[petsc-users] incomplete cholesky with a drop tolerance

[Umut Tabak]

On 07/22/2012 08:17 PM, Jed Brown wrote:
> On Sun, Jul 22, 2012 at 1:11 PM, Umut Tabak <u.tabak at tudelft.nl 
> <mailto:u.tabak at tudelft.nl>> wrote:
>
>     Well, basically, I am not interested in time domain response. What
>     I would like to do is to find the eigenvalues/vectors of the
>     system so it is in the frequency domain. What I was doing it
>     generally is the fact that I first factorize the operator matrix
>     with the normal factorization operation and use it to do multiple
>     solves in my Block Lanczos eigenvalue solver. Then in my
>     performance evaluations I saw that this is the point that I should
>     make faster, then I realized that I could solve this particular
>     system, that is pinned in your words, faster with iterative
>     methods almost %20 percent faster. And this is the reason why I am
>     trying to dig under.
>
>
> How many grid points per wavelength?
I am not sure at the moment I should check it further but the mesh is 
fine enough that this should not be a problem in the frequency range of 
interest.
>
>>         basically the operator is singular however for my problem I
>>         can delete one of the rows of the matrix, for this case, I 
>>         and get a non-singular operator that I can continue my
>>         operations, basically, I am getting a matrix with size n-1,
>>         where original problem size is n.
>>
>>
>>     This is often bad for iterative solvers. See the User's Manual
>>     section on solving singular systems. What is the condition number
>>     of the original operator minus the zero eigenvalue (instead of
>>     "pinning" on point)?
>     This is not clear to me... You mean something like projecting the
>     original operator on the on the zero eigenvector, some kind of a
>     deflation.
>
>
> See the User's Manual section. As long as the preconditioner is 
> stable, convergence is as good as for the nonsingular problem by 
> removing the null space on each iteration.
Ok I will see that part,
Thx.
U.

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