We come back to the prey-predator model we introduced in the opening part. In dimensionless form it is given by the equations
$$ \begin{cases} \dot{x}= x(1-x-y)\\ \dot{y}=\beta(x-\alpha)y \end{cases} $$
where $\alpha,\beta$ are positive parameters. Under this form, the carrying capacity of prey is normalized to be equal to one. We are only interested in the positive quadrant since $x,y$ are interpreted as abundances of populations.
One can check that the positive quadrant is invariant in the sense that, starting from $(x_0,y_0)$ lying inside the positive quadrant, the resulting trajectories will never cross its boundary (which is made of the $x$ and $y$ axes). This is because the boundary of the positive quadrant is the union of five trajectories: the points $(0,0)$ and $(1,0)$, the $y$-axis, and the two intervals $]0, 1[$ and $]1,+\infty[$ of the $x$-axis. And we know that trajectories never cross.
We can easily check that we have at most three fixed points:
$$ (0,0), (1,0), (\alpha,1-\alpha). $$
These fixed points occur at the intersections of the prey and predator nullclines. The third fixed point has a biological meaning if the nullclines $y=1-x$ and $x=\alpha$ do intersect in the positive quadrant; this depends on the value of $\alpha$.The numerical experiment shows the following behaviors:
- the origin is always a saddle;
- If $\alpha>1$, the fixed point $(1,0)$ seems to be a nodal sink, in which case the third fixed point is ruled out (it is outside the positive quadrant). In this case, the predator goes extinct, and the prey population stabilizes at its carrying capacity;
- If $\alpha<1$, the fixed point $(1,0)$ seems to be a saddle whose stable manifold is the positive $x$-axis. The third fixed point is now in the positive quadrant, and seems to be either a spiral or a nodal sink. Thus, if we start with prey and predator densities that are strictly positive, the system settles in a coexistence regime;
- Lastly, for the marginal case $\alpha=1$, the fixed points $(1,0)$ and $(\alpha,1-\alpha)$ coincide. We have the same situation as for $\alpha>1$.
Let’s confirm these observations by making a local analysis about each of these fixed points, so let’s compute the Jacobian matrix at an arbitrary point $x,y$. We find
$$ \begin{pmatrix} 1-2x-y & -x \\ \beta y & \beta(x-\alpha) \end{pmatrix}. $$
At the fixed point $(0,0)$ we have$$ A= \begin{pmatrix} 1 & 0 \\ 0 & -\beta \alpha \end{pmatrix}. $$
It is a saddle since this matrix has eigenvalues $1$ and $-\beta\alpha$. One can check that $(1,0)$ and $(0,1)$ are the corresponding eigenvectors, respectively. Or, we can directly use the fact that$\text{det}(A)=-\beta\alpha$; since it is negative, we know that $(1,0)$ must be a saddle.
At the fixed point $(1,0)$ the Jacobian matrix is
$$ A= \begin{pmatrix} -1 & -1 \\ 0 & \beta(1- \alpha) \end{pmatrix}. $$
We have $\text{det}(A)=\beta(1-\alpha)$. If $\alpha<1$, $\text{det}(A)<0$, thus the fixed point $(1,0)$ is a saddle. If $\alpha>1$, the determinant is $\text{det}(A)>0$, and $\text{tr}(A)=-1+\beta(1-\alpha)<0$, thus the fixed point $(1,0)$ is a sink. We can refine slightly our analysis by computing the discriminant:$$ \text{tr}(A)^2-4\text{det}(A)=\big(1+\beta(\alpha-1)\big)^2-4\beta(\alpha-1)=\big(1-\beta(\alpha-1)\big)^2>0. $$
Thus, the fixed point $(1,0)$ is a nodal sink.Finally, at the fixed point $(\alpha,1-\alpha)$ (assuming that $\alpha<1$), we have
$$ A= \begin{pmatrix} -\alpha & -\alpha \\ \beta(1- \alpha) & 0 \end{pmatrix}. $$
We have $\text{det}(A)=\alpha\beta(1-\alpha)>0$ and $\text{tr}(A)=-\alpha<0$, so the fixed point $(\alpha,1-\alpha)$ is a sink. The sign of the discriminant is given by the sign of $\alpha-4\beta(1-\alpha)$. Thus, we have a nodal sink if $\alpha>\frac{4\beta}{1+4\beta}$, and a spiral sink if $\alpha<\frac{4\beta}{1+4\beta}$.Let’s conclude this section by noting that the numerical experiment shows that the attractiveness of the fixed point $(\alpha,1-\alpha)$, when it exists, is manifestly global, not only local. By `global’ we mean that, whatever is the initial condition $(x_0,y_0)$, provided that $x_0>0,y_0>0$, we have
$$ (x(t),y(t))\to (\alpha,1-\alpha). $$
A linear stability analysis cannot help us proving this observation is true. This can be indeed proven using a powerful tool: Lyapunov functions.