Question 3.6: Use the iterative method to solve [K]{U} = {F} where. [K]=[2...
Use the iterative method to solve [K]{U} = {F} where.
[K]=\begin{bmatrix} 2 & -1 & 0 & 0 & 0 \\ -1 & 2 & -1 & 0 & 0\\ 0 & -1 & 2 & -1 & 0\\ 0 & 0 & -1 & 2 & -1\\ 0 & 0 & 0 & -1 & 2 \end{bmatrix} and \left\{F\right\} =\begin{Bmatrix} 5 \\ 2 \\ 2 \\ 2 \\5 \end{Bmatrix}
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Let \left\{U\right\} ^{(0)} = [1 1 1 1 1] and the terminate tolerance ε = 0.0001.
Using Eq. (3.99) to execute the iterative process, the process to obtain the final solutions is
U^{(k+ 1)}_{i} =\frac{1}{k_{ii} } \left[F_{i}-\sum\limits_{j=1;j\neq i}^{N}{K_{ij}U^{(k)}_{j} } \right] , (i=1,2,…N) (3.99)
Step | \left\{x_{i} \right\} ^{(k)} | Convergence |
0 | [1 1 1 1 1] | |
1 | [3.0000 2.0000 2.0000 2.0000 3.0000] | 3.5000 |
10 | [6.5762 8.6270 9.1523 8.6270 6.5762] | 0.7750 |
20 | [7.6621 10.4369 11.3242 10.4369 7.6621] | 0.1839 |
30 | [7.9198 10.8664 11.8396 10.8664 7.9198] | 0.0436 |
40 | [7.9810 10.9683 11.9619 10.9683 7.9810] | 0.0104 |
50 | [7.9955 10.9925 11.9910 10.9925 7.9955] | 0.0025 |
60 | [7.9989 10.9982 11.9979 10.9982 7.9989] | 5.8325e-04 |
70 | [7.9997 10.9996 11.9995 10.9996 7.9997] | 1.3841e-04 |
73 | [7.9998 10.9997 11.9997 10.9997 7.9998] | 8.9898e-05 |
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