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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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The case n = 2 is particularly important and lends itself to visualization in the x1 x2-plane, called the phase plane. By evaluating Ax at a large number of points and plotting the resulting vectors one obtains a direction field of tangent vectors to solutions of the system of differential equations. A qualitative understanding of the behavior of solutions can usually be gained from a direction field. More precise information results from including in the plot some solution curves, or trajectories. A plot that shows a representative sample of trajectories for a given system is called a phase portrait. Examples of direction fields and phase portraits occur later in this section.
To construct the general solution of the system (1) we proceed by analogy with the treatment of second order linear equations in Section 3.1. Thus we seek solutions of Eq. (1) ofthe form
x = ?ert, (3)
where the exponent r and the constant vector ? are to be determined. Substituting from Eq. (3) for x in the system (1) gives
r?ert = A?ert.
Upon canceling the nonzero scalar factor ert we obtain A? = r?, or
(A - rI)? = 0, (4)
where I is the n x n identity matrix. Thus, to solve the system of differential equations (1), we must solve the system of algebraic equations (4). This latter problem is precisely the one that determines the eigenvalues and eigenvectors of the matrix A. Therefore the vector x given by Eq. (3) is a solution of Eq. (1) provided that r is an eigenvalue and ? an associated eigenvector of the coefficient matrix A.
The following two examples illustrate the solution procedure in the case of 2 x 2 coefficient matrices. We also show how to construct the corresponding phase portraits. Later in the section we return to a further discussion of the general n x n system.
374
Chapter 7. Systems ofFirst Order Linear Equations
EXAMPLE
1
Consider the system
x
1 1 4 1
(5)
Plot a direction field and determine the qualitative behavior of solutions. Then find the general solution and draw several trajectories.
A direction field for this system is shown in Figure 7.5.1. From this figure it is easy to see that a typical solution departs from the neighborhood of the origin and ultimately has a slope of approximately 2 in either the first or third quadrant.
I \ I V
I I ³ \ ³ I ³ ² ³ I ² ³ I
² ³ I
-
Ø /. /
\ ×-» \ \ -V \
\ \ 4 \ \ \
I ³ \
I t v / ³ \
/ I I
t ³ I
2
1 -

/ / / / -/ / / / / / / / / / / / / / / /
/ / / / / ////
/ / / /
/ / / /
/ / / /
/ / / /
/ / / t t / / t t / / ³ t / / f
/ / / / / / / / / / / / / / / / / /
-1 /
/
/
/
/ / / / / / / / / / / /
/ /
/ -1 / /
/ /
/ / . -2 / ? ? ^
{ I "• \
- V ^ N
1
t I f t t 1 t \ t \
? -V •
N N \ \
¦ V \ \ \ \ \ \ \
FIGURE 7.5.1 Direction field for the system (5).
To find solutions explicitly we assume that x = gert, and substitute for x in Eq. (5). We are led to the system of algebraic equations
1 — r 1 4 1 - r
=
(6)
Equations (6) have a nontrivial solution if and only if the determinant of coefficients is zero. Thus allowable values of r are found from the equation
1 — r 1 4 1 - r
= (1 — r )2 — 4
r2 2r 3 0.
(7)
Equation (7) has the roots rl = 3 and r2 = — 1; these are the eigenvalues of the coefficient matrix in Eq. (5). If r = 3, then the system (6) reduces to the single equation
— 2%1 + %2 = 0
Thus %2 = 2?j and the eigenvector corresponding to rl = 3 can be taken as
g(1) =
(8)
(9)
2
7.5 Homogeneous Linear Systems with Constant Coefficients
375
Similarly, corresponding to r2 = — 1, we find that ?2 = — 2f1, so the eigenvector is
?(2) = ( j) • (10)
—2
The corresponding solutions of the differential equation are
x'"(t) = (2) e3', x«'(() = (e-'• (11)
The Wronskian of these solutions is
W[x(1), x(2)](t) =
= —4e2t, (12)
which is never zero. Hence the solutions x(1) and x(2) form a fundamental set, and the general solution of the system (5) is
x = c1x(1)(t) + c2x(2)(t)
= q (Ã)e3t+ci( ^e—t, (13)
^2/ —2f
where c1 and c2 are arbitrary constants.
To visualize the solution (13) it is helpful to consider its graph in the x1 x2-plane for various values of the constants c1 and c2. We start with x = c1x(1)(t), or in scalar form
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