Past exam of the mathematics course of the University of Cambridge / 2021 / iii / Paper 332 / 3 / Solution

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.