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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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Yes, I think that was the intended meaning. It's admittedly sloppy, but it's standard to indicate boundary conditions at infinity that way. –  J. M. Sep 7 '11 at 2:39
    
@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.,$$\lim_{|x|\to\infty}u(x)=\frac{1}{2\pi r}\int_{|y|=r}u(y)ds(y).$$ –  Jack Sep 10 '11 at 1:30

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