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


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


        Nonlocal boundary value problems Cone Fixed point theorem


        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

        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
        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

        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.


        Lemma 2.1[3]. Let y(t) ∈ C([0, 1]), then the problem
        has a unique solution
        where the Green function G(t, s) is
        It is easy to see that

        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
        Obviously, uC2(0, 1) is a solution of Equation 1 if and only if uC(0, 1) satisfies the following nonlinear integral equation

        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
        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
        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
        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

        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

        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
        Thus, it follows from above that

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


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

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

        (H5) There exists a constant 2 ≤ p1 such that
        (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

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

        Proof. From (H4), there exists a 0 < η < ∞ such that
        Choosing R1 ∈ (0, η), set Ω1 = {uE : ||u|| < R1}. We now prove that
        Let uP ∩ ∂Ω1. Since minθt≤1-θu(t) ≥ θ ||u|| and ||u|| = R1, from Equation 3, (H1) and (H3), it follows that

        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
        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
        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
        If uP ∩ ∂Ω2, we have
        From Equations 5, 6, we can prove

        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,
        It is easy to see that (H1) to (H3) hold. We also can have
        Take p1 = 2, then it is clear to see that (H4) and (H5) hold. Since

        then (H6) hold.

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

        (H7) There exists a constant 2 ≤ p1 such that
        (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

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

        Proof. From (H7), there exists η1 > 0 such that
        From (H8), there exists η2 > 0 such that
        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
        If uP ∩ ∂Ω1, we have
        From Equations 8, 9, we can prove

        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
        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
        If uP ∩ ∂Ω2, Since minθt≤1-θu(t) ≥ θ ||u|| and ||u|| = R2, we have
        By Equation 11, (H1) and (H3), it follows that

        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:

        (H12) There exists a constant a > 0 such that

        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

        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
        If http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-5/MediaObjects/13661_2011_Article_5_IEq21_HTML.gif , then

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

        Next, from (H11), there exists d' ∈ (0, a) such that
        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

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

        In fact,
        For uP (σ, a, b), we have
        for all t ∈ [θ, 1 -θ]. Then, we have
        by (H1) to (H3), (H12). In addition, for each uP (θ, a, c) with ||Tu|| > b, we have
        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

        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,
        From a simple computation, we have

        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.


        Authors’ Affiliations

        College of Aeronautics and Astronautics, Nanjing University of Aeronautics and Astronautics
        College of Science, Hohai University
        Department of Mathematics, Nanjing University of Aeronautics and Astronautics


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        © Wang et al; licensee Springer. 2011

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