= Solution
Introduce three positive functions of the unknown front parameter:
$$
P(\lambda)=\frac{e^{-\lambda^2}}{\sqrt\pi\operatorname{erf}\lambda},\quad
Q(\lambda)=\frac{e^{-\lambda^2}}{\sqrt\pi\operatorname{erfc}\lambda},\quad
R(\lambda)=\frac{\sqrt{Le}\,e^{-Le\lambda^2}}{\sqrt\pi\operatorname{erfc}(\sqrt{Le}\lambda)}.
$$
Differentiating the <error function> profiles gives $\theta^{s\prime}=2P\theta_\lambda$, $\theta^{l\prime}=-2Q(\theta_\lambda-1+\mathcal R_\theta)$ and $\mathcal C^{l\prime}=-2R(\theta_\lambda-1)$ at the interface. The three flux conditions are therefore exactly
$$
\boxed{\begin{aligned}
\phi\lambda S&=P\theta_\lambda+Q(\theta_\lambda-1+\mathcal R_\theta),\\
\phi Le\lambda(\theta_\lambda-1+\mathcal R_C)&=R(\theta_\lambda-1),\\
Q(\theta_\lambda-1+\mathcal R_\theta)&=R(\theta_\lambda-1).
\end{aligned}}
$$
These undivided equations are the safest description of the <constant-fraction mushy-layer similarity solution>, including degenerate parameter choices. In the nondegenerate case the third equation first gives
$$
\boxed{\theta_\lambda=1+\frac{Q(\lambda)\mathcal R_\theta}{R(\lambda)-Q(\lambda)}.}
$$
The second then gives the <solid fraction>
$$
\boxed{\phi=\frac{R(\lambda)(\theta_\lambda-1)}{Le\lambda(\theta_\lambda-1+\mathcal R_C)}.}
$$
Use the third equation to replace the $Q$ term in the first, and eliminate $\phi$. The front parameter is a positive root of the single <transcendental equation>
$$
\boxed{\frac{S R(\lambda)(\theta_\lambda-1)}{Le(\theta_\lambda-1+\mathcal R_C)}-P(\lambda)\theta_\lambda-R(\lambda)(\theta_\lambda-1)=0,\qquad
\theta_\lambda=1+\frac{Q(\lambda)\mathcal R_\theta}{R(\lambda)-Q(\lambda)}.}
$$
Then substitute back for $\theta_\lambda,\phi$ and $h(t)=2\lambda\sqrt{\kappa t}$. Since the physical parameters have not been assigned numbers, an implicit scalar equation is the appropriate determination of $\lambda$.
The divided formulas require $R\ne Q$ and $\theta_\lambda-1+\mathcal R_C\ne0$. If either vanishes, return to the boxed three-equation system. For example $Le=1$ gives $R=Q$, so the last equation demands $\mathcal R_\theta=0$; otherwise this proposed <similarity solution> cannot exist. A physical branch must also have $\lambda>0$ and $0\leq\phi\leq1$. The algebra alone does not establish existence or uniqueness of such a branch for arbitrary parameters.
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