So that's what I meant by having extra contributions. I thought you needed
a local_rhs = 0 after/within:
for (unsigned int face_n = 0;
face_n < GeometryInfo<dim>::faces_per_cell;
++face_n)
if (cell->at_boundary(face_n))
{
fe_face_values.reinit(cell, face_n);
In the same way that if you enforce them strongly, it basically ignores the
contributions to the cell vertices that are on the boundary. Because this
is a Dirichlet condition, should we not reset the local_rhs contribution
that have accumulated just prior to the boundary condition? putting
local_rhs = 0 again within the boundary contribution terms seems to
actually do the job.
For Stokes - normal component of the normal stress:
Ok, that's interesting. I seems to be getting a non-zero contribution when
I do:
local_rhs(i) = topstress_values[q] * fe_face_values.normal_vector(q)*
fe_face_values[velocities].value(i,q_point)*
fe_face_values_rock.JxW(q);
where topstress_values is the value of the normal component of the normal
stress.
Is this incorrect??
On Saturday, March 16, 2019 at 12:35:40 AM UTC, Wolfgang Bangerth wrote:
>
>
> Jane,
>
> > Always setting local_rhs = 0 immediately before the below implementation
> > would take into account all cases so that would be the best:
> >
> > for (unsigned int q=0; q<n_face_q_points; ++q)
> > for (unsigned int i=0; i<dofs_per_cell; ++i)
> > local_rhs(i) += -(fe_face_values[velocities].value (i, q) *
> > fe_face_values.normal_vector
> > <
> https://www.dealii.org/current/doxygen/deal.II/classFEValuesBase.html#a130eea0fa89263d93b20521addc830c7>(q)
>
>
> > *
> > boundary_values[q] *
> > fe_face_values.JxW
> > <
> https://www.dealii.org/current/doxygen/deal.II/classFEValuesBase.html#ad097580a2f71878695096cc73b271b9d>(q));
>
>
>
> But it's already there, in the marked line (lkine 494 in the current
> sources):
>
> for (const auto &cell : dof_handler.active_cell_iterators())
> {
> fe_values.reinit(cell);
> local_matrix = 0;
> local_rhs = 0; // ***********************
>
> right_hand_side.value_list(fe_values.get_quadrature_points(),
> rhs_values);
> k_inverse.value_list(fe_values.get_quadrature_points(),
> k_inverse_values);
>
> for (unsigned int q = 0; q < n_q_points; ++q)
> for (unsigned int i = 0; i < dofs_per_cell; ++i)
> ...cell integration...
>
>
> for (unsigned int face_n = 0;
> face_n < GeometryInfo<dim>::faces_per_cell;
> ++face_n)
> if (cell->at_boundary(face_n))
> {
> fe_face_values.reinit(cell, face_n);
>
> pressure_boundary_values.value_list(
> fe_face_values.get_quadrature_points(),
> boundary_values);
>
> for (unsigned int q = 0; q < n_face_q_points; ++q)
> for (unsigned int i = 0; i < dofs_per_cell; ++i)
> local_rhs(i) += ...
>
> So all that is happening is that we accumulate contributions from all
> faces of the cell before we put it all into the global matrix. This is
> the right thing to do.
>
>
> > As for the Stokes case, I am making the analogy for the INHOMOGENEOUS
> > normal component of normal stress boundary condition. \
> > You are right in that the no tangential stress condition requires
> > nothing, but then i have a nonzero value of the normal component of the
> > normal stress which needs to be in the weak form.
> > I can't quite figure out if this is a "dirichlet' condition so I also
> > set local_rhs = 0 before the boundary implementation, or whether it is
> > just an extra addition for the cells on that boundary.
>
> It is correct that the normal component of the normal stress is nonzero.
> But it is multiplied by a test function whose normal component is zero
> (because the normal component of the solution is prescribed). So the
> product of these two terms is zero, and there is no additional
> contribution to the weak form.
>
> Best
> W.
>
> --
> ------------------------------------------------------------------------
> Wolfgang Bangerth email: bang...@colostate.edu
> <javascript:>
> www: http://www.math.colostate.edu/~bangerth/
>
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