| 1 |
You can show that for Jacobi, best choice is ? = 1 i.e. that relaxation strategy is not very useful
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| 2 |
For Gauss-Seidel, then in simple one dimensional case with N grid points, one can show that there is an optimal choice ? = 2 ( 1 - ? / N) i.e. almost equals 2
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| 3 |
Then the number of iterations needed by optimal SOR to get a given error can be shown to be proportional to N not N2 as in Jacobi or Gauss Seidel. So Ratio of Iterations is:
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| 4 |
Jacobi 2 N2/ ?2
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| 5 |
Gauss-Seidel N2/ ?2
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| 6 |
SOR on Gauss-Seidel N/ ?
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