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Combining the lubrication gravity-current flux with vertical imbibition under a draining liquid layer givesThe injected volume is the integral of $h+\varphi l$, not just the surface current thickness. The model neglects horizontal subsurface flow and applies only where the substrate is wetted and the surface current supplies its drainage.

Let $w$ be downward Darcy flux, measured per unit total horizontal area. Neglect the resistance of the displaced phase, capillary pressure, and horizontal subsurface flow. Taking atmospheric pressure as zero, the upper end of the saturated column has pressure $\rho gh$ and its advancing lower end has pressure zero. Across a column of depth $l$, Darcy law therefore givesThe pressure head from the overlying layer supplies $h/l$, and gravity supplies one. Because the new liquid fills a pore volume $\varphi dl$ per unit area, the vertical imbibition under a draining liquid layer obeys $\varphi \dot{l}=w$. Volume conservation gives $h+\varphi l=h_0$, so, until the surface layer is exhausted,The initial condition is understood as a limiting solution of this singular equation. For $0<\varphi <1$, separation of variables givesAt small penetration, expansion of the logarithm yields $t\sim \varphi \nu l^2/(2gk{h}_0)$, henceA sufficient dimensional early-time condition is $t\ll \varphi \nu h_0/(gk)$, for which $l\ll h_0$. The printed $t\ll 1$ needs a chosen time unit or nondimensionalization. In the limiting case $\varphi =1$, $l^2=2gk{h}_{0}t/\nu$ holds exactly until $h=0$. The divergent initial Darcy flux is an idealization of sudden contact with a dry continuum; it cannot describe penetration below the pore scale.

For the spreading surface layer, measure $z$ upwards from the substrate. A thin, slow gravity current has hydrostatic pressure $p=\rho g(h-z)$ and a leading horizontal viscous balance $\nu u_{zz}=g{h}_x$. A no-slip boundary condition at $z=0$ and a stress-free boundary condition at $z=h$ giveThis lubrication gravity-current flux drives horizontal spreading down the surface-pressure gradient. Surface volume conservation is $h_t+q_x=-w$, while conservation in the pores is $\varphi l_t=w$. Combining these balances gives the deep-substrate drainage of a gravity current modelDrainage transfers liquid from the surface layer into the substrate. Indeed $(h+\varphi l{)}_t+q_x=0$.

For definiteness take a one-sided current on $0<x<x_N(t)$, supplied at $x=0$. The imposed volume is per unit span, so $V_0$ has dimensions of area. The boundary conditions and volume conservation constraint areThe source starts from zero injected volume. Initially dry points acquire $l>0$ only when the advancing current reaches them; the formula $h/l$ is not applied outside that region. At the advancing nose these are limiting conditions for a singular drainage equation. For a source feeding two identical sides, each side has half the volume and inlet flux; only the corresponding numerical prefactors change. The integral and inlet-flux conditions are equivalent once the local balance and nose conditions hold.

The physical approximations are lubrication theory, negligible inertia, slowly varying surface height, small pore-scale Reynolds number, and negligible capillary entry pressure. Treating the substrate as vertically draining columns also requires its lateral flow to be negligible and its depth to exceed the penetration depth. The model is valid while surface liquid remains available above each draining column.

For a similarity solution, let $h$ and $l$ scale as $t^a$, and let the current length scale as $t^b$. Since $h/l$ then has no time dependence, drainage requires $a-1=0$. Balancing $h_t$ with the horizontal spreading term gives $a-1=4a-2b$, so $a=1$ and $b=2$. These exponents also give injected volume $t^{a+b}=t^3$, explaining the particular supply law.

Define scaleswhich satisfy $CD=V_0/{\tau }^3$ and $D^2=(g/\nu )C^3$. The cubic-input similarity for a draining gravity current isSubstitution into the two conservation equations giveswhereThus only $K$ and $\varphi$ remain in the dimensionless problem. Its constraints areIntegrating the sum of the two ordinary differential equations verifies that the inlet flux and integral normalization agree. The dimensionless endpoint ${\xi }_N={\xi }_N(K,\varphi )$ is the undetermined multiplicative constant in the length law; solving for it is not required.