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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.

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    You probably have $H^{1/2}(\partial \Omega)$, not $H^{1/2}(\Omega)$. – 2017-01-08
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    @MichałMiśkiewicz: You are totally right. Thank you very much :) – 2017-01-08
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    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. – 2017-01-08
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    @MichałMiśkiewicz: $\partial _\nu v=\left<\nabla u,\nu\right>$. – 2017-01-08
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    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](http://math.stackexchange.com/questions/609372/the-normal-derivative-in-pde-problems-how-is-the-weak-form-defined). – 2017-01-08

1 Answers 1

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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.

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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 ? – 2017-01-08
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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"? – 2017-01-08
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    @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](https://en.wikipedia.org/wiki/Compact_operator#Completely_continuous_operators), so actually it is not necessary to choose another subsequence. I also added some comment before "Comment" - is it sufficient? – 2017-01-08
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    @MichałMiśkiewicz Ah yes, thanks. I didn't understand why $v_n\to 0$ in $L^2$ but I was missing that in fact $v=0$! – 2017-01-08
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    Sorry, but why $\|\nabla v\|=0$ is a contradiction ? – 2017-01-19
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    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 $$\int |\nabla u|^2=\left+\int_\Omega v_n(v_n+f_n),$$ I thougt it was divergence theorem, but it doesn't work. – 2017-01-19
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    1. $\|\nabla v_n\|_{L^2(\Omega)} \to 0$ 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. – 2017-01-20
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    I put question for norm $H^{1/2}$ and $H^{-1/2}$ [here](https://math.stackexchange.com/questions/2309420/what-are-partial-nu-v-n-h-1-2-partial-omega-and-v-n-h1). 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 :-) – 2017-06-04
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    @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. – 2017-06-05
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    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 :-) – 2017-06-06
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    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. – 2017-06-06
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    Thanks a lot, now I got it ! thousand thanks for every thing :-) – 2017-06-07
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    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 ? – 2017-06-16
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    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](https://math.stackexchange.com/questions/2309420/what-are-partial-nu-v-n-h-1-2-partial-omega-and-v-n-h1/2311227#2311227) is more on topic here. – 2017-06-16
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    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 ? – 2017-06-18
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    **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](https://math.stackexchange.com/questions/2110432/prove-v-h1-omega-leq-c-f-l2-omega-v-h1-2-partial?noredirect=1&lq=1) – 2017-06-18
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    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. – 2017-06-18
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    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) – 2017-06-18
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    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. – 2017-06-18
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    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 $H^{1/2}$. – 2017-06-18
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    I think you would learn more by trying to see the things by yourself. By definition, $\| v_n \| \to 0$ is equivalent to $v_n \to 0$. – 2017-06-18
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    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 :-)) – 2017-06-18