= Solution
Let $x$ point downslope and $z$ point normally away from the plane, with $0\leq z\leq h$. Write the parallel <velocity> as $u(z)e_x$ and take $\tau>0$ to be the magnitude of the air's upslope stress. The steady equations are
$$
0=-p_x+\rho g\sin\alpha+\mu u'',
\qquad
0=-p_z-\rho g\cos\alpha,
$$
with <boundary conditions>
$$
u(0)=0,
\qquad
p(h)=p_0,
\qquad
\mu u'(h)=-\tau.
$$
The free-surface condition makes $p_x=0$, and integration gives
$$
\boxed{p(x,z)=p_0+\rho g\cos\alpha\,(h-z)},
$$
$$
\boxed{u(z)=\frac{\rho g\sin\alpha}{\mu}
\left(hz-\frac{z^2}{2}\right)-\frac{\tau z}{\mu}}.
$$
The surface <velocity>, downslope shear exerted by the fluid on the plane, and volume flux per unit width are
$$
\boxed{u_h=\frac h\mu\left(\frac{\rho gh\sin\alpha}{2}-\tau\right)},
$$
$$
\boxed{\tau_0=\mu u'(0)=\rho gh\sin\alpha-\tau},
$$
$$
\boxed{q=\int_0^hu(z)\,dz
=\frac{\rho g h^3\sin\alpha}{3\mu}
-\frac{\tau h^2}{2\mu}}.
$$
Thus the <inclined viscous film with opposing surface shear> reverses in the three senses when
$$
\begin{array}{c|c}
\text{quantity directed upslope}&\text{condition}\\ \hline
u_h&\tau>\frac12\rho gh\sin\alpha\\
q&\tau>\frac23\rho gh\sin\alpha\\
\tau_0&\tau>\rho gh\sin\alpha.
\end{array}
$$
The order of increasing required air stress is therefore
$$
\boxed{\text{surface velocity},\quad\text{volume flux},\quad\text{stress on the plane}}.
$$
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