[petsc-users] ksppreonly question

[Shao-Ching Huang]

On Fri, Sep 21, 2012 at 3:17 PM, Jed Brown <jedbrown at mcs.anl.gov> wrote:
> So, not a Laplacian at all. Rather, a vector operator involving the
> symmetrized gradient of the velocity (or, not symmetrized if you are
> cheating due to nice boundary conditions). But what splitting are you using?

You are right. I loosely called it Laplacian because the non-zero
entries locations are the same as a standard Laplacian. Sorry for the
confusion. To be accurate, it is a Laplacian plus additional diagonal
terms.