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$\textbf{ Statement of Theorem:}$ If $u \in C^2(U)$ is harmonic, then $$u(x) = \frac{1}{m(\partial B(x,r))}\int_{\partial B(x,r)} u dS = \frac{1}{m(B(x,r))}\int_{B(x,r)} u dy$$ for each $B(x,r) \subset U$.

$\textbf{Proof:}$ Let $$ \phi(r) := \frac{1}{m(\partial B(x,r))} u(y) d S(y) = \frac{1}{m(B(0,1))} \int u(x+rz) dS(z)$$ Then, $$ \phi'(r) = \frac{1}{m(\partial B(0,1))} \int_{\partial B(0,1)} Du(x + rz) \cdot z dS(z)$$

Using Green's formula we compute, $$ \phi'(r) = \frac{1}{m(\partial B(x,r))} \int_{\partial B(x,r)} D u(y) \cdot \frac{y-x}{r} dS(y)$$ $$\frac{1}{m(\partial B(x,r))} \int_{ \partial B(x,r)} \frac{\partial u}{\partial \nu} dS(y) \hspace{10mm} (*)$$ where $\nu$ is the outward facing normal vector. $$ = \frac{r}{n} \frac{1}{m(B(x,r))} \int_{B(x,r)} \Delta u(y) dy, \hspace{10mm} (**)$$

$\textbf{Question:}$ What happened between steps $(*)$ and $(**)$. I can see we used the surface area of the ball, however I am not sure how $\frac{\partial u}{\partial \nu}$ turned in to the Laplacian, $\Delta u$.

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    $\begingroup$ What is $\nu$? You have $u(x)$ but nowhere is $\nu$ defined. $\endgroup$ – Winther Apr 18 '15 at 1:54
  • $\begingroup$ If I understand it correctly then $\frac{\partial u}{\partial \nu} = \nabla u \cdot \nu$. $\endgroup$ – Winther Apr 18 '15 at 1:59
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Divergence Theorem : $$ \int_U {\rm div}\ \nabla f =\int_{\partial U} \nabla f\cdot \nu $$ where $\nu$ is an unit outnormal to $\partial U$

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