= Solution
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 therefore
$$
kT''+\rho c_pVT'+q_0e^{-\lambda z}=0,
\qquad T(0)=T_s,\quad T(\infty)=T_\infty.
$$
Take 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> $kT'(0)=\rho LV$.
Write $p=V/\kappa$ and $\Delta=T_s-T_\infty$. For $p\ne\lambda$, solve the constant-coefficient equation to obtain
$$
\boxed{T(z)=T_\infty+\Delta e^{-pz}
+\frac{q_0}{k\lambda(p-\lambda)}(e^{-\lambda z}-e^{-pz}).}
$$
For the resonant case $p=\lambda$, its continuous limit is
$$
\boxed{T(z)=T_\infty+
\left[\Delta+\frac{q_0z}{kp}\right]e^{-pz}.}
$$
An especially transparent way to find $V$ is to integrate the <heat equation> over the half-space. Its far-field gradient vanishes, so
$$
-kT'(0)+\rho c_pV(T_\infty-T_s)+\frac{q_0}\lambda=0.
$$
Using the surface heat balance gives
$$
\boxed{V=\frac{q_0/\lambda}{\rho L+\rho c_p(T_s-T_\infty)}.}
$$
Absorbed 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 is
$$
G=T'(0)=\frac{\rho LV}k>0.
$$
Thus <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 $S=L/(c_p\Delta)$. Then
$$
\frac{T-T_\infty}{\Delta}
=\frac{r(S+1)e^{-x}-(rS+1)e^{-rx}}{r-1},
$$
with limit $[1+(S+1)x]e^{-x}$ when $r=1$. The maximum is at $x_*=[\log((rS+1)/(S+1))]/(r-1)$, or $S/(S+1)$ at $r=1$.
\Image[/past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2002/iii/paper-55-sublimation-temperature.png]
{title=Subsurface temperature maxima under absorbed radiation and the criterion for internal melting}
\b[<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 point, after which an all-solid steady solution is no longer valid and an internal melt region must be modeled.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 is
$$
\hat T(0)+G\hat\eta=\Gamma\alpha^2\hat\eta.
$$
In the quasi-static thermal limit, $\hat T''-\alpha^2\hat T=0$, hence
$$
\hat T(z)=(\Gamma\alpha^2-G)\hat\eta e^{-\alpha z}.
$$
Linearizing the <Stefan condition>, with the base gradient taken constant, gives $\rho L\sigma\hat\eta=k\hat T'(0)$. Therefore
$$
\boxed{\sigma\simeq\frac{k}{\rho L}\alpha(G-\Gamma\alpha^2).}
$$
Positive $G$ 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 $\alpha_* =\sqrt{G/(3\Gamma)}$. This is the <sublimation-front capillary instability>.
The qualifications on wavelength matter. Retaining the perturbation's time dependence and mean recession gives
$$
\kappa\hat T''+V\hat T'-(\kappa\alpha^2+\sigma)\hat T=0,
$$
whose decay constant is $p_\alpha=[V+\sqrt{V^2+4\kappa(\kappa\alpha^2+\sigma)}]/(2\kappa)$. Replacing it by $\alpha$ requires
$$
\boxed{\frac V{\kappa\alpha}\ll1,\qquad
\frac{|\sigma|}{\kappa\alpha^2}\ll1.}
$$
Writing $H=L/c_p$, 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: $\alpha\gg |T_0''(0)|/G$. 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 $kT_0''(0)/(\rho L)$, and the simple quoted dispersion relation need not describe those long waves.
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