Multi-point boundary value problem for first order impulsive integro-differential equations with multi-point jump conditions

  • Chatthai Thaiprayoon1, 3Email author,

    Affiliated with

    • Decha Samana1, 3 and

      Affiliated with

      • Jessada Tariboon2, 3

        Affiliated with

        Boundary Value Problems20122012:38

        DOI: 10.1186/1687-2770-2012-38

        Received: 8 August 2011

        Accepted: 5 April 2012

        Published: 5 April 2012

        Abstract

        In this article we introduce a new definition of impulsive conditions for boundary value problems of first order impulsive integro-differential equations with multi-point boundary conditions. By using the method of lower and upper solutions in reversed order coupled with the monotone iterative technique, we obtain the extremal solutions of the boundary value problem. An example is also discussed to illustrate our results.

        Mathematics Subject Classification 2010: 34B15; 34B37.

        Keywords

        impulsive integro-differential equations multi-point boundary value problem lower and upper solutions monotone iterative technique.

        1 Introduction

        Impulsive differential equations describe processes which have a sudden change of their state at certain moments. Impulse effects are important in many real world applications, such as physics, medicine, biology, control theory, population dynamics, etc. (see, for example [13]). In this article, we consider the following boundary value problem for first order impulsive integro-differential equations (BVP):
        x t = f t , x t , F x t , S x t , t J = 0 , T , t t k , Δ x t k = I k l = 1 c k ρ l k x η l k , k = 1 , 2 , m , x 0 + μ k = 1 m l = 1 c k τ l k x η l k = x T , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-38/MediaObjects/13661_2011_Article_162_Equ1_HTML.gif
        (1.1)
        where fC(J × R3, R), 0 = t0< t1< t2< · · · < t m < tm+1= T,
        F 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%2F1687-2770-2012-38/MediaObjects/13661_2011_Article_162_Equa_HTML.gif

        kC(D, R+), D = {(t, s) ∈ J × J: ts}, hC(J × J, R+). I k C(R, R), Δ x t k = x t k + - x t k - , t k - 1 < η 1 k < η 2 k < < η c k k t k , τ l k , ρ l k 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-38/MediaObjects/13661_2011_Article_162_IEq1_HTML.gif, l = 1, 2, ..., c k , c k N = {1, 2, ...}, k = 1, 2, ..., m, μ ≥ 0.

        The monotone iterative technique coupled with the method of lower and upper solutions is a powerful method used to approximate solutions of several nonlinear problems (see [414]). Boundary value problems for first order impulsive functional differential equations with lower and upper solutions in reversed order have been widely discussed in recent years (see [1520]). However, the discussion of multi-point boundary value problems for first order impulsive functional differential equations is very limited (see [21]). In all articles concerned with applications of the monotone iterative technique to impulsive problems, the authors have assumed that Δx(t k ) = I k (x(t k )), that is a short-term rapid change of the state at impulse point t k depends on the left side of the limit of x(t k ).

        Recently, Tariboon [22] and Liu et al. [23] studied some types of impulsive boundary value problems with the impulsive integral conditions
        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 , k = 1 , 2 , , m . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-38/MediaObjects/13661_2011_Article_162_Equ2_HTML.gif
        (1.2)

        It should be noticed that the terms t k - τ k t k x s d s http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-38/MediaObjects/13661_2011_Article_162_IEq2_HTML.gif and t k - 1 t k - 1 + σ k - 1 x s d s http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-38/MediaObjects/13661_2011_Article_162_IEq3_HTML.gif of impulsive condition (1.2) illustrate the past memory state on [t k - τ k , t k ] before impulse points t k and the history effects after the past impulse points tk- 1on (tk- 1, tk- 1+ σk- 1], respectively.

        The aim of the present article is to discuss the new impulsive multi-point condition
        Δ x t k = I k l = 1 c k ρ l k x η l k = I k ρ 1 k x η 1 k + + ρ l k x η l k + + ρ c k k x η c k k , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-38/MediaObjects/13661_2011_Article_162_Equ3_HTML.gif
        (1.3)

        for t k - 1 < η 1 k < η 2 k < < η c k k t k , k = 1 , 2 , , m http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-38/MediaObjects/13661_2011_Article_162_IEq4_HTML.gif. The new jump conditions mean that a sudden change of the state at impulse point t k depends on the multi-point η l k l = 1 , 2 , , c k http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-38/MediaObjects/13661_2011_Article_162_IEq5_HTML.gif of past states on (tk- 1, t k ]. We note that if c k = 1, η c k k = t k http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-38/MediaObjects/13661_2011_Article_162_IEq6_HTML.gif and ρ c k k = 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-38/MediaObjects/13661_2011_Article_162_IEq7_HTML.gif, then the impulsive condition (1.3) is reduced to the simple impulsive condition Δx(t k ) = I k (x(t k )).

        Firstly, we introduce the definitions of lower and upper solutions and formulate some lemmas which are used in our discussion. In the main results, we obtain the existence of extreme solutions for BVP (1.1) by using the method of lower and upper solutions in reversed order and the monotone iterative technique. Finally, we give an example to illustrate the obtained results.

        2 Preliminaries

        Let J - = J \ {t1, t2, ..., t m }. PC(J, R) = {x: J → R; x(t) is continuous everywhere except for some t k at which x t k - http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-38/MediaObjects/13661_2011_Article_162_IEq8_HTML.gif and x t k + http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-38/MediaObjects/13661_2011_Article_162_IEq9_HTML.gif exist and x t k - = x t k http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-38/MediaObjects/13661_2011_Article_162_IEq10_HTML.gif, k = 1, 2, ..., m}, PC1(J, R) = {xPC(J, R); x'(t) is continuous everywhere except for some t k at which x t k + http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-38/MediaObjects/13661_2011_Article_162_IEq11_HTML.gif and x t k - http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-38/MediaObjects/13661_2011_Article_162_IEq12_HTML.gif exist and x t k - = x t k http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-38/MediaObjects/13661_2011_Article_162_IEq13_HTML.gif}. Let E = PC(J, R) and F = P C 1 J , R http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-38/MediaObjects/13661_2011_Article_162_IEq14_HTML.gif, then E and F http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-38/MediaObjects/13661_2011_Article_162_IEq15_HTML.gif are Banach spaces with the nomes ||x|| E = suptJ|x(t)| and x F = max x E , x E http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-38/MediaObjects/13661_2011_Article_162_IEq16_HTML.gif, respectively. A function x F http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-38/MediaObjects/13661_2011_Article_162_IEq17_HTML.gif is called a solution of BVP (1.1) if it satisfies (1.1).

        Definition 2.1. A function α 0 F http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-38/MediaObjects/13661_2011_Article_162_IEq18_HTML.gif is called a lower solution of BVP (1.1) if:
        α 0 t f t , α 0 t , F α 0 t , S α 0 t , t J - , Δ α 0 t k I k l = 1 c k ρ l k α 0 η l k , k = 1 , 2 , , m , α 0 0 + μ k = 1 m l = 1 c k τ l k α 0 η l k α 0 T . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-38/MediaObjects/13661_2011_Article_162_Equb_HTML.gif
        Analogously, a function β 0 F http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-38/MediaObjects/13661_2011_Article_162_IEq19_HTML.gif is called an upper solution of BVP (1.1) if:
        β 0 t f t , β 0 t , F β 0 t , S β 0 t , t J - , Δ β 0 t k I k l = 1 c k ρ l k β 0 η l k , k = 1 , 2 , , m , β 0 0 + μ k = 1 m l = 1 c k τ l k β 0 η l k β 0 T , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-38/MediaObjects/13661_2011_Article_162_Equc_HTML.gif

        where t k - 1 < η l k t k , ρ l k , τ l k 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-38/MediaObjects/13661_2011_Article_162_IEq20_HTML.gif, l = 1, 2, ..., c k , c k N = {1, 2, ...}, k = 1, 2, ..., m and μ ≥ 0.

        Let us consider the following boundary value problem of a linear impulsive integro-differential equation (BVP):
        x t - M x t = H F x t + K S x t + v t , t J - , Δ x t k = L k l = 1 c k ρ l k x η l k + I k l = 1 c k ρ l k σ η l k - L k l = 1 c k ρ l k σ η l k , k = 1 , 2 , , m , x 0 + μ k = 1 m l = 1 c k τ l k σ η l k = x T , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-38/MediaObjects/13661_2011_Article_162_Equ4_HTML.gif
        (2.1)

        where M > 0, H, K ≥ 0, L k ≥ 0, t k - 1 < η l k t k , τ l k , ρ l k 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-38/MediaObjects/13661_2011_Article_162_IEq21_HTML.gif, l = 1, 2, ..., c k , c k N = {1, 2, ...}, k = 1, 2,..., m are constants and v(t), σ(t) ∈ E.

        Lemma 2.1. x F http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-38/MediaObjects/13661_2011_Article_162_IEq22_HTML.gifis a solution of (2.1) if and only if xE is a solution of the impulsive integral equation
        x t = μ e M t e M T - 1 k = 1 m l = 1 c k τ l k σ η l k - 0 T G t , s P s d s - k = 1 m G t , t k L k l = 1 c k ρ l k x η l k + I k l = 1 c k ρ l k σ η l k - L k l = 1 c k ρ l k σ η l k , t J , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-38/MediaObjects/13661_2011_Article_162_Equ5_HTML.gif
        (2.2)
        where P(t) = H(Fx)(t) + K(Sx)(t) + v(t) and
        G t , s = e M t - s e M T - 1 , 0 s < t T , e M T + t - s e M T - 1 , 0 t s T . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-38/MediaObjects/13661_2011_Article_162_Equd_HTML.gif
        Proof. Assume that x(t) is a solution of BVP (2.1). By using the variation of parameters formula, we get
        x t = x 0 e M t + 0 t e M t - s P s d s + 0 < t k < t e M t - t k L k l = 1 c k ρ l k x η l k + I k l = 1 c k ρ l k σ η l k - L k l = 1 c k ρ l k σ η l k . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-38/MediaObjects/13661_2011_Article_162_Equ6_HTML.gif
        (2.3)
        Putting t = T in (2.3), we have
        x T = x 0 e M T + 0 T e M T - s P s d s + k = 1 m e M T - t k L k l = 1 c k ρ l k x η l k + I k l = 1 c k ρ l k σ η l k - L k l = 1 c k ρ l k σ η l k . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-38/MediaObjects/13661_2011_Article_162_Equ7_HTML.gif
        (2.4)
        From x 0 + μ k = 1 m l = 1 c k τ l k σ η l k = x T http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-38/MediaObjects/13661_2011_Article_162_IEq23_HTML.gif, we obtain
        x 0 = - 1 e M T - 1 - μ k = 1 m l = 1 c k τ l k σ η l k + 0 T e M T - s P s d s + k = 1 m e M T - t k L k l = 1 c k ρ l k x η l k + I k l = 1 c k ρ l k σ η l k - L k l = 1 c k ρ l k σ η l k . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-38/MediaObjects/13661_2011_Article_162_Equ8_HTML.gif
        (2.5)

        Substituting (2.5) into (2.3), we see that xE satisfies (2.2). Hence, x(t) is also the solution of (2.2).

        Conversely, we assume that x(t) is a solution of (2.2). By computing directly, we have
        G t t , s = M e M t - s e M T - 1 , 0 s < t T , M e M T + t - s e M T - 1 , 0 t s T , = M G t , s . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-38/MediaObjects/13661_2011_Article_162_Eque_HTML.gif
        Differentiating (2.2) for t ≠ t k , we obtain
        x t = M x t + H F x t + K S x t + v t . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-38/MediaObjects/13661_2011_Article_162_Equf_HTML.gif
        It is easy to see that
        Δ x t k = L k l = 1 c k ρ l k x η l k + I k l = 1 c k ρ l k σ η l k - L k l = 1 c k ρ l k σ η l k . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-38/MediaObjects/13661_2011_Article_162_Equg_HTML.gif

        Since G(0, s) = G(T, s), then x 0 + μ k = 1 m l = 1 c k τ l k σ η l k = x T http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-38/MediaObjects/13661_2011_Article_162_IEq24_HTML.gif. This completes the proof.   □

        Lemma 2.2. Assume that M > 0, H, K ≥ 0, L k ≥ 0, t k - 1 < η l k t k , ρ l k 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-38/MediaObjects/13661_2011_Article_162_IEq25_HTML.gif, l = 1, 2, ..., c k , c k N = {1, 2, ...}, k = 1, 2, ..., m, and the following inequality holds:
        e M T e M T - 1 0 T H 0 s k s , r d r + K 0 T h s , r d r d s + e M T e M T - 1 k = 1 m L k l = 1 c k ρ l k < 1 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-38/MediaObjects/13661_2011_Article_162_Equ9_HTML.gif
        (2.6)

        Then BVP (2.1) has a unique solution.

        Proof. For any xE, we define an operator A by
        A x t = μ e M t e M T - 1 k = 1 m l = 1 c k τ l k σ η l k - 0 T G t , s H F x s + K S x s + v s d s - k = 1 m G t , t k L k l = 1 c k ρ l k x η l k + I k l = 1 c k ρ l k σ η l k - L k l = 1 c k ρ l k σ η l k , t J , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-38/MediaObjects/13661_2011_Article_162_Equ10_HTML.gif
        (2.7)
        where G(t, s) is defined as in Lemma 2.1. Since max t [ 0 , T ] G t , s = e M T e M T - 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-38/MediaObjects/13661_2011_Article_162_IEq26_HTML.gif, we have for any x, yE, that
        A x - A y E = - 0 T G t , s H 0 s k s , r x r - y r d r + K 0 T h s , r x r - y r d r d s - k = 1 m G t , t k L k l = 1 c k ρ l k x η l k - y η l k e M T e M T - 1 0 T H 0 s k s , r d r + K 0 T h s , r d r d s + e M T e M T - 1 k = 1 m L k l = 1 c k ρ l k x - y E . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-38/MediaObjects/13661_2011_Article_162_Equh_HTML.gif

        From (2.6) and the Banach fixed point theorem, A has a unique fixed point x ¯ E http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-38/MediaObjects/13661_2011_Article_162_IEq27_HTML.gif. By Lemma 2.1, x ¯ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-38/MediaObjects/13661_2011_Article_162_IEq28_HTML.gif is also the unique solution of (2.1).   □

        Lemma 2.3. Assume that x F http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-38/MediaObjects/13661_2011_Article_162_IEq29_HTML.gifsatisfies
        x t M x t + H F x t + K S x t , t J - , Δ x t k L k l = 1 c k ρ l k x η l k , k = 1 , 2 , , m , x 0 x T , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-38/MediaObjects/13661_2011_Article_162_Equ11_HTML.gif
        (2.8)
        where M > 0, H, K ≥ 0, L k ≥ 0, t k - 1 < η l k t k , ρ l k 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-38/MediaObjects/13661_2011_Article_162_IEq30_HTML.gif, l = 1, 2, ..., c k , c k N = {1, 2, ...}, k = 1, 2, ..., m. In addition assume that
        e M T 0 T q s d s + k = 1 m L k l = 1 c k ρ l k e - M t k - η l k 1 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-38/MediaObjects/13661_2011_Article_162_Equ12_HTML.gif
        (2.9)

        where q t = H 0 t k t , s e - M t - s d s + K 0 T h t , s e - M t - s d s http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-38/MediaObjects/13661_2011_Article_162_IEq31_HTML.gif. Then, x(t) ≤ 0 for all tJ.

        Proof. Set u(t) = x(t)e -Mt for tJ , then we have
        u t H 0 t k t , s e - M t - s u s d s + K 0 T h t , s e - M t - s u s d s , t J - , Δ u t k L k l = 1 c k ρ l k e - M t k - η l k u η l k , k = 1 , 2 , , m , u 0 e M T u T . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-38/MediaObjects/13661_2011_Article_162_Equ13_HTML.gif
        (2.10)
        Obviously, the function u(t) and x(t) have the same sign. Suppose, to the contrary, that u(t) > 0 for some tJ. Then, there are two cases:
        1. (i)

          There exists a t*∈ J , such that u(t*) > 0 and u(t) ≥ 0 for all tJ.

           
        2. (ii)

          There exists t*, t *J, such that u(t*) > 0 and u(t *) < 0.

           

        Case (i): Equation (2.10) implies that u'(t) ≥ 0 for tJ - and Δu(t k ) ≥ 0 for k = 1, 2, ..., m. This means that u(t) is nondecreasing in J. Therefore, u(T) ≥ u(t*) > 0 and u(T) ≥ u(0) ≥ u(T)e MT , which is a contradiction.

        Case (ii): Let t* ∈ (t i , ti+1], i ∈ {0, 1, ..., m}, such that u(t*) = inf {u(t): tJ} < 0 and t* ∈ (t j , tj+1], j ∈ {0, 1, ..., m}, such that u(t*) > 0. We first claim that u(0) ≤ 0. Otherwise, if u(0) > 0, then by (2.10), we have
        u t * - u 0 H 0 t * 0 s k s , r e - M s - r u r d r d s + K 0 t * 0 T h s , r e - M s - r u r d r d s + k = 1 i Δ u t k u t * 0 t * q s d s + k = 1 i L k l = 1 c k ρ l k e - M t k - η l k u t * , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-38/MediaObjects/13661_2011_Article_162_Equ14_HTML.gif
        (2.11)

        a contradiction, and so u(0) ≤ 0.

        If t* < t*, then j ≤ i. Integrating the differential inequality in (2.10) from t* to t*, we obtain
        u t * - u t * H t * t * 0 s k s , r e - M s - r u r d r d s + K t * t * 0 T h s , r e - M s - r u r d r d s + k = j + 1 i Δ u t k u t * t * t * q s d s + k = j + 1 i Δ u t k u t * t * t * q s d s + k = j + 1 i L k l = 1 c k ρ l k e - M t k - η l k u η l k u t * 0 T q s d s + k = 1 m L k l = 1 c k ρ l k e - M t k - η l k u t * , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-38/MediaObjects/13661_2011_Article_162_Equi_HTML.gif

        which is a contradiction to u(t*) > 0.

        Now, assume that t*< t*. Since 0 ≥ u(0) ≥ e MT u(T), then u(T) 0. From (2.10), we have
        u T - u t * H t * T 0 s k s , r e - M s - r u r d r d s + K t * T 0 T h s , r e - M s - r u r d r d s + k = j + 1 m Δ u t k u t * t * T q s d s + k = j + 1 m L k l = 1 c k ρ l k e - M t k - η l k , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-38/MediaObjects/13661_2011_Article_162_Equj_HTML.gif
        and u(0) ≥ e MT u(T). In consequence,
        u 0 e M T u T e M T u t * + u t * e M T t * T q s d s + k = j + 1 m L k l = 1 c k ρ l k e - M t k - η l k http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-38/MediaObjects/13661_2011_Article_162_Equ15_HTML.gif
        (2.12)

        can be obtained.

        If t* = 0, then
        u t * e M T u t * + u t * e M T t * T q s d s + k = j + 1 m L k l = 1 c k ρ l k e - M t k - η l k e M T u t * + u t * . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-38/MediaObjects/13661_2011_Article_162_Equk_HTML.gif

        This contradicts the fact that u(t*) > 0.

        If t*> 0, we obtain from (2.11),
        u t * - u t * 0 t * q s d s + k = 1 i L k l = 1 c k ρ l k e - M t k - η l k u 0 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-38/MediaObjects/13661_2011_Article_162_Equl_HTML.gif
        This joint to (2.12) yields
        u t * - u t * 0 t * q s d s + k = 1 i L k l = 1 c k ρ l k e - M t k - η l k e M T u t * + u t * e M T t * T q s d s + k = j + 1 m L k l = 1 c k ρ l k e - M t k - η l k . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-38/MediaObjects/13661_2011_Article_162_Equm_HTML.gif
        Therefore,
        u t * - e M T u t * u t * e M T t * T q s d s + k = j + 1 m L k l = 1 c k ρ l k e - M t k - η l k + u t * 0 t * q s d s + k = 1 i L k l = 1 c k ρ l k e - M t k - η l k u t * e M T t * T q s d s + k = j + 1 m L k l = 1 c k ρ l k e - M t k - η l k + u t * e M T 0 t * q s d s + k = 1 i L k l = 1 c k ρ l k e - M t k - η l k u t * e M T 0 T q s d s + k = 1 m L k l = 1 c k ρ l k e - M t k - η l k u t * . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-38/MediaObjects/13661_2011_Article_162_Equn_HTML.gif

        This is a contradiction and so u(t) ≤ 0 for all tJ. The proof is complete.   □

        3 Main results

        In this section, we are in a position to prove our main results concerning the existence criteria for solutions of BVP (1.1).

        For β 0 , α 0 F http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-38/MediaObjects/13661_2011_Article_162_IEq32_HTML.gif, we denote
        β 0 , α 0 = x F : β 0 t x t α 0 t , t J , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-38/MediaObjects/13661_2011_Article_162_Equo_HTML.gif

        and we write β0α0 if β0(t) ≤ α0(t) for all tJ.

        Theorem 3.1. Let the following conditions hold.

        (H1) The functions α0and β0are lower and upper solutions of BVP (1.1), respectively, such that β0(t) ≤ α0(t) on J.

        (H2) The function fC(J × R3, R) satisfies
        f t , x , y , z - f t , x ¯ , y ¯ , z ¯ M x - x ¯ + H y - y ¯ + K z - z ¯ , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-38/MediaObjects/13661_2011_Article_162_Equp_HTML.gif

        for β 0 t x ¯ t x t α 0 t , F β 0 t y ¯ t y t F α 0 t , S β 0 t z ¯ t z t S α 0 t t J http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-38/MediaObjects/13661_2011_Article_162_IEq33_HTML.gif.

        (H3) The function I k C(R, R) satisfies
        I k l = 1 c k ρ l k x η l k - I k l = 1 c k ρ l k y η l k L k l = 1 c k ρ l k x η l k - y η l k , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-38/MediaObjects/13661_2011_Article_162_Equq_HTML.gif

        whenever β 0 η l k y η l k x η l k α 0 η l k http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-38/MediaObjects/13661_2011_Article_162_IEq34_HTML.gif, l = 1, 2, ..., c k , c k N = {1, 2, ...}, L k ≥ 0, k = 1, 2, ..., m.

        (H4) Inequalities (2.6) and (2.9) hold.

        Then there exist monotone sequences α n , β n F http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-38/MediaObjects/13661_2011_Article_162_IEq35_HTML.gifsuch that lim n→∞ α n (t) = x*(t), lim n→∞ β n (t) = x* (t) uniformly on J and x*, x*are maximal and minimal solutions of BVP (1.1), respectively, such that
        β 0 β 1 β 2 β n x * x x * α n α 2 α 1 α 0 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-38/MediaObjects/13661_2011_Article_162_Equr_HTML.gif

        on J, where x is any solution of BVP (1.1) such that β0 (t) ≤ x(t) ≤ α0(t) on J.

        Proof. For any σ ∈ [β0, α0], we consider BVP (2.1) with
        v t = f t , σ t , F σ t , S σ t - M σ t - H F σ t - K S σ t . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-38/MediaObjects/13661_2011_Article_162_Equs_HTML.gif
        By Lemma 2.2, BVP (2.1) has a unique solution x(t) for tJ. We define an operator A by x = , then the operator A is an operator from [β0, α0] to F http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-38/MediaObjects/13661_2011_Article_162_IEq36_HTML.gif and A has the following properties.
        1. (i)

          β 0 0, 0α 0;

           
        2. (ii)

          For any σ l, σ 2 ∈ [β 0, α 0], σ lσ 2 implies l 2.

           
        To prove (i), set φ = β0 - β1, where β1 = 0. Then from (Hl) and (2.1) for tJ-, we have
        φ t = β 0 t - β 1 t , f t , β 0 t , F β 0 t , S β 0 t - M β 1 t + H F β 1 t + K S β 1 t + f t , β 0 t , F β 0 t , S β 0 t - M β 0 t - H F β 0 t - K S β 0 t = M φ t + H F φ t + K S φ t , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-38/MediaObjects/13661_2011_Article_162_Equt_HTML.gif
        Δ φ t k = Δ β 0 t k - Δ β 1 t k I k l = 1 c k ρ l k β 0 η l k - L k l = 1 c k ρ l k β 1 η l k + I k l = 1 c k ρ l k β 0 η l k - L k l = 1 c k ρ l k β 0 η l k = L k l = 1 c k ρ l k φ η l k , k = 1 , 2 , , m , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-38/MediaObjects/13661_2011_Article_162_Equu_HTML.gif
        and
        φ 0 = β 0 0 - β 1 0 β 0 T - μ k = 1 m l = 1 c k τ l k β 0 η l k - β 1 T + μ k = 1 m l = 1 c k τ l k β 0 η l k = φ T . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-38/MediaObjects/13661_2011_Article_162_Equv_HTML.gif

        By Lemma 2.3, we get that φ(t) ≤ 0 for all tJ , i.e., β00. Similarly, we can prove that 0α0.

        To prove (ii), let ul = l, u2 = 2, where σlσ2 on J and σl, σ2 ∈ [β0, α0]. Set φ = ul - u2. Then for tJ- and by (H2), we obtain
        φ t = u 1 t - u 2 t = M u 1 t + H F u 1 t + K S u 1 t + f t , σ 1 t , F σ 1 t , S σ 1 t - M σ 1 t - H F σ 1 t - K S σ 1 t - M u 2 t + H F u 2 t + K S u 2 t + f t , σ 2 t , F σ 2 t , S σ 2 t - M σ 2 t - H F σ 2 t - K S σ 2 t M u 1 t - u 2 t + H F u 1 - u 2 t + K S u 1 - u 2 t , = M φ t + H F φ t + K S φ t , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-38/MediaObjects/13661_2011_Article_162_Equw_HTML.gif
        and by (H 3);
        Δ φ t k = Δ u 1 t k - Δ u 2 t k = L k l = 1 c k ρ l k u 1 η l k + I k l = 1 c k ρ l k σ 1 η l k - L k l = 1 c k ρ l k σ 1 η l k - L k l = 1 c k ρ l k u 2 η l k + I k l = 1 c k ρ l k σ 2 η l k - L k l = 1 c k ρ l k σ 2 η l k L k l = 1 c k ρ l k u 1 η l k - u 2 η l k = L k l = 1 c k ρ l k φ η l k , k = 1 , 2 , , m . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-38/MediaObjects/13661_2011_Article_162_Equx_HTML.gif
        It is easy to see that
        φ 0 = u 1 0 - u 2 0 = u 1 T - μ k = 1 m l = 1 c k τ l k σ 1 η l k - u 2 T + μ k = 1 m l = 1 c k τ l k σ 2 η l k φ T . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-38/MediaObjects/13661_2011_Article_162_Equy_HTML.gif

        Then by using Lemma 2.3, we have φ(t) ≤ 0, which implies that l2.

        Now, we define the sequences {α n }, {β n } such that αn+l= n and βn+l= n . From (i) and (ii) the sequence {α n }, {β n } satisfy the inequality
        β 0 β 1 β n α n α 1 α 0 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-38/MediaObjects/13661_2011_Article_162_Equz_HTML.gif
        for all nN. Obviously, each α n , β n (n = 1, 2, ...) satisfy
        α n t = M α n t + H F α n t + K S α n t + f t , α n - 1 t , F α n - 1 t , S α n - 1 t - M α n - 1 t - H F α n - 1 t - K S α n - 1 t , t J - , Δ α n t k = L k l = 1 c k ρ l k α n η l k + I k l = 1 c k ρ l k α n - 1 η l k - L k l = 1 c k ρ l k α n - 1 η l k , k = 1 , 2 , , m , α n 0 + μ k = 1 m l = 1 c k τ l k α n - 1 η l k = α n T , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-38/MediaObjects/13661_2011_Article_162_Equaa_HTML.gif
        and
        β n t = M β n t + H F β n t + K S β n t + f t , β n - 1 t , F β n - 1 t , S β n - 1 t - M β n - 1 t - H F β n - 1 t - K S β n - 1 t , t J - , Δ β n t k = L k l = 1 c k ρ l k β n η l k + I k l = 1 c k ρ l k β n - 1 η l k - L k l = 1 c k ρ l k β n - 1 η l k , k = 1 , 2 , , m , β n 0 + μ k = 1 m l = 1 c k τ l k β n - 1 η l k = β n T . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-38/MediaObjects/13661_2011_Article_162_Equab_HTML.gif

        Therefore, there exist x* and x*, such that lim n→∞ β n = x* and lim n→∞ α n = x* uniformly on J. Clearly, x*, x* are solutions of BVP (1.1).

        Finally, we are going to prove that x*, x* are minimal and maximal solutions of BVP (1.1). Assume that x(t) is any solution of BVP (1.1) such that x ∈ [β0, α0] and that there exists a positive integer n such that β n (t) ≤ x(t) ≤ α n (t) on J. Let φ = βn+1- x, then for tJ-,
        φ t = β n + 1 t - x t = M β n + 1 t + H F β n + 1 t + K S β n + 1 t + f t , β n t , F β n t , S β n t - M β n t - H F β n t - K S β n t - f t , x t , F x t , S x t M φ t + H F φ t + K S φ t , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-38/MediaObjects/13661_2011_Article_162_Equac_HTML.gif
        Δ φ t k = Δ β n + 1 t k - Δ x t k = L k l = 1 c k ρ l k β n + 1 η l k + I k l = 1 c k ρ l k β n η l k - L k l = 1 c k ρ l k β n η l k - I k l = 1 c k ρ l k x η l k L k l = 1 c k ρ l k β n + 1 η l k - x η l k = L k l = 1 c k ρ l k φ η l k , k = 1 , 2 , , m , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-38/MediaObjects/13661_2011_Article_162_Equad_HTML.gif
        and
        φ 0 = β n + 1 0 - x 0 = β n + 1 T - μ k = 1 m l = 1 c k τ l k β n η l k - x T + μ k = 1 m l = 1 c k τ l k x η l k φ T . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-38/MediaObjects/13661_2011_Article_162_Equae_HTML.gif

        Then by using Lemma 2.3, we have φ(t) ≤ 0, which implies that βn+1x on J. Similarly we obtain x ≤ αn+1on J. Since β0x ≤ α0 on J , by induction we get β n ≤ × ≤ α n on J for every n. Therefore, x* (t) ≤ x(t) ≤ x*(t) on J by taking n → ∞. 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 BVP
        x t = t 3 1 + x t + 1 54 t 0 t t s x s d s 3 + 1 81 t 2 0 1 t s x s d s 3 , t J = 0 , 1 , t 1 2 , Δ x 1 2 = 1 4 1 5 x 1 10 + 3 10 x 1 5 + 1 10 x 3 10 + 1 5 x 2 5 + 1 5 x 1 2 , k = 1 , x 0 + 1 5 1 5 x 1 5 + 2 5 x 3 10 + 2 5 x 1 2 = x 1 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-38/MediaObjects/13661_2011_Article_162_Equ16_HTML.gif
        (4.1)

        where k(t, s) = h(t, s) = ts, m = 1, t 1 = 1 2 , c 1 = 5 , ρ 1 1 = 1 5 , ρ 2 1 = 3 10 , ρ 3 1 = 1 10 , ρ 4 1 = 1 5 , ρ 5 1 = 1 5 , η 1 1 = 1 10 , η 2 1 = 1 5 , η 3 1 = 3 10 , η 4 1 = 2 5 , η 5 1 = 1 2 , τ 1 1 = 0 , τ 2 1 = 1 5 , τ 3 1 = 2 5 , τ 4 1 = 0 , τ 5 1 = 2 5 , μ = 1 5 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-38/MediaObjects/13661_2011_Article_162_IEq37_HTML.gif.

        Obviously, α0 = 0, β 0 = - 5 , t 0 , 1 2 - 6 , t 1 2 , 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-38/MediaObjects/13661_2011_Article_162_IEq38_HTML.gif are lower and upper solutions for (4.1), respectively, and β0≤ α0.

        Let
        f t , x , y , z = t 3 1 + x + 1 54 t y 3 + 1 81 t 2 z 3 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-38/MediaObjects/13661_2011_Article_162_Equaf_HTML.gif
        Then,
        f t , x , y , z - f t , x ¯ , y ¯ , z ¯ x - x ¯ + 1 2 y - y ¯ + 1 3 z - z ¯ , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-38/MediaObjects/13661_2011_Article_162_Equag_HTML.gif
        where β 0 t x ¯ t x t α 0 t , F β 0 t y ¯ t y t F α 0 t , S β 0 t z ¯ t z t S α 0 t , t J . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-38/MediaObjects/13661_2011_Article_162_IEq39_HTML.gif It is easy to see that
        I 1 l = 1 5 ρ l 1 x η l 1 - I 1 l = 1 5 ρ l 1 y η l 1 = 1 4 l = 1 5 ρ l 1 x η l 1 - y η l 1 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-38/MediaObjects/13661_2011_Article_162_Equah_HTML.gif

        whenever β 0 η l 1 y η l 1 x η l 1 α 0 η l 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-38/MediaObjects/13661_2011_Article_162_IEq40_HTML.gif, l = 1, ..., 5.

        Taking L 1 = 1 4 , M = 1 , H = 1 2 , K = 1 3 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-38/MediaObjects/13661_2011_Article_162_IEq41_HTML.gif, it follows that
        e M T 0 T H 0 s k s , r e - M s - r d r + K 0 T h s , r e - M s - r d r d s + k = 1 m L k e - M t k l = 1 c k ρ l k e M η l k = e 0 1 1 2 0 s sr e - s - r d r + 1 3 0 1 sr e - s - r d r d s + 1 4 e - 1 2 1 5 e 1 10 + 3 10 e 1 5 + 1 10 e 3 10 + 1 5 e 2 5 + 1 5 e 1 2 0 . 9287149 1 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-38/MediaObjects/13661_2011_Article_162_Equai_HTML.gif
        and
        e M T e M T - 1 0 T H 0 s k s , r d r + K 0 T h s , r d r d s + e M T e M T - 1 k = 1 m L k l = 1 c k ρ l k = e e - 1 0 1 1 2 0 s srdr + 1 3 0 1 srdr d s + e e - 1 1 4 1 5 + 3 10 + 1 10 + 1 5 + 1 5 0 . 6261991 < 1 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-38/MediaObjects/13661_2011_Article_162_Equaj_HTML.gif

        Therefore, (4.1) satisfies all conditions of Theorem 3.1. So, BVP (4.1) has minimal and maximal solutions in the segment [β0, α0].

        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)
        Department of Mathematics, Faculty of Applied Science, King Mongkut's University of Technology North Bangkok
        (3)
        Centre of Excellence in Mathematics, CHE

        References

        1. Lakshmikantham V, Bainov DD, Simeonov PS: Theory of Impulsive Differential Equations. World Scientific, Singapore 1989.
        2. Bainov DD, Simeonov PS: Impulsive Differential Equations: Periodic Solutions and Applications. Longman Scientific & Technical, New York 1993.
        3. Samoilenko AM, Perestyuk NA: Impulsive Differential Equations. World Scientific, Singapore 1995.
        4. 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
        5. Zhang F, Li M, Yan J: Nonhomogeneous boundary value problem for first-order impulsive differential equations with delay. Comput Math Appl 2006, 51: 927-936. 10.1016/j.camwa.2005.11.028MathSciNetView Article
        6. Chen L, Sun J: Nonlinear boundary problem of first order impulsive integro-differential equations. J Comput Appl Math 2007, 202: 392-401. 10.1016/j.cam.2005.10.041MathSciNetView Article
        7. 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
        8. Ding W, Xing Y, Han M: Anti-periodic boundary value problems for first order impulsive functional differential equations. Appl Math Comput 2007, 186: 45-53. 10.1016/j.amc.2006.07.087MathSciNetView Article
        9. Yang X, Shen J: Nonlinear boundary value problems for first order impulsive functional differential equations. Appl Math Comput 2007, 189: 1943-1952. 10.1016/j.amc.2006.12.085MathSciNetView Article
        10. 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
        11. Wang X, Zhang J: Impulsive anti-periodic boundary value problem of first-order integro-differential equations. J Comput Appl Math 2010, 234: 3261-3267. 10.1016/j.cam.2010.04.024MathSciNetView Article
        12. 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
        13. Nieto JJ, Rodríguez-López R: Existence and approximation of solutions for nonlinear functional differential equations with periodic boundary value conditions. Comput Appl Math 2000, 40: 433-442. 10.1016/S0898-1221(00)00171-1View Article
        14. 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
        15. 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
        16. 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
        17. Chen L, Sun J: Nonlinear boundary value problem of first order impulsive functional differential equations. J Math Anal Appl 2006, 318: 726-741. 10.1016/j.jmaa.2005.08.012MathSciNetView Article
        18. Chen L, Sun J: Nonlinear boundary value problem for first order impulsive integro-differential equations of mixed type. J Math Anal Appl 2007, 325: 830-842. 10.1016/j.jmaa.2006.01.084MathSciNetView Article
        19. Wang G, Zhang L, Song G: Extremal solutions for the first order impulsive functional differential equations with upper and lower solutions in reversed order. J Comput Appl Math 2010, 235: 325-333. 10.1016/j.cam.2010.06.014MathSciNetView Article
        20. Zhang L: Boundary value problem for first order impulsive functional integro-differential equations. J Comput Appl Math 2011, 235: 2442-2450. 10.1016/j.cam.2010.10.045MathSciNetView Article
        21. Zhang Y, Zhang F: Multi-point boundary value problem of first order impulsive functional differential equations. J Appl Math Comput 2009, 31: 267-278. 10.1007/s12190-008-0209-2MathSciNetView Article
        22. 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
        23. 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

        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.