Hi Wolfgang, thanks for a swift reply.
and apologies, I wasn't very clear.
when I'm imposing dirichlet conditions everywhere, i don't have a condition
for pressure (those were two separate test cases - sorry i was unclear).
and i'm using:
VectorTools::interpolate_boundary_values (dof_handler,
0,
BoundaryValues<dim>(),
constraints,
fe.component_mask(velocities));
where the boundary values are just 1.0 so that u is a constant vector field.
this still gives me the bizarre 22 factor in my pressures solution (which
is just linear in z - the resulting pressure solution is atleast linear),
but the pressure solution aside, both the x and y components of the
velocity are 0.0 ON the boundaries and 1.0 everywhere else.
Yup sorry that was a mistake on my part - my real problem has total stress
conditions so I naively just related it to pressure which isn't correct.
would appreciate any help with why my interpolating boundary values aren't
quite working.
Thanks again
On Wednesday, February 1, 2017 at 1:11:12 PM UTC, Wolfgang Bangerth wrote:
>
>
> > I'm trying to solve a stokes system with some funny boundary conditions
> > so I've stripped the code to make sure I was getting the correct answers
> > to a very simple system as follows:
> > governing equations are:
> > div grad u + grad p = f
> > div u = 0
> > i'm setting u=0 on the boundaries, giving u=0 (or close enough)
> everywhere.
> > setting f = -(0,1), i should get a linear profile with pressure
> > increasing with depth at the same rate going downwards.
>
> Yes. But if you impose Dirichlet boundary values for the velocity all
> around the boundary, then the pressure is only determined up to a
> constant, i.e., the pressure should be a linear function but the offset
> is not determined by the equation.
>
>
> > I'm clearly not understanding how to use
> > VectorTools::interpolate_boundary_values correctly, and I would
> > appreciate some clarification on it.
> >
> > I'm imposing a zero pressure at the top of my rectangle with height of
> > 10, which means I should get 10 at the bottom for pressure.
>
> You can't do that. If you have Dirichlet boundary conditions for the
> velocity, then you can't impose anything on the pressure. In fact, for
> the incompressible Stokes equations, you can never impose any pressure
> boundary conditions.
>
> There is a description of boundary conditions for the Stokes equations
> in the introduction of step-22. Take a look there to see what you can
> and cannot impose for the Stokes equations.
>
> Best
> W.
>
> --
> ------------------------------------------------------------------------
> Wolfgang Bangerth email: [email protected]
> <javascript:>
> www: http://www.math.colostate.edu/~bangerth/
>
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