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I am a beginner of Elliptic PDE. This is really hard for me who do not have a sound foundation in Calculus III. I get stumbled in the following proof, especially the part in the red rectangle. I would be very grateful if you can help explain it in detail. Any other explanation on the proof is also more than welcome. Thank you so much!

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Anna Le
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Well the first part: $$\int_{\partial B_p}\frac{\partial u}{\partial \nu}\,ds=\int_{B_p}\Delta u\,dx=0$$ (I will leave out the brackets for ease of typing) follows directly from the divergence theorem (also called Gauss's Theorem), which in a nutshell is as follows for $U\subset\Bbb R^n$, and some vector field $F$:

$\int_{U}\nabla\cdot F\,dx=\int_{\partial U}F\cdot \nu\,ds$, where $\nu$ is the unit outward normal to $U$. Taking into account that $\Delta u=\nabla\cdot(\nabla u)$, we can see that:

$\int_{B_p}\Delta u\,dx = \int_{B_p}\nabla\cdot(\nabla u)\,dx = \int_{\partial B_p}\nabla u\cdot\nu\,ds$ now $\nabla u\cdot\nu=\frac{\partial u}{\partial\nu}$ (these are just two different notations for the same thing).

Ellya
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  • Hi, thank you so much for your reply. What I don't understand is why x is replace by $y + r\omega $ and $\frac{{\partial u}}{{\partial v}}$ is replaced by $\frac{{\partial u}}{{\partial r}}$. Thank you! – Anna Le Sep 26 '14 at 15:03
  • @AnnaLe this is because we have $\omega = \frac{x-y}{r}$, rearrange this and you obtain $x = y+r\omega$, and $\frac{\partial u}{\partial\nu}$ is the normal derivative, i.e. a derivative taken in the direction that is normal to the domain of integration, which in this case is $B_\rho$. Since $r$ is the radius of the sphere, $r$ is normal to the sphere, and so the normal derivative is the derivative in the direction $r$, so basically $\nu = r$ and $\frac{\partial u}{\partial \nu}=\frac{\partial u}{\partial r}$ – Ellya Sep 26 '14 at 15:12
  • Sorry when I said $r$ is the radius of the sphere, what I meant was $r$ points in the direction of the radius of the sphere. – Ellya Sep 26 '14 at 15:31