Timeline for $H^1$-continuity of Laplace's equation with respect to boundary data
Current License: CC BY-SA 3.0
9 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jun 21, 2016 at 6:31 | comment | added | user35593 | It is also discussed here: mathoverflow.net/questions/96532/… | |
| Jun 20, 2016 at 19:54 | comment | added | Michael Renardy | Yes, with $H^{1/2}$ the inequality holds. | |
| Jun 20, 2016 at 19:51 | comment | added | user35593 | Ok, now I understand your argument. Can we say that the inequality holds with $L^\infty$ replaced by $H^{1/2}$? | |
| Jun 20, 2016 at 19:44 | vote | accept | user35593 | ||
| Jun 20, 2016 at 19:05 | comment | added | Christian Remling | Why was this downvoted? It may be a bit terse, but Michael is making a very valid point. You can also consider concrete examples such $u(z)=\textrm{Re}\exp ((z+1)/(z-1))$ on the unit disk (and approximate $\phi$ in $L^1$ by continuous functions) to see that what you were hoping for is false. | |
| Jun 20, 2016 at 17:39 | comment | added | Michael Renardy | Just to be clear: What I am saying it that if $\phi$ is in $L^\infty$ and not in $H^{1/2}$, then the solution to Laplace's equation (or to any other equation) is not in $H^1$. | |
| Jun 20, 2016 at 17:28 | comment | added | Michael Renardy | Regardless of what equation is satisfied, the regularity of the boundary data does not improve over what is given! | |
| Jun 20, 2016 at 17:15 | comment | added | user35593 | Here we have not just a H1 function but the harmonic map for which we can expect higher regularity. I do not see why it should not be possible. | |
| Jun 20, 2016 at 15:33 | history | answered | Michael Renardy | CC BY-SA 3.0 |