Past exam of the mathematics course of the University of Cambridge / 2018 / iii / Paper 331 / 1

For irrotational flow, ${\mathbf{u}}_j=\nabla {\varphi }_j$. An impermeable stationary horizontal wall has zero normal velocity, so the kinematic boundary condition isNo tangential no-slip condition is imposed on these inviscid fluids.

The interface is material: fluid particles on $F=z-\eta (x,t)=0$ remain there. Its material derivative therefore vanishes, giving $w_j={\eta }_t+u_j{\eta }_x=D_j\eta /Dt$ on the interface. Since $w_j={\partial }_z{\varphi }_j$, this is the kinematic boundary condition for a free-surface graph on each side. The material derivatives use the respective tangential fluid velocities; they enforce the same moving normal boundary.

The Unsteady Bernoulli equation gives $p_j+{\rho }_j[{\partial }_t{\varphi }_j+|\nabla {\varphi }_j{|}^2/2+gz]=C_j(t)$. Absorb the spatially constant functions $C_j$ into the gauges of the velocity potentials. Evaluating at $z=\eta$ and subtracting the equations yields the stated dynamic boundary condition for an inviscid interface. Surface tension supplies the pressure jump, rather than pressure continuity. Thus the first condition enforces material motion, and the second enforces the Bernoulli pressure balance.

Use the linearized boundary condition: evaluate at $z=0$ and discard products of perturbations. The kinematic boundary condition and Young–Laplace equation becomeIncompressible flow and irrotational flow give the Laplace equation for each potential. For a normal mode, the wall conditions selectWith $\eta =B{e}^{i(kx-\omega t)}$, the kinematic conditions give $A_1=i\omega B/[k\sinh (k{L}_1)]$ and $A_2=-i\omega B/[k\sinh (k{L}_2)]$. Hence ${\hat{\varphi }}_1(0)=i\omega B\coth (k{L}_1)/k$ and ${\hat{\varphi }}_2(0)=-i\omega B\coth (k{L}_2)/k$. Substitution into the dynamic condition gives the finite-depth Rayleigh-Taylor dispersion relation:Its denominator is positive. Assuming $g>0$ and $\gamma >0$, negative ${\omega }^2$ therefore produces a growing branch $\omega =i\sigma$ with $\sigma >0$, as in Rayleigh-Taylor instability.

Put $A=g({\rho }_1-{\rho }_2)>0$. In the deep-layer limit, the finite-depth Rayleigh-Taylor dispersion relation gives ${\omega }^2=(\gamma k^3-Ak)/({\rho }_1+{\rho }_2)$. Its sign changes atThe Rayleigh-Taylor instability band is $0<k<k_c$; $k=k_c$ is neutral and $k>k_c$ is oscillatory.

For the growing normal mode in two deep layers,It vanishes at the endpoints of the unstable band and has its unique interior maximum where $A-3\gamma k^2=0$. Using $k_c=\sqrt{A/\gamma }$ givesThis is the maximum exponential growth rate, not the real oscillation frequency.

The stationary-point condition for the squared growth rate is $A-3\gamma k_m^2=0$. Its second derivative is $-6\gamma k_m/({\rho }_1+{\rho }_2)<0$, so

For $L_2\to \infty$, the denominator is ${\rho }_1\coth (k{L}_1)+{\rho }_2>0$. The numerator is unchanged, so the cutoff is exactly independent of depth:This remains exact before taking the thin-layer limit in the finite-depth Rayleigh-Taylor dispersion relation.

Write $A=g({\rho }_1-{\rho }_2)$ and $\epsilon =k_c{L}_1\ll 1$. Throughout the unstable band, $k{L}_1\le \epsilon$ and $\coth (k{L}_1)=1/(k{L}_1)+O(k{L}_1)$. Since ${\rho }_2<{\rho }_1$,The leading polynomial is maximized at $k^2=A/(2\gamma )$, where it equals $L_1{A}^2/(4{\rho }_1\gamma )$. Thus the thin-layer Rayleigh-Taylor growth asymptotics giveThis is a leading asymptotic expression; the exact maximum for a small but nonzero layer depth has corrections.

Maximize the leading thin-layer squared growth rate, $(L_1/{\rho }_1)(A{k}^2-\gamma k^4)$. Its positive stationary point obeys $2Ak-4\gamma k^3=0$ and is the unique interior maximum. ThereforeThe maximizing point stays within the thin-layer regime because $k_m{L}_1\sim k_c{L}_1/\sqrt{2}\ll 1$.

Use superscripts $T$ and $D$ for the thin-upper-layer and two-deep-layer cases. The cutoff is identical, whileThus confinement suppresses the Rayleigh-Taylor instability growth strongly, but moves the most amplified disturbance to a somewhat larger wavenumber. The cutoff depends on the gravity-capillary balance, whereas growth depends also on the inertia of the moving layers.