= Solution
Seek the branch continuous from a uniformly wall-advected film, $H=1+\alpha H_1+O(\alpha^2)$. Expanding either the flux equation or the left-hand side gives
$$
H-\alpha\cos\theta\,H^3=1+\alpha(H_1-\cos\theta)+O(\alpha^2).
$$
Equating this to one yields $H_1=\cos\theta$, hence
$$
\boxed{h(\theta)=\frac QU\left[1+\alpha\cos\theta+O(\alpha^2)\right].}
$$
The film is thicker on the ascending side, where a reduced mean fluid speed must be compensated by greater thickness to carry the same flux. It is thinner on the descending side, where gravity assists the wall motion. The approximation is uniformly small only while $\alpha\ll1$, not near the critical turning point discussed next.
Back to article page
