Past exam of the mathematics course of the University of Cambridge / 2013 / iii / Paper 65 / 3 / b / Solution

Use the positive transport coefficient $D$ from the preceding solution. The porous medium equation here is $h_t=D(h^2{h}_x{)}_x=(D/3)(h^3{)}_{xx}$. A planar pulse conserves the water cross-sectional area $\mathcal{V}=\int hdx$. If the supplied lake volume is a three-dimensional volume $V$, introduce the constant out-of-plane width $B_y$ and take $\mathcal{V}=V/B_y$; alternatively $V$ may be understood as volume per unit span. A two-dimensional model cannot determine an absolute extent from an unspecified three-dimensional volume alone.

For a symmetric localized release, write $h=t^{-a}f(x/t^b)$. Area conservation requires $a=b$, and balancing the PDE powers gives $a+1=3a+2b$, hence $a=b=1/4$. With $\eta =x/t^{1/4}$, the profile equation isIntegrate once, using symmetry and zero central flux: $D{f}^2{f}^{\prime }=-\eta f/4$. Inside the wet region this gives $f^2=C-{\eta }^2/(4D)$. The dry continuation is zero. Write the result asThe normalization follows from ${\int }_{-L}^{L}hdx=\pi HL/2=\mathcal{V}$. This cubic diffusion pulse has finite support $-L<x<L$, total extent $2L$, and spreading speed $\dot{L}=L/(4t)$. Although $h_x$ is singular at the ideal nose, the flux vanishes there and $u=-Dh{h}_x=x/(4t)$ tends to the finite front speed.

For a one-sided pulse on $0<x<L$ with a reflecting boundary at zero and the same area $\mathcal{V}$, replace $\mathcal{V}$ in the displayed full-line formulas by $2\mathcal{V}$. The power laws are unchanged. These are source-type Barenblatt solutions for an instantaneous localized release; a finite initial lake footprint approaches the profile at long times and may require a virtual time origin, rather than matching this singular initial condition exactly.