Transcript Math 260

Ch 7.7: Fundamental Matrices
Suppose that x(1)(t),…, x(n)(t) form a fundamental set of
solutions for x' = P(t)x on  < t < .
The matrix
 x1(1) (t )  x1( n ) (t ) 


Ψ(t )   

 ,
 x (1) (t )  x ( n ) (t ) 
n
 n

whose columns are x(1)(t),…, x(n)(t), is a fundamental matrix
for the system x' = P(t)x. This matrix is nonsingular since its
columns are linearly independent, and hence det  0.
Note also that since x(1)(t),…, x(n)(t) are solutions of x' = P(t)x,
 satisfies the matrix differential equation ' = P(t).
Example 1:
Consider the homogeneous equation x' = Ax below.
 1 1
x
x  
 4 1
In Chapter 7.5, we found the following fundamental
solutions for this system:
 1 3t ( 2)
 1 t
x (t )   e , x (t )   e
 2
  2
(1)
Thus a fundamental matrix for this system is
 e 3t
Ψ (t )   3t
 2e
e t 

t 
 2e 
Fundamental Matrices and General Solution
The general solution of x' = P(t)x
x  c1x(1) (t )   cn x( n)
can be expressed x = (t)c, where c is a constant vector with
components c1,…, cn:
 x1(1) (t )  x1( n ) (t )  c1 

 
x  Ψ(t )c   

   
 x (1) (t )  x ( n ) (t )  c 
n
 n
 n 
Fundamental Matrix & Initial Value Problem
Consider an initial value problem
x' = P(t)x, x(t0) = x0
where  < t0 <  and x0 is a given initial vector.
Now the solution has the form x = (t)c, hence we choose c
so as to satisfy x(t0) = x0.
Recalling (t0) is nonsingular, it follows that
Ψ(t0 )c  x0  c  Ψ1 (t0 )x0
Thus our solution x = (t)c can be expressed as
x  Ψ(t )Ψ1 (t0 )x0
Recall: Theorem 7.4.4
Let
e (1)
1
0
0
 
 
 
0
1
0
  0 , e ( 2 )   0 ,  , e ( n )    
 
 
 


0
0
0
1
 
 
 
Let x(1),…, x(n) be solutions of x' = P(t)x on I:  < t <  that
satisfy the initial conditions
x(1) (t0 )  e(1) , , x( n) (t0 )  e( n) ,   t0  
Then x(1),…, x(n) are fundamental solutions of x' = P(t)x.
Fundamental Matrix & Theorem 7.4.4
Suppose x(1)(t),…, x(n)(t) form the fundamental solutions given
by Thm 7.4.4. Denote the corresponding fundamental matrix
by (t). Then columns of (t) are x(1)(t),…, x(n)(t), and hence
1

0
 (t0 )  


0

0
1

0




0

0
I



1 
Thus -1(t0) = I, and the hence general solution to the
corresponding initial value problem is
x  Φ(t )Φ1 (t0 )x0  Φ(t )x0
It follows that for any fundamental matrix (t),
x  Ψ(t )Ψ1 (t0 )x0  Φ(t )x0  Φ(t )  Ψ(t )Ψ1 (t0 )
The Fundamental Matrix 
and Varying Initial Conditions
Thus when using the fundamental matrix (t), the general
solution to an IVP is
x  Φ(t )Φ1 (t0 )x0  Φ(t )x0
This representation is useful if same system is to be solved for
many different initial conditions, such as a physical system
that can be started from many different initial states.
Also, once (t) has been determined, the solution to each set
of initial conditions can be found by matrix multiplication, as
indicated by the equation above.
Thus (t) represents a linear transformation of the initial
conditions x0 into the solution x(t) at time t.
Example 2: Find (t) for 2 x 2 System
(1 of 5)
Find (t) such that (0) = I for the system below.
 1 1

x
x  
 4 1
Solution: First, we must obtain x(1)(t) and x(2)(t) such that
 1 ( 2)
 0
x (0)   , x (0)   
 0
 1
(1)
We know from previous results that the general solution is
 1 3t
 1 t
x  c1  e  c2  e
 2
  2
Every solution can be expressed in terms of the general
solution, and we use this fact to find x(1)(t) and x(2)(t).
Example 2: Use General Solution (2 of 5)
Thus, to find x(1)(t), express it terms of the general solution
 1 3t
 1 t
x (t )  c1  e  c2  e
 2
  2
(1)
and then find the coefficients c1 and c2.
To do so, use the initial conditions to obtain
 1
 1  1
x (0)  c1    c2     
 2
  2  0
(1)
or equivalently,
1 c1   1
1

    
 2  2  c2   0 
Example 2: Solve for x(1)(t)
(3 of 5)
To find x(1)(t), we therefore solve
1 c1   1
1

    
 2  2  c2   0 
by row reducing the augmented matrix:
1 1  1
1
1  1 1
1  1 0 1 / 2 
1

  
  
  

 2  2 0  0  4  2  0 1 1/ 2  0 1 1/ 2
c1
 1/ 2

c2  1 / 2
Thus
 1 3t 1  t 
1
1




1
1
 e  e 
x (1) (t )   e 3t   e t   2
2 
2  2
2   2
3
t
 e  e t 


Example 2: Solve for x(2)(t)
(4 of 5)
To find x(2)(t), we similarly solve
1 c1   0 
1

    
 2  2  c2   1
by row reducing the augmented matrix:
1 0  1
1 0  1 1
0  1 0
1/ 4
1

  
  
  

 2  2 1  0  4 1  0 1  1 / 4   0 1  1 / 4 
c1
 1/ 4

c2  1 / 4
Thus
 1 3t 1 t 
1  1 3t 1  1 t  4 e  4 e 
( 2)

x (t )   e   e  
1
1
4  2
4   2
 e3t  e t 
2 
2
Example 2: Obtain (t)
(5 of 5)
The columns of (t) are given by x(1)(t) and x(2)(t), and thus
from the previous slide we have
 1 3t 1 t
 e  e
2
Φ(t )   2
 e3t  e t

1 3t 1 t 
e  e 
4
4 
1 3t 1 t 
e  e 
2
2 
Note (t) is more complicated than (t) found in Ex 1.
However, now that we have (t), it is much easier to
determine the solution to any set of initial conditions.
 e 3t
Ψ (t )   3t
 2e
e t 

t 
 2e 
Matrix Exponential Functions
Consider the following two cases:
The solution to x' = ax, x(0) = x0, is x = x0eat, where e0 = 1.
The solution to x' = Ax, x(0) = x0, is x = (t)x0, where (0) = I.
Comparing the form and solution for both of these cases, we
might expect (t) to have an exponential character.
Indeed, it can be shown that (t) = eAt, where

n n
n n

A
t
A
t
e At  
I
n 0 n !
n 1 n !
is a well defined matrix function that has all the usual
properties of an exponential function. See text for details.
Thus the solution to x' = Ax, x(0) = x0, is x = eAtx0.
Example 3: Matrix Exponential Function
Consider the diagonal matrix A below.
1 0

A  
 0 2
Then
 1 0  1 0   1 0  3  1 0  1 0   1 0 

  
, A  

  
,
A  
2
2 
3
 0 2  0 2   0 2 
 0 2  0 2   0 2 
2
In general,
1 0 

A  
n
0 2 
n
Thus
0  n  et
A nt n  1 / n !
 t  
e 
  
n
2 / n !
n 0 n !
n 0  0
0

At
0

2t 
e 
Coupled Systems of Equations
Recall that our constant coefficient homogeneous system
x1  a11 x1  a12 x2    a1n xn

xn  an1 x1  an 2 x2    ann xn ,
written as x' = Ax with
 x1 (t ) 
 a11  a1n 




x(t )    , A      ,
 x (t ) 
a

 n 
 n1  ann 
is a system of coupled equations that must be solved
simultaneously to find all the unknown variables.
Uncoupled Systems & Diagonal Matrices
In contrast, if each equation had only one variable, solved for
independently of other equations, then task would be easier.
In this case our system would have the form
x1  d11 x1  0 x2    0 xn
x2  0 x1  d11 x2    0 xn

xn  0 x1  0 x2    d nn xn ,
or x' = Dx, where D is a diagonal matrix:
 x1 (t ) 


x(t )    ,
 x (t ) 
 n 
 d11

 0
D


 0

0
d 22

0
0 

 0 
  

 d nn 

Uncoupling: Transform Matrix T
In order to explore transforming our given system x' = Ax of
coupled equations into an uncoupled system x' = Dx, where D
is a diagonal matrix, we will use the eigenvectors of A.
Suppose A is n x n with n linearly independent eigenvectors
(1),…, (n), and corresponding eigenvalues 1,…, n.
Define n x n matrices T and D using the eigenvalues &
eigenvectors of A:

  


T      ,
  (1)   ( n ) 
n 
 n
(1)
1
(n)
1
 1 0

 0 2
D
 

0 0

0

 0
  

 n 

Note that T is nonsingular, and hence T-1 exists.
Uncoupling: T-1AT = D
Recall here the definitions of A, T and D:

  
 a11  a1n 




A      , T      ,
  (1)   ( n ) 
a


a
nn 
n 
 n1
 n
(1)
1
(n)
1
 1 0

 0 2
D
 

0 0

0

 0
  

 n 

Then the columns of AT are A(1),…, A(n), and hence
 11(1)  n1( n ) 


AT   

   TD
   (1)    ( n ) 
n n 
 1 n
It follows that T-1AT = D.
Similarity Transformations
Thus, if the eigenvalues and eigenvectors of A are known,
then A can be transformed into a diagonal matrix D, with
T-1AT = D
This process is known as a similarity transformation, and A
is said to be similar to D. Alternatively, we could say that A
is diagonalizable.
 1 0
(1)
(n)



a

a



 11
1n 
1
1




 0 2
A      , T      , D  
 
  (1)   ( n ) 
a



a
nn 
n 
 n1
0 0
 n

0

 0
  

 n 

Similarity Transformations: Hermitian Case
Recall: Our similarity transformation of A has the form
T-1AT = D
where D is diagonal and columns of T are eigenvectors of A.
If A is Hermitian, then A has n linearly independent
orthogonal eigenvectors (1),…, (n), normalized so that
((i), (i)) =1 for i = 1,…, n, and ((i), (k)) = 0 for i  k.
With this selection of eigenvectors, it can be shown that
T-1 = T*. In this case we can write our similarity transform as
T*AT = D
Nondiagonalizable A
Finally, if A is n x n with fewer than n linearly independent
eigenvectors, then there is no matrix T such that T-1AT = D.
In this case, A is not similar to a diagonal matrix and A is not
diagonlizable.
 1 0
(1)
(n)



a

a



 11
1n 
1
1




 0 2
A      , T      , D  
 
  (1)   ( n ) 
a



a
nn 
n 
 n1
0 0
 n

0

 0
  

 n 

Example 4:
Find Transformation Matrix T
(1 of 2)
For the matrix A below, find the similarity transformation
matrix T and show that A can be diagonalized.
 1 1

A  
 4 1
We already know that the eigenvalues are 1 = 3, 2 = -1
with corresponding eigenvectors
 1 ( 2)
 1
ξ (t )   , ξ (t )   
 2
  2
(1)
Thus
1
1
 3 0
, D  

T  
 2  2
 0  1
Example 4: Similarity Transformation
(2 of 2)
To find T-1, augment the identity to T and row reduce:
1
1

2  2
1 0
 
0 1
1 0  1
1
1 0  1 1
1
0
  
  

0 1  0  4  2 1  0 1 1 / 2  1 / 4 
1/ 2
1/ 4
1/ 4
1 / 2
  T 1  

1/ 2 1/ 4 
1 / 2  1 / 4 
Then
1 / 4    1 1  1
1 
1 / 2
  
 
 
T AT  
1 / 2  1 / 4    4 1  2  2  
1 / 4   3  1  3 0 
1 / 2
 
  
  D
 
1 / 2  1 / 4   6 2   0  1
1
Thus A is similar to D, and hence A is diagonalizable.
Fundamental Matrices for Similar Systems
(1 of 3)
Recall our original system of differential equations x' = Ax.
If A is n x n with n linearly independent eigenvectors, then A
is diagonalizable. The eigenvectors form the columns of the
nonsingular transform matrix T, and the eigenvalues are the
corresponding nonzero entries in the diagonal matrix D.
Suppose x satisfies x' = Ax, let y be the n x 1 vector such that
x = Ty. That is, let y be defined by y = T-1x.
Since x' = Ax and T is a constant matrix, we have Ty' = ATy,
and hence y' = T-1ATy = Dy.
Therefore y satisfies y' = Dy, the system similar to x' = Ax.
Both of these systems have fundamental matrices, which we
examine next.
Fundamental Matrix for Diagonal System
(2 of 3)
A fundamental matrix for y' = Dy is given by Q(t) = eDt.
Recalling the definition of eDt, we have
n



  1
Dnt n
Q(t )  
  0
n 0 n !
n 0 
0
 e  1t 0
0 

 0  0 

 t
0
0 e n 


  1t n


0 0 n 
n!
n 0
t
 0  
0

n  n!
0 n 
0





0
0



0


nt n 
0 

n ! 
n 0
Fundamental Matrix for Original System
(3 of 3)
To obtain a fundamental matrix (t) for x' = Ax, recall that the
columns of (t) consist of fundamental solutions x satisfying
x' = Ax. We also know x = Ty, and hence it follows that
t
  1(1) e 1t  1( n ) e  nt 
 1(1)  1( n )  e 1 0
0
 



Ψ  TQ       0  0    

 
 (1)  1t
 nt 
  (1)   ( n )  0
( n )  nt 

0 e

e
 n e 
n 
n
 n
 

The columns of (t) given the expected fundamental solutions
of x' = Ax.
Example 5:
Fundamental Matrices for Similar Systems
We now use the analysis and results of the last few slides.
Applying the transformation x = Ty to x' = Ax below, this
system becomes y' = T-1ATy = Dy:
 1 1
 3 0




 y
x 
x  y  

 4 1
 0  1
A fundamental matrix for y' = Dy is given by Q(t) = eDt:
 e 3t
Q (t )  
 0
0 

t 
e 
Thus a fundamental matrix (t) for x' = Ax is
1 e3t
1

Ψ(t )  TQ  
 2  2  0
0   e 3t
   3t
t 
e   2e
e t 

t 
 2e 