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

Boyce W.E Elementary differential equations 7th edition - Wiley publishing , 2001. - 1310 p.
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 os are a measure of the inhibitory effect the growth of each population has on itself, while the as 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 coexistone must die out.
9.4 Competing Species
501
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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