After rainfall stops, the late finite-domain solution is the <separable aquifer drawdown>
$$
h(x,t)\sim\frac{nL^2}{Kt}F(X),
\qquad X=\frac xL.
$$
The dimensionless profile is determined by
$$
\boxed{(FF')'+F=0,
\qquad F(0)=0,
\qquad F'(1)=0.}
$$
If $C_d=\lim_{X\downarrow0}FF'>0$, then
$$
\boxed{q(0,t)\sim-\frac{n^2L^3}{Kt^2}C_d,}
$$
and
$$
\boxed{F(X)\sim(2C_dX)^{1/2},
\qquad
h(x,t)\sim\frac{nL^2}{Kt}\left(\frac{2C_dx}{L}\right)^{1/2}.}
$$
Thus aquifer thickness decays as $t^{-1}$ and river discharge as $t^{-2}$.

Solution

Past exam of the mathematics course of the University of Cambridge / 2022 / iii / Paper 332 / 1 / a / iii / Solution

After recharge stops on a finite aquifer connected to a river, the late solution of the <Boussinesq equation for an unconfined aquifer> has the form $h\sim t^{-1}F(x/L)$. The profile solves a nonlinear eigenvalue problem, and river discharge decays as $t^{-2}$.

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 Incoming links: Separable aquifer drawdown