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Let nonnegative function $u$ is a solution of $-\Delta u=\lambda u+u|u|^{2^*-2}$ with $u=\frac{\partial u}{\partial\nu}=0$ on $\partial\Omega$, where $\lambda\leq0$, then u vanishes identically in $\Omega$ by the principle of unique continuation. Could anyone explain the principle of unique continuation?

u is smooth on $\bar\Omega$. In fact, I'm reading the book "Vartiational methods" by Struwe. The question is from the proof of 1.3 Theorem in page 171. By Pohozaev's identity, it is proved that the equation $-\Delta u=\lambda u+u|u|^{2^*-2}$ with $u=0$ on $\partial\Omega$ has no nontrivial solutions.

Thank you in advance.

Keith Chen
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  • $u$ is smooth on $\bar\Omega$. In fact, I'm reading "Vartiational methods" by Struwe. The question is from the proof of 1.3 Theorem in page 171. By Pohozaev's identity, it is proved that the equation $-\Delta u=\lambda u+u|u|^{2^*-2}$ with u=0 on ∂Ω has no nontrivial solutions. – Keith Chen Oct 30 '13 at 23:11

1 Answers1

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I am following here the proof given by Struwe in his book cited in the comments by OP.

First consider the case where $\lambda <0$. In page 172, Struwe concludes that $$2|\lambda|\int_\Omega |u|^2+\int_{\partial\Omega}\left|\frac{\partial u}{\partial \nu}\right|^2 x\cdot\nu=0 \tag{1}$$

Because $\Omega$ is a star shaped domain with respect to the origin, we have that $x\cdot\nu >0$ for all $x\in \partial\Omega$. Therefore, both terms of $(1)$ are $0$, which implies in this case that $u=0$ almost everywhere.

If $\lambda =0$, we conclude from $(1)$ that $\frac{\partial u}{\partial \nu}=0$ almost everywhere in the boundary.

If there is a neighborhood $V$ of $\partial\Omega$ such that $u=0$ in $V$ then, by the unique principle continuation (see comments below, Pederson paper), we have that $u=0$ in $\Omega$, hence, we can assume that the boundary of one of the following sets must intersect $\partial\Omega$.

Let $\Omega_+=\{x\in \Omega:\ u(x)>0\}$, $\Omega_-=\{x\in \Omega:\ u(x)<0\}$. Note that

$$ \left\{ \begin{array}{ccc} -\Delta u>0 &\mbox{ in $\Omega_+$} \\ u=0 &\mbox{ in $\partial\Omega_+$} \end{array} \right. $$

$$ \left\{ \begin{array}{ccc} -\Delta u<0 &\mbox{ in $\Omega_-$} \\ u=0 &\mbox{ in $\partial\Omega_-$} \end{array} \right. $$

Suppose for example that it is the set $\Omega_+$, hence, by Hopf's Boundary Lemma, we must have that $\frac{\partial u(x)}{\partial\nu}\neq0$ on $\partial\Omega\cap\partial\Omega_+$, which is an absurd.

From the last paragraph is straightforward to conclude that $u=0$ in all $\Omega$, in fact, here is where we apply some kind on Principle of Unique Continuation, i.e., if $u=0$ in a open set of $\Omega$, then $u=0$ in all $\Omega$.

Remark 1: This paper from Pucci and Serrin also contains your result.

Remark 2: You have said in the first line of your question that $u\geq 0$. In this case, the aswer is immediate, so I think this is a typo?

Tomás
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  • By Hopf's Lemma, it seems that we should assume $u>0$ in $\Omega$, but I cannot prove it. Maybe you can give me some hints. Thanks! – Keith Chen Oct 31 '13 at 02:45
  • Dear @KeithChen, if a function is superharmonic, and if it attains its minimum in the interior of $\Omega$, then it must be constant, i.e. $u=0$, therefore, we can assume without loss of generality that $u>0$ in $\Omega$. You can see this result in Gilbard-Trudinger EDP book chapter 2. Now let me ask you something. In the proof of Theorem 1.3. in Struwe book, they concluded the follwing equality: $$2|\lambda|\int_\Omega |u|^2+\int_{\partial\Omega}\left|\frac{\partial u}{\partial\nu}\right|^2 x\cdot\nu =0$$ Then, they concluded that (continued...) – Tomás Oct 31 '13 at 10:47
  • $\frac{\partial u}{\partial\nu}=0$, because $x\cdot\nu>0$. But if $\frac{\partial u}{\partial\nu}=0$ is zero, hence, we must have $$2|\lambda|\int_\Omega |u|^2=0$$ If $\lambda<0$, we conclude that $u=0$, otherwise, if $\lambda =0$, we use the principle of unique continuation. Is my argument right? What do you think? – Tomás Oct 31 '13 at 10:49
  • Thank you for your kind answer, but I could not get it yet. First, we could not claim that $u$ is superharmonic before we know $u\geq0$. Second, I don't figure out the principle of unique continuation yet since you mentioned it at last. – Keith Chen Nov 01 '13 at 06:40
  • @KeithChen, You have stated in the first line of you question that $u\geq 0$. – Tomás Nov 01 '13 at 12:15
  • Thank you very much. I think I've almost got it. Still, you claim one of the sets $\partial\Omega_-$ and $\partial\Omega_+$ must containt some point of the boundary $\partial\Omega$. I think you use the principle of unique continuation here. That is, if the claim is not true, then $u\equiv 0$ in some neighbourhood of $\partial\Omega$, and $u\equiv0$ follows from the principle of unique continuation. Am I right? In addition, what kind of elliptic equations have this unique continuation principle? – Keith Chen Nov 02 '13 at 04:16
  • Dear @KeithChen, you are right in your reasoning. It seems to me that every solution of a scond order uniformly elliptic equation does have this principle. The idea behind this principle is the maximum or minimum principle. When i was trying to solve your problem, I have made a search in google and found a lot of thing, I suggest you to do tthe same. Later on, I will post here some that I have found. Also, feel free to edit my answer and fill up the details if you want. – Tomás Nov 02 '13 at 11:03
  • Thanks for your response and adivice. I should think it more before editing your answer. I did search this topic in google, but I got few information to solve my question. I look forward to what you'll post here about this topic. – Keith Chen Nov 02 '13 at 12:55
  • Dear @KeithChen, I found the result we need in the following paper Theorem 2, page 7: http://onlinelibrary.wiley.com/doi/10.1002/cpa.3160110104/abstract If you don't have access to it, you can give me your e-mail and I can send one copy to you. There is another paper from Protter which seems to have the result too, But I do not have time to read it. Take a look on it (continued in the next comment): – Tomás Nov 02 '13 at 14:57
  • http://www.google.com/url?sa=t&rct=j&q=&esrc=s&frm=1&source=web&cd=2&ved=0CDwQFjAB&url=http%3A%2F%2Fwww.ams.org%2Fjournals%2Ftran%2F1960-095-01%2FS0002-9947-1960-0113030-3%2FS0002-9947-1960-0113030-3.pdf&ei=RBB1UuC6D8PdkQfdjIC4DQ&usg=AFQjCNHbTQV1OAK7yG_z_n-rZjjwebz3QQ&sig2=iqUaEV44W84d1jOcSVp9vg – Tomás Nov 02 '13 at 14:58
  • Please send a copy of Pederson's paper to my e-mail address [email protected] since I can not access to it, :-(. Thank you very much. – Keith Chen Nov 03 '13 at 11:44
  • @KeithChen, I sent a copy to you. – Tomás Nov 03 '13 at 14:32