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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.

For freezing, the liquid concentration decreases away from the interface, so the liquidus $T_L=-mC$ increases there. Constitutional supercooling means $T_l<T_L$ somewhere ahead of the interface. Since these temperatures agree at the interface, its local onset isFor the similarity solution and salt rejection condition, this is exactlyIt is also the global criterion for this freezing solution when $D<\kappa$ and $T_{\infty }>T_i$: the ratio $T_L^{\prime }(x)/T_l^{\prime }(x)$ is a positive constant times $\exp [-x^2(1/D-1/\kappa )/(4t)]$. For $x\ge a\ge 0$ this ratio decreases with $x$. If the ratio is at most one at the interface, $T_l-T_L$ cannot become negative; if it is greater than one there, a supercooled interval appears immediately ahead. Equality marks onset, not a finite supercooled interval.

For fixed positive $\Delta \theta$ and $\lambda =O(1)>0$, the left side of the inequality is $O(ϵ)$ and the right side is positive of order one, so freezing is constitutionally supercooled for sufficiently small $ϵ$. To locate the boundary between the two regimes, resolve the smaller scale $\Delta \theta =O(ϵ)$, $\lambda =O(ϵ)$. The salt equation and liquidus then giveHere $ϵ\lambda =O({ϵ}^2)$, so the thermal equation improves to $T_i=(T_{\infty }+T_{-\infty })/2+O({ϵ}^2)$; this justifies the second expansion including latent heat. Also $T_{\infty }-T_i={\theta }_{+}+O(ϵ)$ and the left side of the exact criterion is $ϵ{\theta }_{+}/\sqrt{\pi }+O({ϵ}^2)$. Therefore the small-diffusivity constitutional-supercooling threshold isThis printed condition is an asymptotic onset condition, rather than an exact finite-$ϵ$ inequality. In its $O({ϵ}^2)$ transition window the complete algebraic system and exact gradient criterion must be used.

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.

For a steady field in the apparatus frame, the time derivative in the translating crystal frame is $-V{\partial }_z$. The given solute equation therefore becomesSubstituting the density-change flow in a pulled mushy layer reduces this to $(1-\phi )C_z=C{\phi }_z$. Thus the solute conservation in a steadily pulled mush integrates to $(1-\phi )C=C_0$, using the zero solid fraction at the top boundary.

Set $\mathcal{C}=m{C}_0/\Delta T$. The liquidus gives $C=C_0(1-\theta /\mathcal{C})$, soIn particular, $\phi =0$ at $\theta =0$, and immediately above the eutectic temperature, where $\theta =-1$, the mush has $\phi =1/(\mathcal{C}+1)$. It becomes fully solid across the eutectic front below.

Impose the eutectic temperature at $\zeta =0$ in the large-concentration thermal profile of a pulled mush:ThereforeThe thickness is proportional to thermal diffusivity divided by pulling speed. Latent-heat release modifies the exponential decay rate through $\Omega$, while the far-liquid temperature fixes how far the profile can remain below the liquidus.

Measure $z$ upward from the heat exchanger and let the steady ice front be at $z=h>0$. In the exchanger frame, salt in the liquid satisfies the advection-diffusion equationThe decaying solution and the prescribed total salt mass aresoThe linear liquidus condition fixes the interface temperature asA solid layer between the exchanger and the interface can therefore exist only if

Write $\vartheta =T-T_{\infty }$. Heat advection, conduction, and environmental loss giveIts characteristic exponents areBelow the exchanger and above the ice front, boundedness givesIn $0<z<h$, the ice temperature iswhereSubstitution of these fields into the Stefan conditiongives one scalar equation for the steady height $h$, which can be solved numerically. The heat flux may jump at $z=0$ because the exchanger supplies the required localized cooling.

The local equilibrium freezing temperature ahead of the front isBecause $T_l(h)=T_L(h)$, constitutional supercooling begins when the actual liquid-temperature gradient at the interface is smaller than the liquidus gradient:Using the solutions above, this is

At criticality the inequality is an equality. If $F/k\ll V^2/{\kappa }^2$, then $q\sim V/\kappa$ and, for $\kappa >D$,The critical curve is proportional to $S^{-1}$. If $F/k\gg V^2/{\kappa }^2$, put $q\sim \sqrt{F/k}$ to obtainThis curve behaves as $S^{-1/2}$ for small $S$ and approaches $D\sqrt{F/k}$ for large $S$. Supercooling occurs above the corresponding critical curve, where solute rejection steepens the liquidus faster than heat transport raises the actual temperature.

Put ice in $x<a(t)$ and ocean in $x>a(t)$, with $x$ increasing into the ocean. This is a saline Stefan problem. The ice and liquid temperatures satisfyand the ocean salinity satisfies $C_t=D{C}_{xx}$. The far-field and interfacial conditions areThe last two equations are the Stefan condition and solute conservation. Salt is taken to be absent from the ice. The field sketch has two broad thermal boundary layers of width $O(\sqrt{\kappa t})$ and a liquid-side salinity layer of width $O(\sqrt{Dt})$.

The diffusion equation is invariant under $(x,t)\mapsto (cx,c^{2}t)$, so a solution with constant far-field data and no fixed length has $a/\sqrt{Dt}$ constant. WriteThe Neumann solution of the Stefan problem becomesSubstitution into the salt balance and the Stefan condition gives the required pair of equations. With $c_p=k_T/(\rho \kappa )$ andthey are

When $ϵ\ll 1$, heat diffuses much farther than salt. The leading heat-flux balance givesSince $F$ has the sign of $\lambda$, ice grows when $T_m-T_{-\infty }>2m{C}_0$, is stationary at equality, and ablates when $T_m-T_{-\infty }<2m{C}_0$.

During growth, salt rejection gives $C_i>C_0$. Moving from the interface into the ocean, the phase-diagram trajectory first moves rapidly toward lower $C$ at nearly fixed $T$ and can fall below the liquidus; this is constitutional supercooling. During ablation, $C_i<C_0$, so the near-interface trajectory moves toward larger $C$ and into the stable liquid region above the liquidus. Ablation occurs because the heat conducted from the warmer ocean to the interface exceeds the heat that the colder ice can remove; melting and salt diffusion then maintain local liquidus equilibrium.

At times long compared with the encounter time at depth $H$, the fixed depth $H$ is negligible relative to the growing diffusion lengths. Salt obeys the diffusion equation with diffusivity $D$. Withthe self-similar concentration profile in the liquid isIt has $C(h,t)=C_i$ and tends to $C_0$ in the far field.

Because the solid contains no salt, conservation of solute at the moving boundary requires the rejected solute flux from Fick's first law to equal the rate at which the interface sweeps up salt:Substituting the similarity profile givesThis relation determines $C_i$ if $\lambda$ is already known. More generally it must be solved together with the thermal Stefan condition and the interfacial phase-equilibrium relation $T_i=-m{C}_i$ from the liquidus. For $ϵ\ll 1$, salt occupies a much thinner boundary layer than heat and $C_i$ can be much larger than $C_0$.

Before the ice reaches the salty layer, the state moves vertically in a temperature--concentration phase diagram along $C=0$: cooling crosses the pure-water melting point and produces pure ice. After the encounter, the liquid state starts at $(T_{\infty },C_0)$ and approaches the interface stateon the liquidus $T_L=-mC$. The solid remains on the $C=0$ axis.

When $ϵ\ll 1$, the solutal diffusion length $\sqrt{Dt}$ is much smaller than the thermal diffusion length $\sqrt{\kappa t}$. Immediately ahead of the interface, rejected salt makes $C$ fall steeply from $C_i$ to $C_0$, so the local liquidus $T_L=-mC$ rises steeply with depth while the actual temperature changes comparatively slowly. Constitutional supercooling occurs whereverIts local onset criterion at the interface isThe supercooled zone lies in the thin salty boundary layer immediately ahead of the planar front. A small forward protrusion then enters liquid that is already below its local freezing point, so the planar interface is susceptible to a morphological instability.

A saline Stefan problem couples a heat equation, a solute diffusion equation, a Stefan condition, solute conservation, and a concentration-dependent liquidus at a moving solid-liquid boundary. Different thermal and solutal diffusivities produce nested boundary layers.

Salinity measures dissolved salt content. A transport model must specify its units; salt mass fraction is one convenient convention. A dilute solution's ice liquidus is often approximated by $T_L=T_m-mC$. Salt-free ice has zero salinity, and the moving interface's salt balance distinguishes salt rejection during freezing from dilution during melting.

When salt-free ice forms from a saline liquid, the moving phase boundary excludes the dissolved salt. With equal phase mass densities, no bulk flow, liquid salt diffusion coefficient $D$, and interface concentration $C_i$, salt conservation gives the displayed condition, where $x$ increases from solid to liquid. Positive freezing speed $\dot{a}$ therefore requires a negative liquid concentration gradient. During melting the same condition gives dilution. Coupling salt rejection, the liquidus, and the Stefan condition produces the saline Stefan problem.

For a saline Stefan problem with common thermal properties, $ϵ=\sqrt{D/\kappa }\ll 1$, and fixed latent-to-sensible heat ratio, set $T_0=-m{C}_0$, ${\theta }_{-}=T_0-T_{-\infty }$ and ${\theta }_{+}=T_{\infty }-T_0>0$. A finite-$\lambda$ similarity solution has $T_i=(T_{\infty }+T_{-\infty })/2+O(ϵ)$. At the onset of constitutional supercooling, $\lambda =O(ϵ)$ and the error in this mean is $O({ϵ}^2)$. The salt-rejection function for a saline Stefan front then gives $m\lambda C_i=({\theta }_{-}-{\theta }_{+})/(2\sqrt{\pi })+O({ϵ}^2)$, whereas $\sqrt{Dt}{T}_{l,x}(a^{+})=ϵ{\theta }_{+}/\sqrt{\pi }+O({ϵ}^2)$. Comparing the actual temperature gradient to the liquidus gradient gives the displayed onset, with supercooling on the side of greater ${\theta }_{-}$. For a freezing front at $a\ge 0$ and $D<\kappa$, the ratio of these two gradients decreases into the liquid, so their interface comparison also decides whether a supercooled interval occurs anywhere ahead. The formula is an asymptotic criterion; finite-$ϵ$ onset requires the full Stefan condition and salt balance.

The solidus bounds the entirely solid equilibrium region of a phase diagram. For a linear liquidus $T_L(C_l)=-m{C}_l$ and a constant segregation coefficient $C_s=k_D{C}_l$, the corresponding solidus is $T_S(C_s)=-m{C}_s/k_D$. The concentration argument is the solid concentration, not the neighboring liquid concentration.

With pure solid crystals, negligible solute diffusion, and the density-change flow in a pulled mushy layer, steady solute balance gives $(1-\phi )C=C_0$. If $\theta =(T-T_0)/\Delta T$, $\mathcal{C}=m{C}_0/\Delta T$, and the liquidus is $T-T_0=-m(C-C_0)$, then

In a stationary mushy layer with fixed $T=T_E+Gz$, pure solid and local liquidus $C=-T/m$, solute conservation is ${\partial }_t[(1-\varphi )C]=D{\partial }_z[(1-\varphi )C_z]$. Since $C_z=-G/m$ is constant, it reduces to $(z+T_E/G){\varphi }_t=D{\varphi }_z$. The method of characteristics makes the displayed $\eta$ constant and gives $\varphi =f(\eta )$ within the region that remains mush. For negative $T_E$, the characteristics move upward where the equilibrium concentration is positive. Extrapolating a zero-solid-fraction level is not sufficient to determine a boundary when additional flux constraints create a liquid cap or a solid-fraction jump.

A stagnant mushy-layer model has no bulk advection, equal phase mass densities, pure solid, and local equilibrium on the liquidus. With liquid fraction $\varphi$, specific enthalpy $E=c_{p}T+L\varphi$ and bulk solute concentration $B=\varphi C$, neglecting solute diffusion inside the mushy layer gives $T_t+(L/c_p){\varphi }_t=\kappa T_{zz}$ and $(\varphi C{)}_t=0$. The latter fixes composition at each stationary material level; it does not automatically set every level to the initial liquid concentration when solute can diffuse ahead of the growing mush.