= Solution
For a steady field in the apparatus frame, the time derivative in the translating crystal frame is $-V\partial_z$. The given solute equation therefore becomes
$$
[u_z-V(1-\varphi)]C_z=-rVC\varphi_z.
$$
Substituting the <density-change flow in a pulled mushy layer> reduces this to $(1-\varphi)C_z=C\varphi_z$. Thus the <solute conservation in a steadily pulled mush> integrates to $(1-\varphi)C=C_0$, using the zero <solid fraction> at the top boundary.
Set $\mathcal C=mC_0/\Delta T$. The <liquidus> gives $C=C_0(1-\theta/\mathcal C)$, so
$$
\boxed{\varphi=-\frac{\theta}{\mathcal C-\theta},\qquad 1-\varphi=\frac{\mathcal C}{\mathcal C-\theta}.}
$$
In particular, $\varphi=0$ at $\theta=0$, and immediately above the <eutectic temperature>, where $\theta=-1$, the mush has $\varphi=1/(\mathcal C+1)$. It becomes fully solid across the eutectic front below.
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