Appendix: All Solutions to the Problems
895
where, D = {(x, y)| − a < x < a, −b < y < b}; Γ is the boundary of the
rectangular domain D.
Solution It follows from the symmetry of the equation that the solution to find should
be the even function of x and odd function of y, so let the form of the solution be
u n = a 1 + a 2 y + a 3 x
2
+ a 4 y
3
+ a 5 x
2 y + a 6 x
4
+ a 7 x
4 y + a 8 x
2 y
3
+ a 9 y
5
+ · · ·
Taking n = 9, u 9 should satisfy the following boundary conditions
∂u 9
∂ x
x=±a
= ±(2a 3 a + 2a 5 ay + 4a 6 a
3
+ 4a 7 a
3 y + 2a 8 ay
3
) = 0
∂u 9
∂ y
y=±b
= ±(a 2 + 3a 4 b
2
+ a 5 x
2
+ a 7 x
4
+ 3a 8 x
2 b
2
+ 5a 9 b
4
) = 0
Since the coefficients of different powers of x and y are equal to zero, we get
a 5 = a 7 = a 8 = 0, a 3 = −2a 6 a
2
, a 2 = −3a 4 b
2
− 5a 9 b
4
Because u n is the odd function of y, there is a 1 = 0, taking the approximate
solution
u 9 = (−3a 4 b
2
− 5a 9 b
4
)y + (−2a 6 a
2
)x
2
+ a 4 y
3
+ a 6 x
4
+ a 9 y
5
= a 6 (x
4
− 2a
2 x
2
) + a 4 (y
3
− 3b
2 y) + a 9 (y
5
− 5b
4 y)
namely choosing the coordinate functions ϕ 1 = x
4
− 2a
2 x
2 , ϕ 2 = y
3
− 3b
2 y,
ϕ 3 = y
5
− 5b
4 y.
Calculating the two second partial derivatives and the residual
∂
2 u 9
∂ x 2 = (12x
2
− 4a
2
)a 6 ,
∂
2 u 9
∂ y 2 = 6ya 4 + 20y
3 a 9
u 9 − y = (12x
2
− 4a
2
)a 6 + 6ya 4 + 20y
3 a 9 − y
The Galerkin equations are
a
−a
b
−b
((u 9 − y)ϕ i dxdy = 0(i = 1, 2, 3)
First substituting ϕ 1 into the Galerkin equation, integrating, we get
2b
24
7
−
56
5
+
16
3
a
7 a 6 = −
512
105
a
7 ba 6 = 0
Getting a 6 = 0.