[petsc-users] Dealing with 2nd Derivatives on Boundaries

[Jed Brown]

... and you describe a "boundary condition" as "second derivative is zero",
which is not a boundary condition, making your problem ill-posed. (Indeed,
consider the family of problems in which you extend the domain in that
patch and apply _any_ boundary conditions in the extended domain. All of
those solutions are also solutions of your problem with "second derivative
is zero" on your "boundary".)


On Thu, Jan 10, 2013 at 9:44 AM, David Scott <d.scott at ed.ac.uk> wrote:

> All right, I'll say it differently. I wish to solve
>   div.grad phi = 0
> with the boundary conditions that I have described.
>
>
> On 10/01/2013 15:28, Jed Brown wrote:
>
>> Second derivative is not a boundary condition for Poisson; that is the
>> equation satisfied in the interior. Unless you are intentionally
>> attempting to apply a certain kind of outflow boundary condition (i.e.,
>> you're NOT solving Laplace) then there is a problem with your
>> formulation. I suggest you revisit the continuum problem and establish
>> that it is well-posed before concerning yourself with implementation
>> details.
>>
>>
>> On Thu, Jan 10, 2013 at 9:03 AM, David Scott <d.scott at ed.ac.uk
>> <mailto:d.scott at ed.ac.uk>> wrote:
>>
>>     Hello,
>>
>>     I am solving Poisson's equation (actually Laplace's equation in this
>>     simple test case) on a 3D structured grid. The boundary condition in
>>     the first dimension is periodic. In the others there are Von Neumann
>>     conditions except for one surface where the second derivative is
>>     zero. I have specified DMDA_BOUNDARY_NONE in these two dimensions
>>     and deal with the boundary conditions by constructing an appropriate
>>     matrix. Here is an extract from the Fortran code:
>>
>>                    if (j==0) then
>>                      ! Von Neumann boundary conditions on y=0 boundary.
>>                      v(1) = 1
>>                        col(MatStencil_i, 1) = i
>>                        col(MatStencil_j, 1) = j
>>                        col(MatStencil_k, 1) = k
>>                      v(2) = -1
>>                        col(MatStencil_i, 2) = i
>>                        col(MatStencil_j, 2) = j+1
>>                        col(MatStencil_k, 2) = k
>>                      call MatSetValuesStencil(B, 1, row, 2, col, v,
>>     INSERT_VALUES, ierr)
>>                    else if (j==maxl) then
>>                      ! Boundary condition on y=maxl boundary.
>>                      v(1) = 1
>>                        col(MatStencil_i, 1) = i
>>                        col(MatStencil_j, 1) = j
>>                        col(MatStencil_k, 1) = k
>>                      v(2) = -2
>>                        col(MatStencil_i, 2) = i
>>                        col(MatStencil_j, 2) = j-1
>>                        col(MatStencil_k, 2) = k
>>                      v(3) = 1
>>                        col(MatStencil_i, 3) = i
>>                        col(MatStencil_j, 3) = j-2
>>                        col(MatStencil_k, 3) = k
>>                      call MatSetValuesStencil(B, 1, row, 3, col, v,
>>     INSERT_VALUES, ierr)
>>                    else if (k==0) then
>>
>>
>>     Here the second clause deals with the second derivative on the
>> boundary.
>>
>>     In order for this code to work I have to set the stencil width to 2
>>     even though 'j-2' refers to an interior, non-halo
>>     point in the grid. This leads to larger halo swaps than would be
>>     required if a stencil width of 1 could be used.
>>
>>     Is there a better way to encode the problem?
>>
>>     David
>>
>>     --
>>     The University of Edinburgh is a charitable body, registered in
>>     Scotland, with registration number SC005336.
>>
>>
>>
>
> --
> Dr. D. M. Scott
> Applications Consultant
> Edinburgh Parallel Computing Centre
> Tel. 0131 650 5921
>
>
> The University of Edinburgh is a charitable body, registered in
> Scotland, with registration number SC005336.
>
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