= Solution
Use the usual <saline Stefan problem> approximation: no bulk flow, salt-free <ice> with negligible salt transport, constant properties, equal phase densities, and the same <thermal conductivity> $k_T$ and <thermal diffusivity> $\kappa$ in both phases. Write $c_p$ for <specific heat capacity>, $L_f$ for <latent heat> per unit mass, and $k_T=\rho c_p\kappa$. These thermal symmetries are needed for the arithmetic-mean interface <temperature> requested in the paper; unequal phase conductivities would give a weighted balance instead.
Set $\eta=x/(2\sqrt{\kappa t})$, $\xi=x/(2\sqrt{Dt})$, $\epsilon=\sqrt{D/\kappa}$ and $a(t)=2\lambda\sqrt{Dt}$. The <heat equation> and salt <diffusion equation> reduce to $f''+2sf'=0$. Their <similarity solutions>, in terms of the <complementary error function>, are
$$
\begin{aligned}
T_s(x,t)&=T_{-\infty}+(T_i-T_{-\infty})\frac{\operatorname{erfc}(-\eta)}{\operatorname{erfc}(-\epsilon\lambda)},&&x<a(t),\\
T_l(x,t)&=T_\infty+(T_i-T_\infty)\frac{\operatorname{erfc}(\eta)}{\operatorname{erfc}(\epsilon\lambda)},&&x>a(t),\\
C(x,t)&=C_0+(C_i-C_0)\frac{\operatorname{erfc}(\xi)}{\operatorname{erfc}(\lambda)},&&x>a(t).
\end{aligned}
$$
These have the required interface values and far-field limits. At each fixed $x\ne0$ they recover the initial data as $t\downarrow0$. The <liquidus> condition is $T_i=-mC_i$.
<Salt rejection> and the <Stefan condition> give, with <gradients> evaluated on the appropriate sides of the interface,
$$
-D C_x(a^+,t)=\dot a\,C_i,\qquad \rho L_f\dot a=k_T[T_{s,x}(a^-,t)-T_{l,x}(a^+,t)].
$$
For example, <salt rejection> is obtained by differentiating the total salt on a moving liquid interval: the moving lower endpoint removes $\dot a C_i$, which must be supplied by diffusive transport away from the salt-free solid. Substitution, including the <salt-rejection function for a saline Stefan front>, gives the complete algebraic system for the <diffusion-controlled iceberg growth and ablation>:
$$
\boxed{\begin{aligned}
C_i\left[1-\sqrt\pi\lambda e^{\lambda^2}\operatorname{erfc}(\lambda)\right]&=C_0,\\
T_i&=-mC_i,\\
\frac{L_f}{c_p}\epsilon\lambda&=\frac{e^{-\epsilon^2\lambda^2}}{\sqrt\pi}\left[\frac{T_i-T_{-\infty}}{\operatorname{erfc}(-\epsilon\lambda)}-\frac{T_\infty-T_i}{\operatorname{erfc}(\epsilon\lambda)}\right].
\end{aligned}}
$$
The sign of $\lambda$ distinguishes freezing from melting; neither sign should be excluded in the general <similarity solution>.
For $\lambda=O(1)$, $\epsilon\ll1$, and fixed $L_f/c_p$ and far-field <temperatures>, the thermal equation has leading right-hand side $(2T_i-T_{-\infty}-T_\infty)/\sqrt\pi$, while its left-hand side is $O(\epsilon)$. Hence
$$
\boxed{T_i=\frac{T_\infty+T_{-\infty}}2+O(\epsilon).}
$$
This is a leading-order balance, not an exact cancellation of <latent heat>. The resulting $C_i=-T_i/m$ and the first algebraic equation determine the leading $\lambda$. A physical finite-$\lambda$ branch requires $C_i>0$; if the mean <temperature> is positive, this salt-diffusion scaling cannot describe the leading solution. Likewise, a <latent-to-sensible heat ratio> diverging as $1/\epsilon$ changes the leading thermal balance.
Define $\theta_-=-mC_0-T_{-\infty}$ and $\theta_+=T_\infty+mC_0$. Put $T_0=-mC_0$ and $\Delta\theta=\theta_--\theta_+$. Then $T_i=T_0-\Delta\theta/2+O(\epsilon)$, which will determine the freezing and <constitutional supercooling> conditions below.
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