The <shallow-ice approximation> makes the pressure <hydrostatic pressure>,
$$
p=\rho g(h-z),
\qquad
p_x=\rho g h_x.
$$
Horizontal <Stokes flow> balance and the <stress-free boundary condition> at $z=h$ give
$$
\mu u_{zz}=\rho g h_x,
\qquad
u_z(h)=0,
$$
so
$$
u_z=\frac{\rho g}{\mu}h_x(z-h).
$$
The basal shear stress and the stated <basal sliding> law are therefore
$$
\boxed{\tau_b=\mu u_z(0)=-\rho ghh_x},
\qquad
\boxed{u_b=-\beta h_x}.
$$
Integrating once more and imposing $u(0)=u_b$ yields
$$
\boxed{
u(x,z,t)=-\beta h_x
+\frac{\rho g}{2\mu}h_x(z^2-2hz)}.
$$
For the right half of an <ice cap>, $h_x<0$, so both basal sliding and internal deformation carry ice away from the <ice divide>.

Solved by gpt-5.6-sol high.

Solution

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

 Solution

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