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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.
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det (A - rl) = 0 (4)
and the eigenvectors are determined from Eq. (3) up to an arbitrary multiplicative constant.
In Section 2.5 we found that points where the right side of Eq. (1) is zero are of special importance. Such points correspond to constant solutions, or equilibrium solutions, of Eq. (1), and are often called critical points. Similarly, for the system (2), points where Ax = 0 correspond to equilibrium (constant) solutions, and again they are called critical points. We will assume that A is nonsingular, or that det A = 0. It follows that x = 0 is the only critical point of the system (2).
Recall that a solution of Eq. (2) is a vector function x = ) that satisfies the
differential equation. Such a function can be viewed as a parametric representation for a curve in the xxx2-plane. It is often useful to regard this curve as the path, or trajectory, traversed by a moving particle whose velocity dx/dt is specified by the differential equation. The xlx2-plane itself is called the phase plane and a representative set of trajectories is referred to as a phase portrait.
In analyzing the system (2) several different cases must be considered, depending on the nature of the eigenvalues of A. These cases also occurred in Sections 7.5 through 7.8, where we were primarily interested in finding a convenient formula for the general solution. Now our main goal is to characterize the differential equation according to the geometric pattern formed by its trajectories. In each case we discuss the behavior of the trajectories in general and illustrate it with an example. It is important that you become familiar with the types of behavior that the trajectories have for each case, because these are the basic ingredients of the qualitative theory of differential equations.
CASE 1 Real Unequal Eigenvalues of the Same Sign. The general solution of Eq. (2) is
x = clg(1)er1t + c2g(2)er2t, (5)
where rr and r2 are either both positive or both negative. Suppose first that rr < r2 < 0, and that the eigenvectors g(1) and g(2) are as shown in Figure 9.1.1a. It follows from Eq. (5) that x ^ 0 as t regardless of the values of q and c2; in other words, all solutions approach the critical point at the origin as t ^ro. If the solution starts at an
9.1 The Phase Plane: Linear Systems
461
initial point on the line through ?(1), then c2 = 0. Consequently, the solution remains on the line through ?(1) for all t, and approaches the origin as t ^ro. Similarly, if the initial point is on the line through ?(2), then the solution approaches the origin along that line. In the general situation, it is helpful to rewrite Eq. (5) in the form
x = er2t [c1?(1)e(r1 -r2)t + c2?(2)]. (6)
Observe that r1 — r2 < 0. Therefore, as long as c2 = 0, the term c1?(1) exp[(r1 — r2)t] is negligible compared to c2?(2) for t sufficiently large. Thus, as t ^ro, the trajectory not only approaches the origin, it also tends toward the line through ?(2). Hence all solutions approach the critical point tangent to ?(2) except for those solutions that start exactly on the line through ?(1). Several trajectories are sketched in Figure 9.1.1a. Some typical graphs of x1 versus t are shown in Figure 9.1.1b, illustrating that all solutions exhibit exponential decay in time. The behavior of x2 versus t is similar. This type of critical point is called a node or a nodal sink.
FIGURE 9.1.1 A node; rl < r2 < 0. (a) The phase plane. (b) Xj versus t.
If r1 and r2 are both positive, and 0 < r2 < r1 , then the trajectories have the same pattern as in Figure 9.1.1a, but the direction of motion is away from, rather than toward, the critical point at the origin. In this case x1 and x2 grow exponentially as functions of t. Again the critical point is called a node, or (often) a nodal source.
An example of a node occurs in Example 2 of Section 7.5, and its trajectories are shown in Figure 7.5.4.
CASE 2 Real Eigenvalues of Opposite Sign. The general solution of Eq. (2) is
x = c1?(1)er1t + c2?(2)er2t, (7)
where r1 > 0 and r2 < 0. Suppose that the eigenvectors ?(1) and ?(2) are as shown in Figure 9.1.2a. If the solution starts at an initial point on the line through ?(1), then it follows that c2 = 0. Consequently, the solution remains on the line through ?(1) for all t, and since r1 > 0, ||x|| ^ ro as t ^ ro .If the solution starts at an initial point on the line through ?(2), then the situation is similar except that ||x|| ^ 0 as t ^ ro because r2 < 0. Solutions starting at other initial points follow trajectories such as those shown in Figure 9.1.2a. The positive exponential is the dominant term in Eq. (7) for large t, so eventually all these solutions approach infinity asymptotic to the line determined
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