 Solution

Introduce the positive outlet constantFor a draining similarity profile, $J$ is finite and positive. A regular linear profile would have $f{f}^{\prime }\to 0$ and $(f{f}^{\prime }{)}^{\prime }\ge 0$ at the origin, incompatible with the rain-fed aquifer similarity solution $(f{f}^{\prime }{)}^{\prime }(0)=-1$. Instead there is a square-root drainage boundary. At leading order, $(f^2{)}^{\prime }=2f{f}^{\prime }\to 2J$, givingMore explicitly, the rain-fed aquifer similarity solution implies $(f{f}^{\prime }{)}^{\prime }=-1+O(\sqrt{\eta })$, hence $f{f}^{\prime }=J-\eta +O({\eta }^{3/2})$ and $f^2=2J\eta -{\eta }^2+O({\eta }^{5/2})$. The height vanishes but its derivative diverges; their product remains finite.

Use $t_d,x_d,h_d$ temporarily for dimensional quantities. The nondimensionalization in part (a) givesTaking the drainage flux to be positive out of the left-hand end, $Q_0=-q(0,t_d)$, givesThe printed expression in terms of $f(0)$ and $f^{\prime }(0)$ must be understood as this limit; a literal product of their separate endpoint values is undefined. This is discharge per unit transverse width. Very close to the river, the divergent slope marks the limit of the Dupuit approximation, while the outer solution still supplies the leading discharge.