Past exam of the mathematics course of the University of Cambridge / 2018 / iii / Paper 329 / 3

Let $z$ measure distance normal to the plane, with the wall at $z=0$ and the free surface at $z=h$. Take density $\rho$, dynamic viscosity $\mu$, and gravitational acceleration $g$. In lubrication theory, normal momentum balance gives hydrostatic pressure while inertia and streamwise viscous derivatives are smaller. With negligible surface tension and negligible air shear,Apply the no-slip boundary condition $u=v=0$ at $z=0$, and zero leading tangential stress $u_z=v_z=0$ at $z=h$. The velocity profiles areDepth integration gives the volume flux, with $K=\rho g/(3\mu )$:Combining incompressibility with the wall and surface kinematic conditions gives conservation of mass, $h_t+{\partial }_x{q}_x+{\partial }_y{q}_y=0$. Therefore the two-dimensional version of the gravity-driven thin film on an incline satisfiesDirect downslope gravity supplies the advective term, while gradients of the hydrostatic pressure spread the film along the surface. At $\alpha =0$ this reduces to nonlinear gravitational spreading in a horizontal film.

For an axisymmetric viscous gravity current on a cone, the cylindrical radius at slant distance $x$ is $x\cos \alpha$. A surface annulus therefore has area $2\pi x\cos \alpha dx$. The local downslope volume flux is the same as in part (a), but conservation of mass must account for this growing circumference. ThusThis is a leading thin-film description away from the apex, with $0<\alpha <\pi /2$ and thickness small relative to the local curvature and streamwise length scales.

The original PDF uses the long-time criterion $x_N\gg (V\csc \alpha {)}^{1/3}$; the local TeX incorrectly reads $V\cos \alpha$. To see why the PDF criterion is the relevant one, scale $h\sim V/(x_N^2\cos \alpha )$. In the bulk, the hydrostatic-gradient term relative to direct downslope gravity isHence the leading outer evolution, with $\beta =K\sin \alpha$, isFor a fixed-volume similarity solution, let $x_N\propto t^b$ and $h\propto t^a$. Conservation of mass gives $a+2b=0$, while the evolution equation gives $a-1=3a-b$. Thus $b=1/5$, $a=-2/5$. Writing $h=t^{-2/5}H(\xi )$, $\xi =x/t^{1/5}$, yieldsThe positive branch $H=c\sqrt{\xi }$ satisfies this equation when $c^2=1/(5\beta )$. It givesThe ideal outer current has zero thickness beyond its front. Its mass integral isThereforeThe reduced equation is a conservation law after multiplication by $x$. Its front obeys the Rankine-Hugoniot condition, ${\dot{x}}_N=\beta h_N^2=x_N/(5t)$, consistent with the mass-derived front position. The outer profile has a finite jump at the front; the omitted hydrostatic-gradient term resolves a narrow front region. There is also a small apex region where the outer gradient and conical curvature violate the local approximations. Neither region sets the leading conserved-volume scaling above.

For the horizontal film, part (a) gives the lubrication theory equationThe length scale is fixed at $L$, so balancing $h/t$ against $K{h}^4/L^2$ gives $h\sim (L^2/(Kt){)}^{1/3}$. The draining viscous film on a finite horizontal plate therefore has the long-time similarity solutionAn additive shift of the time origin can represent the initial transient without changing the long-time law. The positive profile is even, with $H^{\prime }(0)=0$ and $H(\pm 1)=0$. Substitution yieldsTo integrate, put $q=H^3{H}^{\prime }$ and use $q^{\prime }=qdq/dH/H^3$. ThenFor $0<\eta <1$ the decreasing branch has $q<0$, so inversion givesThe negative half of the profile follows by symmetry.

At the edge, $\eta (0)=1$. Set $u=(s/H_0{)}^5$ in the integral to obtainHere $B$ is the beta function. Thus $H_0^3=10/(3C^2)$, and the centre thickness is

For $H\ll H_0$, the implicit integral remaining between $0$ and $H$ givesConsequently the edge region of a draining viscous film hasIts slope satisfies $|h_x|\sim (h_0/L)(1-x/L{)}^{-3/4}$ up to a constant factor, so it becomes order one atThe local height is then also $O(\ell )$, violating the shallow geometry required by lubrication theory. A full local flow is needed to turn the fluid over the edge.

The imposed zero thickness is nevertheless a consistent leading outer boundary condition: $\ell /h_0\sim (h_0/L{)}^{1/3}\ll 1$, so the unresolved edge height is small compared with the film's bulk height. It should not be interpreted as an exact pointwise prediction inside the edge region. Moreover, $H^3{H}^{\prime }\to -\sqrt{2/15}{H}_0^{5/2}$ at the right edge, so the outward volume flux remains finite even as the outer thickness tends to zero. The singular slope is what allows this outer solution to describe drainage.