MATH 332 - ADVANCED CALCULUS
- Calculus of variations
Question Bank - Set 4
Liberty University
Question 1
Question
Find the extremals of the functional
J[y] = Z1
02y′(x)2+y(x)2dx
subject to the boundary conditions y(0) = 0 and y(1) = 1.
Solution
Step 1: Set up the Euler-Lagrange equation. The Euler-Lagrange equation for
a functional of the form
J[y] = Zb
a
F(x, y, y′)dx
is given by
d
dx ∂F
∂y′−∂F
∂y = 0
In this case, F(x, y, y′)=2y′(x)2+y(x)2. So, we have
d
dx ∂
∂y′(2y′(x)2)−∂
∂y (2y′(x)2) + d
dx ∂
∂y (y(x)2)−∂
∂y (y(x)2) = 0
Simplifying, we get
d
dx (4y′(x)) −2y(x)+2y(x) = 0
4y′′(x)=0
Thus, the extremals must satisfy y′′(x) = 0.
Step 2: Solve the differential equation. Integrating y′′(x) = 0 twice gives
y(x) = Ax +B
Applying the boundary conditions y(0) = 0 and y(1) = 1, we find that A= 1
and B= 0.
Therefore, the extremal that minimizes the functional is y(x) = x.
Question 2
Question
Find the extremals of the functional
J[y] = Z1
0
(y′2−2y)dx
subject to the boundary conditions y(0) = 0 and y(1) = 1.
Solution
Step 1: Define the Lagrangian.
The Lagrangian associated with the given functional J[y] is given by
L(x, y, y′) = y′2−2y+λ1y(0) + λ2(y(1) −1)
where λ1and λ2are the Lagrange multipliers corresponding to the boundary
conditions.
Step 2: Set up the Euler-Lagrange equation.
The Euler-Lagrange equation associated with the given functional is
d
dx ∂L
∂y′−∂L
∂y = 0
This gives us
d
dx (2y′)−(−2) = 0
which simplifies to
2y′′ + 2 = 0
or
y′′ =−1
Step 3: Solve the differential equation for y(x).
Integrating y′′ =−1 twice gives
y′=−x+C1
2
y=−1
2x2+C1x+C2
Applying the boundary conditions y(0) = 0 and y(1) = 1, we have
C2= 0
−1
2+C1+ 0 = 1
C1=3
2
Therefore, the extremal that minimizes the functional is
y(x) = −1
2x2+3
2x
Question 3
Question
Find the extremals of the functional
J[y] = Zb
a
(2y−y′)p1 + y′2dx
subject to the boundary conditions y(a) = 1 and y(b) = 2.
Solution
Step 1: Compute the Euler-Lagrange equation by differentiating the integrand
with respect to yand y′.
d
dx ∂f
∂y′−∂f
∂y = 0
where f(y, y′, x) = (2y−y′)p1 + y′2.
Step 2: Compute the partial derivatives needed for the Euler-Lagrange equa-
tion. ∂f
∂y′= 2y−y′−y(2y−y′)y′
p1 + y′2
and ∂f
∂y = 2p1 + y′2
Step 3: Differentiate ∂f
∂y′with respect to x.
d
dx 2y−y′−y(2y−y′)y′
p1 + y′2!=d
dx (2y−y′)−d
dx y(2y−y′)y′
p1 + y′2!
3
Step 4: Compute the first term in the differentiation.
d
dx (2y−y′)=2y′−y′′
Step 5: Compute the second term in the differentiation.
d
dx y(2y−y′)y′
p1 + y′2!=d
dx 2y2y′−y′2y2
p1 + y′2!
Step 6: Simplify and combine the terms.
2y′−y′′ −4y2y′y′2−y′2y2
(1 + y′2)3/2= 0
Step 7: Integrate the Euler-Lagrange equation subject to the boundary con-
ditions y(a) = 1 and y(b) = 2 to find the extremal y(x).
Question 4
Question
Find the extremals of the functional
J[y] = Z1
0
(y′2−y)dx
subject to the boundary conditions y(0) = 0 and y(1) = 1.
Solution
Step 1: Define the Euler-Lagrange equation.
The Euler-Lagrange equation for a functional of the form
J[y] = Zb
a
F(x, y, y′)dx
is given by
d
dx ∂F
∂y′−∂F
∂y = 0
Step 2: Compute the partial derivatives.
In this case, F(x, y, y′) = y′2−y. Therefore, the partial derivatives are:
∂F
∂y =−1
∂F
∂y′= 2y′
4
d
dx ∂F
∂y′= 2y′′
Step 3: Formulate the Euler-Lagrange equation.
Plugging the partial derivatives into the Euler-Lagrange equation, we get:
2y′′ + 1 = 0
Step 4: Solve the differential equation.
Solving the differential equation 2y′′ + 1 = 0, we find
y′′ =−1
2
Integrating twice gives
y′=−x
2+C1
y=−x2
4+C1x+C2
Step 5: Apply the boundary conditions.
Using the boundary conditions y(0) = 0 and y(1) = 1, we get the system of
equations
C2= 0
−1
4+C1= 1
which implies C1=5
4.
Step 6: Final solution.
Therefore, the extremal that minimizes the functional is
y=−x2
4+5
4x
Question 5
Question
Find the extremals of the functional
J[y] = Z1
0
(y′2−y2)dx
subject to the boundary conditions y(0) = 0 and y(1) = 1.
5
Solution
Step 1: Compute the Euler-Lagrange equation. The Euler-Lagrange equation
for the functional J[y] is given by
d
dx ∂f
∂y′−∂f
∂y = 0
where f=y′2−y2.
So, we have
d
dx (2y′)−(−2y)=0
2y′′ + 2y= 0
Step 2: Solve the differential equation. The general solution to the differen-
tial equation is of the form
y(x) = Acos(x) + Bsin(x)
Step 3: Apply the boundary conditions. Using the boundary conditions
y(0) = 0 and y(1) = 1, we find
y(0) = Acos(0) + Bsin(0) = A= 0
y(1) = Bsin(1) = 1
Thus, B=1
sin(1) .
Step 4: The extremal function. Therefore, the extremal function that mini-
mizes the functional J[y] is
y(x) = sin(x)
sin(1)
Question 6
Question
Find the extremals of the functional
J[y] = Z1
0
(2y′2−2y)dx
subject to the boundary conditions y(0) = 1 and y(1) = 2.
Solution
Step 1: Compute the Euler-Lagrange equation.
The Euler-Lagrange equation for this functional is given by
d
dx ∂f
∂y′−∂f
∂y = 0,
6
where f= 2y′2−2y.
Step 2: Calculate the partial derivatives.
The partial derivative with respect to y′is
∂f
∂y′= 4y′
and the partial derivative with respect to yis
∂f
∂y =−2.
Step 3: Apply the Euler-Lagrange equation.
Substitute the partial derivatives back into the Euler-Lagrange equation to get
d
dx (4y′) + 2 = 0.
Step 4: Solve the differential equation.
This simplifies to 4y′′ +2 = 0, which gives the second-order differential equation
y′′ =−1
2.
Step 5: Solve the differential equation with boundary conditions.
Integrating this twice gives y(x) = −1
4x2+3
4x+5
4.
Step 6: Apply the boundary conditions.
Using the boundary conditions y(0) = 1 and y(1) = 2, we find that the extremal
which minimizes J[y] subject to the given boundary conditions is y(x) = −1
4x2+
3
4x+5
4.
Question 7
Question
Find the function y(x) that minimizes the functional
J[y] = Z1
0
(y′2−4y)dx
subject to the boundary conditions y(0) = 0 and y(1) = 3.
Solution
Step 1: Define the Euler-Lagrange equation. This is given by
d
dx ∂f
∂y′−∂f
∂y = 0
where f=y′2−4y.
7
Step 2: Compute the partial derivatives of fwith respect to yand y′.
∂f
∂y =−4
∂f
∂y′= 2y′
Step 3: Calculate the derivative of ∂f
∂y′with respect to x.
d
dx ∂f
∂y′=d
dx (2y′)=2y′′
Step 4: Set up the Euler-Lagrange equation using the results from Steps 2
and 3.
2y′′ + 4 = 0
Step 5: Solve the differential equation from Step 4 subject to the boundary
conditions y(0) = 0 and y(1) = 3. The general solution to the differential
equation is y(x) = Asin(2x) + Bcos(2x)−2.
Step 6: Apply the boundary conditions to find the values of Aand B.
Using y(0) = 0, we have B−2 = 0, so B= 2. Using y(1) = 3, we have
Asin(2) + 2 cos(2) −2 = 3. Solving for A, we find A≈0.531.
Therefore, the function y(x) that minimizes the given functional is
y(x)≈0.531 sin(2x) + 2 cos(2x)−2
Question 8
Question
Find the extremals of the functional
J[y] = Z1
0
(y′2−y2)dx
subject to the boundary conditions y(0) = 0 and y(1) = 1.
Solution
Step 1: Set up the Euler-Lagrange equation The extremals of the func-
tional J[y] are the solutions to the Euler-Lagrange equation, which is given
by
d
dx ∂f
∂y′−∂f
∂y = 0
where f=y′2−y2.
8
Step 2: Compute the partial derivatives First, compute the partial
derivative ∂f
∂y′:
∂f
∂y′=∂
∂y′(y′2−y2)=2y′
Next, compute the partial derivative ∂f
∂y :
∂f
∂y =∂
∂y (y′2−y2) = −2y
Step 3: Formulate the Euler-Lagrange equation Substitute the partial
derivatives into the Euler-Lagrange equation:
d
dx (2y′)−(−2y)=0
Simplify to obtain:
d
dx (2y′)+2y= 0
Step 4: Solve the differential equation The differential equation to be
solved is:
2y′′ + 2y= 0
which simplifies to:
y′′ +y= 0
Step 5: Find the general solution The general solution to the ODE
y′′ +y= 0 is:
y(x) = c1cos(x) + c2sin(x)
Step 6: Apply boundary conditions Using the boundary conditions
y(0) = 0 and y(1) = 1, we have:
c1= 0
c2sin(1) = 1
Solve the second equation to find c2=1
sin(1) .
Step 7: Final solution Therefore, the extremal that minimizes the func-
tional J[y] is given by:
y(x) = sin(x)
sin(1)
Question 9
Question
Find the extremals of the functional
J[y] = Z1
0
(y2+y′2)dx
subject to the boundary conditions y(0) = 0 and y(1) = 1.
9
Solution
Step 1: Define the Euler-Lagrange equation. The Euler-Lagrange equation for
a functional J[y] = Rb
aF(x, y, y′)dx is given by
d
dx ∂F
∂y′−∂F
∂y = 0
Applying this to the given functional J[y] = R1
0(y2+y′2)dx, we have
d
dx (2y′)−2y= 0
Step 2: Solve the differential equation. Solving the above differential equa-
tion, we get
2y′′ −2y= 0
y′′ −y= 0
The general solution to this differential equation is y(x) = c1ex+c2e−x.
Step 3: Apply the boundary conditions. Using the boundary conditions
y(0) = 0 and y(1) = 1, we have
0 = c1+c2
1 = c1e+c2e−1
Solving these equations simultaneously, we find c1=e−1
e−1and c2=−1
e−1.
Step 4: Find the extremals. Therefore, the extremal that minimizes the
functional J[y] subject to the given boundary conditions is
y(x) = e−1
e−1ex−1
e−1e−x
Question 10
Question
Find the function y(x) that minimizes the functional
J[y] = Z1
0
(y′2−2y)dx
subject to the boundary conditions y(0) = 0 and y(1) = 1.
10
Solution
Step 1: Define the Euler-Lagrange equation. This is given by
d
dx ∂f
∂y′−∂f
∂y = 0
where f(y, y′, x) = y′2−2y.
Step 2: Compute the partial derivatives of fwith respect to yand y′.
∂f
∂y =−2 and ∂f
∂y′= 2y′
Step 3: Differentiate ∂f
∂y′with respect to x.
d
dx ∂f
∂y′=d
dx (2y′)=2y′′
Step 4: Substitute the derivatives back into the Euler-Lagrange equation.
2y′′ + 2 = 0
Step 5: Solve the differential equation y′′ + 1 = 0 subject to the boundary
conditions y(0) = 0 and y(1) = 1. The general solution to the differential
equation is y(x) = Asin(x) + Bcos(x).
Step 6: Apply the boundary conditions to find Aand B. Using y(0) = 0,
we have 0 = B. Using y(1) = 1, we have 1 = Asin(1), which implies A=1
sin(1) .
Step 7: Therefore, the function y(x) that minimizes the functional is
y(x) = 1
sin(1) sin(x)
Question 11
Question
Find the extremals of the functional
J[y] = Z1
0
(y′2+y2)dx
subject to the boundary conditions y(0) = 0 and y(1) = 1.
Solution
Step 1: Define the Lagrangian The Lagrangian for the given functional is defined
as
L(x, y, y′) = y′2+y2
11
Step 2: Set up the Euler-Lagrange equation We need to solve the Euler-
Lagrange equation
d
dx ∂L
∂y′−∂L
∂y = 0
Step 3: Calculate the partial derivatives
∂L
∂y′= 2y′
∂L
∂y = 2y
Step 4: Apply the Euler-Lagrange equation Substitute the partial derivatives
back into the Euler-Lagrange equation to get
d
dx (2y′)−2y= 0
which simplifies to
2y′′ −2y= 0
Step 5: Solve the differential equation The general solution to the differential
equation 2y′′ −2y= 0 is
y(x) = c1ex+c2e−x
Step 6: Apply boundary conditions Using the boundary conditions y(0) = 0
and y(1) = 1, we have
0 = c1+c2
1 = c1e+c2e−1
Solving these equations simultaneously gives
c1=e
e−1, c2=−1
e−1
Therefore, the extremal of the functional J[y] subject to the given boundary
conditions is
y(x) = e
e−1ex−1
e−1e−x
Question 12
Question
Find the function y(x) that minimizes the functional
J[y] = Z1
0
[(y′)2+ (y′′)2−2yy′]dx
subject to the boundary conditions y(0) = 0 and y(1) = 1.
12
Solution
Step 1: Set up the Euler-Lagrange equation. The Euler-Lagrange equation for
minimizing the functional J[y] is given by
d
dx ∂f
∂y′−∂f
∂y = 0
where f= (y′)2+ (y′′)2−2yy′.
Step 2: Compute the partial derivatives. We have
∂f
∂y′= 2y′+ 2y′′
and ∂f
∂y =−2y′
Step 3: Apply the Euler-Lagrange equation. Substituting the partial deriva-
tives into the Euler-Lagrange equation, we get
d
dx (2y′+ 2y′′)+2y′= 0
which simplifies to
2y′′′ + 4y′= 0
Step 4: Solve the differential equation. The general solution to the differen-
tial equation 2y′′′ + 4y′= 0 is
y(x) = c1+c2e−√2x+c3e√2x
Step 5: Apply the boundary conditions. Using the boundary conditions
y(0) = 0 and y(1) = 1, we find c1= 0, c2= 0, and c3=1
e√2.
Therefore, the function that minimizes the functional J[y] subject to the
given boundary conditions is
y(x) = 1
e√2x
Question 13
Question
Consider the functional J[y] = R2
0(y′2−2y)dx where y(0) = 0 and y(2) = 1.
Find the function y(x) that minimizes J[y].
13
Solution
Step 1: Define the Euler-Lagrange equation. The Euler-Lagrange equation for
the functional J[y] is given by:
d
dx ∂F
∂y′−∂F
∂y = 0
where F=y′2−2y.
Step 2: Compute ∂F
∂y′and ∂F
∂y .
∂F
∂y′= 2y′
∂F
∂y =−2
Step 3: Apply the Euler-Lagrange equation.
d
dx (2y′) + 2 = 0
Step 4: Solve the differential equation.
2y′′ =−2 =⇒y′′ =−1
Step 5: Integrate y′′ twice to find y(x). Integrating y′′ =−1 with respect to
xtwice gives:
y′(x) = −x+C1
y(x) = −1
2x2+C1x+C2
Step 6: Apply the boundary conditions y(0) = 0 and y(2) = 1. Putting
x= 0 and x= 2 in y(x) and y′(x) to find C1and C2:
y(0) = C2= 0
y′(2) = −2 + C1= 0 =⇒C1= 2
Step 7: Determine the function that minimizes J[y]. Substitute C1= 2 and
C2= 0 back into y(x) to find the function y(x) that minimizes J[y]:
y(x) = −1
2x2+ 2x
Question 14
Question
Find the extremals of the functional
J[y] = Z1
0
(y′2−y2)dx
subject to the boundary conditions y(0) = 0 and y(1) = 1.
14
Solution
Step 1: Define the Lagrangian function. The Lagrangian function for this prob-
lem is defined as
L(x, y, y′) = y′2−y2
Step 2: Set up the Euler-Lagrange equation. The Euler-Lagrange equation
for this problem is
d
dx ∂L
∂y′−∂L
∂y = 0
Step 3: Calculate the partial derivatives. We first calculate the partial
derivative with respect to y′:
∂L
∂y′= 2y′
and then the partial derivative with respect to y:
∂L
∂y =−2y
Step 4: Apply the Euler-Lagrange equation. Substitute the partial deriva-
tives into the Euler-Lagrange equation:
d
dx (2y′)−(−2y)=0
Simplify this to obtain the differential equation:
d2y
dx2+y= 0
Step 5: Solve the differential equation. The general solution to the differen-
tial equation is
y(x) = c1sin(x) + c2cos(x)
where c1and c2are constants to be determined.
Step 6: Apply the boundary conditions. Using the boundary conditions
y(0) = 0 and y(1) = 1, we find:
y(0) = c2= 0
y(1) = c1sin(1) = 1
So, c1=1
sin(1) .
Therefore, the extremal that minimizes the functional J[y] subject to the
given boundary conditions is
y(x) = 1
sin(1) sin(x)
15
Question 15
Question
Find the extremal for the functional
J[y] = Z1
0
(y′2−y2)dx
subject to the boundary conditions y(0) = 0 and y(1) = 1.
Solution
Step 1: Define the Lagrangian function. Given the functional
J[y] = Z1
0
(y′2−y2)dx
we define the Lagrangian function as
L(x, y, y′) = y′2−y2
Step 2: Apply the Euler-Lagrange equation. The Euler-Lagrange equa-
tion states that the derivative of the Lagrangian with respect to yminus the
derivative of the derivative of the Lagrangian with respect to y′is equal to 0.
Mathematically, this can be expressed as
d
dx ∂L
∂y′−∂L
∂y = 0
Step 3: Calculate the partial derivatives.
∂L
∂y′=∂
∂y′(y′2−y2)=2y′
∂L
∂y =∂
∂y (y′2−y2) = −2y
Step 4: Apply the Euler-Lagrange equation. Substitute the partial deriva-
tives into the Euler-Lagrange equation:
d
dx (2y′)+2y= 0
Step 5: Solve the differential equation. Solving the differential equation gives
us
y′′ −y= 0
The general solution to this differential equation is
y(x) = c1cosh(x) + c2sinh(x)
16
Applying the boundary conditions y(0) = 0 and y(1) = 1, we find c1= 0 and
c2=1
sinh(1) .
Step 6: Find the extremal. Therefore, the extremal for the functional J[y]
subject to the given boundary conditions is
y(x) = sinh(x)
sinh(1)
Question 16
Question
Find the extremals of the functional
J(y) = Z1
0
(y′2−y2)dx
subject to the boundary conditions y(0) = 0 and y(1) = 1.
Solution
Step 1: Define the Euler-Lagrange equation
d
dx ∂F
∂y′−∂F
∂y = 0
where F=y′2−y2.
Step 2: Calculate the partial derivatives of F
∂F
∂y =−2y
∂F
∂y′= 2y′
Step 3: Differentiate ∂F
∂y′with respect to x
d
dx ∂F
∂y′=d
dx (2y′)=2y′′
Step 4: Apply the Euler-Lagrange equation
2y′′ + 2y= 0
Step 5: The general solution of the ODE is
y(x) = c1sin(x) + c2cos(x)
Step 6: Apply the boundary conditions y(0) = 0 and y(1) = 1
y(0) = c2= 0 ⇒c2= 0
17
y(1) = c1sin(1) = 1 ⇒c1=1
sin(1)
Step 7: Therefore, the extremal function that minimizes the functional J(y)
is
y(x) = 1
sin(1) sin(x)
Question 17
Question
Given the functional
J[y] = Z1
0
(y′2−y2)dx
with the boundary conditions y(0) = 0 and y(1) = 1, find the function y(x) that
minimizes J[y].
Solution
Step 1: Define the Euler-Lagrange equation for the given functional: The Euler-
Lagrange equation is given by:
d
dx ∂f
∂y′−∂f
∂y = 0
where f=y′2−y2.
Step 2: Compute the partial derivatives needed for the Euler-Lagrange equa-
tion: ∂f
∂y′= 2y′,d
dx ∂f
∂y′= 2y′′,and ∂f
∂y =−2y
Step 3: Substitute these derivatives into the Euler-Lagrange equation:
2y′′ + 2y= 0
Step 4: Solve the differential equation for y(x): Solving the differential equa-
tion 2y′′ + 2y= 0, we obtain the general solution y(x) = c1cos(x) + c2sin(x).
Step 5: Apply the boundary conditions: Given y(0) = 0, we have c1= 0.
Given y(1) = 1, we have c2sin(1) = 1, so c2=1
sin(1) .
Step 6: Final solution: Therefore, the function that minimizes the functional
J[y] is y(x) = 1
sin(1) sin(x).
18
Question 18
Question
Find the extremal for the functional
J[y] = Z1
0
(y′2−2y)dx
subject to the boundary conditions y(0) = 2 and y(1) = 3.
Solution
To find the extremal for the given functional J[y], we will apply the Euler-
Lagrange equation:
d
dx ∂f
∂y′−∂f
∂y = 0
where f=y′2−2y.
Step 1: Compute ∂f ∂y′and ∂f
∂y
∂f
∂y′= 2y′
∂f
∂y =−2
Step 2: Apply the Euler-Lagrange equation
d
dx (2y′) + 2 = 0
d
dx (2y′) = −2
2y′′ =−2
y′′ =−1
Step 3: Solve the differential equation y′′ =−1 Integrate twice to find
y(x):
y′=−x+C1
y=−1
2x2+C1x+C2
Step 4: Apply boundary conditions Using y(0) = 2 and y(1) = 3:
2 = C2
3 = −1
2+C1+ 2
C1=5
2
Step 5: Final solution Therefore, the extremal for the functional J[y]
subject to the boundary conditions y(0) = 2 and y(1) = 3 is given by
y(x) = −1
2x2+5
2x+ 2
19
Question 19
Question
Consider the functional J[y] = R1
0(y′2−2y)dx subject to the boundary condi-
tions y(0) = 0 and y(1) = 1. Find the function y(x) that minimizes J[y].
Solution
Step 1: Write down the Euler-Lagrange equation for the given functional J[y]:
The Euler-Lagrange equation is given by
d
dx ∂f
∂y′−∂f
∂y = 0
where f(y, y′, x) = y′2−2yin this case.
Step 2: Calculate the partial derivatives:
∂f
∂y =−2,∂f
∂y′= 2y′
d
dx ∂f
∂y′=d
dx (2y′)=2y′′
Step 3: Plugin the derivatives into the Euler-Lagrange equation:
2y′′ + 2 = 0
Step 4: Solve the differential equation:
y′′ =−1
Step 5: Integrate y′′ =−1 twice to find y(x): Integrating twice gives
y′(x) = −x+C1
Integrating again gives
y(x) = −1
2x2+C1x+C2
Step 6: Apply the boundary conditions y(0) = 0 and y(1) = 1: From
y(0) = 0, we get C2= 0. Substituting y(1) = 1, we get
−1
2+C1= 1 =⇒C1=3
2
Step 7: The function that minimizes J[y] with the given boundary condi-
tions is:
y(x) = −1
2x2+3
2x
20
Question 20
Question
Find the function y(x) that minimizes the functional
J[y] = Z1
0
[(y′)2−y]dx
subject to the boundary conditions y(0) = 0 and y(1) = 1.
Solution
Step 1: Compute the Euler-Lagrange equation by differentiating fwith respect
to yand y′and then applying the Euler-Lagrange equation.
Given functional J[y] = R1
0[(y′)2−y]dx, the integrand is f(y, y′, x)=(y′)2−
y. The partial derivative of fwith respect to yis ∂f
∂y =−1 and the partial
derivative with respect to y′is ∂f
∂y′= 2y′.
Applying the Euler-Lagrange equation, we have:
d
dx ∂f
∂y′=∂f
∂y
d
dx (2y′) = −1
2y′′ =−1
y′′ =−1
2
Step 2: Solve the differential equation y′′ =−1
2with the boundary conditions
y(0) = 0 and y(1) = 1.
Integrating y′′ =−1
2twice gives:
y′=−x
2+C1
y=−x2
4+C1x+C2
Applying the boundary conditions:
y(0) = 0 =⇒C2= 0
y(1) = 1 =⇒ − 1
4+C1= 1 =⇒C1=5
4
Thus, the function that minimizes the functional is:
y(x) = −x2
4+5
4x
21
Question 21
Question
Find the extremals of the functional
J[y] = Z1
0
(y′2−y2)dx
subject to the boundary conditions y(0) = 0 and y(1) = 1.
Solution
Step 1: Let’s define the integrand Las L(x, y, y′) = y′2−y2and the Euler-
Lagrange equation is given by
d
dx ∂L
∂y′−∂L
∂y = 0
Step 2: Compute the partial derivatives of Lwith respect to yand y′.
∂L
∂y =−2yand ∂L
∂y′= 2y′
Step 3: Differentiate ∂L
∂y′with respect to x.
d
dx ∂L
∂y′=d
dx (2y′)=2y′′
Step 4: Substituting into the Euler-Lagrange equation, we get
2y′′ + 2y= 0
which simplifies to
y′′ +y= 0
Step 5: The general solution to the differential equation y′′ +y= 0 is
y(x) = c1cos x+c2sin x
Step 6: Apply the boundary conditions y(0) = 0 and y(1) = 1.
y(0) = c1= 0
y(1) = c2sin 1 = 1
Thus, the unique extremal for the given functional is y(x) = sin x.
22
Question 22
Question
Find the extremals of the functional
J[y] = Z2
1
y2−y′2dx
subject to the boundary conditions y(1) = 0 and y(2) = 1.
Solution
To find the extremals of the functional J[y], we need to solve the Euler-Lagrange
equation, which is given by
d
dx ∂f
∂y′−∂f
∂y = 0
where f=y2−y′2.
Step 1: Compute ∂f
∂y and ∂f
∂y′.
∂f
∂y = 2yand ∂f
∂y′=−2y′
Step 2: Compute d
dx ∂f
∂y′.
d
dx ∂f
∂y′=d
dx (−2y′) = −2y′′
Step 3: Set up the Euler-Lagrange equation and solve.
−2y′′ −2y= 0
y′′ +y= 0
Step 4: Find the general solution to the differential equation y′′ +y= 0.
The general solution to this differential equation is of the form y(x) = c1cos(x)+
c2sin(x).
Step 5: Apply the boundary conditions y(1) = 0 and y(2) = 1. Plugging in
x= 1 and y(1) = 0, we get
0 = c1cos(1) + c2sin(1)
Plugging in x= 2 and y(2) = 1, we get
1 = c1cos(2) + c2sin(2)
Step 6: Solve the system of equations for c1and c2. This system of equations
can be solved to find the values of c1and c2.
Step 7: Write down the extremals. The extremals of the given functional
are the solutions to the differential equation found in Step 3 with the appropriate
values of c1and c2.
23
Question 23
Question
Given the functional
J[y] = Z1
0
(y′2−2y)dx
where y(0) = 0 and y(1) = 1, find the function y(x) that minimizes J[y].
Solution
Step 1: Define the Euler-Lagrange equation. The Euler-Lagrange equation for
a functional of the form J[y] = Rb
aF(x, y, y′)dx is given by
d
dx ∂F
∂y′−∂F
∂y = 0
Step 2: Calculate the partial derivatives of F=y′2−2ywith respect to y
and y′.
∂F
∂y =−2 and ∂F
∂y′= 2y′
Step 3: Substitute the partial derivatives into the Euler-Lagrange equation.
d
dx (2y′)−(−2) = 0
Simplify this to get
d
dx (2y′) + 2 = 0
Step 4: Integrate the above equation with respect to x.
2y′+ 2 = C
Where Cis a constant of integration.
Step 5: Solve the differential equation by isolating y′.
2y′=C−2
y′=C
2−1
Step 6: Integrate y′with respect to xto find y(x).
y=Cx
2−x+D
Where Dis another constant of integration.
Step 7: Apply the boundary conditions y(0) = 0 and y(1) = 1. From
y(0) = 0, we have 0 = D. From y(1) = 1, we have 1 = C
2−1. Solving these
equations gives C= 4.
Step 8: Plug in the values of Cand Dto find the minimizing function y(x).
y(x) = 2x−x=x
Therefore, the function that minimizes the given functional is y(x) = x.
24
Question 24
Question
Find the extremal of the functional
J[y] = Z1
0
(y′2+y2)dx
subject to the boundary conditions y(0) = 0 and y(1) = 1.
Solution
Step 1: Define the Lagrangian function L(x, y, y′) = y′2+y2.
Step 2: The Euler-Lagrange equation is given by
d
dx ∂L
∂y′−∂L
∂y = 0
Plugging in the expression for Lgives
d
dx (2y′)−2y= 0
2y′′ −2y= 0
Step 3: Solve the differential equation to find y(x). The solution to the
differential equation y′′ −y= 0 is y(x) = Acos(x) + Bsin(x).
Step 4: Apply the boundary conditions to solve for Aand B. From y(0) = 0,
we have A= 0. From y(1) = 1, we have Bsin(1) = 1, so B=1
sin(1) .
Step 5: Therefore, the extremal of the functional J[y] subject to the given
boundary conditions is
y(x) = sin(x)
sin(1)
Question 25
Question
Minimize the functional
J[y] = Z1
0
(y′2−y2)dx
subject to the boundary conditions y(0) = 0 and y(1) = 1.
25
Solution
Step 1: Compute the Euler-Lagrange equation for the given functional. The
Euler-Lagrange equation is given by
d
dx ∂f
∂y′−∂f
∂y = 0
where f=y′2−y2.
Step 2: Compute the partial derivatives in the Euler-Lagrange equation. We
have ∂f
∂y′=∂
∂y′(y′2−y2)=2y′
∂f
∂y =∂
∂y (y′2−y2) = −2y
Step 3: Apply the Euler-Lagrange equation. Substituting the partial deriva-
tives into the Euler-Lagrange equation gives
d
dx (2y′)+2y= 0
Simplifying further,
2y′′ + 2y= 0
Step 4: Solve the second-order differential equation obtained in Step 3. The
general solution to this differential equation is of the form
y(x) = c1cos(x) + c2sin(x)
Step 5: Use the boundary conditions to find the constants c1and c2. Ap-
plying the boundary conditions y(0) = 0 and y(1) = 1 gives the system of
equations
c1= 0
c1cos(1) + c2sin(1) = 1
Step 6: Solve the system of equations for c1and c2. From the first equation,
c1= 0. Substituting this into the second equation, we get
0 + c2sin(1) = 1
c2=1
sin(1)
Therefore, the unique minimizer of the functional is given by
y(x) = 1
sin(1) sin(x)
26
Question 26
Question
Let J(y) = R2
1(y(x)2+y′(x)2)dx be the functional to be minimized subject to
the boundary conditions y(1) = 1 and y(2) = 2. Find the function y(x) that
minimizes J(y).
Solution
Step 1: We start by writing the Euler-Lagrange equation for the given functional
J(y). The Euler-Lagrange equation is given by:
d
dx ∂f
∂y′−∂f
∂y = 0
where f(y, y′, x) = y2+y′2.
Step 2: Compute the partial derivatives:
∂f
∂y = 2yand ∂f
∂y′= 2y′
Step 3: Compute the derivative of ∂f
∂y′with respect to x:
d
dx ∂f
∂y′=d
dx (2y′)=2y′′
Step 4: Now we can write the Euler-Lagrange equation:
d
dx (2y′)−2y= 0
Step 5: Simplify the equation:
2y′′ −2y= 0
Step 6: Solve the differential equation 2y′′ −2y= 0 by guessing a solution
of the form y(x) = erx:
2r2erx −2erx = 0
Step 7: Simplify the above expression to find the characteristic equation:
r2−1=0
Step 8: Solve the characteristic equation r2−1 = 0 to find the roots r:
r1= 1 and r2=−1
Step 9: The general solution of the differential equation is:
y(x) = c1ex+c2e−x
27
Step 10: Apply the boundary conditions y(1) = 1 and y(2) = 2 to find the
specific solution. We get:
c1e+c2e−1= 1 and c1e2+c2e−2= 2
Step 11: Solve the system of equations to find c1and c2:
c1=e−e−1
e2−e−2and c2=e2−2e+ 1
e2−e−2
Step 12: Therefore, the function that minimizes J(y) is:
y(x) = ex−e−x
e2−e−2+e2−2e+ 1
e2−e−2e−x
Question 27
Question
Find the extremals of the functional
J[y] = Z1
0
(y′2−2y)dx
subject to the boundary conditions y(0) = 0 and y(1) = 1.
Solution
Step 1: Compute the Euler-Lagrange equation. The Euler-Lagrange equation
for the functional J[y] is given by
d
dx ∂f
∂y′−∂f
∂y = 0,
where f=y′2−2y.
Step 2: Compute the partial derivatives. Let’s find the partial derivatives
needed for the Euler-Lagrange equation:
∂f
∂y′=∂
∂y′(y′2−2y) = 2y′,
∂f
∂y =∂
∂y (y′2−2y) = −2.
Step 3: Apply the Euler-Lagrange equation. Substitute the partial deriva-
tives back into the Euler-Lagrange equation:
d
dx (2y′) + 2 = 0.
28
Step 4: Solve the differential equation. Integrating the above equation gives
us:
2y′+C=−2,
where Cis a constant of integration.
Step 5: Apply boundary conditions. Using the boundary conditions y(0) =
0 and y(1) = 1, we find C=−2.
Step 6: Find the extremal function. Thus, we have 2y′−2 = −2, which
simplifies to y′= 0. Integrating once more, we find y(x) = Ax +B.
Step 7: Apply boundary conditions to find Aand B. Using the boundary
conditions y(0) = 0 and y(1) = 1:
0 = A(0) + B⇒B= 0,
1 = A(1) ⇒A= 1.
Step 8: Final Answer. Therefore, the extremal of the functional J[y] subject
to the given boundary conditions is y(x) = x.
Question 28
Question
Find the extremals of the functional
J[y] = Z1
0
(y′2−y)dx
subject to the boundary conditions y(0) = 1 and y(1) = 0.
Solution
To find the extremals of the functional J[y], we need to apply the Euler-Lagrange
equation:
d
dx ∂f
∂y′−∂f
∂y = 0
where f=y′2−y.
Step 1: Calculate ∂f
∂y′and ∂f
∂y .
∂f
∂y′= 2y′and ∂f
∂y =−1
Step 2: Apply the Euler-Lagrange equation to obtain the differential equa-
tion satisfied by y(x):
d
dx (2y′) + 1 = 0
⇒2y′′ + 1 = 0
29
Step 3: Solve the differential equation with the boundary conditions y(0) =
1 and y(1) = 0. The general solution to the differential equation is y(x) =
−1
2x2+cx +d, where cand dare constants to be determined.
Applying the boundary conditions:
y(0) = d= 1
y(1) = −1
2+c+ 1 = 0
c=3
2
Therefore, the extremal function that minimizes the functional is y(x) =
−1
2x2+3
2x+ 1.
Question 29
Question
Find the function y(x) that minimizes the functional
J[y] = Z1
0
(y′2−2y)dx
subject to the boundary conditions y(0) = 0 and y(1) = 1.
Solution
Step 1: Write down the Euler-Lagrange equation for the given functional. The
Euler-Lagrange equation is given by
d
dx ∂f
∂y′−∂f
∂y = 0
where f=y′2−2y.
Step 2: Compute the partial derivatives needed for the Euler-Lagrange
equation:
∂f
∂y′=∂
∂y′(y′2−2y)=2y′
d
dx ∂f
∂y′=d
dx (2y′)=2y′′
∂f
∂y =∂
∂y (y′2−2y) = −2
Step 3: Plug the derivatives back into the Euler-Lagrange equation:
2y′′ + 2 = 0
30
Step 4: Solve the differential equation obtained in Step 3:
y′′ =−1
Integrating once gives
y′=−x+C1
Integrating again gives
y=−1
2x2+C1x+C2
Step 5: Apply the boundary conditions y(0) = 0 and y(1) = 1: From
y(0) = 0, we have
0 = 0 + 0 + C2⇒C2= 0
From y(1) = 1, we have
1 = −1
2(1)2+C1(1) ⇒C1=3
2
Step 6: The function that minimizes the functional is therefore
y(x) = −1
2x2+3
2x
Question 30
Question
Let J[y] = R1
0(y′2−y2)dx where y(0) = 0 and y(1) = 1. Determine the function
y(x) which minimizes J[y].
Solution
Step 1: Define the Euler-Lagrange equation. The Euler-Lagrange equation for
a functional J[y] = Rb
aL(x, y, y′)dx is given by:
d
dx ∂L
∂y′−∂L
∂y = 0
In this case, L(x, y, y′) = y′2−y2.
Step 2: Compute the partial derivatives of L.
∂L
∂y′= 2y′and ∂L
∂y =−2y
Step 3: Substitute the partial derivatives into the Euler-Lagrange equation.
d
dx (2y′)+2y= 0
31
Step 4: Simplify the above differential equation.
2y′′ + 2y= 0
Step 5: Solve the differential equation. The general solution of 2y′′ + 2y= 0
is y(x) = c1cos(x) + c2sin(x), where c1and c2are constants.
Step 6: Use the boundary conditions y(0) = 0 and y(1) = 1 to find c1and
c2. From y(0) = 0, we have c1= 0. From y(1) = 1, we have c2sin(1) = 1 ⇒
c2=1
sin(1) .
Step 7: Substitute c1= 0 and c2=1
sin(1) back into the general solution.
Therefore, the function that minimizes J[y] is y(x) = sin(x)
sin(1) .
Question 31
Question
Let J[y] = R1
0(y′2+y2)dx where y(0) = 0, y(1) = 1. Find the extremal of J[y].
Solution
Step 1: Define the Lagrangian function Las L(x, y, y′) = y′2+y2.
Step 2: Apply the Euler-Lagrange equation d
dx ∂L
∂y′−∂L
∂y = 0.
Step 3: Compute ∂L
∂y′=∂
∂y′(y′2+y2)=2y′.
Step 4: Compute d
dx ∂L
∂y′=d
dx (2y′)=2y′′.
Step 5: Compute ∂L
∂y =∂
∂y (y′2+y2) = 2y.
Step 6: Form the Euler-Lagrange equation: 2y′′ −2y= 0.
Step 7: Solve the differential equation y′′ −y= 0.
Step 8: The general solution to the differential equation is y(x) = C1ex+
C2e−x, where C1and C2are constants.
Step 9: Apply the boundary conditions y(0) = 0 and y(1) = 1 to find the
values of C1and C2.
Step 10: From y(0) = 0, we have C1+C2= 0 =⇒C1=−C2.
Step 11: Substitute C1=−C2into y(1) = 1 to solve for C1and C2.
Step 12: After solving, we find C1=e
e2−1and C2=−1
e2−1.
Step 13: Therefore, the extremal of J[y] is y(x) = e
e2−1ex−1
e2−1e−x.
32
Question 32
Question
Find the extremals for the functional
J[y] = Z1
0
(y′2+y2)dx
subject to the boundary conditions y(0) = 0 and y(1) = 1.
Solution
Step 1: Compute the Euler-Lagrange equation. Let F=y′2+y2in the func-
tional J[y] = R1
0F dx. The Euler-Lagrange equation is given by
d
dx ∂F
∂y′−∂F
∂y = 0.
This yields
d
dx (2y′)−2y= 0
2y′′ −2y= 0.
Step 2: Solve the differential equation. The general solution to the differen-
tial equation is
y(x) = c1cos(x) + c2sin(x).
Step 3: Apply the boundary conditions. Given y(0) = 0 and y(1) = 1, we
have the system of equations
0 = c1,
1 = c2sin(1).
Step 4: Solve for the constants. From the first equation, c1= 0. Then, from
the second equation, c2=1
sin(1) .
Step 5: Final answer. The extremal that minimizes the functional J[y]
subject to the boundary conditions y(0) = 0 and y(1) = 1 is
y(x) = 1
sin(1) sin(x).
Question 33
Question
Find the extremal of the functional
J[y] = Z1
0
(y′2−y2)dx
subject to the boundary conditions y(0) = 0 and y(1) = 1.
33
Solution
Step 1: Define the Lagrangian function The Lagrangian function is given by
L(x, y, y′) = y′2−y2
Step 2: Apply the Euler-Lagrange equation The Euler-Lagrange equation is
given by
d
dx ∂L
∂y′−∂L
∂y = 0
Step 3: Compute the partial derivatives
∂L
∂y′= 2y′and ∂L
∂y =−2y
Step 4: Apply the Euler-Lagrange equation
d
dx (2y′)−(−2y) = 0
d
dx (2y′)+2y= 0
Step 5: Solve the differential equation
2y′′ + 2y= 0
y′′ +y= 0
The general solution to this differential equation is
y(x) = c1cos(x) + c2sin(x)
Step 6: Apply the boundary conditions Using the boundary conditions
y(0) = 0 and y(1) = 1, we have
(c1= 0
c1cos(1) + c2sin(1) = 1
Since c1= 0, we find c2=1
sin(1) .
Therefore, the extremal of the functional is y(x) = sin(x)
sin(1) .
Question 34
Question
Find the extremal corresponding to the functional
J[y] = Z1
0y′2−y2dx
subject to the boundary conditions y(0) = 0 and y(1) = 1.
34
Solution
Step 1: Set up the Euler-Lagrange equation for the given functional. The
Euler-Lagrange equation is given by
d
dx ∂f
∂y′−∂f
∂y = 0
where f(y, y′, x) = y′2−y2.
Step 2: Compute the partial derivatives involved in the Euler-Lagrange
equation. We have:
∂f
∂y′=∂
∂y′(y′2−y2)=2y′
∂f
∂y′
dx =d
dx (2y′)=2y′′
∂f
∂y =∂
∂y (y′2−y2) = −2y
Step 3: Plug the partial derivatives into the Euler-Lagrange equation and
simplify:
d
dx (2y′)+2y= 0
2y′′ + 2y= 0
Step 4: Solve the differential equation 2y′′ +2y= 0 subject to the boundary
conditions y(0) = 0 and y(1) = 1.
Step 5: The general solution to the differential equation is y(x) = c1cos(x)+
c2sin(x).
Step 6: Applying the boundary condition y(0) = 0, we find c1= 0.
Step 7: Applying the boundary condition y(1) = 1, we find y(1) = c2sin(1) =
1, so c2=1
sin(1) .
Step 8: Therefore, the extremal corresponding to the functional J[y] is
y(x) = sin(x)
sin(1) .
Question 35
Question
Consider the functional
J[y] = Z1
0
(2y2+ 2(y′)2−4y)dx
subject to the boundary conditions y(0) = 0 and y(1) = 1. Find the function y
that minimizes J[y].
35
Solution
Step 1: Let’s denote the integrand by L(y, y′, x) = 2y2+ 2(y′)2−4y. Our task
is to minimize the functional J[y] = R1
0L(y, y′, x)dx subject to the boundary
conditions y(0) = 0 and y(1) = 1.
Step 2: We will apply the Euler-Lagrange equation, which states that if y
minimizes J[y], then the function ymust satisfy the differential equation
d
dx ∂L
∂y′−∂L
∂y = 0
Step 3: First, we compute ∂L
∂y′= 4y′and ∂L
∂y = 4y−4. Therefore, the
Euler-Lagrange equation becomes
d
dx (4y′)−(4y−4) = 0
Step 4: Simplifying the above equation, we get
4y′′ −4=0
Step 5: Integrating once, we have
4y′−4x+C1= 0
Step 6: Solving the above differential equation, we find
y′(x) = x+C1
Step 7: Integrating again, we get
y(x) = 1
2x2+C1x+C2
Step 8: Applying the boundary conditions y(0) = 0 and y(1) = 1, we find
C1= 0 and C2= 0.5.
Step 9: Therefore, the function that minimizes J[y] is y(x) = 1
2x2+1
2.
36
Question 15
Question
Find the extremal for the functional
J[y] = Z1
0
(y′2−y2)dx
subject to the boundary conditions y(0) = 0 and y(1) = 1.
Solution
Step 1: Define the Lagrangian function. Given the functional
J[y] = Z1
0
(y′2−y2)dx
we define the Lagrangian function as
L(x, y, y′) = y′2−y2
Step 2: Apply the Euler-Lagrange equation. The Euler-Lagrange equa-
tion states that the derivative of the Lagrangian with respect to yminus the
derivative of the derivative of the Lagrangian with respect to y′is equal to 0.
Mathematically, this can be expressed as
d
dx ∂L
∂y′−∂L
∂y = 0
Step 3: Calculate the partial derivatives.
∂L
∂y′=∂
∂y′(y′2−y2)=2y′
∂L
∂y =∂
∂y (y′2−y2) = −2y
Step 4: Apply the Euler-Lagrange equation. Substitute the partial deriva-
tives into the Euler-Lagrange equation:
d
dx (2y′)+2y= 0
Step 5: Solve the differential equation. Solving the differential equation gives
us
y′′ −y= 0
The general solution to this differential equation is
y(x) = c1cosh(x) + c2sinh(x)
16
Applying the boundary conditions y(0) = 0 and y(1) = 1, we find c1= 0 and
c2=1
sinh(1) .
Step 6: Find the extremal. Therefore, the extremal for the functional J[y]
subject to the given boundary conditions is
y(x) = sinh(x)
sinh(1)
Question 16
Question
Find the extremals of the functional
J(y) = Z1
0
(y′2−y2)dx
subject to the boundary conditions y(0) = 0 and y(1) = 1.
Solution
Step 1: Define the Euler-Lagrange equation
d
dx ∂F
∂y′−∂F
∂y = 0
where F=y′2−y2.
Step 2: Calculate the partial derivatives of F
∂F
∂y =−2y
∂F
∂y′= 2y′
Step 3: Differentiate ∂F
∂y′with respect to x
d
dx ∂F
∂y′=d
dx (2y′)=2y′′
Step 4: Apply the Euler-Lagrange equation
2y′′ + 2y= 0
Step 5: The general solution of the ODE is
y(x) = c1sin(x) + c2cos(x)
Step 6: Apply the boundary conditions y(0) = 0 and y(1) = 1
y(0) = c2= 0 ⇒c2= 0
17
y(1) = c1sin(1) = 1 ⇒c1=1
sin(1)
Step 7: Therefore, the extremal function that minimizes the functional J(y)
is
y(x) = 1
sin(1) sin(x)
Question 17
Question
Given the functional
J[y] = Z1
0
(y′2−y2)dx
with the boundary conditions y(0) = 0 and y(1) = 1, find the function y(x) that
minimizes J[y].
Solution
Step 1: Define the Euler-Lagrange equation for the given functional: The Euler-
Lagrange equation is given by:
d
dx ∂f
∂y′−∂f
∂y = 0
where f=y′2−y2.
Step 2: Compute the partial derivatives needed for the Euler-Lagrange equa-
tion: ∂f
∂y′= 2y′,d
dx ∂f
∂y′= 2y′′,and ∂f
∂y =−2y
Step 3: Substitute these derivatives into the Euler-Lagrange equation:
2y′′ + 2y= 0
Step 4: Solve the differential equation for y(x): Solving the differential equa-
tion 2y′′ + 2y= 0, we obtain the general solution y(x) = c1cos(x) + c2sin(x).
Step 5: Apply the boundary conditions: Given y(0) = 0, we have c1= 0.
Given y(1) = 1, we have c2sin(1) = 1, so c2=1
sin(1) .
Step 6: Final solution: Therefore, the function that minimizes the functional
J[y] is y(x) = 1
sin(1) sin(x).
18
Question 18
Question
Find the extremal for the functional
J[y] = Z1
0
(y′2−2y)dx
subject to the boundary conditions y(0) = 2 and y(1) = 3.
Solution
To find the extremal for the given functional J[y], we will apply the Euler-
Lagrange equation:
d
dx ∂f
∂y′−∂f
∂y = 0
where f=y′2−2y.
Step 1: Compute ∂f ∂y′and ∂f
∂y
∂f
∂y′= 2y′
∂f
∂y =−2
Step 2: Apply the Euler-Lagrange equation
d
dx (2y′) + 2 = 0
d
dx (2y′) = −2
2y′′ =−2
y′′ =−1
Step 3: Solve the differential equation y′′ =−1 Integrate twice to find
y(x):
y′=−x+C1
y=−1
2x2+C1x+C2
Step 4: Apply boundary conditions Using y(0) = 2 and y(1) = 3:
2 = C2
3 = −1
2+C1+ 2
C1=5
2
Step 5: Final solution Therefore, the extremal for the functional J[y]
subject to the boundary conditions y(0) = 2 and y(1) = 3 is given by
y(x) = −1
2x2+5
2x+ 2
19
Question 19
Question
Consider the functional J[y] = R1
0(y′2−2y)dx subject to the boundary condi-
tions y(0) = 0 and y(1) = 1. Find the function y(x) that minimizes J[y].
Solution
Step 1: Write down the Euler-Lagrange equation for the given functional J[y]:
The Euler-Lagrange equation is given by
d
dx ∂f
∂y′−∂f
∂y = 0
where f(y, y′, x) = y′2−2yin this case.
Step 2: Calculate the partial derivatives:
∂f
∂y =−2,∂f
∂y′= 2y′
d
dx ∂f
∂y′=d
dx (2y′)=2y′′
Step 3: Plugin the derivatives into the Euler-Lagrange equation:
2y′′ + 2 = 0
Step 4: Solve the differential equation:
y′′ =−1
Step 5: Integrate y′′ =−1 twice to find y(x): Integrating twice gives
y′(x) = −x+C1
Integrating again gives
y(x) = −1
2x2+C1x+C2
Step 6: Apply the boundary conditions y(0) = 0 and y(1) = 1: From
y(0) = 0, we get C2= 0. Substituting y(1) = 1, we get
−1
2+C1= 1 =⇒C1=3
2
Step 7: The function that minimizes J[y] with the given boundary condi-
tions is:
y(x) = −1
2x2+3
2x
20
Question 20
Question
Find the function y(x) that minimizes the functional
J[y] = Z1
0
[(y′)2−y]dx
subject to the boundary conditions y(0) = 0 and y(1) = 1.
Solution
Step 1: Compute the Euler-Lagrange equation by differentiating fwith respect
to yand y′and then applying the Euler-Lagrange equation.
Given functional J[y] = R1
0[(y′)2−y]dx, the integrand is f(y, y′, x)=(y′)2−
y. The partial derivative of fwith respect to yis ∂f
∂y =−1 and the partial
derivative with respect to y′is ∂f
∂y′= 2y′.
Applying the Euler-Lagrange equation, we have:
d
dx ∂f
∂y′=∂f
∂y
d
dx (2y′) = −1
2y′′ =−1
y′′ =−1
2
Step 2: Solve the differential equation y′′ =−1
2with the boundary conditions
y(0) = 0 and y(1) = 1.
Integrating y′′ =−1
2twice gives:
y′=−x
2+C1
y=−x2
4+C1x+C2
Applying the boundary conditions:
y(0) = 0 =⇒C2= 0
y(1) = 1 =⇒ − 1
4+C1= 1 =⇒C1=5
4
Thus, the function that minimizes the functional is:
y(x) = −x2
4+5
4x
21
Question 21
Question
Find the extremals of the functional
J[y] = Z1
0
(y′2−y2)dx
subject to the boundary conditions y(0) = 0 and y(1) = 1.
Solution
Step 1: Let’s define the integrand Las L(x, y, y′) = y′2−y2and the Euler-
Lagrange equation is given by
d
dx ∂L
∂y′−∂L
∂y = 0
Step 2: Compute the partial derivatives of Lwith respect to yand y′.
∂L
∂y =−2yand ∂L
∂y′= 2y′
Step 3: Differentiate ∂L
∂y′with respect to x.
d
dx ∂L
∂y′=d
dx (2y′)=2y′′
Step 4: Substituting into the Euler-Lagrange equation, we get
2y′′ + 2y= 0
which simplifies to
y′′ +y= 0
Step 5: The general solution to the differential equation y′′ +y= 0 is
y(x) = c1cos x+c2sin x
Step 6: Apply the boundary conditions y(0) = 0 and y(1) = 1.
y(0) = c1= 0
y(1) = c2sin 1 = 1
Thus, the unique extremal for the given functional is y(x) = sin x.
22
Question 22
Question
Find the extremals of the functional
J[y] = Z2
1
y2−y′2dx
subject to the boundary conditions y(1) = 0 and y(2) = 1.
Solution
To find the extremals of the functional J[y], we need to solve the Euler-Lagrange
equation, which is given by
d
dx ∂f
∂y′−∂f
∂y = 0
where f=y2−y′2.
Step 1: Compute ∂f
∂y and ∂f
∂y′.
∂f
∂y = 2yand ∂f
∂y′=−2y′
Step 2: Compute d
dx ∂f
∂y′.
d
dx ∂f
∂y′=d
dx (−2y′) = −2y′′
Step 3: Set up the Euler-Lagrange equation and solve.
−2y′′ −2y= 0
y′′ +y= 0
Step 4: Find the general solution to the differential equation y′′ +y= 0.
The general solution to this differential equation is of the form y(x) = c1cos(x)+
c2sin(x).
Step 5: Apply the boundary conditions y(1) = 0 and y(2) = 1. Plugging in
x= 1 and y(1) = 0, we get
0 = c1cos(1) + c2sin(1)
Plugging in x= 2 and y(2) = 1, we get
1 = c1cos(2) + c2sin(2)
Step 6: Solve the system of equations for c1and c2. This system of equations
can be solved to find the values of c1and c2.
Step 7: Write down the extremals. The extremals of the given functional
are the solutions to the differential equation found in Step 3 with the appropriate
values of c1and c2.
23
Question 23
Question
Given the functional
J[y] = Z1
0
(y′2−2y)dx
where y(0) = 0 and y(1) = 1, find the function y(x) that minimizes J[y].
Solution
Step 1: Define the Euler-Lagrange equation. The Euler-Lagrange equation for
a functional of the form J[y] = Rb
aF(x, y, y′)dx is given by
d
dx ∂F
∂y′−∂F
∂y = 0
Step 2: Calculate the partial derivatives of F=y′2−2ywith respect to y
and y′.
∂F
∂y =−2 and ∂F
∂y′= 2y′
Step 3: Substitute the partial derivatives into the Euler-Lagrange equation.
d
dx (2y′)−(−2) = 0
Simplify this to get
d
dx (2y′) + 2 = 0
Step 4: Integrate the above equation with respect to x.
2y′+ 2 = C
Where Cis a constant of integration.
Step 5: Solve the differential equation by isolating y′.
2y′=C−2
y′=C
2−1
Step 6: Integrate y′with respect to xto find y(x).
y=Cx
2−x+D
Where Dis another constant of integration.
Step 7: Apply the boundary conditions y(0) = 0 and y(1) = 1. From
y(0) = 0, we have 0 = D. From y(1) = 1, we have 1 = C
2−1. Solving these
equations gives C= 4.
Step 8: Plug in the values of Cand Dto find the minimizing function y(x).
y(x) = 2x−x=x
Therefore, the function that minimizes the given functional is y(x) = x.
24
Question 24
Question
Find the extremal of the functional
J[y] = Z1
0
(y′2+y2)dx
subject to the boundary conditions y(0) = 0 and y(1) = 1.
Solution
Step 1: Define the Lagrangian function L(x, y, y′) = y′2+y2.
Step 2: The Euler-Lagrange equation is given by
d
dx ∂L
∂y′−∂L
∂y = 0
Plugging in the expression for Lgives
d
dx (2y′)−2y= 0
2y′′ −2y= 0
Step 3: Solve the differential equation to find y(x). The solution to the
differential equation y′′ −y= 0 is y(x) = Acos(x) + Bsin(x).
Step 4: Apply the boundary conditions to solve for Aand B. From y(0) = 0,
we have A= 0. From y(1) = 1, we have Bsin(1) = 1, so B=1
sin(1) .
Step 5: Therefore, the extremal of the functional J[y] subject to the given
boundary conditions is
y(x) = sin(x)
sin(1)
Question 25
Question
Minimize the functional
J[y] = Z1
0
(y′2−y2)dx
subject to the boundary conditions y(0) = 0 and y(1) = 1.
25
Solution
Step 1: Compute the Euler-Lagrange equation for the given functional. The
Euler-Lagrange equation is given by
d
dx ∂f
∂y′−∂f
∂y = 0
where f=y′2−y2.
Step 2: Compute the partial derivatives in the Euler-Lagrange equation. We
have ∂f
∂y′=∂
∂y′(y′2−y2)=2y′
∂f
∂y =∂
∂y (y′2−y2) = −2y
Step 3: Apply the Euler-Lagrange equation. Substituting the partial deriva-
tives into the Euler-Lagrange equation gives
d
dx (2y′)+2y= 0
Simplifying further,
2y′′ + 2y= 0
Step 4: Solve the second-order differential equation obtained in Step 3. The
general solution to this differential equation is of the form
y(x) = c1cos(x) + c2sin(x)
Step 5: Use the boundary conditions to find the constants c1and c2. Ap-
plying the boundary conditions y(0) = 0 and y(1) = 1 gives the system of
equations
c1= 0
c1cos(1) + c2sin(1) = 1
Step 6: Solve the system of equations for c1and c2. From the first equation,
c1= 0. Substituting this into the second equation, we get
0 + c2sin(1) = 1
c2=1
sin(1)
Therefore, the unique minimizer of the functional is given by
y(x) = 1
sin(1) sin(x)
26
Question 26
Question
Let J(y) = R2
1(y(x)2+y′(x)2)dx be the functional to be minimized subject to
the boundary conditions y(1) = 1 and y(2) = 2. Find the function y(x) that
minimizes J(y).
Solution
Step 1: We start by writing the Euler-Lagrange equation for the given functional
J(y). The Euler-Lagrange equation is given by:
d
dx ∂f
∂y′−∂f
∂y = 0
where f(y, y′, x) = y2+y′2.
Step 2: Compute the partial derivatives:
∂f
∂y = 2yand ∂f
∂y′= 2y′
Step 3: Compute the derivative of ∂f
∂y′with respect to x:
d
dx ∂f
∂y′=d
dx (2y′)=2y′′
Step 4: Now we can write the Euler-Lagrange equation:
d
dx (2y′)−2y= 0
Step 5: Simplify the equation:
2y′′ −2y= 0
Step 6: Solve the differential equation 2y′′ −2y= 0 by guessing a solution
of the form y(x) = erx:
2r2erx −2erx = 0
Step 7: Simplify the above expression to find the characteristic equation:
r2−1=0
Step 8: Solve the characteristic equation r2−1 = 0 to find the roots r:
r1= 1 and r2=−1
Step 9: The general solution of the differential equation is:
y(x) = c1ex+c2e−x
27
Step 10: Apply the boundary conditions y(1) = 1 and y(2) = 2 to find the
specific solution. We get:
c1e+c2e−1= 1 and c1e2+c2e−2= 2
Step 11: Solve the system of equations to find c1and c2:
c1=e−e−1
e2−e−2and c2=e2−2e+ 1
e2−e−2
Step 12: Therefore, the function that minimizes J(y) is:
y(x) = ex−e−x
e2−e−2+e2−2e+ 1
e2−e−2e−x
Question 27
Question
Find the extremals of the functional
J[y] = Z1
0
(y′2−2y)dx
subject to the boundary conditions y(0) = 0 and y(1) = 1.
Solution
Step 1: Compute the Euler-Lagrange equation. The Euler-Lagrange equation
for the functional J[y] is given by
d
dx ∂f
∂y′−∂f
∂y = 0,
where f=y′2−2y.
Step 2: Compute the partial derivatives. Let’s find the partial derivatives
needed for the Euler-Lagrange equation:
∂f
∂y′=∂
∂y′(y′2−2y) = 2y′,
∂f
∂y =∂
∂y (y′2−2y) = −2.
Step 3: Apply the Euler-Lagrange equation. Substitute the partial deriva-
tives back into the Euler-Lagrange equation:
d
dx (2y′) + 2 = 0.
28
Step 4: Solve the differential equation. Integrating the above equation gives
us:
2y′+C=−2,
where Cis a constant of integration.
Step 5: Apply boundary conditions. Using the boundary conditions y(0) =
0 and y(1) = 1, we find C=−2.
Step 6: Find the extremal function. Thus, we have 2y′−2 = −2, which
simplifies to y′= 0. Integrating once more, we find y(x) = Ax +B.
Step 7: Apply boundary conditions to find Aand B. Using the boundary
conditions y(0) = 0 and y(1) = 1:
0 = A(0) + B⇒B= 0,
1 = A(1) ⇒A= 1.
Step 8: Final Answer. Therefore, the extremal of the functional J[y] subject
to the given boundary conditions is y(x) = x.
Question 28
Question
Find the extremals of the functional
J[y] = Z1
0
(y′2−y)dx
subject to the boundary conditions y(0) = 1 and y(1) = 0.
Solution
To find the extremals of the functional J[y], we need to apply the Euler-Lagrange
equation:
d
dx ∂f
∂y′−∂f
∂y = 0
where f=y′2−y.
Step 1: Calculate ∂f
∂y′and ∂f
∂y .
∂f
∂y′= 2y′and ∂f
∂y =−1
Step 2: Apply the Euler-Lagrange equation to obtain the differential equa-
tion satisfied by y(x):
d
dx (2y′) + 1 = 0
⇒2y′′ + 1 = 0
29
Step 3: Solve the differential equation with the boundary conditions y(0) =
1 and y(1) = 0. The general solution to the differential equation is y(x) =
−1
2x2+cx +d, where cand dare constants to be determined.
Applying the boundary conditions:
y(0) = d= 1
y(1) = −1
2+c+ 1 = 0
c=3
2
Therefore, the extremal function that minimizes the functional is y(x) =
−1
2x2+3
2x+ 1.
Question 29
Question
Find the function y(x) that minimizes the functional
J[y] = Z1
0
(y′2−2y)dx
subject to the boundary conditions y(0) = 0 and y(1) = 1.
Solution
Step 1: Write down the Euler-Lagrange equation for the given functional. The
Euler-Lagrange equation is given by
d
dx ∂f
∂y′−∂f
∂y = 0
where f=y′2−2y.
Step 2: Compute the partial derivatives needed for the Euler-Lagrange
equation:
∂f
∂y′=∂
∂y′(y′2−2y)=2y′
d
dx ∂f
∂y′=d
dx (2y′)=2y′′
∂f
∂y =∂
∂y (y′2−2y) = −2
Step 3: Plug the derivatives back into the Euler-Lagrange equation:
2y′′ + 2 = 0
30
Step 4: Solve the differential equation obtained in Step 3:
y′′ =−1
Integrating once gives
y′=−x+C1
Integrating again gives
y=−1
2x2+C1x+C2
Step 5: Apply the boundary conditions y(0) = 0 and y(1) = 1: From
y(0) = 0, we have
0 = 0 + 0 + C2⇒C2= 0
From y(1) = 1, we have
1 = −1
2(1)2+C1(1) ⇒C1=3
2
Step 6: The function that minimizes the functional is therefore
y(x) = −1
2x2+3
2x
Question 30
Question
Let J[y] = R1
0(y′2−y2)dx where y(0) = 0 and y(1) = 1. Determine the function
y(x) which minimizes J[y].
Solution
Step 1: Define the Euler-Lagrange equation. The Euler-Lagrange equation for
a functional J[y] = Rb
aL(x, y, y′)dx is given by:
d
dx ∂L
∂y′−∂L
∂y = 0
In this case, L(x, y, y′) = y′2−y2.
Step 2: Compute the partial derivatives of L.
∂L
∂y′= 2y′and ∂L
∂y =−2y
Step 3: Substitute the partial derivatives into the Euler-Lagrange equation.
d
dx (2y′)+2y= 0
31
Step 4: Simplify the above differential equation.
2y′′ + 2y= 0
Step 5: Solve the differential equation. The general solution of 2y′′ + 2y= 0
is y(x) = c1cos(x) + c2sin(x), where c1and c2are constants.
Step 6: Use the boundary conditions y(0) = 0 and y(1) = 1 to find c1and
c2. From y(0) = 0, we have c1= 0. From y(1) = 1, we have c2sin(1) = 1 ⇒
c2=1
sin(1) .
Step 7: Substitute c1= 0 and c2=1
sin(1) back into the general solution.
Therefore, the function that minimizes J[y] is y(x) = sin(x)
sin(1) .
Question 31
Question
Let J[y] = R1
0(y′2+y2)dx where y(0) = 0, y(1) = 1. Find the extremal of J[y].
Solution
Step 1: Define the Lagrangian function Las L(x, y, y′) = y′2+y2.
Step 2: Apply the Euler-Lagrange equation d
dx ∂L
∂y′−∂L
∂y = 0.
Step 3: Compute ∂L
∂y′=∂
∂y′(y′2+y2)=2y′.
Step 4: Compute d
dx ∂L
∂y′=d
dx (2y′)=2y′′.
Step 5: Compute ∂L
∂y =∂
∂y (y′2+y2) = 2y.
Step 6: Form the Euler-Lagrange equation: 2y′′ −2y= 0.
Step 7: Solve the differential equation y′′ −y= 0.
Step 8: The general solution to the differential equation is y(x) = C1ex+
C2e−x, where C1and C2are constants.
Step 9: Apply the boundary conditions y(0) = 0 and y(1) = 1 to find the
values of C1and C2.
Step 10: From y(0) = 0, we have C1+C2= 0 =⇒C1=−C2.
Step 11: Substitute C1=−C2into y(1) = 1 to solve for C1and C2.
Step 12: After solving, we find C1=e
e2−1and C2=−1
e2−1.
Step 13: Therefore, the extremal of J[y] is y(x) = e
e2−1ex−1
e2−1e−x.
32
Question 32
Question
Find the extremals for the functional
J[y] = Z1
0
(y′2+y2)dx
subject to the boundary conditions y(0) = 0 and y(1) = 1.
Solution
Step 1: Compute the Euler-Lagrange equation. Let F=y′2+y2in the func-
tional J[y] = R1
0F dx. The Euler-Lagrange equation is given by
d
dx ∂F
∂y′−∂F
∂y = 0.
This yields
d
dx (2y′)−2y= 0
2y′′ −2y= 0.
Step 2: Solve the differential equation. The general solution to the differen-
tial equation is
y(x) = c1cos(x) + c2sin(x).
Step 3: Apply the boundary conditions. Given y(0) = 0 and y(1) = 1, we
have the system of equations
0 = c1,
1 = c2sin(1).
Step 4: Solve for the constants. From the first equation, c1= 0. Then, from
the second equation, c2=1
sin(1) .
Step 5: Final answer. The extremal that minimizes the functional J[y]
subject to the boundary conditions y(0) = 0 and y(1) = 1 is
y(x) = 1
sin(1) sin(x).
Question 33
Question
Find the extremal of the functional
J[y] = Z1
0
(y′2−y2)dx
subject to the boundary conditions y(0) = 0 and y(1) = 1.
33
Solution
Step 1: Define the Lagrangian function The Lagrangian function is given by
L(x, y, y′) = y′2−y2
Step 2: Apply the Euler-Lagrange equation The Euler-Lagrange equation is
given by
d
dx ∂L
∂y′−∂L
∂y = 0
Step 3: Compute the partial derivatives
∂L
∂y′= 2y′and ∂L
∂y =−2y
Step 4: Apply the Euler-Lagrange equation
d
dx (2y′)−(−2y) = 0
d
dx (2y′)+2y= 0
Step 5: Solve the differential equation
2y′′ + 2y= 0
y′′ +y= 0
The general solution to this differential equation is
y(x) = c1cos(x) + c2sin(x)
Step 6: Apply the boundary conditions Using the boundary conditions
y(0) = 0 and y(1) = 1, we have
(c1= 0
c1cos(1) + c2sin(1) = 1
Since c1= 0, we find c2=1
sin(1) .
Therefore, the extremal of the functional is y(x) = sin(x)
sin(1) .
Question 34
Question
Find the extremal corresponding to the functional
J[y] = Z1
0y′2−y2dx
subject to the boundary conditions y(0) = 0 and y(1) = 1.
34
Solution
Step 1: Set up the Euler-Lagrange equation for the given functional. The
Euler-Lagrange equation is given by
d
dx ∂f
∂y′−∂f
∂y = 0
where f(y, y′, x) = y′2−y2.
Step 2: Compute the partial derivatives involved in the Euler-Lagrange
equation. We have:
∂f
∂y′=∂
∂y′(y′2−y2)=2y′
∂f
∂y′
dx =d
dx (2y′)=2y′′
∂f
∂y =∂
∂y (y′2−y2) = −2y
Step 3: Plug the partial derivatives into the Euler-Lagrange equation and
simplify:
d
dx (2y′)+2y= 0
2y′′ + 2y= 0
Step 4: Solve the differential equation 2y′′ +2y= 0 subject to the boundary
conditions y(0) = 0 and y(1) = 1.
Step 5: The general solution to the differential equation is y(x) = c1cos(x)+
c2sin(x).
Step 6: Applying the boundary condition y(0) = 0, we find c1= 0.
Step 7: Applying the boundary condition y(1) = 1, we find y(1) = c2sin(1) =
1, so c2=1
sin(1) .
Step 8: Therefore, the extremal corresponding to the functional J[y] is
y(x) = sin(x)
sin(1) .
Question 35
Question
Consider the functional
J[y] = Z1
0
(2y2+ 2(y′)2−4y)dx
subject to the boundary conditions y(0) = 0 and y(1) = 1. Find the function y
that minimizes J[y].
35
Solution
Step 1: Let’s denote the integrand by L(y, y′, x) = 2y2+ 2(y′)2−4y. Our task
is to minimize the functional J[y] = R1
0L(y, y′, x)dx subject to the boundary
conditions y(0) = 0 and y(1) = 1.
Step 2: We will apply the Euler-Lagrange equation, which states that if y
minimizes J[y], then the function ymust satisfy the differential equation
d
dx ∂L
∂y′−∂L
∂y = 0
Step 3: First, we compute ∂L
∂y′= 4y′and ∂L
∂y = 4y−4. Therefore, the
Euler-Lagrange equation becomes
d
dx (4y′)−(4y−4) = 0
Step 4: Simplifying the above equation, we get
4y′′ −4=0
Step 5: Integrating once, we have
4y′−4x+C1= 0
Step 6: Solving the above differential equation, we find
y′(x) = x+C1
Step 7: Integrating again, we get
y(x) = 1
2x2+C1x+C2
Step 8: Applying the boundary conditions y(0) = 0 and y(1) = 1, we find
C1= 0 and C2= 0.5.
Step 9: Therefore, the function that minimizes J[y] is y(x) = 1
2x2+1
2.
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