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

Boundary Value Problems20122012:18

DOI: 10.1186/1687-2770-2012-18

Received: 12 October 2011

Accepted: 15 February 2012

Published: 15 February 2012

Abstract

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

D 0 + β φ p D 0 + α u ( t ) + f ( t , u ( t ) ) = 0 , 0 < t < 1 , u ( 0 ) = 0 , u ( 1 ) + σ D 0 + γ u ( 1 ) = 0 , D 0 + α u ( 0 ) = 0 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-18/MediaObjects/13661_2011_Article_127_Equa_HTML.gif

where D 0 + β , D 0 + α http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-18/MediaObjects/13661_2011_Article_127_IEq1_HTML.gif and D 0 + γ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-18/MediaObjects/13661_2011_Article_127_IEq2_HTML.gif 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 φ p (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.

Keywords

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 [15]). 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, [1028] 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)
D 0 + α u ( t ) = λ f ( u ( t ) ) , t ( 0 , 1 ) , u ( 0 ) + u ( 0 ) = 0 , u ( 1 ) + u ( 1 ) = 0 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-18/MediaObjects/13661_2011_Article_127_Equ1_HTML.gif
(1.1)

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
D 0 + α u ( t ) + a ( t ) f ( u ( t ) ) = 0 , 0 < t < 1 , 1 < α 2 , u ( 0 ) = 0 , u ( 1 ) = 0 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-18/MediaObjects/13661_2011_Article_127_Equb_HTML.gif
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
c D q u ( t ) = f ( t , u ( t ) ) , 1 < q 2 , t J , Δ u ( t k ) = Q k ( u ( t k ) ) , Δ u ( t k ) = I k ( u ( t k ) ) , k = 1 , 2 , . . . p , au ( 0 ) - b u ( 0 ) = x 0 , c u ( 1 ) + d u ( 1 ) = x 1 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-18/MediaObjects/13661_2011_Article_127_Equc_HTML.gif
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]
c D 0 + q u ( t ) = f ( t , u ( t ) , ( K u ) ( t ) , ( H u ) ( t ) ) , 1 < t < 1 , a 1 u ( 0 ) - b 1 u ( 0 ) = d 1 u ( ξ 1 ) , a 2 u ( 1 ) + b 2 u ( 1 ) = u ( ξ 2 ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-18/MediaObjects/13661_2011_Article_127_Equd_HTML.gif
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, [2933] 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
{ ( ϕ p ( u ( t ) ) + f ( t , u ( t ) ) = 0 , t ( 0 , 1 ) , a u ( 0 ) b u ( 0 ) = 0 c u ( 1 ) + u ( 1 ) = 0 , u ( 0 ) = 0 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-18/MediaObjects/13661_2011_Article_127_Equ2_HTML.gif
(1.2)

where φ p (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
D 0 + β φ p D 0 + α u ( t ) + f ( t , u ( t ) ) = 0 , 0 < t < 1 , u ( 0 ) = 0 , u ( 1 ) = au ( ξ ) , D 0 + α u ( 0 ) = 0 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-18/MediaObjects/13661_2011_Article_127_Equ3_HTML.gif
(1.3)

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
D 0 + β φ p D 0 + α u ( t ) + f ( t , u ( t ) ) , 0 < t < 1 , u ( 0 ) = 0 , u ( 1 ) = au ( ξ ) , D 0 + α u ( 0 ) = 0 , D 0 + α u ( 1 ) = b D 0 + α u ( η ) , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-18/MediaObjects/13661_2011_Article_127_Equ4_HTML.gif
(1.4)

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
D 0 + β φ p D 0 + α u ( t ) + f ( t , u ( t ) ) = 0 , 0 < t < 1 , u ( 0 ) = 0 , u ( 1 ) + σ D 0 + γ u ( 1 ) = 0 , D 0 + α u ( 0 ) = 0 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-18/MediaObjects/13661_2011_Article_127_Equ5_HTML.gif
(1.5)

where D 0 + β , D 0 + α http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-18/MediaObjects/13661_2011_Article_127_IEq1_HTML.gif and D 0 + γ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-18/MediaObjects/13661_2011_Article_127_IEq2_HTML.gif 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 φ p (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), D 0 + α u ( 1 ) = b D 0 + α u ( η ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-18/MediaObjects/13661_2011_Article_127_IEq3_HTML.gif show the relations between the derivatives of same order D 0 + μ u ( 1 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-18/MediaObjects/13661_2011_Article_127_IEq4_HTML.gif and D 0 + μ u ( ζ ) ( μ = 0 , α ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-18/MediaObjects/13661_2011_Article_127_IEq5_HTML.gif. By contrast with that, the condition u ( 1 ) + σ D 0 + γ u ( 1 ) = 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-18/MediaObjects/13661_2011_Article_127_IEq6_HTML.gif in BVP (1.5) shows that relation between the derivatives of different order u(1) and D 0 + γ u ( 1 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-18/MediaObjects/13661_2011_Article_127_IEq7_HTML.gif 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
u = max u ( t ) : t I . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-18/MediaObjects/13661_2011_Article_127_Eque_HTML.gif

Define the cone P in C(I, ) as P = {uC(I, ): u(t) ≥ 0, tI}. Let q > 1 satisfy the relation 1 q + 1 p = 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-18/MediaObjects/13661_2011_Article_127_IEq8_HTML.gif, 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
I a + α y ( t ) = 1 Γ ( α ) a t ( t - s ) α - 1 y ( s ) d s , t a , b . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-18/MediaObjects/13661_2011_Article_127_Equf_HTML.gif
Definition 2.2. [6] The Riemann-Liouville fractional derivative of order α > 0 of function y : (a, b] → is given by
D a + α y ( t ) = 1 Γ ( n - α ) d d t n a t y ( s ) ( t - s ) α - n + 1 d s , t a , b , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-18/MediaObjects/13661_2011_Article_127_Equg_HTML.gif

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

Lemma 2.1. [34] Let α > 0. If uC(0, 1) ⋂ L(0, 1) possesses a fractional derivative of order α that belongs to C(0, 1) ⋂ L(0, 1), then
I 0 + α D 0 + α u ( t ) = u ( t ) + c 1 t α - 1 + c 2 t α - 2 + + c n t α - n , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-18/MediaObjects/13661_2011_Article_127_Equh_HTML.gif

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

A function uC(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
D 0 + α u ( t ) + ϕ ( t ) = 0 , 0 < t < 1 , u ( 0 ) = 0 , u ( 1 ) + σ D 0 + γ u ( 1 ) = 0 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-18/MediaObjects/13661_2011_Article_127_Equ6_HTML.gif
(2.1)

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

By Lemma 2.1, we have
u ( t ) = c 1 t α - 1 + c 2 t α - 2 - I 0 + α ϕ ( t ) , t [ 0 , 1 ] . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-18/MediaObjects/13661_2011_Article_127_Equi_HTML.gif
From the boundary condition u(0) = 0, we have c2 = 0, and so
u ( t ) = c 1 t α - 1 - I 0 + α ϕ ( t ) , t [ 0 , 1 ] . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-18/MediaObjects/13661_2011_Article_127_Equ7_HTML.gif
(2.2)
Thus,
D 0 + γ u ( t ) = c 1 Γ ( α ) Γ ( α - γ ) t α - γ - 1 - I 0 + α - γ ϕ ( t ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-18/MediaObjects/13661_2011_Article_127_Equj_HTML.gif
and
u ( 1 ) = c 1 - I 0 + α ϕ ( 1 ) , D 0 + γ u ( 1 ) = c 1 Γ ( α ) Γ ( α - γ ) - I 0 + α - γ ϕ ( 1 ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-18/MediaObjects/13661_2011_Article_127_Equk_HTML.gif
From the boundary condition u ( 1 ) + σ D 0 + γ u ( 1 ) = 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-18/MediaObjects/13661_2011_Article_127_IEq6_HTML.gif, it follows that
1 + σ Γ ( α ) Γ ( α - γ ) c 1 - I 0 + α ϕ ( 1 ) + σ I 0 + α - γ ϕ ( 1 ) = 0 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-18/MediaObjects/13661_2011_Article_127_Equl_HTML.gif
Let δ = 1 + σ Γ ( α ) Γ ( α - γ ) - 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-18/MediaObjects/13661_2011_Article_127_IEq9_HTML.gif. Then
c 1 = δ I 0 + α ϕ ( 1 ) + σ I 0 + α - γ ϕ ( 1 ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-18/MediaObjects/13661_2011_Article_127_Equ8_HTML.gif
(2.3)
Substituting (2.3) into (2.2), we have
u ( t ) = δ I 0 + α ϕ ( 1 ) + σ I 0 + α - γ ϕ ( 1 ) t α - 1 - I 0 + α ϕ ( t ) = δ t α - 1 1 Γ ( α ) 0 1 ( 1 - s ) α - 1 ϕ ( s ) d s + 1 Γ ( α - γ ) σ 0 1 ( 1 - s ) α - γ - 1 ϕ ( s ) d s - 1 Γ ( α ) 0 t ( t - s ) α - 1 ϕ ( s ) d s = 1 Γ ( α ) δ t α - 1 0 1 ( 1 - s ) α - 1 + σ Γ ( α ) Γ ( α - γ ) ( 1 - s ) α - γ - 1 ϕ ( s ) d s - 0 t ( t - s ) α - 1 ϕ ( s ) d s = 1 Γ ( α ) 0 t δ t α - 1 ( 1 - s ) α - 1 + σ Γ ( α ) Γ ( α - γ ) ( 1 - s ) α - γ - 1 - ( t - s ) α - 1 ϕ ( s ) d s + δ t α - 1 t 1 ( 1 - s ) α - 1 + σ Γ ( α ) Γ ( α - γ ) ( 1 - s ) α - γ - 1 ϕ ( s ) d s = 0 1 G ( t , s ) ϕ ( s ) d s , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-18/MediaObjects/13661_2011_Article_127_Equ9_HTML.gif
(2.4)
where
G ( t , s ) = 1 Γ ( α ) g 1 ( t , s ) , 0 s t 1 , g 2 ( t , s ) , 0 t s 1 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-18/MediaObjects/13661_2011_Article_127_Equm_HTML.gif
and
g 1 ( t , s ) = δ t α - 1 ( 1 - s ) α - 1 + σ Γ ( α ) Γ ( α - γ ) ( 1 - s ) α - γ - 1 - ( t - s ) α - 1 , 0 s t , g 2 ( t , s ) = δ t α - 1 ( 1 - s ) α - 1 + σ Γ ( α ) Γ ( α - γ ) ( 1 - s ) α - γ - 1 , t s 1 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-18/MediaObjects/13661_2011_Article_127_Equn_HTML.gif

So, we obtain the following lemma.

Lemma 2.2. The solution of Equation (2.1) is given by
u ( t ) = 0 1 G ( t , s ) ϕ ( s ) d s , t [ 0 , 1 ] . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-18/MediaObjects/13661_2011_Article_127_Equo_HTML.gif

Also, we have the following lemma.

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

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

     
  2. (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.
  1. (ii)

    There are two cases to consider.

     
  2. (1)
    If 0 < st < 1, then
    Γ ( α ) g 1 ( t , s ) = t α - 1 δ ( 1 - s ) α - 1 + σ Γ ( α ) Γ ( α - γ ) ( 1 - s ) α - γ - 1 - 1 - s t α - 1 > t α - 1 δ ( 1 - s ) α - 1 + σ Γ ( α ) Γ ( α - γ ) ( 1 - s ) α - γ - 1 - ( 1 - s ) α - 1 = t α - 1 ( 1 - s ) α - 1 δ 1 + σ Γ ( α ) Γ ( α - γ ) ( 1 - s ) - γ - 1 t α - 1 ( 1 - s ) α - 1 δ 1 + σ Γ ( α ) Γ ( α - γ ) - 1 = 0 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-18/MediaObjects/13661_2011_Article_127_Equp_HTML.gif
     
  3. (2)

    If 0 < ts < 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 η 0 = γ δ σ Γ ( α ) Γ ( α - γ ) 1 α - 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-18/MediaObjects/13661_2011_Article_127_IEq10_HTML.gif. Denote η ( s ) = γ δ σ Γ ( α ) Γ ( α - γ ) s 2 - α http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-18/MediaObjects/13661_2011_Article_127_IEq11_HTML.gif, s ∈ [0, 1]. Set g(s) = G(s, s), s ∈ [0, 1]. From 0 < γ ≤ 1, σ > 0, 1 < α ≤ 2 and δ = 1 + σ Γ ( α ) Γ ( α - γ ) - 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-18/MediaObjects/13661_2011_Article_127_IEq12_HTML.gif, 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
  1. (i)

    G(t, s) ≤ G(s, s),s, t ∈ [0, 1].

     
  2. (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 ≤ st ≤ 1. In this case, since the following relation
g 1 ( t , s ) t = ( α - 1 ) δ t α - 2 ( 1 - s ) α - 1 + σ Γ ( α ) Γ ( α - γ ) ( 1 - s ) α - γ - 1 - ( t - s ) α - 2 ( α - 1 ) δ t α - 2 1 + σ Γ ( α ) Γ ( α - γ ) - ( t - s ) α - 2 = ( α - 1 ) t α - 2 - ( t - s ) α - 2 < 0 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-18/MediaObjects/13661_2011_Article_127_Equq_HTML.gif
holds for 0 < s < t ≤ 1, we have
G ( t , s ) G ( s , s ) , 0 s t 1 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-18/MediaObjects/13661_2011_Article_127_Equr_HTML.gif
Case 2. 0 ≤ ts ≤ 1. In this case, from the expression of g2(t, s), it is easy to see that
G ( t , s ) G ( s , s ) , 0 t s 1 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-18/MediaObjects/13661_2011_Article_127_Equs_HTML.gif
  1. (ii)

    We will consider the following two cases.

     
Case 1. When 0 < sη0, η0t ≤ 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
min t [ η 0 , 1 ] G ( t , s ) = G ( 1 , s ) = g 1 ( 1 , s ) / Γ ( α ) , s 0 , η 0 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-18/MediaObjects/13661_2011_Article_127_Equ10_HTML.gif
(2.5)
and so
min t [ η 0 , 1 ] G ( t , s ) G ( s , s ) = g 1 ( 1 , s ) g ( s ) , for s 0 , η 0 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-18/MediaObjects/13661_2011_Article_127_Equt_HTML.gif
Case 2. η0 < s < 1, η0t ≤ 1.
  1. (a)
    If st, then by similar arguments to (2.5), we also have
    min t [ s , 1 ] G ( t , s ) = G ( 1 , s ) = g 1 ( 1 , s ) / Γ ( α ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-18/MediaObjects/13661_2011_Article_127_Equu_HTML.gif
     
  2. (b)
    If η0ts, then the following relation
    min t [ η 0 , s ] G ( t , s ) = g 2 ( η 0 , s ) / Γ ( α ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-18/MediaObjects/13661_2011_Article_127_Equv_HTML.gif
     

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

To summarize,
min t [ η 0 , 1 ] G ( t , s ) G ( s , s ) min g 1 ( 1 , s ) g ( s ) , g 2 ( η 0 , s ) g ( s ) , for all s ( η 0 , 1 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-18/MediaObjects/13661_2011_Article_127_Equ11_HTML.gif
(2.6)
Now, we shall show that
min t [ η 0 , 1 ] G ( t , s ) G ( s , s ) γ δ σ Γ ( α ) Γ ( α - γ ) s 2 - α , s ( 0 , 1 ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-18/MediaObjects/13661_2011_Article_127_Equ12_HTML.gif
(2.7)
In fact, for s ∈ (0, 1), we have
g 1 ( 1 , s ) = δ [ ( 1 s ) α 1 + σ Γ ( α ) Γ ( α γ ) ( 1 s ) α γ 1 ] ( 1 s ) α 1 = δ [ ( 1 s ) α 1 + σ Γ ( α ) Γ ( α γ ) ( 1 s ) α 1 ] + δ σ Γ ( α ) Γ ( α γ ) [ ( 1 s ) α γ 1 ( 1 s ) α 1 ] ( 1 s ) α 1 = δ [ 1 + σ Γ ( α ) Γ ( α γ ) ] ( 1 s ) α 1 + δ σ Γ ( α ) Γ ( α γ ) ( 1 s ) α γ 1 [ 1 ( 1 s ) γ ) ] ( 1 s ) α 1 = δ σ Γ ( α ) Γ ( α γ ) ( 1 s ) α γ 1 [ 1 ( 1 s ) γ ] > δ σ Γ ( α ) Γ ( α γ ) ( 1 s ) α γ 1 γ s , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-18/MediaObjects/13661_2011_Article_127_Equw_HTML.gif
and so
g 1 ( 1 , s ) g ( s ) > γ δ σ Γ ( α ) Γ ( α - γ ) ( 1 - s ) α - γ - 1 s δ s α - 1 ( 1 - s ) α - 1 + σ Γ ( α ) Γ ( α - γ ) ( 1 - s ) α - γ - 1 = γ σ Γ ( α ) Γ ( α - γ ) s 2 - α ( 1 - s ) γ + σ Γ ( α ) Γ ( α - γ ) > γ σ Γ ( α ) Γ ( α - γ ) s 2 - α 1 + σ Γ ( α ) Γ ( α - γ ) = γ δ σ Γ ( α ) Γ ( α - γ ) s 2 - α , s ( 0 , 1 ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-18/MediaObjects/13661_2011_Article_127_Equ13_HTML.gif
(2.8)
On the other hand, for s ∈ (η0, 1), we have
g 2 ( η 0 , s ) g ( s ) = η 0 α - 1 s 1 - α . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-18/MediaObjects/13661_2011_Article_127_Equ14_HTML.gif
(2.9)
Since η 0 α - 1 = γ δ σ Γ ( α ) Γ ( α - γ ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-18/MediaObjects/13661_2011_Article_127_IEq13_HTML.gif, the equality
η 0 α - 1 s 1 - α = γ δ σ Γ ( α ) Γ ( α - γ ) s 2 - α http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-18/MediaObjects/13661_2011_Article_127_Equx_HTML.gif
holds for s = 1. Thus,
η 0 α - 1 s 1 - α > γ δ σ Γ ( α ) Γ ( α - γ ) s 2 - α , s ( 0 , 1 ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-18/MediaObjects/13661_2011_Article_127_Equ15_HTML.gif
(2.10)
Since 1 < α ≤ 2, it follows from (2.9) that
g 2 ( η 0 , s ) g ( s ) > γ δ σ Γ ( α ) Γ ( α - γ ) s 2 - α , s ( η 0 , 1 ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-18/MediaObjects/13661_2011_Article_127_Equ16_HTML.gif
(2.11)
Hence, from (2.8) and (2.11), we immediately have
min g 1 ( 1 , s ) g ( s ) , g 2 ( η 0 , s ) g ( s ) > γ δ σ Γ ( α ) Γ ( α - γ ) s 2 - α , s ( η 0 , 1 ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-18/MediaObjects/13661_2011_Article_127_Equ17_HTML.gif
(2.12)
Thus, from (2.6) and (2.12 ), it follows that
min t [ η 0 , 1 ] G ( t , s ) G ( s , s ) > γ δ σ Γ ( α ) Γ ( α - γ ) s 2 - α , s ( η 0 , 1 ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-18/MediaObjects/13661_2011_Article_127_Equ18_HTML.gif
(2.13)
Also, by (2.8), the following inequality
g 1 ( 1 , s ) g ( s ) > γ δ σ Γ ( α ) Γ ( α - γ ) s 2 - α , s 0 , η 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-18/MediaObjects/13661_2011_Article_127_Equy_HTML.gif
holds, and therefore
min t [ η 0 , 1 ] G ( t , s ) G ( s , s ) > γ δ σ Γ ( α ) Γ ( α - γ ) s 2 - α , s 0 , η 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-18/MediaObjects/13661_2011_Article_127_Equ19_HTML.gif
(2.14)

from the proof in Case 1.

Summing up the above relations (2.13)-(2.14), we have
min t [ η 0 , 1 ] G ( t , s ) G ( s , s ) > γ δ σ Γ ( α ) Γ ( α - γ ) s 2 - α , s ( 0 , 1 ) , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-18/MediaObjects/13661_2011_Article_127_Equz_HTML.gif
and so
min t [ η 0 , 1 ] G ( t , s ) η ( s ) G ( s , s ) , s [ 0 , 1 ] . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-18/MediaObjects/13661_2011_Article_127_Equaa_HTML.gif

The proof of Lemma 2.4 is complete.

To study BVP (1. 5), we first consider the associated linear BVP
D 0 + β φ p D 0 + α u ( t ) + h ( t ) = 0 , 0 < t < 1 , u ( 0 ) = 0 , u ( 1 ) + σ D 0 + γ u ( 1 ) = 0 , D 0 + α u ( 0 ) = 0 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-18/MediaObjects/13661_2011_Article_127_Equ20_HTML.gif
(2.15)

where hP.

Let w = D 0 + α u , v = φ p ( w ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-18/MediaObjects/13661_2011_Article_127_IEq14_HTML.gif. By Lemma 2.1, the solution of initial value problem
D 0 + β u ( t ) + h ( t ) = 0 , t ( 0 , 1 ) , v ( 0 ) = 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-18/MediaObjects/13661_2011_Article_127_Equab_HTML.gif
is given by
v ( t ) = C 1 t β - 1 - I 0 + β h ( t ) , t 0 , 1 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-18/MediaObjects/13661_2011_Article_127_Equac_HTML.gif
From the relations v(0) = 0, 0 < β ≤ 1, it follows that C1 = 0, and so
v ( t ) = - I 0 + β h ( t ) , t [ 0 , 1 ] . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-18/MediaObjects/13661_2011_Article_127_Equ21_HTML.gif
(2.16)
Noting that D 0 + α u = w , w = φ p - 1 ( v ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-18/MediaObjects/13661_2011_Article_127_IEq15_HTML.gif, from (2.16), we know that the solution of (2.15) satisfies
{ D 0 + α u ( t ) = φ p 1 ( I 0 β h ) ( t ) ) , t ( 0 , 1 ) , u ( 0 ) = 0 , u ( 1 ) + σ D 0 + γ u ( 1 ) = 0. http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-18/MediaObjects/13661_2011_Article_127_Equ22_HTML.gif
(2.17)
By Lemma 2.2, the solution of Equation (2.17) can be written as
u ( t ) = - 0 1 G ( t , s ) φ p - 1 - I 0 β h ( s ) d s , t I . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-18/MediaObjects/13661_2011_Article_127_Equ23_HTML.gif
(2.18)
Since h(s) ≥ 0, s ∈ [0, 1], we have φ p - 1 - I 0 + β h ( s ) = - I 0 + β h ( s ) q - 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-18/MediaObjects/13661_2011_Article_127_IEq16_HTML.gif, s ∈ [0, 1], and so
u ( t ) = 0 1 G ( t , s ) I 0 + β h ( s ) q - 1 d s = 1 ( Γ ( β ) ) q - 1 0 1 G ( t , s ) d s 0 s h ( τ ) ( s - τ ) β - 1 d τ q - 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-18/MediaObjects/13661_2011_Article_127_Equ24_HTML.gif
(2.19)

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

Lemma 2.5. Let hP. Then the solution of Equation (2.15) in P is given by
u ( t ) = 1 ( Γ ( β ) ) q - 1 0 1 G ( t , s ) d s 0 s h ( τ ) ( s - τ ) β - 1 d τ q - 1 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-18/MediaObjects/13661_2011_Article_127_Equad_HTML.gif

We also need the following lemmas to obtain our results.

Lemma 2.6. If a, b ≥ 0, γ > 0, then
( a + b ) γ max { 2 γ - 1 , 1 } ( a γ + b γ ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-18/MediaObjects/13661_2011_Article_127_Equae_HTML.gif

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

Let ϕ(t) = t γ , t ∈ [0, +∞).
  1. (i)
    If γ > 1, then ϕ(t) is convex on (0, +∞), and so
    ϕ 1 2 a + 1 2 b 1 2 ϕ ( a ) + 1 2 ϕ ( b ) , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-18/MediaObjects/13661_2011_Article_127_Equaf_HTML.gif
    .
     
i.e., 1 2 γ ( a + b ) γ 1 2 ( a γ + b γ ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-18/MediaObjects/13661_2011_Article_127_IEq17_HTML.gif. Thus
( a + b ) γ 2 γ - 1 ( a γ + b γ ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-18/MediaObjects/13661_2011_Article_127_Equag_HTML.gif
  1. (ii)
    If 0 < γ < 1, then ϕ(t) is concave on [0, +∞), and so
    ϕ ( a ) = ϕ b a + b 0 + a a + b ( a + b ) b a + b ϕ ( 0 ) + a a + b ϕ ( a + b ) = a a + b ϕ ( a + b ) , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-18/MediaObjects/13661_2011_Article_127_Equah_HTML.gif
    ϕ ( b ) = ϕ b a + b 0 + a a + b ( a + b ) a a + b ϕ ( 0 ) + b a + b ϕ ( a + b ) = b a + b ϕ ( a + b ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-18/MediaObjects/13661_2011_Article_127_Equai_HTML.gif
     
Thus, ϕ(a) + ϕ(b) ≥ ϕ(a + b), namely,
( a + b ) γ a γ + b γ . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-18/MediaObjects/13661_2011_Article_127_Equaj_HTML.gif

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
  1. (i)

    If γ > 1, then |x γ - y γ | ≤ γcγ-1|x - y|,

     
  2. (ii)

    If 0 < γ ≤ 1, then |x γ - y γ | ≤ |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.
  1. (i)
    If γ > 1, then we set ϕ(t) = t γ , t ∈ [0, c]. by virtue of mean value theorem, there exists a ξ ∈ (0, c) such that
    x γ - y γ = γ ξ γ - 1 ( x - y ) γ c γ - 1 ( x - y ) , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-18/MediaObjects/13661_2011_Article_127_Equak_HTML.gif
     
i.e.,
x γ - y γ γ c γ - 1 x - y . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-18/MediaObjects/13661_2011_Article_127_Equal_HTML.gif
  1. (ii)
    If 0 < γ < 1, then by Lemma 2.6, it is easy to see that
    x γ - y γ = ( x - y - y ) γ - y γ ( x - y ) γ + y γ - y γ = ( x - y ) γ , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-18/MediaObjects/13661_2011_Article_127_Equam_HTML.gif
     
and so
x γ - y γ x - y γ . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-18/MediaObjects/13661_2011_Article_127_Equan_HTML.gif

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

Let D = 0 1 G ( s , s ) s β ( q - 1 ) d s , Q = 0 1 η ( s ) G ( s , s ) s β ( q - 1 ) d s http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-18/MediaObjects/13661_2011_Article_127_IEq18_HTML.gif,
l = Γ ( β + 1 ) D max 2 q - 1 , 1 1 1 - q , μ = Γ ( β + 1 ) Q 1 q - 1 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-18/MediaObjects/13661_2011_Article_127_Equao_HTML.gif
By simple calculation, we know that
D = δ Γ ( α ) B α , α + β ( q - 1 ) + σ δ Γ ( α - γ ) B ( α - γ , α + β ( q - 1 ) ) , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-18/MediaObjects/13661_2011_Article_127_Equ25_HTML.gif
(2.20)
Q = γ δ 2 σ Γ ( α - γ ) B α , 2 + β ( q - 1 ) + σ Γ ( α ) Γ ( α - γ ) B α - γ , 2 + β ( q - 1 ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-18/MediaObjects/13661_2011_Article_127_Equ26_HTML.gif
(2.21)

In this article, the following hypotheses will be used.

(H1) fC(I × +, +).

(H2) lim x + ¯ max t I f ( t , x ) x p - 1 < l , lim x 0 + min t I f ( t , x ) x p - 1 > μ . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-18/MediaObjects/13661_2011_Article_127_IEq19_HTML.gif.

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

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
u ( t ) = 1 ( Γ ( β ) ) q - 1 0 1 G ( t , s ) 0 s f ( τ , u ( τ ) ) ( s - τ ) β - 1 d τ q - 1 d s , t I http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-18/MediaObjects/13661_2011_Article_127_Equ27_HTML.gif
(2.22)
has a solution in P. Let c be a positive number, P be a cone and P c = {yP : ∥y∥ ≤ c}. Let α be a nonnegative continuous concave function on P and
P ( α , a , b ) = u P | a α ( u ) , u b . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-18/MediaObjects/13661_2011_Article_127_Equap_HTML.gif

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

Lemma 2.9. [35] Let A : P c ¯ P c ¯ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-18/MediaObjects/13661_2011_Article_127_IEq20_HTML.gif be completely continuous and α be a nonnegative continuous concave function on P such that α(y) ≤ ∥y∥ for all y P c ¯ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-18/MediaObjects/13661_2011_Article_127_IEq21_HTML.gif. Suppose that there exist a, b and d with 0 < a < b < dc such that

(C1) { y P ( α , b , d ) } | α ( y ) > b } 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-18/MediaObjects/13661_2011_Article_127_IEq22_HTML.gif and α(Ay) > b, for all yP(α, b, d);

(C2) ∥Ay∥ < a, for ∥y∥ ≤ a;

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

Then A has at least three fixed points y1, y2, y3 satisfying
y 1 < a , b < α ( y 2 ) , and y 3 > a with α ( y 3 ) < b . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-18/MediaObjects/13661_2011_Article_127_Equaq_HTML.gif

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 u = 1 ( Γ ( β ) ) q - 1 0 1 G ( t , s ) 0 s f ( τ , u ( τ ) ) ( s - τ ) β - 1 d τ q - 1 d s , u P . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-18/MediaObjects/13661_2011_Article_127_Equ28_HTML.gif
(3.1)

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

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

Proof. First, under assumption (H1), it is obvious that APP from Lemma 2.3. Next, we shall show that operator A is completely continuous on P. Let E = 0 1 G ( s , s ) d s http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-18/MediaObjects/13661_2011_Article_127_IEq23_HTML.gif. 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 uB. According to the continuity of f, we have L max f ( t , x ) ( t , x ) I × [ 0 , M ] < + http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-18/MediaObjects/13661_2011_Article_127_IEq24_HTML.gif. Thus, by Lemmas 2.3 and 2.4, it follows that
0 ( A u ) ( t ) L q - 1 ( Γ ( β ) ) q - 1 0 1 G ( s , s ) 0 s ( s - τ ) β - 1 d τ q - 1 d s < L q - 1 ( Γ ( β + 1 ) ) q - 1 0 1 G ( s , s ) d s = L q - 1 ( Γ ( β + 1 ) ) q - 1 E , t I . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-18/MediaObjects/13661_2011_Article_127_Equar_HTML.gif
Thus,
A u L q - 1 ( Γ ( β + 1 ) ) q - 1 E . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-18/MediaObjects/13661_2011_Article_127_Equas_HTML.gif

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, t2I with |t1 - t2| < δ, the following inequality
G ( t 1 , s ) - G ( t 2 , s ) < ε ( Γ ( β + 1 ) ) q - 1 L q - 1 ( β ( q - 1 ) + 1 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-18/MediaObjects/13661_2011_Article_127_Equat_HTML.gif
holds for all sI. Therefore,
A u ( t 1 ) - A u ( t 2 ) 1 ( Γ ( β ) ) q - 1 0 1 G ( t 1 , s ) - G ( t 2 , s ) 0 s f ( τ , u ( τ ) ) ( s - τ ) β - 1 d τ q - 1 d s L q - 1 ( Γ ( β + 1 ) ) q - 1 0 1 s ( q - 1 ) β d s ( Γ ( β + 1 ) ) q - 1 L q - 1 ( β ( q - 1 ) + 1 ) ε = ε . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-18/MediaObjects/13661_2011_Article_127_Equau_HTML.gif

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

Step 2. The operator A is continuous.

Let {u n } be an arbitrary sequence in P with u n u0P. Then exists an L > 0 such that
0 f ( τ , u n ( τ ) ) L , τ [ 0 , 1 ] , n 0 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-18/MediaObjects/13661_2011_Article_127_Equav_HTML.gif
Thus,
0 s f ( τ , u n ( τ ) ) ( s - τ ) β - 1 d τ L 0 s ( s - τ ) β - 1 d τ L β c , s [ 0 , 1 ] . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-18/MediaObjects/13661_2011_Article_127_Equaw_HTML.gif
On the other hand, the uniform continuity of f combined with the fact that ∥u n - u0∥ → 0 yields that there exists a N ≥ 1 such that the following estimate
| f ( τ , u n ( τ ) f ( t , u 0 ( τ ) ) | < ε http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-18/MediaObjects/13661_2011_Article_127_Equax_HTML.gif
holds for nN.
  1. (1)
    If 1 < q ≤ 2, then from Lemma 2.7 (ii), we have
    0 s f ( τ , u n ( τ ) ) ( s - τ ) β - 1 d τ q - 1 - 0 s f ( τ , u 0 ( τ ) ) ( s - τ ) β - 1 d τ q - 1 0 s f ( τ , u n ( τ ) ) - f ( τ , u 0 ( τ ) ) ( s - τ ) β - 1 d τ q - 1 < ε q - 1 1 β q - 1 s β ( q - 1 ) , s [ 0 , 1 ] . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-18/MediaObjects/13661_2011_Article_127_Equay_HTML.gif
     
Hence, by Lemmas 2.3 and 2.4, from (3.1), we obtain
A u n ( t ) - A u 0 ( t ) < ε q - 1 ( Γ ( β + 1 ) ) q - 1 0 1 G ( s , s ) d s = E ( Γ ( β + 1 ) ) q - 1 ε q - 1 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-18/MediaObjects/13661_2011_Article_127_Equaz_HTML.gif
Thus,
A u n - A u 0 E ( Γ ( β + 1 ) ) q - 1 ε q - 1 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-18/MediaObjects/13661_2011_Article_127_Equ29_HTML.gif
(3.2)
  1. (2)
    If q > 2, then from Lemma 2.7 (i), we have
    0 s f ( τ , u n ( τ ) ) ( s - τ ) β - 1 d τ q - 1 - 0 s f ( τ , u 0 ( τ ) ) ( s - τ ) β - 1 d τ q - 1 ( q - 1 ) c q - 2 0 s f ( τ , u n ( τ ) ) - f ( τ , u 0 ( τ ) ) ( s - τ ) β - 1 d τ < q - 1 β c q - 2 s β ε , s [ 0 , 1 ] . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-18/MediaObjects/13661_2011_Article_127_Equba_HTML.gif
     
Thus, we have
A u n ( t ) - A u 0 ( t ) < ( q - 1 ) c q - 2 ε β ( Γ ( β ) ) q - 1 0 1 G ( s , s ) d s = ( q - 1 ) c q - 2 E β ( Γ ( β ) ) q - 1 ε , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-18/MediaObjects/13661_2011_Article_127_Equbb_HTML.gif
and so
A u n - A u 0 ( q - 1 ) c q - 2 E β ( Γ ( β ) ) q - 1 ε . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-18/MediaObjects/13661_2011_Article_127_Equ30_HTML.gif
(3.3)

From (3.2)-(3.3), it follows that ∥Au n - 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 : PP 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
lim x + ¯ max t I f ( t , x ) x p - 1 < l - ε 0 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-18/MediaObjects/13661_2011_Article_127_Equbc_HTML.gif
Therefore, there exists a R0 > 0 such that the inequality
f ( t , x ) < ( l - ε 0 ) x p - 1 , t I http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-18/MediaObjects/13661_2011_Article_127_Equ31_HTML.gif
(3.4)

holds for xR0.

Let M = max ( t , x ) I × [ 0 , R 0 ] f ( t , x ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-18/MediaObjects/13661_2011_Article_127_IEq25_HTML.gif. It follows from (3.4) that
f ( t , x ) ( l - ε 0 ) x p - 1 + M , x + , t I . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-18/MediaObjects/13661_2011_Article_127_Equ32_HTML.gif
(3.5)

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.

Set
D 1 = max { 2 q - 2 , 1 } D ( Γ ( β + 1 ) ) q - 1 , E = D 1 k , G = D 1 M q - 1 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-18/MediaObjects/13661_2011_Article_127_Equ33_HTML.gif
(3.6)
where D is as (2.20). Take R > G E http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-18/MediaObjects/13661_2011_Article_127_IEq26_HTML.gif. Set Ω R = {uP : ∥u∥ < R}. We shall show that the relation
A u μ u , u Ω R μ 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-18/MediaObjects/13661_2011_Article_127_Equ34_HTML.gif
(3.7)

holds.

In fact, if not, then there exists a u0 Ω R and a μ0 ≥ 1 with
μ 0 u 0 = A u 0 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-18/MediaObjects/13661_2011_Article_127_Equbd_HTML.gif
By (3.5), we have
f ( t , u 0 ( t ) ) ( l - ε 0 ) u 0 p - 1 ( t ) + M ( l - ε 0 ) u 0 p - 1 + M = ( l - ε 0 ) R p - 1 + M , t I . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-18/MediaObjects/13661_2011_Article_127_Eqube_HTML.gif
Therefore, in view of Lemmas 2.3, 2.4, from (3.1), it follows that
A u ( t ) ( ( l - ε 0 ) R p - 1 + M ) q - 1 ( Γ ( β + 1 ) ) q - 1 0 1 G ( s , s ) s β ( q - 1 ) d s = D ( Γ ( β + 1 ) ) q - 1 ( l - ε 0 ) R p - 1 + M q - 1 , t I . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-18/MediaObjects/13661_2011_Article_127_Equ35_HTML.gif
(3.8)
Also, keeping in mind that (p - 1)(q - 1) = 1, by Lemma 2.6, we have
( l - ε 0 ) R p - 1 + M q - 1 max 2 q - 2 , 1 ( l - ε 0 ) q - 1 R + M q - 1 < max 2 q - 2 , 1 ( l q - 1 - k ) R + M q - 1 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-18/MediaObjects/13661_2011_Article_127_Equ36_HTML.gif
(3.9)
Hence, from (3.6), (3.8), and (3.9), it follows that
u 0 μ u 0 = A u 0 D 1 l q - 1 - E R + G , t I . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-18/MediaObjects/13661_2011_Article_127_Equ37_HTML.gif
(3.10)
By definition of l, we have D1lq-1= 1. From (3. 10), it follows that R = ∥u0∥ ≤ (1 - E)R + G, and so R G E http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-18/MediaObjects/13661_2011_Article_127_IEq27_HTML.gif, which contradicts the choice of R. Hence, the condition (3.7) holds. By virtue of the fixed point index theorem, we have
i ( A , Ω R , P ) = 1 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-18/MediaObjects/13661_2011_Article_127_Equ38_HTML.gif
(3.11)
Step 2. By (H2), we can choose a ε0 > 0 such that
lim ¯ x 0 + min t I f ( t , x ) x p 1 > μ + ε 0 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-18/MediaObjects/13661_2011_Article_127_Equbf_HTML.gif
Hence, there exists a r1 ∈ (0, r0) such that
f ( t , x ) > ( μ + ε 0 ) x p - 1 , t I , x [ 0 , r 1 ] , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-18/MediaObjects/13661_2011_Article_127_Equ39_HTML.gif
(3.12)

where r0 is given by (H3).

Take 0 < r < min {R, r1}, and set Ω r = {uP : ∥u∥ < r}. Now, we show that
  1. (i)

    inf u Ω r A u > 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-18/MediaObjects/13661_2011_Article_127_IEq28_HTML.gif,

     
  2. (ii)

    Auμu, ∀u Ω r , μ ∈ [0, 1].

     
We first prove that (i) holds. In fact, for any u Ω r , we have 0 ≤ u(t) ≤ r. By (H3), the function f(t, x) is nonincreasing relative to x on [0, r] for any tI, and so
f ( t , u ( t ) ) f ( t , r ) ( μ + ε 0 ) r p - 1 , t [ 0 , 1 ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-18/MediaObjects/13661_2011_Article_127_Equ40_HTML.gif
(3.13)

from (3.12).

Thus, in view of Lemma 2.4 combined with (3.1) and (3.13), we have
A u ( t ) 1 ( Γ ( β ) ) q - 1 0 1 G ( t , s ) 0 s ( μ , + ε 0 ) r p - 1 ( s - τ ) β - 1 d τ q - 1 d s = ( μ + ε 0 ) q - 1 r ( Γ ( β + 1 ) ) q - 1 0 1 G ( t , s ) s β ( q - 1 ) d s ( μ + ε 0 ) q - 1 r ( Γ ( β + 1 ) ) q - 1 0 1 η ( s ) G ( s , s ) s β ( q - 1 ) d s = ( μ + ε 0 ) q - 1 Q ( Γ ( β + 1 ) ) q - 1 r , for all t [ η 0 , 1 ] , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-18/MediaObjects/13661_2011_Article_127_Equbg_HTML.gif
where Q is as (2.21). Consequently,
A u ( μ + ε 0 ) q - 1 Q ( Γ ( β + 1 ) ) q - 1 r c 0 > 0 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-18/MediaObjects/13661_2011_Article_127_Equ41_HTML.gif
(3.14)
Thus inf u Ω r A u c 0 > 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-18/MediaObjects/13661_2011_Article_127_IEq29_HTML.gif.
  1. (ii)
    Suppose on the contrary that there exists a u0 Ω r and μ0 ∈ [0, 1] such that
    μ 0 u 0 = A u 0 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-18/MediaObjects/13661_2011_Article_127_Equbd_HTML.gif
    (3.15)
     
Then, by similar arguments to (3.14), we have
A u 0 B r , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-18/MediaObjects/13661_2011_Article_127_Equ42_HTML.gif
(3.16)

where B = ( μ + ε 0 ) q - 1 Q ( Γ ( β + 1 ) ) q - 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-18/MediaObjects/13661_2011_Article_127_IEq30_HTML.gif.

By (3.15)-(3.16), we obtain
r = u 0 μ 0 u 0 = A u 0 B r . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-18/MediaObjects/13661_2011_Article_127_Equbh_HTML.gif

The hypothesis μ = Γ ( β + 1 ) Q 1 q - 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-18/MediaObjects/13661_2011_Article_127_IEq31_HTML.gif 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
i ( A , Ω R , P ) = 0 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-18/MediaObjects/13661_2011_Article_127_Equ43_HTML.gif
(3.17)
By (3.11) and (3.17), we have
i ( A , Ω R \ Ω r , P ) = 1 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-18/MediaObjects/13661_2011_Article_127_Equbi_HTML.gif

and so, there exists u * Ω R \ Ω ̄ r http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-18/MediaObjects/13661_2011_Article_127_IEq32_HTML.gif 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
0 s f ( τ , u * ( τ ) ) ( s - τ ) β - 1 d τ 0 , s ( 0 , 1 ) , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-18/MediaObjects/13661_2011_Article_127_Equbj_HTML.gif
and so
u * ( t ) = A u * ( t ) > 0 , t ( 0 , 1 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-18/MediaObjects/13661_2011_Article_127_Equ44_HTML.gif
(3.18)

from the fact that G(t, s) > 0 and 0 s f ( τ , u ( τ ) ) ( s - τ ) β - 1 d τ 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-18/MediaObjects/13661_2011_Article_127_IEq33_HTML.gif, 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 M 1 = Γ ( β + 1 ) D 1 q - 1 , M 2 = Γ ( β + 1 ) B 2 1 q - 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-18/MediaObjects/13661_2011_Article_127_IEq34_HTML.gif, where D is as (2.20).

Let
B 2 γ δ 2 σ ( 1 - η 0 ) α + β ( q - 1 ) + 1 Γ ( α - γ ) B ( α , β ( q - 1 ) + 2 ) + ( 1 - η 0 ) - γ σ Γ ( α ) Γ ( α - γ ) B ( α - γ , β ( q - 1 ) + 2 ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-18/MediaObjects/13661_2011_Article_127_Equ45_HTML.gif
(3.19)

Set P r = {uP : ∥u∥ < r}, for r > 0. Let ω ( u ) = min t [ η 0 , 1 ] u ( t ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-18/MediaObjects/13661_2011_Article_127_IEq35_HTML.gif, for uP. 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], tI; f(t, x)≤ l1ap-1, x ∈ [0, a], tI,

(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
u 1 < a , b < min t [ η 0 , 1 ] u 2 ( t ) , a < u 3 , min t [ η 0 , 1 ] u 3 ( t ) < b . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-18/MediaObjects/13661_2011_Article_127_Equbk_HTML.gif
Proof. By Lemmas 2.3 and 2.4, for u P ̄ c http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-18/MediaObjects/13661_2011_Article_127_IEq36_HTML.gif, from (3.1) and condition (D1), it follows that
A u ( t ) ( l 1 ) q - 1 c ( Γ ( β + 1 ) ) q - 1 0 1 G ( s , s ) s β ( q - 1 ) d s = ( l 1 ) q - 1 c ( Γ ( β + 1 ) ) q - 1 D , t I , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-18/MediaObjects/13661_2011_Article_127_Equbl_HTML.gif
and so
A u ( l 1 ) q - 1 D ( Γ ( β + 1 ) ) q - 1 c < c http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-18/MediaObjects/13661_2011_Article_127_Equbm_HTML.gif

from the hypothesis l1 < M1.

Thus, we obtain A : P ̄ c P c http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-18/MediaObjects/13661_2011_Article_127_IEq37_HTML.gif. Similarly, we can also obtain A : P ̄ a P a http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-18/MediaObjects/13661_2011_Article_127_IEq38_HTML.gif by condition (D1). Take u 0 = b + c 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-18/MediaObjects/13661_2011_Article_127_IEq39_HTML.gif. Then ω(u0) > b, and so { u P ( ω , b , c ) | ω ( u ) > b } http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-18/MediaObjects/13661_2011_Article_127_IEq40_HTML.gif.

For any uP(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 u ( t ) = 1 ( Γ ( β ) ) q - 1 0 1 G ( t , s ) ( 0 s f ( τ , u ( τ ) ) ( s - τ ) β - 1 d τ ) q - 1 d s 1 ( Γ ( β ) ) q - 1 0 1 η ( s ) G ( s , s ) 0 s f ( τ , u ( τ ) ) ( s - τ ) β - 1 d τ q - 1 d s 1 ( Γ ( β ) ) q - 1 η 0 1 η ( s ) G ( s , s ) 0 s f ( τ , u ( τ ) ) ( s - τ ) β - 1 d τ q - 1 d s 1 ( Γ ( β ) ) q - 1 η 0 1 η ( s ) G ( s , s ) η 0 s f ( τ , u ( τ ) ) ( s - τ ) β - 1 d τ q - 1 d s l 2 q - 1 b ( Γ ( β + 1 ) ) q - 1 η 0 1 η ( s ) G ( s , s ) ( s - η 0 ) β ( q - 1 ) d s . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-18/MediaObjects/13661_2011_Article_127_Equ46_HTML.gif
(3.20)
Also, by changing the variable θ = s - η 0 1 - η 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-18/MediaObjects/13661_2011_Article_127_IEq41_HTML.gif, we have
η 0 1 η ( s ) G ( s , s ) ( s - η 0 ) β ( q - 1 ) d s = γ δ 2 σ Γ ( α - γ ) η 0 1 s ( 1 - s ) α - 1 + σ Γ ( α ) Γ ( α - γ ) ( 1 - s ) α - γ - 1 ( s - η 0 ) β ( q - 1 ) d s > γ δ 2 σ ( 1 - η 0 ) α + β ( q - 1 ) + 1 Γ ( α - γ ) 0 1 θ 1 + β ( q - 1 ) ( 1 - θ ) α - 1 1 + σ Γ ( α ) Γ ( α - γ ) ( 1 - η 0 ) - γ ( 1 - θ ) - γ d θ = B 2 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-18/MediaObjects/13661_2011_Article_127_Equ47_HTML.gif
(3.21)

where B2 is given by (3.19).

Substituting (3.21) into (3.20), we obtain
A u ( t ) l 2 q - 1 B 2 ( Γ ( β + 1 ) ) q - 1 b , t [ η 0 , 1 ] , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-18/MediaObjects/13661_2011_Article_127_Equbn_HTML.gif

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
u 1 < a , b < min t [ η 0 , 1 ] u 2 ( t ) , a < u 3 , min t [ η 0 , 1 ] u 3 ( t ) < b . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-18/MediaObjects/13661_2011_Article_127_Equbo_HTML.gif

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
D 0 + β φ p D 0 + α u ( t ) + ( t 2 + cos 2 u ) = 0 , 0 < t < 1 , u ( 0 ) = 0 , u ( 1 ) + σ D 0 + γ u ( 1 ) = 0 , D 0 + α u ( 0 ) = 0 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-18/MediaObjects/13661_2011_Article_127_Equ48_HTML.gif
(3.22)

where 1 < α < 2, 0 < β < 1, 0 < γ < 1, 0 ≤ α - γ - 1, σ > 0 and the p-Laplacian operator is defined as φ p (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
D 0 + 1 2 φ 3 2 D 0 + 3 2 u ( t ) + f ( t , u ) = 0 , 0 < t < 1 , u ( 0 ) = 0 , u ( 1 ) + 2 π D 0 + 1 2 u ( 1 ) = 0 , D 0 + 3 2 u ( 0 ) = 0 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-18/MediaObjects/13661_2011_Article_127_Equ49_HTML.gif
(3.23)
where α = 2 3 , β = γ = 1 2 , σ = 2 π , p = 3 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-18/MediaObjects/13661_2011_Article_127_IEq42_HTML.gif relative to Theorem 3.2. With the aid of computation we have that M 1 = π π 4 20 32 + 5 π = 1 . 527 . . . , M 2 = 64 15 π π 4 7 35 + 4 15 1 2 = 3 . 74 . . . , η 0 = 1 16 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-18/MediaObjects/13661_2011_Article_127_IEq43_HTML.gif. Take l1 = 1.5, l2 = 4. Then l1 ∈ (0, M1), l2 ∈ (M2, ∞). Again choosing a = 2 . 25 3 4 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-18/MediaObjects/13661_2011_Article_127_IEq44_HTML.gif, b = 1, c = 196, and setting
f ( t , x ) = ( sin π 90 ( 32 t + 13 ) ) ( x 1 2 + 7 ) , x 1 , + , 8 ( sin π 90 ( 32 t + 13 ) ) x 2 , x 0 , 1 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-18/MediaObjects/13661_2011_Article_127_Equbp_HTML.gif
for t ∈ [0, 1], then we see that f satisfies the following relations:
f ( t , x ) 21 = 1 . 5 × 19 6 1 2 = l 1 c p - 1 , t [ 0 , 1 ] , x [ 0 , 196 ] = [ 0 , c ] , f ( t , x ) 4 = l 2 b p - 1 , t 1 16 , 1 = [ η 0 , 1 ] , x [ 1 , 196 ] = [ b , c ] , f ( t , x ) 3 1 . 5 3 4 = 1 . 5 × 1 . 5 3 2 = l 1 a p - 1 , t [ 0 , 1 ] , t 0 , 2 . 25 3 4 = [ 0 , a ] . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-18/MediaObjects/13661_2011_Article_127_Equbq_HTML.gif
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
u 1 < 2 . 25 3 4 , 1 < min t [ η 0 , 1 ] u 2 ( t ) , 2 . 25 3 4 < u 3 , min t [ η 0 , 1 ] u 3 ( t ) < 1 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-18/MediaObjects/13661_2011_Article_127_Equbr_HTML.gif

Declarations

Acknowledgements

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).

Authors’ Affiliations

(1)
College of Mathematics and Statistics, Hubei Normal University

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© Chai; licensee Springer. 2012

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