= Solution
The <hydrostatic approximation> gives $p=\rho g(h-z)$ above the horizontal floor, up to atmospheric pressure. The horizontal <Stokes flow> equation is $\mu\mathbf u_{zz}=\rho g\nabla h$, with the <no-slip boundary condition> at $z=0$ and zero tangential stress on the <free surface>. Integrating twice gives $\mathbf u=(\rho g/\mu)\nabla h\,(z^2/2-hz)$ and the <lubrication gravity-current flux>
$$
\mathbf q_{\rm lab}=-\frac{\rho g}{3\mu}h^3\nabla h.
$$
For the <squeegee lubrication model>, use $h_0$ vertically, $\ell=\rho gh_0^3/(\mu U)$ horizontally, and $\ell/U$ in time. If $X$ is the laboratory coordinate in the direction of travel, set $x=(X-Ut)/\ell$, $y=Y/\ell$, $t=Ut_{\rm lab}/\ell$, and $h=h_{\rm lab}/h_0$. The blade half-length is $\alpha=L/\ell$. The moving-frame flux is $\mathbf q=-h\mathbf e_x-h^3\nabla h/3$. Conservation of volume gives \b[the dimensionless lubrication equation]
$$
\boxed{h_t-h_x=\frac13\nabla\cdot(h^3\nabla h),\qquad h\to1\quad(x\to\infty).}
$$
Using these scales as representative of the actual flow, <lubrication theory> requires a small aspect ratio and negligible inertia relative to vertical viscous resistance:
$$
\boxed{\frac{h_0}{\ell}=\frac{\mu U}{\rho gh_0^2}\ll1,\qquad\frac{\rho Uh_0^2}{\mu\ell}=\frac{U^2}{gh_0}\ll1.}
$$
The second is the <Reynolds number> based on $h_0$, multiplied by $h_0/\ell$. Surface tension is neglected as assumed. Large piles, rounded gaps, and narrow end regions need their own local slope and inertia checks if these representative scales cease to describe them.
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