An Application: Can cats control a rat population?

Now that we understand a little bit about stability, let's look at an example to see how it might arise. The town of Hamlin has a rat problem. The rats are growing in an explosive, uncontrolled fashion. Rather than a pied piper, the mayor decides to get some cats to control the population. Will this strategy work?
Well, of course, we think it should work if we have enough cats, but this simple question demonstrates the way in which differential equations and stability of equilibrium solutions can be used to understand an issue.

Growth of the rat population
We are told that the rat population is growing in an uncontrolled fashion. A good way to think about this is to suppose that the rate at which the rats are reproducing is proportional to the rat population. That is, if we call the number of rats, then the birth rate of rats is given by where r is some positive constant which represents the number of births per capita per unit time.

The Predation Rate
The other issue is the rate at which the cats eat the rats---we call this the predation rate. It is a bit more interesting to model than the birth rate of the rats because there are several ways to think about it. In fact, three different ways have been proposed by Hollings in the paper The Components of Predation as Revealed by a Study of Small Mammal Predation of the European Pine Sawfly, Canadian Entomology 91: 283-320. Let's take a minute to describe each of the models and then we will choose one to apply to our problem.
We will use to denote the predation rate---the number of rats consumed per unit time.

Hollings Type I Response: This is the simplest possible model. It simply says that the more rats there are, the more the cats eat. In other words, the predation rate is proportional to the rat population:

$P(y) = py$
where p is some positive constant.

Hollings Type II Response: Let's add a new feature to the model above. Realistically, if there are lots of rats, each cat can only eat so many of them. After a while, they become satiated and go to sleep. This means that for large numbers of rats, the predation rate should level off.

$P(y) = P_{max}\frac{y}{a+y}$
where $P_{max}$ and a are positive constants.

Hollings Type III Response: Let's add another new feature. If the population of rats is very low, the cats may have a hard time finding any before they can catch them. This makes there hunting efficiency very low. One way to describe this is:

$P(y) = P_{max}\frac{y^2}{a^2+y^2}$
where again $P_{max}$ and a are positive constants.

Setting up the Model
We will now set up the model for how the rat population behaves assuming the Hollings Type II Response applies. Of course, this is an assumption which would require serious scrutiny were we actually the mayor of Hamlin. However, since we are just pretending, we will not question the assumption. It should be said, however, that this model is useful in many other areas such as chemistry and biochemistry.
Before we begin, let's study the predation rate a bit more carefully. We will take
$P(y) = P_{max} \frac{y}{a+y}$
What is the meaning of these different terms? Notice that:

.
$P(a) = P_{max} \frac{a}{a+a} = \frac{P_{max}}{2}$
$P^\prime(y) = P_{max}\frac{(a+y) - y}{(a+y)^2} = \frac{P_{max}a}{(a+y)^2} > 0$
$P^\prime(0) = \frac{P_{max}}{a}$
When y is very large, we have $a+y \approx y$ so that $P(y) \approx P_{max}$

Putting all this together, we can deduce that the graph is as shown below. The function always increases and the constant a is controlling how fast the function increases. Also, for very large values of the population, the predation rate stabilizes around $P_{max}$ .

Now the changes in the rat population are governed by new births and the predation by cats. That is, the rate of change of the rat population is the difference in the birth rate minus the predation rate. This means that
$\frac{dy}{dt} = ry - P(y) = ry - P_{max} \frac{y}{a+y}$
To understand this differential equation, we would like to understand its equilibrium solutions and whether they are stable or unstable.
Let's first look for equilibrium solutions. This means:
$\frac{dy}{dt} = ry - P_{max}\frac{y}{a+y} = y(r - \frac{P_{max}}{a+y})= 0$
In other words, $\bar{y} = 0$ is an equilibrium solution. Also, when
$r - \frac{P_{max}}{a+y} = 0$
we have an equilibrium solution. This happens when $r = \frac{P_{max}}{a+y}$ or $r(a+y) = P_{max}$ which means that
$\bar{y} = \frac{P_{max}}{r} - a$
is an equilibrium solution. Still some care is needed here: remember that only positive values for the population make sense. This is, this equilibrium solution really only makes sense when $\frac{P_{max}}{r} - a > 0$ or $\frac{P_{max}}{a} > r$ . This will give us two cases to work with depending on whether this condition holds or not.

Case 1: When $\frac{P_{max}}{a} < r$ , we only have one equilibrium solution which has any meaning---namely, $\bar{y} = 0$ . Let's think about this solution and detemrine whether it is stable or not.
Remember, if the differential equation is $\frac{dy}{dt} = f(y)$ , the derivative of f can tell us about stability. This is the Linear Stability Test. In our case, so that $f^\prime(y) = r - P^\prime(y)$ . This means that
$f^\prime(0) = r - P^\prime(0) = r - \frac{P_{max}}{a} > 0$
This tells us that the equilibrium solution $\bar{y} = 0$ is unstable.
Case 2: When $\frac{P_{max}}{a} > r$ , there are two equilibrium solutions. As above, we have
$f^\prime(0) = r - P^\prime(0) = r - \frac{P_{max}}{a} > 0$
which tells us that $\bar{y} = 0$ is a stable equilibrium.
Now the other equilibrium occurs at $\bar{y} = \frac{P_{max}}{r} - a$ . For this value, we have
$\begin{eqnarray*} f^\prime(\frac{P_{max}}{r} - a) & = & r - \frac{P_{max}a}{(a + \frac{P_{max}}{r} - a)^2} \\ & = & r - \frac{r^2a}{P_{max}} \\ & = & r(1-\frac{ra}{P_{max}}) > 0 \end{eqnarray*}$
In other words, this equilibrium solution is stable.

We can interpret these results to derive some conclusions. Let's think about the equilibrium solutions in a different way though. Remember that equilibrium solutions are given by
$f(y) = ry - P(y) = 0$
or . We can understand this geometrically as the intersection of the graphs of and as below.

In this picture, you can adjust the value of $P_{max}$ . In terms of our story, this could correspond to getting more cats, say. This sheds some light on our condition $\frac{P_{max}}{a} > r$ . The term on the left $\frac{P_{max}}{a}$ is the slope of the graph of at the origin. The condition to have an equilbrium solution other than at $\bar{y} = 0$ means that the slope of the graph of at the origin should be larger than the slope of the line. You can see above that this is indeed the condition necessary to generate another intersection point.
When $\frac{P_{max}}{a} < r$ , there is only the equilibrium solution $\bar{y} = 0$ . It is unstable which means that solutions will increase away from . In other words, the rat population will always be increasing. This means that the cats have not controlled the population at all.
When $\frac{P_{max}}{a} > r$ , the equilibrium solution $\bar{y} = 0$ becomes stable and there is another equilibrium solution which is unstable. For values of the population between these two equilibrium solutions, the population will steadily decrease to zero. This means that the rat population has been eliminated!
However, for values of the rat population above this second equilibrium, the population steadily increases and there is still a rat problem. In other words, the cats can only control the rat population provided there are not too many rats to begin with.