Use the <dimensional analysis>
$$
h=h^*H,\qquad
x=\ell X,\qquad
t=\frac{h^*}{\alpha}\,\mathcal T,
\qquad
\ell=\left(\frac{\rho g(h^*)^4}{3\mu\alpha}\right)^{1/2},
$$
and define the dimensionless sliding parameter
$$
\boxed{s=\frac{3\mu\beta}{\rho g(h^*)^2}}.
$$
The dimensionless flux and steady conservation law are
$$
Q=-(H^3+sH)H_X,
\qquad
Q_X=
\begin{cases}
+1,&H>1,\\
-1,&H<1.
\end{cases}
$$
On the right half-cap, let the <snowline> be $X=X_s$ and the nose be $X=X_n$. The zero-flux condition at the <ice divide> gives $Q=X$ in the accumulation region. Continuity of $Q$ gives $Q=2X_s-X$ in the ablation region, and zero nose flux gives
$$
\boxed{X_n=2X_s}.
$$

Introduce the increasing function
$$
\Phi(H)=\frac{H^4}{4}+\frac{sH^2}{2},
\qquad
\Phi'(H)=H^3+sH.
$$
Since $\Phi_X=-Q$, the ablation profile satisfying $H(X_n)=0$ is
$$
\boxed{\Phi(H)=\frac12(2X_s-X)^2},
\qquad X_s\leq X\leq2X_s.
$$
At the snowline $H=1$, so
$$
\boxed{X_s=\sqrt{\frac12+s}},
\qquad
\boxed{X_n=2\sqrt{\frac12+s}}.
$$
In the accumulation region,
$$
\boxed{\Phi(H)=2\Phi(1)-\frac{X^2}{2}},
\qquad 0\leq X\leq X_s,
$$
and the divide thickness $H_0$ is fixed by
$$
\boxed{\Phi(H_0)=2\Phi(1)=\frac12+s}.
$$
These two implicit formulas give a continuous thickness and flux at the snowline. Reflection across $X=0$ gives the full two-dimensional ice cap.

Solved by gpt-5.6-sol high.

Solution

Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 332 / 3 / c / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2026/iii/paper-332/3/c/solution