B2.24

Fluid Dynamics II
Part II, 2002

A thin layer of liquid of kinematic viscosity ν\nu flows under the influence of gravity down a plane inclined at an angle α\alpha to the horizontal (0απ/2)(0 \leq \alpha \leq \pi / 2). With origin OO on the plane, and axes OxO x down the line of steepest slope and OyO y normal to the plane, the free surface is given by y=h(x,t)y=h(x, t), where h/x1|\partial h / \partial x| \ll 1. The pressure distribution in the liquid may be assumed to be hydrostatic. Using the approximations of lubrication theory, show that

ht=g3νx{h3(cosαhxsinα)}.\frac{\partial h}{\partial t}=\frac{g}{3 \nu} \frac{\partial}{\partial x}\left\{h^{3}\left(\cos \alpha \frac{\partial h}{\partial x}-\sin \alpha\right)\right\} .

Now suppose that

h=h0+η(x,t)h=h_{0}+\eta(x, t)

where

η(x,0)=η0ex2/a2\eta(x, 0)=\eta_{0} e^{-x^{2} / a^{2}}

and h0,η0h_{0}, \eta_{0} and aa are constants with η0a,h0\eta_{0} \ll a, h_{0}. Show that, to leading order,

η(x,t)=aη0(a2+4Dt)1/2exp{(xUt)2a2+4Dt}\eta(x, t)=\frac{a \eta_{0}}{\left(a^{2}+4 D t\right)^{1 / 2}} \exp \left\{-\frac{(x-U t)^{2}}{a^{2}+4 D t}\right\}

where UU and DD are constants to be determined.

Explain in physical terms the meaning of this solution.