478
8 Direct Methods of Variational Problems
−
32
45
a 1 (a
5 b
3
+ a
3 b
5
) +
4
9
a
3 b
3 m = 0
Eliminating the common factor a
3 b
3 , and solving for a 1 , we obtain
a 1 =
5m
8(a 2 + b 2 )
Thus, the first approximate solution of the given boundary value problem is
u 1 =
5m
8(a 2 + b 2 )
(a
2
− x
2
)(b
2
− y
2
)
(8.5.20)
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
)(b
2
− y
2
)a 1 + (a
2
− x
2
)(b
2
− y
2
)(x
2
+ y
2
)a 2
∂
2 u 2
∂ x 2 = −2(b
2
− y
2
)a 1 + 2(a
2
− 6x
2
− y
2
)(b
2
− y
2
)a 2
∂
2 u 2
∂ y 2 = −2(a
2
− x
2
)a 1 + 2(a
2
− x
2
)(b
2
− x
2
− 6y
2
)a 2
u 2 + m = −2(a
2
+ b
2
− x
2
− y
2
)a 1 + 2[2a
2 b
2
− (a
2
+ 7b
2
)x
2
+ 12x
2 y
2
− (7a
2
+ b
2
)y
2
+ x
4
+ y
4
]a 2 + m
The Galerkin equations are
⎧
⎪ ⎪ ⎪ ⎨
⎪ ⎪ ⎪ ⎩
a
−a
b
−b
((u 2 + m)ϕ 1 dxdy = 0
a
−a
b
−b
((u 2 + m)ϕ 2 dxdy = 0
Because the integrands are both the even functions, the Galerkin equations can
be simplified to
⎧
⎪ ⎪ ⎪ ⎨
⎪ ⎪ ⎪ ⎩
4
a
0
b
0
((u 2 + m)ϕ 1 dxdy = 0
4
a
0
b
0
((u 2 + m)ϕ 2 dxdy = 0
Integrating the above expressions, and the results are eliminated the common
factor 4a
3 b
3 , the equation are obtained