8.5 The Galerkin Method
477
Example 8.5.4 Find the first and second approximate solutions of the boundary
value problem by the Galerkin method
u = −m
(x, y ∈ D)
u(x, y)| Γ = 0
(8.5.19)
where, D = {(x, y)| − a < x < a, −b < y < b}, Γ is the boundary of the rectangle
domain D, m is a constant.
Solution Taking note that the problem to solve is symmetrical about x, y, choosing
the coordinate function sequence
ϕ 1 , ϕ 1 x
2
, ϕ 1 y
2
, ϕ 1 x
4
, ϕ 1 x
2 y
2
, ϕ 1 y
4
, . . .
In order to find the first and second approximate solutions, choosing the coordinate
functions satisfying the boundary conditions
ϕ 1 (x, y) = (a
2
− x
2
)(b
2
− y
2
)
ϕ 2 (x, y) = (a
2
− x
2
)(b
2
− y
2
)(x
2
+ y
2
)
Find the first approximate solution. Putting
u 1 (x, y) = a 1 ϕ 1 (x, y) = a 1 (a
2
− x
2
)(b
2
− y
2
)
∂
2 u 1
∂ x 2 = −2a 1 (b
2
− y
2
),
∂
2 u 1
∂ y 2 = −2a 1 (a
2
− x
2
)
u 1 + m = −2a 1 (a
2
+ b
2
− x
2
− y
2
) + m
Substituting the above expressions into the Galerkin equation, taking note that
the integrands are both the even functions, we obtain
a
−a
b
−b
[−2a 1 (a
2
+ b
2
− x
2
− y
2
) + m](a
2
− x
2
)(b
2
− y
2
)dxdy
= 4
a
0
b
0
[−2a 1 (a
2
+ b
2
− x
2
− y
2
) + m](a
2
− x
2
)(b
2
− y
2
)dxdy = 0
Eliminating the coefficient 4 before the integral sign, the Galerkin equation can
be simplified to
a
0
b
0
[−2a 1 (a
2
+ b
2
− x
2
− y
2
) + m](a
2
− x
2
)(b
2
− y
2
)dxdy = 0
Integrating, we get