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For the steady positive-velocity branch, conservation of buoyancy flux gives $bhu=F_0$. Substituting into the other two balances givesUse a power-law ansatz $h=C{x}^p$, $u=D{x}^q$. The entrainment balance forces $p=1$ and $C(1+q)=\alpha$. The momentum powers are $x^{2q}$, $x^{-q}$ and $x^{-q}$, so the nonzero inertial branch has $q=0$. Therefore the constant-speed entraining gravity current on a slope isIts positive speed requires $\tan \theta >\alpha /2$ for $\theta <\pi /2$. In the shallow, strictly hyperbolic range this meansThe same steady integral balances have a formal vertical-plume limit at $\theta =\pi /2$, although their time-dependent system then loses strict hyperbolicity. At the lower threshold the assumed positive-flux, finite-speed solution fails.

The line source is singular: $h,uh$ and the momentum flux tend to zero, but $b$ diverges as $x\downarrow 0$. Thus this is a downstream ideal-source solution, valid only far enough away that $b/g\ll 1$ and the Boussinesq approximation and top-hat plume model hold. It does not describe the immediate source region.

For dilute sediment volume concentration $\varphi$, define $G=g({\rho }_p-{\rho }_0)/{\rho }_0$. Neglecting particle volume in the fluid-volume balance,This applies the Boussinesq approximation to the bulk mixture, requiring $G\varphi /g\ll 1$, and uses a deep ambient layer $h\ll H$. A gravity-current front condition closes the unresolved nose. Write it as $\dot{L}=F\sqrt{g^{\prime }h}$, with a prescribed order-one front coefficient $F$. If the chosen Froude number is instead $\mathrm{Fr}=\dot{L}/c$, then $F=\sqrt{2/3}\mathrm{Fr}$.

Use a parabolic-channel sediment-current box model for one side of the dam: initially its length is $L_0$, depth $h_0$, and uniform concentration ${\varphi }_0$. Its conserved fluid volume isFor vertical settling onto the channel sides, the projected deposition width is $b(h)=\sqrt{h}$. Thus the total particle deposition flux is $W_s\varphi \sqrt{h}L$ andThe geometric factor $3/2$ follows from $b(h)/A(h)$; using a rectangular deposition formula would give a different result.

Eliminate time and use $h^{3/2}=h_0^{3/2}{L}_0/L$:The runout length of a gravity current is the limit where suspended concentration vanishes:The model approaches this length asymptotically in time. It assumes negligible entrainment, erosion and water loss, and does not resolve the initial release or nose. If the initially disturbed region supplies both sides, apply the formula to each side using its share of the initial volume and length.

Use a one-sided top-hat plume model with Boussinesq approximation and neglect the solid/liquid density difference in the interfacial volume source. Define the positive ablation speed $U=-\dot{a}$, volume flux per unit wall span $Q=bw$, and kinematic momentum flux $J=b{w}^2$. Let ambient entrainment per height be $E=\epsilon w$, with entrainment coefficient $\epsilon$, and take $\rho ={\rho }_{\infty }[1+{\beta }_C(C-C_{\infty })]$ for the salinity contribution to density. Conservation of volume and vertical momentum, with negligible wall stress and negligible vertical momentum of entering fluid, givesOnly the ocean-facing edge entrains. Using two edges would change the entrainment coefficient in these formulas.

To derive the heat balance, choose liquid specific enthalpy $h_l=c_{p}T+L$ and solid specific enthalpy $h_s=c_p{T}_s$. A horizontal slab receives ambient volume $Edz$ with enthalpy $h_{\infty }=c_p{T}_{\infty }+L$, and solid mass-equivalent volume $Udz$ with enthalpy $h_s$. ThusSubtract $(c_p{T}_{\infty }+L)Q^{\prime }$ and divide by $c_p$:Similarly, the incoming salt flux is $E{C}_{\infty }+U{C}_s$, so $(QC{)}^{\prime }=E{C}_{\infty }+U{C}_s$. Subtracting $C_{\infty }{Q}^{\prime }$ provesBoth anomalies are negative for fresh, cold melt entering the ocean. These balances account for the heat needed to melt the entering ice, rather than treating it as already-liquid water at $T_s$.

Now use $L/c_p\gg T_{\infty }-T_s$, the large latent-to-sensible heat ratio convention of the paper. Put $\Delta C=C_{\infty }-C_s>0$. The stated small plume-deficit limits allow us to replace plume $T,C$ by their ambient values in the diffusive interfacial fluxes. The Stefan condition and interfacial salt balance are thenFor the compositional Péclet number $P\equiv P{e}_c=U\delta /D$, solve the second equation and impose the liquidus:Substitution in the heat balance gives the diffusive glacier-ablation interface lawHere $\Gamma$ is the liquidus slope, unrelated to the surface-energy parameter of the previous question. If $A_T=LD/(c_p\kappa )$, $\Theta =T_{\infty }-T_m$ and $B_T=\Gamma \Delta C$, the physical root isIt is positive when $\Theta +B_T>0$, exactly the condition that the ocean is warmer than its own liquidus. Since $\delta$ and ambient properties are fixed, $U$ is constant with height in this approximation.

For the melting-driven saline wall plume, measure height $Z=z-z_0$ from a virtual origin. The integrated anomaly balances areIn the specified small-deficit regime the meltwater volume is negligible compared with entrainment. Indeed, the solution below gives $(C_{\infty }-C)/\Delta C=4U/(3\epsilon w)$, so a small compositional plume deficit implies $U\ll \epsilon w$. Thus retain $Q^{\prime }=\epsilon w$ for the leading power laws, although $U$ remains essential in the salt and heat sources. Try $b=BZ$, $w=W{Z}^q$. The salt deficit is proportional to $Z^{-q}$, and the momentum balance has powers $Z^{2q}=Z^{1-q}$, hence $q=1/3$. The volume balance gives $(4/3)BW=\epsilon W$, and the momentum coefficient gives $(5/3)B{W}^2=g{\beta }_{C}U\Delta C/W$. ThereforeThese are entrainment-dominated power laws, not an exact solution with the $U$ term retained in the volume equation. They fail close to the virtual origin, where the formal temperature and salinity deficits diverge. Their buoyancy source is distributed melting along the wall, unlike a line plume with a constant buoyancy flux from a localized source.

The profiles rise from $T_i,C_i$ across the laminar boundary layer to the almost-ambient plume values $T,C$, then approach $T_{\infty },C_{\infty }$ across the entraining edge. The sketch uses the linear-gradient approximation implicit in the prescribed diffusive fluxes; it is not an exact normal advection–diffusion profile at large $P$.

For $P\ll 1$,The last formula retains both first-order terms and is appropriate near the no-ablation threshold. Salt diffusion keeps the interface saline, and freezing-point depression allows ice dissolution in saline water even when $T_{\infty }<T_m$. Latent heat is still needed, but the thermodynamic permission to ablate comes from the depressed liquidus; small $P$ does not mean a negligible energy cost.

For $P\gg 1$, $C_i-C_s\sim \Delta C/P$ and $T_i\sim T_m-B_T/P$. Rapid production of fresh melt dilutes the interface, so its temperature approaches the pure-ice melting point. In the usual thermally driven regime $\Theta >0$ and $\Theta \gg B_T/P$,This is predominantly thermal melting. More generally the retained large-$P$ balance is $A_{T}P\simeq \Theta +B_T/P$. For example, $\Theta =0$ gives $P\sim \sqrt{B_T/A_T}$, which can be large while the driving term is still saline freezing-point depression. Thus the melting/dissolution interpretation must also consider the ocean's temperature relative to $T_m$, not the compositional Péclet number alone.

Let $f(y)$ denote the signed vertical component of the Coriolis parameter; it is negative in the southern hemisphere. In the steady, linear, inviscid interior, depth integration of horizontal momentum giveswhere $\mathbf{U}={\int }_{-H}^0(u,v)dz$ and $\Pi ={\int }_{-H}^{0}pdz/{\rho }_0$. Horizontal curl eliminates the pressure. Its left side is $f{\nabla }_h\cdot \mathbf{U}+\beta V$. Incompressibility and impermeable flat top and bottom imply ${\nabla }_h\cdot \mathbf{U}=w(-H)-w(0)=0$, givingThis is Sverdrup balance. It assumes steady small-amplitude motion, a Boussinesq approximation, negligible nonlinear advection, negligible Rayleigh and bottom/lateral drag in the interior, a flat fixed-depth domain and no normal top/bottom flux. The applied surface wind stress remains a forcing, even though interior friction is neglected. With cooling absent, the steady buoyancy equation additionally requires $B_z=0$, if that equation is imposed; the curl balance itself does not depend on the cooling coefficient. In the closed, $x$-independent channel below, depth-integrated $V=0$, so a nonzero wind-stress curl cannot be balanced by an inviscid interior alone.

Steady forced channel. Now take $r>0,\alpha >0$. Put $a=\pi /L$, $s=(z+h{)}_{+}=\max (z+h,0)$ and write ${\tau }_0$ for the positive wind-stress amplitude. The prescribed profiles giveThe buoyancy jumps across the sharp forcing interface in this simplified model; pressure and the circulation below are continuous. Let $P=p_y/{\rho }_0$, $D=f^2+r^2$, and $b_y^{\mathrm{top}}=-a{B}_0\sin ay/(\alpha h)$. Hydrostatic balance gives $P=C(y)+b_y^{\mathrm{top}}s$. The horizontal momentum equations and their solution areThe no-normal-flow conditions imply ${\int }_{-H}^{0}vdz=0$: this integral is independent of $y$ by incompressibility and is zero at the meridional walls. It determines the otherwise free depth-uniform pressure gradient,DefineA meridional overturning streamfunction satisfying $v=-{\Psi }_z$, $w={\Psi }_y$ isIt vanishes on all four boundaries. Together with the expressions for $b,P,u$, this solves all the steady equations. The formula permits variable $f(y)$; freezing it at a negative $f_0$ gives the simple schematic $w=aA\cos ayG(z)$. No unspecified latitude dependence is silently needed for the sketch.

Since $A>0$, upper-layer meridional transport is northward and the deep return is southward. At constant $f$, the circulation rises on the southern side and sinks on the northern side. The meridional velocity changes sign at $z=-h+h^2/(2H)$, slightly above the forcing interface, not exactly at $-h$. The required schematic follows from this derived field:

Forcing comparison. The buoyancy and wind contributions have the same vertical shape and, with these signs, the same circulation sense. Their ratio isThe printed scaling $f{\tau }_0/{\rho }_0\simeq B_0$ must be understood in magnitude, since $f<0$ and both forcing amplitudes are positive. Surface confinement supplies the small factor $h/L$, making direct buoyancy forcing negligible in this circulation model.

The missing physical mechanism in the stratified Southern Ocean is eddy-induced overturning. The wind-driven mean Deacon cell tilts isopycnals and builds available potential energy; baroclinic instability produces mesoscale eddies whose buoyancy transport largely opposes the mean overturning. The residual circulation depends on the small net balance and the diabatic buoyancy transformation. Wind energy also enters mean kinetic energy; eddies redistribute momentum through form stress and transfer energy toward dissipative scales and bottom friction. The displayed linear model already balances its wind input by imposed drag, but omits this mean-advection/eddy buoyancy budget, so the necessity of an opposing eddy circulation is a physical extension, not another mathematical constraint on the four given equations.

Use a top-hat plume model per unit span. Define volume and kinematic momentum fluxes $Q=bw$ and $J=b{w}^2$, and set the pure-ice reduced gravity $g^{\prime }=g({\rho }_0-{\rho }_s)/{\rho }_0>0$. The suspension buoyancy is $g^{\prime }\varphi$. The supplied equations express vertical momentum conservation, conservation of relative enthalpy including latent heat, and conservation of excess salt, respectively. At the release height, prescribe compatible source volume/momentum, salt and enthalpy fluxes, or equivalent source width, velocity, composition and ice fraction; impose ambient entrained water at $(T_0,C_0)$ with no vertical momentum. A pure far-field similarity solution may instead be described using a virtual origin, but its singular near-origin values are not physical inlet conditions.

The missing balance is total mass, equivalently volume under the Boussinesq approximation. In a wall plume, only the exposed outer edge entrains, so the Batchelor entrainment hypothesis $v_e=ew$ givesIf two free edges are intended, replace $e$ by $2e$ everywhere; the numerical coefficient is a geometry convention. The glacier boundary itself has no normal flow. Local phase conversion does not add volume at retained Boussinesq order, and crystals are assumed to share the liquid velocity.

For a box of height $dz$, the upward momentum flux changes by ${\rho }_0[J(z+dz)-J(z)]$. The ambient hydrostatic pressure has been subtracted; the remaining force is the suspension buoyancy ${\rho }_{0}b{g}^{\prime }\varphi dz$. Entrained ambient water carries no vertical momentum, and drag is neglected, giving $J^{\prime }=b{g}^{\prime }\varphi$. Salt entering through the exposed edge is $C_0{Q}^{\prime }dz$, so $(\bar{C}Q{)}^{\prime }=C_0{Q}^{\prime }$ and thereforeThese control-volume derivations account for entrainment rather than treating $Q$ as constant.

Thermodynamic elimination. Let the conserved excess enthalpy and salt fluxes be $E$ and $S$:The ambient satisfies $T_0=-m{C}_0$, while equilibrium in the ice-bearing plume givesSubstitute this temperature in the enthalpy invariant. Exact rearrangement givesThus, at dilute-crystal order,The ice-bearing wall-plume thermodynamic invariant fixes the leading advected ice volume and buoyancy flux from the source deficits. A rising ice-dominated branch requires $I_0>0$. It is not determined from reduced gravity alone. The small-$\varphi$ expansion also presumes finite source fluxes and no cancellation making the discarded terms comparable to $-E-c_{p}mS$.

The three remaining leading equations are $Q^{\prime }=ew$, $(Qw{)}^{\prime }=\mathcal{F}/w$, and $\varphi =I_0/Q$. Seek a similarity form in $Z=z-z_v>0$ with $Q\propto Z^p$, $w\propto Z^q$. Continuity gives $p=q+1$, while momentum gives $p+2q=1$. Therefore $q=0,p=1$. The coefficients obey $e{w}^3=\mathcal{F}$, giving the self-similar ice-bearing wall plumeThe width grows linearly, the speed approaches a constant, and the crystals dilute inversely with height. This is a buoyancy-conserving wall line plume, with buoyancy supplied mainly by the ice fraction rather than a temperature anomaly. The dilute-crystal regime requires $Z\gg w^2/g^{\prime }$ and lies away from the formal virtual-origin singularity. If all source flux deficits vanish, $I_0=0$ and this nontrivial rising branch does not exist. The exact thermodynamic invariants above, rather than their truncated version, are needed near a concentrated source.

Use the nonrotating linear Boussinesq approximation with reference density ${\rho }_0$ and $N^2=-(g/{\rho }_0)d\bar{\rho }/dz>0$:Let $K_h^2=k^2+l^2>0$, $K^2=K_h^2+m^2$ and $E=e^{i(kx+ly+mz-\omega t)}$. The horizontal momentum equations give $u^{\prime }=k{p}^{\prime }/({\rho }_0\omega )$ and $v^{\prime }=l{p}^{\prime }/({\rho }_0\omega )$. Continuity requires $k{u}^{\prime }+l{v}^{\prime }+m{w}^{\prime }=0$. Thus the internal-wave polarization isPhysical perturbations are the real parts. The density is in temporal quadrature with the vertical velocity, because buoyancy responds to the vertical fluid displacement. Substitution into the vertical momentum equation determines the internal gravity wave dispersion relation,A nontrivial propagating wave requires a nonzero horizontal wave vector and a nonzero frequency. The formulas are not to be divided by $K_h^2=0$: incompressibility excludes a nonzero oscillatory vertical velocity for a purely vertical wave vector.

Write $G=g^{\prime }$ for the positive reduced gravity of dense fluid. With $z$ upwards, $G_z>0$ places denser fluid above lighter fluid and permits convective turbulence. A displacement over the radius scale $d$ samples a buoyancy difference of order $d{G}_z$; balancing kinetic energy per mass with buoyant work over $d$ givesThe buoyancy-gradient mixing-length closure then gives an eddy diffusivity $K\sim ud=d^2\sqrt{G_z}$. We use the paper's unit-prefactor convention for this dimensional closure and its effective cross-sectional area $d^2$; order-one mixing constants and the cylinder's geometric factor $\pi$ are suppressed in its displayed evolution equation.

The downward reduced-gravity transport magnitude and its convergence are different quantities:The printed expression labelled a flux is actually the second quantity, a flux convergence per unit height. The dimensions distinguish them: $\mathcal{F}$ has units $L^4{T}^{-3}$, whereas $F_{\mathrm{conv}}$ has units $L^3{T}^{-3}$. A constant positive gradient has nonzero transport but zero convergence, providing a direct counterexample to identifying the derivative with the transport itself.

The upward total buoyancy flux is $QG-\mathcal{F}$. Conservation therefore yieldsThis flux convergence in turbulent buoyancy mixing obtains the displayed evolution equation with the correct transport interpretation.

For the arrested cubic buoyancy profile, an undisturbed region below $-H$ has $G=0$ and no turbulent buoyancy flux. The joining condition is $G(-H)=0$ andIn a steady state, integrating the conservation equation from that front gives $QG=d^4(G_z{)}^{3/2}$. For a nonzero mixed region,and henceThe associated gradient is $Q^2(z+H{)}^2/(9d^8)$, so it matches the required zero-gradient state continuously. We take the mixed interval to start immediately above $-H$; adding a further zero interval would merely relocate the arrest front.

The source buoyancy flux is $B_s=Q_s{g}_s^{\prime }$, with $g_s^{\prime }=g\Delta \rho /\rho$ in the Boussinesq approximation. The stated top-source convention means that this entire flux enters the downward turbulent transport there: $\mathcal{F}(0)=B_s$. Since steady transport also satisfies $\mathcal{F}(0)=QG(0)$,This is the extent in the paper's dimensional-closure normalization. It presupposes positive $Q$; with no upflow there is no finite steady arrest depth. For a mixture of the two original fluids, $G(0)\le g_s^{\prime }$ is necessary, so an extrapolation to $Q_s/Q>1$ cannot represent a physical concentration profile under this top-flux boundary model. A dilute-source regime avoids appreciable source-volume corrections.

Assuming the same scalar-mixing coefficient for a passive tracer, its eddy diffusivity in the mixed region isIt vanishes at the arrest front and increases linearly towards the source. A different turbulent scalar diffusivity ratio would multiply this result by its corresponding constant.

The geometric normalization can be restored without changing the reasoning. With actual area $S=\pi d^2$ and $K=C_m{d}^2\sqrt{G_z}$, put $R=S{C}_m{d}^2$. Then $G=Q^2(z+H{)}^3/(27R^2)$, $H=3R^{2/3}{B}_s^{1/3}/Q$, and $K=Q(z+H)/(3S)$. These formulas show which numerical prefactors are convention-dependent.

Put $P=p/{\rho }_0$ and use the downward buoyancy anomaly $b=g{\rho }^{\prime }/{\rho }_0$. This sign convention is required by the printed potential vorticity. The linear rotating Boussinesq approximation givesLet $D=u_x+v_y$ and $\zeta =v_x-u_y$. Horizontal divergence and curl give $D_t-f\zeta =-{\nabla }_H^{2}P$ and ${\zeta }_t=-fD=f{w}_z$. HenceThus the linear potential vorticity is a time-independent field fixed by the initial data. Differentiate the pressure identity twice. Using ${\zeta }_{tt}=-f(P_{zz}+b_z)$ and $b_{ztt}=-N^2(P_{zz}+b_z)$ gives ${\nabla }^2{P}_{tt}=(N^2-f^2)(P_{zz}+b_z)$. Combining terms,The linear pressure equation for rotating stratified flow has source $Q={\rho }_{0}f{N}^{2}q$. “Arbitrary” means that different initial potential-vorticity fields give different time-independent sources; it is not independent forcing that may vary in time. If buoyancy were instead defined upward as $-g{\rho }^{\prime }/{\rho }_0$, both appearances of its sign would change consistently. The pressure source represents the balanced component alongside propagating inertia-gravity waves.

Rotating Rayleigh-Bénard convection combines buoyancy-driven instability with the Coriolis force. Consider a plane layer of depth $d$ rotating uniformly about the vertical axis, heated from below. Adopt the Boussinesq approximation, fixed boundary temperatures and, for explicit formulas, impermeable stress-free boundary conditions. The conductive state of Rayleigh-Bénard convection is motionless with a linear temperature profile. The dimensionless controls are the Rayleigh number, Prandtl number and Taylor number:Here $\nu$ is kinematic viscosity and $\kappa$ thermal diffusivity. In thermal-diffusion time units, linear perturbations satisfyRotation does no direct mechanical work, since $\mathbf{u}\cdot ({\mathbf{e}}_z\times \mathbf{u})=0$, but couples vertical motion to vertical vorticity and changes the damping and oscillation balance.

For horizontal wavenumber $k$ and vertical mode $n$, put $m=n^2{\pi }^2$ and $a=k^2+m$. Use $w=W\sin (n\pi z)$, $\theta =\Theta \sin (n\pi z)$ and vertical vorticity $\zeta =Z\cos (n\pi z)$ with growth rate $\sigma$. Curling the momentum equation and eliminating pressure givesTheir determinant, without dividing by a possibly zero factor, is the rotating-convection growth-rate polynomialThis makes the linear stability analysis question precise: onset occurs when a root reaches zero real part and all other modes still decay.

A stationary neutral root has $\sigma =0$, givingRotation raises this stationary threshold. An oscillatory neutral root has $\sigma =i\varpi$ with $\varpi \ne 0$. Real-imaginary separation gives the oscillatory neutral curve of rotating convectionThis branch is admissible only when ${\varpi }^2>0$, requiring $P<1$ and sufficiently strong rotation. The restoring Coriolis force coupling permits an inertial/thermal oscillation whose phase-lagged buoyancy can overcome dissipation. At large $P$, temperature and momentum diffusion do not permit that overstability mechanism at primary onset, so the exchange of stabilities is stationary. The actual threshold is the minimum of the stationary and admissible oscillatory curves over all allowed modes, not an arbitrary formal value of $R_o$. Rigid plates require a different vertical eigenproblem and Ekman layers, so the explicit free-slip numbers are not universal.

At large Taylor number, the first vertical mode is selected in the ideal plane layer. Let $t=k^2,m={\pi }^2$. Minimizing the stationary curve givesThus the stationary neutral curve of rotating convection hasThe physical horizontal wavelength is $2\pi d/k_c$, hence decreases as ${\mathrm{Ta}}^{-1/6}$; its prefactor depends on the boundary convention. Thin nearly vertical cells reconcile the strong Coriolis force constraint with viscous and thermal diffusion. The oscillatory minimization replaces the right side of the wavenumber equation by $P^2\mathrm{Ta}m/(1+P{)}^2$, giving the same Taylor number exponent at fixed positive $P$. Where its frequency remains admissible,Equality is $8P^4-P-1=0$, whose positive root is approximately $P_{\ast }=0.6766$. Accordingly, for sufficiently rapid rotation in this free-slip problem, $P<P_{\ast }$ selects oscillatory onset and $P>P_{\ast }$ stationary onset. The weaker condition $P<1$ is only necessary for an oscillatory neutral mode; it does not by itself identify the first instability. Finite Taylor number, finite lateral geometry, allowed discrete wave numbers and plate conditions change the selection.

For the counterpropagating Hopf amplitudes in rotating convection near a simple oscillatory onset, the Hopf bifurcation produces slow complex amplitudes for counterpropagating roll waves. After separating the fast carrier oscillation, symmetry permits the cubic equationswith generally complex coefficients; an $i{\omega }_0{Z}_{\pm }$ term restores the fast frequency if desired. The real parts govern amplitude saturation and the imaginary parts give nonlinear frequency shifts. Write $a_s=\mathrm{Re}{g}_s$, $a_c=\mathrm{Re}{g}_c$. For a travelling wave from a supercritical bifurcation with only one amplitude nonzero, $|Z_{\pm }{|}^2=r/a_s$ with $r,a_s>0$, and the competing wave's growth rate is $r(1-a_c/a_s)$. It is amplitude-stable against that competitor when $a_c>a_s$. A standing wave has equal intensities $r/(a_s+a_c)$; provided this is positive, its intensity-difference mode is stable when $a_s>a_c$. Temporal and spatial phase symmetries leave neutral phase directions, so these are orbital/amplitude stability statements, not decay of every phase displacement.

These coefficients follow from nonlinear interactions and the Fredholm solvability condition obtained by projection onto an adjoint eigenfunction; symmetry alone cannot decide their signs. A negative saturating coefficient gives subcritical bifurcation behavior requiring higher-order terms. Spatial modulation leads to coupled complex Ginzburg–Landau equations with group velocities and diffusion; phase instabilities, mean-flow coupling and differently oriented rolls can destabilize a wave stable in the restricted two-amplitude system. A weakly nonlinear expansion of oscillations therefore predicts travelling waves or standing waves, frequency shifts, modulation and possible secondary mode competition, not a unique universal periodic state.

The Küppers–Lortz instability is a different route to time dependence: it destabilizes steady saturated rolls against oblique roll perturbations. For stationary-roll amplitudes of orientations ${\varphi }_j$, a leading competition system hasA pure roll has $|A_i{|}^2=r/g_0$ with $g_0>0$. An infinitesimal new roll at relative angle $\vartheta$ grows atFor sufficiently strong rotation in appropriate boundary and Prandtl number regimes, some finite oblique angle has $g(\vartheta )<g_0$, so a steady roll is unstable arbitrarily close above its stationary onset. Rotation is handed and allows $g(\vartheta )\ne g(-\vartheta )$, so replacement of one roll by another can favor a definite cyclic sense. Three or more competing orientations can form a heteroclinic cycle; whether it attracts depends on contraction/expansion rates and other modes. Noise, spatially varying domains and modulation can turn this competition into repeated orientation switching and irregular patterns.

The invading rolls are three-dimensional disturbances even when the original straight roll is described by a two-dimensional section. The finite-angle Küppers–Lortz instability mechanism should also be distinguished from the small-angle instability of rotating convection rolls mediated by large-scale mean flow at finite Prandtl number. Numerical thresholds and favored angles depend on mechanical boundaries and material parameters; the essential criterion is the cross-coupling relative to self-saturation. Rotation both changes primary onset and wavelength, and can prevent the resulting steady roll pattern from remaining a stable nonlinear state.

Let $K$ be intrinsic permeability of a porous medium, $\varphi$ the mobile pore fraction, and let $h$ measure thickness normal to the sloping cap. Define cold and reservoir-temperature properties byThe imposed inequality makes ${\delta }_c>0$, so the cold fluid is still buoyant. Assume $K>0$, $\sin \theta >0$, positive densities and viscosities, and the usual positive thermal-expansion and viscosity coefficients. Away from localized fronts, neglect the along-current thickness gradient. Hydrostatic water pressure and Darcy's law then give the upslope Darcy velocitiesTypically $u_h>u_c$: warming increases buoyancy and decreases dynamic viscosity.

Current depths and the moving moving thermal front in a porous current. For this slope-driven porous gravity current, the injected mass flux per unit well length fixes the cold plateau thickness:The moving thermal front in a porous current moves at $V_T=\Gamma u_c$. Since $u_c$ is explicitly a Darcy velocity, the fluid mass stored per unit horizontal area is $\varphi \rho h$. Apply mass conservation across the moving moving thermal front in a porous current:Writing $\gamma =\varphi \Gamma$ and $r=u_c/u_h$, this gives the distal warm depthFor a retarded moving thermal front in a porous current take $0\le \gamma <1$; otherwise this two-plateau geometry needs reconsideration. The factor $(1-\gamma )/(1-\gamma r)$ is important: the distal mass flux is not generally $Q$, because mass is being stored as the cold region replaces the warm region. Setting the two fluxes equal would silently assume a stationary moving thermal front in a porous current.

If the model absorbs porosity into storage and uses pore velocity as its transport speed, the same formulas use $\gamma =\Gamma$. The explicit Darcy convention in the PDF instead gives $\gamma =\varphi \Gamma$; porosity must be specified or absorbed consistently. In the stationary-front limit $\gamma =0$, the two steady depths reduce to $Q/({\rho }_c{u}_c)$ and $Q/({\rho }_h{u}_h)$. In the Boussinesq approximation, use a common reference density in the mass factors while retaining ${\delta }_c,{\delta }_h$ in the driving force.

Leakage thresholds. A normal thickness $h_i$ generates cap overpressure ${\Pi }_i={\delta }_{i}g{h}_i\cos \theta$. Before leakage modifies the current, the two plateau values areFor the caprock leakage threshold $\Delta p>0$, the candidate critical mass-injection rates are thereforeCold-only leakage occurs for $Q_c^{\ast }\le Q<Q_h^{\ast }$, provided $Q_c^{\ast }<Q_h^{\ast }$. Both plateaus can leak once $Q\ge \max (Q_c^{\ast },Q_h^{\ast })$. In the usual common-density approximation, ${\mu }_c>{\mu }_h$ and $r<1$ ensure $Q_c^{\ast }<Q_h^{\ast }$, giving the expected sequence: no leakage, cold-only leakage, then leakage from both cold and warm regions. Without that approximation the ordering must be checked; the given inequality $\Delta \rho >\alpha \Delta T$ alone does not establish it. These are onset criteria computed on the nonleaking current, not a post-leakage mass budget.