[petsc-dev] How do you get RIchardson?

[Matthew Knepley]

On Fri, Sep 16, 2011 at 4:13 PM, Vijay S. Mahadevan <vijay.m at gmail.com>wrote:

> I've always understood nonlinear Richardson to solve a problem A(x) x
> = b as using a linearization and reformulating as
>
> A(x_n) \delta x_n+1 = r_n,  where r_n = b - A(x_n) x_n, \delta x_n+1 =
> x_n+1 - x_n
>
> In essence, when A(x_n) contains the exact A evaluated at x_n, it
> simplifies to the standard Newton iteration. But when A contains only
> parts of the true A, I understand it to be a nonlinear fixed point
> iteration. This is quite often done with multiphysics problems where
> say with two physics nonlinearly coupled to each other, the true
> jacobian operator (exact newton) is
>
> A = [ W X ;
>        Y Z ;]
>
> But with A = [ W 0;
>                     0  Z;]
>
> it still converges, conditionally to the same solution as exact
> newton. Variations for A yield different rates of convergence. When
> A=1, you get the classical Picard iteration that Matt mentioned (?).
>

Not even close.

  Matt


> I like this formulation because it allows the control of including the
> stiff physics and using other algebraic/physics-based preconditioners
> on top of that. I am not sure if this is the standard way of writing
> out nonlinear Richardson or Picard and sorry for adding to the
> confusion ! Just my 2 cents.
>
> Vijay
>
> On Fri, Sep 16, 2011 at 8:35 PM, Jed Brown <jedbrown at mcs.anl.gov> wrote:
> > On Fri, Sep 16, 2011 at 22:14, Matthew Knepley <knepley at gmail.com>
> wrote:
> >>
> >> Water Resources is your standard for mathematical terminology?
> >
> > It's the whole first page of results for each query.
> > More seriously though, what is the problem with
> > x_{n+1} = A(x_n)^{-1} b
> > being a valid fixed-point iteration?
>



-- 
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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