MATH 332 - ADVANCED CALCULUS
- Calculus of variations
Question Bank - Set 3
Liberty University
Question 1
Question
Let y(x) be a function that minimizes the functional
I[y] = Z2
1
(y′2+y2)dx
subject to the boundary conditions y(1) = 1, y(2) = 2. Find the function y(x)
that minimizes the functional.
Solution
Step 1: Formulate the Euler-Lagrange equation
The Euler-Lagrange equation for minimizing the functional I[y] is given by:
d
dx ∂f
∂y′−∂f
∂y = 0
where f(y, y′, x) = y′2+y2in this case.
Step 2: Compute the partial derivatives
First, we 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: Apply the Euler-Lagrange equation
Substitute the derivatives back into the Euler-Lagrange equation:
d
dx(2y′)−2y= 0
Simplify to obtain the differential equation:
2y′′ −2y= 0
Step 4: Solve the differential equation with boundary conditions
The general solution to the differential equation is of the form y(x) =
c1cos(x) + c2sin(x).
Applying the boundary conditions y(1) = 1 and y(2) = 2, we find:
c1cos(1) + c2sin(1) = 1
c1cos(2) + c2sin(2) = 2
Solving these equations simultaneously gives the values of c1and c2. Thus,
the function that minimizes the functional is y(x) = c1cos(x) + c2sin(x).
Question 2
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. Let L(y, y′, x) = y′2−y2. The
Euler-Lagrange equation is given by
d
dx ∂L
∂y′−∂L
∂y = 0
Step 2: Calculate the partial derivatives. We have
∂L
∂y′= 2y′and ∂L
∂y =−2y
Step 3: Use the Euler-Lagrange equation. Substituting the partial deriva-
tives into the Euler-Lagrange equation, we get
d
dx(2y′)−(−2y)=0
2
2y′′ + 2y= 0
Step 4: Solve the differential equation. The general solution to the differen-
tial equation is
y(x) = c1sin(x) + c2cos(x)
Step 5: Apply the boundary conditions. Using y(0) = 0 and y(1) = 1, we
have
0 = c2
1 = c1sin(1)
Thus, c1=1
sin(1) .
Step 6: Final solution. Therefore, the extremal of the functional J[y] is
y(x) = 1
sin(1) sin(x)
Question 3
Question
Find the extremals of the functional
J[y] = Z1
0
(y(x)2+y′(x)2)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 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, we have
F(x, y, y′) = y(x)2+y′(x)2
So, the Euler-Lagrange equation becomes
d
dx ∂
∂y′(y(x)2+y′(x)2)−∂
∂y (y(x)2+y′(x)2)=0
Step 2: Compute the derivatives.
The partial derivatives are:
3
∂
∂y′(y(x)2+y′(x)2)=2y′(x)
∂
∂y (y(x)2+y′(x)2)=2y(x)
d
dx (2y′(x)) = 2y′′(x)
So, the Euler-Lagrange equation simplifies to
2y′′(x)−2y(x)=0
Step 3: Solve the differential equation.
The general solution to the differential equation y′′(x)−y(x) = 0 is
y(x) = c1ex+c2e−x
Step 4: Apply the boundary conditions.
Using the boundary conditions y(0) = 0 and y(1) = 1, we find
0 = c1+c2
1 = c1e+c2e−1
Solving these equations simultaneously, we get c1=e
e2−1and c2=−1
e2−1.
Step 5: Final solution.
Therefore, the extremal of the functional is
y(x) = e
e2−1ex−1
e2−1e−x
Question 4
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
Step 1: Define the Euler-Lagrange equation. Let F(y, y′, x) = y′2−y2be the
integrand in the functional J[y]. The Euler-Lagrange equation states that the
function y(x) that minimizes the functional J[y] must satisfy the equation
d
dx ∂F
∂y′−∂F
∂y = 0
4
Step 2: Compute the partial derivatives. We have ∂F
∂y′= 2y′and ∂F
∂y =−2y.
Therefore, the Euler-Lagrange equation becomes
d
dx(2y′)+2y= 0
Step 3: Solve the differential equation. Rearranging the equation gives us
2y′′ + 2y= 0, which simplifies to y′′ +y= 0. This is a second-order linear
homogeneous differential equation with constant coefficients.
Step 4: Find the general solution. The characteristic equation is r2+ 1 = 0,
which has complex roots r=±i. This implies that the general solution to the
differential equation is
y(x) = c1cos x+c2sin x
Step 5: Apply the boundary conditions. Using y(0) = 0, we have c1= 0.
Using y(1) = 1, we have c2sin 1 = 1, which gives c2=1
sin 1 .
Step 6: Final answer. Thus, the function that minimizes the functional J[y]
subject to the given boundary conditions is
y(x) = sin x
sin 1
Question 5
Question
Find the extremals of the functional
J[y] = Z1
0
(y′2−y2−y)dx
subject to the boundary conditions y(0) = 0 and y(1) = 1.
Solution
Step 1: Define the Lagrangian
L(x, y, y′) = y′2−y2−y
Step 2: Apply the Euler-Lagrange equation The Euler-Lagrange equation is
given by
d
dx ∂L
∂y′=∂L
∂y
Plugging the Lagrangian into the Euler-Lagrange equation, we get
d
dx(2y′) = −2y−1
which simplifies to
2y′′ =−2y−1
5
Step 3: Solve the differential equation The general solution to the homoge-
neous part of the differential equation is
yh(x) = c1cos(x) + c2sin(x)
where c1and c2are constants.
To find the particular solution yp(x), we assume a solution of the form erx.
Substituting this into the differential equation gives us the characteristic equa-
tion:
r2=−1
which has solutions r=±i. Therefore, the particular solution has the form
yp(x) = C1cos(x) + C2sin(x)
where C1and C2are constants.
Step 4: Apply the boundary conditions Using the boundary conditions
y(0) = 0 and y(1) = 1, we find the values of C1and C2.
From y(0) = 0, we have
y(0) = C1= 0
From y(1) = 1, we have
y(1) = C2sin(1) = 1
C2=1
sin(1)
Therefore, the extremal of the functional is
y(x) = 1
sin(1) sin(x)
Question 6
Question
Find the extremals of the functional
J[y] = Z1
0
y2(1 + y′2)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′) = y2(1 + y′2).
6
Step 2: Apply the Euler-Lagrange equation, which states that d
dx ∂L
∂y′=
∂L
∂y .
By applying this equation, we have
d
dx (2yy′)=2yy′.
Step 3: Simplify the equation from Step 2. We have
2yy′′ + 2(y′)2=y.
Step 4: Rearrange the equation from Step 3 to the standard form:
y−2yy′′ −(y′)2= 0.
Step 5: Solve the differential equation subject to the boundary conditions
y(0) = 0 and y(1) = 1.
The solution to the differential equation satisfies the boundary conditions
y(0) = 0 and y(1) = 1 if y(x) = x.
Therefore, the extremals of the functional are the straight lines passing
through the points (0,0) and (1,1).
Question 7
Question
Find the extremals for the functional J[y] = R1
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+λ1y(0) + λ2(y(1) −1)
where λ1, λ2are the Lagrange multipliers.
Step 2: Compute the Euler-Lagrange equation by differentiating the La-
grangian with respect to yand y′and then subtracting the derivative of the
resulting equation with respect to x.
d
dx
∂L
∂y′−∂L
∂y = 0
d
dx(2y′)+2y−λ1δ(x) + λ2δ(x−1) = 0
Integrating by parts in the first term on the right side and using the boundary
conditions, we find
2y−2y′(0) −λ1+λ2= 0
7
2y(1) −2y′(1) + λ2= 0
Step 3: Solve the system of equations obtained in Step 2 using the boundary
conditions to find the extremals. Plugging in y(0) = 0 and y(1) = 1, we get the
following system of equations
−λ1+λ2= 0
2−2y′(1) + λ2= 0
Solving this system, we find λ1= 1 and λ2= 1.
Step 4: Substitute λ1= 1 and λ2= 1 back into the differential equation in
Step 2, we obtain
2y−2y′(0) + 1 = 0
2−2y′(1) + 1 = 0
which can be solved to find the extremals. Therefore, the extremals for the
given functional are y(x) = sin(πx).
Question 8
Question
Consider the functional
J[y] = Z1
0
(y′2+y2)dx
subject to the boundary conditions y(0) = 0, y(1) = 1. Find the function y(x)
that minimizes J[y].
Solution
Step 1: Define the Lagrangian function
L(x, y, y′) = y′2+y2+λ1y(0) + λ2(y(1) −1)
where λ1and λ2are the Lagrange multipliers for the given boundary conditions.
Step 2: Find the Euler-Lagrange equation
∂L
∂y −d
dx ∂L
∂y′= 0
2y−d
dx(2y′) = 0
2y−2y′′ = 0
Step 3: Solve the Euler-Lagrange equation for y(x). The general solution to
the differential equation 2y−2y′′ = 0 is
y(x) = c1ex+c2e−x
8
Step 4: Apply the boundary conditions y(0) = 0 and y(1) = 1 to find the
specific solution. From y(0) = 0, we have c1+c2= 0. From y(1) = 1, we have
c1e+c2/e = 1. Solving these simultaneous equations, we find c1= 1/2 and
c2=−1/2.
Therefore, the function that minimizes J[y] is
y(x) = 1
2ex−1
2e−x
Question 9
Question
Find the extremals of the functional
J[y] = Z1
0
(y′2−2yy′)dx
subject to the boundary conditions y(0) = 1 and y(1) = 2.
Solution
Step 1: Calculate the Euler-Lagrange equation by differentiating the integrand
with respect to yand y′.
d
dx ∂f
∂y′−∂f
∂y = 0
Step 2: Given f(y, y′) = y′2−2yy′, calculate ∂f
∂y′and ∂f
∂y .
∂f
∂y′= 2y′−2yand ∂f
∂y =−2y′
Step 3: Differentiate ∂f
∂y′with respect to xto get d
dx ∂f
∂y′.
d
dx ∂f
∂y′= 2y′′ −2y′
Step 4: Substitute the expressions into the Euler-Lagrange equation and
solve.
2y′′ −2y′+ 2y= 0
Step 5: Solve the differential equation with the given boundary conditions
y(0) = 1 and y(1) = 2.
y(x) = ex
Step 6: Verify if the extremal found is a minimum or maximum by using the
second variation test.
Therefore, the extremal of the given functional is y(x) = ex.
9
Question 10
Question
Find the extremal of the functional
J[y] = Z1
0
y(x)2+y′(x)2dx
subject to the boundary conditions y(0) = 0 and y(1) = 1.
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: d
dx ∂L
∂y′−∂L
∂y = 0.
Step 3: Compute the partial derivatives in the Euler-Lagrange equation:
∂L
∂y = 2yand ∂L
∂y′= 2y′
Step 4: Compute the derivative with respect to xof ∂L
∂y′:
d
dx ∂L
∂y′=d
dx(2y′)=2y′′
Step 5: Apply the Euler-Lagrange equation:
2y′′ −2y= 0
Step 6: Solve the differential equation 2y′′ −2y= 0 with the boundary
conditions y(0) = 0 and y(1) = 1.
Step 7: The general solution to the differential equation is y(x) = c1cos(x)+
c2sin(x).
Step 8: Apply the boundary conditions to find the values of c1and c2:
y(0) = c1= 0 and y(1) = c2sin(1) = 1
Step 9: The solution is y(x) = sin(x).
Question 11
Question
Find the extremals of the functional J[y] = R1
0(y2+y′2)dx subject to the bound-
ary conditions y(0) = 0 and y(1) = 1.
10
Solution
Step 1: Compute the Euler-Lagrange equation. To find the extremals of J[y],
we need to solve the Euler-Lagrange equation:
d
dx ∂f
∂y′−∂f
∂y = 0
where f=y2+y′2. Computing the partial derivatives, we get:
∂f
∂y = 2yand ∂f
∂y′= 2y′
d
dx ∂f
∂y′=d
dx(2y′)=2y′′
So the Euler-Lagrange equation simplifies to:
2y′′ −2y= 0
Step 2: Solve the differential equation. The differential equation 2y′′−2y= 0
simplifies to y′′ −y= 0. The characteristic equation is r2−1 = 0, which has
roots r= 1 and r=−1. Therefore, the general solution to the 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 find:
(c1+c2= 0
c1e+c2e−1= 1
Solving this system of equations gives c1=e
e−1and c2=−1
e−1.
Therefore, the extremal that minimizes 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 extremals for the functional
J[y] = Z1
0
(y′2+y2)dx
subject to the boundary conditions y(0) = 0 and y(1) = 1.
11
Solution
Step 1: Define the Euler-Lagrange equation.
For the functional J[y] = R1
0(y′2+y2)dx, the Euler-Lagrange equation is given
by
d
dx ∂f
∂y′−∂f
∂y = 0
where f(y, y′, x) = y′2+y2.
Step 2: Calculate the partial derivatives.
Let’s calculate the partial derivatives needed for the Euler-Lagrange equation:
∂f
∂y′=∂
∂y′(y′2+y2)=2y′
∂f
∂y =∂
∂y (y′2+y2)=2y
Step 3: Form the Euler-Lagrange equation.
Substitute the partial derivatives into the Euler-Lagrange equation:
d
dx(2y′)−2y= 0
d
dx(2y′)−2y= 0
Step 4: Solve the Euler-Lagrange equation.
Solving the differential equation, we get:
2y′′ −2y= 0
y′′ −y= 0
Step 5: Solve the differential equation with the boundary conditions.
The general solution to the differential equation is y(x) = c1ex+c2e−x. Applying
the boundary conditions y(0) = 0 and y(1) = 1:
c1+c2= 0
c1e+c2e−1= 1
Solving these equations gives c1=e−1
1−e−2and c2=−e
1−e−2.
Step 6: Write down the extremal function.
Therefore, the extremal for the functional is
y(x) = e−1
1−e−2ex−e
1−e−2e−x
12
Question 13
Question
Find the extremum of 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: Set up the Euler-Lagrange equation for the given functional. The
Euler-Lagrange equation for a functional J[y] is given by:
d
dx ∂f
∂y′−∂f
∂y = 0
where f(y, y′, x) = y2+y′2in this case.
Step 2: Compute the partial derivatives. Let f(y, y′, x) = y2+y′2. Then,
∂f
∂y = 2yand ∂f
∂y′= 2y′
d
dx ∂f
∂y′=d
dx(2y′)=2y′′
Step 3: Plug into the Euler-Lagrange equation. Substitute the partial
derivatives into the Euler-Lagrange equation to get:
d
dx(2y′)−2y= 0
2y′′ −2y= 0
Step 4: Solve the differential equation. The solution to the differential
equation 2y′′ −2y= 0 is of the form y(x) = Acos√2x+Bsin√2x.
Step 5: Apply the boundary conditions. Using the boundary conditions
y(0) = 0 and y(1) = 1, we find:
0 = Acos(0) + Bsin(0) ⇒A= 0
1 = Bsin√2⇒B=1
sin√2
Step 6: Determine the extremum. The extremum of the functional J[y] is
given by the function y(x) = sin(√2x)
sin(√2)which minimizes J[y] subject to the given
boundary conditions.
13
Question 14
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) = 2.
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. We have:
∂L
∂y′= 2y′
∂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
Step 5: Solve the differential equation. Integrating the equation gives:
2y′= 2y+C
y′=y+C1
Step 6: Apply the boundary conditions. Using the boundary conditions
y(0) = 1, we find C1= 0. So, the solution is:
y′=y
Step 7: Solve the differential equation. The general solution to the differ-
ential equation y′=yis:
y(x) = Aex
14
Step 8: Apply the boundary condition y(1) = 2. Substitute the boundary
condition to find A:
y(1) = Ae = 2
A= 2/e
Step 9: Final solution. Therefore, the extremal that minimizes the func-
tional J[y] subject to the given boundary conditions is:
y(x) = 2
e·ex= 2ex−1
Question 15
Question
Consider the functional J[y] = R1
0(y′2−y2)dx subject to the boundary condi-
tions y(0) = 1 and y(1) = 2. Find the function y(x) that minimizes J[y].
Solution
Step 1: Write the Euler-Lagrange equation Given the functional J[y] = R1
0(y′2−
y2)dx, 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 The partial derivative ∂f
∂y′is 2y′and
∂f
∂y is −2y.
Step 3: Apply the Euler-Lagrange equation Substitute the partial derivatives
into the Euler-Lagrange equation:
d
dx2y′+ 2y= 0
Step 4: Solve the differential equation Rearranging the equation gives d
dx (2y′) =
−2y. Integrating both sides with respect to xyields
2y′=−2y+C
where Cis the constant of integration.
Step 5: Solve for yTo find y, solve the first-order ordinary differential equa-
tion 2y′=−2y+Cwith the initial conditions y(0) = 1.
Step 6: Apply the boundary conditions Use the boundary conditions y(0) =
1 and y(1) = 2 to solve for the constant of integration C.
Step 7: Finalize the solution After finding the constant of integration C,
substitute it back into the expression for y(x) to get the final function that
minimizes J[y].
15
Question 16
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) = 2.
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: Calculate the partial derivatives. We have:
∂f
∂y′=∂
∂y′(y′2−2y)=2y′
and d
dx ∂f
∂y′=d
dx(2y′)=2y′′
∂f
∂y =∂
∂y (y′2−2y) = −2
Step 3: Apply the Euler-Lagrange equation. Substitute the derivatives into
the Euler-Lagrange equation:
2y′′ + 2 = 0
which simplifies to
y′′ =−1
Step 4: Solve the differential equation. Integrating twice, we get:
y′=−x+C1
y=−x2
2+C1x+C2
Step 5: Apply the boundary conditions. Using the boundary conditions
y(0) = 1 and y(1) = 2, we find:
C2= 1
−1
2+C1+ 1 = 2 ⇒C1=3
2
Step 6: Determine the extremals. The extremal that minimizes the func-
tional is:
y=−x2
2+3
2x+ 1
16
Question 17
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
To find the extremals of the functional J[y], we will use the Euler-Lagrange
equation:
d
dx ∂f
∂y′−∂f
∂y = 0
where f=y′2−y2.
Step 1: Compute ∂f ∂y′and ∂f
∂y .
∂f
∂y′= 2y′
∂f
∂y =−2y
Step 2: Apply the Euler-Lagrange equation. Plugging into the Euler-
Lagrange equation, we get:
d
dx(2y′)+2y= 0
2y′′ + 2y= 0
Step 3: Solve the differential equation. The solution to this differential
equation is of the form y(x) = Asin x+Bcos x. Applying boundary conditions
y(0) = 0 and y(1) = 1:
y(0) = B= 0
y(1) = Asin 1 = 1
A=1
sin 1
Therefore, the extremal of the functional is given by:
y(x) = sin x
sin 1
17
Question 18
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 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, we have
F(x, y, y′) = y′2−2y,
so the Euler-Lagrange equation becomes
d
dx(2y′)−(−2) = 0
which simplifies to
2y′′ + 2 = 0.
Step 2: Solve the differential equation. Solving the differential equation
2y′′ + 2 = 0 gives
y′′ =−1.
Integrating once gives
y′=−x+c1,
where c1is a constant of integration. Integrating again yields
y=−1
2x2+c1x+c2,
where c2is another constant of integration.
Step 3: Apply the boundary conditions. We have y(0) = 0 and y(1) = 1.
Substituting these into the expression for y, we get the system of equations
c2= 0
18
−1
2+c1= 1.
Solving this system gives c1=3
2.
Step 4: Determine the extremals. The extremals are given by the solution
to the differential equation with the values of c1and c2determined from the
boundary conditions. Therefore, the extremal for the functional J[y] is
y=−1
2x2+3
2x.
Question 19
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. 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
In this case, our functional is J[y] = R1
0(y′2−y2)dx, so F(x, y, y′) = y′2−y2.
Step 2: Compute the partial derivatives needed for the Euler-Lagrange
equation. We need to compute ∂F
∂y′and ∂F
∂y .
∂F
∂y′=∂
∂y′(y′2−y2)=2y′
∂F
∂y =∂
∂y (y′2−y2) = −2y
Step 3: Apply the Euler-Lagrange equation. Substitute the derivatives into
the Euler-Lagrange equation:
d
dx (2y′)−(−2y) = 0
Simplify this equation to get:
2y′′ + 2y= 0
19
Step 4: Solve the differential equation. The general solution to 2y′′ +2y= 0
is y(x) = c1cos√2x+c2sin√2x.
Step 5: Apply the boundary conditions. Using the boundary conditions
y(0) = 0 and y(1) = 1, we find:
c1= 0 and c2sin√2= 1
This gives us c2=1
sin(√2).
Step 6: Write the solution to the variational problem. Therefore, the ex-
tremal for the given functional subject to the boundary conditions is:
y(x) = sin√2x
sin√2
Question 20
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 this functional is given by
d
dx ∂f
∂y′−∂f
∂y = 0
where f= (y′)2−2y.
Step 2: Compute the partial derivatives. We compute the partial derivatives
of fwith respect to yand y′:
∂f
∂y =−2 and ∂f
∂y′= 2y′
Step 3: Apply the Euler-Lagrange equation. Substitute the partial deriva-
tives into the Euler-Lagrange equation to get
d
dx(2y′) + 2 = 0
Step 4: Simplify the equation. Simplify the equation to get
2y′′ + 2 = 0
20
Step 5: Solve the differential equation. Solving the differential equation gives
us y(x) = −1
2x2+3
2x.
Step 6: Check the boundary conditions. Using the boundary conditions
y(0) = 0 and y(1) = 1, we find that the extremal that satisfies these conditions
is
y(x) = −1
2x2+3
2x
Therefore, the extremals of the given functional subject to the boundary
conditions are the functions of the form y(x) = −1
2x2+3
2x.
Question 21
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
Applying this to the given functional, we have
d
dx (2y′)−(−2y) = 0
2y′′ + 2y= 0
Step 2: Solve the differential equation. The general solution to the differen-
tial equation 2y′′ + 2y= 0 is
y(x) = c1cos(x) + c2sin(x)
Step 3: Apply the boundary conditions. Using the boundary conditions
y(0) = 0 and y(1) = 1, we get
c1= 0
c2sin(1) = 1
c2=1
sin(1)
Step 4: Final extremal. Therefore, the extremal for the given functional
subject to the boundary conditions is
y(x) = sin(x)
sin(1)
21
Question 22
Question
Find the function y(x) that minimizes 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: Compute the Euler-Lagrange equation. Let f=y′2−y′. The Euler-
Lagrange equation is given by
d
dx ∂f
∂y′−∂f
∂y = 0
d
dx(2y′−1) −(−1) = 0
2y′′ = 1
y′′ =1
2
Step 2: Solve the differential equation. Integrating y′′ =1
2twice gives
y′=x
2+A
y=x2
4+Ax +B
Step 3: Apply the boundary conditions. From y(1) = 0, we have
0 = 1
4+A+B
From y(2) = 1, we have
1 = 1 + 2A+B
Solving these equations simultaneously gives A=−1
2and B=1
4.
Therefore, the function that minimizes the functional is
y(x) = x2
4−1
2x+1
4
22
Question 23
Question
Minimize the functional
J[y] = Z1
0
(y′2−y2)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 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−y2, so
∂f
∂y =−2yand ∂f
∂y′= 2y′
and d
dx (2y′)−(−2y) = 0
2y′′ + 2y= 0
Step 2: Solve the differential equation.
The general solution to the differential equation 2y′′+2y= 0 is y(x) = c1cos(x)+
c2sin(x), where c1and c2are constants.
Applying the boundary conditions, we have:
y(0) = c1= 1 and y(1) = c1cos(1) + c2sin(1) = 2
From c1= 1, we get cos(1) + c2sin(1) = 2.
Step 3: Solve for the remaining constant.
Solving for c2, we have:
c2=2−cos(1)
sin(1)
Hence, the solution to the variational problem is y(x) = cos(x)+ 2−cos(1)
sin(1) sin(x).
23
Question 24
Question
Find the extremals of the functional
J[y] = Z1
0
(y′2−yy′)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 a functional of the form J[y] = Rb
aF(x, y, y′)dx
is given by
d
dx ∂F
∂y′−∂F
∂y = 0
For our functional J[y] = R1
0(y′2−yy′)dx, we have F(x, y, y′) = y′2−yy′.
So,
d
dx ∂F
∂y′−∂F
∂y = 0
d
dx(2y′−y)−(−y′) = 0
2y′′ −y′+y′= 0
2y′′ = 0
y′′ = 0
Step 2: Solve the differential equation.
Integrating y′′ = 0 twice, we get y(x) = Ax +B, where Aand Bare constants.
Step 3: Apply the boundary conditions.
Using the boundary conditions y(0) = 0 and y(1) = 1, we have:
y(0) = A·0 + B= 0 =⇒B= 0
y(1) = A·1 + 0 = 1 =⇒A= 1
Therefore, the extremal of the functional J[y] that satisfies the boundary
conditions is y(x) = x.
Question 25
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.
24
Solution
Step 1: We define the Euler-Lagrange equation for this functional as
∂F
∂y −d
dx ∂F
∂y′= 0
where F(y, y′, x) = y2+y′2.
Step 2: Computing the necessary partial derivatives, we have
∂F
∂y = 2yand ∂F
∂y′= 2y′
Step 3: Differenciating w.r.t. xgives
d
dx ∂F
∂y′=d
dx(2y′)=2y′′
Step 4: Substituting back into the Euler-Lagrange equation, we get
∂F
∂y −d
dx ∂F
∂y′= 0 =⇒2y−2y′′ = 0
Step 5: Rearranging the equation gives the differential equation
y′′ =y
Step 6: The general solutions to this differential equation is
y(x) = Acos x+Bsin x
Step 7: Applying the boundary conditions y(0) = 0 and y(1) = 1, we find
y(0) = A= 0
y(1) = Bsin 1 = 1 =⇒B=1
sin 1
Step 8: Therefore, the extremal that minimizes the functional J[y] is
y(x) = sin x
sin 1
Question 26
Question
Find the extremals of the functional
J[y] = Zx2
x1
(y′2−y2)dx
subject to the boundary conditions y(x1) = Aand y(x2) = B, where Aand B
are constants.
25
Solution
Step 1: Set up the Euler-Lagrange equation. The Euler-Lagrange equation for
the given functional is
d
dx ∂F
∂y′−∂F
∂y = 0
where F=y′2−y2.
Step 2: Compute the partial derivatives. We have
∂F
∂y′= 2y′and ∂F
∂y =−2y
Now, differentiate ∂F
∂y′with respect to x:
d
dx ∂F
∂y′=d
dx(2y′)=2y′′
Step 3: Substitute into the Euler-Lagrange equation. We substitute the
partial derivatives into the Euler-Lagrange equation to get
d
dx(2y′)−(−2y)=0
which simplifies to
2y′′ + 2y= 0
Step 4: Solve the differential equation. The general solution to the differen-
tial equation y′′ +y= 0 is
y(x) = c1cos x+c2sin x
where c1and c2are arbitrary constants.
Step 5: Apply the boundary conditions. Using the boundary conditions
y(x1) = Aand y(x2) = B, we find that
c1cos x1+c2sin x1=A
c1cos x2+c2sin x2=B
Step 6: Solve for c1and c2. Solve the above system of equations to find the
values of c1and c2.
Step 7: Write down the extremals. The extremals of the functional J[y] are
given by
y(x) = c1cos x+c2sin x
where c1and c2are determined by the boundary conditions.
Question 27
Question
Find the extremals for the functional J[y] = R1
0(y′2−2y)dx subject to the
boundary conditions y(0) = 0 and y(1) = 1.
26
Solution
Step 1: Define the Euler-Lagrange equation. This is given by
d
dx ∂f
∂y′−∂f
∂y = 0
where f=y′2−2y.
Step 2: Calculate the partial derivatives of fwith respect to yand y′.
∂f
∂y =∂
∂y (−2y) = −2 and ∂f
∂y′=∂
∂y′(y′2)=2y′
Step 3: Apply the Euler-Lagrange equation to find the extremal.
d
dx(2y′)−(−2) = 0
d
dx(2y′) + 2 = 0
2y′′ + 2 = 0
y′′ =−1
Step 4: Solve the differential equation y′′ =−1.
y′′ =−1 =⇒y′=−x+C1
Integrating with respect to x, we get:
y=−1
2x2+C1x+C2
Step 5: Apply the boundary conditions y(0) = 0 and y(1) = 1 to find C1
and C2. When x= 0:
0 = −1
2(0)2+C1(0) + C2
0 = C2
When x= 1:
1 = −1
2(1)2+C1(1) + 0
1 = −1
2+C1
C1=3
2
Step 6: Substitute C1=3
2and C2= 0 back into the general solution to find
the extremal.
y=−1
2x2+3
2x
So, the extremal for the given functional is y=−1
2x2+3
2x.
27
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: We begin by setting up the Euler-Lagrange equation. Let F=y′2−y2.
The Euler-Lagrange equation is given by
d
dx ∂F
∂y′−∂F
∂y = 0
Plugging in F=y′2−y2, we have
d
dx (2y′)−(−2y) = 0
d
dx(2y′)+2y= 0
2y′′ + 2y= 0
Step 2: The solution to this differential equation is of the form y(x) =
Asin(x) + Bcos(x). Applying the boundary conditions y(0) = 0 and y(1) = 1,
we find
y(0) = B= 0
y(1) = Asin(1) = 1
A=1
sin(1)
Therefore, the extremal of the functional J[y] is y(x) = sin(x)
sin(1) .
Question 29
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.
28
Solution
Step 1: Define 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
Step 2: Plug in the values of a,b,F(x, y, y′) into the Euler-Lagrange equa-
tion. Given J[y] = R1
0(y′2+y2)dx with a= 0 and b= 1, and F(x, y, y′) =
y′2+y2, we have:
d
dx ∂
∂y′(y′2+y2)−∂
∂y (y′2+y2)=0
Step 3: Compute the derivatives and simplify the Euler-Lagrange equation.
Differentiating Fwith respect to y′, we get:
d
dx(2y′)=2y′′
Differentiating Fwith respect to y, we get:
−2y
Substitute these into the Euler-Lagrange equation:
d
dx(2y′)−(−2y)=0
2y′′ + 2y= 0
y′′ +y= 0
Step 4: Find the general solution of the differential equation y′′ +y= 0.
The general solution of the differential equation y′′ +y= 0 is given by:
y(x) = c1cos(x) + c2sin(x)
Step 5: Apply the boundary conditions y(0) = 0 and y(1) = 1 to find the
specific solution. Using the boundary condition y(0) = 0:
c1= 0
Using the boundary condition y(1) = 1:
c1cos(1) + c2sin(1) = 1
c2sin(1) = 1
c2=1
sin(1)
Therefore, the extremal of the functional is y(x) = 1
sin(1) sin(x).
29
Question 30
Question
Find the extremal for the functional
J[y] = Z1
0
(y′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 the given functional J[y] is given by:
d
dx ∂f
∂y′−∂f
∂y = 0
where f(y, y′, x) = y′2−2y.
Step 2: Compute the partial derivatives. We compute the partial derivatives
in the Euler-Lagrange equation:
∂f
∂y′= 2y′and ∂f
∂y =−2
Step 3: Apply the Euler-Lagrange equation. Substitute the partial deriva-
tives into the Euler-Lagrange equation and simplify:
d
dx(2y′) + 2 = 0
2y′′ + 2 = 0
Step 4: Solve the second-order ODE. Solving the ODE 2y′′ + 2 = 0 gives:
y′′ =−1
Step 5: Integrate the ODE twice. Integrating y′′ =−1 twice gives:
y′=−x+c1
y=−1
2x2+c1x+c2
Step 6: Apply the boundary conditions. Using the boundary conditions
y(0) = 1 and y(1) = 2, we have:
y(0) = c2= 1
y(1) = −1
2+c1+ 1 = 2
Step 7: Solve for the constants. Solving −1
2+c1+ 1 = 2 gives c1=5
2.
Step 8: Determine the extremal function. Therefore, the extremal for the
functional J[y] subject to the given boundary conditions is:
y=−1
2x2+5
2x+ 1
30
Question 31
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
To find the extremals of the functional J[y], we need to solve the Euler-Lagrange
equation:
d
dx ∂f
∂y′−∂f
∂y = 0
where f(y, y′, x) = y′2−y.
Step 1: Find ∂f ∂y′:
∂f
∂y′=∂
∂y′(y′2−y) = 2y′
Step 2: Find ddx ∂f
∂y′:
d
dx ∂f
∂y′=d
dx(2y′)=2y′′
Step 3: Find ∂f ∂y:
∂f
∂y =∂
∂y (y′2−y) = −1
Step 4: Set up the Euler-Lagrange equation:
2y′′ + 1 = 0
Step 5: Solve the differential equation: Solving the differential equation
gives us y(x) = Ax2+Bx+C, where A,B, and Care constants to be determined.
Step 6: Apply the boundary conditions: Using y(0) = 0 and y(1) = 1,
we get the system of equations:
C= 0
A+B= 1
Solving the system, we find A= 1 and B= 0.
Step 7: Final solution: Therefore, the extremal for the functional J[y]
subject to the given boundary conditions is y(x) = x2.
31
Question 32
Question
Let J[y] = R1
0(y′2−y2)dx be a functional. Find the function y(x) that minimizes
J[y] subject to the boundary conditions y(0) = 0 and y(1) = 1.
Solution
Step 1: Calculate the Euler-Lagrange equation.
d
dx ∂f
∂y′−∂f
∂y = 0
where f=y′2−y2.
Step 2: Compute the partial derivatives.
∂f
∂y′= 2y′
d
dx ∂f
∂y′= 2y′′
∂f
∂y =−2y
Step 3: Write the Euler-Lagrange equation.
2y′′ + 2y= 0
Step 4: Solve the differential equation. The solution to the differential equa-
tion is a linear combination of sinh xand cosh x. Let y(x) = c1sinh x+c2cosh x.
Step 5: Apply the boundary conditions. Using the boundary conditions
y(0) = 0 and y(1) = 1:
0 = c2
1 = c1sinh 1
c1=1
sinh 1
Step 6: Obtain the function y(x) that minimizes J[y]. Therefore, the func-
tion y(x) that minimizes J[y] subject to the boundary conditions y(0) = 0 and
y(1) = 1 is:
y(x) = sinh x
sinh 1
32
Question 33
Question
Find the extremal for the functional
J[y] = Z1
02y−y′2dx
subject to the boundary conditions y(0) = 1 and y(1) = 2.
Solution
Step 1: Write down 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
Applying this to the given functional J[y] where F(x, y, y′)=2y−y′2:
d
dx ∂F
∂y′−∂F
∂y = 0
d
dx (2 −2y′)−0=0
d
dx(−2y′)=0
−2y′′ = 0
Step 2: Solve the differential equation y′′ = 0. Integrating −2y′′ = 0 once:
−2y′=C1
Integrating again:
−2y=C1x+C2
Now, we apply the boundary conditions y(0) = 1 and y(1) = 2: From y(0) = 1:
−2(0) = C1(0) + C2
C2= 1
From y(1) = 2:
−2(1) = C1(1) + 1
C1=−3
Therefore, the extremal for the given functional is:
y(x) = −3x+ 1
33
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.
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 of F(x, y, y′).
In this case, F(x, y, y′) = y′2−y2, so we have
∂F
∂y =−2y
and ∂F
∂y′= 2y′
Step 3: Apply the Euler-Lagrange equation.
Substitute the partial derivatives into the Euler-Lagrange equation and we get
d
dx(2y′)−(−2y)=0
which simplifies to
2y′′ + 2y= 0
Step 4: Solve the differential equation.
The general solution to this second order linear differential equation is
y(x) = c1cos(x) + c2sin(x)
Step 5: Apply the boundary conditions.
Using the boundary condition y(0) = 0, we have c1= 0.
Using the boundary condition y(1) = 1, we get
1 = c2sin(1)
34
so c2=1
sin(1) .
Step 6: Final solution.
Therefore, the extremals of the functional J[y] subject to the given boundary
conditions are
y(x) = 1
sin(1) sin(x)
Question 35
Question
Find the function y(x) that minimizes the functional
J[y] = Z1
0y′2+y2dx
subject to the boundary conditions y(0) = 1 and y(1) = 2.
Solution
Step 1: Set up the Euler-Lagrange equation. Let F=y′2+y2. The Euler-
Lagrange equation is given by
d
dx ∂F
∂y′−∂F
∂y = 0
Step 2: Compute the partial derivatives.
∂F
∂y′=∂
∂y′(y′2+y2)=2y′
d
dx ∂F
∂y′=d
dx(2y′)=2y′′
Step 3: Compute the partial derivative with respect to y.
∂F
∂y =∂
∂y (y′2+y2)=2y
Step 4: Plug these derivatives into the Euler-Lagrange equation.
2y′′ −2y= 0
Step 5: Solve the differential equation. The general solution to the differen-
tial equation is of the form y(x) = c1cos(x) + c2sin(x).
Step 6: Apply the boundary conditions. Given y(0) = 1 and y(1) = 2, we
have the system of equations
(c1= 1
c1cos(1) + c2sin(1) = 2
35
Question 10
Question
Find the extremal of the functional
J[y] = Z1
0
y(x)2+y′(x)2dx
subject to the boundary conditions y(0) = 0 and y(1) = 1.
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: d
dx ∂L
∂y′−∂L
∂y = 0.
Step 3: Compute the partial derivatives in the Euler-Lagrange equation:
∂L
∂y = 2yand ∂L
∂y′= 2y′
Step 4: Compute the derivative with respect to xof ∂L
∂y′:
d
dx ∂L
∂y′=d
dx(2y′)=2y′′
Step 5: Apply the Euler-Lagrange equation:
2y′′ −2y= 0
Step 6: Solve the differential equation 2y′′ −2y= 0 with the boundary
conditions y(0) = 0 and y(1) = 1.
Step 7: The general solution to the differential equation is y(x) = c1cos(x)+
c2sin(x).
Step 8: Apply the boundary conditions to find the values of c1and c2:
y(0) = c1= 0 and y(1) = c2sin(1) = 1
Step 9: The solution is y(x) = sin(x).
Question 11
Question
Find the extremals of the functional J[y] = R1
0(y2+y′2)dx subject to the bound-
ary conditions y(0) = 0 and y(1) = 1.
10
Solution
Step 1: Compute the Euler-Lagrange equation. To find the extremals of J[y],
we need to solve the Euler-Lagrange equation:
d
dx ∂f
∂y′−∂f
∂y = 0
where f=y2+y′2. Computing the partial derivatives, we get:
∂f
∂y = 2yand ∂f
∂y′= 2y′
d
dx ∂f
∂y′=d
dx(2y′)=2y′′
So the Euler-Lagrange equation simplifies to:
2y′′ −2y= 0
Step 2: Solve the differential equation. The differential equation 2y′′−2y= 0
simplifies to y′′ −y= 0. The characteristic equation is r2−1 = 0, which has
roots r= 1 and r=−1. Therefore, the general solution to the 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 find:
(c1+c2= 0
c1e+c2e−1= 1
Solving this system of equations gives c1=e
e−1and c2=−1
e−1.
Therefore, the extremal that minimizes 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 extremals for the functional
J[y] = Z1
0
(y′2+y2)dx
subject to the boundary conditions y(0) = 0 and y(1) = 1.
11
Solution
Step 1: Define the Euler-Lagrange equation.
For the functional J[y] = R1
0(y′2+y2)dx, the Euler-Lagrange equation is given
by
d
dx ∂f
∂y′−∂f
∂y = 0
where f(y, y′, x) = y′2+y2.
Step 2: Calculate the partial derivatives.
Let’s calculate the partial derivatives needed for the Euler-Lagrange equation:
∂f
∂y′=∂
∂y′(y′2+y2)=2y′
∂f
∂y =∂
∂y (y′2+y2)=2y
Step 3: Form the Euler-Lagrange equation.
Substitute the partial derivatives into the Euler-Lagrange equation:
d
dx(2y′)−2y= 0
d
dx(2y′)−2y= 0
Step 4: Solve the Euler-Lagrange equation.
Solving the differential equation, we get:
2y′′ −2y= 0
y′′ −y= 0
Step 5: Solve the differential equation with the boundary conditions.
The general solution to the differential equation is y(x) = c1ex+c2e−x. Applying
the boundary conditions y(0) = 0 and y(1) = 1:
c1+c2= 0
c1e+c2e−1= 1
Solving these equations gives c1=e−1
1−e−2and c2=−e
1−e−2.
Step 6: Write down the extremal function.
Therefore, the extremal for the functional is
y(x) = e−1
1−e−2ex−e
1−e−2e−x
12
Question 13
Question
Find the extremum of 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: Set up the Euler-Lagrange equation for the given functional. The
Euler-Lagrange equation for a functional J[y] is given by:
d
dx ∂f
∂y′−∂f
∂y = 0
where f(y, y′, x) = y2+y′2in this case.
Step 2: Compute the partial derivatives. Let f(y, y′, x) = y2+y′2. Then,
∂f
∂y = 2yand ∂f
∂y′= 2y′
d
dx ∂f
∂y′=d
dx(2y′)=2y′′
Step 3: Plug into the Euler-Lagrange equation. Substitute the partial
derivatives into the Euler-Lagrange equation to get:
d
dx(2y′)−2y= 0
2y′′ −2y= 0
Step 4: Solve the differential equation. The solution to the differential
equation 2y′′ −2y= 0 is of the form y(x) = Acos√2x+Bsin√2x.
Step 5: Apply the boundary conditions. Using the boundary conditions
y(0) = 0 and y(1) = 1, we find:
0 = Acos(0) + Bsin(0) ⇒A= 0
1 = Bsin√2⇒B=1
sin√2
Step 6: Determine the extremum. The extremum of the functional J[y] is
given by the function y(x) = sin(√2x)
sin(√2)which minimizes J[y] subject to the given
boundary conditions.
13
Question 14
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) = 2.
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. We have:
∂L
∂y′= 2y′
∂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
Step 5: Solve the differential equation. Integrating the equation gives:
2y′= 2y+C
y′=y+C1
Step 6: Apply the boundary conditions. Using the boundary conditions
y(0) = 1, we find C1= 0. So, the solution is:
y′=y
Step 7: Solve the differential equation. The general solution to the differ-
ential equation y′=yis:
y(x) = Aex
14
Step 8: Apply the boundary condition y(1) = 2. Substitute the boundary
condition to find A:
y(1) = Ae = 2
A= 2/e
Step 9: Final solution. Therefore, the extremal that minimizes the func-
tional J[y] subject to the given boundary conditions is:
y(x) = 2
e·ex= 2ex−1
Question 15
Question
Consider the functional J[y] = R1
0(y′2−y2)dx subject to the boundary condi-
tions y(0) = 1 and y(1) = 2. Find the function y(x) that minimizes J[y].
Solution
Step 1: Write the Euler-Lagrange equation Given the functional J[y] = R1
0(y′2−
y2)dx, 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 The partial derivative ∂f
∂y′is 2y′and
∂f
∂y is −2y.
Step 3: Apply the Euler-Lagrange equation Substitute the partial derivatives
into the Euler-Lagrange equation:
d
dx2y′+ 2y= 0
Step 4: Solve the differential equation Rearranging the equation gives d
dx (2y′) =
−2y. Integrating both sides with respect to xyields
2y′=−2y+C
where Cis the constant of integration.
Step 5: Solve for yTo find y, solve the first-order ordinary differential equa-
tion 2y′=−2y+Cwith the initial conditions y(0) = 1.
Step 6: Apply the boundary conditions Use the boundary conditions y(0) =
1 and y(1) = 2 to solve for the constant of integration C.
Step 7: Finalize the solution After finding the constant of integration C,
substitute it back into the expression for y(x) to get the final function that
minimizes J[y].
15
Question 16
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) = 2.
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: Calculate the partial derivatives. We have:
∂f
∂y′=∂
∂y′(y′2−2y)=2y′
and d
dx ∂f
∂y′=d
dx(2y′)=2y′′
∂f
∂y =∂
∂y (y′2−2y) = −2
Step 3: Apply the Euler-Lagrange equation. Substitute the derivatives into
the Euler-Lagrange equation:
2y′′ + 2 = 0
which simplifies to
y′′ =−1
Step 4: Solve the differential equation. Integrating twice, we get:
y′=−x+C1
y=−x2
2+C1x+C2
Step 5: Apply the boundary conditions. Using the boundary conditions
y(0) = 1 and y(1) = 2, we find:
C2= 1
−1
2+C1+ 1 = 2 ⇒C1=3
2
Step 6: Determine the extremals. The extremal that minimizes the func-
tional is:
y=−x2
2+3
2x+ 1
16
Question 17
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
To find the extremals of the functional J[y], we will use the Euler-Lagrange
equation:
d
dx ∂f
∂y′−∂f
∂y = 0
where f=y′2−y2.
Step 1: Compute ∂f ∂y′and ∂f
∂y .
∂f
∂y′= 2y′
∂f
∂y =−2y
Step 2: Apply the Euler-Lagrange equation. Plugging into the Euler-
Lagrange equation, we get:
d
dx(2y′)+2y= 0
2y′′ + 2y= 0
Step 3: Solve the differential equation. The solution to this differential
equation is of the form y(x) = Asin x+Bcos x. Applying boundary conditions
y(0) = 0 and y(1) = 1:
y(0) = B= 0
y(1) = Asin 1 = 1
A=1
sin 1
Therefore, the extremal of the functional is given by:
y(x) = sin x
sin 1
17
Question 18
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 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, we have
F(x, y, y′) = y′2−2y,
so the Euler-Lagrange equation becomes
d
dx(2y′)−(−2) = 0
which simplifies to
2y′′ + 2 = 0.
Step 2: Solve the differential equation. Solving the differential equation
2y′′ + 2 = 0 gives
y′′ =−1.
Integrating once gives
y′=−x+c1,
where c1is a constant of integration. Integrating again yields
y=−1
2x2+c1x+c2,
where c2is another constant of integration.
Step 3: Apply the boundary conditions. We have y(0) = 0 and y(1) = 1.
Substituting these into the expression for y, we get the system of equations
c2= 0
18
−1
2+c1= 1.
Solving this system gives c1=3
2.
Step 4: Determine the extremals. The extremals are given by the solution
to the differential equation with the values of c1and c2determined from the
boundary conditions. Therefore, the extremal for the functional J[y] is
y=−1
2x2+3
2x.
Question 19
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. 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
In this case, our functional is J[y] = R1
0(y′2−y2)dx, so F(x, y, y′) = y′2−y2.
Step 2: Compute the partial derivatives needed for the Euler-Lagrange
equation. We need to compute ∂F
∂y′and ∂F
∂y .
∂F
∂y′=∂
∂y′(y′2−y2)=2y′
∂F
∂y =∂
∂y (y′2−y2) = −2y
Step 3: Apply the Euler-Lagrange equation. Substitute the derivatives into
the Euler-Lagrange equation:
d
dx (2y′)−(−2y) = 0
Simplify this equation to get:
2y′′ + 2y= 0
19
Step 4: Solve the differential equation. The general solution to 2y′′ +2y= 0
is y(x) = c1cos√2x+c2sin√2x.
Step 5: Apply the boundary conditions. Using the boundary conditions
y(0) = 0 and y(1) = 1, we find:
c1= 0 and c2sin√2= 1
This gives us c2=1
sin(√2).
Step 6: Write the solution to the variational problem. Therefore, the ex-
tremal for the given functional subject to the boundary conditions is:
y(x) = sin√2x
sin√2
Question 20
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 this functional is given by
d
dx ∂f
∂y′−∂f
∂y = 0
where f= (y′)2−2y.
Step 2: Compute the partial derivatives. We compute the partial derivatives
of fwith respect to yand y′:
∂f
∂y =−2 and ∂f
∂y′= 2y′
Step 3: Apply the Euler-Lagrange equation. Substitute the partial deriva-
tives into the Euler-Lagrange equation to get
d
dx(2y′) + 2 = 0
Step 4: Simplify the equation. Simplify the equation to get
2y′′ + 2 = 0
20
Step 5: Solve the differential equation. Solving the differential equation gives
us y(x) = −1
2x2+3
2x.
Step 6: Check the boundary conditions. Using the boundary conditions
y(0) = 0 and y(1) = 1, we find that the extremal that satisfies these conditions
is
y(x) = −1
2x2+3
2x
Therefore, the extremals of the given functional subject to the boundary
conditions are the functions of the form y(x) = −1
2x2+3
2x.
Question 21
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
Applying this to the given functional, we have
d
dx (2y′)−(−2y) = 0
2y′′ + 2y= 0
Step 2: Solve the differential equation. The general solution to the differen-
tial equation 2y′′ + 2y= 0 is
y(x) = c1cos(x) + c2sin(x)
Step 3: Apply the boundary conditions. Using the boundary conditions
y(0) = 0 and y(1) = 1, we get
c1= 0
c2sin(1) = 1
c2=1
sin(1)
Step 4: Final extremal. Therefore, the extremal for the given functional
subject to the boundary conditions is
y(x) = sin(x)
sin(1)
21
Question 22
Question
Find the function y(x) that minimizes 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: Compute the Euler-Lagrange equation. Let f=y′2−y′. The Euler-
Lagrange equation is given by
d
dx ∂f
∂y′−∂f
∂y = 0
d
dx(2y′−1) −(−1) = 0
2y′′ = 1
y′′ =1
2
Step 2: Solve the differential equation. Integrating y′′ =1
2twice gives
y′=x
2+A
y=x2
4+Ax +B
Step 3: Apply the boundary conditions. From y(1) = 0, we have
0 = 1
4+A+B
From y(2) = 1, we have
1 = 1 + 2A+B
Solving these equations simultaneously gives A=−1
2and B=1
4.
Therefore, the function that minimizes the functional is
y(x) = x2
4−1
2x+1
4
22
Question 23
Question
Minimize the functional
J[y] = Z1
0
(y′2−y2)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 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−y2, so
∂f
∂y =−2yand ∂f
∂y′= 2y′
and d
dx (2y′)−(−2y) = 0
2y′′ + 2y= 0
Step 2: Solve the differential equation.
The general solution to the differential equation 2y′′+2y= 0 is y(x) = c1cos(x)+
c2sin(x), where c1and c2are constants.
Applying the boundary conditions, we have:
y(0) = c1= 1 and y(1) = c1cos(1) + c2sin(1) = 2
From c1= 1, we get cos(1) + c2sin(1) = 2.
Step 3: Solve for the remaining constant.
Solving for c2, we have:
c2=2−cos(1)
sin(1)
Hence, the solution to the variational problem is y(x) = cos(x)+ 2−cos(1)
sin(1) sin(x).
23
Question 24
Question
Find the extremals of the functional
J[y] = Z1
0
(y′2−yy′)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 a functional of the form J[y] = Rb
aF(x, y, y′)dx
is given by
d
dx ∂F
∂y′−∂F
∂y = 0
For our functional J[y] = R1
0(y′2−yy′)dx, we have F(x, y, y′) = y′2−yy′.
So,
d
dx ∂F
∂y′−∂F
∂y = 0
d
dx(2y′−y)−(−y′) = 0
2y′′ −y′+y′= 0
2y′′ = 0
y′′ = 0
Step 2: Solve the differential equation.
Integrating y′′ = 0 twice, we get y(x) = Ax +B, where Aand Bare constants.
Step 3: Apply the boundary conditions.
Using the boundary conditions y(0) = 0 and y(1) = 1, we have:
y(0) = A·0 + B= 0 =⇒B= 0
y(1) = A·1 + 0 = 1 =⇒A= 1
Therefore, the extremal of the functional J[y] that satisfies the boundary
conditions is y(x) = x.
Question 25
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.
24
Solution
Step 1: We define the Euler-Lagrange equation for this functional as
∂F
∂y −d
dx ∂F
∂y′= 0
where F(y, y′, x) = y2+y′2.
Step 2: Computing the necessary partial derivatives, we have
∂F
∂y = 2yand ∂F
∂y′= 2y′
Step 3: Differenciating w.r.t. xgives
d
dx ∂F
∂y′=d
dx(2y′)=2y′′
Step 4: Substituting back into the Euler-Lagrange equation, we get
∂F
∂y −d
dx ∂F
∂y′= 0 =⇒2y−2y′′ = 0
Step 5: Rearranging the equation gives the differential equation
y′′ =y
Step 6: The general solutions to this differential equation is
y(x) = Acos x+Bsin x
Step 7: Applying the boundary conditions y(0) = 0 and y(1) = 1, we find
y(0) = A= 0
y(1) = Bsin 1 = 1 =⇒B=1
sin 1
Step 8: Therefore, the extremal that minimizes the functional J[y] is
y(x) = sin x
sin 1
Question 26
Question
Find the extremals of the functional
J[y] = Zx2
x1
(y′2−y2)dx
subject to the boundary conditions y(x1) = Aand y(x2) = B, where Aand B
are constants.
25
Solution
Step 1: Set up the Euler-Lagrange equation. The Euler-Lagrange equation for
the given functional is
d
dx ∂F
∂y′−∂F
∂y = 0
where F=y′2−y2.
Step 2: Compute the partial derivatives. We have
∂F
∂y′= 2y′and ∂F
∂y =−2y
Now, differentiate ∂F
∂y′with respect to x:
d
dx ∂F
∂y′=d
dx(2y′)=2y′′
Step 3: Substitute into the Euler-Lagrange equation. We substitute the
partial derivatives into the Euler-Lagrange equation to get
d
dx(2y′)−(−2y)=0
which simplifies to
2y′′ + 2y= 0
Step 4: Solve the differential equation. The general solution to the differen-
tial equation y′′ +y= 0 is
y(x) = c1cos x+c2sin x
where c1and c2are arbitrary constants.
Step 5: Apply the boundary conditions. Using the boundary conditions
y(x1) = Aand y(x2) = B, we find that
c1cos x1+c2sin x1=A
c1cos x2+c2sin x2=B
Step 6: Solve for c1and c2. Solve the above system of equations to find the
values of c1and c2.
Step 7: Write down the extremals. The extremals of the functional J[y] are
given by
y(x) = c1cos x+c2sin x
where c1and c2are determined by the boundary conditions.
Question 27
Question
Find the extremals for the functional J[y] = R1
0(y′2−2y)dx subject to the
boundary conditions y(0) = 0 and y(1) = 1.
26
Solution
Step 1: Define the Euler-Lagrange equation. This is given by
d
dx ∂f
∂y′−∂f
∂y = 0
where f=y′2−2y.
Step 2: Calculate the partial derivatives of fwith respect to yand y′.
∂f
∂y =∂
∂y (−2y) = −2 and ∂f
∂y′=∂
∂y′(y′2)=2y′
Step 3: Apply the Euler-Lagrange equation to find the extremal.
d
dx(2y′)−(−2) = 0
d
dx(2y′) + 2 = 0
2y′′ + 2 = 0
y′′ =−1
Step 4: Solve the differential equation y′′ =−1.
y′′ =−1 =⇒y′=−x+C1
Integrating with respect to x, we get:
y=−1
2x2+C1x+C2
Step 5: Apply the boundary conditions y(0) = 0 and y(1) = 1 to find C1
and C2. When x= 0:
0 = −1
2(0)2+C1(0) + C2
0 = C2
When x= 1:
1 = −1
2(1)2+C1(1) + 0
1 = −1
2+C1
C1=3
2
Step 6: Substitute C1=3
2and C2= 0 back into the general solution to find
the extremal.
y=−1
2x2+3
2x
So, the extremal for the given functional is y=−1
2x2+3
2x.
27
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: We begin by setting up the Euler-Lagrange equation. Let F=y′2−y2.
The Euler-Lagrange equation is given by
d
dx ∂F
∂y′−∂F
∂y = 0
Plugging in F=y′2−y2, we have
d
dx (2y′)−(−2y) = 0
d
dx(2y′)+2y= 0
2y′′ + 2y= 0
Step 2: The solution to this differential equation is of the form y(x) =
Asin(x) + Bcos(x). Applying the boundary conditions y(0) = 0 and y(1) = 1,
we find
y(0) = B= 0
y(1) = Asin(1) = 1
A=1
sin(1)
Therefore, the extremal of the functional J[y] is y(x) = sin(x)
sin(1) .
Question 29
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.
28
Solution
Step 1: Define 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
Step 2: Plug in the values of a,b,F(x, y, y′) into the Euler-Lagrange equa-
tion. Given J[y] = R1
0(y′2+y2)dx with a= 0 and b= 1, and F(x, y, y′) =
y′2+y2, we have:
d
dx ∂
∂y′(y′2+y2)−∂
∂y (y′2+y2)=0
Step 3: Compute the derivatives and simplify the Euler-Lagrange equation.
Differentiating Fwith respect to y′, we get:
d
dx(2y′)=2y′′
Differentiating Fwith respect to y, we get:
−2y
Substitute these into the Euler-Lagrange equation:
d
dx(2y′)−(−2y)=0
2y′′ + 2y= 0
y′′ +y= 0
Step 4: Find the general solution of the differential equation y′′ +y= 0.
The general solution of the differential equation y′′ +y= 0 is given by:
y(x) = c1cos(x) + c2sin(x)
Step 5: Apply the boundary conditions y(0) = 0 and y(1) = 1 to find the
specific solution. Using the boundary condition y(0) = 0:
c1= 0
Using the boundary condition y(1) = 1:
c1cos(1) + c2sin(1) = 1
c2sin(1) = 1
c2=1
sin(1)
Therefore, the extremal of the functional is y(x) = 1
sin(1) sin(x).
29
Question 30
Question
Find the extremal for the functional
J[y] = Z1
0
(y′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 the given functional J[y] is given by:
d
dx ∂f
∂y′−∂f
∂y = 0
where f(y, y′, x) = y′2−2y.
Step 2: Compute the partial derivatives. We compute the partial derivatives
in the Euler-Lagrange equation:
∂f
∂y′= 2y′and ∂f
∂y =−2
Step 3: Apply the Euler-Lagrange equation. Substitute the partial deriva-
tives into the Euler-Lagrange equation and simplify:
d
dx(2y′) + 2 = 0
2y′′ + 2 = 0
Step 4: Solve the second-order ODE. Solving the ODE 2y′′ + 2 = 0 gives:
y′′ =−1
Step 5: Integrate the ODE twice. Integrating y′′ =−1 twice gives:
y′=−x+c1
y=−1
2x2+c1x+c2
Step 6: Apply the boundary conditions. Using the boundary conditions
y(0) = 1 and y(1) = 2, we have:
y(0) = c2= 1
y(1) = −1
2+c1+ 1 = 2
Step 7: Solve for the constants. Solving −1
2+c1+ 1 = 2 gives c1=5
2.
Step 8: Determine the extremal function. Therefore, the extremal for the
functional J[y] subject to the given boundary conditions is:
y=−1
2x2+5
2x+ 1
30
Question 31
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
To find the extremals of the functional J[y], we need to solve the Euler-Lagrange
equation:
d
dx ∂f
∂y′−∂f
∂y = 0
where f(y, y′, x) = y′2−y.
Step 1: Find ∂f ∂y′:
∂f
∂y′=∂
∂y′(y′2−y) = 2y′
Step 2: Find ddx ∂f
∂y′:
d
dx ∂f
∂y′=d
dx(2y′)=2y′′
Step 3: Find ∂f ∂y:
∂f
∂y =∂
∂y (y′2−y) = −1
Step 4: Set up the Euler-Lagrange equation:
2y′′ + 1 = 0
Step 5: Solve the differential equation: Solving the differential equation
gives us y(x) = Ax2+Bx+C, where A,B, and Care constants to be determined.
Step 6: Apply the boundary conditions: Using y(0) = 0 and y(1) = 1,
we get the system of equations:
C= 0
A+B= 1
Solving the system, we find A= 1 and B= 0.
Step 7: Final solution: Therefore, the extremal for the functional J[y]
subject to the given boundary conditions is y(x) = x2.
31
Question 32
Question
Let J[y] = R1
0(y′2−y2)dx be a functional. Find the function y(x) that minimizes
J[y] subject to the boundary conditions y(0) = 0 and y(1) = 1.
Solution
Step 1: Calculate the Euler-Lagrange equation.
d
dx ∂f
∂y′−∂f
∂y = 0
where f=y′2−y2.
Step 2: Compute the partial derivatives.
∂f
∂y′= 2y′
d
dx ∂f
∂y′= 2y′′
∂f
∂y =−2y
Step 3: Write the Euler-Lagrange equation.
2y′′ + 2y= 0
Step 4: Solve the differential equation. The solution to the differential equa-
tion is a linear combination of sinh xand cosh x. Let y(x) = c1sinh x+c2cosh x.
Step 5: Apply the boundary conditions. Using the boundary conditions
y(0) = 0 and y(1) = 1:
0 = c2
1 = c1sinh 1
c1=1
sinh 1
Step 6: Obtain the function y(x) that minimizes J[y]. Therefore, the func-
tion y(x) that minimizes J[y] subject to the boundary conditions y(0) = 0 and
y(1) = 1 is:
y(x) = sinh x
sinh 1
32
Question 33
Question
Find the extremal for the functional
J[y] = Z1
02y−y′2dx
subject to the boundary conditions y(0) = 1 and y(1) = 2.
Solution
Step 1: Write down 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
Applying this to the given functional J[y] where F(x, y, y′)=2y−y′2:
d
dx ∂F
∂y′−∂F
∂y = 0
d
dx (2 −2y′)−0=0
d
dx(−2y′)=0
−2y′′ = 0
Step 2: Solve the differential equation y′′ = 0. Integrating −2y′′ = 0 once:
−2y′=C1
Integrating again:
−2y=C1x+C2
Now, we apply the boundary conditions y(0) = 1 and y(1) = 2: From y(0) = 1:
−2(0) = C1(0) + C2
C2= 1
From y(1) = 2:
−2(1) = C1(1) + 1
C1=−3
Therefore, the extremal for the given functional is:
y(x) = −3x+ 1
33
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.
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 of F(x, y, y′).
In this case, F(x, y, y′) = y′2−y2, so we have
∂F
∂y =−2y
and ∂F
∂y′= 2y′
Step 3: Apply the Euler-Lagrange equation.
Substitute the partial derivatives into the Euler-Lagrange equation and we get
d
dx(2y′)−(−2y)=0
which simplifies to
2y′′ + 2y= 0
Step 4: Solve the differential equation.
The general solution to this second order linear differential equation is
y(x) = c1cos(x) + c2sin(x)
Step 5: Apply the boundary conditions.
Using the boundary condition y(0) = 0, we have c1= 0.
Using the boundary condition y(1) = 1, we get
1 = c2sin(1)
34
so c2=1
sin(1) .
Step 6: Final solution.
Therefore, the extremals of the functional J[y] subject to the given boundary
conditions are
y(x) = 1
sin(1) sin(x)
Question 35
Question
Find the function y(x) that minimizes the functional
J[y] = Z1
0y′2+y2dx
subject to the boundary conditions y(0) = 1 and y(1) = 2.
Solution
Step 1: Set up the Euler-Lagrange equation. Let F=y′2+y2. The Euler-
Lagrange equation is given by
d
dx ∂F
∂y′−∂F
∂y = 0
Step 2: Compute the partial derivatives.
∂F
∂y′=∂
∂y′(y′2+y2)=2y′
d
dx ∂F
∂y′=d
dx(2y′)=2y′′
Step 3: Compute the partial derivative with respect to y.
∂F
∂y =∂
∂y (y′2+y2)=2y
Step 4: Plug these derivatives into the Euler-Lagrange equation.
2y′′ −2y= 0
Step 5: Solve the differential equation. The general solution to the differen-
tial equation is of the form y(x) = c1cos(x) + c2sin(x).
Step 6: Apply the boundary conditions. Given y(0) = 1 and y(1) = 2, we
have the system of equations
(c1= 1
c1cos(1) + c2sin(1) = 2
35
Step 7: Solve the system of equations to find c1and c2. Solving the system
of equations, we find c1= 1 and c2≈1.5574.
Therefore, the function that minimizes the functional J[y] is y(x) = cos(x)+
1.5574 sin(x).
36
