Past exam of the mathematics course of the University of Cambridge / 2019 / iii / Paper 332 / 3

Use the common-density, equal-volume-heat-capacity model implicit in the stated speed law. Let the liquid fraction be $\varphi$ ahead of the melting front and $\varphi +\Delta \varphi$ behind it. Melting adds exactly the liquid needed to fill the newly created pores, so total mass conservation makes Darcy flux continuous. Consequently,Here $u$ denotes Darcy velocity and $v$ the pore-liquid velocity in the stationary rock frame. Treating the pore-space increase as storage without its simultaneous melting source would incorrectly change the Darcy flux.

Relative to the cold unmelted material, the bulk enthalpy increase behind the front is $\rho c_p\Delta T+\rho L\Delta \varphi$: all phases gain sensible heat and the melted ice consumes latent heat. The advective heat-flux difference is $\rho c_{p}U\Delta T$. Applying the Rankine-Hugoniot condition to this energy balance therefore givesso the advection-driven melting front in a porous matrix moves atThe advected latent contribution of the liquid is the same on both sides and cancels; the denominator measures sensible heating plus phase-change energy per unit bulk volume.

Put $k_a=k$ ahead of the front and $k_b=k+\Delta k$ behind. With viscosity $\mu$, Darcy's law gives the planar pressure gradients $p_{a0}^{\prime }=-\mu U/k_a$ and $p_{b0}^{\prime }=-\mu U/k_b$. Write the displacement as $\eta e^{\sigma t+i\alpha y}$ with $\alpha >0$, and take the unperturbed interface to be $x=0$ in its translating frame.

The liquid flux is divergence-free away from the interface, so the pressure perturbations are harmonic. The decaying normal modes areExpanding pressure continuity at the displaced interface and matching normal Darcy flux givesHenceThe common normal-flux perturbation is $\delta u_n=-k_b\alpha B/\mu =\alpha U\Delta k\eta /(k_a+k_b)$. The local enthalpy balance converts it into a front-speed perturbation, $\sigma \eta =\delta u_n/(1+S\Delta \varphi )$. Thus the melting-front instability due to permeability contrast hasFor a signed transverse wavenumber, replace $\alpha$ by $|\alpha |$.

Melting raises permeability of a porous medium. An advancing protrusion therefore offers a less resistive path to hot liquid, attracting larger Darcy flux and receiving more heat. The resulting faster melting reinforces the protrusion: this is a reactive infiltration instability. A permeability decrease would reverse this feedback, while zero contrast gives no growth in this ideal model.

For positive contrast, the growth rate is proportional to transverse wavenumber, so the model has no finite fastest-growing wavelength. The latent-to-sensible heat ratio slows the front but supplies no short-wave cutoff. Finite thermal transport and a resolved melting zone can introduce a length scale; pore geometry eventually limits the continuum description. If interface curvature changes the equilibrium melting temperature through the Gibbs--Thomson relation, that can also oppose short-wave corrugations. The modified dispersion relation must include such effects before a preferred wavelength can be calculated; a length scale alone does not guarantee that every added mechanism selects one.