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!
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).