Timeline for Boundary behavior of $H^2_0(\Omega)$ functions
Current License: CC BY-SA 4.0
10 events
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| Apr 2, 2019 at 22:28 | comment | added | Mateusz Kwaśnicki | I bet the answer is "yes", although in an integral form rather than point-wise. I do not know the details, though. "Higher-order Hardy's inequality" (in Lipschitz domains) seems to be a good search phrase. | |
| Apr 2, 2019 at 22:06 | comment | added | user123672 | @MateuszKwaśnicki I see. Thank you. Then can we say anything at all about the boundary decay of $u \in H^2_0(\Omega)$? | |
| Apr 2, 2019 at 21:58 | comment | added | Mateusz Kwaśnicki | I mean: $\delta < 1/2$. Then $u(x) = (1 - x^2)^{2 - \delta} \mathbb{1}_{(-1,1)}(x)$ satisfies $u, u', u'' \in L^2$, and therefore $u \in H^2(\mathbb{R})$. And since $u = 0$ outside $(-1, 1)$, in fact $u \in H^2_0((-1,1))$. | |
| Apr 2, 2019 at 21:42 | comment | added | user123672 | @MateuszKwaśnicki What do you mean by "if $\delta$ is small enough"? | |
| Apr 2, 2019 at 21:40 | history | edited | user123672 | CC BY-SA 4.0 |
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| Apr 2, 2019 at 21:38 | comment | added | user123672 | @MateuszKwaśnicki You're right: I've edited the question. We can assume whatever smoothness is necessary (although I hope none). | |
| Apr 2, 2019 at 21:38 | comment | added | Mateusz Kwaśnicki | As for the upper bound, $|u(x)| \le C \operatorname{dist}(x, \partial \Omega)^2$, this is not true even in dimension $1$: $(1-x^2)^{2-\delta}$ is in $H_0^2((-1, 1))$ if $\delta > 0$ is small enough. | |
| Apr 2, 2019 at 21:37 | history | edited | user123672 | CC BY-SA 4.0 |
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| Apr 2, 2019 at 21:35 | comment | added | Mateusz Kwaśnicki | Any smoothness assumptions on the boundary? Also, in any case there is no hope for the lower bound (if $\asymp$ stands for the ratio being bounded between two positive constants). | |
| Apr 2, 2019 at 21:31 | history | asked | user123672 | CC BY-SA 4.0 |