On at different values

If at steady state, $x_i \ne x_j$ and both are non-0, then

$$x_i^2-\sigma x_i = x_j^2 -\sigma x_j \left( =-\alpha D^2 \right)$$

There are thus only two possible non-0 steady-state values, noted $\bar{x}_a$ and $\bar{x}_b$, with $\bar{x}_a<\bar{x}_b$. Noting $P(x)=x^2-\sigma x$, and supposing that $\bar{x}_a$ and $\bar{x}_b$ exist, $P'(\bar{x}_a)<0$, ie $\frac{2 \bar{x}_a}{\sigma}<1$.

Consider the Jacobian matrix of the system, reordered so that variables having $\bar{x}_a$ as a value come before those having $\bar{x}_b$ as a value:

$$\begin{pmatrix} \overbrace{\begin{matrix}a_{\phantom{2}} & c & \c... ... \ddots & \vdots e & \cdots & e & _{\phantom{1}} b \end{matrix}\end{pmatrix}$$

With the appropriate eigenvectors, it is easy to show that $b-e$ and $a-c$ are eigenvalues for this matrix, of order $k-1$ and $p-1$. Thus, if $k>1$ and $p>1$, a necessary condition for stability of an equilibrium is $e>b$ and $c>a$. In particular, there can be at most 1 variable having $\bar{x}_a$ as a value.

The first term is positive because the values of $\bar{x}_a$ and $\bar{x}_b$ are symmetrical with respect to $\sigma/2$. The second term is also positive, and the sufficient condition for the instability of the equilibrium is thus met.

Thus, there is no stable equilibrium with non-0 variables having different values.