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Article

# Nontrivial Solutions for a System of Fractional q-Difference Equations Involving q-Integral Boundary Conditions

1
School of Mathematics and Statistics, Suzhou University, Suzhou 234000, Anhui, China
2
School of Mathematics and Information Science, Henan Polytechnic University, Jiaozuo 454000, Henan, China
3
School of Mathematics, Statistics and Applied Mathematics, National University of Ireland, H91 CF50 Galway, Ireland
4
School of Mathematical Sciences, Chongqing Normal University, Chongqing 401331, China
*
Author to whom correspondence should be addressed.
Mathematics 2020, 8(5), 828; https://doi.org/10.3390/math8050828
Received: 24 April 2020 / Revised: 12 May 2020 / Accepted: 13 May 2020 / Published: 20 May 2020
(This article belongs to the Special Issue Mathematical Analysis and Boundary Value Problems)

## Abstract

:
In this paper, we study the existence of nontrivial solutions for a system of fractional q-difference equations involving q-integral boundary conditions, and we use the topological degree to establish our main results by considering the first eigenvalue of some associated linear integral operators.

## 1. Introduction

The initial work of q-difference calculus can be dated back to Jackson [1,2] and for results on fractional q-difference calculus or quantum calculus we refer the reader to [3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,28] and the references therein. For example, in [3,4] the author studied some basic properties of fractional q-fractional integral and differential operators, and used fixed point theorems in cones to investigate the existence of nontrivial solutions for the fractional q-difference equation
$D q α y ( x ) = − f ( x , y ( x ) ) , 0 < x < 1 ,$
with the boundary conditions
$y ( 0 ) = y ( 1 ) = 0 , as α ∈ ( 1 , 2 ] ,$
or
$y ( 0 ) = D q y ( 0 ) = 0 , D q y ( 1 ) = β ≥ 0 , as α ∈ ( 2 , 3 ] .$
In  the author considered the three-point boundary value problem of fractional q-difference equations
$D q α + f ( t , x ( t ) , x ( t ) ) + g ( t , x ( t ) ) = 0 , 0 < t < 1 , x ( 0 ) = D q x ( 0 ) = 0 , D q x ( 1 ) = β D q x ( η ) ,$
where $β η α − 2 ∈ ( 0 , 1 )$, $q ∈ ( 0 , 1 )$, $α ∈ ( 2 , 3 )$ and based on fixed point theorems on mixed monotone operators, some sufficient conditions are used to guarantee the existence and uniqueness of positive solutions for the above problem. In  the authors discussed the following nonhomogeneous boundary value problem with fractional q-derivatives
$( D q α u ) ( t ) + f ( t , u ( t ) ) = 0 , t ∈ ( 0 , 1 ) , u ( 0 ) = D q u ( 0 ) = 0 , γ D q u ( 1 ) + β D q 2 u ( 1 ) = λ ,$
where $q ∈ ( 0 , 1 ) , 2 < α ≤ 3 , γ ≥ 0 , β > 0 ,$ and $λ$ is a parameter. Using the generalized Banach contraction principle and Krasnoselskii’s fixed point theorem, uniqueness, existence, and multiplicity of positive solutions for the above problem were obtained in terms of explicit intervals for the nonhomogeneous term.
Coupled systems of fractional q-difference equations were investigated in [23,24,25,26,27,28] (also see [29,30,31,32,33,34,35]). In  the authors studied the following system of fractional q-difference equations with four-point boundary conditions
$D q α u ( t ) + f ( t , v ( t ) ) = 0 , 0 < t < 1 , D q β v ( t ) + g ( t , u ( t ) ) = 0 , 0 < t < 1 , u ( 0 ) = 0 , u ( 1 ) = γ 1 u η 1 , v ( 0 ) = 0 , v ( 1 ) = γ 2 u η 2 ,$
where $0 < q < 1$, $1 < β ≤ α ≤ 2 , 0 < η 1 , η 2 < 1 , 0 < γ 1 η 1 α − 1 < 1 ,$$0 < γ 2 η 2 β − 1 < 1$ and using the monotone iterative approach they constructed two convergent monotone iterative schemes and obtained two positive solutions for the above problem. In  the authors studied the coupled system of fractional q-integro-difference equations with nonlocal fractional q-integral boundary conditions
$D q α x ( t ) = f t , x ( t ) , I r δ y ( t ) , t ∈ [ 0 , T ] , 1 < α ≤ 2 , D p β y ( t ) = g t , y ( t ) , I z ε x ( t ) , t ∈ [ 0 , T ] , 1 < β ≤ 2 , x ( 0 ) = 0 , λ 1 I m γ x ( η ) = I n κ y ( ξ ) , y ( 0 ) = 0 , λ 2 I h μ y ( θ ) = I k v x ( τ ) ,$
where $0 < p , q , r , z , m , n , h , k < 1$, $η , ξ , θ , τ ∈ ( 0 , T )$, $δ , ε , γ , κ , μ , v > 0 ,$ and $λ 1 , λ 2 ∈ R$ are given constants, $I ϕ ψ$ is the fractional $ϕ$-integral of order $ψ$ with $ϕ ∈ { r , z , m , n , h , k }$ and $ψ ∈ { δ , ε , γ , κ , μ , v }$. Using the Banach contraction principle and the Leray-Schauder alternative, they obtained existence and uniqueness of solutions under some appropriate conditions on $f , g$.
Motivated by the mentioned works above, in this paper we use topological degree theory to study nontrivial solutions for the following system of fractional q-difference equations with q-integral boundary conditions:
$D q α x ( t ) + f 1 ( t , y ( t ) ) = 0 , t ∈ ( 0 , 1 ) , D q α y ( t ) + f 2 ( t , x ( t ) ) = 0 , t ∈ ( 0 , 1 ) , x ( 0 ) = 0 , D q x ( 0 ) = 0 , D q v x ( 1 ) = ∫ 0 1 h ( t ) D q v x ( t ) d q t , y ( 0 ) = 0 , D q y ( 0 ) = 0 , D q v y ( 1 ) = ∫ 0 1 h ( t ) D q v y ( t ) d q t ,$
where $α ∈ ( 2 , 3 ) , v ∈ ( 1 , 2 )$, $D q α$ is the $α$-order Riemann-Liouville’s fractional q-derivative.
Now, we list our assumptions for h, $f i ( i = 1 , 2 )$:
Hypothesis 1 (H1).
$h ≥ 0$and$1 − ∫ 0 1 h ( t ) t α − v − 1 d q t : = A > 0$.
Hypothesis 2 (H2).
$f i ∈ C [ 0 , 1 ] × R , R$.
Hypothesis 3 (H3).
There exist $b i ( t ) , c i ( t ) ∈ C ( [ 0 , 1 ] , R + )$ with $c i ( t ) ≢ 0$ and $K 1 y , K 2 ( x ) ∈ C R , R +$ such that
$f 1 t , y ≥ − b 1 ( t ) − c 1 ( t ) K 1 y , f 2 t , x ≥ − b 2 ( t ) − c 2 ( t ) K 2 x , ∀ x , y ∈ R , t ∈ [ 0 , 1 ] , i = 1 , 2 .$
Hypothesis 4 (H4).
$lim y → + ∞ K 1 y y = 0$, $lim x → + ∞ K 2 x x = 0$.
Hypothesis 5 (H5).
$lim inf y → + ∞ f 1 t , y y > λ 1 ,$$lim inf x → + ∞ f 2 t , x x > λ 1 ,$ uniformly for $t ∈ [ 0 , 1 ]$.
Hypothesis 6 (H6).
$lim sup y → 0 f 1 t , y y < λ 1 ,$$lim sup x → 0 f 2 t , x x < λ 1 ,$ uniformly for $t ∈ [ 0 , 1 ]$, where $λ 1$ is the first eigenvalue of the following eigenvalue problem
$D q α x ( t ) + λ x ( t ) = 0 , t ∈ ( 0 , 1 ) , x ( 0 ) = 0 , D q x ( 0 ) = 0 , D q v x ( 1 ) = ∫ 0 1 h ( t ) D q v x ( t ) d q t ,$
where λ is a parameter, and $α , v , h$ are as in (8).
Finally, we state our main result in this paper:
Theorem 1.
Suppose that (H1)–(H6) hold. Then (8) has at least one nontrivial solution.
Remark 1.
In (H1), the function h is a non-negative function on $[ 0 , 1 ]$ (it also can be a zero function). If $h ≢ 0$ on $[ 0 , 1 ]$, we demand that $∫ 0 1 h ( t ) t α − v − 1 d q t ∈ [ 0 , 1 )$.

## 2. Preliminaries

Let $q ∈ ( 0 , 1 )$ and define
$[ a ] q = 1 − q a 1 − q , a ∈ R .$
The q-analogue of the power function $( a − b ) n$ with $n ∈ N 0$ is
$( a − b ) 0 = 1 , ( a − b ) n = ∏ k = 0 n − 1 a − b q k , n ∈ N , a , b ∈ R .$
More generally, if $α ∈ R ,$ then
$( a − b ) ( α ) = a α ∏ n = 0 ∞ a − b q n a − b q α + n .$
Please note that if $b = 0$ then $a ( α ) = a α .$ The q-gamma function is defined by
$Γ q ( x ) = ( 1 − q ) ( x − 1 ) ( 1 − q ) x − 1 , x ∈ R \ { 0 , − 1 , − 2 , … } ,$
and satisfies $Γ q ( x + 1 ) = [ x ] q Γ q ( x )$.
Definition 1
(see , Definition 2.2). Let $α ≥ 0$ and f be a function defined on $[ 0 , 1 ] .$ The fractional q-integral of the Riemann-Liouville type is $I q 0 f ( x ) = f ( x )$ and
$I q α f ( x ) = 1 Γ q ( α ) ∫ 0 x ( x − q t ) ( α − 1 ) f ( t ) d q t , α > 0 , x ∈ [ 0 , 1 ] .$
The fractional q-derivative of order $α ≥ 0$ is defined by $D q 0 f ( x ) = f ( x )$ and $D q α f ( x ) = D q m I q m − α f ( x )$ for $α > 0 ,$ where m is the smallest integer greater or equal than α.
Lemma 1
(see , Lemma 2.3). Let $α , β ≥ 0$ and f be a function defined on $[ 0 , 1 ] .$ Then, the next formulas hold:
(i) $I q β I q α f ( x ) = I q α + β f ( x )$;
(ii) $D q α I q α f ( x ) = f ( x )$.
Lemma 2
(see , Theorem 2.4). Let $α > 0$ and p be a positive integer. Then, the following equality holds:
$I q α D q p f ( x ) = D q p I q α f ( x ) − ∑ k = 0 p − 1 x α − p + k Γ q ( α + k − p + 1 ) D q k f ( 0 ) .$
Lemma 3.
Suppose that (H1) holds, and $α ∈ ( 2 , 3 ) , v ∈ ( 1 , 2 )$. If $g ∈ C [ 0 , 1 ] ,$ then the following boundary value problem
$D q α x ( t ) + g ( t ) = 0 , t ∈ ( 0 , 1 ) , x ( 0 ) = 0 , D q x ( 0 ) = 0 , D q v x ( 1 ) = ∫ 0 1 h ( t ) D q v x ( t ) d q t ,$
has a unique solution
$x ( t ) = ∫ 0 1 G ( t , q s ) g ( s ) d q s ,$
where
$G ( t , q s ) = G 0 ( t , q s ) + t α − 1 A ∫ 0 1 h ( t ) G 1 ( t , q s ) d q t , G 0 ( t , q s ) = 1 Γ q ( α ) t α − 1 ( 1 − q s ) ( α − v − 1 ) − ( t − q s ) ( α − 1 ) , 0 ≤ s ≤ t ≤ 1 , t α − 1 ( 1 − q s ) ( α − v − 1 ) , 0 ≤ t ≤ s ≤ 1 , G 1 ( t , q s ) = 1 Γ q ( α ) t α − v − 1 ( 1 − q s ) ( α − v − 1 ) − ( t − q s ) ( α − v − 1 ) , 0 ≤ s ≤ t ≤ 1 , t α − v − 1 ( 1 − q s ) ( α − v − 1 ) , 0 ≤ t ≤ s ≤ 1 .$
Proof.
Using Definition 2 and Lemmas 2 and 3 we have
$x ( t ) = c 1 t α − 1 + c 2 t α − 2 + c 3 t α − 3 − I q α g ( t ) , c i ∈ R , i = 1 , 2 , 3 .$
Note from $x ( 0 ) = D q x ( 0 ) = 0$ we have $c 2 = c 3 = 0 .$ Hence,
$x ( t ) = c 1 t α − 1 − I q α g ( t ) ,$
and
$D q v x ( t ) = Γ q ( α ) Γ q ( α − v ) c 1 t α − v − 1 − 1 Γ q ( α − v ) ∫ 0 t ( t − q s ) ( α − v − 1 ) g ( s ) d q s .$
Consequently, we have
$D q v x ( 1 ) = Γ q ( α ) Γ q ( α − v ) c 1 − 1 Γ q ( α − v ) ∫ 0 1 ( 1 − q s ) ( α − v − 1 ) g ( s ) d q s = ∫ 0 1 h ( t ) D q v x ( t ) d q t = Γ q ( α ) Γ q ( α − v ) c 1 ∫ 0 1 h ( t ) t α − v − 1 d q t − 1 Γ q ( α − v ) ∫ 0 1 h ( t ) ∫ 0 t ( t − q s ) ( α − v − 1 ) g ( s ) d q s d q t .$
This, together with (H1), implies that
$c 1 = 1 A Γ q ( α ) ∫ 0 1 ( 1 − q s ) ( α − v − 1 ) g ( s ) d q s − 1 A Γ q ( α ) ∫ 0 1 h ( t ) ∫ 0 t ( t − q s ) ( α − v − 1 ) g ( s ) d q s d q t .$
Thus, we have
$x ( t ) = 1 A Γ q ( α ) ∫ 0 1 t α − 1 ( 1 − q s ) ( α − v − 1 ) g ( s ) d q s − t α − 1 A Γ q ( α ) ∫ 0 1 h ( t ) ∫ 0 t ( t − q s ) ( α − v − 1 ) g ( s ) d q s d q t − 1 Γ q ( α ) ∫ 0 t ( t − q s ) ( α − 1 ) g ( s ) d q s = 1 A Γ q ( α ) ∫ 0 1 t α − 1 ( 1 − q s ) ( α − v − 1 ) g ( s ) d q s − t α − 1 A Γ q ( α ) ∫ 0 1 h ( t ) ∫ 0 t ( t − q s ) ( α − v − 1 ) g ( s ) d q s d q t − 1 Γ q ( α ) ∫ 0 t ( t − q s ) ( α − 1 ) g ( s ) d q s + 1 Γ q ( α ) ∫ 0 1 t α − 1 ( 1 − q s ) ( α − v − 1 ) g ( s ) d q s − 1 Γ q ( α ) ∫ 0 1 t α − 1 ( 1 − q s ) ( α − v − 1 ) g ( s ) d q s = ∫ 0 1 G 0 ( t , q s ) g ( s ) d q s + t α − 1 A Γ q ( α ) ∫ 0 1 h ( t ) t α − v − 1 d q t ∫ 0 1 ( 1 − q s ) ( α − v − 1 ) g ( s ) d q s − t α − 1 A Γ q ( α ) ∫ 0 1 h ( t ) ∫ 0 t ( t − q s ) ( α − v − 1 ) g ( s ) d q s d q t = ∫ 0 1 G 0 ( t , q s ) g ( s ) d q s + t α − 1 A Γ q ( α ) ∫ 0 1 ∫ 0 1 h ( t ) t α − v − 1 ( 1 − q s ) ( α − v − 1 ) d q t − ∫ s 1 h ( t ) ( t − q s ) ( α − v − 1 ) d q t g ( s ) d q s = ∫ 0 1 G 0 ( t , q s ) g ( s ) d q s + t α − 1 A ∫ 0 1 ∫ 0 1 h ( t ) G 1 ( t , q s ) d q t g ( s ) d q s = ∫ 0 1 G ( t , q s ) g ( s ) d q s .$
This completes the proof. □
Lemma 4
(see (, Lemma 2.2), (, Lemma 3.0.7), (, Lemma 2.7)). The functions $G i ( i = 0 , 1 )$ has the following properties
(i)
$G i ( t , q s ) ≥ 0$for$t , s ∈ [ 0 , 1 ]$,
(ii)
$t α − 1 G 0 ( 1 , q s ) ≤ G 0 ( t , q s ) ≤ G 0 ( 1 , q s )$for$t , s ∈ [ 0 , 1 ]$.
Lemma 5.
The function G satisfies
$t α − 1 φ 1 ( q s ) ≤ G ( t , q s ) ≤ φ 1 ( q s ) , for t , s ∈ [ 0 , 1 ] ,$
and
$G ( t , q s ) ≤ t α − 1 φ 2 ( q s ) , for t , s ∈ [ 0 , 1 ] ,$
where
$φ 1 ( s ) = G 0 ( 1 , s ) + 1 A ∫ 0 1 h ( t ) G 1 ( t , s ) d q t , s ∈ [ 0 , 1 ] ,$
and
$φ 2 ( s ) = 1 Γ q ( α ) ( 1 − s ) ( α − v − 1 ) + 1 A ∫ 0 1 h ( t ) G 1 ( t , s ) d q t , s ∈ [ 0 , 1 ] .$
This is the direct result from Lemma 4, so we omit the proof.
Let $E : = C [ 0 , 1 ] , ∥ x ∥ : = max t ∈ [ 0 , 1 ] | x ( t ) |$ and $P : = { x ∈ E :$$x ( t ) ≥ 0 , ∀ t ∈ [ 0 , 1 ] } .$ Then $( E , ∥ · ∥ )$ is a real Banach space and P is a cone on $E .$ Moreover, $E 2 = E × E$ is a Banach space with the norm $∥ ( u , v ) ∥ = ∥ u ∥ + ∥ v ∥$, and $P 2 = P × P$ is a cone on $E 2$. From Lemma 3 we can define operators $T i ( i = 1 , 2 ) : E → E$, and $T : E 2 → E 2$ as follows:
$( T 1 y ) ( t ) : = ∫ 0 1 G ( t , q s ) f 1 ( s , y ( s ) ) d q s ,$
$( T 2 x ) ( t ) : = ∫ 0 1 G ( t , q s ) f 2 ( s , x ( s ) ) d q s ,$
and
$T ( x , y ) ( t ) = ( ( T 1 y ) , ( T 2 x ) ) ( t ) , t ∈ [ 0 , 1 ] , x , y ∈ E ,$
where G is determined in Lemma 3. Please note that $T i ( i = 1 , 2 )$ and T are completely continuous operators, and $( x , y )$ solves (8) if and only if $( x , y )$ is a fixed point of the operator $T .$
In addition, from Lemma 3 we can obtain that (9) is equivalent to
$x ( t ) = λ ∫ 0 1 G ( t , q s ) x ( s ) d q s , t ∈ [ 0 , 1 ] .$
For our purposes, we need to define the operator L by
$( L x ) ( t ) = ∫ 0 1 G ( t , q s ) x ( s ) d q s , t ∈ [ 0 , 1 ] , x ∈ E .$
It is not difficult to prove that $L : E → E$ is a linear completely continuous and $T ( P ) ⊂ P$. From Lemmas 2 and 3 in  we obtain that the spectral radius, denoted by $r ( L )$, is not equal to 0, and L has a positive eigenfunction $φ ∗$ corresponding to its first eigenvalue $λ 1 = ( r ( L ) ) − 1$, i.e., $φ ∗ = λ 1 L φ ∗$.
Lemma 6.
Let $P 0 = { x ∈ P : x ( t ) ≥ t α − 1 ∥ x ∥ , ∀ t ∈ [ 0 , 1 ] }$. Then $L ( P ) ⊂ P 0$.
Proof.
If $x ∈ P$, and from (11) we have
$t α − 1 ∫ 0 1 φ 1 ( q s ) x ( s ) d q s ≤ ∫ 0 1 G ( t , q s ) x ( s ) d q s ≤ ∫ 0 1 φ 1 ( q s ) x ( s ) d q s , for t ∈ [ 0 , 1 ] .$
Therefore, we have
$( L x ) ( t ) ≥ t α − 1 ∫ 0 1 φ 1 ( q s ) x ( s ) d q s ≥ t α − 1 ∥ L x ∥ , for t ∈ [ 0 , 1 ] .$
This completes the proof. □
Remark 2.
From Lemma 6 we have $φ ∗ ∈ P 0$.
We recall the following topological degree theorems, which will play important roles in proving our main results.
Lemma 7
(see (, Theorem A.3.3)). [Let Ω be a bounded open set in a Banach space E, and $T : Ω → E$ a continuous compact operator. If there exists $x 0 ∈ E \ { 0 }$ such that
$x − T x ≠ μ x 0 , ∀ x ∈ ∂ Ω , μ ≥ 0 ,$
then the topological degree $deg ( I − T , Ω , 0 ) = 0$.
Lemma 8
(see (, Lemma 2.5.1)). Let Ω be a bounded open set in a Banach space E with $0 ∈ Ω$, and $T : Ω → E$ a continuous compact operator. If
$T x ≠ μ x , ∀ x ∈ ∂ Ω , μ ≥ 1 ,$
then the topological degree $deg ( I − T , Ω , 0 ) = 1$.

## 3. Proof of Theorem 1

In this section, we present the detailed proof of Theorem 1. From (H6) there exist $ε 0 ∈ ( 0 , λ 1 )$ and $r 1 > 0$ such that
$| f 1 ( t , y ) | ≤ ( λ 1 − ε 0 ) | y | , | f 2 ( t , x ) | ≤ ( λ 1 − ε 0 ) | x | , ∀ t ∈ [ 0 , 1 ] , x , y ∈ R with | x | , | y | ≤ r 1 .$
This implies that
$| ( T 1 y ) ( t ) | ≤ ∫ 0 1 G ( t , q s ) | f 1 ( s , y ( s ) ) | d q s ≤ ( λ 1 − ε 0 ) ∫ 0 1 G ( t , q s ) | y ( s ) | d q s ,$
and
$| ( T 2 x ) ( t ) | ≤ ∫ 0 1 G ( t , q s ) | f 2 ( s , x ( s ) ) | d q s ≤ ( λ 1 − ε 0 ) ∫ 0 1 G ( t , q s ) | x ( s ) | d q s .$
Now we prove that
$( x , y ) ≠ μ T ( x , y ) for all x , y ∈ ∂ B r 1 and μ ∈ [ 0 , 1 ] .$
We argue by contradiction. Suppose there exist $x , y ∈ ∂ B r 1$ and $μ ∈ [ 0 , 1 ]$ such that
$( x , y ) = μ T ( x , y ) .$
Therefore,
$x = μ T 1 y , and y = μ T 2 x .$
This implies that
$| x ( t ) | = μ | ( T 1 y ) ( t ) | ≤ ( λ 1 − ε 0 ) ∫ 0 1 G ( t , q s ) | y ( s ) | d q s , t ∈ [ 0 , 1 ] ,$
and
$| y ( t ) | = μ | ( T 2 x ) ( t ) | ≤ ( λ 1 − ε 0 ) ∫ 0 1 G ( t , q s ) | x ( s ) | d q s , t ∈ [ 0 , 1 ] .$
Consequently, we have
$| x ( t ) | + | y ( t ) | ≤ ( λ 1 − ε 0 ) ∫ 0 1 G ( t , q s ) ( | x ( s ) | + | y ( s ) | ) d q s , t ∈ [ 0 , 1 ] .$
Let $z ( t ) = x ( t ) + y ( t ) .$
Then $z ∈ P$ and
$z ( t ) ≤ ( λ 1 − ε 0 ) ∫ 0 1 G ( t , q s ) z ( s ) d q s = ( λ 1 − ε 0 ) ( L z ) ( t ) , t ∈ [ 0 , 1 ] .$
The nth iteration of this inequality shows that
$z ( t ) ≤ ( λ 1 − ε 0 ) n L n z ( t ) ( n = 1 , 2 , … ) , and then ∥ z ∥ ≤ ( λ 1 − ε 0 ) n L n · ∥ z ∥ , i . e . , 1 ≤ ( λ 1 − ε 0 ) n L n .$
This yields
$1 ≤ ( λ 1 − ε 0 ) lim n → ∞ L n n = ( λ 1 − ε 0 ) r ( L ) = λ 1 − ε 0 λ 1 < 1 ,$
which is a contradiction. Hence, (14) holds. It follows from Lemma 8 that
$deg I − T , B r 1 , 0 = 1 .$
On the other hand, from (H5) there exist $ε 1 > 0$ and $r 2 > 0$ such that
$f 1 ( t , y ) ≥ ( λ 1 + ε 1 ) | y | , f 2 ( t , x ) ≥ ( λ 1 + ε 1 ) | x | , ∀ t ∈ [ 0 , 1 ] , | x | , | y | > r 2 .$
Let $M 1 = max t ∈ [ 0 , 1 ] , | y | ≤ r 2 [ | f 1 ( t , y ) | + ( λ 1 + ε 1 ) | y | ] , M 2 = max t ∈ [ 0 , 1 ] , | x | ≤ r 2 [ | f 2 ( t , x ) | + ( λ 1 + ε 1 ) | x | ]$.
Then
$f 1 ( t , y ) ≥ ( λ 1 + ε 1 ) | y | − M 1 , f 2 ( t , x ) ≥ ( λ 1 + ε 1 ) | x | − M 2 , ∀ t ∈ [ 0 , 1 ] , x , y ∈ R .$
For any given $ε , ε ˜$ with $ε 1 − ∥ c 1 ∥ ε > 0 , ε 1 − ∥ c 2 ∥ ε ˜ > 0 ,$ and from $( H 4 )$ there exists $r 3 > r 2$ such that
$K 1 ( y ) ≤ ε | y | , K 2 ( x ) ≤ ε ˜ | x | , ∀ | x | , | y | > r 3 .$
Let $K 1 ∗ = max | y | ≤ r 3 K 1 ( y )$, and $K 2 ∗ = max | x | ≤ r 3 K 2 ( x )$. Then we have
$K 1 ( y ) ≤ ε | y | + K 1 ∗ , K 2 ( x ) ≤ ε ˜ | x | + K 2 ∗ , ∀ x , y ∈ R .$
Please note that $ε$, $ε ˜$ can be chosen arbitrarily small, so we can let $R 1 > max { r 1 , N 1 , N 2 , N 3 , N 4 }$, where $r 1$ is defined by (14), and
$N 1 = ( 2 ∥ b 1 ∥ + 2 ∥ c 1 ∥ K 1 ∗ + M 1 ) ∫ 0 1 φ 1 ( q s ) d q s 1 2 − ε ∥ c 1 ∥ ∫ 0 1 φ 1 ( q s ) d q s , N 2 = ( 2 ∥ b 2 ∥ + 2 ∥ c 2 ∥ K 2 ∗ + M 2 ) ∫ 0 1 φ 1 ( q s ) d q s 1 2 − ε ˜ ∥ c 2 ∥ ∫ 0 1 φ 1 ( q s ) d q s , N 3 = N 5 ( 2 ∥ b 1 ∥ + 2 ∥ b 2 ∥ + 2 ∥ c 1 ∥ K 1 ∗ + 2 ∥ c 2 ∥ K 2 ∗ + M 1 + M 2 ) ( ε 1 − ∥ c 1 ∥ ε ) − N 5 ( ∥ c 1 ∥ ε + ∥ c 2 ∥ ε ˜ ) , N 4 = N 6 ( 2 ∥ b 1 ∥ + 2 ∥ b 2 ∥ + 2 ∥ c 1 ∥ K 1 ∗ + 2 ∥ c 2 ∥ K 2 ∗ + M 1 + M 2 ) ( ε 1 − ∥ c 2 ∥ ε ˜ ) − N 6 ( ∥ c 1 ∥ ε + ∥ c 2 ∥ ε ˜ ) , N 5 = ( ε 1 − ∥ c 1 ∥ ε ) ∫ 0 1 φ 1 ( q s ) d q s + ( λ 1 + ε 1 − ∥ c 1 ∥ ε ) ∫ 0 1 φ 2 ( q s ) d q s , N 6 = ( ε 1 − ∥ c 2 ∥ ε ˜ ) ∫ 0 1 φ 1 ( q s ) d q s + ( λ 1 + ε 1 − ∥ c 2 ∥ ε ˜ ) ∫ 0 1 φ 2 ( q s ) d q s .$
Now we prove that
$( x , y ) − T ( x , y ) ≠ μ ( φ ∗ , φ ∗ ) , ∀ x , y ∈ ∂ B R 1 , μ ≥ 0 ,$
where $φ ∗$ is the positive eigenfunction of L corresponding to the eigenvalue $λ 1$. We argue by contradiction. Suppose there exist $x , y ∈ ∂ B R 1$ and $μ ≥ 0$ such that
$( x , y ) − T ( x , y ) = μ ( φ ∗ , φ ∗ ) ,$
and thus
$x = T 1 y + μ φ ∗ , y = T 2 x + μ φ ∗ .$
Let
$x ˜ ( t ) = ∫ 0 1 G ( t , q s ) [ 2 b 2 ( s ) + c 2 ( s ) K 2 ( x ( s ) ) + M 2 + ∥ c 2 ∥ K 2 ∗ ] d q s ,$
and
$y ˜ ( t ) = ∫ 0 1 G ( t , q s ) [ 2 b 1 ( s ) + c 1 ( s ) K 1 ( y ( s ) ) + M 1 + ∥ c 1 ∥ K 1 ∗ ] d q s .$
Now we estimate the norms $∥ x ˜ ∥$ and $∥ y ˜ ∥$. Please note that $∥ x ∥ = ∥ y ∥ = R 1$, and from (17) we have
$∥ y ˜ ∥ ≤ ∫ 0 1 φ 1 ( q s ) [ 2 ∥ b 1 ∥ + ∥ c 1 ∥ ( ε ∥ y ∥ + K 1 ∗ ) + M 1 + ∥ c 1 ∥ K 1 ∗ ] d q s = ∫ 0 1 φ 1 ( q s ) d q s × ( 2 ∥ b 1 ∥ + 2 ∥ c 1 ∥ K 1 ∗ + M 1 + ∥ c 1 ∥ ε R 1 ) < 1 2 R 1 ,$
and
$∥ x ˜ ∥ ≤ ∫ 0 1 φ 1 ( q s ) [ 2 ∥ b 2 ∥ + ∥ c 2 ∥ ( ε ˜ ∥ x ∥ + K 2 ∗ ) + M 2 + ∥ c 2 ∥ K 2 ∗ ] d q s = ∫ 0 1 φ 1 ( q s ) d q s × ( 2 ∥ b 2 ∥ + 2 ∥ c 2 ∥ K 2 ∗ + M 2 + ∥ c 2 ∥ ε ˜ R 1 ) < 1 2 R 1 .$
Furthermore, from (H3) and Lemma 6 we have $x ˜ , y ˜ ∈ P 0$. Consequently, we obtain
$x ( t ) + y ˜ ( t ) = ( T 1 y ) ( t ) + μ φ ∗ ( t ) + y ˜ ( t ) = ∫ 0 1 G ( t , q s ) [ f 1 ( s , y ( s ) ) + 2 b 1 ( s ) + c 1 ( s ) K 1 ( y ( s ) ) + M 1 + ∥ c 1 ∥ K 1 ∗ ] d q s + μ φ ∗ ( t ) ,$
and
$y ( t ) + x ˜ ( t ) = ( T 2 x ) ( t ) + μ φ ∗ ( t ) + x ˜ ( t ) = ∫ 0 1 G ( t , q s ) [ f 2 ( s , x ( s ) ) + 2 b 2 ( s ) + c 2 ( s ) K 2 ( x ( s ) ) + M 2 + ∥ c 2 ∥ K 2 ∗ ] d q s + μ φ ∗ ( t ) .$
Using (H3), Lemma 6 and Remark 2 we have
$x + y ˜ ∈ P 0 , y + x ˜ ∈ P 0 .$
Please note that $∥ x ∥ = ∥ y ∥ = R 1$, $y + y ˜ + x ˜ ∈ P 0$, and $x + y ˜ + x ˜ ∈ P 0$. Therefore, we get
$x ( t ) + y ˜ ( t ) + x ˜ ( t ) ≥ t α − 1 ∥ x + y ˜ + x ˜ ∥ ≥ t α − 1 ( ∥ x | | − ∥ y ˜ + x ˜ ∥ ) ≥ t α − 1 [ ∥ x ∥ − ( ∥ y ˜ ∥ + ∥ x ˜ ∥ ) ] ,$
and
$y ( t ) + y ˜ ( t ) + x ˜ ( t ) ≥ t α − 1 ∥ y + y ˜ + x ˜ ∥ ≥ t α − 1 ( ∥ y | | − ∥ y ˜ + x ˜ ∥ ) ≥ t α − 1 [ ∥ y ∥ − ( ∥ y ˜ ∥ + ∥ x ˜ ∥ ) ] .$
Consequently, note that the range value of $R 1$, from (22) and (23) we find
$( ε 1 − ∥ c 1 ∥ ε ) ∫ 0 1 G ( t , q s ) ( y ( s ) + y ˜ ( s ) + x ˜ ( s ) ) d q s − ( λ 1 + ε 1 − ∥ c 1 ∥ ε ) ∫ 0 1 G ( t , q s ) ( y ˜ ( s ) + x ˜ ( s ) ) d q s ≥ ( ε 1 − ∥ c 1 ∥ ε ) ∫ 0 1 G ( t , q s ) s α − 1 ( R 1 − ( ∥ y ˜ ∥ + ∥ x ˜ ∥ ) ) d q s − ( λ 1 + ε 1 − ∥ c 1 ∥ ε ) ∫ 0 1 G ( t , q s ) ∫ 0 1 G ( s , q τ ) [ 2 b 1 ( τ ) + 2 b 2 ( τ ) + c 1 ( τ ) K 1 ( y ( τ ) ) + c 2 ( τ ) K 2 ( x ( τ ) ) + M 1 + M 2 + ∥ c 1 ∥ K 1 ∗ + ∥ c 2 ∥ K 2 ∗ ] d q τ d q s ≥ ( ε 1 − ∥ c 1 ∥ ε ) ∫ 0 1 G ( t , q s ) s α − 1 ( R 1 − ( ∥ y ˜ ∥ + ∥ x ˜ ∥ ) ) d q s − ( λ 1 + ε 1 − ∥ c 1 ∥ ε ) ∫ 0 1 G ( t , q s ) ∫ 0 1 s α − 1 φ 2 ( q τ ) [ 2 ∥ b 1 ∥ + 2 ∥ b 2 ∥ + ∥ c 1 ∥ ( ε ∥ y ∥ + K 1 ∗ ) + ∥ c 2 ∥ ( ε ˜ ∥ x ∥ + K 2 ∗ ) + M 1 + M 2 + ∥ c 1 ∥ K 1 ∗ + ∥ c 2 ∥ K 2 ∗ ] d q τ d q s ≥ 0 ,$
and
$( ε 1 − ∥ c 2 ∥ ε ˜ ) ∫ 0 1 G ( t , q s ) ( x ( s ) + x ˜ ( s ) + y ˜ ( s ) ) d q s − ( λ 1 + ε 1 − ∥ c 2 ∥ ε ˜ ) ∫ 0 1 G ( t , q s ) ( x ˜ ( s ) + y ˜ ( s ) ) d q s ≥ ( ε 1 − ∥ c 2 ∥ ε ˜ ) ∫ 0 1 G ( t , q s ) s α − 1 ( R 1 − ( ∥ y ˜ ∥ + ∥ x ˜ ∥ ) ) d q s − ( λ 1 + ε 1 − ∥ c 2 ∥ ε ˜ ) ∫ 0 1 G ( t , q s ) ∫ 0 1 s α − 1 φ 2 ( q τ ) [ 2 ∥ b 1 ∥ + 2 ∥ b 2 ∥ + ∥ c 1 ∥ ( ε ∥ y ∥ + K 1 ∗ ) + ∥ c 2 ∥ ( ε ˜ ∥ x ∥ + K 2 ∗ ) + M 1 + M 2 + ∥ c 1 ∥ K 1 ∗ + ∥ c 2 ∥ K 2 ∗ ] d q τ d q s ≥ 0 .$
Consequently, from (16) and (17) we have
$∫ 0 1 G ( t , q s ) [ f 1 ( s , y ( s ) ) + 2 b 1 ( s ) + c 1 ( s ) K 1 ( y ( s ) ) + M 1 + ∥ c 1 ∥ K 1 ∗ ] d q s ≥ ∫ 0 1 G ( t , q s ) [ ( λ 1 + ε 1 ) | y ( s ) | − M 1 − b 1 ( s ) − c 1 ( s ) ( ε | y ( s ) | + K 1 ∗ ) + b 1 ( s ) + M 1 + ∥ c 1 ∥ K 1 ∗ ] d q s ≥ ∫ 0 1 G ( t , q s ) [ ( λ 1 + ε 1 ) | y ( s ) | − ∥ c 1 ∥ ( ε | y ( s ) | + K 1 ∗ ) + ∥ c 1 ∥ K 1 ∗ ] d q s = ( λ 1 + ε 1 − ∥ c 1 ∥ ε ) ∫ 0 1 G ( t , q s ) | y ( s ) | d q s ≥ ( λ 1 + ε 1 − ∥ c 1 ∥ ε ) ∫ 0 1 G ( t , q s ) y ( s ) d q s = ( λ 1 + ε 1 − ∥ c 1 ∥ ε ) ∫ 0 1 G ( t , q s ) ( y ( s ) + y ˜ ( s ) + x ˜ ( s ) ) d q s − ( λ 1 + ε 1 − ∥ c 1 ∥ ε ) ∫ 0 1 G ( t , q s ) ( y ˜ ( s ) + x ˜ ( s ) ) d q s ≥ λ 1 L ( y + y ˜ + x ˜ ) ( t ) ≥ λ 1 L ( y + x ˜ ) ( t ) ,$
and
$∫ 0 1 G ( t , q s ) [ f 2 ( s , x ( s ) ) + 2 b 2 ( s ) + c 2 ( s ) K 2 ( x ( s ) ) + M 2 + ∥ c 2 ∥ K 2 ∗ ] d q s ≥ ∫ 0 1 G ( t , q s ) [ ( λ 1 + ε 1 ) | x ( s ) | − M 2 − b 2 ( s ) − c 2 ( s ) ( ε ˜ | x ( s ) | + K 2 ∗ ) + b 2 ( s ) + M 2 + ∥ c 2 ∥ K 2 ∗ ] d q s ≥ ∫ 0 1 G ( t , q s ) [ ( λ 1 + ε 1 ) | x ( s ) | − ∥ c 2 ∥ ( ε ˜ | x ( s ) | + K 2 ∗ ) + ∥ c 2 ∥ K 2 ∗ ] d q s ≥ ( λ 1 + ε 1 − ∥ c 2 ∥ ε ˜ ) ∫ 0 1 G ( t , q s ) x ( s ) d q s = ( λ 1 + ε 1 − ∥ c 2 ∥ ε ˜ ) ∫ 0 1 G ( t , q s ) ( x ( s ) + x ˜ ( s ) + y ˜ ( s ) ) d q s − ( λ 1 + ε 1 − ∥ c 2 ∥ ε ˜ ) ∫ 0 1 G ( t , q s ) ( x ˜ ( s ) + y ˜ ( s ) ) d q s ≥ λ 1 L ( x + x ˜ + y ˜ ) ( t ) ≥ λ 1 L ( x + y ˜ ) ( t ) .$
Now, using (26)–(27) we have
$T 1 y + y ˜ ≥ λ 1 L ( y + x ˜ ) , T 2 x + x ˜ ≥ λ 1 L ( x + y ˜ ) .$
Thus, from (19) we have
$x + y + x ˜ + y ˜ = T 1 y + T 2 x + x ˜ + y ˜ + 2 μ φ ∗ ≥ λ 1 L ( x + y + x ˜ + y ˜ ) + 2 μ φ ∗ ≥ 2 μ φ ∗ .$
Define $μ ∗ = sup S μ : = sup { μ > 0 : x + y + x ˜ + y ˜ ≥ 2 μ φ ∗ }$. Then $S μ ( ≠ ∅ )$ is a limited set, $μ ∗ ≥ μ$ and $x + y + x ˜ + y ˜ ≥ 2 μ ∗ φ ∗$. From $φ ∗ = λ 1 L φ ∗$, we obtain
$λ 1 L ( x + y + x ˜ + y ˜ ) ≥ λ 1 L ( 2 μ ∗ φ ∗ ) = 2 μ ∗ λ 1 L φ ∗ = 2 μ ∗ φ ∗ .$
Hence
$x + y + x ˜ + y ˜ ≥ λ 1 L ( x + y + x ˜ + y ˜ ) + 2 μ φ ∗ ≥ 2 ( μ + μ ∗ ) φ ∗ ,$
which contradicts the definition of $μ ∗$. Therefore, (18) holds, and from Lemma 7 we obtain
$deg ( I − T , B R 1 , 0 ) = 0 .$
Now (15) and (28) together imply that
$deg ( I − T , B R 1 ∖ B ¯ r 1 , 0 ) = deg ( I − T , B R 1 , 0 ) − deg ( I − T , B r 1 , 0 ) = − 1 .$
Therefore the operator T has at least one fixed point in $B R 1 \ B ¯ r 1$. Equivalently, (8) has at least one nontrivial solution. This completes the proof.

## 4. Conclusions

In this paper, we use topological degree to study nontrivial solutions for the system of fractional q-difference Equation (8) with q-integral boundary conditions. There are only a few papers in the literature which consider systems of fractional q-difference equations with q-integral boundary conditions where the nonlinear terms may be unbounded from below. Our main theorem is obtained under some conditions concerning the first eigenvalues corresponding to the relevant linear operators. As a result, our main result generalizes and improves the corresponding ones in the works cited in this paper.

## Author Contributions

Conceptualization, Y.L., J.L., J.X. and D.O.; methodology, Y.L., J.L., J.X. and D.O.; software, Y.L., J.L.; validation, Y.L., J.L., J.X. and D.O.; formal analysis, Y.L., J.L., J.X. and D.O.; investigation, J.L., J.X. and D.O.; resources, Y.L., J.X.; data curation, Y.L., J.X.; writing original draft preparation, J.L., J.X. and D.O.; writing review and editing, Y.L., J.L., J.X. and D.O.; visualization, Y.L., J.X. and D.O.; supervision, J.X. and D.O.; project administration, J.X.; funding acquisition, Y.L., J.X. All authors have read and agreed to the published version of the manuscript.

## Funding

This work is supported by the China Postdoctoral Science Foundation (Grant No. 2019M652348), Technology Research Foundation of Chongqing Educational Committee(Grant No. KJQN201900539), Key Research Funds for the Universities of Henan Province (19A110018, 20B110006), Fundamental Research Funds for the Universities of Henan Province (NSFRF180320) and Henan Polytechnic University Doctor Fund (No. B2016-58), Outstanding Young Foundation of Anhui Provincial Education Department (Grant No. gxyqZD2016339), Natural Science Foundation of Anhui Provincial Education Department (Grant No. KJ2018A0452).

## Conflicts of Interest

The authors declare no conflict of interest.

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MDPI and ACS Style

Li, Y.; Liu, J.; O’Regan, D.; Xu, J. Nontrivial Solutions for a System of Fractional q-Difference Equations Involving q-Integral Boundary Conditions. Mathematics 2020, 8, 828. https://doi.org/10.3390/math8050828

AMA Style

Li Y, Liu J, O’Regan D, Xu J. Nontrivial Solutions for a System of Fractional q-Difference Equations Involving q-Integral Boundary Conditions. Mathematics. 2020; 8(5):828. https://doi.org/10.3390/math8050828

Chicago/Turabian Style

Li, Yaohong, Jie Liu, Donal O’Regan, and Jiafa Xu. 2020. "Nontrivial Solutions for a System of Fractional q-Difference Equations Involving q-Integral Boundary Conditions" Mathematics 8, no. 5: 828. https://doi.org/10.3390/math8050828

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