= Quasistatic evaporation of a receding parabolic drop
{title2=$h=\frac\theta{2s}(s^2-x^2),\quad s_T=-\frac3{2\theta},\quad\overline u=-\frac{\epsilon x}{2s\theta}$}
For a symmetric two-dimensional thin drop with uniform evaporative loss $\epsilon$, a freely receding <contact line> and fixed leading outer <contact angle> $\theta$, quasistatic capillary balance makes the height quadratic. Its volume per transverse length is $2\theta s^2/3$. Since $h_t+(h\overline u)_x=-\epsilon$ and the contact-line flux vanishes, volume loss is $-2\epsilon s$. On slow time $T=\epsilon t$, this gives the stated constant recession speed.
Substitute that speed into $h_t$ and integrate flux from the centre, obtaining $h\overline u=-\epsilon x(s^2-x^2)/(4s^2)$. Division by height gives the displayed inward <velocity>. With a pinned width instead, the same uniform-loss model gives $\theta_t=-3\epsilon/s$ and outward <velocity> $\overline u=\epsilon x/(s\theta)$. Thus the contact-line assumption alone reverses solute transport. A pinned perimeter can produce a <coffee-ring effect>; the receding fixed-angle outer model predicts central transport instead. A strictly no-slip sharp moving <contact line> is singular, so the finite outer angle is a leading quasistatic approximation requiring microscopic regularization at finite speed.
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