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

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