Timeline for A constrained prolongement
Current License: CC BY-SA 3.0
7 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Aug 13, 2012 at 4:54 | comment | added | Yemon Choi | As Leonid Kovalev appears to have given a nice counter-example over MSE, I vote to close the question here. | |
| Aug 13, 2012 at 4:53 | comment | added | Yemon Choi | Cross-posted to MSE: math.stackexchange.com/questions/179109/… | |
| Aug 5, 2012 at 1:34 | comment | added | hardy | If we try to extend the above idea to the case $n=2$ : for $x\in \Omega-\omega,$ we consider the normal to $\partial\omega$ at $\xi$ and passing from $x$. Then we set $ \tilde{\theta}(x)=\theta'(\xi)(x-\xi)+\theta(\xi).$ | |
| Aug 3, 2012 at 16:16 | comment | added | Francois Monard | $W^{2,\infty}$-estimates in dimension $n\ge 2$ are not always true without assuming further regularity on the other terms, so dimension $1$ might not generalize so nicely. As regards your problem, $\tilde\theta$ may only depends on the $\theta$ and $\frac{\partial\theta}{\partial n}$ at the boundary $\partial\omega$. The values inside $\omega$ don't matter so much as long as they match with the traces I mentioned. | |
| Aug 3, 2012 at 14:24 | comment | added | hardy | I sudied the case $n=1,\, \Omega = (a,b)$ and $ \omega=(c,d).$ Then I complete $\theta$ on $\Omega$ by the tangent equtions $\theta'(c)(x-c)+\theta(c), $ for $a<x<c $ and $ \theta'(d)(x-d)+\theta(d),$ for $d<x<b.$ | |
| Aug 3, 2012 at 4:31 | comment | added | Yemon Choi | Which special cases have you already tried to do? What have you already tried doing? See mathoverflow.net/howtoask | |
| Aug 3, 2012 at 3:03 | history | asked | hardy | CC BY-SA 3.0 |