A3.16

Transport Processes
Part II, 2002

(i) A layer of fluid of depth h(x,t)h(x, t), density ρ\rho and viscosity μ\mu sits on top of a rigid horizontal plane at y=0y=0. Gravity gg acts vertically and surface tension is negligible.

Assuming that the horizontal velocity component uu and pressure pp satisfy the lubrication equations

0=px+μuyy0=pyρg,\begin{aligned} &0=-p_{x}+\mu u_{y y} \\ &0=-p_{y}-\rho g, \end{aligned}

together with appropriate boundary conditions at y=0y=0 and y=hy=h (which should be stated), show that hh satisfies the partial differential equation

ht=g3ν(h3hx)x,h_{t}=\frac{g}{3 \nu}\left(h^{3} h_{x}\right)_{x},

where ν=μ/ρ\nu=\mu / \rho.

(ii) A two-dimensional blob of the above fluid has fixed area AA and time-varying width 2X(t)2 X(t), such that

A=X(t)X(t)h(x,t)dxA=\int_{-X(t)}^{X(t)} h(x, t) d x

The blob spreads under gravity.

Use scaling arguments to show that, after an initial transient, X(t)X(t) is proportional to t1/5t^{1 / 5} and h(0,t)h(0, t) is proportional to t1/5t^{-1 / 5}. Hence show that equation ()(*) of Part (i) has a similarity solution of the form

h(x,t)=(A2νgt)1/5H(ξ), where ξ=x(A3gt/ν)1/5h(x, t)=\left(\frac{A^{2} \nu}{g t}\right)^{1 / 5} H(\xi), \quad \text { where } \quad \xi=\frac{x}{\left(A^{3} g t / \nu\right)^{1 / 5}}

and find the differential equation satisfied by H(ξ)H(\xi).

Deduce that

H={[910(ξ02ξ2)]1/3 in ξ0<ξ<ξ00 in ξ>ξ0H= \begin{cases}{\left[\frac{9}{10}\left(\xi_{0}^{2}-\xi^{2}\right)\right]^{1 / 3}} & \text { in }-\xi_{0}<\xi<\xi_{0} \\ 0 & \text { in }|\xi|>\xi_{0}\end{cases}

where

X(t)=ξ0(A3gtν)1/5X(t)=\xi_{0}\left(\frac{A^{3} g t}{\nu}\right)^{1 / 5}

Express ξ0\xi_{0} in terms of the integral

I=11(1u2)1/3duI=\int_{-1}^{1}\left(1-u^{2}\right)^{1 / 3} d u