840
Appendix: All Solutions to the Problems
y
=
λ
2(1 + λ)
, (1 − λ)z
= 0
Obviously, there should be 1 ± λ = 0. Integrating the above two equations two
times respectively, we get
y =
λx
2
4(1 + λ)
+ c 1 x + c 2 , z = c 3 x + c 4
From the boundary conditions y(0) = z(0) = 0, we get c 2 = c 4 = 0; From the
boundary conditions y(1) = z(1) = 1, we get c 1 =
4+3λ
4(1+λ)
, c 3 = 1. Therefore
y =
λx
2
4(1 + λ)
+
4 + 3λ
4(1 + λ)
x, z = x
y
=
λx
2(1 + λ)
+
4 + 3λ
4(1 + λ)
, z
= 1
Substituting the above relations into the isoperimetric condition, we obtain
1
0
λx
2(1 + λ)
+
4 + 3λ
4(1 + λ)
2
−
λx
2
2(1 + λ)
−
4 + 3λ
4(1 + λ)
x − 1
dx = 2
Integrating the above expression and arranging, we get
121λ
2
+ 242λ + 120 = 0
Solve for λ 1 = −
10
11
, λ 2 = −
12
11
. Substituting the two solutions into the y equation,
we get y 1 = −
5
2
x
2
+
7
2
x and y 2 = 3x
2
− 2x.
Substituting the two solutions into the original functional, we get
J [y 1 , z] =
1
0
−5x +
7
2
2
+ 1 − 8x
dx = − 66
J [y 2 , z] =
1
0
[(6x − 2)
2
+ 1 − 8x]dx = 1.5 > −66
The second variation of the functional is
δ
2 J [y, z] =
1
0
[(δy
)
2
+ (δz
)
2
]dx > 0
The second variation of the functional is greater than zero, the functional should
obtain the minimum. It is follows that the result found by y 2 is not the minimum. So
that