[petsc-users] fixed point interations

[Dominik Szczerba]

I think I get it now. Thanks a lot.
On Nov 6, 2011 11:06 PM, "Matthew Knepley" <knepley at gmail.com> wrote:

> 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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