 Solution

The transformed time tends to a finite limit, $s\to {\Omega }^{-1}$. Therefore the symmetric current's maximum distance from the source isand its maximum total width is $2R_{\max }$. With the corrected physical leakage coefficient, this becomes $R_{\max }=[9kbV/(2\varphi \lambda ){]}^{1/3}$. The porous gravity current approaches this limiting distance asymptotically, while its height and mobile fluid volume tend to zero. With no leakage, $\Omega \to 0$, the limiting distance diverges and the ordinary $t^{1/3}$ spreading is recovered.