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

Solution

At the grounding line, hydrostatic flotation and continuity of ice flux give

h (x_{G}) = h_{G} = \frac{ρ _{w}}{ρ} b, q (x_{G}^{-}) = q (x_{G}^{+}) .

The normal stress and longitudinal membrane stress also match onto the floating ice shelf. For Newtonian ice this matching supplies the Newtonian grounding-line flux

q_{G} = \frac{g g ^{'}}{6 ν} h_{G}^{3}, g^{'} = g \frac{ρ _{w} - ρ}{ρ _{w}} .

The accumulation region still has

q = A x

. In the ablation region,

q = A (2 x_{s} - x)

, so

q (x_{G}) = q_{G}

gives

x_{G} = 2 x_{s} - \frac{g g ^{'}}{6 ν A} h_{G}^{3} .

Integrating the grounded flux law from

x_{s}

x_{G}

gives

x_{s}^{2} = \frac{g}{6 ν A} (h_{s}^{4} - h_{G}^{4}) + (\frac{q _{G}}{A})^{2} .

Under the stated limit

A ≪ g^{'} h_{G}^{2} / ν

, the

h_{G}^{4}

term is asymptotically smaller than the grounding-flux term, and hence

x_{s} \sim (\frac{g}{6 ν A})^{1/2} (h_{s}^{4} + \frac{g ^{'}}{6 ν A} h_{G}^{6})^{1/2} .

Finally the accumulation-region profile gives

h_{0}^{4} = h_{s}^{4} + 6 ν A x_{s}^{2} / g

, so

h_{0} \sim (2 h_{s}^{4} + \frac{g ^{'}}{6 ν A} h_{G}^{6})^{1/4} .

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