= Solution
The required boundary model is a fracture that drains after the storm, with $h(0,t)=0$, and a current of finite extent whose flux vanishes at its advancing nose. Introduce <mass density> $\rho$, gravity $g$, constant <porosity> $\phi$ and <dynamic viscosity> $\mu$. Hydrostatic <Darcy law> gives the depth-integrated discharge $q=-k\rho g h h_x/\mu$. Fluid-volume conservation then gives the <Boussinesq equation for an unconfined aquifer>
$$
h_t=\kappa(hh_x)_x,\qquad\kappa=\frac{k\rho g}{\mu\phi}.
$$
The fluid volume per fracture length is $V=\phi\int_0^\infty h\,dx$; omitting the constant $\phi$ does not change its fractional decay rate.
The <conserved first moment of a draining porous current> follows directly:
$$
\frac{d}{dt}\int_0^\infty xh\,dx
=\kappa[xhh_x]_0^\infty-\kappa\int_0^\infty hh_x\,dx
=-\frac\kappa2[h^2]_0^\infty=0.
$$
The finite outlet discharge does not contribute to $[xhh_x]$ at $x=0$. This establishes $D=\int xh\,dx$ constant. An unspecified nonzero fracture head would instead give $\dot D=\kappa h(0,t)^2/2$; the drained boundary is essential.
For the <dipole similarity solution of a draining porous current>, set $h=a t^\alpha f(\eta)$ and $\eta=x/(d t^\beta)$. The conserved first moment requires $\alpha+2\beta=0$, while matching the time powers in the evolution equation gives $\alpha-1=2\alpha-2\beta$. Hence
$$
\boxed{\alpha=-\frac12,\qquad\beta=\frac14}.
$$
Choose the nose at $\eta=1$ and the normalization $\kappa a/d^2=1$. The profile equation becomes
$$
-\frac12f-\frac14\eta f'=(ff')'.
$$
For $f=b\eta^2+c\eta^{1/2}$, the two sides have coefficients $-b,-5c/8$ and $6b^2,15bc/4$, respectively. The nonzero positive profile therefore has $b=-1/6$; its zero at $\eta=1$ gives $c=1/6$. Its normalization is determined by
$$
D=ad^2\int_0^1\eta f(\eta)\,d\eta=\frac{ad^2}{40},\qquad a=\frac{d^2}{\kappa}.
$$
Thus a complete choice of the constants is
$$
\boxed{d=(40\kappa D)^{1/4},\quad a=(40D/\kappa)^{1/2},\quad
b=-\frac16,\quad c=\frac16,\quad
h=\frac{a}{6\sqrt t}\left(\sqrt\eta-\eta^2\right)\ \ (0<\eta<1)}.
$$
Set $h=0$ beyond the nose. The square-root behavior at the outlet allows finite drainage: $ff'\to1/72$ as $\eta\downarrow0$, while $ff'\to0$ at the nose.
Since $\int_0^1f\,d\eta=1/18$, the volume is
$$
\boxed{V(t)=\frac{\phi ad}{18}t^{-1/4},\qquad\frac1V\frac{dV}{dt}=-\frac1{4t}}.
$$
The <drainage volume decay in a dipole porous current> shows that the printed positive rate cannot describe drainage and its factor is also inconsistent with the conserved-moment scaling. This solution supplies a direct counterexample to that rate and the corrected one. If the initial volume $V_0$ is prescribed at the start of a similarity phase, replace $t$ by $t+t_0$ with $t_0=(\phi ad/(18V_0))^4$. A general post-storm initial shape need not be exactly self-similar; this is the dipole similarity profile and its long-time scaling, not an asserted exact profile for every initial condition. The initial volume alone does not specify $D$.
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