Dynamical analysis

0 is a stable steady state. If $x_i(0)=0$, then $\forall t>0, x_i(t)=0$. If $x_i(0)>0$, then $\forall t>0, x_i(t)>0$. One can thus suppose that $\forall i,\forall t\ge 0, x_i(t)>0$. Consider a state in which there is one variable strictly superior to all others (ie, a state not belonging to the line $x_1=x_2=..=x_n$). Suppose without loss of generality that the variable in question is $x_1$. Consider the function

$$f_{1}(x)=\frac{x_1^2}{\alpha D^2 + x_1^2}$$

$$\dot{f_{1}(x)}=2\alpha D x_1\frac{\dot{x_1}\left( D-x_1 \right) - x_1 \Sigma_{i=2}^n \dot{x_i}}{\left( \alpha D^2 + x_1^2 \right)^2}$$

$$\frac{\left( \alpha D^2 + x_1^2 \right)^2}{2\alpha D x_1}\dot{f_{1}(x)}=\dot{x_1} +$$

$$\sigma \Sigma_{i=2}^n \frac{x_1 x_i \left( x_1 - x_i \right)\left... ...i \right)} {\left(\alpha D^2 + x_1^2 \right) \left( \alpha D^2 + x_i^2 \right)}$$

For $\alpha \ge 1/2$, the second term is positive.

We have

$$\frac{\mathrm{d}x_1(t)}{\mathrm{d}t}= \sigma f_{1}(x) - x_1$$

We first consider the case in which $\forall t\ge 0,\forall n\ge j>1, x_1>x_j$.

Suppose that $\sigma f_{1}(0) \ge x(0)$. In this case, $\dot{f_1}\left( 0 \right) > 0$, and $x_1$ and $f$ are strictly increasing functions of time. If $\sigma f_{1}(0)<x_1(0)$, then $\dot{f_1}\left( 0 \right)$ can be negative or positive. In the first case, $x_1$ is decreasing as long as $f_1$ is. If at some time $t_0$ $\sigma f_{1}(t_0) \ge x(t_0)$, then for $t>t_0$, $x_1$ and $f_1$ are increasing functions of time. Thus there can be at most one change in the monotony of $x_1$. Thus $\lim_{t\rightarrow \infty} x_1(t)$ exists. Since $\ddot{x_1}$ exists and is bounded on any trajectory, $\lim_{t\rightarrow \infty} \dot{x_1}(t)=0$. All trajectories thus converge to a steady state where $\forall j>1, x_j=x_1 \mathrm{or} x_j=0$.

If $\exists t \mathrm{st} \forall n>j>1, x_1(t)=x_j(t)$, the system is brought back to one dimension. Note that it is impossible for any variable to outgrow $x_1$.