Past exam of the mathematics course of the University of Cambridge / 2009 / iii / Paper 73 / 1

Choose $x$ upslope and $z$ normally downwards from the impermeable roof, so the buoyant layer occupies $0<z<h(x,t)$. Use the hydrostatic approximation and neglect the ambient return-flow pressure correction of relative order $h/H$. Pressure continuity at the interface gives the pressure inside the buoyant layer asThe along-slope Darcy velocity is thereforeDefine $u=kg\Delta \rho \sin \theta /\mu$. Integrating over depth gives the volume flux per unit width $F=uh-u\cot \theta h{h}_x$. Local fluid-volume conservation, including porosity, is $\varphi h_t+F_x=0$. Hence the inclined porous gravity current satisfiesThe first term is upslope advection driven by the mass density difference; the second is nonlinear hydrostatic spreading. The Darcy velocity $u$ and the corresponding pore velocity $u/\varphi$ must be distinguished.

When the hydrostatic spreading term is small, the depth equation reduces to $h_t+c{h}_x=0$, where $c=u/\varphi$. Its characteristic curves have $x=c(t-t_0)$, and the depth is constant along each characteristic. At the inlet the imposed flux gives $uh(0,t_0)=Q{e}^{-t_0/\tau }$. Therefore $h_0=h(0,0)=Q/u$ in this advective approximation andThis is the advection limit of an inclined porous current, behind its leading nose and after the relevant characteristic has left the inlet. For injection starting into an empty aquifer, the advective support is $0<x<ut/\varphi$.

The qualification “sufficiently large time” must refer to an advection-dominated outer region, not an exact solution of the full nonlinear equation. Substituting this exponential into the full equation leaves a nonzero spreading term. The next part quantifies when it can be neglected and explains the nonuniform leading-edge limit.

For the proposed exponential profile, put $\ell =u\tau /\varphi$. Then $h_x=h/\ell$ and $(h{h}_x{)}_x=2h^2/{\ell }^2$. Thus the exact ratio of the discarded spreading term to the retained advective term isThis is a quantitative validity condition for the advection limit of an inclined porous current. For a specified small tolerance $\eta >0$, at a fixed $x$ it is enough that the inlet characteristic has arrived andIf the logarithm is negative, arrival itself already gives this tolerance. A large-distance estimate comparing the overall current length $ct$ with a height-induced pressure length $h\cot \theta$ is $t\gg \varphi h\cot \theta /u$, but the local gradient criterion above is the sharper test for this particular injection history.

At the advective nose, $x=ut/\varphi$, the exponential factor is one. Therefore large time alone is insufficient for uniform validity: the outer profile also needs $2\varphi h_0\cot \theta /(u\tau )\ll 1$, and its abrupt nose needs an inner spreading layer. This limitation follows by direct substitution, not from a missing constant in the solution.

Indeed, with $\xi =x-ut/\varphi$, the full equation becomes $h_t=D(h{h}_{\xi }{)}_{\xi }$, $D=u\cot \theta /\varphi$. Once a finite-volume injection pulse is effectively over, nonlinear diffusion continues to broaden it. Dimensional balances give width ${\ell }_d\sim (DMt{)}^{1/3}$ and height $M/{\ell }_d$, where $M=\int hd\xi$ is the conserved area. A fixed exponential translating profile cannot be the uniform infinite-time limit of that equation. The printed formula is consequently an outer advection approximation with the displayed smallness condition.

Let $X(t)$ be the advancing nose into previously unoccupied pores. In the advective limit the interior flux is $uh$, while the exterior current thickness and flux are zero. Applying the Rankine-Hugoniot condition to the storage $\varphi h$ givesThis is the leading-edge kinematic boundary condition. The outer depth immediately behind this nose is $h(X(t{)}^{-},t)=h_0$, whereas ahead it is zero. A first-order advection model allows this jump; imposing zero depth on its interior trace would contradict its characteristic solution.

For the full inclined porous gravity current, the nose is instead resolved by a continuous spreading layer with $h(X,t)=0$. Taking the flux divided by depth at that interface gives $\dot{X}=(u/\varphi )(1-\cot \theta h_x{|}_{X^{-}})$, when this one-sided derivative exists. Its leading outer value is $u/\varphi$ only when the nose correction is negligible on the scale of the approximation. The zero-depth condition belongs to that resolved layer, not to the discontinuous outer profile itself.

At a fixed location behind the advancing nose, the depth decreases after passage. A fall $-dh>0$ leaves the immobile fluid volume $s\varphi (-dh)$ per unit length, so total local fluid storage changes by $\varphi dh+s\varphi (-dh)=\varphi (1-s)dh$. The thinning-region equation is thereforeThe inlet flux remains $uh(0,t)=Q{e}^{-t/\tau }$. Applying the method of characteristics with thinning-wave speed $u/[\varphi (1-s)]$ gives the residual-trapping attenuation of a porous current:The nose advances into new pore space, which has no residual fluid yet, so its storage factor is still $\varphi$ and its speed is still $u/\varphi$. Consequently $X(t)=ut/\varphi$ and the interior nose trace becomes $h_0{e}^{-st/\tau }$. Using the faster thinning-wave speed for the leading nose would incorrectly erase the attenuation caused by capillary residual trapping. The broken printed reference “(??)” points to the preceding outer profile.

The mobile fluid volume per unit well length includes the porosity:Integrating the residual-trapping attenuation of a porous current gives, for $0\le s<1$,This is mobile volume in the current; retained fluid outside its mobile thickness is not counted. Since $h_t=-h/\tau$ in the thinning region, the rate of accumulation of immobile retained volume is ${\dot{V}}_r=sV/\tau$. Direct differentiation verifies $\dot{V}+{\dot{V}}_r=Q{e}^{-t/\tau }$, so the mobile-volume expression obeys the overall injection balance.

Without capillary residual trapping, $s=0$ gives $V=Q\tau (1-e^{-t/\tau })$. The limiting expression as $s\to 1$ is $V=Qt{e}^{-t/\tau }$, though the local thinning equation degenerates at exactly $s=1$. Both limits provide useful consistency checks.

For $0<s<1$, differentiation givesIts unique zero solves $e^{-(1-s)t/\tau }=s$. Therefore the mobile-volume peak of an exponentially forced porous current occurs atThe derivative is initially positive and is negative after this time. More explicitly, at the stationary point $V^{\prime \prime }=-Qs{e}^{-st/\tau }/\tau <0$. Thus it is the global positive maximum, withThe claim of a finite maximum requires nonzero capillary residual trapping. At $s=0$ the mobile volume rises monotonically towards $Q\tau$ and the peak time tends to infinity. As $s\to 1$, the limiting peak time is $\tau$ and the limiting maximum is $Q\tau /e$.