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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 Previous << 1 .. 538 539 540 541 542 543 < 544 > 545 546 547 548 549 550 .. 609 >> Next — A2 — (tr A)A + det A
— 0
The eigenvalues of A are the roots A i and A2 of this quadratic equation. We assume A ł — A2. For the eigenvalue A i we can find a corresponding eigenvector
v(i) by solving the vector equation
Av(i) — Aiv(i)
for v(i). In a similar fashion we can find an eigenvector v(2) corresponding to the eigenvalue A2.
Example: Take a look at the system
x' — Ax, A —
" 0 i Xi
1 , x --- _x2_
2
1
1
(i8)
Since
tr A — 0 + 3 — 3 and det A — 0 ¦ 3 — i ¦ (—2) — 2 the characteristic equation is
A2 — (tr A)A + det A — A2 — 3A + 2 — 0
The eigenvalues are Ai — i and A2 — 2. To find an eigenvector v(i) for Ai, let’s solve
0 i
-2 3
— Aiv(i) — v(i)
for v(i). Denoting the components of v(i) by a and â, we have
01 a â a
1 â ---2a + 3â â
3
2
---
1
This gives two equations for a and â:
â — a, —2a + 3â — â
Phase Portraits
99
4
2
0
-2
Figure 6.1: Graphs of five solutions X (t) (left), X2(t) (right) of system (18).
t t
The second equation is equivalent to the first, so we may as well set ŕ = â = 1, which gives us an eigenvector v(1). In a similar way for the eigenvalue X2, we can find an eigenvector v(2) with components ŕ = 1, â = 2. So the general solution of x; = Ax in this case is
x = C1v(1)elit + C2v(2)e^2t
= Ci
et + C2
e2t
or in component form
xi = Ci et + C2e2t x2 = Ci et + 2C2e2t
where Ci and C2 are arbitrary constants.
? Find a formula for the solution of system (18) if xi(0) = 1, x2 (0) = —1. Figure 6.1 shows graphs of x1 (t) and x2(t) where x1 (0) = 1, x2(0) = 0, ±0.5, ±1. Which graphs correspond to x1(0) = 1, x2(0) = —1? What happens as
t ^ +c? As t ^ —c?
¦ Phase Portraits
We can view solutions graphically in several ways. For example, we can draw plots of x1 (t) vs. t, x2(t) vs. t, and so on. These plots are called component plots (see Figure 6.1). Alternatively, we can interpret equations (12) as a set of parametric equations with t as the parameter. Then each specific value of t corresponds to a set of values for x1,... , xn. We can view this set of val-Another term for phase ues as coordinates of a point in x1 x2 ¦ xn-space, called the phase space. (If
space is state space. n = 2 it’s called the phase plane.) For an interval of t-values, the correspond-
ing points form a curve in phase space. This curve is called a phase plot, a trajectory, or an orbit.
100
Chapter 6
Trajectories starting on either line at t = 0 stay on the line.
Eigenvalues of opposite signs imply a saddle.
Phase plots are particularly useful if n = 2. In this case it is often worthwhile to draw several trajectories starting at different initial points on the same set of axes. This produces a phase portrait, which gives us the best possible overall view of the behavior of solutions. Whatever the value of n, the trajectories of system (10) can never intersect because system (10) is autonomous.
If A in system (10) is a 2 x 2 matrix, then it is useful to examine and classify the various cases that can arise. There aren’t many cases when n = 2, but even so these cases give important information about higher-dimensional linear systems, as well as nonlinear systems (see Chapter 7). We won’t consider here the cases where the two eigenvalues are equal, or where one or both of them are zero.
A direction field (or vector field) for an autonomous system when n = 2 is afield of line segments. The slope of the segment at the point (x1, x2) is X2 / x!1. The trajectory through (x1, x2) is tangent to the segment. An arrowhead on the segment shows the direction of the flow. See Figures 6.2-6.5 for examples.
Real Eigenvalues
If the eigenvalues k 1 and k2 are real, the general solution is
x = Civ(1)ek11 + C2V(2) ek2t (19)
where C1 and C2 are arbitrary real constants.
Let’s first look at the case where k1 and k2 have opposite signs, with k1 > 0 and k2 < 0. The term in formula (19) involving k1 dominates as t ^ +tx>, and the term involving k2 dominates as t ^ —ţ. Thus as t ^ +» the trajectories approach the line that goes through the origin and has the same slope as v(1), and as t ^ —ńţ, they approach the line that goes through the origin and has the same slope as v(2). A typical phase portrait for this case is shown in Figure 6.2. The origin is called a saddle point, and it is unstable, since most solutions move away from the point. Previous << 1 .. 538 539 540 541 542 543 < 544 > 545 546 547 548 549 550 .. 609 >> Next 