10.1 – Theory of Linear Systems General form

Lecture 7

Kemal Ahmed 2013-06-03

Test scores up tomorrow. ~50% average for the scantron, but the written problems were worse, like 6/20.

10.1 – Theory of Linear Systems

General form

 

_{ }

_{ }

_{ }

_{ }

 

 

_{ }

_{ }

_{ }

 

_{ }

_{ }

_{ }

1

11 1 12 2 1 1

2

21 1 2 2

1 1

d

... d

d

... d

d

... d

n n

g t

n n

n

n nn n n

x t

a t x a t x a t x f t

t

x t

a t x a t x f t

t

x t

a t x a t x f t

t

   

   

   

Unknowns are: x t₁

   

,x t₂ ,...,x t_n

 

Independent variable is t.

 

 

 

1 2

n

x t

x t X

x t              

←column of unknowns

11 12 1

21 22 2

1 2

... ...

...

n n

n n nn

a a a

a a a

A

a a a

 

 

 



 

 

 

 

 

 

 

 

1 2

system

** ' matrix form

n

f t

f t F t

f t

X A X F

 

 

 

  

 

 

 

x'=ax + b

The system is a system of linear first order differential equations. If F is not a zero vector, then (**) is non-homo system. If F is zero vector, (**) is a homo system.

e.g. 1)

 

 

d

3 4 d

d

5 7 d

3 4 0

5 7 0

3 4

' '

5 7 x

x y

t y

x y

t x t

X A F

y t

X AX X X

 

 

     

_{ } _{ } _{ }



   

 

 

  _{  }



 

e.g. 2)

5

d

3 7 17 d

d

3 7

d d

d

x

y z x

t

y

z x t

t z

z t t

        

Is everything going to be linear? What is the first unknown? Everything is good, except the second equation. So let’s go through it step-by-step: Rearrange equation 2

5

d

3 7 17 d

d

3 7 d

d d

x

y z x

t y

z x t t

z

z t t

       

The first is linear. The second doesn’t have any y’s, so it is also linear, and so is the third one.

YAY! They’re all linear (nothing other than constants in front of the y for dy and no powers for

5

, ,

17 3 7 0

; 7 0 3 ;

0 0 1 17 3 7

' 7 0 3 0 0 1

x y z

x

X y A F t

z _t

x

X y F

z

  

   

 

   

_{ } _{ }  _{ }

     

    _{ }



   

   

_{   } 

   

   

e.g. 3)

Now, let’s verify that something is the solution.

 

 

d

7 d

1 3 '

5 3

y y t y t ce

X X

X e

 

       

e.g. 4)

 

2

1

1 1

t

X t    e 

  is a solution of

1 3 '

5 3

X   _X  

Expand:

 

2

1 ₂

t t

e X t

e





 

  



 

Now, verify that 1 1

1 3 '

5 3

X   _X  

 

2

1 2

2 2

2 2

2 '

2

1 3 2

5 3 2

t t

t t

t t

e X t

e

e e

e e





 

 

 

  

 

   

 



   

 _{ }

    

Therefore, this is verified!

Solution vector

 

 

1

n

x t

X

x t

 

 

  

 

 

, whose entries are differentiable functions that satisfy the system.

Hope that all the entries are continuous functions and if possible, restrain the I to make sure they are continuous.

   

1 0 , 2 0 ,...

x x

For IVP’s, give X(0) or X t

 

₀ , i.e. X₁

 

0 ,X₂

 

0 ,... If they are evaluated at different times, then it is a BVP.

Existence & Uniqueness Theorem

Let the entries, ofA t

   

,F t be continuous functions on a common interval, I, that contain a point,t₀. Then there exists a unique solution of the IVP. Note: this is for IVP’s only. Also, zeroes are continuous functions.

 

   

 

 

0 0

'

X t A t X t F t

X t X

 



 

 

 

 

 

3

7

[0, ) (0, )

1

2

1 5

( ,0) (0, )

(0, ) _{(0, )}

1

3 ₁

'

ln ln

1 7

5

z

t

t

t _t

X t x t

t t _t

X



 



  

 _

  _{  }

  _{  }

 

 

_{ }   

 _{ }

 

  _{ }

 

      

First, check where the discontinuities are and use it to determine your I. It seems the common I is (0,∞) and 7ϵ(0,∞). YAY!

Homogeneous Systems

(*)X’=AX, A = n × n

Superposition principle: let X X₁, ₂,...,X_k(there are k number of solutions) be a set of vector solutions, i.e. X_i' AX i_i, 1,k, of (*) on an interval, I. Then, any linear combination of the set,

1... k

X X , forms a new solution of (*).

1 1 2 2 ... k k

X c X c X  c X is a solution, where are the c’s are constants

e.g. 5)

The sets,

 

 

 

 

 

1 1

1 2 2 2

0 cos

; cos sin

0 cos sin

t

t

X e X t t

t t

 

 

 

 

_{ }  _  _

   _{ } 

   

1 0 1

' 1 1 0

2 0 1

A X X        _{ }   

Claim the matrix,X t

 

c X_{1 1}c X_{2 2}, is also a solution.

 

 

 

 

 

 

 

 

 

 

 

2 1 1

1 2 2 2 2

2 2

2

1 1

1 2 2 2 2

2 2

0 cos

cos sin

0 cos sin

cos cos sin cos sin t t c t

X t c e c t c t

c t c t

c t

c e c t c t

c t c t

        _{ } _  _    _          _   _  _{ }   

 

 

 

 

 

 

2 1 1

1 2 2 2 2

2 2

sin

' sin cos

sin cos

t

c t

X t c e c t c t

c t c t

     _   _  _   

Now, verify this by multiplying A∙X, i.e. (she didn’t do this, I did it for fun)

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

2

1 1 1 1

1 2 2 2 2 1 2 2 2

2 2

2

1 1

1 2 2 2 2

2 2

'

sin 1 0 1 0 cos

sin cos 1 1 0 cos sin

sin cos 2 0 1 0 cos sin

sin

sin cos

sin cos

t t

t

X A X

c t t

c e c t c t c e c t t

c t c t t t

c t

c e c t c t

c t c t

                _{ } _   _ _{ }                 _  _{  } _{ }  _{ }           _{ } _    _   

 

 

 

 

 

 

 

 

 

 

 

 

 

2 1 1

1 2 2 2 2

2 2

2 2

1 1 1 1

1 2 2 2 2 1 2 2 2 2

2 2

1 0 1 0 cos

1 1 0 cos sin

2 0 1 0 cos sin

sin 1 0 1 cos

sin cos 1 1 0 cos sin

sin cos 2 0 1

t

t t

c t

c e c t c t

c t c t

c t c t

c e c t c t c e c t c t

c t c t c

             _{   }        _{ }     _{ }              _{ }  _  _{ }        _  _{   }  

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

2 2 2 2

1 1 1 1

1 2 2 2 2 1 2 2 2 2

2 2 2 2

2

1 1

1 2 2 2 2

2 2

cos sin

sin 1 0 1 cos

sin cos 1 1 0 cos sin

sin cos 2 0 1 cos sin

sin

sin cos

sin cos

t t

t

t c t

c t c t

c e c t c t c e c t c t

c t c t c t c t

c t

c e c t c t

c t c t

       _           _{ } _  _{ }          _  _{  } _{ }          _{ }     _   

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

2 2 2

1 1

2 1 2 2 2 2

2 2 2

2 2

1 1 1 1

1 2 2 2 2 1 2 2 2 2

2 2 2 2

cos cos sin

cos cos sin

2 cos cos sin

sin sin

sin cos sin cos

sin cos sin cos

t

t t

c t c t c t

c t c e c t c t

c t c t c t

c t c t

c e c t c t c e c t c t

c t c t c t c t

Linear Independence and Dependence

Let X X₁, ₂,...X_kbe a set of solution vectors of the homo system,X'AX A,  n n, on an interval, I. We say that the set,



X₁,...X_k



is linearly independent on the interval, I, if there exists constants, c₁,...c_k, of which there must be at least one non-zero constant and the following:

1 1 2 2 ... k k 0

c X c X  c X  for every t ϵ I. Otherwise, they are called linearly-dependent. If the constants are all zero, then it is inconclusive.

   





 

_{ }

 

_{ }

     





 

 

 

, , 0

' '

, , , ' ' ' 0

" " "

f x g x

f x g x

f x g x

f x g x m x

f x g x m x





Criteria for linear independence

Let

11 21 1

1 2

1 2

1 2

; ;

W ...

n n

n n nn

n

X X X

X X X

X X X

X X X

     

     

_{ } _{ } _{ }

     

     

   

e.g. 6)

2 1

6 2

1 1 3 5

t

t

X e

X e



    _         

find the solution of the system

Are X₁ and X₂ linearly independent?





 

1 2

2 6

2 6

1 2

4 4

4

3

W , ₅

5 3

8 0

t t

t t

X X

t t

t

e e

X X _{e e}

e e

e t





   

  

Fundamental Sets of Solutions

Any set,



X₁,...,X_n



(for an n by n matrix, expect n solutions), of n linearly independent solution vectors of the homo system,X'AX A,  n n, on an interval, I, is said to be a

fundamental set of solutions. Once you have a fundamental set of solutions, you can write the

Assume you have a fundamental set of solutions,



X₁,...X_n



, then X c X_{1 1}c X_{2 2}...c Xn n is

the general solution, where the set of constants,c₁,...,c_n, have at least one element that is non-zero.

No-Homo Systems

You can solve the homo equation associated (i.e. complementary solution) and then add a particular solution.

e.g. 7)

3 4 5 6

p

t X

t 

 

 _{ } 

 is a particular solution of:

1 3 12 11 '

5 3 3

t X _ _x_  _



   

I = (−∞, +∞)

Verify that ₁ 1 2 ₂ 3 6

1 5

t t

c

X c  _{ }e c  _{ }e 

    is a solution of

1 3 '

5 3

X   _X   . Claim X Xp Xcis a solution of X’

2 6

1 2

2 6

1 2

2 6

1 2

2 6

1 2

2 6

1 2

2 6

1 2

2 6

1 2

2 6

1 2

3 4 3

5 6 5

3 4 3

5 6 5

3 2 18

'

5 2 3

1 3 3 4 3 1

5 3 5 6 5

p c

t t

t t

t t

t t

t t

t t

t t

t t

X X X

t c e c e

t c e c e

t c e c e

t c e c e

c e c e X

c e c e

t c e c e

t c e c e

















 

   

 

_{   }

   

   

    

 _{   } 

 

   

  

  

 

    

 

_{     } 

  

2 1 3

t

 

 _ 

 

10.2 – Solving Homo Systems

(BTW, I’m not homo-phobic)

(*)X'AX , where A is an n × n matrix, every entry is a constant

1. Find eigenvalues of A,  ₁, ₂,.... Unfortunately you have to go through all the cases 2. Find the associated eigenvectors, K K₁, ₂,...

3. The general solution of (*) on (−∞,∞) is given by: 1 2

1 1 2 2 ...

nt

t t

n n

X c K e c K e  c K e .

e.g. 8)

d

2 3 d

d 2 d

x

x y t

y

x y t

    What are the unknowns? x(t) and y(t).











2

2 3 2 1

det 0

2 3

0 2 1

2 1 6 0

3 4 0

A

A I 



 

 

        



     

  

1 2

4 1    

 

Sub in the eigenvalues. First λ1

1

2 4 3 2 3

2 1 4 2 3

2 3 0

2 3 0

2 3 0 line 1 2 3 0 line 2

2 3

3 2

3 2

a

b

a b

a b

a b

a b

K

 

   



 _   _ 

   



       _             

   

         

2

2 1 3 0

2 2 0

3 3 0

3 2 0

1 1

1

a

b

a b

a b

a b

b

K 

               

   

   

    _  

 

 

4

1 1 1 1

1

2 2 2 2

3 4

2 1 1

1

t

t

K X t K e

K X t K e



 

 

  _{ }    

 

   _{ }  

 

 

 

 

   

1 2

4 1

2

4 4

3 3

3

3

2

1

1

3

W ₂

3 2

5 0

t

t

t t

t t

X t X t

t t

t

X t e

X t e

e e

t _{e e}

e e

e







           _  

 _

  

  

=> The set,



X t₁

 

,X₂

 

t



, forms a fundamental set of solutions General solution:

 

 

 

 

 

 

 

1 1 2 2

4

1 2

1 2

3 1

2 1

t t

X t c X t c X t

c e c e

x t

y t

x t

x t



 

     _{ }  _{ }

      

           

4

1 2

4

1 2

3 2

t t

t t

c e c e

c e c e





  

  _ 

 

x(t) = first row

Repeated Eigenvalues

This is simply the case, when even if you have repeated eigenvalues, you can still find linearly independent eigenvectors for each eigenvalue.

e.g. 9)

1 2 2

' 2 1 2

2 2 1

X X



 

 

 _  _

 _ 

 

Hint: We’ll only have to deal with homogeneous in the exam.

T

A Ais a symmetric matrix <=> All the eigenvalues are real. This says nothing about their multiplicity (i.e. repeats of roots).





3 2

3

det 0

1 2 2

2 1 2 0

2 2 1

1 2 2

2 1 2 0

0 1 1

R R

A I 

  



 



 

 

   

 

 

    

    Expand along the first column:



 

  



 



  









 















 



1 1 2 1

2

2

1 2

3

1 2 2 2

1 1 2 1 0

1 1 1 1

1 1 1 2 1 2 2 2 2 1 0

1 1 1 2 2 2 2 2 2 2 0

1 1 2 2 8 8 0

1 5 0

1 5 



   

     

     

   

 

 



    

     

       

 _        _ _     _  _     _    

 

 _    _  

   

   

It is a big problem that  ₁ ₂because K₁&K₂should be linearly independent.

2 2 2 0

2 2 2 0

2 2 2 0

a

b

c 

    

_{ }    _

    

 _    

2 2 2 0

2 2 2 0

a b c

a b c        2a2b2c0

a – b + c = 0

1 2 3

1 0 1

1 ; 1 ; 0

0 1 1

K K K

     

     

_{ } _{ } _{ }

     _

     

3 3

1 5

1 1 2 2 3 3

1

5 1

1

t t t

K

x c K e c K e c K e 

 

       _{ }    

  

Given n multiplicity, expect (n − 1) to (n + 1) distinct eigenvectors.

Each K is a column. Since the K matrix is

1 0 1 1 1 0 0 1 1

, you can see thatK₁K₂ K₃, you only keep the first 2 K’s. Well, actually, it doesn’t matter which 2 you choose, but it’s probably easier

if you choose those 2.

Complex Eigenvalues

For a size 2 × 2 matrix,

 

1 2

1 1

2 2 1

1 1 2 2

t t

i K

i K K

X t c K e c K e   

  

      

 

Find all real solutions.

 

 

1

cos sin

t t i t

t i t

t

e e

e e

e t i t

  

 

 _{ }





 

 _  _









1 1

1 2

1

1 2 1 1

2 2 1 1

;

K K

i

i i

B K K

B K K

       

 

  

 

 

 

 

 

 

 

1 1 2 2

1 1 2

2 2 2

cos sin memorize!

cos sin

X t c X c X

X t B t B t

X t B t B t

 

 



 



_  _{ }



_  _

e.g. 10)

 

2 8 '

1 2 2 0

1

X X

X

 

 _{ } 

 

    _  







2 2

1 2

2 8

0

1 2

2 2 8 0

4 8 0

4 0

2

2 i

i 



 

 

  



  

    

   

    





























1 2 1 ?

2 2 8 0

1 2 2 0

2 2 8 0

2 2 0

8 8 2 2 0

2 2 0

2 2 0

2 2 0

2 2 0

1

1 2 2

i K

i a

i b

i a b

a i b

a i b

a i b

a i b

a i b

a i b

a

b

i     

    



 _{  }    

    

  

    

  

   

  

  

  



  

 Multiply by complex conjugate 1

2 2i

1 1

4 4

1 _{1 1}

4 4

2 2 8

1 i b

b i

K

i



  

 

 

 _{ } 

1 1 _{1 1}

4 4

2 2 _{1 1}

4 4

1 2

1 2

i K

i

i K

i 



 

  _{  }

 

 

 

   _{  }  

 









1

1 2 1 2

2 4

1 4 1

2 2 2 1

2 4

2

2 4

1 4

2 1 2

1

0 1 2

0

0

i i

B K K

B K K

i



 

 

  _

 

 

  _

 

 

 

 _ 

 

 

  _ 

 

      

 

 

 

 

 

 

 

0

1 1 2

0

2 2 1

1 1 2 2

cos 2 sin 2 cos 2 sin 2

t

t

X t e B t B t

X t e B t B t

X t c X c X

 _  _

 _  _