[petsc-dev] How do you get RIchardson?

[Jed Brown]

On Sat, Sep 17, 2011 at 00:28, Barry Smith <bsmith at mcs.anl.gov> wrote:

>   But this is actually never Newton: if F(x) = A(x) x - b then J_F(x) =
> A(x) + J_A(x)x  so Just using A(x_n) will never give you Newton. But yes is
> commonly done because it doesn't require the horrible computation of J_A(x).


I think Vijay intended to distinguish the A in the solve step from the A in
the residual. Assume the iteration

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

Assume in this case that F(x_n) can be written as A(x_n) x_n - b where
A(x_n) distinguishes some "frozen" coefficients. Usually these are all but
those variables involved in the highest order derivatives.

Now we define J:

J = F_x(x_n), the true Jacobian. This is Newton

J = A(x_n). This is "Picard with solve", the version most commonly used for
PDEs. This is also the "Picard linearization" that is discussed in many
textbooks, especially in CFD.

J = \lambda I. This is gradient descent, aka Richardson or "Matt's Picard".
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