Convergence rate Jacobi/Gauss-Seidel with mesh resolution

Question

In the book

A Multigrid Tutorial - Briggs, Henson. McCormick

in the beginning of Chapter 3, it is mentioned that

...because the convergence factor behaves as 1-$O(h^{2})$, the coarse grid will have a marginally improved convergence rate...

Here, $h$ is the mesh spacing (say from Finite Difference Discretization).

Question

How is this factor of $1-O(h^{2})$ derived ?

Does it mean solution on low resolution grids converges faster than high resolution grids ? But if this is the case then say $h=\frac{1}{2}$, then $1-\frac{1}{4}=\frac{3}{4}$ is the convergence rate.

When $h=\frac{1}{4}$ then the convergence rate is $1-\frac{1}{16}=\frac{15}{16}$.

Clearly the latter is more and hence fine grids have better convergence rates with this approximation. What am I missing ?

Answer 1

The convergence rate that is mentioned here is in the sense that the error in iteration $k$ and $k-1$ are related by $$ \| x^{(k)} - x^\ast\| \le r \| x^{(k-1)} - x^\ast\|, $$ which implies that $$ \| x^{(k)} - x^\ast\| \le r^k \| x^{(0)} - x^\ast\|. $$ For this to converge at all, we need that $r<1$, which the statement you quote provides. But for fast convergence, we need that $r$ is not marginally smaller than one, but ideally close to zero -- say, $r=0.1$, in which case you'd reduce the error by a factor of ten in each iteration. On the other hand, $r=1-ch^2$ is close to one if $h$ is small enough, and so it will take a large number of iterations until the reduction $r^k$ over the initial residual is substantial.