Past exam of the fluids and waves course of the University of Oxford / 2014 / 2

The Milne-Thomson circle theorem constructs a circular impermeable boundary in planar potential flow. Suppose $f(\zeta )$ is an incident complex potential of a two-dimensional potential flow with no singularities on or inside $|\zeta |=R$. Thenis a complex potential of a two-dimensional potential flow outside the circle, with the same exterior singularities and their strengths as the incident flow; the added image singularities lie inside the excluded disk. On $|\zeta |=R$, $R^2/\bar{\zeta }=\zeta$, so $W=f+\bar{f}$ is real. Its imaginary part, the streamfunction, is constant on the circle, proving the no-penetration boundary condition. The theorem gives the usual zero-circulation construction; a separately prescribed circulation can be added by its logarithmic vortex potential.

A uniform flow with physical velocity $(U\cos \alpha ,U\sin \alpha )$ has complex velocity $U{e}^{-i\alpha }$ and incident potential $f(\zeta )=U{e}^{-i\alpha }\zeta$. Apply the Milne-Thomson circle theorem with $R=1$:Indeed, on $\zeta =e^{i\theta }$ it is $W=2U\cos (\theta -\alpha )$, which is real, so the boundary is a streamline. At infinity, $W^{\prime }(\zeta )\to U{e}^{-i\alpha }$, verifying the specified incoming uniform flow.

Parameterize the unit circle by $\zeta =e^{i\theta }$. Under the Joukowski map,Consequently its image is the ellipsewith horizontal and vertical semi-axesBoth are positive because $0<a<1$. The horizontal axis is the major axis. The derivative $dz/d\zeta =1-a^2/{\zeta }^2$ is nonzero for $|\zeta |\ge 1$. Also, equality of two images implies $({\zeta }_1-{\zeta }_2)(1-a^2/({\zeta }_1{\zeta }_2))=0$; two distinct exterior points cannot satisfy the second factor since $|{\zeta }_1{\zeta }_2|>1>a^2$. Thus the exterior map is one-to-one, mapping the exterior of the unit circle conformally onto the exterior of the ellipse.

At large $|\zeta |$, $z=\zeta +O({\zeta }^{-1})$, so the map tends to the identity. Its exterior inverse has the expansion $\zeta =z-a^2/z+O(z^{-3})$. Therefore the physical complex potential of a two-dimensional potential flow is $w(z)=W(\zeta (z))$, withIts leading term is unchanged, so the far-field velocity still has magnitude $U$ and inclination $\alpha$.

The chain rule gives the complex velocity:The parameter $\zeta$ denotes the unique exterior preimage of $z$. If an expression solely in $z$ is desired, takewhere the exterior holomorphic square root is chosen so that $\sqrt{z^2-4a^2}\sim z$ at infinity. Its branch points lie inside the ellipse. This is uniform potential flow around an ellipse.

The complex velocity vanishes where $e^{-i\alpha }{\zeta }^2-e^{i\alpha }=0$, provided its denominator is nonzero. Thus $\zeta =\pm e^{i\alpha }$. Both lie on the unit circle, and ${\zeta }^2-a^2\ne 0$ since $a<1$. Mapping these points into the physical plane givesEquivalently, the two stagnation points are the opposite points $(A\cos \alpha ,B\sin \alpha )$ and $(-A\cos \alpha ,-B\sin \alpha )$ on the ellipse. The conformal map preserves the zeros of the complex velocity here because its derivative is finite and nonzero on the boundary.