# Elementary differential equations 7th edition - Boyce W.E

ISBN 0-471-31999-6

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(a) Write the Lienard equation as a system of two first order equations by introducing the variable y = dx/dt.

(b) Show that (0, 0) is a critical point and that the system is almost linear in the neighborhood of (0, 0).

(c) Show that if c(0) > 0 and g(0) > 0, then the critical point is asymptotically stable, and that if c(0) < 0 or g(0) < 0, then the critical point is unstable.

Hint: Use Taylor series to approximate c and g in the neighborhood of x = 0.

9.4 Competing Species

In this section and the next we explore the application of phase plane analysis to some problems in population dynamics. These problems involve two interacting populations and are extensions of those discussed in Section 2.5, which dealt with a single population. While the equations discussed here are extremely simple, compared to the very complex relationships that exist in nature, it is still possible to acquire some insight into ecological principles from a study of these model problems.

Suppose that in some closed environment there are two similar species competing for a limited food supply; for example, two species of fish in a pond that do not prey on each other, but do compete for the available food. Let x and y be the populations of the two species at time t. As discussed in Section 2.5, we assume that the population

of each of the species, in the absence of the other, is governed by a logistic equation.

Thus

dx/ dt = x (ˆ1 — o 1 x), (1a)

dy/dt = y(ˆz — o2 y), (1b)

respectively, where ˆ1 and e2 are the growth rates of the two populations, and ˆ^/01 and ˆ2/ct2 are their saturation levels. However, when both species are present, each will impinge on the available food supply for the other. In effect, they reduce the growth rates and saturation populations of each other. The simplest expression for reducing the growth rate of species x due to the presence of species y is to replace the growth rate factor ˆ1 — 01 x in Eq. (1a) by ˆ1 — oq x — «1 y, where «1 is a measure of the degree to

492

Chapter 9. Nonlinear Differential Equations and Stability

EXAMPLE

1

which species y interferes with species x. Similarly, in Eq. (1b) we replace e2 — ct2y by e2 — ct2 y — a2 x. Thus we have the system of equations

dx/dt = x(e1 — a1 x — a1 y),

(2)

dy/dt = y(e2 — a2 y — a2x)-

The values of the positive constants ov ai, ˆ2, a2, and a2 depend on the particular species under consideration and in general must be determined from observations. We are interested in solutions of Eqs. (2) for which x and y are nonnegative. In the following two examples we discuss two typical problems in some detail. At the end of the section we return to the general equations (2).

Discuss the qualitative behavior of solutions of the system

dx/dt = x (1 — x — y),

(3)

dy/dt = y (0.75 — y — 0.5x).

We find the critical points by solving the system of algebraic equations

x (1 — x — y) = 0, y(0.75 — y — 0.5x) = 0. (4)

There are four points that satisfy Eqs. (4), namely, (0, 0), (0, 0.75), (1, 0), and (0.5, 0.5); they correspond to equilibrium solutions of the system (3). The first three of these points involve the extinction of one or both species; only the last corresponds to the long-term survival of both species. Other solutions are represented as curves or trajectories in the xy-plane that describe the evolution of the populations in time. To begin to discover their qualitative behavior we can proceed in the following way.

A direction field for the system (3) in the positive quadrant is shown in Figure 9.4.1; the heavy dots in this figure are the critical points or equilibrium solutions. Based on the direction field it appears that the point (0.5, 0.5) attracts other solutions and is therefore asymptotically stable, while the other three critical points are unstable. To confirm these conclusions we can look at the linear approximations near each critical point.

The system (3) is almost linear in the neighborhood of each critical point. There are two ways to obtain the linear system near a critical point (X, Y). First, we can use the substitution x = X + u, y = Y + v in Eqs. (3), retaining only the terms that are linear in u and v. Alternatively, we can use Eq. (13) of Section 9.3, that is,

d M ( Fx (X, Y) Fy (X, Y)x !u

dt W_ (X, Y) Gy(X, Y)J\v

(5)

where, for the system (3),

F(x, y) = x(1 — x — y), G(x, y) = y(0.75 — y — 0.5x). (6)

Thus Eq. (5) becomes

/1 — 2X — Y — X

dt w V —0.5Y 0.75 — 2 Y — 0.5X

(7)

u

v

9.4 Competing Species

493

J'

1 I / 1 / / / / / zZ Z Z Z Z Z z z z Z

1 1 1 I / J / / yz Z z z z Z z Z z z

1 / / / / / z z z z Z Z Z z z

I 1 / / ? / / Z z z z Z Z Z z z

1 / / / y

0 75« 1 1 I. / / / zZ y

s 1 { / / / ^ y

/ —* \ I / / / y

K / \ I ? y y z z Z yy z z z yy

t / zz" Z / ^ y y y Z Z z z z z yy yy

0.5 - / ZZ» Z Z • ^ z—' y y y~ z Z yy yy .z yy yy

/ / z-» Z N y y yy yy Z Z z z z yy yy

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