= Solution
Use the positive transport coefficient $D$ from the preceding solution. The <porous medium equation> here is $h_t=D(h^2h_x)_x=(D/3)(h^3)_{xx}$. A planar pulse conserves the water cross-sectional area $\mathcal V=\int h\,dx$. 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 \b[$a=b=1/4$]. With $\eta=x/t^{1/4}$, the profile equation is
$$
-\frac14(f+\eta f')=D(f^2f')'.
$$
Integrate once, using symmetry and zero central flux: $Df^2f'=-\eta f/4$. Inside the wet region this gives $f^2=C-\eta^2/(4D)$. The dry continuation is zero. Write the result as
$$
\boxed{h(x,t)=H(t)\left[1-\frac{x^2}{L(t)^2}\right]_+^{1/2},\quad
L(t)=2\left(\frac{\mathcal V}{\pi}\right)^{1/2}(Dt)^{1/4},\quad
H(t)=\left(\frac{\mathcal V}{\pi}\right)^{1/2}(Dt)^{-1/4}.}
$$
The normalization follows from $\int_{-L}^Lh\,dx=\pi HL/2=\mathcal V$. 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 $u=-Dhh_x=x/(4t)$ 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.
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