Timeline for Dependence of Neumann eigenvalues on the domain
Current License: CC BY-SA 4.0
11 events
| when toggle format | what | by | license | comment | |
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| Oct 25, 2023 at 8:47 | comment | added | Manuel Cañizares | I was trying to use this generic simplicity to then calculate the derivative of the eigenvalues with respect to perturbation of the domain in a given direction. | |
| Oct 25, 2023 at 8:46 | comment | added | Manuel Cañizares | Does degeneracy then destroy your argument, @username? There is a concept that I haven't yet fully grasped, called generic simplicity (see e.g. this paper, section 5), that kind of says that there is a big amount of domains for which all eigenvalues are simple. | |
| Oct 24, 2023 at 20:20 | comment | added | username | I noticed that I assumed something on the distribution of eigenvalues of the laplacian so I asked if it was true. | |
| Oct 19, 2023 at 6:29 | comment | added | Manuel Cañizares | @ChristianRemling I like that idea, I will try it to make it rigurous, thank you! However, I deliberately omitted a detail of the problem in my question to make it easier and less specific to me. This is, I need the boundary of the domain to include a prescribed small open set (open in the $n-1$ dimensional topology). Therefore in principle one can't just consider concentric balls. Maybe the idea can be adapted though. | |
| Oct 19, 2023 at 5:37 | comment | added | Christian Remling | If we just take balls $B_r(0)$ and increase $r$, we'd have to run into a rather weird scenario to keep a fixed $\lambda$ an eigenvalue throughout. Since the operator on the whole space has essential spectrum $[0,\infty)$, all eigenvalues have to move towards zero (though perhaps not monotone). So $\lambda_n(r)<\lambda$ eventually, and if $\lambda$ stays an eigenvalue, that means that it must become degenerate whenever a $\lambda_n(r)$ crosses $\lambda$. | |
| Oct 17, 2023 at 5:56 | history | edited | Manuel Cañizares |
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| Oct 16, 2023 at 22:11 | comment | added | Manuel Cañizares | I completely agree with you Christian. Maybe the proof will be easy in hindsight, but I haven't been able to put the argument together so far. | |
| Oct 16, 2023 at 21:59 | comment | added | Christian Remling | It's quite annoying that this would be hard to prove since common sense of course suggests that it will be difficult to find a domain for which this is not true. | |
| Oct 16, 2023 at 21:19 | comment | added | Manuel Cañizares | Thank you for your answer. In principle, there isn't any hypothesis for $V$ in that regard. I'm happy if you can write the answer with that hypothesis, nonetheless. Maybe I can work from there to relax that hypothesis. | |
| Oct 16, 2023 at 18:14 | comment | added | username | Is $V\geq 0$, or bounded below? If that is the case, what you suggested works well (I can write it as an answer if you like). If not, it might work as well, but it looks a bit more technical at first sight. | |
| Oct 16, 2023 at 13:22 | history | asked | Manuel Cañizares | CC BY-SA 4.0 |