= Solution
{c}
Introduce the positive outlet constant
$$
J=\lim_{\eta\downarrow0}f(\eta)f'(\eta).
$$
For a draining <similarity profile>, $J$ is finite and positive. A regular linear profile would have $ff'\to0$ and $(ff')'\geq0$ at the origin, incompatible with the <rain-fed aquifer similarity solution> $(ff')'(0)=-1$. Instead there is a <square-root drainage boundary>. At leading order, $(f^2)'=2ff'\to2J$, giving
$$
\boxed{f(\eta)\sim\sqrt{2J\eta},\qquad
f'(\eta)\sim\sqrt{\frac{J}{2\eta}}.}
$$
More explicitly, the <rain-fed aquifer similarity solution> implies $(ff')'=-1+O(\sqrt\eta)$, hence $ff'=J-\eta+O(\eta^{3/2})$ 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) gives
$$
h_d=\frac R\phi t_d f(\eta),\qquad
\eta=\frac{\phi x_d}{\sqrt{KR}\,t_d},\qquad
\partial_{x_d}h_d=\sqrt{\frac RK}f'(\eta).
$$
Taking the drainage flux to be positive out of the left-hand end, $Q_0=-q(0,t_d)$, gives
$$
\boxed{Q_0(t_d)=K\lim_{x_d\downarrow0}h_d\partial_{x_d}h_d
=\frac{\sqrt K\,R^{3/2}}\phi\,t_d
\lim_{\eta\downarrow0}f(\eta)f'(\eta)
=\frac{\sqrt K\,R^{3/2}}\phi Jt_d.}
$$
The printed expression in terms of $f(0)$ and $f'(0)$ 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.
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