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Use constant porosity $\varphi$, isotropic rock permeability of a porous medium $K$, common liquid dynamic viscosity $\mu$ and gravitational acceleration $g$. Treat the two liquids as separated by a sharp interface, with negligible capillary pressure, molecular mixing, ambient-current resistance and inertia. At distances beyond the initial source scale the current is shallow relative to its horizontal length, so its pressure is hydrostatic. The initial release is symmetric about $x=0$ and has actual liquid volume $V$ per unit length of source. Consequently its geometric thickness integral is $V/\varphi$, not $V$.

Below the interface the excess pressure relative to the ambient hydrostatic field is $p^{\prime }=\Delta \rho g(h-z)$, so Darcy law gives horizontal discharge $q_x=-K\Delta \rho g{h}_x/\mu$. The vertically integrated discharge is $Q_x=-K\Delta \rho gh{h}_x/\mu$. Take the specified downward leakage $v=\gamma h$ to mean Darcy flux per bulk horizontal area. Mass conservation then givesIf $v$ is instead defined as interstitial seepage speed, the actual leakage discharge is $\varphi v$ and the corresponding coefficient is $\lambda =\gamma$. Stating the velocity convention removes that otherwise ambiguous porosity factor. For a shale of thickness $d_s$ and permeability $K_s$, a simple hydrostatic leakage model has $\gamma =K_s\Delta \rho g/(\mu d_s)$, if the reservoir below it has the same ambient pressure reference and no capillary entry threshold.

Let $R(t)$ be the half-width of the support; the full end-to-end length will be $L(t)=2R(t)$. Appropriate conditions are symmetry $h_x(0,t)=0$, $h(\pm R,t)=0$, no flux beyond the support, and the moving-front condition $\dot{R}=-\beta h_x(R^{-},t)$. Initially $\varphi h(x,t)dx$ converges weakly to $V{\delta }_0$, expressing the concentrated line release. The equations describe the spreading porous gravity current, not the singular immediate release itself.

With $\lambda =0$, conserved volume and balance of $h_t$ against $\beta (h{h}_x{)}_x$ give width proportional to $t^{1/3}$ and height proportional to $t^{-1/3}$. Write $h=t^{-1/3}f(\eta )$, $\eta =x/t^{1/3}$. Substitution givesIntegrating from the symmetry axis gives $\beta f{f}^{\prime }=-\eta f/3$. Inside the support $f>0$, so $f^{\prime }=-\eta /(3\beta )$ and $f=A-{\eta }^2/(6\beta )$. Thus the compact Barenblatt solution isThe positive-part notation makes $h=0$ outside the current. Volume fixes its remaining constant:If “length” denotes the distance from the source to one front, it is $R$ rather than $L$; both conventions are given explicitly. Differentiating gives $\dot{R}=R/(3t)=-\beta h_x(R^{-},t)$, so the front condition holds. Using $V/2$ on each reflecting half-line gives the same normalization.

For positive leakage define actual retained volume $\mathcal{V}(t)=\varphi \int hdx$. Since the edge thickness and boundary flux vanish, integration of the equation yieldsPut $h=e^{-\lambda t}H(x,\tau )$, where $\tau =(1-e^{-\lambda t})/\lambda$. Because $\dot{\tau }=e^{-\lambda t}$, substitution cancels the leakage term and leaves $H_{\tau }=\beta (H{H}_x{)}_x$. This derives the exponential drainage transform for porous-medium diffusion. Its conserved geometric volume is still $V/\varphi$, soThe full length is again $L(t)=2R(t)$. In particular,At early times $\tau \sim t$, recovering impermeable-bed spreading. At late times, drainage reduces the current thickness and therefore its hydrostatic driving pressure; the effective spreading time saturates at $1/\lambda$. The current approaches a finite horizontal reach while its retained volume and thickness vanish exponentially. This does not mean a positive-volume pool remains at a finite final depth. For small positive $\lambda$ the length scale diverges as ${\lambda }^{-1/3}$, consistently with unbounded spreading when leakage is absent.

For a concentrated line release, take $V$ to be actual fluid volume per unit source length and allow symmetric spreading to both sides of $x=0$. PutSubstitution in the porous medium equation gives the exponential drainage transform for porous-medium diffusion:The Barenblatt solution has conserved area ${\int }_{-\infty }^{\infty }Hdx=V/\varphi$. A quadratic cap satisfies this equation:Indeed, inside the current $H_x=-x/(3Ds)$ and $R\propto s^{1/3}$; differentiating verifies $H_s=D(H{H}_x{)}_x$. Its area is $\int Hdx=2R^3/(9Ds)=V/\varphi$, fixing the prefactor. The flux vanishes at the two moving noses because $H=0$ there. Hence the remaining physical volume and occupied width areFor a one-sided current on a reflecting half-line, replace the $2$ in $R^3$ by $1$ when $V$ denotes the total release into that side. Volume alone does not select this exact solution for an arbitrary finite initial patch; it describes an impulsive line release. Its initial central height is singular, so a finite injection region and the constraint $h\ll w$ regularize the near-source early-time behavior.

Take the boundary to be horizontal, put $z=0$ on it, and let $z=h(x,t)$ be the interface. Write $\Delta \rho =\rho -{\rho }_2>0$. Under the long-thin hydrostatic approximation, the deep ambient fluid has pressure $p_2=p_{\ast }-{\rho }_{2}gz$ to leading order, with negligible horizontal pressure gradient. Matching pressure at the interface gives, inside the dense porous gravity current,The Darcy velocity is superficial volume flux per total area, not pore speed. The horizontal Darcy law and depth-integrated flux therefore giveThe porosity $\varphi$ multiplies storage, so mass conservation is $\varphi h_t+q_x=0$. ConsequentlyThe deep-ambient and long-thin assumptions are what permit neglecting the ambient return-flow pressure variation, vertical inertia, and nonhydrostatic pressure corrections.

For a localized finite release, first use the reflecting half-line $x\ge 0$, so $h_x(0,t)=0$. Let $S=V/\varphi$ be the conserved cross-sectional area of the current; $V$ here is actual pore-fluid volume per unit cross-flow distance. A similarity solution of the porous medium equation has thickness proportional to $t^{-\alpha }$ and extent proportional to $t^{\beta }$. Conservation gives $\alpha =\beta$, and balancing $h_t$ against $(h{h}_x{)}_x$ gives $\alpha +2\beta =1$. Thus $\alpha =\beta =1/3$. PutSubstitution gives $-(F+\xi F^{\prime })/3=(F{F}^{\prime }{)}^{\prime }$. Zero flux at $\xi =0$ allows one integration: $F{F}^{\prime }=-\xi F/3$. In the positive-current region, $F^{\prime }=-\xi /3$, so $F=({\xi }_f^2-{\xi }^2)/6$. The normalization ${\int }_0^{{\xi }_f}Fd\xi =1$ gives ${\xi }_f^3=9$. The one-sided constant-volume porous gravity current is thereforeHere $[a{]}_{+}=\max (a,0)$. Its Darcy flux vanishes at the moving front because $h=0$ there, and its physical volume is $\varphi {\int }_0^{L}hdx=V$. This is the exact concentrated-release Barenblatt solution of the depth equation, physically used after the short initial adjustment to a long-thin current; an arbitrary finite initial patch requires its initial profile as well as its volume, although the same profile describes the usual late spreading regime. If a line release spreads symmetrically in both directions and $V$ denotes the total volume, use the same parabola on $-L<x<L$ with $L^3=9\lambda Vt/(2\varphi )$. If instead volume has already been divided by porosity, replace $V/\varphi$ by that geometric volume.

Drainage at superficial rate $\varphi \Gamma h$ changes mass conservation to $\varphi h_t+q_x=-\varphi \Gamma h$, soFor constant $\Gamma >0$, use the exponential drainage transform for porous-medium diffusionIndeed, $h_t+\Gamma h=e^{-\Gamma t}{\tau }^{\prime }{H}_{\tau }$ and $(h{h}_x{)}_x=e^{-2\Gamma t}(H{H}_x{)}_x$. Choosing ${\tau }^{\prime }=e^{-\Gamma t}$ leaves $H_{\tau }=\lambda (H{H}_x{)}_x$. Substituting the undrained Barenblatt solution givesUse the factor $1/2$ in $L_d^3$ for a total-volume symmetric release, as above. For general initial data the same transform is exact, but $H$ is the corresponding undrained solution rather than automatically a source parabola.

The remaining physical volume is $V(t)=V{e}^{-\Gamma t}$. Early times have $\tau \sim t$, recovering undrained spreading. Late times have $\tau \to 1/\Gamma$, so the front approaches a finite runout while the current drains exponentially:The front slows because drainage reduces the hydrostatic driving head. The finite transformed time also explains why drainage does not guarantee that an arbitrary initial profile reaches its undrained universal large-time shape.