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Let $\Omega $ be a smooth bounded subset of $\mathbb R^d$. Assume that $u=0$ if $u\in H^1(\Omega )$ and $u$ satisfies $$-\Delta u-u=0\ \ in\ \ \Omega \quad \text{and}\quad \partial _\nu u=u=0\ \ on \ \ \partial \Omega .$$

Let $f\in L^2(\Omega )$ and assume that $v$ is solution of $-\Delta v-v=f$ in $\Omega $. Prove that

$$\|v\|_{H^1(\Omega )}\leq C(\|f\|_{L^2(\Omega )}+\|v\|_{H^{1/2}(\partial \Omega )}+\|\partial _\nu v\|_{H^{-1/2}(\partial \Omega )})),$$ where $C$ is independent of $v$ and $f$.

My attempts

My attempts are unfortunately very few... I'm stuck on this problem since yesterday. I tried to make the substitution $w=v-u$, then $w$ is a solution of $\nabla v-v=f$, but it doesn't look to help a lot.

Glitch
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user349449
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  • You probably have $H^{1/2}(\partial \Omega)$, not $H^{1/2}(\Omega)$. – Michał Miśkiewicz Jan 08 '17 at 12:48
  • @MichałMiśkiewicz: You are totally right. Thank you very much :) – user349449 Jan 08 '17 at 13:03
  • The same goes for $\partial_\nu v$, but I'm not sure how to interpret it. Try arguing by contradiction - assume there is a sequence $v_n$ such that $| v_n |{H^1(\Omega)} = 1$, but the norms $|f|{L^2(\Omega )}$, $|v|{H^{1/2}(\partial \Omega )}$, $|\partial\nu v|_{H^{-1/2}(\partial \Omega )}$ tend to zero. – Michał Miśkiewicz Jan 08 '17 at 13:04
  • @MichałMiśkiewicz: $\partial _\nu v=\left<\nabla u,\nu\right>$. – user349449 Jan 08 '17 at 13:04
  • I know the symbol, the problem is to make sense of $\partial_\nu v$ for non-smooth $v$. For arbitrary $v \in H^1(\Omega)$ this is not possible, but here $\Delta v \in L^2(\Omega)$ and you can define $\partial_\nu v$, see this. – Michał Miśkiewicz Jan 08 '17 at 13:20

1 Answers1

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Assume by contradiction that such inequality doesn't hold for any $C$. Then for every $n \in \mathbb{N}$ one can find $v_n \in H^1(\Omega)$ such that

  • $-\Delta v_n - v_n = f_n$,
  • $\| v_n \|_{H^1(\Omega)} = 1$,
  • $\|f_n\|_{L^2(\Omega)}, \|v_n\|_{H^{1/2}(\partial \Omega)}, \|\partial_\nu v_n\|_{H^{-1/2}(\partial \Omega)} < \frac 1n$.

Since $\Omega$ is bounded, one can choose a subsequence so that $v_n \rightharpoonup v$ in $H^1(\Omega)$ (by Banach-Alaoglu theorem) and $v_n \to v$ in $L^2(\Omega)$ (by Rellich-Kondrashov compactness theorem). It is easy to check that $-\Delta v - v = 0$ and $v = \partial_\nu u = 0$ on $\partial \Omega$. By assumption, the only solution is $v=0$. To summarize, we have that $\| v_n \|_{L^2(\Omega)} \to 0$.

What is left is to show a contradiction with $\| v_n \|_{H^1(\Omega)} = 1$. To this end, test the definition of $\partial_\nu v_n$ (given at the bottom) with $v_n$ itself: \begin{align*} \int_\Omega |\nabla v_n|^2 & = \langle \partial_\nu v_n, v_n \rangle - \int_\Omega v_n \Delta v_n \\ & = \langle \partial_\nu v_n, v_n \rangle + \int_\Omega v_n (v_n+f_n) \\ & \xrightarrow{n \to \infty} 0.\end{align*} The convergence follows from $\|\partial_\nu v_n\|_{H^{-1/2}(\partial \Omega)}, \|v_n\|_{H^{1/2}(\partial \Omega)} \to 0$ for the first term and from $\|v_n\|_{L^2(\Omega)}, \|f_n\|_{L^2(\Omega)} \to 0$ for the second. Thus we obtain $\|\nabla v_n\|_{L^2(\Omega)} \to 0$, which together with $\| v_n \|_{L^2(\Omega)} \to 0$ gives a contradiction.

Comment. For regular $v$, the divergence theorem yields the following identity: $$ \int_{\partial \Omega} \varphi \partial_\nu v = \int_\Omega \operatorname{div}(\varphi \nabla v) = \int_\Omega \nabla \varphi \nabla v + \varphi \Delta v \quad \text{for any } \varphi \in C^1(\overline{\Omega}). $$

If $v \in H^1(\Omega)$ and $\Delta v \in L^2(\Omega)$ (as it is in our case), the RHS can be used to define $\langle \partial_\nu v, \varphi \rangle$ for any $\varphi \in H^{1/2}(\partial \Omega)$, that is $$ \langle \partial_\nu v, \varphi \rangle := \int_\Omega \nabla \varphi \nabla v + \varphi \Delta v \quad \text{for } \varphi \in H^{1/2}(\partial \Omega). $$ It can be checked that this doesn't depend on the choice of the extension $\varphi \in H^{1}(\Omega)$. Look up the discussion here.

Michał Miśkiewicz
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  • Thanks a lot for your answer. I'm sorry, but I don't understand why is there such a sequence $v_n$ neither why $v_n\to v$ weakly in $H^1(\Omega )$ and $v_n\to v$ in $L^2(\Omega )$. Could you please detail a little bit ? – user349449 Jan 08 '17 at 15:46
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    @MathBeginner By Banach-Alaoglu, since $(v_n)$ is bounded in $H^1$ it admits a weakly convergent subsequence. From the compact Sobolev embedding $H^1\to L^2$, up to another subsequence you can also suppose that $v_n$ converges strongly in $L^2$. Michal Can you please explain a bit the last line before "Comment"? – Del Jan 08 '17 at 17:52
  • @Del I added your explanation (or at least a part of it) to my answer. Compact operators map weakly convergent sequences to strongly convergent sequences, so actually it is not necessary to choose another subsequence. I also added some comment before "Comment" - is it sufficient? – Michał Miśkiewicz Jan 08 '17 at 21:53
  • @MichałMiśkiewicz Ah yes, thanks. I didn't understand why $v_n\to 0$ in $L^2$ but I was missing that in fact $v=0$! – Del Jan 08 '17 at 21:59
  • Sorry, but why $|\nabla v|=0$ is a contradiction ? – user349449 Jan 19 '17 at 12:46
  • Could you pleas tell me what is $|v_n|{H^{1/2}(\partial \Omega )}$ and $|\partial _nv_n|{H^{-1/2}(\partial \Omega )}$ I really don't understand those norms. I also don't understand how you get $$\int |\nabla u|^2=\left<v_n,\partial \nu v_n\right>+\int\Omega v_n(v_n+f_n),$$ I thougt it was divergence theorem, but it doesn't work. – user349449 Jan 19 '17 at 13:12
  • $|\nabla v_n|{L^2(\Omega)} \to 0$ is a contradiction with $| v_n |{H^1(\Omega)} = 1$. 3. In a way, this equality comes from the divergence theorem. Take the definition of $\langle \partial_\nu v, \varphi \rangle$, plug in $v = \varphi = v_n$, finally use the equation you have for $\Delta v_n$. 2. As for the norms $H^{1/2}$, $H^{-1/2}$, I couldn't explain in the comments. I suggest you post another question about these, but try googling before.
    • – Michał Miśkiewicz Jan 20 '17 at 10:15
  • I put question for norm $H^{1/2}$ and $H^{-1/2}$ here. If you have time, thanks for answering :-) By the way, I don't understand why $\int|\nabla v_n|^2=...\to 0$ come frome the convergence of $|v_n|{H^{1/2}},|f_n|\to 0$ and $|\partial _\nu v_n|{H^{-1/2}}\to 0$, but if you help me to understand better those $H^s$ norme, I'll probably understand better :-) – user349449 Jun 04 '17 at 14:46
  • @MathBeginner I edited the answer once more, giving some more detailed explanations. You need to take the definition of $\partial_\nu v_n$ (or rather $\langle \partial_\nu v_n, \varphi \rangle$) given at the bottom and plug in $\varphi = v_n$. I'm not sure whether this was your doubt though (if not, try to be even more specific).I'll take a look at your other question. – Michał Miśkiewicz Jun 05 '17 at 20:24
  • Thanks for all your answers, I really appreciate :-) There is something I have a doubt : I still don't see in what the fact that $|\partial \nu v_n|{H^{-1/2}(\partial \Omega )}\to 0$ and $|v_n|{H^{1/2}(\partial \Omega )}\to 0$ give us that $<\partial _\nu v_n,v_n>+\int\Omega v_n(v_n+f_n)\to 0$. I'm sorry, and I really appreciate the effort you give to me... some thing is not totally clear, but I'm sure all understand soon thanks to you :-) – user349449 Jun 06 '17 at 20:15
  • The $H^{-1/2}$ norm is defined as the operator norm, so by definition $|\langle \partial_\nu v_n, v_n \rangle | \le |\partial_\nu v_n |{H^{-1/2}} | v_n |{H^{1/2}}$. It's enough if one of the norms is bounded and the other tends to zero, here both norms tend to zero. The justification for the second term comes from convergence in $L^2$, as I indicated. – Michał Miśkiewicz Jun 06 '17 at 21:02
  • Thanks a lot, now I got it ! thousand thanks for every thing :-) – user349449 Jun 07 '17 at 14:55
  • Sorry, I have one more detail to fix. When you write $|\partial n v_n|{H^{-1/2}}$, the $\partial \nu v_n$ must be a functional $H^{-1/2}\to \mathbb R$, no ? Is it true ? And the function si $$\partial _\nu v_n: \varphi\longmapsto \int{\partial \Omega } \partial _\nu v_n \varphi$$ for $\varphi\in H^{1/2}$, right ? – user349449 Jun 16 '17 at 17:45
  • Yes, $\partial_\nu v$ is a functional, i.e., a linear function $H^{1/2} \to \mathbb R$ (I guess that was a misprint). The latter is also true for regular $v$ (e.g. $v \in C^1(\overline{\Omega})$). In general $\partial_\nu v$ is not a function on $\partial \Omega$, so the integral $\int_{\partial \Omega} \partial_\nu v \varphi$ doesn't make sense. That's why we have to use another definition for $\partial_\nu v$. I guess your other question is more on topic here. – Michał Miśkiewicz Jun 16 '17 at 18:40
  • Thanks a lot. A last thing : 1) Why $|v_n|{L^2}\to 0$ ? I see nowhere this hypothesis, so where do we get it ? 2) Is $|\cdot |{H^{\pm 1/2}}$ lower semi continuous ? i.e. that $|v|{H^{1/2}}\leq \liminf{n\to \infty }|v_n|{H^{1/2}}$ and $|\partial _\nu v|{H^{-1/2}}\leq \liminf_{n\to \infty }|\partial \nu v_n|{H^{-1/2}}$ ? If yes, $|v|{H^{1/2}}=|\partial _\nu v|{H^{-1/2}}=0$ is clear, otherwise, it's not that obvious... is it ? – user349449 Jun 18 '17 at 09:41
  • 3) And to finish, could you tell me why $$\int|\nabla v_n|^2=\int |v_n|^2+\int_{\partial \Omega }v_n\partial \nu v_n+\int{\Omega }fv_n$$ implies that $|\nabla v_n|\leq 2|v_n|$ when $n$ big enough ? See here – user349449 Jun 18 '17 at 09:46
  • Ad 1. If you read carefully, we have $v_n \to v$ in $L^2$ and $v=0$. Ad 2. This is way more elementary: $|v_n|{H^{1/2}(\partial \Omega)}, |\partial\nu v_n|_{H^{-1/2}(\partial \Omega)} < \frac 1n$ by assumption, so we have convergence to zero. – Michał Miśkiewicz Jun 18 '17 at 09:46
  • Ok but $|v_n|{H^{1/2}}\to 0$ doesn't imply that $|v|{H^{1/2}}=0$ (except if we have the lower semi continuity) – user349449 Jun 18 '17 at 09:48
  • Since $v_n \rightharpoonup v$ in $H^1(\Omega)$, the traces converge as well: $v_n \rightharpoonup v$ in $H^{1/2}(\Omega)$. On the other hand, $v_n \to 0$ in $H^{1/2}(\Omega)$. Thus $v=0$ on the boundary. Ad 3. This seems unrelated. – Michał Miśkiewicz Jun 18 '17 at 09:52
  • I agree with $v_n\to v$ weakly in $H^{1/2}(\partial \Omega )$, and that $|v_n|_{H^{1/2}}\to 0$, but I don't see why it implies that $v_n\to 0$ strongly in $H^{1/2}$. – user349449 Jun 18 '17 at 09:56
  • I think you would learn more by trying to see the things by yourself. By definition, $| v_n | \to 0$ is equivalent to $v_n \to 0$. – Michał Miśkiewicz Jun 18 '17 at 10:00
  • Something looks strange. Even if $v_n\to 0$ strongly in $H^{1/2}$, nothing tell us that $|v_n|{H^{1/2}}\to |v|{H^{1/2}}$ and thus that $|v|_{H^{1/2}}=0$... (If I could give you 1000 point I would do it, but unfortunately I can't... but I really appreciate all your help :-)) – user349449 Jun 18 '17 at 11:23