Positive solutions of nonhomogeneous boundary value problems for some nonlinear equation with ϕ-Laplacian

  • Liang-Gen Hu1Email author and

    Affiliated with

    • Jing Xu1

      Affiliated with

      Boundary Value Problems20122012:130

      DOI: 10.1186/1687-2770-2012-130

      Received: 10 June 2012

      Accepted: 23 October 2012

      Published: 12 November 2012

      Abstract

      We will consider the nonhomogeneous ϕ-Laplacian differential equation

      { ( ϕ ( u ( t ) ) ) = h ( t ) f ( u ( t ) ) , t ( 0 , T ) , u ( 0 ) = i = 1 k α i u ( η i ) , ϕ ( u ( T ) ) = β , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-130/MediaObjects/13661_2012_Article_231_Equa_HTML.gif

      where ϕ : R ( b , b ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-130/MediaObjects/13661_2012_Article_231_IEq1_HTML.gif ( 0 < b + http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-130/MediaObjects/13661_2012_Article_231_IEq2_HTML.gif) is an increasing homeomorphism such that ϕ ( 0 ) = 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-130/MediaObjects/13661_2012_Article_231_IEq3_HTML.gif, h : [ 0 , T ] R + http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-130/MediaObjects/13661_2012_Article_231_IEq4_HTML.gif and f : R + R + http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-130/MediaObjects/13661_2012_Article_231_IEq5_HTML.gif are continuous, β 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-130/MediaObjects/13661_2012_Article_231_IEq6_HTML.gif and η i ( 0 , T ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-130/MediaObjects/13661_2012_Article_231_IEq7_HTML.gif and α i R http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-130/MediaObjects/13661_2012_Article_231_IEq8_HTML.gif, i = 1 , 2 , , k http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-130/MediaObjects/13661_2012_Article_231_IEq9_HTML.gif. Based on the Krasnosel’skii fixed point theorem, the existence of a positive solution is obtained, even if some of the α i http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-130/MediaObjects/13661_2012_Article_231_IEq10_HTML.gif coefficients are negative. Two examples are also given to illustrate our main results.

      Keywords

      nonhomogeneous ϕ-Laplacian positive solution fixed point negative coefficient

      1 Introduction

      We are concerned with the ϕ-Laplacian differential equation with the nonhomogeneous Dirichlet-Neumann boundary conditions
      { ( ϕ ( u ( t ) ) ) = h ( t ) f ( u ( t ) ) , t ( 0 , T ) , u ( 0 ) = ω ( u ) : = i = 1 k α i u ( η i ) , ϕ ( u ( T ) ) = β , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-130/MediaObjects/13661_2012_Article_231_Equ1_HTML.gif
      (1.1)

      where ϕ : R ( b , b ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-130/MediaObjects/13661_2012_Article_231_IEq1_HTML.gif ( 0 < b + http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-130/MediaObjects/13661_2012_Article_231_IEq11_HTML.gif) is an increasing homeomorphism such that ϕ ( 0 ) = 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-130/MediaObjects/13661_2012_Article_231_IEq3_HTML.gif, h : [ 0 , T ] R + http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-130/MediaObjects/13661_2012_Article_231_IEq4_HTML.gif and f : R + R + http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-130/MediaObjects/13661_2012_Article_231_IEq5_HTML.gif are continuous ( R + = [ 0 , + ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-130/MediaObjects/13661_2012_Article_231_IEq12_HTML.gif), β 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-130/MediaObjects/13661_2012_Article_231_IEq6_HTML.gif, α i R http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-130/MediaObjects/13661_2012_Article_231_IEq8_HTML.gif and η i ( 0 , T ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-130/MediaObjects/13661_2012_Article_231_IEq7_HTML.gif, for i I : = { 1 , 2 , , k } http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-130/MediaObjects/13661_2012_Article_231_IEq13_HTML.gif.

      Boundary value problems, including the ϕ-Laplacian operator, have received a lot of attention with respect to the existence and multiplicity of solutions. Since 2004, with a number of papers, Bereanu and Mawhin have considered such problems with Dirichlet, Neumann or periodic boundary conditions (see [14] and the references therein). In these papers, the various boundary value problems are reduced to the search for fixed points of some nonlinear operators defined on Banach spaces. In particular, they have studied some boundary value problems with nonhomogeneous boundary conditions and obtained the existence of solutions by the use of Schauder’s fixed point theorem (see [3, 4]). Recently, Torres [5] has proved the existence of a solution of a forced Liénard differential equation with ϕ-Laplacian by means of Schauder’s fixed point theorem.

      However, many nonlinear differential equations need to seek the existence of positive solutions because the positive solutions are very meaningful. The existence of positive solutions for homogeneous and nonhomogeneous boundary value problems have been studied by several authors and many interesting results have been obtained (only to mention some of them, see [69], their references and the papers citing them). The problems with negative coefficients for the boundary conditions (see [1012]) often occur in some heat flow problems, the deflection of a beam, and Floquet theory of the beam equation and have been considered by some experts (see [7, 8, 1012]). If the coefficient takes a negative value, then it is sometimes difficult to find an appropriate cone to guarantee the existence of a positive solution of the corresponding differential equation. Comparing with the previous result [24], the cone may be smaller.

      The purpose of this paper is to establish the criteria of the existence of a positive solution to the problem (1.1) by utilizing the Krasnosel’skii fixed point theorem, even if some of the α i http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-130/MediaObjects/13661_2012_Article_231_IEq10_HTML.gif coefficients are negative. The method of proof is inspired by the ideas exposed in [35, 7, 8]. As we will see, our results are new, and the interesting points of those results are the following two aspects: (i) Some of the α i http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-130/MediaObjects/13661_2012_Article_231_IEq10_HTML.gif coefficients appearing in (1.1) are allowed to take a negative value. (ii) The existence of a positive solution for the class of ϕ-Laplacian differential equations with a nonhomogeneous boundary condition is proved. Notice that the existence of a positive solution for the class of ϕ-Laplacian equations has been less studied in the related literature.

      This paper is organized as follows. In Section 2, we give some lemmas, which play an important role in the proof of the main theorem. In Section 3, we obtain the existence of a positive solution to the problem (1.1). Moreover, two examples are also given to illustrate the main results.

      2 Preliminaries and lemmas

      Let X denote the Banach space C ( [ 0 , T ] , R ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-130/MediaObjects/13661_2012_Article_231_IEq14_HTML.gif of continuous functions endowed with the maximum norm u = max t [ 0 , T ] | u ( t ) | http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-130/MediaObjects/13661_2012_Article_231_IEq15_HTML.gif. Define a nonlinear operator S : X X http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-130/MediaObjects/13661_2012_Article_231_IEq16_HTML.gif by
      ( S u ) ( t ) : = i = 1 k α i u ( η i ) + 0 t ϕ 1 ( β + s T h ( τ ) f ( u ( τ ) ) d τ ) d s : = ω ( u ) + 0 t ϕ 1 ( β + s T h ( τ ) f ( u ( τ ) ) d τ ) d s . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-130/MediaObjects/13661_2012_Article_231_Equ2_HTML.gif
      (2.1)
      Deriving on both sides of (2.1) leads to
      ( S u ) ( t ) = ϕ 1 ( β + t T h ( s ) f ( u ( s ) ) d s ) , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-130/MediaObjects/13661_2012_Article_231_Equ3_HTML.gif
      (2.2)
      i.e.,
      ϕ ( ( S u ) ( t ) ) = β + t T h ( s ) f ( u ( s ) ) d s . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-130/MediaObjects/13661_2012_Article_231_Equ4_HTML.gif
      (2.3)
      Again, deriving in (2.3) implies
      ( ϕ ( ( S u ) ( t ) ) ) = h ( t ) f ( u ( t ) ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-130/MediaObjects/13661_2012_Article_231_Equb_HTML.gif

      Moreover, from (2.1) and (2.3), we get that ( S u ) ( 0 ) = ω ( u ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-130/MediaObjects/13661_2012_Article_231_IEq17_HTML.gif and ϕ ( ( S u ) ( T ) ) = β http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-130/MediaObjects/13661_2012_Article_231_IEq18_HTML.gif. Therefore, the existence of a solution for Eq. (1.1) is equivalent to seeking a fixed point of the nonlinear operator S.

      For the sake of convenience, we give the following conditions.
      1. (A)
        Denote Δ : = { i I : α i < 0 } http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-130/MediaObjects/13661_2012_Article_231_IEq19_HTML.gif and : = { i I : α i > 0 } http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-130/MediaObjects/13661_2012_Article_231_IEq20_HTML.gif, and α i http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-130/MediaObjects/13661_2012_Article_231_IEq10_HTML.gif satisfies the conditions
        i = 1 k α i 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-130/MediaObjects/13661_2012_Article_231_Equc_HTML.gif
         
      and
      i α i < 1 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-130/MediaObjects/13661_2012_Article_231_Equd_HTML.gif
      1. (F)

        The function f : R + R + http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-130/MediaObjects/13661_2012_Article_231_IEq5_HTML.gif is continuous and satisfies M 1 f ( u ) M 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-130/MediaObjects/13661_2012_Article_231_IEq21_HTML.gif, for any u 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-130/MediaObjects/13661_2012_Article_231_IEq22_HTML.gif, where 0 < M 1 < M 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-130/MediaObjects/13661_2012_Article_231_IEq23_HTML.gif are two constants.

         
      2. (B)

        β + M 2 0 T h ( t ) d t < b http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-130/MediaObjects/13661_2012_Article_231_IEq24_HTML.gif (for 0 < b < + http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-130/MediaObjects/13661_2012_Article_231_IEq25_HTML.gif).

         
      3. (H)
        There exists a d ( 0 , T ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-130/MediaObjects/13661_2012_Article_231_IEq26_HTML.gif such that h ( d ) > 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-130/MediaObjects/13661_2012_Article_231_IEq27_HTML.gif, and let the inequality
        i Δ α i 0 η i ϕ 1 [ β + M 2 s T h ( τ ) d τ ] d s + i α i 0 η i ϕ 1 [ β + M 1 s T h ( τ ) d τ ] d s 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-130/MediaObjects/13661_2012_Article_231_Eque_HTML.gif
         

      be true.

      For the unbounded ϕ-Laplacian ( b = + http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-130/MediaObjects/13661_2012_Article_231_IEq28_HTML.gif), we obtain the following results.

      Lemma 1 Assume that the conditions (F) and (H) hold, u ( t ) 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-130/MediaObjects/13661_2012_Article_231_IEq29_HTML.gif, ω ( u ) 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-130/MediaObjects/13661_2012_Article_231_IEq30_HTML.gif, and 0 < a < T http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-130/MediaObjects/13661_2012_Article_231_IEq31_HTML.gif. Then there exists a constant γ ( 0 , 1 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-130/MediaObjects/13661_2012_Article_231_IEq32_HTML.gif such that
      min t [ a , T ] ( S u ) ( t ) γ S u . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-130/MediaObjects/13661_2012_Article_231_Equf_HTML.gif
      Proof From the representation (2.2) and the conditions (F)-(H), we have
      ( S u ) ( t ) = ϕ 1 ( β + t T h ( τ ) f ( u ( τ ) ) d τ ) 0 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-130/MediaObjects/13661_2012_Article_231_Equg_HTML.gif
      Again since β 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-130/MediaObjects/13661_2012_Article_231_IEq6_HTML.gif, we get that S u = ( S u ) ( T ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-130/MediaObjects/13661_2012_Article_231_IEq33_HTML.gif. Therefore, applying the condition (F) leads to
      S u = ( S u ) ( T ) = ω ( u ) + 0 T ϕ 1 ( β + s T h ( τ ) f ( u ( τ ) ) d τ ) d s ω ( u ) + 0 T ϕ 1 ( β + M 2 s T h ( τ ) d τ ) d s , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-130/MediaObjects/13661_2012_Article_231_Equh_HTML.gif
      and
      min t [ a , T ] ( S u ) ( t ) = ( S u ) ( a ) = ω ( u ) + 0 a ϕ 1 ( β + s T h ( τ ) f ( u ( τ ) ) d τ ) d s ω ( u ) + 0 a ϕ 1 ( β + M 1 s T h ( τ ) d τ ) d s γ S u , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-130/MediaObjects/13661_2012_Article_231_Equi_HTML.gif
      where
      γ = 0 a ϕ 1 ( β + M 1 s T h ( τ ) d τ ) d s 0 T ϕ 1 ( β + M 2 s T h ( τ ) d τ ) d s . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-130/MediaObjects/13661_2012_Article_231_Equj_HTML.gif

      This completes the proof. □

      Next, let us define a cone by
      P : = { u X : u 0 , ω ( u ) 0 , min t [ a , T ] u ( t ) γ u } . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-130/MediaObjects/13661_2012_Article_231_Equk_HTML.gif

      The definition of the cone is inspired by the results in [7, 8]. To show our main results, the following lemma is essential.

      Lemma 2 Let the conditions (A), (F), and (H) hold and the nonlinear operator S be defined by (2.1). Then S : P P http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-130/MediaObjects/13661_2012_Article_231_IEq34_HTML.gif.

      Proof From the definition of the operator S, we find for any u P http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-130/MediaObjects/13661_2012_Article_231_IEq35_HTML.gif that
      ( S u ) ( t ) = ω ( u ) + 0 t ϕ 1 ( β + s T h ( τ ) f ( u ( τ ) ) d τ ) d s 0 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-130/MediaObjects/13661_2012_Article_231_Equl_HTML.gif
      The conditions (F) and (H) yield
      ω ( S u ) = i = 1 k α i ( S u ) ( η i ) = i = 1 k α i { ω ( u ) + 0 η i ϕ 1 ( β + s T h ( τ ) f ( u ( τ ) ) d τ ) d s } = ω ( u ) ( i = 1 k α i ) + i Δ α i 0 η i ϕ 1 ( β + s T h ( τ ) f ( u ( τ ) ) d τ ) d s + i α i 0 η i ϕ 1 ( β + s T h ( τ ) f ( u ( τ ) ) d τ ) d s i Δ α i 0 η i ϕ 1 ( β + M 2 s T h ( τ ) d τ ) d s + i α i 0 η i ϕ 1 ( β + M 1 s T h ( τ ) d τ ) d s 0 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-130/MediaObjects/13661_2012_Article_231_Equm_HTML.gif
      Further, Lemma 1 shows
      min t [ a , T ] ( S u ) ( t ) γ S u . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-130/MediaObjects/13661_2012_Article_231_Equn_HTML.gif

      Consequently, we get that S : P P http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-130/MediaObjects/13661_2012_Article_231_IEq34_HTML.gif. □

      Remark 1 If the coefficients { α i : i I } http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-130/MediaObjects/13661_2012_Article_231_IEq36_HTML.gif are nonnegative, then the conclusion in Lemma 2 also holds without the hypothesis (H).

      Lemma 3 If 0 < b < + http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-130/MediaObjects/13661_2012_Article_231_IEq25_HTML.gif and, in addition, the assumptions of Lemma  2 and the condition (B) are satisfied, then the conclusions of Lemma  1 and Lemma  2 hold.

      Lemma 4 (See [13])

      Let X be a Banach space and P X http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-130/MediaObjects/13661_2012_Article_231_IEq37_HTML.gif be a cone. Suppose that Ω 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-130/MediaObjects/13661_2012_Article_231_IEq38_HTML.gif and Ω 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-130/MediaObjects/13661_2012_Article_231_IEq39_HTML.gif are bounded open sets contained in X such that 0 Ω 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-130/MediaObjects/13661_2012_Article_231_IEq40_HTML.gif and Ω ¯ 1 Ω 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-130/MediaObjects/13661_2012_Article_231_IEq41_HTML.gif. Suppose further that S : P ( Ω ¯ 2 Ω 1 ) P http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-130/MediaObjects/13661_2012_Article_231_IEq42_HTML.gif is a completely continuous operator. If either
      1. (i)

        S u u http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-130/MediaObjects/13661_2012_Article_231_IEq43_HTML.gif for u P Ω 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-130/MediaObjects/13661_2012_Article_231_IEq44_HTML.gif and S u u http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-130/MediaObjects/13661_2012_Article_231_IEq45_HTML.gif for u P Ω 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-130/MediaObjects/13661_2012_Article_231_IEq46_HTML.gif or

         
      2. (ii)

        S u u http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-130/MediaObjects/13661_2012_Article_231_IEq45_HTML.gif for u P Ω 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-130/MediaObjects/13661_2012_Article_231_IEq44_HTML.gif and S u u http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-130/MediaObjects/13661_2012_Article_231_IEq43_HTML.gif for u P Ω 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-130/MediaObjects/13661_2012_Article_231_IEq46_HTML.gif,

         

      then S has at least one fixed point in P ( Ω ¯ 2 Ω 1 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-130/MediaObjects/13661_2012_Article_231_IEq47_HTML.gif.

      3 The main result

      Theorem 1 Assume that the conditions (A), (F), and (H) hold and b = + http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-130/MediaObjects/13661_2012_Article_231_IEq28_HTML.gif. Then Eq. (1.1) has at least one positive solution.

      Proof Lemma 2 shows that S : P P http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-130/MediaObjects/13661_2012_Article_231_IEq48_HTML.gif. In addition, a standard argument involving the Arzela-Ascoli theorem implies that S is a completely continuous operator.

      Now, we choose a positive constant r 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-130/MediaObjects/13661_2012_Article_231_IEq49_HTML.gif such that
      r 1 0 a ϕ 1 ( β + M 1 s T h ( τ ) d τ ) d s http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-130/MediaObjects/13661_2012_Article_231_Equo_HTML.gif
      and define Ω 1 : = { u X : u < r 1 } http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-130/MediaObjects/13661_2012_Article_231_IEq50_HTML.gif. For any u P Ω 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-130/MediaObjects/13661_2012_Article_231_IEq44_HTML.gif, we get from the condition (F) that
      S u = ( S u ) ( T ) = ω ( u ) + 0 T ϕ 1 ( β + s T h ( τ ) f ( u ( τ ) ) d τ ) d s 0 T ϕ 1 ( β + s T h ( τ ) f ( u ( τ ) ) d τ ) d s 0 a ϕ 1 ( β + M 1 s T h ( τ ) d τ ) d s r 1 = u . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-130/MediaObjects/13661_2012_Article_231_Equp_HTML.gif
      Thus, for any u P Ω 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-130/MediaObjects/13661_2012_Article_231_IEq44_HTML.gif, we find that
      S u u . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-130/MediaObjects/13661_2012_Article_231_Equ5_HTML.gif
      (3.1)
      From the hypothesis (A), we can let i α i = 1 ε http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-130/MediaObjects/13661_2012_Article_231_IEq51_HTML.gif, ε ( 0 , 1 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-130/MediaObjects/13661_2012_Article_231_IEq52_HTML.gif. Next, we choose a positive constant r 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-130/MediaObjects/13661_2012_Article_231_IEq53_HTML.gif such that
      r 2 = max { 1 ε 0 T ϕ 1 ( β + M 2 s T h ( τ ) d τ ) d s , r 1 ε } http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-130/MediaObjects/13661_2012_Article_231_Equq_HTML.gif
      and define Ω 2 : = { u X : u < r 2 } http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-130/MediaObjects/13661_2012_Article_231_IEq54_HTML.gif. Clearly, for any u P Ω 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-130/MediaObjects/13661_2012_Article_231_IEq46_HTML.gif, we obtain
      S u = ( S u ) ( T ) = ω ( u ) + 0 T ϕ 1 ( β + s T h ( τ ) f ( u ( τ ) ) d τ ) d s i α i u + 0 T ϕ 1 ( β + M 2 s T h ( τ ) d τ ) d s i α i u + ε r 2 = r 2 [ i α i + ε ] = r 2 = u . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-130/MediaObjects/13661_2012_Article_231_Equr_HTML.gif
      Then, for any u P Ω 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-130/MediaObjects/13661_2012_Article_231_IEq46_HTML.gif, it implies that
      S u u . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-130/MediaObjects/13661_2012_Article_231_Equ6_HTML.gif
      (3.2)

      Based on Lemma 4, we get from (3.1) and (3.2) that the operator S has at least one fixed point. Thus, it follows that Eq. (1.1) has at least one positive solution. □

      Remark 2 If the coefficients { α i : i I } http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-130/MediaObjects/13661_2012_Article_231_IEq55_HTML.gif are nonnegative, then the condition (A) is replaced with

      (A′) i = 1 k α i ( 0 , 1 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-130/MediaObjects/13661_2012_Article_231_IEq56_HTML.gif.

      Applying the results in Remark 1 and Theorem 1, we get the following result.

      Corollary 1 Assume that the conditions (A′) and (F) hold and b = + http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-130/MediaObjects/13661_2012_Article_231_IEq28_HTML.gif. Then Eq. (1.1) has at least one positive solution.

      If ϕ : R ( b , b ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-130/MediaObjects/13661_2012_Article_231_IEq1_HTML.gif ( 0 < b < + http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-130/MediaObjects/13661_2012_Article_231_IEq25_HTML.gif ), then we have the following result.

      Theorem 2 Assume that the conditions (A), (F), (B), and (H) hold. Then Eq. (1.1) has at least one positive solution.

      Proof Using Lemma 3 and the proof of Theorem 1, we get that the conclusion holds. □

      Example 1 Consider the differential equation
      ( | u ( t ) | 2 u ( t ) ) = t ( e sin u ( t ) + cos u 2 ( t ) + 5 ) , t ( 0 , 4 ) , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-130/MediaObjects/13661_2012_Article_231_Equ7_HTML.gif
      (3.3)
      subjected to the boundary conditions
      u ( 0 ) = 1 8 u ( 1 ) 1 4 u ( 2 ) + 1 2 u ( 3 ) , ϕ ( u ( 4 ) ) = 1 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-130/MediaObjects/13661_2012_Article_231_Equ8_HTML.gif
      (3.4)
      Clearly, we find
      ϕ = ϕ 4 is one  p -Laplacian operator,  p = 4 , h ( t ) = t , f ( u ) = e sin u + cos u 2 + 5 , ω ( u ) = 1 8 u ( 1 ) 1 4 u ( 2 ) + 1 2 u ( 3 ) , α 1 = 1 8 , α 2 = 1 4 , α 3 = 1 2 , i = 1 3 α i = 3 8 and α 1 + α 3 = 5 8 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-130/MediaObjects/13661_2012_Article_231_Equs_HTML.gif
      Computing yields
      http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-130/MediaObjects/13661_2012_Article_231_Equt_HTML.gif

      Therefore, we conclude from Theorem 1 that Eq. (3.3)-(3.4) has at least one positive solution.

      Example 2 Consider the differential equation
      ( u ( t ) 1 + ( u ( t ) ) 2 ) = t 4 ( sin u ( t ) + 2 ) , t ( 0 , 1 ) , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-130/MediaObjects/13661_2012_Article_231_Equ9_HTML.gif
      (3.5)
      subjected to the boundary conditions
      u ( 0 ) = 1 40 u ( 0.2 ) + 1 4 u ( 0.3 ) + 1 2 u ( 0.7 ) , ϕ ( u ( 1 ) ) = 1 4 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-130/MediaObjects/13661_2012_Article_231_Equ10_HTML.gif
      (3.6)
      Obviously, we obtain
      ϕ ( u ) = u 1 + u 2 , h ( t ) = t 4 , f ( u ) = sin u ( t ) + 2 , ω ( u ) = 1 40 u ( 0.2 ) + 1 4 u ( 0.3 ) + 1 2 u ( 0.7 ) , α 1 = 1 40 , α 2 = 1 4 , α 3 = 1 2 , i = 1 3 α i = 29 40 and α 2 + α 3 = 3 4 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-130/MediaObjects/13661_2012_Article_231_Equu_HTML.gif

      It is easy to verify that the conditions (B), (F), and (H) hold. Consequently, we get from Theorem 2 that the equation (3.5)-(3.6) has at least one positive solution.

      Declarations

      Acknowledgements

      The authors would like to thank the anonymous referees very much for helpful comments and suggestions which led to the improvement of the presentation and quality of the work. The work was supported partly by NSFC of Tianyuan Youth Foundation (No.11126125), K.C. Wong Magna Fund of Ningbo University and Ningbo Natural Science Foundation (2012A610031).

      Authors’ Affiliations

      (1)
      Department of Mathematics, Ningbo University

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