Books in black and white
 Books Biology Business Chemistry Computers Culture Economics Fiction Games Guide History Management Mathematical Medicine Mental Fitnes Physics Psychology Scince Sport Technics

# 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.
Previous << 1 .. 263 264 265 266 267 268 < 269 > 270 271 272 273 274 275 .. 609 >> Next

Since the eigenvalues are of opposite sign, the critical point (0.5, 0.5) is a saddle point, and therefore is unstable, as we had surmised earlier. One pair of trajectories approaches
1
0
1
0
v
1
0
3
1
1
1
498
Chapter 9. Nonlinear Differential Equations and Stability
the critical point as t — to; the others depart from it. As they approach the critical point, the entering trajectories are tangent to the line with slope (ë/57 — 3)/8 = 0.5687 determined from the eigenvector g(2).
A phase portrait for the system (21) is shown in Figure 9.4.4. Near each of the critical points the trajectories of the nonlinear system behave as predicted by the corresponding linear approximation. Of particular interest is the pair of trajectories that enter the saddle point. These trajectories form a separatrix that divides the first quadrant into two basins of attraction. Trajectories starting above the separatrix ultimately approach the node at (0, 2), while trajectories starting below the separatrix approach the node at Bad math construct!(1,0)Bad math construct!. If the initial state lies precisely on the separatrix, then the solution (x, y) will approach the saddle point as t — to. However, the slightest perturbation as one follows this trajectory will dislodge the point (x, y) from the separatrix and cause it to approach one of the nodes instead. Thus, in practice, one species will survive the competition and the other will not.
FIGURE 9.4.4 A phase portrait of the system (21).
Examples 1 and 2 show that in some cases the competition between two species leads to an equilibrium state of coexistence, while in other cases the competition results in the eventual extinction of one of the species. To understand more clearly how and why this happens, and to learn how to predict which situation will occur, it is useful to look
9.4 Competing Species
499
again at the general system (2). There are four cases to be considered, depending on the relative orientation of the lines
e1 — a1 x — a1 y = 0 and e2 — a2 y — a2 x = 0,
(34)
as shown in Figure 9.4.5. These lines are called the x and y nullclines, respectively, because x is zero on the first and / is zero on the second. Let (X, Y) denote any critical point in any one of the four cases. As in Examples 1 and 2 the system (2) is almost linear in the neighborhood of this point because the right side of each differential equation is a quadratic polynomial. To study the system (2) in the neighborhood of this critical point we can look at the corresponding linear system obtained from Eq. (13) of Section 9.3,
— /u
dt U
e1 — 2ñò1 X — a1 Y —a2Y
—a1 X ^2 — 2^2 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)
61CT2 — ª2à1
ª2à1 — 4a2
CT1CT2 — a1a2
CT1CT2 — a1a2
FIGURE 9.4.5 The various cases for the competing species system (2).
500
Chapter 9. Nonlinear Differential Equations and Stability
Further, since ˆ1 — à1X — a1Y = 0 and ˆ2 — a2Y — a2X = 0, Eq. (35) immediately reduces to
d /u\ _ /—à1X — a1X\ I u
dt \ v / \ — «2 Y —02 Y / \v
(37)
The eigenvalues of the system (37) are found from the equation
r2 + (a1 X + a2Y)r + (a1a2 — axa2) XY = 0. (38)
Thus
— (aj X + 02 Y) ± Ó (aj X + 02 Y)2 — 4(axa2 — «1^) XY
(39)
If aja2 — aja2 < 0, then the radicand of Eq. (39) is positive and greater than (aj X + a2 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 aj = 1, aj = 1, a2 = 0.25, a2 = 0.75, and aja2 — aja2 = —0.5. j j 2 2 j 2 j 2
On the other hand, if aja2 — aja2 > 0, then the radicand of Eq. (39) is less than (à1 X + a2 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 aj = 1, aj = 1, a2 = 1, a2 = 0.5, and aja2 — aja2 = 0.5.
Let us relate this result to Figures 9.4.5c and 9.4.5d. In Figure 9.4.5c we have
ˆ 1 ˆ2 ˆ1
— > — or ˆ,à, > ˆ?a, and — > — or ˆ2a1 > ˆ,a,. (40)
a1 a2 a2 a1
These inequalities, coupled with the condition that X and Y given by Eqs. (36) be positive, yield the inequality à1à2 < a1a2. Hence in this case the critical point is a saddle point. On the other hand, in Figure 9.4.5 d we have
Previous << 1 .. 263 264 265 266 267 268 < 269 > 270 271 272 273 274 275 .. 609 >> Next