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

Boyce W.E. Elementary Differential Equations and Boundary Value Problems - John Wiley & Sons, 2001. - 1310 p.
Download (direct link): elementarydifferentialequations2001.pdf
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— (x dt\y.
1 0 0 0.75
x2 + xy 0.5xy + y2
(8)
or by setting X = Y = 0 in Eq. (7), we see that near the origin the corresponding linear system is
— (x
dt\y,
The eigenvalues and eigenvectors of the system (9) are
1 0 0 0.75
=
r2 = 0.75,
g(2) =
so the general solution of the system is
e + c2
„0.75 t
(9)
(10)
(11)
Thus the origin is an unstable node of both the linear system (9) and the nonlinear system (8) or (3). In the neighborhood of the origin all trajectories are tangent to the y-axis except for one trajectory that lies along the x-axis.
x = 1, y = 0. This corresponds to a state in which species x survives the competition, but species y does not. The corresponding linear system is
—iu dt U
-1 -1
0 0.25
(12)
0
1
u
v
494
Chapter 9. Nonlinear Differential Equations and Stability
Its eigenvalues and eigenvectors are
r, = -
g(1) =
r2 = 0.25,
g(2) =
5
and its general solution is
c
10
+ c2
_4 ) e0 25t
(13)
(14)
Since the eigenvalues have opposite signs, the point (1, 0) is a saddle point, and hence is an unstable equilibrium point of the linear system (12) and of the nonlinear system (3). The behavior of the trajectories near (1, 0) can be seen from Eq. (14). If c2 = 0, then there is one pair of trajectories that approaches the critical point along the x-axis. All other trajectories depart from the neighborhood of (1, 0).
x = 0, y = 0.75. In this case species j survives but x does not. The analysis is similar to that for the point (1, 0). The corresponding linear system is
d (u
dt \v
The eigenvalues and eigenvectors are
0.25
0
-0.375 -0.75
r1 = 0.25,
g(1) =
8
r2 = -0.75,
g(2) =
so the general solution of Eq. (15) is
c
3
^ e025t + cJl') e-075t
(15)
(16)
(17)
Thus the point (0, 0.75) is also a saddle point. All trajectories leave the neighborhood of this point except one pair that approaches along the j-axis.
x = 0.5, y = 0.5. This critical point corresponds to a mixed equilibrium state, or coexistence, in the competition between the two species. The eigenvalues and eigenvectors of the corresponding linear system
d_ /u\_ /-0.5 -0.5
dt W = 1-0.25 -0.5
are
r1 = (-2 + V2)/4 = -0.146,
r2 = (-2 - V2)/4 = -0.854,
c(1) _ (^2^ .
g(1) =
1
g(2) = ( V2
Therefore the general solution of Eq. (18) is
c
-0.
1
0.146t
+ c2
-0.
1
0.854t
(18)
(19)
(20)
Since both eigenvalues are negative, the critical point (0.5, 0.5) is an asymptotically stable node of the linear system (18) and of the nonlinear system (3). All trajectories approach the critical point as t — to. One pair of trajectories approaches the critical
1
4
0
1
e
v
0
1
9.4 Competing Species
495
point along the line with slope ë/2/2 determined from the eigenvector ?(2). All other trajectories approach the critical point tangent to the line with slope —V2/2 determined from the eigenvector ?(1).
A phase portrait for the system (3) is shown in Figure 9.4.2. By looking closely at the trajectories near each critical point you can see that they behave in the manner predicted by the linear system near that point. In addition, note that the quadratic terms on the right side of Eqs. (3) are all negative. Since for x and ó large and positive these terms are the dominant ones, it follows that far from the origin in the first quadrant both x and ó are negative, that is, the trajectories are directed inward. Thus all trajectories that start at a point (x0, y0) with x0 > 0 and y0 > 0 eventually approach the point (0.5, 0.5).
FIGURE 9.4.2 A phase portrait of the system (3).
EXAMPLE
2
Discuss the qualitative behavior of the solutions of the system
dx/dt = x (1 — x — y),
(21)
dy/dt = ó (0.5 — 0.25 ó — 0.75x),
when x and ó are nonnegative. Observe that this system is also a special case of the system (2) for two competing species.
Once again, there are four critical points, namely, (0, 0), (1, 0), (0, 2), and (0.5, 0.5), corresponding to equilibrium solutions of the system (21). Figure 9.4.3 shows a direction field for the system (21), together with the four critical points. From the direction field it appears that the mixed equilibrium solution (0.5, 0.5) is a saddle point, and therefore unstable, while the points (1, 0) and (0, 2) are asymptotically stable. Thus, for competition described by Eqs. (21), one species will eventually overwhelm the other and drive it to extinction. The surviving species is determined by the initial state of the system. To confirm these conclusions we can look at the linear approximations near each critical point.
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