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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
Download (direct link): elementarydifferentialequat2001.pdf
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a2i + a22 + (aiia22 - ai2a2i) b _ ai2a22 + alla2l
_ 2A ’ _ A '
a2
C= —
+ a12 + (aiia22 - ai2a2i)
2A
where A = (an + a22)(ana22 — al2a21).
(c) Using the result of part (a) show that A > 0 and then show (several steps of algebra are required) that
4 AC - B 2 = (ah + a12 + a21 + a22 )(alla22 — al2a21) + 2(al1 a22 — al2 a21)2 > 0
= A2 '
Thus by Theorem 9.6.4, V is positive definite.
11. In this problem we show that the Liapunov function constructed in the preceding problem is also a Liapunov function for the almost linear system (i). We must show that there is some region containing the origin for which Vis negative definite.
(a) Show that
V(x, y) = -(x2 + y2) + (2Ax + By)F1(x, y) + (Bx + 2Cy)G1(x, y).
(b) Recall that Fj(x, y)/r ^ 0 and Gj(x, y)/r ^ 0 as r = (x2 + y2)1/2 ^ 0. This means that given any e > 0 there exists a circle r = R about the origin such that for 0 < r < R, | F1(x, y)| < er,and | Gj (x, y)| < er. Letting M be the maximum of |2 A|, | B|, and |2C|, show by introducing polar coordinates that R can be chosen so that V(x, y) < 0 for r < R.
Hint: Choose e sufficiently small in terms of M.
12. In this problem we prove a part of Theorem 9.3.2 relating to instability.
(a) Show that if an + a22 > 0 and ana22 — a12a21 > 0, then the critical point (0, 0) of the linear system (ii) is unstable.
(b) The same result holds for the almost linear system (i). As in Problems 10 and 11 construct a positive definite function V such that V(x, y) = x2 + y2 and hence is positive definite, and then invoke Theorem 9.6.2.
9.7 Periodic Solutions and Limit Cycles
In this section we discuss further the possible existence of periodic solutions of second order autonomous systems
x' = f(x). (i)
Such solutions satisfy the relation
x(t + T) = x(t) (2)
522
Chapter 9. Nonlinear Differential Equations and Stability
EXAMPLE
1
for all t and for some nonnegative constant T called the period. The corresponding trajectories are closed curves in the phase plane. Periodic solutions often play an important role in physical problems because they represent phenomena that occur repeatedly. In many situations a periodic solution represents a “final state” toward which all “neighboring” solutions tend as the transients due to the initial conditions die out.
A special case of a periodic solution is a constant solution x = x0, which corresponds to a critical point of the autonomous system. Such a solution is clearly periodic with any period. In this section, when we speak of a periodic solution, we mean a nonconstant periodic solution. In this case the period T is positive and is usually chosen as the smallest positive number for which Eq. (2) is valid.
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
/x\ = ( y + x — x(x2 + yp) . (4)
y) \—x + y — y(xl + 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
\$=(—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 Q, where
x = r cos Q, y = r sin Q, (6)
and r > 0. If we multiply the first of Eqs. (4) by x, the second by y, and add, we then obtain
x^T + yd-y = (x2 + ?) — (x2 + /)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— = r 2(1 — r2). (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.
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