[petsc-dev] rename SNES methods ls, tr etc

[Anton Popov]

On 12/3/12 8:51 PM, Jed Brown wrote:
> On Mon, Dec 3, 2012 at 11:34 AM, Barry Smith <bsmith at mcs.anl.gov 
> <mailto:bsmith at mcs.anl.gov>> wrote:
>
>       What is "a Picard linearization"? As opposed to a non-Picard
>     linearization? Also if you phrase it as in my other email isn't
>     Newton "a Picard linearization"?   You act as if the term "a
>     Picard linearization" has a well defined meaning, but Matt never
>     found it in any book in history.
>
>
Some info on Picard linearization can be found for example in Chapter 2, 
Volume 2 of the  book by Zienkiewicz & Taylor on Finite Elements. Couple 
of equations are given on page 29 (5-th edition), although they're quite 
unclear. Nevertheless it can be helpful.

The idea is that one approximates total nonlinear solution vector (not 
just defect correction) from a linear system with a secant matrix that 
itself depends on the latest solution, and a fixed (at least on a time 
step) right hand side.

When no satisfactory approximation for solution is given, doing couple 
of Picard steps is a good strategy to start with. Then one can switch to 
Newton with/without line search. In general, the advantage of Picard is 
stability at the expense of linear vs. potentially quadratic convergence 
(for Newton, when exact Jacobian and blah-blah-blah is known).

I know, these are just words, it's a bit difficult to generalize it for 
all possible cases.

Anton

> If you have a quasi-linear problem, then you can write the homogenous 
> part of the operator as A(u) u. That A(u) is the Picard linearization. 
> Achi calls it the "principle linearization" in some FAS papers because 
> it's provably all that is necessary in the smoother (the other terms 
> in Newton linearization involve lower frequencies, thus are not needed 
> in the smoother).
>
> Some equations, perhaps most notably the Euler flux, satisfy the 
> "homogeneity property" that F(u) = F'(u) u, i.e., A(u) _is_ the 
> Jacobian, in which case Picard would be equal to Newton. (People don't 
> normally "solve" the flux equation.)

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