= Solution
Let $x$ point down the plane and $z$ normal to it, with velocity $(u,w)$. The two-dimensional incompressible <Navier-Stokes equation> is
$$
u_x+w_z=0,
$$
$$
u_t+uu_x+wu_z
=-\frac1\rho p_x+\nu(u_{xx}+u_{zz})+g\sin\alpha,
$$
$$
w_t+uw_x+ww_z
=-\frac1\rho p_z+\nu(w_{xx}+w_{zz})-g\cos\alpha.
$$
The <lubrication theory> limit has film thickness much smaller than its streamwise length. Transverse viscous derivatives dominate, inertia is negligible at late times, and the leading equations are
$$
0=-\frac1\rho p_x+\nu u_{zz}+g\sin\alpha,
\qquad
0=-\frac1\rho p_z-g\cos\alpha.
$$
At the wall, $u=w=0$ by the <no-slip boundary condition>. At the free surface $z=h(x,t)$, take $p=p_{\rm atm}$ and $u_z=0$, neglecting surface tension. The normal balance gives the <hydrostatic pressure>
$$
p=p_{\rm atm}+\rho g\cos\alpha\,(h-z),
\qquad
\frac{p_x}{\rho}=g\cos\alpha\,h_x.
$$
If the current has thickness scale $H$ and length scale $L$, then
$$
\frac{|p_x|/\rho}{g\sin\alpha}
\sim\frac HL\cot\alpha\ll1.
$$
Thus the streamwise pressure gradient is asymptotically smaller than the downslope body force for a fixed nonzero inclination. The streamwise equation becomes
$$
\nu u_{zz}=-g\sin\alpha.
$$
Using $u(0)=0$ and $u_z(h)=0$ gives
$$
u(z)=\frac{g\sin\alpha}{\nu}
\left(hz-\frac{z^2}{2}\right).
$$
The <volume flux> is
$$
q=\int_0^hu\,dz
=\frac{g\sin\alpha}{3\nu}h^3.
$$
Integrating incompressibility across the film and using the <kinematic boundary condition> gives $h_t+q_x=0$, hence
$$
\boxed{
h_t+\frac{\partial}{\partial x}
\left(\frac{gh^3\sin\alpha}{3\nu}\right)=0
}.
$$
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