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:

$$\{\begin{array}{c}{D}_q^{\alpha }u(t)=f(t,u(t)),t\in (0,T),\hfill \\ u(0)=0,u(T)={\sum }_{i=1}^m{\gamma }_i(I_{q_i}^{{\beta }_i}u){|}_{{\eta }_i}^{{\xi }_i}={\sum }_{i=1}^m{\gamma }_i(I_{q_i}^{{\beta }_i}u({\xi }_i)-I_{q_i}^{{\beta }_i}u({\eta }_i)),\hfill \end{array}$$

(1.1)

where $1<\alpha \le 2$, $0<q,q_i<1$, ${\beta }_i>0$, $0\le {\eta }_i<{\xi }_i\le T$, ${\gamma }_i\in \mathbb{R}$ 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 $f:[0,T]\times \mathbb{R}\to \mathbb{R}$ 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 [5–12] 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 [13–20] 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:

$$u(0)=0,u(T)={\gamma }_1{\int }_{{\eta }_1}^{{\xi }_1}u(s)d_{q_1}s+{\gamma }_2{\int }_{{\eta }_2}^{{\xi }_2}u(s)d_{q_2}s+\cdots +{\gamma }_m{\int }_{{\eta }_m}^{{\xi }_m}u(s)d_{q_m}s.$$

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 $q\in (0,1)$, define

$${[a]}_q=\frac{1-q^a}{1-q},a\in \mathbb{R}.$$

(2.1)

The q-analogue of the power function ${(a-b)}^k$ with $k\in {\mathbb{N}}_0:=\{0,1,2,\dots \}$ is

$${(a-b)}^{(0)}=1,{(a-b)}^{(k)}=\prod _{i=0}^{k-1}(a-b{q}^i),k\in \mathbb{N},a,b\in \mathbb{R}.$$

(2.2)

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

$${(a-b)}^{(\gamma )}=a^{\gamma }\prod _{i=0}^{\mathrm{\infty }}\frac{1-(b/a)q^i}{1-(b/a)q^{\gamma +i}},a\ne 0.$$

(2.3)

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

$${\mathrm{\Gamma }}_q(x)=\frac{{(1-q)}^{(x-1)}}{{(1-q)}^{x-1}},x\in \mathbb{R}\setminus \{0,-1,-2,\dots \}.$$

(2.4)

Obviously, ${\mathrm{\Gamma }}_q(x+1)={[x]}_q{\mathrm{\Gamma }}_q(x)$.

The q-derivative of a function h is defined by

$$(D_{q}h)(x)=\frac{h(x)-h(qx)}{(1-q)x}\text{for }x\ne 0\text{and}(D_{q}h)(0)=\underset{x\to 0}{lim}(D_{q}h)(x),$$

(2.5)

and q-derivatives of higher order are given by

$$\left(D_q^{0}h\right)(x)=h(x)\text{and}\left(D_q^{k}h\right)(x)=D_q\left(D_q^{k-1}h\right)(x),k\in \mathbb{N}.$$

(2.6)

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

$$(I_{q}h)(x)={\int }_0^{x}h(s)d_{q}s=x(1-q)\sum _{i=0}^{\mathrm{\infty }}h\left(x{q}^i\right)q^i,x\in [0,b].$$

(2.7)

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

$${\int }_a^{b}h(s)d_{q}s={\int }_0^{b}h(s)d_{q}s-{\int }_0^{a}h(s)d_{q}s.$$

(2.8)

Similar to derivatives, an operator $I_q^k$ is given by

$$\left(I_q^{0}h\right)(x)=h(x)\text{and}\left(I_q^{k}h\right)(x)=I_q\left(I_q^{k-1}h\right)(x),k\in \mathbb{N}.$$

(2.9)

The fundamental theorem of calculus applies to these operators $D_q$ and $I_q$, i.e.,

$$(D_q{I}_{q}h)(x)=h(x),$$

(2.10)

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

$$(I_q{D}_{q}h)(x)=h(x)-h(0).$$

(2.11)

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

$$\left(I_q^{\nu }h\right)(x)=\frac{1}{{\mathrm{\Gamma }}_q(\nu )}{\int }_0^x{(x-qs)}^{(\nu -1)}h(s)d_{q}s,\nu >0,x\in [0,T].$$

(2.12)

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

$$\left(D_q^{\nu }h\right)(x)=\left(D_q^l{I}_q^{l-\nu }h\right)(x),\nu >0,$$

(2.13)

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

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

$$B_q(x,s)={\int }_0^1{u}^{(x-1)}{(1-qu)}^{(s-1)}{d}_{q}u$$

(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

$$B_q(x,s)=\frac{{\mathrm{\Gamma }}_q(x){\mathrm{\Gamma }}_q(s)}{{\mathrm{\Gamma }}_q(x+s)}.$$

Lemma 2.4 [4]

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

• (1) $(I_q^{\beta }{I}_q^{\alpha }f)(x)=(I_q^{\alpha +\beta }f)(x)$,

• (2) $(D_q^{\alpha }{I}_q^{\alpha }f)(x)=f(x)$.

Lemma 2.5 [22]

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

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

(2.15)

3 Some auxiliary lemmas

Lemma 3.1 Let $\alpha ,\beta >0$ and $0<q<1$. Then we have

$${\int }_0^{\eta }{(\eta -qs)}^{(\alpha -1)}{s}^{(\beta )}{d}_{q}s={\eta }^{\alpha +\beta }{B}_q(\alpha ,\beta +1).$$

(3.1)

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

$$\begin{array}{rcl}{\int }_0^{\eta }{(\eta -qs)}^{(\alpha -1)}{s}^{(\beta )}{d}_{q}s& =& (1-q)\eta \sum _{n=0}^{\mathrm{\infty }}{q}^n{(\eta -q\eta q^n)}^{(\alpha -1)}{\left(\eta q^n\right)}^{\beta }\\ =& (1-q)\eta \sum _{n=0}^{\mathrm{\infty }}{q}^n{\eta }^{\alpha -1}{(1-q{q}^n)}^{(\alpha -1)}{\eta }^{\beta }{q}^{n\beta }\\ =& (1-q){\eta }^{\alpha +\beta }\sum _{n=0}^{\mathrm{\infty }}{q}^n{(1-q{q}^n)}^{(\alpha -1)}{q}^{n\beta }\\ =& {\eta }^{\alpha +\beta }{\int }_0^1{(1-qs)}^{(\alpha -1)}{s}^{(\beta )}{d}_{q}s\\ =& {\eta }^{\alpha +\beta }{B}_q(\alpha ,\beta +1).\end{array}$$

The proof is complete. □

Lemma 3.2 Let $\alpha ,\beta >0$ and $0<p,q<1$. Then we have

$${\int }_0^{\eta }{\int }_0^x{(\eta -px)}^{(\alpha -1)}{(x-qy)}^{(\beta -1)}{d}_{q}y{d}_{p}x=\frac{{\eta }^{\alpha +\beta }}{{[\beta ]}_q}\frac{{\mathrm{\Gamma }}_p(\alpha ){\mathrm{\Gamma }}_p(\beta +1)}{{\mathrm{\Gamma }}_p(\alpha +\beta +1)}.$$

(3.2)

Proof Taking into account Lemma 3.1, we have

$$\begin{array}{c}{\int }_0^{\eta }{\int }_0^x{(\eta -px)}^{(\alpha -1)}{(x-qy)}^{(\beta -1)}{d}_{q}y{d}_{p}x\hfill \\ ={\int }_0^{\eta }{(\eta -px)}^{(\alpha -1)}{\int }_0^x{(x-qy)}^{(\beta -1)}{d}_{q}y{d}_{p}x\hfill \\ =\frac{1}{{[\beta ]}_q}{\int }_0^{\eta }{(\eta -px)}^{(\alpha -1)}{x}^{(\beta )}{d}_{p}x\hfill \\ =\frac{1}{{[\beta ]}_q}{\eta }^{\alpha +\beta }{B}_p(\alpha ,\beta +1)\hfill \\ =\frac{{\eta }^{\alpha +\beta }}{{[\beta ]}_q}\frac{{\mathrm{\Gamma }}_p(\alpha ){\mathrm{\Gamma }}_p(\beta +1)}{{\mathrm{\Gamma }}_p(\alpha +\beta +1)}.\hfill \end{array}$$

This completes the proof. □

For convenience, we set a nonzero constant

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

(3.3)

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

$$D_q^{\alpha }u(t)=y(t),t\in (0,T),1<\alpha \le 2,$$

(3.4)

subject to the multi-strip fractional q-integral condition

$$u(0)=0,u(T)=\sum _{i=1}^m{\gamma }_i\left(I_{q_i}^{{\beta }_i}u\right){|}_{{\eta }_i}^{{\xi }_i}=\sum _{i=1}^m{\gamma }_i(I_{q_i}^{{\beta }_i}u({\xi }_i)-I_{q_i}^{{\beta }_i}u({\eta }_i)),$$

(3.5)

is given by

$$\begin{array}{rcl}u(t)& =& -\frac{t^{\alpha -1}}{\mathrm{\Lambda }}\{{\int }_0^T\frac{{(T-qs)}^{(\alpha -1)}}{{\mathrm{\Gamma }}_q(\alpha )}y(s)d_{q}s\\ -\sum _{i=1}^m\frac{{\gamma }_i}{{\mathrm{\Gamma }}_{q_i}({\beta }_i){\mathrm{\Gamma }}_q(\alpha )}({\int }_0^{{\xi }_i}{\int }_0^s{({\xi }_i-q_{i}s)}^{({\beta }_i-1)}{(s-qx)}^{(\alpha -1)}y(x)d_{q}x{d}_{q_i}s\\ +{\int }_0^{{\eta }_i}{\int }_0^s{({\eta }_i-q_{i}s)}^{({\beta }_i-1)}{(s-qx)}^{(\alpha -1)}y(x)d_{q}x{d}_{q_i}s\left)\right\}\\ +{\int }_0^t\frac{{(t-qs)}^{(\alpha -1)}}{{\mathrm{\Gamma }}_q(\alpha )}y(s)d_{q}s,\end{array}$$

(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

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

Using Lemma 2.5, we obtain

$$u(t)=c_1{t}^{\alpha -1}+c_2{t}^{\alpha -2}+{\int }_0^t\frac{{(t-qs)}^{(\alpha -1)}}{{\mathrm{\Gamma }}_q(\alpha )}y(s)d_{q}s$$

(3.7)

for some constants $c_1,c_2\in \mathbb{R}$. Since $u(0)=0$, 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

$$\begin{array}{rcl}{I}_{q_i}^{{\beta }_i}u({\xi }_i)& =& {\int }_0^{{\xi }_i}\frac{{({\xi }_i-q_{i}s)}^{({\beta }_i-1)}}{{\mathrm{\Gamma }}_{q_i}({\beta }_i)}(c_1{s}^{\alpha -1}+{\int }_0^s\frac{{(s-qx)}^{(\alpha -1)}}{{\mathrm{\Gamma }}_q(\alpha )}y(x)d_{q}x)d_{q_i}s\\ =& \frac{1}{{\mathrm{\Gamma }}_{q_i}({\beta }_i){\mathrm{\Gamma }}_q(\alpha )}{\int }_0^{{\xi }_i}{\int }_0^s{({\xi }_i-q_{i}s)}^{({\beta }_i-1)}{(s-qx)}^{(\alpha -1)}y(x)d_{q}x{d}_{q_i}s\\ +\frac{c_1}{{\mathrm{\Gamma }}_{q_i}({\beta }_i)}{\int }_0^{{\xi }_i}{({\xi }_i-q_{i}s)}^{({\beta }_i-1)}{s}^{\alpha -1}{d}_{q_i}s\\ =& \frac{1}{{\mathrm{\Gamma }}_{q_i}({\beta }_i){\mathrm{\Gamma }}_q(\alpha )}{\int }_0^{{\xi }_i}{\int }_0^s{({\xi }_i-q_{i}s)}^{({\beta }_i-1)}{(s-qx)}^{(\alpha -1)}y(x)d_{q}x{d}_{q_i}s\\ +c_1\frac{{\mathrm{\Gamma }}_{q_i}(\alpha ){\xi }_i^{\alpha +{\beta }_i-1}}{{\mathrm{\Gamma }}_{q_i}(\alpha +{\beta }_i)}.\end{array}$$

(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:

$$\begin{array}{rcl}{c}_1& =& \frac{1}{\mathrm{\Lambda }}\left\{\sum _{i=1}^m\frac{{\gamma }_i}{{\mathrm{\Gamma }}_{q_i}({\beta }_i){\mathrm{\Gamma }}_q(\alpha )}\right({\int }_0^{{\xi }_i}{\int }_0^s{({\xi }_i-q_{i}s)}^{({\beta }_i-1)}{(s-qx)}^{(\alpha -1)}y(x)d_{q}x{d}_{q_i}s\\ -{\int }_0^{{\eta }_i}{\int }_0^s{({\eta }_i-q_{i}s)}^{({\beta }_i-1)}{(s-qx)}^{(\alpha -1)}y(x)d_{q}x{d}_{q_i}s)\\ -{\int }_0^T\frac{{(T-qs)}^{(\alpha -1)}}{{\mathrm{\Gamma }}_q(\alpha )}y(s)d_{q}s\}.\end{array}$$

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

$$\begin{array}{rcl}(\mathcal{A}u)(t)& =& -\frac{t^{\alpha -1}}{\mathrm{\Lambda }}\{{\int }_0^T\frac{{(T-qs)}^{(\alpha -1)}}{{\mathrm{\Gamma }}_q(\alpha )}f(s,u(s))d_{q}s\\ -\sum _{i=1}^m\frac{{\gamma }_i}{{\mathrm{\Gamma }}_{q_i}({\beta }_i){\mathrm{\Gamma }}_q(\alpha )}({\int }_0^{{\xi }_i}{\int }_0^s{({\xi }_i-q_{i}s)}^{({\beta }_i-1)}{(s-qx)}^{(\alpha -1)}f(x,u(x))d_{q}x{d}_{q_i}s\\ +{\int }_0^{{\eta }_i}{\int }_0^s{({\eta }_i-q_{i}s)}^{({\beta }_i-1)}{(s-qx)}^{(\alpha -1)}f(x,u(x))d_{q}x{d}_{q_i}s\left)\right\}\\ +{\int }_0^t\frac{{(t-qs)}^{(\alpha -1)}}{{\mathrm{\Gamma }}_q(\alpha )}f(s,u(s))d_{q}s,\end{array}$$

(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

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

(4.2)

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

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

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

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 ${sup}_{t\in [0,T]}|f(t,0)|=M_0<\mathrm{\infty }$ 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 $B_r=\{u\in \mathcal{C}:\parallel u\parallel \le r\}$. For any $u\in B_r$, and taking into account Lemma 3.2, we have

$$\begin{array}{rcl}\parallel \mathcal{A}u\parallel & \le & \underset{t\in [0,T]}{sup}\left\{\frac{t^{\alpha -1}}{|\mathrm{\Lambda }|}\right\{{\int }_0^T\frac{{(T-qs)}^{(\alpha -1)}}{{\mathrm{\Gamma }}_q(\alpha )}\left|f(s,u(s))\right|d_{q}s\\ +\sum _{i=1}^m\frac{|{\gamma }_i|}{{\mathrm{\Gamma }}_{q_i}({\beta }_i){\mathrm{\Gamma }}_q(\alpha )}({\int }_0^{{\xi }_i}{\int }_0^s{({\xi }_i-q_{i}s)}^{({\beta }_i-1)}{(s-qx)}^{(\alpha -1)}\left|f(x,u(x))\right|d_{q}x{d}_{q_i}s\\ +{\int }_0^{{\eta }_i}{\int }_0^s{({\eta }_i-q_{i}s)}^{({\beta }_i-1)}{(s-qx)}^{(\alpha -1)}\left|f(x,u(x))\right|d_{q}x{d}_{q_i}s\left)\right\}\\ +{\int }_0^t\frac{{(t-qs)}^{(\alpha -1)}}{{\mathrm{\Gamma }}_q(\alpha )}\left|f(s,u(s))\right|d_{q}s\}\\ \le & \frac{T^{\alpha -1}}{|\mathrm{\Lambda }|}\{{\int }_0^T\frac{{(T-qs)}^{(\alpha -1)}}{{\mathrm{\Gamma }}_q(\alpha )}(|f(s,u(s))-f(s,0)|+|f(s,0)|)d_{q}s\\ +\sum _{i=1}^m\frac{|{\gamma }_i|}{{\mathrm{\Gamma }}_{q_i}({\beta }_i){\mathrm{\Gamma }}_q(\alpha )}({\int }_0^{{\xi }_i}{\int }_0^s{({\xi }_i-q_{i}s)}^{({\beta }_i-1)}{(s-qx)}^{(\alpha -1)}\\ \times (|f(x,u(x))-f(x,0)|+|f(x,0)|)d_{q}x{d}_{q_i}s\\ +{\int }_0^{{\eta }_i}{\int }_0^s{({\eta }_i-q_{i}s)}^{({\beta }_i-1)}{(s-qx)}^{(\alpha -1)}\\ \times (|f(x,u(x))-f(x,0)|+|f(x,0)|)d_{q}x{d}_{q_i}s\left)\right\}\\ +{\int }_0^T\frac{{(T-qs)}^{(\alpha -1)}}{{\mathrm{\Gamma }}_q(\alpha )}(|f(s,u(s))-f(s,0)|+|f(s,0)|)d_{q}s\\ \le & (Lr+M_0)\left\{\frac{T^{\alpha -1}}{|\mathrm{\Lambda }|{\mathrm{\Gamma }}_q(\alpha +1)}\right(T^{\alpha }+\sum _{i=1}^m\frac{|{\gamma }_i|{\xi }_i^{{\beta }_i+\alpha }{\mathrm{\Gamma }}_{q_i}(\alpha +1)}{{\mathrm{\Gamma }}_{q_i}(\alpha +{\beta }_i+1)}\\ +\sum _{i=1}^m\frac{|{\gamma }_i|{\eta }_i^{{\beta }_i+\alpha }{\mathrm{\Gamma }}_{q_i}(\alpha +1)}{{\mathrm{\Gamma }}_{q_i}(\alpha +{\beta }_i+1)})+\frac{T^{\alpha }}{{\mathrm{\Gamma }}_q(\alpha +1)}\}\\ =& (Lr+M_0)\mathrm{\Phi }\le r.\end{array}$$

It follows that $\mathcal{A}{B}_r\subset B_r$.

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

$$\begin{array}{c}|\mathcal{A}u(t)-\mathcal{A}v(t)|\hfill \\ \le \frac{T^{\alpha -1}}{|\mathrm{\Lambda }|}\{{\int }_0^T\frac{{(T-qs)}^{(\alpha -1)}}{{\mathrm{\Gamma }}_q(\alpha )}\left(|f(s,u(s))-f(s,v(s))|\right)d_{q}s\hfill \\ +\sum _{i=1}^m\frac{|{\gamma }_i|}{{\mathrm{\Gamma }}_{q_i}({\beta }_i){\mathrm{\Gamma }}_q(\alpha )}({\int }_0^{{\xi }_i}{\int }_0^s{({\xi }_i-q_{i}s)}^{({\beta }_i-1)}{(s-qx)}^{(\alpha -1)}\hfill \\ \times \left(|f(s,u(s))-f(s,v(s))|\right)d_{q}x{d}_{q_i}s\hfill \\ +{\int }_0^{{\eta }_i}{\int }_0^s{({\eta }_i-q_{i}s)}^{({\beta }_i-1)}{(s-qx)}^{(\alpha -1)}\left(|f(s,u(s))-f(s,v(s))|\right)d_{q}x{d}_{q_i}s\left)\right\}\hfill \\ +{\int }_0^T\frac{{(T-qs)}^{(\alpha -1)}}{{\mathrm{\Gamma }}_q(\alpha )}\left(|f(s,u(s))-f(s,v(s))|\right)d_{q}s\hfill \\ \le L\parallel u-v\parallel \left\{\frac{T^{\alpha -1}}{|\mathrm{\Lambda }|{\mathrm{\Gamma }}_q(\alpha +1)}\right(T^{\alpha }+\sum _{i=1}^m\frac{|{\gamma }_i|{\xi }_i^{{\beta }_i+\alpha }{\mathrm{\Gamma }}_{q_i}(\alpha +1)}{{\mathrm{\Gamma }}_{q_i}(\alpha +{\beta }_i+1)}\hfill \\ +\sum _{i=1}^m\frac{|{\gamma }_i|{\eta }_i^{{\beta }_i+\alpha }{\mathrm{\Gamma }}_{q_i}(\alpha +1)}{{\mathrm{\Gamma }}_{q_i}(\alpha +{\beta }_i+1)})+\frac{T^{\alpha }}{{\mathrm{\Gamma }}_q(\alpha +1)}\}\hfill \\ =L\mathrm{\Phi }\parallel u-v\parallel .\hfill \end{array}$$

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

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

If the following condition holds

$$\frac{L}{{\mathrm{\Gamma }}_q(\alpha +1)}(\frac{T^{2\alpha -1}}{|\mathrm{\Lambda }|}+T^{\alpha })<1,$$

(4.4)

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

Proof We define ${sup}_{t\in [0,T]}|\psi (t)|=\parallel \psi \parallel$ 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 $B_R=\{u\in \mathcal{C}:\parallel u\parallel \le R\}$ by

$$\begin{array}{c}({\mathcal{A}}_{1}u)(t)\hfill \\ =\frac{t^{\alpha -1}}{\mathrm{\Lambda }}\sum _{i=1}^m\frac{{\gamma }_i}{{\mathrm{\Gamma }}_{q_i}({\beta }_i){\mathrm{\Gamma }}_q(\alpha )}{\int }_0^{{\xi }_i}{\int }_0^s{({\xi }_i-q_{i}s)}^{({\beta }_i-1)}{(s-qx)}^{(\alpha -1)}f(x,u(x))d_{q}x{d}_{q_i}s\hfill \\ -\frac{t^{\alpha -1}}{\mathrm{\Lambda }}\sum _{i=1}^m\frac{{\gamma }_i}{{\mathrm{\Gamma }}_{q_i}({\beta }_i){\mathrm{\Gamma }}_q(\alpha )}{\int }_0^{{\eta }_i}{\int }_0^s{({\eta }_i-q_{i}s)}^{({\beta }_i-1)}{(s-qx)}^{(\alpha -1)}f(x,u(x))d_{q}x{d}_{q_i}s,\hfill \end{array}$$

and

$$({\mathcal{A}}_{2}u)(t)=-\frac{t^{\alpha -1}}{\mathrm{\Lambda }}{\int }_0^T\frac{{(T-qs)}^{(\alpha -1)}}{{\mathrm{\Gamma }}_q(\alpha )}f(s,u(s))d_{q}s+{\int }_0^t\frac{{(t-qs)}^{(\alpha -1)}}{{\mathrm{\Gamma }}_q(\alpha )}f(s,u(s))d_{q}s.$$

It should be noticed that $\mathcal{A}={\mathcal{A}}_1+{\mathcal{A}}_2$.

For any $u,v\in B_R$, we have

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

Therefore $({\mathcal{A}}_{1}u)+({\mathcal{A}}_{2}v)\in B_R$. 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 ${sup}_{(t,u)\in [0,T]\times B_R}|f(t,u)|=M^{\ast }<\mathrm{\infty }$. For $t_1,t_2\in [0,T]$, $t_2<t_1$ and $u\in B_R$, we have

$$\begin{array}{c}|({\mathcal{A}}_{1}u)(t_1)-({\mathcal{A}}_{1}u)(t_2)|\hfill \\ \le \frac{|t_1^{\alpha -1}-t_2^{\alpha -1}|}{|\mathrm{\Lambda }|}\sum _{i=1}^m\frac{|{\gamma }_i|}{{\mathrm{\Gamma }}_{q_i}({\beta }_i){\mathrm{\Gamma }}_q(\alpha )}\hfill \\ \times {\int }_0^{{\xi }_i}{\int }_0^s{({\xi }_i-q_{i}s)}^{({\beta }_i-1)}{(s-qx)}^{(\alpha -1)}\left|f(x,u(x))\right|d_{q}x{d}_{q_i}s\hfill \\ +\frac{|t_1^{\alpha -1}-t_2^{\alpha -1}|}{|\mathrm{\Lambda }|}\sum _{i=1}^m\frac{|{\gamma }_i|}{{\mathrm{\Gamma }}_{q_i}({\beta }_i){\mathrm{\Gamma }}_q(\alpha )}\hfill \\ \times {\int }_0^{{\eta }_i}{\int }_0^s{({\eta }_i-q_{i}s)}^{({\beta }_i-1)}{(s-qx)}^{(\alpha -1)}\left|f(x,u(x))\right|d_{q}x{d}_{q_i}s\hfill \\ \le M^{\ast }\frac{|t_1^{\alpha -1}-t_2^{\alpha -1}|}{|\mathrm{\Lambda }|{\mathrm{\Gamma }}_q(\alpha +1)}\{\sum _{i=1}^m\frac{|{\gamma }_i|{\xi }_i^{{\beta }_i+\alpha }{\mathrm{\Gamma }}_{q_i}(\alpha +1)}{{\mathrm{\Gamma }}_{q_i}(\alpha +{\beta }_i+1)}+\sum _{i=1}^m\frac{|{\gamma }_i|{\eta }_i^{{\beta }_i+\alpha }{\mathrm{\Gamma }}_{q_i}(\alpha +1)}{{\mathrm{\Gamma }}_{q_i}(\alpha +{\beta }_i+1)}\}.\hfill \end{array}$$

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 $[0,T]$. 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:

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

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, $F(\overline{U})$ is a relatively compact subset of C) map. Then either

• (i) F has a fixed point in $\overline{U}$, or

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

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

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

$$|f(t,u)|\le p(t)\varphi (|u|)\mathit{\text{for each }}(t,u)\in [0,T]\times \mathbb{R};$$

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

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

where Φ is defined by (4.2).

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

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 $B_{\rho }=\{u\in \mathcal{C}:\parallel u\parallel \le \rho \}$ be a bounded ball in $\mathcal{C}$. Then, for $t\in [0,T]$, we have

$$\begin{array}{rcl}|\mathcal{A}u(t)|& \le & \frac{T^{\alpha -1}}{|\mathrm{\Lambda }|}\{{\int }_0^T\frac{{(T-qs)}^{(\alpha -1)}}{{\mathrm{\Gamma }}_q(\alpha )}\left|f(s,u(s))\right|d_{q}s\\ +\sum _{i=1}^m\frac{|{\gamma }_i|}{{\mathrm{\Gamma }}_{q_i}({\beta }_i){\mathrm{\Gamma }}_q(\alpha )}({\int }_0^{{\xi }_i}{\int }_0^s{({\xi }_i-q_{i}s)}^{({\beta }_i-1)}{(s-qx)}^{(\alpha -1)}\left|f(x,u(x))\right|d_{q}x{d}_{q_i}s\\ +{\int }_0^{{\eta }_i}{\int }_0^s{({\eta }_i-q_{i}s)}^{({\beta }_i-1)}{(s-qx)}^{(\alpha -1)}\left|f(x,u(x))\right|d_{q}x{d}_{q_i}s\left)\right\}\\ +{\int }_0^T\frac{{(T-qs)}^{(\alpha -1)}}{{\mathrm{\Gamma }}_q(\alpha )}\left|f(s,u(s))\right|d_{q}s\\ \le & \frac{T^{\alpha -1}}{|\mathrm{\Lambda }|{\mathrm{\Gamma }}_q(\alpha +1)}(\parallel p\parallel \varphi (\parallel u\parallel )T^{\alpha }+\parallel p\parallel \varphi (\parallel u\parallel )\sum _{i=1}^m\frac{|{\gamma }_i|{\xi }_i^{{\beta }_i+\alpha }{\mathrm{\Gamma }}_{q_i}(\alpha +1)}{{\mathrm{\Gamma }}_{q_i}(\alpha +{\beta }_i+1)}\\ +\parallel p\parallel \varphi (\parallel u\parallel )\sum _{i=1}^m\frac{|{\gamma }_i|{\eta }_i^{{\beta }_i+\alpha }{\mathrm{\Gamma }}_{q_i}(\alpha +1)}{{\mathrm{\Gamma }}_{q_i}(\alpha +{\beta }_i+1)})+\parallel p\parallel \varphi (\parallel u\parallel )\frac{T^{\alpha }}{{\mathrm{\Gamma }}_q(\alpha +1)}\\ \le & \frac{\parallel p\parallel \varphi (\rho )T^{\alpha -1}}{|\mathrm{\Lambda }|{\mathrm{\Gamma }}_q(\alpha +1)}(T^{\alpha }+\sum _{i=1}^m\frac{|{\gamma }_i|{\xi }_i^{{\beta }_i+\alpha }{\mathrm{\Gamma }}_{q_i}(\alpha +1)}{{\mathrm{\Gamma }}_{q_i}(\alpha +{\beta }_i+1)}+\sum _{i=1}^m\frac{|{\gamma }_i|{\eta }_i^{{\beta }_i+\alpha }{\mathrm{\Gamma }}_{q_i}(\alpha +1)}{{\mathrm{\Gamma }}_{q_i}(\alpha +{\beta }_i+1)})\\ +\parallel p\parallel \varphi (\rho )\frac{T^{\alpha }}{{\mathrm{\Gamma }}_q(\alpha +1)}\\ :=& K.\end{array}$$

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 ${sup}_{(t,u)\in [0,T]\times B_{\rho }}|f(t,u)|=K^{\ast }<\mathrm{\infty }$, ${\tau }_1,{\tau }_2\in [0,T]$ with ${\tau }_2<{\tau }_1$ and $u\in B_{\rho }$. Then we have

$$\begin{array}{c}|(\mathcal{A}u)({\tau }_1)-(\mathcal{A}u)({\tau }_2)|\hfill \\ \le \frac{|{\tau }_1^{\alpha -1}-{\tau }_2^{\alpha -1}|}{|\mathrm{\Lambda }|}\{{\int }_0^T\frac{{(T-qs)}^{(\alpha -1)}}{{\mathrm{\Gamma }}_q(\alpha )}\left|f(s,u(s))\right|d_{q}s\hfill \\ +\sum _{i=1}^m\frac{|{\gamma }_i|}{{\mathrm{\Gamma }}_{q_i}({\beta }_i){\mathrm{\Gamma }}_q(\alpha )}({\int }_0^{{\xi }_i}{\int }_0^s{({\xi }_i-q_{i}s)}^{({\beta }_i-1)}{(s-qx)}^{(\alpha -1)}\left|f(x,u(x))\right|d_{q}x{d}_{q_i}s\hfill \\ +{\int }_0^{{\eta }_i}{\int }_0^s{({\eta }_i-q_{i}s)}^{({\beta }_i-1)}{(s-qx)}^{(\alpha -1)}\left|f(x,u(x))\right|d_{q}x{d}_{q_i}s\left)\right\}\hfill \\ +|{\int }_0^{{\tau }_1}\frac{{({\tau }_1-qs)}^{(\alpha -1)}}{{\mathrm{\Gamma }}_q(\alpha )}\left|f(s,u(s))\right|d_{q}s-{\int }_0^{{\tau }_2}\frac{{({\tau }_2-qs)}^{(\alpha -1)}}{{\mathrm{\Gamma }}_q(\alpha )}\left|f(s,u(s))\right|d_{q}s|\hfill \\ \le \frac{|{\tau }_1^{\alpha -1}-{\tau }_2^{\alpha -1}|K^{\ast }}{|\mathrm{\Lambda }|{\mathrm{\Gamma }}_q(\alpha +1)}(T^{\alpha }+\sum _{i=1}^m\frac{|{\gamma }_i|{\xi }_i^{{\beta }_i+\alpha }{\mathrm{\Gamma }}_{q_i}(\alpha +1)}{{\mathrm{\Gamma }}_{q_i}(\alpha +{\beta }_i+1)}+\sum _{i=1}^m\frac{|{\gamma }_i|{\eta }_i^{{\beta }_i+\alpha }{\mathrm{\Gamma }}_{q_i}(\alpha +1)}{{\mathrm{\Gamma }}_{q_i}(\alpha +{\beta }_i+1)})\hfill \\ +\frac{|{\tau }_1^{\alpha -1}-{\tau }_2^{\alpha -1}|K^{\ast }}{{\mathrm{\Gamma }}_{q_i}(\alpha +1)}.\hfill \end{array}$$

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 $t\in [0,T]$, and following similar computations as in the first step with (H₃), we have

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

Consequently, we have

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

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

$$U=\{x\in \mathcal{C}:\parallel u\parallel <N\}.$$

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 $\lambda \in (0,1)$. 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 $f:[0,T]\times \mathbb{R}\to \mathbb{R}$ 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

$$|f(t,u)|\le \omega |u|+\mathrm{\Psi }\mathit{\text{for each }}(t,u)\in [0,T]\times \mathbb{R}.$$

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

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

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

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

$$u\ne \theta \mathcal{A}u,\mathrm{\forall }u\in \partial B_{{\rho }^{\ast }},\mathrm{\forall }\theta \in [0,1].$$

(4.5)

Now, we set

$$H(\theta ,u)=\theta \mathcal{A}u,u\in \mathcal{C},\theta \in [0,1].$$

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 $h_{\theta }(u)=u-H(\theta ,u)=u-\theta \mathcal{A}u$ 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

$$\begin{array}{rcl}deg(h_{\theta },B_{{\rho }^{\ast }},0)& =& deg(I-\theta \mathcal{A},B_{{\rho }^{\ast }},0)=deg(h_1,B_{{\rho }^{\ast }},0)\\ =& deg(h_0,B_{{\rho }^{\ast }},0)=deg(I,B_{{\rho }^{\ast }},0)=1\ne 0,0\in B_{{\rho }^{\ast }},\end{array}$$

where I denotes the unit operator. By the nonzero property of Leray-Schauder degree, we have $h_1(u)=u-Au=0$ for at least one $u\in B_{{\rho }^{\ast }}$. Let us assume that $u=\theta \mathcal{A}u$ for some $\theta \in [0,1]$. Then, for all $t\in [0,T]$, we have

$$\begin{array}{rcl}|u(t)|& =& |\theta (\mathcal{A}u)(t)|\\ \le & \frac{T^{\alpha -1}}{|\mathrm{\Lambda }|}\{{\int }_0^T\frac{{(T-qs)}^{(\alpha -1)}}{{\mathrm{\Gamma }}_q(\alpha )}\left|f(s,u(s))\right|d_{q}s\\ +\sum _{i=1}^m\frac{|{\gamma }_i|}{{\mathrm{\Gamma }}_{q_i}({\beta }_i){\mathrm{\Gamma }}_q(\alpha )}({\int }_0^{{\xi }_i}{\int }_0^s{({\xi }_i-q_{i}s)}^{({\beta }_i-1)}{(s-qx)}^{(\alpha -1)}\left|f(x,u(x))\right|d_{q}x{d}_{q_i}s\\ +{\int }_0^{{\eta }_i}{\int }_0^s{({\eta }_i-q_{i}s)}^{({\beta }_i-1)}{(s-qx)}^{(\alpha -1)}\left|f(x,u(x))\right|d_{q}x{d}_{q_i}s\left)\right\}\\ +{\int }_0^T\frac{{(T-qs)}^{(\alpha -1)}}{{\mathrm{\Gamma }}_q(\alpha )}\left|f(s,u(s))\right|d_{q}s\\ \le & (\omega |u|+\mathrm{\Psi })\left\{\frac{T^{\alpha -1}}{|\mathrm{\Lambda }|{\mathrm{\Gamma }}_q(\alpha +1)}\right(T^{\alpha }+\sum _{i=1}^m\frac{|{\gamma }_i|{\xi }_i^{{\beta }_i+\alpha }{B}_{q_i}({\beta }_i,\alpha +1)}{{\mathrm{\Gamma }}_{q_i}({\beta }_i)}\\ +\sum _{i=1}^m\frac{|{\gamma }_i|{\eta }_i^{{\beta }_i+\alpha }{B}_{q_i}({\beta }_i,\alpha +1)}{{\mathrm{\Gamma }}_{q_i}({\beta }_i)})+\frac{T^{\alpha }}{{\mathrm{\Gamma }}_q(\alpha +1)}\}\\ =& (\omega |u|+\mathrm{\Psi })\mathrm{\Phi }.\end{array}$$

Taking norm ${sup}_{t\in [0,T]}|u(t)|=\parallel u\parallel$ 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:

$$\{\begin{array}{c}{D}_{\frac{1}{2}}^{\frac{7}{4}}u(t)=\frac{4|u(t)|}{{(5+t)}^2(1+|u(t)|)},t\in (0,1),\hfill \\ u(0)=0,\hfill \\ u(1)=2(I_{\frac{1}{4}}^{\frac{2}{3}}u){|}_{\frac{1}{6}}^{\frac{1}{5}}+\frac{3}{4}(I_{\frac{1}{2}}^{\frac{4}{5}}u){|}_{\frac{2}{5}}^{\frac{2}{3}}+10(I_{\frac{1}{5}}^{\frac{7}{6}}u){|}_{\frac{1}{2}}^1.\hfill \end{array}$$

(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 $f(t,u)=(4|u(t)|)/({(5+t)}^2(1+|u(t)|))$. Since

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

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

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

Therefore, we get

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

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

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

$$\{\begin{array}{c}{D}_{\frac{5}{8}}^{\frac{4}{3}}u(t)=\frac{1}{u^2+3{\pi }^2}sin(\frac{\pi u}{4})+\frac{1}{20\pi }(5+sin(\pi t)),t\in (0,2),\hfill \\ u(0)=0,\hfill \\ u(2)=(I_{\frac{1}{3}}^{\frac{1}{2}}u){|}_{\frac{1}{5}}^{\frac{1}{4}}+\frac{2}{3}(I_{\frac{1}{2}}^{\frac{6}{5}}u){|}_1^2-3(I_{\frac{1}{8}}^{\frac{3}{2}}u){|}_{\frac{1}{3}}^1+\frac{4}{5}(I_{\frac{1}{4}}^{\frac{3}{4}}u){|}_{\frac{2}{5}}^{\frac{1}{2}}.\hfill \end{array}$$

(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 $f(t,u)=((sin(\pi u/4))/(u^2+3{\pi }^2))+((5+sin(\pi t))/(20\pi ))$. By using the Maple program, we find that

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

Clearly,

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

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

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

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

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

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

(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 $f(t,u)=(({tan}^{-1}(3u))/(12\pi ))+((2|u(t)|)/(1+|u(t)|))$. By using the Maple program, we find that

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

We observe that

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

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 $[0,3]$.

Authors’ information

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