Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 332 (Fluid Dynamics of the Solid Earth)

Put ice in $x<a(t)$ and ocean in $x>a(t)$, with $x$ increasing into the ocean. This is a saline Stefan problem. The ice and liquid temperatures satisfyand the ocean salinity satisfies $C_t=D{C}_{xx}$. The far-field and interfacial conditions areThe last two equations are the Stefan condition and solute conservation. Salt is taken to be absent from the ice. The field sketch has two broad thermal boundary layers of width $O(\sqrt{\kappa t})$ and a liquid-side salinity layer of width $O(\sqrt{Dt})$.

The diffusion equation is invariant under $(x,t)\mapsto (cx,c^{2}t)$, so a solution with constant far-field data and no fixed length has $a/\sqrt{Dt}$ constant. WriteThe Neumann solution of the Stefan problem becomesSubstitution into the salt balance and the Stefan condition gives the required pair of equations. With $c_p=k_T/(\rho \kappa )$ andthey are

When $ϵ\ll 1$, heat diffuses much farther than salt. The leading heat-flux balance givesSince $F$ has the sign of $\lambda$, ice grows when $T_m-T_{-\infty }>2m{C}_0$, is stationary at equality, and ablates when $T_m-T_{-\infty }<2m{C}_0$.

During growth, salt rejection gives $C_i>C_0$. Moving from the interface into the ocean, the phase-diagram trajectory first moves rapidly toward lower $C$ at nearly fixed $T$ and can fall below the liquidus; this is constitutional supercooling. During ablation, $C_i<C_0$, so the near-interface trajectory moves toward larger $C$ and into the stable liquid region above the liquidus. Ablation occurs because the heat conducted from the warmer ocean to the interface exceeds the heat that the colder ice can remove; melting and salt diffusion then maintain local liquidus equilibrium.

Let $K=k\Delta \rho g/\mu$. Hydrostatic pressure continuity at the lower boundary of the light current givesinside the current, so $p_x=\Delta \rho g{h}_x$. Darcy law gives the depth-integrated horizontal flux per unit transverse widthThus the porous gravity current satisfiesaway from the fracture. The boundary and front conditions areAt $x=L$, the pressure excess at the base of the fracture is $\Delta \rho g{h}_L$. Taking upward leakage as positive,This jump is the local mass conservation law for a leaky porous gravity current.

At late times, $q=Q$ to leading order on $0<x<L$, while $Q_l\simeq Q$. Integrating $q=-K(h^2{)}_x/2$ givesLeakage balance and the pressure-driven drop determine

In the far field, $h(L,t)\simeq h_L$ andBalancing the two sides with $h=O(h_L)$ givesMore precisely, setThe self-similar solution is determined byand $x_N=L+{\eta }_N(K{h}_{L}t/\varphi {)}^{1/2}$.

A shelfy stream is grounded ice with nearly depth-independent horizontal velocity. If the lubricating till has thickness $d$ and viscosity ${\mu }_t$, its simple-shear traction is approximately ${\mu }_{t}u/d$. Comparison with ${\tau }_b=\mu u/\lambda$ givesso $\lambda$ has dimensions of length.

Hydrostatic pressure contributes the depth-integrated longitudinal force $-\rho g{h}^2/2$, while the Newtonian extensional stress contributes $4\mu h{u}_x$ by the shallow-shelf approximation. Balancing the change of their sum against basal drag on a slice givesFor a freely floating ice shelf, hydrostatic flotation reduces the gravitational driving by $1-\rho /{\rho }_w$ and removes basal drag:Depth-integrated mass conservation isin the absence of accumulation or ablation.

In a thin stream with horizontal scale $L_x$, the ratio of extensional resistance to basal drag is $O(\lambda h/L_x^2)$, which is small when the thin-film condition $h/L_x\ll 1$ is combined with $\lambda \ll h_G$. For steady flux $hu=q$, neglecting extension givesAt the grounding line, flotation over a bed depth $b$ givesIntegration yieldsThe grounding-line slope is $|h_x(0)|=\mu q/(\rho g\lambda h_G^2)$. Thin-film theory therefore also requiresThis and $\lambda \ll h_G$ are compatible precisely when $\mu q/(\rho g{h}_G^3)\ll 1$.

The total horizontal force resultant in the stream isThe floating shelf equations and its ocean-front traction imply that the force it exerts at the grounding line is the ocean's hydrostatic force, soUsing flotation and the approximate stream solution,Consequentlyand the grounding-line flux-thickness relation is

Because the wavelength is much smaller than the shelf thickness, the ice-shelf corrugation relaxation may be treated as Stokes flow in the half-space $z>0$, with the mean ice-ocean boundary at $z=0$. Letso $u=-{\psi }_z$ and $w={\psi }_x$. Taking the curl of the Stokes equation gives the Biharmonic stream function for planar Stokes flow equation ${\nabla }^4\psi =0$. Decay as $z\to \infty$ removes the growing modes, leavingThusand the perturbation pressure obtained from the Stokes equation is

The linearized zero-shear stress condition at $z=0$ isso $B=kA$. The water is hydrostatic, and displacement of the density interface gives the normal-stress conditionAfter $B=kA$, one has $w_z(0)=0$, soThe kinematic boundary condition ${\eta }_t=w(0)$ gives $\sigma \hat{\eta }=ikA$. Eliminating $A$ yieldsHence a corrugation of wavelength $\lambda =2\pi /k$ decays on the timescaleHydrostatic buoyancy supplies the restoring stress, while viscous deformation over depth $O(k^{-1})$ supplies the resistance. Shorter wavelengths deform a shallower but more strongly sheared layer and therefore have a longer decay time in this gravity-only model. As ice is advected away from the grounding line, long corrugations should disappear first, leaving progressively shorter-wavelength structure farther downstream.