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Positive solutions of Riemann-Stieltjes integral boundary problems for the nonlinear coupling system involving fractional-order differential

Abstract

In this article, we study Riemann-Stieltjes integral boundary value problems of nonlinear fractional functional differential coupling system involving higher-order Caputo fractional derivatives. Some sufficient criteria are obtained for the existence, multiplicity, and nonexistence of positive solutions by applying fixed-point theorems on a convex cone. As applications, some examples are provided to illustrate our main results.

1 Introduction

Fractional differential equations arise in many engineering and scientific disciplines as the mathematical modeling of systems and processes in the fields of physics, chemistry, aerodynamics, electrodynamics of complex medium, polymer rheology, Bode’s analysis of feedback amplifiers, capacitor theory, electrical circuits, electron-analytical chemistry, biology, control theory, fitting of experimental data, and so forth, and involves derivatives of fractional order. Fractional derivatives provide an excellent tool for the description of memory and hereditary properties of various materials and processes. This is the main advantage of fractional differential equations in comparison with classical integer-order models. In consequence, the subject of fractional differential equations is gaining much importance and attention. Especially, there have been many papers focused on boundary value problems of fractional ordinary differential equations (see [116]). Moreover, the boundary value problems with Riemann-Stieltjes integral boundary condition arise in a variety of different areas of applied mathematics and physics (for more comments on Stieltjes integral boundary condition and its importance, we refer the reader to the papers by Webb and Infante [11, 12] and their other related works). For example, blood flow problems, chemical engineering, thermo-elasticity, underground water flow, population dynamics, and so on can be reduced to nonlocal integral boundary problems. Nonlocal boundary value problems of fractional-order differential equations constitute a class of very interesting and important problems. This type of boundary value problems has been investigated in [9, 10, 1316]. To the best of our knowledge, there are only few papers dealing with the existence, multiplicity, and nonexistence of positive solutions of Riemann-Stieltjes integral boundary problems for high-order nonlinear fractional differential coupling system. Therefore, we study the existence, multiplicity, and nonexistence of positive solutions for the following high-order nonlinear fractional differential coupling system (abbreviated by BVPs (1.1)-(1.2) throughout this paper):

{ D 0 + α u ( t ) + f ( t , u ( t ) , v ( t ) , u ( t ) , v ( t ) ) = 0 , t ( 0 , 1 ) , n 1 < α n , D 0 + β v ( t ) + g ( t , u ( t ) , u ( t ) ) = 0 , t ( 0 , 1 ) , m 1 < β m ,
(1.1)

subject to the integral boundary conditions

{ u ( 0 ) = u ( 0 ) = = u ( n 1 ) ( 0 ) = 0 , u ( 1 ) = 0 1 u ( s ) d H ( s ) , v ( 0 ) = v ( 0 ) = = v ( m 1 ) ( 0 ) = 0 , v ( 1 ) = 0 1 v ( s ) d K ( s ) ,
(1.2)

where n,mN, n,m3. D 0 + α , D 0 + β are the Caputo fractional derivatives of order n1<αn, m1<βm. f:[0,1]× [ 0 , + ) 4 [0,+), g:[0,1]× [ 0 , + ) 2 [0,+) are continuous functions. The integrals from (1.2) are Riemann-Stieltjes integrals. H,K:[0,1]R are the function of bounded variation with Δ 1 1 0 1 sdH(s)0 and Δ 2 1 0 1 sdK(s)0. To the best of our knowledge, the study of existence of positive solutions of nonlinear fractional differential system (1.1)-(1.2) has not been done.

The rest of this paper is organized as follows. In Section 2, we recall some useful definitions and properties, and present the properties of the Green’s functions. In Section 3, we give some sufficient conditions for the existence and nonexistence of positive solutions for boundary value problem (1.1)-(1.2). Some examples are also provided to illustrate our main results in Section 4.

2 Preliminaries

For the convenience of the reader, we present here the necessary definitions from fractional calculus theory. These definitions and properties can be found in the recent literature.

Definition 2.1 (see [17, 18])

The Riemann-Liouville fractional integral of order α>0 of a function f:(0,)R is given by

I 0 + α f(t)= 1 Γ ( α ) 0 t ( t s ) α 1 f(s)ds,

provided that the right-hand side is pointwise defined on (0,).

Definition 2.2 (see [17, 18])

The Caputo fractional derivative of order α>0 of a continuous function f:(0,)R is given by

D 0 + α f(t)= 1 Γ ( n α ) 0 t f ( n ) ( s ) ( t s ) α n + 1 ds,

where n1<αn, provided that the right-hand side is pointwise defined on (0,).

Lemma 2.1 (see [17])

Assume thatuC(0,1)L(0,1)with a Caputo fractional derivative of orderα>0that belongs tou C n [0,1], then

I 0 + α D 0 + α u(t)=u(t)+ C 0 + C 1 t++ C n 1 t n 1 ,

for some C i R, i=0,1,,n1, where n is the smallest integer greater than or equal to α.

Here we introduce the following useful fixed-point theorems.

Lemma 2.2 (see [19])

Let E be a Banach space, PEa cone, and Ω 1 , Ω 2 are two bounded open balls of E centered at the origin with0 Ω 1 and Ω ¯ 1 Ω 2 . Suppose thatT:P( Ω ¯ 2 Ω 1 )Pis a completely continuous operator such that either

  1. (i)

    Tuu, uP Ω 1 andTuu, uP Ω 2 , or

  2. (ii)

    Tuu, uP Ω 1 andTuu, uP Ω 2

holds. Then T has at least one fixed point inP( Ω ¯ 2 Ω 1 ).

Let E be a real Banach space with a cone PE. Define a partial order in E as vu if uvP. For uvE, the order interval v,u is defined as v,u={xE:vxu}.

Lemma 2.3 (see [20])

Let P be a normal cone in a real Banach space E, v 0 , u 0 PandT: v 0 , u 0 v 0 , u 0 be an increasing operator. If T is completely continuous, then T has a fixed point u v 0 , u 0 .

Now we present the Green’s functions for system associated with BVPs (1.1)-(1.2).

Lemma 2.4 IfH:[0,1]Ris a function of bounded variation with Δ 1 1 0 1 sdH(s)0andyC([0,1]), then the unique solution of

{ D 0 + α u ( t ) + y ( t ) = 0 , t ( 0 , 1 ) , n 1 < α n , n 3 , u ( 0 ) = u ( 0 ) = = u ( n 1 ) ( 0 ) = 0 , u ( 1 ) = 0 1 u ( s ) d H ( s ) ,
(2.1)

is given by

u(t)= 0 1 G α (t,s)y(s)ds,

where

G α (t,s)= g α (t,s)+ t Δ 1 0 1 g α (τ,s)dH(τ),
(2.2)

and

g α (t,s)= { ( α 1 ) t ( 1 s ) α 2 ( t s ) α 1 Γ ( α ) , 0 s t 1 , ( α 1 ) t ( 1 s ) α 2 Γ ( α ) , 0 t s 1 .
(2.3)

Proof Applying Lemma 2.1, Eq. (2.1) is translated into an equivalent integral equation

u(t)= 1 Γ ( α ) 0 t ( t s ) α 1 y(s)ds+ C 0 + C 1 t++ C n 1 t n 1 .

In the light of u(0)= u (0)== u ( n 1 ) (0)=0, we have C 0 = C 2 == C n 1 =0. From u (1)= 0 1 u(s)dH(s), we deduce

1 Γ ( α ) 0 1 ( α 1 ) ( 1 s ) α 2 y ( s ) d s + C 1 = 0 1 [ 1 Γ ( α ) 0 s ( s τ ) α 1 y ( τ ) d τ + C 1 s ] d H ( s ) ,

namely,

C 1 ( 1 0 1 s d H ( s ) ) = α 1 Γ ( α ) 0 1 ( 1 s ) α 2 y ( s ) d s 1 Γ ( α ) 0 1 ( 0 s ( s τ ) α 1 y ( τ ) d τ ) d H ( s ) ,

which implies

C 1 = α 1 Δ 1 Γ ( α ) 0 1 ( 1 s ) α 2 y ( s ) d s 1 Δ 1 Γ ( α ) 0 1 ( 0 s ( s τ ) α 1 y ( τ ) d τ ) d H ( s ) = α 1 Δ 1 Γ ( α ) 0 1 ( 1 s ) α 2 y ( s ) d s 1 Δ 1 Γ ( α ) 0 1 ( τ 1 ( s τ ) α 1 d H ( s ) ) y ( τ ) d τ = α 1 Δ 1 Γ ( α ) 0 1 ( 1 s ) α 2 y ( s ) d s 1 Δ 1 Γ ( α ) 0 1 ( s 1 ( τ s ) α 1 d H ( τ ) ) y ( s ) d s .

Therefore, the solution of BVPs (2.1) is

u ( t ) = 1 Γ ( α ) 0 t ( t s ) α 1 y ( s ) d s + t Δ 1 Γ ( α ) [ 0 1 ( α 1 ) ( 1 s ) α 2 y ( s ) d s 0 1 ( s 1 ( τ s ) α 1 d H ( τ ) ) y ( s ) d s ] = 1 Γ ( α ) { 0 t [ ( α 1 ) t ( 1 s ) α 2 ( t s ) α 1 ] y ( s ) d s + t 1 ( α 1 ) t ( 1 s ) α 2 y ( s ) d s 0 1 ( α 1 ) t ( 1 s ) α 2 y ( s ) d s + t Δ 1 [ 0 1 ( α 1 ) ( 1 s ) α 2 y ( s ) d s 0 1 ( s 1 ( τ s ) α 1 d H ( τ ) ) y ( s ) d s ] } = 1 Γ ( α ) { 0 t [ ( α 1 ) t ( 1 s ) α 2 ( t s ) α 1 ] y ( s ) d s + t 1 ( α 1 ) t ( 1 s ) α 2 y ( s ) d s 1 Δ 1 ( 1 0 1 τ d H ( τ ) ) 0 1 ( α 1 ) t ( 1 s ) α 2 y ( s ) d s + t Δ 1 [ 0 1 ( α 1 ) ( 1 s ) α 2 y ( s ) d s 0 1 ( s 1 ( τ s ) α 1 d H ( τ ) ) y ( s ) d s ] } = 1 Γ ( α ) { 0 t [ ( α 1 ) t ( 1 s ) α 2 ( t s ) α 1 ] y ( s ) d s + t 1 ( α 1 ) t ( 1 s ) α 2 y ( s ) d s + t Δ 1 [ 0 1 ( 0 1 ( α 1 ) τ ( 1 s ) α 2 d H ( τ ) ) y ( s ) d s 0 1 ( s 1 ( τ s ) α 1 d H ( τ ) ) y ( s ) d s ] } = 1 Γ ( α ) { 0 t [ ( α 1 ) t ( 1 s ) α 2 ( t s ) α 1 ] y ( s ) d s + t 1 ( α 1 ) t ( 1 s ) α 2 y ( s ) d s + t Δ 1 [ 0 1 ( 0 s ( α 1 ) τ ( 1 s ) α 2 d H ( τ ) ) y ( s ) d s + 0 1 ( s 1 [ ( α 1 ) τ ( 1 s ) α 2 ( τ s ) α 1 ] d H ( τ ) ) y ( s ) d s ] } = 0 1 g α ( t , s ) y ( s ) d s + t Δ 1 0 1 ( 0 1 g α ( τ , s ) d H ( τ ) ) y ( s ) d s = 0 1 G α ( t , s ) y ( s ) d s ,

where G α (t,s) and g α (t,s) are defined by (2.2) and (2.3).

Now, we will prove the uniqueness of solution for BVPs (2.1). In fact, let u 1 (t), u 2 (t) are any two solutions of (2.1). Denote w(t)= u 1 (t) u 2 (t), then (2.1) is changed into the following system:

{ D 0 + α w ( t ) = 0 , t ( 0 , 1 ) , n 1 < α n , n 3 , w ( 0 ) = w ( 0 ) = = w ( n 1 ) ( 0 ) = 0 , w ( 1 ) = 0 .

Similar to the above argument, we get w(t)=0, that is u 1 (t)= u 2 (t), which mean that the solution for BVPs (2.1) is unique. The proof is complete. □

Lemma 2.5 IfH:[0,1]Ris a nondecreasing function and Δ 1 >0, we also let G α (t,s) t G α (t,s), g α (t,s) t g α (t,s), then we have the following properties:

  1. (1)

    g α (t,s) t α 1 g α (1,s) t α 1 g α (t,s), for all(t,s)[0,1]×[0,1].

  2. (2)

    G α (t,s) t α 1 J α (s) t α 1 G α (t,s), for all(t,s)[0,1]×[0,1], where J α (s)= g α (1,s)+ 1 Δ 1 0 1 g α (τ,s)dH(τ), s[0,1].

  3. (3)

    G α (t,s)0for all(t,s)[0,1]×[0,1], and for everyθ(0, 1 2 ), we have

    min t [ θ , 1 θ ] g α ( t , s ) γ 1 g α ( 1 , s ) γ 2 g α ( t , s ) , t , s [ 0 , 1 ] , min t [ θ , 1 θ ] G α ( t , s ) γ 1 J α ( s ) γ 2 G α ( t , s ) , t , s [ 0 , 1 ] ,

where γ 1 1 ( 1 θ ) α 2 , γ 2 α 2 α 1 γ 1 .

Proof (1) For t,s[0,1], from (2.3), we have

g α (t,s)= { ( α 1 ) ( 1 s ) α 2 ( α 1 ) ( t s ) α 2 Γ ( α ) , 0 s t 1 , ( α 1 ) ( 1 s ) α 2 Γ ( α ) , 0 t s 1 .
(2.4)

Clearly, g α (t,s)0 for t,s[0,1] which indicates g α (t,s) is increasing with respect to t[0,1]. Therefore, g α (t,s) g α (1,s) for t,s[0,1].

On the other hand, for ts, then

g α ( t , s ) g α ( 1 , s ) = ( α 1 ) t ( 1 s ) α 2 ( t s ) α 1 ( α 1 ) ( 1 s ) α 2 ( 1 s ) α 1 ( α 1 ) t α 1 ( 1 s ) α 2 ( t s ) α 1 ( α 1 ) ( 1 s ) α 2 ( 1 s ) α 1 = t α 1 [ ( α 1 ) ( 1 s ) α 2 ( 1 s t ) α 1 ] ( α 1 ) ( 1 s ) α 2 ( 1 s ) α 1 t α 1 [ ( α 1 ) ( 1 s ) α 2 ( 1 s ) α 1 ] ( α 1 ) ( 1 s ) α 2 ( 1 s ) α 1 = t α 1 .

Thus, for ts, we have

g α ( t , s ) g α ( 1 , s ) = ( α 1 ) t ( 1 s ) α 2 ( α 1 ) ( 1 s ) α 2 ( 1 s ) α 1 ( α 1 ) t ( 1 s ) α 2 ( t s ) α 1 ( α 1 ) ( 1 s ) α 2 ( 1 s ) α 1 t α 1 .

Therefore, g α (t,s) t α 1 g α (1,s) t α 1 g α (t,s), for all (t,s)[0,1]×[0,1].

  1. (2)

    From (2.2), we have

    G α ( t , s ) = g α ( t , s ) + t Δ 1 0 1 g α ( τ , s ) d H ( τ ) t α 1 g α ( 1 , s ) + t α 1 Δ 1 0 1 g α ( τ , s ) d H ( τ ) = t α 1 J α ( s ) ,

where

J α ( s ) = g α ( 1 , s ) + 1 Δ 1 0 1 g α ( τ , s ) d H ( τ ) g α ( t , s ) + t Δ 1 0 1 g α ( τ , s ) d H ( τ ) = G α ( t , s ) .

Therefore, G α (t,s) t α 1 J α (s) t α 1 G α (t,s), for all (t,s)[0,1]×[0,1].

  1. (3)

    From (2.4), for t[0,1], we have 2 g α ( t , s ) t 2 0. Thus, g α (t,s) is decreasing with respect to t[0,1]. Therefore, for θ(0, 1 2 ), we have

    min t [ θ , 1 θ ] g α (t,s) g α (1θ,s) ( α 1 ) ( 1 s ) α 2 ( α 1 ) ( 1 θ s ) α 2 Γ ( α ) .

For any s[0,1], we get

g α ( 1 θ , s ) g α ( s , s ) ( α 1 ) ( 1 s ) α 2 ( α 1 ) ( 1 θ s ) α 2 ( α 1 ) ( 1 s ) α 2 = 1 ( 1 θ 1 s ) α 2 1 ( 1 θ ) α 2 γ 1 .

By

g α ( s , s ) g α ( 1 , s ) = ( α 1 ) ( 1 s ) α 2 ( α 1 ) ( 1 s ) α 2 ( 1 s ) α 1 = α 1 α + s 2 ,

we derive

min t [ θ , 1 θ ] g α ( t , s ) g α ( 1 θ , s ) γ 1 g α ( s , s ) = α 1 α + s 2 γ 1 g α ( 1 , s ) γ 1 g α ( 1 , s ) , g α ( t , s ) g α ( s , s ) = ( α 1 ) ( 1 s ) α 2 Γ ( α ) = α 1 α + s 2 g α ( 1 , s ) α 1 α 2 g α ( 1 , s ) .

Therefore,

min t [ θ , 1 θ ] g α (t,s) γ 1 g α (1,s) α 2 α 1 γ 1 g α ( t , s ) γ 2 g α ( t , s ) , t ,s[0,1].

From (2.2), for t,s[0,1], we have

G α ( t , s ) = g α ( t , s ) + 1 Δ 1 0 1 g α ( τ , s ) d H ( τ ) α 1 α 2 g α ( 1 , s ) + 1 Δ 1 0 1 g α ( τ , s ) d H ( τ ) α 1 α 2 [ g α ( 1 , s ) + 1 Δ 1 0 1 g α ( τ , s ) d H ( τ ) ] = α 1 α 2 J α ( s ) .

According to g α (t,s)0, Δ 1 >0, it is clearly that G α (t,s)0. For t ,s[0,1], we obtain

min t [ θ , 1 θ ] G α ( t , s ) = min t [ θ , 1 θ ] g α ( t , s ) + 1 Δ 1 0 1 g α ( τ , s ) d H ( τ ) γ 1 g α ( 1 , s ) + 1 Δ 1 0 1 g α ( τ , s ) d H ( τ ) γ 1 [ g α ( 1 , s ) + 1 Δ 1 0 1 g α ( τ , s ) d H ( τ ) ] = γ 1 J α ( s ) α 2 α 1 γ 1 G α ( t , s ) = γ 2 G α ( t , s ) .

The proof of Lemma 2.5 is complete. □

From Lemma 2.5, we have the following lemma.

Lemma 2.6 IfH:[0,1]Ris a nondecreasing function and Δ 1 >0, then the Green’s functions G α , G α of BVPs (2.1) are continuous on[0,1]×[0,1]and satisfy G α (t,s), G α (t,s)0for all(t,s)[0,1]×[0,1]. Moreover, ifyC([0,1])satisfiesy(t)0for allt[0,1], then the unique solutionu(t)of BVPs (2.1) satisfiesu(t)0, min t [ θ , 1 θ ] u(t) θ α 1 max t [ 0 , 1 ] u( t ), u (t)= 0 1 G α (t,s)y(s)ds0for allt[0,1]and min t [ θ , 1 θ ] u (t) γ 2 max t [ 0 , 1 ] u ( t ).

We can also formulate similar results as Lemmas 2.4-2.6 above for the fractional differential equation

{ D 0 + β v ( t ) + h ( t ) = 0 , t ( 0 , 1 ) , m 1 < β m , m 3 , v ( 0 ) = v ( 0 ) = = v ( m 1 ) ( 0 ) = 0 , v ( 1 ) = 0 1 v ( s ) d K ( s ) ,
(2.5)

where mN, m3, K:[0,1]R is a nondecreasing function and hC([0,1]). In a similar manner as Δ 1 , γ 1 , γ 2 , g α , g α , G α , G α and J α , we introduce Δ 2 , γ 1 , γ 2 , g β , g β , G β , G β and J β the corresponding constants and functions for BVPs (2.5) defined by Δ 2 1 0 1 sdK(s)0, γ 1 1 ( 1 θ ) β 2 , γ 2 β 2 β 1 γ 1 , G β (t,s)= g β (t,s)+ t Δ 2 0 1 g β (τ,s)dK(τ), G β (t,s) t G β (t,s), g β (t,s) t g β (t,s), J β (s)= g β (1,s)+ 1 Δ 2 0 1 g β (τ,s)dK(τ),

g β (t,s)= { ( β 1 ) t ( 1 s ) β 2 ( t s ) β 1 Γ ( β ) , 0 s t 1 , ( β 1 ) t ( 1 s ) β 2 Γ ( β ) , 0 t s 1 .

3 Existence and nonexistence of positive solutions

In this section, we will discuss the existence and nonexistence of positive solutions to the BVPs (1.1)-(1.2) under various assumptions on f and g.

We present the assumptions that we shall use in the sequel.

(H1) H,K:[0,1]R are nondecreasing functions, Δ 1 =1 0 1 sdH(s)>0, Δ 2 =1 0 1 sdK(s)>0.

(H2) The functions f:[0,1]× [ 0 , ) 4 [0,), g:[0,1]× [ 0 , ) 2 [0,) are continuous and f(t,0,0,0,0)=g(t,0,0)=0 for all t[0,1].

For simplicity, we introduce some important notations as follows:

f 0 = lim sup x + y + z + w 0 + max t [ 0 , 1 ] f ( t , x , y , z , w ) x + y + z + w , g 0 = lim sup x + y 0 + max t [ 0 , 1 ] g ( t , x , y ) x + y , f 0 = lim inf x + y + z + w 0 + min t [ θ , 1 θ ] f ( t , x , y , z , w ) x + y + z + w , g 0 = lim inf x + y 0 + min t [ θ , 1 θ ] g ( t , x , y ) x + y , f = lim sup x + y + z + w + max t [ 0 , 1 ] f ( t , x , y , z , w ) x + y + z + w , g = lim sup x + y + max t [ 0 , 1 ] g ( t , x , y ) x + y , f = lim inf x + y + z + w + min t [ θ , 1 θ ] f ( t , x , y , z , w ) x + y + z + w , g = lim inf x + y + min t [ θ , 1 θ ] g ( t , x , y ) x + y , A 1 = 2 α 3 α 2 0 1 J α ( s ) d s , A 2 = ( θ α 1 + γ 1 ) θ 1 θ J α ( s ) d s , B 1 = 2 β 3 β 2 0 1 J β ( s ) d s , B 2 = γ 0 ( θ β 1 + γ 1 ) θ 1 θ J β ( s ) d s ,

where γ 0 =min{ θ α 1 , γ 2 }.

Let E= C 1 [0,1] be endowed with the norm

u= max t [ 0 , 1 ] | u ( t ) | + max t [ 0 , 1 ] | u ( t ) | = u 0 + u 0 .
(3.1)

Let the cone PE and the operators T:PP be, respectively, defined by

P= { u E : u ( t ) , u ( t ) 0 , min t [ θ , 1 θ ] u ( t ) θ α 1 u 0 , min t [ θ , 1 θ ] u ( t ) γ 2 u 0 }
(3.2)

and

( T u ) ( t ) = 0 1 G α ( t , s ) f ( s , u ( s ) , 0 1 G β ( s , τ ) g ( τ , u ( τ ) , u ( τ ) ) d τ , u ( s ) , 0 1 G β ( s , τ ) g ( τ , u ( τ ) , u ( τ ) ) d τ ) d s , t [ 0 , 1 ] .
(3.3)

It is easy to see that if x(t) is a fixed point of T, then BVPs (1.1)-(1.2) have a pair of solution (u,v) expressed as

{ u ( t ) = x ( t ) , t [ 0 , 1 ] , v ( t ) = 0 1 G β ( t , s ) g ( s , x ( s ) , x ( s ) ) d s , t [ 0 , 1 ] .

Theorem 3.1 Assume that (H1)-(H2) hold. Assume A 1 f 0 < 1 2 < A 2 f , and B 1 g 0 <1< B 2 g . Then BVPs (1.1)-(1.2) have at least a pair of positive solutions(u(t),v(t)).

Proof In view of A 1 f 0 < 1 2 and B 1 g 0 <1, there exists ε 1 >0 such that

A 1 ( f 0 + ε 1 ) 1 2 , B 1 ( g 0 + ε 1 ) 1.
(3.4)

By the definition of f 0 , g 0 , we may choose σ 1 >0 such that, for t[0,1], 0x+yx+y+z+w σ 1 , we have

f(t,x,y,z,w) ( f 0 + ε 1 ) (x+y+z+w),g(t,x,y) ( g 0 + ε 1 ) (x+y).
(3.5)

Let Ω 1 ={uP:u< σ 1 }. Define the operator T: Ω 1 Ω 1 the same as (3.3). We shall prove the theorem through two steps.

Step 1. We assert that T: Ω 1 Ω 1 is completely continuous. In fact, by the definition T, it is easy to see that T is continuous in Ω 1 . It follows from (3.4), (3.5), and Lemma 2.5 that, for any u Ω 1 , s[0,1],

0 1 ( G β ( s , τ ) + G β ( s , τ ) ) g ( τ , u ( τ ) , u ( τ ) ) d τ 0 1 ( J β ( τ ) + β 1 β 2 J β ( τ ) ) ( g 0 + ε 1 ) ( u ( τ ) + u ( τ ) ) d τ ( 2 β 3 ) ( g 0 + ε 1 ) u β 2 0 1 J β ( τ ) d τ < σ 1 ,

which implies that

T u = max 0 t 1 | T u ( t ) | + max 0 t 1 | ( T u ) ( t ) | = 0 1 ( G α ( t , s ) + G α ( t , s ) ) f ( s , u ( s ) , 0 1 G β ( s , τ ) g ( τ , u ( τ ) , u ( τ ) ) d τ , u ( s ) , 0 1 G β ( s , τ ) g ( τ , u ( τ ) , u ( τ ) ) d τ ) d s 0 1 ( J α ( s ) + α 1 α 2 J α ( s ) ) ( f 0 + ε 1 ) ( 0 1 G β ( s , τ ) g ( τ , u ( τ ) , u ( τ ) ) d τ + 0 1 G β ( s , τ ) g ( τ , u ( τ ) , u ( τ ) ) d τ + u ( s ) + u ( s ) ) d s ( 2 α 3 ) ( f 0 + ε 1 ) α 2 0 1 J α ( s ) ( ( 2 β 3 ) ( g 0 + ε 1 ) β 2 × 0 1 J β ( τ ) ( u ( τ ) + u ( τ ) ) d τ + u ) d s 2 α 3 α 2 0 1 J α ( s ) d s ( f 0 + ε 1 ) [ 2 β 3 β 2 0 1 J β ( s ) d s ( g 0 + ε 1 ) + 1 ] u = A 1 ( f 0 + ε 1 ) [ B 1 ( g 0 + ε 1 ) + 1 ] u u < σ 1 .
(3.6)

Thus, we show that T( Ω 1 ) Ω 1 and T( Ω 1 ) is uniformly bounded.

Next, we prove that T: Ω 1 Ω 1 is equicontinuous in [0,1], that is, for any u Ω 1 , t 1 , t 2 [0,1], ϵ>0, δ=δ(ϵ)>0, when | t 1 t 2 |<δ, then |(Tu)( t 1 )(Tu)( t 2 )|<ϵ. Indeed, take δ=δ(ϵ)= ( 2 α 3 ) ϵ ( α 1 ) σ 1 , we have

| ( T u ) ( t 1 ) ( T u ) ( t 2 ) | = | 0 1 ( G α ( t 1 , s ) G α ( t 2 , s ) ) f ( s , u ( s ) , 0 1 G β ( s , τ ) g ( τ , u ( τ ) , u ( τ ) ) d τ , u ( s ) , 0 1 G β ( s , τ ) g ( τ , u ( τ ) , u ( τ ) ) d τ ) d s | 0 1 | G α ( ξ , s ) | | t 1 t 2 | | f ( s , u ( s ) , 0 1 G β ( s , τ ) g ( τ , u ( τ ) , u ( τ ) ) d τ , u ( s ) , 0 1 G β ( s , τ ) g ( τ , u ( τ ) , u ( τ ) ) d τ ) d s | = [ 0 1 G α ( ξ , s ) f ( s , u ( s ) , 0 1 G β ( s , τ ) g ( τ , u ( τ ) , u ( τ ) ) d τ , u ( s ) , 0 1 G β ( s , τ ) g ( τ , u ( τ ) , u ( τ ) ) d τ ) d s ] | t 1 t 2 | [ ( α 1 ) ( f 0 + ε 1 ) α 2 0 1 J α ( s ) ( 0 1 G β ( s , τ ) g ( τ , u ( τ ) , u ( τ ) ) d τ + 0 1 G β ( s , τ ) g ( τ , u ( τ ) , u ( τ ) ) d τ + u ( s ) + u ( s ) ) d s ] | t 1 t 2 | 2 α 3 α 2 0 1 J α ( s ) d s ( f 0 + ε 1 ) × [ 2 β 3 β 2 0 1 J β ( s ) d s ( g 0 + ε 1 ) + 1 ] u α 1 2 α 3 | t 1 t 2 | = A 1 ( f 0 + ε 1 ) [ B 1 ( g 0 + ε 1 ) + 1 ] u α 1 2 α 3 | t 1 t 2 | < ( α 1 ) σ 1 2 α 3 | t 1 t 2 | < ϵ .

Step 2. Now we verify condition (i) or (ii) of Lemma 2.2. In fact, for all uP Ω 1 , s[0,1], similar to the argument of (3.6), we get

Tuu,uP Ω 1 .
(3.7)

On the other hand, since 1 2 < A 2 f and 1< B 2 g , there exists ε 2 >0 such that

A 2 ( f ε 2 ) 1 2 , B 2 ( g ε 2 )1.
(3.8)

By the definition of f , g , we can choose σ 2 > σ 1 such that, for t[θ,1θ], σ 2 x+yx+y+z+w<, we have

f(t,x,y,z,w)( f ε 2 )(x+y),g(t,x,y)( g ε 2 )(x+y).
(3.9)

Let σ 2 =max{ σ 1 , σ 2 γ 0 }= σ 2 γ 0 , where γ 0 =min{ θ α 1 , γ 2 }. Set Ω 2 ={uP:u< σ 2 }. Define the operator T: Ω 2 Ω 2 as (3.3). Similar to the above discussion of T: Ω 1 Ω 1 , we know that T: Ω 2 Ω 2 is completely continuous. Then for t[θ,1θ], uP Ω 2 implies that

u(t)+ u (t) θ α 1 u 0 + γ 2 u 0 min { θ α 1 , γ 2 } ( u 0 + u 0 ) = γ 0 u σ 2 .

It follows from (3.8), (3.9), and Lemma 2.5 that, for any uP Ω 2 , s[θ,1θ], we have

0 1 ( G β ( s , τ ) + G β ( s , τ ) ) g ( τ , u ( τ ) , u ( τ ) ) d τ θ 1 θ ( θ β 1 J β ( τ ) + γ 1 J β ( τ ) ) ( g ε 2 ) ( u ( τ ) + u ( τ ) ) d τ ( g ε 2 ) γ 0 u ( θ β 1 + γ 1 ) θ 1 θ J β ( τ ) d τ = B 2 ( g ε 2 ) u u .
(3.10)

Then, for t[θ,1θ], by (3.8)-(3.10) and Lemma 2.5, we get

T u = max 0 t 1 | T u ( t ) | + max 0 t 1 | ( T u ) ( t ) | = 0 1 ( G α ( t , s ) + G α ( t , s ) ) f ( s , u ( s ) , 0 1 G β ( s , τ ) g ( τ , u ( τ ) , u ( τ ) ) d τ , u ( s ) , 0 1 G β ( s , τ ) g ( τ , u ( τ ) , u ( τ ) ) d τ ) d s θ 1 θ ( θ α 1 J α ( s ) + γ 1 J α ( s ) ) ( f ε 2 ) ( 0 1 G β ( s , τ ) g ( τ , u ( τ ) , u ( τ ) ) d τ + 0 1 G β ( s , τ ) g ( τ , u ( τ ) , u ( τ ) ) d τ + u ( s ) + u ( s ) ) d s A 2 ( f ε 2 ) [ B 2 ( g ε 2 ) + 1 ] u u ,

which implies that

Tuu,uP Ω 2 .
(3.11)

By (3.7), (3.11), and condition (i) of Lemma 2.2, we know that T has at least one fixed point u 1 P( Ω ¯ 2 Ω 1 ). Consequently, BVPs (1.1)-(1.2) have at least a pair of positive solution (u,v)P×P, here u(t)= u 1 (t), v(t)= 0 1 G β (t,s)g(s, u 1 (s), u 1 (s))ds. The proof is complete. □

Similarly, we can get the following theorem.

Theorem 3.2 Assume that (H1)-(H2) hold. Assume A 1 f < 1 2 < A 2 f 0 and B 1 g <1< B 2 g 0 . Then BVPs (1.1)-(1.2) have at least a pair of positive solution.

Theorem 3.3 Assume that (H1)-(H2) hold. If A 2 f 0 > 1 2 , B 2 g 0 >1, A 2 f > 1 2 , B 2 g >1, B 2 g 0 <2, and there exists a constantμ>0such that

max { g ( t , x , y ) : t [ 0 , 1 ] , x + y [ 0 , μ ] } < μ B 1 ;
(3.12)
max { f ( t , x , y , z , w ) : t [ 0 , 1 ] , x + y + z + w [ 0 , μ ] } < μ A 1 .
(3.13)

Then BVPs (1.1)-(1.2) have at least two pairs of positive solutions.

Proof In view of A 2 f 0 >1 and B 2 g 0 >1, there exists ε>0 such that

A 2 ( f 0 ε) 1 2 , B 2 ( g 0 ε)1.
(3.14)

By the definition of f 0 , g 0 , we may choose σ ˆ 1 >0 such that, for t[θ,1θ], 0x+yx+y+z+w σ ˆ 1 , we have

f(t,x,y,z,w)( f 0 ε)(x+y+z+w),g(t,x,y)( g 0 ε)(x+y).
(3.15)

Moreover, from B 2 g 0 <2, take ρ 1 satisfying 0< ρ 1 < B 2 2 B 1 σ ˆ 1 <μ such that

g(t,x,y) g 0 ×(x+y) 2 ρ 1 B 2 ,t[0,1],x+y[0, ρ 1 ].

Set Ω 1 ={uP:u< ρ 1 }. Define the operator T: Ω 1 Ω 1 as (3.3). Similar to the discussion of Theorem 3.1, we know that T: Ω 1 Ω 1 is completely continuous. It follows from (3.14)-(3.15) and Lemma 2.5 that, for any uP Ω 1 , s[0,1],

0 1 ( G β ( s , τ ) + G β ( s , τ ) ) g ( τ , u ( τ ) , u ( τ ) ) d τ 2 ρ 1 B 2 0 1 ( J β ( τ ) + β 1 β 2 J β ( τ ) ) d τ = 2 B 1 B 2 ρ 1 < σ ˆ 1 .
(3.16)

Then, for t[θ,1θ], by (3.14)-(3.16) and Lemma 2.5, we obtain

T u = max 0 t 1 | T u ( t ) | + max 0 t 1 | ( T u ) ( t ) | = 0 1 ( G α ( t , s ) + G α ( t , s ) ) f ( s , u ( s ) , 0 1 G β ( s , τ ) g ( τ , u ( τ ) , u ( τ ) ) d τ , u ( s ) , 0 1 G β ( s , τ ) g ( τ , u ( τ ) , u ( τ ) ) d τ ) d s θ 1 θ ( θ α 1 J α ( s ) + γ 1 J α ( s ) ) ( f 0 ε ) ( 0 1 G β ( s , τ ) g ( τ , u ( τ ) , u ( τ ) ) d τ + 0 1 G β ( s , τ ) g ( τ , u ( τ ) , u ( τ ) ) d τ + u ( s ) + u ( s ) ) d s ( θ α 1 + γ 1 ) θ 1 θ J α ( s ) ( f 0 ε ) × ( θ 1 θ ( G β ( s , τ ) + G β ( s , τ ) ) g ( τ , u ( τ ) , u ( τ ) ) d τ + u ( s ) + u ( s ) ) d s ( θ α 1 + γ 1 ) θ 1 θ J α ( s ) ( f 0 ε ) × ( ( θ β 1 + γ 1 ) θ 1 θ J β ( τ ) ( g 0 ε ) ( u ( τ ) + u ( τ ) ) d τ + u ( s ) + u ( s ) ) d s ( θ α 1 + γ 1 ) θ 1 θ J α ( s ) d s ( f 0 ε ) × [ ( θ β 1 + γ 1 ) θ 1 θ J β ( s ) d s ( g 0 ε ) γ 0 + 1 ] u = A 2 ( f 0 ε ) [ B 2 ( g 0 ε ) + 1 ] u u .

Therefore,

Tuu,uP Ω 1 .
(3.17)

Secondly, according to A 2 f > 1 2 and B 2 g >1, similar to the proof of (3.11), choosing σ 2 >μ, setting Ω 2 ={uP:u< σ 2 } and defining the operator T: Ω 2 Ω 2 as (3.3), we easily get

Tuu,uP Ω 2 .
(3.18)

On the other hand, let Ω 3 ={uP:u<μ}. Define the operator T: Ω 3 Ω 3 as (3.3). Similar to the discussion of Theorem 3.1, we know that T: Ω 3 Ω 3 is completely continuous. Then, for any uP Ω 3 , it follows from (3.12) and (3.13) that

0 1 ( G β ( s , τ ) + G β ( s , τ ) ) g ( τ , u ( τ ) , u ( τ ) ) dτ< μ B 1 0 1 ( J β ( τ ) + β 1 β 2 J β ( τ ) ) dτ=μ,

and

T u = max 0 t 1 | T u ( t ) | + max 0 t 1 | ( T u ) ( t ) | = 0 1 ( G α ( t , s ) + G α ( t , s ) ) f ( s , u ( s ) , 0 1 G β ( s , τ ) g ( τ , u ( τ ) , u ( τ ) ) d τ , u ( s ) , 0 1 G β ( s , τ ) g ( τ , u ( τ ) , u ( τ ) ) d τ ) d s < 0 1 ( J α ( s ) + α 1 α 2 J α ( s ) ) d s μ A 1 = μ = u .

So

Tu<u,uP Ω 3 .
(3.19)

By (3.17), (3.19), and condition (ii) of Lemma 2.2, we know that T has at least a fixed point in u 1 P( Ω ¯ 3 Ω 1 ), that is, ρ 1 u 1 μ. Equations (3.18) and (3.19) together with condition (i) of Lemma 2.2 imply that T has at least one fixed point u 2 P( Ω ¯ 2 Ω 3 ), namely, μ u 2 σ 2 . It is worth noting that ρ 1 <μ< σ 2 , and (3.19) is a strict inequality, that is to say, the operator T has not the fixed point on the boundary Ω 3 . So we conclude that BVPs (1.1)-(1.2) have at least two pairs of positive solutions ( u 1 , v 1 ) and ( u 2 , v 2 ) with the properties of ρ 1 u 1 <μ< u 2 σ 2 and v i (t)= 0 1 G β (t,s)g(s, u i (s), u i (s))ds (i=1,2). The proof is complete. □

Similarly, we get the following theorem.

Theorem 3.4 Assume that (H1)-(H2) hold. Assume A 1 f 0 < 1 2 , B 1 g 0 <1, A 1 f < 1 2 , B 2 g > γ 0 , and there is aη>0such that

min { g ( t , x , y ) : t [ θ , 1 θ ] , x + y [ γ 0 η , ) } > γ 0 2 η B 2 ;
(3.20)
min { f ( t , x , y , z , w ) : t [ θ , 1 θ ] , x + y + z + w [ γ 0 η , ) } > η A 2 .
(3.21)

Then BVPs (1.1)-(1.2) have at least two pairs of positive solutions.

Theorem 3.5 Assume that (H1)-(H2) hold. Further suppose thatf(t,x,y,z,w)andg(t,x,y)are nondecreasing functions with respect to each variable x, y, z, w for eacht[0,1], and there exist u 0 , w 0 satisfyingT u 0 u 0 , T w 0 w 0 for0 u 0 w 0 , 0 u 0 w 0 , 0t1. Then BVPs (1.1)-(1.2) have at least a pair of positive solution( u , v )such that u 0 (t) u (t) w 0 (t), v (t)= 0 1 G β (t,s)g(s, u (s), u (s))ds.

Proof Define the normal cone PE as (3.2) and the operator T:PP as (3.3). By the definition of T, it is easy to show that T is continuous. For any bounded subset Ω={uP:u<R} of P, similar to the proof of (3.6) in Theorem 3.1, we know that T(Ω)ΩP which implies that P is relatively compact set in E. Hence T:PP is completely continuous.

For any u ¯ , v ¯ P defined by (3.2), we define the relationship ≤ on P as v ¯ u ¯ . It is easy to verify that ≤ is a partial order on P. Let u,wP be such that uw, u w , then g(t,u(t), u (t))g(t,w(t), w (t)), for t[0,1]. Thus we have

( T u ) ( t ) = 0 1 G α ( t , s ) f ( s , u ( s ) , 0 1 G β ( s , τ ) g ( τ , u ( τ ) , u ( τ ) ) d τ , u ( s ) , 0 1 G β ( s , τ ) g ( τ , u ( τ ) , u ( τ ) ) d τ ) d s 0 1 G α ( t , s ) f ( s , w ( s ) , 0 1 G β ( s , τ ) g ( τ , w ( τ ) , w ( τ ) ) d τ , w ( s ) , 0 1 G β ( s , τ ) g ( τ , w ( τ ) , w ( τ ) ) d τ ) d s = ( T w ) ( t ) .

Hence T is an increasing operator. By the assumptions T u 0 u 0 , T w 0 w 0 , we have T: u 0 , w 0 u 0 , w 0 . Since T:PP is completely continuous, by Lemma 2.3, T has one fixed point u u 0 , w 0 . Thus BVPs (1.1)-(1.2) have at least a pair of positive solution ( u , v ) such that u 0 (t) u (t) w 0 (t), v (t)= 0 1 G β (t,s)g(s, u (s), u (s))ds. The proof is complete. □

Theorem 3.6 Assume that (H1)-(H2) hold. Assume A 1 f(t,x,y,z,w)< 1 2 (x+y+z+w)and B 1 g(t,x,y)<x+yfort[0,1], 0x+yx+y+z+w<. Then BVPs (1.1)-(1.2) have no monotone positive solution.

Proof Define the cone PE as (3.2), the operator T:PP as (3.3) and the partial order ≤ on P as the proof of Theorem 3.5. By the definition of T, it is easy to show that T is continuous. For any bounded subset Ω={uP:u<R} of P, similar to the proof of (3.6) in Theorem 3.1, we know that T(Ω)ΩP, which implies that P is relatively compact set in E. Hence T:PP is completely continuous.

Suppose on the contrary that u is a monotone positive solution of BVPs (1.1)-(1.2). Then, for t[0,1], we obtain u(t)0, u (t)0, and

u = max 0 t 1 | u ( t ) | + max 0 t 1 | u ( t ) | = 0 1 [ G α ( t , s ) + G α ( t , s ) ] f ( s , u ( s ) , 0 1 G β ( s , τ ) g ( τ , u ( τ ) , u ( τ ) ) d τ , u ( s ) , 0 1 G β ( s , τ ) g ( τ , u ( τ ) , u ( τ ) ) d τ ) d s < 1 2 A 1 0 1 ( J α ( s ) + α 1 α 2 J α ( s ) ) ( 0 1 [ G β ( s , τ ) + G β ( s , τ ) ] × g ( τ , u ( τ ) , u ( τ ) ) d τ + u ( s ) + u ( s ) ) d s < 1 2 A 1 2 α 3 α 2 0 1 J α ( s ) d s [ 1 B 1 0 1 ( J β ( τ ) + β 1 β 2 J β ( τ ) ) [ u ( τ ) + u ( τ ) ] d τ + u ] < 1 2 A 1 2 α 3 α 2 0 1 J α ( s ) d s [ 1 B 1 2 β 3 β 2 0 1 J β ( τ ) d τ + 1 ] u < u ,

which is a contradiction. Then BVPs (1.1)-(1.2) have no monotone positive solution. The proof is complete. □

Similarly, we obtain the following theorem.

Theorem 3.7 Assume that (H1)-(H2) hold. If A 2 f(t,x,y,z,w)> 1 2 (x+y+z+w)and B 2 g(t,x,y)>x+yfort[θ,1θ], 0x+yx+y+z+w<. Then BVPs (1.1)-(1.2) have no monotone positive solution.

4 Illustrative examples

Consider the following coupling system of fractional differential equations:

{ D 0 + 5 2 u ( t ) + f ( t , u ( t ) , v ( t ) , u ( t ) , v ( t ) ) = 0 , t ( 0 , 1 ) , n = 3 , D 0 + 7 2 v ( t ) + g ( t , u ( t ) , u ( t ) ) = 0 , t ( 0 , 1 ) , m = 4 ,
(4.1)

subject to the integral boundary conditions

{ u ( 0 ) = u ( 0 ) = 0 , u ( 1 ) = 0 1 u ( s ) d H ( s ) , v ( 0 ) = v ( 0 ) = v ( 0 ) = 0 , v ( 1 ) = 0 1 v ( s ) d K ( s ) ,
(4.2)

where H(t)= t 2 , K(t)= t 3 for all t[0,1]. Then we obtain

Δ 1 = 1 0 1 s d H ( s ) = 1 2 0 1 s 2 d s = 1 3 > 0 , Δ 2 = 1 0 1 s d K ( s ) = 1 3 0 1 s 3 d s = 1 4 > 0 .

Take θ= 1 4 , for the functions J α and J β , we obtain

J α ( s ) = g α ( 1 , s ) + 1 Δ 1 0 1 g α ( τ , s ) d H ( τ ) = 1 Γ ( α ) { ( α 1 ) ( 1 s ) α 2 ( 1 s ) α 1 + 6 ( 0 1 ( α 1 ) τ 2 ( 1 s ) α 2 d τ s 1 τ ( τ s ) α 1 d τ ) } = 4 3 π [ 9 2 ( 1 s ) 1 2 ( 1 s ) 3 2 24 s + 60 35 ( 1 s ) 5 2 ] , s [ 0 , 1 ] ,

and

J β ( s ) = g β ( 1 , s ) + 1 Δ 2 0 1 g β ( τ , s ) d K ( τ ) = 1 Γ ( β ) { ( β 1 ) ( 1 s ) β 2 ( 1 s ) β 1 + 12 ( 0 1 ( β 1 ) τ 3 ( 1 s ) β 2 d τ s 1 τ 2 ( τ s ) β 1 d τ ) } = 8 15 π [ 10 ( 1 s ) 3 2 ( 1 s ) 5 2 24 ( 8 s 2 + 28 s + 63 ) 693 ( 1 s ) 7 2 ] , s [ 0 , 1 ] .

A simple calculation shows that

A 1 = 2 α 3 α 2 0 1 J α ( s ) d s 6.2186 , A 2 = ( θ α 1 + γ 1 ) θ 1 θ J α ( s ) d s 0.3075 , B 1 = 2 β 3 β 2 0 1 J β ( s ) d s 2.5571 , B 2 = γ 0 ( θ β 1 + γ 1 ) θ 1 θ J β ( s ) d s 0.0080 .

Case 1. Let

f ( t , u , v , u , v ) = 1 16 ( 1 + t ) [ u + v + u + v e u + v + u + v + 50 ( u + v + u + v ) 2 1 + u + v + u + v ] , g ( t , u , u ) = 1 1 + t [ u + u 4 ln [ e + u + u ] + 250 ( u + u ) 2 1 + u + u ] .

Clearly, f(t,0,0,0,0)=g(t,0,0)=0. By a simple computation, we get f 0 = 1 16 , f = 25 14 , g 0 = 1 4 , and g = 1 , 000 7 , which implies that A 1 f 0 0.38866< 1 2 <0.54911 A 2 f and B 1 g 0 0.63928<1<1.14286 B 2 g . Hence BVPs (4.1)-(4.2) have at least a pair of positive solutions by Theorem 3.1.

Case 2. Let

f ( t , u , v , u , v ) = 7 1 + t [ u + v + u + v e u + v + u + v + ( u + v + u + v ) 2 10 5 + u + v + + u + v ] , g ( t , u , u ) = 240 1 + t [ u + u e u + u + ( u + u ) 2 10 6 + u + u ] .

Clearly, f(t,0,0,0,0)=g(t,0,0)=0. By a simple computation, we obtain f 0 = f =4, g 0 = g = 960 7 , and g 0 =240, which shows that A 2 f 0 1.23> 1 2 , A 2 f 1.23> 1 2 , B 2 g 0 1.09714>1, B 2 g 1.09714>1, and B 2 g 0 1.92<2.

Choose μ= 10 3 , we get

max { g ( t , x , y ) : t [ 0 , 1 ] , x + y [ 0 , 10 3 ] } < 240 ( 1 e + 10 6 10 6 + 10 3 ) 328.0513 < μ B 1 391.0680 , max { f ( t , x , y , z , w ) : t [ 0 , 1 ] , x + y + z + w [ 0 , 10 3 ] } < 7 ( 1 e + 10 6 10 5 + 10 3 ) 71.8821 < μ A 1 160.8079 .

Hence BVPs (4.1)-(4.2) have at least two pairs of positive solutions by Theorem 3.3.

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Acknowledgements

The author would like to thank the anonymous referees for their useful and valuable suggestions. This work is supported by the National Natural Sciences Foundation of Peoples Republic of China under Grant (No. 11161025), Yunnan Province natural scientific research fund project (No. 2011FZ058).

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Zhao, K., Gong, P. Positive solutions of Riemann-Stieltjes integral boundary problems for the nonlinear coupling system involving fractional-order differential. Adv Differ Equ 2014, 254 (2014). https://doi.org/10.1186/1687-1847-2014-254

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Keywords

  • coupling fractional differential system
  • positive solutions
  • Riemann-Stieltjes integral BVPs
  • fixed point theorem