Thank you all for the responses.
Colin: you're right no need to specify zero pressure strongly. Yes I'm checking against exact solution U_x = y(1-y), U_y = 0.
Stephan: I think I understand what you mean but can you just check below. I have tried specifying tangential velocity on the boundary strongly - that did not work (solution just didn't anything like the exact). So I tried adding the boundary integral term into the formulation weakly following the reasoning below:
when I include that term, however the L2 norm drops only by a half (and I still get vertical flow near boundary), that is:
without the boundary integral term L2 = 0.00677501099488
with the boundary term L2 = 0.00339740206367
excluding the (d v_j / d x_i ) term by hand gives L2 = 5.86838100481e-14 ...
I must be doing something wrong.
Sent: 23 April 2016 08:56
To: firedrake
Subject: Re: [firedrake] Pressure outlet boundary condition in plane poiseuille flow
On 22/04/16 18:49, Colin Cotter wrote:
> OK, are you trying to match an analytical solution?
>
> I think it depends on whether you integrated by parts in the grad term,
> or the divergence. If you integrated by parts in the grad term, then you
> can get the 0 pressure boundary condition naturally i.e. without setting
> anything.
>
> all the best
> --cjc
To add to that, the natural ("do nothing") boundary condition leads to a
zero stress condition. This is I think also where the difference between
including and excluding the grad-div term comes from. The tangential
component of stress is different (dv/dx vs. dv/dx+du/dy) and is not
sufficiently constrained by the boundary conditions. If you include the
grad-div term the tangential stress is non-zero and so you either have
to explicitly add that as a boundary term to weakly enforce the correct
value of dv/dx+du/dy, or you apply a strong dirichlet on the tangential
component of velocity. Without the grad-div term the "do-nothing" bc for
the tangential stress automatically enforces dv/dx=0.
Cheers
Stephan
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