Boyce/DiPrima 9th ed, Ch 7.6:

Complex EigenvaluesElementary Differential Equations and Boundary Value Problems, 9th edition, by William E. Boyce and Richard C. DiPrima, ©2009 by John Wiley & Sons, Inc.

We consider again a homogeneous system of n first order

linear equations with constant, real coefficients,

22221212

12121111

nn

nn

xaxaxax

xaxaxax

+++=′

+++=′

⋮

…

…

and thus the system can be written as x' = Ax, where

,2211 nnnnnn xaxaxax +++=′ …

⋮

=

=

nnnn

n

n

n aaa

aaa

aaa

tx

tx

tx

t

⋯

⋮⋱⋮⋮

⋯

⋯

⋮

21

22221

11211

2

1

,

)(

)(

)(

)( Ax

Conjugate Eigenvalues and Eigenvectors

We know that x = ξξξξert is a solution of x' = Ax, provided r is

an eigenvalue and ξξξξ is an eigenvector of A.

The eigenvalues r1,…, rn are the roots of det(A-rI) = 0, and

the corresponding eigenvectors satisfy (A-rI)ξξξξ = 0.

If A is real, then the coefficients in the polynomial equation If A is real, then the coefficients in the polynomial equation

det(A-rI) = 0 are real, and hence any complex eigenvalues

must occur in conjugate pairs. Thus if r1 = λ + iµ is an

eigenvalue, then so is r2 = λ - iµ.

The corresponding eigenvectors ξξξξ(1), ξξξξ(2) are conjugates also.

To see this, recall A and I have real entries, and hence

( ) ( ) ( ) 0ξIA0ξIA0ξIA =−⇒=−⇒=− )2(

2

)1(

1

)1(

1 rrr

Conjugate Solutions

It follows from the previous slide that the solutions

corresponding to these eigenvalues and eigenvectors are

conjugates conjugates as well, since

trtree 21 )2()2()1()1( , ξxξx ==

conjugates conjugates as well, since

)1()1()2()2( 22 xξξx === trtree

Example 1: Direction Field (1 of 7)

Consider the homogeneous equation x' = Ax below.

A direction field for this system is given below.

xx

−−

−=′

2/11

12/1

A direction field for this system is given below.

Substituting x = ξξξξert in for x, and rewriting system as

(A-rI)ξξξξ = 0, we obtain

=

−−−

−−

0

0

2/11

12/1

1

1

ξ

ξ

r

r

Example 1: Complex Eigenvalues (2 of 7)

We determine r by solving det(A-rI) = 0. Now

( )4

512/1

2/11

12/122 ++=++=

−−−

−−rrr

r

r

Thus

Therefore the eigenvalues are r1 = -1/2 + i and r2 = -1/2 - i.

ii

r ±−=±−

=−±−

=2

1

2

21

2

)4/5(411 2

Example 1: First Eigenvector (3 of 7)

Eigenvector for r1 = -1/2 + i: Solve

( )

=

⇔

=

−

⇔

=

−−−

−−⇔=−

0101

0

0

2/11

12/1

11

1

1

ξξ

ξ

ξ

ii

r

rr 0ξIA

by row reducing the augmented matrix:

Thus

=

−−

⇔

=

−−

⇔0101 2

1

2

1

ξξ ii

=→

−=→

→

−− i

ii

i

i 1choose

000

01

01

01)1(

2

2)1( ξξξ

ξ

+

=

1

0

0

1)1(

iξ

Example 1: Second Eigenvector (4 of 7)

Eigenvector for r1 = -1/2 - i: Solve

( )

=

−

⇔

=

⇔

=

−−−

−−⇔=−

0101

0

0

2/11

12/1

11

1

1

ξξ

ξ

ξ

ii

r

rr 0ξIA

by row reducing the augmented matrix:

Thus

=

−

−⇔

=

−⇔

0

0

1

1

0

0

1

1

2

1

2

1

ξ

ξ

ξ

ξ

i

i

i

i

−=→

=→

−→

−

−

i

ii

i

i 1choose

000

01

01

01)2(

2

2)2( ξξξ

ξ

−+

=

1

0

0

1)2(

iξ

Example 1: General Solution (5 of 7)

The corresponding solutions x = ξξξξert of x' = Ax are

=

+

=

−=

−

=

−−

−−

tettet

t

tettet

tt

tt

sincos

0sin

1)(

sin

cossin

1

0cos

0

1)(

2/2/

2/2/

v

u

The Wronskian of these two solutions is

Thus u(t) and v(t) are real-valued fundamental solutions of

x' = Ax, with general solution x = c1u + c2v.

=

+

= −−

t

tettet

tt

cos

sincos

1

0sin

0

1)( 2/2/v

[ ] 0cossin

sincos)(,

2/2/

2/2/

)2()1( ≠=−

= −

−−

−−t

tt

tt

etete

tetetW xx

Example 1: Phase Plane (6 of 7)

Given below is the phase plane plot for solutions x, with

Each solution trajectory approaches origin along a spiral path

as t → ∞, since coordinates are products of decaying

+

−=

=

−

−

−

−

te

tec

te

tec

x

x

t

t

t

t

cos

sin

sin

cos2/

2/

22/

2/

1

2

1x

as t → ∞, since coordinates are products of decaying

exponential and sine or cosine factors.

The graph of u passes through (1,0),

since u(0) = (1,0). Similarly, the

graph of v passes through (0,1).

The origin is a spiral point, and

is asymptotically stable.

Example 1: Time Plots (7 of 7)

The general solution is x = c1u + c2v:

As an alternative to phase plane plots, we can graph x1 or x2

as a function of t. A few plots of x are given below, each

+−

+=

=

−−

−−

tectec

tectec

tx

tx

tt

tt

cossin

sincos

)(

)(2/

2

2/

1

2/

2

2/

1

2

1x

1 2

as a function of t. A few plots of x1 are given below, each

one a decaying oscillation as t → ∞.

General Solution

To summarize, suppose r1 = λ + iµ, r2 = λ - iµ, and that

r3,…, rn are all real and distinct eigenvalues of A. Let the

corresponding eigenvectors be

Then the general solution of x' = Ax is

)()4()3()2()1( ,,,,, nii ξξξbaξbaξ …−=+=

Then the general solution of x' = Ax is

where

trn

n

tr necectctc)()3(

3213)()( ξξvux ++++= …

( ) ( )ttetttettt µµµµ λλ cossin)(,sincos)( bavbau +=−=

Real-Valued Solutions

Thus for complex conjugate eigenvalues r1 and r2 , the

corresponding solutions x(1) and x(2) are conjugates also.

To obtain real-valued solutions, use real and imaginary parts

of either x(1) or x(2). To see this, let ξξξξ(1) = a + ib. Then

where

are real valued solutions of x' = Ax, and can be shown to be

linearly independent.

( ) ( ) ( )( ) ( )

)()(

cossinsincos

sincos)1()1(

tit

ttiette

titeie

tt

tti

vu

baba

baξx

+=

++−=

++== +

µµµµ

µµλλ

λµλ

( ) ( ),cossin)(,sincos)( ttetttettt µµµµ λλ bavbau +=−=

Spiral Points, Centers,

Eigenvalues, and Trajectories

In previous example, general solution was

The origin was a spiral point, and was asymptotically stable.

+

−=

=

−

−

−

−

te

tec

te

tec

x

x

t

t

t

t

cos

sin

sin

cos2/

2/

22/

2/

1

2

1x

If real part of complex eigenvalues is positive, then

trajectories spiral away, unbounded, from origin, and hence

origin would be an unstable spiral point.

If real part of complex eigenvalues is zero, then trajectories

circle origin, neither approaching nor departing. Then origin

is called a center and is stable, but not asymptotically stable.

Trajectories periodic in time.

The direction of trajectory motion depends on entries in A.

Example 2:

Second Order System with Parameter (1 of 2)

The system x' = Ax below contains a parameter α.

Substituting x = ξξξξert in for x and rewriting system as

xx

−=′

02

2α

(A-rI)ξξξξ = 0, we obtain

Next, solve for r in terms of α :

=

−−

−

0

0

2

2

1

1

ξ

ξα

r

r

2

1644)(

2

2 22 −±

=⇒+−=+−=−−

− αααα

αrrrrr

r

r

Example 2:

Eigenvalue Analysis (2 of 2)

The eigenvalues are given by the quadratic formula above.

For α < -4, both eigenvalues are real and negative, and hence

origin is asymptotically stable node.

For α > 4, both eigenvalues are real and positive, and hence the

origin is an unstable node.

2

162 −±=

ααr

origin is an unstable node.

For -4 < α < 0, eigenvalues are complex with a negative real

part, and hence origin is asymptotically stable spiral point.

For 0 < α < 4, eigenvalues are complex with a positive real

part, and the origin is an unstable spiral point.

For α = 0, eigenvalues are purely imaginary, origin is a center.

Trajectories closed curves about origin & periodic.

For α = ± 4, eigenvalues real & equal, origin is a node (Ch 7.8)

Second Order Solution Behavior and

Eigenvalues: Three Main Cases

For second order systems, the three main cases are:

Eigenvalues are real and have opposite signs; x = 0 is a saddle point.

Eigenvalues are real, distinct and have same sign; x = 0 is a node.

Eigenvalues are complex with nonzero real part; x = 0 a spiral point.

Other possibilities exist and occur as transitions between two Other possibilities exist and occur as transitions between two

of the cases listed above:

A zero eigenvalue occurs during transition between saddle point and

node. Real and equal eigenvalues occur during transition between

nodes and spiral points. Purely imaginary eigenvalues occur during a

transition between asymptotically stable and unstable spiral points.

a

acbbr

2

42 −±−=

Example 3: Multiple Spring-Mass System (1 of 6)

The equations for the system of two masses and three

springs discussed in Section 7.1, assuming no external

forces, can be expressed as:

)( and)( 232122

2

2

2221212

1

2

1 xkkxkdt

xdmxkxkk

dt

xdm +−=++−=

Given , , the

equations become

' and,',, where

)(' and)('or

)( and)(

24132211

23212422212131

23212222212121

xyxyxyxy

ykkykymykykkym

xkkxkdt

mxkxkkdt

m

====

+−=++−=

+−=++−=

4/15 and,3,1,4/9,2 32121 ===== kkkmm

2142134231 33/4' and,2/32',',' yyyyyyyyyy −=+−===

Example 3: Multiple Spring-Mass System (2 of 6)

Writing the system of equations in matrix form:

AyAyAyAyyyyyy'y'y'y' =

−

−=

0033/4

002/32

1000

0100

2142134231 33/4' and,2/32',',' yyyyyyyyyy −=+−===

Assuming a solution of the form y = ξξξξert , where r must be

an eigenvalue of the matrix A and ξξξξ is the corresponding

eigenvector, the characteristic polynomial of A is

yielding the eigenvalues:

− 0033/4

)4)(1(45 2224 ++=++ rrrr

iriririr 2 and,2,, 4321 −==−==

Example 3: Multiple Spring-Mass System (3 of 6)

For the eigenvalues the correspond-

ing eigenvectors are

AyAyAyAyyyyyy'y'y'y' =

−

−=

0033/4

002/32

1000

0100

iriririr 2 and,2,, 4321 −==−==

−

−=

−

−=

−

−=

=

i

i

i

i

i

i

i

i

8

6

4

3

and,

8

6

4

3

,

2

3

2

3

,

2

3

2

3

)4()3()2()1( ξξξξ

The products yield the complex-valued solutions:

−− iiii 8822

ititee

2)3()1( and ξξ

)()(

2cos8

2cos6

2sin4

2sin3

2sin8

2sin6

2cos4

2cos3

)2sin2(cos

8

6

4

3

)()(

cos2

cos3

sin2

sin3

sin2

sin3

cos2

cos3

)sin(cos

2

3

2

3

)2()2(2)3(

)1()1()1(

tit

t

t

t

t

i

t

t

t

t

tit

i

ie

tit

t

t

t

t

i

t

t

t

t

tit

i

ie

it

it

vu

vu

+=

−

−+

−

−=+

−

−=

+=

+

−

−=+

=

ξ

ξ

Example 3: Multiple Spring-Mass System (4 of 6)

After validating that are linearly

independent, the general solution of the system of equations can be

written as

)(),(,)(),( )2()2()1()1(tttt vuvu

−+

−

−+

+

−=

t

t

t

ct

t

t

ct

t

t

ct

t

t

c2cos6

2sin4

2sin3

2sin6

2cos4

2cos3

cos3

sin2

sin3

sin3

cos2

cos3

4321yyyy

2142134231 33/4' and,2/32',',' yyyyyyyyyy −=+−===

where are arbitrary constants.

Each solution will be periodic with period 2π, so each trajectory is a

closed curve. The first two terms of the solution describe motions with

frequency 1 and period 2π while the second two terms describe

motions with frequency 2 and period π. The motions of the two masses

will be different relative to one another for solutions involving only the

first two terms or the second two terms.

−

− tttt 2cos82sin8cos2sin2

4321 ,,, cccc

Example 3: Multiple Spring-Mass System (5 of 6)

To obtain the fundamental mode of vibration with frequency 1

To obtain the fundamental mode of vibration with frequency 2

’

)0(2)0(3 and)0(2)0(3 when occurs 0 341243 yyyycc ==→==

)0(4)0(3 and)0(4)0(3 when occurs 0 341221 yyyycc −=−=→==

2y←

',' and masses theofmotion therepresent and 141321yyyyyy ==

Plots of and parametric plots (y, y’) are shown for a

selected solution with frequency 1

5 10 15 20t

-3

-2

-1

1

2

3

u H1LHtL

-3 -2 -1 1 2 3y1, y2

-3

-2

-1

1

2

3

y3, y4

),( 31 yy←

),( 42 yy←1y←

2y←

=

=

0

0

2

3

)0(

)0(

)0(

)0(

)0(

4

3

2

1

y

y

y

y

y

Plots of the solutions as functions of time Phase plane plots

21 and yy

Example 3: Multiple Spring-Mass System (6 of 6)

Plots of and parametric plots (y, y’) are shown for a selected

solution with frequency 2

2 4 6 8 10 12t

2

4

uH2L

HtL

↓1y

Plots of the solutions as functions of time

-4 -2 2 4y1, y2

5

10

y3, y4

Phase plane plots

),( 42 yy←

),( 31 yy←

→2y

−=

=

0

0

4

3

)0(

)0(

)0(

)0(

)0(3

2

1

y

y

y

y

yyyy

',' and masses theofmotion therepresent and 141321yyyyyy ==

21 and yy

Plots of and parametric plots (y, y’) are shown for a selected

solution with mixed frequencies satisfying the initial condition stated

-4

-2

2 4 6 8 10 12t

-4

-2

2

4

yHtL

-4 -2 2 4y1, y2

-5

5

y3, y4

),( 42 yy←

),( 31 yy←

1y←→2y

Plots of the solutions as functions of time

Phase plane plots

-5

0)0(4y

−

=

1

1

4

1

)0(yyyy

21 and yy