Transcript Math 260

Ch 7.5: Homogeneous Linear Systems with
Constant Coefficients
We consider here a homogeneous system of n first order linear
equations with constant, real coefficients:
x1  a11 x1  a12 x2    a1n xn
x2  a21 x1  a22 x2    a2 n xn

xn  an1 x1  an 2 x2    ann xn
This system can be written as x' = Ax, where
 x1 (t ) 
 a11



 x2 (t ) 
 a21
x(t )  
, A






 x (t ) 
a
m


 n1
a12
a22

an 2
 a1n 

 a2 n 
  

 ann 
Solving Homogeneous System
To construct a general solution to x' = Ax, assume a solution
of the form x = ert, where the exponent r and the constant
vector  are to be determined.
Substituting x = ert into x' = Ax, we obtain
rξert  Aξert
 rξ  Aξ 
A  rIξ  0
Thus to solve the homogeneous system of differential
equations x' = Ax, we must find the eigenvalues and
eigenvectors of A.
Therefore x = ert is a solution of x' = Ax provided that r is
an eigenvalue and  is an eigenvector of the coefficient
matrix A.
Example 1: Direction Field
(1 of 9)
Consider the homogeneous equation x' = Ax below.
 1 1

x
x  
 4 1
A direction field for this system is given below.
Substituting x = ert in for x, and rewriting system as
(A-rI) = 0, we obtain
1  1   0 
1  r

    
 4 1  r  1   0 
Example 1: Eigenvalues (2 of 9)
Our solution has the form x = ert, where r and  are found
by solving
1  1   0 
1  r

    
 4 1  r  1   0 
Recalling that this is an eigenvalue problem, we determine r
by solving det(A-rI) = 0:
1 r
1
 (1  r ) 2  4  r 2  2r  3  (r  3)(r  1)
4 1 r
Thus r1 = 3 and r2 = -1.
Example 1: First Eigenvector (3 of 9)
Eigenvector for r1 = 3: Solve
1 1   0 
1  3
     
A  rI ξ  0  
 4 1  3   2   0 
1 1   0 
 2

    
 4  2   2   0 
by row reducing the augmented matrix:
1 0   1 1/ 2
 2

  
2
 4  2 0  4
1 / 2 2 
1 / 2 
(1)
  c 
, c
 ξ  
 1 
 2 
1  1 / 2 2
0  1 1/ 2 0
  
  1
0 2
0  0
0 0
 1
(1)
arbitrary choose ξ   
 2
0
0
Example 1: Second Eigenvector (4 of 9)
Eigenvector for r2 = -1: Solve
1 1   0 
1  1
     
A  rI ξ  0  
 4 1  1  2   0 
 2 1 1   0 

    
 4 2   2   0 
by row reducing the augmented matrix:
1  1 / 2 2  0
 2 1 0  1 1/ 2 0  1 1/ 2 0

  
  
  1
0 2  0
2 0 0
0 0
 4 2 0  4
  1 / 2 2 
 1/ 2 
 1
( 2)
( 2)
  c 
, c arbitrary choose ξ   
 ξ  
2   1 
  2

Example 1: General Solution (5 of 9)
The corresponding solutions x = ert of x' = Ax are
 1 3t ( 2)
 1 t
x (t )   e , x (t )   e
 2
  2
(1)
The Wronskian of these two solutions is

W x (1) , x ( 2)

e 3t
(t )  3t
2e
e t
 2t


4
e
0
t
 2e
Thus x(1) and x(2) are fundamental solutions, and the general
solution of x' = Ax is
x(t )  c1x (1) (t )  c2 x ( 2 ) (t )
 1 3 t
 1 t
 c1  e  c2  e
 2
  2
Example 2:
(1 of 9)
Consider the homogeneous equation x' = Ax below.
3
2
x
x  

 2  2
Substituting x = ert in for x, and rewriting system as
(A-rI) = 0, we obtain
3 r

 2

2  1   0 
    
 2  r  1   0 
Example 2: Eigenvalues (2 of 9)
Our solution has the form x = ert, where r and  are found
by solving
3 r

 2

2  1   0 
    
 2  r  1   0 
Recalling that this is an eigenvalue problem, we determine r
by solving det(A-rI) = 0:
3 r
2
2
2r
 (3  r )(2  r )  2  r 2  5r  4  (r  1)(r  4)
Thus r1 = -1 and r2 = -4.
Example 2: First Eigenvector (3 of 9)
Eigenvector for r1 = -1: Solve
  3 1
2  1   0 

     
A  rI ξ  0  
 
2

2

1

 2   0 
2

 2

2  1   0 
    
 1  2   0 
by row reducing the augmented matrix:
2

 2

 ξ (1)
2 0  1  2 / 2 0  1  2 / 2 0







1 0   2
1 0   0
0 0 
 2 / 2 2 
 1
(1)



 choose ξ   

2 
 2

Example 2: Second Eigenvector (4 of 9)
Eigenvector for r2 = -4: Solve
3 4
2  1   0 
     
A  rI ξ  0  
2  2  4   2   0 

 1

 2

by row reducing the augmented matrix:
 1

 2

2 0  1

2 0   0
 choose ξ
( 2)
 2 

 

1


  2 2 
2 0

  ξ ( 2)  




0 0
2

2  1   0 
    
2   2   0 
Example 2: General Solution (5 of 9)
The corresponding solutions x = ert of x' = Ax are
  2   4t
 1 t ( 2)
e
x (t )   e , x (t )  

1
 2


(1)
The Wronskian of these two solutions is


W x (1) , x ( 2 ) (t ) 
e t
 2e 4t
2e  t
e  4t
 3e 5t  0
Thus x(1) and x(2) are fundamental solutions, and the general
solution of x' = Ax is
x(t )  c1x (1) (t )  c2 x ( 2) (t )
  2   4t
 1 t
e
 c1  e  c2 

1
 2


Eigenvalues, Eigenvectors
and Fundamental Solutions
In general, for an n x n real linear system x' = Ax:
All eigenvalues are real and different from each other.
Some eigenvalues occur in complex conjugate pairs.
Some eigenvalues are repeated.
If eigenvalues r1,…, rn are real & different, then there are n
corresponding linearly independent eigenvectors (1),…, (n).
The associated solutions of x' = Ax are
x(1) (t )  ξ (1)er1t ,, x( n) (t )  ξ ( n)ernt
Using Wronskian, it can be shown that these solutions are
linearly independent, and hence form a fundamental set of
solutions. Thus general solution is
x  c1ξ (1)er1t   cnξ ( n)ernt
Hermitian Case: Eigenvalues, Eigenvectors &
Fundamental Solutions
If A is an n x n Hermitian matrix (real and symmetric), then
all eigenvalues r1,…, rn are real, although some may repeat.
In any case, there are n corresponding linearly independent
and orthogonal eigenvectors (1),…, (n). The associated
solutions of x' = Ax are
x(1) (t )  ξ (1)er1t ,, x( n) (t )  ξ ( n)ernt
and form a fundamental set of solutions.
Example 3: Hermitian Matrix
(1 of 3)
Consider the homogeneous equation x' = Ax below.
0 1 1


x   1 0 1  x
1 1 0


The eigenvalues were found previously in Ch 7.3, and were:
r1 = 2, r2 = -1 and r3 = -1.
Corresponding eigenvectors:
ξ (1)
 1
 1
 0
  ( 2 )   ( 3)  
 1, ξ   0  , ξ   1
 1
  1
  1
 
 
 
Example 3: General Solution (2 of 3)
The fundamental solutions are
x (1)
 1
 1
 0
  2 t ( 2 )    t ( 3)    t
  1 e , x   0  e , x   1 e
 1
  1
  1
 
 
 
with general solution
1
 1
 0
  2t
  t
  t
x  c1 1e  c2  0 e  c3  1e
1
  1
  1
 
 
 