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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.

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.

In the frame of a planar interface moving at speed $V$, the steady liquid temperature obeys $-V{T}_z=\kappa T_{zz}$ and approaches $T_{\infty }$. ThusThe Stefan condition, with negligible solid-side heat flux, givesso $T_i-T_{\infty }=L/c_p=S\Delta T$. The kinetic undercooling law then giveswhich requires $S<1$ for advancing solidification.

Scale length with $\kappa /V$, time with $\kappa /V^2$, and temperature with $\Delta T$. The base liquid temperature is ${\Theta }_0=S{e}^{-z}$. For an interface displacement $\zeta e^{i\alpha x+\sigma t}$, write the thermal perturbation asThe heat equation givesLinearizing the Stefan condition giveswhile the kinetic law with stabilizing curvature undercooling givesEliminating $A$ gives the general implicit dispersion relation

At marginal stability $\sigma =0$,and the dispersion relation becomesSince ${\alpha }^2=\lambda (\lambda -1)$ on this curve, the right-hand side is simply $\Gamma {\lambda }^2$. A positive cutoff wavenumber therefore exists exactly whenThe unstable band is $0<\alpha <{\alpha }_c$, whereA sketch of $\Gamma {\lambda }^2$ against the constant $S/(1-S)$ shows no crossing above $\alpha =0$ below threshold and one cutoff above it. As $S\uparrow 1$, ${\lambda }_c$ and ${\alpha }_c$ diverge like $(1-S{)}^{-1/2}$, so the range of the morphological instability of a kinetically limited solidification front becomes unbounded.