[petsc-users] How to use MatSetValuesStencil() with DOF>1

[Lucas Banting]

Yes, I think that makes sense to me. Each cell is at i,j but then you can set the 25 coefficients per node using MatStencil_c from 0 to 4 for both idxm and idxn.

Thanks for your help.

Lucas
________________________________
From: Matthew Knepley <knepley at gmail.com>
Sent: Friday, May 29, 2020 12:47 PM
To: Lucas Banting <bantingl at myumanitoba.ca>
Cc: petsc-users at mcs.anl.gov <petsc-users at mcs.anl.gov>
Subject: Re: [petsc-users] How to use MatSetValuesStencil() with DOF>1

On Fri, May 29, 2020 at 12:16 PM Lucas Banting <bantingl at myumanitoba.ca<mailto:bantingl at myumanitoba.ca>> wrote:
Hello,

I have a structured grid problem with 5 unknowns per cell: U,V,P,T,F, this is a CFD code but I don't think these details are necessary to answer my question.
I was wondering how exactly I should use MatSetValuesStencil() to fill my matrix.
I have 5 degrees of freedom, so I need to set up to 25 coefficients per grid cell. I don't understand how i,j, and c map to all coefficients.

For a 1 degree of freedom system to fill my 9 point stencil I did:

          values(1)= ASW(II,JJ)
          values(2)= AS (II,JJ)
          values(3)= ASE(II,JJ)
          values(4)= AW (II,JJ)
          values(5)= AP (II,JJ)
          values(6)= AE (II,JJ)
          values(7)= ANW(II,JJ)
          values(8)= AN (II,JJ)
          values(9)= ANE(II,JJ)

          idxm(MatStencil_i,1) = II;idxm(MatStencil_j,1) = JJ

          idxn(MatStencil_i,1) = II-1;idxn(MatStencil_j,1) = JJ-1
          idxn(MatStencil_i,2) = II    ;idxn(MatStencil_j,2) = JJ-1
          idxn(MatStencil_i,3) = II+1;idxn(MatStencil_j,3) = JJ-1
          idxn(MatStencil_i,4) = II-1 ;idxn(MatStencil_j,4) = JJ
          idxn(MatStencil_i,5) = II    ;idxn(MatStencil_j,5) = JJ
          idxn(MatStencil_i,6) = II+1;idxn(MatStencil_j,6) = JJ
          idxn(MatStencil_i,7) = II-1 ;idxn(MatStencil_j,7) = JJ+1
          idxn(MatStencil_i,8) = II     ;idxn(MatStencil_j,8) = JJ+1
          idxn(MatStencil_i,9) = II+1;idxn(MatStencil_j,9) = JJ+1
          call MatSetValuesStencil(A,1,idxm,9,idxn,values,INSERT_VALUES,ierr)

Which seemed to work just fine. For 5 degrees of freedom, instead of just having a single AP coefficient for example, I have an AP matrix:

ap_uu
ap_uv
ap_up
ap_ut
ap_uf
ap_vu
ap_vv
ap_vp
ap_vt
ap_vf
ap_pu
ap_pv
ap_pp
ap_pt
ap_pf
ap_tu
ap_tv
ap_tp
ap_tt
ap_tf
ap_fu
ap_fv
ap_fp
ap_ft
ap_ff

In 1 degree of freedom, i and j corresponded to the solution variable in the grid. For multi degree of freedom, I don't understand how values of i, j, and c could distinguish from an ap_uv and ap_vv coefficent for example, wouldn't they both be at i, j ,c=2?
Is there anyway I can use MatSetValuesStencil() to fill in my 9 point stencil with my 5x5 matrix coefficients? To clarify I have nine 5x5 matrices for each cell which correspond to the 5 unknowns per cell.

In MatSetValuesStencil(), I think it is best to think of the Stencils as another way of providing row/col numbers. So each MatStencil corresponds
to some row number. So your ap_uv is some entry in the input matrix, and thus corresponds to some row number (a MatStencil) and some
column number (another MatStencil). Since ap is one point, it appears that

  ap_uv --> row = i,j,0   col = i,j,1
  ap_vv --> row = i,j,1   col = i,j,1

So you line up your MatStencil arguments to match the order of your input. It sounds like you want the field index to be the fastest in your input.

Does that make sense?

  Thanks,

     Matt


Thanks,

Lucas


--
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

https://www.cse.buffalo.edu/~knepley/<http://www.cse.buffalo.edu/~knepley/>
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