# Elementary Differential Equations and Boundary Value Problems - Boyce W.E.

**Download**(direct link)

**:**

**210**> 211 212 213 214 215 216 .. 609 >> Next

where the coefficient matrix A is real-valued. If we seek solutions of the form x = ?ert, then it follows as in Section 7.5 that r must be an eigenvalue and ? a corresponding eigenvector of the coefficient matrix A. Recall that the eigenvalues r1,..., rn of A are the roots of the equation

det(A — rI) = 0, (2)

and that the corresponding eigenvectors satisfy

(A — r I)? = 0. (3)

If A is real, then the coefficients in the polynomial equation (2) for r are real, and any complex eigenvalues must occur in conjugate pairs. For example, if rj = X + ii, where X and i are real, is an eigenvalue of A, then so is r2 = X — ii. Further, the corresponding eigenvectors ?(1) and ?(2) are also complex conjugates. To see that this is so, suppose that rj and ?(1) satisfy

(A — rjI)?(1) = 0. (4)

On taking the complex conjugate of this equation, and noting that A and I are realvalued, we obtain

(A — r jI)?(1) = 0, (5)

where r1 and ?(1) are the complex conjugates of rj and ?(1), respectively. In other

words, r2 = r 1 is also an eigenvalue, and ?(2) = ?(1) is a corresponding eigenvector.

The corresponding solutions

x(1)(t) = ? (1)er1t, x(2)(t) = ? (1)er1t (6)

of the differential equation (1) are then complex conjugates of each other. Therefore, as in Section 3.4, we can find two real-valued solutions ofEq. (1) corresponding to the eigenvalues rj and r2 by taking the real and imaginary parts of x(1) (t) or x(2)(t) given by Eq. (6). 1 ( ) 2

Let us write ?(1) = a + ib, where a and b are real; then we have

x(1)(t) = (a + i b) e(X+i i)t

= (a + ib)eXt(cos lit + i sin fit). (7)

Upon separating x(1) (t) into its real and imaginary parts, we obtain

x(1)(t) = eXt (a cos it — b sin it) + ieXt (a sin it + b cos it). (8)

If we write x(1)(t) = u(t) + i v(t), then the vectors

u(t) = eXt(acos it — b sin it),

Xt (9)

v(t) = e (a sin it + b cos it)

7.6 Complex Eigenvalues

385

are real-valued solutions of Eq. (1). It is possible to show that u and v are linearly independent solutions (see Problem 27).

For example, suppose that r1 = ê + iã, r2 = ê — iã, and that r3,, rn are all real and distinct. Let the corresponding eigenvectors be g(1) = a + ib, g(2) = a — ib, g(3),..., g(n). Then the general solution of Eq. (1) is

x = qu(0 + c2v(t) + c3g {3)er3( + ... + cng (n)ernl, (10)

where u(t) and v(t) are given by Eqs. (9). We emphasize that this analysis applies only if the coefficient matrix A in Eq. (1) is real, for it is only then that complex eigenvalues and eigenvectors must occur in conjugate pairs.

The following examples illustrate the case n = 2, both to simplify the algebraic calculations and to permit easy visualization of the solutions in the phase plane.

Find a fundamental set of real-valued solutions of the system

EXAMPLE

x

1

-1 -1

2/

(11)

and display them graphically.

A direction field for the system (11) is shown in Figure 7.6.1. This plot suggests that the trajectories in the phase plane spiral clockwise toward the origin.

\ \ \ \ \ \

SNW \ \ \ \ \ \

- «4 4 \ \ \ \ \ \ \

XX W \ \ \ \ \ '

.^X \ \ \ \ \ \ '

XX \ \ \ \ \ \

~ 4 \ \ \ I I ,

X \ \ \ I I

Ó Ó s' s’ / S S s'

/ / / s

/ / / / -------1---

-2

\ \

. . \ \ J \ \ \

²Ø

t t t 1 \

1 \ \ N \ \ \ \ \ \ \ \ \ \ \ \ \ N \ \ X

Uk-

\ \ X -

w-t \ xx-

\ Õ-×-\ N V' N X -2 X X-v" \NN'

nVillhii.

/////;// >

**210**> 211 212 213 214 215 216 .. 609 >> Next