# Proof of Green's third identity

I don't completely understand the proof of the following result, which is also called Green's third identity:

Let $D\subset{\mathbb R^m}$ be a bounded domain of class $C^1$ and let $u\in C^2(\bar{D})$ be harmonic in $D$. Then $$u(x)=\int_{\partial D}\bigg\{\frac{\partial u}{\partial\nu}(y)\Phi(x,y)-u(y)\frac{\partial \Phi(x,y)}{\partial\nu}\bigg\}ds(y),\quad x\in D$$ where $$\Phi(x,y):= \begin{cases} \frac{1}{2\pi}\ln\frac{1}{|x-y|},\quad m=2\\ \frac{1}{4\pi}\frac{1}{|x-y|},\quad m=3. \end{cases}$$

Proof. For $x\in D$ choose a sphere $\Omega(x;r):=\{y\in{\mathbb R}^m:|y-x|=r\}$ of radius $r$ such that $\Omega(x;r)\subset D$ and direct the unit normal $\nu$ to $\Omega(x;r)$ into the interior of $\Omega(x;r)$. Apply Green's second identity to the harmonic functions $u$ and $\Phi(x,\cdot)$ in the domain $\{y\in D:|y-x|>r\}$ to obtain $$\int_{\partial D\cup\Omega(x;r)}\bigg\{u(y)\frac{\partial\Phi(x,y)}{\partial\nu(y)}-\frac{\partial u}{\partial\nu}(y)\Phi(x,y)\bigg\}ds(y)=0.$$ Now we have $$\int_{\Omega(x;r)}\bigg\{u(y)\frac{\partial\Phi(x,y)}{\partial\nu(y)}-\frac{\partial u}{\partial\nu}(y)\Phi(x,y)\bigg\}ds(y)=\int_{\partial D}\bigg\{\frac{\partial u}{\partial\nu}(y)\Phi(x,y)-u(y)\frac{\partial\Phi(x,y)}{\partial\nu(y)}\bigg\}ds(y)$$

Here is my question:

How can I show that $$\lim_{r\to 0}\int_{\Omega(x;r)}\bigg\{u(y)\frac{\partial\Phi(x,y)}{\partial\nu(y)}-\frac{\partial u}{\partial\nu}(y)\Phi(x,y)\bigg\}ds(y)=u(x)?$$

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For the second term, since $u\in C^2(\bar{D})$, $\partial u/\partial \nu$ is bounded, say by $M$. Then

$$\begin{eqnarray} \left|\int_{\Omega(x;r)}\frac{\partial u}{\partial\nu}(y)\Phi(x,y)\mathrm ds(y)\right| &\le& M\int_{\Omega(x;r)}|\Phi(x,y)|\mathrm ds(y) \\ &=& Mcr^{m-1}\Phi(r)\;, \end{eqnarray}$$

where $c$ is the surface area of the unit sphere ($2\pi$ and $4\pi$, respectively) and $\Phi(r)$ is the respective function such that $\Phi(x,y)=\Phi(|x-y|)$. This product goes to zero as $r\to0$, since $r\ln r\to0$.

For the first term, we have $\partial\Phi(x,y)/\partial\nu(y)=-\Phi'(r)=1/(cr^{m-1})$, where the minus sign arises because the unit normal is directed into the interior. Thus

$$\begin{eqnarray} \lim_{r\to 0}\int_{\Omega(x;r)}u(y)\frac{\partial\Phi(x,y)}{\partial\nu(y)}ds(y) &=& \lim_{r\to 0}\int_{\Omega(x;r)}\frac{u(y)}{cr^{m-1}}ds(y) \\ &=& \lim_{r\to 0}\frac1c\int_{\Omega(x;r)}u(y)\mathrm d\Omega \\ &=& u(x) \end{eqnarray}$$

(where $\mathrm d\Omega$ denotes integration over the solid angle), since $u$ is continuous.

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+1. Thanks for you nice answer. I don't quite understand the second last equality: $$\int_{\Omega(x;r)}\frac{u(y)}{r^{m-1}}ds(y)=\int_{\Omega(x;r)}u(y)d\Omega.$$ What's "integration over the solid angle"? – Jack Aug 23 '11 at 14:08
I guess you did a scaling and the second integral should be $\int_{\Omega(x,1)}u(y)ds(y)$? – Jack Aug 23 '11 at 14:13
@Jack: Yes, you can write it that way. I was using $\mathrm d\Omega$ and "integration over the solid angle" as they're used e.g. here; I guess "integration over the full solid angle" would be more precise, since "solid angle" can also refer to a specific solid angle less than $4\pi$; also for the $m=2$ case "angle" would be more appropriate. – joriki Aug 23 '11 at 14:24
Fair enough. Thanks for you comment. :-) – Jack Aug 23 '11 at 14:34