One-sided kinetic solidification dispersion ($\sigma (1+S/p)=SV-\Gamma {\alpha }^2-S{V}^2/p$)

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.

The cubic follows by eliminating $\sigma =p^2-Vp-{\alpha }^2$ from one-sided kinetic solidification dispersion, with $V=1-S$. For $\Gamma =0$, the interface branch tends to $\sigma =SV-SV/\alpha +O({\alpha }^{-2})$. For fixed $0<\Gamma <1$, substituting $p=\sqrt{1-\Gamma }\alpha +d$ gives $2d-V=S\Gamma /(1-\Gamma )$, hence $\sigma =-\Gamma {\alpha }^2+[S\Gamma /\sqrt{1-\Gamma }]\alpha +O(1)$. At $\Gamma =1$ the balance is $p^3\sim S{\alpha }^2$, giving $\sigma =-{\alpha }^2+S^{2/3}{\alpha }^{4/3}+o({\alpha }^{4/3})$. For $1<\Gamma <1+2S/V$, $p\to S/(\Gamma -1)>V/2$ and $\sigma =-{\alpha }^2+p^2-Vp+o(1)$. Stronger capillarity can remove the real isolated interface mode from the chosen thermal branch; it does not create growth. These distinctions explain why replacing $p$ by $\alpha$ is not a controlled fixed-capillarity limit.

For $0<S<1$ and $V=1-S$, the translational branch of one-sided kinetic solidification dispersion has $\sigma =(S/V-\Gamma ){\alpha }^2+O({\alpha }^4)$. Expand $p=V+(\sigma +{\alpha }^2)/V+O({\alpha }^4)$ and substitute to prove this limit. At zero growth, $p_0=[V+\sqrt{V^2+4{\alpha }^2}]/2$, and the destabilizing right side has sign ${\alpha }^2(SV/p_0^2-\Gamma )$. For nonnegative real $\sigma$, the residual has derivative at least one, so there is one growing root precisely when $\Gamma <SV/p_0^2$. Thus instability exists exactly when $\Gamma <S/V$, with unstable band ${\alpha }^2<p_c(p_c-V)$, $p_c=\sqrt{SV/\Gamma }$ for $\Gamma >0$. At threshold the next term is $\sigma =-2S{\alpha }^4/V^3+O({\alpha }^6)$. Surface energy removes short-wave growth even when long waves remain unstable.