On q-Difference and Fractional

Equations and Their Solutions

By

Zeinab Sayed Ibrahim Mansour

SUBMITTED FOR THE

REQUIREMENTS FOR THE DEGREE OF

DOCTOR OF PHILOSOPHY

AT

CAIRO UNIVERSITY

GIZA, EGYPT

FEBRUARY 2006

Supervisors

M.H. Annaby M.E.H. Ismail

Cairo University University of Centeral Florida

Table of Contents

Table of Contents iii

Acknowledgements iv

Notations v

Preface vi

I Classical Results 1

1 Basic Cosine and Sine Functions 2

1.1 A q-calculus . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2

1.2 Asymptotics of zeros of cos(z; q) and sin(z; q) . . . . . . . . . . . . . . 10

1.3 Asymptotics of cos(z; q) and sin(z; q) . . . . . . . . . . . . . . . . . . 20

1.4 Basic analogues of a theorem of Polya . . . . . . . . . . . . . . . . . . 30

2 Zeros of Finite q-Hankel Transforms 35

2.1 Introduction and preliminaries . . . . . . . . . . . . . . . . . . . . . . 35

2.2 Main results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 37

2.3 Applications . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 45

II Basic Difference Equations 50

3 Introduction 51

3.1 q-successive approximations . . . . . . . . . . . . . . . . . . . . . . . 51

3.2 q-Initial value problems . . . . . . . . . . . . . . . . . . . . . . . . . 53

3.3 Linear q-difference equations . . . . . . . . . . . . . . . . . . . . . . . 55

3.4 A q-type Wronskian . . . . . . . . . . . . . . . . . . . . . . . . . . . 58

ii

4 Basic Sturm-Liouville Problems 60

4.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 60

4.2 Fundamental solutions . . . . . . . . . . . . . . . . . . . . . . . . . . 63

4.3 The self adjoint problem . . . . . . . . . . . . . . . . . . . . . . . . . 68

4.4 Basic Green’s function . . . . . . . . . . . . . . . . . . . . . . . . . . 74

4.5 Eigenfunction expansions . . . . . . . . . . . . . . . . . . . . . . . . . 79

4.6 Asymptotics of eigenvalues and eigenfunctions . . . . . . . . . . . . . 88

III Basic Fractional Calculus and Equations 107

5 Basic Fractional Calculus 108

5.1 Fractional calculus and equations . . . . . . . . . . . . . . . . . . . . 108

5.2 Some fractional integral operators . . . . . . . . . . . . . . . . . . . . 112

5.3 q-Notations and results . . . . . . . . . . . . . . . . . . . . . . . . . . 118

5.4 Basic Riemann–Liouville fractional calculi . . . . . . . . . . . . . . . 126

6 Basic Mittag-Leffler Functions 141

6.1 Basic Mittag-Leffler function . . . . . . . . . . . . . . . . . . . . . . . 141

6.2 Basic Volterra integral equations . . . . . . . . . . . . . . . . . . . . . 146

6.3 Zeros of basic Mittag-Leffler functions . . . . . . . . . . . . . . . . . . 152

7 Basic Fractional Difference Equations 157

7.1 Basic Riemann–Liouville fractional order systems . . . . . . . . . . . 157

7.2 Basic Caputo fractional order systems . . . . . . . . . . . . . . . . . . 165

7.3 Solutions via a q-Laplace transform . . . . . . . . . . . . . . . . . . . 170

7.4 A q-Cauchy problem . . . . . . . . . . . . . . . . . . . . . . . . . . . 176

Bibliography 185

iii

Acknowledgements

I wish to thank professors Mahmoud Annaby, Cairo University, and Mourad Ismail,

University of Central Florida, for their guidance throughout the preparation of this

thesis. I do not find words of thank that may express my gratitude and appreciation

to my parents for being keen on teaching me all morals needed to achieve success

in my life. Last but not least, I would like to dedicate this thesis to my family, my

husband Abdallah and daughter Maryam, for their love, patience, and understanding

so that they allowed me to spend most of the time on this thesis.

Zeinab S.I. Mansour

February 1, 2006

iv

Notations

N := 0, 1, 2, · · ·

Z+ := 1, 2, 3, · · ·

Z := N⋃−N

Z∗ := Z\ 0

R := (−∞,∞)

R+ := (0,∞)

C is the set of all complex numbers.

f.s. is abbreviation for fundamental set.

RLF is abbreviation for Riemann–Liouville fractional.

v

Preface

This thesis is concerned with the fast growing area of basic special functions and basic

difference equations. It consists of three parts. The first one is concerned with some

basic analogues of classical results. This part contains a self contained q-calculus,

which has considered before in [9, 90]. We also derive some new basic type identities

which will be needed throughout the thesis. The first part contains two new results

concerning zeros of basic functions and q-integral transforms. We study the q-sine

and q-cosine functions, their asymptotics and zeros. Then we derive a q-analogue of

a theorem of George Polya in Chapter 1. In Chapter 2 we study the zeros of a finite

q-Hankel transform. As an application of the results of this chapter we introduce

another q-analogue of the theorem of George Polya of Chapter 1.

Part II is devoted to study q-difference equations. We start this part with Chapter

3 which contains some known results in this subject, see e.g. [8, 9, 90], in particular

existence and uniqueness theorems as well as, q-linear equations. Our new contri-

bution lie in Chapter 4. We establish a q-Sturm–Liouville theory and investigate

the asymptotic behavior of eigenvalues and eigenfunctions of basic Sturm–Liouville

eigenvalue problems.

In the last part we study q-analogues of fractional calculus and fractional differ-

ential equations. In Chapter 5 we give a brief account about the history of fractional

calculi and some results which we are intended in deriving their q-analogues. Basic

Riemann–Liouville fractional calculus introduced in [10,11,13] is studied in a rigorous

analytic way. We derive q-analogues of other types of fractional derivatives, namely

vi

vii

Grunwald–Letinkov derivative and Caputo fractional derivative. q-type fractional dif-

ference equations are introduced in Chapter 7. The solutions of fractional q-difference

equations are also studied in this part. For this task we define q-Mittag-Leffler func-

tions, and study some of their properties. We also solve q-fractional difference equa-

tions with constant coefficients by using q-type Laplace transform.

Part I

Classical Results

1

Chapter 1

Basic Cosine and Sine Functions

This chapter includes some q-notations and results. It starts with a q-calculus and

including some q-notations and functions. We also consider q-sine and q-cosine functions,

some of their properties and their zeros. Then we end up with a basic analogue of a theorem

of George Polya.

1.1 A q-calculus

Throughout this thesis unless otherwise stated q is a positive number less than 1 and by

the word ”basic” we mean q-analogue. In this section we introduce some of the q-notations

and results. We start with the q-shifted factorial, see [53], for a ∈ C,

(a; q)n :=

1, n = 0,

n−1∏i=0

(1− aqi), n = 1, 2, . . . .(1.1.1)

The limit of (a; q)n as n tends to infinity exists and will be denoted by (a; q)∞. Moreover

(a; q)∞ has the following series representation, cf., e.g. [53, p. 11],

(a; q)∞ =∞∑

n=0

(−1)nqn(n−1)

2an

(q; q)n. (1.1.2)

The multiple q-shifted factorial for complex numbers a1, . . . , ak is defined by

(a1, a2, . . . , ak; q)n :=k∏

j=1

(aj ; q)n. (1.1.3)

2

3

We also use the following notations for the q-binomial coefficients[α0

]q

= 1,[αn

]q

=(1− qα)(1− qα−1) . . . (1− qα−n+1)

(q; q)n, n ∈ Z+, α ∈ R. (1.1.4)

For γ ∈ C, aqγ 6= q−n, n ∈ N, we define (a; q)γ to be

(a; q)γ :=(a; q)∞

(aqγ ; q)∞. (1.1.5)

The θ-function is defined for z ∈ C\ 0 , 0 < |q| < 1 to be

θ(z; q) :=∞∑

n=−∞qn2

zn. (1.1.6)

The following identity is introduced by C.G.J. Jacobi in 1829, and it is called Jacobi triple

product identity, see [53]

∞∑n=−∞

qn2zn =

(q2,−qz,−qz−1; q2

)∞, z ∈ C\ 0 , 0 < |q| < 1. (1.1.7)

Therefore θ(z; q) has only real and simple zeros at the pointsq2m+1, m ∈ Z

. The following

formula was obtained by Euler (1748) and by Gauss (1876), cf. [47, p. 67]

m∑n=0

(−1)n[mn

]qqn(n−1)/2un = (u; q)m. (1.1.8)

Let rφs denote the q-Hypergeometric series

rφs

a1, . . . , ar

b1, . . . , bs

∣∣∣∣∣∣∣ q, z =

∞∑n=0

(a1, . . . , ar; q)n

(q, b1, . . . , bs; q)nzn(−q(n−1)/2)n(s+1−r). (1.1.9)

The q-Gamma function, [53,71], is defined by

Γq(z) :=(q; q)∞(qz; q)∞

(1− q)1−z, z ∈ C, 0 < |q| < 1. (1.1.10)

Here we take the principal values of qz and (1−q)1−z. Then Γq(z) is a meromorphic function

with poles at z = −n, n ∈ N. Because Γq(z) has no zeros,1

Γq(z)is an entire function with

zeros at z = −n, n ∈ N. Γq(z) is the unique function that satisfies the functional equation

Γq(z + 1) =1− qz

1− qΓq(z), Γq(1) = 1, (1.1.11)

4

and logarithmically convex. See [14, 53]. Hence, Γq(n) = (q; q)n−1

/(1− q)n−1. The q-Beta

function is defined by

Bq(a, b) :=∫ 1

0xa−1(qx; q)b−1 dqx, a, b > 0, (1.1.12)

where the integration is defined in (1.1.31) below. Using the q-binomial theorem, cf. [14,

p. 488], we have∞∑

n=0

(a; q)n

(q; q)nzn =

(az; q)∞(z; q)∞

; |z| < 1, (1.1.13)

and therefore, cf. [14, p.494] we can prove that

Bq(a, b) =Γq(a)Γq(b)Γq(a+ b)

. (1.1.14)

The third type of q-Bessel functions is defined for z ∈ C by, cf. e.g. [63, 67,68],

J (3)ν (z; q) := zν (qν+1; q)∞

(q; q)∞1φ1

0

qν+1

∣∣∣∣∣∣∣ q; qz2

= zν (qν+1; q)∞

(q; q)∞

∞∑n=0

(−1)nqn(n+1)/2 z2n

(q; q)n(qν+1; q)n

(1.1.15)

This function is called in some literature the Hahn-Exton q-Bessel function, see [78, 112].

It is also called the 1φ1 q-Bessel function, cf. [77]. Since the other types of the q-Bessel

functions, i.e. J(1)ν (·; q), J (2)

ν (·; q), see e.g. [63, 69, 104], will not be used throughout this

thesis we use the notation Jν(·; q) for J (3)ν (·; q). The basic trigonometric functions cos(z; q)

and sin(z; q) are defined on C by

cos(z; q) :=∞∑

n=0

(−1)n qn2

(z(1− q))2n

(q; q)2n

=(q2; q2)∞(q; q2)∞

(zq−1/2(1− q))1/2J−1/2

(z(1− q)/

√q ; q2

),

(1.1.16)

sin(z; q) :=∞∑

n=0

(−1)n qn(n+1)(z(1− q))2n+1

(q; q)2n+1

=(q2; q2)∞(q; q2)∞

(z(1− q))1/2J1/2

(z(1− q); q2

).

(1.1.17)

5

They are q-analogues of the cosine and sine functions, [14, 53]. The basic hyperbolic

trigonometric functions cosh(z; q) and sinh(z; q) are defined for z ∈ C by

cosh(z; q) := cos(iz; q), sinh(z; q) := −i sin(iz; q). (1.1.18)

A q-analogue of the exponential function is the function

E(z; q) :=∞∑

n=0

qn2/4 zn

Γq(n+ 1), z ∈ C. (1.1.19)

One can easily see that

E(iz; q) = cos(z; q) + iq1/4 sin(z; q), (1.1.20)

and

E(|z|; q) = cosh(|z|; q) + q1/4 sin(|z|; q). (1.1.21)

F. H. Jackson [70, 1904] introduced the functions eq(z) and Eq(z) as q-analogues of the

exponential functions, where

Eq(z) :=∞∑

n=0

q(n2) zn

(q; q)n= (−z; q)∞, z ∈ C, (1.1.22)

eq(z) :=∞∑

n=0

zn

(q; q)n, |z| < 1. (1.1.23)

Eq(z) is an entire function with simple zeros at the points −q−n, n ∈ N. Since

eq(z)Eq(−z) ≡ 1, |z| < 1, (1.1.24)

see e.g. [53, 70], then the domain of the function eq(z) can be extended to C by defining

eq(z), z ∈ C, to be

eq(z) :=1

∞∏n=0

(1 + qnz

) . (1.1.25)

Hence, the relation (1.1.24) holds in C, and the function eq(z) has simple poles at the points

−q−n, n ∈ N. Let the basic trigonometric functions sinq z, cosq z, Sinqz, and Cosqz be

6

defined by

sinq z :=eq(iz)− eq(−iz)

2i, cosq z :=

eq(iz)− eq(−iz)2

, |z| < 1, (1.1.26)

Sinqz :=Eq(iz)− Eq(−iz)

2i, Cosqz :=

Eq(iz)− Eq(−iz)2

, z ∈ C. (1.1.27)

In the next section we prove that the functions cos(z; q) and sin(z; q) have only real and

simple zeros, which will be denoted throughout this thesis by ±xm∞m=1, and 0,± ym∞m=1

respectively, where xm and ym, m > 1, are the positive zeros, cf. [7, 28,78].

If µ ∈ R, a subset A of R is called a µ-geometric set if µx ∈ A for all x ∈ A. If a subset

A of R is a µ-geometric, then it contains all geometric sequences xµn∞n=0, x ∈ A. Let

f be a function, real or complex valued, defined on a q-geometric set A. The q-difference

operator is defined by

Dqf(x) :=f(x)− f(qx)

x− qx, x ∈ A\ 0 . (1.1.28)

If 0 ∈ A, the q-derivative at zero is defined by

Dqf(0) := limn→∞

f(xqn)− f(0)xqn

, x ∈ A\ 0 , (1.1.29)

if the limit exists and does not depend on x. In some literature the q-derivative at zero is

defined to be f ′(0) if it exists, cf. [78,112], but the above definition is more suitable for our

approaches. Also the q−1-derivative of a function f(x) at zero is defined by

Dq−1f(0) := limn→−∞

f(xq−n)− f(0)xq−n

= Dqf(0), x ∈ A\ 0 , (1.1.30)

provided that the limit exists and does not depend on x. A right inverse to Dq, the Jackson

q-integration, cf. [72], is∫ x

0f(t) dqt := x(1− q)

∞∑n=0

qnf(xqn), x ∈ A, (1.1.31)

provided that the series converges, and∫ b

af(t) dqt :=

∫ b

0f(t) dqt−

∫ a

0f(t) dqt, a, b ∈ A. (1.1.32)

7

If A is q−1-geometric, then the q-integration over [x,∞), x ∈ A, is defined by∫ ∞

xf(t) dqt :=

∞∑n=1

xq−n(1− q)f(xq−n). (1.1.33)

Lemma 1.1.1. The q-integrals (1.1.31) and (1.1.33) exist if

limn→∞

xqnf(xqn) = 0 and limn→∞

xq−nf(xq−n) = 0,

respectively.

Proof. Based on the facts that, for x ∈ C

limk→∞

xqkf(xqk) = 0 =⇒ ∃α ∈ [0, 1) ∃C > 0, |f(xqk)| 6 C|xqk|−α, k ∈ N, (1.1.34)

limk→∞

xq−kf(xq−k) = 0 =⇒ ∃α ∈ (1,∞) ∃C > 0, |f(xq−k)| 6 C|xq−k|−α, k ∈ N. (1.1.35)

Kac and Cheung [75, p. 68] have proved that if xαf(x) is bounded on [0, a] for some

0 6 α < 1, then∫ x

0f(t) dqt exists for all x ∈ [0, a].

Definition 1.1.1. Let f be a function defined on a q-geometric set A. We say that f is

q-integrable on A if and only if∫ x

0f(t) dqt exists for all x ∈ A.

Definition 1.1.2. A function f which is defined on a q-geometric set A, 0 ∈ A, is said to

be q-regular at zero if

limn→∞

f(xqn) = f(0), for every x ∈ A.

Theorem 1.1.2 below, which we call a fundamental theorem of q-calculus, shows that the

q-integral operator and the q-difference operator are, in certain sense, inverse operations.

Theorem 1.1.2. Let f : [0, b] −→ C be q-regular at zero. Define

F (x) =∫ x

cf(t) dqt, 0 6 c 6 b. (1.1.36)

8

Then F is q-regular at zero. Furthermore, DqF (x) exists for every x ∈ [0, b] and

DqF (x) = f(x) for every x ∈ [0, b].

Conversely if 0 6 a 6 b, then ∫ b

aDqf(t) dqt = f(b)− f(a). (1.1.37)

Proof. See [9].

It is worth mentioning that in Theorem 1.1.2 if f is not q-regular at zero, then∫ b

0Dqf(t) dqt = f(b)− lim

n→∞f(bqn). (1.1.38)

Consequently (1.1.37) would take the form∫ b

aDqf(t) dqt =

[f(b)− lim

n→∞f(bqn)

]−[f(a)− lim

n→∞f(aqn)

]. (1.1.39)

The non-symmetric Leibniz’ rule is

Dq(fg)(x) = g(x)Dqf(x) + f(qx)Dqg(x). (1.1.40)

Relation (1.1.40) can be symmetrized using the relation f(qx) = f(x) − x(1 − q)Dqf(x),

giving the additional term x(1− q)Dqf(x)Dqg(x). Also the rule of q-integration by parts is∫ a

0g(x)Dqf(x) dqx = (fg)(a)− lim

n→∞(fg)(aqn)−

∫ a

0Dqg(x)f(qx) dqx. (1.1.41)

If f, g are q-regular at zero, then limn→∞(fg)(aqn) on the right hand side of (1.1.41) will

be replaced by (fg)(0).

In the following we define Hilbert spaces where a q-analogue of Sturm–Liouville problems

will be considered. Let L2q(0, a) be the space of all complex valued functions defined on [0, a]

such that

‖f‖ :=(∫ a

0|f(x)|2 dqx

)1/2

<∞. (1.1.42)

9

The space L2q(0, a) is a separable Hilbert space with the inner product

〈f, g〉 :=∫ a

0f(x)g(x) dqx, f, g ∈ L2

q(0, a), (1.1.43)

and the orthonormal basis

ϕn(x) =

1√

x(1−q), x = aqn,

0, otherwise,(1.1.44)

n = 0, 1, 2, . . ., cf. [15]. The space L2q

((0, a) × (0, a)

)is the space of all complex valued

functions f(x, t) defined on [0, a]× [0, a] such that

‖f(·, ·)‖2 :=(∫ a

0

∫ a

0|f(x, t)|2 dqxdqt

)1/2

<∞. (1.1.45)

The elements of L2q

((0, a)× (0, a)

)are equivalence classes where f, g are in the same equiv-

alence class if f(aqm, aqn) = g(aqm, aqn), m, n ∈ N. The zero element is the equivalence

class of all functions f(x, t) which satisfy f(aqm, aqn) = 0, for all m,n ∈ N.

Lemma 1.1.3. L2q

((0, a)× (0, a)

)is a separable Hilbert space with the inner product

〈f, g〉2 :=∫ a

0

∫ a

0f(x, t)g(x, t) dqxdqt. (1.1.46)

Proof. Similar to [15, pp. 217-218], L2q

((0, a) × (0, a)

)is a Banach space. To prove

separability, it suffices to prove that

φij(x, t) := φi(x)φj(t), i, j = 1, 2, . . . , (1.1.47)

is an orthonormal basis of L2q

((0, a) × (0, a)

)whenever φi(·)∞i=1 is an orthonormal basis

of L2q(0, a). Indeed,

〈φjk, φmn〉2 =∫ a

0

∫ a

0φj(x)φk(t)φm(x)φn(t) dqx dqt

=∫ a

0φj(x)φm(x) dqx

∫ a

0φk(t)φn(t) dqt = δjmδkn,

10

proving orthogonality. To prove that φij is a basis, we prove that if there exists

f ∈ L2q

((0, a)× (0, a)

)such that 〈f, φij〉2 = 0, then f is the zero element. Indeed,

0 = 〈f, φij〉 =∫ a

0

∫ a

0f(x, t)φi(x)φj(t) dqx dqt

=∫ a

0φj(t)

(∫ a

0f(x, t)φi(x) dqx

)dqt =

∫ a

0h(t)φj(t) dqt.

Thus

h(t) :=∫ a

0f(x, t)φi(x) dqx (1.1.48)

is orthogonal to the φj ’s which implies that h(aqn) = 0, for all n ∈ N. So, f(x, aqn) is

orthogonal to each φi. Consequently, f(aqm, aqn) = 0, for all m,n ∈ N.

By L1q(0, a), a > 0, we mean the Banach space of all complex valued functions defined

on (0, a] such that

‖f‖ :=∫ a

0|f(t)| dqt <∞. (1.1.49)

Let L1q(0, a) denote the space of all functions f defined on (0, a] such that f ∈ L1

q(0, x) for

all x ∈ (qa, a].

1.2 Asymptotics of zeros of cos(z; q) and sin(z; q)

There are studies on the zeros of q-trigonometric functions, see references below. In this

section we establish other formulae for the asymptotic behavior of the zeros of cos(z; q) and

sin(z; q), i.e. xm and ym respectively. Our asymptotic formulae fit with the study for the

eigenvalues of q-Sturm–Liouville problems as well as for deriving a q-analogue of a theorem

of G. Polya. We also state and prove some inequalities for expressions involving xm and

ym. We first state the asymptotic formulae in the existing literature, namely [7, 28]. We

start with the following preliminaries concerning entire functions, cf. e.g [27, 61, 84]. Let

f, g be entire functions , we say that

f(z) = O(g(z)

), as z →∞, (1.2.1)

11

if f(z)/g(z) is bounded in a neighborhood of ∞. On the other hand we write

f(z) ∼ g(z), as z →∞, if limz→∞

|f(z)||g(z)|

= 1. (1.2.2)

Definition 1.2.1. Let f(z) :=∑∞

n=0 anzn be an entire function. The maximum modulus

is defined by

M(r; f) := sup |f(z)| : |z| = r , r > 0. (1.2.3)

The order of f , ρ(f), is defined by, cf. [27, 84]

ρ(f) := lim supr→∞

log logM(r, f)log r

= lim supn→∞

n log nlog a−1

n. (1.2.4)

Theorem 1.2.1. [27] If ρ(f) is finite and is not equal to a positive integer, then f has

infinitely many zeros, or is a polynomial.

One can easily see that Jν(·; q2) is an entire functions of order zero. Therefore, from

Theorem 1.2.1, Jν(·; q2) has infinitely many zeros. Koelink and Swarttouw proved that

Jν(·; q2) has only real and simple zeros, cf. [78, 1994]. Let w(ν)m , ν > −1, denote the positive

zeros of Jν(·; q2) in an increasing order of m ∈ Z+. In [7] L.D. Abreu et al. proved that if

q2ν+2 < (1− q2)2, then

w(ν)m (q) = q−m+2εm(ν), 0 < εm(ν) <

log(1− q2m+2ν

1−q2m

)2 log q

,

∞∑m=0

εm(ν) =log(q2ν+2; q2)∞

4 log q.

(1.2.5)

Moreover w(ν)m ∼ q−m when m → ∞ without the restriction q2ν+2 < (1 − q2)2. Applying

these formulae to the cases ν = ∓1/2 respectively and using (1.1.16) and (1.1.17), we obtain

the following estimates

if q < (1− q2)2, then xm =q−m+1/2+2ε

(−1/2)m

1− q, m > 1, (1.2.6)

if q3 < (1− q2)2, then ym =q−m+2ε

(1/2)m

1− q, m > 1. (1.2.7)

12

Moreover

xm ∼ q−m+1/2

(1− q), ym ∼ q−m

(1− q)as m→∞ (1.2.8)

without further restrictions on q. The results for ym coincide with those of Bustoz and

Cardoso [28]. A study of asymptotic formulae of zeros of another class of basic trigonometric

functions is given in [110, 111]. This section also includes some results involving xm, ym,

which will be needed for the derivation of the asymptotic behavior of eigenvalues of the

basic Sturm–Liouville problems. In the following we introduce some results of Bergweiler

et al. [26], which are essential in our investigations. These results are concerned with the

functional equationl∑

j=0

aj(z)f(cjz) = Q(z), (1.2.9)

where c ∈ C, 0 < |c| < 1, is fixed, Q and the aj ’s are polynomials. Set

pj := deg (aj), and dj := pj − p0, j = 0, 1, . . . , l. (1.2.10)

The Newton–Puiseux diagram associated with equation (1.2.9), P , is defined to be the

convex hull ofl⋃

j=0

(x, y) ∈ R2 : x > j and y 6 dj

. (1.2.11)

Let K > 0 be the number of vertices of P . Thus the vertices of P are (jk, djk), k = 0, . . . ,K,

where

0 = j0 < j1 < . . . < jK 6 l.

It is proved in [26, Theorem 1.1] that if (1.2.9) has a transcendental entire solution, then

there exists j ∈ 1, . . . , l such that dj > 1. In other words, in this case the Newton–Puiseux

diagram of (1.2.9) has at least one vertex which is not (0, 0), i.e. K > 1. For k = 1, . . . ,K

set

σk :=djk

− djk−1

jk − jk−1.

13

Then σ1 > σ2 > . . . > σK > 0. The σk’s are the slopes of the segments which together with

the negative part of the y-axis form the boundary of P . It is shown in [26, Theorem 1.2]

and also in [105] that if f is an entire transcendental solution of (1.2.9), then there exists

k0 ∈ 1, . . . ,K such that

logM(r; f) ∼ σk0

−2 log |c|(log r)2 as r −→∞. (1.2.12)

The following example is a special case of (1.2.9) and is important in our study since the

solutions are nothing but basic trigonometric functions defined in (1.1.16) and (1.1.17).

Example 1.2.1. Consider the functional equation

qy(z) +(− (1 + q) + q2z2(1− q)2

)y(qz) + y(q2z) = 0, z ∈ C. (1.2.13)

The functions cos(z; q), sin(z; q) are transcendental entire solutions of (1.2.13). Using the

previous notations we have

l = 2, a0(z) = q, a1(z) =(− (1 + q) + q2z2(1− q)2

), a2(z) = 1, Q(z) = 0. (1.2.14)

Therefore

p0 = p2 = 0, p1 = 2, d0 = d2 = 0, d1 = 2. (1.2.15)

Hence P is the minimal convex set containing

(x, y) ∈ R2 : x > 0, y 6 0

∪(x, y) ∈ R2 : x > 1, y 6 2

∪(x, y) ∈ R2 : x > 2, y 6 0

. (1.2.16)

Therefore P is the set in R2 whose boundary consists of the negative part of the y-axis, the

line segment y = 2x, 0 6 x 6 1 and the line y = 2, x > 1. This convex hull is illustrated in

Figure 1 below. Clearly P has only two vertices (0, 0) and (1, 2). Thus there is only one line

segment contained in the boundary of P with a slope, σ1, σ1 = 2. So from (1.2.12) it follows

that the functions cos(z; q), sin(z; q) and E(z; q) have the following asymptotic behavior as

r →∞

logM(r; cos(z; q)

), logM

(r; sin(z; q)

), logM(r;E(z; q)) ∼ 1

− log q(log r)2. (1.2.17)

14

Figure 1.1: The Newton-Puiseux diagram P of (1.2.13).

Notice that the asymptotic of E(z; q) is obtained via (1.1.20).

The following lemma is taken from [25] and will be needed in the sequel. It is concerned

with the asymptotic behavior of the θ-function defined in (1.1.6).

Lemma A. Suppose that 0 < |q| < 1 and that z ∈ C, |z| > 1. For m ∈ Z+ let Am be the

annulus defined by

Am :=z ∈ C, q−2m+2 6 |z| < |q|−2m

. (1.2.18)

Then we have, uniformly as m→∞,

log |θ(z; q)| = −(log |z|)2

4 log |q|+ log |1 + q2m−1z|+O(1), z ∈ Am. (1.2.19)

Using the technique of Bergweiler and Hayman [25], we will derive the asymptotic

formulae for the zeros of cos(z; q) and sin(z; q) in the following theorem.

Theorem 1.2.2. If xm are the positive zeros of cos(z; q) and ym are the positive zeros

15

of sin(z; q), then we have for sufficiently large m,

xm = q−m+1/2(1− q)−1(1 +O(qm)

), (1.2.20)

ym = q−m(1− q)−1(1 +O(qm)

). (1.2.21)

Proof. We prove the theorem only for the zeros of cos(z; q) since the proof for the zeros of

sin(z; q) is similar. Let R(z) be the difference

R(z) := cos(z; q)− 1(q; q)∞

θ(−z2(1− q)2; q), z ∈ C\ 0 . (1.2.22)

Hence

R(z) =∞∑

k=−∞δ2kz

2k, δ2k :=

(−1)k qk2

(1− q)2k

(q; q)∞

((q2k+1; q)∞ − 1

), k > 0,

(−1)k+1 qk2

(1− q)2k

(q; q)∞, k < 0.

(1.2.23)

We shall prove that

(q2k+1; q)∞ − 1 = O(q2k) as k −→ +∞. (1.2.24)

Indeed from (1.1.2), we have for k ∈ Z+

∣∣∣(q2k+1; q)∞ − 1∣∣∣ =

∣∣∣∣∣∣∞∑

j=1

(−1)jqj(j−1)/2 q(2k+1)j

(q; q)j

∣∣∣∣∣∣ 6 q2k+1∞∑

j=1

qj(j−1)/2

(q; q)j

6q2k+1

(q; q)∞

∞∑j=1

q(j−1)/2 =q2k+1

(1−√q)(q; q)∞. (1.2.25)

This proves (1.2.24). We conclude from (1.2.23) and (1.2.25) that

|δ2k| 6qk2+2k(1− q)2k

(1−√q)(q; q)2∞, k ∈ Z. (1.2.26)

Thus for z ∈ C\ 0,

|R(z)| 6 1(1−√q)(q; q)2∞

∞∑−∞

qk2+2k(1− q)2k|z2k|

=1

(1−√q)(q; q)2∞θ(q2(1− q)2|z|2; q).

(1.2.27)

16

Taking the logarithm of both sides of (1.2.27) yields

log |R(z)| 6 log θ(q2(1− q)2|z|2; q) + log1

(1−√q)(q; q)2∞, z ∈ C\ 0 . (1.2.28)

We deduce from (1.2.19) that there exists m0 ∈ N and a constant C > 0 such that if

q−2m+2 6 |z| < q−2m, m > m0,

then

−(log |z|)2

4 log q+log |1+q2m−1z|−C 6 log |θ(z; q)| 6 −(log |z|)2

4 log q+log |1+q2m−1z|+C. (1.2.29)

Thus if

q−2m+2 6 q2|z|2(1− q)2 < q−2m, m > m0, (1.2.30)

then replacing z in (1.2.29) by q2(1− q)2|z|2, we obtain for m > m0

log |θ(q2(1− q)2|z|2; q)| 6 −(log |z|2(1− q)2q2)2

4 log q+log

∣∣1 + q2m+1(1− q)2|z|2∣∣+C. (1.2.31)

Since log(1 + x) < x, x > 0, then (1.2.30) implies

log∣∣1 + q2m+1(1− q)2|z|2

∣∣ 6 q2m+1(1− q)2|z|2 6 q−1. (1.2.32)

Substituting in (1.2.31), we obtain for z ∈ C, which satisfies (1.2.30),

log |θ(q2(1− q)2|z|2; q)| 6 −(log |z|(1− q)q)2

log q+ q−1 + C. (1.2.33)

Consequently, by (1.2.28), for z satisfies (1.2.30)

log |R(z)| 6 −(log |z|(1− q)q

)2log q

+ q−1 + C + log1

(1−√q)(q; q)2∞

= −(log |z|(1− q)

)2log q

− log |z|2 + C1,

(1.2.34)

where

C1 := C − log(1−√q)(q; q)2∞ − 2 log(1− q)− log q + q−1. (1.2.35)

17

Notice that (1.2.34) holds uniformly for z satisfies (1.2.30), m > m0 and C1 is positive. Let

Am, m > m0, be the annulus defined by

Am :=z ∈ C : q−2m 6 |z|2(1− q)2 < q−2m−2

. (1.2.36)

The other side of (1.2.29) implies for z ∈ Am

−(log r(1− q)

)2log q

6 log(|θ(−(1− q)2z2; q)|

)− log

∣∣1− q2m+1z2(1− q)2∣∣+ C. (1.2.37)

Therefore by (1.2.34), z ∈ Am,

log |R(z)| 6 log((q; q)−1

∞ |θ(−(1−q)2z2; q)|)−log

∣∣1−q2m+1z2(1− q)2∣∣−log r2+C2, (1.2.38)

where C2 = C + C1 + log(q; q)∞. If z ∈ ∂Am, m > m0, then

|1− q2m+1z2(1− q)2| > q2m+1|z|2(1− q)2 − 1 = q−1(1− q), |z| = q−m−1

1− q, (1.2.39)

|1− q2m+1z2(1− q)2| > 1− q2m+1|z|2(1− q)2 = (1− q), |z| = q−m

1− q. (1.2.40)

That is

log |1− q2m+1z2(1− q)2| > log(1− q), z ∈ ∂Am. (1.2.41)

Substituting in (1.2.38) we obtain

log |R(z)| 6 log((q; q)−1

∞∣∣θ(−(1− q)2z2; q)

∣∣)− log(1− q)− log r2 +C2, z ∈ ∂Am. (1.2.42)

We can also choose m0 sufficiently large such that − log(1− q)− log r2 + C2 < 0. Hence

|R(z)| 6((q; q)−1

∞ |θ(−(1− q)2z2; q)|), z ∈ ∂Am, m > m0. (1.2.43)

Hence an application of Rouche theorem shows that cos(z; q) and θ(−(1−q)2z2; q) have the

same number of zeros in Am. Since θ(−(1−q)2z2; q) has two simple symmetric zeros in Am,

we deduce that cos(z; q) has exactly two zeros in Am. Since cos(z; q) is an even function,

18

then these two zeros are real, symmetric and simple. Assume now that xm, m > m0, is the

positive zero of cos(z; q) in Am. Then from (1.2.22)

|R(xm)| =∣∣((q; q)−1

∞ |θ(−(1− q)2x2m; q)|

)∣∣and we obtain from (1.2.38) that

log∣∣1− q2m+1x2

m(1− q)2∣∣ 6 −2 log xm + C2. (1.2.44)

In other words

log∣∣1− qm+(1/2)xm(1− q)

∣∣+ log∣∣1 + qm+(1/2)xm(1− q)

∣∣ 6 −2 log xm + C2. (1.2.45)

Thus, either

log∣∣1− qm+(1/2)xm(1− q)

∣∣ 6 − log xm +C2

2, (1.2.46)

or

log∣∣1 + qm+(1/2)xm(1− q)

∣∣ 6 − log xm +C2

2. (1.2.47)

We can assume without any loss of generality that (1.2.46) holds. Then for all m > m0

∣∣1− qm+(1/2)xm(1− q)∣∣ = O(x−1

m ). (1.2.48)

This implies that xm ∼ q−m−1/2(1− q)−1 for sufficiently large m. More precisely, we have

xm = q−m−1/2(1− q)−1(1 +O(qm)

), as m→∞. (1.2.49)

Lemma 1.2.3. As m −→∞, we have

∣∣q2k x2m+k

x2m

− 1∣∣ = O(qm), k ∈ N. (1.2.50)

The O-term is uniform in k ∈ N.

19

Proof. From (1.2.20) we see that for m→∞

xm =q−m+1/2

1− q

(1 +O(qm)

). (1.2.51)

Hence there exists a constant C > 0 and m0 ∈ N such that

|qm−1/2(1− q)xm − 1| 6 Cqm, for all m > m0. (1.2.52)

Consequently for all k ∈ N we have

|qm+k−1/2(1− q)xm+k − 1| 6 Cqm+k, for all m > m0. (1.2.53)

We can choose m0 sufficiently large such that 1−Cqm > 0 for all m > m0. So from (1.2.52)

and (1.2.53) we obtain

(1− Cqm+k)2

(1 + Cqm)26 q2k x

2m+k

x2m

6(1 + Cqm+k)2

(1− Cqm)2, m > m0. (1.2.54)

Thus

−C(1 + qk)(2 + Cqm(1− qk))(1 + Cqm)2

6

(q2k x

2m+k

x2m

− 1)q−m 6

C(1 + qk)(2− Cqm(1− qk))(1− Cqm)2

. (1.2.55)

That is−4C(1 + Cqm)

(1 + Cqm)26(q2k x

2m+k

x2m

− 1)q−m 6

4C(1 + Cqm)(1− Cqm)2

, (1.2.56)

holds for all k ∈ N and m > m0. Hence∣∣∣∣∣(q2k x2m+k

x2m

− 1)q−m

∣∣∣∣∣ 6 max

4C(1 + Cqm)(1 + Cqm)2

,4C(1 + Cqm)(1− Cqm)2

=

4C(1 + Cqm)(1− Cqm)2

, (1.2.57)

for all m > m0 and k ∈ N. Thus the sequence(q2k x2

m+k

x2m

− 1)q−m

∞m=1

is uniformly

bounded for k ∈ N since

4C(1 + Cqm)(1− Cqm)2

−→ 4C as m −→∞. (1.2.58)

20

Lemma 1.2.4. For n ∈ N, if z ∈ C and x ∈ R are such that |zx| > q−n

1− q, then for r ∈ Z+,

we haver∏

m=1

x2m + q2n|z|2x2

x2m + |z|2x2

6 q2nrr∏

m=1

(1 + x2m). (1.2.59)

Proof. Assume that z ∈ C, x ∈ R such that |zx| > q−n

1− q, n ∈ N. Then

x2m + q2n|z|2x2

x2m + |z|2x2

− q2n =(1− q2n)x2

m

x2m + |zx|2

6x2

m

|zx|26 q2nx2

m. (1.2.60)

Consequently

r∏m=1

x2m + q2n|z|2x2

x2m + |z|2x2

6r∏

m=1

q2n(1 + x2m) = q2nr

r∏m=1

(1 + x2m), (1.2.61)

proving the lemma.

Similarly we have

Lemma 1.2.5. For n ∈ N, if z ∈ C and y ∈ R are such that |zy| > q−n

1− q, then for r ∈ Z+,

we haver∏

m=1

y2m + q2m|z|2y2

y2m + |z|2y2

6 q2nrr∏

m=1

(1 + y2m). (1.2.62)

1.3 Asymptotics of cos(z; q) and sin(z; q)

In this section, we study the asymptotic behavior of the basic trigonometric functions

cos(z; q), and sin(z; q). The results of this section are contained in [19]. We start our investi-

gations with the following lemma that exhibits a major difference between the trigonometric

functions and their basic analogues.

Lemma 1.3.1. The basic trigonometric functions sin(x; q) and cos(x; q) are not bounded

on R.

21

Proof. Daalhuis in [38] and Chen et al. in [35] proved that Jν(x; q2) has the asymptotic

behavior

Jν(x; q2) ∼|x|ν(x2q2; q2)∞

(q2; q2)∞, as |x| → ∞. (1.3.1)

But (x2; q2)∞ is not bounded on R. Indeed, take

rn := q−n−1 (1 + q)2

, n ∈ Z+. (1.3.2)

Then

q−n < rn < q−n−1, n ∈ Z+. (1.3.3)

Therefore

∞∏k=n+2

(1− r2nq2k) >

∞∏k=n+2

(1− q2k−2n−2) =∞∏

j=1

(1− q2j) = (q2; q2)∞, (1.3.4)

andn−1∏k=0

(1− r−2n q−2k) >

n−1∏k=0

(1− q2n−2k) = (q2; q2)n > (q2; q2)∞. (1.3.5)

But

∣∣(r2n; q2)∞∣∣ = ∞∏

k=0

|1− r2nq2k|

= (r2nq2n − 1)(1− r2nq

2n+2)n−1∏k=0

|1− r2nq2k|

∞∏k=n+2

(1− r2nq2k)

=q−2

16(1 + 2q − 3q2)(3− 2q − q2)r2n

n qn(n−1)n−1∏k=0

(1− r−2n q−2k)

∞∏k=n+2

(1− r2nq2k).

(1.3.6)

Combining equations (1.3.3)–(1.3.6) yields

∣∣(r2n; q2)∞∣∣ > q−2

16(1 + 2q − 3q2)(3− 2q − q2)(q2; q2)2∞q

n2−nr2nn

=q−2

16(1 + 2q − 3q2)(3− 2q − q2)(q2; q2)2∞q

−n2−3n

(1 + q

2

)2n

.

(1.3.7)

But (1 + q

2

)2n

> q2n, for all n ∈ N. (1.3.8)

22

Thus

|(r2n; q2)∞| >q−2

16(1 + 2q − 3q2)(3− 2q − q2)(q2; q2)2∞q

−n2−n. (1.3.9)

Hence ∣∣(r2n; q2)∞∣∣ −→∞ as n −→∞.

This proves that (x2; q2)∞ is not bounded on R. By (1.3.1), Jν(x; q2) is also not bounded on

R. Applying this result to the cases ν = ∓1/2 respectively and using (1.1.16) and (1.1.17),

we deduce that the functions cos(x; q) and sin(x; q) are not bounded on R.

The following two lemmas will be needed in the sequel.

Lemma 1.3.2. For a positive sequence zm∞m=1, if zm → 1 as m→∞, then infm∈N

zm > 0.

Proof. Since zm → 1 as m → ∞, then for 0 < ε < 1, there exists m0 ∈ N such that

|zm−1| < ε for all m > m0. Hence zm > 1−ε for all m > m0. That is infm>m0

zm > 1−ε > 0.

Therefore

infm∈N

zm = minz1, z2, . . . , zm0 , inf

m>m0

zm

> 0.

Lemma 1.3.3. If z ∈ C such that 0 6 |z| < 1, then

∣∣ log |1− z|∣∣ 6 log

11− |z|

. (1.3.10)

Proof. Let z ∈ C such that 0 6 |z| < 1. Since |1− z| > 1− |z|, then we have

log |1− z| > log(1− |z|) = − log1

1− |z|. (1.3.11)

Also |1− z| 6 1 + |z|, then

log |1− z|(1− |z|) 6 log(1− |z2|) 6 0.

Consequently

log |1− z| 6 − log(1− |z|) = log1

(1− |z|). (1.3.12)

23

Combining (1.3.11) and (1.3.12) we obtain (1.3.10).

In Lemma A the asymptotic behavior of log |θ(z; q)| as z →∞ is introduced by studying

the asymptotic behavior of log |θ(z; q)| in the set of annuli Am∞m=1 defined in (1.2.18) when

m is large enough. Similarly we shall study the asymptotic behavior of log | cos(z; q)| and

log | sin(z; q)|. We begin with the q-cosine function. We divide the complex plane except

for a finite disk D with center at zero into an infinite set of annuli Acm

∞m=1, where the

disk D will be determined later. Then we study the asymptotic behavior of log | cos(z; q)|

as z →∞ via the study of its behavior in each Acm, for sufficiently large m. Let

αm := log (xm/xm+1)/ log q, m ∈ Z+. (1.3.13)

Then αm > 0 for all m ∈ Z+. Moreover, from (1.2.20)

αm = 1 +(

log(

1 +O(qm)1 +O(qm+1)

))/log q,

for sufficiently large m. That is limm→∞

αm = 1. Then from Lemma 1.3.2

0 < α := infm∈Z+

αm 6 1. (1.3.14)

We define two sequences, (am)∞m=1, (bm)∞m=1, to be

am :=

αm + α

2, if αm 6= α,

α2 , if αm = α,

(1.3.15)

and

b1 :=α

2, bm+1 :=

αm − α

2, if αm 6= α,

α2 , if αm = α,

(1.3.16)

where m > 1. Then we have

xmq−am = xm+1q

bm+1 , m > 1. (1.3.17)

Thus the set of annuli Acm

∞m=1 where Ac

m is defined to be

Acm :=

z ∈ C : xmq

bm 6 |z| < xmq−am

, m > 1 (1.3.18)

24

Figure 1.2: The annulus Ac2

divide the regionz ∈ C : |z| > qα/2x1

.

In the following we shall study the asymptotic behavior of log | cos(z; q)| in each of the

annuli described above when m is sufficiently large.

Theorem 1.3.4. Let α be the positive number defined in (1.3.14). Assume that |z| > qα/2x1

and let Acm, m > 1, be the annulus defined by (1.3.18). Then we have, uniformly as m→∞,

the asymptotic relation

log | cos(z; q)| = −(log |z|)2

log q+ log |1− z2

x2m

|+O(1), z ∈ Acm. (1.3.19)

Proof. First of all, it should be noted that z ∈ Acm, m > 1, if and only if

|z| = xmqgm(t), gm(t) := −(am + bm)t+ bm, t ∈ [0, 1). (1.3.20)

Let m > 1 and z ∈ Acm be fixed. Then there exists t ∈ [0, 1) such that

r := |z| = xmqgm(t) = xmq

−(am+bm)t+bm . (1.3.21)

25

Since cos(z; q) is an entire function of order zero, then, see e.g. [27],

cos(z; q) =∞∏

n=1

(1− z2

x2n

). (1.3.22)

Thus

log

∣∣∣∣∣ cos(z; q)1− ( z

xm)2

∣∣∣∣∣ =m−1∑n=1

log∣∣∣∣1− z2

x2n

∣∣∣∣+ ∞∑n=m+1

log∣∣∣∣1− z2

x2n

∣∣∣∣=

m−1∑n=1

log | zxn|2 +

m−1∑n=1

log∣∣∣∣1− x2

n

z2

∣∣∣∣+ ∞∑n=m+1

log∣∣∣∣1− z2

x2n

∣∣∣∣ .(1.3.23)

Set

γn := qn− 12 (1− q)xn − 1, n = 1, 2, . . . . (1.3.24)

From (1.2.20), γn = O(qn) as n→∞. Define the constants l, L to be

l := infn∈N

|1 + γn|, L := supn∈N

|1 + γn|. (1.3.25)

Since 1 + γn > 0, for all n ∈ N and 1 + γn → 1 as n→∞, then l > 0. We define constants

K1 and K2 as follows

K1 :=L

l, K2 := q−a; a = sup

n∈Z+

an. (1.3.26)

Then qgm(t) 6 q−am 6 K2 for all t ∈ [0, 1). From (1.3.24), we have for n > m+ 1∣∣∣∣ zxn

∣∣∣∣ = xmqgm(t)

xn=qn−m 1 + γm

1 + γnqgm(t)

6 K1q−amqn−m 6 K1K2q

n−m.

(1.3.27)

Consequently

∞∑n=m+1

log∣∣1− z2

x2n

∣∣ 6∞∑

n=m+1

log

(1 +

∣∣∣∣ zxn

∣∣∣∣2)

6∞∑

n=m+1

∣∣∣∣ zxn

∣∣∣∣26 K2

1K22

∞∑n=m+1

q2(n−m) 6 K21K

22

q2

1− q2. (1.3.28)

As for n = 1, 2, . . . ,m− 1,∣∣∣xn

z

∣∣∣ = xnq−gm(t)

xm= qm−n 1 + γn

1 + γmq−gm(t) 6 K1q

−bmqm−n. (1.3.29)

26

From (1.3.15) and (1.3.16) we obtain

q−bm = qαq−am−1 6 qαq−a = K2qα. (1.3.30)

Combining (1.3.29) and (1.3.30) yields

∣∣∣xn

z

∣∣∣ 6 K1K2qm−n. (1.3.31)

Moreover ∣∣xn

z

∣∣ = xnq−gm(t)

xm6xm−1q

−gm(t)

xm, n = 1, 2, . . . ,m− 1. (1.3.32)

But from (1.3.17)

xm−1

xm= qam−1+bm , and q−gm(t) = qamt−(1−t)bm 6 q−bm .

Then ∣∣∣xn

z

∣∣∣ 6 qam−1 < qα < 1, z ∈ Acm. (1.3.33)

So from (1.3.10) we obtain

∣∣∣∣log∣∣∣∣1− x2

n

z2

∣∣∣∣∣∣∣∣ 6 log1(

1− |x2n

z2 |) = log

1 +∞∑

j=1

|xn

z|2j

= log

(1 +

|xnz |

2

1− |xnz |2

)6

|xnz |

2

1− |xnz |2

6|xn

z |2

1− q2α

6K2

1K22

1− q2αq2(m−n),

(1.3.34)

Therefore ∣∣∣∣∣m−1∑n=1

log∣∣∣∣1− x2

n

z2

∣∣∣∣∣∣∣∣∣ 6 K2

1K22

1− q2α

m−1∑n=1

q2(m−n) 6K2

1K22q

2

(1− q2α)(1− q2). (1.3.35)

Hence from (1.3.23), (1.3.28), and (1.3.35) we obtain

log

∣∣∣∣∣ cos(z; q)1− ( z

xm)2

∣∣∣∣∣ =m−1∑n=1

log | zxn|2 +O(1), as z →∞. (1.3.36)

27

Now

m−1∑n=1

log∣∣∣∣ zxn

∣∣∣∣2 =m−1∑n=1

log |z|2 −m−1∑n=1

log |xn|2

= 2(m− 1) log |z| − 2m−1∑n=1

(−n+ 1/2) log q − 2m−1∑n=1

log |1 + γn|

= 2(m− 1) log |z|+ (m− 1)2 log q − 2m−1∑n=1

log |1 + γn|. (1.3.37)

Since γn = O(qn) as n → ∞, then there exists a constant K3 > 0 and n0 ∈ N such that

|γn| 6 K3qn, for all n > n0. Let

K4 := maxq−1|γ1|, . . . , q−m0+1|γn0−1|,K3

.

Thus |γn| 6 K4qn for all n ∈ Z+, and

m−1∑n=1

log |1 + γn| 6m−1∑n=1

log(1 + |γn|

)6

m−1∑n=1

|γn| 6∞∑

n=1

|γn| 6 K4

∞∑n=1

qn 6 K4q

1− q.

(1.3.38)

Since r = q−m+1/2

(1−q) (1 + γm)qgm(t), then

m− 1 = − log rlog q

+ dm, dm := −1/2 +log(1 + γm)

log q+ gm(t)− log(1− q)

log q. (1.3.39)

A simple computation yields

2(m− 1) log r + (m− 1)2 log q = −(log r)2

log q+ d2

m log q = −(log r)2

log q+O(1), (1.3.40)

as m→∞ since dm∞m=1 is a bounded sequence. Hence the theorem follows by combining

(1.3.36), (1.3.37), and (1.3.40).

Corollary 1.3.5.

M(r; cos(z; q)) = O

((exp

(−(log |z|)2

log q

)), as z −→∞. (1.3.41)

28

Proof. From (1.3.19) we conclude that there exists m0 ∈ N such that for all m > m0 we

have

−(log r)2

log q+ log |1− z2

x2m

| − δ 6 log | cos(z; q)| 6 −(log r)2

log q+ log |1− z2

x2m

|+ δ, z ∈ Acm.

(1.3.42)

From (1.3.18) we obtain

q2bm 6|z|2

x2m

6 q−2am . (1.3.43)

Hence

log∣∣∣∣1− z2

x2m

∣∣∣∣ 6 log∣∣∣∣1 +

|z|2

x2m

∣∣∣∣ 6 log q−2am , m > 1, (1.3.44)

Since the sequence am∞m=1 is bounded, then there exists a constant R > 0 such that

log | cos(z; q)| 6 −(log |z|)2

log q+R, z ∈ Ac

m, m > m0. (1.3.45)

Thus

| cos(z; q)| 6 eRe−(log |z|)2

log q , z ∈ Acm, m > m0, (1.3.46)

proving the corollary.

The same study may be carried out for sin(z; q) and for this purpose we define suitable

annuli. Let

βm := log (ym/ym+1)/ log q, m ∈ Z+. (1.3.47)

Since βm → 1 as m→∞, then

β := infm∈Z+

βm > 0. (1.3.48)

Let (cm)∞m=1 and (dm)∞m=1 be the sequence defined by

cm :=

βm + β

2, if βm 6= β,

β2 , if βm = β,

(1.3.49)

29

and

d1 :=β

2, dm+1 :=

βm − β

2, if βm 6= β,

β2 , if βm = β,

(1.3.50)

where m > 1. Then we have

ymq−cm = ym+1q

dm+1 , m > 1. (1.3.51)

Thus the set of annuli Asm

∞m=1 where As

m is defined to be

Asm :=

z ∈ C : ymq

dm 6 |z| < ymq−cm

, m > 1 (1.3.52)

divide the regionz ∈ C : |z| > qβ/2y1

.

Theorem 1.3.6. Let β be the positive number defined in (1.3.48). Assume that |z| > qβ/2x1.

Let Asm, m > 1, be the annulus defined in (1.3.52). Then we have, uniformly as m→∞

log | sin(z; q)| = −(log |z|)2

log q+ log |1− z2

y2m

|+O(1), z ∈ Asm. (1.3.53)

The proof is similar to that of Theorem 1.3.4, and so it will not be given here.

Also we have the following corollary

Corollary 1.3.7.

M(r; sin(z; q)) = O

((exp

(−(log |z|)2

log q

)), as z −→∞. (1.3.54)

Lemma 1.3.8. The function tanh(|z|; q) :=sinh(|z|; q)cosh(|z|; q)

is bounded on C.

Proof. From (1.3.19) and (1.3.53) we conclude that there exists m0 ∈ N such that for all

m > m0 we have

−(log r)2

log q+ log |1 +

|z|2

x2m

| − δ 6 log | cosh(|z|; q)| 6 −(log r)2

log q+ log |1 +

|z|2

x2m

|+ δ, z ∈ Acm

(1.3.55)

30

and

− (log r)2

log q+log

∣∣∣∣1 +|z|2

y2m

∣∣∣∣−δ 6 log | sinh(|z|; q)| 6 − (log r)2

log q+log

∣∣∣∣1 +|z|2

y2m

∣∣∣∣+δ, z ∈ Asm. (1.3.56)

But if z ∈ Asm, m > 1, then |z|2/y2

m 6 q−2cm , where the sequence cm∞m=1 defined in

(1.3.49) is bounded. Hence there exists a constant L > 0 such that

1 +|z|2

y2m

6 L, for all m ∈ Z+.

Let r0 := maxqbm0xm0 , q

dm0ym0

. Thus by (1.3.55) and (1.3.56) we obtain for all z ∈ C,

|z| > r0sinh(|z|; q)cosh(|z|; q)

6 e2δ|1 +|z|2

y2m

| 6 Le2δ. (1.3.57)

1.4 Basic analogues of a theorem of Polya

In this section we give a q-analogue of Theorem 1.4.1 below of George Polya, cf. [102,103].

These q-analogues are the main results of [18].

Theorem 1.4.1. If the function f ∈ L1(0, 1) is positive and increasing, then

1. the zeros of the entire functions of exponential type

U(z) =∫ 1

0f(t) cos(zt) dt, V (z) =

∫ 1

0f(t) sin(zt) dt (1.4.1)

are real, infinite and simple.

2. U(z) is an even function having no zeros in [0, π2 ), and its positive zeros are situated in

the intervals (πk−π/2, πk+π/2), 1 6 k <∞, one in each. The odd function V (z) has only

one zero z = 0 in [0, π), and its positive zeros are situated in the intervals (πk, π(k + 1)),

1 6 k <∞, one zero in each interval.

31

Theorem 1.4.2. If f ∈ L1q(0, a) and Uf (z) is defined for z ∈ C by

Uf (z) :=∫ 1

0f(t) cos(tz; q) dqt, 0 < q < 1. (1.4.2)

Then Uf (z) has at most a finite number of non real zeros and it has an infinite number of

real zeros ±ζm∞m=1, ζm > 0, such that ζm ∼ xm as m→∞. More precisely

ζm = xm(1 +O(qm)) as m→∞. (1.4.3)

Proof. From the definition of the q-integral (1.1.31) Uf (z) can be written as

Uf (z) = H(z) +R(z), (1.4.4)

where

H(z) := (1− q)f(1) cos(z; q), and R(z) :=∞∑

k=1

qk(1− q)f(qk) cos(qkz; q). (1.4.5)

From (1.3.41) we deduce that there exists constants r0, C > 0 such that

max|z|=r

| cos(z; q)| 6 Ce−(log r)2/log q, for all r > r0. (1.4.6)

Let z ∈ C such that |z| > q−1r0, then

|R(z)| 6 Ce− (log qr)2

log q

∞∑k=1

qk(1− q)|f(qk)| = q−1Ce−(log r)2/ log q

r2

∫ q

0|f(t)| dqt. (1.4.7)

Thus

log |R(z)| 6 C1 − 2 log r −(log r

)2log q

, (1.4.8)

where C1 := log(q−1C

∫ q0 |f(t) dqt

). Let Ac

m, m > 1, be the annuls defined in (1.3.18).

Then

log | cos(z; q)| = −(log r

)2log q

+ log |1− z2

x2m

|+O(1), z ∈ Acm (1.4.9)

uniformly as m → ∞. Hence there exists a constant C2 > 0 and m0 ∈ N such that for all

z ∈ Acm, m > m0 we have

−(log r

)2log q

+ log∣∣1− z2

x2m

∣∣−C2 6 log | cos(z; q)| 6 −(log r

)2log q

+ log∣∣1− z2

x2m

∣∣+C2. (1.4.10)

32

Consequently by (1.4.8)

log |R(z)| 6 C1 + C2 − 2 log r + log | cos(z; q)| − log∣∣1− z2

x2m

∣∣= log |H(z)| − 2 log r − log

∣∣1− z2

x2m

∣∣+ C3, (1.4.11)

where C3 := C1 + C2 − log(1− q)|f(1)|. Let Dm0 be the disk defined by

Dm0 :=z ∈ C : |z| 6 qbm0xm0

. (1.4.12)

Clearly

C := Dm0 ∪ ∪m>m0Am.

If z ∈ ∂Dm0 , i.e. |z| = qbm0xm0 , then by (1.4.11)

log |R(z)| 6 log |H(z)| − 2 log r − log∣∣1− q2bm0 |+ C3. (1.4.13)

We can choose the m0 sufficiently large such that −2 log r− log∣∣1− q2bm0 +C3 < 0. That is

log |R(z)| 6 log |H(z)|, |z| = qbm0xm0 . (1.4.14)

So applying Rouche theorem on Dm0 , we conclude that H(z) and Uf (z) have the same

number of zeros inside Dm0 . As cos(z; q) has exactly 2m0 − 2 zeros inside Dm0 , then Uf (z)

has at most 2m0 − 2 zeros inside Dm0 .

If z ∈ ∂Acm, m > m0, then

∣∣1− z2

x2m

∣∣ > 1− q2bm ; |z| = qbmxm,∣∣1− z2

x2m

∣∣ > q−2am − 1; |z| = q−amxm. (1.4.15)

Thus

log |1− z2

x2m

| > infm>m0

log(1− q2bm) log(q−2am − 1)

, z ∈ ∂Ac

m. (1.4.16)

Since the sequences am∞m=1 and bm∞m=1 are bounded and positive, then there exits a

constant C4 > 0 such that

log |1− z2

x2m

| > C4, z ∈ ∂Acm, m > m0. (1.4.17)

33

That is

log |R(z)| 6 log |H(z)| − 2 log r + C3 − C4. (1.4.18)

Again, we choose the m0 large enough such that if r = |z| ∈ Acm, m > m0, then −2 log r +

C3 − C4 < 0. That is

|R(z)| 6 |H(z)|, z ∈ ∂Acm, m > m0.

Applying Rouche theorem on Acm, m > m0, we conclude that H(z) and Uf (z) have the

same number of zeros inside Acm. As cos(z; q) has two simple symmetric zeros there, we

deduce that the even function Uf (z) has only two real, symmetric and simple zeros inside

Acm, m > m0.

Now we give the asymptotic behavior of such zeros. Let ζm, be a positive zero of Uf (z)

in Acm, then log |H(ζm)| = log |R(ζm)|. Consequently by (1.4.11)

log∣∣∣∣1− ζ2

m

x2m

∣∣∣∣ 6 C3 − 2 log |ζm|, m > m0. (1.4.19)

Thus either

log∣∣∣∣1− ζm

xm

∣∣∣∣ 6 C3

2− log |ζm|, (1.4.20)

or

log∣∣∣∣1 +

ζmxm

∣∣∣∣ 6 C3

2− log |ζm|. (1.4.21)

We can assume without any loss of generality that ζm satisfies (1.4.20), then

∣∣1− ζmxm

∣∣ = O(|ζm|−1), as m→∞. (1.4.22)

Thus ζm ∼ xm for all sufficiently large m. More precisely

ζm = xm(1 +O(qm)), as m→∞. (1.4.23)

Since Uf (z) is even function then ±ζm are both zeros of Uf (z) in Acm. This completes the

proof.

Similarly we have the following theorem when we replace cos(tz; q) by sin(tz; q) in (1.4.2).

34

Theorem 1.4.3. If f ∈ L1q(0, a) and Vf (z) be defined for z ∈ C by

Vf (z) :=∫ 1

0f(t) sin(tz; q) dqt, 0 < q < 1. (1.4.24)

Then Vf (z) has at most a finite number of non real zeros and it has an infinite number

of real zeros ±ηm∞m=1, ηm > 0, for all m ∈ Z+ such that ηm ∼ ym as m → ∞. More

precisely

ηm = ym(1 +O(qm)) as m→∞. (1.4.25)

Chapter 2

Zeros of Finite q-HankelTransforms

In this chapter we prove that the basic finite Hankel transform whose kernel is the third-

type Jackson q-Bessel function has infinitely many real and simple zeros under certain

conditions on q. We also study the asymptotic behavior of the zeros. The obtained results

are applied to investigate the zeros of q-Bessel functions as well as zeros of q-trigonometric

functions. A basic analogue of a theorem of G. Polya (1918) on the zeros of sine and cosine

transformations is given as a special case.

2.1 Introduction and preliminaries

This chapter is devoted to a study the zeros of the q-type finite Hankel transform

F (z) =∫ 1

0f(t)Jν(tz; q2) dqt, z ∈ C, (2.1.1)

where Jν(·; q2) is the third Jackson q-Bessel function defined in (1.1.15), cf. [17]. Throughout

this chapter we assume that ν > −1. Next we state and prove the main results of this

chapter. These results prove that, under certain conditions on f and q, the finite q-Hankel

transform (2.1.1) has only real and simple zeros and there are infinitely many zeros. We

also investigate the distribution of these zeros. Section 2.3 contains some applications of

35

36

the results of Section 2.2. Among these applications we derive another basic analogue of a

theorem of Polya. We also derive in a manner different from the analysis of [7, 28, 63, 112]

the distribution of zeros of q-Bessel and q-trigonometric functions. It is worth mentioning

that q-type Hankel transforms are studied in [80] where the Hankel transform is defined on

(0,∞) simulating the classical situation. The following lemma is needed for the proof of

the main results of this chapter. It is taken from [103, p. 143].

Lemma A. Let g1(z), g2(z), . . . , gn(z), . . . be entire functions which have real zeros only.

If

limn→∞

gn(z) = g(z),

uniformly in any compact subset of C, then g(z) is an entire function with only real zeros.

We use the following convenient notations. For f ∈ L1q(0, 1), we denote by Ak(f) the

q-moments of f , i.e.

Ak(f) :=∫ 1

0tkf(t) dqt, k ∈ N, (2.1.2)

see (1.1.31) for the definition of the q-integration. Let also ck,ν(f), bk,ν(f), ν > −1, k ∈ N,

denote the numbers

ck,ν(f) :=A2k+2(f)

A2k(f)(1− q2k+2+2ν)(1− q2k+2), k ∈ N, (2.1.3)

and

bk,ν(f) :=A2k+3(f)

A2k+1(f)(1− q2k+2+2ν)(1− q2k+2), k ∈ N, (2.1.4)

We start our investigations with the following lemma.

Lemma 2.1.1. Let f ∈ L1q(0, 1) be positive on 0, qn, n ∈ N. Then the numbers

cν,f := infk∈N

ck,ν(f), Cν,f := supk∈N

ck,ν(f), (2.1.5)

and

bν,f := infk∈N

bk,ν(f), Bν,f := supk∈N

bk,ν(f), (2.1.6)

37

exist and they are finite positive numbers.

Proof. We will prove only (2.1.5). The proof of (2.1.6) is similar. First of all it should

be noted that the sequence Ak(f)∞k=0 is positive and strictly decreasing. Using Cauchy-

Schwarz inequality, for k ∈ N, k > 2, we obtain

(A2k(f)

)2 =(∫ 1

0t2kf(t) dqt

)2

=

∞∑j=0

qj(1− q)q2kjf(qj)

2

6∞∑

j=0

qj(1− q)q(2k+2)jf(qj)∞∑

j=0

qj(1− q)q(2k−2)jf(qj)

= A2k+2(f)A2k−2(f).

That is,A2k+2(f)A2k(f)

∞k=0

is an increasing sequence of positive numbers. Since A2k(f)∞k=0

is strictly decreasing, then the above inequality implies

A2(f)A0(f)

6A2k+2(f)A2k(f)

< 1, k = 0, 1, . . . . (2.1.7)

Therefore, for k ∈ N we have

A2(f)A0(f)

6A2k+2(f)A2k(f)

6A2k+2(f)

A2k(f)(1− q2k+2+2ν)(1− q2k+2)= ck,ν(f)

<1

(1− q2k+2ν+2)(1− q2k+2)6

1(1− q2ν+2)(1− q2)

.

(2.1.8)

Inequalities (2.1.8) and the Bolzano–Weierstrass theorem guarantee the existence of the

positive numbers cν,f , Cν,f .

2.2 Main results

In this section we will investigate the zeros of the finite q-Hankel Transforms.

Uν,f (z) := z−ν

∫ 1

0t−νf(t)Jν(tz; q2) dqt, z ∈ C, (2.2.1)

Vν,f (z) := z1−ν

∫ 1

0t1−νf(t)Jν(tz; q2) dqt, z ∈ C. (2.2.2)

38

It is known that Jν(z; q) is an entire function of order zero, cf. (1.1.15). An extensive study

of entire functions of order zero may be found in [87]. In the next lemma we investigate

the growth of Uν,f (z) and Vν,f (z).

Lemma 2.2.1. Let f ∈ L1q(0, 1) be positive on 0, qn, n ∈ N. Then the functions Uν,f (z)

and Vν,f (z) are entire in z of order zero.

Proof. We prove the lemma for Uν,f (z) since the proof for Vν,f (z) is similar. From (1.1.15)

and (1.1.31) we have for z ∈ C,

Uν,f (z) = z−ν

∫ 1

0t−νf(t)Jν(tz; q2) dqt

=(q2ν+2; q2)∞

(q2; q2)∞

∞∑k=0

(−1)kqk2+k z2k

(q2ν+2; q2)k(q2; q2)k

∫ 1

0t2kf(t) dqt

=(q2ν+2; q2)∞

(q2; q2)∞

∞∑k=0

(−1)kqk2+kA2k(f)z2k

(q2ν+2; q2)k(q2; q2)k

=∞∑

k=0

(−1)kakz2k, (2.2.3)

where ak is the sequence of positive numbers

ak :=(q2ν+2; q2)∞

(q2; q2)∞qk2+kA2k(f)

(q2ν+2; q2)k(q2; q2)k, k ∈ N. (2.2.4)

Since the radius of convergence of (2.2.3) is

limk→∞

ak

ak+1= lim

k→∞

A2k

A2k+2q−2k−2(1− q2k+2)(1− q2ν+2k+2)

= limk→∞

A2k

A2k+2q−2k−2 = ∞,

(2.2.5)

Uν,f (z) is entire.

Now we show that Uν,f (z) has order zero. From (1.2.3)

ρ (Uν,f ) = lim supk→∞

k log klog a−1

k

. (2.2.6)

From (2.2.4) we obtain

log1ak

= log(

(q2; q2)∞(q2; q2)k

(q2ν+2k+2; q2)∞A2k(f)

)+ (k2 + k) log

1q, (2.2.7)

39

where we have used(q2ν+2; q2)k

(q2ν+2; q2)∞=(q2ν+2k+2; q2

)∞

−1.

Since Ak(f)∞k=0 is a bounded decreasing sequence, then from the continuity of the loga-

rithmic function, it follows that

log(

(q2; q2)∞(q2; q2)k

(q2ν+2k+2; q2)∞A2k(f)

)−→ log

((q2; q2)2∞A(f)

), as k −→∞,

where A(f) := limk→∞

Ak(f). Recalling that 0 < q < 1, we deduce from (2.2.7)

limk→∞

log 1ak

k log k= lim

k→∞

(k2 + k) log 1q

k log k= ∞. (2.2.8)

Thus ρ (Uν,f ) = 0.

The following is the first main result of this chapter. The second is Theorem 2.2.3 below.

So far we only assumed 0 < q < 1. In our main results as well as those in the existing similar

studies, cf. e.g. [7, 28], more restrictions are imposed on q.

Theorem 2.2.2. Let ν > −1 and f ∈ L1q(0, 1) be positive on 0, qn, n ∈ N. If

q−1(1− q)cν,f

Cν,f> 1, (2.2.9)

then the zeros of Uν,f (z) are real, simple and infinite in number. Moreover, Uν,f (z) is an

even function with no zeros in[0, q−1/

√Cν,f

), and its positive zeros lie in the intervals(

q−r+1/2√Cν,f

,q−r−1/2√Cν,f

), r = 1, 2, 3, . . . , (2.2.10)

one zero in each interval.

Proof. We prove the theorem in three steps.

1. We show that Uν,f (z) has no zeros in BR(0), where R := 1/q√Cν,f and

BR(0) := z ∈ C : |z| 6 R, z 6= ±R .

40

First

0 <a1

a0= q2c0,ν(f) 6 q2Cν,f = R−2,

0 <ak+1

ak= q2k+2ck,ν(f) < q2Cν,f = R−2, k > 1.

(2.2.11)

Therefore, a0 −R2a1 > 0 and ak −R2ak+1 > 0 for all k > 1. If z ∈ BR(0), then by (2.2.3)

∣∣(R2 + z2)Uν,f (z)∣∣ =

∣∣∣∣∣a0R2 −

∞∑k=1

(−1)kz2k(ak−1 −R2ak)

∣∣∣∣∣> a0R

2 −

∣∣∣∣∣∞∑

k=1

(−1)kz2k(ak−1 −R2ak)

∣∣∣∣∣ . (2.2.12)

If z ∈ C is not real, i.e. z = reiθ, θ 6= kπ, k ∈ Z, then for c0, c1 > 0,

|c0 + c1z|2 = c20 + 2c0c1r cos θ + c21r2 < c20 + 2c0c1r + c21r

2 = (c0 + c1|z|)2. (2.2.13)

Consequently |c0 + c1z| < c0 + c1|z|. Thus for non real z ∈ BR(0), we obtain∣∣∣∣∣∞∑

k=1

(−1)kz2k(ak−1 −R2ak)

∣∣∣∣∣6

∣∣z4(a1 − a2R2) + z6(a2 − a3R

2)∣∣+ ∞∑

k=1k 6=2,3

|z|2k(ak−1 −R2ak)

< |z4|((a1 − a2R

2) + |z|2(a2 − a3R2))

+∞∑

k=1k 6=2,3

|z|2k(ak−1 −R2ak) (2.2.14)

=∞∑

k=1

|z|2k(ak−1 −R2ak) 6∞∑

k=1

(R2kak−1 −R2k+2ak)

= a0R2 − lim

m→∞R2m+2am.

From (2.2.11)

m−1∏k=0

ak+1

ak=

m−1∏k=0

q2k+2ck,ν(f) 6m−1∏k=0

q2k+2Cν,f = qm2+mCmν,f . (2.2.15)

Thus

am 6 qm2+mCmν,fa0, R2m+2am 6 qm2−m−2 a0

Cν,f. (2.2.16)

41

That is limm→∞R2m+2am = 0, and therefore we have for a non real z ∈ BR(0), z 6= ±iR,

∣∣(R2 + z2)Uν,f (z)∣∣ > lim

m→∞R2m+2am = 0. (2.2.17)

If z = ±iR, then by (2.2.3)

Uν,f (±iR) =∞∑

k=0

akR2k > 0. (2.2.18)

Thus Uν,f (z) > 0 for all non real z ∈ BR(0). On the other hand if 0 < |z| < R, z is real,

then

∣∣(R2 + z2)Uν,f (z)∣∣ = ∣∣∣∣∣a0R

2 −∞∑

k=1

(−1)kz2k(ak−1 −R2ak)

∣∣∣∣∣> a0R

2 −∞∑

k=1

|z|2k(ak−1 −R2ak)

> a0R2 −

∞∑k=1

R2k(ak−1 −R2ak),

(2.2.19)

since if 0 < αn < βn, and∑∞

n=0 βn <∞, then∑∞

n=0 αn <∑∞

n=0 βn. Moreover

a0R2 −

∞∑k=1

R2k(ak−1 −R2ak) = a0R2 −

(a0R

2 − limm→∞

R2m+2am

)= 0.

Since

Uν,f (0) =(q2ν+2; q2)∞

(q2; q2)∞A0(f) 6= 0,

then∣∣(R2 + z2)Uν,f (z)

∣∣ > 0 for all real z ∈ BR(0). Consequently Uν,f (z) has no zeros in

BR(0).

2. We prove that Uν,f (z) has an infinite number of real zeros only. Let U2n,ν,f (z) be the

polynomial

U2n,ν,f (z) :=(q2ν+2; q2)∞

(q2; q2)∞

2n∑k=0

(−1)k qk2+kA2k(f)(q2ν+2; q2)k(q2; q2)k

z2k, n ∈ N, z ∈ C. (2.2.20)

Obviously U2n,ν,f (z) approaches Uν,f (z) uniformly as n → ∞ on any compact subset of

C. Take zr := q−r−1/2/√Cν,f , r ∈ N. We prove that U2n,ν,f (z) has a zero in the interval

42

(zr−1, zr), r = 1, 2, . . . , 2n. Indeed,

U2n,ν,f (zr) =2n∑

k=0

(−1)kαk(r), (2.2.21)

where αk(r) is the positive sequence

αk(r) :=(q2ν+2; q2)∞

(q2; q2)∞qk2−2kr

Ckν,f (q2; q2)k(q2ν+2; q2)k

A2k(f), k ∈ N. (2.2.22)

If 0 6 k 6 r − 1, then from (2.2.9)

αk+1(r)αk(r)

= q2k−2r+1 ck,ν(f)Cν,f

>cν,f

Cν,fq−1 > (1− q)−1 > 1, (2.2.23)

andαk+1(r)αk(r)

6 q < 1, if k > r. (2.2.24)

From (2.2.23), and (2.2.24), we obtain for r > 1

αr+1(r)αr(r)

+αr−1(r)αr(r)

< q + (1− q) = 1. (2.2.25)

Therefore, we have the following inequality

αr+1(r) < αr(r)− αr−1(r), r = 1, 2, 3, . . . . (2.2.26)

Now we prove that

signU2n,ν,f (zr) = (−1)r, r = 0, 1, . . . , 2n. (2.2.27)

If r = 0, then from (2.2.24)αk+1(0)αk(0)

< 1 for all k > 0. Consequently

U2n,ν,f (z0) =n−1∑j=0

(α2j(0)− α2j+1(0)

)+ α2n(0) > 0.

Take r = 2m, 1 6 m 6 n. Then

U2n,ν,f (zr) =m−1∑k=1

(α2k(r)− α2k−1(r)) +n−1∑

k=m+1

(α2k(r)− α2k+1(r))

+ α0(r) + α2n(r) +(αr(r)− αr−1(r)

)− αr+1(r).

(2.2.28)

43

Hence from (2.2.23), (2.2.24), and (2.2.26),

U2n,ν,f

(zr)

>(αr(r)− αr−1(r)

)− αr+1(r) > 0.

Similarly if r = 2m− 1, 1 6 m 6 n, we have

U2n,ν,f

(zr)

= −m−2∑k=0

(α2k+1(r)− α2k(r)

)−

n∑k=m+1

(α2k−1(r)− α2k(r)

)−(αr(r)− αr−1(r)

)+ αr+1(r) < 0.

(2.2.29)

Thus, U2n,ν,f (z) has at least one zero in each of the intervals (zr−1, zr), r = 1, 2, . . . , 2n,

i.e. it has at least 2n positive zeros. Since U2n,ν,f (z) is an even polynomial of degree 4n,

then it also has 2n negative zeros. So, the 4n zeros of the polynomial U2n,ν,f (z) are all

real and simple. Consequently, from Lemma A, Uν,f (z) has only real zeros. Since αk(r) is

independent of n, the same argument holds when n is replaced by ∞, and noting that all

series converge absolutely, shows that

signUν,f (zr) = (−1)r, r ∈ N. (2.2.30)

proving that Uν,f (z) has an infinite number of real zeros ±w(ν)r , w(ν)

r ∈ (zr, zr+1), r ∈ N.

3. Finally we prove that the zeros of Uν,f (z) are simple. Consider the annulus Dr defined

for r ∈ Z+ by

Dr := z ∈ C : zr−1 6 |z| 6 zr . (2.2.31)

Since Uν,f (z) has no zeros on the boundary ∂Dr of Dr, where |z| = q−r±(1/2)√Cν,f

, then there

exists δ > 0 such that |Uν,f (z)| > δ on ∂Dr. For this δ we can find N0 ∈ N, N0 > r such

that

|U2m,ν,f (z)− Uν,f (z)| 6 δ, m > N0, z ∈ Dr. (2.2.32)

Hence,

|U2m,ν,f (z)− Uν,f (z)| < |Uν,f (z)|, m > N0, z ∈ ∂Dr. (2.2.33)

44

Applying Rouche’s Theorem, we conclude that Uν,f (z) and U2m,ν,f (z), m > N0, have the

same numbers of zeros inside Dr. Since for allm > N0, U2m,ν,f (z) has exactly two symmetric

simple real zeros inside Dr, then so is Uν,f (z). This completes the proof of the theorem.

Remark 2.2.1. Step 2 of the proof of Theorem 2.2.2 includes a proof of the fact that the

function Uν,f has infinitely many zeros. However, according to Theorem 1.2.1 we do not

need to give a proof because ρ (Uν,f ) = 0.

Theorem 2.2.3. Let ν > −1 and f ∈ L1q(0, 1) be positive on 0, qn, n ∈ N. If

q−1(1− q)bν,f

Bν,f> 1, (2.2.34)

then the zeros of Vν,f (z) are real, simple and infinite in number. The odd function Vν,f (z)

has only one zero z = 0 in [0, q−1√Bν,f

), and its positive zeros are situated in the intervals

(q−r+1/2√Bν,f

,q−r−1/2√Bν,f

), r = 1, 2, 3, . . . , (2.2.35)

one zero in each interval.

Proof. The proof is similar to that of the previous theorem.

Corollary 2.2.4. Let ν > −1 and f ∈ L1q(0, 1) be positive and decreasing on 0, qn, n ∈ N.

If

q−1(1− q)2(1− q2)(1− q2+2ν)f2(1)f2(0)

> 1, (2.2.36)

then the zeros of Uν,f (z) are real, simple and infinite in number. Moreover, the positive

zeros lie in the intervals(q−r+1/2

/√Cν,f , q

−r−1/2/√

Cν,f

), r = 1, 2, . . . ,

one zero in each interval. We have the same result if f is increasing and positive on

0, qn, n ∈ N and q satisfies

q−1(1− q)2(1− q2)(1− q2+2ν)f2(0)f2(1)

> 1, (2.2.37)

instead of (2.2.36).

45

Proof. We notice that if

q−1(1− q)lν,f

Lν,f> 1, (2.2.38)

where lν,f and Lν,f are lower and upper bounds of the sequence ck,ν(f)∞k=0 respectively,

then (2.2.9) is satisfied and the results of Theorem 2.2.2 hold. This observation suffices to

prove the corollary because if f is decreasing and positive on 0, qn, n ∈ N, then by (2.1.2)

f(1)(1− q)(1− q2k+1)

6 A2k(f) 6f(0)(1− q)(1− q2k+1)

, k ∈ N. (2.2.39)

Hencef(1)(1− q2k+1)f(0)(1− q2k+3)

6A2k+2(f)A2k(f)

6f(0)(1− q2k+1)f(1)(1− q2k+3)

, k ∈ N. (2.2.40)

Now the sequence (1− q2k+1)

((1− q2k+2)(1− q2k+3)(1− q2k+2ν+2))

∞k=0

is bounded above by 1/((1 − q2)(1 − q2ν+2)

)and it is bounded below by (1 − q). Let lν,f

and Lν,f be the numbers

lν,f =f(1)(1− q)

f(0), Lν,f =

f(0)f(1)(1− q2)(1− q2ν+2)

. (2.2.41)

Then from (2.1.3), lν,f and Lν,f are lower and upper bounds of ck,ν,f respectively. A

direct substitution with lν,f and Lν,f of (2.2.41) in (2.2.38) yields (2.2.36). Similarly if f is

increasing and positive on 0, qn, n ∈ N, then lν,f and Lν,f can be taken to be

lν,f =f(0)(1− q)

f(1), Lν,f =

f(1)f(0)(1− q2)(1− q2ν+2)

.

A similar result holds for Vν,f when f is positive and monotonic on 0, qn, n ∈ N.

2.3 Applications

In this section we introduce some applications of the results of Section 2.2. First we inves-

tigate the zeros of Jν(z; q2) and the q-trigonometric functions.

46

Lemma 2.3.1. If q ∈ (0, 1) satisfies the inequality

q−1(1− q)(1− q2)(1− q2ν+2) > 1, ν > −1, (2.3.1)

then the zeros of the third Jackson q-Bessel function Jν(·; q2) are real, simple, and infinite

in number. Moreover Jν(·; q2) has a zero at z = 0 if and only if ν > 0. The positive zeros

of Jν(·; q2) belong to the intervals(q−r+1/2

√(1− q2ν+2)(1− q2), q−r−1/2

√(1− q2ν+2)(1− q2)

), r = 1, 2, . . . , (2.3.2)

where each interval contains only one zero.

Proof. Applying Theorem 2.2.2 with

f(t) :=

(1− q)−1, t = 1,

0, otherwise,

then Uν,f (z) = z−νJν(z; q2). For k ∈ N, we have

Ak(f) ≡ 1, k ∈ N, Cν,f =1

(1− q2)(1− q2ν+2), cν,f = 1.

Therefore condition (2.2.9) is nothing but (2.3.1) and the lemma is proved.

Remark 2.3.1. In [66] asymptotics of the q-Bessel functions J (1)ν (·; q) and J (2)

ν (·; q) is given.

Also, a study for the zeros of the second Jackson q-Bessel function J(2)ν (x; q) is given in

[60].

Corollary 2.3.2. i. If 0 < q < γ0, where γ0 ≈ 0.429052 is the root of

(1− q)(1− q2)(1− q3)− q, q ∈ (0, 1),

then the zeros of sin(z; q) are real, infinite and simple and the positive zeros belong to the

intervals(q−r+1/2

√(1− q2)(1− q3), q−r−1/2

√(1− q2)(1− q3)

), r = 1, 2, . . . , (2.3.3)

47

and each interval contains only one zero.

ii. If 0 < q < β0, where β0 ≈ 0.38197 is the zero of (1− q)3(1 + q)− q in 0 < q < 1, then

the zeros of cos(z; q) are real and simple and the positive zeros belong to the intervals

(q−r(1− q)

√1 + q, q−r−1(1− q)

√1 + q

), r = 0, 1, 2, . . . , (2.3.4)

and each interval contains only one zero.

Proof. To prove i, substitute with ν = 1/2 in (2.3.1). Then (2.3.1) becomes

q−1(1− q)(1− q2)(1− q)3 > 1

which holds if 0 < q < γ0. Since sin(z; q) has the representation (1.1.17), then applying

Lemma 2.3.1 with ν = 1/2, we conclude that if 0 < q < γ0, the zeros of sin(z; q) are real,

infinite and simple and they lie in the intervals (2.3.3). The proof of ii. follows similarly by

considering ν = −1/2.

Remark 2.3.2. In [78] a study on the zeros of Jν(z; q) has been established. The existence,

simplicity and reality of the zeros of Jν(·; q), ν > −1 was also proved in [78, Theorem 3.4].

Bustoz and Cardoso in [28] and Abreu et al. [7] derived asymptotic formulae for the zeros

of sin(·; q) and cos(·; q) as well as for Jν(·; q), ν > −1 under a condition on q which is less

restrictive than (2.3.9). However our results are more general in another direction because

we study more general problems. This enables us to obtain the distribution of zeros of

combinations of q-functions. For example the zeros ±vr∞r=0 of the q-function

Vt(z) :=∫ 1

0t sin(tz; q) dqt =

1z2

(sin(z; q)− z cos(q−1/2z; q)

), 0 < q < ε0, (2.3.5)

are real, infinite and simple where ε0 ≈ 0.441751 is the zero in 0 < q < 1 of

(1− q)(1− q2)(1− q5)− q.

48

Moreover,

vr ∈(q−r+1/2

√(1− q2)(1− q5), q−r−1/2

√(1− q2)(1− q5)

), r = 0, 1, . . . . (2.3.6)

As another example, define a function f on [0, 1] to be

f(t) :=

tν/(1− q), t = 1, q,

0, otherwise,, (2.3.7)

ν > −1. If q−1(1− q)cν,f

/Cν,f > 1, then the zeros ±ur∞r=0 of the q-function

Uν,f := z−ν

∫ 1

0t−νf(t)Jν(tz; q2) dqt = z−ν

(Jν(z; q2) + qJν(qz; q2)

), (2.3.8)

are real, infinite, simple and ur ∈(q−r+1/2/

√Cν,f , q

−r−1/2/√Cν,f

), r = 0, 1, 2, . . ..

In the following we introduce another q-analogue of Theorem 1.4.1.

Theorem 2.3.3. Let f ∈ L1q(0, 1) be positive on 0, qn, n ∈ N. If

q−1(1− q)c−1/2,f

C−1/2,f> 1, (2.3.9)

then the zeros of the entire function of order zero

Uf (z) :=∫ 1

0f(t) cos(tz; q) dqt, z ∈ C, (2.3.10)

are real, simple and infinite. Moreover Uf (z) is an even function with no zeros in the

interval[0, q−1/2/

(√C−1/2,f (1− q)

)), and its positive zeros lie in the intervals

(q−r/

((1− q)

√C−1/2,f ), q−r−1/

((1− q)

√C−1/2,f )

), r = 0, 1, . . . , (2.3.11)

one zero in each interval.

Proof. From (1.1.16) we conclude that

Uf (z) =(q2; q2)∞(q; q2)∞

U−1/2,f (z(1− q)q−1/2). (2.3.12)

49

Applying Theorems 2.2.2 to U−1/2,f (z(1 − q)q−1/2) where q-satisfies (2.3.9), proves the

theorem.

Similarly, we have the following theorem.

Theorem 2.3.4. Let f ∈ L1q(0, 1) be positive on 0, qn, n ∈ N. If

q−1(1− q)b1/2,f

B1/2,f> 1, (2.3.13)

then the zeros of the entire function of order zero

Vf (z) :=∫ 1

0f(t) sin(tz; q) dqt, z ∈ C, (2.3.14)

are real, simple and infinite. Moreover Vf (z) is an odd function with exactly one zero, z = 0,

in the interval [0, q−1√B1/2,f (1−q)

), and its positive zeros are located in the intervals

(q−r+1/2

(1− q)√B1/2,f

,q−r−1/2

(1− q)√B1/2,f

), r = 0, 1, . . . , (2.3.15)

one in each interval.

A remarkable difference between Polya’s theorem and its basic analogue here is that in

Theorems 2.3.3 and 2.3.4 we do not have the monotonicity condition. However the price

for this are the restrictions (2.3.9) and (2.3.13) on q. This allowed us to prove that Uf (z)

and Vf (z) have no non-real zeros.

Part II

Basic Difference Equations

50

Chapter 3

Introduction

In this introductory chapter we briefly introduce the main concepts and theorems of linear

q-difference equations as stated in [8,9,90]. We start with the existence and uniqueness the-

orem of q-Cauchy problems based on q-analogue of the celebrated successive approximation

method. Then we state some results concerning q-linear difference equations.

3.1 q-successive approximations

In this section we define the q-successive approximations associated with q-Cauchy problems.

This technique is the main method we use to prove the theorems of existence as in [9, 90].

Proofs will not be given here. The interested reader could find them in [9, 90].

Definition 3.1.1. Let r, s, and ni, i = 0, 1, . . . , r, be positive integers and let

N = (n0 + 1) + . . .+ (nr + 1)− 1.

Let Fj(x, y0, y1, . . . , yN ) , j = 0, 1, . . . , s, be real or complex–valued functions, where x is a

real variable lying in some interval I and each yi is a complex variable lying in some region

Di of the complex plane. If there are a sub–interval J of I and functions φi, 0 ≤ i ≤ r,

defined in J such that

51

52

1. φi has ni q-derivatives in J for 0 ≤ i ≤ r ;

2. Dmq φi exists and lies in the region Di for all x in J, 0 ≤ m ≤ ni, and 0 ≤ i ≤ r, for

which the left–hand side in equation (3.1.1) below is defined;

3. for all x in J and 0 ≤ j ≤ s, the following equations hold

Fj

(x, φ0(x), Dqφ0(x), . . . , Dn0

q φ0(x), . . . , φr(x), . . . , Dnrq φr(x)

)= 0; (3.1.1)

then we say that φiri=0 is a solution of the system of the q-difference equations

Fj

(x, y0(x), Dqy0(x), . . . , Dn0

q y0(x), . . . , yr(x), . . . , Dnrq yr(x)

)= 0, (3.1.2)

0 ≤ j ≤ s, valid in J , or that the set φiri=0 satisfies (3.1.2) in J . If there are no such J

and functions φi, we say that the system (3.1.2) has no solutions.

System (3.1.2) is said to be of order n, where n := max06i6p ni. If the functions Fj are

such that the equations (3.1.2) can be solved for the Dniq yi in the form

Dniq yi(x) = fi (x, y0(x), Dqy0(x), . . . , y1(x) . . .) , 0 6 i 6 p, (3.1.3)

or in the form

Dniq yi(x) = fi (qx, y0(qx), Dq,qxy0(qx), . . . , y1(qx) . . .) , 0 6 i 6 p, (3.1.4)

then the systems (3.1.3) and (3.1.4) will be called normal systems. The following systems

are examples of normal first order systems,

Dqyi(x) = fi

(qx, y0(qx), y1(qx), . . . , yp(qx)

), i = 0, 1, . . . , p, (3.1.5)

and

Dqyi(x) = fi

(x, y0(x), y1(x), . . . , yp(x)

), i = 0, 1, . . . , p. (3.1.6)

In a more restrictive manner, first order normal linear systems have been defined in [34].

The existence of a solution of system (3.1.5) or (3.1.6) in a neighborhood of a zero is

53

established in [9, 90] by using a q-analogue of the Picard-Lindelof method of successive

approximations , see e.g. [37, 41]. The following is a description of this q-analogue.

Define sequences of functions φi,m∞m=1, i = 0, 1, . . . , p, by the equations

φi,m(x) =

bi, m = 1

bi +∫ x

0fi (t, φ0,m(t), . . . , φp,m(t)) dqt, m > 2,

(3.1.7)

where bi are constants which lie in Di. It is worth mentioning here that the constants

bipi=0 should be replaced by q-periodic functions if we are looking for solutions which are

not necessarily continuous at zero.

As is stated in Theorem 3.2.2 below under suitable conditions on the fi’s, there are

functions φi and an interval J contained in I and containing zero such that φi,m(x) ⇒ φi(x)

on J as m→∞, where by ”⇒” we mean the uniform convergence. Then, it will be possible

to apply Lemma 3.2.1 below to obtain

φi(x) = bi +∫ x

0fi (t, φ0(t), . . . , φp(t)) dqt. (3.1.8)

Hence φi(0) = bi, and φipi=0 is a solution of (3.1.6). Similarly for (3.1.5).

3.2 q-Initial value problems

Here we state an existence and uniqueness theorem of q-Cauchy problems as treated in

[9, 90].

Definition 3.2.1. Let bi ∈ Di be arbitrary values. By a q-initial value problem in a

neighborhood of zero we mean the problem of finding functions yipi=0, continuous at zero

and satisfying the system (3.1.5) or (3.1.6) and the initial conditions

yi(0) = bi, bi ∈ Di, 0 6 i 6 p. (3.2.1)

The proof of the following lemma is straightforward and will be omitted.

54

Lemma 3.2.1. Let I and J be intervals containing zero, such that J ⊆ I. Let fn, f be

functions defined in I, n ∈ N, such that

limn→∞

fn(t) = f(t), when t ∈ I, and fn ⇒ f on J. (3.2.2)

Then,

limn→∞

∫ x

0fn(t) dqt =

∫ x

0f(t) dqt, for all x ∈ I. (3.2.3)

Theorem 3.2.2. Let I be an interval containing zero and let Er be disks of the form

Er := y ∈ C : |y − br| < β , β > 0, br ∈ C,

and r = 0, 1, . . . , p. Let fi(x, y0, y1, . . . , yp), i = 0, 1, . . . , p, be functions defined on I ×E0×

. . .× Ep for which:

(i) for yr ∈ Er, 0 6 r 6 p, fi(x, y0, y1, . . . , yp) is continuous at x = 0, 0 6 i 6 p,

(ii) there is a positive constant A such that, for x ∈ I and yr, yr ∈ Er, 0 6 r, i 6 p, the

following Lipschiz condition is fulfilled

∣∣fi(x, y0, . . . , yp)− fi(x, y0, . . . , yp)∣∣ 6 A

(|y0 − y0|+ . . .+ |yp − yp|

). (3.2.4)

Then, if 0 is not an end point of I, there exists a positive h such that the q-Cauchy

problem (3.1.6), (3.2.1) has a unique solution valid for |x| ≤ h . Moreover if 0 is the left or

right end point of I, the result holds, except that the interval [−h, h] is replaced by [0, h] or

[−h, 0], respectively

Theorem 3.2.3 (Range of validity). Assume that all conditions of Theorem 3.2.2 are

satisfied with Er = C for all r; r = 0, 1, . . . , p. Then problem (3.1.6), (3.2.1) has a unique

solution valid at least in I ∩(− 1

A(p+1)(1−q) ,1

A(p+1)(1−q)

).

Remark 3.2.1. Theorem 3.2.2 holds for the q-Cauchy problem (3.1.5), (3.2.1), but the so-

lution will be valid throughout the whole interval I if fi’s satisfy the conditions (i), (ii) of

Theorem 3.2.2 with Er = C, 0 6 r, i 6 p.

55

Example 3.2.1. It is known, [53], that both of the functions eq(x) and Eq(x) are q-analogues

of the exponential function. While Eq(x) is entire, eq(x) is analytic only when |x| < 1. They

are defined and related by

eq(x) =1

Eq(x)=:

1∏∞k=0

(1− xqk

) , x ∈ C−q−n, n ∈ N

.

Theorem 3.2.2 and Remark 3.2.1 gave us an explanation since eq(x) and Eq(x) are the

solutions of the q-Cauchy problems

Dqy(x) =1

1− qy(x), y(0) = 1; Dqy(x) =

11− q

y(qx), y(0) = 1, (3.2.5)

respectively. From Theorem 3.2.3 it follows that the solution of the first problem is valid at

most in |x| < 1 and from Remark 3.2.1 the solution of the second one is valid in R.

3.3 Linear q-difference equations

This section contains a brief study of linear q-difference equations of order n, n ∈ Z+.

Complete treatments could be found in [8, 9, 90].

For n ∈ Z+, bi ∈ C, Theorem 3.2.2 can be applied to discuss the existence and uniqueness

of the n-th order q-initial value problemDn

q y(x) = f(x, y(x), Dqy(x), . . . , Dn−1

q y(x))

Di−1q y(0) = bi, bi ∈ C; 1 6 i 6 n .

(3.3.1)

Corollary 3.3.1. Let p, I, Er, br be as in the previous theorem. Let f(x, y0, y1, . . . , yp) be

a function defined on I × E0 × . . .× Ep such that the following conditions are satisfied

(i) For any fixed values of the yr in Er , f(x, y0, y1, . . . , yp) is continuous at zero.

(ii) f satisfies Lipschitz condition with respect to y0, . . . , yp.

56

If 0 is an interior point of I, then there exists a positive h such that q-Cauchy problem

(3.3.1) has a unique solution φ valid for |x| ≤ h. If 0 is a left or right endpoint of I the

result holds, except that the interval |x| ≤ h is replaced by [0, h] or [−h, 0] respectively.

Proof. Assume that 0 is an interior point of I. The q-Cauchy problem (3.3.1) is equivalent

to the first order q-initial value problem

Dqyi(x) = fi(x, y0, . . . , yn−1) yi(0) = bi, ; 1 6 i 6 n, (3.3.2)

in the sense that φin−1i=0 is a solution of (3.3.2) if and only if φ0 is a solution of (3.3.1).

Here fi are the functions

fi(x, y0, . . . , yn−1) =

yi+1, 0 6 i 6 n− 2

f(x, y0, . . . , yn−1), i = n.(3.3.3)

By Theorem (3.2.2) there exists h > 0 such that system (3.3.2) has a unique solution valid

in |x| 6 h.

Corollary 3.3.2. Consider the q-Cauchy problema0(x)Dn

q y(x) + a1(x)Dn−1q y(x) + . . .+ an(x)y(x) = b(x)

Diqφ(0) = bi, 0 6 i 6 n− 1 .

(3.3.4)

Let the aj(x), 0 6 j 6 n, and b(x) be defined on an interval I containing zero such that

a0(x) 6= 0 for all x in I and the functions aj(x), b(x) are continuous at zero and bounded

on I. Then for any complex numbers br, there exists a subinterval J of I containing zero

such that (3.3.4) has a unique solution valid in J .

Proof. Dividing by a0(x), we get the equation

Dnq y(x) = An(x)Dn−1

q y(x) + . . .+A1(x)y(x) +B(x), (3.3.5)

57

where Aj(x) = −aj(x)/a0(x), 1 6 j 6 n, and B(x) = b(x)/a0(x). Equation (3.3.5) is of the

form (3.3.2) with

f(x, y0, . . . , yn−1) = An(x)y0 + . . .+A1(x)yn−1 +B(x).

Since Aj(x) and B(x) are continuous at zero and bounded on I, then f(x, y0, . . . , yp) satisfies

the conditions of Theorem 3.3.1. Hence, there exists a subinterval J of I containing zero

such that equation (3.3.5) has a unique solution valid in J .

Consider the non-homogeneous linear q-difference equation of order n

a0(x)Dnq y(x) + a1(x)Dn−1

q y(x) + . . .+ an(x)y(x) = b(x), x ∈ I, (3.3.6)

for which ai, 0 6 i 6 n, and b are continuous at zero functions defined on I and a0(x) 6= 0

for all x ∈ I . From Corollary 3.3.2 equation (3.3.6) together with the initial conditions

Di−1q y(0) = bi, bi ∈ C, i = 1, . . . , n, (3.3.7)

has a unique solution which is continuous at zero in a subinterval J of I, J = [−h, h], h > 0.

The corresponding n-th order homogeneous linear equation of (3.3.6) is

a0(x)Dnq y(x) + a1(x)Dn−1

q y(x) + . . .+ an(x)y(x) = 0, x ∈ I. (3.3.8)

Let M denote the set of solutions of (3.3.8) valid in a subset J ⊆ I, 0 ∈ J . Then,cf. [8],

M is a linear space over C of dimension n. Also from the existence and uniqueness of the

solutions, if φ ∈ M and Diqφ(0) = 0, 0 6 i 6 n − 1, then φ(x) ≡ 0 on J . Moreover,

Diqφn−1

i=0are continuous at zero for any φ ∈ M . A set of n solutions of (3.3.8) is said to

be a fundamental set (f.s.) for (3.3.8) valid in J or a f.s. of M if it is linearly independent

in J . Moreover, as in differential equations, if bij , 1 ≤ i, j ≤ n, are numbers, and, for each

j, φj is the unique solution of (3.3.8) which satisfies the initial conditions

Di−1q φj(0) = bij , 1 ≤ i ≤ n,

then φjnj=1 is a f.s. of (3.3.8) if and only if det(bij) 6= 0.

58

3.4 A q-type Wronskian

In the following we give a q-analogue of the Wronskian for n-th order q−difference equations.

A study for a second order Wronskian is established by R.F. Swarttouw and H.G. Meijer

in [113]. The results of the present section are also included in [8, 90].

Definition 3.4.1. Let yi, 1 ≤ i ≤ n, be functions defined on a q-geometric set A. The

q-Wronskian of the functions yi which will be denoted by Wq(y1, . . . , yn)(x) is defined to be

Wq(y1, . . . , yn)(x) :=

∣∣∣∣∣∣∣∣∣∣∣∣∣

y1(x) . . . yn(x)

Dqy1(x) . . . Dqyn(x)...

. . ....

Dn−1q y1(x) . . . Dn−1

q yn(x)

∣∣∣∣∣∣∣∣∣∣∣∣∣, (3.4.1)

provided that the derivatives exist in I. For convenience we write Wq(x) instead of

Wq(y1, . . . , yn)(x).

Theorem 3.4.1. If y1, y2, . . . , yn are solutions of (3.3.8) in J ⊆ I, then their q-Wronskian

satisfies the first order q-difference equation

DqWq(x) = −R(x)Wq(x), x ∈ J − 0 ; R(x) =n−1∑k=0

(x− qx)k ak+1(x)a0(x)

. (3.4.2)

The following theorems gives a q-type Liouville’s formula for the q-Wronskian.

Theorem 3.4.2. Suppose that x(1 − q)R(x) 6= −1 when x ∈ J . Then the q-Wronskian of

any set of solutions φini=1 of equation (3.3.8) is given by

Wq(x) = Wq(φ1, . . . , φn)(x) =1∏∞

k=0(1 + x(1− q)qkR(xqk))Wq(0), x ∈ J. (3.4.3)

Corollary 3.4.3. Let φini=1 be a set of solutions of (3.3.8) in some subinterval J of I

which contains zero. Then Wq(x) is either never zero or identically zero in I. The first

case occurs when φini=1 is a fundamental set of (3.3.8) and the second when it is not.

59

Example 3.4.1. In this example we calculate the q-Wronskian of

−1qDq−1Dqy(x) + y(x) = 0, x ∈ R. (3.4.4)

The solutions of (3.4.4) subject to the initial conditions

y(0) = 0, Dqy(0) = 1 and y(0) = 1, Dqy(0) = 0,

are sin(x; q), cos(x; q), x ∈ R, respectively. Since (3.4.4) can be written as

D2qy(x) + qx(1− q)Dqy(x)− qy(x) = 0, (3.4.5)

then a0(x) ≡ 1, a1(x) = qx(1− q) and a2(x) = −q. Thus

R(x) ≡ 0 on R and Wq(x) ≡Wq(0).

But

Wq(0) = Wq

(cos(·; q), sin(·; q)

)(0)

=(cos(x; q) cos(

√qx; q) +

√q sin(x; q) sin(

√qx; q)

)∣∣x=0

= 1.

Then, Wq(x) ≡ 1 for all x ∈ R.

Example 3.4.2. We calculate the q-Wronskian of solutions of the q-difference equation

−D2qy(x) + y(x) = 0, x ∈ R. (3.4.6)

The functions sinq x(1− q), cosq x(1− q), |x|(1− q) < 1, are solutions of (3.4.6) subject

to the initial conditions

y(0) = 0, Dqy(0) = 1 and y(0) = 1, Dqy(0) = 0,

respectively. Here R(x) = x(1− q). So, x(1− q)R(x) 6= −1 for all x in R. Hence,

Wq(x) =Wq(0)∏∞

n=0(1 + q2nx(1− q)2), |x|(1− q) < 1.

But

Wq(0) = Wq

(cosq, sinq

)(0) =

(cosq

2x+ sinq2x)∣∣

x=0= 1. (3.4.7)

Therefore, Wq(x) ≡ 1/∏∞

n=0(1 + q2nx(1− q)2), |x|(1− q) < 1.

Chapter 4

Basic Sturm-Liouville Problems

This chapter is devoted to a study a q-analogue of the Sturm-Liouville eigenvalue prob-

lem. We formulate a formally self adjoint q-difference operator in a Hilbert space. Some of

the properties of the eigenvalues and the eigenfunctions are discussed. The Green’s func-

tion is constructed and the problem in question is inverted into a q-type Fredholm integral

operator with a symmetric kernel. The set of eigenfunctions is shown to be a complete

orthogonal set in the Hilbert space. Examples involving basic trigonometric functions will

be given. A study for the asymptotics of the eigenvalues and eigenfunctions will be given

at the end of this Chapter. The results of this Chapter also exists in [16,19].

4.1 Introduction

Let [a, b] be a finite closed interval in R and let ν be a continuous real-valued function

defined on [a, b]. By a Sturm-Liouville problem we mean the problem of finding a function

y and a number λ in C satisfying the differential equation

Ly := −y′′ + ν(x)y(x) = λy(x), a 6 x 6 b, (4.1.1)

together with the boundary conditions

U1(y) := a1y(a) + a2y′(a) = 0, (4.1.2)

60

61

U2(y) := b1y(b) + b2y′(b) = 0, (4.1.3)

where ai and bi, i = 1, 2 are real numbers for which

|a1|+ |a2| 6= 0 6= |b1|+ |b2|. (4.1.4)

This problem has been extensively studied. It is known that the differential equation (4.1.1)

and the boundary conditions (4.1.2)–(4.1.3) determine a self adjoint operator in L2(a, b).

There is a sequence of real numbers λn∞n=0 with limn→∞ λn = +∞, such that corre-

sponding to each λn there is one and only one linearly independent solution of the problem

(4.1.1)–(4.1.3). The sequence λn∞n=0 is the sequence of eigenvalues and the sequence of

corresponding solutions φn∞n=0 is said to be a sequence of eigenfunctions. One of the

most important properties of these eigenfunctions is that, φn∞n=0 is an orthogonal basis

of L2(a, b). For example, let ν(x) ≡ 0 on [a, b]. If we take a = 0, b = π, a1 = 1, a2 = 0,

b1 = 1, and b2 = 0, we get

λn = n2, φn(x) = sinnx, n = 1, 2, . . . , (4.1.5)

leading to the well known fact that sinnx∞n=1 is a complete orthogonal set of L2(0, π),

while taking a = 0, b = π, a1 = 0, a2 = 1, b1 = 0, and b2 = 1, we get

λn = n2, φn(x) = cosnx, n = 0, 1, 2, . . . , (4.1.6)

which leads to the completeness of cosnx∞n=0 in L2(0, π). We mention here that the

Fourier orthogonal basiseinx

∞n=0

of L2(−π, π) does not arise in this setting but arises

from a simpler situation, namely the first order problem

−iy′ = λy, y(−π) = y(π). (4.1.7)

Among several references for the above mentioned facts we mention the monographs of

Coddington and Levinson [37], Estham [41], Levitan and Sargsjan [85, 86], Marchenko [91]

and Titchmarsh [114].

62

The discrete analogue of the theory outlined above, i.e. when the differential operator

d/dx is replaced by the forward difference operator ∆y(n) = y(n+1)−y(n) or the backward

operator ∇y(n) = y(n)− y(n− 1) where n is a positive integer belonging to a finite set of

integers of the form m,m+ 1,m+ 2, . . . ,m+N, m > 1, is treated in Atkinson’s [23], see

also [73].

The aim of this chapter is to study a basic analogue of Sturm–Liouville systems when

the differential operator is replaced by the q-difference operator Dq, see (1.1.28)–(1.1.29).

In [48, Chapter 5] and [49], a basic Sturm-Liouville system is defined. It is the system

Dq(r(x)Dqy) +(l(x) + λw(x)

)y(qx) = 0, a 6 x 6 b, (4.1.8)

h1y(a) + h2Dqy(a) = 0, (4.1.9)

k1y(b) + k2Dqy(b) = 0. (4.1.10)

where r, l and w are real-valued functions which posses appropriate q-derivatives, h1, h2,

k1, k2 are constants. It is proved [48, pp.164-170] that all eigenvalues of this system are real

and the eigenfunctions satisfy an orthogonality relation, [48, equation(5.1.5)]. Exton, [48],

considered only formal computational aspects to prove an orthogonality relation of some q-

special functions. No attention is paid to several points, which may lead to several mistakes.

First the existence of eigenvalues is not proved and it is not indicated how to determine the

eigenvalues and the eigenfunctions. A basic point here is that if a 6= 0 6= b, then it is not

guaranteed that initial conditions at either a or b determine a unique solution of (4.1.8),

see [9, 90]. The geometric and algebraic simplicity of the eigenvalues, which plays a major

rule in proving the reality of the eigenvalues and the orthogonality of the eigenfunctions

are not proved or even assumed. Moreover the space where the problem is defined is not

specified. If an inner product is defined in the view of [48, equation (5.1.5)], there will

be no orthogonality if h1, h2, k1 and k2 are not real. For more information concerning the

monograph [48], see the Review by Mourad Ismail in [SIAM Rev. 16(1985), pp. 279-281].

63

See also the Review of [49] by Wolfgang Hahn in [Math Rev. 84i:39002 (1982)]. In the

present paper we formulate a self adjoint basic Sturm–Liouville eigenvalue problem in a

Hilbert space. We prove the existence of a sequence of real eigenvalues with no finite limit

points. The associated Green’s function is constructed and the equivalence between the basic

Sturm-Liouville problem and a q-type Fredholm integral operator is proved. Illustrative

examples are given at the end of the paper.

There are several physical models involving q-derivatives, q-integrals, q-functions and

their related problems, see e.g. [36, 51, 52, 55, 108, 109]. Also the problem of expansions of

functions in terms of q-orthogonal functions, which seems to be first discussed by Carmichael

in [33,34] has attracted the work of several authors, see e.g. [6,28,29,64,109,111]. However,

as far we know, the general problem in the present setting has not been studied. At this

point, it is worth mentioning that our work based on the q-difference operator which is

attributed to Jackson, see [69], and a similar study of the Sturm-Liouville systems generated

by the Askey-Wilson divided difference operator, cf. [21] is very much needed. The results

of this chapter except Section 4.6 are contained in [16].

4.2 Fundamental solutions

In this section we investigate the fundamental solutions of the basic Sturm-Liouville equa-

tion

−1qDq−1Dqy(x) + ν(x)y(x) = λy(x), 0 6 x 6 a <∞, λ ∈ C, (4.2.1)

where ν is defined on [0, a] and continuous at zero. Let C2q (0, a) be the space of all functions

y defined on [0, a] such that y, Dqy are continuous at zero. Clearly, C2q (0, a) is a subspace

of the Hilbert space L2q(0, a).

Theorem 4.2.1. For c1, c2 ∈ C, equation (4.2.1) has a unique solution in C2q (0, a) which

64

satisfies

φ(0, λ) = c1, Dq−1φ(0, λ) = c2, λ ∈ C. (4.2.2)

Moreover φ(x, λ) is entire in λ for all x ∈ [0, a].

Proof. The functions

ϕ1(x, λ) = cos(sx; q), and ϕ2(x, λ) =

sin(sx; q)

s , λ 6= 0

x, λ = 0, (4.2.3)

where s :=√λ is defined with respect to the principal branch, is a fundamental set of

1qDq−1Dqy(x) + λy(x) = 0, (4.2.4)

with Wq

(ϕ1(·, λ), ϕ2(·, λ)

)≡ 1 on [0, a], λ ∈ C. For all x ∈ [0, a], λ ∈ C, we define a

sequence ym(·, λ)∞m=1 of successive approximations by

y1(x, λ) = c1ϕ1(x, λ) + c2ϕ2(x, λ), (4.2.5)

ym+1(x, λ) = c1ϕ1(x, λ) + c2ϕ2(x, λ)

+q∫ x

0ϕ2(x, λ)ϕ1(qt, λ)− ϕ1(x, λ)ϕ2(qt, λ) ν(qt) ym(qt, λ) dqt. (4.2.6)

We shall prove that for each fixed λ ∈ C the uniform limit of ym(·, λ)∞m=1 as m → ∞

exists and defines a solution of (4.2.1) and (4.2.2). Let λ ∈ C be fixed. There exist positive

numbers K(λ) and A such that

|ν(x)| 6 A, |ϕi(x, λ)| 6√K(λ)

2; i = 1, 2; x ∈ [0, a]. (4.2.7)

Let K(λ) :=(|c1| + |c2|

)√K(λ)2 . Then, from (4.2.7), |y1(x, λ)| 6 K(λ), for all x ∈ [0, a].

Using mathematical induction, we have

|ym+1(x, λ)− ym(x, λ)| 6 K(λ)qm(m+1)

2(AK(λ)x(1− q))m

(q; q)m, m = 1, 2, . . . . (4.2.8)

Consequently by Weierstrass’ test the series

y1(x, λ) +∞∑

m=1

ym+1(x, λ)− ym(x, λ) (4.2.9)

65

converges uniformly in [0, a]. Since the m-th partial sums of the series is nothing but

ym+1(·, λ), then ym+1(·, λ) approaches a function φ(·, λ) uniformly in [0, a] as m → ∞,

where φ(x, λ) is the sum of the series. We can also prove by induction on m that ym(x, λ)

and Dqym(x, λ) are continuous at zero, where

Dqym+1(x, λ) = c1Dqϕ1(x, λ) + c2Dqϕ2(x, λ)

+q∫ x

0Dqφ2(x, λ)ϕ1(qt, λ)−Dqϕ1(x, λ)ϕ2(qt, λ) ν(qt)ym(qt, λ) dqt,

(4.2.10)

m ∈ Z+. Hence, both φ(·, λ) and Dqφ(·, λ) are continuous at zero. i.e. φ(·, λ) ∈ C2q (0, a).

Because of the uniform convergence, letting m→∞ in (4.2.6) we obtain

φ(x, λ) = c1ϕ1(x, λ) + c2ϕ2(x, λ)

+q∫ x

0ϕ2(x, λ)ϕ1(qt, λ)− ϕ1(x, λ)ϕ2(qt, λ) ν(qt)φ(qt, λ) dqt.(4.2.11)

We prove that φ(·, λ) satisfies (4.2.1) and (4.2.2). Clearly φ(0, λ) = c1 and

Dq−1φ(0, λ) = limn→∞

φ(xqn, λ)− φ(0, λ)xqn

= c1Dq−1φ1(0, λ) + c2Dq−1φ2(0, λ) = c2, (4.2.12)

i.e. φ(·, λ) satisfies (4.2.2). To prove that φ(·, λ) satisfies (4.2.1), We distinguish between

two cases, x 6= 0 and x = 0. If x 6= 0, then from (1.1.28),

Dqφ(x, λ) = c1Dqϕ1(x, λ) + c2Dqϕ2(x, λ)

+ q

∫ x

0Dqφ2(x, λ)ϕ1(qt, λ)−Dqϕ1(x, λ)ϕ2(qt, λ) ν(qt)φ(qt, λ) dqt

(4.2.13)

and hence

−1qDq−1Dqφ(x, λ) =

−1qDq−1Dqϕ1(x, λ)

(c1 −

∫ x

0ϕ2(qt, λ)ν(qt)φ(qt, λ) dqt

)− 1qDq−1Dqϕ2(x, λ)

(c2 +

∫ x

0ϕ1(qt, λ)ν(qt)φ(qt, λ) dqt

)− ν(x)φ(x, λ).

(4.2.14)

Substituting from (4.2.3), (4.2.4) and (4.2.11) in (4.2.14), we conclude that φ(x, λ) satisfies

(4.2.1) for x 6= 0. If x = 0, then (4.2.1) is nothing but

D2qy(0)− qν(0)y(0) = −qλy(0). (4.2.15)

66

From (1.1.29), one can see that

D2qφ(0, λ) = c1D

2qϕ1(0, λ) + c2D

2qϕ2(0, λ) + qν(0)φ(0, λ)

= −c1qλϕ1(0, λ)− c2qλϕ2(0, λ) + qν(0)φ(0, λ)

= −qλφ(0, λ) + qν(0)φ(0, λ). (4.2.16)

Consequently (4.2.1) is satisfied at x = 0.

To prove that problem (4.2.1), (4.2.2) has a unique solution, assume the contrary, that

is ψi(·, λ), i = 1, 2, are two solutions of (4.2.1), (4.2.2). Let

χ(x, λ) = ψ1(x, λ)− ψ2(x, λ), x ∈ [0, a].

Then χ(·, λ) is a solution of (4.2.1) subject to the initial conditions

χ(0, λ) = Dq−1χ(0, λ) = 0.

Applying the q-integration process on (4.2.1) twice yields

χ(x, λ) =∫ x

0(x− t)

(λ− ν(t)

)χ(t, λ) dqt. (4.2.17)

Since χ(x, λ), ν(x) are continuous at zero, then there exist positive numbers Nx,λ, Mx,λ

such that

Nx,λ = supn∈N

|χ(xqn, λ)|, Mx,λ = supn∈N

∣∣λ− ν(xqn)∣∣. (4.2.18)

Again we can prove by mathematical induction on k that

|χ(x, λ)| 6 Nx,λMkx,λq

k2(1− q)2k x2k

(q; q)2k, k ∈ N, x ∈ [0, a]. (4.2.19)

Indeed, if (4.2.19) holds at k ∈ N, then from (4.2.17)

|χ(x, λ)| 6 Nx,λMk+1x,λ qk2 (1− q)2k

(q; q)2k

∫ x

0(x− t)t2k dqt

= Nx,λMk+1x,λ qk2

q2k+1 (1− q)2k+2

(q; q)2k+2x2k+2

= Nx,λMk+1x,λ q(k+1)2 (1− q)2k+2

(q; q)2k+2x2k+2. (4.2.20)

67

Hence (4.2.19) holds at k+1 and therefore (4.2.19) holds for all k ∈ N because from (4.2.18)

it holds at k = 0. Since limk→∞

Mkx,λq

k2(1−q)2k x2k

(q; q)2k= 0, then χ(x, λ) = 0, for all x ∈ [0, a].

This proves the uniqueness. Now, let M > 0 be arbitrary but fixed. To prove that φ(x, λ),

x ∈ [0, a], is entire in λ, it is sufficient to prove that φ(x, λ) is analytic in each disk ΩM ;

ΩM := λ ∈ C : |λ| 6 M. We prove by induction on m that

for all x ∈ [0, a] ym(x, λ) is analytic on ΩM , (4.2.21)

for all λ ∈ ΩM∂

∂λym(x, λ) is continuous at (0, λ). (4.2.22)

Clearly, ϕ1(x, λ), ϕ2(x, λ) are entire functions of λ for x ∈ [0, a]. Moreover∂

∂λφi(x, λ) is

continuous at (0, λ) for each λ ∈ C. Then (4.2.21) and (4.2.22 ) holds at m = 1. Assume

that (4.2.21) and (4.2.22) hold at m > 1. Then for x0 ∈ [0, a], λ0 ∈ ΩM , we obtain

∂ym+1(x0, λ)∂λ

∣∣λ=λ0

= q∂ϕ2(x0, λ)

∂λ

∣∣∣λ=λ0

∫ x0

0ϕ1(qt, λ)ym(qt, λ) dqt

+∂y1(x0, λ)

∂λ

∣∣∣λ=λ0

− q∂ϕ1(x0, λ)

∂λ

∣∣∣λ=λ0

∫ x0

0ϕ2(qt, λ)ym(qt, λ) dqt

+ qϕ2(x0, λ)∂

∂λ

(∫ x0

0ϕ1(qt, λ)ym(qt, λ) dqt

)∣∣∣λ=λ0

− qϕ1(x0, λ)∂

∂λ

(∫ x0

0ϕ2(qt, λ)ym(qt, λ) dqt

)∣∣∣λ=λ0

(4.2.23)

From (4.2.22) we conclude that ∂/∂λ(ϕi(qt, λ)ym(qt, λ)

), i = 1, 2, are continuous at (0, λ0).

Therefore there exist constants C, δ > 0 such that∣∣∣∣∣ ∂∂λ(ϕi(x0qn, λ)ym(x0q

n, λ))∣∣∣∣∣ 6 C, n ∈ N, |λ− λ0| 6 δ. (4.2.24)

Hence

x0(1− q)qn

∣∣∣∣∣ ∂∂λ(ϕi(x0qn+1, λ)ym(x0q

n+1 λ))∣∣∣∣∣ 6 x0A(1− q)qn, n ∈ N,

for all λ in the disk |λ− λ0| 6 δ, i.e. the series corresponding to the q-integrals∫ x0

0

∂

∂λ

(ϕi(qt, λ)ym(qt, λ)

)dqt, i = 1, 2 (4.2.25)

68

are uniformly convergent in a neighborhood of λ = λ0. Thus, we can interchange the

differentiation and integration processes in (4.2.23). Since x0, λ0 are arbitrary, then

∂

∂λym+1(x, λ) =

∂

∂λy1(x, λ)− q

∫ x

0

∂

∂λ

(ϕ2(x, λ)ϕ1(qt, λ)ym(qt, λ)

)ν(qt) dqt

+ q

∫ x

0

∂

∂λ

(ϕ1(x, λ)ϕ2(qt, λ)ym(qt, λ)

)ν(qt) dqt,

(4.2.26)

for all x ∈ [0, a], λ ∈ ΩM . From (4.2.22) the integrals in (4.2.26) are continuous at (0, λ).

Consequently ∂∂λym+1(x, λ) is continuous at (0, λ). Let x0 ∈ [0, a] be arbitrary. Then there

exists B(x0), B(x0) > 0 such that

|ϕi(x0, λ)| 6√B(x0)

2, i = 1, 2, |y1(x, λ)| 6 B(x0), λ ∈ ΩM . (4.2.27)

Finally the use of the mathematical induction yields

|ym+1(x0, λ)− ym(x0, λ)| 6 B(x0)qm(m+1)

2(AB(x0)λ(1− q))m

(q; q)m. (4.2.28)

Consequently the series (4.2.9), with x = x0, converges uniformly in ΩM to φ(x0, λ). Hence

φ(x0, λ) is analytic in ΩM , i.e. it is entire.

4.3 The self adjoint problem

In this section we define a basic Sturm-Liouville problem and prove that it is formally self

adjoint in L2q(0, a). The following lemma which is needed in the sequel indicates that unlike

the classical differential operator d/dx, Dq is neither self adjoint nor skew self adjoint.

Equation (4.3.2) below indicates that the adjoint of Dq is −1qDq−1 .

Lemma 4.3.1. Let f , g in L2q(0, a) be defined on [0, q−1a]. Then, for x ∈ (0, a], we have

(Dqg)(xq−1) = Dq,xq−1g(xq−1) = Dq−1g(x), (4.3.1)

〈Dqf, g〉 = f(a)g(aq−1)− limn→∞

f(aqn)g(aqn−1) +⟨f,−1qDq−1g

⟩, (4.3.2)⟨

− 1qDq−1f, g

⟩= lim

n→∞f(aqn−1)g(aqn)− f(aq−1)g(a) + 〈f,Dqg〉 . (4.3.3)

69

Proof. Relation (4.3.1) follows from

Dq−1g(x) =g(x)− g(q−1x)x(1− q−1)

=g(xq−1)− g(x)xq−1(1− q)

= (Dqg)(xq−1) = Dq,xq−1g(xq−1). (4.3.4)

Using the formula (1.1.41) of q-integration by parts we obtain

〈Dqf, g〉 =∫ a

0Dqf(x)g(x) dqx

= f(a)g(a)− limn→∞

f(aqn)g(aqn)−∫ a

0f(qt)Dqg(t) dqt

= f(a)g(a)− limn→∞

f(aqn)g(aqn)−∫ qa

0f(t)

1qDq−1g(t) dqt

= f(a)g(a)− limn→∞

f(aqn)g(aqn) + aq−1(1− q)f(a)Dq−1g(a)

+∫ a

0f(t)

−1qDq−1g(t) dqt

= f(a)g(aq−1)− limn→∞

f(aqn)g(aqn−1) +⟨f,−1qDqg

⟩, (4.3.5)

proving (4.3.2). Equation (4.3.3) can be proved by a use of (4.3.2).

Now consider the basic Sturm-Liouville problem

`(y) := −1qDq−1Dqy(x) + ν(x)y(x) = λy(x), 0 6 x 6 a <∞, λ ∈ C, (4.3.6)

U1(y) := a11y(0) + a12Dq−1y(0) = 0, (4.3.7)

U2(y) := a21y(a) + a22Dq−1y(a) = 0, (4.3.8)

where ν is a continuous at zero real valued function and aij , i, j ∈ 1, 2 are arbitrary real

numbers such that the rank of the matrix (aij)16i,j62 is 2. Problem (4.3.6)–(4.3.8) is said to

be formally self adjoint if for any functions y and z of C2q (0, a) which satisfy (4.3.7)–(4.3.8),

〈`y, z〉 = 〈y, `z〉 . (4.3.9)

Theorem 4.3.2. The basic Sturm- Liouville eigenvalue problem (4.3.6)–(4.3.8) is formally

self adjoint.

70

Proof. We first prove that for y, z in L2q(0, a), we have the following q-Lagrange’s identity∫ a

0

(`y(x)z(x)− y(x)`z(x)

)dqx = [y, z](a)− lim

n→∞[y, z](aqn), (4.3.10)

where

[y, z](x) := y(x)Dq−1z(x)−Dq−1y(x)z(x). (4.3.11)

Applying (4.3.3) with f(x) = Dqy(x) and g(x) = z(x), we obtain⟨− 1qDq−1Dqy(x), z(x)

⟩= −(Dqy)(aq−1)z(a) + lim

n→∞(Dqy)(aqn−1)z(aqn) +

⟨Dqy,Dqz

⟩= −Dq−1y(a)z(a) + lim

n→∞Dq−1y(aqn)z(aqn) +

⟨Dqy,Dqz

⟩.

(4.3.12)

Applying (4.3.2) with f(x) = y(x), g(x) = Dqz(x),

⟨Dqy,Dqz

⟩= y(a)Dqz(aq−1)− lim

n→∞y(aqn)Dqz(aqn−1) +

⟨y,−1

qDq−1Dqz

⟩= y(a)Dq−1z(a)− lim

n→∞y(aqn)Dq−1z(aqn) +

⟨y,−1

qDq−1Dqz

⟩. (4.3.13)

Therefore,

⟨− 1qDq−1Dqy(x), z(x)

⟩= [y, z](a)− lim

n→∞[y, z](aqn) +

⟨y,−1

qDq−1Dqz

⟩. (4.3.14)

Lagrange’s identity (4.3.10) results from (4.3.14) and the reality of ν(x). Letting y, z in

C2q (0, a) and assuming that they satisfy (4.3.7)–(4.3.8), we obtain

a11y(0) + a12Dq−1y(0) = 0, a11z(0) + a12Dq−1z(0) = 0. (4.3.15)

The continuity of y, z at zero implies that limn→∞[y, z](aqn) = [y, z](0). Then (4.3.14) will

be ⟨− 1qDq−1Dqy(x), z(x)

⟩= [y, z](a)− [y, z](0) +

⟨y,−1

qDq−1Dqz

⟩. (4.3.16)

Since a11 and a12 are not both zero, it follows from (4.3.15) that

[y, z](0) = y(0)Dq−1z(0)−Dq−1y(0)z(0) = 0.

71

Similarly,

[y, z](a) = y(a)Dq−1z(a)−Dq−1y(a)z(a) = 0.

Since ν(x) is real valued, then

〈`(y), z〉 =⟨− 1qDq−1Dqy(x) + ν(x)y(x), z(x)

⟩=

⟨− 1qDq−1Dqy(x), z(x)

⟩+ 〈ν(x)y, z(x)〉

=⟨y,−1

qDq−1Dqz(x)

⟩+⟨y, ν(x)z(x)

⟩= 〈y, `(z)〉 ,

i.e. (4.3.6)–(4.3.8) is a formally self adjoint operator.

A complex number λ∗ is said to be an eigenvalue of the problem (4.3.6)–(4.3.8) if there

is a non trivial solution φ∗ which satisfies the problem at this λ∗. In this case we say that

φ∗ is an eigenfunction of the basic Sturm-Liouville problem corresponding to the eigenvalue

λ∗. The geometric multiplicity of an eigenvalue is defined to be the number of linearly

independent solutions corresponding to it. In particular an eigenvalue is simple if and only

if it has only one linearly independent solution.

Lemma 4.3.3. The eigenvalues and the eigenfunctions of the boundary value problem

(4.3.6)–(4.3.8) have the following properties:

i. The eigenvalues are real.

ii. Eigenfunctions that belong to different eigenvalues are orthogonal.

iii. All eigenvalues are simple from the geometric point of view.

Proof. i. Let λ0 be an eigenvalue with an eigenfunction y0(·). From (4.3.9), we have

〈`(y0), y0〉 = 〈y0, `(y0)〉 . (4.3.17)

But `(y0) = λ0y0, implying

(λ0 − λ0

) ∫ a

0|y(x)|2 dqx = 0. (4.3.18)

72

Since y0(·) is non-trivial then λ0 = λ0, which proves i.

ii. Let λ, µ be two (real) distinct eigenvalues with corresponding eigenfunctions y(·), z(·),

respectively. Again, we make use of (4.3.9) to get

(λ− µ)∫ a

0y(x)z(x) dqx = 0. (4.3.19)

Since λ 6= µ, then y(·) and z(·) are orthogonal, proving ii.

iii. Let λ0 be an eigenvalue with two eigenfunctions y1(·) and y2(·). We prove that they

are linearly dependent by proving that their q-Wronskian vanishes at x = 0, see [8] for a

complete treatment of linear q-difference equations . Indeed,

Wq(y1, y2)(0) = y1(0)Dqy2(0)− y2(0)Dqy1(0) = 0, (4.3.20)

since both y1 and y2 satisfy (4.3.7).

In the following we indicate how to obtain the eigenvalues and the corresponding eigen-

functions. Let φ1(·, λ), φ2(·, λ) be the linearly independent solutions of (4.3.6) determined

by the initial conditions

Dj−1q φi(0, λ) = δij , i, j = 1, 2, λ ∈ C. (4.3.21)

Thus φ1(·, λ) is determined by (4.2.5) by taking c1 = 1, c2 = 0 and φ2(·, λ) is determined

by taking c1 = 0, c2 = 1. Then, every solution of (4.3.6) is of the form

y(x, λ) = A1φ1(x, λ) +A2φ2(x, λ), (4.3.22)

where A1 and A2 does not depend on x. A solution y(·, λ) of (4.3.6) will be an eigenfunction

if it satisfies the boundary conditions (4.3.7)–(4.3.8), i.e. if we can find a non trivial solution

of the linear system

A1U1(φ1) +A2U1(φ2) = 0,

A1U2(φ1) +A2U2(φ2) = 0,(4.3.23)

73

Hence, λ in R is an eigenvalue if and only if

∆(λ) =

∣∣∣∣∣∣∣U1(φ1) U1(φ2)

U2(φ1) U2(φ2)

∣∣∣∣∣∣∣ = 0. (4.3.24)

The function ∆(λ) defined in (4.3.24) is called the characteristic determinant associated

with the basic Sturm- Liouville problem (4.3.6)–(4.3.8). The zeros of ∆(λ) are exactly the

eigenvalues of the problem. Since φ1(x, λ) and φ2(x, λ) are entire in λ for each fixed x in

[0, a], then ∆(λ) is also entire. Thus the eigenvalues of the basic Sturm-Liouville system

(4.3.6)–(4.3.8) are at most countable with no finite limit points. From Lemma 4.3.3 we

know that all eigenvalues are simple from the geometric point of view. We can prove that

the eigenvalues are also simple algebraically, i.e. they are simple zeros of ∆(λ). Indeed, let

θ1(·, λ) and θ2(·, λ) be defined by the relations

θ1(x, λ) := U1(φ2)φ1(x, λ)− U1(φ1)φ2(x, λ),

θ2(x, λ) := U2(φ2)φ1(x, λ)− U2(φ1)φ2(x, λ).(4.3.25)

Hence, θ1(·, λ), θ2(·, λ) are solutions of (4.3.6) such that

θ1(0, λ) = a12, Dq−1θ1(0, λ) = −a11; θ2(a, λ) = a22, Dq−1θ2(a, λ) = −a21. (4.3.26)

One can verify that

Wq

(θ1(·, λ), θ2(·, λ)

)(x, λ) = ∆(λ)Wq

(φ1(·, λ), φ2(·, λ)

)= ∆(λ). (4.3.27)

Let λ0 be an eigenvalue of (4.3.6)–(4.3.8). Then λ0 is a real number and therefore θi(x, λ0)

can be taken to be real valued, i = 1, 2. From (4.3.27), we conclude that θ1(x, λ0), θ2(x, λ0)

are linearly dependent eigenfunctions. So, there exists a non zero constant k0 such that

θ1(x, λ0) = k0θ2(x, λ0). (4.3.28)

From (4.3.25) and (4.3.26)

θ1(a, λ0) = k0a22 = k0θ1(a, λ), Dq−1θ1(a, λ0) = −k0a21 = k0Dq−1θ1(a, λ). (4.3.29)

74

In the q-Lagrange identity (4.3.10), taking y(x) = θ1(x, λ), and z(x) = θ1(x, λ0) implies

(λ− λ0

) ∫ a

0θ1(x, λ)θ1(x, λ0) dqx = θ1(a, λ)Dq−1θ1(a, λ0)−Dq−1θ1(a, λ)θ1(a, λ0)

= k0

(θ1(a, λ)Dq−1θ2(a, λ)− θ2(a, λ)Dq−1θ1(a, λ)

)= k0Wq(θ1(·, λ), θ2(·, λ))(q−1a) = k0∆(λ).

Since ∆(λ) is entire in λ,

∆′(λ0) := limλ→λ0

∆(λ)λ− λ0

=1k0

∫ a

0θ21(x, λ0) dqx 6= 0. (4.3.30)

Therefore λ0 is a simple zero of ∆(λ).

4.4 Basic Green’s function

The q-type Green’s function arises when we seek a solution of the nonhomogeneous equation

−1qDq−1Dqy(x) + −λ+ ν(x) y(x) = f(x), x ∈ [0, a], λ ∈ C, (4.4.1)

which satisfies the boundary conditions (4.3.7)–(4.3.8), where f ∈ L2q(0, a) is given. First,

we note that if λ is not an eigenvalue of the Sturm-Liouville problem (4.3.6)–(4.3.8), then

the solution of (4.4.1), if it exists, would be unique. To see this, assume that χ1(x, λ),

χ2(x, λ) are two solutions of (4.4.1). Then χ1(x, λ) − χ2(x, λ) is a solution of the problem

(4.3.6)–(4.3.8). So, it is identically zero if λ is not an eigenvalue. Another proof of this

assertion is included in the proof of the next theorem.

Theorem 4.4.1. Suppose that λ is not an eigenvalue of (4.3.6)–(4.3.8). Let φ(·, λ) sat-

isfy the q-difference equation (4.4.1) and the boundary conditions (4.3.7)–(4.3.8), where

f ∈ L2q(0, a). Then

φ(x, λ) =∫ a

0G(x, t, λ)f(t) dqt, x ∈ 0, aqm, m ∈ N , (4.4.2)

75

where G(x, t, λ) is the Green’s function of problem (4.3.6)–(4.3.8) and it given by

G(x, t, λ) =−1

∆(λ)

θ2(x, λ)θ1(t, λ) , 0 6 t 6 x,

θ1(x, λ)θ2(t, λ) , x < t 6 a.

(4.4.3)

Conversely the function φ(·, λ) defined by (4.4.2) satisfies (4.4.1) and (4.3.7)–(4.3.8). Green’s

function G(x, t, λ) is unique in the sense that if there exists another function G(x, t, λ) such

that (4.4.2) is satisfied, then

G(x, t, λ) = G(x, t, λ), in L2q

((0, a)× (0, a)

). (4.4.4)

If f is q-regular at zero, then (4.4.2) holds for all x ∈ [0, a].

Proof. Using a q-analogue of the methods of variation of constants, a particular solution

of the non-homogenous equation (4.4.1) may be given by

φ(x, λ) = c1(x)θ1(x, λ) + c2(x)θ2(x, λ), (4.4.5)

where c1(x), c2(x) are solutions of the first order q-difference equations

Dq,xc1(x) =q

∆(λ)θ2(qx, λ)f(qx), Dq,xc2(x) = − q

∆(λ)θ1(qx, λ)f(qx). (4.4.6)

Define the q-geometric set Af by

Af :=x ∈ [0, a] :

∫ x

0θi(qt, λ)f(qt) dqt exists, i = 1, 2

. (4.4.7)

Af is a q-geometric set containing 0, aqm, m ∈ N because f ∈ L2q(0, a). From Holder

inequality we obtain∫ x

0|θi(qt, λ)f(qt)| dqt 6

(∫ x

0|θi(qt, λ)|2 dqt

)1/2(∫ x

0|f(t)|2 dqt

)1/2

, x ∈ Af .

Since f, θi ∈ L2q(0, a), i = 1, 2 and ∆(λ) 6= 0, then Dqci(·), i = 1, 2, are q−integrable on

[0, x] for all x ∈ Af . Hence direct computations lead to the following appropriate solutions

of (4.4.6)

c1(x) = c1(0) +q

∆(λ)

∫ x

0θ2(qt, λ)f(qt) dqt, x ∈ Af (4.4.8)

76

c2(x) = c2(a) +q

∆(λ)

∫ a

xθ1(qt, λ)f(qt) dqt, x ∈ Af . (4.4.9)

That is the general solution of (4.4.1) is given by

φ(x, λ) = c1θ1(x, λ) + c2θ2(x, λ) +q

∆(λ)θ1(x, λ)

∫ x

0θ2(qt, λ)f(qt) dqt

+q

∆(λ)θ2(x, λ)

∫ a

xθ1(qt, λ)f(qt) dqt, (4.4.10)

where x ∈ Af , and c1, c2 are arbitrary constants. Now, we determine c1, c2 for which

φ(x, λ) satisfies (4.3.7)–(4.3.8). It easy to see that

φ(0, λ) = c1θ1(0, λ) +(c2 +

q

∆(λ)

∫ a

0θ1(qt, λ)f(qt) dqt

)θ2(0, λ),

Dq−1φ(0, λ) = limn→∞x∈Af

φ(xqn, λ)− φ(0, λ)xqn

= c1Dq−1θ1(0, λ) +(c2 +

q

∆(λ)

∫ a

0θ1(qt, λ)f(qt) dqt

)Dq−1θ2(0, λ).

The boundary condition a11φ(0, λ) + a12Dq−1φ(0, λ) = 0 implies that

(c2 +

q

∆(λ)

∫ a

0θ1(qt, λ)f(qt) dqt

)Wq(θ1, θ2)(0) = 0. (4.4.11)

Therefore,

c2 =−q

∆(λ)

∫ a

0θ1(qt, λ)f(qt) dqt. (4.4.12)

Hence,

φ(x, λ) = c1θ1(x, λ) +q

∆(λ)

∫ x

0

(θ1(x, λ)θ2(qt, λ)− θ2(x, λ)θ1(qt, λ)

)f(qt) dqt. (4.4.13)

Now we compute φ(a, λ) and Dq−1φ(a, λ). Indeed, from the definition of the q-integration

(1.1.31) and relation (4.4.10)

φ(a, λ) = c1θ1(a, λ) +q

∆(λ)

∫ a

0

(θ1(a, λ)θ2(qt, λ)− θ2(a, λ)θ1(qt, λ)

)f(qt) dqt

= c1θ1(a, λ) +q

∆(λ)

∫ q−1a

0

(θ1(a, λ)θ2(qt, λ)− θ2(a, λ)θ1(qt, λ)

)f(qt) dqt

77

and

Dq−1φ(a, λ) = Dq−1θ1(a, λ)(c1 +

q

∆(λ)

∫ q−1a

0θ2(qt, λ)f(qt) dqt

)− q

∆(λ)Dq−1θ2(a, λ)

∫ q−1a

0θ1(qt, λ)f(qt) dqt.

The boundary condition a21φ2(a, λ) + a22Dq−1φ2(a, λ) = 0 implies(c1 +

q

∆(λ)

∫ q−1a

0θ2(qt, λ)f(qt) dqt

)Wq(θ1, θ2)(a) = 0. (4.4.14)

Hence

c1 =−q

∆(λ)

∫ q−1a

0θ2(qt, λ)f(qt) dq. (4.4.15)

So for x ∈ Af

φ(x, λ) =−q

∆(λ)θ2(x, λ)

∫ x

0θ1(qt, λ)f(qt) dqt−

q

∆(λ)θ1(x, λ)

∫ q−1a

xθ2(qt, λ)f(qt) dqt

=−1

∆(λ)θ2(x, λ)

∫ qx

0θ1(t, λ)f(t) dqt−

1∆(λ)

θ1(x, λ)∫ a

qxθ2(t, λ)f(t) dqt

=−1

∆(λ)θ2(x, λ)

∫ x

0θ1(t, λ)f(t) dqt−

1∆(λ)

θ1(x, λ)∫ a

xθ2(t, λ)f(t) dqt.

proving (4.4.2)–(4.4.3). Conversely, by direct computations, if φ(x, λ) is given by (4.4.2),

then it is a solution of (4.4.1) and satisfies the boundary conditions (4.3.7)–(4.3.8). To prove

the uniqueness, suppose that there exists another function, G(x, t, λ), such that

ψ(x, λ) =∫ a

0G(x, t, λ) f(t) dqt, (4.4.16)

is a solution of (4.4.1) which satisfies (4.3.7)–(4.3.8). For convenience, let

G(x, t, λ) =

G1(x, t, λ), 0 6 t 6 x,

G2(x, t, λ), x 6 t 6 a.

, G(x, t, λ) =

G1(x, t, λ), 0 6 t 6 x,

G2(x, t, λ), x 6 t 6 a.

By subtraction, we see that∫ a

0

G(x, t, λ)− G(x, t, λ)

f(t) dqt = 0, x ∈ 0, aqm, m ∈ N , (4.4.17)

78

holds for all functions f(t) ∈ L2q(0, a). Let us take

f(t) := G(x, t, λ)− G(x, t, λ), x ∈ aqm, m ∈ N .

Then ∫ a

0

∣∣∣G(aqm, t, λ)− G(aqm, t, λ)∣∣∣2 dqt

=∫ aqm

0

∣∣∣G1(aqm, t, λ)− G1(aqm, t, λ)∣∣∣2 dqt+

∫ a

aqm

∣∣∣G2(aqm, t, λ)− G2(aqm, t, λ)∣∣∣2 dqt

= a(1− q)∞∑

n=0

qn∣∣∣G(aqm, aqn, λ)− G(aqm, aqn, λ)

∣∣∣2 = 0 (4.4.18)

Therefore, from (4.4.18) we conclude that

G(aqm, aqn, λ) = G(aqm, aqn, λ), m, n ∈ N.

If f is q-regular at zero, then Af ≡ [0, a] and (4.4.2) will be defined for all x ∈ [0, a].

Theorem 4.4.2. Basic Green’s function has the following properties

(i) G(x, t, λ) is continuous at the point (0, 0).

(ii) G(x, t, λ) = G(t, x, λ).

(iii) For each fixed t ∈ (0, qa], as a function of x, G(x, t, λ) satisfies the q-difference equation

(4.3.6) in the intervals [0, t), (t, a] and it also satisfies the boundary conditions (4.3.7)–

(4.3.8).

(iv) Let λ0 be a zero of ∆(λ). Then λ0 can be a simple pole of the function G(x, t, λ), and

in this case

G(x, t, λ) =−ψ0(x)ψ0(t)

λ− λ0+ G(x, t, λ), (4.4.19)

where G(x, t, λ) is analytic function of λ in a neighborhood of λ0 and ψ0 is a normalized

eigenfunction corresponding to λ0.

79

Proof. (i) Follows from the continuity of θ1(·, λ), θ2(·, λ) at zero for each fixed λ ∈ C and

(ii) is easily checked. Now, we prove (iii). Let t ∈ (0, qa] be fixed. If x ∈ [0, t], then

G(x, t, λ) =1

∆(λ)θ1(x, λ)θ2(t, λ).

So,

`G(x, t, λ) =1

∆(λ)θ2(t, λ)`θ1(x, λ) =

λ

∆(λ)θ2(t, λ)θ1(x, λ) = λG(x, t, λ).

Similarly if x ∈ [t, a]. From (4.3.26) and (4.4.3), we have

a11G(0, t, λ) + a12Dq−1G(0, t, λ) =1

∆(λ)θ2(t, λ)

a11θ1(0, λ) + a12Dq−1θ1(0, λ)

= 0,

a21G(a, t, λ) + a22Dq−1G(a, t, λ) =1

∆(λ)θ1(t, λ)

a21θ1(a, λ) + a22Dq−1θ1(a, λ)

= 0.

(iv) Let λ0 be a pole of G(x, t, λ), and R(x, t) be the residue of G(x, t, λ) at λ = λ0. From

(4.3.28) and (4.3.30), we obtain

R(x, t) = limλ→λ0

(λ− λ0)G(x, t, λ) = k−10 θ1(x, λ0)θ1(t, λ0) lim

λ→λ0

λ− λ0

∆(λ)

= −θ1(x, λ0)θ1(t, λ0)∫ a0 |θ1(u, λ)|2 dqu

= −ψ0(x, λ0)ψ1(t, λ0).

4.5 Eigenfunction expansions

In this section, the existence of a countable sequence of eigenvalues of ` with no finite limit

points will be proved by using the spectral theorem of compact self adjoint operators in

Hilbert spaces, see e.g. [24]. Moreover it will be proved that the corresponding eigenfunctions

form an orthonormal basis of L2q(0, a). We define the operator L : DL → L2

q(0, a) to be

Ly = `y for all y ∈ DL, where DL is the subspace of L2q(0, a) consisting of those complex

valued functions y that satisfies (4.3.7)–(4.3.8) such that Dqy is q-regular at zero and D2qy

lies in L2q(0, a). Thus L is the difference operator generated by the difference expression `

and the boundary conditions (4.3.7)–(4.3.8). By L(y) = λy, we mean that `(y) = λy and

80

y satisfies (4.3.7)–(4.3.8). The operator L has the same eigenvalues of the basic Sturm-

Liouville problem (4.3.6)–(4.3.8). We assume without any loss of generality that λ = 0 is

not an eigenvalue. Thus kerL = 0. From the previous section the solution of the problem

(Ly)(x) = f(x), f ∈ L2q(0, a), (4.5.1)

is given uniquely in L2q(0, a) by

y(x) =∫ a

0G(x, t)f(t) dqt, (4.5.2)

where

G(x, t) = G(x, t, 0) =

cθ1(t)θ2(x), 0 6 t 6 x

cθ1(x)θ2(t), x 6 t 6 a,

, c := − 1Wq(θ1, θ2)

. (4.5.3)

Replacing f by λy in (4.5.1). Then the eigenvalue problem

(Ly)(x) = λy(x), (4.5.4)

is equivalent to the following basic Fredholm integral equation of the second kind

y(x) = λ

∫ a

0G(x, t)y(t) dqt, in L2

q(0, a). (4.5.5)

Let G be the integral operator

G : L2q(0, a) −→ L2

q(0, a), (Gf)(x) =∫ a

0G(x, t)f(t) dqt. (4.5.6)

We prove that

(LG)f = f, f ∈ L2q(0, a). (4.5.7)

We show first that y = Gf ∈ DL. From (4.5.5) and (4.5.6)

y(x) = (Gf)(x) = θ2(x)y1(x) + θ1(x)y2(x), (4.5.8)

81

where

y1(x) = c

∫ x

0θ1(t)f(t) dqt, y2(x) = c

∫ a

xθ2(t)f(t) dqt.

Thus, for all x ∈ Af , cf. (4.4.7),

Dqy(x) = Dqθ2(x)y1(qx) +Dqθ1(x)y2(qx), (4.5.9)

D2qy(x) = −qν(qx)(y)(qx)− qf(qx) ∈ L2

q(0, a). (4.5.10)

Since Dqθi(x, λ), yi(x), i = 1, 2, are q-regular at zero, then so is Dqy and

Dq−1y(0) = Dqy(0) = limn→∞x∈Af

y(xqn)− y(0)xqn

= Dq−1θ2(0)y2(0),

y1(0) = 0, and y2(a) = 0, then

a11y(0) + a12Dq−1y(0) =(a11θ1(0) + a12Dq−1θ1(0)

)y2(0) = 0,

and

a21y(a) + a22Dq−1y(a) =(a21θ2(a) + a22Dq−1θ2(a)

)y1(a) = 0.

Thus y ∈ DL. It follows from (4.5.10) that Ly = (LG)(f) = f . Hence we have established

(4.5.7). Also, we can see that

(GL)(y) = y, y ∈ DL. (4.5.11)

Indeed, replacing f in (4.5.7) by Ly, we get Ly = LGLy. Thus y = GLy since L is assumed

to be injective. It follows from (4.5.7) and (4.5.11) that kerG = 0 and φ is an eigenfunction

of G with eigenvalue µ if and only if φ is an eigenfunction of L with eigenvalue 1/µ.

Theorem 4.5.1. The operator G is compact and self adjoint.

Proof. Let f, h,∈ L2q(0, a). Since G(x, t) is a real valued function defined on [0, a]× [0, a]

and G(x, t) = G(t, x), then for f, h ∈ L2q(0, a),

〈G(f), h〉 =∫ a

0(Gf)(x)h(x) dqx =

∫ a

0

∫ a

0G(x, t)f(t)h(x) dqt dqx

=∫ a

0f(t)

( ∫ a

0G(t, x)h(x) dqx

)dqt = 〈f,G(h)〉 ,

82

i.e. G is self adjoint. Let

φij(x, t) = φi(x)φj(t)∞i,j=1

be an orthonormal basis of L2q

((0, a) × (0, a)

). Consequently G =

∑∞i,j=1

⟨G,φij

⟩φij . For

n ∈ Z+ set Gn =∑n

i,j=1

⟨G,φij

⟩φij , and let Gn be the finite rank integral operator defined

on L2q(0, a) by

Gn(f)(x) :=∫ a

0Gn(x, t)f(t) dqt, in L2

q(0, a). (4.5.12)

Obviously Gn is compact for all n ∈ N. From the Cauchy-Schwarz inequality

‖(G − Gn)(f)‖ =(∫ a

0|(G − Gn)(f)(x)|2 dqt

)1/2

=(∫ a

0

∣∣∣ ∫ a

0(G−Gn)(x, t)(f)(t) dqt

∣∣∣2 dqx

)1/2

6

(∫ a

0

∫ a

0|(G−Gn)(x, t)|2 dqt dqx

)1/2(∫ a

0|f(x)|2 dqx

)1/2

= ‖G−Gn‖2 ‖f‖ ,

then

‖G − Gn‖ 6 ‖G−Gn‖2 → 0 as n→∞.

This completes the proof.

Corollary 4.5.2. The eigenvalues of the operator L form an infinite sequence λk∞k=1 of

real numbers which can be ordered so that

|λ1| 6 |λ2| 6 . . . 6 |λn| 6 . . . →∞ as n→∞.

The set of all normalized eigenfunctions of L form an orthonormal basis for L2q(0, a).

Proof. Since G is a compact self adjoint operator on L2q(0, a), then G has an infinite

sequence of non zero real eigenvalues µn∞n=1, ⊆ R, µn → 0 as n → ∞. Let φn∞n=1

denote an orthonormal set of eigenfunctions corresponding to µn∞n=1. From the spectral

83

theorem of compact self adjoint operators, we have,

G(f) =∞∑

n=0

λn

⟨f, φn

⟩φn. (4.5.13)

Since the eigenvalues λn∞n=1 of the operator L are the reciprocal of those of G, then

|λn| =1|µn|

→ ∞ as n→∞. (4.5.14)

Let y ∈ DL. Then, y = G(f), for some f ∈ L2q(0, a). Consequently,

y =∞∑

n=0

λn 〈f, φn〉φn =∞∑

n=0

µn 〈ly, φn〉φn

=∞∑

n=0

µn 〈y, `φn〉φn =∞∑

n=0

〈y, φn〉φn.

If zero is an eigenvalue of L. Then, we can choose r ∈ R such that r is not an eigenvalue of

L. Now, applying the above result on L − rI in place of L yields the corollary.

Example 4.5.1. Consider the q-Sturm-Liouville boundary value problem

−1qDq−1Dqy(x) = λy(x), (4.5.15)

with the q-Dirichlet conditions

U1(y) = y(0) = 0, U2(y) = y(1) = 0. (4.5.16)

A fundamental set of solutions of (4.5.15) is

φ1(x, λ) = cos(√λx; q), φ2(x, λ) =

sin(√λx; q)√λ

. (4.5.17)

Now, the eigenvalues of problem (4.5.15) are the zeros of the determinant

∆(λ) =

∣∣∣∣∣∣∣U1(φ1) U2(φ1)

U1(φ2) U2(φ2)

∣∣∣∣∣∣∣ = φ2(1, λ) =sin(

√λ; q)√λ

. (4.5.18)

84

Hence, the eigenvalues λm∞m=1 are the zeros of sin(√λ; q). From (1.2.21),

λm =q−2m

(1− q)2(1 +O(qm)

), m > 1, (4.5.19)

for sufficiently large m and the corresponding set of eigenfunctions

sin(√

λmx;q)√λm

∞m=1

is an

orthogonal basis of L2q(0, 1). In the previous notations

θ1(x, λ) =sin(

√λx; q)√λ

, (4.5.20)

and

θ2(x, λ) =sin(

√λ; q)√λ

cos(√λx; q) + cos(

√λ; q)

sin(√λx; q)√λ

. (4.5.21)

So, if λ is not an eigenvalue, Green’s function is given by

G(x, t, λ) =sin(

√λt; q)

sin(√λ; q)

(cos(

√λx; q)

sin(√λ; q)√λ

− cos(√λ; q)

sin(√λx; q)√λ

), (4.5.22)

for 0 6 t 6 x, and

G(x, t, λ) =sin(

√λx; q)

sin(√λ; q)

(cos(

√λt; q)

sin(√λ; q)√λ

− cos(√λ; q)

sin(√λt; q)√λ

), (4.5.23)

for x 6 t 6 1. Since λ = 0 is not an eigenvalue, then Green’s function G(x, t) is nothing

but

G(x, t) = G(x, t, 0) =

t(1− x), 0 6 t 6 x,

x(1− t), x 6 t 6 1.

Hence the boundary value problem (4.5.15)–(4.5.16) is equivalent to the basic Fredholm in-

tegral equation

y(x) = λ

∫ 1

0G(x, t)y(t) dqt. (4.5.24)

Example 4.5.2. Consider the equation (4.5.15) with the q-Neumann boundary conditions

U1(y) = Dq−1y(0) = 0, U2(y) = Dq−1y(1) = 0. (4.5.25)

In this case θ1(x, λ) = cos(√λx; q), and

θ2(x, λ) = cos(√λq−1/2; q) cos(

√λx; q) +

√q sin(

√λq−1/2; q) sin(

√λx; q).

85

Since ∆(λ) =√qλ sin(

√λq−1/2; q). Then λ0 = 0 and for sufficiently large m, the eigenvalue

are given by

λm =q−2m+1

(1− q)2(1 +O(qm)) , m > 1 (4.5.26)

Therefore,1, cos(

√λmx; q)

∞m=1

is an orthogonal basis of L2q(0, 1). If λ is not an eigen-

value, then the Green’s function G(x, t, λ) is defined for x, t ∈ [0, a]× [0, a] by

G(x, t, λ) = − cos(√λt; q)

√qλ sin(

√λq−1/2; q)

(cos(

√λq−1/2; q) cos(

√λx; q)

+√q sin(

√λq−1/2; q) sin(

√λx; q)

),

and

G(x, t, λ) = − cos(√λx; q)

√qλ sin(

√λq−1/2; q)

(cos(

√λq−1/2; q) cos(

√λt; q)

+√q sin(

√λq−1/2; q) sin(

√λt; q)

), x 6 t 6 1.

The operator L associated with problem (4.5.15), (4.5.25) is not invertible since zero is an

eigenvalue.

Example 4.5.3. Consider (4.5.15) with the following boundary conditions

U1(y) = y(0) = 0, U2(y) = y(1) +Dq−1y(1) = 0. (4.5.27)

Then

∆(λ) = φ2(1, λ) +Dq−1φ2(1, λ)

=sin(

√λ; q)√λ

+ cos(√λq−1/2; q).

(4.5.28)

The eigenvalues λm∞m=1 of this boundary value problem are the solutions of the equation

sin(√λ; q)√λ

= − cos(√λq−1/2; q) (4.5.29)

86

and the corresponding eigenfunctions are

sin(√

λmx;q)√λm

∞m=1

. The functions θ1(x, λ) and

θ2(x, λ) are

θ1(x, λ) =sin(

√λx; q)√λ

,

θ2(x, λ) =(cos(

√λq−1/2; q) +

sin(√λ; q)√λ

)cos(

√λx; q)

−(−√λq sin(

√λq−1/2; q) + cos(

√λ; q)

)sin(√λx; q)√λ

.

If λ is not an eigenvalue, then the Green’s function G(x, t, λ) is defined to be

G(x, t, λ) =−1

sin(√λ; q) +

√λ cos(

√λq−1/2; q)

sin(

√λt; q)θ2(x, λ), 0 6 t 6 x,

sin(√λx; q)θ2(t, λ), x 6 t 6 1.

and

G(x, t) =−12

t(2− x), 0 6 t 6 x,

x(2− t), x 6 t 6 1.

Therefore the boundary value problem (4.5.15), (4.5.27) is equivalent to the basic Fredholm

integral equation

y(x) = λ

∫ 1

0G(x, t)y(t) dqt.

Remark 4.5.1. Let r be a function defined on [0, q−1a] and let w be a function defined on

[0, a] such that Dq−1r(0) exists and w is positive on 0, aqn, n ∈ N. The Sturm–Liouville

problem (4.3.6)–(4.3.8) for 0 6 x 6 a <∞, λ ∈ C is

M(y) :=−1qDq−1

(r(x)Dqy(x)

)+ ν(x)y(x) = λw(x)y(x), (4.5.30)

U1(y) := a11y(0) + a12Dq−1y(0) = 0, (4.5.31)

U2(y) := a21y(a) + a22Dq−1y(a) = 0, (4.5.32)

87

where the functions ν and the constants aij , i, j ∈ 1, 2 as in Section 4.3. In this case

we will have the Lagrange identity∫ a

0

((My)(x)z(x)− y(x)(Mz)(x)

)w(x) dqx

= r(aq−1)[y, z](a)− limn→∞

r(aqn−1)[y, z](aqn).(4.5.33)

If y, z, and r are q-regular at zero, then the Lagrange identity (4.5.33) becomes∫ a

0

((My)(x)z(x)− y(x)(Mz)(x)

)w(x) dqx = r(aq−1)[y, z](a)− r(0)[y, z](0). (4.5.34)

The problem (4.5.30)–(4.5.32) is formally self adjoint in L2q

((0, a), w

), where L2

q

((0, a), w

)is the space with the inner product

〈f, g〉w =∫ a

0f(x)g(x)w(x) dqx, f, g ∈ L2

q

((0, a);w

). (4.5.35)

Let φ1(·, λ), φ2(·, λ) be the linearly independent solutions of (4.5.30) subject to the initial

conditions

Dj−1q φi(0, λ) = δij , i, j = 1, 2, λ ∈ C. (4.5.36)

The Green’s function of the problem (4.5.30)–(4.5.32)

−1qDq−1

(r(x)Dqy(x)

)+ −λw(x) + ν(x) y(x) = f(x), f ∈ L2

q

((0, a), w

), (4.5.37)

is given by

G(x, t, λ) =−1

∆(λ)

θ2(x, λ)θ1(t, λ)r(t)

, 0 6 t 6 x,

θ1(x, λ)θ2(t, λ)r(t)

, x < t 6 a,

(4.5.38)

where θi(x, λ) are linearly independent solutions of (4.5.30) in [0, a] such that

U1θ1(·, λ) = 0, U2θ2(·, λ) = 0. (4.5.39)

We define the operator M : DM → L2q

((0, a), w

)by My = My for all y ∈ DM, where

DM is the subspace of L2q

((0, a), w

)consisting of those complex valued functions that

88

satisfy (4.5.31)–(4.5.32) such that(rDqy

)(·) is q-regular at zero and Dq−1

((rDqy)(·)

)∈

L2q

((0, a), w

). An analysis similar work to Section 4.5 shows that the operator M has an

infinite sequence of eigenvalues with no finite limit points and the eigenfunctions form an

orthogonal basis of L2q

((0, a), w

).

4.6 Asymptotics of eigenvalues and eigenfunctions

In this section we investigate the asymptotic behavior of the eigenvalues and the eigenfunc-

tions of the basic Sturm–Liouville problem (4.3.6)–(4.3.8). The results of this section are

contained in [19]. The trigonometric identity

sin z cosw − cos z sinw = sin(z − w), (4.6.1)

where z, w are complex numbers, plays an important role in deriving the asymptotics of the

eigenvalues of the classical Sturm–Liouville problem (4.1.1)–(4.1.3). Unfortunately, there is

no known q-analogue of such a rule. For For z in C and k in Z+, we define the a function

Sk(z; q) by

Sk(z; q) := sin(z; q) cos(qkz; q)− cos(z; q) sin(qkz; q). (4.6.2)

We have the following useful relation which plays a key role in our investigation.

Lemma 4.6.1. For any z ∈ C and k ∈ N, k > 1, we have

Sk(z; q) =k−1∑n=0

(−1)n qn2+nz2n+1

Γq(2n+ 2)(qk−n; q)2n+1. (4.6.3)

Proof. Since the power series expansions of the basic cosine and sine functions defined

respectively in (1.1.16) and (1.1.17) are absolutely convergent, then we obtain

Sk(z; q) =∞∑

n=0

(−1)nqn2+n z2n+1

Γq(2n+ 2)

2n+1∑j=0

(−1)j

[2n+ 1j

]q

qj(j−1)/2q(k−n)j . (4.6.4)

89

Applying (1.1.8) with m = 2n+ 1 and u = qk−n yields

2n+1∑j=0

(−1)j

[2n+ 1j

]q

qj(j−1)/2q(k−n)j = (qk−n; q)2n+1. (4.6.5)

Moreover (qk−n; q)2n+1 = 0 whenever n > k. This completes the proof.

Remark 4.6.1. Identity (4.6.3) is a q-analogue of the result of [115, p.259, Eq. (2)] when

ν = ±1/2.

Lemma 4.6.2. For any k ∈ Z+ we obtain

|Sk(z; q)| 6 sinh(|z|; q) 6 q−1/4E(|z|; q). (4.6.6)

Proof. From (4.6.3) we find that

|Sk(z; q)| 6k−1∑n=0

qn2+n|z|2n+1

Γq(2n+ 2)6

∞∑n=0

qn2+n|z|2n+1

Γq(2n+ 2)= sinh(|z; q|). (4.6.7)

Moreover from (1.1.20), we have

E(|z|; q) = cosh(|z|; q) + q1/4 sinh(|z|; q). (4.6.8)

Consequently

sinh(|z|; q) 6 q−1/4E(|z|; q). (4.6.9)

Thus (4.6.6) follows from combining (4.6.7) and (4.6.9).

Lemma 4.6.3. For k ∈ Z+, x ∈ R, and z ∈ C, if |zx| > q−k

1− q, then for any l ∈ N

∣∣Sk(zx; q)∣∣

l+k∏m=l+1

(1 +

|z|2x2

x2m

) 6qk2

(1− q)2k−1

(q; q)∞

l+k∏m=l+1

x2m, (4.6.10)

where Sk(·; q) is defined in (4.6.3).

90

Proof. From (4.6.3) we obtain

∣∣Sk(zx; q)∣∣

l+k∏m=l+1

(1 +

|z|2x2

x2m

) 6

k−1∑r=0

qr2+r(|zx|)2r+1

(qk−r; q

)2r+1

Γq(2r + 2)l+k∏

m=l+1

(1 +

|z|2x2

x2m

)6

( l+k∏m=l+1

x2m

) k−1∑r=0

qr2+r (|zx|)2r−2k+1

Γq(2r + 2)

=( l+k∏

m=l+1

x2m

) k−1∑j=0

q(k−j)(k−j−1) (|zx|)−2j−1

Γq(2k − 2j). (4.6.11)

If |zx| > q−k

1−q , then |zx|−2j−1q−2jk 6 qk(1− q)2j+1. Therefore

k−1∑j=0

qk2−k−2kj+j(j+1) (|zx|)−2j−1

Γq(2k − 2j)6 qk2

k−1∑j=0

qj(j+1) (1− q)2j+1

Γq(2k − 2j)

6 qk2(1− q)2k

k−1∑j=0

qj(j+1)

(q; q)2k−2j−16qk2

(1− q)2k−1

(q; q)∞.

(4.6.12)

The lemma follows by combining (4.6.11) and (4.6.12).

For a positive integer k let fk be a function defined on C by

fk(zx; q) := Sk(z; q)∞∏

m=1

(1 +

q2k|z|2

x2m

)(1 +

|z|2

x2m

) . (4.6.13)

Theorem 4.6.4. There exists a positive constant h such that for all n ∈ N, if |xz| > q−n

1− q,

where x ∈ R, and z ∈ C, then

|fk(zx; q)| 6 h, k = 1, 2, . . . , n. (4.6.14)

Proof. If x = 0, then fk(zx; q)|x=0 = Sk(0; q) = 0, for all k ∈ N and the lemma holds. In

the following we study the case x 6= 0. From relation (1.2.50) there exists a positive integer

91

m0 and a positive constant A such that

∣∣1− q2kx2m+k

x2m

∣∣ 6 Aqm, k ∈ N, for all m > m0. (4.6.15)

Choosing m0 sufficiently large, from (1.2.20) we deduce that there exists a positive constant

B such that ∣∣qm−1/2(1− q)xm − 1∣∣ 6 Bqm, for all m > m0. (4.6.16)

Set

h(m0) :=

(−B; q

)2∞(−Aqm0+1; q

)∞

(1− q)(q; q)∞

m0∏m=1

(1 + x2m). (4.6.17)

In the following we prove that for any n ∈ N, h(m0) is an upper bound of the functions

fk(|z|x; q), k = 1, 2, . . . , n in the region |zx| > q−n

1−q . Notice that for z ∈ C, x ∈ R, we have

1 +q2k|z|2x2

x2m

6(1 +

|z|2x2

x2m+k

)+ |zx|2

∣∣ 1x2

m+k

− q2k

x2m

∣∣. (4.6.18)

Thus from (4.6.15) we obtain for m > m0,

1 +q2k|z|2x2

x2m

1 + |z|2x2

x2m+k

6 1 +|zx|2

1 + |z|2x2

x2m+k

∣∣∣∣∣ 1x2

m+k

− q2k

x2m

∣∣∣∣∣6 1 +

∣∣∣∣∣1− q2kx2m+k

x2m

∣∣∣∣∣ 6 (1 +Aqm).

(4.6.19)

Hence by (4.6.17)

∞∏m=m0+1

1 + q2k|z|2x2

x2m

1 + |z|2x2

x2m+k

6∞∏

m=m0+1

(1 +Aqm) =(−Aqm0+1; q

)∞. (4.6.20)

Consequently

∞∏m=m0+1

1 +q2k|z|2x2

x2m

1 +|z|2x2

x2m

6(−Aqm0+1; q

)∞

∞∏m=m0+1

1 +|z|2x2

x2m+k

1 +|z|2x2

x2m

=

(−Aqm0+1; q

)∞∏m0+k

m=m0+1

(1 + |z|2x2

x2m

) .(4.6.21)

92

For k ∈ N, x ∈ R, and z ∈ C let

gk(zx; q) := Sk(zx; q)∞∏

m=m0+1

(1 +

q2k|z|2x2

x2m

)(1 +

|z|2x2

x2m

) . (4.6.22)

Thus

fk(zx; q) =( m0∏

m=1

(1 + q2k|z|2x2

x2m

)(1 + |z|2x2

x2m

) )gk(zx; q), k ∈ N, (4.6.23)

From (4.6.21) and Lemma 4.6.3 we conclude that

|gk(zx; q)| 6(−Aqm0+1; q

)∞

∣∣Sk(zx; q)∣∣∏m0+k

m=m0+1

(1 + |z|2x2

x2m

)6

(−Aqm0+1; q

)∞q

k2(1− q)2k−1

(q; q)∞

( m0+k∏m=m0+1

x2m

).

(4.6.24)

Indeed, since |zx| >q−n

1− q, then |zx| >

q−k

1− q, k = 1, 2, . . . , n, then from Lemma 1.2.4 we

obtain

m0∏m=1

1 +q2k|z|2x2

x2m

1 +|z|2x2

x2m

=m0∏

m=1

x2m + q2k|z|2x2

x2m + |z|2x2 6 q2km0

m0∏m=1

(1 + x2m), k = 1, 2, . . . , n. (4.6.25)

Combining (4.6.23), (4.6.24) and (4.6.25), yields

|fk(zx; q)| 6(−Aqm0+1; q

)∞q

k2+2m0k(1− q)2k−1

(q; q)∞

m0+k∏m=m0+1

x2m

m0∏m=1

(1 + x2m), (4.6.26)

k = 1, 2, . . . , n, |zx| > q−n

1− q. From (4.6.16) we have

0 6 x2m 6 q−2m+1(1− q)−2(1 +Bqm)2. (4.6.27)

Thereforem0+k∏

m=m0+1

x2m 6 q−2m0k−k2

(1− q)−2km0+k∏

m=m0+1

(1 +Bqm)2

= q−2m0k−k2(1− q)−2k

( ∞∏m=0

(1 +Bqm))2 = q−2m0k−k2

(1− q)−2k(−B; q)2∞.

(4.6.28)

93

Thus for |zx| > q−n

1− q, formula (4.6.14) is satisfied. Clearly the constant h(m0) defined on

(4.6.17) is independent of n and x. This completes the proof of the theorem.

Lemma 4.6.5. The functions

φ1(x, λ) = cos(sx; q)+

q

s

∫ x

0cos(sx; q) sin(sqt; q)− sin(sx; q) cos(sqt; q) ν(qt)φ1(qt, λ) dqt,

(4.6.29)

φ2(x, λ) =sin(sx; q)

s+

q

s

∫ x

0cos(sx; q) sin(sqt; q)− sin(sx; q) cos(sqt; q) ν(qt)φ2(qt, λ) dqt,

(4.6.30)

form a fundamental set of solutions of (4.3.6) subject to the initial conditions

Dj−1q φi(0, λ) = δij , i, j = 1, 2, (4.6.31)

where s :=√λ is defined with respect to the principal branch.

Proof. Applying the q-analogue of the Picard– Lindelof method of successive approxima-

tions introduced in [8], we conclude that a particular solution φ0 of (4.3.6) is given in terms

of a fundamental set of

−1qDq−1Dqy(x) = λy(x), x ∈ R, (4.6.32)

by the identity

φ0(x, λ) = u1(x)ϕ1(x, λ) + u2(x)ϕ2(x, λ),

where the functions u1 and u2 are solutions of the first order q-difference equations

Dqui(x) =Wq,i(ϕ1, ϕ2)(qx, λ)Wq(ϕ1, ϕ2)(qx, λ)

ν(qx)ψ(x, λ), i = 1, 2, (4.6.33)

where Wq,i(x) is the determinant obtained from Wq(x) by replacing the i-th column by the

column(

01

). The functions

ϕ1(x, λ) = cos(sx; q), ϕ2(x, λ) =sin(sx; q)

s,

94

form a fundamental set of solutions of (4.6.32) with Wq(ϕ1, ϕ2)(x, λ) ≡ 1. Then,

Wq,1(ϕ1, ϕ2)(x, λ) = −ϕ2(x, λ) and Wq,2(φ1, φ2)(x, λ) = ϕ1(x, λ).

Thus, the functions

u1(x, λ) := −∫ x

0ϕ2(qt, λ)ν(qt)φ(qt, λ) dqt, and u2(x, λ) =

∫ x

0ϕ1(qt, λ)ν(qt)φ(qt, λ) dqt,

are solutions of (4.6.33). Since the general solution of (4.3.6) is given by

φ(x, λ) = c1ϕ1(x, λ) + c2ϕ2(x, λ) + φ0(x, λ), (4.6.34)

where c1, c2 are not both zero, then one can easily verify that (4.6.29) and (4.6.30) are a

fundamental set of solutions satisfying (4.6.31).

Computing the q−1-derivatives of φ1(·, λ) and φ2(·, λ) we obtain the following lemma.

Lemma 4.6.6. For x ∈ (0, a] and λ 6= 0, we have

Dq−1φ1(x, λ) := −√qs sin(sq−1/2x; q)+

q

∫ xq−1

0

cos(sq−1/2x; q) cos(sqt; q) +

√q sin(sq−1/2x; q) sin(sqt; q)

ν(qt)φ1(qt, λ) dqt,

(4.6.35)

Dq−1φ2(x, λ) := cos(sq−1/2x; q)+

q

∫ xq−1

0

cos(sq−1/2x; q) cos(sqt; q) +

√q sin(sq−1/2x; q) sin(sqt; q)

ν(qt)φ2(qt, λ) dqt.

(4.6.36)

Theorem 4.6.7. For x ∈ [0, a], λ ∈ C, when |λ| → ∞, we have

φ1(x, λ) = cos(sx; q) +O

(E(|s|x; q)

|s|

), (4.6.37)

φ2(x, λ) =sin(sx; q)

s+O

(E(|s|x; q)|s|2

), (4.6.38)

where for each x ∈ (0, a] the O-terms are uniform on xqn, n ∈ N. Moreover, if ν is

bounded on [0, a], the O-terms (4.6.37)–(4.6.38) will be uniform for all x ∈ [0, a].

95

Proof. Let x ∈ (0, a] and n ∈ N be fixed. Consider the annulus

q−n+1

1− q6 |sx| 6 q−n

1− q.

Then |sx|qj 6 (1− q)−1, for all j > n. Hence from (4.6.3), we obtain for |sx| 6 q−n

1−q

|Sk(sxqj ; q)| 6k−1∑j=0

qj2+j(1− q)−2j−1

Γq(2j + 2)6

∞∑j=0

qj2+j(1− q)−2j−1

Γq(2j + 2)= sinh((1− q)−1; q),

(4.6.39)

j > n. Similarly for j > n,

| cos(sxqj ; q)| 6 cosh((1− q)−1; q), |sx| 6 q−n

1− q. (4.6.40)

From (4.6.3) and (4.6.29) we deduce

φ1(xqj , λ) = cos(sxqj ; q)− q

s

∞∑k=0

xqk+j(1− q)Sk+1(sxqj ; q)ν(xqk+j+1)φ1(xqk+j+1, λ)

= cos(sxqj ; q)− 1s

∞∑k=1

xqk+j(1− q)Sk(sxqj ; q)ν(xqk+j)φ1(xqk+j , λ).

(4.6.41)

Set

ψ1(xqj , λ) :=φ1(xqj , λ)

cosh(|sx|qj ; q)=

φ1(xqj , λ)∞∏

r=1

(1 +

q2j |xs|2

x2r

) , j ∈ N. (4.6.42)

Multiplying both sides of the second equation of (4.6.41) by 1/

cosh(|sx|qj ; q) and noting

that | cos(z; q)| 6 cosh(|z|; q), we obtain for j > n

∣∣ψ1(xqj , λ)∣∣ 6 1 +

sinh((1− q)−1; q)|s|

supj>n

∣∣ψ1(xqj , λ)∣∣×

∞∑k=1

xqk+j(1− q)∞∏

r=1

(x2

r + |sx|2q2j+2k)(

x2r + |sx|2q2j

) |ν(xqk+j |

6 1 +sinh((1− q)−1; q)

|s|supj>n

∣∣ψ1(xqj , λ)∣∣ ∫ x

0|ν(t)| dqt,

(4.6.43)

Thus

supj>n

|ψ1(xqj , λ)|(1− q sinh((1− q)−1; q)

|s|

∫ x

0|ν(t)| dqt

)6 1. (4.6.44)

96

We can choose n large enough such that

ι :=(

1− sinh((1− q)−1; q)|s|

∫ qx

0|ν(t)| dqt

)> 0.

Therefore

|ψ1(xqj , λ)| 6 ι−1, for all j > n. (4.6.45)

Set κ := ι∏∞

r=1

(1 + (1−q)−2

x2r

). Then

|φ1(xqj , λ)| 6 ι∞∏

r=1

(1 +

q2j |xs|2

x2r

)6 ι

∞∏r=1

(1 +

(1− q)−2

x2r

)= κ, (4.6.46)

for all j > n. For j < n, we have

φ1(xqj , λ) = cos(sxqj ; q)− 1s

n−j−1∑k=1

xqk+j(1− q)Sk(sxqj ; q)ν(xqk+j)φ1(xqk+j , λ)

− 1s

∞∑k=n−j

xqk+j(1− q)Sk(sxqj ; q)ν(xqk+j)φ1(xqk+j , λ). (4.6.47)

But from (4.6.46) and (4.6.10) we obtain∣∣∣ ∞∑k=n−j

xqk+j(1− q)Sk(sxqj ; q)ν(xqk+j)φ1(xqk+j , λ)∣∣∣

6 κ

(∫ x

0|ν(t)| dqt

)sinh(|sx|qj ; q).

(4.6.48)

That is

∣∣∣1s

∞∑k=n−j

xqk+j(1− q)Sk(sxqj ; q)ν(xqk+j)φ1(xqk+j , λ)∣∣∣ = O

(1|s|

sinh(|sx|qj ; q)). (4.6.49)

Substituting in (4.6.47) yields

φ1(xqj , λ) = cos(sxqj ; q)− 1s

n−j−1∑k=1

xqk+j(1− q)Sk(sxqj ; q)ν(xqk+j)φ1(xqk+j , λ)

+O(

1|s|

sinh(|sx|qj ; q)).

(4.6.50)

97

Multiplying both sides of (4.6.47) by 1/

cosh(|sx|qj ; q) and using (4.6.42), we obtain∣∣ψ1(xqj , λ)∣∣ 6 1

+1|s|

max16k6n−j−1

∣∣∣ψ1(xqk+j , λ)∣∣∣ n−j−1∑

k=1

xqk+j(1− q)∣∣∣fk(sxqj ; q)ν(xqk+j)

∣∣∣+O

(1|s|

sinh(|sx|qj ; q)cosh(|sx|qj ; q)

),

(4.6.51)

where fk(·; q) is defined in (4.6.13). Since |sx| > q−n+1

1−q , then for j = 1, 2, . . . , n− 1, we have

|sxqj | > q−(n−j−1)

1−q . Hence, from Theorem 4.6.4, there exists a constant h > 0 such that for

all n ∈ N and j ∈ 1, 2, . . . , n− 1, if |sxqj | > q−(n−j−1)

1−q , then

|fk(sxqj ; q)| 6 h, k = 1, 2, . . . , n− j − 1. (4.6.52)

Consequently for j = 1, 2, . . . , n− 1 we get

|ψ1(xqj , λ)| 6 1

+ max1<j<n

|ψ1(xqj , λ)| h|s|

n∑k=j+1

xqk(1− q)|ν(xqk)|+O

(sinh(|sx|qj ; q)|s| cosh(|sx|qj ; q)

)

6 1 + h max1<j<n

|ψ1(xqj , λ)|∫ x

0|ν(t)| dqt+O(

sinh(|sx|qj ; q)|s| cosh(|sx|qj ; q)

).

(4.6.53)

From Lemma 1.3.8 we have

|ψ1(xqj , λ)| 6 1 + h max1<j<n

|ψ1(xqj , λ)|∫ x

0|ν(t)| dqt+O

(1|s|

), (4.6.54)

j = 1, 2, . . . , n− 1. Hence

max16j<n

|ψ1(xqj , λ)|(

1− h

|s|

∫ x

0ν(qt) dqt

)6 1 +O

(1|s|

). (4.6.55)

Again, we choose n large enough such that(1− h

|s|

∫ x

0|ν(qt)| dqt

)> 0. (4.6.56)

Then there exists a constant γ > 0 such that

|φ1(xqj , λ)| 6 γ∞∏

r=1

(1 +

q2j |xs|2

x2r

), j = 1, 2 . . . , n− 1. (4.6.57)

98

Using (4.6.57), (4.6.45), and (4.6.13) we obtain

∣∣∣1s

n−1∑k=1

xqk(1− q)Sk(sx; q)ν(xqk)φ1(xqk, λ)∣∣∣

6 γ∞∏

r=1

(1 +

|xs|2

x2r

) ∣∣∣∣∣1sn−1∑k=1

xqk(1− q)fk(|sx|; q)ν(xqk)

∣∣∣∣∣6γh

|s|

∞∏r=1

(1 +

|xs|2

x2r

) ∫ qx

0|ν(qt)| dqt = O

(E(|sx|; q)

|s|

),

(4.6.58)

and∣∣∣qs

∞∑k=n

xqk(1− q)Sk+1(sx; q)ν(xqk+1)φ1(xqk+1, λ)∣∣∣

6 ι sinh(|sx|; q)∫ qx

0|ν(t)| dqt = O

(E(|sx|; q)

|s|

).

(4.6.59)

Thus ∣∣∣∣∣1s∞∑

k=1

xqk(1− q)Sk(sx; q)ν(xqk)φ1(xqk, λ)

∣∣∣∣∣6 max ι, γ

(∫ qx

0|ν(t)| dqt

)E(|sx|; q)|s|

.

(4.6.60)

One can verify that if we replace x by xqm, m ∈ Z+ in the left-hand side of (4.6.60) we

obtain the same right-hand side of (4.6.60). That is the O-term in (4.6.37) is uniform on

xqm, m ∈ N. If ν is bounded on [0, a], and M := supx∈[0,a] |ν(x)| then we can redefine

the constants ι to be

ι := 1− Ma sinh((1− q)−1; q)|s|

, (4.6.61)

and (4.6.60) would be∣∣∣∣∣1s∞∑

k=1

xqk(1− q)Sk(sx; q)ν(xqk)φ1(xqk, λ)

∣∣∣∣∣ 6 Mamax ι, γ E(|sx|; q)|s|

. (4.6.62)

Thus the O-term is uniform for x ∈ [0, a]. This proves (4.6.37) and (4.6.38) is similarly

proved.

99

Corollary 4.6.8. As |λ| → ∞, we have

φ1(x, λ) = cos(sx; q) +O

(|s|−1 exp

(− (log |sx|)2

log q

)), (4.6.63)

φ2(x, λ) =sin(sx; q)

s+O

(|s|−2 exp

(− (log |sx|)2

log q

)). (4.6.64)

Proof. From (1.3.41) and (1.3.54) we obtain

M(r; sinh(|z|; q)), M(r; cos(|z|; q)) = O

(exp(− (log |z|)2

log q

))as z →∞. Consequently from (1.1.21)

M(r;E(z; q)

)= O

(exp

(−(log |z|)2

log q

))as z −→∞.

Thus for each fixed x ∈ [0, a]

E(|sx|; q) = O

(exp

(− (log |sx|)2

log q

))as |λ| → ∞. (4.6.65)

Thus the big O-terms in (4.6.37) and (4.6.38) can be replaced by the big O-terms in (4.6.63)

and (4.6.64) respectively.

Theorem 4.6.9. Let s =√λ be defined with respect to the principal branch. As |λ| → ∞,

we have

Dq−1φ1(x, λ) = −√qs sin(sq−1/2x; q) +O

(exp

(− (log |sxq−1/2|)2

log q

)), (4.6.66)

Dq−1φ2(x, λ) = cos(sq−1/2x; q) +O

(|s|−1 exp

(− (log |sxq−1/2|)2

log q

)), (4.6.67)

where

Proof. From (4.6.63), we conclude that as |λ| → ∞,

φ1(x, λ) = cos(sx; q) + g(x, λ), |g(x, λ)| = O

(|s|−1 exp

(− (log |sx|)2

log q

)). (4.6.68)

100

Hence there exists a constant R > 0 such that

|g(t, λ)| 6 R

|s|exp

(−(log |s|t)2

log q

), t = xqn, n = −1, 0, s ∈ C\ 0 . (4.6.69)

Consequently

∣∣Dq−1g(x, λ)∣∣ =

∣∣g(xq−1, λ)− g(x, λ)xq−1(1− q)

∣∣6

R

|sx|(1− q)

(exp

(−(log |s|q−1x)2

log q

)+ exp

(−(log |sx|)2

log q

))=

q−3/4R

1− qexp

(−(log |s|q−1/2x)2

log q

)(1 +

1|s|2x2q−1

)

= O

(exp

(−(log |s|q−1/2x)2

log q

)), (4.6.70)

as s → ∞. Since Dq−1 cos(sx; q) = −√qs sin(√qsx; q), then (4.6.66) follows and the proof

of (4.6.67) is similar.

Theorem 4.6.10. As |λ| → ∞

∆(λ) = a12a22

(−√qs sin(sq−1/2a; q) +O

(exp

(−(log |s|q−1/2a)2

log q

)))

+ a11a21

(sin(sa; q)

s+O

(|s|−2 exp

(−(log |s|a)2

log q

)))+ a11a22

(√q cos(sq−1/2a; q) +O

(|s|−1 exp

(−(log |s|q−1/2a)2

log q

)))

− a12a21

(cos(sa; q) +O

(|s|−1 exp

(−(log |s|a)2

log q

))).

(4.6.71)

Proof. From the definition of ∆(λ), cf. (4.3.24), we obtain

∆(λ) = U1(φ1)U2(φ2)− U1(φ2)U2(φ1). (4.6.72)

Since

Dj−1q φi(0, λ) = δij , i, j = 1, 2, (4.6.73)

101

then from (4.3.7) and (4.3.8), we obtain

U1(φ1) = a11, U2(φ2) = a12, (4.6.74)

U2(φ1) = a21φ1(a, λ) + a22D−1q φ1(a, λ), (4.6.75)

U2(φ2) = a21φ2(a, λ) + a22D−1q φ2(a, λ). (4.6.76)

Thus

∆(λ) = a11a21φ2(a, λ) + a11a22Dq−1φ2(a, λ)− a12a22φ1(a, λ)− a12a22Dq−1φ1(a, λ).

Substituting from (4.6.63)–(4.6.64) and (4.6.66)–(4.6.67) when x = a in the previous equa-

tion, we obtain (4.6.71).

It follows from (4.6.71) that ∆(λ) →∞ as λ→ −∞. Consequently ∆(λ) has at most a

finite number of negative eigenvalues because

Theorem 4.6.11. ∆(λ) has an infinite number of positive zeros and may have a finite

number of negative zeros. The positive zeros λm are given asymptotically as m→∞ by

λm =

q

a2y2

m

(1 +O(q

m2 )), a12a22 6= 0;

q

a2x2

m

(1 +O(q

m2 )), a12 = 0;

1a2x2

m

(1 +O(q

m2 )), a22 = 0.

(4.6.77)

Proof. We shall consider the case of a12a22 6= 0 in detail, the proof of the other cases are

similar.

If a12a22 6= 0, then

∆(λ) = −a12a22√qs sin(sq−1/2a; q) +O

(exp

(−(log |s|q−1/2a)2

log q

)). (4.6.78)

Set

∆(λ) = F (λ) +R(λ), (4.6.79)

102

where

F (λ) = −a12a22√qs sin(sq−1/2a; q), R(λ) = O

(exp

(−(log |s|q−1/2a)2

log q

)), (4.6.80)

Therefore there exists a constant c > 0 such that as |λ| → ∞

|R(λ)| 6 c exp(−(log |s|q−1/2a)2

log q

), for all λ ∈ C.

Recall that Asm, m > 1, is the annulus defined in (1.3.52) in terms of the positive ze-

ros ym∞m=1 of sin(z; q) and the sequences cm∞m=1 and dm∞m=1 defined respectively in

(1.3.49) and (1.3.50). Thus from (1.3.53) if saq−1/2 ∈ Asm, m > 1, then

log | sin(sq−1/2a; q)| = −(log |s|q−1/2a

)2log q

+ log∣∣1− qa2s2

y2m

∣∣+O(1) (4.6.81)

as m→∞. So there exists l0 ∈ N, and δ1 > 0 such that∣∣∣∣∣log | sin(sq−1/2a; q)|+(log |s|q−1/2a

)2log q

− log∣∣1− a2q−1s2

y2m

∣∣∣∣∣∣∣ 6 δ1. (4.6.82)

Then

log |R(λ)| 6 log c−(log |s|q−1/2a

)2log q

6 log | sin(sq−1/2a; q)| − log∣∣1− s2a2q−1

y2m

∣∣+ δ1 + log c

6 log |F (λ)| − log |s| − log∣∣1− s2a2q−1

y2m

∣∣+ δ2,

(4.6.83)

where δ2 := δ1 + log c+ log√q|a12a22|. If z ∈ ∂As

m, then

log∣∣1− s2a2q−1

y2m

∣∣ > log(1− q2dm

), |saq−1/2| = ymq

dm ,

log∣∣1− s2a2q−1

y2m

∣∣ > log(q−2cm − 1

), |saq−1/2| = ymq

−cm .

Since the sequences cm∞m=1, dm∞m=1 are bounded and positive, then there exists a con-

stant δ3 > 0 such that

δ3 := infm>1

log(1− q2dm), log(q−2cm − 1)

.

103

Therefore

log∣∣∣∣1− s2a2q

y2m

∣∣∣∣ > δ3.

We choose l0 sufficiently large such that − log |s| − δ3 + δ2 < 0. That is

|R(λ)| 6 |F (λ)|, λ ∈ ∂Asm, m > l0.

An application of Rouche’s theorem shows that ∆(λ) and F (λ) have the same number of

zeros in Asm, m > l0. As sin(sq−1/2a; q) has only two real simple symmetric zeros in As

m,

then ∆(λ) has exactly two zeros there. In the following we show that these zeros are real,

simple and symmetric. If ω is a zero of ∆(λ), then |F (ω)| = |R(ω)| and we deduce from

(4.6.83) that if |ω| is sufficiently large, s :=√ω, ω > 0, then

log∣∣1− a2q−1ω

y2m

∣∣ 6 − log |s|+ δ2. (4.6.84)

Consequently either

log∣∣1− saq−1/2

ym

∣∣ 6 − log |s1/2|+ δ22, (4.6.85)

or

log∣∣1 +

saq−1/2

ym

∣∣ 6 − log |s1/2|+ δ22. (4.6.86)

Thus if (4.6.85) holds, then ∣∣1− saq−1/2

ym

∣∣ = O(|s|−1/2

). (4.6.87)

This implies that s ∼ q1/2a−1ym. More precisely, from (1.2.21), we have

s = q1/2a−1ym

(1 +O(ym)−1/2

)(4.6.88)

Similarly we have if (4.6.86) holds that

−s = q1/2a−1ym

(1 +O(ym)−1/2

). (4.6.89)

Thus ±s are the zeros of ∆(λ) in Asm, s is sufficiently large. On the other hand ∆(λ) has

a finite number of negative zeros, because if ω < 0, then the left hand side of (4.6.71) does

not tends to zero when |w| is large enough.

104

Corollary 4.6.12. As m→∞

√λm =

q−m+1/2

a(1− q)(1 +O(qm/2)

), a12 6= 0;

q−m+1

a(1− q)(1 +O(qm/2)

), a12 = 0;

(4.6.90)

Proof. The proof follows by substituting in (4.6.77) with the asymptotic behavior of xm, ym

given in equations (1.2.20), (1.2.21) respectively.

The following lemma will be needed in the sequel.

Lemma 4.6.13. Let f : C → C be a function. If there exists a constant δ > 0 such that

|f(aqnz)| 6 δE(aqn|z|; q), n ∈ N, a > 0,

then ∫ a

0|f(tz)| dqt = O

(exp

(−(log |z|a)2

log q

))as |z| → ∞. (4.6.91)

Proof. From the definition of the q-integration (1.1.31) we obtain∫ a

0|f(tz)| dqt =

∞∑n=0

aqn(1− q)|f(aqnz)| 6 δ∞∑

n=0

aqn(1− q)E(aqn|z|; q)

6 aδM (ar;E(a|z|; q)) = O

(exp

(−(log |z|a)2

log q

)),

(4.6.92)

as |z| → ∞.

Theorem 4.6.14. The corresponding normalized eigenfunctions Ψm(·)∞m=1 have the

asymptotics

±Ψm(x) =

cos(√λmx; q)( ∫ a

0 cos2(√λmx; q) dqx

)1/2

(1 +O(qm)

), a12 6= 0;

sin(√λmx; q)( ∫ a

0 sin2(√λmx; q) dqx

)1/2

(1 +O(qm)

), a12 = 0,

(4.6.93)

as m→∞.

105

Proof. Since ψm(x) satisfies (4.3.6) with λ = λm, there are numbers c1 and c2 independent

of x such that

ψm(x) = c1φ1(x, λm) + c2φ2(x, λm). (4.6.94)

Since ψm(x) also satisfies the boundary condition (4.3.7), we have

c1a11 + c2a12 = 0. (4.6.95)

Since |c1|+ |c2| > 0, we can write c1 = kma12 and c2 = −kma11, where km 6= 0. Hence

k−1m ψm(x) = a12φ1(x, λm)− a11φ2(x, λm). (4.6.96)

If a12 6= 0, Theorem 4.6.7 with λ = λm gives

k−1m ψm(x) = a12 cos(

√λmx; q) +O

(E(√λmx; q)√λm

)(4.6.97)

as m→∞, where the O-term is uniform on xqm, m ∈ N. Thus

K−2m ψ2

m(x) = a212 cos2(

√λmx; q) +O

(E2(

√λmx; q)√λm

)(4.6.98)

But∣∣cos(

√λmx; q)

∣∣2 6 E2(√λmx; q), for all m ∈ Z+ and x ∈ R. Then∣∣∣∣∫ a

0cos2(

√λmx; q) dqx

∣∣∣∣ = O

(exp

(−2(log a√λm)2

log q))

, (4.6.99)

as m→∞. Applying Lemma 4.6.13 we obtain

K−2m = k−2

m

∫ a

0ψ2

m(x) dqx

= a212

∫ a

0cos2(

√λmx; q) dqx+O

(exp

(−2(log a√λm)2

log q)/√

λm

)= a2

12

(∫ a

0cos2(

√λnx; q) dqx

)(1 +O(

1√λm

)).

106

Hence from (4.6.96)

ψm(x) = ± cos(√λmx; q)(∫ a

0 cos2(√λmx; q) dqx

)1/2

(1 +O(

1√λm

))−1/2 +O(

1√λm

)

= ± cos(√λmx; q)(∫ a

0 cos2(√λmx; q) dqx

)1/2+O(

1√λm

)

= ± cos(√λmx; q)(∫ a

0 cos2(√λmx; q) dqx

)1/2

(1 +O(qm)

).

Similarly, we can prove the second formula of ψm when a12 = 0.

Part III

Basic Fractional Calculus and

Equations

107

Chapter 5

Basic Fractional Calculus

In this chapter we study q-analogue of Riemann–Liouville fractional calculus and fractional

difference equations. We begin with a brief account of the history of fractional calculi and

some results which we are interested in deriving their q-analogues. We give also q-analogues

of the Grunwald–Letinkov fractional derivative and the Caputo fractional derivative at the

end of this chapter.

5.1 Fractional calculus and equations

The following is a brief account about the history of the fractional calculus. This summary

which is taken from [30, 107] contains some important stages in the developments of the

concepts of fractional integrals and derivatives which we are interested in. The idea of

generalizing the notation of differentiation dk/dxk, k ∈ N to non-integer values has begun

with the birth of differential calculus itself. In 1695 L’ Hospital raised the question of the

meaning of dny/dxn if n = 1/2, c.f. [30]. Euler (1738) was the first who observed that the

derivative dpxa/dxp of the power function xa, a ∈ R, has a meaning for a non-integer p,

cf. [107]. S. F. Lacroix [81, 1819] derived the explicit formula

d1/2

dx1/2xa =

Γ(a+ 1)Γ(a+ 1/2)

xa−1/2, a ∈ R. (5.1.1)

108

109

Formula (5.1.1) coincides with the Riemann–Liouville fractional derivative of xa. See equa-

tion (5.2.12) below. J.B.J Fourier 1882 derived the integral representation

f(x) =12π

∫ ∞

−∞f(u) du

∫ ∞

−∞cos t(x− u) dt. (5.1.2)

Then he obtained formally the p-derivative version, p ∈ R, to be

dpf(x)dxp

=12π

∫ ∞

−∞λp dλ

∫ ∞

−∞f(t) cos(λx− tλ+ pπ/2) dt. (5.1.3)

This [30] was the first definition for the derivative of arbitrary order for a function defined by

(5.1.2) which is not necessarily a power function. The proper history of fractional calculus

began with the work of Abel and Liouville. In the two consecutive papers [4,5], Abel solved

the integral equation ∫ x

0

φ(t)(x− t)1−α

dt = f(x), x > 0, 0 < α < 1, (5.1.4)

concretely via the formula

φ(x) =1

1− α

d

dx

∫ x

0

f(t)(x− t)α

dt. (5.1.5)

J. Lutzen [89, p. 314] indicated that Abel never solved the problem of fractional calculus.

He only showed that the solution, found by other means, could be written as a fractional

derivative. Throughout his series of papers published between 1823 and 1837, Liouville

has been considered as the founder of the theory of fractional calculus, see e.g. [30, 88,

107]. Liouville’s first definition of fractional derivative is established for functions having

expansion of the form f(x) =∑∞

k=0 ckeakx. For such functions Liouville’s definition is

Dαf(x) =∞∑

k=0

ckaαk e

akx, α ∈ C. (5.1.6)

The restrictiveness of such a definition is clearly related to the convergence properties of

the series expansion of the differentiated function. He also defined the formula

D−αf(x) =(−1)α

Γ(α)

∫ ∞

0φ(x+ t)tα−1 dt, x ∈ R, <α > 0. (5.1.7)

110

This is now called the Liouville’s form of fractional integration where the factor (−1)α is

omitted. It is worth mentioning that Liouville in 1823 gave the idea of defining the fractional

derivative as a limit of a difference quotient ∆αhf/h

α, where ∆αhf(x) is the right–handed

difference of fractional order

∆αhf(x) =

∞∑j=0

(−1)j

(α

j

)f(x− jh),

(α

j

)=α(α− 1) . . . (α− j + 1)

j!, (5.1.8)

which coincides with the classical r-th right handed difference ∆rhf(x) for α = r ∈ N.

However, Liouville gave no substantial development of this idea. He derived an example

involving Fourier’s formula (5.1.3) for non integer p, based on this idea. He also evalu-

ated fractional derivative of some elementary functions via this approach. This idea was

more deeply considered by Grunwald [56] and Letinkov [83] who established the so called

Grunwald–Letinkov fractional derivative. Riemann in [106] defined the fractional integral

for α ∈ R to be1

Γ(α)

∫ x

0(x− t)α−1φ(t) dt, x > 0. (5.1.9)

which had become one of the main formulae of fractional integration together with Li-

ouville’s construction (5.1.7). Formula (5.1.9) is called the Riemann–Liouville fractional

integral operator of a real-valued function φ(x) defined on (0, T ], T > 0, as is seen in the

next section. Another direction in the history of fractional calculus is connected with the

Cauchy integral formula

f (n)(z) =n!2πi

∫γ

f(t)(t− z)n+1

dt, (5.1.10)

for an analytic function f(z) in the complex plane. The direct extension of this formula to

non-integer values of n was first made by Sonine. In 1917 Weyl, [116], defined fractional

integration for periodic functions as

K(α)± φ ∼

∞∑n=−∞

(±in)−αφneinx, φn =

12π

∫ 2π

0e−intφ(t) dt, φ0 = 0. (5.1.11)

111

It is realized as a convolution

K(α)± φ =

12π

∫ 2π

0Ψα±(x− t)φ(t) dt, (5.1.12)

for a certain special function Ψα±(x). He had shown that the fractional integrals (5.1.11)–

(5.1.12) may be written in the form

Kα+φ =

1Γ(α)

∫ x

−∞

φ(t)(x− t)1−α

dt, Kα−φ =

1Γ(α)

∫ ∞

x

φ(t)(x− t)1−α

dt (5.1.13)

0 < α < 1, provided that the integrals in the last equation of the periodic function φ were

understood to be conventionally convergent ones over an infinite interval. For this reason,

fractional integrals over infinite intervals, especially the integral Kα−φ, were named in many

later papers as Weyl integrals. This is noted by Samko et al. [107] to be a misunderstanding.

The reason as indicated in [107] is that fractional integrals over infinite intervals had first

appeared in the paper of Liouville, whereas Weyl arrived at the integrals in (5.1.13) from

(5.1.11)–(5.1.12), with specific interpretation of these integrals for periodic functions only

satisfying the condition φ0 = 0, cf. [119]. Thus fractional integrals over infinite intervals

considered for arbitrary (non-periodic) functions have no direct relation to Weyl’s ideas.

So, following [107], the fractional integral (5.1.13) are called Liouville fractional integrals,

the first being the left-hand sided and the second is the right-hand sided. M. Caputo has

defined the fractional derivative of order p by

Ca D

αt f(t) =

1Γ(n− α)

∫ t

a(t− τ)n−α−1f (n)(τ) dτ, (5.1.14)

n− 1 < α < n. In the following section, we would exhibit some properties of the Riemann-

Liouville fractional derivative, M. Caputo fractional derivative, Grunwald-Letinkov frac-

tional derivative, and Liouville fractional integral operator. These will be the main proper-

ties and relations of which we will derive the basic analogue.

112

5.2 Some fractional integral operators

This section is devoted to constructing the fractional calculi of which we will derive the

q-analogues. First we denote by AC(n)[a, b], n ∈ N, to the set of functions f which have

continuous derivatives up to order n− 1 on [a, b] with f (n−1) ∈ AC[a, b].

Lemma A. [79] The set AC(n)[a, b] consists of those and only those functions f , which are

represented in the form

f(x) =n−1∑k=0

ck(x− a)k +1

n− 1!

∫ x

a(x− t)n−1φ(t) dt, (5.2.1)

where φ ∈ L1(a, b), and the ck’s, k = 0, 1, . . . , n − 1, are arbitrary constants. Moreover

φ(x) = f (n)(x) a.e., and ck =f (k)(a)k!

, k = 1, 2, . . . , n.

The first fractional integral operator we define in this section is that of Riemann and

Liouville. It is connected to Abel’s integral equation

1Γ(α)

∫ x

a(x− t)α−1φ(t) dt = f(x), x > a, α > 0, f ∈ L1(a, b), (5.2.2)

cf. [107, p. 32].

Theorem B. The Abel integral equation (5.2.2) with 0 < α < 1 has a unique solution in

L1(a, b) if and only if the function f1−α defined by

f1−α(x) :=1

Γ(1− α)

∫ x

a(x− t)−αf(t) dt (5.2.3)

is absolutely continuous on (a, b) and f1−α(a) = 0. Moreover the unique solution φ is given

by

φ(x) =1

Γ(1− α)d

dx

∫ x

a(x− t)−αf(t) dt =

d

dxf1−α(x). (5.2.4)

If f ∈ AC[a, b], then f1−α ∈ AC[a, b] and (5.2.4) becomes

φ(x) =1

Γ(1− α)

[ f(a)(x− a)α

+∫ x

a

f ′(s)(x− s)α

ds]. (5.2.5)

113

The following Cauchy formulae can be considered as the n-th primitive of a function

f ∈ L1(a, b).∫ x

a

∫ xn−1

a. . .

∫ x1

af(t) dtdx1 . . . dxn−1 =

1(n− 1)!

∫ x

a(x− t)n−1f(t) dt, (5.2.6)

∫ b

x

∫ b

xn−1

. . .

∫ b

x1

f(t) dtdx1 . . . dxn−1 =1

(n− 1)!

∫ b

x(t− x)n−1f(t) dt, (5.2.7)

n ∈ Z+. Since the right hand side of (5.2.6) and (5.2.7) exist also for non integer values

of n, the Riemann–Liouville fractional integral can be considered as an extension of (5.2.6)

and (5.2.7) when we replace n by α ∈ R+ and n − 1! by Γ(α). Indeed for α ∈ (0,∞) and

f ∈ L1(a, b) the fractional Riemann–Liouville integral is defined to be

D−αa+ f(x) :=

1Γ(α)

∫ x

a(x− t)α−1f(t) dt, D−α

b− f(x) :=1

Γ(α)

∫ b

x(t− x)α−1f(t) dt, (5.2.8)

x ∈ (a, b) with respect to x = a and x = b respectively. In some literature they are called

left-sided and right-sided Riemann–Liouville fractional integrals respectively. It is known,

see e.g [100,107], that if f ∈ L1(a, b), then both of D−αa+ f and D−α

b− f exist a.e. and they are

L1(a, b)-functions. Moreover for f ∈ L1(a, b), we have

limα→0+

D−αf(x) = f(x) a.e. (5.2.9)

We now pass to the definition of the fractional derivative of arbitrary order. The existence

of the fractional derivative is connected to the solvability of the Abel’s integral equation

(5.2.2). For f ∈ L1(a, b), the left and right sided Riemann–Liouville fractional derivative of

order α, α ∈ R+, are defined formally by

Dαa+f(x) := DkD

−(k−α)a+ f(x) =

1Γ(k − α)

dk

dxk

∫ x

a(x− t)k−α−1f(t) dt, (5.2.10)

and

Dαb−f(x) := (−1)kDkD

−(k−α)b− f(x) =

(−1)k

Γ(k − α)dk

dxk

∫ b

x(t− x)k−α−1f(t) dt, (5.2.11)

114

x ∈ (a, b), k = [α] + 1, respectively. The fractional derivatives Dαa+f , Dα

b−f exist if f ∈

L1(a, b) and D−(k−α)a+ f , D−(k−α)

b− f are in the class AC(k)[a, b], see [107]. For instance the

Riemann–Liouville fractional derivative of xa, a ∈ R is given by

Dαa+x

a =Γ(a+ 1)

Γ(a− α+ 1)xa−α, a ∈ R. (5.2.12)

Now we state some properties of the Riemann–Liouville fractional calculus which we will

derive their q-analogues in Section 5.4. We confine ourselves to the case of the left–sided

Riemann–Liouville fractional calculus since we will study its basic analogue. Let α, β ∈ R+.

If f ∈ L1(a, b), then the semigroup property

D−βa+ D−α

a+ f(x) = D−αa+ D−β

a+f(x) = D−(α+β)a+ f(x), (5.2.13)

holds for almost all x ∈ [a, b]. If f(x) satisfies the conditions

f ∈ L1(a, b) and D−(k−α)a+ f ∈ AC(k)[a, b], k = [α] + 1, (5.2.14)

then Dα−ja+ f ∈ L1(a, b), j = 0, 1, . . . , k. Besides, Dα−j

a+ f ∈ AC(j)[a, b], 1 6 j 6 k − 1.

Moreover, for functions satisfying the appropriate conditions

Dαa+D

−αa+ f(x) = f(x), a.e. (5.2.15)

Dαa+D

−βa+f(x) = D

−(β−α)a+ f(x), a.e. if β > α > 0, (5.2.16)

Dαa+D

−βa+f(x) = Dα−β

a+ f(x), a.e. if α > β > 0. (5.2.17)

Also,

D−αa+D

αa+f(x) = f(x)−

k∑j=1

Dα−ja+ f(t)

∣∣∣∣t=a

(x− a)α−j

Γ(1 + α− j). (5.2.18)

If f ∈ L1(a, b) and, in addition, D−(n−β)a+ f ∈ AC(n)[a, b], n = [β] + 1, then the equality

D−αa+D

βa+f(x) = Dβ−α

a+ f(x)−n∑

j=1

Dβ−j

a+ f(t) ∣∣∣∣

t=a

(x− a)α−j

Γ(1 + α− j)(5.2.19)

115

holds almost everywhere in (a, b) for any α > 0. Finally, the identities

D−αa+D

αa+f(x) = f(x), and D−α

a+Dβa+f(x) = Dβ−α

a+ f(x), (5.2.20)

hold if f ∈ Iαa+(L1) and f ∈ Iβ

a+(L1), respectively, cf. [107], where by Iαa+(L1) we denote

the set of functions f represented by the left sided fractional integral of order α, α > 0, of

a summable function, i.e. f = Iαa+φ, φ ∈ L1(a, b). For proof of these and other properties,

see e.g. [40, 100,107].

The other types of fractional derivatives we discuss here are those of Grunwald–Letinkov

and Caputo. It is known that the nth derivative of a function f is defined for x ∈ (a, b) to

be

f (n)(x) = limh→0

∆nhf(x)hn

, (5.2.21)

where

∆nhf(x) =

1hn

n∑r=0

(−1)r

(n

r

)f(x− rh). (5.2.22)

Starting from this formula, A.K. Grunwald [56] and A.V. Letinkov [83] developed an ap-

proach to fractional differentiation for which the formal definition of the fractional derivative

Dαa+f(x) is the limit

Dαa+f(x) := lim

h→0

∆αhf(x)hα

, α ∈ R+, (5.2.23)

∆αhf(x) being the right-handed difference of fractional order α defined in (5.1.8).

Grunwald [56] gave a formal proof while Letinkov ’s [83] gave a rigorous proof of the fact

that D−α coincides with the Riemann–Liouville fractional integral operator when α > 0 and

f is continuous on [a, b], cf. [82]. The following theorem gives the conditions on which the

Grunwald–Letinkov fractional derivative is nothing but the Riemann–Liouville fractional

derivative.

Theorem C. [100] Assume that f ∈ AC(n)[a, b]. Then for every α, 0 < α < n, the

Riemann–Liouville fractional derivative Dαa+f(t) exists and coincides with the Grunwald–

Letinkov derivative Dαa+f(t). i.e. If 0 6 m − 1 6 α 6 m 6 n, then for a < t < b the

116

following holds:

Dαa+f(t) = Dα

a+f(t) =m−1∑j=0

f (j)(a)(t− a)j−α

j!+

1Γ(m− α)

∫ t

a(t− τ)m−1−αf (m)(τ) dτ.

(5.2.24)

It is known that in applied problems we require definitions of fractional problems which

allow the involvement of initial conditions with physical meanings, i.e. conditions like e.g.

f (j)(a) = bj , j = 1, 2, . . . , k (5.2.25)

where the bj ’s and k ∈ Z+ are given constants. The Riemann–Liouville approach leads to

initial conditions containing the limit values of the Riemann–Liouville fractional derivatives

at the lower end point t = a, for example

limt→a

Dα−ja+ f(t) = cj , j = 1, 2, . . . , k, (5.2.26)

where the cj ’s and k ∈ Z+ are given constants and α is not an integer. It is frequently

stated that the physical meaning of initial conditions of the form (5.2.26) is unclear or

even non existent, see e.g. [107, p. 78]. The requirement for physical interpretation of such

initial conditions was most clearly formulated recently by Diethelm et al. [74]. In [62], N.

Heymans and I. Podlubny show that initial conditions of the form (5.2.26) for Riemann-

Liouville fractional differential equations have physical meaning, and that the corresponding

quantities can be obtained from measurements. However, the problem of interpretation of

initial conditions still remained open. In [31], [32], M. Caputo gave a solution for this

problem when he defined the fractional derivative of order α, Ca D

αt f(t), by the formula

Ca D

αt f(t) =

1Γ(n− α)

∫ t

a(t− τ)n−α−1f (n)(τ) dτ, (5.2.27)

n−1 < α 6 n, n ∈ Z+. The Caputo approach leads to initial conditions of the form (5.2.25)

which has a physical meaning. Recently, several initial value problems based on the Caputo

117

fractional operator have been studied, see e.g. [42,43,99,101]. The relationship between the

Caputo fractional derivative and the Riemann–Liouville Calculus is obviously

Ca D

αt f(t) = D

−(n−α)a+ f (n)(t). (5.2.28)

If the function f(t) has n+ 1 continuous derivatives in [a, b], then cf. [100]

limα→n

Ca D

αt f(t) = f (n)(t), t ∈ [a, b].

The main advantage of the Caputo fractional derivative is that Ca D

αt c = 0, where c is a

constant, while the Riemann-Liouville fractional derivative of a constant c is not equal to

zero. In fact, if n = [α] + 1, then

Dαa+c = DnIn−αc =

c

Γ(n− α+ 1)Dn(x− a)n−α =

c

Γ(1− α)(x− a)−α. (5.2.29)

Since we will establish a method for using the basic Laplace transform defined by [57,

1949] in solving basic fractional equations, we state before closing this section some results

concerning the Laplace transform.

Let f ∈ L1(0, b) for any b > 0. If In−αf ∈ AC(n)[0, b] for any b > 0 such that In−αf(x)

is of exponential order as x→∞. That is there exist real constants K,T, p0 ∈ R+, p0 > 0,

such that

|In−αf(x)| 6 Kep0x if x > T. (5.2.30)

Then the Laplace transform of the Riemann–Liouville fractional derivative Dαa+f(x) exists,

cf. [117], and satisfies the following identity∫ ∞

0e−ptDα

a+f(t) dt = pαF (p)−n−1∑k=0

pkDαa+f(t)|t=0, n− 1 < α 6 n, (5.2.31)

for all <p > p0, cf. [107,117]. Also the Laplace transform of the Caputo derivative Ca D

αt f(x)

exists if f(x) ∈ AC(n)[0, b] for any b > 0 and f(x) is of exponential order as x→∞. In this

118

case we can find p0 > 0 such that the Laplace transform of the Caputo derivative satisfies

the identity ∫ ∞

0e−ptC

0 Dαt f(t) dt = pαF (p)−

n−1∑k=0

pα−k−1f (k)(0), (5.2.32)

for all <p > p0. The Laplace transform method is frequently used for solving applied prob-

lems. So, it is clear that the Laplace transform of the Caputo derivative allows utilization of

initial values of classical integer-order derivative with known physical interpretations, while

the Laplace transform of the Riemann-Liouville fractional derivative allows utilization of

initial conditions of the form

Dα−jf(t)|t=0 = bj , j = 1, 2, . . . , [α] + 1,

which can cause problems without their physical interpretation.

5.3 q-Notations and results

In this section, we introduce q-notations and results which we need throughout this part.

The following definition of q-translation εy is given by M.E.H. Ismail [65]. The q-translation

is defined on monomials by

εyxn := xn(− y/x; q

)n, (5.3.1)

and it is extendeed to polynomials as a linear operator. Thus

εy

(m∑

n=0

fnxn

):=

m∑n=0

fnxn(− y/x; q

)n. (5.3.2)

The q-translation is defined for xa, a ∈ R+ to be

εyxa := xa(− y/x; q

)a. (5.3.3)

Let L1q(0, a) be the space of all functions f defined on [0, a] such that f ∈ L1

q(0, x) for all

x ∈ (qa, a].

119

Lemma 5.3.1. For positive numbers α, β and positive integer n, we haven∑

k=0

[nk

]qqkβ(qα; q)k(qβ ; q)n−k = (qα+β ; q)n. (5.3.4)

Proof. We prove the lemma by induction. Clearly For n = 0, (5.3.4) is true. Assume that

(5.3.4) is true at n. Then, using[n+ 1k

]q

=[

n

k − 1

]q

+ qk[nk

]q, (5.3.5)

cf. e.g [53, p. 25], we obtainn+1∑k=0

[n+ 1k

]q

qkβ(qα; q)k(qβ ; q)n+1−k

= (qβ ; q)n+1 +n∑

k=1

[n

k − 1

]q

qkβ(qα; q)k(qβ ; q)n+1−k

+ q(n+1)β(qα; q)n+1 +n∑

k=1

[nk

]qq(β+1)k(qα; q)k(qβ ; q)n−k

=n∑

k=0

[nk

]qq(k+1)β(qα; q)k+1(qβ ; q)n−k +

n∑k=0

[nk

]qq(β+1)k(qα; q)k(qβ ; q)n+1−k

=(qβ(1− qα) + (1− qβ)

)(qα+β+1; q)n = (qα+β ; q)n+1.

This completes the proof.

Remark 5.3.1. There is another proof of the previous Lemma. This proof depends on Heine’s

q-analogue of Gauss’ summation formula

2φ1(q−n, b; c; q, q) =(c/b, q)n

(c, q)nbn, (5.3.6)

see e.g. [53, p.14]. Indeedn∑

k=0

[nk

]qqkβ(qα; q)k(qβ ; q)n−k = (qβ ; q)n 2φ1(q−n, qα; q1−β−n; q, q)

= (qβ ; q)nqnα (q1−β−α−n; q)n

(q1−β−n; q)n= (qα+β ; q)n.

Corollary 5.3.2. For α, β > 0 and n ∈ Z+, we haven∑

k=0

qkβ(1− q)(qk+1; q)α−1(qn−k+1; q)β−1 = Bq(α, β)(qn+1; q)α+β−1. (5.3.7)

120

Proof. Using equations (1.1.5) and (1.1.10) one can easily see that for any γ > 0, j ∈ N

(qj+1; q)γ−1 =(q; q)∞(qγ ; q)j

(q; q)j(qγ ; q)∞= Γq(γ)(1− q)γ−1 (qγ ; q)j

(q; q)j(5.3.8)

Thus from (5.3.4) we obtain

n∑k=0

qkβ(1− q)(qk+1; q)α−1(qn−k+1; q)β−1

=(1− q)α+β−1Γq(α)Γq(β)

(q; q)n

n∑k=0

qkβ[nk

]q(qα; q)k (qβ; q)n−k

= (1− q)α+β−1Γq(α)Γq(β)(qα+β ; q)n

(q; q)n. (5.3.9)

But from (5.3.8)(qα+β ; q)n

(q; q)n=

(1− q)1−α−β

Γq(α+ β)(qn+1; q)α+β−1. (5.3.10)

By combining (5.3.9) and (5.3.10) we obtain (5.3.7).

The following property is essential in our investigations and it is taken from [90]. This

property indicates that the nth q-derivative Dnq of a function f can be expressed in terms

of its values at the points qjx, j = 0, 1, . . . , n.

Property 5.3.3. Let f be a function defined in an interval I and assume that f has q-

derivatives up to order n in I, n in N. Then, (Dkq f)(x) can be expressed as

Dkq f(x) =

1xk(1− q)k

j=k∑j=0

(−1)j

[k

j

]q

f(qjx)qj(j−1)/2+j(k−j)

, (5.3.11)

for every x in I \ 0, k = 1, 2, . . . , n.

Lemma 5.3.4. Let h(t, x) be a function defined on [0, a]× [0, a], such that

Djq,xh(t, x), j = 0, 1, . . . , k − 1,

are q-integrable functions on [0, a], . If for some x ∈ (0, a] and k ∈ N

h(xqr, xqj) = 0, r = 0, 1, 2, . . . , j − 1; j = 1, 2, . . . , k, (5.3.12)

121

then

Dkq,x

∫ x

0h(t, x) dqt =

∫ x

0Dk

q,xh(t, x) dqt, x ∈ (0, a]. (5.3.13)

Proof. From (5.3.11) we obtain

Dkq,x

∫ x

0h(t, x) dqt =

j=k∑j=0

(−1)j

[k

j

]q

qj(j+1)

2−kj

xk(1− q)k

∫ xqj

0h(t, xqj) dqt. (5.3.14)

From the condition (5.3.12) we find that∫ xqj

0h(t, xqj) dqt =

∫ x

0h(t, xqj) dqt, j = 1, 2, . . . , k. (5.3.15)

Hence

Dkq,x

∫ x

0h(t, x) dqt =

j=k∑j=0

(−1)j

[k

j

]q

qj(j+1)

2−kj

xk(1− q)k

∫ x

0h(t, xqj) dqt

=∫ x

0

j=k∑j=0

(−1)j

[k

j

]q

qj(j+1)

2−kj

xk(1− q)kh(t, xqj)

dqt

=∫ x

0Dk

q,xh(t, x) dqt. (5.3.16)

Lemma 5.3.5. For α, β > 0 and f ∈ L1q(0, x), x > 0, we obtain

xα−1

∫ x

0(qt/x; q)α−1t

β−1

∫ t

0(qu/t; q)β−1f(u) dqu dqt

= Bq(α, β)xα+β−1

∫ x

0

(qt/x; q

)α+β−1

f(t) dqt.

(5.3.17)

Proof. Let α, β > 0 and f ∈ L1q(0, x), x > 0. Then by the definition of the q-integration,

122

cf. (1.1.31) we obtain

xα−1

∫ x

0(qt/x; q)α−1t

β−1

∫ t

0(qu/t; q)β−1f(u) dqu dqt (5.3.18)

= xα+β−1∞∑

n=0

qnβ(qn+1; q)α−1

∞∑m=0

xqn+m(1− q)(qm+1; q)β−1f(xqn+m)

= xα+β−1∞∑

n=0

xqnβ(1− q)(qn+1; q)α−1(qn+1; q)α−1

∞∑k=n

qk(1− q)(qk−n+1; q)β−1f(xqk)

= xα+β−1∞∑

k=1

xqk(1− q)f(xqk)k∑

n=0

qnβ(1− q)(qn+1; q)α−1(qk−n+1; q)β−1. (5.3.19)

Thus from (5.3.7)

xα−1

∫ x

0(qt/x; q)α−1t

β−1

∫ t

0(qu/t; q)β−1f(u) dqu dqt

= xα+β−1Bq

(α, β

) ∞∑k=1

xqk(1− q)(qk+1; q)α+β−1f(xqk)

= xα+β−1Bq

(α, β

) ∫ x

0(qt/x; q)α+β−1 f(t) dqt.

(5.3.20)

Definition 5.3.1. A function f defined on [0, a] is called q−absolutely continuous if for

any ε > 0 and x, y ∈ (0, a] there exists n0 ∈ N such that for any n > n0, if(xqk, yqk)

n

k=n0

are pairwise disjoint and non-intersecting intervals, then we have

n∑j=n0

∣∣f(xqj − f(yqj)∣∣ 6 ε. (5.3.21)

The space of these functions will be denoted by ACq[0, a].

Lemma 5.3.6. If f ∈ ACq[0, a], then f is q-regular at zero.

Proof. Let f ∈ ACq[0, a]. Then given ε > 0 and t, v ∈ (0, a], there exists n0 ∈ N such that

for any n > n0 if(tqk, vqk)

n

k=n0are pairwise disjoint non-intersecting intervals, then

n∑j=n0

∣∣f(tqj − f(vqj)∣∣ 6 ε. (5.3.22)

123

Take v = qt in (5.3.22) and assume that t ∈ (qa, a]. Then

n∑j=n0

∣∣f(tqj)− f(tqj+1)∣∣ 6 ε, n > n0. (5.3.23)

Thus for all m > n > n0, t ∈ (qa, a], we obtain

|f(tqn)− f(tqm)| =

∣∣∣∣∣∣m−1∑j=n

f(tqj)− f(tqj+1)

∣∣∣∣∣∣ < ε. (5.3.24)

Consequently

lt := limj→∞

f(tqj) exists for all t ∈ (qa, a]. (5.3.25)

We prove that lt = lv for all t, v ∈ (qa, a]. Indeed from (5.3.25) we can find j > n0 such

that

|lt − f(tqj)|, |lv − f(vqj)| < ε. (5.3.26)

Then by (5.3.22)

|lt − lv| 6 |lt − f(tqj)|+ |f(tqj)− f(vqj)|+ |f(vqj − lv)| < 3ε. (5.3.27)

Hence lt does not depend on t ∈ (qa, q] as required. Set l := lt, t ∈ (qa, a]. Then If we

redefine f(0) to be l, then f is q−regular at zero.

Remark 5.3.2. Form the previous theorem all the functions in ACq[0, a] are q−regular at

zero. Therefore from now on if f ∈ ACq[0, a], then by f(0) we mean lim n→∞x∈(qa,a]

f(xqn).

Theorem 5.3.7. Let f be a function in the space ACq[0, a]. If t ∈ xqn, n ∈ N, x ∈ (qa, a],

then

f ∈ ACq[0, a] ⇔ f(t) = c+∫ t

0φ(u) dqu, φ ∈ L1

q(0, x), (5.3.28)

where c = f(0) and φ(x) = Dqf(x) for all x ∈ (0, a].

Proof. Assume that f ∈ ACq[0, a]. Fix x ∈ (qa, a] the series∑∞

n=0 f(xqn) − f(xqn+1) is

absolutely convergent and limn→∞ f(xqn) = c. Obviously if t = xqn, x ∈ (qa, a], n ∈ N, we

124

get

f(t) = c+∞∑

k=n

f(xqk)− f(xqk+1) = c+∫ t

0Dqf(u) dqu. (5.3.29)

But ∣∣∣∣∫ t

0Dqf(u) dqu

∣∣∣∣ 6 ∞∑k=0

|f(xqk)− f(xqk+1)| <∞, t = xqn; n ∈ N. (5.3.30)

Hence Dqf ∈ L1q(0, x) for all x ∈ (qa, a]. To prove the converse if f satisfies the identity

in (5.3.28), then Dqf(x) = φ(x) for all x 6= 0. Hence∑∞

n=0 xqn(1 − q)|(Dqf)(xqn)| is

convergent. Hence from Lemma 1.1.1 and relation (1.1.34) there exists αi ∈ (0, 1], i = 1, 2,

and n0 ∈ N such that

|f(xqn)− f(xqn+1)| 6 C|xqn|1−α1 , |f(yqn)− f(yqn+1)| 6 C|yqn|1−α2 , n > n0 (5.3.31)

for some constant C > 0. Thus from (5.3.28) we have

f(xqn)− f(yqn) =∞∑

k=n

f(xqk)− f(xqk+1)−∞∑

k=n

f(yqk)− f(yqk+1). (5.3.32)

Then for all n > n0 we obtain

|f(xqn)− f(yqn)| 6 C

(x1−α1

(1− q1−α1)qn(1−α1) +

y1−α2

(1− q1−α2)qn(1−α2)

), (5.3.33)

Thus

∞∑n=n0

|f(xqn)− f(yqn)| 6 C

(x1−α1

(1− q1−α1)2qn0(1−α1) +

y1−α2

(1− q1−α2)2qn0(1−α2)

). (5.3.34)

We choose n0 large enough such that

C

(x1−α1

(1− q1−α1)2qn0(1−α1) +

y1−α2

(1− q1−α2)2qn0(1−α2)

)< ε.

Hence f ∈ ACq[0, a].

Definition 5.3.2. Let AC(n)q [0, a], n ∈ Z+, be the space of all functions f defined on [0, a]

such that f , Dqf , . . ., Dn−1q f are q-regular at zero and Dn−1

q f(x) ∈ ACq[0, a].

125

When n = 1, we call AC(1)q [0, a] the space of q-absolutely continuous function and we

refer to it by ACq[0, a] for simplicity.

Lemma 5.3.8. A function f : [0, a] → C lies in AC(n)q [0, a] if and only if there exists a

function φ ∈ L1q(0, x), qa < x 6 a such that

f(t) =n−1∑k=0

cktk +

tn−1

Γq(n)

∫ t

0

(qu/t; q

)n−1

φ(u) dqu, (5.3.35)

where t = xqn, x ∈ (qa, a], n ∈ N, ck =Dk

q f(0)Γq(k + 1)

and φ(x) = Dnq f(x) for all x 6= 0.

Proof. The sufficient condition is easy to prove. We shall prove the necessary condition

by using the mathematical induction. The case of n = 1 is proved in Theorem 5.3.7. Now

assume that (5.3.35) holds at n = m. Let f(x) ∈ AC(m+1)q [0, a]. That is, f, . . . , Dm

q f

are q-regular at zero and Dmq f ∈ AC1

q [0, a]. Hence, Dqf ∈ AC(m)q [0, a]. Then, from the

hypothesis of the induction

Dqf(t) =m−1∑j=0

cjtj +

tm−1

Γq(m)

∫ t

0

(qu/t; q

)m−1

φ(u) dqu, (5.3.36)

where t ∈ xqn, n ∈ N, x ∈ (qa, a],

cj = DjqDqf(0)/Γq(j + 1) = Dj+1

q f(0)/Γq(j + 1) and φ(x) = Dm+1q f(x)

for all x 6= 0. Now, replacing t with u and u with v in (5.3.36) and integrating from 0 to t.

Then applying Lemma 5.3.5 yields

f(t) =m∑

k=0

Dkq f(0)

Γq(k + 1)tk +

tm

Γq(m+ 1)

∫ t

0

(qu/t; q

)mDm+1

q f(u) dqu, (5.3.37)

which completes the induction steps.

126

5.4 Basic Riemann–Liouville fractional calculi

In 1966, W.A. Al-Salam [110] defined q-integral operator K−αq by the basic integral

K−αq φ(x) :=

q−12α(α−1)

Γq(α)

∫ ∞

xtα−1

(x/t; q

)α−1

φ(tq1−α) dqt,

K0qφ(x) := φ(x),

(5.4.1)

where α 6= −1,−2, . . .. Al-Salam defined (5.4.1) as a generalization of the following basic

Cauchy formula which is introduced in [12] for a positive integer n via

K−nq φ(x) =

∫ ∞

x

∫ ∞

xn−1

. . .

∫ ∞

x1

φ(t) dqtdqx1 . . . dqxn−1

=q−

12N(N−1)

Γq(N)

∫ ∞

xtN−1 (x/t; q)N−1 φ(tq1−N ) dqt.

(5.4.2)

Using (1.1.33), we can write the equation (5.4.1) as

K−αq φ(x) = q−

12α(α−1)xα(1− q)α

∞∑k=0

(−1)kqk(k−1)

2

[−αk

]q

φ(xq−α−k), (5.4.3)

which is valid for all α. The fractional q-integral operator K−αq is a q-analogue of Liouville

fractional integral, Kα−, defined in (5.1.13). Al-Salam also proved the following semigroup

identity, cf. [13],

Kαq K

βq φ(x) = Kα+β

q φ(x), α, β > 0. (5.4.4)

Moreover, in [13], Al-Salam defined a two parameter basic fractional operator by

Kη,αq f(x) :=

q−ηxη

Γq(α)

∫ ∞

x

(x/t; q

)α−1

t−η−1f(tq1−α) dqt, (5.4.5)

α 6= −1,−2, . . .. This is a q-analogue of the Erdelyi and Sneddon fractional operator,

cf. [44, 46],

Kη,αf(x) =xη

Γ(α)

∫ ∞

x

(t− x)α−1t−η−1f(t) dt. (5.4.6)

By means of (1.1.33), (5.4.5) can be written as

Kη,αq f(x) = (1− q)α

∞∑k=0

(−1)kqkη

[−αk

]q

f(xq−α−k), (5.4.7)

127

valid for all α. Again Al-Salam gave the following semigroup identity for η, α and β in R+.

Kη,αq Kη+α,β

q f(x) = Kη,α+βq f(x). (5.4.8)

In 1968, R.P. Agarwal [10] defined another two parameter family fractional q-operator

corresponding to the basic integral (1.1.31). This operator is defined by

Iη,αq f(x) =

x−η−1

Γq(α)

∫ x

0

(qt/x; q

)α−1

tηf(t) dqt, α 6= −1,−2, . . . (5.4.9)

=(1− q)Γq(α)

∞∑k=0

q(η+1)k(qk+1; q

)α−1

f(xqk). (5.4.10)

valid for all α. He also proved the following semigroup identity when η, λ, and µ are positive

constants.

Iη,λq Iη+λ,µ

q f(x) = Iη,µ+λq = Iη+λ,µ

q Iη,λq f(x) (5.4.11)

= Iη,µq Iµ+η,λ

q f(x) = Iη+µ,λq Iη,µ

q f(x). (5.4.12)

It should be mentioned here that in most of the proofs of [10, 11, 13], the domain where

the fractional integrals and the related properties hold is not determined precisely. A very

restrictive condition might be found in [10, p.366]. A q-analogue of the Riemann-Liouville

fractional(RLF) integral operator is introduced in [11] by Al-Salam through

Iαq f(x) :=

xα−1

Γq(α)

∫ x

0

(qt/x; q

)α−1

f(t) dqt, (5.4.13)

α 6= −1,−2, . . .. This RLF integral is an extension of the basic Cauchy formula, [12],

Inq f(x) =

∫ x

a

∫ xn−1

a. . .

∫ x1

af(t) dqtdqx1 . . . dqxn−1

=xn−1

Γq(n)

∫ x

a

(qt/x; q

)n−1

f(t) dqt.

(5.4.14)

This basic analogue of RLF integral is also given by Agarwal in [10]. In fact Agarwal defined

the q-fractional derivative to be,

Dαq f(x) := I−α

q f(x) =x−α−1

Γq(−α)

∫ x

0

(qt/x; q

)−α−1

f(t) dqt, α > 0. (5.4.15)

128

Using (1.1.31), (5.4.13) reduces to

Iαq f(x) = xα(1− q)α

∞∑n=0

qn (qα; q)n

(q; q)nf(xqn). (5.4.16)

In the following we will derive basic analogue of the results mentioned above concerning

RLF calculus. These results will be used in our investigations below. From now on we

assume that α > 0.

The following Theorem taken from [14, p.491] is needed in the proof of Lemma 5.4.2

below.

Theorem 5.4.1. Suppose λ and µ are real. Then

limq→1−

(qλx; q)∞(qµx; q)∞

= (1− x)µ−λ, (5.4.17)

uniformly on

x ∈ C : |x| 6 1 , if µ > λ, µ+ λ > 1,

and uniformly on compact subsets of x ∈ C : |x| 6 1, x 6= 1 for other choices of λ and µ.

Lemma 5.4.2. Let f be a function defined on [0, a], a > 0. If f is Riemann integrable on

[0, x], then

limq→1−

Iαq f(x) = Iαf(x), x ∈ (0, a], α > 0. (5.4.18)

Proof. From (1.1.13) and (5.4.17), we obtain for 0 < t < x

limq→1−

xα−1(qt/x; q

)α−1

= limq→1−

xα−1

(qt/x; q

)∞(

qαt/x; q)∞

= xα−1(1− t

x)α−1 = (x− t)α−1

(5.4.19)

uniformly for 0 < t < x. Hence given ε > 0 there exists δ > 0 such that for all t ∈qkx, k ∈ N

, we have

∣∣∣xα−1(qt/x; q

)α−1

− (x− t)α−1∣∣∣ 6 ε

/2

129

whenever 0 < 1− q < δ. That is∣∣∣∣Iαq f(x)− 1

Γq(α)

∫ x

0(x− t)α−1f(t) dqt

∣∣∣∣ 6 ε

2Γq(α)(5.4.20)

and since∣∣Iαq f(x)− Iαf(x)

∣∣ 6 ∣∣∣∣Iαq f(x)− 1

Γq(α)

∫ x

0(x− t)α−1f(t) dqt

∣∣∣∣+∣∣∣∣ 1Γq(α)

∫ x

0(x− t)α−1f(t) dqt− Iαf(x)

∣∣∣∣ , (5.4.21)

then the lemma follows if f is Riemann integrable on [0, x].

Lemma 5.4.3. If

limn→∞

qnf(xqn) = 0 for all x ∈ (qa, a], (5.4.22)

then the fractional q-integral of order α exists.

Proof. We apply Lemma 1.1.1 to (5.4.16) and use that limn→∞(qn+1; q)α−1 = 1.

Remark 5.4.1. From the previous lemma we conclude that Iαq f(x) exists if and only if

f ∈ L1q(0, x) for all x ∈ (qa, a], i.e. f ∈ L1

q(0, a).

Lemma 5.4.4. Let f : [0, a] → C be a function. If f ∈ L1q(0, a), then Iα

q f ∈ L1q(0, a).

Proof. Let f be a function defined on [0, a] such that f ∈ L1q(0, a). Then from (5.4.13) we

get for x ∈ (qa, a]∫ x

0|Iα

q f(t)| dqt =1

Γq(α)

∫ x

0tα−1

∣∣∣∣∫ t

0

(qu/t; q

)α−1

f(u) dqu

∣∣∣∣ dqt

61

Γq(α)

∫ x

0tα−1

∫ t

0

(qu/t; q

)α−1

|f(u)| dqu dqt

=xα+1(1− q)2

Γq(α)

∞∑n=0

qnα∞∑

m=0

qn+m(qm+1; q

)α−1

|f(xqn+m)|

=xα+1(1− q)2

Γq(α)

∞∑n=0

qnα∞∑

k=n

qk(qk−n+1; q

)α−1

|f(xqk)|

=xα+1(1− q)

Γq(α)

∞∑k=0

qk|f(xqk)|k∑

n=0

(1− q)qnα(qk−n+1; q

)α−1

. (5.4.23)

130

But from (5.3.7) we obtain

k∑n=0

(1− q)qnα(qk−n+1; q

)α−1

= Bq(1, α)xα+1∞∑

k=0

qk(1− q)(qk+1; q

)α|f(xqk)|

=xα

Γq(α)Bq(1, α)

∫ x

0

(qt/x; q

)α|f(t)| dqt

6xα

Γq(α+ 1)

∫ x

0|f(t)| dqt (5.4.24)

Combining (5.4.23) and (5.4.24) we conclude that Iαq f ∈ L1

q(0, x), for all x ∈ (qa, a]. That

is Iαq f ∈ L1

q(0, a).

Similarly to property (5.2.9) we have the following Lemma

Lemma 5.4.5. Let f ∈ L1q(0, a). Then for x ∈ (qa, a] and t ∈ xqm, m ∈ N

limα→0+

Iαq f(t) = f(t). (5.4.25)

Proof. Let x ∈ (qa, a] and t ∈ xqm, m ∈ N be fixed. From (5.4.16) we have

limα→0+

Iαq f(t) = lim

α→0+tα(1− q)α

∞∑n=0

qn (qα; q)n

(q; q)nf(tqn). (5.4.26)

Since

qn (qα; q)n

(q; q)n|f(tqn)| 6 qn|f(tqn)|, n ∈ N, α ∈ (0, 1], (5.4.27)

then from Weierstrass M -test, the series in (5.4.26) is uniformly convergent for α ∈ (0, 1).

Thus

limα→0+

∞∑n=0

qn (qα; q)n

(q; q)nf(tqn) = f(t). (5.4.28)

Hence

limα→0+

Iαq f(t) = lim

α→0+tα(1− q)α lim

α→0+

∞∑n=0

qn (qα; q)n

(q; q)nf(tqn) = f(t). (5.4.29)

From now on we use the notation I0q f(x) = f(x). The property

Iβq I

αq f(x) = Iα

q Iβq f(x) = I(α+β)

q f(x), α, β > 0 (5.4.30)

131

is a q-analogue of the semigroup property (5.2.13) and it is proved by R.P. Agarwal [10].

As in the case of the RLF derivative, the existence of the fractional q-Riemann-Liouville

derivative holds only for a more restrictive class of functions. For this reason we study a

q-analogue of Abel’s integral equation.

Theorem 5.4.6. The q-Abel integral equation

xα−1

Γq(α)

∫ x

0

(qt/x; q)α−1 φ(t) dqt = f(x), 0 < α < 1, x ∈ (0, a] (5.4.31)

has a unique solution φ(x) ∈ L1q(0, a) if and only if

φ(x) = Dq,xI1−αq f(x), (5.4.32)

and

I1−αq f(x) ∈ ACq[0, a] and I1−α

q f(0) = 0. (5.4.33)

Proof. Assume that (5.4.31) has a solution φ ∈ L1q(0, a). Hence replacing x with t and

t with u in (5.4.31), and multiplying both sides of the equation with x−α(qt/x; q

)−α

, and

then make the q-integration from 0 to x. Then, we have

x−α

Γq(α)

∫ x

0

(qt/x; q

)−α

∫ t

0tα−1

(qu/t; q

)α−1

φ(u) dqu dqt

= x−α

∫ x

0

(qt/x; q

)−αf(t) dqt.

(5.4.34)

But from (5.3.17),

x−α

Γq(α)

∫ x

0

(qt/x; q

)−α

∫ t

0tα−1

(qu/t; q

)α−1

φ(u) dqu dqt

=B(1− α, α)

Γq(α)

∫ x

0φ(t) dqt.

(5.4.35)

Consequently,

Iqφ(x) = I1−αq f(x). (5.4.36)

132

Applying the operator Dq on the two sides of (5.4.36) yields (5.4.32). Since φ ∈ Lq(0, a),

then by Remark 5.3.2

I1−αq f(0) := lim

n→∞x∈(qa,a]

I1−αq f(xqn)

= limn→∞

x∈(qa,a]

Iqφ(xqn) = limn→∞

x∈(qa,a]

∞∑k=n

xqk(1− q)φ(xqk) = 0.(5.4.37)

Moreover from Lemma 5.3.7 I1−αq f ∈ ACq[0, a].

Conversely assume that f satisfies (5.4.33) and φ(x) := Dq,xI1−αq f(x). Then φ ∈

L1q(0, a). We shall prove that φ is a solution of (5.4.31). Indeed let g be the function

defined byxα−1

Γq(α)

∫ x

0

(qt/x; q

)α−1

φ(t) dqt = g(x), x ∈ (0, a]. (5.4.38)

It suffices to show g(x) = f(x), x ∈ (0, a]. Hence φ is a solution of (5.4.38). From the

sufficient part, φ should have the form (5.4.32) with g instead of f . Thus

φ(x) = Dq,xI1−αq f(x) = Dq,xI

1−αq g(x)

= Dq,xx−α

Γq(1− α)

∫ x

0

(qt/x; q

)−αg(t) dqt,

(5.4.39)

for all x ∈ (0, a]. Hence

Dq,x

(I1−αq

(f − g

)(x))

= 0, x > 0. (5.4.40)

Solving equation (5.4.40) we obtain

I1−αq

(f − g

)(x) = c(x), x > 0, (5.4.41)

where c is a q-periodic function. The function I1−αq g(x) ∈ ACq[0, a] because it satisfies

equation (5.4.36) with g(x) on the right hand side. Also, we have I1−αq g(0) = 0 because

(5.4.38) is solvable equation. Thus I1−αq

(f − g

)(x) ∈ ACq[0, a]. So, I1−α

q

(f − g

)is q-regular

at zero such that I1−αq

(f − g

)(0) = 0. Consequently, c ≡ 0. That is,

x−α

Γq(1− α)

∫ x

0

(qt/x; q

)−α

(f(t)− g(t)) dqt = 0, for all x > 0. (5.4.42)

133

The previous is an equation of the form (5.4.31). The uniqueness of its solutions leads to

f(x) = g(x), for all x ∈ (0, a].

From Theorem 5.2.2, the solution φ should be the fractional derivative of order α of f .

Starting from the previous theorem we define the basic fractional derivative as follows.

Definition 5.4.1. Let f be a function defined on [0, a]. For α > 0 the fractional q-derivative

of order α is defined to be

Dαq f(x) := φ(x) = Dk

q Ik−αq f(x), k = [α] + 1, (5.4.43)

provided that

f(x) ∈ L1q(0, a), Ik−α

q f(x) ∈ AC(k)q [0, a]. (5.4.44)

The following lemma gives the relationship between the fractional q-derivative of order

α and I−αq .

Lemma 5.4.7. Let f ∈ L1q(0, a) and α > 0, k = [α] + 1, be such that Ik−α

q f ∈ AC(k)q [0, a].

Then

Dαq f(x) = I−α

q f(x) =x−α−1

Γq(α)

∫ x

0

(qt/x; q

)−α−1

f(t) dqt, x ∈ (0, a]. (5.4.45)

Proof. From (5.4.43) we have

Dαq f(x) = Dk

q Ik−αq f(x) = Dk

q

(xk−α−1

Γq(k − α)

∫ x

0

(qt/x; q

)k−α−1

f(t) dqt

)(5.4.46)

Set

h(t, x) :=xk−α−1

Γq(k − α)(qt/x; q

)k−α−1

f(t). (5.4.47)

Hence

h(xqr, xqj) = 0, r = 0, 1, . . . , j − 1, j = 1, 2, . . . , k. (5.4.48)

Thus applying Lemma 5.3.4 to (5.4.46) we obtain

Dαq f(x) =

∫ x

0Dk

q,x

xk−α−1

Γq(k − α)(qt/x; q

)k−α−1

f(t) dqt

=x−α−1

Γq(−α)

∫ x

0

(qt/x; q

)−α−1

f(t) dqt = I−αq f(t)

(5.4.49)

134

Now we investigate some properties of the fractional q-integral and q-derivative. Similar

to property (5.2.15) we have the following.

Lemma 5.4.8. If f ∈ L1q(0, a), then

Dαq I

αq f(x) = f(x); α > 0, x ∈ (0, a]. (5.4.50)

Proof. It is obvious that if α = n, n ∈ N, then Dnq I

nq f(x) = f(x). For a non integer and

positive α, k − 1 6 α < k, k ∈ Z+, we have by applying the semi group identity (5.4.30)

that

Dαq I

αq f(x) = Dk

q Ik−αq Iα

q f(x) = Dkq I

kq f(x) = f(x), x ∈ (0, a].

The converse of (5.2.18) is the following

Lemma 5.4.9. Let α ∈ R+ and k := [α] + 1. If

f ∈ L1q(0, a) such that Ik−α

q f ∈ ACkq (0, a), (5.4.51)

then

Iαq D

αq f(x) = f(x)−

k∑j=1

xα−j

Γq(α− j + 1)Dα−j

q f(x)∣∣∣x=0

, x ∈ (0, a]. (5.4.52)

Proof. Set

h(t, x) = xα (qt/x; q)αDαq f(t), x ∈ (0, a], 0 < t 6 x. (5.4.53)

Hence h(x, qx) = 0 for all x ∈ (0, a]. So applying Lemma 5.3.4 yields

Iαq D

αq f(x) =

xα−1

Γq(α)

∫ x

0

(qt/x; q

)α−1

Dαq f(t) dqt

= Dq,x

(xα

Γq(α+ 1)

∫ x

0

(qt/x; q

)αDα

q f(t) dqt

).

(5.4.54)

135

Now, applying the q-integration by parts (1.1.41) k times on the last q-integral of (5.4.54),

we get

xα

Γq(α+ 1)

∫ x

0

(qt/x; q

)αDα

q f(t) dqt =xα−1

Γq(α)

∫ x

0

(qt/x; q

)α−1

Dkq I

k−αq f(t) dqt

= −k∑

j=1

xα−j+1

Γq(α− j + 2)Dα−j

q,x f(x)∣∣∣x=0

+xα−k

Γq(α− k + 1)

∫ x

0

(qt/x; q

)α−k

Ik−αq f(t) dqt

= −k∑

j=1

xα−j+1

Γq(α− j + 2)Dα−j

q,x f(x)∣∣∣x=0

+ Iα−k+1q Ik−α

q f(x).

(5.4.55)

Applying the semigroup identity (5.4.30) on (5.4.55) we obtain

xα

Γq(α+ 1)

∫ x

0

(qt/x; q

)αDα

q f(t) dqt = −k∑

j=1

xα−j+1

Γq(α− j + 2)Dα−j

q,x f(x)∣∣∣x=0

+Iqf(x). (5.4.56)

Thus we deduce (5.4.52) by computing the q-derivative of the two sides of (5.4.56) for all

x ∈ (0, a].

From (5.4.52) the equality

Iαq D

αq f(x) = f(x), for all x ∈ (0, a] (5.4.57)

holds if and only if

Dα−jq f(0) = 0, j = 1, 2, . . . , k. (5.4.58)

Lemma 5.4.10. Let f satisfy the conditions

f ∈ L1q(0, a) and Ik−α

q f ∈ AC(k)q [0, a], (5.4.59)

where α > 0 and k := [α] + 1. Then

Dα−jq f(x) ∈ L1

q(0, a), j = 0, 1, 2, . . . , k − 1. (5.4.60)

Moreover

Dα−jq f(x) ∈ AC(j)

q [0, a], j = 1, 2, . . . , k − 1. (5.4.61)

136

Proof. The proof of (5.4.60) follows immediately from (5.4.59) and that of (5.4.61) follows

by noting that

Dα−jq f(x) = Dk−j

q Ik−j−(α−j)q f(x)

= Dk−jq Ik−α

q f(x) ∈ AC(j)q [0, a], 1 6 j 6 k − 1.

(5.4.62)

In the following three lemmas we discuss the results of combining q-fractional integral

and derivative of not necessarily equal orders.

Lemma 5.4.11. If f ∈ L1q(0, a), then

Dαq I

βq f(x) = Iβ−α

q f(x), β > α > 0, x ∈ (0, a]. (5.4.63)

If in addition Dα−βq f(x) exists in (0, a], then

Dαq I

βq f(x) = Dα−β

q f(x), α > β > 0. (5.4.64)

Proof. First, assume that β > α. Then, β = α+ (β − α) and from (5.4.30) we obtain

Dαq I

βq f(x) = Dα

q Iαq I

β−αq f(x) = Iβ−α

q f(x).

Now let β < α and m := [α] + 1 and n := [α − β] + 1. Then n 6 m. Using (5.4.43) and

(5.4.30) we have for x ∈ (0, a]

Dαq I

βq f(x) = Dm

q Im−αq Iβ

q f(x) = Dmq I

m−nq In−α+β

q f(x)

= Dnq I

n−α+βq f(x) = Dα−β

q f(x).(5.4.65)

Lemma 5.4.12. Let f ∈ L1q(0, a) such that In−β

q f ∈ AC(n)q [0, a], where β > 0 and

n := [β] + 1. Then for any α > 0 we have

Iαq D

βq f(x) = D−α+β

q f(x)−n∑

j=1

Dβ−jq f(x)|x=0

xα−j

Γq(α− j + 1), x ∈ (0, a]. (5.4.66)

137

Proof. If α = 0, then Γq(−j + 1) = 0 for all j ∈ N and the identity (5.4.66) holds for any

β > 0. Therefore we shall assume that α > 0. We distinguish between two cases. First if

α > β, then from (5.4.30) and (5.4.52) we obtain

Iαq D

βq f(x) = Iα−β

q

(Iβq D

βq

)f(x)

= Iα−βq

(f(x)−

n∑j=1

xβ−j

Γq(β − j + 1)Dβ−j

q f(x)∣∣∣x=0

)= Iα−β

q f(x)−n∑

j=1

xα−j

Γq(α− j + 1)Dβ−j

q f(x)∣∣∣x=0

,

(5.4.67)

for all x ∈ (0, a]. If β > α, then from (5.4.63) and (5.4.52) we obtain

Iαq D

βq f(x) = Dβ−α

q

(Iβq D

βq f(x)

)= Dβ−α

q

(f(x)−

n∑j=1

xβ−j

Γq(β − j + 1)Dβ−j

q f(x)∣∣∣x=0

)= Dβ−α

q f(x)−n∑

j=1

xα−j

Γq(α− j + 1)Dβ−j

q f(x)∣∣∣x=0

,

(5.4.68)

for all x ∈ (0, a].

Lemma 5.4.13. If f ∈ L1q(0, a) and In−β

q f(x) ∈ AC(n)q [0, a], where β > 0 and n := [β]+ 1,

then

Dαq D

βq f(x) = Dα+β

q f(x)−n∑

j=1

x−α−j

Γq(−α− j + 1)Dβ−j

q f(0), x ∈ (0, a], (5.4.69)

provided that Dα+βq f(x) exists for any α > 0.

Proof. Let x ∈ (0, a]. From (5.4.52) and (5.4.64) we obtain

Dαq D

βq f(x)=Dα+β

q Iβq D

βq f(x)=Dα+β

q

f(x)−n∑

j=1

xβ−j

Γq(β − j + 1)Dβ−j

q f(x)∣∣∣x=0

= Dα+β

q f(x)−n∑

j=1

x−α−j

Γq(−α− j + 1)Dβ−j

q f(x)∣∣∣x=0

.

(5.4.70)

138

Remark 5.4.2. Similar to (5.4.69), if Ik−αq ∈ AC(k)

q [0, a] and Dα+βq f(x) exists, where k :=

[α] + 1, then

DβqD

αq f(x) = Dβ+α

q f(x)−k∑

i=1

Dα−iq f(x)|x=0

x−β−i

Γq(−β − i+ 1), x ∈ (0, a]. (5.4.71)

If

Dβ−jq f(x)

∣∣∣x=0

= 0, j = 1, 2, . . . , n, (5.4.72)

and

Dα−jq f(x)|x=0 = 0, j = 1, 2, . . . , k, (5.4.73)

then we have

Dαq D

βq = Dβ

qDαq = Dα+β

q . (5.4.74)

Similar to property (5.2.24) we have the following

Lemma 5.4.14. Let α > 0, n := [α] + 1. If f ∈ AC(n)q [0, a], then for any x ∈ (0, a], we

have

Dαq f(x) =

n−1∑j=0

x−α+j

Γq(−α+ j + 1)Dj

qf(0) +xn−α−1

Γq(n− α)

∫ x

0

(qt/x; q

)n−α−1

Dnq f(t) dqt. (5.4.75)

Moreover, Dαq f(0) = 0 if and only if Dj

qf(0) = 0, for j = 0, 1, . . . , n− 1.

Proof. Since f ∈ AC(n)q [0, a], then from lemma 5.3.8 we obtain

f(x) =n−1∑k=0

Dkq f(0)

Γq(k + 1)xk +

xn−1

Γq(n)

∫ x

0

(qt/x; q

)n−1

Dnq f(t) dqt, x ∈ (0, a]. (5.4.76)

Consequently,

f(x) =n−1∑k=0

Dkq f(0)

Γq(k + 1)xk + In

q Dnq f(x), x ∈ (0, a]. (5.4.77)

139

Applying the operator Dαq to both sides of (5.4.77) and using (5.4.63) yields

Dαq f(x) =

n−1∑k=0

Dkq f(0)

Γq(k − α+ 1)xk−α +Dα

q Inq D

nq f(x)

=n−1∑k=0

Dkq f(0)

Γq(k − α+ 1)xk−α + In−α

q Dnq f(x)

=n−1∑k=0

Dkq f(0)

Γq(k − α+ 1)xk−α +

xn−α−1

Γq(n− α)

∫ x

0

(qt/x; q

)n−α−1

Dnq f(t) dqt,

(5.4.78)

which is (5.4.75). The rest of the proof arises from (5.4.75).

In the following we define a q-analogue of the Grunwald–Letinkov fractional derivative.

If f is a function defined on [0, a], a > 0, then, from Property 5.3.3, Dkq f(x), k ∈ N, x ∈ (0, a]

is given by

Dkq f(x) =

1xk(1− q)k

j=k∑j=0

(−1)j

[k

j

]q

f(qjx)qj(j−1)/2+j(k−j)

. (5.4.79)

Since the right hand side of (5.4.79) has a meaning when we replace k by any real number

α, we define the fractional operator

Dαq f(x) =

1xα(1− q)α

∞∑j=0

(−1)j

[α

j

]q

f(qjx)qj(j−1)/2+j(α−j)

. (5.4.80)

Clearly Dαq f(x) = Dk

qf(x), if α = k ∈ N because[

kj

]q

= 0 for all j > k. The operator

(5.4.80) is a q-analogue of the Grunwald–Letinkov fractional operator. From (1.1.4)

(−1)j

[α

j

]q

qj(j+1)/2−jα = qj (q−α; q)j

(q; q)j, j ∈ N. (5.4.81)

Then

Dαq f(x) = x−α(1− q)−α

∞∑j=0

(−1)jqj(j+1)/2qαj

[α

j

]q

f(qjx)

= x−α(1− q)−α∞∑

j=0

qj (q−α; q)j

(q; q)jf(qjx).

(5.4.82)

Therefore, from (5.4.16), the fractional operator Dαq , α ∈ R, is the same as Riemann–

Liouville fractional operator Dαq .

Now we introduce a q-analogue of the Caputo fractional derivative defined in (5.2.27).

140

Definition 5.4.2. For α > 0, the Caputo fractional q-derivative of order α is defined to be

cDαq f(x) := Ik−α

q Dkq f(x), k = [α] + 1. (5.4.83)

Clearly cDαq f(x) exists if f ∈ AC(k)

q [0, a]. Obviously cDαq c = 0 for any constant c ∈ C.

We shall study existence and uniqueness of solutions of systems of fractional order of

basic Caputo derivative in the Section 7.2.

Chapter 6

Basic Mittag-Leffler Functions

The two parameter Mittag-Leffler functions play an important role in the theory of fractional

calculus and fractional differential equations, see e.g. [24,40,100]. In fact they are solutions

of fractional integro differential equations. In this chapter we derive a basic analogue of

Mittag-Leffler function and study some of their properties. In particular the zeros of the

basic Mittag-Leffler functions will be investigated.

6.1 Basic Mittag-Leffler function

The special function of the from

Eρ(z) =∞∑

n=0

zn

Γ(nρ + 1)

; ρ > 0, (6.1.1)

and more general functions

Eρ,µ(z) =∞∑

n=0

zn

Γ(nρ + µ)

; ρ > 0, µ ∈ C, (6.1.2)

where z ∈ C, are known as Mittag-Leffler functions, see e.g. [45,65,76]. The function Eρ(z)

which defines a family of exponential functions of one parameter was first introduced by

Mittag-Leffler [97] who in [97,98] investigated some of its properties, while Eρ,µ(z) was first

introduced in [118]. It is an entire function of z of order ρ and type 1. See [39, 40, 54, 100]

141

142

for properties and historical notes. The following group of identities follows directly from

the definition of Eρ,µ(z).

E1/2,1(z) = cosh√z, E1/2,2(z) =

sinh√z√

z, (6.1.3)

E1,1(z) = ez, E1,2(z) =ez − 1z

, (6.1.4)

Eρ,µ(z) =1

Γ(µ)+ zEρ,µ+1/ρ(z), (6.1.5)

Eρ,µ(z) = µEρ,µ+1(z) +z

ρE′ρ,µ+1(z),

′ :=d

dz(6.1.6)

Furthermore, the termwise integration of the power expansion of the function Eρ,µ(z) along

the interval (0, z) leads to the integral relation∫ z

0Eρ,µ(λt

1ρ )tµ−1dt = zµEρ,µ+1(λz1/ρ) (6.1.7)

and its generalization

1Γ(α)

∫ z

0(z − t)α−1Eρ,µ(λt

1ρ )tµ−1 dt = zµ+α−1Eρ,µ+α(λz1/ρ), µ > 0, α > 0. (6.1.8)

In particular, when µ > 0, we have

1Γ(µ)

∫ z

0(z − t)µ−1 cosh

√λt dt = zµE1/2,µ+1(λz

2), (6.1.9)

1Γ(µ)

∫ z

0(z − t)µ−1 sinh

√λt√

λdt = z1+µE1/2,2+µ(λz2), (6.1.10)

1Γ(µ)

∫ z

0(z − t)µ−1eλt dt = zµE1,1+µ(λz), (6.1.11)

cf. [40]. Since we have two major q-exponential functions, namely, eq(z) and Eq(z), then

we will define two q-analogues of the Mittag-Leffler functions. We define

eν,µ(z; q) :=∞∑

n=0

zn

Γq(νn+ µ), |z| < (1− q)−ν ,

Eν,µ(z; q) :=∞∑

n=0

qνn(n−1)

2 zn

Γq(νn+ µ), z ∈ C,

(6.1.12)

143

where ν > 0, µ ∈ C. When µ = 1, eν,1(z; q) and Eν,1(z; q) define a family of q-exponential

functions of one parameter. Another one-parameter family of q-exponential functions is

defined by, cf. [22, 50,65],

E(α)q (z) :=

∞∑k=0

qαk2/2

(q; q)kzk, α ∈ C. (6.1.13)

Similarly to formulae (6.1.3)–(6.1.6), the following formulae follow at once from the defini-

tion of eν,µ(z; q):

e2,1(z; q) = coshq

√z(1− q), e2,2(z; q) =

sinhq√z(1− q)√z

, (6.1.14)

e1,1(z; q) = eq(z(1− q)), e1,2(z; q) =eq(z(1− q)

)− 1

z, (6.1.15)

eν,µ(z; q) =1

Γq(µ)+ zeν+1,µ(z; q), (6.1.16)

qµzDqeν,µ+1(zν ; q) = −1− qµ

1− qeν,µ+1(zν ; q) + eν,µ(zν ; q). (6.1.17)

The following group of formulae follow at once from the definition of Eν,µ(z; q).

E2,1(z; q) = cosh(q−1/2√z; q), E2,2(z; q) =sinh(q−1√z; q)

q−1√z

, (6.1.18)

E1,1(z; q) = Eq(z(1− q)), E1,2(qz; q) =1z(−1 + Eq(z(1− q))), (6.1.19)

Eν,µ(z; q) =1

Γq(µ)+ zEν,ν+µ(qz, q), µ > 0, (6.1.20)

Eν,µ(zν ; q) =1− qµ

1− qEν,µ+1(zν ; q) + qµzDqEν,µ+1(zν ; q). (6.1.21)

Lemma 6.1.1. Let α, µ, β be positive numbers. Then

zα−1

Γq(α)

∫ z

0

(qt/z; q

)α−1

Eβ,µ(λtβ ; q)tµ−1 dqt = zα+µ−1Eβ,µ+α(λzβ; q). (6.1.22)

Proof. Replace Eβ,µ(λtβ; q) by its power series expansion on the left-hand side of (6.1.22).

144

Then

zα−1

Γq(α)

∫ z

0

(qt/z; q

)α−1

Eβ,µ(λtβ; q)tµ−1 dqt

=zα−1

Γq(α)

∫ z

0

(qt/z; q

)α−1

tµ−1

( ∞∑n=0

qn(n−1)β/

2

(λtβ)n

Γq(nβ + µ)

)dqt

=zα−1

Γq(α)

∞∑n=0

qn(n−1)β/

2 λn

Γq(nβ + µ)

∫ z

0

(qt/z; q

)α−1

tnβ+µ−1 dqt. (6.1.23)

Substituting with u := t/z, z 6= 0, in the q-integral of (6.1.23) and using (1.1.12) yields∫ z

0

(qt/z; q

)α−1

tnβ+µ−1 dqt = znβ+µ

∫ 1

0

(qu; q

)α−1

unβ+µ−1 dqu

= znβ+µBq

(α, nβ + µ

).

(6.1.24)

Consequently

zα−1

Γq(α)

∞∑n=0

qn(n−1)β/

2 λn

Γq(nβ + µ)

∫ z

0

(qt/z; q

)α−1

tnβ+µ−1 dqt

=zα−1

Γq(α)

∞∑n=0

qn(n−1)β/

2 λnznβ+µ

Γq(nβ + µ)Bq

(α, nβ + µ

)=

∞∑n=0

qn(n−1)β/

2 λnznβ+µ+α−1

Γq(nβ + µ+ α)= zα+µ−1Eβ,µ+α(λzβ; q),

(6.1.25)

this completes the proof.

Corollary 6.1.2. Let ν, µ, α be positive numbers. Then∫ z

0Eν,µ(tν ; q)tµ−1 dqt = zµEν,µ+1(zν ; q), (6.1.26)

zα−1

Γq(α)

∫ z

0

(qt/z; q

)α−1

cosh(√λt; q) dqt = zαE2,α+1(λz2q; q), (6.1.27)

zα−1

Γq(α)

∫ z

0

(qt/z; q

)α−1

sinh(q−1√λt; q)√

λdqt = qzα+1E2,2+α(q2λz2; q), (6.1.28)

zα−1

Γq(α)

∫ z

0

(qt/z; q

)α−1

Eq(λt(1− q)) dqt = zαE1,α+1(λz; q). (6.1.29)

Proof. Formula (6.1.26) follows from (6.1.22) by taking α = 1. Formula (6.1.27) follows

from (6.1.22) by taking β = 2, µ = 1, and from (6.1.18) by replacing z in E1,2(z; q) by λt2.

145

As for formula (6.1.28) it follows from (6.1.22) by taking β = µ = 2 and using (6.1.18) when

z in E2,2(z; q) is replaced by λt2. Finally formula (6.1.29) follows from (6.1.22) by taking

β = µ = 1 and using (6.1.19) when z is replaced by λt.

Lemma 6.1.3. Eν,µ(z; q) is an entire function of order zero.

Proof. From (1.2.4) we have

ρ(Eν,µ(z; q)

)= lim sup

n→∞

n log nlog(1/|cn|)

, (6.1.30)

where cn is the coefficient of zn in (6.1.12), see [27], i.e.

cn =q

νn(n−1)2

Γq(nν + µ), n ∈ N. (6.1.31)

Hence,

log(

1|cn|

)= −ν

2n(n− 1) log q + log |Γq(nν + µ)| .

From (1.1.2) and the definition of the q−Gamma function, cf. (1.1.10) we obtain for n ∈ N,

ν ∈ R+, and µ ∈ C

log |Γq(nν + µ)| = log∣∣∣∣ (q; q)∞(qnν+µ; q)∞

(1− q)1−nν−µ

∣∣∣∣= log

∣∣∣∣ (q; q)∞(qnν+µ; q)∞

(1− q)1−nν−<µ

∣∣∣∣= log(q; q)∞ + (1− nν −<µ) log(1− q)− log

∣∣(qnν+µ; q)∞∣∣ ,

(6.1.32)

and

log∣∣(qnν+µ; q)∞

∣∣ = log( ∞∏

k=0

∣∣∣1− qnν+µ+k∣∣∣ ) = log

(lim

m→∞

m∏k=0

∣∣∣1− qnν+µ+k∣∣∣ )

= limm→∞

m∑k=0

log∣∣∣1− qnν+µ+k

∣∣∣ = ∞∑k=0

log∣∣∣1− qnν+µ+k

∣∣∣ . (6.1.33)

Since

log∣∣∣1− qnν+µ+k

∣∣∣ 6 log(1 + |qnν+µ+k|

)6 |qnν+µ+k| = qnν+k+<µ, (6.1.34)

146

then∞∑

k=0

log∣∣∣1− qnν+µ+k

∣∣∣ 6 ∞∑k=0

qnν+k+<µ =qnν+<µ

1− q. (6.1.35)

Therefore limn→∞

log |(qnν+µ; q)∞|n log n

= 0. Consequently by (6.1.32) limn→∞

log |Γq(nν + µ)|n log n

= 0.

Since limn→∞

−ν2n(n− 1) log qn log n

= ∞, then limn→∞log(1/|cn|)

n log n = ∞, i.e. ρ(Eν,µ(z; q)

)= 0.

Remark 6.1.1. From Theorem 1.2.1 we conclude that Eν,µ(z; q) has infinitely many zeros.

In Section 6.3 we prove that for specific values of ν and µ, Eν,µ(z; q), 0 < q < 1, may have

only a finite number of complex zeros. Moreover if q satisfies an additional relation, then

the zeros of Eν,µ(z; q) are all real.

6.2 Basic Volterra integral equations

In this section we investigate the solutions of q-analogues of a Volterra integral equation of the

second kind. Actually we introduce basic analogues of the following theorem of [40, p.3].

Theorem 6.2.1. Let ρ > 0 and λ ∈ C. If f ∈ L1(0, a), 0 < a < ∞, then the integral

equation

u(x) = f(x) +λ

Γ(1/ρ)

∫ x

0(x− t)1/ρ−1u(t)dt, x ∈ (0, a) (6.2.1)

has a unique solution in L1(0, a) given by the formula

u(x) = f(x) + λ

∫ x

0(x− t)1/ρ−1Eρ

(λ(x− t)1/ρ; 1/ρ

)f(t) dt, x ∈ (0, a). (6.2.2)

The following theorem gives explicit solutions of a basic Volterra integral equation of

the second kind in terms of the translation operator defined by Ismail [65], cf. (5.3.2).

Theorem 6.2.2. Let ν, a > 0 and λ ∈ C be such that

|λ|aν(1− q)ν < 1 (6.2.3)

147

If f ∈ L1q(0, a), then the q-integral equation

u(x) = f(x) +λxν−1

Γq(ν)

∫ x

0

(qt/x; q

)ν−1

u(t) dqt, x ∈ (0, a], (6.2.4)

has the unique solution

u(x) =∞∑

k=0

λkIkνq f(x) = f(x) + λxν−1

∫ x

0

(qt/x; q

)ν−1

ε−qνteν,ν(λxν ; q)f(t) dqt, (6.2.5)

x ∈ (0, a] in the space L1q(0, a).

Proof. We prove the theorem in four steps.

i. First we prove the existence of a solution by using q-analogue of the method of successive

approximations. Define a sequence um∞m=0 recursively by

u0(x) = f(x),

um(x) = f(x) +λxν−1

Γq(ν)

∫ x

0

(qt/x; q

)ν−1

um−1(t) dqt, x ∈ (0, a], m > 1.(6.2.6)

We prove by mathematical induction that

um(x) =m∑

k=0

λkIkνq f(x), m ∈ N, x ∈ (0, a]. (6.2.7)

Indeed if um satisfies the identity (6.2.7), then from (6.2.6) we obtain

um+1(x) = f(x) +λxν−1

Γq(ν)

∫ x

0

(qt/x; q

)ν−1

m∑k=0

λkIkνq f(t) dqt

= f(x) +m∑

k=0

λk+1 xν−1

Γq(ν)

∫ x

0

(qt/x; q

)ν−1

Ikνq f(t)

= f(x) +m∑

k=0

λk+1Iνq I

kνq f(x), (6.2.8)

for all x ∈ (0, a]. Hence from the semigroup identity (5.4.30) we have, x ∈ (0, a],

um+1(x) = f(x) +m∑

k=0

λk+1Ikν+νq f(x) =

m+1∑k=0

λkIkνq f(x). (6.2.9)

148

Thus (6.2.7) is true for all m ∈ N because it is true at m = 0. Now we prove that the series∑∞k=0 λ

kIkνq f(x) is absolutely convergent for all x ∈ (0, a] and defines a function u(x) on

(0, a]. Take y ∈ (qa, a] and t ∈ yqm, m ∈ N. Since∣∣∣λkIkνq f(t)

∣∣∣ 6 |λ|ktkν−1

Γq(kν)

∞∑r=0

tqr(1− q)(qr+1; q

)kν−1

|f(tqr)|

6|λ|kykν−1

(qν ; q)∞Γq(kν)

∫ y

0|f(u)| dqu,

(6.2.10)

k ∈ Z+, and the series

∞∑k=1

λkykν−1

Γq(kν)= λyν−1

∞∑k=0

λkykν

Γq(kν + ν)= λyν−1eν,ν(λyν ; q),

is convergent for all y 6 a and a satisfies the condition (6.2.3), then the series

∞∑k=0

∣∣∣λkIkνq f(t)

∣∣∣is uniformly convergent on yqm, m ∈ N, y ∈ (qa, a]. That is it is convergent on (0, a].

Thus the function u(x) is defined for x ∈ (0, a] by

u(x) :=∞∑

k=0

λkIkνq f(x). (6.2.11)

ii . Now we prove that u ∈ L1q(0, a), i.e. u ∈ L1

q(0, y) for all y ∈ (qa, a]. Indeed from (6.2.11)

and the semigroup identity (5.4.30) we obtain∫ y

0|u(ξ)| dqξ = Iq|u|(y) 6

∞∑k=0

|λ|k∣∣∣Iνk+1

q f(y)∣∣∣ . (6.2.12)

But from (6.2.10)

∞∑k=0

|λ|k∣∣∣Iνk+1

q f(y)∣∣∣ 6 ∞∑

k=0

|λ|k∫ y

0

∣∣∣Iνkq f(t)

∣∣∣ dqt

6

∫ y0 |f(u)| dqu

(qν ; q)∞

∞∑k=0

|λ|k ykν

|Γq(kν + 1)|

=

∫ y0 |f(u)| dqu

(qν ; q)∞eν,1(|λ|yν ; q) 6

∫ y0 |f(u)| dqu

(qν ; q)∞eν,1(|λ|aν ; q),

(6.2.13)

149

for all y ∈ (qa, a]. Since a satisfies the condition (6.2.3) and f ∈ L1q(0, y), for all y ∈ (qa, a],

then so is u.

iii. To prove the uniqueness assume that u and v are solutions of (6.2.4) in L1q(0, a) valid

in (0, h], h 6 a. Set

z(x) ≡ u(x)− v(x), x ∈ (0, h].

Then, z(x) satisfies the functional equation

z(x) =λxν−1

Γq(ν)

∫ x

0

(qt/x; q

)ν−1

z(t) dqt. (6.2.14)

Since z ∈ L1q(0, a), then z ∈ L1

q(0, h), i.e.

‖z0‖w :=∫ w

0|z(ξ)| dqξ <∞, w ∈ (qh, h]. (6.2.15)

Consequently from (6.2.14) we obtain

|z(τ)| 6 |λ|Γq(ν)(qν ; q)∞

‖z0‖w τν−1, τ ∈ wqm, m ∈ N . (6.2.16)

Applying (6.2.16) to (6.2.14) to estimate the value of z(τ), we get

|z(τ)| 6 |λ|2

Γ2q(ν)(qν ; q)∞

‖z0‖w τ2ν−1Bq(ν, ν) =

|λ|2

Γq(2ν)(qν ; q)∞‖z0‖w τ

2ν−1.

Repeating the estimate n times, we obtain

|z(τ)| 6 |λ|n

Γq(nν)(qν ; q)∞‖z0‖w τ

nν−1, n ∈ Z+, τ ∈ wqm, m ∈ N , w ∈ (qh, h]. (6.2.17)

Since

‖z0‖w

∞∑n=1

|λ|n

Γq(nν)(qν ; q)∞τnν−1 =

λτν−1

(qν ; q)∞‖z0‖w

∞∑n=0

|λ|n

Γq(nν + ν)τnν

=λτν−1

(qν ; q)∞‖z0‖w eν,ν(|λ|τν ; q),

(6.2.18)

then

limn→∞

λn

Γq(nν + ν)τnν = 0, τ ∈ wqm, m ∈ N, w ∈ (qh, h] .

150

Consequently,

z(τ) = 0, τ ∈ wqm, m ∈ N, w ∈ (qh, h] .

That is z(x) = 0 for all x ∈ (0, h] and the uniqueness of the solution follows.

iv. Now we prove that u can be represented by the second identity in (6.2.5). Since

u(x) =∞∑

k=0

Ikνq f(x) = f(x) +

∞∑k=1

λkxkν−1

Γq(kν)

∫ x

0

(qt/x; q

)kν−1

f(t) dqt (6.2.19)

= f(x) + λxν−1∞∑

k=0

λkxkν

Γq(kν + ν)

∫ x

0

(qt/x; q

)kν+ν−1

f(t) dqt (6.2.20)

and

(qt/x; q)kν+ν−1 = (qt/x; q)ν−1(qνt/x; q)kν ,

then

u(x) = f(x) + λxν−1

∫ x

0

( ∞∑k=0

λkxkν

Γq(kν)(qνt/x; q

)kν

)(qt/x; q

)ν−1

f(t) dqt

u(x) = f(x) + λxν−1

∫ x

0

(qt/x; q

)ν−1

ε−qνteν,ν(λxν)f(t) dqt.

(6.2.21)

where the q−translation operator εy is defined in (5.3.3).

Theorem 6.2.3. Let ν, a > 0 and λ ∈ C. If f ∈ L1q(0, a), then the q-integral equation

u(x) = f(x) +λxν−1

Γq(ν)

∫ x

0

(qt/x; q

)ν−1

u(qt) dqt, (6.2.22)

has the unique solution

u(x) = f(x) + λqνxν−1

∫ x

0

(qt/x; q

)ν−1

∞∑j=0

λjqνj(j−1)

2

Γq(νj + ν)(qνt/x; q

)νj−1

f(qj+1t) dqt, (6.2.23)

x ∈ (0, a], in the space L1q(0, a).

151

Proof. First we prove the existence of a solution of (6.2.23). Define a sequence um(x)∞m=0

recursively by

u0(x) = f(x),

um(x) = f(x) +λxν−1

Γq(ν)

∫ x

0

(qt/x; q

)ν−1

um−1(qt) dqt, x ∈ (0, a], m > 1.(6.2.24)

We can prove by mathematical induction that

um(x) =m∑

j=0

λjqj(j−1)

2νIνj

q f(qjx), m ∈ N. (6.2.25)

Indeed from (6.2.24) we obtain

um+1(x) = f(x) + λIνq (f(qx))

+λxν−1

Γq(ν)

∫ x

0

(qt/x; q)ν−1

m∑j=1

λjqj(j−1)ν

2(Ijνq

(f(qjx)

)(x)∣∣∣x=qt

dqt

= f(x) + λIνq

(f(qx)

)+

λxν−1

Γq(ν)

m∑j=1

λjqj(j+1)ν

2 −1

Γq(νj)

∫ x

0

(qt/x; q)ν−1tνj−1

∫ qt

0

(u/t; q)jν−1f(qju) dqu dqt.

But from Lemma 5.3.5∫ x

0(qt/x; q)ν−1t

νj−1

∫ qt

0(u/t; q)jν−1f(qju) dqu dqt

= q

∫ x

0(qt/x; q)ν−1t

νj−1

∫ t

0(qξ/t; q)jν−1f(qj+1ξ) dqξ dqt

= qBq(ν, νj)xνj

∫ x

0(qt/x; q)(j+1)ν−1f(qj+1t) dqt.

(6.2.26)

Thus

um+1(x) = f(x) + λIνq

(f(qx)

)+

m∑j=1

λj+1qj(j+1)ν

2 I(j+1)νq f(qj+1x)

=m+1∑j=0

λjqj(j−1)

2νIνj

q f(qjx).

(6.2.27)

This proves (6.2.25) because it holds at m = 0. Now we prove that the series

m∑j=0

λjqj(j−1)

2νIνj

q f(qjx)

152

is absolutely convergent for all x ∈ (0, a] and defines a function u(x) on (0, a]. Take y ∈

(qa, a] and t ∈ yqm, m ∈ N. Since

(qn+1; q)jν−1 6(qn+1; q)∞(qn+ν ; q)∞

< (qν ; q)−1∞ , n ∈ N, ν > 0, j > 1, (6.2.28)

and f ∈ L1q(0, a), then for all y ∈ (qa, a] and j > 1 we obtain∣∣∣∣Iνj

q

(f(qjy)

)∣∣∣∣ = yνj−1

Γq(νj)

∣∣∣∣∫ y

0

(qt/y; q

)ν j−1

f(qjt) dqt

∣∣∣∣ 6 (qν ; q)−1∞ ‖f‖y

yjν−1

Γq(jν). (6.2.29)

Hence ∣∣∣∣∣∣∞∑

j=1

λjqj(j−1)

2νIνj

q

(f(qjy)

)∣∣∣∣∣∣ 6 ‖f‖y (qν ; q)−1∞

∞∑j=1

|λ|jqj(j−1)

2ν y

jν−1

Γq(jν)

= ‖f‖y (qν ; q)−1∞

∞∑j=0

|λ|j+1qj(j+1)

2ν yjν+ν−1

Γq(jν + ν)

= ‖f‖y (qν ; q)−1∞ |λ|yν−1Eν,ν(qν |λ|yν ; q).

Since the basic Mittag-Leffler function Eν,ν(qν |λ|yν ; q) is convergent for all y ∈ R and λ ∈ C,

then the series∑∞

k=0

∣∣λkIkνq f(t)

∣∣ is uniformly convergent on yqm, m ∈ N, y ∈ (qa, a].

That is it is convergent on (0, a]. Thus the function u(x) is defined for x ∈ (0, a] by

u(x) :=∞∑

j=0

λjqj(j−1)

2νIνj

q f(qjx). (6.2.30)

Then the series u(x) is the limit of (6.2.25) as m→∞. The proof of u is the unique solution

in L1q(0, a) is similar to that of Theorem 6.2.2 and it is omitted.

6.3 Zeros of basic Mittag-Leffler functions

In this section we investigate the zeros of the basic Mittag-Leffler functions Eν,µ(z; q) when

0 < q < 1, and when q satisfies an additional relation. The zeros will be investigated

for specific values for the parameters ν and µ. Actually we shall give an analogue of the

following theorem, cf. [40, p.7].

153

Theorem 6.3.1. Let

Eσ(z; ν) = E1/2(−σ2z; 1 + ν), σ ∈ R+. (6.3.1)

Then

1. If 0 6 ν < 2, then the zeros of εσ(z2; ν) are simple, real and they are symmetric with

respect to the point z = 0.

2. If 0 < ν < 1, the all zeros of the function εσ(z; ν) are situated in the intervals

∆(1)k =

(πσk − π

2σ,π

σk +

π

2σ

), k ∈ Z∗, (6.3.2)

one zero in each interval and if ν = 0, then the zeros of this function are the endpoints

of these intervals.

3. If 1 < ν < 2, then the zeros of Eσ(z2; ν) are situated in the intervals

∆(2)k =

(

πσk,

πσ (k + 1)

), 1 6 k <∞,(

πσ (k − 1), π

σk), −∞ < k 6 −1,

(6.3.3)

one zero in each interval and if ν = 1, then they are the endpoints of these intervals.

Consider the function

Eγ,σ(z; q) := E2,1+γ(−σ2z; q), z ∈ C, σ ∈ R+, (6.3.4)

and recall that ±xn∞n=1 and 0,±yn∞n=1, xn, yn > 0, denote the zeros of cos(z; q) and

sin(z; q). Then we have the following theorem

Theorem 6.3.2. Let 0 6 γ < 2. If γ 6= 0 6= 1. Then

1. For any q ∈ (0, 1) the function Eγ,σ(z2; q) has at most a finite number of complex

zeros and it has an infinite number of real, simple and symmetric zeros,±ξ(γ)

n

∞n=1

,

ξ(γ)n > 0, with the asymptotic behavior

ξ(γ)n =

q1/2xn

σ

(1 +O(qn)

), 0 < γ < 1,

qyn

σ

(1 +O(qn)

), 1 < γ < 2,

(6.3.5)

as n→∞.

154

2. If q-satisfies the condition

q−1(1− q)(1− qγ+1)(1− qγ+2) > 1, γ ∈ (0, 2), γ 6= 1, (6.3.6)

then the zeros of Eγ,σ(z2; q) are real, simple and symmetric with respect to the point

z = 0 such that for n ∈ Z+

ξ(γ)n ∈

(q−n+3/2

√(1− qγ+1)(1− qγ+2)σ(1− q)

,q−n+1/2

√(1− qγ+1)(1− qγ+2)σ(1− q)

), γ ∈ (0, 1),

(q−n+5/2

√(1− qγ+1)(1− qγ+2)σ(1− q)

,q−n+3/2

√(1− qγ+1)(1− qγ+2)σ(1− q)

), γ ∈ (1, 2).

(6.3.7)

3. If γ ∈ 0, 1, then Eγ,σ(z2; q) has only real, simple and symmetric zeros such that

ξ(0)n :=

q1/2xn

σ, ξ(1)n :=

qyn

σ, n = 1, 2, . . . . (6.3.8)

Proof. If γ ∈ 0, 1, then from (6.1.18)

E0,σ(z2; q) = cos(q−1/2σz; q), E1,σ(z2; q) =sin(q−1σz; q)

q−1σz. (6.3.9)

Since the zeros of cos(·; q) and sin(·; q) are real and simple, then the desired statement

follows for γ ∈ 0, 1. From formula (6.1.22) with β = 2, λ = −σ2 we have

E2,µ+α(−σ2z2; q) =z−µ

Γq(α)

∫ z

0

(qt/z; q

)α−1

E2,µ

(− σ2t2; q

)tµ−1 dqt. (6.3.10)

Substitute with u := t/z, z 6= 0, on the q−integral of (6.3.10). Then∫ z

0

(qt/z; q

)α−1

E2,µ

(− σ2t2; q

)tµ−1 dqt = zµ

∫ 1

0(qu; q)α−1E2,µ(−σ2u2z2; q)tµ−1 dqt.

(6.3.11)

Thus (6.3.10) is nothing but

E2,µ+α(−σ2z2; q) =∫ 1

0(qu; q)α−1E2,µ(−σ2u2z2; q)tµ−1 dqt. (6.3.12)

155

If 0 < γ < 1, substituting in (6.3.12) with µ = 1 and α = γ yields

Eγ,σ(z2; q) =1

Γq(γ)

∫ 1

0(qt; q)γ−1 cos(q−1/2σtz; q) dqt. (6.3.13)

If 1 < γ < 2, substituting in (6.3.12) with µ = 2 and α = γ − 1 yields

q−1σzEγ,σ(z2; q) =1

Γq(γ − 1)

∫ 1

0(qt; q)γ−2 sin(q−1σtz; q) dqt. (6.3.14)

For δ ∈ (0, 1) let fδ be the function defined on [0, 1] by

fδ(t) :=(qt; q)δ−1

Γq(δ). (6.3.15)

Sincefδ(qk)fδ(qk+1)

=1− qk+1

1− qk+δ+1> 1, k ∈ N,

then

fδ(1) > fδ(q) > fδ(q2) > . . . > f(qn) > . . . > f(0) = 1.

That is fδ, δ ∈ (0, 1) is increasing and positive on the sequence 0, qn, n ∈ N. Moreover

f ∈ L1q(0, 1) because ∫ 1

0fδ(t) dqt = Bq(δ + 1, 1) =

1− q

1− qδ+1.

Hence applying Theorem 1.4.2 on (6.3.13) and Theorem 1.4.3 on (6.3.14) we conclude that

for any q ∈ (0, 1), Eγ,σ(z2; q) has at most a finite number of non-real zeros, and it has an

infinite number of real, simple and symmetric zeros such thatξ(γ)n

satisfies the asymptotic

properties (6.3.5). From (2.1.3) and (2.1.4)we have

ck,−1/2,fγ−1=

1(1− q2k+γ+1)(1− q2k+γ+2)

, 0 < γ < 1, (6.3.16)

bk,1/2,fγ−2=

1(1− q2k+γ+1)(1− q2k+γ+2)

, 1 < γ < 2, (6.3.17)

for all k ∈ N. Consequently

c−1/2,fγ−1= 1, C−1/2,fγ−1

=1

(1− qγ+1)(1− qγ+2), γ ∈ (0, 1), (6.3.18)

156

b1/2,fγ−2= 1, C−1/2,fγ−2

=1

(1− qγ+1)(1− qγ+2), γ ∈ (1, 2). (6.3.19)

Thus the condition (2.3.9) of Theorem 2.3.3 and condition (2.3.13) of Theorem 2.3.4 are

nothing but the condition (6.3.6). Therefore the proof of the point 2 of the theorem follows

from Theorems 2.3.3 and 2.3.4 of Section 2.3.

Chapter 7

Basic Fractional DifferenceEquations

As in the classical theory of ordinary fractional differential equations, q-difference equations

of fractional order are divided into linear, nonlinear, homogeneous, and inhomogeneous

equations with constant and variable coefficients. This chapter is devoted to study certain

problems of fractional q-difference equations based on the basic Riemann-Liouville fractional

derivative and the basic Caputo fractional derivative. We shall study questions concerning

the solvability of these equations in a certain space of functions. Also we shall solve an equa-

tion of fractional order with constant coefficients by using q−analogue of Laplace transform.

A solution for a fractional integro-differential equation q-Cauchy problem is given.

7.1 Basic Riemann–Liouville fractional order sys-

tems

In this section we deal with two systems of basic Riemann–Liouville fractional derivatives,

namely the system

Dαq yi(x) = fi(x, y1(x), . . . , yn(x)), i = 1, 2, . . . , n, (7.1.1)

157

158

and the system

Dαq yi(x) = fi(qx, y1(qx), . . . , yn(qx)), i = 1, 2, . . . , n, (7.1.2)

where 0 < α < 1. System (7.1.1) together with the initial conditions

Dα−1q yi(x)|x=0 = bi, bj ∈ R, i = 1, 2, . . . , n, (7.1.3)

is called an initial value problem. By a solution of the initial value problem (7.1.1), (7.1.3)

we mean a set of functions yini=1 that satisfies (7.1.1) and the initial conditions (7.1.3)

in an interval I containing zero such that x1−αyi, i = 1, 2, . . . , n, are continuous at zero.

Similarly we can define the solution of the initial value problem (7.1.2)–(7.1.3). We begin

this section with the existence and uniqueness of solutions of the initial value problem

(7.1.1), (7.1.3).

Theorem 7.1.1. Let fi(x, y1, . . . , yn) be functions defined for x ∈ (0, a] and yi in a domain

Gi, satisfying the following conditions

(i) there is a positive constant A such that, for x ∈ (0, a] and yi, yi ∈ Gi, 1 6 i 6 n, the

following Lipschiz’ condition is fulfilled

|fi(x, y1, . . . , yn)− fi(x, y1, . . . , yn)| 6 A (|y1 − y1|+ . . .+ |yn − yn|) . (7.1.4)

(ii) There exists M > 0 such that

|fi(x, y1, . . . , yn)| 6 M, yi ∈ Gi, i = 1, . . . , n, x ∈ (0, a]. (7.1.5)

Let 0 < α < 1 and a, K be such that

0 < a 61

(An)1α (1− q)

and K >Maα

Γq(α+ 1). (7.1.6)

Let Di(a,K) ⊂ Gi be the set of points yi ∈ Gi satisfying the relation

|yi(x)− bixαi−1

Γq(αi)| < K, for all x ∈ (0, a]. (7.1.7)

159

Then, the initial value problem (7.1.1), (7.1.3) has a unique solution yi(x)ni=1 valid in

(0, a], yi ∈ Di(a,K), i = 1, 2, . . . , n.

Proof.

Existence. Define sequences φi,m∞m=1, i = 1, . . . , n, by the following relationships

φi,1(x) =bi x

α−1

Γq(α),

φi,m(x) =xα−1

Γq(α)

(bi +

∫ x

0

(qt/x; q

)α−1

fi

(t, φ1,m−1(t), . . . , φn,m−1(t) dqt

), (7.1.8)

where m > 2. We will show that limm→∞ φi,m(x) exists and gives the required solution

φini=1 of the initial value problem (7.1.1)–(7.1.3). We prove the existence in four steps.

i. We prove by induction on m that

φi,m ∈ Di(a, k), m ∈ Z+, i = 1, 2, . . . , n. (7.1.9)

Clearly, φi,1 ∈ Di(a,K) for i = 1, . . . , n. If we assume that φi,m ∈ Di(a, k) for i = 1, . . . , n,

then by (7.1.5) we obtain

|fi(x, φ1,m(x), . . . , φn,m(x))| 6 M, x ∈ (0, a]. (7.1.10)

Thus by (1.1.12)

|φi,m+1(x)−bi

Γq(α)xα−1| 6

xα−1

Γq(α)

∫ x

0

(qt/x; q

)α−1

|fi(t, φ1,m(t), . . . , φn,m(t))| dqt

6Mxα−1

Γq(α)

∫ x

0

(qt/x; q

)α−1

dqt =M

Γq(α)Bq(α, 1)xα

=Mxα

Γq(α+ 1)6

Maα

Γq(α+ 1)6 K. (7.1.11)

So, φi,m+1 ∈ Di(a,K). This completes the induction steps and prove (7.1.9).

ii. We prove that x1−αφi,m is continuous at zero for all m ∈ N, i = 1, 2, . . . , n. From (7.1.9)

we conclude that ∣∣∣∣φi,m(x)− biΓq(α)

xα−1|∣∣∣∣ 6 K, m ∈ Z+, x ∈ (0, a]. (7.1.12)

160

Consequently ∣∣∣∣x1−αφi,m(x)− biΓq(α)

|∣∣∣∣ 6 Kx1−α, m ∈ Z+, x ∈ (0, a]. (7.1.13)

Therefore x1−αφi,m is continuous at zero for all m ∈ N, i = 1, 2, . . . , n.

iii. We prove by induction on m that

|φi,m+1(x)− φi,m(x)| 6 MBm−1x(m)α

Γq(mα+ 1), B := An, m ∈ Z+. (7.1.14)

From (7.1.11), inequality (7.1.14) is true at m = 1. Assume that (7.1.14) is true at m = k,

that is,

|φi,k+1(x)− φi,k(x)| 6MBk−1xkα

Γq(kα+ 1), i = 1, 2, . . . , n. (7.1.15)

Consequently, we have

|φi,k+2(x)− φi,k+1(x)|

6xα−1

Γq(α)

∫ x

0

(qt/x; q

)α−1

|fi(t, φ1,k+1(t), . . . , φn,k+1(t))− fi(t, φ1,k(t), . . . , φn,k(t))| dqt

6Axα−1

Γq(α)

∫ x

0

(qt/x; q

)α−1

n∑j=1

|φj,k+1(t)− φj,k(t)| dqt

=MBkxα−1

Γq(α)Γq(kα+ 1)

∫ x

0

(qt/x; q

)α−1

tkα dqt

=MBk

Γq(α)Γq(kα+ 1)Bq(α, kα+ 1)x(k+1)α =

BkM

Γq

((k + 1)α+ 1

)x(k+1)α.

That is (7.1.14) is true at m = k + 1 and hence it is true for all m > 1.

iv. We prove that limm→∞ φi,m exists for i = 1, 2, . . . , n, x ∈ (0, a] such that φini=1,

φi(x) := limm→∞

φi,m(x), i = 1, 2, . . . , n, x ∈ (0, a],

defines a solution of (7.1.1), (7.1.3). Consider the infinite series

φi,1(x) +∞∑

m=1

φi,m+1(x)− φi,m(x). (7.1.16)

161

From (7.1.14) we obtain

∞∑m=1

|φi,m+1(x)− φi,m(x)| 6 M

B

∞∑m=1

(Bxα)m

Γq(mα+ 1)

6M

B

∞∑m=0

(Bxα)m

Γq(mα+ 1)

=M

Beα,(Bxα; q) 6

M

Beα,1(Baα; q).

(7.1.17)

But eα,1(Baα; q) is defined only for aαB < (1 − q)−α which is satisfied, cf. (7.1.6). This

means that the series (7.1.16) is uniformly convergent on (0, a] to a function φi, and φi(x) =

limm→∞

φi,m(x). Since x1−αφi,m is continuous at zero for all m ∈ N, i = 1, . . . , n, then so is φi.

Also φi,m ∈ Di(a, k) implies that φi ∈ Di(a, k). The uniform convergence of the sequences

φi,m allows us to let m→∞ in the relationship (7.1.8), this gives that

φi(x) =bi

Γq(α)xα−1 +

xα−1

Γq(α)

∫ x

0

(qt/x; q

)α−1

fi(t, φ1(t), . . . , φn(t)) dqt, x ∈ (0, a]. (7.1.18)

Consequently

I1−αq φi(x) = bi + I1−α

q Iαq fi(x, y1(x), . . . , yn(x)) = bi + Iqfi(x, y1(x), . . . , yn(x)). (7.1.19)

Since fi(x, y1(x), . . . , yn(x)), i = 1, 2, . . . , n, are bounded on (0, a], then I1−αq φi(0) = bi,

i = 1, 2, . . . , n, i.e. φini=1 satisfies the initial conditions (7.1.3).

Uniqueness. To prove the uniqueness assume that ψini=1 is another solution valid in an

interval (0, h], h 6 a. Set

χi(x) := φi(x)− ψi(x), x ∈ (0, h], i = 1, 2, . . . , n, (7.1.20)

and

gi(x) := fi(x, φ1(x), . . . , φn(x))− fi(x, ψ1(x), . . . , ψn(x)), j = 1, 2, . . . , n. (7.1.21)

Hence

Dαq χi(x) = gi(x), j = 1, 2, . . . , n, (7.1.22)

162

then by (5.4.52) we obtain

χi(x) =xα−1

Γq(α)

∫ x

0

(qt/x; q

)α−1

gi(t) dqt. (7.1.23)

Now x1−αχi is continuous at zero for i = 1, 2, . . . , n. Then for each x ∈ (qh, h] there exists

a constant Cx > 0 such that

t1−α|χi(t)| 6Cx

Γq(α), t ∈ xqm, m ∈ N . (7.1.24)

Fix x ∈ (qh, h] and t ∈ xqm, m ∈ N. Hence from (7.1.23) we obtain

|χi(t)| 6Atα−1

Γq(α)

n∑i=1

∫ t

0

(qu/t; q

)α−1

χi(u) dqu

6AnCxt

α−1

Γ2q(α)

∫ t

0uα−1

(qu/t; q

)α−1

dqu

=AnCxt

2α−1

Γ2q(α)

Bq(α, α) =BCx

Γq(2α)t2α−1

(7.1.25)

Repeating the estimates k times we obtain

|χi(t)| 6 CxBk+1tαk+α−1

Γq(αk + α), k ∈ Z+, i = 1, 2, . . . , n. (7.1.26)

SinceBktαk

Γq(αk + α)is the general term of the series of the q-Mittag-Leffler function eα,α(βtα),

βtα 6 βaα < (1− q)−α,

then limk→∞

Bktαk

Γq(αk + α)= 0. Therefore

χi(t) = 0, t ∈ xqm, m ∈ N , x ∈ (qh, h].

That is χ(x) = 0 for all x ∈ (0, h], proving the uniqueness.

As for the initial value problem (7.1.2)–(7.1.3) we have the following theorem

Theorem 7.1.2. Let fi(x, y1, . . . , yn) be functions defined for x ∈ (0, a] and yi in a domain

Gi, satisfying the following conditions

163

(i) there is a positive constant A such that, for x ∈ (0, a] and yr, yi ∈ Gi, 1 6 r, i 6 n, the

following Lipschiz condition is fulfilled

∣∣fi(x, y1, . . . , yn)− fi(x, y1, . . . , yn)∣∣ 6 A

(|y1 − y1|+ . . .+ |yn − yn|

). (7.1.27)

(ii) There exists M > 0 such that

|fi(x, y1, . . . , yn)| 6 M, i = 1, . . . , n, x ∈ (0, a], and yi ∈ Gi.

Let 0 < α < 1 and K be a constant that satisfies K > Maα

Γq(α+1) , and Di(a,K) ⊂ Gi be the

set of points yi ∈ Gi which satisfy that

|yi(x)− bixα−1

Γq(α)| < K, for all x ∈ (0, a]

Then, the initial value problem (7.1.2)–(7.1.3) has a unique solution yini=1 valid in (0, a].

Proof. Consider the system of functions φi,m∞m=1, i = 1, . . . , n,

φi,1(x) =bi x

α−1

Γq(α),

φi,m(x) =xα−1

Γq(α)

(bi +

∫ x

0

(qt/x; q

)α−1

fi

(qt, φ1,m−1(qt), . . . , φn,m−1(qt)

)dqt,

(7.1.28)

where m > 2. We will show that limm→∞ φi,m(x) exists and give the required solution

φini=1 of the initial value problem (7.1.2)–(7.1.3). As in the proof of Theorem 7.1.1 we

can prove by induction on m that φi,m ∈ Di(a, k), m ∈ Z+ and

|φi,m+1(x)− φi,m(x)| 6 MBm−1qm(m−1)α/

2xmα

Γq(mα+ 1), B := An, m > 1, (7.1.29)

for i = 1, . . . , n. According to the estimate (7.1.29), for 0 < x 6 a, the absolute value of

the terms of the series

φi,1(x) +∞∑

m=1

φi,m+1(x)− φi,m(x). (7.1.30)

is less than the corresponding terms of the convergent numeric series

M

B

∞∑m=0

Bmqm(m−1)α/2a(m)α

Γq(mα+ 1)=M

BEα,1(Baα; q).

164

Therefore the series (7.1.30) is uniformly convergent on (0, a] to a function φi, where φi(x) =

limm→∞

φi,m(x). Since x1−αφi,m is continuous at zero for all m ∈ Z+, i = 1, . . . , n, then so

is φi. Also φi,m ∈ Di(a, k) implies that φi ∈ Di(a, k). The uniform convergence of the

sequencesφi,m(x)

, x ∈ (0, a], allows us to let m → ∞ in the relationship (7.1.28), this

gives that

φi(x) =bi

Γq(α)xα−1 +

xα−1

Γq(α)

∫ x

0

(qt/x; q

)α−1

fi(t, φ1(qt), . . . , φn(qt)) dqt.

Consequently

I1−αq φi(x) = bi+I1−α

q Iαq fi(x, y1(qx), . . . , yn(qx)) = bi+Iqfi(x, y1(qx), . . . , yn(qx)). (7.1.31)

Thus I1−αq φi(0) = bi, i = 1, 2, . . . , n, i.e. φin

i=1 satisfies the initial conditions (7.1.3).

The proof of the uniqueness of the solution is similar to that of Theorem 7.1.1 so it is

omitted.

Remark 7.1.1. When α = 1, the initial value problems (7.1.1), (7.1.3) and (7.1.2)–(7.1.3)

are nothing but the initial value problems (3.1.5), (3.2.1) and (3.1.6), (3.2.1), respectively.

See Section 3.2. Also cf. [9, 90].

Example 7.1.1. Using the method of successive approximation, one can prove that the

solution of the initial value problem

Dαq y1(x) = y2(x), Dα

q y2(x) = −y1(x), (7.1.32)

Dα−1q y1(0) = 0, Dα−1

q y2(0) = 1, (7.1.33)

is

y1(x) = x2α−1e2α,2α(−x2α; q), y2(x) = xα−1e2α,α(−x2α; q), |x| < (1− q)−α. (7.1.34)

If we take α = 1, we get

y1(x) = sinq x(1− q), and y2(x) = cosq x(1− q). (7.1.35)

165

Example 7.1.2. Using the successive approximation method, one can prove that the solu-

tion of the initial value problem

Dαq y1(x) = y2(x), Dα

q y2(x) = −q1−2αy1(qαx), (7.1.36)

Dα−1q y1(0) = 0, Dα−1

q y2(0) = 1, (7.1.37)

is

y1(x) = x2α−1E2α,2α(−x2α; q), y2(x) = xα−1E2α,α(−x2α; q), x ∈ R. (7.1.38)

If we take α = 1, we get

y1(x) = q−1 sin(q−1x; q), and y2(x) = cos(q−1/2x; q), x ∈ R. (7.1.39)

7.2 Basic Caputo fractional order systems

In this section we treat two systems of basic Caputo fractional derivatives, namely the

system

cDαq yi(x) = fi(x, y1, . . . , yn), i = 1, 2, . . . , n, (7.2.1)

and the system

cDαq yi(x) = fi(qx, y1(qx), . . . , yn(qx)), i = 1, 2, . . . , n, (7.2.2)

where 0 < α < 1. We call system (7.2.1) together with initial conditions of the form

yi(0) = bi, bj ∈ R, i = 1, 2, . . . , n, (7.2.3)

an initial value problem. Similarly for the system (7.2.2) with the initial conditions (7.2.3).

By a solution of the initial value problem (7.2.1), (7.2.3) we mean a set of functions yini=1

that satisfies (7.2.1) and the initial conditions (7.2.3) in an interval I containing zero such

that yi(x), i = 1, 2, . . . , n, are continuous at zero. Similarly we can define the solution

of the initial value problem (7.2.2)–(7.2.3). Now consider the existence and uniqueness of

solutions of the initial value problem (7.2.1), (7.2.3).

166

Theorem 7.2.1. Let fi(x, y1, . . . , yn) be functions defined for x ∈ [0, a] and yi in a domain

Gi, satisfying the following conditions

(i) there is a positive constant A such that, for x ∈ (0, a] and yi, yi ∈ Gi, 1 6 i 6 n, the

following Lipschiz’ condition is fulfilled

|fi(x, y1, . . . , yn)− fi(x, y1, . . . , yn)| 6 A (|y1 − y1|+ . . .+ |yn − yn|) . (7.2.4)

(ii) There exists M > 0 such that

|fi(t, y1, . . . , yn)| 6 M, i = 1, . . . , n, t ∈ [0, a], and yi ∈ Gi.

For 0 < α 6 1, let a, K be such that

0 < a 61

(An)1α (1− q)

and K >Maα

Γq(α+ 1)

Let Di(a,K) ⊂ Gi be the set of points yi ∈ Gi which satisfy that

|yi(x)− bixα−1

Γq(α)| < K, for all x ∈ [0, a]

Then, the initial value problem (7.2.1), (7.2.3) has a unique solution yi(x)ni=1 valid in

(0, a].

Proof. Define sequences φi,m∞m=1, i = 1. . . . , n, by the following relationships

φi,1(x) = bi,

φi,m(x) = bi +xα−1

Γq(α)

∫ x

0

(qt/x; q

)α−1

fi

(t, φi,m−1(t), . . . , φn,m−1(t) dqt,

(7.2.5)

where m > 2. We will show that limm→∞

φi,m(x) exists and give the required solution φini=1

of the initial value problem (7.2.1), (7.2.3). First we prove by induction on m that

φi,m ∈ Di(a, k), i = 1, . . . , n, m ∈ Z+. (7.2.6)

Clearly, φi,1 ∈ Di(a,K) for i = 1, . . . , n. If we assume that φi,m ∈ Di(a, k) for i = 1, . . . , n,

then

|fi(x, φ1,m(x), . . . , φn,m(x))| 6 M, x ∈ [0, a].

167

Then

|φi,m+1(x)− bi| 6xα−1

Γq(α)

∫ x

0

(qt/x; q

)α−1

|fi(t, φ1,m(t), . . . , φn,m(t))| dqt

6Mxα−1

Γq(α)

∫ x

0

(qt/x; q

)α−1

dqt =M

Γq(α)Bq(α, 1)xα

=Mxα

Γq(α+ 1)(7.2.7)

Thus φi,m is continuous at zero for all m ∈ Z+ and i = 1, 2, . . . , n. Moreover from (7.2.7)

we obtain

|φi,m+1(x)− bi| 6Maα

Γq(α+ 1)6 K. (7.2.8)

So, φi,m ∈ Di(a,K), for all m ∈ N, i = 1, 2, . . . , n. Second, we prove by induction on m

that

|φi,m+1(x)− φi,m(x)| 6 MBm−1xmα

Γq(mα+ 1), B := An, m > 1. (7.2.9)

From (7.2.7), inequality (7.2.9) is true at m = 1. Assume that (7.2.9) is true at m = k,

that is,

|φi,k+1(x)− φi,k(x)| 6MBk−1xkα

Γq(kα+ 1), i = 1, 2, . . . , n. (7.2.10)

Consequently,

|φi,k+2(x)− φi,k+1(x)|

6xα−1

Γq(α)

∫ x

0

(qt/x; q

)α−1

|fi(t, φ1,k+1(t), . . . , φn,k+1(t))− fi(t, φ1,k(t), . . . , φn,k(t))| dqt

6Axα−1

Γq(α)

∫ x

0

(qt/x; q

)α−1

n∑j=1

|φj,k+1(t)− φj,k(t)| dqt

6MBkxα−1

Γq(α)Γq(kα+ 1)qkα

∫ x

0

(qt/x; q

)α−1

tkα dqt

=MBk

Γq(α)Γq(kα+ 1)Bq(α, kα+ 1)x(k+1)α =

BkM

Γq

((k + 1)α+ 1

)x(k+1)α.

Now, consider the infinite series

φi,1(x) +∞∑

m=1

φi,m+1(x)− φi,m(x). (7.2.11)

168

According to the estimate (7.2.9), for 0 < x 6 a, the absolute value of its terms is less than

the corresponding terms of the convergent numeric series

M

B

∞∑m=0

Bmamα

Γq(mα+ 1)=M

Beα,1(Baα; q).

This means that the series (7.2.11) is uniformly convergent on (0, a] to a function φi, and

φi(x) = limm→∞

φi,m(x). Since φi,m is continuous at zero for all m ∈ N, i = 1, . . . , n, then so

is φi. Also φi,m ∈ Di(a, k) implies that φi(x) ∈ Di(a, k). The uniform convergence of the

sequences φi,m allows us to let m→∞ in identity (7.2.5), this gives that

φi(x) = bi +xα−1

Γq(α)

∫ x

0

(qt/x; q

)α−1

fi(t, φ1(t), . . . , φn(t)) dqt.

Obviously the initial conditions (7.2.3) holds. The proof of the uniqueness of the solution

is similar to of that of Theorem 7.1.1 so it is omitted.

Similar to Theorem 7.2.1 we have the following theorem

Theorem 7.2.2. Let fi(x, y1, . . . , yn) be functions defined for x ∈ [0, a] and yi in a domain

Gi, satisfying the following conditions

(i) for fixed x ∈ [0, a], and yi, yi in Gi

|fi(x, y1, . . . , yn)− fi(x, y1, . . . , yn)| 6 A (|y1 − y1|+ . . .+ |yn − yn|) ,

where A > 0 and i = 1, 2, . . . , n.

(ii) There exists M > 0 such that

|fi(t, y1, . . . , yn)| 6 M, i = 1, . . . , n, t ∈ [0, a], and yi ∈ Gi.

Let K be a constant that satisfies K >Maα

Γq(αi + 1), and Di(a,K) ⊂ Gi be the set of points

yi ∈ Gi which satisfy that

|yi(x)− bi| < K, for all x ∈ (0, a].

Then, the initial value problem (7.2.2)–(7.2.3) has a unique solution yini=1 valid in (0, a].

169

Proof. Consider the system of successive functions φi,m∞m=1, i = 1, . . . , n,

φi,1(x) = bi,

φi,m(x) = bi +xα−1

Γq(α)

∫ x

0

(qt/x; q

)α−1

fi

(qt, φi,m−1(qt), . . . , φn,m−1(qt) dqt,

(7.2.12)

where m > 2. We will show that limm→∞ φi,m(x) exists and give the required solution

φini=1 of the initial value problem (7.2.2)–(7.2.3). As in the proof of Theorem 7.2.1 we

can prove by induction on m that φi,m ∈ Di(a, k) for i = 1, . . . , n, m ∈ Z+ and

|φi,m+1(x)− φi,m(x)| 6 MBm−1qm(m−1)α/

2xmα

Γq(mα+ 1), B := An, m > 1. (7.2.13)

According to the estimate (7.2.13), for 0 6 x 6 a, the absolute value of the terms of the

series

φi,1(x) +∞∑

m=1

φi,m+1(x)− φi,m(x). (7.2.14)

is less than the corresponding terms of the convergent numeric series

M

B

∞∑m=0

Bmqm(m−1)α/2amα

Γq(mα+ 1)=M

BEα,1(Baα; q).

This means that the series (7.2.14) is uniformly convergent on (0, a] to a function φi, where

φi(x) = limm→∞

φi,m(x). Since φi,m ∈ L1q(0, a), for all m ∈ N, i = 1, . . . , n, then so is φi. Also

φi,m ∈ Di(a, k) implies that φi(x) ∈ Di(a, k). The uniform convergence of the sequences

φi,m(x), x ∈ (0, a], allows us to let m→∞ in the relationship (7.2.12), this gives that

φi(x) = bi +xα−1

Γq(α)

∫ x

0

(qt/x; q

)α−1

fi(t, φ1(t), . . . , φn(t)) dqt.

The proof of the uniqueness of the solution is similar to that of Theorem 7.1.1 so it is

omitted.

Example 7.2.1. Using the method of successive approximations, one can easily see that

the solution of the initial value problem

cDαq y1(x) = y2(x), cDα

q y2(x) = −y1(x), (7.2.15)

y1(0) = 0, y2(0) = 1, (7.2.16)

170

is

y1(x) = xαe2α,2α+1(−x2α; q), y2(x) = e2α,1(−x2α; q), |x| < (1− q)−α. (7.2.17)

If we take α = 1, we get

y1(x) = sinq x(1− q), y2(x) = cosq x(1− q), |x| < (1− q)−1. (7.2.18)

Example 7.2.2. Using the successive approximation method, one can prove that the solu-

tion of the initial value problem

cDαq y1(x) = y2(x), cDα

q y2(x) = −q−αy1(qαx), (7.2.19)

y1(0) = 0, y2(0) = 1, (7.2.20)

is

y1(x) = xαE2α,α+1(−x2α; q), y2(x) = E2α,1(−x2α; q), x ∈ R. (7.2.21)

If we take α = 1, we get

y1(x) = q sin(q−1x; q), and y2(x) = cos(q−1/2x; q), x ∈ R. (7.2.22)

7.3 Solutions via a q-Laplace transform

In 1949, W. Hahn [57] defined two q-analogues of the classical Laplace transform through

the functional transformations

qLsf(x) = φ(s) =1

1− q

∫ s−1

0Eq(−qsx) f(x) dqx, (7.3.1)

and

qLsf(x) = Φ(s) =1

1− q

∫ ∞

0eq(−sx)f(x) dqx, (7.3.2)

<(s) > 0. In 1961, Abdi [1] studied certain properties of these analogues. In [2] he used

these analogues to solve linear q-difference equations with constant coefficients and certain

171

allied equations. The object of this section is to use (7.3.1) to solve fractional Riemann–

Liouville q-difference equations with constant coefficients. First we establish some properties

of the operator qLs of (7.3.1). For extensive study of the properties of the Laplace transforms

(7.3.1) and (7.3.2), cf. e.g. [1–3, 57–59, 92–96]. Using the definition of the q−integration

(1.1.31), the q-Laplace transform (7.3.1) is nothing but

qLsf(x) =(q; q)∞s

∞∑j=0

qj

(q; q)jf(s−1qj). (7.3.3)

As an example if φr(x) = xr

Γq(r+1) , r > −1, then

qLsφr(x) =(1− q)r

sr+1. (7.3.4)

In [57, equation (9.5)] Hahn defined the convolution of two functions F, G to be

(F ∗G)(x) =t

1− q

∫ 1

0F (tx)G[t− tqx] dqt, (7.3.5)

where G[x− y], for

G(x) :=∞∑

n=0

anxn,

is defined to be

G[x− y] :=∞∑

n=0

an[x− y]n, with [x− y]n := xn(yx

; q)n. (7.3.6)

There is a misprint in (7.3.5) and its correction is

(F ∗G)(x) :=x

1− q

∫ 1

0F (tx)G[x− tqx] dqt. (7.3.7)

Obviously using the definition of q-integration (F ∗G) is nothing but

(F ∗G)(x) =1

1− q

∫ x

0F (t)G[x− qt] dqt. (7.3.8)

From (5.3.2), we use that

G[x− qt] = ε−qtG(x). (7.3.9)

172

This motivated Ismail [65] to define the convolution of two functions F, G through(F ∗G

)=

11− q

∫ x

0F (t)ε−qtG(x) dqt. (7.3.10)

It is proved by W. Hahn [57] that

qLs(F ∗G) = qLsF ∗ qLsG. (7.3.11)

Abdi [1] showed that if Φ(s) = qLsF (x), then

qLsF (ax) =1aΦ(s

a), a 6= 0, (7.3.12)

and

qLsDnq F (x) =

( s

1− q

)nΦ(s)−n∑

m=1

Dn−mq F (0)

sm−1

(1− q)m, n ∈ Z+. (7.3.13)

We shall compute the q-Laplace transform of the fractional q-integral and q-derivative in

the following lemma.

Lemma 7.3.1. If F ∈ L1q(0, a) and Φ(s) := qLsF (x), then

qLsIαq F (x) =

(1− q)α−1

sαΦ(s), α > 0. (7.3.14)

If n− 1 < α < n and In−αq F (x) ∈ AC(n)

q [0, a], then

qLsDαq F (x) =

sα

(1− q)α+1Φ(s)−

n∑m=1

Dα−mq F (0)

sm−1

(1− q)m. (7.3.15)

Proof. Since

Iαq F (x) =

xα−1

Γq(α)∗ F (x) = φα(x) ∗ F (x),

then from (7.3.4) and (7.3.11) we get (7.3.14). Now we prove (7.3.15). From (5.4.43),

(7.3.13), and (7.3.14) we get

qLs

(Dα

q F (x))

= qLs

(Dn

q In−αq F (x)

)=

sn

(1− q)n qLsIn−αq F (x)−

n∑m=1

Dn−mq In−α

q F (0)sm−1

(1− q)m

=sα

(1− q)α+1Φ(s)−

n∑m=1

Dα−mq F (0)

sm−1

(1− q)m,

(7.3.16)

proving the lemma.

173

Remark 7.3.1. It is worth mentioning that if we take α = n in (7.3.15) we do not obtain

(7.3.13). This is because

qLsIn−αq f(x) =

(1− q)n−α−1

sn−α, n− 1 < α < n.

Thus if we let α→ n on the second identity of (7.3.16) we obtain 1/(1− q), and we obtain

formula (7.3.13) with (1− q)n+1 instead of (1− q)n.

Lemma 7.3.2. If qLsf(x) = Φ(s), then the Laplace transform of the Caputo fractional

q-derivative is given by

qLscDα

q f(x) =sα

(1− q)α+1

(Φ(s)−

n−1∑r=0

Drqf(0)

(1− q)r−1

sr+1

). (7.3.17)

Proof. Since cDαq f(x) = Dn

q f(x) ∗ φn−α−1(x), then by (7.3.4) and (7.3.13) we obtain

qLscDα

q f(x) = qLsφn−α−1(x) qLsDnq f(x) =

(1− q)n−α−1

sn−α qLs(Dnq f(x))

=(1− q)n−α−1

sn−α

((

s

1− q)nΦ(s)−

n∑m=1

Dn−mq f(0)

sm−1

(1− q)m

)

=

(sα

(1− q)αΦ(s)−

n−1∑r=0

Drqf(0)

s−r+α−1

(1− q)−r+α+1

). (7.3.18)

From now on by f (k)(u) we mean dk

duk f(u). The next lemma will be needed in the proof

of the main result of this section.

Lemma 7.3.3. Let α, β, a ∈ R+ and k ∈ N. Then

qLs

(tkα+β−1e

(k)α,β

(± atα; q

))=pα−β

1− q

k!(pα ∓ a)k+1

(7.3.19)

Proof. From (6.1.12) we obtain

e(k)α,β(u; q) =

∞∑n=0

n(n− 1) . . . (n− k + 1)Γq(αn+ β)

un−k, |u| < (1− q)−α. (7.3.20)

174

Thus for |atα| < (1− q)−α we have

tkα+β−1e(k)α,β

(± atα

)=

∞∑n=0

n(n− 1) . . . (n− k + 1)Γq(αn+ β)

(±a)n−ktαk+β−1. (7.3.21)

Consequently

qLs

(tkα+β−1e

(k)α,β

(± atα; q

))=

∞∑n=0

n(n− 1) . . . (n− k + 1)(±a)n−k (1− q)αn+β−1

sαn+β

=1

1− qp−kα−β

∞∑n=0

n(n− 1) . . . (n− k + 1)(±ap−α)n−k

=1

1− qp−kα−β d

k

dξk

∞∑n=0

ξn,

(7.3.22)

where ξ := ±ap−α < 1 when we choose |s| > (1− q)a−α−1. But

p−kα−β

1− q

dk

dξk

∞∑n=0

ξn =p−kα−β

1− q

dk

dξk

11− ξ

=p−kα−β

1− q

k!(1− ξ)k+1

=p−kα−β

1− q

k!(1∓ ap−α

)k+1=pα−β

1− q

k!(pα ∓ a)k+1

.

(7.3.23)

Theorem 7.3.4. Let ai ∈ C, an 6= 0, and 1 > βn > βn−1 > . . . > β1 > β0 > 0. Let

F (s) := qLsf exists for |s| > R0, R0 > 0. Consider the n term fractional q-difference

equation

anDβnq y(x) + an−1D

βn−1q y(x) + . . .+ a1D

β1q y(x) + a0D

β0q y(x) = f(x), x > 0 (7.3.24)

If

φ(x) =(

1an

∞∑m=0

(−1)m

m!

∑k0+...+kn−2=m

k0>0,...,kn−2>0

(m

k0, k1, . . . , kn−2

) n−2∏i=0

( ai

an

)ki

x(βn−βn−1)m+βn−∑n−2

i=0 (βi−βn)ki−1e(m)

βn−βn−1,βn−∑n−2

i=0 (βi−βn)ki(−an−1

anxβn−βn−1)

),

(7.3.25)

175

|x(1− q)| <(

anan−1

)1/(βn−βn−1), then the function y(·) defined by

y(x) =n∑

j=0

ajI1−βjq y(0)φ(x) + (1− q)φ(x) ∗ f(x)

=n∑

j=0

ajI1−βjq y(0)φ(x) +

∫ x

0f(t)ε−qtφ(x) dqt,

(7.3.26)

is a solution of (7.3.24) valid in |x(1− q)| <(

anan−1

)1/(βn−βn−1).

Proof. Assume that qLs y(x) = Y (s) and qLs f(x) = F (s) for |s| > R0 > 0. From (7.3.15),

the q-Laplace transform of (7.3.24) is n∑j=0

ajsβj

(1− q)βj+1

Y (s)− 11− q

n∑j=0

ajI1−βjq y(0) = F (s). (7.3.27)

Set

p :=s

1− q, and gn(p) :=

1∑nj=0 ajpβj

. (7.3.28)

We can choose R > R0 such that for |s| > R,∑n

j=0 ajpβj 6= 0 and

|n−2∑j=0

ajpβj | < |an−1p

βn−1 + anpβn |. (7.3.29)

Consequently for |s| > R

Y (s) = cgn(p) + (1− q)F (s)gn(p), c :=n∑

j=0

ajI(1−βj)q y(0). (7.3.30)

Using (7.3.29) we get

gn(p) =1

anpβn + an−1pβn−1 + . . .+ a0pβ0=

(anp

βn + an−1pβn−1

)−1(1 +

∑n−2j=0 ajp

βj

anpβn + an−1pβn−1

)

=∞∑

m=0

(−1)m

n−2∑j=0

ajpβj

m

(anp

βn + an−1pβn−1

)m+1 =∞∑

m=0

(−1)m

anpβn−1

(∑n−2j=0

aj

anpβj−βn−1

)m(an−1

an+ pβn−βn−1

)m+1 .

176

Applying the multinomial theorem on(∑n−2

j=0aj

anpβj−βn−1

)m we obtain

( n−2∑j=0

aj

anpβj−βn−1

)m =∑

k0+...+kn−2=m

k0>0,...,kn−2>0

(m

k0, k1, . . . , kn−2

) n−2∏j=0

( aj

an

)kjp(βj−βn−1)kj

=∑

k0+...+kn−2=m

k0>0,...,kn−2>0

(m

k0, k1, . . . , kn−2

)n−2∏j=0

( aj

an

)kj

p∑n−2

j=0 (βj−βn−1)kj .

(7.3.31)

Thus

gn(p) =∞∑

m=0

(−1)m

an

∑k0+...+kn−2=m

k0>0,...,kn−2>0

(m

k0, k1, . . . , kn−2

) n−2∏j=0

( aj

an

)kj p−βn−1+

∑n−2j=0 (βj−βn)kj(

an−1

an+ pβn−βn−1

)m+1 ,

(7.3.32)

Since the last series in (7.3.32) is absolutely convergent for |s| > R, then using the addition

theorem [1, p.393] and (7.3.19) we obtain gn(p) = qLsφ(x), |s| > R. Consequently from

(7.3.30)

y(x) = cφ(x) + (1− q)f(x) ∗ φ(x),

which is nothing but (7.3.26). This completes the proof of the theorem.

7.4 A q-Cauchy problem

In this section we investigate the existence and uniqueness of solutions of a q-Cauchy prob-

lem defined in terms of a set of real parameters γj3j=0 and a set Lj3

j=0 of q-integro

difference operators of fractional order. This problem is a q-analogue of a Cauchy problem

given by Djrbashian [40].

Let the set of parameters γj3j=0 satisfy the conditions

12< γ0, γ3 6 1, 0 6 γ1, γ2 6 1,

3∑j=0

γj = 3. (7.4.1)

Then obviously

1 < γ0 + γ3 6 2, 1 6 γ1 + γ2 < 2. (7.4.2)

177

We shall also consider the two special cases

γ0 = γ2 = γ3 = 1 when γ1 = 0,

γ0 = γ1 = γ3 = 1 when γ2 = 0.(7.4.3)

We associate a given set of parameters γj3j=0 satisfying (7.4.1) with a set Lj3

j=0 of

q-integro-difference operators of fractional order setting

L0y(x) := D−(1−γ0)q y(x),

L1y(x) := D−(1−γ1)q DqL0y(x) = D−(1−γ1)

q y(x)Dγ0q y(x),

L2y(x) := D−(1−γ2)q DqL1y(x) = D−(1−γ2)

q Dγ1q D

γ0q y(x),

L1/2y(x) := L3y(x) = D−(1−γ3)q DqL2y(x) = D−(1−γ3)

q Dγ2q D

γ1q D

γ0q y(x).

(7.4.4)

In the following we introduce the domain of the definitions of the operators Lj3j=0.

Definition 7.4.1. Let ACγ[0, a], 0 < a < ∞, be the set of functions y(x) satisfying the

conditions

1. y(x) ∈ L1q(0, a).

2. Ljy(x) ∈ AqC[0, a], j = 0, 1, 2.

Obviously if Ljy(x) ∈ AC[0, a], then DqLjy(x) ∈ L1q(0, a), j = 0, 1, 2, and L1/2y(x) ∈

L1q(0, a). Let y(x) ∈ ACγ [0, a] be a given function. We introduce the following constants:

mk(y) = Lky(x)∣∣∣x=0

, k = 0, 1, 2, (7.4.5)

and

µk =k∑

j=0

γj , k = 0, 1, 2. (7.4.6)

One can easily verify that

γj1 + γj2 6 2, j1 6= j2, (7.4.7)

for any pair of numbers of the set γj3j=0.

178

The q-Cauchy Problem which we are interested in is

L1/2y(x) + λy(qx) = 0, 0 < x < a, (7.4.8)

Lky(x)|x=0 = ak, k = 0, 1, 2, (7.4.9)

where ak2k=0 and λ are arbitrary complex numbers.

The following lemmas are essential in our investigation.

Lemma 7.4.1. If y ∈ ACγ[0, a], then

D−2q L1/2y(x) = y(x)−

2∑k=0

mk(y)xµk−1

Γq(µk), for all x ∈ (0, a]. (7.4.10)

Proof. Since L1/2y(x) ∈ L1q(0, a), using (5.4.30) we obtain the following equality for all

x ∈ (0, a]

D−2q L1/2y(x) = D−2

q D−(1−γ3)q DqL2y(x) = D−(2−γ3)

q

D−1

q DqL2y(x)

= D−(2−γ3)q L2y(x)−m2(y) = D−(2−γ3)

q L2y(x)−m2(y)x2−γ3

Γq(3− γ3).

(7.4.11)

Further, we obtain

D−(2−γ3)q L2y(x) = D−(2−γ3)

q

D−(1−γ2)

q Dq,xL1y(x)

= D−(2−γ2−γ3)q

D−1

q DqL1y(x)

= D−(2−γ2−γ3)q L1y(x)−m1(y)

= D−(2−γ2−γ3)q L1y(x)−m1(y)

x2−γ2−γ3

Γq(3− γ2 − γ3). (7.4.12)

Similarly,

D−(2−γ2−γ3)q L1y(x) = D−γ0

q Dq,xL0y(x)

= D−γ0q Dq,x

D−(1−γ0)

q y(x)

= D−γ0q Dγ0

q y(x).(7.4.13)

But

D−γ0q Dγ0

q y(x) = y(x)−D−(1−γ0)q y(x)

∣∣∣x=0

.

179

Therefore,

D−(2−γ2−γ3)q L1y(x) = y(x)−m0(y)

xγ0−1

Γq(γ0), x ∈ (0, a).

This formula together with (7.4.11) and (7.4.12) implies the desired representation (7.4.10).

Lemma 7.4.2. If y(x) ∈ ACγ [0, a], then

L1/2y(x) = D2qy(x)−

2∑k=0

mk(y)xµk−3

Γq(µk − 2). (7.4.14)

Proof. Applying the operation D2q to both sides of identity (7.4.10), we arrive at formula

(7.4.14).

Lemma 7.4.3. The q-Cauchy problem (7.4.8)–(7.4.9) may have only a unique solution in

the class ACγ [0, a].

Proof. If there exist two solutions yj(x) ∈ ACγ [0, a], j=1,2, then their difference y(x) =

y1(x)−y2(x) is of the same class, and it is a solution of the homogeneous q-Cauchy problem

L1/2y(x) + λy(qx) = 0, x ∈ (0, a], Lky(x)|x=0 = mk(y), k = 0, 1, 2. (7.4.15)

By (7.4.10) and (7.4.15), D−2L1/2y(x) = y(x) = −λD−2q y(qx), x ∈ (0, a). In other words,

the function y(x) ∈ L1q(0, a) is a solution of the homogeneous q-Volterra equation

y(x) = −λ∫ x

0(x− qt)y(qt) dqt, x ∈ (0, a]. (7.4.16)

Hence, from Theorem 6.2.3, y(x) = 0 for all x ∈ (0, a].

The next Lemma deals with the function

yµ(x, λ) = xµ−1E2,µ(−λqµ−1x2; q) =∞∑

k=0

(−1)kqk2+(µ−2)k(−λ)k x2k+µ−1

Γq(2k + µ), (7.4.17)

which obviously satisfies the conditions

yµ(x, λ) ∈

L1q(0, a) when, µ > 0,

L2q(0, a) when, µ > 1/2,

(7.4.18)

180

where by L2q(0, a) we mean the set of all functions f defined on (0, a] such that f ∈ L2

q(0, x)

for all x ∈ (qa, a].

Lemma 7.4.4. If

µr =r∑

j=0

γj , r = 0, 1, 2 (7.4.19)

then the functions yµr(·, λ)2r=0 is a fundamental set of solution of the q-Cauchy problem

L1/2y(x) + λy(x) = 0, x ∈ (0, a], (7.4.20)

Lky(x)|x=0 = δk,r ≡

1 when k = r,

0 when k 6= r,k = 0, 1, 2. (7.4.21)

in the space ACγ[0, a].

Proof. We consider three cases

1. If µ = µ0 = γ0, then

L0yµ0(x;λ) =∞∑

k=0

(−λ)kqk2+(µ0−2)k x2k

Γq(1 + 2k),

L1yµ0(x;λ) = −λ∞∑

k=0

(−λ)kq(k+1)2+(µ0−2)(k+1) x2k+2−γ1

Γq(3− γ1 + 2k),

L2yµ0(x;λ) = −λ∞∑

k=0

(−λ)kq(k+1)2+(µ0−2)(k+1) x2k+2−γ1−γ2

Γq(3− γ1 − γ2 + 2k).

Since γ1 + γ2 < 2, and 3− γ1 − γ2 − γ3 = γ0 = µ0, we obtain

L1/2y(x) = L3y(x) = −λ∞∑

k=0

(−λ)kq(k+1)2+(µ0−2)(k+1) x2k+2−γ1−γ2−γ3

Γq(3− γ1 − γ2 − γ3 + 2k)

= −λ∞∑

k=0

(−λ)kqk2+(µ0−2)k (qx)2k+µ0−1

Γq(2k + µ0)= −λy(qx).

That is, yµ0(x;λ) is a solution of the Problem (7.4.20)–(7.4.21).

181

2. If µ = µ1 = γ0 + γ1, then

L0yµ1(x;λ) =∞∑

k=0

(−λ)kqk2+(µ1−2)k x2k+γ1

Γq(2k + γ1 + 1),

L1yµ1(x;λ) =∞∑

k=0

(−λ)kqk2+(µ1−2)k x2k

Γq(2k + 1),

L2yµ1(x;λ) = −λ∞∑

k=0

q(k+1)2+(µ1−2)(k+1) x2k−γ2+2

Γq(3− γ2 + 2k),

L3yµ1(x;λ) = L1/2yµ1(x;λ) = −λ∞∑

k=0

(−λ)kqk2+(µ1−2)k (qx)2k+2−γ2−γ3

Γq(3− γ2 − γ3 + 2k).

(7.4.22)

But in this case, 3− γ2− γ3 = γ0 + γ1 = µ1 according to (7.4.1). Therefore the last identity

may be written in the form

L1/2yµ1(x;λ) + λyµ1(x;λ) = 0, x ∈ (0, a]. (7.4.23)

By formulae (7.4.22), yµ1(x;λ) ∈ ACγ[0, a] and, in addition, this function satisfies (7.4.20)

with initial condition (7.4.21) with r = 1. This completes the proof of the Lemma.

3. if µ = µ2 = γ0 + γ1 + γ2, then

L0yµ2(x;λ) =∞∑

k=0

(−λ)kqk2+(µ2−2)k x2k+γ1+γ2

Γq(2k + γ1 + γ2 + 1),

L1yµ2(x;λ) =∞∑

k=0

(−λ)kqk2+(µ2−2)k x2k+γ2

Γq(2k + γ2 + 1),

L2yµ2(x;λ) =∞∑

k=0

(−λ)kqk2+(µ2−2)k x2k

Γq(2k + 1),

L3yµ2(x;λ) = L1/2yµ2(x;λ) = −λ∞∑

k=0

(−λ)kqk2+(µ2−2)k (qx)2k+2−γ3

Γq(3− γ3 + 2k).

(7.4.24)

But µ2 = 3− γ3 according to (7.4.1) and the last identity may be written in the form

L1/2yµ2(x;λ) + λyµ2(x;λ) = 0, x ∈ (0, a). (7.4.25)

By formulae (7.4.24), yµ2(x;λ) ∈ ACγ [0, a] and yµ2(x;λ) satisfies (7.4.20) with initial con-

dition (7.4.21) with r = 1.

182

Now we are ready to prove the following theorem

Theorem 7.4.5. Let a > 0 be an arbitrary number. Then the function

Y (x; λ) =2∑

j=0

ajyµj (x;λ) ∈ L2q(0, a) (7.4.26)

is the unique solution of the cauchy type problem (7.4.8)–(7.4.9) in the class ACγ [0, a].

Proof. By Lemma 7.4.4,

L1/2Y (x; λ) + λY (qx;λ) =2∑

j=0

aj

L1/2yµj (x;λ) + λY (qx;λ)

= 0, x ∈ (0, a]

and

LkY (x;λ)∣∣x=0

=2∑

j=0

ajLµj (x;λ)∣∣x=0

=2∑

j=0

ajδk,j = ak, k = 0, 1, 2.

The uniqueness of the solution is proved in Lemma 7.4.3.

Finally, we present the forms taken by the operators Lj3j=0, the corresponding Cauchy

type problems and their solutions in the special cases (7.4.3). In the case when γ1 = 0 and

γ0 = γ1 = γ3 = 1,

L0y(x) = y(x), L1y(x) = D−1q Dqy(x) = y(x)− y(0),

L2y(x) = Dqy(x), L1/2y(x) = L3y(x) = D2qy(x),

(7.4.27)

and we have also µ0 = 1, µ1 = 1, µ2 = 2. In the case when γ2 = 0 and γ0 = γ1 = γ3 = 1,

L0y(x) = y(x), L1y(x) = Dqy(x),

L2y(x) = D−1q D2

qy(x) = Dqy(x)−Dqy(0), L1/2y(x) = L3y(x) = D2qy(x),

(7.4.28)

and we also have µ0 = 1, µ1 = 2, µ2 = 2. In both these cases we arrive at the same Cauchy

problem

−1qDq−1Dqy(x) + λy(x) = 0, x ∈ (0, a), y(0) = α, Dqy(0) = β

183

whose solution is

Y (x;λ) = αE2,1(−λx2; q) + βE2,2(−λqx2; q)x

= α cos(√λ

qx; q)

+ β

√q

λsin(√λ

qx; q).

184

Table 7.1:

φ Iαq φ

xβ−1 xα+β−1 Γq(β)

Γq(α+β), <β > 0

eq(λx) xαe1,α+1(λx; q)

Eq(λx) xαE1,α+1(λx; q)

xβ−1eq(λx) xα+β−1(1−q)α(qα+β ;q)∞(qβ ;q)∞ 2φ1(0, q

β; qα+β, q, λx)

xβ−1Eq(λx) xα+β−1(1−q)α(qα+β ;q)∞(qβ ;q)∞ 1φ1(q

β; qα+β, q, λx)

cosq λx xαe2,α+1(−λ2x2), <α > 0

sinq λx λxα+1e2,α+1(−λ2x2), <α > 0

Cosqλx xα

Γq(α+1)1φ2(q2; qα, qα+1; q2,−λ2x2)

Sinqλx xα+1

Γq(α+2)λxq2(1− q)1φ2(q

2; qα, qα+1; q2,−λ2x2)

cos(λx; q) xαE2,α+1(−λ2qx2; q)

sin(λx; q) λxα+1E2,α+2(−λ2q2x2; q)

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Index

C2q (0, a), 63

Eq(z), 5

Eν,µ(z; q), 143

Eρ(z), 141

Jν(·; q), 4

L1q(0, a), 10

L2q(0, a), 8, 63

L2q

((0, a)× (0, a)

), 9

Wq(y1, . . . , yn)(x), 58

Cosqz, 5

Sinqz, 5

cos(z; q), 10

cosq z, 5

AC(n)[a, b], 112

ACq[0, a], 122, 123, 125

AC(n)q [0, a], 125

ACγ[0, a], 177

L1q(0, a), 10, 118, 147

L2q(0, a), 180

µ-geometric set, 6

∼, 11

sin(z; q), 10

sinq z, 5

eq(z), 5

eν,µ(z; q), 143

q-Abel integral equation, 131

q-Bessel functions, 4

q-Beta function, 4

q-Cauchy problem, 56

q-Cauchy problem, 53, 176

q-Gamma function, 3

q-Hankel Transforms, 35

q-Hypergeometric series, 3

q-Laplace transform, 170

q-Wronskian, 58

q-analogues of Mittag-Leffler functions, 142

q-binomial coefficients, 3

q-difference operator, 6

q-integration by parts, 8

q-moments of a function f , 36

q-regular at zero function, 7

q-shifted factorial, 2

q-successive approximations, 51

q-translation , 118

q-type Fredholm integral operator, 60

196

197

Abel’s integral equation, 113

Newton–Puiseux diagram, 12

A fundamental theorem of q-calculus, 7

Abel’s integral equation, 112

Basic Green’s function, 74

Basic Sturm-Liouville problem, 68, 69, 71

Basic Volterra integral equations, 146

Caputo fractional derivative, 139

fractional derivative, 115

Gauss’ summation formula, 119

Grunwald–Letinkov, 110

fractional derivative, 139

Green’s function, 60

Jackson q-integration, 6

Jacobi triple product identity, 3

Linear q-difference equations, 55

Mittag-Leffler functions, 141

Multiple q-shifted factorial, 2

O, 11

Polya, 30, 49

Riemann–Liouville

fractional integral operator, 110

zeros of Eν,µ(z; q), 152