Bounds on $c$ and $\sigma$

Let us note $x_e$ verifying $F(x_e,..,x_e)=0$. The set of such equilibria $x_e$ is defined by

$$x_e+(n-1)x_e^{c+1}=\sigma$$ (5)

$s$ being a strictly positive number, and considering limits when $x_e$ tends to 0 and to infinity, one sees that this equation has at least one solution. There is thus at least one undesirable equilibrium; this equilibrium should be unstable. We have

$$J_{F}(x_e,..,x_e)=\left(\begin{array}{cccc} -1&a&\dots&a\\ a&-1&\dots&a\\ \vdots&\vdots&\vdots&\vdots\\ a&a&\dots&-1 \end{array}\right),$$

where

$$a=\frac{-\sigma c{x}_e^{c-1}}{(1+(n-1)x_e^c)^2}$$

The eigenvalues of $J_{F}(x_e,..,x_e)$ are $-(a+1)$ and $-1-(n-1)a$. Thus, a necessary and sufficient condition for the ss to be unstable is $-a>1$, i.e.

$$\frac{\sigma c{x}_e^{c-1}}{(1+(n-1){x}_e^c)^2}>1$$

Using equation 5, we derive

$$c{x}_e^{c+1}>\sigma$$

Replacing $\sigma$ by its value derived from equation 5, we have

$$c{x}_e^{c+1}>x_e+(n-1){x}_e^{c+1},$$

$${x}_e^{c}(c-(n-1))>1 (x_e\ne 0),$$

from which we derive

$$c>n-1$$

and

$${x}_e^{c}>\frac{1}{c-(n-1)}$$

Combining this last equation with equation 5, we find

$$\sigma>(c-(n-1))^{-\frac{1}{c}}+(n-1)(c-(n-1))^{-\frac{c+1}{c}}$$