We come back to preys and predators. We previously studied a basic model for the interaction of sharks and sardines, namely
$$ \begin{cases} \dot{x}= x(1-x) - xy\\ \dot{y}=\beta y ( x-\alpha) \end{cases} $$
where $x$ represents the density of sardines, $y$ the density of sharks, and $\alpha,\beta$ are positive parameters. We observed and proved that there cannot be limit cycles in this model, which means that the populations of sharks and sardines cannot oscillate periodically in a robust way.A natural question is: what biological factors create limit cycles in a prey-predator model? The inclusion of a more realistic of what ecologists call ‘functional response’ is one of such factor. The functional response is the rate at which each predator captures prey. In the above model, the functional response is a linearly increasing function of prey density, which means that there is no satiety for predators! A simple way to remedy this is to take a functional response of the form
$$ \frac{cx}{a+x} $$
where $a,c$ are positive parameters.
This modification leads to the following model, that we put in a nondimensional form:
$$ \begin{cases} \dot{x}= x\left( 1-\frac{x}{\gamma}\right) - \frac{xy}{1+x}\\ \dot{y}=\beta y \left( \frac{x}{1+x}-\alpha\right) \end{cases} $$
where $\alpha,\beta,\gamma$ are positive parameters. This model was introduced by Rosenzweig and McArthur.Invariance of the positive quadrant. We are only interested in the positive quadrant for the obvious reason that population densities are nonnegative numbers. In order that the model be well-defined, we thus need to check that starting in the positive quadrant, we have $(x(t),y(t))$ remaining in it for all $t>0$. We have
$$ \frac{\dot{x}}{x} = \frac{\text{d}}{\text{d} t}\ln x=1-\frac{x}{\gamma} -\frac{y}{1+x}. $$
Integrating both sides from time $0$ to time $t$, we get$$ x(t)=x_0 \, \exp\left( \int_0^t \left(1-\frac{x(s)}{\gamma} -\frac{y(s)}{1+x(s)}\right) \mathrm{d}s\right). $$
From this we see that is $x_0=0$ then $x(t)=0$ for all $t>0$, and if $x_0>0$, then $x(t)>0$for all $t>0$. We get the same conclusion for $y$. This proves more than the invariance
of the positive quadrant: it proves the invariance of its interior and that of its boundary.
Before playing with the model, let’s determine the nullclines and the fixed points.
Nullclines.
Setting $\dot{x}=0$, we find the $x$-nullclines:
$$ x=0\quad\text{and}\quad y=g(x)=(1+x)\left( 1-\frac{x}{\gamma}\right) $$
where we introduced the function $g(x)$ for later convenience. This function is a parabole whose peak is attained for$$ \frac{\gamma-1}{2}. $$
Setting $\dot{y}=0$, we find the $y$-nullclines:$$ y=0\quad\text{and}\quad x=\frac{\alpha}{1-\alpha}. $$
Fixed points.
For all parameter values, $(0,0)$ and $(\gamma,0)$ are fixed points. The former corresponds to the absence of both populations. The latter corresponds to the absence of predators, whereas preys have density $\gamma$. There is possibly a third fixed point:
$$ (\bar{x}, g(\bar{x})) $$
where$$ \bar{x}=\frac{\alpha}{1-\alpha}. $$
Since population densities are positive numbers, this fixed point is relevant only if $\bar{x}\leq \gamma$. This condition simply means that the nullclines $x=\alpha/(1-\alpha)$ and $y=g(x)$ intersect inside the positive quadrant.We further assume that $\alpha<1$, because otherwise all solutions starting from strictly positive initial densities tend to $(\gamma,0)$.
The reason for that is simple: for any $x>0$, $x/(1+x)<1$, and if $\alpha>1$ then $\dot{y}<0$, whence $y(t)\to 0$ when $t\to+\infty$. Consequently, the equation for $x$ reduces to the logistic equation $\dot{x}=x\left( 1-\frac{x}{\gamma}\right)$, whence $x(t)\to \gamma$ as $t\to+\infty$.