[petsc-dev] How do you get RIchardson?

[Jed Brown]

On Sat, Sep 17, 2011 at 00:20, Matthew Knepley <knepley at gmail.com> wrote:

> My definition is in no way a "strict" subset. Define your nonlinear
> operator to have a solve, and it has what you want.


You want to solve

f(x) = 0

but you have to write

x = g(x)

to apply a fixed point method. So you do, e.g.

x = A(x)^{-1} b

Now we have to come up with a nonlinear problem f(x) = 0 such that

x = x - f(x) = A(x)^{-1} b

Evidently that is f(x) = x - A(x)^{-1} b. Now I have this extra x floating
around just so it can be subtracted. Just because you can transform
something to make a certain choice general doesn't make it so.
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