Existence and multiplicity of positive solutions for a nonlocal differential equation

  • Yunhai Wang1Email author,

    Affiliated with

    • Fanglei Wang2, 3 and

      Affiliated with

      • Yukun An3

        Affiliated with

        Boundary Value Problems20112011:5

        DOI: 10.1186/1687-2770-2011-5

        Received: 21 February 2011

        Accepted: 11 July 2011

        Published: 11 July 2011

        Abstract

        In this paper, the existence and multiplicity results of positive solutions for a nonlocal differential equation are mainly considered.

        Keywords

        Nonlocal boundary value problems Cone Fixed point theorem

        Introduction

        In this paper, we are concerned with the existence and multiplicity of positive solutions for the following nonlinear differential equation with nonlocal boundary value condition
        http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-5/MediaObjects/13661_2011_Article_5_Equ1_HTML.gif
        (1)

        where α, β, γ, δ are nonnegative constants, ρ = αγ + αδ + βγ > 0, q ≥ 1; http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-5/MediaObjects/13661_2011_Article_5_IEq1_HTML.gif , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-5/MediaObjects/13661_2011_Article_5_IEq2_HTML.gif denote the Riemann-Stieltjes integrals.

        Many authors consider the problem
        http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-5/MediaObjects/13661_2011_Article_5_Equ2_HTML.gif
        (2)
        because of the importance in numerous physical models: system of particles in thermodynamical equilibrium interacting via gravitational potential, 2-D fully turbulent behavior of a real flow, one-dimensional fluid flows with rate of strain proportional to a power of stress multiplied by a function of temperature, etc. In [1, 2], the authors use the Kras-noselskii fixed point theorem to obtain one positive solution for the following nonlocal equation with zero Dirichlet boundary condition
        http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-5/MediaObjects/13661_2011_Article_5_Equa_HTML.gif

        when the nonlinearity f is a sublinear or superlinear function in a sense to be established when necessary. Nonlocal BVPs of ordinary differential equations or system arise in a variety of areas of applied mathematics and physics. In recent years, more and more papers were devoted to deal with the existence of positive solutions of nonlocal BVPs (see [39] and references therein). Inspired by the above references, our aim in the present paper is to investigate the existence and multiplicity of positive solutions to Equation 1 using the Krasnosel'skii fixed point theorem and Leggett-Williams fixed point theorem.

        This paper is organized as follows: In Section 2, some preliminaries are given; In Section 3, we give the existence results.

        Preliminaries

        Lemma 2.1[3]. Let y(t) ∈ C([0, 1]), then the problem
        http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-5/MediaObjects/13661_2011_Article_5_Equb_HTML.gif
        has a unique solution
        http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-5/MediaObjects/13661_2011_Article_5_Equc_HTML.gif
        where the Green function G(t, s) is
        http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-5/MediaObjects/13661_2011_Article_5_Equd_HTML.gif
        It is easy to see that
        http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-5/MediaObjects/13661_2011_Article_5_Eque_HTML.gif

        and there exists a http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-5/MediaObjects/13661_2011_Article_5_IEq3_HTML.gif such that G(t, s) ≥ θ G(s, s), θt ≤ 1 - θ, 0 ≤ s ≤ 1.

        For convenience, we assume the following conditions hold throughout this paper:

        (H1) f, g, Φ: R+R+ are continuous and nondecreasing functions, and Φ (0) > 0;

        (H2) φ(t) is an increasing nonconstant function defined on [0, 1] with φ(0) = 0;

        (H3) h(t) does not vanish identically on any subinterval of (0, 1) and satisfies
        http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-5/MediaObjects/13661_2011_Article_5_Equf_HTML.gif
        Obviously, uC2(0, 1) is a solution of Equation 1 if and only if uC(0, 1) satisfies the following nonlinear integral equation
        http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-5/MediaObjects/13661_2011_Article_5_Equg_HTML.gif

        At the end of this section, we state the fixed point theorems, which will be used in Section 3.

        Let E be a real Banach space with norm || · || and PE be a cone in E, P r = {xP : ||x|| < r}(r > 0). Then, http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-5/MediaObjects/13661_2011_Article_5_IEq4_HTML.gif . A map α is said to be a nonnegative continuous concave functional on P if α: P → [0, +∞) is continuous and
        http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-5/MediaObjects/13661_2011_Article_5_Equh_HTML.gif
        for all x, yP and t ∈ [0, 1]. For numbers a, b such that 0 < a < b and α is a nonnegative continuous concave functional on P, we define the convex set
        http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-5/MediaObjects/13661_2011_Article_5_Equi_HTML.gif
        Lemma 2.2[10]. Let http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-5/MediaObjects/13661_2011_Article_5_IEq5_HTML.gif be completely continuous and α be a nonnegative continuous concave functional on P such that α (x) = ||x|| for all http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-5/MediaObjects/13661_2011_Article_5_IEq6_HTML.gif . Suppose there exists 0 < d < a < b = c such that
        1. (i)

          {xP (α, a, b): α (x) > a} ≠ ∅ and α (Ax) > a for xP (α, a, b);

           
        2. (ii)

          ||Ax|| < d for ||x|| ≤ d;

           

        (iii) α(Ax) > a for xP (α, a, c) with ||Ax|| > b.

        Then, A has at least three fixed points x1, x2, x3 satisfying
        http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-5/MediaObjects/13661_2011_Article_5_Equj_HTML.gif
        Lemma 2.3[10]. Let E be a Banach space, and let PE be a closed, convex cone in E, assume Ω1, Ω2 are bounded open subsets of E with http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-5/MediaObjects/13661_2011_Article_5_IEq7_HTML.gif , and http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-5/MediaObjects/13661_2011_Article_5_IEq8_HTML.gif be a completely continuous operator such that either
        1. (i)

          ||Au|| ≤ ||u||, uP ∩ ∂Ω1 and ||Au|| ≥ ||u||, uP ∩ ∂Ω2; or

           
        2. (ii)

          ||Au|| ≥ ||u||, uP ∩ ∂Ω1 and ||Au|| ≤ ||u||, uP ∩ ∂Ω2.

           

        Then, A has a fixed point in http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-5/MediaObjects/13661_2011_Article_5_IEq9_HTML.gif .

        Main result

        Let E = C[0, 1] endowed norm ||u|| = max0≤t≤1|u|, and define the cone PE by
        http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-5/MediaObjects/13661_2011_Article_5_Equk_HTML.gif

        Then, it is easy to prove that E is a Banach space and P is a cone in E.

        Define the operator T: EE by
        http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-5/MediaObjects/13661_2011_Article_5_Equl_HTML.gif

        Lemma 3.1. T: EE is completely continuous, and Te now prove thatPP.

        Proof. For any uP, then from properties of G(t, s), T (u)(t) ≥ 0, t ∈ [0, 1], and it follows from the definition of T that
        http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-5/MediaObjects/13661_2011_Article_5_Equm_HTML.gif
        Thus, it follows from above that
        http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-5/MediaObjects/13661_2011_Article_5_Equn_HTML.gif

        From the above, we conclude that TPP. Also, one can verify that T is completely continuous by the Arzela-Ascoli theorem.   □

        Let
        http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-5/MediaObjects/13661_2011_Article_5_Equo_HTML.gif

        Then, it is clear to see that 0 < lL < L.

        Theorem 3.2. Assume (H1) to (H3) hold. In addition,

        (H4)
        http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-5/MediaObjects/13661_2011_Article_5_Equp_HTML.gif
        (H5) There exists a constant 2 ≤ p1 such that
        http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-5/MediaObjects/13661_2011_Article_5_Equq_HTML.gif
        (H6) There exists a constant p2 with http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-5/MediaObjects/13661_2011_Article_5_IEq10_HTML.gif such that
        http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-5/MediaObjects/13661_2011_Article_5_Equr_HTML.gif

        Then, problem (Equation 1) has one positive solution.

        Proof. From (H4), there exists a 0 < η < ∞ such that
        http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-5/MediaObjects/13661_2011_Article_5_Equ3_HTML.gif
        (3)
        Choosing R1 ∈ (0, η), set Ω1 = {uE : ||u|| < R1}. We now prove that
        http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-5/MediaObjects/13661_2011_Article_5_Equ4_HTML.gif
        (4)
        Let uP ∩ ∂Ω1. Since minθt≤1-θu(t) ≥ θ ||u|| and ||u|| = R1, from Equation 3, (H1) and (H3), it follows that
        http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-5/MediaObjects/13661_2011_Article_5_Equs_HTML.gif

        Then, Equation 4 holds.

        On the other hand, from (H5), there exists http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-5/MediaObjects/13661_2011_Article_5_IEq11_HTML.gif such that
        http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-5/MediaObjects/13661_2011_Article_5_Equ5_HTML.gif
        (5)
        From (H6), there exists http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-5/MediaObjects/13661_2011_Article_5_IEq12_HTML.gif such that
        http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-5/MediaObjects/13661_2011_Article_5_Equ6_HTML.gif
        (6)
        Choosing http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-5/MediaObjects/13661_2011_Article_5_IEq13_HTML.gif , set Ω2 = {uE : ||u|| < R2}. We now prove that
        http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-5/MediaObjects/13661_2011_Article_5_Equ7_HTML.gif
        (7)
        If uP ∩ ∂Ω2, we have
        http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-5/MediaObjects/13661_2011_Article_5_Equt_HTML.gif
        From Equations 5, 6, we can prove
        http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-5/MediaObjects/13661_2011_Article_5_Equu_HTML.gif

        Then, Equation 7 holds.

        Therefore, by Equations 4 and 7 and the second part of Lemma 2.3, T has a fixed point in http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-5/MediaObjects/13661_2011_Article_5_IEq9_HTML.gif , which is a positive solution of Equation 1.   □

        Example. Let q = 2, h(t) = 1, Φ(s) = 2 + s, φ(t) = 2t, http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-5/MediaObjects/13661_2011_Article_5_IEq14_HTML.gif and http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-5/MediaObjects/13661_2011_Article_5_IEq15_HTML.gif , namely,
        http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-5/MediaObjects/13661_2011_Article_5_Equv_HTML.gif
        It is easy to see that (H1) to (H3) hold. We also can have
        http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-5/MediaObjects/13661_2011_Article_5_Equw_HTML.gif
        Take p1 = 2, then it is clear to see that (H4) and (H5) hold. Since
        http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-5/MediaObjects/13661_2011_Article_5_Equx_HTML.gif

        then (H6) hold.

        Theorem 3.3. Assume (H1) to (H3) hold. In addition,

        (H7) There exists a constant 2 ≤ p1 such that
        http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-5/MediaObjects/13661_2011_Article_5_Equy_HTML.gif
        (H8) There exists a constant p2 with http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-5/MediaObjects/13661_2011_Article_5_IEq10_HTML.gif such that
        http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-5/MediaObjects/13661_2011_Article_5_Equz_HTML.gif
        (H9)
        http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-5/MediaObjects/13661_2011_Article_5_Equaa_HTML.gif

        Then, problem (Equation 1) has one positive solution.

        Proof. From (H7), there exists η1 > 0 such that
        http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-5/MediaObjects/13661_2011_Article_5_Equ8_HTML.gif
        (8)
        From (H8), there exists η2 > 0 such that
        http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-5/MediaObjects/13661_2011_Article_5_Equ9_HTML.gif
        (9)
        Choosing http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-5/MediaObjects/13661_2011_Article_5_IEq16_HTML.gif , set Ω1 = {uE : ||u|| < R1}. We now prove that
        http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-5/MediaObjects/13661_2011_Article_5_Equ10_HTML.gif
        (10)
        If uP ∩ ∂Ω1, we have
        http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-5/MediaObjects/13661_2011_Article_5_Equab_HTML.gif
        From Equations 8, 9, we can prove
        http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-5/MediaObjects/13661_2011_Article_5_Equac_HTML.gif

        Then, Equation 10 holds.

        On the other hand, from (H7), there exists http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-5/MediaObjects/13661_2011_Article_5_IEq11_HTML.gif such that
        http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-5/MediaObjects/13661_2011_Article_5_Equ11_HTML.gif
        (11)
        Choosing http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-5/MediaObjects/13661_2011_Article_5_IEq17_HTML.gif , set Ω2 = {uE : ||u|| < R2}. We now prove that
        http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-5/MediaObjects/13661_2011_Article_5_Equ12_HTML.gif
        (12)
        If uP ∩ ∂Ω2, Since minθt≤1-θu(t) ≥ θ ||u|| and ||u|| = R2, we have
        http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-5/MediaObjects/13661_2011_Article_5_Equ13_HTML.gif
        (13)
        By Equation 11, (H1) and (H3), it follows that
        http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-5/MediaObjects/13661_2011_Article_5_Equad_HTML.gif

        Then, Equation 12 holds.

        Therefore, by Equations 10 and 12 and the first part of Lemma 2.3, T has a fixed point in http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-5/MediaObjects/13661_2011_Article_5_IEq9_HTML.gif , which is a positive solution of Equation 1.   □

        Example. Let q = 2, h(t) = t, Φ(s) = 2 + s, φ(t) = 2t, http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-5/MediaObjects/13661_2011_Article_5_IEq18_HTML.gif and g(s) = s2.

        Theorem 3.4. Assume that (H1) to (H3) hold. In addition, φ(1) ≥ 1, and the functions f, g satisfy the following growth conditions:

        (H10)
        http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-5/MediaObjects/13661_2011_Article_5_Equae_HTML.gif
        (H11)
        http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-5/MediaObjects/13661_2011_Article_5_Equaf_HTML.gif
        (H12) There exists a constant a > 0 such that
        http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-5/MediaObjects/13661_2011_Article_5_Equag_HTML.gif

        Then, BVP (Equation 1) has at least three positive solutions.

        Proof. For the sake of applying the Leggett-Williams fixed point theorem, define a functional σ(u) on cone P by
        http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-5/MediaObjects/13661_2011_Article_5_Equah_HTML.gif

        Evidently, σ: PR+ is a nonnegative continuous and concave. Moreover, σ(u) ≤ ||u|| for each uP.

        Now, we verify that the assumption of Lemma 2.2 is satisfied.

        Firstly, it can verify that there exists a positive number c with http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-5/MediaObjects/13661_2011_Article_5_IEq19_HTML.gif such that http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-5/MediaObjects/13661_2011_Article_5_IEq20_HTML.gif .

        By (H10), it is easy to see that there exists τ > 0 such that
        http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-5/MediaObjects/13661_2011_Article_5_Equai_HTML.gif
        Set
        http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-5/MediaObjects/13661_2011_Article_5_Equaj_HTML.gif
        Taking
        http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-5/MediaObjects/13661_2011_Article_5_Equak_HTML.gif
        If http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-5/MediaObjects/13661_2011_Article_5_IEq21_HTML.gif , then
        http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-5/MediaObjects/13661_2011_Article_5_Equal_HTML.gif

        by (H1) to (H3) and (H10).

        Next, from (H11), there exists d' ∈ (0, a) such that
        http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-5/MediaObjects/13661_2011_Article_5_Equam_HTML.gif
        Take http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-5/MediaObjects/13661_2011_Article_5_IEq22_HTML.gif . Then, for each http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-5/MediaObjects/13661_2011_Article_5_IEq23_HTML.gif , we have
        http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-5/MediaObjects/13661_2011_Article_5_Equan_HTML.gif

        Finally, we will show that {uP (σ, a, b): σ(u) > a} ≠ ∅ and σ(Tu) > a for all uP(σ, a, b).

        In fact,
        http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-5/MediaObjects/13661_2011_Article_5_Equao_HTML.gif
        For uP (σ, a, b), we have
        http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-5/MediaObjects/13661_2011_Article_5_Equap_HTML.gif
        for all t ∈ [θ, 1 -θ]. Then, we have
        http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-5/MediaObjects/13661_2011_Article_5_Equaq_HTML.gif
        by (H1) to (H3), (H12). In addition, for each uP (θ, a, c) with ||Tu|| > b, we have
        http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-5/MediaObjects/13661_2011_Article_5_Equar_HTML.gif
        Above all, we know that the conditions of Lemma 2.2 are satisfied. By Lemma 2.2, the operator T has at least three fixed points u i (i = 1, 2, 3) such that
        http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-5/MediaObjects/13661_2011_Article_5_Equas_HTML.gif

        The proof is complete.   □

        Example. Let q = 2, h(t) = t, Φ(s) = 2 + s, φ(t) = 2t, http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-5/MediaObjects/13661_2011_Article_5_IEq24_HTML.gif and, http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-5/MediaObjects/13661_2011_Article_5_IEq25_HTML.gif , namely,
        http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-5/MediaObjects/13661_2011_Article_5_Equat_HTML.gif
        From a simple computation, we have
        http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-5/MediaObjects/13661_2011_Article_5_Equau_HTML.gif

        Then, it is easy to see that (H1) to (H3) and (H10) to (H11) hold. Especially, take a = 1, by http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-5/MediaObjects/13661_2011_Article_5_IEq26_HTML.gif and (H1), then (H12) holds.

        Declarations

        Authors’ Affiliations

        (1)
        College of Aeronautics and Astronautics, Nanjing University of Aeronautics and Astronautics
        (2)
        College of Science, Hohai University
        (3)
        Department of Mathematics, Nanjing University of Aeronautics and Astronautics

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