# $\int_{\partial \Omega}\frac{\partial u}{\partial N}d\sigma$ if $\Delta u = 0$ in $\Omega$

Can you help me please with this problem?

1. What do you know about $\int_{\partial \Omega}\frac{\partial u}{\partial N}d\sigma$ if $\Delta u = 0$ in $\Omega$?
2. Is it always possible to solve an equation $u''=0$ in $[0,1]$ if $u'(0)$ and $u'(1)$ are given?(What is connection to first question?) Is the solution - unique?

Thanks!

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Hint: $\Delta u = \mbox{div} \nabla u$. Do you know the divergence theorem? –  user20266 Jul 1 '12 at 10:02

## 1 Answer

For the first one

From the divergence theorem for the vector field $\text{grad} u$ we get that:

$\displaystyle{ \int_{\Omega} \text{div(grad u}) \nu = \int_{\partial \Omega} < \text{grad}u ,N> \sigma} \quad (\bigstar)$

where $\nu$ is the volume element.

Using now the ypothesis that $0=\Delta u := \text{div(grad u)}$ and that $\displaystyle{\frac{\partial u}{\partial N} := <\text{grad g ,N}>}$

The conclusion now follows from $(\bigstar)$.

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Uhmm -- which conclusion? –  user20266 Jul 1 '12 at 10:38
@passenger Thanks! Can you please explain it in more detailed way? –  Mushka Jul 10 '12 at 6:48
Since $\Delta u =0$ the left hand side of $(\star)$ is $0$ and so does the right hand side which is exactly what you want. –  passenger Jul 10 '12 at 23:55