Null space of discrete Laplace with periodic boundary conditions
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? Thomas
On Tue, Feb 14, 2012 at 09:20, Thomas Witkowski < [email protected]> 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.
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" <[email protected]> wrote:
On Tue, Feb 14, 2012 at 09:20, Thomas Witkowski < [email protected]> 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.
There is no boundary. On Feb 14, 2012 5:47 PM, "Mohammad Mirzadeh" <[email protected]> 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" <[email protected]> wrote:
On Tue, Feb 14, 2012 at 09:20, Thomas Witkowski < [email protected]> 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.
Oh sorry. This is surface Laplacian . My bad On Feb 14, 2012 2:48 PM, "Jed Brown" <[email protected]> wrote:
There is no boundary.
On Feb 14, 2012 5:47 PM, "Mohammad Mirzadeh" <[email protected]> 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" <[email protected]> wrote:
On Tue, Feb 14, 2012 at 09:20, Thomas Witkowski <
[email protected]> 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.
participants (3)
-
Jed Brown -
Mohammad Mirzadeh -
Thomas Witkowski