Download (direct link):
(c) Analyze the nature of the trajectories when a1e2 — a2e1 = J.
11. Consider the system (3) in Example 1 of the text. Recall that this system has an asymptotically stable critical point at (J.5, J.5), corresponding to the stable coexistence of the two population species. Now suppose that immigration or emigration occurs at the constant rates of Sa and S b for the species x and y, respectively. In this case Eqs. (3) are replaced by
dx/dt = x (1 — x — y) + Sa,
dy/dt = y(J.75 — y — J.5x) + Sb.
The question is what effect this has on the location of the stable equilibrium point.
(a) To find the new critical point we must solve the equations
x (1 — x — y) + Sa = J,
y(J.75 — y — J.5x) + Sb = J.
One way to proceed is to assume that x and y are given by power series in the parameter S; thus
x = xJ + x1S + •••> y = yJ + y1S + •••• (iii)
Substitute Eqs. (iii) into Eqs. (ii) and collect terms according to powers of S.
(b) From the constant terms (the terms not involving S) show that xJ = J.5 and yJ = J.5, thus confirming that in the absence of immigration or emigration the critical point is (J.5, J.5).
(c) From the terms that are linear in S show that
x1 = 4a — 4b, y1 = —2a + 4b. (iv)
(d) Suppose that a > J and b > J so that immigration occurs for both species. Show that the resulting equilibrium solution may represent an increase in both populations, or an increase in one but a decrease in the other. Explain intuitively why this is a reasonable result.
> 12. Suppose that a certain pair of competing species are described by the system
dx/dt = x (4 — x — y), dy/dt = y(2 + 2a — y — ax), where a > J is a parameter.
(a) Find the critical points. Note that (2, 2) is a critical point for all values of a.
(b) Determine the nature of the critical point (2, 2) for a = J.75 and for a = 1.25. There is a value of a between J.75 and 1.25 where the nature of the critical point changes abruptly. Denote this value by aJ; it is called a bifurcation point.
(c) Find the approximate linear system near the point (2, 2) in terms of a.
(d) Find the eigenvalues of the linear system in part (c) as functions of a. Then determine
the bifurcation point aJ.
(e) Draw phase portraits near (2, 2) for a = aJ and for values of a slightly less than, and slightly greater than, aJ. Explain how the transition in the phase portrait takes place as a passes through aJ.
> 13. The system
x = —y, y = — ó y — x (x — J.15)(x — 2)
9.5 Predator-Prey Equations 503
results from an approximation to the Hodgkin-Huxley2 equations, which model the transmission of neural impulses along an axon.
(a) Find the critical points and classify them by investigating the approximate linear system near each one.
(b) Draw phase portraits for ó = 0.8 and for y = 1.5.
(c) Consider the trajectory that leaves the critical point (2, 0). Find the value of ó for which this trajectory ultimately approaches the origin as t — to. Draw a phase portrait for this value of ó.
9.5 Predator-Prey Equations
In the preceding section we discussed a model of two species that interact by competing for a common food supply or other natural resource. In this section we investigate the situation in which one species (the predator) preys on the other species (the prey), while the prey lives on a different source of food. For example, consider foxes and rabbits in a closed forest: The foxes prey on the rabbits, the rabbits live on the vegetation in the forest. Other examples are bass in a lake as predators and redear as prey, or ladybugs as predators and aphids as prey. We emphasize again that a model involving only two species cannot fully describe the complex relationships among species that actually occur in nature. Nevertheless, the study of simple models is the first step toward an understanding of more complicated phenomena.
We will denote by x and y the populations of the prey and predator, respectively, at time t. In constructing a model of the interaction of the two species, we make the following assumptions: