Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 329 / 3

Take $x$ downslope, $y$ across the slope, and $z$ normal to the plane. The leading normal momentum balance givesThe tangential lubrication theory equations, with no slip at $z=0$ and zero tangential stress at $z=h$, then give the depth-integrated fluxFor $\alpha \ll 1$, defineDropping tildes, steady mass conservation $\nabla \cdot \mathbf{q}=0$ becomeswith $h\to 1$ as $x\to -\infty$.

With no $y$ dependence, integrate the dimensionless equation once. The upstream condition fixes the constant:Write $h=1+\eta$ upstream. Linearization gives ${\eta }_x=3\eta$, soFor $A>0$, the thickness increases monotonically. At large $h$,Integration gives $x-x_0=h-frac12h^{-2}+O(h^{-5})$, and inversion yieldsIn dimensional variables, $d{h}_{\dim }/d{x}_{\dim }\to \alpha$. The increasing thickness therefore cancels the plane's downward slope, so the free surface becomes asymptotically horizontal. The profile represents the upslope edge of a deep viscous pool or pond held back by an obstruction.

The outward normal from the semicircular barrier is ${\mathbf{e}}_r=(-\cos \varphi ,\sin \varphi )$. The leading radial flux in the thick region isForone has $h_s=-H/\Delta$. Negligible radial flux therefore requiresThis expresses local hydrostatic blocking: the free-surface gradient opposes the downslope gravitational flux. Streamlines arriving from upslope divide at $\varphi =0$ and run around the two sides of the barrier toward $\varphi =\pm \pi /2$.

At leading order the azimuthal flux density is $q_{\varphi }\simeq h^3\sin \varphi$. Its integral across the thick region isThis must equal the upstream unit flux intercepted between the symmetry axis and that polar angle, namely $R\sin \varphi$. HenceCombining this with $H=\Delta \cos \varphi$ gives

Let $\phi =\pi /2-\varphi \to 0$ near a shoulder of the barrier. Since $\cos \varphi \sim \phi$, the solution from part c predictsThus the supposedly thin radial region broadens while its large-thickness assumption eventually weakens, and azimuthal derivatives become singular; the approximation cannot remain uniform at the shoulder.

The radial and azimuthal derivative scales of $h^3$ areThey become comparable when $\Delta \sim R\phi$. Using the expression above,soAt this transition,

Downslope of the barrier, the two side streams turn inward under the transverse hydrostatic pressure gradient. They enclose a thin wake or thickness deficit immediately behind the barrier, while excess fluid initially remains concentrated near its shoulders. Farther downstream, lateral spreading fills the wake and restores the uniform layer.

To estimate the recovery distance, set $h=1+\eta$ with $|\eta |\ll 1$ in the governing equation. In a far wake that varies slowly in $x$,which is a diffusion equation with downslope coordinate acting as time. A transverse disturbance of width $R$ therefore spreads over downslope distanceBefore this scale the two-dimensional wake retains the barrier's cross-slope structure; after it, transverse leveling has mixed that structure across its full width and the approach to the uniform film changes character.