Appendix: All Solutions to the Problems
855
Solution Making the auxiliary functional J
∗
[y] =
1
0 [y
2
+ λ(y − y
2
)]dx, then
the Euler equation of the functional is
λ + 2λy
− 2y = 0
The above equation can be written as
y
=
λ
2(1 − λ)
= c
Integrating twice, we get
y = c 1 x
2
+ c 2 x + c 3
From the boundary conditions y(0) = 0, y(1) =
1
4
, we get c 3 = 0, c 2 =
1
4
− c 1 ,
the extremal curve is y = c 1 x
2
+
1
4
− c 1
x, the derivative is y
= 2c 1 x +
1
4
− c 1 .
Substituting them into the isoperimetric condition, we get
1
0
c 1 x
2
+
1
4
− c 1
x −
2c 1 x +
1
4
− c 1
2
dx =
1
12
Integrating and arranging, we get
16c
2
1 + 8c 1 + 1 = 0
Solving the quadratic equation with one unknown, we obtain
c 1 =
−8 ±
√
64 − 4 × 16 × 1
2 × 16
= −
1
4
Thus getting c 2 =
1
2
, the extremal curve is y =
1
4
(2x − x
2
).
5.16 Find the solution of the minimum problem of the functional J [y] =
1
−1 (y
2
− k
2 y
2
)dx, the boundary conditions are y(−1) = y(1) = 0, the additional
condition is
1
−1 y
2 dx = 1.
Solution Making the auxiliary functional
J
∗
=
1
1
[y
2
+ (λ − k
2
)y
2
]dx
The Euler equation of the functional is
2(λ − k
2
)y − 2y
= 0