Research

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

Chatthai Thaiprayoon13*, Decha Samana13 and Jessada Tariboon23

Author Affiliations

1 Department of Mathematics, Faculty of Science, King Mongkut's Institute of Technology Ladkrabang, Bangkok 10520, Thailand

2 Department of Mathematics, Faculty of Applied Science, King Mongkut's University of Technology North Bangkok, Bangkok 10800, Thailand

3 Centre of Excellence in Mathematics, CHE, Sri Ayutthaya Road, Bangkok 10400, Thailand

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Boundary Value Problems 2012, 2012:38  doi:10.1186/1687-2770-2012-38

 Received: 8 August 2011 Accepted: 5 April 2012 Published: 5 April 2012

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.

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 [1-3]). 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 , (1.1)

where f C(J × R3, R), 0 = t0 < t1 < t2 < · · · < tm < 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 ,

k C(D, R+), D = {(t, s) ∈ J × J: t s}, h C(J × J, R+). Ik 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 , l = 1, 2, ..., ck , ck 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 [4-14]). 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 [15-20]). 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(tk) = Ik (x(tk)), that is a short-term rapid change of the state at impulse point tk depends on the left side of the limit of x(tk).

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 . (1.2)

It should be noticed that the terms t k - τ k t k x s d s and t k - 1 t k - 1 + σ k - 1 x s d s of impulsive condition (1.2) illustrate the past memory state on [tk - τk , tk] before impulse points tk and the history effects after the past impulse points tk-1 on (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 , (1.3)

for t k - 1 < η 1 k < η 2 k < < η c k k t k , k = 1 , 2 , , m . The new jump conditions mean that a sudden change of the state at impulse point tk depends on the multi-point η l k l = 1 , 2 , , c k of past states on (tk-1, tk]. We note that if ck = 1, η c k k = t k and ρ c k k = 1 , then the impulsive condition (1.3) is reduced to the simple impulsive condition Δx(tk) = Ik (x(tk)).

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, ..., tm}. PC(J, R) = {x: J → R; x(t) is continuous everywhere except for some tk at which x t k - and x t k + exist and x t k - = x t k , k = 1, 2, ..., m}, PC1(J, R) = {x PC(J, R); x'(t) is continuous everywhere except for some tk at which x t k + and x t k - exist and x t k - = x t k }. Let E = PC(J, R) and F = P C 1 J , R , then E and F   are Banach spaces with the nomes ||x||E = suptJ |x(t)| and x F = max x E , x E , respectively. A function x F is called a solution of BVP (1.1) if it satisfies (1.1).

Definition 2.1. A function α 0 F 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 .

Analogously, a function β 0 F 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 ,

where t k - 1 < η l k t k , ρ l k , τ l k 0 , l = 1, 2, ..., ck, ck 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 , (2.1)

where M > 0, H, K ≥ 0, Lk ≥ 0, t k - 1 < η l k t k , τ l k , ρ l k 0 , l = 1, 2, ..., ck, ck N = {1, 2, ...}, k = 1, 2,..., m are constants and v(t), σ(t) ∈ E.

Lemma 2.1. x F is a solution of (2.1) if and only if x E 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 , (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 .

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 . (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 . (2.4)

From x 0 + μ k = 1 m l = 1 c k τ l k σ η l k = x T , 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 . (2.5)

Substituting (2.5) into (2.3), we see that x E 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 .

Differentiating (2.2) for t ≠ tk, we obtain

x t = M x t + H F x t + K S x t + v t .

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 .

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 . This completes the proof.   □

Lemma 2.2. Assume that M > 0, H, K ≥ 0, Lk ≥ 0, t k - 1 < η l k t k , ρ l k 0 , l = 1, 2, ..., ck , ck 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 . (2.6)

Then BVP (2.1) has a unique solution.

Proof. For any x E, 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 , (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 , we have for any x, y E, 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 .

From (2.6) and the Banach fixed point theorem, A has a unique fixed point x ¯ E . By Lemma 2.1, x ¯ is also the unique solution of (2.1).   □

Lemma 2.3. Assume that x F satisfies

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 , (2.8)

where M > 0, H, K ≥ 0, Lk ≥ 0, t k - 1 < η l k t k , ρ l k 0 , l = 1, 2, ..., ck , ck 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 , (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 . Then, x(t) ≤ 0 for all t J.

Proof. Set u(t) = x(t)e-Mt for t J , 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 . (2.10)

Obviously, the function u(t) and x(t) have the same sign. Suppose, to the contrary, that u(t) > 0 for some t J. Then, there are two cases:

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

(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 t J- and Δu(tk) ≥ 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)eMT, which is a contradiction.

Case (ii): Let t* ∈ (ti, ti+1], i ∈ {0, 1, ..., m}, such that u(t*) = inf {u(t): t J} < 0 and t* ∈ (tj, 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 * , (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 * ,

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

Now, assume that t* < t*. Since 0 ≥ u(0) ≥ eMTu(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 ,

and u(0) ≥ eMT 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 (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 * .

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 .

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 .

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 * .

This is a contradiction and so u(t) ≤ 0 for all t J. 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 , we denote

β 0 , α 0 = x F : β 0 t x t α 0 t , t J ,

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

Theorem 3.1. Let the following conditions hold.

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

(H2) The function f C(J × R3, R) satisfies

f t , x , y , z - f t , x ¯ , y ¯ , z ¯ M x - x ¯ + H y - y ¯ + K z - z ¯ ,

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 .

(H3) The function Ik 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 ,

whenever β 0 η l k y η l k x η l k α 0 η l k , l = 1, 2, ..., ck , ck N = {1, 2, ...}, Lk ≥ 0, k = 1, 2, ..., m.

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

Then there exist monotone sequences α n , β n F such that limn→∞ αn(t) = x*(t), limn→∞ βn(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 ,

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 .

By Lemma 2.2, BVP (2.1) has a unique solution x(t) for t J. We define an operator A by x = , then the operator A is an operator from [β0, α0] to F   and A has the following properties.

(i) β0 0, 0 α0;

(ii) For any σl, σ2 ∈ [β0, α0], σl σ2 implies l 2.

To prove (i), set φ = β0 - β1, where β1 = 0. Then from (Hl) and (2.1) for t J-, 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 ,

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 ,

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 .

By Lemma 2.3, we get that φ(t) ≤ 0 for all t J , i.e., β0 0. Similarly, we can prove that 0 α0.

To prove (ii), let ul = l, u2 = 2, where σl σ2 on J and σl, σ2 ∈ [β0, α0]. Set φ = ul - u2. Then for t J- 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 ,

and by (H3);

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 .

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 .

Then by using Lemma 2.3, we have φ(t) ≤ 0, which implies that l 2.

Now, we define the sequences {αn}, {βn} such that αn+l = n and βn+l = n. From (i) and (ii) the sequence {αn}, {βn} satisfy the inequality

β 0 β 1 β n α n α 1 α 0 ,

for all n N. Obviously, each αn, βn (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 ,

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 .

Therefore, there exist x* and x*, such that limn→∞ βn = x* and limn→∞ αn = 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 βn(t) ≤ x(t) ≤ αn (t) on J. Let φ = βn+1 - x, then for t J-,

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 ,

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 ,

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 .

Then by using Lemma 2.3, we have φ(t) ≤ 0, which implies that βn+1 x on J. Similarly we obtain x ≤ αn+1 on J. Since β0 x ≤ α0 on J , by induction we get βn ≤ × ≤ αn 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 , (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 .

Obviously, α0 = 0, β 0 = - 5 , t 0 , 1 2 - 6 , t 1 2 , 1 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 .

Then,

f t , x , y , z - f t , x ¯ , y ¯ , z ¯ x - x ¯ + 1 2 y - y ¯ + 1 3 z - z ¯ ,

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 . 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 ,

whenever β 0 η l 1 y η l 1 x η l 1 α 0 η l 1 , l = 1, ..., 5.

Taking L 1 = 1 4 , M = 1 , H = 1 2 , K = 1 3 , 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 ,

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 .

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

Competing interests

The authors declare that they have no competing interests.

Acknowledgements

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

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