Lemma 4.1.14.D17D3298467B The Green function of $B$ is symmetric: $G_ B(x,y)=G_ B(y,x)$ for $x,y\in B$. For $x,y\in B$ with $y\neq 0$, put $y^*=R^2y/|y|^2$. Then for $d\geq 3$,
\[ G_ B(x,y)=\Gamma _ d(x-y)-\left(\frac R{|y|}\right)^{d-2}\Gamma _ d(x-y^*), \]
and for $d=2$,
\[ G_ B(x,y)=\frac1{2\pi }\log \frac{|y|\, |x-y^*|}{R|x-y|}. \]
For $y=0$, $G_ B(x,0)=\Gamma _ d(x)-\Gamma _ d(Re_1)$ if $d\geq 3$ and $G_ B(x,0)=(2\pi )^{-1}\log (R/|x|)$ if $d=2$; here $e_1$ is the first coordinate unit vector. Moreover, for $0<r<R$, these formulas extend the difference $G_ B(x,y)-\Gamma _ d(x-y)$ to a smooth function of $(x,y)$ on the set where $|y|\leq r$ and $|x|<R^2/r$; this range of $x$ contains a neighborhood of $\overline B$.
Proof.
Fix $y\in B\setminus \{ 0\} $. Since $|y^*|=R^2/|y|>R$, the function $x\mapsto \Gamma _ d(x-y^*)$ is harmonic on a neighborhood of $\overline B$. Expanding the square gives
B44D46472BE9
4.1.15
\begin{equation} |y|^2|x-y^*|^2=R^4-2R^2x\mathbin {\cdot }y+|x|^2|y|^2, \end{equation}
and therefore $|y|^2|x-y^*|^2-R^2|x-y|^2=(R^2-|x|^2)(R^2-|y|^2)$. Each displayed expression, as a function of $x$, is the sum of $\Gamma _ d(x-y)$ and a function harmonic near $\overline B$. By proposition 3.1.13, it is therefore harmonic on $B\setminus \{ y\} $ and locally integrable on $B$, with distributional Laplacian $-\delta _ y$ on $B$. Moreover, the last identity shows that $|y|\, |x-y^*|>R|x-y|$ on $B$, with equality on $\partial B$. Hence the expression is positive on $B\setminus \{ y\} $ and tends to zero at every point of $\partial B$. The ball is a regular domain by theorem 3.1.29, and the uniqueness assertion of theorem 3.1.34 identifies the displayed expression with $G_ B(y,x)$ for $x\neq y$. The same argument applies to $y=0$, where the stated expressions differ from $\Gamma _ d(x)$ by constants.
Let $Q(x,y)$ denote the right side of (4.1.15); it is symmetric in $x$ and $y$, and $Q(x,0)=R^4$. The formulas just proved can be written, for all $x,y\in B$ including $y=0$, as
\[ \begin{aligned} G_ B(y,x)& =\Gamma _ d(x-y)-\frac{R^{d-2}Q(x,y)^{(2-d)/2}}{(d-2)s_{d-1}}& & (d\geq 3),\\ G_ B(y,x)& =\Gamma _2(x-y)+\frac1{4\pi }\log \frac{Q(x,y)}{R^2}& & (d=2). \end{aligned} \]
Both right sides are symmetric in $x$ and $y$, so $G_ B(y,x)=G_ B(x,y)$; this also proves the formulas in the order stated. Finally, $Q(x,y)\geq (R^2-|x||y|)^2>0$ when $|y|\leq r$ and $|x|<R^2/r$, so the correction terms $G_ B(x,y)-\Gamma _ d(x-y)$ in the last display are smooth functions of $(x,y)$ there.
$\square$
Pome
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