V-shaped channel gravity current ($(h^2{)}_t=C(h^4{h}_x{)}_x$)

For a fixed half-angle $0<\alpha <\pi /2$, a slow slender gravity current in a V-shaped channel has an almost horizontal free surface across each section. Its area is $h^2\tan \alpha$, and its axial volume flux is $-gQ(\alpha )h^4{h}_x/\nu$. Here $Q$ is the integral of the dimensionless cross-sectional Poisson equation solution: $\Delta \varphi =-1$, with zero velocity on the two rigid slopes and zero normal velocity derivative on the horizontal shear-free surface. Conservation of mass gives $C=gQ/(\nu \tan \alpha )$. This treats the full two-dimensional section and does not require a shallow V angle.

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.

The variable $w=h^2$ converts the shallow V-channel lubrication flux equation into a porous medium equation:Indeed $h^4{h}_x=w^{3/2}{w}_x/2=(w^{5/2}{)}_x/5$. Its fixed-mass source-type solution gives $h\propto t^{-1/7}$ and support length proportional to $t^{2/7}$. Integrating the similarity equation once gives $C{h}^4{h}_x=-2x{h}^2/(7t)$, and hence $h=[3(X(t{)}^2-x^2)/(7Ct){]}_{+}^{1/3}$. The constant-volume similarity in a V-shaped channel fixes $X$ from the positive integral $J={\int }_{-1}^1(1-s^2{)}^{2/3}ds$: $X=(\alpha V/J{)}^{3/7}(7Ct/3{)}^{2/7}$.

The symmetric similarity solution for a localized fixed volume $V$ in a V-shaped channel gravity current has $h(x,t)=[3(X^2-x^2)/(7Ct){]}_{+}^{1/3}$. Integrating the similarity conservation of mass equation once gives $C{h}^4{h}_x=-2x{h}^2/(7t)$, and vanishing thickness at the two noses gives the profile. Its volume fixes $X=[V/(I\tan \alpha ){]}^{3/7}(7Ct/3{)}^{2/7}$, where $I={\int }_{-1}^1(1-s^2{)}^{2/3}ds$. At any fixed nonzero position, the thickness first rises after arrival and then falls; its maximum occurs when $X^2=7x^2/3$.