Next Article in Journal / Special Issue
New Applications of Fractional q-Calculus Operator for a New Subclass of q-Starlike Functions Related with the Cardioid Domain
Previous Article in Journal / Special Issue
Mandelbrot Set as a Particular Julia Set of Fractional Order, Equipotential Lines and External Rays of Mandelbrot and Julia Sets of Fractional Order
 
 
Font Type:
Arial Georgia Verdana
Font Size:
Aa Aa Aa
Line Spacing:
Column Width:
Background:
Article

Positive Solutions for a System of Fractional q-Difference Equations with Multi-Point Boundary Conditions

Department of Mathematics, Gh. Asachi Technical University, 700506 Iasi, Romania
Fractal Fract. 2024, 8(1), 70; https://doi.org/10.3390/fractalfract8010070
Submission received: 31 December 2023 / Revised: 17 January 2024 / Accepted: 19 January 2024 / Published: 21 January 2024

Abstract

:
We explore the existence, uniqueness, and multiplicity of positive solutions to a system of fractional q-difference equations that include fractional q-integrals. This investigation is carried out under coupled multi-point boundary conditions featuring q-derivatives and fractional q-derivatives of various orders. The proofs of our principal findings employ a range of fixed-point theorems, including the Guo–Krasnosel’skii fixed-point theorem, the Leggett–Williams fixed-point theorem, the Schauder fixed-point theorem, and the Banach contraction mapping principle.

1. Introduction

In recent decades, there has been a notable surge in the exploration of nonlocal boundary value problems, including those involving multi-point scenarios, within the realm of ordinary differential or difference equations. This area of research is experiencing rapid growth, spurred not only by theoretical interest but also by the practical applications of modeling various phenomena in engineering, physics, and life sciences. To illustrate, consider systems with feedback controls, such as the steady states of a thermostat. In this context, a second-order ordinary differential equation subject to a three-point boundary condition can capture the dynamics, where a controller at one end adjusts the heat based on the temperature recorded at another point. Another instance is found in the vibrations of a guy wire with a uniform cross-section composed of N parts of varying densities. Such scenarios can be effectively modeled as multi-point boundary value problems (refer to [1]).
In this paper, we will investigate the system of fractional q-difference equations
D q α u ( t ) + f t , u ( t ) , v ( t ) , I q δ 1 u ( t ) , I q γ 1 v ( t ) = 0 , t [ 0 , 1 ] , D q β v ( t ) + g t , u ( t ) , v ( t ) , I q δ 2 u ( t ) , I q γ 2 v ( t ) = 0 , t [ 0 , 1 ] ,
supplemented with the multi-point boundary conditions
D q i u ( 0 ) = 0 , i = 0 , , n 2 , D q ς u ( 1 ) = i = 1 a a i D q ϱ i u ( ξ i ) + i = 1 b b i D q σ i v ( ω i ) , D q i v ( 0 ) = 0 , i = 0 , , m 2 , D q ϑ v ( 1 ) = i = 1 c c i D q η i u ( ζ i ) + i = 1 d d i D q ρ i v ( θ i ) .
Here, q ( 0 , 1 ) , α , β R , α ( n 1 , n ] , β ( m 1 , m ] , n , m N , n , m 3 ; a , b , c , d N , 0 ϱ i ς < α 1 , i = 1 , , a , ς 1 , 0 η i ς , i = 1 , , c , 0 σ i ϑ < β 1 , i = 1 , , b , ϑ 1 , 0 ρ i ϑ , i = 1 , , d ; δ i , γ i > 0 , i = 1 , 2 ; ξ i , ω j , ζ k , θ ι ( 0 , 1 ) , and a i , b j , c k , d ι 0 for i = 1 , , a , j = 1 , , b , k = 1 , , c , ι = 1 , , d ; D q κ is the fractional q-derivative of order κ for κ = α , β , ς , ϑ , ϱ i , σ j , η k , ρ ι , i = 1 , , a , j = 1 , , b , k = 1 , , c , ι = 1 , , d ; D q i is the q-derivative of order i for i = 0 , , n 2 and i = 0 , , m 2 ; and I q κ is the fractional q-integral of order κ for κ = δ i , γ i , i = 1 , 2 .
We will establish conditions on the functions f and g under which problem (1),(2) possesses at least one positive solution. Our proofs will leverage several fixed point theorems, including the Guo–Krasnosel’skii fixed point theorem, the Leggett–Williams fixed point theorem, the Schauder fixed point theorem, and the Banach contraction mapping principle. Subsequently, we will provide references to pertinent papers that are closely related to our investigated problem. In [2], the authors studied the solvability for the system of nonlinear fractional q-difference equations
( D q α u ) ( t ) + P ( t , u ( t ) , v ( t ) , I q ω 1 u ( t ) , I q δ 1 v ( t ) ) = 0 , t ( 0 , 1 ) , ( D q β v ) ( t ) + Q ( t , u ( t ) , v ( t ) , I q ω 2 u ( t ) , I q δ 2 v ( t ) ) = 0 , t ( 0 , 1 ) ,
subject to the coupled nonlocal boundary conditions
D q i u ( 0 ) = 0 , i = 0 , , n 2 , D q ζ 0 u ( 1 ) = 0 1 D q ζ v ( t ) d q H ( t ) , t ( 0 , 1 ) , D q i v ( 0 ) = 0 , i = 0 , , z 2 , D q ξ 0 v ( 1 ) = 0 1 D q ξ u ( t ) d q K ( t ) , t ( 0 , 1 ) ,
where q ( 0 , 1 ) , α , β R , α ( n 1 , n ] , β ( z 1 , z ] , n , z N , n 2 , z 2 , ω i > 0 , δ i > 0 , i = 1 , 2 , ζ [ 0 , β 1 ) , ξ [ 0 , α 1 ) , ζ 0 [ 0 , α 1 ) , ξ 0 [ 0 , β 1 ) , and the integrals in (4) are Riemann–Stieltjes integrals with H - and K -bounded variation functions. By using varied fixed-point theorems, they obtained the existence and uniqueness results for the solutions of problem (3),(4). In [3], the authors examined the existence of solutions for the fractional q-difference equation with nonlinear integral conditions
( C D q α y ) ( t ) = f ( t , y ( t ) ) , for a . e . t [ 0 , T ] , y ( 0 ) y ( 0 ) = 0 T g ( s , y ( s ) d s , y ( T ) + y ( T ) = 0 T h ( s , y ( s ) ) d s ,
where T > 0 , q ( 0 , 1 ) , α ( 1 , 2 ] , and C D q α is the Caputo fractional q-derivative of order α . In the proof of the main result, they utilized measures of noncompactness and the Mönch fixed-point theorem. In [4], the authors investigated the existence, uniqueness, and multiplicity of positive solutions to the fractional q-difference equation with the nonlocal boundary conditions
( D q α u ) ( t ) + f ( t , u ( t ) ) = 0 , t ( 0 , 1 ) , ( D q i u ) ( 0 ) = 0 , i = 0 , , n 2 , ( D q β u ) ( 1 ) = a ( D q β u ) ( η ) ,
where q ( 0 , 1 ) , α ( n 1 , n ] , n > 2 , β [ 1 , n 2 ] , η ( 0 , 1 ) , a [ 0 , 1 ] , and f : [ 0 , 1 ] × [ 0 , ) [ 0 , ) satisfies Caratheodory-type conditions. In the proof of the main theorems, they applied several fixed-point theorems. In [5], by using the Guo–Krasnosel’skii fixed-point theorem, the author analyzed the existence of positive solutions to the fractional q-difference equation with the boundary conditions
( D q α y ) ( t ) = f ( t , y ( t ) ) , t ( 0 , 1 ) , y ( 0 ) = ( D q y ) ( 0 ) = 0 , ( D q y ) ( 1 ) = β ,
where q ( 0 , 1 ) , α ( 2 , 3 ] , β 0 , and f : [ 0 , 1 ] × R R is a nonnegative continuous function. In [6], the author explored the existence of nontrivial solutions to the nonlinear q-fractional boundary value problem
( D q α y ) ( t ) = f ( t , y ( t ) ) , t ( 0 , 1 ) , y ( 0 ) = y ( 1 ) = 0 ,
where q ( 0 , 1 ) , α ( 1 , 2 ] , and f : [ 0 , 1 ] × R R is a nonnegative continuous function. To prove the main finding, he also used the Guo–Krasnosel’skii fixed-point theorem. In [7], the authors proved the existence of solutions to the second-order q-difference equation with the boundary conditions
D q 2 u ( t ) = f ( t , u ( t ) , D q u ( t ) ) , t I , D q u ( 0 ) = 0 , D q u ( 1 ) = α u ( 1 ) ,
where q ( 0 , 1 ) , I = { q n , n N } { 0 , 1 } , f C ( I × R 2 , R ) , and α 0 is a fixed real number. In [8], the authors studied the existence of solutions to the second-order q-difference equation subject to the boundary conditions
D q 2 u ( t ) = f ( t , u ( t ) ) , t I , u ( 0 ) = η u ( T ) , D q u ( 0 ) = η D q u ( T ) ,
where I = [ 0 , T ] q N ¯ , q N ¯ = { q n , n N } { 0 } , T q N ¯ is a fixed constant, η 1 is a fixed real number, and f C ( I × R , R ) . Regarding additional papers that investigate systems of fractional q-difference equations with either coupled or uncoupled boundary conditions or that focus on fractional q-difference equations specifically, we refer to the following papers: [9,10,11,12,13,14,15,16].
The field of q-difference calculus, also known as quantum calculus, traces its origins back to the pioneering work of Jackson ([17,18]). To explore various applications of this discipline, readers are directed to the research of Ernst ([19]). The fractional q-difference calculus originated in the works of Al-Salam ([20]) and Agarwal ([21]). For advancements in this branch, encompassing q-analogs of integral and differential fractional operators, including properties like the fractional Leibniz q-formula, q-analogs of Cauchy’s formula, q-Laplace transform, q-Taylor’s formula, and q-analogs of the Mittag-Leffler function, refer to the papers [22,23,24,25,26].
The novel aspect of our problem (1),(2), compared to (3),(4) from [2], lies in the inclusion of generalized coupled boundary conditions (2) for the system of q-fractional difference equations in Equation (1). In this formulation, the q-fractional derivative of order ς for the unknown function u at the point 1 is contingent upon the q-fractional derivatives of various orders for both functions u and v at different points within the interval ( 0 , 1 ) . Similarly, the q-fractional derivative of order ϑ for the unknown function v at the point 1 is linked to the q-fractional derivatives of distinct orders for functions u and v at diverse points within the interval ( 0 , 1 ) . Furthermore, unlike the approach in the paper [2], we have explored the presence of positive solutions to our specific problem.
Our paper is organized as follows: Section 2 introduces key definitions and properties from q-calculus and fractional q-calculus, along with an existence result for the associated linear problem, the relevant Green functions, and their properties. Section 3 will then present the primary existence results for problem (1),(2), while Section 4 will provide illustrative examples to demonstrate the applicability of our theorems. Finally, Section 5 concludes the paper by summarizing the findings and presenting the overall conclusions.

2. Preliminary Results

In this section, we will introduce certain definitions and properties derived from q-calculus and fractional q-calculus. Additionally, we will outline some auxiliary findings that will play a pivotal role in the subsequent section.
Let q ( 0 , 1 ) . Define the number
[ a ] q = 1 q a 1 q , a R .
Next, to introduce the q-gamma function, we present the q-analog of the power function ( a b ) n with n N { 0 } :
( a b ) ( 0 ) = 1 , ( a b ) ( n ) = k = 0 n 1 ( a b q k ) , n N , a , b R .
For α R , define
( a b ) ( α ) = a α n = 0 1 b a q n n = 0 1 b a q α + n .
If b = 0 , then a ( α ) = a α .
The q-gamma function is defined by
Γ q ( α ) = ( 1 q ) ( α 1 ) ( 1 q ) α 1 = 1 ( 1 q ) α 1 n = 0 ( 1 q n + 1 ) n = 0 ( 1 q n + α ) , α R { 0 , 1 , 2 , } .
This function satisfies the relation Γ q ( α + 1 ) = [ α ] q Γ q ( α ) .
Definition 1.
The q-derivative of a real function f is defined by
( D q f ) ( t ) = f ( t ) f ( q t ) ( 1 q ) t , t 0 ; ( D q f ) ( 0 ) = lim t 0 ( D q f ) ( t ) .
Definition 2.
The q-derivatives of higher order of a real function f are defined by
( D q 0 f ) ( t ) = f ( t ) , ( D q n f ) ( t ) = D q ( D q n 1 f ) ( t ) , n N .
Definition 3.
The q-integral of a function f defined in the interval [ 0 , b ] is defined by
( I q f ) ( t ) = 0 t f ( s ) d q s = t ( 1 q ) n = 0 f ( t q n ) q n , t [ 0 , b ] .
Definition 4.
If a [ 0 , b ] and f is defined in the interval [ 0 , b ] , then its q-integral from a to b is given by
a b f ( s ) d q s = 0 b f ( s ) d q s 0 a f ( s ) d q s .
Definition 5.
The q-integrals of a function f of higher order are defined by
( I q 0 f ) ( t ) = f ( t ) , ( I q n f ) ( t ) = I q ( I q n 1 f ) ( t ) , n N .
The fundamental theorem of q-calculus says that ( D q I q f ) ( t ) = f ( t ) , and if f is continuous at t = 0 , then ( I q D q f ) ( t ) = f ( t ) f ( 0 ) . The properties of the operators D q and I q are presented in [6,27]. Below, we present some properties that will be used later.
Lemma 1.
For a , t , s , α R , and f : [ 0 , b ] R , we have
(i) [ a ( t s ) ] ( α ) = a α ( t s ) ( α ) ;
(ii) t D q ( t s ) ( α ) = [ α ] q ( t s ) ( α 1 ) ;
(iii) If α > 1 , then D q ( t α ) = [ α ] q t α 1 ;
(iv) If α > 0 , then I q ( t α ) = 1 [ α + 1 ] q t α + 1 ;
(v) If α > 0 and c d t , then ( t c ) ( α ) ( t d ) ( α ) ;
(vi)  t D q 0 t f ( t , s ) d q s ( t ) = 0 t t D q f ( t , s ) d q s + f ( q t , t ) , t [ 0 , b ] , where t D q denotes the q-derivative with respect to variable t .
Definition 6
([21]). Let f be a function defined on [ 0 , 1 ] . The fractional q-integral of the Riemann–Liouville type of order α 0 is defined by ( I q 0 f ) ( t ) = f ( t ) and
( I q α f ) ( t ) = 1 Γ q ( α ) 0 t ( t q s ) ( α 1 ) f ( s ) d q s , t [ 0 , 1 ] , α > 0 .
Definition 7
([24]). The fractional q-derivative of the Riemann–Liouville type of order α 0 is defined by ( D q 0 f ) ( t ) = f ( t ) and
( D q α f ) ( t ) = ( D q m I q m α f ) ( t ) , α > 0 ,
where m is the smallest integer greater than or equal to α.
Lemma 2
([21,24]). Let α , β 0 , γ > 0 , ζ [ 0 , γ ) , and λ 0 , and let f be a function defined on [ 0 , 1 ] . Then, the following relations are satisfied:
(a) ( I q β I q α f ) ( t ) = ( I q α + β f ) ( t ) ;
(b) ( D q α I q α f ) ( t ) = f ( t ) ;
(c) If α β > 0 , then ( D q β I q α f ) ( t ) = ( D q β I q β I q α β f ) ( t ) = ( I q α β f ) ( t ) ;
(d) D q ζ ( t γ ) = Γ q ( γ + 1 ) Γ q ( γ ζ + 1 ) t γ ζ ;
(e) I q α ( t λ ) = Γ q ( λ + 1 ) Γ q ( α + λ + 1 ) t α + λ .
Lemma 3
([6]). Let α > 0 and p be a positive integer. Then, the following relation holds:
( I q α D q p f ) ( t ) = ( D q p I q α f ) ( t ) k = 0 p 1 t α p + k Γ q ( α + k p + 1 ) ( D q k f ) ( 0 ) .
Lemma 4.
If w C [ 0 , 1 ] , then for κ > 0 , we have
| I q κ w ( t ) | w Γ q ( κ + 1 ) , t [ 0 , 1 ] ,
where  w = sup t [ 0 , 1 ] | w ( t ) | .
Proof. 
By Definition 6 and Lemma 2 (e) (with α = κ and λ = 0 ), we obtain
| I q κ w ( t ) | = 1 Γ q ( κ ) 0 t ( t q s ) ( κ 1 ) w ( s ) d q s = 1 Γ q ( κ ) t ( 1 q ) n = 0 ( t t q n + 1 ) ( κ 1 ) w ( t q n ) q n w 1 Γ q ( κ ) t ( 1 q ) n = 0 ( t t q n + 1 ) ( κ 1 ) q n = w | ( I q κ 1 ) ( t ) | = w t κ Γ q ( κ + 1 ) w Γ q ( κ + 1 ) , t [ 0 , 1 ] .
In what follows, we will study the linear problem associated with our problem (1),(2). We consider the system of fractional q-difference equations
( D q α u ) ( t ) + h ( t ) = 0 , t [ 0 , 1 ] , ( D q β v ) ( t ) + k ( t ) = 0 , t [ 0 , 1 ] ,
subject to boundary conditions (2), where h , k C [ 0 , 1 ] .
We define
Λ 1 = Γ q ( α ) Γ q ( α ς ) i = 1 a a i Γ q ( α ) Γ q ( α ϱ i ) ξ i α ϱ i 1 , Λ 2 = i = 1 b b i Γ q ( β ) Γ q ( β σ i ) ω i β σ i 1 , Λ 3 = i = 1 c c i Γ q ( α ) Γ q ( α η i ) ζ i α η i 1 , Λ 4 = Γ q ( β ) Γ q ( β ϑ ) i = 1 d d i Γ q ( β ) Γ q ( β ρ i ) θ i β ρ i 1 , Δ = Λ 1 Λ 4 Λ 2 Λ 3 .
Lemma 5.
If Δ 0 , then the solution ( u ( t ) , v ( t ) ) , t [ 0 , 1 ] , of problem (25),(2) is given by
u ( t ) = 1 Γ q ( α ) 0 t ( t q s ) ( α 1 ) h ( s ) d q s + t α 1 Δ ( A Λ 4 + B Λ 2 ) , t [ 0 , 1 ] , v ( t ) = 1 Γ q ( β ) 0 t ( t q s ) ( β 1 ) k ( s ) d q s + t β 1 Δ ( B Λ 1 + A Λ 3 ) , t [ 0 , 1 ] ,
where
A = 1 Γ q ( α ς ) 0 1 ( 1 q s ) ( α ς 1 ) h ( s ) d q s i = 1 a a i Γ q ( α ϱ i ) 0 ξ i ( ξ i q s ) ( α ϱ i 1 ) h ( s ) d q s i = 1 b b i Γ q ( β σ i ) 0 ω i ( ω i q s ) ( β σ i 1 ) k ( s ) d q s , B = 1 Γ q ( β ϑ ) 0 1 ( 1 q s ) ( β ϑ 1 ) k ( s ) d q s i = 1 c c i Γ q ( α η i ) 0 ζ i ( ζ i q s ) ( α η i 1 ) h ( s ) d q s i = 1 d d i Γ q ( β ρ i ) 0 θ i ( θ i q s ) ( β ρ i 1 ) k ( s ) d q s .
Proof. 
We apply I q α and I q β , respectively, to the equations of system (25). Then, by Lemma 2 (a),(b) and Lemma 3, we obtain
u ( t ) = I q α h ( t ) + a ˜ 1 t α 1 + a ˜ 2 t α 2 + + a ˜ n t α n = 1 Γ q ( α ) 0 t ( t q s ) ( α 1 ) h ( s ) d q s + a ˜ 1 t α 1 + a ˜ 2 t α 2 + + a ˜ n t α n , t [ 0 , 1 ] , v ( t ) = I q β k ( t ) + b ˜ 1 t β 1 + b ˜ 2 t β 2 + + b ˜ m t β m = 1 Γ q ( β ) 0 t ( t q s ) ( β 1 ) k ( s ) d q s + b ˜ 1 t β 1 + b ˜ 2 t β 2 + + b ˜ m t β m , t [ 0 , 1 ] ,
for some a ˜ i , b ˜ j R , i = 1 , , n , j = 1 , , m .
From the conditions D q i u ( 0 ) = 0 for i = 0 , , n 2 , and D q j v ( 0 ) = 0 for j = 0 , , m 2 , we deduce that a ˜ i = 0 , i = 2 , , n , and b ˜ j = 0 , j = 2 , , m . So, the solution (29) becomes
u ( t ) = a ˜ 1 t α 1 1 Γ q ( α ) 0 t ( t q s ) ( α 1 ) h ( s ) d q s , t [ 0 , 1 ] , v ( t ) = b ˜ 1 t β 1 1 Γ q ( β ) 0 t ( t q s ) ( β 1 ) k ( s ) d q s , t [ 0 , 1 ] .
On the other hand, we find
D q κ u ( t ) = a ˜ 1 Γ q ( α ) Γ q ( α κ ) t α κ 1 1 Γ q ( α κ ) 0 t ( t q s ) ( α κ 1 ) h ( s ) d q s ,
for κ = ς , ϱ i , η j , i = 1 , , a , j = 1 , , c , and
D q κ v ( t ) = b ˜ 1 Γ q ( β ) Γ q ( β κ ) t β κ 1 1 Γ q ( β κ ) 0 t ( t q s ) ( β κ 1 ) k ( s ) d q s ,
for κ = ϑ , σ i , ρ j , i = 1 , , b , j = 1 , , d .
By using relations (31) and (32), the boundary conditions D q ς u ( 1 ) = i = 1 a a i D q ϱ i u ( ξ i ) + i = 1 b b i D q σ i v ( ω i ) and D q ϑ v ( 1 ) = i = 1 c c i D q η i u ( ζ i ) + i = 1 d d i D q ρ i v ( θ i ) become
a ˜ 1 Γ q ( α ) Γ q ( α ς ) 1 Γ q ( α ς ) 0 1 ( 1 q s ) ( α ς 1 ) h ( s ) d q s = i = 1 a a i a ˜ 1 Γ q ( α ) Γ q ( α ϱ i ) ξ i α ϱ i 1 1 Γ q ( α ϱ i ) 0 ξ i ( ξ i q s ) ( α ϱ i 1 ) h ( s ) d q s + i = 1 b b i b ˜ 1 Γ q ( β ) Γ q ( β σ i ) ω i β σ i 1 1 Γ q ( β σ i ) 0 ω i ( ω i q s ) ( β σ i 1 ) k ( s ) d q s , b ˜ 1 Γ q ( β ) Γ ( β ϑ ) 1 Γ q ( β ϑ ) 0 1 ( 1 q s ) ( β ϑ 1 ) k ( s ) d q s = i = 1 c c i a ˜ 1 Γ q ( α ) Γ q ( α η i ) ζ i α η i 1 1 Γ q ( α η i ) 0 ζ i ( ζ i q s ) ( α η i 1 ) h ( s ) d q s + i = 1 d d i b ˜ 1 Γ q ( β ) Γ q ( β ρ i ) θ i β ρ i 1 1 Γ q ( β ρ i ) 0 θ i ( θ i q s ) ( β ρ i 1 ) k ( s ) d q s .
or
a ˜ 1 Γ q ( α ) Γ q ( α ς ) i = 1 a a i Γ q ( α ) Γ q ( α ϱ i ) ξ i α ϱ i 1 b ˜ 1 i = 1 b b i Γ q ( β ) Γ q ( β σ i ) ω i β σ i 1 = 1 Γ q ( α ς ) 0 1 ( 1 q s ) ( α ς 1 ) h ( s ) d q s i = 1 a a i Γ q ( α ϱ i ) 0 ξ i ( ξ i q s ) ( α ϱ i 1 ) h ( s ) d q s i = 1 b b i Γ q ( β σ i ) 0 ω i ( ω i q s ) ( β σ i 1 ) k ( s ) d q s , a ˜ 1 i = 1 c c i Γ q ( α ) Γ q ( α η i ) ζ i α η i 1 + b ˜ 1 Γ q ( β ) Γ q ( β ϑ ) i = 1 d d i Γ q ( β ) Γ q ( β ρ i ) θ i β ρ i 1 = 1 Γ q ( β ϑ ) 0 1 ( 1 q s ) ( β ϑ 1 ) k ( s ) d q s i = 1 c c i Γ q ( α η i ) 0 ζ i ( ζ i q s ) ( α η i 1 ) h ( s ) d q s i = 1 d d i Γ q ( β ρ i ) 0 θ i ( θ i q s ) ( β ρ i 1 ) k ( s ) d q s .
The determinant of system (34) in the unknowns a ˜ 1 and b ˜ 1 is Δ , which, by the assumption of this lemma, is nonzero. So, system (34) has the unique solution
a ˜ 1 = 1 Δ ( A Λ 4 + B Λ 2 ) , b ˜ 1 = 1 Δ ( B Λ 1 + A Λ 3 ) .
By substituting (35) into (30), we obtain the solution ( u ( t ) , v ( t ) ) , t [ 0 , 1 ] , of problem (25),(2) given by (27). □
Lemma 6.
If Δ 0 , then the solution ( u ( t ) , v ( t ) ) , t [ 0 , 1 ] , of problem (25),(2) can be written as
u ( t ) = 0 1 G 1 ( t , q s ) h ( s ) d q s + 0 1 G 2 ( t , q s ) k ( s ) d q s , t [ 0 , 1 ] , v ( t ) = 0 1 G 3 ( t , q s ) h ( s ) d q s + 0 1 G 4 ( t , q s ) k ( s ) d q s , t [ 0 , 1 ] ,
where
G 1 ( t , s ) = g 1 ( t , s ) + t α 1 Δ Λ 4 i = 1 a a i g 1 i ( ξ i , s ) + Λ 2 i = 1 c c i g 2 i ( ζ i , s ) , G 2 ( t , s ) = t α 1 Δ Λ 4 i = 1 b b i g 3 i ( ω i , s ) + Λ 2 i = 1 d d i g 4 i ( θ i , s ) , G 3 ( t , s ) = t β 1 Δ Λ 3 i = 1 a a i g 1 i ( ξ i , s ) + Λ 1 i = 1 c c i g 2 i ( ζ i , s ) , G 4 ( t , s ) = g 2 ( t , s ) + t β 1 Δ Λ 3 i = 1 b b i g 3 i ( ω i , s ) + Λ 1 i = 1 d d i g 4 i ( θ i , s ) ,
and
g 1 ( t , s ) = 1 Γ q ( α ) t α 1 ( 1 s ) ( α ς 1 ) ( t s ) ( α 1 ) , 0 s t 1 , t α 1 ( 1 s ) ( α ς 1 ) , 0 t s 1 , g 2 ( t , s ) = 1 Γ q ( β ) t β 1 ( 1 s ) ( β ϑ 1 ) ( t s ) ( β 1 ) , 0 s t 1 , t β 1 ( 1 s ) ( β ϑ 1 ) , 0 t s 1 , g 1 i ( t , s ) = 1 Γ q ( α ϱ i ) t α ϱ i 1 ( 1 s ) ( α ς 1 ) ( t s ) ( α ϱ i 1 ) , 0 s t 1 , t α ϱ i 1 ( 1 s ) ( α ς 1 ) , 0 t s 1 , g 2 j ( t , s ) = 1 Γ q ( α η j ) t α η j 1 ( 1 s ) ( α ς 1 ) ( t s ) ( α η j 1 ) , 0 s t 1 , t α η j 1 ( 1 s ) ( α ς 1 ) , 0 t s 1 , g 3 k ( t , s ) = 1 Γ q ( β σ k ) t β σ k 1 ( 1 s ) ( β ϑ 1 ) ( t s ) ( β σ k 1 ) , 0 s t 1 , t β σ k 1 ( 1 s ) ( β ϑ 1 ) , 0 t s 1 , g 4 ι ( t , s ) = 1 Γ q ( β ρ ι ) t β ρ ι 1 ( 1 s ) ( β ϑ 1 ) ( t s ) ( β ρ ι 1 ) , 0 s t 1 , t β ρ ι 1 ( 1 s ) ( β ϑ 1 ) , 0 t s 1 ,
for t , s [ 0 , 1 ] , i = 1 , , a , j = 1 , , c , k = 1 , , b , ι = 1 , , d .
Proof. 
For u ( t ) , we deduce
u ( t ) = 1 Γ q ( α ) 0 t ( t q s ) ( α 1 ) h ( s ) d q s + t α 1 Δ Λ 4 Γ q ( α ς ) 0 1 ( 1 q s ) ( α ς 1 ) h ( s ) d q s t α 1 Δ Λ 4 i = 1 a a i Γ q ( α ϱ i ) 0 ξ i ( ξ i q s ) ( α ϱ i 1 ) h ( s ) d q s t α 1 Δ Λ 2 i = 1 c c i Γ q ( α η i ) 0 ζ i ( ζ i q s ) ( α η i 1 ) h ( s ) d q s + t α 1 Δ Λ 4 i = 1 b b i Γ q ( β σ i ) 0 ω i ( ω i q s ) ( β σ i 1 ) k ( s ) d q s + Λ 2 Γ q ( β ϑ ) 0 1 ( 1 q s ) ( β ϑ 1 ) k ( s ) d q s Λ 2 i = 1 d d i Γ q ( β ρ i ) 0 θ i ( θ i q s ) ( β ρ i 1 ) k ( s ) d q s = t α 1 Γ q ( α ) 0 t ( 1 q s ) ( α ς 1 ) h ( s ) d q s 1 Γ q ( α ) 0 t ( t q s ) ( α 1 ) h ( s ) d q s + 1 Γ q ( α ) t 1 t α 1 ( 1 q s ) ( α ς 1 ) h ( s ) d q s t α 1 Γ q ( α ) 0 1 ( 1 q s ) ( α ς 1 ) h ( s ) d q s + t α 1 Δ Λ 4 Γ q ( α ς ) 0 1 ( 1 q s ) ( α ς 1 ) h ( s ) d q s t α 1 Δ Λ 4 i = 1 a a i Γ q ( α ϱ i ) 0 ξ i ( ξ i q s ) ( α ϱ i 1 ) h ( s ) d q s t α 1 Δ Λ 2 i = 1 c c i Γ q ( α η i ) 0 ζ i ( ζ i q s ) ( α η i 1 ) h ( s ) d q s + t α 1 Δ Λ 4 i = 1 b b i Γ q ( β σ i ) 0 ω i ( ω i q s ) ( β σ i 1 ) k ( s ) d q s + Λ 2 Γ q ( β ϑ ) 0 1 ( 1 q s ) ( β ϑ 1 ) k ( s ) d q s Λ 2 i = 1 d d i Γ q ( β ρ i ) 0 θ i ( θ i q s ) ( β ρ i 1 ) k ( s ) d q s , t [ 0 , 1 ] .
Amplifying the fourth term on the right-hand side of the above relation by Δ = Λ 1 Λ 4 Λ 2 Λ 3 , we obtain
u ( t ) = 1 Γ q ( α ) 0 1 t α 1 ( 1 q s ) ( α ς 1 ) h ( s ) d q s 1 Γ q ( α ) 0 t ( t q s ) ( α 1 ) h ( s ) d q s + t α 1 Δ Γ q ( α ) Γ q ( α ς ) Γ q ( β ) Γ q ( β ϑ ) 1 Γ q ( α ) 0 1 ( 1 q s ) ( α ς 1 ) h ( s ) d q s + Γ q ( β ) Γ q ( β ϑ ) i = 1 a a i Γ q ( α ) Γ q ( α ϱ i ) ξ i α ϱ i 1 1 Γ q ( α ) 0 1 ( 1 q s ) ( α ς 1 ) h ( s ) d q s + Γ q ( α ) Γ q ( α ς ) i = 1 d d i Γ q ( β ) Γ q ( β ρ i ) θ i β ρ i 1 1 Γ q ( α ) 0 1 ( 1 q s ) ( α ς 1 ) h ( s ) d q s i = 1 a a i Γ q ( α ) Γ q ( α ϱ i ) ξ i α ϱ i 1 i = 1 d d i Γ q ( β ) Γ q ( β ρ i ) θ i β ρ i 1 1 Γ q ( α ) 0 1 ( 1 q s ) ( α ς 1 ) h ( s ) d q s + i = 1 c c i Γ q ( α ) Γ q ( α η i ) ζ i α η i 1 i = 1 b b i Γ q ( β ) Γ q ( β σ i ) η i β η i 1 1 Γ q ( α ) 0 1 ( 1 q s ) ( α ς 1 ) h ( s ) d q s + Γ q ( β ) Γ q ( β ϑ ) 1 Γ q ( α ς ) 0 1 ( 1 q s ) ( α ς 1 ) h ( s ) d q s i = 1 d d i Γ q ( β ) Γ q ( β ρ i ) θ i β ρ i 1 1 Γ q ( α ς ) 0 1 ( 1 q s ) ( α ς 1 ) h ( s ) d q s Γ q ( β ) Γ q ( β ϑ ) i = 1 a a i Γ q ( α ϱ i ) 0 ξ i ( ξ i q s ) ( α ϱ i 1 ) h ( s ) d q s + i = 1 d d i Γ q ( β ) Γ q ( β ρ i ) θ i β ρ i 1 i = 1 a a i Γ q ( α ϱ i ) 0 ξ i ( ξ i q s ) ( α ϱ i 1 ) h ( s ) d q s i = 1 b b i Γ q ( β ) Γ q ( β σ i ) η i β η i 1 i = 1 c c i Γ q ( α η i ) 0 ζ i ( ζ i q s ) ( α η i 1 ) h ( s ) d q s + t α 1 Δ Γ q ( β ) Γ q ( β ϑ ) i = 1 b b i Γ q ( β σ i ) 0 ω i ( ω i q s ) ( β σ i 1 ) k ( s ) d q ( s ) + i = 1 d d i Γ q ( β ) Γ q ( β ρ i ) θ i β ρ i 1 i = 1 b b i Γ q ( β σ i ) 0 ω i ( ω i q s ) ( β σ i 1 ) k ( s ) d q s + i = 1 b b i Γ q ( β ) Γ q ( β σ i ) ω i β σ i 1 1 Γ q ( β ϑ ) 0 1 ( 1 q s ) ( β ϑ 1 ) k ( s ) d q s i = 1 b b i Γ q ( β ) Γ q ( β σ i ) ω i β σ i 1 i = 1 d d i Γ q ( β ρ i ) 0 θ i ( θ i q s ) ( β ρ i 1 ) k ( s ) d q s = 1 Γ q ( α ) 0 1 t α 1 ( 1 q s ) ( α ς 1 ) h ( s ) d q s 1 Γ q ( α ) 0 t ( t q s ) ( α 1 ) h ( s ) d q s + t α 1 Δ Γ q ( β ) Γ q ( β ϑ ) i = 1 a a i Γ q ( α ϱ i ) ξ i α ϱ i 1 0 1 ( 1 q s ) ( α ς 1 ) h ( s ) d q s 0 ξ i ( ξ i q s ) ( α ϱ i 1 ) h ( s ) d q s i = 1 d d i Γ q ( β ) Γ q ( β ρ i ) θ i β ρ i 1 i = 1 a a i Γ q ( α ϱ i ) ξ i α ϱ i 1 0 1 ( 1 q s ) ( α ς 1 ) h ( s ) d q s 0 ξ i ( ξ i q s ) ( α ϱ i 1 ) h ( s ) d q s + i = 1 b b i Γ q ( β ) Γ q ( β σ i ) η i β η i 1 i = 1 c c i Γ q ( α η i ) ζ i α η i 1 0 1 ( 1 q s ) ( α ς 1 ) h ( s ) d q s 0 ζ i ( ζ i q s ) ( α η i 1 ) h ( s ) d q s + t α 1 Δ i = 1 b b i Γ q ( β ) Γ q ( β ϑ ) 1 Γ q ( β σ i ) 0 1 ω i β σ i 1 ( 1 q s ) ( β i ϑ 1 ) k ( s ) d q s 0 ω i ( ω i q s ) ( β σ i 1 ) k ( s ) d q s i = 1 d d i Γ q ( β ) Γ q ( β ρ i ) θ i β ρ i 1 i = 1 b b i Γ q ( β σ i ) 0 1 ω i β σ i 1 ( 1 q s ) ( β ϑ 1 ) k ( s ) d q s + i = 1 d d i Γ q ( β ) Γ q ( β ρ i ) θ i β ρ i 1 i = 1 b b i Γ q ( β σ i ) 0 ω i ( ω i q s ) ( β σ i 1 ) k ( s ) d q s + i = 1 b b i Γ q ( β ) Γ q ( β σ i ) ω i β σ i 1 i = 1 d d i Γ q ( β ρ i ) 0 1 θ i β ρ i 1 ( 1 q s ) ( β ϑ 1 ) k ( s ) d q s i = 1 b b i Γ q ( β ) Γ q ( β σ i ) ω i β σ i 1 i = 1 d d i Γ q ( β ρ i ) 0 θ i ( θ i q s ) ( β ρ i 1 ) k ( s ) d q s , t [ 0 , 1 ] .
So, we deduce
u ( t ) = 1 Γ q ( α ) 0 1 t α 1 ( 1 q s ) ( α ς 1 ) h ( s ) d q s 1 Γ q ( α ) 0 t ( t q s ) ( α 1 ) h ( s ) d q s + t α 1 Δ Γ q ( β ) Γ q ( β ϑ ) i = 1 d d i Γ q ( β ) Γ q ( β ρ i ) θ i β ρ i 1 × i = 1 a a i Γ q ( α ϱ i ) ξ i α ϱ i 1 0 1 ( 1 q s ) ( α ς 1 ) h ( s ) d q s 0 ξ i ( ξ i q s ) ( α ϱ i 1 ) h ( s ) d q s + i = 1 b b i Γ q ( β ) Γ q ( β σ i ) η i β η i 1 × i = 1 c c i Γ q ( α η i ) ζ i α η i 1 0 1 ( 1 q s ) ( α ζ 1 ) h ( s ) d q s 0 ζ i ( ζ i q s ) ( α η i 1 ) h ( s ) d q s + t α 1 Δ i = 1 b b i Γ q ( β ) Γ q ( β ϑ ) 1 Γ q ( β σ i ) 0 1 ω i β σ i 1 ( 1 q s ) ( β ϑ 1 ) k ( s ) d q s 0 ω i ( ω i q s ) ( β σ i 1 ) k ( s ) d q s i = 1 d d i Γ q ( β ) Γ q ( β ρ i ) θ i β ρ i 1 × i = 1 b b i Γ q ( β σ i ) 0 1 ω i β σ i 1 ( 1 q s ) ( β ϑ 1 ) k ( s ) d q s 0 ω i ( ω i q s ) ( β σ i 1 ) k ( s ) d q s + i = 1 b b i Γ q ( β ) Γ q ( β σ i ) ω i β σ i 1 × i = 1 d d i Γ q ( β ρ i ) 0 1 θ i β ρ i 1 ( 1 q s ) ( β ϑ 1 ) k ( s ) d q s 0 θ i ( θ i q s ) ( β ρ i 1 ) k ( s ) d q s = 1 Γ q ( α ) 0 1 t α 1 ( 1 q s ) ( α ς 1 ) h ( s ) d q s 1 Γ q ( α ) 0 t ( t q s ) ( α 1 ) h ( s ) d q s + t α 1 Δ Λ 4 i = 1 a a i Γ q ( α ϱ i ) ξ i α ϱ i 1 0 1 ( 1 q s ) ( α ς 1 ) h ( s ) d q s 0 ξ i ( ξ i q s ) ( α ϱ i 1 ) h ( s ) d q s + Λ 2 i = 1 c c i Γ q ( α η i ) ζ i α η i 1 0 1 ( 1 q s ) ( α ς 1 ) h ( s ) d q s 0 ζ i ( ζ i q s ) ( α η i 1 ) h ( s ) d q s + t α 1 Δ Λ 4 i = 1 b b i Γ q ( β σ i ) 0 1 ω i β σ i 1 ( 1 q s ) ( β ϑ 1 ) k ( s ) d q s 0 ω i ( ω i q s ) ( β σ i 1 ) k ( s ) d q s + Λ 2 i = 1 d d i Γ q ( β ρ i ) 0 1 θ i β ρ i 1 ( 1 q s ) ( β ϑ 1 ) k ( s ) d q s 0 θ i ( θ i q s ) ( β ρ i 1 ) k ( s ) d q s .
Therefore, we obtain
u ( t ) = 0 1 g 1 ( t , q s ) + t α 1 Δ Λ 4 i = 1 a a i g 1 i ( ξ i , q s ) + Λ 2 i = 1 c c i g 2 i ( ζ i , q s ) h ( s ) d q s + t α 1 Δ 0 1 Λ 4 i = 1 b b i g 3 i ( ω i , q s ) + Λ 2 i = 1 d d i g 4 i ( θ i , q s ) k ( s ) d q s = 0 1 G 1 ( t , q s ) h ( s ) d q s + 0 1 G 2 ( t , q s ) k ( s ) d q s , t [ 0 , 1 ] ,
where G 1 , G 2 , g 1 , g 1 i , i = 1 , , a , g 2 j , j = 1 , , c , g 3 k , k = 1 , , b , and g 4 ι , ι = 1 , , d , are given by (37) and (38).
In a similar manner, we find
v ( t ) = t β 1 Δ 0 1 Λ 3 i = 1 a a i g 1 i ( ξ i , q s ) + Λ 1 i = 1 c c i g 2 i ( ζ i , q s ) h ( s ) d q s + 0 1 g 2 ( t , q s ) + t β 1 Δ Λ 3 i = 1 b b i g 3 i ( ω i , q s ) + Λ 1 i = 1 d d i g 4 i ( θ i , q s ) k ( s ) d q s = 0 1 G 3 ( t , q s ) h ( s ) d q s + 0 1 G 4 ( t , q s ) k ( s ) d q s , t [ 0 , 1 ] ,
where G 3 , G 4 , and g 2 are given by (37) and (38). So, we deduce the formulas in (36) for the solution ( u ( t ) , v ( t ) ) , t [ 0 , 1 ] , of problem (25),(2). □
With a similar proof to that of Lemma 12 from [4], we obtain the next result.
Lemma 7.
The functions g 1 , g 2 , g 1 i , i = 1 , , a , g 2 j , j = 1 , , c , g 3 k , k = 1 , , b , and g 4 ι , ι = 1 , , d , have the following properties:
(a) g 1 ( t , q s ) 0 , g 2 ( t , q s ) 0 , t , s [ 0 , 1 ] ;
(b) g 1 ( t , q s ) g 1 ( 1 , q s ) , g 2 ( t , q s ) g 2 ( 1 , q s ) , t , s [ 0 , 1 ] ;
(c) g 1 ( t , q s ) t α 1 g 1 ( 1 , q s ) , g 2 ( t , q s ) t β 1 g 2 ( 1 , q s ) , t , s [ 0 , 1 ] ;
(d) g 1 i ( t , q s ) 0 , g 2 j ( t , q s ) 0 , g 3 k ( t , q s ) 0 , g 4 ι ( t , q s ) 0 , t , s [ 0 , 1 ] , i = 1 , , a , j = 1 , , c , k = 1 , , b , ι = 1 , , d .
Remark 1.
If ϖ = q n with n N , then we obtain
min t [ ϖ , 1 ] g 1 ( t , q s ) ϖ α 1 g 1 ( 1 , q s ) , min t [ ϖ , 1 ] g 2 ( t , q s ) ϖ β 1 g 2 ( 1 , q s ) , s [ 0 , 1 ] .
Lemma 8.
If Λ 1 > 0 , Λ 4 > 0 , and Δ > 0 , then the functions G i , i = 1 , , 4 satisfy the following inequalities:
(a) t α 1 G 1 ( 1 , q s ) G 1 ( t , q s ) G 1 ( 1 , q s ) , t , s [ 0 , 1 ] ;
(b) t α 1 G 2 ( 1 , q s ) = G 2 ( t , q s ) G 2 ( 1 , q s ) , t , s [ 0 , 1 ] ;
(c) t β 1 G 3 ( 1 , q s ) = G 3 ( t , q s ) G 3 ( 1 , q s ) , t , s [ 0 , 1 ] ;
(d) t β 1 G 4 ( 1 , q s ) G 4 ( t , q s ) G 4 ( 1 , q s ) , t , s [ 0 , 1 ] .
Proof. 
( a ) ( d ) By using Lemma 7, we deduce the following for all t , s [ 0 , 1 ] :
G 1 ( t , q s ) = g 1 ( t , q s ) + t α 1 Δ Λ 4 i = 1 a a i g 1 i ( ξ i , q s ) + Λ 2 i = 1 c c i g 2 i ( ζ i , q s ) g 1 ( 1 , q s ) + 1 Δ Λ 4 i = 1 a a i g 1 i ( ξ i , q s ) + Λ 2 i = 1 c c i g 2 i ( ζ i , q s ) = G 1 ( 1 , q s ) ; G 1 ( t , q s ) t α 1 g 1 ( 1 , q s ) + t α 1 Δ Λ 4 i = 1 a a i g 1 i ( ξ i , q s ) + Λ 2 i = 1 c c i g 2 i ( ζ i , q s ) = t α 1 G 1 ( 1 , q s ) ; G 2 ( t , q s ) = t α 1 Δ Λ 4 i = 1 b b i g 3 i ( ω i , q s ) + Λ 2 i = 1 d d i g 4 i ( θ i , q s ) 1 Δ Λ 4 i = 1 b b i g 3 i ( ω i , q s ) + Λ 2 i = 1 d d i g 4 i ( θ i , q s ) = G 2 ( 1 , q s ) ; G 2 ( t , q s ) = t α 1 G 2 ( 1 , q s ) ; G 3 ( t , q s ) = t β 1 Δ Λ 3 i = 1 a a i g 1 i ( ξ i , q s ) + Λ 1 i = 1 c c i g 2 i ( ζ i , q s ) 1 Δ Λ 3 i = 1 a a i g 1 i ( ξ i , q s ) + Λ 1 i = 1 c c i g 2 i ( ζ i , q s ) = G 3 ( 1 , q s ) ; G 3 ( t , q s ) = t β 1 G 3 ( 1 , q s ) ; G 4 ( t , q s ) = g 2 ( t , q s ) + t β 1 Δ Λ 3 i = 1 b b i g 3 i ( ω i , q s ) + Λ 1 i = 1 d d i g 4 i ( θ i , q s ) g 2 ( 1 , q s ) + 1 Δ Λ 3 i = 1 b b i g 3 i ( ω i , q s ) + Λ 1 i = 1 d d i g 4 i ( θ i , q s ) = G 4 ( 1 , q s ) ; G 4 ( t , q s ) t β 1 g 2 ( 1 , q s ) + t β 1 Δ Λ 3 i = 1 b b i g 3 i ( ω i , q s ) + Λ 1 i = 1 d d i g 4 i ( θ i , q s ) = t β 1 G 4 ( 1 , q s ) .
  □
Lemma 9.
If  Λ 1 > 0 , Λ 4 > 0 , Δ > 0 , h ( t ) 0 , and  k ( t ) 0  for all  t [ 0 , 1 ] then the solution  ( u ( t ) , v ( t ) ) , t [ 0 , 1 ] of problem (25),(2) satisfies the inequalities  u ( t ) 0  and  v ( t ) 0  for all  t [ 0 , 1 ]  and  u ( t ) t α 1 u ( τ )  and  v ( t ) t β 1 v ( τ )  for all  t , τ [ 0 , 1 ] .
Proof. 
We can easily verify that u ( t ) 0 and v ( t ) 0 for all t [ 0 , 1 ] . In addition, by using Lemma 8, we obtain
u ( t ) = 0 1 G 1 ( t , q s ) h ( s ) d q s + 0 1 G 2 ( t , q s ) k ( s ) d q s 0 1 G 1 ( 1 , q s ) h ( s ) d q s + 0 1 G 2 ( 1 , q s ) k ( s ) d q s , t [ 0 , 1 ] , u ( t ) 0 1 t α 1 G 1 ( 1 , q s ) h ( s ) d q s + 0 1 t α 1 G 2 ( 1 , q s ) k ( s ) d q s = t α 1 0 1 G 1 ( 1 , q s ) h ( s ) d q s + 0 1 G 2 ( 1 , q s ) k ( s ) d q s t α 1 u ( τ ) , t , τ [ 0 , 1 ] , v ( t ) = 0 1 G 3 ( t , q s ) h ( s ) d q s + 0 1 G 4 ( t , q s ) k ( s ) d q s 0 1 G 3 ( 1 , q s ) h ( s ) d q s + 0 1 G 4 ( 1 , q s ) k ( s ) d q s , t [ 0 , 1 ] ,
v ( t ) 0 1 t β 1 G 3 ( 1 , q s ) h ( s ) d q s + 0 1 t β 1 G 4 ( 1 , q s ) k ( s ) d q s = t β 1 0 1 G 3 ( 1 , q s ) h ( s ) d q s + 0 1 G 4 ( 1 , q s ) k ( s ) d q s t β 1 v ( τ ) , t , τ [ 0 , 1 ] .
Remark 2.
By Lemma 9, we find u ( t ) t α 1 u and v ( t ) t β 1 v for all t [ 0 , 1 ] . In addition, by Lemma 8, for ϖ = q n with n N , we have
min t [ ϖ , 1 ] G 1 ( t , q s ) ϖ α 1 G 1 ( 1 , q s ) , s [ 0 , 1 ] ; min t [ ϖ , 1 ] G 2 ( t , q s ) = ϖ α 1 G 2 ( 1 , q s ) , s [ 0 , 1 ] ; min t [ ϖ , 1 ] G 3 ( t , q s ) = ϖ β 1 G 3 ( 1 , q s ) , s [ 0 , 1 ] ; min t [ ϖ , 1 ] G 4 ( t , q s ) ϖ β 1 G 4 ( 1 , q s ) , s [ 0 , 1 ] ,
and so
min t [ ϖ , 1 ] u ( t ) ϖ α 1 u , min t [ ϖ , 1 ] v ( t ) ϖ β 1 v .

3. Main Results

In this section, we will outline the results concerning the existence of positive solutions for our given problem (1),(2).
Initially, we state our primary assumptions:
(H1)
q ( 0 , 1 ) , α , β R , α ( n 1 , n ] , β ( m 1 , m ] , n , m N , n , m 3 ; a , b , c , d N , 0 ϱ i ς < α 1 , i = 1 , , a , ς 1 , 0 η i ς , i = 1 , , c , 0 σ i ϑ < β 1 , i = 1 , , b , ϑ 1 , 0 ρ i ϑ , i = 1 , , d ; δ i , γ i > 0 , i = 1 , 2 ; ξ i , ω j , ζ k , θ ι ( 0 , 1 ) , a i , b j , c k , d ι 0 for i = 1 , , a , j = 1 , , b , k = 1 , , c , ι = 1 , , d ; Λ 1 > 0 , Λ 4 > 0 , Δ > 0 (given by (26)).
(H2)
f , g : [ 0 , 1 ] × R + 4 R + are continuous functions ( R + = [ 0 , ) ).
Problem (1),(2) can be expressed equivalently to the following system of fractional q-integral equations:
u ( t ) = 0 1 G 1 ( t , q s ) f ( s , u ( s ) , v ( s ) , I q δ 1 u ( s ) , I q γ 1 v ( s ) ) d q s + 0 1 G 2 ( t , q s ) g ( s , u ( s ) , v ( s ) , I q δ 2 u ( s ) , I q γ 2 v ( s ) ) d q s , t [ 0 , 1 ] , v ( t ) = 0 1 G 3 ( t , q s ) f ( s , u ( s ) , v ( s ) , I q δ 1 u ( s ) , I q γ 1 v ( s ) ) d q s + 0 1 G 4 ( t , q s ) g ( s , u ( s ) , v ( s ) , I q δ 2 u ( s ) , I q γ 2 v ( s ) ) d q s , t [ 0 , 1 ] .
Let X = C [ 0 , 1 ] be the Banach space endowed with the norm u = sup t [ 0 , 1 ] | u ( t ) | , and let Y = X × X be the Banach space with the norm ( u , v ) Y = u + v . We also introduce the cone P Y by
P = { ( u , v ) Y , u ( t ) t α 1 u , v ( t ) t β 1 v , t [ 0 , 1 ] } .
We now define the operator A : P Y , A ( u , v ) = ( A 1 ( u , v ) , A 2 ( u , v ) ) , for ( u , v ) P , where A 1 , A 2 : P X are given by
A 1 ( u , v ) ( t ) = 0 1 G 1 ( t , q s ) f ( s , u ( s ) , v ( s ) , I q δ 1 u ( s ) , I q γ 1 v ( s ) ) d q s + 0 1 G 2 ( t , q s ) g ( s , u ( s ) , v ( s ) , I q δ 2 u ( s ) , I q γ 2 v ( s ) ) d q s , A 2 ( u , v ) ( t ) = 0 1 G 3 ( t , q s ) f ( s , u ( s ) , v ( s ) , I q δ 1 u ( s ) , I q γ 1 v ( s ) ) d q s + 0 1 G 4 ( t , q s ) g ( s , u ( s ) , v ( s ) , I q δ 2 u ( s ) , I q γ 2 v ( s ) ) d q s ,
for all t [ 0 , 1 ] and ( u , v ) P .
We observe that ( u , v ) constitutes a positive solution to problem (49) (or equivalently, (1),(2)) if and only if it serves as a fixed point for the operator A . Consequently, our subsequent analysis will focus on examining the existence of fixed points for A .
Under assumptions ( H 1 ) and ( H 2 ) , using standard arguments, we deduce that operator A is completely continuous. In addition, by Lemma 9, we obtain
A 1 ( u , v ) ( t ) t α 1 A 1 ( u , v ) , A 2 ( u , v ) ( t ) t β 1 A 1 ( u , v ) , t [ 0 , 1 ] ,
that is, A ( P ) P .
For ϖ = q n , with n N , we introduce the constants
L 1 = 0 1 G 1 ( 1 , q s ) d q s , L 2 = 0 1 G 2 ( 1 , q s ) d q s , L 3 = 0 1 G 3 ( 1 , q s ) d q s , L 4 = 0 1 G 2 ( 1 , q s ) d q s , M 1 = L 1 + L 3 , M 2 = L 2 + L 4 , L ˜ 1 = ϖ 1 G 1 ( 1 , q s ) d q s , L ˜ 2 = ϖ 1 G 2 ( 1 , q s ) d q s , L ˜ 3 = ϖ 1 G 3 ( 1 , q s ) d q s , L ˜ 4 = ϖ 1 G 4 ( 1 , q s ) d q s , M ˜ 1 = ϖ α 1 L ˜ 1 + ϖ β 1 L ˜ 3 , M ˜ 2 = ϖ α 1 L ˜ 2 + ϖ β 1 L ˜ 4 .
We remark that L 2 , L 3 , L ˜ 2 , and L ˜ 3 0 and L 1 ,   L 4 ,   L ˜ 1 ,   L ˜ 4 ,   M 1 ,   M 2 ,   M ˜ 1 ,  and  M ˜ 2 > 0 .
Our initial existence result for positive solutions to problem (1),(2) relies on the Guo–Krasnosel’skii fixed-point theorem (refer to [28]).
Theorem 1.
Let ϖ = q n with n N . Assume that ( H 1 ) and ( H 2 ) are satisfied. In addition, we suppose that there exist two positive constants r 2 > r 1 > 0 and the constants σ ˜ 1 ( 0 , M 1 1 ] , σ ˜ 2 ( 0 , M 2 1 ] , σ ˜ 3 [ M ˜ 1 1 , ) , and σ ˜ 4 [ M ˜ 2 1 , ) such that
  • (H3) f ( t , u , v , x , y ) σ ˜ 3 r 1 2 , t [ ϖ , 1 ] , u , v 0 , u + v r 1 , x 0 , r 1 Γ q ( δ 1 + 1 ) , y 0 , r 1 Γ q ( γ 1 + 1 ) , g ( t , u , v , x , y ) σ ˜ 4 r 1 2 , t [ ϖ , 1 ] , u , v 0 , u + v r 1 , x 0 , r 1 Γ q ( δ 2 + 1 ) , y 0 , r 1 Γ q ( γ 2 + 1 ) ;
  • (H4) f ( t , u , v , x , y ) σ ˜ 1 r 2 2 , t [ 0 , 1 ] , u , v 0 , u + v r 2 , x 0 , r 2 Γ q ( δ 1 + 1 ) , y 0 , r 2 Γ q ( γ 1 + 1 ) , g ( t , u , v , x , y ) σ ˜ 2 r 2 2 , t [ 0 , 1 ] , u , v 0 , u + v r 2 , x 0 , r 2 Γ q ( δ 2 + 1 ) , y 0 , r 2 Γ q ( γ 2 + 1 ) .
Then, problem (1),(2) has at least one positive solution ( u ( t ) , v ( t ) ) , t [ 0 , 1 ] such that  ( u , v ) P  and  r 1 ( u , v ) Y r 2 .
Proof. 
We introduce the set Ω 2 = { ( u , v ) Y , ( u , v ) Y < r 2 } . Then, for ( u , v ) P Ω 2 , we have u + v = r 2 , so u ( t ) + v ( t ) r 2 for all t [ 0 , 1 ] . Because u r 2 and v r 2 , by Lemma 4, we find | I q δ i u ( s ) | r 2 Γ q ( δ i + 1 ) , and | I q γ j v ( s ) | r 2 Γ q ( γ j + 1 ) for all s [ 0 , 1 ] , i , j = 1 , 2 . We define F u v δ 1 γ 1 ( s ) = f ( s , u ( s ) , v ( s ) , I q δ 1 u ( s ) , I q γ 1 v ( s ) ) and G u v δ 2 γ 2 ( s ) = g ( s , u ( s ) , v ( s ) , I q δ 2 u ( s ) , I q γ 2 v ( s ) ) for s [ 0 , 1 ] .
Then, by Lemma 8 and ( H 4 ) , we obtain
A 1 ( u , v ) = sup t [ 0 , 1 ] | A 1 ( u , v ) ( t ) | sup t [ 0 , 1 ] 0 1 G 1 ( t , q s ) F u v δ 1 γ 1 ( s ) d q s + sup t [ 0 , 1 ] 0 1 G 2 ( t , q s ) G u v δ 2 γ 2 ( s ) d q s 0 1 sup t [ 0 , 1 ] G 1 ( t , q s ) F u v δ 1 γ 1 ( s ) d q s + 0 1 sup t [ 0 , 1 ] G 2 ( t , q s ) G u v δ 2 γ 2 ( s ) d q s 0 1 G 1 ( 1 , q s ) F u v δ 1 γ 1 ( s ) d q s + 0 1 G 2 ( 1 , q s ) G u v δ 2 γ 2 ( s ) d q s σ ˜ 1 r 2 2 0 1 G 1 ( 1 , q s ) d q s + σ ˜ 2 r 2 2 0 1 G 2 ( 1 , q s ) d q s = σ ˜ 1 r 2 2 L 1 + σ ˜ 2 r 2 2 L 2 = σ ˜ 1 L 1 2 + σ ˜ 2 L 2 2 r 2 ,
and
A 2 ( u , v ) = sup t [ 0 , 1 ] | A 2 ( u , v ) ( t ) | sup t [ 0 , 1 ] 0 1 G 3 ( t , q s ) F u v δ 1 γ 1 ( s ) d q s + sup t [ 0 , 1 ] 0 1 G 4 ( t , q s ) G u v δ 2 γ 2 ( s ) d q s 0 1 sup t [ 0 , 1 ] G 3 ( t , q s ) F u v δ 1 γ 1 ( s ) d q s + 0 1 sup t [ 0 , 1 ] G 4 ( t , q s ) G u v δ 2 γ 2 ( s ) d q s 0 1 G 3 ( 1 , q s ) F u v δ 1 γ 1 ( s ) d q s + 0 1 G 4 ( 1 , q s ) G u v δ 2 γ 2 ( s ) d q s σ ˜ 1 r 2 2 0 1 G 3 ( 1 , q s ) d q s + σ ˜ 2 r 2 2 0 1 G 4 ( 1 , q s ) d q s = σ ˜ 1 r 2 2 L 3 + σ ˜ 2 r 2 2 L 4 = σ ˜ 1 L 3 2 + σ ˜ 2 L 4 2 r 2 .
Then, we deduce
A ( u , v ) Y = A 1 ( u , v ) + A 2 ( u , v ) σ ˜ 1 ( L 1 + L 3 ) 2 + σ ˜ 2 ( L 2 + L 4 ) 2 r 2 = σ ˜ 1 M 1 2 + σ ˜ 2 M 2 2 r 2 r 2 ,
that is,
A ( u , v ) Y ( u , v ) Y , ( u , v ) P Ω 2 .
Now, we define the set Ω 1 = { ( u , v ) Y , ( u , v ) Y < r 1 } . Then, for ( u , v ) P Ω 1 , we have u + v = r 1 , so u ( t ) + v ( t ) r 1 for all t [ 0 , 1 ] .
Therefore, by Lemma 8 and ( H 3 ) , we obtain
A 1 ( u , v ) = sup t [ 0 , 1 ] | A 1 ( u , v ) ( t ) | = sup t [ 0 , 1 ] 0 1 G 1 ( t , q s ) F u v δ 1 γ 1 ( s ) d q s + 0 1 G 2 ( t , q s ) G u v δ 2 γ 2 ( s ) d q s inf t [ ϖ , 1 ] 0 1 G 1 ( t , q s ) F u v δ 1 γ 1 ( s ) d q s + 0 1 G 2 ( t , q s ) G u v δ 2 γ 2 ( s ) d q s inf t [ ϖ , 1 ] 0 1 G 1 ( t , q s ) F u v δ 1 γ 1 ( s ) d q s + inf t [ ϖ , 1 ] 0 1 G 2 ( t , q s ) G u v δ 2 γ 2 ( s ) d q s 0 1 inf t [ ϖ , 1 ] G 1 ( t , q s ) F u v δ 1 γ 1 ( s ) d q s + 0 1 inf t [ ϖ , 1 ] G 2 ( t , q s ) G u v δ 2 γ 2 ( s ) d q s ϖ 1 ϖ α 1 G 1 ( 1 , q s ) F u v δ 1 γ 1 ( s ) d q s + ϖ 1 ϖ α 1 G 2 ( 1 , q s ) G u v δ 2 γ 2 ( s ) d q s σ ˜ 3 r 1 ϖ α 1 2 ϖ 1 G 1 ( 1 , q s ) d q s + σ ˜ 4 r 1 ϖ α 1 2 ϖ 1 G 2 ( 1 , q s ) d q s = σ ˜ 3 r 1 ϖ α 1 L ˜ 1 2 + σ ˜ 4 r 1 ϖ α 1 L ˜ 2 2 = σ ˜ 3 ϖ α 1 L ˜ 1 2 + σ ˜ 4 ϖ α 1 L ˜ 2 2 r 1 ,
and
A 2 ( u , v ) = sup t [ 0 , 1 ] | A 2 ( u , v ) ( t ) | = sup t [ 0 , 1 ] 0 1 G 3 ( t , q s ) F u v δ 1 γ 1 ( s ) d q s + 0 1 G 4 ( t , q s ) G u v δ 2 γ 2 ( s ) d q s inf t [ ϖ , 1 ] 0 1 G 3 ( t , q s ) F u v δ 1 γ 1 ( s ) d q s + 0 1 G 4 ( t , q s ) G u v δ 2 γ 2 ( s ) d q s inf t [ ϖ , 1 ] 0 1 G 3 ( t , q s ) F u v δ 1 γ 1 ( s ) d q s + inf t [ ϖ , 1 ] 0 1 G 4 ( t , q s ) G u v δ 2 γ 2 ( s ) d q s 0 1 inf t [ ϖ , 1 ] G 3 ( t , q s ) F u v δ 1 γ 1 ( s ) d q s + 0 1 inf t [ ϖ , 1 ] G 4 ( t , q s ) G u v δ 2 γ 2 ( s ) d q s ϖ 1 ϖ β 1 G 3 ( 1 , q s ) F u v δ 1 γ 1 ( s ) d q s + ϖ 1 ϖ β 1 G 4 ( 1 , q s ) G u v δ 2 γ 2 ( s ) d q s σ ˜ 3 r 1 ϖ β 1 2 ϖ 1 G 3 ( 1 , q s ) d q s + σ ˜ 4 r 1 ϖ β 1 2 ϖ 1 G 4 ( 1 , q s ) d q s = σ ˜ 3 r 1 ϖ β 1 L ˜ 3 2 + σ ˜ 4 r 1 ϖ β 1 L ˜ 4 2 = σ ˜ 3 ϖ β 1 L ˜ 3 2 + σ ˜ 4 ϖ β 1 L ˜ 4 2 r 1 .
Then, we conclude
A ( u , v ) Y = A 1 ( u , v ) + A 2 ( u , v ) σ ˜ 3 ( ϖ α 1 L ˜ 1 + ϖ β 1 L ˜ 3 ) 2 + σ ˜ 4 ( ϖ α 1 L ˜ 2 + ϖ β 1 L ˜ 4 ) 2 r 1 = σ ˜ 3 M ˜ 1 2 + σ ˜ 4 M ˜ 2 2 r 1 r 1 ,
and so
A ( u , v ) Y ( u , v ) Y , ( u , v ) P Ω 1 .
As we mentioned before, the operator A is completely continuous. Then, by (57), (61), and the Guo–Krasnosel’skii fixed-point theorem, we deduce that the operator A has a fixed point ( u , v ) P ( Ω ¯ 2 Ω 1 ) , which is a solution of problem (1),(2). This solution satisfies r 1 ( u , v ) Y r 2 , u ( t ) 0 , v ( t ) 0 for all t [ 0 , 1 ] , u ( t ) t α 1 u , v ( t ) t β 1 v for all t [ 0 , 1 ] ; because u + v r 1 , we obtain u   > 0 or v   > 0 , that is, u ( t ) > 0 for all t ( 0 , 1 ] or v ( t ) > 0 for all t ( 0 , 1 ] . □
Subsequently, we will establish the existence of at least three positive solutions to problem (1),(2) using the Leggett–Williams fixed-point theorem (refer to Theorem 3.3 in [29] or Theorem 2.3 in [30]).
Theorem 2.
Let  ϖ = q n  with  n N Assume that  ( H 1 )  and  ( H 2 )  are satisfied. In addition, we suppose that there exist positive constants  0 < r 1 < r 2 < r 3  such that 
  • (H5) f ( t , u , v , x , y ) < r 1 2 M 1 , t [ 0 , 1 ] , u , v 0 , u + v r 1 , x 0 , r 1 Γ q ( δ 1 + 1 ) , y 0 , r 1 Γ q ( γ 1 + 1 ) , g ( t , u , v , x , y ) < r 1 2 M 2 , t [ 0 , 1 ] , u , v 0 , u + v r 1 , x 0 , r 1 Γ q ( δ 2 + 1 ) , y 0 , r 1 Γ q ( γ 2 + 1 ) ;
  • (H6) f ( t , u , v , x , y ) > r 2 2 M ˜ 1 , t [ ϖ , 1 ] , u , v 0 , r 2 u + v r 3 , x 0 , r 3 Γ q ( δ 1 + 1 ) , y 0 , r 3 Γ q ( γ 1 + 1 ) , g ( t , u , v , x , y ) > r 2 2 M ˜ 2 , t [ ϖ , 1 ] , u , v 0 , r 2 u + v r 3 , x 0 , r 3 Γ q ( δ 2 + 1 ) , y 0 , r 3 Γ q ( γ 2 + 1 ) ;
  • (H7) f ( t , u , v , x , y ) r 3 2 M 1 , t [ 0 , 1 ] , u , v 0 , u + v r 3 , x 0 , r 3 Γ q ( δ 1 + 1 ) , y 0 , r 3 Γ q ( γ 1 + 1 ) , g ( t , u , v , x , y ) r 3 2 M 2 , t [ 0 , 1 ] , u , v 0 , u + v r 3 , x 0 , r 3 Γ q ( δ 2 + 1 ) , y 0 , r 3 Γ q ( γ 2 + 1 ) .
Then, problem (1),(2) has at least three positive solutions  ( u i ( t ) , v i ( t ) ) , t [ 0 , 1 ] , i = 1 , , 3 such that  ( u i , v i ) P i = 1 , , 3 ( u 1 , v 1 )   < r 1 ( u 3 , v 3 )   > r 1 inf t [ ϖ , 1 ] ( u 2 ( t ) + v 2 ( t ) ) > r 2 and  inf t [ ϖ , 1 ] ( u 3 ( t ) + v 3 ( t ) ) < r 2 .
Proof. 
Firstly, we prove that A : P ¯ r 3 P ¯ r 3 , where P r 3 = { ( u , v ) P , ( u , v )   < r 3 } . For any ( u , v ) P ¯ r 3 , we have ( u , v ) r 3 , and so u + v r 3 and 0 u ( t ) + v ( t ) r 3 for all t [ 0 , 1 ] . By using ( H 7 ) and Lemma 8, we find
A 1 ( u , v )   = sup t [ 0 , 1 ] | A 1 ( u , v ) ( t ) | 0 1 G 1 ( 1 , q s ) F u v δ 1 γ 1 ( s ) d q s + 0 1 G 2 ( 1 , q s ) G u v δ 2 γ 2 ( s ) d q s r 3 2 M 1 0 1 G 1 ( 1 , q s ) d q s + r 3 2 M 2 0 1 G 2 ( 1 , q s ) d q s = r 3 2 M 1 L 1 + r 3 2 M 2 L 2 = L 1 2 M 1 + L 2 2 M 2 r 3 ,
and
A 2 ( u , v )   = sup t [ 0 , 1 ] | A 2 ( u , v ) ( t ) | 0 1 G 3 ( 1 , q s ) F u v δ 1 γ 1 ( s )