Past exam of the mathematics course of the University of Cambridge / 2001 / iii / Paper 44 / 4

For a stationary solid skeleton, the Darcy velocity $\mathbf{u}$ is liquid volume transport per unit total cross-sectional area, while the pore velocity $\mathbf{v}$ is the intrinsic mean fluid velocity within the connected pore space. ThusFor an incompressible liquid in a saturated region, conservation of mass is ${\varphi }_t+\nabla \cdot \mathbf{u}=0$. With time-independent but spatially varying porosity, it becomesnot generally $\nabla \cdot \mathbf{v}=0$. Neglecting dispersion, a fluid interface is advected by the pore velocity; its mean normal speed is $V_n=\mathbf{v}\cdot \mathbf{n}=\mathbf{u}\cdot \mathbf{n}/\varphi$.

For the thin layer on the inclined plane, the gravitational acceleration components are $g\sin \theta$ downslope and $-g\cos \theta$ normal to the plane. The hydrostatic approximation and constant pressure $p_a$ above the current giveWith permeability of a porous medium $k(z)$, Darcy's law therefore givesHere $u_x,u_y$ denote the components of the Darcy velocity. Integrate over depth, using the prescribed power laws:Both integrals are finite for the positive exponents in the model. Conservation of the stored fluid gives ${\mathcal{V}}_t+{\nabla }_{x,y}\cdot \mathbf{Q}=0$, orFor $0<\theta <\pi /2$, set $X=x\tan \theta$, $Y=y\tan \theta$ andSince $\cos \theta {\tan }^2\theta =\sin \theta \tan \theta$, the depth-dependent inclined porous-current equation isThe longitudinal flux carries fluid downslope, while the thickness-gradient terms spread it. The degenerate horizontal or vertical slope limits need their own coordinate scaling.

Far downslope let the typical height be $H(X)$ and the cross-slope half-width be $W(X)$. Slenderness gives $W\ll X$ and slowly varying longitudinal thickness. In a steady porous gravity current, the longitudinal spreading term is smaller than advection by the ratio $H/X$, while longitudinal diffusion relative to transverse diffusion is of order $(W/X{)}^2$. Thus the leading balance isIntegrating across the finite current, with zero flux through its dry side edges, makes the leading downslope volume flux constant:Balancing the two differential terms givesEliminate $H\propto W^{-1/\beta }$ to find the far-downslope width of a constant-flux porous current:The typical height is correspondingly $H\propto X^{-1/(2\beta +1)}$. For every $\beta >0$, the width exponent is less than one half, so $W/X\to 0$ and $H/X\to 0$, consistent with the discarded longitudinal terms. The porosity exponent $\alpha$ drops out because it enters storage, which has zero time derivative in this steady problem.

For constant porosity and permeability, the equation is $h_T+h_X={\nabla }_{X,Y}\cdot (h{\nabla }_{X,Y}h)$. Under $\tau =T$, $\xi =X-T$, $\eta =Y$, the chain rule gives $h_T=h_{\tau }-h_{\xi }$ and $h_X=h_{\xi }$. Thus the moving-frame porous medium equation isIts conserved mass is $M=\iint hd\xi d\eta$.

There is a source inconsistency in the printed volume formula. Actual liquid volume is $V=P\iint hdxdy$. The stated coordinates have Jacobian determinantso the volume Jacobian for downslope stretched coordinates requiresThe printed factor ${\tan }^2\theta$ in the first expression is its reciprocal and cannot be proved with these coordinates. For example, at $\theta =\pi /6$ an integral equal to one corresponds to actual volume $3P$, whereas the printed formula gives $P/3$. The physical-volume normalization below uses the corrected Jacobian.

Let $r=\sqrt{{\xi }^2+{\eta }^2}$. For the source-type axisymmetric similarity solution, putMass conservation imposes $a=2b$, and balancing $h_{\tau }$ against $\nabla \cdot (h\nabla h)$ gives $a+1=2a+2b$. Hence $a=1/2$ and $b=1/4$. The profile equation isMultiplying by $s$, the left side is $-\frac{1}{4}(s^{2}f{)}^{\prime }$. Integration and regularity at the origin giveWhere $f>0$, divide by $sf$ to obtain $f^{\prime }=-s/4$, soSet the profile to zero beyond this front. The mass normalization isTherefore the two-dimensional Barenblatt profile for quadratic porous-medium diffusion iswhere $[q{]}_{+}=\max (q,0)$. Its radius and central thickness areThe profile is regular at the center, its flux $-h{h}_r$ vanishes at the dry edge, and the front speed $r_N^{\prime }=r_N/(4\tau )$ equals the limiting advective speed $-h_r$ there. Its support shrinks to the release point as $\tau \downarrow 0$ while its mass stays $M$, giving the concentrated initial release in the weak sense.

In physical coordinates its center travels atand $r=\tan \theta \sqrt{(x-x_c{)}^2+y^2}$ with $\tau =\rho gK\sin \theta \tan \theta t/(\mu P)$. If the printed volume expression were instead adopted merely as a definition of a formal parameter $\tilde{V}$, the same profile would use $M=\tilde{V}/(P{\tan }^2\theta )$; that parameter would equal $V{\tan }^4\theta$, not the actual released fluid volume.