= Solution
Write the pole as $(x_0,y_0)$ with $x_0,y_0>0$. The <method of images> uses three reflected poles with signs $-, -, +$. Thus
$$
G(x,y)=\frac1{2\pi}\left[\log|(x,y)-(x_0,y_0)|-\log|(x,y)-(-x_0,y_0)|-\log|(x,y)-(x_0,-y_0)|+\log|(x,y)-(-x_0,-y_0)|\right].
$$
The logarithms cancel in pairs on each axis. Only the original pole is in the interior, so this is the required <Dirichlet Green function in a quadrant>.
On the horizontal boundary $(t,0)$ the outward normal is $-\mathbf e_y$. Differentiating the four logarithms gives
$$
\begin{aligned}
\partial_nG(t,0)&=\frac{y_0}{\pi}\left(\frac1{(t-x_0)^2+y_0^2}-\frac1{(t+x_0)^2+y_0^2}\right)\\
&=\frac{4x_0y_0t}{\pi[(t-x_0)^2+y_0^2][(t+x_0)^2+y_0^2]}.
\end{aligned}
$$
Multiply by the horizontal <Dirichlet boundary condition> $e^{-t^2}$ and integrate to obtain
$$
F(x_0,y_0)=\frac{4x_0y_0}{\pi}\int_0^\infty\frac{te^{-t^2}}{[(t-x_0)^2+y_0^2][(t+x_0)^2+y_0^2]}\,dt.
$$
By exchanging the axes, the vertical boundary contributes $F(y_0,x_0)$. The representation in part (a) consequently gives
$$
\boxed{\phi(x_0,y_0)=F(x_0,y_0)+F(y_0,x_0).}
$$
For the unbounded-domain condition, the same expression is the <Poisson integral> after the conformal <change of variables> $w=(x+iy)^2$ to the upper half-plane, with real-axis data $e^{-|s|}$. This continuous boundary function tends to zero at both ends. Its <Poisson integral> tends to zero as $|w|\to\infty$: split the data into a compact part, whose kernel integral tends to zero, and a tail uniformly smaller than any prescribed positive number. Hence the constructed function has the required decay, and its two boundary traces agree at the corner. The <maximum principle for harmonic functions> on expanding quarter-discs proves uniqueness among decaying solutions. The corner value is obtained by continuity as $1$.
Back to article page
