Appendix: All Solutions to the Problems
839
or
y
− λy = 0
The characteristic equation of the equation is r
2
− λ = 0. Obviously λ = 0. If
λ > 0, then the general solution of the Euler equation is
y = c 1 cosh
√ λx + c 2 sinh
√
λx
and only when c 1 = c 2 = 0, the boundary conditions y(0) = 0, y(π) = 0 can be
satisfied, so the general solution is y = 0, but at this time the isoperimetric condition
π
0 y
2 dx = 1 can not hold. Thus let λ < 0, at the moment the general solution of the
Euler equation is
y = c 1 cos
√
−λx + c 2 sin
√
−λx
From the boundary conditions y(0) = 0, y(π) = 0, we get c 1 = 0, −λ =
n
2 (n = 1, 2, . . .), and y n = c 2 sin nx, substituting it into the isoperimetric condition,
we get
π
0 (c 2 sin nx)
2 dx = c
2
2
π
2
= 1, c 2 = ±
2
π
, so the extremal curve is y n =
±
2
π
sin nx.
In the extremal curve y n = ±
2
π
sin nx passing through point (0,0) and point
(π,0), only two extremal curves satisfy the Jacobi condition, namely y = ±
2
π
sin x.
On the two extremal curves, the functional can reach the minimum
J [y] =
π
0
y
2 dx =
π
0
2
π
cos
2 xdx = 1
5.2 Find the extremal curve of the functional J [y, z] =
1
0 (y
2
+ z
2
− 4xz
− 4z)dx
under the isoperimetric condition
1
0 (y
2
− x y
− z
2
)dx = 2, the boundary
conditions are y(0) = z(0) = 0 and y(1) = z(1) = 1.
Solution Making the auxiliary functional
J
∗
[y] =
1
0
[y
2
+ z
2
− 4xz
− 4z + λ(y
2
− x y
− z
2
)]dx
The Euler equations of the functional are
−2y
− 2λy
+ λ = 0, −2z
+ 4 − 4 + 2λz
= 0
or