= Solution
Take $z$ positive downwards from the roof. Write $T_0=T_m-mC_0$, $C_B=(T_m-T_B)/m$, $\Delta T=T_0-T_B=m(C_B-C_0)>0$ and $\Delta C=C_B-C_0$. The cooled <magma> follows the decreasing <liquidus> from $(C_0,T_0)$ towards $(C_B,T_B)$; the initially superheated liquid starts at $(C_0,T_\infty)$ above that <liquidus>. The solid has zero solute <concentration>. At each <temperature> in the <mushy layer>, a horizontal tie line joins the pure solid to the residual liquid on the <liquidus>. The <lever rule> gives the <liquid fraction> from the bulk <concentration>. Since $T_B>T_E$, this path stops before the <eutectic temperature> and does not cross the <eutectic composition>.
\Image[/past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2016/iii/paper-332-phase-diagram.png]
{title=Liquid cooling, residual-liquid enrichment and a solid–liquid tie line in the stagnant magma mush}
The <porosity> $\phi$ denotes the <liquid fraction>, so $1-\phi$ is the <solid fraction>. Equal phase <mass densities> remove volume-change flow; the stable <density stratification> suppresses <convection>. In the <stagnant mushy-layer model>, define bulk <specific enthalpy> and bulk solute <concentration> by
$$
E=c_pT+L\phi,\qquad B=\phi C+(1-\phi)C_s=\phi C.
$$
The arbitrary reference of <specific enthalpy> has been omitted. The two conservation laws in $0<z<h(t)$ are
$$
\boxed{T_t+\frac L{c_p}\phi_t=\kappa T_{zz},\qquad (\phi C)_t=0,\qquad T=T_m-mC.}
$$
Here $\kappa=k_T/(\rho c_p)$ is the <thermal diffusivity>, with $k_T$ the <thermal conductivity>. The positive $L\phi_t$ term records increased <specific enthalpy> on melting; during freezing it supplies <latent heat>. In the liquid $z>h(t)$, the <heat equation> and solute <diffusion equation> are
$$
T_t=\kappa T_{zz},\qquad C_t=D C_{zz},\qquad \phi=1,
$$
where $D$ is the <solutal diffusivity>. Neglecting solute <diffusion> inside the <mushy layer> does not permit dropping $D C_{zz}$ in the liquid at this stage.
Use $T(0,t)=T_B$, liquid $T\to T_\infty$ and $C\to C_0$ as $z\to\infty$, and initially $h=0$, $T=T_\infty$, $C=C_0$ for $z>0$. These are the half-space boundary conditions appropriate before cooling reaches the bottom of the chamber. For a closed chamber of finite depth $H$, replace the far-field conditions by the specified bottom conditions, for example $T_z=C_z=0$ at an insulated impermeable floor; the half-space <similarity solution> then ceases to apply when the thermal penetration depth approaches $H$.
At $z=h(t)$, let $C_i$ and $T_i=T_m-mC_i$ be the common liquid <concentration> and <temperature>, and let $\phi_i$ be the limiting <porosity> on the mush side. Continuity of $T,C$, the <Rankine-Hugoniot condition> for solute, and the interfacial <conservation of energy> give
$$
-D C_z(h^+,t)=\dot h(1-\phi_i)C_i,\qquad
\kappa[T_z(h^-,t)-T_z(h^+,t)]=\frac L{c_p}\dot h(1-\phi_i).
$$
The first condition has the sign of solute rejection into the liquid. For the sharp-edge <stagnant mushy-layer model> with nonzero liquid <solutal diffusivity>, close the free-boundary problem by <marginal equilibrium at a mush-liquid boundary>:
$$
T_z(h^+,t)=-m C_z(h^+,t),\qquad T(z,t)\ge T_m-mC(z,t)\quad(z>h).
$$
This prevents <constitutional supercooling> immediately ahead of the <mushy layer>. A small <solid fraction> jump at its edge is allowed: imposing $\phi_i=1$ as well as these finite-$D$ conditions would generally overdetermine the reduced model. At the roof, $C=C_B$ follows from the <liquidus>, and no additional solute-flux boundary condition is needed in the diffusion-free mush.
To obtain the large-<concentration> <similarity solution>, put
$$
\theta=\frac{T-T_0}{\Delta T},\quad \theta_\infty=\frac{T_\infty-T_0}{\Delta T}>0,\quad
\mathcal C=\frac{C_0}{\Delta C},\quad \mathcal S=\frac L{c_p\Delta T},\quad
r=\frac D\kappa,\quad \eta=\frac z{2\sqrt{\kappa t}},\quad h=2\lambda\sqrt{\kappa t}.
$$
The positive superheat specifies the physically intended liquid initial state. On a <similarity solution>, $C_i,\phi_i$ are constant, so every newly incorporated level leaves the same constant $B=\phi_iC_i$. Write $C_i=C_0+\Delta C\,\delta$ and $B=\Delta C(\mathcal C+a)$. Then throughout the <mushy layer>
$$
C=C_0-\Delta C\,\theta,\qquad
\phi=\frac{\mathcal C+a}{\mathcal C-\theta},\qquad
\left[1+\mathcal S\frac{\mathcal C+a}{(\mathcal C-\theta)^2}\right]\theta_t=\kappa\theta_{zz}.
$$
For $a,\theta=O(1)$, $\mathcal C\gg1$, and $\mathcal S/\mathcal C=O(1)$, the effective <heat capacity> is constant to relative error $O(\mathcal C^{-1})$. Thus set
$$
\boxed{\Omega=\sqrt{1+\mathcal S/\mathcal C},\qquad \Omega^2\theta_t=\kappa\theta_{zz}.}
$$
Integration of the <similarity solution> <ordinary differential equation> gives the leading analytical profiles
$$
\boxed{\theta_m=-1+(1-\delta)\frac{\operatorname{erf}(\Omega\eta)}{\operatorname{erf}(\Omega\lambda)},\qquad
\theta_l=\theta_\infty-(\theta_\infty+\delta)\frac{\operatorname{erfc}\eta}{\operatorname{erfc}\lambda}.}
$$
The <temperature> is $T=T_0+\Delta T\theta$. In the <mushy layer>, the <concentration> and <porosity> are the displayed $C,\phi$; in the liquid,
$$
\boxed{C_l=C_0+\Delta C\,\delta\frac{\operatorname{erfc}(\eta/\sqrt r)}{\operatorname{erfc}(\lambda/\sqrt r)},\qquad \phi_l=1.}
$$
These retain the liquid solutal <boundary layer>. For completeness, finite-$r$ values of $\delta,a,\lambda$ are determined analytically, to the same large-$\mathcal C$ order, by setting
$$
K(x)=\frac{e^{-x^2}}{\operatorname{erfc}x},\quad
A_r=\frac{\sqrt r K(\lambda/\sqrt r)}{\sqrt\pi\lambda},\quad
\delta=\frac{\theta_\infty K(\lambda)}{K(\lambda/\sqrt r)/\sqrt r-K(\lambda)},\quad
b=\delta A_r,\quad a=\delta-b,
$$
and solving
$$
(1-\delta)\frac{\Omega e^{-\Omega^2\lambda^2}}{\operatorname{erf}(\Omega\lambda)}
-(\theta_\infty+\delta)K(\lambda)
=\sqrt\pi\frac{\mathcal S}{\mathcal C}\lambda b.
$$
The solute jump gives $\phi_i=1-b/(\mathcal C+\delta)$ exactly; replacing $\mathcal C$ by $\mathcal C+\delta$ on the right of the last equation retains that jump's next-order factor, but does not make the linearized thermal profiles exact at finite $\mathcal C$. The branch has $0\le\delta<1$, $0<\phi\le1$. All these profiles are leading asymptotic expressions, not exact solutions of the nonlinear finite-$\mathcal C$ <heat equation>.
For $r\to0$ at fixed positive $\lambda$, the <complementary error function> asymptotic gives $K(\lambda/\sqrt r)\sim\sqrt\pi\lambda/\sqrt r$, hence $\delta=O(r)$, $b=O(r)$, $a\to0$, and $\phi_i\to1$. The result simplifies to
$$
\boxed{\begin{gathered}
\theta_m=-1+\frac{\operatorname{erf}(\Omega\eta)}{\operatorname{erf}(\Omega\lambda)},\qquad
\theta_l=\theta_\infty\left(1-\frac{\operatorname{erfc}\eta}{\operatorname{erfc}\lambda}\right),\\
C_m=C_0-\Delta C\theta_m,\quad \phi_m=\frac{\mathcal C}{\mathcal C-\theta_m},\quad C_l=C_0,\quad h(t)=2\lambda\sqrt{\kappa t}.
\end{gathered}}
$$
In particular, $\phi(0,t)=C_0/C_B=\mathcal C/(\mathcal C+1)$ and $\phi(h^-,t)=1$. There is no finite interfacial <latent heat> jump in this limit: the <latent heat> is released continuously throughout the <mushy layer>. Matching the two <temperature gradients> gives
$$
\frac{\Omega e^{-\Omega^2\lambda^2}}{\operatorname{erf}(\Omega\lambda)}=
\frac{\theta_\infty e^{-\lambda^2}}{\operatorname{erfc}\lambda},\qquad
\boxed{\Omega e^{\lambda^2}\operatorname{erfc}\lambda
=\theta_\infty e^{\Omega^2\lambda^2}\operatorname{erf}(\Omega\lambda).}
$$
The printed final equation is missing the square on $\Omega$ in the exponential. The corrected exponent follows directly from differentiating $\operatorname{erf}(\Omega\eta)$ and is consistent with $\Omega^2=1+\mathcal S/\mathcal C$. The positive root is unique: the mush-side gradient decreases from infinity to zero with $\lambda$, whereas $\theta_\infty K(\lambda)$ increases. The zero-superheat limit is singular and is not described by a finite positive $\lambda$ in this two-region construction.
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