[petsc-users] Most efficient way to set a matrix representing the linearized operator
Timothée Nicolas
timothee.nicolas at gmail.com
Fri Sep 4 19:20:53 CDT 2015
I see, good. Then there should be no problem. Do you also have a suggestion
for questions 2. and 3. ?
Timothée
2015-09-05 3:42 GMT+09:00 Matthew Knepley <knepley at gmail.com>:
> On Fri, Sep 4, 2015 at 11:20 AM, Timothée Nicolas <
> timothee.nicolas at gmail.com> wrote:
>
>> Hi,
>>
>> (short version at the end)
>>
>> I am starting to write the matrices for the preconditioner of my
>> application. It's a physics-based preconditioner (L Chacon Journal of
>> Comput. Phys. 2002, 2003 and Physics of Plasmas 2010 if you want to know
>> more), so I must compute the linearized operators as matrices. I then plan
>> to use KSP Multigrid to invert the resulting linear systems.
>>
>> My question is, what is the most efficient way of setting the matrices
>> for linearized operators in the context of a 3D grid handled with DMDA and
>> several degrees of freedom (8 in my case) ? Take for instance a term like
>> curl(v x B) in the induction equation, which, linearized, gives curl(v0 x .
>> ) for the magnetic part. I think the logical approach is to use vectors of
>> the form (0,...,0,1,0,...,0) and to compute the result of the operator, and
>> store it in the matrix by columns using MatSetValuesStencil. The matrix is
>> created with DMCreateMatrix.
>>
>> Since all the vectors have mostly zeros, by writing by hand what parts
>> you need to compute, the number of evaluations for each vector is only the
>> number of nonzeros in a column, typically 21 in my case (7 point stencil *
>> 3 directions), so the total number of evaluations of the operator should be
>> reasonable and efficient that way.
>>
>> It gives promising results, however I am having trouble handling the
>> regions at the interface between the processors. Because if your vector has
>> its only non zero value on a grid point which is on the boundary between
>> two processors, then it means that one of the non-zero value in the result
>> vector will be living on a ghost point. And to compute the operator at this
>> ghost point, you need an *extra* ghost point, which makes the whole thing
>> quite cumbersome. You could of course increase the stencil_width of the
>> DMDA by one, but it does not seem the good approach.
>>
>
> Why not? The only reason that the stencil width exists is to specify how
> much data from other processors
> is necessary to compute the residual/Jacobian. It sounds like two ghosts
> are needed.
>
> Thanks,
>
> Matt
>
>
>> Instead, I was wondering if it would be possible to write the matrix by
>> column vectors. So I would still create vectors of the form
>> (0,...,0,1,0,...,0) but in Petsc Vec type, then compute the result directly
>> in Petsc Vec, and put this vector in the matrix as a column. But somehow it
>> does not seem really efficient, unless there is a way Petsc knows that the
>> vector you input has mostly zeros, as well as the result vector.
>>
>> So, in short
>>
>> 1. How would you compute a matrix representing an operator in a 3D system
>> with several degrees of freedom in the most efficient way ?
>>
>> And two other question related to this matter:
>>
>> 2. When I do the Multigrid inversion, I will momentarily ignore 3 degrees
>> of freedom (the velocity). What is the more efficient between (i) putting 1
>> everywhere on the diagonal in the velocity part or (ii) somehow create a
>> new DMDA with 8-3 = 5 degrees of freedom to handle the problem separately.
>>
>> 3. When I use DMCreateMatrix, do I have to preallocate the matrix ? I
>> have tried, but got errors like
>> [11]PETSC ERROR: New nonzero at (296,296) caused a malloc
>> Use MatSetOption(A, MAT_NEW_NONZERO_ALLOCATION_ERR, PETSC_FALSE) to turn
>> off this check
>>
>> which could not be prevented by using the advised MatSetOption(A,
>> MAT_NEW_NONZERO_ALLOCATION_ERR, PETSC_FALSE)
>>
>> Sorry for the long question
>>
>> Best
>>
>> Timothée
>>
>
>
>
> --
> What most experimenters take for granted before they begin their
> experiments is infinitely more interesting than any results to which their
> experiments lead.
> -- Norbert Wiener
>
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