Timeline for Monotonicity of the top eigenfunction of the generator of a diffusion
Current License: CC BY-SA 4.0
12 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Feb 18, 2022 at 17:36 | vote | accept | Michal Kotowski | ||
| Feb 18, 2022 at 17:16 | answer | added | Giorgio Metafune | timeline score: 1 | |
| Feb 18, 2022 at 13:58 | comment | added | Giorgio Metafune | I will try to do it, even though it is not a full answer but only a way to get it. It is strange, I never thought it could be true, it is false in an interval beacuse the boundary conditions between $u$ and $u'$ change. It was nice that you asked the question. | |
| Feb 18, 2022 at 13:23 | comment | added | Michal Kotowski | Thanks, it is now clear. If you edit your comments as an answer, I'll accept it. | |
| Feb 18, 2022 at 12:26 | comment | added | Giorgio Metafune | The first eigenvalue is simple and the first eigenfunction is positive is a general fact about second order operators. In your situation the operator $B$ is self-adjoint and with a non-negative potential, the first eigenvalue is characterized by a minumum problem and the first eigenfunction is positive. One should adapt the usual argument for uniformly elliptic operators | |
| Feb 18, 2022 at 12:23 | comment | added | Michal Kotowski | OK, so with your definition we have actually $B = A - V''$, where $V''$ is the operator of multiplication by $V''$. But how do we know a priori that the lowest eigenfunction of B is strictly positive? Otherwise the argument seems to me circular. Do I misunderstand something? | |
| Feb 18, 2022 at 10:32 | comment | added | Giorgio Metafune | The notation is unclear, sorry. I have this in mind: if for example $Au=u''-x^3 u'=\lambda u$, then $u'''-x^3 u''-3x^2 u'=\lambda u'$ so that $B=D^2-x^3D-3x^2$. | |
| Feb 18, 2022 at 10:02 | comment | added | Michal Kotowski | I'm sorry, I have trouble following your derivation - why are you claiming that $Bu' = \lambda u'$? | |
| Feb 17, 2022 at 23:24 | comment | added | Giorgio Metafune | It seems that it can be true. Call $A$ your operator and $B=DA$, where $D$ is the derivative. If $Au=\lambda u$ with $\lambda \neq 0$, then $u$ is not a constant and $Bu'=\lambda u'$ with $u' \neq 0$. Conversely, from this last we get $\lambda u-Au= c$ and then $(\lambda-A)(u-c/\lambda)=0$. Also $u-c/\lambda$ cannot be zero otherwise $u'=0$. Of course we should check that $u,u'$ belong to the right spaces simultaneusly, etc...This means that $\sigma (A)=\sigma (B) \cup \{0\}$ and then the first nonzero eigenvalue of $A$ is the first of $B$ which has a positive eigenfunction, namely $u'$. | |
| Feb 17, 2022 at 19:11 | comment | added | Michal Kotowski | Not much more than wishful thinking, I'd be happy to see a counterexample. | |
| Feb 17, 2022 at 15:45 | comment | added | Giorgio Metafune | Why do you think it should be true? | |
| Feb 17, 2022 at 11:17 | history | asked | Michal Kotowski | CC BY-SA 4.0 |