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Consider the fundamental solution of Laplace's equation: $$\Phi(x,y):= \begin{cases} \frac{1}{2\pi}\ln\frac{1}{|x-y|},\quad x,y\in{\mathbb R}^2\\ \frac{1}{4\pi}\frac{1}{|x-y|},\quad x,y\in{\mathbb R}^3. \end{cases}$$ Let $D\subset{\mathbb R^m} (m=2,3)$ be a bounded domain of class $C^1$. I am thinking about how I can get the following asymptotic behavior: $$\Phi(x,y)= \begin{cases} \frac{1}{2\pi}\ln\frac{1}{|x|}+O(\frac{1}{|x|}),\quad x,y\in{\mathbb R}^2\\ O(\frac{1}{|x|}),\quad x,y\in{\mathbb R}^3 \end{cases}$$ for $|x|\to\infty$, which holds uniformly for all directions $x/|x|$ and all $y\in\partial D$.

According to the definition of the big O notation, for the case in ${\mathbb R}^2$, I think finally I need to show that there exist constants $C,M>0$ such that $$ \frac{1}{2\pi}\bigg(\ln\frac{1}{|x-y|}-\ln\frac{1}{|x|}\bigg)\leq C\frac{1}{|x|} $$ for all $|x|>M$. This boils down to do the estimation $$ \ln\frac{|x|}{|x-y|}\leq \tilde{C}\frac{1}{|x|} $$ which I have no idea how to go on.

Similarly for the case in ${\mathbb R}^3$, one needs to show $$\frac{1}{|x-y|}\leq C\frac{1}{|x|}$$ for all $|x|>M$. This one is simpler: I know how to manipulate it in terms of coordinates.

Here is the question:

How should I do the case in ${\mathbb R}^2$ ?

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$\ln|x - y| = \ln(|x||1 - {y \over x}|) = \ln|x| + \ln|1 - {y \over x}|$ and use $\ln(1 - \epsilon) \sim -\epsilon$ for small $\epsilon$. –  Zarrax Sep 2 '11 at 14:27
    
It seems that we don't need the assumption for the domain $D$. –  Jack Sep 2 '11 at 14:49
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Following Zarrax's comment: for "large enough" $x$, we have $$ \ln\frac{|x|}{|x-y|}=\ln\frac{1}{|1-\frac{|y|}{|x|}|}=-\ln(1-\frac{|y|}{|x|}) $$ Now one can use the Taylor series for $\ln(1-x)$ to do the estimation.

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