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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?

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  • $\begingroup$ 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. $\endgroup$ – knl Feb 4 '16 at 19:07
  • $\begingroup$ 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). $\endgroup$ – Lukas Bystricky Feb 4 '16 at 20:38
  • $\begingroup$ $P_k - P_{-(k-1)}$ (minus indicating discontinuous spaces) may be unstable, but $Q_k - P_{-(k-1)}$ is stable on quadrilaterals. $\endgroup$ – Wolfgang Bangerth Feb 4 '16 at 22:55
  • $\begingroup$ 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. $\endgroup$ – knl Feb 5 '16 at 11:15
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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).

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  • $\begingroup$ What is the problem with the accuracy here? That the P_k pressures do not fit well into the cube meshes and therefore do strange things inside each element, despite being good multipliers in terms of stability? Or does something bad also happen to the velocities? $\endgroup$ – Christian Waluga Feb 5 '16 at 7:36
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    $\begingroup$ @ChristianWaluga: I don't know of any theoretical argument, but my interpretation is that the more pressure shape functions you have, the more constraints the velocity has to satisfy and the less it is able to just minimize the energy. So you get a larger approximation error in the velocity. Of course, if you have too many pressure variables, you'll eventually end up in a situation where the problem can not be solved in a stable way any more at all. $\endgroup$ – Wolfgang Bangerth Feb 6 '16 at 13:52
  • $\begingroup$ So am I correct in assuming that $Q_k - P_{k-1}$ (continuous pressures) is also stable? $\endgroup$ – Lukas Bystricky Feb 6 '16 at 15:46
  • $\begingroup$ @H H: Yes, as far as I remember for k>=2 if the pressures are discontinous (how would you use continuous Pk on cubes?). in detail this should be covered in Boffi/Brezzi/Fortin or other standard books on saddle point problems. $\endgroup$ – Christian Waluga Feb 7 '16 at 8:54
  • $\begingroup$ You can't have continuous $P_{k-1}$ on quadrilaterals. But it's stable with discontinuous $P_{k-1}$. $\endgroup$ – Wolfgang Bangerth Feb 7 '16 at 21:19

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