= Solution
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> $\mathcal M>0$:
$$
\boxed{U=\mathcal M(T_m-\gamma_T\mathcal K-T_i),\qquad
T_i=T_m-\frac U{\mathcal M}-\gamma_T\mathcal K,\qquad
\gamma_T=\frac{T_m\gamma}{\rho L}.}
$$
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 define
$$
U_0=\mathcal M\Delta T,\quad \ell=\kappa/U_0,\quad t_0=\kappa/U_0^2,\quad
\theta=\frac{T-T_\infty}{\Delta T},\quad S=\frac{L}{c_p\Delta T},\quad
\Gamma=\frac{\gamma_T}{\Delta T\ell}.
$$
The 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 $\theta_0''+V\theta_0'=0$, while the bounded solid steady field is constant. Matching a decaying liquid profile to the <Stefan condition> $SV=-\theta_0'(0)$ gives
$$
\boxed{\theta_l(\xi)=S e^{-V\xi}\quad(\xi>0),\qquad \theta_s=S,
\qquad T_i=T_\infty+\frac L{c_p}.}
$$
The <kinetic undercooling> condition is $\theta_i=1-V$. Thus
$$
\boxed{V=1-S>0,\qquad U=\mathcal M\left(T_m-T_\infty-\frac L{c_p}\right)=\frac{\mathcal ML}{c_p}(S^{-1}-1).}
$$
The 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 $A_1e^{-p\xi}e^{\sigma t+i\alpha y}$. The liquid <heat equation> requires
$$
p^2-Vp=\sigma+\alpha^2,\qquad
\boxed{p=\frac{V+\sqrt{V^2+4(\sigma+\alpha^2)}}2.}
$$
For 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 $A_1+\theta_0'(0)\zeta=A_1-SV\zeta$. The <kinetic undercooling> law therefore gives
$$
A_1=(SV-\sigma-\Gamma\alpha^2)\zeta.
$$
Likewise the derivative evaluated at the displaced surface is $\theta_0'(0)+\theta_0''(0)\zeta-pA_1$, with $\theta_0''(0)=SV^2$. The <Stefan condition> at perturbation order reads
$$
S\sigma\zeta=pA_1-SV^2\zeta.
$$
Elimination proves the <one-sided kinetic solidification dispersion>:
$$
\boxed{\sigma\left(1+\frac Sp\right)=SV-\Gamma\alpha^2-\frac{SV^2}{p}.}
$$
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 gives
$$
\sigma\left(1+\frac1{\sqrt{\sigma+\alpha^2}}\right)=-\Gamma\alpha^2.
$$
For $\Gamma>0$ 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. Expand
$$
p=V+\frac{\sigma+\alpha^2}{V}+O(\alpha^4),\qquad
\frac{V^2}{p}=V-\frac{\sigma+\alpha^2}{V}+O(\alpha^4).
$$
Substituting into the <dispersion relation> cancels $S\sigma/V$ on both sides and gives
$$
\boxed{\sigma=\left(\frac SV-\Gamma\right)\alpha^2+O(\alpha^4).}
$$
At 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 is
$$
p^3+(S-V)p^2+[(\Gamma-1)\alpha^2-2SV]p-S\alpha^2+SV^2=0,\qquad
\sigma=p^2-Vp-\alpha^2.
$$
It prevents the uncontrolled replacement $p\simeq\alpha$ when $\sigma$ itself is of order $\alpha^2$. In the usual small-capillarity range the leading results are
$$
\boxed{\Gamma=0:\quad \sigma=SV-\frac{SV}{\alpha}+O(\alpha^{-2}),}
$$
$$
\boxed{0<\Gamma<1:\quad
\sigma=-\Gamma\alpha^2+\frac{S\Gamma}{\sqrt{1-\Gamma}}\alpha+O(1).}
$$
For 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 $\Gamma\ll1$ and $|\sigma|\ll\alpha^2$, the useful reduced approximation is
$$
\sigma\simeq\frac{SV-\Gamma\alpha^2-SV^2/\alpha}{1+S/\alpha};
$$
its 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 $1<\Gamma<1+2S/V$, $p\to p_*\equiv S/(\Gamma-1)>V/2$, so $\sigma=-\alpha^2+p_*^2-Vp_*+O(\alpha^{-2})$. 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 $k\ge0$. 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> becomes
$$
\alpha^2\left(\frac{SV}{p_0^2}-\Gamma\right).
$$
For real $\sigma\ge0$, the residual $F=\sigma(1+S/p)-SV+\Gamma\alpha^2+SV^2/p$ is strictly increasing, since
$$
\frac{dF}{d\sigma}=1+\frac{S(p^2+\alpha^2-V^2)}{p^2(2p-V)}\ge1.
$$
Therefore a unique positive root exists precisely when $\Gamma<SV/p_0^2$. For $0<\Gamma<S/V$, the unstable band is
$$
\boxed{0<\alpha<\alpha_c,\qquad \alpha_c^2=p_c(p_c-V),\qquad p_c=\sqrt{SV/\Gamma}.}
$$
There are no unstable oscillatory roots hidden by this real-root test. Indeed, if $p=a+ib$ and $\operatorname{Re}\sigma\ge0$, the thermal equation implies $a\ge V$. For $b\ne0$, 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 $\sigma$.
\Image[/past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2010/iii/paper-72-instability.png]
{title=Growth-rate curves for a one-sided kinetic freezing front}
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 $S\to1$ 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.
Back to article page
