= Solution
Take $x$ downslope, $y$ across the slope, and $z$ normal to the plane. The leading normal momentum balance gives
$$
p=p_a+\rho g\cos\alpha\,(h-z).
$$
The tangential <lubrication theory> equations, with no slip at $z=0$ and zero tangential stress at $z=h$, then give the depth-integrated flux
$$
\mathbf q=\frac{\rho gh^3}{3\mu}
\left(\sin\alpha\,\mathbf e_x-cos\alpha\,\nabla h\right).
$$
For $\alpha\ll1$, define
$$
\widetilde h=\frac h{h_0},
\qquad
(\widetilde x,\widetilde y)
=\frac\alpha{h_0}(x,y),
\qquad
\widetilde{\mathbf q}
=\frac{3\mu}{\rho g\alpha h_0^3}\mathbf q.
$$
Dropping tildes, steady <mass conservation> $\nabla\mathbin\cdot\mathbf q=0$ becomes
$$
\boxed{\nabla\mathbin\cdot(h^3\mathbf e_x)
=\nabla\mathbin\cdot(h^3\nabla h)},
$$
with $h\to1$ as $x\to-\infty$.
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