Equilibrium existence

$$\forall j, \left(x_{j} - \alpha \right) \left( 1+\Sigma_{i=1}^{n}x_{i}^\mathrm{c} \right)=\sigma x_{j}^{c}$$

If $\forall j, x_{j} = \bar{x}$,

$$n \bar{x}^{c+1} - \left( \sigma + n \alpha \right) \bar{x}^c +\bar{x}-\alpha=0$$ (8)

There is at least one solution, maybe 3 (or 2 in degenerate cases) depending on the parameters. The solutions are noted $\bar{x}_l$, $\bar{x}_u$, and $\bar{x}_h$, with $\bar{x}_l<\bar{x}_i<\bar{x}_h$.

If $f(x)=n x^{c+1} - \left( \sigma + n \alpha \right) x^c +x$, $f'(x)=(c+1) n x^c - c (\sigma + n \alpha)x^{c-1} +1$, $f''(x)=c(c+1)nx^{c-1} - c(c-1)(\sigma+n\alpha)x^{c-2}$. $f''\left(\frac{c-1} {n(c+1)} \left(\sigma + n\alpha\right)\right)=0$. $f'$ takes negative values iff $f'\left(\frac{c-1} {n(c+1)} \left(\sigma + n\alpha\right)\right)<0$, which is a necessary condition for the existence of 3 equilibria with all variables on.

The dynamics of the system constrained to $\forall i, x_i=x$ are defined by

$$\dot{x}=-x + \frac{\sigma x^c}{1+nx^c} + \alpha$$

The sign of $\dot{x}(t)$ is the opposite of that of $f(x(t))$. Because $\bar{x}_u$ is such that $f'(\bar{x}_i)<0$, it is easy to see that the steady state $\bar{x}_u$ is unstable for the constrained system, and thus for the full system.