Past exam of the mathematics course of the University of Cambridge / 2008 / iii / Paper 84 / 1

Assume the dense current lies on the bottom, $\Delta \rho >0$, and its height is small compared with the confined aquifer depth $w$. This permits the hydrostatic approximation and neglect of leading-order resistance from the return flow of the original fluid. The dense-layer pressure anomaly on the basal seal is $\Delta \rho gh$. The along-layer Darcy flux, integrated over the current thickness, and the downward leakage flux are respectivelyVolume conservation requires $\varphi h_t+{\partial }_x{q}_x=-q_{\ell }$. Thus the porous gravity current with hydrostatic leakage through a basal seal obeysHere permeability of a porous medium has units of area, and $\Omega$ has units of inverse time. The original PDF instead prints the sink coefficient $\lambda k\Delta \rho g/(\mu b)$. With $\lambda$ defined there as a permeability, that coefficient has units of area per time and cannot multiply $h$ to give $h_t$. It also omits the division of leakage by porosity. This is a genuine source error, not a Darcy law convention. The remaining parts solve the stated linear-drainage equation for a constant $\Omega >0$; the physically consistent coefficient is the one above. Formally treating the printed coefficient as an independently prescribed inverse-time constant leaves those mathematical solutions unchanged.

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.

The transformed time tends to a finite limit, $s\to {\Omega }^{-1}$. Therefore the symmetric current's maximum distance from the source isand its maximum total width is $2R_{\max }$. With the corrected physical leakage coefficient, this becomes $R_{\max }=[9kbV/(2\varphi \lambda ){]}^{1/3}$. The porous gravity current approaches this limiting distance asymptotically, while its height and mobile fluid volume tend to zero. With no leakage, $\Omega \to 0$, the limiting distance diverges and the ordinary $t^{1/3}$ spreading is recovered.

Write $R=C{s}^{1/3}$, where $C^3=9DV/(2\varphi )$. At a fixed $x$ inside the current,Since $ds/dt>0$, instantaneous constancy of depth is equivalent to ${\partial }_{s}h=0$. Differentiation givesThus the lateral locations at which the depth is neither increasing nor decreasing areAt a fixed nonzero position reached by the front, the height first increases from zero, attains this maximum, and later decreases. The invasion envelope of an exponentially draining porous current is therefore the parametric curveWithin the aquifer, the previously invaded region lies between the basal boundary and this maximum-height envelope, for $|x|<R_{\max }$. The envelope tends to zero height at the limiting lateral nose. Its central singularity belongs to the ideal impulsive release, so it cannot be extended physically above the finite aquifer thickness.