Quasistatic evaporation of a receding parabolic drop ($h=\frac{\theta }{2s}(s^2-x^2),s_T=-\frac{3}{2\theta },\overline{u}=-\frac{ϵx}{2s\theta }$)

For a symmetric two-dimensional thin drop with uniform evaporative loss $ϵ$, 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 $h_t+(h\overline{u}{)}_x=-ϵ$ and the contact-line flux vanishes, volume loss is $-2ϵs$. On slow time $T=ϵt$, this gives the stated constant recession speed.

Substitute that speed into $h_t$ and integrate flux from the centre, obtaining $h\overline{u}=-ϵx(s^2-x^2)/(4s^2)$. Division by height gives the displayed inward velocity. With a pinned width instead, the same uniform-loss model gives ${\theta }_t=-3ϵ/s$ and outward velocity $\overline{u}=ϵx/(s\theta )$. 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.