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Lecture 20
Teodor Burghelea
Department of Mathematics, University of British Columbia
*
I.
MORE SOLVED EXAMPLES FOR HOMOGENEOUS LINEAR SYSTEMS WITH CONSTANT
COEFFICIENTS
Example 1[COMPLEX CONJUGATED EIGENVALUES WITH NON ZERO REAL PARTSPIRAL ]
Solve the following 2
×
2 homogeneous linear system:
→
X
0
(
t
)=

1
2
1

1

1
2

→
X
(
t
)
(1)
The ﬁrst step, as usual, get the eigenvalues of the coeﬃcient matrix
A
=

1
2
1

1

1
2
, that is solve the following
algebraic equation:
λ
2

Tr
(
A
)
λ
+
det
(
A
)=0
⇒
λ
2
+
λ
+
5
4
=0
(2)
The roots of the characteristic equation are:
λ
1
=

1
2
+
i, λ
2
=

1
2

i
(3)
Please note the following:
The roots are complex conjugated and have non zero real part. As you are about to see, in this case
the origin [which is always a critical point for a homogeneous linear system
] is called a
spiral point
.
The solution of the linear system has the following structure:
→
X
(
t
)=
C
1
Re
[
→
v
1
e
λ
1
t
]+
C
2
Re
[
→
v
1
e
λ
2
t
]
(4)
where

→
v
1
is the eigenvector associated to the eigenvalue
λ
1
.
In order to display the ﬁnal solution of the exercise, we only have to calculate the eigenvector
→
v
1
, that is to solve the
*
Electronic address:
bteodor@math.ubc.ca;www.math.ubc.ca\
~
bteodor\math_255.html
Page 2
2
following matrix equation:

1
2
1

1

1
2
v
1
x
v
1
y
=(

1
2
+
i
)
v
1
x
v
1
y
(5)
I will illustrate once more how to solve such a linear algebraic system(
in the future I will skip such trivial calculations,
assuming you have all got the point
.)The system above can be written(just calculate the matrix product in the left
hand side and equate it to the right hand side):
v
1
y
=
iv
1
x

v
1
x
=
iv
1
y
The two equations above are actually identical. Indeed, if you multiply by
i
one of them you obtain the second one.
So, the eigenvector is determined only up to a multiplicative constant:
→
v
1
=
v
1
x
1
i
where
v
1
x
is an arbitrary real number. If one takes
v
1
x
=1
,
the eigenvector is
→
v
1
=
1
i
(6)
Finally, the general solution can be written:

→
X
(
t
)=
C
1
e

t
2
cos
(
t
)

sin
(
t
)
+
C
2
e

t
2
sin
(
t
)
cos
(
t
)
(7)
In order to understand the behavior of this solution, note the following points:
1:
lim
t
→∞
→
X
(
t
)=
0
0
simply because
lim
t
→∞
e

t
2
= 0. So, no matter what the initial condition is (that is to say,
regardless to the values of the constants
C
1
,
2
) after long times, the solution approaches the critical point, the origin.
2:
The components
x
(
t
)
,y
(
t
) of the solution vector
→
X
(
t
) are periodic functions, in this case with period 2
π
. So, the
trajectories are actually spinning around the origin and, because of the remark above, they approach the origin as
the time increases.
THIS IS A TYPICAL BEHAVIOR OF A SPIRAL. FOR THE CASE OF COMPLEX CONJUGATED
EIGENVALUES WITH NON ZERO REAL PART, THE ORIGIN (THE CRITICAL POINT) IS
A SPIRAL. IT IS STABLE (IT APPROACHES THE ORIGIN) IF THE REAL PART OF THE
EIGENVALUES IS NEGATIVE AND UNSTABLE IF THE REAL PART IS POSITIVE.
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