Timeline for Interior Schauder estimates with weights
Current License: CC BY-SA 3.0
13 events
| when toggle format | what | by | license | comment | |
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| Mar 22, 2014 at 14:40 | answer | added | NJK | timeline score: 1 | |
| Feb 19, 2014 at 8:28 | comment | added | username | Did you look at Evans' PDE textbook ? The derivation of $H^2$ estimates is done in details. Using $\sqrt{w}\xi$ instead of $\xi$ in the proof could (maybe) give you an idea. | |
| Feb 18, 2014 at 10:17 | answer | added | Juhana Siljander | timeline score: 3 | |
| Feb 17, 2014 at 3:19 | comment | added | Tomas | @Juhana, thanks for your comments. I will be very glad to see some references about it. | |
| Feb 16, 2014 at 20:42 | comment | added | Juhana Siljander | I think most of the interior regularity theory should work if you are working with a doubling measure $d\mu$ which supports a Poincaré inequality instead of the standard Lebesgue measure. So, while I haven't checked the details, I think you should be able to more or less repeat the standard arguments by just considering the measure $d\mu:=w(x) dx$ instead of $dx$. Of course, if you like to work in whole of $\mathbb{R}^n$, then there might be problems at the infinity. If you are interested in this approach, I may try to find some references. | |
| Jan 2, 2014 at 7:33 | comment | added | Craig | @shanlin. Take $ w=1$ for now and try the above approach. But after integrating the equation put the derivative on $V(x)$ back onto the other terms by integrating by parts. This appears to work at least formally. | |
| Jan 2, 2014 at 1:10 | history | edited | Tomas | CC BY-SA 3.0 |
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| Dec 31, 2013 at 19:54 | comment | added | Tomas | @Craig, thanks for your comments, the assumption on $V$ is just bounded in the paper, so I think perhaps there is a more general way to do this | |
| Dec 31, 2013 at 12:14 | comment | added | Craig | To attempt a proof (lets assume $V=1$) multiply the equation by $ w u \phi^2$ where $ \phi$ a cut off function and integrate and apply Young's inequality. This should get the weighted $L^2$ norm of gradient controlled by weighted $L^2$ norm. To do the next step take a derivative of equation and try same procedure. At first glance it appears this method may not work unless $V(x)$ is assumed nice... | |
| Dec 31, 2013 at 3:07 | history | edited | Tomas | CC BY-SA 3.0 |
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| Dec 29, 2013 at 12:55 | comment | added | Michael Renardy | I think the issue is about behavior at infinity rather than the boundary. | |
| Dec 29, 2013 at 5:52 | comment | added | Craig | Is the result even true? Take $ w(x)=1$. I think its false since you can't expect to have the regularity right to the boundary.. ??? | |
| Dec 29, 2013 at 1:54 | history | asked | Tomas | CC BY-SA 3.0 |