Past exam of the mathematics course of the University of Cambridge / 2022 / iii / Paper 332 (Fluid Dynamics of the Solid Earth)

Let $n=1-\varphi$ be the pore fraction and put $K=k\rho g/\mu$. In the long-wave approximation, the pore pressure is hydrostatic, $p=\rho g(h-z)$, so Darcy's law gives the horizontal volume flux per unit widthLocal mass conservation $n{h}_t+q_x=R$ therefore gives the Boussinesq equation for an unconfined aquiferThe river and drainage-divide conditions are

Before the river disturbance reaches the divide, use the similarity solutionSubstitution gives the nonlinear boundary-value problemIt can be solved numerically. If $C={\lim }_{\eta \downarrow 0}H{H}^{\prime }>0$, then the riverward flux and near-river profile areThe negative sign means flow toward decreasing $x$; the discharge into the river is $-q(0,t)$.

At steady state, $q_x=R$ and $q(L)=0$, soUsing $h(0)=0$ givesNear the river,

After rainfall stops, the late finite-domain solution is the separable aquifer drawdownThe dimensionless profile is determined byIf $C_d={\lim }_{X\downarrow 0}F{F}^{\prime }>0$, thenandThus aquifer thickness decays as $t^{-1}$ and river discharge as $t^{-2}$.

Let ${\rho }_s$ be the grain density and $g^{\prime }=g/g_0$. Vertical force balance for the poroelastic aquifer giveswith compression taken negative in ${\sigma }_e=E({\varphi }_0-\varphi )/{\varphi }_0$. HenceThe zero-effective-stress condition $\varphi (h)={\varphi }_0$ then givesThis identifies the poroelastic compaction length.

The water volume per unit horizontal distance isFor $h\ll {\ell }_c$, a Taylor expansion yieldsStorage conservation is $V_t+q_x=R$. Integrating from river to divide and using $q(L,t)=0$ gives the river influxFor $g^{\prime }=1+ϵe^{i\omega t}$ and a groundwater profile that changes little during one tide, the direct storage oscillation has magnitudeWith typical thickness $H$ and mean recharge discharge ${\overline{Q}}_r=RL$,Using the steady balance $R\sim K{H}^2/L^2$ gives the equivalent estimate $|\delta Q_r|/{\overline{Q}}_r\sim ϵ\omega {\varphi }_0{L}^2/(2{\ell }_{c}K)$.

In the frame of a planar interface moving at speed $V$, the steady liquid temperature obeys $-V{T}_z=\kappa T_{zz}$ and approaches $T_{\infty }$. ThusThe Stefan condition, with negligible solid-side heat flux, givesso $T_i-T_{\infty }=L/c_p=S\Delta T$. The kinetic undercooling law then giveswhich requires $S<1$ for advancing solidification.

Scale length with $\kappa /V$, time with $\kappa /V^2$, and temperature with $\Delta T$. The base liquid temperature is ${\Theta }_0=S{e}^{-z}$. For an interface displacement $\zeta e^{i\alpha x+\sigma t}$, write the thermal perturbation asThe heat equation givesLinearizing the Stefan condition giveswhile the kinetic law with stabilizing curvature undercooling givesEliminating $A$ gives the general implicit dispersion relation

At marginal stability $\sigma =0$,and the dispersion relation becomesSince ${\alpha }^2=\lambda (\lambda -1)$ on this curve, the right-hand side is simply $\Gamma {\lambda }^2$. A positive cutoff wavenumber therefore exists exactly whenThe unstable band is $0<\alpha <{\alpha }_c$, whereA sketch of $\Gamma {\lambda }^2$ against the constant $S/(1-S)$ shows no crossing above $\alpha =0$ below threshold and one cutoff above it. As $S\uparrow 1$, ${\lambda }_c$ and ${\alpha }_c$ diverge like $(1-S{)}^{-1/2}$, so the range of the morphological instability of a kinetically limited solidification front becomes unbounded.

The Newtonian shallow-ice flux on the flat bed isso mass conservation givesAt the ice divide, symmetry gives $h_x(0,t)=q(0,t)=0$. At the terminus, $h(x_N,t)=0$ and the moving-front mass balance applies; in a steady state $q(x_N)=0$. At the snowline, $h(x_s,t)=h_s$ and both $h$ and $q$ are continuous, hence $h_x$ is continuous because $h_s>0$.

In a steady state, $q_x=A$ in the accumulation region and $q_x=-A$ in the ablation region. ThereforeFlux continuity gives $x_N=2x_s$. Integrating the flux law gives the piecewise shapeMatching $h(x_s)=h_s$ yieldsThe largest surface slope is of order $\sqrt{\nu A/g}/h_s$, so the shallow-ice approximation requires

At the grounding line, hydrostatic flotation and continuity of ice flux giveThe normal stress and longitudinal membrane stress also match onto the floating ice shelf. For Newtonian ice this matching supplies the Newtonian grounding-line flux

The accumulation region still has $q=Ax$. In the ablation region, $q=A(2x_s-x)$, so $q(x_G)=q_G$ givesIntegrating the grounded flux law from $x_s$ to $x_G$ givesUnder the stated limit $A\ll g^{\prime }{h}_G^2/\nu$, the $h_G^4$ term is asymptotically smaller than the grounding-flux term, and henceFinally the accumulation-region profile gives $h_0^4=h_s^4+6\nu A{x}_s^2/g$, so