Paper 2, Section II, E

Numerical Analysis | Part II, 2018

The Poisson equation d2udx2=f\frac{d^{2} u}{d x^{2}}=f in the unit interval [0,1][0,1], with u(0)=u(1)=0u(0)=u(1)=0, is discretised with the formula

ui12ui+ui+1=h2fi,1in,u_{i-1}-2 u_{i}+u_{i+1}=h^{2} f_{i}, \quad 1 \leqslant i \leqslant n,

where u0=un+1=0,h=(n+1)1u_{0}=u_{n+1}=0, h=(n+1)^{-1}, the grid points are at x=ihx=i h and uiu(ih)u_{i} \approx u(i h).

(a) Write the above system of equations in the vector form Au=bA \boldsymbol{u}=\boldsymbol{b} and describe the relaxed Jacobi method with relaxation parameter ω\omega for solving this linear system.

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

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

Express HωH_{\omega} in terms of the matrix AA and the relaxation parameter ω\omega. Using the fact that for any n×nn \times n Toeplitz symmetric tridiagonal matrix, the eigenvectors vk(k=1,,n)\boldsymbol{v}_{k}(k=1, \ldots, n) have the form:

vk=(sinikx)i=1n,x=πn+1\boldsymbol{v}_{k}=(\sin i k x)_{i=1}^{n}, \quad x=\frac{\pi}{n+1}

find the eigenvalues λk(A)\lambda_{k}(A) of AA. Hence deduce the eigenvalues λk(ω)\lambda_{k}(\omega) of HωH_{\omega}.

(c) 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 the 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} when nn is large.

(d) Show that, for an appropriate range of ω\omega, the high frequency components ak(ν)a_{k}^{(\nu)} (n+12kn)\left(\frac{n+1}{2} \leqslant k \leqslant n\right) of the error e(ν)e^{(\nu)} tend to zero much faster than the rate obtained in part (c). Determine the optimal parameter ω\omega_{*} which provides the largest supression of the high frequency components per iteration, and find the corresponding attenuation factor μ\mu_{*} assuming nn is large. That is, find 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\frac{n+1}{2} \leqslant k \leqslant n.

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