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 $p=2^\star=\frac{2N}{N-2}$.
定理 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.
证明 . 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 $x\cdot n>0,\forall x \in \Omega
\Rightarrow u=0$.
\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 $x\cdot n>0,\forall x \in \Omega
\Rightarrow u=0$.
这是个我印象最深刻的等式