Past exam of the mathematics course of the University of Cambridge / 2007 / ii / Paper 1 / 36B (Fluid Dynamics II)

For lubrication theory, take layer thickness $H$, horizontal length $L$ and speed $U$. Require aspect ratio $\epsilon =H/L\ll 1$, small surface slopes, inertial ratio $U{H}^2/(\nu L)\ll 1$, and slow evolution $H^2/(\nu T)\ll 1$. The longitudinal viscous term then dominates the longitudinal derivative term, while transverse momentum is hydrostatic.

Here $\epsilon \sim \alpha$ and $U\sim g{H}^2\alpha /\nu$, so sufficient conditions are $\alpha \ll 1$ and $g{H}^3{\alpha }^2/{\nu }^2\ll 1$, together with the slow-time condition. With atmospheric pressure at the surface, $p=p_{\mathrm{atm}}+\rho g\cos \alpha (h-y)$. No slip at the wall and zero surface shear giveDepth-integrated mass conservation, $h_t+q_x=0$, therefore gives the stated equation withA steady linear profile with $h_x=A/B=\tan \alpha$ has zero flux: its free surface is horizontal in laboratory coordinates, so the hydrostatic gradient balances downslope gravity. A constant-depth profile also gives steady through-flow.

For the travelling front of a gravity-driven thin film, set $h=F(\xi )$, $\xi =x-ct$. Integrating and using the dry tip gives $-cF=-A{F}^3+B{F}^3{F}^{\prime }$. The state $F\to h_0$, $F^{\prime }\to 0$ behind the front fixes $\boxed{c=A{h}_0^2}$. Consequently $F^{\prime }=(A/B)(1-h_0^2/F^2)$, and integration with $F(0)=0$ yieldsThis tends to $-\infty$ as $F\uparrow h_0$. Expanding at zero gives $\xi \sim -B{F}^3/(3A{h}_0^2)$, so $\boxed{F\sim (-3c\xi /B{)}^{1/3}.}$ The divergent slope at the formal tip violates the small-slope approximation sufficiently close to the contact line; the interior profile is the lubrication result.