Existence and uniqueness of positive solution to singular fractional differential equations

  • Yongqing Wang1Email author,

    Affiliated with

    • Lishan Liu1, 2 and

      Affiliated with

      • Yonghong Wu2

        Affiliated with

        Boundary Value Problems20122012:81

        DOI: 10.1186/1687-2770-2012-81

        Received: 6 April 2012

        Accepted: 10 July 2012

        Published: 28 July 2012

        Abstract

        In this paper, we discuss the existence and uniqueness of a positive solution to the following singular fractional differential equation with nonlocal boundary value conditions:

        { D 0 + α u ( t ) + f ( t , u ( t ) ) = 0 , 0 < t < 1 , u ( 0 ) = 0 , D 0 + β u ( 1 ) = i = 1 m 2 η i D 0 + β u ( ξ i ) , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-81/MediaObjects/13661_2012_Article_210_Equa_HTML.gif

        where 1 < α 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-81/MediaObjects/13661_2012_Article_210_IEq1_HTML.gif, 0 < β < α 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-81/MediaObjects/13661_2012_Article_210_IEq2_HTML.gif, 0 < ξ 1 < < ξ m 2 < 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-81/MediaObjects/13661_2012_Article_210_IEq3_HTML.gif with i = 1 m 2 η i ξ i α β 1 < 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-81/MediaObjects/13661_2012_Article_210_IEq4_HTML.gif, D 0 + α http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-81/MediaObjects/13661_2012_Article_210_IEq5_HTML.gif is the standard Riemann-Liouville derivative, f may be singular at t = 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-81/MediaObjects/13661_2012_Article_210_IEq6_HTML.gif, t = 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-81/MediaObjects/13661_2012_Article_210_IEq7_HTML.gif, and u = 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-81/MediaObjects/13661_2012_Article_210_IEq8_HTML.gif.

        MSC:34B10, 34B15.

        Keywords

        fractional differential equation positive solution iterative scheme singular boundary value problem

        1 Introduction

        In this paper, we consider the following fractional differential equation:
        { D 0 + α u ( t ) + f ( t , u ( t ) ) = 0 , 0 < t < 1 , u ( 0 ) = 0 , D 0 + β u ( 1 ) = i = 1 m 2 η i D 0 + β u ( ξ i ) , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-81/MediaObjects/13661_2012_Article_210_Equ1_HTML.gif
        (1.1)

        where 1 < α 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-81/MediaObjects/13661_2012_Article_210_IEq9_HTML.gif, 0 < β < α 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-81/MediaObjects/13661_2012_Article_210_IEq2_HTML.gif, 0 < ξ 1 < < ξ m 2 < 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-81/MediaObjects/13661_2012_Article_210_IEq3_HTML.gif with i = 1 m 2 η i ξ i α β 1 < 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-81/MediaObjects/13661_2012_Article_210_IEq4_HTML.gif, D 0 + α http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-81/MediaObjects/13661_2012_Article_210_IEq5_HTML.gif is the standard Riemann-Liouville derivative, f C ( ( 0 , 1 ) × ( 0 , + ) [ 0 , + ) ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-81/MediaObjects/13661_2012_Article_210_IEq10_HTML.gif may be singular at t = 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-81/MediaObjects/13661_2012_Article_210_IEq6_HTML.gif, t = 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-81/MediaObjects/13661_2012_Article_210_IEq7_HTML.gif, and u = 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-81/MediaObjects/13661_2012_Article_210_IEq8_HTML.gif. In this paper, by a positive solution to (1.1), we mean a function u C [ 0 , 1 ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-81/MediaObjects/13661_2012_Article_210_IEq11_HTML.gif which satisfies D 0 + α u L ( 0 , 1 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-81/MediaObjects/13661_2012_Article_210_IEq12_HTML.gif, positive on ( 0 , 1 ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-81/MediaObjects/13661_2012_Article_210_IEq13_HTML.gif and satisfies (1.1).

        Recently, many results were obtained dealing with the existence of solutions for nonlinear fractional differential equations by using the techniques of nonlinear analysis; see [123] and references therein. The multi-point boundary value problems (BVP for short) have provoked a great deal of attention, for example [1319]. In [10], the authors discussed some positive properties of the Green function for Direchlet-type BVP of nonlinear fractional differential equation
        { D 0 + α u ( t ) + f ( t , u ( t ) ) = 0 , 0 < t < 1 , u ( 0 ) = 0 , u ( 1 ) = 0 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-81/MediaObjects/13661_2012_Article_210_Equ2_HTML.gif
        (1.2)

        where 1 < α < 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-81/MediaObjects/13661_2012_Article_210_IEq14_HTML.gif, D 0 + α http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-81/MediaObjects/13661_2012_Article_210_IEq5_HTML.gif is the standard Riemann-Liouville derivative, f C ( [ 0 , 1 ] × [ 0 , + ) [ 0 , + ) ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-81/MediaObjects/13661_2012_Article_210_IEq15_HTML.gif. By using the Krasnosel’skii fixed point theorem, the existence of positive solutions were obtained under suitable conditions on f.

        In [14], the authors investigated the existence and multiplicity of positive solutions by using some fixed point theorems for the fractional differential equation
        { D 0 + α u ( t ) + f ( t , u ( t ) ) = 0 , 0 < t < 1 , u ( 0 ) = 0 , D 0 + β u ( 1 ) = a D 0 + β u ( ξ ) , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-81/MediaObjects/13661_2012_Article_210_Equ3_HTML.gif
        (1.3)

        where 1 < α 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-81/MediaObjects/13661_2012_Article_210_IEq16_HTML.gif, 0 β 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-81/MediaObjects/13661_2012_Article_210_IEq17_HTML.gif, 0 < ξ < 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-81/MediaObjects/13661_2012_Article_210_IEq18_HTML.gif, 0 a 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-81/MediaObjects/13661_2012_Article_210_IEq19_HTML.gif with a ξ α β 2 < 1 β http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-81/MediaObjects/13661_2012_Article_210_IEq20_HTML.gif, 0 α β 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-81/MediaObjects/13661_2012_Article_210_IEq21_HTML.gif, f : [ 0 , 1 ] × [ 0 , + ) [ 0 , + ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-81/MediaObjects/13661_2012_Article_210_IEq22_HTML.gif satisfied Carathéodory type conditions.

        In [20, 21], the authors considered the fractional differential equation given by
        { D 0 + α u ( t ) + f ( t , u , u , , u ( n 2 ) ) = 0 , 0 < t < 1 , n 1 < α n , n 2 , u ( 0 ) = u ( 0 ) = = u ( n 2 ) ( 0 ) = 0 , u ( n 2 ) ( 1 ) = 0 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-81/MediaObjects/13661_2012_Article_210_Equ4_HTML.gif
        (1.4)
        In order to obtain the existence of positive solutions of (1.4), they considered the following fractional differential equation:
        { D 0 + α n + 2 v ( t ) + f ( t , I 0 + n 2 v ( t ) , I 0 + n 3 v ( t ) , , I 0 + 1 v ( t ) , v ( t ) ) = 0 , 0 < t < 1 , v ( 0 ) = v ( 1 ) = 0 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-81/MediaObjects/13661_2012_Article_210_Equ5_HTML.gif
        (1.5)

        In [20], f = q ( t ) ( g + h ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-81/MediaObjects/13661_2012_Article_210_IEq23_HTML.gif, and g, h have different monotone properties. By using the fixed point theorem for the mixed monotone operator, Zhang obtained (1.4) and had a unique positive solution u ( t ) = I 0 + n 2 v ( t ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-81/MediaObjects/13661_2012_Article_210_IEq24_HTML.gif with v Q = : { x ( t ) : 1 M t α n + 1 x ( t ) M t α n + 1 } http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-81/MediaObjects/13661_2012_Article_210_IEq25_HTML.gif. But the results are not true since v ( t ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-81/MediaObjects/13661_2012_Article_210_IEq26_HTML.gif is a positive solution of (1.5), and v ( 1 ) = 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-81/MediaObjects/13661_2012_Article_210_IEq27_HTML.gif. What causes it lies in the unsuitable using of properties of the Green function.

        In [21], f C ( [ 0 , 1 ] × [ 0 , + ) × R n 2 [ 0 , + ) ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-81/MediaObjects/13661_2012_Article_210_IEq28_HTML.gif, f ( t , y 1 , y 2 , , y n 1 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-81/MediaObjects/13661_2012_Article_210_IEq29_HTML.gif is increasing for y i 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-81/MediaObjects/13661_2012_Article_210_IEq30_HTML.gif, i = 1 , 2 , , n 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-81/MediaObjects/13661_2012_Article_210_IEq31_HTML.gif. By using the positive properties of the Green function obtained in [ 10 ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-81/MediaObjects/13661_2012_Article_210_IEq32_HTML.gif and fixed point theory for the u 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-81/MediaObjects/13661_2012_Article_210_IEq33_HTML.gif concave operator, the authors obtained the uniqueness of a positive solution for the BVP (1.4).

        Motivated by the works mentioned above, in this paper we aim to establish the existence and uniqueness of a positive solution to the BVP (1.1). Our work presented in this paper has the following features. Firstly, the BVP (1.1) possesses singularity, that is, f may be singular at t = 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-81/MediaObjects/13661_2012_Article_210_IEq6_HTML.gif, t = 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-81/MediaObjects/13661_2012_Article_210_IEq7_HTML.gif, and u = 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-81/MediaObjects/13661_2012_Article_210_IEq8_HTML.gif. Secondly, we impose weaker positivity conditions on the nonlocal boundary term, that is, some of the coefficients η i http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-81/MediaObjects/13661_2012_Article_210_IEq34_HTML.gif can be negative. Thirdly, the unique positive solution can be approximated by an iterative scheme.

        The rest of the paper is organized as follows. In Section 2, we present some preliminaries and lemmas that will be used to prove our main results. We also develop some new positive properties of the Green function. In Section 3, we discuss the existence and uniqueness of a positive solution of the BVP (1.1), we also give an example to demonstrate the application of our theoretical results.

        2 Preliminaries

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

        Definition 2.1 The fractional integral of order α > 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-81/MediaObjects/13661_2012_Article_210_IEq35_HTML.gif of a function u : ( 0 , + ) R http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-81/MediaObjects/13661_2012_Article_210_IEq36_HTML.gif is given by
        I 0 + α u ( t ) = 1 Γ ( α ) 0 t ( t s ) α 1 u ( s ) d s http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-81/MediaObjects/13661_2012_Article_210_Equb_HTML.gif

        provided the right-hand side is defined pointwise on ( 0 , + ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-81/MediaObjects/13661_2012_Article_210_IEq37_HTML.gif.

        Definition 2.2 The fractional derivative of order α > 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-81/MediaObjects/13661_2012_Article_210_IEq35_HTML.gif of a continuous function u : ( 0 , + ) R http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-81/MediaObjects/13661_2012_Article_210_IEq36_HTML.gif is given by
        D 0 + α u ( t ) = 1 Γ ( n α ) ( d d t ) n 0 t ( t s ) n α 1 u ( s ) d s , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-81/MediaObjects/13661_2012_Article_210_Equc_HTML.gif

        where n = [ α ] + 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-81/MediaObjects/13661_2012_Article_210_IEq38_HTML.gif, [ α ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-81/MediaObjects/13661_2012_Article_210_IEq39_HTML.gif denotes the integral part of the number α, provided the right-hand side is pointwisely defined on ( 0 , + ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-81/MediaObjects/13661_2012_Article_210_IEq37_HTML.gif.

        Definition 2.3 By u L ( 0 , 1 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-81/MediaObjects/13661_2012_Article_210_IEq40_HTML.gif, we mean 0 1 | u ( t ) | d t < http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-81/MediaObjects/13661_2012_Article_210_IEq41_HTML.gif.

        Lemma 2.1 ([3])

        Let α > 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-81/MediaObjects/13661_2012_Article_210_IEq35_HTML.gif. Then the following equality holds for u L ( 0 , 1 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-81/MediaObjects/13661_2012_Article_210_IEq40_HTML.gif, D 0 + α u L ( 0 , 1 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-81/MediaObjects/13661_2012_Article_210_IEq12_HTML.gif,
        I 0 + α D 0 + α u ( t ) = u ( t ) + c 1 t α 1 + c 2 t α 2 + + c n t α n , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-81/MediaObjects/13661_2012_Article_210_Equd_HTML.gif

        where c i R http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-81/MediaObjects/13661_2012_Article_210_IEq42_HTML.gif, i = 1 , 2 , , n http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-81/MediaObjects/13661_2012_Article_210_IEq43_HTML.gif, n 1 < α n http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-81/MediaObjects/13661_2012_Article_210_IEq44_HTML.gif.

        Set
        http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-81/MediaObjects/13661_2012_Article_210_Equ6_HTML.gif
        (2.1)
        http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-81/MediaObjects/13661_2012_Article_210_Equ7_HTML.gif
        (2.2)
        http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-81/MediaObjects/13661_2012_Article_210_Equ8_HTML.gif
        (2.3)
        where
        q ( s ) = p ( s ) p ( 0 ) Γ ( α ) p ( 0 ) ( 1 s ) α β 1 , p ( 0 ) = 1 i = 1 m 2 η i ξ i α β 1 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-81/MediaObjects/13661_2012_Article_210_Equ9_HTML.gif
        (2.4)

        For the convenience in presentation, we here list the assumption to be used throughout the paper.

        ( H 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-81/MediaObjects/13661_2012_Article_210_IEq45_HTML.gif) p ( 0 ) > 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-81/MediaObjects/13661_2012_Article_210_IEq46_HTML.gif, q ( s ) 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-81/MediaObjects/13661_2012_Article_210_IEq47_HTML.gif on [ 0 , 1 ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-81/MediaObjects/13661_2012_Article_210_IEq48_HTML.gif.

        Remark 2.1 If η i = 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-81/MediaObjects/13661_2012_Article_210_IEq49_HTML.gif ( i = 1 , , m 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-81/MediaObjects/13661_2012_Article_210_IEq50_HTML.gif), we have p ( 0 ) = 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-81/MediaObjects/13661_2012_Article_210_IEq51_HTML.gif and q ( s ) 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-81/MediaObjects/13661_2012_Article_210_IEq52_HTML.gif. If η i 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-81/MediaObjects/13661_2012_Article_210_IEq53_HTML.gif ( i = 1 , , m 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-81/MediaObjects/13661_2012_Article_210_IEq54_HTML.gif) and i = 1 m 2 η i ξ i α β 1 < 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-81/MediaObjects/13661_2012_Article_210_IEq4_HTML.gif, we have p ( 0 ) > 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-81/MediaObjects/13661_2012_Article_210_IEq55_HTML.gif and q ( s ) 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-81/MediaObjects/13661_2012_Article_210_IEq47_HTML.gif on [ 0 , 1 ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-81/MediaObjects/13661_2012_Article_210_IEq48_HTML.gif.

        Lemma 2.2 ([14])

        Assume that g L ( 0 , 1 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-81/MediaObjects/13661_2012_Article_210_IEq56_HTML.gif and α > 1 β 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-81/MediaObjects/13661_2012_Article_210_IEq57_HTML.gif. Then
        D 0 + β 0 t ( t s ) α 1 g ( s ) d s = Γ ( α ) Γ ( α β ) 0 t ( t s ) α β 1 g ( s ) d s . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-81/MediaObjects/13661_2012_Article_210_Eque_HTML.gif
        Lemma 2.3 Assume ( H 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-81/MediaObjects/13661_2012_Article_210_IEq45_HTML.gif) holds, and y L ( 0 , 1 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-81/MediaObjects/13661_2012_Article_210_IEq58_HTML.gif. Then the unique solution of the problem
        { D 0 + α u ( t ) + y ( t ) = 0 , 0 < t < 1 , u ( 0 ) = 0 , D 0 + β u ( 1 ) = i = 1 m 2 η i D 0 + β u ( ξ i ) , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-81/MediaObjects/13661_2012_Article_210_Equ10_HTML.gif
        (2.5)
        is
        u ( t ) = 0 1 G ( t , s ) y ( s ) d s , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-81/MediaObjects/13661_2012_Article_210_Equf_HTML.gif

        where G ( t , s ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-81/MediaObjects/13661_2012_Article_210_IEq59_HTML.gif is called the Green function of BVP (2.5).

        Proof From Lemma 2.1, we have the solution of (2.5) given by
        u ( t ) = I 0 + α y ( t ) + c 1 t α 1 + c 2 t α 2 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-81/MediaObjects/13661_2012_Article_210_Equg_HTML.gif
        Consequently,
        u ( t ) = 1 Γ ( α ) 0 t ( t s ) α 1 y ( s ) d s + c 1 t α 1 + c 2 t α 2 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-81/MediaObjects/13661_2012_Article_210_Equh_HTML.gif

        From u ( 0 ) = 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-81/MediaObjects/13661_2012_Article_210_IEq60_HTML.gif, we have c 2 = 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-81/MediaObjects/13661_2012_Article_210_IEq61_HTML.gif.

        By Lemma 2.2, we have
        D 0 + β u ( t ) = 1 Γ ( α β ) 0 t ( t s ) α β 1 y ( s ) d s + c 1 Γ ( α ) Γ ( α β ) t α β 1 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-81/MediaObjects/13661_2012_Article_210_Equi_HTML.gif
        Therefore,
        D 0 + β u ( 1 ) = 1 Γ ( α β ) 0 1 ( 1 s ) α β 1 y ( s ) d s + c 1 Γ ( α ) Γ ( α β ) , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-81/MediaObjects/13661_2012_Article_210_Equj_HTML.gif
        and
        D 0 + β u ( ξ i ) = 1 Γ ( α β ) 0 ξ i ( ξ i s ) α β 1 y ( s ) d s + c 1 Γ ( α ) Γ ( α β ) ξ i α β 1 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-81/MediaObjects/13661_2012_Article_210_Equk_HTML.gif
        By D 0 + β u ( 1 ) = i = 1 m 2 η i D 0 + β u ( ξ i ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-81/MediaObjects/13661_2012_Article_210_IEq62_HTML.gif, we have
        c 1 = 0 1 ( 1 s ) α 1 y ( s ) d s i = 1 m 2 η i 0 ξ i ( ξ i s ) α β 1 y ( s ) d s Γ ( α ) p ( 0 ) = 0 1 ( 1 s ) α β 1 p ( s ) y ( s ) d s Γ ( α ) p ( 0 ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-81/MediaObjects/13661_2012_Article_210_Equl_HTML.gif
        Therefore, the solution of (2.5) is
        u ( t ) = c 1 t α 1 1 Γ ( α ) 0 t ( t s ) α 1 y ( s ) d s = 0 1 G ( t , s ) y ( s ) d s . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-81/MediaObjects/13661_2012_Article_210_Equm_HTML.gif

         □

        Lemma 2.4 The function G 0 ( t , s ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-81/MediaObjects/13661_2012_Article_210_IEq63_HTML.gif has the following properties:
        1. (1)

          G 0 ( t , s ) > 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-81/MediaObjects/13661_2012_Article_210_IEq64_HTML.gif, for t , s ( 0 , 1 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-81/MediaObjects/13661_2012_Article_210_IEq65_HTML.gif;

           
        2. (2)

          Γ ( α ) G 0 ( t , s ) t α 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-81/MediaObjects/13661_2012_Article_210_IEq66_HTML.gif, for t , s [ 0 , 1 ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-81/MediaObjects/13661_2012_Article_210_IEq67_HTML.gif;

           
        3. (3)
          β t α 1 h ( s ) Γ ( α ) G 0 ( t , s ) h ( s ) t α 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-81/MediaObjects/13661_2012_Article_210_IEq68_HTML.gif, for t , s ( 0 , 1 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-81/MediaObjects/13661_2012_Article_210_IEq65_HTML.gif, where
          h ( s ) = s ( 1 s ) α β 1 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-81/MediaObjects/13661_2012_Article_210_Equ11_HTML.gif
          (2.6)
           
        Proof It is obvious that (1), (2) hold. In the following, we will prove (3).
        1. (i)
          When 0 < s t < 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-81/MediaObjects/13661_2012_Article_210_IEq69_HTML.gif, noticing 0 < β < α 1 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-81/MediaObjects/13661_2012_Article_210_IEq70_HTML.gif, we have
          http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-81/MediaObjects/13661_2012_Article_210_Equ12_HTML.gif
          (2.7)
           
        Therefore,
        t α 2 s ( 1 s ) α β 1 ( t α 1 ( 1 s ) α β 1 ( t s ) α 1 ) t α 2 s t α 1 + ( t s ) α 1 = t α 2 ( t s ) + ( t s ) α 1 0 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-81/MediaObjects/13661_2012_Article_210_Equn_HTML.gif
        which implies
        Γ ( α ) G 0 ( t , s ) h ( s ) t α 2 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-81/MediaObjects/13661_2012_Article_210_Equ13_HTML.gif
        (2.8)
        On the other hand, we have
        s { β s + ( 1 s ) β } 0 , s [ 0 , 1 ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-81/MediaObjects/13661_2012_Article_210_Equo_HTML.gif
        Therefore, β s + ( 1 s ) β 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-81/MediaObjects/13661_2012_Article_210_IEq71_HTML.gif, which implies
        [ 1 ( 1 s ) β ] β s . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-81/MediaObjects/13661_2012_Article_210_Equp_HTML.gif
        Then
        Γ ( α ) G 0 ( t , s ) = t α 1 ( 1 s ) α β 1 ( t s ) α 1 t α 1 ( 1 s ) α β 1 ( t s ) β ( t t s ) α β 1 = [ 1 ( 1 s t ) β ] t α 1 ( 1 s ) α β 1 [ 1 ( 1 s ) β ] t α 1 ( 1 s ) α β 1 β t α 1 h ( s ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-81/MediaObjects/13661_2012_Article_210_Equ14_HTML.gif
        (2.9)
        1. (ii)
          When 0 < t s < 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-81/MediaObjects/13661_2012_Article_210_IEq72_HTML.gif, we have
          Γ ( α ) G 0 ( t , s ) = t α 1 ( 1 s ) α β 1 = t α 2 t ( 1 s ) α β 1 t α 2 s ( 1 s ) α β 1 = h ( s ) t α 2 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-81/MediaObjects/13661_2012_Article_210_Equ15_HTML.gif
          (2.10)
           
        On the other hand, clearly we have
        Γ ( α ) G 0 ( t , s ) = t α 1 ( 1 s ) α β 1 β t α 1 h ( s ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-81/MediaObjects/13661_2012_Article_210_Equ16_HTML.gif
        (2.11)

        (2.8)-(2.11) implies (3) holds. □

        By Lemma 2.4 we have the following results.

        Lemma 2.5 Assume ( H 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-81/MediaObjects/13661_2012_Article_210_IEq45_HTML.gif) holds, then the Green function defined by (2.3) satisfies:
        1. (1)

          G ( t , s ) > 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-81/MediaObjects/13661_2012_Article_210_IEq73_HTML.gif, t , s ( 0 , 1 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-81/MediaObjects/13661_2012_Article_210_IEq74_HTML.gif;

           
        2. (2)

          G ( t , s ) t α 1 ( 1 Γ ( α ) + q ( s ) ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-81/MediaObjects/13661_2012_Article_210_IEq75_HTML.gif, t , s [ 0 , 1 ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-81/MediaObjects/13661_2012_Article_210_IEq76_HTML.gif;

           
        3. (3)
          β t α 1 Φ ( s ) G ( t , s ) t α 2 Φ ( s ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-81/MediaObjects/13661_2012_Article_210_IEq77_HTML.gif, t , s ( 0 , 1 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-81/MediaObjects/13661_2012_Article_210_IEq74_HTML.gif, where
          Φ ( s ) = h ( s ) Γ ( α ) + q ( s ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-81/MediaObjects/13661_2012_Article_210_Equq_HTML.gif
           
        Lemma 2.6 Assume ( H 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-81/MediaObjects/13661_2012_Article_210_IEq45_HTML.gif) holds, then the function G ( t , s ) = : t 2 α G ( t , s ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-81/MediaObjects/13661_2012_Article_210_IEq78_HTML.gif satisfies:
        1. (1)

          G ( t , s ) > 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-81/MediaObjects/13661_2012_Article_210_IEq79_HTML.gif, t , s ( 0 , 1 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-81/MediaObjects/13661_2012_Article_210_IEq74_HTML.gif;

           
        2. (2)

          G ( t , s ) t ( 1 Γ ( α ) + q ( s ) ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-81/MediaObjects/13661_2012_Article_210_IEq80_HTML.gif, t , s [ 0 , 1 ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-81/MediaObjects/13661_2012_Article_210_IEq76_HTML.gif;

           
        3. (3)

          β t Φ ( s ) G ( t , s ) Φ ( s ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-81/MediaObjects/13661_2012_Article_210_IEq81_HTML.gif, t , s [ 0 , 1 ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-81/MediaObjects/13661_2012_Article_210_IEq82_HTML.gif.

           

        For convenience, we list here two more assumptions to be used later:

        ( H 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-81/MediaObjects/13661_2012_Article_210_IEq83_HTML.gif) f ( t , u ) = g ( t , u , u ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-81/MediaObjects/13661_2012_Article_210_IEq84_HTML.gif, here g C ( ( 0 , 1 ) × [ 0 , + ) × ( 0 , + ) [ 0 , + ) ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-81/MediaObjects/13661_2012_Article_210_IEq85_HTML.gif, g ( t , u , v ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-81/MediaObjects/13661_2012_Article_210_IEq86_HTML.gif is nondecreasing on u, nonincreasing on v, and there exists μ ( 0 , 1 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-81/MediaObjects/13661_2012_Article_210_IEq87_HTML.gif such that
        g ( t , r u , v r ) r μ g ( t , u , v ) , u , v > 0 , r ( 0 , 1 ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-81/MediaObjects/13661_2012_Article_210_Equ17_HTML.gif
        (2.12)
        ( H 3 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-81/MediaObjects/13661_2012_Article_210_IEq88_HTML.gif)
        0 < 0 1 g ( s , s α 1 , s α 1 ) d s < + . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-81/MediaObjects/13661_2012_Article_210_Equ18_HTML.gif
        (2.13)
        Remark 2.2 Inequality (2.12) is equivalent to
        g ( t , u r , r v ) r μ g ( t , u , v ) , u , v > 0 , r ( 0 , 1 ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-81/MediaObjects/13661_2012_Article_210_Equ19_HTML.gif
        (2.14)
        Let E = C [ 0 , 1 ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-81/MediaObjects/13661_2012_Article_210_IEq89_HTML.gif be endowed with the maximum norm u = max 0 t 1 | u ( t ) | http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-81/MediaObjects/13661_2012_Article_210_IEq90_HTML.gif. Define a cone P by
        P = { u E :  there exists  l u > 0  such that  β u t u ( t ) l u t } . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-81/MediaObjects/13661_2012_Article_210_Equr_HTML.gif
        Let
        A ( u , v ) ( t ) = 0 1 G ( t , s ) g ( s , s α 2 u ( s ) , s α 2 v ( s ) ) d s . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-81/MediaObjects/13661_2012_Article_210_Equ20_HTML.gif
        (2.15)

        Set Q = P { θ } http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-81/MediaObjects/13661_2012_Article_210_IEq91_HTML.gif, where θ is the zero element of E. We have the following lemma.

        Lemma 2.7 Suppose that ( H 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-81/MediaObjects/13661_2012_Article_210_IEq45_HTML.gif)-( H 3 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-81/MediaObjects/13661_2012_Article_210_IEq88_HTML.gif) hold. Then A : Q × Q Q http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-81/MediaObjects/13661_2012_Article_210_IEq92_HTML.gif.

        Proof For any u , v Q http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-81/MediaObjects/13661_2012_Article_210_IEq93_HTML.gif, there exists l 1 , l 2 > 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-81/MediaObjects/13661_2012_Article_210_IEq94_HTML.gif, such that
        β u t u ( t ) l 1 t , β v t v ( t ) l 2 t . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-81/MediaObjects/13661_2012_Article_210_Equs_HTML.gif
        By ( H 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-81/MediaObjects/13661_2012_Article_210_IEq83_HTML.gif), ( H 3 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-81/MediaObjects/13661_2012_Article_210_IEq88_HTML.gif) and (2) of Lemma 2.6, we get
        A ( u , v ) ( t ) = 0 1 G ( t , s ) g ( s , s α 2 u ( s ) , s α 2 v ( s ) ) d s t 0 1 ( 1 Γ ( α ) + q ( s ) ) g ( s , s α 2 u ( s ) , s α 2 v ( s ) ) d s t 0 1 ( 1 Γ ( α ) + q ( s ) ) g ( s , l 1 s α 1 , β v s α 1 ) d s t 0 1 ( 1 Γ ( α ) + q ( s ) ) g ( s , ( 1 + l 1 ) s α 1 , β v ( 1 + β ) ( 1 + v ) s α 1 ) d s L 1 t 0 1 ( 1 Γ ( α ) + q ( s ) ) g ( s , s α 1 , s α 1 ) d s < + , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-81/MediaObjects/13661_2012_Article_210_Equ21_HTML.gif
        (2.16)

        where L 1 = max { ( 1 + l 1 ) μ , ( ( 1 + β ) ( 1 + v ) β v ) μ } http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-81/MediaObjects/13661_2012_Article_210_IEq95_HTML.gif. This implies that A is well defined in Q × Q http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-81/MediaObjects/13661_2012_Article_210_IEq96_HTML.gif.

        On the other hand, by (3) of Lemma 2.6, we have
        http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-81/MediaObjects/13661_2012_Article_210_Equt_HTML.gif

        Therefore, A ( u , v ) ( t ) β A ( u , v ) t http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-81/MediaObjects/13661_2012_Article_210_IEq97_HTML.gif. Combining with (2.16), we have A : Q × Q Q http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-81/MediaObjects/13661_2012_Article_210_IEq92_HTML.gif. □

        Remark 2.3 By ( H 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-81/MediaObjects/13661_2012_Article_210_IEq83_HTML.gif) and (2.15), A is a mixed monotone operator.

        3 Main results

        Theorem 3.1 Suppose that ( H 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-81/MediaObjects/13661_2012_Article_210_IEq45_HTML.gif)-( H 3 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-81/MediaObjects/13661_2012_Article_210_IEq88_HTML.gif) hold. Then the BVP (1.1) has a unique positive solution.

        Proof For any r ( 0 , 1 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-81/MediaObjects/13661_2012_Article_210_IEq98_HTML.gif, by Remark 2.2, we have
        A ( u r , r v ) r μ A ( u , v ) , u , v Q . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-81/MediaObjects/13661_2012_Article_210_Equu_HTML.gif
        For any w Q http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-81/MediaObjects/13661_2012_Article_210_IEq99_HTML.gif, noticing A ( w , w ) Q http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-81/MediaObjects/13661_2012_Article_210_IEq100_HTML.gif, we can choose r 0 ( 0 , 1 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-81/MediaObjects/13661_2012_Article_210_IEq101_HTML.gif small enough such that
        r 0 1 μ w A ( w , w ) r 0 ( 1 μ ) w . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-81/MediaObjects/13661_2012_Article_210_Equ22_HTML.gif
        (3.1)
        Set
        u 0 = r 0 w , v 0 = r 0 1 w . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-81/MediaObjects/13661_2012_Article_210_Equ23_HTML.gif
        (3.2)
        Clearly,
        u 0 , v 0 Q , u 0 v 0 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-81/MediaObjects/13661_2012_Article_210_Equv_HTML.gif
        Let
        u n = A ( u n 1 , v n 1 ) , v n = A ( v n 1 , u n 1 ) , n = 1 , 2 , . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-81/MediaObjects/13661_2012_Article_210_Equ24_HTML.gif
        (3.3)
        It is easy to see that
        u 0 u 1 u n v n v 1 v 0 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-81/MediaObjects/13661_2012_Article_210_Equ25_HTML.gif
        (3.4)
        Noticing
        u 1 = A ( r 0 w , r 0 1 w ) r 0 μ A ( w , w ) , v 1 = A ( r 0 1 w , r 0 w ) r 0 μ A ( w , w ) , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-81/MediaObjects/13661_2012_Article_210_Equw_HTML.gif
        therefore,
        u 1 r 0 2 μ v 1 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-81/MediaObjects/13661_2012_Article_210_Equx_HTML.gif
        Suppose that u n r 0 2 μ n v n http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-81/MediaObjects/13661_2012_Article_210_IEq102_HTML.gif, then v n r 0 2 μ n u n http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-81/MediaObjects/13661_2012_Article_210_IEq103_HTML.gif, and
        u n + 1 = A ( u n , v n ) A ( r 0 2 μ n v n , r 0 2 μ n u n ) r 0 2 μ n + 1 A ( v n , u n ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-81/MediaObjects/13661_2012_Article_210_Equy_HTML.gif
        By induction, we can get
        u n r 0 2 μ n v n , n = 1 , 2 , . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-81/MediaObjects/13661_2012_Article_210_Equ26_HTML.gif
        (3.5)
        By (3.4), (3.5), we have
        0 u n + m u n v n u n ( 1 r 0 2 μ n ) v n ( 1 r 0 2 μ n ) v 0 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-81/MediaObjects/13661_2012_Article_210_Equz_HTML.gif
        which implies { u n } http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-81/MediaObjects/13661_2012_Article_210_IEq104_HTML.gif is a Cauchy sequence. Similarly, { v n } http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-81/MediaObjects/13661_2012_Article_210_IEq105_HTML.gif is a Cauchy sequence. Noticing (3.4), there exist u , v Q http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-81/MediaObjects/13661_2012_Article_210_IEq106_HTML.gif, such that { u n } http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-81/MediaObjects/13661_2012_Article_210_IEq104_HTML.gif converges to u http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-81/MediaObjects/13661_2012_Article_210_IEq107_HTML.gif and { v n } http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-81/MediaObjects/13661_2012_Article_210_IEq105_HTML.gif converges to v http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-81/MediaObjects/13661_2012_Article_210_IEq108_HTML.gif. Moreover,
        u n u v v n , n = 1 , 2 , . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-81/MediaObjects/13661_2012_Article_210_Equ27_HTML.gif
        (3.6)
        (3.5) and (3.6) imply that
        v u v n u n ( 1 r 0 μ n ) v 0 , n = 1 , 2 , . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-81/MediaObjects/13661_2012_Article_210_Equ28_HTML.gif
        (3.7)

        This implies that u = v http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-81/MediaObjects/13661_2012_Article_210_IEq109_HTML.gif.

        By the mixed monotone property of A and (3.6), we have
        A ( u , v ) A ( u n , v n ) = u n + 1 , A ( v , u ) A ( v n , u n ) = v n + 1 , n = 1 , 2 , . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-81/MediaObjects/13661_2012_Article_210_Equaa_HTML.gif
        Let n + http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-81/MediaObjects/13661_2012_Article_210_IEq110_HTML.gif, we get
        u A ( u , v ) = A ( v , u ) v . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-81/MediaObjects/13661_2012_Article_210_Equab_HTML.gif

        Since u = v http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-81/MediaObjects/13661_2012_Article_210_IEq109_HTML.gif, we have u http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-81/MediaObjects/13661_2012_Article_210_IEq107_HTML.gif is a positive fixed point of A.

        In the following, we will prove the positive fixed point of A is unique.

        Suppose u u http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-81/MediaObjects/13661_2012_Article_210_IEq111_HTML.gif is a positive fixed point of A. By Lemma 2.6, we can get u Q http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-81/MediaObjects/13661_2012_Article_210_IEq112_HTML.gif. Let
        r 1 = sup { r ( 0 , 1 ) : r u u r 1 u } . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-81/MediaObjects/13661_2012_Article_210_Equac_HTML.gif
        Then 0 < r 1 < 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-81/MediaObjects/13661_2012_Article_210_IEq113_HTML.gif, and r 1 u u r 1 1 u http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-81/MediaObjects/13661_2012_Article_210_IEq114_HTML.gif. Therefore
        u = A ( u , u ) A ( r 1 u , r 1 1 u ) r 1 μ A ( u , u ) = r 1 μ u , u = A ( u , u ) A ( r 1 1 u , r 1 u ) r 1 μ A ( u , u ) = r 1 μ u . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-81/MediaObjects/13661_2012_Article_210_Equad_HTML.gif

        Thus, r 1 μ u u r 1 μ u http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-81/MediaObjects/13661_2012_Article_210_IEq115_HTML.gif, which contradicts the definition of r 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-81/MediaObjects/13661_2012_Article_210_IEq116_HTML.gif. Consequently, the positive fixed point of A is unique.

        It is clear that y ( t ) = t α 2 u ( t ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-81/MediaObjects/13661_2012_Article_210_IEq117_HTML.gif satisfies
        y ( t ) = 0 1 G ( t , s ) g ( s , y ( s ) , y ( s ) ) d s = 0 1 G ( t , s ) f ( s , y ( s ) ) d s , t ( 0 , 1 ] . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-81/MediaObjects/13661_2012_Article_210_Equae_HTML.gif
        On the other hand, since u Q http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-81/MediaObjects/13661_2012_Article_210_IEq118_HTML.gif, we have β u t u ( t ) l u t http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-81/MediaObjects/13661_2012_Article_210_IEq119_HTML.gif. Then, β u t α 1 y ( t ) l u t α 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-81/MediaObjects/13661_2012_Article_210_IEq120_HTML.gif. By Lemma 2.5 and ( H 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-81/MediaObjects/13661_2012_Article_210_IEq83_HTML.gif), ( H 3 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-81/MediaObjects/13661_2012_Article_210_IEq88_HTML.gif), we can get g ( t , y ( t ) , y ( t ) ) L ( 0 , 1 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-81/MediaObjects/13661_2012_Article_210_IEq121_HTML.gif. Moreover,
        lim t 0 y ( t ) = lim t 0 0 1 G ( t , s ) g ( s , y ( s ) , y ( s ) ) d s t α 1 0 1 ( 1 Γ ( α ) + q ( s ) ) g ( s , y ( s ) , y ( s ) ) d s = 0 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-81/MediaObjects/13661_2012_Article_210_Equaf_HTML.gif

        Lemma 2.3 implies y ( t ) = t α 2 u ( t ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-81/MediaObjects/13661_2012_Article_210_IEq117_HTML.gif is a positive solution of (1.1).

        On the other hand, if y ( t ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-81/MediaObjects/13661_2012_Article_210_IEq122_HTML.gif is a positive solution of (1.1), then
        y ( t ) = 0 1 G ( t , s ) g ( s , y ( s ) , y ( s ) ) d s . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-81/MediaObjects/13661_2012_Article_210_Equag_HTML.gif
        By Lemma 2.5, we have there exists l 1 , l 2 > 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-81/MediaObjects/13661_2012_Article_210_IEq94_HTML.gif such that
        l 2 t α 1 y ( t ) l 1 t α 1 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-81/MediaObjects/13661_2012_Article_210_Equah_HTML.gif
        Set u ( t ) = t 2 α y ( t ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-81/MediaObjects/13661_2012_Article_210_IEq123_HTML.gif, we have
        l 2 t u ( t ) l 1 t , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-81/MediaObjects/13661_2012_Article_210_Equai_HTML.gif
        and
        u ( t ) = 0 1 G ( t , s ) g ( s , s α 2 u ( s ) , s α 2 v ( s ) ) d s , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-81/MediaObjects/13661_2012_Article_210_Equaj_HTML.gif

        which implies u is a positive fixed point of A.

        Then y ( t ) = t α 2 u ( t ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-81/MediaObjects/13661_2012_Article_210_IEq117_HTML.gif is the unique positive solution of the BVP (1.1). □

        Remark 3.1 The unique positive solution y of (1.1) can be approximated by the iterative schemes: for any w Q http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-81/MediaObjects/13661_2012_Article_210_IEq99_HTML.gif, let u 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-81/MediaObjects/13661_2012_Article_210_IEq33_HTML.gif, v 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-81/MediaObjects/13661_2012_Article_210_IEq124_HTML.gif be defined as (3.2) and u n = A ( u n 1 , v n 1 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-81/MediaObjects/13661_2012_Article_210_IEq125_HTML.gif, v n = A ( v n 1 , u n 1 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-81/MediaObjects/13661_2012_Article_210_IEq126_HTML.gif, n = 1 , 2 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-81/MediaObjects/13661_2012_Article_210_IEq127_HTML.gif , then t α 2 u n y http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-81/MediaObjects/13661_2012_Article_210_IEq128_HTML.gif.

        Example 3.1 (A 4-point BVP with coefficients of both signs)

        Consider the following problem:
        { D 0 + 7 4 u ( t ) + f ( t , u ( t ) ) = 0 , t ( 0 , 1 ) , u ( 0 ) = 0 , D 0 + 1 4 u ( 1 ) = D 0 + 1 4 u ( 1 4 ) 1 2 D 0 + 1 4 u ( 4 9 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-81/MediaObjects/13661_2012_Article_210_Equ29_HTML.gif
        (3.8)
        with
        f ( t , x ) = x 1 2 ln t x 1 3 ln ( 1 t ) + x 1 3 ln ( 1 + x 1 2 ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-81/MediaObjects/13661_2012_Article_210_Equak_HTML.gif
        Then
        G 0 ( t , s ) = 1 Γ ( 7 4 ) { t 3 4 ( 1 s ) 1 2 , 0 t s 1 , t 3 4 ( 1 s ) 1 2 ( t s ) 3 4 , 0 s t 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-81/MediaObjects/13661_2012_Article_210_Equal_HTML.gif
        and
        p ( s ) = { 1 ( 1 4 s 1 s ) 1 2 1 2 ( 4 9 s 1 s ) 1 2 , 0 s 1 4 , 1 1 2 ( 4 9 s 1 s ) 1 2 , 1 4 < s 4 9 , 1 , 4 9 < s 1 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-81/MediaObjects/13661_2012_Article_210_Equam_HTML.gif

        By direct calculations, we have p ( 0 ) = 5 6 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-81/MediaObjects/13661_2012_Article_210_IEq129_HTML.gif and q ( s ) 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-81/MediaObjects/13661_2012_Article_210_IEq47_HTML.gif, which implies ( H 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-81/MediaObjects/13661_2012_Article_210_IEq45_HTML.gif) holds.

        Let
        g ( t , x , y ) = x 1 2 ln t y 1 3 ln ( 1 t ) + y 1 3 ln ( 1 + x 1 2 ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-81/MediaObjects/13661_2012_Article_210_Equan_HTML.gif
        Obviously, g C ( ( 0 , 1 ) × [ 0 , + ) × ( 0 , + ) [ 0 , + ) ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-81/MediaObjects/13661_2012_Article_210_IEq130_HTML.gif, g ( t , x , y ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-81/MediaObjects/13661_2012_Article_210_IEq131_HTML.gif is nondecreasing on x and nonincreasing on y. It is easy to see that
        ln ( 1 + r x ) r ln ( 1 + x ) , x 0 , r ( 0 , 1 ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-81/MediaObjects/13661_2012_Article_210_Equ30_HTML.gif
        (3.9)
        Then
        g ( t , r x , y r ) r 5 6 g ( t , x , y ) , x , y > 0 , r ( 0 , 1 ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-81/MediaObjects/13661_2012_Article_210_Equao_HTML.gif

        Therefore ( H 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-81/MediaObjects/13661_2012_Article_210_IEq83_HTML.gif) holds. It is easy to get that ( H 3 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-81/MediaObjects/13661_2012_Article_210_IEq88_HTML.gif) holds. Therefore, the assumptions of Theorem 3.1 are satisfied. Thus Theorem 3.1 ensures that the BVP (3.8) has a unique positive solution.

        Declarations

        Acknowledgements

        The authors are grateful to the anonymous referee for his/her valuable suggestions. The first and second authors were supported financially by the National Natural Science Foundation of China (11071141, 11126231) and Project of Shandong Province Higher Educational Science and Technology Program (J11LA06). The third author was supported financially by the Australia Research Council through an ARC Discovery Project Grant.

        Authors’ Affiliations

        (1)
        School of Mathematical Sciences, Qufu Normal University
        (2)
        Department of Mathematics and Statistics, Curtin University of Technology

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