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2 Variational Problems with Fixed Boundaries
Example 2.6.1 Find the extremal curves of the functional
J [y, z] =
π
2
0
(2yz + y
2
+ z
2
)dx
the boundary conditions are y(0) = 0, y
π
2
= 1, z(0) = 0, z
π
2
= −1.
Solution Let the integrand F = 2yz + y
2
+ z
2 , then the Euler equations of the
functional are
y
− z = 0, z
− y = 0
Solve the second order linear differential equations, derivate twice to the front
equation, eliminate z
, we obtain
y
(4)
− y = 0
In the same way, we obtain
z
(4)
− z = 0
The general solutions are
y = c 1 e
x
+ c 2 e
−x
+ c 3 cos x + c 4 sin x
z = c 1 e
x
+ c 2 e
−x
− c 3 cos x − c 4 sin x
Using the boundary conditions, we obtain
c 1 = c 2 = c 3 = 0, c 4 = 1
So the extremal curves are
y = sin x, z = − sin x
Example 2.6.2 Find the extremal curve of the functional J [y, z] =
π
4
0 (2z − 4y
2
+ y
2
− z
2
)dx, the boundary conditions are y(0) = 0, y
π
4
= 1,
z(0) = 0, z
π
4
= 1.
Solution The Euler equations of the functional are
y
+ 4y = 0
z
+ 1 = 0
The integral results are