# Green’s Formula and Green’s Identity

Suppose $\Omega\subset\R^n$ is a domain with $C^1$ boundary, given a $C^1$ vector field $\mathbf w\in C^1(\bar\Omega)$, we have the divergence theorem
$\int_\Omega\div \mathbf w\rd x=\int_{\pt\Omega}\mathbf w\cdot\n\rd S,$
where $x\in\Omega$ and $\n$ is the outward unit normal vector of $\pt\Omega$.

Problem 1. Show that the divergence theorem also holds for $\mathbf w\in C(\bar \Omega)\cap C^1(\Omega)$.

Remark 1. For any function $u\in C(\bar \Omega)\cap C^1(\Omega)$, define $\mathbf w$ by $\mathbf w\eqdef(0,\ldots,u,\ldots,0)$, i.e., the $i$’s components is $u$ and others are $0$, then the divergence theorem shows that
$\int_\Omega (D_iu)(x)\rd x=\int_{\Omega}\div \mathbf w\rd x=\int_{\pt\Omega}\mathbf w\cdot\n\rd S=\int_{\pt\Omega}u(x)\nu_i\rd S,$
where $\n=(\nu_1,\nu_2,\ldots, \nu_n)$ is the exterior unit normal vector of $\pt\Omega$. This formula can also be obtained by integration by parts.

Now, take $\mathbf w$ as the gradient of a function $u\in C^1(\bar\Omega)\cap C^2(\Omega)$, i.e., $\mathbf w=\D u=\left(\frac{\pt u}{\pt x_1}, \frac{\pt u}{\pt x_2},\ldots,\frac{\pt u}{\pt x_n}\right)$, and since $\div(\D u)=\sum_{i=1}^n\frac{\pt}{\pt x_i}\left(\frac{\pt u}{\pt x_i}\right)=\Delta u$, we obtain
$\int_\Omega\Delta u\rd x=\int_{\pt\Omega}\frac{\pt u}{\pt\n}\rd S\eqdef\int_{\pt\Omega}\D u\cdot\n\rd S.$
If $u,v\in C^1(\bar\Omega)\cap C^2(\Omega)$, take $\mathbf w$ as $u\D v$ and $v\D u$ in the above formula, since $\div(u\D v)=u\Delta v+\D u\D v$, then we have, respectively
\label{eq:1}
\int_\Omega u\Delta v\rd x=\int_{\pt \Omega} u\frac{\pt v}{\pt\n}\rd S-\int_\Omega \D u\cdot\D v\rd x,

\label{eq:2}
\int_\Omega v\Delta u\rd x=\int_{\pt \Omega} v\frac{\pt u}{\pt\n}\rd S-\int_\Omega \D u\cdot\D v\rd x.

Both \eqref{eq:1} and \eqref{eq:2} are called the Green’s first formula. Subtraction \eqref{eq:2} from \eqref{eq:1}, we obtain
\label{eq:3}
\int_\Omega (u\Delta v-v\Delta u)\rd x=\int_{\pt\Omega}\left(u\frac{\pt v}{\pt\n}-v\frac{\pt u}{\pt\n}\right)\rd S,

which is called the Green’s second formula.

Now, we will use Green’s formula to prove Green’s identity, which play an important role in discussion of harmonic functions.

Theorem 1. Suppose $\Omega$ is a bounded domain in $\R^n$ with $C^1$ boundary and $u\in C^1(\bar\Omega)\cap C^2(\Omega)$, then for any $x\in\Omega$ we have the following Green’s identity(also called Green’s representation)
\begin{split}
u(x)&=\int_\Omega\Gamma(x-y)\Delta_y u(y)\rd y\\
\Gamma(x-y)\frac{\pt u}{\pt\n_y}(y)
-u(y)\frac{\pt\Gamma}{\pt \n_y}(x-y)\right)\rd S_y,
\end{split}
where $\n_y$ is the exterior unit normal vector of $\pt\Omega$.

Remark 2. In particular, If $u$ is a harmonic function in $\Omega$, then
$u(x)=\int_{\pt\Omega}\left(u\frac{\pt\Gamma(x-y)}{\pt\n_y} -\Gamma(x-y)\frac{\pt u}{\pt\n_y}\right)\rd S_y,$
called the fundamental integral formula of harmonic functions. It states that for harmonic function $u\in C^1(\pt\Omega)\cap C^2(\Omega)$, its value at any point $x\in\Omega$ can be represented by $u$ and its normal derivative on $\pt\Omega$.

Proof . Firstly, we should note that: for any fixed $x\in\Omega$, $\Gamma(x-y)$ has a singular point at $y=x$. Thus, directly apply the Green’s second formula is not allowed. The idea is replace $\Omega$ by $\Omega\setminus B_r(x)$, $B_r(x)\subsubset\Omega$ is a ball centered at $x$ with radius $r$, and let $r$ limits to $0$ to obtain the desired result.

Just as we have analyzed, fix $x\in\Omega$, take $u=\Gamma=\Gamma(x-\cdot)$ and $v=u$ in \eqref{eq:3}, we get
$\begin{split} \int_{\Omega\setminus B_r(x)}\Gamma\Delta u-u\Delta\Gamma\rd y &=\int_{\pt(\Omega\setminus B_r(x))}\left( \Gamma\frac{\pt u}{\pt\n_y}- u\frac{\pt\Gamma}{\pt\n_y}\right)\rd S_y\\ &=\int_{\pt\Omega}\left( \Gamma\frac{\pt u}{\pt\n_y}- u\frac{\pt\Gamma}{\pt\n_y}\right)\rd S_y +\int_{\pt B_r(x)}\left( \Gamma\frac{\pt u}{\pt\n_y}- u\frac{\pt\Gamma}{\pt\n_y}\right)\rd S_y, \end{split}$
where $\n_y$ is the unit exterior normal vector of $\pt(\Omega\setminus B_r(x))$. Take $r\to 0$, and note that $\Delta\Gamma=0$ in $\Omega\setminus B_r(x)$, we have
\label{eq:4}
\int_\Omega\Gamma\Delta u\rd y=\int_{\pt\Omega}\left(
\Gamma\frac{\pt u}{\pt\n_y}-
u\frac{\pt\Gamma}{\pt\n_y}\right)\rd S_y
+\lim_{r\to 0}\int_{\pt B_r(x)}\left(
\Gamma\frac{\pt u}{\pt\n_y}-
u\frac{\pt\Gamma}{\pt\n_y}\right)\rd S_y.

Since $u$ is $C^2$ in $\bar B_r(x)$, there exists a constant $M>0$, such that $|\Delta u(y)|\leq M$ for any $y\in B_r(x)$. Now,
$\begin{split} \int_{\pt B_r(x)} \Gamma\frac{\pt u}{\pt \n_y}\rd S_y &=\Gamma(r)\int_{\pt B_r(x)}\frac{\pt u}{\pt \n_y}\rd S_y\\ &=-\Gamma(r)\int_{\pt B_r(x)}\frac{\pt u}{\pt\mathbf r_y}\rd S_y\\ &=-\Gamma(r)\int_{B_r(x)}\Delta u\rd y, \end{split}$
and
$\begin{split} \left|\int_{\pt B_r(x)}\Gamma\frac{\pt u}{\pt\n_y}\rd S_y\right| &=\left|-\Gamma(r)\int_{B_r(x)}\Delta u\rd y\right|\\ &\leq\left|\Gamma(r)\right|\int_{B_r(x)}\left|\Delta u\right|\rd y\\ &\leq M\left|\Gamma(r)r^{n-1}\right|\omega_n\\ &=\begin{cases} \frac{M}{2}r|\ln r| ,&n=2,\\ \frac{M}{n(n-2)}r, &n\geq3. \end{cases} \end{split}$
Thus,
\label{eq:5}
\lim_{r\to0}\int_{\pt B_{r}(x)}\Gamma\frac{\pt u}{\pt\n_y}\rd S_y=0.

Lastly,
$\begin{split} \int_{\pt B_r(x)} u\frac{\pt\Gamma}{\pt\n_y}\rd S_y &=\int_{\pt B_r(x)}u \frac{1}{n\omega_n}r^{1-n} \langle -\frac{x-y}{r}, \n_y\rangle\rd S_y\\ &=\int_{\pt B_r(x)}u \frac{1}{n\omega_n}r^{1-n} \langle -\frac{x-y}{r}, \frac{x-y}{r}\rangle\rd S_y\\ &=-\frac{1}{n\omega_n r^{n-1}}\int_{\pt B_r(x)} u\rd S_y, \end{split}$
thus,
\label{eq:6}
\lim_{r\to0}\int_{\pt B_r(x)}u\frac{\pt\Gamma}{\pt\n_y}\rd S_y=-u(x).

Plugs \eqref{eq:6} and \eqref{eq:7} into \eqref{eq:5}, we shall reach the desired result.

Problem 2. Show in detail that \eqref{eq:7} holds.

Remark 3. Comparing with Remark 1, we have
$\int_{\pt\Omega}\frac{\pt\Gamma}{\pt\n_y}(x-y)\rd S_y=1,$
for any $x\in\Omega$. This can be obtained by letting $u\equiv1$ in thm:1.
PDE is analysis, and analysis mainly are inequality, thus the ability of analysis is totally depend on the ability to estimate inequalities.