Existence of positive solutions for nonlinear m-point boundary value problems on time scales

  • Junfang Zhao1Email author,

    Affiliated with

    • Hairong Lian1 and

      Affiliated with

      • Weigao Ge2

        Affiliated with

        Boundary Value Problems20122012:4

        DOI: 10.1186/1687-2770-2012-4

        Received: 4 May 2011

        Accepted: 17 January 2012

        Published: 17 January 2012

        Abstract

        In this article, we study the following m-point boundary value problem on time scales,

        ( ϕ p ( u Δ ( t ) ) ) + h ( t ) f ( t , u ( t ) ) = 0 , t ( 0 , T ) T , u ( 0 ) - δ u Δ ( 0 ) = i = 1 m - 2 β i u Δ ( ξ i ) , u Δ ( T ) = 0 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-4/MediaObjects/13661_2011_Article_110_Equa_HTML.gif

        where T http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-4/MediaObjects/13661_2011_Article_110_IEq1_HTML.gif is a time scale such that 0 , T T , δ , β i > 0 , i = 1 , , m - 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-4/MediaObjects/13661_2011_Article_110_IEq2_HTML.gif, ϕ p (s) = |s| p-2 s,p > 1,hC ld ((0, T), (0, +∞)), and fC([0,+∞), (0,+∞)), 0 < ξ 1 < ξ 2 < < ξ m - 2 < T T http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-4/MediaObjects/13661_2011_Article_110_IEq3_HTML.gif. By using several well-known fixed point theorems in a cone, the existence of at least one, two, or three positive solutions are obtained. Examples are also given in this article.

        AMS Subject Classification: 34B10; 34B18; 39A10.

        Keywords

        positive solutions cone multi-point boundary value problem time scale

        1 Introduction

        The study of dynamic equations on time scales goes back to its founder Hilger [1], and is a new area of still theoretical exploration in mathematics. Motivating the subject is the notion that dynamic equations on time scales can build bridges between continuous and discrete mathematics. Further, the study of time scales has led to several important applications, e.g., in the study of insect population models, neural networks, heat transfer, epidemic models, etc. [2].

        Multipoint boundary value problems of ordinary differential equations (BVPs for short) arise in a variety of different areas of applied mathematics and physics. For example, the vibrations of a guy wire of a uniform cross section and composed of N parts of different densities can be set up as a multi-point boundary value problem [3]. Many problems in the theory of elastic stability can be handled by the method of multi-point problems [4]. Small size bridges are often designed with two supported points, which leads into a standard two-point boundary value condition and large size bridges are sometimes contrived with multi-point supports, which corresponds to a multi-point boundary value condition [5]. The study of multi-point BVPs for linear second-order ordinary differential equations was initiated by Il'in and Moiseev [6]. Since then many authors have studied more general nonlinear multi-point BVPs, and the multi-point BVP on time scales can be seen as a generalization of that in ordinary differential equations.

        Recently, the existence and multiplicity of positive solutions for nonlinear differential equations on time scales have been studied by some authors [711], and there has been some merging of existence of positive solutions to BVPs with p-Laplacian on time scales [1219].

        He [20] studied
        ( ϕ p ( u Δ ( t ) ) ) + a ( t ) f ( t ) = 0 , t ( 0 , T ) T , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-4/MediaObjects/13661_2011_Article_110_Equ1_HTML.gif
        (1.1)
        subject to one of the following boundary conditions
        u ( 0 ) - B 0 ( u Δ ( η ) ) = 0 , u Δ ( T ) = 0 , u Δ ( 0 ) = 0 , u ( T ) - B 1 ( u Δ ( η ) ) = 0 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-4/MediaObjects/13661_2011_Article_110_Equ2_HTML.gif
        (1.2)

        where η ( 0 , T ) T http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-4/MediaObjects/13661_2011_Article_110_IEq4_HTML.gif. By using a double fixed-point theorem, the authors get the existence of at least two positive solutions to BVP (1.1) and (1.2).

        Anderson [21] studied
        - u Δ ( t ) = η a ( t ) f ( u ( t ) ) , t ( t 1 , t n ) T , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-4/MediaObjects/13661_2011_Article_110_Equ3_HTML.gif
        (1.3)
        subject to one of the following boundary conditions
        u ( t 1 ) = i = 2 n - 1 α i u ( t i ) , u Δ ( t n ) = 0 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-4/MediaObjects/13661_2011_Article_110_Equ4_HTML.gif
        (1.4)
        u Δ ( t 1 ) = 0 , u ( t n ) = i = 2 n - 1 α i u ( t i ) , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-4/MediaObjects/13661_2011_Article_110_Equ5_HTML.gif
        (1.5)

        by using a functional-type cone expansion-compression fixed-point theorem, the author gets the existence of at least one positive solution to BVP (1.3), (1.4) and BVP (1.3), (1.5).

        However, to the best of the authors' knowledge, up to now, there are few articles concerned with the existence of m-point boundary value problem with p-Laplacian on time scales. So, in this article, we try to fill this gap. Motivated by the article mentioned above, in this article, we consider the following m-point BVP with one-dimensional p-Laplacian,
        ( ϕ p ( u Δ ( t ) ) ) + h ( t ) f ( t , u ( t ) ) = 0 , t ( 0 , T ) T , u ( 0 ) - δ u Δ ( 0 ) = i = 1 m - 2 β i u Δ ( ξ i ) , u Δ ( T ) = 0 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-4/MediaObjects/13661_2011_Article_110_Equ6_HTML.gif
        (1.6)

        where ϕ p (s) = |s| p-2 s,p > 1,hC ld ((0,T), (0, +∞)), 0 < ξ 1 < ξ 2 < < ξ m - 2 < T T http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-4/MediaObjects/13661_2011_Article_110_IEq5_HTML.gif. δ, β i > 0, i = 1,..., m - 2.

        We will assume throughout

        (S1) h ∈ C ld ((0, T), [0, ∞)) such that 0 T h ( s ) s < http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-4/MediaObjects/13661_2011_Article_110_IEq6_HTML.gif;

        (S2) fC([0, ∞), (0, ∞)), f ≢ 0 on f 0  on  [ 0 , T ] T http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-4/MediaObjects/13661_2011_Article_110_IEq7_HTML.gif;

        (S3) By ϕ q we denote the inverse to ϕ p , where 1 p + 1 q = 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-4/MediaObjects/13661_2011_Article_110_IEq8_HTML.gif;

        (S4) By t ∈ [a, b] we mean that t [ a , b ] T http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-4/MediaObjects/13661_2011_Article_110_IEq9_HTML.gif, where 0 ≤ abT.

        2 Preliminaries

        In this section, we will give some background materials on time scales.

        Definition 2.1. [7, 22] For t < sup T http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-4/MediaObjects/13661_2011_Article_110_IEq10_HTML.gif and t > inf T http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-4/MediaObjects/13661_2011_Article_110_IEq11_HTML.gif, define the forward jump operator σ and the backward jump operator ρ, respectively,
        σ ( t ) = inf { τ T | τ > t } T , ρ ( r ) = sup { τ T | τ < r } T http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-4/MediaObjects/13661_2011_Article_110_Equb_HTML.gif
        for all r , t T http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-4/MediaObjects/13661_2011_Article_110_IEq12_HTML.gif. If σ(t) > t, t is said to be right scattered, and if ρ(r) < r, r is said to be left scattered. If σ(t) = t, t is said to be right dense, and if ρ(r) = r, r is said to be left dense. If T http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-4/MediaObjects/13661_2011_Article_110_IEq1_HTML.gif has a right scattered minimum m, define T κ = T - { m } http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-4/MediaObjects/13661_2011_Article_110_IEq13_HTML.gif; Otherwise set T κ = T http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-4/MediaObjects/13661_2011_Article_110_IEq14_HTML.gif. The backward graininess μ b : T κ 0 + http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-4/MediaObjects/13661_2011_Article_110_IEq15_HTML.gif is defined by
        μ b ( t ) = t - ρ ( t ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-4/MediaObjects/13661_2011_Article_110_Equc_HTML.gif
        If T http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-4/MediaObjects/13661_2011_Article_110_IEq1_HTML.gif has a left scattered maximum M, define T κ = T - { M } http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-4/MediaObjects/13661_2011_Article_110_IEq16_HTML.gif; Otherwise set T κ = T http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-4/MediaObjects/13661_2011_Article_110_IEq17_HTML.gif. The forward graininess μ f : T κ 0 + http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-4/MediaObjects/13661_2011_Article_110_IEq18_HTML.gif is defined by
        μ f ( t ) = σ ( t ) - t . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-4/MediaObjects/13661_2011_Article_110_Equd_HTML.gif
        Definition 2.2. [7, 22] For x : T R http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-4/MediaObjects/13661_2011_Article_110_IEq19_HTML.gif and t T κ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-4/MediaObjects/13661_2011_Article_110_IEq20_HTML.gif, we define the "Δ" derivative of x(t), xΔ(t), to be the number (when it exists), with the property that, for any ε > 0, there is neighborhood U of t such that
        [ x ( σ ( t ) ) - x ( s ) ] - x Δ ( t ) [ σ ( t ) - s ] < ε | σ ( t ) - s | http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-4/MediaObjects/13661_2011_Article_110_Eque_HTML.gif
        for all sU. For x : T R http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-4/MediaObjects/13661_2011_Article_110_IEq19_HTML.gif and t T κ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-4/MediaObjects/13661_2011_Article_110_IEq21_HTML.gif, we define the "∇" derivative of x(t),xΔ (t), to be the number(when it exists), with the property that, for any ε > 0, there is a neighborhood V of t such that
        [ x ( ρ ( t ) ) - x ( s ) ] - x ( t ) [ ρ ( t ) - s ] < ε | ρ ( t ) - s | http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-4/MediaObjects/13661_2011_Article_110_Equf_HTML.gif

        for all sV.

        Definition 2.3. [22] If FΔ (t) = f(t), then we define the "Δ" integral by
        a t f ( s ) Δ s = F ( t ) - F ( a ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-4/MediaObjects/13661_2011_Article_110_Equg_HTML.gif
        If F (t) = f(t), then we define the "∇" integral by
        a t f ( s ) s = F ( t ) - F ( a ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-4/MediaObjects/13661_2011_Article_110_Equh_HTML.gif

        Lemma 2.1. [23]The following formulas hold:

        (i) ( a t f ( t ) Δ s ) Δ = f ( t ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-4/MediaObjects/13661_2011_Article_110_IEq22_HTML.gif,

        (ii) ( a t f ( t ) Δ s ) = f ( ρ ( t ) ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-4/MediaObjects/13661_2011_Article_110_IEq23_HTML.gif,

        (iii) ( a t f ( t ) s ) Δ = f ( σ ( t ) ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-4/MediaObjects/13661_2011_Article_110_IEq24_HTML.gif,

        (iv) ( a t f ( t ) s ) = f ( t ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-4/MediaObjects/13661_2011_Article_110_IEq25_HTML.gif.

        Lemma 2.2. [7, Theorem 1.75 in p. 28] If fC rd and t T κ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-4/MediaObjects/13661_2011_Article_110_IEq20_HTML.gif, then
        t σ ( t ) f ( τ ) Δ τ = μ f ( t ) f ( t ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-4/MediaObjects/13661_2011_Article_110_Equi_HTML.gif

        According to [23, Theorem 1.30 in p. 9], we have the following lemma, which can be proved easily. Here, we omit it.

        Lemma 2.3. Let a , b T http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-4/MediaObjects/13661_2011_Article_110_IEq26_HTML.gifand fC ld .

        (i) If T = http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-4/MediaObjects/13661_2011_Article_110_IEq27_HTML.gif, then
        a b f ( t ) t = a b f ( t ) d t , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-4/MediaObjects/13661_2011_Article_110_Equj_HTML.gif

        where the integral on the right is the usual Riemann integral from calculus.

        (ii) If [a, b] consists of only isolated points, then
        a b f ( t ) t = t ( a , b ] μ b ( t ) f ( t ) , i f a < b , 0 , i f a = b , - t ( a , b ] μ b ( t ) f ( t ) , i f a > b . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-4/MediaObjects/13661_2011_Article_110_Equk_HTML.gif
        (iii) If T = h = { h k : k } http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-4/MediaObjects/13661_2011_Article_110_IEq28_HTML.gif, where h > 0, then
        a b f ( t ) t = k = a h + 1 b h f ( k h ) h , i f a < b , 0 , i f a = b , - k = b h + 1 a h f ( k h ) h , i f a > b . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-4/MediaObjects/13661_2011_Article_110_Equl_HTML.gif
        (iv) If T = http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-4/MediaObjects/13661_2011_Article_110_IEq29_HTML.gif, then
        a b f ( t ) t = t = a + 1 b f ( t ) , i f a < b , 0 , i f a = b , - t = b + 1 a f ( t ) , i f a > b . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-4/MediaObjects/13661_2011_Article_110_Equm_HTML.gif

        In what follows, we list the fixed point theorems that will be used in this article.

        Theorem 2.4. [24]Let E be a Banach space and PE be a cone. Suppose Ω1, Ω2E open and bounded, 0 Ω 1 Ω ¯ 1 Ω 2 Ω ¯ 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-4/MediaObjects/13661_2011_Article_110_IEq30_HTML.gif. Assume A : ( Ω ¯ 2 \ Ω 1 ) P P http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-4/MediaObjects/13661_2011_Article_110_IEq31_HTML.gifis completely continuous. If one of the following conditions holds

        (i) ∥Ax∥ ≤ ∥x∥, ∀x ∈ ∂Ω1P, ∥Ax∥ ≥ ∥x∥, ∀x ∈ ∂Ω2P;

        (ii) ∥Ax∥ ≥ ∥x∥, ∀x ∈ ∂Ω1P, ∥Ax∥ ≤ ∥x∥, ∀x ∈ ∂Ω2P.

        Then, A has a fixed point in ( Ω ¯ 2 \ Ω 1 ) P http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-4/MediaObjects/13661_2011_Article_110_IEq32_HTML.gif.

        Theorem 2.5. [25]Let P be a cone in the real Banach space E. Set
        P ( γ , r ) = { u P , γ ( u ) < r } . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-4/MediaObjects/13661_2011_Article_110_Equn_HTML.gif
        If α and γ are increasing, nonnegative continuous functionals on P, let θ be a nonnegative continuous functional on P with θ(0) = 0 such that for some positive constants r, M,
        γ ( u ) θ ( u ) α ( u ) a n d u M γ ( u ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-4/MediaObjects/13661_2011_Article_110_Equo_HTML.gif
        for all u P ( γ , r ) ¯ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-4/MediaObjects/13661_2011_Article_110_IEq33_HTML.gif. Further, suppose there exists positive numbers a < b < r such that
        θ ( λ u ) λ θ ( u ) f o r a l l 0 λ 1 , u P ( θ , b ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-4/MediaObjects/13661_2011_Article_110_Equp_HTML.gif

        If A : P ( γ , r ) ¯ P http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-4/MediaObjects/13661_2011_Article_110_IEq34_HTML.gif is completely continuous operator satisfying

        (i) γ(Au) > r for all u ∈ ∂P(γ, r);

        (ii) θ(Au) < b for all u∂P(θ, r);

        (iii) P ( α , b ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-4/MediaObjects/13661_2011_Article_110_IEq35_HTML.gifand α(Au) > a for all u∂P(α, a).

        Then, A has at least two fixed points u 1 and u 2 such that
        a < α ( u 1 ) , w i t h θ ( u 1 ) < b , a n d b < θ ( u 2 ) , w i t h γ ( u 1 ) < r , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-4/MediaObjects/13661_2011_Article_110_Equq_HTML.gif

        Let a, b, c be constants, P r = {uP : ∥u∥ < r}, P(ψ, b, d) = {uP : aψ(u), ∥u∥ ≤ b}.

        Theorem 2.6. [26]Let A : P ¯ c P ¯ c http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-4/MediaObjects/13661_2011_Article_110_IEq36_HTML.gifbe a completely continuous map and ψ be a nonnegative continuous concave functional on P such that for u P ¯ c http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-4/MediaObjects/13661_2011_Article_110_IEq37_HTML.gif, there holds ψ(u) ≤ ∥u∥. Suppose there exist a, b, d with 0 < a < b < dc such that

        (i) { u P ( ψ , b , d ) : ψ ( u ) > b } http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-4/MediaObjects/13661_2011_Article_110_IEq38_HTML.gifand ψ(Au) > b for all uP(ψ, b, d);

        (ii) ∥Au∥ < a for all u P ¯ a http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-4/MediaObjects/13661_2011_Article_110_IEq39_HTML.gif;

        (iii) ψ(Au) > b for all uP(ψ, b, d) withAu∥ > d.

        Then, A has at least three fixed points u1,u2, and u3satisfying
        u 1 < a , b < α ( u 2 ) , u 3 > a , a n d u 3 < b . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-4/MediaObjects/13661_2011_Article_110_Equr_HTML.gif
        Let the Banach space E = C ld [0, T] be endowed with the norm ∥u∥ = supt ∈ [0,T]u(t), and cone PE is defined as
        P = { u E , u ( t ) 0  for  t [ 0 , T ]  and  u Δ ( t ) 0  for  t ( 0 , T ) , u Δ ( T ) = 0 } . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-4/MediaObjects/13661_2011_Article_110_Equs_HTML.gif
        It is obvious that ∥u∥ = u(T) for uP. Define A : PE as
        ( A u ) ( t ) = 0 t ϕ q s T h ( τ ) f ( τ , u ( τ ) ) τ Δ s + δ ϕ q 0 T h ( s ) f ( s , u ( s ) ) s + i = 1 m - 2 β i ϕ q ξ i T h ( s ) f ( s , u ( s ) ) s http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-4/MediaObjects/13661_2011_Article_110_Equt_HTML.gif

        for t ∈ [0, T].

        In what follows, we give the main lemmas which are important for getting the main results.

        Lemma 2.7. A : PP is completely continuous.

        Proof. First, we try to prove that A : PP.
        ( A u ) Δ ( t ) = ϕ q t T h ( s ) f ( s , u ( s ) ) s . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-4/MediaObjects/13661_2011_Article_110_Equu_HTML.gif

        Thus, (Au)Δ (T) = 0 and by Lemma 2.1 we have (Au)Δ∇ (t) = -h(t)f(t, u(t)) ≤ 0 for t ∈ (0, T). Consequently, A : PP.

        By standard argument we can prove that A is completely continuous. For more details, see [27]. The proof is complete.

        Lemma 2.8. For uP, there holds u ( t ) t T u f o r t [ 0 , T ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-4/MediaObjects/13661_2011_Article_110_IEq40_HTML.giffor t ∈ [0,T].

        Proof. For uP, we have uΔ∇ (t) ≤ 0, it follows that uΔ (t) is non-increasing. Therefore, for 0 < t < T,
        u ( t ) - u ( 0 ) = 0 t u Δ ( s ) Δ s t u Δ ( t ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-4/MediaObjects/13661_2011_Article_110_Equ7_HTML.gif
        (2.1)
        and
        u ( T ) - u ( t ) = t T u Δ ( s ) Δ s ( T - t ) u Δ ( t ) , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-4/MediaObjects/13661_2011_Article_110_Equ8_HTML.gif
        (2.2)
        thus
        u ( T ) - u ( 0 ) T u Δ ( t ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-4/MediaObjects/13661_2011_Article_110_Equ9_HTML.gif
        (2.3)
        Combining (2.1) and (2.3) we have
        T ( u ( t ) - u ( 0 ) ) T t u Δ ( t ) t ( u ( T ) - u ( 0 ) ) , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-4/MediaObjects/13661_2011_Article_110_Equv_HTML.gif
        as u(0) ≥ 0, it is immediate that
        u ( t ) t u ( T ) + ( T - t ) u ( 0 ) T t T u ( T ) = t T u . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-4/MediaObjects/13661_2011_Article_110_Equw_HTML.gif

        The proof is complete.

        3 Existence of at least one positive solution

        First, we give some notations. Set
        Λ = δ + i = 1 m - 2 β i + T ϕ q 0 T h ( s ) s , B = ξ 1 T δ ϕ q 0 T h ( s ) s + i = 1 m - 2 β i ϕ q ξ i T h ( s ) s + 0 T ϕ q s T h ( τ ) τ Δ s . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-4/MediaObjects/13661_2011_Article_110_Equx_HTML.gif

        Theorem 3.1. Assume in addition to (S 1) and (S 2), the following conditions are satisfied, there exists 0 < r < ξ 1 ρ T < ρ < http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-4/MediaObjects/13661_2011_Article_110_IEq41_HTML.gif such that

        (H1) f ( t , u ) ϕ p ( u Λ ) , f o r t [ 0 , T ] , u [ 0 , r ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-4/MediaObjects/13661_2011_Article_110_IEq42_HTML.gif;

        (H2) f ( t , u ) ϕ p u B , f o r t [ ξ 1 , T ] , u [ ξ 1 ρ T , ρ ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-4/MediaObjects/13661_2011_Article_110_IEq43_HTML.gif.

        Then, BVP (1.6) has at least one positive solution.

        Proof. Cone P is defined as above. By Lemma 2.7 we know that A : PP is completely continuous. Set Ω r = {uE, ∥u∥ < r}. In view of (H1), for u Ω r P,
        A u = ( A u ) ( T ) = δ ϕ q ( 0 T h ( s ) f ( s , u ( s ) ) s ) + i = 1 m 2 β i ϕ q ( ξ i T h ( s ) f ( s , u ( s ) ) s ) + 0 T ϕ q ( s T h ( τ ) f ( τ , u ( τ ) ) τ ) Δ s ( δ + i = 1 m 2 β i + T ) ϕ q ( 0 T ϕ p ( u ( s ) Λ ) h ( s ) s ) u Λ ( δ + i = 1 m 2 β i + T ) ϕ q ( 0 T h ( s ) s ) u , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-4/MediaObjects/13661_2011_Article_110_Equy_HTML.gif

        which means that for u ∈ ∂Ω r P, ||Au|| ≤ ||u||.

        On the other hand, for uP, in view of Lemma 2.8, there holds u ( t ) ξ 1 T u http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-4/MediaObjects/13661_2011_Article_110_IEq44_HTML.gif, for t ∈ [ξ1, T]. Denote Ω ρ = {uE, ∥u∥ < ρ}. Then for u ∈ ∂Ω ρ P, considering (H2), we have
        A u = ( A u ) ( T ) = δ ϕ q 0 T h ( s ) f ( s , u ( s ) ) s + i = 1 m - 2 β i ϕ q ξ i T h ( s ) f ( s , u ( s ) ) s + 0 T ϕ q s T h ( τ ) f ( τ , u ( τ ) ) τ Δ s δ ϕ q 0 T ϕ p u ( s ) B h ( s ) s + i = 1 m - 2 β i ϕ q ξ i T ϕ p u ( s ) B h ( s ) s + 0 T ϕ q s T ϕ p u ( τ ) B h ( τ ) τ Δ s ξ 1 u T B δ ϕ q 0 T h ( s ) s + i = 1 m - 2 β i ϕ q ξ i T h ( s ) s + 0 T ϕ q s T h ( τ ) τ Δ s = u . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-4/MediaObjects/13661_2011_Article_110_Equz_HTML.gif

        which implies that for u Ω ρ P, ∥Au∥ ≥ ∥u∥ Therefore, the immediate result of Theorem 2.4 is that A has at least one fixed point u ∈ (Ω ρ r ) ∩ P. Also, it is obvious that the fixed point of A in cone P is equivalent to the positive solution of BVP (1.6), this yields that BVP (1.6) has at least one positive solution u satisfies r ≤ ∥u∥ ≤ ρ. The proof is complete.

        Here is an example.

        Example 3.2. Let T = 1 , 1 = k = 0 [ 2 k , 2 k + 1 ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-4/MediaObjects/13661_2011_Article_110_IEq45_HTML.gif. Consider the following four point BVP on time scale 1 , 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-4/MediaObjects/13661_2011_Article_110_IEq46_HTML.gif.
        x Δ ( t ) + f ( t , u ( t ) ) = 0 , t [ 0 , T ] T , x ( 0 ) - 2 x Δ ( 0 ) = x Δ ( 2 ) + x Δ ( 3 ) , x Δ ( 4 ) = 0 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-4/MediaObjects/13661_2011_Article_110_Equ10_HTML.gif
        (3.1)
        where
        f ( t , u ) = t u 128 , 0 u 100 , 39 t 512 ( u - 100 ) + 25 t 32 , 100 u 500 , t u 16 , u 500 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-4/MediaObjects/13661_2011_Article_110_Equaa_HTML.gif
        and h(t) = 1, T = 4, ξ1 = 2, ξ2 = 3, δ = 2, β1 = β2 = 1,p = q = 2. In what follows, we try to calculate Λ, B. By Lemmas 2.2 and 2.3, we have
        Λ = δ + i = 1 m - 2 β i + T ϕ q 0 T h ( s ) s = ( 2 + 1 + 1 + 4 ) 0 4 s = 8 × 0 1 d s + 2 3 d s + 1 2 s + 3 4 s = 8 × 0 1 d s + 2 3 d s + ν ( 2 ) × 1 + ν ( 4 ) × 1 = 8 × ( 1 + 1 + 1 + 1 ) = 32 . B = ξ 1 T δ ϕ q 0 T h ( s ) s + i = 1 m - 2 β i ϕ q ξ i T h ( s ) s + 0 T ϕ q s T h ( τ ) τ Δ s = 2 4 2 0 4 s + 2 4 s + 3 4 s + 0 4 s 4 τ Δ s = 1 2 8 + 2 + 1 + 0 1 s 4 τ Δ s + 1 2 2 4 τ Δ s + 2 3 s 4 τ Δ s + 3 4 s 4 τ Δ s = 1 2 11 + 0 1 s 4 τ Δ s + 1 2 s 4 τ Δ s + 2 3 s 4 τ Δ s + 3 4 s 4 τ Δ s , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-4/MediaObjects/13661_2011_Article_110_Equab_HTML.gif
        where
        0 1 s 4 τ Δ s = 0 1 s 1 τ + 1 4 τ Δ s = 0 1 s 1 d τ d s + 0 1 1 2 τ + 2 3 τ + 3 4 τ Δ s = 0 1 s 1 d τ + 2 3 d τ d s + 0 1 1 2 τ + 3 4 τ Δ s = 1 2 + 1 + 1 + 1 = 7 2 , 2 3 s 4 τ Δ s = 2 3 s 3 τ + 3 4 τ Δ s = 2 3 s 3 d τ d s + 2 3 3 4 τ Δ s = 1 2 + 1 = 3 2 , 1 2 s 4 τ Δ s = σ ( 1 ) × 1 4 τ = 3 , 3 4 s 4 τ Δ s = σ ( 3 ) × 3 4 τ = 1 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-4/MediaObjects/13661_2011_Article_110_Equac_HTML.gif

        Thus, B = 1 2 11 + 7 2 + 3 2 + 3 + 1 = 10 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-4/MediaObjects/13661_2011_Article_110_IEq47_HTML.gif. Let r = 100 < 2 4 ρ < ρ = 1000 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-4/MediaObjects/13661_2011_Article_110_IEq48_HTML.gif. Then, we have

        (i) f ( t , u ) 4 u 128 = u 32 = ϕ p ( u Λ ) , f o r t [ 0 , 4 ] , u [ 0 , 100 ] ; http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-4/MediaObjects/13661_2011_Article_110_IEq49_HTML.gif

        (ii) f ( t , u ) 2 u 16 = u 8 > ϕ p ( u B ) , f o r t [ 2 , 4 ] , u [ 500 , 1000 ] . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-4/MediaObjects/13661_2011_Article_110_IEq50_HTML.gif

        Thus, if all the conditions in Theorem 3.1 satisfied, then BVP (3.1) has at least one positive solution lies between 100 and 1000.

        4 Existence of at least two positive solutions

        In this section, we will apply fixed point Theorem 2.5 to prove the existence of at least two positive solutions to the nonlinear BVP (1.6).

        Fix η T http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-4/MediaObjects/13661_2011_Article_110_IEq51_HTML.gif such that
        0 < ξ m - 2 η < T , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-4/MediaObjects/13661_2011_Article_110_Equad_HTML.gif
        and define the increasing, nonnegative, continuous functionals γ, θ,α on P by
        γ ( u ) = min t [ ξ 1 , η ] u ( t ) = u ( ξ 1 ) , θ ( u ) = max t [ 0 , ξ m - 2 ] u ( t ) = u ( ξ m - 2 ) , α ( u ) = min t [ η , T ] u ( t ) = u ( η ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-4/MediaObjects/13661_2011_Article_110_Equae_HTML.gif
        We can see that, for u ∈ P, there holds
        γ ( u ) θ ( u ) α ( u ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-4/MediaObjects/13661_2011_Article_110_Equaf_HTML.gif
        In addition, Lemma 2.8 implies that γ ( u ) = u ( ξ 1 ) ξ 1 T | | u | | http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-4/MediaObjects/13661_2011_Article_110_IEq52_HTML.gif which means that
        | | u | | T ξ 1 γ ( u ) for u P . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-4/MediaObjects/13661_2011_Article_110_Equag_HTML.gif
        We also see that
        θ ( λ u ) = λ θ ( u ) for λ [ 0 , 1 ] , u P ( θ , b ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-4/MediaObjects/13661_2011_Article_110_Equah_HTML.gif
        For convenience, we give some notations,
        K = δ + i = 1 m - 2 β i + ξ m - 2 ϕ q 0 T h ( s ) s , M = δ ϕ q 0 T h ( s ) s + i = 1 m - 2 β i ϕ q ξ i T h ( s ) s + 0 ξ 1 ϕ q s T h ( τ ) τ Δ s , L = δ ϕ q 0 T h ( s ) s + i = 1 m - 2 β i ϕ q ξ i T h ( s ) s + 0 η ϕ q s T h ( τ ) τ Δ s . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-4/MediaObjects/13661_2011_Article_110_Equai_HTML.gif

        Theorem 4.1. Assume in addition to (S 1), (S 2) there exist positive constants a < T η a < b < T ξ m - 2 b < c http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-4/MediaObjects/13661_2011_Article_110_IEq53_HTML.gifsuch that the following conditions hold

        (H3) f(t, u) > ϕ p (c/M) for t ∈ [ξ1,T] u ∈ [c,Tc/ξ1];

        (H4) f(t, u) < ϕ p (b/K) for t ∈ [0,ξm-2], u ∈ [b,Tb/ξm-2];

        (H5) f(t, u) > ϕ p (a/L) for t ∈ [η,T], u ∈ [a,Ta/η].

        Then BVP (1.6) has at least two positive solutions u1and u2such that
        α ( u 1 ) > a , w i t h θ ( u 1 ) < b , a n d b < θ ( u 2 ) , w i t h γ ( u 2 ) < c . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-4/MediaObjects/13661_2011_Article_110_Equ11_HTML.gif
        (4.1)

        Proof. From Lemma 2.7 we know that A : P(γ, c) → P is completely continuous. In what follows, we will prove the result step by step.

        Step one: To verify (i) of theorem 2.5 holds.

        We choose u ∈ ∂P(γ,c), then γ ( u ) = min t [ ξ 1 , η ] u ( t ) = u ( ξ 1 ) = c http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-4/MediaObjects/13661_2011_Article_110_IEq54_HTML.gif. This implies that u(t) ≥ c for t ∈ [ξ1,T], considering that | | u | | T ξ 1 γ ( u ) = T ξ 1 c http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-4/MediaObjects/13661_2011_Article_110_IEq55_HTML.gif, we have
        c u ( t ) T ξ 1 c for t [ ξ 1 , T ] . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-4/MediaObjects/13661_2011_Article_110_Equaj_HTML.gif
        As a consequence of (H3),
        f ( t , u ( t ) ) > ϕ p ( c M ) for t [ ξ 1 , T ] . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-4/MediaObjects/13661_2011_Article_110_Equak_HTML.gif
        Since AuP, we have
        γ ( A u ) = ( A u ) ( ξ 1 ) = δ ϕ q 0 T h ( s ) f ( s , u ( s ) ) s + i = 1 m - 2 β i ϕ q ξ i T h ( s ) f ( s , u ( s ) ) s + 0 ξ 1 ϕ q s T h ( τ ) f ( τ , u ( τ ) ) τ Δ s > c M δ ϕ q 0 T h ( s ) s + i = 1 m - 2 β i ϕ q ξ i T h ( s ) s + 0 ξ 1 ϕ q s T h ( τ ) τ Δ s = c . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-4/MediaObjects/13661_2011_Article_110_Equal_HTML.gif

        Thus, (i) of Theorem 2.5 is satisfied.

        Step two: To verify (ii) of Theorem 2.5 holds.

        Let u∂P(θ,b), then θ ( u ) = max t [ 0 , ξ m - 2 ] = u ( ξ m - 2 ) = b http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-4/MediaObjects/13661_2011_Article_110_IEq56_HTML.gif, this implies that 0 ≤ u(t) ≤ b, t ∈ [0,ξm-2] and since uP, we have ∥u∥ = u(T), note that | | u | | T ξ m - 2 u ( ξ m - 2 ) = T ξ m - 2 θ ( u ) = T ξ m - 2 b http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-4/MediaObjects/13661_2011_Article_110_IEq57_HTML.gif. So,
        0 u ( t ) T ξ m - 2 b for t [ 0 , T ] . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-4/MediaObjects/13661_2011_Article_110_Equam_HTML.gif
        From (H4) we know that f ( t , u ( t ) ) < ϕ p ( b K ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-4/MediaObjects/13661_2011_Article_110_IEq58_HTML.gif for t ∈ [0, ξm-2] and so
        θ ( A u ) = ( A u ) ( ξ m - 2 ) = δ ϕ q 0 T h ( s ) f ( s , u ( s ) ) s + i = 1 m - 2 β i ϕ q ξ i T h ( s ) f ( s , u ( s ) ) s + 0 ξ m - 2 ϕ q s T h ( τ ) f τ , u ( τ ) τ Δ s < b K δ ϕ q 0 T h ( s ) s + i = 1 m - 2 β i ϕ q ξ T h ( s ) s + 0 ξ m - 2 ϕ q s T h ( τ ) τ Δ s < b K δ + i = 1 m - 2 β i + ξ m - 2 ϕ q 0 T h ( s ) s = b . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-4/MediaObjects/13661_2011_Article_110_Equan_HTML.gif

        Thus, (ii) of Theorem 2.5 holds.

        Step three: To verify (iii) of Theorem 2.5 holds.

        Choose u 0 ( t ) = a 2 , t [ 0 , T ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-4/MediaObjects/13661_2011_Article_110_IEq59_HTML.gif, obviously, u0(t) ∈ P(α, a) and α ( u 0 ) = a 2 < a http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-4/MediaObjects/13661_2011_Article_110_IEq60_HTML.gif, thus P ( α , a ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-4/MediaObjects/13661_2011_Article_110_IEq61_HTML.gif.

        Now, let u∂P(α, a), then, α(u) = mint∈[η,T]u(t) = u(η) = a. Recalling that | | u | | T η u ( η ) = T η α ( u ) = T η a http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-4/MediaObjects/13661_2011_Article_110_IEq62_HTML.gif. Thus, we have
        a u ( t ) T η a for t [ η , T ] . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-4/MediaObjects/13661_2011_Article_110_Equao_HTML.gif
        From assumption (H5) we know that
        f ( t , u ( t ) ) > ϕ p a L for t [ η , T ] , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-4/MediaObjects/13661_2011_Article_110_Equap_HTML.gif
        and so
        α ( A u ) = ( A u ) ( η ) = δ ϕ q 0 T h ( s ) f ( s , u ( s ) ) s + i = 1 m - 2 β i ϕ q ξ i T h ( s ) f ( s , u ( s ) ) s + 0 η ϕ q s T h ( τ ) f ( τ , u ( τ ) ) τ Δ s > a L δ ϕ q 0 T h ( s ) s + i = 1 m - 2 β i ϕ q ξ i T h ( s ) s + 0 η ϕ q s T h ( τ ) τ Δ s = a . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-4/MediaObjects/13661_2011_Article_110_Equaq_HTML.gif

        Therefore, all the conditions of Theorem 2.5 are satisfied, thus A has at least two fixed points in P(γ,c), which implies that BVP (1.6) has at least two positive solutions u1,u2 which satisfies (4.1). The proof is complete.

        Example 4.2. Let T = { 2 n , n } { 0 } http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-4/MediaObjects/13661_2011_Article_110_IEq63_HTML.gif. Consider the following four point boundary value problem on time scale T http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-4/MediaObjects/13661_2011_Article_110_IEq1_HTML.gif.
        ( ϕ p ( x Δ ) ) ( t ) + t f ( t , u ( t ) ) = 0 , t [ 0 , 8 ] T , x ( 0 ) - x Δ ( 0 ) = x Δ ( 1 ) + 2 x Δ ( 2 ) , x Δ ( 8 ) = 0 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-4/MediaObjects/13661_2011_Article_110_Equ12_HTML.gif
        (4.2)
        where
        f ( t , u ) = | sin t | + u 10 5 , 0 u 9 . 3 × 10 6 , | sin t | + 93 , 9 . 3 × 10 6 u 4 × 10 8 , | sin t | + 247 u 6 × 10 8 - 215 3 , u 4 × 10 8 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-4/MediaObjects/13661_2011_Article_110_Equar_HTML.gif
        and h(t) = t, T = 8, ξ1 = 1, ξ2 = 2, δ = 1, β1 = 1, β2 = 2,p = 3/2, q = 3. In what follows, we try to calculate K, M, L. By Lemmas 2.2 and 2.3, we have
        K = δ + i = 1 m - 2 β i + ξ m - 2 ϕ q 0 T h ( s ) s = ( 1 + 1 + 2 + 2 ) ϕ q 0 8 s s = 6 × 0 1 s s + 1 2 s s + 2 4 s s + 4 8 s s 2 = 6 × ( ν ( 1 ) × 1 + ν ( 2 ) × 2 + ν ( 4 ) × 4 + ν ( 8 ) × 8 ) 2 = 6 × ( 1 + 2 + 8 + 32 ) 2 = 6 × 1849 = 11094 . M = δ ϕ q 0 T h ( s ) s + i = 1 m - 2 β i ϕ q ξ i T h ( s ) s + 0 ξ i ϕ q s T h ( τ ) τ Δ s = 0 8 s s 2 + 1 8 s s 2 + 2 2 8 s s 2 + 0 1 ϕ q s 8 τ τ Δ s = ( 1 + 2 + 8 + 32 ) 2 + ( 2 + 8 + 32 ) 2 + 2 × ( 8 + 32 ) 2 + 0 8 s s 2 = 2 * ( 1 + 2 + 8 + 32 ) 2 + ( 2 + 8 + 32 ) 2 + 2 × ( 8 + 32 ) 2 = 8662 . L = δ ϕ q 0 T h ( s ) s + i = 1 m - 2 β i ϕ q ξ i T h ( s ) s + 0 η ϕ q s T h ( τ ) τ Δ s = 0 8 s s 2 + 1 8 s s 2 + 2 2 8 s s 2 + 0 4 s 8 h ( τ ) τ 2 Δ s = ( 1 + 2 + 8 + 32 ) 2 + ( 2 + 8 + 32 ) 2 + 2 × ( 8 + 32 ) 2 + 0 1 s 8 h ( τ ) τ 2 Δ s + 1 2 s 8 h ( τ ) τ 2 Δ s + 2 4 s 8 h ( τ ) τ 2 Δ s = 6813 + μ ( 0 ) × 0 8 s s 2 + μ ( 1 ) × 0 8 s s 2 + μ ( 2 ) × 0 8 s s 2 = 6813 + ( 1 + 2 + 8 + 32 ) 2 + ( 2 + 8 + 32 ) 2 + 2 × ( 8 + 32 ) 2 = 13626 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-4/MediaObjects/13661_2011_Article_110_Equas_HTML.gif

        Let a = 106, b = 108, c = 109, then we have

        (i) f ( t , u ) 340 > 10 9 8662 1 / 2 = ϕ p c M http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-4/MediaObjects/13661_2011_Article_110_IEq64_HTML.gif, for t ∈ [1, 8], u ∈ [109, 8 × 109];

        (ii) f ( t , u ) 94 < 10 8 11094 1 / 2 = ϕ p b K http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-4/MediaObjects/13661_2011_Article_110_IEq65_HTML.gif, for t ∈ [0, 2], u ∈ [108, 4 × 108];

        (iii) f ( t , u ) > 9 > 10 6 13326 1 / 2 = ϕ p a L http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-4/MediaObjects/13661_2011_Article_110_IEq66_HTML.gif, for t ∈ [4, 8], u ∈ [106, 2 × 106].

        Thus, if all the conditions in Theorem 4.1 are satisfied, then BVP (4.2) has at least two positive solutions satisfying (4.1).

        5 Existence of at least three positive solutions

        Let ψ ( u ) = min t [ ξ 1 , T ] u ( t ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-4/MediaObjects/13661_2011_Article_110_IEq67_HTML.gif, then 0 < ψ(u) ≤ ∥u∥. Denote
        D = δ ϕ q 0 T h ( s ) s + i = 1 m - 2 β i ϕ q ξ i T h ( s ) s + 0 ξ 1 ϕ q s T h ( τ ) τ Δ s , R = δ + i = 1 m - 2 β i + T ϕ q 0 T h ( s ) s . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-4/MediaObjects/13661_2011_Article_110_Equat_HTML.gif

        In this section, we will use fixed point Theorem 2.6 to get the existence of at least three positive solutions.

        Theorem 5.1. Assume that there exists positive number d, ν, g satisfying d < ν < min ξ 1 T , D R g < g http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-4/MediaObjects/13661_2011_Article_110_IEq68_HTML.gif, such that the following conditions hold.

        (H6) f(t, u) < ϕ p (d/R), t ∈ [0,T],u ∈ [0,d];

        (H7) f(t, u) > ϕ p (ν/D), t ∈ [ξ1, T], u ∈ [ν, Tυ/ξ1];

        (H8) f(t, u) ≤ ϕ p (g/R), t ∈ [0,T],u ∈ [0,g],

        then BVP (1.6) has at least three positive solutions u1, u2, u3satisfying
        | | u 1 | | < d , ψ ( u 2 ) > ν , | | u 3 | | > d , w i t h ψ ( u 3 ) < ν . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-4/MediaObjects/13661_2011_Article_110_Equ13_HTML.gif
        (5.1)

        Proof. From Lemma 2.8 we know that A : PP is completely continuous. Now we only need to show that all the conditions in Theorem 2.6 are satisfied.

        For u P ̄ g , | | u | | g http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-4/MediaObjects/13661_2011_Article_110_IEq69_HTML.gif. By (H8), one has
        | | A u | | = ( A u ) ( T ) = δ ϕ q ( 0 T h ( s ) f ( s , u ( s ) ) s ) + i = 1 m 2 β i ϕ q ( ξ i T h ( s ) f ( s , u ( s ) ) s ) + 0 T ϕ q ( s T h ( τ ) f ( τ , u ( τ ) ) τ ) Δ s g R ( δ + i = 1 m 2 β i + T ) ϕ q ( 0 T h ( s ) s ) = g . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-4/MediaObjects/13661_2011_Article_110_Equau_HTML.gif

        Thus, A : P ̄ g P ̄ g http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-4/MediaObjects/13661_2011_Article_110_IEq70_HTML.gif. Similarly, by (H6), we can prove (ii) of Theorem 2.6 is satisfied.

        In what follows, we try to prove that (i) of theorem 2.6 holds. Choose u 1 ( t ) = T ξ 1 ν , t [ 0 , T ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-4/MediaObjects/13661_2011_Article_110_IEq71_HTML.gif, obviously, ψ(u1) > ν, thus { u P ( ψ , ν , T ν / ξ 1 ) : ψ ( u ) > ν } http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-4/MediaObjects/13661_2011_Article_110_IEq72_HTML.gif. For uP(ψ,ν,Tν/ξ1),
        ψ ( A u ) = ( A u ) ( ξ 1 ) = δ ϕ q 0 T h ( s ) f ( s , u ( s ) ) s + i = 1 m - 2 β i ϕ q ξ i T h ( s ) f ( s , u ( s ) ) s + 0 ξ 1 ϕ q s T h ( τ ) f ( τ , u ( τ ) ) τ Δ s > ν D δ ϕ q 0 T h ( s ) s + i = 1 m - 2 β i ϕ q ξ i T h ( s ) s + 0 ξ 1 ϕ q s T h ( τ ) τ Δ s = ν . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-4/MediaObjects/13661_2011_Article_110_Equav_HTML.gif

        It remains to prove (iii) of Theorem 2.6 holds. For uP(ψ, ν, Tυ/ξ1), with ∥Au∥ > /ξ1, in view of Lemma 2.8, there holds ψ ( A u ) - ( A u ) ( ξ 1 ) ξ 1 T | | A u | | > ν http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-4/MediaObjects/13661_2011_Article_110_IEq73_HTML.gif, which implies that (iii) of Theorem 2.6 holds.

        Therefore, all the conditions in Theorem 2.6 are satisfied. Thus, BVP (1.6) has at least three positive solutions satisfying (5.1). The proof is complete.

        Example 5.2. Let T = [ 0 , 1 ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-4/MediaObjects/13661_2011_Article_110_IEq74_HTML.gif. Consider the following four point boundary value problem on time scale T http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-4/MediaObjects/13661_2011_Article_110_IEq1_HTML.gif.
        ( ϕ p ( x Δ ) ) ( t ) + e t ( t , u ( t ) ) = 0 , t [ 0 , T ] T , x ( 0 ) - 3 x Δ ( 0 ) = 2 x Δ ( 1 / 2 ) + 3 x Δ ( 1 ) , x Δ ( 8 ) = 0 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-4/MediaObjects/13661_2011_Article_110_Equ14_HTML.gif
        (5.2)
        where
        f ( t , u ) = t 20 + u 2 840 3 , 0 u 126 , t 20 + 18 . 9 3 , u 126 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-4/MediaObjects/13661_2011_Article_110_Equaw_HTML.gif
        and h(t) = e t , T = 2, ξ1 = 1/2, ξ2 = 1, δ = 3, β1 = 2, β2 = 3, p = 4, q = 4/3. In what follows, we try to calculate D, R. By Lemmas 2.2 and 2.3, we have
        D = δ ϕ q 0 T h ( s ) s + i = 1 m - 2 β i ϕ q ξ i T h ( s ) s + 0 ξ 1 ϕ q s T h ( τ ) τ Δ s = 3 0 2 e s s 1 / 3 + 2 1 / 2 2 e s s 1 / 3 + 3 1 2 e s s 1 / 3 + s 1 / 2 s 2 e τ τ 1 / 3 Δ s = 3 0 1 e s d s + 1 2 e s s 1 / 3 + 2 1 / 2 1 e s d s + 1 2 e s s 1 / 3 + 3 1 2 e s s 1 / 3 + 0 1 / 2 s 1 e τ d τ + 1 2 e τ τ 1 / 2 Δ s = 3 ( e + e 2 - 1 ) 1 / 3 + 2 ( e + e 2 - e 1 / 2 ) 1 / 3 + 3 e 2 / 3 + 3 4 ( e + e 2 - 1 ) 4 / 3 - 3 4 ( e + e 2 - e 1 / 2 ) 4 / 3 17 . 5216 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-4/MediaObjects/13661_2011_Article_110_Equax_HTML.gif
        R = δ + i = 1 m - 2 β i + T ϕ q 0 T h ( s ) s = ( 3 + 2 + 3 + 2 ) 0 2 e s s 1 / 3 = 10 ( e + e 2 - 1 ) 1 / 3 = 20 . 8832 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-4/MediaObjects/13661_2011_Article_110_Equay_HTML.gif

        Let d = 40, ν = 50, g = 400, then we have

        (i) f(t, u) < 7.027 = (40/20.8832)3 = ϕ p (d/R), for t ∈ [0, 2], u ∈ [0, 40];

        (ii) f(t, u) > 23.2375 = (50/17.5216)3 = ϕ p (ν/D), for t ∈ [1/2, 2], u ∈ [50, 200];

        (iii) f(t, u) < 7027.305 = (400/20.8832)3 = ϕ p (g/R), for t ∈ [0, 2], u ∈ [0, 400].

        Thus, if all the conditions in Theorem 5.1 are satisfied, then BVP (5.2) has at least three positive solutions satisfying (5.1).

        Declarations

        Acknowledgements

        The authors were very grateful to the anonymous referee whose careful reading of the manuscript and valuable comments enhanced presentation of the manuscript. The study was supported by Pre-research project and Excellent Teachers project of the Fundamental Research Funds for the Central Universities (2011YYL079, 2011YXL047).

        Authors’ Affiliations

        (1)
        School of Science, China University of Geosciences
        (2)
        Department of Mathematics, Beijing Institute of Technology

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