= Solution
{c}
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, $\varepsilon=H/\ell_r\ll1$. The axisymmetric <continuity equation> is
$$
\frac1r\partial_r(rv_r)+\partial_zv_z=0.
$$
Away from singular boundary regions, balancing its terms gives $U/\ell_r\sim W/H$, hence $\boxed{W/U\sim\varepsilon\ll1}$. 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 $\varepsilon^{-2}$. The ratio of <convective acceleration> to vertical viscous <acceleration> is
$$
\frac{U^2/\ell_r}{\nu U/H^2}=\frac{UH^2}{\nu\ell_r}.
$$
For 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> to
$$
\boxed{0\approx\mu\partial_z^2\mathbf v-\nabla p-\rho g\widehat{\mathbf z}.}
$$
Thinness 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 $\varepsilon^2$.
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> yields
$$
\boxed{p(r,z,t)=p_0+\gamma\kappa(r,t)+\rho g[h(r,t)-z].}
$$
Consequently $p_r=\gamma\kappa_r+\rho gh_r$, independent of $z$ at this order. The radial <Stokes equation> and <boundary conditions> are
$$
\mu\partial_z^2v_r=p_r,\qquad
v_r(r,0,t)=0,\qquad \partial_zv_r(r,h,t)=0.
$$
The 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 gives
$$
\boxed{\mathbf v_h\approx
\left(\rho gh_r+\gamma\kappa_r\right)\frac{z(z-2h)}{2\mu}\widehat{\mathbf r}.}
$$
This is the horizontal <velocity> only. A smaller vertical <velocity> is generally required by the <continuity equation>.
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