Past exam of the mathematics course of the University of Cambridge / 2018 / ii / Paper 2 / 40E (Numerical Analysis)

With $\mathbf{u}=(u_1,\dots ,u_n{)}^T$ and $\mathbf{b}=h^2(f_1,\dots ,f_n{)}^T$, the system isIts diagonal part is $D=-2I$. The weighted Jacobi method, also called the Relaxed Jacobi method, isComponentwise,with $u_0^{(\nu )}=u_{n+1}^{(\nu )}=0$.

Subtracting the fixed point from the weighted Jacobi method givesPut $x=\pi /(n+1)$. Acting on the supplied sine eigenvector givesThereforeand the corresponding eigenvalues of the iteration matrix are

A stationary iterative method for a linear system converges from every initial vector exactly when the spectral radius of its iteration matrix is less than one. Here $\cos (kx)$ ranges between $\cos x$ and $-\cos x$, so all modes decay for a fixed $n$ exactly whenRequiring this for every grid size givesIndeed, $\omega \le 0$ leaves or amplifies the lowest mode, while any $\omega >1$ makes the highest-mode eigenvalue tend to $1-2\omega <-1$ as $n\to \infty$.

The slowest mode is $k=1$, for which the Taylor expansion isIf the initial error has a nonzero ${\mathbf{v}}_1$ component, that component is multiplied by ${\lambda }_1(\omega {)}^{\nu }$. Hence for a constant $c>0$ independent of $n$,This is the low-frequency obstruction to grid-independent convergence.

For the high-frequency indices $(n+1)/2\le k\le n$, the angle ${\theta }_k=k\pi /(n+1)$ ranges from $\pi /2$ to $\pi -\pi /(n+1)$. Sincethe large-$n$ high-frequency smoothing factor of weighted Jacobi isFor $0<\omega <1$, this is strictly less than one independently of $n$, so high-frequency error decays geometrically much faster than the low-frequency factor $1-O(n^{-2})$.

To minimize ${\mu }_{\omega }$, balance the endpoint magnitudes:Thus the optimal relaxation and attenuation factors areFor finite $n$, retaining the endpoint $\cos (\pi -\pi /(n+1))=-\cos (\pi /(n+1))$ giveswhich tend to $2/3$ and $1/3$.