Past exam of the fluids course of the University of Oxford / 2022 / 3

Each layer has constant mass density, initially negligible velocity and hence zero vorticity. Its Euler equations for an inviscid fluid have a conservative gravitational body force, so Kelvin's circulation theorem preserves zero bulk vorticity. In a simply connected layer, the irrotational flow therefore has ${\mathbf{v}}_i=\nabla {\varphi }_i$. The incompressible flow condition givesThis applies while the ideal layerwise solution is smooth. It allows an interfacial vortex sheet; it does not assert that the whole two-fluid flow has zero vorticity or that a smooth interface must exist for arbitrarily long times.

The rigid wall no-penetration boundary conditions arewith boundedness and ${\partial }_r{\varphi }_i=0$ at $r=0$ for regular axisymmetric fields. For a small interface displacement and small slope, linearization of the kinematic boundary condition about $z=0$ givesThe small-slope condition is an additional requirement for high-wavenumber modes: $|\alpha \eta |\ll 1$, not just $|\eta |\ll h,a$.

Choose the unperturbed hydrostatic pressures as ${\bar{p}}_i(z)=p_{\ast }-{\rho }_{i}gz$. The linearized Bernoulli equation gives the pressure perturbation $p_i^{\prime }=-{\rho }_i{\partial }_t{\varphi }_i$, after absorbing spatially constant gauges into ${\varphi }_i$. Since surface tension and viscous normal stress vanish, pressure continuity on $z=\eta$ givesorThe initial velocity is zero, so ${\eta }_t(r,0)=0$. Separate conservation of mass in each closed layer also requires ${\int }_0^a\eta (r,t)rdr=0$ relative to the initially horizontal interface.

Write $\eta =H{e}^{\sigma t}{J}_0(\alpha r)$ using the Bessel function of the first kind. The wall boundary conditions at $z=\pm h$ selectBecause $J_0^{\prime }(s)=-J_1(s)$, the sidewall no-penetration boundary condition requireswhere $j_{1,n}$ are the positive zeros of $J_1$. These radial eigenfunctions automatically preserve layer volume becauseThe constant radial mode $\alpha =0$ cannot change the interface height in a closed tank: the Laplace equation and the impermeable lids then give zero vertical velocity, and a uniform displacement would change each layer's volume.

The kinematic boundary condition fixesThus the interface values are ${\varphi }_1(0)=\sigma \eta \coth (\alpha h)/\alpha$ and ${\varphi }_2(0)=-\sigma \eta \coth (\alpha h)/\alpha$. Substitution in the normal stress condition gives the finite-depth Rayleigh-Taylor dispersion relation:soHere $\mathcal{A}$ is the Atwood number. Every allowed mode has a positive growth rate and is subject to Rayleigh-Taylor instability. Since $d[\alpha \tanh (\alpha h)]/d\alpha =\tanh (\alpha h)+\alpha h{\mathrm{sech}}^2(\alpha h)>0$, shorter radial scales grow faster. The limits areThere is no fastest finite wavenumber in the ideal sharp-interface model without surface tension, viscosity or diffusion. The zero-initial-velocity condition uses both exponential branches: an initial mode $H{J}_0({\alpha }_{n}r)$ evolves as $H\cosh ({\sigma }_{n,+}t)J_0({\alpha }_{n}r)$. If the mass density ordering were reversed the same calculation would give oscillatory stable modes; equal mass densities give neutral modes.

For an inviscid flow, the force per unit volume is $\mathbf{F}=-\nabla p-\rho g\hat{\mathbf{z}}$. The gravitational acceleration has zero curl. Using the curl product rule,The supplied two-dimensional vorticity equation therefore becomesThe out-of-plane vorticity has no stretching term in this two-dimensional setting. The source is baroclinic vorticity generation: differently directed mass density and pressure gradients generate rotation even from zero initial vorticity.

For an explicit sign check, set ${\omega }_y={\partial }_{z}u-{\partial }_{x}w$, with $z$ upward. A thin smoothed interface has $\rho \simeq \bar{\rho }(z-\eta (x,t))$, with ${\bar{\rho }}^{\prime }>0$. If $P(x,t)$ denotes the total pressure evaluated on the interface, then within the thin layer $p_x+{\eta }_x{p}_z\simeq P_x$ andIntegrating through that layer and taking its thickness to zero gives the vortex sheet strength $\gamma =u_2-u_1$ and its linear evolution:For a growing Fourier mode, the potentials from the preceding calculation with $\alpha$ replaced by $k$ giveThus the right-hand descending slope of a light-fluid crest produces positive ${\omega }_y$, while the left-hand ascending slope produces negative ${\omega }_y$. Their induced velocity raises the light-fluid crest and lowers the heavy-fluid trough, increasing the slope and generating more vorticity. This is the positive feedback underlying Rayleigh-Taylor instability.

One can close this mechanism quantitatively. For $\eta =H(t)\cos (kx)$, the kinematic boundary conditions give $\gamma =2\dot{H}\coth (kh)\sin (kx)$. Combining with the sheet-generation equation givesFor $kh\ll 1$, the lids confine motion throughout both layers and horizontal redistribution is slow, yielding $\sigma \sim \sqrt{\mathcal{A}gh}k$. For $kh\gg 1$, motion is localized within depth $k^{-1}$ on each side. Less fluid must accelerate at shorter wavelength, giving $\sigma \sim \sqrt{\mathcal{A}gk}$. Since $k_n=n\pi /b$, the ideal growth rate increases with $n$ with no short-wave cutoff.

The sharp-interface result requires $k\delta \ll 1$ for the smoothed layer. When $k\delta$ becomes order one, it no longer applies. A smooth adverse mass density gradient has a local overturning time of order $[g{\bar{\rho }}^{\prime }/\bar{\rho }{]}^{-1/2}$ rather than the unbounded sharp-interface rate. Finite molecular diffusion also relaxes small-scale mass density perturbations on a time of order $(D{k}^2{)}^{-1}$, and viscosity, if included, damps small-scale velocity. Weak diffusion motivates the thin-layer picture; it does not justify extrapolating the discontinuous dispersion relation to arbitrary $n$.