Positive solution for a class of singular semipositone fractional differential equations with integral boundary conditions

Boundary Value Problems20122012:123

DOI: 10.1186/1687-2770-2012-123

Received: 20 July 2012

Accepted: 10 October 2012

Published: 24 October 2012

Abstract

In this article, by employing a fixed point theorem in cones, we investigate the existence of a positive solution for a class of singular semipositone fractional differential equations with integral boundary conditions. We also obtain some relations between the solution and Green’s function.

MSC: 26A33, 34B15, 34B16, 34G20.

Keywords

fractional differential equations integral boundary value problem positive solution semipositone cone

1 Introduction

In this article, we consider the existence of a positive solution for the following singular semipositone fractional differential equations:
{ D 0 + α u ( t ) + f ( t , u ( t ) ) = 0 , 0 < t < 1 , u ( 0 ) = u ( 0 ) = u ( 0 ) = 0 , u ( 1 ) = λ 0 η u ( s ) d s , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-123/MediaObjects/13661_2012_Article_226_Equ1_HTML.gif
(1)

where f C [ ( 0 , 1 ) × [ 0 , + ) , ( , + ) ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-123/MediaObjects/13661_2012_Article_226_IEq1_HTML.gif, 3 < α 4 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-123/MediaObjects/13661_2012_Article_226_IEq2_HTML.gif, 0 < η 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-123/MediaObjects/13661_2012_Article_226_IEq3_HTML.gif, 0 λ η α α < 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-123/MediaObjects/13661_2012_Article_226_IEq4_HTML.gif, D 0 + α http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-123/MediaObjects/13661_2012_Article_226_IEq5_HTML.gif is the standard Riemann-Liouville derivative, f ( t , u ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-123/MediaObjects/13661_2012_Article_226_IEq6_HTML.gif may be singular at t = 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-123/MediaObjects/13661_2012_Article_226_IEq7_HTML.gif and/or t = 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-123/MediaObjects/13661_2012_Article_226_IEq8_HTML.gif. Since the nonlinearity f ( t , x ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-123/MediaObjects/13661_2012_Article_226_IEq9_HTML.gif may change sign, the problem studied in this paper is called the semipositone problem in the literature which arises naturally in chemical reactor theory. Up to now, much attention has been attached to the existence of positive solutions for semipositone differential equations and the system of differential equations; see [111] and references therein to name a few.

Boundary value problems with integral boundary conditions for ordinary differential equations arise in different fields of applied mathematics and physics such as heat conduction, chemical engineering, underground water flow, thermo-elasticity, and plasma physics. Moreover, boundary value problems with integral conditions constitute a very interesting and important class of problems. They include two-point, three-point, multi-point, and nonlocal boundary value problems as special cases, which have received much attention from many authors. For boundary value problems with integral boundary conditions and comments on their importance, we refer the reader to the papers by Gallardo [12], Karakostas and Tsamatos [13], Lomtatidze and Malaguti [14], and the references therein.

On the other hand, fractional differential equations have been of great interest for many researchers recently. This is caused both by the intensive development of the theory of fractional calculus itself and by the applications of such constructions in various fields of science and engineering such as control, porous media, electromagnetic, and other fields. For an extensive collection of such results, we refer the readers to the monographs by Samko et al.[15], Podlubny [16] and Kilbas et al.[17]. For the case where α is an integer, a lot of work has been done dealing with local and nonlocal boundary value problems. For example, in [18] Webb studied the n th-order nonlocal BVP
{ u ( n ) ( t ) + a ( t ) f ( t , u ( t ) ) = 0 , 0 < t < 1 , u ( 0 ) = u ( 0 ) = = u ( n 2 ) ( 0 ) = 0 , u ( 1 ) = 0 1 u ( s ) d A ( s ) , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-123/MediaObjects/13661_2012_Article_226_Equa_HTML.gif
where a ( t ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-123/MediaObjects/13661_2012_Article_226_IEq10_HTML.gif can have singularities, and the nonlinearity f satisfies Carathéodory conditions. Under weak assumptions, Webb obtained sharp results on the existence of positive solutions under a suitable condition on f. In [19] Hao et al. consider the n th-order singular nonlocal BVP
{ u ( n ) ( t ) + λ a ( t ) f ( t , u ( t ) ) = 0 , 0 < t < 1 , u ( 0 ) = u ( 0 ) = = u ( n 2 ) ( 0 ) = 0 , u ( 1 ) = 0 1 u ( s ) d A ( s ) , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-123/MediaObjects/13661_2012_Article_226_Equb_HTML.gif

where λ > 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-123/MediaObjects/13661_2012_Article_226_IEq11_HTML.gif is a parameter, a may be singular at t = 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-123/MediaObjects/13661_2012_Article_226_IEq12_HTML.gif and/or t = 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-123/MediaObjects/13661_2012_Article_226_IEq13_HTML.gif, f ( t , x ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-123/MediaObjects/13661_2012_Article_226_IEq14_HTML.gif may also have singularity at x = 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-123/MediaObjects/13661_2012_Article_226_IEq15_HTML.gif.

In two recent papers [20] and [21], by means of the fixed point theory and fixed point index theory, the authors investigated the existence and multiplicity of positive solutions for the following two kinds of fractional differential equations with integral boundary value problems:
{ D 0 + α u ( t ) + q ( t ) f ( t , u ( t ) ) = 0 , 0 < t < 1 , u ( 0 ) = u ( 0 ) = = u ( n 2 ) ( 0 ) = 0 , u ( 1 ) = 0 1 u ( s ) d A ( s ) , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-123/MediaObjects/13661_2012_Article_226_Equc_HTML.gif
and
{ D α C u ( t ) + f ( t , u ( t ) ) = 0 , 0 < t < 1 , u ( 0 ) = u ( 0 ) = 0 , u ( 1 ) = λ 0 1 u ( s ) d s , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-123/MediaObjects/13661_2012_Article_226_Equd_HTML.gif

where 0 < λ < 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-123/MediaObjects/13661_2012_Article_226_IEq16_HTML.gif, D 0 + α http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-123/MediaObjects/13661_2012_Article_226_IEq5_HTML.gif and D α C http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-123/MediaObjects/13661_2012_Article_226_IEq17_HTML.gif are the standard Riemann-Liouville derivative and the Caputo fractional derivative, respectively.

To the author’s knowledge, there are few papers in the literature to consider fractional differential equations with integral boundary value conditions. Motivated by above papers, the purpose of this article is to investigate the existence of positive solutions for the more general fractional differential equations BVP (1). Firstly, we derive corresponding Green’s function known as fractional Green’s function and argue its positivity. Then a fixed point theorem is used to obtain the existence of positive solutions for BVP (1). We also obtain some relations between the solution and Green’s function. From the example given in Section 4, we know that λ in this article may be greater than 2 and η may take the value 1. Therefore, compared with that in [21], BVP (1) considered in this article has a more general form.

The rest of this article is organized as follows. In Section 2, we give some preliminaries and lemmas. The main result is formulated in Section 3, and an example is worked out in Section 4 to illustrate how to use our main result.

2 Preliminaries and several lemmas

Let E = C [ 0 , 1 ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-123/MediaObjects/13661_2012_Article_226_IEq18_HTML.gif, u = max 0 t 1 | u ( t ) | http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-123/MediaObjects/13661_2012_Article_226_IEq19_HTML.gif, then ( E , ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-123/MediaObjects/13661_2012_Article_226_IEq20_HTML.gif is a Banach space. Denote I = [ 0 , 1 ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-123/MediaObjects/13661_2012_Article_226_IEq21_HTML.gif, J = ( 0 , 1 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-123/MediaObjects/13661_2012_Article_226_IEq22_HTML.gif, R + = [ 0 , + ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-123/MediaObjects/13661_2012_Article_226_IEq23_HTML.gif.

For the reader’s convenience, we present some necessary definitions from fractional calculus theory and lemmas. They can be found in the recent literature; see [1417].

Definition 2.1 The Riemann-Liouville fractional integral of order α > 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-123/MediaObjects/13661_2012_Article_226_IEq24_HTML.gif of a function y : ( 0 , ) R http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-123/MediaObjects/13661_2012_Article_226_IEq25_HTML.gif is given by
I 0 + α y ( t ) = 1 Γ ( α ) 0 t ( t s ) α 1 y ( s ) d s http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-123/MediaObjects/13661_2012_Article_226_Eque_HTML.gif

provided the right-hand side is pointwise defined on ( 0 , ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-123/MediaObjects/13661_2012_Article_226_IEq26_HTML.gif.

Definition 2.2 The Riemann-Liouville fractional derivative of order α > 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-123/MediaObjects/13661_2012_Article_226_IEq24_HTML.gif of a continuous function y : ( 0 , ) R http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-123/MediaObjects/13661_2012_Article_226_IEq25_HTML.gif is given by
D 0 + α y ( t ) = 1 Γ ( n α ) ( d d t ) n 0 t y ( s ) ( t s ) α n + 1 y ( s ) d s , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-123/MediaObjects/13661_2012_Article_226_Equf_HTML.gif

where n = [ α ] + 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-123/MediaObjects/13661_2012_Article_226_IEq27_HTML.gif, [ α ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-123/MediaObjects/13661_2012_Article_226_IEq28_HTML.gif denotes the integer part of the number α, provided that the right-hand side is pointwise defined on ( 0 , ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-123/MediaObjects/13661_2012_Article_226_IEq26_HTML.gif.

From the definition of the Riemann-Liouville derivative, we can obtain the statement.

Lemma 2.1 ([17])

Let α > 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-123/MediaObjects/13661_2012_Article_226_IEq24_HTML.gif. If we assume u C ( 0 , 1 ) L ( 0 , 1 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-123/MediaObjects/13661_2012_Article_226_IEq29_HTML.gif, then the fractional differential equation
D 0 + α u ( t ) = 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-123/MediaObjects/13661_2012_Article_226_Equg_HTML.gif

has 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%2F1687-2770-2012-123/MediaObjects/13661_2012_Article_226_IEq30_HTML.gif, C i R http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-123/MediaObjects/13661_2012_Article_226_IEq31_HTML.gif, i = 1 , 2 , , N http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-123/MediaObjects/13661_2012_Article_226_IEq32_HTML.gif, as unique solutions, where N is the smallest integer greater than or equal to α.

Lemma 2.2 ([17])

Assume that u C ( 0 , 1 ) L ( 0 , 1 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-123/MediaObjects/13661_2012_Article_226_IEq29_HTML.gifwith a fractional derivative of order α > 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-123/MediaObjects/13661_2012_Article_226_IEq24_HTML.gifthat belongs to C ( 0 , 1 ) L ( 0 , 1 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-123/MediaObjects/13661_2012_Article_226_IEq33_HTML.gif.

Then
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%2F1687-2770-2012-123/MediaObjects/13661_2012_Article_226_Equh_HTML.gif

for some C i R http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-123/MediaObjects/13661_2012_Article_226_IEq31_HTML.gif, i = 1 , 2 , , N http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-123/MediaObjects/13661_2012_Article_226_IEq32_HTML.gif, where N is the smallest integer greater than or equal to α.

In the following, we present Green’s function of the fractional differential equation boundary value problem.

Lemma 2.3 Given y C [ 0 , 1 ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-123/MediaObjects/13661_2012_Article_226_IEq34_HTML.gif, the problem
{ D 0 + α u ( t ) + y ( t ) = 0 , u ( 0 ) = u ( 0 ) = u ( 0 ) = 0 , u ( 1 ) = λ 0 η u ( s ) d s , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-123/MediaObjects/13661_2012_Article_226_Equ2_HTML.gif
(2)
where 0 < t < 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-123/MediaObjects/13661_2012_Article_226_IEq35_HTML.gif, 3 < α 4 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-123/MediaObjects/13661_2012_Article_226_IEq2_HTML.gif, 0 < η 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-123/MediaObjects/13661_2012_Article_226_IEq36_HTML.gif, 0 λ η α α < 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-123/MediaObjects/13661_2012_Article_226_IEq4_HTML.gif, is equivalent to
u ( t ) = 0 1 G ( t , s ) y ( s ) d s , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-123/MediaObjects/13661_2012_Article_226_Equi_HTML.gif
where
G ( t , s ) = { t α 1 ( 1 s ) α 1 λ α ( η s ) α t α 1 ( 1 λ α η α ) ( t s ) α 1 p ( 0 ) Γ ( α ) , 0 s t 1 , s η ; t α 1 ( 1 s ) α 1 ( 1 λ α η α ) ( t s ) α 1 p ( 0 ) Γ ( α ) , 0 η s t 1 ; t α 1 ( 1 s ) α 1 λ α ( η s ) α t α 1 p ( 0 ) Γ ( α ) , 0 t s η 1 ; t α 1 ( 1 s ) α 1 p ( 0 ) Γ ( α ) , 0 t s 1 , η s . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-123/MediaObjects/13661_2012_Article_226_Equ3_HTML.gif
(3)

Here, p ( s ) : = 1 λ η α α ( 1 s ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-123/MediaObjects/13661_2012_Article_226_IEq37_HTML.gif, G ( t , s ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-123/MediaObjects/13661_2012_Article_226_IEq38_HTML.gifis called the Green function of BVP (2). Obviously, G ( t , s ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-123/MediaObjects/13661_2012_Article_226_IEq38_HTML.gifis continuous on [ 0 , 1 ] × [ 0 , 1 ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-123/MediaObjects/13661_2012_Article_226_IEq39_HTML.gif.

Proof We may apply Lemma 2.2 to reduce (2) to an equivalent integral equation
u ( t ) = I 0 + α y ( t ) + C 1 t α 1 + C 2 t α 2 + C 3 t α 3 + C 4 t α 4 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-123/MediaObjects/13661_2012_Article_226_Equj_HTML.gif
for some C 1 , C 2 , C 3 , C 4 R http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-123/MediaObjects/13661_2012_Article_226_IEq40_HTML.gif. Consequently, the general solution of (2) is
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 + C 3 t α 3 + C 4 t α 4 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-123/MediaObjects/13661_2012_Article_226_Equk_HTML.gif
By u ( 0 ) = u ( 0 ) = u ( 0 ) = 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-123/MediaObjects/13661_2012_Article_226_IEq41_HTML.gif, one gets that C 2 = C 3 = C 4 = 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-123/MediaObjects/13661_2012_Article_226_IEq42_HTML.gif. On the other hand, u ( 1 ) = λ 0 η u ( s ) d s http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-123/MediaObjects/13661_2012_Article_226_IEq43_HTML.gif combining with
http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-123/MediaObjects/13661_2012_Article_226_Equl_HTML.gif
yields
C 1 = 0 1 ( 1 s ) α 1 Γ ( α ) ( 1 λ η α α ) y ( s ) d s λ 0 η ( η s ) α α Γ ( α ) ( 1 λ η α α ) y ( s ) d s . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-123/MediaObjects/13661_2012_Article_226_Equm_HTML.gif
Therefore, the unique solution of the problem (2) is
u ( t ) = 0 t ( t s ) α 1 Γ ( α ) y ( s ) d s + 1 ( 1 λ η α α ) 0 1 ( 1 s ) α 1 t α 1 Γ ( α ) y ( s ) d s 1 ( 1 λ η α α ) 0 η λ α ( η s ) α t α 1 Γ ( α ) y ( s ) d s . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-123/MediaObjects/13661_2012_Article_226_Equn_HTML.gif
For t η http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-123/MediaObjects/13661_2012_Article_226_IEq44_HTML.gif, one has
u ( t ) = 0 t ( t s ) α 1 Γ ( α ) y ( s ) d s + 1 ( 1 λ α η α ) [ ( 0 t + t η + η 1 ) ( 1 s ) α 1 t α 1 Γ ( α ) y ( s ) d s ] λ ( 1 λ α η α ) [ ( 0 t + t η ) 1 α ( η s ) α t α 1 Γ ( α ) y ( s ) d s ] = 0 t t α 1 ( 1 s ) α 1 λ α ( η s ) α t α 1 ( 1 λ α η α ) ( t s ) α 1 ( 1 λ α η α ) Γ ( α ) y ( s ) d s + t η t α 1 ( 1 s ) α 1 λ α ( η s ) α t α 1 ( 1 λ α η α ) Γ ( α ) y ( s ) d s + η 1 t α 1 ( 1 s ) α 1 ( 1 λ α η α ) Γ ( α ) y ( s ) d s = 0 1 G ( t , s ) y ( s ) d s . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-123/MediaObjects/13661_2012_Article_226_Equo_HTML.gif
For t η http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-123/MediaObjects/13661_2012_Article_226_IEq45_HTML.gif, one has
u ( t ) = ( 0 η + η t ) ( t s ) α 1 Γ ( α ) y ( s ) d s + 1 ( 1 λ α η α ) [ ( 0 η + η t + t 1 ) ( 1 s ) α 1 t α 1 Γ ( α ) y ( s ) d s ] λ ( 1 λ α η α ) 0 η 1 α ( η s ) α t α 1 Γ ( α ) y ( s ) d s = 0 η t α 1 ( 1 s ) α 1 λ α ( η s ) α t α 1 ( 1 λ α η α ) ( t s ) α 1 ( 1 λ α η α ) Γ ( α ) y ( s ) d s + η t t α 1 ( 1 s ) α 1 ( 1 λ α η α ) ( t s ) α 1 ( 1 λ α η α ) Γ ( α ) y ( s ) d s + t 1 t α 1 ( 1 s ) α 1 ( 1 λ α η α ) Γ ( α ) y ( s ) d s = 0 1 G ( t , s ) y ( s ) d s . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-123/MediaObjects/13661_2012_Article_226_Equp_HTML.gif

The proof is complete. □

Lemma 2.4 The function G ( t , s ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-123/MediaObjects/13661_2012_Article_226_IEq38_HTML.gifdefined by (3) satisfies

(a1) G ( t , s ) m 1 e ( t ) s ( 1 s ) α 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-123/MediaObjects/13661_2012_Article_226_IEq46_HTML.gif, t , s [ 0 , 1 ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-123/MediaObjects/13661_2012_Article_226_IEq47_HTML.gif;

(a2) G ( t , s ) M 1 e ( t ) ( 1 s ) α 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-123/MediaObjects/13661_2012_Article_226_IEq48_HTML.gif, t , s [ 0 , 1 ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-123/MediaObjects/13661_2012_Article_226_IEq47_HTML.gif;

(a3) G ( t , s ) M 1 s ( 1 s ) α 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-123/MediaObjects/13661_2012_Article_226_IEq49_HTML.gif, t , s [ 0 , 1 ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-123/MediaObjects/13661_2012_Article_226_IEq47_HTML.gif;

(a4) p ( s ) > 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-123/MediaObjects/13661_2012_Article_226_IEq50_HTML.gif, and p ( s ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-123/MediaObjects/13661_2012_Article_226_IEq51_HTML.gifis not decreasing on [ 0 , 1 ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-123/MediaObjects/13661_2012_Article_226_IEq52_HTML.gif;

(a5) G ( t , s ) > 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-123/MediaObjects/13661_2012_Article_226_IEq53_HTML.gif, t , s ( 0 , 1 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-123/MediaObjects/13661_2012_Article_226_IEq54_HTML.gif,

where m 1 = 1 p ( 0 ) Γ ( α ) p ( 0 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-123/MediaObjects/13661_2012_Article_226_IEq55_HTML.gif, M 1 = α 1 Γ ( α ) + 4 λ α η α 1 p ( 0 ) Γ ( α ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-123/MediaObjects/13661_2012_Article_226_IEq56_HTML.gif, e ( t ) = t α 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-123/MediaObjects/13661_2012_Article_226_IEq57_HTML.gif.

Proof For s t http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-123/MediaObjects/13661_2012_Article_226_IEq58_HTML.gif, s η http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-123/MediaObjects/13661_2012_Article_226_IEq59_HTML.gif,
http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-123/MediaObjects/13661_2012_Article_226_Equq_HTML.gif
For η s t http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-123/MediaObjects/13661_2012_Article_226_IEq60_HTML.gif,
http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-123/MediaObjects/13661_2012_Article_226_Equr_HTML.gif
For t s η http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-123/MediaObjects/13661_2012_Article_226_IEq61_HTML.gif,
http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-123/MediaObjects/13661_2012_Article_226_Equs_HTML.gif
For η s http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-123/MediaObjects/13661_2012_Article_226_IEq62_HTML.gif, t s http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-123/MediaObjects/13661_2012_Article_226_IEq63_HTML.gif,
http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-123/MediaObjects/13661_2012_Article_226_Equt_HTML.gif

From above, (a1), (a2), (a3), (a5) are complete. Clearly, (a4) is true. The proof is complete. □

Throughout this article, we adopt the following conditions.

(H1) f C [ ( 0 , 1 ) × [ 0 , + ) , ( , + ) ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-123/MediaObjects/13661_2012_Article_226_IEq1_HTML.gif and there exist a , b C [ J , R + ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-123/MediaObjects/13661_2012_Article_226_IEq64_HTML.gif, h C [ R + , R + ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-123/MediaObjects/13661_2012_Article_226_IEq65_HTML.gif, g ( t , u ) [ J × R + , R + ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-123/MediaObjects/13661_2012_Article_226_IEq66_HTML.gif such that
g ( t , u ) b ( t ) f ( t , u ) a ( t ) h ( u ) , t J , u R + ; http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-123/MediaObjects/13661_2012_Article_226_Equu_HTML.gif

(H2) There exists [ a , b ] I http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-123/MediaObjects/13661_2012_Article_226_IEq67_HTML.gif such that lim u + g ( t , u ) u = + http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-123/MediaObjects/13661_2012_Article_226_IEq68_HTML.gif uniformly for t [ a , b ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-123/MediaObjects/13661_2012_Article_226_IEq69_HTML.gif;

(H3) There exists r > 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-123/MediaObjects/13661_2012_Article_226_IEq70_HTML.gif such that
0 1 ( 1 s ) α 1 b ( s ) d s < m 1 M 1 2 r , M 1 0 1 ( 1 s ) α 1 ( a ( s ) + b ( s ) ) d s < r max 0 u r { h ( u ) , 1 } . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-123/MediaObjects/13661_2012_Article_226_Equv_HTML.gif
Let
http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-123/MediaObjects/13661_2012_Article_226_Equ4_HTML.gif
(4)

Obviously, Q is a cone in a Banach space E and ( E , Q ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-123/MediaObjects/13661_2012_Article_226_IEq71_HTML.gif is an ordering Banach space.

Let
x 0 ( t ) = 0 1 G ( t , s ) b ( s ) d s , t I , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-123/MediaObjects/13661_2012_Article_226_Equ5_HTML.gif
(5)
where b ( t ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-123/MediaObjects/13661_2012_Article_226_IEq72_HTML.gif is defined as that in (H1). It follows from Lemma 2.4 and (H3) that
0 x 0 ( t ) M 1 e ( t ) 0 1 ( 1 s ) α 1 b ( s ) d s < + , t I . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-123/MediaObjects/13661_2012_Article_226_Equ6_HTML.gif
(6)
So, x 0 P http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-123/MediaObjects/13661_2012_Article_226_IEq73_HTML.gif and it satisfies
{ D 0 + α x 0 ( t ) + x 0 ( t ) = 0 , 0 < t < 1 , x 0 ( 0 ) = x 0 ( 0 ) = x 0 ( 0 ) = 0 , x 0 ( 1 ) = λ 0 η x 0 ( s ) d s . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-123/MediaObjects/13661_2012_Article_226_Equ7_HTML.gif
(7)
For any u Q { θ } http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-123/MediaObjects/13661_2012_Article_226_IEq74_HTML.gif, u ( t ) m 1 M 1 u e ( t ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-123/MediaObjects/13661_2012_Article_226_IEq75_HTML.gif, t I http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-123/MediaObjects/13661_2012_Article_226_IEq76_HTML.gif. Consequently, by (6) and Lemma 2.4, we have
x 0 ( t ) M 1 e ( t ) 0 1 ( 1 s ) α 1 b ( s ) d s M 1 u m 1 M 1 e ( t ) 1 u m 1 M 1 0 1 ( 1 s ) α 1 b ( s ) d s u ( t ) u M 1 2 m 1 0 1 ( 1 s ) α 1 b ( s ) d s . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-123/MediaObjects/13661_2012_Article_226_Equ8_HTML.gif
(8)
For any k E http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-123/MediaObjects/13661_2012_Article_226_IEq77_HTML.gif, denote
[ k ( t ) ] + = { k ( t ) , k ( t ) 0 , 0 , k ( t ) < 0 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-123/MediaObjects/13661_2012_Article_226_Equw_HTML.gif
We define an operator A as follows:
( A u ) ( t ) = 0 1 G ( t , s ) ( f ( s , [ u ( s ) x 0 ( s ) ] + ) + b ( s ) ) d s , u P . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-123/MediaObjects/13661_2012_Article_226_Equ9_HTML.gif
(9)

Lemma 2.5 Suppose that ( H 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-123/MediaObjects/13661_2012_Article_226_IEq78_HTML.gif)-( H 3 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-123/MediaObjects/13661_2012_Article_226_IEq79_HTML.gif) hold. Then A : Q Q http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-123/MediaObjects/13661_2012_Article_226_IEq80_HTML.gifis completely continuous.

Proof For any u P http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-123/MediaObjects/13661_2012_Article_226_IEq81_HTML.gif, it is clear that [ u ( s ) x 0 ( s ) ] + u ( s ) u http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-123/MediaObjects/13661_2012_Article_226_IEq82_HTML.gif. By (H1), we get
f ( s , [ u ( s ) x 0 ( s ) ] + ) + b ( s ) a ( s ) h ( [ u ( s ) x 0 ( s ) ] + ) + b ( s ) max 0 r u { h ( r ) , 1 } ( a ( s ) + b ( s ) ) , s J . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-123/MediaObjects/13661_2012_Article_226_Equ10_HTML.gif
(10)
By (10) and Lemma 2.4, we have
http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-123/MediaObjects/13661_2012_Article_226_Equ11_HTML.gif
(11)

which together with (H3) means that operator A defined by (9) is well defined.

Now, we show that A : Q Q http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-123/MediaObjects/13661_2012_Article_226_IEq80_HTML.gif.

For any u Q http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-123/MediaObjects/13661_2012_Article_226_IEq83_HTML.gif, by (H1) we have by (9) and Lemma 2.4 that
( A u ) ( t ) M 1 0 1 s ( 1 s ) α 1 [ f ( s , [ u ( s ) x 0 ( s ) ] + ) + b ( s ) ] d s , t I , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-123/MediaObjects/13661_2012_Article_226_Equx_HTML.gif
which means that
A u M 1 0 1 s ( 1 s ) α 1 [ f ( s , [ u ( s ) x 0 ( s ) ] + ) + b ( s ) ] d s . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-123/MediaObjects/13661_2012_Article_226_Equ12_HTML.gif
(12)
It follows from (12) and Lemma 2.4 that
( A u ) ( t ) = 0 1 G ( t , s ) ( f ( s , [ u ( s ) x 0 ( s ) ] + ) + b ( s ) ) d s m 1 e ( t ) 0 1 s ( 1 s ) α 1 [ f ( s , [ u ( s ) x 0 ( s ) ] + ) + b ( s ) ] d s = m 1 M 1 e ( t ) 0 1 M 1 s ( 1 s ) α 1 [ f ( s , [ u ( s ) x 0 ( s ) ] + ) + b ( s ) ] d s m 1 M 1 e ( t ) A u . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-123/MediaObjects/13661_2012_Article_226_Equy_HTML.gif

Thus, A maps Q into Q.

Finally, we prove that A maps Q into Q is completely continuous.

Let D Q http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-123/MediaObjects/13661_2012_Article_226_IEq84_HTML.gif be any bounded set. Then there exists a constant L 1 > 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-123/MediaObjects/13661_2012_Article_226_IEq85_HTML.gif such that u L 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-123/MediaObjects/13661_2012_Article_226_IEq86_HTML.gif for any u D http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-123/MediaObjects/13661_2012_Article_226_IEq87_HTML.gif. Notice that [ u ( s ) x 0 ( s ) ] + u ( s ) L 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-123/MediaObjects/13661_2012_Article_226_IEq88_HTML.gif, for any u D http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-123/MediaObjects/13661_2012_Article_226_IEq87_HTML.gif, s I http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-123/MediaObjects/13661_2012_Article_226_IEq89_HTML.gif, by (H3) and (11), we have
| ( A u ) ( t ) | M 1 0 1 ( 1 s ) α 1 ( a ( s ) + b ( s ) ) d s max 0 r L 1 { h ( r ) , 1 } < + . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-123/MediaObjects/13661_2012_Article_226_Equz_HTML.gif

Therefore, A ( D ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-123/MediaObjects/13661_2012_Article_226_IEq90_HTML.gif is uniformly bounded.

On the other hand, since G ( t , s ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-123/MediaObjects/13661_2012_Article_226_IEq38_HTML.gif is continuous on [ 0 , 1 ] × [ 0 , 1 ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-123/MediaObjects/13661_2012_Article_226_IEq91_HTML.gif, it is uniformly continuous on [ 0 , 1 ] × [ 0 , 1 ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-123/MediaObjects/13661_2012_Article_226_IEq39_HTML.gif as well. Thus, for fixed s I http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-123/MediaObjects/13661_2012_Article_226_IEq89_HTML.gif and for any ε > 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-123/MediaObjects/13661_2012_Article_226_IEq92_HTML.gif, there exists a constant δ > 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-123/MediaObjects/13661_2012_Article_226_IEq93_HTML.gif such that for any t 1 , t 2 [ 0 , 1 ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-123/MediaObjects/13661_2012_Article_226_IEq94_HTML.gif and | t 1 t 2 | < δ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-123/MediaObjects/13661_2012_Article_226_IEq95_HTML.gif,
| G ( t 1 , s ) G ( t 2 , s ) | < ε max 0 r L 1 { h ( r ) , 1 } 0 1 ( a ( s ) + b ( s ) ) d s . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-123/MediaObjects/13661_2012_Article_226_Equ13_HTML.gif
(13)
Therefore, for any x D http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-123/MediaObjects/13661_2012_Article_226_IEq96_HTML.gif, we get by (10) and (13)
| ( A u ) ( t 1 ) ( A u ) ( t 2 ) | 0 1 | G ( t 1 , s ) G ( t 2 , s ) | ( f ( s , [ u ( s ) x 0 ( s ) ] + ) + b ( s ) ) d s ε max 0 r L 1 { h ( r ) , 1 } 0 1 ( a ( s ) + b ( s ) ) d s 0 1 ( a ( s ) + b ( s ) ) d s max 0 r L 1 { h ( r ) , 1 } = ε , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-123/MediaObjects/13661_2012_Article_226_Equaa_HTML.gif

which implies that the operator A is equicontinuous. Thus, the Ascoli-Arzela theorem guarantees that A ( D ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-123/MediaObjects/13661_2012_Article_226_IEq90_HTML.gif is a relatively compact set.

Let u n , u 0 Q http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-123/MediaObjects/13661_2012_Article_226_IEq97_HTML.gif, u n u 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-123/MediaObjects/13661_2012_Article_226_IEq98_HTML.gif ( n http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-123/MediaObjects/13661_2012_Article_226_IEq99_HTML.gif). Then { u n } http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-123/MediaObjects/13661_2012_Article_226_IEq100_HTML.gif is bounded. Let L 2 = sup { u n , n = 0 , 1 , 2 , } http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-123/MediaObjects/13661_2012_Article_226_IEq101_HTML.gif, by (10), we get
f ( s , [ u n ( s ) x 0 ( s ) ] + ) + b ( s ) ( max 0 r L 2 { h ( r ) , 1 } ) ( a ( s ) + b ( s ) ) , n = 0 , 1 , 2 , . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-123/MediaObjects/13661_2012_Article_226_Equ14_HTML.gif
(14)
By (9), we have
http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-123/MediaObjects/13661_2012_Article_226_Equ15_HTML.gif
(15)

It follows from (14), (15), (H1), (H3), and the Lebesgue dominated convergence theorem that A is continuous. Thus, we have proved the continuity of the operator A. This completes the complete continuity of A. □

To prove the main result, we need the following well-known fixed point theorem.

Lemma 2.6 (Fixed point theorem of cone expansion and compression of norm type [22])

Let Ω 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-123/MediaObjects/13661_2012_Article_226_IEq102_HTML.gifand Ω 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-123/MediaObjects/13661_2012_Article_226_IEq103_HTML.gifbe two bounded open sets in a Banach space E such that θ Ω 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-123/MediaObjects/13661_2012_Article_226_IEq104_HTML.gifand Ω ¯ 1 Ω 2 , A : P ( Ω ¯ 2 Ω 1 ) P http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-123/MediaObjects/13661_2012_Article_226_IEq105_HTML.gifbe a completely continuous operator, where θ denotes the zero element of E and P a cone of E. Suppose that one of the two conditions holds:
  1. (i)

    A u u http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-123/MediaObjects/13661_2012_Article_226_IEq106_HTML.gif, u P Ω 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-123/MediaObjects/13661_2012_Article_226_IEq107_HTML.gif; A u u http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-123/MediaObjects/13661_2012_Article_226_IEq108_HTML.gif, u P Ω 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-123/MediaObjects/13661_2012_Article_226_IEq109_HTML.gif;

     
  2. (ii)

    A u u http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-123/MediaObjects/13661_2012_Article_226_IEq108_HTML.gif, u P Ω 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-123/MediaObjects/13661_2012_Article_226_IEq107_HTML.gif; A u u http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-123/MediaObjects/13661_2012_Article_226_IEq106_HTML.gif, u P Ω 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-123/MediaObjects/13661_2012_Article_226_IEq110_HTML.gif.

     

Then A has a fixed point in P ( Ω ¯ 2 Ω 1 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-123/MediaObjects/13661_2012_Article_226_IEq111_HTML.gif.

3 Main result

Theorem 3.1 Assume that conditions ( H 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-123/MediaObjects/13661_2012_Article_226_IEq78_HTML.gif)-( H 3 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-123/MediaObjects/13661_2012_Article_226_IEq79_HTML.gif) are satisfied. Then the singular semipositone BVP (1) has at least one positive solution ω ( t ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-123/MediaObjects/13661_2012_Article_226_IEq112_HTML.gif. Furthermore, there exist two constants M > m > 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-123/MediaObjects/13661_2012_Article_226_IEq113_HTML.gifsuch that
m e ( t ) ω ( t ) M e ( t ) , t I . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-123/MediaObjects/13661_2012_Article_226_Equ16_HTML.gif
(16)
Proof Firstly, we show that the operator A has a fixed point in Q. Let
Ω r = { u E | u < r } , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-123/MediaObjects/13661_2012_Article_226_Equab_HTML.gif
where r is the same as that defined in (H3). For any u Ω r Q http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-123/MediaObjects/13661_2012_Article_226_IEq114_HTML.gif, by (10) and (12), we have that
( A u ) ( t ) = 0 1 G ( t , s ) ( f ( s , [ u ( s ) x 0 ( s ) ] + ) + b ( s ) ) d s M 1 0 1 s ( 1 s ) α 1 ( a ( s ) + b ( s ) ) d s max 0 u r { h ( u ) , 1 } < + , t J . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-123/MediaObjects/13661_2012_Article_226_Equac_HTML.gif
Therefore,
A u M 1 0 1 s ( 1 s ) α 1 ( a ( s ) + b ( s ) ) d s max 0 u r { h ( u ) , 1 } < r , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-123/MediaObjects/13661_2012_Article_226_Equad_HTML.gif
which together with (H3) implies that
A u r = u , u Ω r Q . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-123/MediaObjects/13661_2012_Article_226_Equ17_HTML.gif
(17)
For [ a , b ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-123/MediaObjects/13661_2012_Article_226_IEq115_HTML.gif in (H2), it is clear that
e ( t ) a α 1 , t [ a , b ] . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-123/MediaObjects/13661_2012_Article_226_Equ18_HTML.gif
(18)
By (H3), we know that there exists a natural number m 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-123/MediaObjects/13661_2012_Article_226_IEq116_HTML.gif big enough such that
0 1 ( 1 s ) α 1 b ( s ) d s < m 0 m 0 + 1 m 1 M 1 2 r . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-123/MediaObjects/13661_2012_Article_226_Equ19_HTML.gif
(19)
Choose
M > M 1 ( m 0 + 1 ) m 1 a α 1 min t [ a , b ] 0 1 G ( t , s ) d s . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-123/MediaObjects/13661_2012_Article_226_Equ20_HTML.gif
(20)
By (H2), we know there exists R 1 > r http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-123/MediaObjects/13661_2012_Article_226_IEq117_HTML.gif such that
g ( t , u ) M u , t [ a , b ] , u R 1 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-123/MediaObjects/13661_2012_Article_226_Equ21_HTML.gif
(21)
Take
R > max { r , 1 , R 1 ( m 0 + 1 ) M 1 a α 1 m 1 } . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-123/MediaObjects/13661_2012_Article_226_Equ22_HTML.gif
(22)
In the following, we are in a position to show that
A u u , u Ω R Q . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-123/MediaObjects/13661_2012_Article_226_Equ23_HTML.gif
(23)
For any u Ω R Q http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-123/MediaObjects/13661_2012_Article_226_IEq118_HTML.gif, by (8) we get
x 0 ( t ) u ( t ) u M 1 2 m 1 0 1 ( 1 s ) α 1 b ( s ) d s , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-123/MediaObjects/13661_2012_Article_226_Equae_HTML.gif
which together with (18), (19), (22), and (H3) implies that
http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-123/MediaObjects/13661_2012_Article_226_Equ24_HTML.gif
(24)
For u Ω R Q http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-123/MediaObjects/13661_2012_Article_226_IEq119_HTML.gif, t [ a , b ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-123/MediaObjects/13661_2012_Article_226_IEq69_HTML.gif, it follows from (H1), (20), (21), (22), and (24) that
( A u ) ( t ) = 0 1 G ( t , s ) ( f ( s , [ u ( s ) x 0 ( s ) ] + ) + b ( s ) ) d s 0 1 G ( t , s ) g ( s , [ u ( s ) x 0 ( s ) ] + ) d s 0 1 G ( t , s ) M ( u ( s ) x 0 ( s ) ) d s M R m 0 + 1 m 1 M 1 a α 1 min t [ a , b ] 0 1 G ( t , s ) d s > R , t [ a , b ] . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-123/MediaObjects/13661_2012_Article_226_Equ25_HTML.gif
(25)
By (25), we know that (23) holds. So, (17), (23), and Lemma 2.6 guarantee that A has at least one fixed point z 0 ( t ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-123/MediaObjects/13661_2012_Article_226_IEq120_HTML.gif in ( Ω ¯ R Ω r ) Q http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-123/MediaObjects/13661_2012_Article_226_IEq121_HTML.gif and r z 0 R http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-123/MediaObjects/13661_2012_Article_226_IEq122_HTML.gif. Furthermore,
z 0 ( t ) = 0 1 G ( t , s ) ( f ( s , [ z 0 ( s ) x 0 ( s ) ] + ) + b ( s ) ) d s , t I . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-123/MediaObjects/13661_2012_Article_226_Equ26_HTML.gif
(26)
By simple computation, we have that
{ D 0 + α z 0 ( t ) + f ( t , [ z 0 ( t ) x 0 ( t ) ] + ) + b ( t ) = 0 , 0 < t < 1 , z 0 ( 0 ) = z 0 ( 0 ) = z 0 ( 0 ) = 0 , z 0 ( 1 ) = λ 0 η z 0 ( s ) d s . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-123/MediaObjects/13661_2012_Article_226_Equ27_HTML.gif
(27)
Secondly, we show BVP (1) has a positive solution. It follows from (8) and the fact z 0 r http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-123/MediaObjects/13661_2012_Article_226_IEq123_HTML.gif that
x 0 ( t ) z 0 ( t ) z 0 M 1 2 m 1 0 1 ( 1 s ) α 1 b ( s ) d s z 0 ( t ) r M 1 2 m 1 0 1 ( 1 s ) α 1 b ( s ) d s , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-123/MediaObjects/13661_2012_Article_226_Equaf_HTML.gif
which combined with (19) implies that
z 0 ( t ) x 0 ( t ) ( 1 1 r M 1 2 m 1 0 1 ( 1 s ) α 1 b ( s ) d s ) z 0 ( t ) 1 m 0 + 1 z 0 ( t ) 0 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-123/MediaObjects/13661_2012_Article_226_Equ28_HTML.gif
(28)
By (27) and (28), we have
{ D 0 + α z 0 ( t ) + f ( t , [ z 0 ( t ) x 0 ( t ) ] ) + b ( t ) = 0 , 0 < t < 1 , z 0 ( 0 ) = z 0 ( 0 ) = z 0 ( 0 ) = 0 , z 0 ( 1 ) = λ 0 η z 0 ( s ) d s . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-123/MediaObjects/13661_2012_Article_226_Equ29_HTML.gif
(29)
Let ω ( t ) = z 0 ( t ) x 0 ( t ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-123/MediaObjects/13661_2012_Article_226_IEq124_HTML.gif, t I http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-123/MediaObjects/13661_2012_Article_226_IEq76_HTML.gif. It follows from (28) and z 0 Q http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-123/MediaObjects/13661_2012_Article_226_IEq125_HTML.gif that
ω ( t ) 1 m 0 + 1 z 0 ( t ) m 1 r ( m 0 + 1 ) M 1 e ( t ) > 0 , t J . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-123/MediaObjects/13661_2012_Article_226_Equ30_HTML.gif
(30)
By (7), (29), and (30), we obtain
{ D 0 + α ω ( t ) + f ( t , ω ( t ) ) = 0 , 0 < t < 1 , ω ( 0 ) = ω ( 0 ) = ω ( 0 ) = 0 , ω ( 1 ) = λ 0 η ω ( s ) d s . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-123/MediaObjects/13661_2012_Article_226_Equag_HTML.gif

Thus, we have proved that ω ( t ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-123/MediaObjects/13661_2012_Article_226_IEq112_HTML.gif is a positive solution for BVP (1).

Finally, we show that (16) holds. From (26) and Lemma 2.4, we know that
z 0 ( t ) M 1 e ( t ) 0 1 ( 1 s ) α 1 ( f ( s , [ z 0 ( s ) x 0 ( s ) ] + ) + b ( s ) ) d s , t I . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-123/MediaObjects/13661_2012_Article_226_Equ31_HTML.gif
(31)

Since ω ( t ) z 0 ( t ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-123/MediaObjects/13661_2012_Article_226_IEq126_HTML.gif, (30), and (31) mean that (16) holds for m = m 1 r ( m 0 + 1 ) M 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-123/MediaObjects/13661_2012_Article_226_IEq127_HTML.gif and M = M 1 0 1 ( 1 s ) α 1 ( f ( s , [ z 0 ( s ) x 0 ( s ) ] + ) + b ( s ) ) d s http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-123/MediaObjects/13661_2012_Article_226_IEq128_HTML.gif holds. This completes the proof of Theorem 3.1. □

4 Example

Consider the following singular semipositone fractional differential equations:
{ D 0 + 7 2 u ( t ) + f ( t , u ( t ) ) = 0 , 0 < t < 1 , u ( 0 ) = u ( 0 ) = u ( 0 ) = 0 , u ( 1 ) = 16 2 0 1 2 u ( s ) d s , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-123/MediaObjects/13661_2012_Article_226_Equ32_HTML.gif
(32)
where f ( t , u ) = t 2 8 1 t u 5 2 + t 1 4 20 ( 1 t ) ( u u 1 2 3 4 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-123/MediaObjects/13661_2012_Article_226_IEq129_HTML.gif. It is clear (32) has the form of (1), where α = 7 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-123/MediaObjects/13661_2012_Article_226_IEq130_HTML.gif, λ = 16 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-123/MediaObjects/13661_2012_Article_226_IEq131_HTML.gif, η = 1 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-123/MediaObjects/13661_2012_Article_226_IEq132_HTML.gif. By simple computation, we know that 0 < λ η α α 0.5714 < 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-123/MediaObjects/13661_2012_Article_226_IEq133_HTML.gif, p ( 0 ) 0.4286 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-123/MediaObjects/13661_2012_Article_226_IEq134_HTML.gif. Let
a ( t ) = t 2 8 1 t + t 1 4 20 ( 1 t ) , b ( t ) = t 1 4 20 ( 1 t ) , g ( t , u ) = t 2 8 1 t u 5 2 , h ( u ) = u 5 2 + u u 1 2 + 1 4 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-123/MediaObjects/13661_2012_Article_226_Equah_HTML.gif
Notice that
1 u u 1 2 3 4 u u 1 2 + 1 4 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-123/MediaObjects/13661_2012_Article_226_Equai_HTML.gif
we have
t 1 4 20 ( 1 t ) t 1 4 20 ( 1 t ) ( u u 1 2 3 4 ) t 1 4 20 ( 1 t ) ( u u 1 2 + 1 4 ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-123/MediaObjects/13661_2012_Article_226_Equ33_HTML.gif
(33)
It follows from the left side of (33) that
b ( t ) f ( t , u ) g ( t , u ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-123/MediaObjects/13661_2012_Article_226_Equ34_HTML.gif
(34)
Considering u u 1 2 + 1 4 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-123/MediaObjects/13661_2012_Article_226_IEq135_HTML.gif, we get
f ( t , u ) t 2 8 1 t u 5 2 + t 1 4 20 ( 1 t ) ( u u 1 2 + 1 4 ) a ( t ) h ( u ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-123/MediaObjects/13661_2012_Article_226_Equ35_HTML.gif
(35)

By (34) and (35) we know (H1) holds. Obviously, (H2) holds for [ a , b ] = [ 1 4 , 3 4 ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-123/MediaObjects/13661_2012_Article_226_IEq136_HTML.gif.

Now, we check (H3). By simple computation, we have m 1 = 0.4012 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-123/MediaObjects/13661_2012_Article_226_IEq137_HTML.gif, M 1 = 3.9619 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-123/MediaObjects/13661_2012_Article_226_IEq138_HTML.gif, m 1 M 1 2 = 0.0256 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-123/MediaObjects/13661_2012_Article_226_IEq139_HTML.gif, 0 1 ( 1 s ) α 1 b ( s ) d s 0.0136 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-123/MediaObjects/13661_2012_Article_226_IEq140_HTML.gif, M 1 0 1 ( 1 s ) α 1 ( a ( s ) + b ( s ) ) d s 0.1245 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-123/MediaObjects/13661_2012_Article_226_IEq141_HTML.gif. Take r = 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-123/MediaObjects/13661_2012_Article_226_IEq142_HTML.gif, then max 0 u r { h ( u ) , 1 } 1.2500 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-123/MediaObjects/13661_2012_Article_226_IEq143_HTML.gif, r max 0 u r { h ( u ) , 1 } 0.8000 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-123/MediaObjects/13661_2012_Article_226_IEq144_HTML.gif. Thus, (H3) is valid. It follows from Theorem 3.1 that BVP (32) has at least one positive solution.

Declarations

Acknowledgements

The author thanks the referee for his/her careful reading of the manuscript and useful suggestions. The project is supported financially by the Foundation for Outstanding Middle-Aged and Young Scientists of Shandong Province (Grant No. BS2010SF004), a Project of Shandong Province Higher Educational Science and Technology Program (Grant No. J10LA53, No. J11LA02), the China Postdoctoral Science Foundation (Grant No. 20110491154) and the National Natural Science Foundation of China (Grant No. 10971179).

Authors’ Affiliations

(1)
School of Mathematics and Statistics, Huazhong University of Science and Technology
(2)
School of Mathematics, Liaocheng University

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