1
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ Do you know integration by parts? $\endgroup$ – Jeff Dec 7 '11 at 22:36
2
$\begingroup$

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 $u\equiv 0$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.