= 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 $P=\delta^2p$. For $Re\,\delta^2\ll1$, the leading bulk equations are $P_Z=0$, $u_{ZZ}=P_x$ and $u_x+W_Z=0$.
The normal-stress condition gives $P=-\delta^3\widehat C\,h_{xx}$ at leading order, and the tangential condition gives $u_Z(h)=0$. Hence
$$
\boxed{\beta=3,\qquad C=\delta^3\widehat C,\qquad P=-Ch_{xx}.}
$$
<No-slip boundary condition>[No slip] at the bed, $u(0)=0$, yields $u=P_x(Z^2/2-hZ)$. Averaging through the depth gives
$$
\boxed{\overline u=-\frac{h^2}{3}P_x=\frac C3h^2h_{xxx}.}
$$
The kinematic condition becomes $h_t+u(h)h_x-W(h)=-\widehat\epsilon/\delta$ at leading order. Integration of continuity from the impermeable bed then gives
$$
\boxed{h_t+(h\overline u)_x=-\epsilon,\qquad\epsilon=\widehat\epsilon/\delta.}
$$
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 $h_{xxx}=3\overline u/(Ch^2)$, 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 is
$$
h(x,T)=\frac\theta{2s(T)}(s(T)^2-x^2),\qquad T=\epsilon t.
$$
Its cross-sectional <volume> is $V=\int_{-s}^sh\,dx=2\theta s^2/3$. The moving endpoints contribute no boundary-height term, and the end <flux> vanishes, giving $V_t=-2\epsilon s$. Consequently
$$
\boxed{\frac{ds}{dT}=-\frac3{2\theta},\qquad s(T)=s_0-\frac{3T}{2\theta},\qquad T_{\rm dry}=\frac{2\theta s_0}3.}
$$
These 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 $s_t=-3\epsilon/(2\theta)$:
$$
h_t=-\frac{3\epsilon}{4}(1+x^2/s^2),\quad
(h\overline u)_x=\frac\epsilon4(-1+3x^2/s^2).
$$
<Symmetry> makes the <flux> zero at the centre, so $h\overline u=-\epsilon x(s^2-x^2)/(4s^2)$. Division by the height gives
$$
\boxed{\overline u=-\frac{\epsilon x}{2s\theta}.}
$$
The 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 $\theta_t=-3\epsilon/s$ and $\overline u=\epsilon x/(s\theta)$, an outward <velocity>. This comparison isolates the boundary assumption even before adding edge-enhanced <evaporation>.
\Image[/past-exam-of-the-topics-in-fluid-mechanics-course-of-the-university-of-oxford/2008-drying-drop-flow.png]
{title=Leading parabolic-drop models predict inward flow for a receding fixed-angle contact line and outward flow for a pinned contact line}
The receding solution is the <quasistatic evaporation of a receding parabolic drop>.
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