$\textbf{ Statement of Theorem:}$ 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, $\Delta u$.

  • 1
    $\begingroup$ 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

Divergence Theorem : $$ \int_U {\rm div}\ \nabla f =\int_{\partial U} \nabla f\cdot \nu $$ where $\nu$ is an unit outnormal to $\partial U$


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.