Chapter 7: Problem 26
The smoothing factor \(\mu^{*}\) for a discrete operator is defined as the worst (i.e., smallest) factor by which high frequency components are reduced in a single relaxation step. For the two-dimensional Laplacian we have discussed throughout this chapter and a basic relaxation scheme, this can be stated as follows. Suppose \(e_{0}\) is the error before a relaxation step associated with a stationary iteration matrix \(T\) and \(\mathbf{e}_{1}\) the error after that step, and write $$ \mathbf{e}_{0}=\sum_{l, m=1}^{N} \alpha_{l, m} \mathbf{v}_{l, m} $$ where \(\left\\{\mathbf{v}_{l, m}\right\\}_{l, m=1}^{N}\) are the eigenvectors of the iteration matrix. Then $$ \mathbf{e}_{1}=\sum_{l, m=1}^{N} \alpha_{l, m} \mu_{l, m} \mathbf{v}_{l, m} $$ where \(\left\\{\mu_{l, m}\right\\}_{l, m=1}^{N}\) are eigenvalues of the iteration matrix. The smoothing factor is thus given by $$ \mu^{*}=\max \left\\{\left|\mu_{l, m}\right|: \frac{N+1}{2} \leq l, m \leq N\right\\} $$ (a) Denote the discrete Laplacian by \(A\) and the iteration matrix for damped Jacobi by \(T_{\omega}\). Confirm that the eigenvectors of \(A\) are the same as the eigenvectors of \(T_{\omega}\) for this scheme. (If you have already worked on Exercise 11 , this should be old news.) (b) Show that the optimal \(\omega\) that gives the smallest smoothing factor over \(0 \leq \omega \leq 1\) for the two-dimensional Laplacian is \(\omega^{*}=\frac{4}{5}\), and find the smoothing factor \(\mu^{*}=\mu^{*}\left(\omega^{*}\right)\) in this case. Note: \(\mu^{*}\) should not depend on the mesh size. (c) Show that Jacobi (i.e., the case \(\omega=1\) ) is not an effective smoother.
Short Answer
Step by step solution
Key Concepts
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