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Periodic boundary value problems for nonlinear first-order impulsive dynamic equations on time scales

Da-Bin Wang

Author Affiliations

Department of Applied Mathematics, Lanzhou University of Technology, Lanzhou, Gansu 730050, People’s Republic of China

Advances in Difference Equations 2012, 2012:12  doi:10.1186/1687-1847-2012-12

The electronic version of this article is the complete one and can be found online at: http://www.advancesindifferenceequations.com/content/2012/1/12

Received:23 August 2011
Accepted:15 February 2012
Published:15 February 2012

© 2012 Wang; 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.


By using the classical fixed point theorem for operators on cone, in this article, some results of one and two positive solutions to a class of nonlinear first-order periodic boundary value problems of impulsive dynamic equations on time scales are obtained. Two examples are given to illustrate the main results in this article.

Mathematics Subject Classification: 39A10; 34B15.

time scale; periodic boundary value problem; positive solution; fixed point; impulsive dynamic equation

1 Introduction

Let T be a time scale, i.e., T is a nonempty closed subset of R. Let 0, T be points in T, an interval (0, T)T denoting time scales interval, that is, (0, T)T: = (0, T) ⋂ T. Other types of intervals are defined similarly.

The theory of impulsive differential equations is emerging as an important area of investigation, since it is a lot richer than the corresponding theory of differential equations without impulse effects. Moreover, such equations may exhibit several real world phenomena in physics, biology, engineering, etc. (see [1-3]). At the same time, the boundary value problems for impulsive differential equations and impulsive difference equations have received much attention [4-18]. On the other hand, recently, the theory of dynamic equations on time scales has become a new important branch (see, for example, [19-21]). Naturally, some authors have focused their attention on the boundary value problems of impulsive dynamic equations on time scales [22-36]. However, to the best of our knowledge, few papers concerning PBVPs of impulsive dynamic equations on time scales with semi-position condition.

In this article, we are concerned with the existence of positive solutions for the following PBVPs of impulsive dynamic equations on time scales with semi-position condition

<a onClick="popup('http://www.advancesindifferenceequations.com/content/2012/1/12/mathml/M1','MathML',630,470);return false;" target="_blank" href="http://www.advancesindifferenceequations.com/content/2012/1/12/mathml/M1">View MathML</a>


where T is an arbitrary time scale, T > 0 is fixed, 0, T T, f C (J × [0, ∞), (-∞, ∞)), Ik C([0, ∞), [0, ∞)), tk ∈ (0, T)T, 0 < t1 < ⋯ < tm < T, and for each k = 1, 2,..., m, <a onClick="popup('http://www.advancesindifferenceequations.com/content/2012/1/12/mathml/M2','MathML',630,470);return false;" target="_blank" href="http://www.advancesindifferenceequations.com/content/2012/1/12/mathml/M2">View MathML</a> and <a onClick="popup('http://www.advancesindifferenceequations.com/content/2012/1/12/mathml/M3','MathML',630,470);return false;" target="_blank" href="http://www.advancesindifferenceequations.com/content/2012/1/12/mathml/M3">View MathML</a> represent the right and left limits of x(t) at t = tk. We always assume the following hypothesis holds (semi-position condition):

(H) There exists a positive number M such that

<a onClick="popup('http://www.advancesindifferenceequations.com/content/2012/1/12/mathml/M4','MathML',630,470);return false;" target="_blank" href="http://www.advancesindifferenceequations.com/content/2012/1/12/mathml/M4">View MathML</a>

By using a fixed point theorem for operators on cone [37], some existence criteria of positive solution to the problem (1.1) are established. We note that for the case T = R and Ik(x) ≡ 0, k = 1, 2,..., m, the problem (1.1) reduces to the problem studied by [38] and for the case Ik(x) ≡ 0, k = 1, 2,..., m, the problem (1.1) reduces to the problem (in the one-dimension case) studied by [39].

In the remainder of this section, we state the following fixed point theorem [37].

Theorem 1.1. Let X be a Banach space and K X be a cone in X. Assume Ω1, Ω2 are bounded open subsets of X with <a onClick="popup('http://www.advancesindifferenceequations.com/content/2012/1/12/mathml/M5','MathML',630,470);return false;" target="_blank" href="http://www.advancesindifferenceequations.com/content/2012/1/12/mathml/M5">View MathML</a> and Φ: <a onClick="popup('http://www.advancesindifferenceequations.com/content/2012/1/12/mathml/M6','MathML',630,470);return false;" target="_blank" href="http://www.advancesindifferenceequations.com/content/2012/1/12/mathml/M6">View MathML</a> is a completely continuous operator. If

(i) There exists u0 K\{0} such that u - Φu ≠ λu0, u K ⋂ ∂ Ω2, λ≥ 0; Φu τu, u K ⋂ ∂Ω1, τ ≥ 1, or

(ii) There exists u0 K\{0} such that u - Φu ≠ λu0, u K ⋂ ∂Ω1, λ≥ 0; Φu τu, u K ⋂ ∂Ω2, τ ≥ 1.

Then Φ has at least one fixed point in <a onClick="popup('http://www.advancesindifferenceequations.com/content/2012/1/12/mathml/M7','MathML',630,470);return false;" target="_blank" href="http://www.advancesindifferenceequations.com/content/2012/1/12/mathml/M7">View MathML</a>.

2 Preliminaries

Throughout the rest of this article, we always assume that the points of impulse tk are right-dense for each k = 1, 2,...,m.

We define

<a onClick="popup('http://www.advancesindifferenceequations.com/content/2012/1/12/mathml/M8','MathML',630,470);return false;" target="_blank" href="http://www.advancesindifferenceequations.com/content/2012/1/12/mathml/M8">View MathML</a>

where xk is the restriction of x to Jk = (tk, tk+1]T ⊂ (0, σ(T)]T, k = 1, 2,..., m and J0 = [0, t1]T, tm +1 = σ(T).


<a onClick="popup('http://www.advancesindifferenceequations.com/content/2012/1/12/mathml/M9','MathML',630,470);return false;" target="_blank" href="http://www.advancesindifferenceequations.com/content/2012/1/12/mathml/M9">View MathML</a>

with the norm <a onClick="popup('http://www.advancesindifferenceequations.com/content/2012/1/12/mathml/M10','MathML',630,470);return false;" target="_blank" href="http://www.advancesindifferenceequations.com/content/2012/1/12/mathml/M10">View MathML</a>, then X is a Banach space.

Lemma 2.1. Suppose M > 0 and h: [0, T]T R is rd-continuous, then x is a solution of

<a onClick="popup('http://www.advancesindifferenceequations.com/content/2012/1/12/mathml/M11','MathML',630,470);return false;" target="_blank" href="http://www.advancesindifferenceequations.com/content/2012/1/12/mathml/M11">View MathML</a>

where <a onClick="popup('http://www.advancesindifferenceequations.com/content/2012/1/12/mathml/M12','MathML',630,470);return false;" target="_blank" href="http://www.advancesindifferenceequations.com/content/2012/1/12/mathml/M12">View MathML</a>

if and only if x is a solution of the boundary value problem

<a onClick="popup('http://www.advancesindifferenceequations.com/content/2012/1/12/mathml/M13','MathML',630,470);return false;" target="_blank" href="http://www.advancesindifferenceequations.com/content/2012/1/12/mathml/M13">View MathML</a>

Proof. Since the proof similar to that of [34, Lemma 3.1], we omit it here.

Lemma 2.2. Let G(t, s) be defined as in Lemma 2.1, then

<a onClick="popup('http://www.advancesindifferenceequations.com/content/2012/1/12/mathml/M14','MathML',630,470);return false;" target="_blank" href="http://www.advancesindifferenceequations.com/content/2012/1/12/mathml/M14">View MathML</a>

Proof. It is obviously, so we omit it here.

Remark 2.1. Let G(t, s) be defined as in Lemma 2.1, then <a onClick="popup('http://www.advancesindifferenceequations.com/content/2012/1/12/mathml/M15','MathML',630,470);return false;" target="_blank" href="http://www.advancesindifferenceequations.com/content/2012/1/12/mathml/M15">View MathML</a>.

For u X, we consider the following problem:

<a onClick="popup('http://www.advancesindifferenceequations.com/content/2012/1/12/mathml/M16','MathML',630,470);return false;" target="_blank" href="http://www.advancesindifferenceequations.com/content/2012/1/12/mathml/M16">View MathML</a>


It follows from Lemma 2.1 that the problem (2.1) has a unique solution:

<a onClick="popup('http://www.advancesindifferenceequations.com/content/2012/1/12/mathml/M17','MathML',630,470);return false;" target="_blank" href="http://www.advancesindifferenceequations.com/content/2012/1/12/mathml/M17">View MathML</a>

where hu(s) = Mu(σ(s)) - f(s, u(σ(s))), s ∈ [0, T]T.

We define an operator Φ: X X by

<a onClick="popup('http://www.advancesindifferenceequations.com/content/2012/1/12/mathml/M18','MathML',630,470);return false;" target="_blank" href="http://www.advancesindifferenceequations.com/content/2012/1/12/mathml/M18">View MathML</a>

It is obvious that fixed points of Φ are solutions of the problem (1.1).

Lemma 2.3. Φ: X X is completely continuous.

Proof. The proof is divided into three steps.

Step 1: To show that Φ: X X is continuous.

Let <a onClick="popup('http://www.advancesindifferenceequations.com/content/2012/1/12/mathml/M19','MathML',630,470);return false;" target="_blank" href="http://www.advancesindifferenceequations.com/content/2012/1/12/mathml/M19">View MathML</a> be a sequence such that un u (n → ∞) in X. Since f(t, u) and Ik(u) are continuous in x, we have

<a onClick="popup('http://www.advancesindifferenceequations.com/content/2012/1/12/mathml/M20','MathML',630,470);return false;" target="_blank" href="http://www.advancesindifferenceequations.com/content/2012/1/12/mathml/M20">View MathML</a>


<a onClick="popup('http://www.advancesindifferenceequations.com/content/2012/1/12/mathml/M21','MathML',630,470);return false;" target="_blank" href="http://www.advancesindifferenceequations.com/content/2012/1/12/mathml/M21">View MathML</a>

which leads to ||Φun - Φu|| → 0 (n → ∞). That is, Φ: X X is continuous.

Step 2: To show that Φ maps bounded sets into bounded sets in X.

Let B X be a bounded set, that is, ∃ r > 0 such that ∀ u B we have ||u|| ≤ r. Then, for any u B, in virtue of the continuities of f(t, u) and Ik(u), there exist c > 0, ck > 0 such that

<a onClick="popup('http://www.advancesindifferenceequations.com/content/2012/1/12/mathml/M22','MathML',630,470);return false;" target="_blank" href="http://www.advancesindifferenceequations.com/content/2012/1/12/mathml/M22">View MathML</a>

We get

<a onClick="popup('http://www.advancesindifferenceequations.com/content/2012/1/12/mathml/M23','MathML',630,470);return false;" target="_blank" href="http://www.advancesindifferenceequations.com/content/2012/1/12/mathml/M23">View MathML</a>

Then we can conclude that Φu is bounded uniformly, and so Φ(B) is a bounded set.

Step 3: To show that Φ maps bounded sets into equicontinuous sets of X.

Let t1, t2 ∈ (tk, tk+1]T ⋂ [0, σ(T)]T, u B, then

<a onClick="popup('http://www.advancesindifferenceequations.com/content/2012/1/12/mathml/M24','MathML',630,470);return false;" target="_blank" href="http://www.advancesindifferenceequations.com/content/2012/1/12/mathml/M24">View MathML</a>

The right-hand side tends to uniformly zero as |t1 - t2| → 0.

Consequently, Steps 1-3 together with the Arzela-Ascoli Theorem shows that Φ: X X is completely continuous.


<a onClick="popup('http://www.advancesindifferenceequations.com/content/2012/1/12/mathml/M25','MathML',630,470);return false;" target="_blank" href="http://www.advancesindifferenceequations.com/content/2012/1/12/mathml/M25">View MathML</a>

where <a onClick="popup('http://www.advancesindifferenceequations.com/content/2012/1/12/mathml/M26','MathML',630,470);return false;" target="_blank" href="http://www.advancesindifferenceequations.com/content/2012/1/12/mathml/M26">View MathML</a>. It is not difficult to verify that K is a cone in X.

From condition (H) and Lemma 2.2, it is easy to obtain following result:

Lemma 2.4. Φ maps K into K.

3 Main results

For convenience, we denote

<a onClick="popup('http://www.advancesindifferenceequations.com/content/2012/1/12/mathml/M27','MathML',630,470);return false;" target="_blank" href="http://www.advancesindifferenceequations.com/content/2012/1/12/mathml/M27">View MathML</a>


<a onClick="popup('http://www.advancesindifferenceequations.com/content/2012/1/12/mathml/M28','MathML',630,470);return false;" target="_blank" href="http://www.advancesindifferenceequations.com/content/2012/1/12/mathml/M28">View MathML</a>

Now we state our main results.

Theorem 3.1. Suppose that

(H1) f0 > 0, f< 0, I0 = 0 for any k; or

(H2) f> 0, f0 < 0, I= 0 for any k.

Then the problem (1.1) has at least one positive solutions.

Proof. Firstly, we assume (H1) holds. Then there exist ε > 0 and β > α > 0 such that

<a onClick="popup('http://www.advancesindifferenceequations.com/content/2012/1/12/mathml/M29','MathML',630,470);return false;" target="_blank" href="http://www.advancesindifferenceequations.com/content/2012/1/12/mathml/M29">View MathML</a>


<a onClick="popup('http://www.advancesindifferenceequations.com/content/2012/1/12/mathml/M30','MathML',630,470);return false;" target="_blank" href="http://www.advancesindifferenceequations.com/content/2012/1/12/mathml/M30">View MathML</a>



<a onClick="popup('http://www.advancesindifferenceequations.com/content/2012/1/12/mathml/M31','MathML',630,470);return false;" target="_blank" href="http://www.advancesindifferenceequations.com/content/2012/1/12/mathml/M31">View MathML</a>


Let Ω1 = {u X: ||u|| < r1}, where r1 = α. Then u K ⋂ ∂Ω1, 0 < δα = δ ||u|| ≤ u(t) ≤ α, in view of (3.1) and (3.2) we have

<a onClick="popup('http://www.advancesindifferenceequations.com/content/2012/1/12/mathml/M32','MathML',630,470);return false;" target="_blank" href="http://www.advancesindifferenceequations.com/content/2012/1/12/mathml/M32">View MathML</a>

which yields ||Φ(u)|| < ||u||.


<a onClick="popup('http://www.advancesindifferenceequations.com/content/2012/1/12/mathml/M33','MathML',630,470);return false;" target="_blank" href="http://www.advancesindifferenceequations.com/content/2012/1/12/mathml/M33">View MathML</a>


On the other hand, let Ω2 = {u X: ||u|| < r2}, where <a onClick="popup('http://www.advancesindifferenceequations.com/content/2012/1/12/mathml/M34','MathML',630,470);return false;" target="_blank" href="http://www.advancesindifferenceequations.com/content/2012/1/12/mathml/M34">View MathML</a>.

Choose u0 = 1, then u0 K\{0}. We assert that

<a onClick="popup('http://www.advancesindifferenceequations.com/content/2012/1/12/mathml/M35','MathML',630,470);return false;" target="_blank" href="http://www.advancesindifferenceequations.com/content/2012/1/12/mathml/M35">View MathML</a>


Suppose on the contrary that there exist <a onClick="popup('http://www.advancesindifferenceequations.com/content/2012/1/12/mathml/M36','MathML',630,470);return false;" target="_blank" href="http://www.advancesindifferenceequations.com/content/2012/1/12/mathml/M36">View MathML</a> and <a onClick="popup('http://www.advancesindifferenceequations.com/content/2012/1/12/mathml/M37','MathML',630,470);return false;" target="_blank" href="http://www.advancesindifferenceequations.com/content/2012/1/12/mathml/M37">View MathML</a> such that

<a onClick="popup('http://www.advancesindifferenceequations.com/content/2012/1/12/mathml/M38','MathML',630,470);return false;" target="_blank" href="http://www.advancesindifferenceequations.com/content/2012/1/12/mathml/M38">View MathML</a>

Let <a onClick="popup('http://www.advancesindifferenceequations.com/content/2012/1/12/mathml/M39','MathML',630,470);return false;" target="_blank" href="http://www.advancesindifferenceequations.com/content/2012/1/12/mathml/M39">View MathML</a>, then <a onClick="popup('http://www.advancesindifferenceequations.com/content/2012/1/12/mathml/M40','MathML',630,470);return false;" target="_blank" href="http://www.advancesindifferenceequations.com/content/2012/1/12/mathml/M40">View MathML</a>, we have from (3.3) that

<a onClick="popup('http://www.advancesindifferenceequations.com/content/2012/1/12/mathml/M41','MathML',630,470);return false;" target="_blank" href="http://www.advancesindifferenceequations.com/content/2012/1/12/mathml/M41">View MathML</a>


<a onClick="popup('http://www.advancesindifferenceequations.com/content/2012/1/12/mathml/M42','MathML',630,470);return false;" target="_blank" href="http://www.advancesindifferenceequations.com/content/2012/1/12/mathml/M42">View MathML</a>

which is a contradiction.

It follows from (3.4), (3.5) and Theorem 1.1 that Φ has a fixed point <a onClick="popup('http://www.advancesindifferenceequations.com/content/2012/1/12/mathml/M43','MathML',630,470);return false;" target="_blank" href="http://www.advancesindifferenceequations.com/content/2012/1/12/mathml/M43">View MathML</a>, and u* is a desired positive solution of the problem (1.1).

Next, suppose that (H2) holds. Then we can choose ε' > 0 and β' > α' > 0 such that

<a onClick="popup('http://www.advancesindifferenceequations.com/content/2012/1/12/mathml/M44','MathML',630,470);return false;" target="_blank" href="http://www.advancesindifferenceequations.com/content/2012/1/12/mathml/M44">View MathML</a>


<a onClick="popup('http://www.advancesindifferenceequations.com/content/2012/1/12/mathml/M45','MathML',630,470);return false;" target="_blank" href="http://www.advancesindifferenceequations.com/content/2012/1/12/mathml/M45">View MathML</a>



<a onClick="popup('http://www.advancesindifferenceequations.com/content/2012/1/12/mathml/M46','MathML',630,470);return false;" target="_blank" href="http://www.advancesindifferenceequations.com/content/2012/1/12/mathml/M46">View MathML</a>


Let Ω3 = {u X: ||u|| < r3}, where r3 = α'. Then for any u K ⋂ ∂Ω3, 0 < δ ||u|| ≤ u(t) ≤ ||u|| = α'.

It is similar to the proof of (3.5), we have

<a onClick="popup('http://www.advancesindifferenceequations.com/content/2012/1/12/mathml/M47','MathML',630,470);return false;" target="_blank" href="http://www.advancesindifferenceequations.com/content/2012/1/12/mathml/M47">View MathML</a>


Let Ω4 = {u X: ||u|| < r4}, where <a onClick="popup('http://www.advancesindifferenceequations.com/content/2012/1/12/mathml/M48','MathML',630,470);return false;" target="_blank" href="http://www.advancesindifferenceequations.com/content/2012/1/12/mathml/M48">View MathML</a>. Then for any u K ⋂ ∂Ω4, u(t) ≥ δ ||u|| = δr4 = β', by (3.6) and (3.7), it is easy to obtain

<a onClick="popup('http://www.advancesindifferenceequations.com/content/2012/1/12/mathml/M49','MathML',630,470);return false;" target="_blank" href="http://www.advancesindifferenceequations.com/content/2012/1/12/mathml/M49">View MathML</a>


It follows from (3.9), (3.10) and Theorem 1.1 that Φ has a fixed point <a onClick="popup('http://www.advancesindifferenceequations.com/content/2012/1/12/mathml/M50','MathML',630,470);return false;" target="_blank" href="http://www.advancesindifferenceequations.com/content/2012/1/12/mathml/M50">View MathML</a>, and u* is a desired positive solution of the problem (1.1).

Theorem 3.2. Suppose that

(H3) f0 < 0, f< 0;

(H4) there exists ρ > 0 such that

<a onClick="popup('http://www.advancesindifferenceequations.com/content/2012/1/12/mathml/M51','MathML',630,470);return false;" target="_blank" href="http://www.advancesindifferenceequations.com/content/2012/1/12/mathml/M51">View MathML</a>


<a onClick="popup('http://www.advancesindifferenceequations.com/content/2012/1/12/mathml/M52','MathML',630,470);return false;" target="_blank" href="http://www.advancesindifferenceequations.com/content/2012/1/12/mathml/M52">View MathML</a>


Then the problem (1.1) has at least two positive solutions.

Proof. By (H3), from the proof of Theorem 3.1, we should know that there exist β" > ρ > α" > 0 such that

<a onClick="popup('http://www.advancesindifferenceequations.com/content/2012/1/12/mathml/M53','MathML',630,470);return false;" target="_blank" href="http://www.advancesindifferenceequations.com/content/2012/1/12/mathml/M53">View MathML</a>


<a onClick="popup('http://www.advancesindifferenceequations.com/content/2012/1/12/mathml/M54','MathML',630,470);return false;" target="_blank" href="http://www.advancesindifferenceequations.com/content/2012/1/12/mathml/M54">View MathML</a>


where Ω5 = {u X: ||u|| < r5}, Ω6 = {u X: ||u|| < r6}, <a onClick="popup('http://www.advancesindifferenceequations.com/content/2012/1/12/mathml/M55','MathML',630,470);return false;" target="_blank" href="http://www.advancesindifferenceequations.com/content/2012/1/12/mathml/M55">View MathML</a>

By (3.11) of (H4), we can choose ε > 0 such that

<a onClick="popup('http://www.advancesindifferenceequations.com/content/2012/1/12/mathml/M56','MathML',630,470);return false;" target="_blank" href="http://www.advancesindifferenceequations.com/content/2012/1/12/mathml/M56">View MathML</a>


Let Ω7 = {u X: ||u|| < ρ}, for any u K ⋂ ∂Ω7, δρ = δ ||u|| ≤ u(t) ≤ ||u|| = ρ, from (3.12) and (3.15), it is similar to the proof of (3.4), we have

<a onClick="popup('http://www.advancesindifferenceequations.com/content/2012/1/12/mathml/M57','MathML',630,470);return false;" target="_blank" href="http://www.advancesindifferenceequations.com/content/2012/1/12/mathml/M57">View MathML</a>


By Theorem 1.1, we conclude that Φ has two fixed points <a onClick="popup('http://www.advancesindifferenceequations.com/content/2012/1/12/mathml/M58','MathML',630,470);return false;" target="_blank" href="http://www.advancesindifferenceequations.com/content/2012/1/12/mathml/M58">View MathML</a> and <a onClick="popup('http://www.advancesindifferenceequations.com/content/2012/1/12/mathml/M59','MathML',630,470);return false;" target="_blank" href="http://www.advancesindifferenceequations.com/content/2012/1/12/mathml/M59">View MathML</a>, and u** and u*** are two positive solution of the problem (1.1).

Similar to Theorem 3.2, we have:

Theorem 3.3. Suppose that

(H4) f0 > 0, f> 0, I0 = 0, I= 0;

(H5) there exists ρ > 0 such that

<a onClick="popup('http://www.advancesindifferenceequations.com/content/2012/1/12/mathml/M60','MathML',630,470);return false;" target="_blank" href="http://www.advancesindifferenceequations.com/content/2012/1/12/mathml/M60">View MathML</a>

Then the problem (1.1) has at least two positive solutions.

4 Examples

Example 4.1. Let T = [0, 1] ∪ [2,3]. We consider the following problem on T

<a onClick="popup('http://www.advancesindifferenceequations.com/content/2012/1/12/mathml/M61','MathML',630,470);return false;" target="_blank" href="http://www.advancesindifferenceequations.com/content/2012/1/12/mathml/M61">View MathML</a>


where T = 3, f(t, x) = x - (t + 1)x2, and I(x) = x2

Let M = 1, then, it is easy to see that

<a onClick="popup('http://www.advancesindifferenceequations.com/content/2012/1/12/mathml/M62','MathML',630,470);return false;" target="_blank" href="http://www.advancesindifferenceequations.com/content/2012/1/12/mathml/M62">View MathML</a>


<a onClick="popup('http://www.advancesindifferenceequations.com/content/2012/1/12/mathml/M63','MathML',630,470);return false;" target="_blank" href="http://www.advancesindifferenceequations.com/content/2012/1/12/mathml/M63">View MathML</a>

Therefore, by Theorem 3.1, it follows that the problem (4.1) has at least one positive solution.

Example 4.2. Let T = [0, 1] ∪ [2,3]. We consider the following problem on T

<a onClick="popup('http://www.advancesindifferenceequations.com/content/2012/1/12/mathml/M64','MathML',630,470);return false;" target="_blank" href="http://www.advancesindifferenceequations.com/content/2012/1/12/mathml/M64">View MathML</a>


where <a onClick="popup('http://www.advancesindifferenceequations.com/content/2012/1/12/mathml/M65','MathML',630,470);return false;" target="_blank" href="http://www.advancesindifferenceequations.com/content/2012/1/12/mathml/M65">View MathML</a>, and I(x) = x2e-x.

Choose M = 1, ρ = 4e2, then <a onClick="popup('http://www.advancesindifferenceequations.com/content/2012/1/12/mathml/M66','MathML',630,470);return false;" target="_blank" href="http://www.advancesindifferenceequations.com/content/2012/1/12/mathml/M66">View MathML</a>, it is easy to see that

<a onClick="popup('http://www.advancesindifferenceequations.com/content/2012/1/12/mathml/M67','MathML',630,470);return false;" target="_blank" href="http://www.advancesindifferenceequations.com/content/2012/1/12/mathml/M67">View MathML</a>


<a onClick="popup('http://www.advancesindifferenceequations.com/content/2012/1/12/mathml/M68','MathML',630,470);return false;" target="_blank" href="http://www.advancesindifferenceequations.com/content/2012/1/12/mathml/M68">View MathML</a>

Therefore, together with Theorem 3.3, it follows that the problem (4.2) has at least two positive solutions.

Competing interests

The author declares that they have no competing interests.


The author thankful to the anonymous referee for his/her helpful suggestions for the improvement of this article. This work is supported by the Excellent Young Teacher Training Program of Lanzhou University of Technology (Q200907)


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