<p>Oh sorry. This is surface Laplacian . My bad <br>
On Feb 14, 2012 2:48 PM, "Jed Brown" <<a href="mailto:jedbrown@mcs.anl.gov">jedbrown@mcs.anl.gov</a>> wrote:<br>
><br>
> There is no boundary.<br>
><br>
> On Feb 14, 2012 5:47 PM, "Mohammad Mirzadeh" <<a href="mailto:mirzadeh@gmail.com">mirzadeh@gmail.com</a>> wrote:<br>
>><br>
>> What do you set on the sphere? If you impose a Dirichlet BC that makes it nonsingular<br>
>><br>
>> Mohammad<br>
>><br>
>> On Feb 14, 2012 7:27 AM, "Jed Brown" <<a href="mailto:jedbrown@mcs.anl.gov">jedbrown@mcs.anl.gov</a>> wrote:<br>
>>><br>
>>> On Tue, Feb 14, 2012 at 09:20, Thomas Witkowski <<a href="mailto:thomas.witkowski@tu-dresden.de">thomas.witkowski@tu-dresden.de</a>> wrote:<br>
>>>><br>
>>>> 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>
>>><br>
>>><br>
>>> 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.</p>