I am reading a paper, and there is a coupled differential equations:
$$\begin{cases} \dot{x} = ax^2-x^3-y-z\\ \dot{y} = (a+\alpha)x^2-y\\ \dot{z} = \mu(bx+c-z) \end{cases} $$
Let the first two equations be $0$, we have $$z = -x^3-\alpha x^2 = f(x)$$
By the above, we can find three equilibrium points: $$N_1(x_1,y_1), O(x_0,y_0), N_2(x_2,y_2)$$
Then the paper use the following characteristic equation to discuss their stability:
$$s^2-\sigma(x_i)s-f'(x_i)=0, \ \ \ i = 0,1,2$$
where divergence $\sigma = -(1-2ax+3x^2)$ and the slope $f'$.
I just know the stability of equilibrium can be found by the phase plots and characteristic equation. However, I only know the characteristic equation of second or higher order ODEs. I have no idea about the above formula and cannot find some useful information by typing key words.
Can anyone please let me know what is the above formula (some lecture notes are fine). Thanks!
Paper link:
https://aip.scitation.org/doi/abs/10.1063/1.3563581 (from p.3 section II. to p.4, you don't need any background of networks.)
They study this system as slow-fast system: they "freeze" $z$-variable, thinking about it as a parameter for $(x, y)$-system. After that they study stability of equilibria of two-dimensional system regular way. Namely, if you have a system $\dot{x} = P(x, y), \; \dot{y} = Q(x, y)$, then eigenvalues of Jacobi matrix $ J = \left ( \begin{array}{cc} P'_x & P'_y \\ Q'_x & Q'_y \end{array} \right )$ computed at equilibrium determine its stability. The characteristic equation for $2 \times 2$ system is always $\lambda^2 - {\rm tr}\, J \cdot\lambda + {\rm det}\, J = 0$ and ${\rm tr}\, J = P'_x + Q'_y = {\rm div}\, (P, Q)$. The final formula from their paper just follows from the used notation.