in the bidimensional case.

![image](g1){width="100.00000%"}

![image](g2){width="100.00000%"}

We proceed now to determine the explicit solutions for $\xi\left(\tau\right)$ and $\eta\left(\tau\right)$ in the three-dimensional case. We will focus on the study of the solution for $\xi$, as the solution for $\eta$ differs only by notation. We will then use $\xi\left(\tau\right)$ and $\eta\left(\tau\right)$ to determine the solution for $\phi\left(\tau\right)$.

Solution by quadrature
======================

The integration of eq. yields $$\int_{0}^{\tau}du=\pm\int_{\xi_{0}}^{\xi}\frac{udu}{\sqrt{\varepsilon u^{6}+2hu^{4}+2\alpha_{1}u^{2}-p_{\phi}^{2}}},
\label{eq:tau_inverse}$$ where the initial fictitious time has been set to zero[^3], $u$ is a dummy integration variable and $\xi_{0}$ is the initial value of $\xi$. Before proceeding, we need to discuss briefly the nature of the sign ambiguity in this formula.

The left-hand side of eq. represents the fictitious time needed by the dynamical system to evolve from the initial coordinate $\xi_0$ to an arbitrary coordinate $\xi$. As pointed out in the previous section, all phase plots are symmetric with respect to the horizontal axis, and thus, along a trajectory in phase space, each coordinate $\xi$ will be visited twice: once with a positive $p_\xi$ coordinate, and once with a negative $p_\xi$ coordinate[^4]. It follows that we can choose either sign in , and the left-hand side will then represent the evolution time along 