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Write $ϵ=1-{\rho }_s/\rho >0$ and $\Delta v=1/{\rho }_s-1/\rho =ϵ/{\rho }_s$. When the ice sphere grows by volume $4\pi a^2\dot{a}dt$, it consumes liquid volume $({\rho }_s/\rho )4\pi a^2\dot{a}dt$. The difference is expelled from the fixed cavity. Thus mass conservation fixes the outward Darcy flux in the surrounding porous medium:The Darcy law $u_r=-(\Pi /\mu ){\partial }_{r}p$ and $p(\infty )=p_m$ implyMatching at the cavity wall, whose internal pressure is uniform, givesThe solid and liquid pressures are equal when the Gibbs--Thomson relation is neglected; nevertheless the coexistence temperature changes with their common pressure because their specific volumes differ. Along coexistence, equality of chemical potentials givesHence the common-pressure melting-point slope is $d{T}_m/dp=-T_m\Delta v/L$. To first order in the pressure change,The $T_m$ multiplying this coefficient is absolute temperature, approximately $273.15\mathrm{K}$ for water; a temperature difference may equally be measured in kelvin or degrees Celsius. Substituting the numerical value $0$ from the Celsius scale into a thermodynamic denominator would be incorrect.

In the quasi-stationary thermal conduction approximation, neglect advective heat transport and use the stated effective conductivity $k$ outside the sphere. The radial Laplace equation gives the bounded solid temperature and the liquid/porous-region temperature:This treats the exterior as a uniform conducting medium, as required by the printed reduction; distinct cavity and porous-medium conductivities would give an additional thermal resistance. There is no solid-side heat flux in the bounded quasi-stationary solution. The Stefan condition therefore becomesEliminate $T_i$ and $p$ to obtainFor $x=a/R$, $t=t_0\tau$ and $t_0=R^2{\rho }_{s}L/(k\Delta T)$, this isThe pressure follows either from the Darcy flux or the Stefan condition:In particular $T_i-T_{\infty }=\Delta TK/(x+K)$: rising pressure progressively reduces the available undercooling. This is spherical freezing inside a porous cavity, whose fixed hydraulic outlet radius $R$ distinguishes it from freezing directly through a porous matrix.

For the idealized zero-radius start, integration givesA finite seed replaces the right side by its difference from the initial value. The reduction ends when the sphere fills the cavity.

For $K\gg 1$, hydraulic drainage is easy and the pressure feedback is small throughout the cavity. The leading expressions areThis is ordinary conduction-controlled spherical solidification, with a completion time asymptotic to $t_0/2$.

For $K\ll 1$, distinguish the initial crossover from the pressure-dominated regime. At $x\ll K$, equivalently $\tau \ll K^2$, the same square-root growth applies and $p-p_m\simeq p^{\ast }x/K$. At $K\ll x\le 1$, equivalently $K^2\ll \tau \lesssim 1/K$,The interface approaches $T_{\infty }$, the pressure approaches the finite coexistence value $p_m+p^{\ast }$, and growth becomes hydraulically limited. The leading completion time is $t_0/(3K)$. The cube-root approximation is not valid at the instant of nucleation, where the initial square-root regime must be retained.

Use the one-sided thermal conduction model encoded by the requested dispersion relation: neglect solid-side perturbation heat flux, take equal phase densities, and let the normal point from solid into liquid. Write the isotropic linear kinetic undercooling law using the solidification attachment mobility $\mathcal{M}>0$:Here $\gamma$ is solid-liquid surface energy, $\mathcal{K}$ is the sum of principal curvatures, and $T_m$ is absolute temperature. Positive curvature convex into liquid lowers the equilibrium freezing temperature through the Gibbs--Thomson relation. The planar base state has $\mathcal{K}=0$.

Set $\Delta T=T_m-T_{\infty }>0$ and defineThe paper's $S$ is the latent-to-sensible heat ratio, the reciprocal of the common Stefan number convention. Below, velocities, lengths and growth rates are dimensionless in these scales. In the travelling coordinate $\xi =X-Vt$, the liquid heat equation gives ${\theta }_0^{\prime \prime }+V{\theta }_0^{\prime }=0$, while the bounded solid steady field is constant. Matching a decaying liquid profile to the Stefan condition $SV=-{\theta }_0^{\prime }(0)$ givesThe kinetic undercooling condition is ${\theta }_i=1-V$. ThusThe speed is constant for $S<1$, determined by $S$ and the attachment coefficient in the specified scales. Thermal diffusivity sets the exponential decay length $\kappa /U$, not the selected speed. Including solid-side perturbation conduction would change the stability equation below.

For linear stability analysis, corrugate the interface to $\xi =\zeta e^{\sigma t+i\alpha y}$ and write the liquid perturbation as $A_1{e}^{-p\xi }{e}^{\sigma t+i\alpha y}$. The liquid heat equation requiresFor this graph the linearized curvature is ${\alpha }^2\zeta e^{\sigma t+i\alpha y}$. Evaluate the temperature on the displaced interface: its perturbation is $A_1+{\theta }_0^{\prime }(0)\zeta =A_1-SV\zeta$. The kinetic undercooling law therefore givesLikewise the derivative evaluated at the displaced surface is ${\theta }_0^{\prime }(0)+{\theta }_0^{\prime \prime }(0)\zeta -p{A}_1$, with ${\theta }_0^{\prime \prime }(0)=S{V}^2$. The Stefan condition at perturbation order readsElimination proves the one-sided kinetic solidification dispersion:The destabilizing term comes from a protrusion encountering colder liquid; capillarity and kinetic relaxation oppose it.

At $S=0$, $V=1$ and the base liquid has no thermal contrast generated by latent heat. The interface obeys $\boxed{\sigma =-\Gamma {\alpha }^2}$: positive surface energy damps all nonzero modes, while $\Gamma =0$ is neutral at this order. Here the thermal perturbation amplitude vanishes, so the auxiliary decay exponent is immaterial to that pure kinetic relaxation. At $S=1$, $V=0$ and the formal limit givesFor $\Gamma >0$ it has a negative real branch $-{\alpha }^2<\sigma <0$ and no growing branch; at $\Gamma =0$ the interface branch is neutral. Strictly, $V=0$ is a singular endpoint of the travelling profile: no finite-width stationary exponential connects ${\theta }_i=1$ to the far-field value zero. These stability statements describe the limit of the travelling-front model.

For $0<S<1$, assume $\sigma =O({\alpha }^2)$ at small wavenumber. ExpandSubstituting into the dispersion relation cancels $S\sigma /V$ on both sides and givesAt the capillary stability threshold for a kinetic freezing front $\Gamma =S/V$, continuing the same expansion gives $\sigma =-2S{\alpha }^4/V^3+O({\alpha }^6)$: the nonzero long waves are damped even at threshold.

For large wavenumber, multiply the dispersion relation by $p$ and eliminate $\sigma$ using the thermal equation. The exact algebraic form isIt prevents the uncontrolled replacement $p\simeq \alpha$ when $\sigma$ itself is of order ${\alpha }^2$. In the usual small-capillarity range the leading results areFor the second result, set $p=\sqrt{1-\Gamma }\alpha +d+O({\alpha }^{-1})$ in the cubic; its quadratic-order balance gives $2d-V=S\Gamma /(1-\Gamma )$. Thus positive capillarity stabilizes sufficiently short waves. If also $\Gamma \ll 1$ and $|\sigma |\ll {\alpha }^2$, the useful reduced approximation isits finite terms should not be treated as a uniform expansion for arbitrary fixed $\Gamma$.

For completeness, large dimensionless capillarity has different balances. At $\Gamma =1$, $p\sim S^{1/3}{\alpha }^{2/3}$ and $\sigma =-{\alpha }^2+S^{2/3}{\alpha }^{4/3}+o({\alpha }^{4/3})$. For $1<\Gamma <1+2S/V$, $p\to p_{\ast }\equiv S/(\Gamma -1)>V/2$, so $\sigma =-{\alpha }^2+p_{\ast }^2-V{p}_{\ast }+O({\alpha }^{-2})$. At $\Gamma =1+2S/V$, the real branch reaches $p=V/2$: for $S<1/3$ its growth rate is $-{\alpha }^2-V^2/4+O({\alpha }^{-4})$, and for $S=1/3$ that endpoint is exact. For $S>1/3$ at this boundary, or for $\Gamma >1+2S/V$, the selected real-$p$ normal mode ceases to exist at sufficiently large wavenumber; the full diffusion problem includes a thermal continuous spectrum and cannot be assigned a real isolated interface growth rate by this ansatz there. Indeed, writing the bulk thermal perturbation as ${\theta }_1=e^{-V\xi /2}u$ reduces its heat equation to $u_t=u_{\xi \xi }-({\alpha }^2+V^2/4)u$, whose half-line diffusion modes have decay rates $-{\alpha }^2-V^2/4-k^2$ for $k\ge 0$. The thermal branch endpoint is $p=V/2$. Loss of a damped isolated interface mode is not a morphological instability.

The approximate instability condition is already clear from the long-wave coefficient: $\Gamma <S/V$, equivalently $S>\Gamma /(1+\Gamma )$. In fact it is exact for growing modes of the present one-sided model. At $\sigma =0$, let $p_0=[V+\sqrt{V^2+4{\alpha }^2}]/2$, so ${\alpha }^2=p_0(p_0-V)$. The right side of the dispersion relation becomesFor real $\sigma \ge 0$, the residual $F=\sigma (1+S/p)-SV+\Gamma {\alpha }^2+S{V}^2/p$ is strictly increasing, sinceTherefore a unique positive root exists precisely when $\Gamma <SV/p_0^2$. For $0<\Gamma <S/V$, the unstable band isThere are no unstable oscillatory roots hidden by this real-root test. Indeed, if $p=a+ib$ and $\mathrm{Re}\sigma \ge 0$, the thermal equation implies $a\ge V$. For $b\ne 0$, the imaginary parts of the two equations would require $2a-V=S(V-\Gamma {\alpha }^2)/|p+S{|}^2\le SV<V$, contradicting $2a-V\ge V$. Thus growing modes have real $p$ and real $\sigma$.

The sketch shows zero capillarity approaching the positive plateau $SV$, weak capillarity giving a positive band followed by short-wave damping, and threshold/stronger capillarity giving damping at every nonzero wavenumber. The zero-wavenumber mode is a translation. As $S\to 1$ with fixed positive $\Gamma$, both the unstable bandwidth and the positive growth rates shrink to zero; the small-wavenumber expansion is not uniform in that limit.

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.

The Maykut-Untersteiner model solves a seasonally forced, one-dimensional sea ice thermodynamics problem with a snow layer above saline ice. Let $z$ increase downward from the upper surface, with snow–ice interface $z=h_s$ and ice bottom $z=h_s+h_i$. Write $q=-k{T}_z$ for conductive heat flux, positive downward, and $I(z)$ for downward shortwave radiation penetrating the material. Local energy conservation givesFor $I=I_0{e}^{-\kappa z}$, the radiation term is $+\kappa I$: absorption heats the interior. Here $c_{\mathrm{eff}}$ is the derivative of specific enthalpy with temperature at prescribed ice salinity, and includes the latent heat of changing brine fraction. Using only pure-ice specific heat capacity would omit this storage. A usual approximation has ${\rho }_i{c}_{\mathrm{eff}}={\rho }_i{c}_{i,0}+a{S}_i/T_C^2$ and $k_i=k_{i,0}+b{S}_i/T_C$, with $T_C<0$ in degrees Celsius and positive, unit-dependent constants $a,b$. Thus warmer saline ice has larger effective heat capacity and lower thermal conductivity. Layerwise constant $k_i$ gives the usual finite-layer heat equation; full differentiation retains its spatial variation. Snow uses its own density and thermal conductivity, and its stronger extinction can make shortwave penetration negligible beneath it.

Define the net atmospheric input to the upper surface bywhere $\alpha$ is albedo, $H_{\uparrow }$ is turbulent sensible heat loss and $E_{\uparrow }$ is turbulent latent heat loss. Their signs can reverse. The outgoing longwave term uses the Stefan–Boltzmann law. Subtract the penetrating shortwave $I_0$, because it heats the interior rather than the infinitesimal surface. If $M_0\ge 0$ is surface ablation speed and ${\rho }_0$ the density of the material currently exposed, the upper boundary obeysBelow the melting point, $M_0=0$ and the flux balance determines $T_0$. Once it reaches $T_m$, the positive excess produces surface ablation; snow melts first, and only afterwards does that ablation remove ice. At the snow–ice interface, absent interfacial phase change,At the base, $T_i=T_f(S_w)$ and the Stefan condition isHere $G_b$ is signed bottom growth speed, and $F_o>0$ is ocean heat flux toward the ice. An upward conductive loss $-q_b>0$ freezes ice; an excess $F_o$ melts it. For a prescribed upward turbulent ocean heat flux, the model needs no separate ocean temperature solution. A water-side closure could write $F_o={\rho }_w{c}_{p,w}{K}_w{T}_{w,z}$ at the boundary in this downward coordinate. Snowfall is an independent mass input: with snowfall mass flux $P_s$, ${\dot{h}}_s=P_s/{\rho }_s-M_s$ in a constant-density approximation. The ice thickness obeys ${\dot{h}}_i=G_b-M_i$. Compaction or refreezing, if added, requires corresponding mass and energy terms rather than counting snowfall directly as ice growth. These boundary conditions and interior equations determine temperature, surface melt and basal growth together.

Under periodically repeated annual forcing, annual sea-ice thermodynamic equilibrium is a periodic annual state, not constant thickness:Winter accretion then balances net annual ablation. “Equilibrium thickness” normally denotes a mean or a specified point in this annual cycle. Different forcing can instead eliminate a perennial-ice solution.

Increasing ocean heat flux reduces bottom growth directly in the Stefan condition, lowering equilibrium thickness and eventually removing perennial ice. The stabilizing winter response is visible in the approximate conductive fluxwhen ice thins, conductive loss and winter regrowth increase. More snow lowers this flux and inhibits winter growth. However, snow also raises albedo and delays summer exposure of dark ice. The competing snow insulation and summer ice protection effects mean an annual snow sensitivity cannot be deduced from winter insulation alone. Under the original forcing, the Maykut-Untersteiner model gives an almost flat annual-thickness response below about $0.7\mathrm{m}$ annual snow depth, followed by strong thickening as snow persists into summer. This particular transition is not a universal threshold for other forcing or snowfall timing. Increasing summer albedo reduces absorbed solar energy, reduces surface and internal melting, and increases equilibrium thickness; sufficiently low albedo removes perennial ice. The original thermodynamic sensitivity calculations establish these three responses.

The three curves are original qualitative sketches, not a new numerical integration or digitization. The snow marker refers to that historical forcing; the unnumbered loss boundaries and ordinate indicate trends rather than calibrated universal thresholds.

For the summer retreat discussed around the time of this examination, the proposed mechanisms act as follows.

These processes overlap: circulation causes export, export changes thickness and open-water fraction, and those changes alter solar absorption. Without specified regional budgets there is no defensible unique percentage partition. For enhanced Beaufort summer basal melt, upper-ocean solar heating and western-Arctic heat pathways are more directly relevant than invoking unmodified heat from a deep, isolated Atlantic layer. ice-mass-balance observations of extreme Beaufort melt identify increased solar absorption in open water as an important source.

Take ${\rho }_i=900\mathrm{kg}{\mathrm{m}}^{-3}$ as an explicit ice-density approximation and use the supplied $L_f=336000\mathrm{J}\mathrm{k}{\mathrm{g}}^{-1}$. The latent heat for $1.5\mathrm{m}$ basal ablation isJune–September contains about $122$ days, soUsing $917\mathrm{kg}{\mathrm{m}}^{-3}$ would give $43.8\mathrm{W}{\mathrm{m}}^{-2}$. This is the net flux used for melting: the gross ocean heat flux can also supply conductive losses and changes in cold content. It greatly exceeds an annual central-pack background of a few $\mathrm{W}{\mathrm{m}}^{-2}$, but is compatible with a summer upper ocean absorbing solar radiation at tens to hundreds of $\mathrm{W}{\mathrm{m}}^{-2}$. For scale, an added open-water fraction $0.3$, a shortwave input $200\mathrm{W}{\mathrm{m}}^{-2}$ and an ice–water albedo contrast $0.6$ produce $0.3\times 200\times 0.6=36\mathrm{W}{\mathrm{m}}^{-2}$ additional area-mean absorption before losses. This is an illustrative budget, not a measurement. Wind mixing and circulation can deliver stored solar or advected heat to the underside.

The ice-albedo feedback closes the warming loop: retreat exposes darker water; extra solar absorption warms the upper ocean; basal and lateral melt increase retreat. The stored heat also delays autumn freeze-up, exposing water to atmospheric heat transfer for longer. Its later release warms the atmosphere, but must not be counted as a second source in addition to the solar energy that originally entered the ocean. In winter, greater heat loss from open water and rapid growth of thin ice partly oppose the summer positive feedback.

Let $K_w$ be the water thermal conductivity, $L_m$ the latent heat of fusion per unit ice mass, and ${\rho }_i$ the ice density. Define $v_m>0$ as the normal ice-retreat speed that enlarges the cavity. With the water thermal boundary layer at $T_s$ on its warm side and $T_m$ at the melting interface, the heat supply per unit interface area is approximatelyThe ice is specified to be uniformly at $T_m$, so no leading sensible-heat flux into colder ice must be subtracted. The Stefan condition is therefore ${\rho }_i{L}_m{v}_m=q_T$, givingEquivalently $K_w={\rho }_w{c}_p{\kappa }_T$ expresses the heat flux in terms of water thermal diffusivity. The imposed temperatures and constant boundary-layer thickness make this melt rate uniform and constant over the wetted interface. Melting adds latent energy demand to the flow; for an isolated finite pulse, maintaining $T_s$ indefinitely would require heat replenishment. The constant-temperature model is consequently an imposed closure, not a prediction of a permanently hot finite water volume.

Use a planar saline Stefan problem with upward coordinate $z$, the initial contact at $z=0$, and phase boundaries $z=a(t)$ and $z=b(t)$ enclosing pure ice, where $a<b$. Fresh liquid occupies $z>b$ and brine occupies $z<a$. We assume negligible bulk flow, equal constant mass density $\rho$, specific heat capacity $c_p$ and thermal diffusivity $\kappa$ in all regions, and zero salt content and salt transport in the ice. Equal densities remove phase-change volume flow; suppressing convection isolates the molecular-transport mechanism. Different material properties would change the numerical coefficients. Set $K=\rho c_p\kappa$ for the common thermal conductivity and $\Delta T=T_m-T_{\infty }>0$. Denote the brine salt diffusion coefficient by $D$.

Use a linear ice liquidus, $T_L(C)=T_m-mC$, with $m>0$ and $C$ the salt mass fraction. Assume the initial brine is a stable liquid, $T_{\infty }>T_L(C_0)$, and all concentrations considered are below the eutectic composition. Thus $R=m{C}_0/\Delta T>1$. These assumptions exclude independent nucleation in a supercooled brine or an eutectic system event, neither of which is specified by the initial data alone. Start with a negligible seed of ice and use local phase equilibrium without interfacial kinetics or curvature. Temperature differences may be in Celsius, since only differences enter this calculation.

WriteThe paper calls $S$ a Stefan number; it is the latent-to-sensible heat ratio, reciprocal to another common Stefan number convention. At the lower phase boundary, define $T_i=T_L(C_i)$ and letThe liquidus relation is $\vartheta =Rc$.

A similarity solution gives an explicit calculation of the two positions. Putand use $u=z/(2\sqrt{\kappa t})$, $v=z/(2\sqrt{Dt})$. The temperatures and salinities areThese error function and complementary error function profiles satisfy the heat equation and the salt diffusion equation, their far-field conditions, and the interfacial temperature conditions. The fresh-water temperature is constant because the interface and the far field are both at $T_m$.

Salt conservation at the moving lower phase boundary requiresFor upward motion this is dilution by melting salt-free ice, rather than salt rejection by freezing brine. Substitution gives the dilution function for a melting saline Stefan front:For $A>0$, $0<c<1$, so the interfacial brine is fresher than the remote liquid.

The Stefan condition at the upper boundary is $\rho L\dot{b}=K{T}_{s,z}(b)$, since the fresh liquid has no temperature gradient. At the lower boundary it isThese signs follow from the jump in enthalpy: the upper front freezes liquid while a positive $\dot{a}$ melts solid. With $E=\mathrm{erf}B-\mathrm{erf}\delta$, they reduce toTogether with $\delta =ϵA$, these equations determine $A,B,\vartheta$, and hence both positions, without discarding the salt-diffusion correction.

For the stated large latent-to-sensible heat ratio, the layer is thin relative to the thermal diffusion length. Expanding the thermal equations for $B,\delta \ll 1$ givesConsequently a useful leading calculation iswhere the positive $A$ is obtained fromThe left side decreases with $A\ge 0$ while the right side increases, and their values at zero and infinity guarantee a unique positive root for large $S$. Thus this leading calculation includes the translation of the lower boundary as well as the increasing thickness.

If the scale separation also obeys $ϵSA\ll 1$, the familiar simpler result iswithThis last simplification needs the logarithmic refinement $ϵS\sqrt{\log (RS)}\ll 1$. The algebraic ordering $ϵ\ll S^{-1}$ alone should not be used to discard an arbitrarily large logarithmic correction; the preceding coupled equations remain the appropriate calculation when that refinement is unavailable. This distinction is captured by large latent heat in a freezing and melting ice layer.

The illustration uses $S=40$, $ϵ=2.5\times 10^{-4}$ and $R=2$. Solving the full equations gives $A=1.73707$, $B=0.0143053$ and $\vartheta =0.0158766$. It shows the ice layer between freshwater and cold brine, with both fronts advancing upward and a strongly diluted lower liquid boundary.

The phase diagram shows the spatial path from remote brine to the lower interface: temperature first changes over $\sqrt{\kappa t}$ at almost unchanged salinity, then salinity changes over its much smaller diffusion scale at nearly constant temperature. The liquid path ends on the liquidus at $(C_i,T_i)$; the solid has $C=0$ and its temperature rises from $T_i$ to $T_m$. The eutectic position is schematic, with $C_E=2C_0$ in the illustration, and is not used in the calculation. A point initially in the fresh layer freezes when $b$ reaches it, cools in the ice, and later melts when $a$ reaches it. The inset indicates this temporal path in the opposite direction through the solid branch and into the liquid.

The mechanism is simultaneous upper freezing and lower melting, with net growth of the ice thickness. Cold brine accepts most of the latent heat released by upper freezing. Salt reaching the lower contact lowers its equilibrium melting temperature, and melting freshens that liquid until its liquidus is close to $T_m$. The heat delivered through the ice supplies the much smaller melting demand there. The remote brine remains liquid even though it is below the pure-water melting temperature, because it lies above its own saline liquidus. Thus “cold” does not by itself decide which phase is stable. These conclusions apply while the layers remain effectively deep and the stated no-convection, pure-ice and phase-equilibrium assumptions hold.

Let $\xi =z-Vt$ increase into the liquid, with the planar solidification front at $\xi =0$. Denote the common thermal conductivity by $K=\rho c_p\kappa$. A bounded travelling temperature profile satisfies $\kappa T^{\prime \prime }+V{T}^{\prime }=0$. In the solid, boundedness as $\xi \to -\infty$ forces $T_s=T_m$; in the liquid,The Stefan condition then givesThus the travelling solidification front in a pure supercooled melt releases just enough latent heat to warm the initially supercooled liquid to $T_m$. There is no separate imposed heat flux at infinity in this bounded travelling-wave problem.

For the interface perturbation write $\xi =\zeta (y,t)=\hat{\zeta }{e}^{i\alpha y+\sigma t}$. Use the Gibbs--Thomson relation for curvature-induced melting-temperature depressionHere $\gamma$ is the solid-liquid surface energy, $T_m$ in this expression is absolute temperature in kelvin, and $\mathcal{K}$ is the sum of principal curvatures, positive for solid convex into liquid. For this graph, $\mathcal{K}=-{\zeta }_{yy}={\alpha }^2\zeta$. If mean curvature is defined as half this sum, $\mathcal{K}=2H$; this fixes the capillary factor unambiguously.

Under the small-wavelength thermal approximation for a solidification front, take harmonic perturbation temperatures that decay away from the interface:This approximates their temperature equation by Laplace's equation, neglecting perturbation time dependence and the moving-frame advection. Expanding the interfacial temperature at the displaced surface givesThe second expression includes displacement through the negative base liquid gradient. Linearizing the Stefan condition also requires evaluation of the base gradient at the displaced surface:Using $S=1$ gives the requested reduced dispersion relation:orThe destabilizing $V\alpha$ term describes a protrusion entering colder liquid. The cubic term expresses stabilization by the curvature-induced temperature decrease.