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In a unit square with prescribed conductive boundary temperatures and impermeable boundaries, the stream function convention $\mathbf{u}=(-{\psi }_z,0,{\psi }_x)$ gives linear equations $\Delta \psi =R{\theta }_x$, ${\theta }_t={\psi }_x+\Delta \theta$, with $\psi =\theta =0$ on all sides. The velocity equation follows by taking the vertical-plane curl of Darcy law, and the temperature equation by linearizing the heat equation about $1-z$. Unlike a laterally unbounded Darcy layer, the conducting side boundaries produce a two-dimensional real marginal eigenspace at $R_c=8{\pi }^2$.

For a slender porous thermal plume, Darcy law gives $w=\beta (T-T_0)$, with $\beta =Kg\alpha /\nu$. If $C_h={\rho }_0{c}_p$ and $F_h=C_h\int w(T-T_0)dx$ is the conserved excess convective heat flux per span, then the kinematic momentum flux is $M=\int w^{2}dx=\beta F_h/C_h$, and the buoyancy flux is $B=\int g\alpha (T-T_0)wdx=g\alpha F_h/C_h$. The hyperbolic-secant porous plume profile has $Q=(36\kappa Mz{)}^{1/3}$. Multiplication of $Q$ and $M$ by ${\rho }_0$ gives physical mass flux and the density-weighted Darcy second moment. Intrinsic pore-velocity momentum flux instead has an additional inverse-porosity factor. Momentum conservation here follows from the heat/velocity proportionality, not from an inertia balance: porous drag balances buoyancy locally.

A Darcy velocity $\mathbf{u}$ driven by pressure in a stationary porous skeleton dissipates mechanical power $\mu |\mathbf{u}{|}^2/\Pi$ per bulk volume. It follows by dotting Darcy law with the discharge: $-\mathbf{u}\cdot \nabla p=\mu |\mathbf{u}{|}^2/\Pi$. Its contribution to a heat equation can be neglected only when it is small relative to the relevant thermal-energy transfer.

Darcy flow is slow flow through a porous medium whose averaged flux obeys the Darcy law. The Darcy velocity is the fluid volume flux per total cross-sectional area; it differs from the velocity averaged only over pore space.

Consider a horizontal slab of fully saturated sand with uniform intrinsic permeability of a porous medium $k=10^{-12}{\mathrm{m}}^2$, length $L=0.50\mathrm{m}$, total cross-sectional area $A=0.020{\mathrm{m}}^2$, and porosity $\varphi =0.25$. Use one Newtonian fluid with mass density $\rho =1000\mathrm{kg}{\mathrm{m}}^{-3}$ and dynamic viscosity $\mu =10^{-3}\mathrm{Pa}\mathrm{s}$. Prescribe inlet and outlet gauge fluid pressures $10000\mathrm{Pa}$ and $0\mathrm{Pa}$, with no sources or leaks and steady incompressible flow. The two ends have the same elevation, so gravity has no component along the slab. Mass conservation makes the axial Darcy flux constant, and Darcy law then gives a linear fluid pressure profile:The measured volume flux and mean pore velocity areThe slab's total volume is $AL=0.010{\mathrm{m}}^3$, of which $\varphi AL=0.0025{\mathrm{m}}^3=2.5\mathrm{L}$ is pore volume. In the ideal uniform advective model, a nonreacting tracer has transit time $L/v=6250\mathrm{s}\simeq 1.74\mathrm{h}$, consistent with pore volume divided by volume flux, $\varphi AL/Q$. The incorrect substitution of Darcy velocity for pore velocity would give $25000\mathrm{s}$, four times too long. Real passages have different path lengths and local speeds; these numbers describe a bulk mean rather than every particle's journey.

As an independent check, gravitational acceleration magnitude $g=9.81\mathrm{m}{\mathrm{s}}^{-2}$ gives hydraulic conductivity $K=9.81\times 10^{-6}\mathrm{m}{\mathrm{s}}^{-1}$, hydraulic head drop $\Delta h=10000/(1000\times 9.81)\mathrm{m}\simeq 1.01937\mathrm{m}$, and $q=K\Delta h/L=2.0\times 10^{-5}\mathrm{m}{\mathrm{s}}^{-1}$ again. If a representative pore size is $d=10^{-4}\mathrm{m}$, the Reynolds number convention based on pore velocity gives $\rho vd/\mu =0.008$, supporting neglect of pore inertia. This size is an illustrative extra measurement, not something determined by porosity and intrinsic permeability of a porous medium alone.

Regenerate the diagram by running python ../darcy-flux-and-pore-velocity.py from _media; it writes its PNG into the current directory.

Darcy law is a linear relation between the bulk Darcy flux and the net driving force per unit volume. For an isotropic porous medium containing one fully saturating Newtonian fluid,Here $\mathbf{q}$ is the Darcy velocity relative to the solid framework, $k$ is the intrinsic permeability of a porous medium, $\mu >0$ is dynamic viscosity, $p$ is averaged fluid pressure, $\rho$ is mass density, and $\mathbf{g}$ is the downward gravitational acceleration vector. The pressure gradient exerts force density $-\nabla p$, while gravity contributes $\rho \mathbf{g}$; the minus sign makes the discharge follow their sum. The symbols $\mathbf{u}$ and $\mathbf{q}$ are both commonly used for Darcy velocity; neither denotes pore velocity here.

With upward coordinate $z$ and positive gravitational acceleration magnitude $g$, $\mathbf{g}=-g{\mathbf{e}}_z$. Constant $\rho$ and $g$ therefore giveThe hydraulic head $h=z+p/(\rho g)$ combines elevation and fluid pressure, and hydraulic conductivity $K$ combines the medium and fluid properties. Hydrostatic pressure satisfies $\nabla p=\rho \mathbf{g}$, so it produces zero Darcy flux even though fluid pressure increases downward. Conversely, equal pressures at the two ends of a vertical column do not imply zero flow: their elevations give different hydraulic heads.

The appeal of Darcy law is that a complicated connected pore geometry enters a macroscopic calculation through a measurable coefficient. It predicts filter discharge and aquifer flow; combined with mass conservation, it gives an equation for the evolving fluid pressure. Its simple linear form requires a small pore-scale Reynolds number, slow adjustment compared with pore-scale inertial relaxation, and a valid continuum approximation. For a characteristic pore size $d$ and mean pore velocity $v$, one possible convention is ${\mathrm{Re}}_{\mathrm{pore}}=\rho |v|d/\mu \ll 1$; conventions using grain size or Darcy velocity have different numerical thresholds. Merely calling flow laminar does not guarantee negligible inertia. Unsaturated pores, several fluid phases, appreciable inertia, or a fluid whose stress is not Newtonian require further modeling. Woessner and Poeter discuss these limits in their authored hydrology text.^[ref]

Henry Darcy's 1856 book Les fontaines publiques de la ville de Dijon was published by Victor Dalmont; the University Grenoble Alpes digital collection verifies this bibliographic record and supplies a digitized copy.^[ref] Woessner and Poeter's account connects Darcy law to experiments on water flowing through sand filters, relating discharge to cross-sectional area, hydraulic head difference and sample length.^[ref] The modern intrinsic permeability of a porous medium notation above should not be read as a quotation of Darcy's original notation.

With permeability $K$, depth $d$, kinematic viscosity $\nu$ and the relevant effective thermal diffusivity ${\kappa }_T$, the Darcy Rayleigh number compares buoyancy-driven Darcy advection to thermal diffusion. It differs from the clear-fluid Rayleigh number by replacing $d^3$ with $Kd$. Effective heat capacity and porosity factors may be absorbed into the definition of ${\kappa }_T$ and the thermal-time scale, and must be kept consistent with the chosen Darcy law convention.

Darcy velocity, also called Darcy flux or specific discharge, is fluid volume flux relative to the solid framework per unit total cross-sectional area. For a uniform axial discharge $Q$ through bulk area $A$,The area includes both solid and voids. Hence Darcy velocity is a superficial flux with units $\mathrm{m}{\mathrm{s}}^{-1}$, even though no fluid travels through the solid part. For a nonuniform discharge, $Q={\int }_A\mathbf{q}\cdot \mathbf{n}dA$, with $\mathbf{n}$ the unit normal; $Q/A$ is the area average of the normal Darcy flux.

If porosity is $ϵ$, the pore-volume-averaged liquid velocity is ${\mathbf{v}}_l$, and the local solid framework velocity is ${\mathbf{v}}_s$, the convention used here is $\mathbf{u}=ϵ({\mathbf{v}}_l-{\mathbf{v}}_s)$. In a stationary, fully saturated framework with all pore volume participating, ${\mathbf{v}}_l=\mathbf{q}/ϵ$. The notation $\varphi =ϵ$ for porosity is equally common. Darcy flow through a homogeneous slab gives $q=2.0\times 10^{-5}\mathrm{m}{\mathrm{s}}^{-1}$ but pore velocity $v=8.0\times 10^{-5}\mathrm{m}{\mathrm{s}}^{-1}$ at $\varphi =0.25$. Keeping these quantities separate lets Darcy law predict filter discharge while pore velocity predicts ideal advective travel time.^[ref]

A dense porous gravity current of height $h$ places excess hydrostatic pressure $\Delta \rho gh$ on a basal seal. For seal permeability of a porous medium $\lambda$ and thickness $b$, Darcy law gives downward flux $\lambda \Delta \rho gh/(\mu b)$. Dividing the stored height-volume balance by porosity $\varphi$ gives a linear drainage term $-\Omega h$. The coefficient has inverse-time units and does not contain an extra factor of the aquifer permeability.

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.

Use the steady slender-plume model implicit in the requested height-only similarity. Put $\theta =T-T_0$ and introduce permeability $K$, dynamic viscosity $\mu ={\rho }_0\nu$, fluid heat capacity per volume $C_h={\rho }_0{c}_p$, and effective transverse thermal diffusivity $\kappa$. These material constants are needed to give dimensional fluxes. Take $u,w$ to be Darcy velocities, so the thermal transport coefficient for the steady equation is $\kappa =k_e/C_h$ when $k_e$ is the effective conductivity of the saturated medium.

The Darcy law with gravity isThe ambient is hydrostatic with $p_{a,z}=-{\rho }_{0}g$. A slender plume's pressure correction has vertical gradient smaller than the buoyancy term by the squared width-to-height ratio. Indeed, incompressibility gives $u/w$ of order width/height, and the horizontal Darcy relation makes the pressure correction of order $\mu w{\text{width}}^2/(K\text{height})$. Its vertical derivative has the additional inverse-height factor. Thus the leading vertical Darcy balance isThe proportionality is to temperature excess $T-T_0$, not absolute $T$; the velocity vanishes in the unheated ambient. This is the local buoyancy/drag balance of a porous thermal plume, rather than a free-fluid inertial momentum balance.

The governing continuity equation and steady advection-diffusion equation, with axial conduction neglected in the slender high-Rayleigh number limit, areChoose the streamfunction convention $u=-{\psi }_z$, $w={\psi }_x$. It enforces continuity identically and givesA change to the opposite streamfunction convention changes the sign of $\psi$, not the physical velocity or temperature.

The upward convected excess heat flux per unit source length isThis is the quantity denoted $F$ in this question; it differs from Question 2's buoyancy-flux symbol. In conservative form the heat equation is ${\partial }_x(u\theta )+{\partial }_z(w\theta )=\kappa {\theta }_{xx}$. Integrate across the plume:The temperature anomaly and its derivative vanish far from the plume; lateral entrainment brings in ambient-temperature fluid and thus no excess heat. There is no heat source or sink above the line source. Transverse conduction redistributes this heat without removing its total, while axial conduction is negligible in the stated approximation. In an unsteady plume a heat-storage term would have to be retained, so the convective flux alone would not in general be independent of height.

Let $\mathcal{J}=\beta F_h/C_h=\int w^{2}dx>0$. It is constant because $w=\beta \theta$. To obtain similarity exponents without guessing a profile, write the width as $\delta \propto z^a$ and central velocity as $W\propto z^b$. The fixed heat flux requires $W^2\delta$ constant, hence $2b+a=0$. Balancing vertical advection, including its continuity-required horizontal component, against transverse conduction gives $W{\delta }^2\propto \kappa z$, hence $b+2a=1$. Thus $a=2/3$ and $b=-1/3$, and the streamfunction scale $W\delta$ is proportional to $z^{1/3}$. A suitable similarity form isThe volume flux per unit span is $Q=\psi (+\infty )-\psi (-\infty )$ and therefore grows as $z^{1/3}$; it tends to zero at the ideal source. Thus the zero-source-volume requirement fixes the absence of an additive source-length scale, not a finite temperature at the singular point.

Substitution into the streamfunction equation givesChoose the scale normalization $AB=6\kappa$ and $f(+\infty )=1$. Reflection symmetry makes $f$ odd, so the relevant conditions are $f(0)=0$, $f^{\prime \prime }(0)=0$, $f^{\prime }(\infty )=0$, with finite $f(\infty )$; the heat input fixes the remaining dimensional scale. The nonlinear similarity equation becomesIntegrate once. Since $f^{\prime },f^{\prime \prime }\to 0$ in the ambient, the integration constant is zero, giving $f^{\prime \prime }+2f{f}^{\prime }=0$. A second integration gives $f^{\prime }+f^2=C$; the normalized far limit fixes $C=1$. Separation with $f(0)=0$ now yields the complete positive-plume profileIt satisfies the symmetry and decay conditions; the solution of the first-order initial-value equation is unique. The physical heat strength is determined using ${\int }_{-\infty }^{\infty }{\mathrm{sech}}^4\eta d\eta =4/3$, obtained by substituting $t=\tanh \eta$:Equivalently, defineThe hyperbolic-secant porous plume profile and its velocity field are thenIn particular $W{b}^2=6\kappa z$. The far horizontal velocities are inward, $u(\pm \infty ,z)=\mp Wb/(3z)$, supplying the entrained ambient volume. They are consistent with $Q^{\prime }=2Wb/(3z)$.

Finally use specific Darcy-flux moments, per unit source length, for the requested flux relationships. The volume flux, kinematic momentum flux and buoyancy flux areThus the physical mass flux is ${\rho }_{0}Q(z)$, and the conventional density-weighted Darcy second moment is ${\rho }_{0}M$. If intrinsic pore-velocity momentum flux is required instead, a uniform porosity $\phi$ gives ${\rho }_{0}M/\phi$; keeping this convention distinct does not change the displayed Darcy moments. These are the conserved momentum flux of a porous thermal plume relationships. Constancy of $M$ follows here from proportionality of temperature and velocity plus heat conservation, not from ignoring the drag on the solid matrix in a total momentum balance.

The ideal source has zero limiting volume flux but divergent central speed and temperature. Consequently the Boussinesq approximation and slender-plume limit describe the region outside a finite source, where $|\alpha \theta |\ll 1$ and $b/z\ll 1$. Indeed $Wz/\kappa =6(z/b{)}^2\gg 1$ is the local high-Rayleigh condition. These limits improve downstream; they do not make the singular similarity field a literal finite-temperature solution at $x=z=0$.

Let $K$ denote isotropic permeability of a porous medium, $\mu$ the fluid dynamic viscosity, ${\rho }_0$ the reference fluid mass density, $\alpha >0$ its volumetric thermal-expansion coefficient and $g$ gravitational acceleration. The Boussinesq approximation uses $\rho ={\rho }_0[1-\alpha (T-T_{\ast })]$ only in the gravitational force. With upward coordinate $z$, hydraulic pressure $P$ and Darcy flux $\mathbf{q}$, the equations arewhere $\theta$ is the perturbation to the conductive temperature. To keep porous heat-capacity conventions explicit, let $C_e$ be effective heat capacity per bulk volume, $C_f={\rho }_0{c}_f$ the liquid volumetric heat capacity, $k_e$ effective thermal conductivity, $\sigma =C_e/C_f$, and $\kappa =k_e/C_f$. The thermal equation is $\sigma T_t+\mathbf{q}\cdot \nabla T=\kappa {\nabla }^{2}T$. If the usual unit-capacity convention is adopted, $\sigma =1$ and $\kappa$ is the thermal diffusivity. Thresholds below do not depend on $\sigma$; it only changes temporal rates.

For a conductive state with upward temperature decrease $\Gamma =-{\overline{T}}_z>0$, linearization giveswhere $w=q_z$. The second equation follows by taking the divergence of Darcy law to eliminate $P$ and then differentiating its vertical component. It is important that the Laplacian on its right is horizontal.

In the horizontal layer, both boundary planes prescribe the temperature and are impermeable, so $\theta =w=0$ at $z=0,h$. Use a normal mode proportional to $e^{st+i\mathbf{k}\cdot \mathbf{x}}\sin (n\pi z/h)$, with horizontal wavenumber $k=|\mathbf{k}|$, $n\ge 1$. Writing $b=n\pi /h$, the momentum relation gives $w=B{k}^2\theta /(k^2+b^2)$, and the thermal relation givesA neutral mode therefore requires $B\Gamma /\kappa =(k^2+b^2{)}^2/k^2$. Since $k^2+2b^2+b^4/k^2\ge 4b^2$, the smallest value is $4{\pi }^2/h^2$, attained at $n=1$ and $k=\pi /h$. This derives the onset of convection in a horizontal Darcy layer rather than simply importing its critical number. ThusEvery small disturbance decays for $T_m<T_{m,\mathrm{crit}}$; at equality the critical roll is neutral. Strict decay has this threshold as a supremum, not an attained maximum. The first roll wavelength is $2h$, compatible with the very large horizontal extent. Cooling from above gives the opposite, stable buoyancy feedback.

For the tall square column, define $a$ as its side length and $H\gg a$ as its height. The insulated vertical walls impose ${\partial }_n\theta =0$, while impermeability imposes zero normal Darcy flux. Away from the remote ends, a nearly height-independent mode has hydrostatic pressure adjustment and $w=B\theta$, with zero cross-sectional mean. The transverse temperature modes arewhere $m,n\ge 0$ but $(m,n)=(0,0)$ is excluded: a closed column cannot support a net vertical discharge. Its horizontally uniform density perturbation is balanced by pressure, rather than driving overturning. The rate of a nonconstant mode isThe smallest allowed eigenvalue is ${\pi }^2/a^2$, so the insulated square-column Darcy convection onset isAgain strict decay requires $\Gamma <{\Gamma }_{\mathrm{crit}}$. This is the leading long-column result. End conditions that maintain the basic gradient require the vertical velocity to turn around near the ends and give small finite-height corrections; their exact coefficient cannot be inferred from insulation of the vertical walls alone. For a separated vertical mode of small wavenumber $k_z$, the neutral expression is $B\Gamma /\kappa =(k_{\perp }^2+k_z^2{)}^2/k_{\perp }^2$, which tends to the displayed value as $H/a\to \infty$.

Just above threshold, warm fluid rises along one side and colder fluid sinks along the opposite side, with horizontal return flow at the top and bottom. The two lowest modes, proportional to $\cos (\pi x/a)$ and $\cos (\pi y/a)$, have the same threshold. A square therefore admits either orientation or a superposition, including diagonal overturning. Linear theory predicts the weak columnar circulation and its degeneracy; it does not select a unique nonlinear planform.

Use constant porosity $\varphi$, isotropic rock permeability of a porous medium $K$, common liquid dynamic viscosity $\mu$ and gravitational acceleration $g$. Treat the two liquids as separated by a sharp interface, with negligible capillary pressure, molecular mixing, ambient-current resistance and inertia. At distances beyond the initial source scale the current is shallow relative to its horizontal length, so its pressure is hydrostatic. The initial release is symmetric about $x=0$ and has actual liquid volume $V$ per unit length of source. Consequently its geometric thickness integral is $V/\varphi$, not $V$.

Below the interface the excess pressure relative to the ambient hydrostatic field is $p^{\prime }=\Delta \rho g(h-z)$, so Darcy law gives horizontal discharge $q_x=-K\Delta \rho g{h}_x/\mu$. The vertically integrated discharge is $Q_x=-K\Delta \rho gh{h}_x/\mu$. Take the specified downward leakage $v=\gamma h$ to mean Darcy flux per bulk horizontal area. Mass conservation then givesIf $v$ is instead defined as interstitial seepage speed, the actual leakage discharge is $\varphi v$ and the corresponding coefficient is $\lambda =\gamma$. Stating the velocity convention removes that otherwise ambiguous porosity factor. For a shale of thickness $d_s$ and permeability $K_s$, a simple hydrostatic leakage model has $\gamma =K_s\Delta \rho g/(\mu d_s)$, if the reservoir below it has the same ambient pressure reference and no capillary entry threshold.

Let $R(t)$ be the half-width of the support; the full end-to-end length will be $L(t)=2R(t)$. Appropriate conditions are symmetry $h_x(0,t)=0$, $h(\pm R,t)=0$, no flux beyond the support, and the moving-front condition $\dot{R}=-\beta h_x(R^{-},t)$. Initially $\varphi h(x,t)dx$ converges weakly to $V{\delta }_0$, expressing the concentrated line release. The equations describe the spreading porous gravity current, not the singular immediate release itself.

With $\lambda =0$, conserved volume and balance of $h_t$ against $\beta (h{h}_x{)}_x$ give width proportional to $t^{1/3}$ and height proportional to $t^{-1/3}$. Write $h=t^{-1/3}f(\eta )$, $\eta =x/t^{1/3}$. Substitution givesIntegrating from the symmetry axis gives $\beta f{f}^{\prime }=-\eta f/3$. Inside the support $f>0$, so $f^{\prime }=-\eta /(3\beta )$ and $f=A-{\eta }^2/(6\beta )$. Thus the compact Barenblatt solution isThe positive-part notation makes $h=0$ outside the current. Volume fixes its remaining constant:If “length” denotes the distance from the source to one front, it is $R$ rather than $L$; both conventions are given explicitly. Differentiating gives $\dot{R}=R/(3t)=-\beta h_x(R^{-},t)$, so the front condition holds. Using $V/2$ on each reflecting half-line gives the same normalization.

For positive leakage define actual retained volume $\mathcal{V}(t)=\varphi \int hdx$. Since the edge thickness and boundary flux vanish, integration of the equation yieldsPut $h=e^{-\lambda t}H(x,\tau )$, where $\tau =(1-e^{-\lambda t})/\lambda$. Because $\dot{\tau }=e^{-\lambda t}$, substitution cancels the leakage term and leaves $H_{\tau }=\beta (H{H}_x{)}_x$. This derives the exponential drainage transform for porous-medium diffusion. Its conserved geometric volume is still $V/\varphi$, soThe full length is again $L(t)=2R(t)$. In particular,At early times $\tau \sim t$, recovering impermeable-bed spreading. At late times, drainage reduces the current thickness and therefore its hydrostatic driving pressure; the effective spreading time saturates at $1/\lambda$. The current approaches a finite horizontal reach while its retained volume and thickness vanish exponentially. This does not mean a positive-volume pool remains at a finite final depth. For small positive $\lambda$ the length scale diverges as ${\lambda }^{-1/3}$, consistently with unbounded spreading when leakage is absent.

Let $z>0$ measure height from the lower edge and $y>0$ distance from one face of the vertical plate. Use the steady Boussinesq approximation, homogeneous isotropic permeability of a porous medium $K$, reference fluid mass density ${\rho }_0$, dynamic viscosity $\mu$, thermal expansion coefficient $\alpha$, effective thermal conductivity $k_e$ and liquid volumetric heat capacity $C_f={\rho }_0{c}_f$. Define $B={\rho }_{0}g\alpha K/\mu$ and $\kappa =k_e/C_f$, consistently with the thermal convention used above. Write $\theta =T-T_a$, where $T_a$ is the uniform ambient temperature.

Interpret $F$ as uniform wall heat-flux density: a plate segment of height $dz$ inputs heat $Fdz$ per unit horizontal span. The cumulative heat input up to height $z$ is therefore $Fz$. If $F$ instead denotes the sum of flux densities into two identically heated faces, replace it by $F/2$ in each face formula before summing. A specification only of total power per span, without its distribution in height, would not determine this uniformly heated-wall problem. We calculate one heated face; identical heating on both faces doubles the integrated fluxes.

Away from the leading edge, suppose the boundary layer is slender, $\delta (z)\ll z$, and porous inertia and longitudinal heat conduction are negligible. Lateral pressure adjustment leaves the vertical Darcy velocity proportional to buoyancy:At the plate, $v=0$ and $-k_e{\theta }_y=F$; in the far field $w,\theta \to 0$. Darcy law permits tangential discharge at an impermeable plate, so imposing a further no-slip condition $w=0$ would be inappropriate in this leading porous model.

Balance $\theta \sim F\delta /k_e$, $w\sim B\theta$ and $w\theta /z\sim \kappa \theta /{\delta }^2$ to obtainThus the warmed layer thickens and accelerates upward. Define a streamfunction by $w={\psi }_y$, $v=-{\psi }_z$ and setThe last two expressions agree because ${\delta }^3=\kappa k_{e}z/(BF)$. Differentiation gives $v=-(\kappa /3\delta )(2f-\eta f^{\prime })$. Substituting in the thermal equation leaves the single nonlinear equation for the uniformly heated vertical plate in a porous medium:The first condition is impermeability; the second fixes the wall heat flux; the third fixes ambient temperature and vertical velocity. Seek the positive-temperature branch with $f^{\prime }\ge 0$ and finite $f(\infty )$. Decay also gives $f^{\prime \prime }(\infty )=0$. The far-field transverse inflow need not vanish: $v_{\infty }=-(2\kappa /3\delta )f(\infty )$ supplies the increasing discharge.

An approximate solution can be obtained without a numerical shooting calculation. Integrate the differential equation over $0<\eta <\infty$. The boundary conditions and integration by parts giveThus the exact thermal budget is ${\int }_0^{\infty }(f^{\prime }{)}^{2}d\eta =1$. Assume an exponential thermal shape, $f=C(1-e^{-\eta /d})$. The wall derivative requires $C/d^2=1$, and the integral condition requires $C^2/(2d)=1$. Hence $d=2^{1/3}$, $C=2^{2/3}$ and the integral exponential profile for porous wall convection isIt satisfies all boundary conditions, has positive decaying temperature, and matches the exact integrated heat budget. It is an integral approximation, not an exact pointwise solution of the differential equation; its mass-flux coefficient should be read accordingly.

Define the volume discharge, kinematic Darcy momentum flux and buoyancy flux per horizontal span byWith these standard discharge conventions the mass flux is $\dot{m}={\rho }_{0}Q$. Their similarity expressions areThe integral identity evaluates two of them exactly:Equivalently $C_f\int w\theta dy=Fz$: all the heat supplied below height $z$ is carried upward through that section. Although these fluxes are exact within the boundary-layer equations, porous drag means that the momentum flux itself is not conserved with height.

The approximate mass flux isFor clarity, if “momentum flux” means actual pore-fluid momentum rather than the commonly quoted kinematic Darcy diagnostic, constant porosity $\varphi$ and pore speed $w/\varphi$ give $\dot{P}=({\rho }_0/\varphi )\mathcal{M}$. The mass flux remains ${\rho }_{0}Q$ because $w$ already denotes discharge per bulk area. Stating both prevents an implicit porosity error. Validity also requires $\alpha \theta \ll 1$ and a sufficiently slender, low-inertia porous boundary layer; the approximation is not the immediate leading-edge conduction field.

Assume the dense current lies on the bottom, $\Delta \rho >0$, and its height is small compared with the confined aquifer depth $w$. This permits the hydrostatic approximation and neglect of leading-order resistance from the return flow of the original fluid. The dense-layer pressure anomaly on the basal seal is $\Delta \rho gh$. The along-layer Darcy flux, integrated over the current thickness, and the downward leakage flux are respectivelyVolume conservation requires $\varphi h_t+{\partial }_x{q}_x=-q_{\ell }$. Thus the porous gravity current with hydrostatic leakage through a basal seal obeysHere permeability of a porous medium has units of area, and $\Omega$ has units of inverse time. The original PDF instead prints the sink coefficient $\lambda k\Delta \rho g/(\mu b)$. With $\lambda$ defined there as a permeability, that coefficient has units of area per time and cannot multiply $h$ to give $h_t$. It also omits the division of leakage by porosity. This is a genuine source error, not a Darcy law convention. The remaining parts solve the stated linear-drainage equation for a constant $\Omega >0$; the physically consistent coefficient is the one above. Formally treating the printed coefficient as an independently prescribed inverse-time constant leaves those mathematical solutions unchanged.

Take $Q$ as the fluid volume flux per unit source length into one side of the aquifer. For a symmetric release with total rate $Q$, use $Q/2$ on each side. Introduce a constant porosity $\varphi$ to distinguish Darcy velocity from the advecting interstitial velocity; if the given flux is already normalized to pore volume, this factor is absorbed in its definition. With a pressure gradient $G=-p_x>0$, Darcy law givesThus, with $\xi =z/h$, the pore velocity and its depth average areThe tracer concentration initially obeys the two-dimensional advection-diffusion equationReflecting transverse boundaries and constant effective molecular diffusion are assumed. No independent radioactive decay rate is supplied; it can be added separately if it is significant over the experiment.

For times long compared with $h^2/D$, transverse diffusion exchanges tracer between fast and slow streamlines. Write $c=C(x,t)+\chi (z)C_x+\cdots$, where $C=\overline{c}$ and $\overline{\chi }=0$. At the leading transverse-correction order,Integration yieldsIts mean and boundary derivatives are zero. Averaging the concentration equation gives $C_t+U{C}_x=[D-\overline{(u-U)\chi }]C_{xx}$. Integration by parts in the cell equation shows that the enhancement is positive:Consequently Taylor dispersion in a parabolic porous-layer velocity profile givesInjection fixes the inlet concentration or solute-flux boundary condition. This depth-averaged equation requires transverse equilibration and longitudinal concentration variations slow enough for the cell expansion. The supplied PDF hint has both a different polynomial and a false numerical value: its integral evaluates exactly to $-83/2520$, not $0.3525$. The cell problem above supplies the coefficient independently.