 Solution

Put $g(z)=s(az)=m{a}^2{z}^2/({\sigma }^2+a^2{z}^2)$. The nullclines areAny steady state has nonnegative coordinates because $g\ge 0$. For $a\ne 0$, $g$ is strictly increasing on the positive half-line. If a steady state had $x>y$, then $x=g(y)<g(x)=y$, a contradiction. The case $y>x$ is identical, so all steady states lie on the diagonal. Write $x=y=z$. Besides $z=0$, the equation $z=g(z)$ givesand henceThus three distinct steady states exist exactly when $|a|m>2\sigma$, with $a\ne 0$ and the positive parameter assumptions stated above. For positive excitatory coupling this is $am>2\sigma$. At equality the two positive states merge; below it only the origin remains. If $a=0$, the origin is again the only steady state.

Both nullclines begin at the origin and approach the limiting activity $m$ along their saturating direction; they are reflections of each other across $x=y$. Flow points to the right where $x<g(y)$ and to the left where $x>g(y)$. It points upwards where $y<g(x)$ and downwards where $y>g(x)$. The plot shows these nullclines and typical trajectories for the parameters of the next part.

The origin and upper diagonal state attract nearby trajectories, while the lower positive state is a saddle. Its stable manifold separates the two attracting regions: trajectories in each basin of attraction approach its stable node, giving bistability. The diagonal is invariant, and on it the scalar activity increases between $z_{-}$ and $z_{+}$ and decreases between zero and $z_{-}$ or above $z_{+}$. Activities remain nonnegative and bounded because $0\le g\le m$. The vector field has constant negative divergence $-2/\tau$, so the Bendixson-Dulac criterion also rules out periodic orbits in the physical quadrant. The network is a switch between persistent activity levels, rather than an oscillator.