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Let $F(\zeta ,t)$ map the unit disk conformally to a planar Darcy flow region, with $F(0,t)=0$ at a point source of flux $2\pi Q$. With zero velocity potential on the boundary, the logarithmic source potential becomes $\varphi =Q\log |\zeta |$. Differentiating the inverse map at fixed physical position gives ${\varphi }_t=-Q\mathrm{Re}(F_t/(\zeta F_{\zeta }))$, while $|\nabla \varphi {|}^2=Q^2/|\zeta F_{\zeta }{|}^2$. The boundary law ${\varphi }_t+|\nabla \varphi {|}^2=0$ therefore gives the displayed equation, for nonzero $Q$.

The vector $\zeta F_{\zeta }d\theta$ is outward normal times arclength on $\zeta =e^{i\theta }$. Consequently the equation gives the area growth law $d|D|/dt=2\pi Q$. For a circular map $F=R(t)\zeta$, it reduces to $R\dot{R}=Q$, and hence $R(t{)}^2=R(0{)}^2+2Qt$.

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.

Hydraulic head is the sum of elevation and pressure head, measured in metres relative to a chosen height datum. For constant mass density $\rho$ and gravitational acceleration magnitude $g$, the pressure head $p/(\rho g)$ is the height of a column of the same fluid supported by fluid pressure $p$. A water-filled open measuring tube records hydraulic head as the elevation of its water level when $p$ is gauge fluid pressure. Changing the common pressure reference or height datum adds a constant to hydraulic head and leaves its driving gradient unchanged.

For water with $\rho =1000\mathrm{kg}{\mathrm{m}}^{-3}$ and $g=9.81\mathrm{m}{\mathrm{s}}^{-2}$, a point at $z=2\mathrm{m}$ with gauge fluid pressure $p=9810\mathrm{Pa}$ has $h=3\mathrm{m}$. A second point at $z=0$ with gauge fluid pressure $29430\mathrm{Pa}$ also has $h=3\mathrm{m}$: these values are compatible with a stationary column under hydrostatic pressure. It is a difference in hydraulic head, rather than a difference in fluid pressure alone, that drives saturated Darcy flow. This form uses constant mass density; one cannot generally pull spatially varying $\rho$ through the gradient.^[ref]

In a thin fluid-filled crack, lubrication theory gives flux per unit depth $q=-h^3{p}_x/(12\mu )$. The surrounding Darcy flow has normal velocity $-k{p}_y/\mu$. Conservation in the crack gives $q_x=-[u_y{]}_{-}^{+}$, yielding the displayed condition. Pressure is continuous across the crack. This reduced model neglects transverse pressure variation and assumes wall slip lengths are small compared with $h$.

Under spherical symmetry, the continuity equation and Darcy's law giveHence $q_r=C/r^2$, and integrating the pressure gradient from infinity givesThe boundary condition at $r=a$ fixes $C={\Pi }_{1}a\Delta p/\eta$. Therefore spherical Darcy injection hasHere $q_r$ is the superficial Darcy velocity, not the pore velocity $q_r/\phi$.

For the comparison, assume geometrically similar pore networks, equal porosity, the same source radius and far boundary geometry, and equal pump efficiency. Dimensional analysis then gives ${\Pi }_1=C_{\mathrm{geom}}{d}^2$ with the same dimensionless factor. ConsequentlyAt a fixed desired volumetric flow rate, the required excess pressure and pumping power areAquifer A requires half the pumping power at equal throughput, or half the energy to inject the same volume over the same time. This compares the excess pressure-volume work dissipated by the Darcy flow; any different baseline compression or other plant costs need separate accounting.

The imposed comparison matters. At fixed $\Delta p$, A takes twice the volumetric flow rate and twice the instantaneous power, but the same excess energy per injected volume. The pore spacing alone does not determine permeability of a porous medium if network geometry or porosity changes; without the geometric assumptions the preference cannot be inferred from $d$ alone.

The Darcy law here is $\mathbf{u}=-k\nabla p/\mu$, with hydrostatic contributions absorbed into $p$ if present. Since $k$ is uniform and the fluid is incompressible, $\nabla \cdot \mathbf{u}=0$ gives $\Delta p=0$ in the porous matrix.

Pressure is uniform across the thin crack at leading order. No-slip lubrication theory gives its flux per unit depth $q=-h^3{p}_x/(12\mu )$. Conservation in a short crack segment says $q_x=-[u_y{]}_{-}^{+}$, while Darcy flow gives $[u_y{]}_{-}^{+}=-(k/\mu )[p_y{]}_{-}^{+}$. ThereforeThe other interfacial condition is pressure continuity, $[p{]}_{-}^{+}=0$. The far pressure gradient is $-\mu U{\mathbf{e}}_x/k$; outside the slit both pressure and normal Darcy flux are continuous. The leading tapered-crack flux vanishes at its endpoints. The assumptions $h\ll a$ and $h\gg \sqrt{k}$ justify transverse pressure uniformity and neglect of porous-wall slip on scale $\sqrt{k}$. Close to the sharp tips the reduced model is an outer approximation, not a uniformly resolved tip solution. This is lubrication boundary condition for a crack in a Darcy medium.

These are planar elliptic coordinates. The curve $\xi =0$ is the slit $[-a,a]$, traced on its upper and lower sides. The curve $\xi =1$ is the ellipse with semiaxes $a\cosh 1$ and $a\sinh 1$. For fixed noncardinal $\eta$, the coordinate curve is the appropriate branch of $x^2/(a^2{\cos }^2\eta )-y^2/(a^2{\sin }^2\eta )=1$. The cardinal angles give the two exterior horizontal rays and the upper/lower vertical rays. The requested original sketch is:

On the upper face $0<\eta <\pi$, the supplied derivatives reduce to ${\partial }_x=-(a\sin \eta {)}^{-1}{\partial }_{\eta }$ and ${\partial }_y=(a\sin \eta {)}^{-1}{\partial }_{\xi }$. The same point on the lower face has angle $2\pi -\eta$, so its ${\partial }_y$ has the opposite sign. ThusInsert this and the tangential derivative into the crack condition and multiply by $12a^2\sin \eta$ to obtainPressure continuity becomes $p(0,\eta )=p(0,2\pi -\eta )$. The uniform-flow background is $p_0=-\mu Ux/k=-(\mu Ua/k)\cosh \xi \cos \eta$. Its derivatives immediately giveThe harmonic equation is $p_{\xi \xi }+p_{\eta \eta }=0$ because the metric multiplier is nonzero outside the slit.

For the specified profile, $h^3=H_0^3\sin \eta$ on the upper face. Its coefficient in the tangential operator is constant, so the imposed cosine mode separates. With $P_0=\mu Ua/k$, write $p=P_{0}F(\xi )\cos \eta$. Harmonicity gives $F^{\prime \prime }=F$, the interfacial condition gives $F^{\prime }(0)=\alpha F(0)$, and the far field fixes the growing coefficient to $-1/2$. ThereforeReflection symmetry makes the two face derivatives equal. Other homogeneous cosine modes decaying as $e^{-m\xi }$ would require $-m=\alpha m^2$ at the slit, impossible for $m>0$ and $\alpha \ge 0$. The constant pressure gauge is immaterial. This justifies the separated crack pressure in elliptic coordinates, not only a candidate satisfying selected conditions.

At the midpoint $F(0)=-1/(1+\alpha )$ and $p_x=-\mu U/[k(1+\alpha )]$. HenceFor $\alpha \ll 1$, $Q\sim H_0^{3}U/(12k)$ and the crack barely perturbs the imposed Darcy pressure gradient. For $\alpha \gg 1$, $Q\sim 2aU$ and the slit is nearly equipotential: its high axial conductance collects flow from the surrounding matrix rather than sustaining a substantial pressure gradient. The parameter compares crack conductance $H_0^3/\mu$ with matrix conductance $ka/\mu$. In fact $q(x)=Q\sqrt{1-x^2/a^2}$, so the tip flux is zero in this outer model.

Subtracting the exact uniform background gives $\delta p=P_0\alpha e^{-\xi }\cos \eta /(1+\alpha )$. Since $r\sim a{e}^{\xi }/2$ and $\eta \sim \theta$ at infinity,The printed far-field coefficient is too large by a factor of two. The factor $1/2$ is forced by the stated coordinate map. An independent check is the large-$\alpha$ solution $p=-(\mu U/k)\mathrm{Re}\sqrt{z^2-a^2}$, whose expansion has precisely $\delta p\sim \mu U{a}^2\cos \theta /(2kr)$. This is the elliptic crack pressure dipole.

The phrase “suitably chosen curve” matters for the pressure-only dissipation integral. Put $C_d=\alpha \mu U{a}^2/[2(1+\alpha )k]$ and use a rectangle with end faces $x=\pm X$ and transverse extent $|y|\le H$, taking $a\ll X\ll H$. The horizontal faces have $n_x=0$. The two end-face integrals of $\delta p{n}_x$ totalThis is a fixed-throughflow normalization: the integrated velocity perturbation on an infinite end face is zero, since it is proportional to ${\partial }_x\int \delta pdy={\partial }_x(\pi C_d\mathrm{sgn}x)=0$. Work on the transverse faces vanishes as $H/X\to \infty$. Thus the requested line functional givesA bare circular cutoff would give $-\pi U{C}_d$ instead; it does not enforce this same throughflow normalization, so its pressure-only term cannot be substituted into the fixed-throughflow dissipation comparison. Keeping the control surface explicit resolves that ambiguity. This is fixed-throughflow dissipation reduction of a conductive crack.

Let $s=\pi \varphi a^2\alpha /(1+\alpha )\ll 1$. The noninteracting average dissipation per area is $\mu U^2/k+\varphi \Delta \mathcal{D}=(\mu U^2/k)(1-s)$. Equating this with $\mu U^2/k_{xx}^{\ast }$ givesOnly the first-order density term is controlled by the noninteraction assumption; the rational form is not an all-orders result. It increases permeability along the aligned cracks and saturates as their conductance becomes large. This is dilute permeability enhancement by aligned cracks.

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.

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.

For constant wall heat-flux density $F>0$, effective thermal conductivity $k_e$, buoyancy mobility $B={\rho }_{0}g\alpha K/\mu$ and thermal transport coefficient $\kappa =k_e/({\rho }_0{c}_f)$, a slender Darcy flow boundary layer has width $\delta =(\kappa k_{e}z/(BF){)}^{1/3}$. Writing $\eta =y/\delta$, $\psi =\kappa zf(\eta )/\delta$ and $T-T_a=(F\delta /k_e)f^{\prime }(\eta )$ gives the displayed equation with $f(0)=0$, $f^{\prime \prime }(0)=-1$ and $f^{\prime }(\infty )=0$. The exact integral is ${\int }_0^{\infty }(f^{\prime }{)}^{2}d\eta =1$. Consequently its kinematic Darcy momentum flux is $\int w^{2}dy=B\kappa Fz/k_e$ and its buoyancy flux is $g\alpha \kappa Fz/k_e$. These are per span on one heated face.