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Let the surface advance downwards at speed $V>0$. Ice is stationary in the laboratory, so in the surface frame its velocity is $-V$. The steady heat equation with volumetric heating is thereforeTake the atmospheric heat flux to be negligible in the stated model. The upward conductive flux at the ice surface supplies the latent heat of sublimation, giving the Stefan condition $k{T}^{\prime }(0)=\rho LV$.

Write $p=V/\kappa$ and $\Delta =T_s-T_{\infty }$. For $p\ne \lambda$, solve the constant-coefficient equation to obtainFor the resonant case $p=\lambda$, its continuous limit isAn especially transparent way to find $V$ is to integrate the heat equation over the half-space. Its far-field gradient vanishes, soUsing the surface heat balance givesAbsorbed radiation supplies both the latent heat and the sensible heat needed to bring incoming cold ice to the surface temperature. This is the radiatively heated sublimation profile.

The surface gradient isThus temperature first rises into the solid, but ultimately falls to the colder far field: it has a subsurface maximum. For a useful dimensionless form, set $x=\lambda z$, $r=p/\lambda$ and $S=L/(c_p\Delta )$. Thenwith limit $[1+(S+1)x]e^{-x}$ when $r=1$. The maximum is at $x_{\ast }=[\log ((rS+1)/(S+1))]/(r-1)$, or $S/(S+1)$ at $r=1$.

Melting begins below the surface if the interior maximum reaches $T_m$. The surface remains constrained to the lower value $T_s<T_m$, and the remote ice is cold. Increasing absorbed heating can raise the interior maximum to the melting point, after which an all-solid steady solution is no longer valid and an internal melt region must be modeled. The maximum, rather than the far-field temperature or surface temperature alone, decides onset.

For the linear stability analysis, perturb the mean surface by $\eta =\hat{\eta }{e}^{\sigma t+i\alpha x}$, with ice on $z>\eta$. Approximate the base temperature locally by $T_s+Gz$. The prescribed radiative source has no perturbation, so the perturbation heat equation is homogeneous. Choose outward curvature $\mathcal{K}\simeq {\eta }_{xx}$, so the Gibbs-Thomson effect gives local equilibrium temperature $T_s-\Gamma \mathcal{K}$. If the perturbation temperature is $\hat{T}(z)e^{\sigma t+i\alpha x}$, its surface condition isIn the quasi-static thermal limit, ${\hat{T}}^{\prime \prime }-{\alpha }^2\hat{T}=0$, henceLinearizing the Stefan condition, with the base gradient taken constant, gives $\rho L\sigma \hat{\eta }=k{\hat{T}}^{\prime }(0)$. ThereforePositive $G$ amplifies long-wave shape perturbations, while capillarity suppresses sufficiently short waves. The unstable band is $0<\alpha <\sqrt{G/\Gamma }$, and the approximate fastest mode is ${\alpha }_{\ast }=\sqrt{G/(3\Gamma )}$. This is the sublimation-front capillary instability.

The qualifications on wavelength matter. Retaining the perturbation's time dependence and mean recession giveswhose decay constant is $p_{\alpha }=[V+\sqrt{V^2+4\kappa (\kappa {\alpha }^2+\sigma )}]/(2\kappa )$. Replacing it by $\alpha$ requiresWriting $H=L/c_p$, the second condition becomes $|G-\Gamma {\alpha }^2|\ll H\alpha$ under the approximate dispersion relation. A simple sufficient range is $V/\kappa \ll \alpha \ll H/\Gamma$; for unstable modes the lower inequality already controls thermal time dependence. Large $S$ alone does not justify arbitrarily small $\alpha$.

Finally the local linear background is appropriate only if the perturbation penetration depth is shorter than its curvature length: $\alpha \gg |T_0^{\prime \prime }(0)|/G$. From the steady equation this ratio is $V/\kappa +\lambda (1+1/S)$. Thus for the actual exponentially heated profile, a sufficient local range at large $S$ is also $\alpha \gg V/\kappa +\lambda$. Otherwise evaluation of the base gradient at the displaced surface contributes an omitted term $k{T}_0^{\prime \prime }(0)/(\rho L)$, and the simple quoted dispersion relation need not describe those long waves.