Principle of unique continuation 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.
 A: 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? 
