## TABLE OF CONTENTS

Acknowledgment i

Certication ii

Approval iii

Introduction v

Dedication vi

1 Preliminaries 2

1.1 Denitions and basic Theorems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2

1.2 Exponential of matrices . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4

2 Basic Theory of Ordinary Dierential Equations 7

2.1 Denitions and basic properties . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7

2.2 Continuous dependence with respect to the initial conditions . . . . . . . . . . . . 11

2.3 Local existence and blowing up phenomena for ODEs . . . . . . . . . . . . . . . . 12

2.4 Variation of constants formula . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15

3 Stability via linearization principle 21

3.1 Denitions and basic results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21

4 Lyapunov functions and LaSalle’s invariance principle 26

4.1 Denitions and basic results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27

4.2 Instability Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29

4.3 How to search for a Lyapunov function (variable gradient method) . . . . . . . . . 30

4.4 LaSalle’s invariance principle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30

4.5 Barbashin and Krasorskii Corollaries . . . . . . . . . . . . . . . . . . . . . . . . . . 32

4.6 Linear systems and linearization . . . . . . . . . . . . . . . . . . . . . . . . . . . . 36

5 More applications 40

5.1 Control design based on lyapunov’s direct method . . . . . . . . . . . . . . . . . . 41

Conclusion 48

Bibliography 48

## CHAPTER ONE

PRELIMINARIES

1.1 Denitions and basic Theorems

In this chapter, we focussed on the basic concepts of the ordinary dierential equations. Also, we

emphasized on relevant theroems in ordinary dierential equations.

Denition 1.1.1 An equation containing only ordinary derivatives of one or more dependent vari-

ables with respect to a single independent variable is called an ordinary dierential equation ODE.

The order of an ODE is the order of the highest derivative in the equation. In symbol, we can

express an n-th order ODE by the form

x(n) = f(t; x; :::; x(n?1)) (1.1.1)

Denition 1.1.2 (Autonomous ODE ) When f is time-independent, then (1.1.1) is said to be

an autonomous ODE. For example,

x0(t) = sin(x(t))

Denition 1.1.3 (Non-autonomous ODE ) When f is time-dependent, then (1.1.1) is said to

be a non autonomous ODE. For example,

x0(t) = (1 + t2)y2(t)

Denition 1.1.4 f : Rn ! Rn is said to be locally Lipschitz, if for all r > 0 there exists k(r) > 0

such that

kf(x) ? f(y)k k(r)kx ? yk; for all x; y 2 B(0; r):

f : Rn ! Rn is said to be Lipschitz, if there exists k > 0 such that

kf(x) ? f(y)k kkx ? yk; for all x; y 2 Rn:

Denition 1.1.5 (Initial value problem (IVP) Let I be an interval containing x0, the follow-

ing problem (

x(n)(t) = f(t; x(t); :::; x(n?1)(t))

x(t0) = x0; x0(t0) = x1; :::; x(n?1)(t0) = xn?1

(1.1.2)

is called an initial value problem (IVP).

x(t0) = x0; x0(t0) = x1; :::; x(n?1)(t0) = xn?1

are called initial condition.

2

Lemma 1.1.6 [9](Gronwall’s Lemma) Let u; v : [a; b] ! R+ be continuous such that there exists

> 0 such that

u(x) +

x

a

u(s)v(s)ds; for all x 2 [a; b]:

Then,

u(x) e

x

a

v(s)ds

; for all x 2 [a; b]:

Proof .

u(x) +

x

a

u(s)v(s)ds

implies that

u(x)

+

x

a

u(s)v(s)ds

v(x):

So,

u(x)v(x)

+

x

a

u(s)v(s)ds

v(x);

which implies that

x

a

u(x)v(x)

+

x

a

u(s)v(s)ds

ds

x

a

v(x)ds:

So, taking exponential of both side we get

u(x) +

x

a

u(s)v(s)ds

x

a

u(s)v(s)ds:

Thus,

u(x) e

x

a

v(s)ds

; x 2 [a; b]:

Corollary 1.1.7 Let u; v : [a; b] ! R+ be continuous such that

u(x)

x

a

u(s)v(s)ds; for all x 2 [a; b]:

Then, u = 0 on [a; b].

Proof . Now,

u(x)

x

a

u(s)v(s)ds

implies that

u(x)

x

a

u(s)v(s)ds

1

n

+

x

a

u(s)v(s)ds; for all n 1:

So, by Gronwall’s lemma,

u(x)

1

n

e

x

a u(s)v(s)ds;

so as

n ! 1; u(x) ! 0:

Thus, u(x) = 0, since u(x) 0. Hence, u = 0 on [a; b].

3

1.2 Exponential of matrices

Denition 1.2.1 Let A 2 Mnn(R), then eA is an n n matrix given by the power series

eA =

1X

k=0

Ak

k!

The series above converges absolutely for all A 2 Mnn(R)

Proof . The n-th partial sum is

Sn =

Xn

k=0

Ak

k!

So, let n > m Then,

Sn ? Sm =

Xn

k=m+1

Ak

k!

:

So,

kSn ? Smk

Xn

k=m+1

kAkk

k!

:

So as

m ! 1; kSn ? Smk ! 0

So, (Sn)n is Cauchy. Thus, converges.

Theorem 1.2.2 [3](Cayley Hamilton Theorem)

Let A 2 Mnn(R) and () = det(I ? A) its characteristic polynomial then

(A) = 0:

Proof . Let A 2 Mnn(R);

() = det(I ? A) = c0 + c1 + c22 + ::: + cnn:

adj(A ? I) = B0 + B1 + B22 + ::: + Bn?2n?2 + Bn?1n?1;

where Bi 2 Mnn(R) for i = 0; 1; 2; :::; n; but, from linear algebra we have that

A?1 =

adj(A)

det(A)

;

where adj(A) denotes the adjugate or classical adjoint of A. So,

det(I ? tA)I = (I ? tA)adj(I ? tA):

(A ? I)(B0 + B1 + B22 + ::: + Bn?2n?2 + Bn?1n?1) = (c0 + c1 + c22 + ::: + cnn)I:

Observe that the entries in adj(I ?tA) are polynomials in of degree at most n?1. So, Bi is the

zero matrix for i = n. Equating the coecients of n on both sides gives

c0I + c1A + c2A2 + ::: + cnAn = 0:

Thus,

(A) = 0:

4

Example 1.2.3 (Application of Cayley Hamilton Theorem)

Find etA for A =

0 1

?1 0

!

Solution:

The characteristic equation is s2 + 1 = 0, and the eigenvalues are 1 = i, and 2 = ?i. So, by

Theorem 1.2.2 we have that,

etA = 0I + 1A;

where we are to nd the value of 0, and 1. So,

eti = cos t + i sin t = 0 + 1i

e?ti = cos t ? i sin t = 0 ? 1i

which implies that 0 = cos t, and 1 = sin t. So,

etA = cos(t)I + sin(t)A =

cos t sin t

?sin t cos t

!

Theorem 1.2.4 [11] Let A;B 2 Mnn(R). Then,

(1) If 0 denotes the zero matrix, then e0 = I, the identity matrix.

(2) If A is invertible, then eABA?1

= AeBA?1.

Proof . Recall that, for all integers s 0, we have (ABA?1)s = ABsA?1. Now,

eABA?1

= I + ABA?1 +

(ABA?1)2

2!

+ :::

= I + ABA?1 +

AB2A?1

2!

+ :::

= A(I + B +

B2

2!

+ :::)A?1

= AeBA?1:

(3) If A is symmetric such that A = AT , then

e(AT ) = (eA)T :

Proof .

eA =

1X

k=0

Ak

k!

:

Then

eAT

=

1X

k=0

(AT )k

k!

=

1X

k=0

(Ak)T

k!

= (

1X

k=0

Ak

k!

)T = (eA)T :

(4) If AB = BA, then

eA+B = eAeB:

Proof .

eAeB = (I + A +

A2

2!

+

A3

3!

+ :::)(I + B +

B2

2!

+

B3

3!

+ :::)

= (

1X

k=0

Ak

k!

)(

1X

j=0

Bj

j!

)

=

1X

k=0

1X

j=0

(A + B)k+j

j!k!

5

Put m = j + k, then j = m ? k then from the binomial theorem that

eAeB =

1X

m=0

1X

k=0

AmBm?k

(m ? k)!k!

=

1X

m=0

Am

m!

1X

k=0

m!

(m ? k)!

Bm?k

k!

=

1X

m=0

(A + B)m

m!

= eA+B:

Theorem 1.2.5 [9]

detA

dt

= AetA = etAA; for t 2 R:

Proof . x(t; x0) = etAx0. Then,

dx(t; x0)

dt

= etAx0A =

1X

k=0

tkAk

k!

x0A = ( lim

n!1

Xn

k=0

tkAk

k!

)x0A

= lim

n!1

Xn

k=0

tkAk+1

k!

x0 = lim

n!1

Xn

k=0

AtkAk

k!

x0 = A

1X

k=0

tkAk

k!

x0 = AetAx0

So,

detA

dt

= AetA = etAA:

Proposition 1.2.6 The solution x(:; x0) of the following linear space

(

x0(t) = Ax(t); t 2 R

x(0) = x0 2 Rn

where A 2 Mnn(R), is given by

x(t; x0) = etAx0:

6

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