 Solution

In lubrication theory, the gap scale $H$ is small compared with the axial scale $L$, so $\delta =H/L\ll 1$ and wall slopes are small. Incompressibility makes the transverse velocity scale $\delta U$. The reduced Reynolds number is $\rho U{H}^2/(\mu L)=\delta {\mathrm{Re}}_H$, and must be small to neglect convection; the transverse viscous diffusion time must also be short compared with the wall-motion time to neglect unsteady inertia. For a Newtonian incompressible fluid with no axial wall motion, the leading equations are $p_y=0$ and $\mu u_{yy}=p_x$.

No slip gives $u(0)=u(h)=0$. Integrating twice and then across the gap givesIntegrate $u_x+v_y=0$ and differentiate the moving-limit flux. The lower wall has $v(0)=0$, and the upper kinematic condition is $v(h)=h_t+u(h)h_x$. These terms cancel to giveFor a travelling gap $h(x-ct)$, this becomes $(q-ch{)}_x=0$, so $q=ch+Q$ with $Q$ independent of position. Zero mean pressure gradient givesthereforeThe positivity of $h$ is assumed so these averages exist. The wall pattern travels without an imposed mean pressure difference; the changing gap and its inverse-cubic hydraulic resistance produce the pumping.