Past exam of the mathematics course of the University of Cambridge / 2022 / iii / Paper 332 / 3

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