\documentclass[reqno]{amsart} \usepackage{hyperref} \AtBeginDocument{{\noindent\small \emph{Electronic Journal of Differential Equations}, Vol. 2016 (2016), No. 17, pp. 1--14.\newline ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu \newline ftp ejde.math.txstate.edu} \thanks{\copyright 2016 Texas State University.} \vspace{9mm}} \begin{document} \title[\hfilneg EJDE-2016/17 q-integral equations of fractional orders] {q-integral equations of fractional orders} \author[M. Jleli, M. Mursaleen, B. Samet \hfil EJDE-2016/17\hfilneg] {Mohamed Jleli, Mohammad Mursaleen, Bessem Samet} \address{Mohamed Jleli \newline Department of Mathematics, College of Science, King Saud University, P.O. Box 2455, Riyadh 11451, Saudi Arabia} \email{jleli@ksu.edu.sa} \address{Mohammad Mursaleen \newline Department of Mathematics, Aligarh Muslim University, Aligarh 202002, India} \email{mursaleenm@gmail.com} \address{Bessem Samet \newline Department of Mathematics, College of Science, King Saud University, P.O. Box 2455, Riyadh 11451, Saudi Arabia} \email{bsamet@ksu.edu.sa} \thanks{Submitted October 2, 2015. Published January 8, 2016.} \subjclass[2010]{31A10, 26A33, 47H08} \keywords{q-fractional integral; existence; measure of non-compactness} \begin{abstract} The aim of this paper is to study the existence of solutions for a class of $q$-integral equations of fractional orders. The techniques in this work are based on the measure of non-compactness argument and a generalized version of Darbo's theorem. An example is presented to illustrate the obtained result. \end{abstract} \maketitle \numberwithin{equation}{section} \newtheorem{theorem}{Theorem}[section] \newtheorem{lemma}[theorem]{Lemma} \newtheorem{proposition}[theorem]{Proposition} \allowdisplaybreaks \section{Introduction} In this paper, we are concerned with the following functional equation \begin{equation}\label{eq} x(t)=F\Big(t,x(a(t)),\frac{f(t,x(b(t)))}{\Gamma_q(\alpha)} \int_0^t (t-qs)^{(\alpha-1)}u(s,x(s))\,d_qs\Big),\quad t\in I, \end{equation} where $\alpha>1$, $q\in (0,1)$, $I=[0,1]$, $f,u:[0,1]\times \mathbb{R}\to \mathbb{R}$, $a,b: I\to I$ and $F: I\times \mathbb{R}\times \mathbb{R}\to \mathbb{R}$. Equation \eqref{eq} can be written as $$ x(t)=F\left(t,x(a(t)),f(t,x(b(t)))I_q^\alpha u(\cdot,x(\cdot))(t)\right),\quad t\in I, $$ where $I_q^\alpha$ is the $q$-fractional integral of order $\alpha$ defined by (see \cite{AR}) $$ I_q^\alpha h (t)=\frac{1}{\Gamma_q(\alpha)} \int_0^t (t-qs)^{(\alpha-1)}h(s)\,d_qs,\quad t\in [0,1]. $$ Using a measure of non-compactness argument combined with a generalized version of Darbo's theorem, we provide sufficient conditions for the existence of at least one solution to \eqref{eq}. We present also some examples to illustrate our obtained result. The measure of non-compactness argument was used by several authors to study the existence of solutions for various classes of integral equations. As examples, we refer the reader to Aghajani et al \cite{A,AGM,AGP}, Bana\'s et al \cite{BC,BANL,BANR}, Bana\'s and Goebel \cite{BG}, Bana\'s and Rzepka \cite{BANRZ}, Bana\'s and Martinon \cite{BANM}, Caballero et al \cite{CABL,CABL2,CABLR}, Darwish \cite{D}, \c{C}akan and Ozdemir \cite{CA}, Dhage and Bellale \cite{DHB}, Mursaleen and Mohiuddine \cite{M}, Mursaleen and Alotaibi \cite{MA}, and the references therein. For other applications of the measure of non-compactness concept, we refer to \cite{BMU,MN}. In \cite{DR}, via a measure of non-compactness concept, Darwish obtained an existence result for the singular integral equation $$ y(t)=f(t)+\frac{y(t)}{\Gamma(\alpha)} \int_0^t \frac{u(s,y(s))}{(t-s)^{1-\alpha}}\,ds,\quad t\in [0,1],\, \alpha>0. $$ The above equation can be written in the form $$ y(t)=f(t)+y(t) I^\alpha u(\cdot,y(\cdot))(t),\quad t\in [0,1], $$ where $I^\alpha$ is the Riemann-Liouville fractional integral of order $\alpha$ defined by (see \cite{S}) $$ I^\alpha h(t)=\frac{1}{\Gamma(\alpha)} \int_0^t \frac{h(s)}{(t-s)^{1-\alpha}}\,ds, \quad t\in [0,1]. $$ Following the above work, there has been a great interest in the study of functional equations involving fractional integrals. In this direction, we refer the reader to \cite{AGB,BE,D2,DD3,BRZ,BO,E} and the references therein. The concept of $q$-calculus (quantum calculus) was introduced by Jackson (see \cite{J1,J2}). This subject is rich in history and has several applications (see \cite{ER,K}). Fractional $q$-difference concept was initiated by Agarwal and by Al-Salam (see \cite{AR,AL}). Because of the considerable progress in the study of fractional differential equations, a great interest appeared from many authors in studying fractional $q$-difference equations (see for examples \cite{AH,AL,F,F2,L} and the references therein). The paper is organized as follows. In Section \ref{s2}, we fix some notations that will be used through this paper, we recall some basic tools on $q$-calculus and we collect some basic definitions and facts on the measure of non-compactness concept. In Section \ref{s3}, we state and prove our main result. Next, we present an illustrative example. \section{Preliminaries}\label{s2} At first, we recall some concepts on fractional $q$-calculus and present additional properties that will be used later. For more details, we refer to \cite{AR,AN,R}. Let $q$ be a positive real number such that $q\neq 1$. For $x\in \mathbb{R}$, the $q$-real number $[x]_q$ is defined by $$ [x]_q=\frac{1-q^x}{1-q}. $$ The $q$-shifted factorial of real number $x$ is defined by $$ (x,q)_0=1,\quad (x,q)_k=\prod_{i=0}^{k-1} (1-xq^i),\quad k=1,2,\dots,\infty. $$ For $(x,y)\in \mathbb{R}^2$, the $q$-analog of $(x-y)^k$ is defined by $$ (x-y)^{(0)}=1,\,\, (x-y)^{(k)}=\prod_{i=0}^{k-1}(x-q^iy),\,\,k=1,2,\dots $$ For $\beta\in \mathbb{R}$, $(x,y)\in \mathbb{R}^2$ and $x\geq 0$, $$ (x-y)^{(\beta)}=x^\beta \prod_{i=0}^\infty \frac{x-yq^i}{x-yq^{\beta+i}}. $$ Note that if $y=0$, then $x^{(\beta)}=x^\beta$. The following inequality (see \cite{F}) will be used later. \begin{lemma}\label{LF} If $\beta> 0$ and $0\leq a \leq b \leq t$, then $$ (t-b)^{(\beta)}\leq (t-a)^{(\beta)}. $$ \end{lemma} The $q$-gamma function is given by $$ \Gamma_q(x)=\frac{(1-q)^{(x-1)}}{(1-q)^{x-1}},\quad x\not\in \{0,-1,-2,\dots\}. $$ We have the following property $$ \Gamma_q(x+1)=[x]_q\Gamma_q(x). $$ Let $f: [0,b]\to \mathbb{R}$ ($b>0$) be a given function. The $q$-integral of the function $f$ is given by $$ I_q f (t)=\int_0^t f(s)\,d_qs =t(1-q)\sum_{n=0}^\infty f(tq^n)q^n,\quad t\in [0,b]. $$ If $c\in [0,b]$, we have $$ \int_c^b f(s)\,d_qs=\int_0^b f(s)\,d_qs-\int_0^c f(s)\,d_qs. $$ \begin{lemma}\label{L1} If $f: [0,1]\to \mathbb{R}$ is a continuous function, then $$ \big|\int_0^t f(s)\,d_qs\big|\leq \int_0^t |f(s)|\,d_qs,\quad t\in [0,1]. $$ \end{lemma} Note that in general, if $0\leq t_1\leq t_2\leq 1$ and $f: [0,1]\to \mathbb{R}$ is a continuous function, the inequality $$ \big|\int_{t_1}^{t_2} f(s)\,d_qs\big|\leq \int_{t_1}^{t_2} |f(s)|\,d_qs $$ is not satisfied. We remark that in many papers dealing with $q$-difference boundary value problems, the use of such inequality yields wrong results. As a counter-example, we refer the reader to \cite[Page.12]{AN}. Let $f: [0,1]\to \mathbb{R}$ be a given function. The fractional $q$-integral of order $\alpha\geq 0$ of the function $f$ is given by $I_q^0 f(t)=f(t)$ and $$ I_q^\alpha f(t)= \frac{1}{\Gamma_q(\alpha)} \int_0^t (t-qs)^{(\alpha-1)}f(s)\,d_qs,\quad t\in [0,1],\, \alpha>0. $$ Note that for $\alpha=1$, we have $$ I_q^1 f(t)=I_q f(t),\quad t\in [0,1]. $$ If $f\equiv 1$, then $$ I_q^\alpha 1 (t)=\frac{1}{\Gamma_q(\alpha+1)}t^\alpha,\quad t\in [0,1]. $$ Now, we recall the notion of measure of non-compactness, which is the main tool used in this paper. Let $\mathbb{E}$ be a Banach space over $\mathbb{R}$ with respect to a certain norm $\|\cdot\|$. We denote by $B_{\mathbb E}$ the set of all nonempty and bounded subsets of $\mathbb{E}$. A mapping $\sigma: B_{\mathbb E}\to [0,\infty)$ is a measure of non-compactness in $\mathbb{E}$ (see \cite{BMU}) if the following conditions are satisfied: \begin{itemize} \item[(i)] for all $M\in B_{\mathbb E}$, we have $$ \sigma(M)=0 \text{ implies } M \text{ is precompact}; $$ \item[(ii)] for every pair $(M_1,M_2)\in B_{\mathbb E}\times B_{\mathbb E}$, we have $$ M_1\subseteq M_2\Longrightarrow \sigma(M_1)\leq \sigma(M_2); $$ \item[(iii)] for every $M\in B_{\mathbb E}$, $$ \sigma(\overline{M})=\sigma(M)=\sigma(\overline{\operatorname{co}} M), $$ where $\overline{\operatorname{co}} M$ denotes the closed convex hull of $M$; \item[(iv)] for every pair $(M_1,M_2)\in B_{\mathbb E}\times B_{\mathbb E}$ and $\lambda\in (0,1)$, we have $$ \sigma(\lambda M_1+(1-\lambda)M_2)\leq \lambda\sigma(M_1)+(1-\lambda)\sigma(M_2); $$ \item[(v)] if $\{M_n\}$ is a sequence of closed and decreasing (w.r.t $\subseteq $) sets in $B_{\mathbb E}$ such that $\sigma(M_n)\to 0$ as $n\to \infty$, then $M_\infty:=\cap_{n=1}^\infty M_n$ is nonempty. \end{itemize} Let $\Lambda$ be the set of functions $\eta: [0,\infty)\to [0,\infty)$ such that \begin{enumerate} %($\Lambda_1$) \item $\eta$ is a non-decreasing function; \item $\eta$ is an upper semi-continuous function; \item $\eta(s)0$. \end{enumerate} For our purpose, we need the following generalized version of Darbo's theorem (see \cite{A}). \begin{lemma}\label{LD} Let $D$ be a nonempty, bounded, closed and convex subset of a Banach space $\mathbb{E}$. Let $T: D\to D$ be a continuous mapping such that $$ \sigma(T M)\leq \eta(\sigma(M)),\quad M\subseteq D, $$ where $\eta\in \Lambda$ and $\sigma$ is a measure of non-compactness in $\mathbb{E}$. Then $T$ has at least one fixed point. \end{lemma} \begin{lemma}\label{CF} Let $\eta_1,\eta_2\in \Lambda$ and $\tau \in (0,1)$. Then the function $\gamma: [0,\infty)\to [0,\infty)$ defined by $$ \gamma(t)=\max\{\eta_1(t),\eta_2(t),\tau\,t\},\quad t\geq 0 $$ belongs to the set $\Lambda$. \end{lemma} \begin{proof} Let $(t,s)\in \mathbb{R}^2$ be such that $0\leq t\leq s$. Since $\eta_1, \eta_2$ are non-decreasing and $\tau\in (0,1)$, we have \begin{gather*} \eta_i(t)\leq \eta_i (s)\leq \gamma(s),\quad i=1,2\,, \\ \tau t\leq \tau s\leq \gamma(s), \end{gather*} which yield $\gamma(t)\leq \gamma(s)$. Therefore, $\gamma$ is a non-decreasing function. Now, for all $s>0$, we have $\eta_i(s)0. $$ On the other hand, it is well-known that the maximum of finitely many upper semi-continuous functions is upper semi-continuous. As consequence, the function $\gamma$ belongs to the set $\Lambda$. \end{proof} In what follows let $\mathbb{E}=C(I;\mathbb{R})$ be the set of real and continuous functions in the compact set $I$. It is well-known that $\mathbb{E}$ is a Banach space with respect to the norm $$ \|x\|=\max\{|x(t)|: t\in I\},\quad x\in \mathbb{E}. $$ Now, let $M\in B_{\mathbb E}$ be fixed. For $(x,\rho)\in M\times (0,\infty)$, we denote by $\omega(x,\rho)$ the modulus of continuity of $x$; that is, $$ \omega(x,\rho)=\sup\{|x(t_1)-x(t_2)|: (t_1,t_2)\in I\times I,\; |t_1-t_2|\leq \rho\}. $$ Further on let us define $$ \omega(M,\rho)=\sup\{\omega(x,\rho): x\in M\}. $$ Define the mapping $\sigma: B_{\mathbb E}\to [0,\infty)$ by $$ \sigma(M)=\lim_{\rho\to 0^+} \omega(M,\rho),\quad M\in B_{\mathbb E}. $$ Then $\sigma$ is a measure of non-compactness in $\mathbb{E}$ (see \cite{BG}). \section{Main result}\label{s3} Define the operator $T$ on $\mathbb{E}=C(I;\mathbb{R})$ by $$ (Tx)(t)=F\Big(t,x(a(t)),\frac{f(t,x(b(t)))}{\Gamma_q(\alpha)} \int_0^t (t-qs)^{(\alpha-1)}u(s,x(s))\,d_qs\Big), \quad (x,t)\in \mathbb{E}\times I. $$ We consider the assumption \begin{itemize} \item[(A1)] The functions $f,u: [0,1]\times \mathbb{R}\to \mathbb{R}$, $a,b: I\to I$ and $F:[0,1]\times\mathbb{R}\times \mathbb{R}\to \mathbb{R}$ are continuous. \end{itemize} \begin{proposition}\label{PR1} Under assumption \rm{(A1)}, the operator $T$ maps $\mathbb{E}$ into itself. \end{proposition} \begin{proof} From assumption (A1), we have just to show that the operator $H$ defined on $\mathbb{E}$ by \begin{equation}\label{HO} (Hx)(t)=\int_0^t (t-qs)^{(\alpha-1)}u(s,x(s))\,d_qs,\quad (x,t)\in \mathbb{E}\times I \end{equation} maps $\mathbb{E}$ into itself. To do this, let us fix $x\in \mathbb{E}$. For all $t\in I$, we have \begin{align*} (Hx)(t) &= \int_0^t (t-qs)^{(\alpha-1)} u(s,x(s))\,d_qs\\ &= t(1-q)\sum_{n=0}^\infty q^n(t-q^{n+1}t)^{(\alpha-1)} u(tq^n,x(tq^n))\\ &= t^\alpha (1-q)\sum_{n=0}^\infty q^n(1-q^{n+1})^{(\alpha-1)} u(tq^n,x(tq^n)). \end{align*} On the other hand, since $00$ and a non-decreasing function $\varphi_F: [0,\infty)\to [0,\infty)$ such that $$ |F(t,x,y)-F(t,z,w)|\leq \varphi_F(|x-z|)+C_F|y-w|,\quad (t,x,y,z,w)\in I\times \mathbb{R}\times \mathbb{R}\times\mathbb{R}\times\mathbb{R}. $$ \item[(A3)] There exists a constant $C_f>0$ such that $$ |f(t,x)-f(t,y)|\leq C_f |x-y|,\quad (t,x,y)\in I\times \mathbb{R}\times \mathbb{R}. $$ \item[(A4)] There exists a non-decreasing and continuous function $\varphi_u: [0,\infty)\to [0,\infty)$ such that \begin{gather*} |u(t,x)-u(t,y)|\leq \varphi_u(|x-y|),\quad (t,x,y)\in I\times \mathbb{R}\times \mathbb{R},\; \varphi_u(t)0,\\ u(t,0)=0,\quad t\in I. \end{gather*} \item[(A5)] There exists $r_0>0$ such that $$ \varphi_F(r_0)+F^*+\frac{C_F(C_f r_0+f^*)\varphi_u(r_0)}{\Gamma_q(\alpha+1)}\leq r_0, $$ where $$ F^*=\max\{|F(t,0,0)|:t\in I\}\quad\text{and}\quad f^*=\max\{|f(t,0)|:t\in I\}. $$ \end{itemize} Let the closed ball of center $0$ and radius $r_0$ be denote by $$ \overline{B(0,r_0)}=\{x\in \mathbb{E}: \|x\|\leq r_0\}. $$ \begin{proposition}\label{PR2} Under assumptions \rm{(A1)--(A5)}, the operator $T$ maps $\overline{B(0,r_0)}$ into itself. \end{proposition} \begin{proof} Let $x\in \overline{B(0,r_0)}$. Using the considered assumptions, for all $t\in I$, we have \begin{align*} &|(Tx)(t)|\\ &\leq \Big|F\Big(t,x(a(t)),\frac{f(t,x(b(t)))}{\Gamma_q(\alpha)} \int_0^t (t-qs)^{(\alpha-1)}u(s,x(s))\,d_qs\Big)-F(t,0,0)\Big|\\ &\quad +|F(t,0,0)| \\ &\leq \varphi_F(|x(a(t))|)+C_F \frac{|f(t,x(b(t)))|}{\Gamma_q(\alpha)} \int_0^t (t-qs)^{(\alpha-1)}|u(s,x(s))|\,d_qs+F^*\\ &\leq \varphi_F(\|x\|) +C_F \frac{|f(t,x(b(t)))-f(t,0)|+|f(t,0)|}{\Gamma_q(\alpha)}\\ &\quad\times \int_0^t (t-qs)^{(\alpha-1)}|u(s,x(s))|\,d_qs +F^*\\ &\leq \varphi_F(\|x\|) +C_F \frac{\left(C_f|x(b(t))|+f^*\right)\varphi_u(\|x\|)}{\Gamma_q(\alpha)} \int_0^t (t-qs)^{(\alpha-1)}\,d_qs +F^*\\ &\leq \varphi_F(\|x\|)+C_F \frac{\left(C_f\|x\|+f^*\right) \varphi_u(\|x\|)}{\Gamma_q(\alpha+1)}t^\alpha+F^*\\ &\leq \varphi_F(r_0)+C_F \frac{\left(C_fr_0+f^*\right) \varphi_u(r_0)}{\Gamma_q(\alpha+1)}+F^*. \end{align*} Therefore, $$ \|Tx\|\leq \varphi_F(r_0)+C_F \frac{\left(C_fr_0+f^*\right) \varphi_u(r_0)}{\Gamma_q(\alpha+1)}+F^*,\quad x\in \overline{B(0,r_0)}. $$ Using the above inequality and assumption (A5), we obtain the desired result. \end{proof} \begin{proposition} \label{PRC} Under assumptions \rm{(A1)--(A5)}, the operator $T$ maps continuously $\overline{B(0,r_0)}$ into itself. \end{proposition} \begin{proof} Define the operators $\gamma_1, \gamma_2$ and $\gamma_3$ on $\mathbb{E}$ by \begin{gather*} (\gamma_1 x)(t)= t,\quad (x,t)\in \mathbb{E}\times I,\\ (\gamma_2 x)(t)= x(a(t)),\quad (x,t)\in \mathbb{E}\times I,\\ (\gamma_3 x)(t)= f(t,x(b(t))),\quad (x,t)\in \mathbb{E}\times I. \end{gather*} Obviously, $\gamma_1: \mathbb{E}\to \mathbb{E}$ is continuous. Moreover, for all $x,y\in \mathbb{E}$, we have $$ |(\gamma_2 x)(t)-(\gamma_2 y)(t)|=|x(a(t))-y(a(t))|\leq \|x-y\|,\quad t\in I, $$ which implies that $$ \|\gamma_2 x-\gamma_2 y\|\leq \|x-y\|,\quad (x,y)\in \mathbb{E}\times \mathbb{E}. $$ Therefore, $\gamma_2$ is uniformly continuous on $\mathbb{E}$. Similarly, for all $x,y\in \mathbb{E}$, for all $t\in I$, we have \begin{align*} |(\gamma_3 x)(t)-(\gamma_3 y)(t)| &=|f(t,x(b(t)))-f(t,y(b(t)))|\\ &\leq C_f |x(b(t))-y(b(t))\|\leq C_f\|x-y\|, \end{align*} which implies $$ \|\gamma_3 x-\gamma_3 y\|\leq C_f\|x-y\|,\quad (x,y)\in \mathbb{E}\times \mathbb{E}. $$ Then $\gamma_3$ is also uniformly continuous on $\mathbb{E}$. So, in order to prove that $T$ is continuous on $\overline{B(0,r_0)}$, we only need to show that the operator $H$ defined by \eqref{HO} is continuous on $\overline{B(0,r_0)}$. To do this, let us consider $\varepsilon>0$ and $(x,y)\in \overline{B(0,r_0)}\times \overline{B(0,r_0)}$ such that $\|x-y\|\leq \varepsilon$. For all $t\in I$, we have \begin{align*} (Hx)(t)-(Hy)(t) &= \int_0^t (t-qs)^{(\alpha-1)}u(s,x(s))\,d_qs -\int_0^t (t-qs)^{(\alpha-1)}u(s,y(s))\,d_qs\\ &= \int_0^t (t-qs)^{(\alpha-1)} \left(u(s,x(s))-u(s,y(s))\right)\,d_qs. \end{align*} Set $$ u_{r_0}(\varepsilon)=\sup\{|u(t,x)-u(t,y)|: t\in I,\, (x,y)\in [-r_0,r_0]\times [-r_0,r_0],\, |x-y|\leq \varepsilon\}, $$ we obtain \[ (Hx)(t)-(Hy)(t) \leq \frac{t^\alpha}{[\alpha]_q} u_{r_0}(\varepsilon)\\ \leq \frac{u_{r_0}(\varepsilon)}{[\alpha]_q}, \] for all $t\in I$. Therefore, $$ \|Hx-Hy\|\leq \frac{u_{r_0}(\varepsilon)}{[\alpha]_q}. $$ Passing to the limit as $\varepsilon\to 0^+$ and using the uniform continuity of $u$ on the compact set $I\times [-r_0,r_0]$, we obtain $$ \lim_{\varepsilon\to 0^+} \frac{u_{r_0}(\varepsilon)}{[\alpha]_q}=0, $$ which completes the proof. \end{proof} To prove our main result, the following additional assumptions are needed. \begin{itemize} \item[(A6)] The function $\varphi_F: [0,\infty)\to [0,\infty)$ is continuous and it satisfies $\varphi_F(s)0$. \item[(A7)] The function $a: I\to I$ satisfies $$ |a(t)-a(s)|\leq \varphi_a(|t-s|),\quad (t,s)\in I\times I, $$ where $\varphi_a: [0,\infty)\to [0,\infty)$ is non-decreasing and $\lim_{t\to 0^+}\varphi_a(t)=0$. \item[(A8)] The function $b: I\to I$ satisfies $$ |b(t)-b(s)|\leq \varphi_b(|t-s|),\quad (t,s)\in I\times I, $$ where $\varphi_b: [0,\infty)\to [0,\infty)$ is non-decreasing and $\lim_{t\to 0^+}\varphi_b(t)=0$. \item[(A9)] We suppose that $$ 0<\varphi_u(r_0)< \frac{\Gamma_q(\alpha+1)}{C_FC_f}\quad\text{and}\quad \frac{C_F}{\Gamma_q(\alpha)}(C_fr_0+f^*)<1. $$ \end{itemize} Our main result is the following. \begin{theorem}\label{T} Under assumptions \rm{(A1)--(A9)}, Equation \eqref{eq} has at least one solution $x^*\in C(I;\mathbb{R})$ satisfying $\|x^*\|\leq r_0$. \end{theorem} \begin{proof} From Proposition \ref{PRC}, we know that $T: \overline{B(0,r_0)}\to \overline{B(0,r_0)}$ is a continuous operator. Now, let $M$ be a nonempty subset of $\overline{B(0,r_0)}$. Let $\rho>0$, $x\in M$ and $(t_1,t_2)\in I\times I$ be such that $|t_1-t_2|\leq \rho$. Without restriction of the generality, we may assume that $t_1\geq t_2$. We have \begin{equation} \begin{aligned} &|(Tx)(t_1)-(Tx)(t_2)|\\ &=\Big|F\Big(t_1,x(a(t_1)),\frac{f(t_1,x(b(t_1)))} {\Gamma_q(\alpha)}\int_0^{t_1} (t_1-qs)^{(\alpha-1)}u(s,x(s))\,d_qs\Big)\\ &\quad -F\Big(t_2,x(a(t_2)),\frac{f(t_2,x(b(t_2)))}{\Gamma_q(\alpha)} \int_0^{t_2} (t_2-qs)^{(\alpha-1)}u(s,x(s))\,d_qs\Big)\Big|\\ &\leq \Big|F\Big(t_1,x(a(t_1)),\frac{f(t_1,x(b(t_1)))} {\Gamma_q(\alpha)}\int_0^{t_1} (t_1-qs)^{(\alpha-1)}u(s,x(s))\,d_qs\Big) \\ &\quad -F\Big(t_2,x(a(t_1)),\frac{f(t_1,x(b(t_1)))} {\Gamma_q(\alpha)}\int_0^{t_1} (t_1-qs)^{(\alpha-1)}u(s,x(s))\,d_qs\Big)\Big|\\ &\quad +\Big|F\Big(t_2,x(a(t_1)),\frac{f(t_1,x(b(t_1)))} {\Gamma_q(\alpha)}\int_0^{t_1} (t_1-qs)^{(\alpha-1)}u(s,x(s))\,d_qs\Big) \\ &\quad -F\Big(t_2,x(a(t_2)),\frac{f(t_2,x(b(t_2)))} {\Gamma_q(\alpha)}\int_0^{t_2} (t_2-qs)^{(\alpha-1)}u(s,x(s))\,d_qs\Big)\Big|\\ &= (I)+(II). \end{aligned} \label{III} \end{equation} \noindent $\bullet$ Estimate for (I). We have \begin{align*} &\Big|\frac{f(t_1,x(b(t_1)))}{\Gamma_q(\alpha)} \int_0^{t_1} (t_1-qs)^{(\alpha-1)}u(s,x(s))\,d_qs\Big|\\ & \leq \frac{|f(t_1,x(b(t_1)))|}{\Gamma_q(\alpha)} \int_0^{t_1} (t_1-qs)^{(\alpha-1)}|u(s,x(s))|\,d_qs\\ &\leq \frac{|f(t_1,x(b(t_1)))-f(t_1,0)|+|f(t_1,0)|} {\Gamma_q(\alpha)}\int_0^{t_1} (t_1-qs)^{(\alpha-1)}\varphi_u(|x(s)|)\,d_qs\\ &\leq \frac{(C_f|x(b(t_1))|+f^*)\varphi_u(\|x\|)} {\Gamma_q(\alpha+1)}t_1^\alpha\\ &\leq \frac{(C_f\|x\|+f^*)\varphi_u(\|x\|)} {\Gamma_q(\alpha+1)} \\ &\leq \frac{(C_fr_0+f^*)\varphi_u(r_0)}{\Gamma_q(\alpha+1)}=D. \end{align*} Set \begin{align*} C(F,\delta) = \sup\big\{&|F(t,x,y)-F(s,x,y)|: (t,s)\in I\times I,\,|t-s|\leq \rho,\\\ &x\in [-r_0,r_0],\,y\in [-D,D]\big\}, \end{align*} we obtain \begin{equation}\label{I} (I)\leq C(F,\delta). \end{equation} \noindent $\bullet$ Estimate for (II). We have \begin{align*} (II) &\leq \varphi_F(|x(a(t_1))-x(a(t_2))|)\\ &\quad + \frac{C_F}{\Gamma_q(\alpha)} \Big|f(t_1,x(b(t_1)))\int_0^{t_1} (t_1-qs)^{(\alpha-1)}u(s,x(s))\,d_qs\\ &\quad - f(t_2,x(b(t_2)))\int_0^{t_2} (t_2-qs)^{(\alpha-1)}u(s,x(s))\,d_qs\Big|. \end{align*} On the other hand, $$ |x(a(t_1))-x(a(t_2))|\leq \omega(x\circ a,\rho), $$ which yields $$ \varphi_F(|x(a(t_1))-x(a(t_2))|)\leq \varphi_F(\omega(x\circ a,\rho)). $$ Now, we have \begin{align*} &\Big|f(t_1,x(b(t_1)))\int_0^{t_1} (t_1-qs)^{(\alpha-1)}u(s,x(s))\,d_qs\\ &\quad -f(t_2,x(b(t_2)))\int_0^{t_2} (t_2-qs)^{(\alpha-1)}u(s,x(s))\,d_qs\Big|\\ &\leq \Big|f(t_1,x(b(t_1)))\int_0^{t_1} (t_1-qs)^{(\alpha-1)}u(s,x(s))\,d_qs \\ &\quad -f(t_2,x(b(t_2)))\int_0^{t_1} (t_1-qs)^{(\alpha-1)}u(s,x(s))\,d_qs\Big|\\ &\quad +\Big|f(t_2,x(b(t_2)))\int_0^{t_1} (t_1-qs)^{(\alpha-1)}u(s,x(s))\,d_qs \\ &\quad -f(t_2,x(b(t_2)))\int_0^{t_2} (t_2-qs)^{(\alpha-1)}u(s,x(s))\,d_qs\Big|\\ &\leq \frac{|f(t_1,x(b(t_1)))-f(t_2,x(b(t_2)))|\varphi_u(\|x\|)}{[\alpha]_q}\\ &\quad +|f(t_2,x(b(t_2)))|\Big|\int_0^{t_1} (t_1-qs)^{(\alpha-1)}u(s,x(s))\,d_qs\\ &\quad -\int_0^{t_2} (t_2-qs)^{(\alpha-1)}u(s,x(s))\,d_qs\Big|\\ &=(III)+(IV). \end{align*} Let us define $$ \omega_f(r_0,\rho)=\sup\{|f(t,x)-f(s,x)|:(t,s)\in I\times I,\,|t-s| \leq \rho,\, x\in [-r_0,r_0]\}. $$ Then \begin{align*} (III)&\leq \frac{\varphi_u(\|x\|)}{[\alpha]_q}|f(t_1,x(b(t_1)))-f(t_1,x(b(t_2)))|\\ &\quad +\frac{\varphi_u(\|x\|)}{[\alpha]_q}|f(t_1,x(b(t_2)))-f(t_2,x(b(t_2)))|\\ &\leq \frac{\left[C_f|x(b(t_1))-x(b(t_2))|+\omega_f(r_0,\rho)\right] \varphi_u(r_0)}{[\alpha]_q}\\ &\leq \frac{\left[C_f\omega(x\circ b,\rho)+\omega_f(r_0,\rho)\right] \varphi_u(r_0)}{[\alpha]_q}. \end{align*} Now, let us estimate $(IV)$. At first, we have \begin{align*} |f(t_2,x(b(t_2)))| &\leq |f(t_2,x(b(t_2)))-f(t_2,0)|+|f(t_2,0)|\\ &\leq C_f |x(b(t_2))|+f^*\leq C_f r_0+f^*. \end{align*} Next, we have \begin{align*} &\Big|\int_0^{t_1} (t_1-qs)^{(\alpha-1)}u(s,x(s))\,d_qs -\int_0^{t_2} (t_2-qs)^{(\alpha-1)}u(s,x(s))\,d_qs\Big|\\ &= (1-q) \sum_{n=0}^\infty q^n(1-q^{n+1})^{(\alpha-1)} \left|t_1^\alpha u(q^nt_1,x(q^nt_1))- t_2^\alpha u(q^nt_2,x(q^nt_2))\right|. \end{align*} We can write \begin{align*} &\left|t_1^\alpha u(q^nt_1,x(q^nt_1))- t_2^\alpha u(q^nt_2,x(q^nt_2))\right|\\ &\leq t_1^\alpha \left|u(q^nt_1,x(q^nt_1))-u(q^nt_1,x(q^nt_2))\right|\\ &\quad +\left|t_1^\alpha u(q^nt_1,x(q^nt_2))-t_2^\alpha u(q^nt_2,x(q^nt_2))\right|\\ &\leq \varphi_u(|x(q^nt_1)-x(q^nt_2)|)+A_\rho\\ &\leq \varphi_u(\omega(x,\rho))+A_\rho,\end{align*} where \begin{align*} A_\rho= \sup\big\{&|\mathcal{N}(\tau,s,x)-\mathcal{N}(\tau',s',x)|: (\tau,s,\tau',s')\in I^4,\, |\tau-\tau'|\leq \rho,\\ & |s-s'|\leq \rho,\,x\in [-r_0,r_0]\big\} \end{align*} and $$ \mathcal{N}(\tau,s,x)=\tau^\alpha u(s,x),\; (\tau,s,x)\in I\times I\times \mathbb{R}. $$ Then, we obtain \begin{align*} &\Big|\int_0^{t_1} (t_1-qs)^{(\alpha-1)}u(s,x(s))\,d_qs -\int_0^{t_2} (t_2-qs)^{(\alpha-1)}u(s,x(s))\,d_qs\Big|\\ &\leq \varphi_u(\omega(x,\rho))+A_\rho. \end{align*} As consequence, we have $$ (IV)\leq (C_fr_0+f^*)(\varphi_u(\omega(x,\rho))+A_\rho). $$ Using the above inequalities, we obtain \begin{align*} (II) &\leq \varphi_F(\omega(x\circ a,\rho)) +\frac{C_F}{\Gamma_q(\alpha)}\Big(\frac{[C_f\omega(x\circ b,\rho) +\omega_f(r_0,\rho)]\varphi_u(r_0)}{[\alpha]_q}\\ &\quad +(C_fr_0+f^*)(\varphi_u(\omega(x,\rho))+A_\rho) \Big). \end{align*} Now, observe that from assumption (A7), we have \begin{align*} \omega(x\circ a,\rho) &= \sup\{|x(a(t))-x(a(s))|: (t,s)\in I\times I,\, |t-s|\leq \rho\}\\ &\leq \sup\{|x(\mu)-x(\nu)|: (\mu,\nu)\in I\times I,\, |\mu-\nu|\leq \varphi_a(\rho)\}\\ &= \omega(x,\varphi_a(\rho)). \end{align*} Similarly, from assumption (A8), we have $$ \omega(x\circ b,\rho)\leq \omega(x,\varphi_b(\rho)). $$ Then \begin{equation} \label{II} \begin{aligned} (II) &\leq \varphi_F(\omega(x,\varphi_a(\rho))) +\frac{C_F}{\Gamma_q(\alpha)}\Big(\frac{[C_f\omega(x,\varphi_b(\rho)) +\omega_f(r_0,\rho)]\varphi_u(r_0)}{[\alpha]_q}\\ &\quad +(C_fr_0+f^*)(\varphi_u(\omega(x,\rho))+A_\rho) \Big). \end{aligned} \end{equation} Next, using \eqref{III}, \eqref{I} and \eqref{II}, we obtain \begin{align*} \omega(Tx,\rho) &\leq C(F,\delta)+\varphi_F(\omega(x,\varphi_a(\rho))) +\frac{C_F}{\Gamma_q(\alpha)}\Big(\frac{[C_f\omega(x,\varphi_b(\rho)) +\omega_f(r_0,\rho)]\varphi_u(r_0)}{[\alpha]_q}\\ &\quad +(C_fr_0+f^*)(\varphi_u(\omega(x,\rho))+A_\rho) \Big), \end{align*} which yields \begin{align*} \omega(TM,\rho) &\leq C(F,\delta)+\varphi_F(\omega(M,\varphi_a(\rho)))\\ &\quad +\frac{C_F}{\Gamma_q(\alpha)}\Big(\frac{[C_f\omega(M,\varphi_b(\rho)) +\omega_f(r_0,\rho)]\varphi_u(r_0)}{[\alpha]_q}\\ &\quad +(C_fr_0+f^*)(\varphi_u(\omega(M,\rho))+A_\rho) \Big). \end{align*} Recall that from assumptions (A7)--(A8), we have $$ \lim_{\rho\to 0^+}\varphi_a(t)=\lim_{\rho\to 0^+}\varphi_b(t)=0. $$ Then passing to the limit as $\rho\to 0^+$ in the above inequality, we obtain $$ \sigma(TM)\leq \varphi_F(\sigma(M)) +\frac{C_F}{\Gamma_q(\alpha)} \Big(\frac{C_f\sigma(M)\varphi_u(r_0)}{[\alpha]_q}+(C_fr_0+f^*)\varphi_u(\sigma(M)) \Big). $$ Therefore, $$ \sigma(TM)\leq \eta(\sigma(M)), $$ where $$ \eta(t)=\max\{\varphi_F(t), L\varphi_u(t),Nt\},\quad t\geq 0, $$ with \[ L=\frac{C_F}{\Gamma_q(\alpha)}(C_fr_0+f^*), \quad N=\frac{C_FC_f}{\Gamma_q(\alpha+1)}\varphi_u(r_0). \] From assumption (A9) and Lemma \ref{CF}, the function $\eta$ belongs also to the set $\Lambda$. Finally, applying Lemma \ref{LD}, we obtain the existence of at least one fixed point of the operator $T$ in $\overline{B(0,r_0)}$, which is a solution to \eqref{eq}. \end{proof} We end the paper with the following illustrative example. Consider the integral equation \begin{equation}\label{ex1} x(t)=\frac{t}{32}+\frac{x(t)}{4}+[\alpha]_q \big(\frac{t}{2}+\frac{x(t)}{4}\big)\int_0^t (t-qs)^{(\alpha-1)} \frac{x(s)}{(2+s^2)}\,d_qs, \end{equation} for $t\in I=[0,1]$, where $\alpha>1$ and $q\in (0,1)$. Observe that \eqref{ex1} can be written in the form \eqref{eq}, where \begin{gather*} a(t)= t,\quad t\in I,\quad b(t)= t,\quad t\in I,\\ F(t,x,y) = \frac{t}{32}+\frac{x}{4}+\Gamma_q(\alpha+1)y,\quad (t,x,y)\in I\times \mathbb{R}\times \mathbb{R},\\ f(t,x) = \frac{t}{2}+\frac{x}{4},\quad (t,x)\in I\times \mathbb{R},\\ u(t,x) = \frac{x}{(2+t^2)},\quad (t,x)\in I \times \mathbb{R}. \end{gather*} Now, let us check that the required assumptions by Theorem \ref{T} are satisfied. \noindent $\bullet$ Assumption (A1). It is trivial. \noindent $\bullet$ Assumption (A2). For all $(t,x,y,z,w)\in I\times\mathbb{R}\times\mathbb{R} \times\mathbb{R}\times\mathbb{R}$, we have \begin{align*} |F(t,x,y)-F(t,z,w)| &= \left|\frac{x}{4}+\Gamma_q(\alpha+1)y -\frac{z}{4}-\Gamma_q(\alpha+1)w\right|\\ &\leq \frac{|x-z|}{4}+\Gamma_q(\alpha+1)|y-w|. \end{align*} Then assumption (A2) is satisfied with \begin{gather*} \varphi_F(t)=\frac{t}{4},\quad t\geq 0,\\ C_F=\Gamma_q(\alpha+1). \end{gather*} \noindent $\bullet$ Assumption (A3). For all $(t,x,y)\in I\times \mathbb{R}\times \mathbb{R}$, we have $$ |f(t,x)-f(t,y)|=\frac{|x-y|}{4}. $$ Then assumption (A3) is satisfied with $C_f=\frac{1}{4}$. \noindent $\bullet$ Assumption (A4). For all $(t,x,y)\in I\times \mathbb{R}\times \mathbb{R}$, we have $$ |u(t,x)-u(t,y)|=\frac{|x-y|}{2+t^2}\leq \frac{|x-y|}{2}. $$ Take $\varphi_u(t)=\frac{t}{2}$, $t\geq 0$, assumption (A4) holds. \noindent $\bullet$ Assumption (A5). At first, in our case, we have $F^*=\frac{1}{32}$ and $f^*=\frac{1}{2}$. Now, the inequality $$ \varphi_F(r_0)+F^*+\frac{C_F(C_f r_0+f^*)\varphi_u(r_0)}{\Gamma_q(\alpha+1)}\leq r_0 $$ is equivalent to $$ r_0^2-4r_0+\frac{1}{4}\leq 0. $$ The above inequality is satisfied for any $r_0\in [\frac{4-\sqrt 15}{2},\frac{4+\sqrt 15}{2}]$. \noindent $\bullet$ Assumptions (A6)--(A8) are trivial. \noindent $\bullet$ Assumption (A9). The inequality $$ 0<\varphi_u(r_0)<\frac{\Gamma_q(\alpha+1)}{C_FC_f} $$ is equivalent to $0