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Take $A>0$ for the repulsive molecular interaction, $p_0$ as the common reference pressure, and let $T_i=T_m-\delta T$ be the ice-film temperature. All ratios involving $T_m$ use its absolute value, approximately $273.15\mathrm{K}$, rather than the numerical Celsius value zero. The premelting film permits water to reach and freeze at the bottom of the ice while the upper surface is displaced upward without freezing there.

For equal phase mass densities, equality of the solid and liquid chemical potentials gives the Clapeyron pressure relation for equal-density phases. Indeed, $d\mu =-sdT+dp/\rho$ and $s_l-s_s=L/T_m$, so expansion around coexistence at $(T_m,p_0)$ yieldsThis is a pressure difference between phases, not a same-pressure Clausius-Clapeyron slope. The molecular disjoining pressure supports this difference:Treat the normal solid load as its effective pressure in this planar model. The gravitational load corresponding to hydrostatic pressure is $p_s=p_0+\rho gh$, so $p_l=p_0+\rho gh-p_T$. Refer the bath pressure to the sheet's upper gravitational datum, or neglect the sheet-scale hydrostatic head. The Darcy flow law then gives the upward supply and the ice growth rateHere $\Pi$ is permeability of a porous medium, $\mu$ is dynamic viscosity, and equal densities identify supplied liquid volume with added ice volume to leading order in $a/h$.

To recover the printed law, use the usual hydraulic control of frost heave idealization: the sheet's top is at $T_m$, the liquid film and ice have a common thermal conductivity $K$, and the flow is slow enough that latent-heat production is small compared with the conductive heat passing through the layer. In addition to a quasi-steady ice temperature field, this needsThe leading heat flux is consequently continuous through the film and ice. Their thermal resistances giveLet $B_T=\rho L\Delta T/T_m$, a pressure scale. The thermomolecular pressure and the molecular law then giveSubstitute this pressure into the Darcy flow law:The first term draws water toward the undercooled ice; the increasing gravitational load eventually cancels it.

Quasi-steady temperature alone is not sufficient to fix this exact prefactor. With film conductivity $K_l$, ice conductivity $K_s$, and nonnegligible latent heat, the appropriate additional balance isTogether with the Darcy equation, this is the implicit growth model. Even in the slow-flow limit, a conductivity ratio $K_s/K_l=r$ changes the leading driving pressure by $r^{3/4}$. Taking $r=2$ is a counterexample to obtaining the printed prefactor from quasi-steadiness alone. Thus the boxed equation is the intended equal-conductivity, thin-film, hydraulically limited model, rather than a consequence of only the stated quasi-steady assumption. This distinction is the heat-balance correction to a premelted-film growth model.

For that reduced equation defineThe scale $h_{\ast }$ balances $F{h}_{\ast }^{-3/4}$ against $\rho g{h}_{\ast }$; $F$ has units of pressure times length to the power $3/4$, so $h_{\ast }$ is a length and $t_{\ast }$ is a time. They yield

Set $u={\eta }^{7/4}$. The equation becomes linear:For initial thickness $\eta (0)={\eta }_0\ge 0$, the explicit solution of gravity-limited frost heave isThe usual zero-initial-thickness sketch usesIts early and late behaviour areFor $0<\eta <1$, the derivative is positive and ${\eta }^{\prime \prime }=(-\frac{3}{4}{\eta }^{-7/4}-1){\eta }^{\prime }<0$, so the curve is increasing and concave downward. It approaches the stable thickness $h_{\ast }$ exponentially. Initial thickness above $h_{\ast }$ instead relaxes downward; at $h_{\ast }$ it is stationary.

The singular slope at zero belongs to the formal reduced solution. Since $a/h\propto h^{-3/4}$, the thin-film assumption eventually fails as $h\to 0$. The early power law therefore describes an intermediate continuum regime after any microscopic startup, not arbitrarily early physical times. A finite positive ${\eta }_0$ supplies a regular initial condition when that regime begins. Gravity limits the ultimate thickness; molecular suction and the Darcy resistance set the growth toward it.

For a stationary, uniformly permeable skeleton, Darcy law gives $\mathbf{u}=-(\Pi /\mu )\nabla p$. Constant-density conservation of mass gives $\nabla \cdot \mathbf{u}=0$, hence the pressure is harmonic:With uniform gravitational acceleration, the same argument applies to pressure after subtracting its hydrostatic part. The spherical calculation below neglects body-force variation and uses the excess pressure relative to the far field.

Set $ϵ=1-{\rho }_s/\rho >0$ and $\beta =\varphi ϵ$. When a shell freezes, its pore mass decreases at rate $4\pi a^2\varphi (\rho -{\rho }_s)\dot{a}$; the difference must leave as liquid. Since Darcy velocity is volume discharge per total area, not pore-fluid speed, the interfacial and radial balances giveThe positive sign means outward expulsion by the expansion on freezing. Integrating Darcy law with $p(\infty )=0$ givesThere is no liquid Darcy velocity in the ice-filled region.

To state a closed Stefan problem, let $k_{-},k_{+}$ be the effective thermal conductivities of the frozen and unfrozen composites and $C_{-},C_{+}$ their volumetric heat capacities. If the matrix mass density is ${\rho }_m$, the common specific heat gives $C_{-}=c_p[(1-\varphi ){\rho }_m+\varphi {\rho }_s]$ and $C_{+}=c_p[(1-\varphi ){\rho }_m+\varphi \rho ]$. These material coefficients are not numerically specified. With $\mathcal{L}=\varphi {\rho }_{s}L$, the thermal equations areThe exterior equation retains advected sensible heat. A more detailed mechanical-energy model adds the Darcy dissipation source $\mu u^2/\Pi$; its smallness at leading order is checked below. At the origin require ${\partial }_r{T}_{-}(0)=0$, and at infinity require $T_{+}\to T_{\infty }$, $p\to 0$. At the phase boundary impose temperature continuity, phase equilibrium and the Stefan condition:These equations, the radial liquid-mass balance and Darcy law, together with an initial radius and compatible initial temperature fields, specify the moving-boundary model. No curvature correction is imposed. A vanishing initial seed corresponds to the leading growth law below with $a(0)=0$.

Write $\Delta =T_m-T_{\infty }$ and $S=L/(c_p\Delta )$, the latent-to-sensible heat ratio. A conductive interfacial balance gives $a\dot{a}\lesssim k_{+}\Delta /\mathcal{L}$. On length scale $a$, the ratio of thermal diffusion time to growth time is consequently bounded byIt is small for $S\gg 1$ with fixed porosity and comparable constituent densities and thermal properties. Very small porosity requires the sharper condition $C_{\pm }\Delta /(\varphi {\rho }_{s}L)\ll 1$; the bare heat-capacity ratio alone should not be used uniformly as $\varphi \to 0$. The advective-to-conductive ratio isEven if mechanical heating is retained, its size relative to latent-heat release is $p_i\beta /\mathcal{L}=(T_m-T_i)/T_m\le \Delta /T_m$. For fixed water properties this is also small as $S\to \infty$, consistently with the linearized Clausius-Clapeyron relation. Thus unsteadiness, thermal advection and mechanical heating are higher-order effects. The local leading temperature fields satisfy Laplace's equation. The approximation describes radii of order $a$; the distant transient diffusion tail supplies the far-field matching and need not be globally steady.

Regularity at zero and the interface/far-field values giveThe Stefan condition becomes $\mathcal{L}a\dot{a}=k_{+}(T_i-T_{\infty })$. Introduce the dimensionless pressure-feedback factorCombining this heat balance with pressure-dependent phase equilibrium yields the growth and interface values:This is pressure-limited spherical freezing in a porous medium. The interface values are constant during this leading solution, although $p(r,t)$ changes at a fixed exterior radius.

As permeability of a porous medium decreases, $\Lambda$ increases: the interface pressure rises toward the finite value $\Delta /\gamma$, while the interface temperature falls toward $T_{\infty }$. Hydraulic resistance suppresses freezing through pressure-induced freezing-point depression. In the limit $\Pi \to 0$, $a\dot{a}\sim \Pi \Delta /(\gamma \mu \beta )$; a zero-seed radius grows as $\sqrt{\Pi t}$ and tends to zero at fixed time. A finite seed instead hardly changes. At exactly zero permeability an incompressible rigid model cannot expel the expansion volume, so sustained growth stops; deformation or compressibility would require a different model.

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.

Take $z$ positive downwards from the roof. Write $T_0=T_m-m{C}_0$, $C_B=(T_m-T_B)/m$, $\Delta T=T_0-T_B=m(C_B-C_0)>0$ and $\Delta C=C_B-C_0$. The cooled magma follows the decreasing liquidus from $(C_0,T_0)$ towards $(C_B,T_B)$; the initially superheated liquid starts at $(C_0,T_{\infty })$ above that liquidus. The solid has zero solute concentration. At each temperature in the mushy layer, a horizontal tie line joins the pure solid to the residual liquid on the liquidus. The lever rule gives the liquid fraction from the bulk concentration. Since $T_B>T_E$, this path stops before the eutectic temperature and does not cross the eutectic composition.

The porosity $\varphi$ denotes the liquid fraction, so $1-\varphi$ is the solid fraction. Equal phase mass densities remove volume-change flow; the stable density stratification suppresses convection. In the stagnant mushy-layer model, define bulk specific enthalpy and bulk solute concentration byThe arbitrary reference of specific enthalpy has been omitted. The two conservation laws in $0<z<h(t)$ areHere $\kappa =k_T/(\rho c_p)$ is the thermal diffusivity, with $k_T$ the thermal conductivity. The positive $L{\varphi }_t$ term records increased specific enthalpy on melting; during freezing it supplies latent heat. In the liquid $z>h(t)$, the heat equation and solute diffusion equation arewhere $D$ is the solutal diffusivity. Neglecting solute diffusion inside the mushy layer does not permit dropping $D{C}_{zz}$ in the liquid at this stage.

Use $T(0,t)=T_B$, liquid $T\to T_{\infty }$ and $C\to C_0$ as $z\to \infty$, and initially $h=0$, $T=T_{\infty }$, $C=C_0$ for $z>0$. These are the half-space boundary conditions appropriate before cooling reaches the bottom of the chamber. For a closed chamber of finite depth $H$, replace the far-field conditions by the specified bottom conditions, for example $T_z=C_z=0$ at an insulated impermeable floor; the half-space similarity solution then ceases to apply when the thermal penetration depth approaches $H$.

At $z=h(t)$, let $C_i$ and $T_i=T_m-m{C}_i$ be the common liquid concentration and temperature, and let ${\varphi }_i$ be the limiting porosity on the mush side. Continuity of $T,C$, the Rankine-Hugoniot condition for solute, and the interfacial conservation of energy giveThe first condition has the sign of solute rejection into the liquid. For the sharp-edge stagnant mushy-layer model with nonzero liquid solutal diffusivity, close the free-boundary problem by marginal equilibrium at a mush-liquid boundary:This prevents constitutional supercooling immediately ahead of the mushy layer. A small solid fraction jump at its edge is allowed: imposing ${\varphi }_i=1$ as well as these finite-$D$ conditions would generally overdetermine the reduced model. At the roof, $C=C_B$ follows from the liquidus, and no additional solute-flux boundary condition is needed in the diffusion-free mush.

To obtain the large-concentration similarity solution, putThe positive superheat specifies the physically intended liquid initial state. On a similarity solution, $C_i,{\varphi }_i$ are constant, so every newly incorporated level leaves the same constant $B={\varphi }_i{C}_i$. Write $C_i=C_0+\Delta C\delta$ and $B=\Delta C(\mathcal{C}+a)$. Then throughout the mushy layerFor $a,\theta =O(1)$, $\mathcal{C}\gg 1$, and $\mathcal{S}/\mathcal{C}=O(1)$, the effective heat capacity is constant to relative error $O({\mathcal{C}}^{-1})$. Thus setIntegration of the similarity solution ordinary differential equation gives the leading analytical profilesThe temperature is $T=T_0+\Delta T\theta$. In the mushy layer, the concentration and porosity are the displayed $C,\varphi$; in the liquid,These retain the liquid solutal boundary layer. For completeness, finite-$r$ values of $\delta ,a,\lambda$ are determined analytically, to the same large-$\mathcal{C}$ order, by settingand solvingThe solute jump gives ${\varphi }_i=1-b/(\mathcal{C}+\delta )$ exactly; replacing $\mathcal{C}$ by $\mathcal{C}+\delta$ on the right of the last equation retains that jump's next-order factor, but does not make the linearized thermal profiles exact at finite $\mathcal{C}$. The branch has $0\le \delta <1$, $0<\varphi \le 1$. All these profiles are leading asymptotic expressions, not exact solutions of the nonlinear finite-$\mathcal{C}$ heat equation.

For $r\to 0$ at fixed positive $\lambda$, the complementary error function asymptotic gives $K(\lambda /\sqrt{r})\sim \sqrt{\pi }\lambda /\sqrt{r}$, hence $\delta =O(r)$, $b=O(r)$, $a\to 0$, and ${\varphi }_i\to 1$. The result simplifies toIn particular, $\varphi (0,t)=C_0/C_B=\mathcal{C}/(\mathcal{C}+1)$ and $\varphi (h^{-},t)=1$. There is no finite interfacial latent heat jump in this limit: the latent heat is released continuously throughout the mushy layer. Matching the two temperature gradients givesThe printed final equation is missing the square on $\Omega$ in the exponential. The corrected exponent follows directly from differentiating $\mathrm{erf}(\Omega \eta )$ and is consistent with ${\Omega }^2=1+\mathcal{S}/\mathcal{C}$. The positive root is unique: the mush-side gradient decreases from infinity to zero with $\lambda$, whereas ${\theta }_{\infty }K(\lambda )$ increases. The zero-superheat limit is singular and is not described by a finite positive $\lambda$ in this two-region construction.

Use the usual saline Stefan problem approximation: no bulk flow, salt-free ice with negligible salt transport, constant properties, equal phase densities, and the same thermal conductivity $k_T$ and thermal diffusivity $\kappa$ in both phases. Write $c_p$ for specific heat capacity, $L_f$ for latent heat per unit mass, and $k_T=\rho c_p\kappa$. These thermal symmetries are needed for the arithmetic-mean interface temperature requested in the paper; unequal phase conductivities would give a weighted balance instead.

Set $\eta =x/(2\sqrt{\kappa t})$, $\xi =x/(2\sqrt{Dt})$, $ϵ=\sqrt{D/\kappa }$ and $a(t)=2\lambda \sqrt{Dt}$. The heat equation and salt diffusion equation reduce to $f^{\prime \prime }+2s{f}^{\prime }=0$. Their similarity solutions, in terms of the complementary error function, areThese have the required interface values and far-field limits. At each fixed $x\ne 0$ they recover the initial data as $t\downarrow 0$. The liquidus condition is $T_i=-m{C}_i$.

Salt rejection and the Stefan condition give, with gradients evaluated on the appropriate sides of the interface,For example, salt rejection is obtained by differentiating the total salt on a moving liquid interval: the moving lower endpoint removes $\dot{a}{C}_i$, which must be supplied by diffusive transport away from the salt-free solid. Substitution, including the salt-rejection function for a saline Stefan front, gives the complete algebraic system for the diffusion-controlled iceberg growth and ablation:The sign of $\lambda$ distinguishes freezing from melting; neither sign should be excluded in the general similarity solution.

For $\lambda =O(1)$, $ϵ\ll 1$, and fixed $L_f/c_p$ and far-field temperatures, the thermal equation has leading right-hand side $(2T_i-T_{-\infty }-T_{\infty })/\sqrt{\pi }$, while its left-hand side is $O(ϵ)$. HenceThis is a leading-order balance, not an exact cancellation of latent heat. The resulting $C_i=-T_i/m$ and the first algebraic equation determine the leading $\lambda$. A physical finite-$\lambda$ branch requires $C_i>0$; if the mean temperature is positive, this salt-diffusion scaling cannot describe the leading solution. Likewise, a latent-to-sensible heat ratio diverging as $1/ϵ$ changes the leading thermal balance.

Define ${\theta }_{-}=-m{C}_0-T_{-\infty }$ and ${\theta }_{+}=T_{\infty }+m{C}_0$. Put $T_0=-m{C}_0$ and $\Delta \theta ={\theta }_{-}-{\theta }_{+}$. Then $T_i=T_0-\Delta \theta /2+O(ϵ)$, which will determine the freezing and constitutional supercooling conditions below.

Use a quasistatic surface energy balance with negligible heat storage in the surface or ice, a linear ice temperature profile, and a fixed freezing temperature at the ice-ocean interface. These are the assumptions behind the single-category sea-ice thermodynamic model used below. Write $\theta =t/\tau$, $F_O={\lambda }_O(T_O-T_m)>0$ andHere ${\lambda }_A$ and ${\lambda }_O$ are heat transfer coefficients, $k$ is thermal conductivity, $\kappa =k/(\rho c_p)$ is thermal diffusivity, and $\mathcal{S}$ is the inverse Stefan number. Here $L$ is latent heat per unit mass. The thermal $k$ denotes conductivity; the previous questions use $k$ for permeability.

While there is open water, the upward ocean heat supply balances the loss to air: ${\lambda }_O(T_O-T_s)={\lambda }_A(T_s-T_A)$. ThusFreezing begins when this surface reaches $T_m$ on the cooling part of the annual cycle, which requires $r<1$ and givesFor $r\ge 1$ the surface never cools strictly below freezing and no positive ice thickness develops in this model. The limit $r\to 0$ gives $t_1\to \pi \tau /2$.

With ice present and a cold surface, thermal conduction and Newton's law of cooling give $k(T_m-T_s)/h={\lambda }_A(T_s-T_A)$. The surface temperature isDuring surface melting the temperature is capped at $T_s=T_m$; the ice is not permitted to follow the above formula to a temperature higher than its melting point. The upward conductive heat flux during the cold phase isThe Stefan condition at the ocean interface gives $\rho L\dot{h}=F_C-F_O$. Under the permitted approximation of negligible ocean heat during growth,Starting from $h(t_1)=0$, integration yields the implicit growth lawThe leading maximum occurs when the air returns to freezing from below:The equality for $h_m$ is within the ocean-heat-neglected growth model, with the freezing time set by the open-water balance. If $aH\gg 1$, conductive resistance through the ice dominates the atmospheric boundary resistance at maximum thickness. ConsequentlyThis reproduces the displayed large-conductance limit. A sufficient condition for the quasistatic temperature assumption is $h_m^2/\kappa \ll \tau$; large $\mathcal{S}$ ensures this in the thick-ice scaling.

The approximation concerning ocean heat needs care. Its ratio to atmospheric heat removal during the cold phase isIt is uniformly small away from the beginnings and ends of winter providedFor the thick-ice limit this requires both $r\ll 1$ and $raH\ll 1$. This also makes the integrated ocean heat small relative to the latent heat of the maximum ice column. It is not possible to neglect $F_O$ pointwise throughout the closed growth interval: at the actual freezing onset it balances the atmospheric removal, and at the true maximum the growth rate is zero because the two fluxes again balance. The ocean-heat correction to seasonal ice growth therefore gives narrow endpoint corrections. In the full model the maximum is slightly earlier than $3\pi \tau /2$, withThus $t_2=3\pi \tau /2-O(\tau r[1+a{h}_m])$ in the stated approximation, rather than an exact maximum time for finite ocean heat.

For $3\pi \tau /2\le t\le 2\pi \tau$, while ice survives, both the warm air and ocean melt it. The surface is at $T_m$, and the total Stefan condition isIntegrating from the leading maximum givesuntil its first zero. The physical thickness is then zero for the rest of this warm interval and the open-water temperature applies. Therefore at the end of the first year,Positive ice at that time is equivalent, in the specified approximation, toThis establishes the algebraic criterion printed in the paper.

There is, however, an actual qualification to the paper's word “perennial.” This inequality tests peak-temperature sea-ice survival, since $t=2\pi \tau$ is the next atmospheric temperature maximum. Surface melting continues until at least $t=5\pi \tau /2$, so positive thickness at the maximum does not guarantee survival throughout the summer. In the same approximation, survival through that full warm half-cycle would require the stronger inequalityIn fact, starting ice-free under the specified zero-mean sinusoidal forcing, even the zero-ocean-heat upper bound on winter growth cannot meet it. In that limit $t_1=\pi \tau /2$ andsince $h_m^2/2>0$. The expression on the right is already the atmospheric melt over the entire subsequent warm half-cycle. Positive ocean heat can only reduce the winter ice and increase summer melt. Thus the printed condition is a first-year peak-temperature survival test, not a sufficient criterion for literal year-round ice in this model. For example, with $aH=1/2$ and very small positive $r$, it gives $h_m/H\simeq 2(\sqrt{2}-1)\simeq 0.828$, while first-year melt is approximately $0.5H$ but full warm-season melt is approximately $H$. Ice is present at the year's endpoint and disappears later that same summer. No reinterpretation of the printed inequality as guaranteed perennial cover is justified without an additional seasonal assumption.