= Shallow V-channel lubrication flux
{title2=$Q=-\rho gh^4h_x/(6\mu\alpha)$}
For a shallow V-channel with floor $z=\alpha|y|$, $0<\alpha\ll1$, let the <free surface> be $z=h(x,t)$. <Hydrostatic pressure> gives $p_x=\rho g h_x$. A vertical strip has thickness $H_y=h-\alpha|y|$ and, with no slip below and zero shear above, flux $-\rho gH_y^3h_x/(3\mu)$. Integrating across $|y|<h/\alpha$ gives
$$
Q=-\frac{\rho g}{6\mu\alpha}h^4h_x,\qquad A=\frac{h^2}{\alpha}.
$$
Conservation of volume is $(h^2)_t=C(h^4h_x)_x$, $C=\rho g/(6\mu)$. Here $\alpha$ is the wall slope to the horizontal, not the opening half-angle measured from the vertical in a general V-section. Vertical shear dominates transverse shear by the shallow-slope approximation.
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