Timeline for Continuous vs discontinuous pressure elements in fluid flow problems
Current License: CC BY-SA 3.0
8 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Dec 3, 2016 at 22:53 | vote | accept | Lukas Bystricky | ||
| Feb 5, 2016 at 11:15 | comment | added | knl | 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. | |
| Feb 5, 2016 at 1:44 | history | tweeted | twitter.com/StackSciComp/status/695422690505547777 | ||
| Feb 4, 2016 at 22:59 | answer | added | Wolfgang Bangerth | timeline score: 4 | |
| Feb 4, 2016 at 22:55 | comment | added | Wolfgang Bangerth | $P_k - P_{-(k-1)}$ (minus indicating discontinuous spaces) may be unstable, but $Q_k - P_{-(k-1)}$ is stable on quadrilaterals. | |
| Feb 4, 2016 at 20:38 | comment | added | Lukas Bystricky | 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). | |
| Feb 4, 2016 at 19:07 | comment | added | knl | 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. | |
| Feb 4, 2016 at 16:29 | history | asked | Lukas Bystricky | CC BY-SA 3.0 |