*3.1. Case 1:* −1 < *x* < 1

From Equation (5), we have *g*(*x*) = *m*0*x*. At the equilibrium point, we get *y* = *x*˙ = 0, *z* = *y*˙ = 0, and

$$-\mathfrak{x} - z + m\_0 \mathfrak{x} + \mathfrak{a}^2 \mathfrak{e}^{y/\mathfrak{a}} = \dot{z} = 0.$$

Thus, the equilibrium for Case 1 is given by *<sup>E</sup>*<sup>1</sup> = (*x*1, *<sup>y</sup>*1, *<sup>z</sup>*1)=( *<sup>α</sup>*<sup>2</sup> 1−*m*<sup>0</sup> , 0, 0). When *α* tends to 0, the equilibrium point reaches the origin (0, 0, 0).

The system (6) can be put in the vector form

$$X'(t) = AX(t) + B(t),\tag{7}$$

where

$$X(t) = \begin{bmatrix} x(t) \\ y(t) \\ z(t) \end{bmatrix}, \quad B(t) = \begin{bmatrix} 0 \\ 0 \\ a^2 e^{y(t)/a} \end{bmatrix}, \quad A = \begin{bmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ m\_0 - 1 & 0 & -1 \end{bmatrix}.$$

*3.2. Case 2:* <sup>−</sup>*<sup>v</sup> <sup>f</sup> ve x* -−1

In this case, we have *g*(*x*) = *m*1*x* − *m*<sup>0</sup> + *m*1. At the equilibrium point, we obtain *y* = *x*˙ = 0, *z* = *y*˙ = 0, and

$$-\mathbf{x} - z + m\_1 \mathbf{x} - m\_0 + m\_1 + \alpha^2 e^{y/a} = \dot{z} = 0.$$

Thus, the equilibrium for Case 2 is given by

$$E\_2 = (x\_{2"} y\_{2"} z\_2) \ = (\frac{\alpha^2 - m\_0 + m\_1}{1 - m\_1}, 0, 0) \ .$$

When *<sup>α</sup>* <sup>→</sup> 0, we have (*x*2, *<sup>y</sup>*2, *<sup>z</sup>*2) reaches the point ( <sup>−</sup>*m*0+*m*<sup>1</sup> <sup>1</sup>−*m*<sup>1</sup> , 0, 0). The system (6) can be put in the vector form (7) where

$$A = \begin{bmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ m\_1 - 1 & 0 & -1 \end{bmatrix}, \quad B(t) = \begin{bmatrix} 0 \\ 0 \\ -m\_0 + m\_1 + a^2 e^{y(t)/a} \end{bmatrix}.$$

*3.3. Case 3:* 1 *x <sup>v</sup> <sup>f</sup> ve*

We have *g*(*x*) = *m*1*x* + *m*<sup>0</sup> − *m*<sup>1</sup> and, thus, the equilibrium point is given by

$$E\_3 = (x\_{3\prime}y\_{3\prime}z\_3)\_- = (\frac{\alpha^2 + m\_0 - m\_1}{1 - m\_1}, 0, 0).$$

When *<sup>α</sup>* <sup>→</sup> 0, we have (*x*3, *<sup>y</sup>*3, *<sup>z</sup>*3) reaches ( *<sup>m</sup>*0−*m*<sup>1</sup> <sup>1</sup>−*m*<sup>1</sup> , 0, 0). We also have the vector form (7) where the Jacobian matrix *A* is the same as that in the previous case, and

$$B(t) = \begin{bmatrix} 0 \\ 0 \\ m\_0 - m\_1 + a^2 e^{y(t)/\alpha} \end{bmatrix}.$$

Thus, the system (6) has three colinear equilibrium points on the *X*-axis. Note that when *α* → 0, we have that the points *E*<sup>2</sup> and *E*<sup>3</sup> are opposite to each other with respect to the origin *E*1. This observation shows the symmetry of the equilibrium points.
