Timeline for Weak divergence implies weak differentiability of components?
Current License: CC BY-SA 3.0
6 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Oct 29, 2012 at 7:04 | vote | accept | Beni Bogosel | ||
| Oct 28, 2012 at 21:22 | answer | added | Bazin | timeline score: 2 | |
| Oct 23, 2012 at 19:25 | comment | added | Delio Mugnolo | as Rbega says: if it is weakly differentiable (i.e., $\sigma\in H^1(\Omega)$, then it clearly also has a weak divergence, but (generally) not viceversa. i do not even see why elliptic regularity applies here. what one can generally say is that if $u\in H^1(\Omega)$ is a weak solution of $\Delta u=f\in L^2(\Omega)$, then $u\in H^2_{loc}(\Omega)$. | |
| Oct 23, 2012 at 17:35 | comment | added | Beni Bogosel | That is what I was thinking, but I wanted another confirmation. | |
| Oct 23, 2012 at 17:14 | comment | added | Rbega | If I understood your comment at the end what was actually shown was that a function $u\in H^1(\Omega)$ satisfied $\Delta u=f\in L^2(\Omega)$ weakly. This is a standard consequence of "elliptic regularity". What you are asking can't possibly be true since you have no control over most of the derivatives of each component of $\sigma$. | |
| Oct 23, 2012 at 17:01 | history | asked | Beni Bogosel | CC BY-SA 3.0 |