Timeline for Question about Neumann eigenvalues on manifolds
Current License: CC BY-SA 4.0
13 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Apr 17 at 20:53 | comment | added | Neal | @Student I was thinking this is the case as well and am quite surprised. | |
| Apr 17 at 19:00 | comment | added | Student | the first eigenvalue of the lune is just 2 and strangely it is independent of the angle. It corresponds to the case when k=0 and l=1 above. | |
| Apr 14 at 13:19 | comment | added | Student | that makes sense but the order of the Legendre function is now fractional so I am not sure how to work with it, but I think that in general, the eigenfunctions should be $\cos\left(\frac{k \pi}{\beta}\theta\right)P^{\frac{k \pi}{\beta}}_l(\cos(r))$ and $\lambda = l (l+1),$ do you agree? | |
| Apr 12 at 22:47 | comment | added | Neal | @Student after separating variables $u=R\Theta$, Neumann conditions mean that the latitudinal function $\Theta(\theta) = \cos(k\pi\theta/\beta)$ with $k=0,1,2...$. So the longitudinal function $R$ can be a Legendre function of order zero, as well as positive order. | |
| Apr 12 at 18:11 | comment | added | Student | but by repeating the ODE argument it seems that they should be the same, what do you think? | |
| Apr 12 at 18:09 | comment | added | Student | that is also the first Dirichlet eigenvalue, see Section 2 here link.springer.com/article/10.1007/s10455-021-09797-y | |
| Apr 12 at 17:26 | comment | added | Neal | @Student you sure that's not the first Dirichlet eigenvalue? | |
| Apr 12 at 6:07 | history | bounty ended | CommunityBot | ||
| Apr 12 at 3:30 | comment | added | Student | The first eigenvalue of this lune is $\frac{\pi}{2\epsilon}(\frac{\pi}{2\epsilon}+1)$ which blows up as $\epsilon\to 0,$ so I am not sure if this argument would work. | |
| Apr 11 at 14:15 | vote | accept | Student | ||
| Apr 11 at 14:15 | |||||
| Apr 11 at 3:46 | comment | added | Neal | Good catch. I think this is fixed by considering a "banana" between two geodesics that intersect at the equator on the boundary of $\mathbb{S}^+$, andhave an angle of $2\epsilon$ between them. This domain's convex (a geodesic connecting two points must be the "long way round" since a shorter path along the bdry can be found), and the estimate is modified with inner integral limits dependent on $s$. I think since integrands are all positive, the upper bound will carry through. If I have time later this week I'll check this and update my answer more carefully. But lmk if you see a flaw here. | |
| Apr 9 at 21:30 | comment | added | Student | Thank you for your answer, but I am not sure if the thin strip is a geodesically convex subset of the hemisphere, what do you think? | |
| Apr 9 at 3:07 | history | answered | Neal | CC BY-SA 4.0 |