Past exam of the fluids and waves course of the University of Oxford / 2017 / 3 / b

Treat the free surface as the material level set $F(x,y,t)=y-\eta (x,t)=0$. The kinematic boundary condition $DF/Dt=0$, with $\mathbf{u}=\nabla \varphi$, givesThis says that fluid initially on the free surface remains on it.

For constant mass density $\rho$, the Euler momentum equation under gravity is $D\mathbf{u}/Dt=-\nabla p/\rho -g{\mathbf{e}}_y$. Since the flow is irrotational, $(\mathbf{u}\cdot \nabla )\mathbf{u}=\nabla (|\mathbf{u}{|}^2/2)$. Integration in space therefore gives the Unsteady Bernoulli equationAt the free surface, $p=P_a$ and $y=\eta$. Adding a function $G(t)$ to the velocity potential leaves the velocity field unchanged; choose $G^{\prime }=P_a/\rho -B(t)$. With this choice the surface dynamic boundary condition for an inviscid interface becomesThe constant atmospheric pressure has been absorbed into this choice of potential.

Linearize about fluid at rest, with mean free surface $y=0$ and mean bottom $y=-d$. Regard $a$, $\eta$ and the perturbation velocity potential as first-order quantities. A Taylor expansion from $y=\eta$ to $y=0$ produces products of first-order quantities, as do ${\varphi }_x{\eta }_x$ and $|\nabla \varphi {|}^2$. Discarding these second-order terms gives the linearized free-surface boundary conditionsEquivalently, differentiate the second condition in time and substitute the first to obtain ${\varphi }_{tt}+g{\varphi }_y=0$ at $y=0$.

The bottom is also impermeable and moving. Its exact kinematic boundary condition follows from $D(y-h)/Dt=0$:Here $h_t=-akc\cos (k(x-ct))$, while ${\varphi }_x{h}_x$ and the change from evaluating ${\varphi }_y$ at $h$ to evaluating it at $-d$ are second order. HenceThe interior Laplace equation remains ${\varphi }_{xx}+{\varphi }_{yy}=0$ on $-d<y<0$. This linearization requires small slopes and small resulting excursions as well as $a\ll d$; close to resonance a small prescribed bottom displacement can produce a response too large for it.

Seek the periodic response of a surface gravity wave forced by a traveling bottom, with the same wavenumber $k$ and angular frequency $kc$ as the bottom. The linear equations have constant coefficients in $x,t$, so they preserve each such harmonic. There is no dissipative phase lag in this model; away from resonance, the surface response is in phase or in opposition with the bottom. Thus write $\eta =b\sin \theta$, $\theta =k(x-ct)$, with a signed coefficient $b$. Its physical amplitude is $|b|$. Arbitrary additional free surface gravity waves depend on initial data; the paper's expression selects the forced traveling component.

For $k>0$, $c\ne 0$, use $\varphi =F(y)\cos \theta$. The Laplace equation gives $F^{\prime \prime }-k^{2}F=0$. At $y=0$, the two linearized free-surface boundary conditions implyThese determine both constants in the vertical solution:Its vertical derivative at the bottom isMatching it to $-akc\cos \theta$ and multiplying by $c$ givesand thereforeAn equivalent expression for the velocity potential, which makes the bottom condition immediate, isLet ${\omega }_f=kc$ be the forcing angular frequency, and let ${\omega }_0^2=gk\tanh (kd)$ be the natural dispersion relation from Airy wave theory. ThenThe denominator vanishes at $c^2=(g/k)\tanh (kd)$, when the forcing travels at the free-wave phase velocity. For nonzero forcing there is no bounded periodic forced solution at resonance; the undamped linear response develops secular growth. Below that speed $b<0$, giving a phase reversal, while above it $b>0$. At $c=0$ the bottom is stationary: the rest solution has $\varphi =0$ and $\eta =0$, consistent with the zero limit of $b$, though the formula containing $1/c$ must be interpreted through that limit.