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\begin{document}


\begin{center}
Representation Formula
\end{center}

\begin{theorem}
If $\Delta u=0$ then 
\begin{equation*}
u(x_{0})=\frac{1}{4\pi }\diint\limits_{\partial D}\left( \frac{1}{|x-x_{0}|}%
\frac{\partial u}{\partial n}-u(x)\frac{\partial }{\partial n}(\frac{1}{%
|x-x_{0}|})\right) dS
\end{equation*}
\end{theorem}

\begin{proof}
By Green's second identity 
\begin{equation*}
\diiint \left( u\Delta v-v\Delta u\right) dV=\diint\limits_{\partial
D}\left( u(x)\frac{\partial v}{\partial n}-v(x)\frac{\partial u}{\partial n}%
\right) dS
\end{equation*}

Choose%
\begin{equation*}
v(x)=-\frac{1}{4\pi |x-x_{0}|}\sim \frac{1}{r}
\end{equation*}

Then%
\begin{equation*}
\Delta u=0\text{ and }\Delta v=0\text{ except at }x=x_{0}
\end{equation*}

Consider a region $D_{\epsilon }=D-$circle of radius $r$ about x$_{0}.$

So%
\begin{equation*}
\diint\limits_{\partial D_{\epsilon }}\left( u(x)\frac{\partial }{\partial n}%
\left( \frac{1}{r}\right) -\frac{1}{r}\frac{\partial u}{\partial n}\right)
dS=0
\end{equation*}
\end{proof}

\newpage

\begin{center}
Green's function
\end{center}

\bigskip

Consider%
\begin{align*}
-\Delta u &= 0\qquad \text{in }D \\
u &= f\qquad \text{on }\partial D
\end{align*}

\begin{definition}
$G(x,x_{0})$is a Green's function for $-\Delta $ in $D$ and $x_{0}\in D$ if

\begin{enumerate}
\item $G(x)\in C^{2}$

\item $-\Delta G(x,x_{0})=0$\quad $x\in D\quad x\neq x_{0}$

\item $G(x,x_{0})=0\quad x\in \partial D$

\item $\widetilde{G}=G(x,x_{0})+\frac{1}{4\pi |x-x_{0}|}$ is finite at $%
x_{0} $ and so $\Delta \widetilde{G}=0$ everywhere in $D$
\end{enumerate}
\end{definition}

\bigskip

\begin{theorem}
$G(x,x_{0})$ exists and is unique
\end{theorem}

\bigskip

\begin{theorem}
\begin{equation*}
u(x_{0})=\diint\limits_{D}u(x)\frac{\partial G(x,x_{0})}{\partial n}dS
\end{equation*}
\end{theorem}

\begin{proof}
Using the second Green's identity%
\begin{equation*}
\diiint \left( u\Delta H-H\Delta u\right) dV=\diint\limits_{\partial
D}\left( u(x)\frac{\partial H}{\partial n}-H(x)\frac{\partial u}{\partial n}%
\right) dS
\end{equation*}

Let $\Delta u=0$ and $\Delta H=0$ then%
\begin{equation*}
(\ast )\qquad 0=\diint\limits_{\partial D}\left( u(x)\frac{\partial H}{%
\partial n}-H(x)\frac{\partial u}{\partial n}\right) dS
\end{equation*}

Define 
\begin{eqnarray*}
v &=&-\frac{1}{4\pi |x-x_{0}|} \\
\text{so\quad }\Delta v &=&\delta (x-x_{0})
\end{eqnarray*}

Then 
\begin{equation*}
(\ast \ast )\qquad u(x_{0})=\diint\limits_{\partial D}\left( u(x)\frac{%
\partial v}{\partial n}-v(x)\frac{\partial u}{\partial n}\right) dS
\end{equation*}

Choose%
\begin{equation*}
G(x,x_{0})=H(x)+v(x)
\end{equation*}

adding $(\ast )$ and $(\ast \ast )$ we get%
\begin{equation*}
u(x_{0})=\diint\limits_{\partial D}\left( u(x)\frac{\partial G}{\partial n}%
-G(x)\frac{\partial u}{\partial n}\right) dS
\end{equation*}

but $G=0$ on the boundary. So

\begin{equation*}
u(x_{0})=\diint\limits_{\partial D}u(x)\frac{\partial G}{\partial n}dS
\end{equation*}
\end{proof}

\begin{theorem}
\begin{equation*}
G(x,x_0) = G(x_0,x)
\end{equation*}
\end{theorem}

\noindent For a general region $G$ is difficult to find

\end{document}
