Past exam of the options II course of the University of Oxford / 2010 / K (Fluid Dynamics and Waves)

The Milne-Thomson circle theorem says that if a complex potential $f(z)$ has no singularities on or inside $|z|=a$, then a circular impermeable boundary can be introduced by takingThe new singularities are the reflected ones inside the circle. On $|z|=a$, $a^2/z=\overline{z}$, so $W(z)=f(z)+\overline{f(z)}$ is real; the stream function is constant and there is no normal velocity. At infinity the added term tends to $f^{\ast }(0)$, a harmless potential constant, so it preserves the far-field flow. This proves the theorem with zero added net circulation about the cylinder. A logarithmic term may then specify a different cylinder circulation.

For $f(z)=-i\Gamma \log (z-b)/(2\pi )$, put $c=a^2/b$. The reflected term iswith locally consistent logarithm branches. The circle-theorem solution has zero cylinder circulation: its image vortex at $c$ and its central vortex cancel. To impose circulation $-\Gamma$, add $i\Gamma \log z/(2\pi )$. The result isIt consists of the physical vortex and an opposite image vortex inside the obstacle. On the circle, $|z-b|=(b/a)|z-c|$, so its stream function is indeed constant.

The complex velocity isInside the positively oriented obstacle contour, $q^2$ has only the double pole at $c$. Its residue isThe residue theorem and the supplied Blasius force theorem therefore giveConsequentlyThe force points toward the held vortex, independent of the sign of its circulation.

The residue calculation is the force on a circular cylinder from a held exterior vortex.

The conformal map $\zeta =z^3$ sends the wedge $0<\arg z<\pi /3$ to the upper half-plane. The physical vortex maps to $\zeta =c^3$; an opposite image vortex at ${\overline{c}}^{,3}$ makes the real axis a streamline. Pulling this potential back givesSince $z^3-c^3=(z-c)(z^2+cz+c^2)$, splitting the first logarithm, up to an additive constant, gives exactly the required form. The image construction enforces zero normal velocity on both rays.

To find the vortex motion, hold $c$ fixed while differentiating with respect to $z$, remove the self-induced singular term $-i\Gamma /(2\pi (z-c))$, and then evaluate at $z=c$. The remaining complex velocity isA complex velocity $u-iv$ equals $d\overline{c}/dt$, rather than $dc/dt$. ThusThe overbar on the source derivative is essential to the trajectory calculation.

Write $c=r{e}^{i\theta }$. Since $c^3-{\overline{c}}^{,3}=2i{r}^3\sin (3\theta )$, multiply the equation by $e^{i\theta }$ to obtainTaking real and imaginary parts givesThereforeproving the trajectory invariantThe calculation also covers $\Gamma =0$, when the vortex position is stationary.

This is the sixty-degree case of a point vortex in a wedge.