Past exam of the options II course of the University of Oxford / 2006 / K (Fluid Dynamics and Waves)

Use the convention that a positive vortex has counterclockwise circulation, with complex potential of a two-dimensional potential flow $\Gamma \log (z-z_1)/(2\pi i)$. To make both coordinate axes streamlines, place images of strength $-\Gamma$ at ${\bar{z}}_1$ and $-{\bar{z}}_1$, and strength $+\Gamma$ at $-z_1$. The resulting potential isAdditive constants and consistent logarithm branches do not affect velocity. The quotient has unit modulus on either wall, so the streamfunction is constant there and there is no normal flow.

The vortex moves with the regular fluid velocity after removing its own singular field. Since $W^{\prime }=u-iv$, evaluating the three image contributions at $z_1$ gives exactlyThe overall minus sign and the two reflected-term minus signs here agree with the source scan.

Write $z_1=x+iy$, with $x,y>0$. Taking real and imaginary parts yieldsDividing gives $dy/dx=-y^3/x^3$, which integrates toPut $\ell =ab/\sqrt{a^2+b^2}$. The first-quadrant branch has vertical and horizontal asymptotes $x=\ell$ and $y=\ell$. Its motion is to the right and downward: from large $y$ near the vertical asymptote toward large $x$ near the horizontal one. Both coordinates stay strictly positive, so no wall is crossed.

The exact free-surface kinematic condition is ${\eta }_t+{\varphi }_x{\eta }_x={\varphi }_y$ evaluated at $y=\eta$. Linearizing about a flat surface at rest drops products of small quantities and evaluates at $y=0$, giving ${\varphi }_y={\eta }_t$. Bernoulli's equation with constant atmospheric pressure gives ${\varphi }_t+g\eta =0$ to the same order. The rigid bed has no normal flow, hence ${\varphi }_y=0$ at $y=-h$.

In the fluid, incompressibility and the velocity-potential representation give $\Delta \varphi =0$. For $k>0$, takeIt is harmonic, satisfies the bed condition and the linear dynamic condition. At the surface its vertical derivative is $Agk\tanh (kh)\sin (kx-\omega t)/\omega$, while ${\eta }_t=A\omega \sin (kx-\omega t)$. Therefore the kinematic condition is equivalent toFinally $\tanh q=q-q^3/3+O(q^5)$. Taking the square root givesThe first term is nondispersive shallow-water speed; the next term is the leading finite-depth dispersion correction.