$$\label{z}
\zeta = \zeta(u, v).$$ With this assumption the first equation in (\[Rxx, Ryy, Rzz\]) reduces to $$\label{equation for lambda in the self-similar solution}
2 \; \frac{\lambda_{\zeta \zeta}}{\lambda_{\zeta}} + \left(\lambda_{\zeta} + \mu_{\zeta} + \sigma_{\zeta}\right) + 2\; \frac{{\zeta}_{u v}}{{\zeta}_{u}{\zeta}_{v}} = 0.$$ The assumed symmetry requires $\left({{\zeta}_{u v}}/{{\zeta}_{u}{\zeta}_{v}}\right)$ to be some function of $\zeta$, say $ Z(\zeta) = \left({{\zeta}_{u v}}/{{\zeta}_{u}{\zeta}_{v}}\right)$. Integrating we get $$\label{lambdaz}
\lambda_{\zeta} = 2 \alpha \; e^{- \left(\lambda + \mu + \sigma\right)/2} e^{- \int{Z(\zeta) d\zeta}} \equiv 2 \alpha \left(\frac{f_{\zeta}}{f}\right),$$ where $\alpha$ is an arbitrary constant of integration. Similar equations, with new constants, e.g., $\beta$ and $\gamma$, are obtained for $\mu$ and $\sigma$ by means of a cyclic transformation. What this means is that $$\label{lambdaz, muz, sigma z}
\frac{\lambda_{\zeta}}{2 \alpha} = \frac{\mu_{\zeta}}{2 \beta} = \frac{\sigma_{\zeta}}{2 \gamma} = \frac{f_{\zeta}}{f},$$ which upon integration yields $$\label{elambda, emu, esigma}
e^{\lambda}  = C_{1}f^{2 \alpha}, \;\;\;e^{\mu} = C_{2} f^{2 \beta}, \;\;\;e^{\sigma} = C_{3} f^{2 \gamma},$$ where $C_{1}$, $C_{2}$, $C_{3}$ are constants of integration. A single differential equation for $f(\zeta) = f(u, v)$ can be easily obtained by substituting (\[elambda, emu, esigma\]) into any (\[Rxx, Ryy, Rzz\]), namely $$\label{equation for f}
f f_{u v}  + \left(a - 1\right) f_{u} f_{v} = 0.$$

[*Notation*]{}: Here and henceforth we 