Existence of symmetric positive solutions for a multipoint boundary value problem with sign-changing nonlinearity on time scales

  • Fatma Tokmak1 and

    Affiliated with

    • Ilkay Yaslan Karaca1Email author

      Affiliated with

      Boundary Value Problems20132013:52

      DOI: 10.1186/1687-2770-2013-52

      Received: 30 September 2012

      Accepted: 12 February 2013

      Published: 14 March 2013

      Abstract

      In this paper, we make use of the four functionals fixed point theorem to verify the existence of at least one symmetric positive solution of a second-order m-point boundary value problem on time scales such that the considered equation admits a nonlinear term f whose sign is allowed to change. The discussed problem involves both an increasing homeomorphism and homomorphism, which generalizes the p-Laplacian operator. An example which supports our theoretical results is also indicated.

      MSC:34B10, 39A10.

      Keywords

      symmetric positive solution fixed-point theorem time scales m-point boundary value problem increasing homeomorphism and homomorphism

      1 Introduction

      The theory of time scales was introduced by Stefan Hilger [1] in his PhD thesis in 1988 in order to unify continuous and discrete analysis. This theory was developed by Agarwal, Bohner, Peterson, Henderson, Avery, etc. [25]. Some preliminary definitions and theorems on time scales can be found in books [3, 4] which are excellent references for calculus of time scales.

      There have been extensive studies on a boundary value problem (BVP) with sign-changing nonlinearity on time scales by using the fixed point theorem on cones. See [6, 7] and references therein. In [8], Feng, Pang and Ge discussed the existence of triple symmetric positive solutions by applying the fixed point theorem of functional type in a cone.

      In [9], Ji, Bai and Ge studied the following singular multipoint boundary value problem:
      http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-52/MediaObjects/13661_2012_Article_345_Equa_HTML.gif

      where 0 < ξ 1 < ξ 2 < < ξ m 2 < 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-52/MediaObjects/13661_2012_Article_345_IEq1_HTML.gif, 0 < η 1 < η 2 < < η m 2 < 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-52/MediaObjects/13661_2012_Article_345_IEq2_HTML.gif, ξ i < η i http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-52/MediaObjects/13661_2012_Article_345_IEq3_HTML.gif, α i > 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-52/MediaObjects/13661_2012_Article_345_IEq4_HTML.gif for i = 1 , 2 , , m 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-52/MediaObjects/13661_2012_Article_345_IEq5_HTML.gif. By using fixed point index theory [10] and the Legget-Williams fixed point theorem [11], sufficient conditions for the existence of countably many positive solutions are established.

      Sun, Wang and Fan [12] studied the nonlocal boundary value problem with p-Laplacian of the form
      http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-52/MediaObjects/13661_2012_Article_345_Equb_HTML.gif

      where 0 t 1 < ξ 1 < ξ 2 < < ξ m 2 < t m http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-52/MediaObjects/13661_2012_Article_345_IEq6_HTML.gif and t 1 < η 1 < η 2 < < η n < t m < + http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-52/MediaObjects/13661_2012_Article_345_IEq7_HTML.gif and ϵ i > 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-52/MediaObjects/13661_2012_Article_345_IEq8_HTML.gif, θ j 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-52/MediaObjects/13661_2012_Article_345_IEq9_HTML.gif for i = 1 , 2 , , m 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-52/MediaObjects/13661_2012_Article_345_IEq5_HTML.gif and j = 1 , 2 , , n http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-52/MediaObjects/13661_2012_Article_345_IEq10_HTML.gif. By using the four functionals fixed point theorem and five functionals fixed point theorem, they obtained the existence criteria of at least one positive solution and three positive solutions.

      Inspired by the mentioned works, in this paper we consider the following m-point boundary value problem with an increasing homeomorphism and homomorphism:
      http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-52/MediaObjects/13661_2012_Article_345_Equ1_HTML.gif
      (1.1)
      http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-52/MediaObjects/13661_2012_Article_345_Equ2_HTML.gif
      (1.2)
      where T http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-52/MediaObjects/13661_2012_Article_345_IEq11_HTML.gif is a time scale, ϕ : R R http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-52/MediaObjects/13661_2012_Article_345_IEq12_HTML.gif is an increasing homeomorphism and homomorphism with ϕ ( 0 ) = 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-52/MediaObjects/13661_2012_Article_345_IEq13_HTML.gif. A projection ϕ : R R http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-52/MediaObjects/13661_2012_Article_345_IEq14_HTML.gif is called an increasing homeomorphism and homomorphism if the following conditions are satisfied:
      1. (i)

        If x y http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-52/MediaObjects/13661_2012_Article_345_IEq15_HTML.gif, then ϕ ( x ) ϕ ( y ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-52/MediaObjects/13661_2012_Article_345_IEq16_HTML.gif for all x , y R http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-52/MediaObjects/13661_2012_Article_345_IEq17_HTML.gif;

         
      2. (ii)

        ϕ is a continuous bijection and its inverse mapping is also continuous;

         
      3. (iii)

        ϕ ( x y ) = ϕ ( x ) ϕ ( y ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-52/MediaObjects/13661_2012_Article_345_IEq18_HTML.gif for all x , y R http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-52/MediaObjects/13661_2012_Article_345_IEq19_HTML.gif, where R = ( , ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-52/MediaObjects/13661_2012_Article_345_IEq20_HTML.gif.

         

      We assume that the following conditions are satisfied:

      1. (H1)

        α i 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-52/MediaObjects/13661_2012_Article_345_IEq21_HTML.gif, i = 1 m 2 α i < 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-52/MediaObjects/13661_2012_Article_345_IEq22_HTML.gif, 0 < ξ 1 < ξ 2 < < ξ m 2 < 1 / 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-52/MediaObjects/13661_2012_Article_345_IEq23_HTML.gif, ξ i + η i = 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-52/MediaObjects/13661_2012_Article_345_IEq24_HTML.gif, i = 1 , 2 , , m 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-52/MediaObjects/13661_2012_Article_345_IEq5_HTML.gif, 1 / 2 T http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-52/MediaObjects/13661_2012_Article_345_IEq25_HTML.gif;

         
      2. (H2)

        f C ( [ 0 , 1 ] T × [ 0 , ) × ( , ) , ( , ) ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-52/MediaObjects/13661_2012_Article_345_IEq26_HTML.gif is symmetric on [ 0 , 1 ] T http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-52/MediaObjects/13661_2012_Article_345_IEq27_HTML.gif (i.e., f ( t , u , v ) = f ( 1 t , u , v ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-52/MediaObjects/13661_2012_Article_345_IEq28_HTML.gif for t [ 0 , 1 ] T http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-52/MediaObjects/13661_2012_Article_345_IEq29_HTML.gif);

         
      3. (H3)

        h C l d ( [ 0 , 1 ] T , [ 0 , ) ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-52/MediaObjects/13661_2012_Article_345_IEq30_HTML.gif symmetric on [ 0 , 1 ] T http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-52/MediaObjects/13661_2012_Article_345_IEq27_HTML.gif (i.e., h ( t ) = h ( 1 t ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-52/MediaObjects/13661_2012_Article_345_IEq31_HTML.gif for t [ 0 , 1 ] T http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-52/MediaObjects/13661_2012_Article_345_IEq32_HTML.gif) and h ( t ) 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-52/MediaObjects/13661_2012_Article_345_IEq33_HTML.gif on any subinterval of [ 0 , 1 ] T http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-52/MediaObjects/13661_2012_Article_345_IEq27_HTML.gif.

         

      By using four functionals fixed point theorem [5], we establish the existence of at least one symmetric positive solution for BVP (1.1)-(1.2). In particular, the nonlinear term f ( t , x ( t ) , x Δ ( t ) ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-52/MediaObjects/13661_2012_Article_345_IEq34_HTML.gif is allowed to change sign. The remainder of this paper is organized as follows. Section 2 is devoted to some preliminary lemmas. We give and prove our main result in Section 3. Section 4 contains an illustrative example. To the best of our knowledge, symmetric positive solutions for multipoint BVP for an increasing homeomorphism and homomorphism with sign-changing nonlinearity on time scales by using four functionals fixed point theorem [5] have not been considered till now. In this paper, we intend to fill in such gaps in the literature.

      In this paper, a symmetric positive solution x of (1.1) and (1.2) means a solution of (1.1) and (1.2) satisfying x > 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-52/MediaObjects/13661_2012_Article_345_IEq35_HTML.gif and x ( t ) = x ( 1 t ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-52/MediaObjects/13661_2012_Article_345_IEq36_HTML.gif, t [ 0 , 1 ] T http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-52/MediaObjects/13661_2012_Article_345_IEq37_HTML.gif.

      2 Preliminaries

      To prove the main result in this paper, we will employ several lemmas. These lemmas are based on the BVP
      http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-52/MediaObjects/13661_2012_Article_345_Equ3_HTML.gif
      (2.1)
      http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-52/MediaObjects/13661_2012_Article_345_Equ4_HTML.gif
      (2.2)
      Lemma 2.1 If condition (H1) holds, then for y C l d [ 0 , 1 ] T http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-52/MediaObjects/13661_2012_Article_345_IEq38_HTML.gif, boundary value problem (2.1) and (2.2) has a unique solution x ( t ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-52/MediaObjects/13661_2012_Article_345_IEq39_HTML.gif
      x ( t ) = i = 1 m 2 α i 0 η i ϕ 1 ( A y 0 s y ( τ ) τ ) Δ s + ϕ 1 ( A y 0 1 y ( τ ) τ ) i = 1 m 2 α i + 0 t ϕ 1 ( A y 0 s y ( τ ) τ ) Δ s http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-52/MediaObjects/13661_2012_Article_345_Equ5_HTML.gif
      (2.3)
      or
      x ( t ) = ϕ 1 ( A y ) + i = 1 m 2 α i ξ i 1 ϕ 1 ( A y 0 s y ( τ ) τ ) Δ s i = 1 m 2 α i t 1 ϕ 1 ( A y 0 s y ( τ ) τ ) Δ s , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-52/MediaObjects/13661_2012_Article_345_Equ6_HTML.gif
      (2.4)
      where A y http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-52/MediaObjects/13661_2012_Article_345_IEq40_HTML.gif satisfies
      0 1 ϕ 1 ( A y 0 s y ( τ ) τ ) Δ s = 0 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-52/MediaObjects/13661_2012_Article_345_Equ7_HTML.gif
      (2.5)
      Proof Suppose x is a solution of BVP (2.1), (2.2). Integrating (2.1) from 0 to t, we have
      x Δ ( t ) = ϕ 1 ( A y 0 t y ( τ ) τ ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-52/MediaObjects/13661_2012_Article_345_Equ8_HTML.gif
      (2.6)
      Integrating (2.6) from 0 to t, we get
      x ( t ) = x ( 0 ) + 0 t ϕ 1 ( A y 0 s y ( τ ) τ ) Δ s , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-52/MediaObjects/13661_2012_Article_345_Equc_HTML.gif
      or integrating the same equation from t to 1, we achieve
      x ( t ) = x ( 1 ) t 1 ϕ 1 ( A y 0 s y ( τ ) τ ) Δ s . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-52/MediaObjects/13661_2012_Article_345_Equd_HTML.gif
      Using boundary condition (2.2), we get
      x ( t ) = i = 1 m 2 α i 0 η i ϕ 1 ( A y 0 s y ( τ ) τ ) Δ s + ϕ 1 ( A y 0 1 y ( τ ) τ ) i = 1 m 2 α i + 0 t ϕ 1 ( A y 0 s y ( τ ) τ ) Δ s http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-52/MediaObjects/13661_2012_Article_345_Equ9_HTML.gif
      (2.7)
      or
      x ( t ) = ϕ 1 ( A y ) + i = 1 m 2 α i ξ i 1 ϕ 1 ( A y 0 s y ( τ ) τ ) Δ s i = 1 m 2 α i t 1 ϕ 1 ( A y 0 s y ( τ ) τ ) Δ s , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-52/MediaObjects/13661_2012_Article_345_Equ10_HTML.gif
      (2.8)

      where A y http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-52/MediaObjects/13661_2012_Article_345_IEq40_HTML.gif satisfies (2.5).

      On the other hand, it is easy to verify that if x is the solution of (2.3) or (2.4), then x is a solution of BVP (2.1), (2.2). The proof is accomplished. □

      Lemma 2.2 If y ( t ) C l d [ 0 , 1 ] T http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-52/MediaObjects/13661_2012_Article_345_IEq41_HTML.gif is nonnegative on [ 0 , 1 ] T http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-52/MediaObjects/13661_2012_Article_345_IEq27_HTML.gif and y ( t ) 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-52/MediaObjects/13661_2012_Article_345_IEq42_HTML.gif on any subinterval of [ 0 , 1 ] T http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-52/MediaObjects/13661_2012_Article_345_IEq27_HTML.gif, then there exists a unique A y ( , + ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-52/MediaObjects/13661_2012_Article_345_IEq43_HTML.gif satisfying (2.5). Moreover, there is a unique σ y ( 0 , 1 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-52/MediaObjects/13661_2012_Article_345_IEq44_HTML.gif such that A y = 0 σ y y ( τ ) τ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-52/MediaObjects/13661_2012_Article_345_IEq45_HTML.gif.

      Proof For any y ( t ) (2.1) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-52/MediaObjects/13661_2012_Article_345_IEq46_HTML.gif, define
      H y ( c ) = 0 1 ϕ 1 ( c 0 s y ( τ ) τ ) Δ s . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-52/MediaObjects/13661_2012_Article_345_Eque_HTML.gif
      So, H y : R R http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-52/MediaObjects/13661_2012_Article_345_IEq47_HTML.gif is continuous and strictly increasing. It is easy to see that
      H y ( 0 ) < 0 , H y ( 0 1 y ( τ ) τ ) > 0 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-52/MediaObjects/13661_2012_Article_345_Equf_HTML.gif
      Therefore there exists a unique A y ( 0 , 0 1 y ( τ ) τ ) ( , + ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-52/MediaObjects/13661_2012_Article_345_IEq48_HTML.gif satisfying (2.5). Furthermore, let
      F ( t ) = 0 t y ( τ ) τ . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-52/MediaObjects/13661_2012_Article_345_Equg_HTML.gif
      Then F ( t ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-52/MediaObjects/13661_2012_Article_345_IEq49_HTML.gif is continuous and strictly increasing on [ 0 , 1 ] T http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-52/MediaObjects/13661_2012_Article_345_IEq27_HTML.gif and F ( 0 ) = 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-52/MediaObjects/13661_2012_Article_345_IEq50_HTML.gif, F ( 1 ) = 0 1 y ( τ ) τ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-52/MediaObjects/13661_2012_Article_345_IEq51_HTML.gif. Thus
      0 = F ( 0 ) < A y < F ( 1 ) = 0 1 y ( τ ) τ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-52/MediaObjects/13661_2012_Article_345_Equh_HTML.gif

      implies that there exists a unique σ y ( 0 , 1 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-52/MediaObjects/13661_2012_Article_345_IEq44_HTML.gif such that A y = 0 σ y y ( τ ) τ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-52/MediaObjects/13661_2012_Article_345_IEq52_HTML.gif. Lemma is proved. □

      Remark 2.1 By Lemmas 2.1 and 2.2, the unique solution of BVP (2.1), (2.2) can be rewritten in the form
      x ( t ) = { 1 i = 1 m 2 α i [ i = 1 m 2 α i 0 η i ϕ 1 ( s σ y y ( τ ) τ ) Δ s ϕ 1 ( σ y 1 y ( τ ) τ ) ] + 0 t ϕ 1 ( s σ y y ( τ ) τ ) Δ s , 0 t σ y , 1 i = 1 m 2 α i [ i = 1 m 2 α i ξ i 1 ϕ 1 ( σ y s y ( τ ) τ ) Δ s ϕ 1 ( 0 σ y y ( τ ) τ ) ] + t 1 ϕ 1 ( σ y s y ( τ ) τ ) Δ s , σ y t 1 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-52/MediaObjects/13661_2012_Article_345_Equ11_HTML.gif
      (2.9)
      Lemma 2.3 Let (H1) hold. If y C l d Δ [ 0 , 1 ] T http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-52/MediaObjects/13661_2012_Article_345_IEq53_HTML.gif is nonnegative on [ 0 , 1 ] T http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-52/MediaObjects/13661_2012_Article_345_IEq27_HTML.gif and y ( t ) 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-52/MediaObjects/13661_2012_Article_345_IEq42_HTML.gif on any subinterval of [ 0 , 1 ] T http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-52/MediaObjects/13661_2012_Article_345_IEq27_HTML.gif, then the unique solution x ( t ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-52/MediaObjects/13661_2012_Article_345_IEq39_HTML.gif of BVP (2.1)-(2.2) has the following properties:
      1. (i)

        x ( t ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-52/MediaObjects/13661_2012_Article_345_IEq39_HTML.gif is concave on [ 0 , 1 ] T http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-52/MediaObjects/13661_2012_Article_345_IEq27_HTML.gif,

         
      2. (ii)

        x ( t ) > 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-52/MediaObjects/13661_2012_Article_345_IEq54_HTML.gif,

         
      3. (iii)

        there exists a unique t 0 ( 0 , 1 ) T http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-52/MediaObjects/13661_2012_Article_345_IEq55_HTML.gif such that x Δ ( t 0 ) = 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-52/MediaObjects/13661_2012_Article_345_IEq56_HTML.gif,

         
      4. (iv)

        σ y = t 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-52/MediaObjects/13661_2012_Article_345_IEq57_HTML.gif.

         
      Proof Suppose that x ( t ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-52/MediaObjects/13661_2012_Article_345_IEq39_HTML.gif is a solution of BVP (2.1)-(2.2), then
      1. (i)

        ( ϕ ( x Δ ) ) ( t ) = y ( t ) 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-52/MediaObjects/13661_2012_Article_345_IEq58_HTML.gif, ϕ ( x Δ ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-52/MediaObjects/13661_2012_Article_345_IEq59_HTML.gif is nonincreasing so x Δ ( t ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-52/MediaObjects/13661_2012_Article_345_IEq60_HTML.gif is nonincreasing. This implies that x ( t ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-52/MediaObjects/13661_2012_Article_345_IEq39_HTML.gif is concave.

         
      2. (ii)
        We have x Δ ( 0 ) = i = 1 m 2 α i x ( ξ i ) = ϕ 1 ( A y ) > 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-52/MediaObjects/13661_2012_Article_345_IEq61_HTML.gif and x Δ ( 1 ) = ϕ 1 ( A y 0 1 y ( s ) ( s ) ) < 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-52/MediaObjects/13661_2012_Article_345_IEq62_HTML.gif. Furthermore, we get
        http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-52/MediaObjects/13661_2012_Article_345_Equi_HTML.gif
         
      If we continue like this, we have
      α m 2 x ( ξ m 2 ) α m 2 x ( 0 ) = α m 2 0 ξ m 2 x Δ ( s ) Δ s α m 2 ξ m 2 x Δ ( 0 ) = α m 2 ξ m 2 i = 1 m 2 α i x ( ξ i ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-52/MediaObjects/13661_2012_Article_345_Equj_HTML.gif
      Using (H1), we obtain
      i = 1 m 2 α i x ( ξ i ) i = 1 m 2 α i x ( 0 ) i = 1 m 2 α i x ( ξ i ) i = 1 m 2 α i ξ i < i = 1 m 2 α i x ( ξ i ) , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-52/MediaObjects/13661_2012_Article_345_Equk_HTML.gif
      which implies that x ( 0 ) > 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-52/MediaObjects/13661_2012_Article_345_IEq63_HTML.gif. Similarly,
      http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-52/MediaObjects/13661_2012_Article_345_Equl_HTML.gif
      If we continue in this way, we attain that
      α m 2 x ( 1 ) α m 2 x ( η m 2 ) = α m 2 η m 2 1 x Δ ( s ) Δ s α m 2 ( 1 η m 2 ) x Δ ( 1 ) = α m 2 ( 1 η m 2 ) i = 1 m 2 α i x ( η i ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-52/MediaObjects/13661_2012_Article_345_Equm_HTML.gif
      Using (H1), we have i = 1 m 2 α i x ( 1 ) > 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-52/MediaObjects/13661_2012_Article_345_IEq64_HTML.gif, x ( 1 ) > 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-52/MediaObjects/13661_2012_Article_345_IEq65_HTML.gif. Therefore, we get x ( t ) > 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-52/MediaObjects/13661_2012_Article_345_IEq54_HTML.gif, t [ 0 , 1 ] T http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-52/MediaObjects/13661_2012_Article_345_IEq66_HTML.gif.
      1. (iii)

        x Δ ( 0 ) = i = 1 m 2 α i x ( ξ i ) > 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-52/MediaObjects/13661_2012_Article_345_IEq67_HTML.gif, x Δ ( 1 ) = i = 1 m 2 α i x ( η i ) < 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-52/MediaObjects/13661_2012_Article_345_IEq68_HTML.gif imply that there is a t 0 ( 0 , 1 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-52/MediaObjects/13661_2012_Article_345_IEq69_HTML.gif such that x Δ ( t 0 ) = 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-52/MediaObjects/13661_2012_Article_345_IEq56_HTML.gif.

         
      If there exist t 1 , t 2 ( 0 , 1 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-52/MediaObjects/13661_2012_Article_345_IEq70_HTML.gif, t 1 < t 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-52/MediaObjects/13661_2012_Article_345_IEq71_HTML.gif, such that x Δ ( t 1 ) = 0 = x Δ ( t 2 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-52/MediaObjects/13661_2012_Article_345_IEq72_HTML.gif, then
      0 = ϕ ( x Δ ( t 2 ) ) ϕ ( x Δ ( t 1 ) ) = t 1 t 2 y ( τ ) τ < 0 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-52/MediaObjects/13661_2012_Article_345_Equn_HTML.gif
      which is a contradiction.
      1. (iv)

        From Lemmas 2.1 and 2.2, we have x Δ ( t ) = ϕ 1 ( t σ y y ( τ ) τ ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-52/MediaObjects/13661_2012_Article_345_IEq73_HTML.gif. Hence we obtain that x Δ ( σ y ) = x Δ ( t 0 ) = 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-52/MediaObjects/13661_2012_Article_345_IEq74_HTML.gif. This implies σ y = t 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-52/MediaObjects/13661_2012_Article_345_IEq57_HTML.gif.

         

      The lemma is proved. □

      Lemma 2.4 If y ( t ) C l d [ 0 , 1 ] T http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-52/MediaObjects/13661_2012_Article_345_IEq41_HTML.gif is symmetric nonnegative on [ 0 , 1 ] T http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-52/MediaObjects/13661_2012_Article_345_IEq27_HTML.gif and y ( t ) 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-52/MediaObjects/13661_2012_Article_345_IEq42_HTML.gif on any subinterval of [ 0 , 1 ] T http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-52/MediaObjects/13661_2012_Article_345_IEq27_HTML.gif, then the unique solution x ( t ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-52/MediaObjects/13661_2012_Article_345_IEq39_HTML.gif of (2.1), (2.2) is concave and symmetric with x ( t ) 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-52/MediaObjects/13661_2012_Article_345_IEq75_HTML.gif on [ 0 , 1 ] T http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-52/MediaObjects/13661_2012_Article_345_IEq27_HTML.gif.

      Proof Clearly, x ( t ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-52/MediaObjects/13661_2012_Article_345_IEq39_HTML.gif is concave and x ( t ) 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-52/MediaObjects/13661_2012_Article_345_IEq75_HTML.gif from Lemma 2.3. We show that x ( t ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-52/MediaObjects/13661_2012_Article_345_IEq39_HTML.gif is symmetric on [ 0 , 1 ] T http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-52/MediaObjects/13661_2012_Article_345_IEq27_HTML.gif. For the symmetry of y ( t ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-52/MediaObjects/13661_2012_Article_345_IEq76_HTML.gif, it is easy to see that H y ( 0 1 / 2 y ( τ ) τ ) = 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-52/MediaObjects/13661_2012_Article_345_IEq77_HTML.gif, i.e., σ y = 1 / 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-52/MediaObjects/13661_2012_Article_345_IEq78_HTML.gif. Therefore, from (2.9) and for t [ 0 , 1 / 2 ] T http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-52/MediaObjects/13661_2012_Article_345_IEq79_HTML.gif, by the transformation τ = 1 τ ˆ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-52/MediaObjects/13661_2012_Article_345_IEq80_HTML.gif, we have
      x ( t ) = i = 1 m 2 α i 0 η i ϕ 1 ( s 1 / 2 y ( τ ) τ ) Δ s + ϕ 1 ( 1 / 2 1 y ( τ ) τ ) i = 1 m 2 α i + 0 t ϕ 1 ( s 1 / 2 y ( τ ) τ ) Δ s = i = 1 m 2 α i 0 η i ϕ 1 ( 1 s 1 / 2 y ( τ ˆ ) τ ˆ ) Δ s + ϕ 1 ( 0 1 / 2 y ( τ ˆ ) τ ˆ ) i = 1 m 2 α i 0 t ϕ 1 ( 1 s 1 / 2 y ( τ ˆ ) τ ˆ ) Δ s . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-52/MediaObjects/13661_2012_Article_345_Equo_HTML.gif
      Again, let s = 1 s ˆ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-52/MediaObjects/13661_2012_Article_345_IEq81_HTML.gif. Then
      x ( t ) = i = 1 m 2 α i 1 1 η i ϕ 1 ( s ˆ 1 / 2 y ( τ ˆ ) τ ˆ ) Δ s ˆ + ϕ 1 ( 0 1 / 2 y ( τ ˆ ) τ ˆ ) i = 1 m 2 α i + 1 1 t ϕ 1 ( s ˆ 1 / 2 y ( τ ˆ ) τ ˆ ) Δ s ˆ = i = 1 m 2 α i ξ i 1 ϕ 1 ( 1 / 2 s y ( τ ) τ ) Δ s + ϕ 1 ( 0 1 / 2 y ( τ ) τ ) i = 1 m 2 α i + 1 t 1 ϕ 1 ( 1 / 2 s y ( τ ) τ ) Δ s = x ( 1 t ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-52/MediaObjects/13661_2012_Article_345_Equp_HTML.gif

      So, x ( t ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-52/MediaObjects/13661_2012_Article_345_IEq39_HTML.gif is symmetric on [ 0 , 1 ] T http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-52/MediaObjects/13661_2012_Article_345_IEq27_HTML.gif. The proof is accomplished. □

      Let E = C l d Δ [ 0 , 1 ] T http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-52/MediaObjects/13661_2012_Article_345_IEq82_HTML.gif. Then E is a Banach space with the norm
      x = max { sup t [ 0 , 1 ] T | x ( t ) | , sup t [ 0 , 1 ] T | x Δ ( t ) | } . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-52/MediaObjects/13661_2012_Article_345_Equq_HTML.gif
      We define two cones by
      http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-52/MediaObjects/13661_2012_Article_345_Equr_HTML.gif
      Define the operator F : P E http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-52/MediaObjects/13661_2012_Article_345_IEq83_HTML.gif by
      ( F x ) ( t ) = { 1 i = 1 m 2 α i i = 1 m 2 α i 0 η i ϕ 1 ( s 1 / 2 h ( τ ) f ( τ , x ( τ ) , x Δ ( τ ) ) τ ) Δ s + 1 i = 1 m 2 α i ϕ 1 ( 1 / 2 1 h ( τ ) f ( τ , x ( τ ) , x Δ ( τ ) ) τ ) + 0 t ϕ 1 ( s 1 / 2 h ( τ ) f ( τ , x ( τ ) , x Δ ( τ ) ) τ ) Δ s , 0 t 1 / 2 , 1 i = 1 m 2 α i ϕ 1 ( 0 1 / 2 h ( τ ) f ( τ , x ( τ ) , x Δ ( τ ) ) τ ) 1 i = 1 m 2 α i i = 1 m 2 α i ξ i 1 ϕ 1 ( 1 / 2 s h ( τ ) f ( τ , x ( τ ) , x Δ ( τ ) ) τ ) Δ s + t 1 ϕ 1 ( 1 / 2 s h ( τ ) f ( τ , x ( τ ) , x Δ ( τ ) ) τ ) Δ s , 1 / 2 t 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-52/MediaObjects/13661_2012_Article_345_Equs_HTML.gif
      and T : K E http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-52/MediaObjects/13661_2012_Article_345_IEq84_HTML.gif as follows:
      ( T x ) ( t ) = { 1 i = 1 m 2 α i i = 1 m 2 α i 0 η i ϕ 1 ( s 1 / 2 h ( τ ) f + ( τ , x ( τ ) , x Δ ( τ ) ) τ ) Δ s + 1 i = 1 m 2 α i ϕ 1 ( 1 / 2 1 h ( τ ) f + ( τ , x ( τ ) , x Δ ( τ ) ) τ ) + 0 t ϕ 1 ( s 1 / 2 h ( τ ) f + ( τ , x ( τ ) , x Δ ( τ ) ) τ ) Δ s , 0 t 1 / 2 , 1 i = 1 m 2 α i ϕ 1 ( 0 1 / 2 h ( τ ) f + ( τ , x ( τ ) , x Δ ( τ ) ) τ ) 1 i = 1 m 2 α i i = 1 m 2 α i ξ i 1 ϕ 1 ( 1 / 2 s h ( τ ) f + ( τ , x ( τ ) , x Δ ( τ ) ) τ ) Δ s + t 1 ϕ 1 ( 1 / 2 s h ( τ ) f + ( τ , x ( τ ) , x Δ ( τ ) ) τ ) Δ s , 1 / 2 t 1 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-52/MediaObjects/13661_2012_Article_345_Equt_HTML.gif

      where f + ( t , x ( t ) , x Δ ( t ) ) = max { f ( t , x ( t ) , x Δ ( t ) ) , 0 } http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-52/MediaObjects/13661_2012_Article_345_IEq85_HTML.gif. Obviously, x is a solution of BVP (2.1)-(2.2) if and only if x is a fixed point of the operator F.

      Lemma 2.5 If (H1) holds, then sup t [ 0 , 1 ] T x ( t ) M sup t [ 0 , 1 ] T | x Δ ( t ) | http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-52/MediaObjects/13661_2012_Article_345_IEq86_HTML.gif for x K http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-52/MediaObjects/13661_2012_Article_345_IEq87_HTML.gif, where
      M = 1 + 1 i = 1 m 2 α i . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-52/MediaObjects/13661_2012_Article_345_Equ12_HTML.gif
      (2.10)
      Proof For x K http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-52/MediaObjects/13661_2012_Article_345_IEq87_HTML.gif, one arrives at
      0 = x ( 1 ) x ( 0 ) x ( ξ i ) x ( 0 ) ξ i , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-52/MediaObjects/13661_2012_Article_345_Equu_HTML.gif
      i.e., x ( ξ i ) x ( 0 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-52/MediaObjects/13661_2012_Article_345_IEq88_HTML.gif. Hence,
      i = 1 m 2 α i x ( ξ i ) i = 1 m 2 α i x ( 0 ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-52/MediaObjects/13661_2012_Article_345_Equv_HTML.gif
      By x Δ ( 0 ) = i = 1 m 2 α i x ( ξ i ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-52/MediaObjects/13661_2012_Article_345_IEq89_HTML.gif, we get
      x ( 0 ) 1 i = 1 m 2 α i x Δ ( 0 ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-52/MediaObjects/13661_2012_Article_345_Equw_HTML.gif
      Hence
      x ( t ) = 0 t x Δ ( s ) Δ s + x ( 0 ) t x Δ ( 0 ) + x ( 0 ) t x Δ ( 0 ) + 1 i = 1 m 2 α i x Δ ( 0 ) ( 1 + 1 i = 1 m 2 α i ) x Δ ( 0 ) = M x Δ ( 0 ) , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-52/MediaObjects/13661_2012_Article_345_Equx_HTML.gif
      i.e.,
      sup t [ 0 , 1 ] T x ( t ) M x Δ ( 0 ) = M sup t [ 0 , 1 ] T x Δ ( t ) M sup t [ 0 , 1 ] T | x Δ ( t ) | . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-52/MediaObjects/13661_2012_Article_345_Equy_HTML.gif

      The proof is finalized. □

      From Lemma 2.5, we obtain
      x = max { sup t [ 0 , 1 ] T | x ( t ) | , sup t [ 0 , 1 ] T | x Δ ( t ) | } max { M sup t [ 0 , 1 ] T | x Δ ( t ) | , sup t [ 0 , 1 ] T | x Δ ( t ) | } M sup t [ 0 , 1 ] T | x Δ ( t ) | . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-52/MediaObjects/13661_2012_Article_345_Equz_HTML.gif

      Lemma 2.6 Suppose that (H1)-(H3) hold, then T : K K http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-52/MediaObjects/13661_2012_Article_345_IEq90_HTML.gif is completely continuous.

      Proof Let x K http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-52/MediaObjects/13661_2012_Article_345_IEq87_HTML.gif. According to the definition of T and Lemma 2.3, it follows that ( ϕ ( ( T x ) Δ ) ) ( t ) 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-52/MediaObjects/13661_2012_Article_345_IEq91_HTML.gif, which implies the concavity of ( T x ) ( t ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-52/MediaObjects/13661_2012_Article_345_IEq92_HTML.gif on [ 0 , 1 ] T http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-52/MediaObjects/13661_2012_Article_345_IEq27_HTML.gif. On the other hand, from the definition of f and h, ( T x ) ( t ) = ( T x ) ( 1 t ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-52/MediaObjects/13661_2012_Article_345_IEq93_HTML.gif holds for t [ 0 , 1 / 2 ] T http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-52/MediaObjects/13661_2012_Article_345_IEq94_HTML.gif, i.e., Tx is symmetric on [ 0 , 1 ] T http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-52/MediaObjects/13661_2012_Article_345_IEq27_HTML.gif. So, T K K http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-52/MediaObjects/13661_2012_Article_345_IEq95_HTML.gif. By applying the Arzela-Ascoli theorem on time scales, we can obtain that T ( K ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-52/MediaObjects/13661_2012_Article_345_IEq96_HTML.gif is relatively compact. In view of the Lebesgue convergence theorem on time scales, it is obvious that T is continuous. Hence, T : K K http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-52/MediaObjects/13661_2012_Article_345_IEq97_HTML.gif is a completely continuous operator. The proof is completed. □

      3 Existence of one symmetric positive solution

      Let α and Ψ be nonnegative continuous concave functionals on P, and let β and θ be nonnegative continuous convex functionals on P, then for positive numbers r, j, n and R, we define the sets:
      http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-52/MediaObjects/13661_2012_Article_345_Equ13_HTML.gif
      (3.1)

      Theorem 3.1 [5]

      If P is a cone in a real Banach space E, α and Ψ are nonnegative continuous concave functionals on P, β and θ are nonnegative continuous convex functionals on P and there exist positive numbers r, j, n and R such that
      A : Q ( α , β , r , R ) P http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-52/MediaObjects/13661_2012_Article_345_Equaa_HTML.gif
      is a completely continuous operator, and Q ( α , β , r , R ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-52/MediaObjects/13661_2012_Article_345_IEq98_HTML.gif is a bounded set. If
      1. (i)

        { x U ( Ψ , j ) : β ( x ) < R } { x V ( θ , n ) : r < α ( x ) } http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-52/MediaObjects/13661_2012_Article_345_IEq99_HTML.gif;

         
      2. (ii)

        α ( A x ) r http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-52/MediaObjects/13661_2012_Article_345_IEq100_HTML.gif for all x Q ( α , β , r , R ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-52/MediaObjects/13661_2012_Article_345_IEq101_HTML.gif, with α ( x ) = r http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-52/MediaObjects/13661_2012_Article_345_IEq102_HTML.gif and n < θ ( A x ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-52/MediaObjects/13661_2012_Article_345_IEq103_HTML.gif;

         
      3. (iii)

        α ( A x ) r http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-52/MediaObjects/13661_2012_Article_345_IEq100_HTML.gif for all x V ( θ , n ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-52/MediaObjects/13661_2012_Article_345_IEq104_HTML.gif, with α ( x ) = r http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-52/MediaObjects/13661_2012_Article_345_IEq102_HTML.gif;

         
      4. (iv)

        β ( A x ) R http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-52/MediaObjects/13661_2012_Article_345_IEq105_HTML.gif for all x Q ( α , β , r , R ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-52/MediaObjects/13661_2012_Article_345_IEq101_HTML.gif, with β ( x ) = R http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-52/MediaObjects/13661_2012_Article_345_IEq106_HTML.gif and Ψ ( A x ) < j http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-52/MediaObjects/13661_2012_Article_345_IEq107_HTML.gif;

         
      5. (v)

        β ( A x ) R http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-52/MediaObjects/13661_2012_Article_345_IEq105_HTML.gif for all x U ( Ψ , j ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-52/MediaObjects/13661_2012_Article_345_IEq108_HTML.gif, with β ( x ) = R http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-52/MediaObjects/13661_2012_Article_345_IEq106_HTML.gif.

         

      Then A has a fixed point x in Q ( α , β , r , R ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-52/MediaObjects/13661_2012_Article_345_IEq98_HTML.gif.

      Suppose ω , z T http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-52/MediaObjects/13661_2012_Article_345_IEq109_HTML.gif with 0 < ω < z < 1 / 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-52/MediaObjects/13661_2012_Article_345_IEq110_HTML.gif. For the convenience, we take the notations
      Ω = w z h ( τ ) τ , Λ = 0 1 h ( τ ) τ , L = i = 1 m 2 α i 1 i = 1 m 2 α i ξ i η i http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-52/MediaObjects/13661_2012_Article_345_Equab_HTML.gif
      and define the maps
      α ( x ) = min t [ ω , z ] T x ( t ) , θ ( x ) = max t [ 0 , 1 ] T x ( t ) , β ( x ) = Ψ ( x ) = sup t [ 0 , 1 ] T | x Δ ( t ) | http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-52/MediaObjects/13661_2012_Article_345_Equ14_HTML.gif
      (3.2)

      and let Q ( α , β , r , R ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-52/MediaObjects/13661_2012_Article_345_IEq98_HTML.gif, U ( Ψ , j ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-52/MediaObjects/13661_2012_Article_345_IEq111_HTML.gif and V ( θ , n ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-52/MediaObjects/13661_2012_Article_345_IEq112_HTML.gif be defined by (3.1).

      Theorem 3.2 Assume (H1)-(H3) hold. If there exist constants r, j, n, R with max { r 2 ω , R } n http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-52/MediaObjects/13661_2012_Article_345_IEq113_HTML.gif, max { L + 1 L j , L + 1 L ω ( 1 ω ) + 1 r } < R http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-52/MediaObjects/13661_2012_Article_345_IEq114_HTML.gif and suppose that f satisfies the following conditions:

      1. (C1)

        f ( t , u , v ) 1 Ω ϕ ( r ω ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-52/MediaObjects/13661_2012_Article_345_IEq115_HTML.gif for ( t , u , v ) [ ω , z ] T × [ r , n ] × [ R , R ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-52/MediaObjects/13661_2012_Article_345_IEq116_HTML.gif;

         
      2. (C2)

        f ( t , u , v ) 1 Λ ϕ ( R ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-52/MediaObjects/13661_2012_Article_345_IEq117_HTML.gif for ( t , u , v ) [ 0 , 1 ] T × [ 0 , M R ] × [ j , R ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-52/MediaObjects/13661_2012_Article_345_IEq118_HTML.gif;

         
      3. (C3)

        f ( t , u , v ) 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-52/MediaObjects/13661_2012_Article_345_IEq119_HTML.gif for ( t , u , v ) [ 0 , 1 ] T × [ 0 , M R ] × [ R , R ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-52/MediaObjects/13661_2012_Article_345_IEq120_HTML.gif.

         
      Then BVP (1.1)-(1.2) has at least one symmetric positive solution x P http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-52/MediaObjects/13661_2012_Article_345_IEq121_HTML.gif such that
      min t [ ω , z ] T x ( t ) r , max t [ 0 , 1 ] T x ( t ) R . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-52/MediaObjects/13661_2012_Article_345_Equac_HTML.gif

      Proof Boundary value problem (1.1)-(1.2) has a solution x = x ( t ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-52/MediaObjects/13661_2012_Article_345_IEq122_HTML.gif if and only if x solves the operator equation x = T x http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-52/MediaObjects/13661_2012_Article_345_IEq123_HTML.gif. Thus we set out to verify that the operator T satisfies four functionals fixed point theorem, which will prove the existence of a fixed point of T.

      We first show that Q ( α , β , r , R ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-52/MediaObjects/13661_2012_Article_345_IEq98_HTML.gif is bounded and T : Q ( α , β , r , R ) K http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-52/MediaObjects/13661_2012_Article_345_IEq124_HTML.gif is completely continuous. For all x Q ( α , β , r , R ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-52/MediaObjects/13661_2012_Article_345_IEq101_HTML.gif with Lemma 2.5, we have
      x M sup t [ 0 , 1 ] T | x Δ ( t ) | = M β ( x ) M R , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-52/MediaObjects/13661_2012_Article_345_Equad_HTML.gif

      which means that Q ( α , β , r , R ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-52/MediaObjects/13661_2012_Article_345_IEq98_HTML.gif is a bounded set. According to Lemma 2.6, it is clear that T : Q ( α , β , r , R ) K http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-52/MediaObjects/13661_2012_Article_345_IEq124_HTML.gif is completely continuous.

      Let
      x 0 = R L + 1 ( L t ( 1 t ) + 1 ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-52/MediaObjects/13661_2012_Article_345_Equae_HTML.gif
      Clearly, x 0 K http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-52/MediaObjects/13661_2012_Article_345_IEq125_HTML.gif. By direct calculation,
      http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-52/MediaObjects/13661_2012_Article_345_Equaf_HTML.gif

      So, x 0 { x U ( Ψ , j ) : β ( x ) < R } { x V ( θ , n ) : r < α ( x ) } http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-52/MediaObjects/13661_2012_Article_345_IEq126_HTML.gif, which means that (i) in Theorem 3.1 is satisfied.

      For all x Q ( α , β , r , R ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-52/MediaObjects/13661_2012_Article_345_IEq101_HTML.gif, with α ( x ) = r http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-52/MediaObjects/13661_2012_Article_345_IEq102_HTML.gif and n < θ ( T x ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-52/MediaObjects/13661_2012_Article_345_IEq127_HTML.gif, we have from concavity
      α ( T x ) = T x ( ω ) ω 1 / 2 T x ( 1 / 2 ) = 2 ω θ ( T x ) > 2 ω n r . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-52/MediaObjects/13661_2012_Article_345_Equag_HTML.gif

      So, α ( T x ) > r http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-52/MediaObjects/13661_2012_Article_345_IEq128_HTML.gif. Hence (ii) in Theorem 3.1 is fulfilled.

      For all x V ( θ , n ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-52/MediaObjects/13661_2012_Article_345_IEq104_HTML.gif, with α ( x ) = r http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-52/MediaObjects/13661_2012_Article_345_IEq102_HTML.gif,
      α ( T x ) = min t [ ω , z ] T T x ( t ) = ( T x ) ( ω ) = 1 i = 1 m 2 α i i = 1 m 2 α i 0 η i ϕ 1 ( s 1 / 2 h ( τ ) f + ( τ , x ( τ ) , x Δ ( τ ) ) τ ) Δ s + 1 i = 1 m 2 α i ϕ 1 ( 1 / 2 1 h ( τ ) f + ( τ , x ( τ ) , x Δ ( τ ) ) τ ) + 0 ω ϕ 1 ( s 1 / 2 h ( τ ) f + ( τ , x ( τ ) , x Δ ( τ ) ) τ ) Δ s 0 ω ϕ 1 ( ω 1 / 2 h ( τ ) f + ( τ , x ( τ ) , x Δ ( τ ) ) τ ) Δ s = ω ϕ 1 ( ω 1 / 2 h ( τ ) f + ( τ , x ( τ ) , x Δ ( τ ) ) τ ) ω ϕ 1 ( ω z h ( τ ) f + ( τ , x ( τ ) , x Δ ( τ ) ) τ ) ω ϕ 1 ( 1 Ω ϕ ( r ω ) ω z h ( τ ) τ ) = r http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-52/MediaObjects/13661_2012_Article_345_Equah_HTML.gif
      and for all x U ( Ψ , j ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-52/MediaObjects/13661_2012_Article_345_IEq108_HTML.gif, with β ( x ) = R http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-52/MediaObjects/13661_2012_Article_345_IEq106_HTML.gif,
      β ( T x ) = max t [ 0 , 1 ] T | ( T x ) Δ ( t ) | = ( T x ) Δ ( 0 ) = ϕ 1 ( 0 1 / 2 h ( τ ) f + ( τ , x ( τ ) , x Δ ( τ ) ) τ ) ϕ 1 ( 0 1 h ( τ ) f + ( τ , x ( τ ) , x Δ ( τ ) ) τ ) ϕ 1 ( 1 Λ ϕ ( R ) 0 1 h ( τ ) τ ) = R . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-52/MediaObjects/13661_2012_Article_345_Equai_HTML.gif

      Thus (iii) and (v) in Theorem 3.1 hold true. We finally prove that (iv) in Theorem 3.1 holds.

      For all x Q ( α , β , r , R ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-52/MediaObjects/13661_2012_Article_345_IEq101_HTML.gif, with β ( x ) = R http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-52/MediaObjects/13661_2012_Article_345_IEq106_HTML.gif and Ψ ( T x ) < j http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-52/MediaObjects/13661_2012_Article_345_IEq129_HTML.gif, we have
      β ( T x ) = Ψ ( T x ) < j < L L + 1 R < R . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-52/MediaObjects/13661_2012_Article_345_Equaj_HTML.gif

      Thus, all conditions of Theorem 3.1 are satisfied. T has a fixed point x in Q ( α , β , r , R ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-52/MediaObjects/13661_2012_Article_345_IEq98_HTML.gif. Clearly, 0 x ( t ) M R http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-52/MediaObjects/13661_2012_Article_345_IEq130_HTML.gif, t [ 0 , 1 ] T http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-52/MediaObjects/13661_2012_Article_345_IEq37_HTML.gif. By condition (C3), we have f ( t , x ( t ) , x Δ ( t ) ) 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-52/MediaObjects/13661_2012_Article_345_IEq131_HTML.gif, t [ 0 , 1 ] T http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-52/MediaObjects/13661_2012_Article_345_IEq37_HTML.gif, that is, f + ( t , x ( t ) , x Δ ( t ) ) = f ( t , x ( t ) , x Δ ( t ) ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-52/MediaObjects/13661_2012_Article_345_IEq132_HTML.gif. Hence, F x = T x http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-52/MediaObjects/13661_2012_Article_345_IEq133_HTML.gif. This means that x is a fixed point of the operator F. Therefore, BVP (1.1)-(1.2) has at least one symmetric positive solution. The proof is completed. □

      4 An example

      Example 4.1 Let T = [ 0 , 1 3 ] { 1 2 } [ 2 3 , 1 ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-52/MediaObjects/13661_2012_Article_345_IEq134_HTML.gif. If we choose m = 3 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-52/MediaObjects/13661_2012_Article_345_IEq135_HTML.gif, ξ 1 = 1 5 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-52/MediaObjects/13661_2012_Article_345_IEq136_HTML.gif, η 1 = 4 5 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-52/MediaObjects/13661_2012_Article_345_IEq137_HTML.gif, α 1 = 1 3 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-52/MediaObjects/13661_2012_Article_345_IEq138_HTML.gif, h ( t ) = 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-52/MediaObjects/13661_2012_Article_345_IEq139_HTML.gif in boundary value problem (1.1)-(1.2), then we have the following BVP on time scale T http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-52/MediaObjects/13661_2012_Article_345_IEq11_HTML.gif:
      { ( ϕ ( x Δ ) ) ( t ) + f ( t , x ( t ) , x Δ ( t ) ) = 0 , t [ 0 , 1 ] T , x Δ ( 0 ) 1 3 x ( 1 5 ) = 0 , x Δ ( 1 ) + 1 3 x ( 4 5 ) = 0 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-52/MediaObjects/13661_2012_Article_345_Equ15_HTML.gif
      (4.1)
      where ϕ ( x ) = x http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-52/MediaObjects/13661_2012_Article_345_IEq140_HTML.gif,
      http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-52/MediaObjects/13661_2012_Article_345_Equ16_HTML.gif
      (4.2)
      Set ω = 1 5 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-52/MediaObjects/13661_2012_Article_345_IEq141_HTML.gif, z = 1 4 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-52/MediaObjects/13661_2012_Article_345_IEq142_HTML.gif. By simple calculation, we get
      Ω = 1 20 , Λ = 1 , L = 25 71 , M = 4 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-52/MediaObjects/13661_2012_Article_345_Equak_HTML.gif
      Choose r = 1 100 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-52/MediaObjects/13661_2012_Article_345_IEq143_HTML.gif, n = 10 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-52/MediaObjects/13661_2012_Article_345_IEq144_HTML.gif, j = 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-52/MediaObjects/13661_2012_Article_345_IEq145_HTML.gif and R = 8 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-52/MediaObjects/13661_2012_Article_345_IEq146_HTML.gif. It is easy to check that max { 1 40 , 8 } 10 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-52/MediaObjects/13661_2012_Article_345_IEq147_HTML.gif, max { 192 25 , 8 625 } < 8 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-52/MediaObjects/13661_2012_Article_345_IEq148_HTML.gif.
      1. (1)

        f ( t , x ( t ) , x Δ ( t ) ) 76 75 1 Ω ϕ ( r ω ) = 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-52/MediaObjects/13661_2012_Article_345_IEq149_HTML.gif for ( t , x ( t ) , x Δ ( t ) ) [ 1 5 , 1 4 ] T × [ 1 100 , 10 ] × [ 8 , 8 ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-52/MediaObjects/13661_2012_Article_345_IEq150_HTML.gif;

         
      2. (2)

        f ( t , x ( t ) , x Δ ( t ) ) 2 1 Λ ϕ ( R ) = 8 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-52/MediaObjects/13661_2012_Article_345_IEq151_HTML.gif for ( t , x ( t ) , x Δ ( t ) ) [ 0 , 1 ] T × [ 0 , 32 ] × [ 2 , 8 ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-52/MediaObjects/13661_2012_Article_345_IEq152_HTML.gif;

         
      3. (3)

        f ( t , x ( t ) , x Δ ( t ) ) 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-52/MediaObjects/13661_2012_Article_345_IEq131_HTML.gif for ( t , x ( t ) , x Δ ( t ) ) [ 0 , 1 ] T × [ 0 , 32 ] × [ 8 , 8 ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-52/MediaObjects/13661_2012_Article_345_IEq153_HTML.gif.

         
      So, all conditions of Theorem 3.2 hold. Thus, by Theorem 3.2, BVP (4.1) has at least one symmetric positive solution x such that
      min t [ 1 5 , 1 4 ] T x ( t ) 1 100 , max t [ 0 , 1 ] T x ( t ) 8 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-52/MediaObjects/13661_2012_Article_345_Equal_HTML.gif

      Declarations

      Authors’ Affiliations

      (1)
      Department of Mathematics, Ege University

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      Copyright

      © Tokmak and Karaca; licensee Springer 2013

      This article is published under license to BioMed Central Ltd. This is an Open Access article distributed under the terms of the Creative Commons Attribution License (http://​creativecommons.​org/​licenses/​by/​2.​0), which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.