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I learned the extension of Green's formula to unbounded domains in Kress's Linear Integral Equations (p.71):

Theorem 6.10 Assume that $D$ is a bounded domain of class $C^1$ with a connected boundary $\partial D$ and outward unit normal $\nu$ and let $u\in C^2({\mathbb R}^m\setminus D)$ be a bounded harmonic function. Then $ u(x)=u_{\infty}+\int_{\partial D}\bigg\{u(y)\frac{\partial\Phi(x,y)}{\partial\nu(y)}-\frac{\partial u}{\partial \nu}\Phi(x,y)\bigg\}ds(y) $ for $x\in{\mathbb R}^m\setminus D$ and some constant $u_{\infty}$. For $m=2$, in addition, $ \int_{\partial D}\frac{\partial u}{\partial \nu}ds = 0 $ and the mean value property at infinity $ u_{\infty}=\frac{1}{2\pi r}\int_{|y|=r}u(y)ds(y) $ for sufficiently large $r$ is satisfied.

Here are my questions:

  • Since the expression $u(\infty)$ may be meaningless, how should I understand the constant $u_{\infty}$ here?
  • [EDITED: ] Should the last sentence be interpreted as $ \exists M>0, ~\text{such that}, ~\forall r>M,~ u_{\infty}=\frac{1}{2\pi r}\int_{|y|=r}u(y)ds(y)? $
  • Do we have $ \lim_{|x|\to\infty}u(x)=u_{\infty} $ [EDITED:] i.e., $ \lim_{|x|\to\infty}\int_{\partial D}\bigg\{u(y)\frac{\partial\Phi(x,y)}{\partial\nu(y)}-\frac{\partial u}{\partial \nu}\Phi(x,y)\bigg\}ds(y)=0 $ in this theorem?
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    @J.M.: I agree with you. But I think it is not that obvious to see that $\lim_{|x|\to\infty}u(x)=u_{\infty}$, i.e.,$\lim_{|x|\to\infty}u(x)=\frac{1}{2\pi r}\int_{|y|=r}u(y)ds(y).$2011-09-10

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