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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
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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 = + i, r2 = i, 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
s' s / S S s'
/ / / s
/ / / / -------1---
-2
\ \
. . \ \ J \ \ \

t t t 1 \
1 \ \ N \ \ \ \ \ \ \ \ \ \ \ \ \ N \ \ X
Uk-
\ \ X -
w-t \ xx-
\ --\ N V' N X -2 X X-v" \NN'
nVillhii.
/////;// >
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