= Solution
In the frame of a planar interface moving at speed $V$, the steady liquid temperature obeys $-VT_z=\kappa T_{zz}$ and approaches $T_\infty$. Thus
$$
T=T_\infty+(T_i-T_\infty)e^{-Vz/\kappa}.
$$
The <Stefan condition>, with negligible solid-side heat flux, gives
$$
LV=c_p(T_i-T_\infty)V,
$$
so $T_i-T_\infty=L/c_p=S\Delta T$. The <kinetic undercooling> law then gives
$$
\boxed{V=G(T_m-T_i)=G\Delta T(1-S),}
$$
which 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 $\Theta_0=Se^{-z}$. For an interface displacement $\zeta e^{i\alpha x+\sigma t}$, write the thermal perturbation as
$$
\theta=Ae^{-\lambda z+i\alpha x+\sigma t}.
$$
The <heat equation> gives
$$
\boxed{\lambda^2-\lambda-\alpha^2-\sigma=0,
\qquad
\lambda=\frac12\left[1+\sqrt{1+4(\alpha^2+\sigma)}\right].}
$$
Linearizing the <Stefan condition> gives
$$
\lambda A=S(1+\sigma)\zeta,
$$
while the kinetic law with stabilizing curvature undercooling gives
$$
A=\left[S-(1-S)(\sigma+\Gamma\alpha^2)\right]\zeta,
\qquad
\Gamma=\frac{\gamma G}{\kappa}.
$$
Eliminating $A$ gives the general implicit <dispersion relation>
$$
\boxed{
\frac{S(1+\sigma)}{\lambda}
=S-(1-S)(\sigma+\Gamma\alpha^2),
\qquad
\lambda^2-\lambda=\alpha^2+\sigma.}
$$
At marginal stability $\sigma=0$,
$$
\lambda=\frac12(1+\sqrt{1+4\alpha^2}),
$$
and the dispersion relation becomes
$$
\boxed{\frac S{1-S}=\Gamma\frac{\lambda}{\lambda-1}\alpha^2.}
$$
Since $\alpha^2=\lambda(\lambda-1)$ on this curve, the right-hand side is simply $\Gamma\lambda^2$. A positive cutoff wavenumber therefore exists exactly when
$$
\frac S{\Gamma(1-S)}>1,
\qquad\text{that is,}\qquad
\boxed{S>\frac{\Gamma}{1+\Gamma}.}
$$
The unstable band is $0<\alpha<\alpha_c$, where
$$
\lambda_c=\left[\frac S{\Gamma(1-S)}\right]^{1/2},
\qquad
\alpha_c^2=\lambda_c(\lambda_c-1).
$$
A 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 $S\uparrow1$, $\lambda_c$ and $\alpha_c$ diverge like $(1-S)^{-1/2}$, so the range of the <morphological instability of a kinetically limited solidification front> becomes unbounded.
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