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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?

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1 Answer 1

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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.

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    $\begingroup$ You could also just use that $\int_\Omega \vert \nabla u\vert^2\,dx=-\int_\Omega u \Delta u \, dx + \int_{\partial \Omega} u \frac {\partial u}{\partial n} \, dS =0.$ $\endgroup$
    – JackT
    Jun 19, 2023 at 21:29
  • $\begingroup$ 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 $\dfrac{\partial u}{\partial n}=\nabla u.n=0\nRightarrow u\equiv const.$) $\endgroup$
    – Berban
    Jun 20, 2023 at 5:27
  • $\begingroup$ Yes Really, thank you. $\endgroup$
    – Berban
    Jun 20, 2023 at 6:28

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