On the existence of solutions for a multi-singular pointwise defined fractional q-integro-differential equation

Abstract

By using the Caputo type and the Riemann–Liouville type fractional q-derivative, we investigate the existence of solutions for a multi-term pointwise defined fractional q-integro-differential equation with some boundary value conditions. In fact, we give some results by considering different conditions and using some classical fixed point techniques and the Lebesgue dominated convergence theorem.

1 Introduction

It is known that the subject of q-difference equations was introduced by Jackson in 1910 [1]. After that, some researchers studied q-difference equations [2–20]. On the other hand, many modern works on integro-differential equations by using different views and fractional derivatives have been published recently, and young researchers could use the main idea of the works for their works (see, for example, [21–50]).

In 2012, Ahmad et al. studied the existence and uniqueness of solutions for the fractional q-difference equation ${}^{c}D_{q}^{ \alpha }u(t)= T ( t, u(t) )$ with boundary conditions $\alpha _{1} u(0) - \beta _{1} D_{q} u(0) = \gamma _{1} u(\eta _{1})$ and $\alpha _{2} u(1) - \beta _{2} D_{q} u(1) = \gamma _{2} u(\eta _{2})$, where $\alpha \in (1, 2]$, $\alpha _{i}$, $\beta _{i}$, $\gamma _{i}$, $\eta _{i}$ are real numbers for $i=1,2$ and $T \in C(J \times \mathbb{R}, \mathbb{R})$ [6]. In 2013, Zhao et al. reviewed the q-integral problem $(D_{q}^{\alpha }u)(t) + f(t, u(t) )=0$ with boundary conditions $u(1)=\mu I_{q}^{\beta }u(\eta )$ and $u(0)=0$ for almost all $t \in (0,1)$, where $q \in (0,1)$, $\alpha \in (1, 2]$, $\beta \in (0, 2]$, $\eta \in (0,1)$, μ is a positive real number, $D_{q}^{\alpha }$ is the q-derivative of Riemann–Liouville and real-valued continuous map u defined on $I \times [0, \infty )$ [15]. In 2014, Ahmad et al. investigated the problem

$$ {}^{c}D^{\beta }_{q} \bigl( {}^{c}D^{\gamma }_{q} + \lambda \bigr) u(t) = p f \bigl(t, u(t)\bigr) + k I_{q}^{\xi }g\bigl(t, u(t)\bigr), $$

with boundary conditions $\alpha _{1} u(0) - \beta _{1} (t^{(1-\gamma )} D_{q} u(0)) |_{t=0}= \sigma _{1} u(\eta _{1})$ and $\alpha _{2} u(1) + \beta _{2} D_{q} u(1)= \sigma _{2} u(\eta _{2})$, where $t, q \in [0,1]$, ${}^{c}D_{q}^{\beta }$ is the fractional Caputo q-derivative, $0 < \beta$, $\gamma \leq 1$, $I_{q}^{\xi }(\cdot)$ denotes the Riemann–Liouville integral with $\xi \in (0, 1)$, f and g are given continuous functions, λ and p, k are real constants, $\alpha _{i}, \beta _{i}, \sigma _{i}\in \mathbb{R}$ and $\eta _{i} \in (0, 1)$ for $i=1,2$ [5]. In 2017, Wang considered the existence of uniqueness and nonexistence of positive solution for fractional differential equations $D_{0^{+}}^{\sigma }x(t) + f(t, x(t))=0$ for $t \in (0,1)$ under conditions the $x(0) = x'(0) = \cdots = x^{(n-2)} (0) =0$ and $D_{0^{+}}^{\alpha }x(1) = \int _{0}^{b} \mu (t) D_{0^{+}}^{\beta }x(t) \,\mathrm{d}t$, where $n-1< \sigma \leq n$, $n \geq 3$, $\alpha \in (0,1)$,

$$ \varGamma (\sigma - \alpha ) \int _{0}^{b} \mu (t) t^{\sigma - \beta -1} \,\mathrm{d}t < \varGamma (\sigma - \beta ), $$

$b \in (0, 1]$, $D_{0^{+}}^{\sigma }$, $D_{0^{+}}^{\alpha }$, $D_{0^{+}} ^{\beta }$ are the standard Riemann–Liouville derivatives, $f : (0,1) \times [0, \infty ) \to [0, \infty )$ is continuous and $\mu (t) \in L^{1} ([0,1])$ is nonnegative [51]. Also, in 2018 he investigated the existence and multiplicity of positive solutions for the fractional differential equation $D_{0^{+}}^{\sigma }x(t) + f(t, x(t))=0$ for $t \in (0,1)$ under the conjugate type integral boundary conditions $x(0) = x'(0) = \cdots = x^{(n-2)} (0) =0$ and $D_{0^{+}}^{\alpha }x(1) = \int _{0}^{b} \mu (t) D_{0^{+}}^{\beta }x(t) \,\mathrm{d}V(t)$, where $D_{0^{+}}^{\alpha }$, $D_{0^{+}}^{ \beta }$ are the standard Riemann–Liouville derivatives, $n \geq 3$, $\alpha \in (0,1)$, $0 \leq \beta < \sigma -1$, $b \in (0, 1]$, $f(t, x)$ may be singular at $t=0, 1$ and $x=0$, $\mu (t) \in L^{1} [0,1] \cap C(0,1)$ is nonnegative, $\int _{0}^{b} \mu (t) t^{\sigma - \beta -1} \,\mathrm{d}V(t)$ denotes the Riemann–Stieltjes integral, in which V has bounded variation [52].

In 2019, Samei et al. reviewed the existence of solutions for some multi-term q-integro-differential equations with non-separated and initial boundary conditions [12]. Also, Ntouyas et al. [10], by applying definitions of the fractional q-derivative of the Caputo type and the fractional q-integral of the Riemann–Liouville type, studied the existence and uniqueness of solutions for multi-term nonlinear fractional q-integro-differential equations under some boundary conditions

$$ {}^{c}D_{q}^{\alpha } x(t) = w \bigl( t, x(t), ( \varphi _{1} x) (t), ( \varphi _{2} x) (t), {}^{c}D_{q}^{ \beta _{1}} x(t), {}^{c}D_{q}^{\beta _{2}} x(t), \ldots , {}^{c}D_{q}^{ \beta _{n}}x(t) \bigr). $$

In 2020, Liang et al. investigated the existence of solutions for nonlinear problems regular and singular fractional q-differential equation

$$ {}^{c}D_{q}^{\alpha }f(t) = w \bigl(t, f(t), f'(t), {}^{c}D_{q}^{ \beta }f(t) \bigr), $$

with conditions $f(0) = c_{1} f(1)$, $f'(0)= c_{2} {}^{c}D_{q}^{ \beta } f (1)$, and $f^{(k)}(0) = 0$ for $2\leq k \leq n-1$, here $n-1 < \alpha < n$ with $n\geq 3$, $\beta , q , c_{1}\in (0,1)$, $c_{2} \in (0, \varGamma _{q} (2- \beta ))$, function w is an $L^{\kappa }$-Carathéodory, $w(t, x_{1}, x_{2}, x_{3})$ may be singular, and ${}^{c}D_{q}^{\alpha }$ is the fractional Caputo type q-derivative [17]. Also, they discussed the existence of solutions for the fractional q-derivative inclusions

$$ {}^{c}D_{q}^{\alpha }x(t) \in F \bigl( t, x(t), x'(t), {}^{c}D_{q} ^{\beta }x(t) \bigr), $$

$x(0) + x'(0) + {}^{c}D_{q}^{\beta }x(0) = \int _{0}^{\eta _{1}} x(s) \,\mathrm{d}s$, and $x(1) + x'(1) + {}^{c}D_{q}^{\beta }x(1) = \int _{0}^{\eta _{2}} x(s) \,\mathrm{d}s$ for any t in I and $q, \eta _{1}, \eta _{2}, \beta \in (0,1)$, where F maps $I\times \mathbb{R}^{3}$ into $2^{\mathbb{R}}$ is a compact-valued multifunction and ${}^{c}D_{q}^{\alpha }$ is the fractional Caputo type q-derivative operator of order $\alpha \in (1, 2]$, and

$$ \varGamma _{q} (2- \beta ) \bigl(\eta ^{2} \nu - \nu ^{2} \eta - \eta ^{2} + \nu ^{2} + 4\eta - 2\nu -2\bigr) + 2(1-\eta ) \neq 0 $$

such that $\alpha -\beta >1$ [14]. Similar results have been presented in other studies [12, 13, 19, 20, 37].

By using the main idea of [41, 42, 53], we are going to investigate the multi-singular fractional q-integro-differential pointwise defined equation

$$ D_{q}^{\alpha } u(t)= \omega \bigl(t, u(t), u'(t), D_{q}^{\beta _{1}} u(t), I_{q}^{\beta _{2}} u(t) \bigr) $$

(1)

under two distinct boundary conditions

$$ \begin{aligned} &u'(0) = u(a),\qquad u(1) =\int _{0}^{b} u(r) \,\mathrm{d}r, \quad \alpha \in [2,3), \\ &u'(0) = u(a),\qquad u(1) = \int _{0}^{b} u(r) \,\mathrm{d}r, \quad \alpha \in [3,\infty ), \end{aligned} $$

(2)

and $u^{(j)}(0)=0$ for $j=2,\ldots ,[\alpha ]-1$, where $t \in \overline{J}=[0,1]$, $u \in \mathcal{B}=C^{1}(\overline{J})$, α, $\beta _{1}$, $\beta _{2}$ belong to $[2,\infty )$, $J=(0,1)$, $(1, \infty )$, $a, b \in J$, $D_{q}^{\alpha }$ is the Caputo fractional q-derivative of order α, and $\omega : \overline{J} \times \mathbb{R}^{4} \to \mathbb{R}$ is a function such that $\omega (t, \cdot, \cdot, \cdot, \cdot)$ is singular at some points $t\in \overline{J}$.

2 Preliminaries

Here, we recall some basic notion, lemmas, and theorems which are used in the subsequent sections. Let $q \in (0,1)$ and $a \in \mathbb{R}$. Define $[a]_{q}=\frac{1-q^{a}}{1-q}$ [1]. The power function $(x-y)_{q}^{n}$ with $n \in \mathbb{N}_{0}$ is defined by $(x-y)_{q}^{(n)}= \prod_{k=0}^{n-1} (x - yq^{k})$ for $n\geq 1$ and $(x-y)_{q}^{(0)}=1$, where x and y are real numbers and $\mathbb{N}_{0} := \{ 0\} \cup \mathbb{N}$ [2]. Also, for $\alpha \in \mathbb{R}$ and $a \neq 0$, we have

$$ (x-y)_{q}^{(\alpha )}= x^{\alpha }\prod _{k=0}^{\infty }\bigl(x-yq^{k}\bigr) \big/ \bigl(x - yq^{\alpha + k}\bigr). $$

If $y=0$, then it is clear that $x^{(\alpha )}= x^{\alpha }$ (Algorithm 1). The q-gamma function is given by $\varGamma _{q}(z) = (1-q)^{(z-1)} / (1-q)^{z -1}$, where $z \in \mathbb{R} \backslash \{0, -1, -2, \ldots \}$ [1]. Note that $\varGamma _{q} (z+1) = [z]_{q} \varGamma _{q} (z)$. The value of q-gamma function is $\varGamma _{q}(z)$ for input values q and z with counting the number of sentences n in summation by simplifying analysis (see Tables 1–3). For this design, we prepare a pseudo-code description of the technique for estimating q-gamma function of order n, which is shown in Algorithm 2. The q-derivative of function f is defined by $(D_{q} f)(x) = \frac{f(x) - f(qx)}{(1- q)x}$ and $(D_{q} f)(0) = \lim_{x \to 0} (D_{q} f)(x)$, which is shown in Algorithm 3 [2]. Also, the higher order q-derivative of a function f is defined by $(D_{q}^{n} f)(x) = D_{q}(D_{q}^{n-1} f)(x)$ for all $n \geq 1$, where $(D_{q}^{0} f)(x) = f(x)$ [2, 3]. The q-integral of a function f defined on $[0,b]$ is defined by

$$ I_{q} f(x) = \int _{0}^{x} f(s) \,\mathrm{d}_{q} s = x(1- q) \sum_{k=0} ^{\infty } q^{k} f\bigl(x q^{k}\bigr) $$

for $0 \leq x \leq b$, provided the series is absolutely convergent [2, 3]. The q-derivative of function f is defined by $(D_{q} f)(x) = \frac{f(x) - f(qx)}{(1- q)x}$ and $(D_{q} f)(0) = \lim_{x \to 0} (D_{q} f)(x)$, which is shown in Algorithm 3 [2, 3]. If $a \in [0, b]$, then

$$ \int _{a}^{b} f(u) \,\mathrm{d}_{q} u = (1-q) \sum_{k=0}^{\infty } q ^{k} \bigl[ b f\bigl(b q^{k}\bigr) - a f\bigl(a q^{k}\bigr) \bigr], $$

whenever the series exists [2, 3]. The operator $I_{q}^{n}$ is given by $(I_{q}^{0} h)(x) = h(x)$ and $(I_{q}^{n} h)(x) = (I_{q} (I_{q}^{n-1} h)) (x)$ for $n \geq 1$ and $g \in C([0,b])$ [2, 3]. It has been proved that $(D_{q} (I_{q} f))(x) = f(x)$ and $(I_{q} (D_{q} f))(x) = f(x) - f(0)$ whenever f is continuous at $x =0$ [2, 3]. The fractional Riemann–Liouville type q-integral of the function f on J for $\alpha \geq 0$ is defined by $(I_{q}^{0} f)(t) = f(t)$ and

$$ \bigl(I_{q}^{\alpha }f\bigr) (t) = \frac{1}{\varGamma _{q}(\alpha )} \int _{0}^{t} (t- qs)^{(\alpha - 1)} f(s) \,\mathrm{d}_{q}s $$

for $t \in J$ and $\alpha >0$ [9]. Also, the Caputo fractional q-derivative of a function f is defined by

$$ \begin{aligned}[b] \bigl( {}^{c}D_{q}^{\alpha }f \bigr) (t) & = \bigl( I_{q}^{[\alpha ]-\alpha }\bigl( D_{q}^{[\alpha ]} f\bigr) \bigr) (t) \\ & = \frac{1}{\varGamma _{q} ([\alpha ]-\alpha )} \int _{0} ^{t} (t- qs)^{ ([\alpha ]-\alpha -1 )} \bigl( D_{q}^{[ \alpha ]} f \bigr) (s) \,\mathrm{d}_{q}s, \end{aligned} $$

(3)

where $t \in J$ and $\alpha >0$ ([9]). It has been proved that $( I_{q}^{\beta } (I_{q}^{\alpha } f)) (x) = ( I_{q}^{ \alpha + \beta } f) (x)$ and $(D_{q}^{\alpha } (I_{q}^{\alpha } f) ) (x)= f(x)$, where $\alpha , \beta \geq 0$ ([9]). By using Algorithm 2, we can calculate $(I_{q}^{\alpha }f)(x)$, which is shown in Algorithm 4.

Algorithm 1

The proposed method for calculated $(a-b)_{q}^{(\alpha )}$

Algorithm 2

The proposed method for calculated $\varGamma _{q}(x)$

Algorithm 3

The proposed method for calculated $(D_{q} f)(x)$

Algorithm 4

The proposed method for calculated $(I_{q}^{\alpha}f)(x)$

Table 1 Some numerical results for calculation of $\varGamma _{q}(x)$ with $q=\frac{1}{8}$, which is constant, for $x=9.5, 65, 110, 780$ in Algorithm 2

Table 2 Some numerical results for calculation of $\varGamma _{q}(x)$ with $q=\frac{1}{8}, \frac{1}{2}, \frac{4}{5}, \frac{8}{9}$ for $x=9.5$ of Algorithm 2

Table 3 Some numerical results for calculation of $\varGamma _{q}(x)$ with $q=\frac{1}{8}, \frac{1}{2}, \frac{4}{5}, \frac{8}{9}$ for $x=110$ of Algorithm 2

We say f is multi-singular when it is singular at more than one point t. Also, we say that $D_{q}^{\alpha } u(t) + h(t)=0$ is a pointwise defined equation on J̅ if there exists a set $E \subset \overline{J}$ such that the measure of $E^{c}$ is zero and the equation holds on E. In this paper, we use $\|\cdot\|_{1}$, $\|\cdot\|$ and $\Vert w \Vert _{*} = \max \{\| w\|, \|w'\| \}$ as the norm of $\overline{\mathcal{L}} =L^{1}(\overline{J})$, the sup norm $\overline{\mathcal{A}}=C(\overline{J})$, and the norm of $\overline{ \mathcal{B}}= C^{1}(\overline{J})$, respectively. Let Ψ be the family of nondecreasing functions $\psi : [0, \infty ) \to [0,\infty )$ such that $\sum_{n=1}^{\infty } \psi ^{n}(t) < \infty$ for all $t> 0$ [54]. One can check that $\psi (t)< t$ for all $t>0$ [54]. Let $T : \mathcal{X} \to \mathcal{X}$ and $\alpha : \mathcal{X} \times \mathcal{X} \to [0,\infty )$ be two maps. Then T is called an α-admissible map whenever $\alpha ( x, y) \geq 1$ implies $\alpha (Tx,Ty) \geq 1$ [55]. Let $(\mathcal{X}, \rho )$ be a complete metric space, $\psi \in \varPsi$, and $\alpha : \mathcal{X} \times \mathcal{X} \to [0, \infty )$ be a map. A self-map $T : \mathcal{X} \to \mathcal{X}$ is called an α-ψ-contraction whenever $\alpha ( s, t) \rho ( Ts, Tt ) \leq \psi ( \rho ( s, t))$ for all $s, t \in \mathcal{X}$ [55]. We need the following results.

Lemma 1

([56])

Suppose that$0< n-1\leq \alpha < n$and$u \in \overline{\mathcal{A}} \cap \overline{\mathcal{L}}$. Then$I_{q}^{\alpha } D_{q}^{\alpha } u(t)= u(t)+ \sum_{i=0}^{n-1} c_{i} t ^{i}$for some constants$c_{i} \in \mathbb{R}$.

Lemma 2

([55])

Let$(\mathcal{X}, \rho )$be a complete metric space, $\psi \in \varPsi$, $\alpha : \mathcal{X} \times \mathcal{X} \to [0,\infty )$be a map, and$T : \mathcal{X} \to \mathcal{X}$be anα-admissibleα-ψ-contraction. ThenThas a fixed point wheneverTis continuous and there exists$x_{0} \in \mathcal{X}$such that$\alpha ( x_{0}, Tx_{0}) \geq 1$.

3 Main results

First, we state and prove the following key results.

Lemma 3

Let$\alpha \geq 2$, $a, b \in J$, and$v_{0} \in L^{1}(\overline{J})$. Then$v(t)= \int ^{1}_{0} G_{q}(t,s) v_{0}(s) \,\mathrm{d}s$is a solution for the pointwise defined problem$D_{q}^{\alpha }u(t) + v_{0}(t) = 0$with boundary conditions (2), where

$$ G_{q}(t,s)= G_{q}^{0}(t, s) + \frac{1}{ 1 - b } \int _{0}^{b} G_{q} ^{0}(t,s) \,\mathrm{d}t $$

(4)

and

$$ G_{q}^{0}(t,s) = \textstyle\begin{cases} \frac{1}{\varGamma _{q}(\alpha ) } [ (1 - qs)^{(\alpha - 1)} - ( t- qs)^{( \alpha - 1)} & \\ \quad {} + (1- t) (a- qs)^{(\alpha -1)} ], & 0\leq s \leq t \leq 1, s\leq a, \\ \frac{( 1 -qs)^{(\alpha - 1)} - ( t - qs)^{(\alpha -1)}}{\varGamma _{q}( \alpha )}, & 0\leq a\leq s \leq t \leq 1, \\ \frac{(1-qs)^{(\alpha - 1)} + (1-t)( a- qs)^{(\alpha -1)}}{\varGamma _{q}( \alpha )}, & 0\leq t \leq s \leq a \leq 1, \\ \frac{(1 - qs)^{(\alpha - 1)}}{\varGamma _{q}(\alpha )},& 0\leq t \leq s \leq 1, a \leq s. \end{cases} $$

Proof

Let $E \subset \overline{J}$ be such that the equation $D_{q}^{\alpha }u(t) + v_{0}(t) = 0$ holds for all $t\in E$ and the measure of $E^{c}$ is zero. Choose $v \in \overline{\mathcal{A }}\cap \overline{ \mathcal{L}}$ such that $v =v_{0}$ on E. If $v_{0} \in \overline{\mathcal{ A }}$ is a solution for the pointwise defined problem, then we put $v(t) = -D_{q}^{\alpha } u_{0}(t)$ for all $t \in \overline{J}$. Note that $v \in \overline{\mathcal{A}} \cap \overline{ \mathcal{L}}$ and $v =v_{0}|_{E}$. Also, we have

$$\begin{aligned} I_{q}^{\alpha }\bigl(v_{0}(t)\bigr) &= \frac{1}{\varGamma _{q}(\alpha )} \int _{0} ^{t} (t - qs)^{(\alpha - 1)} v_{0}(s) \,\mathrm{d}_{q}s \\ & = \frac{1}{\varGamma _{q}( \alpha )} \biggl[ \int _{[0,t] \cap E} ( t -qs)^{( \alpha - 1)} v_{0}(s) \,\mathrm{d}_{q}s \\ &\quad {} + \int _{[0,t] \cap E^{c}} ( t -qs)^{(\alpha - 1)} v_{0}(s) \,\mathrm{d}_{q}s \biggr] \\ & = \frac{1}{\varGamma _{q}( \alpha )} \int _{[0,t] \cap E} ( t - qs)^{( \alpha -1)} v(s) \,\mathrm{d}_{q}s \\ & = \frac{1}{\varGamma _{q}(\alpha )} \biggl[ \int _{[0,t] \cap E} ( t - qs)^{( \alpha - 1)} v(s) \,\mathrm{d}_{q}s \\ &\quad {} + \int _{[0,t]\cap E^{c}} ( t - qs)^{(\alpha - 1)} v(s) \,\mathrm{d}_{q}s \biggr] \\ & = \frac{1}{\varGamma _{q}(\alpha )} \int _{0}^{t} ( t - qs )^{(\alpha -1)} v(s) \,\mathrm{d}_{q}s = I_{q}^{\alpha }\bigl(v(t) \bigr) \end{aligned}$$

for each $t\in E$. Let $t\in E^{c} \backslash \{0\}$. Choose $\{ t_{n}\}$ in E such that $t_{n} \to t^{-}$. Hence,

$$\begin{aligned} I_{q}^{\alpha } \bigl( v_{0}(t)\bigr) & = \frac{1}{\varGamma _{q}(\alpha )} \int _{0}^{t} ( t -qs)^{(\alpha -1)} v_{0}(s) \,\mathrm{d}_{q}s \\ & = \lim_{n\to \infty } \frac{1}{\varGamma _{q}(\alpha )} \int _{0}^{t _{n}} (t_{n} - qs)^{(\alpha - 1)} v_{0}(s) \,\mathrm{d}_{q}s = \lim _{n\to \infty } I_{q}^{\alpha } \bigl( v_{0} (t_{n}) \bigr) \\ & = \lim_{n\to \infty } I_{q}^{\alpha } \bigl( v ( t_{n} ) \bigr) = \lim_{ n\to \infty } \frac{1}{\varGamma _{q}(\alpha )} \int _{0}^{t_{n}} (t _{n} - q s)^{(\alpha -1)} v(s) \,\mathrm{d}_{q}s \\ & = \frac{1}{\varGamma _{q}(\alpha )} \int _{0}^{t} (t - qs )^{(\alpha -1)} v_{0}(s) \,\mathrm{d}_{q}s =I_{q}^{\alpha } \bigl( v(t)\bigr). \end{aligned}$$

If $t=0 \in E^{c}$, then $I_{q}^{\alpha }( v_{0}(t)) = I_{q}^{\alpha } ( v(t))= 0$, and so $I_{q}^{\alpha }( v_{0}(t)) = I_{q}^{\alpha } ( w(t))$ for all $t \in \overline{J}$. Thus, $I_{q}^{\alpha } ( D_{q} ^{\alpha }u(t)) = I_{q}^{\alpha }( - v_{0}(t))$ for each $t \in \overline{J}$ whenever $D_{q}^{\alpha }u(t) + v_{0}(t) = 0$ for $t \in E$. Hence, $I_{q}^{\alpha }(D^{\alpha }u(t)) = I_{q}^{\alpha }( - v(t))$ on J̅. By employing the boundary conditions and Lemma 1, we can conclude that

$$ u(t)= - \frac{1}{ \varGamma _{q}(\alpha )} \int ^{t}_{0} ( t -qs)^{(\alpha - 1)} v(s) + c_{0} + c_{1} t. $$

Since $u'(0)= u(a)$, $c_{1} = - I_{q}^{\alpha }v(a)$, and so $u(t)= -I_{q}^{\alpha }v(t) + c_{0} - I_{q}^{\alpha }v(a)$. Hence,

$$ \int ^{b}_{0} u(s) \,\mathrm{d}s = u(1)= - I_{q}^{\alpha }v(1) + c_{0} - I_{q}^{\alpha }v(a). $$

So $c_{0}= \int ^{b}_{0} u(s) \,\mathrm{d}s + I_{q}^{\alpha }v(1) + I _{q}^{\alpha }v(a)$. Thus,

$$ u(t) = -I_{q}^{\alpha }v(t) - t I_{q}^{\alpha }v(a)+ \int ^{b}_{0} u(s) \,\mathrm{d}_{q}s + I_{q}^{\alpha }v(1) + I_{q}^{\alpha }v(a). $$

Put $h(t) = -I_{q}^{\alpha }v(t) + (1- t) I_{q}^{\alpha }v(a) + I_{q} ^{\alpha }v(1)$. Then we get

$$ u(t)= h(t)+ \int ^{b}_{0} u(s) \,\mathrm{d}s. $$

(5)

We consider two cases. If $t\geq a$, then

$$\begin{aligned} h(t) & = - \frac{1}{\varGamma _{q}(\alpha )} \biggl[ \int ^{a}_{0} ( t -qs)^{( \alpha - 1)} v(s) \,\mathrm{d}_{q} s \\ &\quad {} + \int ^{t}_{a} ( t -qs)^{(\alpha - 1)} v(s) \,\mathrm{d} _{q}s \biggr] \\ &\quad {} - \frac{t}{\varGamma _{q}(\alpha )} \int ^{a}_{0} (a- qs)^{( \alpha - 1)} v(s) \,\mathrm{d}_{q}s \\ &\quad {} + \frac{1}{\varGamma _{q}( \alpha )} \biggl[ \int ^{a}_{0} ( 1 - qs)^{( \alpha - 1)} v(s) \,\mathrm{d}_{q}s \\ &\quad {} + \int ^{t}_{a} ( 1 - qs)^{(\alpha - 1)} v(s) \,\mathrm{d} _{q}s + \int ^{1}_{t} ( 1 - qs)^{(\alpha - 1)} v(s) \,\mathrm{d}_{q}s \biggr] \\ &\quad {} + \frac{1}{\varGamma _{q}(\alpha )} \int ^{a}_{0} ( a- qs )^{( \alpha - 1)} v(s) \,\mathrm{d}_{q}s \\ & = \frac{1}{ \varGamma _{q}(\alpha )} \int ^{a}_{0} \bigl[ (1 - sq)^{( \alpha -1)} -(t - qs)^{(\alpha -1)} \\ &\quad {} + (1-t) (a - qs)^{(\alpha -1)} \bigr]v(s) \,\mathrm{d}_{q}s \\ &\quad {} + \frac{1}{\varGamma _{q}(\alpha )} \int ^{t}_{a} \bigl[ ( 1 -qs)^{( \alpha -1)} -( t -qs)^{(\alpha -1)} \bigr] v(s) \,\mathrm{d}_{q}s \\ &\quad {} + \frac{1}{ \varGamma _{q}(\alpha )} \int ^{1}_{t} (1-qs)^{ ( \alpha -1)} v(s) \,\mathrm{d}_{q}s. \end{aligned}$$

(6)

If $t\leq a$, then we have

$$\begin{aligned} h(t) &= -\frac{1}{\varGamma _{q}(\alpha )} \int ^{t}_{0}( t - qs)^{(\alpha - 1)} v(s) \,\mathrm{d}_{q}s \\ &\quad {} - \frac{t}{\varGamma _{q}(\alpha )} \biggl[ \int ^{t}_{0} ( a - qs)^{( \alpha - 1)} v(s) \,\mathrm{d}_{q}s \\ &\quad {} + \int ^{a}_{t} (a- qs)^{(\alpha - 1)} v(s) \,\mathrm{d}_{q}s \biggr] \\ &\quad {} + \frac{1}{\varGamma _{q}(\alpha )} \biggl[ \int ^{t}_{0} ( 1 -qs)^{( \alpha - 1)} v(s) \,\mathrm{d}_{q}s \\ &\quad {} + \int ^{a}_{t} ( 1 -qs)^{(\alpha - 1)} v(s) \,\mathrm{d}_{q}s + \int ^{1}_{a} ( 1 -qs)^{(\alpha - 1)} v(s) \,\mathrm{d}_{q}s \biggr] \\ &\quad {} + \frac{1}{\varGamma _{q}(\alpha )} \biggl[ \int ^{t}_{0} (a - qs)^{( \alpha - 1)} v(s) \,\mathrm{d}_{q}s \\ &\quad {} + \int ^{a}_{t} (a - qs)^{(\alpha - 1)} v(s) \,\mathrm{d}_{q}s \biggr] \\ & = \frac{1 }{\varGamma _{q}(\alpha )} \int ^{t}_{0} \bigl[ ( 1 -qs)^{( \alpha -1)} -( t -qs)^{(\alpha -1)} \\ &\quad {} + (1- t ) (a- qs)^{(\alpha -1)}+ \bigr] v(s) \,\mathrm{d}_{q}s \\ &\quad {} + \frac{1 }{\varGamma _{q}(\alpha )} \int ^{a}_{t} \bigl[ ( 1 -qs)^{( \alpha -1)}- t (a- qs)^{(\alpha -1)} \\ &\quad {} + (a- qs)^{(\alpha -1)} \bigr] v(s) \,\mathrm{d}_{q}s \\ &\quad {} + \frac{1}{ \varGamma _{q}(\alpha )} \int ^{1}_{a} ( 1-qs)^{( \alpha - 1)} v(s) \,\mathrm{d}_{q}s. \end{aligned}$$

(7)

Thus, equations (6) and (7) imply that $h(t) = \int _{0}^{1} G_{q}^{0}(t,s) v(s) \,\mathrm{d}_{q}s$, and by entering $h(t)$ in equation (5), we see that

$$ u(t) = \int _{0}^{1} G_{q}^{0}(t,s) v(s) \,\mathrm{d}_{q}s + \int _{0} ^{b} u(s) \,\mathrm{d}s. $$

This implies that

$$\begin{aligned} \int _{0}^{b} u(t) \,\mathrm{d}t & = \int _{0}^{b} \int _{0}^{1} G_{q}^{0}(t,s) v(s) \,\mathrm{d}_{q}s \,\mathrm{d}t + \int _{0}^{b} \int _{0}^{b} u(s) \,\mathrm{d}s \,\mathrm{d}t \\ & = \int _{0}^{1} \biggl[ \int _{0}^{b} G_{q}^{0}(t,s) \,\mathrm{d}t \biggr] v(s) \,\mathrm{d}_{q}s + b \int _{0}^{b} u(s) \,\mathrm{d}s. \end{aligned}$$

Thus, $(1 - b ) \int _{0}^{b} u(t) \,\mathrm{d}t = \int _{0}^{1} [ \int _{0}^{b} G_{q}^{0}(t,s) \,\mathrm{d}t ] v(s) \,\mathrm{d} _{q}s$, and so

$$ \int _{0}^{b} u(t) \,\mathrm{d}t = \int _{0}^{1} \frac{1}{ 1-b} \biggl[ \int _{0} ^{b} G_{q}^{0} (t, s) \,\mathrm{d}t \biggr] v(s) \,\mathrm{d}_{q}s. $$

Hence,

$$\begin{aligned} u(t) &= \int _{0}^{1} G_{q}^{0}(t,s) v(s) \,\mathrm{d}_{q}s + \int _{0}^{1} \frac{1}{ 1 -b } \biggl[ \int _{0}^{b} G_{q}^{0}(t,s) \,\mathrm{d}t \biggr] v(s) \,\mathrm{d}_{q}s \\ & = \int _{0}^{1} \biggl[ G_{q}^{0}(t,s) + \frac{1}{ 1 - b } \int _{0} ^{b} G_{q}^{0}(t,s) \,\mathrm{d}t \biggr] v(s) \,\mathrm{d}_{q}s \\ & = \int _{0}^{1} G_{q}(t,s) v(s) \,\mathrm{d}_{q}s = \int _{0}^{1} G _{q}(t,s) v_{0}(s) \,\mathrm{d}_{q}s. \end{aligned}$$

This completes the proof. □

Lemma 4

Let$G_{q}(t,s)$be given in Lemma 3. Then

$$ 0 \leq G_{q}(t,s) \leq A_{1}(\alpha , b) (1-qs)^{(\alpha - 1)}, $$

$\vert \frac{\partial G_{q}}{\partial t}(t,s) \vert \leq A_{2}( \alpha , b) (1-qs)^{(\alpha - 1)}$, where

$$\begin{aligned} A_{1}(\alpha , b) & = \frac{3}{ (1 - b) \varGamma _{q}(\alpha ) }, \\ A_{2}(\alpha , b) & = \frac{2}{ (1 - b ) \varGamma _{q}(\alpha -1)}, \end{aligned}$$

and finally

$$ 0 \leq \frac{(1-qs)^{(\alpha - 1)}}{\varGamma _{q}(\alpha )} \biggl[ - t + \frac{ 2 -b^{2}}{ 2 (1 - b)} \biggr] \leq G_{q}(t,s) $$

(8)

for$t, s \in \overline{J}$.

Proof

We consider some cases. If $0 \leq s \leq t \leq 1$ and $s \leq a$, then $(a - qs)^{\alpha - 1} \geq t(a - qs)^{(\alpha - 1)}$ and $(1 - qs)^{( \alpha - 1)} \geq ( t - qs)^{(\alpha - 1)}$. Hence,

$$ (1 - qs)^{(\alpha - 1)} + (1-t) ( a -qs)^{(\alpha - 1)} -(t - qs)^{( \alpha - 1)} \geq 0 $$

and so $G_{q}^{0}(t,s) \geq 0$. Thus, $G_{q}(t,s) \geq 0$. In other cases, the proof is easy. One can see that $G_{q}^{0}(t,s) \leq 3 (1- qs)^{(\alpha -1)}$ for each $t,s \in \overline{J}$, and so

$$\begin{aligned} G_{q}(t,s) & \leq 3( 1 - q s)^{( \alpha -1)} + \frac{1}{( 1 -b) \varGamma _{q}(\alpha ) } \int _{0}^{b} 3 ( 1 -qs)^{(\alpha -1)} \,\mathrm{d}t \\ & = 3( 1 - qs)^{(\alpha -1)} + \frac{ 3b ( 1 - qs)^{ (\alpha -1)}}{ (1- b) \varGamma _{q}(\alpha ) } \\ & = \frac{3( 1 -qs)^{(\alpha -1) }}{ (1 - b) \varGamma _{q}(\alpha )} = A _{1}(\alpha , b) (1- qs)^{(\alpha -1)}. \end{aligned}$$

From q-Green function $G_{q}^{0}(t,s)$ be given in Lemma 3, since

$$ \frac{\partial G_{q}^{0}(t,s)}{\partial t} = \textstyle\begin{cases} \frac{- ( t - qs)^{(\alpha -2)} -(a - qs)^{(\alpha -1)} }{ \varGamma _{q}( \alpha - 1)},& 0\leq s \leq t \leq 1, s\leq a, \\ \frac{ - ( t -qs)^{(\alpha -2)}}{\varGamma _{q}(\alpha - 1)},& 0 \leq s \leq a \leq t \leq 1, \\ \frac{-( a- qs)^{(\alpha -1)}}{\varGamma _{q}(\alpha )},& 0\leq t \leq s \leq a \leq 1, \\ 0, & 0\leq t \leq s \leq 1, a\leq s, \end{cases} $$

we have

$$ \biggl\vert \frac{\partial G_{q}^{0}(t,s)}{\partial t} \biggr\vert \leq \frac{(t-qs)^{( \alpha -1)} +(a- qs)^{ (\alpha -1)}}{\varGamma _{q}( \alpha - 1)} \leq \frac{ 2 (1 - qs)^{(\alpha -1)}}{ \varGamma _{q}(\alpha - 1)}, $$

and so

$$\begin{aligned} \biggl\vert \frac{\partial G_{q} (t,s)}{\partial t} \biggr\vert & \leq \frac{ 2 (1-qs)^{ \alpha - 1}}{\varGamma _{q}(\alpha - 1)} + \frac{ 2 b (1 -qs)^{(\alpha - 1)}}{ 1 - b} \\ & =\frac{ 2 ( 1 -qs)^{(\alpha -1)}}{\varGamma _{q}(\alpha - 1)} \biggl[ 1 + \frac{b}{1 -b} \biggr] \\ & = \frac{2(1-qs)^{(\alpha -1)}}{(1 -b) \varGamma _{q}(\alpha - 1)} = A _{2}(\alpha , b) (1-qs)^{(\alpha -1)}. \end{aligned}$$

If $0< s< t< 1$ and $s\leq a$, then $t - st >0$, and so $t ( 1 -qs) - s + t>0$. Hence, $s- t < (1- qs)t$ and $t(1-qs) > t-qs$. Since $t<1$ and $\alpha \geq 2$,

$$ \biggl( \frac{1-qs}{t-qs} \biggr)^{(\alpha -1)} > \biggl( \frac{1}{t} \biggr)^{(\alpha -1)} > \frac{1}{t}, $$

and so

$$\begin{aligned} &(1 - qs)^{(\alpha -1)} - ( t-qs)^{(\alpha -1)}+ (1- t) (a-qs)^{( \alpha - 1)} \\ &\quad > (1-qs)^{(\alpha -1)} -t(1-qs)^{(\alpha -1)} + ( 1 - t) ( a- qs)^{( \alpha -1)} \\ &\quad = (1 - t) \bigl((1- qs)^{(\alpha -1)} +(a-qs)^{(\alpha -1)} \bigr) \\ &\quad \geq (1-t) (1-qs)^{(\alpha -1)}. \end{aligned}$$

Thus, $G_{q}^{0}(t,s) > (1-t) (1-qs)^{(\alpha -1)}$. If $0< s\leq a < t <1$, then

$$\begin{aligned} -(t-qs)^{(\alpha -1)} +( 1 - qs)^{(\alpha -1)} &> - t ( 1-qs)^{( \alpha -1)}+ (1-qs)^{(\alpha -1)} \\ & =(1-t) (1-qs)^{(\alpha -1)}, \end{aligned}$$

and so $G_{q}^{0}(t,s) > \frac{ (1- t)(1-qs)^{(\alpha -1)}}{ \varGamma _{q}(\alpha )}$. Hence,

$$\begin{aligned} G_{q}(t,s) &\geq \frac{1}{\varGamma _{q}(\alpha )} \biggl[ (1- t) (1 - qs)^{( \alpha -1)} + \frac{1}{1 -b} \int _{0}^{b} ( 1 - t) ( 1 -qs)^{(\alpha -1)} \,\mathrm{d}t \biggr] \\ & = \frac{1}{\varGamma _{q}(\alpha )} \biggl[(1-t) (1 - qs)^{ (\alpha -1)} + \frac{(1-qs)^{(\alpha -1)}}{ 1 - b} \biggl( b-\frac{b^{2}}{2} \biggr) \biggr] \\ & = \frac{(1 - qs)^{(\alpha -1)}}{\varGamma _{q} ( \alpha )} \biggl[ - t + \frac{ 2 -b^{2}}{2( 1 - b)} \biggr] \geq 0, \end{aligned}$$

and so inequality (8) holds. □

Consider the self-map $T : \overline{\mathcal{B}} \to \overline{ \mathcal{B}}$ defined by

$$ T_{u}(t) = \int _{0}^{1} G_{q}(t,s) \omega \bigl(s, u(s), u'(s), D_{q} ^{\beta _{1}} u(s), I_{q}^{\beta } u(s) \bigr) \,\mathrm{d}s, $$

(9)

where $G_{q}(t,s)$ is the q-Green function in Lemma 3. By applying Lemma 3, one can easily see that the fractional q-integro-differential equation (1) has a solution if and only if T has a fixed point.

Here, we provide our first result about the existence of solutions for problem (1).

Theorem 5

Assume that the mapTis defined by equation (9) and$\omega : \overline{J} \times \overline{\mathcal{A}}^{4} \to \mathbb{R}$is a singular function at some points$t \in \overline{J}$, $\mu _{1}, \ldots , \mu _{4} \in \overline{\mathcal{L}}$are some nonnegative real-valued maps. Then fractional differential pointwise defined equation (1) under boundary conditions (2) has a solution whenever the following assumptions hold:

• (1) The functionωsatisfies the contraction condition

  $$ \bigl\vert \omega ( t, u_{1}, \ldots , u_{4}) -\omega (t, v_{1}, \ldots , v _{4}) \bigr\vert \leq \sum _{i=1}^{4} \mu _{i}(t) \Vert u_{i} - v_{i} \Vert $$

  for all$u_{1}, \ldots , u_{4}, v_{1},\ldots ,v_{4} \in \overline{ \mathcal{B}}$and$t \in \overline{J}$.

• (2) There exist a natural number$k_{0}$, some functions$\gamma _{1}, \ldots , \gamma _{k_{0}} \in \overline{\mathcal{L}}$, $\varTheta _{1}, \ldots , \varTheta _{k_{0}} : \mathbb{R}^{4} \to [0,\infty )$, nonnegative maps$\gamma _{1}, \ldots , \gamma _{k_{0}}$, and nonnegative and nondecreasing maps in their all components$\varTheta _{1}, \ldots , \varTheta _{k_{0}}$such that

  $$ \bigl\vert \omega ( t,u_{1},\ldots , u_{4}) \bigr\vert \leq \sum_{i=1}^{k _{0}} \gamma _{i}(t) \varTheta _{i} (u_{1}, \ldots , u_{4}) $$

  for all$(u_{1},\ldots , u_{4}) \in \overline{\mathcal{B}}^{4}$and$t\in \overline{J}$and$\lim_{w\to \infty } \frac{ \varTheta _{i} (w,w,w,w)}{w} = \eta _{0}$, where$\eta _{0}$is a nonnegative real number with

  $$ 0 \leq \eta _{0} \leq \frac{m_{0}}{M(\alpha , b) \sum_{i =1}^{k_{0}} \Vert \gamma _{i} \Vert + \delta _{0}} $$

  for some$\delta _{0} > 0$, $M(\alpha , b)= \max \{ A_{1}(\alpha , b), A_{2}(\alpha , b)\}$, and

  $$ m_{0}= \min \bigl\{ 1, \varGamma _{q}(2 -\beta _{1}), \varGamma _{q}(\beta _{2} +1) \bigr\} . $$

• (3) We have

  $$ \tau (\alpha , b)= \biggl[\hat{\mu }_{1} + \hat{\mu }_{2} + \frac{ \hat{\mu }_{3}}{ \varGamma _{q}( 2 -\beta _{1})} + \frac{ \hat{\mu }_{4}}{ \varGamma _{q}(\beta _{2} + 1 )} \biggr] M( \alpha , b) < 1, $$

  where$\hat{\mu }_{i}= \int _{0}^{1} (1-qs)^{(\alpha - 1)}\mu _{i}(s) \,\mathrm{d}_{q}s = \varGamma _{q}(\alpha ) I_{q}^{\alpha }\mu _{i}(1)$.

Proof

Let $u_{1}, u_{2} \in \overline{\mathcal{B}}$ and t belong to J̅. Then we obtain

$$\begin{aligned} \bigl\vert T_{ u_{1}}(t) - T_{u_{2}}(t) \bigr\vert &\leq \int _{0}^{1} G_{q}(t,s) \bigl\vert \omega \bigl(u_{1}(s), u'_{1}(s), D_{q}^{\beta _{1}} u_{1} (s), I _{q}^{\beta _{2}} u_{1}(s) \bigr) \\ &\quad {} - \omega \bigl( u_{2} (s), u'_{2}(s), D_{q}^{\beta _{1}} u_{2}(s), I_{q}^{\beta _{2}} u_{2} (s) \bigr) \bigr\vert \,\mathrm{d}_{q}s \\ & \leq \int _{0}^{1} A_{1}(\alpha , b) ( 1 -qs)^{(\alpha -1)} \bigl( \mu _{1}(s) \Vert u_{1} - u_{2} \Vert + \mu _{2}(s) \bigl\Vert u'_{1} - u'_{2} \bigr\Vert \\ &\quad {} + \mu _{3} (s) \bigl\Vert D_{q}^{\beta _{1}} u_{1} - D_{q}^{\beta _{1}} u _{2} \bigr\Vert - \mu _{4}(s) \bigl\Vert I_{q}^{\beta _{2}} u_{1} - I_{q}^{\beta _{2}} u _{2} \bigr\Vert \bigr) \,\mathrm{d}_{q}s. \end{aligned}$$

Since

$$\begin{aligned} \bigl\vert I_{q}^{\beta _{2}} u(t) \bigr\vert & \leq \frac{1}{ \varGamma _{q}(\beta _{2})} \int _{0}^{t} (t-qs)^{(\beta _{2})} \bigl\vert u(s) \bigr\vert \,\mathrm{d}_{q}s \\ & \leq \frac{ \Vert u \Vert }{ \varGamma _{q}(\beta _{2})} \biggl( \frac{1}{\beta _{2}} \bigl[ (t-qs)^{(\beta _{2})}|_{0}^{t} \bigr] \biggr) = \frac{ \Vert u \Vert }{ \varGamma _{q}(\beta _{2} + 1)} t^{\beta _{2}}, \end{aligned}$$

$\varGamma _{q}(\beta _{2} + 1) \|I_{q}^{\beta _{2}} u\| \leq \| u\|$. Hence,

$$ \varGamma _{q} (\beta _{2} + 1) \bigl\Vert I_{q}^{\beta _{2}} u_{1} - I_{q}^{\beta _{2}} u_{2} \bigr\Vert = \varGamma _{q} (\beta _{2} + 1) \bigl\Vert I_{q}^{\beta _{2}}(u_{1} - u_{2}) \bigr\Vert \leq \Vert u_{1} - u_{2} \Vert . $$

Similarly, one can conclude that $\varGamma _{q}( 2 - \beta _{1})\|D_{q} ^{\beta _{1}} u_{1} - D_{q}^{\beta _{1}} u_{2}\| \leq \| u_{1} - u_{2} \|$ for each $u_{1}, u_{2} \in \overline{\mathcal{B}}$. Therefore

$$\begin{aligned} \bigl\vert T_{u_{1}}(t) - T_{u_{2}}(t) \bigr\vert &\leq A_{1}(\alpha , b) \int _{0}^{1} \biggl[\mu _{1} (s) \Vert u_{1} - u_{2} \Vert + \mu _{2} (s) \bigl\Vert u'_{1} - u'_{2} \bigr\Vert \\ &\quad {} + \mu _{3} (s) \frac{ \Vert u'_{1} - u'_{2} \Vert }{ \varGamma _{q}(2 - \beta _{1})} + \mu _{4}(s) \frac{ \Vert u_{1} - u_{2} \Vert }{ \varGamma _{q}(\beta _{2} + 1)} \biggr] ( 1 -qs)^{(\alpha -1)} \,\mathrm{d}_{q}s \\ & = A_{1}(\alpha , b) \int _{0}^{1} \biggl[ \biggl( \mu _{1}(s)+ \frac{ \mu _{4} (s) }{ \varGamma _{q}(\beta _{2} + 1 )} \biggr) ( 1 -qs)^{(\alpha -1)} \Vert u_{1} - u_{2} \Vert \\ &\quad {} + \biggl( \mu _{2} (s) + \frac{ \mu _{3}(s)}{ \varGamma _{q}(2- \beta _{1})} \biggr) ( 1 -qs)^{(\alpha -1)} \bigl\Vert u'_{1} - u'_{2} \bigr\Vert \biggr] \,\mathrm{d}_{q}s \\ & \leq A_{1}(\alpha , b) \Vert u_{1} - u_{2} \Vert _{*} \int _{0}^{1} \biggl[ \mu _{1} (s) + \mu _{2} (s) \\ &\quad {} + \frac{ \mu _{3}(s)}{ \varGamma _{q}( 2 -\beta _{1})} + \frac{ \mu _{4}(s)}{ \varGamma _{q}( \beta _{2} + 1)} \biggr]( 1 -qs)^{(\alpha -1)} \,\mathrm{d}_{q}s \\ & = A_{1}(\alpha , b) \biggl[ \hat{\mu }_{1} + \hat{ \mu }_{2} + \frac{ \hat{\mu }_{3}}{ \varGamma _{q}(2 - \beta _{1})} + \frac{\hat{\mu }_{4}}{ \varGamma _{q}( \beta _{2} + 1)} \biggr] \Vert u_{1} - u_{2} \Vert _{*}. \end{aligned}$$

(10)

Also, we have

$$\begin{aligned} \bigl\vert T'_{u_{1}}(t) - T'_{u_{2}}(t) \bigr\vert &\leq \int _{0}^{1} \biggl\vert \frac{ \partial G_{q}(t,s)}{ \partial t} \biggr\vert | \omega \bigl( s, u_{1}(s), u_{1}(s), D_{q}^{\beta _{1}} u_{1}(s), I_{q}^{\beta _{2}} u_{1}(s) \bigr) \\ &\quad {} - \omega \bigl(s, u_{2}(s), u_{2}(s), D_{q}^{\beta _{1}} u_{2}(s), I_{q}^{\beta _{2}} u_{2}(s)\bigr) | \,\mathrm{d}_{q}s \\ & \leq \int _{0}^{1} A_{2}(\alpha , b) ( 1 -qs)^{(\alpha -1)} \bigl[ \mu _{1} (s) \Vert u_{1} - u_{2} \Vert + \mu _{2}(s) \bigl\Vert u'_{1} - u'_{2} \bigr\Vert \\ &\quad {} + \mu _{3} (s) \bigl\Vert D_{q}^{\beta _{1}} u_{1} -D_{q}^{\beta _{1}} u _{2} \bigr\Vert + \mu _{1}(s) \bigl\Vert I_{q}^{\beta _{2}} u_{1} - I_{q}^{\beta _{2}} u _{2} \bigr\Vert \bigr] \,\mathrm{d}_{q}s \\ & \leq A_{2}(\alpha , b) \int _{0}^{1} \biggl[ \mu _{1}(s) \Vert u_{1} - u _{2} \Vert + \mu _{2} (s) \bigl\Vert u'_{1} - u'_{2} \bigr\Vert \\ &\quad {} + \mu _{3} (s) \frac{ \Vert u'_{1} - u'_{2} \Vert }{ \varGamma _{q}( 2 - \beta _{1})} + \mu _{4}(s) \frac{ \Vert u_{1} - u_{2} \Vert }{\varGamma _{q}(\beta _{2} + 1)} \biggr] (1-qs)^{(\alpha -1) } \,\mathrm{d}_{q}s \\ & = A_{2}(\alpha , b) \biggl[ \hat{\mu }_{1} + \hat{ \mu }_{2} + \frac{ \hat{\mu }_{3}}{\varGamma _{q}(2 - \beta _{1})} + \frac{\hat{\mu }_{4}}{ \varGamma _{q}( \beta _{2} + 1)} \biggr] \Vert u_{1} - u_{2} \Vert _{*}. \end{aligned}$$

(11)

By using (10) and (11), we obtain

$$\begin{aligned} \Vert T_{u_{1}} - T_{u_{2}} \Vert _{*} & = \max \bigl\{ \Vert T_{u_{1}} - T _{u_{2}} \Vert , \bigl\Vert T'_{u_{1}} - T'_{u_{2}} \bigr\Vert \bigr\} \\ & \leq M(\alpha , b) \biggl[\hat{\mu }_{1} + \hat{\mu }_{2} + \frac{ \hat{\mu }_{3}}{ \varGamma _{q}(2 - \beta _{1})} + \frac{ \hat{\mu }_{4}}{ \varGamma _{q}(\beta _{2} + 1)} \biggr] \Vert u_{1} - u_{2} \Vert _{*}. \end{aligned}$$

Hence, $\| T_{u_{1}} - T_{u_{2}} \|_{*} \to 0$ as $\| u_{1} - u_{2} \|_{*} \to 0$, and so T is continuous. Since $\eta _{0} M(\alpha , b) \sum_{i=1}^{k_{0}} \| \gamma _{i}\| < m_{0}$, we can choose $\varepsilon _{0} > 0$ such that

$$ (\eta _{0} + \varepsilon _{0}) M(\alpha , b) \sum _{i=1}^{k_{0}} \Vert \gamma _{i} \Vert < m_{0}. $$

Since $\frac{\varTheta (w,w,w,w)}{w} \to \eta _{0}$ as $w \to \infty$, there exists $r = \Delta (\varepsilon _{0})>0$ such that $\frac{\varTheta ( w, w, w, w)}{w} < \eta _{0} + \varepsilon _{0}$ for all $w \geq \Delta (\varepsilon _{0})$. So

$$ \varTheta (w,w,w,w) < (\eta _{0} + \varepsilon _{0})w $$

(12)

for $w\geq \Delta (\varepsilon _{0})$. Put $B_{r} = \{ u\in \overline{ \mathcal{B}} : \| u\|_{*} < r \}$ and define $\alpha : \overline{ \mathcal{B}}^{2} \to [0, \infty )$ by $\alpha (u, v) = 1$ whenever $u, v \in B_{r}$ and $\alpha (u, v)=0$ otherwise. If $\alpha (u, v) \geq 1$, then $\|u\|_{*}$ and $\|v\|_{*}$ are less than r. Let $t \in \overline{J}$. Then we obtain

$$\begin{aligned} \bigl\vert T_{u}(t) \bigr\vert & \leq \int _{0}^{1} G_{q}(t,s) \bigl\vert \omega \bigl( s, u(s), u'(s), D_{q}^{\beta _{1}} u(s), I_{q}^{\beta _{2}} u(s) \bigr) \bigr\vert \,\mathrm{d}_{q}s \\ & \leq A_{1}(\alpha , b) \int _{0}^{1} (1 -qs)^{(\alpha - 1)} \\ &\quad{} \times \sum_{i=1}^{k_{0}} \gamma _{i} (s) \varTheta _{i} \bigl(u(s), u'(s), D_{q}^{\beta _{1}} u(s), I_{q}^{\beta _{2}} u(s) \bigr) \,\mathrm{d}_{q}s \\ & \leq A_{1}(\alpha , b) \sum_{i=1}^{k_{0}} \int _{0}^{1} (1 -qs)^{( \alpha - 1) } \gamma _{i} (s) \\ &\quad{} \times \varTheta _{i} \biggl( \Vert u \Vert , \bigl\Vert u'_{1} \bigr\Vert , \frac{ \Vert u' \Vert }{ \varGamma _{q}( 2 -\beta _{1})}, \frac{ \Vert u \Vert }{ \varGamma _{q}(\beta _{2} + 1)} \biggr) \,\mathrm{d}_{q}s \\ & \leq A_{1}(\alpha , b) \sum_{i=1}^{k_{0}} \varTheta _{i} \biggl( r, r, \frac{r}{ \varGamma _{q}(2-\beta _{1})}, \frac{ r}{\varGamma _{q}(\beta _{2} + 1)} \biggr) \\ &\quad{} \times \int _{0}^{1} \gamma _{i}(s) \sup ( 1 -qs)^{(\alpha - 1)} \,\mathrm{d}_{q}s \\ & \leq A_{1}(\alpha , b) \sum_{i=1}^{k_{0}} \varTheta _{i} \biggl( \frac{r}{ m_{0}}, \frac{r}{m_{0}}, \frac{r}{ m_{0}}, \frac{r}{m_{0}} \biggr) \Vert \gamma _{i} \Vert _{1}, \end{aligned}$$

where

$$ m_{0} = \min \bigl\{ 1, \varGamma _{q}(\beta _{2} +1), \varGamma _{q}(2 - \beta _{1}) \bigr\} = \min \bigl\{ \varGamma _{q}(\beta _{2} +1), \varGamma _{q}(2- \beta _{1}) \bigr\} . $$

Since $r >m_{0} r$, by using (12) we obtain

$$ \varTheta _{i} \biggl( \frac{r}{m_{0}}, \frac{r}{m_{0}}, \frac{r}{m_{0}}, \frac{r}{m _{0}} \biggr)< (\eta _{0} + \varepsilon _{0}) \frac{r}{m_{0}}, $$

and so

$$\begin{aligned} \bigl\vert T_{u}(t) \bigr\vert & \leq A_{1}( \alpha , b) \sum_{i=1}^{k_{0}} \frac{r}{m _{0}} (\eta _{0} + \varepsilon _{0}) \Vert \gamma _{i} \Vert _{1} \\ & = (\eta _{0} + \varepsilon _{0}) \biggl[ \frac{A_{1}(\alpha , b) \sum_{i=1}^{k_{0}} \Vert \gamma _{i} \Vert _{1} }{m_{0}} \biggr]r < r. \end{aligned}$$

Hence, $\|T_{u}\| \leq r$. Also, one can conclude that

$$\begin{aligned} \bigl\vert T'_{u}(t) \bigr\vert & \leq \int _{0}^{1} \biggl\vert \frac{\partial G_{q}( t,s)}{ \partial t} \biggr\vert \bigl\vert \omega \bigl( s, u(s), u'(s), D_{q}^{\beta _{1}} u(s), I_{q}^{\beta _{2}} u(s) \bigr) \bigr\vert \,\mathrm{d}_{q}s \\ & \leq A_{2}(\alpha , b) \int _{0}^{1} ( 1 -qs)^{(\alpha - 1)} \\ &\quad{} \times \sum_{i=1}^{k_{0}} \gamma _{i}(s) \varTheta _{i} \bigl( u(s), u'(s), D_{q}^{\beta _{1}} u(s), I_{q}^{\beta _{2}} u(s) \bigr) \,\mathrm{d}_{q}s \\ & \leq A_{2}(\alpha , b) \Biggl[\sum_{i=1}^{k_{0}} \varTheta _{i} \biggl( r, r, \frac{ r}{ \varGamma _{q}(2-\beta _{1})}, \frac{r}{\varGamma _{q}(\beta _{2} +1)} \biggr) \Biggr] \\ &\quad{} \times \int _{0}^{1} (1-qs)^{(\alpha - 1)} \gamma _{i}(s) \,\mathrm{d}_{q}s \\ &\leq A_{2}(\alpha , b) \sum_{i=1}^{k_{0}} \varTheta _{i} \biggl( \frac{r}{m _{0}},\frac{r}{m_{0}}, \frac{ r}{m_{0}}, \frac{ r }{ m_{0}} \biggr) \int _{0}^{1} \Vert \gamma _{i} \Vert _{1} \\ & \leq (\eta _{0} + \varepsilon _{0}) \biggl[ \frac{ A_{2}(\alpha , b) \sum_{i=1}^{k_{0}} \Vert \gamma _{i} \Vert _{1} }{m_{0}} \biggr] r < r \end{aligned}$$

and $\|T_{u}\|_{*} = \max \{ \|T_{u}\|,\|T'_{u}\|_{*} \} \leq r$. This implies that $T_{u}$ and so $T_{v} \in B_{r}$, that is, $\alpha ( T _{u}, T_{v}) \geq 1$. Thus, T is α-admissible. Since $B_{r} \neq \emptyset$, there exists $u_{0} \in B_{r}$ such that $T_{u_{0}} \in B_{r}$. Hence, $\alpha ( u_{0}, T_{u_{0}} ) \geq 1$. Put $\psi (t) = \tau (\alpha , b) t$ for each $t \in [0, \infty )$, here $\tau (\alpha , b) <1$. Since

$$ \sum_{n=1}^{\infty } \psi ^{n}(t) = \sum_{n=1}^{\infty } \tau (\alpha , b)^{n} t = \biggl( \frac{ \tau (\alpha , b) }{1 - \tau (\alpha , b) } \biggr) t < \infty $$

and $\psi : [0,\infty ) \to [0,\infty )$ is nondecreasing, we get $\psi \in \varPsi$. Note that

$$\begin{aligned} \bigl\vert T_{u}(t)- T_{v}(t) \bigr\vert & \leq \int _{0}^{1} G_{q}(t,s) \bigl\vert \omega \bigl(s, u(s), u'(s), D_{q}^{ \beta _{1}} u(s), I_{q}^{\beta _{2}} u(s) \bigr) \\ &\quad {} - \omega \bigl( s, v(s), v'(s), D_{q}^{ \beta _{1}} v(s), I _{q}^{\beta _{2}} v (s) \bigr) \bigr\vert \,\mathrm{d}_{q}s \\ & \leq A_{1}(\alpha , b) \biggl[ \hat{\mu }_{1} + \hat{ \mu }_{2} + \frac{ \hat{\mu }_{3}}{ \varGamma _{q}( 2 - \beta _{1})} + \frac{\hat{\mu }_{4}}{ \varGamma _{q}(\beta _{2} +1)} \biggr] \Vert u - v \Vert _{*}, \end{aligned}$$

and so

$$ \Vert T_{u} - T_{v} \Vert \leq A_{1}(\alpha , b) \biggl[ \hat{\mu }_{1} + \hat{\mu }_{2} + \frac{ \hat{\mu }_{3}}{ \varGamma _{q}(2 - \beta _{1})} + \frac{ \hat{\mu }_{4}}{ \varGamma _{q}( \beta _{2} +1)} \biggr] \Vert u - v \Vert _{*}. $$

(13)

Also,

$$\begin{aligned} \bigl\vert T'_{u}(t)- T'_{v}(t) \bigr\vert & \leq \int _{0}^{1} \biggl\vert \frac{ \partial G _{q} (t,s)}{ \partial t} \biggr\vert \bigl\vert \omega \bigl(s, u(s), u'(s), D _{q}^{\beta _{1}} u(s), I_{q}^{\beta _{2}} u(s) \bigr) \\ &\quad {} - \omega \bigl( s, v(s), v'(s), D_{q}^{\beta _{1}} v(s), I_{q} ^{\beta _{2}} v(s) \bigr) \bigr\vert \,\mathrm{d}_{q}s \\ & \leq A_{2}(\alpha , b) \biggl[\hat{\mu }_{1} + \hat{ \mu }_{2} + \frac{ \hat{\mu }_{3}}{ \varGamma _{q}(2- \beta _{1})} + \frac{\hat{\mu }_{4}}{ \varGamma _{q}(\beta _{2} + 1)} \biggr] \Vert u - v \Vert _{*}. \end{aligned}$$

This implies

$$ \bigl\Vert T'_{u}- T'_{v} \bigr\Vert \leq A_{2}(\alpha , b) \biggl[ \hat{\mu }_{1} + \hat{\mu }_{2} + \frac{ \hat{\mu }_{3} }{ \varGamma _{q}(2 -\beta _{1}) } + \frac{ \hat{\mu }_{4} }{ \varGamma _{q}(\beta _{2} +1) } \biggr] \Vert u - v \Vert _{*}. $$

(14)

Thus, equations (13) and (14) imply that

$$\begin{aligned} \Vert T_{u}- T_{v} \Vert _{*} & \leq M(\alpha , b) \biggl[ \hat{\mu }_{1} + \hat{\mu }_{2} + \frac{ \hat{\mu }_{3} }{ \varGamma _{q}( 2 -\beta _{1})} + \frac{ \hat{ \mu }_{4}}{ \varGamma _{q}(\beta _{2} +1)} \biggr] \Vert u - v \Vert _{*} \\ & = \tau (\alpha , b) \Vert u - v \Vert _{*} = \psi \bigl( \Vert u - v \Vert _{*} \bigr) \end{aligned}$$

for u and v in $B_{r}$. Hence, $\alpha (u, v) \|T_{u} - T_{v}\| _{*} \leq \psi ( \rho (u, v))$ for each $u, v \in \overline{ \mathcal{B}}$. By using Lemma 2, T has a fixed point, which is a solution for problem (1). □

Theorem 6

Letωbe a real-valued function on$\overline{J} \times \overline{ \mathcal{A}}^{4}$. Then the pointwise defined problem (1) with boundary conditions (2) has a solution whenever the following assumptions hold:

• (1) There exist natural numbers$k_{1}$, some maps$\varTheta _{1}, \ldots , \varTheta _{k_{1}}:\mathbb{R}^{4} \to \mathbb{R}$which are nondecreasing in their all components, $\varTheta _{i}(w, w, w, w) \geq 0$for all$w\geq 0$and$\frac{ \varTheta _{i} (w, w, w, w)}{w} \to \eta _{i}$as$w \to 0^{+}$for some$\eta _{i} \in [0, 1)$ ($i=1, \ldots , k_{1}$), and there are some nonnegative real-valued functions$\mu _{1}, \ldots , \mu _{k_{1}} : \overline{J} \to [0, \infty )$such that

  $$\begin{aligned} &\bigl\vert \omega (t, u_{1},u_{2}, u_{3}, u_{4}) - \omega (t, v_{1}, v_{2}, v _{3}, v_{4}) \bigr\vert \\ &\quad \leq \sum_{i=1}^{k_{1}} \mu _{i} (t) \varTheta _{i}( u_{1} - v_{1}, u _{2}- v_{2}, u_{3}- v_{3}, u_{4}- v_{4}) \end{aligned}$$

  for all$u_{1}, \ldots , u_{4}, v_{1},\ldots , v_{4} \in \overline{ \mathcal{B} }$with$u_{j} \geq v_{j} \geq 0$ ($j=1,\ldots ,4$) and$t \in \overline{J}$.

• (2) If$M(\alpha , b)= \max \{ A_{1}(\alpha , b), A_{2}(\alpha , b)\}$, then

  $$ M(\alpha , b) \sum_{i=1}^{k_{1}} \bigl\Vert (1-qt)^{(\alpha -1)} \mu _{i} \bigr\Vert _{1} \leq 1. $$

Proof

Since $\lim_{w \to 0^{+}} \frac{\varTheta _{i}(w, w, w, w)}{ w} = \eta _{i} < 1$ for $i=1,\ldots , k_{1}$, for each $\varepsilon _{i} > 0$ there exist $\delta _{i} = \delta (\varepsilon _{i}) > 0$ such that $\frac{w}{m_{0}} \in (0, \delta _{i})$ implies

$$ \varTheta _{i} \biggl( \frac{w}{m_{0}}, \ldots , \frac{w}{ m_{0}} \biggr) \leq (\eta _{i} + \varepsilon _{i}) \frac{w}{m_{0}}, $$

where $m_{0} = \min \{ \varGamma _{q}(2 -\beta _{1}), \varGamma _{q} (\beta _{2} + 1)\}$. Let $\varepsilon ^{0}_{i}$ be such that $\eta _{i} + \varepsilon ^{0}_{i}< 1$ and $\delta ^{0}_{i} = \delta ( \varepsilon ^{0}_{i})$. Put $\eta = \max \{ \eta _{1}, \ldots , \eta _{k_{1}}\}$, $\varepsilon _{0} = \min \{ \varepsilon ^{0}_{1}, \ldots , \varepsilon ^{0}_{k_{1}}\}$, and $\delta = \min \{ \delta ^{0}_{1} , \ldots , \delta ^{0}_{ k_{1}}, \varepsilon _{0} \}$. Thus, $\eta + \varepsilon _{0} < 1$ and

$$ \varTheta _{i} \biggl(\frac{w}{ m_{0}}, \ldots , \frac{w}{m_{0}} \biggr) < ( \eta _{i} + \varepsilon _{0}) \frac{ w}{m_{0}} $$

for $\frac{w}{ m_{0}} \in (0, \delta )$ and $1 \leq i \leq k_{1}$. Also,

$$ \varTheta _{i} \biggl(\frac{ \delta }{ m_{0}}, \ldots , \frac{ \delta }{ m _{0}} \biggr) < ( \eta + \varepsilon _{0} ) \frac{ \delta }{ m_{0}} m _{0} = (\eta + \varepsilon _{0} ) \delta \leq (\eta + \varepsilon _{0}) \varepsilon _{0}. $$

Now, we define the map $\alpha : \overline{\mathcal{B}} \times \overline{ \mathcal{B}} \to [ 0,\infty )$ by $\alpha (u, v) =1$ whenever $\| u - v\|_{*} \leq \delta$ and $\alpha (u, v)=0$ otherwise. If $\alpha (u, v) \geq 1$, then $\| u - v\|_{*} \leq \delta$, and so

$$\begin{aligned} \bigl\vert T_{u}(t) - T_{v}(t) \bigr\vert & \leq \int _{0}^{1} G_{q}(t,s) \bigl\vert \omega \bigl(s, u(s), u'(s), D_{q}^{\beta _{1}} u(s), I_{q}^{\beta _{2}} u( s) \bigr) \\ &\quad {} - \omega \bigl( s, v(s), v'(s), D_{q}^{\beta _{1}} v(s), I_{q} ^{\beta _{2}} v(s) \bigr) \bigr\vert \,\mathrm{d}_{q}s \\ &\leq \int _{0}^{1} G_{q}(t,s) \sum _{i=1}^{k_{1}} \mu _{i}(s) \\ &\quad{} \times \bigl\vert \varTheta _{i} \bigl( (u - v) (s), (u - v)'(s), D_{q} ^{\beta _{1}}(u - v) (s), I_{q}^{\beta _{2}} (u - v) (s) \bigr) \bigr\vert \,\mathrm{d}_{q}s \\ & \leq A_{1}(\alpha , b) \int _{0}^{1} (1-qs)^{(\alpha -1)} \sum _{i=1} ^{k_{1}} \mu _{i}(s) \\ &\quad{} \times \bigl\vert \varTheta _{i} \bigl( \Vert u -v \Vert , \bigl\Vert ( u - v)' \bigr\Vert , \bigl\Vert D _{q}^{\beta _{1}}( u - v) \bigr\Vert , \bigl\Vert I_{q}^{\beta _{2}} (u - v) \bigr\Vert \bigr) \bigr\vert \,\mathrm{d}_{q}s \\ & \leq A_{1}(\alpha , b) \sum_{i=1}^{k_{1}} \biggl( \int _{0}^{1} (1 - qs)^{( \alpha -1)} \mu _{i}(s) \,\mathrm{d}_{q}s \biggr) \\ &\quad{} \times \varTheta _{i} \biggl(\delta , \delta , \frac{ \delta }{ \varGamma _{q}( 2 -\beta _{1})}, \frac{ \delta }{\varGamma _{q}(\beta _{2} + 1)} \biggr) \\ &\leq A_{1}(\alpha , b) \sum_{i=1}^{k_{1}} \biggl( \int _{0}^{1} (1-qs)^{( \alpha -1)} \mu _{i}(s) \,\mathrm{d}_{q}s \biggr) \\ &\quad{} \times \varTheta _{i} \biggl(\frac{\delta }{ m_{0}}, \frac{\delta }{m_{0}}, \frac{\delta }{m_{0}}, \frac{\delta }{m_{0}} \biggr) \\ & \leq A_{1}(\alpha , b) \sum_{i=1}^{k_{1}} \bigl\Vert (1-qt)^{(\alpha -1)} \mu _{i} \bigr\Vert _{1} ( \eta +\varepsilon _{0}) \delta \\ &\leq A_{1}(\alpha , b) \sum_{i=1}^{k_{1}} \bigl\Vert (1-qt)^{(\alpha -1)} \mu _{i} \bigr\Vert _{1} \delta \leq \delta . \end{aligned}$$

(15)

Hence, $\|T_{u} - T_{v}\|_{*} \leq \delta$, which implies $\alpha (T _{u}, T_{v})=1$. If

$$ \lambda = M(\alpha , b) \sum_{i=1}^{k_{1}} \bigl\Vert (1-qt)^{(\alpha - 1)} \mu _{i} \bigr\Vert _{1} (\eta + \varepsilon _{0}), $$

then by using the assumption we get $\lambda <1$. If $\psi (t) = \lambda t$, then $\psi \in \varPsi$. If $\|u - v\| \leq \delta$, then

$$ \Vert T_{u}- T_{v} \Vert \leq A_{1}(\alpha , b) \sum_{i=1}^{k_{1}} \bigl\Vert (1-qt)^{( \alpha -1)} \mu _{i} \bigr\Vert _{1} ( \eta + \varepsilon _{0}) \Vert u - v \Vert _{*} \leq \lambda \Vert u - v \Vert _{*} $$

(16)

and

$$ \bigl\Vert T'_{u} - T'_{v} \bigr\Vert \leq A_{2}(\alpha , b) \sum_{i=1}^{ k_{1}} \bigl\Vert (1 - qt)^{(\alpha -1)} \mu _{i} \bigr\Vert _{1} (\eta + \varepsilon _{0}) \Vert u - v \Vert _{*} \leq \lambda \Vert u - v \Vert _{*}. $$

(17)

Thus, from inequality (16), we have $\|T_{u} - T_{v}\|_{*} \leq \lambda \|u - v\|_{*}= \psi (\| u - v\|_{*})$ and so $\alpha ( u, v) \|T_{u} - T_{v}\|_{*} \leq \psi ( \|u - v\|_{*})$ for each $u, v\in \overline{\mathcal{B}}$. Let $\varepsilon > 0$ be given. Since $\varTheta _{i} (\frac{w}{m_{0}}, \ldots , \frac{w}{m_{0}}) \to 0$ as $w \to 0^{+}$, for each $i=1, \ldots , k_{1}$, there exists $\delta _{i} > 0$ such that $\varTheta _{i} ( \frac{w}{ m_{0}}, \ldots , \frac{w}{ m_{0} }) < \frac{\varepsilon }{k'}$ for all $0 < w \leq \delta _{i}$, where

$$ \lambda ' = \Biggl[M(\alpha , b) \sum _{i=1}^{k_{1}} \bigl\Vert (1-qt)^{(\alpha -1)} \mu _{i} \bigr\Vert _{1} \Biggr] + 1. $$

If $\delta = \min \{\delta _{i}: 1\leq i \leq k_{1} \}$, then $\varTheta _{i} (\frac{w}{m_{0}}, \ldots , \frac{w}{m_{0}}) < \frac{ \varepsilon }{\lambda '}$ for all $w \in (0, \delta ]$ and $i=1, \ldots , k_{1}$. If $u_{n} \to u$, then there exists a natural number $n_{0}$ such that $\|u_{n} - u\|_{*} < \delta$ for all $n \geq n_{0}$. Hence,

$$\begin{aligned} &\bigl\vert T_{u_{n}}(t) - T_{u}(t) \bigr\vert \\ &\quad \leq \int _{0}^{1} G_{q}(t,s) \bigl\vert \omega \bigl(s, u_{n}(s), u'_{n}(s), D_{q}^{\beta _{1}} u_{n}(s), I_{q}^{\beta _{2}} u_{n}(s)\bigr) \\ &\qquad {} - \omega \bigl(s, u(s), u'(s), D_{q}^{\beta _{2}} u(s), I_{q} ^{\beta _{2}} u(s) \bigr) \bigr\vert \,\mathrm{d}s \\ &\quad \leq \int _{0}^{1} G_{q}(t,s) \sum _{i=1}^{k_{1}} \mu _{i}(s) \\ &\qquad{} \times \varTheta _{i} \bigl( \Vert u_{n} -u \Vert , \bigl\Vert (u_{n} - u)' \bigr\Vert , \bigl\Vert D _{q}^{\beta _{1}}( u_{n} -u) \bigr\Vert , \bigl\Vert I_{q}^{\beta _{2}} (u_{n}- u) \bigr\Vert \bigr) \,\mathrm{d}s \\ &\quad \leq A_{1}(\alpha , b) \sum_{i=1}^{k_{1}} \biggl[ \varTheta _{i} \biggl( \delta , \delta , \frac{\delta }{ \varGamma _{q}( 2- \beta _{1})}, \frac{ \delta }{\varGamma _{q}(\beta _{2} +1)} \biggr) \\ &\qquad{} \times \int _{0}^{1} (1-qs)^{(\alpha -1)} \mu _{i}(s) \,\mathrm{d}_{q}s \biggr] \\ & \quad \leq A_{1}(\alpha , b) \sum_{i=1}^{k_{1}} \biggl[ \varTheta _{i} \biggl( \frac{ \delta }{ m_{0}}, \frac{ \delta }{m_{0}}, \frac{\delta }{ m _{0}}, \frac{\delta }{ m_{0}} \biggr) \\ &\qquad{} \times \int _{0}^{1} (1-qs)^{(\alpha -1)} \mu _{i} (s) \,\mathrm{d}_{q}s \biggr] \\ & \quad \leq A_{1}(\alpha , b) \frac{\varepsilon }{k'} \sum _{i=1}^{k_{1}} \bigl\Vert (1 - qt)^{(\alpha -1)} \mu _{i} \bigr\Vert _{1} < \varepsilon \end{aligned}$$

(18)

for $t \in \overline{J}$ and $n \geq n_{0}$. By repeating the similar method, we obtain

$$\begin{aligned} \bigl\vert T'_{u_{n}}(t) - T'_{u}(t) \bigr\vert & \leq \int _{0}^{1} \biggl\vert \frac{ \partial G_{q} (t,s)}{\partial t} \biggr\vert \bigl\vert \omega \bigl(s, u_{n}(s), u'_{n}(s), D_{q}^{\beta _{1}} u_{n}(s), I_{q}^{\beta _{2}} u_{n}(s) \bigr) \\ &\quad {} - \omega \bigl( s, u(s), u'(s), D_{q}^{\beta } u(s), I_{q} ^{\beta _{2}} u(s) \bigr) \bigr\vert \,\mathrm{d}_{q}s \\ & \leq A_{2}(\alpha , b) \frac{\varepsilon }{k'} \sum _{i=1}^{k_{1}} \bigl\Vert ( 1 -qt)^{(\alpha -1)} \mu _{i} \bigr\Vert _{1} < \varepsilon \end{aligned}$$

(19)

for all t in J̅ and $n \geq n_{0}$. Now, (18) and (19) imply that $\|T_{u_{n}} - T_{u} \| \leq \varepsilon$ and $\|T'_{u_{n}} - T'_{u} \| \leq \varepsilon$, respectively, for $n \geq n_{0}$. Thus,

$$ \Vert T_{u_{n}} - T_{u} \Vert _{*} = \max \bigl\{ \Vert T_{u_{n}} - T_{u} \Vert , \bigl\Vert T'_{u _{n}} - T'_{u} \bigr\Vert \bigr\} \leq \varepsilon $$

for $n \geq n_{0}$, and so $T_{u_{n}} \to T_{u}$ as $u_{n} \to u$. Indeed, T is continuous. Now, we show that there exists $u_{0} \in \overline{\mathcal{B}}$ such that $\alpha (u_{0} , T_{u_{0}}) = 1$. In this way, we have to show that $\|T_{u_{0}}- u_{0}\| \leq \delta$ for some $u_{0}\in \overline{\mathcal{B}}$. Let $r_{0} > 0$ be a fixed real number. Since $\varTheta _{i}(w, w, w, w) \to 0$ as $w \to 0^{+}$, for each $\varepsilon > 0$, there exists $n = n(\varepsilon )$ such $0 < w \leq \frac{ r_{0}}{n}$ implies $\varTheta _{i}( w, w, w, w) \leq \varepsilon$ for all $1 \leq i \leq k_{1}$. Hence, $\varTheta _{i}(\frac{ r_{0}}{ n }, \ldots , \frac{r_{0}}{n}) \leq \varepsilon$. Put

$$ M = \max \biggl\{ \frac{1}{\varGamma _{q}( 2 - \beta _{1})}, \frac{1}{\varGamma _{q}(\beta _{2} +1) }, 1 \biggr\} = \max \biggl\{ \frac{1}{ \varGamma _{q}(2- \beta _{1})}, \frac{ 1}{ \varGamma _{q}(\beta _{2} +1)} \biggr\} $$

and choose $\varepsilon _{M}$ such that

$$ \sum_{i=1}^{k_{1}} \bigl\Vert (1-qt)^{(\alpha -1)} \mu _{i} \bigr\Vert _{1} \varepsilon _{M} M(\alpha , b) < \delta . $$

Take $n_{1} = n(\varepsilon _{M})$ and choose a natural number $n_{2}$ such that

$$ \sum_{i=1}^{k_{1}} \bigl\Vert ( 1 -qt)^{(\alpha -1)} \mu _{i} \bigr\Vert _{1} \varepsilon _{M} M(\alpha , b) \leq \delta - \frac{1}{n_{2}}. $$

If $n_{0} = \max \{n_{1}, n_{2}\}$, then $\varTheta _{i} ( \frac{M}{n _{0}}, \ldots , \frac{M}{n_{0}} ) \leq \varepsilon _{M}$ for $1=1, \ldots , k_{1}$. Define

$$ u_{0} (t)= \textstyle\begin{cases} 0,& t \leq \frac{1}{n_{0} +1}, \\ \frac{6 n_{0}^{2}}{6 n_{0}^{2} + 5n_{0} +2} \\ \quad {} \times [ \frac{t^{3}}{3} - \frac{2n_{0} +1}{2n_{0} (n_{0} +1)} t^{2} + \frac{t}{n_{0} (n_{0} +1)} ] +\frac{1}{n_{0} +2},& \frac{1}{n_{0} +1} < t < \frac{1}{n_{0}}, \\ \frac{1}{n_{0}}, & \frac{1}{n_{0}} \leq t. \end{cases} $$

One can easily see that $u_{0} \in \overline{\mathcal{A}}$ and $u(t) \in [0, \frac{1}{n_{0}} ]$, $u_{0} ( \frac{1}{n _{0} + 1} )= 0$ and

$$ u'_{0} (t)= \textstyle\begin{cases} 0, & t \leq \frac{1}{n_{0} +1}, \\ \frac{6 n_{0}^{2}}{6 n_{0}^{2} + 5n_{0} +2} [ t^{2} - \frac{2n _{0} +1}{n_{0} (n_{0} +1)} t + \frac{1}{n_{0} (n_{0} +1)} ], & \frac{1}{n_{0} +1} < t < \frac{1}{n_{0}}, \\ 0, & \frac{1}{n_{0}} \leq t. \end{cases} $$

Hence, $u'_{0}$ belongs to $\overline{\mathcal{A}}$ and $u'_{0}(\frac{1}{n _{0}+ 1} )= u'_{0}(\frac{1}{n_{0}}) = 0$. Thus, $u_{0} \in \overline{ \mathcal{B}}$. Also, we have

$$\begin{aligned} u'_{0}(t) & \leq \frac{6 n_{0}^{2}}{6 n_{0}^{2} + 5n_{0} +2} \biggl( \frac{1}{n ^{2}_{0}} + \frac{1}{n^{4}_{0}(n_{0} +1)} - \frac{ 2n_{0} +1}{ 2n_{0} (n_{0}+1)^{2}} \biggr) \\ & \leq 1 \times \frac{1}{n_{0}} \times \biggl( \frac{1}{n_{0}} + \frac{1}{n ^{3}_{0}(n_{0} +1)} - \frac{2n_{0} }{ (n_{0}+1)^{2}} \biggr) \leq \frac{1}{n _{0}}, \end{aligned}$$

and so $n_{0} \|u_{n_{0}} \|_{*} \leq 1$. This implies that

$$\begin{aligned} &\bigl\vert T_{u_{0}}(t) - u_{0}(t) \bigr\vert \\ &\quad = \biggl\vert \int _{0}^{1} G_{q}(t,s) \omega \bigl( s, u_{0}(s), u'_{0}(s), D_{q}^{\beta _{1}} u_{0}(s), I_{q}^{\beta _{2}} u_{0}(s) \bigr) \,\mathrm{d}_{q}s - u_{0}(s) \biggr\vert \\ &\quad \leq \int _{0}^{1} G_{q}(t,s) \bigl\vert \omega \bigl(s, u_{0}(s), u'_{0}(s), D_{q}^{\beta } u_{0}(s), I_{q}^{\beta _{2}} u_{0}(s) \bigr) \,\mathrm{d}_{q}s \bigr\vert + \frac{1}{n_{0}} \\ &\quad \leq A_{1}(\alpha , b) \int _{0}^{1} ( 1 -qs)^{(\alpha -1)} \\ &\qquad{} \times \Biggl[ \sum_{i=1}^{k_{1}} \mu _{i}(s) \varTheta _{i} \biggl( \Vert u_{0} \Vert , \bigl\Vert u'_{0} \bigr\Vert , \frac{ \Vert u'_{0} \Vert }{ \varGamma _{q}(2 - \beta _{1})}, \frac{ \Vert u_{0} \Vert }{ \varGamma _{q}(\beta _{2} +1)} \biggr) \Biggr] \,\mathrm{d}_{q}s + \frac{1}{n_{0}} \\ &\quad \leq A_{1} (\alpha , b) \sum_{i=1}^{k_{1}} \biggl[ \biggl( \int _{0} ^{1} (1-qs)^{(\alpha -1)} \mu _{i}(s) \,\mathrm{d}_{q}s \biggr) \\ &\qquad{} \times \varTheta _{i} \biggl( \frac{r_{0}}{n_{0}}, \frac{r_{0}}{n _{0}}, \frac{r_{0}}{n_{0}}, \frac{r_{0}}{n_{0}} \biggr) \biggr] + \frac{1}{n _{0}} \\ &\quad \leq A_{1}(\alpha , b) \Biggl[\sum_{i=1}^{k_{1}} \bigl\Vert ( 1 -qt)^{( \alpha -1)} \mu _{i} \bigr\Vert \Biggr] \varepsilon _{M} + \frac{1}{n_{0}} \leq \delta \end{aligned}$$

(20)

for $t \in \overline{J}$, and so $\|T_{u_{0}} -u_{0} \| \leq \delta$. By using a similar method, we get

$$\begin{aligned} \bigl\vert \bigl( T_{u_{0}}(t) - u_{0}(t) \bigr)' \bigr\vert &= \bigl\vert T'_{u_{0}}(t) - u'_{0}(t) \bigr\vert \\ & \leq \int _{0}^{1} \biggl\vert \frac{\partial G_{q}}{\partial t} (t,s) \biggr\vert \bigl\vert \omega \bigl( s, u_{0}(s), u'_{0}(s), D_{q}^{\beta _{1}} u _{0} (s), I_{q}^{\beta _{2}} u_{0}(s) \bigr) \,\mathrm{d}_{q}s \bigr\vert + \frac{1}{n_{0}} \\ & \leq A_{2}(\alpha , b) \Biggl[ \sum_{i=1}^{k_{1}} \bigl\Vert (1-qt)^{( \alpha -1)} \mu _{i} \bigr\Vert \Biggr] \varepsilon _{M} + \frac{1}{n_{0}} \leq \delta , \end{aligned}$$

(21)

and so $\|(T_{u_{0}} -u_{0})' \| \leq \delta$. Hence, from equations (20) and (21), we obtain $\|T_{u_{0}} - u_{0} \|_{*} = \max \{\|T_{u_{0}} - u_{0} \|, \|(T_{u_{0}} - u_{0})' \| \} \leq \delta$. Thus, $\alpha (u_{0}, T_{u_{0}}) = 1$. Now, by using Lemma 2, the map T has a fixed point, which is a solution for the multi-singular fractional q-problem (1). □

Now, we present our final result.

Theorem 7

Assume that$\omega : \overline{J} \times \overline{ \mathcal{B} } ^{4} \to [0, \infty ]$is such that$\omega ( t, u_{1}, u_{2}, u_{3}, u_{4}) < \infty$for all$u_{1}, u_{2}, u_{3}, u_{4}\in \overline{ \mathcal{B}}$and$t\in E$, where$E^{c}$is a null subset ofJ̅, that is, the measure of$E^{c}$is zero, the map$\omega (t, u_{1}, u_{2}, u_{3}, u_{4})$is continuous with respect to the components$u_{1}$, $u_{2}$, $u_{3}$, and$u_{4}$for all$t\in E$. Then the pointwise defined problem (1) with boundary conditions (2) has a solution whenever the following assumptions hold:

• (1) There exist a natural number$n_{1}\geq 1$and some maps$\mu _{1}, \ldots , \mu _{n_{1}} : \overline{J} \to [0,\infty )$such that$\mu _{1}, \ldots , \mu _{n_{1}} \in \overline{\mathcal{L}}$, the maps$F_{1}, \ldots , F_{n_{1}} : \mathbb{R}^{4}\to [0,\infty )$and$\varOmega : \mathbb{R}^{4} \to [0,\infty )$so that

  $$\begin{aligned}& \Vert \varOmega \Vert _{1}^{*} =\sup _{x \in \overline{\mathcal{L}} } \int _{0} ^{1} \varOmega \bigl( u(t), u(t), u(t), u(t) \bigr) \,\mathrm{d}t < \infty , \\& \Vert F_{i} \Vert _{\infty } = \sup _{w \in \mathbb{R}} \bigl\{ F_{i}( w, w, w, w)\bigr\} < \infty \end{aligned}$$

  for$i=1,\ldots , n_{1}$and

  $$ \bigl\vert \omega ( t, u_{1}, u_{2}, u_{3}, u_{4}) \bigr\vert \leq \sum _{i=1}^{n_{1}} \mu _{i}(t) F_{i} (u_{1}, u_{2}, u_{3}, u_{4}) + \varOmega (u_{1}, u_{2}, u_{3}, u_{4}) $$

  for$u_{1}, \ldots , u_{4} \in \overline{\mathcal{B}}$and$t \in \overline{J}$.

• (2) There exist some maps$\psi :\mathbb{R}^{4} \to [0,\infty )$and$h : \overline{J} \to [0,\infty )$such that$\| h \|_{1}^{L} = \varGamma _{q}(\alpha ) I_{q}^{\alpha }h(1) < \infty$and$h(t) \psi (u _{1}, \ldots , u_{4}) \leq \omega (t, u_{1}, \ldots , u_{4})$for$u_{1}, \ldots , u_{4} \in \overline{\mathcal{B}}$and$t \in \overline{J}$.

• (3) There exist$\gamma _{1}, \gamma _{2}, \gamma _{3}, \gamma _{4} \in \overline{\mathcal{L}}$and$\phi : [0, \infty ) \to [0, \infty )$such that

  $$ M( \alpha , b) \sum_{i=1}^{4} \Vert \gamma _{i} \Vert _{1} < 1, $$

  $\phi _{m_{0}} \in \varPsi$and

  $$ \bigl\vert \omega ( t, u_{1}, \ldots , u_{4}) - w(t, v_{1}, \ldots , v _{4}) \bigr\vert \leq \sum _{i=1}^{4} \gamma _{i}(t) \phi \bigl( \Vert u_{i} - v _{i} \Vert \bigr) $$

  for all$(u_{1}, \ldots , u_{4})$and$(v_{1}, \ldots , v_{4}) \in \overline{ \mathcal{B}}^{4}$with$\| u_{i}\|, \|v_{i}\| \in [\delta _{1}, \delta _{2}]$, where$\phi _{\lambda }(z):= \phi (\frac{z}{\lambda })$for all$\lambda \in (0, \infty )$,

  $$ \Vert \psi \Vert _{m}:= \min \bigl\{ \psi (u_{1}, \ldots , u_{4}): (u_{1}, \ldots , u_{4}) \in \mathbb{R}^{4} \bigr\} , $$

  $2\delta _{1} \varGamma _{q}(\alpha ) (1- \alpha ) \leq \| \psi \|_{m} \|h \|_{1}^{L} ( 4 - \alpha ^{2} - 2\alpha )$and

  $$ \delta _{2} \geq M(\alpha ,b) \Biggl( \sum _{i=1}^{ n_{1}} \Vert F_{i} \Vert _{\infty } \Vert \mu _{i} \Vert _{1} + \Vert \varOmega \Vert _{1}^{*} \Biggr). $$

Proof

Let $\{u_{n}\}$ be a sequence such that $\|u_{n} - u\|_{*} \to 0$. Then $u_{n} \to u$ and $u'_{n} \to u'$. By using the inequalities $\varGamma _{q}(2 - \beta _{1}) \|D_{q}^{\beta _{1}}( u_{n} - u) \| \leq \|(u _{n} -u)' \|$ and

$$ \varGamma _{q}(\beta _{2} + 1) \bigl\Vert I_{q}^{\beta _{2}}( u_{n} -u) \bigr\Vert \leq \Vert u _{n} -u \Vert , $$

we get $D_{q}^{\beta } u_{n} \to D_{q}^{\beta } u$ and $I_{q}^{\beta _{2}} u_{n} \to I_{q}^{\beta _{2}} u$. Since $\omega (t, u_{1}, \ldots , u_{4})$ is continuous with respect to $u_{1}, \ldots , u_{4}$ for all $t \in E$, we can conclude that

$$ \omega \bigl( t, u_{n}, u'_{n}, D_{q}^{\beta _{1}} u_{n}, I_{q}^{\beta _{2}} u_{n} \bigr) \to \omega \bigl( t, u, u', D_{q}^{\beta _{1}} u, I_{q}^{\beta _{2}} u \bigr) $$

for $t\in E$. Let $u \in \overline{\mathcal{B}}$ be given and $t \in \overline{J}$. Then we have

$$\begin{aligned} \bigl\vert T_{u}(t) \bigr\vert & \leq A_{1}( \alpha , b) \Biggl[ \int _{0}^{1} ( 1 -qs)^{ \alpha -1} \Biggl( \Biggl[ \sum_{i=1}^{n_{1}} \mu _{i}(s) F_{i} \bigl( u(s), u'(s), D_{q}^{\beta _{1}} u(s), I_{q}^{\beta _{2}} u(s) \bigr) \Biggr] \\ &\quad {} + \varOmega \bigl( u(s), u'(s), D_{q}^{\beta _{1}} u(s), I_{q} ^{\beta _{2}} u(s) \bigr) \Biggr) \,\mathrm{d}_{q}s \Biggr]. \end{aligned}$$

If $u_{M}(s) := \max \{u(s),u'(s), D_{q}^{\beta _{1}} u(s), I_{q}^{ \beta _{2}} u(s)\}$, then $u_{M} \in \overline{\mathcal{A}}$, and so

$$\begin{aligned} \bigl\vert T_{u}(t) \bigr\vert & \leq A_{1}( \alpha , b) \Biggl[ \sum_{i=1}^{n_{1}} F_{i} \biggl( \Vert u \Vert , \bigl\Vert u' \bigr\Vert , \frac{ \Vert u' \Vert }{ \varGamma _{q}( 2 - \beta _{1})}, \frac{ \Vert u \Vert }{ \varGamma _{q}(\beta _{2} + 1)} \biggr) \Biggr] \int _{0}^{1} \mu _{i}(s) \,\mathrm{d}s \\ &\quad {} + \int _{0}^{1} \varOmega \bigl( u_{M}(s), u_{M}(s), u_{M}(s), u_{M}(s)\bigr) \,\mathrm{d}s \\ &\leq A_{1}(\alpha , b) \Biggl[ \sum_{i=1}^{n_{1} } \Vert F_{i} \Vert _{\infty } \Vert \mu _{i} \Vert + \Vert \varOmega \Vert _{1}^{*} \Biggr]. \end{aligned}$$

Similarly, one can see that $|T'_{u}(t)| \leq A_{2}(\alpha , b) [\sum_{i=1}^{n_{1} } \|F_{i}\|_{\infty } \| \mu _{i}\| + \|\varOmega \|_{1} ^{*} ]$. Thus,

$$ \Vert T_{u} \Vert _{*} \leq M( \alpha , b) \Biggl[ \sum_{i=1}^{n_{1}} \Vert F_{i} \Vert _{\infty } \Vert \mu _{i} \Vert + \Vert \varOmega \Vert _{1}^{*} \Biggr] < \infty $$

(22)

for all $u \in \overline{\mathcal{B}}$. By using the Lebesgue dominated convergence theorem, we conclude that

$$\begin{aligned} T_{u_{n}}(t) & = \int _{0}^{1} G_{q}(t,s) \omega \bigl(s, u_{n}(s), u _{n}'(s), D_{q}^{\beta _{1}} u_{n}(s), I_{q}^{\beta _{2}} u_{n}(s) \bigr) \,\mathrm{d}_{q}s \\ & \to \int _{0}^{1} G_{q}(t,s) \omega \bigl( s, u(s), u'(s), D _{q}^{\beta _{1}} u(s), I_{q}^{\beta _{2}} u(s) \bigr) \,\mathrm{d}_{q}s = T_{u}(t) \end{aligned}$$

for t belonging to J̅, and so the self-map T on $\overline{\mathcal{B}}$ is continuous. Define the map $\alpha : \overline{ \mathcal{B}}^{2} \to [0,\infty )$ by $\alpha (u, v)=1$ whenever $\| u\|_{*}, \| v\|_{*} \in [\delta _{1}, \delta _{2}]$, $\alpha (u, v)=0$, otherwise. If $\alpha (u, v) \geq 1$, then $\| u\|_{*}, \| v\| _{*} \in [\delta _{1}, \delta _{2}]$, and so

$$\begin{aligned} \bigl\vert T_{u}(t) \bigr\vert & = \biggl\vert \int _{0}^{1} G_{q}(t,s) \omega \bigl(s, u(s), u'(s), D_{q}^{\beta _{1}} u(s), I_{q}^{\beta _{2}} u(s) \bigr) \,\mathrm{d}_{q}s \biggr\vert \\ & \geq \int _{0}^{1} \frac{ (1-qs)^{b-1}}{ \varGamma _{q}(\alpha ) } \biggl[-t + \frac{ 2 -b^{2}}{ 2 ( 1 - b)} \biggr] h(s) \\ &\quad{} \times \psi \bigl[ u(s), u'(s), D_{q}^{\beta _{1}} u(s), I_{q} ^{\beta _{2}} u(s) \bigr] \,\mathrm{d}_{q}s \\ &\geq \Vert \psi \Vert _{m} \biggl[\frac{-t}{ \varGamma _{q}(\alpha )} \int _{0} ^{1} (1- qs)^{ \alpha - 1} h(s) \,\mathrm{d}_{q}s \\ &\quad {} + \frac{ 2 -b^{2}}{ 2 (1 - b) } \int _{0}^{1} (1-qs)^{(b-1)} h(s) \,\mathrm{d}_{q}s \biggr] \\ &\geq \Vert \psi \Vert _{m} \Vert h \Vert _{1}^{L} \biggl[ \frac{1}{ \varGamma _{q}( \alpha )} + \frac{ 2 -b^{2}}{ 2 (1 - b)} \biggr] \\ & = \Vert \psi \Vert _{m} \Vert h \Vert _{1}^{L} \biggl[\frac{4 - b^{2} - 2 b}{2 \varGamma _{q}(\alpha ) (1- b)} \biggr] \end{aligned}$$

for $t \in \overline{J}$. Thus, $2 \varGamma _{q}(\alpha ) (1 - b) \|T _{u}\| \geq \|\psi \|_{m} \|h\|_{1}^{L} (4 - b^{2} - 2b)$, and so

$$ \Vert T_{u} \Vert _{*} := \max \bigl\{ \Vert T_{u} \Vert , \bigl\Vert T'_{u} \bigr\Vert \bigr\} \geq \frac{ \Vert \psi \Vert _{m} \Vert h \Vert _{1}^{L} (4 - b^{2} - 2b)}{2 \varGamma _{q}( \alpha ) ( 1 - b)} \geq \delta _{1}. $$

By using (22), we obtain

$$ \Vert T_{x} \Vert _{*} \leq M(\alpha , b) \Biggl( \sum_{i=1}^{n_{1}} \Vert F_{i} \Vert _{\infty } \Vert \mu _{i} \Vert _{1} + \Vert \varOmega \Vert _{1}^{*} \Biggr) \leq \delta _{2}, $$

and so $\alpha (T_{u}, F_{v}) \geq 1$. If $u_{0} \in [\delta _{1}, \delta _{2}]$, then it is easy to check that $\alpha (T_{u_{0}}, u_{0}) \geq 1$. Let $u, v \in [\delta _{1}, \delta _{2}]$. Then

$$\begin{aligned} &\alpha (u, v) \bigl\vert T_{u}(t) - T_{v}(t) \bigr\vert \\ &\quad \leq \int _{0}^{1} G _{q}(t,s) \bigl\vert \omega \bigl( s, u(s), u'(s), D_{q}^{\beta _{1}} u(s), I _{q}^{\beta _{2}} u(s) \bigr) \\ &\qquad {} - \omega \bigl( s, v(s), v'(s), D_{q}^{\beta _{1}} v(s), I_{q} ^{\beta _{2}} v(s) \bigr) \bigr\vert \,\mathrm{d}_{q}s \\ &\quad \leq A_{1}(\alpha , b) \int _{0}^{1} (1-qs)^{(\alpha -1)} \bigl( \gamma _{1}(s) \phi \bigl( \vert u - v \vert \bigr) + \gamma _{2}(s) \phi \bigl( \bigl\vert u' - v' \bigr\vert \bigr) \\ &\qquad {} + \gamma _{3}(s) \phi \bigl( \bigl\vert D_{q}^{\beta _{1}} u - D_{q}^{ \beta _{1}} v \bigr\vert \bigr) + \gamma _{4}(s) \phi \bigl( \bigl\vert I_{q}^{\beta _{2}} u- I_{q}^{\beta _{2}} v \bigr\vert \bigr) \bigr) \,\mathrm{d}_{q}s \\ & \quad \leq A_{1}(\alpha , b) \int _{0}^{1} (1-qs)^{(\alpha -1)} \biggl( \gamma _{1}(s) \phi \bigl( \Vert u- v \Vert \bigr) + \gamma _{2}(s) \phi \bigl( \bigl\Vert u' - v' \bigr\Vert \bigr) \\ &\qquad {} + \gamma _{3}(s) \phi \biggl( \frac{ \Vert u' - v' \Vert }{ \varGamma _{q}(2 - \beta _{1})} \biggr) + \gamma _{4}(s) \phi \biggl( \frac{ \Vert u- v \Vert }{\varGamma _{q}(\beta _{2}+ 1)} \biggr) \biggr) \,\mathrm{d}_{q}s \\ &\quad \leq A_{1}(\alpha , b) \int _{0}^{1} (1-qs)^{(\alpha -1)} \phi \biggl( \frac{ \Vert u- v \Vert _{*}}{ m_{0}} \biggr) \Biggl[ \sum_{i=1}^{4} \gamma _{i}(s) \Biggr] \,\mathrm{d}_{q}s \\ &\quad \leq A_{1}(\alpha , b) \phi \biggl( \frac{ \Vert u- v \Vert _{*}}{ m_{0}} \biggr) \sum_{i=1}^{4} \int _{0}^{1} ( 1 -qs)^{(\alpha -1)} \gamma _{i}(s) \,\mathrm{d}_{q}s \\ &\quad \leq \sum_{i=1}^{4} \Vert \gamma _{i} \Vert _{1} \phi \biggl( \frac{ \Vert u - v \Vert _{*}}{ m_{0}} \biggr) \leq \phi _{\gamma _{0}}\bigl( \Vert u - v \Vert _{*} \bigr). \end{aligned}$$

(23)

Similarly, we conclude that

$$ \alpha (u, v) \bigl\Vert T'_{u} - T'_{v} \bigr\Vert \leq \phi _{m_{0}} \bigl( \Vert u - v \Vert _{*}\bigr). $$

(24)

Now, (23) and (24) imply that $\alpha (u , v) \| T _{u} - T_{v} \|$ and $\alpha ( u, v) \| T'_{u} - T'_{v} \|_{*}$ are less than or equal to $\phi _{ m_{0}}( \|u - v\|_{*})$ for all $u, v\in \overline{\mathcal{B}}$. By using Theorem 2, the self-map T has a fixed point, which is a solution for problem (1). □

4 Examples and algorithms for the problem

Here, we provide some examples to illustrate our main results. In this way, we give a computational technique for checking problem (1). We need to present that a simplified analysis could be executed on values of the q-gamma function. To this aim, we consider a pseudo-code description of the method for calculation of the q-gamma function of order n in Algorithms 2, 3, 4, and 5 (for more details, see the link https://en.wikipedia.org/wiki/Q-gamma_function).

Algorithm 5

The proposed method for calculated $\int _{a}^{b} f(r) \,\mathrm{d}_{q} r$

Example 1

Consider the following pointwise defined problem, similar to (1):

$$ D_{q}^{\frac{7}{2}} u(t) + \frac{1}{96 \sqrt{\pi } [ g(t)]^{ \frac{1}{3}}} \bigl( \Vert u \Vert + \bigl\Vert u' \bigr\Vert + \bigl\Vert D_{q}^{\frac{2}{3}} u \bigr\Vert + \bigl\Vert I _{q}^{\frac{1}{2}} u \bigr\Vert \bigr) = 0 $$

(25)

with boundary conditions $u'(0)=u(\frac{5}{6})$, $u(1)=\int _{0}^{ \frac{1}{2}} u(s) \,\mathrm{d}s$, and $u''(0)=0$, where $g(t) = 0$ whenever $t \in \overline{J} \cap Q$, $g(t) = t$ whenever $t \in J \cap Q^{c}$. Let $\alpha = \frac{7}{2}$, $\beta _{1}= \frac{2}{3}$, $\beta _{2} = \frac{1}{3}$, $a= \frac{5}{6}$, and $b= \frac{1}{2}$. Then we have

$$\begin{aligned} M(\alpha , b) &= \max \bigl\{ A_{1}(\alpha , b), A_{2}(\alpha , b)\bigr\} \\ & = \max \biggl\{ \frac{3}{(1-b) \varGamma _{q}(\alpha )}, \frac{2}{(1-b) \varGamma _{q}(\alpha -1)} \biggr\} \\ &= \max \biggl\{ \frac{3}{( 1 -\frac{1}{2}) \varGamma _{q}(\frac{7}{2})}, \frac{2}{( 1- \frac{1}{2}) \varGamma _{q}(\frac{5}{2})} \biggr\} \end{aligned}$$

(26)

and

$$ m_{0} = \min \bigl\{ \varGamma _{q}( 2 -\beta _{1}), \varGamma _{q}(\beta _{2} +1) \bigr\} = \min \biggl\{ \varGamma _{q} \biggl( \frac{4}{3} \biggr), \varGamma _{q} \biggl(\frac{3}{2} \biggr) \biggr\} . $$

(27)

Tables 4 and 5 show the values of $M(\alpha , b)$ in equation (26) and $m_{0}$ in equation (27), respectively. Put

$$ \omega \bigl( t, u(t), u'(t), D_{q}^{\beta _{1}} u(t), I_{q}^{\beta _{2}} u(t)\bigr) := \frac{1}{96 \sqrt{\pi } [ g(t)]^{\frac{1}{3}}} \bigl( \Vert u \Vert + \bigl\Vert u' \bigr\Vert + \bigl\Vert D_{q}^{\frac{2}{3}} u \bigr\Vert + \bigl\Vert I_{q}^{\frac{1}{2}} u \bigr\Vert \bigr), $$

$\mu _{1}(t)= \mu _{2}(t) =\mu _{3}(t) =\mu _{4}(t) = \mu (t)= \frac{1}{96\sqrt{ \pi } t^{\frac{1}{3}}}$, $\gamma _{1}(t) = \gamma _{2}(t) = \gamma _{3}(t) =\gamma _{4}(t) =\gamma (t)= \frac{1}{96\sqrt{\pi } t^{\frac{1}{3}}}$ and $\varTheta _{i} (u_{1}, u_{2}, u_{3}, u_{4} ):= \|u_{i}\|$ for $i=1,\ldots ,4$. Then $\|\mu \|_{1} = \|\gamma \|_{1} = \frac{1}{96\sqrt{ \pi }(1-\frac{1}{3})} = \frac{1}{64\sqrt{\pi } }$,

$$\begin{aligned} &\bigl\vert \omega ( t, u_{1}, u_{2}, u_{3}, u_{4}) - \omega (t, v_{1}, v_{2}, v _{3}, v_{4} ) \bigr\vert \\ &\quad = \frac{1}{96\sqrt{\pi } ( g(t) )^{ \frac{1}{3}}} \sum_{i=1}^{4} \Vert u_{i} \Vert - \Vert v_{i} \Vert \\ &\quad \leq \frac{1}{ 96 \sqrt{\pi } (g(t))^{ \frac{1}{3}}} \sum_{i=1} ^{4} \Vert u_{i} - v_{i} \Vert \\ &\quad = \frac{1}{96\sqrt{\pi } t^{\frac{1}{3}}} \sum_{i=1}^{4} \Vert u _{i} - v_{i} \Vert \\ &\quad = \sum_{i=1}^{4} \mu (t) \Vert u_{i} - v_{i} \Vert , \end{aligned}$$

and

$$\begin{aligned} \bigl\vert \omega ( t, u_{1}, u_{2}, u_{3}, u_{4} ) \bigr\vert &= \frac{1}{96\sqrt{ \pi } (g(t))^{\frac{1}{3}} } \sum_{i=1}^{4} \Vert u_{i} \Vert \\ & = \frac{1}{96\sqrt{\pi }t^{\frac{1}{3}}} \sum_{i=1}^{4} \Vert u_{i} \Vert \\ & = \sum_{i=1}^{4} \gamma (t) \varTheta _{i}( u_{1}, u_{2}, u_{3}, u_{4} ) \end{aligned}$$

for all $(u_{1}, u_{2}, u_{3}, u_{4})$ and $(v_{1}, v_{2}, v_{3}, v _{4}) \in \overline{\mathcal{B}}^{4}$ and $t \in \overline{J}$. On the other hand, we have $\lim_{w \to \infty } \frac{\varTheta _{i}(w, w, w, w)}{w} = 1$, and by equations (26) and (27), we get

$$ \varLambda = \frac{m_{0}}{ M(\alpha , b) \sum_{i=1}^{4} \Vert \gamma _{i}(t) \Vert _{1}} = \frac{m_{0}}{ M(\alpha , b) \sum_{i=1}^{4} \Vert \gamma (t) \Vert _{1}} > 1. $$

(28)

Table 6 shows the values of Λ. Choose $\delta _{0}>0$ such that $m_{0} \geq M( \alpha , b) \sum_{i=1}^{4} \| \gamma \|_{1}+ \delta _{0}$. Since

$$ \hat{\mu }= \int _{0}^{1} (1-qs)^{(\frac{7}{2}-1)} \mu (s) \,\mathrm{d}_{q}s \leq \int _{0}^{1} \mu (s) \,\mathrm{d}s = \Vert \mu \Vert _{1} = \frac{1}{64 \sqrt{\pi }}, $$

we obtain

$$\begin{aligned} \tau (\alpha , b) & = \biggl[\hat{\mu } + \hat{\mu } + \frac{ \hat{\mu }}{\varGamma _{q}(2 - \beta _{1})} + \frac{ \hat{\mu }}{ \varGamma _{q}(\beta _{2} +1)} \biggr] M( \alpha , b) \\ & \leq \frac{1}{64 \sqrt{\pi }} \biggl[1+ 1 + \frac{1}{ \varGamma _{q}( \frac{4}{3}) } + \frac{1}{ \varGamma _{q}(\frac{3}{2})} \biggr] M( \alpha , b) < 1. \end{aligned}$$

(29)

Table 7 shows the values of $\tau (\alpha , b)$. Now, by using Theorem 5, the pointwise defined problem (25) has a solution.

Table 4 Some numerical results of $M(\alpha , b)$ in equation (26) in Example 1 for $q=\frac{1}{7}, \frac{1}{2}, \frac{8}{9}$

Table 5 Some numerical results of $m_{0}$ in equation (27) in Example 1 for $q=\frac{1}{7}, \frac{1}{2}, \frac{8}{9}$

Table 6 Some numerical results of $\varLambda >1$ in equation (28) in Example 1 for $q=\frac{1}{7}, \frac{1}{2}, \frac{8}{9}$

Table 7 Some numerical results of $\tau (\alpha , b)<1$ in equation (29) in Example 1 for $q=\frac{1}{7}, \frac{1}{2}, \frac{8}{9}$. Note that $(a)=\varGamma _{q}(\beta _{2}+1)$, $(b)= \varGamma _{q}(2-\beta _{1})$, and $(c) = M(\alpha , b)$

Example 2

Consider the fractional q-integro-differential equation

$$ D_{q}^{\frac{5}{2}} u(t) + \frac{0.09}{ t^{\frac{1}{4}} ( t - \frac{1}{3})^{ \frac{1}{8}}} \biggl[1 - \biggl( \frac{3}{5} \biggr)^{ \frac{2}{5} ( u(t) + u'(t) + D_{q}^{ \frac{1}{4}} u(t) + I_{q} ^{\frac{1}{5}} u(t) ) } \biggr]= 0 $$

(30)

for $t\in \overline{J}$, with boundary condition $u'(0) =u( \frac{1}{2})$ and $u(1)= \int ^{ \frac{1}{7}}_{0} u(s) \,\mathrm{d}s$. Put $\alpha = \frac{5}{2}$, $\beta _{1}=\frac{1}{4}$, $\beta _{2}= \frac{1}{5}$, $a = \frac{1}{2}$, $b= \frac{1}{7}$, and $k_{1}=1$. Note that

$$\begin{aligned} M(\alpha , b) &= \max \biggl\{ \frac{3}{(1-b) \varGamma _{q}(\alpha )}, \frac{2}{(1-b) \varGamma _{q}(\alpha -1)} \biggr\} \\ & = \max \biggl\{ \frac{3}{ ( 1 -\frac{1}{7}) \varGamma _{q}( \frac{5}{2})}, \frac{2}{ (1 -\frac{1}{7}) \varGamma _{q}( \frac{3}{2})} \biggr\} . \end{aligned}$$

(31)

Table 8 shows the values of $M(\alpha , b)$ in equation (31). Define

$$ \omega (t, u_{1}, \ldots , u_{4})= \frac{ 0.09}{ t^{\frac{1}{4}} ( t - \frac{1}{3} )^{\frac{1}{8}}} \biggl[ 1 - \biggl(\frac{3}{5} \biggr)^{\frac{2}{5} (u_{1} + \cdots + u_{4})} \biggr], $$

$\mu (t)= \frac{0.09}{ t^{\frac{1}{4}} ( t -\frac{1}{3})^{\frac{1}{8}}}$, and

$$ \varTheta (u_{1}, \ldots , u_{4})=1 - \biggl( \frac{3}{5} \biggr)^{ \frac{2}{5} ( u_{1}+ \cdots + u_{4})}. $$

One can easily see that Θ is nondecreasing in all its components and $\varTheta (w, w, w, w) \geq 0$ for all $w \geq 0$. Assume that $(u_{1}, \ldots , u_{4})$ and $(v_{1}, \ldots , v_{4})$ belong to $\overline{\mathcal{B}}^{4}$ and $u_{i} \geq v_{i} \geq 0$ for $i = 1, \ldots , 4$. Since $( \frac{3}{5} )^{v_{i}} \geq (\frac{3}{5})^{u_{i}}$,

$$ \biggl( \frac{3}{5} \biggr)^{v_{i}} \biggl[ \biggl( \frac{3}{5} \biggr) ^{v_{i}} - \biggl( \frac{3}{5} \biggr)^{u_{i}} \biggr] \leq \biggl(\frac{3}{5} \biggr)^{v_{i}} - \biggl( \frac{3}{5} \biggr) ^{u_{i}}, $$

and so

$$ \biggl(\frac{3}{5} \biggr)^{v_{i}} \biggl[ \biggl( \frac{3}{5} \biggr) ^{v_{i}} - \biggl(\frac{3}{5} \biggr)^{u_{i}} \biggr] \leq \biggl(\frac{3}{5} \biggr)^{v_{i}} \biggl[1 - \biggl(\frac{3}{5} \biggr)^{u_{i} - v_{i}} \biggr]. $$

Thus,

$$ \biggl[ 1 - \biggl(\frac{3}{5} \biggr)^{u_{i} } \biggr] - \biggl[ 1 - \biggl(\frac{3}{5} \biggr)^{v_{i}} \biggr] \leq 1 - \biggl(\frac{3}{5} \biggr)^{ u_{i} -v_{i}} . $$

By replacing $u_{i}$, $v_{i}$ with $\frac{2}{5}\sum_{i=1}^{4} u_{i}$, $\frac{2}{5} \sum_{i=1}^{4} v_{i}$, respectively, we get

$$ \biggl[1 - \biggl(\frac{3}{5} \biggr)^{\frac{2}{5} \sum _{i=1}^{4} u _{i} } \biggr] - \biggl[1 - \biggl(\frac{3}{5} \biggr)^{ \frac{2}{5} \sum _{i=1}^{4} v_{i}} \biggr] \leq 1 - \biggl(\frac{3}{5} \biggr)^{\frac{2}{5} (\sum _{i=1}^{4} u_{i} - v_{i})}. $$

Hence, $\varTheta (u_{1}, \ldots , u_{4}) - \varTheta (v_{1}, \ldots , v _{4}) \leq \varTheta (u_{1} - v_{1}, \ldots , u_{4} - v_{4})$. On the other hand, we have

$$ \omega ( t, u_{1}, \ldots , u_{4}) - \omega (t, v_{1}, \ldots , v_{4}) \leq \mu (t) \varTheta (u_{1} - v_{1}, \ldots , u_{4} - v_{4}) $$

and

$$ \lim_{w \to 0^{+}} \frac{\varTheta (w, w, w, w)}{w} = \lim \limits _{w \to 0^{+}} \frac{1 - (\frac{3}{5})^{4\times \frac{2}{5} w}}{w} = - 4 \biggl(\frac{2}{5} \biggr) \biggl[ \ln \biggl( \frac{3}{5} \biggr) \biggr] = 0.81 < 1. $$

Now, by using Theorem 6, the fractional q-integro-differential pointwise defined equation (30) has a solution.

Table 8 Some numerical results of $M(\alpha , b)$ in equation (31) in Example 2 for $q=\frac{1}{7}, \frac{1}{2}, \frac{8}{9}$

Example 3

Consider the fractional q-integro-differential equation

$$ D_{q}^{\frac{5}{2}} u(t) + \frac{0.01}{g(t)} F \bigl( u(t), u'(t), D _{q}^{\frac{1}{3}} u(t), I_{q}^{\frac{2}{3}} u(t) \bigr)+ 2 = 0\quad (t \in \overline{J}) $$

(32)

with boundary condition $u'(0) = u(\frac{1}{2})$ and $u(1)= \int ^{\frac{4}{5}}_{0} u(s) \,\mathrm{d}s$, where $g: \overline{J} \to [0,\infty )$ is defined by $g(t) =0$ whenever $t \in Q \cap \overline{J}$ and $g(t) = \sqrt{t}$ whenever $t \in Q^{c} \cap \overline{J}$ and the map $F: \mathbb{R}^{4} \to [0,\infty )$ is defined by

$$ F(u_{1}, \ldots , u_{4}) = \textstyle\begin{cases} \frac{1}{2} \sum_{i=1}^{4} \frac{ \Vert u_{i} \Vert }{ 1 + \Vert u_{i} \Vert }, & u _{1}, \ldots , u_{4} \in [0, 29], \\ \vert \sin (u_{1} + \cdots + u_{4}) \vert , & u_{1}, \ldots , u_{4} \in (-\infty , 0), \\ \frac{-29}{48} ( \frac{1 }{4} \sum_{i=1}^{4} u_{i} - 24 ), & u_{1}, \ldots , u_{4} \in [29, 30], \\ 0,& \text{otherwise}. \end{cases} $$

Put $\alpha = \frac{5}{2}$, $\beta _{1}=\frac{1}{3}$, $\beta _{2}= \frac{2}{3}$, $a= \frac{1}{2}$, and $b= \frac{4}{5}$. Then we have

$$\begin{aligned} M(\alpha , b) & = \max \biggl\{ \frac{3}{( 1 -b) \varGamma _{q}(\alpha )}, \frac{2}{( 1 -b) \varGamma _{q}(\alpha -1) } \biggr\} \\ & = \max \biggl\{ \frac{3}{ ( 1 -\frac{4}{5}) \varGamma _{q}( \frac{5}{2})}, \frac{2}{(1 -\frac{4}{5}) \varGamma _{q}(\frac{3}{2})} \biggr\} \end{aligned}$$

(33)

and

$$ m_{0} = \min \bigl\{ \varGamma _{q}(2- \beta _{1}), \varGamma _{q}( \beta _{2} + 1) \bigr\} =\min \biggl\{ \varGamma _{q} \biggl( \frac{5}{3} \biggr), \varGamma _{q} \biggl(\frac{5}{3} \biggr) \biggr\} = \varGamma _{q} \biggl( \frac{5}{3} \biggr). $$

(34)

Tables 9 and 10 show the values of $M(\alpha , b)$ and $m_{0}$ in equations (33) and (34), respectively. By simple checking, we can see that $\|F\|_{\infty } = 1$. Let $n_{1}=1$. Define the maps $\psi ( u_{1}, \ldots , u_{4}):= F( u_{1}, \ldots , u_{4})$, $\mu (t)= h(t)= \gamma (t) := 0.01t$, and $\phi (t) =\frac{1}{2} t$. If $\varOmega (u_{1}, \ldots , u _{4})=2$,

$$ \omega ( t, u_{1}, \ldots , u_{4})= \frac{0.01}{g(t)} T(u_{1}, \ldots , u _{4})+ 2, $$

and $[\delta _{1} , \delta _{2}] =[0, 29]$, then $\omega (t ,u_{1}, u _{2}, u_{3}, u_{4}) < \infty$ for $u_{1}, \ldots , u_{4} \in \overline{ \mathcal{B}}$ and $t\in E: = Q^{c}\cap \overline{J}$, $\omega ( t, u _{1}, u_{2}, u_{3}, u_{4})$ is continuous with respect to the components $u_{1}$, $u_{2}$, $u_{3}$, and $u_{4}$ for all $t\in E$,

$$ \omega ( t, u_{1}, \ldots , u_{4} )= \frac{1}{g(t)} F( u_{1}, \ldots , u _{4})+ 2 \geq h(t) \psi ( u_{1}, \ldots , u_{4}) $$

and

$$\begin{aligned} \bigl\vert \omega ( t, u_{1} , \ldots , u_{4}) - \omega (t, v_{1}, \ldots , v _{4}) \bigr\vert & \leq \sum_{i=1}^{4} \frac{1}{2} \times \frac{0.01}{ \sqrt{t}} \Vert u_{i} - v_{i} \Vert \\ & = \sum_{i=1}^{4} \gamma (t) \phi \bigl( \Vert u_{i} - v_{i} \Vert \bigr) \end{aligned}$$

for all $(u_{1}, \ldots , u_{4})$, $(v_{1}, \ldots , v_{4})\in \overline{ \mathcal{B}}^{4}$ and $t \in \overline{J}$. Note that $\phi _{m_{0}} \in \varPsi$,

$$\begin{aligned}& M(\alpha , b) \sum_{i=1}^{4} \Vert \gamma _{i} \Vert _{1} = 4 \times 0.005 \times M(\alpha , b) = 0.02 \times M(\alpha , b) < 1, \end{aligned}$$

(35)

$$\begin{aligned}& \frac{ \Vert \psi \Vert _{m} \Vert h \Vert _{1}^{L} (4- \alpha ^{2} - 2\alpha ) }{2 \varGamma _{q}(\alpha ) (1 - \alpha )} \geq 0 = \delta _{1}, \end{aligned}$$

(36)

and

$$ M(\alpha , b) \bigl( \Vert F \Vert _{\infty } \Vert \mu \Vert _{1} + \Vert \varOmega \Vert _{1} ^{*} \bigr) \leq (1 \times 0.02 + 2 )M(\alpha , b) \leq 29 = \delta _{2}. $$

(37)

Table 10 shows the values of equation (37). Now, by using Theorem 7, the fractional q-integro-differential pointwise defined equation (32) has a solution.

Table 9 Some numerical results of $M(\alpha , b)$ in equation (33) in Example 3 for $q=\frac{1}{7}, \frac{1}{2}, \frac{8}{9}$

Table 10 Some numerical results of $M(\alpha , b)$ in equation (33) in Example 3 for $q=\frac{1}{7}, \frac{1}{2}, \frac{8}{9}$. Note that $(a) = m_{0}$, $(b) = 0.02 \times M(\alpha , b)$, and $(c) =2.02 \times M(\alpha , b)$