[petsc-users] fixed point interations

[Matthew Knepley]

On Sun, Nov 6, 2011 at 9:34 PM, Dominik Szczerba <dominik at itis.ethz.ch>wrote:

> On Sun, Nov 6, 2011 at 6:59 PM, Matthew Knepley <knepley at gmail.com> wrote:
> > On Sun, Nov 6, 2011 at 5:52 PM, Dominik Szczerba <dominik at itis.ethz.ch>
> > wrote:
> >>
> >> >>> I want to start small by porting a very simple code using fixed
> point
> >> >>> iterations as follows: A(x)x = b(x) is approximated as A(x0)x =
> b(x0),
> >> >>> then solved by KSP for x, then x0 is updated to x, then repeat until
> >> >>> convergence.
> >> >
> >> > Run the usual "Newton" methods with A(x) in place of the true
> Jacobian.
> >>
> >> When I substitute A(x) into eq. 5.2 I get:
> >>
> >> A(x) dx = -F(x) (1)
> >> A(x) dx = -A(x) x + b(x) (2)
> >> A(x) dx + A(x) x = b(x) (3)
> >> A(x) (x+dx) = b(x) (4)
> >>
> >> My questions:
> >>
> >> * Will the procedure somehow optimally group the two A(x) terms into
> >> one, as in 3-4? This requires knowledge, will this be efficiently
> >> handled?
> >
> > There is no grouping. You solve for dx and do a vector addition.
> >
> >>
> >> * I am solving for x+dx, while eq. 5.3 solves for dx. Is this, and
> >> how, correctly handled? Should I somehow disable the update myself?
> >
> > Do not do any update yourself, just give the correct A at each iteration
> in
> > your FormJacobian routine.
> >    Matt
>
> OK, no manual update, this is clear now. What is still not clear is
> that by substituting A for F' I arrive at an equation in x+dx (my eq.
> 4), and not dx (Petsc eq. 5.3)...


Newton's equation is for dx. Then you add that to x to get the next guess.
This
is described in any book on numerical analysis, e.g. Henrici.

   Matt


>
> Dominik
>

-- 
What most experimenters take for granted before they begin their
experiments is infinitely more interesting than any results to which their
experiments lead.
-- Norbert Wiener
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