Past exam of the fluids course of the University of Oxford / 2021 / 3 / a / Solution

Let $H$ be the typical thickness, ${\ell }_r$ the radial variation length, $U$ the horizontal speed, and $W$ the vertical speed. For a thin smooth profile, $\epsilon =H/{\ell }_r\ll 1$. The axisymmetric continuity equation isAway from singular boundary regions, balancing its terms gives $U/{\ell }_r\sim W/H$, hence $\boxed{W/U\sim \epsilon \ll 1}$. Near $r=0$, regularity gives $v_r=O(r)$ and the same conclusion follows without a divergent $v_r/r$ term.

Vertical viscous derivatives dominate horizontal ones by ${\epsilon }^{-2}$. The ratio of convective acceleration to vertical viscous acceleration isFor thickness-evolution time $T$, the local inertial ratio is $H^2/(\nu T)$. When both ratios are small, lubrication theory reduces the Navier-Stokes equation toThinness must therefore accompany a slow enough motion; it is not a justification for dropping arbitrarily rapid imposed acceleration. The small vertical velocity also makes the leading vertical balance hydrostatic balance, $p_z=-\rho g$. The neglected normal viscous stress at the interface is smaller than the characteristic lubrication pressure by order ${\epsilon }^2$.

Use the source's curvature sign: $p_{\mathrm{air}}-p_{\mathrm{liquid}}=-\gamma \kappa$. The Young–Laplace equation and atmospheric pressure $p_0$ thus give $p(r,h,t)=p_0+\gamma \kappa$. Integrating the hydrostatic balance yieldsConsequently $p_r=\gamma {\kappa }_r+\rho g{h}_r$, independent of $z$ at this order. The radial Stokes equation and boundary conditions areThe first is the no-slip boundary condition; the second is the leading stress-free boundary condition, since the atmosphere has negligible dynamic viscosity and the surface tension is constant. Integrating twice givesThis is the horizontal velocity only. A smaller vertical velocity is generally required by the continuity equation.