Past exam of the Probability, statistics and options II course of the University of Oxford / 2010 / K / K4 / Solution

The conformal map $\zeta =z^3$ sends the wedge to the upper half-plane. A fluid vortex at $c^3$ there has an opposite image at ${\bar{c}}^3$, so the pulled-back potential isThe two boundary rays map to the real axis, where the imaginary part is constant, giving the required impermeability. Factoring $z^3-c^3=(z-c)(z^2+cz+c^2)$ gives the printed form up to branch constants.

The fluid vortex moves with the regular part of the complex velocity, excluding its own singular term. Since $w^{\prime }=u-iv$, its motion is the conjugated derivative. Differentiating the remaining terms and setting $z=c$ givesThe overbar on the time derivative is essential.

Write $c=r{e}^{i\theta }$. Using $c^3-{\bar{c}}^3=2i{r}^3\sin 3\theta$ and multiplying the velocity equation by $e^{i\theta }$ givesThus $\dot{r}=3\Gamma \cot 3\theta /(4\pi r)$ and $\dot{\theta }=-\Gamma /(4\pi r^2)$. Direct differentiation now yieldsHence $\boxed{r\sin (3\theta )=\text{constant}}$ along the trajectory while it stays inside the wedge.