= Solution
Put $c=2\alpha/\pi$. In the <right half-plane>,
$$
\Phi_u=-\frac{cv}{u^2+v^2},\quad
\Phi_v=\frac{cu}{u^2+v^2},\quad
\Phi_{uu}=\frac{2cuv}{(u^2+v^2)^2},\quad
\Phi_{vv}=-\frac{2cuv}{(u^2+v^2)^2}.
$$
The last two cancel, so $\boxed{\Phi_{uu}+\Phi_{vv}=0}$. The possible singular point $(0,0)$ lies outside the open domain $u>0$.
For $\alpha\ne0$, a <level curve> of value $\Phi_0$ has $v/u=\tan(\pi\Phi_0/(2\alpha))$. These are rays from the origin, restricted to the <right half-plane>. For positive $\alpha$, the values increase with angle from $-\alpha$ on the lower <boundary> to $+\alpha$ on the upper <boundary>. If $\alpha=0$, the <function> is identically zero. The left panel of the figure in (c) shows these ray-shaped <level curves>.
Back to article page
