= Solution
Write $R=C s^{1/3}$, where $C^3=9DV/(2\phi)$. At a fixed $x$ inside the current,
$$
h=\frac{1-\Omega s}{6D}
\left(C^2s^{-1/3}-x^2s^{-1}\right).
$$
Since $ds/dt>0$, instantaneous constancy of depth is equivalent to $\partial_s h=0$. Differentiation gives
$$
0=-\frac{C^2}{3}s^{-4/3}(1+2\Omega s)+x^2s^{-2}.
$$
Thus the lateral locations at which the depth is neither increasing nor decreasing are
$$
\boxed{x_*(t)=\pm R(t)\sqrt{\frac{1+2\Omega s(t)}3}
=\pm R(t)\sqrt{1-\frac23e^{-\Omega t}}.}
$$
At 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 curve
$$
\boxed{|x_*|=C s^{1/3}\sqrt{\frac{1+2\Omega s}{3}},
\qquad h_*=\frac{C^2}{9D}s^{-1/3}(1-\Omega s)^2,
\quad 0<s<\Omega^{-1}.}
$$
Within 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.
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