<div class="gmail_quote">On Tue, Feb 14, 2012 at 09:20, Thomas Witkowski <span dir="ltr"><<a href="mailto:thomas.witkowski@tu-dresden.de">thomas.witkowski@tu-dresden.de</a>></span> wrote:<br><blockquote class="gmail_quote" style="margin:0 0 0 .8ex;border-left:1px #ccc solid;padding-left:1ex">
I discretize the Laplace operator (using finite element) on the unit square equipped with periodic boundary conditions on all four edges. Is it correct that the null space is still constant? I wounder, because when I run the same code on a sphere (so a 2D surface embedded in 3D), the resulting matrix is non-singular. I thought, that both cases should be somehow equal with respect to the null space?<br>
</blockquote><div><br></div><div>The continuum operators for both cases have a constant null space, so if either is nonsingular in your finite element code, it's a discretization problem.</div></div>