Timeline for Positive first eigenvalue; operator satisfies maximum principle
Current License: CC BY-SA 4.0
7 events
| when toggle format | what | by | license | comment | |
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| Dec 17, 2021 at 12:03 | comment | added | Math604 | thanks for the comment. It would take me a while to find the paper (I actually was looking at the paper a long time ago and didn't understand it then...but only recently thought I wanted to learn it since I might want to apply it). Ya I agree that it seems problematic for general $\phi$ but maybe for that particular one it works. If I can find the reference I will post it. | |
| Dec 17, 2021 at 11:40 | comment | added | Mateusz Kwaśnicki | If $f$ is increasing, i.e. if $f'(u) \geqslant 0$, then the maximum principle fails for the linearised equation $-\Delta \phi - f'(u) \phi = 0$: if we set $\phi = 1$ on the boundary, then $\phi > 1$ in the interior, unless $f'(u) = 0$ a.e. in $\Omega$. Of course, this does not necessarily mean that the maximum principle fails for the particular choice of $\phi$ such as $\phi = u_{x_i}$. Can you give us a reference to the paper where you found this argument? | |
| Dec 17, 2021 at 7:36 | answer | added | username | timeline score: 1 | |
| Dec 17, 2021 at 7:19 | comment | added | Math604 | so in general f is increasing... its really coming from a Gelfand problem like $ -\Delta u = \lambda f(u)$ and the minimal solution is stable. Typical $f$ are $ f( u ) = e^u$ and $ f(u) = (u+1)^p$ for $p>1$. In all reality I don't overly care about this particular problem. Its this maximum principle idea that I would like to understand to be able to apply it in some other places. | |
| Dec 17, 2021 at 7:06 | comment | added | Mateusz Kwaśnicki | @username: If $f$ were decreasing, $u$ would automatically be stable, so I suppose it is not. :-) | |
| Dec 17, 2021 at 7:03 | comment | added | username | $f$ is decreasing? | |
| Dec 17, 2021 at 4:47 | history | asked | Math604 | CC BY-SA 4.0 |