# Elementary Differential Equations and Boundary Value Problems - Boyce W.E.

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492

Chapter 9. Nonlinear Differential Equations and Stability

EXAMPLE

1

which species ó interferes with species x. Similarly, in Eq. (1b) we replace e2 — a2ó by e2 — o2ó — a2x. Thus we have the system of equations

dx/dt = x (ej — CTj x — al y), (2)

(2)

dy/dt = y(e2 — a2 ó — a2x ).

The values of the positive constants e±, ox, ax, e2, ct2, 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 ó 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).

(3)

Discuss the qualitative behavior of solutions of the system

dx/dt = x (1 — x — y), dy/dt = ó (0.75 — ó — 0.5x).

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

x (1 — x — y) = 0, y(0.75 — ó — 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, ó = 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 (u\-( Fx(X’Y) Fy(X YA (u\ (5)

dt W = [gx(X, Y) Gy(X, Y)) (v) ’ (5)

where, for the system (3),

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

Thus Eq. (5) becomes

d (u\ (1-2X-Y -X

dt W V — 0.5Y 0.75 — 2Y — 0.5XI (7)

9.4 Competing Species

493

y

1 I / I / / / / / S' *< jS

1 I I / J / / S S ³/ ? S' jS

I ¯ / / / / Ó JS S' S' jS

I I / / ? / / Ó S' S jS

I / / / Ó Ó

0 75 I 1 I / / / Ó

^ s, 1 { / / / ^ Ó

/ ---* I / / / S

Ê / \ ³ ? ó *s JS jS JtS jS

t / 's. / ........ jS jS ---

0.5 - / è X» S' S' • ^ *- ....... *- *<¦ ¦*-" Jr** Mr-

/ / X» S' N +- ...... ¦*-

t / / S' S' ^ / N

t / / / S' è /

t / / Z' S' Ó f

0.25 - Ó S' ó ó /» / \ N

è / ó S' s* ó Ó ó ÷

è ó

1 1

0 0.25 0.50 0.75 , x

1

FIGURE 9.4.1 Critical points and direction field for the system (3).

x = 0, y = 0. This critical point corresponds to a state in which both species die as a result of their competition. By rewriting the system (3) in the form

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