I have the following problem:
Let $ u $ be in $ C^2(\Omega) $ and in $ C^1(\overline{\Omega}) $, where $ \Omega $ is a normal bounded domain in $ R^n $, and suppose that
\begin{equation*}
\begin{split}
\Delta u&=0 ~~ \text{in}~~\Omega\\
\dfrac{\partial u}{\partial n}&=0 ~~ \text{in}~~\partial\Omega
\end{split}
\end{equation*}
Then show that $ u $ is constant in $ \overline{\Omega} $.
Here I try like this:
From Green's first identity,
\begin{equation*}
\int_{\Omega}[u\Delta w+(\nabla u).(\nabla w)]dv=\int_{\partial\Omega}u\dfrac{\partial w}{\partial n}d\sigma
\end{equation*}
taking $ u=1 $ and $ w=u $, I have
\begin{equation*}
\int_{\Omega}\Delta udv=\int_{\partial\Omega}\dfrac{\partial u}{\partial n}d\sigma
\end{equation*}
And using the fact that
$$ \overline{\Omega}=\Omega\cup\partial\Omega$$
And
\begin{equation*}
\int_{\overline{\Omega}}f=\int_{\Omega}f+\int_{\partial\Omega}f-\int_{\Omega\cap\partial\Omega}f
\end{equation*}
I got
\begin{equation*}
\int_{\overline{\Omega}}\Delta udv=\int_{\partial\Omega}\dfrac{\partial u}{\partial n}d\sigma-\int_{\Omega}0.\nabla udv+\int_{\partial\Omega}\Delta udv-\int_{\Omega\cap\partial\Omega}\Delta udv
\end{equation*}
which is reduced to
\begin{equation*}
\int_{\overline{\Omega}}\Delta udv=\int_{\partial\Omega}\Delta udv
\end{equation*}
Now in order to be $ u $ constant on $ \overline{\Omega} $, the integrand $ \Delta u $ must be 0 (?!). To be that the integral on the right hand side should vanish. But how can I show that? Any help?
Best Answer
Let $u$ be a solution to \begin{align} \Delta u&=0 \,\,\, \text{in}\,\,\,\Omega, \tag{1a} \\ \dfrac{\partial u}{\partial n}&=0 \,\,\,\text{in}\,\,\,\partial\Omega. \tag{1b} \end{align} Then $$ \int_{\Omega}\Delta (u^2)\,dv=\int_{\partial\Omega}\frac{\partial u^2} {\partial n}\,d\sigma=\int_{\partial\Omega}2u\,\frac{\partial u} {\partial n}\,d\sigma=0. \tag{2} $$ On the other hand, $\Delta (u^2)=2[u\Delta u+(\nabla u)^2]=2(\nabla u)^2$, so $(2)$ implies $$ \int_{\Omega}(\nabla u)^2\,dv=0, \tag{3} $$ from which follows that $u$ is constant.