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Appendix: All Solutions to the Problems
c
2
1 + (1 + c 2 )
2
= 1, c
2
1 + (−1 + c 2 )
2
= 1
Solving for c 1 = c 2 = 0. Thus the extremal curve is x
2
+ y
2
= 1, the parametric
forms are x = cos t, y = sin t.
5.4 Under the isoperimetric condition
1
0 y
2 dx = 2, find the extremal curve of the
functional J [y] =
1
0 (y
2
+ x
2
)dx, the boundary conditions are y(0) = y(1) = 0.
Solution Making the auxiliary functional J
∗
[y] =
1
0 (y
2
+ x
2
+ λy
2
)dx, then
F = y
2
+ x
2
+ λy
2
The Euler equation of the functional is
2λy − 2y
= 0
or
y
− λy = 0
When λ ≥ 0, the general solution of the Euler equation is
y = c 1 x + c 2 (λ = 0)
or
y = c 1 e
−
√ λx
+ c 2 e
√
λx
(λ > 0)
From the boundary conditions y(0) = y(1) = 0, we get c 1 = c 2 = 0, y = 0. The
solution does not satisfy the given isoperimetric condition, there is λ < 0. Thus the
above Euler equation becomes to
y
+ |λ|y = 0
The general solution of the Euler equation is
y = c 1 cos(
|λ|x) + c 2 sin(
|λ|x)
From the boundary conditions y(0) = 0, 得c 1 = 0. From the boundary conditions
y(1) = 0, we get
√ |λ| = ±nπ, where n is integer. Therefore y = ±c 2 sin(n π x).
Substituting it into the isoperimetric condition, there is
1
0
c
2
2 sin
2
(n π x)dx = c
2
2
1
0
1
2
[1 − cos(n π x)]dx =
c
2
2
2
= 2