768
Appendix: All Solutions to the Problems
Solution The Euler equation of the functional is
−
1
y 2 + 2
x y
y 3 +
1
y 2 − 2
x y
y 3 = 0
namely 0 = 0, the variational problem is meaningless. It is also seen from that the
integral is the linear function of y
that the variational problem is meaningless.
2.19 Discuss whether the variational problem of the functional J [y] =
π
2
0
x sin y +
1
2
x
2 y
cos y
dx makes sense, the boundary conditions are y(0) = 0,
y
π
2
=
π
2
.
Solution Let the integrand F = x sin y +
1
2
x
2 y
cos y, the Euler equation of the
functional is
x cos y −
x
2 y
2
sin y − x cos y +
x
2 y
2
sin y = 0
namely 0 = 0, the variational problem is meaningless. In fact, the integral of the
functional can be written as
J [y] =
π
2
0
x sin ydx +
x
2
2
cos ydy
=
π
2
0
d
x
2
2
sin y
=
π
2
8
The integral result is a constant and has nothing to do with the integral path, the
variational problem loses its meaning.
2.20 Discuss the variational problem of the functional J [y] =
π
2
0 (y
2
+ y
sin 2x)dx,
the boundary conditions are y(0) = k, y
π
2
= −1. What value does k take its
extremal curve exists? and calculate the extremal curve.
Solution Let the integrand F = y
2
+ y
sin 2x, the Euler equation of the functional
is
2y − 2 cos 2x = 0
or
y = cos 2x
It can be seen from the above equation that when x = 0时, y = 1, so only when
k = 1, there is the extremal curve y = cos 2x, when k = 1, the extremal curve does
not exist.