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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

The electronic version of this article is the complete one and can be found online at: http://www.boundaryvalueproblems.com/content/2012/1/38


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

© 2012 Thaiprayoon et al; licensee Springer.

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):

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/38/mathml/M1','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/38/mathml/M1">View MathML</a>

(1.1)

where f C(J × R3, R), 0 = t0 < t1 < t2 < · · · < tm < tm+1 = T,

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/38/mathml/M2','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/38/mathml/M2">View MathML</a>

k C(D, R+), D = {(t, s) ∈ J × J: t s}, h C(J × J, R+). Ik C(R, R), <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/38/mathml/M3','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/38/mathml/M3">View MathML</a>, 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

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/38/mathml/M4','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/38/mathml/M4">View MathML</a>

(1.2)

It should be noticed that the terms <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/38/mathml/M5','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/38/mathml/M5">View MathML</a> and <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/38/mathml/M6','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/38/mathml/M6">View MathML</a> 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

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/38/mathml/M7','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/38/mathml/M7">View MathML</a>

(1.3)

for <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/38/mathml/M8','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/38/mathml/M8">View MathML</a>. The new jump conditions mean that a sudden change of the state at impulse point tk depends on the multi-point <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/38/mathml/M9','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/38/mathml/M9">View MathML</a> of past states on (tk-1, tk]. We note that if ck = 1, <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/38/mathml/M10','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/38/mathml/M10">View MathML</a> and <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/38/mathml/M11','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/38/mathml/M11">View MathML</a>, 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 <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/38/mathml/M12','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/38/mathml/M12">View MathML</a> and <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/38/mathml/M13','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/38/mathml/M13">View MathML</a> exist and <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/38/mathml/M14','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/38/mathml/M14">View MathML</a>, k = 1, 2, ..., m}, PC1(J, R) = {x PC(J, R); x'(t) is continuous everywhere except for some tk at which <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/38/mathml/M15','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/38/mathml/M15">View MathML</a> and <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/38/mathml/M16','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/38/mathml/M16">View MathML</a> exist and <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/38/mathml/M17','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/38/mathml/M17">View MathML</a>}. Let E = PC(J, R) and <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/38/mathml/M18','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/38/mathml/M18">View MathML</a>, then E and <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/38/mathml/M19','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/38/mathml/M19">View MathML</a> are Banach spaces with the nomes ||x||E = suptJ |x(t)| and <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/38/mathml/M20','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/38/mathml/M20">View MathML</a>, respectively. A function <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/38/mathml/M21','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/38/mathml/M21">View MathML</a> is called a solution of BVP (1.1) if it satisfies (1.1).

Definition 2.1. A function <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/38/mathml/M22','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/38/mathml/M22">View MathML</a> is called a lower solution of BVP (1.1) if:

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/38/mathml/M23','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/38/mathml/M23">View MathML</a>

Analogously, a function <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/38/mathml/M24','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/38/mathml/M24">View MathML</a> is called an upper solution of BVP (1.1) if:

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/38/mathml/M25','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/38/mathml/M25">View MathML</a>

where <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/38/mathml/M26','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/38/mathml/M26">View MathML</a>, 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):

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/38/mathml/M27','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/38/mathml/M27">View MathML</a>

(2.1)

where M > 0, H, K ≥ 0, Lk ≥ 0, <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/38/mathml/M28','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/38/mathml/M28">View MathML</a>, l = 1, 2, ..., ck, ck N = {1, 2, ...}, k = 1, 2,..., m are constants and v(t), σ(t) ∈ E.

Lemma 2.1. <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/38/mathml/M29','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/38/mathml/M29">View MathML</a>is a solution of (2.1) if and only if x E is a solution of the impulsive integral equation

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/38/mathml/M30','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/38/mathml/M30">View MathML</a>

(2.2)

where P(t) = H(Fx)(t) + K(Sx)(t) + v(t) and

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/38/mathml/M31','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/38/mathml/M31">View MathML</a>

Proof. Assume that x(t) is a solution of BVP (2.1). By using the variation of parameters formula, we get

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/38/mathml/M32','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/38/mathml/M32">View MathML</a>

(2.3)

Putting t = T in (2.3), we have

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/38/mathml/M33','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/38/mathml/M33">View MathML</a>

(2.4)

From <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/38/mathml/M34','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/38/mathml/M34">View MathML</a>, we obtain

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/38/mathml/M35','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/38/mathml/M35">View MathML</a>

(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

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/38/mathml/M36','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/38/mathml/M36">View MathML</a>

Differentiating (2.2) for t ≠ tk, we obtain

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/38/mathml/M37','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/38/mathml/M37">View MathML</a>

It is easy to see that

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/38/mathml/M38','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/38/mathml/M38">View MathML</a>

Since G(0, s) = G(T, s), then <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/38/mathml/M39','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/38/mathml/M39">View MathML</a>. This completes the proof.   □

Lemma 2.2. Assume that M > 0, H, K ≥ 0, Lk ≥ 0,<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/38/mathml/M40','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/38/mathml/M40">View MathML</a>, l = 1, 2, ..., ck , ck N = {1, 2, ...}, k = 1, 2, ..., m, and the following inequality holds:

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/38/mathml/M41','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/38/mathml/M41">View MathML</a>

(2.6)

Then BVP (2.1) has a unique solution.

Proof. For any x E, we define an operator A by

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/38/mathml/M42','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/38/mathml/M42">View MathML</a>

(2.7)

where G(t, s) is defined as in Lemma 2.1. Since <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/38/mathml/M43','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/38/mathml/M43">View MathML</a>, we have for any x, y E, that

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/38/mathml/M44','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/38/mathml/M44">View MathML</a>

From (2.6) and the Banach fixed point theorem, A has a unique fixed point <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/38/mathml/M45','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/38/mathml/M45">View MathML</a>. By Lemma 2.1, <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/38/mathml/M46','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/38/mathml/M46">View MathML</a> is also the unique solution of (2.1).   □

Lemma 2.3. Assume that <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/38/mathml/M47','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/38/mathml/M47">View MathML</a>satisfies

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/38/mathml/M48','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/38/mathml/M48">View MathML</a>

(2.8)

where M > 0, H, K ≥ 0, Lk ≥ 0,<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/38/mathml/M49','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/38/mathml/M49">View MathML</a>, l = 1, 2, ..., ck , ck N = {1, 2, ...}, k = 1, 2, ..., m. In addition assume that

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/38/mathml/M50','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/38/mathml/M50">View MathML</a>

(2.9)

where <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/38/mathml/M51','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/38/mathml/M51">View MathML</a>. Then, x(t) ≤ 0 for all t J.

Proof. Set u(t) = x(t)e-Mt for t J , then we have

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/38/mathml/M52','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/38/mathml/M52">View MathML</a>

(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

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/38/mathml/M53','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/38/mathml/M53">View MathML</a>

(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

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/38/mathml/M54','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/38/mathml/M54">View MathML</a>

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

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/38/mathml/M55','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/38/mathml/M55">View MathML</a>

and u(0) ≥ eMT u(T). In consequence,

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/38/mathml/M56','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/38/mathml/M56">View MathML</a>

(2.12)

can be obtained.

If t* = 0, then

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/38/mathml/M57','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/38/mathml/M57">View MathML</a>

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

If t* > 0, we obtain from (2.11),

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/38/mathml/M58','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/38/mathml/M58">View MathML</a>

This joint to (2.12) yields

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/38/mathml/M59','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/38/mathml/M59">View MathML</a>

Therefore,

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/38/mathml/M60','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/38/mathml/M60">View MathML</a>

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 <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/38/mathml/M61','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/38/mathml/M61">View MathML</a>, we denote

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/38/mathml/M62','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/38/mathml/M62">View MathML</a>

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

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/38/mathml/M63','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/38/mathml/M63">View MathML</a>

for <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/38/mathml/M64','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/38/mathml/M64">View MathML</a>.

(H3) The function Ik C(R, R) satisfies

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/38/mathml/M65','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/38/mathml/M65">View MathML</a>

whenever <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/38/mathml/M66','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/38/mathml/M66">View MathML</a>, 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 <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/38/mathml/M67','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/38/mathml/M67">View MathML</a>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

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/38/mathml/M68','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/38/mathml/M68">View MathML</a>

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

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/38/mathml/M69','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/38/mathml/M69">View MathML</a>

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 <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/38/mathml/M70','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/38/mathml/M70">View MathML</a> 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

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/38/mathml/M71','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/38/mathml/M71">View MathML</a>

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/38/mathml/M72','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/38/mathml/M72">View MathML</a>

and

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/38/mathml/M73','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/38/mathml/M73">View MathML</a>

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

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/38/mathml/M74','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/38/mathml/M74">View MathML</a>

and by (H3);

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/38/mathml/M75','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/38/mathml/M75">View MathML</a>

It is easy to see that

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/38/mathml/M76','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/38/mathml/M76">View MathML</a>

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

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/38/mathml/M77','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/38/mathml/M77">View MathML</a>

for all n N. Obviously, each αn, βn (n = 1, 2, ...) satisfy

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/38/mathml/M78','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/38/mathml/M78">View MathML</a>

and

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/38/mathml/M79','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/38/mathml/M79">View MathML</a>

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-,

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/38/mathml/M80','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/38/mathml/M80">View MathML</a>

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/38/mathml/M81','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/38/mathml/M81">View MathML</a>

and

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/38/mathml/M82','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/38/mathml/M82">View MathML</a>

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

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/38/mathml/M83','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/38/mathml/M83">View MathML</a>

(4.1)

where k(t, s) = h(t, s) = ts, m = 1, <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/38/mathml/M84','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/38/mathml/M84">View MathML</a>.

Obviously, α0 = 0, <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/38/mathml/M85','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/38/mathml/M85">View MathML</a> are lower and upper solutions for (4.1), respectively, and β0 ≤ α0.

Let

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/38/mathml/M86','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/38/mathml/M86">View MathML</a>

Then,

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/38/mathml/M87','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/38/mathml/M87">View MathML</a>

where <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/38/mathml/M88','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/38/mathml/M88">View MathML</a> It is easy to see that

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/38/mathml/M89','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/38/mathml/M89">View MathML</a>

whenever <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/38/mathml/M90','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/38/mathml/M90">View MathML</a>, l = 1, ..., 5.

Taking <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/38/mathml/M91','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/38/mathml/M91">View MathML</a>, it follows that

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/38/mathml/M92','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/38/mathml/M92">View MathML</a>

and

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/38/mathml/M93','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/38/mathml/M93">View MathML</a>

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.

Authors' contributions

All authors contributed equally in this article. They read and approved the final manuscript.

Acknowledgements

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

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