8.5 The Galerkin Method
481
Find the second approximate solution, putting
u 2 (x, y) = a 1 ϕ 1 (x, y) + a 2 ϕ 2 (x, y) = (a
2 − x
2 )
2 (b
2 − y
2 )
2 a 1
+ (a
2 − x
2 )
2 (b
2 − y
2 )
2 (x
2 + y
2 )a 2 = (a
4 − 2a
2 x
2 + x
4 )(b
4 − 2b
2 y
2 + y
4 )a 1
+ (a
4 x
2 − 2a
2 x
4 + x
6 )(b
4 − 2b
2 y
2 + y
4 )a 2
+ (a
4 − 2a
2 x
2 + x
4 )(b
4 y
2 − 2b
2 y
4 + y
6 )a 2
Computing the various partial derivatives
∂
4 u 2
∂ x 4 = 24(b
4
− 2b
2 y
2
+ y
4
)a 1 + 24(15x
2
− 2a
2
)(b
4
− 2b
2 y
2
+ y
4
)a 2
+ 24(b
4 y
2
− 2b
2 y
4
+ y
6
)a 2
∂
4 u 2
∂ y 4 = 24(a
4
− 2a
2 x
2
+ x
4
)a 1 + 24(15y
2
− 2b
2
)(a
4
− 2a
2 x
2
+ x
4
)a 2
+ 24(a
4 x
2
− 2a
2 x
4
+ x
6
)a 2
∂
4 u 2
∂ x 2 ∂ y 2 = 16(3x
2
− a
2
)(3y
2
− b
2
)a 1 + 8(a
4
− 12a
2 x
2
+ 15x
4
)(3y
2
− b
2
)a 2
+ 8(b
4
− 12b
2 y
2
+ 15y
4
)(3x
2
− a
2
)a 2
Substituting the above various equations into the following Galerkin equations
a
0
b
0 ((
2 u 2 − m)(a
2
− x
2
)
2
(b
2
− y
2
)
2 dxdy = 0
a
0
b
0 ((
2 u 2 − m)(a
2
− x
2
)
2
(b
2
− y
2
)
2
(x
2
+ y
2
)dxdy = 0
The integral result of the first Galerkin equation is
a
5 b
9 64
225
×
128
7
a 1 + a
7 b
9 64
225
×
128
7 × 7
a 2 + a
5 b
11 64
225
×
128
7 × 11
a 2
+ a
9 b
5 64
225
×
128
7
a 1 + a
9 b
7 64
225
×
128
7 × 7
a 2 + a
11 b
5 64
225
×
128
7 × 11
a 2
+ a
7 b
7 64
225
×
128 × 4
7 × 7
a 1 − a
5 b
5 64
225
m = 0
The integral result of the second Galerkin equation is
a
7 b
9 64
105
×
128
105
a 1 + a
5 b
11 64
105
×
128
5 × 33
a 1 + a
9 b
9 64
105
×
128
35
a 2
+ a
7 b
11 64
105
×
128 × 2
35 × 33
a 2 + a
5 b
13 64
105
×
128
5 × 143
a 2