[petsc-dev] How do you get RIchardson?

[Jed Brown]

On Fri, Sep 16, 2011 at 23:21, Matthew Knepley <knepley at gmail.com> wrote:

> 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.
>

>From Barry's description at the top of this thread:

 x^{n+1}   = x^{n}  - lambda F(x^{n})


This looks oddly similar to

 x^{n+1}   = x^{n}  - J(x^n)^{-1} F(x^{n})

I wonder where I've seen that before.
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