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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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Recall that the solutions of the linear autonomous system
x; = Ax (3)
are periodic if and only if the eigenvalues of A are pure imaginary. In this case the critical point at the origin is a center, as discussed in Section 9.1. We emphasize that if the eigenvalues of A are pure imaginary, then every solution of the linear system (3) is periodic; while if the eigenvalues are not pure imaginary, then there are no (nonconstant) periodic solutions. The predator-prey equations discussed in Section 9.5, although nonlinear, behave similarly: All solutions in the first quadrant are periodic. The following example illustrates a different way in which periodic solutions of nonlinear autonomous systems can occur.
Discuss the solutions of the system
(xY = ( y+xx(x2 + yp). (4)
y) \x + y y(x2 + y2))
It is not difficult to show that (0, 0) is the only critical point of the system (4), and also that the system is almost linear in the neighborhood of the origin. The corresponding linear system
y)'=(1 1)(y) <5>
has eigenvalues 1 i. Therefore the origin is an unstable spiral point for both the linear system (5) and the nonlinear system (4). Thus any solution that starts near the origin in the phase plane will spiral away from the origin. Since there are no other critical points, we might think that all solutions of Eqs. (4) correspond to trajectories that spiral out to infinity. However, we now show that this is incorrect because far away from the origin the trajectories are directed inward.
It is convenient to introduce polar coordinates r and , where
x = r cos , y = r sin , (6)
and r > 0. If we multiply the first of Eqs. (4) by x, the second by y, and add, we then obtain
x^x + ydJL = (x2 + y2) (x2 + y2)2. (7)
dt dt
9.7 Periodic Solutions and Limit Cycles
523
Since r2 = x2 + y2 and r (dr/dt) = x(dx/dt) + y( dy/dt), it follows from Eq. (7) that
dr
r = r2(1 - r ). (8)
dt
This is similar to the equations discussed in Section 2.5. The critical points (for r > 0) are the origin and the point r = 1 , which corresponds to the unit circle in the phase
plane. From Eq. (8) it follows that dr/dt > 0 if r < 1 and dr/dt < 0 if r > 1. Thus,
inside the unit circle the trajectories are directed outward, while outside the unit circle they are directed inward. Apparently, the circle r = 1 is a limiting trajectory for this system.
To determine an equation for we multiply the first of Eqs. (4) by y, the second by x, and subtract, obtaining
/ - x& = x2 + y2. (9)
J dt dt J w
Upon calculating dx/dt and dy/dt from Eqs. (6), we find that the left side of Eq. (9) is -r2(de/dt), so Eq. (9) reduces to
d
= -1. (10)
dt v '
The system of equations (8), (10) for r and is equivalent to the original system (4). One solution of the system (8), (10) is
r = 1, = -t + tQ, (11)
where tQ is an arbitrary constant. As t increases, a point satisfying Eqs. (11) moves clockwise around the unit circle. Thus the autonomous system (4) has a periodic solution. Other solutions can be obtained by solving Eq. (8) by separation of variables; if r = 0 and r = 1, then
dr
---------T = dt. (12)
r(1 - r2)
Equation (12) can be solved by using partial fractions to rewrite the left side and then integrating. By performing these calculations we find that the solution of Eqs. (10) and (12) is
r = , 1 =, = -t + tQ, (13)
1 + cQ e-2t
where cQ and tQ are arbitrary constants. The solution (13) also contains the solution
(11), which is obtained by setting cQ = 0 in the first of Eqs. (13).
The solution satisfying the initial conditions r = p, = a at t = 0 is given by
r = , ' , = -(t - a). (14)
/1 + [(1/P2) - 1]e-2t
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