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Positive solutions for boundary value problem of fractional differential equation with p-Laplacian operator

Guoqing Chai

Author affiliations

College of Mathematics and Statistics, Hubei Normal University, Hubei 435002, P.R. China

Citation and License

Boundary Value Problems 2012, 2012:18  doi:10.1186/1687-2770-2012-18

The electronic version of this article is the complete one and can be found online at: http://www.boundaryvalueproblems.com/content/2012/1/18

Received:12 October 2011
Accepted:15 February 2012
Published:15 February 2012

© 2012 Chai; 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.


In this article, the author investigates the existence and multiplicity of positive solutions for boundary value problem of fractional differential equation with p-Laplacian operator

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/18/mathml/M1','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/18/mathml/M1">View MathML</a>

where <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/18/mathml/M2','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/18/mathml/M2">View MathML</a> and <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/18/mathml/M3','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/18/mathml/M3">View MathML</a> are the standard Riemann-Liouville derivatives with 1 < α ≤ 2, 0 < β ≤ 1, 0 < γ ≤ 1, 0 ≤ α - γ - 1, the constant σ is a positive number and p-Laplacian operator is defined as φp(s) = |s|p-2s, p > 1. By means of the fixed point theorem on cones, some existence and multiplicity results of positive solutions are obtained.

2010 Mathematical Subject Classification: 34A08; 34B18.

fractional differential equations; fixed point index; p-Laplacian operator; positive solution; multiplicity of solutions

1 Introduction

Differential equations of fractional order have been recently proved to be valuable tools in the modeling of many phenomena in various fields of science and engineering. Indeed, we can find numerous applications in viscoelasticity, electrochemistry, control, porous media, electromagnetism, etc. (see [1-5]). There has been a significant development in the study of fractional differential equations in recent years, see the monographs of Kilbas et al. [6], Lakshmikantham et al. [7], Podlubny [4], Samko et al. [8], and the survey by Agarwal et al. [9].

For some recent contributions on fractional differential equations, see for example, [10-28] and the references therein. Especially, in [15], by means of Guo-Krasnosel'skiĭ's fixed point theorem, Zhao et al. investigated the existence of positive solutions for the nonlinear fractional boundary value problem (BVP for short)

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/18/mathml/M4','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/18/mathml/M4">View MathML</a>


where 1 < α ≤ 2, f : [0, +∞) → (0, +∞).

In [16], relying on the Krasnosel'skiĭ's fixed point theorem as well as on the Leggett-Williams fixed point theorem, Kaufmann and Mboumi discussed the existence of positive solutions for the following fractional BVP

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/18/mathml/M5','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/18/mathml/M5">View MathML</a>

In [17], by applying Altman's fixed point theorem and Leray-Schauder' fixed point theorem, Wang obtained the existence and uniqueness of solutions for the following BVP of nonlinear impulsive differential equations of fractional order q

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/18/mathml/M6','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/18/mathml/M6">View MathML</a>

In [18], relying on the contraction mapping principle and the Krasnosel'skiĭ's fixed point theorem, Zhou and Chu discussed the existence of solutions for a nonlinear multi-point BVP of integro-differential equations of fractional order q ∈ (1, 2]

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/18/mathml/M7','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/18/mathml/M7">View MathML</a>

On the other hand, integer-order p-Laplacian boundary value problems have been widely studied owing to its importance in theory and application of mathematics and physics, see for example, [29-33] and the references therein. Especially, in [29], by using the fixed point index method, Yang and Yan investigated the existence of positive solution for the third-order Sturm-Liouville boundary value problems with p-Laplacian operator

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/18/mathml/M8','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/18/mathml/M8">View MathML</a>


where φp(s) = |s|p-2s.

However, there are few articles dealing with the existence of solutions to boundary value problems for fractional differential equation with p-Laplacian operator. In [24], the authors investigated the nonlinear nonlocal problem

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/18/mathml/M9','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/18/mathml/M9">View MathML</a>


where 0 < β ≤ 1, 1 < α ≤ 2, 0 ≤ a ≤ 1, 0 < ξ < 1. By using Krasnosel'skiĭ's fixed point theorem and Leggett-Williams theorem, some sufficient conditions for the existence of positive solutions to the above BVP are obtained.

In [25], by using upper and lower solutions method, under suitable monotone conditions, the authors investigated the existence of positive solutions to the following nonlocal problem

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/18/mathml/M10','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/18/mathml/M10">View MathML</a>


where 1 < α, β ≤ 2, 0 ≤ a, b ≤ 1, 0 < ξ, η < 1.

No contribution exists, as far as we know, concerning the existence of solutions for the fractional differential equation with p-Laplacian operator

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/18/mathml/M11','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/18/mathml/M11">View MathML</a>


where <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/18/mathml/M2','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/18/mathml/M2">View MathML</a> and <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/18/mathml/M3','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/18/mathml/M3">View MathML</a> are the standard Riemann-Liouville derivative with 1 < α ≤ 2, 0 < β ≤ 1, 0 < γ ≤ 1, 0 ≤ α - γ - 1, the constant σ is a positive number, the p-Laplacian operator is defined as φp(s) = |s|p-2s, p > 1, and function f is assumed to satisfy certain conditions, which will be specified later. To obtain the existence and multiplicity of positive solutions to BVP (1.5), the fixed point theorem on cones will be applied.

It is worth emphasizing that our work presented in this article has the following features which are different from those in [24,25]. Firstly, BVP (1.5) is an important two point BVP. When γ = 1, the boundary value conditions in (1.5) reduce to u(0) = 0, u(1) + σu'(1) = 0, which are the well-known Sturm-Liouville boundary value conditions u(0) + bu'(0) = 0, u(1) + σu'(1) = 0 (such as BVP (1.1)) with b = 0. It is a well-known fact that the boundary value problems with Sturm-Liouville boundary value conditions for integral order differential equations have important physical and applied background and have been studied in many literatures, while BVPs (1.3) and (1.4) are the nonlocal boundary value problems, which are not able to substitute BVP (1.5). Secondly, when α = 2, β = 1, γ = 1, then BVP (1.5) reduces to BVP (1.2) with b = 0. So, BVP (1.5) is an important generalization of BVP (1.2) from integral order to fractional order. Thirdly, in BVPs (1.3) or (1.4), the boundary value conditions u(1) = au(ξ), <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/18/mathml/M12','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/18/mathml/M12">View MathML</a> show the relations between the derivatives of same order <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/18/mathml/M13','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/18/mathml/M13">View MathML</a> and <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/18/mathml/M14','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/18/mathml/M14">View MathML</a>. By contrast with that, the condition <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/18/mathml/M15','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/18/mathml/M15">View MathML</a> in BVP (1.5) shows that relation between the derivatives of different order u(1) and <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/18/mathml/M16','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/18/mathml/M16">View MathML</a> is regarded as the derivative value of zero order of u at t = 1), which brings about more difficulties in deducing the property of green's function than the former. Finally, order α + β satisfies that 2 < α + β ≤ 4 in BVP (1.4), while order α + β satisfies that 1 < α + β ≤ 3 in BVP (1.5). In the case for α, β taking integral numbers, the BVPs (1.5) and (1.4) are the third-order BVP and the fourth-order BVP, respectively. So, BVP (1.5) differs essentially from BVP (1.4). In addition, the conditions imposed in present paper are easily verified.

The organization of this article is as follows. In Section 2, we present some necessary definitions and preliminary results that will be used to prove our main results. In Section 3, we put forward and prove our main results. Finally, we will give two examples to demonstrate our main results.

2 Preliminaries

In this section, we introduce some preliminary facts which are used throughout this article.

Let ℕ be the set of positive integers, be the set of real numbers and + be the set of nonnegative real numbers. Let I = [0, 1]. Denote by C(I, ) the Banach space of all continuous functions from I into with the norm

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/18/mathml/M17','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/18/mathml/M17">View MathML</a>

Define the cone P in C(I, ) as P = {u C(I, ): u(t) ≥ 0, t I}. Let q > 1 satisfy the relation <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/18/mathml/M18','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/18/mathml/M18">View MathML</a>, where p is given by (1. 5).

Definition 2.1. [6] The Riemann-Liouville fractional integral of order α > 0 of a function y : (a, b] → is given by

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/18/mathml/M19','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/18/mathml/M19">View MathML</a>

Definition 2.2. [6] The Riemann-Liouville fractional derivative of order α > 0 of function y : (a, b] → is given by

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/18/mathml/M20','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/18/mathml/M20">View MathML</a>

where n = [α] + 1 and [α] denotes the integer part of α.

Lemma 2.1. [34] Let α > 0. If u C(0, 1) ⋂ L(0, 1) possesses a fractional derivative of order α that belongs to C(0, 1) ⋂ L(0, 1), then

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/18/mathml/M21','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/18/mathml/M21">View MathML</a>

for some ci , i = 1, 2,..., n, where n = [α] + 1.

A function u C(I, ) is called a nonnegative solution of BVP (1.5), if u ≥ 0 on [0, 1] and satisfies (1.5). Moreover, if u(t) > 0, t ∈ (0, 1), then u is said to be a positive solution of BVP (1.5).

For forthcoming analysis, we first consider the following fractional differential equation

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/18/mathml/M22','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/18/mathml/M22">View MathML</a>


where α, γ, σ are given by (1.5) and ϕ C(I, ).

By Lemma 2.1, we have

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/18/mathml/M23','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/18/mathml/M23">View MathML</a>

From the boundary condition u(0) = 0, we have c2 = 0, and so

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/18/mathml/M24','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/18/mathml/M24">View MathML</a>



<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/18/mathml/M25','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/18/mathml/M25">View MathML</a>


<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/18/mathml/M26','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/18/mathml/M26">View MathML</a>

From the boundary condition <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/18/mathml/M15','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/18/mathml/M15">View MathML</a>, it follows that

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/18/mathml/M27','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/18/mathml/M27">View MathML</a>

Let <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/18/mathml/M28','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/18/mathml/M28">View MathML</a>. Then

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/18/mathml/M29','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/18/mathml/M29">View MathML</a>


Substituting (2.3) into (2.2), we have

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/18/mathml/M30','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/18/mathml/M30">View MathML</a>



<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/18/mathml/M31','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/18/mathml/M31">View MathML</a>


<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/18/mathml/M32','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/18/mathml/M32">View MathML</a>

So, we obtain the following lemma.

Lemma 2.2. The solution of Equation (2.1) is given by

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/18/mathml/M33','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/18/mathml/M33">View MathML</a>

Also, we have the following lemma.

Lemma 2.3. The Green's function G(t, s) has the following properties

(i) G(t, s) is continuous on [0, 1] × [0, 1],

(ii) G(t, s) > 0, s, t ∈ (0, 1).

Proof. (i) Owing to the fact 1 < α ≤ 2, 0 < γ ≤ 1, 0 ≤ α - γ - 1, from the expression of G, it is easy to see that conclusion (i) of Lemma 2.3 is true.

(ii) There are two cases to consider.

(1) If 0 < s t < 1, then

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/18/mathml/M34','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/18/mathml/M34">View MathML</a>

(2) If 0 < t s < 1, then conclusion (ii) of Lemma 2.3 is obviously true from the expression of G.

We need to introduce some notations for the forthcoming discussion.

Let <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/18/mathml/M35','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/18/mathml/M35">View MathML</a>. Denote <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/18/mathml/M36','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/18/mathml/M36">View MathML</a>, s ∈ [0, 1]. Set g(s) = G(s, s), s ∈ [0, 1]. From 0 < γ ≤ 1, σ > 0, 1 < α ≤ 2 and <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/18/mathml/M37','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/18/mathml/M37">View MathML</a>, we know that η0 ∈ (0, 1).

The following lemma is fundamental in this article.

Lemma 2.4. The Green's function G has the properties

(i) G(t, s) ≤ G(s, s),s, t ∈ [0, 1].

(ii) G(t, s) ≥ η(s)G(s, s), t ∈ [η0, 1], s ∈ [0, 1].

Proof. (i) There are two cases to consider.

Case 1. 0 ≤ s t ≤ 1. In this case, since the following relation

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/18/mathml/M38','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/18/mathml/M38">View MathML</a>

holds for 0 < s < t ≤ 1, we have

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/18/mathml/M39','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/18/mathml/M39">View MathML</a>

Case 2. 0 ≤ t s ≤ 1. In this case, from the expression of g2(t, s), it is easy to see that

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/18/mathml/M40','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/18/mathml/M40">View MathML</a>

(ii) We will consider the following two cases.

Case 1. When 0 < s η0, η0 t ≤ 1, then from the above argument in (i) of proof, we know that g1(t, s) is decreasing with respect to t on [η0, 1]. Thus

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/18/mathml/M41','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/18/mathml/M41">View MathML</a>


and so

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/18/mathml/M42','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/18/mathml/M42">View MathML</a>

Case 2. η0 < s < 1, η0 t ≤ 1.

(a) If s t, then by similar arguments to (2.5), we also have

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/18/mathml/M43','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/18/mathml/M43">View MathML</a>

(b) If η0 t s, then the following relation

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/18/mathml/M44','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/18/mathml/M44">View MathML</a>

holds in view of the expression of g2(t, s).

To summarize,

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/18/mathml/M45','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/18/mathml/M45">View MathML</a>


Now, we shall show that

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/18/mathml/M46','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/18/mathml/M46">View MathML</a>


In fact, for s ∈ (0, 1), we have

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/18/mathml/M47','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/18/mathml/M47">View MathML</a>

and so

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/18/mathml/M48','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/18/mathml/M48">View MathML</a>


On the other hand, for s ∈ (η0, 1), we have

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/18/mathml/M49','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/18/mathml/M49">View MathML</a>


Since <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/18/mathml/M50','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/18/mathml/M50">View MathML</a>, the equality

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/18/mathml/M51','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/18/mathml/M51">View MathML</a>

holds for s = 1. Thus,

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/18/mathml/M52','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/18/mathml/M52">View MathML</a>


Since 1 < α ≤ 2, it follows from (2.9) that

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/18/mathml/M53','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/18/mathml/M53">View MathML</a>


Hence, from (2.8) and (2.11), we immediately have

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/18/mathml/M54','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/18/mathml/M54">View MathML</a>


Thus, from (2.6) and (2.12 ), it follows that

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/18/mathml/M55','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/18/mathml/M55">View MathML</a>


Also, by (2.8), the following inequality

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/18/mathml/M56','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/18/mathml/M56">View MathML</a>

holds, and therefore

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/18/mathml/M57','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/18/mathml/M57">View MathML</a>


from the proof in Case 1.

Summing up the above relations (2.13)-(2.14), we have

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/18/mathml/M58','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/18/mathml/M58">View MathML</a>

and so

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/18/mathml/M59','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/18/mathml/M59">View MathML</a>

The proof of Lemma 2.4 is complete.

To study BVP (1. 5), we first consider the associated linear BVP

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/18/mathml/M60','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/18/mathml/M60">View MathML</a>


where h P.

Let <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/18/mathml/M61','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/18/mathml/M61">View MathML</a>. By Lemma 2.1, the solution of initial value problem

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/18/mathml/M62','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/18/mathml/M62">View MathML</a>

is given by

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/18/mathml/M63','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/18/mathml/M63">View MathML</a>

From the relations v(0) = 0, 0 < β ≤ 1, it follows that C1 = 0, and so

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/18/mathml/M64','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/18/mathml/M64">View MathML</a>


Noting that <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/18/mathml/M65','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/18/mathml/M65">View MathML</a>, from (2.16), we know that the solution of (2.15) satisfies

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/18/mathml/M66','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/18/mathml/M66">View MathML</a>


By Lemma 2.2, the solution of Equation (2.17) can be written as

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/18/mathml/M67','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/18/mathml/M67">View MathML</a>


Since h(s) ≥ 0, s ∈ [0, 1], we have <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/18/mathml/M68','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/18/mathml/M68">View MathML</a>, s ∈ [0, 1], and so

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/18/mathml/M69','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/18/mathml/M69">View MathML</a>


from (2.18). Thus, by Lemma 2.3, we have obtained the following lemma.

Lemma 2.5. Let h P. Then the solution of Equation (2.15) in P is given by

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/18/mathml/M70','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/18/mathml/M70">View MathML</a>

We also need the following lemmas to obtain our results.

Lemma 2.6. If a, b ≥ 0, γ > 0, then

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/18/mathml/M71','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/18/mathml/M71">View MathML</a>

Proof. Obviously, without loss of generality, we can assume that 0 < a < b, γ ≠ 1.

Let ϕ(t) = tγ, t ∈ [0, +∞).

(i) If γ > 1, then ϕ(t) is convex on (0, +∞), and so

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/18/mathml/M72','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/18/mathml/M72">View MathML</a>


i.e., <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/18/mathml/M73','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/18/mathml/M73">View MathML</a>. Thus

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/18/mathml/M74','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/18/mathml/M74">View MathML</a>

(ii) If 0 < γ < 1, then ϕ(t) is concave on [0, +∞), and so

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/18/mathml/M75','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/18/mathml/M75">View MathML</a>

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/18/mathml/M76','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/18/mathml/M76">View MathML</a>

Thus, ϕ(a) + ϕ(b) ≥ ϕ(a + b), namely,

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/18/mathml/M77','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/18/mathml/M77">View MathML</a>

By (i), (ii) above, we know that the conclusion of Lemma 2.6 is true.

Lemma 2.7. Let c > 0, γ > 0. For any x, y ∈ [0, c], we have that

(i) If γ > 1, then |xγ - yγ| ≤ γcγ-1 |x - y|,

(ii) If 0 < γ ≤ 1, then |xγ - yγ| ≤ |x - y|γ.

Proof. Obviously, without loss of generality, we can assume that 0 < y < x since the variables x and y are symmetrical in the above inequality.

(i) If γ > 1, then we set ϕ(t) = tγ, t ∈ [0, c]. by virtue of mean value theorem, there exists a ξ ∈ (0, c) such that

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/18/mathml/M78','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/18/mathml/M78">View MathML</a>


<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/18/mathml/M79','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/18/mathml/M79">View MathML</a>

(ii) If 0 < γ < 1, then by Lemma 2.6, it is easy to see that

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/18/mathml/M80','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/18/mathml/M80">View MathML</a>

and so

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/18/mathml/M81','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/18/mathml/M81">View MathML</a>

Now we introduce some notations, which will be used in the sequel.

Let <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/18/mathml/M82','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/18/mathml/M82">View MathML</a>,

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/18/mathml/M83','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/18/mathml/M83">View MathML</a>

By simple calculation, we know that

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/18/mathml/M84','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/18/mathml/M84">View MathML</a>


<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/18/mathml/M85','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/18/mathml/M85">View MathML</a>


In this article, the following hypotheses will be used.

(H1) f C(I × +, +).

(H2) <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/18/mathml/M86','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/18/mathml/M86">View MathML</a>.

(H3) There exists a r0 > 0 such that f(t, x) is nonincreasing relative to x on [0, r0] for any fixed t I.

By Lemma 2.5, it is easy to know that the following lemma is true.

Lemma 2.8. If (H1) holds, then BVP (1.5) has a nonnegative solution if and only if the integral equation

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/18/mathml/M87','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/18/mathml/M87">View MathML</a>


has a solution in P. Let c be a positive number, P be a cone and Pc = {y P : ∥y∥ ≤ c}. Let α be a nonnegative continuous concave function on P and

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/18/mathml/M88','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/18/mathml/M88">View MathML</a>

We will use the following lemma to obtain the multiplicity results of positive solutions.

Lemma 2.9. [35] Let <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/18/mathml/M89','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/18/mathml/M89">View MathML</a> be completely continuous and α be a nonnegative continuous concave function on P such that α(y) ≤ ∥y∥ for all <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/18/mathml/M90','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/18/mathml/M90">View MathML</a>. Suppose that there exist a, b and d with 0 < a < b < d c such that

(C1) <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/18/mathml/M91','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/18/mathml/M91">View MathML</a> and α(Ay) > b, for all y P(α, b, d);

(C2) ∥Ay∥ < a, for ∥y∥ ≤ a;

(C3) α(Ay) > b, for y P(α, b, c) with ∥Ay∥ > d.

Then A has at least three fixed points y1, y2, y3 satisfying

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/18/mathml/M92','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/18/mathml/M92">View MathML</a>

3 Main results

In this section, our objective is to establish existence and multiplicity of positive solution to the BVP (1.5). To this end, we first define the operator on P as

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/18/mathml/M93','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/18/mathml/M93">View MathML</a>


The properties of the operator A are given in the following lemma.

Lemma 3.1. Let (H1) hold. Then A : P P is completely continuous.

Proof. First, under assumption (H1), it is obvious that AP P from Lemma 2.3. Next, we shall show that operator A is completely continuous on P. Let <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/18/mathml/M94','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/18/mathml/M94">View MathML</a>. The following proof will be divided into two steps.

Step 1. We shall show that the operator A is compact on P.

Let B be an arbitrary bounded set in P. Then exists an M > 0 such that ∥u∥ ≤ M for all u B. According to the continuity of f, we have <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/18/mathml/M95','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/18/mathml/M95">View MathML</a>. Thus, by Lemmas 2.3 and 2.4, it follows that

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/18/mathml/M96','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/18/mathml/M96">View MathML</a>


<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/18/mathml/M97','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/18/mathml/M97">View MathML</a>

That is, the set AB is uniformly bounded.

On the other hand, the uniform continuity of G(t, s) on I × I implies that for arbitrary ε > 0, there exists a δ > 0 such that whenever t1, t2 I with |t1 - t2| < δ, the following inequality

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/18/mathml/M98','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/18/mathml/M98">View MathML</a>

holds for all s I. Therefore,

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/18/mathml/M99','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/18/mathml/M99">View MathML</a>

Thus, AB is equicontinuous. Consequently, the operator is compact on P by Arzelà-Ascoli theorem.

Step 2. The operator A is continuous.

Let {un} be an arbitrary sequence in P with un u0 P. Then exists an L > 0 such that

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/18/mathml/M100','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/18/mathml/M100">View MathML</a>


<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/18/mathml/M101','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/18/mathml/M101">View MathML</a>

On the other hand, the uniform continuity of f combined with the fact that ∥un - u0∥ → 0 yields that there exists a N ≥ 1 such that the following estimate

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/18/mathml/M102','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/18/mathml/M102">View MathML</a>

holds for n N.

(1) If 1 < q ≤ 2, then from Lemma 2.7 (ii), we have

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/18/mathml/M103','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/18/mathml/M103">View MathML</a>

Hence, by Lemmas 2.3 and 2.4, from (3.1), we obtain

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/18/mathml/M104','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/18/mathml/M104">View MathML</a>


<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/18/mathml/M105','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/18/mathml/M105">View MathML</a>


(2) If q > 2, then from Lemma 2.7 (i), we have

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/18/mathml/M106','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/18/mathml/M106">View MathML</a>

Thus, we have

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/18/mathml/M107','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/18/mathml/M107">View MathML</a>

and so

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/18/mathml/M108','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/18/mathml/M108">View MathML</a>


From (3.2)-(3.3), it follows that ∥Aun - Au0∥ → 0(n → ∞).

Summing up the above analysis, we obtain that the operator A is completely continuous on P.

We are now in a position to state and prove the first theorem in this article.

Theorem 3.1. Let (H1), (H2), and (H3) hold. Then BVP (1.5) has at least one positive solution.

Proof. By Lemma 2.8, it is easy to know that BVP (1.5) has a nonnegative solution if and only if the operator A has a fixed point in P. Also, we know that A : P P is completely continuous by Lemma 3.1.

The following proof is divided into two steps.

Step 1. From (H2), we can choose a ε0 ∈ (0, l) such that

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/18/mathml/M109','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/18/mathml/M109">View MathML</a>

Therefore, there exists a R0 > 0 such that the inequality

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/18/mathml/M110','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/18/mathml/M110">View MathML</a>


holds for x R0.

Let <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/18/mathml/M111','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/18/mathml/M111">View MathML</a>. It follows from (3.4) that

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/18/mathml/M112','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/18/mathml/M112">View MathML</a>


From the fact that (l - ε0)q-1 < lq-1, we can choose a k > 0 such that (l - ε0)q-1 < lq-1 - k.


<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/18/mathml/M113','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/18/mathml/M113">View MathML</a>


where D is as (2.20). Take <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/18/mathml/M114','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/18/mathml/M114">View MathML</a>. Set ΩR = {u P : ∥u∥ < R}. We shall show that the relation

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/18/mathml/M115','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/18/mathml/M115">View MathML</a>



In fact, if not, then there exists a u0 ΩR and a μ0 ≥ 1 with

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/18/mathml/M116','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/18/mathml/M116">View MathML</a>

By (3.5), we have

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/18/mathml/M117','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/18/mathml/M117">View MathML</a>

Therefore, in view of Lemmas 2.3, 2.4, from (3.1), it follows that

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/18/mathml/M118','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/18/mathml/M118">View MathML</a>


Also, keeping in mind that (p - 1)(q - 1) = 1, by Lemma 2.6, we have

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/18/mathml/M119','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/18/mathml/M119">View MathML</a>


Hence, from (3.6), (3.8), and (3.9), it follows that

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/18/mathml/M120','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/18/mathml/M120">View MathML</a>


By definition of l, we have D1lq-1 = 1. From (3. 10), it follows that R = ∥u0∥ ≤ (1 - E)R + G, and so <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/18/mathml/M121','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/18/mathml/M121">View MathML</a>, which contradicts the choice of R. Hence, the condition (3.7) holds. By virtue of the fixed point index theorem, we have

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/18/mathml/M122','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/18/mathml/M122">View MathML</a>


Step 2. By (H2), we can choose a ε0 > 0 such that

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/18/mathml/M123','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/18/mathml/M123">View MathML</a>

Hence, there exists a r1 ∈ (0, r0) such that

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/18/mathml/M124','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/18/mathml/M124">View MathML</a>


where r0 is given by (H3).

Take 0 < r < min {R, r1}, and set Ωr = {u P : ∥u∥ < r}. Now, we show that

(i) <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/18/mathml/M125','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/18/mathml/M125">View MathML</a>,

(ii) Au μu, ∀u Ωr, μ ∈ [0, 1].

We first prove that (i) holds. In fact, for any u Ωr, we have 0 ≤ u(t) ≤ r. By (H3), the function f(t, x) is nonincreasing relative to x on [0, r] for any t I, and so

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/18/mathml/M126','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/18/mathml/M126">View MathML</a>


from (3.12).

Thus, in view of Lemma 2.4 combined with (3.1) and (3.13), we have

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/18/mathml/M127','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/18/mathml/M127">View MathML</a>

where Q is as (2.21). Consequently,

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/18/mathml/M128','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/18/mathml/M128">View MathML</a>


Thus <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/18/mathml/M129','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/18/mathml/M129">View MathML</a>.

(ii) Suppose on the contrary that there exists a u0 Ωr and μ0 ∈ [0, 1] such that

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/18/mathml/M116','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/18/mathml/M116">View MathML</a>


Then, by similar arguments to (3.14), we have

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/18/mathml/M130','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/18/mathml/M130">View MathML</a>


where <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/18/mathml/M131','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/18/mathml/M131">View MathML</a>.

By (3.15)-(3.16), we obtain

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/18/mathml/M132','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/18/mathml/M132">View MathML</a>

The hypothesis <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/18/mathml/M133','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/18/mathml/M133">View MathML</a> implies that B > 1, and so r > r from above inequality, which is a contradiction. That means that (ii) holds.

Hence, applying fixed point index theorem, we have

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/18/mathml/M134','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/18/mathml/M134">View MathML</a>


By (3.11) and (3.17), we have

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/18/mathml/M135','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/18/mathml/M135">View MathML</a>

and so, there exists <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/18/mathml/M136','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/18/mathml/M136">View MathML</a> with Au* = u*, ∥u*∥ > r. Hence, u* is a nonnegative solution of BVP (1.5) satisfying ∥u*∥ > r. Now, we show that u*(t) > 0, t ∈ (0, 1).

In fact, since ∥u*∥ > r, u* P, G(t, s) > 0, t, s ∈ (0, 1), from (3.1), we have

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/18/mathml/M137','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/18/mathml/M137">View MathML</a>

and so

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/18/mathml/M138','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/18/mathml/M138">View MathML</a>


from the fact that G(t, s) > 0 and <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/18/mathml/M139','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/18/mathml/M139">View MathML</a>, s ∈ [0, 1]. That is, u* is a positive solution of BVP (1.5).

The proof is complete.

Now, we state another theorem in this article. First, let me introduce some notations which will be used in the sequel.

Let <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/18/mathml/M140','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/18/mathml/M140">View MathML</a>, where D is as (2.20).


<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/18/mathml/M141','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/18/mathml/M141">View MathML</a>


Set Pr = {u P : ∥u∥ < r}, for r > 0. Let <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/18/mathml/M142','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/18/mathml/M142">View MathML</a>, for u P. Obviously, ω is a nonnegative continuous concave functional on P.

Theorem 3.2. Let (H1) hold. Assume that there exist constants a, b, c, l1, l2 with 0 < a < b < c and l1 ∈ (0, M1), l2 ∈ (M2, ∞) such that

(D1) f(t, x) ≤ l1cp-1, x ∈ [0,c], t I; f(t, x)≤ l1ap-1, x ∈ [0, a], t I,

(D2) f(t, x) ≥ l1bp-1, x ∈ [b,c], t ∈ [η0, 1].

Then BVP (1.5) has at least one nonnegative solution u1 and two positive solutions u2, u3 with

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/18/mathml/M143','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/18/mathml/M143">View MathML</a>

Proof. By Lemmas 2.3 and 2.4, for <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/18/mathml/M144','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/18/mathml/M144">View MathML</a>, from (3.1) and condition (D1), it follows that

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/18/mathml/M145','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/18/mathml/M145">View MathML</a>

and so

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/18/mathml/M146','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/18/mathml/M146">View MathML</a>

from the hypothesis l1 < M1.

Thus, we obtain <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/18/mathml/M147','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/18/mathml/M147">View MathML</a>. Similarly, we can also obtain <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/18/mathml/M148','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/18/mathml/M148">View MathML</a> by condition (D1). Take <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/18/mathml/M149','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/18/mathml/M149">View MathML</a>. Then ω(u0) > b, and so <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/18/mathml/M150','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/18/mathml/M150">View MathML</a>.

For any u P(ω, b, c), we have that u(t) ≥ b, t ∈ [η0, 1] and ∥u∥ ≤ c. Consequently, by Lemma 2.3, 2.4 and the formula (3.1), for any t ∈ [η0, 1], it follows from condition (D2) that

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/18/mathml/M151','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/18/mathml/M151">View MathML</a>


Also, by changing the variable <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/18/mathml/M152','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/18/mathml/M152">View MathML</a>, we have

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/18/mathml/M153','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/18/mathml/M153">View MathML</a>


where B2 is given by (3.19).

Substituting (3.21) into (3.20), we obtain

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/18/mathml/M154','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/18/mathml/M154">View MathML</a>

and so ω(Au) > b from the hypothesis l2 > M2.

Summing up the above analysis, we know that all the conditions of Lemma 2.9 with c = d are satisfied, and so BVP (1.5) has at least three nonnegative solutions u1, u2, u3 with

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/18/mathml/M155','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/18/mathml/M155">View MathML</a>

By similar argument to (3.18), we can deduce that u2 and u3 are two positive solutions.

The proof is complete.

Example 3.1. Consider the following BVP

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/18/mathml/M156','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/18/mathml/M156">View MathML</a>


where 1 < α < 2, 0 < β < 1, 0 < γ < 1, 0 ≤ α - γ - 1, σ > 0 and the p-Laplacian operator is defined as φp(s) = |s|p-2s, p > 1.

It is easy to verify that all assumptions of Theorem 3.1 are satisfied. Hence, by the conclusion of Theorem 3.1, BVP (3.22) has at least one positive solution on [0, 1].

Example 3.2. Consider the following BVP

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/18/mathml/M157','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/18/mathml/M157">View MathML</a>


where <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/18/mathml/M158','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/18/mathml/M158">View MathML</a> relative to Theorem 3.2. With the aid of computation we have that <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/18/mathml/M159','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/18/mathml/M159">View MathML</a>. Take l1 = 1.5, l2 = 4. Then l1 ∈ (0, M1), l2 ∈ (M2, ∞). Again choosing <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/18/mathml/M160','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/18/mathml/M160">View MathML</a>, b = 1, c = 196, and setting

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/18/mathml/M161','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/18/mathml/M161">View MathML</a>

for t ∈ [0, 1], then we see that f satisfies the following relations:

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/18/mathml/M162','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/18/mathml/M162">View MathML</a>

So, all the assumptions of Theorem 3.2 are satisfied. By Theorem 3.2, we arrive at BVP (3.23) has at least one nonnegative solution u1 and two positive solutions u2, u3 with

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/18/mathml/M163','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/18/mathml/M163">View MathML</a>

Competing interests

The author declares that they have no competing interests.


The author sincerely thanks the anonymous referees for their valuable suggestions and comments which have greatly helped improve this article. Supported by the Natural Science Foundation of Educational Committee of Hubei Province (D200722002).


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