MATH 332 - ADVANCED CALCULUS
- Calculus of variations
Question Bank - Set 1
Liberty University
Question 1
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: Calculate 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, y′, x) = y′2−y2.
Step 2: Compute the partial derivatives of fwith respect to y′and y.
∂f
∂y′=∂
∂y′(y′2−y2)=2y′
∂f
∂y =∂
∂y (y′2−y2) = −2y
Step 3: Plug the derivatives into the Euler-Lagrange equation and solve.
d
dx (2y′)−(−2y)=0
2y′′ + 2y= 0
Step 4: Solve the resulting second-order differential equation. The general
solution to the above differential equation is
y(x) = c1cos(x) + c2sin(x)
Step 5: Apply the boundary conditions y(0) = 0 and y(1) = 1.
y(0) = c1cos(0) + c2sin(0) = c1= 0
y(1) = c2sin(1) = 1 =⇒c2=1
sin(1)
Therefore, the extremal for the functional J[y] subject to the given boundary
conditions is
y(x) = sin(x)
sin(1)
Question 2
Question
Find the extremal for the functional
J[y] = Z1
0
(y′(x))2+y(x)2dx
subject to the boundary conditions y(0) = 1 and y(1) = 2.
Solution
Step 1: Define the Lagrangian function as L(x, y, y′)=(y′(x))2+y(x)2.
Step 2: Apply the Euler-Lagrange equation, which states that
d
dx ∂L
∂y′−∂L
∂y = 0
to find the extremal.
Step 3: Compute ∂L
∂y′=∂
∂y′((y′(x))2+y(x)2)=2y′(x) and ∂L
∂y =∂
∂y ((y′(x))2+
y(x)2)=2y(x).
Step 4: Differentiate ∂L
∂y′with respect to x:
d
dx ∂L
∂y′=d
dx (2y′(x)) = 2y′′(x)
Step 5: Plug ∂L
∂y′and ∂L
∂y into the Euler-Lagrange equation:
d
dx ∂L
∂y′−∂L
∂y = 2y′′(x)−2y(x) = 0
2
Step 6: Solve the differential equation y′′(x)−y(x) = 0 with the boundary
conditions y(0) = 1 and y(1) = 2.
Step 7: The general solution to the differential equation is y(x) = c1ex+
c2e−x.
Step 8: Apply the boundary conditions to find c1and c2:
y(0) = c1+c2= 1
y(1) = c1e+c2e−1= 2
Step 9: Solve the system of equations to find c1and c2.
Step 10: The extremal for the functional is the function y(x) that satisfies
the Euler-Lagrange equation and the boundary conditions.
Question 3
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: Define the Euler-Lagrange equation.
The Euler-Lagrange equation for the 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′2−2y, so we have
d
dx ∂
∂y′(y′2−2y)−∂
∂y (y′2−2y) = 0.
Step 2: Compute the partial derivatives.
Let’s first compute ∂
∂y′(y′2−2y) and ∂
∂y (y′2−2y).
∂
∂y′(y′2−2y) = 2y′,
∂
∂y (y′2−2y) = −2.
Step 3: Apply the Euler-Lagrange equation and solve for y(x).
Substitute the partial derivatives into the Euler-Lagrange equation:
d
dx (2y′) + 2 = 0.
3
Solving this differential equation gives us y′′ = 0, which implies y′=c1and
y=c1x+c2.
Applying the boundary conditions y(0) = 0 and y(1) = 1, we find c2= 0
and c1= 1.
Therefore, the function that minimizes the functional is y(x) = x.
Question 4
Question
Find the extremals for the functional J[y] = R1
0(y′2−y)dx subject to the bound-
ary conditions y(0) = 0 and y(1) = 1.
Solution
Step 1: Consider the Euler-Lagrange equation:
d
dx ∂f
∂y′−∂f
∂y = 0
where f(y, y′, x) = y′2−y.
Step 2: Compute the partial derivatives of f:
∂f
∂y′= 2y′and ∂f
∂y =−1
Step 3: Substitute these derivatives into the Euler-Lagrange equation to
obtain: d
dx (2y′) + 1 = 0
Simplify this equation to get:
2y′′ = 0
Step 4: Solve the differential equation 2y′′ = 0 to find the general solution
for y(x):
y(x) = Ax +B
Step 5: Apply the boundary conditions y(0) = 0 and y(1) = 1:
y(0) = A·0 + B= 0 =⇒B= 0
y(1) = A·1 = 1 =⇒A= 1
Step 6: Therefore, the extremal that minimizes the functional J[y] is y(x) =
x.
4
Question 5
Question
Find the extremal for the functional
J[y] = Z2
1
(y′2−y)dx
subject to the boundary conditions y(1) = 0 and y(2) = 1.
Solution
Step 1: Set up the Euler-Lagrange equation:
d
dx ∂f
∂y′−∂f
∂y = 0
where f=y′2−y.
Step 2: Compute the partial derivatives:
∂f
∂y′= 2y′and ∂f
∂y =−1
Step 3: Compute the derivative with respect to xof ∂f
∂y′:
d
dx (2y′)=2y′′
Step 4: Set up the Euler-Lagrange equation:
2y′′ + 1 = 0
Step 5: Solve the differential equation 2y′′ + 1 = 0 subject to the boundary
conditions y(1) = 0 and y(2) = 1.
Step 6: Integrating once gives:
y′=−1
2x+c1
Step 7: Integrating again gives:
y=−1
4x2+c1x+c2
Step 8: Apply the boundary conditions y(1) = 0 and y(2) = 1 to solve for
c1and c2.
Step 9: Substituting x= 1 and y= 0 gives:
0 = −1
4+c1+c2
5
Step 10: Substituting x= 2 and y= 1 gives:
1 = −1+2c1+c2
Step 11: Solve the system of equations to find c1and c2.
Step 12: Plug c1and c2back into the equation for yto find the extremal of
the functional J[y].
Question 6
Question
Find the extremals of the functional
J[y] = Z1
0
(2y′2−y2)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] = Rb
aF(x, y, y′)dx is given by
d
dx ∂F
∂y′−∂F
∂y = 0
In this case, F(x, y, y′)=2y′2−y2. Therefore, the Euler-Lagrange equation
becomes d
dx ∂(2y′2−y2)
∂y′−∂(2y′2−y2)
∂y = 0
d
dx (4y′)−(−2y)=0
4y′′ + 2y= 0
Step 2: Solve the differential equation: The solution to the differential equa-
tion y′′ +1
2y= 0 is given by
y(x) = c1cos √2
2x!+c2sin √2
2x!
Step 3: Apply the boundary conditions: Using the boundary conditions
y(0) = 0 and y(1) = 1, we can solve for the constants c1and c2.
y(0) = c1= 0
6
y(1) = 0 + c2sin √2
2!= 1
c2=1
sin √2
2
Therefore, the extremal of the functional is given by
y(x) = 1
sin √2
2sin √2
2x!
Question 7
Question
Find the function y(x) that minimizes the functional
J[y] = Z1
0
(y2+y′2)dx
subject to the boundary conditions y(0) = 0 and y(1) = 1.
Solution
Step 1: Compute the Euler-Lagrange equation for this functional. The Euler-
Lagrange equation is given by
d
dx ∂f
∂y′−∂f
∂y = 0,
where f=y2+y′2.
Step 2: Compute the partial derivatives involved in the Euler-Lagrange equa-
tion: ∂f
∂y = 2y,
∂f
∂y′= 2y′.
Therefore,
d
dx ∂f
∂y′=d
dx (2y′)=2y′′,
Step 3: Substitute the partial derivatives back into the Euler-Lagrange equa-
tion:
2y′′ −2y= 0
y′′ −y= 0.
7
Step 4: Solve the differential equation y′′ −y= 0 with the boundary condi-
tions y(0) = 0 and y(1) = 1. The general solution to the differential equation
is
y(x) = c1cos x+c2sin x.
Step 5: Apply the boundary conditions to find c1and c2: Using y(0) = 0:
c1cos 0 + c2sin 0 = 0
c1= 0.
Using y(1) = 1:
c1cos 1 + c2sin 1 = 1.
Step 6: Since c1= 0, we have
c2sin 1 = 1
c2=1
sin 1.
Therefore, the function y(x) that minimizes the functional J[y] is given by
y(x) = sin x
sin 1 .
Question 8
Question
Find the extremal for the functional
J[y] = Z1
0
(2y−y′)2dx
subject to the boundary conditions y(0) = 0 and y(1) = 1.
Solution
Step 1: Compute the functional derivative of J[y] with respect to y. The func-
tional derivative is defined as
δJ[y]
δy =d
dx ∂f
∂y′−∂f
∂y
where f= (2y−y′)2. Compute the partial derivatives as follows:
∂f
∂y′= 2(2y−y′)
∂f
∂y = 0
8
Step 2: Calculate the derivative of ∂f
∂y′with respect to x:
d
dx ∂f
∂y′=d
dx (2(2y−y′)) = 2(2y′−y′′)
Step 3: Substitute the partial derivatives back into the expression for the
functional derivative: δJ[y]
δy = 2(2y′−y′′)
Step 4: Set the functional derivative equal to 0 to find the extremal:
2(2y′−y′′) = 0
Step 5: Solve the differential equation 2y′−y′′ = 0. The general solution is
y(x) = c1e−x+c2e2x
Step 6: Apply the boundary conditions y(0) = 0 and y(1) = 1 to determine
the values of c1and c2: From y(0) = 0, we have c1+c2= 0. From y(1) = 1, we
get c1e−1+c2e2= 1.
Step 7: Solve the system of equations c1+c2= 0 and c1e−1+c2e2= 1 to
find the values of c1and c2.
Step 8: Substitute the values of c1and c2back into the general solution y(x)
to get the extremal for the functional J[y].
Question 9
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. Let F=y′2−2y. The Euler-
Lagrange equation for this functional is given by:
d
dx ∂F
∂y′−∂F
∂y = 0
Differentiating Fwith respect to y′gives:
∂F
∂y′= 2y′
9
Differentiating again with respect to xgives:
d
dx ∂F
∂y′=d
dx (2y′)=2y′′
Differentiating Fwith respect to ygives:
∂F
∂y =−2
Therefore, the Euler-Lagrange equation becomes:
2y′′ + 2 = 0
Step 2: Solve the differential equation. Solving the differential equation
2y′′ + 2 = 0, we get:
y′′ =−1
Integrating once gives:
y′=−x+C1
Integrating again gives:
y=−1
2x2+C1x+C2
Step 3: Apply boundary conditions. Using the boundary conditions y(0) = 0
and y(1) = 1, we have:
0 = 0 + 0 + C2
1 = −1
2+C1+C2
Solving these equations simultaneously, we find C1=3
2and C2= 0.
Therefore, the extremal that minimizes the functional J[y] is given by:
y=−1
2x2+3
2x
Question 10
Question
Find the extremals of the functional
J[y] = Z1
0
(y′2−y2)dx
subject to the boundary conditions y(0) = 1 and y(1) = e.
10
Solution
Step 1: Define the Lagrangian The Lagrangian is defined as
L(x, y, y′) = y′2−y2
Step 2: Set up 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
Step 5: Solve the differential equation
2y′′ + 2y= 0
This is a second-order linear homogeneous differential equation with character-
istic polynomial r2+ 1 = 0. The solutions are y(x) = Asin(x) + Bcos(x).
Step 6: Apply the boundary conditions Using the boundary conditions
y(0) = 1 and y(1) = e, we have:
(B= 1
Asin(1) + Bcos(1) = e
Substitute B= 1 into the second equation to find A.
Question 11
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) = 2.
11
Solution
Step 1: Define the Lagrangian The Lagrangian is defined as
L(x, y, y′) = y′2−y2
Step 2: Set up the Euler-Lagrange equation The Euler-Lagrange equation
for the functional J[y] is given by
d
dx ∂L
∂y′−∂L
∂y = 0
Step 3: Compute ∂L
∂y′and ∂L
∂y We have
∂L
∂y′=∂
∂y′(y′2−y2)=2y′
and ∂L
∂y =∂
∂y (y′2−y2) = −2y
Step 4: Apply the Euler-Lagrange equation Substitute ∂L
∂y′and ∂L
∂y into the
Euler-Lagrange equation to get
d
dx (2y′)+2y= 0
Step 5: Solve the differential equation This differential equation can be re-
arranged to
y′′ −y= 0
which has the general solution
y(x) = c1ex+c2e−x
where c1and c2are constants to be determined.
Step 6: Apply the boundary conditions Using the boundary conditions
y(0) = 0 and y(1) = 2, we get the equations:
c1+c2= 0
c1e+c2e−1= 2
Solving these equations gives c1=eand c2=−e.
Therefore, the extremal for the given functional subject to the boundary
conditions is
y(x) = ex−e−x
12
Question 12
Question
Consider the functional
J[y] = Z1
0
(y′)2+y2dx
subject to the boundary conditions 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 minimizing the functional
J[y]:
d
dx ∂f
∂y′−∂f
∂y = 0
where f= (y′)2+y2.
Step 2: Compute the partial derivatives:
∂f
∂y′= 2y′and ∂f
∂y = 2y
Step 3: Apply the Euler-Lagrange equation:
d
dx (2y′)−2y= 0
Step 4: Simplify the above expression:
2y′′ −2y= 0
Step 5: Solve the differential equation. The general solution is:
y(x) = c1cos(x) + c2sin(x)
Step 6: Apply the boundary conditions y(0) = 0 and y(1) = 1:
c1= 0 and c2sin(1) = 1
c2=1
sin(1)
Step 7: Therefore, the function that minimizes the functional J[y] is:
y(x) = 1
sin(1) sin(x)
13
Question 13
Question
Find the extremals of the functional
J[y] = Z1
0
(y′2−y−y2)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, y′, x) = y′2−y−y2.
Step 2: Compute the partial derivatives. First, we compute the derivative
with respect to y′:
∂f
∂y′=∂
∂y′(y′2−y−y2)=2y′.
Next, we compute the derivative with respect to y:
∂f
∂y =∂
∂y (y′2−y−y2) = −1−2y.
Step 3: Apply the Euler-Lagrange equation. Substitute the partial deriva-
tives into the Euler-Lagrange equation to obtain
d
dx (2y′)−(−1−2y)=0.
Step 4: Simplify the equation. Simplify the above equation to get
2y′′ + 2y+ 1 = 0.
Step 5: Solve the differential equation. The general solution to the differen-
tial equation 2y′′ + 2y+ 1 = 0 is
y(x) = c1cos x+c2sin x−1
2,
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 get the system of equations:
(c1−1
2= 0
c1cos 1 + c2sin 1 −1
2= 1
14
Solving this system, we find c1=1
2and c2=2 sin 1
cos 1 .
Thus, the extremal of the functional is
y(x) = 1
2cos x+2 sin 1
cos 1 sin x−1
2.
Question 14
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 Euler-Lagrange equation. The Euler-Lagrange equation for
the given functional J[y] is given by
d
dx ∂f
∂y′−∂f
∂y = 0
where f=y′2−y2.
Step 2: Calculate the partial derivatives. Calculating the partial deriva-
tives, we have:
∂f
∂y =−2y
∂f
∂y′= 2y′
d
dx ∂f
∂y′=d
dx (2y′)=2y′′
Step 3: Apply the Euler-Lagrange equation. Substitute the partial deriva-
tives into the Euler-Lagrange equation to get:
d
dx (2y′)+2y= 0
Step 4: Solve the differential equation. The differential equation simplifies
to 2y′′ + 2y= 0, which simplifies further to y′′ +y= 0. The general solution to
this homogeneous second-order differential equation is
y(x) = c1cos(x) + c2sin(x)
Step 5: Apply the boundary conditions. Using the boundary conditions
y(0) = 0 and y(1) = 1, we find:
c1= 0
15
c2sin(1) = 1
Thus, c2=1
sin(1) .
Step 6: Final answer. Therefore, the extremal of the functional J[y] subject
to the given boundary conditions is
y(x) = 1
sin(1) sin(x)
Question 15
Question
Let I[y] = R1
0(y′2−y2)dx where y(0) = 0, y(1) = 1, and y′(0) = 2. Determine
the function y(x) that minimizes I[y] using the calculus of variations.
Solution
Step 1: Set up the Euler-Lagrange equation. Given I[y] = R1
0(y′2−y2)dx, the
corresponding Lagrange density is L(y, y′, x) = y′2−y2. The Euler-Lagrange
equation is then given by:
d
dx ∂L
∂y′−∂L
∂y = 0
Step 2: Compute the partial derivatives. We have:
∂L
∂y′= 2y′and ∂L
∂y =−2y
Step 3: Apply the Euler-Lagrange equation. Substituting the partial deriva-
tives into the Euler-Lagrange equation and simplifying, we obtain:
d
dx (2y′)+2y= 0
Step 4: Solve the differential equation. The equation simplifies to 2y′′ +2y=
0, which has the general solution y(x) = c1cos(x) + c2sin(x).
Step 5: Apply the boundary conditions. Using the boundary conditions
y(0) = 0 and y(1) = 1, we find that c1= 0 and c2= 1.
Step 6: Determine y(x). Thus, the function that minimizes I[y] is y(x) =
sin(x).
Question 16
Question
Find the extremals of the functional
J[y] = Z1
0
(y′2−y2)dx
16
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 the functional J[y] = Rb
aF(x, y, y′)dx is given
by:
d
dx ∂F
∂y′−∂F
∂y = 0
Step 2: Write the Euler-Lagrange equation for the given functional.
For J[y] = R1
0(y′2−y2)dx,F(x, y, y′) = y′2−y2. Thus, the Euler-Lagrange
equation becomes:
d
dx (2y′)+2y= 0
Step 3: Solve the Euler-Lagrange equation.
Differentiating 2y′with respect to x, we get:
2y′′ + 2y= 0
This is a second-order linear homogeneous differential equation. The general
solution of this differential equation is:
y(x) = c1cos(x) + c2sin(x)
Step 4: Apply the boundary conditions.
Given that y(0) = 0 and y(1) = 1, we have:
0 = c1
1 = c2sin(1)
Hence, the extremal of the functional is y(x) = sin(x).
Question 17
Question
Find the function y(x) that minimizes the functional
J[y] = Z1
0
(y′2+y2)dx
subject to the boundary conditions y(0) = 0 and y(1) = 1.
17
Solution
Step 1: Define the Euler-Lagrange equation.
The Euler-Lagrange equation for minimizing the functional J[y] = Rb
aF(x, y, y′)dx
is given by
d
dx ∂F
∂y′−∂F
∂y = 0
Step 2: Apply the Euler-Lagrange equation to the given functional.
For the given functional J[y] = R1
0(y′2+y2)dx, we have
F(x, y, y′) = y′2+y2
Therefore,
∂F
∂y = 2yand ∂F
∂y′= 2y′
Now, the Euler-Lagrange equation becomes
d
dx (2y′)−2y= 0
2y′′ −2y= 0
Step 3: Solve the differential equation above.
The solution to the differential equation 2y′′ −2y= 0 is of the form y(x) =
c1cos(x) + c2sin(x).
Step 4: Apply the boundary conditions.
Given that y(0) = 0 and y(1) = 1, we have c1= 0 (from the first boundary
condition) and c2sin(1) = 1 (from the second boundary condition). Thus,
c2=1
sin(1) .
Therefore, the function that minimizes the given functional is
y(x) = sin(x)
sin(1)
Question 18
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) = 0.
18
Solution
To find the extremals, we need to solve 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+y2in this case.
Step 1: Compute ∂f
∂y′and ∂f
∂y .
∂f
∂y′= 2y′,∂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.
2y′′ −2y= 0
Step 4: Solve the differential equation with the boundary conditions y(0) =
0 and y(1) = 0. The general solution to the differential equation is
y(x) = c1sin(x) + c2cos(x)
Applying the boundary conditions:
y(0) = c2= 0
y(1) = c1sin(1) = 0
Since c1cannot be 0, we must have sin(1) = 0, which gives the constraint
c1=kπ where k∈Z.
Therefore, the extremals of the functional are of the form
y(x) = kπ sin(x)
where k∈Z.
Question 19
Question
Find the extremals of the functional
J[y] = Z1
0
(y′2+ 2y)dx
subject to the boundary conditions y(0) = 1 and y(1) = −1.
19
Solution
Step 1: Define the Lagrangian
The Lagrangian L(x, y, y′) for this functional is given by
L(x, y, y′) = y′2+ 2y
Step 2: Set up the Euler-Lagrange equation
The Euler-Lagrange equation is given by
d
dx ∂L
∂y′−∂L
∂y = 0
Plugging in the Lagrangian L(x, y, y′) = y′2+ 2y, we have
d
dx (2y′)−2=0
Step 3: Solve the Euler-Lagrange equation
Integrating the Euler-Lagrange equation with respect to x, we get
2y′−2 = c
where cis an arbitrary constant.
Step 4: Impose the boundary conditions
Using the boundary conditions y(0) = 1 and y(1) = −1, we find that c= 2.
Step 5: Solve for y(x)
Solving the differential equation 2y′−2 = 2 with initial condition y(0) = 1, we
get
y(x) = −x2+x+ 1
Step 6: Confirm the extremals
Plugging the extremal y(x) = −x2+x+ 1 back into the functional J[y], we
calculate
J[y] = Z1
0
((−2x+ 1)2+ 2(−x2+x+ 1)) dx =−11
6
Therefore, the extremal of the functional is y(x) = −x2+x+ 1.
Question 20
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.
20
Solution
Step 1: Compute the Euler-Lagrange equation. The Euler-Lagrange equation
for the 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. Therefore, we have
d
dx ∂L
∂y′−∂L
∂y =d
dx (2y′)−(−2y)=2y′′ + 2y= 0.
Step 2: Solve the differential equation. The general solution to 2y′′ + 2y= 0
is y(x) = c1cos(x)+c2sin(x). Next, we apply the boundary conditions y(0) = 0
and y(1) = 1: y(0) = c1= 0 ⇒y(x) = c2sin(x). y(1) = c2sin(1) = 1 ⇒
c2=1
sin(1) ⇒y(x) = sin(x)
sin(1) .
Therefore, the extremal for the functional is y(x) = sin(x)
sin(1) .
Question 21
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: Set up the Euler-Lagrange equation for the given functional. The
Euler-Lagrange equation for the functional J[y] = Rb
aF(x, y, y′)dx is given by
d
dx ∂F
∂y′−∂F
∂y = 0
In this case, F(x, y, y′) = y′2−y, so
d
dx (2y′)−(−1) = 0
Thus, we have
2y′′ = 1
Step 2: Solve the differential equation that we obtained in Step 1. Inte-
grating both sides with respect to x, we get
2y′=x+C1
21
Integrating again, we have
y=1
2x2+C1x+C2
Step 3: Apply the boundary conditions to find the constants C1and C2.
Using the boundary conditions y(0) = 0 and y(1) = 1, we get
0 = 0 + 0 + C2⇒C2= 0
1 = 1
2+C1⇒C1=1
2
Step 4: Finalize the solution by substituting the values of C1and C2into
the general solution. Therefore, the function that minimizes the given functional
subject to the given boundary conditions is
y(x) = 1
2x2+1
2x
Question 22
Question
Find the extremals of the functional
J[y] = Z1
0
(y2+y′2)dx
subject to the boundary conditions y(0) = 1 and y(1) = 2.
Solution
Step 1: Write down the Euler-Lagrange equation. The Euler-Lagrange equation
for the given functional is
d
dx ∂f
∂y′−∂f
∂y = 0
where f=y2+y′2.
Step 2: Compute the partial derivatives.
∂f
∂y = 2y
∂f
∂y′= 2y′
Step 3: Find the derivative of ∂f
∂y′with respect to x.
d
dx ∂f
∂y′= 2y′′
22
Step 4: Apply the Euler-Lagrange equation. Substitute the partial deriva-
tives into the Euler-Lagrange equation, we get
d
dx (2y′)−2y= 0
which simplifies to
2y′′ −2y= 0
Step 5: Solve the differential equation. The general solution to the differ-
ential equation is given by y(x) = Acos(x) + Bsin(x), where Aand Bare
constants to be determined.
Step 6: Apply the boundary conditions. Using the boundary conditions
y(0) = 1 and y(1) = 2, we find
y(0) = A= 1
y(1) = Acos(1) + Bsin(1) = 2
Substitute A= 1 into the second equation and solve for Bto find B=2−cos(1)
sin(1) .
Step 7: Write down the extremal solution. The extremal function that
minimizes the given functional is
y(x) = cos(x) + 2−cos(1)
sin(1) sin(x)
Question 23
Question
Let y(x) be a function that minimizes the functional
J[y] = Z2
1
(y′2−2y2)dx
subject to the boundary conditions y(1) = 3 and y(2) = −1. Find the function
y(x) that minimizes J[y].
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, y′, x) = y′2−2y2.
Step 2: Compute the partial derivatives. We have
∂f
∂y′=∂
∂y′(y′2−2y2)=2y′,
23
and ∂f
∂y =∂
∂y (y′2−2y2) = −4y.
Step 3: Substitute into the Euler-Lagrange equation. Plugging the partial
derivatives into the Euler-Lagrange equation, we get
d
dx (2y′)−(−4y)=0
⇒d
dx (2y′)+4y= 0
⇒2y′′ + 4y= 0.
Step 4: Solve the differential equation. The solution to the differential equa-
tion 2y′′ + 4y= 0 is of the form y(x) = Asin(2x) + Bcos(2x), where Aand B
are constants to be determined.
Step 5: Apply the boundary conditions. Using the boundary conditions
y(1) = 3 and y(2) = −1, we find
3 = Asin(2) + Bcos(2)
−1 = Asin(4) + Bcos(4).
Step 6: Solve for Aand B. Solving the system of equations, we find A=
3 cos(4)+sin(4)
sin(2) sin(4)−2 cos(2) cos(4) and B=−3 sin(4)+cos(4)
sin(2) sin(4)−2 cos(2) cos(4) .
Therefore, the function y(x) that minimizes the given functional is y(x) =
3 cos(4)+sin(4)
sin(2) sin(4)−2 cos(2) cos(4) sin(2x) + −3 sin(4)+cos(4)
sin(2) sin(4)−2 cos(2) cos(4) cos(2x).
Question 24
Question
Find the function y(x) that minimizes the functional
J[y] = Z1
0
(y′2−y2)dx
subject to the boundary conditions y(0) = 0 and y(1) = 1.
Solution
To find the function y(x) that minimizes the functional J[y], we will use the
Euler-Lagrange equation. Let L(y, y′, x) = y′2−y2be the integrand of J[y].
Step 1: Compute ∂L
∂y and d
dx ∂L
∂y′.
∂L
∂y =−2yand d
dx ∂L
∂y′=d
dx (2y′) = 2y′′
24
Step 2: Apply the Euler-Lagrange equation d
dx ∂L
∂y′−∂L
∂y = 0.
2y′′ + 2y= 0
Step 3: Solve the second-order differential equation 2y′′ + 2y= 0 with
boundary conditions y(0) = 0 and y(1) = 1. The general solution is y(x) =
c1sin(x) + c2cos(x).
Applying the boundary conditions, we have:
y(0) = 0 =⇒c2= 0
y(1) = 1 =⇒c1sin(1) = 1 =⇒c1=1
sin(1)
Thus, the function y(x) that minimizes the functional J[y] is y(x) = 1
sin(1) sin(x).
Question 25
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 L(x, y, y′) as L(x, y, y′) = y′2−y2.
Step 2: Apply the Euler-Lagrange equation
d
dx ∂L
∂y′−∂L
∂y = 0
to find the extremal.
Step 3: Calculate the partial derivatives:
∂L
∂y =−2yand ∂L
∂y′= 2y′
Step 4: Compute the derivative of ∂L
∂y′with respect to x:
d
dx ∂L
∂y′=d
dx (2y′)=2y′′
Step 5: Substitute the derivatives into the Euler-Lagrange equation:
2y′′ + 2y= 0
25
Step 6: Solve the differential equation subject to the boundary conditions
y(0) = 0 and y(1) = 1.
The general solution to the differential equation is given by
y(x) = c1cos(x) + c2sin(x)
Step 7: Apply the boundary conditions to find the values of c1and c2.
From y(0) = 0, we have c1= 0.
Substitute c1= 0 into the general solution:
y(x) = c2sin(x)
From y(1) = 1, we have c2sin(1) = 1, which implies c2=1
sin(1) .
Therefore, the extremal for the given functional is
y(x) = 1
sin(1) sin(x)
Question 26
Question
Find the function y(x) that extremizes the functional
J[y] = Z1
0y2+y′2dx
subject to the boundary conditions y(0) = 1 and y(1) = 2.
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, y′, x) = y2+y′2.
Step 2: Find the partial derivatives. We have
∂f
∂y = 2y, and ∂f
∂y′= 2y′.
Now, differentiate ∂f
∂y′with respect to x:
d
dx ∂f
∂y′= 2y′′.
26
Therefore, the Euler-Lagrange equation becomes
2y′′ −2y= 0.
Step 3: Solve the differential equation. The general solution to the differen-
tial equation y′′ −y= 0 is given by
y(x) = c1ex+c2e−x,
where c1and c2are constants to be determined.
Step 4: Apply the boundary conditions. Using the boundary conditions
y(0) = 1 and y(1) = 2, we find
y(0) = c1+c2= 1
y(1) = c1e+c2e−1= 2
Step 5: Solve for c1and c2. By solving the system of equations from step 4,
we find c1=2e
e+1 and c2=2
e+1 .
Step 6: Final solution. Therefore, the function y(x) that extremizes the
functional J[y] subject to the given boundary conditions is
y(x) = 2e
e+ 1ex+2
e+ 1e−x.
Question 27
Question
Find the extremal for 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 by finding the derivative of the
integrand with respect to yand the derivative of the integrand with respect to
y′.
d
dx ∂f
∂y′−∂f
∂y = 0
Step 2: Let f(y, y′, x) = y′2−2y. Then, compute the partial derivatives.
∂f
∂y =−2,∂f
∂y′= 2y′
27
Step 3: Compute d
dx ∂f
∂y′.
d
dx (2y′)=2y′′
Step 4: Set up the Euler-Lagrange equation:
2y′′ + 2 = 0
Step 5: Solve the differential equation with the boundary conditions y(0) = 0
and y(1) = 1.
y′′ =−1 =⇒y′=−x+C1, y =−1
2x2+C1x+C2
Step 6: Use the boundary conditions y(0) = 0 and y(1) = 1 to find C1and
C2.
0 = C2,1 = −1
2+C1=⇒C1=3
2
Step 7: Therefore, the extremal for the given functional is y=−1
2x2+3
2x.
Question 28
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: Compute the Euler-Lagrange equation by taking the derivative with
respect to yand y′of the integrand in the functional J[y]:
d
dx
∂f
∂y′−∂f
∂y = 0
d
dx (2y′)−(−2y)=0
d
dx (2y′)+2y= 0
2y′′ + 2y= 0
28
Step 2: Solve the differential equation 2y′′ + 2y= 0 with the boundary con-
ditions y(0) = 0 and y(1) = 1. The general solution to the differential equation
is given by y(x) = Asin(x) + Bcos(x). Applying the boundary conditions:
y(0) = 0 ⇒B= 0
y(1) = 1 ⇒Asin(1) = 1 ⇒A=1
sin(1)
Therefore, the extremal that minimizes the functional J[y] is given by y(x) =
1
sin(1) sin(x).
Question 29
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: Set up the Euler-Lagrange equation: The Euler-Lagrange equation for
the functional J[y] = Rb
aF(x, y, y′)dx is given by
d
dx ∂F
∂y′−∂F
∂y = 0
In this case, we have F(x, y, y′) = y′2−y2, so
d
dx ∂F
∂y′−∂F
∂y =d
dx (2y′)−(−2y) = d
dx (2y′)+2y= 0
Step 2: Solve the Euler-Lagrange equation: Solving the Euler-Lagrange
equation gives us
2y′′ + 2y= 0
which simplifies to
y′′ +y= 0
Step 3: Find the general solution of the differential equation: The general
solution of y′′ +y= 0 is
y(x) = Acos x+Bsin x
where Aand Bare constants to be determined.
29
Step 4: Apply the boundary conditions: From the boundary conditions
y(0) = 0 and y(1) = 1, we have
A= 0
Bsin 1 = 1
which gives B=1
sin 1 .
Therefore, the extremal for the functional is
y(x) = sin x
sin 1
Question 30
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. Let L(y, y′, x) = y′2−ybe the
integrand of the functional J[y].
Step 2: Compute ∂L
∂y −d
dx ∂L
∂y′= 0.
∂L
∂y =−1,∂L
∂y′= 2y′
d
dx ∂L
∂y′=d
dx (2y′)=2y′′
Euler-Lagrange equation: −1−2y′′ = 0
Step 3: Solve the differential equation y′′ =−1
2. Integrate twice to obtain
the general solution: y(x) = −1
4x2+1
2x+c1x+c2
Step 4: Apply the boundary conditions y(0) = 0 and y(1) = 1.
y(0) = c2= 0
y(1) = −1
4+1
2+c1= 1
c1=3
4
Step 5: The extremal is the function y(x) = −1
4x2+1
2x+3
4x.
Therefore, the extremal of the functional J[y] subject to the given boundary
conditions is y(x) = −1
4x2+1
2x+3
4x.
30
Question 31
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: Calculate the Euler-Lagrange equation. Given the functional
J[y] = Z1
0
(y′2−y2)dx
we have the Euler-Lagrange equation
d
dx ∂f
∂y′−∂f
∂y = 0
where f=f(x, y, y′) = y′2−y2.
Step 2: Compute the partial derivatives. We have
∂f
∂y′= 2y′,∂f
∂y =−2y
and d
dx ∂f
∂y′=d
dx (2y′)=2y′′
Step 3: Write down the Euler-Lagrange equation. Substituting into the
Euler-Lagrange equation, we have
2y′′ + 2y= 0
Step 4: Solve the differential equation. The general solution to this differ-
ential equation is
y(x) = c1cos(x) + c2sin(x)
Step 5: Apply the boundary conditions. From y(0) = 0 and y(1) = 1, we
find
c1= 0
and
c2sin(1) = 1
which gives c2=1
sin(1) .
Therefore, the extremal for the given functional subject to the boundary
conditions is
y(x) = sin(x)
sin(1)
31
Question 32
Question
Find the extremals of the functional
J[y] = Z1
0
(y′)2−y2dx
subject to the boundary conditions y(0) = 1 and y(1) = 0.
Solution
Step 1: Let’s denote by L(y, y′, x)=(y′)2−y2the integrand of the functional
J[y].
Step 2: The Euler-Lagrange equation is given by d
dx ∂L
∂y′−∂L
∂y = 0.
Step 3: First, let’s calculate ∂L
∂y′:
∂L
∂y′= 2y′
Step 4: Then, we find d
dx ∂L
∂y′:
d
dx ∂L
∂y′= 2y′′
Step 5: Next, let’s compute ∂L
∂y :
∂L
∂y =−2y
Step 6: Substituting these derivatives into the Euler-Lagrange equation, we
get:
2y′′ + 2y= 0
Step 7: The general solution to this differential equation is of the form
y(x) = c1cos(x) + c2sin(x).
Step 8: Now, we apply the boundary conditions y(0) = 1 and y(1) = 0 to
find the values of c1and c2.
Step 9: From y(0) = 1, we have c1= 1.
Step 10: Substituting c1= 1 into y(1) = 0, we find c2=−1.
Step 11: Therefore, the extremal that minimizes the functional J[y] under
the given boundary conditions is y(x) = cos(x)−sin(x).
32
Question 33
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) = 0.
Solution
Step 1: Define the Euler-Lagrange equation. The Euler-Lagrange equation for
the 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, so the Euler-Lagrange equation becomes
d
dx (2y′)+2y= 0
Step 2: Solve the Euler-Lagrange equation. Since the Euler-Lagrange equa-
tion is a second-order ordinary differential equation, we can rewrite it as
y′′ +y= 0
The general solution of this differential equation is
y(x) = c1sin x+c2cos x
Step 3: Apply the boundary conditions. We have y(0) = 0 and y(1) = 0,
which gives us the following equations:
(0 = c2
0 = c1sin 1
Since c2= 0, the second equation becomes c1sin 1 = 0, which implies c1= 0
since sin 1 = 0.
Step 4: Conclusion. The only extremal for the functional J[y] subject to
the given boundary conditions is the identically zero function, i.e., y(x) = 0.
Question 34
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.
33
Solution
Step 1: Define the Lagrangian for the functional J[y] as
L(x, y, y′) = y′2+y2+λ1(0)y+λ2(1)(y−1)
where λ1and λ2are the Lagrange multipliers.
Step 2: Compute the Euler-Lagrange equation for y:
d
dx ∂L
∂y′−∂L
∂y = 0
This gives
d
dx (2y′)−2y= 0.
Step 3: Solve the differential equation obtained in Step 2:
2y′′ −2y= 0
y′′ −y= 0
The general solution to this differential equation is
y(x) = c1ex+c2e−x
Step 4: Apply the boundary conditions y(0) = 0 and y(1) = 1: From
y(0) = 0, we have c1+c2= 0. From y(1) = 1, we have c1e+c2e−1= 1.
Step 5: Solve the system of equations in Step 4 to find the constants c1and
c2: Solving the system, we find c1=e
e−1and c2=−1
e−1.
Therefore, the extremal for the given functional is
y(x) = e
e−1ex−1
e−1e−x
Question 35
Question
Let J[y] = R1
0(y′2−y2)dx where ysatisfies the boundary conditions y(0) = 0
and y(1) = 1. Find the function ythat minimizes J[y].
Solution
Step 1: Let’s start by setting up the Euler-Lagrange equation for this variational
problem. The Euler-Lagrange equation is given by:
d
dx ∂f
∂y′−∂f
∂y = 0
where f(x, y, y′) = y′2−y2.
34
2y′′ + 2y= 0
Step 4: Solve the resulting second-order differential equation. The general
solution to the above differential equation is
y(x) = c1cos(x) + c2sin(x)
Step 5: Apply the boundary conditions y(0) = 0 and y(1) = 1.
y(0) = c1cos(0) + c2sin(0) = c1= 0
y(1) = c2sin(1) = 1 =⇒c2=1
sin(1)
Therefore, the extremal for the functional J[y] subject to the given boundary
conditions is
y(x) = sin(x)
sin(1)
Question 2
Question
Find the extremal for the functional
J[y] = Z1
0
(y′(x))2+y(x)2dx
subject to the boundary conditions y(0) = 1 and y(1) = 2.
Solution
Step 1: Define the Lagrangian function as L(x, y, y′)=(y′(x))2+y(x)2.
Step 2: Apply the Euler-Lagrange equation, which states that
d
dx ∂L
∂y′−∂L
∂y = 0
to find the extremal.
Step 3: Compute ∂L
∂y′=∂
∂y′((y′(x))2+y(x)2)=2y′(x) and ∂L
∂y =∂
∂y ((y′(x))2+
y(x)2)=2y(x).
Step 4: Differentiate ∂L
∂y′with respect to x:
d
dx ∂L
∂y′=d
dx (2y′(x)) = 2y′′(x)
Step 5: Plug ∂L
∂y′and ∂L
∂y into the Euler-Lagrange equation:
d
dx ∂L
∂y′−∂L
∂y = 2y′′(x)−2y(x) = 0
2
Step 6: Solve the differential equation y′′(x)−y(x) = 0 with the boundary
conditions y(0) = 1 and y(1) = 2.
Step 7: The general solution to the differential equation is y(x) = c1ex+
c2e−x.
Step 8: Apply the boundary conditions to find c1and c2:
y(0) = c1+c2= 1
y(1) = c1e+c2e−1= 2
Step 9: Solve the system of equations to find c1and c2.
Step 10: The extremal for the functional is the function y(x) that satisfies
the Euler-Lagrange equation and the boundary conditions.
Question 3
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: Define the Euler-Lagrange equation.
The Euler-Lagrange equation for the 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′2−2y, so we have
d
dx ∂
∂y′(y′2−2y)−∂
∂y (y′2−2y) = 0.
Step 2: Compute the partial derivatives.
Let’s first compute ∂
∂y′(y′2−2y) and ∂
∂y (y′2−2y).
∂
∂y′(y′2−2y) = 2y′,
∂
∂y (y′2−2y) = −2.
Step 3: Apply the Euler-Lagrange equation and solve for y(x).
Substitute the partial derivatives into the Euler-Lagrange equation:
d
dx (2y′) + 2 = 0.
3
Solving this differential equation gives us y′′ = 0, which implies y′=c1and
y=c1x+c2.
Applying the boundary conditions y(0) = 0 and y(1) = 1, we find c2= 0
and c1= 1.
Therefore, the function that minimizes the functional is y(x) = x.
Question 4
Question
Find the extremals for the functional J[y] = R1
0(y′2−y)dx subject to the bound-
ary conditions y(0) = 0 and y(1) = 1.
Solution
Step 1: Consider the Euler-Lagrange equation:
d
dx ∂f
∂y′−∂f
∂y = 0
where f(y, y′, x) = y′2−y.
Step 2: Compute the partial derivatives of f:
∂f
∂y′= 2y′and ∂f
∂y =−1
Step 3: Substitute these derivatives into the Euler-Lagrange equation to
obtain: d
dx (2y′) + 1 = 0
Simplify this equation to get:
2y′′ = 0
Step 4: Solve the differential equation 2y′′ = 0 to find the general solution
for y(x):
y(x) = Ax +B
Step 5: Apply the boundary conditions y(0) = 0 and y(1) = 1:
y(0) = A·0 + B= 0 =⇒B= 0
y(1) = A·1 = 1 =⇒A= 1
Step 6: Therefore, the extremal that minimizes the functional J[y] is y(x) =
x.
4
Question 5
Question
Find the extremal for the functional
J[y] = Z2
1
(y′2−y)dx
subject to the boundary conditions y(1) = 0 and y(2) = 1.
Solution
Step 1: Set up the Euler-Lagrange equation:
d
dx ∂f
∂y′−∂f
∂y = 0
where f=y′2−y.
Step 2: Compute the partial derivatives:
∂f
∂y′= 2y′and ∂f
∂y =−1
Step 3: Compute the derivative with respect to xof ∂f
∂y′:
d
dx (2y′)=2y′′
Step 4: Set up the Euler-Lagrange equation:
2y′′ + 1 = 0
Step 5: Solve the differential equation 2y′′ + 1 = 0 subject to the boundary
conditions y(1) = 0 and y(2) = 1.
Step 6: Integrating once gives:
y′=−1
2x+c1
Step 7: Integrating again gives:
y=−1
4x2+c1x+c2
Step 8: Apply the boundary conditions y(1) = 0 and y(2) = 1 to solve for
c1and c2.
Step 9: Substituting x= 1 and y= 0 gives:
0 = −1
4+c1+c2
5
Step 10: Substituting x= 2 and y= 1 gives:
1 = −1+2c1+c2
Step 11: Solve the system of equations to find c1and c2.
Step 12: Plug c1and c2back into the equation for yto find the extremal of
the functional J[y].
Question 6
Question
Find the extremals of the functional
J[y] = Z1
0
(2y′2−y2)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] = Rb
aF(x, y, y′)dx is given by
d
dx ∂F
∂y′−∂F
∂y = 0
In this case, F(x, y, y′)=2y′2−y2. Therefore, the Euler-Lagrange equation
becomes d
dx ∂(2y′2−y2)
∂y′−∂(2y′2−y2)
∂y = 0
d
dx (4y′)−(−2y)=0
4y′′ + 2y= 0
Step 2: Solve the differential equation: The solution to the differential equa-
tion y′′ +1
2y= 0 is given by
y(x) = c1cos √2
2x!+c2sin √2
2x!
Step 3: Apply the boundary conditions: Using the boundary conditions
y(0) = 0 and y(1) = 1, we can solve for the constants c1and c2.
y(0) = c1= 0
6
y(1) = 0 + c2sin √2
2!= 1
c2=1
sin √2
2
Therefore, the extremal of the functional is given by
y(x) = 1
sin √2
2sin √2
2x!
Question 7
Question
Find the function y(x) that minimizes the functional
J[y] = Z1
0
(y2+y′2)dx
subject to the boundary conditions y(0) = 0 and y(1) = 1.
Solution
Step 1: Compute the Euler-Lagrange equation for this functional. The Euler-
Lagrange equation is given by
d
dx ∂f
∂y′−∂f
∂y = 0,
where f=y2+y′2.
Step 2: Compute the partial derivatives involved in the Euler-Lagrange equa-
tion: ∂f
∂y = 2y,
∂f
∂y′= 2y′.
Therefore,
d
dx ∂f
∂y′=d
dx (2y′)=2y′′,
Step 3: Substitute the partial derivatives back into the Euler-Lagrange equa-
tion:
2y′′ −2y= 0
y′′ −y= 0.
7
Step 4: Solve the differential equation y′′ −y= 0 with the boundary condi-
tions y(0) = 0 and y(1) = 1. The general solution to the differential equation
is
y(x) = c1cos x+c2sin x.
Step 5: Apply the boundary conditions to find c1and c2: Using y(0) = 0:
c1cos 0 + c2sin 0 = 0
c1= 0.
Using y(1) = 1:
c1cos 1 + c2sin 1 = 1.
Step 6: Since c1= 0, we have
c2sin 1 = 1
c2=1
sin 1.
Therefore, the function y(x) that minimizes the functional J[y] is given by
y(x) = sin x
sin 1 .
Question 8
Question
Find the extremal for the functional
J[y] = Z1
0
(2y−y′)2dx
subject to the boundary conditions y(0) = 0 and y(1) = 1.
Solution
Step 1: Compute the functional derivative of J[y] with respect to y. The func-
tional derivative is defined as
δJ[y]
δy =d
dx ∂f
∂y′−∂f
∂y
where f= (2y−y′)2. Compute the partial derivatives as follows:
∂f
∂y′= 2(2y−y′)
∂f
∂y = 0
8
Step 2: Calculate the derivative of ∂f
∂y′with respect to x:
d
dx ∂f
∂y′=d
dx (2(2y−y′)) = 2(2y′−y′′)
Step 3: Substitute the partial derivatives back into the expression for the
functional derivative: δJ[y]
δy = 2(2y′−y′′)
Step 4: Set the functional derivative equal to 0 to find the extremal:
2(2y′−y′′) = 0
Step 5: Solve the differential equation 2y′−y′′ = 0. The general solution is
y(x) = c1e−x+c2e2x
Step 6: Apply the boundary conditions y(0) = 0 and y(1) = 1 to determine
the values of c1and c2: From y(0) = 0, we have c1+c2= 0. From y(1) = 1, we
get c1e−1+c2e2= 1.
Step 7: Solve the system of equations c1+c2= 0 and c1e−1+c2e2= 1 to
find the values of c1and c2.
Step 8: Substitute the values of c1and c2back into the general solution y(x)
to get the extremal for the functional J[y].
Question 9
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. Let F=y′2−2y. The Euler-
Lagrange equation for this functional is given by:
d
dx ∂F
∂y′−∂F
∂y = 0
Differentiating Fwith respect to y′gives:
∂F
∂y′= 2y′
9
Differentiating again with respect to xgives:
d
dx ∂F
∂y′=d
dx (2y′)=2y′′
Differentiating Fwith respect to ygives:
∂F
∂y =−2
Therefore, the Euler-Lagrange equation becomes:
2y′′ + 2 = 0
Step 2: Solve the differential equation. Solving the differential equation
2y′′ + 2 = 0, we get:
y′′ =−1
Integrating once gives:
y′=−x+C1
Integrating again gives:
y=−1
2x2+C1x+C2
Step 3: Apply boundary conditions. Using the boundary conditions y(0) = 0
and y(1) = 1, we have:
0 = 0 + 0 + C2
1 = −1
2+C1+C2
Solving these equations simultaneously, we find C1=3
2and C2= 0.
Therefore, the extremal that minimizes the functional J[y] is given by:
y=−1
2x2+3
2x
Question 10
Question
Find the extremals of the functional
J[y] = Z1
0
(y′2−y2)dx
subject to the boundary conditions y(0) = 1 and y(1) = e.
10
Solution
Step 1: Define the Lagrangian The Lagrangian is defined as
L(x, y, y′) = y′2−y2
Step 2: Set up 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
Step 5: Solve the differential equation
2y′′ + 2y= 0
This is a second-order linear homogeneous differential equation with character-
istic polynomial r2+ 1 = 0. The solutions are y(x) = Asin(x) + Bcos(x).
Step 6: Apply the boundary conditions Using the boundary conditions
y(0) = 1 and y(1) = e, we have:
(B= 1
Asin(1) + Bcos(1) = e
Substitute B= 1 into the second equation to find A.
Question 11
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) = 2.
11
Solution
Step 1: Define the Lagrangian The Lagrangian is defined as
L(x, y, y′) = y′2−y2
Step 2: Set up the Euler-Lagrange equation The Euler-Lagrange equation
for the functional J[y] is given by
d
dx ∂L
∂y′−∂L
∂y = 0
Step 3: Compute ∂L
∂y′and ∂L
∂y We have
∂L
∂y′=∂
∂y′(y′2−y2)=2y′
and ∂L
∂y =∂
∂y (y′2−y2) = −2y
Step 4: Apply the Euler-Lagrange equation Substitute ∂L
∂y′and ∂L
∂y into the
Euler-Lagrange equation to get
d
dx (2y′)+2y= 0
Step 5: Solve the differential equation This differential equation can be re-
arranged to
y′′ −y= 0
which has the general solution
y(x) = c1ex+c2e−x
where c1and c2are constants to be determined.
Step 6: Apply the boundary conditions Using the boundary conditions
y(0) = 0 and y(1) = 2, we get the equations:
c1+c2= 0
c1e+c2e−1= 2
Solving these equations gives c1=eand c2=−e.
Therefore, the extremal for the given functional subject to the boundary
conditions is
y(x) = ex−e−x
12
Question 12
Question
Consider the functional
J[y] = Z1
0
(y′)2+y2dx
subject to the boundary conditions 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 minimizing the functional
J[y]:
d
dx ∂f
∂y′−∂f
∂y = 0
where f= (y′)2+y2.
Step 2: Compute the partial derivatives:
∂f
∂y′= 2y′and ∂f
∂y = 2y
Step 3: Apply the Euler-Lagrange equation:
d
dx (2y′)−2y= 0
Step 4: Simplify the above expression:
2y′′ −2y= 0
Step 5: Solve the differential equation. The general solution is:
y(x) = c1cos(x) + c2sin(x)
Step 6: Apply the boundary conditions y(0) = 0 and y(1) = 1:
c1= 0 and c2sin(1) = 1
c2=1
sin(1)
Step 7: Therefore, the function that minimizes the functional J[y] is:
y(x) = 1
sin(1) sin(x)
13
Question 13
Question
Find the extremals of the functional
J[y] = Z1
0
(y′2−y−y2)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, y′, x) = y′2−y−y2.
Step 2: Compute the partial derivatives. First, we compute the derivative
with respect to y′:
∂f
∂y′=∂
∂y′(y′2−y−y2)=2y′.
Next, we compute the derivative with respect to y:
∂f
∂y =∂
∂y (y′2−y−y2) = −1−2y.
Step 3: Apply the Euler-Lagrange equation. Substitute the partial deriva-
tives into the Euler-Lagrange equation to obtain
d
dx (2y′)−(−1−2y)=0.
Step 4: Simplify the equation. Simplify the above equation to get
2y′′ + 2y+ 1 = 0.
Step 5: Solve the differential equation. The general solution to the differen-
tial equation 2y′′ + 2y+ 1 = 0 is
y(x) = c1cos x+c2sin x−1
2,
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 get the system of equations:
(c1−1
2= 0
c1cos 1 + c2sin 1 −1
2= 1
14
Solving this system, we find c1=1
2and c2=2 sin 1
cos 1 .
Thus, the extremal of the functional is
y(x) = 1
2cos x+2 sin 1
cos 1 sin x−1
2.
Question 14
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 Euler-Lagrange equation. The Euler-Lagrange equation for
the given functional J[y] is given by
d
dx ∂f
∂y′−∂f
∂y = 0
where f=y′2−y2.
Step 2: Calculate the partial derivatives. Calculating the partial deriva-
tives, we have:
∂f
∂y =−2y
∂f
∂y′= 2y′
d
dx ∂f
∂y′=d
dx (2y′)=2y′′
Step 3: Apply the Euler-Lagrange equation. Substitute the partial deriva-
tives into the Euler-Lagrange equation to get:
d
dx (2y′)+2y= 0
Step 4: Solve the differential equation. The differential equation simplifies
to 2y′′ + 2y= 0, which simplifies further to y′′ +y= 0. The general solution to
this homogeneous second-order differential equation is
y(x) = c1cos(x) + c2sin(x)
Step 5: Apply the boundary conditions. Using the boundary conditions
y(0) = 0 and y(1) = 1, we find:
c1= 0
15
c2sin(1) = 1
Thus, c2=1
sin(1) .
Step 6: Final answer. Therefore, the extremal of the functional J[y] subject
to the given boundary conditions is
y(x) = 1
sin(1) sin(x)
Question 15
Question
Let I[y] = R1
0(y′2−y2)dx where y(0) = 0, y(1) = 1, and y′(0) = 2. Determine
the function y(x) that minimizes I[y] using the calculus of variations.
Solution
Step 1: Set up the Euler-Lagrange equation. Given I[y] = R1
0(y′2−y2)dx, the
corresponding Lagrange density is L(y, y′, x) = y′2−y2. The Euler-Lagrange
equation is then given by:
d
dx ∂L
∂y′−∂L
∂y = 0
Step 2: Compute the partial derivatives. We have:
∂L
∂y′= 2y′and ∂L
∂y =−2y
Step 3: Apply the Euler-Lagrange equation. Substituting the partial deriva-
tives into the Euler-Lagrange equation and simplifying, we obtain:
d
dx (2y′)+2y= 0
Step 4: Solve the differential equation. The equation simplifies to 2y′′ +2y=
0, which has the general solution y(x) = c1cos(x) + c2sin(x).
Step 5: Apply the boundary conditions. Using the boundary conditions
y(0) = 0 and y(1) = 1, we find that c1= 0 and c2= 1.
Step 6: Determine y(x). Thus, the function that minimizes I[y] is y(x) =
sin(x).
Question 16
Question
Find the extremals of the functional
J[y] = Z1
0
(y′2−y2)dx
16
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 the functional J[y] = Rb
aF(x, y, y′)dx is given
by:
d
dx ∂F
∂y′−∂F
∂y = 0
Step 2: Write the Euler-Lagrange equation for the given functional.
For J[y] = R1
0(y′2−y2)dx,F(x, y, y′) = y′2−y2. Thus, the Euler-Lagrange
equation becomes:
d
dx (2y′)+2y= 0
Step 3: Solve the Euler-Lagrange equation.
Differentiating 2y′with respect to x, we get:
2y′′ + 2y= 0
This is a second-order linear homogeneous differential equation. The general
solution of this differential equation is:
y(x) = c1cos(x) + c2sin(x)
Step 4: Apply the boundary conditions.
Given that y(0) = 0 and y(1) = 1, we have:
0 = c1
1 = c2sin(1)
Hence, the extremal of the functional is y(x) = sin(x).
Question 17
Question
Find the function y(x) that minimizes the functional
J[y] = Z1
0
(y′2+y2)dx
subject to the boundary conditions y(0) = 0 and y(1) = 1.
17
Solution
Step 1: Define the Euler-Lagrange equation.
The Euler-Lagrange equation for minimizing the functional J[y] = Rb
aF(x, y, y′)dx
is given by
d
dx ∂F
∂y′−∂F
∂y = 0
Step 2: Apply the Euler-Lagrange equation to the given functional.
For the given functional J[y] = R1
0(y′2+y2)dx, we have
F(x, y, y′) = y′2+y2
Therefore,
∂F
∂y = 2yand ∂F
∂y′= 2y′
Now, the Euler-Lagrange equation becomes
d
dx (2y′)−2y= 0
2y′′ −2y= 0
Step 3: Solve the differential equation above.
The solution to the differential equation 2y′′ −2y= 0 is of the form y(x) =
c1cos(x) + c2sin(x).
Step 4: Apply the boundary conditions.
Given that y(0) = 0 and y(1) = 1, we have c1= 0 (from the first boundary
condition) and c2sin(1) = 1 (from the second boundary condition). Thus,
c2=1
sin(1) .
Therefore, the function that minimizes the given functional is
y(x) = sin(x)
sin(1)
Question 18
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) = 0.
18
Solution
To find the extremals, we need to solve 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+y2in this case.
Step 1: Compute ∂f
∂y′and ∂f
∂y .
∂f
∂y′= 2y′,∂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.
2y′′ −2y= 0
Step 4: Solve the differential equation with the boundary conditions y(0) =
0 and y(1) = 0. The general solution to the differential equation is
y(x) = c1sin(x) + c2cos(x)
Applying the boundary conditions:
y(0) = c2= 0
y(1) = c1sin(1) = 0
Since c1cannot be 0, we must have sin(1) = 0, which gives the constraint
c1=kπ where k∈Z.
Therefore, the extremals of the functional are of the form
y(x) = kπ sin(x)
where k∈Z.
Question 19
Question
Find the extremals of the functional
J[y] = Z1
0
(y′2+ 2y)dx
subject to the boundary conditions y(0) = 1 and y(1) = −1.
19
Solution
Step 1: Define the Lagrangian
The Lagrangian L(x, y, y′) for this functional is given by
L(x, y, y′) = y′2+ 2y
Step 2: Set up the Euler-Lagrange equation
The Euler-Lagrange equation is given by
d
dx ∂L
∂y′−∂L
∂y = 0
Plugging in the Lagrangian L(x, y, y′) = y′2+ 2y, we have
d
dx (2y′)−2=0
Step 3: Solve the Euler-Lagrange equation
Integrating the Euler-Lagrange equation with respect to x, we get
2y′−2 = c
where cis an arbitrary constant.
Step 4: Impose the boundary conditions
Using the boundary conditions y(0) = 1 and y(1) = −1, we find that c= 2.
Step 5: Solve for y(x)
Solving the differential equation 2y′−2 = 2 with initial condition y(0) = 1, we
get
y(x) = −x2+x+ 1
Step 6: Confirm the extremals
Plugging the extremal y(x) = −x2+x+ 1 back into the functional J[y], we
calculate
J[y] = Z1
0
((−2x+ 1)2+ 2(−x2+x+ 1)) dx =−11
6
Therefore, the extremal of the functional is y(x) = −x2+x+ 1.
Question 20
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.
20
Solution
Step 1: Compute the Euler-Lagrange equation. The Euler-Lagrange equation
for the 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. Therefore, we have
d
dx ∂L
∂y′−∂L
∂y =d
dx (2y′)−(−2y)=2y′′ + 2y= 0.
Step 2: Solve the differential equation. The general solution to 2y′′ + 2y= 0
is y(x) = c1cos(x)+c2sin(x). Next, we apply the boundary conditions y(0) = 0
and y(1) = 1: y(0) = c1= 0 ⇒y(x) = c2sin(x). y(1) = c2sin(1) = 1 ⇒
c2=1
sin(1) ⇒y(x) = sin(x)
sin(1) .
Therefore, the extremal for the functional is y(x) = sin(x)
sin(1) .
Question 21
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: Set up the Euler-Lagrange equation for the given functional. The
Euler-Lagrange equation for the functional J[y] = Rb
aF(x, y, y′)dx is given by
d
dx ∂F
∂y′−∂F
∂y = 0
In this case, F(x, y, y′) = y′2−y, so
d
dx (2y′)−(−1) = 0
Thus, we have
2y′′ = 1
Step 2: Solve the differential equation that we obtained in Step 1. Inte-
grating both sides with respect to x, we get
2y′=x+C1
21
Integrating again, we have
y=1
2x2+C1x+C2
Step 3: Apply the boundary conditions to find the constants C1and C2.
Using the boundary conditions y(0) = 0 and y(1) = 1, we get
0 = 0 + 0 + C2⇒C2= 0
1 = 1
2+C1⇒C1=1
2
Step 4: Finalize the solution by substituting the values of C1and C2into
the general solution. Therefore, the function that minimizes the given functional
subject to the given boundary conditions is
y(x) = 1
2x2+1
2x
Question 22
Question
Find the extremals of the functional
J[y] = Z1
0
(y2+y′2)dx
subject to the boundary conditions y(0) = 1 and y(1) = 2.
Solution
Step 1: Write down the Euler-Lagrange equation. The Euler-Lagrange equation
for the given functional is
d
dx ∂f
∂y′−∂f
∂y = 0
where f=y2+y′2.
Step 2: Compute the partial derivatives.
∂f
∂y = 2y
∂f
∂y′= 2y′
Step 3: Find the derivative of ∂f
∂y′with respect to x.
d
dx ∂f
∂y′= 2y′′
22
Step 4: Apply the Euler-Lagrange equation. Substitute the partial deriva-
tives into the Euler-Lagrange equation, we get
d
dx (2y′)−2y= 0
which simplifies to
2y′′ −2y= 0
Step 5: Solve the differential equation. The general solution to the differ-
ential equation is given by y(x) = Acos(x) + Bsin(x), where Aand Bare
constants to be determined.
Step 6: Apply the boundary conditions. Using the boundary conditions
y(0) = 1 and y(1) = 2, we find
y(0) = A= 1
y(1) = Acos(1) + Bsin(1) = 2
Substitute A= 1 into the second equation and solve for Bto find B=2−cos(1)
sin(1) .
Step 7: Write down the extremal solution. The extremal function that
minimizes the given functional is
y(x) = cos(x) + 2−cos(1)
sin(1) sin(x)
Question 23
Question
Let y(x) be a function that minimizes the functional
J[y] = Z2
1
(y′2−2y2)dx
subject to the boundary conditions y(1) = 3 and y(2) = −1. Find the function
y(x) that minimizes J[y].
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, y′, x) = y′2−2y2.
Step 2: Compute the partial derivatives. We have
∂f
∂y′=∂
∂y′(y′2−2y2)=2y′,
23
and ∂f
∂y =∂
∂y (y′2−2y2) = −4y.
Step 3: Substitute into the Euler-Lagrange equation. Plugging the partial
derivatives into the Euler-Lagrange equation, we get
d
dx (2y′)−(−4y)=0
⇒d
dx (2y′)+4y= 0
⇒2y′′ + 4y= 0.
Step 4: Solve the differential equation. The solution to the differential equa-
tion 2y′′ + 4y= 0 is of the form y(x) = Asin(2x) + Bcos(2x), where Aand B
are constants to be determined.
Step 5: Apply the boundary conditions. Using the boundary conditions
y(1) = 3 and y(2) = −1, we find
3 = Asin(2) + Bcos(2)
−1 = Asin(4) + Bcos(4).
Step 6: Solve for Aand B. Solving the system of equations, we find A=
3 cos(4)+sin(4)
sin(2) sin(4)−2 cos(2) cos(4) and B=−3 sin(4)+cos(4)
sin(2) sin(4)−2 cos(2) cos(4) .
Therefore, the function y(x) that minimizes the given functional is y(x) =
3 cos(4)+sin(4)
sin(2) sin(4)−2 cos(2) cos(4) sin(2x) + −3 sin(4)+cos(4)
sin(2) sin(4)−2 cos(2) cos(4) cos(2x).
Question 24
Question
Find the function y(x) that minimizes the functional
J[y] = Z1
0
(y′2−y2)dx
subject to the boundary conditions y(0) = 0 and y(1) = 1.
Solution
To find the function y(x) that minimizes the functional J[y], we will use the
Euler-Lagrange equation. Let L(y, y′, x) = y′2−y2be the integrand of J[y].
Step 1: Compute ∂L
∂y and d
dx ∂L
∂y′.
∂L
∂y =−2yand d
dx ∂L
∂y′=d
dx (2y′) = 2y′′
24
Step 2: Apply the Euler-Lagrange equation d
dx ∂L
∂y′−∂L
∂y = 0.
2y′′ + 2y= 0
Step 3: Solve the second-order differential equation 2y′′ + 2y= 0 with
boundary conditions y(0) = 0 and y(1) = 1. The general solution is y(x) =
c1sin(x) + c2cos(x).
Applying the boundary conditions, we have:
y(0) = 0 =⇒c2= 0
y(1) = 1 =⇒c1sin(1) = 1 =⇒c1=1
sin(1)
Thus, the function y(x) that minimizes the functional J[y] is y(x) = 1
sin(1) sin(x).
Question 25
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 L(x, y, y′) as L(x, y, y′) = y′2−y2.
Step 2: Apply the Euler-Lagrange equation
d
dx ∂L
∂y′−∂L
∂y = 0
to find the extremal.
Step 3: Calculate the partial derivatives:
∂L
∂y =−2yand ∂L
∂y′= 2y′
Step 4: Compute the derivative of ∂L
∂y′with respect to x:
d
dx ∂L
∂y′=d
dx (2y′)=2y′′
Step 5: Substitute the derivatives into the Euler-Lagrange equation:
2y′′ + 2y= 0
25
Step 6: Solve the differential equation subject to the boundary conditions
y(0) = 0 and y(1) = 1.
The general solution to the differential equation is given by
y(x) = c1cos(x) + c2sin(x)
Step 7: Apply the boundary conditions to find the values of c1and c2.
From y(0) = 0, we have c1= 0.
Substitute c1= 0 into the general solution:
y(x) = c2sin(x)
From y(1) = 1, we have c2sin(1) = 1, which implies c2=1
sin(1) .
Therefore, the extremal for the given functional is
y(x) = 1
sin(1) sin(x)
Question 26
Question
Find the function y(x) that extremizes the functional
J[y] = Z1
0y2+y′2dx
subject to the boundary conditions y(0) = 1 and y(1) = 2.
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, y′, x) = y2+y′2.
Step 2: Find the partial derivatives. We have
∂f
∂y = 2y, and ∂f
∂y′= 2y′.
Now, differentiate ∂f
∂y′with respect to x:
d
dx ∂f
∂y′= 2y′′.
26
Therefore, the Euler-Lagrange equation becomes
2y′′ −2y= 0.
Step 3: Solve the differential equation. The general solution to the differen-
tial equation y′′ −y= 0 is given by
y(x) = c1ex+c2e−x,
where c1and c2are constants to be determined.
Step 4: Apply the boundary conditions. Using the boundary conditions
y(0) = 1 and y(1) = 2, we find
y(0) = c1+c2= 1
y(1) = c1e+c2e−1= 2
Step 5: Solve for c1and c2. By solving the system of equations from step 4,
we find c1=2e
e+1 and c2=2
e+1 .
Step 6: Final solution. Therefore, the function y(x) that extremizes the
functional J[y] subject to the given boundary conditions is
y(x) = 2e
e+ 1ex+2
e+ 1e−x.
Question 27
Question
Find the extremal for 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 by finding the derivative of the
integrand with respect to yand the derivative of the integrand with respect to
y′.
d
dx ∂f
∂y′−∂f
∂y = 0
Step 2: Let f(y, y′, x) = y′2−2y. Then, compute the partial derivatives.
∂f
∂y =−2,∂f
∂y′= 2y′
27
Step 3: Compute d
dx ∂f
∂y′.
d
dx (2y′)=2y′′
Step 4: Set up the Euler-Lagrange equation:
2y′′ + 2 = 0
Step 5: Solve the differential equation with the boundary conditions y(0) = 0
and y(1) = 1.
y′′ =−1 =⇒y′=−x+C1, y =−1
2x2+C1x+C2
Step 6: Use the boundary conditions y(0) = 0 and y(1) = 1 to find C1and
C2.
0 = C2,1 = −1
2+C1=⇒C1=3
2
Step 7: Therefore, the extremal for the given functional is y=−1
2x2+3
2x.
Question 28
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: Compute the Euler-Lagrange equation by taking the derivative with
respect to yand y′of the integrand in the functional J[y]:
d
dx
∂f
∂y′−∂f
∂y = 0
d
dx (2y′)−(−2y)=0
d
dx (2y′)+2y= 0
2y′′ + 2y= 0
28
Step 2: Solve the differential equation 2y′′ + 2y= 0 with the boundary con-
ditions y(0) = 0 and y(1) = 1. The general solution to the differential equation
is given by y(x) = Asin(x) + Bcos(x). Applying the boundary conditions:
y(0) = 0 ⇒B= 0
y(1) = 1 ⇒Asin(1) = 1 ⇒A=1
sin(1)
Therefore, the extremal that minimizes the functional J[y] is given by y(x) =
1
sin(1) sin(x).
Question 29
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: Set up the Euler-Lagrange equation: The Euler-Lagrange equation for
the functional J[y] = Rb
aF(x, y, y′)dx is given by
d
dx ∂F
∂y′−∂F
∂y = 0
In this case, we have F(x, y, y′) = y′2−y2, so
d
dx ∂F
∂y′−∂F
∂y =d
dx (2y′)−(−2y) = d
dx (2y′)+2y= 0
Step 2: Solve the Euler-Lagrange equation: Solving the Euler-Lagrange
equation gives us
2y′′ + 2y= 0
which simplifies to
y′′ +y= 0
Step 3: Find the general solution of the differential equation: The general
solution of y′′ +y= 0 is
y(x) = Acos x+Bsin x
where Aand Bare constants to be determined.
29
Step 4: Apply the boundary conditions: From the boundary conditions
y(0) = 0 and y(1) = 1, we have
A= 0
Bsin 1 = 1
which gives B=1
sin 1 .
Therefore, the extremal for the functional is
y(x) = sin x
sin 1
Question 30
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. Let L(y, y′, x) = y′2−ybe the
integrand of the functional J[y].
Step 2: Compute ∂L
∂y −d
dx ∂L
∂y′= 0.
∂L
∂y =−1,∂L
∂y′= 2y′
d
dx ∂L
∂y′=d
dx (2y′)=2y′′
Euler-Lagrange equation: −1−2y′′ = 0
Step 3: Solve the differential equation y′′ =−1
2. Integrate twice to obtain
the general solution: y(x) = −1
4x2+1
2x+c1x+c2
Step 4: Apply the boundary conditions y(0) = 0 and y(1) = 1.
y(0) = c2= 0
y(1) = −1
4+1
2+c1= 1
c1=3
4
Step 5: The extremal is the function y(x) = −1
4x2+1
2x+3
4x.
Therefore, the extremal of the functional J[y] subject to the given boundary
conditions is y(x) = −1
4x2+1
2x+3
4x.
30
Question 31
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: Calculate the Euler-Lagrange equation. Given the functional
J[y] = Z1
0
(y′2−y2)dx
we have the Euler-Lagrange equation
d
dx ∂f
∂y′−∂f
∂y = 0
where f=f(x, y, y′) = y′2−y2.
Step 2: Compute the partial derivatives. We have
∂f
∂y′= 2y′,∂f
∂y =−2y
and d
dx ∂f
∂y′=d
dx (2y′)=2y′′
Step 3: Write down the Euler-Lagrange equation. Substituting into the
Euler-Lagrange equation, we have
2y′′ + 2y= 0
Step 4: Solve the differential equation. The general solution to this differ-
ential equation is
y(x) = c1cos(x) + c2sin(x)
Step 5: Apply the boundary conditions. From y(0) = 0 and y(1) = 1, we
find
c1= 0
and
c2sin(1) = 1
which gives c2=1
sin(1) .
Therefore, the extremal for the given functional subject to the boundary
conditions is
y(x) = sin(x)
sin(1)
31
Question 32
Question
Find the extremals of the functional
J[y] = Z1
0
(y′)2−y2dx
subject to the boundary conditions y(0) = 1 and y(1) = 0.
Solution
Step 1: Let’s denote by L(y, y′, x)=(y′)2−y2the integrand of the functional
J[y].
Step 2: The Euler-Lagrange equation is given by d
dx ∂L
∂y′−∂L
∂y = 0.
Step 3: First, let’s calculate ∂L
∂y′:
∂L
∂y′= 2y′
Step 4: Then, we find d
dx ∂L
∂y′:
d
dx ∂L
∂y′= 2y′′
Step 5: Next, let’s compute ∂L
∂y :
∂L
∂y =−2y
Step 6: Substituting these derivatives into the Euler-Lagrange equation, we
get:
2y′′ + 2y= 0
Step 7: The general solution to this differential equation is of the form
y(x) = c1cos(x) + c2sin(x).
Step 8: Now, we apply the boundary conditions y(0) = 1 and y(1) = 0 to
find the values of c1and c2.
Step 9: From y(0) = 1, we have c1= 1.
Step 10: Substituting c1= 1 into y(1) = 0, we find c2=−1.
Step 11: Therefore, the extremal that minimizes the functional J[y] under
the given boundary conditions is y(x) = cos(x)−sin(x).
32
Question 33
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) = 0.
Solution
Step 1: Define the Euler-Lagrange equation. The Euler-Lagrange equation for
the 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, so the Euler-Lagrange equation becomes
d
dx (2y′)+2y= 0
Step 2: Solve the Euler-Lagrange equation. Since the Euler-Lagrange equa-
tion is a second-order ordinary differential equation, we can rewrite it as
y′′ +y= 0
The general solution of this differential equation is
y(x) = c1sin x+c2cos x
Step 3: Apply the boundary conditions. We have y(0) = 0 and y(1) = 0,
which gives us the following equations:
(0 = c2
0 = c1sin 1
Since c2= 0, the second equation becomes c1sin 1 = 0, which implies c1= 0
since sin 1 = 0.
Step 4: Conclusion. The only extremal for the functional J[y] subject to
the given boundary conditions is the identically zero function, i.e., y(x) = 0.
Question 34
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.
33
Solution
Step 1: Define the Lagrangian for the functional J[y] as
L(x, y, y′) = y′2+y2+λ1(0)y+λ2(1)(y−1)
where λ1and λ2are the Lagrange multipliers.
Step 2: Compute the Euler-Lagrange equation for y:
d
dx ∂L
∂y′−∂L
∂y = 0
This gives
d
dx (2y′)−2y= 0.
Step 3: Solve the differential equation obtained in Step 2:
2y′′ −2y= 0
y′′ −y= 0
The general solution to this differential equation is
y(x) = c1ex+c2e−x
Step 4: Apply the boundary conditions y(0) = 0 and y(1) = 1: From
y(0) = 0, we have c1+c2= 0. From y(1) = 1, we have c1e+c2e−1= 1.
Step 5: Solve the system of equations in Step 4 to find the constants c1and
c2: Solving the system, we find c1=e
e−1and c2=−1
e−1.
Therefore, the extremal for the given functional is
y(x) = e
e−1ex−1
e−1e−x
Question 35
Question
Let J[y] = R1
0(y′2−y2)dx where ysatisfies the boundary conditions y(0) = 0
and y(1) = 1. Find the function ythat minimizes J[y].
Solution
Step 1: Let’s start by setting up the Euler-Lagrange equation for this variational
problem. The Euler-Lagrange equation is given by:
d
dx ∂f
∂y′−∂f
∂y = 0
where f(x, y, y′) = y′2−y2.
34
Step 2: Compute the partial derivatives involved:
∂f
∂y′= 2y′,∂f
∂y =−2y
Step 3: Differentiate ∂f
∂y′with respect to x:
d
dx ∂f
∂y′= 2y′′
Step 4: Now we have the Euler-Lagrange equation in the form:
2y′′ + 2y= 0
Step 5: Solve the differential equation with the boundary conditions y(0) = 0
and y(1) = 1.
Step 6: The general solution to the differential equation is:
y(x) = c1sin(x) + c2cos(x)
Step 7: Apply the boundary conditions y(0) = 0 and y(1) = 1 to find c1and
c2.
Step 8: From y(0) = 0, we have:
c2= 0
Step 9: Substituting c2= 0 into the general solution, we get:
y(x) = c1sin(x)
Step 10: Then, from y(1) = 1, we have:
c1sin(1) = 1 =⇒c1=1
sin(1)
Step 11: Therefore, the function ythat minimizes J[y] is:
y(x) = sin(x)
sin(1)
35
