Timeline for Taylor expansion of error - Finite elements approximation
Current License: CC BY-SA 3.0
14 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Aug 9, 2023 at 20:13 | comment | added | Wolfgang Bangerth | @feynman That's not true. People have proven error estimates for FD schemes since at least the 1970s. | |
| Aug 8, 2023 at 12:56 | comment | added | feynman | I mean in general for any FD schemes no error inequalities like those in FE ones are seen | |
| Aug 8, 2023 at 12:26 | comment | added | Wolfgang Bangerth | @feynman I still don't understand the question. Why should people write papers about the detailed behavior of schemes that are useless? | |
| Aug 7, 2023 at 13:05 | comment | added | feynman | sorry the question could be rephrased as: as long as the PDE is of high regularity there's such an error inequality for FE convergence, but for the same PDE FD schemes (like forward Euler) could diverge and nobody mentions such error inequalities for FD. Why is that? | |
| Aug 7, 2023 at 10:29 | comment | added | Wolfgang Bangerth | @feynman I don't understand the question. No "reasonable" FD scheme diverges; no reasonable FE scheme diverges either. If a scheme diverges, then just don't use it -- it's not good for anything. The same is true for a scheme that converges, but to something that is not the solution. | |
| Aug 6, 2023 at 20:24 | comment | added | feynman | @WolfgangBangerth thank you. Then if a finite difference (FD) scheme diverges, the above Taylor expansion also diverges for a FD. But why does this divergence issue never happen to a FE? | |
| Aug 6, 2023 at 12:18 | comment | added | Wolfgang Bangerth | @feynman I see no reason why this argument couldn't be applied to finite differences just as well. But note that I made no assumption that a scheme actually converges: $\lambda(h=0)$ is simply the limit as the mesh is made finer and finer. I'm not assuming that $\lambda(0)=\lambda_\text{exact}$. | |
| Aug 5, 2023 at 21:00 | comment | added | feynman | @WolfgangBangerth Thank you very for this great answer. A follow up question, in that case why does this Taylor expansion argument not apply to finite differences, for otherwise all finite differences can converge? | |
| Mar 5, 2018 at 2:13 | comment | added | Wolfgang Bangerth | @BeniBogosel -- nope, it's just my intuition, backed by 20 years of experience. | |
| Mar 3, 2018 at 13:06 | comment | added | Beni Bogosel | Do you have any reference in mind for your statement: "if this dependence is continuous, then you can do a Taylor expansion around 0"? Thank you. | |
| Feb 24, 2018 at 21:30 | vote | accept | Beni Bogosel | ||
| Feb 24, 2018 at 20:43 | comment | added | Wolfgang Bangerth | @BeniBogosel -- yes, at least if the eigenvalue is single (i.e., has multiplicity one) I would expect that to be true. | |
| Feb 23, 2018 at 12:10 | comment | added | Beni Bogosel | Thank you for your answer. In my case the function $\lambda(h)$ is the first eigenvalue of an elliptic operator (Laplace-Beltrami). It verifies something like $-\Delta_h u_h = \lambda_h u_h$, where $u_h$ is non-negative (the first eigenfunction) and $L^2$ normalized. I guess in this case $\lambda(h)$ should be regular enough for the development to work... | |
| Feb 22, 2018 at 23:45 | history | answered | Wolfgang Bangerth | CC BY-SA 3.0 |