Past exam of the Probability, statistics and options II course of the University of Oxford / 2011 / K (Fluid Dynamics and Waves)

On the circle put $\zeta =a{e}^{i\theta }$. The Joukowski map gives $z=2a\cos \theta$, so its image is the slit $S=[-2a,2a]$. Away from that circle, the inverse equation is ${\zeta }^2-z\zeta +a^2=0$. For $z\notin S$, neither root has modulus $a$, and their product is $a^2$, so exactly one root has modulus greater than $a$. Selecting that root defines the exterior inverse.

More explicitly, define $s(z)=\sqrt{z^2-4a^2}$ to be the branch analytic on $\mathbb{C}\setminus S$ normalized by $s(z)/z\to 1$ at infinity. It is positive for real $z>2a$ and negative for real $z<-2a$. ThenExistence and analyticity of this branch also follow from the exterior inverse: $J^{\prime }(\zeta )=1-a^2/{\zeta }^2$ is nonzero for $|\zeta |>a$, so the selected simple root varies analytically; $s=2g-z$ has the stated normalization. The exterior map is injective because $J({\zeta }_1)=J({\zeta }_2)$ with distinct roots would imply ${\zeta }_1{\zeta }_2=a^2$, impossible for two exterior points. Thus $g$ is a conformal bijection throughout $\mathbb{C}\setminus S$.

The proposed potential is a uniform stream plus its cylinder doublet and the circulation term. On $\zeta =a{e}^{i\theta }$, the first two terms sum to the real quantity $2Ua\cos (\theta -\alpha )$, while the imaginary part of the logarithmic term is the constant $-\Gamma \log a/(2\pi )$. Thus the cylinder is a streamline. At infinity the complex velocity tends to $U{e}^{-i\alpha }$ and the circulation is $\Gamma$, verifying both far-field conditions. Logarithms may be chosen locally; their derivative gives a single-valued velocity.

Composition therefore gives the plate potentialAt the edge $z=2a$, corresponding to $\zeta =a$, the physical complex velocity is $W^{\prime }(\zeta )/J^{\prime }(\zeta )$. Since the denominator has a simple zero, finiteness requireswhich is exactlyWith this cancellation, the limiting ratio is $W^{\prime \prime }(a)/J^{\prime \prime }(a)$. The Kutta value of the circulation gives $W^{\prime \prime }(a)=2U\cos \alpha /a$ and $J^{\prime \prime }(a)=2/a$. Hence the finite edge velocity is

The material surface is $F(x,y,t)=y-\eta (x,t)=0$. Its material derivative vanishes, giving the exact kinematic conditionFor irrotational flow $u=\nabla \varphi$, the Euler equations and $(u\cdot \nabla )u=\nabla (|u{|}^2/2)$ integrate spatially to ${\varphi }_t+|\nabla \varphi {|}^2/2+gy+p/\rho =C(t)$. Adding a suitable function of time to $\varphi$ changes no velocity and makes the right side $p_a/\rho$. ThusTake $p_a$ as the air-side pressure and use the given capillary pressure jump on the fluid side. At first order in wave amplitude, the product ${\varphi }_x{\eta }_x$ and kinetic-energy term are quadratic, the curvature denominator is one to the required order, and replacing $y=\eta$ by $y=0$ changes first-order quantities only at second order. ConsequentlyFor a harmonic wave $\eta =A\cos (kx-\omega t)$, the harmonic potential decaying with depth isIt satisfies the kinematic condition. Substituting into the dynamic condition givesThe phase speed is $c=\omega /k$ and the group velocity of a narrow wave packet is $c_g=d\omega /dk$. For the positive-frequency branch and $\sigma =T/\rho$,For positive surface tension this ratio exceeds one exactly when $\sigma k^2>g$, proving