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Article

Positive Solutions of a Singular Fractional Boundary Value Problem with r-Laplacian Operators

1
Department of Computer Science and Engineering, Gh. Asachi Technical University, 700050 Iasi, Romania
2
Department of Mathematics, Gh. Asachi Technical University, 700506 Iasi, Romania
*
Author to whom correspondence should be addressed.
Fractal Fract. 2022, 6(1), 18; https://doi.org/10.3390/fractalfract6010018
Submission received: 18 November 2021 / Revised: 27 December 2021 / Accepted: 29 December 2021 / Published: 30 December 2021

Abstract

:
We investigate the existence and multiplicity of positive solutions for a system of Riemann–Liouville fractional differential equations with r-Laplacian operators and nonnegative singular nonlinearities depending on fractional integrals, supplemented with nonlocal uncoupled boundary conditions which contain Riemann–Stieltjes integrals and various fractional derivatives. In the proof of our main results we apply the Guo–Krasnosel’skii fixed point theorem of cone expansion and compression of norm type.

1. Introduction

We consider the system of fractional differential equations with r 1 -Laplacian and r 2 -Laplacian operators
D 0 + γ 1 φ r 1 D 0 + δ 1 u ( τ ) = f τ , u ( τ ) , v ( τ ) , I 0 + σ 1 u ( τ ) , I 0 + σ 2 v ( τ ) , τ ( 0 , 1 ) , D 0 + γ 2 φ r 2 D 0 + δ 2 v ( τ ) = g τ , u ( τ ) , v ( τ ) , I 0 + ς 1 u ( τ ) , I 0 + ς 2 v ( τ ) , τ ( 0 , 1 ) ,
subject to the uncoupled nonlocal boundary conditions
u ( i ) ( 0 ) = 0 , i = 0 , , p 2 , D 0 + δ 1 u ( 0 ) = 0 , φ r 1 ( D 0 + δ 1 u ( 1 ) ) = 0 1 φ r 1 ( D 0 + δ 1 u ( η ) ) d H 0 ( η ) , D 0 + α 0 u ( 1 ) = k = 1 n 0 1 D 0 + α k u ( η ) d H k ( η ) , v ( i ) ( 0 ) = 0 , i = 0 , , q 2 , D 0 + δ 2 v ( 0 ) = 0 , φ r 2 ( D 0 + δ 2 v ( 1 ) ) = 0 1 φ r 2 ( D 0 + δ 2 v ( η ) ) d K 0 ( η ) , D 0 + β 0 v ( 1 ) = k = 1 m 0 1 D 0 + β k v ( η ) d K k ( η ) ,
where γ 1 , γ 2 ( 1 , 2 ] , δ 1 ( p 1 , p ] , p N , p 3 , δ 2 ( q 1 , q ] , q N , q 3 , n , m N , σ 1 , σ 2 , ς 1 , ς 2 > 0 , α k R , k = 0 , , n , 0 α 1 < α 2 < < α n α 0 < δ 1 1 , α 0 1 , β k R , k = 0 , , m , 0 β 1 < β 2 < < β m β 0 < δ 2 1 , β 0 1 , φ r i ( η ) = | η | r i 2 η , φ r i 1 = φ ϱ i , ϱ i = r i r i 1 , i = 1 , 2 , r i > 1 , i = 1 , 2 , f , g : ( 0 , 1 ) × R + 4 R + are continuous functions, singular at τ = 0 and/or τ = 1 , ( R + = [ 0 , ) ), I 0 + κ is the Riemann–Liouville fractional integral of order κ (for κ = σ 1 , σ 2 , ς 1 , ς 2 ), D 0 + κ is the Riemann-Liouville fractional derivative of order κ (for κ = γ 1 , δ 1 , γ 2 , δ 2 , α 0 , , α n , β 0 , , β m ), and the integrals from the boundary conditions (2) are Riemann–Stieltjes integrals with H i : [ 0 , 1 ] R , i = 0 , , n and K i : [ 0 , 1 ] R , i = 0 , , m functions of bounded variation.
We give in this paper various conditions for the functions f and g such that problems (1) and (2) have at least one or two positive solutions. From a positive solution of (1) and (2) we understand a pair of functions ( u , v ) ( C ( [ 0 , 1 ] , R + ) ) 2 satisfying the system (1) and the boundary conditions (2), with u ( τ ) > 0 for all τ ( 0 , 1 ] or v ( τ ) > 0 for all τ ( 0 , 1 ] . In the proof of our main results we use the Guo–Krasnosel’skii fixed point theorem of cone expansion and compression of norm type. We now present some recent results which are connected with our problem. In [1], the authors studied the existence of multiple positive solutions for the system of nonlinear fractional differential equations with a p-Laplacian operator
D 0 + β 1 ( φ p 1 ( D 0 + α 1 x ( τ ) ) ) = f ( τ , x ( τ ) , y ( τ ) ) , τ ( 0 , 1 ) , D 0 + β 2 ( φ p 2 ( D 0 + α 2 y ( τ ) ) ) = g ( τ , x ( τ ) , y ( τ ) ) , τ ( 0 , 1 ) ,
supplemented with the uncoupled boundary conditions
x ( 0 ) = 0 , D 0 + γ 1 x ( 1 ) = i = 1 m 2 ξ 1 i D 0 + γ 1 x ( η 1 i ) , D 0 + α 1 x ( 0 ) = 0 , φ p 1 ( D 0 + α 1 x ( 1 ) ) = i = 1 m 2 ζ 1 i φ p 1 ( D 0 + α 1 x ( η 1 i ) ) , y ( 0 ) = 0 , D 0 + γ 2 y ( 1 ) = i = 1 m 2 ξ 2 i D 0 + γ 2 y ( η 2 i ) , D 0 + α 2 y ( 0 ) = 0 , φ p 2 ( D 0 + α 2 y ( 1 ) ) = i = 1 m 2 ζ 2 i φ p 2 ( D 0 + α 2 y ( η 2 i ) ) ,
where α i , β i ( 1 , 2 ] , γ i ( 0 , 1 ] , α i + β i ( 3 , 4 ] , α i > γ i + 1 , i = 1 , 2 , ξ 1 i , η 1 i , ζ 1 i , ξ 2 i , η 2 i , ζ 2 i ( 0 , 1 ) for i = 1 , , m 2 , and f and g are nonnegative and nonsingular functions. In the proof of the existence results they use the Leray–Schauder alternative theorem, the Leggett–Williams fixed point theorem and the Avery–Henderson fixed point theorem. In [2], the authors investigated the existence and multiplicity of positive solutions for the system of fractional differential equations with ϱ 1 -Laplacian and ϱ 2 -Laplacian operators
D 0 + γ 1 ( φ ϱ 1 ( D 0 + δ 1 x ( τ ) ) ) + f ( τ , x ( τ ) , y ( τ ) ) = 0 , τ ( 0 , 1 ) , D 0 + γ 2 ( φ ϱ 2 ( D 0 + δ 2 y ( τ ) ) ) + g ( τ , x ( τ ) , y ( τ ) ) = 0 , τ ( 0 , 1 ) ,
subject to the uncoupled nonlocal boundary conditions
x ( j ) ( 0 ) = 0 , j = 0 , , p 2 ; D 0 + δ 1 x ( 0 ) = 0 , D 0 + α 0 x ( 1 ) = i = 1 n 0 1 D 0 + α i x ( τ ) d H i ( τ ) , y ( j ) ( 0 ) = 0 , j = 0 , , q 2 ; D 0 + δ 2 y ( 0 ) = 0 , D 0 + β 0 y ( 1 ) = i = 1 m 0 1 D 0 + β i y ( τ ) d K i ( τ ) ,
where γ 1 , γ 2 ( 0 , 1 ] , δ 1 ( p 1 , p ] , δ 2 ( q 1 , q ] , p , q N , p , q 3 , n , m N , α i R for all i = 0 , 1 , , n , 0 α 1 < α 2 < < α n α 0 < δ 1 1 , α 0 1 , β i R for all i = 0 , 1 , , m , 0 β 1 < β 2 < < β m β 0 < δ 2 1 , β 0 1 , ϱ 1 , ϱ 2 > 1 , the functions f and g are nonnegative and continuous, and they may be singular at τ = 0 and/or τ = 1 , and H i , i = 1 , , n and K j , j = 1 , , m are functions of bounded variation. In the proof of the main existence results they applied the Guo–Krasnosel’skii fixed point theorem. In [3], the authors studied the existence and nonexistence of positive solutions for the system (3) with two positive parameters λ and μ , supplemented with the coupled nonlocal boundary conditions
x ( j ) ( 0 ) = 0 , j = 0 , , p 2 ; D 0 + δ 1 x ( 0 ) = 0 , D 0 + α 0 x ( 1 ) = i = 1 n 0 1 D 0 + α i y ( τ ) d H i ( τ ) , y ( j ) ( 0 ) = 0 , j = 0 , , q 2 ; D 0 + δ 2 y ( 0 ) = 0 , D 0 + β 0 y ( 1 ) = i = 1 m 0 1 D 0 + β i x ( τ ) d K i ( τ ) ,
where n , m N , α i R for all i = 0 , , n , 0 α 1 < α 2 < < α n β 0 < δ 2 1 , β 0 1 , β i R for all i = 0 , , m , 0 β 1 < β 2 < < β m α 0 < δ 1 1 , α 0 1 , the functions f , g C ( [ 0 , 1 ] × R + × R + , R + ) , and the functions H i , i = 1 , , n and K j , j = 1 , , m are bounded variation functions. They presented sufficient conditions on the functions f and g, and intervals for the parameters λ and μ such that the problem (3) with these parameters and (4) has positive solutions. In [4], by using the Guo–Krasnosel’skii fixed point theorem, the authors investigated the existence and multiplicity of positive solutions for the nonlinear singular fractional differential equation
D 0 + α w ( τ ) + f ( τ , w ( τ ) , D 0 + α 1 w ( τ ) , , D 0 + α n 2 w ( τ ) ) = 0 , τ ( 0 , 1 ) ,
with the nonlocal boundary conditions
w ( 0 ) = D 0 + γ 1 w ( 0 ) = = D 0 + γ n 2 w ( 0 ) = 0 , D 0 + β 1 w ( 1 ) = 0 η h ( τ ) D 0 + β 2 w ( τ ) d A ( τ ) + 0 1 a ( τ ) D 0 + β 3 w ( τ ) d A ( τ ) ,
where α ( n 1 , n ] , n 3 , α k , γ k ( k 1 , k ] , k = 1 , , n 2 , α γ j ( n j 1 , n j ] , j = 1 , , n 2 , α α n 2 1 ( 1 , 2 ] , γ n 2 α n 2 , β 1 β 2 , β 1 β 3 , α β i + 1 , β i α n 2 + 1 , i = 1 , 2 , 3 , β 1 n 1 , the function f : ( 0 , 1 ) × R + n 1 R + is continuous, a , h C ( ( 0 , 1 ) , R + ) , and A is a function of bounded variation. In [5], the authors studied the existence of a unique positive solution for a system of three Caputo fractional equations with ( p , q , r ) -Laplacian operators subject to two-point boundary conditions, by using an n-fixed point theorem of ternary operators in partially ordered complete metric spaces. By relying on the properties of the Kuratowski noncompactness measure and the Sadovskii fixed point theorem; in [6], the authors obtained new existence results for the solutions of a Riemann–Liouville fractional differential equation with a p-Laplacian operator in a Banach space, supplemented with multi-point boundary conditions with fractional derivatives. In [7], the authors investigated the existence of solutions for a mixed fractional differential equation with p ( t ) -Laplacian operator and two-point boundary conditions at resonance, by applying the continuation theorem of coincidence degree theory. By using the Leggett–Williams fixed-point theorem, the authors studied in [8] the multiplicity of positive solutions for a Riemann–Liouville fractional differential equation with a p-Laplacian operator, subject to four-point boundary conditions. In [9], the authors established suitable criteria for the existence of positive solutions for a Riemann–Liouville fractional equation with a p-Laplacian operator and infinite-point boundary value conditions, by using the Krasnosel’skii fixed point theorem and Avery–Peterson fixed point theorem. By applying the Guo–Krasnosel’skii fixed point theorem the authors investigated in [10] the existence, multiplicity and the nonexistence of positive solutions for a mixed fractional differential equation with a generalized p-Laplacian operator and a positive parameter, supplemented with two-point boundary conditions. We also mention some recent monographs devoted to the investigation of boundary value problems for fractional differential equations and systems with many examples and applications, namely [11,12,13,14,15].
So in comparison with the above papers, the new characteristics of our problem (1) and (2) consist in a combination between the fractional orders γ 1 , γ 2 ( 1 , 2 ] with the arbitrary fractional orders δ 1 , δ 2 , the existence of the fractional integral terms in equations of (1), and the general uncoupled nonlocal boundary conditions with Riemann–Stieltjes integrals and fractional derivatives. In addition, one of its special feature is the singularity of the nonlinearities from the system (1), that is f , g become unbounded in the vicinity of 0 and/or 1 in the first variable (see Assumption ( I 2 ) in Section 3).
The structure of this paper is as follows. In Section 2, some preliminary results including the properties of the Green functions associated to our problem (1) and (2) are presented. In Section 3 we discuss the existence and multiplicity of positive solutions for (1) and (2). Then two examples to illustrate our obtained theorems are given in Section 4, and Section 5 contains the conclusions for this paper.

2. Preliminary Results

We consider the fractional differential equation
D 0 + γ 1 φ r 1 D 0 + δ 1 u ( τ ) = x ( τ ) , τ ( 0 , 1 ) ,
where x C ( 0 , 1 ) L 1 ( 0 , 1 ) , with the boundary conditions
u ( i ) ( 0 ) = 0 , i = 0 , , p 2 , D 0 + δ 1 u ( 0 ) = 0 , φ r 1 ( D 0 + δ 1 u ( 1 ) ) = 0 1 φ r 1 ( D 0 + δ 1 u ( η ) ) d H 0 ( η ) , D 0 + α 0 u ( 1 ) = k = 1 n 0 1 D 0 + α k u ( η ) d H k ( η ) .
We denote by
a 1 = 1 0 1 η γ 1 1 d H 0 ( η ) , a 2 = Γ ( δ 1 ) Γ ( δ 1 α 0 ) i = 1 n Γ ( δ 1 ) Γ ( δ 1 α i ) 0 1 η δ 1 α i 1 d H i ( η ) .
Lemma 1.
If a 1 0 and a 2 0 , then the unique solution u C [ 0 , 1 ] of problem (5) and (6) is given by
u ( τ ) = 0 1 G 2 ( τ , η ) φ ϱ 1 0 1 G 1 ( η , ϑ ) x ( ϑ ) d ϑ d η , τ [ 0 , 1 ] ,
where
G 1 ( τ , η ) = g 1 ( τ , η ) + τ γ 1 1 a 1 0 1 g 1 ( ϑ , η ) d H 0 ( ϑ ) , ( τ , η ) [ 0 , 1 ] × [ 0 , 1 ] ,
with
g 1 ( τ , η ) = 1 Γ ( γ 1 ) τ γ 1 1 ( 1 η ) γ 1 1 ( τ η ) γ 1 1 , 0 η τ 1 , τ γ 1 1 ( 1 η ) γ 1 1 , 0 τ η 1 ,
and
G 2 ( τ , η ) = g 2 ( τ , η ) + τ δ 1 1 a 2 i = 1 n 0 1 g 2 i ( ϑ , η ) d H i ( ϑ ) , ( τ , η ) [ 0 , 1 ] × [ 0 , 1 ] ,
with
g 2 ( τ , η ) = 1 Γ ( δ 1 ) τ δ 1 1 ( 1 η ) δ 1 α 0 1 ( τ η ) δ 1 1 , 0 η τ 1 , τ δ 1 1 ( 1 η ) δ 1 α 0 1 , 0 τ η 1 , g 2 i ( τ , η ) = 1 Γ ( δ 1 α i ) τ δ 1 α i 1 ( 1 η ) δ 1 α 0 1 ( τ η ) δ 1 α i 1 , 0 η τ 1 , τ δ 1 α i 1 ( 1 η ) δ 1 α 0 1 , 0 τ η 1 , i = 1 , , n .
Proof. 
We denote by φ r 1 ( D 0 + δ 1 u ( τ ) ) = ϕ 1 ( τ ) , τ ( 0 , 1 ) . Hence problems (5) and (6) are equivalent to the following two boundary value problems
( I ) D 0 + γ 1 ϕ 1 ( τ ) = x ( τ ) , τ ( 0 , 1 ) , ϕ 1 ( 0 ) = 0 , ϕ 1 ( 1 ) = 0 1 ϕ 1 ( η ) d H 0 ( η ) ,
and
( II ) D 0 + δ 1 u ( τ ) = φ ϱ 1 ( ϕ 1 ( τ ) ) , τ ( 0 , 1 ) , u ( j ) ( 0 ) = 0 , j = 0 , , p 2 , D 0 + α 0 u ( 1 ) = k = 1 n 0 1 D 0 + α k u ( η ) d H k ( η ) .
By using Lemma 4.1.5 from [14], the unique solution ϕ 1 C [ 0 , 1 ] of problem ( I ) is
ϕ 1 ( τ ) = 0 1 G 1 ( τ , ϑ ) x ( ϑ ) d ϑ , τ [ 0 , 1 ] ,
where G 1 is given by (9). By using Lemma 2.4.2 from [12], the unique solution u C [ 0 , 1 ] of problem ( I I ) is
u ( τ ) = 0 1 G 2 ( τ , η ) φ ϱ 1 ( ϕ 1 ( η ) ) d η , τ [ 0 , 1 ] ,
where G 2 is given by (11). Combining the relations (13) and (14) we obtain the solution u of problem (5) and (6) which is given by relation (8). □
We consider now the fractional differential equation
D 0 + γ 2 φ r 2 D 0 + δ 2 v ( τ ) = y ( τ ) , τ ( 0 , 1 ) ,
where y C ( 0 , 1 ) L 1 ( 0 , 1 ) , with the boundary conditions
v ( i ) ( 0 ) = 0 , i = 0 , , q 2 , D 0 + δ 2 v ( 0 ) = 0 , φ r 2 ( D 0 + δ 2 v ( 1 ) ) = 0 1 φ r 2 ( D 0 + δ 2 v ( η ) ) d K 0 ( η ) , D 0 + β 0 v ( 1 ) = k = 1 m 0 1 D 0 + β k v ( η ) d K k ( η ) .
We denote by
b 1 = 1 0 1 η γ 2 1 d K 0 ( η ) , b 2 = Γ ( δ 2 ) Γ ( δ 2 β 0 ) i = 1 m Γ ( δ 2 ) Γ ( δ 2 β i ) 0 1 η δ 2 β i 1 d K i ( η ) .
Similar to Lemma 1 we obtain the next result.
Lemma 2.
If b 1 0 and b 2 0 , then the unique solution v C [ 0 , 1 ] of problem (15) and (16) is given by
v ( τ ) = 0 1 G 4 ( τ , η ) φ ϱ 2 0 1 G 3 ( η , ϑ ) y ( ϑ ) d ϑ d η , τ [ 0 , 1 ] ,
where
G 3 ( τ , η ) = g 3 ( τ , η ) + τ γ 2 1 b 1 0 1 g 3 ( ϑ , η ) d K 0 ( ϑ ) , ( τ , η ) [ 0 , 1 ] × [ 0 , 1 ] ,
with
g 3 ( τ , η ) = 1 Γ ( γ 2 ) τ γ 2 1 ( 1 η ) γ 2 1 ( τ η ) γ 2 1 , 0 η τ 1 , τ γ 2 1 ( 1 η ) γ 2 1 , 0 τ η 1 ,
and
G 4 ( τ , η ) = g 4 ( τ , η ) + τ δ 2 1 b 2 i = 1 m 0 1 g 4 i ( ϑ , η ) d K i ( ϑ ) , ( τ , η ) [ 0 , 1 ] × [ 0 , 1 ] ,
with
g 4 ( τ , η ) = 1 Γ ( δ 2 ) τ δ 2 1 ( 1 η ) δ 2 β 0 1 ( τ η ) δ 2 1 , 0 η τ 1 , τ δ 2 1 ( 1 η ) δ 2 β 0 1 , 0 τ η 1 , g 4 i ( τ , η ) = 1 Γ ( δ 2 β i ) τ δ 2 β i 1 ( 1 η ) δ 2 β 0 1 ( τ η ) δ 2 β i 1 , 0 η τ 1 , τ δ 2 β i 1 ( 1 η ) δ 2 β 0 1 , 0 τ η 1 , i = 1 , , m .
Lemma 3.
We assume that a 1 , a 2 , b 1 , b 2 > 0 , H i , i = 0 , , n , and K j , j = 0 , , m are nondecreasing functions. Then the functions G i , i = 1 , , 4 given by (9), (11), (19) and (21) have the properties
(a) G i : [ 0 , 1 ] × [ 0 , 1 ] [ 0 , ) , i = 1 , , 4 are continuous functions;
(b) G 1 ( τ , η ) J 1 ( η ) , ( τ , η ) [ 0 , 1 ] × [ 0 , 1 ] , where
J 1 ( η ) = h 1 ( η ) + 1 a 1 0 1 g 1 ( ϑ , η ) d H 0 ( ϑ ) , η [ 0 , 1 ] ,
with h 1 ( η ) = 1 Γ ( γ 1 ) ( 1 η ) γ 1 1 , η [ 0 , 1 ] ;
(c) G 2 ( τ , η ) J 2 ( η ) , ( τ , η ) [ 0 , 1 ] × [ 0 , 1 ] , where
J 2 ( η ) = h 2 ( η ) + 1 a 2 i = 1 n 0 1 g 2 i ( ϑ , η ) d H i ( ϑ ) , η [ 0 , 1 ] ,
with h 2 ( η ) = 1 Γ ( δ 1 ) ( 1 η ) δ 1 α 0 1 ( 1 ( 1 η ) α 0 ) , η [ 0 , 1 ] ;
(d) G 2 ( τ , η ) τ δ 1 1 J 2 ( η ) , ( τ , η ) [ 0 , 1 ] × [ 0 , 1 ] ;
(e) G 3 ( τ , η ) J 3 ( η ) , ( τ , η ) [ 0 , 1 ] × [ 0 , 1 ] , where
J 3 ( η ) = h 3 ( η ) + 1 b 1 0 1 g 3 ( ϑ , η ) d K 0 ( ϑ ) , η [ 0 , 1 ] ,
with h 3 ( η ) = 1 Γ ( γ 2 ) ( 1 η ) γ 2 1 , η [ 0 , 1 ] ;
(f) G 4 ( τ , η ) J 4 ( η ) , ( τ , η ) [ 0 , 1 ] × [ 0 , 1 ] , where
J 4 ( η ) = h 4 ( η ) + 1 b 2 i = 1 m 0 1 g 4 i ( ϑ , η ) d K i ( ϑ ) , η [ 0 , 1 ] ,
with h 4 ( η ) = 1 Γ ( δ 2 ) ( 1 η ) δ 2 β 0 1 ( 1 ( 1 η ) β 0 ) , η [ 0 , 1 ] ;
(g) G 4 ( τ , η ) τ δ 2 1 J 4 ( η ) , ( τ , η ) [ 0 , 1 ] × [ 0 , 1 ] .
Proof. 
(a) Based on the continuity of functions g 1 , g 2 , g 2 i , i = 1 , , n , g 3 , g 4 , g 4 i , i = 1 , , m (given by (10), (12), (20) and (22)), we obtain that the functions G i , i = 1 , , 4 are continuous.
(b) By the definition of g 1 we find
G 1 ( τ , η ) 1 Γ ( γ 1 ) ( 1 η ) γ 1 1 + 1 a 1 0 1 g 1 ( ϑ , η ) d H 0 ( ϑ ) = h 1 ( η ) + 1 a 1 0 1 g 1 ( ϑ , η ) d H 0 ( ϑ ) = J 1 ( η ) , τ , η [ 0 , 1 ] .
(c–d) Using our assumptions and the properties of function g 2 from Lemma 2.1.3 from [12], namely g 2 ( τ , η ) 1 Γ ( δ 1 ) ( 1 η ) δ 1 α 0 1 ( 1 ( 1 η ) α 0 ) = h 2 ( η ) and g 2 ( τ , η ) τ δ 1 1 h 2 ( η ) for all τ , η [ 0 , 1 ] , we deduce
G 2 ( τ , η ) h 2 ( η ) + 1 a 2 i = 1 n 0 1 g 2 i ( ϑ , η ) d H i ( ϑ ) = J 2 ( η ) , G 2 ( τ , η ) τ δ 1 1 h 2 ( η ) + 1 a 2 i = 1 n g 2 i ( ϑ , η ) d H i ( ϑ ) = τ δ 1 1 J 2 ( η ) , τ , η [ 0 , 1 ] .
(e) By the definition of g 3 we obtain
G 3 ( τ , η ) 1 Γ ( γ 2 ) ( 1 η ) γ 2 1 + 1 b 1 0 1 g 3 ( ϑ , η ) d K 0 ( ϑ ) = h 3 ( η ) + 1 b 1 0 1 g 3 ( ϑ , η ) d K 0 ( ϑ ) = J 3 ( η ) , τ , η [ 0 , 1 ] .
(f–g) Using the assumptions of this lemma and the properties of function g 4 from Lemma 2.1.3 from [12], namely g 4 ( τ , η ) 1 Γ ( δ 2 ) ( 1 η ) δ 2 β 0 1 ( 1 ( 1 η ) β 0 ) = h 4 ( η ) and g 4 ( τ , η ) τ δ 2 1 h 4 ( η ) for all τ , η [ 0 , 1 ] , we find
G 4 ( τ , η ) h 4 ( η ) + 1 b 2 i = 1 m 0 1 g 4 i ( ϑ , η ) d K i ( ϑ ) = J 4 ( η ) , G 4 ( τ , η ) τ δ 2 1 h 4 ( η ) + 1 b 2 i = 1 m g 4 i ( ϑ , η ) d K i ( ϑ ) = τ δ 2 1 J 4 ( η ) , τ , η [ 0 , 1 ] .
Lemma 4.
We assume that a 1 , a 2 , b 1 , b 2 > 0 , H i , i = 0 , , n , and K j , j = 0 , , m are nondecreasing functions, x , y C ( 0 , 1 ) L 1 ( 0 , 1 ) with x ( τ ) 0 , y ( τ ) 0 for all τ ( 0 , 1 ) . Then the solutions u and v of problems (5), (6) and (15), (16), respectively, satisfy the inequalities u ( τ ) 0 , v ( τ ) 0 for all τ [ 0 , 1 ] and u ( τ ) τ δ 1 1 u ( s ) and v ( τ ) τ δ 2 1 v ( s ) for all τ , s [ 0 , 1 ] .
Proof. 
Based on the assumptions of this lemma, we obtain that the solutions u and v of problems (5), (6) and (15), (16), respectively, are nonnegative, that is u ( τ ) 0 , v ( τ ) 0 for all τ [ 0 , 1 ] . In addition, by using Lemma 3, we deduce
u ( τ ) τ δ 1 1 0 1 J 2 ( η ) φ ϱ 1 0 1 G 1 ( η , ϑ ) x ( ϑ ) d ϑ d η τ δ 1 1 0 1 G 2 ( s , η ) φ ϱ 1 0 1 G 1 ( η , ϑ ) x ( ϑ ) d ϑ d η = τ δ 1 1 u ( s ) , v ( τ ) τ δ 2 1 0 1 J 4 ( η ) φ ϱ 2 0 1 G 3 ( η , ϑ ) y ( ϑ ) d ϑ d η τ δ 2 1 0 1 G 4 ( s , η ) φ ϱ 2 0 1 G 3 ( η , ϑ ) y ( ϑ ) d ϑ d η = τ δ 2 1 v ( s ) ,
for all τ , s [ 0 , 1 ] . □
We present finally in this section the Guo–Krasnosel’skii fixed point theorem, which we will use in the proofs of our main results.
Theorem 1.
([16]). Let X be a real Banach space with the norm · , and let C X be a cone in X . Assume Ω 1 and Ω 2 are bounded open subsets of X with 0 Ω 1 , Ω ¯ 1 Ω 2 and let A : C ( Ω ¯ 2 Ω 1 ) C be a completely continuous operator such that, either
(i) A u u , u C Ω 1 , and A u u , u C Ω 2 ; or
(ii) A u u , u C Ω 1 , and A u u , u C Ω 2 .
Then A has at least one fixed point in C ( Ω ¯ 2 Ω 1 ) .

3. Existence of Positive Solutions

According to Lemmas 1 and 2, the pair of functions ( u , v ) is a solution of problem (1) and (2) if and only if ( u , v ) is a solution of the system
u ( τ ) = 0 1 G 2 ( τ , ζ ) φ ϱ 1 0 1 G 1 ( ζ , ϑ ) f ϑ , u ( ϑ ) , v ( ϑ ) , I 0 + σ 1 u ( ϑ ) , I 0 + σ 2 v ( ϑ ) d ϑ d ζ , v ( τ ) = 0 1 G 4 ( τ , ζ ) φ ϱ 2 0 1 G 3 ( ζ , ϑ ) g ϑ , u ( ϑ ) , v ( ϑ ) , I 0 + ς 1 u ( ϑ ) , I 0 + ς 2 v ( ϑ ) d ϑ d ζ ,
for all τ [ 0 , 1 ] . We introduce the Banach space X = C [ 0 , 1 ] with supreme norm u = sup τ [ 0 , 1 ] | u ( τ ) | , and the Banach space Y = X × X with the norm ( u , v ) Y = u + v . We define the cone
P = { ( u , v ) Y , u ( τ ) 0 , v ( τ ) 0 , τ [ 0 , 1 ] } .
We also define the operators A 1 , A 2 : Y X and A : Y Y by
A 1 ( u , v ) ( τ ) = 0 1 G 2 ( τ , ζ ) φ ϱ 1 0 1 G 1 ( ζ , ϑ ) f ϑ , u ( ϑ ) , v ( ϑ ) , I 0 + σ 1 u ( ϑ ) , I 0 + σ 2 v ( ϑ ) d ϑ d ζ , A 2 ( u , v ) ( τ ) = 0 1 G 4 ( τ , ζ ) φ ϱ 2 0 1 G 3 ( ζ , ϑ ) g ϑ , u ( ϑ ) , v ( ϑ ) , I 0 + ς 1 u ( ϑ ) , I 0 + ς 2 v ( ϑ ) d ϑ d ζ ,
for τ [ 0 , 1 ] and ( u , v ) Y , and A ( u , v ) = ( A 1 ( u , v ) , A 2 ( u , v ) ) , ( u , v ) Y . We see that ( u , v ) is a solution of problem (1) and (2) if and only if ( u , v ) is a fixed point of operator A .
We introduce now the basic assumptions that we will use in this section.
( I 1 )
γ 1 , γ 2 ( 1 , 2 ] , δ 1 ( p 1 , p ] , p N , p 3 , δ 2 ( q 1 , q ] , q N , q 3 , n , m N , σ 1 , σ 2 , ς 1 , ς 2 > 0 , α j R , j = 0 , , n , 0 α 1 < α 2 < < α n α 0 < δ 1 1 , α 0 1 , β j R , j = 0 , , m , 0 β 1 < β 2 < < β m β 0 < δ 2 1 , β 0 1 , φ r i ( τ ) = | τ | r i 2 τ , φ r i 1 = φ ϱ i , ϱ i = r i r i 1 , i = 1 , 2 , r i > 1 , i = 1 , 2 , H i : [ 0 , 1 ] R , i = 0 , , n , and K j : [ 0 , 1 ] R , j = 0 , , m are nondecreasing functions, a 1 , a 2 , b 1 , b 2 > 0 (given by (7) and (17)).
( I 2 )
The functions f , g C ( ( 0 , 1 ) × R + 4 , R + ) and there exist the functions ψ 1 , ψ 2 C ( ( 0 , 1 ) , R + ) and χ 1 , χ 2 C ( [ 0 , 1 ] × R + 4 , R + ) with Λ 1 = 0 1 ( 1 τ ) γ 1 1 ψ 1 ( τ ) d τ ( 0 , ) , Λ 2 = 0 1 ( 1 τ ) γ 2 1 ψ 2 ( τ ) d τ ( 0 , ) , such that
f ( τ , z 1 , z 2 , z 3 , z 4 ) ψ 1 ( τ ) χ 1 ( τ , z 1 , z 2 , z 3 , z 4 ) , g ( τ , z 1 , z 2 , z 3 , z 4 ) ψ 2 ( τ ) χ 2 ( τ , z 1 , z 2 , z 3 , z 4 ) ,
for any τ ( 0 , 1 ) , z i R + , i = 1 , , 4 .
Lemma 5.
We assume that assumptions ( I 1 ) and ( I 2 ) are satisfied. Then operator A : P P is completely continuous.
Proof. 
We denote by M 1 = 0 1 J 1 ( η ) ψ 1 ( η ) d η , M 2 = 0 1 J 3 ( η ) ψ 2 ( η ) d η . By using ( I 2 ) and Lemma 3, we deduce that M 1 > 0 and M 2 > 0 . In addition we find
M 1 = 0 1 J 1 ( η ) ψ 1 ( η ) d η = 0 1 h 1 ( η ) + 1 a 1 0 1 g 1 ( ζ , η ) d H 0 ( ζ ) ψ 1 ( η ) d η 1 Γ ( γ 1 ) 0 1 ( 1 η ) γ 1 1 ψ 1 ( η ) d η + 1 a 1 0 1 0 1 1 Γ ( γ 1 ) ζ γ 1 1 ( 1 η ) γ 1 1 d H 0 ( ζ ) ψ 1 ( η ) d η = 1 + 1 a 1 0 1 ζ γ 1 1 d H 0 ( ζ ) 1 Γ ( γ 1 ) 0 1 ( 1 η ) γ 1 1 ψ 1 ( η ) d η < , M 2 = 0 1 J 3 ( η ) ψ 2 ( η ) d η = 0 1 h 3 ( η ) + 1 b 1 0 1 g 3 ( ζ , η ) d K 0 ( ζ ) ψ 2 ( η ) d η 1 Γ ( γ 2 ) 0 1 ( 1 η ) γ 2 1 ψ 2 ( η ) d η + 1 b 1 0 1 0 1 1 Γ ( γ 2 ) ζ γ 2 1 ( 1 η ) γ 2 1 d K 0 ( ζ ) ψ 2 ( η ) d η = 1 + 1 b 1 0 1 ζ γ 2 1 d K 0 ( ζ ) 1 Γ ( γ 2 ) 0 1 ( 1 η ) γ 2 1 ψ 2 ( η ) d η < .
Also, by Lemma 3 we conclude that A maps P into P .
We will prove that A maps bounded sets into relatively compact sets. Let E P be an arbitrary bounded set. Then there exists Ξ 1 > 0 such that ( u , v ) Y Ξ 1 for all ( u , v ) E . By the continuity of χ 1 and χ 2 , we deduce that there exists Ξ 2 > 0 such that Ξ 2 = max { sup τ [ 0 , 1 ] , z i [ 0 , ω ] , i = 1 , , 4 χ 1 ( τ , z 1 , z 2 , z 3 , z 4 ) , sup τ [ 0 , 1 ] , z i [ 0 , ω ] , i = 1 , , 4 χ 2 ( τ , z 1 , z 2 , z 3 , z 4 ) } , where ω = Ξ 1 max 1 , 1 Γ ( σ 1 + 1 ) , 1 Γ ( σ 2 + 1 ) , 1 Γ ( ς 1 + 1 ) , 1 Γ ( ς 2 + 1 ) . Based on the inequality | I 0 + ξ w ( η ) | w Γ ( ξ + 1 ) , for ξ > 0 and w C [ 0 , 1 ] , and by Lemma 3, we find for any ( u , v ) E and η [ 0 , 1 ]
A 1 ( u , v ) ( η ) 0 1 J 2 ( ζ ) φ ϱ 1 0 1 J 1 ( τ ) ψ 1 ( τ ) χ 1 ( τ , u ( τ ) , v ( τ ) , I 0 + σ 1 u ( τ ) , I 0 + σ 2 v ( τ ) ) d τ d ζ Ξ 2 ϱ 1 1 φ ϱ 1 0 1 J 1 ( τ ) ψ 1 ( τ ) d τ 0 1 J 2 ( ζ ) d ζ = M 1 ϱ 1 1 Ξ 2 ϱ 1 1 M 3 , A 2 ( u , v ) ( η ) 0 1 J 4 ( ζ ) φ ϱ 2 0 1 J 3 ( τ ) ψ 2 ( τ ) χ 2 ( τ , u ( τ ) , v ( τ ) , I 0 + ς 1 u ( τ ) , I 0 + ς 2 v ( τ ) ) d τ d ζ Ξ 2 ϱ 2 1 φ ϱ 2 0 1 J 3 ( τ ) ψ 2 ( τ ) d τ 0 1 J 4 ( ζ ) d ζ = M 2 ϱ 2 1 Ξ 2 ϱ 2 1 M 4 ,
where M 3 = 0 1 J 2 ( ζ ) d ζ and M 4 = 0 1 J 4 ( ζ ) d ζ .
Then A 1 ( u , v ) M 1 ϱ 1 1 Ξ 2 ϱ 1 1 M 3 , A 2 ( u , v ) M 2 ϱ 2 1 Ξ 2 ϱ 2 1 M 4 for all ( u , v ) E , and A ( u , v ) Y M 1 ϱ 1 1 Ξ 2 ϱ 1 1 M 3 + M 2 ϱ 2 1 Ξ 2 ϱ 2 1 M 4 for all ( u , v ) E , that is A 1 ( E ) , A 2 ( E ) and A ( E ) are bounded.
We will show that A ( E ) is equicontinuous. By using Lemma 1, for ( u , v ) E and η [ 0 , 1 ] we obtain
A 1 ( u , v ) ( η ) = 0 1 g 2 ( η , ζ ) + η δ 1 1 a 2 i = 1 n 0 1 g 2 i ( τ , ζ ) d H i ( τ ) φ ϱ 1 0 1 G 1 ( ζ , ϑ ) × f ( ϑ , u ( ϑ ) , v ( ϑ ) , I 0 + σ 1 u ( ϑ ) , I 0 + σ 2 v ( ϑ ) ) d ϑ d ζ = 0 η 1 Γ ( δ 1 ) [ η δ 1 1 ( 1 ζ ) δ 1 α 0 1 ( η ζ ) δ 1 1 ] × φ ϱ 1 0 1 G 1 ( ζ , ϑ ) f ( ϑ , u ( ϑ ) , v ( ϑ ) , I 0 + σ 1 u ( ϑ ) , I 0 + σ 2 v ( ϑ ) ) d ϑ d ζ + η 1 1 Γ ( δ 1 ) η δ 1 1 ( 1 ζ ) δ 1 α 0 1 φ ϱ 1 0 1 G 1 ( ζ , ϑ ) f ( ϑ , u ( ϑ ) , v ( ϑ ) , I 0 + σ 1 u ( ϑ ) , I 0 + σ 2 v ( ϑ ) ) d ϑ d ζ + η δ 1 1 a 2 0 1 i = 1 n 0 1 g 2 i ( τ , ζ ) d H i ( τ ) φ ϱ 1 0 1 G 1 ( ζ , ϑ ) f ( ϑ , u ( ϑ ) , v ( ϑ ) , I 0 + σ 1 u ( ϑ ) , I 0 + σ 2 v ( ϑ ) ) d ϑ d ζ .
Then for any η ( 0 , 1 ) we deduce
( A 1 ( u , v ) ) ( η ) = 0 η 1 Γ ( δ 1 ) [ ( δ 1 1 ) η δ 1 2 ( 1 ζ ) δ 1 α 0 1 ( δ 1 1 ) ( η ζ ) δ 1 2 ] × φ ϱ 1 0 1 G 1 ( ζ , ϑ ) f ( ϑ , u ( ϑ ) , v ( ϑ ) , I 0 + σ 1 u ( ϑ ) , I 0 + σ 2 v ( ϑ ) ) d ϑ d ζ + η 1 1 Γ ( δ 1 ) ( δ 1 1 ) η δ 1 2 ( 1 ζ ) δ 1 α 0 1 φ ϱ 1 0 1 G 1 ( ζ , ϑ ) f ( ϑ , u ( ϑ ) , v ( ϑ ) , I 0 + σ 1 u ( ϑ ) , I 0 + σ 2 v ( ϑ ) ) d ϑ d ζ + ( δ 1 1 ) η δ 1 2 a 2 0 1 i = 1 n 0 1 g 2 i ( τ , ζ ) d H i ( τ ) φ ϱ 1 0 1 G 1 ( ζ , ϑ ) f ( ϑ , u ( ϑ ) , v ( ϑ ) , I 0 + σ 1 u ( ϑ ) , I 0 + σ 2 v ( ϑ ) ) d ϑ d ζ .
So for any η ( 0 , 1 ) we find
| ( A 1 ( u , v ) ) ( η ) | 1 Γ ( δ 1 1 ) 0 η [ η δ 1 2 ( 1 ζ ) δ 1 α 0 1 + ( η ζ ) δ 1 2 ] × φ ϱ 1 0 1 J 1 ( ϑ ) ψ 1 ( ϑ ) χ 1 ( ϑ , u ( ϑ ) , v ( ϑ ) , I 0 + σ 1 u ( ϑ ) , I 0 + σ 2 v ( ϑ ) ) d ϑ d ζ + 1 Γ ( δ 1 1 ) η 1 η δ 1 2 ( 1 ζ ) δ 1 α 0 1 φ ϱ 1 0 1 J 1 ( ϑ ) ψ 1 ( ϑ ) χ 1 ( ϑ , u ( ϑ ) , v ( ϑ ) , I 0 + σ 1 u ( ϑ ) , I 0 + σ 2 v ( ϑ ) ) d ϑ d ζ + ( δ 1 1 ) η δ 1 2 a 2 0 1 i = 1 n 0 1 g 2 i ( τ , ζ ) d H i ( τ ) φ ϱ 1 0 1 J 1 ( ϑ ) χ 1 ( ϑ , u ( ϑ ) , v ( ϑ ) , I 0 + σ 1 u ( ϑ ) , I 0 + σ 2 v ( ϑ ) ) d ϑ d ζ Ξ 2 ϱ 1 1 M 1 ϱ 1 1 1 Γ ( δ 1 1 ) 0 η [ η δ 1 2 ( 1 ζ ) δ 1 α 0 1 + ( η ζ ) δ 1 2 ] d ζ + 1 Γ ( δ 1 1 ) η 1 η δ 1 2 ( 1 ζ ) δ 1 α 0 1 d ζ + ( δ 1 1 ) η δ 1 2 a 2 0 1 i = 1 n 0 1 g 2 i ( τ , ζ ) d H i ( τ ) d ζ .
Therefore, for η ( 0 , 1 ) we obtain
| ( A 1 ( u , v ) ) ( η ) | Ξ 2 ϱ 1 1 M 1 ϱ 1 1 1 Γ ( δ 1 1 ) η δ 1 2 δ 1 α 0 + η δ 1 1 δ 1 1 + ( δ 1 1 ) η δ 1 2 a 2 0 1 i = 1 n 0 1 1 Γ ( δ 1 α i ) ( 1 ζ ) δ 1 α 0 1 d ζ τ δ 1 α i 1 d H i ( τ ) = Ξ 2 ϱ 1 1 M 1 ϱ 1 1 1 Γ ( δ 1 1 ) η δ 1 2 δ 1 α 0 + η δ 1 1 δ 1 1 + ( δ 1 1 ) η δ 1 2 a 2 ( δ 1 α 0 ) i = 1 n 1 Γ ( δ 1 α i ) × 0 1 τ δ 1 α i 1 d H i ( τ ) .
We denote by
Θ 0 ( η ) = 1 Γ ( δ 1 1 ) η δ 1 2 δ 1 α 0 + η δ 1 1 δ 1 1 + ( δ 1 1 ) η δ 1 2 a 2 ( δ 1 α 0 ) i = 1 n 1 Γ ( δ 1 α i ) 0 1 τ δ 1 α i 1 d H i ( τ ) , η ( 0 , 1 ) .
This function Θ 0 L 1 ( 0 , 1 ) , because
0 1 Θ 0 ( η ) d η = 1 Γ ( δ 1 ) 1 δ 1 α 0 + 1 δ 1 + 1 a 2 ( δ 1 α 0 ) × i = 1 n 1 Γ ( δ 1 α i ) 0 1 τ δ 1 α i 1 d H i ( τ ) < .
Then for any s 1 , s 2 [ 0 , 1 ] with s 1 < s 2 and ( u , v ) E , by (23) and (24) we conclude
| A 1 ( u , v ) ( s 1 ) A 1 ( u , v ) ( s 2 ) | = s 1 s 2 ( A 1 ( u , v ) ) ( τ ) d τ Ξ 2 ϱ 1 1 M 1 ϱ 1 1 s 1 s 2 Θ 0 ( τ ) d τ .
By (24) and (25), we deduce that A 1 ( E ) is equicontinuous. By a similar method, we find that A 2 ( E ) is also equicontinuous, and then A ( E ) is equicontinuous too. Using the Arzela–Ascoli theorem, we conclude that A 1 ( E ) and A 2 ( E ) are relatively compact sets, and so A ( E ) is also relatively compact. In addition, we can show that A 1 , A 2 and A are continuous on P (see Lemma 1.4.1 from [14]). Hence, A is a completely continuous operator on P . □
We define now the cone
P 0 = { ( u , v ) P , u ( η ) η δ 1 1 u , v ( η ) η δ 2 1 v , η [ 0 , 1 ] } .
Under the assumptions ( I 1 ) and ( I 2 ) , by using Lemma 4, we deduce that A ( P ) P 0 , and so A | P 0 : P 0 P 0 (denoted again by A ) is also a completely continuous operator. For θ > 0 we denote by B θ the open ball centered at zero of radius θ , and by B ¯ θ and B θ its closure and its boundary, respectively.
We also denote by M 1 = 0 1 J 1 ( τ ) ψ 1 ( τ ) d τ , M 2 = 0 1 J 3 ( τ ) ψ 2 ( τ ) d τ , M 3 = 0 1 J 2 ( τ ) d τ , M 4 = 0 1 J 4 ( τ ) d τ , and for θ 1 , θ 2 ( 0 , 1 ) , θ 1 < θ 2 , M 5 = θ 1 θ 2 J 2 ( ζ ) θ 1 ζ G 1 ( ζ , τ ) d τ ϱ 1 1 d ζ , M 6 = θ 1 θ 2 J 4 ( ζ ) θ 1 ζ G 3 ( ζ , τ ) d τ ϱ 2 1 d ζ .
Theorem 2.
We suppose that assumptions ( I 1 ) , ( I 2 ) ,
( I 3 )
There exist c i 0 , i = 1 , , 4 with i = 1 4 c i > 0 , d i 0 , i = 1 , , 4 with i = 1 4 d i > 0 , and μ 1 1 , μ 2 1 such that
χ 10 = lim sup i = 1 4 c i z i 0 max η [ 0 , 1 ] χ 1 ( η , z 1 , z 2 , z 3 , z 4 ) φ r 1 ( ( c 1 z 1 + c 2 z 2 + c 3 z 3 + c 4 z 4 ) μ 1 ) < l 1 ,
and
χ 20 = lim sup i = 1 4 d i z i 0 max η [ 0 , 1 ] χ 2 ( η , z 1 , z 2 , z 3 , z 4 ) φ r 2 ( ( d 1 z 1 + d 2 z 2 + d 3 z 3 + d 4 z 4 ) μ 2 ) < l 2 ,
where l 1 = ( 2 r 1 1 M 1 M 3 r 1 1 ρ 1 μ 1 ( r 1 1 ) ) 1 , l 2 = ( 2 r 2 1 M 2 M 4 r 2 1 ρ 2 μ 2 ( r 2 1 ) ) 1 , with ρ 1 = 2 max c 1 , c 2 , c 3 Γ ( σ 1 + 1 ) , c 4 Γ ( σ 2 + 1 ) , ρ 2 = 2 max d 1 , d 2 , d 3 Γ ( ς 1 + 1 ) , d 4 Γ ( ς 2 + 1 ) ;
( I 4 )
There exist p i 0 , i = 1 , , 4 with i = 1 4 p i > 0 , q i 0 , i = 1 , , 4 with i = 1 4 q i > 0 , θ 1 , θ 2 ( 0 , 1 ) , θ 1 < θ 2 and λ 1 > 1 , λ 2 > 1 such that
f = lim inf i = 1 4 p i z i min η [ θ 1 , θ 2 ] f ( η , z 1 , z 2 , z 3 , z 4 ) φ r 1 ( p 1 z 1 + p 2 z 2 + p 3 z 3 + p 4 z 4 ) > l 3 ,
or
g = lim inf i = 1 4 q i z i min η [ θ 1 , θ 2 ] g ( η , z 1 , z 2 , z 3 , z 4 ) φ r 2 ( q 1 z 1 + q 2 z 2 + q 3 z 3 + q 4 z 4 ) > l 4 ,
where l 3 = λ 1 ( 2 ρ 3 M 5 θ 1 δ 1 1 ) 1 r 1 , l 4 = λ 2 ( 2 ρ 4 M 6 θ 1 δ 2 1 ) 1 r 2 with ρ 3 = min p 1 θ 1 δ 1 1 , p 2 θ 1 δ 2 1 , p 3 θ 1 σ 1 + δ 1 1 Γ ( δ 1 ) Γ ( δ 1 + σ 1 ) , p 4 θ 1 σ 2 + δ 2 1 Γ ( δ 2 ) Γ ( δ 2 + σ 2 ) , ρ 4 = min q 1 θ 1 δ 1 1 , q 2 θ 1 δ 2 1 , q 3 θ 1 ς 1 + δ 1 1 Γ ( δ 1 ) Γ ( δ 1 + ς 1 ) , q 4 θ 1 ς 2 + δ 2 1 Γ ( δ 2 ) Γ ( δ 2 + ς 2 ) ,
hold. Then there exists a positive solution ( u ( τ ) , v ( τ ) ) , τ [ 0 , 1 ] of problems (1) and (2).
Proof. 
By ( I 3 ) there exists R ( 0 , 1 ) such that
χ 1 ( η , z 1 , z 2 , z 3 , z 4 ) l 1 φ r 1 ( ( c 1 z 1 + c 2 z 2 + c 3 z 3 + c 4 z 4 ) μ 1 ) , χ 2 ( η , z 1 , z 2 , z 3 , z 4 ) l 2 φ r 2 ( ( d 1 z 1 + d 2 z 2 + d 3 z 3 + d 4 z 4 ) μ 2 ) ,
for all η [ 0 , 1 ] , z i 0 , i = 1 , , 4 with i = 1 4 c i z i R and i = 1 4 d i z i R . We define R 1 min { R / ρ 1 , R / ρ 2 , R } . For any ( u , v ) B ¯ R 1 P and ζ [ 0 , 1 ] we have
c 1 u ( ζ ) + c 2 v ( ζ ) + c 3 I 0 + σ 1 u ( ζ ) + c 4 I 0 + σ 2 v ( ζ ) 2 max c 1 , c 2 , c 3 Γ ( σ 1 + 1 ) , c 4 Γ ( σ 2 + 1 ) ( u , v ) Y = ρ 1 ( u , v ) Y ρ 1 R 1 R , d 1 u ( ζ ) + d 2 v ( ζ ) + d 3 I 0 + ς 1 u ( ζ ) + d 4 I 0 + ς 2 v ( ζ ) 2 max d 1 , d 2 , d 3 Γ ( ς 1 + 1 ) , d 4 Γ ( ς 2 + 1 ) ( u , v ) Y = ρ 2 ( u , v ) Y ρ 2 R 1 R .
Then by (26) and Lemma 3, for any ( u , v ) B R 1 P 0 and η [ 0 , 1 ] , we deduce
( A 1 ( u , v ) ) ( η ) 0 1 J 2 ( ζ ) φ ϱ 1 0 1 J 1 ( ϑ ) f ( ϑ , u ( ϑ ) , v ( ϑ ) , I 0 + σ 1 u ( ϑ ) , I 0 + σ 2 v ( ϑ ) ) d ϑ d ζ = M 3 φ ϱ 1 0 1 J 1 ( ϑ ) f ( ϑ , u ( ϑ ) , v ( ϑ ) , I 0 + σ 1 u ( ϑ ) , I 0 + σ 2 v ( ϑ ) ) d ϑ M 3 φ ϱ 1 0 1 J 1 ( ϑ ) ψ 1 ( ϑ ) χ 1 ( ϑ , u ( ϑ ) , v ( ϑ ) , I 0 + σ 1 u ( ϑ ) , I 0 + σ 2 v ( ϑ ) ) d ϑ M 3 φ ϱ 1 0 1 J 1 ( ϑ ) ψ 1 ( ϑ ) l 1 φ r 1 ( c 1 u ( ϑ ) + c 2 v ( ϑ ) + c 3 I 0 + σ 1 u ( ϑ ) + c 4 I 0 + σ 2 v ( ϑ ) ) μ 1 d ϑ M 3 φ ϱ 1 φ r 1 ( ( ρ 1 ( u , v ) Y ) μ 1 ) φ ϱ 1 ( l 1 ) φ ϱ 1 ( M 1 ) = M 3 M 1 ϱ 1 1 l 1 ϱ 1 1 ρ 1 μ 1 ( u , v ) Y μ 1 M 3 M 1 ϱ 1 1 l 1 ϱ 1 1 ρ 1 μ 1 ( u , v ) Y = 1 2 ( u , v ) Y , ( A 2 ( u , v ) ) ( η ) 0 1 J 4 ( ζ ) φ ϱ 2 0 1 J 3 ( ϑ ) g ( ϑ , u ( ϑ ) , v ( ϑ ) , I 0 + ς 1 u ( ϑ ) , I 0 + ς 2 v ( ϑ ) ) d ϑ d ζ = M 4 φ ϱ 2 0 1 J 3 ( ϑ ) g ( ϑ , u ( ϑ ) , v ( ϑ ) , I 0 + ς 1 u ( ϑ ) , I 0 + ς 2 v ( ϑ ) ) d ϑ M 4 φ ϱ 2 0 1 J 3 ( ϑ ) ψ 2 ( ϑ ) χ 2 ( ϑ , u ( ϑ ) , v ( ϑ ) , I 0 + ς 1 u ( ϑ ) , I 0 + ς 2 v ( ϑ ) ) d ϑ M 4 φ ϱ 2 0 1 J 3 ( ϑ ) ψ 2 ( ϑ ) l 2 φ r 2 ( d 1 u ( ϑ ) + d 2 v ( ϑ ) + d 3 I 0 + ς 1 u ( ϑ ) + d 4 I 0 + ς 2 v ( ϑ ) ) μ 2 d ϑ M 4 φ ϱ 2 φ r 2 ( ( ρ 2 ( u , v ) Y ) μ 2 ) φ ϱ 2 ( l 2 ) φ ϱ 2 ( M 2 ) = M 4 M 2 ϱ 2 1 l 2 ϱ 2 1 ρ 2 μ 2 ( u , v ) Y μ 2 M 4 M 2 ϱ 2 1 l 2 ϱ 2 1 ρ 2 μ 2 ( u , v ) Y = 1 2 ( u , v ) Y .
Then we conclude that
A ( u , v ) Y = A 1 ( u , v ) + A 2 ( u , v ) ( u , v ) Y , ( u , v ) B R 1 P 0 .
Now we suppose in ( I 4 ) that f > l 3 (in a similar manner we study the case g > l 4 ). Then there exists C 1 > 0 such that
f ( η , z 1 , z 2 , z 3 , z 4 ) l 3 φ r 1 ( p 1 z 1 + p 2 z 2 + p 3 z 3 + p 4 z 4 ) C 1 ,
for all η [ θ 1 , θ 2 ] and z i 0 , i = 1 , , 4 . By definition of I 0 + σ 1 , for any ( u , v ) P 0 and η [ 0 , 1 ] we have
I 0 + σ 1 u ( η ) = 1 Γ ( σ 1 ) 0 η ( η ζ ) σ 1 1 u ( ζ ) d ζ 1 Γ ( σ 1 ) 0 η ( η ζ ) σ 1 1 ζ δ 1 1 u d ζ = ζ = η y u Γ ( σ 1 ) 0 1 ( η η y ) σ 1 1 η δ 1 1 y δ 1 1 η d y = u Γ ( σ 1 ) η σ 1 + δ 1 1 0 1 y δ 1 1 ( 1 y ) σ 1 1 d y = u Γ ( σ 1 ) η σ 1 + δ 1 1 B ( δ 1 , σ 1 ) = u η σ 1 + δ 1 1 Γ ( δ 1 ) Γ ( δ 1 + σ 1 ) ,
and in a similar way
I 0 + σ 2 v ( η ) v η σ 2 + δ 2 1 Γ ( δ 2 ) Γ ( δ 2 + σ 2 ) ,
where B ( p , q ) is the first Euler function. Then by using (28) and (29), for any ( u , v ) P 0 and η [ θ 1 , θ 2 ] we obtain
( A 1 ( u , v ) ) ( η ) θ 1 θ 2 G 2 ( η , ζ ) φ ϱ 1 θ 1 ζ G 1 ( ζ , ϑ ) f ( ϑ , u ( ϑ ) , v ( ϑ ) , I 0 + σ 1 u ( ϑ ) , I 0 + σ 2 v ( ϑ ) ) d ϑ d ζ θ 1 δ 1 1 θ 1 θ 2 J 2 ( ζ ) θ 1 ζ G 1 ( ζ , ϑ ) l 3 ( p 1 u ( ϑ ) + p 2 v ( ϑ ) + p 3 I 0 + σ 1 u ( ϑ ) + p 4 I 0 + σ 2 v ( ϑ ) ) r 1 1 C 1 d ϑ ϱ 1 1 d ζ θ 1 δ 1 1 θ 1 θ 2 J 2 ( ζ ) θ 1 ζ G 1 ( ζ , ϑ ) l 3 p 1 θ 1 δ 1 1 u + p 2 θ 1 δ 2 1 v + p 3 θ 1 σ 1 + δ 1 1 Γ ( δ 1 ) Γ ( δ 1 + σ 1 ) u + p 4 θ 1 σ 2 + δ 2 1 Γ ( δ 2 ) Γ ( δ 2 + σ 2 ) v r 1 1 C 1 d ϑ ϱ 1 1 d ζ θ 1 δ 1 1 θ 1 θ 2 J 2 ( ζ ) θ 1 ζ G 1 ( ζ , ϑ ) l 3 min p 1 θ 1 δ 1 1 , p 2 θ 1 δ 2 1 , p 3 θ 1 σ 1 + δ 1 1 Γ ( δ 1 ) Γ ( δ 1 + σ 1 ) , p 4 θ 1 σ 2 + δ 2 1 Γ ( δ 2 ) Γ ( δ 2 + σ 2 ) 2 ( u , v ) Y r 1 1 C 1 d ϑ ϱ 1 1 d ζ = θ 1 δ 1 1 θ 1 θ 2 J 2 ( ζ ) θ 1 ζ G 1 ( ζ , ϑ ) l 3 2 ρ 3 ( u , v ) Y r 1 1 C 1 d ϑ ϱ 1 1 d ζ = M 5 θ 1 δ 1 1 l 3 2 ρ 3 ( u , v ) Y r 1 1 C 1 ϱ 1 1 = M 5 r 1 1 θ 1 ( δ 1 1 ) ( r 1 1 ) l 3 2 r 1 1 ρ 3 r 1 1 ( u , v ) Y r 1 1 M 5 r 1 1 θ 1 ( δ 1 1 ) ( r 1 1 ) C 1 ϱ 1 1 = λ 1 ( u , v ) Y r 1 1 C 2 ϱ 1 1 , C 2 = M 5 r 1 1 θ 1 ( δ 1 1 ) ( r 1 1 ) C 1 .
Then we deduce
A ( u , v ) Y A 1 ( u , v ) | A 1 ( u , v ) ( θ 1 ) | λ 1 ( u , v ) Y r 1 1 C 2 ϱ 1 1 , ( u , v ) P 0 .
We choose R 2 max 1 , C 2 ϱ 1 1 / ( λ 1 1 ) ϱ 1 1 and we obtain
A ( u , v ) Y ( u , v ) Y , ( u , v ) B R 2 P 0 .
By Lemma 5, (27), (30) and Theorem 1 (i), we conclude that A has a fixed point ( u , v ) ( B ¯ R 2 B R 1 ) P 0 , so R 1 ( u , v ) Y R 2 , and u ( τ ) τ δ 1 1 u and v ( τ ) τ δ 2 1 v for all τ [ 0 , 1 ] . Then u > 0 or v > 0 , that is u ( τ ) > 0 for all τ ( 0 , 1 ] or v ( τ ) > 0 for all τ ( 0 , 1 ] . Hence ( u ( τ ) , v ( τ ) ) , τ [ 0 , 1 ] is a positive solution of problem (1) and (2). □
Theorem 3.
We suppose that assumptions ( I 1 ) , ( I 2 ) ,
( I 5 )
There exist e i 0 , i = 1 , , 4 with i = 1 4 e i > 0 , k i 0 , i = 1 , , 4 with i = 1 4 k i > 0 such that
χ 1 = lim sup i = 1 4 e i z i max η [ 0 , 1 ] χ 1 ( η , z 1 , z 2 , z 3 , z 4 ) φ r 1 ( e 1 z 1 + e 2 z 2 + e 3 z 3 + e 4 z 4 ) < m 1 ,
and
χ 2 = lim sup i = 1 4 k i z i max η [ 0 , 1 ] χ 2 ( η , z 1 , z 2 , z 3 , z 4 ) φ r 2 ( k 1 z 1 + k 2 z 2 + k 3 z 3 + k 4 z 4 ) < m 2 ,
where m 1 < min { 1 / ( 2 M 1 ( ξ 1 M 3 ) r 1 1 ) , 1 / ( M 1 ( 2 ξ 1 M 3 ) r 1 1 ) } , m 2 < min { 1 / ( 2 M 2 ( ξ 2 M 4 ) r 2 1 ) , 1 / ( M 2 ( 2 ξ 2 M 4 ) r 2 1 ) } with ξ 1 = 2 max e 1 , e 2 , e 3 Γ ( σ 1 + 1 ) , e 4 Γ ( σ 2 + 1 ) , ξ 2 = 2 max k 1 , k 2 , k 3 Γ ( ς 1 + 1 ) , k 4 Γ ( ς 2 + 1 ) ;
( I 6 )
There exist s i 0 , i = 1 , , 4 with i = 1 4 s i > 0 , t i 0 , i = 1 , , 4 with i = 1 4 t i > 0 , θ 1 , θ 2 ( 0 , 1 ) , θ 1 < θ 2 and ν 1 ( 0 , 1 ] , ν 2 ( 0 , 1 ] , λ 3 1 , λ 4 1 such that
f 0 = lim inf i = 1 4 s i z i 0 min η [ θ 1 , θ 2 ] f ( η , z 1 , z 2 , z 3 , z 4 ) φ r 1 ( ( s 1 z 1 + s 2 z 2 + s 3 z 3 + s 4 z 4 ) ν 1 ) > m 3 ,
or
g 0 = lim inf i = 1 4 t i z i 0 min η [ θ 1 , θ 2 ] g ( η , z 1 , z 2 , z 3 , z 4 ) φ r 2 ( ( t 1 z 1 + t 2 z 2 + t 3 z 3 + t 4 z 4 ) ν 2 ) > m 4 ,
where m 3 = λ 3 r 1 1 ( M 5 2 ν 1 ξ 3 ν 1 θ 1 δ 1 1 ) 1 r 1 , m 4 = λ 4 r 2 1 ( M 6 2 ν 2 ξ 4 ν 2 θ 1 δ 2 1 ) 1 r 2 , with ξ 3 = min s 1 θ 1 δ 1 1 , s 2 θ 1 δ 2 1 , s 3 θ 1 σ 1 + δ 1 1 Γ ( δ 1 ) Γ ( δ 1 + σ 1 ) , s 4 θ 1 σ 2 + δ 2 1 Γ ( δ 2 ) Γ ( δ 2 + σ 2 ) , ξ 4 = min t 1 θ 1 δ 1 1 , t 2 θ 1 δ 2 1 , t 3 θ 1 ς 1 + δ 1 1 Γ ( δ 1 ) Γ ( δ 1 + ς 1 ) , t 4 θ 1 ς 2 + δ 2 1 Γ ( δ 2 ) Γ ( δ 2 + ς 2 ) ,
hold. Then there exists a positive solution ( u ( τ ) , v ( τ ) ) , τ [ 0 , 1 ] of problem (1) and (2).
Proof. 
From ( I 5 ) there exist C 3 > 0 , C 4 > 0 such that
χ 1 ( η , z 1 , z 2 , z 3 , z 4 ) m 1 φ r 1 ( e 1 z 1 + e 2 z 2 + e 3 z 3 + e 4 z 4 ) + C 3 , χ 2 ( η , z 1 , z 2 , z 3 , z 4 ) m 2 φ r 2 ( k 1 z 1 + k 2 z 2 + k 3 z 3 + k 4 z 4 ) + C 4 ,
for any η [ 0 , 1 ] and z i 0 , i = 1 , , 4 . By using ( I 2 ) and (31) for any ( u , v ) P 0 and η [ 0 , 1 ] we find
A 1 ( u , v ) ( η ) 0 1 J 2 ( ζ ) φ ϱ 1 0 1 J 1 ( ϑ ) f ( ϑ , u ( ϑ ) , v ( ϑ ) , I 0 + σ 1 u ( ϑ ) , I 0 + σ 2 v ( ϑ ) ) d ϑ d ζ M 3 φ ϱ 1 0 1 J 1 ( ϑ ) ψ 1 ( ϑ ) χ 1 ( ϑ , u ( ϑ ) , v ( ϑ ) , I 0 + σ 1 u ( ϑ ) , I 0 + σ 2 v ( ϑ ) ) d ϑ M 3 φ ϱ 1 0 1 J 1 ( ϑ ) ψ 1 ( ϑ ) m 1 φ r 1 e 1 u ( ϑ ) + e 2 v ( ϑ ) + e 3 I 0 + σ 1 u ( ϑ ) + e 4 I 0 + σ 2 v ( ϑ ) + C 3 d ϑ M 3 φ ϱ 1 0 1 J 1 ( ϑ ) ψ 1 ( ϑ ) m 1 e 1 u + e 2 v + e 3 u Γ ( σ 1 + 1 ) + e 4 v Γ ( σ 2 + 1 ) r 1 1 + C 3 d ϑ M 3 φ ϱ 1 m 1 max e 1 , e 2 , e 3 Γ ( σ 1 + 1 ) , e 4 Γ ( σ 2 + 1 ) 2 ( u , v ) Y r 1 1 + C 3 × 0 1 J 1 ( ϑ ) ψ 1 ( ϑ ) d ϑ ϱ 1 1 = M 1 ϱ 1 1 M 3 m 1 ξ 1 r 1 1 ( u , v ) Y r 1 1 + C 3 ϱ 1 1 ,
and
A 2 ( u , v ) ( η ) 0 1 J 4 ( ζ ) φ ϱ 2 0 1 J 3 ( ϑ ) g ( ϑ , u ( ϑ ) , v ( ϑ ) , I 0 + ς 1 u ( ϑ ) , I 0 + ς 2 v ( ϑ ) ) d ϑ d ζ M 4 φ ϱ 2 0 1 J 3 ( ϑ ) ψ 2 ( ϑ ) χ 2 ( ϑ , u ( ϑ ) , v ( ϑ ) , I 0 + ς 1 u ( ϑ ) , I 0 + ς 2 v ( ϑ ) ) d ϑ M 4 φ ϱ 2 0 1 J 3 ( ϑ ) ψ 2 ( ϑ ) m 2 φ r 2 k 1 u ( ϑ ) + k 2 v ( ϑ ) + k 3 I 0 + ς 1 u ( ϑ ) + k 4 I 0 + ς 2 v ( ϑ ) + C 4 d ϑ M 4 φ ϱ 2 0 1 J 3 ( ϑ ) ψ 2 ( ϑ ) m 2 k 1 u + k 2 v + k 3 u Γ ( ς 1 + 1 ) + k 4 v Γ ( ς 2 + 1 ) r 2 1 + C 4 d ϑ M 4 φ ϱ 2 m 2 max k 1 , k 2 , k 3 Γ ( ς 1 + 1 ) , k 4 Γ ( ς 2 + 1 ) 2 ( u , v ) Y r 2 1 + C 4 × 0 1 J 3 ( ϑ ) ψ 2 ( ϑ ) d ϑ ϱ 2 1 = M 2 ϱ 2 1 M 4 m 2 ξ 2 r 2 1 ( u , v ) Y r 2 1 + C 4 ϱ 2 1 .
Then we obtain
A 1 ( u , v ) M 1 ϱ 1 1 M 3 m 1 ξ 1 r 1 1 ( u , v ) Y r 1 1 + C 3 ϱ 1 1 , A 2 ( u , v ) M 2 ϱ 2 1 M 4 m 2 ξ 2 r 2 1 ( u , v ) Y r 2 1 + C 4 ϱ 2 1 ,
and so
A ( u , v ) Y M 1 ϱ 1 1 M 3 m 1 ξ 1 r 1 1 ( u , v ) Y r 1 1 + C 3 ϱ 1 1 + M 2 ϱ 2 1 M 4 m 2 ξ 2 r 2 1 ( u , v ) Y r 2 1 + C 4 ϱ 2 1 ,
for all ( u , v ) P 0 . We choose
R 3 max 1 , M 1 ϱ 1 1 M 3 2 ϱ 1 2 C 3 ϱ 1 1 + M 2 ϱ 2 1 M 4 2 ϱ 2 2 C 4 ϱ 2 1 1 ( M 1 ϱ 1 1 M 3 2 ϱ 1 2 m 1 ϱ 1 1 ξ 1 + M 2 ϱ 2 1 M 4 2 ϱ 2 2 m 2 ϱ 2 1 ξ 2 ) , M 1 ϱ 1 1 M 3 C 3 ϱ 1 1 + M 2 ϱ 2 1 M 4 C 4 ϱ 2 1 1 ( M 1 ϱ 1 1 M 3 m 1 ϱ 1 1 ξ 1 + M 2 ϱ 2 1 M 4 m 2 ϱ 2 1 ξ 2 ) , M 1 ϱ 1 1 M 3 C 3 ϱ 1 1 + M 2 ϱ 2 1 M 4 2 ϱ 2 2 C 4 ϱ 2 1 1 ( M 1 ϱ 1 1 M 3 m 1 ϱ 1 1 ξ 1 + M 2 ϱ 2 1 M 4 2 ϱ 2 2 m 2 ϱ 2 1 ξ 2 ) ,
M 1 ϱ 1 1 M 3 2 ϱ 1 2 C 3 ϱ 1 1 + M 2 ϱ 2 1 M 4 C 4 ϱ 2 1 1 ( M 1 ϱ 1 1 M 3 2 ϱ 1 2 m 1 ϱ 1 1 ξ 1 + M 2 ϱ 2 1 M 4 m 2 ϱ 2 1 ξ 2 ) ,
and then we conclude
A ( u , v ) Y ( u , v ) Y , ( u , v ) B R 3 P 0 .
The above number R 3 was chosen based on the inequalities ( x + y ) ϖ 2 ϖ 1 ( x ϖ + y ϖ ) for ϖ 1 and x , y 0 , and ( x + y ) ϖ x ϖ + y ϖ for ϖ ( 0 , 1 ] and x , y 0 . Here ϖ = ϱ 1 1 or ϱ 2 1 . We prove the inequality (33) in one case, namely ϱ 1 [ 2 , ) and ϱ 2 [ 2 , ) . In this case, by using (32) and the relations M 1 ϱ 1 1 M 3 2 ϱ 1 2 m 1 ϱ 1 1 ξ 1 < 1 / 2 and M 2 ϱ 2 1 M 4 2 ϱ 2 2 m 2 ϱ 2 1 ξ 2 < 1 / 2 (from the inequalities for m 1 and m 2 in ( I 5 ) ) we have the inequalities
M 1 ϱ 1 1 M 3 ( m 1 ξ 1 r 1 1 R 3 r 1 1 + C 3 ) ϱ 1 1 + M 2 ϱ 2 1 M 4 ( m 2 ξ 2 r 2 1 R 3 r 2 1 + C 4 ) ϱ 2 1 M 1 ϱ 1 1 M 3 2 ϱ 1 2 ( m 1 ϱ 1 1 ξ 1 R 3 + C 3 ϱ 1 1 ) + M 2 ϱ 2 1 M 4 2 ϱ 2 2 ( m 2 ϱ 2 1 ξ 2 R 3 + C 4 ϱ 2 1 ) = ( M 1 ϱ 1 1 M 3 2 ϱ 1 2 m 1 ϱ 1 1 ξ 1 + M 2 ϱ 2 1 M 4 2 ϱ 2 2 m 2 ϱ 2 1 ξ 2 ) R 3 + ( M 1 ϱ 1 1 M 3 2 ϱ 1 2 C 3 ϱ 1 1 + M 2 ϱ 2 1 M 4 2 ϱ 2 2 C 4 ϱ 2 1 ) R 3 .
In a similar manner we consider the cases ϱ 1 ( 1 , 2 ] and ϱ 2 ( 1 , 2 ] ; ϱ 1 [ 2 , ) and ϱ 2 ( 1 , 2 ] ; ϱ 1 ( 1 , 2 ] and ϱ 2 [ 2 , ) .
In ( I 6 ) , we suppose that g 0 > m 4 (in a similar manner we can study the case f 0 > m 3 ). We deduce that there exists R ˜ 4 ( 0 , 1 ] such that
g ( η , z 1 , z 2 , z 3 , z 4 ) m 4 φ r 2 ( ( t 1 z 1 + t 2 z 2 + t 3 z 3 + t 4 z 4 ) ν 2 ) ,
for all η [ θ 1 , θ 2 ] , z i 0 , i = 1 , , 4 , i = 1 4 t i z i R ˜ 4 . We take R 4 min { R ˜ 4 / ξ ˜ 4 , R ˜ 4 } , where ξ ˜ 4 = 2 max t 1 , t 2 , t 3 Γ ( ς 1 + 1 ) , t 4 Γ ( ς 2 + 1 ) . Then for any ( u , v ) B ¯ R 4 P and η [ 0 , 1 ] we have
t 1 u ( ζ ) + t 2 v ( ζ ) + t 3 I 0 + ς u ( ζ ) + t 4 I 0 + ς v ( ζ ) t 1 u + t 2 v + t 3 u Γ ( ς 1 + 1 ) + t 4 v Γ ( ς 2 + 1 ) max t 1 , t 2 , t 3 Γ ( ς 1 + 1 ) , t 4 Γ ( ς 2 + 1 ) 2 ( u , v ) Y = ξ ˜ 4 ( u , v ) Y ξ ˜ 4 R 4 R ˜ 4 .
Therefore by using (34) and (29), we obtain for any ( u , v ) B R 4 P 0 and η [ θ 1 , θ 2 ]
A 2 ( u , v ) ( η ) θ 1 θ 2 G 4 ( η , ζ ) φ ϱ 2 θ 1 ζ G 3 ( ζ , ϑ ) g ( ϑ , u ( ϑ ) , v ( ϑ ) , I 0 + ς 1 u ( ϑ ) , I 0 + ς 2 v ( ϑ ) ) d ϑ d ζ θ 1 δ 2 1 θ 1 θ 2 J 4 ( ζ ) φ ϱ 2 θ 1 ζ G 3 ( ζ , ϑ ) m 4 φ r 2 ( t 1 u ( ϑ ) + t 2 v ( ϑ ) + t 3 I 0 + ς 1 u ( ϑ ) + t 4 I 0 + ς 2 v ( ϑ ) ν 2 d ϑ d ζ θ 1 δ 2 1 θ 1 θ 2 J 4 ( ζ ) φ ϱ 2 θ 1 ζ G 3 ( ζ , ϑ ) m 4 t 1 θ 1 δ 1 1 u + t 2 θ 1 δ 2 1 v + + t 3 θ 1 ς 1 + δ 1 1 Γ ( δ 1 ) Γ ( δ 1 + ς 1 ) u + t 4 θ 1 ς 2 + δ 2 1 Γ ( δ 2 ) Γ ( δ 2 + ς 2 ) v ν 2 ( r 2 1 ) d ϑ d ζ θ 1 δ 2 1 θ 1 θ 2 J 4 ( ζ ) θ 1 ζ G 3 ( ζ , ϑ ) m 4 2 ξ 4 ( u , v ) Y ν 2 ( r 2 1 ) d ϑ ϱ 2 1 d ζ = θ 1 δ 2 1 m 4 ϱ 2 1 ( 2 ξ 4 ) ν 2 ( ϱ 2 1 ) ( r 2 1 ) ( u , v ) Y ν 2 θ 1 θ 2 J 4 ( ζ ) θ 1 ζ G 3 ( ζ , ϑ ) d ϑ ϱ 2 1 d ζ = M 6 θ 1 δ 2 1 m 4 ϱ 2 1 2 ν 2 ξ 4 ν 2 ( u , v ) Y ν 2 = λ 4 ( u , v ) Y ν 2 ( u , v ) Y ν 2 ( u , v ) Y .
Then we deduce A 2 ( u , v ) ( u , v ) Y and then
A ( u , v ) Y ( u , v ) Y , ( u , v ) B R 4 P 0 .
From Lemma 5, (33), (35) and Theorem 1 (ii), we conclude that A has a fixed point ( u , v ) ( B ¯ R 3 B R 4 ) P 0 , so R 4 ( u , v ) Y R 3 , which is a positive solution of problem (1) and (2). □
Theorem 4.
We suppose that assumptions ( I 1 ) , ( I 2 ) , ( I 4 ) and ( I 6 ) hold. In addition, the functions ψ i and χ i , i = 1 , 2 satisfy the condition
( I 17 )
M 3 M 1 ϱ 1 1 D 0 ϱ 1 1 < 1 2 , M 4 M 2 ϱ 2 1 D 0 ϱ 2 1 < 1 2 , where
D 0 = max max η [ 0 , 1 ] , z i [ 0 , ω 0 ] , i = 1 , , 4 χ 1 ( η , z 1 , z 2 , z 3 , z 4 ) , max η [ 0 , 1 ] , z i [ 0 , ω 0 ] , i = 1 , , 4 χ 2 ( η , z 1 , z 2 , z 3 , z 4 ) ,
with ω 0 = max 1 , 1 Γ ( σ 1 + 1 ) , 1 Γ ( σ 2 + 1 ) , 1 Γ ( ς 1 + 1 ) , 1 Γ ( ς 2 + 1 ) .
Then there exist two positive solutions ( u 1 ( τ ) , v 1 ( τ ) ) , ( u 2 ( τ ) , v 2 ( τ ) ) , τ [ 0 , 1 ] of problem (1) and (2).
Proof. 
Under assumptions ( I 1 ) , ( I 2 ) and ( I 4 ) , Theorem 2 gives us the existence of R 2 > 1 such that
A ( u , v ) Y ( u , v ) Y , ( u , v ) B R 2 P 0 .
Under assumptions ( I 1 ) , ( I 2 ) and ( I 6 ) , Theorem 3 gives us the existence of R 4 < 1 such that
A ( u , v ) Y ( u , v ) Y , ( u , v ) B R 4 P 0 .
Now we consider the set B 1 = { ( u , v ) Y , ( u , v ) Y < 1 } . By ( I 7 ) , for any ( u , v ) B 1 P 0 and η [ 0 , 1 ] , we obtain
A 1 ( u , v ) ( η ) 0 1 J 2 ( ζ ) φ ϱ 1 0 1 J 1 ( ϑ ) ψ 1 ( ϑ ) χ 1 ( ϑ , u ( ϑ ) , v ( ϑ ) , I 0 + σ 1 u ( ϑ ) , I 0 + σ 2 v ( ϑ ) ) d ϑ d ζ D 0 ϱ 1 1 0 1 J 2 ( ζ ) d ζ 0 1 J 1 ( ϑ ) ψ 1 ( ϑ ) d ϑ ϱ 1 1 = M 3 D 0 ϱ 1 1 M 1 ϱ 1 1 < 1 2 , A 2 ( u , v ) ( η ) 0 1 J 4 ( ζ ) φ ϱ 2 0 1 J 3 ( ϑ ) ψ 2 ( ϑ ) χ 2 ( ϑ , u ( ϑ ) , v ( ϑ ) , I 0 + ς 1 u ( ϑ ) , I 0 + ς 2 v ( ϑ ) ) d ϑ d ζ D 0 ϱ 2 1 0 1 J 4 ( ζ ) d ζ 0 1 J 3 ( ϑ ) ψ 2 ( ϑ ) d ϑ ϱ 2 1 = M 4 D 0 ϱ 2 1 M 2 ϱ 2 1 < 1 2 .
Then A i ( u , v ) < 1 / 2 for all ( u , v ) B 1 P 0 , i = 1 , 2 . Hence
A ( u , v ) Y = A 1 ( u , v ) + A 2 ( u , v ) < 1 = ( u , v ) Y , ( u , v ) B 1 P 0 .
So from (36), (38) and Theorem 1, we deduce that problem (1) and (2) has one positive solution ( u 1 , v 1 ) P 0 with 1 < ( u 1 , v 1 ) Y R 2 . From (37) and (38) and the Guo–Krasnosel’skii fixed point theorem, we conclude that problem (1) and (2) have another positive solution ( u 2 , v 2 ) P 0 with R 4 ( u 2 , v 2 ) Y < 1 . Then problem (1) and (2) have at least two positive solutions ( u 1 ( τ ) , v 1 ( τ ) ) , ( u 2 ( τ ) , v 2 ( τ ) ) , τ [ 0 , 1 ] . □

4. Examples

Let γ 1 = 3 / 2 , γ 2 = 7 / 6 , p = 4 , q = 3 , δ 1 = 10 / 3 , δ 2 = 12 / 5 , σ 1 = 2 / 5 , σ 2 = 29 / 7 , ς 1 = 11 / 9 , ς 2 = 21 / 4 , n = 2 , m = 1 , α 0 = 13 / 8 , α 1 = 5 / 7 , α 2 = 3 / 4 , β 0 = 10 / 9 , β 1 = 7 / 8 , r 1 = 17 / 4 , r 2 = 25 / 8 , ϱ 1 = 17 / 13 , ϱ 2 = 25 / 17 , H 0 ( t ) = { 2 / 7 , t [ 0 , 3 / 4 ) ; 11 / 4 , t [ 3 / 4 , 1 ] } , H 1 ( t ) = t / 2 , t [ 0 , 1 ] , H 2 ( t ) = { 1 / 2 , t [ 0 , 1 / 2 ) ; 13 / 10 , t [ 1 / 2 , 1 ] } , K 0 ( t ) = 4 t / 9 , t [ 0 , 1 ] , K 1 ( t ) = { 1 / 4 , t [ 0 , 1 / 3 ) ; 29 / 20 ,
t [ 1 / 3 , 1 ] } .
We consider the system of fractional differential equations
D 0 + 3 / 2 φ 17 / 4 D 0 + 10 / 3 u ( τ ) = f ( τ , u ( τ ) , v ( τ ) , I 0 + 2 / 5 u ( τ ) , I 0 + 29 / 7 v ( τ ) ) , τ ( 0 , 1 ) , D 0 + 7 / 6 φ 25 / 8 D 0 + 12 / 5 v ( τ ) = g ( τ , u ( τ ) , v ( τ ) , I 0 + 11 / 9 u ( τ ) , I 0 + 21 / 4 v ( τ ) ) , τ ( 0 , 1 ) ,
with the boundary conditions
u ( 0 ) = u ( 0 ) = u ( 0 ) = 0 , D 0 + 10 / 3 u ( 0 ) = 0 , D 0 + 10 / 3 u ( 1 ) = 1 2 4 / 13 D 0 + 10 / 3 u 3 4 , D 0 + 13 / 8 u ( 1 ) = 1 2 0 1 D 0 + 5 / 7 u ( η ) d η + 4 5 D 0 + 3 / 4 u 1 2 , v ( 0 ) = v ( 0 ) = 0 , D 0 + 12 / 5 v ( 0 ) = 0 , φ 25 / 8 D 0 + 12 / 5 v ( 1 ) = 4 9 0 1 φ 25 / 8 D 0 + 12 / 5 v ( η ) d η , D 0 + 10 / 9 v ( 1 ) = 6 5 D 0 + 7 / 8 v 1 3 .
We have here a 1 0.56698729 > 0 , a 2 2.16111947 > 0 , b 1 0.61904762 > 0 , b 2 0.43774133 > 0 . So, assumption ( I 1 ) is satisfied. We also obtain
g 1 ( τ , η ) = 1 Γ ( 3 / 2 ) τ 1 / 2 ( 1 η ) 1 / 2 ( τ η ) 1 / 2 , 0 η τ 1 , τ 1 / 2 ( 1 η ) 1 / 2 , 0 τ η 1 ,
g 2 ( τ , η ) = 1 Γ ( 10 / 3 ) τ 7 / 3 ( 1 η ) 17 / 24 ( τ η ) 7 / 3 , 0 η τ 1 , τ 7 / 3 ( 1 η ) 17 / 24 , 0 τ η 1 ,
g 21 ( τ , η ) = 1 Γ ( 55 / 21 ) τ 34 / 21 ( 1 η ) 17 / 24 ( τ η ) 34 / 21 , 0 η τ 1 , τ 34 / 21 ( 1 η ) 17 / 24 , 0 τ η 1 ,
g 22 ( τ , η ) = 1 Γ ( 31 / 12 ) τ 19 / 12 ( 1 η ) 17 / 24 ( τ η ) 19 / 12 , 0 η τ 1 , τ 19 / 12 ( 1 η ) 17 / 24 , 0 τ η 1 ,
g 3 ( τ , η ) = 1 Γ ( 7 / 6 ) τ 1 / 6 ( 1 η ) 1 / 6 ( τ η ) 1 / 6 , 0 η τ 1 , τ 1 / 6 ( 1 η ) 1 / 6 , 0 τ η 1 ,
g 4 ( τ , η ) = 1 Γ ( 12 / 5 ) τ 7 / 5 ( 1 η ) 13 / 45 ( τ η ) 7 / 5 , 0 η τ 1 , τ 7 / 5 ( 1 η ) 13 / 45 , 0 τ η 1 ,
g 41 ( τ , η ) = 1 Γ ( 61 / 40 ) τ 21 / 40 ( 1 η ) 13 / 45 ( τ η ) 21 / 40 , 0 η τ 1 , τ 21 / 40 ( 1 η ) 13 / 45 , 0 τ η 1 ,
𝒢 1 ( τ , η ) = g 1 ( τ , η ) + τ 1 / 2 2 a 1 g 1 ( 3 4 , η ) ,   ( τ , η ) [ 0 ,   1 ] × [ 0 ,   1 ] ,
𝒢 2 ( τ , η ) = g 2 ( τ , η ) + τ 7 / 3 a 2 ( 1 2 0 1 g 21 ( v , η ) d v + 4 5 g 22 ( 1 2 , η ) ) ,   ( τ , η ) [ 0 ,   1 ] × [ 0 ,   1 ] ,
𝒢 3 ( τ , η ) = g 3 ( τ , η ) + 4 τ 1 / 6 9 b 2 0 1 g 3 ( v , η ) d v + ,   ( τ , η ) [ 0 ,   1 ] × [ 0 ,   1 ] ,
𝒢 4 ( τ , η ) = g 4 ( τ , η ) + 6 τ 7 / 5 5 b 2 g 41 ( 1 3 , η ) ,   ( τ , η ) [ 0 ,   1 ] × [ 0 ,   1 ] ,
h 1 ( η ) = 1 Γ ( 3 / 2 ) ( 1 η ) 1 / 2 ,   h 2 ( η ) = 1 Γ ( 10 / 2 ) ( 1 η ) 17 / 24 ( 1 ( 1 η ) 13 / 8 ) ,   η [ 0 ,   1 ] .
h 3 ( η ) = 1 Γ ( 7 / 6 ) ( 1 η ) 1 / 6 ,   h 4 ( η ) = 1 Γ ( 12 / 5 ) ( 1 η ) 13 / 45 ( 1 ( 1 η ) 10 / 9 ) ,   η [ 0 ,   1 ] .
Besides we deduce
J 1 ( ζ ) = h 1 ( ζ ) + 1 2 a 1 Γ ( 3 / 2 ) 3 4 1 / 2 ( 1 ζ ) 1 / 2 3 4 ζ 1 / 2 , 0 ζ 3 4 , h 1 ( ζ ) + 1 2 a 1 Γ ( 3 / 2 ) 3 4 1 / 2 ( 1 ζ ) 1 / 2 , 3 4 < ζ 1 ,
J 2 ( ζ ) = h 2 ( ζ ) + 1 a 2 1 2 Γ ( 76 / 21 ) ( 1 ζ ) 17 / 24 ( 1 ζ ) 55 / 21 + 4 5 Γ ( 31 / 12 ) 1 2 19 / 12 ( 1 ζ ) 17 / 24 1 2 ζ 19 / 12 , 0 ζ 1 2 , h 2 ( ζ ) + 1 a 2 1 2 Γ ( 76 / 21 ) ( 1 ζ ) 17 / 24 ( 1 ζ ) 55 / 21 + 4 5 Γ ( 31 / 12 ) 1 2 19 / 12 ( 1 ζ ) 17 / 24 , 1 2 < ζ 1 ,
J 3 ( ζ ) = h 3 ( ζ ) + 4 9 b 1 Γ ( 13 / 6 ) ( 1 ζ ) 1 / 6 ( 1 ζ ) 7 / 6 ,   ζ [ 0 ,   1 ] ,
J 4 ( ζ ) = h 4 ( ζ ) + 6 5 b 2 Γ ( 61 / 40 ) 1 3 21 / 40 ( 1 ζ ) 13 / 45 1 3 ζ 21 / 40 , 0 ζ 1 3 , h 4 ( ζ ) + 6 5 b 2 Γ ( 61 / 40 ) 1 3 21 / 40 ( 1 ζ ) 13 / 45 , 1 3 < ζ 1 .
.
Example 1.
We consider the functions
f ( η , z 1 , z 2 , z 3 , z 4 ) = ( 2 z 1 + z 2 + 5 z 3 + 7 z 4 ) 13 a / 4 η κ 1 ( 1 η ) κ 2 , g ( η , z 1 , z 2 , z 3 , z 4 ) = ( 3 z 1 + 8 z 2 + 2 z 3 + 9 z 4 ) 17 b / 8 η κ 3 ( 1 η ) κ 4 ,
for η ( 0 , 1 ) , z i 0 , i = 1 , , 4 , where a > 1 , b > 1 , κ 1 ( 0 , 1 ) , κ 2 ( 0 , 3 / 2 ) , κ 3 ( 0 , 1 ) , κ 4 ( 0 , 7 / 6 ) . Here ψ 1 ( η ) = 1 η κ 1 ( 1 η ) κ 2 , ψ 2 ( η ) = 1 η κ 3 ( 1 η ) κ 4 for η ( 0 , 1 ) , χ 1 ( η , z 1 , z 2 , z 3 , z 4 ) = ( 2 z 1 + z 2 + 5 z 3 + 7 z 4 ) 13 a / 4 and χ 2 ( η , z 1 , z 2 , z 3 , z 4 ) = ( 3 z 1 + 8 z 2 + 2 z 3 + 9 z 4 ) 17 b / 8 for η [ 0 , 1 ] , z i 0 , i = 1 , , 4 . We also find Λ 1 = 0 1 ( 1 τ ) 1 / 2 ψ 1 ( τ ) d τ = B 1 κ 1 , 3 2 κ 2 ( 0 , ) , Λ 2 = 0 1 ( 1 τ ) 1 / 6 ψ 2 ( τ ) d τ = B 1 κ 3 , 7 6 κ 4 ( 0 , ) . Then assumption ( I 2 ) is also satisfied. Moreover, in ( I 3 ) , for c 1 = 2 , c 2 = 1 , c 3 = 5 , c 4 = 7 , μ 1 = 1 , d 1 = 3 , d 2 = 8 , d 3 = 2 , d 4 = 9 , μ 2 = 1 , we obtain χ 10 = 0 , χ 20 = 0 . In ( I 4 ) , for [ θ 1 , θ 2 ] ( 0 , 1 ) , p 1 = 2 , p 2 = 1 , p 3 = 5 , p 4 = 7 , we have f = . By Theorem 2, we deduce that there exists a positive solution ( u ( τ ) , v ( τ ) ) , τ [ 0 , 1 ] of problems (39) and (40) with the nonlinearities (41).
Example 2.
We consider the functions
f ( η , z 1 , z 2 , z 3 , z 4 ) = s 0 ( η + 2 ) ( η 2 + 6 ) η 2 3 1 4 z 1 + 1 3 z 2 + z 3 + 1 2 z 4 ω 1 + 1 4 z 1 + 1 3 z 2 + z 3 + 1 2 z 4 ω 2 , η ( 0 , 1 ] , z i 0 , i = 1 , , 4 , g ( η , z 1 , z 2 , z 3 , z 4 ) = t 0 ( 3 + sin η ) ( η + 2 ) 4 ( 1 η ) 3 5 e z 1 + ln ( z 2 + z 3 + 1 ) + z 4 ω 3 , η [ 0 , 1 ) , z i 0 , i = 1 , , 4 ,
where s 0 > 0 , t 0 > 0 , ω 1 > 13 4 , ω 2 0 , 13 4 , ω 3 > 0 . Here, we have ψ 1 ( η ) = 1 η 2 3 , η ( 0 , 1 ] , χ 1 ( η , z 1 , z 2 , z 3 , z 4 ) = s 0 ( η + 2 ) ( η 2 + 6 ) 1 4 z 1 + 1 3 z 2 + z 3 + 1 2 z 4 ω 1 + 1 4 z 1 + 1 3 z 2 + z 3 + 1 2 z 4 ω 2 , η [ 0 , 1 ] , z i 0 , i = 1 , , 4 , ψ 2 ( η ) = 1 ( 1 η ) 3 5 , η [ 0 , 1 ) , χ 2 ( η , z 1 , z 2 , z 3 , z 4 ) = t 0 ( 3 + sin η ) ( η + 2 ) 4 e z 1 + ln ( z 2 + z 3 + 1 ) + z 4 ω 3 , η [ 0 , 1 ] , z i 0 , i = 1 , , 4 . We find Λ 1 = 0 1 ( 1 τ ) 1 / 2 1 τ 2 3 d τ = B 1 3 , 3 2 ( 0 , ) , Λ 2 = 0 1 ( 1 τ ) 1 / 6 1 ( 1 τ ) 3 5 d τ = 30 17 ( 0 , ) . Then assumption ( I 2 ) is satisfied. For [ θ 1 , θ 2 ] ( 0 , 1 ) , p 1 = 1 / 4 , p 2 = 1 / 3 , p 3 = 1 , p 4 = 1 / 2 , we obtain f = , and for s 1 = 1 / 4 , s 2 = 1 / 3 , s 3 = 1 , s 4 = 1 / 2 and ν 1 4 ω 2 13 , 1 , we have f 0 = . So assumptions ( I 4 ) and ( I 6 ) are satisfied. Then after some computations, we deduce M 1 = 0 1 J 1 ( τ ) ψ 1 ( τ ) d τ 3.04682891 , M 2 = 0 1 J 3 ( τ ) ψ 2 ( τ ) d τ 2.64937892 , M 3 = 0 1 J 2 ( τ ) d τ 0.15582207 , M 4 = 0 1 J 4 ( τ ) d τ 1.25629509 . In addition, we obtain that ω 0 = 1 Γ ( 7 / 5 ) 1.12706049 , D 0 = max { 3 s 0 7 25 12 ω 0 ω 1 + 25 12 ω 0 ω 2 , t 0 m 0 [ e ω 0 + ln ( 2 ω 0 + 1 ) + ω 0 ω 3 ] } , with m 0 = max η [ 0 , 1 ] 3 + sin η ( η + 2 ) 4 3.0123699 . If
s 0 < min 7 3 ( 2 M 3 ) 13 / 4 M 1 ( 25 ω 0 / 12 ) ω 1 + ( 25 ω 0 / 12 ) ω 2 , 7 3 ( 2 M 4 ) 17 / 8 M 2 ( 25 ω 0 / 12 ) ω 1 + ( 25 ω 0 / 12 ) ω 2 , t 0 < min 1 ( 2 M 3 ) 13 / 4 M 1 m 0 e ω 0 + ln ( 2 ω 0 + 1 ) + ω 0 ω 3 , 1 ( 2 M 4 ) 17 / 8 M 2 m 0 e ω 0 + ln ( 2 ω 0 + 1 ) + ω 0 ω 3 ,
then the inequalities M 3 M 1 4 / 13 D 0 4 / 13 < 1 2 , M 4 M 2 8 / 17 D 0 8 / 17 < 1 2 are satisfied (that is, assumption ( I 7 ) is satisfied). For example, if ω 1 = 4 , ω 2 = 2 , ω 3 = 3 , and s 0 0.0034 and t 0 0.0031 , then the above inequalities are satisfied. By Theorem 4, we conclude that problem (39) and (40) with the nonlinearities (42) has at least two positive solutions ( u 1 ( τ ) , v 1 ( τ ) ) , ( u 2 ( τ ) , v 2 ( τ ) ) , τ [ 0 , 1 ] .

5. Conclusions

In this paper we investigate the system of Riemann–Liouville fractional differential Equations (1) with r 1 -Laplacian and r 2 -Laplacian operators and fractional integral terms, subject to the uncoupled boundary conditions (2) which contain Riemann–Stieltjes integrals and fractional derivatives of various orders. The nonlinearities f and g from the system are nonnegative functions and they may be singular at τ = 0 and/or τ = 1 . First we present the Green functions associated to our problem (1) and (2) and some of their properties. Then we give various conditions for the functions f and g such that (1) and (2) has at least one or two positive solutions. In the proof of our main results we use the Guo–Krasnosel’skii fixed point theorem of cone expansion and compression of norm type. We finally present two examples for illustrating the obtained existence theorems.

Author Contributions

Conceptualization, R.L.; formal analysis, A.T. and R.L.; methodology, A.T. and R.L. All authors have read and agreed to the published version of the manuscript.

Funding

This research received no external funding.

Institutional Review Board Statement

Not applicable.

Informed Consent Statement

Not applicable.

Data Availability Statement

Not applicable.

Acknowledgments

The authors thank the referees for their valuable comments and suggestions.

Conflicts of Interest

The authors declare no conflict of interest.

References

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Tudorache, A.; Luca, R. Positive Solutions of a Singular Fractional Boundary Value Problem with r-Laplacian Operators. Fractal Fract. 2022, 6, 18. https://doi.org/10.3390/fractalfract6010018

AMA Style

Tudorache A, Luca R. Positive Solutions of a Singular Fractional Boundary Value Problem with r-Laplacian Operators. Fractal and Fractional. 2022; 6(1):18. https://doi.org/10.3390/fractalfract6010018

Chicago/Turabian Style

Tudorache, Alexandru, and Rodica Luca. 2022. "Positive Solutions of a Singular Fractional Boundary Value Problem with r-Laplacian Operators" Fractal and Fractional 6, no. 1: 18. https://doi.org/10.3390/fractalfract6010018

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