How to efficiently change just the diagonal vector in a matrix at every time step

Ben Tay zonexo at
Fri May 9 07:33:21 CDT 2008

I have a matrix and I inserted all the relevant values during the 1st 
step. I'll then solve it. For the subsequent steps, I only need to 
change the diagonal vector of the matrix before solving. I wonder how I 
can do it efficiently. Of course, the RHS vector also change but I've 
not included them here.

I set these at the 1st step:

call KSPSetOperators(ksp_semi_x,A_semi_x,A_semi_x,SAME_NONZERO_PATTERN,ierr)

call KSPGetPC(ksp_semi_x,pc_semi_x,ierr)


    call KSPSetType(ksp_semi_x,ksptype,ierr)

    ptype = PCILU

    call PCSetType(pc_semi_x,ptype,ierr)

    call KSPSetFromOptions(ksp_semi_x,ierr)

    call KSPSetInitialGuessNonzero(ksp_semi_x,PETSC_TRUE,ierr)



and what I did at the subsequent steps is:

do II=1,total
    call MatSetValues(A_semi_x,1,II,1,II,new_value,INSERT_VALUES,ierr)

end do

call MatAssemblyBegin(A_semi_x,MAT_FINAL_ASSEMBLY,ierr)

call MatAssemblyEnd(A_semi_x,MAT_FINAL_ASSEMBLY,ierr)

call KSPSolve(ksp_semi_x,b_rhs_semi_x,xx_semi_x,ierr)

I realise that the answers are slightly different as compared to calling 
all the options such as KSPSetType, KSPSetFromOptions, KSPSetTolerances 
at every time step. Should that be so? Is this the best way?

Also, I can let the matrix be equal at every time step by fixing the 
delta_time. However, it may give stability problems. I wonder how 
expensive is these type of value changing and assembly for a matrix?

Thank you very much.


More information about the petsc-users mailing list