Past exam of the mathematics course of the University of Cambridge / 2017 / iii / Paper 332 / 2

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 $\eta =x/(2\sqrt{\kappa t})$, $\xi =x/(2\sqrt{Dt})$, $ϵ=\sqrt{D/\kappa }$ and $a(t)=2\lambda \sqrt{Dt}$. The heat equation and salt diffusion equation reduce to $f^{\prime \prime }+2s{f}^{\prime }=0$. Their similarity solutions, in terms of the complementary error function, areThese have the required interface values and far-field limits. At each fixed $x\ne 0$ they recover the initial data as $t\downarrow 0$. The liquidus condition is $T_i=-m{C}_i$.

Salt rejection and the Stefan condition give, with gradients evaluated on the appropriate sides of the interface,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:The sign of $\lambda$ distinguishes freezing from melting; neither sign should be excluded in the general similarity solution.

For $\lambda =O(1)$, $ϵ\ll 1$, 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 $O(ϵ)$. HenceThis 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 $1/ϵ$ changes the leading thermal balance.

Define ${\theta }_{-}=-m{C}_0-T_{-\infty }$ and ${\theta }_{+}=T_{\infty }+m{C}_0$. Put $T_0=-m{C}_0$ and $\Delta \theta ={\theta }_{-}-{\theta }_{+}$. Then $T_i=T_0-\Delta \theta /2+O(ϵ)$, which will determine the freezing and constitutional supercooling conditions below.

Let $E(\lambda )=\sqrt{\pi }\lambda e^{{\lambda }^2}\mathrm{erfc}(\lambda )$. The salt equation in the saline Stefan problem gives $C_i-C_0=E(\lambda )C_i$. Since the complementary error function is positive on the real axis and $C_i>0$, the signs of $C_i-C_0$ and $\lambda$ agree. Thus freezing is equivalent to $C_i>C_0$, or $T_i<T_0=-m{C}_0$. Using the leading thermal balance,at leading order away from a vanishing temperature difference. In fact the exact onset is also ${\theta }_{-}={\theta }_{+}$: at $\lambda =0$ the salt balance forces $C_i=C_0$, and the Stefan condition then forces $T_{-\infty }+T_{\infty }=2T_0$. The solution with $\lambda =0$ has a stationary interface and balancing conductive heat fluxes, despite nontrivial diffusion of the initial temperature discontinuity. For completeness, the exact freezing criterion follows without relying on that local crossing. By the salt-rejection function for a saline Stefan front, $T_i(\lambda )$ decreases continuously from $T_0$ to $-\infty$ as positive $\lambda$ increases. If $T_{-\infty }<T_0$, let ${\lambda }_s$ be where $T_i=T_{-\infty }$. On $[0,{\lambda }_s]$, the solid-side thermal term decreases, the liquid-side term increases, and the latent-heat term increases: the factors $e^{-{ϵ}^2{\lambda }^2}/\mathrm{erfc}(-ϵ\lambda )$ and $e^{-{ϵ}^2{\lambda }^2}/\mathrm{erfc}(ϵ\lambda )$ decrease and increase respectively. These monotonicities follow from the positive integral representation of $e^{v^2}\mathrm{erfc}v$. The thermal residual starts at $({\theta }_{-}-{\theta }_{+})/\sqrt{\pi }$ and is negative at ${\lambda }_s$, so a positive root exists uniquely precisely when ${\theta }_{-}>{\theta }_{+}$. Beyond ${\lambda }_s$ both thermal contributions to freezing are negative, so no further root is possible. If $T_{-\infty }\ge T_0$, the same sign argument excludes every positive root.

For freezing, the liquid concentration decreases away from the interface, so the liquidus $T_L=-mC$ increases there. Constitutional supercooling means $T_l<T_L$ somewhere ahead of the interface. Since these temperatures agree at the interface, its local onset isFor the similarity solution and salt rejection condition, this is exactlyIt is also the global criterion for this freezing solution when $D<\kappa$ and $T_{\infty }>T_i$: the ratio $T_L^{\prime }(x)/T_l^{\prime }(x)$ is a positive constant times $\exp [-x^2(1/D-1/\kappa )/(4t)]$. For $x\ge a\ge 0$ this ratio decreases with $x$. If the ratio is at most one at the interface, $T_l-T_L$ cannot become negative; if it is greater than one there, a supercooled interval appears immediately ahead. Equality marks onset, not a finite supercooled interval.

For fixed positive $\Delta \theta$ and $\lambda =O(1)>0$, the left side of the inequality is $O(ϵ)$ and the right side is positive of order one, so freezing is constitutionally supercooled for sufficiently small $ϵ$. To locate the boundary between the two regimes, resolve the smaller scale $\Delta \theta =O(ϵ)$, $\lambda =O(ϵ)$. The salt equation and liquidus then giveHere $ϵ\lambda =O({ϵ}^2)$, so the thermal equation improves to $T_i=(T_{\infty }+T_{-\infty })/2+O({ϵ}^2)$; this justifies the second expansion including latent heat. Also $T_{\infty }-T_i={\theta }_{+}+O(ϵ)$ and the left side of the exact criterion is $ϵ{\theta }_{+}/\sqrt{\pi }+O({ϵ}^2)$. Therefore the small-diffusivity constitutional-supercooling threshold isThis printed condition is an asymptotic onset condition, rather than an exact finite-$ϵ$ inequality. In its $O({ϵ}^2)$ transition window the complete algebraic system and exact gradient criterion must be used.