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This article is part of the series A Tribute to Professor Ivan Kiguradze.

Open Access Research

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

Johnny Henderson1 and Rodica Luca2*

Author Affiliations

1 Department of Mathematics, Baylor University, Waco, TX, 76798-7328, USA

2 Department of Mathematics, Gh. Asachi Technical University, Iasi, 700506, Romania

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Boundary Value Problems 2014, 2014:60  doi:10.1186/1687-2770-2014-60


The electronic version of this article is the complete one and can be found online at: http://www.boundaryvalueproblems.com/content/2014/1/60


Received:8 January 2014
Accepted:11 March 2014
Published:20 March 2014

© 2014 Henderson and Luca; licensee Springer.

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

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.

Keywords:
Riemann-Liouville fractional differential equation; integral boundary conditions; positive solutions

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 [1]) and some theorems from the fixed point index theory (from [2] and [3]). 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 [4] (the nonsingular case) and [5] (the singular case). In [6], 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 [7-13]).

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 ) d s , 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 d t , 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 d s , 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 ([10])

Let α > 0 and n = α + 1 for α N and n = α for α N ; that is, nis the smallest integer greater than or equal toα. Then the solutions of the fractional differential equation D 0 + α u ( t ) = 0 , 0 < t < 1 , are

u ( t ) = c 1 t α 1 + c 2 t α 2 + + c n t α n , 0 < t < 1 ,

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

Lemma 2.2 ([6,10])

Let α > 0 , nbe the smallest integer greater than or equal toα ( 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 < t < 1 , are

u ( t ) = 1 Γ ( α ) 0 t ( t s ) α 1 y ( s ) d s + c 1 t α 1 + + c n t α n , 0 < t < 1 ,

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 < t < 1 , n 1 < α n , (1)

with the integral boundary conditions

u ( 0 ) = u ( 0 ) = = u ( n 2 ) ( 0 ) = 0 , u ( 1 ) = 0 1 u ( s ) d H ( s ) , (2)

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

Lemma 2.3If H : [ 0 , 1 ] R is a function of bounded variation, Δ 1 = 1 0 1 s α 1 d H ( 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 ) d s + 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 ) d s .

Now, by condition u ( 1 ) = 0 1 u ( s ) d H ( s ) , we deduce

c 1 1 Γ ( α ) 0 1 ( 1 s ) α 1 y ( s ) d s = 0 1 [ c 1 s α 1 1 Γ ( α ) 0 s ( s τ ) α 1 y ( τ ) d τ ] d H ( 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 2.4Under 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 ) d H ( 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 ) d s for all t [ 0 , 1 ] . □

Lemma 2.5 ([6])

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

(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 ] .

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

(c) For any c ( 0 , 1 / 2 ) ,

min t [ c , 1 c ] g 1 ( t , s ) γ 1 g 1 ( θ 1 ( s ) , s ) for all  s [ 0 , 1 ] ,

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 2.6If 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 solutionuof 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 2.7Assume 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:

(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 ) d H ( τ ) , s [ 0 , 1 ] .

(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 2.8Assume 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 ) d s γ 1 0 1 J 1 ( s ) y ( s ) d s γ 1 0 1 G 1 ( t , s ) y ( s ) d s = 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 < t < 1 , m 1 < β m , (6)

with the integral boundary conditions

v ( 0 ) = v ( 0 ) = = v ( m 2 ) ( 0 ) = 0 , v ( 1 ) = 0 1 v ( s ) d K ( 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 d H ( s ) > 0 , Δ 2 = 1 0 1 s β 1 d K ( 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

( 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 , t [ 0 , 1 ] ,

and B : P X , C : P X by

( B u ) ( t ) = 0 1 G 1 ( t , s ) u ( s ) d s , ( C u ) ( t ) = 0 1 G 2 ( t , s ) u ( s ) d s , t [ 0 , 1 ] .

Under the assumptions (H1) and (H2), using also Lemma 2.6, it is easy to see that , ℬ, 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 3.1Assume that (H1)-(H2) hold. If the functionsfandgalso 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 [4], 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

( A u ) ( 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 ) d s C 3 , t [ 0 , 1 ] , (9)

where C 3 = C 2 c 1 c J 1 ( s ) d s .

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 = B y P 0 and v = C y 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 ) d s = ( B y 0 ) ( t ) 0 , t [ 0 , 1 ] , with y 0 ( t ) = 1 for all t [ 0 , 1 ] . We define the set

M = { u P ;  there exists  λ 0  such that  u = A u + λ u 0 } .

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 ) = ( A u ) ( t ) + λ u 0 ( t ) , t [ 0 , 1 ] . From the definition of u 0 , we have

u ( t ) = ( A u ) ( 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 ( F u ) ( t ) = f ( t , 0 1 G 2 ( t , s ) g ( s , u ( s ) ) d s ) . 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 ) d τ > 0 , m 2 = c 1 c ( J 2 ( τ ) ) p d τ > 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 A u + λ u 0 , u B L P , λ 0 .

From [2], 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 ) d s > 0 , M 2 = 0 1 J 2 ( s ) d s > 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 ) ) d s ε 1 0 1 J 2 ( s ) ( u ( s ) ) 1 / q d s 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 A u u / 2 for all u B δ 1 P . From [2], 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 < L .

Let v 1 ( t ) = 0 1 G 2 ( t , s ) g ( s , u 1 ( s ) ) d s . 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 ) ) d s = 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 [4], we also obtain the following results for our problem (S)-(BC).

Theorem 3.2Assume that (H1)-(H2) hold. If the functionsfandgalso 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 3.3Assume that (H1)-(H3), and (H6) hold. If the functionsfandgalso satisfy the condition:

(H7) For each t [ 0 , 1 ] , f ( t , u ) and g ( t , u ) are nondecreasing with respect tou, 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 ) d t < , 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 ) d s .

Lemma 4.1Assume 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 ) d s and β ˜ = 0 1 J 2 ( s ) p 2 ( s ) d s . 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 ) d t | M ¯ 3 t 1 t 2 μ ( t ) d t . (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 4.1Assume that ( H 1 ˜ )-( H 2 ˜ ) hold. If the functionsfandgalso 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 [5], 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 ) d s and θ ¯ 2 = c 1 c J 2 ( s ) d s . 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 ) d s . 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 ) ) d τ 0 1 J 2 ( τ ) p 2 ( τ ) q 2 ( u ( τ ) ) d τ 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 ) ) d s . The proof of Theorem 4.1 is completed. □

Using similar arguments as those used in the proof of Theorem 2 in [5] (see also [14] for a particular case of the problem studied in [5]), we also obtain the following result for our problem (S)-(BC).

Theorem 4.2Assume that ( H 1 ˜ )-( H 2 ˜ ) hold. If the functionsfandgalso 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 ) d H ( s ) = 2 u ( 1 4 ) + u ( 3 4 ) and 0 1 v ( s ) d K ( s ) = 3 0 1 s 2 v ( s ) d s .

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 d H ( 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 d K ( s ) = 1 3 0 1 s 7 / 3 d s = 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 >