Greetings fellow members,

I'm trying to implement a Discontinuous Galerkin scheme for a Stokes flow (Poiseuille). While I get very satisfactory results on the velocity, I'm suprised with negative values for the pressure. I know this is due to imposing a of zero-value for the mean value of the pressure so as to retrieve uniqueness (theoretically), but how to retrieve the "physically meaningful" value ? I tried adding numerous terms to the formulation (of the form integral over interior/boundary alpha * jump(p) * jump(q), alpha * jump(dn(p)) * jump(dn(q)) ..) with no success. Any idea ?

Thank you, and best regards.


1 Answer 1


In the Stokes equations, there is no "physically reasonable" pressure. The pressure is only defined up to an additive constant. Every pressure you move up or down by a constant is physically reasonable.

In practice, we often want to compare solutions to what we get from experiments, however, and in those cases we normalize so that not the average pressure is zero, but the average along the top surface is zero.

  • $\begingroup$ For Poiseuille flow, wouldn't making the average at outflow boundary zero be more meaningful? I believe this is the canonical way since it leads to positive pressure at inflow. But I also saw many variations like making the average pressure zero at inflow boundary, or prescribing a pressure value at a point in the channel. $\endgroup$ Commented Apr 27 at 5:35
  • $\begingroup$ @AbdullahAliSivas Sure. Any kind of convention of the sort "average is zero at some part of the boundary" makes sense, depending on the application. $\endgroup$ Commented Apr 27 at 7:07
  • $\begingroup$ Thank you for the replies. I need the pressure to be 400 at inflow and 0 at outflow. I'm using nitsche's method to enforce boundary conditions. What I get is 200 at inflow and -200 at outflow. In a way I need to move the pressure up by 200.. Any idea how to enforce this ? A reference maybe ? $\endgroup$
    – Aubium
    Commented Apr 28 at 0:05
  • $\begingroup$ @Aubium You let the solver do its thing, get whatever pressure it produces, and then you add a constant to the pressure in such a way that you get what you want at in- and out-flow. $\endgroup$ Commented Apr 28 at 1:54

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