Past exam of the fluids and waves course of the University of Oxford / 2021 / 2

The Joukowski map is a holomorphic function for $z\ne 0$, withA holomorphic function is a local conformal map wherever its derivative is nonzero. Hence the two finite nonzero critical points areThey lie on the boundary of the fluid region. At $z=0$ the displayed formula has a pole and is undefined as a map to the finite complex plane; zero is outside the fluid region. Thus $g$ is locally conformal everywhere in the fluid.

For $z=x+iy$,Both factors are positive in the fluid region, so its image is contained in the upper half-plane. The boundary correspondence is also explicit: on the upper unit semicircle, $z=e^{i\theta }$ gives $g(z)=2\cos \theta \in [-2,2]$, while the real rays $x>1$ and $x<-1$ map onto the rays $\zeta >2$ and $\zeta <-2$.

To prove the image is the whole upper half-plane, fix any $\mathrm{Im}\zeta >0$. The equation $g(z)=\zeta$ isIts roots have product one. Neither root can have unit modulus, since that would make $\zeta =z+1/z$ real. Exactly one root therefore has modulus greater than one. For that root the formula for $\mathrm{Im}g$ forces $y>0$, so it belongs to the fluid region. Moreover, if $g(z_1)=g(z_2)$, thenTwo distinct roots in the fluid region would require $z_1{z}_2=1$, impossible when both moduli exceed one. Thus the Joukowski map of the upper exterior of a unit disk is a bijection from the fluid region onto the upper half-plane.

The quadratic formula gives two algebraic possibilities. To select one consistent holomorphic square root, setwhere each factor is the principal square root. Both arguments lie in the upper half-plane, so neither crosses its branch cut. This choice satisfies $S(\zeta {)}^2={\zeta }^2-4$ and $S(\zeta )\sim \zeta$ at large $\zeta$ in the upper half-plane. The required inverse function isFor $\zeta =iR$, $R>0$, this gives $z=i(R+\sqrt{R^2+4})/2$, whose modulus exceeds one. As $\zeta$ varies continuously in the upper half-plane, this root cannot cross $|z|=1$, since that would make $\zeta$ real. Hence the selected root always lies in the fluid region. The other root is its reciprocal and lies inside the unit disk. Simply writing the principal square root of ${\zeta }^2-4$ would not specify this branch correctly throughout the upper half-plane.

The line vortex is mapped toIts imaginary part is positive because $c$ lies in the fluid region. A conformal map preserves the circulation of the local line vortex, so its strength in the mapped complex plane is still $\Gamma$.

Take positive $\Gamma$ to mean counterclockwise circulation. A line vortex at ${\zeta }_0$ has complex potential of a two-dimensional potential flow $\Gamma \log (\zeta -{\zeta }_0)/(2\pi i)$. The method of images imposes the no-penetration boundary condition on the real axis by placing an image line vortex of strength $-\Gamma$ at ${\bar{\zeta }}_0$. ThusFor real $\zeta$, the numerator and denominator have equal modulus, so the streamfunction isThe real axis is therefore a streamline. The image line vortex lies outside the mapped fluid region, while the complex velocitydecays as $|\zeta |\to \infty$. This selects the flow due to the line vortex and the boundary, without superposing an imposed background velocity field.

The physical complex potential of a two-dimensional potential flow is $w(z)=W(g(z))$. Applying the chain rule gives the complex velocityThe physical velocity is $(\mathrm{Re}{w}^{\prime },-\mathrm{Im}{w}^{\prime })$. A useful equivalent expression follows fromFactoring the image expression in the same way makes the $-1/z$ terms cancel, leavingThis represents the physical line vortex at $c$, an image line vortex of the same strength at $c^{-1}$, and image point vortices of opposite strength at $\bar{c}$ and ${\bar{c}}^{-1}$. All three image locations are outside the fluid region. Both boundary pieces inherit the constant streamfunction of the real axis in the mapped complex plane.

A free point vortex is advected by the regular part of the velocity field at its position. Its own singular complex velocity $\Gamma /[2\pi i(z-c)]$ must first be subtracted. The image expression therefore givesEquivalently, $\dot{c}$ is the complex conjugate of the boxed expression. This is the instantaneous velocity, evaluated with the current value of $c$.

To see why mapping the image velocity alone would miss a term, expand the mapped physical singularity:Thus the same point-vortex drift under a conformal map can be writtenThe finite term involving $g^{\prime \prime }$ is part of the physical regular velocity field.

If $c=id$ with $d>1$, direct substitution into the image expression givesThe bracket is strictly positive, and the result is real. Hence a positive line vortex initially moves horizontally to the right, a negative line vortex moves to the left, and the initial vertical velocity is zero. As $d\to \infty$, the leading term $\Gamma /(4\pi d)$ is the familiar drift of a point vortex above a straight wall.