On Sep 13, 2013, at 6:21 AM, Bishesh Khanal <[email protected]> wrote:
> Dear all,
> I'm trying to reuse an operator for the linear system multiple times with
> different RHS.
> But it seems that the RHS is not changed in the KSP object. Here is the code
> to illustrate what I'm doing:
>
> Vec mX[3], mB[3]; //Solving Ax=b for 3 different RHS.
> ...
> ierr = KSPSetComputeOperators(mKsp,computeMatrix,this);CHKERRQ(ierr);
> ierr = KSPSetDM(mKsp,mDa);CHKERRQ(ierr);
Move the line up above the previous line
> ierr = KSPSetFromOptions(mKsp);CHKERRQ(ierr);
>
> for(int i = 0; i<3; ++i) {
> mBComponent = i; //sets different component for RHS.
> ierr = KSPSetComputeRHS(mKsp,computeRhs,this);CHKERRQ(ierr);
Put the above line in the setup code above, not in the loop
> ierr = KSPSetDM(mKsp,mDa);CHKERRQ(ierr);
Remove the line above
> ierr = KSPSolve(mKsp,NULL,NULL);CHKERRQ(ierr);
> ierr = KSPGetSolution(mKsp,&mX[i]);CHKERRQ(ierr);
> ierr = KSPGetRhs(mKsp,&mB[i]);CHKERRQ(ierr);
> }
After you make this changes it should work. If it does not then run with
-start_in_debugger noxterm and put a break point in computeRHS, make sure it is
being called.
>
> What I expect is to have three different solutions in mX array for the
> different Rhs computed by computeRhs function. However, the result when I see
> using VecView is that the Rhs mB[0], mB[1] and mB[2] are all identical (which
> should not have been the case). And mX also has three identical vectors as
> solution.
> What am I doing wrong here ?
>
>
> Thanks,
> Bishesh
>
