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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
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