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The dimensionless density contrast $({\rho }_h-{\rho }_l)/({\rho }_h+{\rho }_l)$, conventionally using heavy and light densities. In the inviscid, zero-surface-tension heavy-over-light Rayleigh-Taylor instability, the growth rate is $s=\sqrt{\mathcal{A}gk}$. State the signed ordering if labels are not heavy/light.

More suspended heavy particles above fewer particles puts denser fluid above lighter fluid. Such a particle suspension may develop Rayleigh-Taylor instability and overturning in addition to one-dimensional kinematic sedimentation. A rarefaction solution alone does not establish hydrodynamic stability.

For a deep inviscid liquid above a gas of negligible inertia, an inverted free surface has dispersion relation ${\omega }^2=q(\sigma q^2/\rho -g)$, $q=|k|$. The Rayleigh-Taylor instability affects $0<q<k_c=\sqrt{\rho g/\sigma }$. A bounded aperture can suppress it if its smallest allowed wavenumber is at least $k_c$. The capillary length is ${\ell }_c=k_c^{-1}$ and the marginal wavelength is $2\pi {\ell }_c$. A pinned one-dimensional span of width $L$ with a half-sine lowest mode instead has $k_{min}=\pi /L$, giving $L\le \pi {\ell }_c$. A numerical maximum container size therefore also requires aperture geometry and contact-line boundary conditions; the marginal wavelength alone is not a universal diameter.

For two resting incompressible inviscid layers between impermeable horizontal walls, the upper layer has density ${\rho }_1$ and depth $L_1$, and the lower layer has ${\rho }_2<{\rho }_1$ and depth $L_2$. Solving the Laplace equation for the velocity potential and imposing the linearized kinematic and capillary pressure conditions gives the displayed dispersion relation. With positive surface tension $\gamma$, exponential Rayleigh-Taylor instability occurs for $0<k<k_c$, where $k_c=\sqrt{g({\rho }_1-{\rho }_2)/\gamma }$. The depths change the growth rate, but not this cutoff.

For two inviscid layers of equal depth $h$ between rigid lids, with lower density ${\rho }_2$ and upper density ${\rho }_1$, a horizontal mode of wavenumber $k$ satisfiesThe sign reverses when the heavier fluid is above, producing the Rayleigh-Taylor instability.

For the density jump of a magnetic pressure discontinuity in an isothermal atmosphere, with $0<|B_{+}|<|B_{-}|$, the fastest Rayleigh-Taylor instability has $m=0$ and $|k|=K_{\ast }=(B_{-}^2-B_{+}^2)/(4B_{+}^{2}H)$. Its vertical decay length is $K_{\ast }^{-1}$. Hence localization relative to both common scale heights requires $4B_{+}^2\ll B_{-}^2-B_{+}^2$. This also makes its growth speed small compared with the isothermal sound speed. If $B_{+}=0$, the ideal sharp-interface dispersion relation instead has unbounded growth as $|k|\to \infty$: there is no finite maximizing mode.

Two uniform horizontal magnetic fields on either side of a horizontal interface exert no bulk force, but their magnetic pressures differ. For a common-temperature isothermal atmosphere with sound-speed squared $c_T^2=\mathcal{R}{T}_0$ and gravity $g$, hydrostatic equilibrium gives $p_i=c_T^2{\rho }_i$ and $p_i(z)=p_i(0)e^{-z/H}$, $H=c_T^2/g$. Continuity of magnetohydrodynamic total pressure gives the displayed density jump. A stronger field below means denser gas above, supplying a Rayleigh-Taylor instability even though each half-space is hydrostatic.

// Target: continuum-mechanics.bigb

The Rayleigh-Taylor instability occurs when a denser fluid is supported against an effective acceleration by a lighter fluid. Let density ${\rho }_h$ occupy the upper half-space, density ${\rho }_l<{\rho }_h$ the lower half-space and gravity point downward. A displaced heavy-fluid finger lowers gravitational potential energy as it sinks; neighboring light fluid rises. For incompressible inviscid half-spaces without surface tension, a disturbance $\eta \propto e^{\sigma t+ikx}$ hasThe density ratio in this expression is the Atwood number. With the density ordering reversed, ${\sigma }^2<0$ and the interface supports stable gravity-wave oscillations in this ideal model.

Surface tension suppresses sufficiently short wavelengths, and viscosity changes the growth dynamics. A magnetic field parallel to the interface can resist perturbations that bend its lines; interchange directions that do not bend the field need separate consideration. In astrophysics, effective deceleration can make dense ejecta or a swept-up shell supported by a lower-density hot interior unstable. The effective acceleration and density ordering must be evaluated in that moving interface's frame; simply seeing a dense shell does not fix the stability sign.

A local thermal imbalance can amplify a gas perturbation even without an initial displacement of an interface. Let $\mathcal{L}(\rho ,T)$ denote net cooling per unit mass, including heating with a negative sign, and suppose $\mathcal{L}({\rho }_0,T_0)=0$. When the sound-crossing time is short compared with the thermal evolution time, pressure equilibrates and an ideal gas perturbation is approximately isobaric. The linearized thermal equation isThus a negative constant-pressure derivative produces isobaric thermal instability: a cooler parcel experiences a net loss that cools it further, while a warmer parcel heats further. For example, isobaric thermal stability of a heated ideal gas with $\mathcal{L}=A\rho T^{\alpha }-H$ gives $\sigma =(1-\alpha )H/(c_p{T}_0)$ and instability for $\alpha <1$. Different heating laws and fixed-density perturbations give different criteria; thermal conduction damps sufficiently short-scale temperature contrasts. Such feedback helps a gas separate into denser cool and rarer warm regions when stable thermal branches exist.

The Rayleigh-Taylor instability instead releases gravitational potential energy when a heavier fluid lies above a lighter fluid in a downward gravitational field. For two inviscid incompressible half-spaces with no surface tension or magnetic field, a horizontal perturbation of wavenumber $k$ hasThe Atwood number measures the density contrast. A descending heavy-fluid finger and an ascending light-fluid bubble both lower the total gravitational potential energy, increasing the original displacement. Interchanging the layers gives an oscillatory stable interface. An accelerating or decelerating astrophysical shell has an analogous effective acceleration, allowing mixing at suitable ejecta interfaces in a supernova remnant. Finite interface width, viscosity, surface tension and magnetic tension modify the short-wavelength behavior; the ideal sharp-interface formula does not select a finite fastest-growing wavelength.

The Rayleigh-Taylor instability extracts gravitational potential energy when a heavy fluid is supported above a lighter one. A small downward excursion of heavy material and upward excursion of light material can grow instead of experiencing restoring buoyancy. The Kelvin-Helmholtz instability instead extracts kinetic energy from a velocity difference across an interface or shear layer. The interface displacement and its induced flow reinforce one another by tapping the background shear; a reversed density ordering is not required.

A useful common model has two inviscid, incompressible, semi-infinite fluids separated at $z=0$, with lower density/velocity ${\rho }_1,U_1$, upper density/velocity ${\rho }_2,U_2$, downward gravity $g$, and surface tension $\sigma$. For a disturbance proportional to $e^{i(kx-\omega t)}$, with $k>0$, the two-fluid interfacial gravity-shear dispersion relation isA negative right side gives exponential growth. With no shear or surface tension and ${\rho }_2>{\rho }_1$, the growth rate of the Rayleigh-Taylor instability is $s=\sqrt{\mathcal{A}gk}$, where $\mathcal{A}=({\rho }_2-{\rho }_1)/({\rho }_2+{\rho }_1)$ is the Atwood number. With equal densities and no restoring force, the growth rate of the Kelvin-Helmholtz instability is $s=k|U_1-U_2|/2$. Heavy-over-light stratification drives Rayleigh-Taylor growth; velocity shear drives Kelvin-Helmholtz growth. A heavy lower fluid supplies gravitational stabilization, and surface tension stabilizes sufficiently short waves in this ideal interface model.

Finite shear-layer thickness, viscosity, compressibility and magnetic tension can change the growth rates and unstable wavelength range. In particular, unlimited growth with increasing $k$ is a limitation of the zero-thickness, zero-surface-tension idealization. Density-interface overturning and jet-boundary mixing are appropriate physical applications of the two mechanisms, but the nonlinear mixed state is not determined by the linear growth rate alone.

For an inverted free surface, gravity promotes a displacement instead of restoring it. The surface tension term still resists curvature. Hence the gravity-capillary wave relation becomes, with $q=|k|$,For $0<q<k_c$, where $k_c=\sqrt{\rho g/\sigma }$, ${\omega }^2<0$. Writing the displacement as $\eta \propto e^{st+ikx}$ gives the Rayleigh-Taylor instability growth rateShorter wavelengths, $q>k_c$, remain oscillatory because surface tension dominates. The marginal scale is determined by the capillary length,A sufficiently small aperture excludes the long unstable modes: this is finite-aperture stabilization of an inverted liquid surface. The estimate requested is therefore a container opening of order one centimetre, with $1.7\mathrm{cm}$ the marginal full wavelength rather than a geometry-independent maximum diameter.

The exact largest opening cannot be fixed from the stated infinite-surface dispersion relation alone. It requires the aperture shape and contact-line boundary conditions. For example, a one-dimensional span of width $L$ whose contact line is pinned at both ends has a half-sine lowest mode with $q_{\min }=\pi /L$. In that simple model,Different geometries change the numerical factor. This is a small-disturbance stability estimate with negligible air inertia and no leakage or contact-line failure; a guarantee that a real inverted container retains its water needs those additional boundary details.

Write $\eta =H{e}^{\sigma t}{J}_0(\alpha r)$ using the Bessel function of the first kind. The wall boundary conditions at $z=\pm h$ selectBecause $J_0^{\prime }(s)=-J_1(s)$, the sidewall no-penetration boundary condition requireswhere $j_{1,n}$ are the positive zeros of $J_1$. These radial eigenfunctions automatically preserve layer volume becauseThe constant radial mode $\alpha =0$ cannot change the interface height in a closed tank: the Laplace equation and the impermeable lids then give zero vertical velocity, and a uniform displacement would change each layer's volume.

The kinematic boundary condition fixesThus the interface values are ${\varphi }_1(0)=\sigma \eta \coth (\alpha h)/\alpha$ and ${\varphi }_2(0)=-\sigma \eta \coth (\alpha h)/\alpha$. Substitution in the normal stress condition gives the finite-depth Rayleigh-Taylor dispersion relation:soHere $\mathcal{A}$ is the Atwood number. Every allowed mode has a positive growth rate and is subject to Rayleigh-Taylor instability. Since $d[\alpha \tanh (\alpha h)]/d\alpha =\tanh (\alpha h)+\alpha h{\mathrm{sech}}^2(\alpha h)>0$, shorter radial scales grow faster. The limits areThere is no fastest finite wavenumber in the ideal sharp-interface model without surface tension, viscosity or diffusion. The zero-initial-velocity condition uses both exponential branches: an initial mode $H{J}_0({\alpha }_{n}r)$ evolves as $H\cosh ({\sigma }_{n,+}t)J_0({\alpha }_{n}r)$. If the mass density ordering were reversed the same calculation would give oscillatory stable modes; equal mass densities give neutral modes.

For an inviscid flow, the force per unit volume is $\mathbf{F}=-\nabla p-\rho g\hat{\mathbf{z}}$. The gravitational acceleration has zero curl. Using the curl product rule,The supplied two-dimensional vorticity equation therefore becomesThe out-of-plane vorticity has no stretching term in this two-dimensional setting. The source is baroclinic vorticity generation: differently directed mass density and pressure gradients generate rotation even from zero initial vorticity.

For an explicit sign check, set ${\omega }_y={\partial }_{z}u-{\partial }_{x}w$, with $z$ upward. A thin smoothed interface has $\rho \simeq \bar{\rho }(z-\eta (x,t))$, with ${\bar{\rho }}^{\prime }>0$. If $P(x,t)$ denotes the total pressure evaluated on the interface, then within the thin layer $p_x+{\eta }_x{p}_z\simeq P_x$ andIntegrating through that layer and taking its thickness to zero gives the vortex sheet strength $\gamma =u_2-u_1$ and its linear evolution:For a growing Fourier mode, the potentials from the preceding calculation with $\alpha$ replaced by $k$ giveThus the right-hand descending slope of a light-fluid crest produces positive ${\omega }_y$, while the left-hand ascending slope produces negative ${\omega }_y$. Their induced velocity raises the light-fluid crest and lowers the heavy-fluid trough, increasing the slope and generating more vorticity. This is the positive feedback underlying Rayleigh-Taylor instability.

One can close this mechanism quantitatively. For $\eta =H(t)\cos (kx)$, the kinematic boundary conditions give $\gamma =2\dot{H}\coth (kh)\sin (kx)$. Combining with the sheet-generation equation givesFor $kh\ll 1$, the lids confine motion throughout both layers and horizontal redistribution is slow, yielding $\sigma \sim \sqrt{\mathcal{A}gh}k$. For $kh\gg 1$, motion is localized within depth $k^{-1}$ on each side. Less fluid must accelerate at shorter wavelength, giving $\sigma \sim \sqrt{\mathcal{A}gk}$. Since $k_n=n\pi /b$, the ideal growth rate increases with $n$ with no short-wave cutoff.

The sharp-interface result requires $k\delta \ll 1$ for the smoothed layer. When $k\delta$ becomes order one, it no longer applies. A smooth adverse mass density gradient has a local overturning time of order $[g{\bar{\rho }}^{\prime }/\bar{\rho }{]}^{-1/2}$ rather than the unbounded sharp-interface rate. Finite molecular diffusion also relaxes small-scale mass density perturbations on a time of order $(D{k}^2{)}^{-1}$, and viscosity, if included, damps small-scale velocity. Weak diffusion motivates the thin-layer picture; it does not justify extrapolating the discontinuous dispersion relation to arbitrary $n$.

Use the standard ideal-fluid assumptions underlying the stated relation: both fluids are incompressible, inviscid and unbounded away from the interface, with no surface tension. An irrotational flow has $u_j=\nabla {\varphi }_j$, and incompressibility gives $\Delta {\varphi }_j=0$ in each fluid. The unsteady Bernoulli equation iswhere the arbitrary time functions can be absorbed into the potentials. On $z=\zeta (x,t)$ the material-interface and normal-stress conditions areDisturbance velocities decay as $z\to \pm \infty$. The equilibrium pressures are $p_{\ast }-{\rho }_{j}gz$.

Linearize at the resting flat interface. The equations become $\Delta {\varphi }_j=0$, ${\zeta }_t={\varphi }_{j,z}$ on $z=0$, andFor $k>0$, write $\zeta =\eta e^{i(kx-\omega t)}$. Decay selects ${\varphi }_1=A_1{e}^{-kz}{e}^{i(kx-\omega t)}$ above and ${\varphi }_2=A_2{e}^{kz}{e}^{i(kx-\omega t)}$ below. The kinematic conditions give $A_1=i\omega \eta /k$ and $A_2=-i\omega \eta /k$. Substituting into the dynamic condition givesFor a signed wave number replace $k$ by $|k|$. A heavier lower fluid gives stable interfacial gravity waves; a heavier upper fluid gives ${\omega }^2<0$ and Rayleigh-Taylor instability. The density ordering is not silently assumed positive in the derivation.

Write $c_T^2=\mathcal{R}{T}_0$ for the squared isothermal sound speed, distinguishing the gas constant $\mathcal{R}$ from a Rayleigh number, and put $H=c_T^2/g$. A uniform magnetic field has no bulk Lorentz force, so hydrostatic equilibrium and the ideal gas equation giveAt the interface, it is magnetohydrodynamic total pressure, not gas pressure alone, that is continuous:Define ${\Delta }_B=B_1^2-B_2^2>0$. The magnetic pressure discontinuity in an isothermal atmosphere therefore producesIn particular, the upper gas is denser at the interface. Both halves have the same scale height $H$, despite their different pressure normalizations.

For the disturbance calculation, write ${\rho }_{-}=\rho$, ${\rho }_{+}=\rho +\delta \rho$, where $\delta \rho ={\Delta }_B/(2{\mu }_0{c}_T^2)$, and ${\Sigma }_{\rho }={\rho }_{-}+{\rho }_{+}$. Work to leading order in the prescribed short-vertical-scale approximation: the bulk motion is locally incompressible potential flow, while the isothermal equation determines the equilibrium density jump. Its scale separation and small compressive correction will be checked at the maximizing mode.

Let the interface displacement be $h{e}^{st+ikx+imy}$ and $K=\sqrt{k^2+m^2}>0$. Using displacement rather than velocity potentials, write ${\mathbf{\xi }}_i=\nabla {\chi }_i$ and ${\mathbf{u}}_i=s{\mathbf{\xi }}_i$. Incompressibility and irrotational flow give $\Delta {\chi }_i=0$. Decay away from the interface and continuity of its normal displacement fixwith the common horizontal and temporal exponential understood. Both satisfy ${\partial }_z{\chi }_i(0)=h$. Thus the disturbance has vertical extent of order $K^{-1}$.

In ideal magnetohydrodynamics, magnetic flux freezing gives ${\mathbf{b}}_i=({\mathbf{B}}_i\cdot \nabla ){\mathbf{\xi }}_i$ for these locally incompressible displacements. If ${\pi }_i=p_i^{\prime }+{\mathbf{B}}_i\cdot {\mathbf{b}}_i/{\mu }_0$ is the total-pressure perturbation, linear momentum isSince $({\mathbf{B}}_i\cdot \nabla {)}^2=-({\mathbf{k}}_h\cdot {\mathbf{B}}_i{)}^2$ on the mode, integration in each half-space yieldsAt the displaced interface, continuity of Lagrangian magnetohydrodynamic total pressure requires ${\pi }_1-{\rho }_{-}gh={\pi }_2-{\rho }_{+}gh$. Inserting the two potentials gives the orthogonal-field magnetic Rayleigh-Taylor dispersion relation:The first term is the destabilizing Rayleigh-Taylor instability of the heavier upper gas; the second is stabilizing magnetic tension. A negative right-hand side gives an oscillatory mode rather than growth.

For fixed $K$, the magnetic penalty isIt is minimized by $m=0$: the wavevector is along the weaker upper magnetic field and perpendicular to the stronger lower field. If $B_2\ne 0$, define $A=g\delta \rho /{\Sigma }_{\rho }$ and $C=B_2^2/({\mu }_0{\Sigma }_{\rho })$. The growth law for this orientation is $s^2=AK-C{K}^2$, a concave quadratic with maximum at $K_{\ast }=A/(2C)$. HenceFor this orientation the unstable interval is $0<K<2K_{\ast }$; the maximum lies halfway along it.

The localization condition for the fastest orthogonal-field interface mode follows directly from the vertical decay length:This is the condition for ${\ell }_{\ast }\ll H$ on both sides. It also controls the locally incompressible approximation, sinceAcoustic adjustment is then fast compared with growth, so bulk density perturbations are higher-order corrections. If this localization condition fails, the constant-background approximation cannot be used to identify the physical fastest mode.

Finally, the printed assumptions allow $B_2=0$. In that case $m=0$ eliminates magnetic tension on both sides and gives $s^2=g\delta \rho |k|/{\Sigma }_{\rho }$. There is no finite maximum growth rate or maximizing wavenumber in the ideal sharp-interface model: growth is unbounded as $|k|\to \infty$. The finite maximum and localization formula above therefore require $0<|B_2|<|B_1|$.

In the growth curves, $\theta$ is the angle of the horizontal wavevector from the $x$-axis. The illustrated ratio $B_1^2/B_2^2=101$ gives $K_{\ast }H=25$, and shows how even a small rotation toward the stronger magnetic field lowers the maximum growth.

Incompressibility makes each perturbation velocity potential a harmonic function. A horizontal mode $e^{i(kx+\ell y)+st}$ has $q=\sqrt{k^2+{\ell }^2}>0$. Decay away from the interface selects ${\phi }_1=A_1{e}^{qz}{e}^{i(kx+\ell y)+st}$ below and ${\phi }_2=A_2{e}^{-qz}{e}^{i(kx+\ell y)+st}$ above. The kinematic boundary condition givesInsert these into the dynamic boundary condition for an inviscid interface and cancel a nonzero mode amplitude to obtainWriting ${\rho }_T={\rho }_1+{\rho }_2$ and $\bar{U}=({\rho }_1{U}_1+{\rho }_2{U}_2)/{\rho }_T$, completion of the square givesConsequentlyThe imaginary advection term does not change growth rates. A mode grows exponentially precisely when the radicand is positive, equivalentlyIf the upper fluid is heavier, every nonzero wavevector is unstable even without shear: this is Rayleigh-Taylor instability. If the lower fluid is heavier, gravity stabilizes sufficiently long waves, but any nonzero velocity difference destabilizes sufficiently short waves with $\ell =0$: this is Kelvin-Helmholtz instability. For equal densities, shear destabilizes modes with $k\ne 0$. A negative radicand gives purely imaginary frequencies and oscillatory modes; equality is the marginal threshold for exponential growth.

Let $z$ measure distance inward from the surface. While $a\ll R$, curvature is negligible and this is a planar two-phase Stefan problem. Use a common reference mass density $\rho$, specific heat capacity $c_p$, thermal conductivity $k$ and thermal diffusivity $\kappa =k/(\rho c_p)$; retain the mass density differences only in the subsequent buoyancy estimates. Define $\Delta T=T_m-T_s>0$, $\Theta =\overline{T}-T_m\ge 0$, and $\vartheta =\Theta /\Delta T$. The paper's parameter $S=L/(c_p\Delta T)$ is the latent-to-sensible heat ratio, reciprocal to the more usual Stefan number.

For superheated two-phase Stefan similarity, solving $T_t=\kappa T_{zz}$ with $T(0)=T_s$, $T(a)=T_m$ and $T(\infty )=\overline{T}$ gives the Neumann solution of the Stefan problem. With $\eta =z/(2\sqrt{\kappa t})$,The solid expression holds for $0<z<a$ and the liquid expression for $z>a$. The interface gradients inserted in the Stefan condition determine the positive constant $\lambda$:The right-hand bracket decreases from infinity and crosses zero, while the equivalent left-hand side $S\sqrt{\pi }\lambda e^{{\lambda }^2}$ increases, proving uniqueness. The liquid-side flux subtracts from the roof cooling: it is heat brought to the interface by the superheated interior.

For large $S$, $\lambda \ll 1$. Keeping the superheat term while using $\mathrm{erf}\lambda \simeq 2\lambda /\sqrt{\pi }$ and $\mathrm{erfc}\lambda \simeq 1$ yields $S{\lambda }^2+(\vartheta /\sqrt{\pi })\lambda \simeq 1/2$. Hence the small-interface-parameter growth with liquid superheat approximation isThe solid temperature is approximately linear across its thin crust, while the liquid thermal boundary layer has thickness of order $\sqrt{\kappa t}$. For fixed superheat and $S\gg 1$, $A\simeq \sqrt{2/S}$; for $\vartheta \gg \sqrt{S}$, $A\simeq \sqrt{\pi }/\vartheta$. This retains the strong suppression of growth by a very hot interior.

To estimate overturn, take local gravitational acceleration $g$ at the roof. The liquid's mass density contrast is $\Delta {\rho }_l={\rho }_f\alpha \Theta$, while the solid contrast is $\Delta {\rho }_s={\rho }_s-{\rho }_f$. A liquid thermal layer of thickness $\delta$ has Rayleigh number $\Delta {\rho }_{l}g{\delta }^3/({\mu }_l\kappa )$. As $\delta \sim \sqrt{\kappa t}$, it reaches its critical value after a time scaling asIn viscous overturn of a dense crust, an unstably stratified viscous crust has a Rayleigh-Taylor instability growth time of order ${\mu }_s/(\Delta {\rho }_{s}ga)$. Setting this equal to its age, with $a=A\sqrt{\kappa t}$, givesThese are formation-to-overturn times. A specified critical liquid Rayleigh number ${\mathrm{Ra}}_c$ multiplies ${\tau }_l$ by ${\mathrm{Ra}}_c^{2/3}$; logarithmic amplification and geometrical constants also affect the crust estimate. They are omitted for the requested order-of-magnitude scaling.

The given numbers yield $S\simeq 2.54$, $\Delta {\rho }_s=500\mathrm{kg}{\mathrm{m}}^{-3}$ and $\Delta {\rho }_l=100\vartheta \mathrm{kg}{\mathrm{m}}^{-3}$. ThusLiquid thermal overturn is enormously faster than crustal overturn throughout the stated range. Even a critical-Rayleigh prefactor of order $10^2$ does not change that ordering. Since $S\simeq 2.54$ is not asymptotically large, low-superheat coefficients from the permitted approximation are only approximate. At the upper end $\alpha \Theta =1.25$, so the linear thermal-density approximation also ceases to be quantitatively credible; that endpoint is a formal scaling estimate under the supplied constant-property model. At a common age the instantaneous viscous instability-time ratio instead scales as $5A10^{-18}/\vartheta$; it should not be confused with the above formation-time ratio.

For a boundary-layer renewal model, the liquid layer reaches ${\delta }_l\sim \sqrt{\kappa {\tau }_l}$. Its cycle-averaged thermal flux is consequentlyIf a critical Rayleigh number is retained, insert a factor ${\mathrm{Ra}}_c^{-1/3}$. Averaging the diffusive $t^{-1/2}$ flux over a cycle only changes an order-one coefficient. The crust reachesso its average solid-volume flux per roof area is the crust-removal fluxThe corresponding mass flux is ${\rho }_s{J}_s$ and latent-heat flux is ${\rho }_{s}L{J}_s$. This independent-cycle estimate is a scale for a crust that reaches overturn thickness and is then renewed. Because the liquid renews much more rapidly, in a hot interior its average heat flux can limit actual crust growth: the consistent interfacial budget is $\rho L\dot{a}\simeq k\Delta T/a-F_l$. A stalled crust would not realize repeated solid cycles. In a superheated interior, detached solid may also remelt; permanent core deposition is most directly justified in the late regime requested next.

At late times $\Theta =0$, liquid thermal buoyancy vanishes and $A^2\kappa \simeq 2\kappa /S=2k\Delta T/(\rho L)$. ThusAssuming thin renewed crust, efficient sinking and constant asteroid radius, conservation of mass gives $4\pi b^2\dot{b}\simeq 4\pi R^2{J}_0$ under the permitted equal-density approximation. Therefore the asteroidal core growth by crustal sinking law isFor a self-gravitating approximately uniform sphere, $g\simeq 4\pi G\rho R/3$, so $J_0\propto R^{1/3}$ and a zero-seed core scales as $b\propto R^{7/9}(t-t_0{)}^{1/3}$. This law applies while the thin-crust renewal model remains valid and liquid remains available; it cannot continue beyond complete solidification. The specific numerical lifetime cannot be inferred without the absolute solid viscosity and conductivity.

Put $a={\varphi }_1$, $b={\varphi }_2=a/2$, $z_m=4H/5$, $L=H-z_m$ and $D=(1-b)z_m$. The quadratic hindered-settling flux is convex. At the initial internal interface its lower particle volume fraction is smaller than its upper particle volume fraction, so characteristic curves diverge: the entropy solution has a settling rarefaction fan, not a compression shock. Within this fan,The two bounding straight shocks initially haveThe upper front's nominal meeting with the fan's upper edge occurs atThe lower front meets the fan's lower edge atThe lower meeting is always physical in the stated range. The upper meeting need not happen before the suspension disappears; this distinction matters for sufficiently small $a$.

For the curved settling fronts within a rarefaction fan, the Rankine-Hugoniot condition gives ${\dot{z}}_u=u_s(\varphi -1)$ at the clear-fluid front and ${\dot{z}}_d=u_s\varphi$ at the deposit front. Substitution of the fan particle volume fraction and matching to the straight sections givesThe first curve applies after $t_{c1}$ if that meeting occurs, and the second after $t_{c2}$ until it leaves the fan or deposition ends. If both fronts meet inside the fan, their intersection isThe last surviving characteristic of a settling fan joins $(z_m,0)$ to $(h_c,t_c)$. It has constant particle volume fractionIt is swallowed simultaneously by the clear-fluid and deposit shocks at final deposition. This requires $b\le {\varphi }_{\ast }\le a$, which in the given geometry is equivalent toEquality places the last characteristic curve on the upper fan edge. In the generic fan-intersection regime $a>a_{\ast }$, both nominal meetings and the two curved fronts are realized.

For $0<a<a_{\ast }$ there is early extinction of a settling rarefaction fan. The deposit shock reaches the upper fan edge before the clear-fluid front does, atIt then travels through the remaining uniform particle volume fraction $a$ with speed $u_{s}a$. The upper front retains $V_1$ until the final intersection. Since the final bed height is still $h_c=3aH/5$, the actual stopping time isIn this regime $(h_{c1},t_{c1})$ is only an extrapolated intersection; for $a<1/6$ its extrapolated height is even negative. The line joining the initial internal interface to the final point would have a particle volume fraction greater than $a$, so it is not a characteristic curve emitted by that fan. The single fan-characteristic curve stopping-time construction does not apply to the entire printed interval $0<a<1/2$; the missing subregime must be included.