Paper 1, Section II, B

Numerical Analysis | Part II, 2009

(i) Define the Jacobi method with relaxation for solving the linear system Ax=bA \boldsymbol{x}=\boldsymbol{b}.

(ii) For x\boldsymbol{x}^{*} and x(ν)\boldsymbol{x}^{(\nu)} being the exact and the iterated solution, respectively, let e(ν):=x(ν)xe^{(\nu)}:=\boldsymbol{x}^{(\nu)}-\boldsymbol{x}^{*} be the error and HωH_{\omega} the iteration matrix, so that

e(ν+1)=Hωe(ν)e^{(\nu+1)}=H_{\omega} e^{(\nu)}

Express HωH_{\omega} in terms of the matrix AA, its diagonal part DD and the relaxation parameter ω\omega, and find the eigenvectors vk\boldsymbol{v}_{k} and the eigenvalues λk(ω)\lambda_{k}(\omega) of HωH_{\omega} for the n×nn \times n tridiagonal matrix

A=[2112112112]A=\left[\begin{array}{rrrrr} 2 & -1 & & & \\ -1 & 2 & -1 & & \\ & \ddots & \ddots & \ddots & \\ & & -1 & 2 & -1 \\ & & & -1 & 2 \end{array}\right]

[Hint: The eigenvectors and eigenvalues of AA are

(uk)i=sinkiπn+1,i=1,,n,λk(A)=4sin2kπ2(n+1),k=1,,n.]\left.\left(\boldsymbol{u}_{k}\right)_{i}=\sin \frac{k i \pi}{n+1}, \quad i=1, \ldots, n, \quad \lambda_{k}(A)=4 \sin ^{2} \frac{k \pi}{2(n+1)}, \quad k=1, \ldots, n .\right]

(iii) For AA as above, let

e(ν)=k=1nak(ν)vk\boldsymbol{e}^{(\nu)}=\sum_{k=1}^{n} a_{k}^{(\nu)} \boldsymbol{v}_{k}

be the expansion of the error with respect to the eigenvectors (vk)\left(\boldsymbol{v}_{k}\right) of HωH_{\omega}.

Find the range of parameter ω\omega which provides convergence of the method for any nn, and prove that, for any such ω\omega, the rate of convergence e(ν)0e^{(\nu)} \rightarrow 0 is not faster than (1c/n2)ν\left(1-c / n^{2}\right)^{\nu}.

(iv) Show that, for some ω\omega, the high frequency components (n+12kn)\left(\frac{n+1}{2} \leqslant k \leqslant n\right) of the error e(ν)e^{(\nu)} tend to zero much faster. Determine the optimal parameter ω\omega_{*} which provides the largest suppression of the high frequency components per iteration, and find the corresponding attenuation factor μ\mu_{*} (i.e. the least μω\mu_{\omega} such that ak(ν+1)μωak(ν)\left|a_{k}^{(\nu+1)}\right| \leqslant \mu_{\omega}\left|a_{k}^{(\nu)}\right| for n+12kn)\left.\frac{n+1}{2} \leqslant k \leqslant n\right).

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