# Continuous vs discontinuous pressure elements in fluid flow problems

When solving fluid flow problems using finite elements you typically end up requiring that $p\in L^2$.

For Darcy flow problems, a popular choice of elements is the Raviart-Thomas element and discontinuous pressure elements.

For Stokes/Navier-Stokes flow a popular choice of elements is the Taylor-Hood element which has piecewise continuous pressure elements. This is actually more continuity than is required, because for a function to be in $L^2$ it does not have to be continuous.

Is there some reason that discontinuous pressure elements are preferred (or necessary) for Darcy flow problems, but not for Stokes/Navier-Stokes flow problems?

• If I recall correctly then $P_k-P_{k-1}$ (with discontinuous pressure) is stable only for $k\geq4$. If you use continuous pressure, then $P_2-P_1$, i.e. Taylor-Hood, works. Thus, one might not want to use such a large polynomial degree for example due to reduced regularity. – knl Feb 4 '16 at 19:07
• That may be true, but I should have mentioned that in both cases pressure is typically just linear elements (continuous for Stokes, discontinuous for Darcy). – Lukas Bystricky Feb 4 '16 at 20:38
• $P_k - P_{-(k-1)}$ (minus indicating discontinuous spaces) may be unstable, but $Q_k - P_{-(k-1)}$ is stable on quadrilaterals. – Wolfgang Bangerth Feb 4 '16 at 22:55
• Yeah, I checked from my references that $Q_2 - P_{-1}$ (as you call it) is stable. However, $Q_1-P_0$ is not. A classic counterexample of a non-unique pressure is the checkerboard mode with rectangular mesh. – knl Feb 5 '16 at 11:15

Using discontinuous pressure spaces has the advantage that the solution is cellwise conservative (because you can test the equation $\nabla \cdot u=0$ with the characteristic function of each element). Thus, there is an advantage for using the $Q_k - P_{-(k-1)}$ element on quadrilateral/hexahedra. People do indeed use this element in practice.
On the other hand, this implies 3 pressure degrees per cell (in 2d) as opposed to roughly one per cell for the $Q_1$ element. So it's more expensive. But a practical observation is that it is not more accurate, but in fact often less accurate than just the regular Taylor-Hood element. The choice therefore comes to "better accuracy and cheaper" (Taylor-Hood) vs "locally conservative but more expensive" (discontinuous pressures).
• So am I correct in assuming that $Q_k - P_{k-1}$ (continuous pressures) is also stable? – Lukas Bystricky Feb 6 '16 at 15:46
• You can't have continuous $P_{k-1}$ on quadrilaterals. But it's stable with discontinuous $P_{k-1}$. – Wolfgang Bangerth Feb 7 '16 at 21:19