Past exam of the mathematics course of the University of Cambridge / 2008 / ii / Paper 2 / 36A / Solution

The lubrication approximation requires depth $h$ small compared with the horizontal variation scale $L$, small bottom slopes, and negligible inertia on the scale $\rho U{h}^2/(\mu L)\ll 1$. Treat the prescribed flat surface as steady and impermeable to leading order. The leading Stokes equations give $p_z=0$ and $\mu {\partial }_z^2{\mathbf{u}}_H={\nabla }_{H}p$. No slip at $z=-h$ and the imposed shear $\mu {\partial }_z{\mathbf{u}}_H{|}_0=\mathbf{S}$ yieldIntegration and depth-integrated incompressibility now giveThe surface value is ${\mathbf{u}}_0=-h^2{\nabla }_{H}p/(2\mu )+h\mathbf{S}/\mu$. Eliminate the pressure gradient with the flux relation to obtain

For the circular container, write $\mathbf{q}=\nabla \psi \times \hat{z}=({\psi }_y,-{\psi }_x)$. Curl the relation $\mu \mathbf{q}/h^3=-\nabla p/3+\mathbf{S}/(2h)$ for $\mathbf{S}=(S_0,0)$. Its vertical component givesThe wall condition $\mathbf{q}\cdot \mathbf{n}=0$ says $\psi$ is constant on the boundary; choose that constant zero. If $h=h_0$, the equation is Laplace's equation with zero Dirichlet data. The strong maximum principle for harmonic functions gives $\psi =0$ and $\mathbf{q}=0$, though there is a sheared return flow through the depth. Its surface velocity is $\boxed{{\mathbf{u}}_0=(h_0{S}_0/(4\mu ),0)}$.

For $h=h_0(1+\epsilon y/a)$, the first-order equation is $h_0^{-3}\Delta \psi =-\epsilon S_0/(2\mu h_{0}a)+O({\epsilon }^2)$. The zero-boundary quadratic solution isThus $\mathbf{q}=2\epsilon C(y,-x)+O({\epsilon }^2)$. Inserting this and the varying depth into the surface formula givesThe circulation is first order in the bottom slope, while its wall-normal flux remains zero.