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The conserved first moment of a draining porous current fixes $\alpha +2\beta =0$ in a self-similar solution $h=a{t}^{\alpha }f(x/(d{t}^{\beta }))$. The Boussinesq equation for an unconfined aquifer supplies $\alpha =2\beta -1$, giving $\alpha =-1/2$, $\beta =1/4$. Normalize $\kappa a=d^2$ and the nose at $\eta =1$. Then $-f/2-\eta f^{\prime }/4=(f{f}^{\prime }{)}^{\prime }$ is solved byThe square-root outlet has finite drainage flux and the nose has zero flux. This is a particular similarity solution and its scaling family, not an exact description of arbitrary initial data.

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 $\varphi$ and dynamic viscosity $\mu$. Hydrostatic Darcy law gives the depth-integrated discharge $q=-k\rho gh{h}_x/\mu$. Fluid-volume conservation then gives the Boussinesq equation for an unconfined aquiferThe fluid volume per fracture length is $V=\varphi {\int }_0^{\infty }hdx$; omitting the constant $\varphi$ does not change its fractional decay rate.

The conserved first moment of a draining porous current follows directly:The finite outlet discharge does not contribute to $[xh{h}_x]$ at $x=0$. This establishes $D=\int xhdx$ 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 $\alpha -1=2\alpha -2\beta$. HenceChoose the nose at $\eta =1$ and the normalization $\kappa a/d^2=1$. The profile equation becomesFor $f=b{\eta }^2+c{\eta }^{1/2}$, 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 byThus a complete choice of the constants isSet $h=0$ beyond the nose. The square-root behavior at the outlet allows finite drainage: $f{f}^{\prime }\to 1/72$ as $\eta \downarrow 0$, while $f{f}^{\prime }\to 0$ at the nose.

Since ${\int }_0^{1}fd\eta =1/18$, the volume isThe 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 $t_0=(\varphi ad/(18V_0){)}^4$. 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 $D$.

Let $n=1-\varphi$ be the pore fraction and put $K=k\rho g/\mu$. In the long-wave approximation, the pore pressure is hydrostatic, $p=\rho g(h-z)$, so Darcy's law gives the horizontal volume flux per unit widthLocal mass conservation $n{h}_t+q_x=R$ therefore gives the Boussinesq equation for an unconfined aquiferThe river and drainage-divide conditions are

The hydrostatic pressure gives $p_x=\rho g{h}_x$, independent of depth. Thus Darcy's law gives $u=-K{h}_x$ and the horizontal volume flux per unit width isThe coefficient $K$ is the hydraulic conductivity, with units of velocity. For a column of unit horizontal area, the stored water volume is $\varphi h$, and rainfall supplies volume flux $R$. The local mass conservation equation is $\varphi h_t+q_x=R$. Therefore the Boussinesq equation for an unconfined aquifer isThis assumes rainfall is entirely converted into recharge and neglects unsaturated storage and losses before the water reaches the water table.

Set $x=L\tilde{x}$, $h=H\tilde{h}$, and $t=T\tilde{t}$. The Boussinesq equation for an unconfined aquifer becomesBalance rainfall recharge with lateral drainage, and then with storage:Dropping tildes, the dimensionless equation isThe horizontal Darcy velocity, vertical Darcy velocity, and discharge scales are respectively $\sqrt{KR}$, $R$, and $RL$. Thus the vertical thickness scale is set by rainfall balanced against hydraulic conductivity over the finite drainage length $L$, rather than by porosity. The latter determines the storage time $T$. The Dupuit approximation requires $H/L=\sqrt{R/K}\ll 1$ for these scales. The initially dry initial condition is $h(x,0)=0$.

With $R=0$, substitute $h=X(x)T(t)$ into the dimensional Boussinesq equation for an unconfined aquifer:For a positive draining profile, take $\lambda >0$. Normalize $T(t^{\ast })=1$, so $X$ has the dimensions of height. The temporal ordinary differential equation integrates toThe spatial ordinary differential equation and boundary conditions areHere $\lambda$ has units of inverse length with this normalization. The scale freedom between $X$ and $T$ is fixed by the chosen value $T(t^{\ast })=1$; the amplitude of $X$, and hence $\lambda$, must be matched to the available water.

This separable aquifer drawdown is exact if the inherited height profile has the separated spatial shape. For a general rainfall-generated initial condition, it describes the late drought regime; it cannot reproduce an arbitrary profile at the instant rainfall stops merely by choosing one temporal constant.

For the dimensionless rainfall-driven Boussinesq equation for an unconfined aquifer, $h_t=(h{h}_x{)}_x+1$, balancing every time power in $h=t^{a}f(x/t^b)$ gives $a=b=1$. The similarity profile obeys $(f{f}^{\prime }{)}^{\prime }+\eta f^{\prime }-f+1=0$. A river drained to base level gives $f(0)=0$, while undrained storage far away gives $f(\infty )=1$. The solution approximates a finite aquifer before its drainage-affected length reaches the far boundary. If $J={\lim }_{\eta \downarrow 0}f{f}^{\prime }>0$, the dimensional drainage flux is $\sqrt{K}{R}^{3/2}Jt/\varphi$.

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 $t^{-2}$.