# Imposing total pressure over surface in FEM

I am trying to solve Stokes problem using Finite element method.

My question is how to impose that total pressure over the surface is zero to remove the constant pressure mode?

You must add an additional equation to your system. It is: $$\int_{\Gamma_p}{p\,d\sigma}=0\implies \sum_{p_i\in \Gamma_p}p_i\int_{\Gamma_{pi}}{\phi(t)\,dt}=0$$

Where $\phi(t)$ is the basis function of $p$ along the parameterised boundary described with the parameter $t$ in the element $\Gamma_{pi}$. Remember that $\Gamma_p=\cup{ \Gamma_{pi}}$

For Finite Element formulation for fluid flow, it is sufficient to impose zero-pressure on one of the nodes.

Usually, this node is chosen on the outflow boundary. If the outflow boundary does not exist, then pick a node far away from the region(s) of steep pressure gradients.

• How is my answer not helpful? Can the person who downvoted my answer care to explain? Otherwise, I have to assume that there are many ignorant users here who are just wanna be experts. Aug 13 '17 at 21:59
• It wasn't me. Your answer is OK -- that's how you can eliminate the constant in practice. Although, you will not get a zero mean (which was part of the question too). Your second remark can be very misleading: Typically, if you have outflow, the pressure is completely defined. If you then try to pin it, you will run into troubles... (That's why I don't upvote either)
– Jan
Sep 10 '17 at 4:27
• Unless otherwise specified, the pressure at the outflow is prescribed as zero. For mixed-Galerkin FE formulation, we don't have to do anything for this case because the contribution of traction term to the RHS is zero. That is why it is sufficient to fix one pressure node on the outflow. For the cases with non-zero pressure at the outflow, one can fix a pressure node (at the outflow) with the specified non-zero value to avoid numerical issues. I didn't write this explicitly as I did not want to spoon feed. Sep 10 '17 at 16:49