 Solution

Stretch $z=\delta Z$ and write $f=\delta h$, $w=\delta W$, with horizontal velocity and time left order one. The longitudinal viscous term is ${\delta }^{-2}{u}_{ZZ}$, so set $P={\delta }^{2}p$. For $Re{\delta }^2\ll 1$, the leading bulk equations are $P_Z=0$, $u_{ZZ}=P_x$ and $u_x+W_Z=0$.

The normal-stress condition gives $P=-{\delta }^3\hat{C}{h}_{xx}$ at leading order, and the tangential condition gives $u_Z(h)=0$. HenceNo slip at the bed, $u(0)=0$, yields $u=P_x(Z^2/2-hZ)$. Averaging through the depth givesThe kinematic condition becomes $h_t+u(h)h_x-W(h)=-\widehat{ϵ}/\delta$ at leading order. Integration of continuity from the impermeable bed then gives

A nonzero contact speed or evaporative flux is singular in the strictly no-slip sharp-contact-line model with mobility proportional to $h^3$. A finite nonzero limiting $\overline{u}$ would require $h_{xxx}=3\overline{u}/(C{h}^2)$, which is singular for a linearly vanishing height and generally incompatible with a prescribed finite slope at the exact line. Thus a dynamic finite-contact-angle condition needs microscopic regularization, such as slip or a precursor film. In the slow-drying limit a fixed contact angle can instead be used for the leading quasistatic outer profile; it is not an exact regular finite-rate contact-line solution.

At leading quasistatic order $h_{xxx}=0$, so the symmetric profile satisfying the imposed outer angle isIts cross-sectional volume is $V={\int }_{-s}^{s}hdx=2\theta s^2/3$. The moving endpoints contribute no boundary-height term, and the end flux vanishes, giving $V_t=-2ϵs$. ConsequentlyThese are leading slow-time results up to extinction; the approximation need not remain uniform at the final vanishing scale.

To recover the small velocity, differentiate the quasistatic profile using $s_t=-3ϵ/(2\theta )$:Symmetry makes the flux zero at the centre, so $h\overline{u}=-ϵx(s^2-x^2)/(4s^2)$. Division by the height givesThe flow is inward on both sides, suggesting transport of suspended matter toward a central deposit rather than a perimeter ring. Coffee commonly leaves the opposite, ring-like pattern. A possible missing ingredient is contact-line pinning: holding the perimeter fixed makes evaporation require outward replenishment, and stronger evaporation near the edge reinforces that transport. Receding constant-angle and pinned contact lines therefore have qualitatively different deposit mechanisms.

For comparison, the same uniform-evaporation parabolic model with fixed $s$ gives ${\theta }_t=-3ϵ/s$ and $\overline{u}=ϵx/(s\theta )$, an outward velocity. This comparison isolates the boundary assumption even before adding edge-enhanced evaporation.

The receding solution is the quasistatic evaporation of a receding parabolic drop.