# Solution of Neumann problem for Laplace equation.

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?

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.
• You could also just use that $\int_\Omega \vert \nabla u\vert^2\,dx=-\int_\Omega u \Delta u \, dx + \int_{\partial \Omega} u \frac {\partial u}{\partial n} \, dS =0.$ Jun 19, 2023 at 21:29
• But what about on $\partial\Omega$? to be $u$ constant on $\bar{\Omega}$, it should also be on the boundary. $\dfrac{\partial u}{\partial n}=0$ in $\partial\Omega$ doesn't imply $u$ is constant, i think. (as $\dfrac{\partial u}{\partial n}=\nabla u.n=0\nRightarrow u\equiv const.$) Jun 20, 2023 at 5:27