Study of new class of q-fractional integral operator

TL;DR

This paper introduces a new class of q-fractional integral operators using an iterated Cauchy integral approach, generalizing existing q-analogues of fractional differential operators and exploring their properties.

Contribution

It presents a novel q-fractional integral operator that extends previous operators and encompasses the q-analogue of Hadamard fractional differential operator.

Findings

The new operator generalizes existing q-fractional operators.

Properties of the new q-fractional integral operator are investigated.

The operator covers the q-analogue of Hadamard fractional differential operator.

Abstract

We study the new class of q-fractional integral operator. In the aid of iterated Cauchy integral approach to fractional integral operator, we applied t^pf(t) instead of f(t) in these integrals and with parameter p a new class of q-fractional integral operator is introduced. Recently the q-analogue of fractional differential integral operator is studied.[8][4][9][10][13][14]All of these operators are q-analogue of Riemann fractional differential operator. The new class of introduced operator generalize all these defined operator and can be cover the q-analogue of Hadamard fractional differential operator. Some properties of this operator is investigated.

Equations

$${\left[a\right]}_q=\frac{1-q^a}{1-q}\left(q\ne 1\right);{\left[0\right]}_q!=1;{\left[n\right]}_q!={\left[n\right]}_q{\left[n-1\right]}_{q}n\in \mathbb{N},a\in \mathbb{C}.$$

$${\left(a;q\right)}_0$$

$${\left(a;q\right)}_{\infty }$$

$$\left(\begin{array}[c]{c}n\\ k\end{array}\right)_{q}=\frac{\left(q;q\right)_{n}}{\left(q;q\right)_{n-k}\left(q;q\right)_{k}},$$

$$e_q\left(z\right)$$

$$E_q(z)$$

$$\int f(x)d_{q}x=(1-q)x\sum _{i=0}^{\infty }{q}^{i}f(q^{i}x)$$

$$\underset{0}{\overset{b}{\int }}f(x)d_{q}x$$

$$\underset{a}{\overset{b}{\int }}f(x)d_{q}x$$

$$\underset{a}{\overset{b}{\int }}g(x)D_{q}f(x)d_{q}x=\left(g(b)f(b-g(a)f(a)\right)-\underset{a}{\overset{b}{\int }}f(qx)D_{q}g(x)d_{q}x$$

$$\underset{0}{\overset{x}{\int }}\underset{0}{\overset{v}{\int }}f(s)d_{q}s{d}_{q}v=\underset{0}{\overset{x}{\int }}\underset{qs}{\overset{x}{\int }}f(s)d_{q}v{d}_{q}s$$

$$\underset{0}{\overset{x}{\int }}\underset{qs}{\overset{x}{\int }}f(s)d_{q}v{d}_{q}s=\underset{0}{\overset{x}{\int }}\left(x-qs\right)f(s)d_{q}s=x(1-q)\sum _{i=0}^{\infty }{q}^{i}f(q^{i}x)\left(x-q^{i+1}x\right)=x^2(1-q{)}^2\sum _{i=0}^{\infty }{q}^{i}f(q^{i}x)\left(\sum _{j=0}^{\infty }{q}^j\right)$$

$$\underset{0}{\overset{x}{\int }}\underset{0}{\overset{v}{\int }}f(s)d_{q}s{d}_{q}v=x(1-q)\sum _{i=0}^{\infty }{q}^i\underset{0}{\overset{x{q}^i}{\int }}f(s)d_{q}s=x^2(1-q{)}^2\sum _{i=0}^{\infty }\sum _{j=0}^{\infty }{q}^{i+2j}f(q^{i+j}x)$$

$${\Gamma }_q(t)=\underset{0}{\overset{\frac{1}{1-q}}{\int }}{x}^{t-1}{E}_q(-qx)d_{q}x$$

$${\Gamma }_q(t+1)=-\underset{0}{\overset{\frac{1}{1-q}}{\int }}{x}^t{d}_q{E}_q(-x)=0+{\left[t\right]}_q\underset{0}{\overset{\frac{1}{1-q}}{\int }}{x}^{t-1}{E}_q(-qx)d_{q}x={\left[t\right]}_q{\Gamma }_q(t)$$

$$D_q\text{vskip}12.0ptplus4.0ptminus4.0pt(f(q^{-1}x))=-f(x)$$

$$f(x)=\frac{f(0)}{{\prod }_{k=0}^{\infty }\left(1+\left(q-1\right)q{-}^{k}x\right)}=f(0)e_{q^{-1}}\left(x\right)=f(0)E_q(x)$$

$${\tilde{\gamma }}_q^{(A)}(t+1)=q^{-t}{\left[t\right]}_q{\tilde{\gamma }}_q^{(A)}(t)$$

$$(x-a{)}^{\left(\alpha \right)}=x^{\alpha }\prod _{k=0}^{\infty }\frac{\left(1-\frac{x}{a}{q}^k\right)}{\left(1-\frac{x}{a}{q}^{k+\alpha }\right)}=\frac{x^{\alpha }{\left(\frac{x}{a};q\right)}_{\infty }}{{\left(q^{\alpha }\frac{x}{a};q\right)}_{\infty }}$$

$${\Gamma }_q(t)=\frac{(1-q{)}^{\left(t-1\right)}}{(1-q{)}^{t-1}}$$

$$(x-y{)}_{q^{p+1}}^{\left(\alpha \right)}=x^{\alpha }\prod _{k=0}^{\infty }\frac{\left(x-y{\left(q^{p+1}\right)}^k\right)}{\left(x-y{\left(q^{p+1}\right)}^{k+\alpha }\right)}=\frac{x^{\alpha }{\left(\frac{y}{x};q^{p+1}\right)}_{\infty }}{{\left(q^{\alpha \left(p+1\right)}\frac{y}{x};q^{p+1}\right)}_{\infty }}$$

$${\left[p+1\right]}^{\left(\alpha \right)}$$

$$=\frac{{\left(1-q^{p+1}\right)}_{q^{p+1}}^{\left(\alpha \right)}}{{\left(1-q\right)}^{\left(\alpha \right)}}=\frac{{\left(1-q^{p+1}\right)}_{q^{p+1}}^{\left(\alpha \right)}}{{\left(1-q\right)}^{\alpha }{\Gamma }_q(\alpha +1)}=\frac{{\Gamma }_{q^{p+1}}(\alpha +1)}{{\Gamma }_q(\alpha +1)}{\left({\left[p+1\right]}_q\right)}^{\alpha }$$

$${\beta }_q(t,s)=\underset{0}{\overset{1}{\int }}{x}^{t-1}{\left(1-qx\right)}_q^{s-1}{d}_{q}x=\frac{{\Gamma }_q(t){\Gamma }_q(s)}{{\Gamma }_q(t+s)}$$

$$\underset{a}{\overset{x}{\int }}{dt}_1\underset{a}{\overset{t_1}{\int }}{dt}_2...\underset{a}{\overset{t_{n-1}}{\int }}f(t_n){dt}_n=\frac{1}{\Gamma (n)}\underset{a}{\overset{x}{\int }}{\left(x-t\right)}^{n-1}f(t)dt$$

$${}_a{I}^{\alpha }(f(x))=\frac{1}{\Gamma (\alpha )}\underset{a}{\overset{x}{\int }}{\left(x-t\right)}^{\alpha -1}f(t)dt$$

$$\underset{a}{\overset{x}{\int }}\frac{1}{t_1}{dt}_1\underset{a}{\overset{t_1}{\int }}\frac{1}{t_2}{dt}_2...\underset{a}{\overset{t_{n-1}}{\int }}\frac{1}{t_n}f(t_n){dt}_n=\frac{1}{\Gamma (n)}\underset{a}{\overset{x}{\int }}{\left(Log\left(\frac{x}{t}\right)\right)}^{n-1}f(t)\frac{dt}{t}$$

$${}_a{J}^{\alpha }(f(x))=\frac{1}{\Gamma (\alpha )}\underset{a}{\overset{x}{\int }}{\left(Log\left(\frac{x}{t}\right)\right)}^{\alpha -1}f(t)\frac{dt}{t}$$

$$I_{q,a}^{\alpha }(f(x))=\frac{1}{{\Gamma }_q(\alpha )}\underset{a}{\overset{x}{\int }}{K}_q(t,x)f(t)d_{q}t$$

$$J_{p,q}^2\left(f(a)\right)=\underset{0}{\overset{a}{\int }}\underset{0}{\overset{x}{\int }}{x}^p{y}^{p}f(y)d_q{yd}_{q}x=\underset{0}{\overset{a}{\int }}\underset{qy}{\overset{a}{\int }}{x}^p{y}^{p}f(y)d_q{xd}_{q}y=\frac{1}{{\left[p+1\right]}_q}\underset{0}{\overset{a}{\int }}{y}^{p}f(y)\left[a^{p+1}-q^{p+1}{y}^{p+1}\right]d_{q}y$$

$$J_{p,q}^3\left(f(a)\right)$$

$$=\frac{1}{{\left[p+1\right]}_q{\left[2p+2\right]}_q}\underset{0}{\overset{a}{\int }}{z}^{p}f(z)\left[{\left(a^{p+1}\right)}^2-\frac{{\left[2p+2\right]}_q}{{\left[p+1\right]}_q}{\left(zq\right)}^{p+1}{a}^{p+1}+\left(\frac{{\left[2p+2\right]}_q}{{\left[p+1\right]}_q}-1\right){\left({\left(zq\right)}^{p+1}\right)}^2\right]d_{q}z$$

$$J_{p,q}^3\left(f(a)\right)==\frac{1}{{\left[p+1\right]}_q{\left[2p+2\right]}_q}\underset{0}{\overset{a}{\int }}{z}^{p}f(z)\left(a^{p+1}-{\left(zq\right)}^{p+1}\right)\left(a^{p+1}-{\left(zq\right)}^{p+1}{q}^{p+1}\right)d_{q}z$$

Full text

Keyword: q-fractional differential integral operator, fractional calculus, Hadamard fractional differential operator

1 Introduction

Fractional calculus has a long history and has recently gone through a period of rapid development. When Jackson (1908)[16] defined $q$-differential operator, $q-$calculus became a bridge between mathematics and physics. It has a lot of applications in different mathematical areas such as combinatorics, number theory, basic hypergeometric functions and other sciences: quantum theory, mechanics, theory of relativity, capacitor theory, electrical circuits, particle physics, viscoelastic, electro analytical chemistry, neurology, diffusion systems, control theory and statistics; The $q-$Riemann-Liouville fractional integral operator was introduced by Al-Salam [12], from that time few $q$-analogues of Riemann operator were studied. [9][2][8][4]

On the other hand, recent studies on fractional differential equations indicate that a variety of interesting and important results concerning existence and uniqueness of solutions, stability properties of solutions, and analytic and numerical methods of solutions for these equations have been obtained, and the surge for investigating more and more results is underway. Several real world problems were modeled in the aid of using fractional calculus. Nowadays, fractional-order differential equations can be traced in a variety of applications such as diffusion processes, biomathematics, thermo-elasticity, etc.[18]. However, most of the work on the topic is based on Riemann-Liouville, and Caputo-type fractional differential equations. $q$-analogue of this operator is defined [12] and application of this operator is investigated.[4][8][10] Another kind of fractional derivatives that appears side by side to Riemann-Liouville and Caputo derivatives in the literature is the fractional derivative due to Hadamard, introduced in 1892 [6], which contains logarithmic function of arbitrary exponent in the kernel of the integral appearing in its definition. In a paper from 1751, Leonhard Euler (1707–1783) introduced the series which can be assumed as a $q-$analogue of logarithm function. [15]Since that time, a lot of mathematician have tried to define $q-$logarithm function. Because of dificulty of working with these function, there is no $q-$analogue of Hadamard fractional differential integral operator. Hadamard-type integrals arise in the formulation of many problems in mechanics such as in fracture analysis. For details and applications of Hadamard fractional derivative and integral, we refer the reader to a new book that gathered all of these applications.[20]

In this paper, new $q$-integral operator is introduced, Some properties and relations are investigated. In fact, a parameter is used to generalize the Riemann operator and a new class of $q-$fractional difference operator is introduced. In the first section, let us introduce some familar concepts of q-calculus. Most of these definitions and concepts are available in [1] and [3]. We use $[n]_{q}$ as a $q-$analogue of any complex number. Naturally, we can define $\left[n\right]_{q}!$ as

[TABLE]

The $q$-shifted factorial and $q$-polynomial coefficient are defined by

[TABLE]

[TABLE]

In the standard approach to $q-$calculus, two exponential function are used more.The first and second identities of Euler leads to two $q-$exponentials function as follow

[TABLE]

Let for some $0\leq\alpha<1,$ the function $|f(x)x^{\alpha}|$ is bounded on the interval $\left(0,A\right]$, then Jakson integral defines as [1]

[TABLE]

converges to a function $F(x)$ on $\left(0,A\right],$ which is a $q-$antiderivative of $f(x)$. Suppose $0<a<b$, the definite $q-$integral is defined as

[TABLE]

$q-$analogue of integral by part can be written as

[TABLE]

In addition, we can interchange the order of double q-integral by

[TABLE]

Actually this interchange of order is true, since

[TABLE]

In addition, the left side can be written as

[TABLE]

Let $i+j=t$ to see that this releation is true. $q-$analogue of Gamma function is defined as

[TABLE]

Several $q$-exponential functions for different purposes were defined. [5]Natural question is appeard about this definition of q-exponential function, Why did we use $E_{q}(x)$ in the definition of q-Gamma function? Next lemma answer this question.

Lemma 1

For given q-Gamma function $\Gamma_{q}(t+1)=\left[t\right]_{q}\Gamma_{q}(t)$ and this relation is true only where we use $E_{q}(-qx)$ instead of q-exponential function.

Proof. Use q-integral by part and the fact that $D_{q}$

$(E_{q}(-x))=-E_{q}(-qx)$ to find the recurrence formula for q-gamma function.

[TABLE]

Now if we assume that $f(x)=E_{q}(x)$, to have the above property, function $f(x)$ should satisfies the following q-difference equation

[TABLE]

If we apply q-derivative formula for this, we can see that $f(x)\left[1+(q-1)x\right]=f(q^{-1}x).$ If we rewrite this equation for $f(q^{-j}x)$ and tend $j$ to infinity, then we have

[TABLE]

Which shows the uniqueness of $E_{q}(-qx)$ in this definition.

Actually, the authors in [2] define another version of $q$-Gamma function. In that paper, Authors defined $q$-Gamma function such that the classic results satisfied. This definition is based on $e_{q}\left(x\right)$ and as we mentioned it in the last lemma, the recurence formula is not given the same terms like in lemma. In fact,

[TABLE]

Where $\widetilde{\gamma}_{q}^{(A)}(t)=\int\limits_{0}^{\frac{\infty}{A\left(1-q\right)}}\mathit{x}^{t-1}e_{q}(-x)d_{q}x$. The limit boundary of integration is changed also and author used the interesting function as a kind of weight and make the equivalent definition for $q$-gamma function.

$q$-shifted factorial may extend to the following definition

[TABLE]

We can rewrite the $q$-Gamma function by using this definition as[2]

[TABLE]

Let us generalize definition (7) to the following form

[TABLE]

In addition, we define another version of $q-$analogue of exponent as

[TABLE]

In addition, $q-$Beta function is defined as

[TABLE]

2 Itterated q-integral to approach new class of operators

There are several approaches to fractional differential operators. One of demonstration methods of fractional differential equation is using the itterated Cauchy integrals. The Riemann–Liouville fractional integral is a generalization of the following itterated Cauchy integral:

[TABLE]

In the aid of this formula, for any positive real value $0<\alpha,$ we have

[TABLE]

If we put $\frac{1}{t_{i}}$ in the chain of integration, then we can reach to Hadamard operator. The related itterated integral can be written as

[TABLE]

Therefore, Hadamard fractional integral can be written as [6]

[TABLE]

The author in [7] assumed $t_{i}^{p}$ in the chain of integration and reached to the general formula for fractional integral operator.There are four different models of $q$-analogues of Riemann-Liouville fractional integral operators. No one has investigated the Hadamard type and there is not any $q$-analogue of this operator. First let us rewrite the definition of $q-$fractional integral operator in all introduced forms. In fact, for $\alpha\geq 0$ and $f:\left[a,b\right]\rightarrow\mathbb{R},$ the $\alpha$ order fractional $q$-integral of a function $f(x)$ is defined by

[TABLE]

The kernel of this integral is defined as $K_{q}(t,x)=(x-qt)^{\left(\alpha-1\right)}$ [9], $K_{q}(t,x)=(x-qt)_{q}^{\alpha-1}$ [2], $K_{q}(t,x)=x^{\alpha-1}(\frac{qt}{x};q)_{\alpha-1}$ [4], and $K_{q}(t,x)=(x-qt)_{\alpha-1}$ [8]. There is a good discussion about the form of the kernel function at [10].

In this section, we investigate the new definition for q-integral operator. In the aid of definition of Hadamard operator as itterated integral for the case $\alpha\in\mathbb{N}.$ We rewrite this integrals by using Jackson integral instead of Riemann integral. We will use the technique of interchanging the order of q-integral which is mentioned in the last section. So for $n=2,$ we have:

[TABLE]

Now Consider the case $n=3$, then we have

[TABLE]

we can simplify this with the little calculation as

[TABLE]

For case $n=4$, we have

[TABLE]

Now the little calculation, shows that

[TABLE]

We can easily see that for any natural number $k\in\mathbb{N},$ in the aid of induction we have:

[TABLE]

This relation for itteration integral motivate us to define $q$-analogue of integral operator as follow

Definition 2

for $\alpha>0$ and $x>0,$ if $f(x)$ satisfies the condistion of existance for following Jackson integral, we define $q$-fractional integral as

[TABLE]

Remark 3

First, we should clear some identities. For any natural number $k\in\mathbb{N}$ we have:

[TABLE]

In addition, this definition is really the unification of $q$-analogue of Reimann and Hadamard integral operator. For instance, let $q\rightarrow 1^{-}$ then we have

[TABLE]

This is exactly as the same as operator which is introduced at [7]. If we let $p\rightarrow-1^{+}$ and use Hopital, we have

[TABLE]

When $p=0$, we arrive at the well-known $q-$fractional Reimann  integral [9].

Remark 4

The new operator can be expanded as

[TABLE]

In addition, we may express the $q$-Reimann type operator for $q^{p+1}$ as follow

[TABLE]

These expressions shows that the new operator is not only the simple modification of the available operator.

3 Some properties of new q-fractional integral operator

In this section, we study some familar properties of fractional integral operator as semi-group properties of this operator. This property is essentially useful, because to solve related $q-$difference equation we should apply this property. In addition, we will define inverse operator as $q$-fractional derivative and at the end, properties of these operators will be studied. In this procedure, we study some useful identities and relations as well. $q-$fractional Reimann integral operators were extensively investigated in several resources.[9] In the aid of Hine’s transform for q-hypergeometric functions, useful identities were introduced and a lot of identities were studied.[9] [4] We start with the following lemma from [9]. This relation is important and make a rule like beta function. Most of $q$-analogue of ordinary cases can be interpreted in the aid of this lemma, We will introduce the sequence of identities to achive semi-group property.

Lemma 5

For $\alpha,\beta,\mu\in\mathbb{R}^{+}$, the following identity is valid[9]

[TABLE]

Remark 6

We mention that the terms at the doniminator of summation is not related to $t$ and we can take it out from the summation. Now, we modify the lemma to prepare the suitable conditions for using in our operator. For this reason, let us substitute $q$ by $q^{p+1}$ in the last lemma. In the aid of definition that is mentioned by identity (9), we have :

[TABLE]

In this step, let us calculate the following q-integral by using the last remark.

Lemma 7

Following Jackson integral for real positive $\alpha$ and $\lambda>-1$ holds true:

[TABLE]

Proof.* In the aid of definition of Jackson integral, left hand side of this inequality can be written as*

[TABLE]

Here, the second integral is zero, because for some $i\in\mathbb{N}$ the factor $(\left(aq^{i}\right)^{p+1}-a^{p+1})_{q^{p+1}}^{\left(\lambda\right)}=0$ and we can expand the integral as

[TABLE]

Now expand the first integral and use the last remark to see

[TABLE]

Remark 8

put $a=0$ in the last integral to find $J_{p,q}^{\alpha}\left(f(t)\right)$ where $f(t)=t^{\lambda(p+1)}$. In the aid of last lemma we have

[TABLE]

In addition, we interpret logarithm function by limit of expression in remark 3. Hadamard integral operator has the following property: [11]

[TABLE]

On the other hand, we have this limit

[TABLE]

Now, in the aid of last lemma, we derive to q-analogue of the property in [11]

[TABLE]

Proposition 9

The given q-fractional integral operator has semi-group property. Means

[TABLE]

Proof.* Write the left hand side of this identity as*

[TABLE]

Now apply the last lemma to have

[TABLE]

**

Definition 10

Let $\alpha\geq 0$ and $n=\left\lfloor\alpha\right\rfloor+1$ means $n$ is the smallest integer such that $n\geq\alpha$ and $p>0$. The corresponding generalized $q-$fractional derivatives is defined by

[TABLE]

if the integral does exist.

**

Now we can related the defined q-derivative and q-integral operator as follow

[TABLE]

It is easy to see that $(x^{-p}D_{q})^{n}(J_{p,q}^{n}f)(t)=f(t)$. We can prove it by induction. For instance, let us consider the case that $0<\alpha<1$ in next proposition:

Proposition 11

Assume that $0<\alpha<1$ and $p>0$ and integral does exist, then the following identity holds

[TABLE]

Proof.* Direct calculation of the identity in the aid of lemma (5) shows that*

[TABLE]

**

In this paper, we defined class of generalized $q-$fractional difference integral operator and the inverse operator also is defined. A few properties of these operators were investigated, but still there are a lot of identities and formulae related to this operator which can be studied. $q$-calculus is the world of mathematics without limit and the introduced operator can be make a rule as a part of these objects.