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For saturated Darcy flow through a long dam occupying a channel of width $\alpha z^2$, let hydraulic conductivity be $K=k\rho g/\mu$ and free-surface height $h(x)$. The Dupuit approximation gives $Q=-K\alpha h^3{h}_x/3$. Fixed heads $h(0)=H$ and $h(L)=\eta$ give $h^4=H^4-(H^4-{\eta }^4)x/L$ and the displayed steady discharge. Porosity relates pore velocity to superficial Darcy velocity but does not independently multiply this discharge when $k$ is intrinsic permeability. For depth-dependent permeability, integrate the horizontal conductivity over the saturated cross-section first.

Assume quasi-steady saturated Darcy flow, hydrostatic pore pressure, a long shallow barrier, isotropic intrinsic permeability of a porous medium $k$, an impermeable lined base, and negligible capillarity. Maintain upstream and downstream free-surface heads at $H$ and $\eta <H$. The Dupuit approximation neglects vertical velocity except in small entrance/exit regions. Let $\mu$ be the fluid's dynamic viscosity and $K=k\rho g/\mu =kg/\nu$ its hydraulic conductivity.

Since the pressure is $p=\rho g(h(x)-z)$, the superficial Darcy velocity is $v_D=-K{h}_x$, uniform with depth at fixed $x$. Integrating over the printed channel width givesThe steady discharge is constant, so integration with the two head boundary values gives Dupuit flow in a channel with cubic wetted area:Here $Q$ is total volume per time, including the changing channel width. The pore-space mean speed is $v_p=v_D/\varphi =Q/[\varphi A(h)]$. Thus porosity affects storage and pore transit time but does not separately multiply this steady discharge when $k$ is the intrinsic permeability in Darcy's law. If “permeability” is instead supplied as hydraulic conductivity, use that value directly for $K$. An exact seepage-face/free-boundary calculation near the outlet would go beyond the shallow Dupuit model.

Use the Dupuit approximation: the saturated region is shallow enough that hydrostatic pressure is $p=\rho g(h-z)$ and flow is predominantly horizontal. Darcy's law then gives the horizontal Darcy velocity $u_x=-u_b(1+\beta z)h_x$, where $u_b=k_0\rho g/\mu$. Integrating through the saturated depth gives the volume flux per unit widthThe stored water volume per unit area is $\varphi h$, so mass conservation yields the unconfined aquifer with depth-dependent permeability equationDefine the positive river discharge by $Q(t)=-q(0,t)$. A useful check on all subsequent results is the integrated mass conservation law

Let $H\ll L$ be the characteristic saturated thickness and let $\delta =H/L$. Scale pressure by $\rho gH$. The horizontal part of Darcy's law gives the scaleBy incompressibility, $w\sim \delta u\sim K{\delta }^2$. The vertical part of Darcy's law isIts left side is smaller than the scale $K$ by ${\delta }^2$. Therefore $p_z+\rho g=O(\rho g{\delta }^2)$, and the leading pressure is hydrostatic pressure:Here the water table has pressure $p_a$. This is the Dupuit approximation, valid away from the small region near an outlet where the surface slope need not remain small.

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 $h(x,0)=0$.

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.

A zero-height outlet can carry finite nonzero volume flux per unit width in a nonlinear Darcy flow model with $q=-Kh{h}_x$. If $h{h}_x\to J>0$ as $x\downarrow 0$, then $(h^2{)}^{\prime }\to 2J$, and integration gives $h\sim \sqrt{2Jx}$ and $h_x\sim \sqrt{J/(2x)}$. Thus the height vanishes while its slope diverges; the flux must be evaluated as a limit of the product. The divergent slope also identifies a region where the outer Dupuit approximation eventually needs a local correction.

With permeability of a porous medium $k(z)=k_0(1+\beta z)$, the Dupuit approximation givesHere $\varphi$ is porosity and $R$ is recharge. The low-height limit has mobility proportional to $h$, whereas large heights have mobility proportional to $h^2$. This changes both the forced filling similarity for power-law diffusion and the separable draining profile for power-law diffusion.

The water table is the upper surface of the saturated zone in an unconfined aquifer, where liquid pressure equals atmospheric pressure in the simplest drainage model. Under the Dupuit approximation, its height determines hydrostatic pressure throughout the saturated region and therefore the lateral Darcy flux.