Timeline for A differential inequality involving gradient and laplacian
Current License: CC BY-SA 4.0
13 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jul 6, 2020 at 14:15 | vote | accept | tituf | ||
| Jul 5, 2020 at 12:10 | history | bounty ended | tituf | ||
| Jul 3, 2020 at 19:06 | comment | added | Piero D'Ancona | Maybe, but I do not see how to use this extra freedom | |
| Jul 2, 2020 at 21:59 | comment | added | tituf | If $U$ is allowed to depend on $\alpha$, is everything easier? | |
| Jul 2, 2020 at 18:22 | comment | added | Piero D'Ancona | You could try with a multiplier of the form $\phi(x,V)$. The derivatives w.r.to $x$ should be of lower order, that is, small compared to the main terms. For instance of polynomial growth in $x$ | |
| Jul 2, 2020 at 17:23 | comment | added | tituf | If I weaken the request and ask the inequality to hold only for $0<\alpha<\alpha_0$, then $\Delta V\leq \theta |\nabla V|^2$ for some $\theta<\alpha_0^{-1}$ is a sufficient hypothesis. This can be seen by choosing $U=V$. Do you see any further possibility when only $0<\alpha<\alpha_0$ is considered? | |
| Jul 1, 2020 at 21:46 | comment | added | Piero D'Ancona | It seems difficult, since you want your inequality to hold for arbitrarily large $\alpha$, but I can not exclude it | |
| Jul 1, 2020 at 15:40 | comment | added | tituf | Thank you for your answer. Taking $U=\phi(V)$ is great, but is it possible to weaken the hypothesis on $V$, for example $\Delta V \leq \theta |\nabla V|^2$ for some fixed $\theta$? | |
| Jun 30, 2020 at 18:01 | history | edited | Piero D'Ancona | CC BY-SA 4.0 |
edited body
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| Jun 30, 2020 at 8:13 | history | undeleted | Piero D'Ancona | ||
| Jun 30, 2020 at 8:13 | history | edited | Piero D'Ancona | CC BY-SA 4.0 |
added 140 characters in body
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| Jun 30, 2020 at 8:07 | history | deleted | Piero D'Ancona | via Vote | |
| Jun 30, 2020 at 8:04 | history | answered | Piero D'Ancona | CC BY-SA 4.0 |