= Solution
For <lubrication theory>, take layer thickness $H$, horizontal length $L$ and speed $U$. Require aspect ratio $\varepsilon=H/L\ll1$, small surface slopes, inertial ratio $UH^2/(\nu L)\ll1$, and slow evolution $H^2/(\nu T)\ll1$. The longitudinal viscous term then dominates the longitudinal derivative term, while transverse momentum is hydrostatic.
Here $\varepsilon\sim\alpha$ and $U\sim gH^2\alpha/\nu$, so sufficient conditions are $\alpha\ll1$ and $gH^3\alpha^2/\nu^2\ll1$, together with the slow-time condition. With atmospheric pressure at the surface, $p=p_{\rm atm}+\rho g\cos\alpha(h-y)$. No slip at the wall and zero surface shear give
$$
u=\frac g\nu(\sin\alpha-\cos\alpha\,h_x)(hy-y^2/2),\qquad
q=\frac{gh^3}{3\nu}(\sin\alpha-\cos\alpha\,h_x).
$$
Depth-integrated <mass conservation>, $h_t+q_x=0$, therefore gives the stated equation with
$$
\boxed{A=\frac{g\sin\alpha}{3\nu}\sim\frac{g\alpha}{3\nu},\qquad B=\frac{g\cos\alpha}{3\nu}\sim\frac g{3\nu}.}
$$
A steady linear profile with $h_x=A/B=\tan\alpha$ has zero flux: its free surface is horizontal in laboratory coordinates, so the hydrostatic gradient balances downslope gravity. A constant-depth profile also gives steady through-flow.
For the <travelling front of a gravity-driven thin film>, set $h=F(\xi)$, $\xi=x-ct$. Integrating and using the dry tip gives $-cF=-AF^3+BF^3F'$. The state $F\to h_0$, $F'\to0$ behind the front fixes $\boxed{c=Ah_0^2}$. Consequently $F'=(A/B)(1-h_0^2/F^2)$, and integration with $F(0)=0$ yields
$$
\boxed{\xi=\frac BA\left[F+\frac{h_0}{2}\log\frac{h_0-F}{h_0+F}\right].}
$$
This tends to $-\infty$ as $F\uparrow h_0$. Expanding at zero gives $\xi\sim-BF^3/(3Ah_0^2)$, so $\boxed{F\sim(-3c\xi/B)^{1/3}.}$ The divergent slope at the formal tip violates the small-slope approximation sufficiently close to the <contact line>; the interior profile is the lubrication result.
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