Timeline for Asymptotic formula for fractional Laplacian
Current License: CC BY-SA 4.0
23 events
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| Apr 26, 2021 at 22:55 | comment | added | Jun | Also, about $1_{\Omega^c}$ not being in the domain of $(-\Delta)^s$: I know that usually one requires a zero condition on $\Omega^c$, but haven't nonhomogeneous fractional problems been studied? | |
| Apr 26, 2021 at 22:43 | comment | added | Jun | I see. And then $u^\epsilon/\epsilon \to g(x)$. But I'm maybe missing a piece: from $$ \epsilon^{-2} w + \epsilon^{-1} (-\Delta)^s w(x) = \epsilon^{-1} g(x) $$ we also get $w/\epsilon \to 0$ instead of $\to g(x)$? | |
| Apr 26, 2021 at 22:19 | comment | added | Mateusz Kwaśnicki | We do have $\lim_{\epsilon \to 0^+} u^\epsilon(x) = 0$, just as in the classical case. It is only the rate of convergence that changes: exponential of $-1/\epsilon$ in the classical case, and linear in $\epsilon$ in the non-local case. | |
| Apr 26, 2021 at 22:09 | comment | added | Jun | I see. Thanks! By the way, is an $\epsilon^{-1}$ missing in front of $g(x)$? Otherwise (formally), we would get $w = 0$ in the limit | |
| Apr 26, 2021 at 20:42 | comment | added | Mateusz Kwaśnicki | Yes, that's right. (I would not call this "change of variables", though.) A word of caution: neither $w$ nor $\mathbb{1}_{\Omega^c}$ belongs to any of the usual domains of $(-\Delta)^s$ (at least not if $s > \tfrac12$). I'm not experienced in applying PDE methods, but I suppose making all of this rigorous is not straightforward. | |
| Apr 26, 2021 at 20:14 | comment | added | Jun | Thanks! That's perfect. What is the change of variables that leads from the PDE for $u$ to the one for $w$? I guess $w = u - 1_{\Omega^c}$, if so is $g(x) = (-\Delta)^s 1_{\Omega^c} $? | |
| Apr 26, 2021 at 19:52 | comment | added | Mateusz Kwaśnicki | There you go. This is very informal — I could add more details if you like when I have time, but this will not be before schools in Poland are open again. :-) | |
| Apr 26, 2021 at 19:51 | history | edited | Mateusz Kwaśnicki | CC BY-SA 4.0 |
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| Apr 26, 2021 at 15:29 | comment | added | Jun | Thank you so much for all your help. One thing remains unclear to me: at the PDE level, at least formally, how does one obtain this result? This is related to the question mathoverflow.net/questions/390439/… (and to some extent to mathoverflow.net/questions/390207/…) | |
| Apr 26, 2021 at 15:05 | vote | accept | Jun | ||
| Apr 21, 2021 at 17:08 | comment | added | Mateusz Kwaśnicki | This does not give exact asymptotics, just a two-sided bound. Asymptotics for a given point should be much easier, but as I said before, I do not have a reference. You may like to check papers by Bogdan, Byczkowski and co-authors, and by Chen, Song and co-authors, from 1997–2005. | |
| Apr 21, 2021 at 17:05 | comment | added | Mateusz Kwaśnicki | I had no time to think about that, but it looks like one can get a two-sided bound by using two-sided estimates of the heat kernel $p_t^D(x,y)$; note that $\mathbb E^x e^{-k\tau_D} = \int_0^\infty \int_D e^{-kt} p_t^D(x,y) dy dt$. For the bounds on $p_t^D$, see DOI:10.4171/JEMS/231 and DOI:10.1214/10-AOP532. | |
| Apr 21, 2021 at 9:12 | comment | added | Jun | Hello. In the end, did you find the reference for the estimate on $\mathbb E^x e^{-k\tau_D}$? | |
| Apr 15, 2021 at 7:39 | comment | added | Jun | Thanks! But as $\epsilon \to 0$, we formally get that the limit of $u^\epsilon/\epsilon$ should solve $u= 0$ in $\Omega$ and $u=\infty$ in $\mathbb R^N \setminus \Omega$. Is this right? | |
| Apr 14, 2021 at 10:34 | comment | added | Mateusz Kwaśnicki | @Jun: I believe if you read what you just wrote carefully, you will realise right away that $I(x) > 0$. :-) | |
| Apr 14, 2021 at 10:05 | comment | added | Jun | Thanks for the update. Is it then true that $I(x) = \int_{\mathbb R^N \setminus \Omega} |x - z|^{-N-2s} dz $ satisfies $I(x) = 0$ for $x \in \Omega$ and $I(x) = \infty$ for $x \in \mathbb R^N \setminus \Omega$? | |
| Apr 14, 2021 at 9:20 | history | edited | Mateusz Kwaśnicki | CC BY-SA 4.0 |
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| Apr 14, 2021 at 8:36 | comment | added | Mateusz Kwaśnicki | I cannot parse your question: the only solution of $\lambda u = 0$ is of course $u = 0$, is it not? By the way, I realise the "$\dist(x,\partial \Omega)$" term is not quite correct, I'll update the answer momentarily. | |
| Apr 14, 2021 at 8:21 | comment | added | Jun | Thanks! Formally, is it clear that $\lambda^{-1}\mathrm{dist}(x,\partial\Omega)^{-2s} $ solves the limit equation $\lambda u = 0$ in $\Omega$ and $u=\infty$ in $\mathbb R^N \setminus \Omega$? | |
| Apr 10, 2021 at 9:18 | comment | added | Mateusz Kwaśnicki | I think this works as expected: $u^\epsilon / \epsilon$ is equal to $1/\epsilon$ in the complement of $\Omega$, and this goes to infinity as $\epsilon \to 0^+$. | |
| Apr 10, 2021 at 8:54 | comment | added | Jun | Thanks! Isn't it also strange that the right-hand side in the conjecture does not seem to satisfy the boundary condition $u^\epsilon =1$ in $\mathbb R^N \setminus \Omega$? | |
| Apr 10, 2021 at 8:00 | comment | added | Mateusz Kwaśnicki | Time permitting, I'll think about references. As $s \to 1$, the constant $C(s)$ will converge to zero, so in a sense we do recover the local case in a very weak form. :-) This is one of the numerous results where there is a "phase transition" at $s = 1$. The simplest one (and one that is closely related to this question) is the estimate of the heat kernel: for a fixed $t$, it is $(1+|x|)^{-N-2s}$ when $s < 1$, but $\exp(-|x|^2)$ when $s = 1$. | |
| Apr 9, 2021 at 21:22 | history | answered | Mateusz Kwaśnicki | CC BY-SA 4.0 |