2.4 The Euler Equations of the Simplest Functional
117
J [y] =
2
1
2
x 2 + x
2 4
x 4
dx =
2
1
6
x 2 dx = 3
Example 2.4.2 Find the extremal curve of the functional J [y] =
π
2
0 (y
2
− y
2
)dx,
the boundary conditions are y(0) = 0, y
π
2
= 1.
Solution Let the integrand F = y
2
− y
2 , the Euler equation of the functional is
F y −
d
dx
F y = −2y − 2y
= 0
namely y
+y = 0, the general solution is y = c 1 cos x +c 2 sin x. Using the boundary
conditions, we obtain c 1 = 0, c 2 = 1. So the extremal curve is y = sin x.
Example 2.4.3 Find the extremal curve of the functional J [y] =
2π
0 (y
2
− y
2
)dx,
the boundary conditions are y(0) = 1, y(2π) = 1.
Solution It can be seen from Example 2.4.2 that the general solution is y = c 1 cos x+
c 2 sin x. Using the boundary conditions, we obtain c 1 = 1, c 2 = c, where, c is an
arbitrary constant, the extremal curve is y = cos x + c sin x, Thus the variational
problem of the functional has an infinite number of solutions.
It is observed from Examples 2.4.2 and 2.4.3 that the extremal curve of a functional is not only related to the integrand, also related to the integral interval and the
boundary conditions.
Example 2.4.4 Find the extremal curve of the functional J [y] =
2
1 (y
2
− 2x y)dx,
the boundary conditions are y(1) = 0, y(2) = −1.
Solution The Euler equation of the functional is
y
+ x = 0
Integrating twice, we obtain
y = −
x
3
6
+ c 1 x + c 2
According to the boundary conditions y(1) = 0, y(2) = −1, solve for c 1 =
1
6
,
c 2 = 0, thus the extremal curve is
y =
x
6
(1 − x
2
)