Past exam of the mathematics course of the University of Cambridge / 2014 / iii / Paper 73 / 2 / i / Solution

With zero surfactant flux and positive concentration, the nondiffusive surface velocity is zero. The reduced equations become ${\Gamma }_X=6Q/H^2$, $H_X=-12Q/H^3$. Integrating, with the prescribed initial height and concentration, givesAt the far end, the imposed concentration drop fixes $H(1{)}^2=H_0^2+4\delta$, henceThis steady surfactant film with zero surface flux rises to the right while its concentration decreases. The liquid flux is negative, despite the rightward Marangoni traction: the adverse hydrostatic gradient is strong enough to immobilize the surface and drive the interior leftwards.

In dimensionless height $Y=y/h_{\ast }$, the velocity profile isIt vanishes at both boundaries and is negative in the interior when $\delta >0$. Its integral is $Q=-H^3{H}_X/12$. For $\delta =0$, the nondegenerate film is flat and stationary. If its initial thickness is zero, the formal $X^{1/4}$ edge has an unbounded slope, so the lubrication approximation applies away from that edge rather than at the exact dry point.