MA342H: Homework #3 solutions
1. Find the critical points of
J(y) =
Z
0
π/2 2
′
2
y (x) − y(x) − 4y(x) dx
subject to the constraints y(0) = 0 and y(π/2) = 1.
First of all, we need to solve the Euler-Lagrange equation
d
Ly ′ = Ly
dx
=⇒
2y ′′ (x) = −2y(x) − 4
=⇒
y ′′ (x) + y(x) = −2.
This is a second-order linear equation with constant coefficients whose solution is
y(x) = c1 sin x + c2 cos x − 2.
To ensure that the boundary conditions hold, we need to ensure that
0 = y(0) = c2 − 2,
1 = y(π/2) = c1 − 2.
This gives c2 = 2 and c1 = 3, so the only critical point is y(x) = 3 sin x + 2 cos x − 2.
2. Find the critical points of
J(y) =
Z
3
1
p
1 + y ′ (x)2 dx
x
subject to the constraints y(1) = 3 and y(3) = 5.
Since the Lagrangian is independent of y, the Euler-Lagrange equation gives
d
Ly ′ = Ly = 0
dx
=⇒
y ′ (x)
x
p
1 + y ′ (x)2
= Ly′ = c.
We square both sides and then rearrange terms to find that
y ′ (x)2 = c2 x2 + c2 x2 y ′ (x)2
=⇒
y ′ (x) = √
cx
.
1 − c 2 x2
This is a separable equation that we can simply integrate to get
√
1 − c 2 x2
1 − c 2 x2
y(x) = −
+ a =⇒ (y − a)2 =
= c−2 − x2 .
2
c
c
According to the given boundary conditions, we must also have
(3 − a)2 + 1 = c−2 = (5 − a)2 + 9
=⇒
=⇒
a2 − 6a + 10 = a2 − 10a + 34
4a = 24.
In particular, a = 6 and c−2 = 10, so the only critical point is
√
√
√
1 − c 2 x2
y(x) = a −
= a − c−2 − x2 = 6 − 10 − x2 .
c
3. Find the critical points of
J(y) =
subject to the constraints
Z 1
x · y(x) dx = 1,
Z
1
y ′ (x)2 dx
0
y(0) = 0,
y(1) = 1.
0
The critical points of I(y) =
R1
0
L(x, y, y ′ ) = xy
x · y(x) dx satisfy the Euler-Lagrange equation
=⇒
d
Ly ′ = Ly
dx
=⇒
0 = x,
so this functional has no critical points. The critical points of J(y) − λI(y) satisfy
L(x, y, y ′ ) = (y ′ )2 − λxy
=⇒
d
Ly ′ = Ly
dx
=⇒
2y ′′ (x) = −λx.
Once we now integrate this equation twice, we find that
y ′′ (x) = −
λx
2
=⇒
y ′ (x) = −
λx2
+a
4
=⇒
y(x) = −
λx3
+ ax + b.
12
The condition y(0) = 0 implies that b = 0 and the condition y(1) = 1 implies that
1=−
λ
+a
12
=⇒
λ = 12(a − 1).
In view of the integral condition I(y) = 1, we must also have
Z 1
λ
λx4
a
a−1 a
2a + 3
2
−
1=
+ ax dx = − + = −
+ =
.
12
60 3
5
3
15
0
This gives a = 6 and λ = 60, so the only critical point is y(x) = −5x3 + 6x.
4. Find the critical points of
J(y) =
Z 1
′
2
′
y (x) + y(x)y (x) + 4y(x) dx.
0
First of all, we need to solve the Euler-Lagrange equation
d
Ly ′ = Ly
dx
=⇒
2y ′′ (x) + y ′ (x) = y ′ (x) + 4
=⇒
y ′′ (x) = 2.
This equation can be easily integrated twice to give
y ′′ (x) = 2
=⇒
y ′ (x) = 2x + a
=⇒
y(x) = x2 + ax + b.
In view of the natural boundary conditions, we must also have
0 = Ly′ = 2y ′ (x) + y(x) = 2(2x + a) + x2 + ax + b
when x = 0, 1. This gives 2a + b = 0 = 3a + b + 5, so it easily follows that
a = −5
=⇒
b = −2a = 10
=⇒
y(x) = x2 − 5x + 10.