Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2011 / Paper 2 / 19 / a / Solution

Put $c=2\alpha /\pi$. In the right half-plane,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 $\alpha \ne 0$, 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.