Timeline for Gradient estimates for a boundary value problem
Current License: CC BY-SA 4.0
13 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Feb 14, 2021 at 6:25 | answer | added | Connor Mooney | timeline score: 2 | |
| Feb 13, 2021 at 22:49 | answer | added | Mateusz Kwaśnicki | timeline score: 3 | |
| Feb 13, 2021 at 20:20 | comment | added | Harish | @MateuszKwaśnicki Since $w=0$ in the inner boundary and with the same argument applied not to gradient but to the value of $w$ we get $\int_{\partial B_r}P(x,y)\varphi \,d\sigma=0$ for every boundary data $\varphi$. This give us $P(x,y)=0$ on $\partial B_r$. Which should not be true! I think the Newtonian potential estimates cannot be applied on the boundary points. | |
| Feb 13, 2021 at 20:09 | comment | added | Mateusz Kwaśnicki | Everything scales nicely, so you are free to choose $r = 1$ with no loss of generality. | |
| Feb 13, 2021 at 18:22 | comment | added | Harish | @MateuszKwaśnicki I am not sure if that intiution can be taken that far, because $\delta(x)\delta(y)$ when differentiated w.r.t. normal vectors at boundary give 1. So we should have been left with just integral $\int_{\partial B_r} \varphi$ on RHS. but it is avergage integral divided by $r$. | |
| Feb 13, 2021 at 18:15 | comment | added | Mateusz Kwaśnicki | Yes, that is what I mean. (And, of course, this only works for $\varphi \ge 0$; otherwise, the integral of $\varphi$ can well be zero, while the LHS is positive for some $x$ unless $\varphi$ is identically zero). | |
| Feb 13, 2021 at 18:05 | comment | added | Harish | @MateuszKwaśnicki I see, hence considering $x$ in in the inner sphere and $y$ in the outer sphere, we can say $\partial _{\nu(y)}\delta (y) \approx 1 $, and makes no affect in the integration. We are left with $\int _{\partial B_r} \varphi$. what could explain the avergae integral? | |
| Feb 13, 2021 at 17:50 | comment | added | Mateusz Kwaśnicki | I may have not been clear enough, sorry: in the expression for $w(x)$, $x$ is in the interior, but in the expression for $\nabla w(x)$, $x$ is on $\partial B_{kr}$. And for smooth enough domains it is known that $G(x,y) \approx \delta(x) \delta(y)$ when $x$ and $y$ are not too close (and $\delta(x)$ is the distance to the boundary). | |
| Feb 13, 2021 at 17:42 | comment | added | Harish | @MateuszKwaśnicki Thank you, but these are internal estimates, do such estimate also hold true on the boundary of interior sphere? (that is on $\partial B_{kr}$)? | |
| Feb 13, 2021 at 17:35 | comment | added | Mateusz Kwaśnicki | If $G(x,y)$ is the Green function for $B_r \setminus B_{kr}$, then $w(x) = \int_{\partial B_r} P(x, y) \varphi(y) \sigma(dy)$ and $|\nabla w(x)| = |\partial_n w(x)| = |\int_{\partial B_r} Q(x, y) \varphi(y) \sigma(dy)|$, where $P(x,y) = \partial_{n(y)} G(x, y)$ is the Poisson kernel and $Q(x, y) = \partial_{n(x)} \partial_{n(y)} G(x,y)$ is the normal derivative of the Green function with respect to both variables. It remains to note that $Q(x,y)$ is bounded (by known estimates of the Green function, for example). | |
| Feb 13, 2021 at 17:30 | comment | added | Harish | @Math604 yes, constant $C$ can depend on $k$. | |
| Feb 13, 2021 at 16:43 | comment | added | Math604 | You should try a scaling argument to remove the $r$ . Can the $C$ depend on $k$ ? (I assume it can otherwise its probably false) | |
| Feb 13, 2021 at 15:41 | history | asked | Harish | CC BY-SA 4.0 |