Theory and Modern Applications

# Multi-strip fractional q-integral boundary value problems for nonlinear fractional q-difference equations

## Abstract

In this article, we study the existence and uniqueness of solutions for multi-strip fractional q-integral boundary value problems of nonlinear fractional q-difference equations. By using the Banach contraction principle, Krasnoselskii’s fixed point theorem, Leray-Schauder’s nonlinear alternative and Leray-Schauder degree theory some interesting results are obtained. Some examples are presented to illustrate the results.

MSC:34A08, 34B18, 39A13.

## 1 Introduction

In this article, we investigate the following nonlinear fractional q-difference equation for multi-strip fractional q-integral boundary condition:

$\left\{\begin{array}{c}{D}_{q}^{\alpha }u\left(t\right)=f\left(t,u\left(t\right)\right),\phantom{\rule{1em}{0ex}}t\in \left(0,T\right),\hfill \\ u\left(0\right)=0,\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}u\left(T\right)={\sum }_{i=1}^{m}{\gamma }_{i}\left({I}_{{q}_{i}}^{{\beta }_{i}}u\right){|}_{{\eta }_{i}}^{{\xi }_{i}}={\sum }_{i=1}^{m}{\gamma }_{i}\left({I}_{{q}_{i}}^{{\beta }_{i}}u\left({\xi }_{i}\right)-{I}_{{q}_{i}}^{{\beta }_{i}}u\left({\eta }_{i}\right)\right),\hfill \end{array}$
(1.1)

where $1<\alpha \le 2$, $0, ${\beta }_{i}>0$, $0\le {\eta }_{i}<{\xi }_{i}\le T$, ${\gamma }_{i}\in \mathbb{R}$ for all $i=1,2,\dots ,m$ are given constants, ${D}_{q}^{\alpha }$ is the fractional q-derivative of Riemann-Liouville type of order α, ${I}_{{q}_{i}}^{{\beta }_{i}}$ is the fractional ${q}_{i}$-integral of order ${\beta }_{i}$ and $f:\left[0,T\right]×\mathbb{R}\to \mathbb{R}$ is a continuous function.

q-Difference calculus or quantum calculus was initiated by Jackson [1]. Basic definitions and properties of quantum calculus can be found in the book [2]. The fractional q-difference calculus had its origin in the works by Al-Salam [3] and Agarwal [4]. For some recent work on the subject, we refer to [512] and the references cited therein.

Strip conditions appear in the mathematical modeling of certain real world problems. For motivation, discussion on multi-strip boundary conditions, examples and a consistent bibliography on these problems, we refer to the papers [1320] and the references therein. As it is pointed out in [20], the boundary condition in (1.1) can be interpreted in the sense that a controller at the right-end of the considered interval is influenced by a discrete distribution of finite many nonintersecting strips of arbitrary length expressed in terms of fractional integral boundary conditions.

The significance of investigating problem (1.1) is that the multi-strip fractional q-integral boundary condition is very general and includes many conditions as special cases. In particular, if ${\beta }_{i}=1$ for $i=1,2,\dots ,m$, then the condition of (1.1) is reduced to the multi-strip q-integral condition as follows:

$u\left(0\right)=0,\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}u\left(T\right)={\gamma }_{1}{\int }_{{\eta }_{1}}^{{\xi }_{1}}u\left(s\right)\phantom{\rule{0.2em}{0ex}}{d}_{{q}_{1}}s+{\gamma }_{2}{\int }_{{\eta }_{2}}^{{\xi }_{2}}u\left(s\right)\phantom{\rule{0.2em}{0ex}}{d}_{{q}_{2}}s+\cdots +{\gamma }_{m}{\int }_{{\eta }_{m}}^{{\xi }_{m}}u\left(s\right)\phantom{\rule{0.2em}{0ex}}{d}_{{q}_{m}}s.$

Moreover, we emphasize that we have different quantum numbers and as far as we know this is new in the literature.

The rest of the paper is organized as follows. In Section 2 we briefly give some basic notations, definitions and lemmas. In Section 3 we collect some auxiliary results needed in the proofs of our main results. Section 4 contains the main results concerning existence and uniqueness results for problem (1.1), which are shown by applying the Banach contraction principle, Krasnoselskii’s fixed point theorem, Leray-Schauder’s nonlinear alternative and Leray-Schauder degree theory. Some examples are presented in Section 5 to illustrate the results.

## 2 Preliminaries

To make this paper self-contained, below we recall some known facts on fractional q-calculus. The presentation here can be found in, for example, [21, 22].

For $q\in \left(0,1\right)$, define

${\left[a\right]}_{q}=\frac{1-{q}^{a}}{1-q},\phantom{\rule{1em}{0ex}}a\in \mathbb{R}.$
(2.1)

The q-analogue of the power function ${\left(a-b\right)}^{k}$ with $k\in {\mathbb{N}}_{0}:=\left\{0,1,2,\dots \right\}$ is

${\left(a-b\right)}^{\left(0\right)}=1,\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}{\left(a-b\right)}^{\left(k\right)}=\prod _{i=0}^{k-1}\left(a-b{q}^{i}\right),\phantom{\rule{1em}{0ex}}k\in \mathbb{N},a,b\in \mathbb{R}.$
(2.2)

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

${\left(a-b\right)}^{\left(\gamma \right)}={a}^{\gamma }\prod _{i=0}^{\mathrm{\infty }}\frac{1-\left(b/a\right){q}^{i}}{1-\left(b/a\right){q}^{\gamma +i}},\phantom{\rule{1em}{0ex}}a\ne 0.$
(2.3)

Note if $b=0$, then ${a}^{\left(\gamma \right)}={a}^{\gamma }$. We also use the notation ${0}^{\left(\gamma \right)}=0$ for $\gamma >0$. The q-gamma function is defined by

${\mathrm{\Gamma }}_{q}\left(x\right)=\frac{{\left(1-q\right)}^{\left(x-1\right)}}{{\left(1-q\right)}^{x-1}},\phantom{\rule{1em}{0ex}}x\in \mathbb{R}\setminus \left\{0,-1,-2,\dots \right\}.$
(2.4)

Obviously, ${\mathrm{\Gamma }}_{q}\left(x+1\right)={\left[x\right]}_{q}{\mathrm{\Gamma }}_{q}\left(x\right)$.

The q-derivative of a function h is defined by

(2.5)

and q-derivatives of higher order are given by

$\left({D}_{q}^{0}h\right)\left(x\right)=h\left(x\right)\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}\left({D}_{q}^{k}h\right)\left(x\right)={D}_{q}\left({D}_{q}^{k-1}h\right)\left(x\right),\phantom{\rule{1em}{0ex}}k\in \mathbb{N}.$
(2.6)

The q-integral of a function h defined on the interval $\left[0,b\right]$ is given by

$\left({I}_{q}h\right)\left(x\right)={\int }_{0}^{x}h\left(s\right)\phantom{\rule{0.2em}{0ex}}{d}_{q}s=x\left(1-q\right)\sum _{i=0}^{\mathrm{\infty }}h\left(x{q}^{i}\right){q}^{i},\phantom{\rule{1em}{0ex}}x\in \left[0,b\right].$
(2.7)

If $a\in \left[0,b\right]$ and h is defined in the interval $\left[0,b\right]$, then its integral from a to b is defined by

${\int }_{a}^{b}h\left(s\right)\phantom{\rule{0.2em}{0ex}}{d}_{q}s={\int }_{0}^{b}h\left(s\right)\phantom{\rule{0.2em}{0ex}}{d}_{q}s-{\int }_{0}^{a}h\left(s\right)\phantom{\rule{0.2em}{0ex}}{d}_{q}s.$
(2.8)

Similar to derivatives, an operator ${I}_{q}^{k}$ is given by

$\left({I}_{q}^{0}h\right)\left(x\right)=h\left(x\right)\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}\left({I}_{q}^{k}h\right)\left(x\right)={I}_{q}\left({I}_{q}^{k-1}h\right)\left(x\right),\phantom{\rule{1em}{0ex}}k\in \mathbb{N}.$
(2.9)

The fundamental theorem of calculus applies to these operators ${D}_{q}$ and ${I}_{q}$, i.e.,

$\left({D}_{q}{I}_{q}h\right)\left(x\right)=h\left(x\right),$
(2.10)

and if h is continuous at $x=0$, then

$\left({I}_{q}{D}_{q}h\right)\left(x\right)=h\left(x\right)-h\left(0\right).$
(2.11)

Definition 2.1 Let $\nu \ge 0$ and h be a function defined on $\left[0,T\right]$. The fractional q-integral of Riemann-Liouville type is given by $\left({I}_{q}^{0}h\right)\left(x\right)=h\left(x\right)$ and

$\left({I}_{q}^{\nu }h\right)\left(x\right)=\frac{1}{{\mathrm{\Gamma }}_{q}\left(\nu \right)}{\int }_{0}^{x}{\left(x-qs\right)}^{\left(\nu -1\right)}h\left(s\right)\phantom{\rule{0.2em}{0ex}}{d}_{q}s,\phantom{\rule{1em}{0ex}}\nu >0,x\in \left[0,T\right].$
(2.12)

Definition 2.2 The fractional q-derivative of Riemann-Liouville type of order $\nu \ge 0$ is defined by $\left({D}_{q}^{0}h\right)\left(x\right)=h\left(x\right)$ and

$\left({D}_{q}^{\nu }h\right)\left(x\right)=\left({D}_{q}^{l}{I}_{q}^{l-\nu }h\right)\left(x\right),\phantom{\rule{1em}{0ex}}\nu >0,$
(2.13)

where l is the smallest integer greater than or equal to ν.

Definition 2.3 For any $x,s>0$,

${B}_{q}\left(x,s\right)={\int }_{0}^{1}{u}^{\left(x-1\right)}{\left(1-qu\right)}^{\left(s-1\right)}\phantom{\rule{0.2em}{0ex}}{d}_{q}u$
(2.14)

is called the q-beta function.

From [2], the expression of q-beta function in terms of the q-gamma function can be written as

${B}_{q}\left(x,s\right)=\frac{{\mathrm{\Gamma }}_{q}\left(x\right){\mathrm{\Gamma }}_{q}\left(s\right)}{{\mathrm{\Gamma }}_{q}\left(x+s\right)}.$

Lemma 2.4 [4]

Let $\alpha ,\beta \ge 0$ and f be a function defined in $\left[0,T\right]$. Then the following formulas hold:

1. (1)

$\left({I}_{q}^{\beta }{I}_{q}^{\alpha }f\right)\left(x\right)=\left({I}_{q}^{\alpha +\beta }f\right)\left(x\right)$,

2. (2)

$\left({D}_{q}^{\alpha }{I}_{q}^{\alpha }f\right)\left(x\right)=f\left(x\right)$.

Lemma 2.5 [22]

Let $\alpha >0$ and ν be a positive integer. Then the following equality holds:

$\left({I}_{q}^{\alpha }{D}_{q}^{\nu }f\right)\left(x\right)=\left({D}_{q}^{\nu }{I}_{q}^{\alpha }f\right)\left(x\right)-\sum _{k=0}^{\nu -1}\frac{{x}^{\alpha -\nu +k}}{{\mathrm{\Gamma }}_{q}\left(\alpha +k-\nu +1\right)}\left({D}_{q}^{k}f\right)\left(0\right).$
(2.15)

## 3 Some auxiliary lemmas

Lemma 3.1 Let $\alpha ,\beta >0$ and $0. Then we have

${\int }_{0}^{\eta }{\left(\eta -qs\right)}^{\left(\alpha -1\right)}{s}^{\left(\beta \right)}\phantom{\rule{0.2em}{0ex}}{d}_{q}s={\eta }^{\alpha +\beta }{B}_{q}\left(\alpha ,\beta +1\right).$
(3.1)

Proof Using the definitions of q-analogue of power function and q-beta function, we have

$\begin{array}{rcl}{\int }_{0}^{\eta }{\left(\eta -qs\right)}^{\left(\alpha -1\right)}{s}^{\left(\beta \right)}\phantom{\rule{0.2em}{0ex}}{d}_{q}s& =& \left(1-q\right)\eta \sum _{n=0}^{\mathrm{\infty }}{q}^{n}{\left(\eta -q\eta {q}^{n}\right)}^{\left(\alpha -1\right)}{\left(\eta {q}^{n}\right)}^{\beta }\\ =& \left(1-q\right)\eta \sum _{n=0}^{\mathrm{\infty }}{q}^{n}{\eta }^{\alpha -1}{\left(1-q{q}^{n}\right)}^{\left(\alpha -1\right)}{\eta }^{\beta }{q}^{n\beta }\\ =& \left(1-q\right){\eta }^{\alpha +\beta }\sum _{n=0}^{\mathrm{\infty }}{q}^{n}{\left(1-q{q}^{n}\right)}^{\left(\alpha -1\right)}{q}^{n\beta }\\ =& {\eta }^{\alpha +\beta }{\int }_{0}^{1}{\left(1-qs\right)}^{\left(\alpha -1\right)}{s}^{\left(\beta \right)}\phantom{\rule{0.2em}{0ex}}{d}_{q}s\\ =& {\eta }^{\alpha +\beta }{B}_{q}\left(\alpha ,\beta +1\right).\end{array}$

The proof is complete. □

Lemma 3.2 Let $\alpha ,\beta >0$ and $0. Then we have

${\int }_{0}^{\eta }{\int }_{0}^{x}{\left(\eta -px\right)}^{\left(\alpha -1\right)}{\left(x-qy\right)}^{\left(\beta -1\right)}\phantom{\rule{0.2em}{0ex}}{d}_{q}y\phantom{\rule{0.2em}{0ex}}{d}_{p}x=\frac{{\eta }^{\alpha +\beta }}{{\left[\beta \right]}_{q}}\frac{{\mathrm{\Gamma }}_{p}\left(\alpha \right){\mathrm{\Gamma }}_{p}\left(\beta +1\right)}{{\mathrm{\Gamma }}_{p}\left(\alpha +\beta +1\right)}.$
(3.2)

Proof Taking into account Lemma 3.1, we have

$\begin{array}{c}{\int }_{0}^{\eta }{\int }_{0}^{x}{\left(\eta -px\right)}^{\left(\alpha -1\right)}{\left(x-qy\right)}^{\left(\beta -1\right)}\phantom{\rule{0.2em}{0ex}}{d}_{q}y\phantom{\rule{0.2em}{0ex}}{d}_{p}x\hfill \\ \phantom{\rule{1em}{0ex}}={\int }_{0}^{\eta }{\left(\eta -px\right)}^{\left(\alpha -1\right)}{\int }_{0}^{x}{\left(x-qy\right)}^{\left(\beta -1\right)}\phantom{\rule{0.2em}{0ex}}{d}_{q}y\phantom{\rule{0.2em}{0ex}}{d}_{p}x\hfill \\ \phantom{\rule{1em}{0ex}}=\frac{1}{{\left[\beta \right]}_{q}}{\int }_{0}^{\eta }{\left(\eta -px\right)}^{\left(\alpha -1\right)}{x}^{\left(\beta \right)}\phantom{\rule{0.2em}{0ex}}{d}_{p}x\hfill \\ \phantom{\rule{1em}{0ex}}=\frac{1}{{\left[\beta \right]}_{q}}{\eta }^{\alpha +\beta }{B}_{p}\left(\alpha ,\beta +1\right)\hfill \\ \phantom{\rule{1em}{0ex}}=\frac{{\eta }^{\alpha +\beta }}{{\left[\beta \right]}_{q}}\frac{{\mathrm{\Gamma }}_{p}\left(\alpha \right){\mathrm{\Gamma }}_{p}\left(\beta +1\right)}{{\mathrm{\Gamma }}_{p}\left(\alpha +\beta +1\right)}.\hfill \end{array}$

This completes the proof. □

For convenience, we set a nonzero constant

$\mathrm{\Lambda }={T}^{\alpha -1}-\sum _{i=1}^{m}\frac{{\gamma }_{i}{\mathrm{\Gamma }}_{{q}_{i}}\left(\alpha \right)}{{\mathrm{\Gamma }}_{{q}_{i}}\left(\alpha +{\beta }_{i}\right)}\left({\xi }_{i}^{\alpha +{\beta }_{i}-1}-{\eta }_{i}^{\alpha +{\beta }_{i}-1}\right).$
(3.3)

Lemma 3.3 Let ${\beta }_{i}>0$, $0, ${\gamma }_{i}\in \mathbb{R}$, ${\eta }_{i},{\xi }_{i}\in \left(0,T\right)$ and ${\eta }_{i}<{\xi }_{i}$ for all $i=1,2,\dots ,m$. Then, for a given $y\in C\left(\left[0,1\right],\mathbb{R}\right)$, the unique solution of the linear q-difference equation

${D}_{q}^{\alpha }u\left(t\right)=y\left(t\right),\phantom{\rule{1em}{0ex}}t\in \left(0,T\right),1<\alpha \le 2,$
(3.4)

subject to the multi-strip fractional q-integral condition

$u\left(0\right)=0,\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}u\left(T\right)=\sum _{i=1}^{m}{\gamma }_{i}\left({I}_{{q}_{i}}^{{\beta }_{i}}u\right){|}_{{\eta }_{i}}^{{\xi }_{i}}=\sum _{i=1}^{m}{\gamma }_{i}\left({I}_{{q}_{i}}^{{\beta }_{i}}u\left({\xi }_{i}\right)-{I}_{{q}_{i}}^{{\beta }_{i}}u\left({\eta }_{i}\right)\right),$
(3.5)

is given by

$\begin{array}{rcl}u\left(t\right)& =& -\frac{{t}^{\alpha -1}}{\mathrm{\Lambda }}\left\{{\int }_{0}^{T}\frac{{\left(T-qs\right)}^{\left(\alpha -1\right)}}{{\mathrm{\Gamma }}_{q}\left(\alpha \right)}y\left(s\right)\phantom{\rule{0.2em}{0ex}}{d}_{q}s\\ -\sum _{i=1}^{m}\frac{{\gamma }_{i}}{{\mathrm{\Gamma }}_{{q}_{i}}\left({\beta }_{i}\right){\mathrm{\Gamma }}_{q}\left(\alpha \right)}\left({\int }_{0}^{{\xi }_{i}}{\int }_{0}^{s}{\left({\xi }_{i}-{q}_{i}s\right)}^{\left({\beta }_{i}-1\right)}{\left(s-qx\right)}^{\left(\alpha -1\right)}y\left(x\right)\phantom{\rule{0.2em}{0ex}}{d}_{q}x\phantom{\rule{0.2em}{0ex}}{d}_{{q}_{i}}s\\ +{\int }_{0}^{{\eta }_{i}}{\int }_{0}^{s}{\left({\eta }_{i}-{q}_{i}s\right)}^{\left({\beta }_{i}-1\right)}{\left(s-qx\right)}^{\left(\alpha -1\right)}y\left(x\right)\phantom{\rule{0.2em}{0ex}}{d}_{q}x\phantom{\rule{0.2em}{0ex}}{d}_{{q}_{i}}s\right)\right\}\\ +{\int }_{0}^{t}\frac{{\left(t-qs\right)}^{\left(\alpha -1\right)}}{{\mathrm{\Gamma }}_{q}\left(\alpha \right)}y\left(s\right)\phantom{\rule{0.2em}{0ex}}{d}_{q}s,\end{array}$
(3.6)

where Λ is defined by (3.3).

Proof Since $1<\alpha \le 2$, we take $n=2$. In view of Definition 2.2 and Lemma 2.4, the linear q-difference equation (3.4) can be written as

$\left({I}_{q}^{\alpha }{D}_{q}^{2}{I}_{q}^{2-\alpha }u\right)\left(t\right)=\left({I}_{q}^{\alpha }y\right)\left(t\right).$

Using Lemma 2.5, we obtain

$u\left(t\right)={c}_{1}{t}^{\alpha -1}+{c}_{2}{t}^{\alpha -2}+{\int }_{0}^{t}\frac{{\left(t-qs\right)}^{\left(\alpha -1\right)}}{{\mathrm{\Gamma }}_{q}\left(\alpha \right)}y\left(s\right)\phantom{\rule{0.2em}{0ex}}{d}_{q}s$
(3.7)

for some constants ${c}_{1},{c}_{2}\in \mathbb{R}$. Since $u\left(0\right)=0$, we get ${c}_{2}=0$.

Applying the Riemann-Liouville fractional ${q}_{i}$-integral of order ${\beta }_{i}>0$ with ${c}_{2}=0$ for (3.7) and taking into account Lemma 3.1, we have

$\begin{array}{rcl}{I}_{{q}_{i}}^{{\beta }_{i}}u\left({\xi }_{i}\right)& =& {\int }_{0}^{{\xi }_{i}}\frac{{\left({\xi }_{i}-{q}_{i}s\right)}^{\left({\beta }_{i}-1\right)}}{{\mathrm{\Gamma }}_{{q}_{i}}\left({\beta }_{i}\right)}\left({c}_{1}{s}^{\alpha -1}+{\int }_{0}^{s}\frac{{\left(s-qx\right)}^{\left(\alpha -1\right)}}{{\mathrm{\Gamma }}_{q}\left(\alpha \right)}y\left(x\right)\phantom{\rule{0.2em}{0ex}}{d}_{q}x\right)\phantom{\rule{0.2em}{0ex}}{d}_{{q}_{i}}s\\ =& \frac{1}{{\mathrm{\Gamma }}_{{q}_{i}}\left({\beta }_{i}\right){\mathrm{\Gamma }}_{q}\left(\alpha \right)}{\int }_{0}^{{\xi }_{i}}{\int }_{0}^{s}{\left({\xi }_{i}-{q}_{i}s\right)}^{\left({\beta }_{i}-1\right)}{\left(s-qx\right)}^{\left(\alpha -1\right)}y\left(x\right)\phantom{\rule{0.2em}{0ex}}{d}_{q}x\phantom{\rule{0.2em}{0ex}}{d}_{{q}_{i}}s\\ +\frac{{c}_{1}}{{\mathrm{\Gamma }}_{{q}_{i}}\left({\beta }_{i}\right)}{\int }_{0}^{{\xi }_{i}}{\left({\xi }_{i}-{q}_{i}s\right)}^{\left({\beta }_{i}-1\right)}{s}^{\alpha -1}\phantom{\rule{0.2em}{0ex}}{d}_{{q}_{i}}s\\ =& \frac{1}{{\mathrm{\Gamma }}_{{q}_{i}}\left({\beta }_{i}\right){\mathrm{\Gamma }}_{q}\left(\alpha \right)}{\int }_{0}^{{\xi }_{i}}{\int }_{0}^{s}{\left({\xi }_{i}-{q}_{i}s\right)}^{\left({\beta }_{i}-1\right)}{\left(s-qx\right)}^{\left(\alpha -1\right)}y\left(x\right)\phantom{\rule{0.2em}{0ex}}{d}_{q}x\phantom{\rule{0.2em}{0ex}}{d}_{{q}_{i}}s\\ +{c}_{1}\frac{{\mathrm{\Gamma }}_{{q}_{i}}\left(\alpha \right){\xi }_{i}^{\alpha +{\beta }_{i}-1}}{{\mathrm{\Gamma }}_{{q}_{i}}\left(\alpha +{\beta }_{i}\right)}.\end{array}$
(3.8)

Repeating the above process with $t={\eta }_{i}$ and using the second condition of (3.5), we get a constant ${c}_{1}$ as follows:

$\begin{array}{rcl}{c}_{1}& =& \frac{1}{\mathrm{\Lambda }}\left\{\sum _{i=1}^{m}\frac{{\gamma }_{i}}{{\mathrm{\Gamma }}_{{q}_{i}}\left({\beta }_{i}\right){\mathrm{\Gamma }}_{q}\left(\alpha \right)}\left({\int }_{0}^{{\xi }_{i}}{\int }_{0}^{s}{\left({\xi }_{i}-{q}_{i}s\right)}^{\left({\beta }_{i}-1\right)}{\left(s-qx\right)}^{\left(\alpha -1\right)}y\left(x\right)\phantom{\rule{0.2em}{0ex}}{d}_{q}x\phantom{\rule{0.2em}{0ex}}{d}_{{q}_{i}}s\\ -{\int }_{0}^{{\eta }_{i}}{\int }_{0}^{s}{\left({\eta }_{i}-{q}_{i}s\right)}^{\left({\beta }_{i}-1\right)}{\left(s-qx\right)}^{\left(\alpha -1\right)}y\left(x\right)\phantom{\rule{0.2em}{0ex}}{d}_{q}x\phantom{\rule{0.2em}{0ex}}{d}_{{q}_{i}}s\right)\\ -{\int }_{0}^{T}\frac{{\left(T-qs\right)}^{\left(\alpha -1\right)}}{{\mathrm{\Gamma }}_{q}\left(\alpha \right)}y\left(s\right)\phantom{\rule{0.2em}{0ex}}{d}_{q}s\right\}.\end{array}$
(3.9)

Substituting the values of constants ${c}_{1}$ and ${c}_{2}$ in the linear solution (3.7), the desired result in (3.6) is obtained. □

## 4 Main results

Let $\mathcal{C}=C\left(\left[0,T\right],\mathbb{R}\right)$ denote the Banach space of all continuous functions from $\left[0,T\right]$ to endowed with the supremum norm defined by $\parallel u\parallel ={sup}_{t\in \left[0,T\right]}|u\left(t\right)|$. In view of Lemma 3.3, we define an operator $\mathcal{A}:\mathcal{C}\to \mathcal{C}$ by

$\begin{array}{rcl}\left(\mathcal{A}u\right)\left(t\right)& =& -\frac{{t}^{\alpha -1}}{\mathrm{\Lambda }}\left\{{\int }_{0}^{T}\frac{{\left(T-qs\right)}^{\left(\alpha -1\right)}}{{\mathrm{\Gamma }}_{q}\left(\alpha \right)}f\left(s,u\left(s\right)\right)\phantom{\rule{0.2em}{0ex}}{d}_{q}s\\ -\sum _{i=1}^{m}\frac{{\gamma }_{i}}{{\mathrm{\Gamma }}_{{q}_{i}}\left({\beta }_{i}\right){\mathrm{\Gamma }}_{q}\left(\alpha \right)}\left({\int }_{0}^{{\xi }_{i}}{\int }_{0}^{s}{\left({\xi }_{i}-{q}_{i}s\right)}^{\left({\beta }_{i}-1\right)}{\left(s-qx\right)}^{\left(\alpha -1\right)}f\left(x,u\left(x\right)\right)\phantom{\rule{0.2em}{0ex}}{d}_{q}x\phantom{\rule{0.2em}{0ex}}{d}_{{q}_{i}}s\\ +{\int }_{0}^{{\eta }_{i}}{\int }_{0}^{s}{\left({\eta }_{i}-{q}_{i}s\right)}^{\left({\beta }_{i}-1\right)}{\left(s-qx\right)}^{\left(\alpha -1\right)}f\left(x,u\left(x\right)\right)\phantom{\rule{0.2em}{0ex}}{d}_{q}x\phantom{\rule{0.2em}{0ex}}{d}_{{q}_{i}}s\right)\right\}\\ +{\int }_{0}^{t}\frac{{\left(t-qs\right)}^{\left(\alpha -1\right)}}{{\mathrm{\Gamma }}_{q}\left(\alpha \right)}f\left(s,u\left(s\right)\right)\phantom{\rule{0.2em}{0ex}}{d}_{q}s,\end{array}$
(4.1)

with $\mathrm{\Lambda }\ne 0$. It should be noticed that problem (1.1) has solutions if and only if the operator $\mathcal{A}$ has fixed points.

For the sake of convenience, we put

$\begin{array}{rcl}\mathrm{\Phi }& =& \frac{{T}^{\alpha -1}}{|\mathrm{\Lambda }|{\mathrm{\Gamma }}_{q}\left(\alpha +1\right)}\left({T}^{\alpha }+\sum _{i=1}^{m}\frac{|{\gamma }_{i}|{\xi }_{i}^{{\beta }_{i}+\alpha }{\mathrm{\Gamma }}_{{q}_{i}}\left(\alpha +1\right)}{{\mathrm{\Gamma }}_{{q}_{i}}\left(\alpha +{\beta }_{i}+1\right)}+\sum _{i=1}^{m}\frac{|{\gamma }_{i}|{\eta }_{i}^{{\beta }_{i}+\alpha }{\mathrm{\Gamma }}_{{q}_{i}}\left(\alpha +1\right)}{{\mathrm{\Gamma }}_{{q}_{i}}\left(\alpha +{\beta }_{i}+1\right)}\right)\\ +\frac{{T}^{\alpha }}{{\mathrm{\Gamma }}_{q}\left(\alpha +1\right)}.\end{array}$
(4.2)

The first existence and uniqueness result is based on the Banach contraction mapping principle.

Theorem 4.1 Let $f:\left[0,T\right]×\mathbb{R}\to \mathbb{R}$ be a continuous function satisfying the assumption

(H1) there exists a constant $L>0$ such that $|f\left(t,u\right)-f\left(t,v\right)|\le L|u-v|$ for each $t\in \left[0,T\right]$ and $u,v\in \mathbb{R}$.

If

$L\mathrm{\Phi }<1,$
(4.3)

where a constant Φ is given by (4.2), then the multi-strip boundary value problem (1.1) has a unique solution on $\left[0,T\right]$.

Proof We transform problem (1.1) into a fixed point problem, $u=\mathcal{A}u$, where the operator $\mathcal{A}$ is defined by (4.1). Applying the Banach contraction mapping principle, we will show that the operator $\mathcal{A}$ has a fixed point which is a unique solution of problem (1.1).

Setting ${sup}_{t\in \left[0,T\right]}|f\left(t,0\right)|={M}_{0}<\mathrm{\infty }$ and choosing

$r\ge \frac{{M}_{0}\mathrm{\Phi }}{1-L\mathrm{\Phi }},$

with L Φ satisfying (4.3), we will show that $\mathcal{A}{B}_{r}\subset {B}_{r}$, where the set ${B}_{r}=\left\{u\in \mathcal{C}:\parallel u\parallel \le r\right\}$. For any $u\in {B}_{r}$, and taking into account Lemma 3.2, we have

$\begin{array}{rcl}\parallel \mathcal{A}u\parallel & \le & \underset{t\in \left[0,T\right]}{sup}\left\{\frac{{t}^{\alpha -1}}{|\mathrm{\Lambda }|}\left\{{\int }_{0}^{T}\frac{{\left(T-qs\right)}^{\left(\alpha -1\right)}}{{\mathrm{\Gamma }}_{q}\left(\alpha \right)}|f\left(s,u\left(s\right)\right)|\phantom{\rule{0.2em}{0ex}}{d}_{q}s\\ +\sum _{i=1}^{m}\frac{|{\gamma }_{i}|}{{\mathrm{\Gamma }}_{{q}_{i}}\left({\beta }_{i}\right){\mathrm{\Gamma }}_{q}\left(\alpha \right)}\left({\int }_{0}^{{\xi }_{i}}{\int }_{0}^{s}{\left({\xi }_{i}-{q}_{i}s\right)}^{\left({\beta }_{i}-1\right)}{\left(s-qx\right)}^{\left(\alpha -1\right)}|f\left(x,u\left(x\right)\right)|\phantom{\rule{0.2em}{0ex}}{d}_{q}x\phantom{\rule{0.2em}{0ex}}{d}_{{q}_{i}}s\\ +{\int }_{0}^{{\eta }_{i}}{\int }_{0}^{s}{\left({\eta }_{i}-{q}_{i}s\right)}^{\left({\beta }_{i}-1\right)}{\left(s-qx\right)}^{\left(\alpha -1\right)}|f\left(x,u\left(x\right)\right)|\phantom{\rule{0.2em}{0ex}}{d}_{q}x\phantom{\rule{0.2em}{0ex}}{d}_{{q}_{i}}s\right)\right\}\\ +{\int }_{0}^{t}\frac{{\left(t-qs\right)}^{\left(\alpha -1\right)}}{{\mathrm{\Gamma }}_{q}\left(\alpha \right)}|f\left(s,u\left(s\right)\right)|\phantom{\rule{0.2em}{0ex}}{d}_{q}s\right\}\\ \le & \frac{{T}^{\alpha -1}}{|\mathrm{\Lambda }|}\left\{{\int }_{0}^{T}\frac{{\left(T-qs\right)}^{\left(\alpha -1\right)}}{{\mathrm{\Gamma }}_{q}\left(\alpha \right)}\left(|f\left(s,u\left(s\right)\right)-f\left(s,0\right)|+|f\left(s,0\right)|\right)\phantom{\rule{0.2em}{0ex}}{d}_{q}s\\ +\sum _{i=1}^{m}\frac{|{\gamma }_{i}|}{{\mathrm{\Gamma }}_{{q}_{i}}\left({\beta }_{i}\right){\mathrm{\Gamma }}_{q}\left(\alpha \right)}\left({\int }_{0}^{{\xi }_{i}}{\int }_{0}^{s}{\left({\xi }_{i}-{q}_{i}s\right)}^{\left({\beta }_{i}-1\right)}{\left(s-qx\right)}^{\left(\alpha -1\right)}\\ ×\left(|f\left(x,u\left(x\right)\right)-f\left(x,0\right)|+|f\left(x,0\right)|\right)\phantom{\rule{0.2em}{0ex}}{d}_{q}x\phantom{\rule{0.2em}{0ex}}{d}_{{q}_{i}}s\\ +{\int }_{0}^{{\eta }_{i}}{\int }_{0}^{s}{\left({\eta }_{i}-{q}_{i}s\right)}^{\left({\beta }_{i}-1\right)}{\left(s-qx\right)}^{\left(\alpha -1\right)}\\ ×\left(|f\left(x,u\left(x\right)\right)-f\left(x,0\right)|+|f\left(x,0\right)|\right)\phantom{\rule{0.2em}{0ex}}{d}_{q}x\phantom{\rule{0.2em}{0ex}}{d}_{{q}_{i}}s\right)\right\}\\ +{\int }_{0}^{T}\frac{{\left(T-qs\right)}^{\left(\alpha -1\right)}}{{\mathrm{\Gamma }}_{q}\left(\alpha \right)}\left(|f\left(s,u\left(s\right)\right)-f\left(s,0\right)|+|f\left(s,0\right)|\right)\phantom{\rule{0.2em}{0ex}}{d}_{q}s\\ \le & \left(Lr+{M}_{0}\right)\left\{\frac{{T}^{\alpha -1}}{|\mathrm{\Lambda }|{\mathrm{\Gamma }}_{q}\left(\alpha +1\right)}\left({T}^{\alpha }+\sum _{i=1}^{m}\frac{|{\gamma }_{i}|{\xi }_{i}^{{\beta }_{i}+\alpha }{\mathrm{\Gamma }}_{{q}_{i}}\left(\alpha +1\right)}{{\mathrm{\Gamma }}_{{q}_{i}}\left(\alpha +{\beta }_{i}+1\right)}\\ +\sum _{i=1}^{m}\frac{|{\gamma }_{i}|{\eta }_{i}^{{\beta }_{i}+\alpha }{\mathrm{\Gamma }}_{{q}_{i}}\left(\alpha +1\right)}{{\mathrm{\Gamma }}_{{q}_{i}}\left(\alpha +{\beta }_{i}+1\right)}\right)+\frac{{T}^{\alpha }}{{\mathrm{\Gamma }}_{q}\left(\alpha +1\right)}\right\}\\ =& \left(Lr+{M}_{0}\right)\mathrm{\Phi }\le r.\end{array}$

It follows that $\mathcal{A}{B}_{r}\subset {B}_{r}$.

For $u,v\in \mathcal{C}$ and for each $t\in \left[0,T\right]$, we have

$\begin{array}{c}|\mathcal{A}u\left(t\right)-\mathcal{A}v\left(t\right)|\hfill \\ \phantom{\rule{1em}{0ex}}\le \frac{{T}^{\alpha -1}}{|\mathrm{\Lambda }|}\left\{{\int }_{0}^{T}\frac{{\left(T-qs\right)}^{\left(\alpha -1\right)}}{{\mathrm{\Gamma }}_{q}\left(\alpha \right)}\left(|f\left(s,u\left(s\right)\right)-f\left(s,v\left(s\right)\right)|\right)\phantom{\rule{0.2em}{0ex}}{d}_{q}s\hfill \\ \phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}+\sum _{i=1}^{m}\frac{|{\gamma }_{i}|}{{\mathrm{\Gamma }}_{{q}_{i}}\left({\beta }_{i}\right){\mathrm{\Gamma }}_{q}\left(\alpha \right)}\left({\int }_{0}^{{\xi }_{i}}{\int }_{0}^{s}{\left({\xi }_{i}-{q}_{i}s\right)}^{\left({\beta }_{i}-1\right)}{\left(s-qx\right)}^{\left(\alpha -1\right)}\hfill \\ \phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}×\left(|f\left(s,u\left(s\right)\right)-f\left(s,v\left(s\right)\right)|\right)\phantom{\rule{0.2em}{0ex}}{d}_{q}x\phantom{\rule{0.2em}{0ex}}{d}_{{q}_{i}}s\hfill \\ \phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}+{\int }_{0}^{{\eta }_{i}}{\int }_{0}^{s}{\left({\eta }_{i}-{q}_{i}s\right)}^{\left({\beta }_{i}-1\right)}{\left(s-qx\right)}^{\left(\alpha -1\right)}\left(|f\left(s,u\left(s\right)\right)-f\left(s,v\left(s\right)\right)|\right)\phantom{\rule{0.2em}{0ex}}{d}_{q}x\phantom{\rule{0.2em}{0ex}}{d}_{{q}_{i}}s\right)\right\}\hfill \\ \phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}+{\int }_{0}^{T}\frac{{\left(T-qs\right)}^{\left(\alpha -1\right)}}{{\mathrm{\Gamma }}_{q}\left(\alpha \right)}\left(|f\left(s,u\left(s\right)\right)-f\left(s,v\left(s\right)\right)|\right)\phantom{\rule{0.2em}{0ex}}{d}_{q}s\hfill \\ \phantom{\rule{1em}{0ex}}\le L\parallel u-v\parallel \left\{\frac{{T}^{\alpha -1}}{|\mathrm{\Lambda }|{\mathrm{\Gamma }}_{q}\left(\alpha +1\right)}\left({T}^{\alpha }+\sum _{i=1}^{m}\frac{|{\gamma }_{i}|{\xi }_{i}^{{\beta }_{i}+\alpha }{\mathrm{\Gamma }}_{{q}_{i}}\left(\alpha +1\right)}{{\mathrm{\Gamma }}_{{q}_{i}}\left(\alpha +{\beta }_{i}+1\right)}\hfill \\ \phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}+\sum _{i=1}^{m}\frac{|{\gamma }_{i}|{\eta }_{i}^{{\beta }_{i}+\alpha }{\mathrm{\Gamma }}_{{q}_{i}}\left(\alpha +1\right)}{{\mathrm{\Gamma }}_{{q}_{i}}\left(\alpha +{\beta }_{i}+1\right)}\right)+\frac{{T}^{\alpha }}{{\mathrm{\Gamma }}_{q}\left(\alpha +1\right)}\right\}\hfill \\ \phantom{\rule{1em}{0ex}}=L\mathrm{\Phi }\parallel u-v\parallel .\hfill \end{array}$

The above result leads to $\parallel \mathcal{A}u-\mathcal{A}v\parallel \le L\mathrm{\Phi }\parallel u-v\parallel$. As $L\mathrm{\Phi }<1$, by (4.3), therefore $\mathcal{A}$ is a contraction. Hence, by the Banach contraction mapping principle, we deduce that $\mathcal{A}$ has a fixed point which is the unique solution of problem (1.1). □

Next, we prove the existence of at least one solution by using Krasnoselskii’s fixed point theorem.

Lemma 4.2 (Krasnoselskii’s fixed point theorem [23])

Let M be a closed, bounded, convex and nonempty subset of a Banach space X. Let A, B be the operators such that (a) $Ax+By\in M$ whenever $x,y\in M$; (b) A is compact and continuous; (c) B is a contraction mapping. Then there exists $z\in M$ such that $z=Az+Bz$.

Theorem 4.3 Assume that $f:\left[0,T\right]×\mathbb{R}\to \mathbb{R}$ is a continuous function satisfying assumption (H1). In addition, we suppose that

(H2) $|f\left(t,u\right)|\le \psi \left(t\right)$, $\mathrm{\forall }\left(t,u\right)\in \left[0,T\right]×\mathbb{R}$ and $\psi \in C\left(\left[0,T\right],{\mathbb{R}}^{+}\right)$.

If the following condition holds

$\frac{L}{{\mathrm{\Gamma }}_{q}\left(\alpha +1\right)}\left(\frac{{T}^{2\alpha -1}}{|\mathrm{\Lambda }|}+{T}^{\alpha }\right)<1,$
(4.4)

then the multi-strip boundary value problem (1.1) has at least one solution on $\left[0,T\right]$.

Proof We define ${sup}_{t\in \left[0,T\right]}|\psi \left(t\right)|=\parallel \psi \parallel$ and choose a suitable constant R such that

$R\ge \parallel \psi \parallel \mathrm{\Phi },$

where Φ is defined by (4.2). Furthermore, we define the operators ${\mathcal{A}}_{1}$ and ${\mathcal{A}}_{2}$ on ${B}_{R}=\left\{u\in \mathcal{C}:\parallel u\parallel \le R\right\}$ by

$\begin{array}{c}\left({\mathcal{A}}_{1}u\right)\left(t\right)\hfill \\ \phantom{\rule{1em}{0ex}}=\frac{{t}^{\alpha -1}}{\mathrm{\Lambda }}\sum _{i=1}^{m}\frac{{\gamma }_{i}}{{\mathrm{\Gamma }}_{{q}_{i}}\left({\beta }_{i}\right){\mathrm{\Gamma }}_{q}\left(\alpha \right)}{\int }_{0}^{{\xi }_{i}}{\int }_{0}^{s}{\left({\xi }_{i}-{q}_{i}s\right)}^{\left({\beta }_{i}-1\right)}{\left(s-qx\right)}^{\left(\alpha -1\right)}f\left(x,u\left(x\right)\right)\phantom{\rule{0.2em}{0ex}}{d}_{q}x\phantom{\rule{0.2em}{0ex}}{d}_{{q}_{i}}s\hfill \\ \phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}-\frac{{t}^{\alpha -1}}{\mathrm{\Lambda }}\sum _{i=1}^{m}\frac{{\gamma }_{i}}{{\mathrm{\Gamma }}_{{q}_{i}}\left({\beta }_{i}\right){\mathrm{\Gamma }}_{q}\left(\alpha \right)}{\int }_{0}^{{\eta }_{i}}{\int }_{0}^{s}{\left({\eta }_{i}-{q}_{i}s\right)}^{\left({\beta }_{i}-1\right)}{\left(s-qx\right)}^{\left(\alpha -1\right)}f\left(x,u\left(x\right)\right)\phantom{\rule{0.2em}{0ex}}{d}_{q}x\phantom{\rule{0.2em}{0ex}}{d}_{{q}_{i}}s,\hfill \end{array}$

and

$\left({\mathcal{A}}_{2}u\right)\left(t\right)=-\frac{{t}^{\alpha -1}}{\mathrm{\Lambda }}{\int }_{0}^{T}\frac{{\left(T-qs\right)}^{\left(\alpha -1\right)}}{{\mathrm{\Gamma }}_{q}\left(\alpha \right)}f\left(s,u\left(s\right)\right)\phantom{\rule{0.2em}{0ex}}{d}_{q}s+{\int }_{0}^{t}\frac{{\left(t-qs\right)}^{\left(\alpha -1\right)}}{{\mathrm{\Gamma }}_{q}\left(\alpha \right)}f\left(s,u\left(s\right)\right)\phantom{\rule{0.2em}{0ex}}{d}_{q}s.$

It should be noticed that $\mathcal{A}={\mathcal{A}}_{1}+{\mathcal{A}}_{2}$.

For any $u,v\in {B}_{R}$, we have

$\begin{array}{rcl}\parallel {\mathcal{A}}_{1}u+{\mathcal{A}}_{2}v\parallel & \le & \parallel \psi \parallel \left\{\frac{{T}^{\alpha -1}}{|\mathrm{\Lambda }|{\mathrm{\Gamma }}_{q}\left(\alpha +1\right)}\left({T}^{\alpha }+\sum _{i=1}^{m}\frac{|{\gamma }_{i}|{\xi }_{i}^{{\beta }_{i}+\alpha }{B}_{{q}_{i}}\left({\beta }_{i},\alpha +1\right)}{{\mathrm{\Gamma }}_{{q}_{i}}\left({\beta }_{i}\right)}\\ +\sum _{i=1}^{m}\frac{|{\gamma }_{i}|{\eta }_{i}^{{\beta }_{i}+\alpha }{B}_{{q}_{i}}\left({\beta }_{i},\alpha +1\right)}{{\mathrm{\Gamma }}_{{q}_{i}}\left({\beta }_{i}\right)}\right)+\frac{{T}^{\alpha }}{{\mathrm{\Gamma }}_{q}\left(\alpha +1\right)}\right\}\\ =& \parallel \psi \parallel \mathrm{\Phi }\\ \le & R.\end{array}$

Therefore $\left({\mathcal{A}}_{1}u\right)+\left({\mathcal{A}}_{2}v\right)\in {B}_{R}$. Obviously, condition (4.4) implies that ${\mathcal{A}}_{2}$ is a contraction mapping.

Finally, we will show that ${\mathcal{A}}_{1}$ is compact and continuous. The continuity of f coupled with assumption (H2) implies that the operator ${\mathcal{A}}_{1}$ is continuous and uniformly bounded on ${B}_{R}$. We define ${sup}_{\left(t,u\right)\in \left[0,T\right]×{B}_{R}}|f\left(t,u\right)|={M}^{\ast }<\mathrm{\infty }$. For ${t}_{1},{t}_{2}\in \left[0,T\right]$, ${t}_{2}<{t}_{1}$ and $u\in {B}_{R}$, we have

$\begin{array}{c}|\left({\mathcal{A}}_{1}u\right)\left({t}_{1}\right)-\left({\mathcal{A}}_{1}u\right)\left({t}_{2}\right)|\hfill \\ \phantom{\rule{1em}{0ex}}\le \frac{|{t}_{1}^{\alpha -1}-{t}_{2}^{\alpha -1}|}{|\mathrm{\Lambda }|}\sum _{i=1}^{m}\frac{|{\gamma }_{i}|}{{\mathrm{\Gamma }}_{{q}_{i}}\left({\beta }_{i}\right){\mathrm{\Gamma }}_{q}\left(\alpha \right)}\hfill \\ \phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}×{\int }_{0}^{{\xi }_{i}}{\int }_{0}^{s}{\left({\xi }_{i}-{q}_{i}s\right)}^{\left({\beta }_{i}-1\right)}{\left(s-qx\right)}^{\left(\alpha -1\right)}|f\left(x,u\left(x\right)\right)|\phantom{\rule{0.2em}{0ex}}{d}_{q}x\phantom{\rule{0.2em}{0ex}}{d}_{{q}_{i}}s\hfill \\ \phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}+\frac{|{t}_{1}^{\alpha -1}-{t}_{2}^{\alpha -1}|}{|\mathrm{\Lambda }|}\sum _{i=1}^{m}\frac{|{\gamma }_{i}|}{{\mathrm{\Gamma }}_{{q}_{i}}\left({\beta }_{i}\right){\mathrm{\Gamma }}_{q}\left(\alpha \right)}\hfill \\ \phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}×{\int }_{0}^{{\eta }_{i}}{\int }_{0}^{s}{\left({\eta }_{i}-{q}_{i}s\right)}^{\left({\beta }_{i}-1\right)}{\left(s-qx\right)}^{\left(\alpha -1\right)}|f\left(x,u\left(x\right)\right)|\phantom{\rule{0.2em}{0ex}}{d}_{q}x\phantom{\rule{0.2em}{0ex}}{d}_{{q}_{i}}s\hfill \\ \phantom{\rule{1em}{0ex}}\le {M}^{\ast }\frac{|{t}_{1}^{\alpha -1}-{t}_{2}^{\alpha -1}|}{|\mathrm{\Lambda }|{\mathrm{\Gamma }}_{q}\left(\alpha +1\right)}\left\{\sum _{i=1}^{m}\frac{|{\gamma }_{i}|{\xi }_{i}^{{\beta }_{i}+\alpha }{\mathrm{\Gamma }}_{{q}_{i}}\left(\alpha +1\right)}{{\mathrm{\Gamma }}_{{q}_{i}}\left(\alpha +{\beta }_{i}+1\right)}+\sum _{i=1}^{m}\frac{|{\gamma }_{i}|{\eta }_{i}^{{\beta }_{i}+\alpha }{\mathrm{\Gamma }}_{{q}_{i}}\left(\alpha +1\right)}{{\mathrm{\Gamma }}_{{q}_{i}}\left(\alpha +{\beta }_{i}+1\right)}\right\}.\hfill \end{array}$

Actually, as $|{t}_{1}-{t}_{2}|\to 0$ the right-hand side of the above inequality tends to zero independently of u. So ${\mathcal{A}}_{1}$ is relatively compact on ${B}_{R}$. Therefore, by the Arzelá-Ascoli theorem, ${\mathcal{A}}_{1}$ is compact on ${B}_{R}$. Thus all the assumptions of Lemma 4.2 are satisfied. Thus, the boundary value problem (1.1) has at least one solution on $\left[0,T\right]$. The proof is complete. □

Remark 4.4 In the above theorem we can interchange the roles of the operators ${\mathcal{A}}_{1}$ and ${\mathcal{A}}_{2}$ to obtain the second result replacing (4.4) by the following condition:

$\frac{L{T}^{\alpha -1}}{|\mathrm{\Lambda }|{\mathrm{\Gamma }}_{q}\left(\alpha +1\right)}\left(\sum _{i=1}^{m}\frac{|{\gamma }_{i}|{\xi }_{i}^{{\beta }_{i}+\alpha }{\mathrm{\Gamma }}_{{q}_{i}}\left(\alpha +1\right)}{{\mathrm{\Gamma }}_{{q}_{i}}\left(\alpha +{\beta }_{i}+1\right)}+\sum _{i=1}^{m}\frac{|{\gamma }_{i}|{\eta }_{i}^{{\beta }_{i}+\alpha }{\mathrm{\Gamma }}_{{q}_{i}}\left(\alpha +1\right)}{{\mathrm{\Gamma }}_{{q}_{i}}\left(\alpha +{\beta }_{i}+1\right)}\right)<1.$

Now, our third existence result is based on Leray-Schauder’s nonlinear alternative.

Lemma 4.5 (Nonlinear alternative for single-valued maps [24])

Let E be a Banach space, C be a closed, convex subset of E, U be an open subset of C and $0\in U$. Suppose that $F:\overline{U}\to C$ is a continuous, compact (that is, $F\left(\overline{U}\right)$ is a relatively compact subset of C) map. Then either

1. (i)

F has a fixed point in $\overline{U}$, or

2. (ii)

there is $u\in \partial U$ (the boundary of U in C) and $\lambda \in \left(0,1\right)$ with $u=\lambda F\left(u\right)$.

Theorem 4.6 Assume that $f:\left[0,T\right]×\mathbb{R}\to \mathbb{R}$ is a continuous function. In addition we suppose that:

(H3) there exist a continuous nondecreasing function $\varphi :\left[0,\mathrm{\infty }\right)\to \left(0,\mathrm{\infty }\right)$ and a function $p\in C\left(\left[0,T\right],{\mathbb{R}}^{+}\right)$ such that

(H4) there exists a constant $N>0$ such that

$\frac{N}{\parallel p\parallel \varphi \left(N\right)\mathrm{\Phi }}>1,$

where Φ is defined by (4.2).

Then the multi-strip boundary value problem (1.1) has at least one solution on $\left[0,T\right]$.

Proof Firstly, we will show that the operator $\mathcal{A}$ defined by (4.1) maps bounded sets (balls) into bounded sets in $\mathcal{C}$. For a positive number ρ, let ${B}_{\rho }=\left\{u\in \mathcal{C}:\parallel u\parallel \le \rho \right\}$ be a bounded ball in $\mathcal{C}$. Then, for $t\in \left[0,T\right]$, we have

$\begin{array}{rcl}|\mathcal{A}u\left(t\right)|& \le & \frac{{T}^{\alpha -1}}{|\mathrm{\Lambda }|}\left\{{\int }_{0}^{T}\frac{{\left(T-qs\right)}^{\left(\alpha -1\right)}}{{\mathrm{\Gamma }}_{q}\left(\alpha \right)}|f\left(s,u\left(s\right)\right)|\phantom{\rule{0.2em}{0ex}}{d}_{q}s\\ +\sum _{i=1}^{m}\frac{|{\gamma }_{i}|}{{\mathrm{\Gamma }}_{{q}_{i}}\left({\beta }_{i}\right){\mathrm{\Gamma }}_{q}\left(\alpha \right)}\left({\int }_{0}^{{\xi }_{i}}{\int }_{0}^{s}{\left({\xi }_{i}-{q}_{i}s\right)}^{\left({\beta }_{i}-1\right)}{\left(s-qx\right)}^{\left(\alpha -1\right)}|f\left(x,u\left(x\right)\right)|\phantom{\rule{0.2em}{0ex}}{d}_{q}x\phantom{\rule{0.2em}{0ex}}{d}_{{q}_{i}}s\\ +{\int }_{0}^{{\eta }_{i}}{\int }_{0}^{s}{\left({\eta }_{i}-{q}_{i}s\right)}^{\left({\beta }_{i}-1\right)}{\left(s-qx\right)}^{\left(\alpha -1\right)}|f\left(x,u\left(x\right)\right)|\phantom{\rule{0.2em}{0ex}}{d}_{q}x\phantom{\rule{0.2em}{0ex}}{d}_{{q}_{i}}s\right)\right\}\\ +{\int }_{0}^{T}\frac{{\left(T-qs\right)}^{\left(\alpha -1\right)}}{{\mathrm{\Gamma }}_{q}\left(\alpha \right)}|f\left(s,u\left(s\right)\right)|\phantom{\rule{0.2em}{0ex}}{d}_{q}s\\ \le & \frac{{T}^{\alpha -1}}{|\mathrm{\Lambda }|{\mathrm{\Gamma }}_{q}\left(\alpha +1\right)}\left(\parallel p\parallel \varphi \left(\parallel u\parallel \right){T}^{\alpha }+\parallel p\parallel \varphi \left(\parallel u\parallel \right)\sum _{i=1}^{m}\frac{|{\gamma }_{i}|{\xi }_{i}^{{\beta }_{i}+\alpha }{\mathrm{\Gamma }}_{{q}_{i}}\left(\alpha +1\right)}{{\mathrm{\Gamma }}_{{q}_{i}}\left(\alpha +{\beta }_{i}+1\right)}\\ +\parallel p\parallel \varphi \left(\parallel u\parallel \right)\sum _{i=1}^{m}\frac{|{\gamma }_{i}|{\eta }_{i}^{{\beta }_{i}+\alpha }{\mathrm{\Gamma }}_{{q}_{i}}\left(\alpha +1\right)}{{\mathrm{\Gamma }}_{{q}_{i}}\left(\alpha +{\beta }_{i}+1\right)}\right)+\parallel p\parallel \varphi \left(\parallel u\parallel \right)\frac{{T}^{\alpha }}{{\mathrm{\Gamma }}_{q}\left(\alpha +1\right)}\\ \le & \frac{\parallel p\parallel \varphi \left(\rho \right){T}^{\alpha -1}}{|\mathrm{\Lambda }|{\mathrm{\Gamma }}_{q}\left(\alpha +1\right)}\left({T}^{\alpha }+\sum _{i=1}^{m}\frac{|{\gamma }_{i}|{\xi }_{i}^{{\beta }_{i}+\alpha }{\mathrm{\Gamma }}_{{q}_{i}}\left(\alpha +1\right)}{{\mathrm{\Gamma }}_{{q}_{i}}\left(\alpha +{\beta }_{i}+1\right)}+\sum _{i=1}^{m}\frac{|{\gamma }_{i}|{\eta }_{i}^{{\beta }_{i}+\alpha }{\mathrm{\Gamma }}_{{q}_{i}}\left(\alpha +1\right)}{{\mathrm{\Gamma }}_{{q}_{i}}\left(\alpha +{\beta }_{i}+1\right)}\right)\\ +\parallel p\parallel \varphi \left(\rho \right)\frac{{T}^{\alpha }}{{\mathrm{\Gamma }}_{q}\left(\alpha +1\right)}\\ :=& K.\end{array}$

Therefore, we deduce that $\parallel \mathcal{A}u\parallel \le K$.

Secondly, we will show that $\mathcal{A}$ maps bounded sets into equicontinuous sets of $\mathcal{C}$. Let ${sup}_{\left(t,u\right)\in \left[0,T\right]×{B}_{\rho }}|f\left(t,u\right)|={K}^{\ast }<\mathrm{\infty }$, ${\tau }_{1},{\tau }_{2}\in \left[0,T\right]$ with ${\tau }_{2}<{\tau }_{1}$ and $u\in {B}_{\rho }$. Then we have

$\begin{array}{c}|\left(\mathcal{A}u\right)\left({\tau }_{1}\right)-\left(\mathcal{A}u\right)\left({\tau }_{2}\right)|\hfill \\ \phantom{\rule{1em}{0ex}}\le \frac{|{\tau }_{1}^{\alpha -1}-{\tau }_{2}^{\alpha -1}|}{|\mathrm{\Lambda }|}\left\{{\int }_{0}^{T}\frac{{\left(T-qs\right)}^{\left(\alpha -1\right)}}{{\mathrm{\Gamma }}_{q}\left(\alpha \right)}|f\left(s,u\left(s\right)\right)|\phantom{\rule{0.2em}{0ex}}{d}_{q}s\hfill \\ \phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}+\sum _{i=1}^{m}\frac{|{\gamma }_{i}|}{{\mathrm{\Gamma }}_{{q}_{i}}\left({\beta }_{i}\right){\mathrm{\Gamma }}_{q}\left(\alpha \right)}\left({\int }_{0}^{{\xi }_{i}}{\int }_{0}^{s}{\left({\xi }_{i}-{q}_{i}s\right)}^{\left({\beta }_{i}-1\right)}{\left(s-qx\right)}^{\left(\alpha -1\right)}|f\left(x,u\left(x\right)\right)|\phantom{\rule{0.2em}{0ex}}{d}_{q}x\phantom{\rule{0.2em}{0ex}}{d}_{{q}_{i}}s\hfill \\ \phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}+{\int }_{0}^{{\eta }_{i}}{\int }_{0}^{s}{\left({\eta }_{i}-{q}_{i}s\right)}^{\left({\beta }_{i}-1\right)}{\left(s-qx\right)}^{\left(\alpha -1\right)}|f\left(x,u\left(x\right)\right)|\phantom{\rule{0.2em}{0ex}}{d}_{q}x\phantom{\rule{0.2em}{0ex}}{d}_{{q}_{i}}s\right)\right\}\hfill \\ \phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}+|{\int }_{0}^{{\tau }_{1}}\frac{{\left({\tau }_{1}-qs\right)}^{\left(\alpha -1\right)}}{{\mathrm{\Gamma }}_{q}\left(\alpha \right)}|f\left(s,u\left(s\right)\right)|\phantom{\rule{0.2em}{0ex}}{d}_{q}s-{\int }_{0}^{{\tau }_{2}}\frac{{\left({\tau }_{2}-qs\right)}^{\left(\alpha -1\right)}}{{\mathrm{\Gamma }}_{q}\left(\alpha \right)}|f\left(s,u\left(s\right)\right)|\phantom{\rule{0.2em}{0ex}}{d}_{q}s|\hfill \\ \phantom{\rule{1em}{0ex}}\le \frac{|{\tau }_{1}^{\alpha -1}-{\tau }_{2}^{\alpha -1}|{K}^{\ast }}{|\mathrm{\Lambda }|{\mathrm{\Gamma }}_{q}\left(\alpha +1\right)}\left({T}^{\alpha }+\sum _{i=1}^{m}\frac{|{\gamma }_{i}|{\xi }_{i}^{{\beta }_{i}+\alpha }{\mathrm{\Gamma }}_{{q}_{i}}\left(\alpha +1\right)}{{\mathrm{\Gamma }}_{{q}_{i}}\left(\alpha +{\beta }_{i}+1\right)}+\sum _{i=1}^{m}\frac{|{\gamma }_{i}|{\eta }_{i}^{{\beta }_{i}+\alpha }{\mathrm{\Gamma }}_{{q}_{i}}\left(\alpha +1\right)}{{\mathrm{\Gamma }}_{{q}_{i}}\left(\alpha +{\beta }_{i}+1\right)}\right)\hfill \\ \phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}+\frac{|{\tau }_{1}^{\alpha -1}-{\tau }_{2}^{\alpha -1}|{K}^{\ast }}{{\mathrm{\Gamma }}_{{q}_{i}}\left(\alpha +1\right)}.\hfill \end{array}$

Obviously, the right-hand side of the above inequality tends to zero independently of $x\in {B}_{\rho }$ as ${\tau }_{2}\to {\tau }_{1}$. Therefore it follows by the Arzelá-Ascoli theorem that $\mathcal{A}:\mathcal{C}\to \mathcal{C}$ is completely continuous.

Let u be a solution of problem (1.1). Then, for $t\in \left[0,T\right]$, and following similar computations as in the first step with (H3), we have

$\begin{array}{rcl}\parallel u\parallel & \le & \frac{{T}^{\alpha -1}}{|\mathrm{\Lambda }|{\mathrm{\Gamma }}_{q}\left(\alpha +1\right)}\left(\parallel p\parallel \varphi \left(\parallel u\parallel \right){T}^{\alpha }+\parallel p\parallel \varphi \left(\parallel u\parallel \right)\sum _{i=1}^{m}\frac{|{\gamma }_{i}|{\xi }_{i}^{{\beta }_{i}+\alpha }{\mathrm{\Gamma }}_{{q}_{i}}\left(\alpha +1\right)}{{\mathrm{\Gamma }}_{{q}_{i}}\left(\alpha +{\beta }_{i}+1\right)}\\ +\parallel p\parallel \varphi \left(\parallel u\parallel \right)\sum _{i=1}^{m}\frac{|{\gamma }_{i}|{\eta }_{i}^{{\beta }_{i}+\alpha }{\mathrm{\Gamma }}_{{q}_{i}}\left(\alpha +1\right)}{{\mathrm{\Gamma }}_{{q}_{i}}\left(\alpha +{\beta }_{i}+1\right)}\right)+\parallel p\parallel \varphi \left(\parallel u\parallel \right)\frac{{T}^{\alpha }}{{\mathrm{\Gamma }}_{q}\left(\alpha +1\right)}\\ =& \parallel p\parallel \varphi \left(\parallel u\parallel \right)\mathrm{\Phi }.\end{array}$

Consequently, we have

$\frac{\parallel u\parallel }{\parallel p\parallel \varphi \left(\parallel u\parallel \right)\mathrm{\Phi }}\le 1.$

In view of (H4), there exists a constant $N>0$ such that $\parallel u\parallel \ne N$. Let us set

$U=\left\{x\in \mathcal{C}:\parallel u\parallel

Note that the operator $\mathcal{A}:\overline{U}\to \mathcal{C}$ is continuous and completely continuous. From the choice of U, there is no $u\in \partial U$ such that $u=\lambda \mathcal{A}u$ for some $\lambda \in \left(0,1\right)$. Consequently, by nonlinear alternative of Leray-Schauder type (Lemma 4.5), we deduce that $\mathcal{A}$ has a fixed point in $\overline{U}$, which is a solution of the boundary value problem (1.1). This completes the proof. □

As the forth result, we prove the existence of solutions of (1.1) by using Leray-Schauder degree theory.

Theorem 4.7 Let $f:\left[0,T\right]×\mathbb{R}\to \mathbb{R}$ be a continuous function. Assume that

(H5) there exist constants $0\le \omega <{\mathrm{\Phi }}^{-1}$, where Φ are given by (4.2), and $\mathrm{\Psi }>0$ such that

Then the multi-strip boundary value problem (1.1) has at least one solution on $\left[0,T\right]$.

Proof Let $\mathcal{A}$ be the operator defined by (4.1). We will prove that there exists at least one solution $u\in \mathcal{C}$ of the operator equation $u=\mathcal{A}u$.

Setting a ball ${B}_{{\rho }^{\ast }}\subset \mathcal{C}$, where a constant radius ${\rho }^{\ast }>0$, by

${B}_{{\rho }^{\ast }}=\left\{u\in \mathcal{C}:\underset{t\in \left[0,T\right]}{sup}|u\left(t\right)|<{\rho }^{\ast }\right\},$

it is sufficient to show that $\mathcal{A}:{\overline{B}}_{{\rho }^{\ast }}\to \mathcal{C}$ satisfies

$u\ne \theta \mathcal{A}u,\phantom{\rule{1em}{0ex}}\mathrm{\forall }u\in \partial {B}_{{\rho }^{\ast }},\mathrm{\forall }\theta \in \left[0,1\right].$
(4.5)

Now, we set

$H\left(\theta ,u\right)=\theta \mathcal{A}u,\phantom{\rule{1em}{0ex}}u\in \mathcal{C},\theta \in \left[0,1\right].$

As shown in Theorem 4.6, we have that the operator $\mathcal{A}$ is continuous, uniformly bounded and equicontinuous. Then, by the Arzelá-Ascoli theorem, a continuous map ${h}_{\theta }\left(u\right)=u-H\left(\theta ,u\right)=u-\theta \mathcal{A}u$ is completely continuous. If (4.5) holds, then the following Leray-Schauder degrees are well defined. From the homotopy invariance of topological degree, it follows that

$\begin{array}{rcl}deg\left({h}_{\theta },{B}_{{\rho }^{\ast }},0\right)& =& deg\left(I-\theta \mathcal{A},{B}_{{\rho }^{\ast }},0\right)=deg\left({h}_{1},{B}_{{\rho }^{\ast }},0\right)\\ =& deg\left({h}_{0},{B}_{{\rho }^{\ast }},0\right)=deg\left(I,{B}_{{\rho }^{\ast }},0\right)=1\ne 0,\phantom{\rule{1em}{0ex}}0\in {B}_{{\rho }^{\ast }},\end{array}$

where I denotes the unit operator. By the nonzero property of Leray-Schauder degree, we have ${h}_{1}\left(u\right)=u-Au=0$ for at least one $u\in {B}_{{\rho }^{\ast }}$. Let us assume that $u=\theta \mathcal{A}u$ for some $\theta \in \left[0,1\right]$. Then, for all $t\in \left[0,T\right]$, we have

$\begin{array}{rcl}|u\left(t\right)|& =& |\theta \left(\mathcal{A}u\right)\left(t\right)|\\ \le & \frac{{T}^{\alpha -1}}{|\mathrm{\Lambda }|}\left\{{\int }_{0}^{T}\frac{{\left(T-qs\right)}^{\left(\alpha -1\right)}}{{\mathrm{\Gamma }}_{q}\left(\alpha \right)}|f\left(s,u\left(s\right)\right)|\phantom{\rule{0.2em}{0ex}}{d}_{q}s\\ +\sum _{i=1}^{m}\frac{|{\gamma }_{i}|}{{\mathrm{\Gamma }}_{{q}_{i}}\left({\beta }_{i}\right){\mathrm{\Gamma }}_{q}\left(\alpha \right)}\left({\int }_{0}^{{\xi }_{i}}{\int }_{0}^{s}{\left({\xi }_{i}-{q}_{i}s\right)}^{\left({\beta }_{i}-1\right)}{\left(s-qx\right)}^{\left(\alpha -1\right)}|f\left(x,u\left(x\right)\right)|\phantom{\rule{0.2em}{0ex}}{d}_{q}x\phantom{\rule{0.2em}{0ex}}{d}_{{q}_{i}}s\\ +{\int }_{0}^{{\eta }_{i}}{\int }_{0}^{s}{\left({\eta }_{i}-{q}_{i}s\right)}^{\left({\beta }_{i}-1\right)}{\left(s-qx\right)}^{\left(\alpha -1\right)}|f\left(x,u\left(x\right)\right)|\phantom{\rule{0.2em}{0ex}}{d}_{q}x\phantom{\rule{0.2em}{0ex}}{d}_{{q}_{i}}s\right)\right\}\\ +{\int }_{0}^{T}\frac{{\left(T-qs\right)}^{\left(\alpha -1\right)}}{{\mathrm{\Gamma }}_{q}\left(\alpha \right)}|f\left(s,u\left(s\right)\right)|\phantom{\rule{0.2em}{0ex}}{d}_{q}s\\ \le & \left(\omega |u|+\mathrm{\Psi }\right)\left\{\frac{{T}^{\alpha -1}}{|\mathrm{\Lambda }|{\mathrm{\Gamma }}_{q}\left(\alpha +1\right)}\left({T}^{\alpha }+\sum _{i=1}^{m}\frac{|{\gamma }_{i}|{\xi }_{i}^{{\beta }_{i}+\alpha }{B}_{{q}_{i}}\left({\beta }_{i},\alpha +1\right)}{{\mathrm{\Gamma }}_{{q}_{i}}\left({\beta }_{i}\right)}\\ +\sum _{i=1}^{m}\frac{|{\gamma }_{i}|{\eta }_{i}^{{\beta }_{i}+\alpha }{B}_{{q}_{i}}\left({\beta }_{i},\alpha +1\right)}{{\mathrm{\Gamma }}_{{q}_{i}}\left({\beta }_{i}\right)}\right)+\frac{{T}^{\alpha }}{{\mathrm{\Gamma }}_{q}\left(\alpha +1\right)}\right\}\\ =& \left(\omega |u|+\mathrm{\Psi }\right)\mathrm{\Phi }.\end{array}$

Taking norm ${sup}_{t\in \left[0,T\right]}|u\left(t\right)|=\parallel u\parallel$ and solving for $\parallel u\parallel$, we get

$\parallel u\parallel \le \frac{\mathrm{\Psi }\mathrm{\Phi }}{1-\omega \mathrm{\Phi }}.$

Choosing ${\rho }^{\ast }=\frac{\mathrm{\Psi }\mathrm{\Phi }}{1-\omega \mathrm{\Phi }}+1$, then we deduce that (4.5) holds. This completes the proof. □

## 5 Examples

In this section, we present some examples to illustrate our results.

Example 5.1 Consider the following multi-strip fractional q-integral boundary value problem:

$\left\{\begin{array}{c}{D}_{\frac{1}{2}}^{\frac{7}{4}}u\left(t\right)=\frac{4|u\left(t\right)|}{{\left(5+t\right)}^{2}\left(1+|u\left(t\right)|\right)},\phantom{\rule{1em}{0ex}}t\in \left(0,1\right),\hfill \\ u\left(0\right)=0,\hfill \\ u\left(1\right)=2\left({I}_{\frac{1}{4}}^{\frac{2}{3}}u\right){|}_{\frac{1}{6}}^{\frac{1}{5}}+\frac{3}{4}\left({I}_{\frac{1}{2}}^{\frac{4}{5}}u\right){|}_{\frac{2}{5}}^{\frac{2}{3}}+10\left({I}_{\frac{1}{5}}^{\frac{7}{6}}u\right){|}_{\frac{1}{2}}^{1}.\hfill \end{array}$
(5.1)

Here $\alpha =7/4$, $q=1/2$, $T=1$, $m=3$, ${\gamma }_{1}=2$, ${\gamma }_{2}=3/4$, ${\gamma }_{3}=10$, ${\beta }_{1}=2/3$, ${\beta }_{2}=4/5$, ${\beta }_{3}=7/6$, ${q}_{1}=1/4$, ${q}_{2}=1/2$, ${q}_{3}=1/5$, ${\xi }_{1}=1/5$, ${\xi }_{2}=2/3$, ${\xi }_{3}=1$, ${\eta }_{1}=1/6$, ${\eta }_{2}=2/5$, ${\eta }_{3}=1/2$ and $f\left(t,u\right)=\left(4|u\left(t\right)|\right)/\left({\left(5+t\right)}^{2}\left(1+|u\left(t\right)|\right)\right)$. Since

$|f\left(t,u\right)-f\left(t,v\right)|\le \frac{4}{25}|u-v|,$

then (H1) is satisfied with $L=4/25$. Using the Maple program, we find that

$\begin{array}{c}\mathrm{\Lambda }={T}^{\alpha -1}-\sum _{i=1}^{m}\frac{{\gamma }_{i}{\mathrm{\Gamma }}_{{q}_{i}}\left(\alpha \right)}{{\mathrm{\Gamma }}_{{q}_{i}}\left(\alpha +{\beta }_{i}\right)}\left({\xi }_{i}^{\alpha +{\beta }_{i}-1}-{\eta }_{i}^{\alpha +{\beta }_{i}-1}\right)\hfill \\ \phantom{\mathrm{\Lambda }}\approx -5.259895840,\hfill \\ \mathrm{\Phi }=\frac{{T}^{\alpha -1}}{|\mathrm{\Lambda }|{\mathrm{\Gamma }}_{q}\left(\alpha +1\right)}\left({T}^{\alpha }+\sum _{i=1}^{m}\frac{|{\gamma }_{i}|{\xi }_{i}^{{\beta }_{i}+\alpha }{\mathrm{\Gamma }}_{{q}_{i}}\left(\alpha +1\right)}{{\mathrm{\Gamma }}_{{q}_{i}}\left(\alpha +{\beta }_{i}+1\right)}+\sum _{i=1}^{m}\frac{|{\gamma }_{i}|{\eta }_{i}^{{\beta }_{i}+\alpha }{\mathrm{\Gamma }}_{{q}_{i}}\left(\alpha +1\right)}{{\mathrm{\Gamma }}_{{q}_{i}}\left(\alpha +{\beta }_{i}+1\right)}\right)\hfill \\ \phantom{\mathrm{\Phi }=}+\frac{{T}^{\alpha }}{{\mathrm{\Gamma }}_{q}\left(\alpha +1\right)}\hfill \\ \phantom{\mathrm{\Phi }}\approx 2.200354723.\hfill \end{array}$

Therefore, we get

$L\mathrm{\Phi }=\frac{4}{25}\left(2.200354723\right)\approx 0.352056756<1.$

Hence, by Theorem 4.1, the boundary value problem (5.1) has a unique solution on $\left[0,1\right]$.

Example 5.2 Consider the following multi-strip fractional q-integral boundary value problem:

$\left\{\begin{array}{c}{D}_{\frac{5}{8}}^{\frac{4}{3}}u\left(t\right)=\frac{1}{{u}^{2}+3{\pi }^{2}}sin\left(\frac{\pi u}{4}\right)+\frac{1}{20\pi }\left(5+sin\left(\pi t\right)\right),\phantom{\rule{1em}{0ex}}t\in \left(0,2\right),\hfill \\ u\left(0\right)=0,\hfill \\ u\left(2\right)=\left({I}_{\frac{1}{3}}^{\frac{1}{2}}u\right){|}_{\frac{1}{5}}^{\frac{1}{4}}+\frac{2}{3}\left({I}_{\frac{1}{2}}^{\frac{6}{5}}u\right){|}_{1}^{2}-3\left({I}_{\frac{1}{8}}^{\frac{3}{2}}u\right){|}_{\frac{1}{3}}^{1}+\frac{4}{5}\left({I}_{\frac{1}{4}}^{\frac{3}{4}}u\right){|}_{\frac{2}{5}}^{\frac{1}{2}}.\hfill \end{array}$
(5.2)

Here $\alpha =4/3$, $q=5/8$, $T=2$, $m=4$, ${\gamma }_{1}=1$, ${\gamma }_{2}=2/3$, ${\gamma }_{3}=-3$, ${\gamma }_{4}=4/5$, ${\beta }_{1}=1/2$, ${\beta }_{2}=6/5$, ${\beta }_{3}=3/2$, ${\beta }_{4}=3/4$, ${q}_{1}=1/3$, ${q}_{2}=1/2$, ${q}_{3}=1/8$, ${q}_{4}=1/4$, ${\xi }_{1}=1/4$, ${\xi }_{2}=2$, ${\xi }_{3}=1$, ${\xi }_{4}=1/2$, ${\eta }_{1}=1/5$, ${\eta }_{2}=1$, ${\eta }_{3}=1/3$, ${\eta }_{4}=2/5$ and $f\left(t,u\right)=\left(\left(sin\left(\pi u/4\right)\right)/\left({u}^{2}+3{\pi }^{2}\right)\right)+\left(\left(5+sin\left(\pi t\right)\right)/\left(20\pi \right)\right)$. By using the Maple program, we find that

$\begin{array}{c}\mathrm{\Lambda }={T}^{\alpha -1}-\sum _{i=1}^{m}\frac{{\gamma }_{i}{\mathrm{\Gamma }}_{{q}_{i}}\left(\alpha \right)}{{\mathrm{\Gamma }}_{{q}_{i}}\left(\alpha +{\beta }_{i}\right)}\left({\xi }_{i}^{\alpha +{\beta }_{i}-1}-{\eta }_{i}^{\alpha +{\beta }_{i}-1}\right)\hfill \\ \phantom{\mathrm{\Lambda }}\approx 2.448357686,\hfill \\ \mathrm{\Phi }=\frac{{T}^{\alpha -1}}{|\mathrm{\Lambda }|{\mathrm{\Gamma }}_{q}\left(\alpha +1\right)}\left({T}^{\alpha }+\sum _{i=1}^{m}\frac{|{\gamma }_{i}|{\xi }_{i}^{{\beta }_{i}+\alpha }{\mathrm{\Gamma }}_{{q}_{i}}\left(\alpha +1\right)}{{\mathrm{\Gamma }}_{{q}_{i}}\left(\alpha +{\beta }_{i}+1\right)}+\sum _{i=1}^{m}\frac{|{\gamma }_{i}|{\eta }_{i}^{{\beta }_{i}+\alpha }{\mathrm{\Gamma }}_{{q}_{i}}\left(\alpha +1\right)}{{\mathrm{\Gamma }}_{{q}_{i}}\left(\alpha +{\beta }_{i}+1\right)}\right)\hfill \\ \phantom{\mathrm{\Phi }=}+\frac{{T}^{\alpha }}{{\mathrm{\Gamma }}_{q}\left(\alpha +1\right)}\hfill \\ \phantom{\mathrm{\Phi }}\approx 5.843174987.\hfill \end{array}$

Clearly,

$|f\left(t,u\right)|=|\frac{1}{{u}^{2}+3{\pi }^{2}}sin\left(\frac{\pi u}{4}\right)+\frac{5+sin\left(\pi t\right)}{20\pi }|\le \left(5+sin\left(\pi t\right)\right)\left(\frac{5|u|+3}{60\pi }\right).$

Choosing $p\left(t\right)=5+sin\left(\pi t\right)$ and $\psi \left(|u|\right)=\left(5|u|+3\right)/\left(60\pi \right)$, we can show that

$\frac{N}{\left(6\right)\left(\frac{5N+3}{60\pi }\right)\left(3.493621065\right)}>1,$

which implies that $N>7.967778981$. Hence, by Theorem 4.6, the boundary value problem (5.2) has at least one solution on $\left[0,2\right]$.

Example 5.3 Consider the following multi-strip fractional q-integral boundary value problem:

$\left\{\begin{array}{c}{D}_{\frac{3}{8}}^{\frac{7}{5}}u\left(t\right)=\frac{1}{12\pi }{tan}^{-1}\left(3u\right)+\frac{2|u\left(t\right)|}{1+|u\left(t\right)|},\phantom{\rule{1em}{0ex}}t\in \left(0,3\right),\hfill \\ u\left(0\right)=0,\hfill \\ u\left(3\right)=-2\left({I}_{\frac{1}{2}}^{\frac{6}{5}}u\right){|}_{\frac{3}{2}}^{\frac{5}{3}}+11\left({I}_{\frac{1}{4}}^{\frac{7}{3}}u\right){|}_{\frac{1}{5}}^{\frac{7}{4}}+\left(\frac{9}{5}\right)\left({I}_{\frac{2}{3}}^{\frac{6}{7}}u\right){|}_{\frac{5}{4}}^{\frac{5}{2}}\hfill \\ \phantom{u\left(3\right)=}+5\left({I}_{\frac{2}{5}}^{\frac{3}{4}}u\right){|}_{\frac{3}{7}}^{\frac{1}{2}}+\left(\frac{4}{7}\right)\left({I}_{\frac{3}{5}}^{\frac{2}{3}}u\right){|}_{\frac{1}{3}}^{\frac{9}{5}}.\hfill \end{array}$
(5.3)

Here $\alpha =7/5$, $q=3/8$, $T=3$, $m=5$, ${\gamma }_{1}=-2$, ${\gamma }_{2}=11$, ${\gamma }_{3}=9/5$, ${\gamma }_{4}=5$, ${\gamma }_{5}=4/7$, ${\beta }_{1}=6/5$, ${\beta }_{2}=7/3$, ${\beta }_{3}=6/7$, ${\beta }_{4}=3/4$, ${\beta }_{5}=2/3$, ${q}_{1}=1/2$, ${q}_{2}=1/4$, ${q}_{3}=2/3$, ${q}_{4}=2/5$, ${q}_{5}=3/5$, ${\xi }_{1}=5/3$, ${\xi }_{2}=7/4$, ${\xi }_{3}=5/2$, ${\xi }_{4}=1/2$, ${\xi }_{5}=9/5$, ${\eta }_{1}=3/2$, ${\eta }_{2}=1/5$, ${\eta }_{3}=5/4$, ${\eta }_{4}=3/7$, ${\eta }_{5}=1/3$ and $f\left(t,u\right)=\left(\left({tan}^{-1}\left(3u\right)\right)/\left(12\pi \right)\right)+\left(\left(2|u\left(t\right)|\right)/\left(1+|u\left(t\right)|\right)\right)$. By using the Maple program, we find that

$\begin{array}{c}\mathrm{\Lambda }={T}^{\alpha -1}-\sum _{i=1}^{m}\frac{{\gamma }_{i}{\mathrm{\Gamma }}_{{q}_{i}}\left(\alpha \right)}{{\mathrm{\Gamma }}_{{q}_{i}}\left(\alpha +{\beta }_{i}\right)}\left({\xi }_{i}^{\alpha +{\beta }_{i}-1}-{\eta }_{i}^{\alpha +{\beta }_{i}-1}\right)\hfill \\ \phantom{\mathrm{\Lambda }}\approx -33.26381181,\hfill \\ \mathrm{\Phi }=\frac{{T}^{\alpha -1}}{|\mathrm{\Lambda }|{\mathrm{\Gamma }}_{q}\left(\alpha +1\right)}\left({T}^{\alpha }+\sum _{i=1}^{m}\frac{|{\gamma }_{i}|{\xi }_{i}^{{\beta }_{i}+\alpha }{\mathrm{\Gamma }}_{{q}_{i}}\left(\alpha +1\right)}{{\mathrm{\Gamma }}_{{q}_{i}}\left(\alpha +{\beta }_{i}+1\right)}+\sum _{i=1}^{m}\frac{|{\gamma }_{i}|{\eta }_{i}^{{\beta }_{i}+\alpha }{\mathrm{\Gamma }}_{{q}_{i}}\left(\alpha +1\right)}{{\mathrm{\Gamma }}_{{q}_{i}}\left(\alpha +{\beta }_{i}+1\right)}\right)\hfill \\ \phantom{\mathrm{\Phi }=}+\frac{{T}^{\alpha }}{{\mathrm{\Gamma }}_{q}\left(\alpha +1\right)}\hfill \\ \phantom{\mathrm{\Phi }}\approx 7.22159847.\hfill \end{array}$

We observe that

$|f\left(t,u\right)|=|\frac{1}{12\pi }{tan}^{-1}\left(3u\right)+\frac{2|u\left(t\right)|}{1+|u\left(t\right)|}|\le \frac{|u|}{4\pi }+2.$

Therefore, we have $\mathrm{\Psi }=2$ and

$\omega =1/4\pi <{\mathrm{\Phi }}^{-1}=0.13847350.$

Hence, by Theorem 4.7, the boundary value problem (5.3) has at least one solution on $\left[0,3\right]$.

## Authors’ information

The fourth author is a member of Nonlinear Analysis and Applied Mathematics (NAAM)-Research Group at King Abdulaziz University, Jeddah, Saudi Arabia.

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## Acknowledgements

The research of J. Tariboon and S. Asawasamrit is supported by King Mongkut’s University of Technology North Bangkok, Thailand.

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### Authors’ contributions

All authors contributed equally to the writing of this paper. All authors read and approved the final manuscript.

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Pongarm, N., Asawasamrit, S., Tariboon, J. et al. Multi-strip fractional q-integral boundary value problems for nonlinear fractional q-difference equations. Adv Differ Equ 2014, 193 (2014). https://doi.org/10.1186/1687-1847-2014-193