We study here the simplest model describing two populations that
compete fot the same limited food source or in some way inhibit each other’s growth. For example, competition may be for territory which is directly related to food ressources.
The model is given by
$$ \begin{cases} \dot{x}=r_1 \, x\left(1-\frac{x}{K_1}-\frac{b_{12}y}{K_1}\right) \\ \dot{y}=r_2\, y\left(1-\frac{y}{K_2}-\frac{b_{21}x}{K_2}\right) \end{cases} $$
where $r_1,r_2,K_1,K_2,b_{12},b_{21}$ are positive parameters.In the absence of the other, each population have logistic growth. The parameters $b_{12}$ and $b_{21}$ measure the competitive effect of population $2$ on population $1$, and vice versa. They are generally not equal.
A first step in analyzing this system is to nondimensionalise it. We obtain the following equations, this time with only three parameters:
$$ \begin{cases} \dot{x}=x\, (1-x-a_{12}y) \\ \dot{y}=\rho y\, (1-y-a_{21}x)\, . \end{cases} $$
(More details on getting this form below.)For the same reasons as for the previous model, the positive quadrant is invariant. There are always
three fixed points, namely
$$ (0,0), (1,0), (0,1). $$
The crucial part of the nullclines are the straight lines
$$ 1-x- a_{12} y=0,\,1-y-a_{21} x=0. $$
When they intersect in the positive quadrant, we get a fourth fixed point, namely
$$ (\bar{x},\bar{y})=\left(\frac{1-a_{12}}{1-a_{12}a_{21}},\frac{1-a_{21}}{1-a_{12}a_{21}} \right)\, \cdot $$
(We assume that $a_{12}a_{21}\neq 1$.)
You can observe three qualitatively different regimes:
- When the above straight lines do not cross, which means that either
$$1/a_{12}>1,1/a_{21}<1$$
or the other way around. In the first case, population $2$ tends to extinction, and population $1$ tends to its carrying capacity. In the second, things go the other way around. Thus, one of the populations is dominant. - When the above straight lines cross, and $1/a_{12}>1,1/a_{21}>1$. The fixed point $(\bar{x},\bar{y})$ seems to attract all solutions starting inside the positive quadrant. We have a stable coexistence.
- When the above straight lines cross, and $1/a_{12}<1,1/a_{21}<1$. The fixed point $(\bar{x},\bar{y})$ becomes repulsive: it is unstable to small perturbations. Moreover, each of the fixed points $(1,0)$ and $(0,1)$ has a domain of attraction. One can infer that there is a curve, a separatrix, which divides the positive quadrant into two nonoverlapping regions. It is tempting to assume that the fixed point $(\bar{x},\bar{y})$ is a saddle whose unstable manifold is the separatrix.
The conclusion is that we have a bistable regime: depending on the initial condition, one or the other population gets eliminated.
We can confirm part of the above informations by linearizing around each fixed point. This is left to the reader as an exercise. When the fixed point $(\bar{x},\bar{y})$ exists and is attractive, all what we can say by using the linearized system is that, locally, it is a sink. However, the digital experiment shows clearly that $(\bar{x},\bar{y})$ is globally attractive. We shall see later on how to confirm mathematically this using a Lyapunov function.
More on nondimensionalisation. The starting point is to measure each population abundance by comparing it to the corresponding carrying capacity. So, we make the rescaling $x\to x/K_1$ and $y\to y/K_2$. Then, consider the new time $r_1 t$, and set
$$\rho=\frac{r_2}{r_1},\, a_{12}=b_{12}\frac{K_2}{K_1},\, a_{21}=b_{21}\frac{K_1}{K_2}.$$