# 18.1 Predator Prey Models

Suppose we have two sorts of animals. Type $$A$$ animals eat type $$B$$ ones. We suppose that in the absence of type $$B$$ animals, type $$A$$ ones will not get enough to eat and will die off or move away to avoid doing so. On the other hand we assume that in the absence of type $$A$$ animals, type $$B$$ ones will have a better chance to survive and will experience population growth.

Let $$A$$ represent the population in our area of type $$A$$ animals. In the absence of type $$B$$ creatures the simplest model for these is that change of $$A$$ population is a negative multiple of $$A$$ itself:

$\frac{dA}{dt} = -cA$

Similarly, in the absence of $$A$$ creatures, the simples model for behavior of the population $$B$$ of $$B$$ types will show increase, and obey, for some $$r$$,

$\frac{dB}{dt} = rB$

The interaction between $$A$$ and $$B$$ must be $$0$$ if either $$A$$ or $$B$$ is $$0$$. The simplest interaction model is then that the contribution to the change per unit time in $$A$$ from the presence of $$B$$ is $$sAB$$ for some $$s$$, and the effect on $$B$$ is $$-qAB$$ for some $$q$$.

Our simplest model for this situation then has the form

$\frac{dA}{dt} = -cA + sAB$

and

$\frac{dB}{dt} = rB - qAB$

So what can we say about the behavior of these populations in this model?

We can first look for steady state solutions. These happen when both derivatives are $$0$$. When that happens the populations of the two species stays the same. Such solutions are called fixed points of the equations.

Here this happens when $$cA = sAB$$ and $$rB = qA$$ for which $$sB = c$$, and $$r = qA$$.

A fixed point solution is said to be stable if small deviations in $$A$$ or $$B$$ tend to die out and disappear or at least donâ€™t spiral outward. You can investigate the stability of this fixed point, for values of $$c$$, $$s$$, $$r$$ and $$q$$ of your choosing, by "integrating" these equations numerically. Start with a value of $$A$$ and $$B$$ slightly off from the fixed point and move forward in time, finding both $$A(t)$$ and $$B(t)$$ exactly like one does in an integral (discussed in Chapter 14), using the left hand rule.

What tends to happen here in the $$(A, B)$$ plane, is that solutions generally spiral in toward the fixed point.

Given one orbit in this plane, no other orbit can cross it, because at a common point the derivatives would be the same in both, which would mean that the orbits were identical afterward.

Exercise: Set this up with a spreadsheet, and check on this conclusion.

You can see qualitatively what happens if, for example, the population $$B$$ is suddenly reduced from its fixed point value. This causes a reduction in $$\frac{dA}{dt}$$ from its $$0$$ value at the fixed point, so that the population $$A$$ decreases. This in turn causes an increase in $$B$$. Thus, if $$B$$ is the vertical coordinate used in charting the orbit, starting beneath the fixed point, the orbit moves counterclockwise around it.

You do not even need a spreadsheet to see how solutions behave. Given a starting point $$(A,B)$$, you can draw an arrow from it, pointing in the direction whose tangent is the ratio of the $$A$$ derivative to $$B$$ derivative at that point. Then choose a point a small distance along that arrow, and repeat these steps. You will generate an approximate orbit of the system in the $$(A,B)$$ plane.

How this plays out depends considerably on how fast the two populations recover from their reductions.

An interesting case is one in which flies are the prey and birds the predators that eat them. Reducing the fly population significantly, reduces the bird population as well, as noted above. However, fly populations recover relatively quickly, in on the order of weeks, while bird populations take on the order of years to recover. Thus bird populations tend to decline for a short time, but rise again to their fixed point value only slowly. This means that fly populations increase for quite a long time and spend much of that time at levels well above the fixed point value. Thus a campaign for killing flies is not a productive way to reduce fly populations except very temporarily.