Past exam of the mathematics course of the University of Cambridge / 2021 / iii / Paper 332 (Fluid Dynamics of the Solid Earth)

Let the unperturbed interface rise at pore velocity $U/\varphi$, and write its displacement asIn each fluid, Darcy's law and incompressible flow implyThe decaying pressure perturbations are proportional to $e^{-\alpha |z|}$. Continuity of normal velocity, the kinematic boundary condition $\varphi {\eta }_t=u_n^{\prime }$, and continuity of pressure giveThus a less mobile displaced fluid, ${\mu }_2>{\mu }_1$, and a denser fluid above a lighter one both drive the Saffman–Taylor instability.

For immiscible fluids, the Young–Laplace equation adds the pressure jump $-\gamma {\eta }_{xx}$. The dispersion relation becomeswhereIf $A>0$, the unstable band is $0<\alpha <\sqrt{A/C}$ and differentiation gives

Define the signed characteristic buoyancy velocitySince ${\mu }_2/({\mu }_1+{\mu }_2)=(1+M)/2$ and $Ca={\mu }_{2}U/\gamma$,This expression applies when the quantity in the final parentheses is positive.

Every growth curve starts at the origin. For equal densities its initial slope is proportional to $M$; for ${\rho }_2>{\rho }_1$ buoyancy shifts that slope upward, while for ${\rho }_2<{\rho }_1$ it shifts it downward. When $A>0$, the curve rises to one positive maximum and then crosses zero before its stabilizing $-{\alpha }^3$ capillary tail. When $A\le 0$, every nonzero wavenumber decays.

For a prescribed nonzero wavenumber, neutral stability requiresorIn the quasistatic limit $U\to 0$, viscosity contrast disappears and this reduces to the capillary Rayleigh-Taylor instability threshold ${\rho }_2-{\rho }_1=\gamma {\alpha }^2/g$. Without surface tension, neutral stability in that limit simply requires equal densities.

LetThe Neumann solution of the Stefan problem in the ice and substrate isandContinuity of heat flux at the contact gives, with $r=\sqrt{k_s/k}$,and therefore

At the ice–water interface, the water is isothermal at $T_m$. The Stefan condition $\rho L\dot{h}=k{T}_{i,z}(h^{-})$ givesEliminating $T_0$ yields the implicit equationwhich determines $\lambda$ and hence $h(t)=2\lambda \sqrt{\kappa t}$.

If $k_s\gg k$, then $r\gg 1$, $T_0\sim T_s$, andThe highly conducting substrate acts as a reservoir fixed near its initial cold temperature, giving the usual one-phase Stefan problem.

If $k_s\ll k$, then $r\ll 1$, $T_0\sim T_m$, and $\lambda \sim r/(\sqrt{\pi }S)$. ConsequentlyHere heat removal through the poorly conducting substrate is rate limiting, and only a small temperature drop is needed across the much more conducting ice.

Measure $z$ upward from the heat exchanger and let the steady ice front be at $z=h>0$. In the exchanger frame, salt in the liquid satisfies the advection-diffusion equationThe decaying solution and the prescribed total salt mass aresoThe linear liquidus condition fixes the interface temperature asA solid layer between the exchanger and the interface can therefore exist only if

Write $\vartheta =T-T_{\infty }$. Heat advection, conduction, and environmental loss giveIts characteristic exponents areBelow the exchanger and above the ice front, boundedness givesIn $0<z<h$, the ice temperature iswhereSubstitution of these fields into the Stefan conditiongives one scalar equation for the steady height $h$, which can be solved numerically. The heat flux may jump at $z=0$ because the exchanger supplies the required localized cooling.

The local equilibrium freezing temperature ahead of the front isBecause $T_l(h)=T_L(h)$, constitutional supercooling begins when the actual liquid-temperature gradient at the interface is smaller than the liquidus gradient:Using the solutions above, this is

At criticality the inequality is an equality. If $F/k\ll V^2/{\kappa }^2$, then $q\sim V/\kappa$ and, for $\kappa >D$,The critical curve is proportional to $S^{-1}$. If $F/k\gg V^2/{\kappa }^2$, put $q\sim \sqrt{F/k}$ to obtainThis curve behaves as $S^{-1/2}$ for small $S$ and approaches $D\sqrt{F/k}$ for large $S$. Supercooling occurs above the corresponding critical curve, where solute rejection steepens the liquidus faster than heat transport raises the actual temperature.

Take the rock at $z=0$, the till–ice interface at $z=h$, and the ice surface at $z=h+H$. The common leading hydrostatic pressure has horizontal gradientThe thin-film momentum equations areImpose no slip at the rock, continuity of velocity and shear stress at $z=h$, and zero shear at the ice surface. Expanding for $h/H\ll 1$ gives the till flux and ice fluxThe basal shear stress isApplying mass conservation separately to the ice and till, with erosion source $E{\tau }_0$, givesand

In a steady state $q_i=q_0$. Balancing the two terms in the till equation over length $L$ givesand henceindependently of $q_0$. Without substantial lubrication, the usual shallow-ice balance gives $H^4\sim \mu q_{0}L/(\rho g)$. ThereforeThe sheet is essentially unlubricated when $\boxed{\mathcal{E}\ll 1}$.

For $\mathcal{E}\gg 1$, the sliding term dominates the ice flux:Integrating the steady till equation from $x=0$, where $H=H_0$ and $h=0$, givesEliminating $H_x$ between these equations yieldsWith $F=H/H_0$ and $\xi =x/L$, a second integration giveswhere the physical branch decreases from $F(0)=1$ to $F(1)=0$. The length condition determinesUsing this relation in the till thickness givesThus $H$ decreases monotonically from $H_0$ to zero. The till starts at zero, rises to one interior maximum at $F=1/\sqrt{3}$, and returns to zero at the margin.