Past exam of the mathematics course of the University of Cambridge / 2009 / ii / Paper 1 / 39B / iii / Solution

For a fixed $n$, convergence for every initial error is equivalent to $|1-\omega t_k|<1$ for all $k$, where $t_k=1-\cos (k\pi /(n+1))>0$. ThusA single parameter works for every finite $n$ precisely when $\boxed{0<\omega \le 1}$: the largest $t_k$ tends to two as $n\to \infty$, so every $\omega >1$ eventually fails, whereas $\omega =1$ has all eigenvalues strictly between $-1$ and one for each finite grid.

For such $\omega$ the lowest-frequency eigenvalue satisfies, using $1-\cos x\le x^2/2$,An initial error proportional to $v_1$ retains exactly the factor ${\lambda }_1^{\nu }$. Hence the worst-case spectral radius cannot improve on a factor of the form $(1-c/n^2{)}^{\nu }$, with $c$ independent of grid size, and reducing a slow-mode error by a fixed fraction takes order $n^2$ iterations. This is a worst-case rate assertion, not a lower bound for every individual initial error: a vector containing only faster modes, or the zero vector, need not exhibit the slow mode.