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 | |
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| Dec 4, 2016 at 3:24 | history | edited | Wolfgang Bangerth | CC BY-SA 3.0 |
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| Dec 3, 2016 at 22:53 | vote | accept | Lukas Bystricky | ||
| Feb 7, 2016 at 21:19 | comment | added | Wolfgang Bangerth | You can't have continuous $P_{k-1}$ on quadrilaterals. But it's stable with discontinuous $P_{k-1}$. | |
| Feb 7, 2016 at 8:54 | comment | added | Christian Waluga | @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. | |
| Feb 6, 2016 at 15:46 | comment | added | Lukas Bystricky | So am I correct in assuming that $Q_k - P_{k-1}$ (continuous pressures) is also stable? | |
| Feb 6, 2016 at 13:52 | comment | added | Wolfgang Bangerth | @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. | |
| Feb 5, 2016 at 7:36 | comment | added | Christian Waluga | 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? | |
| Feb 4, 2016 at 22:59 | history | answered | Wolfgang Bangerth | CC BY-SA 3.0 |