$V=2\rho_c\frac{e^{-\sqrt{3}\phi}}{(1+e^{-\sqrt{3}\phi})^2}$, and choosing the solution (\[solution\]) which leads to the background (\[background\]) mimicking the same background as a pressureless fluid, one can see that the function $\frac{\theta''}{\theta}$, which is symmetric with respect to the contracting and expanding phase, satisfies that in the contracting regime is increasing from $\rho=0$, where it vanishes, to $\rho=\frac{\rho_c}{2}$, and decreasing from $\rho=\frac{\rho_c}{2}$ to the bounce, which occurs at $\rho=\rho_c$, and where one has $\left|\frac{\theta''}{\theta}\right|=\frac{5}{4} \rho_c\cong 315$, which, as we will see in Section VI, is bigger than the pivot scale $k_*\sim 5\times 10^{-33}a_E\ll \rho_c$. Then, the pivot scale, which occurs at a very early time, namely $t=-t_H$, reenters to the Hubble radius at time $t=t_H$, so in this case the equation (\[long\]) holds from $-t_H$ to $t_H$ and the equation (\[short\]) holds at very early and late times.

The solution of (\[long\]) is, then, $$\begin{aligned}
\label{z2}
u^{(0)}_k=C_1(k)\theta+C_2(k)\theta\int_{\tau_I}^{\tau}\frac{1}{\theta^2}d\bar\tau  \Longrightarrow \Phi_k^{(0)}=\frac{H}{2a\theta}u_k^{(0)}
=C_1(k)\frac{{\mathcal H}}{2a^2}+C_2(k)\left(1-\frac{\mathcal H}{\mathcal H_I}\left(\frac{a_I}{a} \right)^2-\frac{{\mathcal H}\int_{\tau_I}^{\tau} a^2d\bar\tau}{a^2}\right), \end{aligned}$$ where the sub-index $I$ refers to an early time where GR holds, and the coefficients $C_1(k)$ and $C_2(k)$ are obtained by matching both expressions of $u_k$ in (\[z1\]) and (\[z2\]) when $k^2\tau^2\ll 1$.

Once this solution has been obtained, we consider the $i-0$ equation, which can be written 