Appendix: All Solutions to the Problems
887
∂ J
2∂a 1
=
11
30
a 1 +
11
60
a 2 −
1
12
= 0
∂ J
2∂a 2
=
11
60
a 1 +
1
7
a 2 −
1
20
= 0
Solving the above equations, we get
a 1 =
69
473
, a 2 =
77
473
Thus, the approximate solution can be obtained
y 2 =
69
473
(x
2
− x) +
77
473
(x
3
− x
2
) =
x
473
(77x
2
− 8x − 69)
At the point of x = 0.5, the analytic solution is y = −0.005659056, the approximate solution is y 2 = −0.005681818, the absolute error is y 2 −y = −2.276247×10
5 ,
the relative error is
y 2 −y
y
× 100 = 0.4022339%.
8.8 Using the Ritz method to find the approximate solution of the following boundary
value problem
y
+ (1 + x
2
)y + 1 = 0 (−1 < x < 1)
y(1) = 0, y(−1) = 0
Solution Let the form of approximate solution be
y n = (1 − x
2
)(a 1 + a 2 x
2
+ · · · + a n x
2n
)
Taking the second approximate solution
y 2 = (1 − x
2
)(a 1 + a 2 x
2
)
The corresponding functional of the original is
J =
1
−1
[y
2
− (1 + x
2
)y
2
− 2y]dx
Substituting y 2 into the functional and integrating, then deriving a 1 and a 2
respectively, we get
∂ J
∂a 1
= 4 ×
76
105
a 1 +
52
315
a 2 −
2
3
= 0
∂ J
∂a 2
= 4 ×
52
315
a 1 +
1324
3465
a 2 −
2
15
= 0
Solving the above equations, we get a 1 =
3969
4252
, a 2 = −
231
4252
. Therefore, the second
approximate solution is