890
Appendix: All Solutions to the Problems
u 1 = a 1 (x)(y
2
− k
2 x
2
)
The integrand of the functional can be expressed as
∂u 1
∂ x
2
+
∂u 1
∂ y
2
− 2u 1 = (y
4
− 2k
2 x
2 y
2
+ k
4 x
4
)a
2
1
+ 4(k
4 x
2
+ y
2
)a
2
1 + 4(k
4 x
3
− k
2 x y
2
)a 1 a 1
+ 2(k
2 x
2
− y
2
)a 1
Integrating y in the functional, we get
J [a 1 ] =
b
a
kx
−kx
∂u 1
∂ x
2
+
∂u 1
∂ y
2
− 2u 1
dxdy
=
b
a
16
15
k
5 x
5 a
2
1 +
8k
5
+
8
3
k
3
x
3 a
2
1 +
16
3
k
5 x
4 a 1 a 1
+
8
3
k
3 x
3 a 1
dx
The Euler equation of the functional is
2
5
k
2 x
2 a 1
+ 2k
2 xa 1
+ (k
2
− 1)a 1 =
1
2
This linear ordinary differential equation is the Euler equation. It is transformed
into the standard form
x
2 a 1
+ 5xa 1
+
5
2k 2 (k
2
− 1)a 1 =
5
4k 2
Let x = e
T or t = ln x, and using d to express the derivative with respect to t,
then the original equation can be converted into
d(d − 1)a 1 + 5da 1 +
5
2k 2 (k
2
− 1)a 1 =
5
4k 2
The characteristic equation of the equation is
r (r − 1) + 5r +
5
2k 2 (k
2
− 1) = 0
or
r
2
+ 4r +
5
2k 2 (k
2
− 1) = 0
The two roots of the characteristic equation can be obtained