Existence and multiplicity of positive solutions for a system of fractional boundary value problems

• 914 Accesses

• 29 Citations

Abstract

We study the existence and multiplicity of positive solutions for a system of nonlinear Riemann-Liouville fractional differential equations, subject to integral boundary conditions. The nonsingular and singular cases for the nonlinearities are investigated.

MSC:34A08, 45G15.

1 Introduction

We consider the system of nonlinear ordinary fractional differential equations

$(S) { D 0 + α u ( t ) + f ( t , v ( t ) ) = 0 , t ∈ ( 0 , 1 ) , n − 1 < α ≤ n , D 0 + β v ( t ) + g ( t , u ( t ) ) = 0 , t ∈ ( 0 , 1 ) , m − 1 < β ≤ m ,$

with the integral boundary conditions

$(BC) { u ( 0 ) = u ′ ( 0 ) = ⋯ = u ( n − 2 ) ( 0 ) = 0 , u ( 1 ) = ∫ 0 1 u ( s ) d H ( s ) , v ( 0 ) = v ′ ( 0 ) = ⋯ = v ( m − 2 ) ( 0 ) = 0 , v ( 1 ) = ∫ 0 1 v ( s ) d K ( s ) ,$

where $n,m∈N$, $n,m≥2$, $D 0 + α$ and $D 0 + β$ denote the Riemann-Liouville derivatives of orders α and β, respectively, and the integrals from (BC) are Riemann-Stieltjes integrals.

Under sufficient conditions on functions f and g, which can be nonsingular or singular in the points $t=0$ and/or $t=1$, we study the existence and multiplicity of positive solutions of problem (S)-(BC). We use the Guo-Krasnosel’skii fixed point theorem (see ) and some theorems from the fixed point index theory (from  and ). By a positive solution of problem (S)-(BC) we mean a pair of functions $(u,v)∈C([0,1])×C([0,1])$ satisfying (S) and (BC) with $u(t)≥0$, $v(t)≥0$ for all $t∈[0,1]$ and $sup t ∈ [ 0 , 1 ] u(t)>0$, $sup t ∈ [ 0 , 1 ] v(t)>0$. The system (S) with $α=n$, $β=m$ and the boundary conditions (BC) where H and K are scale functions (that is, multi-point boundary conditions) has been investigated in  (the nonsingular case) and  (the singular case). In , the authors give sufficient conditions for λ, μ, f, and g such that the system

$( S 1 ) { D 0 + α u ( t ) + λ f ( t , u ( t ) , v ( t ) ) = 0 , t ∈ ( 0 , 1 ) , n − 1 < α ≤ n , D 0 + β v ( t ) + μ g ( t , u ( t ) , v ( t ) ) = 0 , t ∈ ( 0 , 1 ) , m − 1 < β ≤ m ,$

with the boundary conditions (BC) with H and K scale functions, has positive solutions ($u(t)≥0$, $v(t)≥0$ for all $t∈[0,1]$, and $(u,v)≠(0,0)$).

Fractional differential equations describe many phenomena in various fields of engineering and scientific disciplines such as physics, biophysics, chemistry, biology, economics, control theory, signal and image processing, aerodynamics, viscoelasticity, electromagnetics, and so on (see ).

In Section 2, we present the necessary definitions and properties from the fractional calculus theory and some auxiliary results dealing with a nonlocal boundary value problem for fractional differential equations. In Section 3, we give some existence and multiplicity results for positive solutions with respect to a cone for our problem (S)-(BC), where f and g are nonsingular functions. The case when f and g are singular at $t=0$ and/or $t=1$ is studied in Section 4. Finally, in Section 5, we present two examples which illustrate our main results.

2 Preliminaries and auxiliary results

We present here the definitions, some lemmas from the theory of fractional calculus and some auxiliary results that will be used to prove our main theorems.

Definition 2.1 The (left-sided) fractional integral of order $α>0$ of a function $f:(0,∞)→R$ is given by

$( I 0 + α f ) (t)= 1 Γ ( α ) ∫ 0 t ( t − s ) α − 1 f(s)ds,t>0,$

provided the right-hand side is pointwise defined on $(0,∞)$, where $Γ(α)$ is the Euler gamma function defined by $Γ(α)= ∫ 0 ∞ t α − 1 e − t dt$, $α>0$.

Definition 2.2 The Riemann-Liouville fractional derivative of order $α≥0$ for a function $f:(0,∞)→R$ is given by

$( D 0 + α f ) (t)= ( d d t ) n ( I 0 + n − α f ) (t)= 1 Γ ( n − α ) ( d d t ) n ∫ 0 t f ( s ) ( t − s ) α − n + 1 ds,t>0,$

where $n=〚α〛+1$, provided that the right-hand side is pointwise defined on $(0,∞)$.

The notation $〚α〛$ stands for the largest integer not greater than α. We also denote the Riemann-Liouville fractional derivative of f by $D 0 + α f(t)$. If $α=m∈N$ then $D 0 + m f(t)= f ( m ) (t)$ for $t>0$, and if $α=0$ then $D 0 + 0 f(t)=f(t)$ for $t>0$.

Lemma 2.1 ()

Let $α>0$ and $n=〚α〛+1$ for $α∉N$ and $n=α$ for $α∈N$; that is, n is the smallest integer greater than or equal to α. Then the solutions of the fractional differential equation $D 0 + α u(t)=0$, $0, are

$u(t)= c 1 t α − 1 + c 2 t α − 2 +⋯+ c n t α − n ,0

where $c 1 , c 2 ,…, c n$ are arbitrary real constants.

Lemma 2.2 ([6, 10])

Let $α>0$, n be the smallest integer greater than or equal to α ($n−1<α≤n$) and $y∈ L 1 (0,1)$. The solutions of the fractional equation $D 0 + α u(t)+y(t)=0$, $0, are

$u(t)=− 1 Γ ( α ) ∫ 0 t ( t − s ) α − 1 y(s)ds+ c 1 t α − 1 +⋯+ c n t α − n ,0

where $c 1 , c 2 ,…, c n$ are arbitrary real constants.

We consider now the fractional differential equation

$D 0 + α u(t)+y(t)=0,0
(1)

with the integral boundary conditions

$u(0)= u ′ (0)=⋯= u ( n − 2 ) (0)=0,u(1)= ∫ 0 1 u(s)dH(s),$
(2)

where $n∈N$, $n≥2$, and $H:[0,1]→R$ is a function of the bounded variation.

Lemma 2.3 If $H:[0,1]→R$ is a function of bounded variation, $Δ 1 =1− ∫ 0 1 s α − 1 dH(s)≠0$ and $y∈C([0,1])$, then the solution of problem (1)-(2) is given by

$u ( t ) = − 1 Γ ( α ) ∫ 0 t ( t − s ) α − 1 y ( s ) d s + t α − 1 Δ 1 Γ ( α ) [ ∫ 0 1 ( 1 − s ) α − 1 y ( s ) d s − ∫ 0 1 ( ∫ s 1 ( τ − s ) α − 1 d H ( τ ) ) y ( s ) d s ] , 0 ≤ t ≤ 1 .$
(3)

Proof By Lemma 2.2, the solutions of equation (1) are

$u(t)=− 1 Γ ( α ) ∫ 0 t ( t − s ) α − 1 y(s)ds+ c 1 t α − 1 +⋯+ c n t α − n ,$

where $c 1 ,…, c n ∈R$. By using the conditions $u(0)= u ′ (0)=⋯= u ( n − 2 ) (0)=0$, we obtain $c 2 =⋯= c n =0$. Then we conclude

$u(t)= c 1 t α − 1 − 1 Γ ( α ) ∫ 0 t ( t − s ) α − 1 y(s)ds.$

Now, by condition $u(1)= ∫ 0 1 u(s)dH(s)$, we deduce

$c 1 − 1 Γ ( α ) ∫ 0 1 ( 1 − s ) α − 1 y(s)ds= ∫ 0 1 [ c 1 s α − 1 − 1 Γ ( α ) ∫ 0 s ( s − τ ) α − 1 y ( τ ) d τ ] dH(s)$

or

$c 1 ( 1 − ∫ 0 1 s α − 1 d H ( s ) ) = 1 Γ ( α ) ∫ 0 1 ( 1 − s ) α − 1 y ( s ) d s − 1 Γ ( α ) ∫ 0 1 ( ∫ 0 s ( s − τ ) α − 1 y ( τ ) d τ ) d H ( s ) .$

So, we obtain

$c 1 = 1 Δ 1 Γ ( α ) ∫ 0 1 ( 1 − s ) α − 1 y ( s ) d s − 1 Δ 1 Γ ( α ) ∫ 0 1 ( ∫ 0 s ( s − τ ) α − 1 y ( τ ) d τ ) d H ( s ) = 1 Δ 1 Γ ( α ) ∫ 0 1 ( 1 − s ) α − 1 y ( s ) d s − 1 Δ 1 Γ ( α ) ∫ 0 1 ( ∫ τ 1 ( s − τ ) α − 1 d H ( s ) ) y ( τ ) d τ = 1 Δ 1 Γ ( α ) ∫ 0 1 ( 1 − s ) α − 1 y ( s ) d s − 1 Δ 1 Γ ( α ) ∫ 0 1 ( ∫ s 1 ( τ − s ) α − 1 d H ( τ ) ) y ( s ) d s .$

Therefore, we get the expression (3) for the solution of problem (1)-(2). □

Lemma 2.4 Under the assumptions of Lemma  2.3, the Green’s function for the boundary value problem (1)-(2) is given by

$G 1 (t,s)= g 1 (t,s)+ t α − 1 Δ 1 ∫ 0 1 g 1 (τ,s)dH(τ),(t,s)∈[0,1]×[0,1],$
(4)

where

$g 1 (t,s)= 1 Γ ( α ) { t α − 1 ( 1 − s ) α − 1 − ( t − s ) α − 1 , 0 ≤ s ≤ t ≤ 1 , t α − 1 ( 1 − s ) α − 1 , 0 ≤ t ≤ s ≤ 1 .$
(5)

Proof By Lemma 2.3 and relation (3), we conclude

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

where $g 1$ and $G 1$ are given in (5) and (4), respectively. Hence $u(t)= ∫ 0 1 G 1 (t,s)y(s)ds$ for all $t∈[0,1]$. □

Lemma 2.5 ()

The function $g 1$ given by (5) has the properties:

1. (a)

$g 1 :[0,1]×[0,1]→ R +$ is a continuous function and $g 1 (t,s)≥0$ for all $(t,s)∈[0,1]×[0,1]$.

2. (b)

$g 1 (t,s)≤ g 1 ( θ 1 (s),s)$, for all $(t,s)∈[0,1]×[0,1]$.

3. (c)

For any $c∈(0,1/2)$,

where $γ 1 = c α − 1$, $θ 1 (s)=s$ if $1<α≤2$ ($n=2$), and

$θ 1 (s)= { s 1 − ( 1 − s ) α − 1 α − 2 , s ∈ ( 0 , 1 ] , α − 2 α − 1 , s = 0 ,$

if $n−1<α≤n$, $n≥3$.

Lemma 2.6 If $H:[0,1]→R$ is a nondecreasing function and $Δ 1 >0$, then the Green’s function $G 1$ of the problem (1)-(2) is continuous on $[0,1]×[0,1]$ and satisfies $G 1 (t,s)≥0$ for all $(t,s)∈[0,1]×[0,1]$. Moreover, if $y∈C([0,1])$ satisfies $y(t)≥0$ for all $t∈[0,1]$, then the unique solution u of problem (1)-(2) satisfies $u(t)≥0$ for all $t∈[0,1]$.

Proof By using the assumptions of this lemma, we have $G 1 (t,s)≥0$ for all $(t,s)∈[0,1]×[0,1]$, and so $u(t)≥0$ for all $t∈[0,1]$. □

Lemma 2.7 Assume that $H:[0,1]→R$ is a nondecreasing function and $Δ 1 >0$. Then the Green’s function $G 1$ of the problem (1)-(2) satisfies the inequalities:

1. (a)

$G 1 (t,s)≤ J 1 (s)$, $∀(t,s)∈[0,1]×[0,1]$, where

$J 1 (s)= g 1 ( θ 1 ( s ) , s ) + 1 Δ 1 ∫ 0 1 g 1 (τ,s)dH(τ),s∈[0,1].$
2. (b)

For every $c∈(0,1/2)$, we have

$min t ∈ [ c , 1 − c ] G 1 (t,s)≥ γ 1 J 1 (s)≥ γ 1 G 1 ( t ′ , s ) ,∀ t ′ ,s∈[0,1].$

Proof The first inequality (a) is evident. For part (b), for $c∈(0,1/2)$ and $t∈[c,1−c]$, $t ′ ,s∈[0,1]$, we deduce

$G 1 ( t , s ) ≥ c α − 1 g 1 ( θ 1 ( s ) , s ) + c α − 1 Δ 1 ∫ 0 1 g 1 ( τ , s ) d H ( τ ) = c α − 1 ( g 1 ( θ 1 ( s ) , s ) + 1 Δ 1 ∫ 0 1 g 1 ( τ , s ) d H ( τ ) ) = γ 1 J 1 ( s ) ≥ γ 1 G 1 ( t ′ , s ) .$

Therefore, we obtain the inequalities (b) of this lemma. □

Lemma 2.8 Assume that $H:[0,1]→R$ is a nondecreasing function and $Δ 1 >0$, $c∈(0,1/2)$, and $y∈C([0,1])$, $y(t)≥0$ for all $t∈[0,1]$. Then the solution $u(t)$, $t∈[0,1]$ of problem (1)-(2) satisfies the inequality $min t ∈ [ c , 1 − c ] u(t)≥ γ 1 max t ′ ∈ [ 0 , 1 ] u( t ′ )$.

Proof For $c∈(0,1/2)$, $t∈[c,1−c]$, $t ′ ∈[0,1]$, we have

$u(t)= ∫ 0 1 G 1 (t,s)y(s)ds≥ γ 1 ∫ 0 1 J 1 (s)y(s)ds≥ γ 1 ∫ 0 1 G 1 ( t ′ , s ) y(s)ds= γ 1 u ( t ′ ) .$

Then we deduce the conclusion of this lemma. □

We can also formulate similar results as Lemmas 2.3-2.8 above for the fractional differential equation

$D 0 + β v(t)+h(t)=0,0
(6)

with the integral boundary conditions

$v(0)= v ′ (0)=⋯= v ( m − 2 ) (0)=0,v(1)= ∫ 0 1 v(s)dK(s),$
(7)

where $m∈N$, $m≥2$, $K:[0,1]→R$ is a nondecreasing function and $h∈C([0,1])$. We denote by $Δ 2$, $γ 2$, $g 2$, $θ 2$, $G 2$, and $J 2$ the corresponding constants and functions for the problem (6)-(7) defined in a similar manner as $Δ 1$, $γ 1$, $g 1$, $θ 1$, $G 1$, and $J 1$, respectively.

3 The nonsingular case

In this section, we investigate the existence and multiplicity of positive solutions for our problem (S)-(BC) under various assumptions on nonsingular functions f and g.

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

(H1) $H,K:[0,1]→R$ are nondecreasing functions, $Δ 1 =1− ∫ 0 1 s α − 1 dH(s)>0$, $Δ 2 =1− ∫ 0 1 s β − 1 dK(s)>0$.

(H2) The functions $f,g:[0,1]×[0,∞)→[0,∞)$ are continuous and $f(t,0)=g(t,0)=0$ for all $t∈[0,1]$.

A pair of functions $(u,v)∈C([0,1])×C([0,1])$ is a solution for our problem (S)-(BC) if and only if $(u,v)∈C([0,1])×C([0,1])$ is a solution for the nonlinear integral system

${ u ( t ) = ∫ 0 1 G 1 ( t , s ) f ( s , ∫ 0 1 G 2 ( s , τ ) g ( τ , u ( τ ) ) d τ ) d s , t ∈ [ 0 , 1 ] , v ( t ) = ∫ 0 1 G 2 ( t , s ) g ( s , u ( s ) ) d s , t ∈ [ 0 , 1 ] .$

We consider the Banach space $X=C([0,1])$ with supremum norm $∥⋅∥$ and define the cone $P⊂X$ by $P={u∈X,u(t)≥0,∀t∈[0,1]}$.

We also define the operators $A:P→X$ by

$(Au)(t)= ∫ 0 1 G 1 (t,s)f ( s , ∫ 0 1 G 2 ( s , τ ) g ( τ , u ( τ ) ) d τ ) ds,t∈[0,1],$

and $B:P→X$, $C:P→X$ by

$(Bu)(t)= ∫ 0 1 G 1 (t,s)u(s)ds,(Cu)(t)= ∫ 0 1 G 2 (t,s)u(s)ds,t∈[0,1].$

Under the assumptions (H1) and (H2), using also Lemma 2.6, it is easy to see that , , and are completely continuous from P to P. Thus the existence and multiplicity of positive solutions of the system (S)-(BC) are equivalent to the existence and multiplicity of fixed points of the operator .

Theorem 3.1 Assume that (H1)-(H2) hold. If the functions f and g also satisfy the conditions:

(H3) There exist positive constants $p∈(0,1]$ and $c∈(0,1/2)$ such that

$( i ) f ∞ i = lim inf u → ∞ inf t ∈ [ c , 1 − c ] f ( t , u ) u p ∈ ( 0 , ∞ ] ; ( ii ) g ∞ i = lim inf u → ∞ inf t ∈ [ c , 1 − c ] g ( t , u ) u 1 / p = ∞ .$

(H4) There exists a positive constant $q∈(0,∞)$ such that

$( i ) f 0 s = lim sup u → 0 + sup t ∈ [ 0 , 1 ] f ( t , u ) u q ∈ [ 0 , ∞ ) ; ( ii ) g 0 s = lim sup u → 0 + sup t ∈ [ 0 , 1 ] g ( t , u ) u 1 / q = 0 ,$

then the problem (S)-(BC) has at least one positive solution $(u(t),v(t))$, $t∈[0,1]$.

Proof Because the proof of the theorem is similar to that of Theorem 3.1 from , we will sketch some parts of it. From assumption (i) of (H3), we deduce that there exist $C 1 , C 2 >0$ such that

$f(t,u)≥ C 1 u p − C 2 ,∀(t,u)∈[c,1−c]×[0,∞).$
(8)

Then for $u∈P$, by using (8), Lemma 2.6, and Lemma 2.7, we obtain after some computations

$(Au)(t)≥ C 1 ∫ c 1 − c G 1 (t,s) ( ∫ c 1 − c ( G 2 ( s , τ ) ) p ( g ( τ , u ( τ ) ) ) p d τ ) ds− C 3 ,∀t∈[0,1],$
(9)

where $C 3 = C 2 ∫ c 1 − c J 1 (s)ds$.

For c given in (H3), we define the cone $P 0 ={u∈P, inf t ∈ [ c , 1 − c ] u(t)≥γ∥u∥}$, where $γ=min{ γ 1 , γ 2 }$. From our assumptions and Lemma 2.8, for any $y∈P$, we can easily show that $u=By∈ P 0$ and $v=Cy∈ P 0$, that is, $B(P)⊂ P 0$ and $C(P)⊂ P 0$.

We now consider the function $u 0 (t)= ∫ 0 1 G 1 (t,s)ds=(B y 0 )(t)≥0$, $t∈[0,1]$, with $y 0 (t)=1$ for all $t∈[0,1]$. We define the set

We will show that $M⊂ P 0$ and M is a bounded subset of X. If $u∈M$, then there exists $λ≥0$ such that $u(t)=(Au)(t)+λ u 0 (t)$, $t∈[0,1]$. From the definition of $u 0$, we have

$u(t)=(Au)(t)+λ(B y 0 )(t)=B ( F u ( t ) ) +λ(B y 0 )(t)=B ( F u ( t ) + λ y 0 ( t ) ) ∈ P 0 ,$

where $F:P→P$ is defined by $(Fu)(t)=f(t, ∫ 0 1 G 2 (t,s)g(s,u(s))ds)$. Therefore, $M⊂ P 0$, and from the definition of $P 0$, we get

$∥u∥≤ 1 γ inf t ∈ [ c , 1 − c ] u(t),∀u∈M.$
(10)

From (ii) of assumption (H3), we conclude that for $ε 0 = ( 2 / ( C 1 m 1 m 2 γ 1 γ 2 p ) ) 1 / p >0$ there exists $C 4 >0$ such that

$( g ( t , u ) ) p ≥ ε 0 p u− C 4 ,∀(t,u)∈[c,1−c]×[0,∞),$
(11)

where $m 1 = ∫ c 1 − c J 1 (τ)dτ>0$, $m 2 = ∫ c 1 − c ( J 2 ( τ ) ) p dτ>0$.

For $u∈M$ and $t∈[c,1−c]$, by using Lemma 2.7 and the relations (9) and (11), it follows that

$u ( t ) = ( A u ) ( t ) + λ u 0 ( t ) ≥ ( A u ) ( t ) ≥ C 1 γ 1 γ 2 p ∫ c 1 − c J 1 ( s ) ( ∫ c 1 − c ( J 2 ( τ ) ) p ( ε 0 p u ( τ ) − C 4 ) d τ ) d s − C 3 ≥ C 1 ε 0 p γ 1 γ 2 p ( ∫ c 1 − c J 1 ( s ) d s ) ( ∫ c 1 − c ( J 2 ( τ ) ) p u ( τ ) d τ ) − C 5 ≥ C 1 ε 0 p γ 1 γ 2 p ( ∫ c 1 − c J 1 ( s ) d s ) ( ∫ c 1 − c ( J 2 ( τ ) ) p d τ ) ⋅ inf τ ∈ [ c , 1 − c ] u ( τ ) − C 5 = 2 inf τ ∈ [ c , 1 − c ] u ( τ ) − C 5 ,$

where $C 5 = C 3 + C 1 C 4 m 1 m 2 γ 1 γ 2 p >0$.

Hence, $inf t ∈ [ c , 1 − c ] u(t)≥2 inf t ∈ [ c , 1 − c ] u(t)− C 5$, and so

$inf t ∈ [ c , 1 − c ] u(t)≤ C 5 ,∀u∈M.$
(12)

Now from relations (10) and (12), one obtains $∥u∥≤( inf t ∈ [ c , 1 − c ] u(t))/γ≤ C 5 /γ$, for all $u∈M$, that is, M is a bounded subset of X.

Besides, there exists a sufficiently large $L>0$ such that

$u≠Au+λ u 0 ,∀u∈∂ B L ∩P,∀λ≥0.$

From , we deduce that the fixed point index of the operator over $B L ∩P$ with respect to P is

$i(A, B L ∩P,P)=0.$
(13)

Next, from assumption (H4), we conclude that there exist $M 0 >0$ and $δ 1 ∈(0,1)$ such that

$f ( t , u ) ≤ M 0 u q , ∀ ( t , u ) ∈ [ 0 , 1 ] × [ 0 , 1 ] ; g ( t , u ) ≤ ε 1 u 1 / q , ∀ ( t , u ) ∈ [ 0 , 1 ] × [ 0 , δ 1 ] ,$
(14)

where $ε 1 =min{1/ M 2 , ( 1 / ( 2 M 0 M 1 M 2 q ) ) 1 / q }>0$, $M 1 = ∫ 0 1 J 1 (s)ds>0$, $M 2 = ∫ 0 1 J 2 (s)ds>0$. Hence, for any $u∈ B ¯ δ 1 ∩P$ and $t∈[0,1]$, we obtain

$∫ 0 1 G 2 (t,s)g ( s , u ( s ) ) ds≤ ε 1 ∫ 0 1 J 2 (s) ( u ( s ) ) 1 / q ds≤ ε 1 M 2 ∥ u ∥ 1 / q ≤1.$
(15)

Therefore, by (14) and (15), we deduce that for any $u∈ B ¯ δ 1 ∩P$ and $t∈[0,1]$

$( A u ) ( t ) ≤ M 0 ∫ 0 1 G 1 ( t , s ) ( ∫ 0 1 G 2 ( s , τ ) g ( τ , u ( τ ) ) d τ ) q d s ≤ M 0 ε 1 q M 2 q ∥ u ∥ ∫ 0 1 J 1 ( s ) d s = M 0 ε 1 q M 1 M 2 q ∥ u ∥ ≤ 1 2 ∥ u ∥ .$

This implies that $∥Au∥≤∥u∥/2$ for all $u∈∂ B δ 1 ∩P$. From , we conclude that the fixed point index of the operator over $B δ 1 ∩P$ with respect to P is

$i(A, B δ 1 ∩P,P)=1.$
(16)

Combining (13) and (16), we obtain

$i ( A , ( B L ∖ B ¯ δ 1 ) ∩ P , P ) =i(A, B L ∩P,P)−i(A, B δ 1 ∩P,P)=−1.$

We deduce that has at least one fixed point $u 1 ∈( B L ∖ B ¯ δ 1 )∩P$, that is, $δ 1 <∥ u 1 ∥.

Let $v 1 (t)= ∫ 0 1 G 2 (t,s)g(s, u 1 (s))ds$. Then $( u 1 , v 1 )∈P×P$ is a solution of (S)-(BC). In addition $∥ v 1 ∥>0$. Indeed, if we suppose that $v 1 (t)=0$, for all $t∈[0,1]$, then by using (H2) we have $f(s, v 1 (s))=f(s,0)=0$, for all $s∈[0,1]$. This implies $u 1 (t)= ∫ 0 1 G 1 (t,s)f(s, v 1 (s))ds=0$, for all $t∈[0,1]$, which contradicts $∥ u 1 ∥>0$. The proof of Theorem 3.1 is completed. □

Using similar arguments as those used in the proofs of Theorem 3.2 and Theorem 3.3 in , we also obtain the following results for our problem (S)-(BC).

Theorem 3.2 Assume that (H1)-(H2) hold. If the functions f and g also satisfy the conditions:

(H5) There exists a positive constant $r∈(0,∞)$ such that

$( i ) f ∞ s = lim sup u → ∞ sup t ∈ [ 0 , 1 ] f ( t , u ) u r ∈ [ 0 , ∞ ) ; ( ii ) g ∞ s = lim sup u → ∞ sup t ∈ [ 0 , 1 ] g ( t , u ) u 1 / r = 0 .$

(H6) There exists $c∈(0,1/2)$ such that

$( i ) f 0 i = lim inf u → 0 + inf t ∈ [ c , 1 − c ] f ( t , u ) u ∈ ( 0 , ∞ ] ; ( ii ) g 0 i = lim inf u → 0 + inf t ∈ [ c , 1 − c ] g ( t , u ) u = ∞ ,$

then the problem (S)-(BC) has at least one positive solution $(u(t),v(t))$, $t∈[0,1]$.

Theorem 3.3 Assume that (H1)-(H3), and (H6) hold. If the functions f and g also satisfy the condition:

(H7) For each $t∈[0,1]$, $f(t,u)$ and $g(t,u)$ are nondecreasing with respect to u, and there exists a constant $N>0$ such that

$f ( t , m 0 ∫ 0 1 g ( s , N ) d s ) < N m 0 ,∀t∈[0,1],$

where $m 0 =max{ K 1 , K 2 }$, $K 1 = max s ∈ [ 0 , 1 ] J 1 (s)$, $K 2 = max s ∈ [ 0 , 1 ] J 2 (s)$, and $J 1$, $J 2$ are defined in Section  2, then the problem (S)-(BC) has at least two positive solutions $( u 1 (t), v 1 (t))$, $( u 2 (t), v 2 (t))$, $t∈[0,1]$.

4 The singular case

In this section, we investigate the existence of positive solutions for our problem (S)-(BC) under various assumptions on functions f and g which may be singular at $t=0$ and/or $t=1$.

The basic assumptions used here are the following.

($H 1 ˜$) ≡ (H1).

($H 2 ˜$) The functions $f,g∈C((0,1)× R + , R + )$ and there exist $p i ∈C((0,1), R + )$, $q i ∈C( R + , R + )$, $i=1,2$, with $0< ∫ 0 1 p i (t)dt<∞$, $i=1,2$, $q 1 (0)=0$, $q 2 (0)=0$ such that

$f(t,x)≤ p 1 (t) q 1 (x),g(t,x)≤ p 2 (t) q 2 (x),∀t∈(0,1),x∈ R + .$

We consider the Banach space $X=C([0,1])$ with supremum norm and define the cone $P⊂X$ by $P={u∈X,u(t)≥0,∀t∈[0,1]}$. We also define the operator $A ˜ :P→X$ by

$( A ˜ u)(t)= ∫ 0 1 G 1 (t,s)f ( s , ∫ 0 1 G 2 ( s , τ ) g ( τ , u ( τ ) ) d τ ) ds.$

Lemma 4.1 Assume that ($H 1 ˜$)-($H 2 ˜$) hold. Then $A ˜ :P→P$ is completely continuous.

Proof We denote by $α ˜ = ∫ 0 1 J 1 (s) p 1 (s)ds$ and $β ˜ = ∫ 0 1 J 2 (s) p 2 (s)ds$. Using ($H 2 ˜$), we deduce that $0< α ˜ <∞$ and $0< β ˜ <∞$. By Lemma 2.6 and the corresponding lemma for $G 2$, we see that $A ˜$ maps P into P.

We shall prove that $A ˜$ maps bounded sets into relatively compact sets. Suppose $D⊂P$ is an arbitrary bounded set. Then there exists $M ¯ 1 >0$ such that $∥u∥≤ M ¯ 1$ for all $u∈D$. By using ($H 2 ˜$) and Lemma 2.7, we obtain $∥ A ˜ u∥≤ α ˜ M ¯ 3$ for all $u∈D$, where $M ¯ 3 = sup x ∈ [ 0 , β ˜ M ¯ 2 ] q 1 (x)$, and $M ¯ 2 = sup x ∈ [ 0 , M ¯ 1 ] q 2 (x)$. In what follows, we shall prove that $A ˜ (D)$ is equicontinuous. By using Lemma 2.4, we have

$( A ˜ u ) ( t ) = ∫ 0 1 G 1 ( t , s ) f ( s , ∫ 0 1 G 2 ( s , τ ) g ( τ , u ( τ ) ) d τ ) d s = ∫ 0 1 [ g 1 ( t , s ) + t α − 1 Δ 1 ∫ 0 1 g 1 ( τ , s ) d H ( τ ) ] f ( s , ∫ 0 1 G 2 ( s , τ ) g ( τ , u ( τ ) ) d τ ) d s = 1 Γ ( α ) ∫ 0 t [ t α − 1 ( 1 − s ) α − 1 − ( t − s ) α − 1 ] f ( s , ∫ 0 1 G 2 ( s , τ ) g ( τ , u ( τ ) ) d τ ) d s + 1 Γ ( α ) ∫ t 1 t α − 1 ( 1 − s ) α − 1 f ( s , ∫ 0 1 G 2 ( s , τ ) g ( τ , u ( τ ) ) d τ ) d s + t α − 1 Δ 1 ∫ 0 1 ( ∫ 0 1 g 1 ( τ , s ) d H ( τ ) ) × f ( s , ∫ 0 1 G 2 ( s , τ ) g ( τ , u ( τ ) ) d τ ) d s , ∀ t ∈ [ 0 , 1 ] .$

Therefore, for any $t∈(0,1)$, we obtain

$( A ˜ u ) ′ ( t ) = 1 Γ ( α ) ∫ 0 t [ ( α − 1 ) t α − 2 ( 1 − s ) α − 1 − ( α − 1 ) ( t − s ) α − 2 ] × f ( s , ∫ 0 1 G 2 ( s , τ ) g ( τ , u ( τ ) ) d τ ) d s + 1 Γ ( α ) ∫ t 1 ( α − 1 ) t α − 2 ( 1 − s ) α − 1 f ( s , ∫ 0 1 G 2 ( s , τ ) g ( τ , u ( τ ) ) d τ ) d s + ( α − 1 ) t α − 2 Δ 1 ∫ 0 1 ( ∫ 0 1 g 1 ( τ , s ) d H ( τ ) ) f ( s , ∫ 0 1 G 2 ( s , τ ) g ( τ , u ( τ ) ) d τ ) d s .$

So, for any $t∈(0,1)$, we deduce

$| ( A ˜ u ) ′ ( t ) | ≤ 1 Γ ( α − 1 ) ∫ 0 t [ t α − 2 ( 1 − s ) α − 1 + ( t − s ) α − 2 ] p 1 ( s ) × q 1 ( ∫ 0 1 G 2 ( s , τ ) g ( τ , u ( τ ) ) d τ ) d s + 1 Γ ( α − 1 ) ∫ t 1 t α − 2 ( 1 − s ) α − 1 p 1 ( s ) q 1 ( ∫ 0 1 G 2 ( s , τ ) g ( τ , u ( τ ) ) d τ ) d s + ( α − 1 ) t α − 2 Δ 1 ∫ 0 1 ( ∫ 0 1 g 1 ( τ , s ) d H ( τ ) ) p 1 ( s ) × q 1 ( ∫ 0 1 G 2 ( s , τ ) g ( τ , u ( τ ) ) d τ ) d s ≤ M ¯ 3 ( 1 Γ ( α − 1 ) ∫ 0 t [ t α − 2 ( 1 − s ) α − 1 + ( t − s ) α − 2 ] p 1 ( s ) d s + 1 Γ ( α − 1 ) ∫ t 1 t α − 2 ( 1 − s ) α − 1 p 1 ( s ) d s + ( α − 1 ) t α − 2 Δ 1 ∫ 0 1 ( ∫ 0 1 g 1 ( τ , s ) d H ( τ ) ) p 1 ( s ) d s ) .$
(17)

We denote

$h ( t ) = 1 Γ ( α − 1 ) ∫ 0 t [ t α − 2 ( 1 − s ) α − 1 + ( t − s ) α − 2 ] p 1 ( s ) d s h ( t ) = + 1 Γ ( α − 1 ) ∫ t 1 t α − 2 ( 1 − s ) α − 1 p 1 ( s ) d s , μ ( t ) = h ( t ) + ( α − 1 ) t α − 2 Δ 1 ∫ 0 1 ( ∫ 0 1 g 1 ( τ , s ) d H ( τ ) ) p 1 ( s ) d s , t ∈ ( 0 , 1 ) .$

For the integral of the function h, by exchanging the order of integration, we obtain

$∫ 0 1 h ( t ) d t = 1 Γ ( α − 1 ) ∫ 0 1 ( ∫ 0 t [ t α − 2 ( 1 − s ) α − 1 + ( t − s ) α − 2 ] p 1 ( s ) d s ) d t + 1 Γ ( α − 1 ) ∫ 0 1 ( ∫ t 1 t α − 2 ( 1 − s ) α − 1 p 1 ( s ) d s ) d t = 1 Γ ( α − 1 ) ∫ 0 1 ( ∫ s 1 [ t α − 2 ( 1 − s ) α − 1 + ( t − s ) α − 2 ] p 1 ( s ) d t ) d s + 1 Γ ( α − 1 ) ∫ 0 1 ( ∫ 0 s t α − 2 ( 1 − s ) α − 1 p 1 ( s ) d t ) d s = 2 Γ ( α ) ∫ 0 1 ( 1 − s ) α − 1 p 1 ( s ) d s < ∞ .$

For the integral of the function μ, we have

$∫ 0 1 μ ( t ) d t = ∫ 0 1 h ( t ) d t + α − 1 Δ 1 ( ∫ 0 1 ( ∫ 0 1 g 1 ( τ , s ) d H ( τ ) ) p 1 ( s ) d s ) ( ∫ 0 1 t α − 2 d t ) ≤ 2 Γ ( α ) ∫ 0 1 ( 1 − s ) α − 1 p 1 ( s ) + H ( 1 ) − H ( 0 ) Δ 1 ∫ 0 1 g 1 ( θ 1 ( s ) , s ) p 1 ( s ) d s ≤ 1 Γ ( α ) ( 2 + H ( 1 ) − H ( 0 ) Δ 1 ) ∫ 0 1 ( 1 − s ) α − 1 p 1 ( s ) d s < ∞ .$
(18)

We deduce that $μ∈ L 1 (0,1)$. Thus for any given $t 1 , t 2 ∈[0,1]$ with $t 1 ≤ t 2$ and $u∈D$, by (17), we conclude

$|( A ˜ u)( t 1 )−( A ˜ u)( t 2 )|=| ∫ t 1 t 2 ( A ˜ u ) ′ (t)dt|≤ M ¯ 3 ∫ t 1 t 2 μ(t)dt.$
(19)

From (18), (19), and the absolute continuity of the integral function, we find that $A ˜ (D)$ is equicontinuous. By the Ascoli-Arzelà theorem, we deduce that $A ˜ (D)$ is relatively compact. Therefore $A ˜$ is a compact operator. Besides, we can easily show that $A ˜$ is continuous on P. Hence $A ˜ :P→P$ is completely continuous. □

Theorem 4.1 Assume that ($H 1 ˜$)-($H 2 ˜$) hold. If the functions f and g also satisfy the conditions:

($H 3 ˜$) There exist $α 1 , α 2 ∈(0,∞)$ with $α 1 α 2 ≤1$ such that

$(i) q 1 ∞ s = lim sup x → ∞ q 1 ( x ) x α 1 ∈[0,∞);(ii) q 2 ∞ s = lim sup x → ∞ q 2 ( x ) x α 2 =0.$

($H 4 ˜$) There exist $β 1 , β 2 ∈(0,∞)$ with $β 1 β 2 ≤1$ and $c∈(0, 1 2 )$ such that

$(i) f ˜ 0 i = lim inf x → 0 + inf t ∈ [ c , 1 − c ] f ( t , x ) x β 1 ∈(0,∞];(ii) g ˜ 0 i = lim inf x → 0 + inf t ∈ [ c , 1 − c ] g ( t , x ) x β 2 =∞,$

then the problem (S)-(BC) has at least one positive solution $(u(t),v(t))$, $t∈[0,1]$.

Proof Because the proof of this theorem is similar to that of Theorem 3 in , we will sketch some parts of it. For c given in ($H 4 ˜$), we consider the cone $P 0 ={u∈X,u(t)≥0,∀t∈[0,1], min t ∈ [ c , 1 − c ] u(t)≥γ∥u∥}$, where $γ=min{ γ 1 , γ 2 }$. Under assumptions ($H 1 ˜$)-($H 2 ˜$), we obtain $A ˜ (P)⊂ P 0$. By ($H 3 ˜$), we deduce that there exist $C 6 , C 7 , C 8 >0$ and $ε 2 ∈(0, ( 2 α 1 C 6 α ˜ β ˜ α 1 ) − 1 / α 1 )$ such that

$q 1 (x)≤ C 6 x α 1 + C 7 , q 2 (x)≤ ε 2 x α 2 + C 8 ,∀x∈[0,∞).$
(20)

By using (20) and ($H 2 ˜$), for any $u∈ P 0$, we conclude

$( A ˜ u ) ( t ) ≤ ∫ 0 1 G 1 ( t , s ) p 1 ( s ) q 1 ( ∫ 0 1 G 2 ( s , τ ) g ( τ , u ( τ ) ) d τ ) d s ≤ C 6 ∫ 0 1 G 1 ( t , s ) p 1 ( s ) ( ∫ 0 1 G 2 ( s , τ ) g ( τ , u ( τ ) ) d τ ) α 1 d s + C 7 ∫ 0 1 J 1 ( s ) p 1 ( s ) d s ≤ C 6 ∫ 0 1 J 1 ( s ) p 1 ( s ) [ ∫ 0 1 G 2 ( s , τ ) p 2 ( τ ) ( ε 2 ( u ( τ ) ) α 2 + C 8 ) d τ ] α 1 d s + α ˜ C 7 ≤ C 6 ( ∫ 0 1 J 1 ( s ) p 1 ( s ) d s ) ( ∫ 0 1 J 2 ( τ ) p 2 ( τ ) d τ ) α 1 ( ε 2 ∥ u ∥ α 2 + C 8 ) α 1 + α ˜ C 7 ≤ C 6 2 α 1 ε 2 α 1 α ˜ β ˜ α 1 ∥ u ∥ α 1 α 2 + C 6 2 α 1 α ˜ β ˜ α 1 C 8 α 1 + α ˜ C 7 , ∀ t ∈ [ 0 , 1 ] .$

By the definition of $ε 2$, we can choose sufficiently large $R 1 >0$ such that

$∥ A ˜ u∥≤∥u∥,∀u∈∂ B R 1 ∩ P 0 .$
(21)

From ($H 4 ˜$), we deduce that there exist positive constants $C 9 >0$, $x 1 >0$, and $ε 3 ≥ ( 1 / ( C 9 γ 1 γ 2 β 1 γ β 1 β 2 θ ¯ 1 θ ¯ 2 β 1 ) ) 1 / β 1$ such that

$f(t,x)≥ C 9 x β 1 ,g(t,x)≥ ε 3 x β 2 ,∀(t,x)∈[c,1−c]×[0, x 1 ],$
(22)

where $θ ¯ 1 = ∫ c 1 − c J 1 (s)ds$ and $θ ¯ 2 = ∫ c 1 − c J 2 (s)ds$. From the assumption $q 2 (0)=0$ and the continuity of $q 2$, we conclude that there exists sufficiently small $ε 4 ∈(0,min{ x 1 ,1})$ such that $q 2 (x)≤ β ˜ − 1 x 1$ for all $x∈[0, ε 4 ]$, where $β ˜ = ∫ 0 1 J 2 (s) p 2 (s)ds$. Therefore for any $u∈∂ B ε 4 ∩ P 0$ and $s∈[0,1]$, we have

$∫ 0 1 G 2 (s,τ)g ( τ , u ( τ ) ) dτ≤ ∫ 0 1 J 2 (τ) p 2 (τ) q 2 ( u ( τ ) ) dτ≤ x 1 .$
(23)

By (22), (23), Lemma 2.7, and Lemma 2.8, for any $u∈∂ B ε 4 ∩ P 0$ and $t∈[c,1−c]$, we obtain

$( A ˜ u ) ( t ) ≥ C 9 ∫ c 1 − c G 1 ( t , s ) ( ∫ c 1 − c G 2 ( s , τ ) g ( τ , u ( τ ) ) d τ ) β 1 d s ≥ C 9 ∫ c 1 − c G 1 ( t , s ) ( ε 3 ∫ c 1 − c G 2 ( s , τ ) ( u ( τ ) ) β 2 d τ ) β 1 d s ≥ C 9 γ 1 ∫ c 1 − c J 1 ( s ) ( ( ε 3 γ 2 ) β 1 ( ∫ c 1 − c J 2 ( τ ) ( u ( τ ) ) β 2 d τ ) β 1 ) d s ≥ C 9 γ 1 γ 2 β 1 ε 3 β 1 γ β 1 β 2 θ ¯ 1 θ ¯ 2 β 1 ∥ u ∥ β 1 β 2 ≥ ∥ u ∥ β 1 β 2 ≥ ∥ u ∥ .$

Therefore

$∥ A ˜ u∥≥∥u∥,∀u∈∂ B ε 4 ∩ P 0 .$
(24)

By (21), (24), and the Guo-Krasnosel’skii fixed point theorem, we deduce that $A ˜$ has at least one fixed point $u 2 ∈( B ¯ R 1 ∖ B ε 4 )∩ P 0$. Then our problem (S)-(BC) has at least one positive solution $( u 2 , v 2 )∈ P 0 × P 0$ where $v 2 (t)= ∫ 0 1 G 2 (t,s)g(s, u 2 (s))ds$. The proof of Theorem 4.1 is completed. □

Using similar arguments as those used in the proof of Theorem 2 in  (see also  for a particular case of the problem studied in ), we also obtain the following result for our problem (S)-(BC).

Theorem 4.2 Assume that ($H 1 ˜$)-($H 2 ˜$) hold. If the functions f and g also satisfy the conditions:

($H 5 ˜$) There exist $r 1 , r 2 ∈(0,∞)$ with $r 1 r 2 ≥1$ such that

$(i) q 10 s = lim sup x → 0 + q 1 ( x ) x r 1 ∈[0,∞);(ii) q 20 s = lim sup x → 0 + q 2 ( x ) x r 2 =0.$

($H 6 ˜$) There exist $l 1 , l 2 ∈(0,∞)$ with $l 1 l 2 ≥1$ and $c∈(0, 1 2 )$ such that

$( i ) f ˜ ∞ i = lim inf x → ∞ inf t ∈ [ c , 1 − c ] f ( t , x ) x l 1 ∈ ( 0 , ∞ ] ; ( ii ) g ˜ ∞ i = lim inf x → ∞ inf t ∈ [ c , 1 − c ] g ( t , x ) x l 2 = ∞ ,$

then the problem (S)-(BC) has at least one positive solution $(u(t),v(t))$, $t∈[0,1]$.

5 Examples

Let $α=5/2$ ($n=3$), $β=4/3$ ($m=2$),

$H(t)= { 0 , t ∈ [ 0 , 1 / 4 ) , 2 , t ∈ [ 1 / 4 , 3 / 4 ) , 3 , t ∈ [ 3 / 4 , 1 ] ,$

and $K(t)= t 3$ for all $t∈[0,1]$. Then $∫ 0 1 u(s)dH(s)=2u( 1 4 )+u( 3 4 )$ and $∫ 0 1 v(s)dK(s)=3 ∫ 0 1 s 2 v(s)ds$.

We consider the system of fractional differential equations

$( S 0 ) { D 0 + 5 / 2 u ( t ) + f ( t , v ( t ) ) = 0 , t ∈ ( 0 , 1 ) , D 0 + 4 / 3 v ( t ) + g ( t , u ( t ) ) = 0 , t ∈ ( 0 , 1 ) ,$

with the boundary conditions

$( BC 0 ) { u ( 0 ) = u ′ ( 0 ) = 0 , u ( 1 ) = 2 u ( 1 4 ) + u ( 3 4 ) , v ( 0 ) = 0 , v ( 1 ) = 3 ∫ 0 1 s 2 v ( s ) d s .$

Then we obtain $Δ 1 =1− ∫ 0 1 s α − 1 dH(s)=1−2 ( 1 4 ) 3 / 2 − ( 3 4 ) 3 / 2 = 3 ( 2 − 3 ) 8 ≈0.1005>0$, $Δ 2 =1− ∫ 0 1 s β − 1 dK(s)=1−3 ∫ 0 1 s 7 / 3 ds= 1 10 =0.1>0$. We also deduce

$g 1 ( t , s ) = 4 3 π { t 3 / 2 ( 1 − s ) 3 / 2 − ( t − s ) 3 / 2 , 0 ≤ s ≤ t ≤ 1 , t 3 / 2 ( 1 − s ) 3 / 2 , 0 ≤ t ≤ s ≤ 1 , g 2 ( t , s ) = 1 Γ ( 4 / 3 ) { t 1 / 3 ( 1 − s ) 1 / 3 − ( t − s ) 1 / 3 , 0 ≤ s ≤ t ≤ 1 , t 1 / 3 ( 1 − s ) 1 / 3 , 0 ≤ t ≤ s ≤ 1 ,$

$θ 1 (s)= 1 3 − 3 s + s 2$ and $θ 2 (s)=s$ for all $s∈[0,1]$.

For the functions $J 1$ and $J 2$, we obtain

$J 1 ( s ) = g 1 ( θ 1 ( s ) , s ) + 1 Δ 1 ∫ 0 1 g 1 ( τ , s ) d H ( τ ) = g 1 ( 1 3 − 3 s + s 2 , s ) + 1 Δ 1 [ 2 g 1 ( 1 4 , s ) + g 1 ( 3 4 , s ) ] = { 4 3 π { s ( 1 − s ) 3 / 2 ( 3 − 3 s + s 2 ) 1 / 2 + 1 3 ( 2 − 3 ) [ 2 ( ( 1 − s ) 3 / 2 − ( 1 − 4 s ) 3 / 2 ) + ( 3 3 ( 1 − s ) 3 / 2 − ( 3 − 4 s ) 3 / 2 ) ] } , 0 ≤ s < 1 4 , 4 3 π { s ( 1 − s ) 3 / 2 ( 3 − 3 s + s 2 ) 1 / 2 + 1 3 ( 2 − 3 ) [ 2 ( 1 − s ) 3 / 2 + ( 3 3 ( 1 − s ) 3 / 2 − ( 3 − 4 s ) 3 / 2 ) ] } , 1 4 ≤ s < 3 4 , 4 3 π [ s ( 1 − s ) 3 / 2 ( 3 − 3 s + s 2 ) 1 / 2 + ( 2 + 3 3 ) ( 1 − s ) 3 / 2 3 ( 2 − 3 ) ] , 3 4 ≤ s ≤ 1$

and

$J 2 ( s ) = g 2 ( θ 2 ( s ) , s ) + 1 Δ 2 ∫ 0 1 g 2 ( τ , s ) d K ( τ ) = g 2 ( s , s ) + 3 Δ 2 ∫ 0 1 τ 2 g 2 ( τ , s ) d τ = 1 Γ ( 4 / 3 ) ( s 1 / 3 ( 1 − s ) 1 / 3 + 9 ( 1 − s ) 1 / 3 − 9 ( 1 − s ) 10 / 3 − 180 7 s ( 1 − s ) 7 / 3 − 45 2 s 2 ( 1 − s ) 4 / 3 ) , s ∈ [ 0 , 1 ] .$

Example 1 We consider the functions

$f(t,u)=a ( u α 0 + u β 0 ) ,g(t,u)=b ( u γ 0 + u δ 0 ) ,t∈[0,1],u≥0,$

where $α 0 >1$, $0< β 0 <1$, $γ 0 >2$, $0< δ 0 <1$, $a,b>0$. We have $K 1 = max s ∈ [ 0 , 1 ] J 1 (s)≈1.8076209$, $K 2 = max s ∈ [ 0 , 1 ] J 2 (s)≈4.16640785$. Then $m 0 =max{ K 1 , K 2 }= K 2$. The functions $f(t,u)$ and $g(t,u)$ are nondecreasing with respect to u, for any $t∈[0,1]$, and for $p=1/2$ and $c∈(0,1/2)$ the assumptions (H3) and (H6) are satisfied; indeed we obtain

$f ∞ i = lim u → ∞ a ( u α 0 + u β 0 ) u 1 / 2 = ∞ , g ∞ i = lim u → ∞ b ( u γ 0 + u δ 0 ) u 2 = ∞ , f 0 i = lim u → 0 + a ( u α 0 + u β 0 ) u = ∞ , g 0 i = lim u → 0 + b ( u γ 0 + u δ 0 ) u = ∞ .$

We take $N=1$ and then $∫ 0 1 g(s,1)ds=2b$ and $f(t,2b m 0 )=a[ ( 2 b m 0 ) α 0 + ( 2 b m 0 ) β 0 ]$. If $a[ ( 2 b m 0 ) α 0 + ( 2 b m 0 ) β 0 ]< 1 m 0 ⇔a[ m 0 α 0 + 1 ( 2 b ) α 0 + m 0 β 0 + 1 ( 2 b ) β 0 ]<1$, then the assumption (H7) is satisfied. For example, if $α 0 =3/2$, $β 0 =1/3$, $b=1/2$, and $a< 1 m 0 5 / 2 + m 0 4 / 3$ (e.g. $a≤0.0237$), then the above inequality is satisfied. By Theorem 3.3, we deduce that the problem ($S 0$)-($BC 0$) has at least two positive solutions.

Example 2 We consider the functions

$f(t,x)= x a t ζ 1 ( 1 − t ) ρ 1 ,g(t,x)= x b t ζ 2 ( 1 − t ) ρ 2 ,$

with $a,b>1$ and $ζ 1 , ρ 1 , ζ 2 , ρ 2 ∈(0,1)$. Here $f(t,x)= p 1 (t) q 1 (x)$ and $g(t,x)= p 2 (t) q 2 (x)$, where

$p 1 (t)= 1 t ζ 1 ( 1 − t ) ρ 1 , p 2 (t)= 1 t ζ 2 ( 1 − t ) ρ 2 , q 1 (x)= x a , q 2 (x)= x b .$

We have $0< ∫ 0 1 p 1 (s)ds<∞$, $0< ∫ 0 1 p 2 (s)ds<∞$.

In ($H 5 ˜$), for $r 1 , $r 2 and $r 1 r 2 ≥1$, we obtain

$lim sup x → 0 + q 1 ( x ) x r 1 =0, lim sup x → 0 + q 2 ( x ) x r 2 =0.$

In ($H 6 ˜$), for $l 1 , $l 2 , $l 1 l 2 ≥1$, and $c∈(0, 1 2 )$, we have

$lim inf x → ∞ inf t ∈ [ c , 1 − c ] f ( t , x ) x l 1 =∞, lim inf x → ∞ inf t ∈ [ c , 1 − c ] g ( t , x ) x l 2 =∞.$

For example, if $a=3/2$, $b=2$, $r 1 =1$, $r 2 =3/2$, $l 1 =1$, $l 2 =3/2$, the above conditions are satisfied. Then, by Theorem 4.2, we deduce that the problem ($S 0$)-($BC 0$) has at least one positive solution.

References

1. 1.

Guo D, Lakshmikantham V: Nonlinear Problems in Abstract Cones. Academic Press, New York; 1988.

2. 2.

Amann H: Fixed point equations and nonlinear eigenvalue problems in ordered Banach spaces. SIAM Rev. 1976, 18: 620-709. 10.1137/1018114

3. 3.

Zhou Y, Xu Y: Positive solutions of three-point boundary value problems for systems of nonlinear second order ordinary differential equations. J. Math. Anal. Appl. 2006, 320: 578-590. 10.1016/j.jmaa.2005.07.014

4. 4.

Henderson J, Luca R: Existence and multiplicity for positive solutions of a system of higher-order multi-point boundary value problems. Nonlinear Differ. Equ. Appl. 2013, 20(3):1035-1054. 10.1007/s00030-012-0195-9

5. 5.

Henderson J, Luca R: Positive solutions for singular systems of higher-order multi-point boundary value problems. Math. Model. Anal. 2013, 18(3):309-324. 10.3846/13926292.2013.804009

6. 6.

Henderson J, Luca R: Positive solutions for a system of nonlocal fractional boundary value problems. Fract. Calc. Appl. Anal. 2013, 16(4):985-1008.

7. 7.

Baleanu D, Diethelm K, Scalas E, Trujillo JJ Series on Complexity, Nonlinearity and Chaos. In Fractional Calculus Models and Numerical Methods. World Scientific, Boston; 2012.

8. 8.

Das S: Functional Fractional Calculus for System Identification and Controls. Springer, New York; 2008.

9. 9.

Graef JR, Kong L, Kong Q, Wang M: Uniqueness of positive solutions of fractional boundary value problems with non-homogeneous integral boundary conditions. Fract. Calc. Appl. Anal. 2012, 15(3):509-528.

10. 10.

Kilbas AA, Srivastava HM, Trujillo JJ North-Holland Mathematics Studies 204. In Theory and Applications of Fractional Differential Equations. Elsevier, Amsterdam; 2006.

11. 11.

Podlubny I: Fractional Differential Equations. Academic Press, San Diego; 1999.

12. 12.

Sabatier J, Agrawal OP, Machado JAT (Eds): Advances in Fractional Calculus: Theoretical Developments and Applications in Physics and Engineering. Springer, Dordrecht; 2007.

13. 13.

Samko SG, Kilbas AA, Marichev OI: Fractional Integrals and Derivatives: Theory and Applications. Gordon & Breach, Yverdon; 1993.

14. 14.

Liu B, Liu L, Wu Y: Positive solutions for singular systems of three-point boundary value problems. Comput. Math. Appl. 2007, 53: 1429-1438. 10.1016/j.camwa.2006.07.014

Acknowledgements

The work of R Luca was supported by the CNCS grant PN-II-ID-PCE-2011-3-0557, Romania.

Author information

Correspondence to Rodica Luca.

Competing interests

The authors declare that they have no competing interests.

Authors’ contributions

Both authors contributed equally to this paper. Both authors read and approved the final manuscript.

Rights and permissions

Open Access This article is distributed under the terms of the Creative Commons Attribution 2.0 International License (https://creativecommons.org/licenses/by/2.0), which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

Reprints and Permissions 