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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 $b-a\propto \sqrt{t}$.

Use $\ell =L/c_p$, $ϵ=\sqrt{D_s/D_l}$ and $\delta =\sqrt{D_l/\kappa }$, with $0<k_D<1$. Equal reference densities and constant material properties are implicit in these diffusion balances. Put $a(t)=2\lambda \sqrt{D_{l}t}$ and $T_a=-m{C}_a$. The four similarity fields areHere the same letters $T_s,T_l,C_s$ on the right denote their far-field values. Substitution into the heat and solute diffusion equations verifies the fields; their interface values and far-field limits verify every boundary condition. The concentration has an equilibrium jump from $k_D{C}_a$ to $C_a$, whereas temperature is continuous.

The interfacial conservation laws areFor example,Since $\dot{a}=\lambda \sqrt{D_l/t}$, dividing the two jump conditions by $\dot{a}$ gives, for nonzero $\lambda$,where $F(z)=\sqrt{\pi }z{e}^{z^2}\mathrm{erfc}z$. At $\lambda =0$, use the undivided flux conditions or the continuous limit, rather than divide by zero. The liquidus relation and the segregation coefficient also give the solidus $T_S(C)=-mC/k_D$.

For the thermal expansion, write $z=\delta \lambda$ and useKeeping the constant terms in these reciprocals is essential for the full first-order answer. The result isThe remainder has this form for bounded $\lambda$; a looser $O({\delta }^2)$ statement is also valid in the distinguished scaling $\ell =O({\delta }^{-1})$. Dropping the temperature-difference term while claiming first-order accuracy for fixed $\ell$ would be incorrect.

Solving the solute balance algebraically givesFor fixed positive $\lambda$, $F(-\lambda /ϵ)\to -\infty$, and $C_a\to 0$, with a positive exponentially small correction. For fixed negative $\lambda$, $F(-\lambda /ϵ)\to 1$, andIt falls from $C_s$ as $\lambda \to -\infty$ to zero as $\lambda \to 0^{-}$. The outer positive branch is zero. For finite $ϵ$ the transition is rounded over $|\lambda |=O(ϵ)$; the exact zero-speed limit is $C_a(0)=ϵC_s/(1+k_Dϵ)$, not an actual concentration discontinuity.

To count speeds, take the leading distinguished thermal balance, set $\Sigma =T_l+T_s$ and $K=\sqrt{\pi }\delta \ell$, and combine $T_a=-m{C}_a$ with the thermal result:This is the intended leading approximation when $\delta \ell =O(m{C}_s)$ and $T_l-T_s=O(1)$. The full first-order expression instead replaces $K$ by $\sqrt{\pi }\delta [\ell +2(T_l-T_s)/\pi ]$. Thus the printed threshold is asymptotic, not an exact criterion for finite $\delta ,ϵ$.

For $r=-\lambda >0$, put $H(r)=\sqrt{\pi }r{e}^{r^2}\mathrm{erfc}(-r)$. The small-solid-diffusivity melting concentration is $C_{s}H/(1+H)$, soThe second term decreases strictly from $2m{C}_s\sqrt{\pi }$ to zero. One can check the strict decrease without guessing its shape. Set $A(r)=\sqrt{\pi }{e}^{r^2}\mathrm{erfc}(-r)$, so $A^{\prime }=2rA+2$, $A\ge \sqrt{\pi }+2r$, and $H=rA$. Direct differentiation givesThe bracket increases with $A$ on $r\ge 0$. Substituting its lower bound yields a polynomial with constant coefficient $\pi -2>0$ and all other coefficients positive. Hence $[H^{\prime }/(1+H{)}^2{]}^{\prime }<0$, as claimed.

If $\delta \ell >2m{C}_s$, then $K>2m{C}_s\sqrt{\pi }$ and $\Sigma (r)$ increases from zero to infinity. The freezing branch has $\Sigma =-K\lambda <0$ and is also monotone. Together these give one speed for each imposed sum within the approximation. If $0<\delta \ell <2m{C}_s$, the melting branch initially decreases, has exactly one minimum and then increases to infinity. For sums between that minimum and zero there are two melting speeds plus one freezing speed: three solutions. At the lower fold there are two distinct speeds, one being a double solution. At zero the limiting outer branches join; finite solid diffusivity rounds the join into a nearby fold. Away from the fold values the other sums have one solution. At the threshold itself the initial slope vanishes; it is the transition between these cases.

For $\Sigma >0$, the solution is on the melting branch $\lambda <0$. Its thermal layers have scale $\sqrt{\kappa t}=\sqrt{D_{l}t}/\delta$, much larger than the liquid-solute scale $\sqrt{D_{l}t}$. Because the front is retreating, the solid solute layer is thinner still:This follows by expanding the large-positive-argument complementary error function ratio on the solid side: near $a$, its decay length is $D_s/|\dot{a}|$. Using simply $\sqrt{D_{s}t}$ would miss the retreating-front compression of that layer.

Across this narrow layer, solid concentration drops from almost $C_s$ to $k_D{C}_a$, while temperature stays almost $T_a$. At the interface $T_a=T_S(k_D{C}_a)$. Moving slightly into the solid increases its concentration toward $C_s$, lowering its solidus to $-m{C}_s/k_D$, whereas the temperature has barely changed. Since $C_a<C_s$ and $k_D<1$, $T_a=-m{C}_a>-m{C}_s/k_D$. Thus an adjacent region can satisfy $T>T_S(C)$: constitutional superheating. Its existence is especially clear when far-field solid is initially stable but heat diffuses across the narrow composition layer.

The phase-diagram path goes from $(C_s,T_s)$ through a broad thermal rise, then nearly horizontally toward $(k_D{C}_a,T_a)$ across the thin solid diffusion layer. At the interface it traverses the isothermal equilibrium tie line to $(C_a,T_a)$. On the liquid side concentration falls to zero over the liquid-solute layer, while the remaining thermal adjustment to $T_l$ takes place over the much wider heat layer. The path of the depleted solid approaches the solidus from its superheated side. That region should melt at additional sites or develop a two-phase mushy layer; the original single-boundary similarity construction is then a mathematical branch rather than a physically stable description of the entire sample.

Take $z$ upward from the unperturbed interface in the apparatus frame, with liquid at $z>0$ and pulling velocity $-V{\mathbf{e}}_z$. The primary ice contains no salt. We use the dilute, equal-density approximation and neglect fluid motion relative to the mould. The initial salt inventory per unit horizontal area, with a common density factor suppressed, is $N=h{C}_0$. Its conservation is essential: the far liquid has zero concentration, so an imposed nonzero far-field composition would describe a different experiment.

The steady advection-diffusion equation is $-V{C}^{\prime }=D{C}^{\prime \prime }$. Decay into the pure water and interfacial salt rejection giveSince ${\int }_0^{\infty }C(z)dz=D{C}_I/V=N$, the finite-inventory directional salt rejection solution isThe decay length is $D/V$. Local liquidus equilibrium givesThus the thermal apparatus selects the interface position at this isotherm. The stated speed inequality is precisely $C_I<C_E$, so the interface is warmer than the eutectic temperature, $T_I>T_E=-m{C}_E$. These are steady values after the initial finite saline layer has rearranged; the final exponential distribution is not confined to the original depth $h$.

For $Vh{C}_0/D>C_E$, a steady pure-ice front would require an interfacial composition beyond the eutectic composition. That branch is unavailable. During the transient, rejected brine concentrates until it reaches $C_E$; secondary salt-bearing crystals can then form and the residual brine freezes as an eutectic system. Salt is incorporated into the product rather than remaining indefinitely in a super-eutectic exponential layer. The detailed pattern can include brine pockets and a nonplanar freezing region; it depends on the thermal history and the morphological stability. The speed inequality alone does not decide that latter stability.

For the stability calculation use an interface displacement $\eta =\hat{\eta }{e}^{ikx+st}$ and a liquid perturbation $c=\hat{c}{e}^{ikx+st-qz}$. Here $k>0$ is the wavenumber, $s$ is the temporal rate and $\Re q>0$ gives spatial decay. The frozen thermal field is $T_I+Gz$. Define $\Gamma ={\gamma }_{sl}{T}_m/(\rho L)>0$, where ${\gamma }_{sl}$ is solid-liquid surface energy, $T_m$ is absolute pure-solvent melting temperature, $\rho$ is mass density and $L$ is specific latent heat. For a solid protrusion into liquid the curvature is $\mathcal{K}=-{\eta }_{xx}=k^2\eta$, and the Gibbs-Thomson effect gives local equilibrium $T=-mC-\Gamma \mathcal{K}$. We assume isotropic surface energy and instantaneous interface kinetics, so no kinetic undercooling term is present.

Linearizing the equilibrium condition at the displaced interface givesThe perturbation advection-diffusion equation givesThe local solute balance at a moving graph, to first order, isIts perturbation is $(V-Dq)\hat{c}+s{C}_I\hat{\eta }=0$, because the terms $D\eta C^{\prime \prime }(0)$ and $V\eta C^{\prime }(0)$ cancel. Consequently the full frozen-temperature solutal interface dispersion isNo quasistatic approximation to solute diffusion was made here. For a growing mode the unique admissible liquid root isWriting $b=(G+\Gamma k^2)/(m{C}_I)$ and eliminating $s$ gives $(q-V/D)(q-V/D+b)=k^2$. The root continuously connected to the growing interface branch is thereforeIn particular, unstable wavelengths exist exactly when $mV{C}_I/D>G$, and their band isThis is the constitutional supercooling criterion: immediately above the plane, the local liquidus rises at rate $-m{C}^{\prime }(0)=mV{C}_I/D$, so it overtakes the imposed thermal gradient when that rate exceeds $G$. The Gibbs-Thomson effect cuts off short waves. At long wavelength $s\sim D(V/D-b)k^2/b$ for positive $G$. The exactly uniform inventory-changing displacement is not an instability of a fixed-inventory experiment. On stable branches the full diffusion problem also admits relaxation transients; the root selection above is sufficient to identify every exponentially growing interface mode.

If one additionally neglects the perturbation time derivative in the liquid diffusion equation, the alternative reduced rate isIt has the same onset and cutoff, but need not have the same rate away from the slow-growth regime. The frozen-temperature approximation by itself does not justify this extra reduction.

After onset, protrusions reach more weakly saline liquid and advance faster, while grooves accumulate rejected brine. The planar interface develops cells and, as growth becomes sufficiently strong, crystal dendrites with concentrated liquid between them. Nonlinear competition changes the selected spacing; linear rates alone cannot fix the final geometry. The interdendritic liquid forms a mushy layer, concentrates toward the eutectic composition, and ultimately freezes into a salt-bearing eutectic product when it reaches $T_E$. This replaces the pure planar salt-rejection picture with trapping and freezing of brine between crystals.

Work in the frame of the steadily advancing mush, so the rigid primary-crystal skeleton moves with $-V{\mathbf{e}}_z$. Let $\varphi$ be solid fraction, $C$ the pore-liquid concentration, $\mathbf{q}=(u,w)$ the Darcy flux relative to that skeleton, and $K$ the common thermal conductivity. Set $\kappa =K/(\rho c_p)$, the thermal diffusivity. The actual pore velocity is $\mathbf{q}/(1-\varphi )-V{\mathbf{e}}_z$. Primary crystals are taken to exclude solute; the residual pore liquid freezes into a two-solid eutectic product at the bottom. These are the standard binary mushy layer assumptions implicit in using a single liquidus and an eutectic floor.

It is convenient to use hydraulic pressure $P=p+\rho gz$, absorbing the constant-density gravitational term. With equal phase properties, no solute diffusion, negligible fluid inertia and local liquidus equilibrium, a complete mush model isThe last equation follows from conservation of energy with mixture specific enthalpy $c_{p}T-L\varphi$. Sensible heat is transported by the total volume flux $\mathbf{q}-V{\mathbf{e}}_z$; the primary solid carries the latent-enthalpy flux $VL\varphi {\mathbf{e}}_z$. In particular the sign of the latent-heat term is fixed by the downward-moving skeleton.

At the liquid roof, imposeHere $P_0=p_0+\rho gh$. The gradient condition matches heat flux into the imposed liquid field, since there is no phase-fraction jump at this roof and the phase conductivities are equal. The pore velocity there equals the relative Darcy velocity. At the floor,There is no liquid penetration into fully solid material. The unspecified solid fraction at the floor is determined by the solute balance, not set to one: remaining eutectic liquid freezes as it crosses the floor. If $C_s$ denotes the bulk concentration of the final solid product, the floor solute balance and Stefan condition areMaintaining the floor at $T_E$ permits the solid-side bath to remove this final latent heat. Equivalently its required gradient is $T_z(0^{-})=T_z(0^{+})+\rho LV[1-\varphi (0)]/K$. No additional imposed solid-side gradient is specified. Laterally we take the quadratic-pressure strain solution, without sidewalls; finite sidewalls would require additional flow conditions and generally break this horizontal invariance.

For uniform permeability of a porous medium, incompressible flow gives ${\nabla }^{2}P=0$. The roof pressure and zero normal flow at the floor are satisfied byHorizontal outflow is balanced by downward inflow. The parameter $\alpha$ is an inverse time; $E^2$ in the printed pressure condition already has dimensions of pressure divided by length squared, so no extra density factor belongs in this definition.

With $T=T(z)$, $C=C(z)$ and $\varphi =\varphi (z)$, the $x$-dependence disappears from both scalar balances. Although $u$ is nonzero, its horizontal divergence cancels $w_z$ in $\nabla \cdot (\mathbf{q}C)$. The exact strain-driven steady mush equations becomeTogether with $T(0)=T_E$, $T(h)=T_0$, $T^{\prime }(h)=G$ and $\varphi (h)=0$, they determine the scalar fields and the unknown thickness. This reduction proves that the equations admit horizontally invariant scalar solutions despite the imposed straining flow. An equivalent useful form, obtained by differentiating the solute equation, isThis exact form also shows why one must not silently discard latent heat in an arbitrary strong-flow calculation.

Define $\Delta T=T_0-T_E>0$, $R=\alpha h/V$ and $S=L/(c_p\Delta T)$. For the stated regular ordering $R\gg S\gg 1$, strain transport dominates skeleton transport and the bounded phase-change enthalpy contribution. More precisely, at fixed finite thermal strain number $\alpha h^2/\kappa$, the dimensionless energy equation isso its skeleton and integrated latent-heat corrections are small in $1/R$ and $S/R$ respectively. The leading strain-dominated mush temperature profile obeys $\kappa T^{\prime \prime }+\alpha z{T}^{\prime }=0$. Put $\delta =\sqrt{2\kappa /\alpha }$ and $H=h/\delta$. Integrating first for $T^{\prime }$ and then for $T$ givesThe error function profile follows from integrating a Gaussian, with $\mathrm{erf}x=(2/\sqrt{\pi }){\int }_0^x{e}^{-s^2}ds$. Matching the roof gradient gives the requested transcendental thickness relationFor prescribed $G>0$ and $\Delta T>0$ the right side decreases continuously from infinity to zero as $H$ increases, so it fixes one positive leading-order thickness. The resulting $h$ must still satisfy the assumed ordering and the phase-fraction constraints.

To find the product composition, integrate the exact solute balance from the roof to an arbitrary depth. With $f=1-\varphi$,At $z=0$, the exact floor composition is thereforeThe minus sign has a direct physical meaning: the horizontal outflow removes more salt than arrives at the same depth from a uniform-$C_0$ liquid, leaving a depleted solid product. Using the leading error function profile andyieldsThe full solid fraction follows by substituting this same profile into the preceding integral formula for $f(z)$. It has $\varphi (h)=0$ and $1-\varphi (0)=C_s/C_E$. A physical branch requires $0\le C_s\le C_E$ and $0\le \varphi \le 1$. In this monotone profile, $C_s\ge 0$ ensures a nonnegative liquid fraction everywhere; it is not permissible to interpret a negative formal result as a material composition.

The error-function reduction is a regular leading approximation, not a uniform statement for every possible scaling of the thermal strain number. If $\alpha h^2/\kappa$ becomes singularly large along the ordering limit, a thin basal layer can amplify the discarded terms. The printed two inequalities alone do not specify that extra scaling. In such a regime the displayed exact scalar equations and boundary conditions, rather than an unqualified zero-latent-heat profile, provide the appropriate thickness and composition calculation.

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.

Take $G,V,D,m,C_0>0$ and $T_E<0$, with the liquidus referenced to the pure-water melting point. The solid-liquid segregation coefficient $k_D=0$ means that newly formed ice contains no salt. A steady sharp front producing only pure ice cannot remove the incoming solute flux. More explicitly, in the liquid the steady solute flux is $J_C=-VC-D{C}_z=-V{C}_0$, whereas in pure solid it is zero. Interfacial salt conservation would require $V{C}_0=0$, a contradiction. An equilibrium mushy layer avoids this by carrying concentrated pore liquid down to the eutectic region, where a salt-bearing solid assemblage can form.

Use equal densities and common downward translation velocity $-V$. Write $q=1-\varphi$ for the liquid fraction; $\varphi$ here is the solid fraction. Local liquidus equilibrium in the mush requiresAbove a stationary roof $h_0$, the liquid advection-diffusion equation is $-V{C}_z=D{C}_{zz}$, with solutionThe marginal equilibrium at a mush-liquid boundary condition is $-m{C}_l^{\prime }(h_0)=G$, hence $C_i=C_0+GD/(mV)$. Matching the temperature and liquidus at the roof givesA positive equilibrium mush requires $(-m{C}_0-T_E)/G>D/V$; otherwise this positive-thickness construction does not apply.

Including solute diffusion in the liquid pore space, the steady solute flux in the mush isThus $q[C+(D/V)C_z]=C_0$. Inserting $C_z=-G/m$ and the value of $h_0$ yieldsThis is the diffusive steadily pulled equilibrium mush. Omitting the diffusive contribution would give a different solid-fraction distribution.