[petsc-users] fixed point interations

[Matthew Knepley]

On Sun, Nov 6, 2011 at 9:56 PM, Dominik Szczerba <dominik at itis.ethz.ch>wrote:

> On Sun, Nov 6, 2011 at 10:40 PM, Matthew Knepley <knepley at gmail.com>
> wrote:
> > 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
>
> I understand that, but I am asking something else... when taking F' =
> A I arrive at an equation in x+dx, which is not Newton equation. What
> is wrong in this picture?


Keep the residual on the rhs, not b

   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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