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