Ch 7.1: Introduction to Systems of First Order Linear Equations

A system of simultaneous first order ordinary differential equations has the general form

where each xk is a function of t. If each Fk is a linear function of x1, x2, …, xn, then the system of equations is said to be linear, otherwise it is nonlinear. Systems of higher order differential equations can similarly be defined.

),,,(

),,,(),,,(

21

2122

2111

nnn

n

n

xxxtFx

xxxtFxxxxtFx

K

M

K

K

=′

=′=′

Example 1The motion of a spring-mass system from Section 3.8 was described by the equation

This second order equation can be converted into a system of first order equations by letting x1 = u and x2 = u'. Thus

or

0)(192)(16)( =+′+′′ tututu

019216 122

21

=++′=′

xxxxx

122

21

19216 xxxxx

−−=′=′

Nth Order ODEs and Linear 1st Order SystemsThe method illustrated in previous example can be used to transform an arbitrary nth order equation

into a system of n first order equations, first by defining

Then

( ))1()( ,,,,, −′′′= nn yyyytFy K

)1(321 ,,,, −=′′=′== n

n yxyxyxyx K

),,,( 21

1

32

21

nn

nn

xxxtFxxx

xxxx

K

M

=′=′

=′=′

−

Solutions of First Order SystemsA system of simultaneous first order ordinary differential equations has the general form

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

that are differentiable on I and satisfy the system of equations 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 φφφ === K

00

0202

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

Example 2The equation

can be written as system of first order equations by letting x1 = y and x2 = y'. Thus

A solution to this system is

which is a parametric descriptionfor the unit circle.

π20,0 <<=+′′ tyy

12

21

xxxx−=′

=′

π20),cos(),sin( 21 <<== ttxtx

Theorem 7.1.1Suppose F1,…, Fn and ∂F1/∂x1,…, ∂F1/∂xn,…, ∂Fn/∂ x1,…, ∂Fn/∂xn, are continuous in the region R of t x1 x2…xn-space defined by α < t < β, α1 < x1 < β1, …, αn < xn < βn, 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.

( )002

010 ,,,, nxxxt K

)(,),(),( 2211 txtxtx nn φφφ === K

),,,(

),,,(),,,(

21

2122

2111

nnn

n

n

xxxtFx

xxxtFxxxxtFx

K

M

K

K

=′

=′=′

Linear SystemsIf each Fk is a linear function of x1, x2, …, xn, then the system of equations has the general form

If each of the gk(t) is zero on I, then the system is homogeneous, otherwise it is nonhomogeneous.

)()()()(

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

2211

222221212

112121111

tgxtpxtpxtpx

tgxtpxtpxtpxtgxtpxtpxtpx

nnnnnnn

nn

nn

++++=′

++++=′++++=′

K

M

K

K

Theorem 7.1.2Suppose p11, p12,…, pnn, g1,…, gn are continuous on an interval I: α < t < β with t0 in I, and let

prescribe the initial conditions. Then there exists a unique solution

that satisfies the IVP, and exists throughout I.

002

01 ,,, nxxx K

)(,),(),( 2211 txtxtx nn φφφ === K

)()()()(

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

2211

222221212

112121111

tgxtpxtpxtpx

tgxtpxtpxtpxtgxtpxtpxtpx

nnnnnnn

nn

nn

++++=′

++++=′++++=′

K

M

K

K

Ch 7.2: Review of MatricesFor theoretical and computation reasons, we review results of matrix theory in this section and the next. A matrix A is an m x n rectangular array of elements, arranged in m rows and n columns, denoted

Some examples of 2 x 2 matrices are given below:

( )⎟⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜⎜

⎝

⎛

==

mnmm

n

n

ji

aaa

aaaaaa

a

L

MOMM

L

L

21

22221

11211

A

⎟⎟⎠

⎞⎜⎜⎝

⎛−+−

=⎟⎟⎠

⎞⎜⎜⎝

⎛=⎟⎟

⎠

⎞⎜⎜⎝

⎛=

iii

CB7654231

,4231

,4321

A

TransposeThe transpose of A = (aij) is AT = (aji).

For example,

⎟⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜⎜

⎝

⎛

=⇒

⎟⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜⎜

⎝

⎛

=

mnnn

m

m

T

mnmm

n

n

aaa

aaaaaa

aaa

aaaaaa

L

MOMM

L

L

L

MOMM

L

L

21

22212

12111

21

22221

11211

AA

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛=⇒⎟⎟

⎠

⎞⎜⎜⎝

⎛=⎟⎟

⎠

⎞⎜⎜⎝

⎛=⇒⎟⎟

⎠

⎞⎜⎜⎝

⎛=

635241

654321

,4231

4321 TT BBAA

ConjugateThe conjugate of A = (aij) is A = (aij).

For example,

⎟⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜⎜

⎝

⎛

=⇒

⎟⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜⎜

⎝

⎛

=

mnmm

n

n

mnmm

n

n

aaa

aaaaaa

aaa

aaaaaa

L

MOMM

L

L

L

MOMM

L

L

21

22221

11211

21

22221

11211

AA

⎟⎟⎠

⎞⎜⎜⎝

⎛+

−=⇒⎟⎟

⎠

⎞⎜⎜⎝

⎛−

+=

443321

443321

ii

ii

AA

AdjointThe adjoint of A is AT , and is denoted by A*

For example,

⎟⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜⎜

⎝

⎛

=⇒

⎟⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜⎜

⎝

⎛

=

mnnn

m

m

mnmm

n

n

aaa

aaaaaa

aaa

aaaaaa

L

MOMM

L

L

L

MOMM

L

L

21

22212

12111

*

21

22221

11211

AA

⎟⎟⎠

⎞⎜⎜⎝

⎛−

+=⇒⎟⎟

⎠

⎞⎜⎜⎝

⎛−

+=

432431

443321 *

ii

ii

AA

Square MatricesA square matrix A has the same number of rows and columns. That is, A is n x n. In this case, A is said to have order n.

For example,

⎟⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜⎜

⎝

⎛

=

nnnn

n

n

aaa

aaaaaa

L

MOMM

L

L

21

22221

11211

A

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛=⎟⎟

⎠

⎞⎜⎜⎝

⎛=

987654321

,4321

BA

VectorsA column vector x is an n x 1 matrix. For example,

A row vector x is a 1 x n matrix. For example,

Note here that y = xT, and that in general, if x is a column vector x, then xT is a row vector.

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛=

321

x

( )321=y

The Zero MatrixThe zero matrix is defined to be 0 = (0), whose dimensions depend on the context. For example,

K,000000

,000000

,0000

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛=⎟⎟

⎠

⎞⎜⎜⎝

⎛=⎟⎟

⎠

⎞⎜⎜⎝

⎛= 000

Matrix EqualityTwo matrices A = (aij) and B = (bij) are equal if aij = bij for all i and j. For example,

BABA =⇒⎟⎟⎠

⎞⎜⎜⎝

⎛=⎟⎟

⎠

⎞⎜⎜⎝

⎛=

4321

,4321

Matrix – Scalar MultiplicationThe product of a matrix A = (aij) and a constant k is defined to be kA = (kaij). For example,

⎟⎟⎠

⎞⎜⎜⎝

⎛−−−−−−

=−⇒⎟⎟⎠

⎞⎜⎜⎝

⎛=

30252015105

5654321

AA

Matrix Addition and SubtractionThe sum of two m x n matrices A = (aij) and B = (bij) is defined to be A + B = (aij + bij). For example,

The difference of two m x n matrices A = (aij) and B = (bij) is defined to be A - B = (aij - bij). For example,

⎟⎟⎠

⎞⎜⎜⎝

⎛=+⇒⎟⎟

⎠

⎞⎜⎜⎝

⎛=⎟⎟

⎠

⎞⎜⎜⎝

⎛=

121086

8765

,4321

BABA

⎟⎟⎠

⎞⎜⎜⎝

⎛−−−−

=−⇒⎟⎟⎠

⎞⎜⎜⎝

⎛=⎟⎟

⎠

⎞⎜⎜⎝

⎛=

4444

8765

,4321

BABA

Matrix MultiplicationThe product of an m x n matrix A = (aij) and an n x rmatrix B = (bij) is defined to be the matrix C = (cij), where

Examples (note AB does not necessarily equal BA):

∑=

=n

kkjikij bac

1

⎟⎟⎠

⎞⎜⎜⎝

⎛=⎟⎟

⎠

⎞⎜⎜⎝

⎛−+++−+++

=⇒⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛

−=⎟⎟

⎠

⎞⎜⎜⎝

⎛=

⎟⎟⎠

⎞⎜⎜⎝

⎛=⎟⎟

⎠

⎞⎜⎜⎝

⎛++++

=⇒

⎟⎟⎠

⎞⎜⎜⎝

⎛=⎟⎟

⎠

⎞⎜⎜⎝

⎛++++

=⇒⎟⎟⎠

⎞⎜⎜⎝

⎛=⎟⎟

⎠

⎞⎜⎜⎝

⎛=

41715

61000512340023

102103

,654321

20141410

16412212291

2511115

169838341

4231

,4321

CDDC

BA

ABBA

Vector MultiplicationThe dot product of two n x 1 vectors x & y is defined as

The inner product of two n x 1 vectors x & y is defined as

Example:

∑=

=n

kji

T yx1

yx

( ) ∑=

==n

kji

T yx,1

yxyx

( ) iiii,

iiiiii

iT

T

2118)55)(3()32)(2()1)(1(

912)55)(3()32)(2()1)(1(55321

,321

+=−+++−==⇒

+−=++−+−=⇒⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛

+−−

=⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛=

yxyx

yxyx

Vector LengthThe length of an n x 1 vector x is defined as

Note here that we have used the fact that if x = a + bi, then

Example:

( )2/1

1

22/1

1

2/1 || ⎥⎦

⎤⎢⎣

⎡=⎥

⎦

⎤⎢⎣

⎡== ∑∑

==

n

kk

n

kkk xxx,xxx

( )

( ) 3016941

)43)(43()2)(2()1)(1(43

21

2/1

=+++=

−+++==⇒⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛

+= ii,

ixxxx

( )( ) 222 xbabiabiaxx =+=−+=⋅

OrthogonalityTwo n x 1 vectors x & y are orthogonal if (x,y) = 0. Example:

( ) 0)1)(3()4)(2()11)(1(14

11

321

=−+−+=⇒⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛

−−=

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛= yxyx ,

Identity MatrixThe multiplicative identity matrix I is an n x n matrix given by

For any square matrix A, it follows that AI = IA = A. The dimensions of I depend on the context. For example,

⎟⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜⎜

⎝

⎛

=

100

010001

L

MOMM

L

L

I

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛=

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛=⎟⎟

⎠

⎞⎜⎜⎝

⎛=⎟⎟

⎠

⎞⎜⎜⎝

⎛⎟⎟⎠

⎞⎜⎜⎝

⎛=

987654321

987654321

100010001

,4321

1001

4321

IBAI

Inverse MatrixA square matrix A is nonsingular, or invertible, if there exists a matrix B such that that AB = BA = I. Otherwise Ais singular. The matrix B, if it exists, is unique and is denoted by A-1

and is called the inverse of A. It turns out that A-1 exists iff detA ≠ 0, and A-1 can be found using row reduction (also called Gaussian elimination) on the augmented matrix (A|I), see example on next slide. The three elementary row operations:

Interchange two rows.Multiply a row by a nonzero scalar.Add a multiple of one row to another row.

Example: Finding the Identity Matrix (1 of 2)

Use row reduction to find the inverse of the matrix A below, if it exists.

Solution: If possible, use elementary row operations to reduce (A|I),

such that the left side is the identity matrix, for then the right side will be A-1. (See next slide.)

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛

−=

834301210

A

( ) ,100834010301001210

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛

−=IA

Example: Finding the Identity Matrix (2 of 2)

Thus

( )

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛

−−−

−−→

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛

−−−→

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛

−→

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛

−−−→

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛

−→

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛

−=

2/122/31001420102/372/9001

143200142010010301

143200001210010301

140430001210010301

100834001210010301

100834010301001210

IA

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛

−−−

−−=−

2/122/31422/372/9

1A

Matrix FunctionsThe 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

tatatatatata

t

tx

txtx

t

mnmm

n

n

m L

MOMM

L

L

MAx

⎟⎠⎞⎜

⎝⎛=⎟⎟

⎠

⎞⎜⎜⎝

⎛= ∫∫

b

a ij

b

a

ij dttadttdt

dadtd )()(, AA

Example & Differentiation RulesExample:

Many of the rules from calculus apply in this setting. For example: ( )

( )

( )⎟⎠⎞

⎜⎝⎛+⎟

⎠⎞

⎜⎝⎛=

+=+

=

dtd

dtd

dtd

dtd

dtd

dtd

dtd

dtd

BABAAB

BABA

CACCA matrixconstant a is where,

⎟⎟⎠

⎞⎜⎜⎝

⎛

−=⇒

⎟⎟⎠

⎞⎜⎜⎝

⎛−

=⇒⎟⎟⎠

⎞⎜⎜⎝

⎛=

∫ πππ

410

)(

,0sin

cos64cos

sin3)(

3

0

2

dtt

ttt

dtd

ttt

t

A

AA

Ch 7.3: Systems of Linear Equations, Linear Independence, Eigenvalues

A system of n linear equations in n variables,

can be expressed as a matrix equation Ax = b:

If b = 0, then system is homogeneous; otherwise it is nonhomogeneous.

⎟⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜⎜

⎝

⎛

=

⎟⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜⎜

⎝

⎛

⎟⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜⎜

⎝

⎛

nnnnnn

n

n

b

bb

x

xx

aaa

aaaaaa

MM

L

MOMM

L

L

2

1

2

1

,2,1,

,22,21,2

,12,11,1

,,22,11,

2,222,211,2

1,122,111,1

nnnnnn

nn

nn

bxaxaxa

bxaxaxabxaxaxa

=+++

=+++

=+++

L

M

L

L

Nonsingular CaseIf the coefficient matrix A is nonsingular, then it is invertible and we can solve Ax = b as follows:

This solution is therefore unique. Also, if b = 0, it follows that the unique solution to Ax = 0 is x = A-10 = 0. Thus if A is nonsingular, then the only solution to Ax = 0 is the trivial solution x = 0.

bAxbAIxbAAxAbAx 1111 −−−− =⇒=⇒=⇒=

Example 1: Nonsingular Case (1 of 3)

From a previous example, we know that the matrix A below is nonsingular with inverse as given.

Using the definition of matrix multiplication, it follows that the only solution of Ax = 0 is x = 0:

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛

−−−

−−=

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛

−= −

2/122/31422/372/9

,834301210

1AA

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛=

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛

−−−

−−== −

000

000

2/122/31422/372/9

10Ax

Example 1: Nonsingular Case (2 of 3)

Now let’s solve the nonhomogeneous linear system Ax = bbelow using A-1:

This system of equations can be written as Ax = b, where

Then

08342301220

321

321

321

=+−−=++

=++

xxxxxxxxx

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛−−

=⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛−

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛

−−−

−−== −

71223

022

2/122/31422/372/9

1bAx

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛−=

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛=

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛

−=

022

,,834301210

3

2

1

bxAxxx

Example 1: Nonsingular Case (3 of 3)

Alternatively, we could solve the nonhomogeneous linear system Ax = b below using row reduction.

To do so, form the augmented matrix (A|b) and reduce, using elementary row operations.

( )

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛−−

=→==+−=+

→⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛ −→

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛ −→

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛

−−

−→

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛

−

−→

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛

−−=

71223

72223

710022102301

1420022102301

843022102301

083422102301

083423012210

3

32

31

x

bA

xxxxx

08342301220

321

321

321

=+−−=++

=++

xxxxxxxxx

Singular CaseIf the coefficient matrix A is singular, then A-1 does not exist, and either a solution to Ax = b does not exist, or there is more than one solution (not unique). Further, the homogeneous system Ax = 0 has more than one solution. That is, in addition to the trivial solution x = 0, there are infinitely many nontrivial solutions.The nonhomogeneous case Ax = b has no solution unless (b, y) = 0, for all vectors y satisfying A*y = 0, where A* is the adjoint of A. In this case, Ax = b has solutions (infinitely many), each of the form x = x(0) + ξ, where x(0) is a particular solution of Ax = b, and ξ is any solution of Ax = 0.

Example 2: Singular Case (1 of 3)

Solve the nonhomogeneous linear system Ax = b below using row reduction.

To do so, form the augmented matrix (A|b) and reduce, using elementary row operations.

( )

soln no1015312

100015301121

400015301121

353015301121

6106015301121

154506511121

3

32

321

→==+=−−

→⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛ −−→

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛

−

−−→

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛

−

−−→

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛

−

−−→

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛

−−−

−−=

xxxxxx

bA

154506511121

321

321

321

−=+−=++−=−−

xxxxxxxxx

Example 2: Singular Case (2 of 3)

Solve the nonhomogeneous linear system Ax = b below using row reduction.

Reduce the augmented matrix (A|b) as follows:

( )

072

27

21000

530121

25

21530

530121

51060530121

545651121

123

123

12

1

13

12

1

13

12

1

3

2

1

=−−→

⎟⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜⎜

⎝

⎛

−−

+−−

→

⎟⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜⎜

⎝

⎛

−

+−−

→

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛

−+

−−→

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛

−−

−−=

bbb

bbb

bbb

bb

bbb

bbbbb

bbb

bA

3321

2321

1321

545651121

bxxxbxxxbxxx

=+−=++−=−−

Example 2: Singular Case (3 of 3)

From the previous slide, we require

Suppose

Then the reduced augmented matrix (A|b) becomes:

ξxxxx +=⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛

−+⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛==→

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛−−

+⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛=→

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛−−

=→

==+=−−

→

⎟⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜⎜

⎝

⎛

−−

+−−

)0(3

3

3

3

3

32

321

123

12

1

357

001

13/53/7

001

3/53/71

00053112

27

21000

530121

cxx

xx

xxxxxx

bbb

bbb

072 123 =−− bbb

5,1,1 321 =−== bbb

Linear Dependence and IndependenceA set of vectors x(1), x(2),…, x(n) is linearly dependent if there exists scalars c1, c2,…, cn, not all zero, such that

If the only solution of

is c1= c2 = …= cn = 0, then x(1), x(2),…, x(n) is linearly independent.

0xxx =+++ )()2(2

)1(1

nnccc L

0xxx =+++ )()2(2

)1(1

nnccc L

Example 3: Linear Independence (1 of 2)

Determine whether the following vectors are linear dependent or linearly independent.

We need to solve

or

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛=

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛

−⇔

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛=

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛+⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛

−+⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛

000

834301210

000

832

301

410

3

2

1

21

ccc

ccc

0xxx =++ )3(3

)2(2

)1(1 ccc

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛=

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛

−=

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛=

832

,301

,410

)3()2()1( xxx

Example 3: Linear Independence (2 of 2)

We thus reduce the augmented matrix (A|b), as before.

Thus the only solution is c1= c2 = …= cn = 0, and therefore the original vectors are linearly independent.

( )

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛=→

==+=+

→

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛→

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛

−=

000

00203

010002100301

083403010210

3

32

31

c

bA

ccccc

Example 4: Linear Dependence (1 of 2)

Determine whether the following vectors are linear dependent or linearly independent.

We need to solve

or

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛−=

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛

−

−=

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛−=

561

,452

,511

)3()2()1( xxx

0xxx =++ )3(3

)2(2

)1(1 ccc

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛=

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛

−−

−−⇔

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛=

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛−+⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛

−

−+⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛−

000

545651121

000

561

452

511

3

2

1

321

ccc

ccc

Example 4: Linear Dependence (2 of 2)

We thus reduce the augmented matrix (A|b), as before.

Thus the original vectors are linearly dependent, with

( )

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛

−=→

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛−−

=→==+=−−

→

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛ −−→

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛

−−

−−=

357

3/53/7

00053012

000005300121

054506510121

3

3

3

3

32

321

kc

cc

cccccc

cc

bA

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛=

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛−−⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛

−

−+⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛−

000

561

3452

5511

7

Linear Independence and InvertibilityConsider the previous two examples:

The first matrix was known to be nonsingular, and its column vectors were linearly independent. The second matrix was known to be singular, and its column vectors were linearly dependent.

This is true in general: the columns (or rows) of A are linearly independent iff A is nonsingular iff A-1 exists.Also, A is nonsingular iff detA ≠ 0, hence columns (or rows) of A are linearly independent iff detA ≠ 0.Further, if A = BC, then det(C) = det(A)det(B). Thus if the columns (or rows) of A and B are linearly independent, then the columns (or rows) of C are also.

Linear Dependence & Vector FunctionsNow consider vector functions x(1)(t), x(2)(t),…, x(n)(t), where

As before, x(1)(t), x(2)(t),…, x(n)(t) is linearly dependent on I if there exists scalars c1, c2,…, cn, not all zero, such that

Otherwise x(1)(t), x(2)(t),…, x(n)(t) is linearly independent on ISee text for more discussion on this.

( ) ( )βα ,,,,2,1,

)(

)()(

)(

)(

)(2

)(1

=∈=

⎟⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜⎜

⎝

⎛

= Itnk

tx

txtx

t

km

k

k

k KM

x

Ittctctc nn ∈=+++ allfor ,)()()( )()2(

2)1(

1 0xxx L

Eigenvalues and EigenvectorsThe eqn. Ax = y can be viewed as a linear transformation that maps (or transforms) x into a new vector y. Nonzero vectors x that transform into multiples of themselves are important in many applications. Thus we solve Ax = λx or equivalently, (A-λI)x = 0. This equation has a nonzero solution if we choose λ such that det(A-λI) = 0. Such values of λ are called eigenvalues of A, and the nonzero solutions x are called eigenvectors.

Example 5: Eigenvalues (1 of 3)

Find the eigenvalues and eigenvectors of the matrix A.

Solution: Choose λ such that det(A-λI) = 0, as follows.

⎟⎟⎠

⎞⎜⎜⎝

⎛−

=6332

A

( )

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

7,373214

336263

32det

0001

6332

detdet

2

−==⇒+−=−+=

−−−−=

⎟⎟⎠

⎞⎜⎜⎝

⎛−−

−=

⎟⎟⎠

⎞⎜⎜⎝

⎛⎟⎟⎠

⎞⎜⎜⎝

⎛−⎟⎟⎠

⎞⎜⎜⎝

⎛−

=−

λλλλλλ

λλλ

λ

λλIA

Example 5: First Eigenvector (2 of 3)

To find the eigenvectors of the matrix A, we need to solve (A-λI)x = 0 for λ = 3 and λ = -7. Eigenvector for λ = 3: Solve

by row reducing the augmented matrix:

( ) ⎟⎟⎠

⎞⎜⎜⎝

⎛=⎟⎟

⎠

⎞⎜⎜⎝

⎛⎟⎟⎠

⎞⎜⎜⎝

⎛−

−⇔⎟⎟

⎠

⎞⎜⎜⎝

⎛=⎟⎟

⎠

⎞⎜⎜⎝

⎛⎟⎟⎠

⎞⎜⎜⎝

⎛−−

−⇔=−

00

9331

00

363332

2

1

2

1

xx

xx

0xIA λ

⎟⎟⎠

⎞⎜⎜⎝

⎛=→⎟⎟

⎠

⎞⎜⎜⎝

⎛=⎟⎟

⎠

⎞⎜⎜⎝

⎛=→

==−

→⎟⎟⎠

⎞⎜⎜⎝

⎛ −→⎟⎟

⎠

⎞⎜⎜⎝

⎛−−

→⎟⎟⎠

⎞⎜⎜⎝

⎛−

−

13

choosearbitrary,133

00031

000031

093031

093031

)1(

2

2)1(

2

21

xx ccxx

xxx

Example 5: Second Eigenvector (3 of 3)

Eigenvector for λ = -7: Solve

by row reducing the augmented matrix:

( ) ⎟⎟⎠

⎞⎜⎜⎝

⎛=⎟⎟

⎠

⎞⎜⎜⎝

⎛⎟⎟⎠

⎞⎜⎜⎝

⎛⇔⎟⎟

⎠

⎞⎜⎜⎝

⎛=⎟⎟

⎠

⎞⎜⎜⎝

⎛⎟⎟⎠

⎞⎜⎜⎝

⎛+−

+⇔=−

00

1339

00

763372

2

1

2

1

xx

xx

0xIA λ

⎟⎟⎠

⎞⎜⎜⎝

⎛−=→⎟⎟

⎠

⎞⎜⎜⎝

⎛−=⎟⎟

⎠

⎞⎜⎜⎝

⎛−=→

==+

→⎟⎟⎠

⎞⎜⎜⎝

⎛→⎟⎟

⎠

⎞⎜⎜⎝

⎛→⎟⎟

⎠

⎞⎜⎜⎝

⎛

31

choosearbitrary,1

3/13/1

0003/11

00003/11

01303/11

013039

)2(

2

2)2(

2

21

xx ccxx

xxx

Normalized EigenvectorsFrom the previous example, we see that eigenvectors are determined up to a nonzero multiplicative constant. If this constant is specified in some particular way, then the eigenvector is said to be normalized. For example, eigenvectors are sometimes normalized by choosing the constant so that ||x|| = (x, x)½ = 1.

Algebraic and Geometric MultiplicityIn finding the eigenvalues λ of an n x n matrix A, we solvedet(A-λI) = 0. Since this involves finding the determinant of an n x nmatrix, the problem reduces to finding roots of an nth degree polynomial. Denote these roots, or eigenvalues, by λ1, λ2, …, λn. If an eigenvalue is repeated m times, then its algebraic multiplicity is m. Each eigenvalue has at least one eigenvector, and a eigenvalue of algebraic multiplicity m may have q linearly independent eigevectors, 1 ≤ q ≤ m, and q is called the geometric multiplicity of the eigenvalue.

Eigenvectors and Linear IndependenceIf an eigenvalue λ has algebraic multiplicity 1, then it is said to be simple, and the geometric multiplicity is 1 also. If each eigenvalue of an n x n matrix A is simple, then Ahas n distinct eigenvalues. It can be shown that the neigenvectors corresponding to these eigenvalues are linearly independent. If an eigenvalue has one or more repeated eigenvalues, then there may be fewer than n linearly independent eigenvectors since for each repeated eigenvalue, we may have q < m. This may lead to complications in solving systems of differential equations.

Example 6: Eigenvalues (1 of 5)

Find the eigenvalues and eigenvectors of the matrix A.

Solution: Choose λ such that det(A-λI) = 0, as follows.

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛=

011101110

A

( )

1,1,2)1)(2(

23

111111

detdet

221

2

3

−=−==⇒+−=

++−=

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛

−−

−=−

λλλλλ

λλ

λλ

λλIA

Example 6: First Eigenvector (2 of 5)

Eigenvector for λ = 2: Solve (A-λI)x = 0, as follows.

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛=→

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛=

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛=→

==−=−

→⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛−−

→⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛−−

→

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛

−−

−→

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛

−−

−→

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛

−−

−

111

choosearbitrary,111

00011011

000001100101

000001100211

033003300211

011201210211

021101210112

)1(

3

3

3)1(

3

32

31

xx ccxxx

xxxxx

Example 6: 2nd and 3rd Eigenvectors (3 of 5)

Eigenvector for λ = -1: Solve (A-λI)x = 0, as follows.

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛

−=

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛

−=→

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛−+⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛−=

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛ −−=→

===++

→⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛→

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛

110

, 101

choose

arbitrary,where,101

011

00000111

000000000111

011101110111

)3()2(

3232

3

2

32)2(

3

2

321

xx

x xxxxxx

xx

xx

xxx

Example 6: Eigenvectors of A (4 of 5)

Thus three eigenvectors of A are

where x(2), x(3) correspond to the double eigenvalue λ = - 1.It can be shown that x(1), x(2), x(3) are linearly independent. Hence A is a 3 x 3 symmetric matrix (A = AT ) with 3 real eigenvalues and 3 linearly independent eigenvectors.

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛

−=

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛

−=

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛=

110

, 101

,111

)3()2()1( xxx

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛=

011101110

A

Example 6: Eigenvectors of A (5 of 5)

Note that we could have we had chosen

Then the eigenvectors are orthogonal, since

Thus A is a 3 x 3 symmetric matrix with 3 real eigenvalues and 3 linearly independent orthogonal eigenvectors.

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛−=

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛

−=

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛=

121

, 101

,111

)3()2()1( xxx

( ) ( ) ( ) 0,,0,,0, )3()2()3()1()2()1( === xxxxxx

Hermitian MatricesA self-adjoint, or Hermitian matrix, satisfies A = A*, where we recall that A* = AT . Thus for a Hermitian matrix, aij = aji. Note that if A has real entries and is symmetric (see last example), then A is Hermitian. An n x n Hermitian matrix A has the following properties:

All eigenvalues of A are real.There exists a full set of n linearly independent eigenvectors of A.If x(1) and x(2) are eigenvectors that correspond to different eigenvalues of A, then x(1) and x(2) are orthogonal. Corresponding to an eigenvalue of algebraic multiplicity m, it is possible to choose m mutually orthogonal eigenvectors, and hence A has a full set of n linearly independent orthogonal eigenvectors.

Ch 7.4: Basic Theory of Systems of First Order Linear EquationsThe general theory of a system of n first order linear 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

tgxtpxtpxtpxtgxtpxtpxtpx

nnnnnnn

nn

nn

++++=′

++++=′++++=′

K

M

K

K

⎟⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜⎜

⎝

⎛

=

⎟⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜⎜

⎝

⎛

=

⎟⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜⎜

⎝

⎛

=

)()()(

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

)(,

)(

)()(

)(,

)(

)()(

)(

21

22221

11211

2

1

2

1

tptptp

tptptptptptp

t

tg

tgtg

t

tx

txtx

t

nnnn

n

n

nn L

MOMM

L

L

MMPgx

Vector Solutions of an ODE SystemA vector x = φ(t) is a solution of x' = P(t)x + g(t) if the components of x,

satisfy the system of equations on I: α < t < β.

For comparison, recall that x' = P(t)x + g(t) represents our system of equations

Assuming P and g continuous on I, such a solution exists by Theorem 7.1.2.

),(,),(),( 2211 txtxtx nn φφφ === K

)()()()(

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

2211

222221212

112121111

tgxtpxtpxtpx

tgxtpxtpxtpxtgxtpxtpxtpx

nnnnnnn

nn

nn

++++=′

++++=′++++=′

K

M

K

K

Example 1Consider the homogeneous equation x' = P(t)x below, with the solutions x as indicated.

To see that x is a solution, substitute x into the equation and perform the indicated operations:

tt

t

eee

t 33

3

21

2)(;

1411

⎟⎟⎠

⎞⎜⎜⎝

⎛=⎟⎟

⎠

⎞⎜⎜⎝

⎛=⎟⎟

⎠

⎞⎜⎜⎝

⎛=′ xxx

xx ′=⎟⎟⎠

⎞⎜⎜⎝

⎛=⎟⎟

⎠

⎞⎜⎜⎝

⎛=⎟⎟

⎠

⎞⎜⎜⎝

⎛⎟⎟⎠

⎞⎜⎜⎝

⎛=⎟⎟

⎠

⎞⎜⎜⎝

⎛t

t

t

t

t

t

ee

ee

ee

3

3

3

3

3

3

23

63

21411

1411

Homogeneous Case; Vector Function NotationAs in Chapters 3 and 4, we first examine the general homogeneous equation x' = P(t)x.Also, the following notation for the vector functions x(1), x(2),…, x(k),… will be used:

KM

KMM

,

)(

)()(

)(,,

)(

)()(

)(,

)(

)()(

)( 2

1

)(

2

22

12

)2(

1

21

11

)1(

⎟⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜⎜

⎝

⎛

=

⎟⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜⎜

⎝

⎛

=

⎟⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜⎜

⎝

⎛

=

tx

txtx

t

tx

txtx

t

tx

txtx

t

nn

n

n

k

nn

xxx

Theorem 7.4.1If the vector functions x(1) and x(2) are solutions of the system x' = P(t)x, then the linear combination c1x(1) + c2x(2) is also a solution for any constants c1 and c2.

Note: By repeatedly applying the result of this theorem, it can be seen that every finite linear combination

of solutions x(1), x(2),…, x(k) is itself a solution to x' = P(t)x. )()()( )()2(

2)1(

1 tctctc kk xxxx +++= K

Example 2Consider the homogeneous equation x' = P(t)x below, with the two solutions x(1) and x(2) as indicated.

Then x = c1x(1) + c2x(2) is also a solution:

⎟⎟⎠

⎞⎜⎜⎝

⎛

−=⎟⎟

⎠

⎞⎜⎜⎝

⎛=⎟⎟

⎠

⎞⎜⎜⎝

⎛=′

−

−

t

t

t

t

ee

tee

t2

)(,2

)(;1411 )2(

3

3)1( xxxx

x

x

′=⎟⎟⎠

⎞⎜⎜⎝

⎛−+⎟⎟⎠

⎞⎜⎜⎝

⎛=

⎟⎟⎠

⎞⎜⎜⎝

⎛−+⎟⎟⎠

⎞⎜⎜⎝

⎛=

⎟⎟⎠

⎞⎜⎜⎝

⎛

−⎟⎟⎠

⎞⎜⎜⎝

⎛+⎟⎟⎠

⎞⎜⎜⎝

⎛⎟⎟⎠

⎞⎜⎜⎝

⎛=⎟⎟

⎠

⎞⎜⎜⎝

⎛

−

−

−

−

−

−

t

t

t

t

t

t

t

t

t

t

t

t

ee

cee

c

ecec

ecec

ecec

ecec

263

263

21411

21411

1411

23

3

1

2

23

1

31

2

23

1

31

Theorem 7.4.2If x(1), x(2),…, x(n) are linearly independent solutions of the system x' = P(t)x for each point in I: α < t < β, then each solution x = φ(t) can be expressed uniquely in the form

If solutions x(1),…, x(n) are linearly independent for each point in I: α < t < β, then they are fundamental solutions on I, and the general solution is given by

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

)1(1 tctctc n

nxxxx +++= K

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

)1(1 tctctc n

nxxxx +++= K

The Wronskian and Linear IndependenceThe proof of Thm 7.4.2 uses the fact that if x(1), x(2),…, x(n)

are linearly independent on I, then detX(t) ≠ 0 on I, where

The Wronskian of x(1),…, x(n) is defined as W[x(1),…, x(n)](t) = detX(t).

It follows that W[x(1),…, x(n)](t) ≠ 0 on I iff x(1),…, x(n) are linearly independent for each point in I.

,)()(

)()()(

1

111

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛=

txtx

txtxt

nnn

n

L

MOM

L

X

Theorem 7.4.3If x(1), x(2),…, x(n) are solutions of the system x' = P(t)x on I: α < t < β, then the Wronskian W[x(1),…, x(n)](t) is either identically zero on I or else is never zero on I.

This result enables us to determine whether a given set of solutions x(1), x(2),…, x(n) are fundamental solutions byevaluating W[x(1),…, x(n)](t) at any point t in α < t < β.

Theorem 7.4.4Let

Let x(1), x(2),…, x(n) be solutions of the system x' = P(t)x,α < t < β, that satisfy the initial conditions

respectively, where t0 is any point in α < t < β. Thenx(1), x(2),…, x(n) are fundamental solutions of x' = P(t)x.

⎟⎟⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜⎜⎜

⎝

⎛

=

⎟⎟⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜⎜⎜

⎝

⎛

=

⎟⎟⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜⎜⎜

⎝

⎛

=

10

00

,,

0

010

,

0

001

)()2()1( MK

MM

neee

,)(,,)( )(0

)()1(0

)1( nn tt exex == K

Ch 7.5: Homogeneous Linear Systems with Constant CoefficientsWe consider here a homogeneous system of n first order linear equations with constant, real coefficients:

This system can be written as x' = Ax, wherennnnnn

nn

nn

xaxaxax

xaxaxaxxaxaxax

+++=′

+++=′+++=′

K

M

K

K

2211

22221212

12121111

⎟⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜⎜

⎝

⎛

=

⎟⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜⎜

⎝

⎛

=

nnnn

n

n

m aaa

aaaaaa

tx

txtx

t

L

MOMM

L

L

M

21

22221

11211

2

1

,

)(

)()(

)( Ax

Equilibrium SolutionsNote that if n = 1, then the system reduces to

Recall that x = 0 is the only equilibrium solution if a ≠ 0. Further, x = 0 is an asymptotically stable solution if a < 0, since other solutions approach x = 0 in this case. Also, x = 0 is an unstable solution if a > 0, since other solutions depart from x = 0 in this case. For n > 1, equilibrium solutions are similarly found by solving Ax = 0. We assume detA ≠ 0, so that x = 0 is the only solution. Determining whether x = 0 is asymptotically stable or unstable is an important question here as well.

atetxaxx =⇒=′ )(

Phase PlaneWhen n = 2, then the system reduces to

This case can be visualized in the x1x2-plane, which is called the phase plane. In the phase plane, a direction field can be obtained by evaluating Ax at many points and plotting the resulting vectors, which will be tangent to solution vectors. A plot that shows representative solution trajectories is called a phase portrait. Examples of phase planes, directions fields and phase portraits will be given later in this section.

2221212

2121111

xaxaxxaxax

+=′+=′

Solving Homogeneous SystemTo construct a general solution to x' = Ax, assume a solution of the form x = ξert, where the exponent r and the constant vector ξ are to be determined. Substituting x = ξert into x' = Ax, we obtain

Thus to solve the homogeneous system of differential equations x' = Ax, we must find the eigenvalues and eigenvectors of A.Therefore x = ξert is a solution of x' = Ax provided that r is an eigenvalue and ξ is an eigenvector of the coefficient matrix A.

( ) 0ξIAAξξAξξ =−⇔=⇔= rreer rtrt

Example 1: Direction Field (1 of 9)

Consider the homogeneous equation x' = Ax below.

A direction field for this system is given below.Substituting x = ξert in for x, and rewriting system as (A-rI)ξ = 0, we obtain

xx ⎟⎟⎠

⎞⎜⎜⎝

⎛=′

1411

⎟⎟⎠

⎞⎜⎜⎝

⎛=⎟⎟

⎠

⎞⎜⎜⎝

⎛⎟⎟⎠

⎞⎜⎜⎝

⎛−

−00

1411

1

1

ξξ

rr

Example 1: Eigenvalues (2 of 9)

Our solution has the form x = ξert, where r and ξ are found by solving

Recalling that this is an eigenvalue problem, we determine rby solving det(A-rI) = 0:

Thus r1 = 3 and r2 = -1.

⎟⎟⎠

⎞⎜⎜⎝

⎛=⎟⎟

⎠

⎞⎜⎜⎝

⎛⎟⎟⎠

⎞⎜⎜⎝

⎛−

−00

1411

1

1

ξξ

rr

)1)(3(324)1(14

11 22 +−=−−=−−=−

−rrrrr

rr

Example 1: First Eigenvector (3 of 9)

Eigenvector for r1 = 3: Solve

by row reducing the augmented matrix:

( ) ⎟⎟⎠

⎞⎜⎜⎝

⎛=⎟⎟

⎠

⎞⎜⎜⎝

⎛⎟⎟⎠

⎞⎜⎜⎝

⎛−

−⇔⎟⎟

⎠

⎞⎜⎜⎝

⎛=⎟⎟

⎠

⎞⎜⎜⎝

⎛⎟⎟⎠

⎞⎜⎜⎝

⎛−

−⇔=−

00

2412

00

314131

2

1

2

1

ξξ

ξξ

0ξIA r

⎟⎟⎠

⎞⎜⎜⎝

⎛=→⎟⎟

⎠

⎞⎜⎜⎝

⎛=⎟⎟

⎠

⎞⎜⎜⎝

⎛=→

==−

→⎟⎟⎠

⎞⎜⎜⎝

⎛ −→⎟⎟

⎠

⎞⎜⎜⎝

⎛−

−→⎟⎟

⎠

⎞⎜⎜⎝

⎛−

−

21

choosearbitrary,12/12/1

0002/11

00002/11

02402/11

024012

)1(

2

2)1(

2

21

ξξ ccξξ

ξξξ

Example 1: Second Eigenvector (4 of 9)

Eigenvector for r2 = -1: Solve

by row reducing the augmented matrix:

( ) ⎟⎟⎠

⎞⎜⎜⎝

⎛=⎟⎟

⎠

⎞⎜⎜⎝

⎛⎟⎟⎠

⎞⎜⎜⎝

⎛⇔⎟⎟

⎠

⎞⎜⎜⎝

⎛=⎟⎟

⎠

⎞⎜⎜⎝

⎛⎟⎟⎠

⎞⎜⎜⎝

⎛+

+⇔=−

00

2412

00

114111

2

1

2

1

ξξ

ξξ

0ξIA r

⎟⎟⎠

⎞⎜⎜⎝

⎛−

=→⎟⎟⎠

⎞⎜⎜⎝

⎛−=⎟⎟

⎠

⎞⎜⎜⎝

⎛−=→

==+

→⎟⎟⎠

⎞⎜⎜⎝

⎛→⎟⎟

⎠

⎞⎜⎜⎝

⎛→⎟⎟

⎠

⎞⎜⎜⎝

⎛

21

choosearbitrary,1

2/12/1

0002/11

00002/11

02402/11

024012

)2(

2

2)2(

2

21

ξξ ccξξ

ξξξ

Example 1: General Solution (5 of 9)

The corresponding solutions x = ξert of x' = Ax are

The Wronskian of these two solutions is

Thus x(1) and x(2) are fundamental solutions, and the general solution of x' = Ax is

tt etet −⎟⎟⎠

⎞⎜⎜⎝

⎛−

=⎟⎟⎠

⎞⎜⎜⎝

⎛=

21

)(,21

)( )2(3)1( xx

[ ] 0422

)(, 23

3)2()1( ≠−=

−= −

−

−t

tt

tt

eeeee

tW xx

tt ecec

tctct

−⎟⎟⎠

⎞⎜⎜⎝

⎛−

+⎟⎟⎠

⎞⎜⎜⎝

⎛=

+=

21

21

)()()(

23

1

)2(2

)1(1 xxx

Example 1: Phase Plane for x(1) (6 of 9)

To visualize solution, consider first x = c1x(1):

Now

Thus x(1) lies along the straight line x2 = 2x1, which is the line through origin in direction of first eigenvector ξ(1)

If solution is trajectory of particle, with position given by (x1, x2), then it is in Q1 when c1 > 0, and in Q3 when c1 < 0. In either case, particle moves away from origin as t increases.

ttt ecxecxecxx

t 312

311

31

2

1)1( 2,21

)( ==⇔⎟⎟⎠

⎞⎜⎜⎝

⎛=⎟⎟

⎠

⎞⎜⎜⎝

⎛=x

121

2

1

13312

311 2

22, xx

cx

cxeecxecx ttt =⇔==⇔==

Example 1: Phase Plane for x(2) (7 of 9)

Next, consider x = c2x(2):

Then x(2) lies along the straight line x2 = -2x1, which is the line through origin in direction of 2nd eigenvector ξ(2)

If solution is trajectory of particle, with position given by (x1, x2), then it is in Q4 when c2 > 0, and in Q2 when c2 < 0. In either case, particle moves towards origin as t increases.

ttt ecxecxecxx

t −−− −==⇔⎟⎟⎠

⎞⎜⎜⎝

⎛−

=⎟⎟⎠

⎞⎜⎜⎝

⎛= 22212

2

1)2( 2,21

)(x

Example 1: Phase Plane for General Solution (8 of 9)

The general solution is x = c1x(1) + c2x(2):

As t →∞, c1x(1) is dominant and c2x(2) becomes negligible. Thus, for c1 ≠ 0, all solutions asymptotically approach the line x2 = 2x1 as t →∞. Similarly, for c2 ≠ 0, all solutions asymptotically approach the line x2 = -2x1 as t → - ∞. The origin is a saddle point,and is unstable. See graph.

tt ecect −⎟⎟⎠

⎞⎜⎜⎝

⎛−

+⎟⎟⎠

⎞⎜⎜⎝

⎛=

21

21

)( 23

1x

Example 1: Time Plots for General Solution (9 of 9)

The general solution is x = c1x(1) + c2x(2):

As an alternative to phase plane plots, we can graph x1 or x2as a function of t. A few plots of x1 are given below. Note that when c1 = 0, x1(t) = c2e-t → 0 as t →∞. Otherwise, x1(t) = c1e3t + c2e-t grows unbounded as t →∞. Graphs of x2 are similarly obtained.

⎟⎟⎠

⎞⎜⎜⎝

⎛

−+

=⎟⎟⎠

⎞⎜⎜⎝

⎛⇔⎟⎟

⎠

⎞⎜⎜⎝

⎛−

+⎟⎟⎠

⎞⎜⎜⎝

⎛=

−

−−

tt

tttt

ecececec

txtx

ecect2

31

23

1

2

12

31 22)(

)(21

21

)(x

Example 2: Direction Field (1 of 9)

Consider the homogeneous equation x' = Ax below.

A direction field for this system is given below.Substituting x = ξert in for x, and rewriting system as (A-rI)ξ = 0, we obtain

xx ⎟⎟⎠

⎞⎜⎜⎝

⎛

−−

=′2223

⎟⎟⎠

⎞⎜⎜⎝

⎛=⎟⎟

⎠

⎞⎜⎜⎝

⎛⎟⎟⎠

⎞⎜⎜⎝

⎛

−−−−

00

2223

1

1

ξξ

rr

Example 2: Eigenvalues (2 of 9)

Our solution has the form x = ξert, where r and ξ are found by solving

Recalling that this is an eigenvalue problem, we determine rby solving det(A-rI) = 0:

Thus r1 = -1 and r2 = -4.

)4)(1(452)2)(3(22

23 2 ++=++=−−−−−=−−

−− rrrrrrr

r

⎟⎟⎠

⎞⎜⎜⎝

⎛=⎟⎟

⎠

⎞⎜⎜⎝

⎛⎟⎟⎠

⎞⎜⎜⎝

⎛

−−−−

00

2223

1

1

ξξ

rr

Example 2: First Eigenvector (3 of 9)

Eigenvector for r1 = -1: Solve

by row reducing the augmented matrix:

( ) ⎟⎟⎠

⎞⎜⎜⎝

⎛=⎟⎟

⎠

⎞⎜⎜⎝

⎛⎟⎟⎠

⎞⎜⎜⎝

⎛

−−

⇔⎟⎟⎠

⎞⎜⎜⎝

⎛=⎟⎟

⎠

⎞⎜⎜⎝

⎛⎟⎟⎠

⎞⎜⎜⎝

⎛

+−+−

⇔=−00

1222

00

122213

2

1

2

1

ξξ

ξξ

0ξIA r

⎟⎟⎠

⎞⎜⎜⎝

⎛=→⎟⎟

⎠

⎞⎜⎜⎝

⎛=→

⎟⎟⎠

⎞⎜⎜⎝

⎛ −→⎟⎟

⎠

⎞⎜⎜⎝

⎛

−−

→⎟⎟⎠

⎞⎜⎜⎝

⎛

−−

21

choose2/2

00002/21

01202/21

012022

)1(

2

2)1( ξξξξ

Example 2: Second Eigenvector (4 of 9)

Eigenvector for r2 = -4: Solve

by row reducing the augmented matrix:

( ) ⎟⎟⎠

⎞⎜⎜⎝

⎛=⎟⎟

⎠

⎞⎜⎜⎝

⎛⎟⎟⎠

⎞⎜⎜⎝

⎛⇔⎟⎟

⎠

⎞⎜⎜⎝

⎛=⎟⎟

⎠

⎞⎜⎜⎝

⎛⎟⎟⎠

⎞⎜⎜⎝

⎛

+−+−

⇔=−00

2221

00

422243

2

1

2

1

ξξ

ξξ

0ξIA r

⎟⎟⎠

⎞⎜⎜⎝

⎛−=→

⎟⎟⎠

⎞⎜⎜⎝

⎛−=→⎟⎟

⎠

⎞⎜⎜⎝

⎛→⎟⎟

⎠

⎞⎜⎜⎝

⎛

12choose

2000021

022021

)2(

2

2)2(

ξ

ξξξ

Example 2: General Solution (5 of 9)

The corresponding solutions x = ξert of x' = Ax are

The Wronskian of these two solutions is

Thus x(1) and x(2) are fundamental solutions, and the general solution of x' = Ax is

tt etet 4)2()1(

12)(,

21

)( −−

⎟⎟⎠

⎞⎜⎜⎝

⎛−=⎟⎟

⎠

⎞⎜⎜⎝

⎛= xx

[ ] 032

2)(, 54

4)2()1( ≠=

−= −

−−

−−t

tt

tt

eeeeetW xx

tt ecec

tctct

421

)2(2

)1(1

12

21

)()()(

−−

⎟⎟⎠

⎞⎜⎜⎝

⎛−+⎟⎟

⎠

⎞⎜⎜⎝

⎛=

+= xxx

Example 2: Phase Plane for x(1) (6 of 9)

To visualize solution, consider first x = c1x(1):

Now

Thus x(1) lies along the straight line x2 = 2½x1, which is the line through origin in direction of first eigenvector ξ(1)

If solution is trajectory of particle, with position given by (x1, x2), then it is in Q1 when c1 > 0, and in Q3 when c1 < 0. In either case, particle moves towards origin as t increases.

ttt ecxecxecxx

t −−− ==⇔⎟⎟⎠

⎞⎜⎜⎝

⎛=⎟⎟

⎠

⎞⎜⎜⎝

⎛= 12111

2

1)1( 2,21

)(x

121

2

1

11211 2

22, xx

cx

cxeecxecx ttt =⇔==⇔== −−−

Example 2: Phase Plane for x(2) (7 of 9)

Next, consider x = c2x(2):

Then x(2) lies along the straight line x2 = -2½x1, which is the line through origin in direction of 2nd eigenvector ξ(2)

If solution is trajectory of particle, with position given by (x1, x2), then it is in Q4 when c2 > 0, and in Q2 when c2 < 0. In either case, particle moves towards origin as t increases.

ttt ecxecxecxx

t 422

421

42

2

1)2( ,212)( −−− =−=⇔⎟⎟⎠

⎞⎜⎜⎝

⎛−=⎟⎟

⎠

⎞⎜⎜⎝

⎛=x

Example 2: Phase Plane for General Solution (8 of 9)

The general solution is x = c1x(1) + c2x(2):

As t →∞, c1x(1) is dominant and c2x(2) becomes negligible. Thus, for c1 ≠ 0, all solutions asymptotically approach origin along the line x2 = 2½x1 as t →∞. Similarly, all solutions are unbounded as t → - ∞. The origin is a node, and is asymptotically stable.

tt etet 4)2()1(

12)(,

21

)( −−

⎟⎟⎠

⎞⎜⎜⎝

⎛−=⎟⎟

⎠

⎞⎜⎜⎝

⎛= xx

Example 2: Time Plots for General Solution (9 of 9)

The general solution is x = c1x(1) + c2x(2):

As an alternative to phase plane plots, we can graph x1 or x2as a function of t. A few plots of x1 are given below. Graphs of x2 are similarly obtained.

⎟⎟⎠

⎞⎜⎜⎝

⎛

+−

=⎟⎟⎠

⎞⎜⎜⎝

⎛⇔⎟⎟

⎠

⎞⎜⎜⎝

⎛−+⎟⎟

⎠

⎞⎜⎜⎝

⎛=

−−

−−−−

tt

tttt

ecececec

txtx

ecect4

21

421

2

1421 2

2)()(

12

21

)(x

2 x 2 Case: Real Eigenvalues, Saddle Points and Nodes

The previous two examples demonstrate the two main cases for a 2 x 2 real system with real and different eigenvalues:

Both eigenvalues have opposite signs, in which case origin is a saddle point and is unstable.Both eigenvalues have the same sign, in which case origin is a node, and is asymptotically stable if the eigenvalues are negative and unstable if the eigenvalues are positive.

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 ncorresponding linearly independent eigenvectors ξ(1),…, ξ(n). The associated solutions of x' = Ax are

Using Wronskian, it can be shown that these solutions are linearly independent, and hence form a fundamental set of solutions. Thus general solution is

trnntr netet )()()1()1( )(,,)( 1 ξxξx == K

trnn

tr necec )()1(1

1 ξξx ++= K

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 ξ(1),…, ξ(n). The associated solutions of x' = Ax are

and form a fundamental set of solutions.

trnntr netet )()()1()1( )(,,)( 1 ξxξx == K

Example 3: Hermitian Matrix (1 of 3)

Consider the homogeneous equation x' = Ax below.

The eigenvalues were found previously in Ch 7.3, and were: r1 = 2, r2 = -1 and r3 = -1.

Corresponding eigenvectors:

xx⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛=′

011101110

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛

−=

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛

−=

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛=

110

, 101

,111

)3()2()1( ξξξ

Example 3: General Solution (2 of 3)

The fundamental solutions are

with general solution

ttt eee −−

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛

−=

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛

−=

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛=

110

, 101

,111

)3()2(2)1( xxx

ttt ececec −−

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛

−+

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛

−+

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛=

110

101

111

322

1x

Example 3: General Solution Behavior (3 of 3)

The general solution is x = c1x(1) + c2x(2) + c3x(3):

As t →∞, c1x(1) is dominant and c2x(2) , c3x(3) become negligible. Thus, for c1 ≠ 0, all solns x become unbounded as t →∞,while for c1 = 0, all solns x → 0 as t →∞.The initial points that cause c1 = 0 are those that lie in plane determined by ξ(2) and ξ(3). Thus solutions that start in this plane approach origin as t →∞.

ttt ececec −−

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛

−+

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛

−+

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛=

110

101

111

322

1x

Complex Eigenvalues and Fundamental SolnsIf some of the eigenvalues r1,…, rn occur in complex conjugate pairs, but otherwise are different, then there are still n corresponding linearly independent solutions

which form a fundamental set of solutions. Some may be complex-valued, but real-valued solutions may be derived from them. This situation will be examined in Ch 7.6.

If the coefficient matrix A is complex, then complex eigenvalues need not occur in conjugate pairs, but solutions will still have the above form (if the eigenvalues are distinct) and these solutions may be complex-valued.

,)(,,)( )()()1()1( 1 trnntr netet ξxξx == K

Repeated Eigenvalues and Fundamental SolnsIf some of the eigenvalues r1,…, rn are repeated, then there may not be n corresponding linearly independent solutions of the form

In order to obtain a fundamental set of solutions, it may be necessary to seek additional solutions of another form. This situation is analogous to that for an nth order linear equation with constant coefficients, in which case a repeated root gave rise solutions of the form

This case of repeated eigenvalues is examined in Section 7.8.

trnntr netet )()()1()1( )(,,)( 1 ξxξx == K

K,,, 2 rtrtrt ettee

Ch 7.6: Complex EigenvaluesWe consider again a homogeneous system of n first order linear equations with constant, real coefficients,

and thus the system can be written as x' = Ax, where

,2211

22221212

12121111

nnnnnn

nn

nn

xaxaxax

xaxaxaxxaxaxax

+++=′

+++=′+++=′

K

M

K

K

⎟⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜⎜

⎝

⎛

=

⎟⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜⎜

⎝

⎛

=

nnnn

n

n

n aaa

aaaaaa

tx

txtx

t

L

MOMM

L

L

M

21

22221

11211

2

1

,

)(

)()(

)( Ax

Conjugate Eigenvalues and EigenvectorsWe know that x = ξert is a solution of x' = Ax, provided r is an eigenvalue and ξ is an eigenvector of A.The eigenvalues r1,…, rn are the roots of det(A-rI) = 0, and the corresponding eigenvectors satisfy (A-rI)ξ = 0. If A is real, then the coefficients in the polynomial equation det(A-rI) = 0 are real, and hence any complex eigenvalues must occur in conjugate pairs. Thus if r1 = λ + iμ is an eigenvalue, then so is r2 = λ - iμ.The corresponding eigenvectors ξ(1), ξ(2) are conjugates also.To see this, recall A and I have real entries, and hence( ) ( ) ( ) 0ξIA0ξIA0ξIA =−⇒=−⇒=− )2(

2)1(

1)1(

1 rrr

Conjugate SolutionsIt follows from the previous slide that the solutions

corresponding to these eigenvalues and eigenvectors are conjugates conjugates as well, since

)1()1()2()2( 22 xξξx === trtr ee

trtr ee 21 )2()2()1()1( , ξxξx ==

Real-Valued SolutionsThus for complex conjugate eigenvalues r1 and r2 , the corresponding solutions x(1) and x(2) are conjugates also.To obtain real-valued solutions, use real and imaginary parts of either x(1) or x(2). To see this, let ξ(1) = a + ib. Then

where

are real valued solutions of x' = Ax, and can be shown to be linearly independent.

( ) ( ) ( )( ) ( )

)()(cossinsincos

sincos)1()1(

titttiette

titeiett

tti

vubaba

baξx

+=++−=

++== +

μμμμ

μμλλ

λμλ

( ) ( ),cossin)(,sincos)( ttetttet tt μμμμ λλ bavbau +=−=

General SolutionTo summarize, suppose r1 = λ + iμ, r2 = λ - iμ, and that r3,…, rn are all real and distinct eigenvalues of A. Let the corresponding eigenvectors be

Then the general solution of x' = Ax is

where

trnn

tr necectctc )()3(321

3)()( ξξvux ++++= K

)()4()3()2()1( ,,,,, nii ξξξbaξbaξ K−=+=

( ) ( )ttetttet tt μμμμ λλ cossin)(,sincos)( bavbau +=−=

Example 1: Direction Field (1 of 7)

Consider the homogeneous equation x' = Ax below.

A direction field for this system is given below.Substituting x = ξert in for x, and rewriting system as (A-rI)ξ = 0, we obtain

xx ⎟⎟⎠

⎞⎜⎜⎝

⎛−−

−=′

2/1112/1

⎟⎟⎠

⎞⎜⎜⎝

⎛=⎟⎟

⎠

⎞⎜⎜⎝

⎛⎟⎟⎠

⎞⎜⎜⎝

⎛−−−

−−00

2/1112/1

1

1

ξξ

rr

Example 1: Complex Eigenvalues (2 of 7)

We determine r by solving det(A-rI) = 0. Now

Thus

Therefore the eigenvalues are r1 = -1/2 + i and r2 = -1/2 - i.

( )4512/1

2/1112/1 22 ++=++=−−−

−−rrr

rr

iir ±−=±−

=−±−

=21

221

2)4/5(411 2

Example 1: First Eigenvector (3 of 7)

Eigenvector for r1 = -1/2 + i: Solve

by row reducing the augmented matrix:

Thus

( )

⎟⎟⎠

⎞⎜⎜⎝

⎛=⎟⎟

⎠

⎞⎜⎜⎝

⎛⎟⎟⎠

⎞⎜⎜⎝

⎛−−

⇔⎟⎟⎠

⎞⎜⎜⎝

⎛=⎟⎟

⎠

⎞⎜⎜⎝

⎛⎟⎟⎠

⎞⎜⎜⎝

⎛−−

−⇔

⎟⎟⎠

⎞⎜⎜⎝

⎛=⎟⎟

⎠

⎞⎜⎜⎝

⎛⎟⎟⎠

⎞⎜⎜⎝

⎛−−−

−−⇔=−

00

11

00

11

00

2/1112/1

2

1

2

1

1

1

ξξ

ξξ

ξξ

ii

ii

rr

r 0ξIA

⎟⎟⎠

⎞⎜⎜⎝

⎛=→⎟⎟

⎠

⎞⎜⎜⎝

⎛−=→⎟⎟

⎠

⎞⎜⎜⎝

⎛→⎟⎟

⎠

⎞⎜⎜⎝

⎛−− i

iiii 1

choose00001

0101 )1(

2

2)1( ξξξξ

⎟⎟⎠

⎞⎜⎜⎝

⎛+⎟⎟⎠

⎞⎜⎜⎝

⎛=

10

01)1( iξ

Example 1: Second Eigenvector (4 of 7)

Eigenvector for r1 = -1/2 - i: Solve

by row reducing the augmented matrix:

Thus

( )

⎟⎟⎠

⎞⎜⎜⎝

⎛=⎟⎟

⎠

⎞⎜⎜⎝

⎛⎟⎟⎠

⎞⎜⎜⎝

⎛−

−⇔⎟⎟

⎠

⎞⎜⎜⎝

⎛=⎟⎟

⎠

⎞⎜⎜⎝

⎛⎟⎟⎠

⎞⎜⎜⎝

⎛−

⇔

⎟⎟⎠

⎞⎜⎜⎝

⎛=⎟⎟

⎠

⎞⎜⎜⎝

⎛⎟⎟⎠

⎞⎜⎜⎝

⎛−−−

−−⇔=−

00

11

00

11

00

2/1112/1

2

1

2

1

1

1

ξξ

ξξ

ξξ

ii

ii

rr

r 0ξIA

⎟⎟⎠

⎞⎜⎜⎝

⎛−

=→⎟⎟⎠

⎞⎜⎜⎝

⎛=→⎟⎟

⎠

⎞⎜⎜⎝

⎛ −→⎟⎟

⎠

⎞⎜⎜⎝

⎛−

−i

iiii 1

choose00001

0101 )2(

2

2)2( ξξξξ

⎟⎟⎠

⎞⎜⎜⎝

⎛−

+⎟⎟⎠

⎞⎜⎜⎝

⎛=

10

01)2( iξ

Example 1: General Solution (5 of 7)

The corresponding solutions x = ξert of x' = Ax are

The Wronskian of these two solutions is

Thus u(t) and v(t) are real-valued fundamental solutions of x' = Ax, with general solution x = c1u + c2v.

⎟⎟⎠

⎞⎜⎜⎝

⎛=⎥

⎦

⎤⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛+⎟⎟

⎠

⎞⎜⎜⎝

⎛=

⎟⎟⎠

⎞⎜⎜⎝

⎛−

=⎥⎦

⎤⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛−⎟⎟

⎠

⎞⎜⎜⎝

⎛=

−−

−−

tt

ettet

tt

ettet

tt

tt

cossin

cos10

sin01

)(

sincos

sin10

cos01

)(

2/2/

2/2/

v

u

[ ] 0cossinsincos

)(, 2/2/

2/2/)2()1( ≠=

−= −

−−

−−t

tt

tt

etetetete

tW xx

Example 1: Phase Plane (6 of 7)

Given below is the phase plane plot for solutions x, with

Each solution trajectory approaches origin along a spiral path as t →∞, since coordinates are products of decaying exponential and sine or cosine factors. The graph of u passes through (1,0), since u(0) = (1,0). Similarly, the graph of v passes through (0,1). The origin is a spiral point, and is asymptotically stable.

⎟⎟⎠

⎞⎜⎜⎝

⎛+⎟⎟⎠

⎞⎜⎜⎝

⎛

−=⎟⎟

⎠

⎞⎜⎜⎝

⎛=

−

−

−

−

tete

ctete

cxx

t

t

t

t

cossin

sincos

2/

2/

22/

2/

12

1x

Example 1: Time Plots (7 of 7)

The general solution is x = c1u + c2v:

As an alternative to phase plane plots, we can graph x1 or x2as a function of t. A few plots of x1 are given below, each one a decaying oscillation as t →∞.

⎟⎟⎠

⎞⎜⎜⎝

⎛

+−+

=⎟⎟⎠

⎞⎜⎜⎝

⎛=

−−

−−

tectectectec

txtx

tt

tt

cossinsincos

)()(

2/2

2/1

2/2

2/1

2

1x

Spiral Points, Centers, Eigenvalues, and TrajectoriesIn previous example, general solution was

The origin was a spiral point, and was asymptotically stable. If real part of complex eigenvalues is positive, then trajectories spiral away, unbounded, from origin, and hence origin would be an unstable spiral point. If real part of complex eigenvalues is zero, then trajectories circle origin, neither approaching nor departing. Then origin is called a center and is stable, but not asymptotically stable. Trajectories periodic in time. The direction of trajectory motion depends on entries in A.

⎟⎟⎠

⎞⎜⎜⎝

⎛+⎟⎟⎠

⎞⎜⎜⎝

⎛

−=⎟⎟

⎠

⎞⎜⎜⎝

⎛=

−

−

−

−

tete

ctete

cxx

t

t

t

t

cossin

sincos

2/

2/

22/

2/

12

1x

Example 2: Second Order System with Parameter (1 of 2)

The system x' = Ax below contains a parameter α.

Substituting x = ξert in for x and rewriting system as (A-rI)ξ = 0, we obtain

Next, solve for r in terms of α :

xx ⎟⎟⎠

⎞⎜⎜⎝

⎛−

=′022α

⎟⎟⎠

⎞⎜⎜⎝

⎛=⎟⎟

⎠

⎞⎜⎜⎝

⎛⎟⎟⎠

⎞⎜⎜⎝

⎛−−

−00

22

1

1

ξξα

rr

21644)(

22 2

2 −±=⇒+−=+−=

−−− αααα

αrrrrr

rr

Example 2: Eigenvalue Analysis (2 of 2)

The eigenvalues are given by the quadratic formula above.For α < -4, both eigenvalues are real and negative, and hence origin is asymptotically stable node. For α > 4, both eigenvalues are real and positive, and hence the origin is an unstable node.For -4 < α < 0, eigenvalues are complex with a negative real part, and hence origin is asymptotically stable spiral point.For 0 < α < 4, eigenvalues are complex with a positive real part, and the origin is an unstable spiral point.For α = 0, eigenvalues are purely imaginary, origin is a center. Trajectories closed curves about origin & periodic.For α = ± 4, eigenvalues real & equal, origin is a node (Ch 7.8)

2162 −±

=ααr

Second Order Solution Behavior and Eigenvalues: Three Main CasesFor second order systems, the three main cases are:

Eigenvalues are real and have opposite signs; x = 0 is a saddle point.Eigenvalues are real, distinct and have same sign; x = 0 is a node.Eigenvalues are complex with nonzero real part; x = 0 a spiral point.

Other possibilities exist and occur as transitions between two of the cases listed above:

A zero eigenvalue occurs during transition between saddle point and node. Real and equal eigenvalues occur during transition between nodes and spiral points. Purely imaginary eigenvalues occur during a transition between asymptotically stable and unstable spiral points.

aacbbr

242 −±−

=

Ch 7.7: Fundamental MatricesSuppose that x(1)(t),…, x(n)(t) form a fundamental set of solutions for x' = P(t)x on α < t < β. The matrix

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 detΨ ≠ 0. Note also that since x(1)(t),…, x(n)(t) are solutions of x' = P(t)x, Ψ satisfies the matrix differential equation Ψ' = P(t)Ψ.

,)()(

)()()(

)()1(

)(1

)1(1

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛

=txtx

txtxt

nnn

n

L

MOM

L

Ψ

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

In Chapter 7.5, we found the following fundamental solutions for this system:

Thus a fundamental matrix for this system is

xx ⎟⎟⎠

⎞⎜⎜⎝

⎛=′

1411

tt etet −⎟⎟⎠

⎞⎜⎜⎝

⎛−

=⎟⎟⎠

⎞⎜⎜⎝

⎛=

21

)(,21

)( )2(3)1( xx

⎟⎟⎠

⎞⎜⎜⎝

⎛

−=

−

−

tt

tt

eeee

t22

)( 3

3

Ψ

Fundamental Matrices and General SolutionThe general solution of x' = P(t)x

can be expressed x = Ψ(t)c, where c is a constant vector with components c1,…, cn:

)()1(1 )( n

nctc xxx ++= L

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛

==

nn

nn

n

c

c

txtx

txtxt M

L

MOM

L 1

)()1(

)(1

)1(1

)()(

)()()( cΨx

Fundamental Matrix & Initial Value ProblemConsider an initial value problem

x' = P(t)x, x(t0) = x0

where α < t0 < β and x0 is a given initial vector.Now the solution has the form x = Ψ(t)c, hence we choose cso as to satisfy x(t0) = x0. Recalling Ψ(t0) is nonsingular, it follows that

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

00

100 )()( xΨcxcΨ tt −=⇒=

00

1 )()( xΨΨx tt −=

Recall: Theorem 7.4.4Let

Let x(1),…, x(n) be solutions of x' = P(t)x on I: α < t < β that satisfy the initial conditions

Then x(1),…, x(n) are fundamental solutions of x' = P(t)x.

⎟⎟⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜⎜⎜

⎝

⎛

=

⎟⎟⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜⎜⎜

⎝

⎛

=

⎟⎟⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜⎜⎜

⎝

⎛

=

10

00

,,

0

010

,

0

001

)()2()1( MK

MM

neee

βα <<== 0)(

0)()1(

0)1( ,)(,,)( ttt nn exex K

Fundamental Matrix & Theorem 7.4.4Suppose x(1)(t),…, x(n)(t) form the fundamental solutions given by Thm 7.4.4. Denote the corresponding fundamental matrix by Φ(t). Then columns of Φ(t) are x(1)(t),…, x(n)(t), and hence

Thus Φ-1(t0) = I, and the hence general solution to the corresponding initial value problem is

It follows that for any fundamental matrix Ψ(t),

I=

⎟⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜⎜

⎝

⎛

=Φ

100

010001

)( 0

L

MOMM

L

L

t

000

1 )()()( xΦxΦΦx ttt == −

)()()()()()( 0100

01 tttttt −− =⇒== ΨΨΦxΦxΨΨx

The Fundamental Matrix Φand Varying Initial ConditionsThus when using the fundamental matrix Φ(t), the general solution to an IVP is

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 Φ(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 Φ(t) represents a linear transformation of the initial conditions x0 into the solution x(t) at time t.

000

1 )()()( xΦxΦΦx ttt == −

Example 2: Find Φ(t) for 2 x 2 System (1 of 5)

Find Φ(t) such that Φ(0) = I for the system below.

Solution: First, we must obtain x(1)(t) and x(2)(t) such that

We know from previous results that the general solution is

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).

xx ⎟⎟⎠

⎞⎜⎜⎝

⎛=′

1411

tt ecec −⎟⎟⎠

⎞⎜⎜⎝

⎛−

+⎟⎟⎠

⎞⎜⎜⎝

⎛=

21

21

23

1x

⎟⎟⎠

⎞⎜⎜⎝

⎛=⎟⎟

⎠

⎞⎜⎜⎝

⎛=

10

)0(,01

)0( )2()1( xx

Example 2: Use General Solution (2 of 5)

Thus, to find x(1)(t), express it terms of the general solution

and then find the coefficients c1 and c2. To do so, use the initial conditions to obtain

or equivalently,

tt ecect −⎟⎟⎠

⎞⎜⎜⎝

⎛−

+⎟⎟⎠

⎞⎜⎜⎝

⎛=

21

21

)( 23

1)1(x

⎟⎟⎠

⎞⎜⎜⎝

⎛=⎟⎟

⎠

⎞⎜⎜⎝

⎛−

+⎟⎟⎠

⎞⎜⎜⎝

⎛=

01

21

21

)0( 21)1( ccx

⎟⎟⎠

⎞⎜⎜⎝

⎛=⎟⎟

⎠

⎞⎜⎜⎝

⎛⎟⎟⎠

⎞⎜⎜⎝

⎛− 0

12211

2

1

cc

Example 2: Solve for x(1)(t) (3 of 5)

To find x(1)(t), we therefore solve

by row reducing the augmented matrix:

Thus2/12/1

2/1102/101

2/110111

240111

022111

2

1

==

→

⎟⎟⎠

⎞⎜⎜⎝

⎛→⎟⎟

⎠

⎞⎜⎜⎝

⎛→⎟⎟

⎠

⎞⎜⎜⎝

⎛−−

→⎟⎟⎠

⎞⎜⎜⎝

⎛−

cc

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛

−

+=⎟⎟⎠

⎞⎜⎜⎝

⎛−

+⎟⎟⎠

⎞⎜⎜⎝

⎛=

−

−−

tt

tttt

ee

eeeet3

33)1( 2

121

21

21

21

21)(x

⎟⎟⎠

⎞⎜⎜⎝

⎛=⎟⎟

⎠

⎞⎜⎜⎝

⎛⎟⎟⎠

⎞⎜⎜⎝

⎛− 0

12211

2

1

cc

Example 2: Solve for x(2)(t) (4 of 5)

To find x(2)(t), we similarly solve

by row reducing the augmented matrix:

Thus4/14/1

4/1104/101

4/110011

140011

122011

2

1

−==

→

⎟⎟⎠

⎞⎜⎜⎝

⎛−

→⎟⎟⎠

⎞⎜⎜⎝

⎛−

→⎟⎟⎠

⎞⎜⎜⎝

⎛−

→⎟⎟⎠

⎞⎜⎜⎝

⎛−

cc

⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

+

−=⎟⎟

⎠

⎞⎜⎜⎝

⎛−

−⎟⎟⎠

⎞⎜⎜⎝

⎛=

−

−

−

tt

tt

tt

ee

eeeet

21

21

41

41

21

41

21

41)(

3

3

3)2(x

⎟⎟⎠

⎞⎜⎜⎝

⎛=⎟⎟

⎠

⎞⎜⎜⎝

⎛⎟⎟⎠

⎞⎜⎜⎝

⎛− 1

02211

2

1

cc

Example 2: Obtain Φ(t) (5 of 5)

The columns of Φ(t) are given by x(1)(t) and x(2)(t), and thus from the previous slide we have

Note Φ(t) is more complicated than Ψ(t) found in Ex 1. However, now that we have Φ(t), it is much easier to determine the solution to any set of initial conditions.

⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

+−

−+=

−−

−−

tttt

tttt

eeee

eeeet

21

21

41

41

21

21

)(33

33

Φ

⎟⎟⎠

⎞⎜⎜⎝

⎛

−=

−

−

tt

tt

eeee

t22

)(3

3

Ψ

Matrix Exponential FunctionsConsider 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 = Φ(t)x0, where Φ(0) = I.

Comparing the form and solution for both of these cases, we might expect Φ(t) to have an exponential character. Indeed, it can be shown that Φ(t) = eAt, where

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.

∑∑∞

=

∞

=

+==10 !! n

nn

n

nnt

nt

nte AIAA

Example 3: Matrix Exponential FunctionConsider the diagonal matrix A below.

Then

In general,

Thus

⎟⎟⎠

⎞⎜⎜⎝

⎛=⎟⎟

⎠

⎞⎜⎜⎝

⎛== ∑∑

∞

=

∞

=t

t

n

nn

n

nnt

ee

tn

nn

te2

00 00

!/200!/1

!AA

⎟⎟⎠

⎞⎜⎜⎝

⎛=

2001

A

K,2001

2001

2001

,2001

2001

2001

323

22

⎟⎟⎠

⎞⎜⎜⎝

⎛=⎟⎟

⎠

⎞⎜⎜⎝

⎛⎟⎟⎠

⎞⎜⎜⎝

⎛=⎟⎟

⎠

⎞⎜⎜⎝

⎛=⎟⎟

⎠

⎞⎜⎜⎝

⎛⎟⎟⎠

⎞⎜⎜⎝

⎛= AA

⎟⎟⎠

⎞⎜⎜⎝

⎛= n

n

2001

A

Coupled Systems of EquationsRecall that our constant coefficient homogeneous system

written as x' = Ax with

is a system of coupled equations that must be solved simultaneously to find all the unknown variables.

,2211

12121111

nnnnnn

nn

xaxaxax

xaxaxax

+++=′

+++=′

K

M

K

,,)(

)()(

1

1111

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛=

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛=

nnn

n

n aa

aa

tx

txt

L

MOM

L

M Ax

Uncoupled Systems & Diagonal MatricesIn 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

or x' = Dx, where D is a diagonal matrix:

,00

0000

21

21112

21111

nnnn

n

n

xdxxx

xxdxxxxxdx

+++=′

+++=′+++=′

K

M

K

K

⎟⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜⎜

⎝

⎛

=⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛=

nnn d

dd

tx

txt

L

MOMM

L

L

M

00

0000

,)(

)()( 22

111

Dx

Uncoupling: Transform Matrix TIn order to explore transforming our given system x' = Ax of coupled equations into an uncoupled system x' = Dx, where Dis a diagonal matrix, we will use the eigenvectors of A.Suppose A is n x n with n linearly independent eigenvectors ξ(1),…, ξ(n), and corresponding eigenvalues λ1,…, λn.Define n x n matrices T and D using the eigenvalues & eigenvectors of A:

Note that T is nonsingular, and hence T-1 exists.

⎟⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜⎜

⎝

⎛

=⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛

=

n

nnn

n

λ

λλ

ξξ

ξξ

L

MOMM

L

L

L

MOM

L

00

0000

, 2

1

)()1(

)(1

)1(1

DT

Uncoupling: T-1AT = DRecall here the definitions of A, T and D:

Then the columns of AT are Aξ(1),…, Aξ(n), and hence

It follows that T-1AT = D.

⎟⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜⎜

⎝

⎛

=⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛

=⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛=

n

nnn

n

nnn

n

aa

aa

λ

λλ

ξξ

ξξ

L

MOMM

L

L

L

MOM

L

L

MOM

L

00

0000

,, 2

1

)()1(

)(1

)1(1

1

111

DTA

TDAT =⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛

=)()1(

1

)(1

)1(11

nnnn

nn

ξλξλ

ξλξλ

L

MOM

L

Similarity TransformationsThus, if the eigenvalues and eigenvectors of A are known, then A can be transformed into a diagonal matrix D, with

T-1AT = DThis process is known as a similarity transformation, and Ais said to be similar to D. Alternatively, we could say that Ais diagonalizable.

⎟⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜⎜

⎝

⎛

=⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛

=⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛=

n

nnn

n

nnn

n

aa

aa

λ

λλ

ξξ

ξξ

L

MOMM

L

L

L

MOM

L

L

MOM

L

00

0000

,, 2

1

)()1(

)(1

)1(1

1

111

DTA

Similarity Transformations: Hermitian CaseRecall: Our similarity transformation of A has the form

T-1AT = Dwhere D is diagonal and columns of T are eigenvectors of A. If A is Hermitian, then A has n linearly independent orthogonal eigenvectors ξ(1),…, ξ(n), normalized so that (ξ(i), ξ(i)) =1 for i = 1,…, n, and (ξ(i), ξ(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 AFinally, 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.

⎟⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜⎜

⎝

⎛

=⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛

=⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛=

n

nnn

n

nnn

n

aa

aa

λ

λλ

ξξ

ξξ

L

MOMM

L

L

L

MOM

L

L

MOM

L

00

0000

,, 2

1

)()1(

)(1

)1(1

1

111

DTA

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.

We already know that the eigenvalues are λ1 = 3, λ2 = -1 with corresponding eigenvectors

Thus

⎟⎟⎠

⎞⎜⎜⎝

⎛=

1411

A

⎟⎟⎠

⎞⎜⎜⎝

⎛−

=⎟⎟⎠

⎞⎜⎜⎝

⎛=

21

)(,21

)( )2()1( tt ξξ

⎟⎟⎠

⎞⎜⎜⎝

⎛−

=⎟⎟⎠

⎞⎜⎜⎝

⎛−

=1003

,2211

DT

Example 4: Similarity Transformation (2 of 2)

To find T-1, augment the identity to T and row reduce:

Then

Thus A is similar to D, and hence A is diagonalizable.

⎟⎟⎠

⎞⎜⎜⎝

⎛−

=→⎟⎟⎠

⎞⎜⎜⎝

⎛−

→

⎟⎟⎠

⎞⎜⎜⎝

⎛−

→⎟⎟⎠

⎞⎜⎜⎝

⎛−−

→⎟⎟⎠

⎞⎜⎜⎝

⎛−

−

4/12/14/12/1

4/12/1104/12/101

4/12/1100111

12400111

10220111

1T

D

ATT

=⎟⎟⎠

⎞⎜⎜⎝

⎛−

=⎟⎟⎠

⎞⎜⎜⎝

⎛ −⎟⎟⎠

⎞⎜⎜⎝

⎛−

=

⎥⎦

⎤⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛−⎟⎟

⎠

⎞⎜⎜⎝

⎛⎟⎟⎠

⎞⎜⎜⎝

⎛−

=−

1003

2613

4/12/14/12/1

2211

1411

4/12/14/12/11

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 Ais 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

( )

( )

⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

=

⎟⎟⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜⎜⎜

⎝

⎛

=⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛

==

∑

∑∑∑

∞

=

∞

=∞

=

∞

=

t

t

n

nn

n

n

n

n

nn

n

n

nn

ne

e

nt

nt

nt

ntt

λ

λ

λ

λ

λ

λ

000000

!00

00

00!

!00

0000

!)(

1

0

0

1

0

1

0

O

OODQ

Fundamental Matrix for Original System (3 of 3)

To obtain a fundamental matrix Ψ(t) for x' = Ax, recall that the columns of Ψ(t) consist of fundamental solutions x satisfying x' = Ax. We also know x = Ty, and hence it follows that

The columns of Ψ(t) given the expected fundamental solutions of x' = Ax.

⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

=⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛

==tn

nt

n

tnt

t

t

nnn

n

n

n

n ee

ee

e

e

λλ

λλ

λ

λ

ξξ

ξξ

ξξ

ξξ

)()1(

)(1

)1(1

)()1(

)(1

)1(1

1

11

000000

L

MOM

L

O

L

MOM

L

TQΨ

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:

A fundamental matrix for y' = Dy is given by Q(t) = eDt:

Thus a fundamental matrix Ψ(t) for x' = Ax is

yyxx ⎟⎟⎠

⎞⎜⎜⎝

⎛−

=′⇒⎟⎟⎠

⎞⎜⎜⎝

⎛=′

1003

1411

⎟⎟⎠

⎞⎜⎜⎝

⎛=

−t

t

ee

t0

0)(

3

Q

⎟⎟⎠

⎞⎜⎜⎝

⎛

−=⎟⎟

⎠

⎞⎜⎜⎝

⎛⎟⎟⎠

⎞⎜⎜⎝

⎛−

==−

−

− tt

tt

t

t

eeee

ee

t220

02211

)( 3

33

TQΨ

Ch 7.8: Repeated EigenvaluesWe consider again a homogeneous system of n first order linear equations with constant real coefficients x' = Ax.If the eigenvalues r1,…, rn of A are real and different, then there are n linearly independent eigenvectors ξ(1),…, ξ(n), and nlinearly independent solutions of the form

If some of the eigenvalues r1,…, rn are repeated, then there may not be n corresponding linearly independent solutions of the above form.In this case, we will seek additional solutions that are products of polynomials and exponential functions.

trnntr netet )()()1()1( )(,,)( 1 ξxξx == K

Example 1: Direction Field (1 of 12)

Consider the homogeneous equation x' = Ax below.

A direction field for this system is given below.Substituting x = ξert in for x, and rewriting system as (A-rI)ξ = 0, we obtain

xx ⎟⎟⎠

⎞⎜⎜⎝

⎛ −=′

3111

⎟⎟⎠

⎞⎜⎜⎝

⎛=⎟⎟

⎠

⎞⎜⎜⎝

⎛⎟⎟⎠

⎞⎜⎜⎝

⎛−−−

00

3111

1

1

ξξ

rr

Example 1: Eigenvalues (2 of 12)

Solutions have the form x = ξert, where r and ξ satisfy

To determine r, solve det(A-rI) = 0:

Thus r1 = 2 and r2 = 2.

22 )2(441)3)(1(31

11−=+−=+−−=

−−−

rrrrrr

r

⎟⎟⎠

⎞⎜⎜⎝

⎛=⎟⎟

⎠

⎞⎜⎜⎝

⎛⎟⎟⎠

⎞⎜⎜⎝

⎛−−−

00

3111

1

1

ξξ

rr

Example 1: Eigenvectors (3 of 12)

To find the eigenvectors, we solve

by row reducing the augmented matrix:

Thus there is only one eigenvector for the repeated eigenvalue r = 2.

( ) ⎟⎟⎠

⎞⎜⎜⎝

⎛=⎟⎟

⎠

⎞⎜⎜⎝

⎛⎟⎟⎠

⎞⎜⎜⎝

⎛ −−⇔⎟⎟

⎠

⎞⎜⎜⎝

⎛=⎟⎟

⎠

⎞⎜⎜⎝

⎛⎟⎟⎠

⎞⎜⎜⎝

⎛−−−

⇔=−00

1111

00

231121

2

1

2

1

ξξ

ξξ

0ξIA r

⎟⎟⎠

⎞⎜⎜⎝

⎛−

=→⎟⎟⎠

⎞⎜⎜⎝

⎛−=→

==+

→⎟⎟⎠

⎞⎜⎜⎝

⎛→⎟⎟

⎠

⎞⎜⎜⎝

⎛→⎟⎟

⎠

⎞⎜⎜⎝

⎛ −−

11

choose

00011

000011

011011

011011

)1(

2

2)1(

2

21

ξξξξ

ξξξ

Example 1: First Solution; andSecond Solution, First Attempt (4 of 12)

The corresponding solution x = ξert of x' = Ax is

Since there is no second solution of the form x = ξert, we need to try a different form. Based on methods for second order linear equations in Ch 3.5, we first try x = ξte2t. Substituting x = ξte2t into x' = Ax, we obtain

or

tet 2)1(

11

)( ⎟⎟⎠

⎞⎜⎜⎝

⎛−

=x

ttt tetee 222 2 Aξξξ =+

02 222 =−+ ttt teete Aξξξ

Example 1: Second Solution, Second Attempt (5 of 12)

From the previous slide, we have

In order for this equation to be satisfied for all t, it is necessary for the coefficients of te2t and e2t to both be zero.From the e2t term, we see that ξ = 0, and hence there is no nonzero solution of the form x = ξte2t.Since te2t and e2t appear in the above equation, we next consider a solution of the form

02 222 =−+ ttt teete Aξξξ

tt ete 22 ηξx +=

Example 1: Second Solution and its Defining Matrix Equations (6 of 12)

Substituting x = ξte2t + ηe2t into x' = Ax, we obtain

or

Equating coefficients yields Aξ = 2ξ and Aη = ξ + 2η, or

The first equation is satisfied if ξ is an eigenvector of Acorresponding to the eigenvalue r = 2. Thus

( )ttttt eteetee 22222 22 ηξAηξξ +=++

( ) tttt eteete 2222 22 AηAξηξξ +=++

( ) ( ) ξηIA0ξIA =−=− 2and2

⎟⎟⎠

⎞⎜⎜⎝

⎛−

=11

ξ

Example 1: Solving for Second Solution (7 of 12)

Recall that

Thus to solve (A – 2I)η = ξ for η, we row reduce the corresponding augmented matrix:

⎟⎟⎠

⎞⎜⎜⎝

⎛−

+⎟⎟⎠

⎞⎜⎜⎝

⎛−

=→⎟⎟⎠

⎞⎜⎜⎝

⎛−−

=→

−−=→⎟⎟⎠

⎞⎜⎜⎝

⎛ −→⎟⎟

⎠

⎞⎜⎜⎝

⎛−−

→⎟⎟⎠

⎞⎜⎜⎝

⎛−

−−

11

10

1

1000111

111111

111111

1

1

12

kηηη

η

ηη

⎟⎟⎠

⎞⎜⎜⎝

⎛−

=⎟⎟⎠

⎞⎜⎜⎝

⎛ −=

11

,3111

ξA

Example 1: Second Solution (8 of 12)

Our second solution x = ξte2t + ηe2t is now

Recalling that the first solution was

we see that our second solution is simply

since the last term of third term of x is a multiple of x(1).

ttt ekete 222

11

10

11

⎟⎟⎠

⎞⎜⎜⎝

⎛−

+⎟⎟⎠

⎞⎜⎜⎝

⎛−

+⎟⎟⎠

⎞⎜⎜⎝

⎛−

=x

,11

)( 2)1( tet ⎟⎟⎠

⎞⎜⎜⎝

⎛−

=x

,10

11

)( 22)2( tt etet ⎟⎟⎠

⎞⎜⎜⎝

⎛−

+⎟⎟⎠

⎞⎜⎜⎝

⎛−

=x

Example 1: General Solution (9 of 12)

The two solutions of x' = Ax are

The Wronskian of these two solutions is

Thus x(1) and x(2) are fundamental solutions, and the general solution of x' = Ax is

[ ] 0)(, 4222

22)2()1( ≠−=

−−= t

ttt

tt

eetee

teetW xx

⎥⎦

⎤⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛−

+⎟⎟⎠

⎞⎜⎜⎝

⎛−

+⎟⎟⎠

⎞⎜⎜⎝

⎛−

=

+=

ttt etecec

tctct

222

21

)2(2

)1(1

10

11

11

)()()( xxx

ttt etetet 22)2(2)1(

10

11

)(,11

)( ⎟⎟⎠

⎞⎜⎜⎝

⎛−

+⎟⎟⎠

⎞⎜⎜⎝

⎛−

=⎟⎟⎠

⎞⎜⎜⎝

⎛−

= xx

Example 1: Phase Plane (10 of 12)

The general solution is

Thus x is unbounded as t →∞, and x → 0 as t → -∞. Further, it can be shown that as t → -∞, x → 0 asymptotic to the line x2 = -x1 determined by the first eigenvector. Similarly, as t →∞, x is asymptotic to a line parallel to x2 = -x1.

⎥⎦

⎤⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛−

+⎟⎟⎠

⎞⎜⎜⎝

⎛−

+⎟⎟⎠

⎞⎜⎜⎝

⎛−

= ttt etecect 222

21 1

011

11

)(x

Example 1: Phase Plane (11 of 12)

The origin is an improper node, and is unstable. See graph. The pattern of trajectories is typical for two repeated eigenvalues with only one eigenvector. If the eigenvalues are negative, then the trajectories are similar but are traversed in the inward direction. In this case the origin is an asymptotically stable improper node.

Example 1: Time Plots for General Solution (12 of 12)

Time plots for x1(t) are given below, where we note that the general solution x can be written as follows.

⎟⎟⎠

⎞⎜⎜⎝

⎛

−+−+

=⎟⎟⎠

⎞⎜⎜⎝

⎛⇔

⎥⎦

⎤⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛−

+⎟⎟⎠

⎞⎜⎜⎝

⎛−

+⎟⎟⎠

⎞⎜⎜⎝

⎛−

=

tt

tt

ttt

tececctecec

txtx

etecect

22

221

22

21

2

1

222

21

)()()(

10

11

11

)(x

General Case for Double EigenvaluesSuppose the system x' = Ax has a double eigenvalue r = ρand a single corresponding eigenvector ξ.The first solution is

x(1) = ξeρ t, where ξ satisfies (A-ρI)ξ = 0. As in Example 1, the second solution has the form

where ξ is as above and η satisfies (A-ρI)η = ξ.

Since ρ is an eigenvalue, det(A-ρI) = 0, and (A-ρI)η = bdoes not have a solution for all b. However, it can be shown that (A-ρI)η = ξ always has a solution. The vector η is called a generalized eigenvector.

tt ete ρρ ηξx +=)2(

Example 2: Fundamental Matrix Ψ (1 of 2)

Recall that a fundamental matrix Ψ(t) for x' = Ax has linearly independent solution for its columns. In Example 1, our system x' = Ax was

and the two solutions we found were

Thus the corresponding fundamental matrix is

⎟⎟⎠

⎞⎜⎜⎝

⎛−−−

=⎟⎟⎠

⎞⎜⎜⎝

⎛

−−−=

111

)( 2222

22

tt

eetee

teet t

ttt

tt

Ψ

ttt etetet 22)2(2)1(

10

11

)(,11

)( ⎟⎟⎠

⎞⎜⎜⎝

⎛−

+⎟⎟⎠

⎞⎜⎜⎝

⎛−

=⎟⎟⎠

⎞⎜⎜⎝

⎛−

= xx

xx ⎟⎟⎠

⎞⎜⎜⎝

⎛ −=′

3111

Example 2: Fundamental Matrix Φ (2 of 2)

The fundamental matrix Φ(t) that satisfies Φ(0) = I can be found using Φ(t) = Ψ(t)Ψ-1(0), where

where Ψ-1(0) is found as follows:

Thus

,1101

)0(,1101

)0( 1⎟⎟⎠

⎞⎜⎜⎝

⎛−−

=⎟⎟⎠

⎞⎜⎜⎝

⎛−−

= −ΨΨ

⎟⎟⎠

⎞⎜⎜⎝

⎛−−

→⎟⎟⎠

⎞⎜⎜⎝

⎛−

→⎟⎟⎠

⎞⎜⎜⎝

⎛−− 1110

010111100101

10110101

⎟⎟⎠

⎞⎜⎜⎝

⎛+−−

=⎟⎟⎠

⎞⎜⎜⎝

⎛−−⎟⎟

⎠

⎞⎜⎜⎝

⎛−−−

=1

11101

111

)( 22

tttt

ett

et ttΦ

Jordan FormsIf A is n x n with n linearly independent eigenvectors, then Acan be diagonalized using a similarity transform T-1AT = D.The transform matrix T consisted of eigenvectors of A, and the diagonal entries of D consisted of the eigenvalues of A. In the case of repeated eigenvalues and fewer than n linearly independent eigenvectors, A can be transformed into a nearly diagonal matrix J, called the Jordan form of A, with

T-1AT = J.

Example 3: Transform Matrix (1 of 2)

In Example 1, our system x' = Ax was

with eigenvalues r1 = 2 and r2 = 2 and eigenvectors

Choosing k = 0, the transform matrix T formed from the two eigenvectors ξ and η is

xx ⎟⎟⎠

⎞⎜⎜⎝

⎛ −=′

3111

⎟⎟⎠

⎞⎜⎜⎝

⎛−

+⎟⎟⎠

⎞⎜⎜⎝

⎛−

=⎟⎟⎠

⎞⎜⎜⎝

⎛−

=11

10

,11

kηξ

⎟⎟⎠

⎞⎜⎜⎝

⎛−−

=1101

T

Example 3: Jordan Form (2 of 2)

The Jordan form J of A is defined by T-1AT = J.Now

and hence

Note that the eigenvalues of A, r1 = 2 and r2 = 2, are on the main diagonal of J, and that there is a 1 directly above the second eigenvalue. This pattern is typical of Jordan forms.

⎟⎟⎠

⎞⎜⎜⎝

⎛−−

=⎟⎟⎠

⎞⎜⎜⎝

⎛−−

= −

1101

,1101 1TT

⎟⎟⎠

⎞⎜⎜⎝

⎛=⎥

⎦

⎤⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛−−⎟⎟

⎠

⎞⎜⎜⎝

⎛ −⎟⎟⎠

⎞⎜⎜⎝

⎛−−

== −

2012

1101

3111

11011ATTJ

Ch 7.9: Nonhomogeneous Linear SystemsThe 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

tgxtpxtpxtpxtgxtpxtpxtpx

nnnnnnn

nn

nn

++++=′

++++=′++++=′

K

M

K

K

⎟⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜⎜

⎝

⎛

=

⎟⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜⎜

⎝

⎛

=

⎟⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜⎜

⎝

⎛

=

)()()(

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

)(,

)(

)()(

)(,

)(

)()(

)(

21

22221

11211

2

1

2

1

tptptp

tptptptptptp

t

tg

tgtg

t

tx

txtx

t

nnnn

n

n

nn L

MOMM

L

L

MMPgx

General SolutionThe general solution of x' = P(t)x + g(t) on I: α < t < β has the form

where

is the general solution of the homogeneous system x' = P(t)xand v(t) is a particular solution of the nonhomogeneous system x' = P(t)x + g(t).

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

)1(1 ttctctc n

n vxxxx ++++= K

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

)1(1 tctctc n

nxxx +++ K

DiagonalizationSuppose x' = Ax + g(t), where A is an n x n diagonalizable constant matrix. Let T be the nonsingular transform matrix whose columns are the eigenvectors of A, and D the diagonal matrix whose diagonal entries are the corresponding eigenvalues of A.Suppose x satisfies x' = Ax, let y be defined by x = Ty. Substituting x = Ty into x' = Ax, we obtain

Ty' = ATy + g(t).or y' = T-1ATy + T-1g(t)or y' = Dy + h(t), where h(t) = T-1g(t).Note that if we can solve diagonal system y' = Dy + h(t) for y, then x = Ty is a solution to the original system.

Solving Diagonal SystemNow y' = Dy + h(t) is a diagonal system of the form

where r1,…, rn are the eigenvalues of A. Thus y' = Dy + h(t) is an uncoupled system of n linear first order equations in the unknowns yk(t), which can be isolated

and solved separately, using methods of Section 2.1:

⎟⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜⎜

⎝

⎛

+

⎟⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜⎜

⎝

⎛

⎟⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜⎜

⎝

⎛

=

⎟⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜⎜

⎝

⎛

′

′′

⇔

⎪⎪⎭

⎪⎪⎬

⎫

++++=′

++++=′++++=′

nnnnnnn

n

n

h

hh

y

yy

r

rr

y

yy

thyryyy

thyyryythyyyry

MM

L

MOMM

L

L

M

K

M

K

K

2

1

2

1

2

1

2

1

121

22212

12111

00

0000

)(00

)(00)(00

nkthyry kkk ,,1),(1 K=+=′

nkecdssheeyt

t

trkk

srtrk

kkk ,,1,)(0

K=+= ∫ −

Solving Original SystemThe solution y to y' = Dy + h(t) has components

For this solution vector y, the solution to the original system x' = Ax + g(t) is then x = Ty. Recall that T is the nonsingular transform matrix whose columns are the eigenvectors of A. Thus, when multiplied by T, the second term on right side of yk produces general solution of homogeneous equation, while the integral term of yk produces a particular solution of nonhomogeneous system.

nkecdssheeyt

t

trkk

srtrk

kkk ,,1,)(0

K=+= ∫ −

Example 1: General Solution of Homogeneous Case (1 of 5)

Consider the nonhomogeneous system x' = Ax + g below.

Note: A is a Hermitian matrix, since it is real and symmetric.The eigenvalues of A are r1 = -3 and r2 = -1, with corresponding eigenvectors

The general solution of the homogeneous system is then

)(3

22112

tt

e t

gAxxx +=⎟⎟⎠

⎞⎜⎜⎝

⎛+⎟⎟

⎠

⎞⎜⎜⎝

⎛−

−=′

−

⎟⎟⎠

⎞⎜⎜⎝

⎛=⎟⎟

⎠

⎞⎜⎜⎝

⎛−

=11

,11 )2()1( ξξ

tt ecect −−⎟⎟⎠

⎞⎜⎜⎝

⎛+⎟⎟

⎠

⎞⎜⎜⎝

⎛−

=11

11

)( 23

1x

Example 1: Transformation Matrix (2 of 5)

Consider next the transformation matrix T of eigenvectors. Using a Section 7.7 comment, and A Hermitian, we haveT-1 = T* = TT, provided we normalize ξ(1)and ξ(2) so that (ξ(1), ξ(1)) = 1 and (ξ(2), ξ(2)) = 1. Thus normalize as follows:

Then for this choice of eigenvectors,

⎟⎟⎠

⎞⎜⎜⎝

⎛=⎟⎟

⎠

⎞⎜⎜⎝

⎛+

=

⎟⎟⎠

⎞⎜⎜⎝

⎛−

=⎟⎟⎠

⎞⎜⎜⎝

⎛−−−+

=

11

21

11

)1)(1()1)(1(1

,11

21

11

)1)(1()1)(1(1

)2(

)1(

ξ

ξ

⎟⎟⎠

⎞⎜⎜⎝

⎛ −=⎟⎟

⎠

⎞⎜⎜⎝

⎛−

= −

1111

21,

1111

21 1TT

Example 1: Diagonal System and its Solution (3 of 5)

Under the transformation x = Ty, we obtain the diagonal system y' = Dy + T-1g(t):

Then, using methods of Section 2.1,

⎟⎟⎠

⎞⎜⎜⎝

⎛

+−

+⎟⎟⎠

⎞⎜⎜⎝

⎛−−

=

⎟⎟⎠

⎞⎜⎜⎝

⎛⎟⎟⎠

⎞⎜⎜⎝

⎛ −+⎟⎟

⎠

⎞⎜⎜⎝

⎛⎟⎟⎠

⎞⎜⎜⎝

⎛−

−=⎟⎟

⎠

⎞⎜⎜⎝

⎛′′

−

−

−

tete

yy

te

yy

yy

t

t

t

3232

213

32

1111

21

1003

2

1

2

1

2

1

( ) ttt

ttt

ectteyteyy

ecteyteyy

−−−

−−−

+−+=⇒+=+′

+⎟⎠⎞

⎜⎝⎛ −−=⇒−=+′

2222

31111

12

322

32

91

323

22

2323

Example 1: Transform Back to Original System (4 of 5)

We next use the transformation x = Ty to obtain the solution to the original system x' = Ax + g(t):

( )

2,

2,

352

21

34

21

123

61

221

1111

1111

21

22

11

23

1

23

1

2

31

2

1

2

1

ckcktetekek

tetekek

ektte

ekte

yy

xx

ttt

ttt

tt

tt

==

⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

+−+⎟⎠⎞

⎜⎝⎛ −+−

+−+⎟⎠⎞

⎜⎝⎛ ++

=

⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

+−+

+⎟⎠⎞

⎜⎝⎛ −−

⎟⎟⎠

⎞⎜⎜⎝

⎛−

=⎟⎟⎠

⎞⎜⎜⎝

⎛⎟⎟⎠

⎞⎜⎜⎝

⎛−

=⎟⎟⎠

⎞⎜⎜⎝

⎛

−−−

−−−

−−

−−

Example 1: Solution of Original System (5 of 5)

Simplifying further, the solution x can be written as

Note that the first two terms on right side form the general solution to homogeneous system, while the remaining terms are a particular solution to nonhomogeneous system.

⎟⎟⎠

⎞⎜⎜⎝

⎛−⎟⎟

⎠

⎞⎜⎜⎝

⎛+⎟⎟

⎠

⎞⎜⎜⎝

⎛+⎟⎟

⎠

⎞⎜⎜⎝

⎛−

+⎟⎟⎠

⎞⎜⎜⎝

⎛+⎟⎟

⎠

⎞⎜⎜⎝

⎛−

=

⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

+−+⎟⎠⎞

⎜⎝⎛ −+−

+−+⎟⎠⎞

⎜⎝⎛ ++

=⎟⎟⎠

⎞⎜⎜⎝

⎛

−−−−

−−−

−−−

54

31

21

11

11

21

11

11

352

21

34

21

23

1

23

1

23

1

2

1

tteeekek

tetekek

tetekek

xx

tttt

ttt

ttt

Nondiagonal CaseIf A cannot be diagonalized, (repeated eigenvalues and a shortage of eigenvectors), then it can be transformed to its Jordan form J, which is nearly diagonal. In this case the differential equations are not totally uncoupled, because some rows of J have two nonzero entries: an eigenvalue in diagonal position, and a 1 in adjacent position to the right of diagonal position. However, the equations for y1,…, yn can still be solved consecutively, starting with yn. Then the solution x to original system can be found using x = Ty.

Undetermined CoefficientsA second way of solving x' = P(t)x + g(t) is the method of undetermined coefficients. Assume P is a constant matrix, and that the components of g are polynomial, exponential or sinusoidal functions, or sums or products of these. The procedure for choosing the form of solution is usually directly analogous to that given in Section 3.6. The main difference arises when g(t) has the form ueλt,where λ is a simple eigenvalue of P. In this case, g(t) matches solution form of homogeneous system x' = P(t)x, and as a result, it is necessary to take nonhomogeneous solution to be of the form ateλt + beλt. This form differs from the Section 3.6 analog, ateλt.

Example 2: Undetermined Coefficients (1 of 5)

Consider again the nonhomogeneous system x' = Ax + g:

Assume a particular solution of the form

where the vector coefficents a, b, c, d are to be determined. Since r = -1 is an eigenvalue of A, it is necessary to include both ate-t and be-t, as mentioned on the previous slide.

tet

e tt

⎟⎟⎠

⎞⎜⎜⎝

⎛+⎟⎟

⎠

⎞⎜⎜⎝

⎛+⎟⎟

⎠

⎞⎜⎜⎝

⎛−

−=⎟⎟

⎠

⎞⎜⎜⎝

⎛+⎟⎟

⎠

⎞⎜⎜⎝

⎛−

−=′ −

−

30

02

2112

32

2112

xxx

dcbav +++= −− tetet tt)(

Example 2: Matrix Equations for Coefficients (2 of 5)

Substituting

in for x in our nonhomogeneous system x' = Ax + g,

we obtain

Equating coefficients, we conclude that

,30

02

2112

te t⎟⎟⎠

⎞⎜⎜⎝

⎛+⎟⎟

⎠

⎞⎜⎜⎝

⎛+⎟⎟

⎠

⎞⎜⎜⎝

⎛−

−=′ −xx

dcbav +++= −− tetet tt)(

cAdAcbaAbaAa =⎟⎟⎠

⎞⎜⎜⎝

⎛−=⎟⎟

⎠

⎞⎜⎜⎝

⎛−−=−= ,

30

,02

,

( ) teteteete ttttt⎟⎟⎠

⎞⎜⎜⎝

⎛+⎟⎟

⎠

⎞⎜⎜⎝

⎛++++=+−+− −−−−−

30

02

AdAcAbAacbaa

Example 2: Solving Matrix Equation for a (3 of 5)

Our matrix equations for the coefficients are:

From the first equation, we see that a is an eigenvector of Acorresponding to eigenvalue r = -1, and hence has the form

We will see on the next slide that α = 1, and hence

cAdAcbaAbaAa =⎟⎟⎠

⎞⎜⎜⎝

⎛−=⎟⎟

⎠

⎞⎜⎜⎝

⎛−−=−= ,

30

,02

,

⎟⎟⎠

⎞⎜⎜⎝

⎛=

αα

a

⎟⎟⎠

⎞⎜⎜⎝

⎛=

11

a

Example 2: Solving Matrix Equation for b (4 of 5)

Our matrix equations for the coefficients are:

Substituting aT = (α,α) into second equation,

Thus α = 1, and solving for b, we obtain

cAdAcbaAbaAa =⎟⎟⎠

⎞⎜⎜⎝

⎛−=⎟⎟

⎠

⎞⎜⎜⎝

⎛−−=−= ,

30

,02

,

⎟⎟⎠

⎞⎜⎜⎝

⎛−

=⎟⎟⎠

⎞⎜⎜⎝

⎛⎟⎟⎠

⎞⎜⎜⎝

⎛ −⇔⎟⎟

⎠

⎞⎜⎜⎝

⎛ −=⎟⎟

⎠

⎞⎜⎜⎝

⎛⎟⎟⎠

⎞⎜⎜⎝

⎛−

−⇔

⎟⎟⎠

⎞⎜⎜⎝

⎛ −=⎟⎟

⎠

⎞⎜⎜⎝

⎛⎥⎦

⎤⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛+⎟⎟⎠

⎞⎜⎜⎝

⎛−

−⇔⎟⎟

⎠

⎞⎜⎜⎝

⎛−⎟⎟⎠

⎞⎜⎜⎝

⎛ −=

100112

1111

21001

21122

2

1

2

1

2

1

2

1

αα

αα

αα

αα

bb

bb

bb

bb

Ab

⎟⎟⎠

⎞⎜⎜⎝

⎛−

=→=→⎟⎟⎠

⎞⎜⎜⎝

⎛−⎟⎟⎠

⎞⎜⎜⎝

⎛=

10

0 choose10

11

bb kk

Example 2: Particular Solution (5 of 5)

Our matrix equations for the coefficients are:

Solving third equation for c, and then fourth equation for d, it is straightforward to obtain cT = (1, 2), dT = (-4/3, -5/3). Thus our particular solution of x' = Ax + g is

Comparing this to the result obtained in Example 1, we see that both particular solutions would be the same if we had chosen k = ½ for b on previous slide, instead of k = 0.

cAdAcbaAbaAa =⎟⎟⎠

⎞⎜⎜⎝

⎛−=⎟⎟

⎠

⎞⎜⎜⎝

⎛−−=−= ,

30

,02

,

⎟⎟⎠

⎞⎜⎜⎝

⎛−⎟⎟

⎠

⎞⎜⎜⎝

⎛+⎟⎟

⎠

⎞⎜⎜⎝

⎛−⎟⎟

⎠

⎞⎜⎜⎝

⎛= −−

54

31

21

10

11

)( tetet ttv

Variation of Parameters: PreliminariesA more general way of solving x' = P(t)x + g(t) is the method of variation of parameters. Assume P(t) and g(t) are continuous on α < t < β, and let Ψ(t) be a fundamental matrix for the homogeneous system. Recall that the columns of Ψ are linearly independent solutions of x' = P(t)x, and hence Ψ(t) is invertible on the interval α < t < β, and also Ψ'(t) = P(t)Ψ(t). Next, recall that the solution of the homogeneous system can be expressed as x = Ψ(t)c. Analogous to Section 3.7, assume the particular solution of the nonhomogeneous system has the form x = Ψ(t)u(t),where u(t) is a vector to be found.

Variation of Parameters: SolutionWe assume a particular solution of the form x = Ψ(t)u(t). Substituting this into x' = P(t)x + g(t), we obtain

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

u'(t) = Ψ-1(t)g(t)Thus

where the vector c is an arbitrary constant of integration.The general solution to x' = P(t)x + g(t) is therefore

cΨu += ∫ − dttgtt )()()( 1

( ) arbitrary,,)()()()( 11

1

βα∈+= ∫ − tdsssttt

tgΨΨcΨx

Variation of Parameters: Initial Value ProblemFor 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

Alternatively, recall that the fundamental matrix Φ(t) satisfies Φ(t0) = I, and hence the general solution is

In practice, it may be easier to row reduce matrices and solve necessary equations than to compute Ψ-1(t) and substitute into equations. See next example.

∫ −− +=t

tdsssttt

0

)()()()()( 1)0(0

1 gΨΨxΨΨx

∫ −+=t

tdssstt

0

)()()()( 1)0( gΨΦxΦx

Example 3: Variation of Parameters (1 of 3)

Consider again the nonhomogeneous system x' = Ax + g:

We have previously found general solution to homogeneous case, with corresponding fundamental matrix:

Using variation of parameters method, our solution is given by x = Ψ(t)u(t), where u(t) satisfies Ψ(t)u'(t) = g(t), or

tet

e tt

⎟⎟⎠

⎞⎜⎜⎝

⎛+⎟⎟

⎠

⎞⎜⎜⎝

⎛+⎟⎟

⎠

⎞⎜⎜⎝

⎛−

−=⎟⎟

⎠

⎞⎜⎜⎝

⎛+⎟⎟

⎠

⎞⎜⎜⎝

⎛−

−=′ −

−

30

02

2112

32

2112

xxx

⎟⎟⎠

⎞⎜⎜⎝

⎛

−=

−−

−−

tt

tt

eeee

t 3

3

)(Ψ

⎟⎟⎠

⎞⎜⎜⎝

⎛=⎟⎟

⎠

⎞⎜⎜⎝

⎛′′

⎟⎟⎠

⎞⎜⎜⎝

⎛

−

−

−−

−−

te

uu

eeee t

tt

tt

32

2

13

3

Example 3: Solving for u(t) (2 of 3)

Solving Ψ(t)u'(t) = g(t) by row reduction,

It follows that

2/312/3

2/31102/301

2/302/30

2/302

32202

32

2

321

32

33

3

3

3

t

tt

t

tt

tt

tt

tt

ttt

tt

ttt

tt

ttt

teuteeu

tetee

teetee

teeeee

teeeee

teeeee

+=′−=′

→⎟⎟⎠

⎞⎜⎜⎝

⎛

+−

→

⎟⎟⎠

⎞⎜⎜⎝

⎛

+−

→⎟⎟⎠

⎞⎜⎜⎝

⎛

+→

⎟⎟⎠

⎞⎜⎜⎝

⎛

+→⎟⎟

⎠

⎞⎜⎜⎝

⎛

−

−−

−−

−−

−−−

−−

−−−

−−

−−−

⎟⎟⎠

⎞⎜⎜⎝

⎛

+−+++−

=⎟⎟⎠

⎞⎜⎜⎝

⎛=

2

1332

2

1

2/32/36/2/2/

)(cetet

ceteeuu

t tt

ttt

u

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

Now x(t) = Ψ(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

1332

3

3

2/32/36/2/2/

cetetcetee

eeee

tt

ttt

tt

tt

x

⎟⎟⎠

⎞⎜⎜⎝

⎛−⎟⎟

⎠

⎞⎜⎜⎝

⎛+⎟⎟

⎠

⎞⎜⎜⎝

⎛−

+⎟⎟⎠

⎞⎜⎜⎝

⎛+⎟⎟

⎠

⎞⎜⎜⎝

⎛+⎟⎟

⎠

⎞⎜⎜⎝

⎛−

= −−−−

54

31

21

11

21

11

11

11

23

1 teteecec ttttx

Laplace TransformsThe Laplace transform can be used to solve systems of equations. Here, the transform of a vector is the vector of component transforms, denoted by X(s):

Then by extending Theorem 6.2.1, we obtain

{ } { }{ } )(

)()(

)()(

)(2

1

2

1 stxLtxL

txtx

LtL Xx =⎟⎟⎠

⎞⎜⎜⎝

⎛=

⎭⎬⎫

⎩⎨⎧

⎟⎟⎠

⎞⎜⎜⎝

⎛=

{ } )0()()( xXx −=′ sstL

Example 4: Laplace Transform (1 of 5)

Consider again the nonhomogeneous system x' = Ax + g:

Taking the Laplace transform of each term, we obtain

where G(s) is the transform of g(t), and is given by

⎟⎟⎠

⎞⎜⎜⎝

⎛+⎟⎟

⎠

⎞⎜⎜⎝

⎛−

−=′

−

te t

32

2112

xx

( )⎟⎟⎠

⎞⎜⎜⎝

⎛ += 23

12)(

ss

sG

)()()0()( ssss GAXxX +=−

Example 4: Transfer Matrix (2 of 5)

Our transformed equation is

If we take x(0) = 0, then the above equation becomes

or

Solving for X(s), we obtain

The matrix (sI – A)-1 is called the transfer matrix.

)()()0()( ssss GAXxX +=−

)()()( ssss GAXX +=

( ) )()( 1 sss GAIX −−=

( ) )()( sss GXAI =−

Example 4: Finding Transfer Matrix (3 of 5)

Then

Solving for (sI – A)-1, we obtain

( ) ⎟⎟⎠

⎞⎜⎜⎝

⎛+−−+

=−⇒⎟⎟⎠

⎞⎜⎜⎝

⎛−

−=

2112

2112

ss

s AIA

( ) ⎟⎟⎠

⎞⎜⎜⎝

⎛+

+++

=− −

2112

)3)(1(11

ss

sss AI

Example 4: Transfer Matrix (4 of 5)

Next, X(s) = (sI – A)-1G(s), and hence

or

⎟⎟⎠

⎞⎜⎜⎝

⎛ +⎟⎟⎠

⎞⎜⎜⎝

⎛+

+++

= 23)1(2

2112

)3)(1(1)(

ss

ss

sssX

( )( ) ( )

( ) ( ) ⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

+++

+++

+++

+++

=

)3(1)2(3

)3(12

)3(13

)3(122

)(22

22

ssss

ss

ssssss

sX

Example 4: Transfer Matrix (5 of 5)

Thus

To solve for x(t) = L-1{X(s)}, use partial fraction expansions of both components of X(s), and then Table 6.2.1 to obtain:

Since we assumed x(0) = 0, this solution differs slightly from the previous particular solutions.

( )( ) ( )

( ) ( ) ⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

+++

+++

+++

+++

=

)3(1)2(3

)3(12

)3(13

)3(122

)(22

22

ssss

ss

ssssss

sX

⎟⎟⎠

⎞⎜⎜⎝

⎛−⎟⎟

⎠

⎞⎜⎜⎝

⎛+⎟⎟

⎠

⎞⎜⎜⎝

⎛+⎟⎟

⎠

⎞⎜⎜⎝

⎛+⎟⎟

⎠

⎞⎜⎜⎝

⎛−

−= −−−

54

31

21

11

12

11

32 3 tteee tttx

Summary (1 of 2)

The method of undetermined coefficients requires no integration but is limited in scope and may involve several sets of algebraic equations.Diagonalization requires finding inverse of transformation matrix and solving uncoupled first order linear equations. When coefficient matrix is Hermitian, the inverse of transformation matrix can be found without calculation, which is very helpful for large systems.The Laplace transform method involves matrix inversion, matrix multiplication, and inverse transforms. This method is particularly useful for problems with discontinuous or impulsive forcing functions.

Summary (2 of 2)

Variation of parameters is the most general method, but it involves solving linear algebraic equations with variable coefficients, integration, and matrix multiplication, and hence may be the most computationally complicated method. For many small systems with constant coefficients, all of these methods work well, and there may be little reason to select one over another.