D gx xs ys x gy xs y s

and we now compare Eq. (2.46) with Eq. (2.37) to determine the values of A, B, C, and D (note that they are not arbitrary but must match the various partial derivatives in Eq. (2.46)), from which we can determine the stability characteristics of the steady states.

To help make the preceding more concrete, we will first consider an example, then an exercise. A very simple model for competition between two types or species x(t) and y(t) is dx

— — y( 1+ a — y — ax) dt where a is a parameter, which we assume to be positive. From the form of these equations, we see that the presence of x increases the rate of change of x and that the presence of both x and y decreases the rate of change of x (and vice versa for y). We say that x andy are auto-catalysts for themselves and anti-catalysts for the other type. This thinking underlay the work of Sir F. C. Frank in his study of spontaneous asymmetric synthesis (see Connections); Eq. (2.47) is also a simple analog of the Lotka-Volterra competition equations, in which the competition is symmetric.

We find the steady states of Eq. (2.47) by setting dx/dt = 0 and dy/dt = 0. For the former, we find that x = 0or x + ay = 1 + a. For the latter, we find thaty = 0 or y + ax = 1 + a. Thus, (1, 1) is a steady state. Before conducting an eigenvalue analysis, we use the isoclines (or more properly, the nullclines, lines on which dx/dt = 0 or dy/dt = 0) of the differential equations to understand properties of the solution. These are shown in Figure 2.16. The steady state (1, 1) can be either a node (if a < 1) or a saddle point (if a > 1). When a = 1, the two isoclines sit on top of each other and the system is structurally unstable. Note also that, because x and y are interchangeable in the two equations, the line y = x is a solution of the equations - points on the line y = x move towards (1, 1), regardless of whether it is a node or a saddle point.

We can now conduct the eigenvalue analysis. In this case f (x,y) = x(1 + a — x — ay) = x(1 + a) — x — axy and g(x, y) = y(1 + a) — y — axy. The partial derivatives are thus f = 1 + a — 2x — ay, f =—ax, gx =—ay, and gy = 1 + a — 2y — ax, and we evaluate these at (1, 1) in order to obtain A, B, C, and D, so that A = 1 + a — 2 — a = —1, B = —a, C =—a, and D = 1 + a — 2 — a = —1. We substitute this into Eq. (2.42) and find that = — 1 ± Va From the eigenvalue analysis, we reach the same conclusion as from the phase plane analysis - that (1,1) is either a stable node or saddle point, depending upon the value of a. Thus, in this case, the eigenvalue analysis told us little that we could not understand from the phase plane. Here's an example where it tells us much more.

Figure 2.16. The isocline analysis of the equations for spontaneous asymmetric synthesis/symmetric competition. In panels (a) and (b), I show the separate isoclines for dx/dt = 0 and dy/dt = 0 and the flow of points in the phase plane. When these are put together, the resulting phase plane shows either a stable node at (1,1) (panel c) or a saddle point (panel d).

Figure 2.16. The isocline analysis of the equations for spontaneous asymmetric synthesis/symmetric competition. In panels (a) and (b), I show the separate isoclines for dx/dt = 0 and dy/dt = 0 and the flow of points in the phase plane. When these are put together, the resulting phase plane shows either a stable node at (1,1) (panel c) or a saddle point (panel d).

Consider the following predator (P)-prey (V) system dV / V\ dP

Assume that the biomass of each is measured in numbers of individuals (but treated as a continuous variable) and time is measured in years. It might be helpful for what follows to think of rabbits and foxes as the victims and predators. It might also be helpful, especially for parts (a) and (b), to convert to per capita growth rates. (a) What are the units of all the parameters? (b) Interpret the biology of both predator and prey. What must be true about the relationship between b and c if the system is mammalian predators such as rabbits and foxes? (c) Conduct an isocline analysis. Note: there are two cases, depending upon the relationship between K and m/c. Be sure to get both of them and carefully think about what each means. (d) Classify the steady states of the system according to their eigenvalues. What does the eigenvalue calculation tell you that the isocline analysis did not? Once again, there are two cases that require careful interpretation. (e) What happens to the eigenvalues as Kn?

What does this mean for the dynamics of the system? If you want more after this, do the same kind of analysis for the mutualism described in Eq. (2.35). We will return to this exercise, in a different context, in Chapter 6.

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