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

ISBN 0-471-31999-6

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(u

dt U

1 2a1 X a1 Y a2Y

a1X 2 2a2 Y a2 X

(35)

We now use Eq. (35) to determine the conditions under which the model described by Eqs. (2) permits the coexistence of the two species x and y. Of the four possible cases shown in Figure 9.4.5 coexistence is possible only in cases (c) and (d). In these cases the nonzero values of X and Y are readily obtained by solving the algebraic equations (34); the result is

X

Y

(36)

1 a2 2a 1

2a1 1a2

a1a2 a1a2

a1a2 a1a2

FIGURE 9.4.5 The various cases for the competing species system (2).

500

Chapter 9. Nonlinear Differential Equations and Stability

Further, since Cj ojX aj Y = 0 and e2 ct2Y a2X = 0, Eq. (35) immediately reduces to

d /u\ _ /oqX ajX\ ! u

dt \ v / \ a2 Y O2 Y J \ v

(37)

The eigenvalues of the system (37) are found from the equation

r2 + (oj X + 02Y)r + (ctjCT2 aja2) XY = 0. (38)

Thus

(ct, X + o2 Y) ± ,/(ct, X + o2 Y)2 4(0,0, a1a2)XY ru =---------1-------2-----^^----------------------------------------------------------- . (39)

Ifo1o2 aja2 < 0, then the radicand of Eq. (39) is positive and greater than (oj X + 02 Y)2. Thus the eigenvalues are real and of opposite sign. Consequently, the critical point (X, Y) is an (unstable) saddle point, and coexistence is not possible. This is the case in Example 2, where Oj = 1, aj = 1, = 0.25, a2 = 0.75, and OjCT2 a^ =

0.5.

On the other hand, if o^ a^ > 0, then the radicand of Eq. (39) is less than (oj X + 02 Y)2. Thus the eigenvalues are real, negative, and unequal, or complex with negative real part. A straightforward analysis of the radicand of Eq. (39) shows that the eigenvalues cannot be complex (see Problem 7). Thus the critical point is an asymptotically stable node, and sustained coexistence is possible. This is illustrated by Example 1, where Oj = 1, aj = 1, o2 = 1, a2 = 0.5, and o^ a^ = 0.5.

Let us relate this result to Figures 9.4.5c and 9.4.5d. In Figure 9.4.5c we have

61 62 62 61

> or > 62oj and > or 62ax > exo2. (40)

o1 a2 o2 a1

These inequalities, coupled with the condition that X and Y given by Eqs. (36) be positive, yield the inequality o^ < a^. Hence in this case the critical point is a saddle point. On the other hand, in Figure 9.4.5dwe have

61 62 62 61

< or 6^2 < 62oj and < or C2aj < 6jo2. (41)

o1 a2 o2 a1

Now the condition that X and Y are positive yields o^ > a^. Hence the critical point is asymptotically stable. For this case we can also show that the other critical points (0, 0), (6jlo1, 0), and (0, 62/o2) are unstable. Thus for any positive initial values of x and y the two populations approach the equilibrium state of coexistence given by Eqs. (36).

Equations (2) provide the biological interpretation of the result that coexistence occurs or not depending on whether o^ aja2 is positive or negative. The os are a measure of the inhibitory effect the growth of each population has on itself, while the as are a measure of the inhibiting effect the growth of each population has on the other species. Thus, when o^ > a^, interaction (competition) is weak and the species can coexist; when o^ < a^, interaction (competition) is strong and the species cannot coexistone must die out.

9.4 Competing Species

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PROBLEMS Each of Problems 1 through 6 can be interpreted as describing the interaction of two species

with populations x and y. In each of these problems carry out the following steps.

(a) Draw a direction field and describe how solutions seem to behave.

(b) Find the critical points.

(c) For each critical point find the corresponding linear system. Find the eigenvalues and eigenvectors of the linear system; classify each critical point as to type, and determine whether it is asymptotically stable, stable, or unstable.

(d) Sketch the trajectories in the neighborhood of each critical point.

(e) Compute and plot enough trajectories of the given system to show clearly the behavior of the solutions.

(f) Determine the limiting behavior of x and y as t and interpret the results in terms of

the populations of the two species.

? 1. dx/dt = x (1.5 x 0.5/) ? 2. dx/dt = x (1.5 x 0.5y)

dy/dt = y(2 y 0.75x) dy/dt = y(2 0.5 y 1.5x)

? 3. dx/dt = x (1.5 0.5x y) ? 4. dx/dt = x (1.5 0.5x y)

dy/dt = y(2 y 1.125x) dy/dt = y(0.75 y 0.125x)

? 5. dx/dt = x (1 x y) ? 6. dx/dt = x (1 x + 0.5 y)

dy/dt = y(1.5 y x) dy/dt = y(2.5 1.5y + 0.25x)

7. Show that

(o1 X + a2 Y)2 4(o1o2 a1 a2)XY = (o1 X a2 Y)2 + 4a1a2XY.

Hence conclude that the eigenvalues given by Eq. (39) can never be complex.

8. Two species of fish that compete with each other for food, but do not prey on each other, are bluegill and redear. Suppose that a pond is stocked with bluegill and redear and let x and y be the populations of bluegill and redear, respectively, at time t. Suppose further that the competition is modeled by the equations

dx/dt = x (1 o1 x a1 y),

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