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Use the Newtonian fluid stress tensor $\sigma =-pI+2\mu e$, where $e=(\nabla u+\nabla u^T)/2$. The Stokes equations for a point force on the fluid are $\nabla \cdot u=0$ and $\nabla \cdot \sigma +F\delta (x)=0$. Put $r=|x|$. Away from the origin, a harmonic vector potential proportional to $F/r$ in the Unscaled Papkovich–Neuber representation gives the rotationally covariant decaying solution. Specifically, take $\Phi =-F/(8\pi \mu r)$ and $\chi =0$. Differentiation yieldsThis satisfies the homogeneous Stokes equations for $r>0$. Its stress isTo verify the point-force normalization, on a sphere $r=\epsilon$ the traction is $\sigma n=-3(F\cdot n)n/(4\pi {\epsilon }^2)$. Since ${\int }_{S^2}{n}_i{n}_{j}d\Omega =4\pi {\delta }_{ij}/3$, its surface integral is $-F$. Thus integrating momentum balance over the small ball gives exactly the applied force $F$, fixing the normalization of the Stokeslet. Its radial velocity is $(F\cdot n)/(4\pi \mu r)$, soThere is no net volume or mass source at the singularity. The kernels have the parity $J(-x)=J(x)$ and $K(-x)=-K(x)$.

For two Stokes flows with the same viscosity and body forces $f^{(1)},f^{(2)}$, the Lorentz reciprocal theorem including body forces isIndeed, apply the divergence theorem to $u_i^{(1)}{\sigma }_{ij}^{(2)}-u_i^{(2)}{\sigma }_{ij}^{(1)}$. The cross-gradient terms cancel: incompressibility removes each pressure contraction and the remaining terms are both $2\mu e^{(1)}:e^{(2)}$. The two stress divergences give the body-force terms shown.

Choose flow 1 to be the desired body-force-free field and flow 2 to be a Stokeslet with arbitrary force vector $q$ at $y$. For $y\in V$, the body-force term is $-q\cdot u(y)$. Rearrange the reciprocal identity and use the parity of $J,K$; equality for every $q$ givesFor $y$ outside the volume there is no point force in $V$, so the same two integrals sum to zero. To obtain the smooth boundary value, exclude a small hemisphere about $y$, use continuity of $u$ to replace it by $u(y)$ on that hemisphere, and take the radius to zero. A full small sphere contributes $u(y)$; a hemisphere contributes half of it. Equivalently, the local solid-angle factor is $2\pi /(4\pi )$. The result is the boundary integral representation of Stokes flow with right side $\frac{1}{2}u(y)$ on a smooth boundary, and with the double layer interpreted by its boundary principal value. A nonsmooth corner would have a different solid-angle coefficient.

For body-force-free Stokes fields with the same viscosity,To prove this Lorentz reciprocal theorem, integrate the divergence of the difference of the two cross-work fluxes. Both stress divergences are zero; incompressibility and symmetry of the Newtonian stress reduce each cross-gradient contraction to $2\mu e^{(1)}:e^{(2)}$. These are equal, so the divergence integrates to zero. The argument applies to an exterior domain by first cutting it off and taking the outer radius to infinity; the decaying velocity and stress make the outer boundary contribution vanish.

Define the generalized force as force and torque exerted by the rigid body on the fluid. With a resting far field and no-slip boundary conditions, $u=U+\Omega \times x$ on the body, and linearity givesFor two such motions the reciprocal theorem becomesSince the six-component velocities can be chosen independently, $\boxed{\mathsf{R}={\mathsf{R}}^T}$. The ordinary work identity givesfor every nonzero rigid-body motion. If the integral were zero, $e=0$ would make the connected exterior flow an infinitesimal rigid motion. Its decay at infinity requires that motion to be zero, which is incompatible with a nonzero no-slip body velocity. Thus $\mathsf{R}$ is positive definite. Fluid-on-body forces have the opposite sign; their map is $-\mathsf{R}$, not a positive-definite resistance matrix.

Write $s_{\varphi }=\sin \varphi$ and $c_{\varphi }=\cos \varphi$. The helix has $|dX/d\theta |=b\sqrt{1+{\tan }^2\varphi }=b\sec \varphi$, soIts unit tangent is $X^{\prime }=c_{\varphi }{e}_{\theta }+s_{\varphi }{e}_z$, where $e_{\theta }=(-\sin \theta ,\cos \theta ,0)$. The given slender-body force density is the force per arc length exerted on the fluid. For pure axial translation, $V=U{e}_z$, it givesFor pure rotation, $V=b\Omega e_{\theta }$, it givesThe axial torque per unit arc length is $b{f}_{\theta }$. Integrating along the wire gives the axial resistance matrix of a slender helixThus translation gives $(F_z,G_z)=(\mathcal{A}U,\mathcal{B}U)$, and rotation gives $(\mathcal{B}\Omega ,\mathcal{D}\Omega )$. These are the requested axial components; the axial swimmer approximation does not require proving statements about the other components. The equality of the two off-diagonal coefficients checks reciprocity, andFor the organism, retain the specified entrainment approximation: the head moves with the local large-scale fluid translation, and its angular velocity $\Omega -\omega$ differs from the local fluid rotation $\Omega$ by $-\omega$. Put $D_0=8\pi \mu a^3$. Its torque on the fluid is therefore $-D_0\omega$, and its force is neglected. Neutral buoyancy and absence of external torque giveSolving,This is the entrained-head approximation for a helical microswimmer, which differs from treating the head as an isolated sphere in otherwise stationary fluid. The flagellum's angular speed is also $\Omega =D_0\omega (1+c_{\varphi }^2)/(CL{b}^2)$.

At fixed $b,L,a,\omega ,ϵ$, the speed is maximal at $\varphi =\pi /4$ and tends to zero at either limiting pitch. At fixed other parameters it decreases as $1/L$: the longer flagellum has more resistance at the fixed torque supplied in this model. Holding the number of turns fixed instead would make $L$ change with pitch, so that is a different comparison.

The motor exerts equal and opposite torques of magnitude $D_0\omega$ on flagellum and head, whose relative angular speed is $\omega$. Therefore its rate of working isEquivalently, the work delivered to the fluid is $G_{\mathrm{flag}}\Omega +G_{\mathrm{head}}(\Omega -\omega )=D_0\omega \Omega -D_0\omega (\Omega -\omega )$. Translation contributes no net work because the total force vanishes. This explains the independence from $U$ and $\Omega$, without assigning an isolated-fluid drag law to the entrained head.

The material derivative follows surfactant carried by the moving interface. The term $-C{\nabla }_s\cdot u_s$ is dilution or concentration by tangential area expansion or contraction. The term $-C(u\cdot n){\nabla }_s\cdot n$ is the additional area change produced by normal motion of a curved interface. The term $D_s{\nabla }_s^{2}C$ represents surface diffusion, while $-k(C-C_0)$ relaxes the concentration towards a reservoir value through exchange. All these terms refer to concentration per surface area, as in surfactant transport with exchange relaxation.

In the steady bubble frame, the spherical interface has $u\cdot n=0$. Write $C=C_0+C^{\prime }$ with uniform $C_0$ and regard both $u$ and $C^{\prime }$ as first order in the rising speed. Advection of $C^{\prime }$ and its product with surface dilation are second order. Hence the linear steady equation isSpherical symmetry and linearity leave only the tangential projection of the imposed translation vector: $u_s=A{I}_{s}U$. An azimuthal term $n\times U$ would change sign under reflection and is excluded for this non-chiral rising bubble. The degree-one forcing does not excite other spherical-harmonic degrees in the linear problem.

On $r=a$, differentiating $n=x/r$ givesEach component of $n$ is a degree-one spherical harmonic, soOne may also obtain the coefficient from $n\cdot n=1$: its surface Laplacian gives $n\cdot {\nabla }_s^{2}n=-|{\nabla }_{s}n{|}^2=-2/a^2$, while rotational symmetry makes the vector Laplacian parallel to $n$. Expanding $I_{s}U=U-n(U\cdot n)$ then yieldsThe candidate $C^{\prime }=BU\cdot n$ therefore solves the linear transport equation ifIts spherical mean is zero. This is the dipolar surfactant distribution. Assuming $k{a}^2+2D_s>0$, its consistency condition isDiffusion or exchange must restore uniform concentration faster than the interfacial flow redistributes it: this holds with a small surface Péclet number, fast exchange, or sufficiently immobilized surface velocity. These same conditions justify neglecting the perturbation's advection. Small capillary number and sufficiently small buoyancy-induced shape distortion are separate requirements for a spherical interface.

Take $+$ to mean the outer liquid, $-$ the bubble interior and $n$ outward from the bubble. The general interfacial stress balance with variable surface tension isFor $\gamma ={\gamma }_0-{\gamma }_1{C}^{\prime }$, its tangential part is $-{\nabla }_s\gamma ={\gamma }_{1}B{I}_{s}U/a$. Thus the requested form isThis positive parameter measures the strength of the restoring Marangoni stress relative to viscous stress, including the relaxation of concentration by diffusion and exchange.

The far velocity $-U$ is produced by the constant harmonic vector potential $\Phi =U$. The two independent decaying degree-one modes are the force-carrying vector potential $\alpha aU/r$ and the scalar dipole $\chi =\beta a^{3}U\cdot \nabla (1/r)$. They are harmonic outside the bubble and have precisely the symmetry of uniform translation. The Unscaled Papkovich–Neuber representation gives, at arbitrary radius,The velocity at $r=a$ agrees with the printed expression. However, the original PDF's traction formula has a factor-of-two error. Direct differentiation with $\sigma =-pI+\mu (\nabla u+\nabla u^T)$ givesnot the printed prefactor $12\mu /a$. This is traction of translating-sphere Papkovich–Neuber potentials. The $\alpha$ term follows immediately from the Stokeslet stress, and the scalar-potential part gives $6\mu \beta [U-3(U\cdot n)n]/a$; neither contribution contains the extra factor two.

With this consistent traction, no penetration requires $\beta -\alpha =1/2$, tangential velocity requires $A=-(1+\alpha +\beta )$, and tangential stress balance requires $\beta =A\lambda$. Solving,For comparison, taking the printed $12\mu /a$ literally would instead giveThose solve the supplied algebraic pair but not the Newtonian stress calculated from the supplied potentials. The factor-six result is the physically consistent translating surfactant-coated bubble solution.

As $\lambda \to 0$, $A\to -1/2$ and $\beta \to 0$: the clean bubble has a mobile interface and zero tangential traction. As $\lambda \to \infty$, $A\to 0$, $\alpha \to -3/4$ and $\beta \to -1/4$: the interface becomes immobilized and the exterior flow approaches that past a no-slip rigid sphere. The limiting tangential traction is $-3\mu I_{s}U/(2a)$, a useful check on the correct normalization.

The normal dynamic traction isapart from an arbitrary constant pressure. In the normal stress balance on a translating bubble, its degree-one part is balanced by the buoyant hydrostatic pressure difference and by the variation $-2{\gamma }_1{C}^{\prime }/a$ of normal capillary stress. A uniform internal gas pressure balances the constant Laplace-pressure term but cannot by itself balance this angular variation. If $U$ is upwards and $\Delta \rho$ is liquid minus bubble mass density, the degree-one balance can be writtenIt fixes $U=\Delta \rho g{a}^2(1+2\lambda )/[3\mu (1+3\lambda )]$, tending to the clean-bubble and rigid-sphere rise speeds. A degree-one radial displacement translates a sphere rather than changing its curvature, so a supposed dipolar shape distortion cannot replace this force balance.

Let $ϵ=1-\alpha \ll 1$, and use local distance $\rho =a\theta$ from the closest pole. The gap isHorizontal rotation has nonzero wall speed $V_s\sim |{\Omega }^{\prime }|a$ at this pole. In the broad region with $h=O(\Delta )$, shear is $O(\mu |{\Omega }^{\prime }|a/\Delta )$, the area is $O(a^2)$ and the lever arm is $O(a)$. Its torque is therefore $O(\mu |{\Omega }^{\prime }|a^4/\Delta )$. The smallest gap patch has radius $O(a\sqrt{ϵ})$ and area $O(a^2ϵ)$; its shear is $O(\mu |{\Omega }^{\prime }|a/(ϵ\Delta ))$. It gives the same bounded torque scale.

These are the two estimates requested in the paper. They omit an intermediate annulus, so the printed uniform $O(\mu |{\Omega }^{\prime }|a^4/\Delta )$ assertion is not correct as $ϵ\to 0$. For $a\sqrt{ϵ}\ll \rho \ll a$, $h\sim \Delta {\rho }^2/a^2$. An annular shear-torque contribution scales asIntegrating between the inner and outer scales produces $\log (1/ϵ)$.

A viscous dissipation lower bound makes this conclusion independent of possible signs of local tractions. A fluid velocity changing by $V_s$ between two no-slip walls dissipates at least $\mu V_s^2/h$ per unit area, by the Cauchy-Schwarz inequality in the cross-gap coordinate. The lubrication velocity profile gives more explicitlywith no cross term. The second term is nonnegative. Thus the annulus entails a logarithmic divergence in total power and hence in $|G^{\prime }|$, since the centre is held fixed and all boundary work is $G^{\prime }{\Omega }^{\prime }$.

The horizontal rotation in a nearly touching spherical shell calculation also gives the corrected force scale. Choose local $x$ along $v_s$ and put $V_s=a{\Omega }^{\prime }$ with its signed value. At leading order the stationary-gap Reynolds lubrication equation isFor $h=h_0+(x^2+y^2)/(2R)$ it has the particular pressurewhich satisfies the equation by direct differentiation. The pressure itself gives no torque about the centre of an exactly spherical inner body, because its normal is radial. It does contribute to tangential shear through the pressure gradient.

For any fixed local outer cutoff, the logarithmically divergent integrals areThe resisting tangential traction is $\mu V_s/h+(h/2)p_x$. Integration by parts gives torque magnitude $a\mu |V_s|(I_1+3I_2/(5R))$ to logarithmic order. Consequently, at leading order in $\Delta /a$,The strict corrected order is $\Theta ((\mu |{\Omega }^{\prime }|a^4/\Delta )\log (1/ϵ))$; the paper's scale is the bounded contribution of each of its two nominated regions, not the full asymptotic answer.

For the horizontal force, pressure dominates shear by a factor $a/\Delta$. Its horizontal projection on the inner sphere has normal component $n_x\simeq x/a$, giving $|F_p|\sim (6\mu |V_s|/(5a))I_2$. Hence the required holding-force magnitude isThe broad region and smallest patch each supply $O(\mu |{\Omega }^{\prime }|a^4/{\Delta }^2)$; the intermediate annulus again supplies the extra logarithm. The shear force is smaller by $\Delta /a$. These expressions assume the completely filled lubrication film of the problem.

The vertical holding force is zero. The geometry is invariant under a half-turn about the vertical axis. That operation reverses the horizontal angular velocity but leaves a vertical force component unchanged. Linearity would reverse that force component too, forcing it to vanish. Equivalently, the induced pressure has the first azimuthal harmonic, whose vertical-force integral over azimuth is zero. The horizontal holding force is the negative of the fluid's horizontal force on the body.

Axisymmetric rotation about the offset axis gives wall velocity $v_{\phi }=a\Omega \sin \theta$. Its azimuthal Couette flow has no pressure-driving divergence, so pressure remains constant. The fluid-on-inner-sphere tangential traction is $-\mu a\Omega \sin \theta /h$. Multiply it by the moment arm $a\sin \theta$ and integrate:Putting $t=\alpha z$ and using the second supplied identity with its generic parameter equal to $\alpha$ gives the axial rotational resistance of eccentric nested spheres:The continuous limit at $\alpha =0$ is $-8\pi \mu \Omega a^4/(3\Delta )$, consistent with a concentric narrow spherical shell. At $\alpha \to 1$, the torque tends to $-4\pi \mu \Omega a^4/\Delta$: it remains finite because the rotational wall speed vanishes at the closest pole.

Set $V=\Delta \dot{\alpha }$, positive downwards. Then $h_t=-V\cos \theta$. The leading axisymmetric lubrication flux is $q_{\theta }=-h^3{p}_{\theta }/(12\mu a)$. Wall Couette flux from the translating sphere is smaller by $\Delta /a$ than the squeeze flux and does not affect this leading pressure. The spherical Reynolds lubrication equation isRegularity at the pole gives $q_{\theta }=(aV/2)\sin \theta$, soThe apparent singularity at $\alpha =0$ is a constant-pressure gauge term: subtract $3\mu a^{2}V/(\alpha {\Delta }^3)$ before taking the limit. The remaining pressure tends to $6\mu a^{2}V\cos \theta /{\Delta }^3$.

To leading lubrication order the force is the pressure integral. Its downward component on the inner sphere isThe original PDF's first integration hint is incorrect: the rational term needs $1-b^2$, not $(1-b{)}^2$. From the antiderivative $1/(1-t)+\log (1-t)$,Thus the squeeze resistance of eccentric nested spheres givesIt opposes the velocity for either sign of $\alpha$. The continuous concentric limit is $F_z=-8\pi \mu a^{4}V/{\Delta }^3$, since the bracket is $4{\alpha }^3/3+O({\alpha }^5)$. As $\alpha \to 1$, its leading magnitude is $6\pi \mu a^4|V|/[{\Delta }^3(1-\alpha )]$.

Let the tangential velocity scale be $U$, gap scale $H$ and along-gap length $L\gg H$. The tangential Stokes equation balances $P/L$ with $\mu U/H^2$, while wall shear is $\tau \sim \mu U/H$. ThereforeThis is lubrication pressure dominates shear stress; pressure force projections may nevertheless be small in a particular direction or may give no torque on a spherical surface.

Write the signed centre displacement along the downward axis as $d=\alpha \Delta$. A point $an$ on the inner sphere has distance $|an+d{e}_z|=a+d\cos \theta +O({\Delta }^2/a)$ from the outer centre. The leading normal gap to radius $a+\Delta$ is consequentlyThe local-curvature corrections are smaller by $\Delta /a$. This approximation also follows by solving the outer-sphere intersection along the ray $n$ from the inner centre and expanding its positive root.

The zeroth law of turbulence states that at fixed outer velocity $u$ and integral scale of turbulence $\ell$, the mean viscous dissipation per unit mass approaches a nonzero finite value as the Reynolds number $\mathrm{Re}=u\ell /\nu$ becomes large. In an established forward energy cascade, this gives $ϵ\sim C_{ϵ}{u}^3/\ell$, with $C_{ϵ}$ of order one. The small-scale velocity gradients become large enough to offset decreasing kinematic viscosity; this is the turbulent dissipation anomaly.

At the Kolmogorov microscales, the local Reynolds number is of order one, $v\eta /\nu \sim 1$, and the viscous dissipation is $ϵ\sim \nu (v/\eta {)}^2$. Solving these balances givesSubstituting the outer-scale estimate, and omitting order-one constants, givesThe associated eddy turnover time is $\eta /v=(\nu /ϵ{)}^{1/2}$.

For a Burgers vortex, integrating the axial vorticity gives the azimuthal velocity:The swirl contribution to the viscous dissipation per unit length and per unit density isFor the equality, expand the two squares using ${\omega }_z=u_{\theta }^{\prime }+u_{\theta }/r$; their integrated difference is proportional to $[u_{\theta }^2{]}_0^{\infty }$, which vanishes. Substitution of the Gaussian vorticity profile givesThis excess dissipation of a Burgers vortex is independent of kinematic viscosity at fixed strain $\alpha$ and circulation $\Gamma$. Multiply by the mass density if a dimensional power per length is wanted.

The qualification "excess" matters. The imposed uniform strain itself has viscous dissipation density $3\nu {\alpha }^2$ per unit mass, so its integral over an infinite cross-section diverges. One must subtract that background or restrict to a finite cross-section. In a core-sized area, its contribution is of order ${\nu }^2\alpha$, smaller than the swirl contribution by order $(\nu /\Gamma {)}^2$ as $\Gamma /\nu \to \infty$.

Meanwhile $\delta \propto {\nu }^{1/2}$ shrinks and the central vorticity $\Gamma /(\pi {\delta }^2)$ grows. The finite dissipation becomes concentrated in a narrow tube: a useful local model of internal intermittency and the turbulent dissipation anomaly, though not a proof that an entire turbulent flow consists of such vortices.

Let $u^2$ denote one-component velocity variance, $F=u^{2}f$ the longitudinal velocity correlation, and $S_p=⟨(\Delta v{)}^p⟩$ the signed longitudinal structure functions. Homogeneity gives $F=u^2-S_2/2$, while the third-order convention here is $u^{3}K=S_3/6$. Thus the Kármán-Howarth equation becomesThe mean kinetic energy per unit mass is $3u^2/2$, so its decay gives ${\partial }_t{u}^2=-2ϵ/3$. In the universal equilibrium range, the small-scale variation ${\partial }_t{S}_2$ is negligible at leading order, hence ${\partial }_{t}F\simeq -2ϵ/3$. Integrating from zero and using regularity givesThe Kolmogorov equation for structure functions is thereforeFor finite local unsteadiness the right side has the additional term $-3r^{-4}{\int }_0^r{s}^4{\partial }_t{S}_2(s,t)ds$; dropping it is the local-equilibrium approximation used here.

In the inertial range the viscous term is also negligible, leaving the Kolmogorov four-fifths law, $S_3(r)=-4ϵr/5$. Its coefficient and sign follow from the exact energy balance and Kármán-Howarth equation, rather than from dimensional similarity. It supplies a firm third-order constraint on Kolmogorov 1941 theory and measures the forward energy cascade, while leaving the second- and higher-order exponents open to internal intermittency corrections.

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.