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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.

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.