Equilibrium existence

The steady-state equations are

$$\forall j, \bar{x}_{j} \left( 1+\Sigma_{i=1}^{n}\bar{x}_{i}^\mathrm{c} \right)=\sigma \bar{x}_{j}^{c},$$

ie

$$\bar{x}_{j}^{c-1}=\frac{1}{\sigma}\left( 1+\Sigma_{i=1}^{n}\bar{x}_{i}^\mathrm{c} \right) \mathrm{or} \bar{x}_j=0$$

Re-arranging the first equation,

$$\frac{1}{\sigma}\bar{x}_j^c -\bar{x}_{j}^{c-1}=- \frac{1}{\sigma}\left( 1+\Sigma_{i\ne j}\bar{x}_{i}^\mathrm{c} \right)$$

Let $f(x)=\frac{1}{\sigma}x^c -x^{c-1}$. Then $f'(x)=\frac{c}{\sigma}x^{c-1}-(c-1)x^{c-2}$. $f'(x)<0$ iff $\frac{c}{\sigma}x<c-1$. The minimal value of $f$ over the positive real set is $f(\frac{c-1}{c}\sigma)=\frac{1}{\sigma}\left( \frac{c-1}{c}\sigma \right) ^{c}... ...1}{c} - 1 \right)=\sigma^{c-1} \left( \frac{c-1}{c} \right) ^{c-1} \frac{-1}{c}$.

The equilibria studied here are such that only $\bar{x}_j$ is non-0, for some $j$. There are either 0 or 2 solutions, 2 iff

$$\sigma^{c-1} \left( \frac{c-1}{c} \right) ^{c-1} \frac{1}{c} > \frac{1}{\sigma}$$

$$\sigma^{c} > c \left( \frac{c}{c-1} \right) ^{c-1}$$

$$\ln{\sigma} > \frac{\ln{c}+\ln{\left( \frac{c}{c-1} \right) ^{c-1}}}{c}$$ (4)

$\ln{\left( \frac{c}{c-1} \right) ^{c-1}}$ is an increasing function of $c$, and $\lim_{c \to \infty} \ln{\left( \frac{c}{c-1} \right) ^{c-1}} =1$. $\frac{ln{c}}{c}$ is decreasing for $c>e \simeq 2.7$. The right-hand side of equation 4 has a maximum for $c=2$, of about $0.7$, matched by $\sigma=2$. Thus, for $\sigma \ge 2,$ there are two equilibria. Both large $c$ and large $\sigma$ are favourable to the existence of an equilibrium with one variable dominating all others.