the modified Friedmann equation for synchronous observers $$\begin{aligned}
\label{modfriedman}
\rho=3H^2+\bar f -3H\bar f_{\chi},\end{aligned}$$ which depicts a curve on the plane $(H,\rho)$. Finally, taking its temporal derivative and using the conservation equation (\[conserb\]) in the form $\dot{\rho}=-3H(P+\rho)$, one finds the Raychaudhuri equation $$\begin{aligned}
\label{raychauduri}
\left(1-  \frac{3}{2}\bar f_{\chi\chi} \right)\dot{H}=-\frac{1}{2}(\rho+P).\end{aligned}$$

A different way to obtain such equations is to directly consider the metric for synchronous observers $ds^2=-dt^2+a^2\delta_{ij}dx^idx^j$ and use the equations $0-0$ and $i-i$, which also leads to the modified Friedmann and Raychaudhuri equations.

Therefore, given a curve $\rho=\bar g(3H)=\bar g(\bar\chi)$ on the plane $(H,\rho)$, so as to obtain the corresponding $\bar f(\bar\chi)$ theory one has to solve the first-order differential equation $$\begin{aligned}
\bar\chi \bar f_{\chi}-\bar f-\frac{1}{3}\bar\chi^2+g(\bar\chi)=0,\end{aligned}$$ whose solution is $$\begin{aligned}
\label{formula}
\bar f(\bar\chi)=+\frac{1}{3}\bar\chi^2-\bar\chi\int\frac{g(\bar\chi)}{\bar\chi^2}d\bar\chi.\end{aligned}$$

Note that, in order to obtain a bouncing background, a necessary condition is to choose a curve on the plane $(H,\rho)$ cutting two or more times the axis $H=0$. One of those points has to be $(0,0)$ because at low energy densities a viable bouncing background has to approach to GR, i.e., the chosen curve has to approach to the parabola $H^2=\frac{\rho}{3}$ and the other points cutting the axis $H=0$ have the form $(0, \rho_i)$ with $\rho_i>0$.

The simplest 