Timeline for What is the spectrum of this differential operator?
Current License: CC BY-SA 4.0
20 events
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| Sep 23 at 20:06 | comment | added | Peter A | Ironically, that is exactly my original form. I scaled by u to get a self adjoint operator, which I thought would make it easier | |
| Sep 23 at 19:23 | comment | added | Bertoldo Baccalà | Another rather trivial remark: if you substitute $g:=uf$ you end up with an equation $g''=u^{-2}\lambda g$, which is in a standard Sturm-Liouville format. Finding explicit solutions will very much depend on the specific choice of $u$. | |
| Sep 23 at 14:17 | comment | added | Peter A | @Willie Wong thank you, yes good point. For the purpose of this question we should assume u is always greater than a constant. Better ignore the case where it can go to zero for now. Mistake on my part to introduce it here | |
| Sep 23 at 13:48 | comment | added | Willie Wong | @PeterA: if $u$ vanishes to zero linearly as $x$ tends to zero, then $1/u$ is not square integrable and cannot be an eigenfunction. | |
| Sep 23 at 9:06 | comment | added | Peter A | Yes 1 / u is certainly admissible | |
| Sep 23 at 6:25 | comment | added | Bertoldo Baccalà | With the inner product you mentioned, you are formally looking for critical points of the Rayleigh quotient $ \frac{\int [\partial_x (uf)]^2 \,dx}{\int f^2\,dx}$, but be cautious with the integrability and BCs. This gives some intuition on the obvious fact that $1/u$ is an eigenfunction to $\lambda=0$ if this is an admissible function in the game. | |
| Sep 22 at 22:11 | comment | added | Peter A | @Willie Wong Yes the domain is the real numbers. The function u can go to zero roughly linearly as x goes to zero in some cases of interest, and it can go to infinity as x goes to infinity. But I am also interested in the strictly positive and bounded case | |
| Sep 22 at 22:01 | comment | added | Steven Stadnicki | If it helps any, two smooth $u()$ that satisfy your conditions and might be analyzable are $u(x)=1-\frac1{2(x^2+1)}$ and $u(x)=1-\frac12e^{-x^2}$. | |
| Sep 22 at 21:51 | comment | added | Willie Wong | I assume the domain is $x\in \mathbb{R}$? Is the function $u$ bounded? Is it bounded away from zero uniformly? If it is allow to approach zero, what is the asymptotic rate? | |
| Sep 22 at 21:12 | history | edited | Peter A | CC BY-SA 4.0 |
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| Sep 22 at 20:05 | comment | added | Carlo Beenakker | if $u(x)=x$, eigenfunctions are $f(x)=x^q$ with eigenvalue $q(q+1)$ | |
| Sep 22 at 19:37 | history | edited | Peter A | CC BY-SA 4.0 |
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| Sep 22 at 9:15 | comment | added | Peter A | Yes, thank you, I have now fixed the brackets | |
| Sep 22 at 9:14 | history | edited | Peter A | CC BY-SA 4.0 |
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| Sep 22 at 8:12 | comment | added | Bertoldo Baccalà | Just for clarity, are there parentheses missing after the derivative? | |
| Sep 22 at 6:27 | history | edited | Peter A | CC BY-SA 4.0 |
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| Sep 21 at 22:33 | comment | added | Peter A | I have tried to clarify this as best I can. Thie background is it will help me to solve a related problem numerically, so I am looking for smooth functions that don't explode, but more generally anything that is known about this type of problem | |
| Sep 21 at 22:29 | history | edited | Peter A | CC BY-SA 4.0 |
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| Sep 21 at 21:28 | comment | added | paul garrett | What space of functions...? Any boundary conditions...? Even things that "look" self-adjoint (some people call them "formally self-adjoint") can fail to have any self-adjoint extensions at all, or have infinitely-many, ... Can you clarify? | |
| Sep 21 at 21:19 | history | asked | Peter A | CC BY-SA 4.0 |