Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 352 / 1 / b

Write ${\tau }_d=\rho gH\sin \theta$. For unidirectional velocity $\mathbf{u}=(u(y,z),0,0)$, the nontrivial force balances areFor the power-law fluid,At the free surface $z=H$, $p=p_{\mathrm{atm}}$ and ${\tau }_{xz}=0$. At the bed $z=0$, continuity of tangential velocity and traction gives the local sliding conditionbecause the stated inequalities keep the downhill bed yielded. The velocity and horizontal shear traction are continuous across $y=0$, and the flow tends to uniform states as $y\to \pm \infty$.

When horizontal shear dominates, $u=U(y)+O(ϵ)$ andIntegrating downhill momentum through $0<z<H$ givesThe Bingham sliding law then yields the ODEDefineThenThe speed decreases monotonically across the material transition, with a smooth margin layer centered at $y=0$.

In either half-plane, let $U_{\infty }$ denote its far-field value. Multiplying the ODE by $U^{\prime }$ and integrating from the far field givesSince the profile decreases,Continuity of $U^{\prime }$ at $y=0$ gives

For $n=1$, let $\ell =A^{-1}=\sqrt{hHK/\mu }$ and $\Delta U=U_{-}-U_{+}$. ThenThe Newtonian disturbance therefore has exponential tails.

For $n<1$ and $y>0$, put $W=U-U_{+}$. Integration givesThe corresponding expression on $y<0$ follows by reflection about $(0,U_0)$. A shear thinning power-law material has algebraic rather than exponential margin tails. Measurements of how rapidly an ice-stream speed approaches its far-field values could therefore distinguish an approximately Newtonian rheology from power-law shear thinning and estimate $n$.

If $\delta$ is the cross-stream transition width and $\Delta U$ its speed change, dominant horizontal shear requiresBalancing the two terms in the reduced ODE gives the scaleThe approximation is strongest in the margin where $U^{\prime }$ is large, for a thin basal layer $h\ll H$ and parameters making this lateral scale short compared with the scale on which vertical shear changes the plug velocity. It fails sufficiently far into either uniform region because $U^{\prime }\to 0$ while the $O(ϵ)$ vertical shear needed to transmit the driving stress remains. It can also fail in narrow neighborhoods where neglected free-surface, bed-transition, or three-dimensional effects vary on the same scale.

For $n>1$, the exponent $m=2/(n+1)$ in $U^{\prime }=-A|U-U_{\infty }{|}^m$ is below one. Its integral reaches $U=U_{\infty }$ at a finite distance, after which the constant far-field solution can be attached. The ideal power law therefore predicts a compactly supported transition with finite-width edges rather than the algebraic tails found for $n<1$. Near those edges horizontal shear vanishes and the neglected vertical or regularizing physics becomes important, so the sharp termination should not be interpreted literally.