Pohozave’s equality and non-existence results

This lecture note is totally based on professor Guo Yuxia’s lecture of  non-linear functional analysis.

Let $\Omega$ be a bounded domain in $R^{N}$, $N\geq 3$,we consider the following N-B problem:
$\left\{\begin{array}{rll} -\Delta u+\lambda u&=u|u|^{p-2},&\quad\text{in }\Omega;\\ u&>0,&\quad\text{in }\Omega; \\ u&=0,&\quad\text{on }\partial \Omega.\tag{A} \end{array}\right.$
where $p=2^\star=\frac{2N}{N-2}$.

Theorem 1. Suppose $\Omega \neq R^{N}$ is smooth (possibly unbounded) domain in $R^{N}$,$N\geq 3$, which is strictly star shaped with respect to the origin in $R^{N}$,$\lambda \leq 0$,then any solution $u\in H_{0}^{1}(\Omega)$ of the problem (A) vanishes.

Proof . Firstly, we need the following
\begin{lem}
Let g:$R\rightarrow R$ be continuous with primitive $G(u)=\int_{0}^{u}g(v)dv$, $u \in C^{2}(\Omega)\cap C^{1}(\Omega)$ be a solution of the problem:
$\left\{\begin{array}{rll} -\Delta u&=g(u), &\quad\text{in }\Omega;\\ u&=0,&\quad \text{on }\partial \Omega; \end{array}\right.$
in a domain in $R^{N}(N\geq 3)$, then there holds
$\frac{N-2}{2}\int_{\Omega}|\nabla u|^{2}dx-N\int_{\Omega}G(u)dx+\int_{\partial \Omega}\frac{1}{2}|\frac{\partial u}{\partial \nu}|^{2}x\cdot n d\sigma=0.$
\end{lem}
Let $G(u)=\frac{\lambda u^{2}}{2}+\frac{|u|^{2^{\star}}}{2^{\star}}$, by standard elliptic discussion, any weak solution of (A) is smooth on $\bar{\Omega}$, hence by Lemma 2, we infer that
$\frac{N-2}{2}\int_{\Omega}|\nabla u|^{2}dx-N\int_{\Omega}G(u)dx+\int_{\partial \Omega }\frac{1}{2}|\frac{\partial u}{\partial \nu}|^{2}x\cdot n d\sigma=0,$
i.e.,
$\int_{\Omega}|\nabla u|^{2}dx-2^{\star}\int_{\Omega}G(u)dx+\int_{\partial \Omega}\frac{1}{N-2}|\frac{\partial u}{\partial \nu}|^{2}x\cdot n d\sigma=0.$
Thus,
$\int_{\Omega}(|\nabla u|^{2}-|u|^{2^{\star}})dx+\frac{2^{\star}}{2}\lambda\int_{\Omega}|u|^{2}dx+\int_{\partial \Omega}\frac{1}{N-2}|\frac{\partial u}{\partial \nu}|^{2}x\cdot n d\sigma=0.$
On the other hand, testing the equation with $u$, we have
$\int_{\Omega}|\nabla u|^{2}-\lambda u^{2}-|u|^{2^{\star}}dx=0,$
hence
$2\lambda\int_{\Omega}|u|^{2}dx+\int_{\partial \Omega}|\frac{\partial u}{\partial \nu}|^{2}x\cdot n d\sigma=0.$
Since $\Omega$ is strictly star shaped with respect to $x\cdot n>0,\forall x \in \Omega \Rightarrow u=0$.

1. 天天学习，好好向上

这是个我印象最深刻的等式