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For an ice layer between freshwater and cold brine, set $ϵ=\sqrt{D/\kappa }$ and use solutal similarity positions $a=2{\lambda }_a\sqrt{Dt}$, $b=2{\lambda }_b\sqrt{Dt}$. A linear ice temperature, two Stefan conditions and interfacial salt conservation determine both negative interface constants. As $ϵ\to 0$ with fixed latent-to-sensible heat ratio $S$, ${\lambda }_b=O(1)$ but ${\lambda }_a=O({ϵ}^{-1})$. Freezing above occurs over the thermal diffusion scale, while dilution-driven melting below has the much shorter solutal scale.

For $\mathcal{C}\gg 1$ and latent-to-sensible heat ratio $S\gg 1$ with $S/\mathcal{C}=O(1)$, put $\Omega =1+S/(r\mathcal{C})$, $\zeta =Vz/\kappa$, and $H=Vh/\kappa$. The steady heat equation becomes ${\theta }^{\prime \prime }+r\Omega {\theta }^{\prime }=0$ in the mush and ${\theta }^{\prime \prime }+r{\theta }^{\prime }=0$ in the liquid. Matching temperature and heat flux at $\zeta =H$, with $\theta (0)=-1$ and liquid temperature ${\theta }_{\infty }>0$ far above, givesThe latent heat modifies the mush's effective thermal decay rate and therefore its thickness.

Use undercooling $\Delta T=T_m-T_{\infty }$, solidification attachment mobility $\mathcal{M}$, velocity scale $U_0=\mathcal{M}\Delta T$, length $\ell =\kappa /U_0$ and time $\kappa /U_0^2$. Let $S=L/(c_p\Delta T)$ be the latent-to-sensible heat ratio, $\theta =(T-T_{\infty })/\Delta T$, and $\Gamma ={\gamma }_T/(\Delta T\ell )$. In the one-sided liquid-diffusion model, a travelling planar front has $\theta =S{e}^{-V\xi }$ and $V=1-S$. For a corrugation $\zeta e^{\sigma t+i\alpha y}$, the decaying thermal perturbation is $A_1{e}^{-p\xi }$, with $p^2-Vp=\sigma +{\alpha }^2$ and the branch $p=[V+\sqrt{V^2+4(\sigma +{\alpha }^2)}]/2$. The displaced kinetic condition gives $A_1=(SV-\sigma -\Gamma {\alpha }^2)\zeta$. The displaced Stefan condition gives $S\sigma \zeta =p{A}_1-S{V}^2\zeta$. Eliminating $A_1$ proves the displayed dispersion relation. Neglect of solid-side perturbation heat flux is part of this model.

Write $H=L/c_p$, $k=\rho c_p\kappa$ and $ϵ=\sqrt{D/\kappa }$. The requested model neglects density differences and uses the same thermal transport coefficients in the interfacial balances. Let $C_b=C(b(t),t)$ and $T_b=T_m-m{C}_b$. With no imposed heat gradient in the fresh liquid, its temperature remains $T_m$. The prescribed linear temperature in the ice isThe salt and temperature fields in the lower liquid satisfy diffusion equations, $C_t=D{C}_{zz}$ and $T_t=\kappa T_{zz}$. For constant interfacial values and the proposed similarity solution, their decaying solutions areThese follow by substituting a variable $z/(2\sqrt{\chi t})$ into the diffusion equation, which gives $U^{\prime \prime }+2\eta U^{\prime }=0$; integrating gives the complementary error function. Their values at $b$ and at infinity give the required boundary data.

For the upward-moving fresh-water/ice boundary, the Stefan condition is $\rho L\dot{a}=k{T}_{i,z}$: negative $\dot{a}$ means fresh water freezes. At the lower boundary the signed heat balance isSalt cannot enter the pure ice. With no bulk liquid motion its interfacial conservation law isThe derivatives and interface velocities areDefine $F(x)=\sqrt{\pi }x{e}^{x^2}\mathrm{erfc}x$. Substitution into all three balances givesTogether with $T_b=T_m-m{C}_b$, these determine the interfacial constants. In particular $C_b=C_0/[1-F({\lambda }_b)]$.

The physical branch has ${\lambda }_a<{\lambda }_b<0$. Since $F$ is negative at negative arguments, the salt relation gives $0<C_b<C_0$: lower-surface melting dilutes the brine. Combining the two heat balances eliminates the ice gradient and givesThus $T_m>T_b>T_{\infty }>T_L(C_0)$. The interface is at local liquidus equilibrium with diluted brine, while the undiluted far-field brine remains above its own liquidus. The similarity model describes a layer starting with zero thickness, or the long-time limit when an initial finite thickness no longer sets the dominant scale.

For the small-diffusivity-ratio limit, put $A=ϵ{\lambda }_a$ and $\Delta =T_m-T_{\infty }$, keeping $S=H/\Delta$ and the imposed concentrations fixed. Seek $A=O(1)$ and ${\lambda }_b=O(1)$. Then $F(ϵ{\lambda }_b)\sim \sqrt{\pi }ϵ{\lambda }_b$ and ${\lambda }_b-{\lambda }_a\sim -{\lambda }_a$. The upper heat balance givesIn the lower balance the two terms of order ${ϵ}^{-1}$ must cancel, givingConsequentlyand $T_b-T_{\infty }\sim -\sqrt{\pi }HA$. Adding the two temperature differences yields $2A^2-\sqrt{\pi }A=1/S$. The negative root isThe resulting $C_b=(T_m-T_b)/m$ lies strictly between zero and $C_0$. The salt relation therefore determines a finite negative ${\lambda }_b$, confirming its $O(1)$ scaling. The top interface travels on the thermal scale $\sqrt{\kappa t}$, whereas the lower melting interface travels on the shorter solutal scale $\sqrt{Dt}$. This is the brine-ice sandwich with disparate diffusion scales; $S$ is the latent-to-sensible heat ratio, the reciprocal of the usual sensible-to-latent Stefan number convention.

The sketches show constant fresh-water temperature, a declining linear ice temperature, and a much broader thermal than solutal boundary layer in the brine. Concentration is zero in freshwater and pure ice, jumps to $C_b$ across the lower interface, and rises to $C_0$. Accordingly the local liquidus is $T_m$ on the salt-free side, jumps to $T_b$ and falls rapidly to $T_L(C_0)$. In the phase diagram, the brine starts on the liquidus at $(C_b,T_b)$ and ends above it at $(C_0,T_{\infty })$; the pure ice has concentration zero, and the interface tie-line joins it to the diluted brine.

Physically, the cold saline liquid permits freshwater to freeze above, but salt lowers the liquidus and allows the existing ice to melt into brine below even though that brine is colder than $T_m$. Dilution raises the lower interfacial freezing temperature. Heat released by upper freezing supplies the lower melting and the conductive loss to the colder deep brine. With $|{\lambda }_a|\gg |{\lambda }_b|$, upper freezing exceeds lower melting and the ice layer thickens as $b-a\propto \sqrt{t}$.

Let $a(t)$ be the particle radius and $\Delta T=T_s-T_{\infty }>0$ the supercooling. Use spherical symmetry, constant thermal properties, equal phase densities $\rho$, and negligible fluid advection. Let $\mathcal{L}$ be latent heat per unit mass, $c_p$ the liquid specific heat, $k_{\ell }$ its thermal conductivity and $\kappa =k_{\ell }/(\rho c_p)$ its thermal diffusivity. The exterior temperature satisfies the radial heat equationwith $T(r,0)=T_{\infty }$ for $r>a_0$, $a(0)=a_0$ and $T(r,t)\to T_{\infty }$ as $r\to \infty$.

The Gibbs-Thomson effect gives curvature-induced melting-temperature depression. For a spherical interface writewhere $\gamma$ is solid-liquid interfacial energy per area and the factor two is the sphere's curvature $2/a$. Here $T_s$ in the thermodynamic formula is an absolute temperature. The notation $\Gamma$ also permits a given measured capillary coefficient without requiring its microscopic derivation.

In the one-phase Stefan problem, the solid has no internal heat flux and the Stefan condition isThe sign follows from heat balance: positive outward liquid heat flux releases latent heat as the solid grows. If the solid's thermal storage is retained, its temperature $T^{(s)}$ instead satisfies its own radial heat equation, ${\partial }_r{T}^{(s)}(0,t)=0$, and $T^{(s)}(a^{-},t)=T(a^{+},t)$. The two-phase condition is $\rho \mathcal{L}\dot{a}=k_s{T}_r^{(s)}(a^{-},t)-k_{\ell }{T}_r(a^{+},t)$. It reduces to the one-phase condition in the quasistatic limit because a regular harmonic temperature inside a sphere is spatially constant. An initial solid temperature profile is additional data for a fully time-dependent two-phase problem.

Two reciprocal Stefan number conventions are common. The large parameter appropriate to the requested slow-interface limit is the latent-to-sensible heat ratioIt measures the energy required for phase change relative to cooling or heating by the supercooling. The often-used sensible-to-latent definition is $\mathrm{Ste}=c_p\Delta T/\mathcal{L}=1/S$. Thus “large Stefan number” here means large $S$, not large $\mathrm{Ste}$ in that other convention.

The thermal diffusion time for scale $a_0$ is $a_0^2/\kappa$, whereas the interface time is $S{a}_0^2/\kappa$. More explicitly, with $R=r/a_0$, $\theta =(T-T_{\infty })/\Delta T$ and slow time $\tau =\kappa t/(S{a}_0^2)$, the liquid equation isFor $S\gg 1$, temperature therefore equilibrates before the radius changes appreciably. The quasistatic spherical Stefan problem with capillarity has a harmonic exterior temperature, fixed by its interface and far-field values:The small initial diffusion adjustment is omitted at this order. Inserting $T_r(a)=-\Delta T/a+\Gamma /a^2$ into the Stefan condition givesThis is the critical radius for a supercooled spherical seed. For $a_0>a_c$, the interface is warmer than the liquid and loses heat, driving growth. For $0<a_0<a_c$, its curvature-depressed interface temperature is colder than the liquid, so heat enters and the solid melts. At $a=a_c$, the uniform temperature $T_{\infty }$ supplies an equilibrium radius. It is unstable, since the radius-velocity derivative there is $\kappa /(S{a}_c^2)>0$. For a growing particle with $a\gg a_c$, $a^2$ increases asymptotically as $2\kappa t/S$.

For a subcritical particle the radius decreases monotonically and remains below $a_0$. HenceIt cannot remain positive beyond $S{a}_0^3/[3\kappa (a_c-a_0)]$, proving finite-time extinction of a subcritical spherical seed. The exact extinction time in the reduced model is obtained by separating variables:With ${\eta }_0=a_0/a_c\in (0,1)$, this isNear disappearance, the capillary term dominates and $\dot{a}\sim -\kappa a_c/(S{a}^2)$, so $\boxed{a^3\sim 3\kappa a_c(t_e-t)/S}$. For $a_0\ll a_c$, the lifetime is $t_e\sim S{a}_0^3/(3\kappa a_c)$; it diverges logarithmically as $a_0\to a_c^{-}$, consistently with the unstable equilibrium. The threshold and lifetime above are conclusions of the specified large-$S$ quasistatic continuum model, rather than an assertion that its approximation is uniform through every finite-$S$ shrinking-interface transient.

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.

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.

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.