diff equation 6 sys 2011fall

8/12/2019 Diff Equation 6 Sys 2011FALL

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S.l.dr.ing.mat. Alina BogoiDifferential Equations

POLITEHNICA University of Bucharest

Faculty of Aerospace Engineering

CHAPTER 5

Systems of First Order

Linear Equations


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Outline Eigenvalues and Eigenvectors

Homogeneous Systems of Equationswith Constant Coefficients

Nonhomogeneous Systems ofEquations: Method of Variation of

Parameters


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Diff_Eq_7_2011 3

Eigenvalues and Eigenvectors

DefinitionLetAbe an n

n matrix. A scalar

is called an eigenvalue

ofA

if there exists a nonzero vector x

in Rn such that

Ax =

x.The vector x

is called an eigenvector

corresponding to .

Figure 6.1


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Eigenvalue and Eigenvector:

Ch6: Eigenvalues and Eigenvectors

:Example

The number is said to be an eigenvalue of the nxn matrix Aprovided there exists a nonzero vector v such that

v is called an eigenvector of the matrix A. v is associated with theeigenvalue

vAv =

=

=

1

2

v,22

65

:consider 1A

= 2

3

v2


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Characteristic Equation:


( )P A I =

:Example

It is a polynomial of order n. ( A is nxn)

= 42

75

)Aa

=

02

80

)Ab

=

20

32)Ac

=

51516

264

003

)Ad

Find the characteristic ploynomial

=

322

102

124

)Ae


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:Example

Eigenvalues: Eigenvalues of A are the roots of the characteristic equation

=

42

75A

=

02

80A

=

20

32A

=

51516

264

003

A

Find all eigenvalues

Eigenvector:

:Example

=42

75A

=02

80A

=

20

32A

=

51516

264

003

A

Find all eigenvectors

6)( 2 = P 16)( 2 +=P 2)2()( =P )1)(3()( = P

:theosolution tzero-nonaisitif

eigenvaluewith theassociatedeigevectoranis

v [ ]0IA


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:Example

=

322

102

124

A

Find the charact. Poly.

Remark:

Find all eigenvalues

For higher-degree polynomial is not easy to factor. DONOT

expand but try to find a common factor

Remarks:Aofeigenvalueanis0= 0=A

singular

A


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Eigenspace:

[ ]0-AsystemtheofspacesolutionThe =

Let be an eigenvalue of the matrix A .W = eigenspace of A associated with

= { all eigenvector of A associated with } U { zero vector}

=

322

102124

A

:Example 2=

[ ][ ]201

011

2

1

==

v

v

3=

[ ]1113=v

},{ 212 vvspanW = }{ 33 vspanW =2)dim( 2 =W

1)dim( 3 =W


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Example 1Find the eigenvalues and eigenvectors of the matrix

=

53

64A

Let us first derive the characteristic polynomial of A.We getSolution

=

=

53

64

10

01

53

642IA

218)5)(4( 22 =+= IA

We now solve the characteristic equation ofA.

1or20)1)(2(022 ==+=


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= 20=

2

1

33

66

x

x

This leads to the system of equations

Thus the eigenvectors ofA corresponding to

= 2 are

nonzero vectors of the form

033

066

21

21

=+

=

xx

xx

1

1

r


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Diff_Eq_7_2011 11

= 10=

2

1

63

63

x

x

This leads to the system of equations

Thus the eigenvectors ofA corresponding to

= 1

are

nonzero vectors of the form

063

063

21

21

=+

=

xx

xx

1

2s


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Example 2Find the eigenvalues and eigenvectors of the matrix

=

222

254

245

A

The matrixA

I3

is obtained by subtracting

from

the diagonal elements ofA.Thus

Solution

=

222

254245

3IA

2

2

)1)(10()1)(10)(1(

]1011)[1(]8)2)(9)[(1(

242

294

001

==+==

=


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Diff_Eq_7_2011 13

= 10

We get

0

0x

=

=

3

2

1

3

822254

245

)10(

xx

x

I

1

2

2

r


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Diff_Eq_7_2011 14

= 1

0

0x

=

=

3

2

1

3

122

244

244

)1(

x

x

x

IA

The solution to this system of equations can be shown to be

x1

=

s

t,x2

= s, andx3

= 2t, where s and t are scalars.

Thus the eigenspace of = 1 is the space of vectors of theform.

ts

ts

2


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Separating the parameters s and t, we can write

Thus the eigenspace of

= 1 is a two-dimensional subspace of

R2

with basis

+

=

2

0

1

0

1

1

2

ts

t

s

ts

0

0

1

,

0

1

1

.

Thus =10 has multiplicity 1, while =1 has multiplicity 2

in this example.


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Theorem

LetAbe an n

n symmetric matrix.

(a) All the eigenvalues ofA are real numbers.(b)

The dimensional of an eigenspace ofA is the multiplicity of

the eigenvalues as a root of the characteristic equation.

(c) The eigenspaces ofA are orthogonal.(d)

A has n linearly independent eigenvectors.


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Matrix Functions

The elements of a matrix can be functions of a real

variable. In this case, we write

Such a matrix is continuous at a point, or on an

interval

(a, b), if each element is continuous there. Similarly

with differentiation and integration:

=

=

)()()(

)()()(

)()()(

)(,

)(

)(

)(

)(

21

22221

11211

2

1

tatata

tatata

tatata

t

tx

tx

tx

t

mnmm

n

n

m L

MOMM

L

L

M Ax

=

= b

a ij

b

a

ij

dttadttdt

da

dt

d)()(, A

A


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Introduction to Systems of First OrderLinear Equations

A system of simultaneous first order ordinary

differential equations has the general form

eachxk =xk( t).

),,,(

),,,(

),,,(

21

2122

2111

nnn

n

n

xxxtFx

xxxtFx

xxxtFx

K

MK

K

=

=

=


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Solutions of First Order Systems

A system of first order ordinary differential equations

It has a solution on I: < t < if there exists n functions

that are differentiable on I and satisfy the system ofequations at all points t in I.

Initial conditions may also be prescribed to give an IVP:

).,,,(

),,,(

21

2111

nnn

n

xxxtFx

xxxtFx

K

M

K

=

=

)(,),(),( 2211 txtxtx nn === K

00

0202

0101 )(,,)(,)( nn xtxxtxxtx === K


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Theorem

Suppose F1,, Fn and F1/x1,, F1/xn,, Fn/ x1,, Fn/xn, are continuous in the region R definedby < t < ,

1

< x1

< 1

, , n

< xn

< n

, and let the

point

be contained in R. Then in some interval (t0 - h, t0 +h) there exists a unique solution

that satisfies the IVP.

00

2

0

10 ,,,, nxxxt K

)(,),(),( 2211 txtxtx nn === K

),,,(

),,,(

),,,(

21

2122

2111

nnn

n

n

xxxtFx

xxxtFx

xxxtFx

K

M

K

K

=

=

=


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Linear Systems

If each Fk is a linear function ofx1,x2,

,xn, then the system of equations hasthe general form

Otherwise it is nonlinear.

)()()()(

)()()()()()()()(

2211

222221212

112121111

tgxtpxtpxtpx

tgxtpxtpxtpxtgxtpxtpxtpx

nnnnnnn

nn

nn

++++=

++++=++++=

KM

KK


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General Solution The general solution of x' = P(t)x + g(t) on I: < t <

has the form

where

is the general solution of the homogeneous system

x' = P(t)x

and v(t) is a particular solution of thenonhomogeneous system x' = P(t)x + g(t).

)()()()( )()2(2)1(

1 ttctctc n

n vxxxx ++++= K

)()()( )()2(2

)1(

1

tctctc n

n

xxx +++ K


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Nth Order ODEs reduces to a Linear 1st

Order Systems An arbitrary nth order equation

Is transformed into a system of n first order

equations, by defining

)1()( ,,,,, = nn yyyytFy K

)1(

321 ,,,, ==== nn yxyxyxyx K

1 2

2 3

1 2( , , , )n n

x x

x x

x F t x x x

==

=M

K


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Practical Importance:

High-order

System Converted

First-order

System

Example: tyyyy sin5'2''3''' =++

'''

3

2

1

yxyx

yx

==

=

'''''''

''

3

2

1

yxyx

yx

==

=

txxxxxx

xx

sin523''

'

1233

32

21

++==

=First-order System

Linear Systems of Differential Equations

Li S t f Diff ti l E ti


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Example: System of DE

)3sin(4022''

26''

tyxy

yxx

+=

+=

Independent variable: tdependent variables: x, y

2nd order

=

LL

LL

)(

)(

ty

tx

Solutions

Example:

yxy

yxx

22'

26'

=

+=

First-order

ytxy

ytxx

222'''

26''

=

+=

3rd-order zxz

zyxy

zyxx

=+=

+=

'

'

2'

First-order

Order of the system




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Example:

'

'

4

3

2

1

yx

yx

xx

xx

=

=

=

=

Transform into first-order system

yxy

xyx

)1(''

)1(''

=

=

314

43

132

21

)1('

'

)1('

'

xxx

xx

xxx

xx

=

=

=

=

Example:

'

'

4

3

2

1

yx

yx

xxxx

=

=

==

Transform into first-order system

tyxy

yxx

+=

+=

''

''

txxx

xx

xxx

xx

+=

=

+==

314

43

312

21

'

'

'

'

+

=

tx

x

x

x

x

x

x

x

0

0

0

0101

1000

0101

0010

'

'

'

'

4

3

2

1

4

3

2

1



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Example:

yxyyxx

22'26'

=+=

3

2

22'

26'

yxy

yxx

=+=

zxz

zyxy

zyxx

=

+=

+=

'

'

2'

Linear system

yzxzzyxy

zxyxx

=+=

+=

''

2'

zexz

zyxy

zytxx

t=+=

+=

''

2' 2



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1 11 1 12 2 1 1

2 21 1 22 2 2 2

1 1 2 2

( ) ( ) ( ) ( ) ( )

( ) ( ) ( ) ( ) ( )

( ) ( ) ( ) ( ) ( )

n n

n n

n n n nn n n

x t a t x a t x a t x f t



= + + + +

= + + + +

= + + + +

KK

KK

M

KK

Matrix Form:

1 11 12 1 1 1

1 21 22 2 1 2

1 1 2 1 1

( ) ( ) ( ) ( )

( ) ( ) ( ) ( )

( ) ( ) ( ) ( )

n

n

n n nn

x a t a t a t x f t

x a t a t a t x f t

x a t a t a t x f t

= +

K

K

M M M M M M

L

System of linear first-order DE

'X A X F= +



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'X A X F= +

If F=0 homogeneous system

If F 0 non-homogeneous system

Therorem ( Existence of a Unique Solution)

0

all entries of ( ) are cont on

all entries of ( ) are cont on

t I

t I

F t I

0 0

' ( ) ( ) (*)

( )

X A t X F t

X t X

= +

=

There exists a uniquesolution of IVP(*)



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Example:

yxytyxx

22'26'

=++=

zxz

zyxy

zyxx

=

+=

+=

'

'

2'

Homog and Non-homg

ttezexz

zyxy

tzytxx

2

22

'

'

2'

+=

+=

+=

+

=

022

26

'

' t

y

x

y

x

=

zy

x

zy

x

101111

121

''

'

+

=

tt e

t

zy

x

e

t

zy

x

2

22

001

111

121

''

'

homognon

'

+= FAXX

homog

' AXX =

homognon

'

+= FAXX


System of Equations: First-Order Linear


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Differential Equations - Substitution

Consider the 2x2 system of linear homogeneous differential

equations (with constant coefficients)

x'(t) = ax(t) + by(t)

y'(t) = cx(t) + dy(t)

We can solve this system using what we know:

1. Isolatey(t) in the first equation =>y(t) =x'(t)/b

ax(t)/b.

2. Differentiate thisy(t) equation =>y'(t) =x"(t)/b

ax'(t)/b.

3. Solve forx(t) andy'(t)

=>x"(t)

(a + d)x'(t) + (ad

bc)x(t) = 0.

4. Go back to step 1. Solve fory(t) in terms ofx'(t) andx(t).



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Differential Equations - SubstitutionExample:

x'(t) = 2x(t) +y(t)

y'(t) = 4x(t)

3y(t).

Solution:

x"(t) +x'(t) 2x(t) = 0.

x(t) =Aet +Be2t.

y(t) = Aet

4Be2t.



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Therorem ( Principle of Superposition)

solutionsnareLet 21 n,X,, XX L

XX ='Consider the sys of DE: (*)

1 1 2 2 also solutionn nc X c X c X + + +L

Example:

=

y

x

y

x

13

24

'

'

=

=

t

t

t

t

e

etX

e

etX

5

5

22

2

1

2)(,

3)(

are both

solutions of (*)

(*)

+

=

t

t

t

t

e

ec

e

ectX

5

5

22

2

23

2

3)( solution of (*)

DEF ( Wronskian)

solutionsnareLet 21 n,X,, XX L


nnn

n

n

xx

xx

XXXW

L

MOM

L

L

1

111

21

),,,( =

their wronskian is the nxn determinant

Example:

=

y

x

y

x

13

24

'

'

=

=

t

t

t

t

e

etXe

etX5

5

22

2

12)(,

3)(

Find W(X1,X2)

(*)




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Example: Consider the first-order linearsystem of DE

=

y

x

y

x

13

24

'

'

=

=

t

t

t

t

e

etX

e

etX

5

5

22

2

1

2)(,

3)(

Verify that the vector functions

are both solutions of (*)

(*)


First-order Systems


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First order Systems

THM ( Wronskian)

solutionsnareLet21 n

,X,, XX L


1 2( , , , ) 0nW X X X L

Example:

=

y

x

y

x

13

24

'

'

=

=

t

t

t

t

e

etX

e

etX

5

5

22

2

1

2)(,

3)(

Linearly dependent or independent ??

(*)

tindependenlinearly

21 n,X,, XX L

First-order Systems


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THM ( general solution for Homog)

indeplinare21 n,X,, XX L

AXX ='Consider the sys of DE: (*)

Example:

=

y

x

y

x

13

24

'

'

=

=

t

t

t

t

e

etX

e

etX

5

5

22

2

1

2)(,

3)(

Find the general solution for (*)

(*)

nnXcXcXctX +++= L2211)(solutionsare

21 n

,X,, XX LThe general sol for (*) is

Example:

=

y

x

y

x

13

24

'

'

Solve IVP

(*)

=

1

1

)0(

)0(

y

x

First order Systems

First-order Systems


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THM ( general solution for non-Homog)

indeplinare21 n,X,, XX L

FAXX +='Consider the sys of DE: (*)

Example:

++

=

t

t

y

x

y

x

3

14

13

24

'

'

=

=

t

t

t

t

e

etX

e

etX

5

5

22

2

1

2)(,

3)(

Find the general solution for (*)

(*)

nnc XcXcXctXwhere +++= L2211)(

(**)solutionsare21 n,X,, XX L The general sol for (*) is

Example: Solve IVP

=

1

1

)0(

)0(

y

x

AXX =' (**)

(*)forsolparticularpX)()()( tXtXtX pc +=

Sol for Homog

=

0)(

ttXP Particular sol for non-Homog

++

=

t

t

y

x

y

x

3

14

13

24

'

'

y

How to solve the system of ODE


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How to solve the system of ODE

System of Linear First-Order DE

(constant Coeff) 'X AX=

Distinct realEigenvalues

repeated realEigenvalues

complexEigenvalues

System of Linear First-Order DE

(Non-homog)'X AX F= +

Variation of

ParametersEigenv

alueM

ethod

Eigenvalue Method


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Diff_Eq_7_2011 41

g

Our Goal:

Example: Solve:

'X AX=Solve the Homog linear system

=

y

x

y

x

22

26

'

'

=

z

y

x

z

y

x

101

111

121

'

'

'

Method:Find all eigenvalues of the matrix A1 n ,,, 21 L

Distinct real

Eigenvalues

repeated real

Eigenvalues

complex

Eigenvalues

Solution: 1) Find n linearly independent solutions

2) The general solution is: their linear combination

nXXX ,,, 21 L


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Eigenvalue Method


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Diff_Eq_7_2011 43

1 ,123 11 1 veX

t=22

2 veX t= n

t

n veX n=

4 nnXcXcXctX +++= L2211)(

n,2 LLL

,1v ,2v nvLLL

LLL

:Example

=

4275A

2

7,

1

1

3,2 ==

Solve: 'X AX=

=1

1)( 21t

etX

=2

7)( 32t

etX

2211)( XcXctX +=The general solution:

+

= t

t

t

t

e

e

ce

e

ctX 3

3

22

2

12

7

)(

g

Eigenvalue Method


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Diff_Eq_7_2011 44

1 ,1

23 11 1 veX t= 22 2 veX t= ntn veX n=

4 nn

XcXcXctX +++= L2211

)(

n,2 LLL,1v ,2v nvLLL

LLL

:Example Solve: 'X AX=

=

3

1

0

)(1tX

=

5

2

0

)(2t

etX

332211)( XcXcXctX ++=The general solution:

+

+

=t

t

t

t

e

e

c

e

ecctX3

3

321

2

0

5

2

0

3

1

0

)(

=

51516

264

003

A

20

1

,5

2

0

,3

1

0

3,1,0 ===

=

2

0

1

)( 32t

etX

g

Eigenvalue Method


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Diff_Eq_7_2011 45

'X AX=

Method:Find all eigenvalues of the matrix A1

i =2,1

Complex conjugate eigenvalues

Find an eigenvector for21v

solution is:

31

1 veX t=

Two-lin independent solutions are:4

i +=1Complex vector

1)]sin()[cos( vtite t +=1

)(ve

ti+=

)real(1 XX = )Imag(2 XX =

Two-lin independent solutions are:52211)( XcXctX +=

Eigenvalue Method


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Diff_Eq_7_2011 46


)(1

tX )(2

tX

2211)( XcXctX +=

The general solution:

=

02

80A i4=

i4=

=

1

2iv

=

1

2)( 4

ietX

ti

+=

1

2)4sin4(cos

itit

+

+=

tit

tti

4sin4cos

4sin24cos2i

t

t

t

t

+

=

4sin

4cos2

4cos

4sin2

+

=

t

tc

t

tctX

4sin

4cos2

4cos

4sin2)( 21

Eigenvalue Method


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Diff_Eq_7_2011 47

'X AX=Method:

Find all eigenvalues of the matrix A1i =2,1

Complex conjugate

eigenvalues

Find an eigenvector for2iBBv 211 +=

Two-lin independent solutions are:

3

i +=1

Two-lin independent solutions are:42211)( XcXctX +=

1 1 2

2 2 1

[ cos sin ]

[ cos sin ]

t

t

X B t B t e

X B t B t e

=

= +

Eigenvalue Method


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Diff_Eq_7_2011 48


)(1

tX )(2

tX

=

02

80A i4=

i4=

=

1

2iv

=

1

2)( 4

ietX

ti

+=

1

2)4sin4(cos

itit

+

+=

tit

tti

4sin4cos

4sin24cos2i

t

t

t

t

+

=

4sin

4cos2

4cos

4sin2

i4=

=

1

2iv i

+

=

0

2

1

0

iBB 21+=

1 1 2

2 2 1

[ cos sin ]

[ cos sin ]

t

t

X B t B t e

X B t B t e

=

= +

ttX 4sin

0

24cos

1

01

= ttX 4sin

1

04cos

0

21

+

=

Multiple Eigenvalue Solution


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Diff_Eq_7_2011 49

, , , repeated real eigenvalues K 'X A X=

:Example 1, 1,5= Solve:

1 2 2

' 2 1 2

2 2 1

X X

=

3

1

1 for =5

1

K

=

Lin. indep eigenvectors

nvvv ,,, 21 L

One single eigenvector

v

n

t

n

tt veXveXveX === ,,, 2211 L


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Another method to compute: Generalized eigenvectors



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Diff_Eq_7_2011 52

, , , repeated real eigenvalues K

KKKKKKKKKKKKKK

[ ] 10 vIA [ ] 21 vvIA

[ ] 32 vvIA

[ ] kk vvIA 1

{ }kvvv ,,, 21 L

Chain of generalized

eigenvectors

Compute:)( IA 2)( IA LL kIA )(

k

k

vIA 0)(

Solve:

Compute:

kk vIAv )(1 =

LL

32 )( vIAv =

21 )( vIAv =

{ }kvvv ,,, 21 L

Chain of

generalized

eigenvectors

:Example Find all generalized eigenvectors:

1,11,=

12 )( = kk vIAv

=

010

122

001

A




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Diff_Eq_7_2011 53

Compute:)( IA 2)( IA LL

kIA )(

k

kvIA 0)(

Solve:

Compute:

kk vIAv )(1 =

LL

32 )( vIAv =

21 )( vIAv =

{ }kvvv ,,, 21 LChain of

generalized

eigenvectors

:Example Find all generalized eigenvectors:

1,11,=

12 )( = kk vIAv

=

010

122

001

A

:Solution[ ]

=

0110

0112

0000

0IA

0110

0001

000023 RR+

22

1R

1 lin indep

eigenvector

Length of

chain =3{ }321 ,, vvv

=

110

112

000

)( IA

=

002

002

000

)( 2IA

=

000

000

000

)( 3IA

=00

1

3v

=

=

02

0

00

1

110112

000

2v

=

=

22

0

02

0

110112

000

1v


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Diff_Eq_7_2011 54

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

whose columns are x(1)(t),, x(n)(t), is a fundamental

matrix for the system x' = P(t)x. This matrix isnonsingular since its columns are linearly independent,

and hence det

0.

,

)()(

)()(

)()()1(

)(

1

)1(

1

=

txtx

txtx

tn

nn

n

L

MOM

L


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Diff_Eq_7_2011 55

Example 1: Consider the homogeneous equation x' = Ax below.

We found the following fundamental solutions for

this system:

Thus a fundamental matrix for this system is

xx

= 14

11

ttetet

=

=

2

1)(,

2

1)( )2(3)1( xx

=

tt

tt

ee

eet

22)(

3

3

Fundamental Matrices and General


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Diff_Eq_7_2011 56

Fundamental Matrices and General

Solution The general solution of x' = P(t)x

can be expressed x = (t)c, where c is a constantvector with components c1,, cn:

)()1(

1 )(

n

nctc xxx ++= L

==n

n

nn

n

c

c

txtx

txtx

t ML

MOM

L 1

)()1(

)(

1

)1(

1

)()(

)()(

)( c

x

Fundamental Matrix & Initial Value


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Diff_Eq_7_2011 57

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 wechoose c so as to satisfy x(t

0) = x0.

Recalling (t0) is nonsingular, it follows that

Thus our solution x = (t)c can be expressed as

0

0

10

0 )()( xcxc tt ==

0

0

1 )()( xx tt =

Nonhomogeneous Linear Systems


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Diff_Eq_7_2011 58

The general theory of a nonhomogeneous system of

equations

parallels that of a single nth order linear equation.

This system can be written as x' = P(t)x + g(t), where

)()()()(

)()()()(

)()()()(

2211

222221212

112121111

tgxtpxtpxtpx

tgxtpxtpxtpx

tgxtpxtpxtpx

nnnnnnn

nn

nn

++++=

++++=

++++=

K

M

K

K

=

=

=

)()()(

)()()(

)()()(

)(,

)(

)(

)(

)(,

)(

)(

)(

)(

21

22221

11211

2

1

2

1

tptptp

tptptp

tptptp

t

tg

tg

tg

t

tx

tx

tx

t

nnnn

n

n

nn L

MOMM

L

L

MM Pgx


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Variation of Parameters: Solution


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Diff_Eq_7_2011 60

We assume a particular solution of the form x =(t)u(t).

Substituting this into x' = P(t)x + g(t), we obtain

'(t)u(t) + (t)u'(t) = P(t)(t)u(t) + g(t) Since '(t) = P(t)(t), the above equation simplifies

to

u'(t) = -1(t)g(t)

Thus

where the vector c is an arbitrary constant of

integration.

cu += dttgtt )()()( 1

( ) arbitrary,,)()()()( 111

+= tdssstt

t

tgcx

Variation of Parameters: Initial Value


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Diff_Eq_7_2011 61

Problem For an initial value problem

x' = P(t)x + g(t),x(t0) = x

(0),

the general solution to x' = P(t)x + g(t)is

+=

t

tdsssttt

0

)()()()()( 1)0(01 gxx


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Diff_Eq_7_2011 62

Example :Variation of Parameters Consider

Fundamental matrix:

(t)u'(t) = g(t), or

2 1 2

1 2 3

te

t

= + x x

=

tt

tt

ee

eet3

3)(

=

t

e

u

u

ee

ee t

tt

tt

3

2

2

1

3

3

22 121

21 11

-1

-det

a aAa aA

=


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Diff_Eq_7_2011 63

Example 3: Solving for u(t) Solving (t)u'(t) = g(t) by row reduction,

It follows that

2 3

1

2

3 / 2

1 3 / 2

t t

t

u e te

u te

=

= +

++

++=

=

2

1

332

2

1

2/32/3

6/2/2/)(

cetet

cetee

u

ut

tt

ttt

u


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Diff_Eq_7_2011 64

Example 3: Solving for x(t) (3 of 3) Now x(t) = (t)u(t), and hence we multiply

to obtain, after collecting terms and simplifying,

Note that this is the same solution as in Example 1.

++++

=

2

1

332

3

3

2/32/36/2/2/

cetetcetee

eeee

tt

ttt

tt

tt

x

+

+

+

+

=

5

4

3

1

2

1

1

1

2

1

1

1

1

1

1

12

3

1 teteecec tttt

x


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Diff_Eq_7_2011 65

Example 3: Solving for x(t) (3 of 3)

t

egt

=

1 06 -1

A =

t

e

t3-3 1

2 -4

A =

2

4

4

4

t

t

eg

e

=

3 -1

-1 3A

=

1(0)

1x

=

Solve the nonhomogeneous system: X'=AX+g

(c)

(a)

(b)


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diff equation 6 sys 2011fall

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