+ \frac{\kappa}{2} \phi^2 +
\frac{\lambda}{6} \phi^3 =0,
\label{phi2}$$ $$m_v^2 V_0 - g_v \rho + 2 \Lambda_v g_v^2\, V_0\, g_{\rho}^2
b_0^2 =0,
\label{V02}$$ $$m_\rho^2 b_0 -\frac{g_\rho}{2} \rho_3 + 2 \Lambda_v
g_{\rho}^2 b_0 g_v^2 V_0^2 = 0,
\label{b02}$$

$${\cal E}= \frac{1}{\pi^2} \sum_i \int_0^{k_{Fi}}
p^2 dp \sqrt{{\mathbf p}^2 + {M^*}^2}$$ $$+\frac{1}{2} \left[m_s^2 \phi^2 - m_v^2 V_0^2 - m_\rho^2 b_0^2 \right]
+ g_v V_0 \rho + \frac{g_\rho}{2} \rho_3 b_0$$ $$+\frac{\kappa}{6} \phi^3 + \frac{\lambda}{24} \phi^4 
-\Lambda_v g_v^2 V_0^2 g_\rho^2 b_0^2.$$ and $$P=\frac{1}{3 \pi^2} \sum_i \int_0^{k_{Fi}} \frac{p^4 dp}{\epsilon}
- \frac{m_{s}^2}{2} \phi^2 + \frac{m_{v}^2}{2} V_0^2
+ \frac{m_{\rho}^2}{2} b_0^2$$ - \^3 - \^4 + \_v g\_v\^2 V\_0\^2 g\_\^2 b\_0\^2.

Based on the geometrical construction and Gibbs conditions for phase coexistence, i.e., the pressure and both chemical potentials are equal in both phases, we build the binodal section given in Fig. \[binodais\]. Notice that we have defined the proton fraction of the system as $$y_p = \frac{\rho_p}{\rho}.$$ The binodal section yields the boundary conditions which we need. For the same pressure, two points, with different proton fractions are found. For each of these points, the meson fields and the densities are well defined and used as the initial and boundary conditions in eqs. (\[elphi\]-\[elA0\]), which are then solved. Once 