Existence results of positive solutions for boundary value problems of fractional differential equations

Boundary Value Problems20132013:109

DOI: 10.1186/1687-2770-2013-109

Received: 15 October 2012

Accepted: 15 April 2013

Published: 29 April 2013

Abstract

In this paper, we are concerned with the following fractional equation:

D 0 + α C u ( t ) = f ( t , u ( t ) , u ( t ) ) , t ( 0 , 1 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_Equa_HTML.gif

with the boundary value conditions

u ( 1 ) = u ( 1 ) = 0 , δ u ( 0 ) = u ( 1 ) , γ u ( 0 ) = u ( 1 ) , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_Equb_HTML.gif

where D 0 + α C http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq1_HTML.gif is the standard Caputo derivative with 3 < α 4 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq2_HTML.gif and δ, γ are constants with δ > 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq3_HTML.gif, γ > 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq4_HTML.gif. By applying a new fixed point theorem on cone and Krasnoselskii’s fixed point theorem, some existence results of positive solution are obtained.

MSC: 34A08, 34B15, 34B18.

Keywords

fractional differential equations existence results fixed point theorem positive solution

1 Introduction

In this paper, we are concerned with the existence of positive solutions for the fractional equation
D 0 + α C u ( t ) = f ( t , u ( t ) , u ( t ) ) , t ( 0 , 1 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_Equ1_HTML.gif
(1.1)
with the boundary value conditions
u ( 1 ) = u ( 1 ) = 0 , δ u ( 0 ) = u ( 1 ) , γ u ( 0 ) = u ( 1 ) , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_Equ2_HTML.gif
(1.2)

where D 0 + α C http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq1_HTML.gif is the standard Caputo derivative with 3 < α 4 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq2_HTML.gif and δ, γ are constants with δ > 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq3_HTML.gif, γ > 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq4_HTML.gif.

Differential equations of fractional order have recently proved to be valuable tools in the modeling of many phenomena in various fields of science and engineering. Indeed, we can find numerous applications in viscoelasticity, electrochemistry, control, porous media, electromagnetism, etc. (see [15]). There has been a significant development in the study of fractional differential equations and inclusions in recent years, see the monographs of Podlubny [5], Kilbas et al. [6], Lakshmikantham et al. [7], Samko et al. [8], Diethelm [9], and the survey by Agarwal et al. [10]. For some recent contributions on fractional differential equations, see [925] and the references therein.

On the other hand, it is well known that the fourth-order boundary value problem describes the deformations of an elastic beam in equilibrium state. Owing to its importance in physics, the existence of solutions to this problem has been studied by many authors; see, for example, [2630] and references therein. Recently, there have been a few papers dealing with the existence of solutions for fractional equations of order α ( 3 , 4 ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq5_HTML.gif.

In [14], Xu et al. discussed the problem
D 0 + α u ( t ) = f ( t , u ( t ) ) , t ( 0 , 1 ) , u ( 0 ) = u ( 0 ) = u ( 1 ) = u ( 1 ) = 0 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_Equc_HTML.gif

where 3 < α 4 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq2_HTML.gif and f C ( [ 0 , 1 ] × ( 0 , ) ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq6_HTML.gif is nonnegative, D 0 + α http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq7_HTML.gif is the Riemann-Liouville fractional derivative of order α. The existence results of positive solutions are obtained by applying the Leray-Schauder nonlinear alternative theorem.

In [15], Liang and Zhang studied the following nonlinear fractional boundary value problem:
D 0 + α u ( t ) = f ( t , u ( t ) ) , t ( 0 , 1 ) , u ( 0 ) = u ( 0 ) = u ( 0 ) = u ( 1 ) = 0 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_Equd_HTML.gif

where 3 < α 4 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq2_HTML.gif, f ( t , u ) C ( [ 0 , 1 ] × [ 0 , ) , [ 0 , ) ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq8_HTML.gif is nondecreasing relative to u, D 0 + α http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq7_HTML.gif is the Riemann-Liouville fractional derivative of order α. By means of the lower and upper solution method and fixed point theorems, some results on the existence of positive solutions were obtained.

In [16], Agarwal and Ahmad studied the solvability of the following anti-periodic boundary value problem for a nonlinear fractional differential equation:
{ D 0 + α u ( t ) = f ( t , u ( t ) ) , t ( 0 , T ) , u ( 0 ) + u ( T ) = 0 , u ( 0 ) + u ( T ) = 0 , u ( 0 ) + u ( T ) = 0 , u ( 0 ) + u ( T ) = 0 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_Eque_HTML.gif

where 3 < α 4 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq2_HTML.gif. The existence results were obtained by the nonlinear alternative theorem.

Inspired by above work, the author will be concerned with the boundary value problem (BVP for short in the sequel) (1.1)-(1.2). To the best of our knowledge, no contribution exists concerning the existence of solutions for BVP (1.1)-(1.2). In the present paper, by applying a new fixed point theorem on cone and Krasnoselskii’s fixed point theorem, some existence results of positive solution for BVP (1.1)-(1.2) are obtained. It is worth to point out that the results in this paper are also new even for α = 4 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq9_HTML.gif relative to the corresponding literature with regard to the fourth-order boundary value problem. In addition, the conditions imposed in this paper are easily verified.

The organization of this paper is as follows. In Section 2, we present some necessary definitions and preliminary results that will be used to prove our main results. In Section 3, we put forward and prove our main results. Finally, we give two examples to demonstrate our main results.

2 Preliminaries

In this section, we introduce some preliminary facts which are useful throughout this paper.

Let ℕ be the set of positive integers, ℝ be the set of real numbers, R + = [ 0 , + ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq10_HTML.gif, and R = ( , 0 ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq11_HTML.gif. Let I = [ 0 , 1 ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq12_HTML.gif. Denote by C 1 ( I , R ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq13_HTML.gif the Banach space endowed with the norm u 1 = u 0 + u 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq14_HTML.gif, where u 0 = max t I | u ( t ) | http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq15_HTML.gif for u C ( I , R ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq16_HTML.gif.

Definition 2.1 [6]

The Riemann-Liouville fractional integral of order α > 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq17_HTML.gif of a function y : ( a , b ] R http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq18_HTML.gif is given by
I a + α y ( t ) = 1 Γ ( α ) 0 t ( t s ) α 1 y ( s ) d s , t ( a , b ] . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_Equf_HTML.gif

Definition 2.2 [6]

The Riemann-Liouville fractional derivative of order α > 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq17_HTML.gif of a function y : ( a , b ] R http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq18_HTML.gif is given by
D a + α y ( t ) = 1 Γ ( n α ) ( d d t ) n 0 t y ( s ) ( t s ) α n + 1 d s , t ( a , b ] , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_Equg_HTML.gif

where n = [ α ] + 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq19_HTML.gif, [ α ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq20_HTML.gif denotes the integer part of α.

Definition 2.3 [6]

The Caputo fractional derivative of order α > 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq17_HTML.gif of a function y on ( a , b ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq21_HTML.gif is defined via the above Riemann-Liouville derivatives by
( C D a + α y ) ( x ) = ( D a + α [ y ( t ) k = 0 n 1 y ( k ) ( a ) k ! ( t a ) k ] ) ( x ) , x ( a , b ] . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_Equh_HTML.gif

Lemma 2.1 [6]

Let α > 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq17_HTML.gif and y C [ a , b ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq22_HTML.gif. Then
( C D a + α I a + α y ) ( x ) = y ( x ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_Equi_HTML.gif

holds on [ a , b ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq23_HTML.gif.

Lemma 2.2 [17]

Let n N http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq24_HTML.gif with n 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq25_HTML.gif, α ( n 1 , n ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq26_HTML.gif. If y C n 1 [ a , b ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq27_HTML.gif and D a + α C y C ( a , b ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq28_HTML.gif, then
I a + α C D a + α y ( t ) = y ( t ) k = 0 n 1 y ( k ) ( a ) k ! ( t a ) k http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_Equj_HTML.gif

holds on ( a , b ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq29_HTML.gif.

For convenience, we first list some hypotheses which will be used throughout this paper.

( H 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq30_HTML.gif) f C ( I × R + × R , R + ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq31_HTML.gif.

( H 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq32_HTML.gif) 3 < α 4 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq2_HTML.gif, γ > 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq4_HTML.gif, δ > 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq3_HTML.gif.

For h C [ 0 , 1 ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq33_HTML.gif, consider the following BVP:
D 0 + α C u ( t ) = h ( t ) , t ( 0 , 1 ) , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_Equ3_HTML.gif
(2.1)
u ( 1 ) = u ( 1 ) = 0 , δ u ( 0 ) = u ( 1 ) , γ u ( 0 ) = u ( 1 ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_Equ4_HTML.gif
(2.2)

We have the following lemma, which is important in this paper.

Lemma 2.3 Let ( H 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq32_HTML.gif) hold. Then u C 3 [ 0 , 1 ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq34_HTML.gif is a solution of BVP (2.1)-(2.2) iff u C [ 0 , 1 ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq35_HTML.gif has the expression as follows:
u ( t ) = 0 1 G ( t , s ) h ( s ) d s , t I , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_Equ5_HTML.gif
(2.3)
where
G ( t , s ) = { G 1 ( t , s ) , 0 s t < 1 , G 2 ( t , s ) , 0 t s < 1 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_Equ6_HTML.gif
(2.4)
and
G 1 ( t , s ) = [ ( t s ) α 1 ( 1 s ) α 1 ] Γ ( α ) + ( 1 t ) ( 1 s ) α 2 Γ ( α 1 ) G 1 ( t , s ) = + ( 1 t ) 2 ( 1 s ) α 3 2 ( δ 1 ) Γ ( α 2 ) + ( 1 t ) 2 [ 1 + 2 δ + ( δ 1 ) t ] 6 ( γ 1 ) ( δ 1 ) Γ ( α 3 ) ( 1 s ) α 4 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_Equ7_HTML.gif
(2.5)
G 2 ( t , s ) = ( 1 s ) α 1 Γ ( α ) + ( 1 t ) ( 1 s ) α 2 Γ ( α 1 ) + ( 1 t ) 2 ( 1 s ) α 3 2 ( δ 1 ) Γ ( α 2 ) G 2 ( t , s ) = + ( 1 t ) 2 [ 1 + 2 δ + ( δ 1 ) t ] 6 ( γ 1 ) ( δ 1 ) Γ ( α 3 ) ( 1 s ) α 4 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_Equ8_HTML.gif
(2.6)
Proof Let u C 3 [ 0 , 1 ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq34_HTML.gif be a solution of (2.1)-(2.2). Then by Lemma 2.2, we have
u ( t ) = c 0 + c 1 t + c 2 t 2 + c 3 t 3 + I 0 + α h ( t ) , t I , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_Equ9_HTML.gif
(2.7)
and so
u ( t ) = c 1 + 2 c 2 t + 3 c 3 t 2 + I 0 + α 1 h ( t ) , t I , u ( t ) = 2 c 2 + 6 c 3 t + I 0 + α 2 h ( t ) , t I , u ( t ) = 6 c 3 + I 0 + α 3 h ( t ) , t I . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_Equk_HTML.gif
Thus, by the boundary value condition (2.2), we can obtain
c 0 + c 1 + c 2 + c 3 + I 0 + α h ( 1 ) = 0 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_Equ10_HTML.gif
(2.8)
c 1 + 2 c 2 + 3 c 3 + I 0 + α 1 h ( 1 ) = 0 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_Equ11_HTML.gif
(2.9)
2 ( 1 δ ) c 2 + 6 c 3 + I 0 + α 2 h ( 1 ) = 0 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_Equ12_HTML.gif
(2.10)
6 ( 1 γ ) c 3 + I 0 + α 3 h ( 1 ) = 0 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_Equ13_HTML.gif
(2.11)
From (2.11), we have
c 3 = 1 6 ( γ 1 ) I 0 + α 3 h ( 1 ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_Equ14_HTML.gif
(2.12)
Substituting (2.12) into (2.10), we get
c 2 = 1 2 ( γ 1 ) ( δ 1 ) I 0 + α 3 h ( 1 ) + 1 2 ( δ 1 ) I 0 + α 2 h ( 1 ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_Equ15_HTML.gif
(2.13)
So, by (2.13), (2.12), and (2.9), we have
c 1 = 1 + δ 2 ( γ 1 ) ( δ 1 ) I 0 + α 3 h ( 1 ) 1 δ 1 I 0 + α 2 h ( 1 ) I 0 + α 1 h ( 1 ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_Equ16_HTML.gif
(2.14)
Thus, from (2.8), we have
c 0 = 1 + 2 δ 6 ( γ 1 ) ( δ 1 ) I 0 + α 3 h ( 1 ) + 1 2 ( δ 1 ) I 0 + α 2 h ( 1 ) + I 0 + α 1 h ( 1 ) I 0 + α h ( 1 ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_Equ17_HTML.gif
(2.15)
Hence, from (2.7) together with (2.12)-(2.15), it follows that
u ( t ) = c 0 + c 1 t + c 2 t 2 + c 3 t 3 + I 0 + α h ( t ) = I 0 + α h ( 1 ) + ( 1 t ) I 0 + α 1 h ( 1 ) + 1 2 ( δ 1 ) ( 1 t ) 2 I 0 + α 2 h ( 1 ) + [ ( 2 δ + 1 ) 3 ( 1 + δ ) t + 3 t 2 + ( δ 1 ) t 3 ] 6 ( γ 1 ) ( δ 1 ) I 0 + α 3 h ( 1 ) + I 0 + α h ( t ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_Equ18_HTML.gif
(2.16)
Noticing that
( 2 δ + 1 ) 3 ( 1 + δ ) t + 3 t 2 + ( δ 1 ) t 3 = ( 1 t ) 2 [ 1 + 2 δ + ( δ 1 ) t ] , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_Equl_HTML.gif
by Definition 2.1, we have
u ( t ) = 0 t [ ( t s ) α 1 ( 1 s ) α 1 Γ ( α ) + ( 1 t ) ( 1 s ) α 2 Γ ( α 1 ) + ( 1 t ) 2 ( 1 s ) α 3 2 ( δ 1 ) Γ ( α 2 ) + ( 1 t ) 2 [ 1 + 2 δ + ( δ 1 ) t ] 6 ( γ 1 ) ( δ 1 ) Γ ( α 3 ) ( 1 s ) α 4 ] h ( s ) d s + t 1 [ ( 1 s ) α 1 Γ ( α ) + ( 1 t ) ( 1 s ) α 2 Γ ( α 1 ) + ( 1 t ) 2 ( 1 s ) α 3 2 ( δ 1 ) Γ ( α 2 ) + ( 1 t ) 2 [ 1 + 2 δ + ( δ 1 ) t ] 6 ( γ 1 ) ( δ 1 ) Γ ( α 3 ) ( 1 s ) α 4 ] h ( s ) d s = 0 1 G ( t , s ) h ( s ) d s , t I . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_Equm_HTML.gif
Conversely, if u has the expression (2.3), then from the fact that h C [ 0 , 1 ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq36_HTML.gif, we can easily verify that
u ( t ) = 0 t [ ( t s ) α 2 ( 1 s ) α 2 Γ ( α 1 ) + ( t 1 ) ( 1 s ) α 3 ( δ 1 ) Γ ( α 2 ) + ( t 1 ) [ 1 + δ + ( δ 1 ) t ] 2 ( γ 1 ) ( δ 1 ) Γ ( α 3 ) ] h ( s ) d s u ( t ) = + t 1 [ ( 1 s ) α 2 Γ ( α 1 ) + ( t 1 ) ( 1 s ) α 3 ( δ 1 ) Γ ( α 2 ) + ( t 1 ) [ 1 + δ + ( δ 1 ) t ] 2 ( γ 1 ) ( δ 1 ) Γ ( α 3 ) ] h ( s ) d s , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_Equ19_HTML.gif
(2.17)
u ( t ) = 0 t [ ( t s ) α 3 Γ ( α 2 ) + ( 1 s ) α 3 ( δ 1 ) Γ ( α 2 ) + 1 + ( δ 1 ) t ( γ 1 ) ( δ 1 ) Γ ( α 3 ) ] h ( s ) d s u ( t ) = + t 1 [ ( 1 s ) α 3 ( δ 1 ) Γ ( α 2 ) + 1 + ( δ 1 ) t ( γ 1 ) ( δ 1 ) Γ ( α 3 ) ] h ( s ) d s , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_Equ20_HTML.gif
(2.18)
u ( t ) = 0 t [ ( t s ) α 4 Γ ( α 3 ) + 1 ( γ 1 ) Γ ( α 3 ) ] h ( s ) d s + t 1 1 ( γ 1 ) Γ ( α 3 ) h ( s ) d s http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_Equ21_HTML.gif
(2.19)

hold for t I http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq37_HTML.gif, and u satisfies the boundary condition (2.2).

Again, from (2.16) and Lemma 2.1, we have that D 0 + α C x ( t ) = h ( t ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq38_HTML.gif, t ( 0 , 1 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq39_HTML.gif. In addition, noting that h C [ 0 , 1 ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq36_HTML.gif, it is easy to see that x C 3 [ 0 , 1 ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq40_HTML.gif from (2.19). □

For the forthcoming analysis, we need to introduce some new notations.

Let η 1 = min { 2 α 5 3 ( α 1 ) , 4 ( 1 + 2 δ ) 27 δ } http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq41_HTML.gif, and η 2 = min { 2 ( α 2 ) 3 ( α + δ 3 ) , 1 + δ 3 δ } http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq42_HTML.gif. Denote
w 1 ( s ) = ( 1 s ) α 2 Γ ( α 1 ) + ( 1 s ) α 3 2 ( δ 1 ) Γ ( α 2 ) + δ ( 1 s ) α 4 2 ( γ 1 ) ( δ 1 ) Γ ( α 3 ) , s [ 0 , 1 ) , w 2 ( s ) = ( α + δ 3 ) ( 1 s ) α 3 ( α 2 ) ( δ 1 ) Γ ( α 2 ) + δ ( 1 s ) α 4 ( γ 1 ) ( δ 1 ) Γ ( α 3 ) , s [ 0 , 1 ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_Equn_HTML.gif

It is easy to verify that η 1 , η 2 ( 0 , 1 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq43_HTML.gif, and w 1 , w 2 C ( [ 0 , 1 ) , ( 0 , + ) ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq44_HTML.gif noting that 3 < α 4 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq2_HTML.gif, γ > 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq4_HTML.gif, δ > 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq3_HTML.gif.

We also need the following lemma, which will play an important role in obtaining our main results in Section 3.

Lemma 2.4 Under the assumption ( H 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq32_HTML.gif), Green’s function G ( t , s ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq45_HTML.gif has the following properties:

(1) G is continuous on [ 0 , 1 ) × [ 0 , 1 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq46_HTML.gif;

(2) G ( t , s ) > 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq47_HTML.gif, t , s [ 0 , 1 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq48_HTML.gif; G t ( t , s ) < 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq49_HTML.gif, t , s [ 0 , 1 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq48_HTML.gif, s t http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq50_HTML.gif;

(3) G ( t , s ) w 1 ( s ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq51_HTML.gif, t , s [ 0 , 1 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq48_HTML.gif; G ( t , s ) η 1 w 1 ( s ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq52_HTML.gif, t [ 1 6 , 1 3 ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq53_HTML.gif, s [ 0 , 1 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq54_HTML.gif;

0 < G t ( t , s ) w 2 ( s ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq55_HTML.gif, t , s [ 0 , 1 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq48_HTML.gif, s t http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq50_HTML.gif; G t ( t , s ) η 2 w 2 ( s ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq56_HTML.gif, t [ 1 6 , 1 3 ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq53_HTML.gif, s [ 0 , 1 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq54_HTML.gif, s t http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq50_HTML.gif.

Proof (1) Observing the expression of Green’s function given by (2.4)-(2.6), the conclusion (1) of Lemma 2.4 is obvious.

(2) We first show that G t ( t , s ) < 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq49_HTML.gif, t , s [ 0 , 1 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq48_HTML.gif, s t http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq50_HTML.gif.

In fact, if 0 s < t < 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq57_HTML.gif, then by (2.5) we have
G 1 t = [ ( t s ) α 2 ( 1 s ) α 2 ] Γ ( α 1 ) ( 1 t ) ( 1 s ) α 3 ( δ 1 ) Γ ( α 2 ) ( 1 t ) [ 1 + δ + ( δ 1 ) t ] 2 ( γ 1 ) ( δ 1 ) Γ ( α 3 ) ( 1 s ) α 4 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_Equ22_HTML.gif
(2.20)

Owing to the fact that 3 < α 4 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq2_HTML.gif and 0 s < t < 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq57_HTML.gif, we have that ( t s ) α 2 < ( 1 s ) α 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq58_HTML.gif. Thus, we immediately obtain that G 1 t < 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq59_HTML.gif for 0 s < t < 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq57_HTML.gif from (2.20) together with the condition δ > 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq3_HTML.gif, γ > 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq4_HTML.gif.

Similarly, we can deduce that
G 2 t = ( 1 s ) α 2 Γ ( α 1 ) ( 1 t ) ( 1 s ) α 3 ( δ 1 ) Γ ( α 2 ) ( 1 t ) [ 1 + δ + ( δ 1 ) t ] 2 ( γ 1 ) ( δ 1 ) Γ ( α 3 ) ( 1 s ) α 4 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_Equ23_HTML.gif
(2.21)

for 0 t < s < 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq60_HTML.gif, and so G 2 t < 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq61_HTML.gif for 0 t < s < 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq60_HTML.gif.

To summarize, G t < 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq62_HTML.gif for all s , t [ 0 , 1 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq63_HTML.gif with s t http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq50_HTML.gif.

Now, since G ( 1 , s ) = G 1 ( 1 , s ) = 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq64_HTML.gif for 0 s < 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq65_HTML.gif, and G t < 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq62_HTML.gif for t , s [ 0 , 1 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq66_HTML.gif with t s http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq67_HTML.gif, it follows that G ( t , s ) > 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq47_HTML.gif for all t , s [ 0 , 1 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq48_HTML.gif.

(3) The proof is divided into four steps.

Step 1. We show that G ( t , s ) w 1 ( s ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq51_HTML.gif for t , s [ 0 , 1 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq48_HTML.gif.

(i) If 0 s t < 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq68_HTML.gif, then by (2.5) and the assumption that 3 < α 4 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq2_HTML.gif, γ > 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq4_HTML.gif, and δ > 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq3_HTML.gif, we have
G 1 ( t , s ) ( 1 t ) ( 1 s ) α 2 Γ ( α 1 ) + ( 1 t ) 2 ( 1 s ) α 3 2 ( δ 1 ) Γ ( α 2 ) + ( 1 t ) 2 [ 1 + 2 δ + ( δ 1 ) t ] 6 ( γ 1 ) ( δ 1 ) Γ ( α 3 ) ( 1 s ) α 4 < ( 1 s ) α 2 Γ ( α 1 ) + ( 1 s ) α 3 2 ( δ 1 ) Γ ( α 2 ) + δ ( 1 s ) α 4 2 ( γ 1 ) ( δ 1 ) Γ ( α 3 ) = Δ w 1 ( s ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_Equ24_HTML.gif
(2.22)
(ii) If 0 t s < 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq69_HTML.gif, then by an argument similar to (2.22), we have
G 2 ( t , s ) w 1 ( s ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_Equo_HTML.gif
Summing up the above analysis (i)-(ii), we obtain
G ( t , s ) w 1 ( s ) for  t , s [ 0 , 1 ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_Equp_HTML.gif

Step 2. We show that G ( t , s ) η 1 w 1 ( s ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq70_HTML.gif for t [ 1 6 , 1 3 ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq53_HTML.gif and s [ 0 , 1 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq54_HTML.gif.

In fact, if 0 s t http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq71_HTML.gif and t [ 1 6 , 1 3 ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq53_HTML.gif, then by (2.5) combined with the assumption that 3 < α 4 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq2_HTML.gif, γ > 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq4_HTML.gif, δ > 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq3_HTML.gif, we have
G 1 ( t , s ) ( 1 s ) α 1 Γ ( α ) + ( 1 t ) ( 1 s ) α 2 Γ ( α 1 ) + ( 1 t ) 2 ( 1 s ) α 3 2 ( δ 1 ) Γ ( α 2 ) + ( 1 t ) 2 [ 1 + 2 δ + ( δ 1 ) t ] 6 ( γ 1 ) ( δ 1 ) Γ ( α 3 ) ( 1 s ) α 4 ( 1 s ) α 2 Γ ( α ) + 2 ( 1 s ) α 2 3 Γ ( α 1 ) + 2 ( 1 s ) α 3 9 ( δ 1 ) Γ ( α 2 ) + 2 ( 1 + 2 δ ) ( 1 s ) α 4 27 ( γ 1 ) ( δ 1 ) Γ ( α 3 ) = ( 2 α 5 ) 3 ( α 1 ) ( 1 s ) α 2 Γ ( α 1 ) + 4 9 ( 1 s ) α 3 2 ( δ 1 ) Γ ( α 2 ) + 4 ( 1 + 2 δ ) 27 δ δ ( 1 s ) α 4 2 ( γ 1 ) ( δ 1 ) Γ ( α 3 ) η 1 w 1 ( s ) , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_Equ25_HTML.gif
(2.23)

where η 1 = min { 2 α 5 3 ( α 1 ) , 4 9 , 4 ( 1 + 2 δ ) 27 δ } = min { 2 α 5 3 ( α 1 ) , 4 ( 1 + 2 δ ) 27 δ } http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq72_HTML.gif, because 2 α 5 3 ( α 1 ) < 4 9 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq73_HTML.gif.

If t [ 1 6 , 1 3 ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq53_HTML.gif with t s < 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq74_HTML.gif, then by an argument similar to (2.23), we have
G 2 ( t , s ) η 1 w 1 ( s ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_Equ26_HTML.gif
(2.24)

So, by (2.23)-(2.24), we have

G ( t , s ) η 1 w 1 ( s ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq52_HTML.gif for s [ 0 , 1 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq54_HTML.gif, and t [ 1 6 , 1 3 ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq53_HTML.gif.

Step 3. Now, we show that
0 < G ( t , s ) t w 2 ( s ) for  s , t [ 0 , 1 ) , s t . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_Equq_HTML.gif
(i) If 0 s < t < 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq57_HTML.gif, then by (2.20) and keeping in mind that 3 < α 4 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq2_HTML.gif, γ > 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq4_HTML.gif, δ > 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq3_HTML.gif, it follows that
G 1 t = ( t s ) α 2 Γ ( α 1 ) + ( 1 s ) α 2 Γ ( α 1 ) + ( 1 t ) ( 1 s ) α 3 ( δ 1 ) Γ ( α 2 ) + ( 1 t ) [ 1 + δ + ( δ 1 ) t ] 2 ( γ 1 ) ( δ 1 ) Γ ( α 3 ) ( 1 s ) α 4 ( 1 s ) α 2 Γ ( α 1 ) + ( 1 s ) α 3 ( δ 1 ) Γ ( α 2 ) + δ ( 1 s ) α 4 ( γ 1 ) ( δ 1 ) Γ ( α 3 ) ( 1 s ) α 3 Γ ( α 1 ) + ( 1 s ) α 3 ( δ 1 ) Γ ( α 2 ) + δ ( 1 s ) α 4 ( γ 1 ) ( δ 1 ) Γ ( α 3 ) = ( α + δ 3 ) ( 1 s ) α 3 ( α 2 ) ( δ 1 ) Γ ( α 2 ) + δ ( 1 s ) α 4 ( γ 1 ) ( δ 1 ) Γ ( α 3 ) = Δ w 2 ( s ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_Equ27_HTML.gif
(2.25)
(ii) If 0 t < s < 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq60_HTML.gif, then by an argument similar to (2.25), from (2.21), we have
G 2 ( t , s ) t = ( 1 s ) α 2 Γ ( α 1 ) + ( 1 t ) ( 1 s ) α 3 ( δ 1 ) Γ ( α 2 ) + ( 1 t ) [ 1 + δ + ( δ 1 ) t ] 2 ( γ 1 ) ( δ 1 ) Γ ( α 3 ) ( 1 s ) α 4 w 2 ( s ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_Equr_HTML.gif
Summing up the above analysis (i)-(ii), and noting Step 2 of the proof as before, it follows that
0 < G ( t , s ) t w 2 ( s ) for  s , t [ 0 , 1 ) , s t . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_Equs_HTML.gif
Step 4. It remains to show that
G ( t , s ) t η 2 w 2 ( s ) for  t [ 1 6 , 1 3 ] , s [ 0 , 1 )  with  s t . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_Equt_HTML.gif
(i) If 0 s < t < 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq57_HTML.gif, then by (2.20) and the fact that ( 1 s ) α 2 > ( t s ) α 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq75_HTML.gif, we know that the relations
G 1 t > ( 1 t ) ( 1 s ) α 3 ( δ 1 ) Γ ( α 2 ) + ( 1 t ) [ 1 + δ + ( δ 1 ) t ] 2 ( γ 1 ) ( δ 1 ) Γ ( α 3 ) ( 1 s ) α 4 2 3 ( 1 s ) α 3 ( δ 1 ) Γ ( α 2 ) + ( 1 + δ ) ( 1 s ) α 4 3 ( γ 1 ) ( δ 1 ) Γ ( α 3 ) = 2 ( α 2 ) 3 ( α + δ 3 ) ( α + δ 3 ) ( 1 s ) α 3 ( α 2 ) ( δ 1 ) Γ ( α 2 ) + 1 + δ 3 δ δ ( 1 s ) α 4 ( γ 1 ) ( δ 1 ) Γ ( α 3 ) η 2 w 2 ( s ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_Equu_HTML.gif
hold for t [ 1 6 , 1 3 ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq76_HTML.gif, s [ 0 , 1 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq54_HTML.gif with s < t http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq77_HTML.gif, where η 2 = min { 2 ( α 2 ) 3 ( α + δ 3 ) , 1 + δ 3 δ } http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq78_HTML.gif, and
w 2 ( s ) = ( α + δ 3 ) ( 1 s ) α 3 ( α 2 ) ( δ 1 ) Γ ( α 2 ) + δ ( 1 s ) α 4 ( γ 1 ) ( δ 1 ) Γ ( α 3 ) , s [ 0 , 1 ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_Equv_HTML.gif
Similarly, we can obtain that
G 2 t η 2 w 2 ( s ) for  t [ 1 6 , 1 3 ]  with  t < s < 1 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_Equw_HTML.gif

The proof is complete. □

Now, we introduce a cone as follows:
P = { u C 1 [ I , R ] : u ( t ) 0 , u ( t ) 0 , t I ; u ( 1 ) = 0 ; u ( t ) η 1 u 0 , u ( t ) η 2 u 0 , t [ 1 6 , 1 3 ] } . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_Equx_HTML.gif

It is easy to check that the above set P is a cone in the space C 1 [ I , R ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq79_HTML.gif, which will be used in the sequel.

We define an operator T on P as follows:
T u = 0 1 G ( t , s ) f ( s , u ( s ) , u ( s ) ) d s , u P . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_Equ28_HTML.gif
(2.26)
Obviously, under the assumption ( H 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq30_HTML.gif)-( H 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq32_HTML.gif), the operator T is well defined. Moreover,
( T u ) ( t ) = 0 1 G t ( t , s ) f ( s , u ( s ) , u ( s ) ) d s , t I , u P , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_Equ29_HTML.gif
(2.27)

where G t ( t , s ) = t G 1 ( t , s ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq80_HTML.gif, 0 s < t < 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq57_HTML.gif, G t ( t , s ) = t G 2 ( t , s ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq81_HTML.gif, 0 t < s < 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq60_HTML.gif, and t G 1 ( t , s ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq82_HTML.gif, t G 2 ( t , s ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq83_HTML.gif are given by (2.20)-(2.21), respectively.

A function u C 3 [ 0 , 1 ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq84_HTML.gif is a positive solution of BVP (1.1)-(1.2) if u ( t ) 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq85_HTML.gif, t I http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq37_HTML.gif, u 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq86_HTML.gif, and u satisfies BVP (1.1)-(1.2).

By Lemma 2.3, it is easy to know that a function u C 3 [ 0 , 1 ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq84_HTML.gif is a positive solution of BVP (1.1)-(1.2) iff u P http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq87_HTML.gif is a nonzero fixed point of T. So, we can focus on seeking the existence of a nonzero fixed point of T in P.

Finally, for the remainder of this section, we give the following two theorems, which are fundamental in the proof of our main results.

Let X be a Banach space, and let P X http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq88_HTML.gif be a cone. Suppose that the functions α, β satisfy the following condition:

(D) α , β : X R + http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq89_HTML.gif are continuous convex functionals satisfying α ( λ u ) = | λ | α ( u ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq90_HTML.gif, β ( λ u ) = | λ | ( u ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq91_HTML.gif for u X http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq92_HTML.gif, λ R http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq93_HTML.gif; u k max { α ( u ) , β ( u ) } http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq94_HTML.gif for u X http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq92_HTML.gif, and α ( u 1 ) α ( u 2 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq95_HTML.gif for u 1 , u 2 P http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq96_HTML.gif with u 1 u 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq97_HTML.gif, where k > 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq98_HTML.gif is a constant.

Lemma 2.5 [31]

Assume that r 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq99_HTML.gif, r 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq100_HTML.gif, L are constants with r 2 > r 1 > 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq101_HTML.gif, L > 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq102_HTML.gif, and
Ω i = { u X : α ( u ) < r i , β ( u ) < L } , i = 1 , 2 are two bounded open sets in X . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_Equy_HTML.gif

Set D i = { u X : α ( u ) = r i } http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq103_HTML.gif. Suppose that T : P P http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq104_HTML.gif is a completely continuous operator satisfying

( C 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq105_HTML.gif) α ( T u ) < r 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq106_HTML.gif, u D 1 P http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq107_HTML.gif; α ( T u ) > r 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq108_HTML.gif, u D 2 P http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq109_HTML.gif;

( C 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq110_HTML.gif) β ( T u ) < L http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq111_HTML.gif, u P http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq87_HTML.gif;

( C 3 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq112_HTML.gif) there is a p ( Ω 2 P ) { 0 } http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq113_HTML.gif such that α ( p ) 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq114_HTML.gif and α ( u + λ p ) α u http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq115_HTML.gif for all u P http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq87_HTML.gif and λ 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq116_HTML.gif.

Then T has at least one fixed point in ( Ω 2 Ω ¯ 1 ) P http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq117_HTML.gif.

Lemma 2.6 [32]

Assume that Ω 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq118_HTML.gif, Ω 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq119_HTML.gif are two open subsets of X with 0 Ω 1 Ω ¯ 1 Ω 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq120_HTML.gif, and let T : P P http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq104_HTML.gif be a completely continuous operator such that either

(i) T u u http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq121_HTML.gif, u P Ω 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq122_HTML.gif; T u u http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq123_HTML.gif, u P Ω 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq124_HTML.gif, or

(ii) T u u http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq123_HTML.gif, u P Ω 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq122_HTML.gif; T u u http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq125_HTML.gif, u P Ω 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq124_HTML.gif.

Then T has a fixed point in P ( Ω ¯ 1 Ω 1 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq126_HTML.gif.

3 Main results

We first prove the following lemma to obtain our main results.

Lemma 3.1 Suppose that ( H 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq30_HTML.gif)-( H 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq32_HTML.gif) hold. Then the operator T defined by (2.26) maps P into P, and T is completely continuous.

Proof It is well known that the norms u 1 = u 0 + u 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq127_HTML.gif and u 2 = max { u 0 , u 0 } http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq128_HTML.gif are equivalent on C 1 [ 0 , 1 ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq129_HTML.gif. So, we can consider that the Banach space C 1 [ 0 , 1 ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq129_HTML.gif is equipped with the norm u 2 = max { u 0 , u 0 } http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq130_HTML.gif in the following proof.

For any u P http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq87_HTML.gif, in view of the conclusion (1)-(2) of Lemma 2.4 and the hypotheses ( H 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq131_HTML.gif)-( H 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq32_HTML.gif), it is easy to see that T u C 1 ( I , R ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq132_HTML.gif, ( T u ) ( t ) 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq133_HTML.gif, t I http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq37_HTML.gif, and ( T u ) ( t ) 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq134_HTML.gif, t I http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq37_HTML.gif observing (2.26)-(2.27). Moreover, the conclusion (3) of Lemma 2.4 implies that
( T u ) ( t ) = 0 1 G ( t , s ) f ( s , u ( s ) , u ( s ) ) d s 0 1 w 1 ( s ) f ( s , u ( s ) , u ( s ) ) d s , t I , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_Equ30_HTML.gif
(3.1)
and
( T u ) ( t ) = 0 1 G ( t , s ) f ( s , u ( s ) , u ( s ) ) d s η 1 0 1 w 1 ( s ) f ( s , u ( s ) , u ( s ) ) d s http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_Equ31_HTML.gif
(3.2)

for t [ 1 6 , 1 3 ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq76_HTML.gif.

From (3.1)-(3.2), it follows that ( T u ) ( t ) η 1 ( T u ) ( τ ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq135_HTML.gif, t [ 1 6 , 1 3 ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq76_HTML.gif, τ I http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq136_HTML.gif. Thus,
( T u ) ( t ) η 1 T u 0 , t [ 1 6 , 1 3 ] . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_Equz_HTML.gif
Similarly, we can obtain
( T u ) ( t ) η 2 ( T u ) , t [ 1 6 , 1 3 ] . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_Equaa_HTML.gif

In addition, ( T u ) ( 1 ) = 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq137_HTML.gif. Thus, T : P P http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq104_HTML.gif.

Now, we show that the operator T is compact on P.

In fact, let U be an arbitrary bounded set in P. Then there exists a positive number L such that u 2 L http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq138_HTML.gif for all u U http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq139_HTML.gif, and so M > 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq140_HTML.gif such that 0 f ( t , u ( s ) , u ( s ) ) M http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq141_HTML.gif, t I http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq37_HTML.gif for all u U http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq139_HTML.gif.

In terms of Lemma 2.4, it follows from (2.26)-(2.27) that
0 ( T u ) ( t ) = 0 1 G ( t , s ) f ( s , u ( s ) , u ( s ) ) d s M 0 1 w 1 ( s ) d s , t I , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_Equ32_HTML.gif
(3.3)
0 ( T u ) ( t ) = 0 1 ( G t ( t , s ) f ( s , u ( s ) , u ( s ) ) d s M 0 1 w 2 ( s ) d s , t I . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_Equ33_HTML.gif
(3.4)

Because the functions w 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq142_HTML.gif and w 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq143_HTML.gif are integrable on I, the formulae (3.3)-(3.4) yield that T u 0 M 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq144_HTML.gif, ( T u ) 0 M 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq145_HTML.gif, where M 0 = M max { 0 1 w 1 ( s ) d s , 0 1 w 2 ( s ) d s } http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq146_HTML.gif. So, T u 1 M 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq147_HTML.gif. That is, TU is uniformly bounded.

On the other hand, for any t 1 , t 2 I http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq148_HTML.gif with t 1 < t 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq149_HTML.gif, by setting h = f ( t , u , u ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq150_HTML.gif, the formula (2.26) implies that
| ( T u ) ( t 2 ) ( T u ) ( t 1 ) | = | 0 t 1 ( G 1 ( t 2 , s ) G 1 ( t 1 , s ) ) h ( s ) d s + t 1 t 2 ( G 1 ( t 2 , s ) G 2 ( t 1 , s ) ) h ( s ) d s + t 2 1 ( G 2 ( t 2 , s ) G 2 ( t 1 , s ) ) h ( s ) d s | M 0 t 1 | G 1 ( t 2 , s ) G 1 ( t 1 , s ) | d s + 2 M t 1 t 2 w 1 ( s ) d s + M t 2 1 | G 2 ( t 2 , s ) G 2 ( t 1 , s ) | d s . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_Equ34_HTML.gif
(3.5)
According to (2.4)-(2.5) and by applying the mean value theorem, we have
| G 1 ( t 2 , s ) G 1 ( t 1 , s ) | [ 2 Γ ( α 1 ) + 1 ( δ 1 ) Γ ( α 2 ) + δ ( 1 s ) α 4 ( γ 1 ) ( δ 1 ) Γ ( α 3 ) ] | t 2 t 1 | , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_Equab_HTML.gif
and so
0 t 1 | G 1 ( t 2 , s ) G 1 ( t 1 , s ) | d s N 1 | t 2 t 1 | , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_Equ35_HTML.gif
(3.6)

where N 1 = 2 Γ ( α 1 ) + 1 ( δ 1 ) Γ ( α 2 ) + δ ( γ 1 ) ( δ 1 ) Γ ( α 2 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq151_HTML.gif.

Similarly, there is another constant N 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq152_HTML.gif such that
t 2 1 | G 2 ( t 2 , s ) G 2 ( t 1 , s ) | d s N 2 | t 1 t 2 | . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_Equ36_HTML.gif
(3.7)
Again, because the function w 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq142_HTML.gif is integrable on I, the absolute continuity of integral of w 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq153_HTML.gif on [ t 1 , t 2 ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq154_HTML.gif ensures that there exists a constant N 3 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq155_HTML.gif such that
t 1 t 2 w 1 ( s ) d s < N 3 ( t 2 t 1 ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_Equ37_HTML.gif
(3.8)
So, (3.5) together with (3.6)-(3.8) implies that there exists a constant N such that the inequality
| ( T u ) ( t 2 ) ( T u ) ( t 1 ) | N ( t 2 t 1 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_Equac_HTML.gif

holds for any u U http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq139_HTML.gif and t 1 , t 2 I http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq148_HTML.gif with t 1 < t 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq149_HTML.gif. That is, the set TU is equicontinuous.

Similarly, we can deduce that the set { ( T u ) | u U } http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq156_HTML.gif is also equicontinuous in terms of (2.27).

So, as a consequence of the Arzelà-Ascoli theorem, we have that TU is a compact set.

Now, we come to prove the operator T is continuous on P.

Let { u n } http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq157_HTML.gif be an arbitrary sequence in P with u n u 0 P http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq158_HTML.gif. Then there exists an L > 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq102_HTML.gif such that
u n ( t ) [ 0 , L ] , u n ( t ) [ L , 0 ] , t I , n = 0 , 1 , 2 , . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_Equad_HTML.gif
According to the uniform continuity of f on I × [ 0 , L ] × [ L , 0 ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq159_HTML.gif, for an arbitrary number ε > 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq160_HTML.gif, there is a number N 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq161_HTML.gif such that
| f ( t , u n ( t ) , u n ( t ) ) f ( t , u 0 ( t ) , u 0 ( t ) ) | < ε http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_Equ38_HTML.gif
(3.9)

for all t I http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq37_HTML.gif, whenever n N http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq162_HTML.gif.

Thus, in view of Lemma 2.4, from (2.26)-(2.27) and (3.9), it follows that
| ( T u n ) ( t ) ( T u 0 ) ( t ) | ε 0 1 w 1 ( s ) d s , t I , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_Equae_HTML.gif
and
| ( T u n ) ( t ) ( T u ) ( t ) | ε 0 1 w 2 ( s ) d s , t I , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_Equaf_HTML.gif

whenever n N http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq162_HTML.gif. That is, T is continuous on P. □

We are now in a position to state and prove the first theorem in the article. Let constants k 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq163_HTML.gif, k 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq164_HTML.gif satisfy k 1 ( 0 , ( 0 1 w 2 ( s ) d s ) 1 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq165_HTML.gif, k 2 > ( η 2 1 6 1 3 w 2 ( s ) d s ) 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq166_HTML.gif.

Theorem 3.1 Suppose that ( H 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq30_HTML.gif)-( H 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq32_HTML.gif) hold. In addition, there are two constants r 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq99_HTML.gif, r 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq100_HTML.gif with 1 6 η 1 η 2 r 2 > r 1 > 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq167_HTML.gif such that f satisfies the following condition:

( H 3 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq168_HTML.gif) f ( t , x , y ) k 1 r 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq169_HTML.gif, for ( t , x , y ) I × [ 0 , r 1 ] × [ r 1 , 0 ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq170_HTML.gif;

f ( t , x , y ) k 2 r 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq171_HTML.gif, for ( t , x , y ) I × [ 1 6 η 1 η 2 r 2 , r 2 ] × [ r 2 , η 2 r 2 ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq172_HTML.gif.

Then BVP (1.1)-(1.2) has at least one positive solution u satisfying r 1 < max t I | u ( t ) | < r 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq173_HTML.gif and max t I | u ( t ) | < r 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq174_HTML.gif.

Proof We already know that T : P P http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq104_HTML.gif is completely continuous by Lemma 3.1.

Let α ( u ) = max t I | u ( t ) | http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq175_HTML.gif, β ( u ) = max t I | u ( t ) | http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq176_HTML.gif for u P http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq87_HTML.gif. It is easy to verify that the functions α, β satisfy the condition (D).

Choose a constant L large enough so that L > max { M 0 1 w 1 ( s ) d s , r 2 } http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq177_HTML.gif, where M = max ( t , x , y ) I × J × Δ f ( t , x , y ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq178_HTML.gif, and J = [ 0 , r 2 ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq179_HTML.gif, Δ = [ r 2 , 0 ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq180_HTML.gif. Set D i = { u C 1 ( I , R ) : α ( u ) = r i } http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq181_HTML.gif, Ω i = { u C 1 ( I , R ) : α ( u ) < r i , β ( u ) < L } http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq182_HTML.gif, i = 1 , 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq183_HTML.gif. Define the function f ¯ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq184_HTML.gif on I × R + × R http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq185_HTML.gif as f ¯ ( t , x , y ) = f ( t , ϕ ( x ) , φ ( y ) ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq186_HTML.gif, ( t , x , y ) I × R + × R http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq187_HTML.gif, where ϕ ( x ) = min { x , r 2 } http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq188_HTML.gif, φ ( y ) = max { y , r 2 } http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq189_HTML.gif.

Consider the following ancillary BVP:
{ D 0 + α C x ( t ) = f ¯ ( t , x ( t ) , x ( t ) ) , t ( 0.1 ) , x ( 1 ) = x ( 1 ) = 0 , δ x ( 0 ) = x ( 1 ) , γ x ( 0 ) = x ( 1 ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_Equ39_HTML.gif
(3.10)

Obviously, the function f ¯ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq184_HTML.gif is continuous on I × R + × R http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq190_HTML.gif according to the continuity of f. Thus, by an argument similar to that in Lemma 3.1, the operator T ¯ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq191_HTML.gif given by T ¯ u = 0 1 G ( t , s ) f ¯ ( s , u ( s ) , u ( s ) ) d s http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq192_HTML.gif is also completely continuous on P and maps P into P.

We will prove that T has at least one nonzero fixed point in P by applying Lemma 2.5. The approach is divided into four steps.

Step 1. We first show that
u 0 u 0 , u ( t ) 1 6 η 1 η 2 u 0 , t [ 1 6 , 1 3 ] for any  u P . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_Equ40_HTML.gif
(3.11)

In fact, for any u P http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq87_HTML.gif, owing to the condition u ( 1 ) = 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq193_HTML.gif, we have that u ( t ) = t 1 u ( s ) d s http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq194_HTML.gif, t I http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq37_HTML.gif, and so u 0 u 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq195_HTML.gif.

On the other hand, applying the mean value theorem, for any u P http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq87_HTML.gif, we have that u 0 | u ( 1 3 ) u ( 1 6 ) | = 1 6 | u ( ξ ) | http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq196_HTML.gif for some ξ ( 1 6 , 1 3 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq197_HTML.gif. Therefore, we have that u 0 1 6 η 2 u 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq198_HTML.gif from the fact that u ( ξ ) η 2 u 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq199_HTML.gif, ξ [ 1 6 , 1 3 ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq200_HTML.gif because u P http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq87_HTML.gif. So,
u ( t ) η 1 u 0 1 6 η 1 η 2 u 0 , t [ 1 6 , 1 3 ] , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_Equag_HTML.gif

keeping in mind that u ( t ) η 1 u 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq201_HTML.gif, t [ 1 6 , 1 3 ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq53_HTML.gif for u P http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq87_HTML.gif.

Step 2. Now, we come to verify that the conditions corresponding to ( C 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq202_HTML.gif) in Lemma 2.5 hold.

For any u P http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq87_HTML.gif with α ( u ) = r 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq203_HTML.gif, we have that u ( t ) 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq85_HTML.gif, u ( t ) 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq204_HTML.gif, t I http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq37_HTML.gif, and u 0 = r 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq205_HTML.gif. Thus, in view of (3.11), we have that 0 u ( t ) r 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq206_HTML.gif, r 1 u ( t ) 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq207_HTML.gif, t I http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq37_HTML.gif. So, according to ( H 3 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq168_HTML.gif), we have
f ¯ ( t , u ( t ) , u ( t ) ) = f ( t , u ( t ) , u ( t ) ) k 1 r 1 , t I . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_Equah_HTML.gif
Thus, from (2.27) and Lemma 2.4, it follows that
( T ¯ u ) ( t ) = 0 1 G t ( t , s ) f ( s , u ( s ) , u ( s ) ) d s k 1 r 1 0 1 ( G t ( t , s ) ) d s k 1 r 1 0 1 w 2 ( s ) d s , t I . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_Equai_HTML.gif

Thus, max t I | ( T ¯ u ) ( t ) | k 1 r 1 0 1 w 2 ( s ) d s < r 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq208_HTML.gif, noting that the assumption k 1 < ( 0 1 w 2 ( s ) d s ) 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq209_HTML.gif. That is, α ( T u ) < r 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq106_HTML.gif.

For any u P http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq210_HTML.gif with α ( u ) = r 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq211_HTML.gif, then u ( t ) 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq85_HTML.gif, u ( t ) 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq212_HTML.gif, t I http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq37_HTML.gif, and max t I | u ( t ) | = r 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq213_HTML.gif. Thus, from (3.11), we obtain
1 6 η 1 η 2 r 2 u ( t ) r 2 , r 2 u ( t ) η 2 r 2 , t [ 1 6 , 1 3 ] , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_Equaj_HTML.gif
and so
f ¯ ( t , u ( t ) , u ( t ) ) = f ( t , u ( t ) , u ( t ) ) k 2 r 2 , t [ 1 6 , 1 3 ] , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_Equak_HTML.gif
from the condition ( H 3 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq168_HTML.gif). Therefore, in view of Lemma 2.4, we have
( T ¯ u ) ( t ) = 0 1 G t ( t , s ) f ( s , u ( s ) , u ( s ) ) d s 1 6 1 3 G t ( t , s ) f ( s , u ( s ) , u ( s ) ) d s k 2 r 2 1 6 1 3 ( G t ( t , s ) ) d s k 2 r 2 η 2 1 6 1 3 w 2 ( s ) d s , t I . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_Equal_HTML.gif

Thus, max t I | T u ( t ) | k 2 η 2 r 2 1 6 1 3 w 2 ( s ) d s > r 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq214_HTML.gif, noting that k 2 > ( η 2 1 6 1 3 w 2 ( s ) d s ) 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq215_HTML.gif. That is, α ( T u ) > r 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq108_HTML.gif.

Step 3. We verify that the conditions corresponding to ( C 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq216_HTML.gif) in Lemma 2.5 hold.

For any u P http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq87_HTML.gif, owing to the fact that 0 ϕ ( u ( t ) ) r 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq217_HTML.gif, r 2 φ ( u ( t ) ) 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq218_HTML.gif, from the meaning of M, we have immediately that
f ¯ ( t , u ( t ) , u ( t ) ) = f ( t , ϕ ( u ( t ) ) , φ ( u ( t ) ) ) M , t I . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_Equam_HTML.gif
Thus,
( T ¯ u ) ( t ) = 0 1 G ( t , s ) f ¯ ( s , u ( s ) , u ( s ) ) d s M 0 1 G ( t , s ) d s M 0 1 w 1 ( s ) d s , t I . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_Equan_HTML.gif

Hence, ( T ¯ u ) 0 M 0 1 w 1 ( s ) d s http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq219_HTML.gif, and so ( T ¯ u ) 0 < L http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq220_HTML.gif from the choice of L. Thus, β ( T ¯ u ) < L http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq221_HTML.gif.

Step 4. Finally, take p = σ 0 1 G ( t , s ) d s http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq222_HTML.gif with 0 < σ < r 2 ( 0 1 w 2 ( s ) d s ) 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq223_HTML.gif. Then, by an argument similar to that in Lemma 3.1, we can know that p P http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq224_HTML.gif. Moreover, 0 p ( t ) = σ 0 1 ( G t ( t , s ) ) d s σ 0 1 w 2 ( s ) d s < r 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq225_HTML.gif, t I http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq37_HTML.gif, and 0 p ( t ) < r 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq226_HTML.gif, t I http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq37_HTML.gif from (3.11). Thus, p ( Ω 2 P ) { 0 } http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq227_HTML.gif, α ( p ) 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq114_HTML.gif. Again,
α ( u + λ p ) = max t I | u + λ p ( t ) | = max ( u ( t ) + λ ( p ( t ) ) λ max ( u ( t ) ) = λ α ( u ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_Equao_HTML.gif

for any u P http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq87_HTML.gif, λ 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq116_HTML.gif.

So, the conditions corresponding to ( C 3 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq112_HTML.gif) in Lemma 2.5 hold.

Summing up the above steps 1-4 and applying Lemma 2.5, we obtain that BVP (3.10) has at least one positive solution u ( Ω 2 Ω ¯ 1 ) P http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq228_HTML.gif. That is, r 1 < u 0 < r 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq229_HTML.gif. u 0 < L http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq230_HTML.gif, and so u 0 < r 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq231_HTML.gif from the fact that r 2 < L http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq232_HTML.gif and u 0 u 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq233_HTML.gif by (3.13). Thus, 0 u ( t ) < r 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq234_HTML.gif, t I http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq37_HTML.gif, 0 u ( t ) < r 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq235_HTML.gif, t I http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq37_HTML.gif, and u 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq236_HTML.gif. Hence, f ¯ ( t , u ( t ) , u ( t ) ) = f ( t , u ( t ) , u ( t ) ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq237_HTML.gif, t I http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq37_HTML.gif, and so u is a positive solution of BVP (1.1)-(1.2). The proof is complete. □

Now, we state another theorem in this paper. Let us begin with introducing some notations.

Let B = R + × R http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq238_HTML.gif. Denote f = lim inf x + | y | min t I f ( t , x , y ) x + | y | http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq239_HTML.gif with ( x , y ) B http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq240_HTML.gif, and f 0 = lim sup x + | y | 0 max t I f ( t , x , y ) x + | y | http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq241_HTML.gif with ( x , y ) B http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq242_HTML.gif. Put λ 1 = ( η 0 μ 1 ) 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq243_HTML.gif, λ 2 = μ 2 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq244_HTML.gif, where μ 1 = 1 6 1 3 [ G ( 0 , s ) G t ( 0 , s ) ] d s http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq245_HTML.gif, μ 2 = 0 1 ( w 1 ( s ) + w 2 ( s ) ) d s http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq246_HTML.gif, η 0 = min { η 1 , η 2 } http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq247_HTML.gif, and η i http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq248_HTML.gif ( i = 1 , 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq183_HTML.gif) are given in Lemma 2.4.

Theorem 3.2 Assume that ( H 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq30_HTML.gif)-( H 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq32_HTML.gif) hold. If f > λ 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq249_HTML.gif, f 0 < λ 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq250_HTML.gif, then BVP (1.1)-(1.2) has at least one positive solution.

Proof As described in the proof of Theorem 3.1, T : P P http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq104_HTML.gif is completely continuous. Again, from f > λ 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq249_HTML.gif, it follows that there exists an R 1 > 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq251_HTML.gif such that
f ( t , x , y ) > λ 1 ( x + | y | ) , t I http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_Equ41_HTML.gif
(3.12)

holds when x + | y | R 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq252_HTML.gif with x 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq253_HTML.gif, y 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq254_HTML.gif.

Take R η 0 1 R 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq255_HTML.gif. Set Ω R = { u C 1 [ 0 , 1 ] : u 1 < R } http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq256_HTML.gif. Now, we show that the following relation holds:
T u 1 u 1 , u Ω R P . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_Equ42_HTML.gif
(3.13)
In fact, for any u Ω R P http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq257_HTML.gif, we have that u P http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq87_HTML.gif with u 1 = R http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq258_HTML.gif. Then
u ( t ) 0 , u ( t ) 0 , t I ; u ( t ) η 1 u 0 , u ( t ) η 2 u 0 , t [ 1 6 , 1 3 ] . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_Equap_HTML.gif
Thus,
u ( t ) + | u ( t ) | η 1 u 0 + η 2 u 0 η 0 ( u 0 + u 0 ) = η 0 R R 1 , for  t [ 1 6 , 1 3 ] . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_Equ43_HTML.gif
(3.14)
Thus, from (3.12), (3.14), it follows that
f ( t , u ( t ) , u ( t ) ) > λ 1 η 0 R , t [ 1 6 , 1 3 ] . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_Equ44_HTML.gif
(3.15)
Hence, (2.26) together with (3.15) implies that
( T u ) ( t ) = 0 1 G ( t , s ) f ( s , u ( s ) , u ( s ) ) d s 1 6 1 3 G ( t , s ) f ( s , u ( s ) , u ( s ) ) d s λ 1 η 0 R 1 6 1 3 G ( t , s ) d s , t I . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_Equaq_HTML.gif
So,
T u 0 ( T u ) ( 0 ) λ 1 η 0 R 1 6 1 3 G ( 0 , s ) d s . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_Equ45_HTML.gif
(3.16)
Similarly, we can obtain
( T u ) 0 ( T u ) ( 0 ) λ 1 η 0 R 1 6 1 3 [ G t ( 0 , s ) ] d s . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_Equ46_HTML.gif
(3.17)
Therefore, noting that λ 1 = ( η 0 μ 1 ) 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq243_HTML.gif, from (3.16)-(3.17), we have
T u 1 = T u 0 + ( T u ) 0 λ 1 η 0 R 1 6 1 3 [ G ( 0 , s ) G t ( 0 , s ) ] d s = λ 1 η 0 μ 1 R = R = u 1 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_Equar_HTML.gif

So, the relation (3.13) holds.

Now, from f 0 < λ 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq250_HTML.gif, it follows that there exists an r 2 > 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq259_HTML.gif such that
f ( t , x , y ) < λ 2 ( x + | y | ) , t I , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_Equ47_HTML.gif
(3.18)

whenever x + | y | r 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq260_HTML.gif with x 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq253_HTML.gif, y 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq254_HTML.gif.

Take 0 < r < min { r 2 , R } http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq261_HTML.gif. Set Ω r = { u C 1 [ 0 , 1 ] : u 1 < r } http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq262_HTML.gif. We prove the following relation holds:
T u 1 u 1 , u Ω r P . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_Equ48_HTML.gif
(3.19)
In fact, for any u Ω r P http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq263_HTML.gif, we have u P http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq87_HTML.gif and u 1 = r http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq264_HTML.gif. Thus, u ( t ) 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq85_HTML.gif, u ( t ) 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq204_HTML.gif, t I http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq37_HTML.gif, and u ( t ) + | u ( t ) | r http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq265_HTML.gif, t I http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq37_HTML.gif. So, by (3.18), it follows that
f ( t , u ( t ) , u ( t ) ) < λ 2 ( u ( t ) + | u ( t ) | ) λ 2 r . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_Equas_HTML.gif
Therefore, from (2.26) and in view of Lemma 2.4, we have
( T u ) ( t ) = 0 1 G ( t , s ) f ( s , u ( s ) , u ( s ) ) d s λ 2 r 0 1 G ( t , s ) d s λ 2 r 0 1 w 1 ( s ) d s , t I . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_Equ49_HTML.gif
(3.20)
So,
T u 0 λ 2 r 0 1 w 1 ( s ) d s . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_Equ50_HTML.gif
(3.21)
Similarly, we can obtain
( T u ) 0 λ 2 r 0 1 w 2 ( s ) d s . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_Equ51_HTML.gif
(3.22)
Consequently, noting that λ 2 = μ 2 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq244_HTML.gif, from (3.21)-(3.22), it follows that
T u 1 = T u 0 + ( T u ) 0 λ 2 r 0 1 ( w 1 ( s ) + w 2 ( s ) ) d s = r λ 2 μ 2 = r = u 1 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_Equat_HTML.gif

So, the relation (3.19) holds.

Summing up (3.13) and (3.19), applying Lemma 2.6, the operator T has at least one fixed point u ( Ω ¯ 2 Ω 1 ) P http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq266_HTML.gif. Thus u is a positive solution of BVP (1.1)-(1.2). The proof is complete. □

Example 3.1 Consider the following BVP:
{ D 0 + α C x ( t ) = f ( t , x ( t ) , x ( t ) ) , t ( 0 , 1 ) , x ( 1 ) = x ( 1 ) = 0 , δ x ( 0 ) = x ( 1 ) , γ x ( 0 ) = x ( 1 ) , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_Equ52_HTML.gif
(3.23)
where 3 < α 4 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq2_HTML.gif, γ > 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq4_HTML.gif, δ > 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq3_HTML.gif, and f is given by
f ( t , x , y ) = e t [ a 1 ( sin x ) 6 + a 2 ( y ) 4 ] , ( t , x , y ) I × R + × R , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_Equau_HTML.gif

where constants a 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq267_HTML.gif, a 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq268_HTML.gif are two positive numbers. Then BVP (3.23) has at least one positive solution.

In fact, assume that the notations k 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq163_HTML.gif, k 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq269_HTML.gif, η 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq270_HTML.gif, and η 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq271_HTML.gif are described in Theorem 3.1. Take 0 < r 1 < min { 1 , k 1 e ( a 1 + a 2 ) } http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq272_HTML.gif, r 2 > max { k 2 a 2 η 2 2 , 6 η 1 η 2 } http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq273_HTML.gif. Then the inequality
f ( t , x , y ) e [ a 1 x 3 + a 2 y 2 ] e [ a 1 + a 2 ] r 1 2 k 1 r 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_Equav_HTML.gif
holds for ( t , x , y ) I × [ 0 , r 1 ] × [ r 1 , 0 ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq170_HTML.gif, and the inequality
f ( t , x , y ) a 2 y 2 a 2 η 2 2 r 2 2 k 2 r 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_Equaw_HTML.gif

holds for ( t , x , y ) I × [ 0 , + ) × [ r 2 , η 2 r 2 ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq274_HTML.gif.

So, by Theorem 3.1, BVP (3.23) has at least one positive solution.

Example 3.2 Consider the following BVP:
{ D 0 + α C x ( t ) = f ( t , x ( t ) , x ( t ) ) , t ( 0 , 1 ) , x ( 1 ) = x ( 1 ) = 0 , δ x ( 0 ) = x ( 1 ) , γ x ( 0 ) = x ( 1 ) , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_Equ53_HTML.gif
(3.24)
where 3 < α 4 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq2_HTML.gif, γ > 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq4_HTML.gif, δ > 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq3_HTML.gif, and f is given by
f ( t , x , y ) = e t [ a 1 ( x + | y | ) β + a 2 ( sin x y ) 2 ] , ( t , x , y ) I × R + × R , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_Equax_HTML.gif

where constants a 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq267_HTML.gif, a 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq268_HTML.gif are two positive numbers and a constant β > 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq275_HTML.gif. Then BVP (3.24) has at least one positive solution.

In fact, observing that f = + http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq276_HTML.gif, f 0 = 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-109/MediaObjects/13661_2012_Article_360_IEq277_HTML.gif, the conclusion follows from Theorem 3.2.

Declarations

Acknowledgements

The author sincerely thanks the anonymous referees for their valuable suggestions and comments which have greatly helped improve this article. Article is supported by the Natural Science Foundation of Hubei Provincial Education Department (D20102502).

Authors’ Affiliations

(1)
College of Mathematics and Statistics, Hubei Normal University

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© Chai; licensee Springer. 2013

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