control, 129–130 Numerical integration, 16

formulas computing work with, 515–518 free-falling bungee jumper,

488–489 higher-order Newton-Cotes

formulas, 507–508 multiple integrals, 512–514 Newton-Cotes formulas, 492–494 open methods, 512 Simpson’s rules, 501–507 trapezoidal rule, 494–501

of functions adaptive quadrature, 537–540 Gauss quadrature, 530–537 Romberg integration, 525–530 root-mean-square current,

540–543 Numerical methods, 1, 2, 9, 13, 104,

136. See also Bisection method conservation laws in engineering and

science, 12–13 devices and types of balances, 14 real drag, 17–19

Nyquist frequency, 419

O Octal, 107 ODE. See Ordinary differential

equations (ODE) ode113 function, 618 ode15s function, 632 ode23 function, 617, 620–621 ode23s function, 632 ode23t function, 632 ode23tb function, 632 ode45 function, 618, 620, 637, 655

Ohm’s law, 241, 541, 552 One-dimensional optimization, 202–211

fminbnd function, 210–211 golden-section search, 203–208 parabolic interpolation, 209–210

One-point iteration. See Simple fixed-point iteration

One-step methods, 581 Open integration formulas, 487, 494 Open methods, 143, 164 Operation counting, 258–261 Optimization, 16, 136, 199

elevation as function of time, 199 equilibrium and minimum potential

energy, 213–215 mathematical perspective, 199 multidimensional, 202, 211–213 one-dimensional, 202–211 optimum analytically by root

location, 200–201 single variable function, 200

Optimum analytically by root location, 200–201

optimset function, 182–183 Ordinary differential equations (ODE),

16, 573, 574, 647 dependent variable, 573–574 first-order ODEs, 574, 576 initial-value problem, 577 Runge-Kutta techniques, 577 shooting method

for linear ODE, 652–654 for nonlinear ODEs, 656–658

solution, 575 Ordinary frequency, 407 Oscillations, 447–449 Overdetermined systems, 238 Overflow error, 109 Overrelaxation, 309

P Parabola. See Second-order polynomial Parabolic interpolation, 209–210 Parameters, 5, 6

passing, 85–87 Partial derivatives, 550, 559–560 Partial differential equation (PDE),

574. See also Ordinary differential equations (ODE)

Partial pivoting, 261 Passed function, 82 Passing functions

anonymous functions, 81–82 building and implementing function

function, 84–85 function functions, 82–83 to M-files, 81

Passing parameters, 85 approach for, 86–87

Pause command, 76 pchip. See Piecewise cubic Hermite

interpolation (pchip) PDE. See Partial differential equation

(PDE) Per-step truncation error estimation, 628 Percent relative error, 102 Periodic function, 405 Permutation matrix, 233 Phase angle, 407, 407 Phase shift, 407 Phase-plane plots, 605, 609 Phasor, 413 Piecewise cubic Hermite interpolation

(pchip), 468, 471 Piecewise cubic spline interpolation, 470 Piecewise interpolation. See also Multi-

dimensional interpolation; Polyno- mial interpolation heat transfer, 476–480 interp1 function, 470–473 in MATLAB, 468 spline function, 468–470

Pipe friction, 186–190 Pivot equation, 256 Pivoting, 261–264

determinant evaluation with gauss elimination, 263–264

LU factorization with, 280–282 MATLAB M-file, 262–263

Pliny’s intermittent fountain, 635–639 plot3 function, 608 Point-slope method. See Euler’s method polyfit functions, 433

polynomial regression implemention with, 394

Polynomial interpolation, 430. See also Multidimensional interpolation examples, 431 extrapolation, 445–447

cha97962_Index_687-704.indd 694 06/11/16 3:39 pm

INDEX 695

inverse interpolation, 444–445 Lagrange interpolating polynomial,

441–444 Newton interpolating polynomial,

433–441 oscillations, 447–449

Polynomial(s), 183–186 MATLAB function, 184–186 method, 328, 329–331 regression, 345, 385–389

implementation, 394 with MATLAB, 392–393

polyval functions, 433 Positional notation, 107 Posttest loop, 76 Potential energy, 213 Power equation, 366

fitting data with, 368–370 Power method, 333–334

for highest eigenvalue, 334–336 Power spectrum, 423

with MATLAB, 424 Preallocation of memory, 74–75 Precision, 100–101, 111–112 Predator-prey

equations, 618 models and chaos, 604–609

Predictor equation, 588 Predictor-corrector approach, 589 Pretest loop, 75 Primary function, 61 Problem solving, 1

real drag, 17–19 Programming with MATLAB

bungee jumper velocity, 87–90 input-output, 61–65 M-files, 54 nesting and indentation, 79–81 passing functions to M-files, 81–87 structured programming, 65–78

Propagated truncation error, 583–584 Proportionality, 291

Q QR factorization, 394 quadadapt function, 537–539 Quadratic

convergence, 170 interpolation, 435–437

polynomial. See Second-order polynomial

splines, 459–462 Quadrature, 489

R Ralston’s method, 595 Random numbers, 353

generation, 345 rand function, 354–356 randn function, 356–358

Range, 111, 349 Rate equations, 573 Regression, 16 Relative tolerance (RelTol), 618 Relaxation, 309–310

Gauss-Seidel method with, 311–312 RelTol. See Relative tolerance (RelTol) Residual, 358

Reverse-wrap-around order, 421, 423

Reynolds number, 17 Richardson extrapolation, 524, 525–526,

555–557 example, 526–527 higher-order corrections, 527

RK methods. See Runge-Kutta methods (RK methods)

RK-Fehlberg methods, 617 rk4sys function, 602–603, 605 Romberg integration, 487, 508, 524,

525, 528. See also Adaptive quadra- ture; Gauss quadrature algorithm, 527–530 M-file to implement, 530 Richardson extrapolation, 525–527

Roots, 135, 164, 165, 184–186 bisection method, 146–152 Brent’s method, 176–181 in engineering and science, 139–140 false position, 152–155 graphical methods, 140–141 greenhouse gases and rainwater,

156–159 MATLAB function, 181–183 Newton-Raphson method, 169–174 pipe friction, 186–190 polynomials, 183–186 root-location techniques, 224 root-mean-square current, 540–543

Secant methods, 174–176 simple fixed-point iteration, 165–169

roots function, 184–186 round function, 40 Roundoff errors, 3, 106, 583. See also

Truncation errors arithmetic manipulations of computer

numbers, 112–114 computer number representation,

106–112 in numerical differentiation,

127–129 Row, 229

row-sum norm, 294 vectors, 229

Runge-Kutta methods (RK methods), 577, 578, 581, 587, 593. See also Adaptive Runge-Kutta methods; Euler’s method classical fourth-order, 595–598 increment function, 593 rk4sys function, 602–603 second-order, 594–595 systems of equations, 600–601,

603–604 types of, 594

Runge’s function, 447, 468 comparison, 448–449, 469–470

S S-shape, 374 Saturation-growth-rate equation, 367 Scalars, 29–30 Script files, 54–55 Secant methods, 174–176. See also

Brent’s method Second finite divided difference, 437 Second forward finite difference, 125 Second-order equation, 574 Second-order polynomial, 435 Second-order Runge-Kutta methods,

594–595 Sequential search, 458 Shooting method, 651

with derivative boundary conditions, 654–656

for linear ODE, 652–654 for nonlinear ODEs, 656–658 trial-and-error approach, 651

Sigmoid, 374

cha97962_Index_687-704.indd 695 06/11/16 3:39 pm

696 INDEX

Signed magnitude method, 107 Signum function, 580 Simple fixed-point iteration, 165–169,

313–314 Simpson’s rules, 501, 529

Simpson’s 1∕3 rule, 486, 501–502, 535, 541 composite Simpson’s 1∕3 rule,

503–505 single application, 502–503

Simpson’s 3∕8 rule, 486, 505–507, 535

Simulation, 353–358 MATLAB function, 353–358

Simulink®, 677–684 Simultaneous equations, determin-

ing polynomial coefficients with, 431–432

single-line if structure, 66 Singular value decomposition, 394 Sinusoid, 405

least-squares fit of, 408–411 sinusoidal functions, 405

curve fitting with, 405–411 Smearing, 114 SOR. See Successive overrelaxation

(SOR) Spectral norm, 294 Splines, 453

cubic, 462–468 drafting technique, 455 functions, 453 heat transfer, 476–480 to higher-order interpolating

polynomials, 454 interpolation, 345 linear, 455–459 in MATLAB, 468–470 quadratic, 459–462 spline function, 468–470

Square matrices, 230 Standard deviation, 349 Steady-state calculation, 12 Stefan-Boltzmann law, 59 Step halving approach, 616 Stiff system, 628 Stiffness, 628

Euler’s method, 629–630 explicit and implicit Euler method,

630–632

MATLAB application, 634–635 MATLAB functions for, 632–634 Pliny’s intermittent fountain,

635–639 stiff solution of single ODE, 629

Stimulus-response computations, 290–291

Stokes drag, 17 Stopping criterion, 103 String functions, 35 Structured programming, 65

animation, 77–78 decisions, 65–72 loops, 72–76

Subfunctions, 60–61 Subtractive cancellation, 113 Successive overrelaxation (SOR), 309 Successive substitution. See Simple

fixed-point iteration sum function, 40 Sunspots, 425–426 Superposition, 291 Swamee-Jain equation, 187, 189 Switch structure, 70–71 Symmetric matrix, 230 Systems of equations, 598

Euler’s method, 598–600 Runge-Kutta methods, 600–604

T Table lookup, 458–459 Taylor series, 114–118, 552–553

to estimating truncation errors, 120–121

expansion, 118, 314, 584 remainder for, 119–120

Taylor theorem, 114, 117 Terminal velocity, 9 Thermocline, 476 Third-order polynomial, 462 Three-point Gauss-Legendre formula,

536–537 Time domain, 414–415 Time plane, 415 Time series, 405 Time-consuming evaluation, 208 Time-variable, 12 Top-down design, 79 Torricelli’s law, 635–636

Total numerical error, 125 control of numerical errors,

129–130 error analysis of numerical differen-

tiation, 126–129 trade-off between roundoff and

truncation error, 126 Transient variable, 12 Transpose, 233, 234, 283 Transposition matrix, 233 trap function, 499–501 Trapezoidal rule, 486, 494, 495, 529,

531, 541, 592, 593 composite trapezoidal rule, 497–499 error of, 495 integral evaluation by, 532 MATLAB M-file function, trap,

499–501 single application of, 495–497 with unequal segments, 508

trapuneq function, 509 trapz functions, 510 Trend analysis, 344 Trial and error technique, 135, 651 Tridiag function, 266 Tridiagonal matrices, 225, 231 Tridiagonal systems, 264

MATLAB M-file, 266 solution of, 265

Truncation errors, 3, 114, 583–584. See also Roundoff errors numerical differentiation, 121–125 in numerical differentiation, 127–129 remainder for Taylor series

expansion, 119–120 Taylor series, 114–118 Taylor series to estimating, 120–121

Two-dimensional interpolation, 473 implementing by first applying

one-dimensional linear interpolation, 474

Two-point Gauss-Legendre formula, 533–536, 543 weighting factors and function

arguments, 536 2s complement technique, 108

U Uncertainty. See Imprecision Unconditionally stable approach, 630

cha97962_Index_687-704.indd 696 06/11/16 3:39 pm

INDEX 697

Underdetermined systems, 238 Underrelaxation, 309 Undetermined coefficients method,

531–533 Unequally spaced data, derivatives of,

557–558 Unimodal, 204 Upper triangular matrix, 230, 280 Upper triangular system, 256

V van der Pol equation, 632–633 Vandermonde matrices, 432

Variable argument list, 71–72 Variable scope, 57–58 Variance, 349 Vectorization, 74 Vectors, 31–32, 293–294

vector fields, visualizations of, 565–567

vector-matrix multiplication, 38 Visualization

two-dimensional function, 212 vector fields, 565–567

Voltage rules, 240, 241 Voltages in circuits, 240–243 Volume integral, 492

W while structure, 75 while... break structure, 75–76 Wolf sunspot number, 425 Word, 106

X xmin function, 213

Z Zeros

of equation, 135 zero-order approximation, 115

cha97962_Index_687-704.indd 697 06/11/16 3:39 pm

Cover
Title
Copyright
Contents
About the author
Preface
Part One: Modeling, Computers, and Error Analysis

1.1 Motivation
1.2 Part Organization
Chapter 1 Mathematical Modeling, Numerical Methods, and Problem Solving

1.1 A Simple Mathematical Model
1.2 Conservation Laws in Engineering and Science
1.3 Numerical Methods Covered in This Book
1.4 Case Study: It's a Real Drag
Problems

Chapter 2 MATLAB Fundamentals

2.1 The MATLAB Environment
2.2 Assignment
2.3 Mathematical Operations
2.4 Use of Built-In Functions
2.5 Graphics
2.6 Other Resources
2.7 Case Study: Exploratory Data Analysis
Problems

Chapter 3 Programming with MATLAB

3.1 M-Files
3.2 Input-Output
3.3 Structured Programming
3.4 Nesting and Indentation
3.5 Passing Functions to M-Files
3.6 Case Study: Bungee Jumper Velocity
Problems

Chapter 4 Roundoff and Truncation Errors

4.1 Errors
4.2 Roundoff Errors
4.3 Truncation Errors
4.4 Total Numerical Error
4.5 Blunders, Model Errors, and Data Uncertainty
Problems

Part Two: Roots and Optimization

2.1 Overview
2.2 Part Organization
Chapter 5 Roots: Bracketing Methods

5.1 Roots in Engineering and Science
5.2 Graphical Methods
5.3 Bracketing Methods and Initial Guesses
5.4 Bisection
5.5 False Position
5.6 Case Study: Greenhouse Gases and Rainwater
Problems

Chapter 6 Roots: Open Methods

6.1 Simple Fixed-Point Iteration
6.2 Newton-Raphson
6.3 Secant Methods
6.4 Brent's Method
6.5 MATLAB Function: fzero
6.6 Polynomials
6.7 Case Study: Pipe Friction
Problems

Chapter 7 Optimization

7.1 Introduction and Background
7.2 One-Dimensional Optimization
7.3 Multidimensional Optimization
7.4 Case Study: Equilibrium and Minimum Potential Energy
Problems

Part Three: Linear Systems

3.1 Overview
3.2 Part Organization
Chapter 8 Linear Algebraic Equations and Matrices

8.1 Matrix Algebra Overview
8.2 Solving Linear Algebraic Equations with MATLAB
8.3 Case Study: Currents and Voltages in Circuits
Problems

Chapter 9 Gauss Elimination

9.1 Solving Small Numbers of Equations
9.2 Naive Gauss Elimination
9.3 Pivoting
9.4 Tridiagonal Systems
9.5 Case Study: Model of a Heated Rod
Problems

Chapter 10 LU Factorization

10.1 Overview of LU Factorization
10.2 Gauss Elimination as LU Factorization
10.3 Cholesky Factorization
10.4 MATLAB Left Division
Problems

Chapter 11 Matrix Inverse and Condition

11.1 The Matrix Inverse
11.2 Error Analysis and System Condition
11.3 Case Study: Indoor Air Pollution
Problems

Chapter 12 Iterative Methods

12.1 Linear Systems: Gauss-Seidel
12.2 Nonlinear Systems
12.3 Case Study: Chemical Reactions
Problems

Chapter 13 Eigenvalues

13.1 Mathematical Background
13.2 Physical Background
13.3 The Power Method
13.4 MATLAB Function: eig
13.5 Case Study: Eigenvalues and Earthquakes
Problems

Part Four: Curve Fitting

4.1 Overview
4.2 Part Organization
Chapter 14 Linear Regression

14.1 Statistics Review
14.2 Random Numbers and Simulation
14.3 Linear Least-Squares Regression
14.4 Linearization of Nonlinear Relationships
14.5 Computer Applications
14.6 Case Study: Enzyme Kinetics
Problems

Chapter 15 General Linear Least-Squares and Nonlinear Regression

15.1 Polynomial Regression
15.2 Multiple Linear Regression
15.3 General Linear Least Squares
15.4 QR Factorization and the Backslash Operator
15.5 Nonlinear Regression
15.6 Case Study: Fitting Experimental Data
Problems

Chapter 16 Fourier Analysis

16.1 Curve Fitting with Sinusoidal Functions
16.2 Continuous Fourier Series
16.3 Frequency and Time Domains
16.4 Fourier Integral and Transform
16.5 Discrete Fourier Transform (DFT)
16.6 The Power Spectrum
16.7 Case Study: Sunspots
Problems

Chapter 17 Polynomial Interpolation

17.1 Introduction to Interpolation
17.2 Newton Interpolating Polynomial
17.3 Lagrange Interpolating Polynomial
17.4 Inverse Interpolation
17.5 Extrapolation and Oscillations
Problems

Chapter 18 Splines and Piecewise Interpolation

18.1 Introduction to Splines
18.2 Linear Splines
18.3 Quadratic Splines
18.4 Cubic Splines
18.5 Piecewise Interpolation in MATLAB
18.6 Multidimensional Interpolation
18.7 Case Study: Heat Transfer
Problems

Part Five: Integration and Differentiation

5.1 Overview
5.2 Part Organization
Chapter 19 Numerical Integration Formulas

19.1 Introduction and Background
19.2 Newton-Cotes Formulas
19.3 The Trapezoidal Rule
19.4 Simpson's Rules
19.5 Higher-Order Newton-Cotes Formulas
19.6 Integration with Unequal Segments
19.7 Open Methods
19.8 Multiple Integrals
19.9 Case Study: Computing Work with Numerical Integration
Problems

Chapter 20 Numerical Integration of Functions

20.1 Introduction
20.2 Romberg Integration
20.3 Gauss Quadrature
20.4 Adaptive Quadrature
20.5 Case Study: Root-Mean-Square Current
Problems

Chapter 21 Numerical Differentiation

21.1 Introduction and Background
21.2 High-Accuracy Differentiation Formulas
21.3 Richardson Extrapolation
21.4 Derivatives of Unequally Spaced Data
21.5 Derivatives and Integrals for Data with Errors
21.6 Partial Derivatives
21.7 Numerical Differentiation with MATLAB
21.8 Case Study: Visualizing Fields
Problems

Part six: Ordinary Differential Equations

6.1 Overview
6.2 Part Organization
Chapter 22 Initial-Value Problems

22.1 Overview
22.2 Euler's Method
22.3 Improvements of Euler's Method
22.4 Runge-Kutta Methods
22.5 Systems of Equations
22.6 Case Study: Predator-Prey Models and Chaos
Problems

Chapter 23 Adaptive Methods and Stiff Systems

23.1 Adaptive Runge-Kutta Methods
23.2 Multistep Methods
23.3 Stiffness
23.4 MATLAB Application: Bungee Jumper with Cord
23.5 Case Study: Pliny's Intermittent Fountain
Problems

Chapter 24 Boundary-Value Problems

24.1 Introduction and Background
24.2 The Shooting Method
24.3 Finite-Difference Methods
24.4 MATLAB Function: bvp4c
Problems

Appendix A: Matlab Built-In Functions
Appendix B: Matlab M-File Functions
Appendix C: Introduction To Simulink
Bibliography
Index

1. 2017-02-02T00:38:44+0000
2. Preflight Ticket Signature