Timeline for Eigenvalues and eigenfunctions of the Laplace operator on entire plane
Current License: CC BY-SA 4.0
20 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jan 29, 2022 at 18:13 | comment | added | UserA | @GiorgioMetafune Thank you very much! Email sent. | |
| Jan 29, 2022 at 18:02 | comment | added | Giorgio Metafune | I checked better the book. Some papers are in italian but this one is in English. If you need I can scan and send a copy via mail. Then you should write to me via mail to [email protected] | |
| Jan 29, 2022 at 18:00 | comment | added | UserA | @GiorgioMetafune I searched quite a lot and was not able to find Symposia Mathematica vol VII, Academic Press 1971 on Google. Do you have any idea where I can find it? | |
| Jan 29, 2022 at 13:23 | comment | added | Giorgio Metafune | I doubt, I have my personal copy | |
| Jan 29, 2022 at 13:16 | vote | accept | UserA | ||
| Jan 29, 2022 at 12:54 | comment | added | UserA | @GiorgioMetafune I only found the Italian version on Springer. Any other place where I can download the paper? | |
| Jan 29, 2022 at 12:52 | comment | added | Giorgio Metafune | @JochenGlueck Since you are interested I go on. If $\Delta u+k^2 u=0$ ($K \neq 0$) and $u$ is radial, then setting $u(r)=r^{(1-n)/2}w(r)$ you get a Bessel equation whose solutions at infinity oscillate lile $\sin$. Then the asympotics for $v$ at inifinity is like $r^{(1-n)/2}$ which is in $L^p$ for $p>2n(n-1)$. | |
| Jan 29, 2022 at 12:51 | comment | added | UserA | @JochenGlueck neither was I and this is actually a remarkable property! On another another, I think I made a mistake by calling the trivial pair $(\lambda, \hat u)=(0,0)$ an eingenpair. | |
| Jan 29, 2022 at 12:48 | comment | added | Giorgio Metafune | It is in English | |
| Jan 29, 2022 at 12:47 | comment | added | UserA | @GiorgioMetafune does the paper you referenced have an English translation? | |
| Jan 29, 2022 at 12:31 | comment | added | Jochen Glueck | @GiorgioMetafune: Thanks a lot for your reply, and for the explanation of how to construct the eigenfunctions! Quite embarrassingly for me, I really wasn't aware (before reading your post) that $\Delta$ has any point spectrum on $L^p$ for sufficiently large $p$) | |
| Jan 29, 2022 at 12:27 | history | edited | Giorgio Metafune | CC BY-SA 4.0 |
added 8 characters in body
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| Jan 29, 2022 at 12:26 | comment | added | Giorgio Metafune | @JochenGlueck Jochen you are right, the argument works for the spectrum minus 0. Of course, there is a way of avoiding semigroups. For $l\lambda$ not a negative number (and not 0), write formally the inverse through the Fourier transform and then use Mikhlin to check the boundedness. If $\lambda<0$ one shows it is an approximate eigenvlaue (the functions $e^{iax}$ should be an approximate eigenvector). For the point spectrum, one first show that if there is an eigenfuntions, then there is a radial one, by averaging, and then ends with a Bessel equation whose asymptotic is known. | |
| Jan 29, 2022 at 12:23 | comment | added | Jochen Glueck | @UserA: I'm not sure I understand your question. What I wrote doesn't contradict your computation, does it? | |
| Jan 29, 2022 at 12:14 | comment | added | UserA | @JochenGlueck I think I computed the Fourier transform properly, right? So where does this problem come from? | |
| Jan 29, 2022 at 12:09 | comment | added | Jochen Glueck | Hmm, maybe I'm misunderstanding something - but $0$ cannot be an eigenvalue of $\Delta$ for any $p < \infty$ since the heat semigroup converges strongly to $0$ on $L^p$ (as $t \to \infty$) for every $p \in (1,\infty)$. | |
| Jan 29, 2022 at 10:59 | comment | added | UserA | How is the heat semi-group defined? | |
| Jan 29, 2022 at 10:37 | comment | added | Giorgio Metafune | The full spectrum is in any case $(-\infty, 0]$, for every $p$. The inclusion in the half line follows from semigroup theory, since the heat semigroup is bounded holomorphic of angle $\pi/2$, the reverse inclusion by showing that negative numbers are approximate eigenvalues. | |
| Jan 29, 2022 at 10:31 | comment | added | UserA | Thank you very much for the reference. However, it remains to show that for $p>2n/(n-2)$ (so in our case $p>4$) and $\Delta:W^{m,p}\to L^p$ , the spectrum is still given by $\sigma(\Delta)=(\infty,0]$. I have a proof of this using the Fourier transform that works, but only for $1\leq p\leq 2$. However, in our case, we need to study the spectrum for $p>4$! | |
| Jan 29, 2022 at 10:18 | history | answered | Giorgio Metafune | CC BY-SA 4.0 |