 Shallow V-channel lubrication flux

For a shallow V-channel with floor $z=\alpha |y|$, $0<\alpha \ll 1$, 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 $-\rho g{H}_y^3{h}_x/(3\mu )$. Integrating across $|y|<h/\alpha$ givesConservation of volume is $(h^2{)}_t=C(h^4{h}_x{)}_x$, $C=\rho g/(6\mu )$. 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.