= Solution
Write
$$
S=(1-\alpha)F_r,
\qquad
F_*=\epsilon\sigma T_m^4-S.
$$
The quasi-steady model is the coupled pair
$$
\epsilon\sigma T_s^4
=S+\frac ah\log\frac{T_m}{T_s},
\qquad
\rho L\dot h=\frac ah\log\frac{T_m}{T_s}-F_w.
$$
For a thin shell, $T_s$ is close to $T_m$. A first-order <asymptotic expansion> gives
$$
\log\frac{T_m}{T_s}\sim\frac{T_m-T_s}{T_m},
$$
and therefore
$$
\boxed{T_s\sim T_m-\frac{hT_m}{a}F_*},
\qquad h\ll \frac a{F_w}.
$$
To leading order $F_c\sim F_*$, so an initially freezing shell grows linearly:
$$
\boxed{h(t)\sim h(0)+\frac{F_*-F_w}{\rho L}t}.
$$
For a thick shell the conductive correction in the surface balance is small. The <radiative equilibrium> temperature and its first correction are
$$
T_e=\left(\frac{S}{\epsilon\sigma}\right)^{1/4},
\qquad
\boxed{T_s\sim T_e+
\frac{a\log(T_m/T_e)}
{4\epsilon\sigma T_e^3h}}.
$$
Thus, with $A=a\log(T_m/T_e)$,
$$
\boxed{\rho L\dot h\sim\frac Ah-F_w},
\qquad
h_\infty\sim\frac A{F_w}.
$$
When $F_w$ is negligible over an intermediate range, this <ordinary differential equation> gives the square-root growth law
$$
h^2-h_0^2\sim\frac{2A}{\rho L}(t-t_0).
$$
Retaining $F_w$, its implicit solution is
$$
t-t_0=\frac{\rho L}{F_w}
\left[
h_0-h+h_\infty
\log\frac{h_\infty-h_0}{h_\infty-h}
\right].
$$
Consequently a sketch of $h(t)$ starts approximately linearly, crosses to square-root growth, and approaches $h_\infty$ with <exponential decay> of $h_\infty-h$. A shell placed above $h_\infty$ instead thins because the basal oceanic heat flux exceeds the conductive loss.
Solved by gpt-5.6-sol high.
Back to article page
