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# 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:

${ D q α u ( t ) = f ( t , u ( t ) ) , t ∈ ( 0 , T ) , u ( 0 ) = 0 , u ( T ) = ∑ i = 1 m γ i ( I q i β i u ) | η i ξ i = ∑ i = 1 m γ i ( I q i β i u ( ξ i ) − I q i β i u ( η i ) ) ,$
(1.1)

where $1<α≤2$, $0, $β i >0$, $0≤ η i < ξ i ≤T$, $γ i ∈R$ for all $i=1,2,…,m$ are given constants, $D q α$ is the fractional q-derivative of Riemann-Liouville type of order α, $I q i β i$ is the fractional $q i$-integral of order $β i$ and $f:[0,T]×R→R$ 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 $β i =1$ for $i=1,2,…,m$, then the condition of (1.1) is reduced to the multi-strip q-integral condition as follows:

$u(0)=0,u(T)= γ 1 ∫ η 1 ξ 1 u(s) d q 1 s+ γ 2 ∫ η 2 ξ 2 u(s) d q 2 s+⋯+ γ m ∫ η m ξ m u(s) d q m s.$

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 $q∈(0,1)$, define

$[ a ] q = 1 − q a 1 − q ,a∈R.$
(2.1)

The q-analogue of the power function $( a − b ) k$ with $k∈ N 0 :={0,1,2,…}$ is

$( a − b ) ( 0 ) =1, ( a − b ) ( k ) = ∏ i = 0 k − 1 ( a − b q i ) ,k∈N,a,b∈R.$
(2.2)

More generally, if $γ∈R$, then

$( a − b ) ( γ ) = a γ ∏ i = 0 ∞ 1 − ( b / a ) q i 1 − ( b / a ) q γ + i ,a≠0.$
(2.3)

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

$Γ q (x)= ( 1 − q ) ( x − 1 ) ( 1 − q ) x − 1 ,x∈R∖{0,−1,−2,…}.$
(2.4)

Obviously, $Γ q (x+1)= [ x ] q Γ q (x)$.

The q-derivative of a function h is defined by

(2.5)

and q-derivatives of higher order are given by

$( D q 0 h ) (x)=h(x)and ( D q k h ) (x)= D q ( D q k − 1 h ) (x),k∈N.$
(2.6)

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

$( I q h)(x)= ∫ 0 x h(s) d q s=x(1−q) ∑ i = 0 ∞ h ( x q i ) q i ,x∈[0,b].$
(2.7)

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

$∫ a b h(s) d q s= ∫ 0 b h(s) d q s− ∫ 0 a h(s) d q s.$
(2.8)

Similar to derivatives, an operator $I q k$ is given by

$( I q 0 h ) (x)=h(x)and ( I q k h ) (x)= I q ( I q k − 1 h ) (x),k∈N.$
(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 $ν≥0$ 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

$( I q ν h ) (x)= 1 Γ q ( ν ) ∫ 0 x ( x − q s ) ( ν − 1 ) h(s) d q s,ν>0,x∈[0,T].$
(2.12)

Definition 2.2 The fractional q-derivative of Riemann-Liouville type of order $ν≥0$ is defined by $( D q 0 h)(x)=h(x)$ and

$( D q ν h ) (x)= ( D q l I q l − ν h ) (x),ν>0,$
(2.13)

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

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

$B q (x,s)= ∫ 0 1 u ( x − 1 ) ( 1 − q u ) ( s − 1 ) d q u$
(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

$B q (x,s)= Γ q ( x ) Γ q ( s ) Γ q ( x + s ) .$

Lemma 2.4 [4]

Let $α,β≥0$ and f be a function defined in $[0,T]$. Then the following formulas hold:

1. (1)

$( I q β I q α f)(x)=( I q α + β f)(x)$,

2. (2)

$( D q α I q α f)(x)=f(x)$.

Lemma 2.5 [22]

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

$( I q α D q ν f ) (x)= ( D q ν I q α f ) (x)− ∑ k = 0 ν − 1 x α − ν + k Γ q ( α + k − ν + 1 ) ( D q k f ) (0).$
(2.15)

## 3 Some auxiliary lemmas

Lemma 3.1 Let $α,β>0$ and $0. Then we have

$∫ 0 η ( η − q s ) ( α − 1 ) s ( β ) d q s= η α + β B q (α,β+1).$
(3.1)

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

$∫ 0 η ( η − q s ) ( α − 1 ) s ( β ) d q s = ( 1 − q ) η ∑ n = 0 ∞ q n ( η − q η q n ) ( α − 1 ) ( η q n ) β = ( 1 − q ) η ∑ n = 0 ∞ q n η α − 1 ( 1 − q q n ) ( α − 1 ) η β q n β = ( 1 − q ) η α + β ∑ n = 0 ∞ q n ( 1 − q q n ) ( α − 1 ) q n β = η α + β ∫ 0 1 ( 1 − q s ) ( α − 1 ) s ( β ) d q s = η α + β B q ( α , β + 1 ) .$

The proof is complete. □

Lemma 3.2 Let $α,β>0$ and $0. Then we have

$∫ 0 η ∫ 0 x ( η − p x ) ( α − 1 ) ( x − q y ) ( β − 1 ) d q y d p x= η α + β [ β ] q Γ p ( α ) Γ p ( β + 1 ) Γ p ( α + β + 1 ) .$
(3.2)

Proof Taking into account Lemma 3.1, we have

$∫ 0 η ∫ 0 x ( η − p x ) ( α − 1 ) ( x − q y ) ( β − 1 ) d q y d p x = ∫ 0 η ( η − p x ) ( α − 1 ) ∫ 0 x ( x − q y ) ( β − 1 ) d q y d p x = 1 [ β ] q ∫ 0 η ( η − p x ) ( α − 1 ) x ( β ) d p x = 1 [ β ] q η α + β B p ( α , β + 1 ) = η α + β [ β ] q Γ p ( α ) Γ p ( β + 1 ) Γ p ( α + β + 1 ) .$

This completes the proof. □

For convenience, we set a nonzero constant

$Λ= T α − 1 − ∑ i = 1 m γ i Γ q i ( α ) Γ q i ( α + β i ) ( ξ i α + β i − 1 − η i α + β i − 1 ) .$
(3.3)

Lemma 3.3 Let $β i >0$, $0, $γ i ∈R$, $η i , ξ i ∈(0,T)$ and $η i < ξ i$ for all $i=1,2,…,m$. Then, for a given $y∈C([0,1],R)$, the unique solution of the linear q-difference equation

$D q α u(t)=y(t),t∈(0,T),1<α≤2,$
(3.4)

subject to the multi-strip fractional q-integral condition

$u(0)=0,u(T)= ∑ i = 1 m γ i ( I q i β i u ) | η i ξ i = ∑ i = 1 m γ i ( I q i β i u ( ξ i ) − I q i β i u ( η i ) ) ,$
(3.5)

is given by

$u ( t ) = − t α − 1 Λ { ∫ 0 T ( T − q s ) ( α − 1 ) Γ q ( α ) y ( s ) d q s − ∑ i = 1 m γ i Γ q i ( β i ) Γ q ( α ) ( ∫ 0 ξ i ∫ 0 s ( ξ i − q i s ) ( β i − 1 ) ( s − q x ) ( α − 1 ) y ( x ) d q x d q i s + ∫ 0 η i ∫ 0 s ( η i − q i s ) ( β i − 1 ) ( s − q x ) ( α − 1 ) y ( x ) d q x d q i s ) } + ∫ 0 t ( t − q s ) ( α − 1 ) Γ q ( α ) y ( s ) d q s ,$
(3.6)

where Λ is defined by (3.3).

Proof Since $1<α≤2$, 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

$( I q α D q 2 I q 2 − α u ) (t)= ( I q α y ) (t).$

Using Lemma 2.5, we obtain

$u(t)= c 1 t α − 1 + c 2 t α − 2 + ∫ 0 t ( t − q s ) ( α − 1 ) Γ q ( α ) y(s) d q s$
(3.7)

for some constants $c 1 , c 2 ∈R$. Since $u(0)=0$, we get $c 2 =0$.

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

$I q i β i u ( ξ i ) = ∫ 0 ξ i ( ξ i − q i s ) ( β i − 1 ) Γ q i ( β i ) ( c 1 s α − 1 + ∫ 0 s ( s − q x ) ( α − 1 ) Γ q ( α ) y ( x ) d q x ) d q i s = 1 Γ q i ( β i ) Γ q ( α ) ∫ 0 ξ i ∫ 0 s ( ξ i − q i s ) ( β i − 1 ) ( s − q x ) ( α − 1 ) y ( x ) d q x d q i s + c 1 Γ q i ( β i ) ∫ 0 ξ i ( ξ i − q i s ) ( β i − 1 ) s α − 1 d q i s = 1 Γ q i ( β i ) Γ q ( α ) ∫ 0 ξ i ∫ 0 s ( ξ i − q i s ) ( β i − 1 ) ( s − q x ) ( α − 1 ) y ( x ) d q x d q i s + c 1 Γ q i ( α ) ξ i α + β i − 1 Γ q i ( α + β i ) .$
(3.8)

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

$c 1 = 1 Λ { ∑ i = 1 m γ i Γ q i ( β i ) Γ q ( α ) ( ∫ 0 ξ i ∫ 0 s ( ξ i − q i s ) ( β i − 1 ) ( s − q x ) ( α − 1 ) y ( x ) d q x d q i s − ∫ 0 η i ∫ 0 s ( η i − q i s ) ( β i − 1 ) ( s − q x ) ( α − 1 ) y ( x ) d q x d q i s ) − ∫ 0 T ( T − q s ) ( α − 1 ) Γ q ( α ) y ( s ) d q s } .$
(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 $C=C([0,T],R)$ denote the Banach space of all continuous functions from $[0,T]$ to endowed with the supremum norm defined by $∥u∥= sup t ∈ [ 0 , T ] |u(t)|$. In view of Lemma 3.3, we define an operator $A:C→C$ by

$( A u ) ( t ) = − t α − 1 Λ { ∫ 0 T ( T − q s ) ( α − 1 ) Γ q ( α ) f ( s , u ( s ) ) d q s − ∑ i = 1 m γ i Γ q i ( β i ) Γ q ( α ) ( ∫ 0 ξ i ∫ 0 s ( ξ i − q i s ) ( β i − 1 ) ( s − q x ) ( α − 1 ) f ( x , u ( x ) ) d q x d q i s + ∫ 0 η i ∫ 0 s ( η i − q i s ) ( β i − 1 ) ( s − q x ) ( α − 1 ) f ( x , u ( x ) ) d q x d q i s ) } + ∫ 0 t ( t − q s ) ( α − 1 ) Γ q ( α ) f ( s , u ( s ) ) d q s ,$
(4.1)

with $Λ≠0$. It should be noticed that problem (1.1) has solutions if and only if the operator $A$ has fixed points.

For the sake of convenience, we put

$Φ = T α − 1 | Λ | Γ q ( α + 1 ) ( T α + ∑ i = 1 m | γ i | ξ i β i + α Γ q i ( α + 1 ) Γ q i ( α + β i + 1 ) + ∑ i = 1 m | γ i | η i β i + α Γ q i ( α + 1 ) Γ q i ( α + β i + 1 ) ) + T α Γ q ( α + 1 ) .$
(4.2)

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

Theorem 4.1 Let $f:[0,T]×R→R$ be a continuous function satisfying the assumption

(H1) there exists a constant $L>0$ such that $|f(t,u)−f(t,v)|≤L|u−v|$ for each $t∈[0,T]$ and $u,v∈R$.

If

$LΦ<1,$
(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, $u=Au$, where the operator $A$ is defined by (4.1). Applying the Banach contraction mapping principle, we will show that the operator $A$ has a fixed point which is a unique solution of problem (1.1).

Setting $sup t ∈ [ 0 , T ] |f(t,0)|= M 0 <∞$ and choosing

$r≥ M 0 Φ 1 − L Φ ,$

with L Φ satisfying (4.3), we will show that $A B r ⊂ B r$, where the set $B r ={u∈C:∥u∥≤r}$. For any $u∈ B r$, and taking into account Lemma 3.2, we have

$∥ A u ∥ ≤ sup t ∈ [ 0 , T ] { t α − 1 | Λ | { ∫ 0 T ( T − q s ) ( α − 1 ) Γ q ( α ) | f ( s , u ( s ) ) | d q s + ∑ i = 1 m | γ i | Γ q i ( β i ) Γ q ( α ) ( ∫ 0 ξ i ∫ 0 s ( ξ i − q i s ) ( β i − 1 ) ( s − q x ) ( α − 1 ) | f ( x , u ( x ) ) | d q x d q i s + ∫ 0 η i ∫ 0 s ( η i − q i s ) ( β i − 1 ) ( s − q x ) ( α − 1 ) | f ( x , u ( x ) ) | d q x d q i s ) } + ∫ 0 t ( t − q s ) ( α − 1 ) Γ q ( α ) | f ( s , u ( s ) ) | d q s } ≤ T α − 1 | Λ | { ∫ 0 T ( T − q s ) ( α − 1 ) Γ q ( α ) ( | f ( s , u ( s ) ) − f ( s , 0 ) | + | f ( s , 0 ) | ) d q s + ∑ i = 1 m | γ i | Γ q i ( β i ) Γ q ( α ) ( ∫ 0 ξ i ∫ 0 s ( ξ i − q i s ) ( β i − 1 ) ( s − q x ) ( α − 1 ) × ( | f ( x , u ( x ) ) − f ( x , 0 ) | + | f ( x , 0 ) | ) d q x d q i s + ∫ 0 η i ∫ 0 s ( η i − q i s ) ( β i − 1 ) ( s − q x ) ( α − 1 ) × ( | f ( x , u ( x ) ) − f ( x , 0 ) | + | f ( x , 0 ) | ) d q x d q i s ) } + ∫ 0 T ( T − q s ) ( α − 1 ) Γ q ( α ) ( | f ( s , u ( s ) ) − f ( s , 0 ) | + | f ( s , 0 ) | ) d q s ≤ ( L r + M 0 ) { T α − 1 | Λ | Γ q ( α + 1 ) ( T α + ∑ i = 1 m | γ i | ξ i β i + α Γ q i ( α + 1 ) Γ q i ( α + β i + 1 ) + ∑ i = 1 m | γ i | η i β i + α Γ q i ( α + 1 ) Γ q i ( α + β i + 1 ) ) + T α Γ q ( α + 1 ) } = ( L r + M 0 ) Φ ≤ r .$

It follows that $A B r ⊂ B r$.

For $u,v∈C$ and for each $t∈[0,T]$, we have

$| A u ( t ) − A v ( t ) | ≤ T α − 1 | Λ | { ∫ 0 T ( T − q s ) ( α − 1 ) Γ q ( α ) ( | f ( s , u ( s ) ) − f ( s , v ( s ) ) | ) d q s + ∑ i = 1 m | γ i | Γ q i ( β i ) Γ q ( α ) ( ∫ 0 ξ i ∫ 0 s ( ξ i − q i s ) ( β i − 1 ) ( s − q x ) ( α − 1 ) × ( | f ( s , u ( s ) ) − f ( s , v ( s ) ) | ) d q x d q i s + ∫ 0 η i ∫ 0 s ( η i − q i s ) ( β i − 1 ) ( s − q x ) ( α − 1 ) ( | f ( s , u ( s ) ) − f ( s , v ( s ) ) | ) d q x d q i s ) } + ∫ 0 T ( T − q s ) ( α − 1 ) Γ q ( α ) ( | f ( s , u ( s ) ) − f ( s , v ( s ) ) | ) d q s ≤ L ∥ u − v ∥ { T α − 1 | Λ | Γ q ( α + 1 ) ( T α + ∑ i = 1 m | γ i | ξ i β i + α Γ q i ( α + 1 ) Γ q i ( α + β i + 1 ) + ∑ i = 1 m | γ i | η i β i + α Γ q i ( α + 1 ) Γ q i ( α + β i + 1 ) ) + T α Γ q ( α + 1 ) } = L Φ ∥ u − v ∥ .$

The above result leads to $∥Au−Av∥≤LΦ∥u−v∥$. As $LΦ<1$, by (4.3), therefore $A$ is a contraction. Hence, by the Banach contraction mapping principle, we deduce that $A$ 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) $Ax+By∈M$ whenever $x,y∈M$; (b) A is compact and continuous; (c) B is a contraction mapping. Then there exists $z∈M$ such that $z=Az+Bz$.

Theorem 4.3 Assume that $f:[0,T]×R→R$ is a continuous function satisfying assumption (H1). In addition, we suppose that

(H2) $|f(t,u)|≤ψ(t)$, $∀(t,u)∈[0,T]×R$ and $ψ∈C([0,T], R + )$.

If the following condition holds

$L Γ q ( α + 1 ) ( T 2 α − 1 | Λ | + T α ) <1,$
(4.4)

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

Proof We define $sup t ∈ [ 0 , T ] |ψ(t)|=∥ψ∥$ and choose a suitable constant R such that

$R≥∥ψ∥Φ,$

where Φ is defined by (4.2). Furthermore, we define the operators $A 1$ and $A 2$ on $B R ={u∈C:∥u∥≤R}$ by

$( A 1 u ) ( t ) = t α − 1 Λ ∑ i = 1 m γ i Γ q i ( β i ) Γ q ( α ) ∫ 0 ξ i ∫ 0 s ( ξ i − q i s ) ( β i − 1 ) ( s − q x ) ( α − 1 ) f ( x , u ( x ) ) d q x d q i s − t α − 1 Λ ∑ i = 1 m γ i Γ q i ( β i ) Γ q ( α ) ∫ 0 η i ∫ 0 s ( η i − q i s ) ( β i − 1 ) ( s − q x ) ( α − 1 ) f ( x , u ( x ) ) d q x d q i s ,$

and

$( A 2 u)(t)=− t α − 1 Λ ∫ 0 T ( T − q s ) ( α − 1 ) Γ q ( α ) f ( s , u ( s ) ) d q s+ ∫ 0 t ( t − q s ) ( α − 1 ) Γ q ( α ) f ( s , u ( s ) ) d q s.$

It should be noticed that $A= A 1 + A 2$.

For any $u,v∈ B R$, we have

$∥ A 1 u + A 2 v ∥ ≤ ∥ ψ ∥ { T α − 1 | Λ | Γ q ( α + 1 ) ( T α + ∑ i = 1 m | γ i | ξ i β i + α B q i ( β i , α + 1 ) Γ q i ( β i ) + ∑ i = 1 m | γ i | η i β i + α B q i ( β i , α + 1 ) Γ q i ( β i ) ) + T α Γ q ( α + 1 ) } = ∥ ψ ∥ Φ ≤ R .$

Therefore $( A 1 u)+( A 2 v)∈ B R$. Obviously, condition (4.4) implies that $A 2$ is a contraction mapping.

Finally, we will show that $A 1$ is compact and continuous. The continuity of f coupled with assumption (H2) implies that the operator $A 1$ is continuous and uniformly bounded on $B R$. We define $sup ( t , u ) ∈ [ 0 , T ] × B R |f(t,u)|= M ∗ <∞$. For $t 1 , t 2 ∈[0,T]$, $t 2 < t 1$ and $u∈ B R$, we have

$| ( A 1 u ) ( t 1 ) − ( A 1 u ) ( t 2 ) | ≤ | t 1 α − 1 − t 2 α − 1 | | Λ | ∑ i = 1 m | γ i | Γ q i ( β i ) Γ q ( α ) × ∫ 0 ξ i ∫ 0 s ( ξ i − q i s ) ( β i − 1 ) ( s − q x ) ( α − 1 ) | f ( x , u ( x ) ) | d q x d q i s + | t 1 α − 1 − t 2 α − 1 | | Λ | ∑ i = 1 m | γ i | Γ q i ( β i ) Γ q ( α ) × ∫ 0 η i ∫ 0 s ( η i − q i s ) ( β i − 1 ) ( s − q x ) ( α − 1 ) | f ( x , u ( x ) ) | d q x d q i s ≤ M ∗ | t 1 α − 1 − t 2 α − 1 | | Λ | Γ q ( α + 1 ) { ∑ i = 1 m | γ i | ξ i β i + α Γ q i ( α + 1 ) Γ q i ( α + β i + 1 ) + ∑ i = 1 m | γ i | η i β i + α Γ q i ( α + 1 ) Γ q i ( α + β i + 1 ) } .$

Actually, as $| t 1 − t 2 |→0$ the right-hand side of the above inequality tends to zero independently of u. So $A 1$ is relatively compact on $B R$. Therefore, by the Arzelá-Ascoli theorem, $A 1$ 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 $A 1$ and $A 2$ to obtain the second result replacing (4.4) by the following condition:

$L T α − 1 | Λ | Γ q ( α + 1 ) ( ∑ i = 1 m | γ i | ξ i β i + α Γ q i ( α + 1 ) Γ q i ( α + β i + 1 ) + ∑ i = 1 m | γ i | η i β i + α Γ q i ( α + 1 ) Γ q i ( α + β i + 1 ) ) <1.$

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 $0∈U$. Suppose that $F: U ¯ →C$ is a continuous, compact (that is, $F( U ¯ )$ is a relatively compact subset of C) map. Then either

1. (i)

F has a fixed point in $U ¯$, or

2. (ii)

there is $u∈∂U$ (the boundary of U in C) and $λ∈(0,1)$ with $u=λF(u)$.

Theorem 4.6 Assume that $f:[0,T]×R→R$ is a continuous function. In addition we suppose that:

(H3) there exist a continuous nondecreasing function $ϕ:[0,∞)→(0,∞)$ and a function $p∈C([0,T], R + )$ such that

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

$N ∥ p ∥ ϕ ( N ) Φ >1,$

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 $A$ defined by (4.1) maps bounded sets (balls) into bounded sets in $C$. For a positive number ρ, let $B ρ ={u∈C:∥u∥≤ρ}$ be a bounded ball in $C$. Then, for $t∈[0,T]$, we have

$| A u ( t ) | ≤ T α − 1 | Λ | { ∫ 0 T ( T − q s ) ( α − 1 ) Γ q ( α ) | f ( s , u ( s ) ) | d q s + ∑ i = 1 m | γ i | Γ q i ( β i ) Γ q ( α ) ( ∫ 0 ξ i ∫ 0 s ( ξ i − q i s ) ( β i − 1 ) ( s − q x ) ( α − 1 ) | f ( x , u ( x ) ) | d q x d q i s + ∫ 0 η i ∫ 0 s ( η i − q i s ) ( β i − 1 ) ( s − q x ) ( α − 1 ) | f ( x , u ( x ) ) | d q x d q i s ) } + ∫ 0 T ( T − q s ) ( α − 1 ) Γ q ( α ) | f ( s , u ( s ) ) | d q s ≤ T α − 1 | Λ | Γ q ( α + 1 ) ( ∥ p ∥ ϕ ( ∥ u ∥ ) T α + ∥ p ∥ ϕ ( ∥ u ∥ ) ∑ i = 1 m | γ i | ξ i β i + α Γ q i ( α + 1 ) Γ q i ( α + β i + 1 ) + ∥ p ∥ ϕ ( ∥ u ∥ ) ∑ i = 1 m | γ i | η i β i + α Γ q i ( α + 1 ) Γ q i ( α + β i + 1 ) ) + ∥ p ∥ ϕ ( ∥ u ∥ ) T α Γ q ( α + 1 ) ≤ ∥ p ∥ ϕ ( ρ ) T α − 1 | Λ | Γ q ( α + 1 ) ( T α + ∑ i = 1 m | γ i | ξ i β i + α Γ q i ( α + 1 ) Γ q i ( α + β i + 1 ) + ∑ i = 1 m | γ i | η i β i + α Γ q i ( α + 1 ) Γ q i ( α + β i + 1 ) ) + ∥ p ∥ ϕ ( ρ ) T α Γ q ( α + 1 ) : = K .$

Therefore, we deduce that $∥Au∥≤K$.

Secondly, we will show that $A$ maps bounded sets into equicontinuous sets of $C$. Let $sup ( t , u ) ∈ [ 0 , T ] × B ρ |f(t,u)|= K ∗ <∞$, $τ 1 , τ 2 ∈[0,T]$ with $τ 2 < τ 1$ and $u∈ B ρ$. Then we have

$| ( A u ) ( τ 1 ) − ( A u ) ( τ 2 ) | ≤ | τ 1 α − 1 − τ 2 α − 1 | | Λ | { ∫ 0 T ( T − q s ) ( α − 1 ) Γ q ( α ) | f ( s , u ( s ) ) | d q s + ∑ i = 1 m | γ i | Γ q i ( β i ) Γ q ( α ) ( ∫ 0 ξ i ∫ 0 s ( ξ i − q i s ) ( β i − 1 ) ( s − q x ) ( α − 1 ) | f ( x , u ( x ) ) | d q x d q i s + ∫ 0 η i ∫ 0 s ( η i − q i s ) ( β i − 1 ) ( s − q x ) ( α − 1 ) | f ( x , u ( x ) ) | d q x d q i s ) } + | ∫ 0 τ 1 ( τ 1 − q s ) ( α − 1 ) Γ q ( α ) | f ( s , u ( s ) ) | d q s − ∫ 0 τ 2 ( τ 2 − q s ) ( α − 1 ) Γ q ( α ) | f ( s , u ( s ) ) | d q s | ≤ | τ 1 α − 1 − τ 2 α − 1 | K ∗ | Λ | Γ q ( α + 1 ) ( T α + ∑ i = 1 m | γ i | ξ i β i + α Γ q i ( α + 1 ) Γ q i ( α + β i + 1 ) + ∑ i = 1 m | γ i | η i β i + α Γ q i ( α + 1 ) Γ q i ( α + β i + 1 ) ) + | τ 1 α − 1 − τ 2 α − 1 | K ∗ Γ q i ( α + 1 ) .$

Obviously, the right-hand side of the above inequality tends to zero independently of $x∈ B ρ$ as $τ 2 → τ 1$. Therefore it follows by the Arzelá-Ascoli theorem that $A:C→C$ is completely continuous.

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

$∥ u ∥ ≤ T α − 1 | Λ | Γ q ( α + 1 ) ( ∥ p ∥ ϕ ( ∥ u ∥ ) T α + ∥ p ∥ ϕ ( ∥ u ∥ ) ∑ i = 1 m | γ i | ξ i β i + α Γ q i ( α + 1 ) Γ q i ( α + β i + 1 ) + ∥ p ∥ ϕ ( ∥ u ∥ ) ∑ i = 1 m | γ i | η i β i + α Γ q i ( α + 1 ) Γ q i ( α + β i + 1 ) ) + ∥ p ∥ ϕ ( ∥ u ∥ ) T α Γ q ( α + 1 ) = ∥ p ∥ ϕ ( ∥ u ∥ ) Φ .$

Consequently, we have

$∥ u ∥ ∥ p ∥ ϕ ( ∥ u ∥ ) Φ ≤1.$

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

$U= { x ∈ C : ∥ u ∥ < N } .$

Note that the operator $A: U ¯ →C$ is continuous and completely continuous. From the choice of U, there is no $u∈∂U$ such that $u=λAu$ for some $λ∈(0,1)$. Consequently, by nonlinear alternative of Leray-Schauder type (Lemma 4.5), we deduce that $A$ has a fixed point in $U ¯$, 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 $f:[0,T]×R→R$ be a continuous function. Assume that

(H5) there exist constants $0≤ω< Φ − 1$, where Φ are given by (4.2), and $Ψ>0$ such that

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

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

Setting a ball $B ρ ∗ ⊂C$, where a constant radius $ρ ∗ >0$, by

$B ρ ∗ = { u ∈ C : sup t ∈ [ 0 , T ] | u ( t ) | < ρ ∗ } ,$

it is sufficient to show that $A: B ¯ ρ ∗ →C$ satisfies

$u≠θAu,∀u∈∂ B ρ ∗ ,∀θ∈[0,1].$
(4.5)

Now, we set

$H(θ,u)=θAu,u∈C,θ∈[0,1].$

As shown in Theorem 4.6, we have that the operator $A$ is continuous, uniformly bounded and equicontinuous. Then, by the Arzelá-Ascoli theorem, a continuous map $h θ (u)=u−H(θ,u)=u−θAu$ 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

$deg ( h θ , B ρ ∗ , 0 ) = deg ( I − θ A , B ρ ∗ , 0 ) = deg ( h 1 , B ρ ∗ , 0 ) = deg ( h 0 , B ρ ∗ , 0 ) = deg ( I , B ρ ∗ , 0 ) = 1 ≠ 0 , 0 ∈ B ρ ∗ ,$

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 $u∈ B ρ ∗$. Let us assume that $u=θAu$ for some $θ∈[0,1]$. Then, for all $t∈[0,T]$, we have

$| u ( t ) | = | θ ( A u ) ( t ) | ≤ T α − 1 | Λ | { ∫ 0 T ( T − q s ) ( α − 1 ) Γ q ( α ) | f ( s , u ( s ) ) | d q s + ∑ i = 1 m | γ i | Γ q i ( β i ) Γ q ( α ) ( ∫ 0 ξ i ∫ 0 s ( ξ i − q i s ) ( β i − 1 ) ( s − q x ) ( α − 1 ) | f ( x , u ( x ) ) | d q x d q i s + ∫ 0 η i ∫ 0 s ( η i − q i s ) ( β i − 1 ) ( s − q x ) ( α − 1 ) | f ( x , u ( x ) ) | d q x d q i s ) } + ∫ 0 T ( T − q s ) ( α − 1 ) Γ q ( α ) | f ( s , u ( s ) ) | d q s ≤ ( ω | u | + Ψ ) { T α − 1 | Λ | Γ q ( α + 1 ) ( T α + ∑ i = 1 m | γ i | ξ i β i + α B q i ( β i , α + 1 ) Γ q i ( β i ) + ∑ i = 1 m | γ i | η i β i + α B q i ( β i , α + 1 ) Γ q i ( β i ) ) + T α Γ q ( α + 1 ) } = ( ω | u | + Ψ ) Φ .$

Taking norm $sup t ∈ [ 0 , T ] |u(t)|=∥u∥$ and solving for $∥u∥$, we get

$∥u∥≤ Ψ Φ 1 − ω Φ .$

Choosing $ρ ∗ = Ψ Φ 1 − ω Φ +1$, 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:

${ D 1 2 7 4 u ( t ) = 4 | u ( t ) | ( 5 + t ) 2 ( 1 + | u ( t ) | ) , t ∈ ( 0 , 1 ) , u ( 0 ) = 0 , u ( 1 ) = 2 ( I 1 4 2 3 u ) | 1 6 1 5 + 3 4 ( I 1 2 4 5 u ) | 2 5 2 3 + 10 ( I 1 5 7 6 u ) | 1 2 1 .$
(5.1)

Here $α=7/4$, $q=1/2$, $T=1$, $m=3$, $γ 1 =2$, $γ 2 =3/4$, $γ 3 =10$, $β 1 =2/3$, $β 2 =4/5$, $β 3 =7/6$, $q 1 =1/4$, $q 2 =1/2$, $q 3 =1/5$, $ξ 1 =1/5$, $ξ 2 =2/3$, $ξ 3 =1$, $η 1 =1/6$, $η 2 =2/5$, $η 3 =1/2$ and $f(t,u)=(4|u(t)|)/( ( 5 + t ) 2 (1+|u(t)|))$. Since

$| f ( t , u ) − f ( t , v ) | ≤ 4 25 |u−v|,$

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

$Λ = T α − 1 − ∑ i = 1 m γ i Γ q i ( α ) Γ q i ( α + β i ) ( ξ i α + β i − 1 − η i α + β i − 1 ) Λ ≈ − 5.259895840 , Φ = T α − 1 | Λ | Γ q ( α + 1 ) ( T α + ∑ i = 1 m | γ i | ξ i β i + α Γ q i ( α + 1 ) Γ q i ( α + β i + 1 ) + ∑ i = 1 m | γ i | η i β i + α Γ q i ( α + 1 ) Γ q i ( α + β i + 1 ) ) Φ = + T α Γ q ( α + 1 ) Φ ≈ 2.200354723 .$

Therefore, we get

$LΦ= 4 25 (2.200354723)≈0.352056756<1.$

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:

${ D 5 8 4 3 u ( t ) = 1 u 2 + 3 π 2 sin ( π u 4 ) + 1 20 π ( 5 + sin ( π t ) ) , t ∈ ( 0 , 2 ) , u ( 0 ) = 0 , u ( 2 ) = ( I 1 3 1 2 u ) | 1 5 1 4 + 2 3 ( I 1 2 6 5 u ) | 1 2 − 3 ( I 1 8 3 2 u ) | 1 3 1 + 4 5 ( I 1 4 3 4 u ) | 2 5 1 2 .$
(5.2)

Here $α=4/3$, $q=5/8$, $T=2$, $m=4$, $γ 1 =1$, $γ 2 =2/3$, $γ 3 =−3$, $γ 4 =4/5$, $β 1 =1/2$, $β 2 =6/5$, $β 3 =3/2$, $β 4 =3/4$, $q 1 =1/3$, $q 2 =1/2$, $q 3 =1/8$, $q 4 =1/4$, $ξ 1 =1/4$, $ξ 2 =2$, $ξ 3 =1$, $ξ 4 =1/2$, $η 1 =1/5$, $η 2 =1$, $η 3 =1/3$, $η 4 =2/5$ and $f(t,u)=((sin(πu/4))/( u 2 +3 π 2 ))+((5+sin(πt))/(20π))$. By using the Maple program, we find that

$Λ = T α − 1 − ∑ i = 1 m γ i Γ q i ( α ) Γ q i ( α + β i ) ( ξ i α + β i − 1 − η i α + β i − 1 ) Λ ≈ 2.448357686 , Φ = T α − 1 | Λ | Γ q ( α + 1 ) ( T α + ∑ i = 1 m | γ i | ξ i β i + α Γ q i ( α + 1 ) Γ q i ( α + β i + 1 ) + ∑ i = 1 m | γ i | η i β i + α Γ q i ( α + 1 ) Γ q i ( α + β i + 1 ) ) Φ = + T α Γ q ( α + 1 ) Φ ≈ 5.843174987 .$

Clearly,

$| f ( t , u ) | =| 1 u 2 + 3 π 2 sin ( π u 4 ) + 5 + sin ( π t ) 20 π |≤ ( 5 + sin ( π t ) ) ( 5 | u | + 3 60 π ) .$

Choosing $p(t)=5+sin(πt)$ and $ψ(|u|)=(5|u|+3)/(60π)$, we can show that

$N ( 6 ) ( 5 N + 3 60 π ) ( 3.493621065 ) >1,$

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:

${ D 3 8 7 5 u ( t ) = 1 12 π tan − 1 ( 3 u ) + 2 | u ( t ) | 1 + | u ( t ) | , t ∈ ( 0 , 3 ) , u ( 0 ) = 0 , u ( 3 ) = − 2 ( I 1 2 6 5 u ) | 3 2 5 3 + 11 ( I 1 4 7 3 u ) | 1 5 7 4 + ( 9 5 ) ( I 2 3 6 7 u ) | 5 4 5 2 u ( 3 ) = + 5 ( I 2 5 3 4 u ) | 3 7 1 2 + ( 4 7 ) ( I 3 5 2 3 u ) | 1 3 9 5 .$
(5.3)

Here $α=7/5$, $q=3/8$, $T=3$, $m=5$, $γ 1 =−2$, $γ 2 =11$, $γ 3 =9/5$, $γ 4 =5$, $γ 5 =4/7$, $β 1 =6/5$, $β 2 =7/3$, $β 3 =6/7$, $β 4 =3/4$, $β 5 =2/3$, $q 1 =1/2$, $q 2 =1/4$, $q 3 =2/3$, $q 4 =2/5$, $q 5 =3/5$, $ξ 1 =5/3$, $ξ 2 =7/4$, $ξ 3 =5/2$, $ξ 4 =1/2$, $ξ 5 =9/5$, $η 1 =3/2$, $η 2 =1/5$, $η 3 =5/4$, $η 4 =3/7$, $η 5 =1/3$ and $f(t,u)=(( tan − 1 (3u))/(12π))+((2|u(t)|)/(1+|u(t)|))$. By using the Maple program, we find that

$Λ = T α − 1 − ∑ i = 1 m γ i Γ q i ( α ) Γ q i ( α + β i ) ( ξ i α + β i − 1 − η i α + β i − 1 ) Λ ≈ − 33.26381181 , Φ = T α − 1 | Λ | Γ q ( α + 1 ) ( T α + ∑ i = 1 m | γ i | ξ i β i + α Γ q i ( α + 1 ) Γ q i ( α + β i + 1 ) + ∑ i = 1 m | γ i | η i β i + α Γ q i ( α + 1 ) Γ q i ( α + β i + 1 ) ) Φ = + T α Γ q ( α + 1 ) Φ ≈ 7.22159847 .$

We observe that

$| f ( t , u ) | =| 1 12 π tan − 1 (3u)+ 2 | u ( t ) | 1 + | u ( t ) | |≤ | u | 4 π +2.$

Therefore, we have $Ψ=2$ and

$ω=1/4π< Φ − 1 =0.13847350.$

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.

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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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### Competing interests

The authors declare that they have no competing interests.

### 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