Past exam of the mathematics course of the University of Cambridge / 2009 / ib / Paper 4 / 18D (Fluid Dynamics)

For small irrotational surface gravity waves, write the velocity field as $\nabla \varphi$. Incompressibility gives ${\nabla }^2\varphi =0$. The rigid walls have ${\varphi }_x=0$ at $x=0,a$, ${\varphi }_y=0$ at $y=0,b$, and ${\varphi }_z=0$ at $z=-h$. At $z=0$, the linear kinematic boundary condition and pressure condition areSeparate variables for a rectangular standing surface-gravity mode:The horizontal cosine factors and vertical hyperbolic cosine satisfy all rigid-wall conditions. Put $\eta =A(t)\cos (m\pi x/a)\cos (n\pi y/b)$ and $\lambda =k\tanh kh$. The free-surface conditions give $\dot{A}=\lambda B$ and $\dot{B}=-gA$, henceBoth integers are nonnegative and not both zero. The zero pair would be a spatially uniform displacement, excluded by fixed fluid volume; a spatially constant potential is only a gauge mode.

The applied pressure changes the dynamic condition to ${\varphi }_t+g\eta =-p/\rho$. For its specified spatial mode, the two amplitude equations becomeWhen $\Omega \ne \omega$, the corresponding particular oscillation isFree oscillations with amplitudes $C\cos \omega t+D\sin \omega t$ may be added. If the mode starts from rest at $t=0$, its total amplitude instead isNear resonance, the response grows too large for the linearization. At exact resonance the rest-start limit is $A=(\epsilon h/2)(\omega t\cos \omega t-\sin \omega t)$, with secular growth. Dissipation or nonlinear effects would be required to limit it.