= Solution
Let $z$ measure distance normal to the plane, with the wall at $z=0$ and the <free surface> at $z=h$. Take density $\rho$, dynamic viscosity $\mu$, and gravitational acceleration $g$. In <lubrication theory>, normal momentum balance gives <hydrostatic pressure> while inertia and streamwise viscous derivatives are smaller. With negligible <surface tension> and negligible air shear,
$$
p=p_{\mathrm{atm}}+\rho g\cos\alpha\,(h-z),\qquad
\mu u_{zz}=\rho g(\cos\alpha\,h_x-\sin\alpha),\qquad
\mu v_{zz}=\rho g\cos\alpha\,h_y.
$$
Apply the <no-slip boundary condition> $u=v=0$ at $z=0$, and zero leading tangential stress $u_z=v_z=0$ at $z=h$. The velocity profiles are
$$
u=\frac{\rho g}{\mu}(\sin\alpha-\cos\alpha\,h_x)(hz-z^2/2),\qquad
v=-\frac{\rho g}{\mu}\cos\alpha\,h_y(hz-z^2/2).
$$
Depth integration gives the <volume flux>, with $K=\rho g/(3\mu)$:
$$
q_x=Kh^3(\sin\alpha-\cos\alpha\,h_x),\qquad
q_y=-Kh^3\cos\alpha\,h_y.
$$
Combining <incompressibility> with the wall and surface kinematic conditions gives <conservation of mass>, $h_t+\partial_xq_x+\partial_yq_y=0$. Therefore the two-dimensional version of the <gravity-driven thin film on an incline> satisfies
$$
\boxed{h_t+K\sin\alpha\,\partial_x(h^3)
=K\cos\alpha\left[\partial_x(h^3h_x)+\partial_y(h^3h_y)\right].}
$$
Direct downslope gravity supplies the advective term, while gradients of the <hydrostatic pressure> spread the film along the surface. At $\alpha=0$ this reduces to nonlinear gravitational spreading in a horizontal film.
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