*4.1. Mathematical Analysis of the System*

For the case of −1 < *x* < 1, the equilibrium points of the system are given by

$$E\_1 = (\frac{\mathfrak{a}^2}{1 - m\_0}, 0, 0) = (5.77838 \times 10^{-4}, 0, 0).$$

In this case, we reach the system *X* (*t*) = *AX*(*t*) + *B*(*t*) where

$$A = \begin{bmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ -1.1768 & 0 & -1 \end{bmatrix}, \quad B(t) = \begin{bmatrix} 0 \\ 0 \\ \alpha^2 e^{y(t)/\alpha} \end{bmatrix}.$$

The eigenvalues of the system associated with the equilibrium point *E*<sup>1</sup> are the solutions of the cubic equation

$$
\lambda^3 + \lambda^2 + 1.1768 = 0.
$$

We get the following eigenvalues

$$
\lambda\_1 = -1.51364, \quad \lambda\_2 = 0.25682 + 0.84351i, \quad \lambda\_3 = 0.25682 - 0.84351i.
$$

For the case −*vf* /*ve x* -−1, we can obtain the equilibrium point

$$E\_2 = (\frac{\alpha^2 - m\_0 + m\_1}{1 - m\_1}, 0, 0) \\ = (-0.969422, 0, 0)$$

associated with eigenvalues

$$
\lambda\_1 = -1.72307, \quad \lambda\_2 = 0.36154 + 1.05603i, \quad \lambda\_3 = 0.36154 - 1.05603i.
$$

For the case 1 *x vf* /*ve*, the system has the equilibrium point *E*<sup>3</sup> = (0.452152, 0, 0). Note that *E*<sup>2</sup> and *E*<sup>3</sup> have the same eigenvalues since their associated matrices are the same.

From the signs of real/imaginary parts of the associated eigenvalues, we conclude that the three equilibrium points *E*1, *E*2, *E*<sup>3</sup> are saddle foci. Hence, the proposed circuit has a chaotic behavior.
