Appendix: All Solutions to the Problems
769
2.21 Discuss the extremum of the functional J [y] =
x 1
x 0
[2x y + (x
2
+ e
y
)y
]dx, the
boundary conditions are y(x 0 ) = y 0 , y(x 1 ) = y 1 .
Solution The integrand linearly depends on y
, there are
∂ M
∂ y
= 2x,
∂ N
∂ x
2x,
∂ M
∂ y
−
∂ N
∂ x
≡ 0
and the integral
J [y] =
x 1
x 0
[2x y + (x
2
+ e
y
)y
]dx =
x 1
x 0
d(x
2 y + e
y
) = x
2
1 y 1 + e
y 1 − x
2
0 y 0 − e
y 0
has nothing to do with the integral path, it only depends on the boundary conditions.
The variational problem does not make sense.
2.22 Discuss the extremum of the functional J [y] =
1
0 (e
y
+ x y
)dx, the boundary
conditions are y(0) = 0, y(1) = a.
Solution The Euler equation of the functional is
e
y
− 1 = 0
It is see from the boundary conditions y(0) = 0, y(1) = a that when a = 0,
y = 0, when a = 0, there is no smooth extremal curve. Moreover the integrand is
the linear function of y
, the variational problem loses its meaning.
2.23 Discuss the extremum of the functional J [y] =
x 1
x 0
(y
2
+ 2x yy
)dx, the
boundary conditions are y(x 0 ) = y 0 , y(x 1 ) = y 1 .
Solution Since the integrand F = y
2
+ 2x yy
, namely F is the linear function of
y
, the integral has nothing to do with the integral path, the variational problem does
not make sense.
2.24 Discuss the extremum of the functional J [y] =
1
0 (x y + y
2
− 2y
2 y
)dx, the
boundary conditions are y(0) = 1, y(1) = 2.
Solution Since the integrand F = x y + y
2
− 2y
2 y
, namely F is the linear function
of y
, in continuous function class the extremum can not reach.
2.25 Find the extremal curve of the functional J [y] =
x 1
x 0
y
(1 + x
2 y
)dx.
Solution Let the integrand F = y
(1 + x
2 y
), because F does not depend on y, the
Euler equation of the functional has the first integral