New concepts of fractional Hahn’s $q,\omega$-derivative of Riemann–Liouville type and Caputo type and applications

Abstract

In this paper, we give the definitions of $q,\omega$-exponential function and $q,\omega$-gamma function. New concepts of fractional Hahn’s $q,\omega$-derivative of Riemann–Liouville type and Caputo type are introduced, meanwhile we discuss some properties. As applications of the new concepts, we give the existence result of positive solutions for boundary value problem of fractional $q,\omega$-derivatives equations. We also definite certain $q,\omega$-Mittag-Leffler function by solving the initial value problem of fractional $q,\omega$-equations.

1 Introduction

The quantum calculus began with Jackson in the early twentieth century. The q-calculus appeared as a connection between mathematics and physics. It has a lot of applications in different mathematical areas such as number theory, combinatorics, and other sciences quantum theory. The book by Kac and Cheung [1] covers many of the fundamental aspects of quantum calculus. In recent years, the topic of q-calculus has attracted the attention of several researchers and a variety of new results can be found in papers [2–6].

The fractional q-calculus has developed into a popular mathematical modeling tool for many real world phenomena. Early developments for fractional q-calculus can be found in the work of Al-Salam [7] and Agarwal [8]. Moreover, the author in [9] generalizes the notions of fractional q-integral and q-derivative by introducing variable lower limit of integration.

Quantum difference operators have also been receiving an increasing interest due to their applications. In [10], Hahn introduced the quantum difference operator $D_{q,\omega }$, where $q\in\,]0,1[$ and $\omega >0$ are fixed. The Hahn operator unifies the two most well-known quantum difference operators: the Jackson q-difference derivative $D_{q}$, where $q\in\,]0,1[$; and the forward difference $\triangle_{\omega }$, where $\omega >0$. Hahn’s difference operator has been applied successfully in the construction of families of orthogonal polynomials as well as in approximation problems. The right inverse of Hahn’s difference operator was introduced by Aldwoah [11] who defined the right inverse of $D_{q,\omega }$ in terms of both the Jackson q-integral containing the right inverse of $D_{q}$ and the Nörlund sum involving the right inverse of $\triangle_{\omega }$. Malinowska and Torres [12] introduced the Hahn quantum variational calculus, while Malinowska and Martins [13] studied the generalized transversality conditions for the Hahn quantum variational calculus. In [14], some concepts of fractional quantum calculus were introduced in terms of a q-shifting operator ${}_{a}\!\Phi_{q}(m)=qm+(1-q)a$. As a research motivation, we redefine the fractional Hahn $q,\omega$-calculus with a q-shifting operator on the interval $[a,b]$.

In this paper, we recall some basic concepts on fractional q-calculus, which will be used in the later section. After preliminaries, in the third section the definitions of $q,\omega$-exponential function and $q,\omega$-gamma function are presented. In the fourth section we discuss some properties of $q,\omega$-calculus. After giving the basic properties, the fractional Hahn $q,\omega$-derivative of Riemann–Liouville type is introduced in the fifth section. Then we change the order of operators, we give the fractional $q,\omega$-derivative of Caputo type in the sixth section. And then, we consider boundary value problems of fractional $q,\omega$-derivative equations. We give the corresponding Green’s function of the boundary value problem and its properties. By using Krasnoselskii’s fixed point theorem, an existence result of positive solutions to the boundary value problem is enunciated. Finally, we define a certain $q,\omega$-Mittag-Leffler function by solving the initial value problem of fractional $q,\omega$-equations.

Let $q\in\,]0,1[$ and $\omega \in\,]0,+\infty [$ be given. Define $\omega_{0}=\frac{\omega }{1-q}$. Throughout the paper, we assume I to be an interval of $\mathbb{R}$ containing $\omega_{0}$.

Definition 1.1

(Hahn’s difference operator [10])

Let $f: I\rightarrow \mathbb{R}$. Hahn’s difference operator is defined by

$$\begin{aligned} D_{q,\omega }f(t):= \textstyle\begin{cases} \frac{f(qt+\omega )-f(t)}{(qt+\omega )-t},& \mbox{if }t\neq\omega_{0}, \\ f'(t),&\mbox{if } t=\omega_{0}, \end{cases}\displaystyle \end{aligned}$$

provided that f is differentiable at $\omega_{0}$.

Definition 1.2

([12])

Let I be a closed interval of $\mathbb{R}$ such that $\omega_{0}$, $a,b\in I$. For $f : I\rightarrow \mathbb{R}$, we define the $q,\omega$-integral of f from a to b by

$$ \int_{a}^{b}f(t)\,{d_{q,\omega }t}:= \int_{\omega_{0}}^{b}f(t)\,{d_{q, \omega }t}- \int_{\omega_{0}}^{a}f(t)\,{d_{q,\omega }t}, $$

(1)

where

$$\begin{aligned} \int_{\omega_{0}}^{x}f(t)\,{d_{q,\omega }t}:=\bigl(x(1-q)- \omega \bigr)\sum_{k=0} ^{\infty }q^{k}f \bigl(xq^{k}+[k]_{q,\omega }\bigr),\quad x\in I, \end{aligned}$$

(2)

with $[k]_{q,\omega }:=\frac{\omega (1-q^{k})}{1-q}$ for $k\in\mathbb{N}_{0}=\mathbb{N}\cup \{0\}$, provided that the series converges at $x=a$ and $x=b$. In this case, f is called $q,\omega$-integrable on $[a,b]$. We say that f is $q,\omega$-integrable over I iff it is $q,\omega$-integrable over $[a,b]$ for all $a,b\in I$.

Lemma 1.3

([12])

Assume $f : I\rightarrow \mathbb{R}$ to be continuous at $\omega_{0}$. Define

$$\begin{aligned} F(x):= \int_{\omega_{0}}^{x}f(t)\,{d_{q,\omega }t}. \end{aligned}$$

Then F is continuous at $\omega_{0}$. Furthermore, $D_{q,w}F(x)$ exists for every $x\in I$ and $D_{q,w}F(x)=f(x)$. Conversely,

$$ \int_{a}^{b}D_{q,w}F(x)\,{d_{q,\omega }t}=f(b)-f(a) $$

(3)

for all $a,b\in I$.

Lemma 1.4

([12])

If $f,g : I\rightarrow \mathbb{R}$ are continuous at $\omega_{0}$, then

$$ \int_{a}^{b}f(t)D_{q,w}g(t)\,{d_{q,\omega }t}=f(t)g(t)|_{t=a}^{t=b}- \int _{a}^{b}D_{q,w}\bigl(f(t)\bigr)g(qt+ \omega )\,{d_{q,\omega }t} $$

(4)

for all $a,b\in I$. This is the integration by parts formula of the Jackson–Nörlund integral.

2 Preliminary

To make this paper self-contained, below we recall some known facts on fractional q-calculus. The presentation here can be found in, for example, [14].

For $q\in\,]0,1[$, define

$$\begin{aligned}{} [m]_{q}=\frac{1-q^{m}}{1-q},\quad m\in \mathbb{R}. \end{aligned}$$

The q-analog of the power function $(n-m)^{k}$ with $k\in \mathbb{N}_{0}:= \{0,1,2,\ldots \}$ is

$$\begin{aligned} (n-m)^{(0)}=1, \qquad (n-m)^{(k)}=\prod_{i=0}^{k-1} \bigl(n-mq^{i}\bigr),\quad k\in \mathbb{N},n,m\in \mathbb{R}. \end{aligned}$$

More generally, if $\gamma \in \mathbb{R}$, then

$$\begin{aligned} (n-m)^{(\gamma )}=\prod_{i=0}^{\infty } \frac{n-mq^{i}}{n-mq^{\gamma +i}}, \quad n\neq 0. \end{aligned}$$

Note if $m=0$, then $n^{(\gamma )}=n^{\gamma }$. We also use $0^{(\gamma )}=0$ for $\gamma >0$. The q-gamma function is defined by

$$ \Gamma_{q}(t)=\frac{(1-q)^{(t-1)}}{(1-q)^{t-1}}, \quad t\in \mathbb{R} \backslash \{0,-1,-2,\ldots \}. $$

(5)

Obviously, $\Gamma_{q}(t+1)=[t]_{q}\Gamma_{q}(t)$.

Lemma 2.1

([9])

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

$$ \sum_{n=0}^{\infty }\frac{(1-\mu q^{1-n})^{(\alpha -1)}(1-q^{1+n})^{( \beta -1)}}{(1-q)^{(\alpha -1)}(1-q)^{(\beta -1)}}q^{\alpha n}= \frac{(1- \mu q)^{(\alpha +\beta -1)}}{(1-q)^{(\alpha +\beta -1)}}. $$

(6)

3 The $q,\omega$-exponential function and the $q,\omega$-gamma function

Let us define a q-shifting operator as

$$\begin{aligned} {}_{a}\!\Phi_{q}(m)=qm+(1-q)a. \end{aligned}$$

For any positive integer k, we have

$$\begin{aligned} {}_{a}\!\Phi_{q}^{k}(m)={}_{a}\!\Phi_{q}^{k-1}\bigl({}_{a}\!\Phi_{q}(m) \bigr) \quad \mbox{and} \quad {}_{a}\!\Phi_{q}^{0}(m)=m. \end{aligned}$$

We also define the new power of q-shifting operator as

$$\begin{aligned} (n-m)_{a}^{(0)}=1, (n-m)_{a}^{(k)}=\prod _{i=0}^{k-1}\bigl(n-{}_{a}\!\Phi_{q} ^{i}(m) \bigr),\quad k\in \mathbb{N}\cup \{\infty \}. \end{aligned}$$

More generally, if $\gamma \in \mathbb{R}$, then

$$ (n-m)_{a}^{(\gamma )}=\prod_{i=0}^{\infty } \frac{n-{}_{a}\!\Phi_{q}^{i}(m)}{n-{}_{a}\!\Phi_{q}^{i+\gamma }(m)}, $$

(7)

with ${}_{a}\!\Phi_{q}^{\gamma }(m)=q^{\gamma }m+(1-q^{\gamma })a$, $\gamma \in \mathbb{R}$.

For any $\gamma ,n\in \mathbb{R}$, we have

$$ \bigl(n-{}_{c}\!\Phi_{q}(s)\bigr)_{c}^{(\gamma )}=(n-c)^{\gamma } \prod_{i=0}^{ \infty }\frac{1-q^{1+i}(\frac{s-c}{n-c})}{1-q^{1+i+\gamma }( \frac{s-c}{n-c})}. $$

(8)

Using Definition 1.1 and (7), we may obtain the very useful examples of the $q,\omega$-derivatives of the following functions:

$$\begin{aligned} &{}_{x}D_{q,\omega }(x-a)_{\omega_{0}}^{(\alpha )}=[\alpha ]_{q}(x-a)_{ \omega_{0}}^{(\alpha -1)}, \\ &{}_{t}D_{q,\omega }(x-t)_{\omega_{0}}^{(\alpha )}=-[\alpha ]_{q}\bigl(x-{}_{\omega_{0}}\!\Phi_{q}(t) \bigr)_{\omega_{0}}^{(\alpha -1)}. \end{aligned}$$

Definition 3.1

The $q,\omega$-exponential function is defined as

$$\begin{aligned} e_{q,\omega }(t)=\prod_{n=0}^{\infty } \frac{1-{}_{\omega_{0}}\!\Phi_{q} ^{n}(t)}{1-\omega_{0}},\quad e_{q,\omega }(\omega_{0})=1. \end{aligned}$$

Note that $e_{q,\omega }(1)=0$ and

$$\begin{aligned} D_{q,\omega }e_{q,\omega }(t)=-\frac{1}{(1-\omega_{0})(1-q)}e_{q, \omega } \bigl({}_{\omega_{0}}\!\Phi_{q}(t)\bigr). \end{aligned}$$

We are now in a position to give the integral representation of the $q,\omega$-gamma function.

Let $\alpha \in \mathbb{R}\backslash \{0,-1,-2,\ldots \}$. Definite the $q,\omega$-gamma function by

$$\begin{aligned} \Gamma_{q,\omega }(\alpha )= \int_{\omega_{0}}^{1} \biggl(\frac{1}{(1-\omega_{0})(1-q)} \biggr)^{\alpha }\bigl(t-{}_{\omega_{0}}\!\Phi_{q}(\omega_{0})\bigr)_{ \omega_{0}}^{(\alpha -1)} e_{q,\omega }\bigl({}_{\omega_{0}}\!\Phi_{q}(t)\bigr) \,{d_{q,\omega }t}. \end{aligned}$$

Lemma 3.2

For $\alpha \in \mathbb{R}$,

$$\begin{aligned} \Gamma_{q,\omega }(\alpha +1)=[\alpha ]_{q}\Gamma_{q,\omega }( \alpha ),\quad \Gamma_{q,\omega }(1)=1. \end{aligned}$$

For any positive integer k,

$$\begin{aligned} \Gamma_{q,\omega }(k+1)=[k]!=\frac{(1-q^{k})}{(1-q)} \frac{(1-q^{k-1})}{(1-q)}\cdots \frac{(1-q^{2})}{(1-q)} \frac{(1-q)}{(1-q)}. \end{aligned}$$

Proof

By Lemma 1.4 and Definition 3.1, we can get

$$\begin{aligned} &\Gamma_{q,\omega }(\alpha +1) \\ &\quad = \int_{\omega_{0}}^{1} \biggl(\frac{1}{(1-\omega_{0})(1-q)} \biggr)^{ \alpha +1}\bigl(t-{}_{\omega_{0}}\!\Phi_{q}( \omega_{0})\bigr)_{\omega_{0}}^{(\alpha )} e_{q,\omega } \bigl({}_{\omega_{0}}\!\Phi_{q}(t)\bigr)\,{d_{q,\omega }t} \\ &\quad = \int_{\omega_{0}}^{1} \biggl(\frac{1}{(1-\omega_{0})(1-q)} \biggr)^{ \alpha +1}\bigl(t-{}_{\omega_{0}}\!\Phi_{q}( \omega_{0})\bigr)_{\omega_{0}}^{(\alpha )} \bigl[-D_{q,\omega }(1- \omega_{0}) (1-q)e_{q,\omega }(t) \bigr] \,{d_{q,\omega }t} \\ &\quad =- \biggl(\frac{1}{(1-\omega_{0})(1-q)} \biggr)^{\alpha } \bigl[ \bigl(t-{}_{\omega_{0}}\!\Phi_{q}(\omega_{0}) \bigr)_{\omega_{0}}^{(\alpha )}e_{q,\omega }(t) \bigr] | _{\omega_{0}}^{1} \\ &\quad \quad + \int_{\omega_{0}}^{1} \biggl(\frac{1}{(1-\omega_{0})(1-q)} \biggr)^{ \alpha }e_{q,\omega }\bigl({}_{\omega_{0}}\!\Phi_{q}(t) \bigr)D_{q,\omega }\bigl(t-{}_{\omega_{0}}\!\Phi_{q}(\omega_{0})\bigr)_{\omega_{0}}^{(\alpha )}\,{d_{q,\omega }t} \\ &\quad =[\alpha ]_{q}\int_{\omega_{0}}^{1} \biggl( \frac{1}{(1-\omega_{0})(1-q)} \biggr)^{\alpha }\bigl(t-{}_{\omega_{0}}\!\Phi _{q}(\omega_{0})\bigr)_{\omega_{0}}^{(\alpha -1)}e_{q,\omega } \bigl({}_{\omega_{0}}\!\Phi_{q}(t)\bigr)\,{d_{q,\omega }t} \\ &\quad =[\alpha ]_{q}\Gamma_{q,\omega }(\alpha ), \end{aligned}$$

and

$$\begin{aligned} \Gamma_{q,\omega }(1)&= \int_{\omega_{0}}^{1} \biggl(\frac{1}{(1-\omega _{0})(1-q)} \biggr)e_{q,\omega }\bigl({}_{\omega_{0}}\!\Phi_{q}(t) \bigr)\,{d_{q,\omega }t} \\ & = \int_{\omega_{0}}^{1} \biggl(\frac{1}{(1-\omega_{0})(1-q)} \biggr) \bigl[-D_{q,\omega }(1-\omega_{0}) (1-q)e_{q,\omega }(t) \bigr] \,{d_{q,\omega }t} \\ & =-e_{q,\omega }(t) | _{\omega_{0}}^{1}=-e_{q,\omega }(1)+e_{q, \omega }( \omega_{0})=1. \end{aligned}$$

 □

Remark 3.3

In [15], authors introduced the q-gamma function as

$$\begin{aligned} \Gamma_{q}(\alpha )=\frac{1}{1-q} \int_{0}^{1} \biggl(\frac{t}{1-q} \biggr)^{\alpha -1}e_{q}(qt)\,{d_{q}t}, \end{aligned}$$

where $e_{q}(t)=\prod_{n=0}^{\infty }(1-q^{n}t)$.

To see that $\Gamma_{q,\omega }(\alpha )=\Gamma_{q}(\alpha )$, in fact

$$\begin{aligned} &\Gamma_{q,\omega }(\alpha ) \\ &= \biggl(\frac{1}{(1-\omega_{0})(1-q)} \biggr)^{\alpha } \int_{\omega _{0}}^{1}\bigl(t-{}_{\omega_{0}}\,\Phi_{q}(\omega_{0})\bigr)_{\omega_{0}}^{(\alpha -1)} e_{q,\omega }\bigl({}_{\omega_{0}}\!\Phi_{q}(t) \bigr)\,{d_{q,\omega }t} \\ &= \biggl(\frac{1}{(1-\omega_{0})(1-q)} \biggr)^{\alpha }\bigl((1-q)-\omega \bigr) \sum_{k=0}^{\infty }q^{k} \bigl({}_{\omega_{0}}\!\Phi_{q}^{k}(1)-{}_{\omega_{0}}\!\Phi_{q}(\omega_{0})\bigr)_{\omega_{0}}^{(\alpha -1)}e_{q,\omega } \bigl({}_{\omega _{0}}\!\Phi_{q}^{k+1}(1)\bigr) \\ &= \biggl(\frac{1}{(1-\omega_{0})(1-q)} \biggr)^{\alpha -1}\sum _{k=0} ^{\infty }q^{k}\bigl(q^{k} \bigr)^{\alpha -1}(1-\omega_{0})^{\alpha -1}\prod _{n=0} ^{\infty }\bigl(1-q^{n+k+1}\bigr) \\ &=\sum_{k=0}^{\infty }q^{k} \biggl( \frac{q^{k}}{1-q} \biggr)^{\alpha -1} \prod_{n=0}^{\infty } \bigl(1-q^{n+k+1}\bigr) \\ &=\frac{1}{1-q}(1-q)\sum_{k=0}^{\infty }q^{k} \biggl( \frac{q^{k}}{1-q} \biggr)^{\alpha -1}e_{q} \bigl(q^{k+1}\bigr) \\ &=\frac{1}{1-q} \int_{0}^{1} \biggl(\frac{t}{1-q} \biggr)^{\alpha -1}e _{q}(qt)\,{d_{q}t}= \Gamma_{q}(\alpha ). \end{aligned}$$

4 Concepts of fractional Hahn’s quantum integration

In the section, we recall some basic concepts of fractional Hahn’s quantum integration.

Definition 4.1

([14])

Let $\nu \geq 0$ and f be a function defined on $[c,b]$. The fractional q-integral of Riemann–Liouville type is given by $({}_{c}I_{q}^{0}f)(t)=h(t)$ and

$$\begin{aligned} \bigl({}_{c}I_{q}^{\nu }f\bigr) (t)= \frac{1}{\Gamma_{q}(\nu )} \int_{c}^{t}\bigl(t-{}_{c}\,\Phi_{q}(s)\bigr)_{c}^{(\nu -1)}h(s)\,{{}_{c}d_{q}s}, \quad \nu >0,t\in [c,b]. \end{aligned}$$

Lemma 4.2

([14])

Let $\alpha , \beta \in \mathbb{R}^{+}$, and f be a continuous function on $[c,b]$, $c\geq 0$. The Riemann–Liouville fractional q-integral has the following semi-group property:

$$ {{}_{c}I_{q}^{\beta }} {{}_{c}I_{q}^{\alpha }}f(t)={{}_{c}I_{q}^{\alpha }} {{}_{c}I_{q}^{\beta }}f(t)={}_{c}I_{q}^{\alpha +\beta }f(t). $$

(9)

Remark $c=\omega_{0}$, we have

$$\begin{aligned} \bigl({}_{\omega_{0}}I_{q}^{\nu }f\bigr) (t)= \frac{1}{\Gamma_{q}(\nu )} \int_{\omega_{0}}^{t}\bigl(t-{}_{\omega_{0}}\!\Phi_{q}(s)\bigr)_{\omega_{0}}^{( \nu -1)}f(s){_{\omega_{0}}d_{q}s}, \quad \nu >0,t\in [c,b], \end{aligned}$$

(10)

and

$$\begin{aligned} {_{\omega_{0}}I_{q}^{\beta }} {_{\omega_{0}}I_{q}^{\alpha }}f(t)={_{ \omega_{0}}I_{q}^{\alpha }} {_{\omega_{0}}I_{q}^{\beta }}f(t)={}_{\omega _{0}}I_{q}^{\alpha +\beta }f(t). \end{aligned}$$

(11)

Definition 4.3

([16])

Let $\nu \geq 0$ and f be a function defined on $[a,b]$. Hahn’s fractional integration of Riemann–Liouville type is given by $({}_{a}I_{q}^{0}f)(t)=f(t)$ and

$$\begin{aligned} \bigl({}_{a}I_{q,\omega }^{\nu }f\bigr) (t)= \frac{1}{\Gamma_{q}(\nu )} \int_{a}^{t}\bigl(t-{}_{\omega_{0}}\!\Phi_{q}(s)\bigr)_{\omega_{0}}^{(\nu -1)}f(s)\,{d_{q,\omega }s}, \quad \nu >0,t\in [a,b]. \end{aligned}$$

(12)

Lemma 4.4

For $\alpha , \beta \in \mathbb{R}^{+}$, the following is valid:

$$ \int_{\omega_{0}}^{a}\bigl(x-{}_{\omega_{0}}\!\Phi_{q}(t)\bigr)_{\omega_{0}}^{(\beta -1)}\bigl({{}_{a}I_{q,\omega }^{\alpha }}f \bigr) (t)\,{d_{q,\omega }t}=0 \quad (\omega_{0}< a< x< b). $$

Proof

Using (2), we have

$$\begin{aligned} &\bigl({}_{a}I_{q,\omega }^{\alpha }f\bigr) \bigl(aq^{n}+[n]_{q,\omega }\bigr) \\ &\quad =\frac{1}{\Gamma_{q}(\alpha )} \int_{a}^{aq^{n}+[n]_{q,\omega }}\bigl(aq ^{n}+[n]_{q,\omega }-{}_{\omega_{0}}\!\Phi_{q}(t)\bigr)_{\omega_{0}}^{(\beta -1)}f(t) \,{d_{q,\omega }t} \\ &\quad =\frac{1}{\Gamma_{q}(\alpha )}\bigl[(1-q) \bigl(aq^{n}+[n]_{q,\omega } \bigr)-\omega\bigr] \\ &\quad \quad {} \times \sum_{k=0}^{\infty }q^{k} \bigl(aq^{n}+[n]_{q,\omega }-{}_{\omega_{0}}\!\Phi_{q} \bigl(aq^{n+k}+[n+k]_{q,\omega }\bigr)\bigr)_{\omega_{0}}^{(\beta -1)}f \bigl(aq ^{n+k}+[n+k]_{q,\omega }\bigr) \\ & \quad \quad {}-\frac{1}{\Gamma_{q}(\alpha )}\bigl[(1-q)a-\omega \bigr] \\ & \quad \quad {}\times \sum_{k=0}^{\infty }q^{k} \bigl(aq^{n}+[n]_{q,\omega }-{}_{\omega_{0}}\!\Phi_{q} \bigl(aq^{k}+[k]_{q,\omega }\bigr)\bigr)_{\omega_{0}}^{(\beta -1)}f \bigl(aq^{k}+[k]_{q, \omega }\bigr) \\ &\quad =-\frac{1}{\Gamma_{q}(\alpha )}\bigl[(1-q)a-\omega \bigr] \\ & \quad \quad {}\times \sum_{k=0}^{n-1}q^{k} \bigl(aq^{n}+[n]_{q,\omega }-{}_{\omega_{0}}\!\Phi _{q} \bigl(aq^{k}+[k]_{q,\omega }\bigr)\bigr)_{\omega_{0}}^{(\beta -1)}f \bigl(aq^{k}+[k]_{q, \omega }\bigr)=0, \end{aligned}$$

since

$$\begin{aligned} &\bigl(aq^{n}+[n]_{q,\omega }-{}_{\omega_{0}}\!\Phi_{q}\bigl(aq^{k}+[k]_{q,\omega }\bigr) \bigr)_{\omega_{0}}^{(\beta -1)} \\ &\quad =\frac{\prod_{i=0}^{\infty }(aq^{n}+[n]_{q,\omega }-{}_{\omega_{0}}\!\Phi_{q}^{i}({}_{\omega_{0}}\!\Phi_{q}(aq^{k}+[k]_{q,\omega })))}{\prod_{i=0}^{\infty }(aq^{n}+[n]_{q,\omega }-{}_{\omega_{0}}\!\Phi_{q}^{i+ \nu }({}_{\omega_{0}}\!\Phi_{q}(aq^{k}+[k]_{q,\omega })))}=0 \end{aligned}$$

for $0\leq k\leq n-1$.

Then, according to the definition of $q,\omega$-integral, it follows

$$\begin{aligned} &\int_{\omega_{0}}^{a}\bigl(x-{}_{\omega_{0}}\!\Phi_{q}(t)\bigr)_{\omega_{0}}^{( \beta -1)}\bigl({{}_{a}I_{q,\omega }^{\alpha }}f \bigr) (t)\,{d_{q,\omega }t} \\ &\quad =\bigl[(1-q)a-\omega \bigr]\sum_{k=0}^{\infty }q^{k} \bigl(x-{}_{\omega_{0}}\!\Phi_{q}\bigl(aq ^{k}+[k]_{q,\omega } \bigr)\bigr)_{\omega_{0}}^{(\beta -1)}\bigl({}_{a}I_{q,\omega } ^{\alpha }f\bigr) \bigl(aq^{k}+[k]_{q,\omega }\bigr) \\ &\quad =0. \end{aligned}$$

 □

Theorem 4.5

Let $\alpha ,\beta \in \mathbb{R}^{+}$. Hahn’s fractional integration has the following semi-group property:

$$\begin{aligned} \bigl({}_{a}I_{q,\omega }^{\beta }\,{{}_{a}I_{q,\omega }^{\alpha }}f \bigr) (x)=\bigl({}_{a}I _{q,\omega }^{\alpha +\beta }f\bigr) (x)\quad (\omega_{0}< a< x< b). \end{aligned}$$

Proof

By previous Lemma 4.4, we have

$$\begin{aligned} &\bigl({}_{a}I_{q,\omega }^{\beta }\,{{}_{a}I_{q,\omega }^{\alpha }}f\bigr) (x) \\ &\quad =\frac{1}{ \Gamma_{q}(\beta )} \int_{\omega_{0}}^{x}\bigl(x-{}_{\omega_{0}}\!\Phi_{q}(t)\bigr)_{ \omega_{0}}^{(\beta -1)}\bigl({{}_{a}I_{q,\omega }^{\alpha }}f \bigr) (t)\,{d_{q, \omega }t} \\ &\quad =\frac{1}{\Gamma_{q}(\alpha )\Gamma_{q}(\beta )} \int_{\omega_{0}} ^{x}\bigl(x-{}_{\omega_{0}}\!\Phi_{q}(t)\bigr)_{\omega_{0}}^{(\beta -1)} \int_{\omega_{0}}^{t} \bigl(t-{}_{\omega_{0}}\!\Phi_{q}(u)\bigr)_{\omega_{0}}^{( \alpha -1)}f(u)\,{d_{q,\omega }u} \,{d_{q,\omega }t} \\ &\quad \quad {}-\frac{1}{\Gamma_{q}(\alpha )\Gamma_{q}(\beta )} \int_{\omega_{0}} ^{x}\bigl(x-{}_{\omega_{0}}\!\Phi_{q}(t)\bigr)_{\omega_{0}}^{(\beta -1)} \int_{\omega_{0}}^{a} \bigl(t-{}_{\omega_{0}}\!\Phi_{q}(u)\bigr)_{\omega_{0}}^{( \alpha -1)}f(u)\,{d_{q,\omega }u} \,{d_{q,\omega }t}. \end{aligned}$$

Using Lemma 4.2, we conclude that

$$\begin{aligned} \bigl({}_{a}I_{q,\omega }^{\beta }\,{{}_{a}I_{q,\omega }^{\alpha }}f \bigr) (x)&=\bigl({}_{ \omega_{0}}I_{q,\omega }^{\alpha +\beta }f\bigr) (x)- \frac{1}{\Gamma_{q}( \alpha )\Gamma_{q}(\beta )} \\ &\quad {} \times \int_{\omega_{0}}^{x}\bigl(x-{}_{\omega_{0}}\!\Phi_{q}(t)\bigr)_{\omega_{0}} ^{(\beta -1)} \int_{\omega_{0}}^{a}\bigl(t-{}_{\omega_{0}}\!\Phi_{q}(u)\bigr)_{ \omega_{0}}^{(\alpha -1)}f(u)\,{d_{q,\omega }u} \,{d_{q,\omega }t}. \end{aligned}$$

Furthermore, we can write

$$\begin{aligned} &\bigl({}_{a}I_{q,\omega }^{\beta }\,{{}_{a}I_{q,\omega }^{\alpha }}f \bigr) (x) \\ &\quad =\bigl({}_{a}I _{q,\omega }^{\alpha +\beta }f\bigr) (x)+ \frac{1}{\Gamma_{q}(\alpha + \beta )} \int_{\omega_{0}}^{a}\bigl(x-{}_{\omega_{0}}\!\Phi_{q}(t)\bigr)_{\omega_{0}} ^{(\alpha +\beta -1)}f(t)\,{d_{q,\omega }t} \\ &\quad \quad {} -\frac{1}{\Gamma_{q}(\alpha )\Gamma_{q}(\beta )}\int_{\omega_{0}}^{x}\bigl(x-{}_{\omega_{0}}\!\Phi_{q}(t)\bigr)_{\omega_{0}}^{(\beta -1)} \int_{\omega_{0}} ^{a}\bigl(t-{}_{\omega_{0}}\!\Phi_{q}(u)\bigr)_{\omega_{0}}^{(\alpha -1)}f(u)\,{d_{q,\omega }u}\,{d_{q,\omega }t}. \end{aligned}$$

Wherefrom it follows

$$\begin{aligned} \bigl({}_{a}I_{q,\omega }^{\beta }\,{{}_{a}I_{q,\omega }^{\alpha }}f \bigr) (x)=\bigl({}_{a}I _{q,\omega }^{\alpha +\beta }f\bigr) (x)+ \bigl[(1-q)a-\omega \bigr]\sum_{k=0}^{ \infty }c_{j}q^{k}f \bigl(aq^{k}+[k]_{q,\omega }\bigr), \end{aligned}$$

with

$$\begin{aligned} c_{j}&=\frac{(x-{}_{\omega_{0}}\!\Phi_{q}(aq^{k}+[k]_{q,\omega }))_{ \omega_{0}}^{(\alpha +\beta -1)}}{\Gamma_{q}(\alpha +\beta )}-\frac{(1-q)x- \omega }{\Gamma_{q}(\alpha )\Gamma_{q}(\beta )} \\ &\quad {}\times \sum_{n=0}^{\infty }q^{n} \bigl(x-{}_{\omega_{0}}\!\Phi_{q}\bigl(xq^{n}+[n]_{q, \omega } \bigr)\bigr)_{\omega_{0}}^{(\beta -1)} \bigl(xq^{n}+[n]_{q,\omega }-{}_{\omega _{0}}\!\Phi_{q}^{k+1}(a)\bigr)_{\omega_{0}}^{(\alpha -1)}. \end{aligned}$$

By using formulas (5) and (7), we get

$$\begin{aligned} c_{j}&=\bigl((x-\omega_{0}) (1-q)\bigr)^{\alpha +\beta -1} \\ & \quad {}\times \Biggl[\frac{(1-\frac{a-\omega_{0}}{x-\omega_{0}}q^{k+1})^{( \alpha +\beta -1)}}{(1-q)^{(\alpha +\beta -1)}} -\sum_{n=0}^{\infty } \frac{(1-q ^{n+1})^{(\beta -1)}(1-\frac{a-\omega_{0}}{x-\omega_{0}}q^{k-n+1})^{( \alpha -1)}}{(1-q)^{(\beta -1)}(1-q)^{(\alpha -1)}}q^{n\alpha } \Biggr]. \end{aligned}$$

Putting $\mu =q^{k}\frac{a-\omega_{0}}{x-\omega_{0}}$ into Lemma 2.1, we see that $c_{j}=0$ for all $j\in \mathbb{N}$, which completes the proof. □

Lemma 4.6

For $\alpha \in \mathbb{R}^{+}$, $\lambda \in (-1,\infty )$, the following is valid:

$$\begin{aligned} _{a}I_{q,\omega }^{\alpha } \bigl((x-a)_{\omega_{0}}^{(\lambda )}\bigr)=\frac{ \Gamma_{q}(\lambda +1)}{\Gamma_{q}(\alpha +\lambda +1)}(x-a)_{\omega _{0}}^{(\alpha +\lambda )}\quad (\omega_{0}< a< x< b). \end{aligned}$$

(13)

Proof

For $\lambda \neq 0$, according to Definition 4.3, we have

$$\begin{aligned} {}_{a}I_{q,\omega }^{\alpha }\bigl((x-a)_{\omega_{0}}^{(\lambda )} \bigr)&=\frac{1}{ \Gamma_{q}(\alpha )} \biggl[ \int_{\omega_{0}}^{x}\bigl(x-{}_{\omega_{0}}\!\Phi _{q}(t)\bigr)_{\omega_{0}}^{(\alpha -1)}(t-a)_{\omega_{0}}^{(\lambda )} \,{d_{q,\omega }t} \\ &\quad {} - \int_{\omega_{0}}^{a}\bigl(x-{}_{\omega_{0}}\!\Phi_{q}(t)\bigr)_{\omega_{0}}^{( \alpha -1)}(t-a)_{\omega_{0}}^{(\lambda )}\,{d_{q,\omega }t} \biggr]. \end{aligned}$$

Also, the following is valid:

$$\begin{aligned} & \int_{\omega_{0}}^{a}\bigl(x-{}_{\omega_{0}}\!\Phi_{q}(t)\bigr)_{\omega_{0}}^{( \alpha -1)}(t-a)_{\omega_{0}}^{(\lambda )}\,{d_{q,\omega }t} \\ &\quad =\bigl[(1-q)a-\omega \bigr]\sum_{k=0}^{\infty }q^{k} \bigl(x-{}_{\omega_{0}}\!\Phi_{q}\bigl(aq ^{k}+[k]_{q,\omega } \bigr)\bigr)_{\omega_{0}}^{(\alpha -1)}\bigl(aq^{k}+[k]_{q, \omega }-a \bigr)_{\omega_{0}}^{(\lambda )} \\ &\quad =\bigl[(1-q)a-\omega \bigr]\sum_{k=0}^{\infty }q^{k} \bigl(x-{}_{\omega_{0}}\!\Phi_{q} ^{k+1}(a) \bigr)_{\omega_{0}}^{(\alpha -1)}\bigl({}_{\omega_{0}}\!\Phi_{q}^{k}(a)-a \bigr)_{ \omega_{0}}^{(\lambda )}=0. \end{aligned}$$

Therefrom, using Lemma 2.1, we get

$$\begin{aligned} & \int_{\omega_{0}}^{x}\bigl(x-{}_{\omega_{0}}\!\Phi_{q}(t)\bigr)_{\omega_{0}}^{( \alpha -1)}(t-a)_{\omega_{0}}^{(\lambda )}\,{d_{q,\omega }t} \\ &\quad =\bigl[(1-q)x-\omega \bigr]\sum_{k=0}^{\infty }q^{k} \bigl(x-{}_{\omega_{0}}\!\Phi_{q} ^{k+1}(x) \bigr)_{\omega_{0}}^{(\alpha -1)}\bigl({}_{\omega_{0}}\!\Phi_{q}^{k}(x)-a \bigr)_{ \omega_{0}}^{(\lambda )} \\ &\quad =(1-q) (x-\omega_{0})^{\alpha +\lambda }(1-q)^{(\alpha -1)}(1-q)^{( \lambda )} \sum_{k=0}^{\infty } \frac{(1-q^{k+1})^{(\alpha -1)}(1-q ^{-k}\frac{a-\omega_{0}}{x-\omega_{0}})^{(\lambda )}}{(1-q)^{(\alpha -1)}(1-q)^{(\lambda )}}q^{(\lambda +1)k} \\ &\quad =(1-q) (x-\omega_{0})^{\alpha +\lambda }(1-q)^{(\alpha -1)}(1-q)^{( \lambda )} \frac{(1-\frac{a-\omega_{0}}{x-\omega_{0}})^{(\alpha + \lambda )}}{(1-q)^{(\alpha +\lambda )}} \\ &\quad =(1-q)\frac{(1-q)^{(\alpha -1)}(1-q)^{(\lambda )}}{(1-q)^{(\alpha + \lambda )}}(x-a)_{\omega_{0}}^{(\alpha +\lambda )}. \end{aligned}$$

Using (5), we obtain the required formula (13).

Particularly, for $\lambda =0$, since the $q,\omega$-derivative over the variable t is

$$\begin{aligned} {}_{t}D_{q,\omega }(x-t)_{\omega_{0}}^{(\alpha )}=-[\alpha ]_{q}\bigl(x-{}_{ \omega_{0}}\!\Phi_{q}(t) \bigr)_{\omega_{0}}^{(\alpha -1)} \end{aligned}$$

(14)

and using Lemma 1.3, we have

$$\begin{aligned} \bigl({}_{a}I_{q,\omega }^{\alpha }1\bigr) (x)= \frac{1}{\Gamma_{q}(\alpha )} \int _{a}^{x}\bigl(x-{}_{\omega_{0}}\!\Phi_{q}(t)\bigr)_{\omega_{0}}^{(\alpha -1)}\,{d_{q, \omega }t}= \frac{1}{\Gamma_{q}(\alpha +1)}(x-a)_{\omega_{0}}^{(\alpha )}. \end{aligned}$$

 □

5 The fractional $q,\omega$-derivative of Riemann–Liouville type

We define the fractional $q,\omega$-derivative of Riemann–Liouville type by

$$\begin{aligned} \bigl({}_{a}D_{q,\omega }^{\alpha }f\bigr) (x)=\textstyle\begin{cases} ({}_{a}I_{q,\omega }^{-\alpha }f)(x),& \alpha < 0, \\ f(x),& \alpha =0, \\ (D_{q,\omega }^{\lceil \alpha \rceil }\,{{}_{a}I_{q,\omega }^{\lceil \alpha \rceil -\alpha }}f)(x),& \alpha >0, \end{cases}\displaystyle \end{aligned}$$

(15)

where $\lceil \alpha \rceil$ denotes the smallest integer greater or equal to α.

Theorem 5.1

For $\alpha \in \mathbb{R}^{+}$, $\lambda \in (-1,\infty )$, the following is valid:

$$\begin{aligned} _{a}D_{q,\omega }^{\alpha }\bigl((x-a)_{\omega_{0}}^{(\lambda )} \bigr)=\frac{ \Gamma_{q}(\lambda +1)}{\Gamma_{q}(\lambda -\alpha +1)}(x-a)_{\omega _{0}}^{(\lambda -\alpha )},\quad ( \omega_{0}< a< x< b). \end{aligned}$$

Proof

For $\lambda \neq 0$, according to (15) and Lemma 4.6, we have

$$\begin{aligned} _{a}D_{q,\omega }^{\alpha }\bigl((x-a)_{\omega_{0}}^{(\lambda )} \bigr)&=D_{q, \omega }^{\lceil \alpha \rceil }\,{{}_{a}I_{q,\omega }^{\lceil \alpha \rceil -\alpha }}(x-a)_{\omega_{0}}^{(\lambda )} \\ & =D_{q,\omega }^{\lceil \alpha \rceil } \biggl(\frac{\Gamma_{q}(\lambda +1)}{\Gamma_{q}(\lambda +\lceil \alpha \rceil -\alpha +1)}(x-a)_{ \omega_{0}}^{(\lambda +\lceil \alpha \rceil -\alpha )} \biggr) \\ & =\frac{\Gamma_{q}(\lambda +1)}{\Gamma_{q}(\lambda +\lceil \alpha \rceil -\alpha +1)}D_{q,\omega }^{\lceil \alpha \rceil -1}\bigl[\lambda + \lceil \alpha \rceil -\alpha \bigr]_{q} (x-a)_{\omega_{0}}^{(\lambda + \lceil \alpha \rceil -\alpha -1)} \\ &=\frac{\Gamma_{q}(\lambda +1)}{\Gamma_{q}(\lambda +\lceil \alpha \rceil -\alpha +1)}\bigl[\lambda +\lceil \alpha \rceil -\alpha \bigr]_{q}\bigl[ \lambda +\lceil \alpha \rceil -\alpha -1 \bigr]_{q}\cdots \\ & \quad {} \times \bigl[\lambda +\lceil \alpha \rceil -\alpha -\bigl(\lceil \alpha \rceil -1\bigr)\bigr]_{q}(x-a)_{\omega_{0}}^{(\lambda -\alpha )} \\ & =\frac{\Gamma_{q}(\lambda +1)}{\Gamma_{q}(\lambda -\alpha +1)}(x-a)_{ \omega_{0}}^{(\lambda -\alpha )}. \end{aligned}$$

 □

Lemma 5.2

For $\alpha \in \mathbb{R}\setminus \mathbb{N}_{0}$, the following is valid:

$$\begin{aligned} \bigl(D_{q,\omega }\,{{}_{a}D_{q,\omega }^{\alpha }}f\bigr) (x)=\bigl({}_{a}D_{q,\omega } ^{\alpha +1}f\bigr) (x),\quad ( \omega_{0}< a< x< b). \end{aligned}$$

(16)

Proof

We will consider three cases. For $\alpha \leq -1$, according to Theorem 4.5, we have

$$\begin{aligned} \bigl(D_{q,\omega }\,{{}_{a}D_{q,\omega }^{\alpha }}f\bigr) (x)&=\bigl(D_{q,\omega }\,{{}_{a}I_{q,\omega }^{-\alpha }}f \bigr) (x)=\bigl(D_{q,\omega }\,{{}_{a}I_{q,\omega } ^{1-\alpha -1}}f\bigr) (x) \\ & =\bigl(D_{q,\omega }{I_{q,\omega }}\,{{}_{a}I_{q,\omega }^{-\alpha -1}}f \bigr) (x)=\bigl({}_{a}I _{q,\omega }^{-(\alpha +1)}f\bigr) (x) \\ & =\bigl({}_{a}D_{q,\omega }^{\alpha +1}f\bigr) (x). \end{aligned}$$

In the case $-1<\alpha <0$, i.e., $0<\alpha +1<1$, we obtain

$$\begin{aligned} \bigl(D_{q,\omega }\,{{}_{a}D_{q,\omega }^{\alpha }}f\bigr) (x)=\bigl(D_{q,\omega }\,{{}_{a}I_{q,\omega }^{-\alpha }}f \bigr) (x)=\bigl(D_{q,\omega }\,{{}_{a}I_{q,\omega } ^{1-\alpha -1}}f\bigr) (x)=\bigl({}_{a}D_{q,\omega }^{\alpha +1}f \bigr) (x). \end{aligned}$$

For $\alpha >0$, we get

$$\begin{aligned} \bigl(D_{q,\omega }\,{{}_{a}D_{q,\omega }^{\alpha }}f\bigr) (x)&=\bigl(D_{q,\omega } {D_{q,\omega }^{\lceil \alpha \rceil }\,{{}_{a}I_{q,\omega }^{\lceil \alpha \rceil -\alpha }}}f \bigr) (x) \\ & =\bigl({D_{q,\omega }^{\lceil \alpha \rceil +1}\,{{}_{a}I_{q,\omega }^{\lceil \alpha \rceil -\alpha }}}f \bigr) (x)=\bigl({}_{a}D_{q,\omega }^{\alpha +1}f\bigr) (x). \end{aligned}$$

 □

Theorem 5.3

For $\alpha \in \mathbb{R}\setminus \mathbb{N}_{0}$, the following is valid:

$$\begin{aligned} \bigl(D_{q,\omega }\,{{}_{a}D_{q,\omega }^{\alpha }}f\bigr) (x)-\bigl({}_{a}D_{q,\omega } ^{\alpha }{D_{q,\omega }}f \bigr) (x)=\frac{f(a)}{\Gamma_{q}(-\alpha )}(x-a)_{ \omega_{0}}^{(-\alpha -1)}\quad ( \omega_{0}< a< x< b). \end{aligned}$$

Proof

We will use Lemma 1.3, Theorem 4.5, and Lemma 4.6 to prove the statement. Let us consider two cases. If $\alpha <0$, then

$$\begin{aligned} \bigl(D_{q,\omega }\,{{}_{a}D_{q,\omega }^{\alpha }}f\bigr) (x)&=\bigl(D_{q,\omega }\,{{}_{a}I_{q,\omega }^{-\alpha }}f \bigr) (x)=D_{q,\omega }\,{{}_{a}I_{q,\omega } ^{-\alpha }} \bigl(({}_{a}I_{q,\omega }{D_{q,\omega }}f) (x)+f(a)\bigr) \\ & =\bigl(D_{q,\omega }\,{{}_{a}I_{q,\omega }^{-\alpha }}\,{{}_{a}I_{q,\omega }} {D _{q,\omega }}f\bigr) (x)+f(a) \bigl(D_{q,\omega }\,{{}_{a}I_{q,\omega }^{-\alpha }}1\bigr) (x) \\ & =\bigl(D_{q,\omega }\,{{}_{a}I_{q,\omega }^{-\alpha +1}} {D_{q,\omega }}f\bigr) (x)+f(a)D _{q,\omega }\frac{(x-a)_{\omega_{0}}^{(-\alpha )}}{\Gamma_{q}(-\alpha +1)} \\ & =\bigl(D_{q,\omega }\,{{}_{a}I_{q,\omega }}\,{{}_{a}I_{q,\omega }^{-\alpha }} {D _{q,\omega }}f\bigr) (x)+f(a)\frac{[-\alpha ]_{q}(x-a)_{\omega_{0}}^{(- \alpha -1)}}{\Gamma_{q}(-\alpha +1)} \\ &=\bigl({}_{a}D_{q,\omega }^{\alpha }{D_{q,\omega }}f\bigr) (x)+\frac{f(a)}{\Gamma _{q}(-\alpha )}(x-a)_{\omega_{0}}^{(-\alpha -1)}. \end{aligned}$$

If $\alpha >0$, there exists $l\in \mathbb{N}_{0}$ such that $\alpha \in (l,l+1)$. Then, applying a similar procedure, we get

$$\begin{aligned} \bigl(D_{q,\omega }\,{{}_{a}D_{q,\omega }^{\alpha }}f\bigr) (x)&=\bigl(D_{q,\omega } {D_{q,\omega }^{l+1}}\,{{}_{a}I_{q,\omega }^{l+1-\alpha }}f\bigr) (x) \\ &=D_{q,\omega }^{l+2}\,{{}_{a}I_{q,\omega }^{l+1-\alpha }} \bigl(({}_{a}I_{q, \omega }{D_{q,\omega }}f) (x)+f(a)\bigr) \\ & =\bigl(D_{q,\omega }^{l+1}{D_{q,\omega }}\,{{}_{a}I_{q,\omega }}\,{{}_{a}I_{q, \omega }^{l+1-\alpha }} {D_{q,\omega }}f\bigr) (x) \\ &\quad {} +\frac{f(a)}{\Gamma_{q}(l+2-\alpha )}D_{q,\omega }^{l+1} \bigl((x-a)_{\omega _{0}}^{(l+1-\alpha )}\bigr) \\ & =\bigl({}_{a}D_{q,\omega }^{\alpha }{D_{q,\omega }}f \bigr) (x)+\frac{f(a)}{\Gamma_{q}(-\alpha )}(x-a)_{\omega_{0}}^{(-\alpha -1)}. \end{aligned}$$

 □

Lemma 5.4

Let $f(x)$ be a function defined on an interval $(\omega_{0},b)$ and $\alpha \in \mathbb{R}^{+}$. Then the following is valid:

$$\begin{aligned} \bigl({}_{a}D_{q,\omega }^{\alpha }\,{{}_{a}I_{q,\omega }^{\alpha }}f \bigr) (x)=f(x)\quad (\omega_{0}< a< x< b). \end{aligned}$$

Proof

For $\alpha >0$, we have

$$\begin{aligned} \bigl({}_{a}D_{q,\omega }^{\alpha }\,{{}_{a}I_{q,\omega }^{\alpha }}f \bigr) (x)&=\bigl(D _{q,\omega }^{\lceil \alpha \rceil }\,{{}_{a}I_{q,\omega }^{\lceil \alpha \rceil -\alpha }}\,{{}_{a}I_{q,\omega }^{\alpha }}f\bigr) (x) = \bigl(D_{q,\omega } ^{\lceil \alpha \rceil }\,{{}_{a}I_{q,\omega }^{\lceil \alpha \rceil - \alpha +\alpha }}f \bigr) (x) \\ &=\bigl({}_{a}D_{q,\omega }^{\lceil \alpha \rceil }\,{{}_{a}I_{q,\omega }^{\lceil \alpha \rceil }}f \bigr) (x)=f(x). \end{aligned}$$

 □

Lemma 5.5

Let $\alpha \in (0,1)$, then

$$\begin{aligned} \bigl({}_{a}I_{q,\omega }^{\alpha }\,{{}_{a}D_{q,\omega }^{\alpha }}f \bigr) (x)=f(x)+K(a) (x-a)_{ \omega_{0}}^{(\alpha -1)}\quad (\omega_{0}< a< x< b), \end{aligned}$$

where $K(a)$ does not depend on x.

Proof

Let $A(x)=({}_{a}I_{q,\omega }^{\alpha }{{}_{a}D_{q,\omega }^{\alpha }}f)(x)-f(x)$. Applying ${}_{a}D_{q,\omega }^{\alpha }$ to both sides of the above expression and using Lemma 5.4, we get

$$\begin{aligned} \bigl({}_{a}D_{q,\omega }^{\alpha }A\bigr) (x)&= \bigl({}_{a}D_{q,\omega }^{\alpha }\,{{}_{a}I_{q,\omega }^{\alpha }} {{}_{a}D_{q,\omega }^{\alpha }}f\bigr) (x)- \bigl({}{}_{a}D_{q,\omega }^{\alpha }f\bigr) (x) \\ &=\bigl(\bigl({}_{a}D_{q,\omega }^{\alpha }\,{{}_{a}I_{q,\omega }^{\alpha }} \bigr){}_{a}D_{q,\omega }^{\alpha }f\bigr) (x)-\bigl({}_{a}D_{q,\omega }^{\alpha }f\bigr) (x)=0. \end{aligned}$$

On the other hand, according to Lemma 4.6, we obtain

$$\begin{aligned} &{}_{a}D_{q,\omega }^{\alpha }(x-a)_{\omega_{0}}^{(\alpha -1)}=D_{q, \omega }\,{{}_{a}I_{q,\omega }^{1-\alpha }}(x-a)_{\omega_{0}}^{(\alpha -1)}= \Gamma_{q}(\alpha ) (D_{q,\omega }1) (x)=0. \end{aligned}$$

Hence, we conclude that $A(x)$ is a function of the form

$$\begin{aligned} A(x)=K(a) (x-a)_{\omega_{0}}^{(\alpha -1)}. \end{aligned}$$

 □

Theorem 5.6

Let $\alpha \in (N-1,N]$. Then, for some constants $c_{i}\in \mathbb{R},i=1,2,\ldots ,N$, the following equality holds:

$$\begin{aligned} \bigl({}_{a}I_{q,\omega }^{\alpha }\,{{}_{a}D_{q,\omega }^{\alpha }}f \bigr) (x)=f(x)+c _{1}(x-a)_{\omega_{0}}^{(\alpha -1)}+c_{2}(x-a)_{\omega_{0}}^{(\alpha -2)}+ \cdots +c_{N}(x-a)_{\omega_{0}}^{(\alpha -N)}. \end{aligned}$$

Proof

By Lemma 5.5, we have

$$\begin{aligned} \bigl({}_{a}I_{q,\omega }^{\alpha }\,{{}_{a}D_{q,\omega }^{\alpha }}f \bigr) (x)&=\bigl({}_{a}I _{q,\omega }^{\alpha }{D_{q,\omega }^{N}}\,{{}_{a}I_{q,\omega }^{N-\alpha }}f\bigr) (x) \\ & =\bigl({}_{a}I_{q,\omega }^{\alpha -1}\,{{}_{a}I_{q,\omega }} {D_{q,\omega }} {D_{q,\omega }^{N-1}}\,{{}_{a}I_{q,\omega }^{N-\alpha }}f \bigr) (x) \\ & =\bigl({}_{a}I_{q,\omega }^{\alpha -1}{D_{q,\omega }^{N-1}}\,{{}_{a}I_{q,\omega }^{N-\alpha }}f\bigr) (x)-D_{q,\omega }^{N-1}\,{{}_{a}I_{q,\omega }^{N-\alpha }}f(a) \bigl({}_{a}I _{q,\omega }^{\alpha -1}1\bigr) (x) \\ & =\bigl({}_{a}I_{q,\omega }^{\alpha -1}{D_{q,\omega }^{N-1}}\,{{}_{a}I_{q,\omega }^{N-\alpha }}f\bigr) (x)-\frac{D_{q,\omega }^{N-1}\,{{}_{a}I_{q,\omega }^{N- \alpha }}f(a)}{\Gamma_{q}(\alpha )}(x-a)_{\omega_{0}}^{(\alpha -1)} \\ & =\bigl({}_{a}I_{q,\omega }^{\alpha -2}{D_{q,\omega }^{N-2}}\,{{}_{a}I_{q,\omega }^{N-\alpha }}f\bigr) (x)-\frac{D_{q,\omega }^{N-2}\,{{}_{a}I_{q,\omega }^{N- \alpha }}f(a)}{\Gamma_{q}(\alpha -1)}(x-a)_{\omega_{0}}^{(\alpha -2)} \\ &\quad {} -\frac{D_{q,\omega }^{N-1}\,{{}_{a}I_{q,\omega }^{N-\alpha }}f(a)}{\Gamma _{q}(\alpha )}(x-a)_{\omega_{0}}^{(\alpha -1)} \\ & =\cdots \\ & =\bigl({}_{a}I_{q,\omega }^{\alpha -N+1}\,{{}_{a}D_{q,\omega }^{\alpha -N+1}}f \bigr) (x)-\frac{D _{q,\omega }\,{{}_{a}I_{q,\omega }^{N-\alpha }}f(a)}{\Gamma_{q}(\alpha -N+2)}(x-a)_{ \omega_{0}}^{(\alpha -N+1)} \\ & \quad {}-\cdots -\frac{D_{q,\omega }^{N-1}\,{{}_{a}I_{q,\omega }^{N- \alpha }}f(a)}{\Gamma_{q}(\alpha )}(x-a)_{\omega_{0}}^{(\alpha -1)} \\ & =f(x)+c_{1}(x-a)_{\omega_{0}}^{(\alpha -1)}+\cdots +c_{N}(x-a)_{\omega_{0}}^{(\alpha -N)}. \end{aligned}$$

 □

Lemma 5.7

$$\begin{aligned} D_{q,\omega } \int_{\omega_{0}}^{t}f(t,s)\,d_{q,\omega }s= \int_{\omega _{0}}^{t}D_{q,\omega }f(t,s)\,d_{q,\omega }s+f(qt+ \omega ,t). \end{aligned}$$

Proof

Using (2) and Definition 1.1, we have

$$\begin{aligned} &D_{q,\omega } \int_{\omega_{0}}^{t}f(t,s)\,d_{q,\omega }s \\ &\quad =D_{q,\omega }\Biggl[\bigl(t(1-q)-\omega \bigr)\sum _{k=0}^{\infty }q^{k}f\bigl(t,q^{k}t+[k]_{q, \omega } \bigr)\Biggr] \\ &\quad =\frac{1}{qt+\omega -t} \Biggl[\bigl((qt+\omega ) (1-q)-\omega \bigr)\sum _{k=0} ^{\infty }q^{k}f\bigl(qt+\omega ,(qt+\omega )q^{k}+[k]_{q,\omega }\bigr) \\ &\quad \quad {}-\bigl(t(1-q)-\omega \bigr)\sum_{k=0}^{\infty }q^{k}f \bigl(t,tq^{k}+[k]_{q,\omega }\bigr) \Biggr] \\ &\quad =-\sum_{k=0}^{\infty }q^{k+1}f \bigl(qt+\omega ,(qt+\omega )q^{k}+[k]_{q, \omega }\bigr)+\sum _{k=0}^{\infty }q^{k}f \bigl(t,tq^{k}+[k]_{q,\omega }\bigr) \\ &\quad =-\sum_{k=0}^{\infty }q^{k}f \bigl(qt+\omega ,(qt+\omega )q^{k-1}+[k-1]_{q, \omega }\bigr)+\sum _{k=0}^{\infty }q^{k}f \bigl(t,tq^{k}+[k]_{q,\omega }\bigr) \\ &\quad \quad {}+f(qt+\omega ,t) \\ &\quad =\bigl(t(1-q)-\omega \bigr) \sum_{k=0}^{\infty }q^{k} \frac{f(qt+\omega ,tq ^{k}+[k]_{q,\omega })-f(t,tq^{k}+[k]_{q,\omega })}{qt+\omega -t} \\ &\quad \quad {} +f(qt+\omega ,t) \\ &\quad = \int_{\omega_{0}}^{t}D_{q,\omega }f(t,s)\,d_{q,\omega }s+f(qt+ \omega ,t). \end{aligned}$$

 □

Theorem 5.8

If $f(t)$ is defined and finite, then for $\nu >0$ with $N-1<\nu \leq N$,

$$\begin{aligned} \bigl(D_{q,\omega }^{\nu }f\bigr) (t)=\frac{1}{\Gamma_{q}(-\nu )} \int_{a}^{t}\bigl(t-{}_{\omega_{0}}\!\Phi_{q}(s)\bigr)_{\omega_{0}}^{(-\nu -1)}f(s)\,d_{q,\omega }s. \end{aligned}$$

Proof

Using Definition 4.3 and Lemma 5.7, we have

$$\begin{aligned} &\bigl(D_{q,\omega }^{\nu }f\bigr) (t)=D_{q,\omega }^{N}\,{{}_{a}I_{q,\omega }^{N- \nu }}f(t) \\ &\quad =D_{q,\omega }^{N} \biggl(\frac{1}{\Gamma_{q}(N-\nu )} \int_{a}^{t}\bigl(t-{}_{ \omega_{0}}\!\Phi_{q}(s)\bigr)_{\omega_{0}}^{(N-\nu -1)}f(s)\,d_{q,\omega }s \biggr) \\ & \quad =D_{q,\omega }^{N-1}{D_{q,\omega }} \biggl( \frac{1}{\Gamma_{q}(N-\nu )} \int_{a}^{t}\bigl(t-{}_{\omega_{0}}\!\Phi_{q}(s)\bigr)_{\omega_{0}}^{(N- \nu -1)}f(s)\,d_{q,\omega }s \biggr) \\ & \quad =D_{q,\omega }^{N-1} \biggl[\frac{1}{\Gamma_{q}(N-\nu )} \int_{a}^{t}D _{q,\omega }\bigl(t-{}_{\omega_{0}}\!\Phi_{q}(s)\bigr)_{\omega_{0}}^{(N-\nu -1)}f(s)\,d_{q,\omega }s \\ & \quad \quad {} +\bigl(qt+\omega -{}_{\omega_{0}}\!\Phi_{q}(t) \bigr)_{\omega_{0}}^{(N-\nu -1)}f(t) \biggr] \\ &\quad =D_{q,\omega }^{N-1}\frac{[N-\nu -1]_{q}}{\Gamma_{q}(N-\nu )} \int_{a} ^{t}\bigl(t-{}_{\omega_{0}}\!\Phi_{q}(s)\bigr)_{\omega_{0}}^{(N-\nu -2)}f(s)\,d_{q,\omega }s\\ &\quad =D_{q,\omega }^{N-1}\frac{1}{\Gamma_{q}(N-\nu -1)}\int_{a}^{t}\bigl(t-{}_{ \omega_{0}}\!\Phi_{q}(s)\bigr)_{\omega_{0}}^{(N-\nu -2)}f(s)\,d_{q,\omega }s\\ &\quad =\cdots \\ & \quad =\frac{1}{\Gamma_{q}(-\nu )}\int_{a}^{t}\bigl(t-{}_{\omega_{0}}\!\Phi_{q}(s)\bigr)_{\omega_{0}}^{(-\nu -1)}f(s)\,d_{q,\omega }s. \end{aligned}$$

 □

6 The fractional $q,\omega$-derivative of Caputo type

If we change the order of operators, we can introduce another type of fractional $q,\omega$-derivative.

The fractional $q,\omega$-derivative of Caputo type is

$$\begin{aligned} \bigl({}_{a}^{c}D_{q,\omega }^{\alpha }f \bigr) (x)= \textstyle\begin{cases} ({}_{a}I_{q,\omega }^{-\alpha }f)(x),& \alpha < 0, \\ f(x),& \alpha =0, \\ ({}_{a}I_{q,\omega }^{\lceil \alpha \rceil -\alpha } D_{q,\omega }^{\lceil \alpha \rceil }f)(x),& \alpha >0, \end{cases}\displaystyle \end{aligned}$$

(17)

where $\lceil \alpha \rceil$ denotes the smallest integer greater or equal to α.

Theorem 6.1

For $\alpha \in \mathbb{R}\setminus \mathbb{N}_{0}$ and $\omega_{0}< a< x< b$, the following is valid:

$$\begin{aligned} \bigl({}_{a}^{c}D_{q,\omega }^{\alpha +1}f \bigr) (x)-\bigl({}_{a}^{c}D_{q,\omega }^{\alpha }{D_{q,\omega }}f \bigr) (x)= \textstyle\begin{cases} \frac{f(a)}{\Gamma_{q}(-\alpha )}(x-a)_{\omega_{0}}^{(-\alpha -1)},& \alpha \leq -1, \\ 0,& \alpha >-1. \end{cases}\displaystyle \end{aligned}$$

(18)

Proof

Clearly, (18) holds for $\alpha =-1$. Next, we will consider three cases.

Case 1. $\alpha <-1$, according to (17) and (3), we have

$$\begin{aligned} \bigl({}_{a}^{c}D_{q,\omega }^{\alpha +1}f\bigr) (x)&=\bigl({}_{a}I_{q,\omega }^{-\alpha -1}f\bigr) (x)={}_{a}I_{q,\omega }^{-\alpha -1}\bigl(({}_{a}I_{q,\omega }{D_{q, \omega }}f) (x)+f(a)\bigr) \\ & =\bigl({}_{a}I_{q,\omega }^{-\alpha }{D_{q,\omega }}f \bigr) (x)+f(a) \bigl({}_{a}I_{q, \omega }^{-\alpha -1}1\bigr) (x) \\ & =\bigl({}_{a}^{c}D_{q,\omega }^{\alpha }{D_{q,\omega }}f \bigr) (x)+\frac{f(a)}{ \Gamma_{q}(-\alpha )}(x-a)_{\omega_{0}}^{(-\alpha -1)}. \end{aligned}$$

Case 2. $-1<\alpha \leq 0$, we obtain

$$\begin{aligned} \bigl({}_{a}^{c}D_{q,\omega }^{\alpha +1}f\bigr) (x)&=\bigl({}_{a}I_{q,\omega }^{1-( \alpha +1)}{D_{q,\omega }}f \bigr) (x) \\ &=\bigl({}_{a}I_{q,\omega }^{-\alpha }{D_{q, \omega }}f \bigr) (x)=\bigl({}_{a}^{c}D_{q,\omega }^{\alpha }{D_{q,\omega }}f \bigr) (x). \end{aligned}$$

Case 3. $\alpha >0$, we assume $\alpha =n+\varepsilon$, $n\in \mathbb{N}_{0}$, $0<\varepsilon <1$, then $\alpha +1\in (n+1,n+2)$, so we obtain

$$\begin{aligned} \bigl({}_{a}^{c}D_{q,\omega }^{\alpha +1}f\bigr) (x)&=\bigl({}_{a}I_{q,\omega }^{1- \varepsilon }{D_{q,\omega }^{n+2}}f \bigr) (x) =\bigl({}_{a}I_{q,\omega }^{1-\varepsilon }{D_{q,\omega }^{n+1}} {D_{q,\omega }}f\bigr) (x) =\bigl({}_{a}^{c}D_{q,\omega }^{\alpha }{D_{q,\omega }}f \bigr) (x). \end{aligned}$$

 □

Theorem 6.2

For $\alpha \in \mathbb{R}\setminus \mathbb{N}_{0}$ and $\omega_{0}< a< x< b$, the following is valid:

$$\begin{aligned} \bigl(D_{q,\omega }\,{{}_{a}^{c}D_{q,\omega }^{\alpha }}f \bigr) (x)-\bigl({}_{a}^{c}D_{q, \omega }^{\alpha +1}f \bigr) (x)= \textstyle\begin{cases} 0,& \alpha\leq -1, \\ \frac{D_{q,\omega }^{\lceil \alpha \rceil }f(a)}{\Gamma_{q}(\lceil\alpha \rceil -\alpha )}(x-a)_{\omega_{0}}^{(-\alpha -1)},& \alpha >-1. \end{cases}\displaystyle \end{aligned}$$

(19)

Proof

We will consider two cases.

Case 1. $\alpha <0$, using Lemma 1.3, Lemma 4.6, and (17), we obtain

$$\begin{aligned} \bigl(D_{q,\omega }\,{{}_{a}^{c}D_{q,\omega }^{\alpha }}f \bigr) (x)&=\bigl(D_{q,\omega }\,{{}_{a}I_{q,\omega }^{-\alpha }}f \bigr) (x)=D_{q,\omega }\,{{}_{a}I_{q,\omega } ^{-\alpha }} \bigl(({}_{a}I_{q,\omega }{D_{q,\omega }}f) (x)+f(a)\bigr) \\ &=\bigl(D_{q,\omega }\,{{}_{a}I_{q,\omega }}\,{{}_{a}I_{q,\omega }^{-\alpha }} {D _{q,\omega }}f\bigr) (x)+f(a)\frac{[-\alpha ]_{q}(x-a)_{\omega_{0}}^{(- \alpha -1)}}{\Gamma_{q}(-\alpha +1)} \\ &=\bigl({}_{a}^{c}D_{q,\omega }^{\alpha +1}f\bigr) (x)+\frac{f(a)}{\Gamma_{q}(- \alpha )}(x-a)_{\omega_{0}}^{(-\alpha -1)}. \end{aligned}$$

By Theorem 5.1, the required equalities are valid both for $\alpha\leq -1$ and $-1<\alpha <0$.

Case 2. $\alpha >0$, we assume $\alpha =n+\varepsilon$, $n\in \mathbb{N}_{0}$, $0<\varepsilon <1$, then $\alpha +1\in (n+1,n+2)$, by Lemma 1.3, Lemma 4.6, and (17), we obtain

$$\begin{aligned} \bigl(D_{q,\omega }\,{{}_{a}^{c}D_{q,\omega }^{\alpha }}f \bigr) (x)&=\bigl(D_{q,\omega }\,{{}_{a}I_{q,\omega }^{1-\varepsilon }} {D_{q,\omega }^{n+1}}f\bigr) (x) \\ & =\bigl(D_{q,\omega }\,{{}_{a}I_{q,\omega }^{2-\varepsilon }} {D_{q,\omega } ^{n+2}}f\bigr) (x)+D_{q,\omega }^{n+1}f(a) \bigl(D_{q,\omega }\,{{}_{a}I_{q,\omega } ^{1-\varepsilon }}1\bigr) (x) \\ & =\bigl({}_{a}^{c}D_{q,\omega }^{\alpha +1}f\bigr) (x)+\frac{D_{q,\omega }^{n+1}f(a)}{ \Gamma_{q}(n+1-\alpha )}(x-a)_{\omega_{0}}^{(n-\alpha )}. \end{aligned}$$

 □

Theorem 6.3

Let $\alpha \in (N-1,N]$. Then, for some constants $c_{i}\in \mathbb{R},i=1,2,\ldots ,N-1$, the following equality holds:

$$\begin{aligned} \bigl({}_{a}I_{q,\omega }^{\alpha }\,{{}_{a}^{c}D_{q,\omega }^{\alpha }}f \bigr) (x)=f(x)+c _{0}+c_{1}(x-a)_{\omega_{0}}^{(1)}+c_{2}(x-a)_{\omega_{0}}^{(2)}+ \cdots +c_{N-1}(x-a)_{\omega_{0}}^{(N-1)}. \end{aligned}$$

Proof

By (17) and Lemma 1.3, Lemma 4.6, we have

$$\begin{aligned} \bigl({}_{a}I_{q,\omega }^{\alpha }\,{{}_{a}^{c}D_{q,\omega }^{\alpha }}f \bigr) (x)&=\bigl({}_{a}I _{q,\omega }^{\alpha }\,{{}_{a}I_{q,\omega }^{N-\alpha }} {D_{q,\omega } ^{N}}f\bigr) (x)=\bigl({}_{a}I_{q,\omega }^{N}{D_{q,\omega }^{N}}f \bigr) (x) \\ & ={}_{a}I_{q,\omega }^{N-1}\bigl(\bigl({D_{q,\omega }^{N-1}}f \bigr) (x)-\bigl({D_{q,\omega } ^{N-1}}f\bigr) (a)\bigr) \\ & =\bigl({}_{a}I_{q,\omega }^{N-1}{D_{q,\omega }^{N-1}}f \bigr) (x)-\frac{D_{q,\omega }^{N-1}f(a)}{\Gamma_{q}(N)}(x-a)_{\omega_{0}}^{(N-1)} \\ & =\bigl({}_{a}I_{q,\omega }^{N-2}{D_{q,\omega }^{N-2}}f \bigr) (x)-\frac{D_{q,\omega }^{N-2}f(a)}{\Gamma_{q}(N-1)}(x-a)_{\omega_{0}}^{(N-2)} \\ &\quad {} -\frac{D_{q,\omega }^{N-1}f(a)}{\Gamma_{q}(N)}(x-a)_{\omega_{0}}^{(N-1)} \\ & =\cdots \\ & =f(x)-\sum_{k=0}^{N-1} \frac{D_{q,\omega }^{k}f(a)}{\Gamma_{q}(k+1)}(x-a)_{ \omega_{0}}^{(k)}. \end{aligned}$$

 □

7 The application

In this section, we deal with the nonlocal $q,\omega$-integral boundary value problem of nonlinear fractional $q,\omega$-derivative equations:

$$\begin{aligned}& \bigl({}_{a}D_{q,\omega }^{\alpha }u\bigr) (t)+f \bigl(qt+\omega ,u(qt+\omega )\bigr)=0, \end{aligned}$$

(20)

$$\begin{aligned}& u(a)=0,\qquad u(b)=\mu \bigl({}_{a}I_{q,\omega }^{\beta }u \bigr) (\eta ), \end{aligned}$$

(21)

where $q\in (0,1)$, $1<\alpha \leq 2$, $0<\beta \leq 2$, $0<\eta <1$, and $\mu >0$ is a parameter, ${}_{a}D_{q,\omega }^{\alpha }$ is the Riemann–Liouville $q,\omega$-derivative of Hahn quantum operator type of order α, $f:[a,b]\times \mathbb{R}^{+}\longrightarrow \mathbb{R}^{+}$ is continuous.

Lemma 7.1

Let $M=\Gamma_{q}(\alpha +\beta )(b-a)_{\omega_{0}}^{(\alpha -1)}- \mu \Gamma_{q}(\alpha )(\eta -a)_{\omega_{0}}^{(\alpha +\beta -1)}>0$. Then, for given $y\in C[a,b]$, the unique solution of the boundary value problem

$$\begin{aligned} \bigl({}_{a}D_{q,\omega }^{\alpha }u\bigr) (t)+y(qt+\omega )=0, \quad 1< \alpha \leq 2, \end{aligned}$$

subject to the boundary condition

$$\begin{aligned} u(a)=0,\qquad u(b)=\mu \bigl({}_{a}I_{q,\omega }^{\beta }u \bigr) (\eta ),\quad 0< \beta \leq 2, 0< \eta < 1, \end{aligned}$$

is given by

$$\begin{aligned} u(t)= \int_{a}^{b}G\bigl(t,{}_{\omega_{0}}\!\Phi_{q}(s)\bigr)y(qs+\omega )\,{d_{q,\omega }s},\quad t\in [a,b], \end{aligned}$$

where

$$\begin{aligned}& G\bigl(t,{}_{\omega_{0}}\!\Phi_{q}(s)\bigr)=g\bigl(t,{}_{\omega_{0}}\!\Phi_{q}(s)\bigr)+\frac{ \mu (t-a)_{\omega_{0}}^{(\alpha -1)}}{M}H\bigl(\eta ,{}_{\omega_{0}}\!\Phi _{q}(s)\bigr), \\& g\bigl(t,{}_{\omega_{0}}\!\Phi_{q}(s)\bigr)=\frac{1}{\Gamma_{q}(\alpha )} \textstyle\begin{cases} {\frac{(t-a)_{\omega_{0}}^{(\alpha -1)}}{(b-a)_{\omega_{0}}^{(\alpha -1)}}}(b-{}_{\omega_{0}}\!\Phi_{q}(s))_{\omega_{0}}^{(\alpha -1)} -(t-{}_{ \omega_{0}}\!\Phi_{q}(s))_{\omega_{0}}^{(\alpha -1)}, \\ \hphantom{{\frac{(t-a)_{\omega_{0}}^{(\alpha -1)}}{(b-a)_{\omega_{0}}^{(\alpha -1)}}}(b-{}_{\omega_{0}}\!\Phi_{q}(s))_{\omega_{0}}^{(\alpha -1)},\quad }{} a\leq_{\omega_{0}}\!\Phi_{q}(s)\leq t\leq b, \\ {\frac{(t-a)_{\omega_{0}}^{(\alpha -1)}}{(b-a)_{\omega_{0}}^{(\alpha -1)}}}(b-{}_{\omega_{0}}\!\Phi_{q}(s))_{\omega_{0}}^{(\alpha -1)},\quad a\leq t\leq_{\omega_{0}}\!\Phi_{q}(s)\leq b, \end{cases}\displaystyle \\& H\bigl(\eta ,{}_{\omega_{0}}\!\Phi_{q}(s)\bigr)= \textstyle\begin{cases} {\frac{(\eta -a)_{\omega_{0}}^{(\alpha +\beta -1)}}{(b-a)_{\omega_{0}} ^{(\alpha -1)}}}(b-{}_{\omega_{0}}\!\Phi_{q}(s))_{\omega_{0}}^{(\alpha -1)}-(\eta -{}_{\omega_{0}}\!\Phi_{q}(s))_{\omega_{0}}^{(\alpha +\beta -1)}, \\ \hphantom{{\frac{(\eta -a)_{\omega_{0}}^{(\alpha +\beta -1)}}{(b-a)_{\omega_{0}} ^{(\alpha -1)}}}(b-{}_{\omega_{0}}\!\Phi_{q}(s))_{\omega_{0}}^{(\alpha -1)}, \quad }{}a\leq_{\omega_{0}}\!\Phi_{q}(s)\leq \eta \leq b, \\ {\frac{(\eta -a)_{\omega_{0}}^{(\alpha +\beta -1)}}{(b-a)_{\omega_{0}} ^{(\alpha -1)}}}(b-{}_{\omega_{0}}\!\Phi_{q}(s))_{\omega_{0}}^{(\alpha -1)}, \quad a\leq \eta \leq_{\omega_{0}}\!\Phi_{q}(s)\leq b. \end{cases}\displaystyle \end{aligned}$$

Proof

In view of Theorem 5.6,

$$\begin{aligned} u(t)=c_{1}(t-a)_{\omega_{0}}^{(\alpha -1)}+c_{2}(t-a)_{\omega_{0}} ^{(\alpha -2)}-\frac{1}{\Gamma_{q}(\alpha )} \int_{a}^{t}\bigl(t-{}_{\omega _{0}}\!\Phi_{q}(s)\bigr)_{\omega_{0}}^{(\alpha -1)}y(qs+\omega )\,{d_{q,\omega }s}, \end{aligned}$$

where $c_{1}, c_{2}$ does not depend on t, $t\in [a,b]$. Since $u(a)=0$, we have $c_{2}=0$. From the boundary condition $u(b)=\mu ({}_{a}I _{q,\omega }^{\beta }u)(\eta )$, we get

$$\begin{aligned} c_{1}&=\frac{\Gamma_{q}(\alpha +\beta )}{M} \biggl[ \frac{1}{\Gamma _{q}(\alpha )} \int_{a}^{b}\bigl(b-{}_{\omega_{0}}\!\Phi_{q}(s)\bigr)_{\omega_{0}} ^{(\alpha -1)}y(qs+\omega )\,{d_{q,\omega }s} \\ &\quad {}-\frac{\mu }{\Gamma_{q}(\alpha +\beta )} \int_{a}^{\eta }\bigl(\eta -{}_{ \omega_{0}}\!\Phi_{q}(s)\bigr)_{\omega_{0}}^{(\alpha +\beta -1)}y(qs+\omega )\,{d_{q,\omega }s} \biggr]. \end{aligned}$$

Hence

$$\begin{aligned} u(t)&= \frac{\Gamma_{q}(\alpha +\beta )(t-a)_{\omega_{0}}^{(\alpha -1)}}{M} \biggl[ \frac{1}{\Gamma_{q}(\alpha )} \int_{a}^{b}\bigl(b-{}_{\omega_{0}}\!\Phi _{q}(s)\bigr)_{\omega_{0}}^{(\alpha -1)}y(qs+\omega )\,{d_{q,\omega }s} \\ &\quad {} -\frac{\mu }{\Gamma_{q}(\alpha +\beta )} \int_{a}^{\eta }\bigl(\eta -{}_{ \omega_{0}}\!\Phi_{q}(s)\bigr)_{\omega_{0}}^{(\alpha +\beta -1)}y(qs+\omega )\,{d_{q,\omega }s} \biggr] \\ &\quad {} -\frac{1}{\Gamma_{q}(\alpha )} \int_{a}^{t}\bigl(t-{}_{\omega_{0}}\!\Phi_{q}(s)\bigr)_{ \omega_{0}}^{(\alpha -1)}y(qs+\omega )\,{d_{q,\omega }s} \\ & =\int_{a}^{b}g\bigl(t,{}_{\omega_{0}}\!\Phi_{q}(s)\bigr)y(qs+\omega )\,{d_{q,\omega}s} \\ &\quad {}+\frac{\mu (t-a)_{\omega_{0}}^{(\alpha -1)}}{M} \int_{a}^{b}H\bigl( \eta ,{}_{\omega_{0}}\!\Phi_{q}(s)\bigr)y(qs+\omega )\,{d_{q,\omega }s} \\ & = \int_{a}^{b}G\bigl(t,{}_{\omega_{0}}\!\Phi_{q}(s)\bigr)y(qs+\omega )\,{d_{q,\omega }s}. \end{aligned}$$

 □

Lemma 7.2

The functions $g(t, {}_{\omega_{0}}\!\Phi_{q}(s))$ and $H(\eta ,{}_{\omega _{0}}\!\Phi_{q}(s))$ satisfy the following properties:

• (a) $g(t, {}_{\omega_{0}}\!\Phi_{q}(s))\geq 0$, $g(t,{}_{\omega_{0}}\!\Phi_{q}(s))\leq g({}_{\omega_{0}}\!\Phi_{q}(s),{}_{\omega_{0}}\!\Phi_{q}(s))$, $a\leq t, {}_{\omega_{0}}\!\Phi_{q}(s)\leq b$.

• (b) $H(\eta , {}_{\omega_{0}}\!\Phi_{q}(s))\geq 0$, $a\leq {}_{\omega_{0}}\!\Phi_{q}(s),\eta \leq b$.

Proof

Let

$$\begin{aligned}& g_{1}\bigl(t, _{\omega_{0}}\!\Phi_{q}(s)\bigr)= \frac{1}{\Gamma_{q}(\alpha )} \biggl[\frac{(t-a)_{\omega_{0}}^{(\alpha -1)}}{(b-a)_{\omega_{0}}^{( \alpha -1)}} \bigl(b-{}_{\omega_{0}}\!\Phi_{q}(s)\bigr)_{\omega_{0}}^{(\alpha -1)}-\bigl(t-{}_{ \omega_{0}}\!\Phi_{q}(s)\bigr)_{\omega_{0}}^{(\alpha -1)} \biggr], \\& g_{2}\bigl(t, _{\omega_{0}}\!\Phi_{q}(s)\bigr)= \frac{1}{\Gamma_{q}(\alpha )} \biggl[\frac{(t-a)_{\omega_{0}}^{(\alpha -1)}}{(b-a)_{\omega_{0}}^{( \alpha -1)}} \bigl(b-{}_{\omega_{0}}\!\Phi_{q}(s)\bigr)_{\omega_{0}}^{(\alpha -1)} \biggr]. \end{aligned}$$

We can easily get $g_{2}(t, {}_{\omega_{0}}\!\Phi_{q}(s))\geq 0$, according to property (7), we have

$$\begin{aligned} &\frac{(t-a)_{\omega_{0}}^{(\alpha -1)}}{(b-a)_{\omega_{0}}^{(\alpha -1)}} \bigl(b-{}_{\omega_{0}}\!\Phi_{q}(s) \bigr)_{\omega_{0}}^{(\alpha -1)}-\bigl(t-{}_{ \omega_{0}}\!\Phi_{q}(s)\bigr)_{\omega_{0}}^{(\alpha -1)} \\ &\quad =\prod_{i=0}^{\infty }\frac{(t-{}_{\omega_{0}}\!\Phi_{q}^{i}(a))}{(t-{}_{ \omega_{0}}\!\Phi_{q}^{i+\alpha -1}(a))} \frac{(b-{}_{\omega_{0}}\!\Phi_{q} ^{i+\alpha -1}(a))}{(b-{}_{\omega_{0}}\!\Phi_{q}^{i}(a))}\times \prod_{i=0} ^{\infty } \frac{(b-{}_{\omega_{0}}\!\Phi_{q}^{i+1}(s))}{(b-{}_{\omega_{0}}\!\Phi_{q}^{i+\alpha }(s))} \\ &\quad \quad {}-\prod_{i=0}^{\infty } \frac{(t-{}_{\omega_{0}}\!\Phi_{q}^{i+1}(s))}{(t-{}_{ \omega_{0}}\!\Phi_{q}^{i+\alpha }(s))}. \end{aligned}$$

Since

$$\begin{aligned} & \biggl(\frac{(b-{}_{\omega_{0}}\!\Phi_{q}^{i+1}(s))}{(b-{}_{\omega_{0}}\!\Phi_{q}^{i+\alpha }(s))}\frac{(t-{}_{\omega_{0}}\!\Phi_{q}^{i+\alpha }(s))}{(t-{}_{ \omega_{0}}\!\Phi_{q}^{i+1}(s))} \biggr)_{s}^{'} \\ &\quad =\frac{1}{(b-{}_{\omega_{0}}\!\Phi_{q}^{i+\alpha }(s))^{2}(t-{}_{\omega _{0}}\!\Phi_{q}^{i+1}(s))^{2}} \bigl[ \bigl( -q^{i+1}\bigl(t-{}_{\omega_{0}}\!\Phi_{q}^{i+\alpha }(s)\bigr)-q^{i+\alpha }\bigl(b-{}_{\omega_{0}}\!\Phi_{q}^{i+1}(s)\bigr) \bigr) \\ &\quad \quad {}\times \bigl(b-{}_{\omega_{0}}\!\Phi_{q}^{i+\alpha }(s)\bigr) \bigl(t-{}_{\omega_{0}}\!\Phi_{q}^{i+1}(s)\bigr) - \bigl( -q^{i+\alpha }\bigl(t-{}_{\omega_{0}}\!\Phi_{q}^{i+1}(s)\bigr)-q^{i+1}\bigl(b-{}_{\omega_{0}}\!\Phi_{q}^{i+\alpha }(s)\bigr) \bigr) \\ &\quad \quad {} \times \bigl(b-{}_{\omega_{0}}\!\Phi_{q}^{i+1}(s)\bigr) \bigl(t-{}_{\omega_{0}}\!\Phi_{q} ^{i+\alpha }(s)\bigr)\bigr] \\ &\quad =\frac{1}{(b-{}_{\omega_{0}}\!\Phi_{q}^{i+\alpha }(s))^{2}(t-{}_{\omega _{0}}\!\Phi_{q}^{i+1}(s))^{2}} \bigl[-q^{i+1}t^{2}b+q^{i+\alpha }t^{2}b-q ^{i+\alpha }b^{2}t+q^{i+1}b^{2}t \\ &\quad \quad {} +_{\omega_{0}}\!\Phi_{q}^{i+\alpha }(s){q}^{i+1}t^{2}-{}_{\omega_{0}}\!\Phi_{q}^{i+1}(s){q}^{i+\alpha }t^{2}+_{\omega_{0}}\!\Phi_{q}^{i+\alpha }(s){q}^{i+1}b^{2}-{}_{\omega_{0}}\!\Phi_{q}^{i+1}(s){q}^{i+\alpha }b^{2} \\ &\quad \quad {} +_{\omega_{0}}\!\Phi_{q}^{i+\alpha }(s)_{\omega_{0}}\!\Phi_{q}^{i+ \alpha }(s){q}^{i+1}(b-t)-{}_{\omega_{0}}\!\Phi_{q}^{i+1}(s)_{\omega_{0}}\!\Phi_{q}^{i+1}(s){q}^{i+\alpha }(b-t) \bigr] \\ &\quad =\frac{1}{(b-{}_{\omega_{0}}\!\Phi_{q}^{i+\alpha }(s))^{2}(t-{}_{\omega _{0}}\!\Phi_{q}^{i+1}(s))^{2}} \bigl[ \bigl({q}^{i+1}-{q}^{i+\alpha } \bigr) \bigl(b^{2}t-t ^{2}b\bigr)+\bigl({q}^{i+1}-{q}^{i+\alpha } \bigr)\omega_{0}t^{2} \\ &\quad \quad {} +\bigl({q}^{i+1}-{q}^{i+\alpha }\bigr) \omega_{0}b^{2}+\bigl({q}^{i+1}-{q}^{i+ \alpha } \bigr) \bigl(\omega_{0}^{2}b-\omega_{0}^{2}t \bigr) \bigr] >0, \end{aligned}$$

we may see that function $\frac{(b-{}_{\omega_{0}}\!\Phi_{q}^{i+1}(s))}{(b-{}_{ \omega_{0}}\!\Phi_{q}^{i+\alpha }(s))}\frac{(t-{}_{\omega_{0}}\!\Phi_{q} ^{i+\alpha }(s))}{(t-{}_{\omega_{0}}\!\Phi_{q}^{i+1}(s))}$ on s is non-decreasing. Thus, we can get

$$\begin{aligned} &\frac{(t-{}_{\omega_{0}}\!\Phi_{q}^{i}(a))(b-{}_{\omega_{0}}\!\Phi_{q}^{i+ \alpha -1}(a))}{(t-{}_{\omega_{0}}\!\Phi_{q}^{i+\alpha -1}(a))(b-{}_{\omega _{0}}\!\Phi_{q}^{i}(a))} \frac{(b-{}_{\omega_{0}}\!\Phi_{q}^{i}({}_{\omega _{0}}\!\Phi_{q}(s)))(t-{}_{\omega_{0}}\!\Phi_{q}^{i+\alpha -1}({}_{\omega_{0}}\!\Phi_{q}(s)))}{(b-{}_{\omega_{0}}\!\Phi_{q}^{i+\alpha -1}({}_{\omega_{0}}\!\Phi_{q}(s)))(t-{}_{\omega_{0}}\!\Phi_{q}^{i}({}_{\omega_{0}}\!\Phi_{q}(s)))} \\ &\quad \geq \frac{(t-{}_{\omega_{0}}\!\Phi_{q}^{i}(a))(b-{}_{\omega_{0}}\!\Phi_{q} ^{i+\alpha -1}(a))}{(t-{}_{\omega_{0}}\!\Phi_{q}^{i+\alpha -1}(a))(b-{}_{ \omega_{0}}\!\Phi_{q}^{i}(a))} \frac{(b-{}_{\omega_{0}}\!\Phi_{q}^{i}(a))(t-{}_{\omega_{0}}\!\Phi_{q}^{i+\alpha -1}(a))}{(b-{}_{\omega_{0}}\!\Phi_{q}^{i+ \alpha -1}(a))(t-{}_{\omega_{0}}\!\Phi_{q}^{i}(a))} =1, \end{aligned}$$

so $g_{1}(t,{}_{\omega_{0}}\!\Phi_{q}(s))\geq 0$, we can get $g(t,{}_{ \omega_{0}}\!\Phi_{q}(s))\geq 0$. Clearly $g(t,{}_{\omega_{0}}\!\Phi_{q}(s)) \leq g({}_{\omega_{0}}\!\Phi_{q}(s),{}_{\omega_{0}}\!\Phi_{q}(s))$ and $H(\eta , {}_{\omega_{0}}\!\Phi_{q}(s))\geq 0$ holds trivially. □

According to the property of being non-increasing of $(t-{}_{\omega_{0}}\!\Phi_{q}(s))_{\omega_{0}}^{(\alpha -1)}$ on s and Lemma 7.2, we may easily obtain Lemma 7.3 as follows.

Lemma 7.3

The function $G(t,{}_{\omega_{0}}\!\Phi_{q}(s))$ satisfies the following properties:

• (a) G is a continuous function and $G(t,{}_{\omega_{0}}\!\Phi_{q}(s))\geq 0$, $(t,{}_{\omega_{0}}\!\Phi_{q}(s))\in [a,b]\times [a,b]$.

• (b) There exists a positive function $\rho \in C(a,b)$, $(a,+\infty )$ such that $\max_{a\leq t\leq b} G(t,{}_{\omega_{0}}\!\Phi_{q}(s))\leq g({}_{\omega_{0}}\!\Phi_{q}(s),{}_{\omega_{0}}\!\Phi_{q}(s)) +\frac{\mu }{M}H(\eta , _{\omega_{0}}\!\Phi_{q}(s))=:\rho ({}_{\omega_{0}}\!\Phi_{q}(s))$, ${}_{\omega_{0}}\!\Phi_{q}(s), s \in (a,b)$.

Lemma 7.4

(Krasnoselskii [12])

Let $\mathbb{E}$ be a Banach space, and let $P\subset \mathbb{E}$ be a cone. Assume $\Omega_{1},\Omega_{2}$ are open subsets of $\mathbb{E}$ with $\theta \in \Omega_{1}\subset \overline{\Omega }_{1}\subset \Omega_{2}$, and let $T:P\cap (\overline{\Omega } _{2}\setminus \Omega_{1})\longrightarrow P$ be a completely continuous operator such that

$$\begin{aligned} \Vert Tu \Vert \geq \Vert u \Vert ,\quad u\in P\cap \partial \Omega_{1}\quad \textit{and} \quad \Vert Tu \Vert \leq \Vert u \Vert ,\quad u \in P\cap \partial \Omega_{2}. \end{aligned}$$

Then T has at least one fixed point in $P\cap (\overline{\Omega } _{2}\setminus \Omega_{1})$.

Let $X=C[a,b]$ be a Banach space endowed with the norm $\Vert u \Vert _{X}=\max_{a\leq t\leq b}|u(t)|$. Define the cone $P\subset \{u\in X:u(t)\geq 0,a\leq t\leq b\}$.

Define the operator $T:P\longrightarrow X$ as follows:

$$\begin{aligned} (T u) (t)= \int_{a}^{b}G\bigl(t,{}_{\omega_{0}}\!\Phi_{q}(s)\bigr)f\bigl(qt+\omega ,u(qt+ \omega )\bigr)\,{d_{q,\omega }s}. \end{aligned}$$

(22)

It follows from the non-negativeness and continuity of G and f that the operator $T:P\longrightarrow X$ satisfies $T(P)\subset P$ and is completely continuous.

Theorem 7.5

Let $f(t,u)$ be a nonnegative continuous function on $[a,b]\times \mathbb{R}^{+}$. In addition, we assume that:

$(H_{1})$ :

There exists a positive constant $r_{1}$ such that

$$\begin{aligned} f(t,u)\geq \kappa r_{1},\quad \textit{for }(t,u)\in [\tau_{1},\tau_{2}]\times [a,r _{1}], \end{aligned}$$

where $a\leq \tau_{1}\leq \tau_{2}\leq b$ and

$$\begin{aligned} \kappa \geq \biggl[ \int_{\tau_{1}}^{\tau_{2}} \biggl(g\bigl({}_{\omega_{0}}\!\Phi_{q}(s),{}_{\omega_{0}}\!\Phi_{q}(s)\bigr)+ \frac{\mu (s-a)_{\omega_{0}} ^{(\alpha -1)}}{M}H\bigl(\eta ,{}_{\omega_{0}}\!\Phi_{q}(s)\bigr) \biggr)\,{d_{q, \omega }s} \biggr]^{-1}. \end{aligned}$$

$(H_{2})$ :

There exists a positive constant $r_{2}$ with $r_{2}>r_{1}$ such that

$$\begin{aligned} f(t,u)\leq Lr_{2},\quad \textit{for } (t,u)\in [a,b]\times [a,r_{2}], \end{aligned}$$

where

$$\begin{aligned} L= \biggl[ \int_{a}^{b} \biggl(\frac{(b-{}_{\omega_{0}}\!\Phi_{q}(s))_{\omega _{0}}^{(\alpha -1)}}{\Gamma_{q}(\alpha )(b-a)_{\omega_{0}}^{(\alpha -1)}} + \frac{\mu }{M}H\bigl(\eta ,{}_{\omega_{0}}\!\Phi_{q}(s)\bigr) \biggr)\,{d_{q,\omega }s} \biggr]^{-1}. \end{aligned}$$

Then the boundary value problem (20), (21) has at least one positive solution $u_{0}$ satisfying $a\leq r_{1}\leq \Vert u_{0} \Vert _{X} \leq r_{2}$.

Proof

By Lemma 7.2, we obtain $\max_{a\leq t\leq b} g(t, {}_{\omega_{0}}\!\Phi_{q}(s))= g({}_{\omega_{0}}\!\Phi_{q}(s),{}_{\omega_{0}}\!\Phi_{q}(s))$. Let $\Omega_{1}=\{u\in X:\Vert u \Vert _{X}< r _{1}\}$. For any $u\in P\cap \partial \Omega_{1}$, according to $(H_{1})$, we have

$$\begin{aligned} \Vert Tu \Vert _{X}&=\max_{a\leq t\leq b} \biggl[ \int_{a}^{b}g\bigl(t, {}_{\omega_{0}}\!\Phi_{q}(s)\bigr)f\bigl(qs+\omega ,u(qs+\omega )\bigr)\,{d_{q,\omega }s} \\ &\quad {} + \int_{a}^{b}\frac{\mu (t-a)_{\omega_{0}}^{(\alpha -1)}}{M}H\bigl(\eta ,{}_{\omega_{0}}\!\Phi_{q}(s)\bigr)f\bigl(qs+\omega ,u(qs+\omega ) \bigr)\,{d_{q,\omega }s} \biggr] \\ & \geq \int_{a}^{b} \biggl[g\bigl({}_{\omega_{0}}\!\Phi_{q}(s),{}_{\omega_{0}}\!\Phi _{q}(s)\bigr)+ \frac{\mu (s-a)_{\omega_{0}}^{(\alpha -1)}}{M} H\bigl(\eta ,{}_{\omega_{0}}\!\Phi_{q}(s)\bigr) \biggr] \\ &\quad {}\times f\bigl(qs+\omega ,u(qs+\omega )\bigr)\,{d_{q,\omega }s} \\ & \geq \kappa r_{1} \int_{\tau_{1}}^{\tau_{2}} \biggl[g\bigl({}_{\omega_{0}}\!\Phi _{q}(s),{}_{\omega_{0}}\!\Phi_{q}(s)\bigr)+ \frac{\mu (s-a)_{\omega_{0}}^{( \alpha -1)}}{M} H\bigl(\eta ,{}_{\omega_{0}}\!\Phi_{q}(s)\bigr) \biggr]\,{d_{q,\omega }s} \\ & =r_{1}=\Vert u \Vert _{X}. \end{aligned}$$

Let $\Omega_{2}=\{u\in X:\Vert u \Vert _{X}< r_{2}\}$. For any $u\in P\cap \partial \Omega_{2}$, by $(H_{2})$, we have

$$\begin{aligned} \Vert Tu \Vert _{X}&=\max_{a\leq t\leq b} \int_{a}^{b}G\bigl(t,{}_{ \omega_{0}}\!\Phi_{q}(s)\bigr)f\bigl(qs+\omega ,u(qs+\omega )\bigr)\,{d_{q,\omega }s} \\ & \leq Lr_{2} \int_{a}^{b}\rho \bigl({}_{\omega_{0}}\!\Phi_{q}(s)\bigr)\,{d_{q,\omega }s} \\ &\leq \Vert u \Vert _{X}=r_{2}. \end{aligned}$$

Now, an application of Lemma 7.4 concludes the proof. □

8 Mittag-Leffler function

Example 8.1

Let $0<\alpha \leq 1$ and consider the fractional $q,\omega$-difference equation

$$\begin{aligned} \bigl(^{c}D_{q,\omega }^{\alpha }y\bigr) (qt+\omega )= \lambda y(t)+f(t),\quad y(0)=a _{0}. \end{aligned}$$

If we apply ${}_{a}I_{q,\omega }^{\alpha }$ on the equation, we see that

$$\begin{aligned} y(t)=a_{0}+\lambda_{a}I_{q,\omega }^{\alpha }y \bigl({}_{\omega_{0}}\!\Phi_{q} ^{-1}(t) \bigr)+_{a}I_{q,\omega }^{\alpha }g(t), \end{aligned}$$

where $g(t)=f({}_{\omega_{0}}\!\Phi_{q}^{-1}(t))$.

To obtain an explicit clear solution, we apply the method of successive approximation. Set $y_{0}(t)=a_{0}$ and

$$\begin{aligned} y_{m}(t)=a_{0}+\lambda_{a}I_{q,\omega }^{\alpha }y_{m-1} \bigl({}_{\omega_{0}}\!\Phi_{q}^{-1}(t) \bigr)+_{a}I_{q,\omega }^{\alpha }g(t),\quad m=1,2,\ldots. \end{aligned}$$

For $m=1$, we have

$$\begin{aligned}& y_{1}(t) =a_{0}+\frac{\lambda }{\Gamma_{q}(\alpha )} \int^{{}_{\omega_{0}}{\Phi_{q}^{-1}(t)}}_{a}\bigl({}_{\omega_{0}}\!\Phi_{q}^{-1}(t)-{}_{\omega_{0}}\!\Phi_{q}(s) \bigr)_{\omega_{0}}^{(\alpha -1)}y_{0}(s)\,{d_{q,\omega }s} +_{a}I _{q,\omega }^{\alpha }g(t) \\& \hphantom{y_{1}(t) } =a_{0} \biggl[1+\frac{\lambda }{\Gamma_{q}(\alpha +1)} \bigl({}_{\omega_{0}}\!\Phi_{q}^{-1}(t)-a \bigr)_{\omega_{0}}^{(\alpha )} \biggr]+_{a}I_{q,\omega } ^{\alpha }g(t), \\& y_{2}(t)=a_{0} \biggl[1+\frac{\lambda }{\Gamma_{q}(\alpha +1)} \bigl({}_{ \omega_{0}}\!\Phi_{q}^{-1}(t)-a \bigr)_{\omega_{0}}^{(\alpha )} +\frac{\lambda ^{2}}{\Gamma_{q}(2\alpha +1)}\bigl({}_{\omega_{0}}\!\Phi_{q}^{-1}(t)-a\bigr)_{\omega _{0}}^{(2\alpha )} \biggr] \\& \hphantom{y_{2}(t) =}{}+\lambda_{a}I_{q,\omega }^{\alpha } \bigl({}_{a}I_{q,\omega }^{\alpha }g\bigr) \bigl({}_{ \omega_{0}}\!\Phi_{q}^{-1}(t)\bigr)+_{a}I_{q,\omega }^{\alpha }g(t). \end{aligned}$$

If we proceed inductively and let $m\rightarrow \infty$, we obtain the solution

$$\begin{aligned} y(t)&=a_{0} \Biggl[ 1+\sum_{k=1}^{\infty } \frac{\lambda^{k}}{\Gamma _{q}(k\alpha +1)}\bigl({}_{\omega_{0}}\!\Phi_{q}^{-1}(t)-a \bigr)_{\omega_{0}}^{(k \alpha )} \Biggr] \\ &\quad {}+ \int_{a}^{t}\sum_{k=0}^{\infty } \frac{\lambda^{k}(t-{}_{\omega_{0}}\!\Phi _{q}(s))_{\omega_{0}}^{(k\alpha +\alpha -1)}}{\Gamma_{q}(k\alpha + \alpha )} f\bigl({}_{\omega_{0}}\!\Phi_{q}^{-1}(t) \bigr)\,{d_{q,\omega }s}. \end{aligned}$$

Definition 8.2

For $z,z_{0}\in \mathbb{C}$, the Mittag-Leffler function for fractional $q,\omega$-difference equation is defined by

$$\begin{aligned} E_{\alpha ,\beta }(\lambda ,z-z_{0})=\sum _{k=0}^{\infty }\frac{\lambda ^{k}(z-z_{0})^{(k\alpha +\beta -1)}}{\Gamma_{q}(k\alpha +\beta )}. \end{aligned}$$

When $\beta =1$, we simply use $E_{\alpha }(\lambda ,z-z_{0})=E_{ \alpha ,1}(\lambda ,z-z_{0})$.

According to Definition 8.2 above, the solution of the fractional $q,\omega$-difference equation in Example 8.1 is expressed by

$$\begin{aligned} y(t)=a_{0}E_{\alpha }\bigl(\lambda ,t-{}_{\omega_{0}}\!\Phi_{q}(a)\bigr)+ \int_{a} ^{t}E_{\alpha ,\alpha }\bigl(\lambda ,t-{}_{\omega_{0}}\!\Phi_{q}(s)\bigr)f\bigl({}_{\omega_{0}}\!\Phi_{q}^{-1}(s)\bigr)\,{d_{q,\omega }s}. \end{aligned}$$

9 Conclusion

The paper [16] focuses on essentials of fractional calculus on a special discrete time scale forming the background for the so-called $(q,h)$-calculus, which can be reduced to the quantum calculus (the case $h=0$) or to the difference calculus (the case $q=h=1$). From this point, in this article, we give a more general extension. The fractional $q,\omega$-calculus is defined in the set of real numbers. The main advantage of our results is that the corresponding properties can deal with the fractional $q,\omega$-calculus equation, the existence result of the solutions is solved, which is important in physical systems. We trust that the field here initiated will prove to be fruitful for further research.