Positive solutions of Sturm-Liouville boundary value problems for singular nonlinear second-order impulsive integro-differential equation in Banach spaces

Boundary Value Problems20122012:86

DOI: 10.1186/1687-2770-2012-86

Received: 5 June 2012

Accepted: 23 July 2012

Published: 6 August 2012

Abstract

In this work, we investigate the existence of positive solutions of Sturm-Liouville boundary value problems for singular nonlinear second-order impulsive integro differential equation in a real Banach space. Some new existence results of positive solutions are established by applying fixed-point index theory together with comparison theorem. Some discussions and an example are given to demonstrate the applications of our main results.

MSC:34B15, 34B25, 45J05.

Keywords

measure of non-comparison positive solution boundary value problem impulsive integro-differential equation

1 Introduction

In this paper, we study the existence of positive solutions to second-order singular nonlinear impulsive integro-differential equation of the form:
{ y + h ( t ) f ( t , y ( t ) , y ( t ) , ( T y ) ( t ) , ( S y ) ( t ) ) = 0 , t J , t t k , Δ y | t = t k = I k ( y ( t k ) ) , k = 1 , , m , Δ y | t = t k = I ¯ k ( y ( t k ) , y ( t k ) ) , k = 1 , , m , α y ( 0 ) β y ( 0 ) = 0 , γ y ( 1 ) + δ y ( 1 ) = 0 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-86/MediaObjects/13661_2012_Article_185_Equ1_HTML.gif
(1.1)
where α , β , γ , δ 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-86/MediaObjects/13661_2012_Article_185_IEq1_HTML.gif, ρ = β γ + α γ + α δ > 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-86/MediaObjects/13661_2012_Article_185_IEq2_HTML.gif, I = [ 0 , 1 ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-86/MediaObjects/13661_2012_Article_185_IEq3_HTML.gif, J = ( 0 , 1 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-86/MediaObjects/13661_2012_Article_185_IEq4_HTML.gif, 0 < t 1 < t 2 < < t m < 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-86/MediaObjects/13661_2012_Article_185_IEq5_HTML.gif, J = J { t 1 , t 2 , , t m } http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-86/MediaObjects/13661_2012_Article_185_IEq6_HTML.gif, J ¯ = [ 0 , 1 ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-86/MediaObjects/13661_2012_Article_185_IEq7_HTML.gif, J 0 = ( 0 , t 1 ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-86/MediaObjects/13661_2012_Article_185_IEq8_HTML.gif, J k = ( t k , t k + 1 ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-86/MediaObjects/13661_2012_Article_185_IEq9_HTML.gif, k = 1 , , m 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-86/MediaObjects/13661_2012_Article_185_IEq10_HTML.gif, J m = ( t m , 1 ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-86/MediaObjects/13661_2012_Article_185_IEq11_HTML.gif, f C [ J × P × P × P × P , P ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-86/MediaObjects/13661_2012_Article_185_IEq12_HTML.gif, and P is a positive cone in E. θ is a zero element of E, I k C [ P , P ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-86/MediaObjects/13661_2012_Article_185_IEq13_HTML.gif, I ¯ k C [ P , P ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-86/MediaObjects/13661_2012_Article_185_IEq14_HTML.gif, and
( T y ) ( t ) = 0 t K ( t , s ) y ( s ) d s , ( S y ) ( t ) = 0 1 H ( t , s ) y ( s ) d s , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-86/MediaObjects/13661_2012_Article_185_Equ2_HTML.gif
(1.2)
in which K C [ D , J ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-86/MediaObjects/13661_2012_Article_185_IEq15_HTML.gif, D = { ( t , s ) J × J : t s } http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-86/MediaObjects/13661_2012_Article_185_IEq16_HTML.gif, H C [ J × J , J ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-86/MediaObjects/13661_2012_Article_185_IEq17_HTML.gif, and K 0 = max { K ( t , s ) : ( t , s ) D } http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-86/MediaObjects/13661_2012_Article_185_IEq18_HTML.gif, H 0 = max { H ( t , s ) : ( t , s ) D } http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-86/MediaObjects/13661_2012_Article_185_IEq19_HTML.gif. Δ y | t = t k http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-86/MediaObjects/13661_2012_Article_185_IEq20_HTML.gif and Δ y | t = t k http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-86/MediaObjects/13661_2012_Article_185_IEq21_HTML.gif denote the jump of y ( t ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-86/MediaObjects/13661_2012_Article_185_IEq22_HTML.gif and y ( t ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-86/MediaObjects/13661_2012_Article_185_IEq23_HTML.gif at t = t k http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-86/MediaObjects/13661_2012_Article_185_IEq24_HTML.gif, i.e.,
Δ y | t = t k = y ( t k + ) y ( t k ) , Δ y | t = t k = y ( t k + ) y ( t k ) , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-86/MediaObjects/13661_2012_Article_185_Equa_HTML.gif

where y ( t k + ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-86/MediaObjects/13661_2012_Article_185_IEq25_HTML.gif, y ( t k + ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-86/MediaObjects/13661_2012_Article_185_IEq26_HTML.gif and y ( t k ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-86/MediaObjects/13661_2012_Article_185_IEq27_HTML.gif, y ( t k ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-86/MediaObjects/13661_2012_Article_185_IEq28_HTML.gif represent the right-hand limit and left-hand limit of y ( t ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-86/MediaObjects/13661_2012_Article_185_IEq22_HTML.gif and y ( t ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-86/MediaObjects/13661_2012_Article_185_IEq23_HTML.gif at t = t k http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-86/MediaObjects/13661_2012_Article_185_IEq24_HTML.gif, respectively. h C ( J , R + ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-86/MediaObjects/13661_2012_Article_185_IEq29_HTML.gif and may be singular at t = 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-86/MediaObjects/13661_2012_Article_185_IEq30_HTML.gif and/or t = 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-86/MediaObjects/13661_2012_Article_185_IEq31_HTML.gif.

Boundary value problems for impulsive differential equations arise from many nonlinear problems in sciences, such as physics, population dynamics, biotechnology, and economics etc. (see [1, 2, 414, 1618]). As it is well known that impulsive differential equations contain jumps and/or impulses which are main characteristic feature in computational biology. Over the past 15 years, a significant advance has been achieved in theory of impulsive differential equations. However, the corresponding theory of impulsive integro-differential equations in Banach spaces does not develop rapidly. Recently, Guo [58] established the existence of a solution, multiple solutions and extremal solutions for nonlinear impulsive integro-differential equations with nonsingular argument in Banach spaces. The main tools of Guo [58] are the Schauder fixed-point theorem, fixed-point index theory, upper and lower solutions together with the monotone iterative technique, respectively. The conditions of the Kuratowski measure of non-compactness in Guo [58] play an important role in the proof of the results. But all kinds of compactness type conditions is difficult to verify in abstract spaces. As a result, it is an interesting and important problem to remove or weak compactness type conditions.

Inspired and motivated greatly by the above works, the aim of the paper is to consider the existence of positive solutions for the boundary value problem (1.1) under simpler conditions. The main results of problem (1.1) are obtained by making use of fixed-point index theory and fixed-point theorem. More specifically, in the proof of these theorems, we construct a special cone for strict set contraction operator. Our main results in essence improve and generalize the corresponding results of Guo [58]. Moreover, our method is different from those in Guo [58].

The rest of the paper is organized as follows: In Section 2, we present some known results and introduce conditions to be used in the next section. The main theorem formulated and proved in Section 3. Finally, in Section 4, some discussions and an example for singular nonlinear integro-differential equations are presented to demonstrate the application of the main results.

2 Preliminaries and lemmas

In this section, we shall state some necessary definitions and preliminaries results.

Definition 2.1 Let E be a real Banach space. A nonempty closed set P E http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-86/MediaObjects/13661_2012_Article_185_IEq32_HTML.gif is called a cone if it satisfies the following two conditions:
  1. (1)

    x P http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-86/MediaObjects/13661_2012_Article_185_IEq33_HTML.gif, λ > 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-86/MediaObjects/13661_2012_Article_185_IEq34_HTML.gif implies λ x P http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-86/MediaObjects/13661_2012_Article_185_IEq35_HTML.gif;

     
  2. (2)

    x P http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-86/MediaObjects/13661_2012_Article_185_IEq33_HTML.gif, x P http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-86/MediaObjects/13661_2012_Article_185_IEq36_HTML.gif implies x = 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-86/MediaObjects/13661_2012_Article_185_IEq37_HTML.gif.

     

A cone is said solid if it contains interior points, P θ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-86/MediaObjects/13661_2012_Article_185_IEq38_HTML.gif. A cone P is called to be generating if E = P P http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-86/MediaObjects/13661_2012_Article_185_IEq39_HTML.gif, i.e., every element y E http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-86/MediaObjects/13661_2012_Article_185_IEq40_HTML.gif can be represented in the form y = x z http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-86/MediaObjects/13661_2012_Article_185_IEq41_HTML.gif, where x , z P http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-86/MediaObjects/13661_2012_Article_185_IEq42_HTML.gif. A cone P in E induces a partial ordering in E given by u v http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-86/MediaObjects/13661_2012_Article_185_IEq43_HTML.gif if v u P http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-86/MediaObjects/13661_2012_Article_185_IEq44_HTML.gif. If u v http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-86/MediaObjects/13661_2012_Article_185_IEq43_HTML.gif and u v http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-86/MediaObjects/13661_2012_Article_185_IEq45_HTML.gif, we write u < v http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-86/MediaObjects/13661_2012_Article_185_IEq46_HTML.gif; if cone P is solid and v u P http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-86/MediaObjects/13661_2012_Article_185_IEq47_HTML.gif, we write u v http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-86/MediaObjects/13661_2012_Article_185_IEq48_HTML.gif.

Definition 2.2 A cone P E http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-86/MediaObjects/13661_2012_Article_185_IEq32_HTML.gif is said to be normal if there exists a positive constant N such that x + y N http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-86/MediaObjects/13661_2012_Article_185_IEq49_HTML.gif, x , y P http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-86/MediaObjects/13661_2012_Article_185_IEq50_HTML.gif, x = 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-86/MediaObjects/13661_2012_Article_185_IEq51_HTML.gif, y = 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-86/MediaObjects/13661_2012_Article_185_IEq52_HTML.gif.

Definition 2.3 Let E be a metric space and S be a bounded subset of E. The measure of non-compactness ϒ ( S ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-86/MediaObjects/13661_2012_Article_185_IEq53_HTML.gif of S is defined by
ϒ ( S ) = inf { δ > 0 : S  admits a finite cover by sets of diameter  δ } . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-86/MediaObjects/13661_2012_Article_185_Equb_HTML.gif

Definition 2.4 An operator B : D E http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-86/MediaObjects/13661_2012_Article_185_IEq54_HTML.gif is said to be completely continuous if it is continuous and compact. B is called a k-set-contraction ( k 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-86/MediaObjects/13661_2012_Article_185_IEq55_HTML.gif) if it is continuous, bounded and ϒ ( B ( S ) ) k ϒ ( S ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-86/MediaObjects/13661_2012_Article_185_IEq56_HTML.gif for any bounded set S D http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-86/MediaObjects/13661_2012_Article_185_IEq57_HTML.gif, where ϒ ( S ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-86/MediaObjects/13661_2012_Article_185_IEq53_HTML.gif denotes the measure of noncompactness of S.

A k-set-contraction is called a strict-set contraction if k < 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-86/MediaObjects/13661_2012_Article_185_IEq58_HTML.gif. An operator B is said to be condensing if it is continuous, bounded, and ϒ ( B ( S ) ) < ϒ ( S ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-86/MediaObjects/13661_2012_Article_185_IEq59_HTML.gif for any bounded set S D http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-86/MediaObjects/13661_2012_Article_185_IEq60_HTML.gif with ϒ ( S ) > 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-86/MediaObjects/13661_2012_Article_185_IEq61_HTML.gif.

Obviously, if B is a strict-set contraction, then B is a condensing mapping, and if operator B is completely continuous, then B is a strict-set contraction.

It is well known that y C 2 ( 0 , 1 ) C [ 0 , 1 ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-86/MediaObjects/13661_2012_Article_185_IEq62_HTML.gif is a solution of the problem (1.1) if and only if x C [ 0 , 1 ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-86/MediaObjects/13661_2012_Article_185_IEq63_HTML.gif is a solution of the following nonlinear integral equation:
y ( t ) = 0 1 G ( t , s ) h ( s ) f ( s , y ( s ) , y ( s ) , ( T y ) ( s ) , ( S y ) ( s ) ) d s + 0 < t k < t [ I k ( y ( t k ) ) ( t t k ) I ¯ k ( y ( t k ) , y ( t k ) ) ] + 1 ρ ( α t + β ) { k = 1 m [ γ ( 1 t k ) I ¯ k ( y ( t k ) , y ( t k ) ) + δ I ¯ k ( y ( t k ) , y ( t k ) ) γ I k ( y ( t k ) ) ] } , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-86/MediaObjects/13661_2012_Article_185_Equc_HTML.gif
where
G ( t , s ) = 1 ρ { ( γ + δ γ t ) ( β + α s ) , 0 s t 1 , ( β + α t ) ( γ + δ γ s ) , 0 t s 1 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-86/MediaObjects/13661_2012_Article_185_Equ3_HTML.gif
(2.1)

where ρ = γ β + α γ + α δ > 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-86/MediaObjects/13661_2012_Article_185_IEq64_HTML.gif. In what follows, we write J 1 = [ 0 , t 1 ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-86/MediaObjects/13661_2012_Article_185_IEq65_HTML.gif, J k = ( t k 1 , t k ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-86/MediaObjects/13661_2012_Article_185_IEq66_HTML.gif ( k = 1 , 2 , , m http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-86/MediaObjects/13661_2012_Article_185_IEq67_HTML.gif), J m = ( t m , 1 ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-86/MediaObjects/13661_2012_Article_185_IEq11_HTML.gif. By making use of (2.1), we can prove that G ( t , s ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-86/MediaObjects/13661_2012_Article_185_IEq68_HTML.gif has the following properties.

Proposition 2.1 0 G ( t , s ) G ( s , s ) 1 ρ ( α + δ ) ( α + β ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-86/MediaObjects/13661_2012_Article_185_IEq69_HTML.gif, t , s [ 0 , 1 ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-86/MediaObjects/13661_2012_Article_185_IEq70_HTML.gif.

Proposition 2.2 0 σ G ( t , s ) G ( t , s ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-86/MediaObjects/13661_2012_Article_185_IEq71_HTML.gif, t , s [ a , b ] [ 0 , 1 ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-86/MediaObjects/13661_2012_Article_185_IEq72_HTML.gif, where a ( 0 , t 1 ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-86/MediaObjects/13661_2012_Article_185_IEq73_HTML.gif, b [ t m , 1 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-86/MediaObjects/13661_2012_Article_185_IEq74_HTML.gifand
0 < σ = min { β + α a α + β , δ + ( 1 b ) γ γ + δ } < 1 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-86/MediaObjects/13661_2012_Article_185_Equ4_HTML.gif
(2.2)

Let P C [ J , E ] = { y : y  is a map from  J  into  E  such that  y ( t )  is continuous at  t t k , left continuous at  t = t k  and  y ( t k + )  exists for  k = 1 , , m }  and  P C [ J , P ] = { y P C [ J , E ] : y ( t ) θ } http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-86/MediaObjects/13661_2012_Article_185_IEq75_HTML.gif. It is easy to verify P C [ J , E ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-86/MediaObjects/13661_2012_Article_185_IEq76_HTML.gif is a Banach space with norm y P C = sup t J y ( t ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-86/MediaObjects/13661_2012_Article_185_IEq77_HTML.gif. Obviously, P C [ J , P ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-86/MediaObjects/13661_2012_Article_185_IEq78_HTML.gif is a cone in Banach space P C [ J , E ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-86/MediaObjects/13661_2012_Article_185_IEq79_HTML.gif.

Let P C 1 [ J , E ] = { y : y  is a map from  J  into  E  such that  y ( t )  exist and is continuous at  t t k , y ( t )  left continuous at  t = t k , and  y ( t k + ) , y ( t k )  exist for  k = 1 , , m } http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-86/MediaObjects/13661_2012_Article_185_IEq80_HTML.gif, P C 1 [ J , P ] = { y P C 1 [ J , E ] : y ( t ) θ , y ( t ) θ } http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-86/MediaObjects/13661_2012_Article_185_IEq81_HTML.gif. It is easy to see that P C 1 [ J , E ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-86/MediaObjects/13661_2012_Article_185_IEq82_HTML.gif is a Banach space with the norm y 1 = max { y P C , y P C } http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-86/MediaObjects/13661_2012_Article_185_IEq83_HTML.gif. Evidently, y 1 y P C + y P C http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-86/MediaObjects/13661_2012_Article_185_IEq84_HTML.gif and P C 1 [ J , P ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-86/MediaObjects/13661_2012_Article_185_IEq85_HTML.gif is a cone in Banach space P C 1 [ J , E ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-86/MediaObjects/13661_2012_Article_185_IEq82_HTML.gif. For any y P C 1 [ J , E ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-86/MediaObjects/13661_2012_Article_185_IEq86_HTML.gif, by making use of the mean value theorem y ( t k ) y ( t k h ) h c o ¯ { y ( t ) : t k h < t < t k } http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-86/MediaObjects/13661_2012_Article_185_IEq87_HTML.gif ( h > 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-86/MediaObjects/13661_2012_Article_185_IEq88_HTML.gif), obviously we see that y ( t k ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-86/MediaObjects/13661_2012_Article_185_IEq89_HTML.gif exists and y ( t k ) = lim h 0 y ( t k ) y ( t k h ) h http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-86/MediaObjects/13661_2012_Article_185_IEq90_HTML.gif.

Let K = { y P C [ J ¯ , P ] : y σ y , t [ a , b ] } http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-86/MediaObjects/13661_2012_Article_185_IEq91_HTML.gif. For any 0 < r < R < + http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-86/MediaObjects/13661_2012_Article_185_IEq92_HTML.gif, let K r = { y K : y < r } http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-86/MediaObjects/13661_2012_Article_185_IEq93_HTML.gif, K r = { y K : y = r } http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-86/MediaObjects/13661_2012_Article_185_IEq94_HTML.gif, K ¯ r , R = { y K : r y R } http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-86/MediaObjects/13661_2012_Article_185_IEq95_HTML.gif.

A map y P C 1 [ J , E ] C 2 [ J , E ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-86/MediaObjects/13661_2012_Article_185_IEq96_HTML.gif is called a nonnegative solution of problem (1.1) if y ( t ) θ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-86/MediaObjects/13661_2012_Article_185_IEq97_HTML.gif, y ( t ) θ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-86/MediaObjects/13661_2012_Article_185_IEq98_HTML.gif for t J http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-86/MediaObjects/13661_2012_Article_185_IEq99_HTML.gif and y ( t ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-86/MediaObjects/13661_2012_Article_185_IEq22_HTML.gif satisfies problem (1.1). An operator y P C 1 [ J , E ] C 2 [ J , E ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-86/MediaObjects/13661_2012_Article_185_IEq100_HTML.gif is called a positive solution of problem (1.1) if y is a nonnegative solution of problem (1.1) and y ( t ) θ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-86/MediaObjects/13661_2012_Article_185_IEq101_HTML.gif.

For convenience and simplicity in the following discussion, we denote
http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-86/MediaObjects/13661_2012_Article_185_Equd_HTML.gif

where ν denote 0 or ∞.

To establish the existence of multiple positive solutions in E of problem (1.1), let us list the following assumptions, which will stand throughout the paper:

(H1) f C ( J × P 4 , P ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-86/MediaObjects/13661_2012_Article_185_IEq102_HTML.gif, h C ( J , P ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-86/MediaObjects/13661_2012_Article_185_IEq103_HTML.gif and
f ( t , y 1 , y 2 , y 3 , y 4 ) a ( t ) + i = 1 4 b i ( t ) y i , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-86/MediaObjects/13661_2012_Article_185_Equ5_HTML.gif
(2.3)
where a ( t ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-86/MediaObjects/13661_2012_Article_185_IEq104_HTML.gif and b i ( t ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-86/MediaObjects/13661_2012_Article_185_IEq105_HTML.gif are Lebesgue integrable functionals on J ( i = 1 , 2 , 3 , 4 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-86/MediaObjects/13661_2012_Article_185_IEq106_HTML.gif) and satisfying
http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-86/MediaObjects/13661_2012_Article_185_Eque_HTML.gif
(H2) I k C ( P , P ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-86/MediaObjects/13661_2012_Article_185_IEq107_HTML.gif, I ¯ k C ( P , P ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-86/MediaObjects/13661_2012_Article_185_IEq108_HTML.gif and there exist positive constants c k http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-86/MediaObjects/13661_2012_Article_185_IEq109_HTML.gif, c k http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-86/MediaObjects/13661_2012_Article_185_IEq110_HTML.gif and c ¯ k http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-86/MediaObjects/13661_2012_Article_185_IEq111_HTML.gif ( k = 1 , , m http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-86/MediaObjects/13661_2012_Article_185_IEq112_HTML.gif) satisfying
k = 1 m ( c k + c k + c ¯ k ) + 1 ρ ( α + β ) k = 1 m [ ( γ + δ ) ( c k + c ¯ k ) + γ c k ] < 1 4 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-86/MediaObjects/13661_2012_Article_185_Equ6_HTML.gif
(2.4)
such that
I k ( y ) c k y , I ¯ k ( y 1 , y 2 ) c k y 1 + c ¯ k y 2 , t J , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-86/MediaObjects/13661_2012_Article_185_Equf_HTML.gif

(H3) for any bounded set B i E http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-86/MediaObjects/13661_2012_Article_185_IEq113_HTML.gif, i = 1 , 2 , 3 , 4 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-86/MediaObjects/13661_2012_Article_185_IEq106_HTML.gif, f ( t , B 1 , B 2 , B 3 , B 4 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-86/MediaObjects/13661_2012_Article_185_IEq114_HTML.gif and I k ( B 1 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-86/MediaObjects/13661_2012_Article_185_IEq115_HTML.gif together with I ¯ k ( B 1 , B 2 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-86/MediaObjects/13661_2012_Article_185_IEq116_HTML.gif are relatively compact sets,

(H4) f 0 > m http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-86/MediaObjects/13661_2012_Article_185_IEq117_HTML.gif,

(H5) f < m http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-86/MediaObjects/13661_2012_Article_185_IEq118_HTML.gif, where
m = max { ( σ a b G ( s , s ) h ( s ) d s ) 1 , ( a b G t ( t , s ) h ( s ) d s ) 1 , 1 ρ ( α + β ) ( γ + δ ) 0 1 h ( s ) a ( s ) d s } . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-86/MediaObjects/13661_2012_Article_185_Equ7_HTML.gif
(2.5)
We shall reduce problem (1.1) to an integral equation in E. To this end, we first consider operator A : K P C [ J ¯ , E ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-86/MediaObjects/13661_2012_Article_185_IEq119_HTML.gif defined by
( A y ) ( t ) = 0 1 G ( t , s ) h ( s ) f ( s , y ( s ) , y ( s ) , ( T y ) ( s ) , ( S y ) ( s ) ) d s + 0 < t k < t [ I k ( y ( t k ) ) ( t t k ) I ¯ k ( y ( t k ) , y ( t k ) ) ] + 1 ρ ( α t + β ) { k = 1 m ( γ ( 1 t k ) + δ ) I ¯ k ( y ( t k ) , y ( t k ) ) γ k = 1 m I k ( y ( t k ) ) } . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-86/MediaObjects/13661_2012_Article_185_Equ8_HTML.gif
(2.6)
Lemma 2.1 y P C 1 [ J ¯ , E ] C 2 [ J , E ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-86/MediaObjects/13661_2012_Article_185_IEq120_HTML.gifis a solution of problem (1.1) if and only if y P C 1 [ J ¯ , E ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-86/MediaObjects/13661_2012_Article_185_IEq121_HTML.gifis a solution of the following impulsive integral equation:
y ( t ) = 0 1 G ( t , s ) h ( s ) f ( s , y ( s ) , y ( s ) , ( T y ) ( s ) , ( S y ) ( s ) ) d s + 0 < t k < t [ I k ( y ( t k ) ) ( t t k ) I ¯ k ( y ( t k ) , y ( t k ) ) ] + 1 ρ ( α t + β ) k = 1 m [ ( γ ( 1 t k ) + δ ) I ¯ k ( y ( t k ) , y ( t k ) ) γ I k ( y ( t k ) ) ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-86/MediaObjects/13661_2012_Article_185_Equ9_HTML.gif
(2.7)

i.e., y is a fixed point of operator A defined by (2.6) in P C 1 [ J , E ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-86/MediaObjects/13661_2012_Article_185_IEq122_HTML.gif.

Proof First suppose that y P C 1 [ J , E ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-86/MediaObjects/13661_2012_Article_185_IEq123_HTML.gif is a solution of problem (1.1). It is easy to see by the integration of problem (1.1) that
y ( t ) = y ( 0 ) 0 t f ( s , y ( s ) , y ( s ) , ( T y ) ( s ) , ( S y ) ( s ) ) d s + 0 < t k < t [ y ( t k + ) y ( t k ) ] = y ( 0 ) 0 t f ( s , y ( s ) , y ( s ) , ( T y ) ( s ) , ( S y ) ( s ) ) d s 0 < t k < t I ¯ k ( y ( t k ) , y ( t k ) ) , t J . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-86/MediaObjects/13661_2012_Article_185_Equ10_HTML.gif
(2.8)
Integrate again, we get
y ( t ) = y ( 0 ) + y ( 0 ) t 0 t ( t s ) f ( s , y ( s ) , y ( s ) , ( T y ) ( s ) , ( S y ) ( s ) ) d s + 0 < t k < t [ I k ( y ( t k ) ) ( t t k ) I ¯ k ( y ( t k ) , y ( t k ) ) ] , t J . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-86/MediaObjects/13661_2012_Article_185_Equ11_HTML.gif
(2.9)
Letting t = 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-86/MediaObjects/13661_2012_Article_185_IEq31_HTML.gif in (2.8) and (2.9), we find that
y ( 1 ) = y ( 0 ) 0 1 f ( s , y ( s ) , y ( s ) , ( T y ) ( s ) , ( S y ) ( s ) ) d s k = 1 m I ¯ k ( y ( t k ) , y ( t k ) ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-86/MediaObjects/13661_2012_Article_185_Equ12_HTML.gif
(2.10)
y ( 1 ) = y ( 0 ) + y ( 0 ) 0 1 ( 1 s ) f ( s , y ( s ) , y ( s ) , ( T y ) ( s ) , ( S y ) ( s ) ) d s + k = 1 m [ I k ( y ( t k ) ) ( 1 t k ) I ¯ k ( y ( t k ) , y ( t k ) ) ] . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-86/MediaObjects/13661_2012_Article_185_Equ13_HTML.gif
(2.11)
Since
y ( 0 ) = β α y ( 0 ) , y ( 1 ) = δ γ { y ( 0 ) 0 1 f ( s , y ( s ) , y ( s ) , ( T y ) ( s ) , ( S y ) ( s ) ) d s k = 1 m I ¯ k ( y ( t k ) , y ( t k ) ) } . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-86/MediaObjects/13661_2012_Article_185_Equ14_HTML.gif
(2.12)
We get
y ( 0 ) = α ρ { δ 0 1 f ( s , y ( s ) , y ( s ) , ( T y ) ( s ) , ( S y ) ( s ) ) d s + γ 0 1 ( 1 s ) f ( s , y ( s ) , y ( s ) , ( T y ) ( s ) , ( S y ) ( s ) ) d s + k = 1 m [ ( γ ( 1 t k ) + δ ) I ¯ k ( y ( t k ) , y ( t k ) ) γ I k ( y ( t k ) ) ] } , t J . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-86/MediaObjects/13661_2012_Article_185_Equ15_HTML.gif
(2.13)
Substituting (2.12) and (2.13) into (2.9), we obtain
y ( t ) = 1 ρ ( α t + β ) { 0 1 ( γ ( 1 s ) + δ ) f ( s , y ( s ) , y ( s ) , ( T y ) ( s ) , ( S y ) ( s ) ) d s k = 1 m [ γ I k ( y ( t k ) ) ( γ ( 1 t k ) + δ ) I ¯ k ( y ( t k ) , y ( t k ) ) ] } 0 t γ ( t s ) f ( s , y ( s ) , y ( s ) , ( T y ) ( s ) , ( S y ) ( s ) ) d s + 0 < t k < t [ I k ( y ( t k ) ) ( t t k ) I ¯ k ( y ( t k ) , y ( t k ) ) ] = 1 ρ 0 t [ ( α t + β ) ( γ ( 1 s ) + δ ) ρ ( t s ) ] f ( s , y ( s ) , y ( s ) , ( T y ) ( s ) , ( S y ) ( s ) ) d s + 1 ρ ( α t + β ) t 1 ( γ ( 1 s ) + δ ) f ( s , y ( s ) , y ( s ) , ( T y ) ( s ) , ( S y ) ( s ) ) d s + 0 < t k < t [ I k ( y ( t k ) ) ( t t k ) I ¯ k ( y ( t k ) , y ( t k ) ) ] + 1 ρ ( α t + β ) k = 1 m [ ( γ ( 1 t k ) + δ ) I ¯ k ( y ( t k ) , y ( t k ) ) γ I k ( y ( t k ) ) ] = 0 1 G ( t , s ) h ( s ) f ( s , y ( s ) , y ( s ) , ( T y ) ( s ) , ( S y ) ( s ) ) d s + 0 < t k < t [ I k ( y ( t k ) ) ( t t k ) I ¯ k ( y ( t k ) , y ( t k ) ) ] + 1 ρ ( α t + β ) k = 1 m [ ( γ ( 1 t k ) + δ ) I ¯ k ( y ( t k ) , y ( t k ) ) γ I k ( y ( t k ) ) ] . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-86/MediaObjects/13661_2012_Article_185_Equg_HTML.gif
Conversely, if y P C 1 [ J , E ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-86/MediaObjects/13661_2012_Article_185_IEq86_HTML.gif is a solution of the integral equation (2.7). Evidently, Δ y | t = t k = I k ( y ( t k ) ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-86/MediaObjects/13661_2012_Article_185_IEq124_HTML.gif ( k = 1 , , m http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-86/MediaObjects/13661_2012_Article_185_IEq125_HTML.gif). For t = t k http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-86/MediaObjects/13661_2012_Article_185_IEq24_HTML.gif, direct differentiation of the integral equation (2.7) implies
y ( t ) = 1 ρ { γ 0 t ( α s + β ) f ( s , y ( s ) , y ( s ) , ( T y ) ( s ) , ( S y ) ( s ) ) d s + α t 1 ( γ ( 1 s ) + δ ) f ( s , y ( s ) , y ( s ) , ( T y ) ( s ) , ( S y ) ( s ) ) d s 0 < t k < t I ¯ k ( y ( t k ) , y ( t k ) ) } α ρ k = 1 m [ γ I k ( y ( t k ) ) ( γ ( 1 t k ) + δ ) I ¯ k ( y ( t k ) , y ( t k ) ) ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-86/MediaObjects/13661_2012_Article_185_Equh_HTML.gif

and y = f ( t , y ( t ) , y ( t ) , ( T y ) ( t ) , ( S y ) ( t ) ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-86/MediaObjects/13661_2012_Article_185_IEq126_HTML.gif. So y C 2 [ J , E ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-86/MediaObjects/13661_2012_Article_185_IEq127_HTML.gif and Δ y | t = t k = I ¯ k ( y ( t k ) , y ( t k ) ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-86/MediaObjects/13661_2012_Article_185_IEq128_HTML.gif ( k = 1 , , m http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-86/MediaObjects/13661_2012_Article_185_IEq129_HTML.gif). It is easy to verify that α y ( 0 ) β y ( 0 ) = 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-86/MediaObjects/13661_2012_Article_185_IEq130_HTML.gif and γ y ( 1 ) + δ y ( 1 ) = 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-86/MediaObjects/13661_2012_Article_185_IEq131_HTML.gif. The proof is complete. □

Thanks to (2.1), we know that
G t ( t , s ) = 1 ρ { γ ( β + α s ) , 0 s t 1 , α ( γ + δ γ s ) , 0 t s 1 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-86/MediaObjects/13661_2012_Article_185_Equi_HTML.gif

In the following, let w = sup t , s J , t s | G t ( t , s ) | http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-86/MediaObjects/13661_2012_Article_185_IEq132_HTML.gif. For B P C 1 [ J , E ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-86/MediaObjects/13661_2012_Article_185_IEq133_HTML.gif, we denote B = { y : y B } P C [ J , E ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-86/MediaObjects/13661_2012_Article_185_IEq134_HTML.gif, B ( t ) = { y ( t ) : y B } E http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-86/MediaObjects/13661_2012_Article_185_IEq135_HTML.gif and B ( t ) = { y ( t ) : y B } E http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-86/MediaObjects/13661_2012_Article_185_IEq136_HTML.gif ( t J http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-86/MediaObjects/13661_2012_Article_185_IEq99_HTML.gif).

Lemma 2.2 ([12])

Let D P C 1 [ J , E ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-86/MediaObjects/13661_2012_Article_185_IEq137_HTML.gifbe a bounded set. Suppose that D ( t ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-86/MediaObjects/13661_2012_Article_185_IEq138_HTML.gifis equi-continuous on each J k http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-86/MediaObjects/13661_2012_Article_185_IEq139_HTML.gif ( k = 1 , , m http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-86/MediaObjects/13661_2012_Article_185_IEq112_HTML.gif). Then
ϒ P C 1 ( D ) = max { sup t J ϒ ( D ( t ) ) , sup t J ϒ ( D ( t ) ) } , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-86/MediaObjects/13661_2012_Article_185_Equj_HTML.gif

where ϒ and ϒ P C 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-86/MediaObjects/13661_2012_Article_185_IEq140_HTML.gifdenote the Kuratowski measures of noncompactness of bounded sets in E and P C 1 [ J , E ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-86/MediaObjects/13661_2012_Article_185_IEq82_HTML.gif, respectively.

Lemma 2.3 ([15])

Let H P C [ J , E ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-86/MediaObjects/13661_2012_Article_185_IEq141_HTML.gifbe bounded equicontinuous, then ϒ ( H ( t ) ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-86/MediaObjects/13661_2012_Article_185_IEq142_HTML.gifis continuous on J and
ϒ ( { J y ( t ) d t : y H } ) J ϒ ( H ( t ) ) d t . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-86/MediaObjects/13661_2012_Article_185_Equk_HTML.gif

Lemma 2.4 ([15])

H P C 1 [ J , E ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-86/MediaObjects/13661_2012_Article_185_IEq143_HTML.gifis relatively compact if and only if each element y ( t ) H http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-86/MediaObjects/13661_2012_Article_185_IEq144_HTML.gifand y ( t ) H http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-86/MediaObjects/13661_2012_Article_185_IEq145_HTML.gifare uniformly bounded and equicontinuous on each J k http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-86/MediaObjects/13661_2012_Article_185_IEq139_HTML.gif ( k = 1 , , m http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-86/MediaObjects/13661_2012_Article_185_IEq112_HTML.gif).

Lemma 2.5 ([15])

Let E be a Banach space and H C [ J , E ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-86/MediaObjects/13661_2012_Article_185_IEq146_HTML.gifif H is countable and there exists φ L [ J , R + ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-86/MediaObjects/13661_2012_Article_185_IEq147_HTML.gifsuch that y ( t ) φ ( t ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-86/MediaObjects/13661_2012_Article_185_IEq148_HTML.gif, t J http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-86/MediaObjects/13661_2012_Article_185_IEq99_HTML.gif, y H http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-86/MediaObjects/13661_2012_Article_185_IEq149_HTML.gif. Then ϒ ( { y ( t ) : y H } ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-86/MediaObjects/13661_2012_Article_185_IEq150_HTML.gifis integrable on J, and
ϒ ( { J y ( t ) d t : y H } ) 2 J ϒ { y ( t ) : y H } d t . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-86/MediaObjects/13661_2012_Article_185_Equl_HTML.gif

Lemma 2.6 A ( K ) K http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-86/MediaObjects/13661_2012_Article_185_IEq151_HTML.gif.

Proof For any y K http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-86/MediaObjects/13661_2012_Article_185_IEq152_HTML.gif, from Proposition 2.1 and (2.6), we obtain
A y 0 1 G ( s , s ) h ( s ) f ( s , y ( s ) , y ( s ) , ( T y ) ( s ) , ( S y ) ( s ) ) d s + 0 < t k < t [ I k ( y ( t k ) ) ( t t k ) I ¯ k ( y ( t k ) , y ( t k ) ) ] + 1 ρ ( α t + β ) { k = 1 m [ ( γ ( 1 t k ) + δ ) I ¯ k ( y ( t k ) , y ( t k ) ) ] k = 1 m γ I k ( y ( t k ) ) } . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-86/MediaObjects/13661_2012_Article_185_Equm_HTML.gif
On the other hand, for any t [ a , b ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-86/MediaObjects/13661_2012_Article_185_IEq153_HTML.gif, by (2.6) and Proposition 2.2, we know that
A y ( t ) = 0 1 G ( t , s ) h ( s ) f ( s , y ( s ) , y ( s ) , ( T y ) ( s ) , ( S y ) ( s ) ) d s + 0 < t k < t [ I k ( y ( t k ) ) ( t t k ) I ¯ k ( y ( t k ) , y ( t k ) ) ] + 1 ρ ( α t + β ) k = 1 m [ ( γ ( 1 t k ) + δ ) I ¯ k ( y ( t k ) , y ( t k ) ) γ I k ( y ( t k ) ) ] σ 0 1 G ( s , s ) h ( s ) f ( s , y ( s ) , y ( s ) , ( T y ) ( s ) , ( S y ) ( s ) ) d s + 0 < t k < t [ I k ( y ( t k ) ) ( t t k ) I ¯ k ( y ( t k ) , y ( t k ) ) ] + 1 ρ ( α t + β ) { k = 1 m [ ( γ ( 1 t k ) + δ ) I ¯ k ( y ( t k ) , y ( t k ) ) γ I k ( y ( t k ) ) ] } σ A y . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-86/MediaObjects/13661_2012_Article_185_Equn_HTML.gif

Hence, A ( K ) K http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-86/MediaObjects/13661_2012_Article_185_IEq151_HTML.gif. □

Lemma 2.7 Suppose that (H1) and (H3) hold. Then A : K K http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-86/MediaObjects/13661_2012_Article_185_IEq154_HTML.gifis completely continuous.

Proof Firstly, we show that A : K K http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-86/MediaObjects/13661_2012_Article_185_IEq154_HTML.gif is continuous. Assume that y n , y 0 K http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-86/MediaObjects/13661_2012_Article_185_IEq155_HTML.gif and y n y 0 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-86/MediaObjects/13661_2012_Article_185_IEq156_HTML.gif, y n y 0 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-86/MediaObjects/13661_2012_Article_185_IEq157_HTML.gif ( n http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-86/MediaObjects/13661_2012_Article_185_IEq158_HTML.gif). Since f C ( J ¯ × P × P × P × P , P ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-86/MediaObjects/13661_2012_Article_185_IEq159_HTML.gif, I k C ( J , P ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-86/MediaObjects/13661_2012_Article_185_IEq160_HTML.gif and I ¯ k C ( J , P ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-86/MediaObjects/13661_2012_Article_185_IEq161_HTML.gif, then
http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-86/MediaObjects/13661_2012_Article_185_Equ16_HTML.gif
(2.14)
and
lim n I ¯ k ( y n ( t k ) , y n ( t k ) ) I ¯ k ( y 0 ( t k ) , y 0 ( t k ) ) = 0 , k = 1 , , m . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-86/MediaObjects/13661_2012_Article_185_Equ17_HTML.gif
(2.15)
Thus, for any t J http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-86/MediaObjects/13661_2012_Article_185_IEq99_HTML.gif, from the Lebesgue dominated convergence theorem together with (2.14) and (2.15), we know that
http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-86/MediaObjects/13661_2012_Article_185_Equo_HTML.gif

Hence, A : K K http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-86/MediaObjects/13661_2012_Article_185_IEq154_HTML.gif is continuous.

Let B K http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-86/MediaObjects/13661_2012_Article_185_IEq162_HTML.gif be any bounded set, then there exists a positive constant R 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-86/MediaObjects/13661_2012_Article_185_IEq163_HTML.gif such that y 1 R 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-86/MediaObjects/13661_2012_Article_185_IEq164_HTML.gif. Thus, for any y B http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-86/MediaObjects/13661_2012_Article_185_IEq165_HTML.gif, t ( 0 , 1 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-86/MediaObjects/13661_2012_Article_185_IEq166_HTML.gif, we know that
| ( A y ) ( t ) | = 1 ρ | { γ 0 t ( α s + β ) h ( s ) f ( s , y ( s ) , y ( s ) , ( T y ) ( s ) , ( S y ) ( s ) ) d s + α t 1 ( γ ( 1 s ) + δ ) h ( s ) f ( s , y ( s ) , y ( s ) , ( T y ) ( s ) , ( S y ) ( s ) ) d s } 0 < t k < t I ¯ k ( y ( t k ) , y ( t k ) ) + α ρ k = 1 m [ ( γ ( 1 t k ) + δ ) I ¯ k ( y ( t k ) , y ( t k ) ) γ I k ( y ( t k ) ) ] | . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-86/MediaObjects/13661_2012_Article_185_Equp_HTML.gif
Therefore,
| ( A y ) ( t ) | 1 ρ { γ 0 t ( α s + β ) h ( s ) | f ( s , y ( s ) , y ( s ) , ( T y ) ( s ) , ( S y ) ( s ) ) | d s + α t 1 ( γ ( 1 s ) + δ ) h ( s ) | f ( s , y ( s ) , y ( s ) , ( T y ) ( s ) , ( S y ) ( s ) ) | d s } + 0 < t k < t | I ¯ k ( y ( t k ) , y ( t k ) ) | + α ρ k = 1 m [ ( γ ( 1 t k ) + δ ) I ¯ k ( y ( t k ) , y ( t k ) ) + γ | I k ( y ( t k ) ) | ] 1 ρ { γ 0 t ( α s + β ) h ( s ) ( a ( s ) + ( i = 1 2 b i ( s ) + K 0 b 3 ( s ) + H 0 b 4 ( s ) ) y 1 ) d s + α t 1 ( γ ( 1 s ) + δ ) h ( s ) ( a ( s ) + ( i = 1 2 b i ( s ) + K 0 b 3 ( s ) + H 0 b 4 ( s ) ) y ( s ) 1 ) d s } + k = 1 m ( c k + c ¯ k ) y ( s ) 1 + α ρ k = 1 m [ ( γ ( 1 t k ) + δ ) ( c k + c ¯ k ) + γ c k ] y ( s ) 1 1 ρ { γ 0 t ( α s + β ) h ( s ) ( a ( s ) + ( i = 1 2 b i ( s ) + K 0 b 3 ( s ) + H 0 b 4 ( s ) ) R 0 ) d s + α t 1 ( γ ( 1 s ) + δ ) h ( s ) ( a ( s ) + ( i = 1 2 b i ( s ) + K 0 b 3 ( s ) + H 0 b 4 ( s ) ) R 0 ) d s } + k = 1 m ( c k + c ¯ k ) R 0 + α ρ k = 1 m [ ( γ ( 1 t k ) + δ ) ( c k + c ¯ k ) + γ c k ] R 0 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-86/MediaObjects/13661_2012_Article_185_Equ18_HTML.gif
(2.16)
t J http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-86/MediaObjects/13661_2012_Article_185_IEq167_HTML.gif, t t k http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-86/MediaObjects/13661_2012_Article_185_IEq168_HTML.gif, k = 1 , , m http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-86/MediaObjects/13661_2012_Article_185_IEq112_HTML.gif. Let
http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-86/MediaObjects/13661_2012_Article_185_Equq_HTML.gif
Integrating ψ ( t ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-86/MediaObjects/13661_2012_Article_185_IEq169_HTML.gif from 0 to 1 and exchanging integral sequence, then
0 1 ψ ( t ) d t = 0 1 0 s α ( γ + δ γ s ) h ( s ) ( a ( s ) + ( i = 1 2 b i ( s ) + K 0 b 3 ( s ) + H 0 b 4 ( s ) ) R 0 ) d t d s γ 0 1 s 1 ( β + α s ) h ( s ) ( a ( s ) + ( i = 1 2 b i ( s ) + K 0 b 3 ( s ) + H 0 b 4 ( s ) ) R 0 ) d t d s α 0 1 s ( γ + δ γ s ) h ( s ) ( a ( s ) + ( i = 1 2 b i ( s ) + K 0 b 3 ( s ) + H 0 b 4 ( s ) ) R 0 ) d s γ 0 1 ( β + α s ) h ( s ) ( a ( s ) + ( i = 1 2 b i ( s ) + K 0 b 3 ( s ) + H 0 b 4 ( s ) ) R 0 ) d s ρ 0 1 G ( s , s ) h ( s ) ( a ( s ) + ( i = 1 2 b i ( s ) + K 0 b 3 ( s ) + H 0 b 4 ( s ) ) R 0 ) d s < + . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-86/MediaObjects/13661_2012_Article_185_Equ19_HTML.gif
(2.17)
Thus, by (H1) and (2.17), we have ψ ( t ) L 1 ( J ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-86/MediaObjects/13661_2012_Article_185_IEq170_HTML.gif. Hence, for any 0 t 1 t 2 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-86/MediaObjects/13661_2012_Article_185_IEq171_HTML.gif and for all y E http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-86/MediaObjects/13661_2012_Article_185_IEq40_HTML.gif, from (2.16), we know that
( A y ) ( t 1 ) ( A y ) ( t 2 ) = | t 1 t 2 ( A y ) ( t ) d t | t 1 t 2 ( ψ ( t ) + M 1 ) d t . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-86/MediaObjects/13661_2012_Article_185_Equ20_HTML.gif
(2.18)

From (2.17), (2.18), and the absolutely continuity of integral function, we see that A ( B ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-86/MediaObjects/13661_2012_Article_185_IEq172_HTML.gif is equicontinuous.

On the other hand, for any y B http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-86/MediaObjects/13661_2012_Article_185_IEq165_HTML.gif and t J http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-86/MediaObjects/13661_2012_Article_185_IEq99_HTML.gif, we know that
| ( A y ) ( t ) | = | 0 1 G ( t , s ) h ( s ) f ( s , y ( s ) , y ( s ) , ( T y ) ( s ) , ( S y ) ( s ) ) d s + 0 < t k < t [ I k ( y ( t k ) ) ( t t k ) I ¯ k ( y ( t k ) , y ( t k ) ) ] + 1 ρ ( α t + β ) { k = 1 m [ ( γ ( 1 t k ) + δ ) I ¯ k ( y ( t k ) , y ( t k ) ) γ I k ( y ( t k ) ) ] } | 0 1 G ( s , s ) h ( s ) ( a ( s ) + ( i = 1 2 b i ( s ) + K 0 b 3 ( s ) + H 0 b 4 ( s ) ) R 0 ) d s + k = 1 m ( c k + c ¯ k ) R 0 ) + α ρ k = 1 m [ ( γ ( 1 t k ) + δ ) ( c k + c ¯ k ) + γ c k ] R 0 < + . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-86/MediaObjects/13661_2012_Article_185_Equr_HTML.gif
Therefore, A ( B ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-86/MediaObjects/13661_2012_Article_185_IEq172_HTML.gif is uniformly bounded. By virtue of Lemma 2.3 and (H3), we know that
http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-86/MediaObjects/13661_2012_Article_185_Equs_HTML.gif

So, ϒ ( A ( B ) ) = 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-86/MediaObjects/13661_2012_Article_185_IEq173_HTML.gif. Therefore, A is compact. To sum up, the conclusion of Lemma 2.7 follows. □

The main tools of the paper are the following well-known fixed-point index theorems (see [24]).

Lemma 2.8 Let A : K K http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-86/MediaObjects/13661_2012_Article_185_IEq174_HTML.gifbe a completely continuous mapping and A y y http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-86/MediaObjects/13661_2012_Article_185_IEq175_HTML.giffor y K r http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-86/MediaObjects/13661_2012_Article_185_IEq176_HTML.gif. Thus, we have the following conclusions:
  1. (i)

    If y A y http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-86/MediaObjects/13661_2012_Article_185_IEq177_HTML.gif for y K r http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-86/MediaObjects/13661_2012_Article_185_IEq176_HTML.gif, then i ( A , K r , K ) = 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-86/MediaObjects/13661_2012_Article_185_IEq178_HTML.gif.

     
  2. (ii)

    If y A y http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-86/MediaObjects/13661_2012_Article_185_IEq179_HTML.gif for y K r http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-86/MediaObjects/13661_2012_Article_185_IEq176_HTML.gif, then i ( A , K r , K ) = 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-86/MediaObjects/13661_2012_Article_185_IEq180_HTML.gif.

     

3 Main results

In this section, we establish the existence of positive solutions for problem (1.1) by making use of Lemma 2.8.

Theorem 3.1 Suppose that (H1)-(H4) hold. Then problem (1.1) has at least one positive solution.

Proof From (H4), there exists ε > 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-86/MediaObjects/13661_2012_Article_185_IEq181_HTML.gif such that f 0 > m + ε http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-86/MediaObjects/13661_2012_Article_185_IEq182_HTML.gif and also there exists r > 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-86/MediaObjects/13661_2012_Article_185_IEq183_HTML.gif such that for any 0 < i = 1 4 y i r http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-86/MediaObjects/13661_2012_Article_185_IEq184_HTML.gif and t [ a , b ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-86/MediaObjects/13661_2012_Article_185_IEq185_HTML.gif, we have
f ( t , y 1 , y 2 , y 3 , y 4 ) ( m + ε ) i = 1 4 y i . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-86/MediaObjects/13661_2012_Article_185_Equ21_HTML.gif
(3.1)
Set K r = { y K : y 1 < r } http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-86/MediaObjects/13661_2012_Article_185_IEq186_HTML.gif. Then for any y K r K http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-86/MediaObjects/13661_2012_Article_185_IEq187_HTML.gif, by virtue of (3.1), we know that
http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-86/MediaObjects/13661_2012_Article_185_Equt_HTML.gif
So ( A y ) ( t ) 1 r http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-86/MediaObjects/13661_2012_Article_185_IEq188_HTML.gif. Therefore,
i ( A , K r K , K ) = 0 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-86/MediaObjects/13661_2012_Article_185_Equ22_HTML.gif
(3.2)
Let R > max { 4 m , 2 r } http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-86/MediaObjects/13661_2012_Article_185_IEq189_HTML.gif. Then K R http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-86/MediaObjects/13661_2012_Article_185_IEq190_HTML.gif is a bounded open subsets in E, and so for any y K R K http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-86/MediaObjects/13661_2012_Article_185_IEq191_HTML.gif and t J http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-86/MediaObjects/13661_2012_Article_185_IEq99_HTML.gif, we obtain
http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-86/MediaObjects/13661_2012_Article_185_Equu_HTML.gif
Hence, A y 1 < R http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-86/MediaObjects/13661_2012_Article_185_IEq192_HTML.gif. Therefore,
i ( A , K R K , K ) = 0 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-86/MediaObjects/13661_2012_Article_185_Equ23_HTML.gif
(3.3)
From (3.2) and (3.3), we get
i ( A , ( K R K ) ( K ¯ r K ) , K ) = i ( A , K R K , K ) i ( A , K r K , K ) = 1 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-86/MediaObjects/13661_2012_Article_185_Equv_HTML.gif

Therefore, A has at least one fixed point on ( K R K ) ( K ¯ r K ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-86/MediaObjects/13661_2012_Article_185_IEq193_HTML.gif. Consequently, problem (1.1) has at least one positive solution. □

Theorem 3.2 Suppose that (H1)∼(H3) and (H5) are satisfied. Then problem (1.1) has at least one positive solution.

Proof From (H5), we can choose ε 1 > 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-86/MediaObjects/13661_2012_Article_185_IEq194_HTML.gif such that f > m + ε http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-86/MediaObjects/13661_2012_Article_185_IEq195_HTML.gif and also there exists R 1 > 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-86/MediaObjects/13661_2012_Article_185_IEq196_HTML.gif such that for any i = 1 4 y i R 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-86/MediaObjects/13661_2012_Article_185_IEq197_HTML.gif and t [ a , b ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-86/MediaObjects/13661_2012_Article_185_IEq185_HTML.gif, we have
f ( t , y 1 , y 2 , y 3 , y 4 ) ( m + ε 1 ) i = 1 4 y i . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-86/MediaObjects/13661_2012_Article_185_Equ24_HTML.gif
(3.4)
Let R > R 1 σ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-86/MediaObjects/13661_2012_Article_185_IEq198_HTML.gif. By virtue of (3.4), we know that
f ( t , y ( t ) , y ( t ) , ( T y ) ( t ) , ( S y ) ( t ) ) ( m + ε 1 ) ( y + y + T y + S y ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-86/MediaObjects/13661_2012_Article_185_Equ25_HTML.gif
(3.5)
Set K R = { y K : y 1 < R } http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-86/MediaObjects/13661_2012_Article_185_IEq199_HTML.gif. Then for any y K R K http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-86/MediaObjects/13661_2012_Article_185_IEq200_HTML.gif, by virtue of (3.5), we know that
http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-86/MediaObjects/13661_2012_Article_185_Equw_HTML.gif
So, ( A y ) ( t ) 1 r http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-86/MediaObjects/13661_2012_Article_185_IEq188_HTML.gif. Therefore,
i ( A , K R K , K ) = 0 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-86/MediaObjects/13661_2012_Article_185_Equ26_HTML.gif
(3.6)
By the same method as the selection of r http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-86/MediaObjects/13661_2012_Article_185_IEq201_HTML.gif in Theorem 3.1, we can obtain r < R http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-86/MediaObjects/13661_2012_Article_185_IEq202_HTML.gif satisfying
i ( A , K r K , K ) = 1 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-86/MediaObjects/13661_2012_Article_185_Equ27_HTML.gif
(3.7)
According to (3.6) and (3.7), we get
i ( A , ( K R K ) ( K ¯ r K ) , K ) = i ( A , K R K , K ) i ( A , K r K , K ) = 1 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-86/MediaObjects/13661_2012_Article_185_Equx_HTML.gif

Therefore, A has at least one fixed point on ( K R K ) ( K ¯ r K ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-86/MediaObjects/13661_2012_Article_185_IEq203_HTML.gif. Consequently, problem (1.1) has at least one positive solution. The proof is complete. □

4 Concerned results and applications

In this section, we deal with a special case of the problem (1.1). The method is just similar to what we have done in Section 3, so we omit the proof of some main results of the section. Case F ( t , x ( t ) , x ( t ) ) = f ( t , x ( t ) , x ( t ) , ( A x ) ( t ) , ( B x ) ( t ) ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-86/MediaObjects/13661_2012_Article_185_IEq204_HTML.gif is treated in the following theorem. Under the case, the problem (1.1) reduces to the following boundary value problems:
{ y ( t ) + h ( t ) F ( t , y ( t ) , y ( t ) ) = θ , t J , t t k , Δ y | t = t k = I k ( y ( t k ) ) , Δ y | t = t k = I ¯ k ( y ( t k ) , y ( t k ) ) , k = 1 , , m , α y ( 0 ) β y ( 0 ) = 0 , γ y ( 1 ) + δ y ( 1 ) = 0 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-86/MediaObjects/13661_2012_Article_185_Equ28_HTML.gif
(4.1)

where F C ( J × P × P , P ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-86/MediaObjects/13661_2012_Article_185_IEq205_HTML.gif, h C ( J ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-86/MediaObjects/13661_2012_Article_185_IEq206_HTML.gif.

Theorem 4.1 Assume that (H2) holds, and the following conditions are satisfied:

(C1) F C ( J × P 2 , P ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-86/MediaObjects/13661_2012_Article_185_IEq207_HTML.gif, h C ( J , P ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-86/MediaObjects/13661_2012_Article_185_IEq103_HTML.gifand
F ( t , y 1 , y 2 ) a ( t ) + i = 1 2 b i ( t ) y i , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-86/MediaObjects/13661_2012_Article_185_Equy_HTML.gif
where a ( t ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-86/MediaObjects/13661_2012_Article_185_IEq104_HTML.gifand b i ( t ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-86/MediaObjects/13661_2012_Article_185_IEq105_HTML.gifare Lebesgue integrable functionals on J ( i = 1 , 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-86/MediaObjects/13661_2012_Article_185_IEq208_HTML.gif) and satisfying
http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-86/MediaObjects/13661_2012_Article_185_Equz_HTML.gif

(C2) for any bounded set B i E http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-86/MediaObjects/13661_2012_Article_185_IEq113_HTML.gif ( i = 1 , 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-86/MediaObjects/13661_2012_Article_185_IEq208_HTML.gif), F ( t , B 1 , B 2 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-86/MediaObjects/13661_2012_Article_185_IEq209_HTML.gifand I k ( B 1 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-86/MediaObjects/13661_2012_Article_185_IEq115_HTML.giftogether with I ¯ k ( B 1 , B 2 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-86/MediaObjects/13661_2012_Article_185_IEq210_HTML.gifare relatively compact sets.

(C3) lim y 1 + y 2 0 F ( t , y 1 , y 2 ) y 1 + y 2 > m http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-86/MediaObjects/13661_2012_Article_185_IEq211_HTML.gif, where m is defined by (2.4). Then the problem (4.1) has at least one positive solution.

Theorem 4.2 Assume that (H2) and (C1)∼(C2) hold, and the following condition is satisfied:

(C4) lim y 1 + y 2 F ( t , y 1 , y 2 ) y 1 + y 2 > m http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-86/MediaObjects/13661_2012_Article_185_IEq212_HTML.gif, where m is defined by (2.4). Then the problem (4.1) has at least one positive solution.

To illustrate how our main results can be used in practice, we present an example.

Example 4.1 Consider the following boundary value problem for scalar second-order impulsive integro-differential equation:
{ y ( t ) = π sin t ( ln ( 3 + t 2 ) + t 3 y + t 0 t e t s y d s + 0 1 e 2 s y d s ) 720 t ( 1 + y + 0 t e t s y ( s ) d s + 0 1 e 2 s y ( s ) d s ) 2 , t t 1 , Δ y | t 1 = 1 3 = 1 120 y 3 ( 1 3 ) , Δ y | t 1 = 1 3 = 1 120 ( y 3 ( 1 3 ) + y ( 1 3 ) ) ( y ( 1 3 ) + y ( 1 3 ) ) 2 , α y ( 0 ) β y ( 0 ) = 0 , γ y ( 1 ) + δ y ( 1 ) = 0 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-86/MediaObjects/13661_2012_Article_185_Equ29_HTML.gif
(4.2)

5 Conclusion

The problem (4.2) has at least one positive solution y ( t ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-86/MediaObjects/13661_2012_Article_185_IEq213_HTML.gif.

For t J http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-86/MediaObjects/13661_2012_Article_185_IEq99_HTML.gif, t t 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-86/MediaObjects/13661_2012_Article_185_IEq214_HTML.gif, let h ( t ) = π 720 t http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-86/MediaObjects/13661_2012_Article_185_IEq215_HTML.gif,
F ( t , y , y , ( T y ) ( t ) , ( S y ) ( t ) ) = sin t ( ln ( 3 + t 2 ) + t 3 y + t 0 t e t s y d s + 0 1 e 2 s y d s ) ( 1 + y + 0 t e t s y ( s ) d s + 0 1 e 2 s y ( s ) d s ) 2 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-86/MediaObjects/13661_2012_Article_185_Equaa_HTML.gif
Choose a ( t ) = sin t http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-86/MediaObjects/13661_2012_Article_185_IEq216_HTML.gif. By simple computation, we know that
http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-86/MediaObjects/13661_2012_Article_185_Equab_HTML.gif

Then conditions (H1)∼(H4) are satisfied. Therefore, by Theorem 3.1, the problem (4.2) has at least one positive solution.

Remark 5.1 In [12], by requiring that f satisfies some noncompact measure conditions and P is a normal cone, Guo established the existence of positive solutions for initial value problem. In the paper, we impose some weaker condition on f, we obtain the positive solution of the problem (1.1).

Remark 5.2 For the special case when the problem (1.1) has no singularities and J = [ 0 , a ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-86/MediaObjects/13661_2012_Article_185_IEq217_HTML.gif, our results still hold. Obviously, our theorems generalize and improve the results in [912].

Declarations

Acknowledgements

The author is very grateful to Professor Lishan Liu and Professor R. P. Agarwal for their making many valuable comments. The author would like to express her thanks to the editor of the journal and the anonymous referees for their carefully reading of the first draft of the manuscript and making many helpful comments and suggestions which improved the presentation of the paper. The author was supported financially by the Foundation of Shanghai Municipal Education Commission (Grant Nos. DYL201105).

Authors’ Affiliations

(1)
School of Mathematical Sciences, Fudan University
(2)
Department of Mathematics, Shanghai Normal University

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