= Solution
Integrating $Y''+2a\eta Y'=0$ once gives $Y'=K e^{-a\eta^2}$. A second integration expresses the solution using the <error function> when the value at zero is prescribed, and the <complementary error function> when decay at infinity is prescribed. Write $\theta_\lambda=\theta^s(\lambda)$ and take $\lambda>0$, $Le>0$. The wall condition and the interface value give
$$
\boxed{\theta^s(\eta)=\mathcal C^s(\eta)=\theta_\lambda\frac{\operatorname{erf}\eta}{\operatorname{erf}\lambda},\qquad0\leq\eta\leq\lambda.}
$$
For the exterior liquid, continuity at the interface fixes $\theta^l(\lambda)=\theta_\lambda-1+\mathcal R_\theta$ and $\mathcal C^l(\lambda)=\theta_\lambda-1$. Thus the two decaying <similarity solutions> are
$$
\boxed{\theta^l(\eta)=(\theta_\lambda-1+\mathcal R_\theta)\frac{\operatorname{erfc}\eta}{\operatorname{erfc}\lambda},\qquad
\mathcal C^l(\eta)=(\theta_\lambda-1)\frac{\operatorname{erfc}(\sqrt{Le}\eta)}{\operatorname{erfc}(\sqrt{Le}\lambda)},\quad\eta\geq\lambda.}
$$
The factor $\sqrt{Le}$ is required because solute and heat diffuse on different length scales. Both denominators are positive for a finite positive $\lambda$. These formulas use the differential equations, wall/far-field data and continuity; the remaining three flux conditions determine their unknown constants.
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