Past exam of the mathematics course of the University of Cambridge / 2023 / ib / Paper 4 / 16C (Fluid Dynamics)

Write the free surface as the zero setBecause this is a material surface, the kinematic boundary condition is $DS/Dt=0$. Thus, on $z=\eta$,Thereforeat the free surface. This is the kinematic boundary condition for a free-surface graph.

For irrotational flow, $\mathbf{u}=\nabla \varphi$. Incompressibility gives the Laplace equationThe rigid side walls impose no penetration:and decay in the infinitely deep fluid requires $\nabla \varphi \to 0$ as $z\to -\infty$.

Linearizing the kinematic condition from part (a) about $z=0$ givesThe Unsteady Bernoulli equation, evaluated at the surface and linearized about hydrostatic equilibrium, gives the dynamic conditionThese are the linearized free-surface boundary conditions. Initial values for $\eta$ and $\varphi$ specify the transient; below we also record the initially quiescent solution.

Forthe normal derivative at $x=0,L$ contains $\sin (\pi x/L)$ and the normal derivative at $y=0,L$ contains $\sin (2\pi y/L)$. Both therefore vanish at the side walls.

At $z=0$, ${\varphi }_t$ has exactly the same horizontal dependence as the imposed pressure. The dynamic boundary condition is consequently consistent provided $\eta$ has that same horizontal mode. This is the rectangular standing surface-gravity mode selected by the forcing.

Substitution into ${\nabla }^2\varphi =0$ givesSetThen $Z=A{e}^{kz}+B{e}^{-kz}$. Since the fluid occupies $z<0$, decay as $z\to -\infty$ forces $B=0$. Absorbing $A$ into $F$ and choosing $Z(0)=1$ givesThis is the usual exponential depth dependence of a deep-water gravity wave.

With $Z^{\prime }(0)=k$, the linearized kinematic condition givesHence

The dynamic condition reduces toDifferentiate $H^{\prime }=kF$ and define the natural frequencyThen the surface amplitude obeys the forced harmonic oscillatorFor $\omega \ne \Omega$, its general solution and the corresponding potential amplitude are

If the fluid is initially quiescent, $H(0)=F(0)=0$, so

The denominators in part (v) vanish when the forcing frequency equals the natural surface gravity wave frequencyAt this value, the initially quiescent resonant solution isBoth contain a term growing linearly in time. This is resonance: the pressure forcing repeatedly supplies energy in phase with the box's $(1,2)$ standing-wave mode. In the ideal inviscid linear model there is no damping or nonlinear saturation, so its amplitude is unbounded. This is the resonance of a pressure-forced rectangular surface-gravity mode.