Determine time spent by orbits near an equilibrium Consider the system of ODEs
\begin{align*}
\dot{x}_1 = x_1(1-x_1-\alpha x_2-\beta x_3)\\
\dot{x}_2 = x_2(1-\beta x_1 - x_2 - \alpha x_3)\\
\dot{x}_3 = x_3(1-\alpha x_1 - \beta x_2 - x_3)
\end{align*} for $0 < \beta < 1 < \alpha$ and $\alpha + \beta > 2$. I am trying to show that the time intervals which a solution spends near the rest points $(1,0,0), (0,1,0), (0,0,1), (1,0,0),...$ increase geometrically with factor $(\alpha - 1)/(1-\beta) > 1$. I'm not sure where to start on this problem. I first tried finding an expression for $dt$ so that I could integrate over a region of time where a solution is near any of the aforementioned points, but the dimensionality of the problem makes this impossible (this method works great for one-dimensional problems).
So then I tried "reducing" the system, e.g. in a neighborhood of $(1,0,0)$, $x_2 \approx 0$ and $x_3 \approx 0$, so that the corresponding derivatives are negligible, leaving $\dot{x}_1 \approx x_1(1-x_1)$, implying the dynamics near the equilibrium are well-approximated by a 1-D system. Employing the aforementioned technique for this new system I still came up short, and feel as though this isn't the intended approach. Any help is greatly appreciated.
 A: Disclaimer: this is not a solution. It provides

*

*a graphical representation and


*exhibits an exceptional solution which does not enter into the general scheme.
A general remark: This system of differential equations is a kind of generalization of the one-dimensional logistic differential equation (see here).
Let us assume that the initial values are such that
$$x_1(0) > 0, \ \ x_2(0)>0, \ \ x_3(0)>0$$
(Otherwise there is divergence).
I have written a Matlab program (using a Runge-Kutta method) for the numerical computation of a certain number of trajectories for a certain common time range $[0,T]$. They all remain into the first octant ; their terminal positions (for $t=T$) is materialized by a small red circle, evidencing the "attracting power" of points $(1,0,0)$, $(0,1,0)$ and $(0,0,1)$.

The trajectory singled out in red follows the line $x=y=z$ ; it corresponds to an exceptional solution.
Indeed, let us look for a common equation
$$\varphi(t):=x(t)=y(t)=z(t) \tag{1}$$
with
$$\varphi(t)>0 \ \text{and} \ \varphi'(t)<0$$
If we plug (1) in any of the 3 equations, we get the same differential equation:
$$\varphi'(t)=\varphi(t)(1- \gamma \varphi(t)) \ \ \text{where} \ \gamma:=1+\alpha+\beta \tag{2}$$
which is a classical logistic differential equation with solution:
$$\varphi(t)=\dfrac{1}{\gamma +c e^{-t}}$$
where $c$ is an arbitrary constant.
We can remark that
$$\lim _{t \to \infty}\varphi(t)=\dfrac{1}{\gamma}$$
The limit point of this trajectory $(\varphi(t), \varphi(t), \varphi(t))$ is therefore $(\dfrac{1}{\gamma},\dfrac{1}{\gamma},\dfrac{1}{\gamma})$.
Final remark: I am wondering if looking for solutions of the form
$$x(t)=X(t)+\varphi(t), y(t)=Y(t)+\varphi(t), z(t)=Z(t)+\varphi(t)$$ wouldn't bring a simplification...
