Timeline for Strong positivity of Neumann Laplacian
Current License: CC BY-SA 4.0
26 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Dec 21, 2022 at 22:23 | comment | added | Bogdan | Thanks a lot Mr. Metafune! Now it is all clear. | |
| Dec 21, 2022 at 15:49 | comment | added | Giorgio Metafune | Both arguments work. If you use the parabolic Sobolev embedding you get continuity in both variables (and you need $p>(N+2)/2$, as you wrote). But also the semigroup approach works. For $t>0$ the map $t \to u(t, \cdot)$ is continuous from the open half line to $W^{2,p}$, hence to $C(\bar{\Omega})$ for $p>N/2$ which gives continuity in both variables. | |
| Dec 21, 2022 at 6:42 | comment | added | Bogdan | Source: Daners, Medina - Abstract evolution equations, periodic problems and applications, page 240, Theorem A3.14). Am I right? | |
| Dec 21, 2022 at 6:42 | comment | added | Bogdan | @GiorgioMetafune Thanks a lot for the reference! Now I see. However there is another problem. Using Sobolev embedding for $p>N/2$ we get $W^{2,p}(\Omega)\hookrightarrow C(\overline{\Omega})$ will give indeed $u(t,\cdot)\geq \epsilon_t>0$. But that does not mean that there is some $\epsilon>0$ such that $u>\epsilon$ for any $(t,x)$. I found an interesting embedding result for the parabolic Sobolev space $W^{2,1}_p((0,T)\times\Omega)\hookrightarrow C(\overline{\Omega}\times [0,T])$, but it requires $p>(N+2)/2$ | |
| Dec 20, 2022 at 21:58 | comment | added | Giorgio Metafune | You find the version for parabolic equations in Chapter 2 of "Partial Differential equations of parabolic type" by A. Friedman. See in particular Section 5, Theorem 14. | |
| Dec 20, 2022 at 20:37 | comment | added | Bogdan | @GiorgioMetafune But the Hopf lemma works just for elliptic PDE's, isn't it? How can we apply it for parabolic equations? I did not find such a version in the classical references on PDE's. So my question is how is Hopf lemma applied here (what version, what hypotheses requires)? | |
| Oct 20, 2022 at 12:45 | comment | added | MaoWao | @GiorgioMetafune Thank you. The Hopf lemma was the ingredient I didn't see (and of course that's where things fail for Dirichlet boundary conditions). | |
| Oct 20, 2022 at 11:06 | comment | added | Giorgio Metafune | @MaoWao At the boundary I use Hopf lemma. Assume that $f \geq 0$, so that $u \geq 0$ and regular up to the boundary. If $u(t_0,x_0)=0$, with $x_0 \in \partial \Omega$, then $(t_0,x_0)$ is a global minimum and$\frac{\partial u}{\partial \nu} (t_0,x_0)=0$ by the boundary condition. But then Hopf boundary point lemma implies that $u$ is a constant. Please, let me know if something is not convincing. | |
| Oct 20, 2022 at 10:43 | comment | added | MaoWao | @GiorgioMetafune How do you obtain that $u(t,\cdot)$ is positive on the closure of $\Omega$? In general, I don't understand which part of your argument uses properties that do not hold for Dirichlet boundary conditions. Could you elaborate? | |
| Oct 19, 2022 at 11:49 | comment | added | Bogdan | Now I see. Thank you very much! | |
| Oct 19, 2022 at 7:22 | comment | added | Giorgio Metafune | $u$ is continuous only for $t>0$, unless $f$ is continuous, too. The approach I have in mind is based on regularity up to the boundary and I need $\partial \Omega \in C^2$. The semigroup is analytic in the whole scale of $L^p$ with domain included into $W^{2,p}$. Choosing $p>N/2$, by Sobolev embedding one obtains thjat $u(t, \cdot)$ is continuous. A standard reference is the bool by A. Lunardi "Analytic semigroups and optimal regularity in parabolic problems". the approach described by @Jochen Glueck is surely more efficient. | |
| Oct 19, 2022 at 3:32 | history | edited | Bogdan | CC BY-SA 4.0 |
added 1 character in body
|
| Oct 19, 2022 at 3:22 | comment | added | Bogdan | @GiorgioMetafune Please, can you suggest me some results/references where I can find more about this regularizing effect? I do not see why $u$ has to be continuous on $\overline{Q_T}$... Thanks! | |
| Oct 18, 2022 at 18:19 | vote | accept | Bogdan | ||
| Oct 18, 2022 at 18:19 | comment | added | Jochen Glueck | @GiorgioMetafune: There's a small issue with this type of argument, though: it only works if one has a regularizing effect up to the boundary. This if fine if one has smooth boundary (as assumed in the question), but for rough boundaries it causes problems (but this can be solved by a more abstract approach, as outlined in my answer). | |
| Oct 18, 2022 at 18:16 | answer | added | Jochen Glueck | timeline score: 7 | |
| Oct 18, 2022 at 17:18 | comment | added | Giorgio Metafune | @bogdan Yes, true. One has to add also a regularizing effect of the semigroup for positive $t$ . In this case the solution is continuous, positive at every point and then the infimum is positive. However, this argumet does not give any quantittave bound. | |
| Oct 18, 2022 at 8:42 | comment | added | Jochen Glueck | For rather general situations (e.g. rough boundary conditions where Hopf's boundary lemma doesn't apply) combining irreducibility with ultracontractivity and a bit of abstract nonsense does the trick. I'll explain the details in an answer as soon as I have a bit more time left. | |
| Oct 18, 2022 at 8:29 | comment | added | MaoWao | Minor remark: Such a result depends on the choice of boundary conditions. For Dirichlet boundary conditions, it is false, while the "positivity improving" property that comes from irreducibility still holds. | |
| Oct 18, 2022 at 8:16 | comment | added | Bogdan | But irreducibility does not imply that $\operatorname{ess\ inf} u>0$ on $Q_T$... | |
| Oct 18, 2022 at 7:58 | comment | added | Giorgio Metafune | As pointed out in the other comments, one possibility is to use the strong maximum principle (see Ch. 2, Sec. 5 in the book of Friedman "Partial differential equations of parabolic type"). A functoipnal analytic approach, which uses "irreducuibility" and form methods, can be found in the book by Ouhabaz "Analysis of the heat equation in domains", see Chapter 4. | |
| Oct 18, 2022 at 7:07 | comment | added | Hannes | $\Omega$ should also be connected for such a statement to possibly be true, right? | |
| Oct 18, 2022 at 6:59 | comment | added | leo monsaingeon | Strong positivity follows from the strong maximum principle, doesn't it? | |
| Oct 17, 2022 at 20:10 | history | edited | Bogdan | CC BY-SA 4.0 |
deleted 4 characters in body
|
| Oct 17, 2022 at 18:15 | history | edited | Bogdan | CC BY-SA 4.0 |
added 35 characters in body
|
| Oct 17, 2022 at 18:09 | history | asked | Bogdan | CC BY-SA 4.0 |