= Solution
In <lubrication theory>, the gap scale $H$ is small compared with the axial scale $L$, so $\delta=H/L\ll1$ and wall slopes are small. <Incompressibility> makes the transverse velocity scale $\delta U$. The reduced <Reynolds number> is $\rho UH^2/(\mu L)=\delta\operatorname{Re}_H$, 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 gives
$$
\boxed{u=\frac{p_x}{2\mu}y(y-h),\qquad q=\int_0^hu\,dy=-\frac{h^3}{12\mu}p_x.}
$$
Integrate $u_x+v_y=0$ 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 give
$$
\boxed{h_t+q_x=0.}
$$
For 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 gives
$$
0=\langle p_x\rangle=-12\mu\left[c\langle h^{-2}\rangle+Q\langle h^{-3}\rangle\right],
$$
therefore
$$
\boxed{\langle q\rangle=c\left[\langle h\rangle-\frac{\langle h^{-2}\rangle}{\langle h^{-3}\rangle}\right].}
$$
The 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.
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