Equilibrium existence

Variables zero at the steady state can be discarded from the analysis. If $k$ variables are non-0, and are all equal, to $\bar{x} \ne 0$,

$$\bar{x}^{2}\left(1 + k^2\alpha \right) + \bar{x}\left( 2k\alpha - \sigma \right) + \alpha = 0$$ (10)

Solutions are

$$\frac{\sigma -2k\alpha \pm \sqrt{\sigma^2-4\alpha\left(1+k\sigma\right)}}{2\left(1+k^{2} \alpha\right)}$$

A sufficient and necessary condition for the existence is

$$4 \alpha \frac{k\sigma + 1}{\sigma^2} < 1$$

It will be shown below that, at a stable steady-state, there is at most 1 non-0 variable which can be different from other non-0 variables. If there is such a variable, equal to $y$, the equation for the value of other variables becomes

$$\bar{x}^{2}\left(1 + k^2\alpha \right) + \bar{x}\left( 2k\alpha \left( 1+y \right) - \sigma \right) +$$

$$\alpha \left(1 + y \right)^2= 0$$ (11)

Solutions are

$$\frac{\sigma -2k\alpha\left(1+y\right) \pm \sqrt{\sigma^2-4\alpha\left(1+k\sigma+y\right)\left(1+y\right)}} {2\left(1+k^{2} \alpha\right)}$$

and the condition for a solution to exist

$$4 \alpha \left(1+y \right) \frac{k\sigma + 1+y}{\sigma^2} < 1$$

The solutions for $y$ are

$$\frac{\sigma -2\alpha\left(1+k\bar{x}\right) \pm \sqrt{\sigma^2-4... ...eft(1+\sigma+k\bar{x}\right)\left(1+k\bar{x}\right)}} {2\left(1+\alpha\right)}$$