 Corner sink in a semi-infinite channel

For unit channel width, incompressible flow towards a corner withdrawal point has $\psi =0$ on the lower wall and $\psi =-m$ on the upper and left walls, with $\psi \to -my$ far downstream. For irrotational flow, separation of variables gives the displayed harmonic correction: the Fourier sine coefficients of $-m(1-y)$ are $-2m/(n\pi )$. Near the corner, $\psi \sim -2m\theta /\pi$ and radial velocity is $-2m/(\pi r)$, whose integrated inward flux across a quarter-circle is $m$.