Conformable Dynamic Equations on Time Scales [1 ed.] 0367517019, 9780367517014

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Conformable Dynamic Equations on Time Scales

Conformable Dynamic Equations on Time Scales

Douglas R. Anderson Concordia College

Svetlin G. Georgiev Sorbonne University

First edition published 2020 by CRC Press 6000 Broken Sound Parkway NW, Suite 300, Boca Raton, FL 33487-2742 and by CRC Press 2 Park Square, Milton Park, Abingdon, Oxon, OX14 4RN  c 2020 Douglas R. Anderson, Svetlin G. Georgiev CRC Press is an imprint of Taylor & Francis Group, LLC Reasonable efforts have been made to publish reliable data and information, but the author and publisher cannot assume responsibility for the validity of all materials or the consequences of their use. The authors and publishers have attempted to trace the copyright holders of all material reproduced in this publication and apologize to copyright holders if permission to publish in this form has not been obtained. If any copyright material has not been acknowledged please write and let us know so we may rectify in any future reprint. Except as permitted under U.S. Copyright Law, no part of this book may be reprinted, reproduced, transmitted, or utilized in any form by any electronic, mechanical, or other means, now known or hereafter invented, including photocopying, microfilming, and recording, or in any information storage or retrieval system, without written permission from the publishers. For permission to photocopy or use material electronically from this work, access www.copyright.com or contact the Copyright Clearance Center, Inc. (CCC), 222 Rosewood Drive, Danvers, MA 01923, 978-750-8400. For works that are not available on CCC please contact [email protected] Trademark notice: Product or corporate names may be trademarks or registered trademarks, and are used only for identification and explanation without intent to infringe. ISBN: 978-0-367-51701-4 (hbk) ISBN: 978-1-003-05740-6 (ebk) Visit the Taylor & Francis Web site at http://www.taylorandfrancis.com and the CRC Press Web site at http://www.crcpress.com

Contents Preface

ix

C HAPTER 1  Conformable Dynamic Calculus on Time Scales

1

1.1

INTRODUCTION

1

1.2

CONFORMABLE DIFFERENTIATION

2

1.3

CONFORMABLE REGRESSIVE FUNCTIONS

17

1.4

THE CONFORMABLE EXPONENTIAL FUNCTION

23

1.5

CONFORMABLE HYPERBOLIC AND TRIGONOMETRIC FUNCTIONS

26

1.6

THE CONFORMABLE LOGARITHM FUNCTION

31

1.7

CONFORMABLE INTEGRATION

38

1.8

TAYLOR’S FORMULA

45

1.9

CALCULUS FOR THE NABLA CONFORMABLE DERIVATIVE

48

1.10 CONFORMABLE PARTIAL DERIVATIVES

51

1.11 ADVANCED PRACTICAL PROBLEMS

58

1.12 NOTES AND REFERENCES

59

C HAPTER 2  First-Order Linear Dynamic Equations

61

2.1

LINEAR FIRST-ORDER DYNAMIC EQUATIONS

61

2.2

CONFORMABLE BERNOULLI EQUATIONS

76

2.3

CONFORMABLE RICCATI EQUATIONS

82

2.4

CONFORMABLE LOGISTIC EQUATIONS

87

2.5

ADVANCED PRACTICAL PROBLEMS

90

C HAPTER 3  Conformable Dynamic Systems on Time Scales 3.1

STRUCTURE OF CONFORMABLE DYNAMIC SYSTEMS ON TIME SCALES

93

93

v

vi  Contents

3.2

CONSTANT COEFFICIENTS

124

3.3

ADVANCED PRACTICAL PROBLEMS

137

C HAPTER 4  Linear Conformable Inequalities

139

4.1

CONFORMABLE GRONWALL INEQUALITY

139

4.2

CONFORMABLE VOLTERRA-TYPE INTEGRAL INEQUALITIES

148

4.3

CONFORMABLE INEQUALITIES OF GAMIDOV AND RODRIGUES

153

4.4

SIMULTANEOUS CONFORMABLE INTEGRAL INEQUALITIES

156

4.5

CONFORMABLE PACHPATTE’S INEQUALITIES

157

4.6

A CONFORMABLE INTEGRO-DYNAMIC INEQUALITY

162

C HAPTER 5  Cauchy-Type Problem for a Class of Nonlinear Conformable Dynamic Equations

165

5.1

EXISTENCE AND UNIQUENESS OF SOLUTIONS

165

5.2

THE DEPENDENCY OF THE SOLUTION UPON THE INITIAL DATA

169

5.3

LYAPUNOV FUNCTIONS

170

5.4

BOUNDEDNESS OF SOLUTIONS

172

5.5

EXPONENTIAL STABILITY

178

5.6

ADVANCED PRACTICAL PROBLEMS

183

C HAPTER 6  Higher-Order Linear Conformable Dynamic Equations with Constant Coefficients 6.1

HOMOGENEOUS HIGHER-ORDER LINEAR CONFORMABLE DYNAMIC EQUATIONS WITH CONSTANT COEFFICIENTS

185

185

6.2

NONHOMOGENEOUS HIGHER-ORDER LINEAR CONFORMABLE DYNAMIC EQUATIONS WITH CONSTANT COEFFICIENTS 190

6.3

ADVANCED PRACTICAL PROBLEMS

C HAPTER 7  Second-Order Conformable Dynamic Equations

194

197

7.1

HOMOGENEOUS SECOND-ORDER LINEAR CONFORMABLE DYNAMIC EQUATIONS 197

7.2

REDUCTION OF ORDER

212

Contents  vii

7.3

METHOD OF FACTORING

218

7.4

NONCONSTANT COEFFICIENTS

225

7.5

CONFORMABLE EULER-CAUCHY EQUATIONS

231

7.6

VARIATION OF PARAMETERS

241

7.7

ADVANCED PRACTICAL PROBLEMS

245

C HAPTER 8  Second-Order Self-Adjoint Conformable Dynamic Equations

249

8.1

SELF-ADJOINT DYNAMIC EQUATIONS

249

8.2

REDUCTION-OF-ORDER THEOREMS

260

8.3

DOMINANT AND RECESSIVE SOLUTIONS

262

8.4

RICCATI EQUATION

273

8.5

CAUCHY FUNCTION AND VARIATION OF CONSTANTS FORMULA

276

BOUNDARY VALUE PROBLEMS AND GREEN FUNCTIONS

278

8.6.1

Conjugate Problem and Disconjugacy

282

8.6.2

Right Focal Problem

286

8.6.3

Periodic Problem

288

8.6

C HAPTER 9  The Conformable Laplace Transform

291

9.1

DEFINITION AND PROPERTIES

291

9.2

DECAY OF THE EXPONENTIAL FUNCTION

299

9.3

CONVERGENCE OF THE CONFORMABLE LAPLACE TRANSFORM

304

9.4

APPLICATIONS TO IVPS

310

9.5

ADVANCED PRACTICAL PROBLEMS

313

A PPENDIX A  Derivatives on Banach Spaces

315

A.1

REMAINDERS

315

A.2

´ DEFINITION AND UNIQUENESS OF THE FRECHET DERIVATIVE ˆ THE GATEAUX DERIVATIVE

317

A.3

324

viii  Contents

A PPENDIX B  A Chain Rule B.1 B.2

MEASURE CHAINS ¨ POTZSCHE’S CHAIN RULE

327 327 329

Bibliography

333

Index

335

Preface The concept of derivatives of non-integer order, known as fractional derivatives, first appeared in the letter between L’Hˆopital and Leibniz in which the question of a halforder derivative was posed. Since then, many formulations of fractional derivatives have appeared. Recently, a new definition of fractional derivative, named “fractional conformable derivative”, has been introduced. This new fractional derivative is compatible with the classical derivative and it has attracted attention in domains such as mechanics, electronics and anomalous diffusion. This book is devoted to the qualitative theory of conformable dynamic equations on time scales. It summarizes the most recent contributions in this area, and vastly expands on them to create a comprehensive theory developed exclusively for this book. Except for a few sections in Chapter 1, the results here are presented for the first time. As a result, the book is intended for researchers who work on dynamic calculus on time scales and its applications. There are nine chapters in this book, all with new content not found in journals. In Chapter 1 we introduce the concept of conformable dynamic differentiation and integration, and deduce some of the properties of these operations. Also in this chapter, we define basic elementary functions, such as conformable dynamic exponential, trigonometric, hyperbolic, and logarithmic functions, and we obtain a Taylor formula. Chapter 2 is devoted to first-order linear dynamic equations on time scales. In particular, we discuss conformable dynamic versions of Bernoulli, Riccati, and logistic equations. In Chapter 3, we investigate the structure of conformable dynamic systems with variable and constant coefficients. Chapter 4 deals with linear conformable inequalities such as Gronwall, Gamidov, Pachpatte, and Volterra-type inequalities. Chapter 5 is devoted to a Cauchytype problem for some classes of nonlinear conformable dynamic equations. For this class, we prove the existence and uniqueness of solutions, and investigate the dependency of solutions on initial data. Lyapunov functions are also discussed, including some criteria for the boundedness and exponential stability of solutions. Higher-order linear dynamic equations with constant coefficients are investigated in Chapter 6. Chapter 7 deals with second-order conformable dynamic equations, and second-order self-adjoint conformable dynamic equations are investigated in Chapter 8. Chapter 9 is devoted to the Laplace transform, including its properties and applications. This book is addressed to a wide audience of specialists such as mathematicians, physicists, engineers, and biologists. It can be used as a textbook at the graduate level and as a reference book for several disciplines. The authors welcome any suggestions for improvement of the text. Douglas R. Anderson, Svetlin G. Georgiev Moorhead, Paris, August 2019 ix

CHAPTER

1

Conformable Dynamic Calculus on Time Scales

1.1

INTRODUCTION

Motivated by a proportional-derivative (PD) controller from control theory, in this book we introduce a conformable dynamic calculus on time scales whose differential operator is modeled after a PD controller. This proportional derivative Dα of order α ∈ [0, 1], where D0 is the identity operator, and D1 is the classical time scales delta derivative operator, will be used to construct a new calculus that is then explored extensively. Let T be a time scale with forward jump operator and delta differentiation operator σ and ∆, respectively. Also, let α ∈ [0, 1]. Throughout this monograph we suppose (A1) k0 , k1 : [0, 1] × T → [0, ∞) are continuous functions such that lim k1 (α,t) = 1,

α→0+

lim k0 (α,t) = 0,

α→0+

k1 (α,t) 6= 0,

lim k1 (α,t) = 0,

t ∈ T,

lim k0 (α,t) = 1,

t ∈ T,

α→1−

α→1−

α ∈ [0, 1),

t ∈ T,

k0 (α,t) 6= 0,

α ∈ (0, 1],

t ∈ T.

Such functions k1 and k0 exist. For instance, k1 (α,t) = (1 − α) 1 + t 2


α

k1 (α,t) = (1 − α)|t|α ,

k0 (α,t) = α|t|1−α ,

k1 (α,t) = 1 − α,

,

k0 (α,t) = α 1 + t 2

k0 (α,t) = α,

k1 (α,t) = (1 − α)3α ,

t ∈ T,

k0 (α,t) = α31−α ,


1−α

t ∈ T,

,

t ∈ T,

α ∈ [0, 1],

α ∈ [0, 1],

α ∈ [0, 1], t ∈ T,

α ∈ [0, 1],

1

2  Conformable Dynamic Equations on Time Scales

 π k1 (α,t) = cos α |t|α , 2

 π k0 (α,t) = sin α |t|1−α , 2

t ∈ T,

α ∈ [0, 1],

satisfy (A1).

1.2

CONFORMABLE DIFFERENTIATION

In control theory particularly, a proportional-derivative controller for controller output u at time t with two tuning parameters has the algorithm u(t) = κ p E(t) + κd

d E(t), dt

where κ p is the proportional gain, κd is the derivative gain, and E is the error between the state variable and the process variable. This is the impetus for the definition and results to follow. Definition 1.2.1 Suppose that f is ∆-differentiable at some t ∈ Tκ . The conformable ∆derivative of f at t is defined by Dα f (t) = k1 (α,t) f (t) + k0 (α,t) f ∆ (t).

(1.1)

Here k1 is a type of the proportional gain k p , k0 is a type of the derivative gain kd , f is the error, and Dα f (t) is the controller output.

Remark 1.2.2 We have D0 f (t) = f (t),

D1 f (t) = f ∆ (t).

Remark 1.2.3 Suppose that f is ∆-differentiable at t ∈ Tκ . 1. Let α ∈ [0, 1). Then k1 (α,t) f (t) = Dα f (t) − k0 (α,t) f ∆ (t), whereupon f (t) =

1 k0 (α,t) ∆ Dα f (t) − f (t). k1 (α,t) k1 (α,t)

2. Let α ∈ (0, 1]. Then k0 (α,t) f ∆ (t) = Dα f (t) − k1 (α,t) f (t), whereupon f ∆ (t) =

1 k1 (α,t) Dα f (t) − f (t). k0 (α,t) k0 (α,t)

Conformable Dynamic Calculus on Time Scales  3

3. Let α ∈ (0, 1). Then   Dα f (t) = k1 (α,t) f σ (t) − µ(t) f ∆ (t) + k0 (α,t) f ∆ (t), whereupon k1 (α,t) f σ (t) = Dα f (t) + µ(t)k1 (α,t) f ∆ (t) − k0 (α,t) f ∆ (t) = Dα f (t) + (µ(t)k1 (α,t) − k0 (α,t)) f ∆ (t), and

  1 k0 (α,t) α f (t) = D f (t) + µ(t) − f ∆ (t). k1 (α,t) k1 (α,t) σ

Example 1.2.4 Let T = 2Z, k1 (α,t) = (1 − α)t 2α ,

k0 (α,t) = αt 2(1−α) ,

α ∈ [0, 1],

t ∈ T,

and f (t) = t 3 − t 2 + 2t + 3,

t ∈ T.

In Fig. 1.1 is shown the function f . 1 ´ 106

800 000

600 000 Out[1]=

400 000

200 000

0

Figure 1.1

10

20

30

f (t) for t ∈ [0, 100] in Example 1.2.4

We will find 1

D 2 f (t),

t ∈ T.

40

50

4  Conformable Dynamic Equations on Time Scales

We have σ (t) = t + 2, f ∆ (t) = (σ (t))2 + tσ (t) + t 2 − σ (t) − t + 2 = (t + 2)2 + t(t + 2) + t 2 − t − 2 − t + 2 = t 2 + 4t + 4 + t 2 + 2t + t 2 − 2t = 3t 2 + 4t + 4,

t ∈ T.

Fig. 1.2 shows the delta derivative f ∆ . 30 000

25 000

20 000

Out[5]= 15 000

10 000

5000

0

Figure 1.2

10

20

30

40

f ∆ (t) for t ∈ [0, 100] in Example 1.2.4

Next, 

1 2

D f (t) = k1

1 ,t 2



 f (t) + k0

1 ,t 2

 f ∆ (t)

=


1
1 3 2 t t − t + 2t + 3 + t 3t 2 + 4t + 4 2 2

=


1 3 2 t t − t + 2t + 3 + 3t 2 + 4t + 4 2

50

Conformable Dynamic Calculus on Time Scales  5

=


1 3 t t + 2t 2 + 6t + 7 , 2

t ∈ T. 1

The Fig. 1.3 shows the conformable delta derivative D 2 f . 5 ´ 107

4 ´ 107

3 ´ 107 Out[6]=

2 ´ 107

1 ´ 107

0

Figure 1.3

10

20

30

40

1

D 2 f (t) for t ∈ [0, 100] in Example 1.2.4

Example 1.2.5 Let T = 2N0 , k1 (α,t) = (1 − α)t α , f (t) = t 3 + t,

k0 (α,t) = αt 1−α ,

α ∈ [0, 1],

t ∈ T.

We will find 1

D 4 f (t),

t ∈ T.

Here σ (t) = 2t,

t ∈ T.

Then f ∆ (t) = (σ (t))2 + tσ (t) + t 2 + 1 = (2t)2 + 2t 2 + t 2 + 1

t ∈ T,

50

6  Conformable Dynamic Equations on Time Scales

= 4t 2 + 2t 2 + t 2 + 1 = 7t 2 + 1,

t ∈ T.

Hence, 

1 4

D f (t) = k1

1 ,t 4



 f (t) + k0

1 ,t 4

 f ∆ (t)

=


3 1 3
1 3 t 4 t + t + t 4 7t 2 + 1 4 4

=

3 13 3 5 7 11 1 3 t 4 + t4 + t 4 + t4, 4 4 4 4

t ∈ T.

Theorem 1.2.6 Let f and g be ∆-differentiable at t ∈ Tκ . Then 1. Dα ( f + g)(t) = Dα f (t) + Dα g(t), 2. Dα (a f )(t) = aDα f (t) for any a ∈ R, 3. Dα ( f g)(t) = (Dα f (t)) g(t) + f σ (t) (Dα g(t)) −k1 (α,t) f σ (t)g(t) = (Dα f (t)) gσ (t) + f (t) (Dα g(t)) −k1 (α,t) f (t)gσ (t), 4.

  f g(t)Dα f (t) − f (t)Dα g(t) f (t) D (t) = + k1 (α,t) σ g g(t)g (t) g(t) α

provided that g(t)gσ (t) 6= 0. Proof 1.2.7

1. We have Dα ( f + g)(t) = k1 (α,t)( f + g)(t) + k0 (α,t)( f + g)∆ (t) = k1 (α,t) f (t) + k0 (α,t) f ∆ (t)

Conformable Dynamic Calculus on Time Scales  7

+k1 (α,t)g(t) + k0 (α,t)g∆ (t) = Dα f (t) + Dα g(t). 2. We have Dα (a f )(t) = k1 (α,t)(a f )(t) + k0 (α,t)(a f )∆ (t)   = a k1 (α,t) f (t) + k0 (α,t) f ∆ (t) = aDα f (t). 3. We have Dα ( f g)(t) = k1 (α,t)( f g)(t) + k0 (α,t)( f g)∆ (t) = k1 (α,t) f (t)g(t) + k0 (α,t) f ∆ (t)g(t) +k0 (α,t) f σ (t)g∆ (t) =

  k1 (α,t) f (t) + k0 (α,t) f ∆ (t) g(t)   + k1 (α,t)g(t) + k0 (α,t)g∆ (t) f σ (t) −k1 (α,t) f σ (t)g(t)

= (Dα f (t)) g(t) + (Dα g(t)) f σ (t) −k1 (α,t) f σ (t)g(t) = k1 (α,t) f (t)g(t) + k0 (α,t) f ∆ (t)gσ (t) +k0 (α,t) f (t)g∆ (t) =

  k1 (α,t)g(t) + k0 (α,t)g∆ (t) f (t)   ∆ + k1 (α,t) f (t) + k0 (α,t) f (t) gσ (t)

8  Conformable Dynamic Equations on Time Scales

−k1 (α,t) f (t)gσ (t) = (Dα f (t)) gσ (t) + f (t)Dα g(t) −k1 (α,t) f (t)gσ (t). 4. We have Dα

 ∆   f (t) f f (t) = k1 (α,t) + k0 (α,t) (t) g g(t) g = k1 (α,t)

f (t) f ∆ (t)g(t) − f (t)g∆ (t) + k0 (α,t) g(t) g(t)gσ (t)

= k1 (α,t)

f (t) f ∆ (t) + k0 (α,t) σ g(t) g (t)

−

f (t)

k0 (α,t)g gσ (t)g(t)

∆

(t)

f (t) g(t)  1  + σ k1 (α,t) f (t) + k0 (α,t) f ∆ (t) g (t)

= k1 (α,t)

−

 f (t)  ∆ k (α,t)g(t) + k (α,t)g (t) 1 0 g(t)gσ (t)

=

f (t) f (t) Dα f (t) − Dα g(t) + k1 (α,t) σ σ g (t) g(t)g (t) g(t)

=

g(t)Dα f (t) − f (t)Dα g(t) f (t) + k1 (α,t) . σ g(t)g (t) g(t) 

This completes the proof. Example 1.2.8 Let T = Z, k1 (α,t) = (1 − α)(1 + t 2 )α , f (t) = t 2 + t,

k0 (α,t) = α(1 + t 2 )1−α ,

g(t) = t 3 + t + 1,

t ∈ T.

We will find   f D (t), g

t ∈ T.

σ (t) = t + 1,

t ∈ T.

1 2

Here

α ∈ [0, 1],

t ∈ T,

Conformable Dynamic Calculus on Time Scales  9

Then f ∆ (t) = σ (t) + t + 1 = t +1+t +1 = 2t + 2, g∆ (t) = (σ (t))2 + tσ (t) + t 2 + 1 = (t + 1)2 + t(t + 1) + t 2 + 1 = t 2 + 2t + 1 + t 2 + t + t 2 + 1 = 3t 2 + 3t + 2, 

1 2

D f (t) = k1

1 ,t 2



 f (t) + k0

1 ,t 2

 f ∆ (t)

=

1p 1p 1 + t 2 (t 2 + t) + 1 + t 2 (2t + 2) 2 2

=

1p 1 + t 2 (t 2 + t + 2t + 2) 2

1p 1 + t 2 (t 2 + 3t + 2), 2     1 1 1 D 2 g(t) = k1 ,t g(t) + k0 ,t g∆ (t) 2 2 =

=


1p
1p 1 + t2 t3 + t + 1 + 1 + t 2 3t 2 + 3t + 2 2 2

=


1p 1 + t 2 t 3 + t + 1 + 3t 2 + 3t + 2 2

=


1p 1 + t 2 t 3 + 3t 2 + 4t + 3 , 2

gσ (t) = (σ (t))3 + σ (t) + 1 = (t + 1)3 + t + 1 + 1

10  Conformable Dynamic Equations on Time Scales

= t 3 + 3t 2 + 3t + 1 + t + 2 = t 3 + 3t 2 + 4t + 3, 1

  f (t) = D g 1 2

1

g(t)D 2 f (t) − f (t)D 2 g(t) g(t)gσ (t)   1 f (t) +k1 ,t 2 g(t) √ (t 3 + t + 1) 12 1 + t 2 (t 2 + 3t + 2) = (t 3 + t + 1)(t 3 + 3t 2 + 4t + 3) √ (t 2 + t) 12 1 + t 2 (t 3 + 3t 2 + 4t + 3) − (t 3 + t + 1)(t 3 + 3t 2 + 4t + 3) 1p t2 + t 1 + t2 3 2 t +t +1 √  1 + t2 = t 5 + 3t 4 + 2t 3 + t 3 + 3t 2 + 2t + t 2 2 (t 3 + t + 1) (t 3 + 3t 2 + 4t + 3)  +3t + 2 − t 5 − 3t 4 − 4t 3 − 3t 2 − t 4 − 3t 3 − 4t 2 − 3t +

1p t2 + t 1 + t2 3 2 t +t +1 √
1 + t 2 −t 4 − 4t 3 − 3t 2 + 2t + 2 2 (t 3 + t + 1) (t 3 + 3t 2 + 4t + 3)

+

=

t2 + t 1p 1 + t2 3 2 t +t +1 √  4  −t − 4t 3 − 3t 2 + 2t + 2 2 1 + t2 +t +t = 2 (t 3 + t + 1) t 3 + 3t 2 + 4t + 3 √  1 + t2 = −t 4 − 4t 3 − 3t 2 + 2t + 2 + t 5 2 (t 3 + t + 1) (t 3 + 3t 2 + 4t + 3)  +3t 4 + 4t 3 + 3t 2 + t 4 + 3t 3 + 4t 2 + 3t +

=

√
1 + t 2 t 5 + 3t 4 + 3t 3 + 4t 2 + 5t + 2 , 2 (t 3 + t + 1) (t 3 + 3t 2 + 4t + 3)

t ∈ T.

Example 1.2.9 Let T = 2N0 , k1 (α,t) = (1 − α)t 4α ,

k0 (α,t) = αt 4(1−α) ,

α ∈ [0, 1],

t ∈ T,

Conformable Dynamic Calculus on Time Scales  11

f (t) = t 2 − t,

g(t) = t 2 + 2t + 3,

t ∈ T.

We will find 1 4

D ( f g)(t) and

D

1 2



 f (t), f +g

Here σ (t) = 2t,

t ∈ T.

We have f ∆ (t) = σ (t) + t − 1 = 2t + t − 1 = 3t − 1, g∆ (t) = σ (t) + t + 2 = 2t + t + 2 = 3t + 2, f σ (t) = (σ (t))2 − σ (t) = (2t)2 − 2t = 4t 2 − 2t, gσ (t) = (σ (t))2 + 2σ (t) + 3 = (2t)2 + 2(2t) + 3 = 4t 2 + 4t + 3, 

1 4

D f (t) = k1 =

1 ,t 4



 f (t) + k0

1 ,t 4

3 2
1 3 t t − t + t (3t − 1) 4 4

 f ∆ (t)

t ∈ T.

12  Conformable Dynamic Equations on Time Scales

=

1 2 t (3t − 3 + 3t 2 − t) 4

1 2 2 t (3t + 2t − 3), 4     1 1 1 4 D g(t) = k1 ,t g(t) + k0 ,t g∆ (t) 4 4 =

=


1 3 2 t t + 2t + 3 + t 3 (3t + 2) 4 4

=


1 t 3t 2 + 6t + 9 + 3t 3 + 2t 2 4


1 t 3t 3 + 5t 2 + 6t + 9 , 4  1   1  1 D 4 f (t) g(t) + f σ (t) D 4 g(t) D 4 ( f g)(t) = =

 −k1 =

1 ,t 4

 f σ (t)g(t)



1 2 2 t 3t + 2t − 3 t 2 + 2t + 3 4 + 4t 2 − 2t


1
t 3t 3 + 5t 2 + 6t + 9 4



3 − t 4t 2 − 2t t 2 + 2t + 3 4 =



1 2 2 t 3t + 2t − 3 t 2 + 2t + 3 4
1 + t 2 (4t − 2) 3t 3 + 5t 2 + 6t + 9 4
1 − t 2 (12t − 6) t 2 + 2t + 3 4

=

1 2 t 3t 4 + 6t 3 + 9t 2 + 2t 3 + 4t 2 + 6t − 3t 2 − 6t − 9 4 +12t 4 + 20t 3 + 24t 2 + 36t − 6t 3 − 10t 2 − 12t − 18 ! −12t 3 − 24t 2 − 36t + 6t 2 + 12t + 18

Conformable Dynamic Calculus on Time Scales  13

! 1 2 t 15t 4 + 10t 3 + 6t 2 − 9 , = 4     1 1 1 D 2 f (t) = k1 ,t f (t) + k0 ,t f ∆ (t) 2 2 =

1 1 2 2 t (t − t) + t 2 (3t − 1) 2 2

=

1 2 2 t (t − t + 3t − 1) 2

=

1 2 2 t (t + 2t − 1) 2

=

1 4 3 1 2 t +t − t , 2 2

1

D 2 g(t) =

1 2 1 t g(t) + t 2 g∆ (t) 2 2

=

1 2 2 1 t (t + 2t + 3) + t 2 (3t + 2) 2 2

=

1 2 2 t (t + 5t + 5), 2

f σ (t) + gσ (t) = (σ (t))2 − σ (t) + (σ (t))2 + 2σ (t) + 3 = 2 (σ (t))2 + σ (t) + 3 = 8t 2 + 2t + 3,

1

D2





f (t) = f +g

 1  1 1 ( f (t) + g(t))D 2 f (t) − f (t) D 2 f (t) + D 2 g(t) ( f (t) + g(t)) ( f σ (t) + gσ (t))   1 f (t) +k1 ,t 2 f (t) + g(t)

   1 1 4 3 1 2 2 = (2t + t + 3) t +t − t (2t 2 + t + 3)(8t 2 + 2t + 3) 2 2  
1 4 3 1 2 1 4 5 3 5 2  − t2 − t t +t − t + t + t + t 2 2 2 2 2 1 t2 − t + t2 2 2 2t + t + 3

14  Conformable Dynamic Equations on Time Scales

=

 1 1 3 1 t 6 + 2t 5 − t 4 + t 5 + t 4 − t 3 + t 4 2 2 (2t + t + 3)(8t + 2t + 3) 2 2 2   3 7 +3t 3 − t 2 − (t 2 − t) t 4 + t 3 + 2t 2 2 2 +

=

t4 − t3 2(2t 2 + t + 3)

 5 3 5 3 1 t6 + t5 + t4 + t3 − t2 − t6 2 2 (2t + t + 3)(8t + 2t + 3) 2 2 2 2  7 7 − t 5 − 2t 4 + t 5 + t 4 + 2t 3 2 2 +

t4 − t3 2(2t 2 + t + 3)

=

3t 4 + 92 t 3 − 32 t 2 t4 − t3 + (2t 2 + t + 3)(8t 2 + 2t + 3) 2(2t 2 + t + 3)

=

6t 4 + 9t 3 − 3t 2 + (8t 2 + 2t + 3)(t 4 − t 3 ) 2(2t 2 + t + 3)(8t 2 + 2t + 3)

=

6t 4 + 9t 3 − 3t 2 + 8t 6 − 8t 5 + 2t 5 − 2t 4 + 3t 4 − 3t 3 2(2t 2 + t + 3)(8t 2 + 2t + 3)

=

8t 6 − 6t 5 + 7t 4 + 6t 3 − 3t 2 , 2(2t 2 + t + 3)(8t 2 + 2t + 3)

t ∈ T.

Example 1.2.10 Let T = 2N0 , k1 (α,t) = (1 − α)t 4α , f (t) = t,

k0 (α,t) = αt 4(1−α) ,

α ∈ [0, 1],

t ∈ T.

Then σ (t) = 2t, f ∆ (t) = 1, 1 2

D f (t) = k1



1 ,t 2



 f (t) + k0

1 ,t 2

 f ∆ (t)

t ∈ T,

Conformable Dynamic Calculus on Time Scales  15

= =  1 ∆ D 2 f (t) = =

1 3 1 2 t + t 2 2 1 3 2 (t + t ), 2  1 (σ (t))2 + tσ (t) + t 2 + σ (t) + t 2 1 2 (4t + 2t 2 + t 2 + 2t + t) 2

1 2 (7t + 3t), 2      1   ∆ 1 1 1 1 1 D 4 D 2 f (t) = k1 ,t D 2 f (t) + k0 ,t D 2 f (t) 4 4     3 1 3 2 1 3 1 2 = t (t + t ) + t (7t + 3t) 4 2 4 2 =

3 4 3 1 (t + t ) + (7t 5 + 3t 4 ) 8 8  1 5 = 7t + 3t 4 + 3t 4 + 3t 3 8  1 5 7t + 6t 4 + 3t 3 , = 8     1 1 1 D 4 f (t) = k1 ,t f (t) + k0 ,t f ∆ (t) 4 4 =

=

3 2 1 3 t + t 4 4

=

1 3 (t + 3t 2 ), 4

 1 ∆ D 4 f (t) = =


1 (σ (t))2 + tσ (t) + t 2 + 3σ (t) + 3t 4
1 2 4t + 2t 2 + t 2 + 6t + 3t 4

1 2 (7t + 9t), 4      1  ∆ 1 1 1 1 1 D 2 D 4 f (t) = k1 ,t D 4 f (t) + k0 ,t D 4 f (t) 2 2 =

16  Conformable Dynamic Equations on Time Scales



   1 3 1 2 1 2 2 (t + 3t ) + t (7t + 9t) 4 2 4

=

1 2 t 2

=


1 2 3 t t + 3t 2 + 7t 2 + 9t 8

=


1 2 3 t t + 10t 2 + 9t , 8

t ∈ T.

Therefore  1   1  1 1 D 4 D 2 f (t) 6= D 2 D 4 f (t),

t ∈ T. 2

Remark 1.2.11 Assume α, β ∈ [0, 1], k1 , and k0 are ∆-differentiable at t ∈ Tκ , and f : 2 T → R is twice ∆-differentiable at t ∈ Tκ . Then, in the general case, we have   Dα Dβ f (t) 6= Dβ (Dα f ) (t). n−1

Definition 1.2.12 Let k1 , k0 be n − 1-times ∆-differentiable at t ∈ Tκ , f : T → R be n n-times ∆-differentiable at t ∈ Tκ , n ∈ N. Then we define      n (Dα )n f (t) = Dα Dα . . . Dα f . . . (t), t ∈ Tκ . {z } | n

Remark 1.2.13 Note that in the general case, we have (Dα )n f (t) 6= Dnα f (t),

n

t ∈ Tκ ,

if nα ∈ (0, 1]. Example 1.2.14 Let T = 2N0 , k1 (α,t) = (1 − α)t 4α , f (t) = t,

k0 (α,t) = αt 4(1−α) ,

t ∈ T.

We have σ (t) = 2t,

t ∈ T.

By the previous example, we have f ∆ (t) = 1, 1

D 4 f (t) =  1 ∆ D 4 f (t) =

3 2 1 3 t + t , 4 4 9 7 t + t 2, 4 4

α ∈ [0, 1],

t ∈ T,

Conformable Dynamic Calculus on Time Scales  17

1

D 2 f (t) =

1 3 1 2 t + t , 2 2

t ∈ T.

Then      1  ∆ 1 1 1 1 D D 4 f (t) = k1 ,t D 4 f (t) + k0 ,t D 4 f (t) 4 4     1 9 7 3 3 2 1 3 t t + t + t3 t + t2 = 4 4 4 4 4 4 1 4

=


1 3 t 3(3 + t) + 9t + 7t 2 16

=


1 3 t 9 + 3t + 9t + 7t 2 16

=


1 3 t 9 + 12t + 7t 2 , 16

t ∈ T.

Consequently,  1  1 1 D 4 D 4 f (t) 6= D 2 f (t),

1.3

t ∈ T.

CONFORMABLE REGRESSIVE FUNCTIONS

Definition 1.3.1 We say that a function f : T → R is a conformable regressive function if k0 (α,t) − µ(t)k1 (α,t) 6= 0 and k0 (α,t) + µ(t) ( f (t) − k1 (α,t)) 6= 0 for any α ∈ (0, 1] and for any t ∈ T. The set of all conformable regressive functions on T will be denoted by Rc . Definition 1.3.2 For f , g ∈ Rc , we define “conformable circle plus” ⊕c as follows: ( f ⊕c g) (t) =

( f (t) + g(t) − k1 (α,t)) k0 (α,t) + µ(t) ( f (t) − k1 (α,t)) (g(t) − k1 (α,t)) , k0 (α,t)

t ∈ T, α ∈ (0, 1]. Remark 1.3.3 When α = 1, we have Rc = R

and

⊕c = ⊕.

Theorem 1.3.4 We have that (Rc , ⊕c ) is an Abelian group.

18  Conformable Dynamic Equations on Time Scales

Proof 1.3.5 Let f , g, h ∈ Rc be arbitrarily chosen. Then k0 + µ (( f ⊕c g) − k1 ) = k0 +

µ (( f + g − k1 )k0 + µ( f − k1 )(g − k1 ) − k1 k0 ) k0

=


1 2 k0 + µ (( f − k1 )(k0 + µ(g − k1 )) + gk0 − k1 k0 ) k0

=

1 (k0 (k0 + µ(g − k1 )) + µ( f − k1 )(k0 + µ(g − k1 ))) k0

=

1 (k0 + µ( f − k1 )) (k0 + µ(g − k1 )) k0

6= 0 on

T,

i.e., f ⊕c g ∈ Rc . Also,   k0 ( f − k1 ) k0 + µ − + k1 − k1 k0 + µ( f − k1 ) = k0 −

µk0 ( f − k1 ) k0 + µ( f − k1 )

=

k02 + µk0 ( f − k1 ) − µk0 ( f − k1 ) k0 + µ( f − k1 )

=

k02 k0 + µ( f − k1 )

6= 0 on

T,

i.e., −

k0 ( f − k1 ) + k1 ∈ Rc . k0 + µ( f − k1 )

We have   k0 ( f − k1 ) f ⊕c − + k1 k0 + µ( f − k1 )   1 k0 ( f − k1 ) = f− + k1 − k1 k0 k0 k0 + µ( f − k1 )    k0 ( f − k1 ) + k1 − k1 ( f − k1 ) +µ − k0 + µ( f − k1 )

Conformable Dynamic Calculus on Time Scales  19

  k0 ( f − k1 ) k0 ( f − k1 ) f− k0 − µ ( f − k1 ) k0 + µ( f − k1 ) k0 + µ( f − k1 )   k02 ( f − k1 ) 1 µk0 ( f − k1 )2 f k0 − − k0 k0 + µ( f − k1 ) k0 + µ( f − k1 )   1 k0 ( f − k1 ) (k0 + µ( f − k1 )) f k0 − k0 k0 + µ( f − k1 ) 1 k0

=

= =



1 ( f k0 − k0 f + k0 k1 ) k0

=

= k1 , i.e., the conformable addition inverse of f is −

k0 ( f − k1 ) + k1 . k0 + µ( f − k1 )

Next, k0 + µ(k1 − k1 ) = k0 6= 0,

for

α ∈ (0, 1] and t ∈ T,

i.e., k1 ∈ Rc , and f ⊕c k1 =

( f + k1 − k1 )k0 + µ( f − k1 )(k1 − k1 ) k0

=

f k0 k0

=

f

=

(k1 + f − k1 )k0 + µ(k1 − k1 )( f − k1 ) k0

= k1 ⊕c f , i.e., k1 is the conformable additive identity for ⊕c . Next, ( f ⊕c g) ⊕c h µ ( f ⊕c g − k1 ) (h − k1 ) k0   µ = f + g − k1 + ( f − k1 )(g − k1 ) + h − k1 k0 = (( f ⊕c g) + h − k1 ) +

20  Conformable Dynamic Equations on Time Scales

µ + k0 =



 µ f + g − k1 + ( f − k1 )(g − k1 ) − k1 (h − k1 ) k0

f + g + h − 2k1 +

µ ( f − k1 )(g − k1 ) k0

+

µ µ ( f − k1 )(h − k1 ) + (g − k1 )(h − k1 ) k0 k0

+

µ2 ( f − k1 )(g − k1 )(h − k1 ) on k02

T,

and f ⊕c (g ⊕c h) µ ( f − k1 ) (g ⊕c h − k1 ) k0   µ = f + g + h − k1 + (g − k1 )(h − k1 ) − k1 k0   µ µ + ( f − k1 ) g + h − k1 + (g − k1 )(h − k1 ) − k1 k0 k0

= ( f + (g ⊕c h) − k1 ) +

=

f + g + h − 2k1 + +

µ µ (g − k1 )(h − k1 ) + (g − k1 )( f − k1 ) k0 k0

µ µ2 ( f − k1 )(h − k1 ) + 2 ( f − k1 )(g − k1 )(h − k1 ) on T. k0 k0

Consequently, ( f ⊕c g) ⊕c h = f ⊕c (g ⊕c h)

on

T.

Also, f ⊕c g = ( f + g − k1 ) +

µ ( f − k1 )(g − k1 ) k0

= (g + f − k1 ) +

µ (g − k1 )( f − k1 ) k0

= g ⊕c f This completes the proof.

on T. 

Definition 1.3.6 Let f ∈ Rc . We define the conformable addition inverse of f under the operation c as follows k0 ( f − k1 ) c f = − + k1 . k0 + µ( f − k1 )

Conformable Dynamic Calculus on Time Scales  21

For f ∈ Rc , we have k0 ( c f − k1 ) + k1 k0 + µ( c f − k1 )   k0 ( f −k1 ) k0 − k +µ( f −k ) + k1 − k1 1  + k1 0 = − k0 ( f −k1 ) k0 + µ − k +µ( f −k ) + k1 − k1

c ( c f ) = −

0

1

k02 ( f

=

k02 + µk0 ( f

=

f − k1 + k1

=

f.

− k1 ) + k1 − k1 ) − µk0 ( f − k1 )

Definition 1.3.7 Let f , g ∈ Rc . We define “conformable circle minus” subtraction c as follows: f c g = f ⊕c ( c g) . For f , g ∈ Rc , we have f c g =

f ⊕c ( c g)

=

f + ( c g) − k1 + µ

=

f−

+

( f − k1 )( c g − k1 ) k0

k0 (g − k1 ) + k1 − k1 k0 + µ(g − k1 )   0 (g−k1 ) µ( f − k1 ) − k k+µ(g−k + k − k 1 1 ) 0

1

k0

=

f−

k0 (g − k1 ) k0 ( f − k1 )(g − k1 ) −µ k0 + µ(g − k1 ) k0 (k0 + µ(g − k1 ))

=

f−

k0 (g − k1 ) ( f − k1 )(g − k1 ) −µ k0 + µ(g − k1 ) k0 + µ(g − k1 )

=

f k0 + µ f (g − k1 ) − (k0 + µ( f − k1 ))(g − k1 ) k0 + µ(g − k1 )

=

f k0 + (g − k1 )(µ f − k0 − µ f + µk1 ) k0 + µ(g − k1 )

=

f k0 − (g − k1 )(k0 − µk1 ) k0 + µ(g − k1 )

22  Conformable Dynamic Equations on Time Scales

=

f k0 − gk0 + µgk1 + k0 k1 − µk12 k0 + µ(g − k1 )

=

( f − g)k0 + k1 (k0 + µ(g − k1 )) k0 + µ(g − k1 )

on

T,

and ( f c g) − k1 = =

( f − g)k0 + k1 − k1 k0 + µ(g − k1 ) ( f − g)k0 k0 + µ(g − k1 )

on

T,

and k0 + µ (( f c g) − k1 ) = k0 + µ

k0 ( f − g) k0 + µ(g − k1 )

 = k0 1 +

µ f − µg k0 + µ(g − k1 )



= k0

k0 + µg − µk1 + µ f − µg k0 + µ(g − k1 )

= k0

k0 + µ( f − k1 ) k0 + µ(g − k1 )

6= 0

on

T,

i.e., f c g ∈ Rc . Definition 1.3.8 Let f ∈ Rc . The conformable generalized square of f is defined as follows 2 f = − f ( c f ) .

For f ∈ Rc , we have 2 f = − f ( c f )

  k0 ( f − k1 ) = −f − + k1 k0 + µ( f − k1 )   k0 ( f − k1 ) = f − k1 k0 + µ( f − k1 ) =

f

k0 ( f − k1 ) − k0 k1 − µk1 ( f − k1 ) k0 + µ( f − k1 )

=

f

(k0 − µk1 )( f − k1 ) − k0 k1 , k0 + µ( f − k1 )

t ∈ Tκ .

Conformable Dynamic Calculus on Time Scales  23

Theorem 1.3.9 Let f ∈ Rc . Then 2 2 ( c f ) = f .

Proof 1.3.10 We have 2 ( c f ) = − ( c f ) ( c ( c f ))

= − f ( c f ) 2 f .

=



This completes the proof.

1.4

THE CONFORMABLE EXPONENTIAL FUNCTION

Definition 1.4.1 Suppose that α ∈ (0, 1], p ∈ Rc . For t,t0 ∈ T, we define the conformable exponential function as follows E p (t,t0 ) = e p−k1 (t,t0 ). k0

We have Dα E p (t,t0 ) = k1 (α,t)e p−k1 (t,t0 ) + k0 (α,t)e∆p−k1 (t,t0 ) k0

k0

= k1 (α,t)e p−k1 (t,t0 ) + k0 (α,t) k0

p(t) − k1 (α,t) e p−k1 (t,t0 ) k0 (α,t) k0

= p(t)e p−k1 (t,t0 ) k0

= p(t)E p (t,t0 ),

t ∈ Tκ .

Below we will list some of the properties of the conformable exponential function. Theorem 1.4.2 (Semigroup Property) The exponential function defined above in Definition 1.8 satisfies E p (t, s)E p (s, r) = E p (t, r), t, s, r ∈ T.

Proof 1.4.3 We have E p (t, s)E p (s, r) = e p−k1 (t, s)e p−k1 (s, r) k0

k0

= e p−k1 (t, r) k0

= E p (t, r), This completes the proof.

t, s, r ∈ T. 

24  Conformable Dynamic Equations on Time Scales

Theorem 1.4.4 We have Ek1 (t,t0 ) = 1,

E p (t,t) = 1,

t,t0 ∈ T.

Proof 1.4.5 We have Ek1 (t,t0 ) = e0 (t,t0 ) = 1 and E p (t,t) = e p−k1 (t,t) = 1,

t,t0 ∈ T.

k0



This completes the proof. Theorem 1.4.6 We have   p(t) − k1 (α,t) E p (σ (t),t0 ) = 1 + µ(t) E p (t,t0 ), k0 (α,t)

t,t0 ∈ T.

Proof 1.4.7 Using the definition of the exponential function, E p (σ (t),t0 ) = e p−k1 (σ (t),t0 ) k0

  p(t) − k1 (α,t) = 1 + µ(t) e p−k1 (t,t0 ) k0 (α,t) k0   p(t) − k1 (α,t) = 1 + µ(t) E p (t,t0 ), t,t0 ∈ T. k0 (α,t) 

This completes the proof. Theorem 1.4.8 Let p ∈ Rc . Then E p (t,t0 ) = E c p (t0 ,t),

α ∈ (0, 1],

t0 ,t ∈ T.

Proof 1.4.9 We have E p (t,t0 ) = e p−k1 (t,t0 ) k0

= e

  (t ,t), p−k 0 k 1

α ∈ (0, 1],

t0 ,t ∈ T.

0

Note that 

p − k1 k0

 (t) =

p(t)−k1 (α,t) k0 (α,t) − 1 (α,t) 1 + µ(t) p(t)−k k0 (α,t)

= −

p(t) − k1 (α,t) , k0 (α,t) + µ(t) (p(t) − k1 (α,t))

α ∈ (0, 1],

t ∈ T.

Conformable Dynamic Calculus on Time Scales  25

Next,   1 k0 (α,t)(p(t) − k1 (α,t)) ( c p)(t) − k1 (α,t) = − − k1 (α,t) + k1 (α,t) k0 (α,t) k0 (α,t) k0 (α,t) + µ(t)(p(t) − k1 (α,t)) =−

p(t) − k1 (α,t) , k0 (α,t) + µ(t)(p(t) − k1 (α,t))

α ∈ (0, 1],

t ∈ T.

Therefore E p (t,t0 ) = E c p (t0 ,t),

t,t0 ∈ T. 

This completes the proof. Theorem 1.4.10 Let f , g ∈ Rc . Then E f (t,t0 )Eg (t,t0 ) = E f ⊕c g (t,t0 ),

α ∈ (0, 1],

t,t0 ∈ T.

Proof 1.4.11 We have E f (t,t0 )Eg (t,t0 ) = e f −k1 (t,t0 )e g−k1 (t,t0 ) k0

k0

 (t,t ),  g−k 0 ⊕ k 1

= e f −k1  k0

α ∈ (0, 1],

t,t0 ∈ T.

0

Note that 

f − k1 k0





g − k1 ⊕ k0

+µ(t)

 (t) =

f (t) − k1 (α,t) g(t) − k1 (α,t) + k0 (α,t) k0 (α,t)

( f (t) − k1 (α,t))(g(t) − k1 (α,t)) , (k0 (α,t))2

α ∈ (0, 1],

t ∈ T,

and ( f ⊕c g)(t) − k1 (α,t) k0 (α,t)

=

1  f (t) + g(t) − k1 (α,t) k0 (α,t) +

=

 µ(t) ( f (t) − k1 (α,t))(g(t) − k1 (α,t)) − k1 (α,t) k0 (α,t)

f (t) − k1 (α,t) g(t) − k1 (α,t) + k0 (α,t) k0 (α,t) +µ(t)

( f (t) − k1 (α,t))(g(t) − k1 (α,t)) , (k0 (α,t))2

α ∈ (0, 1], t ∈ T. Therefore     f − k1 g − k1 ( f ⊕c g)(t) − k1 (α,t) ⊕ (t) = , k0 k0 k0 (α,t)

α ∈ (0, 1],

t ∈ T.

Hence, E f (t,t0 )Eg (t,t0 ) = E f ⊕c g (t,t0 ), This completes the proof.

α ∈ (0, 1],

t,t0 ∈ T. 

26  Conformable Dynamic Equations on Time Scales

Theorem 1.4.12 Let f , g ∈ Rc . Then E f (t,t0 ) = E f c g (t,t0 ), Eg (t,t0 )

α ∈ (0, 1],

t,t0 ∈ T.

Proof 1.4.13 Using Theorem 1.4.8 and Theorem 1.4.10, we get E f (t,t0 ) Eg (t,t0 )

= E f (t,t0 )E c g (t,t0 ) = E f c g (t,t0 ),

α ∈ (0, 1],

t,t0 ∈ T. 

This completes the proof.

1.5

CONFORMABLE HYPERBOLIC AND TRIGONOMETRIC FUNCTIONS

Definition 1.5.1 Let ± f ∈ Rc . Define the conformable hyperbolic functions Cosh f and Sinh f by E f + E− f E f − E− f Cosh f = and Sinh f = . 2 2 Theorem 1.5.2 Let ± f ∈ Rc . Then Dα Cosh f Cosh2f − Sinh2f

=

f Sinh f ,

Dα Sinh f = fCosh f ,

= Eg ,

where g =

f ⊕c (− f )

= −k1 −


µ 2 f − k12 . k0

Proof 1.5.3 We have  α

D Cosh f

= D

α

E f + E− f 2



=

Dα E f + Dα E− f 2

=

f E f − f E− f 2

Conformable Dynamic Calculus on Time Scales  27

E f − E− f 2

=

f

=

f Sinh f , 

α

D Sinh f

Cosh2f

− Sinh2f

= D

α

E f − E− f 2



=

Dα E f − Dα E− f 2

=

f E f + f E− f 2

=

f

=

fCosh f ,

E f + E− f 2

   E f + E− f 2 E f − E− f 2 = − 2 2 1 2 2 E + 2E f E− f + E− = f 4 f  2 −E 2f + 2E f E− f − E− f 

= E f E− f = E f ⊕c (− f ) . Note that f ⊕c (− f ) =

( f − f − k1 )k0 + µ( f − k1 )(− f − k1 ) k0

= −

k0 k1 + µ( f − k1 )( f + k1 ) k0

= −k1 −


µ 2 f − k12 k0

= g. Therefore Cosh2f − Sinh2f = Eg . This completes the proof.



28  Conformable Dynamic Equations on Time Scales

Definition 1.5.4 Suppose that f ± g ∈ Rc . Define the conformable hyperbolic functions Ch f g and Sh f g as follows Ch f g =

E f +g + E f −g 2

and

Sh f g =

E f +g − E f −g . 2

Theorem 1.5.5 Let f ± g ∈ Rc . Then Dα Ch f g =

fCh f g + gSh f g ,

Dα Sh f g = gCh f g + f Sh f g ,

Ch2f g − Sh2f g = E( f +g)⊕c ( f −g) . Proof 1.5.6 We have  α

D Ch f g = D

α

E f +g + E f −g 2



=

Dα E f +g + Dα E f −g 2

=

( f + g)E f +g + ( f − g)E f −g 2

=

f

=

fCh f g + gSh f g ,

E f +g + E f −g E f +g − E f −g +g 2 2

 α

D Sh f g = D

α

E f +g − E f −g 2



=

Dα E f +g − Dα E f −g 2

=

( f + g)E f +g − ( f − g)E f −g 2

=

f

=

f Sh f g + gCh f g ,

E f +g − E f −g E f +g + E f −g +g 2 2

Conformable Dynamic Calculus on Time Scales  29

Ch2f g − Sh2f g

   E f +g − E f −g 2 E f +g + E f −g 2 − = 2 2 1 2 = E + 2E f +g E f −g + E 2f −g 4 f +g  −E 2f +g + 2E f +g E f −g − E 2f −g 

= E f +g E f −g = E( f +g)⊕c ( f −g) . 

This completes the proof.

Definition 1.5.7 Let ±i f ∈ Rc . Define the conformable trigonometric functions Cos f and Sin f as follows: Ei f + E−i f Ei f − E−i f Cos f = , Sin f = . 2 2i

Theorem 1.5.8 Let ±i f ∈ Rc . Then Dα Cos f

= − f Sin f ,

Cos2f + Sin2f

Dα Sin f = fCos f ,

= E(i f )⊕c (−i f ) .

Proof 1.5.9 We have  α

D Cos f

α

= D

Ei f + E−i f 2



=

Dα Ei f + Dα E−i f 2

=

i f Ei f − i f E−i f 2

=

f

E−i f − Ei f 2i

= − f Sin f ,

30  Conformable Dynamic Equations on Time Scales

 α

D Sin f

Cos2f

+ Sin2f

α

= D

Ei f − E−i f 2i



=

Dα Ei f − Dα E−i f 2i

=

i f Ei f + i f E−i f 2i

=

f

=

fCos f ,

Ei f + E−i f 2

   Ei f − E−i f 2 Ei f + E−i f 2 + = 2 2i 1 2 2 = E + 2Ei f E−i f + E−i f 4 if  2 −Ei2f + 2Ei f E−i f − E−i f 

= Ei f E−i f = E(i f )⊕c (−i f ) . 

This completes the proof.

Definition 1.5.10 Let f ± ig ∈ Rc . Define the conformable trigonometric follows: C f g and S f g as follows: E f +ig − E f −ig E f +ig + E f −ig , Sfg = . Cfg = 2 2i

Theorem 1.5.11 Let f ± ig ∈ Rc . Then Dα C f g =

fC f g − gS f g ,

Dα S f g = gC f g + f S f g ,

C2f g + S2f g = E( f +ig)⊕c ( f −ig) . Proof 1.5.12 We have  α

α

D Cfg = D

E f +ig + E f −ig 2



Conformable Dynamic Calculus on Time Scales  31

=

Dα E f +ig + Dα E f −ig 2

=

( f + ig)E f +ig + ( f − ig)E f −ig 2

=

f

=

fC f g − gS f g ,

E f +ig − E f −ig E f +ig + E f −ig −g 2 2i

 α

α

D Sfg = D

E f +ig − E f −ig 2i



=

Dα E f +ig − Dα E f −ig 2i

=

( f + ig)E f +ig − ( f − ig)E f −ig 2i

=

f

E f +ig − E f −ig E f +ig + E f −ig +g 2i 2

= gC f g + f S f g , C2f g + S2f g

   E f +ig − E f −ig 2 E f +ig + E f −ig 2 + = 2 2i 1 2 = E + 2E f +ig E f −ig + E 2f −ig 4 f +ig  −E 2f +ig + 2E f +ig E f −ig − E 2f −ig 

= E f +ig E f −ig = E( f +ig)⊕c ( f −ig) . This completes the proof.

1.6



THE CONFORMABLE LOGARITHM FUNCTION

We now set the foundation for offering a definition of conformable logarithms on time scales. This definition will be of a multi-valued function.

32  Conformable Dynamic Equations on Time Scales

Definition 1.6.1 (Conformable Cylinder Transformation) Let α ∈ (0, 1], and fix t ∈ T. For h > 0, define the multi-valued conformable cylinder transformation ζhc : Cch → C by     z − k1 (α,t) 1   for h 6= 0  log 1 + h k0 (α,t) ζhc (z) = h (1.2) z − k1 (α,t)   for h = 0,  k0 (α,t) where C is the set of complex numbers, Cch is given by   k0 (α,t) c , Ch = z ∈ C : z 6= k1 (α,t) − h

(1.3)

and log is the multi-valued complex logarithm function. Lemma 1.6.2 Fix α ∈ (0, 1]. Let f , g : T → C be ∆-differentiable functions with f , g 6= 0 on T, and let the multi-valued conformable cylinder transformation ζ c be given by (1.2). Then, for fixed τ ∈ Tκ ,  α    α   α  D f D f (τ) D g(τ) Dα g c c c ζµ(τ) ⊕c (τ) = ζµ(τ) + ζµ(τ) . f g f (τ) g(τ) Proof 1.6.3 First, note that the simple useful formula f σ = f + µ f ∆ (suppressing the variable) implies Dα ( f g) fg

= = = =

f σ Dα g + gDα f − k1 (α, ·) f σ g fg
α ∆ f + µ f (D g − k1 (α, ·)g) Dα f + fg f  α  α α ∆ D g D f D g f + − k1 (α, ·) + µ − k1 (α, ·) f g f g Dα f Dα g ⊕c . f g

It follows that for fixed τ ∈ Tκ ,  α   D f Dα g c ζµ(τ) ⊕c (τ) f g  α  D ( f g)(τ) c = ζµ(τ) ( f g)(τ)    ( f g)∆ (τ) 1   log 1 + µ(τ) for µ(τ) 6= 0  µ(τ)  ( f g)(τ)  = α 1 D ( f g)(τ)   − k1 (α, τ) for µ(τ) = 0  k0 (α, τ) ( f g)(τ)    1 ( f g)σ (τ)   log for µ(τ) 6= 0  µ(τ)  ( f g)(τ)  = 1 Dα f (τ) k0 (α, τ) f σ (τ)g∆ (τ)   + − k1 (α, τ) for µ(τ) = 0  k0 (α, τ) f (τ) ( f g)(τ)

Conformable Dynamic Calculus on Time Scales  33

 σ   σ   f (τ) 1 g (τ) 1   log + log for µ(τ) 6= 0  µ(τ)  f (τ) µ(τ)  g(τ) = 1 Dα f (τ) g∆ (τ)   − k1 (α, τ) + for µ(τ) = 0  k0 (α, τ) f (τ) g(τ)      1 ( f + µ f ∆ )(τ) 1 (g + µg∆ )(τ)   log + log for µ(τ) 6= 0  µ(τ)  f (τ) µ(τ) g(τ)    = 1 Dα f (τ) 1 Dα g(τ)   − k1 (α, τ) + − k1 (α, τ) for µ(τ) = 0  k0 (α, τ) f (τ) k0 (α, τ) g(τ)   α   α D g(τ) D f (τ) c c + ζµ(τ) . = ζµ(τ) f (τ) g(τ) 

This completes the proof. Definition 1.6.4 Fix α ∈ (0, 1]. For β ∈ R and f ∈ R(β ), define the operation c via β c f := k1 (α, ·) + β ( f − k1 (α, ·))

Z 1

1+µ 0



 β −1 f − k1 (α, ·) h dh. k0 (α, ·)

(1.4)

Lemma 1.6.5 Let α ∈ (0, 1], β ∈ R, and p : T → C be such that p(t) 6= 0 for all t ∈ T, and if β 6∈ N, then p(t)pσ (t) > 0 for all t ∈ T. For the multi-valued conformable cylinder transformation ζ c given by (1.2), we have
 α  Dα pβ D p = β c p pβ on Tκ . Proof 1.6.6 Using P¨otzsche’s chain rule, we have (suppressing the variable) Dα (pβ ) = k1 (α, ·)pβ + k0 (α, ·)(pβ )∆ Z 1 β −1 = k1 (α, ·)pβ + β k0 (α, ·)p∆ dh p + µ p∆ h 0    h i 1 ∆ β β k (α, ·)pβ + β k (α, ·)p∆ (p + µ p ) − p for µ(τ) 6= 0 1 0 β µ p∆ =  k1 (α, ·)pβ + β k0 (α, ·)p∆ pβ −1 for µ(τ) = 0  " #  β  p∆ k0 (α, ·)   1+µ −1 for µ(τ) 6= 0 k1 (α, ·) + µ p = pβ  p∆   k1 (α, ·) + β k0 (α, ·) for µ(τ) = 0 p  " #  α β  D p − k (α, ·)p k (α, ·)  1 0  1+µ −1 for µ(τ) 6= 0 k1 (α, ·) + µ k0 (α, ·)p β = p  α   D p − k1 (α, ·)p   for µ(τ) = 0. k1 (α, ·) + β k0 (α, ·) k0 (α, ·)p

34  Conformable Dynamic Equations on Time Scales

Therefore,  " #  α β  D p − k (α, ·)p k (α, ·)  1 k (α, ·) + 0 1+µ −1 for µ(τ) 6= 0 Dα (pβ )  1 µ k0 (α, ·)p =   α  pβ D p − k1 (α, ·)p   for µ(τ) = 0. k1 (α, ·) + β p By Definition 1.6.4,  α  D p β c p ! !β −1 Z 1 Dα p Dα p p − k1 (α, ·) = k1 (α, ·) + β − k1 (α, ·) 1+µ h dh p k0 (α, ·) 0  " #  α β  k (α, ·) D p − k (α, ·)p  0 1  1+µ −1 for µ(τ) 6= 0 k1 (α, ·) + µ k (α, ·)p =  α  0  D p − k1 (α, ·)p   for µ(τ) = 0. k1 (α, ·) + β p 

Hence,  β c

Dα p p



Dα (pβ ) , pβ

=



and the proof is complete.

Lemma 1.6.7 Let α ∈ (0, 1], β ∈ R, and f ∈ R(β ). For the multi-valued conformable cylinder transformation ζ c given by (1.2) and for fixed τ ∈ T, c c ( f (τ)). ((β c f )(τ)) = β ζµ(τ) ζµ(τ)

Proof 1.6.8 Assume α ∈ (0, 1], β ∈ R, and f ∈ R(β ). Fix τ ∈ Tκ . If µ(τ) 6= 0, then k0 (α, τ) + µ(τ) (β c f ) (τ) − µ(τ)k1 (α, τ) = k0 (α, τ) + µ(τ)β ( f (τ) − k1 (α, τ))  f (τ) − k1 (α, τ) = k0 (α, τ) 1 + µ(τ) k0 (α, τ)

Z 1

1 + µ(τ) 0 β



 β −1 f (τ) − k1 (α, τ) h dh k0 (α, τ)

.

Therefore, after dividing by k0 (α, τ), we have   µ(τ) f (τ) − k1 (α, τ) β 1+ ((β c f ) (τ) − k1 (α, τ)) = 1 + µ(τ) k0 (α, τ) k0 (α, τ) for any τ ∈ Tκ . It follows that for fixed τ ∈ Tκ , c ζµ(τ) ((β c f ) (τ))

Conformable Dynamic Calculus on Time Scales  35

=

=

= =

     β f − k (α, ·) 1  c 1  log 1 + µ(τ) (τ) for µ(τ) 6= 0  µ(τ) k0 (α, ·)   β c f − k1 (α, ·)   (τ) for µ(τ) = 0  k0 (α, ·)     1 f (τ) − k1 (α, τ) β   log 1 + µ(τ) for µ(τ) 6= 0 µ(τ) k0 (α, τ)    f − k1 (α, ·)  β (τ) for µ(τ) = 0 k0 (α, ·)    1 f (τ) − k1 (α, τ)   log 1 + µ(τ) for µ(τ) 6= 0  µ(τ) k0 (α, τ)   β f − k1 (α, ·)   (τ) for µ(τ) = 0  k0 (α, ·) c β ζµ(τ) ( f (τ)). 

This completes the proof.

Definition 1.6.9 (Logarithm Function) For a ∆-differentiable function p : T → C with p 6= 0 on T, the multi-valued conformable logarithm function on time scales is given by  α  Z t D p(τ) c c ` p (t, s) = ζµ(τ) ∆τ for s,t ∈ T, p(τ) s where ζhc (z) is the multi-valued cylinder transformation given in (1.2). Define the principal logarithm on time scales to be  α  Z t D p(τ) c Lcp (t, s) = ξµ(τ) ∆τ for s,t ∈ T, p(τ) s where ξhc (z) is the single-valued cylinder transformation given by     1 z − k1 (α,t)   for h 6= 0  Log 1 + h k0 (α,t) ξhc (z) = h z − k1 (α,t)   for h = 0,  k0 (α,t)

(1.5)

where Log is the principal logarithm. Using the definition of the multi-valued conformable logarithm on time scales given above, we establish the following properties. Theorem 1.6.10 Let p : T → C be a ∆-differentiable function with p 6= 0 on T. Then, for s,t ∈ T, we have
exp Lcp (t, s) = E Dα p (t, s). p

Proof 1.6.11 Let p : T → C be a ∆-differentiable function with p 6= 0 on T. Then, for s,t ∈ T, we have  α  Z t D p(τ) c c ∆τ. L p (t, s) = ξµ(τ) p(τ) s Now exponentiate both sides and use the definition of the exponential function E p (t, s). 

36  Conformable Dynamic Equations on Time Scales

Corollary 1.6.12 Let p ∈ Rc and s,t ∈ T. Then   exp LEc p (t, s) = E p (t, s). Theorem 1.6.13 (Logarithm of Product & Quotient) Let f , g : T → C be ∆-differentiable functions with f , g 6= 0 on T. Then, for s,t ∈ T, we have `cf g (t, s) = `cf (t, s) + `cg (t, s) and `cf (t, s) = `cf (t, s) − `cg (t, s). g

Proof 1.6.14 Let f , g : T → R be ∆-differentiable functions with f , g 6= 0 on T. Then, for s,t ∈ T, we have via Lemma 1.6.2 and its proof that   α Z t D ( f g)(τ) c c ∆τ ` f g (t, s) = ζµ(τ) ( f g)(τ) s  α   Z t D f Dα g c = ζµ(τ) ⊕c (τ) ∆τ f g s    α  Z t Z t α c D f (τ) c D g(τ) = ζµ ∆τ + ζµ ∆τ f (τ) g(τ) s s = `cf (t, s) + `cg (t, s). In a similar manner, `cf (t, s) = g

= = = This completes the proof.

   Dα gf (τ) c     ∆τ ζµ(τ) f s (τ) g    α Z t D f Dα g c c (τ) ∆τ ζµ(τ) f g s    α  α Z t Z t c D f (τ) c D g(τ) ∆τ − ζµ ∆τ ζµ f (τ) g(τ) s s `cf (t, s) − `cg (t, s). Z t





Theorem 1.6.15 Let α ∈ (0, 1], β ∈ R, and p : T → C be a ∆-differentiable function such that p(t) 6= 0 for all t ∈ T, and if β 6∈ N, then p(t)pσ (t) > 0 for all t ∈ T. Then, for s,t ∈ T, we have `cpβ (t, s) = β `cp (t, s). Proof 1.6.16 For the multi-valued cylinder transformation ζ c given by (1.2) and for fixed τ ∈ Tκ ,     α  Dα p D p(τ) c c ζµ(τ) β c (τ) = β ζµ(τ) p p(τ)

Conformable Dynamic Calculus on Time Scales  37

using Lemma 1.6.7. Moreover, by Lemma 1.6.5, we have  β c

Dα p p




Dα pβ . = pβ

Consequently, `cpβ (t, s) = = = =

!
α pβ (τ) D c ζµ(τ) ∆τ pβ (τ) s    Z t Dα p c (τ) ∆τ ζµ(τ) β c p s  α  Z t D p(τ) c β ζµ(τ) ∆τ p(τ) s β `cp (t, s). Z t

This ends the proof.



Theorem 1.6.17 Let p : T → R be a ∆-differentiable function with p 6= 0 on T. Then, for s,t ∈ T, we have  σ   p (t) 1   log for µ(t) 6= 0 
µ(t) p(t) Dα `cp (t, s) = k1 (α,t)`cp (t, s) + k0 (α,t) p∆ (t)    for µ(t) = 0, p(t) where ∆-differentiation is with respect to t. Proof 1.6.18 Using the definition of the conformable logarithm and applying the Dα operator with respect to t,   α
D p(t) c Dα `cp (t, s) = k1 (α,t)`cp (t, s) + k0 (α,t)ζµ(t) p(t)    ∆ (t) 1 p   log 1 + µ(t) for µ(t) 6= 0  µ(t) p(t) c = k1 (α,t)` p (t, s) + k0 (α,t) p∆ (t)    for µ(t) = 0. p(t) Now substitute µ p∆ = pσ − p. This ends the proof. Example 1.6.19 Let t ∈ T with t 6= 0, and set p(t) = t. For s ∈ T, we have    1 σ (t)   log for µ(t) 6= 0
t Dα `cp (t, s) = k1 (α,t)`cp (t, s) + k0 (α,t) µ(t)  1 for µ(t) = 0, t



38  Conformable Dynamic Equations on Time Scales

where ∆-differentiation is with respect to t. Thus,   t  k (α,t) 0   + for T = R k (α,t) log 1   s t      
t k0 (α,t) h + log 1 + for T = hZ Dα `cp (t, s) = k1 (α,t) log s h t       k (α,t) log t + k0 (α,t) log(q)  for T = qN0 , 1 s (q − 1)t where h > 0 and q > 1.

1.7

CONFORMABLE INTEGRATION

Let a, b ∈ T, a < b. Suppose that k0 (α,t) − µ(t)k1 (α,t) 6= 0,

α ∈ (0, 1],

t ∈ T.

(1.6)

Definition 1.7.1 Let f ∈ Crd (T) and assume (1.6) holds. Define a conformable antiderivative via Z Dα f (t)∆α t = f (t) + cE0 (t,t0 ), c ∈ R, t ∈ T. Define the conformable ∆-integral of f over [a, b] as follows Z t a

Z t

f (s)∆α,t s =

f (s) a

E0 (t, σ (s)) ∆s, k0 (α, s)

t ∈ [a, b].

Theorem 1.7.2 Let α ∈ (0, 1], f ∈ Crd (T) and (1.6) hold. Then Z t  α D f (s)∆α,t s = f (t), t ∈ [a, b]κ . a

Proof 1.7.3 We have  Z t  Z t α f (s)∆α,t s f (s)∆α,t s = k1 (α,t) D a

a

∆ E0 (t, σ (s)) +k0 (α,t) f (s) ∆s k0 (α, s) a Z t  = k1 (α,t) f (s)∆α,t s t

Z

a

+k0 (α,t) f (t)

E0 (σ (t), σ (t)) k0 (α,t)

Z t

+k0 (α,t)

f (s) a

E0∆ (t, σ (s)) ∆s k0 (α, s)

Conformable Dynamic Calculus on Time Scales  39

t

Z = k1 (α,t)

 f (s)∆α,t s + f (t)

a

e∆t k1 (t, σ (s))

Z t

+k0 (α,t)

f (s)

−k

0

k0 (α, s) Z t  = k1 (α,t) f (s)∆α,t s + f (t)

∆s

a

a

−k1 (α,t) =

f (t),

Z a

t

 E0 (t, σ (s)) f (s) ∆s k0 (α, s)

t ∈ [a, b]κ . 

This completes the proof. Example 1.7.4 Let T = Z, k1 (α,t) = (1 − α)t 2α , t2 + 1 , t2 + 3

f (t) =

k0 (α,t) = αt 2(1−α) ,

α ∈ (0, 1],

t ∈ T,

t ∈ T.

We will compute Z 10 0

f (s)∆α,10 s

for αt 2(1−α) − (1 − α)t 2α 6= 0,

t ∈ [0, 10].

Here σ (t) = t + 1, t ∈ T,

µ(t) = 1,

E0 (10, σ (s)) = e− k1 (10, σ (s)) k0

R 10

= e

1 σ (s) µ(τ)

  k (α,τ) Log 1−µ(τ) k1 (α,τ) ∆τ 0

=

    R 1 Log 1−µ(τ) k1 (α,τ) ∆τ+R 10 1 Log 1−µ(τ) k1 (α,τ) ∆τ − sσ (s) µ(τ) s k0 (α,τ) µ(τ) k0 (α,τ) e

=

    k (α,s) k (α,l) − Log 1− k1 (α,s) ∑9l=s Log 1− k1 (α,l) 0 0 e e

40  Conformable Dynamic Equations on Time Scales

=

9

1 1 − kk1 (α,s) (α,s) 0

  k1 (α, l) ∏ 1 − k0 (α, l) l=s

=

9 k0 (α, s) k0 (α, l) − k1 (α, l) ∏ k0 (α, s) − k1 (α, s) l=s k0 (α, l)

=

9 αl 2(1−α) − (1 − α)l 2α αs2(1−α) , ∏ −(1 − α)s2α + αs2(1−α) l=s αl 2(1−α)

s ∈ [0, 9].

Then Z 10 0

f (s)∆α,10 s =

Z 10 2 s + 1 E0 (10, σ (s))

s2 + 3

0

k0 (α, s)

∆s

9

=

s2 + 1 E0 (10, σ (s)) ∑ 2 k0 (α, s) s=0 s + 3

=

∑

9

 s2 + 1

αs2(1−α) s2 + 3 αs2(1−α) −(1 − α)s2α + αs2(1−α)

s=0

1

9

αl 2(1−α) − (1 − α)l 2α αl 2(1−α) l=s

×∏ 9

=

9 αl 2(1−α) − (1 − α)l 2α s2 + 1
∏ αl 2(1−α) (s2 + 3) −(1 − α)s2α + αs2(1−α) l=s

∑

s=0

for αt 2(1−α) − (1 − α)t 2α 6= 0,

t ∈ [0, 10].

Example 1.7.5 Let T = 2N0 , k1 (α,t) = (1 − α)t 4α , f (t) = t 2 ,

k0 (α,t) = αt 4(1−α) ,

α ∈ (0, 1],

t ∈ T.

We will compute Z 16 1

f (t)∆α,16 s

for α − (1 − α)t 8α−3 6= 0, We have σ (t) = 2t, µ(t) = t,

α ∈ (0, 1],

t ∈ T.

t ∈ T,

!

Conformable Dynamic Calculus on Time Scales  41

E0 (16, σ (s)) = e− k1 (t, σ (s)) k0

R 16

= e

1 σ (s) µ(τ)

  k (α,τ) Log 1−µ(τ) k1 (α,τ) ∆τ 0

  R 1 Log 1−µ(τ) k1 (α,τ) ∆τ − sσ (s) µ(τ) k0 (α,τ) e

=

R 16

×e

s

  k (α,τ) Log 1−µ(τ) k1 (α,τ) ∆τ 0

1 µ(τ)

= e− Log(1−

1−α s8α−3 α

8

×e∑l=s Log(1−

)

1−α l 8α−3 α

)

 8  α 1 − α 8α−3 l , 1− α − (1 − α)s8α−3 ∏ α l=s

= α ∈ (0, 1], s ∈ [1, 8]. Then Z 16 1

Z 16

f (s)∆α,16 s =

f (s) 1

Z 16

=

s2

1

=

=

1 α

Z 16

1 α

8

1

∑

E0 (16, σ (s)) ∆s k0 (α, s)

E0 (16, σ (s)) ∆s αs4−4α s4α−2 E0 (16, σ (s))∆s
s4α−1 E0 (16, σ (s))

s=1

! 8  α 1 − α = ∑ s4α−1 α − (1 − α)s8α−3 ∏ 1 − α l 8α−3 s=1 l=s  ! 8 8 s4α−1 1 − α 8α−3 1− = ∑ l , 8α−3 ∏ α s=1 α − (1 − α)s l=s 1 α

8

α ∈ (0, 1], for α − (1 − α)t 8α−3 6= 0,

α ∈ (0, 1],

t ∈ [1, 16].

Theorem 1.7.6 Let α ∈ (0, 1], f ∈ Crd (T) and (1.6) hold. Then Z t a

Dα f (s)∆α,t s = f (t) − f (a)E0 (t, a),

t ∈ [a, b]κ .

42  Conformable Dynamic Equations on Time Scales

Proof 1.7.7 We have Z t a

α

D f (s)∆α,t s = Z t

= a

Z t

 E (t, σ (s)) 0 ∆s k1 (α, s) f (s) + k0 (α, s) f ∆ (s) k0 (α, s)

a

E0 (t, σ (s)) k1 (α, s) f (s) ∆s + k0 (α, s)

Z t

=

k1 (α, s) f (s)

a

−

Z t a

k1 (α, s) f (s)

a

−

Z t a

−k

−

a

−

E0 (t, σ (s)) ∆s + f (t) − f (a)E0 (t, a) k0 (α, s)

k1 (α,s) k0 (α,s) e− k1 (t, s)∆s f (s) k0 a 1 − µ(s) kk1 (α,s) 0 (α,s)

k1 (α, s) f (s)

E0 (t, σ (s)) ∆s + f (t) − f (a)E0 (t, a) k0 (α, s)

Z t a

=

E0 (t, σ (s)) ∆s + f (t) − f (a)E0 (t, a) k0 (α, s)

Z t

Z t

=

s=t E0 (t, σ (s)) ∆s + f (s)E0 (t, s) k0 (α, s) s=a

0

k1 (α, s) f (s)

a

f ∆ (s)E0 (t, σ (s))∆s

f (s)e∆sk1 (t, s)∆s

Z t

=

a

f (s)E0∆s (t, s)∆s

Z t

=

Z t

k1 (α, s) f (s)

E0 (t, σ (s)) ∆s k0 (α, s)

f (t) − f (a)E0 (t, a),

t ∈ [a, b]κ . 

This completes the proof. Theorem 1.7.8 (Integration by Parts) Let f , g ∈ Crd (T) and assume (1.6) holds. Then Z t a

Dα f (s)g(σ (s))∆α,t s =

f (t)g(t) − f (a)g(a)E0 (t, a) −

Z t a

Z t a

Dα f (s)g(s)∆α,t s =

( f (s)Dα g(s) − k1 (α, s) f (s)g(σ (s))) ∆α,t s,

f (t)g(t) − f (a)g(a)E0 (t, a) −

Z t a

( f (σ (s))Dα g(s) − k1 (α, s) f (σ (s))g(s))

× ∆α,t s,

t ∈ [a, b]κ .

Conformable Dynamic Calculus on Time Scales  43

Proof 1.7.9 We have Z t a

Dα f (s)g(σ (s))∆α,t s =

Z t a

Z t

= a

−

(Dα ( f g)(s) − f (s)Dα g(s) + k1 (α, s) f (s)g(σ (s))) ∆α,t s Dα ( f g)(s)∆α,t s

Z t a

=

( f (s)Dα g(s) − k1 (α, s) f (s)g(σ (s))) ∆α,t s

f (t)g(t) − f (a)g(a)E0 (t, a) −

Z t a

( f (s)Dα g(s) − k1 (α, s) f (s)g(σ (s))) ∆α,t s,

t ∈ [a, b]κ .

As above, one can prove the second part of the assertion. This completes the proof.



Theorem 1.7.10 Let f : T × T → R, f ∈ Crd (T × T), f (t, ·), Dtα f (t, ·) ∈ Crd (T), f (·,t) ∈ Crd1 (T) for any t ∈ [a, b]κ , and (1.6) hold. Then Z t  Z t Z t α Dt f (t, s)∆α,t s = Dtα f (t, s)∆α,t s − k1 (α,t) f (σ (t), s)∆α,t s a

a

a

t ∈ [a, b]κ .

+ f (σ (t),t),

Proof 1.7.11 We have Z t  Z t  α D f (t, s)∆α,t s = k1 (α,t) f (t, s)∆α,t s a

a

t

Z +k0 (α,t)

a t

Z = k1 (α,t)

∆t f (t, s)∆α,t s

 f (t, s)∆α,t s

a

∆t E0 (t, σ (s)) +k0 (α,t) f (t, s) ∆s k0 (α, s) a Z t  = k1 (α,t) f (t, s)∆α,t s t

Z

a

t

Z +k0 (α,t)

a

Z t

+k0 (α,t)

 E0 (t, σ (s)) f (t, s) ∆s k0 (α, s) ∆t

a

! E0∆t (t, σ (s)) f (σ (t), s) ∆s k0 (α, s)

44  Conformable Dynamic Equations on Time Scales

+k0 (α,t) f (σ (t),t)

E0 (σ (t), σ (t)) k0 (α,t)

Z t

=

 k1 (α,t) f (t, s) + k0 (α,t) f ∆t (t, s) ∆α,t s

a

e∆t k1 (t, σ (s))

 Z  t

+k0 (α,t) 

f (σ (t), s)

−k

a

0

k0 (α, s)

  ∆s

+ f (σ (t),t) Z t

= a

Dtα f (t, s)∆α,t s

−k1 (α,t)

Z t

f (σ (t), s)∆α,t s

a

t ∈ [a, b]κ .

+ f (σ (t),t),



This completes the proof. Theorem 1.7.12 Let f ∈ Crd (T) and (1.6) hold. Then Z t a

f (s)∆α,t s = −E0 (t, a)

Z a

f (s)∆α,a s,

t

t ∈ [a, b]κ .

Proof 1.7.13 We have Z t a

Z t

f (s)∆α,t s =

f (s) a

= −

E0 (t, σ (s)) ∆s k0 (α, s)

Z a

f (s)

E0 (t, σ (s)) ∆s k0 (α, s)

f (s)

E0 (t, a)E0 (a, σ (s)) ∆s k0 (α, s)

t

= −

Z a t

= −E0 (t, a) = −E0 (t, a)

Z a

f (s) t

E0 (a, σ (s)) ∆s k0 (α, s)

Z a t

f (s)∆α,a s. 

This completes the proof. Theorem 1.7.14 Let f ∈ Crd (T) and (1.6) hold. Then Z t a

Z c

f (s)∆α,t s = E0 (t, c)

a

Z t

f (s)∆α,c s +

c

f (s)∆α,t s,

c,t ∈ [a, b]κ .

Conformable Dynamic Calculus on Time Scales  45

Proof 1.7.15 We have Z t a

Z t

f (s)∆α,t s =

f (s)

E0 (t, σ (s)) ∆s k0 (α, s)

f (s)

E0 (t, c)E0 (c, σ (s)) ∆s k0 (α, s)

a

Z c

= a

Z t

+

f (s) c

E0 (t, σ (s)) ∆s k0 (α, s)

Z c

= E0 (t, c)

a

Z t

f (s)∆α,c s +

c

f (s)∆α,t s,

c,t ∈ [a, b]κ . 

This completes the proof.

1.8

TAYLOR’S FORMULA

Definition 1.8.1 Suppose that α ∈ (0, 1] and (1.6) hold. Define g0 (t, s) = 1, gn (t, s) =

Z t gn−1 (σ (τ), s)k0 (α, τ)E0 (σ (τ),t)

k0 (α, τ) − µ(τ)k1 (α, τ)

s

∆α,t τ.

We have gn (t, s) =

Z t gn−1 (σ (τ), s)k0 (α, τ)E0 (σ (τ),t)E0 (t, σ (τ)) s

Z t

= s

(k0 (α, τ) − µ(τ)k1 (α, τ)) k0 (α, τ)

∆τ

gn−1 (σ (τ), s) ∆τ, k0 (α, τ) − µ(τ)k1 (α, τ)

whereupon g∆n t (t, s) =

gn−1 (σ (t), s) , k0 (α,t) − µ(t)k1 (α,t)

or gn−1 (σ (t), s) = k0 (α,t)g∆n t (t, s) − µ(t)g∆n t (t, s)k1 (α,t) = k0 (α,t)g∆n t (t, s) − (gn (σ (t), s) − gn (t, s)) k1 (α,t) = k0 (α,t)g∆n t (t, s) + gn (t, s)k1 (α,t) −k1 (α,t)gn (σ (t), s)

46  Conformable Dynamic Equations on Time Scales

= Dtα gn (t, s) − k1 (α,t)gn (σ (t), s), Dtα g0 (t, s) − k1 (α,t) = k1 (α,t) − k1 (α,t) = 0,

s,t ∈ T.

Theorem 1.8.2 Let α ∈ (0, 1] and (1.6) hold. If f is n-times conformable ∆-differentiable on T, then !   n−1
Dα ∑ (−1)k (Dα )k f gk = (−1)n−1 (Dα )n f gσn−1 . (1.7) k=0

Proof 1.8.3 We have n−1

D

α

  α k (−1) (D ) f gk ∑ k

!

n−1

=

k=0

   (Dα )k f gk

k=0

n−1

=

∑ (−1)k Dα

∑ (−1)k

       (Dα )k+1 f gσk + (Dα )k f Dα gk − k1 (Dα )k f gσk

k=0 n−1

=

∑ (−1)k

     (Dα )k+1 f gσk + (Dα )k f (Dα gk − k1 gσk )

k=0 n−1

=

∑ (−1)k

    n−1 (Dα )k+1 f gσk + ∑ (−1)k (Dα )k f gσk−1 + f (Dα g0 − k1 )

k=0

k=1

 
= (Dα f ) gσ0 − (Dα )2 f gσ1 + · · · + (−1)n−1 (Dα )n f gσn−1     − (Dα f ) gσ0 + (Dα )2 f gσ1 + · · · + (−1)n−1 (Dα )n−1 f gσn−2
= (−1)n−1 (Dα )n f gσn−1 . 

This completes the proof.

Theorem 1.8.4 (Taylor’s Formula) Let α ∈ (0, 1] and (1.6) hold. If f is n-times conformable ∆-differentiable on T, then n−1

f (t) =

  k α k (D ) f (s)gk (s,t)E0 (t, s) (−1) ∑

k=0

+(−1)n−1

(1.8) Z t s

(Dα )n f (τ)gn−1 (σ (τ),t)∆α,t τ,


s,t ∈ T.

Conformable Dynamic Calculus on Time Scales  47

Proof 1.8.5 We integrate (1.7) from s to t and we get Z t s


(−1)n−1 (Dα )n f (τ)gn−1 (σ (τ),t)∆α,t τ Z t

=

!   ∑ (−1)k (Dα )k f (τ)gk (τ,t) ∆α,t τ

n−1

Dα

s

k=0

n−1

=

  k α k (−1) (D ) f (t)gk (t,t) ∑

k=0

  n−1 − ∑ (−1)k (Dα )k f (s)gk (s,t)E0 (t, s) k=0

=

  n−1 f (t) − ∑ (−1)k (Dα )k f (s)gk (s,t)E0 (t, s),

s,t ∈ T,

k=0



whereupon we get (1.8). This completes the proof. Now we define the polynomials h0 (t, s) = E0 (t, s), Z t

hn (t, s) =

s

hn−1 (τ, s)∆α,t τ,

α ∈ (0, 1],

s,t ∈ T.

Note that h0 (s, s) = 1, Dα h0 (t, s) = 0, Dα hn (t, s) = hn−1 (t, s),

n ∈ N,

s,t ∈ T.

Theorem 1.8.6 Suppose that α ∈ (0, 1] and assume (1.6) holds. Then hn (t, s) = (−1)n gn (s,t)E0 (t, s),

s,t ∈ T,

n ∈ N.

Proof 1.8.7 Note that (Dtα )k hn (t, s) = hn−k (t, s),

s,t ∈ T,

k ∈ T,

0 ≤ k ≤ n,

α ∈ (0, 1].

0 ≤ k ≤ n − 1,

α ∈ (0, 1],

Hence, (Dtα )k hn (s, s) = 0,

s ∈ T,

k ∈ N,

48  Conformable Dynamic Equations on Time Scales

(Dtα )n hn (s, s) = 1, (Dtα )n+1 hn (t, s) = Dtα h0 (t, s) t, s ∈ T,

= 0,

α ∈ (0, 1].

From here and from Taylor’s formula (1.8), we get n

  k α k (−1) (D ) h n (s, s)gk (s,t)E0 (t, s) ∑

hn (t, s) =

k=0

+(−1)n

Z t s

n−1

∑ (−1)k

=

 (Dα )n+1 hn (τ, s)gn (σ (τ),t)∆α,t τ

  (Dα )k hn (s, s)gk (s,t)E0 (t, s)

k=0

+(−1)n gn (s,t)E0 (t, s) = (−1)n gn (s,t)E0 (t, s),

s,t ∈ T,

α ∈ (0, 1],

n ∈ N. 

This completes the proof.

Corollary 1.8.8 (Taylor’s Formula) Let α ∈ (0, 1] and assume (1.6) holds. If f is n-times conformable ∆-differentiable on T, then f (t) =

n−1 

 (Dα )k f (s)hk (t, s)

∑

k=0

Z t

+ s

1.9


(Dα )n f (τ)hn−1 (t, σ (τ)) (E0 (t, σ (τ)))−1 ∆α,t τ,

s,t ∈ T.

CALCULUS FOR THE NABLA CONFORMABLE DERIVATIVE

Suppose that k0 and k1 satisfy (A1). In this section we consider the nabla version of the conformable derivative. In particular, the nabla conformable derivative on time scales is given by b α f (t) = k1 (α,t) f (t) + k0 (α,t) f ∇ (t), D t ∈ Tκ (1.9) provided the right-hand side exists at t, where f ∇ is the time scale nabla derivative. We continue with the next important definition, which establishes a type of exponential function for derivative (1.9). We assume throughout that the reader is familiar with time scale notation, and the basics of nabla derivatives on time scales found in [1]. In particular, recall the graininess function ν(t) = t − ρ(t), where ρ is the backward jump operator ρ(t) := sup{s ∈ T : s < t}, and the nabla derivative [4] of f at t, denoted f ∇ (t), to be the

Conformable Dynamic Calculus on Time Scales  49

number (provided it exists) with the property that given any ε > 0, there is a neighborhood U of t such that | f (ρ(t)) − f (s) − f ∇ (t)[ρ(t) − s]| ≤ ε|ρ(t) − s| for all s ∈ U. Definition 1.9.1 (ν-Regressive) A function f : T → R is conformable ν-regressive if and only if k0 (α,t) − ν(t) ( f (t) − k1 (α,t)) 6= 0 for any α ∈ (0, 1] and for any t ∈ T. The set of all conformable ν-regressive functions on cc . T will be denoted by R Remark 1.9.2 We will often need to divide by the expression k0 + νk1 . Since all three functions are non-negative, and for most of the development to follow, we assume α ∈ (0, 1], we then have k0 (α,t) + ν(t)k1 (α,t) 6= 0 for all t ∈ Tk for all α ∈ (0, 1]. cc , the nabla conformable circle plus is given by Definition 1.9.3 For f , g ∈ R b c g)(t) = f (t) + g(t) − k1 (α,t) − ν(t) (f⊕

( f (t) − k1 (α,t))(g(t) − k1 (α,t)) , k0 (α,t)

and the nabla conformable circle minus is given by b c f (t) = k1 (α,t) −

k0 (α,t)( f (t) − k1 (α,t)) . k0 (α,t) − ν(t)( f (t) − k1 (α,t))

The following theorem is straightforward.   cc , ⊕ b c is an Abelian group. Theorem 1.9.4 The set with binary operation R Definition 1.9.5 (Nabla Conformable Exponential Function) Let α ∈ (0, 1], the points cc . Then the nabla conformable exponential function s,t ∈ T, and let the function p ∈ R b with respect to Dα in (1.9) is defined to be Ebp (t, s) := ebp−k1 (t, s),

(1.10)

k0

where eb is the nabla time scale exponential [1, Section 3]. Note that Ebp satisfies b α Ebp (t, s) = p(t)Ebp (t, s), D

t ∈ Tk ,

s ∈ T,

(1.11)

using (1.9). The following properties of the nabla exponential function are useful direct results of (1.10).

50  Conformable Dynamic Equations on Time Scales

b α satisfy (1.9), let p, q ∈ Lemma 1.9.6 (Nabla Exponential Function Properties) Let D cc , and let t, s, r ∈ T. For the nabla conformable exponential function given in (1.10), the R following properties hold. (i) Ebp (t,t) ≡ 1. (ii) Ebp (t, s)Ebp (s, r) = Ebp (t, r). (iii)

1 Ebp (t, s)

= Ebp (s,t) = Eb b c p (t, s).

(iv) Ebk1 (t, s) ≡ 1 ≡ Eb b c k1 (t, s).   p(t) − k (α,t) 1 (v) Ebp (ρ(t), s) = 1 − ν(t) Ebp (t, s). k0 (α,t) (vi) Ebp (t, s)Ebq (t, s) = Ebp⊕b c q (t, s). (vii)

Ebp (t, s) b = E p b c q (t, s). Ebq (t, s)

(viii) If p is nabla differentiable and

bα p D cc , then ∈R p p(t) Eb Dbα p  (t, s) = . p(s) p

Theorem 1.9.7 (Fundamental Theorem of Integral Calculus) Let α ∈ (0, 1]. Suppose b α is integrable on [a, b]T . Then f : [a, b]T → R is differentiable on [a, b]T and D Z t a

b b α [ f (s)] E0 (t, ρ(s)) ∇s = f (t) − f (a)Eb0 (t, a), D k0 (α, s)

and

"Z bα D

a

t

# Eb0 (t, ρ(s)) f (s) ∇s = f (t). k0 (α, s)

cc and Eb0 is well defined. The equalities then Proof 1.9.8 Since k0 + νk1 6= 0, we have 0 ∈ R follow.  b α satisfy (1.9) for α ∈ [0, 1]. Let the functions Lemma 1.9.9 (Basic Derivatives) Let D f , g : T → R be nabla differentiable as needed. Moreover, define f ρ (t) := f (ρ(t)). Then b α c = ck1 (α,t) for all constants c ∈ R; 1. D

Conformable Dynamic Calculus on Time Scales  51

b α [a f + bg] = aD b α [ f ] + bD b α [g] for all a, b ∈ R; 2. D b α [ f g] = f D b α [g] + gρ D b α [ f ] − f gρ k1 ; 3. D b α [ f g] = gD bα [ f ] + f ρ D b α [g] − f ρ gk1 ; 4. D b b b α [ f /g] = gDα [ f ] − f Dα [g] + f k1 . 5. D gρ g g Theorem 1.9.10 (Equivalence of Exponential Functions) Let α ∈ (0, 1]. If f is continuρ ρ ous and κ0 + ν( f ρ − κ1 ) 6= 0, then E f (t, s) = Eb

κ0 ( f −κ1 )ρ ρ κ0 +ν( f −κ1 )ρ

+κ1

(t, s).

If g is continuous and κ0σ − µ( f σ − κ1σ ) 6= 0, then Ebg (t, s) = E

κ0 (g−κ1 )σ κ0σ −µ(g−κ1 )σ

+κ1

(t, s).

Proof 1.9.11 It is known on time scales (α = 1) that if a function p is continuous and regressive, then e p (t, s) = eb pρ (t, s). 1+ν pρ

Hence, suppressing the arguments, and letting F := E f = eF = eb

Fρ 1+νF ρ

f − κ1 , we see that κ0

= Eb κ0 F ρ

1+νF ρ

+κ1

,

cc . We which yields the first equation. We must also check that the resulting base of Eb is in R have # ! " ρ κ0 κ0 κ0 ( f − κ1 )ρ κ0 − ν = − κ + κ 6= 0, 1 1 ρ ρ κ0 + ν( f − κ1 )ρ κ0 + ν( f − κ1 )ρ and the result holds. Next, if a function q is continuous and ν-regressive, then ebq (t, s) = e

qσ 1−µqσ

(t, s).

Similar calculations then yield the second equation and the appropriate regressivity. This completes the proof. 

1.10

CONFORMABLE PARTIAL DERIVATIVES

Definition 1.10.1 Let u : T×T → C be delta differentiable with respect to the first variable or with respect to the second variable at some point (t, s) ∈ T × T. Define the conformable ∆-partial derivative with respect to the first variable or the conformable ∆-partial derivative with respect to the second variable at (t, s) as follows:

52  Conformable Dynamic Equations on Time Scales

Dtα u(t, s) = k1 (α,t)u(t, s) + k0 (α,t)ut∆ (t, s), Dαs u(t, s) = k1 (α, s)u(t, s) + k0 (α, s)u∆s (t, s), respectively. Example 1.10.2 Let T = Z, k1 (α,t) = (1 − α)t α , u(t, s) = s2 + st,

k0 (α,t) = αt 1−α ,

α ∈ (0, 1],

(t, s) ∈ T × T.

We will find 1

1

Dt4 u(t, s),

Ds4 u(t, s),

(t, s) ∈ T × T.

We have σ (t) = t + 1,  k1  k0

1 ,t 4



1 ,t 4



=

3 1 t4, 4

=

1 3 t4, 4

t ∈ T,

t ∈ T,

ut∆ (t, s) = s, u∆s (t, s) = σ (s) + s + t = s+1+s+t = 2s + t + 1,

(t, s) ∈ T × T.

Then 

1 4

Dt u(t, s) = k1 =

   1 1 ,t u(t, s) + k0 ,t ut∆ (t, s) 4 4


1 3 3 1 2 t 4 s + st + t 4 s 4 4 1

 1 t4s  3s + 3t + t 2 , 4     1 1 1 Ds4 u(t, s) = k1 , s u(t, s) + k0 , s u∆s (t, s) 4 4 =

t ∈ T,

Conformable Dynamic Calculus on Time Scales  53

=


1 3 3 1 2 s 4 s + st + s 4 (2s + t + 1) 4 4

=

1  3 1 1 s4  2 3s + 3st + 2s 2 + ts 2 + s 2 , 4

(t, s) ∈ T × T.

Exercise 1.10.3 Let T = 2N0 , k1 (α,t) = (1 − α)t 2α ,

k0 (α,t) = αt 2(1−α) ,

u(t, s) = s3 + 3s2t + st 2 + t 3 , Find

t ∈ T,

(t, s) ∈ T| × T.

1

1

Dt2 u(t, s),

α ∈ (0, 1],

1

Dt3 u(t, s),

Ds4 u(t, s),

(t, s) ∈ T × T.

Theorem 1.10.4 Let u, v : T × T → C be delta differentiable with respect to the first variable at some point (t, s) ∈ T × T. Then 1. Dtα (u + v)(t, s) = Dtα u(t, s) + Dtα v(t, s), 2. Dtα (au)(t, s) = aDtα u(t, s). Proof 1.10.5

1. We have Dtα (u + v)(t, s) = k1 (α,t)(u + v)(t, s) + k0 (α,t)(u + v)t∆ (t, s) = k1 (α,t)u(t, s) + k1 (α,t)v(t, s) +k0 (α,t)ut∆ (t, s) + k1 (α,t)vt∆ (t, s) = Dtα u(t, s) + Dtα v(t, s).

2. We have Dtα (au)(t, s) = k1 (α,t)(au)(t, s) + k0 (α,t)(au)t∆ (t, s) = ak1 (α,t)u(t, s) + ak0 (α,t)ut∆ (t, s) = aDtα u(t, s). This completes the proof.



54  Conformable Dynamic Equations on Time Scales

Theorem 1.10.6 Let u, v : T × T → C be delta differentiable with respect to the first variable at some point (t, s) ∈ T × T. Then Dtα (uv)(t, s) = (Dtα u(t, s)) v(t, s) + u(σ (t), s)Dtα v(t, s) −k1 (α,t)u(σ (t), s)v(t, s) = (Dtα u(t, s)) v(σ (t), s) + u(t, s)Dtα v(t, s) −k1 (α,t)u(t, s)v(σ (t), s). Proof 1.10.7 We have Dtα (uv)(t, s) = k1 (α,t)u(t, s)v(t, s) + k0 (α,t)(uv)t∆ (t, s) = k1 (α,t)u(t, s)v(t, s) +k0 (α,t)ut∆ (t, s)v(t, s) +k0 (α,t)u(σ (t), s)vt∆ (t, s) = v(t, s)Dtα u(t, s) +k0 (α,t)u(σ (t), s)vt∆ (t, s) +k1 (α,t)u(σ (t), s)v(t, s) −k1 (α,t)u(σ (t), s)v(t, s) = v(t, s)Dtα u(t, s) + u(σ (t), s)Dtα v(t, s) −k1 (α,t)u(σ (t), s)v(t, s) = k1 (α,t)u(t, s)v(t, s) +k0 (α,t)ut∆ (t, s)v(σ (t), s) +k0 (α,t)u(t, s)vt∆ (t, s)

Conformable Dynamic Calculus on Time Scales  55

= u(t, s)Dtα v(t, s) +k0 (α,t)ut∆ (t, s)v(σ (t), s) +k1 (α,t)u(t, s)v(σ (t), s) −k1 (α,t)u(t, s)v(σ (t), s) = u(t, s)Dtα v(t, s) + v(σ (t), s)Dtα u(t, s) −k1 (α,t)u(t, s)v(σ (t), s). 

This completes the proof.

Theorem 1.10.8 Suppose that u, v : T × T → C are delta differentiable with respect to the first variable at some point (t, s) ∈ T × T. Then u v(t, s)Dtα u(t, s) − u(t, s)Dtα v(t, s) (t, s) = Dtα v v(t, s)v(σ (t), s) +k1 (α,t)

u(t, s) v(t, s)

provided that v(t, s)v(σ (t), s) 6= 0. Proof 1.10.9 We have u  u ∆ u(t, s) Dtα (t, s) = k1 (α,t) + k0 (α,t) (t, s) v v(t, s) v t = k1 (α,t)

u(t, s) v(t, s)

+k0 (α,t)

=

ut∆ (t, s)v(t, s) − u(t, s)vt∆ (t, s) v(t, s)v(σ (t), s)

1 k1 (α,t)u(t, s)v(σ (t), s) v(t, s)v(σ (t), s) +k0 (α,t)ut∆ (t, s)v(t, s) ! −k0 (α,t)u(t, s)vt∆ (t, s)

=

1 k1 (α,t)u(t, s)v(σ (t), s) v(t, s)v(σ (t), s)

56  Conformable Dynamic Equations on Time Scales

+k0 (α,t)ut∆ (t, s)v(t, s) +k1 (α,t)u(t, s)v(t, s) −k1 (α,t)u(t, s)v(t, s) ! −k0 (α,t)u(t, s)vt∆ (t, s)

=

1 k1 (α,t)u(t, s)v(σ (t), s) v(t, s)v(σ (t), s) ! +v(t, s)Dtα u(t, s) − u(t, s)Dtα v(t, s)

=

v(t, s)Dtα u(t, s) − u(t, s)Dtα v(t, s) v(t, s)v(σ (t), s) +k1 (α,t)

u(t, s) . v(t, s) 

This completes the proof. Example 1.10.10 Let T = 2N0 , k1 (α,t) = (1 − α)t 2α ,

k0 (α,t) = αt 2(1−α) ,

u(t, s) = 1 + ts, v(t, s) = t 2 + s2 ,

(t, s) ∈ T × T.

We will find 1

Dt2

u v

(t, s),

(t, s) ∈ T × T.

We have σ (t) = 2t, ut∆ (t, s) = s, vt∆ (t, s) = σ (t) + t = 2t + t = 3t,

α ∈ (0, 1],

t ∈ T,

Conformable Dynamic Calculus on Time Scales  57

 k1 

1 ,t 2



1 ,t 2



1 t, 2

=

1 t, 2     1 1 1 Dt2 u(t, s) = k1 ,t u(t, s) + k0 ,t ut∆ (t, s) 2 2

k0

=

1 1 t(1 + ts) + t(3t) 2 2

=

1 t(1 + 3t + st), 2     1 1 1 2 Dt v(t, s) = k1 ,t v(t, s) + k0 ,t vt∆ (t, s) 2 2 =

=

1 2 2 1 t(t + s ) + t(3t) 2 2

=

1 2 t(t + 3t + s2 ), 2

v(σ (t), s) = (σ (t))2 + s2 = (2t)2 + s2 = 4t 2 + s2 ,

(t, s) ∈ T × T.

Then 1 2

Dt

  1 1 2 2 (t, s) = (t + s ) t(1 + st + 3t) v (t 2 + s2 )(4t 2 + s2 ) 2  ! 1 2 1 1 + ts 2 −(1 + ts) t(t + 3t + s ) + t 2 2 2 2 t +s

u

=

1 t 2 + st 3 + 3t 3 + s2 + s3t + 3s2t 2(t 2 + s2 )(4t 2 + s2 ) −t 2 − 3t − s2 − t 3 s − 3t 2 s − ts2 ! 2

2

+(1 + ts)(4t + s ) = (t, s) ∈ T × T.

t 2(t 2 + s2 )(4t 2 + s2 )


3t 3 + 3s2t − 3t 2 s − 3t + 4t 2 + s2 + 4t 3 s + ts3 ,

58  Conformable Dynamic Equations on Time Scales

Exercise 1.10.11 Let T = 2Z, k1 (α,t) = (1 − α)t 3α ,

k0 (α,t) = αt 3(1−α) ,

α ∈ (0, 1],

t ∈ T,

u(t, s) = t 3 + 3ts2 + s3 , v(t, s) = t 4 + s4 , Find

(t, s) ∈ T × T. 1

Dt5

1.11

u v

(t, s) ∈ T × T.

(t, s),

ADVANCED PRACTICAL PROBLEMS

Problem 1.11.1 Let T = 2N0 ,  π 2 k1 (α,t) = sin (1 − α)t α , 2 α ∈ [0, 1], t ∈ T,

 π 2 k0 (α,t) = t 1−α cos (1 − α)t α , 2

f (t) = t 2 ,

Find

1

D 3 f (t),

t ∈ T. t ∈ T.

Problem 1.11.2 Let T = 2N0 , k1 (α,t) = (1 − α)t α , f (t) = t 3 + 3t,

k0 (α,t) = αt 1−α ,

g(t) = t 2 + t,

α ∈ [0, 1],

t ∈ T,

t ∈ T.

Find 1

1. D 2 ( f + g)(t), 1

2. D 2 ( f g − g2 )(t),   1 f −g 3. D 4 (t), f +g   1 f 4. D 6 (t). g Problem 1.11.3 Let T = 2N0 , k1 (α,t) = (1 − α)(1 + t)α , f (t) = t 2 ,

t ∈ T.

k0 (α,t) = α(1 + t)1−α ,

α ∈ [0, 1],

t ∈ T,

Conformable Dynamic Calculus on Time Scales  59

Find 1

1. D 6 f (t),  1  1 2. D 6 D 6 f (t),  1  1  1 3. D 6 D 6 D 6 f (t). Problem 1.11.4 Let T = 4Z, k1 (α,t) = (1 − α) 1 + 2t 2 f (t) = t 2 + t,


α

k0 (α,t) = α 1 + 2t 2

,


1−α

,

α ∈ (0, 1],

α ∈ (0, 1],

t ∈ [0, 40].

t ∈ T,

t ∈ T.

Compute Z 40

f (s)∆α,40 s

0

for α 1 + 2t 2


1−α

− 4(1 − α) 1 + 2t 2


α

6= 0,

Problem 1.11.5 Let T = qN0 , q > 1. Find g1 (t, s),

g2 (t, s),

h1 (t, s),

t, s ∈ T.

h2 (t, s),

Problem 1.11.6 Let T = 3N0 , k0 (α,t) = αt 1−α ,

k1 (α,t) = (1 − α)t α , u(t, s) = (s + t)2 + t 4 , Find

t ∈ T,

(t, s) ∈ T × T. 1

1

Dt2 u(t, s),

α ∈ (0, 1],

1

Ds3 u(t, s),

Ds4 u(t, s).

Problem 1.11.7 Let T = 3N0 , k1 (α,t) = (1 − α)t 2α ,

k0 (α,t) = αt 4(1−α) ,

α ∈ (0, 1],

t ∈ T,

u(t, s) = 2t 2 + 3ts2 + 4s2 , v(t, s) = 3t 2 + 5s2 , Find

1

Dt3

1.12

(t, s) ∈ T × T.

u v

(t, s),

(t, s) ∈ T × T.

NOTES AND REFERENCES

This chapter introduces the concepts of conformable delta (Hilger) and nabla derivatives on time scales, and some of their properties. Results in this chapter include the basic conformable delta derivative, the conformable exponential function, the conformable logarithm function, conformable trigonometric and hyperbolic functions, the conformable delta integral and integral rules and Taylor’s formula.

CHAPTER

2

First-Order Linear Dynamic Equations

Suppose that T is a time scale with forward jump operator and delta differentiation operator σ and ∆, respectively. Let α ∈ (0, 1], k0 and k1 satisfy (A1), and assume (1.6) holds.

2.1

LINEAR FIRST-ORDER DYNAMIC EQUATIONS

Let p, q ∈ Crd (T). Consider the IVP Dα y = (p(t) + k1 (α,t)) y + q(t),

t ∈ Tκ ,

y(t0 ) = y0 ,

(2.1) (2.2)

where t0 ∈ T, y0 ∈ R. Theorem 2.1.1 Suppose that k0 (α,t) + µ(t)p(t) 6= 0,

α ∈ (0, 1],

t ∈ T.

Then the problem (2.1), (2.2) has a unique solution represented in the form Z t

y(t) = y0 E p+k1 (t,t0 ) +

t0

where g=

q(s)Eg (σ (s),t)∆α,t s,

s,t ∈ Tκ ,

(2.3)

p(µk1 − k0 ) . k0 + µ p

Proof 2.1.2 Let t ∈ Tκ . We multiply both sides of equation (2.1) by Eg (σ (t),t0 ) to get (Dα y(t)) Eg (σ (t),t0 ) = (p(t) + k1 (α,t)) y(t)Eg (σ (t),t0 ) + q(t)Eg (σ (t),t0 ).

(2.4)

Note that   g(t) − k1 (α,t) p(t)Eg (σ (t),t0 ) = p(t) 1 + µ(t) Eg (t,t0 ) k0 (α,t) 61

62  Conformable Dynamic Equations on Time Scales

 = p(t) 1 + µ(t)

p(t)(µ(t)k1 (α,t)−k0 (α,t)) k0 (α,t)+µ(t)p(t)

− k1 (α,t)

k0 (α,t)

  Eg (t,t0 )

  µ(t)p(t)k1 (α,t) − p(t)k0 (α,t) − k0 (α,t)k1 (α,t) − µ(t)p(t)k1 (α,t) = p(t) 1 + µ(t) k0 (α,t) (k0 (α,t) + µ(t)p(t)) ×Eg (t,t0 )

= p(t)

(k0 (α,t))2 + µ(t)k0 (α,t)p(t) − µ(t)k0 (α,t)p(t) − µ(t)k0 (α,t)k1 (α,t) Eg (t,t0 ) k0 (α,t) (k0 (α,t) + µ(t)p(t))

= p(t)

k0 (α,t) − µ(t)k1 (α,t) Eg (t,t0 ) k0 (α,t) + µ(t)p(t)

= −g(t)Eg (t,t0 ) = −Dα Eg (t,t0 ),

t ∈ Tκ .

Then (2.4) takes the form (Dα y(t)) Eg (σ (t),t0 ) + (Dα Eg (t,t0 )) y(t) − k1 (α,t)Eg (σ (t),t0 )y(t) = q(t)Eg (σ (t),t0 ), t ∈ Tκ , or Dα (yEg (·,t0 )) (t) = q(t)Eg (σ (t),t0 ),

t ∈ Tκ .

We integrate the last equality from t0 to t and we get Z t t0

Dα (yEg (·,t0 )) (s)∆α,t s =

Z t t0

q(s)Eg (σ (s),t0 )∆α,t s,

or y(t)Eg (t,t0 ) − y(t0 )Eg (t0 ,t0 )E0 (t,t0 ) = or

Z t t0

q(s)Eg (σ (s),t0 )∆α,t s,

Z t

y(t)Eg (t,t0 ) = y0 E0 (t,t0 ) +

t0

t ∈ Tκ ,

q(s)Eg (σ (s),t0 )∆α,t s,

t ∈ Tκ ,

t ∈ Tκ ,

or E0 (t,t0 ) y(t) = y0 + Eg (t,t0 )

Z t t0

q(s)Eg (σ (s),t0 )Eg (t0 ,t)∆α,t s Z t

= y0 E0 c g (t,t0 ) +

t0

q(s)Eg (σ (s),t)∆α,t s,

We have g − k1 =

p(µk1 − k0 ) − k1 k0 + µ p

t ∈ Tκ .

First-Order Linear Dynamic Equations  63

=

p(µk1 − k0 ) − k0 k1 − µ pk1 k0 + µ p

=

µ pk1 − pk0 − k0 k1 − µ pk1 k0 + µ p

=

−pk0 − k0 k1 , k0 + µ p

c g = − c g − k1 = −

k0 (g − k1 ) + k1 , k0 + µ(g − k1 ) k0 (g − k1 ) k0 + µ(g − k1 )

=

k02 (p+k1 ) k0 +µ p 1) k0 − µkk00(p+k +µ p

=

k0 (p + k1 ) k0 + µ p − µ p − µk1

=

k0 (p + k1 ) , k0 − µk1

0 c g = 0 ⊕c ( c g) = 0 + ( c g) − k1 + µ = ( c g) − k1 − = ( c g − k1 ) =

(0 − k1 )( c g − k1 ) k0

µk1 ( c g − k1 ) k0

k0 − µk1 k0

k0 (p + k1 ) k0 − µk1 k0 − µk1 k0

= p + k1 . Consequently, Z t

y(t) = y0 E p+k1 (t,t0 ) +

t0

q(s)Eg (σ (s),t)∆α,t s,

We differentiate with respect to t the last equality and we get
Dα y(t) = y0 Dα E p+k1 (·,t0 ) (t)

t ∈ Tκ .

64  Conformable Dynamic Equations on Time Scales

 α

+D

1 Eg (t,t0 )

Z t

 q(s)Eg (σ (s),t0 )∆α,t s

t0

= (p(t) + k1 (α,t)) y0 E p+k1 (t,t0 )  α

+D +

1 Eg (t,t0 )

Z

t

t0

q(s)Eg (σ (s),t0 )∆α,t s

1 q(t)Eg (σ (t),t0 ) Eg (σ (t),t0 )

−k1 (α,t)

1 Eg (σ (t),t0 )

Z t t0

q(s)Eg (σ (s),t0 )∆α,t s

= (p(t) + k1 (α,t)) y0 E p+k1 (t,t0 ) + q(t) +

Eg (t,t0 )k1 (α,t) − g(t)Eg (t,t0 ) Eg (t,t0 )Eg (σ (t),t0 )

+k1 (α,t) −

1 Eg (t,t0 )

k1 (α,t) Eg (σ (t),t0 )

Z t t0

q(s)Eg (σ (s),t0 )∆α,t s

Z t t0

q(s)Eg (σ (s),t0 )∆α,t s

Z t t0

q(s)Eg (σ (s),t0 )∆α,t s

= (p(t) + k1 (α,t)) y0 E p+k1 (t,t0 ) + q(t) g(t) − Eg (σ (t),t0 ) +k1 (α,t)

Z t t0

q(s)Eg (σ (s),t0 )∆α,t s

1 Eg (t,t0 )

Z t t0

q(s)Eg (σ (s),t0 )∆α,t s

= (p(t) + k1 (α,t)) y0 E p+k1 (t,t0 ) + q(t) 1 +k1 (α,t) Eg (t,t0 )

Z t t0

q(s)Eg (σ (s),t0 )∆α,t s

g(t)  − g(t)−k1 (α,t) 1 + µ(t) k (α,t) Eg (t,t0 )

Z t t0

q(s)Eg (σ (s),t0 )∆α,t s

0

= (p(t) + k1 (α,t)) y0 E p+k1 (t,t0 ) + q(t) +k1 (α,t)

1 Eg (t,t0 )

Z t t0

q(s)Eg (σ (s),t0 )∆α,t s

First-Order Linear Dynamic Equations  65

−

Z t

g(t) 1 (α,t) 1 + µ(t) g(t)−k k (α,t)

t0

0

q(s)Eg (σ (s),t)∆α,t s

= (p(t) + k1 (α,t)) y0 E p+k1 (t,t0 ) + q(t) +k1 (α,t) −

1 Eg (t,t0 )

Z t t0

q(s)Eg (σ (s),t0 )∆α,t s

g(t)k0 (α,t) k0 (α,t) + µ(t) (g(t) − k1 (α,t))

Z t t0

q(s)Eg (σ (s),t)∆α,t s,

Note that gk0 − k0 + µ(g − k1 )

=

p(µk1 −k0 ) k0 +µ p k0 − 1) k0 − µ k0k(p+k 0 +µ p

=

p(µk1 −k0 ) k0 +µ p − 1) 1 − µ(p+k k0 +µ p

= −

p(µk1 − k0 ) k0 − µk1

= p. Therefore Dα y(t) = (p(t) + k1 (α,t)) y0 E p+k1 (t,t0 ) + q(t) Z t

+p(t) t0

+k1 (α,t)

q(s)Eg (σ (s),t)∆α,t s 1 Eg (t,t0 )

Z t t0

q(s)Eg (σ (s),t0 )∆α,t s

= (p(t) + k1 (α,t)) y(t) + q(t), ,

t ∈ Tκ .

Assume that the IVP (2.1), (2.2) has two solutions y1 and y2 . We set v = y1 − y2 . Then v is a solution of the IVP Dα v = (p(t) + k1 (α,t)) v, v(t0 ) = 0.

t ∈ Tκ .

66  Conformable Dynamic Equations on Time Scales

Hence, and considering (2.3), we get that v = 0 on

T,

i.e., y1 = y2

on

T. 

This completes the proof. Remark 2.1.3 Let p ∈ Crd (T)

\

Dα y

Rc and q ∈ Crd (T). Consider the IVP t ∈ Tκ ,

= p(t)y + q(t),

(2.5) y(t0 ) = y0 , where t0 ∈ T, y0 ∈ R. We can rewrite the equation in the form Dα y = (p(t) − k1 (α,t) + k1 (α,t)) y + q(t),

t ∈ Tκ .

Because p ∈ Rc , we have 1 + µ(t) (p(t) − k1 (α,t)) 6= 0,

α ∈ (0, 1],

t ∈ Tκ .

Therefore, the solution y of the IVP (2.5) can be represented in the form Z t

y(t) = y0 E p (t,t0 ) + where g=

t0

q(s)Eg (σ (s),t)∆α,t s,

t ∈ Tκ ,

(p − k1 )(µk1 − k0 ) . k0 + µ(p − k1 )

Example 2.1.4 Let T = 2N0 , 4

k1 (α,t) = (1 − α)t 3 α ,

8

k0 (α,t) = αt 3 α ,

α ∈ (0, 1],

Consider the IVP 3

D4 y =

1 ty + t 2 + 1, 2

y(1) = 2. Here σ (t) = 2t, µ(t) = t, p(t) =

1 t, 2

t ∈ T,

t ∈ T.

First-Order Linear Dynamic Equations  67

q(t) = t 2 + 1,

t ∈ T.

Note that (p − k1 )(µk1 − k0 ) − k1 k0 + µ(p − k1 )

g − k1 =

µk1 (p − k1 ) − k0 (p − k1 ) − k0 k1 − µk1 (p − k1 ) k0 + µ(p − k1 )

=

1+µ

= −

pk0 , k0 + µ(p − k1 )

g − k1 k0

= −

p , k0 + µ(p − k1 )

g − k1 k0

= 1−

µp k0 + µ(p − k1 )

k0 − µk1 . k0 + µ(p − k1 )

= Then 1 + µ(t)

g(t) − k1 k0

3 4 ,t

3 ,t
4




3 4 ,t

k0

=

3 4 ,t




− µ(t)k1

3 4 ,t




+ µ(t) p(t) − k1
3 2 1 4t − t 4t
3 2 1 1 4t + t 2t − 4t

k0

=

3 2 4t − 3 2 4t −

=

= 1,

1 2 4t 1 2 4t

t ∈ T.

Hence, Eg (σ (s),t) = e g−k1 (σ (s),t) k0

R σ (s) t

= e

1 µ(τ)

log 1+µ(τ)

R σ (s) 1 log 1∆τ

= et = 1,

τ

s,t ∈ T,

g(τ)−k1 34 ,τ k0 34 ,τ

( ) ( )

! ∆τ


3 4 ,t





68  Conformable Dynamic Equations on Time Scales 3 4 ,t
− 3 k0 4 ,t

k1

1 t 1 = − 34 2 = − , 3t 4t




t ∈ T,

and E0 (t, σ (s)) = e− k1 (t, σ (s)) k0

Rt

= e = e = e

= e

1 σ (s) µ(τ)

Rt

1 σ (s) τ

k1 1−µ(τ) k0

log

( 34 ,τ ) ( 34 ,τ )

! ∆τ

log(1− 13 )∆τ

  R R − sσ (s) τ1 ∆τ+ st τ1 ∆τ

(log 32 )

  t 2 1 −1+∑l=s

(log 32 )

t s≤ , 2

t, s ∈ T,

,

and p(t) − k1 k0 34 ,t

3 ,t
4


=

1 1 2t − 4t 3 2 4t

=

1 4t 3 2 4t

=

1 , 3t

E p (t,t0 ) = e p−k1 (t,t0 ) k0

Rt

1 t0 µ(τ)

= e

log

p(τ)−k1 34 ,τ 1+µ(τ) k0 34 ,τ

( ) ( )

! ∆τ

Rt 1 log(1+ 13 )∆τ

= e t0 τ

log 4 = e( 3 )

Rt 1 t τ ∆τ 0

t

= e

2 1 (log 43 ) ∑l=t 0

,

t t0 ≤ , 2

t0 ,t ∈ T.

Therefore Z t

y(t) = y(1)E p (t, 1) +

1

q(s)Eg (σ (s),t)∆α,t s

Z t

= 2E p (t, 1) +

1

4 = 2E p (t, 1) + 3

q(s)Eg (σ (s),t)

Z t 2 s +1 1

s2

E0 (t, σ (s))
∆s k0 34 , s

Eg (σ (s),t)E0 (t, σ (s))∆s

First-Order Linear Dynamic Equations  69 t

4 2 s2 + 1 Eg (σ (s),t)E0 (t, σ (s)) = 2E p (t, 1) + ∑ 3 s=1 s 4

t 2

= 2e(log 3 ) ∑l=1 1 t 2

+

4 ∑ 3 s=1

s2 + 1 s

e

  t 2 1 −1+∑l=s

(log 23 )

t ≥ 2.

,

If t = 2k , k ∈ N, then   k−1 1 log 4 2 y 2k = 2e( 3 ) ∑l=20 k−1

4 2 4m + 1 (log 23 ) + ∑ e 3 2m =20 2m

  k−1 −1+∑2l=2m 1

k−1

4

= 2ek(log 3 ) +

4 2 4m + 1 (log 2 )(k−m−1) 3 e 3 2m∑ 2m =20

 k   k−1 4 4 2 4m + 1 2 k−m−1 = 2 + . 3 3 2m∑ 2m 3 =20 This ends the example. Now we consider the conformable dynamic equation Dα y = (−p(t) + k1 (α,t)) yσ + q(t),

t ∈ Tκ ,

(2.6)

subject to the initial condition (2.2), where p, q ∈ Crd (T), p ∈ Rc , t0 ∈ T, y0 ∈ R. Theorem 2.1.5 The problem (2.6), (2.2) has a unique solution represented in the form Z t

y(t) = y0 Eg (t,t0 ) + where g(t) = −

t0

q(s)E p (s,t)∆α,t s,

t ∈ Tκ ,

(p(t) − k1 (α,t))(k0 (α,t) − µ(t)k1 (α,t)) , k0 (α,t) + µ(t)(p(t) − k1 (α,t))

(2.7)

t ∈ T.

Proof 2.1.6 We multiply both sides of equation (2.6) by E p (t,t0 ) to get (Dα y(t)) E p (t,t0 ) + p(t)E p (t,t0 )yσ (t) − k1 (α,t)E p (t,t0 )yσ (t) = q(t)E p (t,t0 ), or Dα (yE p (·,t0 )) (t) = q(t)E p (t,t0 ),

t ∈ Tκ .

t ∈ Tκ ,

70  Conformable Dynamic Equations on Time Scales

The last equality we integrate from t0 to t and we obtain y(t)E p (t,t0 ) − y(t0 )E p (t0 ,t0 )E0 (t,t0 ) = or

Z t t0

q(s)E p (s,t0 )∆α,t s,

Z t

q(s)E p (s,t0 )∆α,t s,

t ∈ Tκ ,

q(s)E p (s,t0 )E p (t0 ,t)∆α,t s,

t ∈ Tκ ,

y(t)E p (t,t0 ) = y0 E0 (t,t0 ) + or

E0 (t,t0 ) y(t) = y0 + E p (t,t0 )

t0

Z t t0

t ∈ Tκ ,

or

Z t

y(t) = y0 E0 c p (t,t0 ) +

t0

t ∈ Tκ .

q(s)E p (s,t)∆α,t s,

Note that c p = − c p − k1 = −

k0 (p − k1 ) + k1 , k0 + µ(p − k1 ) k0 (p − k1 ) , k0 + µ(p − k1 )

0 ⊕c ( c p) = 0 + ( c p) − k1 + µ = ( c p − k1 ) − µk1 = ( c p − k1 ) = −

(0 − k1 )( c p − k1 ) k0

( c p − k1 ) k0

k0 − µk1 k0

(p − k1 )(k0 − µk1 ) k0 + µ(p − k1 )

= g. Therefore Z t

y(t) = y0 Eg (t,t0 ) +

t0

q(s)E p (s,t)∆α,t s,

t ∈ Tκ .

Now we differentiate with respect to t both sides of the last equality and we obtain   Z t 1 α α α D y(t) = y0 D Eg (t,t0 ) + D q(s)E p (s,t0 )∆α,t s E p (t,t0 ) t0 = y0 g(t)Eg (t,t0 )  α

+D

1 E p (t,t0 )

 Z

σ (t)

t0

 q(s)E p (s,t0 )∆α,t s

First-Order Linear Dynamic Equations  71

1 Dα + E p (t,t0 ) k1 (α,t) − E p (t,t0 )

Z

t

t0

 q(s)E p (s,t0 )∆α,t s

Z σ (t) t0

q(s)E p (s,t0 )∆α,t s

= y0 g(t)Eg (t,t0 ) 

k1 (α,t)E p (t,t0 ) − p(t)E p (t,t0 ) k1 (α,t) + + E p (t,t0 )E p (σ (t),t0 ) E p (t,t0 ) +

q(t)E p (t,t0 ) k1 (α,t) − E p (t,t0 ) E p (t,t0 )

 Z

σ (t)

t0

 q(s)E p (s,t0 )∆α,t s

Z σ (t) t0

q(s)E p (s,t0 )∆α,t s

= y0 g(t)Eg (t,t0 ) + q(t) +

k1 (α,t) − p(t) E p (σ (t),t0 )

= y0 g(t) +

q(s)E p (s,t0 )∆α,t s

t0

k0 (α,t) Eg (σ (t),t0 ) + q(t) k0 (α,t) + µ(t)(g(t) − k1 (α,t))

k1 (α,t) − p(t) E p (σ (t),t0 )

= y0 g(t)

Z σ (t)

Z σ (t)

q(s)E p (s,t0 )∆α,t s

t0

k0 (α,t) Eg (σ (t),t0 ) + q(t) k0 (α,t) + µ(t)(g(t) − k1 (α,t))

+ (k1 (α,t) − p(t))

Z σ (t) t0

q(s)E p (s, σ (t))∆α,t s,

t ∈ Tκ .

Note that g − k1 = −

(p − k1 )(k0 − µk1 ) − k1 k0 + µ(p − k1 )

= −

k0 (p − k1 ) − µk1 (p − k1 ) + k1 k0 + µk1 (p − k1 ) k0 + µ(p − k1 )

= −

k0 p , k0 + µ(p − k1 )

k0 + µ(g − k1 ) = k0 −

µk0 p k0 + µ(p − k1 )

=

k0 (k0 + µ p − µk1 − µ p) k0 + µ(p − k1 )

=

k0 (k0 − µk1 ) , k0 + µ(p − k1 )

72  Conformable Dynamic Equations on Time Scales

gk0 k0 + µ(g − k1 )

= −

k0 (p−k1 )(k0 −µk1 ) k0 +µ(p−k1 ) k0 (k0 −µk1 ) k0 +µ(p−k1 )

= k1 − p. Therefore Dα y(t) = y0 (k1 (α,t) − p(t))Eg (σ (t),t0 ) + q(t) +(k1 (α,t) − p(t))

Z σ (t)

q(s)E p (s, σ (t))∆α,t s

t0

= (k1 (α,t) − p(t))yσ (t) + q(t),

t ∈ Tκ .

Assume that the IVP (2.6), (2.2) has two solutions y1 and y2 . Let v = y1 − y2 . Then v solves the IVP Dα v = (−p(t) + k1 (α,t)) vσ ,

t ∈ Tκ ,

v(t0 ) = 0. Hence, and considering (2.7), we get v = 0 on

Tκ ,

i.e., y1 = y2

on

Tκ . 

This completes the proof. Remark 2.1.7 Consider the IVP Dα y

= p(t)yσ + q(t),

t ∈ Tκ , (2.8)

y(t0 ) = y0 . Suppose that t0 ∈ T, y0 ∈ R, p, q ∈ Crd (T) and k0 (α,t) − µ(t)p(t) 6= 0,

t ∈ T,

α ∈ (0, 1].

The IVP (2.8) can be rewritten in the form Dα y = − ((−p(t) + k1 (α,t)) − k1 (α,t)) yσ + q(t), y(t0 ) = y0 .

t ∈ Tκ ,

First-Order Linear Dynamic Equations  73

Note that g = −

p(k0 − µk1 ) (−p + k1 − k1 )(k0 − µk1 ) = . k0 + µ(−p + k1 − k1 ) k0 − µ p

Therefore the solution of the IVP (2.8) can be represented in the form Z t

y(t) = y0 Eg (t,t0 ) +  Example 2.1.8 Let T =

t0

t ∈ Tκ .

q(s)E−p+k1 (s,t)∆α,t s,

 [ 1 N {0} and 2

k1 (α,t) = (1 − α)t 4α ,

k0 (α,t) = αt 4(1−α) ,

t ∈ T,

α ∈ (0, 1].

Consider the IVP 1

D2 y =

1 2 σ t y + t 3, 8

t ∈ T,

y(0) = 1. Here 1 σ (t) = t + , 2 µ(t) =  k1  k0

1 ,t 2



1 ,t 2



1 , 2

=

1 2 t , 2

=

1 2 t , 2

p(t) =

1 2 t , 8

q(t) = t 3 ,

t ∈ T.

Then g(t) =

=

1 1 2 ,t
− µ(t)k1 2 ,t k0 12 ,t − µ(t)p(t)




p(t) k0

1 2 1 2 8t 2t − 1 2 1 2t − 2

1 2

1 2 2 t
1 2 8t









74  Conformable Dynamic Equations on Time Scales

g(t) − k1 k0 12 ,t 1 + µ(t)

g(t) − k1

1 ,t
2




1 ,t
2




=

1 2 1 2 1 2 8t 2t − 4t 1 2 1 2 2 t − 16 t

=

1 2 1 2 8t 4t 7 2 16 t

=

1 2 t , 14

=

1 2 1 2 14 t − 2 t 1 2 2t

= 1−

k0 12 ,t

−p(t) + k1 12 ,t − k1 12 ,t
k0 21 ,t

−p(t) + k1 21 ,t − k1 12 ,t
1 + µ(t) k0 21 ,t
k1 12 ,t
− k0 12 ,t
k1 12 ,t
1 − µ(t) k0 12 ,t

1 2







=

6 2 t − 14 1 2 2t

  6 3 4 = 1− = , 7 7 7

1 2 p(t) 1 8t
= − =− , 1 1 2 4 k0 2 ,t 2t   1 7 1 1 = 1− = , = 1− 2 4 8 8

= −

1 2 t

= − 21

2t

= 1−

2

= −1,

1 1 = , 2 2

t ∈ T.

Hence, Eg (t, 0) = e g−k1 (t, 0) k0

Rt

= e = e

1 0 µ(s)

log

Rt

4 0 2 log 7 ∆s

t− 12

= e∑s=0

log 47

4

= e2t log 7  2t 4 = , 7 E−p+k1 (s,t) = e− p (s,t) k0

6 =− , 7

g(s)−k1 12 ,s 1+µ(s) k0 12 ,s

( ) ( )

! ∆s

First-Order Linear Dynamic Equations  75 !

= e

Rs 1 t µ(y) log 1−µ(y)

Rs

= et

= e−

p(y) k0 12 ,y

∆y

( )

2 log(1− 81 )∆y

Rt s

2 log 78 ∆y

t− 1

= e

− ∑l=s2 (log 78 )

7

= e−2(t−s)(log 8 )  2(t−s) 8 , = 7 E0 (t, σ (s)) = e− k1 (t, σ (s)) k0

= e = e = e

k1 12 ,y 1−µ(y) k0 21 ,y

( ) ( )

Rt

log

Rt

(1− 12 )∆y

1 σ (s) µ(y)

! ∆y

σ (s) 2 log

Rt

1 σ (s) log 4 ∆y

= e−

R σ (s) s

R log 14 ∆y+ st log 41 ∆y

1

t− 12

= e− log 2 +∑l=s 1

log 12

1

= e− log 2 +2(t−s) log 2 t−s

= elog 2−log 4 2

= elog 4t−s =

2 22t−2s

= 21+2s−2t ,

s, .t ∈ T,

s ≤ t.

76  Conformable Dynamic Equations on Time Scales

Then the solution of the considered IVP Z t

y(t) = Eg (t, 0) +

q(s)E−p+k1 (s,t)∆α,t s

0

Z t

= Eg (t, 0) +

0

q(s)E−p+k1 (s,t)

E0 (t, σ (s))
∆s k0 12 , s

 2t Z t  2(t−s) 8 1 1 4 + s3 ∆s = 2t−2s−1 1 2 7 7 2 0 2s  2t Z t 6t−6s 1 2 4 = + 2 s 2(t−s) 2t−2s−1 ∆s 7 2 0 7  2t Z t 4t−4s+1 4 2 = + 2 s 2(t−s) ∆s 7 7 0  2t t− 12 4t−4s+ 1 4 2 = + ∑ s 2(t−s) 7 s=0 7   2t 1 4    if t =   2   7 =  2t t− 12 4t−4s+1   4 2    + ∑ s 2(t−s) ,   7 7 s= 1

t ≥ 1,

t ∈ T.

2

This ends the example.

2.2

CONFORMABLE BERNOULLI EQUATIONS

Consider the following IVP Dα y = −p(t)yσ + k1 (α,t)y + q(t)yyσ ,

t ∈ Tκ ,

y(t0 ) = y0 ,

(2.9) (2.10)

where p, q ∈ Crd (T), k0 (α,t) + µ(t)p(t) 6= 0,

α ∈ (0, 1],

t ∈ T,

t0 ∈ T, y0 ∈ R. Definition 2.2.1 The equation (2.9) will be called the conformable Bernoulli equation. Suppose that y0 6= 0 and y(t) 6= 0, yσ (t) 6= 0, for some t ∈ Tκ , and y is a solution of the problem (2.9), (2.10). Then Dα y 1 1 = −p(t) + k1 (α,t) σ + q(t), σ yy y y

t ∈ Tκ .

(2.11)

First-Order Linear Dynamic Equations  77

Let

1 v= . y

Then   1 D v = D y α

α

= and

k1 y − Dα y k1 + yyσ y

Dα y k1 k1 = σ + − Dα v yyσ y y

on

Tκ .

From the equation (2.11), we get k1 (α,t) k1 (α,t) 1 k1 (α,t) + − Dα v(t) = −p(t) + σ + q(t), σ y (t) y(t) y(t) y (t) or −Dα v(t) = − (p(t) + k1 (α,t))

1 + q(t), y(t)

t ∈ Tκ ,

t ∈ Tκ ,

or Dα v(t) = (p(t) + k1 (α,t)) v(t) − q(t),

t ∈ Tκ .

Also, v(t0 ) =

1 . y0

By Theorem 2.1.1, we get that 1 v(t) = E p+k1 (t,t0 ) − y0 where g(t) =

Z t t0

q(s)Eg (σ (s),t)∆α,t s,

p(t)(µ(t)k1 (α,t) − k0 (α,t)) , k0 (α,t) + µ(t)p(t)

t ∈ Tκ .

Consequently, y(t) =

y0 , E p+k1 (t,t0 ) − y0 t0 q(s)Eg (σ (s),t)∆α,t s Rt

t ∈ Tκ .

By the above computations, we conclude that when we search for a nontrivial solution of equation (2.9), we can reduce equation (2.9) to a first-order linear conformable dynamic equation. Example 2.2.2 Let T = 3Z, k1 (α,t) = (1 − α)(1 + t 2 )3α ,

k0 (α,t) = α(1 + t 2 )3(1−α) ,

α ∈ (0, 1],

t ∈ T.

78  Conformable Dynamic Equations on Time Scales

Consider the equation 1 1 2 D 3 y = − tyσ + (1 + t 2 )y + tyyσ , 3 3

t ∈ Tκ .

Here σ (t) = t + 3, µ(t) = 3, p(t) =

1 t, 3

q(t) = t,  k1  k0

1 ,t 3



1 ,t 3



=

2 (1 + t 2 ), 3

=

1 (1 + t 2 )2 , 3

t ∈ T.

Let y be a solution of the considered equation and y(t) 6= 0, yσ (t) 6= 0. Then, we get 1

D3 y 1 1 2 1 = − t + (1 + t 2 ) σ + t, yyσ 3 y 3 y We set

t ∈ Tκ .

1 v= . y

Hence,   1 D v(t) = D (t) y

1 k1 13 ,t y(t) − D 3 y(t) k1 13 ,t = + y(t)yσ (t) y(t) 1 3

1 3

=

1 2 2 3 3 (1 + t )y(t) − D y(t) σ y(t)y (t)

=

2 1 D 3 y(t) (1 + t 2 ) σ − 3 y (t) y(t)yσ (t)

+

2 2 3 (1 + t )

y(t)

1

2 1 + (1 + t 2 ) , 3 y(t) and

t ∈ Tκ ,

1

1 D 3 y(t) 2 1 2 1 = (1 + t 2 ) σ − D 3 v(t) + (1 + t 2 ) , y(t)yσ (t) 3 y (t) 3 y(t)

t ∈ Tκ .

First-Order Linear Dynamic Equations  79

Therefore 1 2 1 2 1 1 1 2 1 (1 + t 2 ) σ − D 3 v(t) + (1 + t 2 ) =− t + (1 + t 2 ) σ , 3 y (t) 3 y(t) 3 y(t) 3 y (t)

or 1

−D 3 v(t) = − or 1

D 3 v(t) =


1 2 2t + t + 2 v(t) + t, 3


1 2 2t + +2 v(t) − t, 3

t ∈ Tκ ,

t ∈ Tκ , t ∈ Tκ .

Exercise 2.2.3 Let T = 2N0 , k1 (α,t) = (1 − α)(1 + t 4 )4α ,

k0 (α,t) = α(1 + t 4 )4(1−α) ,

α ∈ (0, 1],

t ∈ T.

Reduce the following conformable Bernoulli equations to first-order linear conformable dynamic equations. 1.

2.

3.

1 3 D 4 y = 2yσ + (1 + t 4 )y + t 2 yyσ , 4

t ∈ Tκ ,

1 1 D 2 y = −3yσ + (1 + t 4 )2 y + (1 + t + t 2 )yyσ , 2

1 2 1 t2 + 6 D 6 y = (1 + t 2 )yσ + (1 + t 4 ) 3 y + 2 yyσ , 6 t +t +1

t ∈ Tκ ,

t ∈ Tκ .

This ends the exercise. Now we consider the equation Dα y = p(t)y + k1 (α,t)yσ + q(t)yyσ ,

t ∈ Tκ ,

(2.12)

where p and q satisfy the conditions as above, subject to the initial condition (2.10). Suppose that y0 6= 0, y(t) 6= 0, yσ (t) 6= 0 for some t ∈ Tκ , and y is a solution of the equation (2.12). Then Dα y(t) 1 1 = p(t) σ + k1 (α,t) + q(t), y(t)yσ (t) y (t) y(t) or Dα y(t) 1 1 − = −p(t) σ − k1 (α,t) − q(t), y(t)yσ (t) y (t) y(t) or Dα y(t) k1 (α,t) 1 k1 (α,t) − + = (−p(t) + k1 (α,t)) σ − q(t). σ σ y (t) y(t)y (t) y(t) y (t) Let v(t) =

1 . y(t)

80  Conformable Dynamic Equations on Time Scales

Thus we get Dα v(t) = (−p(t) + k1 (α,t)) vσ (t) − q(t). Hence, and considering Theorem 2.1.5, we obtain v(t) = where g(t) = −

1 Eg (t,t0 ) − y0

Z t t0

q(s)E p (s,t)∆α,t s,

(p(t) − k1 (α,t))(k0 (α,t) − µ(t)k1 (α,t)) . k0 (α,t) + µ(t)(p(t) − k1 (α,t))

Therefore y(t) =

y0 . Eg (t,t0 ) − y0 t0 q(s)E p (s,t)∆α,t s Rt

Example 2.2.4 Let T = 2N0 , k1 (α,t) = (1 − α)t 2α ,

k0 (α,t) = αt 2(1−α) ,

α ∈ (0, 1],

t ∈ T.

Consider the equation 1 3 1 D 4 y = (t 2 + t)y + t 2 yσ + (t 2 + 1)yyσ , 4

t ∈ Tκ .

Here σ (t) = 2t, µ(t) = t, p(t) = t 2 + t, q(t) = t 2 + 1, α =

t ∈ T,

1 . 4

Let y be a solution of the considered equation, y(t) 6= 0, yσ (t) 6= 0 for some t ∈ Tκ . Then 1


1 D 4 y(t) 3 1 1 = t2 + t σ + t 2 + t 2 + 1. σ y(t)y (t) y (t) 4 y(t) We set v(t) =

1 . y(t)

Then 1

1

D 4 v(t) = D 4



1 y(t)



First-Order Linear Dynamic Equations  81

=

k1

1

y(t) − D 4 y(t) k1 14 ,t + y(t)yσ (t) y(t)

1 4 ,t







=

1 3 12 4 4 t y(t) − D y(t) σ y(t)y (t)

=

3 1 1 D 4 y(t) 3 1 t2 σ − + t 2 v(t) σ 4 y (t) y(t)y (t) 4

=

3 1 σ 3 1 D 4 y(t) t 2 v (t) + t 2 v(t) − . 4 4 y(t)yσ (t)

+

3 21 4t

y(t)

1

1

Therefore 1

1 3 1 3 1 D 4 y(t) = t 2 vσ (t) + t 2 v(t) − D 4 v(t) σ y(t)y (t) 4 4

and or

or


1 3 1 σ 3 1 3 1 t 2 v (t) + t 2 v(t) − D 4 v(t) = t 2 + t vσ (t) + t 2 v(t) + t 2 + 1, 4 4 4   1 3 1 −D 4 v(t) = t 2 + t − t 2 vσ (t) + t 2 + 1, 4   3 1 σ 2 D v(t) = −t − t + t 2 v (t) − t 2 − 1. 4 1 4

Exercise 2.2.5 Let T = Z, k1 (α,t) = (1 − α)(1 + t 6 )α ,

k0 (α,t) = α(1 + t 6 )1−α ,

α ∈ (0, 1],

t ∈ T.

Reduce the following conformable Bernoulli’s equations to first-order linear conformable dynamic equations. 1.

2.

3.

1 1 1 D 2 y = ty + (1 + t 6 ) 2 yσ + yyσ , 2

t ∈ Tκ ,


1 1 2 D 3 y = 1 + t 2 + t 4 y + (1 + t 6 ) 3 yσ + (t + 1)yyσ , 3 1 1 3 D 4 y = (1 + t)y + (1 + t 6 ) 4 yσ + (t 2 + 1)yyσ , 4

t ∈ Tκ ,

t ∈ Tκ .

82  Conformable Dynamic Equations on Time Scales

2.3

CONFORMABLE RICCATI EQUATIONS

Consider the equation Dα y = −p(t)yσ + k1 (α,t)y + q(t)yyσ + f (t),

t ∈ Tκ ,

(2.13)

where p, q ∈ Crd (T) and k0 (α,t) + µ(t)p(t) 6= 0,

α ∈ (0, 1],

t ∈ T,

f ∈ Crd (T). Definition 2.3.1 The equation (2.13) is called the conformable Riccati equation. Suppose that y p is a particular solution of the considered equation. Set y = z + yp. We get Dα z + Dα y p = −p(t)zσ − p(t)yσp +k1 (α,t)z + k1 (α,t)y p +q(t)(z + y p )(zσ + yσp ) + f (t) = −p(t)zσ − p(t)yσp +k1 (α,t)z + k1 (α,t)y p + q(t)zzσ + q(t)y p zσ +q(t)zyσp + q(t)y p yσp + f (t) = − (p(t) − q(t)y p ) zσ + (k1 (α,t) + q(t)yσp )z + q(t)zzσ −p(t)yσp + k1 (α,t)y p + q(t)y p yσp + f (t),

t ∈ Tκ .

Hence, Dα z = −(p(t) − q(t)y p )zσ + (k1 (α,t) + q(t)yσp )z + q(t)zzσ , i.e., if we know a particular solution of equation (2.13), then we can reduce it to a Bernoulli equation. Example 2.3.2 Let T = 2N0 , k1 (α,t) = (1 − α)t 2α ,

k0 (α,t) = αt 2(1−α) ,

α ∈ (0, 1],

t ∈ T.

First-Order Linear Dynamic Equations  83

Consider the equation 1 1 3 D 2 y = −(t 2 + t)yσ + ty + tyyσ − 4t 5 + 4t 4 + 4t 3 + t 2 , 2 2

Here σ (t) = 2t, µ(t) = t,  k1  k0

1 ,t 2



1 ,t 2



=

1 t, 2

=

1 t, 2

p(t) = t 2 + t, q(t) = t, 3 f (t) = −4t 5 + 4t 4 + 4t 3 + t 2 , 2

t ∈ T.

We will prove that y p (t) = t 2 ,

t ∈ Tκ ,

is a particular solution of the considered equation. We have y∆p (t) = σ (t) + t = 2t + t = 3t, 

1 2

D y p (t) = k1

t ∈ Tκ ,    1 1 ,t y p (t) + k0 ,t y∆p (t) 2 2

=

1 2 1 t(t ) + t(3t) 2 2

=

1 3 3 2 t + t , 2 2

yσp (t) = (σ (t))2

t ∈ Tκ ,

t ∈ Tκ .

84  Conformable Dynamic Equations on Time Scales

= (2t)2 = 4t 2 ,

t ∈ Tκ ,

and 3 1 −(t 2 + t)yσp (t) + ty p (t) + ty p (t)yσp (t) − 4t 5 + 4t 4 + 4t 3 + t 2 2 2 3 1 = −(t 2 + t)(4t 2 ) + t(t 2 ) + t(t 2 )(4t 2 ) − 4t 5 + 4t 4 + 4t 3 + t 2 2 2 1 3 = −4t 4 − 4t 3 + t 3 + 4t 5 − 4t 5 + 4t 4 + 4t 3 + t 2 2 2 =

1 3 3 2 t + t , 2 2

t ∈ Tκ .

Therefore y p is a particular solution of the considered equation. Note that p(t) − q(t)y p (t) = t 2 + t − t(t 2 ) = t 2 + t − t 3,  k1

 1 ,t + q(t)yσp (t) = 2 =

t ∈ Tκ ,

1 t + t(4t 2 ) 2 1 t + 4t 3 , 2

t ∈ Tκ .

Hence, the considered equation can be reduced to the following Bernoulli equation   1 1 3 2 3 σ 2 t + 4t z + tzzσ , t ∈ Tκ . D z = −(t + t − t )z + 2 Exercise 2.3.3 Let T = 2N0 , k1 (α,t) = (1 − α)(1 + t 2 )2α ,

k0 (α,t) = α(1 + t 2 )2(1−α) ,

α ∈ (0, 1],

Prove that y p (t) = 1,

t ∈ T,

is a particular solution to the following Riccati equation 1 1 D 2 y = −yσ + (1 + t 2 )y + t 2 yyσ + 1 − t 2 , 2

t ∈ Tκ .

Reduce this equation to Bernoulli’s equation. This ends the exercise.

t ∈ T.

First-Order Linear Dynamic Equations  85

Now we consider the equation Dα y = p(t)y + k1 (α,t)yσ + q(t)yyσ + f (t),

t ∈ Tκ ,

(2.14)

where p, q, and f satisfy the conditions as above. Definition 2.3.4 The equation (2.14) is called the alternative conformable Riccati equation. Let y p be a particular solution of the equation (2.14). Set y(t) = z(t) + y p (t),

t ∈ T.

Then we get Dα z(t) + Dα y p (t) = p(t)z(t) + p(t)y p (t) +k1 (α,t)zσ (t) + k1 (α,t)yσp (t)
+q(t) (z(t) + y p (t)) zσ (t) + yσp (t) + f (t) = p(t)z(t) + p(t)y p (t) +k1 (α,t)zσ (t) + k1 (α,t)yσp (t) +q(t)z(t)zσ (t) + q(t)z(t)yσp (t) +q(t)y p (t)zσ (t) + q(t)y p (t)yσp (t) + f (t) =


p(t) + q(t)yσp (t) z(t) + (q(t)y p (t) + k1 (α,t)) zσ (t) +q(t)z(t)zσ (t) +p(t)y p (t) + k1 (α,t)yσp (t) +q(t)y p (t)yσp (t) + f (t),

t ∈ Tκ .

86  Conformable Dynamic Equations on Time Scales

Hence, Dα z(t) =


p(t) + q(t)yσp (t) z(t) + (q(t)y p (t) + k1 (α,t)) zσ (t) +q(t)z(t)zσ (t),

t ∈ Tκ ,

i.e., if we know a particular solution of equation (2.14), then we can reduce it to a conformable Bernoulli equation. Example 2.3.5 Let T = 2Z, k1 (α,t) = (1 − α)t 4α ,

k0 (α,t) = αt 4(1−α) ,

α ∈ (0, 1],

Consider the equation 1 1 3 7 D 4 y = t 2 y + tyσ + yyσ − t 2 − t, 4 4 2

t ∈ Tκ .

We will prove that y p (t) = t,

t ∈ T,

is its particular solution. We have σ (t) = t + 2, yσp (t) = σ (t) = t + 2,  k1  k0

1 ,t 4



1 ,t 4



=

3 t, 4

=

1 3 t , 4

y∆p (t) = 1, p(t) =

1 2 t , 4

q(t) = 1, 7 f (t) = −t 2 − t, t ∈ T, 2     1 1 1 D 4 y p (t) = k1 ,t y p (t) + k0 ,t y∆p (t) 4 4 =

3 2 1 3 t + t , 4 4

t ∈ Tκ .

t ∈ T.

First-Order Linear Dynamic Equations  87

Then 1 2 3 7 t y p (t) + ty p (t) + y p (t)yσp (t) − t 2 − t 4 4 2 =

1 3 3 7 t + t(t + 2) + t(t + 2) − t 2 − t 4 4 2

=

7 1 3 3 2 3 t + t + t + t 2 + 2t − t 2 − t 4 4 2 2

=

1 3 3 2 t + t , 4 4

t ∈ T.

Therefore y p (t) = t, t ∈ T, is a particular solution of the considered equation. Note that p(t) + q(t)yσp (t) =  q(t)y p (t) + k1

1 ,t 4



1 2 t + t + 2, 4

3 = t+ t 4 =

7 t, 4

t ∈ T.

The considered equation is reduced to the following Bernoulli equation   1 1 2 7 4 t + t + 2 z + tzσ + zzσ , t ∈ Tκ . D z= 4 4 This ends the example. Exercise 2.3.6 Let T = 2N0 , k1 (α,t) = (1 − α)t 6α ,

k0 (α,t) = αt 6(1−α) ,

α ∈ (0, 1],

t ∈ T.

Prove that y p (t) = t 2 ,

t ∈ T,

is a particular solution of the equation 1 5 1 5 D 6 y = y + tyσ + yyσ + t 6 − 4t 4 − t 3 − t 2 , 6 2 2

Reduce this equation to Bernoulli’s equation.

2.4

CONFORMABLE LOGISTIC EQUATIONS

Let u solve the linear equation Dα u = pu + q,

t ∈ Tκ .

88  Conformable Dynamic Equations on Time Scales

and let y = 1/u. Then α

D y = = = = =

  1 D u   1 uk1 − Dα u + k1 σ uu u   p σ σ k1 (y + y) − yy +q y k1 y − yσ (p + qy − k1 )   (k0 − µk1 )y + µDα y (p + qy − k1 ) . k1 y − k0 α

Collecting the Dα y terms on the left, we get (k0 + µ(p + qy − k1 )) Dα y = k0 k1 y − (k0 − µk1 )(p + qy − k1 )y. Now, assuming that p + qy ∈ Rc , we see that   k0 k1 − (k0 − µk1 )(p + qy − k1 ) α D y= y, k0 + µ(p + qy − k1 ) which can be written using the circle minus notation as the conformable logistic equation on time scales Dα y = [ c (p + qy)] y, p + qy ∈ Rc . (2.15) In like manner, let v solve the linear equation Dα v = (k1 − p)vσ + q, and let x = 1/v. Then   1 D x = D v α

α

  1 = k1 (xσ + x) − xxσ −(p − k1 ) σ + q x σ = px − x (qx − k1 )   (k0 − µk1 )x + µDα x = px − (qx − k1 ) . k0 Rearranging terms, we arrive at (k0 + µ(qx − k1 )) Dα x = k0 px − (k0 − µk1 )(qx − k1 )x. Assuming that qx ∈ Rc , we have   k0 p − (k0 − µk1 )(qx − k1 ) α D x= x, k0 + µ(qx − k1 )

First-Order Linear Dynamic Equations  89

which can also be rewritten as Dα x = [p c (qx)] x,

qx ∈ Rc .

(2.16)

This equation will also be called a conformable logistic equation on time scales. Throughout this section we will assume p ∈ Rc and q is right-dense continuous. If u 6= 0 and y = 1/u, then y is a solution to (2.15). To check the regressivity condition for (p + qy), note that k0 + µ(p + qy − k1 ) =

k0 u + µ(pu + q) − k1 µy . u

Recalling that Dα u = pu + q, we have k0 + µ(p + qy − k1 ) =

k0 u + µDα u − k1 µy k0 u + k0 uσ − k0 u k0 uσ = = 6= 0, u u u

putting (p + qy) ∈ Rc . Likewise, if v 6= 0 and x = 1/v, then x is a solution to (2.16), and k0 + µ(qx − k1 ) =

(k0 − µk1 )v + µq . v

Recalling that Dα v = (k1 − p)vσ + q, we have k0 + µ(qx − k1 ) =

k0 vσ − µDα v + µq vσ = [k0 + µ(p − k1 )] 6= 0 v v

by the regressivity of p, so that (qx) ∈ Rc . This leads to the following result. Theorem 2.4.1 Suppose p ∈ Rc and q is right-dense continuous. 1. Let y0 6= 0. If Z t

u(t) = u0 E p (t,t0 ) +

t0

q(s)E f (σ (s),t)∆α,t s 6= 0

for all t ∈ T, then y(t) = 1/u(t) solves (2.15) with y(t0 ) = y0 , where f=

(p − k1 )(µk1 − k0 ) . k0 + µ(p − k1 )

2. Let x0 6= 0. If

Z t

v(t) = v0 E f (t,t0 ) +

t0

q(s)E p (s,t)∆α,t s 6= 0

for all t ∈ T, then x(t) = 1/v(t) solves (2.16) with x(t0 ) = x0 . Proof 2.4.2 Using Remark 2.1 to solve Dα u = pu + q for u, and using Theorem 2.2 to solve Dα v = (k1 − p)vσ + q for v, we can obtain the above solutions of the conformable logistic equations (2.15) and (2.16), as stated in the theorem. This completes the proof. 

90  Conformable Dynamic Equations on Time Scales

2.5

ADVANCED PRACTICAL PROBLEMS

Problem 2.5.1 Let T = 3N0 , k1 (α,t) = (1 − α)t 2α ,

k0 (α,t) = αt 2(1−α) ,

t ∈ T,

α ∈ (0, 1].

Find the solution of the IVP 1

D 4 y(t) =

  3 1 t 3 + t 2 y(t) + t 2 + t, 4

t ∈ T,

y(1) = 3. Problem 2.5.2 Let T = 2N0 , k1 (α,t) = (1 − α)(1 + t 2 )α ,

k0 (α,t) = α(1 + t 2 )1−α ,

α ∈ (0, 1],

t ∈ T.

Find the solution of the IVP 1

D 4 y = (t + 2)yσ + t 2 ,

t ∈ T,

t > 128,

y(128) = 1. Problem 2.5.3 Let T = 3N0 , k1 (α,t) = (1 − α)(1 + t)α ,

k0 (α,t) = α(1 + t)1−α ,

α ∈ (0, 1],

t ∈ T.

Reduce the following conformable Bernoulli equations to first-order linear conformable dynamic equations. 1.

2.

3.

1 1 1 D 2 y = (2 + t)yσ + (1 + t) 2 y + t 2 yyσ , 2

t ∈ T,

2 2 1 D 3 y = (3 − t 2 )yσ + (1 + t) 3 y + tyyσ , 3

t ∈ T,

3 3 1 D 4 y = (t 2 + t + 1)yσ + (1 + t) 4 y + (1 + t)yyσ , 4

t ∈ T.

Problem 2.5.4 Let T = 3N0 , k1 (α,t) = (1 − α)t 3α ,

k0 (α,t) = αt 3(1−α) ,

α ∈ (0, 1],

t ∈ T.

Reduce the following conformable Bernoulli equations to first-order linear conformable dynamic equations. 1.

1 1 3 D 2 y = (t 2 + 2t)y + t 2 yσ + tyyσ 2

t ∈ T,

First-Order Linear Dynamic Equations  91

2.

3.

1 3 3 D 4 y = (1 + t)y + t 4 yσ + t 2 yyσ , 4

t ∈ T,

1 7 3 D 8 y = (1 + t 2 )y + t 8 yσ + (t + 1)yyσ , 8

t ∈ T.

Problem 2.5.5 Let T = 3N0 , k1 (α,t) = (1 − α)t 4α ,

k0 (α,t) = αt 4(1−α) ,

α ∈ (0, 1],

t ∈ T.

Prove that y p (t) = t,

t ∈ T,

is a particular solution of the equation 1 3 11 D 4 y = −2tyσ + ty + tyyσ − t 3 + 6t 2 , 4 4

t ∈ T.

Reduce this equation to Bernoulli’s equation. Problem 2.5.6 Let T = 3N0 , k1 (α,t) = (1 − α)t 2α ,

k0 (α,t) = αt 2(1−α) ,

α ∈ (0, 1],

Prove that y p (t) = t is a particular solution to the equation 1 3 3 13 D 4 y = ty + tyσ + t 2 yyσ − 3t 4 − t 2 + t 2 , 4 4

Reduce this equation to Bernoulli’s equation.

t ∈ T.

t ∈ T.

CHAPTER

3

Conformable Dynamic Systems on Time Scales

Suppose that T is a time scale with forward jump operator and delta differentiation operator σ and ∆, respectively. Let α ∈ (0, 1], k0 and k1 satisfy (A1), and assume (1.6) holds.

3.1

STRUCTURE OF CONFORMABLE DYNAMIC SYSTEMS ON TIME SCALES

Suppose that A is an m × n matrix, A = (ai j )1≤i≤m,1≤ j≤n , shortly A + (ai j ), ai j : T → R, 1 ≤ i ≤ m, 1 ≤ j ≤ n. Definition 3.1.1 We say that A is conformable ∆-differentiable at t ∈ Tκ , if each entry is conformable ∆-differentiable at t and Dα A = (Dα ai j ) .

Example 3.1.2 Let T = 2N0 , k1 (α,t) = (1 − α)t 4α ,

k0 (α,t) = αt 4(1−α) ,

and

t2 + t t t − 1 t3

 A(t) =

α ∈ (0, 1],

t ∈ T,

 ,

t ∈ T.

We will find 1

D 4 A(t),

t ∈ T.

Here σ (t) = 2t,  k1

1 ,t 4

 =

3 t, 4 93

94  Conformable Dynamic Equations on Time Scales

 k0

1 ,t 4

 =

1 3 t , 4

a11 (t) = t 2 + t, a12 (t) = t, a21 (t) = t − 1, a22 (t) = t 3 ,

t ∈ T.

Then a∆11 (t) = σ (t) + t + 1 = 2t + t + 1 = 3t + 1, a∆12 (t) = 1, a∆21 (t) = 1, a∆22 (t) = (σ (t))2 + tσ (t) + t 2 = (2t)2 + 2t 2 + t 2 = 4t 2 + 3t 2 = 7t 2 ,

t ∈ T.

Hence, 

1 4

D a11 (t) = k1

   P1 1 ,t a11 (t) + k0 ,t a∆11 (t) 4 4

=

1 3 2 t(t + t) + t 3 (3t + 1) 4 4

=

3 3 3 2 3 4 1 3 t + t + t + t 4 4 4 4

Conformable Dynamic Systems on Time Scales  95

3 4 3 3 2 t +t + t , 4 4     1 1 1 F 4 a12 (t) = k1 ,t a12 (t) + k0 ,t a∆12 (t) 4 4 =

3 2 1 3 t + t , 4 4     1 1 1 ,t a21 (t) + k0 ,t a∆21 (t) D 4 a21 (t) = k1 4 4 =

=

1 3 t(t − 1) + t 3 4 4

3 2 3 1 t − t + t 3, 4 4 4     1 1 1 4 ,t a22 (t) + k0 ,t a∆22 (t) D a22 (t) = k1 4 4 =

=

3 4 1 3 2 t + t (7t ) 4 4

=

7 5 3 4 t + t , 4 4

t ∈ T.

Therefore 

3 4 3 3 2 t +t + t 1  4 D 4 A(t) =  14 3 3 t3 + t2 − t 4 4 4

 1 3 3 2 t + t  4 4 7 5 3 4 , t + t 4 4

t ∈ T.

Exercise 3.1.3 Let T = Z, k1 (α,t) = (1 − α)t 2α , and

 A(t) =

Find

k0 (α,t) = αt 2(1−α) , t 2 + 2t + 4 t 3 − t t 2 − 3t t2 1

D 3 A(t),

α ∈ (0, 1],

t ∈ T,

 ,

t ∈ T.

t ∈ T.

Definition 3.1.4 If A is conformable ∆-differentiable at t ∈ Tκ , then we define Aσ (t) = (aσi j (t)). Theorem 3.1.5 If A is conformable ∆-differentiable at t ∈ Tκ , then   µ(t) µ(t) α Aσ (t) = 1 − k1 (α,t) A(t) + D A(t), α ∈ (0, 1]. k0 (α,t) k0 (α,t)

96  Conformable Dynamic Equations on Time Scales

Proof 3.1.6 We have Aσ (t) = (aσi j (t))    µ(t) µ(t) α = 1− k1 (α,t) ai j (t) + D ai j (t) k0 (α,t) ai j (t)   µ(t) µ(t) α = 1− k1 (α,t) (ai j (t)) + (D ai j (t)) k0 (α,t) ai j (t)   µ(t) α µ(t) k1 (α,t) A(t) + D A(t). = 1− k0 (α,t) ai j (t) 

This completes the proof. Below we suppose that B = (bi j )1≤i≤m,1≤ j≤n , where bi j : T → R, 1 ≤ i ≤ m, 1 ≤ j ≤ n. Theorem 3.1.7 Let A and B be conformable ∆-differentiable at t ∈ Tκ . Then Dα (A + B)(t) = Dα A(t) + Dα B(t),

α ∈ (0, 1].

Proof 3.1.8 We have Dα (A + B)(t) = (Dα (ai j + bi j ) (t)) = (Dα ai j (t) + Dα bi j (t)) = (Dα ai j (t)) + (Dα bi j (t)) = Dα A(t) + Dα B(t). 

This completes the proof. Theorem 3.1.9 Let a ∈ R and A be conformable ∆-differentiable at t ∈ Tκ . Then Dα (aA)(t) = aDα A(t). Proof 3.1.10 We have Dα (aA)(t) = Dα (aai j )(t) = (Dα (aai j )) (t) = (aDα ai j ) (t)

Conformable Dynamic Systems on Time Scales  97

= a (Dα ai j ) (t) = aDα A(t). 

This completes the proof. Theorem 3.1.11 Let m = n and A, B be conformable ∆-differentiable at t ∈ Tκ . Then Dα (AB)(t) = Dα A(t)B(t) + Aσ (t)Dα B(t) − k1 (α,t)Aσ (t)B(t) = Dα A(t)Bσ (t) + A(t)Dα B(t) − k1 (α,t)A(t)Bσ (t). Proof 3.1.12 We have !

n

Dα (AB)(t) = Dα

(t)

∑ ail bl j l=1

!!

n

=

Dα

(t)

∑ ail bl j l=1

!

n

=

∑D

α

(ail bl j ) (t)

l=1 n

=

n

∑ (D

α

ail ) bl j + ∑

l=1

l=1

=

∑ (Dα ail ) bl j

∑

k1 (α, ·)aσil bl j

(t)

l=1

!

n

!

n

aσil Dα bl j −

!

n

(t) +

l=1

∑ aσil Dα bl j

(t)

l=1

!

n

−k1 (α,t)

∑ aσil bl j

(t)

l=1

= Dα A(t)B(t) + Aσ (t)B(t) − k1 (α,t)Aσ (t)B(t) n

=

n

∑ (D

α

ail ) bσl j +

l=1

=

∑ ail D

bl j − ∑

l=1 α

∑ (D

ai j ) bσl j

(t)

!

n

(t) +

l=1

α

∑ ail D

bl j (t)

l=1 n

−k1 (α,t)

k1 (α, ·)ail bσl j

l=1

!

n

!

n α

∑ ail bσl j

! (t)

l=1

= Dα A(t)Bσ (t) + A(t)Dα B(t) − k1 (α,t)A(t)Bσ (t). This completes the proof.



98  Conformable Dynamic Equations on Time Scales

Example 3.1.13 Let T = 2N0 , k1 (α,t) = (1 − α)t 4α , and

 A(t) =

k0 (α,t) = αt 4(1−α) ,

t 2t t +1 1



 ,

B(t) =

α ∈ (0, 1],

t 1 2 3

t ∈ T,

 ,

t ∈ T.

We will find 1

t ∈ T.

D 2 (AB)(t), We have

σ (t) = 2t,  k1  k0

1 ,t 2



1 ,t 2



=

1 2 t , 2

=

1 2 t , 2

t ∈ T.

Let t ∈ T.

C(t) = A(t)B(t), Then  C(t) =

t 2t t +1 1



t 1 2 3



 =

We have c11 (t) = t 2 + 4t, c12 (t) = 7t, c21 (t) = t 2 + t + 2, c22 (t) = t + 4, c∆11 (t) = σ (t) + t + 4 = 2t + t + 4 = 3t + 4, c∆12 (t) = 7,

t 2 + 4t 7t t2 + t + 2 t + 4

 ,

t ∈ T.

Conformable Dynamic Systems on Time Scales  99

c∆21 (t) = σ (t) + t + 1 = 2t + t + 1 = 3t + 1, c∆22 (t) = 1, 

1 2

D c11 (t) = k1

   1 1 ,t c11 (t) + k0 ,t c∆11 (t) 2 2

=


1 1 2 2 t t + 4t + t 2 (3t + 4) 2 2

=

3 1 4 t + 2t 3 + t 3 + 2t 2 2 2

1 4 7 3 t + t + 2t 2 , 2 2     1 1 1 2 D c12 (t) = k1 ,t c12 (t) + k0 ,t c∆12 (t) 2 2 =

7 3 7 2 t + t , 2 2     1 1 1 2 D c21 (t) = k1 ,t c21 (t) + k0 ,t c∆21 (t) 2 2 =

=


1 1 2 2 t t + t + 2 + t 2 (3t + 1) 2 2

=

1 4 1 3 2 3 3 1 2 t + t +t + t + t 2 2 2 2

1 4 3 t + 2t 3 + t 2 , 2 2     1 1 1 D 2 cx22 (t) = k1 ,t c22 (t) + k0 ,t c∆22 (t) 2 2 =

=

1 2 1 t (t + 4) + t 3 2 2

=

1 3 1 t + 2t 2 + t 2 2 2

100  Conformable Dynamic Equations on Time Scales

=

1 3 5 2 t + t , 2 2

t ∈ T.

Therefore 1

1

D 2 (AB)(t) = D 2 C(t) 

1 4 7 3 2  2 t + 2 t + 2t =  1 3 t 4 + 2t 3 + t 2 2 2 Next, a11 (t) = t, a12 (t) = 2t, a21 (t) = t + 1, a22 (t) = 1, b11 (t) = t, b12 (t) = 1, b21 (t) = 2, b22 (t) = 3, a∆11 (t) = 1, a∆12 (t) = 2, a∆21 (t) = 1, a∆22 (t) = 0, b∆11 (t) = 1, b∆12 (t) = 0,

 7 3 7 2 t + t  2 2 1 3 5 2 , t + t 2 2

t ∈ T.

Conformable Dynamic Systems on Time Scales  101

b∆21 (t) = 0, b∆22 (t) = 0, 

1 2

D a11 (t) = k1

   1 1 ,t a11 (t) + k0 ,t a∆11 (t) 2 2

1 3 1 2 t + t , 2 2     1 1 1 D 2 a12 (t) = k1 ,t a12 (t) + k0 ,t a∆12 (t) 2 2 =

= t 3 + t 2, 

1 2

D a21 (t) = k1 =

   1 1 ,t a21 (t) + k0 ,t a∆21 (t) 2 2

1 1 2 t (t + 1) + t 2 2 2

1 3 2 t +t , 2     1 1 1 2 D a22 (t) = k1 ,t a22 )(t) + k0 ,t a∆22 (t) 2 2 =

1 2 t , 2     1 1 1 D 2 b11 (t) = k1 ,t b11 (t) + k0 ,t b∆11 (t) 2 2 =

1 3 1 2 t + t , 2 2     1 1 1 D 2 b12 (t) = k1 ,t b12 (t) + k0 ,t b∆12 (t) 2 2 =

1 2 t , 2     1 1 1 D 2 b21 (t) = k1 ,t b21 (t) + k0 ,t b∆21 (t) 2 2 =

= t 2,

102  Conformable Dynamic Equations on Time Scales



1 2

D b22 (t) = k1 =

   1 1 ,t b22 (t) + k0 ,t b∆22 (t) 2 2

3 2 t , 2

bσ11 (t) = σ (t) = 2t, bσ12 (t) = 1, bσ21 (t) = 2, bσ22 (t) = 3. Hence, 

1

D 2 A(t)

1

D 2 B(t)

Bσ (t)

1 2

σ

D A(t)B (t)

1

A(t)D 2 B(t)

 1 3 1 2 3 2 t + t t + t   2 =  21 , 1 t3 + t2 t2 2 2   1 3 1 2 1 2 t  t + 2t 2  =  2 3 2 , t2 t 2   2t 1 = , 2 3   1 3 1 2 3 2   t + 2 t t + t  2t 1 =  21 1 2  2 3 t3 + t2 t 2 2   7 2 4 3 2 7 3 t + t   t + 3t + 2t 2 2 =  1 3 5 2 , 4 3 2 t + t t + 2t + t 2 2  1 3 1 2 1 2   t t 2t  2 t + 2 t 2 =  3 t +1 1 t2 t2 2   1 4 5 3 7 3 t + t t   2 2 =  12 , 3 1 t4 + t3 + t2 t 3 + 2t 2 2 2 2



  

Conformable Dynamic Systems on Time Scales  103



4

3

2

1 1  t + 3t + 2t D 2 A(t)Bσ (t) + A(t)D 2 B(t) =  t 4 + 2t 3 + t 2

 7 3 7 2 t + t  2 2 1 3 5 2  t + t 2 2



1 4 5 3 7 3 t  2t + 2t 2 + 1 3 1 t4 + t3 + t2 t 3 + 2t 2 2 2 2  3 4 11 3 7 2 2 3  2 t + 2 t + 2t 7t + 2 t  3 5 9 t 4 + 3t 3 + t 2 t 3 + t 2 2 2 2    t 2t 2t 1 t +1 1 2 3   2t 2 + 4t 7t , 2t 2 + 2t + 2 t + 4   1 2 2t 2 + 4t 7t t 2t 2 + 2t + 2 t + 4 2   7 3 4 3 t t + 2t   2 ,  4 3 2 1 3 2 t +t +t t + 2t 2

=

A(t)Bσ (t) =

=  k1

 1 ,t A(t)Bσ (t) = 2

=

     ,

and 1 2

D A(t)B   =    = 



 1 (t) + A(t)D B(t) − k1 ,t A(t)Bσ (t) 2   3 4 11 3 7 t + t + 2t 2 7t 3 + t 2   t 4 + 2t 3 2 2 2 3 4 5 2 9 2 − 4 3 2 3 3 t +t +t t + 3t + t t + t 2 2 2  1 4 7 3 7 2 2 7 3 t + t + 2t t + t  2 2 2 2 1 4 3 1 3 5 2  t + 2t 3 + t 2 t + t 2 2 2 2 1 2

σ

1

= D 2 (AB)(t),

 7 3 t  2  1 3 t + 2t 2 2

t ∈ T.

This ends the example. Exercise 3.1.14 Let T = 3N0 , k1 (α,t) = (1 − α)(1 + t)3α , and

 A(t) =

k0 (α,t) = α(1 + t)3(1−α) ,

t 2t + 3 t +4 t



 ,

B(t) =

t2 1 −1 t

α ∈ (0, 1],  ,

t ∈ T.

t ∈ T,

104  Conformable Dynamic Equations on Time Scales

Find

1

t ∈ T.

D 3 (AB)(t),

Theorem 3.1.15 Let m = n, and assume A−1 exists on T. Then
σ (Aσ )−1 = A−1 on T. Proof 3.1.16 For any t ∈ T we have A(t)A−1 (t) = I,

t ∈ T.

Then Aσ (t) A−1 whereupon


σ

(t) = I,

(Aσ )−1 (t) = A−1


σ

(t),

t ∈ T. 

This completes the proof. Example 3.1.17 Let T = lN0 , l > 0,   t +1 t +2 A(t) = , 1 t +3

t ∈ T.

Then  σ

A (t) =

σ (t) + 1 σ (t) + 2 1 σ (t) + 3



 = (Aσ )−1 (t) =

 t +l +1 t +l +2 , 1 t +l +3   1 t + l + 3 −t − l − 2 , −1 t +l +1 (t + l)(t + l + 3) + 1

t ∈ T.

Next, 1 A (t) = t(t + 3) + 1 −1



t + 3 −t − 2 −1 t +1

 ,

t ∈ T,

whereupon A


−1 σ

(t) = =

Consequently, This ends the example.

 σ (t) + 3 −σ (t) − 2 −1 σ (t) + 1   1 t + l + 3 −t − l − 2 , −1 t +l +1 (t + l)(t + l + 3) + 1 1 σ (t)(σ (t) + 3) + 1



(Aσ )−1 (t) = (A−1 )σ (t),

t ∈ T.

t ∈ T.

Conformable Dynamic Systems on Time Scales  105

Exercise 3.1.18 Let T = 2N0 and  1  t +2 t +1  A(t) =  1 , t2 + 1 t +2 

t ∈ T.

Prove that (Aσ )−1 (t) = (A−1 )σ (t),

t ∈ T.

Theorem 3.1.19 Let m = n, A be conformable ∆-differentiable at t ∈ Tκ and assume A−1 , (Aσ )−1 exist at t. Then Dα A−1 (t) = −A−1 (t) (Dα A(t)) (Aσ )−1 (t) + k1 (α,t) (Aσ )−1 (t), Dα A−1 (t) = (Aσ )−1 (t) (Dα A(t)) A−1 (t) + k1 (α,t)A−1 (t). Proof 3.1.20 We have
I = AA−1 (t). Then
O = Dα AA−1 (t) = (Dα A(t)) A−1 (t) + Aσ (t)Dα A−1 (t) − k1 (α,t)Aσ (t)A−1 (t) = (Dα A(t)) (Aσ )−1 (t) + A(t)Dα A1 (t) − k1 (α,t)A(t) (Aσ )−1 (t). Hence, A(t)Dα A−1 (t) = − (Dα A(t)) (Aσ )−1 (t) + k1 (α,t)A(t) (Aσ )−1 (t), Aσ (t)Dα A−1 (t) = − (Dα A(t)) A−1 (t) + k1 (α,t)Aσ (t)A−1 (t), whereupon we get the desired result. Example 3.1.21 Let m = n, A and B be conformable ∆ - differentiable at t ∈ Tκ . Then
Dα AB−1 (t) = (Dα A(t)) B−1 (t) + Aσ (t)Dα B−1 (t) −k1 (α,t)Aσ (t)B−1 (t) = (Dα A(t)) B−1 (t)   +Aσ (t) − (Bσ )−1 (t)Dα B(t)B−1 (t) + k1 (α,t)B−1 (t)



106  Conformable Dynamic Equations on Time Scales

−k1 (α,t)Aσ (t)B−1 (t) =

  Dα A(t) − Aσ (t) (Bσ )−1 (t)Dα B(t) B−1 (t).

This completes the example. Definition 3.1.22 We say that the matrix A is rd-continuous on T if each entry of A is rdcontinuous. The class of such rd-continuous m × n matrix-valued functions on T is denoted by Crd = Crd (T) = Crd (T, R m×n ). Below we suppose that A and B are n × n matrix-valued functions. Definition 3.1.23 We say that the matrix A is conformable regressive with respect to T provided k0 I + µ(A − k1 I) is invertible for all t ∈ Tκ , and k0 − µk1 6= 0 on T. The class of such regressive and rdcontinuous functions is denoted, similar to the scalar case, by Rc = Rc (T) = Rc (T, Rn×n ).

Theorem 3.1.24 The matrix-valued function A is conformable regressive if the eigenvalues λ j of A are conformable regressive for all 1 ≤ i ≤ n and (1.6) holds. Proof 3.1.25 Let j ∈ {1, . . . , n} be arbitrarily chosen, and let λ j (t) be an eigenvalue corresponding to the eigenfunction y j (t). Then (k0 + µ(λ j − k1 )) y j = k0 Iy j + µ(λ j y j − k1 Iy j ) = k0 Iy j + µ(Ay j − k1 Iy j ) = (k0 I + µ(A j − k1 I)) y j , 

whereupon it follows the desired result. This completes the proof.

Theorem 3.1.26 Let A be a 2×2 matrix-valued function. Then A is conformable regressive on T if and only if k02 + µ 2 k1 + µ(k0 − µk1 )trA + µ 2 det A 6= 0 on and (1.6) holds.

T

Conformable Dynamic Systems on Time Scales  107

Proof 3.1.27 Let

 A=



a11 a12 a21 a22

.

Then  k0 I + µ(A − k1 I) =  =  =

k0 0 0 k0



k0 0 0 k0



 +µ  +µ

a11 a12 a21 a22



 −

a11 − k1 a12 a21 a22 − k1

k0 + µ(a11 − k1 ) µa12 µa21 k0 + µ(a22 − k1 )

k1 0 0 k1 



 .

Hence,  det (k0 I + µ(A − k1 I)) = det

k0 + µ(a11 − k1 ) µa12 µa21 k0 + µ(a22 − k1 )



= (k0 + µ(a11 − k1 )) (k0 + µ(a22 − k1 )) − µ 2 a21 a12 = k02 + k0 µ(a22 − k1 ) + k0 µ(a11 − k1 ) +µ 2 (a11 − k1 )(a22 − k1 ) − µ 2 a21 a12 = k02 + k0 µa22 − k0 k1 µ + k0 µa11 − k0 k1 µ
+µ 2 a11 a22 − k1 a11 − k1 a22 + k12 − µ 2 a21 a12 = k02 + k0 µ(a11 + a22 ) + µ 2 (a11 a22 − a21 a12 ) −µ 2 k1 (a11 + a22 ) + µ 2 k12 = k02 + k0 µtrA + µ 2 detA − µ 2 k1trA + µ 2 k12 = k02 + µ 2 k1 + µ(k0 − µk1 )trA + µ 2 detA. This completes the proof. Definition 3.1.28 Let A and B be regressive on T. Then we define A ⊕c B = (A + B − k1 I) + µk0−1 (A − k1 I)(B − k1 ),



108  Conformable Dynamic Equations on Time Scales

c A = −k0 (k0 I + µ(A − k1 I))−1 (A − k1 I), A c B = A ⊕ ( c B).

Theorem 3.1.29 (Rc , ⊕c ) is a group. Proof 3.1.30 Let A, B < C ∈ Rc . Then k0 − µk1 6= 0 on

T,

and (ko I + µ(A − k1 I))−1 ,

(ko I + µ(B − k1 I))−1 ,

(ko I + µ(C − k1 I))−1

exist. We have k0 I + µ(A ⊕c B − k1 I) = k0 I + µ k0−1 ((A + B − k1 I)k0 + µ(A − k1 I)(B − k1 I)) − k1 I




= k0−1 k02 I + µ ((A + B − k1 I)k0 + µ(A − k1 I)(B − k1 I)) − µk0 k1 I




= k0−1 k02 I + µk0 (A − k1 I) + µk0 B + µ(A − k1 I)(B − k1 I) − µk0 k1 I





= k0−1 k02 I + µk0 (A − k1 I) + µk0 (B − k1 I) + µ(A − k1 I)(B − k1 I) = k0−1 (k0 (k0 I + µ(A − k1 I)) + µ(k0 I + µ(A − k1 I))(B − k1 I)) = k0−1 (k0 I + µ(A − k1 I)) (kI + µ(B − k1 I)) , whereupon we conclude that k0 I + µ(A⊕c B−k1 I) is an invertible matrix on T and A⊕c B ∈ Rc . Also, k0 I + µ(k1 I − k1 I) = k0 I is invertible and hence, k1 I ∈ Rc , and (k1 I) ⊕c A = k0−1 ((k1 I + A − k1 I)k0 + µ(k1 I − k1 I)(A − k1 I)) = k0−1 k0 A = A,

Conformable Dynamic Systems on Time Scales  109

A ⊕c (k1 I) = k0−1 ((A + k1 I − k1 I)k0 + µ(A − k1 I)(k1 I − k1 I)) = k0−1 k0 A = A. Consequently, k1 I is the conformable additive identity for ⊕c . Note that   k0 I + µ −k0 (k0 I + µ(A − k1 I))−1 (A − k1 I) + k1 I − k1 I = k0 I − µk0 (k0 I + µ(A − k1 I))−1 (A − k1 I)
= (k0 I + µ(A − k1 I))−1 k02 I + µk0 (A − k1 I) − µk0 (A − k1 I) = k02 (k0 I + µ(A − k1 I))−1 , i.e., −k0 (k0 I + µ(A − k1 I))−1 (A − k1 I) + k1 I ∈ Rc . Next,   A ⊕c −k0 (k0 I + µ(A − k1 I))−1 (A − k1 I) + k1 I 0

   = k0−1 k0 A − k0 (k0 I + µ(A − k1 I))−1 (A − k1 I) + k1 I − k1 I   +µ(A − k1 I) −k0 (k0 I + µ(A − k1 I))−1 (A − k1 I) + k1 I − k1 I    = k0−1 k0 A − k0 (k0 I + µ(A − k1 I))−12 (A − k1 I)  −µk0 (A − k1 I) (k0 I + µ(A − k1 I))−1 (A − k1 I) =

 Ak0 (k0 I + µ(A − k1 I))−1 (A − k1 I)  −µ(A − k1 I) (k0 I + µ(A − K1 I))−1 (A − k1 I)

= A − (k0 + µ(A − k1 I)) (k0 I + µ(A − k1 I))−1 (A − k1 I)

110  Conformable Dynamic Equations on Time Scales

= A − A + k1 I = k1 I and   −k0 (k0 I + µ(A − k1 I))−1 (A − k1 I) + k1 I ⊕c A = −k0 (k0 I + µ(A − k1 I))−1 (A − k1 I) + k1 I + A − k1 I   +µk0−1 −k0 (k0 I + µ(A − k1 I))−1 (A − k1 I) + k1 I + k1 I − k1 I (A − k1 I) = −k0 (k0 I + µ(A − k1 I))−1 (A − k1 I) + A −µ (k0 I + µ(A − k1 I))−1 (A − k1 I)(A − k1 I) = − (k0 I + µ(A − k1 I))−1 (k0 I + µ(A − k1 I)) (A − k1 I) + A = −A + k1 I + A = k1 I, i.e., the conformable additive inverse of the matrix A is the matrix −k0 (k0 I + µ(A − k1 I))−1 (A − k1 I) + k1 I. Let P = A ⊕c B,

Q = B ⊕c C.

Then P = A + B − k1 I + µk0−1 (A − k1 I)(B − k1 I), Q = B +C − k1 I + µk0−1 (B − k1 I)(C − k1 I), and P ⊕c C = P +C − k1 I + µk0−1 (P − k1 I)(C − k1 I) = A + B − k1 I + µk0−1 (A − k1 I)(B − k1 I) +C − k1 I
+µk0−1 A + B − k1 I + µk01 (A − k1 I)(B − k1 I) − k1 I (C − k1 I)

Conformable Dynamic Systems on Time Scales  111

= A + B +C − 2k1 I + µk0−1 (A − k1 I)(B − k1 I) +µk0−1 (A − k1 I)(C − k1 I) + µk0−1 (B − k1 I)(C − k1 I) + µk0−1


2

(A − k1 I)(B − k1 I)(C − k1 I),

A ⊕c Q = A + Q − k1 I + µk)−1 (A − k1 I)(Q − k1 I) = A + B +C − k1 I + µk0−1 (B − k1 I)(C − k1 I) − k1 I +µk)−1 B +C − k1 I + µk0−1 (B − k1 I)(C − k1 I) − k1 I




= A + B +C − 2k1 I + µk0−1 (B − k1 I)(C − k1 I) +µk0−1 (A − k1 I)(B − k1 I) + µk0−1 (A − k1 I)(C − k1 I) + µk0−1


2

(A − k1 I)(B − k1 I)(C − k1 I).

Therefore A ⊕c Q = P ⊕c C or A ⊕c (B ⊕c C) = (A ⊕c B) ⊕c C. 

This completes the proof. Example 3.1.31 Let T = 2N0 , k0 (α,t) = αt 1−α ,

k1 (α,t) = (1 − α)t α , and

 A(t) =

1 t t +1 3



 ,

B(t) =

α ∈ (0, 1],

−1 t t 2

 ,

0

t ∈ T,

t ∈ T.

We will find (A ⊕c B)(t),

t ∈ T.

Here µ(t) = t,

t ∈ T.

Then  B(t) − k1 (α,t)I =

−1 t t 2



 − (1 − α)t α

1 0 0 1



112  Conformable Dynamic Equations on Time Scales





−1 t t 2



(1 − α)t α 0



−1 − (1 − α)t α t

t 2 − (1 − α)t α

0 = − (1 − α)t α   −1 − (1 − α)t α t = , t 2 − (1 − α)t α     1 t 1 0 α A(t) − k1 (α,t)I = − (1 − α)t t +1 3 0 1   1 − (1 − α)t α t = , t +1 3 − (1 − α)t α  (A(t) − k1 (α,t)I) (B(t) − k1 (α,t)I) =

 ×

 =

1 − (1 − α)t α t +1

t 3 − (1 − α)t α





−1 + (1 − α)2t 2α + t 2 3t − 2(1 − α)t 1+α 2t − 1 − (1 − α)t α t 2 + t + 6 − (1 − α)t α − (1 − α)2t 2α

 ,

µ(t) (k0 (α,t))−1 (A(t) − k1 (α,t)I) (B(t) − k1 (α,t)I)

=



t αt 1−α

−1 + (1 − α)2t 2α + t 2 3t − 2(1 − α)t 1+α 2t − 1 − (1 − α)t α t 2 + t + 6 − (1 − α)t α − (1 − α)2t 2α

1 α (1 − α)2 3α 1 2+α t + t + t −  α α =  α 2 1+α 1 α 1 t − t − (1 − α)t 2α α α α 





 3 1+α 2 1+2α t − (1 − α)t  α α ,
1 2 1 − α α 1 − α 2α (1 − α)2 3α α t +t +6 t − t − t − t α α α α

   −1 t 1 0 α A(t) + B(t) − k1 (α,t)I = + − (1 − α)t t 2 0 1     0 2t (α − 1)t α 0 = + 2t + 1 5 0 (α − 1)t α   (α − 1)t α 2t = , 2t + 1 5 + (α − 1)t α  (A ⊕c B)(t) =

(α − 1)t α 2t + 1

1 t t +1 3

2t 5 + (α − 1)t α







Conformable Dynamic Systems on Time Scales  113

1 α (1 − α)2 3α 1 2+α t + t  −α t + α α + 2 1+α 1 α 1 t − t − (1 − α)t 2α α α α

 3 1+α 2 t − (1 − α)t 1+2α  α α
α 1 − α α 1 − α 2α (1 − α)2 3α  1 2 t +t +6 t − t − t − t α α α α



=

   1 α (1 − α)2 3α 1 2+α t + t + t α − 1 −  α α α   2 1+α 1 α 1 − α 2α 1 + 2t + t − t − t α α α

 3 1+α 2(1 − α) 1+2α t − t  α α ,  2  t +t +6 3 2(1 − α) α 1+α 1+2α +α −1 t +5+ t − t α α α 2t +

α ∈ (0, 1], t ∈ T. This ends the example. Exercise 3.1.32 Let T = 2N0 , k1 (α,t) = (1 − α)(1 + t)α , and

 A(t) =

1 t t 2

k0 (α,t) = α(1 + t)1−α , 

 ,

B(t) =

t 3t 1 4

α ∈ (0, 1],

t ∈ T,

 ,

t ∈ T.

Find 1. (A ⊕c B)(t), t ∈ T, 2. ( c A)(t), t ∈ T, 3. ( c B)(t), t ∈ T. With A we will denote the conjugate matrix of the matrix A. With AT we will denote the
T transpose matrix of the matrix A, and A∗ = A will be the conjugate transpose matrix of the matrix A. Theorem 3.1.33 Let A ∈ Rc . Then A∗ ∈ Rc . Proof 3.1.34 We have (k0 I + µ(A − k1 I)) (k0 I + µ(A − k1 I))−1 = I

on

T.

(k0 I + µ(A − k1 I)) (k0 I + µ(A − k1 I))−1 = I

on

T,

Hence, and

 ∗ (k0 I + µ(A − k1 I))−1 (k0 I + µ(A∗ − k1 I)) = I

on

T,

or (k0 I + µ(A∗ − k1 I))−1 (k0 I + µ(A∗ − k1 I)) = I

on

T.

Therefore k0 I + µ(A∗ − k1 I) is invertible on T. This completes the proof.



114  Conformable Dynamic Equations on Time Scales

Theorem 3.1.35 Let A ∈ Rc . Then ( c A)∗ = c A∗ . Proof 3.1.36 We have  ∗ −k0 (k0 I + µ(A − k1 I))−1 (A − k1 I) + k1 I

( c A)∗ =

= −k0 (A∗ − k1 I) (k0 I + µ(A∗ − k1 I))−1 + k1 I = −k0 (k0 I + µ(A∗ − k1 I))−1 (A∗ − k1 I) + k1 I = c A∗ . 

This completes the proof. Theorem 3.1.37 Let A, B ∈ Rc . Then (A ⊕c B)∗ = B∗ ⊕c A∗ . Proof 3.1.38 We have (A ⊕c B)∗ =


∗ A + B − k1 I + µk0−1 (A − k1 I) (B − k1 I)

= B∗ + A∗ − k1 I + µk0−1 (B∗ − k1 I) (A∗ − k1 I) = B∗ ⊕c A∗ . 

This completes the proof.

Definition 3.1.39 (Conformable Matrix Exponential Function) Let A ∈ Rc and t0 ∈ T. The unique solution of the IVP Dα Y = A(t)Y,

Y (t0 ) = I,

is called the conformable matrix exponential function. It is denoted by EA (·,t0 ). Theorem 3.1.40 Let A, B ∈ Rc and t, s, r ∈ T. Then: 1. EA (t,t) = I, 2. The matrix composition with sigma is given by    µ(t) k1 (α,t) I+ A(t) EA (t, s), EA (σ (t), s) = 1 − µ(t) k0 (α,t) k0 (α,t)

Conformable Dynamic Systems on Time Scales  115

3. The inverse matrix is given by EA (s,t) = (EA (t, s))−1 = (EC (t, s))∗ , where C = k0 (−A∗ + k1 I) (k0 I + µ (A∗ − kI ))−1 , 4. EA (t, r) = EA (t, s)EA (s, r), 5. EA (t, s)EB (t, s) = EA⊕c B (t, s) if A and B commute. Proof 3.1.41

1. Consider the IVP Dα Y = A(t)Y,

Y (s) = I.

By the definition of EA (·, s) we have EA (s, s) = I. 2. We have   µ(t) α µ(t) k1 (α,t) EA (t, s) + D EA (t, s) EA (σ (t), s) = 1− k0 (α,t) k0 (α,t)   µ(t) µ(t) k1 (α,t) EA (t, s) + A(t)EA (t, s) = 1− k0 (α,t) k0 (α,t)    µ(t) µ(t) = 1− k1 (α,t) I + A(t) EA (t, s). k0 (α,t) k0 (α,t) 3. Let

 ∗ Z(t) = (EA (t, s))−1 .

Then  ∗ Dα Z(t) = − (EA (t, s))−1 Dα EA (t, s) (EA (σ (t), s))−1 + k1 (α,t) (EA (σ (t), s))−1 =

 ∗ − (EA (t, s))−1 A(t)EA (t, s) (EA (σ (t), s))−1 + k1 (α,t) (EA (σ (t), s))−1

=

 ∗ (−A(t) + k1 (α,t)I) (EA (σ (t), s))−1

−1 !∗ 1 (k0 (α,t)I + µ(t) (A(t) − k1 (α,t)I)) EA (t, s) = (−A(t) + k1 (α,t)I) k0 (α,t)  ∗ = k0 (α,t) (−A(t) + k1 (α,t)I) (k0 (α,t)I + µ(t) (A(t) − k1 (α,t)I))−1 (EA (t, s))−1 

 ∗ = k0 (α,t) (−A∗ (t) + k1 (α,t)I) (k0 (α,t)I + µ(t) (A∗ (t) − k1 (α,t)I))−1 (EA (t, s))−1

116  Conformable Dynamic Equations on Time Scales = k0 (α,t) (−A∗ (t) + k1 (α,t)I) (k0 (α,t)I + µ(t) (A∗ (t) − k1 (α,t)I))−1 Z(t).

Hence,

∗  (EA (t, s))−1 = EC (t, s)

or (EA (t, s))−1 = EC∗ (t, s). 4. Let Z(t) = EA (t, s)EA (s, r). Then Dα Z(t) = Dα EA (t, s)EA (s, r) = A(t)EA (t, s)EA (s, r) = A(t)Z(t) and Z(r) = I. Therefore EA (t, r) = EA (t, s)EA (s, r). 5. Let Z(t) = EA (t, s)EB (t, s). Then Dα Z(t) = Dα (EA (t, s)EB (t, s)) = Dα EA (t, s)EB (t, s) + EA (σ (t), s)Dα EB (t, s) −k1 (α, s)EA (σ (t), s)EB (t, s) = A(t)EA (t, s)EB (t, s) + EA (σ (t), s)B(t)EB (t, s) −k1 (α, s)EA (σ (t), s)EB (t, s) = A(t)EA (t, s)EB (t, s) +

   µ(t) k1 (α,t) I+ A(t) B(t)EA (t, s)EB (t, s) 1 − µ(t) k0 (α,t) k0 (α,t)

Conformable Dynamic Systems on Time Scales  117

−k1 (α,t)A(t)EA (t, s)EB (t, s)    k1 (α,t) µ(t) + 1 − µ(t) I+ A(t) EA (t, s)EB (t, s) k0 (α,t) k0 (α,t) = A(t)EA (t, s)EB (t, s)   µ(t) (A(t) − k1 (α,t)I) B(t) EA (t, s)EB (t, s) + B(t) + k0 (α,t) −

k1 (α,t) (k0 (α,t)I + µ(t) (A(t) − k1 (α,t)I)) EA (t, s)EB (t, s) k0 (α,t)

  µ(t) = A(t) + B(t) − k1 (α,t)I + (A(t) − k1 (α,t)I) (B(t) − k1 (α,t)I) k0 (α,t) ×EA (t, s)EB (t, s) = (A ⊕c B) (t)EA (t, s)EB (t, s) = (A ⊕c B) (t)Z(t). Therefore EA⊕c B (t, s) = EA (t, s)EB (t, s). 

This completes the proof. Theorem 3.1.42 Let A ∈ Rc and s,t ∈ T. Then Dtα (EA (s,t)) = (−A(t) + k1 (α,t)I) EA (s, σ (t)),

α ∈ (0, 1].

Proof 3.1.43 We have Dtα (EA (s,t)) = Dα ((EA (t, s)) −1) = Dtα (EC (t, s))∗ , where C = k0 (−A∗ + k1 I) (k0 I + µ (A∗ − k1 I))−1 . Then Dtα (EA (s,t)) = (Dtα (EC (t, s)))∗ = (CEC (t, s))∗ = (EC (t, s))∗ C∗

118  Conformable Dynamic Equations on Time Scales

= (EC (t, s))∗ k0 (−A(t) + k1 (α,t)I) (k0 (α,t)I + µ(t)(A(t) − k1 (α,t)I))−1 = k0 (−A(t) + k1 (α,t)I) (k0 (α,t)I + µ(t)(A(t) − k1 (α,t)I))−1 EA (s,t) = (−A(t) + k1 (α,t)I) EA (s, σ (t)). 

This completes the proof.

Theorem 3.1.44 (Variation of Constants) Let A ∈ Rc and q : T → Rn be rd-continuous. Let also, t0 ∈ T, y0 ∈ Rn and assume k0 I + µA is invertible on T. Then the IVP Dα y = A(t)y + q(t),

y(t0 ) = y0 ,

(3.1)

has a unique solution y : T → Rn and its solution is given by Z t

y(t) = y0 EA (t,t0 ) +

t0

EB (σ (s),t)q(s)∆α,t s,

(3.2)

where B = (µk1 − k0 )A (k0 I + µA)−1 . 1. Let y : T → Rn be given by (3.2). Then

Proof 3.1.45

Z t

y(t) = y0 EA (t,t0 ) + EB (t0 ,t)

t0

EB (σ (s),t0 )q(s)∆α,t s,

t ∈ T,

and   Z t D y(t) = y0 D EA (t,t0 ) + D EB (t0 ,t) EB (σ (s),t0 )q(s)∆α,t s α

α

α

t0

= y0 A(t)ES (t,t0 ) + (Dα EB (t0 ,t))

Z t t0

EB (σ (s),t0 )q(s)∆α,t s

+EB (t0 , σ (t))EB (σ (t),t0 )q(t) −k1 (α,t)EB (t0 , σ (t))

Z t t0

EB (σ (s),t0 )q(s)∆α,t s

= y0 A(t)EA (t,t0 ) Z t

+ (−B(t) + k1 (α,t)I) EB (t0 , σ (t))

t0

EB (σ (s),t0 )q(s)∆α,t s

Conformable Dynamic Systems on Time Scales  119

+q(t) −k1 (α,t)EB (t0 , σ (t))

Z t t0

EB (σ (s),t0 )q(s)∆α,t s

= y0 A(t)EA (t,t0 ) −B(t)EB (t0 , σ (t)) +q(t),

Z t t0

EB (σ (s),t0 )q(s)∆α,t s

t ∈ T.

Note that EB (t0 , σ (t)) = k0 (α,t) (k0 (α,t)I + µ(t)(B(t) − k1 (α,t)I)) EB (t0 ,t) and (k0 (α,t)I + µ(t)A(t)) B(t) = (µ(t)k1 (α,t)I − k0 (α,t)I) A(t), whereupon k0 (α,t)B(t) = A(t) (µ(t)k1 (α,t)I − k0 (α,t)I − µ(t)B(t)) = −A(t) (k0 (α,t)I − µ(t)k1 (α,t)I + µ(t)B(t)) and k0 (α,t)B(t) (k0 (α,t)I − µ(t)k1 (α,t)I + µ(t)B(t))−1 = −A(t),

t ∈ T.

Hence, −B(t)EB (t0 , σ (t)) = A(t)EB (t0 ,t),

t ∈ T.

Therefore Z t

α

D y(t) = y0 A(t)EA (t,t0 ) + A(t)EB (t0 ,t)

t0

EB (σ (s),t0 )q(s)∆α,t s + q(t)

  Z t = A(t) y0 EA (t,t0 ) + EB (σ (s),t)q(s)∆α,t s + q(t) t0

= A(t)y(t) + q(t),

t ∈ T.

Also y(t0 ) = y0 . Therefore y satisfies (3.1).

120  Conformable Dynamic Equations on Time Scales

2. Suppose that y1 and y2 are two solutions of the IVP (3.1). Let z = y1 − y2

on

T.

Then z is the solution of the IVP Dα z = A(t)z,

z(t0 ) = 0.

Therefore z = 0 on T, or y1 = y2 on T. This completes the proof.



Example 3.1.46 Suppose that A ∈ Rc , k0 I − µA and k0 I − µB are invertible on T, where B = (k0 − µ = k1 ) (k0 I − µA)−1 A. Also, assume q : T → Rn is rd-continuous. Consider the IVP Dα y = A(t)yσ + q(t), Since

y(t0 ) = y0 .

  µ(t) µ(t) α y (t) = 1 − k1 (α,t) y(t) + D y(t), k0 (α,t) k0 (α,t) σ

t ∈ T,

we get   µ(t) µ(t) k1 (α,t) A(t)y(t) + A(t)Dα y(t) D y(t) = 1− k0 (α,t) k0 (α,t) α

+q(t),

t ∈ T,

or (k0 (α,t)I − µ(t)A(t)) Dα y(t) = (k0 (α,t) − µ(t)k1 (α,t)) A(t)y(t) + k0 (α,t)q(t), or Dα y(t) = (k0 (α,t) − µ(t)k1 (α,t)) (k0 (α,t)I − µ(t)A(t))−1 A(t)y(t) +k0 (α,t) (k0 (α,t)I − µ(t)A(t))−1 q(t) = B(t)y(t) + k0 (α,t) (k0 (α,t)I − µ(t)A(t))−1 q(t), Hence, and considering Theorem 3.1.44, we find Z t

y(t) = y0 EB (t,t0 ) +

t0

EC (σ (s),t)p(s)∆α,t s,

where C = (µk1 − k0 )B(k0 I + µB)−1

on T

and p = k0 (k0 I − µA)−1 q This ends the example.

on T.

t ∈ T,

t ∈ T.

t ∈ T,

Conformable Dynamic Systems on Time Scales  121

Exercise 3.1.47 Let T = Z and
α k1 (α,t) = (1 − α) 1 + t 2 ,

k0 (α,t) = α 1 + t 2

Find the solution of the IVP (3.1), where     1 t t A(t) = , q(t) = , 2 1 2t + 1


1−α

α ∈ (0, 1],

,

 y0 =

1 0

t ∈ T.

 ,

t ∈ T.

Theorem 3.1.48 Let A ∈ Rc be a 2 × 2 matrix and assume that X is a solution of the system Dα X = A(t)X. Then X satisfies Liouville’s formula det X(t) = EC (t,t0 ) det X(t0 ),

t ∈ Tκ ,

det X(t) = EC (t,t0 ) det X(t0 ),

t ∈ Tκ ,

where where   µ(t) µ(t) µ(t) det A(t) − k1 (α,t) 1 − k1 (α,t) + trA(t) , C(t) = trA(t) + k0 (α,t) k0 (α,t) k0 (α,t) t ∈ T. Proof 3.1.49 Let  A(t) =

a11 (t) a12 (t) a21 (t) a22 (t)

Then

 α

D X(t) = and



Dα x11 (t) Dα x12 (t) Dα x21 (t) Dα x22 (t)



 ,

X(t) =

Dα x11 (t) Dα x12 (t) Dα x21 (t) Dα x22 (t) 

 =

x11 (t) x12 (t) x21 (t) x22 (t)

 .



a11 (t) a12 (t) a21 (t) a22 (t)

on 

Tκ

x11 (t) x12 (t) x21 (t) x22 (t)

on Tκ . Therefore we get the system  α   D x11 (t) = a11 (t)x11 (t) + a12 (t)x21 (t)      α   D x12 (t) = a11 (t)x12 (t) + a12 (t)x22 (t)   Dα x21 (t) = a21 (t)x11 (t) + a22 (t)x21 (t)        α D x22 (t) = a21 (t)x12 (t) + a22 (t)x22 (t),

t ∈ Tκ .

Let p11 (t) =

  µ(t) α µ(t) k1 (α,t) x11 (t) + D x11 (t), 1− k0 (α,t) k0 (α,t)



122  Conformable Dynamic Equations on Time Scales

  µ(t) µ(t) α p12 (t) = 1− k1 (α,t) x12 (t) + D x12 (t), k0 (α,t) k0 (α,t) p21 (t) = Dα x21 (t), p22 (t) = Dα x22 (t),   µ(t) α µ(t) k1 (α,t) x11 (t) + D x11 (t), q11 (t) = 1− k0 (α,t) k0 (α,t)   µ(t) µ(t) α q12 (t) = 1− k1 (α,t) x12 (t) + D x12 (t), k0 (α,t) k0 (α,t) q21 (t) = x21 (t), q22 (t) = x22 (t), and

 P(t) =

p11 (t) p12 (t) p21 (t) p22 (t)

t ∈ Tκ , 

 ,

Q(t) =

q11 (t) q12 (t) q21 (t) q22 (t)

 ,

t ∈ Tκ .

Next, det(X(t)) = x11 (t)x22 (t) − x12 (t)x21 (t),

t ∈ T,

and Dα (det(X(t))) = Dα (x11 (t)x22 (t)) − Dα (x12 (t)x21 (t)) σ σ = Dα x11 (t)x22 (t) + x11 (t)Dα x22 (t) − k1 (α,t)x11 (t)x22 (t) σ σ −Dα x12 (t)x21 (t) − x12 (t)Dα x21 (t) + k1 (α,t)x12 (t)x21 (t)



   σ σ Dα x11 (t) Dα x12 (t) x11 (t) x12 (t) = det + det x21 (t) x22 (t) Dα x21 (t) Dα x22 (t)  σ  σ x11 (t) x12 (t) −k1 (α,t) det x21 (t) x22 (t)   a11 (t)x11 (t) + a12 (t)x21 (t) a11 (t)x12 (t) + a12 (t)x22 (t) = det x21 (t) x22 (t) + det P(t) − k1 (α,t) det Q(t)

Conformable Dynamic Systems on Time Scales  123

    µ(t) x11 (t) x12 (t) k1 (α,t) det = a11 (t) det X(t) + 1 − Dα x21 (t) Dα x22 (t) k0 (α,t)  α  µ(t) D x11 (t) Dα x12 (t) det + Dα x21 (t) Dα x22 (t) k0 (α,t)   µ(t) −k1 (α,t) 1 − k1 (α,t) det X(t) k0 (α,t)  α  µ(t) D x11 (t) Dα x12 (t) −k1 (α,t) x21 (t) x22 (t) k0 (α,t)    µ(t) = a11 (t) − k1 (α,t) 1 − k1 (α,t) det X(t) k0 (α,t)   µ(t) + 1− k1 (α,t) k0 (α,t)   x11 (t) x12 (t) × det a21 (t)x11 (t) + a22 (t)x21 (t) a21 (t)x12 (t) + a22 (t)x22 (t)   µ(t) a11 (t)x11 (t) + a12 (t)x21 (t) a11 (t)x11 (t) + a22 (t)x21 (t) −k1 (α,t) x21 (t) x22 (t) k0 (α,t) µ(t) (Dα x11 (t)Dα x22 (t) − Dα x12 (t)Dα x21 (t)) k0 (α,t)    µ(t) k1 (α,t) det X(t) = a11 (t) − k1 (α,t) 1 − k0 (α,t)   µ(t) +a22 (t) 1 − k1 (α,t) det X(t) k0 (α,t) +

−a11 (t)k1 (α,t)

µ(t) det X(t) k0 (α,t)

µ(t)  (a11 (t)x11 (t) + a12 (t)x21 (t)) (a21 (t)x12 (t) + a22 (t)x22 (t)) k0 (α,t)  − (a11 (t)x12 (t) + a12 (t)x22 (t)) (a21 (t)x11 (t) + a22 (t)x21 (t))

+

=

  µ(t) a11 (t) + a22 (t) − k1 (α,t) 1 − k1 (α,t) k0 (α,t) +a22 (t) +

µ(t) µ(t)  + a11 (t) det X(t) k0 (α,t) k0 (α,t)

µ(t)  a11 (t)a21 (t)x11 (t)x22 (t) + a11 (t)a22 (t)x11 (t)x22 (t) k0 (α,t)

124  Conformable Dynamic Equations on Time Scales

+a12 (t)a21 (t)x21 (t)x12 (t) + a12 (t)a22 (t)x21 (t)x22 (t) −a11 (t)a21 (t)x11 (t)x12 (t) − a11 (t)a22 (t)x12 (t)x21 (t)  −a12 (t)a21 (t)x11 (t)x22 (t) − a12 (t)a22 (t)x21 (t)x22 (t) =

  µ(t) k1 (α,t) (a11 (t) + a22 (t) − k1 (α,t) (1 − k0 (α,t) +a22 (t) +

=

µ(t) µ(t)  + a11 (t) det X(t) k0 (α,t) k0 (α,t)

µ(t) (a11 (t)a22 (t) det X(t) − a12 (t)a22 (t) det X(t)) k0 (α,t)

 µ(t) det A(t) trA(t) + k0 (α,t)   µ(t) µ(t) µ(t)  det X(t), −k1 (α,t) 1 − k1 (α,t) + a22 (t) + a11 (t) k0 (α,t) k0 (α,t) k0 (α,t)

t ∈ Tκ , i.e., Dα (det X(t)) = C(t) det X(t),

t ∈ Tκ .

det X(t) = EC (t,t0 ) det X(t0 ),

t ∈ Tκ .

Therefore 

This completes the proof.

3.2

CONSTANT COEFFICIENTS

In this section we suppose that A is an n × n constant matrix and A ∈ Rc . Let t0 ∈ T. Consider the system Dα x = Ax. (3.3)

Theorem 3.2.1 Let u and v be solutions of (3.3). Then x = au + bv, is a solution of the system (3.3). Proof 3.2.2 We have Dα u(t) = Au(t),

a, b ∈ C,

Conformable Dynamic Systems on Time Scales  125

Dα v(t) = Bv(t), Dα x(t) = Dα (au + bv)(t) = aDα u(t) + bDα v(t) = aAu(t) + bAv(t) = A(au(t)) + A(bv(t)) = A(au(t) + bv(t)) = Ax(t),

t ∈ Tκ . 

This completes the proof. Theorem 3.2.3 Let λ , ξ be an eigenpair of A. Then x(t) = Eλ (t,t0 )ξ ,

t ∈ Tκ ,

is a solution of the system (3.3). Proof 3.2.4 We have Aξ = λ ξ . Then Dα x(t) = Dα (Eλ (t,t0 )ξ ) = Dα (Eλ (t,t0 )) ξ = λ Eλ (t,t0 )ξ = Eλ (t,t0 )(λ ξ ) = Eλ (t,t0 )Aξ = A (Eλ (t,t0 )ξ ) = Ax(t), This completes the proof.

t ∈ Tκ . 

126  Conformable Dynamic Equations on Time Scales

Example 3.2.5 Consider the system  α  D x1 (t) = −3x1 − 2x2 

Dα x2 (t) = 3x1 + 4x2 .

Here

 A=

−3 −2 3 4



−3 − λ 3

−2 4−λ

.

Then  0 = det



= (λ − 4)(λ + 3) + 6 = λ2 −λ −6 and λ1 = 3,

λ2 = −2.

The considered system is conformable regressive for any time scale for which −2 ∈ Rc . Note that     −2 1 ξ1 = , ξ2 = −3 1 are eigenvectors corresponding to λ1 and λ2 , respectively. Therefore x(t) = c1 E3 (t,t0 )ξ1 + c2 E−2 (t,t0 )ξ2  = c1 E3 (t,t0 )

1 −3



 + c2 E−2 (t,t0 )

−2 1

 ,

where c1 and c2 are real constants, is a solution of the considered system for any time scale for which −2 ∈ Rc . Example 3.2.6 Consider the system  α D x1 (t) = x1 (t) − x2 (t)      Dα x2 (t) = −x1 (t) + 2x2 (t) − x3 (t)      α D x3 (t) = −x2 (t) + x3 (t). Here



 1 −1 0 A =  −1 2 −1  . 0 −1 1

Conformable Dynamic Systems on Time Scales  127

Then 0 = det(A − λ I) 

1−λ = det  −1 0

−1 2−λ −1

 0 −1  1−λ

= −(λ − 1)2 (λ − 2) + (λ − 1) + (λ − 1) = (λ − 1) (−(λ − 1)(λ − 2) + 2) = (λ − 1) −λ 2 + 3λ




= −λ (λ − 1)(λ − 3). Therefore λ1 = 0,

λ2 = 1,

λ3 = 3.

Note that the matrix A is conformable regressive for any time scale and       1 1 1 ξ1 =  1  , ξ2 =  0  , ξ3 =  −2  1 −1 1 are eigenvectors corresponding to λ1 , λ2 and λ3 , respectively. Consequently, x(t) = c1 E0 (t,t0 )ξ1 + c2 E1 (t,t0 )ξ2 + c3 E3 (t,t0 )ξ3      1 1 1 = c1 E0 (t,t0 )  1  + c2 E1 (t,t0 )  0  + c3 E3 (t,t0 )  −2  , 1 −1 1 

where c1 , c2 and c3 are constants, is a general solution of the considered system. Example 3.2.7 Consider the system  α D x1 (t) = −x1 (t) + x2 (t) + x3 (t)        α   D x2 (t) = x2 (t) − x3 (t) + x4 (t)   Dα x3 (t) = 2x3 (t) − 2x4 (t)        α D x4 (t) = 3x4 (t).

128  Conformable Dynamic Equations on Time Scales

Here



−1  0 A=  0 0

 1 1 0 1 −1 1  . 0 2 −2  0 0 3

Then 0 = det(A − λ I) 

−1 − λ  0 = det   0 0

1 1−λ 0 0

1 −1 2−λ 0

 0 1   −2  3−λ

= (λ + 1)(λ − 1)(λ − 2)(λ − 3) and λ1 = −1,

λ2 = 1,

λ3 = 2,

λ4 = 3.

The matrix A is conformable regressive for any time scale for which −1 ∈ Rc . Note that         0 0 0 1  0   0   1   0         ξ1 =   0  , ξ2 =  0  , ξ3 =  1  and ξ4 =  0  1 0 0 0 are eigenvectors corresponding to λ1 , λ2 , λ3 and λ4 , respectively. Consequently, x(t) = c1 E−1 (t,t0 )ξ1 + c2 E1 (t,t0 )ξ2 + c3 E2 (t,t0 )ξ3 + c4 E5 (t,t0 )ξ4    0 1  1   0     = c1 E−1 (t,t0 )   0  + c2 E1 (t,t0 )  0  0 0     0 0  0   0     +c3 E2 (t,t0 )   1  + c4 E3 (t,t0 )  0  , 0 1 

where c1 , c2 , c3 and c4 are real constants, is a general solution of the considered system. Exercise 3.2.8 Find a general solution of the system  α  D x1 (t) = x2 (t) 

Dα x2 (t) = x1 (t).

Conformable Dynamic Systems on Time Scales  129

Answer.

 x(t) = c1 E1 (t,t0 )

1 1



 + c2 E−1 (t,t0 )

1 −1

 ,

where c1 , c2 ∈ R, for any time scale for which −1 ∈ Rc . Theorem 3.2.9 Assume that A ∈ Rc . If t ∈ Tκ ,

x(t) = u(t) + iv(t),

is a complex vector-valued solution of the system (3.3), where u(t) and v(t) are real vectorvalued functions on T, then u(t) and v(t) are real vector-valued solutions of the system (3.3) on T. Proof 3.2.10 We have Dα x(t) = A(t)x(t) = A(t) (u(t) + iv(t)) = A(t)u(t) + iA(t)v(t) = Dα u(t) + iDα v(t),

t ∈ Tκ .

Equating real and imaginary parts, we get Dα u(t) = A(t)u(t), Dα v(t) = A(t)v(t),

t ∈ Tκ . 

This completes the proof. Example 3.2.11 Consider the system  α  D x1 (t) = x1 (t) + x2 (t) 

Dα x2 (t) = −x1 (t) + x2 (t).

Here

 A=

1 1 −1 1

 .

Then 0 = det(A − λ I)  = det

1−λ −1

1 1−λ



130  Conformable Dynamic Equations on Time Scales

= (λ − 1)2 + 1 = λ 2 − 2λ + 1 + 1 = λ 2 − 2λ + 2, whereupon λ1,2 = 1 ± i. Note that

 ξ=

1 i



is an eigenvector corresponding to the eigenvalue λ = 1 + i. We have   1 x(t) = E1+i (t,t0 ) i   1  = E1 (t,t0 ) Cos 1 (t,t0 ) + iSin 1 (t,t0 ) 1+µ 1+µ i ! !! iSin 1 (t,t0 ) Cos 1 (t,t0 ) 1+µ 1+µ + = E1 (t,t0 ) iCos 1 (t,t0 ) −Sin 1 (t,t0 ) 1+µ

1+µ

Cos 1 (t,t0 )

!

1µ

= E1 (t,t0 )

−Sin 1 (t,t0 )

Sin + iE1 (t,t0 )

1 1+µ

Cos

1µ

!

(t,t0 )

1 1+µ

.

(t,t0 )

Consequently, Cos E1 (t,t0 )

1 1+µ

−Sin

(t,t0 )

1 1+µ

(t,t0 )

!

Sin and

E1 (t,t0 )

are solutions of the considered system. Therefore ! Cos 1 (t,t0 ) 1+µ x(t) = c1 E1 (t,t0 ) + c2 E1 (t,t0 ) −Sin 1 (t,t0 )

1 1+µ

Cos

1 1+µ

Sin

where c1 , c2 ∈ R, is a general solution of the considered system. Example 3.2.12 Consider the system  α D x1 (t) = x2 (t)      Dα x2 (t) = x3 (t)      α D x3 (t) = 2x1 (t) − 4x2 (t) + 3x3 (t).

!

(t,t0 )

1 1+µ

Cos

1+µ

(t,t0 )

1 1+µ

(t,t0 ) (t,t0 )

! ,

Conformable Dynamic Systems on Time Scales  131

Here

 0 1 0 A =  0 0 1 . 2 −4 3 

Then 0 = det(A − λ I) 

−λ = det  0 2

 1 0 −λ 1  −4 3 − λ

= −λ 2 (λ − 3) + 2 − 4λ = − λ 3 − 3λ 2 + 4λ − 2




= −(λ − 1)(λ 2 − 2λ + 2), whereupon λ1 = 1, Note that

λ2,3 = 1 ± i.  1 and ξ2 =  1 + i  2i 

 1 ξ1 =  1  1 

are eigenvectors corresponding to the eigenvalues λ1 = 1 and λ2 = 1 + i, respectively. Note that     1 1   E1+i (t,t0 )  1 + i  = E1 (t,t0 ) Cos 1 (t,t0 ) + iSin 1 (t,t0 )  1 + i  1+µ 1+µ 2i 2i     Cos 1 (t,t0 ) Sin 1 (t,t0 ) 1+µ 1+µ      (1 + i)Cos 1 (t,t0 )  + i  (1 + i)Sin 1 (t,t0 )  = E1 (t,t0 )      1+µ 1+µ 2iCos 1 (t,t0 ) 2iSin 1 (t,t0 ) 1+µ



Cos

1 1+µ

1+µ

(t,t0 )





iSin

1 1+µ



(t,t0 )

     (1 + i)Cos 1 (t,t0 )  +  (−1 + i)Sin 1 (t,t0 )  = E1 (t,t0 )      1+µ 1+µ 2iCos 1 (t,t0 ) −2Sin 1 (t,t0 ) 1+µ



Cos

1+µ

1 1+µ

(t,t0 )





Sin

1 1+µ

(t,t0 )



     Cos 1 (t,t0 ) − Sin 1 (t,t0 )  + i  Cos 1 (t,t0 ) + Sin 1 (t,t0 )  . = E1 (t,t0 )      1+µ 1+µ 1+µ 1+µ −2Sin 1 (t,t0 ) 2Cos 1 (t,t0 ) 1+µ

1+µ

132  Conformable Dynamic Equations on Time Scales

Consequently,   Cos 1 (t,t0 ) 1 1+µ   Cos 1 (t,t0 ) − Sin 1 (t,t0 ) x(t) = E1 (t,t0 ) c1  1  + c2   1+µ 1+µ 1 −2Sin 1 (t,t0 ) 

   

1+µ



 Sin 1 (t,t0 ) 1+µ   Cos 1 (t,t0 ) + Sin 1 (t,t0 )  , +c3    1+µ 1+µ 2Cos 1 (t,t0 ) 1+µ

where c1 , c2 , c3 ∈ R, is a general solution of the considered system. Exercise 3.2.13 Find a general solution of the system  α D x1 (t) = x1 (t) − 2x2 (t) + x3 (t)      Dα x2 (t) = −x1 (t) + x3 (t)      α D x3 (t) = x1 (t) − 2x2 (t) + x3 (t). Theorem 3.2.14 (Conformable Putzer Algorithm) Let A ∈ Rc be a constant n × n matrix and t0 ∈ T. If λ1 , λ2 , . . ., λn are the eigenvalues of A, then n−1

EA (t,t0 ) =

∑ rk+1 (t)Pk , k=0

where



 r1 (t)   r(t) =  ...  rn (t)

is the solution of the IVP     D r=   α

λ1 0 0 1 λ2 0 0 1 λ3 .. .. .. . . . 0 0 0

... ... ... .. .

0 0 0 .. .

     r,  

    r(t0 ) =   

. . . λn

0

P0 = I, Pk+1 = (A − λk+1 I)Pk ,

1 0 0 .. .

0 ≤ k ≤ n − 1.

    ,  

(3.4)

Conformable Dynamic Systems on Time Scales  133

Proof 3.2.15 Since A is conformable regressive, we have that all eigenvalues of A are conformable regressive. Therefore the IVP (3.4) has a unique solution. We set n−1

∑ rk+1 (t)Pk .

X(t) =

k=0

We have P1 = (A − λ1 I)P0 = (A − λ1 I), P2 = (A − λ2 I)P1 = (A − λ2 I)(A − λ1 I), .. . Pn = (A − λn I)Pn−1 = (A − λn I) . . . (A − λ1 I) = 0. Therefore n−1

Dα X(t) =

∑ Dα rk+1 (t)Pk , k=0 n−1

Dα X(t) − AX(t) =

n−1

∑ Dα rk+1 (t)Pk − A ∑ rk+1 (t)Pk k=0

k=0 n−1

= Dα r1 (t)P0 + ∑ Dα rk+1 (t)Pk k=1 n−1

−A ∑ rk+1 (t)Pk k=0 n−1

= λ1 r1 (t)P0 + ∑ (rk (t) + λk+1 rk+1 (t)) Pk k=1 n−1

− ∑ rk+1 (t)APk k=1

134  Conformable Dynamic Equations on Time Scales n−1

=

∑ rk (t)Pk + λ1 r1 (t)P0 k=1 n−1

n−1

+ ∑ λk+1 rk+1 (t)Pk − ∑ rk+1 (t)APk k=0

k=1 n−1

n−1

=

∑ rk (t)Pk + ∑ λk+1 rk+1 (t)Pk k=0

k=1 n−1

− ∑ rk+1 (t)APk k=0 n−1

=

=

n−1

∑ rk (t)Pk − ∑ (A − λk+1 I)rk+1 (t)Pk k=1

k=0

n−1

n−1

∑ rk (t)Pk − ∑ rk+1 (t)Pk+1 k=1

k=0

= −rn (t)Pn t ∈ Tκ .

= 0, Also,

n−1

X(t0 ) =

∑ rk+1 (t0 )Pk k=0

= r1 (t0 )P0 = I. 

This completes the proof. Example 3.2.16 Consider the system  α D x1 (t) = 2x1 (t) + x2 (t) + 2x3 (t)      Dα x2 (t) = 4x1 (t) + 2x2 (t) + 4x3 (t)      α D x3 (t) = 2x1 (t) + x2 (t) + 2x3 (t), t ∈ Tκ . Here



 2 1 2 A =  4 2 4 . 2 1 2

Then 0 = det(A − λ I)

Conformable Dynamic Systems on Time Scales  135



2−λ  4 = det 2

 2 4  2−λ

1 2−λ 1

= −(λ − 2)3 + 8 + 8 + 4(λ − 2) + 4(λ − 2) + 4(λ − 2) = −(λ − 2)3 + 12(λ − 2) + 16 = − λ 3 − 6λ 2 + 12λ − 8 − 12λ + 24 − 16 = − λ 3 − 6λ 2







= −λ 2 (λ − 6), whereupon λ1 = 0,

λ2 = 0,

λ3 = 6.

Consider the IVPs Dα r1 (t) = 0,

r1 (t0 ) = 1,

Dα r2 (t) = r1 (t),

r2 (t0 ) = 0,

Dα r3 (t) = r2 (t) + 6r3 (t),

r3 (t0 ) = 0.

We have r1 (t) = E0 (t,t0 ), Dα r2 (t) = E0 (t,t0 ), Then

t ∈ Tκ r2 (t0 ) = 0.

Z t

r2 (t) = where f (t) =

t0

E f (σ (s),t)E0 (s,t0 )∆α,t s,

t ∈ Tκ ,

k1 (α,t)(k0 (α,t) − µ(t)k1 (α,t)) , k0 (α,t) − µ(t)k1 (α,t)

t ∈ T,

and Dα r3 (t) = r2 (t) + 6r3 (t),

r3 (t0 ) = 0.

Therefore Z t

r3 (t) =

t0

Eg (σ (s),t)r2 (s)∆α,t s,

t ∈ Tκ ,

136  Conformable Dynamic Equations on Time Scales

where g(t) = −

(6 − k1 (α,t))(k0 (α,t) − µ(t)k1 (α,t)) , k0 (α,t) + µ(t)(6 − k1 (α,t))

t ∈ T.

Next, 

P0

 1 0 0 =  0 1 0 , 0 0 1

P1 = (A − λ1 I)P0 = AP0 = A  2 1 2 =  4 2 4 , 2 1 2 

P2 = (A − λ1 I)(A − λ2 I) = A2 I = A2 

2 1 2  4 2 4 = 2 1 2  12 6 =  24 12 12 6

 2 1 2  4 2 4  2 1 2  12 24  , 12 

P3 = 0. Therefore EA (t,t0 ) = r1 (t)P0 + r2 (t)P1 + r3 (t)P2 

   1 0 0 2 1 2 = r1 (t)  0 1 0  + r2 (t)  4 2 4  0 0 1 2 1 2   12 6 12  +r3 (t) 24 12 24  , t ∈ Tκ 12 6 12

Conformable Dynamic Systems on Time Scales  137

and



   x1 (t) c1  x2 (t)  = EA (t,t0 )  c2  , x3 (t) c3

t ∈ Tκ ,

where c1 , c2 , c3 ∈ R, is a general solution of the considered system. This ends the example. Exercise 3.2.17 Using Putzer’s algorithm, find EA (t,t0 ), where 1.

 A=

2.

3.3

1 2 −1 3

 ,

 1 −1 1 0 2 . A= 1 −1 1 1 

ADVANCED PRACTICAL PROBLEMS

Problem 3.3.1 Let T = 3N0 , k1 (α,t) = (1 − α)(1 + t)α , k0 (α,t) = α(1 + t)1−α , α ∈ (0, 1],   t +1 2 t − 3t  A(t) =  t 2 + 1 , t ∈ T. t3 + t2 t4 − t Find

1

D 2 A(t),

t ∈ T,

t ∈ T.

Problem 3.3.2 Let T = 4N0 , k1 (α,t) = (1 − α)(1 + t 2 )3α , and

 A(t) =

k0 (α,t) = α(1 + t 2 )3(1−α) ,

2t + 5 t t t +2

Find



 ,

1

B(t) =

D 4 (AB)(t),

t2 − t t 1 2

t ∈ T.

Problem 3.3.3 Let T = 4N0 and 

1 A(t) =  t + 2 t +1

 1 2t + 3  , t +2

t ∈ T.

Prove that (Aσ )−1 (t) = (A−1 )σ (t),

t ∈ T.

α ∈ (0, 1],  ,

t ∈ T.

t ∈ T,

138  Conformable Dynamic Equations on Time Scales

Problem 3.3.4 Let T = 3N0 , k1 (α,t) = (1 − α)(1 + t)2α , and

 A(t) =

1 + t t2 t +3 2

k0 (α,t) = α(1 + t)2(1−α) ,



 ,

B(t) =

α ∈ (0, 1],

t + 2 3t 2 + t + 1 t + 11 4t

t ∈ T,

 ,

t ∈ T.

Find 1. (A ⊕c B)(t), t ∈ T, 2. ( c A)(t), t ∈ T, 3. ( c B)(t), t ∈ T. Problem 3.3.5 Let T = 2N0 and k1 (α,t) = (1 − α)t 2α ,

k0 (α,t) = αt 2−2α ,

Find the solution of the IVP (3.1), where     t +4 t t2 A(t) = , q(t) = , t 2 3

α ∈ (0, 1],

 y0 =

2 −1

Problem 3.3.6 Find a general solution of the system  α  D x1 (t) = −x1 (t) + x2 (t) 

Dα x2 (t) = x1 (t) − 3x2 (t).

Problem 3.3.7 Find a general solution of the system  α D x1 (t) = −x1 (t) + 2x2 (t)3 x3 (t) + x4 (t)        α   D x2 (t) = x1 (t) − 3x2 (t) + x4 (T )   Dα x3 (t) = −x1 (t) + x2 (t) + x3 (t) + x4 (t)        α D x4 (t) = x1 (t) + x4 (t). Problem 3.3.8 Using Putzer’s algorithm, find EA (t,t0 ), where 1.

 A=

2.



−1 4 1 3

 ,

 1 1 2 A =  −1 3 −2  . 4 −1 −1

t ∈ T.

 ,

t ∈ T.

CHAPTER

4

Linear Conformable Inequalities

Suppose that T is a time scale with forward jump operator and delta differentiation operator σ and ∆, respectively. Let α ∈ (0, 1], k0 and k1 satisfy (A1), and assume (1.6) holds. Also, let a, b ∈ T, a < b, and J = [a, b] be a time scale interval. If A ⊂ T, we define the sets n Rc+ = f ∈ Rc : k0 (α,t) + µ(t) ( f (t) − k1 (α,t)) > 0 f or

Rc+ (A) =

n

t ∈ T,

o α ∈ (0, 1] ,

f ∈ R(A) : k0 (α,t) + µ(t) ( f (t) − k1 (α,t)) > 0

f or

4.1

all

all

t ∈ A,

o α ∈ (0, 1] .

CONFORMABLE GRONWALL INEQUALITY

Theorem 4.1.1 (Conformable Gronwall Inequality) Let f , p, q ∈ Crd (J) be nonnegative functions such that k0 (α,t) − µ(t)(− f (t)p(t) + k1 (α,t)) 6= 0, and

t ∈ J,

k0 (α,t) − µ(t)k1 (α,t) > 0 t ∈ J. k0 (α,t) − µ(t)(−p(t) f (t) + k1 (α,t))

Also, let x ∈ Crd (J) be a nonnegative function. Then the inequality x(t) ≤ f (t)

Z t a

p(s)x(s)∆α,t s + q(t),

t ∈ J,

(4.1) 139

140  Conformable Dynamic Equations on Time Scales

implies the inequality x(t) ≤ f (t)

Z t a

where g(t) =

p(s)q(s)Eg (σ (s),t)∆α,t s + q(t),

t ∈ J,

(− f (t)p(t) + k1 (α,t))(k0 (α,t) − µ(t)k1 (α,t)) , k0 (α,t) − µ(t)(− f (t)p(t) + k1 (α,t))

Proof 4.1.2 Let

Z t

y(t) = a

p(s)x(s)∆α,t s,

t ∈ J.

t ∈ J.

Then, using (4.1), we get x(t) ≤ f (t)y(t) + q(t), We have

t ∈ J.

(4.2)

Dα y(t) = p(t)x(t) ≤ p(t)( f (t)y(t) + q(t))

(4.3)

= p(t) f (t)y(t) + p(t)q(t),

t ∈ J,

and y(a) = 0. Note that Eg (t, a) > 0,

Eg (σ (t), s) > 0,

t ∈ J.

We multiply both sides of the inequality (4.3) by Eg (σ (t), a) and we get Eg (σ (t), a)Dα y(t) ≤ p(t) f (t)Eg (σ (t), a)y(t) (4.4) +p(t)q(t)Eg (σ (t), a),

t ∈ J.

Observe that g(t)Eg (t, a) − k1 (α,t)Eg (σ (t), a)   g(t) − k1 (α,t) = g(t)Eg (t, a) − k1 (α,t) 1 + µ(t) Eg (t, a) k0 (α,t)    g(t) − k1 (α,t) = g(t) − k1 (α,t) 1 + µ(t) Eg (t, a), t ∈ J, k0 (α,t) and g(t) − k1 (α,t) =

(− f (t)p(t) + k1 (α,t))(k0 (α,t) − µ(t)k1 (α,t)) − k1 (α,t) k0 (α,t) − µ(t)(− f (t)p(t) + k1 (α,t))

Linear Conformable Inequalities  141

=

 1 − f (t)p(t)k0 (α,t) k0 (α,t) − µ(t)(− f (t)p(t) + k1 (α,t)) +µ(t) f (t)p(t)k1 (α,t) + k0 (α,t)k1 (α,t) − µ(t)(k0 (α,t))2 2



−k0 (α,t)k1 (α,t) − µ(t) f (t)p(t)k1 (α,t) + µ(t)(k1 (α,t)) = −

f (t)p(t)k0 (α,t) , k0 (α,t) − µ(t)(− f (t)p(t) + k1 (α,t))

t ∈ J,

and g(t)Eg (t, a) − k1 (α,t)Eg (σ (t), a)   g(t) − k1 (α,t) = g(t)Eg (t, a) − k1 (α,t) 1 + µ(t) Eg (t, a) k0 (α,t)   g(t) − k1 (α,t) = g(t) − k1 (α,t) − µ(t)k1 (α,t) Eg (t, a) k0 (α,t) k0 (α,t) − µ(t)k1 (α,t) Eg (t, a) k0 (α,t)   k0 (α,t) − µ(t)k1 (α,t) f (t)p(t)k0 (α,t) Eg (t, a) = − k0 (α,t) − µ(t)(− f (t)p(t) + k1 (α,t)) k0 (α,t) = (g(t) − k1 (α,t))

= −

f (t)p(t)(k0 (α,t) − µ(t)k1 (α,t)) Eg (t, a), k0 (α,t) − µ(t)(− f (t)p(t) + k1 (α,t))

t ∈ J,

and   g(t) − k1 (α,t) − f (t)p(t)Eg (σ (t), a) = − f (t)p(t) 1 + µ(t) Eg (t, a) k0 (α,t)   µ(t) f (t)p(t)k0 (α,t) = − f (t)p(t) 1 − Eg (t, a) k0 (α,t) (k0 (α,t) − µ(t)(− f (t)p(t) + k1 (α,t)))   µ(t) f (t)p(t) = − f (t)p(t) 1 − Eg (t, a) k0 (α,t) − µ(t)(− f (t)p(t) + k1 (α,t)) = − f (t)p(t) = −

k0 (α,t) + µ(t) f (t)p(t) − µ(t)k1 (α,t) − µ(t) f (t)p(t) Eg (t, a) k0 (α,t) − µ(t)(− f (t)p(t) + k1 (α,t))

f (t)p(t) (k0 (α,t) − µ(t)k1 (α,t)) Eg (t, a), k0 (α,t) − µ(t) (− f (t)p(t) + k1 (α,t))

t ∈ J.

Therefore g(t)Eg (t, a) − k1 (α,t)Eg (σ (t), a) = − f (t)p(t)Eg (σ (t), a),

t ∈ J.

142  Conformable Dynamic Equations on Time Scales

Hence, and considering (4.4), we get Eg (σ (t), a)Dα y(t) ≤ (−g(t)Eg (t, a) + k1 (α,t)Eg (σ (t), a)) y(t) +p(t)q(t)Eg (σ (t), a),

t ∈ J,

or Eg (σ (t), a)Dα y(t) + g(t)Eg (t, a)y(t) − k1 (α,t)Eg (σ (t), a)y(t) ≤ p(t)q(t)Eg (σ (t), a),

t ∈ J,

or Dα (Eg (t, a)y(t)) ≤ p(t)q(t)Eg (σ (t), a),

t ∈ J.

From the last inequality, we obtain y(t)Eg (t, a) ≤

Z t a

p(s)q(s)Eg (σ (s), a)∆α,t s,

t ∈ J,

whereupon y(t) ≤ Eg (a,t)

Z t a

p(s)q(s)Eg (σ (s), a)∆α,t s

Z t

= a

p(s)q(s)Eg (σ (s),t)∆α,t s,

t ∈ J.

Hence, and considering (4.2), we obtain x(t) ≤ ≤

f (t)y(t) + q(t) Z t

f (t) a

p(s)q(s)Eg (σ (s),t)∆α,t s + q(t),

t ∈ J. 

This completes the proof. Theorem 4.1.3 Let f , g, h, p ∈ Crd (J) be nonnegative functions, gh ∈ Rc (J), and k0 (α,t) − µ(t)k1 (α,t) > 0, k0 (α,t) − µ(t)(−h(t)g(t) + k1 (α,t))

t ∈ J.

Then the inequality x(t) ≤ f (t) + g(t)

Z t a

(h(s)x(s) + p(s))∆α,t s,

t ∈ J,

implies the inequality Z t

(h(s) f (s) + p(s))Eq (σ (s),t)∆α,t s,

t ∈ J,

(−h(t)g(t) + k1 (α,t))(k0 (α,t) − µ(t)k1 (α,t)) , k0 (α,t) − µ(t)(−h(t)g(t) + k1 (α,t))

t ∈ J.

x(t) ≤ f (t) + g(t)

a

where q(t) =

Linear Conformable Inequalities  143

Proof 4.1.4 Note that Eq (σ (t), a) > 0,

t ∈ J.

(h(s)x(s) + p(s))∆α,t s,

t ∈ J.

Eq (t, a) > 0, Let

Z t

y(t) = a

Then y(a) = 0 and x(t) ≤ f (t) + g(t)y(t),

t ∈ J.

(4.5)

Also, Dα y(t) = h(t)x(t) + p(t) ≤ h(t)( f (t) + g(t)y(t)) + p(t) = h(t) f (t) + p(t) + h(t)g(t)y(t),

t ∈ J,

whereupon Eq (σ (t), a)Dα y(t) ≤ (h(t) f (t) + p(t))Eq (σ (t), a) + h(t)g(t)y(t)Eq (σ (t), a), t ∈ J. Observe that −q(t)Eq (t, a) + k1 (α,t)Eq (σ (t), a) = = = =

  q(t) − k1 (α,t) Eq (t, a) −q(t)Eq (t, a) + k1 (α,t) 1 + µ(t) k0 (α,t)   q(t) − k1 (α,t) Eq (t, a) −(q(t) − k1 (α,t)) + µ(t)k1 (α,t) k0 (α,t)   µ(t)k1 (α,t) (q(t) − k1 (α,t)) − 1 Eq (t, a) k0 (α,t)   µ(t)k1 (α,t) − k0 (α,t) (q(t) − k1 (α,t)) Eq (t, a), t ∈ J, k0 (α,t)

and (−h(t)g(t) + k1 (α,t))(k0 (α,t) − µ(t)k1 (α,t)) − k1 (α,t) k0 (α,t) − µ(t)(−h(t)g(t) + k1 (α,t))  1 − h(t)g(t)k0 (α,t) k0 (α,t) − µ(t)(−h(t)g(t) + k1 (α,t))

q(t) − k1 (α,t) = =

+h(t)g(t)µ(t)k1 (α,t) + k1 (α,t)k0 (α,t) − µ(t)(k1 (α,t))2

144  Conformable Dynamic Equations on Time Scales 2



−k1 (α,t)k0 (α,t) − µ(t)k1 (α,t)h(t)g(t) + µ(t)(k1 (α,t)) = −

h(t)g(t)k0 (α,t) , k0 (α,t) − µ(t)(−h(t)g(t) + k1 (α,t))

t ∈ J.

Thus, −q(t)Eq (t, a) + k1 (α,t)Eq (σ (t), a) h(t)g(t)k0 (α,t) = − k0 (α,t) − µ(t)(−h(t)g(t) + k1 (α,t)) = −



 µ(t)k1 (α,t) − k0 (α,t) Eq (t, a) k0 (α,t)

h(t)g(t)(µ(t)k1 (α,t) − k0 (α,t)) Eq (t, a), k0 (α,t) − µ(t)(−h(t)g(t) + k1 (α,t))

t ∈ J.

Next,   q(t) − k1 (α,t) h(t)g(t)Eq (t, a) = h(t)g(t) 1 + µ(t) Eq (t, a) k0 (α,t)   k0 (α,t) + µ(t)q(t) − µ(t)k1 (α,t) Eq (t, a) = h(t)g(t) k0 (α,t)  h(t)g(t) k0 (α,t) − µ(t)k1 (α,t) = k0 (α,t)  (−h(t)g(t) + k1 (α,t))(k0 (α,t) − µ(t)k1 (α,t)) +µ(t) Eq (t, a) k0 (α,t) − µ(t)(−h(t)g(t) + k1 (α,t))  h(t)g(t)(k0 (α,t) − µ(t)k1 (α,t)) = k0 (α,t) k0 (α,t)(k0 (α,t) − µ(t)(−h(t)g(t) + k1 (α,t)))  −µ(t)(−h(t)g(t) + k1 (α,t)) + µ(t)(−h(t)g(t) + k1 (α,t)) Eq (t, a) =

h(t)g(t)(k0 (α,t) − µ(t)k1 (α,t)) Eq (t, a), (k0 (α,t) − µ(t)(−h(t)g(t) + k1 (α,t)))

t ∈ J.

Therefore −q(t)Eq (t, a) + k1 (α,t)Eq (σ (t), a) = h(t)g(t)Eq (σ (t), a),

t ∈ J.

Hence, Eq (σ (t), a)Dα y(t) ≤ −q(t)y(t)Eq (t, a)+k1 (α,t)Eq (σ (t), a)+(h(t) f (t)+ p(t))Eq (σ (t), a), t ∈ J, or Eq (σ (t), a)Dα y(t) + q(t)y(t)Eq (t, a) − k1 (α,t)y(t) Eq (σ (t), a) ≤ (h(t) f (t) + p(t))Eq (σ (t), a),

Linear Conformable Inequalities  145

t ∈ J, or Dα (Eq (t, a)y(t)) ≤ (h(t) f (t) + p(t)) Eq (σ (t), a),

t ∈ J.

Therefore Eq (t, a)y(t) ≤

Z t a

t ∈ J,

(h(s) f (s) + p(s)) Eq (σ (s), a)∆α,t s,

and y(t) ≤ Eq (a,t)

Z t a

(h(s) f (s) + p(s)) Eq (σ (s), a)∆α,t s

Z t

= a

t ∈ J.

(h(s) f (s) + p(s)) Eq (σ (s),t)∆α,t s,

Hence, and considering (4.5), we get x(t) ≤ f (t) + g(t)

Z t a

(h(s) f (s) + p(s)) Eq (σ (s),t)∆α,t s,

t ∈ J. 

This completes the proof.

Theorem 4.1.5 Let 0 ∈ Rc+ , assume y, h and v are rd-continuous nonnegative functions on J, and assume k is an rd-continuous positive function on J such that h(x)k0 (α, x) k1 (α, x) − > 0, k0 (α, x) + µ(x)(k(t)h(x) − k1 (α, x)) k(t)

a ≤ x ≤ t ≤ b.

If k(t) y(t) ≥ v(x) − E0 (x,t)

Z t x

 h(s) k1 (α, s) ∆α,t s, − k0 (α, s) + µ(s)(k(t)h(s) − k1 (α, s)) k(t)

a ≤ x ≤ t ≤ b, then y(t) ≥ Ek(t)h (x,t)v(x), where Ek(t)h (x,t) = e

a ≤ x ≤ t ≤ b,

  Rx 1 k(t)h(s)−k1 (α,s) ∆s t µ(s) log 1+µ(s) k (α,s) 0

,

a ≤ x ≤ t ≤ b. Proof 4.1.6 Since 0 ∈ Rc+ , we have k0 (α, x) − µ(x)k1 (α, x) > 0,

a ≤ x ≤ t ≤ b,

whereupon k0 (α,t) + µ(x)(k(t)h(x) − k1 (α, x)) > 0,

a ≤ x ≤ t ≤ b.

Let z(x) = y(t) +

k(t) E0 (x,t)

Z t x

 h(s) k1 (α, s) − ∆α,t s, k0 (α, s) + µ(s)(k(t)h(s) − k1 (α, s)) k(t)

146  Conformable Dynamic Equations on Time Scales

a ≤ x ≤ t ≤ b. Then z(x) ≥ v(x),

a ≤ x ≤ t ≤ b,

and z(x) = y(t) − k(t)

Z x t

 h(s) k1 (α, s) − ∆α,x s, k0 (α, s) + µ(s)(k(t)h(s) − k1 (α, s)) k(t)

a ≤ x ≤ t ≤ b, and 

 h(x)k0 (α, x) k1 (α, x) D z(x) = −k(t) − v(x) k0 (α, x) + µ(x)(k(t)h(x) − k1 (α, x)) k(t)   h(x)k0 (α, x) k1 (α, x) ≥ −k(t) − z(x), k0 (α, x) + µ(x)(k(t)h(x) − k1 (α, x)) k(t) α

a ≤ x ≤ t ≤ b. Hence, Ek(t)h (σ (x),t)Dα z(x) ≥ −k(t)

h(x)k0 (α, x)Ek(t)h (σ (x),t) z(x) k0 (α, x) + µ(x)(k(t)h(x) − k1 (α, x))

+k1 (α, x)z(x)Ek(t)h (σ (x),t) = −k(t)h(x)Ek(t)h (x,t)z(x) +k1 (α, x)Ek(t)h (σ (x),t)z(x), a ≤ x ≤ t ≤ b, or Ek(t)h (σ (x),t)Dα z(x) + k(t)h(x)Ek(t)h (x,t)z(x) − k1 (α, x)Ek(t)h (σ (x),t)z(x) ≥ 0, a ≤ x ≤ t. Therefore
Dαx Ek(t)h (·,t)z(·) (x) ≥ 0,

a ≤ x ≤ t ≤ b.

From the last inequality, we obtain z(t) ≥ Ek(t)h (x,t)z(x),

a ≤ x ≤ t ≤ b.

Since z(t) = y(t), z(x) ≥ v(x),

a ≤ x ≤ t ≤ b,

we get y(t) ≥ Ek(t)h (x,t)v(x), This completes the proof.

a ≤ x ≤ t ≤ b. 

Linear Conformable Inequalities  147

Theorem 4.1.7 (Gronwall-Type Inequality) Let 0 ∈ Rc+ , y, f , g : J → R be nonnegative rd-continuous functions on J, − f ∈ Rc+ . If Dα y(t) ≤ f (t)yσ (t) + g(t), then y(t) ≤ y(a)E0 c (− f ) (t, a) +

t ∈ J,

Z t

E− f (s,t)g(s)∆α,t s,

a

t ∈ J.

Proof 4.1.8 We have E− f (t, a) > 0,

t ∈ J.

Then Dα y(t)E− f (t, a) − f (t)(E− f (t, a)yσ (t) ≤ E− f (t, a)g(t),

t ∈ J,

whereupon Dα y(t)E− f (t, a) − f (t)(E− f (t, a)yσ (t) − k1 (α,t)E− f (t, a)yσ (t) ≤ E− f (t, a)g(t),

t ∈ J,

or
Dα yE− f (·, a) (t) ≤ E− f (t, a)g(t), Hence, y(t)E− f (t, a) − y(a)E0 (t, a) ≤ or y(t)E− f (t, a) ≤ y(a)E0 (t, a) + and

E0 (t, a) + y(t) ≤ y(a) E− f (t, a)

t ∈ J.

Z t a

E− f (s, a)g(s)∆α,t s,

t ∈ J,

E− f (s, a)g(s)∆α,t s,

t ∈ J,

Z t a

Z t a

E− f (a,t)E− f (s, a)g(s)∆α,t s,

or y(t) ≤ y(a)E0 c (− f ) (t, a) +

Z t a

E− f (s,t)g(s)∆α,t s,

t ∈ J,

t ∈ J. 

This completes the proof.

Theorem 4.1.9 (Gronwall-Type Inequality) Let 0 ∈ Rc+ , y, f , g : J → R be nonnegative rd-continuous functions on J, − f ∈ Rc+ . If Dα y(t) ≤ f (t)y(t) + g(t),

t ∈ J,

then y(t) ≤ y(a)E0 c (−h) (t, a) +

Z t g(s)(k0 (α, s) − µ(s)k1 (α, s)) a

k0 (α, s) − µ(s)( f (s) + k1 (α, s))

where h(t) =

E− f (s,t)∆α,t s,

k0 (α,t) f (t) , k0 (α,t) − µ(t)( f (t) + k1 (α,t))

t ∈ J.

t ∈ J,

148  Conformable Dynamic Equations on Time Scales

Proof 4.1.10 We have Dα y(t) ≤

f (t)y(t) + g(t) k0 (α,t) f (t) yσ (t) k0 (α,t) − µ(t)k1 (α,t)

=

−

µ(t) f (t) Dα y(t) + g(t), k0 (α,t) − µ(t)k1 (α,t)

t ∈ J,

whereupon k0 (α,t) − µ(t)( f (t) + k1 (α,t)) α k0 (α,t) f (t) D y(t) ≤ yσ (t) + g(t), k0 (α,t) − µ(t)k1 (α,t) k0 (α,t) − µ(t)k1 (α,t)

t ∈ J,

or Dα y(t) ≤

k0 (α,t) f (t) yσ (t) k0 (α,t) − µ(t)( f (t) + k1 (α,t)) +

g(t)(k0 (α,t) − µ(t)k1 (α,t)) k0 (α,t) − µ(t)( f (t) + k1 (α,t))

= h(t)yσ (t) +

g(t)(k0 (α,t) − µ(t)k1 (α,t)) , k0 (α,t) − µ(t)( f (t) + k1 (α,t))

t ∈ J.

From the last inequality and from Theorem 4.1.7, we get the desired result. This completes the proof. 

4.2

CONFORMABLE VOLTERRA-TYPE INTEGRAL INEQUALITIES

Theorem 4.2.1 Let x, f ∈ Crd (J), k ∈ Crd (J × J) be nonnegative on J × J, and x(t) ≤ f (t) +

Z t a

k(t, s)x(s)∆α,t s,

t ∈ J.

(4.6)

Let also, k1 (t, s) = k(t, s), Z t

kl (t, s) =

a

k(t, s1 )kl−1 (s1 , s)∆s1 ,

l ∈ N,

Suppose that ∞

H(t, s) =

∑ kn (t, s)

n=1

l ≥ 2,

a ≤ s ≤ t ≤ b.

Linear Conformable Inequalities  149

is a uniformly convergent series on a ≤ s ≤ t ≤ b. Then x(t) ≤ f (t) +

Z t a

H(t, s) f (s)∆α,t s,

t ∈ J.

(4.7)

Proof 4.2.2 We have x(t) ≤

Z t

f (t) + a

≤

Z t

f (t) + a

k(t, s)x(s)∆α,t s   Z s k(t, s) f (s) + k(s, s1 )x(s1 )∆α,s s1 ∆α,t s a

Z t

=

f (t) + a

k(t, s) f (s)∆α,t s

Z tZ s

+ a

k(t, s)k(s, s1 )x(s1 )∆α,s s1 ∆α,t s

a

Z t

=

f (t) + a

+

k(t, s) f (s)∆α,t s

Z t Z s1 a

a

 k(t, s)k(s, s1 )∆α,s1 s x(s1 )∆α,t s1

Z t

=

f (t) + a

Z t

k1 (t, s) f (s)∆α,t s +

a

k2 (t, s)x(s)∆α,t s,

t ∈ J.

Hence, for t ∈ J, we get x(t) ≤

Z t

f (t) + a

≤

Z t

f (t) + a

Z s

+ a

k(t, s)x(s)∆α,t s Z s  k(t, s) f (s) + k1 (s, s1 ) f (s1 )∆α,s s1 a

 k2 (s, s1 )x(s1 )∆α,t s1 ∆α,t s Z t

=

f (t) + a

k(t, s) f (s)∆α,t s

Z tZ s

+ a

k(t, s)k1 (s, s1 ) f (s1 )∆α,s s1 ∆α,t s

a

Z tZ s

+ a

a

k(t, s)k2 (s, s1 )x(s1 )∆α,s s1 ∆α,t s

Z t

=

f (t) + a

k1 (t, s) f (s)∆α,t s

150  Conformable Dynamic Equations on Time Scales

+

Z t Z s1 a

a

Z t Z

s1

+ a

a

 k(t, s)k1 (s, s1 )∆α,s1 s f (s1 )∆α,t s1  k(t, s)k2 (s, s1 )∆α,s1 s x(s1 )∆α,t s1

Z t

=

f (t) +

Z t

k1 (t, s) f (s)∆α,t s +

a

a

k2 (t, s) f (s)∆α,t s

Z t

+

k3 (t, s)x(s)∆α,t s

a

Z t

=

f (t) + a

(k1 (t, s) + k2 (t, s)) f (s)∆α,t s

Z t

+

k3 (t, s)x(s)∆α,t s.

a

Assume that n

x(t) ≤ f (t) + ∑

Z t

Z t

kl (t, s) f (s)∆α,t s +

l=1 a

a

kn+1 (t, s)x(s)∆α,t s,

t ∈ J,

kn+2 (t, s)x(s)∆α,t s,

t ∈ J.

for some n ∈ N. We will prove that n+1 Z t

x(t) ≤ f (t) + ∑

Z t

kl (t, s) f (s)∆α,t s +

l=1 a

a

Really, for t ∈ J, we have x(t) ≤

Z t

f (t) + a

≤

Z t

f (t) + a

Z s

+ a

k(t, s)x(s)∆α,t s  n Z s k(t, s) f (s) + ∑ kl (s, s1 ) f (s1 )∆α,s s1 l=1 a

 kn+1 (s, s1 )x(s1 )∆α,s s1 ∆α,t s Z t

=

f (t) + a n

k(t, s) f (s)∆α,t s

Z tZ s

+∑

l=1 a

a

k(t, s)kl (s, s1 ) f (s1 )∆α,s s1 ∆α,t s

Z tZ s

+ a

a

k(t, s)kn+1 (s, s1 )x(s1 )∆α,s s1 ∆α,t s

Z t

=

f (t) + a

k(t, s) f (s)∆α,t s

(4.8)

Linear Conformable Inequalities  151 n

Z t Z s1

+∑ +

 k(t, s)kl (s, s1 )∆α,s1 s f (s1 )∆α,t s1

l=1 a

a

Z t Z s1

 k(t, s)kn+1 (s, s1 )∆α,s1 s x(s1 )∆α,t s1

a

a

n

Z t

=

f (t) +

Z t

k1 (t, s) f (s)∆α,t s + ∑

a

l=1 a

kl+1 (t, s) f (s)∆α,t s

Z t

+

kn+2 (t, s)x(s)∆α,t s

a

Z t

=

f (t) +

k1 (t, s) f (s)∆α,t s

a n+1 Z t

+∑

l=2 a

kl (t, s) f (s)∆α,t s

Z t

+ a

kn+2 (t, s)x(s)∆α,t s n+1 Z t

=

f (t) + ∑

l=1 a

Z t

kl (t, s) f (s)∆α,t s +

a

kn+2 (t, s)x(s)∆α,t s.

Therefore (4.8) holds for any n ∈ N. Letting n → ∞ into (4.8) and using that H(t, s) = ∞

∑ kn (t, s) is a uniformly convergent series on a ≤ s ≤ t ≤ b, we get the inequality (4.7).

n=1



This completes the proof.

Theorem 4.2.3 Assume that 0 ∈ Rc+ . Let k : J × J → R be a nonnegative continuous function which is nondecreasing with respect to its first argument and −k ∈ Rc+ . Also, let c be a nonnegative constant. If y(t) ≤ c +

Z t a

k(t, s)k0 (α, s) y(s)∆α,t s, k0 (α, s) − µ(s)(k(t, s) + k1 (α, s))

t ∈ J,

then y(t) ≤ cE0 c (−k(b,·)) (t, a),

t ∈ J.

Proof 4.2.4 Since k is nondecreasing with respect to its first argument, we have k(t, s) ≤ k(b, s), k(t, s) + k1 (α, s) ≤ k(b, s) + k1 (α, s), −µ(s)(k(t, s) + k1 (α, s)) ≥ −µ(s)(k(b, s) + k1 (α, s)),

152  Conformable Dynamic Equations on Time Scales

k0 (α, s) − µ(s)(k(t, s) + k1 (α, s)) ≥ k0 (α, s) − µ(s)(k(b, s) + k1 (α, s)), t, s ∈ J, and k(b, s)k0 (α, s) k(t, s)k0 (α, s) ≤ , k0 (α, s) − µ(s)(k(t, s) + k1 (α, s)) k0 (α, s) − µ(s)(k(b, s) + k1 (α, s))

t, s ∈ J.

From here, y(t) ≤ c +

Z t a

k(b, s)k0 (α, s) y(s)∆α,t s, k0 (α, s) − µ(s)(k(b, s) + k1 (α, s))

t ∈ J.

Define the function z : J → R as follows. Z t

z(t) = c + a

k(b, s)k0 (α, s) y(s)∆α,t s, k0 (α, s) − µ(s)(k(b, s) + k1 (α, s))

t ∈ J.

Then y(t) ≤ z(t),

t ∈ J,

(4.9)

and Dα z(t) ≤

k(b,t)k0 (α,t) y(t) k0 (α,t) − µ(t)(k(b,t) + k1 (α,t))

≤

k(b,t)k1 (α,t) z(t), k0 (α,t) − µ(t)(k(b,t) + k1 (α,t))

t ∈ J.

Hence, E−k(b,·) (σ (t), a)Dα z(t) −

k(b,t)k0 (α,t)E−k(b,·) (σ (t), a) z(t) ≤ 0, k0 (α,t) − µ(t)(k(b,t) + k1 (α,t))

t ∈ J,

or E−k(b,·) (σ (t), a)Dα z(t) − k(b,t)E−k(b,·) (t, a)z(t) ≤ 0,

t ∈ J,

whereupon E−k(b,·) (σ (t), a)Dα z(t)−k(b,t)E−k(b,·) (t, a)z(t)−k1 (α,t)E−k(b,·) (σ (t), a)z(t) ≤ 0, Therefore
Dα E−k(b,·) (t, a)z(t) ≤ 0,

t ∈ J,

and E−k(b,·) (t, a)z(t) − z(a)E0 (t, a) ≤ 0,

t ∈ J,

or E−k(b,·) (t, a)z(t) ≤ cE0 (t, a), or z(t) ≤ c

E0 (t, a) , E−k(b,·) (t, a)

t ∈ J,

t ∈ J,

t ∈ J.

Linear Conformable Inequalities  153

or z(t) ≤ cE0 c (−k(b,·)) (t, a),

t ∈ J.

From the last inequality and from (4.9), we get y(t) ≤ cE0 c (−k(b,·)) (t, a),

t ∈ J. 

This completes the proof.

Theorem 4.2.5 Let 0 ∈ Rc+ , k be as in Theorem 4.2.3, and g : J → R be a positive rdcontinuous and nondecreasing function on J. If y(t) ≤ g(t) +

Z t

k(t, s)k0 (α, s) y(s)∆α,t s, k0 (α, s) − µ(s)(k(t, s) + k1 (α, s))

a

t ∈ J,

(4.10)

then y(t) ≤ g(t)E0 c (−k(b,·)) (t, a),

t ∈ J.

Proof 4.2.6 By the inequality (4.10), using that g is a nondecreasing function on J, we obtain y(t) g(t)

≤ 1+

Z t a

≤ 1+

Z t a

y(s) k(t, s)k0 (α, s) ∆α,t s k0 (α, s) − µ(s)(k(t, s) + k1 (α, s)) g(t) y(s) k(t, s)k0 (α, s) ∆α,t s, k0 (α, s) − µ(s)(k(t, s) + k1 (α, s)) g(s)

Let x(t) = Then x(t) ≤ 1 +

Z t a

y(t) , g(t)

t ∈ J.

t ∈ J.

k(t, s)k0 (α, s) x(s)∆α,t s, k0 (α, s) − µ(s)(k(t, s) + k1 (α, s))

t ∈ J.

Hence, and considering Theorem 4.2.3, we get x(t) ≤ E0 c (−k(b,·)) (t, a),

t ∈ J,

whereupon y(t) ≤ g(t)E0 c (−k(b,·)) (t, a),

t ∈ J. 

This completes the proof.

4.3

CONFORMABLE INEQUALITIES OF GAMIDOV AND RODRIGUES

Theorem 4.3.1 Let 0 ∈ Rc+ , f , gi , hi , i ∈ {1, . . . , n}, be nonnegative rd-continuous functions on J, and let g(t) =

sup i∈{1,...,n}

{gi (t)} ,

154  Conformable Dynamic Equations on Time Scales n

h(t) =

∑ hi (t),

t ∈ J.

i=1

Also, let gh ∈ Rc+ . If n

y(t) ≤ f (t) + ∑ gi (t)

Z t a

i=1

then y(t) ≤ f (t) + g(t)

p(t) =

t ∈ J,

Z t

h(s) f (s)E p (σ (s),t)∆α,t s,

a

where

hi (s)y(s)∆α,t s,

t ∈ J,

(−g(t)h(t) + k1 (α,t))(k0 (α,t) − µ(t)k1 (α,t)) , k0 (α,t) + µ(t)(g(t)h(t) − k1 (α,t))

t ∈ J.

Proof 4.3.2 We have y(t) ≤

Z t

n

a

i=1

f (t) + g(t)

!

∑ hi (s)

y(s)∆α,t s

Z t

=

f (t) + g(t) a

h(s)y(s)∆α,t s,

t ∈ J.

Hence, by Theorem 4.1.1, we get the desired result. This completes the proof.



Theorem 4.3.3 Let 0 ∈ Rc+ , y, f , g1 , g2 , hi , i ∈ {1, . . . , n}, be nonnegative rd-continuous functions on J, h1 g1 ∈ Rc+ , a = t1 ≤ t2 ≤ . . . ≤ tn = b, ci , i ∈ {1, . . . , n}, be nonnegative constants, n

f1 (t) =

f (t) + g2 (t) ∑ mi , i=2

p(t) =

(−g1 (t)h1 (t) + k1 (α,t))(k0 (α,t) − µ(t)k1 (α,t)) , k0 (α,t) + µ(t)(g1 (t)h1 (t) − k1 (α,t)) Z t

q1 (t) =

f1 (t) + g1 (t)

t1

h1 (s) f (s)E p (σ (s),t)∆α,t s,

Z t

q2 (t) =

t1

h1 (s)g2 (s)E p (σ (s),t)∆α,t s,

n

1 >

Z ti

∑ ci

t1

i=2

hi (s)q2 (s)∆α,ti s,

n

1 − ∑ ci

M =

i=2

n

×

∑ ci

i=2

!−1

Z ti t1

Z ti t1

hi (s)q2 (s)∆α,ti s !

hi (s)q1 (s)∆α,t s ,

t ∈ J.

Linear Conformable Inequalities  155

If Z t

y(t) ≤

f (t) + g1 (t)

t1

n

h1 (s)y(s)∆α,t s

Z ti

+g2 (t) ∑ ci

t ∈ J,

hi (s)y(s)∆α,t s,

t1

i=2

then y(t) ≤ q1 (t) + Mq2 (t),

t ∈ J.

Proof 4.3.4 Set Z ti

mi = ci

t1

i ∈ {1, . . . , n},

hi (s)y(s)∆α,ti s,

i ∈ {1, . . . , n}.

Then n

y(t) ≤

Z t

f (t) + g2 (t) ∑ mi + g1 (t) i=2

t1

h1 (s)y(s)∆α,t s

Z t

=

f1 (t) + g1 (t)

t1

t ∈ J.

h1 (s)y(s)∆α,t s,

Hence, and considering Theorem 4.1.1, we get Z t

y(t) ≤

f1 (t) + g1 (t)

t1

h1 (s) f1 (s)E p (σ (s),t)∆α,t s n

Z t

=

f1 (t) + g1 (t)

t1

h1 (s)

!

f (s) + g2 (s) ∑ mi E p (σ (s),t)∆α,t s i=2

Z t

=

f1 (t) + g1 (t) n

t1

h1 (s) f (s)E p (σ (s),t)∆α,t s

Z t

+g1 (t) ∑ mi i=2

t1

h1 (s)g2 (s)E p (σ (s),t)∆α,t s

n

= q1 (t) + ∑ mi q2 (t),

t ∈ J.

i=2

Since n

∑ mi

n

=

i=2

=

=

∑ ci

Z ti

i=2

t1

n

Z ti

∑ ci

i=2

t1

n

Z ti

∑ ci

i=2

t1

hi (s)y(s)∆α,ti s n

hi (s)

!

∑ mi q2 (s) + q1 (s)

i=2

hi (s)q1 (s)∆α,ti s

∆α,ti s

156  Conformable Dynamic Equations on Time Scales

n

n

+ ∑ mi

∑ ci

hi (s)q2 (s)∆α,ti s ,

t1

i=2

i=2

!

Z ti

whereupon n

n

∑ mi 1 − ∑ ci

i=2

!

Z ti t1

i=2

hi (s)q2 (s)∆α,ti s

n

Z ti

= ∑ ci i=2

hi (s)q1 (s)∆α,ti s

t1

and n

∑ mi

n

1 − ∑ ci

=

i=2

i=2

n

×

t1

hi (s)q2 (s)∆α,ti s !

Z ti

∑ ci

i=2

!−1

Z ti

t1

hi (s)q1 (s)∆α,t s

= M. Therefore y(t) ≤ q1 (t) + Mq2 (t),

t ∈ J. 

This completes the proof.

4.4

SIMULTANEOUS CONFORMABLE INTEGRAL INEQUALITIES

Theorem 4.4.1 Let 0 ∈ Rc+ , x, y, a, b, p, hi : J → R, i ∈ {1, 2, 3, 4}, be nonnegative rdcontinuous functions on J, h(t) = max{h1 (t) + h3 (t), h2 (t) + h4 (t)}, c(t) = a(t) + b(t), g(t) =

(−p(t)h(t) + k1 (α,t))(k0 (α,t) − µ(t)k1 (α,t)) , k0 (α,t) + µ(t)(p(t)h(t) − k1 (α,t))

t ∈ J.

If x(t) ≤ a(t) + p(t)

Z

t

h1 (s)x(s)∆α,t s +

a

y(t) ≤ a(t) + p(t)

Z a

Z t

t

a

Z t

h3 (s)x(s)∆α,t s +

a

 h2 (s)y(s)∆α,t s  h4 (s)y(s)∆α,t s ,

then x(t) ≤ c(t) + p(t)

Z t a

y(t) ≤ c(t) + p(t)

h(s)c(s)Eg (σ (s),t)∆α,t s,

Z t a

h(s)c(s)Eg (σ (s),t)∆α,t s,

t ∈ J.

t ∈ J,

Linear Conformable Inequalities  157

Proof 4.4.2 Let z(t) = x(t) + y(t),

t ∈ J.

Then z(t) = x(t) + y(t) Z t ≤ a(t) + b(t) + p(t) (h1 (s) + h3 (s))x(s)∆α,t s a

Z t

+ a

 (h2 (s) + h4 (s))y(s)∆α,t s

≤ c(t) + p(t)

t

Z

Z t

h(s)x(s)∆α,t s +

a

Z t

= c(t) + p(t)

h(s)z(s)∆α,t s,

a

a

 h(s)y(s)∆α,t s

t ∈ J.

Hence, and considering Theorem 4.1.1, it follows that z(t) ≤ c(t) + p(t)

Z t a

h(s)c(s)Eg (σ (s),t)∆α,t s,

t ∈ J. 

This completes the proof.

4.5

CONFORMABLE PACHPATTE’S INEQUALITIES

Theorem 4.5.1 Let 0 ∈ Rc+ , y0 ≥ 0, y, f , g : J → R be nonnegative rd-continuous functions on J and −( f + g) ∈ Rc+ . If y(t) ≤ y0 +

Z t a

f (s)y(s)∆α,t s

Z t

+

Z f (s)

 g(τ)y(τ)∆α,s τ ∆α,t s,

a

a

then

s

  Z t y(t) ≤ y0 E0 (t, a) + f (s)E0 c (−h) (s, a)∆α,t s ,

t ∈ J,

t ∈ J,

a

where h(t) =

k0 (α,t)( f (t) + g(t)) , k0 (α,t) − µ(t)( f (t) + g(t) + k1 (α,t))

t ∈ J.

Proof 4.5.2 Let Z t

z(t) = y0 +

a

f (s)y(s)∆α,t s

Z t

+

Z f (s)

a

a

s

 g(τ)y(τ)∆α,s τ ∆α,t s,

t ∈ J.

158  Conformable Dynamic Equations on Time Scales

Then y(t) ≤ z(t),

t ∈ J,

and Z t

α

D z(t) =

f (t)y(t) + f (t) a

g(s)y(s)∆α,t s

=

  Z t f (t) y(t) + g(s)y(s)∆α,t s

≤

  Z t f (t) z(t) + g(s)z(s)∆α,t s ,

a

t ∈ J.

a

Let

Z t

w(t) = z(t) +

t ∈ J.

g(s)z(s)∆α,t s,

a

Then w(a) = z(a) = y0 , Dα z(t) ≤

f (t)w(t),

Dα w(t) = Dα z(t) + g(t)z(t) ≤

f (t)w(t) + g(t)w(t) t ∈ J.

= ( f (t) + g(t))w(t),

Hence, by the Gronwall inequality and Theorem 4.1.7, we get w(t) ≤ w(a)E0 c (−h) (t, a) = y0 E0 c (−h) (t, a),

t ∈ J.

Then Dα z(t) ≤

f (t)w(t)

≤ y0 f (t)E0 c (−h) (t, a),

t ∈ J.

Now we integrate the last inequality from a to t and we find z(t) ≤ z(a)E0 (t, a) + y0

Z t a

f (s)E0 c (−h) (s, a)∆α,t s

Linear Conformable Inequalities  159

Z t

= y0 E0 (t, a) + y0 and

f (s)E0 c (−h) (s, a)∆α,t s,

a

  Z t f (s)E0 c (−h) (s, a)∆α,t s , y(t) ≤ y0 1 +

t ∈ J,

t ∈ J.

a



This completes the proof.

Theorem 4.5.3 Let 0 ∈ Rc+ , f , g, h, y : J → R be nonnegative rd-continuous functions on J, y0 ≥ 0 and −( f + g + h) ∈ Rc+ . If y(t) ≤ y0 +

Z t a

f (s)y(s)∆α,t s

Z t

+

Z f (s)

Z t

+

Z

 g(τ)y(τ)∆α,s τ ∆α,st

a

a

s

f (s)

Z



τ

g(τ) a

a

then

s



h(l)y(l)∆α,τ l ∆α,s τ ∆α,t s,

a

  Z t y(t) ≤ y0 E0 (t, a) + f (s)E0 c (−p) (s, a)∆α,t s ,

t ∈ J,

t ∈ J,

a

where p(t) = f (t) + g(t) + h(t),

t ∈ J.

Proof 4.5.4 Let Z t

z(t) = y0 +

a

f (s)y(s)∆α,t s

Z t

+

Z

s

f (s) a

a

Z t

+

Z f (s)

a

 g(τ)y(τ)∆α,s τ ∆α,st

s

Z g(τ)

a

a



τ

h(l)y(l)∆α,τ l ∆α,s τ ∆α,t s,

Then y(t) ≤ z(t),

t ∈ J,

z(a) = y0 , and Dα z(t) =



Z t

f (t)y(t) + f (t) a

g(s)y(s)∆α,t s

t ∈ J.

160  Conformable Dynamic Equations on Time Scales Z t

Z

+ f (t)

g(s) a

=

s

a

 h(τ)y(τ)∆α,s τ ∆α,st

 Z t f (t) y(t) + g(s)y(s)∆α,t s a

Z t

+

g(s) a

≤

s

Z a

  h(τ)y(τ)∆α,s τ ∆α,st

 Z t f (t) z(t) + g(s)z(s)∆α,t s a

Z t

+

s

Z g(s)

a

a



 h(τ)z(τ)∆α,s τ ∆α,st ,

t ∈ J.

Let Z t

w(t) = z(t) + a

g(s)z(s)∆α,t s

Z t

+

Z g(s) a

a

s

 h(τ)z(τ)∆α,s τ ∆α,st,

t ∈ J.

Then w(a) = z(a) = y0 , z(t) ≤ w(t), Dα z(t) ≤

f (t)w(t),

Dα w(t) = Dα z(t) + g(t)z(t) Z t

+g(t) a

≤

h(s)z(s)∆α,t s

f (t)w(t) + g(t)w(t) Z t

+g(t) a

h(s)w(s)∆α,t s

  Z t ≤ ( f (t) + g(t)) w(t) + h(s)w(s)∆α,t s , a

Let

Z t

x(t) = w(t) + a

h(s)w(s)∆α,t s,

t ∈ J.

t ∈ J.

Linear Conformable Inequalities  161

Then x(a) = w(a) = y0 , w(t) ≤ x(t), Dα w(t) ≤ ( f (t) + g(t))x(t), Dα x(t) = Dα w(t) + h(t)w(t) ≤ ( f (t) + g(t))w(t) + h(t)w(t) t ∈ J.

= ( f (t) + g(t) + h(t))w(t),

Consequently, by the Gronwall inequality and Theorem 4.1.9, we get x(t) ≤ y0 E0 c (−p) (t, a),

t ∈ J.

Next, w(t) ≤ x(t) ≤ y0 E0 c (−p) (t, a),

t ∈ J,

and Dα z(t) ≤ h(t)w(t) ≤ y0 f (t)E0 c (−p) (t, a),

t ∈ J.

We integrate the last inequality from a to t and we obtain z(t) ≤ z(a)E0 (t, a) + y0

Z t a

f (s)E0 c (−p) (s, a)∆α,t s

  Z t = y0 E0 (t, a) + f (s)E0 c (−p) (s, a)∆α,t s ,

t ∈ J,

a

and y(t) ≤ z(t)   Z t f (s)E0 c (−p) (s, a)∆α,t s , = y0 E0 (t, a) +

t ∈ J.

a

This completes the proof.



162  Conformable Dynamic Equations on Time Scales

4.6

A CONFORMABLE INTEGRO-DYNAMIC INEQUALITY

Theorem 4.6.1 Let a1 , b1 , c1 , y, Dα y : J → R be nonnegative rd-continuous functions on J, b1 ≥ 1 on J, −b1 (c1 + 1) ∈ Rc+ ,   Z t f (t) = c1 (t) y(a)E0 (t, a) + a1 (s)∆α,t s + a1 (t) , a

k0 (α,t)b1 (t)(c1 (t) + 1) , k0 (α,t) − µ(t)(b1 (t)(c1 (t) + 1) + k1 (α,t))

h(t) =

t ∈ J.

If Dα y(t) ≤ a1 (t) + b1 (t)

Z t a

a1 (s) (y(s) + Dα y(s)) ∆α,t s,

t ∈ J,

then Dα y(t) ≤ a1 (t) Z t

+b1 (t)

a

f (s)(k0 (α, s) − µ(s)k1 (α, s)) E−h (s,t)∆α,t s, k0 (α, s) − µ(s)(b1 (s)(c1 (s) + 1) + k1 (α, s))

y(t) ≤ y(a)E0 (t, a) + Z t

+ a

Z t

Z τ

b1 (τ)

a

a

a1 (s)∆α,t s

f (s)(k0 (α, s) − µ(s)k1 (α, s)) E−h (s, τ)∆α,τ s∆α,s τ, k0 (α, s) − µ(s)(b1 (s)(c1 (s) + 1) + k1 (α, s))

t ∈ J. Proof 4.6.2 Let

Z t

z(t) = a

c1 (s) (y(s) + Dα y(s)) ∆α,t s,

t ∈ J.

Then z(a) = 0, Dα y(t) ≤ a1 (t) + b1 (t)z(t),

t ∈ J.

Hence, integrating from a to t, we find y(t) ≤ y(a)E0 (t, a) +

Z t a

a1 (s)∆α,t s

Z t

+ a

b1 (s)z(s)∆α,t s,

By (4.11), we get Dα z(t) ≤ c1 (t) (y(t) + Dα y(t))

t ∈ J.

(4.11)

Linear Conformable Inequalities  163

 Z t ≤ c1 (t) y(a)E0 (t, a) + a1 (s)∆α,t s a

Z t

+ a

 b1 (s)z(s)∆α,t s + a1 (t) + b1 (t)z(t)

  Z t = c1 (t) y(a)E0 (t, a) + a1 (s)∆α,t s + a1 (t) a

  Z t +c1 (t) b1 (t)z(t) + b1 (s)z(s)∆α,t s a

=

  Z t f (t) + c1 (t) b1 (t)z(t) + b1 (s)z(s)∆α,t s ,

t ∈ J.

a

Set

Z t

w(t) = z(t) + a

b1 (s)z(s)∆α,t s,

t ∈ J.

Then z(t) ≤ w(t), Dα z(t) ≤

f (t) + c1 (t)b1 (t)w(t),

t ∈ J,

w(a) = z(a) = 0 and Dα w(t) = Dα z(t) + b1 (t)z(t) ≤

f (t) + c1 (t)b1 (t)w(t) + b(t)w(t)

=

f (t) + b1 (t)(c1 (t) + 1)w(t),

t ∈ J.

Thus, by the Gronwall inequality and Theorem 4.1.9, it follows that w(t) ≤

Z t a

t ∈ J. Then z(t) ≤ w(t)

f (s)(k0 (α, s) − µ(s)k1 (α, s)) E−h (s,t)∆α,t s, k0 (α, s) − µ(s)(b1 (s)(c1 (s) + 1) + k1 (α, s))

164  Conformable Dynamic Equations on Time Scales

≤

Z t a

f (s)(k0 (α, s) − µ(s)k1 (α, s)) E−h (s,t)∆α,t s, k0 (α, s) − µ(s)(b1 (s)(c1 (s) + 1) + k1 (α, s))

Dα y(t) ≤ a(t) Z t

+ a

f (s)(k0 (α, s) − µ(s)k1 (α, s)) E−h (s,t)∆α,t s, k0 (α, s) − µ(s)(b1 (s)(c1 (s) + 1) + k1 (α, s))

t ∈ J,

and y(t) ≤ y(a)E0 (t, a) + Z t

+ a

Z τ

b1 (τ)

a

Z t a

a1 (s)∆α,t s

f (s)(k0 (α, s) − µ(s)k1 (α, s)) E−h (s, τ)∆α,τ s∆α,s τ, k0 (α, s) − µ(s)(b1 (s)(c1 (s) + 1) + k1 (α, s))

t ∈ J. This completes the proof.



CHAPTER

5

Cauchy-Type Problem for a Class of Nonlinear Conformable Dynamic Equations

Throughout this chapter, suppose that T is a time scale with forward jump operator and delta differentiation operator σ and ∆, respectively. Let α ∈ (0, 1], k0 and k1 satisfy (A1), and assume k0 (α,t) − µ(t)k1 (α,t) > 0, α ∈ (0, 1], t ∈ T, holds. Let t0 ∈ T and y0 ∈ R.

5.1

EXISTENCE AND UNIQUENESS OF SOLUTIONS

In this section we will investigate the Cauchy problem Dα y(t) = f (t, y(t)),

t > t0 ,

(5.1)

y(t0 ) = y0 ,

(5.2)

where (H) f : T × R → R, f ∈ C (T × R),

Z t t0

f (s, z)∆α,t s exists for any z ∈ R and for any t ≥ t0 ,

| f (t, s)| ≤ M for any t ≥ t0 and for any s ∈ R and for some positive constant M, | f (t, y1 ) − f (t, y2 )| ≤ L|y1 − y2 | for any t ≥ t0 and for any y1 , y2 ∈ R. Firstly, we will note that the IVP (5.1), (5.2) is equivalent to the equation Z t

y(t) = y0 E0 (t,t0 ) +

t0

f (s, y(s))∆α,t s,

t ≥ t0 .

165

166  Conformable Dynamic Equations on Time Scales

Theorem 5.1.1 Suppose that f satisfies (H). Then the problem (5.1), (5.2) has a unique solution y ∈ C 1 ([t0 , ∞)). Proof 5.1.2 Let a ∈ T, a > t0 , a < ∞, be arbitrarily chosen and fixed. Let K be a positive constant such that (t − t0 )l hl (t,t0 ) ≤ K , l ∈ N, t ∈ [t0 , a]. l! Consider the IVP Dα y(t) = f (t, y(t)), t ∈ [t0 , a], (5.3) y(t0 ) = y0 .

(5.4)

We define the sequence {yl (t)}l∈N , t ∈ [t0 , a], as follows y0 (t) = y0 E0 (t,t0 ), Z t

yl (t) = y0 E0 (t,t0 ) +

t0

f (s, yl−1 (s))∆α,t s,

t ∈ [t0 , a],

l ∈ N.

We have Z t |y1 (t) − y0 (t)| = y0 E0 (t,t0 ) + f (s, y0 (s))∆α,t s − y0 E0 (t,t0 ) t0

Z t = d(s, y0 (s))∆α,t s t 0

≤

Z t t0

≤ M

| f (s, y0 (s))|∆α,t s

Z t t0

∆α,t s

= Mh1 (t,t0 ),

t ∈ [t0 , a].

Assume that |yl−1 (t) − yl−2 (t)| ≤ MLl−2 hl−1 (t,t0 ),

t ∈ [t0 , a],

l ∈ N,

for some l ∈ N, l ≥ 2. We will prove that |yl (t) − yl−1 (t)| ≤ MLl−1 hl (t,t0 ),

t ∈ [t0 , a].

Really, Z t |yl (t) − yl−1 (t)| = y0 E0 (t,t0 ) + f (s, yl−1 (s))∆α,t s t0

−y0 E0 (t,t0 ) −

Z t t0

f (s, yl−2 (s))∆α,t s

(5.5)

Cauchy-Type Problem for a Class of Nonlinear Conformable Dynamic Equations  167

Z t Z t = f (s, yl−1 (s))∆α,t s − f (s, yl−2 (s))∆α,t s t t 0

0

Z t = ( f (s, yl−1 (s)) − f (s, yl−2 (s))) ∆α,t s t 0

≤

Z t t0

≤ L

| f (s, yl−1 (s)) − f (s, yl−2 (s))| ∆α,t s

Z t t0

≤ ML

|yl−1 (s) − yl−2 (s)|∆α,t s

l−1

Z t t0

hl−1 (s,t0 )∆α,t s

≤ MLl−1 hl (t,t0 ),

t ∈ [t0 , a].

Therefore (5.5) holds for any l ∈ N, l ≥ 1, and for any t ∈ [t0 , a]. Note that ∞ lim (yl (t) − y0 (t)) = ∑ (yl (t) − yl−1 (t)) l→∞ l=1 ∞

≤

∑ |yl (t) − yl−1 (t)| l=1

≤

M ∞ l ∑ L hl (t,t0 ) L l=1

≤

M ∞ l ∑ L hl (a,t0 ) L l=1

M ∞ (L(a − t0 ))l ∑ L l=1 l! M ≤ K eL(a−t0 ) L ≤ K

< ∞,

t ∈ [t0 , a].

Consequently, ∞

∑ (yl (t) − yl−1 (t)) l=1

is uniformly convergent on [t0 , a]. Hence, there exists ∞

lim (yl (t) − y0 (t)) =

l→∞

∑ (yl (t) − yl−1 (t)) , l=1

From here, there exists y(t) = lim yl (t), l→∞

t ∈ [t0 , a],

t ∈ [t0 , a].

168  Conformable Dynamic Equations on Time Scales

and y satisfies (5.3), (5.4). Now we assume that the problem (5.3), (5.4) has two solutions y and z. We have Z t

z(t) = y0 E0 (t,t0 ) +

f (s, z(s))∆α,t s,

t0

t ∈ [t0 , a].

Note that |y0 (t) − z(t)| = y0 E0 (t,t0 ) − y0 E0 (t,t0 ) −

Z t t0

f (s, z(s))∆α,t s

Z t = f (s, z(s))∆α,t s t 0

≤

Z t t0

| f (s, z(s))|∆α,t s

≤ bMh1 (t,t0 ),

t ∈ [t0 , a].

Assume that |yl−1 (t) − z(t)| ≤ MLl−1 hl (t,t0 ),

t ∈ [t0 , a],

for some l ∈ N. We will prove that |yl (t) − z(t)| ≤ MLl hl+1 (t,t0 ),

t ∈ [t0 , a].

Really, we have Z t |yl (t) − z(t)| = y0 E0 (t,t0 ) + f (s, yl−1 (s))∆α,t s t0

−0y0 E0 (t,t0 ) −

Z t t0

f (s, z(s))∆α,t s

Z t = ( f (s, yl−1 (s)) − f (s, z(s))) ∆α,t s t 0

≤

Z t t0

≤ L

| f (s, yl−1 (s)) − f (s, z(s))|∆α,t s

Z t t0

≤ MLl

|yl−1 (s) − z(s)|∆α,t s Z t t0

hl (s,t0 )∆α,t s

= MLl hl+1 (t,t0 ),

t ∈ [t0 , a].

(5.6)

Cauchy-Type Problem for a Class of Nonlinear Conformable Dynamic Equations  169

Therefore (5.6) holds for any l ∈ N. Because lim MLl hl+1 (t,t0 ) = 0,

t ∈ [t0 , a],

lim (yl (t) − z(t)) = 0,

t ∈ [t0 , a].

l→∞

we get l→∞

From here y(t) = z(t),

t ∈ [t0 , a].

In this way we get that the problem (5.3), (5.4) has a unique solution in C 1 ([t0 , a]), which we will denote by y1 . Now we consider the problem Dα y(t) = f (t, y(t)),

t ∈ [a, 2a],

y(a) = y1 (a).

(5.7)

(5.8)

As in the above, we get that the problem (5.7), (5.8) has a unique solution y2 ∈ C 1 ([a, 2a]), and so on. Then  1 y (t) i f t ∈ [t0 , a]       y2 (t) i f t ∈ [a, 2a] y(t) =       .. . is the unique solution of the problem (5.1), (5.2) that belongs to C 1 ([t0 , ∞)). This completes the proof. 

5.2

THE DEPENDENCY OF THE SOLUTION UPON THE INITIAL DATA

Let z0 ∈ R. Consider the IVPs Dα y(t) = f (t, y(t)),

t ∈ [t0 , a], (5.9)

y(t0 ) = y0 and

Dα z(t) = f (t, z(t)),

t ∈ [t0 , a], (5.10)

z(t0 ) = z0 , where f satisfies (H1) and k0 (α,t) − µ(t)(k1 (α,t) − L) > 0, Let P be a positive constant such that E0 (t,t0 ) ≤ P,

t ∈ [t0 , a].

t ∈ [t0 , a].

170  Conformable Dynamic Equations on Time Scales

We have Z t

y(t) = y0 E0 (t,t0 ) +

f (s, y(s))∆α,t s,

t0

Z t

z(t) = y0 E0 (t,t0 ) +

f (s, z(s))∆α,t s,

t0

t ∈ [t0 , a].

Then Z t |y(t) − z(t)| = y0 E0 (t,t0 ) + f (s, y(s))∆α,t s t0

−z0 E0 (t,t0 ) −

Z t t0

f (s, z(s))∆α,t s

Z t = (y0 − z0 )E0 (t,t0 ) + ( f (s, y(s)) − f (s, z(s))) ∆α,t s t0

Z t ≤ |y0 − z0 |E0 (t,t0 ) + ( f (s, y(s)) − f (s, z(s))) ∆α,t s t 0

≤ |y0 − z0 |E0 (t,t0 ) + ≤ P|y0 − z0 | + L

Z t t0

Z t t0

| f (s, y(s)) − f (s, z(s))| ∆α,t s

|y(s) − z(s)|∆α,t s,

t ∈ [t0 , a].

From the last inequality and from Theorem 4.1.1, we get   Zt |y(t) − z(t)| ≤ P|y0 − z0 | L Eg (σ (s),t)∆α,t s + 1 , t0

where g(t) =

5.3

(−L + k1 (α,t))(k0 (α,t) − µ(t)k1 (α,t)) , k0 (α,t) − µ(t)(k1 (α,t) − L)

t ∈ [t0 , a],

t ∈ [t0 , a].

LYAPUNOV FUNCTIONS

Let 0 ∈ Rc . Suppose that x is a solution of equation (5.3) and let V : R → R be a continuously differentiable function. By P¨otzsche’s chain rule (see the appendix of this book), we get (V (x(t)))∆ =

Z 1

  V 0 x(t) + hµ(t)x∆ (t) dhx∆ (t)

0

Z 1

  V x(t) + hµ(t)

 1 k1 (α,t) α = D x(t) − x(t) dh k0 (α,t) k0 (α,t) 0   1 k1 (α,t) α × D x(t) − x(t) k0 (α,t) k0 (α,t) 0

Cauchy-Type Problem for a Class of Nonlinear Conformable Dynamic Equations  171

Z 1

 k0 (α,t) − hµ(t)k1 (α,t) hµ(t) α = V x(t) + D x(t) dh k0 (α,t) k0 (α,t) 0   k1 (α,t) 1 α D x(t) − x(t) , t ∈ Tκ , × k0 (α,t) k0 (α,t) 0



whereupon ∆

k0 (α,t)(V (x(t)))

Z 1



 k0 (α,t) − hµ(t)k1 (α,t) hµ(t) α = V x(t) + D x(t) dh k0 (α,t) k0 (α,t) 0   k1 (α,t) α × D x(t) − x(t) k0 (α,t)  Z 1  k0 (α,t) − hµ(t)k1 (α,t) hµ(t) = V0 x(t) + f (t, x(t)) dh k0 (α,t) k0 (α,t) 0   k1 (α,t) × f (t, x(t)) − x(t) , t ∈ Tκ , k0 (α,t) 0

and Z 1

 k0 (α,t) − hµ(t)k1 (α,t) hµ(t) D (V (x(t))) = x(t) + f (t, x(t)) dh V k0 (α,t) k0 (α,t) 0   k1 (α,t) x(t) + k1 (α,t)V (x(t)), t ∈ Tκ . × f (t, x(t)) − k0 (α,t) α

0



This motivates us to define Dα V : T × R → R as follows  Z 1  hµ(t) α 0 k0 (α,t) − hµ(t)k1 (α,t) D (V (t, x)) = x(t) + f (t, x) dh V k0 (α,t) k0 (α,t) 0   k1 (α,t) × f (t, x) − x(t) + k1 (α,t)V (x), t ∈ Tκ . k0 (α,t) Let t ∈ Tκ . If µ(t) = 0, then  k0 (α,t) − hµ(t)k1 (α,t) x(t) dh D (V (t, x)) = V k0 (α,t)   k1 (α,t) x(t) + k1 (α,t)V (x). × f (t, x) − k0 (α,t) 0

α



If µ(t) 6= 0 and f (t, x) 6= 0, then    k0 (α,t) k0 (α,t) − hµ(t)k1 (α,t) hµ(t) D (V (t, x)) = V x(t) + f (t, x) µ(t) f (t, x) k0 (α,t) k0 (α,t)    k0 (α,t) − hµ(t)k1 (α,t) k1 (α,t) −V x f (t, x) − x k0 (α,t) k0 (α,t) α

+k1 (α,t)V (x).

172  Conformable Dynamic Equations on Time Scales

Example 5.3.1 Let V (x) = x. Then  Dα (V (t, x)) =

f (t, x) −

 k1 (α,t) x + k1 (α,t)x. k0 (α,t)

Exercise 5.3.2 Let V (x) = x2 . Find Dα (V (t, x)).

5.4

BOUNDEDNESS OF SOLUTIONS

Definition 5.4.1 We say that a solution y of the IVP (5.3), (5.4) is uniformly bounded if there exists a constant C = C(y0 ), which does not depend on t0 , such that kyk = sup |y(t)| ≤ C. t∈[t0 ,∞)

Theorem 5.4.2 Let 0, 1 ∈ Rc+ , and assume there exists a Lyapunov function V : R → [0, ∞) such that for all (t, y) ∈ [t0 , ∞) × R we have V (y) → ∞,

as kyk∞,

V (y)

≤

φ (kyk),

Dα (V (t, y))

≤

ψ(kyk) + L

and

k0 (α,t) + µ(t)(1 − k1 (α,t)) ψ φ −1 (V (y)) + L k0 (α,t)   k0 (α,t) + µ(t)(1 − k1 (α,t)) +V (y) 1 − k1 (α,t) ≤ γ, k0 (α,t) where φ , ψ are functions such that φ : [0, ∞) → [0, ∞), ψ : [0, ∞) → (−∞, 0], ψ is nonincreasing and φ −1 exists, L and γ are nonnegative constants. Then any solution of the IVP (5.3), (5.4), defined on [t0 , ∞), is uniformly bounded. Proof 5.4.3 Let y be a solution of the IVP (5.3), (5.4), defined on [t0 , ∞). Then Dα (V (t, y)E1 (t,t0 )) = Dα (V (t, y))E1σ (t,t0 ) +V (y(t))E1 (t,t0 ) −k1 (α,t)V (y(t))E1σ (t,t0 ) = Dα (V (t, y))

k0 (α,t) + µ(t)(1 − k1 (α,t)) E1 (t,t0 ) k0 (α,t)

+V (y(t))E1 (t,t0 )

Cauchy-Type Problem for a Class of Nonlinear Conformable Dynamic Equations  173

−

k1 (α,t)(k0 (α,t) + µ(t)(1 − k1 (α,t))) V (y(t))E1 (t,t0 ) k0 (α,t)

≤ (ψ(kyk) + L)

k0 (α,t) + µ(t)(1 − k1 (α,t)) E1 (t,t0 ) k0 (α,t)

+V (y(t))E1 (t,t0 ) −k1 (α,t) ≤

k0 (α,t) + µ(t)(1 − k1 (α,t)) V (y(t))E1 (t,t0 ) k0 (α,t)



k0 (α,t) + µ(t)(1 − k1 (α,t)) ψ φ −1 (V (y(t))) + L E1 (t,t0 ) k0 (α,t) +V (y(t))E1 (t,t0 ) −k1 (α,t)

k0 (α,t) + µ(t)(1 − k1 (α,t)) V (y(t))E1 (t,t0 ) k0 (α,t)



=



k0 (α,t) + µ(t)(1 − k1 (α,t)) ψ φ −1 (V (y(t))) + L k0 (α,t)   k0 (α,t) + µ(t)(1 − k1 (α,t)) +V (y(t)) 1 − k1 (α,t) E1 (t,t0 ) k0 (α,t)

≤ γE1 (t,t0 ),

t ∈ [t0 , ∞).

Hence, V (y(t))E1 (t,t0 ) ≤ V (y0 )E0 (t,t0 ) + γE1 (t,t0 ), and V (y(t)) ≤ V (y0 )

E0 (t,t0 ) + γ, E1 (t,t0 )

t ∈ [t0 , ∞),

t ∈ [t0 , ∞), t ∈ [t0 , ∞).

Since 0, 1 ∈ Rc+ , we have E0 (t,t0 ) E1 (t,t0 )

e− k1 (t,t0 ) =

k0

e 1−k1 (t,t0 ) k0

Rt

e

=

1 t0 µ(τ)

  k (α,τ) log 1−µ(τ) k1 (α,τ) ∆τ 0

  1−k1 (α,τ) 1 ∆τ t0 µ(τ) log 1+µ(τ) k0 (α,τ)

Rt

e

Rt

= e

1 t0 µ(τ)

≤ 1,

  k0 (α,τ)−µ(τ)k1 (α,τ) log k (α,τ)+µ(τ)(1−k ∆τ (α,τ)) 0

t ∈ [t0 , ∞).

1

(5.11)

174  Conformable Dynamic Equations on Time Scales

Then, by (5.11), we get V (y(t)) ≤ V (y0 ) + γ,

t ∈ [t0 , ∞).

Because V (y) → ∞, as kyk → ∞, by the last inequality, we get that there exists a constant C > 0, depending on V (y0 ), γ and L only, such that kyk ≤ C. 

This completes the proof.

Theorem 5.4.4 Let 0, 1 ∈ Rc+ , and assume there exists a Lyapunov function V : R → [0, ∞) such that for all (t, y) ∈ [t0 , ∞) × R, we have V (y) → ∞,

as kyk → ∞,

(5.12)

Dα (V (t, y)) ≤ −λ1 kykr + L, V (y) ≤ λ2 kykq , and  −

 λ1 r q

r q

(V (y)) + L

λ2

k0 (α,t) + µ(t)(1 − k1 (α,t)) k0 (α,t)

  k0 (α,t) + µ(t)(1 − k1 (α,t)) +V (y) 1 − k1 (α,t) ≤ γ, k0 (α,t) where λ1 , λ2 , r, q are positive constants, L and γ are nonnegative constants. Then any solution of the IVP (5.3), (5.4), defined on [t0 , ∞), is uniformly bounded. Proof 5.4.5 Let y be a solution of the IVP (5.3), (5.4), defined on [t0 , ∞). We have Dα (V (t, y)E1 (t,t0 )) = Dα (V (t, y))E1σ (t,t0 ) +V (y(t))E1 (t,t0 ) −k1 (α,t)V (y(t))E1σ (t,t0 ) = Dα (V (t, y))

k0 (α,t) + µ(t)(1 − k1 (α,t)) E1 (t,t0 ) k0 (α,t)

+V (y(t))E1 (t,t0 ) −k1 (α,t)

k0 (α,t) + µ(t)(1 − k1 (α,t)) V (y(t))E1 (t,t0 ) k0 (α,t)

≤ (−λ1 kykr + L)

k0 (α,t) + µ(t)(1 − k1 (α,t)) E1 (t,t0 ) k0 (α,t)

Cauchy-Type Problem for a Class of Nonlinear Conformable Dynamic Equations  175

  k0 (α,t) + µ(t)(1 − k1 (α,t)) +V (y(t)) 1 − k1 (α,t) E1 (t,t0 ) k0 (α,t)   r λ1 k0 (α,t) + µ(t)(1 − k1 (α,t)) ≤ − r (V (y(t))) q + L E1 (t,t0 ) k0 (α,t) λq 2

  k0 (α,t) + µ(t)(1 − k1 (α,t)) +V (y(t)) 1 − k1 (α,t) E1 (t,t0 ) k0 (α,t)   r − λ1r (V (y(t))) q + L k0 (α,t) + µ(t)(1 − k1 (α,t)) ≤ k0 (α,t) λq 2

  k0 (α,t) + µ(t)(1 − k1 (α,t)) +V (y(t)) 1 − k1 (α,t) E1 (t,t0 ) k0 (α,t) ≤ γE1 (t,t0 ),

t ∈ [t0 , ∞).

Now, as in the proof of Theorem 5.4.2, we get V (y(t)) ≤ V (y0 ) + γ,

t ∈ [t0 , ∞),

(5.13)

and there exists a constant C > 0, depending on V (y0 ), γ and L only, such that kyk ≤ C. 

This completes the proof.

Theorem 5.4.6 Assume that all conditions of Theorem 5.4.4 are satisfied with (5.12) replaced by λ3 kyk p ≤ V (y), (5.14) where λ3 and p are positive constants. Then any solution of the IVP (5.3), (5.4), defined on [t0 , ∞), is uniformly bounded and  kyk ≤

V (y0 ) + γ λ3

1

p

.

Proof 5.4.7 Let y be a solution of the IVP (5.3), (5.4), defined on [t0 , ∞). By (5.13), we get V (y(t)) ≤ V (y0 ) + γ,

t ∈ [t0 , ∞).

Hence, and considering (5.14), we arrive at λ3 kyk p ≤ V (y0 ) + γ, whereupon kyk p ≤

V (y0 ) + γ λ3

176  Conformable Dynamic Equations on Time Scales

and  kyk ≤

V (y0 ) + γ λ3

1

p

. 

This completes the proof.

Theorem 5.4.8 Let 0, λ1 ∈ Rc+ and there exists a Lyapunov function V : R → [0, ∞) such that for all (t, y) ∈ [t0 , ∞) × R we have V (y) → ∞, Dα (V (t, y))

≤

as kyk∞,

−λ1V (y) + L

and (−λ1V (y(t)) + L)

k0 (α,t) + µ(t)(λ1 − k1 (α,t)) k0 (α,t)

  k0 (α,t) + µ(t)(λ1 − k1 (α,t)) +V (y(t)) λ1 − k1 (α,t) k0 (α,t) where L and γ are nonnegative constants. Then any solution of the IVP (5.3), (5.4), defined on [t0 , ∞), is uniformly bounded. Proof 5.4.9 Let y be a solution of the IVP (5.3), (5.4), defined on [t0 , ∞). Then
Dα V (t, y)Eλ1 (t,t0 ) = Dα (V (t, y))Eλσ1 (t,t0 ) + λ1V (y(t))Eλ1 (t,t0 ) −k1 (α,t)V (y(t))Eλσ1 (t,t0 ) = Dα (V (t, y))

k0 (α,t) + µ(t)(λ1 − k1 (α,t)) Eλ1 (t,t0 ) k0 (α,t)

+λ1V (y(t))Eλ1 (t,t0 ) −

k1 (α,t)(k0 (α,t) + µ(t)(λ1 − k1 (α,t))) V (y(t))Eλ1 (t,t0 ) k0 (α,t)

≤ (−λ1V (y(t)) + L)

k0 (α,t) + µ(t)(λ1 − k1 (α,t)) Eλ1 (t,t0 ) k0 (α,t)

+λ1V (y(t))Eλ1 (t,t0 ) −k1 (α,t)

k0 (α,t) + µ(t)(λ1 − k1 (α,t)) V (y(t))Eλ1 (t,t0 ) k0 (α,t)

 =

(−λ1V (y(t)) + L)

k0 (α,t) + µ(t)(λ1 − k1 (α,t)) k0 (α,t)

Cauchy-Type Problem for a Class of Nonlinear Conformable Dynamic Equations  177

  k0 (α,t) + µ(t)(λ1 − k1 (α,t)) Eλ1 (t,t0 ) +V (y(t)) λ1 − k1 (α,t) k0 (α,t) ≤ γEλ1 (t,t0 ),

t ∈ [t0 , ∞).

Consequently, V (y(t))Eλ1 (t,t0 ) ≤ V (y0 )E0 (t,t0 ) + γEλ1 (t,t0 ), and V (y(t)) ≤ V (y0 )

E0 (t,t0 ) + γ, Eλ1 (t,t0 )

t ∈ [t0 , ∞),

t ∈ [t0 , ∞),

t ∈ [t0 , ∞).

(5.15)

Now, using that 0, λ1 ∈ Rc+ , we obtain E0 (t,t0 ) Eλ1 (t,t0 )

e− k1 (t,t0 ) =

k0

e λ1 −k1 (t,t0 ) k0

Rt

e

=

1 t0 µ(τ)

  k (α,τ) log 1−µ(τ) k1 (α,τ) ∆τ 0

  λ1 −k1 (α,τ) 1 ∆τ t0 µ(τ) log 1+µ(τ) k0 (α,τ)

Rt

e

Rt

= e

1 t0 µ(τ)

≤ 1,

  k0 (α,τ)−µ(τ)k1 (α,τ) ∆τ log k (α,τ)+µ(τ)(λ −k (α,τ)) 0

1

1

t ∈ [t0 , ∞).

From here and (5.15), we get V (y(t)) ≤ V (y0 ) + γ,

t ∈ [t0 , ∞).

(5.16)

Since V (y) → ∞, as kyk → ∞, by the last inequality, we conclude that there exists a constant C > 0, depending on V (y0 ), γ and L only, such that kyk ≤ C. 

This completes the proof. Theorem 5.4.10 Assume that all conditions of Theorem 5.4.8 are satisfied with V (y) → ∞,

as

kyk → ∞

replaced by λ2 kyk p ≤ V (y),

(5.17)

where λ2 and p are positive constants. Then any solution of the IVP (5.3), (5.4), defined on [t0 , ∞), is uniformly bounded and  kyk ≤

V (y0 ) + γ λ2

1

p

.

178  Conformable Dynamic Equations on Time Scales

Proof 5.4.11 Let y be a solution of the IVP (5.3), (5.4), defined on [t0 , ∞). By (5.16), we get V (y(t)) ≤ V (y0 ) + γ, t ∈ [t0 , ∞). Hence, and considering (5.17), we arrive at λ2 kyk p ≤ V (y0 ) + γ, whereupon kyk p ≤

V (y0 ) + γ λ2

and  kyk ≤

V (y0 ) + γ λ2

1

p

. 

This completes the proof. Exercise 5.4.12 Let T = 2Z, k1 (α,t) = (1 − α)(1 + t 2 )4α ,

k0 (α,t) = α(1 + t 2 )4(1−α) ,

α ∈ (0, 1],

t ∈ T.

Consider the IVP 1

1

D 4 y = ay 3 + by,

t > 0,

y(0) = 2, where a and b are constants. If 1. if V (x) = x, 2. V (x) = x2 , find conditions for the constants a and b so that any solution of the considered IVP, defined on [0, ∞), is uniformly bounded.

5.5

EXPONENTIAL STABILITY

In this section, in addition we suppose that f (t, 0) = 0, t ∈ [t0 , ∞). Definition 5.5.1 We say that the trivial solution of equation (5.3) is exponentially stable if there exist positive constants d and M and a nonnegative constant C such that for any solution of the IVP (5.3), (5.4) we have |y(t)| ≤ C(y0 ,t0 ) (E c M (t,t0 ))d ,

t ∈ [t0 , ∞).

Cauchy-Type Problem for a Class of Nonlinear Conformable Dynamic Equations  179

Theorem 5.5.2 Let 0, M ∈ Rc+ , M > 0, and assume there exists a Lyapunov function V : R → [0, ∞) such that for all (t, y) ∈ [t0 , ∞) × R we have W (kyk) ≤ V (y) ≤ φ (kyk), Dα (V (t, y)) ≤ ψ(kyk) + L and

k0 (α,t) + µ(t)(M ⊕c 0 − k1 (α,t)) ψ φ −1 (V (y)) + L k0 (α,t)   k0 (α,t) + µ(t)(M ⊕c 0 − k1 (α,t)) ≤ −(M ⊕c 0), +V (y) 1 − k1 (α,t) k0 (α,t) where φ , ψ are functions such that W, φ : [0, ∞) → [0, ∞), ψ : [0, ∞) → (−∞, 0], ψ is nonincreasing and φ and W are strictly increasing, L is a nonnegative constant. Then any solution of the IVP (5.3) satisfies kyk ≤ W −1 ((V (y0 ) + 1)E c M (t,t0 )) ,

t ∈ [t0 , ∞).

Proof 5.5.3 Let y be a solution of the IVP (5.3), (5.4), defined on [t0 , ∞). Then σ Dα (V (t, y)EM⊕c 0 (t,t0 )) = Dα (V (t, y))EM⊕ (t,t0 ) +V (y(t))EM⊕c 0 (t,t0 ) c0 σ −k1 (α,t)V (y(t))EM⊕ (t,t0 ) c0

= Dα (V (t, y))

k0 (α,t) + µ(t)(M ⊕c 0 − k1 (α,t)) EM⊕c 0 (t,t0 ) k0 (α,t)

+(M ⊕c 0)V (y(t))EM⊕c 0 (t,t0 ) −

k1 (α,t)(k0 (α,t) + µ(t)(M ⊕c 0 − k1 (α,t))) V (y(t))EM⊕c 0 (t,t0 ) k0 (α,t)

≤ (ψ(kyk) + L)

k0 (α,t) + µ(t)(M ⊕c 0 − k1 (α,t)) EM⊕c 0 (t,t0 ) k0 (α,t)

+(M ⊕c 0)V (y(t))EM⊕c 0 (t,t0 ) −k1 (α,t) ≤

k0 (α,t) + µ(t)(M ⊕c 0 − k1 (α,t)) V (y(t))EM⊕c 0 (t,t0 ) k0 (α,t)



k0 (α,t) + µ(t)(M ⊕c 0 − k1 (α,t)) ψ φ −1 (V (y(t))) + L EM⊕c 0 (t,t0 ) k0 (α,t)

180  Conformable Dynamic Equations on Time Scales

+(M ⊕c 0)V (y(t))EM⊕c 0 (t,t0 ) −k1 (α,t)

k0 (α,t) + µ(t)(M ⊕c 0 − k1 (α,t)) V (y(t))EM⊕c 0 (t,t0 ) k0 (α,t)

 =



k0 (α,t) + µ(t)(M ⊕c 0 − k1 (α,t)) ψ φ −1 (V (y(t))) + L k0 (α,t)   k0 (α,t) + µ(t)(M ⊕c 0 − k1 (α,t)) +V (y(t)) M ⊕c 0 − k1 (α,t) EM⊕c 0 (t,t0 ) k0 (α,t)

≤ −(M ⊕c 0)EM⊕c 0 (t,t0 ),

t ∈ [t0 , ∞).

Hence, V (y(t))EM⊕c 0 (t,t0 ) ≤ V (y0 )E0 (t,t0 ) − EM⊕c 0 (t,t0 ) + E0 (t,t0 ) ≤ (V (y0 ) + 1)E0 (t,t0 ),

t ∈ [t0 , ∞),

and V (y(t)) ≤ (V (y0 ) + 1)E c M (t,t0 ),

t ∈ [t0 , ∞),

t ∈ [t0 , ∞).

Now, using that V (y(t)) ≥ W (kyk), we get kyk ≤ W −1 ((V (y0 ) + 1)E c M (t,t0 )) ,

t ∈ [t0 , ∞). 

This completes the proof.

Theorem 5.5.4 Let 0, M ∈ Rc+ , and assume there exists a Lyapunov function V : R → [0, ∞) such that for all (t, y) ∈ [t0 , ∞) × R, we have Dα (V (t, y)) ≤ −λ1 kykr + L, λ3 kyk p ≤ V (y) ≤ λ2 kykq , and 



− λ1r (V (y)) q + L k0 (α,t) + µ(t)(M ⊕c 0 − k1 (α,t)) k0 (α,t) λ2q   k0 (α,t) + µ(t)(M ⊕c −k1 (α,t)) +V (y) M ⊕c 0 − k1 (α,t) ≤ −(M ⊕c 0), k0 (α,t) r

where λ1 , λ2 , r, p, q are positive constants, L is a nonnegative constant. Then the trivial solution of equation (5.3) is exponentially stable.

Cauchy-Type Problem for a Class of Nonlinear Conformable Dynamic Equations  181

Proof 5.5.5 Let y be a solution of the IVP (5.3), (5.4), defined on [t0 , ∞). We have σ Dα (V (t, y)EM⊕c 0 (t,t0 )) = Dα (V (t, y))EM⊕ (t,t0 ) +V (y(t))EM⊕c 0 (t,t0 ) c0 σ −k1 (α,t)V (y(t))EM⊕ (t,t0 ) c0

= Dα (V (t, y))

k0 (α,t) + µ(t)(M ⊕c 0 − k1 (α,t)) EM⊕c 0 (t,t0 ) k0 (α,t)

+(M ⊕c 0)V (y(t))EM⊕c 0 (t,t0 ) −k1 (α,t)

k0 (α,t) + µ(t)(M ⊕c 0 − k1 (α,t)) V (y(t))EM⊕c 0 (t,t0 ) k0 (α,t)

k0 (α,t) + µ(t)(M ⊕c 0 − k1 (α,t)) EM⊕c 0 (t,t0 ) k0 (α,t)   k0 (α,t) + µ(t)(1 − k1 (α,t)) +V (y(t)) M ⊕c 0 − k1 (α,t) EM⊕c 0 (t,t0 ) k0 (α,t)   r k0 (α,t) + µ(t)(M ⊕c 0 − k1 (α,t)) λ1 EM⊕c 0 (t,t0 ) ≤ − r (V (y(t))) q + L k0 (α,t) λq

≤ (−λ1 kykr + L)

2

  k0 (α,t) + µ(t)(M ⊕c 0 − k1 (α,t)) +V (y(t)) M ⊕c 0 − k1 (α,t) EM⊕c 0 (t,t0 ) k0 (α,t)   r − λ1r (V (y(t))) q + L k0 (α,t) + µ(t)(M ⊕c 0 − k1 (α,t)) ≤ k0 (α,t) λq 2

  k0 (α,t) + µ(t)(M ⊕c 0 − k1 (α,t)) EM⊕c 0 (t,t0 ) +V (y(t)) M ⊕c 0 − k1 (α,t) k0 (α,t) ≤ −(M ⊕c 0)EM⊕c 0 (t,t0 ),

t ∈ [t0 , ∞).

Now, as in the proof of Theorem 5.5.2, we get V (y(t)) ≤ (V (y0 ) + 1)e c M (t,t0 ),

t ∈ [t0 , ∞),

1 p V (y0 ) + 1 e c M (t,t0 ) , λ3

t ∈ [t0 , ∞).

and  y(t) ≤



This completes the proof.

Theorem 5.5.6 Let 0, λ1 ∈ Rc+ , and assume there exists a Lyapunov function V : R → [0, ∞) such that for all (t, y) ∈ [t0 , ∞) × R we have λ kyk p ≤ V (y),

182  Conformable Dynamic Equations on Time Scales

Dα (V (t, y)) ≤ −λ1V (y) + L and (−λ1V (y(t)) + L)

k0 (α,t) + µ(t)(λ1 − k1 (α,t)) k0 (α,t)

  k0 (α,t) + µ(t)(λ1 − k1 (α,t)) ≤ −(λ1 ⊕c 0) +V (y(t)) λ1 − k1 (α,t) k0 (α,t) where L is a nonnegative constant. Then the trivial solution of equation (5.3) is exponentially stable. Proof 5.5.7 Let y be a solution of the IVP (5.3), (5.4), defined on [t0 , ∞). Then
Dα V (t, y)Eλ1 ⊕c 0 (t,t0 ) = Dα (V (t, y))Eλσ1 ⊕c 0 (t,t0 ) + (λ1 ⊕c 0)V (y(t))Eλ1 ⊕c 0 (t,t0 ) −k1 (α,t)V (y(t))Eλσ1 ⊕c 0 (t,t0 )

= Dα (V (t, y))

k0 (α,t) + µ(t)(λ1 ⊕c 0 − k1 (α,t)) Eλ1 ⊕c 0 (t,t0 ) k0 (α,t)

+(λ1 ⊕c 0)V (y(t))Eλ1 ⊕c 0 (t,t0 ) −

k1 (α,t)(k0 (α,t) + µ(t)(λ1 ⊕c 0 − k1 (α,t))) V (y(t))Eλ1 ⊕c 0 (t,t0 ) k0 (α,t)

≤ (−λ1V (y(t)) + L)

k0 (α,t) + µ(t)(λ1 ⊕c 0 − k1 (α,t)) Eλ1 ⊕c 0 (t,t0 ) k0 (α,t)

+(λ1 ⊕c 0)V (y(t))Eλ1 ⊕c 0 (t,t0 ) −k1 (α,t)

k0 (α,t) + µ(t)(λ1 ⊕c 0 − k1 (α,t)) V (y(t))Eλ1 ⊕c 0 (t,t0 ) k0 (α,t)

 =

(−λ1V (y(t)) + L)

k0 (α,t) + µ(t)(λ1 ⊕c 0 − k1 (α,t)) k0 (α,t)

  k0 (α,t) + µ(t)(λ1 ⊕c 0 − k1 (α,t)) +V (y(t)) λ1 ⊕c 0 − k1 (α,t) Eλ1 ⊕c 0 (t,t0 ) k0 (α,t) ≤ −(λ1 ⊕c 0)Eλ1 ⊕c 0 (t,t0 ),

t ∈ [t0 , ∞).

Consequently, V (y(t)) ≤ (V (y0 ) + 1)E c λ1 (t,t0 ),

t ∈ [t0 , ∞),

Cauchy-Type Problem for a Class of Nonlinear Conformable Dynamic Equations  183

and  |y(t)| ≤

1 p V (0) + 1 E c λ1 (t,t0 ) , λ

t ∈ [t0 , ∞). 

This completes the proof. Exercise 5.5.8 Let T = 4N0 , k1 (α,t) = (1 − α)t 8α ,

k0 (α,t) = αt 8(1−α) ,

α ∈ (0, 1],

t ∈ T.

Consider the IVP 1

1

D7 y = t +y3 +

at + b , 1 + y8

t > 1,

y(1) = 1, where a, b are constants. If 1. V (x) = x, 2. V (x) = x2 , find conditions for the constants a, b so that the trivial solution is exponentially stable.

5.6

ADVANCED PRACTICAL PROBLEMS

Problem 5.6.1 Let V (x) = ax4 , where a > 0. Find Dα (V (t, x)). Problem 5.6.2 Let

Z x

V (x) =

p(s)ds, 0

where p : R → [0, ∞) is a continuous function. Find Dα (V (t, x)). Problem 5.6.3 Let T = 2N0 , k1 (α,t) = (1 − α)(1 + t + t 2 )2α ,

k0 (α,t) = α(1 + t + t 2 )2(1−α) ,

α ∈ (0, 1],

Consider the IVP 1

1

1

D 6 y = 1 + t + (a + b)ty 7 + (ct + d)y 3 , y(1) = 1, where a, b, c, d are constants. If 1. V (x) = x, 2. V (x) = x2 ,

t > 1,

t ∈ T.

184  Conformable Dynamic Equations on Time Scales

find conditions for the constants a, b, c, d so that any solution of the considered IVP, defined on [1, ∞), is uniformly bounded. Problem 5.6.4 Let T = 3N0 , k1 (α,t) = (1 − α)(1 + t 4 )6α ,

k0 (α,t) = α(1 + t 4 )6(1−α) ,

α ∈ (0, 1],

t ∈ T.

Consider the IVP 1 b , D 9 y = at + p 1 + y2

t > 1,

y(1) = 4, where a, b are constants. If 1. V (x) = x, 2. V (x) = x2 , find conditions for the constants a, b so that the trivial solution is exponentially stable.

CHAPTER

6

Higher-Order Linear Conformable Dynamic Equations with Constant Coefficients

Suppose that T is a time scale with forward jump operator and delta differentiation operator σ and ∆, respectively. Let α ∈ (0, 1], n ∈ N, k0 and k1 satisfy (A1), k0 (α, ·), k1 (α, ·) ∈ Crdn−1 (T) for any α ∈ (0, 1]. Also, assume (1.6) holds and t0 ∈ T.

6.1

HOMOGENEOUS HIGHER-ORDER LINEAR CONFORMABLE DYNAMIC EQUATIONS WITH CONSTANT COEFFICIENTS

Consider the equation (Dα )n y + a1 (Dα )n−1 y + · · · + an−1 Dα y + an y = 0

on Tκ

n

(6.1)

where al ∈ R, l ∈ {1, . . . , n}. Definition 6.1.1 The equation (6.1) is called a homogeneous nth-order linear conformable dynamic equation with constant coefficients. Definition 6.1.2 A function y ∈ Crdn (T) that satisfies equation (6.1) will be called a solution of equation (6.1). Theorem 6.1.3 Let y1 , y2 ∈ Crdn (T) be two solutions of equation (6.1). Then b1 y1 + b2 y2 is a solution of equation (6.1) for any b1 , b2 ∈ R. Proof 6.1.4 We have (Dα )n (b1 y1 + b2 y2 ) + a1 (Dα )n−1 (b1 y1 + b2 y2 ) + · · · + an−1 Dα (b1 y1 + b2 y2 ) 185

186  Conformable Dynamic Equations on Time Scales

+an (b1 y1 + b2 y2 ) =

 
(Dα )n (b1 y1 ) + (Dα )n (b2 y2 ) + a1 (Dα )n−1 (b1 y1 ) + (Dα )n−1 (b2 y2 ) + · · · + an−1 (Dα (b1 y1 ) + Dα (b2 y2 )) + an (b1 y1 + b2 y2 )

=

 
b1 (Dα )n y1 + b2 (Dα )n y2 + a1 b1 (Dα )n−1 y1 + b2 (Dα )n−1 y2 + · · · + an−1 (b1 Dα y1 + b2 Dα y2 ) + an (b1 y1 + b2 y2 )

  = b1 (Dα )n y1 + a1 (Dα )n−1 y1 + · · · + an−1 Dα y1 + an y1   +b2 (Dα )n y2 + a1 (Dα )n−1 y2 + · · · + an−1 Dα y2 + an y2 = 0 on

n

Tκ . 

This completes the proof. Now, we will search a solution of equation (6.1) in the form y(t) = Eλ (t,t0 ),

n

t ∈ Tκ ,

where λ is a constant. We have Dα y(t) = Dα Eλ (t,t0 ) = λ Eλ (t,t0 ), (Dα )2 y(t) = (Dα )2 Eλ (t,t0 ) = λ Dα Eλ (t,t0 ) = λ 2 Eλ (t,t0 ), .. . (Dα )n y(t) = (Dα )n Eλ (t,t0 ) = λ n−1 Dα Eλ (t,t0 ) = λ n Eλ (t,t0 ),

n

t ∈ Tκ .

Higher-Order Linear Conformable Dynamic Equations with Constant Coefficients  187

Thus, using (6.1), we get λ n Eλ (t,t0 ) + a1 λ n−1 Eλ (t,t0 ) + · · · + an−1 λ Eλ (t,t0 ) + an Eλ (t,t0 ) = 0, or

n

t ∈ Tκ ,

n


λ n + a1 λ n−1 + · · · + an−1 λ + an Eλ (t,t0 ) = 0,

t ∈ Tκ .

This leads to the following definition. Definition 6.1.5 The equation λ n + a1 λ n−1 + · · · + an−1 λ + an = 0

(6.2)

is called the characteristic equation of equation (6.1). Example 6.1.6 The characteristic equation of the equation 3

(Dα )3 y + 3 (Dα )2 y + 2Dα y + y = 0,

t ∈ Tκ ,

is λ 3 + 3λ 2 + 2λ + 1 = 0. Exercise 6.1.7 Find the characteristic equations for each of the following equations. 1.

4

(Dα )4 y − 2 (Dα )3 y + Dα y = 0, 2. (Dα )5 y − y = 0,

t ∈ Tκ , 3

t ∈ Tκ ,

3.

3

(Dα )3 y − 2 (Dα )2 y − Dα y − 10y = 0,

t ∈ Tκ ,

4.

3

(Dα )3 y + 7 (Dα )2 y + 12Dα y − 13y = 0,

t ∈ Tκ ,

5. (Dα )4 y − (Dα )2 y − 5Dα y − 18y = 0,

4

t ∈ Tκ .

Let λ1 , λ2 , . . ., λn be the roots of the characteristic equation (6.2). Then equation (6.1) can be rewritten in the following way (Dα − λ1 ) (Dα − λ2 ) . . . (Dα − λn ) y = 0 We set (Dα − λ2 ) . . . (Dα − λn ) y = y1

on

(Dα − λ1 ) y1 = 0

n

Then we get

on Tκ ,

n

on Tκ . n

Tκ .

188  Conformable Dynamic Equations on Time Scales

which is a first-order linear conformable equation. Let y1 be its solution. Then we set (Dα − λ3 ) . . . (Dα − λn ) y = y2

on

(Dα − λ2 ) y2 = y1

n

and we obtain

Tκ

n

on Tκ .

We solve the last equation and so on. Example 6.1.8 Let T = 2N0 , k1 (α,t) = (1 − α)t 4α ,

k0 (α,t) = αt 4(1−α) ,

α ∈ (0, 1],

t ∈ T.

Consider the equation  1 2 D 4 y − y = 0. The characteristic equation is λ 2 − 1 = 0. Its roots are λ2 = −1.

λ1 = 1,

The given equation we can rewritten in the form  1  1  D 4 − 1 D 4 + 1 y = 0. We set

 1  D 4 + 1 y = y1 .

Thus we get 1

D 4 y1 − y1 = 0. Its solution is given by y1 (t) = c1 E1 (t,t0 ),

t ∈ T.

Here c1 is a constant. Hence, 1

D 4 y + y = c1 E1 (t,t0 ), or

1

D 4 y = −y + c1 E1 (t,t0 ), Note that σ (t) = 2t, µ(t) = t,  k0

1 ,t 4

 =

1 3 t , 4

t ∈ T.

(6.3)

Higher-Order Linear Conformable Dynamic Equations with Constant Coefficients  189

 k1

1 ,t 4

 =

3 t, 4

t ∈ T.

The equation (6.3) is a first-order linear conformable dynamic equation. Here p(t) = −1, q(t) = c1 E1 (t,t0 ),

t ∈ T.

Then g(t) =

=

− k0 14 ,t

k0 41 ,t + µ(t) p(t) − k1 14 ,t

−1 − 43 t 34 t 2 − 14 t 3
1 3 3 4 t + t −1 − 4 t p(t) − k1

1 4
,t





µ(t)k1

= −

(4 + 3t)(3t 2 − t 3 ) 4(t 3 − 4t − 3t 2 )

= −

(3t + 4)(3t − t 2 ) , 4(t 2 − 3t − 4)

1 4 ,t

t ∈ T,








t 6= 4.

Therefore Z t

y(t) = c2 E p (t,t0 ) + c1

t0

E1 (s,t0 )Eg (σ (s),t)∆ 1 ,t s, 4

t ∈ T,

t 6= 4,

is a solution of the considered equation. Here c2 is a constant. This ends the example. Exercise 6.1.9 Let T = 3Z,  α k1 (α,t) = (1 − α) 1 + t 6 ,

 1−α , k0 (α,t) = α 1 + t 6

Find a solution for each of the following equations. 1.

2.

3.

4.

5.

 1 3 D 4 y − y = 0,

t ∈ T,

 1 2 1 D 2 y − 3D 2 y + 2y = 0,

t ∈ T,

 1 2 1 D 3 y − 6D 3 y + 5y = 0,

t ∈ T,

 1 4 D 2 y − y = 0,  1 3 1 D 3 y − D 3 y = 0,

t ∈ T,

t ∈ T.

α ∈ (0, 1],

t ∈ T.

190  Conformable Dynamic Equations on Time Scales

6.2

NONHOMOGENEOUS HIGHER-ORDER LINEAR CONFORMABLE DYNAMIC EQUATIONS WITH CONSTANT COEFFICIENTS

Consider the equation (Dα )n y + a1 (Dα )n−1 y + · · · + an−1 Dα y + an y = f (t),

n

t ∈ Tκ ,

(6.4)

where al ∈ R, l ∈ {1, . . . , n}, f ∈ Crd (T). Theorem 6.2.1 Let f (t) = g(t) + ih(t),

t ∈ T,

where g, h ∈ Crd (T), g, h : T → R. Suppose that y(t) = u(t) + iv(t),

n

t ∈ Tκ ,

where u, v ∈ Crdn (T), u, v : T → R, is a solution of equation (6.4). Then u and v are solutions of the equations n

(Dα )n y + a1 (Dα )n−1 y + · · · + an−1 Dα y + an y = g(t),

t ∈ Tκ .

(Dα )n y + a1 (Dα )n−1 y + · · · + an−1 Dα y + an y = h(t),

t ∈ Tκ ,

n

respectively. Proof 6.2.2 We have g(t) + ih(t) = (Dα )n (u + iv)(t) + a1 (Dα )n−1 (u + iv)(t) + · · · + an−1 Dα (u + iv)(t) + an (u + iv)(t)   = (Dα ) (u(t) + iv(t)) + a1 (Dα )n−1 u(t) + i (Dα )n−1 v(t) + · · · + an−1 Dα (u(t) + iv(t)) + an (u(t) + iv(t)) = (Dα )n u(t) + a1 (Dα )n−1 u(t) + · · · + an−1 Dα u(t) + an u(t)   +i (Dα )n v(t) + a1 (Dα )n−1 v(t) + · · · + an−1 Dα v(t) + an v(t) , n

t ∈ Tκ . Hence, equating the real and imaginary parts of both sides of the last equation, we get n

(Dα )n u(t) + a1 (Dα )n−1 u(t) + · · · + an−1 Dα u(t) + an u(t) = g(t),

t ∈ Tκ .

(Dα )n v(t) + a1 (Dα )n−1 v(t) + · · · + an−1 Dα v(t) + an v(t) = h(t),

t ∈ Tκ .

This completes the proof.

n



Higher-Order Linear Conformable Dynamic Equations with Constant Coefficients  191

Let λ1 , . . ., λn be the roots of equation (6.2). Then we can rewrite equation (6.4) in the form (Dα − λ1 ) (Dα − λ2 ) . . . (Dα − λn ) y = f (t),

n

t ∈ Tκ .

Let y be a solution of equation (6.4). We set y1 = (Dα − λ2 ) . . . (Dα − λn ) y,

n

t ∈ Tκ .

Then we get Dα y1 − λ1 y1 = f (t),

n

t ∈ Tκ ,

which is a first-order linear conformable dynamic equation. Let y1 be its solution. Then we set n y2 = (Dα − λ3 ) . . . (Dα − λn ) y, t ∈ Tκ and we obtain Dα y2 − λ2 y2 = y1 ,

n

t ∈ Tκ ,

and so on. Example 6.2.3 Let T = Z, k1 (α,t) = (1 − α)(1 + t 2 )α ,

k0 (α,t) = α(1 + t 2 )1−α ,

α ∈ (0, 1],

Consider the equation  1 2  1 3 1 D 4 y + 2 D 4 y − D 4 y − 2y = t,

t ∈ T.

The characteristic equation is λ 3 + 2λ 2 − λ − 2 = 0 or λ 2 (λ + 2) − (λ + 2) = 0, or (λ 2 − 1)(λ + 2) = 0, or (λ − 1)(λ + 1)(λ + 2) = 0. Therefore λ1 = 1,

λ2 = −1,

λ3 = −2.

The considered equation can be rewritten in the following form.  1  1  1  D 4 − 1 D 4 + 1 D 4 + 2 y = t, t ∈ T. Let

 1  1  y1 = D 4 + 1 D 4 + 2 y.

t ∈ T.

192  Conformable Dynamic Equations on Time Scales

Hence, we get the equation 1

D 4 y1 − y1 = t,

t ∈ T,

or 1

D 4 y1 = y1 + t,

t ∈ T.

Here µ(t) = 1, σ (t) = t + 1, p(t) = 1, q(t) = t,  k0  k1

1 ,t 4



1 ,t 4



=

3 1 (1 + t 2 ) 4 , 4

=

1 3 (1 + t 2 ) 4 , 4

g(t) =

=



− k0 14 ,t

k0 14 ,t + µ(t) p(t) − k1 14 ,t    1 3 2 ) 14 − 1 (1 + t 2 ) 34 (1 + t 1 − 34 (1 + t 2 ) 4 4 4 p(t) − k1

1 4
,t





µ(t)k1

3 1 2 4 4 (1 + t )

1 4 ,t




1

+ 1 − 34 (1 + t 2 ) 4

Therefore Z t

y1 (t) = c1 E p (t,t0 ) +

t0

sEg (σ (s),t)∆ 1 ,t s, 4

where c1 is a constant. Now we set  1  D 4 + 2 y = y2 . Then

1

D 4 y2 + y2 = y1 (t), or

1

D 4 y2 = −y2 + y1 (t), Here p1 (t) = −1,

t ∈ T, t ∈ T.

t ∈ T,

,

t ∈ T.

Higher-Order Linear Conformable Dynamic Equations with Constant Coefficients  193

q1 (t) = y1 (t), g1 (t) =

=

1 ,t 4




− k0 14 ,t

k0 14 ,t + µ(t) p1 (t) − k1 14 ,t    1 3 2 ) 14 − 1 (1 + t 2 ) 34 −1 − 43 (1 + t 2 ) 4 (1 + t 4 4 p1 (t) − k1





µ(t)k1

3 1 2 4 4 (1 + t )

1 4 ,t




1

− 1 − 34 (1 + t 2 ) 4

,

t ∈ T.

Therefore Z t

y2 (t) = c2 E p1 (t,t0 ) +

t0

y1 (s)Eg1 (σ (s),t)∆ 1 ,t s, 4

t ∈ T,

where c2 is a constant. Now we get the equation 1

D 4 y = −2y + y2 (t),

t ∈ T.

Here p2 (t) = −2, q2 (t) = y2 (t), g2 (t) =

=



− k0 14 ,t

k0 14 ,t + µ(t) p2 (t) − k1 14 ,t    1 3 2 ) 14 − 1 (1 + t 2 ) 34 −2 − 43 (1 + t 2 ) 4 (1 + t 4 4 p2 (t) − k1

1 ,t
4





µ(t)k1

3 1 2 4 4 (1 + t )

1 4 ,t




1

− 2 − 34 (1 + t 2 ) 4

,

t ∈ T.

Consequently, Z t

y(t) = c3 E p2 (t,t0 ) +

t0

y2 (s)Eg2 (σ (s),t)∆ 1 ,t s, 4

t ∈ T,

where c3 is a constant, is a solution. This ends the example. Exercise 6.2.4 Let T = 4Z, k1 (α,t) = (1 − α) 1 + t 4


4α

,

k0 (α,t) = α(1 + t 4 )4(1−α) ,

Find a solution for each of the following equations. 1.

2.

 1 3 D 4 y + y = t + 1,

t ∈ T,

 1 2 1 D 2 y − 4D 2 y + 3y = t 2 ,

t ∈ T,

α ∈ (0, 1],

t ∈ T.

194  Conformable Dynamic Equations on Time Scales

3.

 1 2 1 D 3 y − 8D 3 y + 7y = t + 2,

4.

5.

6.3

 1 4 D 2 y − 16y = t 2 − t,

t ∈ T,

t ∈ T,

 1 2  1 3 1 D 3 y − 4 D 3 y + 3D 3 y = t,

t ∈ T.

ADVANCED PRACTICAL PROBLEMS

Problem 6.3.1 Find the characteristic equations for each of the following equations. 1. (Dα )3 y − (Dα )2 y − 2Dα y − 3y = 0, 2. (Dα )4 y − Dα y = 0, 3. (Dα )10 y − (Dα )8 y − (Dα )4 y − y = 0. 4. (Dα )5 y − 3 (Dα )3 y − y = 0, 5. (Dα )2 y − 4Dα y + y = 0. Problem 6.3.2 Let T = 3N0 , k1 (α,t) = (1 − α)t 6α ,

k0 (α,t) = αt 6(1−α) ,

α ∈ (0, 1],

Find a solution for each of the following equations. 1.

2.

3.

4.

 1 3 D 4 y − y = 0,

t ∈ T,

 1 2 1 D 2 y − 3D 2 y + 2y = 0,

t ∈ T,

 1 2 1 D 3 y − 6D 3 y + 5y = 0,

t ∈ T,

 1 4 D 2 y − y = 0,

t ∈ T,

t ∈ T.

Higher-Order Linear Conformable Dynamic Equations with Constant Coefficients  195

5.

 1 3 1 D 3 y − D 3 y = 0,

t ∈ T.

Problem 6.3.3 Let T = 3N0 , k1 (α,t) = (1 − α)t 3α ,

k0 (α,t) = αt 3(1−α) ,

α ∈ (0, 1],

Find a solution for each of the following equations. 1.

2.

3.

4.

5.

 1 3 D 4 y + y = t + 1,

t ∈ T,

 1 2 1 D 2 y − 4D 2 y + 3y = t 2 ,

t ∈ T,

 1 2 1 D 3 y − 8D 3 y + 7y = t + 2,  1 4 D 2 y − 16y = t 2 − t,

t ∈ T,

t ∈ T,

 1 2  1 3 1 D 3 y − 4 D 3 y + 3D 3 y = t,

t ∈ T.

t ∈ T.

CHAPTER

7

Second-Order Conformable Dynamic Equations

Suppose that T is a time scale with forward jump operator and delta differentiation operator σ and ∆, respectively. Let α ∈ (0, 1], k0 and k1 satisfy (A1), k0 (α, ·), k1 (α, ·) ∈ Crd1 (Tκ ). Also, assume (1.6) holds and t0 ∈ T.

7.1

HOMOGENEOUS SECOND-ORDER LINEAR CONFORMABLE DYNAMIC EQUATIONS

Consider the equation (Dα )2 y + a(t)Dα y + b(t)y = 0,

2

t ∈ Tκ ,

(7.1)

where a, b ∈ Crd (T). Definition 7.1.1 A function y ∈ Crd2 (T) that satisfies equation (7.1) will be called a solution of equation (7.1). Theorem 7.1.2 Let y1 and y2 be solutions of equation (7.1). Then py1 + qy2 is a solution of equation (7.1) for any p, q ∈ R. Proof 7.1.3 We have (Dα )2 y1 + a(t)Dα y1 + b(t)y1 = 0, (Dα )2 y2 + a(t)Dα y2 + b(t)y2 = 0,

2

t ∈ Tκ .

Hence, (Dα )2 (py1 + qy2 ) + a(t)Dα (py1 + qy2 ) + b(t)(py1 + qy2 )

197

198  Conformable Dynamic Equations on Time Scales

= p (Dα )2 y1 + q (Dα )2 y2 + pa(t)Dα y1 + qa(t)Dα y2 +pb(t)y1 + qb(t)y2   = p (Dα )2 y1 + a(t)Dα y1 + b(t)y1   +q (Dα )2 y2 + a(t)Dα y2 + b(t)y2 = 0,

2

t ∈ Tκ . 

This completes the proof.

Definition 7.1.4 For any two functions y1 , y2 ∈ Crd1 (T) we define the conformable Wronskian by   y1 y2 W (y1 , y2 ) = det . Dα y1 Dα y2

Definition 7.1.5 We say that two solutions y1 and y2 of equation (7.1) form a fundamental set of solutions for (7.1) if any t ∈ Tκ .

W (y1 , y2 )(t) 6= 0 for

Definition 7.1.6 With Rc+ we will denote the set of all functions f : T → R such that k0 + µ( f − k1 ) > 0, Example 7.1.7 Let a, b ∈ R be such that √ −a ± a2 − 4b ∈ Rc+ , 2

k0 − µk1 6= 0

on T.

a2 − 4b 6= 0 on

T.

Such T, k0 and k1 exist. Really, let T = Z, k0 (α,t) = α, and

k1 (α,t) = 1 − α, 1 a= , 4

α ∈ (0, 1], 3 α= . 4

b = 0,

Then  k0

3 ,t 4

 =

3 , 4

t ∈ T,

Second-Order Conformable Dynamic Equations  199

 k1

3 ,t 4

 = 1− =

 k0

 3 ,t + µ(t) 4

√  ! 3 −a − a2 − 4b − k1 ,t = 2 4 =

 k0

 3 ,t + µ(t) 4

− 14 − 41 1 − 2 4   3 1 1 − + 4 4 4 3 + 4

3 1 − 4 2

=

1 , 4

√  ! 3 −a + a2 − 4b − k1 ,t = 2 4

√ −a + a2 − 4b , g1 = 2

1 , 4

=

= We set

3 4

3 1 − 4 4 1 , 2

t ∈ T.

√ −a − a2 − 4b g2 = . 2

Note that (Dα )2 Eg1 + aDα Eg1 + bEg1  !2 √ √ 2 2 −a + a − 4b −a + a − 4b +a + b Eg1 =  2 2 

=

= = 0

! √ √ a2 − 2a a2 − 4b + a2 − 4b −a + a2 − 4b +a + b Eg 1 4 2 ! √ √ a2 − a a2 − 4b − 2b −a2 + a a2 − 4b + + b Eg 1 2 2 on

2

Tκ ,

and (Dα )2 Eg2 + aDα Eg2 + bEg2

!

200  Conformable Dynamic Equations on Time Scales

 !2 √ √ 2 2 −a − a − 4b −a − a − 4b =  +a + b Eg2 2 2 

=

= = 0

! √ √ a2 + 2a a2 − 4b + a2 − 4b −a − a2 − 4b +a + b Eg 2 4 2 ! √ √ a2 + a a2 − 4b − 2b −a2 − a a2 − 4b + + b Eg 2 2 2 2

Tκ ,

on

i.e., Eg1 and Eg2 are solutions of (7.1). Next,  W (y1 , y2 )(t) = det

Eg 1 Eg2 g1 Eg1 g2 Eg2



= (g2 − g1 )Eg1 Eg2 p = − a2 − 4bEg1 Eg2 6= 0

on Tκ .

Therefore Eg1 and Eg2 form a fundamental system for (7.1). This ends the example. Theorem 7.1.8 Let y1 and y2 be solutions of equation (7.1). Then  σ  k0 − µk1 y1 yσ2 W (y1 , y2 ) = det . D α y1 D α y2 k0 Proof 7.1.9 We have  σ  y1 yσ2 det = Dα y1 Dα y2 = = = = = This completes the proof.

1 det k0



k0 yσ1 k0 yσ2 Dα y1 Dα y2

1 det k0



k0 y1 + k0 µy∆1 k0 y2 + k0 µy∆2 Dα y1 Dα y2







(k0 − k1 µ)y1 + µDα y1 (k0 − k1 µ)y2 + µDα y2 D α y1 D α y2   1 (k0 − k1 µ)y1 (k0 − k1 µ)y2 det Dα y1 Dα y2 k0   k0 − k1 µ y1 y2 det Dα y1 Dα y2 k0 1 det k0



k0 − k1 µ W (y1 , y2 ). k0 

Second-Order Conformable Dynamic Equations  201

Theorem 7.1.10 Let y1 and y2 be solutions of equation (7.1). Then   yσ1 yσ2 k1 (k0 − µk1 ) α D W (y1 , y2 ) = det − W (y1 , y2 ). (Dα )2 y1 (Dα )2 y2 k0 Proof 7.1.11 We have   D W (y1 , y2 ) = D det α

α

y1 y2 α D y1 D α y2



= Dα (y1 Dα y2 − y2 Dα y1 ) = Dα (y1 Dα y2 ) − Dα (y2 Dα y1 ) = (Dα y1 ) (Dα y2 ) + yσ1 (Dα )2 y2 − k1 yσ1 Dα y2 − (Dα y2 ) (Dα y1 ) − yσ2 (Dα )2 y1 + k1 yσ2 Dα y1  = det  = det

yσ1 yσ2 α 2 α 2 (D ) y1 (D ) y2



yσ1 yσ2 (Dα )2 y1 (Dα )2 y2



 − k1 det −

yσ1 yσ2 α D y1 Dα y2



k1 (k0 − µk1 ) W (y1 , y2 ). k0 

This completes the proof. Theorem 7.1.12 Let y1 and y2 be solutions of equation (7.1). Then   k0 − µk1 k1 (k0 − µk1 ) bµ α + − W (y1 , y2 ). D W (y1 , y2 ) = − a k0 k0 k0 Proof 7.1.13 Note that (Dα )2 y1 = −aDα y1 − by1 , (Dα )2 y2 = −aDα y2 − by2 . Then, using Theorem 7.1.8 and Theorem 7.1.10, we get   yσ1 yσ2 k1 (k0 − µk1 ) α D W (y1 , y2 ) = det − W (y1 , y2 ) α 2 α 2 (D ) y1 (D ) y2 k0   k1 (k0 − µk1 ) yσ1 yσ2 = det − W (y1 , y2 ) α α −aD y1 − by1 −aD y2 − by2 k0    σ  yσ1 yσ2 y1 yσ2 = det + det −aDα y1 −aDα y2 −by1 −by2

202  Conformable Dynamic Equations on Time Scales

k1 (k0 − µk1 ) W (y1 , y2 ) k0  σ    y1 yσ2 y1 + µy∆1 y2 + µy∆2 = −a det − b det Dα y1 Dα y2 y1 y2 −

−

k1 (k0 − µk1 ) W (y1 , y2 ) k0

k0 − µk1 bµ = −a W (y1 , y2 ) − det k0 k0



k0 y∆1 k0 y∆2 y1 y2



k1 (k0 − µk1 ) W (y1 , y2 ) k0   k0 − µk1 k1 (k0 − µk1 ) = − a W (y1 , y2 ) + k0 k0   α bµ D y1 D α y2 − det y1 y2 k0   k0 − µk1 k1 (k0 − µk1 ) bµ = − a + W (y1 , y2 ) + W (y1 , y2 ) k0 k0 k0   k0 − µk1 k1 (k0 − µk1 ) bµ = − a + − W (y1 , y2 ). k0 k0 k0 −



This completes the proof. Theorem 7.1.14 (Abel’s Formula) Let y1 and y2 be solutions of equation (7.1). Then W (y1 , y2 )(t) = E p (t,t0 )W (y1 , y2 )(t0 ), where

  k0 − µk1 k1 (k0 − µk1 ) bµ p=− a + − . k0 k0 k0

Proof 7.1.15 By Theorem 7.1.12, we obtain Dα W (y1 , y2 ) = pW (y1 , y2 ). 

Hence, we get the assertion. This completes the proof. We will now introduce hyperbolic functions. Consider the equation (Dα )2 − γ 2 y = 0,

2

t ∈ Tκ ,

(7.2)

Second-Order Conformable Dynamic Equations  203

where γ ∈ R, γ > 0 and ±γ ∈ Rc . Let y1 (t) = Coshγ (t,t0 ),

2

y2 (t) = Sinhγ (t,t0 ). t ∈ Tκ .

Then Dα y1 (t) = γSinhγ (t,t0 ), (Dα )2 y1 (t) = γ 2Coshγ (t,t0 ), Dα y2 (t) = γCoshγ (t,t0 ), 2

(Dα )2 y2 (t) = γ 2 Sinhγ (t,t0 ),

t ∈ Tκ .

Hence, (Dα )2 y1 (t) − γ 2 y1 (t) = γ 2Coshγ (t,t0 ) − γ 2Coshγ (t,t0 ) = 0,

2

t ∈ Tκ ,

and (Dα )2 y2 (t) − γ 2 y2 (t) = γ 2 Sinhγ (t,t0 ) − γ 2 Sinhγ (t,t0 ) = 0,

2

t ∈ Tκ .

Consequently, y1 and y2 are solutions of equation (7.2). Note that γ ⊕c (−γ) = =

(γ − γ − k1 )k0 + µ(γ − k1 )(−γ − k1 ) k0 −k0 k1 − µ(γ − k1 )(γ + k1 ) k0

µ (γ − k1 )(γ + k1 ), k0   µ k0 + µ (γ ⊕c (−γ) − k1 ) = k0 + µ −k1 − (γ − k1 )(γ + k1 ) − k1 k0 = −k1 −

= k0 − 2µk1 −

µ2 (γ − k1 )(γ + k1 ) k0

= k0 − 2µk1 −

µ2 2 µ2 2 γ + k1 k0 k0

=

k02 − 2µk0 k1 − µ 2 γ 2 + µ 2 k12 k0

(7.3)

204  Conformable Dynamic Equations on Time Scales

=

(k0 − µk1 )2 − µ 2 γ 2 k0

=

(k0 − µk1 − µγ)(k0 − µk1 + µγ) k0

6= 0,

on

T.

Hence, 

y1 (t) y2 (t) Dα y1 (t) Dα y2 (t)



Coshγ (t,t0 ) Sinhγ (t,t0 ) γSinhγ (t,t0 ) γCoshγ (t,t0 )

W (y1 , y2 )(t) = det = det






2
2 = γ Coshγ (t,t0 ) − γ Sinhγ (t,t0 ) = γ




2 
2 Coshγ (t,t0 ) − Sinhγ (t,t0 )

= γEγ⊕c (−γ) (t,t0 ) 6= 0,

2

t ∈ Tκ .

Therefore, y1 and y2 , defined by (7.3), form a fundamental set of solutions of equation (7.2) and y(t) = c1 y1 (t) + c2 y2 (t) = c1Coshγ (t,t0 ) + c2 Sinhγ (t,t0 ),

2

t ∈ Tκ ,

where c1 , c2 ∈ R, is the general solution of equation (7.2). Now we consider the equation
2 (Dα )2 y − 2 f Dα y + f 2 − g2 y = 0, t ∈ Tκ , (7.4) where f , g ∈ R, g 6= 0, f ± g ∈ Rc . Let y1 (t) = Ch f g (t,t0 ),

y2 (t) = Sh f g (t,t0 ),

Note that Dα y1 (t) = Dα Ch f g (t,t0 ) =

fCh f g (t,t0 ) + gSh f g (t,t0 ),

2

t ∈ Tκ .

(7.5)

Second-Order Conformable Dynamic Equations  205

(Dα )2 y1 (t) = =

f Dα Ch f g (t,t0 ) + gDα Sh f g (t,t0 )
f fCh f g (t,t0 ) + gSh f g (t,t0 )
+g gCh f g (t,t0 ) + f Sh f g (t,t0 )

=


f 2 + g2 Ch f g (t,t0 ) + 2 f gSh f g (t,t0 ),

Dα y2 (t) = Dα Sh f g (t,t0 ) = gCh f g (t,t0 ) + f Sh f g (t,t0 ), (Dα )2 y2 (t) = gDα Ch f g (t,t0 ) + f Dα Sh f g (t,t0 )
= g fCh f g (t,t0 ) + gSh f g (t,t0 )
+ f gCh f g (t,t0 ) + f Sh f g (t,t0 ) =


f 2 + g2 Sh f g (t,t0 ) + 2 f gCh f g (t,t0 ),

Hence,
(Dα )2 y1 (t) − 2 f Dα y1 (t) + f 2 − g2 y1 (t) =


f 2 + g2 Ch f g (t,t0 ) + 2 f gSh f g (t,t0 )
−2 f fCh f g (t,t0 ) + gSh f g (t,t0 )
+ f 2 − g2 Ch f g (t,t0 )

= 0,

2

t ∈ Tκ ,

and
(Dα )2 y2 (t) − 2 f Dα y2 (t) + f 2 − g2 y2 (t) =


f 2 + g2 Sh f g (t,t0 ) + 2 f gCh f g (t,t0 )
−2 f gCh f g (t,t0 ) + f Sh f g (t,t0 )

2

t ∈ Tκ .

206  Conformable Dynamic Equations on Time Scales


+ f 2 − g2 Sh f g (t,t0 ) 2

t ∈ Tκ .

= 0,

Therefore y1 and y2 , defined by (7.5), are solutions of equation (7.4). Also, ( f + g) ⊕c ( f − g) = =

(2 f − k1 )k0 + µ( f + g − k1 )( f − g − k1 ) k0
(2 f − k1 )k0 + µ f 2 − 2 f k1 + k12 − g2 on k0

2

Tκ ,

and k0 + µ (( f + g) ⊕c ( f − g) − k1 ) = k0 + µ

!
(2 f − k1 )k0 + µ f 2 − 2 f k1 + k12 + g2 − k1 k0

=

k02 + 2µ f k0 − 2µk0 k1 + µ 2 f 2 − 2µ 2 f k1 + µ 2 k12 − µ 2 g2 k0

=

(k0 + µ f )2 − 2µk0 k1 + 2µ 2 f k1 + µ 2 k12 − µ 2 g2 k0

=

(k0 + µ f )2 − 2µk1 (k0 + µ f ) + µ 2 k12 − µ 2 g2 k0

=

(k0 + µ( f − k1 ))2 − µ 2 g2 k0

=

(k0 + µ( f − g − k1 )) (k0 + µ( f + g − k1 )) k0

6= 0

2

on Tκ .

Next, 

y1 (t) y2 (t) α D y1 (t) Dα y2 (t)



Ch f g (t,t0 ) Sh f g (t,t0 ) fCh f g (t,t0 ) + gSh f g (t,t0 ) gCh f g (t,t0 ) + f Sh f g (t,t0 )

W (y1 , y2 )(t) = =




2 = g Ch f g (t,t0 ) + fCh f g (t,t0 )Sh f g (t,t0 )
2 − fCh f g (t,t0 )Sh f g (t,t0 ) − g Sh f g (t,t0 )



Second-Order Conformable Dynamic Equations  207

= g




2
2  Ch f g (t,t0 ) − Sh f g (t,t0 )

= gE( f +g)⊕c ( f −g) (t,t0 ) 6= 0,

2

t ∈ Tκ .

Consequently, y1 and y2 , defined by (7.5), form a fundamental set of solutions of equation (7.4) and y(t) = c1 y1 (t) + c2 y2 (t) = c1Ch f g (t,t0 ) + c2 Sh f g (t,t0 ), where c1 , c2 ∈ R, is its general solution. Now we consider the equation (Dα )2 y + f 2 y = 0,

2

t ∈ Tκ ,

(7.6)

where f ∈ R, f 6= 0, ±i f ∈ Rc . Let y1 (t) = Cos f (t,t0 ),

y2 (t) = Sin f (t,t0 ),

2

t ∈ Tκ .

Then Dα y1 (t) = Dα Cos f (t,t0 ) = − f Sin f (t,t0 ), (Dα )2 y1 (t) = − f Dα Sin f (t,t0 ) = − f 2Cos f (t,t0 ), Dα y2 (t) = Dα Sin f (t,t0 ) = (Dα )2 y2 (t) =

fCos f (t,t0 ), f Dα Cos f (t,t0 )

= − f 2 Sin f (t,t0 ),

2

t ∈ Tκ .

Hence, (Dα )2 y1 (t) + f 2 y1 (t) = − f 2Cos f (t,t0 ) + f 2Cos f (t,t0 ) = 0,

(7.7)

208  Conformable Dynamic Equations on Time Scales

(Dα )2 y2 (t) + f 2 y2 (t) = − f 2 Sin f (t,t0 ) + f 2 Sin f (t,t0 ) 2

t ∈ Tκ .

= 0,

Therefore, the functions y1 and y2 , defined by (7.7), are solutions of equation (7.6). Next, −k1 k0 − µ(i f − k1 )(i f + k1 ) , k0 !
−k1 k0 + µ f 2 + k12 − k1 k0 + µ ((i f ) ⊕c (−i f ) − k1 ) = k0 + µ k0 (i f ) ⊕c (−i f ) =

=

k02 − 2µk0 k1 + µ 2 k12 + µ 2 f 2 k0

=

(k0 − µk1 )2 − (iµ f )2 k0

=

(k0 − µ(k1 + i f )) (k0 − µ(k1 − i f )) k0

=

(k0 + µ(−i f − k1 )) (k0 + µ(i f − k1 )) k0

6= 0

2

on Tκ ,

and 

y1 (t) y2 (t) α D y1 (t) Dα y2 (t)



Cos f (t,t0 ) − f Sin f (t,t0 )

W (y1 , y2 ) = det



Sin f (t,t0 ) = det fCos f (t,t0 ) 
2 
2 = f Cos f (t,t0 ) + Sin f (t,t0 ) =



f E(i f )⊕c (−i f ) (t,t0 )

6= 0

2

on Tκ .

Consequently, the functions y1 and y2 , defined by (7.7), form a fundamental set of solutions of equation (7.6) and y(t) = c1 y1 (t) + c2 y2 (t) = c1Cos f (t,t0 ) + c2 Sin f (t,t0 ),

Second-Order Conformable Dynamic Equations  209

where c1 , c2 ∈ R, is its general solution. Now we consider the equation
(Dα )2 y − 2 f Dα y + f 2 + g2 y = 0,

2

t ∈ Tκ ,

(7.8)

where f , g ∈ R, f ± ig ∈ Rc , g 6= 0. Let y1 (t) = C f g (t,t0 ),

y2 (t) = S f g (t,t0 ),

2

t ∈ Tκ .

(7.9)

We have Dα y1 (t) = Dα C f g (t,t0 ) = (Dα )2 y1 (t) = =

fC f g (t,t0 ) − gS f g (t,t0 ), f Dα C f g (t,t0 ) − gDα S f g (t,t0 )
f fC f g (t,t0 ) − gS f g (t,t0 )
−g gC f g (t,t0 ) + f S f g (t,t0 )

=


f 2 − g2 C f g (t,t0 ) − 2 f gS f g (t,t0 ),

Dα y2 (t) = Dα S f g (t,t0 ) = gC f g (t,t0 ) + f S f g (t,t0 ), (Dα )2 y2 (t) = gDα C f g (t,t0 ) + f Dα S f g (t,t0 )
= g fC f g (t,t0 ) − gS f g (t,t0 )
+ f gC f g (t,t0 ) + f S f g (t,t0 ) =


f 2 − g2 S f g (t,t0 ) + 2 f gC f g (t,t0 ),

Hence,
(Dα )2 y1 (t) − 2 f Dα y1 (t) + f 2 + g2 y1 (t)

2

t ∈ Tκ .

210  Conformable Dynamic Equations on Time Scales

=


f 2 − g2 C f g (t,t0 ) − 2 f gS f g (t,t0 ) −2 f 2C f g (t,t0 ) + 2 f gS f g (t,t0 )
+ f 2 + g2 C f g (t,t0 )

= 0,

2

t ∈ Tκ ,

and
(Dα )2 y2 (t) − 2 f Dα y2 (t) + f 2 + g2 y2 (t) =


f 2 − g2 S f g (t,t0 ) + 2 f gC f g (t,t0 ) −2 f gC f g (t,t0 ) − 2 f 2 S f g (t,t0 )
+ f 2 + g2 S f g (t,t0 )

= 0,

2

t ∈ Tκ .

Therefore, the functions y1 and y2 , defined by (7.9), are solutions of equation (7.8). Next, ( f + ig) ⊕c ( f − ig) = =

=

(2 f − k1 )k0 + µ( f + ig − k1 )( f − ig − k1 ) k0
(2 f − k1 )k0 + µ ( f − k1 )2 + g2 k0
(2 f − k1 )k0 + µ f 2 − 2 f k1 + k12 + g2 on k0

2

Tκ ,

and k0 + µ (( f + ig) ⊕c ( f − ig) − k1 ) = k0 + µ

!
(2 f − k1 )k0 + µ f 2 − 2 f k1 + k12 + g2 − k1 k0

=

k02 − 2µk0 k1 + 2µ f k0 + µ 2 f 2 − 2µ 2 f k1 + µ 2 k12 + µ 2 g2 k0

=

(k0 + µ f )2 − 2µk1 (k0 + µ f ) + µ 2 k12 + µ 2 g2 k0

=

(k0 + µ( f − k1 ))2 + µ 2 g2 k0

Second-Order Conformable Dynamic Equations  211

=

(k0 + µ( f + ig − k1 )) (k0 + µ( f − ig − k1 )) k0

6= 0

2

on Tκ .

Also, 

y1 (t) y2 (t) Dα y1 (t) Dα y2 (t)



C f g (t,t0 ) S f g (t,t0 ) fC f g (t,t0 ) − gS f g (t,t0 ) gC f g (t,t0 ) + f S f g (t,t0 )

W (y1 , y2 )(t) = det = det






2 = g C f g (t,t0 ) + fC f g (t,t0 )S f g (t,t0 )
2 − fC f g (t,t0 )S f g (t,t0 ) + g S f g (t,t0 ) = g




2 
2 C f g (t,t0 ) + S f g (t,t0 )

= gE( f +ig)⊕c ( f −ig) (t,t0 ) 6= 0,

2

t ∈ Tκ .

Consequently, the functions y1 and y2 , defined by (7.9), form a fundamental set of solutions for equation (7.8) and y(t) = c1 y1 (t) + c2 y2 (t) = c1C f g (t,t0 ) + c2 S f g (t,t0 ),

2

t ∈ Tκ ,

where c1 , c2 ∈ R, is its general solution. Exercise 7.1.16 Find the general solution for each of the following equations. 2

1. (Dα )2 y − 9y = 0, t ∈ Tκ , 2

2. (Dα )2 y − 36y = 0, t ∈ Tκ , 2

3. (Dα )2 y + 4y = 0, t ∈ Tκ , 2

4. 9 (Dα )2 y + y = 0, t ∈ Tκ , 2

5. (Dα )2 y − 4Dα y − 12y = 0, t ∈ Tκ , 2

6. (Dα )2 y − Dα y + y = 0, t ∈ Tκ .

212  Conformable Dynamic Equations on Time Scales

7.2

REDUCTION OF ORDER

In this section we suppose that the graininess function µ is delta differentiable on Tκ . Consider the equation (Dα )2 y − 2γDα y + γ 2 y = 0,

2

t ∈ Tκ ,

(7.10)

where γ ∈ R, γ > 0, γ ∈ Rc . Note that 2

t ∈ Tκ ,

y1 (t) = Eγ (t,t0 ),

is a solution of equation (7.10). We will search for another solution of equation (7.10) in the form 2 y2 (t) = x(t)Eγ (t,t0 ), t ∈ Tκ , where x ∈ Crd2 (T) will be determined so that y1 and y2 are linearly independent solutions of equation (7.10). To this end, note that
Dα y2 (t) = Dα x(t)Eγ (t,t0 ) = (Dα x(t)) Eγσ (t,t0 ) + x(t)Dα Eγ (t,t0 ) −k1 (α,t)x(t)Eγσ (t,t0 )   γ − k1 (α,t) Eγ (t,t0 ) = (D x(t)) 1 + µ(t) k0 (α,t) α

+γx(t)Eγ (t,t0 )   γ − k1 (α,t) −k1 (α,t) 1 + µ(t) x(t)Eγ (t,t0 ) k0 (α,t)   γ − k1 (α,t) α = (D x(t) − k1 (α,t)x(t)) 1 + µ(t) Eγ (t,t0 ) k0 (α,t) +γx(t)Eγ (t,t0 ),

2

t ∈ Tκ ,

and     γ − k1 (α,t) α (D ) y2 (t) = D (D x(t) − k1 (α,t)x(t)) 1 + µ(t) Eγ (t,t0 ) k0 (α,t) α 2

α


+γDα x(t)Eγ (t,t0 )    γ − k1 (α,t) α = D (D x(t) − k1 (α,t)x(t)) 1 + µ(t) Eγσ (t,t0 ) k0 (α,t) α

Second-Order Conformable Dynamic Equations  213

  γ − k1 (α,t) Dα Eγ (t,t0 ) + (D x(t) − k1 (α,t)x(t)) 1 + µ(t) k0 (α,t)   γ − k1 (α,t) α −k1 (α,t) (D x(t) − k1 (α,t)x(t)) 1 + µ(t) Eγσ (t,t0 ) k0 (α,t) α

+γ (Dα x(t)) Eγσ (t,t0 ) + γx(t)Dα Eγ (t,t0 ) −γx(t)k1 (α,t)Eγσ (t,t0 )    γ − k1 (α,t) α = D (D x(t) − k1 (α,t)x(t)) 1 + µ(t) k0 (α,t)   γ − k1 (α,t) × 1 + µ(t) Eγ (t,t0 ) k0 (α,t)   γ − k1 (α,t) α Eγ (t,t0 ) +γ (D x(t) − k1 (α,t)x(t)) 1 + µ(t) k0 (α,t)   γ − k1 (α,t) 2 α Eγ (t,t0 ) −k1 (α,t) (D x(t) − k1 (α,t)x(t)) 1 + µ(t) k0 (α,t)   γ − k1 (α,t) α +γ (D x(t)) 1 + µ(t) Eγ (t,t0 ) k0 (α,t) α

+γ 2 x(t)Eγ (t,t0 )   γ − k1 (α,t) −γx(t)k1 (α,t) 1 + µ(t) Eγ (t,t0 ) k0 (α,t)    γ − k1 (α,t) = Dα (Dα x(t) − k1 (α,t)x(t)) 1 + µ(t) k0 (α,t)   γ − k1 (α,t) Eγ (t,t0 ) × 1 + µ(t) k0 (α,t)   γ − k1 (α,t) α +2γ (D x(t) − k1 (α,t)x(t)) 1 + µ(t) Eγ (t,t0 ) k0 (α,t)   γ − k1 (α,t) 2 α −k1 (α,t) (D x(t) − k1 (α,t)x(t)) 1 + µ(t) Eγ (t,t0 ) k0 (α,t) +γ 2 x(t)Eγ (t,t0 ),

2

t ∈ Tκ .

Hence, 0 = (Dα )2 y2 (t) − 2γDα y2 (t) + γ 2 y2 (t)

214  Conformable Dynamic Equations on Time Scales

   γ − k1 (α,t) α = D (D x(t) − k1 (α,t)x(t)) 1 + µ(t) k0 (α,t)   γ − k1 (α,t) × 1 + µ(t) Eγ (t,t0 ) k0 (α,t)   γ − k1 (α,t) Eγ (t,t0 ) +2γ (Dα x(t) − k1 (α,t)x(t)) 1 + µ(t) k0 (α,t)   γ − k1 (α,t) 2 α −k1 (α,t) (D x(t) − k1 (α,t)x(t)) 1 + µ(t) Eγ (t,t0 ) k0 (α,t) α

+γ 2 x(t)Eγ (t,t0 )   γ − k1 (α,t) −2γ (Dα x(t) − k1 (α,t)x(t)) 1 + µ(t) Eγ (t,t0 ) k0 (α,t) −2γ 2 x(t)Eγ (t,t0 ) + γ 2 x(t)Eγ (t,t0 )    γ − k1 (α,t) α = D (D x(t) − k1 (α,t)x(t)) 1 + µ(t) k0 (α,t)   γ − k1 (α,t) Eγ (t,t0 ) × 1 + µ(t) k0 (α,t)   γ − k1 (α,t) 2 α −k1 (α,t) (D x(t) − k1 (α,t)x(t)) 1 + µ(t) Eγ (t,t0 ), k0 (α,t) α

2

t ∈ Tκ . Set   γ − k1 (α,t) v(t) = (D x(t) − k1 (α,t)x(t)) 1 + µ(t) , k0 (α,t) α

2

t ∈ Tκ .

Then   γ − k1 (α,t) 0 = (D v(t)) 1 + µ(t) Eγ (t,t0 ) k0 (α,t)   γ − k1 (α,t) −k1 (α,t)v(t) 1 + µ(t) Eγ (t,t0 ) k0 (α,t)   γ − k1 (α,t) α = (D v(t) − k1 (α,t)v(t)) 1 + µ(t) Eγ (t,t0 ), k0 (α,t) α

Hence, Dα v(t) = k1 (α,t)v(t),

2

t ∈ Tκ ,

and v(t) = c1 Ek1 (t,t0 ) = c1 ,

2

t ∈ Tκ ,

2

t ∈ Tκ .

Second-Order Conformable Dynamic Equations  215

where c1 is a constant. Take c1 = 1. Then 2

t ∈ Tκ ,

v(t) = 1, and

  γ − k1 (α,t) = 1, (D x(t) − k1 (α,t)x(t)) 1 + µ(t) k0 (α,t)

t ∈ Tκ ,

k0 (α,t) , k0 (α,t) + µ(t)(γ − k1 (α,t))

t ∈ Tκ .

2

α

or Dα x(t) = k1 (α,t)x(t) +

2

Consequently, Z t

x(t) = c2 Ek1 (t,t0 ) + Z t

= c2 +

t0

t0

k0 (α, s) E0 (σ (s),t)∆α,t s k0 (α, s) + µ(s)(γ − k1 (α, s))

k0 (α, s) E0 (σ (s),t)∆α,t s, k0 (α, s) + µ(s)(γ − k1 (α, s))

2

t ∈ Tκ , where c2 is a constant. Take c2 = 0. Then Z t

k0 (α, s) E0 (σ (s),t)∆α,t s, k0 (α, s) + µ(s)(γ − k1 (α, s))

x(t) = t0

and

Z y2 (t) =

t

t0

2

t ∈ Tκ ,

 k0 (α, s) E0 (σ (s),t)∆α,t s Eγ (t,t0 ), k0 (α, s) + µ(s)(γ − k1 (α, s))

κ2

t ∈ T . By the above computations, we have Dα y2 (t) = Eγ (t,t0 ) + γx(t)Eγ (t,t0 ) = (1 + γx(t))Eγ (t,t0 ),

2

t ∈ Tκ .

From here, 

y1 (t) y2 (t) Dα y1 (t) Dα y2 (t)



Eγ (t,t0 ) x(t)Eγ (t,t0 ) γEγ (t,t0 ) (1 + γx(t))Eγ (t,t0 )

W (y1 , y2 )(t) = det = det






2
2 = (1 + γx(t)) Eγ (t,t0 ) − γx(t) Eγ (t,t0 ) =


2 Eγ (t,t0 )

6= 0,

2

t ∈ Tκ .

216  Conformable Dynamic Equations on Time Scales

Consequently, y1 and y2 form a fundamental set of solutions of equation (7.10). Therefore the general solution of equation (7.10) is given by the expression y(t) = c1 y1 (t) + c2 y2 (t) = c1 Eγ (t,t0 ) Z

t

 k0 (α, s) E0 (σ (s),t)∆α,t s Eγ (t,t0 ) +c2 t0 k0 (α, s) + µ(s)(γ − k1 (α, s)) Z t   k0 (α, s) = c1 + c2 E0 (σ (s),t)∆α,t s Eγ (t,t0 ), t0 k0 (α, s) + µ(s)(γ − k1 (α, s)) 2

t ∈ Tκ , where c1 and c2 are constants. Example 7.2.1 Let T = N and k1 (α,t) = (1 − α)t 2α ,

k0 (α,t) = αt 2(1−α) ,

t ∈ T,

Consider the equation  1 2 1 D 2 y y − 2D 2 y + y = 0,

t ∈ T.

The characteristic equation is λ 2 − 2λ + 1 = 0. Then λ1 = λ2 = 1. Here µ(t) = 1,  k1  k0

1 ,t 2



1 ,t 2



=

1 t, 2

=

1 t, 2

t ∈ T.

We have y1 (t) = E1 (t, 1) Rt

= e

1 1 µ(τ)

log

1−k1 12 ,τ 1+µ(τ) k0 21 ,τ

( ) ( )

! ∆τ

α ∈ (0, 1].

Second-Order Conformable Dynamic Equations  217   1− 1 τ 1+ 1 2 ∆τ

Rt

= e

1 log

Rt

= e

(1+ 2−τ τ )∆τ

1 log

Rt

= e

2τ

1 log

τ+2−τ ∆τ τ

t−1

2

= e∑τ=1 log τ t−1

=

2

∏ τ,

t ∈ T,

t ≥ 2,

τ=1

is a solution of the considered equation. Next, !
Z t k0 12 , s

E0 (σ (s),t)∆α,t s E1 (t, 1) y2 (t) = 1 1 1 k0 2 , s + 1 − k1 2 , s !
Z t k0 12 , s E0 (t, σ (s))

E0 (σ (s),t)
∆s E1 (t, 1) = 1 1 k0 12 , s 1 k0 2 , s + 1 − k1 2 , s ! Z t 1

∆s E1 (t, 1) = 1 1 1 k0 2 , s + 1 − k1 2 , s ! Z t 1 = ∆s E1 (t, 1) 1 1 1 2s+1− 2s Z t  = ∆s E1 (t, 1) 1

= (t − 1)E1 (t, 1) t−1 2 = (t − 1) ∏ , τ τ=1

t ∈ T,

t ≥ 2.

Hence, the general solution of the considered equation is y(t) = c1 y1 (t) + c2 y2 (t) t−1

t−1 2 2 + c2 (t − 1) ∏ τ=1 τ τ=1 τ

= c1 ∏

t−1

= (c1 − c2 + c2t)

2 τ |tau=1

∏

218  Conformable Dynamic Equations on Time Scales t−1 2 = (c3 + c2t) ∏ , τ=1 τ

where c1 and c2 are constants, c3 = c1 − c2 . Here ends the example. Exercise 7.2.2 Let T = 2N0 , k1 (α,t) = (1 − α)t 4α ,

k0 (α,t) = αt 4(1−α) ,

α ∈ (0, 1],

t ∈ T.

Find a fundamental set of solutions for each of the following equations.  1 2 1 1. D 3 y − 2D 3 y + y = 0, t ∈ T, 2.

 1 2 1 D 4 y − 6D 4 y + 9y = 0, t ∈ T,

3.

 1 2 1 D 2 y − 4D 2 y + 4y = 0, t ∈ T,

 1 2 1 1 D 2 y − D 2 y + y = 0, t ∈ T, 4  1 2 1 5. D 6 y − 8D 6 y + 16y = 0, t ∈ T. 4.

7.3

METHOD OF FACTORING

Consider the equation 2

Dα (Dα y − py) (t) − q(t) (Dα y(t) − p(t)y(t)) = 0,

t ∈ Tκ ,

where p ∈ Crd1 (T), q ∈ Crd (T), p, q ∈ Rc . Set v(t) = Dα y(t) − p(t)y(t),

2

t ∈ Tκ .

Then

2

Dα v(t) = q(t)v(t),

t ∈ Tκ .

v(t) = c1 Eq (t,t0 ),

r ∈ Tκ ,

Hence,

2

where c1 is a constant. From here, Dα y(t) = p(t)y(t) + c1 Eq (t,t0 ), Then

2

t ∈ Tκ .

Z t

y(t) = c2 E p (t,t0 ) + c1

t0

Eq (s,t0 )Eg (σ (s),t)∆α,t s,

2

t ∈ Tκ ,

where c2 is a constant and g(t) =

(p(t) − k1 (α,t))(µ(t)k1 (α,t) − k0 (α,t)) , k0 (α,t) + µ(t)(p(t) − k1 (α,t))

is the general solution of equation (7.11).

t ∈ T,

(7.11)

Second-Order Conformable Dynamic Equations  219

Example 7.3.1 Let T = N0 , k1 (α,t) = 1 − α,

α ∈ (0, 1],

k0 (α,t) = α,

Consider the equation  3 2 3 D 4 y − 2D 4 y + y = 0,

t ∈ T.

Here σ (t) = t + 1, µ(t) = 1, α =  k1

 k0

3 ,t 4

3 ,t 4

3 , 4

 = 1− =

1 , 4

=

3 , 4



3 4

t ∈ T.

We can rewrite the given equation in the form  3   3  3 D 4 D 4 y − y − D 4 y − y = 0, Set

3

v(t) = D 4 y(t) − y(t), Then

3

t ∈ T.

t ∈ T.

t ∈ T,

D 4 v(t) = v(t), and v(t) = c1 E1 (t, 0) Rt

= c1 e

1 0 µ(s)

Rt

= c1 e

log

1−k1 34 ,t 1+µ(s) k0 34 ,t

( ) ( )

  1− 1 1+ 3 4 ∆s

0 log

4

! ∆s

t ∈ T.

220  Conformable Dynamic Equations on Time Scales

= c1 et log 2 = c1 2t ,

t ∈ T,

where c1 is a constant. Hence, 3

t ∈ T,

D 4 y(t) = y(t) + v(t), and Z t

y(t) = c2 E1 (t, 0) +

0

v(s)Eg (σ (s),t)∆α,t s

Z t

= c2 E1 (t, 0) +

0

v(s)Eg (σ (s),t)

4 = c2 E1 (t, 0) + 3

Z t

4 3

Z t

= c2 E1 (t, 0) +

0

0

E0 (t, σ (s))
∆s k0 34 , s

v(s)Eg (σ (s),t)E c 0 (σ (s),t)∆s v(s)Eg⊕c ( c 0) (σ (s),t)∆s

= c2 E1 (t, 0) 4 + 3

Z t 0

(g ⊕c ( c 0))(s) − k1
v(s) 1 + µ(s) k0 34 , s


!

3 4,s

Eg⊕c ( c 0) (s,t)∆s,

where c2 is a constant, g(s) =

=

=

3 3 3 4 , s
µ(s)k1 4 , s − k0

4 , s k0 43 , s + µ(s) 1 − k1 34 , s





1 − k1 1 − 14 3 4 3 4

1 4




− 34







+ 1 − 41
1

−2 3 2

1 = − , 4 c 0 =

k0

=

3 4 3 4

=

3 16 1 2

=

3 , 8

3 s 4 ,


3 4,s
− k1 34 , s




k0

3 4,s 1 4




− 14

k1








Second-Order Conformable Dynamic Equations  221

− 41 + 38 − 14

(g ⊕c ( c 0)) (s) =

+ − 14 − 14




3 8

− 14




1 3 − 16 − 32 3 4 5 − 32

=

3 4,s

4

3 4

=

(g ⊕c ( c 0)) (s) − k1
1 + µ(s) k0 34 , s


3

3 4

= −

20 96

= −

5 , 24


= 1+

5 − 14 − 24 3 4

= 1−

11 24 3 4

= 1−

11 18

7 , 18

=

s ∈ T,

and Rs

Eg⊕c ( c 0) (s,t) = e = e−

Rt s

1 t µ(τ)

log 1+µ(τ)

(g⊕c ( c 0))(τ)−k1 34 ,τ k0 34 ,τ

( )

( )

7 ∆τ log 18

7

= e−(t−s) log 18  =

18 7

t−s ,

t, s ∈ T,



7 18

s ≤ t.

Then Z

t



18 v(s) 7 0  t−s 14 t−1 18 = c2 E1 (t, 0) + ∑ v(s) 7 27 s=0

4 y(t) = c2 E1 (t, 0) + 3

t−s ∆s

! ∆τ

222  Conformable Dynamic Equations on Time Scales

14 t−1 s 2t−s s2t−2s c1 ∑ 2 27 s=0 7t−s   14 t t−1 9 t−s t = c2 2 + c1 2 ∑ 27 s=0 7  t t−1  s 18 7 t = c2 2 + c3 , ∑ 7 s=0 9 = c2 2t +

where c3 =

14 c1 . This ends the example. 27

Theorem 7.3.2 Let f , g ∈ Crd (T). Consider the equation (Dα )2 y + f (t)Dα y + g(t)y = 0,

2

t ∈ Tκ .

(7.12)

If either one of the two conditions (i) f (t) = −pσ (t) − q(t), g(t) = −Dα p(t) + k1 (α,t)pσ (t) + q(t)p(t),

α ∈ (0, 1],

(ii) f (t) = −p − q(t), g(t) = pq(t),

2

t ∈ Tκ ,

where p is a constant, is satisfied, then equation (7.12) can be written in the factored form (7.11). Proof 7.3.3

1. Suppose (i). Then 0 = (Dα )2 y(t) + (−pσ (t) − q(t)) Dα y(t) + (−Dα p(t) + k1 (α,t)pσ (t) + q(t)p(t)) y(t) = (Dα )2 y(t) − (pσ (t)Dα y(t) + Dα p(t)y(t) − k1 (α,t)pσ (t)y(t)) −q(t)Dα y(t) + q(t)p(t)y(t)

2

t ∈ Tκ ,

Second-Order Conformable Dynamic Equations  223

= (Dα )2 y(t) − Dα (py)(t) − q(t) (Dα y(t) − p(t)y(t)) = Dα (Dα y − py) (t) − q(t) (Dα y(t) − p(t)y(t)) ,

2

t ∈ Tκ .

2. Suppose (ii). Then 0 = (Dα )2 y(t) + (−p − q(t)) Dα y(t) + pq(t)y(t) = (Dα )2 y(t) − pDα y(t) − q(t)Dα y(t) + pq(t)y(t) = Dα (Dα y(t) − py(t)) − q(t) (Dα y(t) − py(t)) ,

2

t ∈ Tκ . 

This completes the proof. Example 7.3.4 Let T = 2N0 , k1 (α,t) = (1 − α)t 4α ,

k0 (α,t) = αt 4(1−α) ,

α ∈ (0, 1],

Consider the equation    1 2 1 13 3 3 4 2 4 4 y − (t + 4t )D y + t − t y = 0, D 4 4 Here 1 α = , 4   1 3 ,t = t, k1 4 4   1 1 3 ,t = t , k0 4 4 σ (t) = 2t, f (t) = −t − 4t 2 , g(t) =

13 3 3 4 t − t , 4 4

Take p(t) = t 2 , q(t) = t,

t ∈ T.

t ∈ T.

2

t ∈ Tκ .

t ∈ T.

224  Conformable Dynamic Equations on Time Scales

Then p∆ (t) = σ (t) + t = 2t + t = 3t, pσ (t) = (σ (t))2 = (2t)2 = 4t 2 , 

1 4

D p(t) = k1

   1 1 ,t p(t) + k0 ,t p∆ (t) 4 4

=

3 2
1 3 t t + t (3t) 4 4

=

3 3 3 4 t + t , 4 4

−pσ (t) − q(t) = −4t 2 − t =

f (t),

t ∈ T,

and 1

−D 4 p(t) + k1



 1 3 3 ,t pσ (t) + q(t)p(t) = − t 3 − t 4 4 4 4
3 + t 4t 2 + t 3 4 =

13 3 3 4 t − t 4 4

= g(t),

t ∈ T.

Hence, and considering Theorem 7.3.2, we conclude that the considered equation can be written in the factored form  1   1  1 D 4 D 4 y − t 2 y − t D 4 y − t 2 y = 0, t ∈ T. This ends the example.

Second-Order Conformable Dynamic Equations  225

Exercise 7.3.5 Let T = 3N0 and k1 (α,t) = (1 − α)t 3α ,

k0 (α,t) = αt 3(1−α) ,

α ∈ (0, 1],

t ∈ T.

Prove that the equation  1 2
1
D 3 y − 3t + t 2 D 3 y + t 2 + t 3 y = 0,

t ∈ T,

can be written in the factored form (7.11).

7.4

NONCONSTANT COEFFICIENTS

Consider the equation   2 (Dα )2 y − q (t) + k1 (α,t)q yσ − k1 (α,t)qy = 0,

(7.13)

2

t ∈ Tκ , where q ∈ Rc is a constant, q 6= 0, (k0 (α,t) − µ(t)k1 (α,t))2 − (µ(t))2 k1 (α,t)q 6= 0,

α ∈ (0, 1],

t ∈ T.

We will show that y1 (t) = Eq (t,t0 ),

t ∈ T,

is a solution of equation (7.13). We have 2 q (t) =

2 q (t) + k1 (α,t)q =

qk0 (α,t)(q − k1 (α,t)) − k1 (α,t)q, k0 (α,t) + µ(t)(q − k1 (α,t)) qk0 (α,t)(q − k1 (α,t)) k0 (α,t) + µ(t)(q − k1 (α,t))

  2 q (t) + k1 (α,t)q Eqσ (t,t0 ) = q(q − k1 (α,t))Eq (t,t0 ), Dα Eq (t,t0 ) = qEq (t,t0 ), (Dα )2 Eq (t,t0 ) = qDα Eq (t,t0 ) = q2 Eq (t,t0 ),

2

t ∈ Tκ .

Hence,   2 (Dα )2 y − q (t) + k1 (α,t)q yσ − k1 (α,t)qy
= q2 Eq (t,t0 ) − q2 − k1 (α,t)q Eq (t,t0 )

226  Conformable Dynamic Equations on Time Scales

−k1 (α,t)qEq (t,t0 ) = 0,

2

t ∈ Tκ .

Note that yσ

= y + µy∆  µk µ 1 k1 y + k0 y∆ − y k0 k0   2 µ µk1 y + Dα y on Tκ . = 1− k0 k0

= y+

Then equation (7.13) takes the form 0 = (Dα )2 y(t) −

qk0 (α,t)(q − k1 (α,t)) k0 (α,t) + µ(t)(q − k1 (α,t))



 k0 (α,t) − µ(t)k1 (α,t) µ(t) α × y(t) + D y(t) k0 (α,t) k0 (α,t) −k1 (α,t)qy(t) = (Dα )2 y(t) −

qµ(t)(q − k1 (α,t)) Dα y(t) k0 (α,t) + µ(t)(q − k1 (α,t))

  q(q − k1 (α,t))(k0 (α,t) − µ(t)k1 (α,t)) + − − k1 (α,t)q y(t), k0 (α,t) + µ(t)(q − k1 (α,t)) 2

t ∈ Tκ . Let a(t) = −

qµ(t)(q − k1 (α,t) , k0 (α,t) + µ(t)(q − k1 (α,t))

b(t) = −

q(q − k1 (α,t))(k0 (α,t) − µ(t)k1 (α,t)) − k1 (α,t)q, k0 (α,t) + µ(t)(q − k1 (α,t))

t ∈ T. Therefore, equation (7.13) can be rewritten in the form (Dα )2 y + a(t)Dα y + b(t)y = 0,

2

t ∈ Tκ .

We will find another solution y2 of equation (7.13) using Abel’s formula, so that y2 (t0 ) = 0,

Dα y2 (t0 ) = 1.

We have  W (y1 , y2 )(t0 ) = det

y1 (t0 ) y2 (t0 ) Dα y1 (t0 ) Dα y2 (t0 )



Second-Order Conformable Dynamic Equations  227

 = det



1 0 q 1

= 1 6= 0, Let p=−

t ∈ Tκ .

(a + k1 )(k0 − µk1 ) bµ + , k0 k0

on T.

Note that p = − −

p − k1 k0

1+µ

p − k1 k0

  1) − kqµ(q−k + k 1 (k0 − µk1 ) µ(q−k ) 0

1

k0 µq(q − k1 )(k0 − µk1 ) µqk1 − k0 (k0 + µ(q − k1 )) k0

= −

k1 (k0 − µk1 ) µqk1 − k0 k0

= −

k1 (k0 + µ(q − k1 )) , k0

= −

k1 (k0 + µ(q − k1 )) k1 − k0 k02

= −

k1 (2k0 + µ(q − k1 )) , k02

= 1−

µk1 (2k0 + µ(q − k1 )) k02

=

k02 − 2µk0 k1 + µ 2 k12 − µ 2 k1 q k02

=

(k0 − µk1 )2 − µ 2 k1 q k02

6= 0

on T.

From here, it follows that E p (t,t0 ) 6= 0,

t ∈ T,

and W (y1 , y2 )(t) 6= 0,

t ∈ Tκ .

228  Conformable Dynamic Equations on Time Scales

Then, using Abel’s formula, we obtain E p (t,t0 )W (y1 , y2 )(t0 ) = W (y1 , y2 )(t),

t ∈ Tκ ,

or 

y1 (t) y2 (t) α α D y1 (t) D y2 3(t)





Eq (t,t0 ) y2 (t) qEq (t,t0 ) Dα y2 (t)



E p (t,t0 ) = det = det

= Eq (t,t0 ) (Dα y2 (t) − qy2 (t)) ,

t ∈ Tκ .

Therefore Dα y2 (t) = qy2 (t) + E p (t,t0 )E c q (t,t0 ) 2

t ∈ Tκ .

= qy2 (t) + E p c q (t,t0 ), Let g(t) =

(q − k1 (α,t))(µ(t)k1 (α,t) − k0 (α,t)) , k0 (α,t) + µ(t)(q − k1 (α,t))

Then

Z t

y2 (t) =

t0

E p c q (s,t0 )Eg (σ (s),t)∆α,t s,

t ∈ T. 2

t ∈ Tκ .

Therefore a general solution of equation (7.13) is given by the expression y(t) = c1 Eq (t,t0 ) Z t

+c2

t0

2

t ∈ Tκ ,

E p c q (s,t0 )Eg (σ (s),t)∆α,t s,

where c1 and c2 are constants. Exercise 7.4.1 Let T = Z, k1 (α,t) = 1 − α,

k0 (α,t) = α,

α ∈ (0, 1],

Find a general solution of the equation  1 2   2 D 3 y − 3 (t) + 2 yσ − 2y = 0,

t ∈ T.

2

t ∈ Tκ .

Now we consider the equation (Dα )2 y + f (t)Dα y + g(t)y = 0,

2

t ∈ Tκ ,

where f , g ∈ Crd (T) and (k0 (α,t) − µ(t)k1 (α,t))2 − f (t)µ(t) (k0 (α,t) − µ(t)k1 (α,t)) +g(t) (µ(t))2 6= 0,

t ∈ T,

α ∈ (0, 1].

(7.14)

Second-Order Conformable Dynamic Equations  229

Suppose that z ∈ Crd1 (T), z ∈ Rc , satisfies the equation Dα z + zσ z − k1 (α,t)zσ + a(t)z + b(t) = 0,

t ∈ Tκ .

We will show that y1 (t) = Ez (t,t0 ),

t ∈ T,

is a solution of equation (7.14). We have Dα y1 (t) = z(t)Ez (t,t0 ), (Dα )2 y1 (t) = Dα z(t)Ez (t,t0 ) + zσ (t)z(t)Ez (t,t0 ) −k1 (α,t)zσ (t)Ez (t,t0 ) = (Dα z(t) + zσ (t)z(t) − k1 (α,t)zσ (t)) Ez (t,t0 ), Then (Dα )2 y1 (t) + f (t)Dα y1 (t) + g(t)y1 (t) = (Dα z(t) + zσ (t)z(t) − k1 (α,t)zσ (t)) Ez (t,t0 ) + f (t)z(t)Ez (t,t0 ) + g(t)Ez (t,t0 ) = (Dα z(t) + zσ (t)z(t) − k1 (α,t)zσ (t) + f (t)z(t) + g(t)) Ez (t,t0 ) = 0,

2

t ∈ Tκ .

Let h(t) = − +

( f (t) + k1 (α,t))(k0 (α,t) − µ(t)k1 (α,t)) k0 (α,t) g(t)µ(t) , k0 (α,t)

t ∈ T,

α ∈ (0, 1].

Then h − k1 k0

=

−( f + k1 )(k0 − µk1 ) + gµ − k0 k1 k02

=

− f k0 + f µk1 + gµ − 2k0 k1 + µk12 , k02

2

t ∈ Tκ .

230  Conformable Dynamic Equations on Time Scales

1+µ

h − k1 k0

=

k02 − 2µk0 k1 + µ 2 k12 − f µ(k0 − µk1 ) + gµ 2 k02

=

(k0 − µk1 )2 − f µ(k0 − µk1 ) + gµ k02

6= 0,

α ∈ (0, 1],

t ∈ T.

Now we will find another solution y2 of equation (7.14) so that y2 (t0 ) = 0,

Dα y2 (t0 ) = 1,

and y1 and y2 are linearly independent. By Abel’s formula, we have W (y1 , y2 )(t) = Eh (t,t0 )W (y1 , y2 )(t0 ) 

y1 (t) y2 (t) = Eh (t,t0 )det α D y1 (t) Dα y2 (t)   1 0 = Eh (t,t0 )det z(t0 ) 1



= Eh (t,t0 ) 6= 0,

t ∈ Tκ .

Hence, 

y1 (t) y2 (t) α D y1 (t) Dα y2 (t)



Ez (t,t0 ) y2 (t) z(t)Ez (t,t0 ) Dα y2 (t)

Eh (t,t0 ) = det = det





= (Dα y2 (t) − z(t)y2 (t)) Ez (t,t0 ),

t ∈ Tκ ,

or Dα y2 (t) = z(t)y2 (t) + Eh c z (t,t0 ), Then

t ∈ Tκ .

Z t

y2 (t) = where h1 =

t0

Eh c z (s,t0 )Eh1 (σ (s),t)∆α,t s,

(z − k1 )(µk1 − k0 ) , k0 + µ(z − k1 )

α ∈ (0, 1],

t ∈ Tκ ,

t ∈ T.

Second-Order Conformable Dynamic Equations  231

We have h1 − k1 = =

(z − k1 )(µk1 − k0 ) − k1 k0 − µk1 (z − k1 ) k0 + µ(z − k1 ) µzk1 − zk0 − µk12 + k1 k0 − k1 k0 − µk1 z + µk12 k0 + µ(z − k1 )

= − 1+µ

h1 − k1 k0

zk0 , k0 + µ(z − k1 )

= 1−

µz k0 + µ(z − k1 )

=

k0 + µ(z − k1 ) − µz k0 + µ(z − k1 )

=

k0 − µk1 k0 + µ(z − k1 )

6= 0,

α ∈ (0, 1],

t ∈ T.

Consequently, h1 ∈ Rc and a general solution of equation (7.14) is given by y(t) = c1 y1 (t) + c2 y2 (t) = c1 Ez (t,t0 ) Z t

+c2

7.5

t0

Eh c z (s,t0 )Eh1 (σ (s),t)∆α,t s,

2

t ∈ Tκ .

CONFORMABLE EULER-CAUCHY EQUATIONS

In this section we suppose that T ⊆ (0, ∞). Definition 7.5.1 Let a, b ∈ R. The equations of the form tσ (t) (Dα )2 y + (a + k0 (α,t) − µ(t)k1 (α,t))tDα y + by = 0, will be called conformable Euler-Cauchy equations. Theorem 7.5.2 Let a2 − 4b 6= 0, √ −a ± a2 − 4b ∈ Rc , 2t and

t ∈ T,

√ −a ± a2 − 4b λ1,2 = . 2 Then E λ1 (t,t0 ) and E λ2 (t,t0 ) are solutions of equation (7.15). t

t

2

t ∈ Tκ ,

(7.15)

232  Conformable Dynamic Equations on Time Scales

Proof 7.5.3 Let y(t) = E λ (t,t0 ), t

t ∈ T,

where λ = λ1 or λ = λ2 . Note that λ 2 + aλ + b = 0. Then Dα y(t) =

λ E λ (t,t0 ) t t λ y(t), t

t ∈ Tκ ,

tDα y(t) = λ y(t),

t ∈ Tκ ,

=



α 2

(D ) y(t) = λ D

α

y(t) t



= λ

tDα y(t) − y(t)Dα t y(t) + λ k1 (α,t) tσ (t) t

= λ

tDα y(t) − ty(t)k1 (α,t) − k0 (α,t)y(t) tσ (t)

+λ k1 (α,t)

y(t) , t

2

t ∈ Tκ ,

tσ (t) (Dα )2 y(t) = λ (tDα y(t) − ty(t)k1 (α,t) − k0 (α,t)y(t)) +λ k1 (α,t)σ (t)y(t)

= λ λ − tk1 (α,t) − k0 (α,t) ! +k1 (α,t)σ (t) y(t),

2

t ∈ Tκ .

Therefore tσ (t) (Dα )2 y(t) + (a + tk1 (α,t) + k0 (α,t) − σ (t)k1 (α,t))tDα y(t) + by(t) = λ 2 y(t) + λ (−tk1 (α,t) − k0 (α,t) + σ (t)k1 (α,t)) y(t) +λ (a + tk1 (α,t) + k0 (α,t) − σ (t)k1 (α,t)) y(t)

Second-Order Conformable Dynamic Equations  233

+by(t)
λ 2 + aλ + b y(t)

=

= 0,

2

t ∈ Tκ . 

This completes the proof. Theorem 7.5.4 Let a, b, λ1 and λ2 be as in Theorem 7.5.2. Then E λ1 (t,t0 ),

t ∈ T,

E λ2 (t,t0 ),

t

t

form a fundamental set of solutions for the Euler-Cauchy equation (7.15). Proof 7.5.5 Let y1 (t) = E λ1 (t,t0 ),

t ∈ T.

y2 (t) = E λ2 (t,t0 ),

t

t

We have  W (y1 , y2 )(t) = det 

y1 (t) y2 (t) α D y1 (t) Dα y2 (t) E λ1 (t,t0 ) t

= det  λ1 E λ (t,t0 ) t t1



E λ2 (t,t0 )



t  λ2 E λ2 (t,t0 ) t t

=

λ2 λ1 E λ1 (t,t0 )E λ2 (t,t0 ) − E λ1 (t,t0 )E λ2 (t,t0 ) t t t t t t

=

λ2 − λ1 E λ1 (t,t0 )E λ2 (t,t0 ) t t t

6= 0,

t ∈ Tκ . 

This completes the proof. Example 7.5.6 Let T = N, k1 (α,t) = (1 − α)(1 + t 2 )α ,

k0 (α,t) = α(1 + t 2 )1−α ,

t ∈ T,

Consider the equation  1 2 1 t(t + 1) D 2 y + 3tD 2 y + 2y = 0, Here α =

1 , 2

t ∈ T.

α ∈ (0, 1].

234  Conformable Dynamic Equations on Time Scales

a = 3, b = 2,  k1  k0

1 ,t 2



1 ,t 2



=

1 1 (1 + t 2 ) 2 , 2

=

1 1 (1 + t 2 ) 2 , 2

t ∈ T.

Then  a + tk1

     1 1 1 ,t + k0 ,t − σ (t)k1 ,t 2 2 2

1 1 1 1 1 t = 3 + (1 + t 2 ) 2 + (1 + t 2 ) 2 − (t + 1) (1 + t 2 ) 2 2 2 2

= 3,

t ∈ T.

Therefore √ −3 + 9 − 8 −3 + 1 = = −1, 2 2 √ −3 − 9 − 8 −3 − 1 = = −2, 2 2

λ1 = λ2 = and

E− 1 (t,t0 ), t

E− 2 (t,t0 ), t

t ∈ T,

form a fundamental set of solutions of the considered equation. Exercise 7.5.7 Let T = 2N0 , k1 (α,t) = (1 − α)t α ,

k0 (α,t) = αt 1−α ,

α ∈ (0, 1],

t ∈ T.

Find a fundamental set of solutions for each of the following equations. 1.

   1 2 1 1 3 1 1 2 2 2 2t D y + 4 − t + t tD 2 y + 3y = 0, 2 2

t ∈ T,

   1 2 1 2 4 1 2 2t D 3 y + 5 − t 3 + t 3 tD 3 y + 4y = 0, 3 3

t ∈ T,

   1 2 1 3 5 1 3 2t 2 D 4 y + 7 − t 4 + t 4 tD 4 y + 6y = 0, 4 4

t ∈ T.

2

2.

2

3.

Second-Order Conformable Dynamic Equations  235

Now we suppose that a 6= 0 and λ a = − ∈ Rc . t 2t

a2 − 4b = 0, Then

a = −2λ , a2 4

b =

4λ 2 4

=

= λ 2. The equation (7.15) can be rewritten in the form tσ (t) (Dα )2 y + (−2λ + k0 (α,t) − µ(t)k1 (α,t))tDα y + λ 2 y = 0, Let

Z t

g(t) = 1 + t0

λ 1 (α,s) s + µ(s) λ −sk k (α,s)

E0 (σ (s),t)∆α,t s,

2

t ∈ Tκ .

t,t0 ∈ T.

0

Note that Dα g(t) = Dα 1 



Z t

+Dα 

t0

λ 1 (α,s) s + µ(s) λ −sk k0 (α,s)

E0 (σ (s),t)∆α,t s

= k1 (α,t) Z t

+

t0 s + µ(s) λ −sk1 (α,s) k0 (α,s) Z t λ

−k1 (α,t)

+

λ

t0

Dtα E0 (σ (s),t)∆α,t s|

1 (α,s) s + µ(s) λ −sk k (α,s) 0

λ 1 (α,t) t + µ(t) λ −tk k (α,t)

E0 (σ (t), σ (t))

0

= k1 (α,t) +

E0 (σ (s), σ (t))∆α,t s

λ 1 (α,t) t + µ(t) λ −tk k (α,t) 0

(7.16)

236  Conformable Dynamic Equations on Time Scales Z t

+

k1 (α,t)E0 (t, σ (s)) E0 (t, σ (s))E0 (σ (t), σ (s))

λ

t0 s + µ(s) λ −sk1 (α,s) k0 (α,s)

! 1 +k1 (α,t) ∆α,t s E0 (t, σ (s)) −k1 (α,t)

Z t t0

λ 1 (α,s) s + µ(s) λ −sk k (α,s)

E0 (σ (s), σ (t))∆α,t s

0

λ

= k1 (α,t) +

1 (α,t) t + µ(t) λ −tk k (α,t) 0

Z t

+k1 (α,t)

t0

λ 1 (α,s) s + µ(s) λ −sk k (α,s) 0

Z t

+k1 (α,t)

−k1 (α,t)

t0

λ 1 (α,s) s + µ(s) λ −sk k (α,s)

λ 1 (α,s) s + µ(s) λ −sk k (α,s)

E0 (σ (s), σ (t))∆α,t s

0





Z t

= k1 (α,t) 1 +

+

E0 (σ (s),t)∆α,t s

0

Z t t0

E0 (σ (s), σ (t))∆α,t s

t0

λ

E0 (σ (s),t)∆α,t s

1 (α,s) s + µ(s) λ −sk k (α,s) 0

λ 1 (α,t) t + µ(t) λ −tk k (α,t) 0

= k1 (α,t)g(t) +

λ 1 (α,t) t + µ(t) λ −tk k (α,t)

t ∈ Tκ .

,

0

Consequently, g satisfies the equation Dα g = k1 (α,t)g +

λ 1 (α,t) t + µ(t) λ −tk k0 (α,t)

,

Now we will prove that y(t) = E λ (t,t0 )g(t), t

t ∈ T,

satisfies equation (7.16). Indeed, we have   Dα y(t) = Dα E λ (t,t0 )g(t) t

=

λ E λ (t,t0 )g(t) + E λ (σ (t),t0 )Dα g(t) t t t

t ∈ Tκ .

Second-Order Conformable Dynamic Equations  237

−k1 (α,t)E λ (σ (t),t0 )g(t) t

=

λ E λ (t,t0 )g(t) + E λ (σ (t),t0 ) (Dα g(t) − k1 (α,t)g(t)) t t t

=

λ E λ (t,t0 )g(t) t t λ t

− k1 (α,t) + 1 + µ(t) k0 (α,t) =

! E λ (t,t0 ) (Dα g(t) − k1 (α,t)g(t)) t

λ λ E λ (t,t0 )g(t) + E λ (t,t0 ) t t t t

λ λ y(t) + E λ (t,t0 ), t t t   λ λ (Dα )2 y(t) = Dα y(t) + E λ (t,t0 ) t t t =

 α

= λD = λ

E λ (t,t0 )

 y(t) + λ Dα t

!

t

t

y(t) tDα y(t) − y(t)Dα t + λ k1 (α,t) tσ (t) t tDα E λ (t,t0 ) − E λ (t,t0 )Dα t t

+λ

t

tσ (t) E λ (t,t0 )

+λ k1 (α,t) t



= λ

t

t

λ λ t y(t) + t E λt

 (t,t0 ) − k1 (α,t)ty(t) − k0 (α,t)y(t) tσ (t)

+λ

+λ

k1 (α,t)y(t) t t λt E λ (t,t0 ) − k1 (α,t)tE λ (t,t0 ) − k0 (α,t)E λ (t,t0 ) t

t

t

tσ (t) E λ (t,t0 )

+λ k1 (α,t)

t

t

λ y(t) + λ E λ (t,t0 ) − k1 (α,t)ty(t) − k0 (α,t)y(t) = λ

t

tσ (t)

238  Conformable Dynamic Equations on Time Scales

+λ

k1 (α,t)y(t) t λ E λ (t,t0 ) − tk1 (α,t)E λ (t,t0 ) − k0 (α,t)E λ (t,t0 ) t

+λ

t

t

tσ (t) E λ (t,t0 )

+λ k1 (α,t)

t

t

2

t ∈ Tκ .

,

Therefore tσ (t) (Dα )2 y(t) + (−2λ + tk1 (α,t) + k0 (α,t) − σ (t)k1 (α,t))tDα y(t) + λ 2 y(t)   = λ λ y(t) + λ E λ (t,t0 ) − k1 (α,t)ty(t) − k0 (α,t)y(t) t

+λ σ (t)k1 (α,t)y(t)   +λ λ E λ (t,t0 ) − tk1 (α,t)E λ (t,t0 ) − k0 (α,t)E λ (t,t0 ) t

t

t

+λ σ (t)k1 (α,t)E λ (t,t0 ) t

  + (−2λ + tk1 (α,t) + k0 (α,t) − σ (t)k1 (α,t)) λ y(t) + λ E λ (t,t0 ) t

+λ 2 y(t) = λ (λ − tk1 (α,t) − k0 (α,t)) E λ (t,t0 ) t

+λ σ (t)k1 (α,t)E λ (t,t0 ) + λ 2 E λ (t,t0 ) t

t

+λ (−2λ + tk1 (α,t) + k0 (α,t) − σ (t)k1 (α,t)) E λ (t,t0 ) t

= 0,

2

t ∈ Tκ .

Let y1 (t) = E λ (t,t0 ), t

y2 (t) = g(t)E λ (t,t0 ), t

Then Dα y1 (t) = =

λ E λ (t,t0 ) t t λ y1 (t), t

t ∈ T.

Second-Order Conformable Dynamic Equations  239

λ λ y2 (t) + E λ (t,t0 ) t t t

Dα y2 (t) =

λ λ y1 (t) + y2 (t), t t

=

t,t0 ∈ T,

and  W (y1 , y2 )(t) = det

y1 (t) y2 (t) Dα y1 (t) Dα y2 (t)





 y1 (t) y2 (t)  = det  λ λ λ y1 (t) y2 (t) + y1 (t) t t t   y1 (t) y2 (t)  = det  λ 0 y1 (t) t = =

λ (y1 (t))2 t 2 λ E λ (t,t0 ) t t

6= 0, i.e., y1 (t) = E λ (t,t0 ), t

y2 (t) = g(t)E λ (t,t0 ), t

t ∈ T,

form a fundamental set of solutions for equation (7.16). Then the general solution of equation (7.16) is given by y(t) = (c1 + c2 g(t)) E λ (t,t0 ), t

t ∈ T.

Example 7.5.8 Let T = 2N0 , k1 (α,t) = (1 − α)t 4α ,

k0 (α,t) = αt 4(1−α) ,

α ∈ (0, 1],

Consider the equation    1 2 1 1 2 1 3 2 2t D 4 y + 2 − t + t tD 4 y + y = 0, 4 4 Here σ (t) = 2t,  k1

1 ,t 4

 =

3 t, 4

t ≥ 4.

t ∈ T.

240  Conformable Dynamic Equations on Time Scales

 k0

1 ,t 4



1 3 t , 4

=

a = 2, b = 1. Then λ = −1, and g(t) = 1 −

Z t 4

= 1−

Z t 4

= 1−

Z t 4

= 1−

s−s

Z t 4

= 1−

1 1+ 34 s2 1 s3 4

1 s−

4+3s2 s2

E0 (2s,t)∆ 1 ,t s

E0 (2s,t)∆ 1 ,t s

s2 E0 (t, 2s) E0 (2s,t) 1 3 ∆s 3 2 s − 3s − 4 4s

Z t

4

4

s(s3 − 3s2 − 4)

4 , 3 − 3s2 − 4 s s=4

= 1− ∑

E− 1 (t, 4) = e − 1 − 3 t (t, 4) t 4 1 t3 4

= e

− 4+3t 4

2

(t, 4)

t

  Rt 1 4+3s2 ∆s 4 s log 1− 3

= e

s

 3 2  Rt 1 s −3s −4 ∆s 4 s log 3

= e

s

  t 2 log s3 −3s2 −4 ∑s=4 3

= e

s

t 2

=

∏

s=4



4

s2 E0 (2s,t)∆ 1 ,t s 4 s3 − 3s2 − 4

t 2

t

4

 s3 − 3s2 − 4 . s3

∆s

Second-Order Conformable Dynamic Equations  241

Consequently, 

 t 3 2 2 4  ∏ s − 3s − 4 y(t) = c1 + c2 − c2 ∑ 3 2 s3 s=4 s − 3s − 4 s=4 t 2



 t 3 2 2 4  ∏ s − 3s − 4 = c3 + c4 ∑ 3 2 s3 s=4 s − 3s − 4 s=4 t 2

is the general solution, where c1 and c2 are constants, c3 = c1 + c2 , c4 = −c2 . This ends the example. Exercise 7.5.9 Let T = 3N0 , k1 (α,t) = (1 − α)(1 + t)α ,

k0 (α,t) = α(1 + t)1−α ,

α ∈ (0, 1],

t ∈ T.

Find the general solution of each of the following equations. 1.

   1 2 1 1 1 1 3t D 2 y + 6 − t(1 + t) 2 + (1 + t) 2 tD 2 y + 9y = 0, 2 2

2.

2

t ∈ Tκ ,

   1 2 1 3 1 3 1 4 4 4 3t D y + 8 − t(1 + t) + (1 + t) tD 4 y + 16y = 0, 2 4

t ∈ Tκ ,

   1 2 1 1 2 4 1 3t D 3 y + 10 − t(1 + t) 3 + (1 + t) 3 tD 3 y + 25y = 0, 3 3

t ∈ Tκ .

2

2

3. 2

7.6

2

VARIATION OF PARAMETERS

Consider the equation (Dα )2 y + a(t)Dα y + b(t)y = f (t),

(7.17)

where a, b, f ∈ Crd (T). Suppose that y1 and y2 form a fundamental set of solutions for the corresponding homogeneous equation (7.1). We will search for a solution of equation (7.17) of the form 2 y(t) = p(t)y1 (t) + q(t)y2 (t), t ∈ Tκ , where the functions p and q will be determined below. We have Dα y(t) = Dα (p(t)y1 (t)) + Dα (q(t)y2 (t)) = (Dα p(t)) yσ1 (t) + p(t)Dα y1 (t) −k1 (α,t)p(t)yσ1 (t)

242  Conformable Dynamic Equations on Time Scales

+ (Dα q(t)) yσ2 (t) + q(t)Dα y2 (t) −k1 (α,t)q(t)yσ2 (t) = p(t)Dα y1 (t) + q(t)Dα y2 (t),

2

t ∈ Tκ ,

provided p and q satisfy (Dα p(t)) yσ1 (t) + (Dα q(t)) yσ2 (t) = k1 (α,t)p(t)yσ1 (t) (7.18) κ2

+k1 (α,t)q(t)yσ2 (t),

t ∈T .

Next, (Dα )2 y(t) = Dα (p(t)Dα y1 (t)) + Dα (q(t)Dα y2 (t)) = p(t) (Dα )2 y1 (t) + (Dα p(t)) (Dα y1 )σ (t) −k1 (α,t)p(t) (Dα y1 )σ (t) +q(t) (Dα )2 y2 (t) + (Dα q(t)) (Dα y2 )σ (t) −k1 (α,t)q(t) (Dα y2 )σ (t),

2

t ∈ Tκ .

Then (Dα )2 y(t) + a(t)Dα y(t) + b(t)y(t)   = p(t) (Dα )2 y1 (t) + a(t)Dα y1 (t) + b(t)y1 (t)   +q(t) (Dα )2 y2 (t) + a(t)Dα y2 (t) +b (t)y2 (t) + (Dα p(t)) (Dα y1 )σ (t) − k1 (α,t)p(t) (Dα y1 )σ (t) + (Dα q(t)) (Dα y2 )σ (t) − k1 (α,t)q(t) (Dα y2 )σ (t) = (Dα p(t)) (Dα y1 )σ (t) − k1 (α,t)p(t) (Dα y1 )σ (t) + (Dα q(t)) (Dα y2 )σ (t) − k1 (α,t)q(t) (Dα y2 )σ (t) =

f (t),

2

t ∈ Tκ .

Second-Order Conformable Dynamic Equations  243

Thus, using (7.18), for p and q we get the system  (Dα p(t)) yσ1 (t) + (Dα q(t)) yσ2 (t)         

= k1 (α,t)p(t)yσ1 (t) +k1 (α,t)q(t)yσ2 (t)

  (Dα p(t)) (Dα y1 )σ (t) + (Dα q(t)) (Dα y2 )σ (t) =       

f (t) + k1 (α,t)p(t) (Dα y1 )σ (t) 2

+k1 (α,t)q(t) (Dα y2 )σ (t),

t ∈ Tκ .

From the last system, we get  yσ2 (t)  α  D p(t) = k (α,t)p(t) − f (t)  1   (W (y1 , y2 ))σ (t)      Dα q(t) = k1 (α,t)q(t) + f (t)

yσ1 (t) , (W (y1 , y2 ))σ (t)

2

t ∈ Tκ .

We take  Z t yσ2 (s)   p(t) = E (t,t ) − f (s) E0 (σ (s),t)∆α,t s  0 k1   (W (y1 , y2 ))σ (s) t0 Z t     f (s)  q(t) = Ek1 (t,t0 ) + t0

yσ1 (s) E0 (σ (s),t)∆α,t s, (W (y1 , y2 ))σ (s)

2

t,t0 ∈ Tκ .

Consequently, y(t) = c1 y1 (t) + c2 y2 (t)  Z t + Ek1 (t,t0 ) − f (s)

 yσ2 (s) E0 (σ (s),t)∆α,t s y1 (t) (W (y1 , y2 ))σ (s) t0   Z t yσ1 (s) + Ek1 (t,t0 ) + E0 (σ (s),t)∆α,t s y2 (t), f (s) (W (y1 , y2 ))σ (s) t0

2

t,t0 ∈ Tκ ,

where c1 and c2 are constants, is the general solution of equation (7.17). Example 7.6.1 Let T = 2N0 , k1 (α,t) = (1 − α)t 2α ,

k0 (α,t) = αt 2(1−α) ,

α ∈ (0, 1],

t ∈ T.

Consider the equation    1 2 1 1 2 1 2t D 2 y + −3 − t + t tD 2 y + 2y = 2t 3 , 2 2 2

Here σ (t) = 2t,

t ∈ T.

(7.19)

244  Conformable Dynamic Equations on Time Scales

 k1  k0

1 ,t 2



1 ,t 2



=

1 t, 2

=

1 t, 2

t ∈ T.

Note that the equation    1 2 1 1 2 1 2t D 2 y + −3 − t + t tD 2 y + 2y = 0, 2 2 2

t ∈ T,

(7.20)

is an Euler-Cauchy equation. Then a = −3,

b = 2.

The functions y1 (t) = E 1 (t,t0 ) and t

y2 (t) = E 2 (t,t0 ), t

t,t0 ∈ T,

form a fundamental set of solutions for equation (7.20). We have   E 1 (t,t0 ) E 2 (t,t0 ) t t  W (y1 , y2 )(t) = det  1 2 E 1 (t,t0 ) E 2 (t,t0 ) t t t t 2 1 = E 1 (t,t0 )E 2 (t,t0 ) − E 1 (t,t0 )E 2 (t,t0 ) t t t t t t 1 E 1 (t,t0 )E 2 (t,t0 ). t t t

=

Note that equation (7.19) can be rewritten in the form  1 2 −6 − t 2 + t 1 1 D2 y+ D 2 y + 2 y = t, 2 4t t

t ∈ T.

Here p(t) = Ek1 (t,t0 ) −

Z t t0

Z t

q(t) = Ek1 (t,t0 ) +

t0

s E0 (2s,t)∆α,t s, E 1 (2s,t0 ) t

s E0 (2s,t)∆α,t s E 2 (2s,t0 ) t

Hence, y(t) = (c1 + p(t))E 1 (t,t0 ) + (c2 + q(t))E 2 (2t,t0 ), t

t

t,t0 ∈ T,

where c1 and c2 are constants, is a general solution of equation (7.19). This ends the example.

Second-Order Conformable Dynamic Equations  245

Exercise 7.6.2 Let T = 3N0 , k1 (α,t) = (1 − α)(1 + t)α ,

k0 (α,t) = α(1 + t)1−α ,

α ∈ (0, 1],

t ∈ T.

Find the general solution of each of the following equations. 1.

   1 2 1 1 1 1 3t D 2 y + 6 − t(1 + t) 2 + (1 + t) 2 tD 2 y + 9y = t 2 − 1, 2 2

2.

   1 2 1 3 1 3 1 4 4 4 3t D y + 8 − t(1 + t) + (1 + t) tD 4 y + 16y = t 2 + 2t + 3, 2 4 2

3.    1 2 1 1 2 4 1 3t D 3 y + 10 − t(1 + t) 3 + (1 + t) 3 tD 3 y + 25y = t 3 + 3t 2 + t + 1. 3 3 2

7.7

ADVANCED PRACTICAL PROBLEMS

Problem 7.7.1 Find a general solution for each of the following equations. 2

1. (Dα )2 y − 121y = 0, t ∈ Tκ , 2

2. (Dα )2 y + 9y, t ∈ Tκ , 2

3. (Dα )2 y + 3Dα y − 2y = 0, t ∈ Tκ , 2

4. (Dα )2 y + Dα y + 5y, t ∈ Tκ . Problem 7.7.2 Let T = N20 , k1 (α,t) = (1 − α)t 3α ,

k0 (α,t) = αt 3(1−α) ,

α ∈ (0, 1],

Find a fundamental set of solutions for each of the following equations.  1 2 1 2 1. D 3 y − 6D 2 y + 9y = 0, t ∈ Tκ , 2.

 1 2 1 2 D 4 y − 10D 4 y + 25y = 0, t ∈ Tκ ,

3.

 1 2 1 2 D 8 − 2D 8 y + y = 0, t ∈ Tκ ,

4.

 1 2 1 2 D 3 y − 8D 3 y + 16y = 0, t ∈ Tκ ,

5.

 1 2 2 2 1 1 D 2 y − D 2 y + y = 0, t ∈ Tκ . 3 9

t ∈ T.

246  Conformable Dynamic Equations on Time Scales

Problem 7.7.3 Let T = 2Z, k1 (α,t) = (1 − α)t 5α ,

k0 (α,t) = αt 5(1−α) ,

α ∈ (0, 1],

t ∈ T.

Prove that the equation    1 2 1 1 4 2 8 5 5 D − (2t + 2)D y + − t + t + t y = 0, 5 5

2

t ∈ Tκ ,

can be written in the factored form (7.11). Problem 7.7.4 Let T = Z, k1 (α,t) = 1 − α,

k0 (α,t) = α,

α ∈ (0, 1],

t ∈ T.

Find a general solution of the equation    1 2 2 (t) + 4 yσ − 4y = 0, D 5 y − 5

2

t ∈ Tκ .

Problem 7.7.5 Let T = 3N0 , k1 (α,t) = (1 − α)t α ,

k0 (α,t) = αt 1−α ,

α ∈ (0, 1],

t ∈ T.

Find a fundamental set of solutions for each of the following equations. 1.

   1 2 3 1 1 1 2 2 2 3t D y + 9 − t + t tD 2 y + 8y = 0, 2 2

2.

   1 2 1 4 4 1 2 3 3 3 y + 12 − t + t tD 3 y + 11y = 0, 3t D 3 3

t ∈ T,

   1 2 1 3 5 1 3 3t D 4 y + 15 − t 4 + t 4 tD 4 y + 14y = 0, 2 4

t ∈ T.

2

3.

t ∈ T,

2

Problem 7.7.6 Let T = 4N0 , k1 (α,t) = (1 − α)(1 + t)α ,

k0 (α,t) = α(1 + t)1−α ,

α ∈ (0, 1],

t ∈ T.

Find a general solution for each of the following equations. 1.

   1 2 1 1 1 3 1 2 2 2 4t D y + 6 − t(1 + t) + (1 + t) tD 2 y + 9y = 0, 2 2 2

2.

   1 2 1 1 3 9 1 4t D 4 y + 8 − t(1 + t) 4 + (1 + t) 4 tD 4 y + 16y = 0, 4 4 2

Second-Order Conformable Dynamic Equations  247

3.

   1 2 1 2 1 1 4t D 3 y + 10 − 2t(1 + t) 3 + (1 + t) 3 tD 3 y + 25y = 0. 3 2

Problem 7.7.7 Let T = 4N0 , k1 (α,t) = (1 − α)(1 + t)α ,

k0 (α,t) = α(1 + t)1−α ,

α ∈ (0, 1],

t ∈ T.

Find a general solution for each of the following equations. 1.

   1 2 1 1 1 3 1 2 2 2 y + −6 − t(1 + t) + (1 + t) tD 2 y + 9y = t + 10, 4t D 2 2 2

2.

   1 2 1 1 3 9 1 t2 + 1 4 4 4 4t D y + −8 − t(1 + t) + (1 + t) tD 4 y + 16y = 4 , 4 4 t +1 2

3.    1 2 1 2 1 1 4t D 3 y + −10 − 2t(1 + t) 3 + (1 + t) 3 tD 3 y + 25y = t 4 + t 3 + t 2 + t + 1. 3 2

CHAPTER

8

Second-Order Self-Adjoint Conformable Dynamic Equations

8.1

SELF-ADJOINT DYNAMIC EQUATIONS

Let p be continuous and q be ld-continuous functions on some time-scale interval I ⊆ Tκκ with p(t) > 0 for all t ∈ I . In this section we are concerned with the conformable secondorder (formally) self-adjoint homogeneous dynamic equation Mc x = 0, or for a continuous function h, the associated non-homogeneous equation Mc x = h,

b α [pDα x] (t) + q(t)x(t), Mc x(t) := D

(8.1)

b α using (1.9) for k0 and k1 satisfying for t ∈ I , α ∈ (0, 1]. We interpret Dα using (1.1) and D (A1). Note that one could also study the equation h i b α x (t) + q(t)x(t) Lc = h, Lc x(t) := Dα pD for t ∈ I , α ∈ (0, 1], though here we will concentrate on (1.1). Definition 8.1.1 Let D denote the set of all functions x : T → R defined on I such that x∆ is b α [pDα x] is ld-continuous on I . Then x is a solution of the homogeneous continuous and D equation Mc x = 0 for Mc in (8.1) on I provided x ∈ D and Mc x(t) = 0 for all t ∈ I . We next state a theorem concerning the existence-uniqueness of solutions of initial value problems for the non-homogeneous self-adjoint equation Mc x = h. Theorem 8.1.2 Assume k0 , k1 satisfy (A1). Let α ∈ (0, 1], t0 ∈ I , and let Dα be as in b α be as in (1.9). Assume p is continuous and q, h are ld-continuous on I with (1.1) and D p(t) 6= 0, and suppose x0 , x1 ∈ R are given constants. Then the initial value problem Mc x = h(t),

x(t0 ) = x0 ,

Dα x(t0 ) = x1

has a unique solution that exists on all of I . 249

250  Conformable Dynamic Equations on Time Scales

Proof 8.1.3 The idea of the proof is to use the induction principle for time scales. This is a straightforward modification of the proof in [4, Theorem 3.1]. Definition 8.1.4 (Wronskian) Let k0 , k1 : [0, 1] × I → [0, ∞) be continuous and satisfy (A1). If x, y : I → R are differentiable on I , then the conformable Wronskian of x and y is given by   x(t) y(t) W (x, y)(t) = det for t ∈ I , (8.2) Dα x(t) Dα y(t) for Dα given in (1.1). Theorem 8.1.5 (Conformable Lagrange Identity) Let Mc be given as in (8.1) and α ∈ (0, 1]. Assume ρ ρ t ∈I. k0 (α,t) − ν(t)k1 (α,t) 6= 0, If x, y ∈ D, then ρ

x(Mc y) − y(Mc x) =

ρ

k0 − νk1 ρ k0

!

ρ

νk1 ) b α [pW (x, y)] + k1 (k0 + D pW (x, y), ρ k0

t ∈ Tκκ ,

for Mc given in (8.1). Equivalently, for t, b ∈ Tκκ we have   pW (x, y)(t) bα = x(Mc y) − y (Mc x) . E0 (t, b)D E0 (t, b)

(8.3)

Proof 8.1.6 Let x, y ∈ D. Then x, y ∈ D implies that Dα x and in particular x∆ is continuous, so that x∇ = x∆ρ ; likewise y∇ = y∆ρ . Using the product rule from Lemma 1.9.9 we have (suppressing all arguments) b α [pW (x, y)] = D b α [xpDα y − ypDα x] D   b α x [pDα y]ρ − k1 x[pDα y]ρ b α [pDα y] + D = xD   b α y [pDα x]ρ + k1 y[pDα x]ρ b α [pDα x] − D −yD     = x(Mc y) − y(Mc x) + [pDα y]ρ k0 x∇ − [pDα x]ρ k0 y∇ h iρ = x(Mc y) − y(Mc x) + k0 x∆ pDα y − y∆ pDα x h iρ = x(Mc y) − y(Mc x) + k0 x∆ pk1 y − y∆ pk1 x  ρ k1 = x(Mc y) − y(Mc x) − k0 [pW (x, y)]ρ k0 on Tκκ . Note that for any nabla differentiable f , bα f . k0 f ρ = (k0 + νk1 ) f − ν D It follows that  ρ h i b α [pW ] = x(Mc y) − y(Mc x) − k1 b α (pW ) . D (k0 + νk1 )pW − ν D k0

Second-Order Self-Adjoint Conformable Dynamic Equations  251

b α [pW ] terms on the left-hand side, we have Putting the D   ρ   ρ k1 b α [pW ] = x(Mc y) − y(Mc x) − k1 1−ν D (k0 + νk1 )pW, k0 k0 which is the first equation in the theorem; this can be rewritten as ! ρ ρ k k1 (k0 + νk1 ) 0 b [x(Mc y) − y(Mc x)] − ρ Dα [pW ] = ρ ρ pW. ρ k0 − νk1 k0 − νk1 Now let

(8.4)

ρ

ζ (t) := k1 (α,t) −

k0 (α,t)k1 (α,t) . ρ ρ k0 (α,t) − ν(t)k1 (α,t)

(8.5)

Observe that E0 = Ebζ by the equivalence of exponentials, and ρ ρ Ebζ k0 − νk1 . ρ ρ = k0 Ebζ

Additionally, by the quotient rule we see that " # i Ebζ h pW (x, y)(t) bα b α (pW ) − ζ pW + k1 pW. Ebζ (t, b)D = ρ D Ebζ (t, b) Eb

(8.6)

ζ

Substitution of (8.4) into (8.6) yields # "  Ebζ pW (x, y)(t) bα = Ebζ (t, b)D ρ Ebζ (t, b) Eb

! ρ ρ k0 k1 (k0 + νk1 ) [x(M y) − y(M x)] − c c ρ ρ ρ ρ pW k0 − νk1 k0 − νk1 ζ  −ζ pW + k1 pW

= x(Mc y) − y(Mc x), and thus (8.3) holds as well after recalling Ebζ = E0 . Remark 8.1.7 Consider ζ defined in (8.5). If T = R, then ν ≡ 0 and ρ(t) = t for all t ∈ T, hence ζ (t) = 0 for all t ∈ I . Similarly, for any time scale T, if α = 1, then k1 ≡ 0 and k0 ≡ 1, thus ζ (t) = 0 for all t ∈ I . Definition 8.1.8 (Inner Product) Let α ∈ (0, 1], let Eb0 be as given in (1.10), and let ζ be as in (8.5), with ρ ρ k0 (α,t) − ν(t)k1 (α,t) 6= 0, t ∈I. Define the (weighted) inner product of rd-continuous functions f , g ∈ C(I ) on [a, b]T ⊆ I to be h f , gi =

Z b f (t)g(t)Eb0 (b, ρ(t)) a

Ebζ (t, b)k0 (α,t)

∇t =

Z b f (t)g(t) a

Ebζ (t, b)

∇α,bt,

∇α,bt :=

Eb0 (b, ρ(t))∇t . (8.7) k0 (α,t)

252  Conformable Dynamic Equations on Time Scales

As Ebζ = E0 by the equivalence of exponential functions, this inner product can also be written as h f , gi =

Z b f (t)g(t)Eb0 (b, ρ(t)) a

E0 (t, b)k0 (α,t)

∇t =

Z b f (t)g(t) a

E0 (t, b)

∇α,bt,

∇α,bt :=

Eb0 (b, ρ(t))∇t . k0 (α,t)

Corollary 8.1.9 (Green’s Formula; Self-Adjoint Operator) Let Mc be given as in (8.1) and α ∈ (0, 1]. Assume ρ

ρ

k0 (α,t) − ν(t)k1 (α,t) 6= 0,

t ∈I.

If x, y ∈ D, then Green’s formula hx, Mc yi − hMc x, yi = p(b)W (x, y)(b) −

p(a)W (x, y)(a) E0 (a, b)Eb0 (a, b)

(8.8)

holds. Moreover, the operator Mc is formally self-adjoint with respect to the inner product (8.7); that is, the identity hx, Mc yi = hMc x, yi holds if and only if x, y ∈ D and x, y satisfy the self-adjoint boundary conditions p(b)W (x, y)(b) =

p(a)W (x, y)(a) , E0 (a, b)Eb0 (a, b)

(8.9)

where we have used the conformable Wronskian matrix on time scales from Definition 8.1.4. Proof 8.1.10 From Theorem 8.1.5 we have the Lagrange identity (8.3) given by " # pW (x, y)(t) bα Ebζ (t, b)D = x(Mc y) − y (Mc x) , Ebζ (t, b) with ζ defined in (8.5). If we multiply both sides of this equation by Eb0 (b, ρ(t)) Ebζ (t, b)k0 (α,t) and integrate from a to b we obtain " # Z b pW (x, y)(t) bα D ∇α,bt = hx, Mc yi − hMc x, yi. a Ebζ (t, b) By Theorem 1.9.7 we have " # Z b pW (x, y)(t) pW (x, y)(b) pW (x, y)(a) b bα D ∇α,bt = − E0 (b, a), a Ebζ (t, b) Ebζ (b, b) Ebζ (a, b) so that Green’s formula (8.8) holds. Thus hx, Mc yi = hMc x, yi if and only if x, y ∈ D satisfy the self-adjoint boundary conditions (8.9). This completes the proof. 

Second-Order Self-Adjoint Conformable Dynamic Equations  253

Corollary 8.1.11 (Abel’s Formula) Let Mc be given as in (8.1) and α ∈ (0, 1]. Assume ρ

ρ

k0 (α,t) − ν(t)k1 (α,t) 6= 0,

t ∈I.

If x, y ∈ D are solutions of Mc x = 0 on I , then for fixed a ∈ I the Wronskian satisfies W (x, y)(t) =

p(b)W (x, y)(b) (8.9) p(a)W (x, y)(a) = p(t)E0 (b,t)Eb0 (b,t) p(t)E0 (a,t)Eb0 (a,t)

for t ∈ I . Proof 8.1.12 As in (8.3) and in the proof of Corollary 8.1.9, for x, y ∈ D we have   pW (x, y)(t) b = x(Mc y) − y (Mc x) . E0 (t, b)Dα E0 (t, b) If x, y are solutions of (8.1) on I , then Mc x = 0 = Mc y and   pW (x, y)(t) b E0 (t, b)Dα = 0. E0 (t, b) As a result,  bα D so that

 pW (x, y)(t) = 0, E0 (t, b)

p(t)W (x, y)(t) = cEb0 (t, b), E0 (t, b)

where c is the constant c = p(b)W (x, y)(b). Consequently, W (x, y)(t) =

p(b)W (x, y)(b) (8.9) p(a)W (x, y)(a) = , p(t)E0 (b,t)Eb0 (b,t) p(t)E0 (a,t)Eb0 (a,t) 

completing the proof. Corollary 8.1.13 Let Mc be given as in (8.1) and α ∈ (0, 1]. Assume ρ

ρ

k0 (α,t) − ν(t)k1 (α,t) 6= 0,

t ∈I.

If x, y ∈ D are solutions of (8.1) on I , then either W (x, y)(t) = 0 for all t ∈ I , or W (x, y)(t) 6= 0 for all t ∈ I . Theorem 8.1.14 Equation (8.1) on I has two linearly independent solutions, and every solution of (8.1) on I is a linear combination of these two solutions. Theorem 8.1.15 (Converse of Abel’s Formula) Let Mc be given as in (8.1) and α ∈ (0, 1]. Assume ρ ρ k0 (α,t) − ν(t)k1 (α,t) 6= 0, t ∈I. Let x be a solution of Mc x = 0 on I such that x 6= 0 on I . If y ∈ D satisfies c W (x, y)(t) = p(t)E0 (b,t)Eb0 (b,t) for some constant c ∈ R, then y is also a solution of Mc x = 0.

254  Conformable Dynamic Equations on Time Scales

Proof 8.1.16 Suppose that x is a solution of Mc x = 0 such that x 6= 0 on I , and assume y ∈ D satisfies p(t)W (x, y)(t) = cE0 (t, b)Eb0 (t, b) for some constant c ∈ R. By the Lagrange identity (Theorem 8.1.5) and Lagrange’s formula (8.3) we have for t ∈ I that   h i p(t)W (x, y)(t) b b α cEb0 (t, b) ≡ 0; x(Mc y)(t) − y(Mc x)(t) = E0 (t, b)Dα = E0 (t, b)D E0 (t, b) since Mc x = 0, this yields x(Mc y)(t) ≡ 0 for t ∈ I . As x 6= 0 on I , (Mc y)(t) ≡ 0 on I . Thus, y is also a solution of Mc x = 0. This completes the proof.  Now, consider a few other second-order conformable dynamic equations, such as M1 x = 0,

b α [Dα x] + p1 [Dα x]ρ + p2 x, M1 x := D

(8.10)

b α [Dα x] + a1 Dα x + a2 x, M2 x := D

(8.11)

M2 x = 0,

where p j , a j : T → R are ld-continuous for j = 1, 2. Let DM denote the set of all functions b α -differentiable on Tκκ , and x : T → R such that x is Dα -differentiable on Tκ , (Dα x) is D α κ b α [D x] is ld-continuous on Tκ . We say x is a solution of M j x = 0 on T provided x ∈ DM D and M j x = 0 on Tκκ for j = 1, 2. + d Theorem 8.1.17 Let α ∈ (0, 1]. If p2 is ld-continuous and (p1 + k1 ) ∈ R c , then the dynamic equation (8.10) can be written in self-adjoint form (8.1), with

p(t) = Ebp1 +k1 (t,t0 ) and

q(t) = p2 (t)Ebp1 +k1 (t,t0 ).

Moreover, in this case, x is a solution of (8.10) if and only if x is a solution of Mc x = 0 for the self-adjoint Mc operator in (8.1). Proof 8.1.18 If p = Ebp1 +k1 and q = p2 Ebp1 +k1 , then (suppressing the arguments) i h b α [pDα x] + qx = D b α Ebp +k Dα x + p2 Ebp +k x Mc x = D 1 1 1 1 b α [Dα x] + (p1 + k1 )Ebp +k [Dα x]ρ = Ebp1 +k1 D 1 1 −k1 Ebp1 +k1 [Dα x]ρ + p2 Ebp1 +k1 x o n b α [Dα x] + (p1 + k1 ) [Dα x]ρ − k1 [Dα x]ρ + p2 x = Ebp1 +k1 D = Ebp1 +k1 M1 x. d + bp +k > 0, and thus Mc x = 0 if and only if M1 x = 0. As (p1 + k1 ) ∈ R c , we know that E 1 1 This completes the proof.  + d Theorem 8.1.19 Let α ∈ (0, 1]. If a2 is ld-continuous and (−a1 ) ∈ R c , then the dynamic equation (8.11) can be written in self-adjoint form (8.1), with

p(t) = Ebγ (t,t0 ) and where

ρ q(t) = a2 (t)Ebγ (t,t0 ),

(k0 + νk1 )(a1 + k1 ) . k0 + ν(a1 + k1 ) Moreover, in this case, x is a solution of (8.11) if and only if x is a solution of Mc x = 0 for the self-adjoint Mc operator in (8.1). γ :=

Second-Order Self-Adjoint Conformable Dynamic Equations  255 + d Proof 8.1.20 Since (−a1 ) ∈ R c , we have

k0 + ν(a1 + k1 ) = k0 − ν(−a1 − k1 ) > 0, so that γ is a well-defined function. Also,   k0 + νk1 γ − k1 = > 0, 1−ν k0 k0 + ν(a1 + k1 ) + d bγ and thus p and q well defined as well. If p = Ebγ and q = a2 Ebγρ , putting γ ∈ R c , making E then h i b α [pDα x] + qx = D b α Ebγ Dα x + a2 Ebγρ x Mc x = D ρb α bγ Dα x − k1 Ebγρ Dα x + a2 Ebγρ x = Ebγ D α [D x] + γ E     γ ρ b α [Dα x] +   Dα x − k1 Dα x + a2 x = Ebγ D   1 1 − ν γ−k k0 ρ = Ebγ M2 x. + d bγρ > 0, and thus Mc x = 0 if and only if M2 x = 0. This completes As γ ∈ R c we know that E the proof. 

Remark 8.1.21 In the next theorem, we will assume for constants a, b ∈ R that (k0 (α,t) − ν(t)k1 (α,t) − aν(t))(k0 (α,t) − ν(t)k1 (α,t)) + bν 2 (t) 6= 0, ρ

ρ

ρ

ρ

t ∈ Tκ .

If α = 1, then k0 ≡ 1 and k1 ≡ 0, and this assumption becomes 1 − aν(t) + bν 2 (t) 6= 0,

t ∈ Tκ ,

which is the standard assumption for second-order nabla equations on time scales. If T = R, the assumption is merely k0 (α,t) 6= 0, which is always true for α ∈ (0, 1]. Remark 8.1.22 Consider the assumption ρ

ρ

k0 (α,t) − ν(t)k1 (α,t) 6= 0,

t ∈ Tκ .

If t is a left-scattered or a left-dense-right-dense point, then this is clearly equivalent to the regressivity assumption k0 (α,t) − µ(t)k1 (α,t) 6= 0,

t ∈ Tκ .

Suppose t is a left-dense right-scattered point. Then σ (t) is a left-scattered point, and ρ

ρ

k0 (α, σ (t)) − ν(σ (t))k1 (α, σ (t)) 6= 0 implies k0 (α,t) − µ(t)k1 (α,t) 6= 0.

256  Conformable Dynamic Equations on Time Scales

Theorem 8.1.23 Let a, b ∈ R be given constants such that λ j ∈ R solves λ 2 + aλ + b = 0 for j = 1, 2, with λ1 6= λ2 . Assume ρ

ρ

k0 (α,t) − ν(t)k1 (α,t) 6= 0,

t ∈ Tκ

and (k0 (α,t) − ν(t)k1 (α,t) − aν(t))(k0 (α,t) − ν(t)k1 (α,t)) + bν 2 (t) 6= 0, ρ

ρ

ρ

ρ

Let p(t) := Ebβ (t,t0 ), where β :=

q(t) :=

t ∈ Tκ .

b(k0 (α,t) + ν(t)k1 (α,t)) b Eβ (t,t0 ), ρ ρ k0 (α,t) − ν(t)k1 (α,t)

  ρ ρ ρ (k0 + νk1 ) (a + k1 )(k0 − νk1 ) − bν ρ

ρ

ρ

(k0 − νk1 )k0

.

Then the general solution of Mc x = 0 for Mc in (8.1) with p, q as given here, is x(t) = c1 Eλ1 + c2 Eλ2 . Proof 8.1.24 Let a, b ∈ R be given constants such that λ j ∈ R solves λ 2 + aλ + b = 0 for j = 1, 2, with λ1 6= λ2 . Then a = −(λ1 + λ2 ) and b = λ1 λ2 . The regressivity conditions we need to check are k0 − µk1 6= 0, k0 + µ(λ − k1 ) 6= 0. By Remark 8.1.22, the assumption ρ

ρ

k0 − νk1 6= 0 implies

k0 − µk1 6= 0

on Tκ . Moreover,
ρ  ρ ρ
ρ ρ ρ  ρ  ρ k0 + ν(λ1 − k1 ) k0 + ν(λ2 − k1 ) = k0 − νk1 − aν k0 − νk1 + bν 2 6= 0 on Tκ by assumption. It follows that  ρ ρ  k0 + ν(λ1 − k1 ) 6= 0,

 ρ ρ  k0 + ν(λ2 − k1 ) 6= 0,

and thus [k0 + µ(λ1 − k1 )] 6= 0,

[k0 + µ(λ2 − k1 )] 6= 0,

putting λ1 , λ2 ∈ Rc . Hence, Eλ1 and Eλ2 are well defined. We will show that x = Eλ is a solution of Mc x = 0 for p and q as in the statement of the theorem, where λ represents, without loss of generality, either λ1 or λ2 . To this end, note that Dα x = λ Eλ , so that h i b α [pDα x] = λ D b α Ebβ Eλ D " # b α Ebβ Eb = λD k (λ −k )ρ 1 k1 + ρ 0 k0 +ν(λ −k1 )ρ

Second-Order Self-Adjoint Conformable Dynamic Equations  257

= λ

bc β⊕

k0 (λ − k1 )ρ k1 + ρ k0 + ν(λ − k1 )ρ

!! Ebβ Eλ .

Consequently, !! ρ k (λ − k ) 0 1 b α [pDα x] + qx = λ β ⊕ b c k1 + ρ D Ebβ Eλ k0 + ν(λ − k1 )ρ ! b(k0 + νk1 ) b Eβ Eλ + ρ ρ k0 − νk1 ! (k0 + νk1 )(λ 2 + aλ + b) b = Eβ Eλ ρ k0 + ν(λ − k1 )ρ = 0 as λ 2 + aλ + b = 0. Hence, x j = Eλ j is a solution for j = 1, 2. Since λ1 6= λ2 , the Wron
skian satisfies W Eλ1 , Eλ2 = (λ2 − λ1 )Eλ1 ⊕c λ2 6= 0, and the two solutions are linearly independent. This completes the proof.  Example 8.1.25 Let a = −8 and b = 15, T = hZ for h > 0, k0 ≡ α, k1 ≡ 1 − α, and let h α ∈ (0, 1] such that α 6= . Clearly λ1 = 3 and λ2 = 5 solve λ 2 + aλ + b = 0, with 1+h λ1 6= λ2 . Checking the regressivity conditions, ρ

ρ

k0 (α,t) − ν(t)k1 (α,t) = α − h(1 − α) 6= 0 as α 6=

h , and 1+h (k0 (α,t) − ν(t)k1 (α,t) − aν(t))(k0 (α,t) − ν(t)k1 (α,t)) + bν 2 (t) ρ

ρ

ρ

ρ

= (α − h(1 − α) + 8h)(α − h(1 − α)) + 15h2 > 0. Let p(t) := Ebβ (t,t0 ), where β :=

q(t) :=

15(α + h(1 − α)) b Eβ (t,t0 ), α − h(1 − α)

(α + h(1 − α)) [(−7 − α)(α − h(1 − α)) − 15h] . (α − h(1 − α))α

Then the general solution of Mc x = 0 for Mc in (8.1) with p, q as given here, is x(t) = c1 E3 (t,t0 ) + c2 E5 (t,t0 ). b In particular, if α = 1/5 and h = 1 for example, h then 1β i= 89 and q = −25E89 , and x = b 1 Eb89 D 5 x − 25Eb89 x = 0. c1 E3 + c2 E5 solves the self-adjoint equation D 5

258  Conformable Dynamic Equations on Time Scales

Lemma 8.1.26 Assume k0 (α,t) − µ(t)k1 (α,t) 6= 0,

t ∈ Tκ .

If f : T → R is ∆-differentiable, then 1 α D [E0 f ] = Dα f − k1 f = k0 f ∆ . E0σ

(8.12)

Similarly, if f : T → R is ∇-differentiable, then 1 b b ∇ b ρ Dα [E0 f ] = Dα f − k1 f = k0 f . b E

(8.13)

0

Theorem 8.1.27 Let a, b ∈ R with 1 − aν(t) + bν 2 (t) 6= 0, t0 ∈ T, and assume k0 (α,t) − µ(t)k1 (α,t) 6= 0,

t ∈ Tκ .

Let p(t) :=

Eb(a−bν)(k0 +νk1 ) (t,t0 ) , k0 (α,t)E0σ (t,t0 )

  k1 (α,t) k0 (α,t)bEb(a−bν)(k0 +νk1 ) (t,t0 ) q(t) := 1 + ν(t) . k0 (α,t) E0 (t,t0 )

If λ is a solution of the characteristic equation λ 2 + aλ + b = 0, then a solution of Mc x = 0 for Mc in (8.1) with p, q as given here, is x(t) = Eλ (k0 −µk1 ) (t,t0 ). Proof 8.1.28 Let p, q be as given in the statement of the theorem, and let x(t) = Eλ (k0 −µk1 ) (t,t0 ) = E0 (t,t0 )E(k0 λ +k1 ) (t,t0 ). Then Dα x(t) = λ (k0 − µk1 )(t)x(t) implies Eb(a−bν)(k0 +νk1 ) (t,t0 ) λ (k0 − µk1 )(t)Eλ (k0 −µk1 ) (t,t0 ) k0 (α,t)E0σ (t,t0 ) λ Eb(a−bν)(k0 +νk1 ) (t,t0 ) = Eλ (k0 −µk1 ) (t,t0 ) E0 (t,t0 ) = λ Eb(a−bν)(k0 +νk1 ) (t,t0 )E(k0 λ +k1 ) (t,t0 ) = λ Eb0 (t,t0 )Eb(k (a−bν)+k ) (t,t0 )E(k λ +k ) (t,t0 ).

p(t)Dα x(t) =

0

1

0

1

Let e p and ebp be the delta and nabla exponential functions on time scales, respectively. Then p(t)Dα x(t) = λ Eb0 (t,t0 )b e(a−bν) (t,t0 )eλ (t,t0 )

Second-Order Self-Adjoint Conformable Dynamic Equations  259

= λ Eb0 (t,t0 )b e(a−bν) (t,t0 )b e

λ 1+λ ν

(t,t0 )

= λ Eb0 (t,t0 )b e( a−bν+λ ) (t,t0 ). 1+λ ν

As a result, b α [p(t)Dα x] (t) D

= (8.13)

=

  b α Eb0 (t,t0 )b λD e( a−bν+λ ) (t,t0 ) 1+λ ν   a − bν + λ ρ b eb( a−bν+λ ) (t,t0 ). λ E0 (t,t0 )k0 (α,t) 1+λν 1+λ ν

Moreover,   k1 (α,t) k0 (α,t)bEb(a−bν)(k0 +νk1 ) (t,t0 ) Eλ (k0 −µk1 ) (t,t0 ) q(t)x(t) = 1 + ν(t) k0 (α,t) E0 (t,t0 )   k1 (α,t) = 1 + ν(t) k0 (α,t)bEb0 (t,t0 )Eb(k0 (a−bν)+k1 ) (t,t0 )E(k0 λ +k1 ) (t,t0 ) k0 (α,t) = bEb0σ (t,t0 )k0 (α,t)Eb(k0 (a−bν)+k1 ) (t,t0 )E(k0 λ +k1 ) (t,t0 ) = bEb0σ (t,t0 )k0 (α,t)b e( a−bν+λ ) (t,t0 ). 1+λ ν It follows that     a − bν + λ ρ λ e( a−bν+λ ) (t,t0 ) + b Eb0 (t,t0 )k0 (α,t)b 1+λν 1+λ ν  2  λ + aλ + b bρ = E0 (t,t0 )k0 (α,t)b e( a−bν+λ ) (t,t0 ) 1+λν 1+λ ν = 0,

b α [p(t)Dα x] (t) + q(t)x(t) = D

as λ is a solution of the characteristic equation. This completes the proof.



Example 8.1.29 Assume k0 (α,t) − µ(t)k1 (α,t) 6= 0,

t ∈ Tκ .

Let ω > 0 and t0 ∈ T. In the previous theorem, Theorem 8.1.27, let a = 0 and b = ω 2 ; clearly 1 + ω 2 ν 2 (t) 6= 0. Let Eb(−ω 2 ν)(k0 +νk1 ) (t,t0 )

  k1 (α,t) k0 (α,t)bEb(−ω 2 ν)(k0 +νk1 ) (t,t0 ) p(t) := , q(t) := 1 + ν(t) . k0 (α,t)E0σ (t,t0 ) k0 (α,t) E0 (t,t0 ) (8.14) If λ is a solution of the characteristic equation λ 2 + ω 2 = 0, then a solution of Mc x = 0 for Mc in (8.1) with p, q as given here, is x(t) = Eλ (k0 −µk1 ) (t,t0 )

260  Conformable Dynamic Equations on Time Scales

= E0 (t,t0 )E(k0 λ +k1 ) (t,t0 ) = E0 (t,t0 )eλ (t,t0 ) = E0 (t,t0 )e±iω (t,t0 ). Recall the trigonometric functions on time scales defined by cosω =

eiω + e−iω , 2

sinω =

eiω − e−iω , 2i

where e p is the exponential function on time scales. It follows that Mc x = 0 has the general solution x(t) = c1 E0 (t,t0 ) cosω (t,t0 ) + c2 E0 (t,t0 ) sinω (t,t0 ) for p, q given in (8.14). This ends the example.

8.2

REDUCTION-OF-ORDER THEOREMS

In this section we establish two related reduction-of-order theorems for the conformable self-adjoint dynamic equation Mc x = 0 for Mc given in (8.1). Theorem 8.2.1 (Reduction of Order I) Let Mc be given as in (8.1) and α ∈ (0, 1]. Assume ρ

ρ

k0 (α,t) − ν(t)k1 (α,t) 6= 0,

t ∈I.

Let t0 ∈ I , and assume x is a solution of Mc x = 0 with x 6= 0 on I . Then for t ∈ I , Z t

y(t) = x(t) t0

Z t E0 (s,t0 )Eb0 (s,t0 ) E0 (s,t0 )E0 (σ (s),t)Eb0 (s,t0 ) ∆s = x(t) ∆α,t s p(s)x(s)xσ (s)k0 (α, s) p(s)x(s)xσ (s) t0

defines a second linearly independent solution y of Mc x = 0 on I . Proof 8.2.2 By the converse of Abel’s formula, Theorem 8.1.15, if y ∈ DR satisfies W (x, y)(t) =

c p(t)E0 (b,t)Eb0 (b,t)

for some constant c ∈ R, then y is also a solution of Mc x = 0. Checking the Wronskian, we have W (x, y) = xDα y − yDα x = x[k1 y + k0 y∆ ] − y[k1 x + k0 x∆ ] = k0 xy∆ − k0 yx∆ # " Z t b b E (s,t ) E (s,t ) E (t,t ) E (t,t ) 0 0 0 0 0 0 0 0 = k0 x x σ + x∆ ∆s − k0 yx∆ σ pxxσ k0 t0 p(s)x(s)x (s)k0 (α, s) =

Z t E0 (t,t0 )Eb0 (t,t0 ) E0 (s,t0 )Eb0 (s,t0 ) ∆ + k0 x x ∆s − k0 x∆ y σ p(t) t0 p(s)x(s)x (s)k0 (α, s)

Second-Order Self-Adjoint Conformable Dynamic Equations  261

E0 (t,t0 )Eb0 (t,t0 ) , p(t)

=

and the condition holds with c = E0 (t0 , b)Eb0 (t0 , b). To show that y ∈ DR , one can modify the argument made by Messer in Chapter 4 (page 98) of Bohner and Peterson’s 2003 monograph to the conformable setting. Thus, by Theorem 8.1.15, y is also a linearly independent solution of Mc x = 0. This completes the proof.  Theorem 8.2.3 (Reduction of Order II) Let Mc be given as in (8.1) and α ∈ (0, 1]. Assume ρ ρ t ∈I. k0 (α,t) − ν(t)k1 (α,t) 6= 0, Let t0 ∈ I , and assume x is a solution of Mc x = 0 with x 6= 0 on I . Then y is a second linearly independent solution of Mc x = 0 if and only if y satisfies the first-order equation Dα [y/x](t) − k1 (t)[y/x](t) =

cE0 (t,t0 )Eb0 (t,t0 ) , p(t)x(t)xσ (t)

t ∈I,

t ≥ t0 ,

(8.15)

for some constant c ∈ R if and only if y is of the form Z t

y(t) = c1 x(t) + c2 x(t)

t0

E0 (s,t0 )Eb0 (s,t0 ) ∆s p(s)x(s)xσ (s)k0 (α, s)

(8.16)

for t ∈ I with t ≥ t0 , where c1 , c2 ∈ R are constants. In the latter case, c1 = y(t0 )/x(t0 ),

c2 = p(t0 )W (x, y)(t0 ).

(8.17)

Proof 8.2.4 Assume x is a solution of Mc x = 0 with x 6= 0 on I . Let y be any solution of Mc x = 0; we must show y is of the form (8.16). Using the Wronskian, set c2 := p(t0 )W (x, y)(t0 ). By Abel’s formula, Corollary 8.1.11 with a = t0 , we must have x(t)Dα y(t) − y(t)Dα x(t) = W (x, y)(t) =

c2 p(t)E0 (t0 ,t)Eb0 (t0 ,t)

,

t ∈I,

so that xDα y − yDα x y c2 E0 (t,t0 )Eb0 (t,t0 ) y + k1 = + k1 . σ σ xx x pxx x Thus, we see that Dα [y/x] =

c2 E0 (t,t0 )Eb0 (t,t0 ) y + k1 pxxσ x

(8.18)

and y satisfies (8.15). Note that the left-hand side of (8.15) simplifies to k0 [y/x]∆ ; divide it by k0 and use the straight ∆-integral on time scales on both sides from t0 to t to get y(t)/x(t) − y(t0 )/x(t0 ) = c2

Z t t0

E0 (s,t0 )Eb0 (s,t0 ) ∆s. p(s)x(s)xσ (s)k0 (α, s)

262  Conformable Dynamic Equations on Time Scales

Recovering y from this yields Z t

y(t) = c1 x(t) + c2 x(t)

t0

E0 (s,t0 )Eb0 (s,t0 ) ∆s p(s)x(s)xσ (s)k0 (α, s)

provided c1 = y(t0 )/x(t0 ). Conversely, assume y is given by (8.16). By Theorem 8.2.1 and linearity, y is a solution of Mc x = 0 on I for t ≥ t0 . Setting t = t0 in (8.16) leads to c1 in (8.17). By Abel’s formula, Corollary 8.1.11 with a = t0 , the Wronskian satisfies W (x, y)(t) =

p(t0 )W (x, y)(t0 ) ; p(t)E0 (t0 ,t)Eb0 (t0 ,t)

to calculate W (x, y)(t0 ), we use (8.16) to obtain W (x, y)(t0 ) = (xDα y − yDα x)(t0 ) =

c2 . p(t0 ) 

This ends the proof.

8.3

DOMINANT AND RECESSIVE SOLUTIONS

In this section we lay the groundwork for further exploration of the non-homogeneous equation (8.1) by introducing the P´olya factorization for the conformable self-adjoint dynamic equation Mc x = 0, which in turn leads to a variation of parameters result for Mc x = h and to the notion of dominant and recessive solutions. Again we assume throughout that the coefficient function p satisfies p > 0. We begin, though, with a discussion of oscillation and disconjugacy for the homogeneous equation. Definition 8.3.1 A function x ∈ DR has a generalized zero at t ∈ I if x(t) = 0 or, if t is a left-scattered point and xρ (t)x(t) < 0. Definition 8.3.2 The homogeneous self-adjoint equation (8.1), namely b α [pDα ] (t) + q(t)x(t) = 0, Mc x(t) = D

(8.19)

is disconjugate on a time-scale interval [a, b]T if and only if the following hold. 1. If x is a nontrivial solution of (8.19) with x(a) = 0, then x has no generalized zeros in (a, b]T . 2. If x is a nontrivial solution of (8.19) with x(a) 6= 0, then x has at most one generalized zero in (a, b]T . Definition 8.3.3 Let ω = sup T; if ω < ∞, assume ρ(ω) = ω. Let a ∈ T. Then (8.19) is oscillatory if and only if every nontrivial real-valued solution has infinitely many generalized zeros in [a, ω)T . Equation (8.19) is non-oscillatory on [a, ω)T if and only if it is not oscillatory on [a, ω)T .

Second-Order Self-Adjoint Conformable Dynamic Equations  263

Lemma 8.3.4 Let ω = sup T; if ω < ∞, assume ρ(ω) = ω. Let a ∈ T. Then if (8.19) is non-oscillatory on [a, ω)T , there exists t0 ∈ T with t0 ≥ a such that (8.19) has a positive solution on [t0 , ω)T . Proof 8.3.5 If (8.19) is non-oscillatory on [a, ω)T , then there exists a nontrivial solution y of (8.19) that has only finitely many generalized zeros in [a, ω)T . Set τ := max{t ∈ T : y has a generalized zero at t}. Then for any t0 ∈ T with t0 > τ, either y > 0 on [t0 , ω)T , or −y > 0 on [t0 , ω)T . This completes the proof.  Theorem 8.3.6 (Sturm Separation) Assume k0 (α,t) − µ(t)k1 (α,t) > 0 for t ∈ Tκ . Let x and y be linearly independent solutions of (8.19) on T. Then x and y have no common zeros in Tκ . If x has a zero at t1 ∈ T and a generalized zero at t2 ∈ T with t2 > t1 , then y has a generalized zero in (t1 ,t2 ]T . If x has generalized zeros at t1 ,t2 ∈ T with t2 > t1 , then y has a generalized zero in [t1 ,t2 ]T . Proof 8.3.7 If x and y share a common zero at t0 ∈ Tκ , then their Wronskian satisfies W (x, y)(t0 ) = 0, making x and y linearly dependent, a contradiction. If x has a zero at t1 ∈ T and a generalized zero at t2 ∈ T with t2 > t1 , then without loss of generality, assume t2 > σ (t1 ) is the first generalized zero to the right of t1 , and that x(t) > 0 on (t1 ,t2 )T with x(t2 ) ≤ 0. Assume y is a linearly independent solution of (8.19) on T with no generalized zeros in (t1 ,t2 ]T ; without loss of generality, y(t) > 0 for t ∈ [t1 ,t2 ]T . Note that on Tκ , W (y, x) = yDα x − xDα y = y(k1 x + k0 x∆ ) − x(k1 y + k0 y∆ ) = k0 (yx∆ − xy∆ ). It follows that for t ∈ [t1 ,t2 ]T ,  ∆ x (t) = y

yx∆ − xy∆ (t) yyσ W (y, x) (t) = yyσ k0 cE0 (t, a)Eb0 (t, a) = p(t)y(t)yσ (t)k0 (α,t)

is of one sign on [t1 ,t2 )T , since y(t) > 0 for t ∈ [t1 ,t2 ]T , Eb0 (t, a), p(t), k0 (α,t) are all positive on T, and the assumption k0 (α,t) − µ(t)k1 (α,t) > 0 for t ∈ Tκ guarantees E0 (t, a) > 0 on T as well. Consequently, (x/y) is monotone on [t1 ,t2 ]T . For any τ ∈ (t1 ,t2 )T , x(t1 ) = 0 and y(t1 )

x(τ) > 0, y(τ)

while

x(t2 ) ≤ 0, y(t2 )

a contradiction of monotonicity. Thus y must have a generalized zero in (t1 ,t2 ]T . If x has generalized zeros at t1 ,t2 ∈ T with t2 > t1 , then without loss of generality, assume t2 > σ (t1 ) is the first generalized zero to the right of t1 . If x(t1 ) = 0, this has been dealt with in the previous case, so x(t1 ) 6= 0, but t1 must be a left-scattered point; assume without loss of

264  Conformable Dynamic Equations on Time Scales

generality that x(t) > 0 on [t1 ,t2 )T , with xρ (t1 ) < 0 and x(t2 ) ≤ 0. Assume y is a linearly independent solution of (8.19) on T with no generalized zeros in [t1 ,t2 )T ; without loss of generality, y(t) > 0 for t ∈ [t1 ,t2 ]T and yρ (t1 ) > 0. As in the previous case, (x/y) is monotone on [ρ(t1 ),t2 ]T . However, xρ (t1 ) < 0 and yρ (t1 )

x(t1 ) > 0, y(t1 )

while

x(t2 ) ≤ 0, y(t2 )

a contradiction of monotonicity. Thus y must have a generalized zero in [t1 ,t2 ]T . This completes the proof.  Theorem 8.3.8 (Disconjugacy) Assume k0 (α,t) − µ(t)k1 (α,t) > 0 for t ∈ Tκ . If (8.19) has a positive solution on a time-scale interval I ⊂ T, then (8.19) is disconjugate on I . Conversely, if (8.19) is disconjugate on [ρ(a), σ (b)]T for some a, b ∈ Tκκ with a < b, then (8.19) has a positive solution on [ρ(a), σ (b)]T . Proof 8.3.9 Assume (8.19) has a positive solution x on the interval I ⊂ T. If (8.19) is not disconjugate on I , then it has a nontrivial solution y with at least two generalized zeros in I . So, without loss of generality, there exist t1 ,t2 ∈ I such that (t1 ,t2 )T 6= 0/ with y(t1 ) ≤ 0,

y(t2 ) ≤ 0,

y(t) > 0,

t ∈ (t1 ,t2 )T .

It follows that for t ∈ [t1 ,t2 ]T ,  y ∆ x

(t) = =

xy∆ − yx∆ (t) xxσ cE0 (t, a)Eb0 (t, a) p(t)x(t)xσ (t)k0 (α,t)

is of one sign on [t1 ,t2 )T , since x(t) > 0 for t ∈ [t1 ,t2 ]T . Consequently, (y/x) is monotone on [t1 ,t2 ]T . For any τ ∈ (t1 ,t2 )T , y(t1 ) ≤ 0 and x(t1 )

y(τ) > 0, x(τ)

while

y(t2 ) ≤ 0, x(t2 )

a contradiction of monotonicity. Thus (8.19) is disconjugate on I . Conversely, assume (8.19) is disconjugate on the compact interval [ρ(a), σ (b)]T for some a, b ∈ Tκκ with a < b. Let u, y be the unique solutions of (8.19) satisfying uρ (a) = 0, Dα uρ (a) = k0 (α, ρ(a)),

yσ (b) = 0, Dα y(b) = −k0 (α, b),

respectively. Note that u then satisfies u∆ρ (a) = 1, while y satisfies y(b) = 0 and y∆ (b) = −1 µ(b)k0 (α, b) > 0 with yσ (b) = 0 if b is a if b is a left-dense point, and y(b) = k0 (α, b) − µ(b)k1 (α, b) left-scattered point. Since (8.19) is disconjugate on [ρ(a), σ (b)]T , we have u(t) > 0 for t ∈ (ρ(a), σ (b)]T and y(t) > 0 for t ∈ [ρ(a), σ (b))T . If we take x = u + y, then x is a positive solution of (8.19) on [ρ(a), σ (b)]T . This completes the proof. 

Second-Order Self-Adjoint Conformable Dynamic Equations  265

Theorem 8.3.10 (P´olya Factorization) Let Mc be as in (8.1), and assume Mc x = 0 has a positive solution x > 0 on an interval [a, b)T with b ≤ ∞. If k0 (α,t) − µ(t)k1 (α,t) > 0,

t ∈ [a, b)T

then for any function y ∈ D we have a P´olya factorization of Mc y on [a, b)T given by b α {ϕ2 Dα [ϕ1 y]} Mc y = ϕ1 D

on [a, b)T ,

where ϕ1 (t) :=

E0 (t, a) > 0, x(t)

ϕ2 (t) :=

p(t)x(t)xσ (t) p(t) = > 0. σ E0 (t, a)E0 (t, a) ϕ1 (t)ϕ1σ (t)

Proof 8.3.11 Recall from (8.12) that 1 α D [E0 f ] = Dα f − k1 f , E0σ a fact that we will use in the calculations below. Assume x > 0 is a positive solution of Mc x = 0 on [a, b)T , and let y ∈ D. Then   p Thm 8.1.5 E0 (·, a) b α α Mc y = Dα [xD y − yD x] x E0 (·, a)    σ 1 y pxx α α bα = ϕ1 D D y− σ D x E0 (·, a) xσ xx    h i σ pxx y α y b = ϕ1 Dα D − k1 E0 (·, a) x x    σ pxx 1 (8.12) α bα = ϕ1 D D [E0 (·, a)y/x] E0 (·, a) E0σ (·, a) b α {ϕ2 Dα [ϕ1 y]} , = ϕ1 D for ϕ1 and ϕ2 as defined in the statement of the theorem. This completes the proof.



Theorem 8.3.12 (Variation of Parameters) Assume k0 (α,t) − µ(t)k1 (α,t) > 0,

t ∈ [a, ∞)T .

Let h be a continuous function defined on [a, ∞)T , and let Mc be given as in (8.1). If the homogeneous equation Mc x = 0 has a positive solution x > 0 on [a, ∞)T , then the nonhomogeneous equation Mc y = h has a solution y given by y(t) =

y(a)x(t) + p(a)W (x, y)(a)x(t) x(a) Z t

+x(t) a

E0 (s, a) (pxxσ )(s)k0 (α, s)

Z t E0 (s, a)Eb0 (s, a)

∆s (pxxσ )(s)k0 (α, s) Z s x(ω)h(ω)Eb0 (s, ρ(ω)) a

a

E0 (ω, a)k0 (α, ω)

! ∇ω ∆s.

266  Conformable Dynamic Equations on Time Scales

Proof 8.3.13 Let y ∈ D be defined on [a, ∞)T , and assume x > 0 is a positive solution of Mc x = 0 on [a, ∞)T . As in Theorem 8.3.10, we use the P´olya factorization of Mc y to get    E0 (ω, a) b p(ω)x(ω)xσ (ω) α E0 (ω, a) h(ω) = Mc y(ω) = Dα D y(ω) . x(ω) E0 (ω, a)E0σ (ω, a) x(ω) Multiplying by Eb0 (t, ρ(ω))E0 (a, ω)x(ω)/k0 (α, ω) and integrating from a to t we arrive, via the fundamental theorem of conformable ∇ calculus, at Z t x(ω)h(ω)Eb0 (t, ρ(ω))

∇ω E0 (ω, a)k0 (α, ω)    Z t p(ω)x(ω)xσ (ω) α E0 (ω, a) Eb0 (t, ρ(ω)) b = Dα D y(ω) ∇ω E0 (ω, a)E0σ (ω, a) x(ω) k0 (α, ω) a   p(t)x(t)xσ (t) α E0 (t, a) p(a)x(a)xσ (a) = − y(t) W (x, y)(a)Eb0 (t, a) D E0 (t, a)E0σ (t, a) x(t) x(a)xσ (a)   p(t)x(t)xσ (t) α E0 (t, a) = D y(t) − p(a)W (x, y)(a)Eb0 (t, a) E0 (t, a)E0σ (t, a) x(t) a

using (8.12), since x > 0 is a solution. This leads to  Dα

 E0 (t, σ (s)) E0 (·, a) y (s) ∆s = x k0 (α, s)

E0 (t, a)p(a) W (x, y)(a)E0 (s, a)Eb0 (s, a)∆s (pxxσ )(s)k0 (α, s) E0 (t, a)E0 (s, a) + (pxxσ )(s)k0 (α, s)

Z s x(ω)h(ω)Eb0 (s, ρ(ω)) a

E0 (ω, a)k0 (α, ω)

! ∇ω ∆s.

Integrating this from a to t and using the fundamental theorem of conformable ∆ calculus yields y(t) y(a) − x(t) x(a)

= p(a)W (x, y)(a)

Z t E0 (s, a)Eb0 (s, a) a

Z t

+ a

(pxxσ )(s)k0 (α, s)

E0 (s, a) σ (pxx )(s)k0 (α, s)

∆s

Z s x(ω)h(ω)Eb0 (s, ρ(ω)) a

E0 (ω, a)k0 (α, ω)

! ∇ω ∆s,

which can be rewritten in the form for y given in the statement of the theorem. Clearly the right-hand side of the form of y above reduces to y(a) at a, and since x > 0 is a solution the conformable derivative reduces to Dα y(a) at a. This completes the proof.  Corollary 8.3.14 Let h be a continuous function defined on [a, ∞)T , and let Mc be given as in (8.1). Assume k0 (α,t) − µ(t)k1 (α,t) > 0, t ∈ [a, b)T . If the homogeneous matrix equation (8.1) has a positive solution x > 0 on [a, ∞)T , then the non-homogeneous initial value problem Mc y = h, has a unique solution.

y(a) = ya ,

Dα y(a) = y0a

(8.20)

Second-Order Self-Adjoint Conformable Dynamic Equations  267

Proof 8.3.15 By Theorem 8.3.12, the non-homogeneous initial value problem (8.20) has a solution. Suppose y1 and y2 both solve (8.20). Then x = y1 − y2 solves the homogeneous initial value problem Mc x = 0, x(a) = 0, Dα x(a) = 0; by Theorem 8.1.2, this has only the trivial solution x ≡ 0, and thus y1 = y2 is unique. This completes the proof.  Theorem 8.3.16 (Trench Divergence) Let Mc be as in (8.1). Let a ∈ T, and b := sup T; if b < ∞, assume ρ(b) = b. Also assume k0 (α,t) − µ(t)k1 (α,t) > 0, and let δ2 (t) =

t ∈ [a, b)T ,

E0 (t, a)E0σ (t, a) . p(t)x(t)xσ (t)

If Mc x = 0 has a positive solution x on an interval [a, b)T with b ≤ ∞, then either Z b a

Z b E0 (t, a)Eb0 (t, a) ∆t = δ2 (t)Eb0 (t, a)∆α,at = ∞, p(t)x(t)xσ (t)k0 (α,t) a

or Z b a

δ2 (t)Eb0 (t, a)  R  ∆α,at = ∞. b b b0 (s, a)∆α,a s b0 (s, a)∆α,a s δ (s) E δ (s) E 2 2 t σ (t)

R

Proof 8.3.17 Since Mc x = 0 has a positive solution x on [a, b)T , for any y ∈ D we have a P´olya factorization of Mc y on [a, b)T given by b α {ϕ2 Dα [ϕ1 y]} Mc y = ϕ1 D where ϕ1 (t) :=

E0 (t, a) > 0, x(t)

ϕ2 (t) :=

on

[a, b)T ,

p(t)x(t)xσ (t) 1 > 0. = σ E0 (t, a)E0 (t, a) δ2 (t)

Either Z b δ2 (s)Eb0 (s, a)E0 (a, σ (s))

k0 (α, s)

a

Z b

∆s = a

δ2 (s)Eb0 (s, a)∆α,a s = ∞,

or Z b δ2 (s)Eb0 (s, a)E0 (a, σ (s))

k0 (α, s)

a

Z b

∆s = a

δ2 (s)Eb0 (s, a)∆α,a s < ∞.

If (8.21) holds, set δ2 (t)Eb0 (t, a)  R  b b b b t δ2 (s)E0 (s, a)∆α,a s σ (t) δ2 (s)E0 (s, a)∆α,a s

β2 (t) = R

for t ∈ [a, b)T . Then by (8.21) we have Z b a

Z c

β2 (t)∆α,at = lim

c→b−

a

δ2 (t)E0 (t, a)  R  ∆α,at b b b b t δ2 (s)E0 (s, a)∆α,a s σ (t) δ2 (s)E0 (s, a)∆α,a s b

R

(8.21)

268  Conformable Dynamic Equations on Time Scales Z c

= lim

c→b− a

δ2 (t)E0 (t, a)E0 (a, σ (t)) ∆t  R  σ (t) δ2 (s)Eb0 (s,a)E0 (a,σ (s)) t δ2 (s)Eb0 (s,a)E0 (a,σ (s)) k (α,t) 0 ∆s ∆s b b k0 (α,s) k0 (α,s) b

R

Z c

E0 (t, a)δ2 (t)Eb0 (t, a) ∆t  R  σ (t) δ2 (s)Eb0 (s,a)E0 (σ (t),σ (s)) t δ2 (s)Eb0 (s,a)E0 (t,σ (s)) k (α,t) c→b− a ∆s ∆s 0 b b k0 (α,s) k0 (α,s) # ( " Z c −E0 (t, a) = lim Dα R t c→b− a δ (s) Eb0 (s, a)∆α,t s b 2   ∆t 1 −k1 (α,t) R b δ2 (s)Eb0 (s,a)E0 (a,σ (s)) ∆s  k0 (α,t) t k0 (α,s)    Z c  −E (t, a) 1 0  = lim E0σ (t, a)Dα  R σ t δ2 (s)Eb0 (s,a)E0 (t,σ (s)) c→b− a E0 (t, a)  ∆s b k0 (α,s)   ∆t 1 −k1 (α,t)E0σ (t, a) R b δ2 (s)Eb0 (s,a)E0 (a,σ (s)) ∆s  k0 (α,t) t k0 (α,s) ( ) Z c 1 ∆t 1 α = lim D E0 (t, a) R b σ b k (α,t) c→b− a E0 (t, a) 0 δ (s) E (s, a)∆ s α,a 0 t 2 ( ! Z c 1 = lim Dα R b − b c→b a t δ2 (s)E0 (s, a)∆α,a s !) ∆t 1 −k1 (α,t) R b b k (α,t) 0 δ (s) E (s, a)∆ s α,a 0 t 2  !∆  Z c  ∆t 1 = lim k0 (α,t) R b b  k0 (α,t) c→b− a  t δ2 (s)E0 (s, a)∆α,a s ! c 1 = lim R b b t=a c→b− t δ2 (s)E0 (s, a)∆α,a s = ∞. = lim

R



This completes the proof.

Theorem 8.3.18 (Dominant and Recessive Solutions) Let Mc be as in (8.1). Also assume k0 (α,t) − µ(t)k1 (α,t) > 0,

t ∈ [a, b)T .

If Mc x = 0 is non-oscillatory on an interval [a, b)T ⊂ (0, ∞)T with b ≤ ∞, then there is a positive solution u, called a recessive solution at b, such that for any second linearly independent solution v, called a dominant solution at b, u(t) = 0, − t→b v(t)

Z b

lim

a

E0 (t, a)Eb0 (t, a) ∆t = ∞, p(t)u(t)uσ (t)k0 (α,t)

Z b

and t0

E0 (s, a)Eb0 (s, a) ∆s < ∞, p(s)v(s)vσ (s)k0 (α, s)

where t0 < b is sufficiently close. Furthermore p(t)Dα u(t) p(t)Dα v(t) > v(t) u(t)

(8.22)

Second-Order Self-Adjoint Conformable Dynamic Equations  269

for t < b sufficiently close. Moreover, the recessive solution is unique up to multiplication by a nonzero constant. Proof 8.3.19 Since we are assuming that Mc x = 0 is non-oscillatory on [a, b)T , without loss of generality, it has a positive solution x on [a, b)T . Case I. Suppose for this solution x > 0 that Z b a

E0 (s, a)Eb0 (s, a) ∆s = p(s)x(s)xσ (s)k0 (α, s)

Z b E0 (s, a)E0σ (s, a) b a

p(s)x(s)xσ (s)

E0 (s, a)∆α,a s = ∞.

Let u = x, and let v(t) = u(t)

Z t E0 (s, a)E0σ (s, a) a

p(s)u(s)uσ (s)

Eb0 (s, a)∆α,a s > 0.

Then v is a second linearly independent solution of Mc x = 0 on [a, b)T by reduction of order, with v(a) = 0. Note that 1

u(t) = lim R t→b− v(t) t→b− t lim

E0 (s,a)E0σ (s,a) b a p(s)u(s)uσ (s) E0 (s, a)∆α,a s

=0

and Z b a

E0 (t, a)Eb0 (t, a) ∆t = ∞ p(t)u(t)uσ (t)k0 (α,t)

by the assumption on the integral in this case. Pick t0 ∈ [a, b)T so that v(t) 6= 0 on [t0 , b)T , which is possible by the positivity of the solution u = x. For t ∈ [t0 , b)T , by reduction of order hui u(t) −E0 (t, a)Eb0 (t, a) Dα (t) − k1 (t) = , v v(t) p(t)v(t)vσ (t) but we also have h i h u(t) 1 ui α u α (t) = D (t) − k1 (t) , D E (·, a) 0 σ E0 (t, a) v v v(t) so that a little manipulation yields h u i E0 (t, σ (s)) −E0 (t, a)E0 (s, a)Eb0 (s, a) Dα E0 (·, a) (s) = v k0 (α, s) p(s)v(s)vσ (s)k0 (α, s) for t ∈ [t0 , b)T . Integrating both sides of this last equality from t0 to t we obtain u(t) u(t0 ) − = v(t) v(t0 )

Z t t0

−E0 (s, a)Eb0 (s, a) ∆s. p(s)v(s)vσ (s)k0 (α, s)

Letting t → b− we get Z b t0

E0 (s, a)Eb0 (s, a) u(t0 ) ∆s = < ∞. p(s)v(s)vσ (s)k0 (α, s) v(t0 )

270  Conformable Dynamic Equations on Time Scales

Finally, for t ∈ [t0 , b)T , p(t)Dα v(t) p(t)Dα u(t) p(t)W (u, v)(t) E0 (t, a)Eb0 (t, a) − = = > 0. v(t) u(t) u(t)v(t) u(t)v(t) Case II. Suppose for the solution x > 0 that Z b a

Z b E0 (s, a)E0σ (s, a) b E0 (s, a)Eb0 (s, a) ∆s = E0 (s, a)∆α,a s < ∞. p(s)x(s)xσ (s)k0 (α, s) p(s)x(s)xσ (s) a

Then for δ2 (t) = Z b a

E0 (t, a)E0σ (t, a) , p(t)x(t)xσ (t)

δ2 (t)Eb0 (t, a)  R  ∆α,at = ∞ b b b b δ (s) E (s, a)∆ s δ (s) E (s, a)∆ s α,a α,a 0 0 t 2 σ (t) 2

R

by the Trench divergence theorem. If we let Z b

u(t) = x(t) t

δ2 (s)Eb0 (s, a)∆α,a s > 0,

then u ∈ D and we have a P´olya factorization of Mc u on [a, b)T given by b α {ϕ2 Dα [ϕ1 u]} Mc u = ϕ1 D where ϕ1 (t) :=

E0 (t, a) > 0, x(t)

ϕ2 (t) :=

on

[a, b)T ,

p(t)x(t)xσ (t) 1 = > 0. σ E0 (t, a)E0 (t, a) δ2 (t)

Thus, 

α

D [ϕ1 u]

= = (8.12)

=

Z b

 D δ2 (s)Eb0 (s, a)∆α,a s t   Z b α b D E0 (t, a) δ2 (s)E0 (s, a)∆α,a s t  Z b   Z b δ2 (s)Eb0 (s, a)∆α,a s − k1 (α,t) δ2 (s)Eb0 (s, a)∆α,a s E0σ (t, a) Dα α

E0 (t, a) x(t) x(t)

t

=

E0σ (t, a)k0 (α,t)

=

−δ2 (t)Eb0 (t, a).

This leads to

t

Z t

b

∆ b δ2 (s)E0 (s, a)∆α,a s

n o b α {ϕ2 Dα [ϕ1 u]} = ϕ1 D b α −Eb0 (t, a) = 0, Mc u = ϕ1 D

Second-Order Self-Adjoint Conformable Dynamic Equations  271

so that u is a positive solution of Mc x = 0 on [a, b)T . Let Z t

v0 (t) = u(t)

a

δ2 (s)Eb0 (s, a)  R  ∆α,a s > 0. b b b b δ (η) E (η, a)∆ η δ (η) E (η, a)∆ η α,a α,a 0 0 s 2 σ (s) 2

R

Then Dα

hv i 0

u

(t) − k1 (α,t)

v  0

u

(t) = k0 (α,t) =

=

 v ∆ 0

u

(t)

δ2 (t)E0 (t, a)E0 (a, σ (t))  R  b b b b t δ2 (η)E0 (η, a)∆α,a η σ (t) δ2 (η)E0 (η, a)∆α,a η b

R

E0 (t, a)Eb0 (t, a) , p(t)u(t)uσ (t)

which by reduction of order and (8.15) makes v0 a solution of Mc x = 0 on [a, b)T . Note that lim

t→b−

1

u(t) = lim v0 (t) t→b− R t

=0

δ2 (s) Eb0 (s,a)

 ∆α,a s R a Rb ( s δ2 (η)Eb0 (η,a)∆α,a η ) σb(s) δ2 (η)Eb0 (η,a)∆α,a η

since we are assuming the integral diverges to positive infinity in this case. Moreover, using the quotient rule from Lemma 1.9.9 we have E0 (t, a)Eb0 (t, a) p(t)u(t)uσ (t)

= Dα

hv i 0

(t) − k1 (α,t)

u W (u, v0 )(t) = . u(t)uσ (t)

v  0

u

(t) (8.23)

Using (8.12) we have hv i h v0 (t) 1 v0 i 0 Dα (t) − k1 (α,t) = σ Dα E0 (·, a) (t), u u(t) E0 (t, a) u so that h 1 v0 i W (u, v0 )(t) E0 (t, a)Eb0 (t, a) α E (·, a) (t) = D = . 0 E0σ (t, a) u u(t)uσ (t) p(t)u(t)uσ (t) Integrating both sides of this last equality from a to t via Theorem 1.9.7, we obtain v0 (t) = 0< u(t)

Z t a

E0 (s, a)Eb0 (s, a) ∆s. p(s)u(s)uσ (s)k0 (α, s)

Letting t → b− we get one of the desired results, namely Z b a

E0 (s, a)Eb0 (s, a) ∆s = ∞. p(s)u(s)uσ (s)k0 (α, s)

Suppose v is any solution of Mc x = 0 such that u and v are linearly independent. Then v(t) = c1 u(t) + c2 v0 (t),

272  Conformable Dynamic Equations on Time Scales

where c2 6= 0. It follows that u(t) u(t) lim = lim = lim − − t→b v(t) t→b c1 u(t) + c2 v0 (t) t→b− c1

u(t) v0 (t) u(t) v0 (t) + c2

= 0.

Again let v be a fixed solution of (8.1) such that u and v are linearly independent, but this time, pick t0 ∈ [a, b)T so that v(t) 6= 0 on [t0 , b)T , which is possible by the non-oscillatory nature of Mc x = 0. Then for t ∈ [t0 , b)T , similar to the calculations above in (8.23), we have Dα

hui v

(t) − k1 (α,t)

h 1 ui α D E (·, a) (t) 0 E0σ (t, a) v W (v, u)(t) = v(t)vσ (t) c2 E0 (t, a)Eb0 (t, a) , = p(t)v(t)vσ (t)

u(t) v(t)

=

where c2 6= 0. Integrating both sides of this last equality from t0 to t we obtain u(t) u(t0 ) − = c2 v(t) v(t0 )

Z t t0

E0 (s, a)Eb0 (s, a) ∆s. p(s)v(s)vσ (s)k0 (α, s)

Letting t → b− we get another one of the desired results, that is, Z b t0

E0 (s, a)Eb0 (s, a) ∆s < ∞. p(s)v(s)vσ (s)k0 (α, s)

We now show that (8.22) holds for t < b sufficiently close. Above we saw that v(t) 6= 0 on [t0 , b)T . Note that the expression p(t)Dα v(t) v(t) is the same if v(t) is replaced by −v(t). Hence without loss of generality we can assume v > 0 on [t0 , b)T . For t ∈ [t0 , b)T , consider p(t)Dα v(t) p(t)Dα u(t) p(t)W (u, v)(t) c3 E0 (t,t0 )Eb0 (t,t0 ) − = = , v(t) u(t) v(t)u(t) v(t)u(t) where c3 =

p(t)W (u, v)(t) E0 (t,t0 )Eb0 (t,t0 )

is a (nonzero) constant by Abel’s formula. It remains to show that c3 > 0. Since lim

t→b−

v(t) = ∞, u(t)

the ordinary ∆-derivative (v/u)∆ > 0 near b. Consequently, we see that  v ∆ (t) 0 < k0 (α,t) u

Second-Order Self-Adjoint Conformable Dynamic Equations  273

= Dα

hvi u

W (u, v)(t) = u(t)uσ (t)

=

v(t) u(t) c3 E0 (t,t0 )Eb0 (t,t0 )

(t) − k1 (α,t)

p(t)u(t)uσ (t)

,

and we get the desired result that c3 > 0. This completes the proof.



Remark 8.3.20 The proof of the previous theorem is non-standard and new on all time scales. Usually, one uses the P´olya factorization of Mc x to derive a Trench factorization, and then proceeds from there to prove the theorem on dominant and recessive solutions. As shown above, however, the proof is possible even if a Trench factorization cannot be found; all one needs is the P´olya factorization and what we have called the Trench divergence, proved earlier. Thus we get the same result with a weaker assumption.

8.4

RICCATI EQUATION

Now we will consider the associated conformable nabla Riccati dynamic equation Rz = 0,

bα z + q + where Rz := D

k0 zρ (z − k1 p)ρ , ρ k0 pρ + ν(z − k1 p)ρ

(8.24)

where p is continuous and q is ld-continuous on I such that p(t) > 0 for all t ∈ I . Also, assume throughout that α ∈ (0, 1]. b α z is Definition 8.4.1 Denote by DR the set of all differentiable functions z such that D ρ ρ ρ ld-continuous and k0 p + ν(z − k1 p) > 0 on I . A function z ∈ DR is a solution of (8.24) on I if and only if Rz(t) = 0 for all t ∈ I . Remark 8.4.2 If α = 1, then k0 ≡ 1, k1 ≡ 0, and (8.24) simplifies to bα z + q + Rz := D

(zρ )2 pρ + νzρ

for any time scale T, while the positivity condition is ρ

k0 pρ + ν(z − k1 p)ρ = pρ + νzρ > 0, both of which agree with the classic time scales form. If T = R, then ν ≡ 0, ρ(t) = t, and (8.24) reduces to z(z − k1 p) , Rz := Dα z + q + p while the positivity condition is ρ

k0 pρ + ν(z − k1 p)ρ = k0 p > 0, both of which agree with [2, 3].

274  Conformable Dynamic Equations on Time Scales

Example 8.4.3 Let T = R, α ∈ (0, 1], and let k0 , k1 satisfy (A1). Furthermore, let θ− , θ+ ∈ R with θ− < 0 < θ+ such that Z θ−

h1 (θ− , 0) :=

0

1 π ds = − , k0 (α, s) 2

Z θ+

h1 (θ+ , 0) :=

0

1 π ds = . k0 (α, s) 2

Then the solution of the initial value problem Rz = 0,

Rz := Dα z + 1 + z2 − k1 z,

z(0) = 0

is given by z(t) = − tan (h1 (t, 0)) ,

t ∈ I := (θ− , θ+ ) .

Here we have solved (8.24) for T = R with p ≡ q ≡ 1.

4

Theorem 8.4.4 (Factorization Theorem) Let Mc be as in (8.1), R as in (8.24), and let x ∈ DR such that x has no generalized zeros in I . If z is defined by z=

pDα x x

(8.25)

on I , then z ∈ DR and Mc x = xRz on I . Proof 8.4.5 Assume x ∈ DR such that x has no generalized zeros in I , and let z have the form (8.25). Then (8.25)

z − k1 p =

pDα x p k0 p ∆ − k1 p = (Dα x − k1 x) = x . x x x

Taking ρ of both sides and using the fact that x ∈ DR implies x∆ρ − x∇ ,   k0 p ρ ∇ ρ (z − k1 p) = x . x Multiplying by ν and using the nabla formula x − xρ = νx∇ , we have       k0 p ρ ∇ k0 p ρ k0 p ρ ρ ρ ρ ν(z − k1 p) = νx = (x − x ) = x − k0 pρ . x x x Consequently, ρ k0 pρ

 ρ

+ ν(z − k1 p) =

k0 p x

ρ x>0

on I since x has no generalized zeros in I . Thus, z ∈ DR . Note that (8.25) also implies zx = k1 x + k0 x∆ . p If we solve for x∆ and take ρ of both sides, we have   z − k1 p ρ ρ x∇ = x . k0 p

Second-Order Self-Adjoint Conformable Dynamic Equations  275

Using the nabla formula xρ = x − νx∇ , we have   z − k1 p ρ ∇ x = (x − νx∇ ); k0 p rearranging and resolving for x∇ yields x∇ =

x(z − k1 p)ρ . (k0 p)ρ + ν(z − k1 p)ρ

Then Mc x

= (8.25)

= = =

= (8.24)

=

b α [pDα x] + qx D b α [zx] + qx D b x + xD b α z − k1 zρ x + qx zρ D  α   bα z + q x zρ k1 x + k0 x∇ − k1 x + D     x(z − k1 p)ρ ρ bα z + q x k0 z D + (k0 p)ρ + ν(z − k1 p)ρ xRz

on I . This completes the proof.



Theorem 8.4.6 The self-adjoint equation Mc x = 0 for Mc in (8.1) has a positive solution on T if and only if the Riccati equation (8.24) has a solution z on Tκ . Proof 8.4.7 First assume Mc x = 0 has a solution x ∈ DR with x > 0 on T, and let z have the Riccati substitution form (8.25). By Theorem 8.4.4 we have z ∈ DR and xRz = Mc x = 0 on T, making z a solution of (8.24) on Tκ , since x positive means x has no generalized zeros. Conversely, let z be a solution of the Riccati equation (8.24) on Tκ . Then z ∈ DR and hence     z/p − k1 ρ ρ ρ ρ ρ 0 < k0 p + ν(z − k1 p) = (k0 p) 1 + ν k0 on Tκκ , and z is continuous on Tκ . In particular, (z/p) is continuous, and the conformable exponential E pz = e z/p−k1 = eb  z/p−k1 ρ = Eb  z/p−k1 ρ k0

k  0  z/p−k1 ρ 1+ν k0

k0

 1+ν

k0  z/p−k1 ρ k0

+k1

exists and is well defined. Let t0 ∈ T, and let x(t) = E pz (t,t0 ),

t ∈ T.

Then x is continuous and positive on T. Note also that Dα x = zx/p is continuous, and if we solve this equation for z we get (8.25). Furthermore, b α [pDα x] (8.25) b α [zx] = zρ D b α x + xD b α z − k1 zρ x D = D

276  Conformable Dynamic Equations on Time Scales

is ld-continuous on Tκκ , putting x ∈ DR . Moreover,   b α [pDα x] = zρ k0 x∇ + xD b α z = k0 zρ  D 

z/p−k1 k0

1+ν



ρ

z/p−k1 k0

  b α z = −qx ρ  x + xD

using (8.24), since Rz = 0. This shows that x is a solution of Mc x = 0 on T, completing the proof. 

8.5

CAUCHY FUNCTION AND VARIATION OF CONSTANTS FORMULA

In this section we discuss the Cauchy function and derive a variation of constants formula for the conformable non-homogeneous self-adjoint dynamic equation Mc x = h for Mc given in (8.1), where we assume h is a continuous function on some interval I ⊆ Tκκ . The following theorem follows from a standard argument, using the linearity of Dα in (1.1) b α in (1.9), and Theorem 8.1.2. and D Theorem 8.5.1 If x1 and x2 are linearly independent solutions of the conformable homogeneous equation Mc x = 0 on I , and y is a particular solution of the conformable nonhomogeneous equation Mc x = h on I , then x = c1 x1 + c2 x2 + y is a general solution of Mc x = h for constants c1 , c2 ∈ R. Definition 8.5.2 (Cauchy Function) Let Mc be as in (8.1). A function x : I × I → R is the Cauchy function for Mc x = 0 provided for each fixed s ∈ I , x(·, s) is the solution of the initial value problem Mc x(·, s) = 0,

x(s, s) = 0,

Dα x(ρ(s), s) =

1 . pρ (s)

It is easy to verify the following example. Example 8.5.3 If q, h ≡ 0 in (8.1), then the Cauchy function for b α [pDα x] (t) = 0 D is given by x(t, s) =

Z t b E0 (τ, ρ(s)) s

for all t, s ∈ I .

p(τ)

∆α,t τ 4

For Mc in (8.1), a formula for the Cauchy function for Mc x = 0 is given in the next theorem.

Second-Order Self-Adjoint Conformable Dynamic Equations  277

Theorem 8.5.4 If u and v are linearly independent solutions of Mc x = 0 for Mc in (8.1), then the Cauchy function x(t, s) for Mc x = 0 is given by x(t, s) =

u(s)v(t) − v(s)u(t) pρ (s)[u(s)Dα vρ (s) − v(s)Dα uρ (s)]

for t, s ∈ I .

(8.26)

Proof 8.5.5 Let y(t, s) be defined by the right-hand side of equation (8.26). Then note that for each fixed s, y(·, s) is a linear combination of the solutions u and v and as such is a solution of (8.1). Clearly y(s, s) = 0. Also note that u(s)Dα v(t) − v(s)Dα u(t) , pρ (s)[u(s)Dα vρ (s) − v(s)Dα uρ (s)]

Dα y(t, s) = from which we have Dα y(ρ(s), s) =

u(s)Dα vρ (s) − v(s)Dα uρ (s) 1 = ρ . ρ α ρ α ρ p (s)[u(s)D v (s) − v(s)D u (s)] p (s)

From the uniqueness of solutions of initial value problems (Theorem 8.1.2) we have that for each fixed s, x(t, s) = y(t, s), 

which gives us the desired result.

Theorem 8.5.6 (Variation of Constants Formula) Assume h is continuous on I and a ∈ I . Let x(t, s) be the Cauchy function for Mc x = 0 for Mc in (8.1). Then x(t) =

Z t x(t, s)h(s) a

k0 (α, s)

t ∈I

∇s,

(8.27)

is the solution of the initial value problem Mc x = h(t),

x(a) = 0,

Dα x(a) = 0.

Proof 8.5.7 In the proof we will use the fact from time scales calculus that if f , f ∆ , and f ∇ are continuous, then Z t ∆ Z t f ∆ (t, s)∇s + f (σ (t), σ (t)) f (t, s)∇s = a

a

and Z

t

∇ Z t f (t, s)∇s = f ∇ (t, s)∇s + f (ρ(t),t). a

a

Let x(t, s) be the Cauchy function for Mc x = 0, and set x(t) =

Z t x(t, s)h(s) a

k0 (α, s)

∇s.

Note that x(a) = 0. Taking the conformable derivative Dα of x, Dα x(t) = k1 x + k0 x∆

278  Conformable Dynamic Equations on Time Scales

Z = k1 (t)x(t) + k0

t

a

Z t α D [x(t, s)]h(s)

=

a

k0 (α, s)

x∆ (t, s)h(s) x(σ (t), σ (t))hσ (t) ∇s + k0 (α, s) k0 (α, σ (t))



∇s

since the Cauchy function satisfies x(σ (t), σ (t)) = 0. Note that in the integral, Dα denotes the derivative with respect to the first variable t; thus Dα x(a) = 0. Multiply by p on both sides to get Z t p(t)Dα [x(t, s)]h(s) ∇s. (8.28) (pDα x)(t) = k0 (α, s) a Then b α [pDα x] D

= (8.28)

=

k1 (pDα x) + k0 (pDα x)∇ Z t k1 (t)p(t)Dα [x(t, s)]h(s)

Z t k0 (t) (pDα [x(·, s)])∇ (t)h(s)

∇s + k0 (α, s) a ρ α +p (t)D x(ρ(t),t)h(t) Z t b α [pDα x(·, s)] (t) h(s) ∇s + h(t) D k0 (α, s) a Z t h(s) ∇s h(t) − q(t)x(t, s) k a 0 (α, s) h(t) − q(t)x(t), a

= = =

k0 (α, s)

∇s

by all of the properties of the Cauchy function. Consequently, Mc x(t) = h(t). This completes the proof. 

8.6

BOUNDARY VALUE PROBLEMS AND GREEN FUNCTIONS

In this section we are concerned with Green functions for a general two-point boundary value problem (abbreviated by BVP) for Mc x = 0, Mc given in (8.1). Also assume k0 (α,t) − µ(t)k1 (α,t) > 0,

t ∈ [a, b)T

throughout this section. Theorem 8.6.1 (Existence and Uniqueness of Solutions for General Two-Point BVPs) Assume that the homogeneous boundary value problem Mc x = 0,

ξ x(a) − β Dα x(a) = γx(b) + δ Dα x(b) = 0

(8.29)

has only the trivial solution. Then the nonhomogeneous BVP Mc x = h(t),

ξ x(a) − β Dα x(a) = A,

γx(b) + δ Dα x(b) = B,

(8.30)

where A and B are given constants and h is continuous, has a unique solution. Proof 8.6.2 The proof is similar to the classical (α = 1) case and thus is omitted.



Second-Order Self-Adjoint Conformable Dynamic Equations  279

In the next example we give a BVP of the type (8.29) which does not have just the trivial solution. In Example 8.6.4 we give a necessary and sufficient condition for some boundary value problems of the form (8.29) to have only the trivial solution. Example 8.6.3 Find all solutions of the BVP b α [pDα x](t) = 0, D

Dα x(a) = Dα x(b) = 0,

t ∈ (a, b)T ,

(8.31)

where a < b for a ∈ Tκ and b ∈ Tκ . The BVP (8.31) is equivalent to a BVP of the form (8.29) if we take q(t) ≡ 0, ξ = γ = 0, and β = δ = 1. A general solution of the boundary value problem in (8.31) is x(t) = c1 E0 (t, a) + c2

Z t b E0 (s, a) a

p(s)

∆α,t s,

t ∈ [a, b]T ,

(8.32)

and the boundary conditions lead to the equations Dα x(a) =

c2 Eb0 (a, a) = 0 and p(a)

Dα x(b) =

c2 Eb0 (b, a) = 0. p(b)

Thus c2 = 0, and x(t) = c1 E0 (t, a) solves (8.31) for any constant c1 ∈ R.

4

Example 8.6.4 Let D = ξγ

Z b b E0 (s, a) a

p(s)

∆α,b s +

β γE0 (b, a) ξ δ Eb0 (b, a) + . p(a) p(b)

Using (8.32), it is straightforward to show that the BVP (8.29) with q ≡ 0 has only the trivial solution if and only if D 6= 0. In particular, D = 0 for the BVP in Example 8.6.3 and thus (8.31) has nontrivial solutions, as seen above. 4 Theorem 8.6.5 (Green Function for General Two-Point BVPs) Assume that the BVP (8.29) has only the trivial solution. For each fixed s ∈ [a, b]T , let u(·, s) be the unique solution of the BVP Mc u(·, s) = 0,

ξ u(a, s) − β Dα u(a, s) = 0,

γu(b, s) + δ Dα u(b, s) = −γx(b, s) − δ Dα x(b, s),

(8.33)

where x(t, s) is the Cauchy function (8.26) for Mc x = 0, Mc in (8.1). Define the Green function G : [a, b]T × [a, b]T → R for the BVP (8.29) by ( u(t, s) : a ≤ t ≤ s ≤ b, G(t, s) = v(t, s) : a ≤ s ≤ t ≤ b, where v(t, s) := u(t, s) + x(t, s) for t, s ∈ [a, b]T . Then for each fixed s ∈ [a, b]T , v(·, s) is a solution of Mc x = 0 and satisfies the second boundary condition in (8.29). If h is continuous, then Z b G(t, s)h(s) x(t) := ∇s k0 (α, s) a is the solution of the nonhomogeneous BVP (8.30) with A = B = 0.

280  Conformable Dynamic Equations on Time Scales

Proof 8.6.6 The existence and uniqueness of u(t, s) is guaranteed by Theorem 8.6.1. Since for each fixed s ∈ [a, b]T , u(·, s) and x(·, s) are solutions of Mc x = 0, Mc in (8.1), we have for each fixed s ∈ [a, b]T that v(·, s) = u(·, s) + x(·, s) is also a solution of Mc x = 0. It follows from (8.33) that v(·, s) satisfies the second boundary condition in (8.29) for each fixed s ∈ [a, b]T . First note that x(t) =

Z b G(t, s)h(s)

∇s k0 (α, s) Z b Z t G(t, s)h(s) G(t, s)h(s) ∇s + ∇s k0 (α, s) t a k0 (α, s) Z t Z b v(t, s)h(s) u(t, s)h(s) ∇s + ∇s k (α, s) k0 (α, s) a t 0 Z t Z b h(s) u(t, s)h(s) [u(t, s) + x(t, s)] ∇s + ∇s k0 (α, s) k0 (α, s) a t Z t Z b x(t, s)h(s) u(t, s)h(s) ∇s + ∇s k0 (α, s) a k0 (α, s) a Z b u(t, s)h(s) ∇s + w(t), k0 (α, s) a a

= = = = =

where, by the variation of constants formula in Theorem 8.5.6, w is the solution of the IVP Mc w = h(t),

w(a) = Dα w(a) = 0.

(8.34)

It follows that Z b

Mc x(t) =

a

Mc u(t, s)

h(s) ∇s + Mc w(t) = Mc w(t) = h(t). k0 (α, s)

Hence x is a solution of the nonhomogeneous equation Mc x = h(t). It remains to show that x satisfies the two boundary conditions in (8.29). Now ξ x(a) − β Dα x(a) =

Z b

[ξ u(a, s) − β Dα u(a, s)]

a

h(s) ∇s = 0, k0 (α, s)

since for each fixed s, u(·, s) satisfies the first boundary condition in (8.29) and w satisfies (8.34). Hence x satisfies the first boundary condition in (8.29). From earlier in this proof, t h(s) h(s) u(t, s) ∇s + x(t, s) ∇s k0 (α, s) k0 (α, s) a a Z b Z b h(s) h(s) v(t, s) ∇s − x(t, s) ∇s k0 (α, s) k0 (α, s) a t Z b Z t h(s) h(s) ∇s + x(t, s) ∇s v(t, s) k0 (α, s) k0 (α, s) a b Z b h(s) v(t, s) ∇s + y(t), k a 0 (α, s)

Z b

x(t) = = = =

Z

Second-Order Self-Adjoint Conformable Dynamic Equations  281

where, by the variation of constants formula in Theorem 8.5.6, y solves Mc y = h(t),

y(b) = Dα y(b) = 0.

(8.35)

Then Z b

α

γx(b) + δ D x(b) =

[γv(b, s) + δ Dα v(b, s)]

a

h(s) ∇s = 0, k0 (α, s)

since for each fixed s, v(·, s) satisfies the second boundary condition in (8.29) and y satisfies (8.35). Hence x satisfies the second boundary condition in (8.29). The uniqueness of solutions of the nonhomogeneous BVP (8.30) (with A = B = 0) follows from Theorem 8.6.1. This completes the proof.  Instead of the Cauchy function approach as above, in the next theorem we find another form of the Green function for the BVP (8.29); of particular interest is a symmetry condition on the square [a, b]T × [a, b]T satisfied by the Green function. Theorem 8.6.7 Assume that the BVP (8.29) has only the trivial solution. Let φ be the solution of the IVP Mc φ = 0, φ (a) = β , Dα φ (a) = ξ , and let ψ be the solution of Mc ψ = 0,

ψ(b) = δ ,

Dα ψ(b) = −γ.

Then the Green function for the BVP (8.29) is given by ( φ (t)ψ(s) : a ≤ t ≤ s ≤ b, 1 G(t, s) = ρ p (s) [φ (s)Dα ψ ρ (s) − ψ(s)Dα φ ρ (s)] ψ(t)φ (s) : a ≤ s ≤ t ≤ b.

(8.36)

Proof 8.6.8 Let φ and ψ be as stated in the theorem. We use Theorem 8.6.5 to prove that G defined by (8.36) is the Green function for the BVP (8.29). Note that ξ φ (a) − β Dα φ (a) = ξ β − β ξ = 0 and γψ(b) + δ Dα ψ(b) = γδ − δ γ = 0. Hence φ and ψ satisfy the first and second boundary conditions in (8.29), respectively. Let u(t, s) :=

φ (t)ψ(s) pρ (s) [φ (s)Dα ψ ρ (s) − ψ(s)Dα φ ρ (s)]

and v(t, s) :=

ψ(t)φ (s) pρ (s) [φ (s)Dα ψ ρ (s) − ψ(s)Dα φ ρ (s)]

for t ∈ [a, b]T , s ∈ [a, b]T . Note that for each fixed s ∈ [a, b]T , u(·, s) and v(·, s) are solutions of Mc x = 0, Mc in (8.1), on [a, b]T . Also for each fixed s ∈ [a, b]T , ξ u(a, s) − β Dα u(a, s) =

ψ(s)[ξ φ (a) − β Dα φ (a)] =0 pρ (s) [φ (s)Dα ψ ρ (s) − ψ(s)Dα φ ρ (s)]

282  Conformable Dynamic Equations on Time Scales

and γv(b, s) + δ Dα v(b, s) =

φ (s)[γψ(b) + δ Dα ψ(b)] = 0. pρ (s) [φ (s)Dα ψ ρ (s) − ψ(s)Dα φ ρ (s)]

Hence for each fixed s ∈ [a, b]T , u(·, s) and v(·, s) satisfy the first and second boundary conditions in (8.29), respectively. Let χ(t, s) := v(t, s) − u(t, s) =

ψ(t)φ (s) − φ (t)ψ(s) pρ (s) [φ (s)Dα ψ ρ (s) − ψ(s)Dα φ ρ (s)]

.

It follows that for each fixed s, χ(·, s) is a solution of Mc x = 0, χ(s, s) = 0, and Dα χ(ρ(s), s) =

φ (s)Dα ψ ρ (s) − ψ(s)Dα φ ρ (s) 1 = ρ . ρ α ρ α ρ p (s) [φ (s)D ψ (s) − ψ(s)D φ (s)] p (s)

Consequently, χ(t, s) = x(t, s) is the Cauchy function for Mc x = 0, and we have v(t, s) = u(t, s) + x(t, s). It remains to prove that for each fixed s, u(·, s) satisfies (8.33). To see this, consider γu(b, s) + δ Dα u(b, s) = γv(b, s) + δ Dα v(b, s) − [γx(b, s) + δ Dα x(b, s)] = −γx(b, s) − δ Dα x(b, s). Hence by Theorem 8.6.5, G(t, s) defined by (8.36) is the Green function for (8.29). This completes the proof.  8.6.1

Conjugate Problem and Disconjugacy

In this subsection we examine Theorem 8.6.5 and Theorem 8.6.7 in more detail, in particular for the special case where the boundary conditions (8.33) are conjugate (also known as Dirichlet) boundary conditions. Corollary 8.6.9 (Green Function for the Conjugate Problem) Assume the BVP Mc x = 0,

x(a) = x(b) = 0,

(8.37)

has only the trivial solution. Let x(t, s) be the Cauchy function for Mc x = 0, Mc in (8.1). For each fixed s ∈ I , let u(·, s) be the unique solution of the BVP Mc u(·, s) = 0, Then

u(a, s) = 0,

u(b, s) = −x(b, s).

( u(t, s) : t ≤ s, G(t, s) = v(t, s) : t ≥ s,

where v(t, s) = u(t, s) + x(t, s), is the Green function for the BVP (8.37). Moreover, for each fixed s ∈ [a, b]T , v(·, s) is a solution of Mc x = 0 and v(b, s) = 0. Proof 8.6.10 This corollary follows from Theorem 8.6.5 with ξ = γ = 1 and β = δ = 0. 

Second-Order Self-Adjoint Conformable Dynamic Equations  283

Corollary 8.6.11 The Green function for the BVP (8.37) with q ≡ 0 is given by Z Z b b t E b0 (τ, a) E0 (τ, ρ(s))   ∆α,t τ ∆α,b τ : a ≤ t ≤ s ≤ b,  −1 p(τ) p(τ) s a G(t, s) = R b Z s Z b b b E0 (τ,a)  E0 (τ, ρ(s))  Eb0 (τ, a) ∆ τ a p(τ) ∆α,b τ  ∆α,b τ : a ≤ s ≤ t ≤ b. α,t p(τ) p(τ) a t Proof 8.6.12 It is easy to check that the BVP b α [pDα x](t) = 0, D

x(a) = x(b) = 0,

has only the trivial solution from a modification of Example 8.6.3. By Example 8.5.3, the Cauchy function for b α [pDα x](t) = 0 D (8.38) is given by x(t, s) =

Z t b E0 (τ, ρ(s))

p(τ)

s

∆α,t τ.

By Corollary 8.6.9, u(·, s) from the statement of Corollary 8.6.9 solves (8.38) for each fixed s ∈ [a, b]T and satisfies u(a, s) = 0 and

u(b, s) = −x(b, s) = −

Z b b E0 (τ, ρ(s))

p(τ)

s

∆α,b τ.

Since x1 (t) = E0 (t, a) and

x2 (t) =

Z t b E0 (τ, a)

p(τ)

a

∆α,t τ

are solutions of (8.38), u(t, s) = ξ (s)E0 (t, a) + β (s)

Z t b E0 (τ, a)

p(τ)

a

∆α,t τ.

Using the boundary conditions (8.39), it can be shown that Z t b E0 (τ, a)

−1

u(t, s) = R b b E0 (τ,a) a

p(τ)

∆α,b τ

a

p(τ)

∆α,t τ

Z b b E0 (τ, ρ(s)) s

p(τ)

∆α,b τ.

Hence G(t, s) has the desired form for t ≤ s. By Corollary 8.6.9 for t ≥ s, G(t, s) = x(t, s) + u(t, s). Therefore for t ≥ s,

G(t, s) =

Z t b E0 (τ, ρ(s)) s

p(τ)

R t Eb0 (τ,a)

∆α,t τ −

a

p(τ)

∆α,t τ

R b Eb0 (τ,ρ(s)) s

R b Eb0 (τ,a) a

p(τ)

p(τ)

∆α,b τ

∆α,b τ

(8.39)

284  Conformable Dynamic Equations on Time Scales R t Eb0 (τ,ρ(s))

=

s

p(τ)

R b Eb0 (τ,a)

∆α,t τ

a

∆α,b τ −

p(τ)

R t Eb0 (τ,ρ(s)) s

p(τ)

R b Eb0 (τ,a)

∆α,t τ

a

s

p(τ)

R s Eb0 (τ,a)

∆α,t τ

a

p(τ)

a

R t Eb0 (τ,a) s

p(τ)

∆α,b τ

R s Eb0 (τ,ρ(s)) a

p(τ)

s

p(τ)

∆α,b τ

∆α,t τ

R b Eb0 (τ,a) s

p(τ)

∆α,b τ

∆α,b τ a

p(τ)

∆α,t τ −

p(τ)

R s Eb0 (τ,ρ(s))

R b Eb0 (τ,a)

R b Eb0 (τ,ρ(s))

∆α,b τ a

p(τ)

∆α,b τ −

p(τ)

∆α,t τ

R t Eb0 (τ,ρ(s))

R b Eb0 (τ,a) R t Eb0 (τ,ρ(s))

=

p(τ)

∆α,b τ −

p(τ)

a

=

a

R b Eb0 (τ,a) a

=

R t Eb0 (τ,a)

p(τ)

∆α,t τ

R b Eb0 (τ,a) s

p(τ)

∆α,b τ

∆α,b τ

R s Eb0 (τ,ρ(s)) a

p(τ)

∆α,t τ

R b Eb0 (τ,a) s

p(τ)

∆α,b τ

R b Eb0 (τ,a)

a p(τ) ∆α,b τ  R b  (τ,a) Z s b − tb E0p(τ) ∆α,b τ E0 (τ, ρ(s))  ∆α,t τ  R b = b E0 (τ,a) p(τ) a ∆ τ α,b a p(τ)

R s Eb0 (τ,a)

= −

a

p(τ)

∆α,t τ

R b Eb0 (τ,ρ(s)) t

R b Eb0 (τ,a) a

p(τ)

p(τ)

∆α,b τ

∆α,b τ 

which is the desired result. The following corollary follows immediately from Corollary 8.6.11 by taking   τ −1 α p (τ) = Eb0 (a, τ)D . E0 (b, τ)

(8.40)

Corollary 8.6.13 Let a, b ∈ T with a < b, and let p be given via (8.40). Then the Green function for the conjugate BVP b α [pDα x] (t) = 0, D

x(a) = x(b) = 0

is given by −Eb0 (a, ρ(s)) G(t, s) = (b − a)E0 (b,t)

( (t − a)(b − s) : a ≤ t ≤ s ≤ b, (s − a)(b − t) : a ≤ s ≤ t ≤ b.

Proof 8.6.14 By Corollary 8.6.11, the Green function for the BVP (8.37) with q ≡ 0 is given by Z Z b b t E b0 (τ, a) E0 (τ, ρ(s))   ∆α,t τ ∆α,b τ : a ≤ t ≤ s ≤ b,  −1 p(τ) p(τ) a s G(t, s) = R b Z s Z b b b E0 (τ,a)  E0 (τ, ρ(s))  Eb0 (τ, a) ∆ τ a p(τ) ∆α,b τ  ∆α,b τ : a ≤ s ≤ t ≤ b, α,t p(τ) p(τ) a t which can be rewritten as Z Z b b t E b0 (τ, ρ(s)) E0 (τ, a)   ∆α,t τ ∆α,b τ  −1 p(τ) p(τ) a s G(t, s) = R b Z s b Z b b b E0 (τ,a)  E0 (τ, a)  E0 (τ, ρ(s)) ∆ τ a p(τ) ∆α,b τ  ∆α,b τ α,t p(τ) p(τ) a t

: a ≤ t ≤ s ≤ b, :a≤s≤t ≤b

Second-Order Self-Adjoint Conformable Dynamic Equations  285

by the semi-group property of the exponential function. Using (8.40) and the fundamental theorem of calculus, we see that   Z b b Z b E0 (τ, a) τ α b b ∆α,b τ = E0 (τ, a)E0 (a, τ)D ∆α,b τ p(τ) E0 (b, τ) a a b a = − E0 (b, a) E0 (b, b) E0 (b, a) = b − a; similarly Z b b E0 (τ, a) s

p(τ)

∆α,b τ = b − s

and Z b b E0 (τ, a) t

p(τ)

∆α,b τ = b − t.

Also, Z t b E0 (τ, ρ(s)) a

p(τ)

Z t

∆α,t τ = = = =

  τ α b b E0 (τ, ρ(s))E0 (a, τ)D ∆α,t τ E0 (b, τ) a   Z t τ Eb0 (a, ρ(s)) Dα ∆α,t τ E0 (b, τ) a   t a b E0 (a, ρ(s)) − E0 (t, a) E0 (b,t) E0 (b, a) Eb0 (a, ρ(s)) (t − a), E0 (b,t)

and Z s b E0 (τ, ρ(s)) a

p(τ)

Z s



 τ ∆α,t τ D E0 (b, τ) a   Z s τ = Eb0 (a, ρ(s))E0 (t, s) Dα ∆α,s τ E0 (b, τ) a   s a b = E0 (a, ρ(s))E0 (t, s) − E0 (s, a) E0 (b, s) E0 (b, a) Eb0 (a, ρ(s)) = (s − a). E0 (b,t)

∆α,t τ = Eb0 (a, ρ(s))

α



This concludes the proof.

The following example follows immediately from Corollary 8.6.11 by taking T = R and p(τ) ≡ 1. Example 8.6.15 Let T = R and dα τ :=

dτ . Then the Green function for the BVP k0 (α, τ)

Dα Dα x = 0,

x(a) = x(b) = 0

286  Conformable Dynamic Equations on Time Scales

is given by Z t Z b  −E0 (t, s)  a 1dα τ s 1dα τ Z Z b G(t, s) = R b  s a 1dα τ  1dα τ 1dα τ a

: a ≤ t ≤ s ≤ b, : a ≤ s ≤ t ≤ b.

t

For example, if k1 (t) = (1−α)(ωt)α and k0 (t) = α(ωt)1−α for α ∈ (0, 1] and ω,t ∈ (0, ∞), then   Z t t α − aα − 1−α2 ω 2α−1 (t 2α −s2α ) 2α 1dα τ = 2 1−α and E0 (t, s) = e α ω a for a, s ∈ (0, ∞), and   − 1−α2 ω 2α−1 (t 2α −s2α )

−e 2α G(t, s) = α 2 ω 1−α (bα − aα )

( (t α − aα ) (bα − sα ) : a ≤ t ≤ s ≤ b, (sα − aα ) (bα − t α ) : a ≤ s ≤ t ≤ b.

The proofs of the following two theorems are similar to their classical (α = 1) counterparts and thus are omitted. Theorem 8.6.16 Let Mc be as in (8.1). If p > 0 and Mc x = 0 is disconjugate on [a, b]T , then the Green function for the conjugate BVP (8.37) exists and satisfies G(t, s) < 0 for t, s ∈ (a, b)T . Theorem 8.6.17 (Comparison Theorem for Conjugate BVPs) Let Mc be as in (8.1). Let p > 0 and assume that Mc x = 0 is disconjugate on [a, b]T . If u, v ∈ D satisfy Mc u(t) ≤ Mc v(t) for all t ∈ [a, b]T ,

u(a) ≥ v(a),

u(b) ≥ v(b),

then u(t) ≥ v(t) for all t ∈ [a, b]T . 8.6.2

Right Focal Problem

Similar to the subsection above on the conjugate boundary conditions and disconjugacy, here we examine Theorem 8.6.5 and Theorem 8.6.7 for the special case where the boundary conditions (8.33) are right focal boundary conditions, namely the boundary value problem Mc x = 0,

x(a) = Dα x(b) = 0.

(8.41)

Corollary 8.6.18 (Green Function for Focal BVPs) Assume that the BVP (8.41) has only the trivial solution. For each fixed s ∈ [a, b]T , let u(·, s) be the solution of the BVP Mc u(·, s) = 0,

u(a, s) = 0,

Dα u(b, s) = −Dα x(b, s),

where x(t, s) is the Cauchy function of Mc x = 0. Then ( u(t, s) :a≤t ≤s≤b G(t, s) = u(t, s) + x(t, s) : a ≤ s ≤ t ≤ b is the Green function for the right focal BVP (8.41).

Second-Order Self-Adjoint Conformable Dynamic Equations  287

Proof 8.6.19 This follows from Theorem 8.6.5 with ξ = δ = 1 and β = γ = 0.



Corollary 8.6.20 The Green function for the focal BVP (8.41) with q ≡ 0 is given by  Z t E b0 (τ, ρ(s))   ∆α,t τ : a ≤ t ≤ s ≤ b, − p(τ) a G(t, s) = Z s b  E0 (τ, ρ(s))  − ∆α,t τ : a ≤ s ≤ t ≤ b. p(τ) a Proof 8.6.21 It is easy to see that (8.41) with q ≡ 0 has only the trivial solution. Hence we can apply Corollary 8.6.18 to find the focal Green function G(t, s). For t ≤ s, G(t, s) = u(t, s), where for each fixed s, u(·, s) solves the BVP Mc u(·, s) = 0,

u(a, s) = 0,

Dα u(b, s) = −Dα x(b, s),

and where x(t, s) is the Cauchy function for (8.38). Since Mc u(·, s) = 0 and u(a, s) = 0, we have that u(·, s) must have the form u(t, s) = A(s)

Z t b E0 (τ, a) a

p(τ)

∆α,t τ

for some function A. The boundary condition at b is Dα u(b, s) = −Dα x(b, s) implies that −Eb0 (b, ρ(s)) A(s) b E0 (b, a) = , p(b) p(b) which yields A(s) = −Eb0 (a, ρ(s)). Overall, we get that u(t, s) = −

Z t b E0 (τ, ρ(s)) a

p(τ)

∆α,t τ,

which is the desired expression for G(t, s) if t ≤ s. If t ≥ s, then G(t, s) = u(t, s) + x(t, s) Z t b Z t b E0 (τ, ρ(s)) E0 (τ, ρ(s)) = − ∆α,t τ + ∆α,t τ p(τ) p(τ) s a Z s b E0 (τ, ρ(s)) = − ∆α,t τ. p(τ) a This completes the proof.



288  Conformable Dynamic Equations on Time Scales

dτ . Then the Green function for the focal k0 (α, τ) BVP (8.41) with p(t) ≡ 1 and q ≡ 0 is given by Z t   1dα τ : a ≤ t ≤ s ≤ b, G(t, s) = −E0 (t, s) Zas   1dα τ : a ≤ s ≤ t ≤ b. Example 8.6.22 Let T = R and dα τ =

a

For example, if k1 (t) = (1−α)(ωt)α and k0 (t) = α(ωt)1−α for α ∈ (0, 1] and ω,t ∈ (0, ∞), then   Z t t α − aα − 1−α2 ω 2α−1 (t 2α −s2α ) 1dα τ = 2 1−α and e0 (t, s) = e 2α α ω a for a, s ∈ (0, ∞), and   − 1−α2 ω 2α−1 (t 2α −s2α )

G(t, s) =

−e

2α

α 2 ω 1−α

( t α − aα sα − aα

: t ≤ s, : t ≥ s.

Corollary 8.6.23 Let a, b ∈ T with a < b, and let p be given via (8.40). Then the Green function for the BVP b α [pDα x] (t) = 0, D

x(a) = Dα x(b) = 0

is given by ( −Eb0 (a, ρ(s)) t − a : a ≤ t ≤ s ≤ b, G(t, s) = E0 (b,t) s − a : a ≤ s ≤ t ≤ b. 8.6.3

Periodic Problem

Finally we consider periodic boundary conditions. In traditional (α = 1 and T = R) calculus, the geometry of periodicity means returning to the same values in the sense that they lie along the same horizontal straight line (geodesic with slope zero). Considering the context of the derivative (8.1), we investigate the periodic BVP Mc x = 0,

x(a) = E0 (a, b)x(b),

Dα x(a) = E0 (a, b)Dα x(b).

(8.42)

Theorem 8.6.24 Let Mc be as in (8.1). Assume that the homogeneous periodic BVP (8.42) has only the trivial solution. Then for t ∈ (a, b)T , the nonhomogeneous BVP Mc x = h(t),

x(a) − E0 (a, b)x(b) = A,

Dα x(a) − E0 (a, b)Dα x(b) = B,

where A and B are given constants and h is continuous, has a unique solution. Proof 8.6.25 Let x1 and x2 be linearly independent solutions of Mc x = 0. Then x(t) = c1 x1 (t) + c2 x2 (t)

(8.43)

Second-Order Self-Adjoint Conformable Dynamic Equations  289

is a general solution of Mc x = 0. Note that x satisfies the boundary conditions in (8.42) if and only if c1 and c2 are constants satisfying   c1 M =0 c2 with

 M=

x1 (a) − E0 (a, b)x1 (b) x2 (a) − E0 (a, b)x2 (b) Dα x1 (a) − E0 (a, b)Dα x1 (b) Dα x2 (a) − E0 (a, b)Dα x2 (b)

 .

Since we are assuming that (8.42) has only the trivial solution, it follows that c1 = c2 = 0 is the unique solution of the above linear system. Hence det M 6= 0.

(8.44)

Now we show that (8.43) has a unique solution. Let u0 be a fixed solution of Mc u = h(t). Then a general solution of Mc u = h(t) is given by u(t) = a1 x1 (t) + a2 x2 (t) + u0 (t). It follows that u satisfies the boundary conditions in (8.43) if and only if a1 and a2 are constants satisfying the system of equations     a1 A − u0 (a) + E0 (a, b)u0 (b) M = . a2 B − Dα u0 (a) + E0 (a, b)Dα u0 (b) This system has a unique solution because of (8.44), and hence (8.43) has a unique solution. This completes the proof.  Theorem 8.6.26 (Green Function for Periodic BVPs) Assume that the homogeneous BVP (8.42) has only the trivial solution. For each fixed s ∈ I , let u(·, s) be the solution of the BVP   Mc u(·, s) = 0 u(a, s) = E0 (a, b) [u(b, s) + x(b, s)] (8.45)  α α α D u(a, s) = E0 (a, b) [D u(b, s) + D x(b, s)] , where x(t, s) is the Cauchy function for Mc x = 0, Mc as in (8.1). Define ( u(t, s) : t ≤ s, G(t, s) := v(t, s) : t ≥ s,

(8.46)

where v(t, s) := u(t, s) + x(t, s). If h is continuous, then Z b

x(t) :=

G(t, s) a

h(s) ∇s k0 (α, s)

is the unique solution of the nonhomogeneous periodic BVP (8.43) with A = B = 0. Furthermore, for each fixed s ∈ [a, b]T , v(·, s) is a solution of Mc x = 0, and u(a, s) = E0 (a, b)v(b, s),

Dα u(a, s) = E0 (a, b)Dα v(b, s).

290  Conformable Dynamic Equations on Time Scales

Proof 8.6.27 The existence and uniqueness of u(t, s) is guaranteed by Theorem 8.6.24. Since v(t, s) = u(t, s) + x(t, s), we have for each fixed s that v(·, s) is a solution of Mc x = 0. Using the boundary conditions in (8.45), it is easy to see that for each fixed s, u(a, s) = E0 (a, b)v(b, s) and Dα u(a, s) = E0 (a, b)Dα v(b, s). Let G be as in (8.46) and notice that Z b

h(s) ∇s k0 (α, s) a Z b Z t h(s) h(s) = u(t, s) ∇s + x(t, s) ∇s k (α, s) k a a 0 0 (α, s) Z b h(s) ∇s + z(t), = u(t, s) k0 (α, s) a

x(t) =

G(t, s)

where, by the variation of constants formula in Theorem 8.5.6, z solves Mc z = h(t), Hence

Z b

Mc x(t) =

a

Mc u(t, s)

z(a) = Dα z(a) = 0.

h(s) ∇s + Mc z(t) = Mc z(t) = h(t). k0 (α, s)

Thus x is a solution of Mc x = h(t). Note that Z b

b h(s) h(s) ∇s + z(a) = E0 (a, b)v(b, s) ∇s k0 (α, s) k0 (α, s) a a Z b h(s) ∇s = E0 (a, b)x(b), = E0 (a, b)G(b, s) k0 (α, s) a

x(a) =

Z

u(a, s)

and Z b

h(s) ∇s + Dα z(a) k (α, s) a 0 Z b h(s) = E0 (a, b)Dα v(b, s) ∇s k0 (α, s) a Z b h(s) = E0 (a, b)Dα G(b, s) ∇s k a 0 (α, s) = E0 (a, b)Dα x(b).

Dα x(a) =

Dα u(a, s)

Hence, x satisfies the periodic boundary conditions in (8.42). This completes the proof. 

CHAPTER

9

The Conformable Laplace Transform

Suppose that T is a time scale with forward jump operator and delta differentiation operator σ and ∆, respectively. Let α ∈ (0, 1], k0 and k1 satisfy (A1), k0 ∈ Crd1 (T) and |E0 (∞, 0)| < ∞.

9.1

DEFINITION AND PROPERTIES

Let h ∈ Crd1 (T) and g ∈ Rc be such that zhσ Egσ (·, 0) = −gEg (·, 0) Dα h − zhhσ + (z − k1 )hσ − k1 h = 0

(9.1)

h(0) = 1 for z ∈ Hc (h), where Hc (h) consists of all complex numbers z ∈ Rc for which z − k1 ∈ Rc and k0 + hσ z(µ − k1 ) 6= 0. Remark 9.1.1 Note that there exists a unique h ∈ Crd1 (T) that satisfies the second equation and the third equation of the system (9.1). Hence, there exists a unique g ∈ Rc that satisfies the first equation of (9.1). Remark 9.1.2 If α = 1, then h = 1 and g = z. Remark 9.1.3 Note that from the first equation (9.1), we get   g − k1 = −g. zhσ 1 + µ k0 Definition 9.1.4 Assume that f : T → C is regulated. Then the conformable Laplace transform of f is defined by Lc ( f )(z) =

Z ∞ 0

f (t)hσ (t)Egσ (t, 0)∆α,∞t 291

292  Conformable Dynamic Equations on Time Scales

for z ∈ Dc ( f ), where Dc ( f ) consists of all complex numbers z ∈ Hc (h) for which the improper integral exists. Theorem 9.1.5 Let f , g : T → C be regulated functions, a, b ∈ C. Then Lc (a f + bg)(z) = aLc ( f )(z) + bLc (g)(z), z ∈ Dc ( f )

\

Dc (g).

Proof 9.1.6 We have Lc (a f + bg)(z) =

Z ∞

(a f (t) + bg(t)) hσ (t)Egσ (t, 0)∆α,∞t

0

Z ∞

a f (t)hσ (t)Egσ (t, 0)∆α,∞t

= 0

Z ∞

+ 0

Z ∞

bg(t)hσ (t)Egσ (t, 0)∆α,∞t f (t)hσ (t)Egσ (t, 0)∆α,∞t

= a 0

Z ∞

+b 0

g(t)hσ (t)Egσ (t, 0)∆α,∞t

= aLc ( f )(z) + bLc (g)(z),

z ∈ Dc ( f )

Dc (g). 

This completes the proof. Example 9.1.7 We will compute Lc (1)(z),

z ∈ Dc (1).

We have Lc (1)(z) =

Z ∞ 0

hσ (t)Egσ (t, 0)∆α,∞t

1 z

Z ∞

1 = − z

Z ∞

= −

0

0

gEg (t, 0)∆α,∞t Dα Eg (t, 0)∆α,∞t

1 = − (Eg (∞, 0) − Eg (0, 0)E0 (∞, 0)) z =

1 E0 (∞, 0) z

for all z ∈ Dc (1) for which Eg (∞, 0) = 0. This ends the example.

\

The Conformable Laplace Transform  293

Theorem 9.1.8 Let n ∈ N, f : T → C be such that (Dα )k f , k ∈ {0, 1, . . . , n}, are regulated. Then
Lc (Dα )n f (z) = zn Lc ( f )(z)  −E0 (∞, 0) f (0)zn−1 + Dα f (0)zn−2 + · · ·

(9.2)

 + (Dα )n−1 f (0) for any z ∈ Dc ( f )

\

Dc (Dα ( f ))

\

...

\

Dc (Dα )n f




(9.3)

for which   lim (Dα )k f (t)h(t)Eg (t, 0) = 0,

k ∈ {0, . . . , n − 1}.

t→∞

(9.4)

Proof 9.1.9 We will use the principle of the mathematical induction. 1. Let n = 1. Using integration by parts and the definition for h and g, we get Lc (D f ) (z) = α

Z ∞ 0

Dα f (t)hσ (t)Egσ (t, 0)∆α,∞t

= lim ( f (t)h(t)Eg (t, 0)) − f (0)h(0)Eg (0, 0)E0 (∞, 0) t→∞

−

Z ∞ 0


f (t)Dα (hEg (·, 0)) (t) − k1 (α,t) f (t)hσ (t)Egσ (t, 0) ∆α,∞t

= − f (0)E0 (∞, 0)

−

Z ∞ 0

f (t) Dα h(t)Egσ (t, 0) + h(t)Dα Eg (t, 0) !

−k1 (α,t)h(t)Egσ (t, 0) ! −k1 (α,t) f (t)h

σ

(t)Egσ (t, 0)

∆α,∞t

= − f (0)E0 (∞, 0)

−

Z ∞ 0

f (t)Dα h(t)Egσ (t, 0) + f (t)h(t)g(t)Eg (t, 0)

−k1 (α,t) f (t)h(t)Egσ (t, 0)

294  Conformable Dynamic Equations on Time Scales

! −k1 (α,t) f (t)hσ (t)Egσ (t, 0) ∆α,∞t = − f (0)E0 (∞, 0)

−

Z ∞ 0

f (t) Dα h(t) − zh(t)hσ (t) − k1 (α,t)h(t) !

−k1 (α,t)hσ (t) Egσ (t, 0)∆α,∞t = − f (0)E0 (∞, 0) + z

Z ∞ 0

f (t)hσ (t)Egσ (t, 0)∆α,∞t

= − f (0)E0 (∞, 0) + zLc ( f )(z) for any z ∈ Dc ( f ) for which lim ( f (t)h(t)Eg (t, 0)) = 0.

t→∞

2. Assume that (9.2) holds for any z that satisfies (9.3) and (9.4). 3. We will prove that   Lc (Dα )n+1 f (z) = zn+1 Lc ( f )(z)  −E0 (∞, 0) f (0)zn + Dα f (0)zn−1 + · · ·  + (D ) f (0) α n

for any z ∈ Dc ( f )

\

Dc (Dα ( f ))

\

...

\

  Dc (Dα )n+1 f

(9.5)

k ∈ {0, . . . , n}.

(9.6)

for which   lim (Dα )k f (t)h(t)Eg (t, 0) = 0,

t→∞

Really, using (9.2), we get  

Lc (Dα )n+1 f (z) = Lc Dα (Dα )n f
= zLc (Dα )n f (z) − (Dα )n f (0)E0 (∞, 0) = zn+1 Lc ( f )(z)

The Conformable Laplace Transform  295

 −E0 (∞, 0) f (0)zn + Dα f (0)zn−1 α n−1

+ · · · + (D )

 f (0)z

− (Dα )n f (0)E0 (∞, 0) = zn+1 Lc ( f )(z)  −E0 (∞, 0) f (0)zn + Dα f (0)zn−1 α n−1

+ · · · + (D )

 f (0)z + (D ) f (0) α n

for any z which satisfies (9.5) and (9.6). This completes the proof.



Theorem 9.1.10 Assume that f : T → C is regulated. If Z t

F(t) = 0

then

for those z ∈ Dc (F)

t ∈ T,

f (s)∆α,t s,

1 Lc (F)(z) = Lc ( f )(z) z \

Dc ( f ) for which lim (h(t)F(t)Eg (t, 0)) = 0.

t→∞

(9.7)

Proof 9.1.11 Note that Dα F(t) = f (t),

t ∈ T.

Hence, by Theorem 9.1.8, we get Lc ( f )(z) = Lc (Dα F(t)) = −F(0)E0 (∞, 0) + zLc (F)(z) = zLc (F)(z) for those z ∈ Dc (F)

\

Dc ( f ) for which (9.7) holds. From here, 1 Lc (F)(z) = Lc ( f )(z) z

for those z ∈ Dc (F)

\

Dc ( f ) for which (9.7) holds. This completes the proof.



296  Conformable Dynamic Equations on Time Scales

Theorem 9.1.12 We have Lc (hn (·, 0)) (z) =

E0 (∞, 0) , zn+1

n ∈ N0 ,

(9.8)

for those z ∈ Dc (hn (·, 0)) for which lim (hn (t, 0)h(t)Eg (t, 0)) = 0.

t→∞

(9.9)

Proof 9.1.13 We will use the principle of the mathematical induction. 1. Let n = 0. Then t ∈ T.

h0 (t, 0) = E0 (t, 0), Hence, and considering Theorem 9.1.8, we get 0 = Lc (Dα E0 (·, 0)) (z)

= zLc (E0 (·, 0))(z) − E0 (0, 0)E0 (∞, 0) = zLc (E0 (·, 0))(z) − E0 (∞, 0) for those z ∈ Dc (E0 (·, 0)) for which lim (h(t)h0 (t, 0)Eg (t, 0)) = 0.

t→∞

(9.10)

Hence, E0 (∞, 0) z for those z ∈ Dc (h0 (·, 0)) for which (9.10) holds. Lc (h0 (·, 0))(z) =

2. Assume (9.8) for some n ∈ N and for those z ∈ Dc (hn (·, 0)) for which (9.9) holds. 3. We will prove that Lc (hn+1 (·, 0)) (z) =

E0 (∞, 0) zn+2

(9.11)

for those z ∈ Dc (hn+1 (·, 0)) for which lim (hn+1 (t, 0)h(t)Eg (t, 0)) = 0.

t→∞

Really, using that Dα hn+1 (t, 0) = hn (t, 0),

t ∈ T,

and using Theorem 9.1.8, we get E0 (∞, 0) zn+1

= Lc (hn (·, 0)) (z) = Lc (Dα hn+1 (·, 0)) (z)

(9.12)

The Conformable Laplace Transform  297

= zLc (hn+1 (·, 0)) (z) −hn+1 (0, 0)E0 (∞, 0) = zLc (hn+1 (·, 0)) (z) for those z ∈ Dc (hn+1 (·, 0)) for which (9.12) holds. Hence, we get (9.11) for those z ∈ Dc (hn+1 (·, 0)) for which (9.12) holds. This completes the proof.  Example 9.1.14 Let f ∈ Rc be a constant. We will compute
Lc E f (·, 0) (z). Note that Dα E f (t, 0) = f E f (t, 0),

t ∈ Tκ .

Then, using Theorem 9.1.8, we get

f Lc E f (·, 0) (z) = Lc f E f (·, 0)
= Lc Dα E f (·, 0) (z)
= zLc E f (·, 0) (z) −E f (0, 0)E0 (∞, 0) for those z ∈ Dc (E f (·, 0)) for which
lim E f (t, 0)h(t)Eg (t, 0) = 0.

t→∞

Hence,
(z − f )Lc E f (·, 0) (z) = E0 (∞, 0) or Lc (E f (·, 0)) =

E0 (∞, 0) z− f

for those z ∈ Dc (E f (·, 0)) for which
lim E f (t, 0)h(t)Eg (t, 0) = 0.

t→∞

Example 9.1.15 Let f ∈ Rc be a constant. We will compute
Lc Cosh f (·, 0) (z).

298  Conformable Dynamic Equations on Time Scales

We have Lc (Coshc (·, 0)) (z) = Lc

for those z ∈ Dc (E f (·, 0))



 E f (·, 0) + E− f (·, 0) (z) 2

=



1 1 Lc E f (·, 0) (z) + Lc E− f (·, 0) (z) 2 2

=

E0 (∞, 0) E0 (∞, 0) + 2(z − f ) 2(z + f )

=

zE0 (∞, 0) z2 − f 2

Dc (E− f (·, 0)) for which

lim E f (t, 0)h(t)Eg (t, 0) = lim E− f (t, 0)h(t)Eg (t, 0) = 0. \

t→∞

t→∞

Example 9.1.16 Let f ∈ Rc be a constant. We will compute
Lc Sinh f (·, 0) (z). We have Lc (Sinhc (·, 0)) (z) = Lc

for those z ∈ Dc (E f (·, 0))



 E f (·, 0) − E− f (·, 0) (z) 2

=



1 1 Lc E f (·, 0) (z) − Lc E− f (·, 0) (z) 2 2

=

E0 (∞, 0) E0 (∞, 0) − 2(z − f ) 2(z + f )

=

f E0 (∞, 0) z2 − f 2

Dc (E− f (·, 0)) for which

lim E f (t, 0)h(t)Eg (t, 0) = lim E− f (t, 0)h(t)Eg (t, 0) = 0.

t→∞

\

t→∞

Exercise 9.1.17 Let f , g ∈ Rc be constants. Under “suitable” assumptions find the following.
1. Lc E f (·, 0) +Ch f g (·, 0) (z),
2. Lc 2E f (·, 0) − 3E f +g (·, 0) + E f −g (·, 0) (z),   3. Lc Sh f g (·, 0) − 3E f 2 (·, 0) (z),
4. Lc Cos f (·, 0) (z),
5. Lc Sin f (·, 0) (z).

The Conformable Laplace Transform  299

9.2

DECAY OF THE EXPONENTIAL FUNCTION

For λ ∈ R(T), define Z t

mλ (t, s) =

s

1 ∆τ, 1 + µ(τ)λ

s,t ∈ T.

Theorem 9.2.1 Let s ∈ T and λ ∈ R + ([s, ∞)). Then 1. m∆λ t (t, s) ≥ 0 for any t ∈ [s, ∞). 2. lim mλ (t, s) = ∞. t→∞

1. Since λ ∈ R + ([s, ∞)), we have

Proof 9.2.2

t ∈ [s, ∞).

1 + µ(t)λ > 0, Hence, mλ∆t (t, s) =

1 1 + µ(t)λ t ∈ [s, ∞).

> 0, 2. We consider the following two cases. (a) Let sup µ(T) < ∞. Then Z t

mλ (t, s)

= s

≥

Z t s

≥

1 ∆τ 1 + λ µ(τ) 1 ∆τ 1 + µ(τ)|λ |

t −s 1 + |λ | sup µ(T)

→ ∞,

as t → ∞.

(b) Let sup µ(T) = ∞. Since λ ∈ R + ([s, ∞)), we conclude that λ ≥ 0. i. Let λ = 0. Then mλ (t, s)

= t −s → ∞,

as t → ∞.

ii. Let λ > 0. Take a sequence {ξk }k∈N ⊂ [s, ∞) such that the sequence {µ(ξk )}k∈N is increasing and divergent. Hence, Z t 1 mλ (t, s) = ∆τ s 1 + λ µ(τ)

300  Conformable Dynamic Equations on Time Scales

Z σ (ξk )

≥

∑ σ (ξk ) ≤ t ξk ≥ s

=

∑ σ (ξk ) ≤ t ξk ≥ s

Since

ξk

1 ∆τ 1 + λ µ(τ)

µ(ξk ) . 1 + λ µ(ξk )

µ(ξk ) 1 = , k→∞ 1 + λ µ(ξk ) λ lim

we get lim mλ (t, s) ≥ lim

t→∞

µ(ξk ) 1 + λ µ(ξk )

∑

t→∞

σ (ξk ) ≤ t ξk ≥ s

= ∞. 

This completes the proof. Theorem 9.2.3 Let s ∈ T and f ∈ R + ([s, ∞)). Then 0 < e f (t, s) < e

Rt s

f (τ)∆τ

,

t ∈ [s, ∞).

Proof 9.2.4 Observe that log(1 +µ (τ) f (τ)) µ(τ)

=

µ(τ) f (τ) − (µ(τ) f (τ) − log(1 + µ(τ) f (τ))) µ(τ)

≤

f (τ),

τ ∈ [s,t],

t ∈ [s, ∞).

Then 0 < e f (t, s) R t log(1+µ(τ) f (τ))

= e ≤ e

s

Rt s

µ(τ)

f (τ)∆τ

,

∆τ

t ∈ [s, ∞). 

This completes the proof. For h > 0 define   1 Ch = z ∈ C : z 6= − h

and

Rh = Ch

\

R,

The Conformable Laplace Transform  301

and C0 = C, R0 = R. For h ≥ 0 define the functions Ψh : Ch × Rh → R and Reh : Ch → R as follows  1 + hλ 1  if h > 0   log h 1 + hz Ψh (z, λ ) =    λ − Re(z) if h = 0,  1   (|1 + hz| − 1) if h > 0 h Reh (z) =   Re(z) if h = 0.

Theorem 9.2.5 Let h ≥ 0 and λ ∈ Rh . Then: 1. Ψh (z, λ ) = Ψh (Reh (z), λ ) for any z ∈ Ch . 2. Suppose 1 + hλ > 0 and x1 , x2 ∈ R, 1 + hx1 > 0, 1 + hx2 > 0. If x1 < x2 , then Ψh (x1 , λ ) > Ψh (x2 , λ ). Proof 9.2.6

1. Let h = 0. Then Ψh (z, λ ) = λ − Re(z), Reh (z) = Re(z), Ψh (Reh (z), λ ) = λ − Re(Reh (z)) = λ − Re(z).

Thus, Ψh (z, λ ) = Ψh (Reh (z), λ ). Let h > 0. Then Ψh (Reh (z), λ ) = = =

1 + hλ 1 log h 1 + hReh (z) 1 1 + hλ log h 1 + |1 + hz| − 1 1 + hλ 1 log h 1 + hz

= Ψh (z, λ ).

302  Conformable Dynamic Equations on Time Scales

2. Let h = 0. Then Ψh (x1 , λ ) = λ − x1 > λ − x2 = Ψh (x2 , λ ). Suppose that h > 0. Then Ψh (x1 , λ ) =

1 + hλ 1 log h 1 + hx1

=

1 + hλ 1 log h 1 + hx1

>

1 + hλ 1 log h 1 + hx2

= Ψh (x2 , λ ). 

This completes the proof. For h ≥ 0 and z ∈ Ch , define  1   log |1 + hz| if h > 0 h ψh (z) =   Re(z) if h = 0. Define the minimal graininess function µ∗ : T → [0, ∞) by µ∗ (s) = inf µ(τ), τ∈[s,∞)

s ∈ T.

For h ≥ 0 and λ ∈ R, define Ch (λ ) = {z ∈ Ch : Reh (z) > λ }.

Theorem 9.2.7 Let s ∈ T and λ ∈0 R + ([s, ∞)). Then, for any z ∈ Cµ∗ (s) (λ ), we have the following properties. 1. |eλ z (t, s)| ≤ eλ Reµ∗ (s) (s) (t, s), 2. lim eλ Reµ∗ (s) (z) (t, s) = 0. t→∞

3. lim eλ z (t, s) = 0. t→∞

t ∈ [s, ∞).

The Conformable Laplace Transform  303

Proof 9.2.8

1. Observe that

ψµ(τ) ((λ z)(τ)) =

=

   

1 log |1 + µ(τ)((λ z)(τ))| µ(τ)

      

Re(λ − z) if µ(τ) = 0 1 + µ(τ)λ 1 if log µ(τ) 1 + µ(τ)z

  

λ − Re(z) if

if

µ(τ) > 0

µ(τ) > 0

µ(τ) = 0

= Ψµ(τ) (z, λ ). Then, using Theorem 9.2.5, we get |eλ z (t, s)| = e = e = e ≤ e

Rt

ψµ(τ) ((λ z)(τ))∆τ

Rt

Ψµ(τ) (z,λ )∆τ

Rt

Ψµ(τ) (Reµ(τ) (z),λ )∆τ

Rt

Ψµ(s) (Reµ∗ (τ) (z),λ )∆τ

s

s

s

s

= eλ Reµ∗ (s) (z) (t, s),

t ∈ [s, ∞).

2. We have λ − Reµ∗ (s) (z) 1 + µ(t)Reµ∗ (s) (z)


λ Reµ∗ (s) (z) (t) =

λ − Reµ∗ (s) (z)

=

1 + λ µ(t) + µ(t) Reµ∗ (s) (z) − λ




and
1 + µ(t) λ Reµ∗ (s) (z) (t) = 1 + =


µ(t) λ − Reµ∗ (s) (z) 1 + λ µ(t) + µ(t) Reµ∗ (s) (z) − λ

1 + λ µ(t)
. 1 + λ µ(t) + µ(t) Reµ∗ (s) (z) − λ

Therefore λ Reµ∗ (s) (z) < 0 and λ Reµ∗ (s) (z) ∈ R + ([s, ∞)). Hence, and considering Theorem 9.2.3, it follows that t eλ Reµ∗ (s) (z) (t, s) ≤ e s (λ Reµ∗ (s) (z))(τ)∆τ

R




304  Conformable Dynamic Equations on Time Scales

(λ −Reµ∗ (s) (z)) = e = e

Rt

∆τ s 1+µ(τ)Re µ∗ (s) (z)

−(Reµ∗ (s) (z)−λ )mRe

µ∗ (s) (z)

(t,s)

,

t ∈ [s, ∞).

Since z ∈ Cµ∗ (s) (λ ) and Reµ∗ (s) (z) ∈ R + ([s, ∞)), we have lim mReµ∗ (s) (z) (t, s) = ∞

t→∞

and lim eλ Reµ∗ (s) (z) (t, s) = 0.

t→∞

3. By 1) and 2), we get lim eλ z (t, s) ≤ lim eλ Reµ∗ (s) (z) (t, s) t→∞

t→∞

= 0. 

This completes the proof.

9.3

CONVERGENCE OF THE CONFORMABLE LAPLACE TRANSFORM

Definition 9.3.1 A function f ∈ Crd (T) is said to be of conformable exponential order β on [0, ∞) if there exist β ∈ R + ([0, ∞)) and a constant K > 0 such that f (t)hσ (t)e g−k (σ (t), 0)e k (∞, σ (t)) ≤ K eβ z (t, 0) k0 (α,t) 1 − 1 k0

k0

for any t ∈ [0, ∞) and z ∈ C.

Theorem 9.3.2 Let f ∈ Crd ([0, ∞)) be of conformable exponential order β . Then the conformable Laplace transform Lc ( f ) exists on Cµ∗ (0) (β ) and converges absolutely. In the case µ∗ (0) > 0, we have lim Lc ( f )(z) = 0. |z|→∞

Proof 9.3.3 Using Theorem 9.2.7, we have Z t |Lc ( f )(z)| = lim f (τ)hσ (τ)Egσ (τ, 0)∆α,∞ τ t→∞

0

Z t E0 (∞, σ (τ)) σ σ = lim f (τ)h (τ)Eg (τ, 0) ∆τ t→∞ 0 k0 (α, τ)

The Conformable Laplace Transform  305

Z t e− k1 (∞, σ (τ)) k0 ∆τ = lim f (τ)hσ (τ)e g−k1 (σ (τ), 0) t→∞ 0 k0 (α, τ) k0 e− k1 (∞, σ (τ)) Z t k0 σ ∆τ ≤ lim f (τ)h (τ)e g−k1 (σ (τ), 0) t→∞ 0 k0 (α, τ) k0 Z t eβ z (τ, 0) ∆τ ≤ K lim t→∞ 0 |1 + µ(τ)z| ≤ K lim

Z t e β −Reµ

∗ (0)

t→∞ 0

= = =

(z)(τ, 0)

1 + µ(τ)Reµ∗ (0) (z)

K lim β − Reµ∗ (0) (z) t→∞

Z t

K lim β − Reµ∗ (0) (z) t→∞

Z t

K lim β − Reµ∗ (0) (z) t→∞

Z t

0

0

0

∆τ

β − Reµ∗ (0) (z) e (τ, 0)∆τ 1 + µ(τ)Reµ∗ (0) (z) β Reµ∗ (0) (z)
β Reµ∗ (0) (z)(z) eβ Reµ∗ (0) (z) (τ, 0)∆τ e∆β Reµ

∗ (0) (z)

(τ, 0)∆τ

=

  K lim 1 − eβ Reµ∗ (0) (z) (t, 0) Reµ∗ (0) (z) − β t→∞

=

K , Reµ∗ (0) (z) − β

z ∈ Cµ∗ (0) (β ).

Let µ∗ (0) > 0. Note that |z| → ∞ implies Reµ∗ (0) (z) → ∞ and K lim Lc ( f ) ≤ lim = 0. |z|→∞ |z|→∞ Reµ∗ (0) (z) − β 

This completes the proof.

Theorem 9.3.4 Let f ∈ Crd ([0, ∞)) be of conformable exponential order β . Then the conformable Laplace transform Lc ( f ) converges uniformly on Cµ∗ (0) (γ) for any γ > β . Proof 9.3.5 By the proof of Theorem 9.3.2, it follows that |Lc ( f )(z)| ≤ ≤

K Reµ∗ (0) (z) − β K . γ −β

Hence, for any ε > 0 there exists an r ∈ [0, ∞) such that Z ∞ σ σ f (τ)h (τ)Eg (τ, 0)∆α,∞ τ < ε t

for any t ∈ [r, ∞) and for any z ∈ Cµ∗ (0) (γ). This completes the proof.



306  Conformable Dynamic Equations on Time Scales

Theorem 9.3.6 (Inversion Formula) Suppose 0 < µmin ≤ µ(t) ≤ µmax < ∞ and a1 ≤ |k0 (α,t) − µ(t)k1 (α,t)| ≤ b1 ,

t ∈ T,

for some positive constants a1 and b1 . If f : T → R is regulated and Z a+i∞ a−i∞

|Lc ( f )(z)||dz| < ∞

and the poles of Lc ( f )(z) are regressive constants {z1 , . . . , zn } of finite order, then   n k0 (α,t) 1 f (t) = ∑ Resz=z j h(t)Eg (t, 0) Lc ( f )(z) , t ∈ T, E0 (∞, σ (t)) j=1 for any z ∈ C with Re(z) > a. Proof 9.3.7 Firstly, we will find an expression for h(·)Eg (·, 0). We have Dα (h(·)Eg (·, 0)) (t) = Dα h(t)Egσ (t, 0) + h(t)g(t)Eg (t, 0) (9.13) −k1 (α,t)h(t)Egσ (t, 0),

κ

t ∈T .

By the second equation of (9.1), we obtain 0 = Dα h(t)Egσ (t, 0) − zh(t)hσ (t)Egσ (t, 0) (9.14) +(z − k1 (α,t))hσ (t)Egσ (t, 0) − k1 (α,t)h(t)Egσ (t, 0),

t ∈ Tκ .

By the first equation of (9.1), we get zh(t)hσ (t)Egσ (t, 0) = −h(t)g(t)Eg (t, 0),

t ∈ Tκ .

Hence, employing (9.14) and (9.13), we find 0 = Dα h(t)Egσ (t, 0) + h(t)g(t)Eg (t, 0) +(z − k1 (α,t))hσ (t)Egσ (t, 0) − k1 (α,t)h(t)Egσ (t, 0) = Dα (h(·)Eg (·, 0)) (t) + (z − k1 (α,t))hσ (t)Egσ (t, 0),

t ∈ Tκ ,

or Dα (h(·), Eg (·, 0)) (t) = −(z − k1 (α,t))hσ (t)Egσ (t, 0),

t ∈ Tκ .

Let r(t) = z =

(k1 (α,t) − z)(k0 (α,t) − µ(t)k1 (α,t)) , k0 (α,t) − µ(t)(k1 (α,t) − z) 1 (r(t) − k1 (α,t)), k0 (α,t)

t ∈ T.

The Conformable Laplace Transform  307

Then z = = =

1 (r(t) − k1 (α,t)) k0 (α,t)   1 (k1 (α,t) − z)(k0 (α,t) − µ(t)k1 (α,t)) − k1 (α,t) k0 (α,t) k0 (α,t) − µ(t)(k1 (α,t) − z)  1 k0 (α,t)k1 (α,t) − µ(t)(k1 (α,t))2 k0 (α,t)(k0 (α,t) − µ(t)(k1 (α,t) − z)) −zk0 (α,t) + zµ(t)k1 (α,t) − k0 (α,t)k1 (α,t)  2 +µ(t)(k1 (α,t)) − zµ(t)k1 (α,t)

= −

zk0 (α,t) k0 (α,t)(k0 (α,t) − µ(t)(k1 (α,t) − z))

z , k0 (α,t) − µ(t)(k1 (α,t) − z)   z 1 + µ(t)z = 1 + µ(t) − k0 (α,t) − µ(t)(k1 (α,t) − z) = −

=

k0 (α,t) − µ(t)k1 (α,t) + zµ(t) − zµ(t) k0 (α,t) − µ(t)(k1 (α,t) − z)

=

k0 (α,t) − µ(t)k1 (α,t) , k0 (α,t) − µ(t)(k1 (α,t) − z)

z = −

z 1 + µ(t)z

=

z k0 (α,t)−µ(t)(k1 (α,t)−z) k0 (α,t)−µ(t)k1 (α,t) k0 (α,t)−µ(t)(k1 (α,t)−z)

=

z , k0 (α,t) − µ(t)k1 (α,t)

t ∈ T.

Denote e z = z,

t ∈ T.

Therefore h(t)Eg (t, 0) = Er (t, 0) = e r−k1 (t, 0) k0

= ez (t, 0) = e ez (t, 0),

t ∈ T.

308  Conformable Dynamic Equations on Time Scales

Hence, Z ∞

Lc ( f )(z) =

0

f (t)hσ (t)Egσ (t, 0)

E0 (∞, σ (t)) ∆t k0 (α,t)

E0 (∞, σ (t)) e (t, 0)∆t k0 (α,t) ez 0   E0 (∞, σ (·)) = L f (·) (e z). k0 (α, ·) Z ∞

=

f (t)

Hence, by the inverse formula for the Laplace transform on time scales, we get f (t)

E0 (∞, σ (t)) k0 (α,t)

n

∑ Resez=ez j eez (t, 0)L ( f )(ez)

=

j=1 n

1

∑ Resz=z j h(t)Eg (t, 0) Lc ( f )(z),

=

t ∈ T,

j=1

for any z ∈ C with Re(z) > a, whereupon f (t) =

k0 (α,t) E0 (∞, σ (t))

n

∑ Resz=z j

j=1



 1 Lc ( f )(z) , h(t)Eg (t, 0)

t ∈ T,

for any z ∈ C with Re(z) > a. This completes the proof.



Theorem 9.3.8 (Uniqueness of the Inverse) If the functions f , g : T → R are regulated and have the same Laplace transform, then f = g a.e. Proof 9.3.9 We have Lc ( f )(z) = Lc (g)(z). By the proof of Theorem 9.3.6, it follows that     E0 (∞, σ (·)) E0 (∞, σ (·)) L f (·) (e z) = L g(·) (e z). k0 (α, ·) k0 (α, ·) By the uniqueness of the inverse of the Laplace transform on time scales, it follows f (·)

E0 (∞, σ (·)) E0 (∞, σ (·)) = g(·) k0 (α, ·) k0 (α, ·)

a.e.,

from where f = g a.e. This completes the proof.



Definition 9.3.10 Suppose 0 < µmin ≤ µ(t) ≤ µmax < ∞ and a1 ≤ |k0 (α,t) − µ(t)k1 (α,t)| ≤ b1 ,

t ∈ T,

The Conformable Laplace Transform  309

for some positive constants a1 and b1 . If f1 , f2 : T → R are regulated and Z a+i∞ a−i∞

|Lc ( f1 )(z)Lc ( f2 )(z)||dz| < ∞

and the poles of Lc ( f1 )(z)Lc ( f2 )(z) are regressive constants {z1 , . . . , zn } of finite order, then we say that f1 and f2 is a conformable Laplace pair.

Definition 9.3.11 Let f1 , f2 : T → R be a conformable Laplace pair. Then we define the conformable convolution of f1 and f2 by   n k0 (α,t) 1 ( f1 ?c f2 )(t) = ∑ Resz=z j h(t)Eg (t, 0) Lc ( f1 )(z)Lc ( f2 )(z) , t ∈ T, E0 (∞, σ (t)) j=1 for any z ∈ C with Re(z) > a. If f1 , f2 : T → R is a conformable Laplace pair, then ( f1 ?c f2 )(t) = ( f2 ?c f1 )(t),

t ∈ T,

and Lc ( f1 ?c f2 )(z) = Lc ( f1 )(z)Lc ( f2 )(z) for any z ∈ C with Re(z) > a. This ends the example. Exercise 9.3.12 Let f1 , f2 , f3 : T → R be regulated and f1 and f2 , f1 and f3 , f2 and f3 be conformable Laplace pairs such that the poles of Lc ( f1 )(z)Lc ( f2 )(z), Lc ( f1 )(z)Lc ( f3 )(z) and Lc ( f2 )(z)Lc ( f3 )(z) are regressive constants {z1 , . . . , zn } of finite order. Prove that ( f1 ?c f2 ) ?c f3 = f1 ?c ( f2 ?c f3 ) . Theorem 9.3.13 Let f1 , f2 : T → R be a conformable Laplace pair. If f1 is conformable ∆-differentiable, then Dα ( f1 ?c f2 )(t) = (Dα f1 ?c f2 ) (t) + E0 (∞, 0) f1 (0) f2 . If f2 is conformable ∆-differentiable, then Dα ( f1 ?c f2 )(t) = ( f1 ?c Dα f2 ) (t) + E0 (∞, 0) f1 f2 (0). Proof 9.3.14 Suppose that f1 is conformable ∆-differentiable. Then Lc (Dα ( f1 ?c f2 )) (z) = zLc ( f1 ?c f2 )(z) − E0 (∞)( f1 ?c f2 )(0) = zLc ( f1 )(z)Lc ( f2 )(z) − (E1 (∞, 0) f1 (0)) Lc ( f2 )(z)

310  Conformable Dynamic Equations on Time Scales

+ (E0 (∞, 0) f1 (0)) Lc ( f2 )(z) = (zLc ( f1 )(z) − E0 (∞, 0) f1 (0)) Lc ( f2 )(z) +E0 (∞, 0) f1 (0)Lc ( f2 )(z) = Lc (Dα f1 ) (z)Lc ( f2 )(z) +Lc (E0 (∞, 0) f1 (0) f2 ) (z) = Lc (Dα f1 ?c f2 ) (z) +Lc (E0 (∞, 0) f1 (0) f2 ) (z) = Lc (Dα f1 ?c f2 + E0 (∞, 0) f1 (0) f2 ) , whereupon Dα ( f1 ?c f2 )(t) = (Dα f1 ?c f2 ) (t) + E0 (∞, 0) f1 (0) f2 . The case when f2 is conformable ∆-differentiable we leave to the reader as an exercise. This completes the proof. 

9.4

APPLICATIONS TO IVPS

Consider the following IVP (Dα )n y + an−1 (Dα )n−1 y + · · · + a1 Dα y + a0 y = f (t),

t > 0,

(9.15)

(Dα )n−1 y(0) = bn−1 , .. . (9.16) α

D y(0)

= b1 ,

y(0)

= b0 ,

where ai , bi ∈ C, i ∈ {1, . . . , n − 1}, f : T → R is regulated. Take the conformable Laplace transform of both sides of equation (9.15) and using the initial conditions (9.16), we get   Lc ( f )(z) = Lc (Dα )n y + an−1 (Dα )n−1 y + · · · + a1 Dα y + a0 y (z)  
= Lc (Dα )n y (z) + an−1 Lc (Dα )n−1 y (z) + · · · + a1 Lc (Dα y) (z) + a0 Lc (y)(z)

The Conformable Laplace Transform  311

  = zn Lc (y)(z) − E0 (∞, 0) y(0)zn−1 + Dα y(0)zn−2 + · · · + (Dα )n−1 y(0) +an−1 zn−1 Lc (y)(z) − E0 (∞, 0)   y(0)zn−2 + Dα y(0)zn−3 + · · · + (Dα )n−2 y(0) +··· +a1 (zLc (y)(z) − E0 (∞, 0)y(0)) +a0 Lc (y)(z) =


zn + an−1 zn−1 + · · · + a1 z + a0 Lc (y)(z) −E0 (∞, 0) b0 zn−1 + b1 zn−2 + · · · + bn−2 z + bn−1




−E0 (∞, 0) b0 zn−2 + b1 zn−3 + · · · + bn−3 z + bn−2




−··· −E0 (∞, 0)b0 =


zn + an−1 zn−1 + · · · + a1 z + a0 Lc (y)(z)  −E0 (∞, 0) b0 zn−1 + (b1 + b2 )zn−2 + (b2 + b1 + b0 )zn−3 + · · · + (bn−2 + bn−1 + · · · + b1 + b0 )  +bn−1 + bn−2 + · · · + b1 + b0 .

Let  l(z) = E0 (∞, 0) b0 zn−1 + (b1 + b2 )zn−2 + (b2 + b1 + b0 )zn−3 + · · · + (bn−2 + bn−1 + · · · + b1 + b0 )  +bn−1 + bn−2 + · · · + b1 + b0 .

312  Conformable Dynamic Equations on Time Scales

Then
zn + an−1 zn−1 + · · · + a1 z + a0 Lc (y)(z) = Lc ( f )(z) + l(z) or Lc (y)(z)(y)(z) =

1 zn + an−1 zn−1 + · · · + a1 z + a0

(Lc ( f )(z) + l(z)) .

Hence, y(t) = Lc−1



 1 (Lc ( f )(z) + l(z)) , zn + an−1 zn−1 + · · · + a1 z + a0

t ≥ 0.

Example 9.4.1 Consider the IVP (Dα )2 y + 3Dα y + 2y = 1,

t > 0,

Dα y(0) = 1, y(0) = 0. We take the conformable Laplace transform of both sides of the considered equation and using the initial conditions, we find   Lc (1)(z) = Lc (Dα )2 y + 3Dα y + 2y (z)   = Lc (Dα )2 y (z) + 3Lc (Dα y) (z) + 2Lc (y)(z) = z2 Lc (y)(z) − E0 (∞, 0) + 3zLc (y)(z) + 2Lc (y)(z) = (z2 + 3z + 2)Lc (y)(z) − E0 (∞, 0) or or

E0 (∞, 0) + E0 (∞, 0) = (z2 + 3z + 2)Lc (y)(z), z   1 1 Lc (y)(z) = E0 (∞, 0) + z(z + 1)(z + 2) (z + 1)(z + 2)   1 1 1 1 1 = E0 (∞, 0) − + + − 2z z + 1 2(z + 2) z + 1 z + 2   1 1 = E0 (∞, 0) − . 2z 2(z + 2)

Thus, y(t) = This ends the example.

1 1 − E−2 (t, 0), 2 2

t ≥ 0.

The Conformable Laplace Transform  313

Exercise 9.4.2 Using the conformable Laplace transform, find a solution of the following IVP (Dα )2 y + 5Dα y + 6y = 3 + 2E1 (t, 0),

t > 0,

Dα y(0) = −11, y(0) = 1.

9.5

ADVANCED PRACTICAL PROBLEMS

Problem 9.5.1 Let f , g ∈ Rc be constants. Under “suitable” assumptions find the following.
1. Lc C f g (·, 0) (z),
2. Lc S f g (·, 0) (z),
3. Lc Cos f (·, 0) + 4S f g (·, 0) − 5E f +g (·, 0) (z),
4. Lc Sin f (·, 0) − 3Cos f (·, 0) (z),
5. Lc Sin f (·, 0) − 4h11 (·, 0) (z). Problem 9.5.2 Let T = 3N0 , k1 (α,t) = (1 − α)t α , u(t, s) = (s + t)2 + t 4 , Find

k0 (α,t) = αt 1−α ,

t ∈ T,

(t, s) ∈ T × T. 1

1

Dt2 u(t, s),

α ∈ (0, 1],

Ds3 u(t, s),

1

Ds4 u(t, s).

Problem 9.5.3 Using the conformable Laplace transform, find solutions of the following IVPs: 1. (Dα )2 y − 7Dα y + 12y = 2E1 (t, 0) + 3Cosh2 (t, 0), Dα y(0) = 2, y(0) = −1.

t > 0,

314  Conformable Dynamic Equations on Time Scales

2. (Dα )2 y − 9y = E2 (t, 0) − Sinh2 (t, 0),

t > 0,

Dα y(0) = 0, y(0) = 1. 3. (Dα )3 y − 6 (Dα )2 y + 11Dα y − 6y = E4 (t, 0) − Sinh1 (t, 0) + 3Cosh7 (t, 0), (Dα )2 y(0) = 4, Dα y(0) = 1, y(0) = 1.

t > 0,

APPENDIX

A

Derivatives on Banach Spaces

A.1

REMAINDERS

Let X and Y be normed spaces. With o(X,Y ) we will denote the set of all maps r : X → Y for which there is some map α : X → Y such that 1. r(x) = α(x)kxk for all x ∈ X, 2. α(0) = 0, 3. α is continuous at 0. Definition A.1.1 The elements of o(X,Y ) will be called remainders.

Exercise A.1.2 Prove that o(X,Y ) is a vector space. Definition A.1.3 Let f : X → Y be a function and x0 ∈ X. We say that f is stable at x0 if there are some ε > 0 and some c > 0 such that kx − x0 k ≤ ε implies k f (x − x0 )k ≤ ckx − x0 k. Example A.1.4 Let T : X → Y be a linear bounded operator. Then kT (x − 0)k = kT (x)k ≤ kT kkxk,

x ∈ X.

Hence, T is stable at 0. Theorem A.1.5 Let X, Y , Z and W be normed spaces, r ∈ o(X,Y ), and assume f : W → X is stable at 0, g : Y → Z is stable at 0. Then r ◦ f ∈ o(W,Y ) and g ◦ r ∈ o(X, Z). 315

316  Conformable Dynamic Equations on Time Scales

Proof A.1.6 Since r ∈ o(X,Y ), then there is a map α : X → Y such that r(x) = α(x)kxk,

x ∈ X,

α(0) = 0 and α is continuous at 0. Define β : W → Y such that  k f (w)k   α( f (w)) if w = 6 0,  kwk β (w) =    0 if w = 0, w ∈ W . Since f : W → Z is stable at 0, then there are constants ε > 0 and c > 0 such that kwk ≤ ε implies that k f (w)k ≤ ckwk. Hence, k f (0)k = 0 and

f (0) = 0.

Next, β (0) = 0 and if w 6= 0, kwk ≤ ε, we get kβ (w)k =

k f (w)k kα( f (w))k kwk

≤ ckα( f (w))k. From here, using that f (w) → 0

as

w→0

and α( f (w)) → 0 as

w → 0,

we get β (w) → 0 as

w → 0.

Therefore β : W → Y is continuous at 0. Also, • if w = 0, then β (0) = 0, r ◦ f (0) = α( f (0))k f (0)k = 0 = β (0). • if w 6= 0, then r ◦ f (w) = α( f (w))k f (w)k

Derivatives on Banach Spaces  317

=

kwkβ (w) k f (w)k k f (w)k

= kwkβ (w). Therefore r ◦ f ∈ o(W,Y ). Since g : Y → Z is stable at 0, then there are constants ε1 > 0 and c1 > 0 such that kwk ≤ ε1 implies kg(w)k ≤ c1 kwk. Define γ : X → Y by  g(kxkα(x))   if  kxk γ(x) =    0 if x = 0.

x 6= 0,

Then g(kxkα(x)) = kxkγ(x),

x ∈ X.

For x 6= 0, x ∈ X, we have kγ(x)k = ≤

kg(kxkα(x))k kxk c1 kxkα(x) kxk

= c1 α(x). Then γ(x) → 0 as

x → 0,

x ∈ X.

Also, g ◦ r(x) = g(r(x)) = g(α(x)kxk) = γ(x)kxk, This completes the proof.

A.2

x ∈ X. 

´ DEFINITION AND UNIQUENESS OF THE FRECHET DERIVATIVE

Suppose that X and Y are normed spaces, U is an open subset of X and x0 ∈ U. With L (X,Y ) we will denote the vector space of all linear bounded operators from X to Y .

318  Conformable Dynamic Equations on Time Scales

Definition A.2.1 We say that a function f : X → Y is Fr´echet differentiable at x0 if there is some L ∈ L (X,Y ) and r ∈ o(X,Y ) such that f (x) = f (x0 ) + L(x − x0 ) + r(x − x0 ),

x ∈ U.

The operator L will be called the Fr´echet derivative of the function f at x0 . We will write D f (x0 ) = L. Suppose that L1 , L2 ∈ L (X,Y ) and r1 , r2 ∈ o(X,Y ) are such that f (x) =

f (x0 ) + L1 (x − x0 ) + r1 (x − x0 ),

f (x) =

f (x0 ) + L2 (x − x0 ) + r2 (x − x0 ),

x ∈ U.

Then f (x0 ) + L1 (x − x0 ) + r1 (x − x0 ) = f (x0 ) + L2 (x − x0 ) + r2 (x − x0 ),

x ∈ U,

or L1 (x − x0 ) − L2 (x − x0 ) = r2 (x − x0 ) − r1 (x − x0 ),

x ∈ U.

Also, let α1 , α2 : X → Y be such that r1 (x) = kxkα2 (x),

r2 (x) = kxkα1 (x),

α1 (0) = α2 (0) = 0, α1 and α2 are continuous at 0. Then L1 (x − x0 ) − L2 (x − x0 ) = kx − x0 kα1 (x − x0 ) − kx − x0 kα2 (x − x0 ) = kx − x0 k (α1 (x − x0 ) − α2 (x − x0 )) ,

x ∈ U.

Let x ∈ X be arbitrarily chosen. Then there is some h > 0 such that for all |t| ≤ h we have x0 + tx ∈ U. Hence, L1 (tx) − L2 (tx) = ktxk (α1 (tx) − α2 (tx)) or t (L1 (x) − L2 (x)) = |t|kxk (α1 (tx) − α2 (tx)) , or L1 (x) − L2 (x)

=

sign(t)kxk (α1 (tx) − α2 (tx))

→ 0

as t → 0.

Because x ∈ X was arbitrarily chosen, we conclude that L1 = L2 and r1 = r2 .

Derivatives on Banach Spaces  319

Definition A.2.2 We denote by C 1 (U,Y ) the set of all functions f : U → Y that are Fr´echet differentiable at each point of U and D f : U → L (X,Y ) is continuous. We denote by C 2 (U,Y ) the set of all functions f ∈ C 1 (U,Y ) such that D f : U → L (X,Y ) is Fr´echet differentiable at each point of U and D(D f ) : U → L (X, L (X,Y )) is continuous.

Theorem A.2.3 Let f1 , f2 : U → Y be Fr´echet differentiable at x0 and a, b ∈ R. Then a f1 + b f2 is Fr´echet differentiable at x0 . Proof A.2.4 Let r1 , r2 ∈ o(X,Y ) be such that f1 (x) =

f1 (x0 ) + D f1 (x0 )(x − x0 ) + r1 (x − x0 ),

f2 (x) =

f2 (x0 ) + D f2 (x0 )(x − x0 ) + r2 (x − x0 ),

x ∈ U.

Hence, (a f1 + b f2 )(x) = a ( f1 (x0 ) + D f1 (x0 )(x − x0 ) + r1 (x − x0 )) +b ( f2 (x0 ) + D f2 (x0 )(x − x0 ) + r2 (x − x0 )) = a f1 (x0 ) + b f2 (x0 ) + (aD f1 (x0 ) + bD f2 (x0 )) (x − x0 ) + (ar1 (x − x0 ) + br2 (x − x0 )) ,

x ∈ U.

Note that ar1 + br2 ∈ o(X,Y ). This completes the proof.



Theorem A.2.5 A function f : U → Y is Fr´echet differentiable at x0 if and only if there is some function F : U → L (X,Y ) that is continuous at x0 and for which f (x) − f (x0 ) = F(x)(x − x0 ), Proof A.2.6 x0 and

x ∈ U.

1. Suppose that there is a function F : U → L (X,Y ) that is continuous at f (x) − f (x0 ) = F(x)(x − x0 ),

x ∈ U.

Then f (x) − f (x0 ) = F(x)(x − x0 ) − F(x0 )(x − x0 ) + F(x0 )(x − x0 ) = F(x0 )(x − x0 ) + r(x − x0 ),

320  Conformable Dynamic Equations on Time Scales

where r(x) =

  (F(x + x0 ) − F(x0 )) (x) for 

0 for

x + x0 ∈ U,

x + x0 ∈ / U.

Define

α(x) =

 (F(x + x0 ) − F(x0 )) (x)     kxk   

for

x + x0 ∈ U,

x 6= 0,

0 for x + x0 ∈ / U,        0 for x = 0.

Then x ∈ X.

r(x) = α(x)kxk,

Let ε > 0 be arbitrarily chosen. Since F : U → L (X,Y ) is continuous at x0 , there exists some δ > 0 for which kxk < δ implies k (F(x + x0 ) − F(x0 )) (x)k ≤ kF(x + x0 ) − F(x0 )kkxk < εkxk. Therefore |α(x)| < ε for kxk < δ , i.e., α is continuous at 0. From here, we conclude that r ∈ o(X,Y ) and F(x0 ) = D f (x0 ). 2. Suppose that f is Fr´echet differentiable at x0 . Then there is some r ∈ o(X,Y ) such that f (x) = f (x0 ) + D f (x0 )(x − x0 ) + r(x − x0 ), x ∈ U, where D f (x0 ) ∈ L (X,Y ). Since r ∈ o(X,Y ), there is some α : X → Y such that r(x)

=

α(x)kxk,

α(0)

=

0,

α(x) → 0 as

x → 0.

By the Hahn-Banach extension theorem, it follows that there is some λx ∈ X ∗ such that λx x = kxk and |λx v| ≤ kvk,

v ∈ X.

Derivatives on Banach Spaces  321

Then r(x) = (λx x)α(x),

x ∈ X,

and
f (x) = f (x0 ) + D f (x0 )(x − x0 ) + λx−x0 (x − x0 ) α(x − x0 ), Let F : U → L (X,Y ) be defined as follows
F(x)(v) = D f (x0 )(v) + λx−x0 v α(x − x0 ),

x ∈ U,

x ∈ U.

v ∈ X.

We have f (x) = r(x − x0 ) = =

f (x0 ) + F(x)(x − x0 ),

x ∈ U,


λx−x0 (x − x0 ) α(x − x0 ) f (x) − f (x0 ) − D f (x0 )(x − x0 )

= F(x)(x − x0 ) − D f (x0 )(x − x0 ),

x ∈ U.

Note that
kF(x)(v) − F(x0 )(v)k = kD f (x0 )(v) + λx−x0 v α(x − x0 ) − D f (x0 )(v)k
= k λx−x0 v α(x − x0 )k = |λx−x0 v|kα(x − x0 )k ≤ kvkkα(x − x0 )k,

x ∈ U,

v ∈ X.

Then kF(x) − F(x0 )k ≤ kα(x − x0 )k,

x ∈ U.

Consequently, F is continuous at x0 . This completes the proof.



Theorem A.2.7 Let Z be a normed space, assume f : U → Z is Fr´echet differentiable at x0 , g : f (U) → Z is Fr´echet differentiable at f (x0 ). Then g ◦ f : U → Z is Fr´echet differentiable at x0 and D(g ◦ f )(x0 ) = Dg( f (x0 )) ◦ D f (x0 ). Proof A.2.8 Let y0 =

f (x0 ),

L1 = D f (x0 ), L2 = Dg(y0 ).

322  Conformable Dynamic Equations on Time Scales

There exist r1 ∈ o(X,Y ), r2 ∈ o(Y, Z) such that f (x0 ) + L1 (x − x0 ) + r1 (x − x0 ),

f (x) =

g(y) = g(y0 ) + L2 (y − y0 ) + r2 (y − y0 ),

x ∈ U, y ∈ f (U).

Hence, g( f (x)) = g( f (x0 )) + L2 ( f (x) − y0 ) + r2 ( f (x) − y0 ) = g(y0 ) + L2 (L1 (x − x0 ) + r1 (x − x0 )) +r2 (L1 (x − x0 ) + r1 (x − x0 )) = g(y0 ) + L2 (L1 (x − x0 )) + L2 (r1 (x − x0 )) +r2 (L1 (x − x0 ) + r1 (x − x0 )) ,

x ∈ U.

Define r3 : X → Z as follows r3 (x) = r2 (L1 (x) + r1 (x)),

x ∈ U.

Fix c > kL1 k and we represent r1 as follows r1 (x) = α1 (x)kxk,

x ∈ U.

We have that α1 : X → Y , α1 (0) = 0 and α1 is continuous at 0. Then there exists some δ > 0 such that if kxk < δ , then kα1 (x)k < c − kL1 k. Hence, if kxk < δ , then kr1 (x)k ≤ (c − kL1 k)kxk. Then, kxk < δ implies kL1 (x) + r1 (x)k ≤ kL1 (x)k + kr1 (x)k ≤ kL1 kkxk + (c − kL1 k) kxk = ckxk.

Derivatives on Banach Spaces  323

Then x → L1 (x) + r1 (x) is stable at 0. Hence, by Theorem A.1.5, we get r3 ∈ o(X, Z). Define r : X → Z as follows r = L1 ◦ r1 + r3 . We have r ∈ o(X, Z) and g ◦ f (r) = g ◦ f (x0 ) + L2 ◦ L1 (x − x0 ) + r(x − x0 ),

x ∈ U.

Since L1 ∈ L (X,Y ), L2 ∈ L (Y, Z), we have L2 ◦ L1 ∈ L (X, Z). Therefore g ◦ f is Fr´echet differentiable at x0 and L2 ◦ L1 = Dg(y0 ) ◦ D f (x0 ) = Dg( f (x0 )) ◦ D f (x0 ). 

This completes the proof.

Theorem A.2.9 Let f1 , f2 : U → R be Fr´echet differentiable at x0 . Then f1 · f2 is Fr´echet differentiable at x0 and D( f1 · f2 )(x0 ) = f2 (x0 )D f1 (x0 ) + f1 (x0 )D f2 (x0 ). Proof A.2.10 Let r1 , r2 ∈ o(X, R) be such that f1 (x) =

f1 (x0 ) + D f1 (x0 )(x − x0 ) + r1 (x − x0 ),

f2 (x) =

f2 (x0 ) + D f2 (x0 )(x − x0 ) + r2 (x − x0 ),

x ∈ U.

Hence, f1 (x) f2 (x) = ( f1 (x0 ) + D f1 (x0 )(x − x0 ) + r1 (x − x0 )) × ( f2 (x0 ) + D f2 (x0 )(x − x0 ) + r2 (x − x0 )) =

f1 (x0 ) f2 (x0 ) + f1 (x0 )D f2 (x0 )(x − x0 ) + f2 (x0 )D f1 (x0 )(x − x0 ) + f1 (x0 )r2 (x − x0 ) + D f1 (x0 )(x − x0 )D f2 (x0 )(x − x0 ) +D f1 (x0 )(x − x0 )r2 (x − x0 ) + r1 (x − x0 ) f2 (x0 ) +D f2 (x0 )(x − x0 )r1 (x − x0 ) + r1 (x − x0 )r2 (x − x0 ),

x ∈ U.

324  Conformable Dynamic Equations on Time Scales

Let r : X → R be defined as follows r(x) =

f1 (x0 )r2 (x) + D f1 (x0 )xD f2 (x0 )x +D f1 (x0 )xr2 (x) + r1 (x) f2 (x0 ) +D f2 (x0 )xr1 (x) + r1 (x)r2 (x),

x ∈ U.

Then f1 (x) f2 (x) =

f1 (x0 ) f2 (x0 ) + f1 (x0 )D f2 (x0 )(x − x0 ) + f2 (x0 )D f1 (x0 )(x − x0 ) + r(x − x0 ),

x ∈ U.

Note that |D f1 (x0 )xD f2 (x0 )x| ≤ kD f1 (x0 )kkD f2 (x0 )kkxk2 , Define α : X → R as follows  D f1 (x0 )xD f2 (x0 )x   ,  kxk α(x) =    0, x = 0.

x ∈ U,

x ∈ U.

x 6= 0,

Then D f1 (x0 )xD f2 (x0 )x = α(x)kxk, |α(x)| = ≤

x ∈ U,

|D f1 (x0 )xD f2 (x0 )x| kxk kD f1 (x0 )kkD f2 (x0 )kkxk2 kxk

= kD f1 (x0 )kkD f2 (x0 )kkxk,

x ∈ U,

x 6= 0.

Then α(x) → 0 as

x → 0.

From here, r ∈ o(X, R). This completes the proof.

A.3

ˆ THE GATEAUX DERIVATIVE

Let X and Y be normed spaces and U be an open subset of X. Let also, x0 ∈ U. Definition A.3.1 Let f : U → Y . If there is some T ∈ L (X,Y ) such that lim

t→0

f (x0 + tv) − f (x0 ) = Tv t



Derivatives on Banach Spaces  325

for any v ∈ X, we say that f is Gˆateaux differentiable at x0 . We write f 0 (x0 ) = T . If f is Gateaux ˆ differentiable at any point of U, then we say that f is Gateaux ˆ differentiable on U.

Example A.3.2 Let f : R2 → R be defined as follows  x4    6 1 3 for (x1 , x2 ) 6= (0, 0), x1 + x2 f (x1 , x2 ) =    0 for (x1 , x2 ) = (0, 0). Let v = (v1 , v2 ) ∈ R2 , (v1 , v2 ) 6= (0, 0), be arbitrarily chosen. We have, for t 6= 0, f (0 + tv) =

=

lim

t→0

f (0 + tv) − f (0) t

t 4 v41 t 6 v61 + t 3 v32 tv41 , t 3 v61 + v32 tv41
t→0 t t 3 v6 + v3 2 1

= lim

v41 t→0 t 3 v6 + v3 2 1

= lim

=

v41 . v32

Therefore f 0 (0, 0)(v1 , v2 ) =

v41 , v32

(v1 , v2 ) ∈ R2 ,

(v1 , v2 ) 6= (0, 0).

This ends the example. Theorem A.3.3 If f : U → Y is Fr´echet differentiable at x0 , then it is Gˆateaux differentiable at x0 . Proof A.3.4 Since f : U → Y is Fr´echet differentiable at x0 , then there is some r ∈ o(X,Y ) such that f (x) = f (x0 ) + D f (x0 )(x − x0 ) + r(x − x0 ), x ∈ U, and r(x) = α(x)kxk,

x ∈ X,

where α : X → Y , α(0) = 0, α is continuous at 0. Then, for v ∈ X and t ∈ R, |t| small enough, we have f (x0 + tv) − f (x0 ) t

=

D f (x0 )(tv) + r(tv) t

326  Conformable Dynamic Equations on Time Scales

=

tD f (x0 )(v) + |t|kvkα(tv) t

=

D f (x0 )(v) + sign(t)kvkα(tv)

→ D f (x0 )(v) as t → 0. This completes the proof.



APPENDIX

B

A Chain Rule

B.1

MEASURE CHAINS

Let T be some set of real numbers. Definition B.1.1 A triple (T, ≤, ν) is called a measure chain provided it satisfies the following axioms. (A1) The relation “≤” satisfies, for r, s,t ∈ T, 1. t ≤ t (reflexive), 2. if t ≤ r and r ≤ s, then t ≤ s (transitive), 3. if t ≤ r and r ≤ t, then t = r (antisymmetric), 4. either r ≤ s or s ≤ r (total). (A2) Any nonvoid subset of T which is bounded above has a least upper bound, i.e., the measure chain (T, ≤) is conditionally complete. (A3) The mapping ν : T × T → R has the following properties, for r, s,t ∈ T. 1. ν(r, s) + ν(s,t) = ν(r,t) (cocycle property), 2. if r > s, then ν(r, s) > 0 (strong isotony), 3. ν is continuous (continuity).

Example B.1.2 Let T be any nonvoid closed subset of real numbers, “≤” is the usual order relation between real numbers and ν(r, s) = r − s,

r, s ∈ T.

Definition B.1.3 The forward jump operator σ and the backward jump operator ρ are defined as follows. σ (t) = inf{s ∈ T : s > t},

ρ(t) = sup{s ∈ T : s < t}, 327

328  Conformable Dynamic Equations on Time Scales

where σ (t) = t

if t = max T,

ρ(t) = t

if t = min T.

The graininess function is defined as follows. µ(t) = ν(σ (t),t),

t ∈ T.

The notions left-scattered, left-dense, right-scattered, right-dense, isolated, and Tκ are defined as in the case of time scales. Definition B.1.4 Let X be a Banach space with a norm k · k. We say that f : T → X is differentiable at t ∈ T if there exists f ∆ (t) ∈ X such that for any ε > 0 there exists a neighborhood U of t such that k f (σ (t)) − f (s) − f ∆ (t)ν(σ (t), s)k ≤ ε|ν(σ (t), s)| for all s ∈ U. In this case f ∆ (t) is said to be a derivative of f at t. Theorem B.1.5 We have ν ∆ (·,t) = 1,

t ∈ T.

Proof B.1.6 Let t ∈ T. Let also, ε > 0 be arbitrarily chosen and U is a neighborhood of t. Then ν(σ (t), s) + ν(s,t) = ν(σ (t),t),

s ∈ T,

and |ν(σ (t),t) − ν(s,t) − ν(σ (t), s)| = |ν(σ (t),t) − ν(σ (t),t)| = 0 ≤ ε|ν(σ (t), s)|, for any s ∈ U. This completes the proof. As in the case of time scales, one can prove the following assertion. Theorem B.1.7 Let f , g : T → X and t ∈ T. 1. If t ∈ Tκ , then f has at most one derivative at t. 2. If f is differentiable at t, then f is continuous at t.



A Chain Rule  329

3. If f is continuous at t and t is right-scattered, then f is differentiable at t and f ∆ (t) =

f (σ (t)) − f (t) . µ(t)

4. If f and g are differentiable at t ∈ Tκ and α, β ∈ R, then α f + β g is differentiable at t and (α f + β g)∆ (t) = α f ∆ (t) + β g∆ (t). 5. If f and g are differentiable at t ∈ Tκ and “·” is bilinear and continuous, then f · g is differentiable at t and ( f · g)∆ (t) = f ∆ (t) · g(t) + f (σ (t)) · g∆ (t). 6. If f and g are differentiable at t ∈ Tκ and g is algebraically invertible, then f · g−1 is differentiable at t with  
∆
f · g−1 (t) = f ∆ (t) − f · g−1 (t) · g∆ (t) · g−1 (σ (t)).

B.2

¨ POTZSCHE’S CHAIN RULE

Throughout this section we suppose that (T, ≤, ν) is a measure chain with forward jump operator σ and graininess µ. Assume that X and Y are Banach spaces and we will write k · k for the norms of X and Y . For a function f : T × X → Y and x0 ∈ X, we denote the delta derivative of t → f (t, x0 ) by ∆1 f (·, x0 ), and for a t0 ∈ T we denote the Fr´echet derivative of x → f (t0 , x) by D2 f (t0 , ·), provided these derivatives exist. Theorem B.2.1 ( P¨otzsche’s Chain Rule) For some fixed t0 ∈ Tκ , let g : T → X, f : T × X → Y be functions such that g, f (·, g(t0 )) are differentiable[ at t0 , and let U ⊆ T be a neighborhood of t0 such that f (t, ·) is differentiable for t ∈ U {σ (t0 )}, D2 f (σ (t0 ), ·) is continuous on the line segment {g(t0 ) + hµ(t0 )g∆ (t0 ) ∈ X : h ∈ [0, 1]} and D2 f is continuous at (t0 , g(t0 )). Then the composition function F : T → Y , F(t) = f (t, g(t)) is differentiable at t0 with derivative F ∆ (t0 ) = ∆1 f (t0 , g(t0 )) Z + 0

1

 ∆

D2 f (σ (t0 ), g(t0 ) + hµ(t0 )g (t0 ))dh g∆ (t0 ).

Proof B.2.2 Let U0 ⊆ U be a neighborhood of t0 such that µ(t0 ) ≤ |ν(t, σ (t0 ))| for t ∈ U0 . Let Φ(t, h) = D2 f (t, g(t0 ) + h(g(t) − g(t0 ))),

t ∈ U0 ,

h ∈ [0, 1].

330  Conformable Dynamic Equations on Time Scales

Note that there exists a constant C > 0 such that kΦ(σ (t0 ), h) − Φ(t0 , h)k ≤ C|ν(t, σ (t0 ))| for t ∈ U0 ,

h ∈ [0, 1].

Let ε > 0 be arbitrarily chosen. We choose ε1 > 0, ε2 > 0 small enough such that   Z 1   Φ(σ (t0 ), h)dhk + ε2 ε1 + 2kg∆ (t0 )k ≤ ε. ε1 1 +Ck 0

Since g and f (·, g(t0 )) are differentiable at t0 , there exists a neighborhood U1 ⊆ U0 of t0 such that kg(t) − g(t0 )k ≤ ε1 , kg(t) − g(σ (t0 )) − ν(t, σ (t0 ))g∆ (t0 )k ≤ ε1 |ν(t, σ (t0 ))|, k f (t, g(t0 )) − f (σ (t0 ), g(t0 )) − ν(t, σ (t0 ))∆1 f (t0 , g(t0 ))k ≤ ε1 |ν(t, σ (t0 ))| for t ∈ U1 . Hence, kg(t) − g(t0 )k = kg(t) − g(σ (t0 )) − ν(t, σ (t0 ))g∆ (t0 ) + g∆ (t0 )ν(t, σ (t0 )) +g(σ (t0 )) − g(t0 )k ≤ kg(t) − g(σ (t0 )) − ν(t, σ (t0 ))g∆ (t0 )k +kg∆ (t0 )k|ν(t, σ (t0 ))| + kg(σ (t0 )) − g(t0 )k ≤ ε1 |ν(t, σ (t0 ))| + kg∆ (t0 )k|ν(t, σ (t0 ))| +kg∆ (t0 )kµ(t0 ) =

  ε1 + kg∆ (t0 )k |ν(t, σ (t0 ))| + kg∆ (t0 )kµ(t0 )

≤

  ε1 + 2kg∆ (t0 )k |ν(t, σ (t0 ))|,

t ∈ U1 .

Since g is continuous at t0 and D2 f is continuous at (t0 , g(t0 )), there exists a neighborhood U2 ⊆ Uof t0 so that kΦ(t, h) − Φ(t0 , h)k ≤ ε2

for t ∈ U2 ,

h ∈ [0, 1].

Hence,   Z 1 kF(t) − F(σ (t0 )) − ν(t, σ (t0 )) ∆1 f (t0 , g(t0 )) + Φ(σ (t0 ), h)dhg∆ (t0 ) k 0

A Chain Rule  331

= k f (t, g(t)) − f (σ (t0 ), g(σ (t0 ))) − f (σ (t0 ), g(t0 )) + f (σ (t0 ), g(t0 )) − f (t, g(t0 )) + f (t, g(t0 )) −ν(t, σ (t0 ))∆1 f (t0 , g(t0 )) −ν(t, σ (t0 )) −

Z 1 0

Z 1

+ 0

Z 1 0

Φ(σ (t0 ), h)dhg∆ (t0 )

Φ(σ (t0 ), h)dh(g(t) − g(t0 )) Φ(σ (t0 ), h)dh(g(t) − g(t0 ))k

≤ k f (t, g(t0 )) − f (σ (t0 ), g(t0 )) − ν(t, σ (t0 ))∆1 f (t0 , g(t0 ))k Z 1

+k 0

  Φ(σ (t0 ), h)dh g(t) − g(t0 ) − ν(t, σ (t0 ))g∆ (t0 ) k

+k f (t, g(t)) − f (t, g(t0 )) − ( f (σ (t0 ), g(σ (t0 ))) − f (σ (t0 ), g(t0 ))) −

Z 1 0

Φ(σ (t0 ), h)dh(g(t) − g(t0 ))k

≤ k f (t, g(t0 )) − f (σ (t0 ), g(t0 )) − ν(t, σ (t0 ))∆1 f (t0 , g(t0 ))k Z 1

+k 0

Z 1

+k 0

Φ(σ (t0 ), h)dhkkg(t) − g(t0 ) − ν(t, σ (t0 ))g∆ (t0 )k (Φ(t, h) − Φ(σ (t0 ), h)) dh(g(t) − g(t0 ))k

≤ k f (t, g(t0 )) − f (σ (t0 ), g(t0 )) − ν(t, σ (t0 ))∆1 f (t0 , g(t0 ))k Z 1

+k 0

Z 1

+k 0

Z 1

+k 0

Φ(σ (t0 ), h)dhkkg(t) − g(t0 ) − ν(t, σ (t0 ))g∆ (t0 )k (Φ(t, h) − Φ(t0 , h)) dhkkg(t) − g(t0 )k (Φ(t0 , h) − Φ(σ (t0 ), h)) dhkkg(t) − g(t0 )k

332  Conformable Dynamic Equations on Time Scales

≤ ε1 |ν(t, σ (t0 ))| + ε1 |ν(t, σ (t0 ))|k

Z 1 0

Φ(σ (t0 ), h)dhk

  ∆ +ε2 ε1 + 2kg (t0 )k |ν(t, σ (t0 ))| +ε1C|ν(t, σ (t0 ))|  Z 1   Φ(σ (t0 ), h)dhk = ε1 1 +C + k 0

  +ε2 ε1 + 2kg∆ (t0 )k |ν(t, σ (t0 ))| ≤ ε|ν(t, σ (t0 ))|, This completes the proof.

t ∈ U1

\

U2 . 

Bibliography [1] D. R. Anderson, J. Bullock, L. Erbe, A. Peterson and H. N. Tran. Nabla Dynamic Equations on Time Scales, Pan American Mathematical Journal, 13:1 (2003) 1–47. [2] D. R. Anderson and D. J. Ulness. Newly Defined Conformable Derivatives, Advances in Dynamical Systems and Applications, Vol. 10, Number 2, pp. 109-137, 2015. [3] D. R. Anderson. Second-Order Self-Adjoint Differential Equations Using a Proportional-Derivative Controller, Communications on Applied Nonlinear Analysis, Vol. 24(2017), Number 1, 17-48. [4] F. M. Atici and G. Sh. Guseinov. On Green’s Functions and Positive Solutions for Boundary Value Problems on Time Scales, J. Comput. Appl. Math., Vol. 141, Issues 1-2, (2002) 75–99. [5] B. Bayour, A. Hammoudi and D. F. M. Torres. A Truly Conformable Calculus on Time Scales, Global and Stochastic Analysis, Vol. 5, No. 1, June (2018), 1-14. [6] M. Bohner and A. Peterson. Dynamic Equations on Time Scales: An Introduction with Applications, Birkh¨auser, Boston, 2001. [7] M. Bohner and A. Peterson(editors). Advances in Dynamic Equations on Time Scales, Birkh¨auser, Boston, 2003. [8] M. Bohner and S. Georgiev. Multivariable Dynamic Calculus on Time Scales, Springer, 2017. [9] S. Georgiev. Integral Equations on Time Scales, Springer 2016. [10] M.R.S. Rahmat. A New Definition of Conformable Fractional Derivative on Arbitrary Time Scales, Advances in Difference Equations, 2019.

333

Index A c B, 108 A ⊕c B, 108 C f g , 30 Ch f g , 28 Cos f , 29 Cosh f , 26 S f g , 30 Sh f g , 28 Sin f , 29 Sinh f , 26 C 1 (U,Y ), 319 C 2 (U,Y ), 319 R + , 139 Rc , 17 c A, 108 ⊕c , 17 Abel’s Formula, 202 Alternative Conformable Riccati’s Equation, 85 boundary value problem general two point, 278, 279 periodic, 288 Cauchy function second-order equation, 277, 282 comparison theorem for BVPs, 286 Conformable ∆-Derivative, 2 Conformable ∆-Differentiable Matrix, 93 Conformable ∆-Integration, 38 Conformable Addition Inverse, 21 Conformable Bernoulli’s Equation, 76 Conformable Circle Minus, 21 Conformable Circle Plus Addition, 17 Conformable Convolution, 309 Conformable Euler-Cauchy Equation, 231 Conformable Exponential Function, 23

Conformable Generalized Square, 22 Conformable Hyperbolic Functions, 26, 28 Conformable Laplace Pair, 309 Conformable Partial Derivative, 52 Conformable Putzer’s Algorithm, 132 Conformable Regressive Function, 17 Conformable Regressive Matrix, 106 Conformable Riccati’s Equation, 82 Conformable Trigonometric Functions, 29, 30 Conformable Wronskian, 198 dominant solution, 268 existence uniqueness theorem BVP, 278, 288 self-adjoint problem, 249 Exponentially Stable Solution, 178 Fr´echet Derivative, 318 Function of Conformable Exponential Order, 304 Fundamental System, 198 Gˆateaux Derivative, 325 Graininess Function, 328 Green function conjugate problem, 282, 285 focal BVP, 286 general two-point BVP, 279, 281 periodic BVP, 289 Gronwall’s Type Inequality, 147, 148 Integration by Parts, 43 Jump Operator, 328 Measure Chain, 327 periodic boundary conditions, 288

335

336  Index

rd-continuous Matrix, 106 recessive solution, 268 Remainder, 315

Taylor’s Formula, 46, 48 Trench divergence, 267 Uniformly Bounded Solution, 172

Semigroup Property, 23