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I am having problems with the following question, any and all help is appreciated.

Suppose $\Delta u = 0$ in $D$

$$\displaystyle\frac{du}{d\eta} +au = 0$$ on $\partial D$ where $D$ is a bounded domain in $\mathbb{R}^3$ for which the divergence theorem holds, vector $\eta$ is a unit outward normal on $\partial D$ and $a >0$. Show that $u \equiv 0$.

I need help starting this though any help towards completing it would also be helpful. Thank you.

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Do you know integration by parts? –  Jeff Dec 7 '11 at 22:36

1 Answer 1

up vote 2 down vote accepted

Since $\Delta u=0$ in $D$, by divergence theorem, we have $$0=\int_Du\Delta u=-\int_{D}|\nabla u|^2+\int_{\partial D}u\frac{du}{d\eta}.$$ Since $\displaystyle\frac{du}{d\eta} +au = 0$ on $\partial D$, the right hand side becomes $$0=-\int_{D}|\nabla u|^2-a\int_{\partial D}u^2.$$ Since $a>0$, the above equality implies that $\nabla u\equiv 0$ in $D$, i.e. $u$ is a constant. Since $u\equiv 0$ on $\partial D$ by the above equality, we conclude that $u\equiv 0$.

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