LECTURE 24: EIGENVALUE METHOD October 31, 2014 y(t) and ~

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LECTURE 24: EIGENVALUE METHOD
MINGFENG ZHAO
October 31, 2014
Theorem 1. Let A be a 2 × 2 matrix, ~y (t) and ~z(t) be two linearly independent solutions to ~x0 = A~x, then the general
solution to ~x0 = A~x is
~x(t) = C1 ~y (t) + C2 ~z(t).
Theorem 2 (Eigenvalue Method). Let λ be an eigenvalue of A and ~v be an eigenvector corresponding to λ, then
~x(t) = eλt~v is a solution to ~x0 = A~x.
For find eigenvalues of A, we need to solve the characteristic polynomial of A, that is:


a b
 = (λ − a)(λ − d) − bc = λ2 − (a + d)λ + (ad − bc) = 0.
det (λI2 − A) = det 
c d
Decoupled system

Definition 1. Let A = 
a
b
c
d

, we say ~x0 = A~x + f~(t) is a decoupled system if bc = 0.
For a decoupled system, we can solve x1 and x2 separately. For example, if b = 0, then we have
(1)
x01
= ax1 + f1 (t)
(2)
x02
= cx1 + dx2 + f2 (t).
For the first equation (1), it’s a first order linear equation of x1 , by using the method of variation of parameters,
we can solve x1 from (1). Then plug x1 (t) into (2), we get a first order linear equation of x2 , by using the method of
variation of parameters, we can solve x2 .

 x0 = x1 ,
1
.
Example 1. Solve the system
 x0 = x − x .
1
2
2
1
2
MINGFENG ZHAO
C1 t
Since x01 = x1 , then x1 (t) = C1 et . Since x02 = x1 − x2 = −x2 + C1 et , solve x2 , we get x2 =
e + C2 e−t . Then the
2

 x0 = x1 ,
1
solution to the system
is:
 x0 = x − x .
1
2
2

~x(t) = 

C 1 et
C1 t
2 e
+ C2 e
−t
.
Eigenvalue method with distinct real eigenvalues
Theorem 3. Let A be a 2 × 2 matrix with two distinct real eigenvalues λ1 and λ2 , and ~v1 and ~v2 are eigenvectors
corresponding to λ1 and λ2 , respectively, then the general solution to ~x0 = A~x is:
~x = C1 eλ1 t~v1 + C2 eλ2 t~v2 .
Example 2. Find the general solution to the system:

Let ~x(t) = 
x1 (t)


, A = 
x2 (t)
2
1
0
1
x01
=
2x1 + x2
x02
=
x2 .

, then we need to solve
~x0 = A~x.

2 1
 and corresponding eigenvectors.
First, we need to find the eigenvalues of A = 
0 1





2 1
λ−2
−1
 = det 
 = (λ − 2)(λ − 1) = 0, then
Since det λI2 − 
0 1
0
λ−1

λ1 = 1,
and λ2 = 2.
I. When λ1 = 1, let’s solve

2
1
0
1
1
1


x1



=
x2
x1

.
x2
That is,



x1

0
0
x2


=
0
0

.
LECTURE 24: EIGENVALUE METHOD
3
Then we know that

x1



 = x1 
x2

So 

1
.
−1

1
 is an eigenvector corresponding to λ1 = 1.
−1
II. When λ1 = 2, let’s solve





2 1
x1
x1


 = 2
.
0 1
x2
x2
That is,


0
1
0
−1

x1



0
=
x2

.
0
Then we know that

x1



 = x1 
x2

So 
1

1
.
0

 is an eigenvector corresponding to λ1 = 2.
0

In summary, the eigenvalues and corresponding eigenvectors of 

λ1 = 1 with 
1
−1
2
1
0
1


 are:

 and λ2 = 2 with 
1

.
0
Therefore, the general solution to ~x0 = A~x is:

x1 (t)

x2 (t)


 = C 1 et 
1
−1


 + C2 e2t 
1


=
0
C1 et + C2 e2t
t
−C1 e

.
Eigenvalue method with distinct complex eigenvalues
Theorem 4. Let A be a 2 × 2 matrix with two distinct complex eigenvalues λ1 and λ2 , and ~v1 and ~v2 are eigenvectors
corresponding to λ1 and λ2 , respectively, then λ2 = λ1 , ~v2 = ~v1 , and the general solution to ~x0 = A~x is:
~x = C1 Re eλ1 t~v1 + C2 Im eλ1 t~v1 .
4
MINGFENG ZHAO
Recall the Euler’s identity:
ea+ib = ea cos(b) + iea sin(b).

1
1
−1
1
λ−1
−1

1
λ−1
Example 3. Find the general solution to the system ~x0 = 

First, let’s find the eigenvalues of A = 
1
1
−1
1

 ~x.

, then

det (λI2 − A) = det 
 = (λ − 1)2 + 1 = 0.
Then
and λ2 = 1 − i.
λ1 = 1 + i,
For λ = 1 + i, let’s solve A~v = λ~v , that is,

1

−1

1
x1

1


x1
 = (1 + i) 
x2

.
x2
So we get

−1
i


x1

1
i


0
=
x2

.
0
So we have

x1



 = x1 
x2

That is, 
1
1

.
i


 is an eigenvalue corresponding to 1 + i, then e(1+i)t 
i
identity, we have
1

 is a solution to ~x0 = A~x. By Euler’s
i

e(1+i)t 
1
i


=
e(1+i)t
ie
(1+i)t


=

et cos(t) + iet sin(t)
t
t

=
i [e cos(t) + ie sin(t)]
et cos(t)
t
−e sin(t)
Therefore, the general solution to ~x0 = A~x is:

~x(t) = C1 
et cos(t)
t
−e sin(t)


 + C2 
et sin(t)
t
e cos(t)

.


 + i
et sin(t)
t
e cos(t)

.
LECTURE 24: EIGENVALUE METHOD
5
Department of Mathematics, The University of British Columbia, Room 121, 1984 Mathematics Road, Vancouver, B.C.
Canada V6T 1Z2
E-mail address: [email protected]
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