Periodic boundary value problems for second-order impulsive integro-differential equations with integral jump conditions

  • Chatthai Thaiprayoon1, 2Email author,

    Affiliated with

    • Decha Samana1, 2 and

      Affiliated with

      • Jessada Tariboon2, 3

        Affiliated with

        Boundary Value Problems20122012:122

        DOI: 10.1186/1687-2770-2012-122

        Received: 24 June 2012

        Accepted: 11 October 2012

        Published: 24 October 2012

        Abstract

        This paper is concerned with the existence of extremal solutions of periodic boundary value problems for second-order impulsive integro-differential equations with integral jump conditions. We introduce a new definition of lower and upper solutions with integral jump conditions and prove some new maximum principles. The method of lower and upper solutions and the monotone iterative technique are used.

        MSC: 34B37, 34K10, 34K45.

        Keywords

        impulsive integro-differential equation lower and upper solutions periodic boundary value problem monotone iterative technique

        1 Introduction

        Differential equations which have impulse effects describe many evolution processes that abruptly change their state at a certain moment. In recent years, impulsive differential equations have become more important tools in some mathematical models of real processes and phenomena studied in physics, biotechnology, chemical technology, population dynamics and economics; see [15]. Many papers have been published about existence analysis of periodic boundary value problems of first and second order for impulsive ordinary or functional or integro-differential equations. We refer the readers to the papers [629]. More recent works on existence results of impulsive problems with integral boundary conditions can be found in [3035] and the reference therein. This literature has lead to significant development of a general theory for impulsive differential equations.

        The monotone iterative technique coupled with the method of upper and lower solutions has been used to study the existence of extremal solutions of periodic boundary value problems for second-order impulsive equations; see, for example, [3641]. This method has been also used to study abstract nonlinear problems; see [42]. However, in most of these papers concerned with applications of the monotone iterative technique to second-order periodic boundary value problems with impulses, the authors assume that the jump conditions at impulse point t k http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-122/MediaObjects/13661_2012_Article_229_IEq1_HTML.gif of solution values and the derivative of solution values depend on the left-hand limits of solutions or the slope of solutions themselves, such as Δ x ( t k ) = I k ( x ( t k ) ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-122/MediaObjects/13661_2012_Article_229_IEq2_HTML.gif, Δ x ( t k ) = I k ( x ( t k ) ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-122/MediaObjects/13661_2012_Article_229_IEq3_HTML.gif, Δ x ( t k ) = I k ( x ( t k ) ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-122/MediaObjects/13661_2012_Article_229_IEq4_HTML.gif, Δ x ( t k ) = I k ( x ( t k ) ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-122/MediaObjects/13661_2012_Article_229_IEq5_HTML.gif.

        In this paper, we consider the periodic boundary value problem for second-order impulsive integro-differential equation (PBVP) with integral jump conditions:
        { x ( t ) = f ( t , x ( t ) , ( K x ) ( t ) , ( S x ) ( t ) ) , t J = [ 0 , T ] , t t k , Δ x ( t k ) = I k ( t k δ k t k ε k x ( s ) d s ) , k = 1 , 2 , , m , Δ x ( t k ) = I k ( t k τ k t k σ k x ( s ) d s ) , k = 1 , 2 , , m , x ( 0 ) = x ( T ) , x ( 0 ) = x ( T ) , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-122/MediaObjects/13661_2012_Article_229_Equ1_HTML.gif
        (1.1)
        where 0 = t 0 < t 1 < t 2 < < t k < < t m < t m + 1 = T http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-122/MediaObjects/13661_2012_Article_229_IEq6_HTML.gif, f : J × R 3 R http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-122/MediaObjects/13661_2012_Article_229_IEq7_HTML.gif is continuous everywhere except at { t k } × R 3 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-122/MediaObjects/13661_2012_Article_229_IEq8_HTML.gif, f ( t k + , x , y , z ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-122/MediaObjects/13661_2012_Article_229_IEq9_HTML.gif, f ( t k , x , y , z ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-122/MediaObjects/13661_2012_Article_229_IEq10_HTML.gif exist, f ( t k , x , y , z ) = f ( t k , x , y , z ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-122/MediaObjects/13661_2012_Article_229_IEq11_HTML.gif, I k C ( R , R ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-122/MediaObjects/13661_2012_Article_229_IEq12_HTML.gif, I k C ( R , R ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-122/MediaObjects/13661_2012_Article_229_IEq13_HTML.gif, Δ x ( t k ) = x ( t k + ) x ( t k ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-122/MediaObjects/13661_2012_Article_229_IEq14_HTML.gif, Δ x ( t k ) = x ( t k + ) x ( t k ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-122/MediaObjects/13661_2012_Article_229_IEq15_HTML.gif, 0 ε k δ k t k t k 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-122/MediaObjects/13661_2012_Article_229_IEq16_HTML.gif, 0 σ k τ k t k t k 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-122/MediaObjects/13661_2012_Article_229_IEq17_HTML.gif, k = 1 , 2 , , m http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-122/MediaObjects/13661_2012_Article_229_IEq18_HTML.gif,
        ( K x ) ( t ) = 0 t k ( t , s ) x ( s ) d s , ( S x ) ( t ) = 0 T h ( t , s ) x ( s ) d s , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-122/MediaObjects/13661_2012_Article_229_Equa_HTML.gif

        k ( t , s ) C ( D , R + ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-122/MediaObjects/13661_2012_Article_229_IEq19_HTML.gif, h ( t , s ) C ( J × J , R + ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-122/MediaObjects/13661_2012_Article_229_IEq20_HTML.gif, D = { ( t , s ) R 2 , 0 s t T } http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-122/MediaObjects/13661_2012_Article_229_IEq21_HTML.gif, R + = [ 0 , + ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-122/MediaObjects/13661_2012_Article_229_IEq22_HTML.gif, k 0 = max { k ( t , s ) : ( t , s ) D } http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-122/MediaObjects/13661_2012_Article_229_IEq23_HTML.gif, h 0 = max { h ( t , s ) : ( t , s ) J × J } http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-122/MediaObjects/13661_2012_Article_229_IEq24_HTML.gif.

        In [43, 44], the authors discussed some kinds of first-order impulsive problems with the integral jump condition
        Δ x ( t k ) = I k ( t k τ k t k x ( s ) d s t k 1 t k 1 + σ k 1 x ( s ) d s ) , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-122/MediaObjects/13661_2012_Article_229_Equ2_HTML.gif
        (1.2)
        where 0 < σ k 1 ( t k t k 1 ) / 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-122/MediaObjects/13661_2012_Article_229_IEq25_HTML.gif, 0 τ k 1 ( t k t k 1 ) / 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-122/MediaObjects/13661_2012_Article_229_IEq26_HTML.gif, k = 1 , 2 , , m http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-122/MediaObjects/13661_2012_Article_229_IEq27_HTML.gif. We note that the jump condition (1.2) depends on functionals of path history before impulse points t k http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-122/MediaObjects/13661_2012_Article_229_IEq1_HTML.gif and after the past impulse points t k 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-122/MediaObjects/13661_2012_Article_229_IEq28_HTML.gif. The aim of our research is to deal with the integral jump conditions
        Δ x ( t k ) = I k ( t k δ k t k ε k x ( s ) d s ) , Δ x ( t k ) = I k ( t k τ k t k σ k x ( s ) d s ) , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-122/MediaObjects/13661_2012_Article_229_Equ3_HTML.gif
        (1.3)

        where 0 ε k δ k t k t k 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-122/MediaObjects/13661_2012_Article_229_IEq16_HTML.gif, 0 σ k τ k t k t k 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-122/MediaObjects/13661_2012_Article_229_IEq29_HTML.gif, k = 1 , 2 , , m http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-122/MediaObjects/13661_2012_Article_229_IEq18_HTML.gif. The integral jump condition (1.3) means that a sudden change of solution values and the derivative of solution values at impulse point t k http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-122/MediaObjects/13661_2012_Article_229_IEq1_HTML.gif depend on the area under the curves of x ( t ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-122/MediaObjects/13661_2012_Article_229_IEq30_HTML.gif and x ( t ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-122/MediaObjects/13661_2012_Article_229_IEq31_HTML.gif between t = t k δ k http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-122/MediaObjects/13661_2012_Article_229_IEq32_HTML.gif to t = t k ε k http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-122/MediaObjects/13661_2012_Article_229_IEq33_HTML.gif and t = t k τ k http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-122/MediaObjects/13661_2012_Article_229_IEq34_HTML.gif to t = t k σ k http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-122/MediaObjects/13661_2012_Article_229_IEq35_HTML.gif, respectively. It should be noticed that the impulsive effects of PBVP (1.1) have memory of the past states.

        This paper is organized as follows. Firstly, we introduce a new concept of lower and upper solutions. After that, we establish some new comparison principles and discuss the existence and uniqueness of the solutions for second-order impulsive integro-differential equations with integral jump conditions. By using the method of upper and lower solutions and the monotone iterative technique, we obtain the existence of an extreme solution of PBVP (1.1). Finally, we give an example to illustrate the obtained results.

        2 Preliminaries

        Let J = J { t 1 , t 2 , , t m } http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-122/MediaObjects/13661_2012_Article_229_IEq36_HTML.gif, J 0 = [ t 0 , t 1 ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-122/MediaObjects/13661_2012_Article_229_IEq37_HTML.gif, J k = ( t k , t k + 1 ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-122/MediaObjects/13661_2012_Article_229_IEq38_HTML.gif for k = 1 , 2 , , m http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-122/MediaObjects/13661_2012_Article_229_IEq39_HTML.gif. Let P C ( J , R ) = { x : J R ; x ( t )  is continuous everywhere except for some  t k  at which  x ( t k + )  and  x ( t k )  exist and x ( t k ) = x ( t k ) , k = 1 , 2 , , m } http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-122/MediaObjects/13661_2012_Article_229_IEq40_HTML.gif, and P C 1 ( J , R ) = { x P C ( J , R ) ; x ( t )  is continuous everywhere except for some  t k  at which  x ( t k + )  and  x ( t k )  exist and  x ( t k ) = x ( t k ) , k = 1 , 2 , , m } http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-122/MediaObjects/13661_2012_Article_229_IEq41_HTML.gif. P C ( J , R ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-122/MediaObjects/13661_2012_Article_229_IEq42_HTML.gif and P C 1 ( J , R ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-122/MediaObjects/13661_2012_Article_229_IEq43_HTML.gif are Banach spaces with the norms x P C = sup { x ( t ) : t J } http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-122/MediaObjects/13661_2012_Article_229_IEq44_HTML.gif and x P C 1 = max { x P C , x P C } http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-122/MediaObjects/13661_2012_Article_229_IEq45_HTML.gif. Let E = P C 1 ( J , R ) C 2 ( J , R ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-122/MediaObjects/13661_2012_Article_229_IEq46_HTML.gif. A function x E http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-122/MediaObjects/13661_2012_Article_229_IEq47_HTML.gif is called a solution of PBVP (1.1) if it satisfies (1.1).

        Definition 2.1 We say that the functions α 0 , β 0 E http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-122/MediaObjects/13661_2012_Article_229_IEq48_HTML.gif are lower and upper solutions of PBVP (1.1), respectively, if there exist M > 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-122/MediaObjects/13661_2012_Article_229_IEq49_HTML.gif, N 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-122/MediaObjects/13661_2012_Article_229_IEq50_HTML.gif, L 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-122/MediaObjects/13661_2012_Article_229_IEq51_HTML.gif, L k 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-122/MediaObjects/13661_2012_Article_229_IEq52_HTML.gif, L k 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-122/MediaObjects/13661_2012_Article_229_IEq53_HTML.gif, 0 ε k δ k t k t k 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-122/MediaObjects/13661_2012_Article_229_IEq16_HTML.gif, 0 σ k τ k t k t k 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-122/MediaObjects/13661_2012_Article_229_IEq17_HTML.gif, such that
        { α 0 ( t ) f ( t , α 0 ( t ) , ( K α 0 ) ( t ) , ( S α 0 ) ( t ) ) + a ( t ) , t J , Δ α 0 ( t k ) = I k ( t k δ k t k ε k α 0 ( s ) d s ) + m k , k = 1 , 2 , , m , Δ α 0 ( t k ) I k ( t k τ k t k σ k α 0 ( s ) d s ) + l k , k = 1 , 2 , , m , α 0 ( 0 ) = α 0 ( T ) , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-122/MediaObjects/13661_2012_Article_229_Equb_HTML.gif
        where
        a ( t ) = { 0 if  α 0 ( 0 ) α 0 ( T ) , [ α 0 ( T ) α 0 ( 0 ) ] T [ 2 + M [ t T t 2 ] + N 0 t k ( t , s ) ( s T s 2 ) d s + L 0 T h ( t , s ) ( s T s 2 ) d s ] if  α 0 ( 0 ) < α 0 ( T ) , m k = { 0 if  α 0 ( 0 ) α 0 ( T ) , L k [ α 0 ( T ) α 0 ( 0 ) ] T t k δ k t k ε k ( T 2 s ) d s if  α 0 ( 0 ) < α 0 ( T ) , l k = { 0 if  α 0 ( 0 ) α 0 ( T ) , L k [ α 0 ( T ) α 0 ( 0 ) ] T t k τ k t k σ k ( s T s 2 ) d s if  α 0 ( 0 ) < α 0 ( T ) , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-122/MediaObjects/13661_2012_Article_229_Equc_HTML.gif
        and
        { β 0 ( t ) f ( t , β 0 ( t ) , ( K β 0 ) ( t ) , ( S β 0 ) ( t ) ) b ( t ) , t J , Δ β 0 ( t k ) = I k ( t k δ k t k ε k β 0 ( s ) d s ) m k , k = 1 , 2 , , m , Δ β 0 ( t k ) I k ( t k τ k t k σ k β 0 ( s ) d s ) l k , k = 1 , 2 , , m , β 0 ( 0 ) = β 0 ( T ) , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-122/MediaObjects/13661_2012_Article_229_Equd_HTML.gif
        where
        b ( t ) = { 0 if  β 0 ( 0 ) β 0 ( T ) , [ β 0 ( 0 ) β 0 ( T ) ] T [ 2 + M [ t T t 2 ] + N 0 t k ( t , s ) ( s T s 2 ) d s + L 0 T h ( t , s ) ( s T s 2 ) d s ] if  β 0 ( 0 ) > β 0 ( T ) , m k = { 0 if  β 0 ( 0 ) β 0 ( T ) , L k [ β 0 ( 0 ) β 0 ( T ) ] T t k δ k t k ε k ( T 2 s ) d s if  β 0 ( 0 ) > β 0 ( T ) , l k = { 0 if  β 0 ( 0 ) β 0 ( T ) , L k [ β 0 ( 0 ) β 0 ( T ) ] T t k τ k t k σ k ( s T s 2 ) d s if  β 0 ( 0 ) > β 0 ( T ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-122/MediaObjects/13661_2012_Article_229_Eque_HTML.gif

        Now we are in the position to establish some new comparison principles which play an important role in the monotone iterative technique.

        Lemma 2.1 Assume that x E http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-122/MediaObjects/13661_2012_Article_229_IEq47_HTML.gif satisfies
        { x ( t ) M x ( t ) + N 0 t k ( t , s ) x ( s ) d s + L 0 T h ( t , s ) x ( s ) d s , t J , Δ x ( t k ) = L k t k δ k t k ε k x ( s ) d s , k = 1 , 2 , , m , Δ x ( t k ) L k t k τ k t k σ k x ( s ) d s , k = 1 , 2 , , m , x ( 0 ) = x ( T ) , x ( 0 ) x ( T ) , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-122/MediaObjects/13661_2012_Article_229_Equ4_HTML.gif
        (2.1)
        where M > 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-122/MediaObjects/13661_2012_Article_229_IEq49_HTML.gif, N 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-122/MediaObjects/13661_2012_Article_229_IEq50_HTML.gif, L 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-122/MediaObjects/13661_2012_Article_229_IEq51_HTML.gif, L k 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-122/MediaObjects/13661_2012_Article_229_IEq52_HTML.gif, L k 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-122/MediaObjects/13661_2012_Article_229_IEq53_HTML.gifare constants and 0 ε k δ k t k t k 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-122/MediaObjects/13661_2012_Article_229_IEq16_HTML.gif, 0 σ k τ k t k t k 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-122/MediaObjects/13661_2012_Article_229_IEq17_HTML.gif, k = 1 , 2 , , m http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-122/MediaObjects/13661_2012_Article_229_IEq54_HTML.gif, and they satisfy
        [ k = 1 m L k ( δ k ε k ) + T ] [ k = 1 m L k ( τ k σ k ) + ( M + N k 0 T + L h 0 T ) T ] 1 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-122/MediaObjects/13661_2012_Article_229_Equ5_HTML.gif
        (2.2)

        Then x ( t ) 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-122/MediaObjects/13661_2012_Article_229_IEq55_HTML.gif, t J http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-122/MediaObjects/13661_2012_Article_229_IEq56_HTML.gif.

        Proof Suppose, to the contrary, that x ( t ) > 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-122/MediaObjects/13661_2012_Article_229_IEq57_HTML.gif for some t J http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-122/MediaObjects/13661_2012_Article_229_IEq56_HTML.gif. We divide the proof into two cases:

        Case (i). There exists a t ˜ J http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-122/MediaObjects/13661_2012_Article_229_IEq58_HTML.gif such that x ( t ˜ ) > 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-122/MediaObjects/13661_2012_Article_229_IEq59_HTML.gif and x ( t ) 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-122/MediaObjects/13661_2012_Article_229_IEq60_HTML.gif for all t J http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-122/MediaObjects/13661_2012_Article_229_IEq56_HTML.gif.

        From (2.1), we have x ( t ) 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-122/MediaObjects/13661_2012_Article_229_IEq61_HTML.gif for t J http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-122/MediaObjects/13661_2012_Article_229_IEq62_HTML.gif. Since Δ x ( t k ) L k t k τ k t k σ k x ( s ) d s 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-122/MediaObjects/13661_2012_Article_229_IEq63_HTML.gif, then x ( t ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-122/MediaObjects/13661_2012_Article_229_IEq30_HTML.gif is nondecreasing in t J http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-122/MediaObjects/13661_2012_Article_229_IEq56_HTML.gif and so x ( 0 ) x ( T ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-122/MediaObjects/13661_2012_Article_229_IEq64_HTML.gif. However, by (2.1) x ( 0 ) x ( T ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-122/MediaObjects/13661_2012_Article_229_IEq65_HTML.gif, then x ( 0 ) = x ( T ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-122/MediaObjects/13661_2012_Article_229_IEq66_HTML.gif, which implies x ( t ) =  constant http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-122/MediaObjects/13661_2012_Article_229_IEq67_HTML.gif for all t J http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-122/MediaObjects/13661_2012_Article_229_IEq56_HTML.gif. Thus, 0 = x ( t ) M x ( t ˜ ) > 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-122/MediaObjects/13661_2012_Article_229_IEq68_HTML.gif, a contradiction.

        Case (ii). There exists t , t J http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-122/MediaObjects/13661_2012_Article_229_IEq69_HTML.gif such that x ( t ) > 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-122/MediaObjects/13661_2012_Article_229_IEq70_HTML.gif, x ( t ) < 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-122/MediaObjects/13661_2012_Article_229_IEq71_HTML.gif.

        Let inf t J x ( t ) = λ < 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-122/MediaObjects/13661_2012_Article_229_IEq72_HTML.gif, then there exists t J i http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-122/MediaObjects/13661_2012_Article_229_IEq73_HTML.gif, for some i { 0 , 1 , , m } http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-122/MediaObjects/13661_2012_Article_229_IEq74_HTML.gif, such that x ( t ) = λ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-122/MediaObjects/13661_2012_Article_229_IEq75_HTML.gif or x ( t i + ) = λ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-122/MediaObjects/13661_2012_Article_229_IEq76_HTML.gif. Without loss of generality, we only consider x ( t ) = λ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-122/MediaObjects/13661_2012_Article_229_IEq75_HTML.gif. For the case x ( t i + ) = λ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-122/MediaObjects/13661_2012_Article_229_IEq76_HTML.gif the proof is similar. It follows that
        http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-122/MediaObjects/13661_2012_Article_229_Equf_HTML.gif

        If x ( t ) > 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-122/MediaObjects/13661_2012_Article_229_IEq77_HTML.gif for all t J http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-122/MediaObjects/13661_2012_Article_229_IEq56_HTML.gif, then Δ x ( t k ) = L k t k δ k t k ε k x ( s ) d s 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-122/MediaObjects/13661_2012_Article_229_IEq78_HTML.gif, k = 1 , 2 , , m http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-122/MediaObjects/13661_2012_Article_229_IEq79_HTML.gif. Hence, x ( t ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-122/MediaObjects/13661_2012_Article_229_IEq31_HTML.gif is strictly increasing on J, which contradicts x ( 0 ) = x ( T ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-122/MediaObjects/13661_2012_Article_229_IEq80_HTML.gif. Then there exists a t ¯ J http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-122/MediaObjects/13661_2012_Article_229_IEq81_HTML.gif such that x ( t ¯ ) 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-122/MediaObjects/13661_2012_Article_229_IEq82_HTML.gif.

        Let t ¯ J j http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-122/MediaObjects/13661_2012_Article_229_IEq83_HTML.gif, j { 0 , 1 , , m } http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-122/MediaObjects/13661_2012_Article_229_IEq84_HTML.gif. By mean value theorem, we have
        http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-122/MediaObjects/13661_2012_Article_229_Equg_HTML.gif
        Summing up the above inequalities, we obtain
        x ( 0 ) x ( t ¯ ) + λ [ k = 1 j L k ( τ k σ k ) + ( M + N k 0 T + L h 0 T ) ( t ¯ t 0 ) ] λ [ k = 1 j L k ( τ k σ k ) + ( M + N k 0 T + L h 0 T ) t ¯ ] . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-122/MediaObjects/13661_2012_Article_229_Equ6_HTML.gif
        (2.3)
        Let t J h http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-122/MediaObjects/13661_2012_Article_229_IEq85_HTML.gif, h { 0 , 1 , , m } http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-122/MediaObjects/13661_2012_Article_229_IEq86_HTML.gif. If t t ¯ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-122/MediaObjects/13661_2012_Article_229_IEq87_HTML.gif by using the method to get (2.3), then we have
        x ( t ) x ( t ¯ ) + λ [ k = h + 1 j L k ( τ k σ k ) + ( M + N k 0 T + L h 0 T ) ( t ¯ t ) ] λ [ k = 1 m L k ( τ k σ k ) + ( M + N k 0 T + L h 0 T ) T ] . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-122/MediaObjects/13661_2012_Article_229_Equh_HTML.gif
        If t > t ¯ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-122/MediaObjects/13661_2012_Article_229_IEq88_HTML.gif, then the above method together with (2.1), (2.3) implies that
        x ( t ) x ( T ) + λ [ k = h + 1 m L k ( τ k σ k ) + ( M + N k 0 T + L h 0 T ) ( T t ) ] x ( 0 ) + λ [ k = h + 1 m L k ( τ k σ k ) + ( M + N k 0 T + L h 0 T ) ( T t ) ] λ [ k = 1 j L k ( τ k σ k ) + ( M + N k 0 T + L h 0 T ) t ¯ ] + λ [ k = h + 1 m L k ( τ k σ k ) + ( M + N k 0 T + L h 0 T ) ( T t ) ] λ [ k = 1 m L k ( τ k σ k ) + ( M + N k 0 T + L h 0 T ) T ] . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-122/MediaObjects/13661_2012_Article_229_Equi_HTML.gif
        Thus,
        x ( t ) λ [ k = 1 m L k ( τ k σ k ) + ( M + N k 0 T + L h 0 T ) T ] , t J . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-122/MediaObjects/13661_2012_Article_229_Equj_HTML.gif
        Let t J r http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-122/MediaObjects/13661_2012_Article_229_IEq89_HTML.gif for some r { 0 , 1 , , m } http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-122/MediaObjects/13661_2012_Article_229_IEq90_HTML.gif. We first assume that t < t http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-122/MediaObjects/13661_2012_Article_229_IEq91_HTML.gif, then i < r http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-122/MediaObjects/13661_2012_Article_229_IEq92_HTML.gif. By the mean value theorem, we have
        http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-122/MediaObjects/13661_2012_Article_229_Equk_HTML.gif
        Summing up, we get
        0 < x ( t ) λ + λ [ k = 1 m L k ( δ k ε k ) + T ] [ k = 1 m L k ( τ k σ k ) + ( M + N k 0 T + L h 0 T ) T ] . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-122/MediaObjects/13661_2012_Article_229_Equl_HTML.gif
        Hence,
        [ k = 1 m L k ( δ k ε k ) + T ] [ k = 1 m L k ( τ k σ k ) + ( M + N k 0 T + L h 0 T ) T ] > 1 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-122/MediaObjects/13661_2012_Article_229_Equm_HTML.gif

        which contradicts (2.2).

        For the case t < t http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-122/MediaObjects/13661_2012_Article_229_IEq93_HTML.gif, the proof is similar, and thus we omit it. This completes the proof. □

        Lemma 2.2 Assume that x E http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-122/MediaObjects/13661_2012_Article_229_IEq47_HTML.gif satisfies
        http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-122/MediaObjects/13661_2012_Article_229_Equn_HTML.gif

        where M > 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-122/MediaObjects/13661_2012_Article_229_IEq49_HTML.gif, N 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-122/MediaObjects/13661_2012_Article_229_IEq50_HTML.gif, L 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-122/MediaObjects/13661_2012_Article_229_IEq51_HTML.gif, L k 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-122/MediaObjects/13661_2012_Article_229_IEq52_HTML.gif, L k 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-122/MediaObjects/13661_2012_Article_229_IEq53_HTML.gifare constants and 0 ε k δ k t k t k 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-122/MediaObjects/13661_2012_Article_229_IEq16_HTML.gif, 0 σ k τ k t k t k 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-122/MediaObjects/13661_2012_Article_229_IEq17_HTML.gif, k = 1 , 2 , , m http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-122/MediaObjects/13661_2012_Article_229_IEq54_HTML.gif, and they satisfy (2.2). Then x ( t ) 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-122/MediaObjects/13661_2012_Article_229_IEq55_HTML.giffor all t J http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-122/MediaObjects/13661_2012_Article_229_IEq56_HTML.gif.

        Proof Let u ( t ) = [ t T t 2 T ] [ x ( T ) x ( 0 ) ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-122/MediaObjects/13661_2012_Article_229_IEq94_HTML.gif, t J http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-122/MediaObjects/13661_2012_Article_229_IEq56_HTML.gif, and define
        w ( t ) = x ( t ) + u ( t ) = x ( t ) + [ t T t 2 T ] [ x ( T ) x ( 0 ) ] . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-122/MediaObjects/13661_2012_Article_229_Equo_HTML.gif
        Note that u ( 0 ) = u ( T ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-122/MediaObjects/13661_2012_Article_229_IEq95_HTML.gif, u ( t ) 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-122/MediaObjects/13661_2012_Article_229_IEq96_HTML.gif for t J http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-122/MediaObjects/13661_2012_Article_229_IEq97_HTML.gif. If we prove that w 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-122/MediaObjects/13661_2012_Article_229_IEq98_HTML.gif, then x ( t ) x ( t ) + u ( t ) 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-122/MediaObjects/13661_2012_Article_229_IEq99_HTML.gif and the proof is complete. Since u ( t ) = [ T 2 t T ] [ x ( T ) x ( 0 ) ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-122/MediaObjects/13661_2012_Article_229_IEq100_HTML.gif, then we get
        w ( 0 ) = x ( 0 ) + u ( 0 ) = x ( T ) + u ( T ) = w ( T ) , w ( 0 ) = x ( 0 ) + u ( 0 ) = x ( 0 ) + x ( T ) x ( 0 ) = x ( T ) , w ( T ) = x ( T ) + u ( T ) = x ( T ) x ( T ) + x ( 0 ) = x ( 0 ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-122/MediaObjects/13661_2012_Article_229_Equp_HTML.gif
        Hence, w ( 0 ) > w ( T ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-122/MediaObjects/13661_2012_Article_229_IEq101_HTML.gif. Indeed, for k = 1 , 2 , , m http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-122/MediaObjects/13661_2012_Article_229_IEq27_HTML.gif,
        Δ w ( t k ) = Δ x ( t k ) + Δ u ( t k ) = L k t k δ k t k ε k x ( s ) d s + L k [ x ( T ) x ( 0 ) ] T t k δ k t k ε k T 2 s d s = L k t k δ k t k ε k x ( s ) + u ( s ) d s = L k t k δ k t k ε k w ( s ) d s , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-122/MediaObjects/13661_2012_Article_229_Equq_HTML.gif
        and
        Δ w ( t k ) = Δ x ( t k ) + Δ u ( t k ) L k t k τ k t k σ k x ( s ) d s + L k [ x ( T ) x ( 0 ) ] T t k τ k t k σ k s T s 2 d s L k t k τ k t k σ k x ( s ) + u ( s ) d s L k t k τ k t k σ k w ( s ) d s . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-122/MediaObjects/13661_2012_Article_229_Equr_HTML.gif
        Meanwhile, for t t k http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-122/MediaObjects/13661_2012_Article_229_IEq102_HTML.gif, t J http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-122/MediaObjects/13661_2012_Article_229_IEq97_HTML.gif,
        w ( t ) M w ( t ) N 0 t k ( t , s ) w ( s ) d s L 0 T h ( t , s ) w ( s ) d s = x ( t ) M x ( t ) M t T [ x ( T ) x ( 0 ) ] N 0 t k ( t , s ) x ( s ) d s L 0 T h ( t , s ) x ( s ) d s [ x ( T ) x ( 0 ) ] T [ 2 + M [ t T t 2 ] + N 0 t k ( t , s ) ( s T s 2 ) d s + L 0 T h ( t , s ) ( s T s 2 ) d s ] 0 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-122/MediaObjects/13661_2012_Article_229_Equs_HTML.gif

        Then by Lemma 2.1, we get w ( t ) 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-122/MediaObjects/13661_2012_Article_229_IEq103_HTML.gif for all t J http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-122/MediaObjects/13661_2012_Article_229_IEq56_HTML.gif, which implies that x ( t ) 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-122/MediaObjects/13661_2012_Article_229_IEq55_HTML.gif, t J http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-122/MediaObjects/13661_2012_Article_229_IEq56_HTML.gif. □

        Consider the linear PBVP
        { x ( t ) = M x ( t ) + N 0 t k ( t , s ) x ( s ) d s + L 0 T h ( t , s ) x ( s ) d s g ( t ) , t J , Δ x ( t k ) = L k t k δ k t k ε k x ( s ) d s + γ k , k = 1 , 2 , , m , Δ x ( t k ) = L k t k τ k t k σ k x ( s ) d s + λ k , k = 1 , 2 , , m , x ( 0 ) = x ( T ) , x ( 0 ) = x ( T ) , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-122/MediaObjects/13661_2012_Article_229_Equ7_HTML.gif
        (2.4)

        where constants M > 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-122/MediaObjects/13661_2012_Article_229_IEq49_HTML.gif, N 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-122/MediaObjects/13661_2012_Article_229_IEq50_HTML.gif, L 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-122/MediaObjects/13661_2012_Article_229_IEq51_HTML.gif, L k 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-122/MediaObjects/13661_2012_Article_229_IEq52_HTML.gif, L k 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-122/MediaObjects/13661_2012_Article_229_IEq53_HTML.gif, γ k http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-122/MediaObjects/13661_2012_Article_229_IEq104_HTML.gif, λ k http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-122/MediaObjects/13661_2012_Article_229_IEq105_HTML.gif are constants and g P C ( J , R ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-122/MediaObjects/13661_2012_Article_229_IEq106_HTML.gif, 0 ε k δ k t k t k 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-122/MediaObjects/13661_2012_Article_229_IEq16_HTML.gif, 0 σ k τ k t k t k 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-122/MediaObjects/13661_2012_Article_229_IEq17_HTML.gif, k = 1 , 2 , , m http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-122/MediaObjects/13661_2012_Article_229_IEq54_HTML.gif.

        Lemma 2.3 x E http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-122/MediaObjects/13661_2012_Article_229_IEq107_HTML.gifis a solution of (2.4) if and only if x P C 1 ( J , R ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-122/MediaObjects/13661_2012_Article_229_IEq108_HTML.gifis a solution of the following impulsive integral equation:
        x ( t ) = 0 T G 1 ( t , s ) [ g ( s ) N 0 s k ( s , r ) x ( r ) d r L 0 T h ( s , r ) x ( r ) d r ] d s + k = 1 m [ G 1 ( t , t k ) ( L k t k τ k t k σ k x ( s ) d s + λ k ) + G 2 ( t , t k ) ( L k t k δ k t k ε k x ( s ) d s + γ k ) ] , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-122/MediaObjects/13661_2012_Article_229_Equ8_HTML.gif
        (2.5)
        where
        G 1 ( t , s ) = [ 2 M ( e M T 1 ) ] 1 { e M ( T t + s ) + e M ( t s ) , 0 s < t T , e M ( T + t s ) + e M ( s t ) , 0 t s T , G 2 ( t , s ) = [ 2 ( e M T 1 ) ] 1 { e M ( T t + s ) e M ( t s ) , 0 s < t T , e M ( T + t s ) + e M ( s t ) , 0 t s T . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-122/MediaObjects/13661_2012_Article_229_Equt_HTML.gif

        This proof is similar to the proof of Lemma 2.1 in [36], and we omit it.

        Lemma 2.4 Let M > 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-122/MediaObjects/13661_2012_Article_229_IEq49_HTML.gif, N 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-122/MediaObjects/13661_2012_Article_229_IEq50_HTML.gif, L 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-122/MediaObjects/13661_2012_Article_229_IEq51_HTML.gif, L k 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-122/MediaObjects/13661_2012_Article_229_IEq52_HTML.gif, L k 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-122/MediaObjects/13661_2012_Article_229_IEq53_HTML.gifare constants and 0 ε k δ k t k t k 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-122/MediaObjects/13661_2012_Article_229_IEq16_HTML.gif, 0 σ k τ k t k t k 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-122/MediaObjects/13661_2012_Article_229_IEq17_HTML.gif, k = 1 , 2 , , m http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-122/MediaObjects/13661_2012_Article_229_IEq54_HTML.gif. If
        http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-122/MediaObjects/13661_2012_Article_229_Equ9_HTML.gif
        (2.6)
        http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-122/MediaObjects/13661_2012_Article_229_Equ10_HTML.gif
        (2.7)

        then (2.4) has a unique solution x in E.

        Proof For any x E http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-122/MediaObjects/13661_2012_Article_229_IEq109_HTML.gif, we define an operator F by
        ( F x ) ( t ) = 0 T G 1 ( t , s ) [ g ( s ) N 0 s k ( s , r ) x ( r ) d r L 0 T h ( s , r ) x ( r ) d r ] d s + k = 1 m [ G 1 ( t , t k ) ( L k t k τ k t k σ k x ( s ) d s + λ k ) + G 2 ( t , t k ) ( L k t k δ k t k ε k x ( s ) d s + γ k ) ] , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-122/MediaObjects/13661_2012_Article_229_Equu_HTML.gif
        where G 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-122/MediaObjects/13661_2012_Article_229_IEq110_HTML.gif, G 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-122/MediaObjects/13661_2012_Article_229_IEq111_HTML.gif are given by Lemma 2.3. Then F x P C 1 ( J , R ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-122/MediaObjects/13661_2012_Article_229_IEq112_HTML.gif and
        ( F x ) ( t ) = 0 T G 2 ( t , s ) [ g ( s ) N 0 s k ( s , r ) x ( r ) d r L 0 T h ( s , r ) x ( r ) d r ] d s + k = 1 m [ G 2 ( t , t k ) ( L k t k τ k t k σ k x ( s ) d s + λ k ) M G 1 ( t , t k ) ( L k t k δ k t k ε k x ( s ) d s + γ k ) ] . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-122/MediaObjects/13661_2012_Article_229_Equv_HTML.gif
        By computing directly, we have
        max ( t , s ) J × J { G 1 ( t , s ) } = 1 + e M T 2 M ( e M T 1 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-122/MediaObjects/13661_2012_Article_229_Equw_HTML.gif
        and
        max ( t , s ) J × J { G 2 ( t , s ) } = 1 2 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-122/MediaObjects/13661_2012_Article_229_Equx_HTML.gif
        On the other hand, for x , y P C 1 ( J , R ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-122/MediaObjects/13661_2012_Article_229_IEq113_HTML.gif, we have
        ( F x ) ( F y ) P C = sup t J | ( F x ) ( t ) ( F y ) ( t ) | = sup t J | 0 T G 1 ( t , s ) [ N 0 s k ( s , r ) [ x ( r ) y ( r ) ] d r L 0 T h ( s , r ) [ x ( r ) y ( r ) ] d r ] d s + k = 1 m [ G 1 ( t , t k ) ( L k t k τ k t k σ k [ x ( s ) y ( s ) ] d s ) + G 2 ( t , t k ) ( L k t k δ k t k ε k [ x ( s ) y ( s ) ] d s ) ] | sup t J 0 T G 1 ( t , s ) [ N | x ( s ) y ( s ) | 0 s k ( s , r ) d r + L | x ( s ) y ( s ) | 0 T h ( s , r ) d r ] d s + k = 1 m [ G 1 ( t , t k ) L k ( τ k σ k ) | x ( t ) y ( t ) | + G 2 ( t , t k ) L k ( δ k ε k ) | x ( t ) y ( t ) | ] x y P C 1 [ sup t J 0 T G 1 ( t , s ) [ N 0 s k ( s , r ) d r + L 0 T h ( s , r ) d r ] d s + k = 1 m [ G 1 ( t , t k ) L k ( τ k σ k ) + G 2 ( t , t k ) L k ( δ k ε k ) ] ] ψ x y P C 1 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-122/MediaObjects/13661_2012_Article_229_Equy_HTML.gif
        Similarly,
        ( F x ) ( F y ) P C = sup t J | 0 T G 2 ( t , s ) [ N 0 s k ( s , r ) [ x ( r ) y ( r ) ] d r L 0 T h ( s , r ) [ x ( r ) y ( r ) ] d r ] d s + k = 1 m [ G 2 ( t , t k ) ( L k t k τ k t k σ k [ x ( s ) y ( s ) ] d s ) M G 1 ( t , t k ) ( L k t k δ k t k ε k [ x ( s ) y ( s ) ] d s ) ] | sup t J 0 T G 2 ( t , s ) [ N | x ( s ) y ( s ) | 0 s k ( s , r ) d r + L | x ( s ) y ( s ) | 0 T h ( s , r ) d r ] d s + k = 1 m [ G 2 ( t , t k ) L k ( τ k σ k ) | x ( t ) y ( t ) | + M G 1 ( t , t k ) L k ( δ k ε k ) | x ( s ) y ( s ) | ] x y P C 1 [ sup t J 0 T G 2 ( t , s ) [ N 0 s k ( s , r ) d r + L 0 T h ( s , r ) d r ] d s + k = 1 m [ G 2 ( t , t k ) L k ( τ k σ k ) + M G 1 ( t , t k ) L k ( δ k ε k ) ] ] μ x y P C 1 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-122/MediaObjects/13661_2012_Article_229_Equz_HTML.gif
        Thus,
        ( F x ) ( F y ) P C 1 max { ψ , μ } x y P C 1 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-122/MediaObjects/13661_2012_Article_229_Equaa_HTML.gif

        By the Banach fixed-point theorem, F has a unique fixed point x P C 1 ( J , R ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-122/MediaObjects/13661_2012_Article_229_IEq114_HTML.gif, and by Lemma 2.3, x http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-122/MediaObjects/13661_2012_Article_229_IEq115_HTML.gif is also the unique solution of (2.4). This completes the proof. □

        3 Main results

        In this section, we establish existence criteria for solutions of PBVP (1.1) by the method of lower and upper solutions and the monotone iterative technique. For α 0 , β 0 E http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-122/MediaObjects/13661_2012_Article_229_IEq48_HTML.gif, we write α 0 β 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-122/MediaObjects/13661_2012_Article_229_IEq116_HTML.gif if α 0 ( t ) β 0 ( t ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-122/MediaObjects/13661_2012_Article_229_IEq117_HTML.gif for all t J http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-122/MediaObjects/13661_2012_Article_229_IEq56_HTML.gif. In such a case, we denote [ α 0 , β 0 ] = { x E : α 0 ( t ) x ( t ) β 0 ( t ) , t J } http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-122/MediaObjects/13661_2012_Article_229_IEq118_HTML.gif.

        Theorem 3.1 Suppose that the following conditions hold:

        (H1) α 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-122/MediaObjects/13661_2012_Article_229_IEq119_HTML.gifand β 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-122/MediaObjects/13661_2012_Article_229_IEq120_HTML.gifare lower and upper solutions for PBVP (1.1), respectively, such that α 0 β 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-122/MediaObjects/13661_2012_Article_229_IEq121_HTML.gif.

        (H2) The function f satisfies
        f ( t , x 2 , y 2 , z 2 ) f ( t , x 1 , y 1 , z 1 ) M ( x 2 x 1 ) + N ( y 2 y 1 ) + L ( z 2 z 1 ) , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-122/MediaObjects/13661_2012_Article_229_Equab_HTML.gif

        for all t J http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-122/MediaObjects/13661_2012_Article_229_IEq56_HTML.gif, α 0 ( t ) x 1 x 2 β 0 ( t ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-122/MediaObjects/13661_2012_Article_229_IEq122_HTML.gif, ( K α 0 ) ( t ) y 1 y 2 ( K β 0 ) ( t ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-122/MediaObjects/13661_2012_Article_229_IEq123_HTML.gif, ( S α 0 ) ( t ) z 1 z 2 ( S β 0 ) ( t ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-122/MediaObjects/13661_2012_Article_229_IEq124_HTML.gif.

        (H3) M > 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-122/MediaObjects/13661_2012_Article_229_IEq49_HTML.gif, N 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-122/MediaObjects/13661_2012_Article_229_IEq50_HTML.gif, L 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-122/MediaObjects/13661_2012_Article_229_IEq51_HTML.gif, L k 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-122/MediaObjects/13661_2012_Article_229_IEq52_HTML.gif, L k 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-122/MediaObjects/13661_2012_Article_229_IEq125_HTML.gifare constants, and 0 ε k δ k t k t k 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-122/MediaObjects/13661_2012_Article_229_IEq16_HTML.gif, 0 σ k τ k t k t k 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-122/MediaObjects/13661_2012_Article_229_IEq17_HTML.gif, k = 1 , 2 , , m http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-122/MediaObjects/13661_2012_Article_229_IEq79_HTML.gif, and they satisfy (2.2), (2.6) and (2.7).

        (H4) The functions I k http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-122/MediaObjects/13661_2012_Article_229_IEq126_HTML.gif, I k http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-122/MediaObjects/13661_2012_Article_229_IEq127_HTML.gifsatisfy
        I k ( t k δ k t k ε k x ( s ) d s ) I k ( t k δ k t k ε k y ( s ) d s ) = L k t k δ k t k ε k x ( s ) y ( s ) d s , I k ( t k τ k t k σ k x ( s ) d s ) I k ( t k τ k t k σ k y ( s ) d s ) L k t k τ k t k σ k x ( s ) y ( s ) d s , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-122/MediaObjects/13661_2012_Article_229_Equac_HTML.gif

        where t k τ k t k σ k α 0 ( s ) d s t k τ k t k σ k y ( s ) d s t k τ k t k σ k x ( s ) d s t k τ k t k σ k β 0 ( s ) d s http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-122/MediaObjects/13661_2012_Article_229_IEq128_HTML.gif, 0 σ k τ k t k t k 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-122/MediaObjects/13661_2012_Article_229_IEq17_HTML.gif, 0 ε k δ k t k t k 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-122/MediaObjects/13661_2012_Article_229_IEq16_HTML.gif, k = 1 , 2 , , m http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-122/MediaObjects/13661_2012_Article_229_IEq129_HTML.gif.

        Then there exist monotone sequences { α n } , { β n } E http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-122/MediaObjects/13661_2012_Article_229_IEq130_HTML.gifwhich converge in E to the extreme solutions of PBVP (1.1) in [ α 0 , β 0 ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-122/MediaObjects/13661_2012_Article_229_IEq131_HTML.gif, respectively.

        Proof For any η [ α 0 , β 0 ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-122/MediaObjects/13661_2012_Article_229_IEq132_HTML.gif, we consider linear PBVP (2.4) with
        g ( t ) = f ( t , η ( t ) , ( K η ) ( t ) , ( S η ) ( t ) ) M η ( t ) N 0 t k ( t , s ) η ( s ) d s L 0 T h ( t , s ) η ( s ) d s , γ k = I k ( t k δ k t k ε k η ( s ) d s ) L k t k δ k t k ε k η ( s ) d s , k = 1 , 2 , , m , λ k = I k ( t k τ k t k σ k η ( s ) d s ) L k t k τ k t k σ k η ( s ) d s , k = 1 , 2 , , m . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-122/MediaObjects/13661_2012_Article_229_Equad_HTML.gif

        By Lemma 2.4, PBVP (2.4) has a unique solution x E http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-122/MediaObjects/13661_2012_Article_229_IEq47_HTML.gif. We define an operator A from [ α 0 , β 0 ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-122/MediaObjects/13661_2012_Article_229_IEq133_HTML.gif to E by x ( t ) = A η ( t ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-122/MediaObjects/13661_2012_Article_229_IEq134_HTML.gif. We complete the proof in four steps.

        Step 1. We claim that α 0 A α 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-122/MediaObjects/13661_2012_Article_229_IEq135_HTML.gif and A β 0 β 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-122/MediaObjects/13661_2012_Article_229_IEq136_HTML.gif. We only prove α 0 A α 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-122/MediaObjects/13661_2012_Article_229_IEq135_HTML.gif since the second inequality can be proved in a similar manner.

        Let α 1 = A α 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-122/MediaObjects/13661_2012_Article_229_IEq137_HTML.gif and p = α 0 α 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-122/MediaObjects/13661_2012_Article_229_IEq138_HTML.gif. Then α 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-122/MediaObjects/13661_2012_Article_229_IEq139_HTML.gif satisfies
        http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-122/MediaObjects/13661_2012_Article_229_Equae_HTML.gif

        We finish Step 1 in two cases.

        Case 1. α 0 ( 0 ) α 0 ( T ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-122/MediaObjects/13661_2012_Article_229_IEq140_HTML.gif, which implies that
        a ( t ) = 0 , α 0 ( t ) f ( t , α 0 ( t ) , ( K α 0 ) ( t ) , ( S α 0 ) ( t ) ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-122/MediaObjects/13661_2012_Article_229_Equaf_HTML.gif
        As α 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-122/MediaObjects/13661_2012_Article_229_IEq119_HTML.gif is a lower solution of PBVP (1.1), then for t J http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-122/MediaObjects/13661_2012_Article_229_IEq62_HTML.gif,
        p ( t ) M p ( t ) N ( K p ) ( t ) L ( S p ) ( t ) = α 0 ( t ) M α 0 ( t ) N ( K α 0 ) ( t ) L ( S α 0 ) ( t ) α 1 ( t ) + M α 1 ( t ) + N ( K α 1 ) ( t ) + L ( S α 1 ) ( t ) f ( t , α 0 ( t ) , ( K α 0 ) ( t ) , ( S α 0 ) ( t ) ) M α 0 ( t ) N ( K α 0 ) ( t ) L ( S α 0 ) ( t ) f ( t , α 0 ( t ) , ( K α 0 ) ( t ) , ( S α 0 ) ( t ) ) + M α 0 ( t ) + N ( K α 0 ) ( t ) + L ( S α 0 ) ( t ) = 0 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-122/MediaObjects/13661_2012_Article_229_Equag_HTML.gif
        and
        http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-122/MediaObjects/13661_2012_Article_229_Equah_HTML.gif

        Then by Lemma 2.1, p ( t ) 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-122/MediaObjects/13661_2012_Article_229_IEq141_HTML.gif, which implies that α 0 ( t ) A α 0 ( t ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-122/MediaObjects/13661_2012_Article_229_IEq142_HTML.gif, i.e., α 0 A α 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-122/MediaObjects/13661_2012_Article_229_IEq135_HTML.gif.

        Case 2. α 0 ( 0 ) < α 0 ( T ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-122/MediaObjects/13661_2012_Article_229_IEq143_HTML.gif, which implies that
        a ( t ) = [ α 0 ( T ) α 0 ( 0 ) ] T [ 2 + M [ t T t 2 ] + N 0 t k ( t , s ) ( s T s 2 ) d s + L 0 T h ( t , s ) ( s T s 2 ) d s ] . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-122/MediaObjects/13661_2012_Article_229_Equai_HTML.gif
        Hence,
        p ( t ) M p ( t ) N ( K p ) ( t ) L ( S p ) ( t ) [ p ( T ) p ( 0 ) ] T [ 2 + M [ t T t 2 ] + N 0 t k ( t , s ) ( s T s 2 ) d s + L 0 T h ( t , s ) ( s T s 2 ) d s ] = α 0 ( t ) M α 0 ( t ) N ( K α 0 ) ( t ) L ( S α 0 ) ( t ) [ α 0 ( T ) α 0 ( 0 ) ] T [ 2 + M [ t T t 2 ] + N 0 t k ( t , s ) ( s T s 2 ) d s + L 0 T h ( t , s ) ( s T s 2 ) d s ] α 1 ( t ) + M α 1 ( t ) + N ( K α 1 ) ( t ) + L ( S α 1 ) ( t ) f ( t , α 0 ( t ) , ( K α 0 ) ( t ) , ( S α 0 ) ( t ) ) M α 0 ( t ) N ( K α 0 ) ( t ) L ( S α 0 ) ( t ) f ( t , α 0 ( t ) , ( K α 0 ) ( t ) , ( S α 0 ) ( t ) ) + M α 0 ( t ) + N ( K α 0 ) ( t ) + L ( S α 0 ) ( t ) = 0 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-122/MediaObjects/13661_2012_Article_229_Equaj_HTML.gif
        and
        Δ p ( t k ) = Δ α 0 ( t k ) Δ α 1 ( t k ) = I k ( t k δ k t k ε k α 0 ( s ) d s ) + L k [ α 0 ( T ) α 0 ( 0 ) ] T t k δ k t k ε k T 2 s d s L k t k δ k t k ε k α 1 ( s ) d s I k ( t k δ k t k ε k α 0 ( s ) d s ) + L k t k δ k t k ε k α 0 ( s ) d s = L k t k δ k t k ε k p ( s ) d s + L k [ p ( T ) p ( 0 ) ] T t k δ k t k ε k T 2 s d s , k = 1 , 2 , , m , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-122/MediaObjects/13661_2012_Article_229_Equak_HTML.gif
        and
        Δ p ( t k ) = Δ α 0 ( t k ) α 1 ( t k ) I k ( t k τ k t k σ k α 0 ( s ) d s ) + L k [ α 0 ( T ) α 0 ( 0 ) ] T t k τ k t k σ k s T s 2 d s L k t k τ k t k σ k α 1 ( s ) d s I k ( t k τ k t k σ k α 0 ( s ) d s ) + L k t k τ k t k σ k α 0 ( s ) d s = L k t k τ k t k σ k p ( s ) d s + L k [ p ( T ) p ( 0 ) ] T t k τ k t k σ k s T s 2 d s , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-122/MediaObjects/13661_2012_Article_229_Equal_HTML.gif
        k = 1 , 2 , , m http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-122/MediaObjects/13661_2012_Article_229_IEq144_HTML.gif, and
        p ( 0 ) = p ( T ) , p ( 0 ) = α 0 ( 0 ) α 1 ( 0 ) < α 0 ( T ) α 1 ( T ) = p ( T ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-122/MediaObjects/13661_2012_Article_229_Equam_HTML.gif

        Then by Lemma 2.2, p ( t ) 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-122/MediaObjects/13661_2012_Article_229_IEq141_HTML.gif, which implies α 0 ( t ) A α 0 ( t ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-122/MediaObjects/13661_2012_Article_229_IEq142_HTML.gif, i.e., α 0 A α 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-122/MediaObjects/13661_2012_Article_229_IEq135_HTML.gif.

        Step 2. We prove that if α 0 η 1 η 2 β 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-122/MediaObjects/13661_2012_Article_229_IEq145_HTML.gif, then A η 1 A η 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-122/MediaObjects/13661_2012_Article_229_IEq146_HTML.gif.

        Let η 1 = A η 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-122/MediaObjects/13661_2012_Article_229_IEq147_HTML.gif, η 2 = A η 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-122/MediaObjects/13661_2012_Article_229_IEq148_HTML.gif, and p = η 1 η 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-122/MediaObjects/13661_2012_Article_229_IEq149_HTML.gif, then for t J http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-122/MediaObjects/13661_2012_Article_229_IEq62_HTML.gif, and by (H2), we obtain
        p ( t ) M p ( t ) N ( K p ) ( t ) L ( S p ) ( t ) = f ( t , η 1 ( t ) , ( K η 1 ) ( t ) , ( S η 1 ) ( t ) ) M η 1 ( t ) N ( K η 1 ) ( t ) L ( S η 1 ) ( t ) f ( t , η 2 ( t ) , ( K η 2 ) ( t ) , ( S η 2 ) ( t ) ) + M η 2 ( t ) + N ( K η 2 ) ( t ) + L ( S η 2 ) ( t ) 0 ( by (H ) 2 ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-122/MediaObjects/13661_2012_Article_229_Equan_HTML.gif
        From (H3), we obtain
        Δ p ( t k ) = L k t k δ k t k ε k p ( s ) d s , k = 1 , 2 , , m , Δ p ( t k ) L k t k τ k t k σ k p ( s ) d s , k = 1 , 2 , , m , p ( 0 ) = p ( T ) , p ( 0 ) = p ( T ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-122/MediaObjects/13661_2012_Article_229_Equao_HTML.gif

        Applying Lemma 2.1, we get p ( t ) 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-122/MediaObjects/13661_2012_Article_229_IEq141_HTML.gif, which implies A η 1 A η 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-122/MediaObjects/13661_2012_Article_229_IEq146_HTML.gif.

        Step 3. We show that PBVP (1.1) has solutions.

        Let α n = A α n 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-122/MediaObjects/13661_2012_Article_229_IEq150_HTML.gif, β n = A β n 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-122/MediaObjects/13661_2012_Article_229_IEq151_HTML.gif, n = 1 , 2 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-122/MediaObjects/13661_2012_Article_229_IEq152_HTML.gif . Following the first two steps, we have
        α 0 α 1 α n β n β 1 β 0 , n N . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-122/MediaObjects/13661_2012_Article_229_Equap_HTML.gif
        Obviously, each α i http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-122/MediaObjects/13661_2012_Article_229_IEq153_HTML.gif, β i http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-122/MediaObjects/13661_2012_Article_229_IEq154_HTML.gif ( i = 1 , 2 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-122/MediaObjects/13661_2012_Article_229_IEq155_HTML.gif) satisfies
        http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-122/MediaObjects/13661_2012_Article_229_Equaq_HTML.gif
        and
        http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-122/MediaObjects/13661_2012_Article_229_Equar_HTML.gif
        Thus, there exist x http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-122/MediaObjects/13661_2012_Article_229_IEq156_HTML.gif and x http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-122/MediaObjects/13661_2012_Article_229_IEq115_HTML.gif such that
        lim i α i ( t ) = x ( t ) , lim i β i ( t ) = x ( t ) , uniformly on  t J . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-122/MediaObjects/13661_2012_Article_229_Equas_HTML.gif

        Clearly, x http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-122/MediaObjects/13661_2012_Article_229_IEq156_HTML.gif, x http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-122/MediaObjects/13661_2012_Article_229_IEq115_HTML.gif satisfy PBVP (1.1).

        Step 4. We show that x http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-122/MediaObjects/13661_2012_Article_229_IEq156_HTML.gif, x http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-122/MediaObjects/13661_2012_Article_229_IEq115_HTML.gif are extreme solutions of PBVP (1.1).

        Let x ( t ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-122/MediaObjects/13661_2012_Article_229_IEq31_HTML.gif be any solution of PBVP (1.1), which satisfies α 0 ( t ) x ( t ) β 0 ( t ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-122/MediaObjects/13661_2012_Article_229_IEq157_HTML.gif, t J http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-122/MediaObjects/13661_2012_Article_229_IEq56_HTML.gif. Suppose that there exists a positive integer n such that for t J http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-122/MediaObjects/13661_2012_Article_229_IEq56_HTML.gif, α n ( t ) x ( t ) β n ( t ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-122/MediaObjects/13661_2012_Article_229_IEq158_HTML.gif. Setting p ( t ) = α n + 1 ( t ) x ( t ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-122/MediaObjects/13661_2012_Article_229_IEq159_HTML.gif, then for t J http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-122/MediaObjects/13661_2012_Article_229_IEq62_HTML.gif,
        p ( t ) = α n + 1 x ( t ) = M α n + 1 ( t ) + N ( K α n + 1 ) ( t ) + L ( S α n + 1 ) ( t ) + f ( t , α n ( t ) , ( K α n ) ( t ) , ( S α n ) ( t ) ) M α n ( t ) N ( K α n ) ( t ) L ( S α n ) ( t ) f ( t , x ( t ) , ( K x ) ( t ) , ( S x ) ( t ) ) = M α n + 1 ( t ) + N ( K α n + 1 ) ( t ) + L ( S α n + 1 ) ( t ) M x ( t ) N ( K x ) ( t ) L ( S x ) ( t ) + f ( t , α n ( t ) , ( K α n ) ( t ) , ( S α n ) ( t ) ) f ( t , x ( t ) , ( K x ) ( t ) , ( S x ) ( t ) ) M ( α n ( t ) x ( t ) ) N ( K ( α n ) ( t ) ( K x ) ( t ) ) L ( ( S α n ) ( t ) ( S x ) ( t ) ) M p ( t ) + N ( K p ) ( t ) + L ( S p ) ( t ) , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-122/MediaObjects/13661_2012_Article_229_Equat_HTML.gif
        and
        Δ p ( t k ) = Δ α n + 1 ( t k ) Δ x ( t k ) = L k t k δ k t k ε k p ( s ) d s , k = 1 , 2 , , m , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-122/MediaObjects/13661_2012_Article_229_Equau_HTML.gif
        and
        Δ p ( t k ) = Δ α n + 1 ( t k ) Δ x ( t k ) = L k t k τ k t k σ k α n + 1 ( s ) d s + I k ( t k τ k t k σ k α n ( s ) d s ) L k t k τ k t k σ k α n ( s ) d s I k ( t k τ k t k σ k x ( s ) d s ) L k t k τ k t k σ k α n + 1 ( s ) d s + L k t k τ k t k σ k α n ( s ) d s L k t k τ k t k σ k α n ( s ) d s L k t k τ k t k σ k x ( s ) d s = L k t k τ k t k σ k p ( s ) d s , k = 1 , 2 , , m , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-122/MediaObjects/13661_2012_Article_229_Equav_HTML.gif
        and
        p ( 0 ) = p ( T ) , p ( 0 ) = p ( T ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-122/MediaObjects/13661_2012_Article_229_Equaw_HTML.gif

        Still by Lemma 2.1, we have for all t J http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-122/MediaObjects/13661_2012_Article_229_IEq56_HTML.gif, p ( t ) 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-122/MediaObjects/13661_2012_Article_229_IEq160_HTML.gif, i.e., α n + 1 ( t ) x ( t ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-122/MediaObjects/13661_2012_Article_229_IEq161_HTML.gif. Similarly, we can prove that x ( t ) β n + 1 ( t ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-122/MediaObjects/13661_2012_Article_229_IEq162_HTML.gif, t J http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-122/MediaObjects/13661_2012_Article_229_IEq56_HTML.gif. Therefore, α n + 1 ( t ) x ( t ) β n + 1 ( t ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-122/MediaObjects/13661_2012_Article_229_IEq163_HTML.gif, for all t J http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-122/MediaObjects/13661_2012_Article_229_IEq56_HTML.gif, which implies x ( t ) x ( t ) x ( t ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-122/MediaObjects/13661_2012_Article_229_IEq164_HTML.gif. The proof is complete. □

        4 An example

        In this section, in order to illustrate our results, we consider an example.

        Example 4.1 Consider the following PBVP:
        { u ( t ) = 1 4 t 3 ( u ( t ) 2 ) + 5 18 [ 0 t t 2 s 4 u ( s ) d s ] 2 u ( t ) = + 1 8 [ 0 1 t 3 s 2 u ( s ) d s ] 2 , t J = [ 0 , 1 ] , t 1 2 , Δ u ( 1 2 ) = 1 2 1 6 3 10 u ( s ) d s , k = 1 , Δ u ( 1 2 ) = 1 3 1 10 3 10 u ( s ) d s , k = 1 , u ( 0 ) = u ( 1 ) , u ( 0 ) = u ( 1 ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-122/MediaObjects/13661_2012_Article_229_Equ11_HTML.gif
        (4.1)

        Set k ( t , s ) = t 2 s 4 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-122/MediaObjects/13661_2012_Article_229_IEq165_HTML.gif, h ( t , s ) = t 3 s 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-122/MediaObjects/13661_2012_Article_229_IEq166_HTML.gif, m = 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-122/MediaObjects/13661_2012_Article_229_IEq167_HTML.gif, t 1 = 1 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-122/MediaObjects/13661_2012_Article_229_IEq168_HTML.gif, δ 1 = 1 3 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-122/MediaObjects/13661_2012_Article_229_IEq169_HTML.gif, ε 1 = 1 5 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-122/MediaObjects/13661_2012_Article_229_IEq170_HTML.gif, τ 1 = 2 5 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-122/MediaObjects/13661_2012_Article_229_IEq171_HTML.gif, σ 1 = 1 5 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-122/MediaObjects/13661_2012_Article_229_IEq172_HTML.gif, T = 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-122/MediaObjects/13661_2012_Article_229_IEq173_HTML.gif. Obviously, α 0 = 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-122/MediaObjects/13661_2012_Article_229_IEq174_HTML.gif, β 0 = 3 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-122/MediaObjects/13661_2012_Article_229_IEq175_HTML.gif are lower and upper solutions for (4.1), respectively, and α 0 β 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-122/MediaObjects/13661_2012_Article_229_IEq116_HTML.gif.

        Let
        f ( t , x 1 , y 1 , z 1 ) = 1 4 t 3 ( x 1 2 ) + 5 18 y 1 2 + 1 8 z 1 2 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-122/MediaObjects/13661_2012_Article_229_Equax_HTML.gif
        we have
        f ( t , x 2 , y 2 , z 2 ) f ( t , x 1 , y 1 , z 1 ) 1 4 ( x 2 x 1 ) + 1 3 ( y 2 y 1 ) + 1 4 ( z 2 z 1 ) , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-122/MediaObjects/13661_2012_Article_229_Equay_HTML.gif
        where α ( t ) x 1 x 2 β ( t ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-122/MediaObjects/13661_2012_Article_229_IEq176_HTML.gif, ( K α ) ( t ) y 1 y 2 ( K β ) ( t ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-122/MediaObjects/13661_2012_Article_229_IEq177_HTML.gif, ( S α ) ( t ) z 1 z 2 ( S β ) ( t ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-122/MediaObjects/13661_2012_Article_229_IEq178_HTML.gif, t J http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-122/MediaObjects/13661_2012_Article_229_IEq56_HTML.gif. It is easy to see that
        I 1 ( 1 6 3 10 x ( s ) d s ) I 1 ( 1 6 3 10 y ( s ) d s ) = 1 2 1 6 3 10 x ( s ) y ( s ) d s , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-122/MediaObjects/13661_2012_Article_229_Equaz_HTML.gif
        and
        I 1 ( 1 10 3 10 x ( s ) d s ) I 1 ( 1 10 3 10 y ( s ) d s ) = 1 3 1 10 3 10 x ( s ) y ( s ) d s , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-122/MediaObjects/13661_2012_Article_229_Equba_HTML.gif

        whenever t k τ k t k σ k α ( s ) d s t k τ k t k σ k y ( s ) d s t k τ k t k σ k x ( s ) d s t k τ k t k σ k β ( s ) d s http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-122/MediaObjects/13661_2012_Article_229_IEq179_HTML.gif, k = 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-122/MediaObjects/13661_2012_Article_229_IEq180_HTML.gif.

        Taking M = 1 4 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-122/MediaObjects/13661_2012_Article_229_IEq181_HTML.gif, N = 1 3 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-122/MediaObjects/13661_2012_Article_229_IEq182_HTML.gif, L = 1 4 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-122/MediaObjects/13661_2012_Article_229_IEq183_HTML.gif, L 1 = 1 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-122/MediaObjects/13661_2012_Article_229_IEq184_HTML.gif, L 1 = 1 3 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-122/MediaObjects/13661_2012_Article_229_IEq185_HTML.gif, it follows that
        [ k = 1 m L k ( δ k ε k ) + T ] [ k = 1 m L k ( τ k σ k ) + ( M + N k 0 T + L h 0 T ) T ] = [ 1 2 ( 1 3 1 5 ) + 1 ] [ 1 3 ( 2 5 1 5 ) + ( 1 4 + ( 1 3 ) ( 1 ) ( 1 ) + ( 1 4 ) ( 1 ) ( 1 ) ) 1 ] = 24 25 1 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-122/MediaObjects/13661_2012_Article_229_Equbb_HTML.gif
        and
        ψ = 1 + e M T 2 M ( e M T 1 ) × [ 0 T [ N 0 s k ( s , r ) d r + L 0 T h ( s , r ) d r ] d s + k = 1 m L k ( τ k σ k ) ] + 1 2 k = 1 m L k ( δ k ε k ) = 1 + e 1 2 2 ( 1 2 ) ( e 1 2 1 ) [ 0 1 [ 1 3 0 s s 2 r 4 d r + 1 4 0 1 s 3 r 2 d r ] d s + 1 3 ( 2 5 1 5 ) ] + ( 1 2 ) ( 1 2 ) ( 1 3 1 5 ) 0.4246196990 < 1 , μ = 1 2 [ 0 T [ N 0 s k ( s , r ) d r + L 0 T h ( s , r ) d r ] d s + k = 1 m L k ( τ k σ k ) ] + M ( 1 + e M T ) 2 ( e M T 1 ) k = 1 m L k ( δ k ε k ) = 1 2 [ 0 1 [ 1 3 0 s s 2 r 4 d r + 1 4 0 1 s 3 r 2 d r ] d s + 1 3 ( 2 5 1 5 ) ] + 1 2 ( 1 + e 1 2 ) 2 ( e 1 2 1 ) ( 1 2 ) ( 1 3 1 5 ) 0.1159664694 < 1 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-122/MediaObjects/13661_2012_Article_229_Equbc_HTML.gif

        Therefore, (4.1) satisfies all the conditions of Theorem 3.1. So, PBVP (4.1) has minimal and maximal solutions in the segment [ α 0 , β 0 ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-122/MediaObjects/13661_2012_Article_229_IEq133_HTML.gif.

        Substituting α 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-122/MediaObjects/13661_2012_Article_229_IEq119_HTML.gif, β 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-122/MediaObjects/13661_2012_Article_229_IEq120_HTML.gif into monotone iterative scheme, we obtain
        { α 1 ( t ) 1 4 α 1 ( t ) 1 3 0 t t 2 s 4 α 1 ( s ) d s 1 4 0 1 t 3 s 2 α 1 ( s ) d s = 1 2 t 3 , J = [ 0 , 1 ] , t 1 2 , Δ α 1 ( 1 2 ) = 1 2 1 6 3 10 α 1 ( s ) d s , k = 1 , Δ α 1 ( 1 2 ) = 1 3 1 10 3 10 α 1 ( s ) d s , k = 1 , α 1 ( 0 ) = α 1 ( T ) , α 1 ( 0 ) = α 1 ( T ) , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-122/MediaObjects/13661_2012_Article_229_Equ12_HTML.gif
        (4.2)
        and
        { β 1 ( t ) 1 4 β 1 ( t ) 1 3 0 t t 2 s 4 β 1 ( s ) d s 1 4 0 1 t 3 s 2 β 1 ( s ) d s = 1 10 t 14 1 5 t 7 + 1 8 t 6 3 4 , J = [ 0 , 1 ] , t 1 2 , Δ β 1 ( 1 2 ) = 1 2 1 6 3 10 β 1 ( s ) d s , k = 1 , Δ β 1 ( 1 2 ) = 1 3 1 10 3 10 β 1 ( s ) d s , k = 1 , β 1 ( 0 ) = β 1 ( T ) , β 1 ( 0 ) = β 1 ( T ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-122/MediaObjects/13661_2012_Article_229_Equ13_HTML.gif
        (4.3)
        After using the variational iteration method [45] for (4.2), (4.3), the approximate solutions for α 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-122/MediaObjects/13661_2012_Article_229_IEq139_HTML.gif and β 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-122/MediaObjects/13661_2012_Article_229_IEq186_HTML.gif can be illustrated as Figure 1 and Figure 2, respectively.
        http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-122/MediaObjects/13661_2012_Article_229_Fig1_HTML.jpg
        Figure 1

        Time history of α 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-122/MediaObjects/13661_2012_Article_229_IEq187_HTML.gif .

        http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-122/MediaObjects/13661_2012_Article_229_Fig2_HTML.jpg
        Figure 2

        Time history of β 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-122/MediaObjects/13661_2012_Article_229_IEq188_HTML.gif .

        Declarations

        Acknowledgements

        This research is supported by the Centre of Excellence in Mathematics, the Commission on Higher Education, Thailand.

        Authors’ Affiliations

        (1)
        Department of Mathematics, Faculty of Science, King Mongkut’s Institute of Technology Ladkrabang
        (2)
        Centre of Excellence in Mathematics, CHE
        (3)
        Department of Mathematics, Faculty of Applied Science, King Mongkut’s University of Technology North Bangkok

        References

        1. Bainov DD, Simeonov PS: Impulsive Differential Equations: Periodic Solutions and Applications. Longman, Harlow; 1993.
        2. Bainov DD, Simeonov PS: Impulsive Differential Equations: Asymptotic Properties of the Solutions. World Scientific, Singapore; 1995.
        3. Benchohra M, Henderson J, Ntouyas S: Impulsive Differential Equations and Inclusions. Hindawi Publishing, New York; 2006.View Article
        4. Lakshmikantham V, Bainov DD, Simeonov PS: Theory of Impulsive Differential Equations. World Scientific, Singapore; 1989.View Article
        5. Samoilenko AM, Perestyuk NA: Impulsive Differential Equations. World Scientific, Singapore; 1995.
        6. Ahmad B, Alsaedi A: Existence of solutions for anti-periodic boundary value problems of nonlinear impulsive functional integro-differential equations of mixed type. Nonlinear Anal. Hybrid Syst. 2009, 3: 501-509. 10.1016/j.nahs.2009.03.007MathSciNetView Article
        7. Boucherif A, Fiagbedzi YA: Periodic solutions of nonlinear integrodifferential equations. Dyn. Contin. Discrete Impuls. Syst. Ser. A, Math. Anal. 2007, 14: 165-175.MathSciNet
        8. Chen J, Tisdell CC, Yuan R: On the solvability of periodic boundary value problems with impulse. J. Math. Anal. Appl. 2007, 331: 902-912. 10.1016/j.jmaa.2006.09.021MathSciNetView Article
        9. Ding W, Mi J, Han M: Periodic boundary value problems for the first order impulsive functional differential equations. Appl. Math. Comput. 2005, 165: 433-446. 10.1016/j.amc.2004.06.022MathSciNetView Article
        10. He Z, Ge W: Periodic boundary value problem for first order impulsive delay differential equations. Appl. Math. Comput. 1999, 104: 51-63. 10.1016/S0096-3003(98)10059-0MathSciNetView Article
        11. He Z, He X: Monotone iterative technique for impulsive integro-differential equations with periodic boundary conditions. Comput. Math. Appl. 2004, 48: 73-84. 10.1016/j.camwa.2004.01.005MathSciNetView Article
        12. He Z, He X: Periodic boundary value problems for first order impulsive integro-differential equations of mixed type. J. Math. Anal. Appl. 2004, 296: 8-20. 10.1016/j.jmaa.2003.12.047MathSciNetView Article
        13. He Z, Yu J: Periodic boundary value problem for first-order impulsive functional differential equations. J. Comput. Appl. Math. 2002, 138: 205-217. 10.1016/S0377-0427(01)00381-8MathSciNetView Article
        14. He Z, Yu J: Periodic boundary value problem for first-order impulsive ordinary differential equations. J. Math. Anal. Appl. 2002, 272: 67-78. 10.1016/S0022-247X(02)00133-6MathSciNetView Article
        15. Huseynov A: Positive solutions of a nonlinear impulsive equation with periodic boundary conditions. Appl. Math. Comput. 2010, 217: 247-259. 10.1016/j.amc.2010.05.055MathSciNetView Article
        16. Li J, Nieto JJ, Shen J: Impulsive periodic boundary value problems of first-order differential equations. J. Math. Anal. Appl. 2007, 325: 226-236. 10.1016/j.jmaa.2005.04.005MathSciNetView Article
        17. Li J, Shen J: Periodic boundary value problems for delay differential equations with impulses. J. Comput. Appl. Math. 2006, 193: 563-573. 10.1016/j.cam.2005.05.037MathSciNetView Article
        18. Li J, Shen J: Periodic boundary value problems for impulsive integro-differential equations of mixed type. Appl. Math. Comput. 2006, 183: 890-902. 10.1016/j.amc.2006.06.037MathSciNetView Article
        19. Liang R, Shen J: Periodic boundary value problem for the first order impulsive functional differential equations. J. Comput. Appl. Math. 2007, 202: 498-510. 10.1016/j.cam.2006.03.017MathSciNetView Article
        20. Liu Y: Further results on periodic boundary value problems for nonlinear first order impulsive functional differential equations. J. Math. Anal. Appl. 2007, 327: 435-452. 10.1016/j.jmaa.2006.01.027MathSciNetView Article
        21. Liu Y: Periodic boundary value problems for first order functional differential equations with impulse. J. Comput. Appl. Math. 2009, 223: 27-39. 10.1016/j.cam.2007.12.015MathSciNetView Article
        22. Luo Z, Jing Z: Periodic boundary value problem for first-order impulsive functional differential equations. Comput. Math. Appl. 2008, 55: 2094-2107. 10.1016/j.camwa.2007.08.036MathSciNetView Article
        23. Nieto JJ: Periodic boundary value problems for first-order impulsive ordinary differential equations. Nonlinear Anal. 2002, 51: 1223-1232. 10.1016/S0362-546X(01)00889-6MathSciNetView Article
        24. Nieto JJ, Rodríguez-López R: Remarks on periodic boundary value problems for functional differential equations. J. Comput. Appl. Math. 2003, 158: 339-353. 10.1016/S0377-0427(03)00452-7MathSciNetView Article
        25. Nieto JJ, Rodríguez-López R: Periodic boundary value problem for non-Lipschitzian impulsive functional differential equations. J. Math. Anal. Appl. 2006, 318: 593-610. 10.1016/j.jmaa.2005.06.014MathSciNetView Article
        26. Tian Y, Jiang D, Ge W: Multiple positive solutions of periodic boundary value problems for second order impulsive differential equations. Appl. Math. Comput. 2008, 200: 123-132. 10.1016/j.amc.2007.10.052MathSciNetView Article
        27. Zhang D, Dai B: Infinitely many solutions for a class of nonlinear impulsive differential equations with periodic boundary conditions. Comput. Math. Appl. 2011, 61: 3153-3160. 10.1016/j.camwa.2011.04.003MathSciNetView Article
        28. Zhang N, Dai B, Qian X: Periodic solutions for a class of higher-dimension functional differential equations with impulses. Nonlinear Anal. 2008, 68: 629-638. 10.1016/j.na.2006.11.024MathSciNetView Article
        29. Zhao A, Bai Z: Existence of solutions to first-order impulsive periodic boundary value problems. Nonlinear Anal. 2009, 71: 1970-1977. 10.1016/j.na.2009.01.036MathSciNetView Article
        30. Han J, Liu Y, Zhao J: Integral boundary value problems for first order nonlinear impulsive functional integro-differential equations. Appl. Math. Comput. 2012, 218: 5002-5009. 10.1016/j.amc.2011.10.067MathSciNetView Article
        31. Hao X, Liu L, Wu Y: Positive solutions for second order impulsive differential equations with integral boundary conditions. Commun. Nonlinear Sci. Numer. Simul. 2011, 16: 101-111. 10.1016/j.cnsns.2010.04.007MathSciNetView Article
        32. Jankowski T: Positive solutions for second order impulsive differential equations involving Stieltjes integral conditions. Nonlinear Anal. 2011, 74: 3775-3785. 10.1016/j.na.2011.03.022MathSciNetView Article
        33. Song G, Zhao Y, Sun X: Integral boundary value problems for first order impulsive integro-differential equations of mixed type. J. Comput. Appl. Math. 2011, 235: 2928-2935. 10.1016/j.cam.2010.12.007MathSciNetView Article
        34. Wang G, Zhang L, Song G: New existence results and comparison principles for impulsive integral boundary value problem with lower and upper solutions in reversed order. Adv. Differ. Equ. 2011., 2011: Article ID 783726
        35. Zhang X, Feng M, Ge W: Existence of solutions of boundary value problems with integral boundary conditions for second-order impulsive integro-differential equations in Banach spaces. J. Comput. Appl. Math. 2010, 233: 1915-1926. 10.1016/j.cam.2009.07.060MathSciNetView Article
        36. Ding W, Han M: Periodic boundary value problem for the second order impulsive functional differential equations. Appl. Math. Comput. 2004, 155: 709-726. 10.1016/S0096-3003(03)00811-7MathSciNetView Article
        37. Ding W, Han M, Mi J: Periodic boundary value problem for the second-order impulsive functional differential equations. Comput. Math. Appl. 2005, 50: 491-507. 10.1016/j.camwa.2005.03.010MathSciNetView Article
        38. Li J: Periodic boundary value problems for second-order impulsive integro-differential equations. Appl. Math. Comput. 2008, 198: 317-325. 10.1016/j.amc.2007.08.079MathSciNetView Article
        39. Liang R, Shen J: Periodic boundary value problem for second-order impulsive functional differential equations. Appl. Math. Comput. 2007, 193: 560-571. 10.1016/j.amc.2007.03.072MathSciNetView Article
        40. Yang X, Shen J: Periodic boundary value problems for second-order impulsive integro-differential equations. J. Comput. Appl. Math. 2007, 209: 176-186. 10.1016/j.cam.2006.10.082MathSciNetView Article
        41. Yao M, Zhao A, Yan J: Periodic boundary value problems of second-order impulsive differential equations. Nonlinear Anal. 2009, 70: 262-273. 10.1016/j.na.2007.11.050MathSciNetView Article
        42. Nieto JJ: An abstract monotone iterative technique. Nonlinear Anal., Theory Methods Appl. 1997, 28: 1923-1933. 10.1016/S0362-546X(97)89710-6View Article
        43. Tariboon J: Boundary value problems for first order functional differential equations with impulsive integral conditions. J. Comput. Appl. Math. 2010, 234: 2411-2419. 10.1016/j.cam.2010.03.007MathSciNetView Article
        44. Liu Z, Han J, Fang L: Integral boundary value problems for first order integro-differential equations with impulsive integral conditions. Comput. Math. Appl. 2011, 61: 3035-3043. 10.1016/j.camwa.2011.03.094MathSciNetView Article
        45. Wazwaz AM: Linear and Nonlinear Integral Equations: Methods and Applications. Springer, New York; 2011.View Article

        Copyright

        © Thaiprayoon et al.; licensee Springer 2012

        This article is published under license to BioMed Central Ltd. This is an Open Access article distributed under the terms of the Creative Commons Attribution License (http://​creativecommons.​org/​licenses/​by/​2.​0), which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.