## Tag Archives: green function

### Green function for annular region

Problem 2.5 Determine the Green’s function for the annular region bounded by two concenteric spheres in ${\mathbb{R}^n}$

Proof: Suppose the annular region is ${\Omega=B_R(0)/B_\rho(0)}$ and denote ${\lambda=\rho/R<1}$ and ${\displaystyle \bar{x}=\frac{R^2x}{|x|^2}}$, and ${\displaystyle \tilde{x}=\frac{r^2x}{|x|^2}}$.
Firstly

$\displaystyle h_1=\Gamma(x-y)-\Gamma(\frac{|x|}{R}|\bar{x}-y|)\quad x,y\in \Omega, x\neq y$

is a harmonic function in ${\Omega}$ and ${h_1|_{\partial B_R}=0}$ but not necessarily 0 on ${\partial B_\rho}$.

$\displaystyle h_2=\Gamma(\frac{|x|}{r}|\tilde{x}-y|)-\Gamma(\frac{|x|}{R}\frac{|\bar{x}|}{r}|\lambda^2x-y|)=\Gamma(\frac{|x|}{r}|\tilde{x}-y|)-\Gamma(\lambda^{-1}|\lambda^2x-y|)$

Then ${h_1-h_2}$ is harmonic in ${\Omega}$ and vanish on ${\partial B_\rho}$. Similarly construct

$\displaystyle h_3=\Gamma(\lambda|\lambda^{-2}x-y|)-\Gamma(\lambda\frac{|x|}{R}|\lambda^{-2}\bar{x}-y|)$

Then ${h_1-h_2+h_3}$ is harmonic in ${\Omega}$ and vanish on ${\partial B_R}$. We hope that

$\displaystyle h=h_1-h_2+h_3-h_4+h_5-\cdots$

is convergent, then ${h}$ is harmonic in ${\Omega}$ and vanish on both boundaries. Formally we get

$\displaystyle h=\Gamma(|x-y|)+\sum\limits_{k=1}^{\infty}[\Gamma(\lambda^{-k}|\lambda^{2k}x-y|)+\Gamma(\lambda^{k}|\lambda^{-2k}x-y|)]$

$\displaystyle -\sum\limits_{k=0}^{\infty}[\Gamma(\frac{|x|}{R}\lambda^{k}|\lambda^{-2k}\bar{x}-y|)+\Gamma(\frac{|x|}{r}\lambda^{-k}|\lambda^{2k}\tilde{x}-y|)]$

${h}$ is well defined, because the terms of infinite sum are geometric series, it is unifomly convergent when ${x\neq y}$. And very term is harmonic, thus ${h}$ is harmonic when ${x\neq y}$. One can verify that ${h}$ is 0 on the ${\partial \Omega}$.

$\Box$

### Properties of Green function

As we all know, Green’s function $G=\Gamma+h$, where $\Gamma$ is the fundamental solution of Laplacian operator

$\displaystyle \Gamma(x,y)=\begin{cases}\frac{1}{n(2-n)\omega_n}|x-y|^{2-n}, n>2\\\frac{1}{2\pi}\log|x-y|, n=2.\end{cases}$

and $h(x,y)$ is solved for each $x\in \Omega$ by

$\displaystyle \begin{cases}\Delta h(x,\cdot)=0\text{ in }\Omega\\h(x,\cdot)=-\Gamma(x,\cdot) \text{ on }\partial \Omega \end{cases}$

Then for any $u\in C^1(\overline{\Omega})\cap C^2(\Omega)$, $G$ satisfies

$\displaystyle u(x)=\int_{\partial \Omega}u\frac{\partial G}{\partial \nu}ds+\int_{\partial G}G(x,y)\Delta u(y)dy$

$\mathbf{Problem:}$ $\Omega\subset \mathbb{R}^n$ is bounded. $G(x,y)$ is the first kind of Green function of $\Omega$.

1. $G(x,y)<0$ for all $x,y\in \Omega$, $x\neq y$.
2. Let $f$ be a bounded and locally integrable function in $\Omega$. Then

$\displaystyle \int_\Omega G(x,y)f(y)dy\to 0$ as $x\to \partial \Omega$.

$\mathbf{Lemma:}$ Let $u$ be a non-negative harmonic function in $\Omega$, $y\in \Omega$ such that $B_{4R}(y)\subset \Omega$, then

$\displaystyle \sup\limits_{B_R(y)}u\leq 3^n\inf\limits_{B_R(y)}u$

We omit the proof of this lemma. The importance of this lemma is sup is bounded by inf with a constant independent of radius.

$\mathbf{Proof:}$ Fix $x\in \Omega$. Denote $u_x(\cdot)=-G(x,\cdot)$. Recall that $u_x(\cdot)$ is a harmonic function in $\Omega\backslash{x}$,  and $u_x(y)\to \infty$ as $y\to x$, there exists  $B_r(x)\subset \Omega$ with $0 such that $\exists \,y_0\in\partial B_r(x)$ and $u_x(y_0)=\epsilon$.

we can cover $\partial B_r(x)$ by finitely many balls centered at $\partial B_r(x)$ with radius $\frac{r}{8}$. By using the lemma iteratively, we know that $u_x(y)\leq M(n)\epsilon$ on the $\partial B_r(x)$.

Since $u_x$ is a harmonic function in $\Omega\backslash B_r(x)$ and $u_x(y)=0$ on $\partial \Omega$, by the maximum principle

$0 in $\Omega\backslash B_r(x)$.

This implies $\displaystyle \int_\Omega -G(x,y)dy=\int_{\Omega\backslash B_r(x)}u_xdy+\int_{B_r(x)}-G(x,y)dy\leq M\epsilon |\Omega|+\int_{B_r(x)}-G(x,y)dy$

When $x\to \partial \Omega$, we must have $r\to 0$, then $\displaystyle \int_{B_r(x)}G(x,y)dy\to 0$, because $G(x,y)=O(|x-y|^{2-n})$.

So $\displaystyle 0\leq \overline{\lim\limits_{x\to \partial \Omega}}\int_\Omega -G(x,y)dy\leq M\epsilon |\Omega|$. Let $\epsilon \to 0$. We have

$\displaystyle \int_\Omega G(x,y)dy\to 0$ as $x\to \partial \Omega$.

This implies when $x\to \partial \Omega$.

$\displaystyle \left|\int_\Omega G(x,y)f(y)dy\right|\leq ||f||_\infty \int_\Omega |G(x,y)|dy=-||f||_\infty \int_\Omega G(x,y)dy\to 0$

$\text{Q.E.D}\hfill \square$

$\mathbf{Remark:}$ GT’s chapter 2.3.