On Fri, May 25, 2012 at 9:06 AM, Mark F. Adams <[email protected]>wrote:
On May 25, 2012, at 9:42 AM, Jed Brown wrote:
The high end of the GS preconditioned operator is still high frequency. If it wasn't, then GS would be a spectrally equivalent preconditioner.
Huh? If I damp Jacobi on the 3-point stencil with 0.5 then the high frequency is _not_ the "high end of the preconditioned operator". It is asymptotically 0. Does that mean it is spectrally equivalent?
When I said "high" frequency, I didn't mean "highest" frequency. The low end of the spectrum (that you can't capture) is relatively unperturbed by local smoothers. So let's look at a damped Jacobi preconditioner. Suppose D = [diag(A)]^{-1}. If you weight it by w=0.5 or whatever, the Chebyshev(2) error propagation operator still looks like (I - a w D A) (I - b w D A) where a and b come from the target interval and we build eigenvalue estimates using K = w D A, so we'll produce exactly the same polynomial as w=1. We need better visualization for modes, but if the preconditioned operator K = P^{-1}A has maximum eigenvalue of 1, the second order Chebyshev polynomial targeting [0.1, 1.1] is about (1 - 0.25 K) (1 - 0.95 K). Thus, if P^{-1} perfectly corrects the high energy mode, we will use more than 0.95 of that correction. Please correct the above reasoning if I've messed up.