= Solution
The transformed time tends to a finite limit, $s\to\Omega^{-1}$. Therefore the symmetric current's maximum distance from the source is
$$
\boxed{R_{\max}=\left(\frac{9DV}{2\phi\Omega}\right)^{1/3},}
$$
and its maximum total width is $2R_{\max}$. With the corrected physical leakage coefficient, this becomes $R_{\max}=[9kbV/(2\phi\lambda)]^{1/3}$. The <porous gravity current> approaches this limiting distance asymptotically, while its height and mobile fluid volume tend to zero. With no leakage, $\Omega\to0$, the limiting distance diverges and the ordinary $t^{1/3}$ spreading is recovered.
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