Multiple positive solutions for a fourth-order integral boundary value problem on time scales

  • Yongkun Li1Email author and

    Affiliated with

    • Yanshou Dong1

      Affiliated with

      Boundary Value Problems20112011:59

      DOI: 10.1186/1687-2770-2011-59

      Received: 9 November 2011

      Accepted: 29 December 2011

      Published: 29 December 2011

      Abstract

      In this article, we investigate the multiplicity of positive solutions for a fourth-order system of integral boundary value problem on time scales. The existence of multiple positive solutions for the system is obtained by using the fixed point theorem of cone expansion and compression type due to Krasnosel'skill. To demonstrate the applications of our results, an example is also given in the article.

      Keywords

      positive solutions fixed points integral boundary conditions time scales

      1 Introduction

      Boundary value problem (BVP) for ordinary differential equations arise in different areas of applied mathematics and physics and so on, the existence and multiplicity of positive solutions for such problems have become an important area of investigation in recent years, lots of significant results have been established by using upper and lower solution arguments, fixed point indexes, fixed point theorems and so on (see [18] and the references therein). Especially, the existence of positive solutions of nonlinear BVP with integral boundary conditions has been extensively studied by many authors (see [918] and the references therein).

      However, the corresponding results for BVP with integral boundary conditions on time scales are still very few [1921]. In this article, we discuss the multiple positive solutions for the following fourth-order system of integral BVP with a parameter on time scales
      x ( 4 Δ ) ( t ) + λ f ( t , x ( t ) , x Δ ( t ) , x Δ Δ ( t ) , y ( t ) , y Δ ( t ) , y Δ Δ ( t ) ) = 0 , t ( 0 , σ ( T ) ) T , y ( 4 Δ ) ( t ) + μ g ( t , x ( t ) , x Δ ( t ) , x Δ Δ ( t ) , y ( t ) , y Δ ( t ) , y Δ Δ ( t ) ) = 0 , t ( 0 , σ ( T ) ) T , x ( 0 ) = x Δ ( 0 ) = 0 , y ( 0 ) = y Δ ( 0 ) = 0 , a 1 x Δ Δ ( 0 ) - b 1 x Δ Δ Δ ( 0 ) = 0 σ ( T ) x Δ Δ ( s ) A 1 ( s ) Δ s , c 1 x Δ Δ ( σ ( T ) ) + d 1 x Δ Δ Δ ( σ ( T ) ) = 0 σ ( T ) x Δ Δ ( s ) B 1 ( s ) Δ s , a 2 y Δ Δ ( 0 ) - b 2 y Δ Δ Δ ( 0 ) = 0 σ ( T ) y Δ Δ ( s ) A 2 ( s ) Δ s , c 2 y Δ Δ ( σ ( T ) ) + d 2 y Δ Δ Δ ( σ ( T ) ) = 0 σ ( T ) y Δ Δ ( s ) B 2 ( s ) Δ s , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-59/MediaObjects/13661_2011_Article_104_Equ1_HTML.gif
      (1.1)

      where a i , b i , c i , d i ≥ 0, and ρ i = a i c i σ(T) + a i d i + b i c i > 0(i = 1, 2), 0 < λ, μ < +∞, f, g C ( ( 0 , σ ( T ) ) T × ( + ) 6 , + ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-59/MediaObjects/13661_2011_Article_104_IEq1_HTML.gif, ℝ+ = [0, +∞), A i and B i are nonnegative and rd-continuous on [ 0 , σ ( T ) ] T ( i = 1 , 2 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-59/MediaObjects/13661_2011_Article_104_IEq2_HTML.gif.

      The main purpose of this article is to establish some sufficient conditions for the existence of at least two positive solutions for system (1.1) by using the fixed point theorem of cone expansion and compression type. This article is organized as follows. In Section 2, some useful lemmas are established. In Section 3, by using the fixed point theorem of cone expansion and compression type, we establish sufficient conditions for the existence of at least two positive solutions for system (1.1). An illustrative example is given in Section 4.

      2 Preliminaries

      In this section, we will provide several foundational definitions and results from the calculus on time scales and give some lemmas which are used in the proof of our main results.

      A time scale T http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-59/MediaObjects/13661_2011_Article_104_IEq3_HTML.gif is a nonempty closed subset of the real numbers ℝ.

      Definition 2.1. [22] For t T http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-59/MediaObjects/13661_2011_Article_104_IEq4_HTML.gif, we define the forward jump operator σ : T T http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-59/MediaObjects/13661_2011_Article_104_IEq5_HTML.gif by σ ( t ) = inf { τ T : τ > t } http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-59/MediaObjects/13661_2011_Article_104_IEq6_HTML.gif, while the backward jump operator ρ : T T http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-59/MediaObjects/13661_2011_Article_104_IEq7_HTML.gif by ρ ( t ) = sup { τ T : τ < t } http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-59/MediaObjects/13661_2011_Article_104_IEq8_HTML.gif.

      In this definition, we put inf = sup T http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-59/MediaObjects/13661_2011_Article_104_IEq9_HTML.gif and sup = inf T http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-59/MediaObjects/13661_2011_Article_104_IEq10_HTML.gif, where ∅, denotes the empty set. If σ(t) > t, we say that t is right-scattered, while if ρ(t) < t, we say that t is left-scattered. Also, if t < sup T http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-59/MediaObjects/13661_2011_Article_104_IEq11_HTML.gif and σ(t) = t, then t is called right-dense, and if t > inf T http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-59/MediaObjects/13661_2011_Article_104_IEq12_HTML.gif and ρ(t) = t, then t is called left-dense. We also need, below, the set T k http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-59/MediaObjects/13661_2011_Article_104_IEq13_HTML.gif, which is derived from the time scale T http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-59/MediaObjects/13661_2011_Article_104_IEq14_HTML.gif as follows: if T http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-59/MediaObjects/13661_2011_Article_104_IEq3_HTML.gif has a left-scattered maximum m, then T k = T - m http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-59/MediaObjects/13661_2011_Article_104_IEq15_HTML.gif. Otherwise, T k = T http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-59/MediaObjects/13661_2011_Article_104_IEq16_HTML.gif.

      Definition 2.2. [22] Assume that x : T http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-59/MediaObjects/13661_2011_Article_104_IEq17_HTML.gif is a function and let t T k http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-59/MediaObjects/13661_2011_Article_104_IEq18_HTML.gif. Then x is called differentiable at t T http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-59/MediaObjects/13661_2011_Article_104_IEq19_HTML.gif if there exists a θ ∈ ℝ such that for any given ε > 0, there is an open neighborhood U of t such that
      | x ( σ ( t ) ) - x ( s ) - x Δ ( t ) | σ ( t ) - s | | ε | σ ( t ) - s | , s U . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-59/MediaObjects/13661_2011_Article_104_Equa_HTML.gif

      In this case, xΔ(t) is called the delta derivative of x at t. The second derivative of x(t) is defined by xΔΔ(t) = (xΔ)Δ(t).

      In a similar way, we can obtain the fourth-order derivative of x(t) is defined by x(4Δ)(t) = (((xΔ)Δ)Δ)Δ(t).

      Definition 2.3. [22] A function f : T R http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-59/MediaObjects/13661_2011_Article_104_IEq20_HTML.gif is called rd-continuous provided it is continuous at right-dense points in T http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-59/MediaObjects/13661_2011_Article_104_IEq3_HTML.gif and its left-sided limits exist at left-dense points in T http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-59/MediaObjects/13661_2011_Article_104_IEq3_HTML.gif. The set of rd-continuous functions f : T R http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-59/MediaObjects/13661_2011_Article_104_IEq21_HTML.gif will be denoted by C r d ( T ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-59/MediaObjects/13661_2011_Article_104_IEq22_HTML.gif.

      Definition 2.4. [22] A function F : T http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-59/MediaObjects/13661_2011_Article_104_IEq23_HTML.gif is called a delta-antiderivative of f : T R http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-59/MediaObjects/13661_2011_Article_104_IEq24_HTML.gif provide FΔ(t) = f(t) holds for all t T k http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-59/MediaObjects/13661_2011_Article_104_IEq25_HTML.gif. In this case we define the integral of f by
      a t f ( s ) = F ( t ) - F ( a ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-59/MediaObjects/13661_2011_Article_104_Equb_HTML.gif
      For convenience, we denote I = [ 0 , σ ( T ) ] T http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-59/MediaObjects/13661_2011_Article_104_IEq26_HTML.gif, I = ( 0 , σ ( T ) ) T http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-59/MediaObjects/13661_2011_Article_104_IEq27_HTML.gif and for i = 1, 2, we set
      D 1 i = Q 1 i 1 - P 1 i , D 2 i = P 2 i 1 - Q 2 i , K 1 i = 1 1 - P 1 i , K 2 i = 1 1 - Q 2 i , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-59/MediaObjects/13661_2011_Article_104_Equc_HTML.gif
      where
      P 1 i = 0 σ ( T ) B i ( s ) a i s + b i ρ i Δ s , P 2 i = 0 σ ( T ) A i ( s ) a i s + b i ρ i Δ s , Q 1 i = 0 σ ( T ) B i ( s ) d i + c i ( σ ( T ) - s ) ρ i Δ s , Q 2 i = 0 σ ( T ) A i ( s ) d i + c i ( σ ( T ) - s ) ρ i Δ s . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-59/MediaObjects/13661_2011_Article_104_Equd_HTML.gif
      To establish the existence of multiple positive solutions of system (1.1), let us list the following assumptions:
      ( H 1 ) P j i , Q j i [ 0 , 1 ) , D 11 D 21 [ 0 , 1 ) , D 21 D 22 [ 0 , 1 ) , j , i = 1 , 2 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-59/MediaObjects/13661_2011_Article_104_Eque_HTML.gif
      In order to overcome the difficulty due to the dependence of f, g on derivatives, we first consider the following second-order nonlinear system
      u Δ Δ ( t ) + λ f ( t , A 2 u , A 1 u , A 0 u , A 2 v , A 1 v , A 0 v ) = 0 , t ( 0 , σ ( T ) ) T , v Δ Δ ( t ) + μ g ( t , A 2 u , A 1 u , A 0 u , A 2 v , A 1 v , A 0 v ) = 0 , t ( 0 , σ ( T ) ) T , a 1 u ( 0 ) - b 1 u Δ ( 0 ) = 0 σ ( T ) u ( s ) A 1 ( s ) Δ s , c 1 u ( σ ( T ) ) + d 1 u Δ ( σ ( T ) ) = 0 σ ( T ) u ( s ) B 1 ( s ) Δ s , a 2 v ( 0 ) - b 2 v Δ ( 0 ) = 0 σ ( T ) v ( s ) A 2 ( s ) Δ s , c 2 v ( σ ( T ) ) + d 2 v Δ ( σ ( T ) ) = 0 σ ( T ) v ( s ) B 2 ( s ) Δ s , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-59/MediaObjects/13661_2011_Article_104_Equ2_HTML.gif
      (2.1)
      where A0 is the identity operator, and
      A i u ( t ) = 0 t ( t - σ ( s ) ) i - 1 u ( s ) Δ s , A i v ( t ) = 0 t ( t - σ ( s ) ) i - 1 v ( s ) Δ s , i = 1 , 2 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-59/MediaObjects/13661_2011_Article_104_Equ3_HTML.gif
      (2.2)

      For the proof of our main results, we will make use of the following lemmas.

      Lemma 2.1. The fourth-order system (1.1) has a solution (x, y) if and only if the nonlinear system (2.1) has a solution (u, v).

      Proof. If (x, y) is a solution of the fourth-order system (1.1), let u(t) = xΔΔ(t), v(t) = yΔΔ(t), then it follows from the boundary conditions of system (1.1) that
      A 1 u ( t ) = x Δ ( t ) , A 2 u ( t ) = x ( t ) , A 1 v ( t ) = y Δ ( t ) , A 2 v ( t ) = y ( t ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-59/MediaObjects/13661_2011_Article_104_Equf_HTML.gif

      Thus (u, v) = (xΔΔ(t), yΔΔ(t)) is a solution of the nonlinear system (2.1).

      Conversely, if (u, v) is a solution of the nonlinear system (2.1), let x(t) = A2u(t), y(t) = A2v(t), then we have
      x Δ ( t ) = A 1 u ( t ) , x Δ Δ ( t ) = u ( t ) , y Δ ( t ) = A 1 v ( t ) , y Δ Δ ( t ) = v ( t ) , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-59/MediaObjects/13661_2011_Article_104_Equg_HTML.gif
      which imply that
      x ( 0 ) = 0 , x Δ ( 0 ) = 0 , y ( 0 ) = 0 , y Δ ( 0 ) = 0 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-59/MediaObjects/13661_2011_Article_104_Equh_HTML.gif

      Consequently, (x, y) = (A2u(t), A2v(t)) is a solution of the fourth-order system (1.1). This completes the proof.

      Lemma 2.2. Assume that D11D21 ≠ 1 holds. Then for any h1C(I', ℝ+), the following BVP
      u Δ Δ ( t ) + h 1 ( t ) = 0 , t ( 0 , σ ( T ) ) T , a 1 u ( 0 ) - b 1 u Δ ( 0 ) = 0 σ ( T ) u ( s ) A 1 ( s ) Δ s , c 1 u ( σ ( T ) ) + d 1 u Δ ( σ ( T ) ) = 0 σ ( T ) u ( s ) B 1 ( s ) Δ s http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-59/MediaObjects/13661_2011_Article_104_Equ4_HTML.gif
      (2.3)
      has a solution
      u ( t ) = 0 σ ( T ) H 1 ( t , s ) h 1 ( s ) Δ s , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-59/MediaObjects/13661_2011_Article_104_Equi_HTML.gif
      where
      H 1 ( t , s ) = G 1 ( t , s ) + r 1 ( t ) 0 σ ( T ) B 1 ( τ ) G 1 ( τ , s ) Δ τ + r 2 ( t ) 0 σ ( T ) A 1 ( τ ) G 1 ( τ , s ) Δ τ , G 1 ( t , s ) = 1 ρ 1 ( a 1 σ ( s ) + b 1 ) [ d 1 + c 1 ( σ ( T ) - t ) ] , σ ( s ) < t , ( a 1 t + b 1 ) [ d 1 + c 1 ( σ ( T ) - σ ( s ) ) ] , t σ ( s ) , r 11 ( t ) = K 11 ( a 1 t + b 1 ) + K 11 D 21 [ d 1 + c 1 ( σ ( T ) - t ) ] ρ 1 ( 1 - D 11 D 21 ) , r 21 ( t ) = K 21 D 11 ( a 1 t + b 1 ) + K 21 [ d 1 + c 1 ( σ ( T ) - t ) ] ρ 1 ( 1 - D 11 D 21 ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-59/MediaObjects/13661_2011_Article_104_Equj_HTML.gif
      Proof. First suppose that u is a solution of system (2.3). It is easy to see by integration of BVP(2.3) that
      u Δ ( t ) = u Δ ( 0 ) - 0 t h 1 ( s ) Δ s . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-59/MediaObjects/13661_2011_Article_104_Equ5_HTML.gif
      (2.4)
      Integrating again, we can obtain
      u ( t ) = u ( 0 ) + t u Δ ( 0 ) - 0 t ( t - σ ( s ) ) h 1 ( s ) Δ s . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-59/MediaObjects/13661_2011_Article_104_Equ6_HTML.gif
      (2.5)
      Let t = σ(T) in (2.4) and (2.5), we obtain
      u Δ ( σ ( T ) ) = u Δ ( 0 ) - 0 σ ( T ) h 1 ( s ) Δ s , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-59/MediaObjects/13661_2011_Article_104_Equ7_HTML.gif
      (2.6)
      u ( σ ( T ) ) = u ( 0 ) + σ ( T ) u Δ ( 0 ) - 0 σ ( T ) ( σ ( T ) - σ ( s ) ) h 1 ( s ) Δ s . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-59/MediaObjects/13661_2011_Article_104_Equ8_HTML.gif
      (2.7)
      Substituting (2.6) and (2.7) into the second boundary value condition of system (2.3), we obtain
      c 1 u ( 0 ) + ( c 1 σ ( T ) + d 1 ) u Δ ( 0 ) = 0 σ ( T ) [ d 1 + c 1 ( σ ( T ) - σ ( s ) ) ] h 1 ( s ) Δ s + 0 σ ( T ) u ( s ) B 1 ( s ) Δ s . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-59/MediaObjects/13661_2011_Article_104_Equ9_HTML.gif
      (2.8)
      From (2.8) and the first boundary value condition of system (2.3), we have
      u Δ ( 0 ) = a 1 ρ 1 0 σ ( T ) [ d 1 + c 1 ( σ ( T ) - σ ( s ) ) ] h 1 ( s ) Δ s + 0 σ ( T ) u ( s ) B 1 ( s ) Δ s http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-59/MediaObjects/13661_2011_Article_104_Equ10_HTML.gif
      (2.9)
      - c 1 a 1 0 σ ( T ) u ( s ) A 1 ( s ) Δ s , u ( 0 ) = b 1 ρ 1 0 σ ( T ) [ d 1 + c 1 ( σ ( T ) - σ ( s ) ) ] h 1 ( s ) Δ s + 0 σ ( T ) u ( s ) B 1 ( s ) Δ s - c 1 a 1 0 σ ( T ) u ( s ) A 1 ( s ) Δ s + 1 a 1 0 σ ( T ) u ( s ) A 1 ( s ) Δ s . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-59/MediaObjects/13661_2011_Article_104_Equ11_HTML.gif
      (2.10)
      Substituting (2.9) and (2.10) into (2.5), we have
      u ( t ) = 0 σ ( T ) G 1 ( t , s ) h 1 ( s ) Δ s + a 1 t + b 1 ρ 1 0 σ ( T ) u ( s ) B 1 ( s ) Δ s + d 1 + c 1 ( σ ( T ) - t ) ρ 1 0 σ ( T ) u ( s ) A 1 ( s ) Δ s . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-59/MediaObjects/13661_2011_Article_104_Equ12_HTML.gif
      (2.11)
      By (2.11), we get
      0 σ ( T ) u ( s ) B 1 ( s ) Δ s = 1 1 - P 11 0 σ ( T ) B 1 ( s ) 0 σ ( T ) G 1 ( s , τ ) h 1 ( τ ) Δ τ Δ s + Q 11 1 - P 11 0 σ ( T ) u ( s ) A 1 ( s ) Δ s , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-59/MediaObjects/13661_2011_Article_104_Equ13_HTML.gif
      (2.12)
      0 σ ( T ) u ( s ) A 1 ( s ) Δ s = 1 1 - Q 21 0 σ ( T ) B 1 ( s ) 0 σ ( T ) G 1 ( s , τ ) h 1 ( τ ) Δ τ Δ s + P 21 1 - Q 21 0 σ ( T ) u ( s ) B 1 ( s ) Δ s . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-59/MediaObjects/13661_2011_Article_104_Equ14_HTML.gif
      (2.13)
      By (2.12) and (2.13), we get
      0 σ ( T ) u ( s ) A 1 ( s ) Δ s = K 11 D 21 1 - D 11 D 21 0 σ ( T ) B 1 ( s ) 0 σ ( T ) G 1 ( s , τ ) h 1 ( τ ) Δ τ Δ s + K 21 1 - D 11 D 21 0 σ ( T ) A 1 ( s ) 0 σ ( T ) G 1 ( s , τ ) h 1 ( τ ) Δ τ Δ s , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-59/MediaObjects/13661_2011_Article_104_Equ15_HTML.gif
      (2.14)
      0 σ ( T ) u ( s ) B 1 ( s ) Δ s = K 11 1 - D 11 D 21 0 T B 1 ( s ) 0 T G 1 ( s , τ ) h 1 ( τ ) Δ τ Δ s + K 21 D 11 1 - D 11 D 21 0 σ ( T ) A 1 ( s ) 0 σ ( T ) G 1 ( s , τ ) h 1 ( τ ) Δ τ Δ s . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-59/MediaObjects/13661_2011_Article_104_Equ16_HTML.gif
      (2.15)
      Substituting (2.14) and (2.15) into (2.11), we have
      u ( t ) = 0 σ ( T ) G 1 ( t , s ) h 1 ( s ) Δ s + K 11 ( a 1 t + b 1 ) + K 11 D 21 [ d 1 + c 1 ( σ ( T ) - t ) ] ρ 1 ( 1 - D 11 D 21 ) 0 σ ( T ) B 1 ( s ) 0 σ ( T ) G 1 ( s , τ ) h 1 ( τ ) Δ τ Δ s + K 21 D 11 ( a 1 t + b 1 ) + K 21 [ d 1 + c 1 ( σ ( T ) - t ) ] ρ 1 ( 1 - D 11 D 21 ) 0 σ ( T ) A 1 ( s ) 0 σ ( T ) G 1 ( s , τ ) h 1 ( τ ) Δ τ Δ s = 0 σ ( T ) G 1 ( t , s ) h 1 ( s ) Δ s + r 11 ( t ) 0 σ ( T ) B 1 ( s ) 0 σ ( T ) G 1 ( s , τ ) h 1 ( τ ) Δ τ Δ s + r 21 ( t ) 0 σ ( T ) A 1 ( s ) 0 σ ( T ) G 1 ( s , τ ) h 1 ( τ ) Δ τ Δ s = 0 σ ( T ) H 1 ( t , s ) h 1 ( s ) Δ s . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-59/MediaObjects/13661_2011_Article_104_Equ17_HTML.gif
      (2.16)
      Conversely, suppose u ( t ) = 0 σ ( T ) H 1 ( t , s ) h 1 ( s ) Δ s http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-59/MediaObjects/13661_2011_Article_104_IEq28_HTML.gif, then
      u ( t ) = 0 σ ( T ) G 1 ( t , s ) h 1 ( s ) Δ s + r 11 ( t ) 0 σ ( T ) B 1 ( s ) 0 σ ( T ) G 1 ( s , τ ) h 1 ( τ ) Δ τ Δ s + r 21 ( t ) 0 σ ( T ) A 1 ( s ) 0 σ ( T ) G 1 ( s , τ ) h 1 ( τ ) Δ τ Δ s . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-59/MediaObjects/13661_2011_Article_104_Equ18_HTML.gif
      (2.17)
      Direct differentiation of (2.17) implies
      u Δ ( t ) = 1 ρ 1 a 1 t σ ( T ) [ d 1 + c 1 ( σ ( T ) - σ ( s ) ) ] h 1 ( s ) Δ s - c 1 0 t ( a 1 σ ( s ) + b 1 ) h 1 ( s ) s + a 1 K 11 - c 1 K 11 D 21 ρ 1 ( 1 - D 11 D 21 ) 0 σ ( T ) B 1 ( s ) 0 σ ( T ) G 1 ( s , τ ) h 1 ( τ ) Δ τ Δ s + a 1 K 21 D 11 - c 1 K 21 ρ 1 ( 1 - D 11 D 21 ) 0 σ ( T ) A 1 ( s ) 0 σ ( T ) G 1 ( s , τ ) h 1 ( τ ) Δ τ Δ s http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-59/MediaObjects/13661_2011_Article_104_Equk_HTML.gif
      and
      u Δ Δ ( t ) = - h 1 ( t ) , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-59/MediaObjects/13661_2011_Article_104_Equl_HTML.gif
      and it is easy to verify that
      a 1 u ( 0 ) - b 1 u Δ ( 0 ) = 0 σ ( T ) u ( s ) A 1 ( s ) Δ s , c 1 u ( σ ( T ) ) + d 1 u Δ ( σ ( T ) ) = 0 σ ( T ) u ( s ) B 1 ( s ) Δ s . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-59/MediaObjects/13661_2011_Article_104_Equm_HTML.gif

      This completes the proof.

      Lemma 2.3. Assume that D12D22 ≠ 1 holds. Then for any h2C(I', ℝ+), the following BVP
      v Δ Δ ( t ) + h 2 ( t ) = 0 , t ( 0 , σ ( T ) ) T , a 2 v ( 0 ) - b 2 v Δ ( 0 ) = 0 σ ( T ) v ( s ) A 2 ( s ) Δ s , c 2 v ( σ ( T ) ) + d 2 v Δ ( σ ( T ) ) = 0 σ ( T ) v ( s ) B 2 ( s ) Δ s http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-59/MediaObjects/13661_2011_Article_104_Equn_HTML.gif
      has a solution
      v ( t ) = 0 σ ( T ) H 2 ( t , s ) h 2 ( s ) Δ s , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-59/MediaObjects/13661_2011_Article_104_Equo_HTML.gif
      where
      H 2 ( t , s ) = G 2 ( t , s ) + r 12 ( t ) 0 σ ( T ) B 2 ( τ ) G 2 ( τ , s ) Δ τ + r 22 ( t ) 0 σ ( T ) A 2 ( τ ) G 2 ( τ , s ) Δ τ , G 2 ( t , s ) = 1 ρ 2 ( a 2 σ ( s ) + b 2 ) [ d 2 + c 2 ( σ ( T ) - t ) ] , σ ( s ) < t , ( a 2 t + b 2 ) [ d 2 + c 2 ( σ ( T ) - σ ( s ) ) ] , t σ ( s ) , r 12 ( t ) = K 12 ( a 2 t + b 2 ) + K 12 D 22 [ d 2 + c 2 ( σ ( T ) - t ) ] ρ 2 ( 1 - D 12 D 22 ) , r 22 ( t ) = K 22 D 12 ( a 2 t + b 2 ) + K 22 [ d 2 + c 2 ( σ ( T ) - t ) ] ρ 2 ( 1 - D 12 D 22 ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-59/MediaObjects/13661_2011_Article_104_Equp_HTML.gif

      Proof. The proof is similar to that of Lemma 2.2 and will omit it here.

      Lemma 2.4. Suppose that (H1) is satisfied, for all t, sI and i = 1, 2, we have
      1. (i)

        G i (t, s) > 0, H i (t, s) > 0,

         
      2. (ii)

        Lim i G i (σ(s), s) ≤ H i (t, s) ≤ M i G i (σ(s), s),

         
      3. (iii)

        mG i (σ(s), s) ≤ H i (t, s) ≤ MG i (σ(s), s),

         
      where
      M i = 1 + r 1 i ¯ 0 σ ( T ) B i ( τ ) Δ τ + r 2 i ¯ 0 σ ( T ) A i ( τ ) Δ τ , r j i ¯ = max 0 t 1 r i ( t ) , m i = 1 + r 1 i ¯ ( t ) 0 σ ( T ) B i ( τ ) Δ τ + r 2 i ¯ ( t ) 0 σ ( T ) A i ( τ ) Δ τ , r j i ¯ = min 0 t 1 r i ( t ) , M = max { M 1 , M 2 } , m = min { L 1 m 1 , L 2 m 2 } , L i = min d i d i + c i , b i a i + b i , i , j = 1 , 2 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-59/MediaObjects/13661_2011_Article_104_Equq_HTML.gif
      Proof. It is easy to verify that G i (t, s) > 0, H i (t, s) > 0 and G i (t, s) ≤ G i (σ(s), s), for all t, sI. Since
      G i ( t , s ) G i ( σ ( s ) , s ) = d i + c z ( σ ( T ) - t ) d i + c z ( σ ( T ) - σ ( s ) ) , σ ( s ) < t , a i t + b i a i σ ( s ) + b i , σ ( s ) t . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-59/MediaObjects/13661_2011_Article_104_Equr_HTML.gif
      Thus G i (t, s)/G i (σ(s), s) ≥ L i and we have
      G i ( t , s ) L i G i ( σ ( s ) , s ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-59/MediaObjects/13661_2011_Article_104_Equs_HTML.gif
      On the one hand, from the definition of L i and m i , for all t, sI, we have
      H i ( t , s ) = G i ( t , s ) + r 1 i ( t ) 0 σ ( T ) B i ( τ ) G i ( τ , s ) Δ τ + r 2 i ( t ) 0 σ ( T ) A i ( τ ) G i ( τ , s ) Δ τ L i G i ( σ ( s ) , s ) 1 + r 1 i ( t ) 0 σ ( T ) B i ( τ ) Δ τ + r 2 i ( t ) 0 σ ( T ) A i ( τ ) Δ τ L i m i G i ( σ ( s ) , s ) , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-59/MediaObjects/13661_2011_Article_104_Equt_HTML.gif
      and on the other hand, we obtain easily that from the definition of M i , for all t, sI,
      H i ( t , s ) G i ( σ ( s ) , s ) + r 1 i ( t ) 0 σ ( T ) B i ( τ ) G i ( σ ( s ) , s ) Δ τ + r 2 i ( t ) 0 σ ( T ) A i ( τ ) G i ( σ ( s ) , s ) Δ τ M i G i ( σ ( s ) , s ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-59/MediaObjects/13661_2011_Article_104_Equu_HTML.gif

      Finally, it is easy to verify that mG i (σ(s), s) ≤ H i (t, s) ≤ MG i (σ(s), s). This completes the proof.

      Lemma 2.5. [23] Let E be a Banach space and P be a cone in E. Assume that Ω1 and Ω2 are bounded open subsets of E, such that 0 ∈ Ω1, Ω 1 ¯ Ω 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-59/MediaObjects/13661_2011_Article_104_IEq29_HTML.gif, and let T : P ( Ω 2 ¯ \ Ω 1 ) P http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-59/MediaObjects/13661_2011_Article_104_IEq30_HTML.gif be a completely continuous operator such that either

      (i) ||Tu|| ≤ ||u||, ∀uP ∩ ∂Ω1 and ||Tu|| ≥ ||u||, ∀uP ∩ ∂Ω2, or

      (ii) ||Tu|| ≥ ||u||, ∀uP ∩ ∂Ω1 and ||Tu|| ≤ ||u||, ∀uP ∩ ∂Ω2

      holds. Then T has a fixed point in P ( Ω 2 ¯ \ Ω 1 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-59/MediaObjects/13661_2011_Article_104_IEq31_HTML.gif.

      To obtain the existence of positive solutions for system (2.1), we construct a cone P in the Banach space Q = C(I, ℝ+) × C(I, ℝ+) equipped with the norm | | ( u , v ) | | = | | u | | + | | v | | = max t I | u | + max t I | v | http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-59/MediaObjects/13661_2011_Article_104_IEq32_HTML.gif by
      P = ( u , v ) Q | u ( t ) 0 , v ( t ) 0 , min t I ( u ( t ) + v ( t ) ) m M | | ( u , v ) | | . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-59/MediaObjects/13661_2011_Article_104_Equv_HTML.gif

      It is easy to see that P is a cone in Q.

      Define two operators T λ , T μ : PC(I, ℝ+) by
      T λ ( u , v ) ( t ) = λ 0 T H 1 ( t , s ) f ( t , A 2 u , A 1 u , A 0 u , A 2 v , A 1 v , A 0 v ) Δ s , t I , T μ ( u , v ) ( t ) = μ 0 T H 2 ( t , s ) g ( t , A 2 u , A 1 u , A 0 u , A 2 v , A 1 v , A 0 v ) Δ s , t I . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-59/MediaObjects/13661_2011_Article_104_Equw_HTML.gif
      Then we can define an operator T : PC(I, ℝ+) by
      T ( u , v ) = ( T λ ( u , v ) , T μ ( u , v ) ) , ( u , v ) P . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-59/MediaObjects/13661_2011_Article_104_Equx_HTML.gif

      Lemma 2.6. Let (H1) hold. Then T : PP is completely continuous.

      Proof. Firstly, we prove that T : PP. In fact, for all (u, v) ∈ P and tI, by Lemma 2.4(i) and (H1), it is obvious that T λ (u, v)(t) > 0, T μ (u, v)(t) > 0. In addition, we have
      T λ ( u , v ) ( t ) = λ 0 σ ( T ) H 1 ( t , s ) f ( t , A 2 u , A 1 u , A 0 u , A 2 v , A 1 v , A 0 v ) Δ s λ M 0 σ ( T ) G 1 ( σ ( s ) , s ) f ( t , A 2 u , A 1 u , A 0 u , A 2 v , A 1 v , A 0 v ) Δ s , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-59/MediaObjects/13661_2011_Article_104_Equ19_HTML.gif
      (2.18)
      which implies | | T λ ( u , v ) | | λ M 0 σ ( T ) G 1 ( σ ( s ) , s ) f ( t , A 2 u , A 1 u , A 0 u , A 2 v , A 1 v , A 0 v ) Δ s http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-59/MediaObjects/13661_2011_Article_104_IEq33_HTML.gif. And we have
      T λ ( u , v ) ( t ) λ L 1 m 1 0 σ ( T ) G 1 ( σ ( s ) , s ) f ( t , A 2 u , A 1 u , A 0 u , A 2 v , A 1 v , A 0 v ) Δ s m M | | T λ ( u , v ) | | . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-59/MediaObjects/13661_2011_Article_104_Equy_HTML.gif
      In a similar way,
      T μ ( u , v ) ( t ) m M | | T μ ( u , v ) | | . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-59/MediaObjects/13661_2011_Article_104_Equz_HTML.gif
      Therefore,
      min t I ( T λ ( u , v ) ( t ) + T μ ( u , v ) ( t ) ) m M | | T λ ( u , v ) | | + m M | | T μ ( u , v ) | | = m M | | T λ ( u , v ) , T μ ( u , v ) | | . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-59/MediaObjects/13661_2011_Article_104_Equaa_HTML.gif

      This shows that T : PP.

      Secondly, we prove that T is continuous and compact, respectively. Let {(u k , v k )} ∈ P be any sequence of functions with lim k ( u k , v k ) = ( u , v ) P http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-59/MediaObjects/13661_2011_Article_104_IEq34_HTML.gif,
      | T λ ( u k , v k ) ( t ) - T λ ( u , v ) ( t ) | λ M 1 sup t I | f ( t , A 2 u k , A 1 u k , A 0 u k , A 2 v k , A 1 v k , A 0 v k ) - f ( t , A 2 u , A 1 u , A 0 u , A 2 v , A 1 v , A 0 v ) | 0 σ ( T ) G 1 ( σ ( s ) , s ) Δ s , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-59/MediaObjects/13661_2011_Article_104_Equab_HTML.gif

      from the continuity of f, we know that ||T λ (u k , v k ) - T λ (u, v)|| → 0 as k → ∞. Hence T λ is continuous.

      T λ is compact provided that it maps bounded sets into relatively compact sets. Let f ̄ = sup t I | f ( t , A 2 u , A 1 u , A 0 u , A 2 v , A 1 v , A 0 v ) | http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-59/MediaObjects/13661_2011_Article_104_IEq35_HTML.gif, and let Ω be any bounded subset of P, then there exists r > 0 such that ||(u, v)|| ≤ r for all (u, v) ∈ Ω. Obviously, from (2.16), we know that
      T λ ( u , v ) ( t ) λ M f ̄ 0 σ ( T ) G 1 ( σ ( s ) , s ) Δ s , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-59/MediaObjects/13661_2011_Article_104_Equac_HTML.gif
      so, T λ Ω is bounded for all (u, v) ∈ Ω. Moreover, let
      L 1 = λ f ̄ ρ 1 a 1 0 σ ( T ) [ d 1 + c 1 ( σ ( T ) - σ ( s ) ) ] Δ s + c 1 0 σ ( T ) ( a 1 σ ( s ) + b 1 ) Δ s + λ f ̄ | a 1 K 11 - c 1 K 11 D 21 | ρ 1 ( 1 - D 11 D 21 ) 0 σ ( T ) B 1 ( s ) 0 σ ( T ) G 1 ( s , τ ) Δ τ Δ s + λ f ̄ | a 1 K 21 D 11 - c 1 K 21 | ρ 1 ( 1 - D 11 D 21 ) 0 σ ( T ) A 1 ( s ) 0 σ ( T ) G 1 ( s , τ ) Δ τ Δ s . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-59/MediaObjects/13661_2011_Article_104_Equad_HTML.gif
      We have
      | T λ ( u , v ) Δ ( t ) | λ ρ 1 a 1 t σ ( T ) [ d 1 + c 1 ( σ ( T ) - σ ( s ) ) ] f ( s , A 2 u , A 1 u , A 0 u , A 2 v , A 1 v , A 0 v ) Δ s - c 1 0 t ( a 1 σ ( s ) + b 1 ) f ( s , A 2 u , A 1 u , A 0 u , A 2 v , A 1 v , A 0 v ) Δ s + λ | a 1 K 11 - c 1 K 11 D 21 | ρ 1 ( 1 - D 11 D 21 ) 0 σ ( T ) B 1 ( s ) 0 σ ( T ) G 1 ( s , τ ) f ( s , A 2 u , A 1 u , A 0 u , A 2 v , A 1 v , A 0 v ) Δ τ Δ s + λ | a 1 K 21 D 11 - c 1 K 21 | ρ 1 ( 1 - D 11 D 21 ) 0 σ ( T ) A 1 ( s ) 0 σ ( T ) G 1 ( s , τ ) f ( s , A 2 u , A 1 u , A 0 u , A 2 v , A 1 v , A 0 v ) Δ τ Δ s L 1 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-59/MediaObjects/13661_2011_Article_104_Equae_HTML.gif
      Thus, for any (u, v) ∈ Ω and ∀ε > 0, let δ = ε L 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-59/MediaObjects/13661_2011_Article_104_IEq36_HTML.gif, then for t1, t2I, |t1 - t2| < δ, we have
      | T λ ( u , v ) ( t 1 ) - T λ ( u , v ) ( t 2 ) | L 1 | t 1 - t 2 | < ε . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-59/MediaObjects/13661_2011_Article_104_Equaf_HTML.gif

      So, for all (u, v) ∈ Ω, T λ Ω is equicontinuous. By Ascoli-Arzela theorem, we obtain that T λ : PP is completely continuous. In a similar way, we can prove that T μ : PP is completely continuous. Therefore, T : PP is completely continuous. This completes the proof.

      3 Main results

      In this section, we will give our main results on multiplicity of positive solutions of system (1.1). In the following, for convenience, we set
      f β = lim i = 1 6 | φ i | β inf t I f ( t , φ 1 , , φ 6 ) q 1 ( t ) i = 1 6 | φ i | , f α = lim i = 1 6 | φ i | α sup t I f ( t , φ 1 , , φ 6 ) q 2 ( t ) i = 1 6 | φ i | , g β = lim i = 1 6 | φ i | β inf t I f ( t , φ 1 , , φ 6 ) q 3 ( t ) i = 1 6 | φ i | , g α = lim i = 1 6 | φ i | α sup t I f ( t , φ 1 , , φ 6 ) q 4 ( t ) i = 1 6 | φ i | , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-59/MediaObjects/13661_2011_Article_104_Equag_HTML.gif
      where q i (t), q j (t) ∈ C rd (I', ℝ+) satisfy
      0 < 0 σ ( T ) G 1 ( σ ( s ) , s ) q i ( s ) Δ s < + ( i = 1 , 2 ) , 0 < 0 σ ( T ) G 2 ( σ ( s ) , s ) q j ( s ) Δ s < + ( j = 3 , 4 ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-59/MediaObjects/13661_2011_Article_104_Equah_HTML.gif

      Theorem 3.1. Assume that (H1) holds. Assume further that

      (H2) there exist a constant R > 0, and two functions p i (t) ∈ C rd (I, R+) satisfying 0 < 0 σ ( T ) G i ( σ ( s ) , s ) p i ( s ) Δ s < + ( i = 1 , 2 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-59/MediaObjects/13661_2011_Article_104_IEq37_HTML.gif such that
      f ( t , φ 1 , , φ 6 ) R p 1 ( t ) , t I , 0 < i = 1 6 | φ i | R , g ( t , φ 1 , , φ 6 ) R p 2 ( t ) , t I , 0 < i = 1 6 | φ i | R , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-59/MediaObjects/13661_2011_Article_104_Equai_HTML.gif

      and one of the folloeing conditions is satisfied

      (E1) λ ( M 3 f 0 , M 4 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-59/MediaObjects/13661_2011_Article_104_IEq38_HTML.gif, μ ( N 3 g , N 4 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-59/MediaObjects/13661_2011_Article_104_IEq39_HTML.gif,

      (E2) λ ( M 3 f , M 4 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-59/MediaObjects/13661_2011_Article_104_IEq40_HTML.gif, μ ( N 3 g 0 , N 4 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-59/MediaObjects/13661_2011_Article_104_IEq41_HTML.gif,

      (E3) λ ( M 3 F α , M 4 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-59/MediaObjects/13661_2011_Article_104_IEq42_HTML.gif, μ ∈ (0, N4),

      (E4) λ ∈ (0, M4), μ ( N 3 G α , N 4 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-59/MediaObjects/13661_2011_Article_104_IEq43_HTML.gif,

      where
      M 3 = m 2 M 0 σ ( T ) G 1 ( σ ( s ) , s ) q 1 ( s ) Δ s - 1 , M 4 = O 1 M N 0 σ ( T ) G 1 ( σ ( s ) , s ) p 1 ( s ) Δ s - 1 , N 3 = m 2 M 0 σ ( T ) G 2 ( σ ( s ) , s ) q 3 ( s ) Δ s - 1 , N 4 = o 2 M N 0 σ ( T ) G 2 ( σ ( s ) , s ) p 2 ( s ) Δ s - 1 , F α = min { f 0 , f } , G α = { g 0 , g } , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-59/MediaObjects/13661_2011_Article_104_Equaj_HTML.gif

      O1, O2 satisfy 1 O 1 + 1 O 2 1 , N = 1 + σ ( T ) + ( σ ( T ) ) 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-59/MediaObjects/13661_2011_Article_104_IEq44_HTML.gif. Then system (1.1) has at least two positive solutions.

      Proof. We only prove the case in which (H2) and (E1) hold, the other case can be proved similarly. Firstly, from (2.2), we have
      i = 0 2 A i u ( t ) + i = 0 2 A i v ( t ) | | ( u , v ) | | + σ ( T ) | | ( u , v ) | | + ( σ ( T ) ) 2 | | ( u , v ) | | = N | | ( u , v ) | | . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-59/MediaObjects/13661_2011_Article_104_Equ20_HTML.gif
      (3.1)
      Take R 1 = R N http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-59/MediaObjects/13661_2011_Article_104_IEq45_HTML.gif, and let Ω1 = {(u, v) ∈ Q; ||(u, v)|| < R1}. For any tI, (u, v) ∈ ∂Ω1P, it follows from λ < M4, μ < N4 and (H2) that
      T λ ( u , v ) ( t ) = λ 0 σ ( T ) H 1 ( t , s ) f ( s , A 2 u , A 1 u , A 0 u , A 2 v , A 1 v , A 0 v ) Δ s M 4 M 0 σ ( T ) G 1 ( σ ( s ) , s ) f ( s , A 2 u , A 1 u , A 0 u , A 2 v , A 1 v , A 0 v ) Δ s M 4 M R 0 σ ( T ) G 1 ( σ ( s ) , s ) p 1 ( s ) Δ s = N M 4 M R 1 0 σ ( T ) G 1 ( σ ( s ) , s ) p 1 ( s ) Δ s 1 O 1 R 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-59/MediaObjects/13661_2011_Article_104_Equak_HTML.gif
      and
      T μ ( u , v ) ( t ) = μ 0 σ ( T ) H 2 ( t , s ) g ( s , A 2 u , A 1 u , A 0 u , A 2 v , A 1 v , A 0 v ) Δ s N 4 M 0 σ ( T ) G 2 ( σ ( s ) , s ) g ( s , A 2 u , A 1 u , A 0 u , A 2 v , A 1 v , A 0 v ) Δ s N 4 M R 0 σ ( T ) G 2 ( σ ( s ) , s ) p 2 ( s ) Δ s = N 4 M N R 1 0 σ ( T ) G 2 ( σ ( s ) , s ) p 2 ( s ) Δ s 1 O 2 R 1 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-59/MediaObjects/13661_2011_Article_104_Equal_HTML.gif
      Consequently, for any (u, v) ∈ ∂Ω1P, we have
      | | T ( u , v ) | | = | | T λ ( u , v ) | | + | | T μ ( u , v ) | | < 1 O 1 R 1 + 1 O 2 R 1 R 1 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-59/MediaObjects/13661_2011_Article_104_Equ21_HTML.gif
      (3.2)
      Second, from λ > M 3 f 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-59/MediaObjects/13661_2011_Article_104_IEq46_HTML.gif, we can choose ε1 > 0 such that λf0 > M3 + ε1, then there exists 0 < l1 < NR1 such that for any i = 1 6 | φ i | < l 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-59/MediaObjects/13661_2011_Article_104_IEq47_HTML.gif and tI,
      f ( t , φ 1 , , φ 6 ) M 3 + ε 1 λ q 1 ( t ) i = 1 6 | φ i | . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-59/MediaObjects/13661_2011_Article_104_Equam_HTML.gif
      And since
      i = 0 2 A i u ( t ) + i = 0 2 A i v ( t ) u ( t ) + v ( t ) m M | | ( u , v ) | | . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-59/MediaObjects/13661_2011_Article_104_Equ22_HTML.gif
      (3.3)
      Take
      R 2 = l 1 N < R 1 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-59/MediaObjects/13661_2011_Article_104_Equan_HTML.gif
      For all (u, v) ∈ Ω2P, where Ω2 = {(u, v) ∈ Q; ||(u, v)|| < R2}, we have
      i = 0 2 A i u ( t ) + i = 0 2 A i v ( t ) u ( t ) + v ( t ) m M R 2 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-59/MediaObjects/13661_2011_Article_104_Equao_HTML.gif
      Thus, for all (u, v) ∈ Ω2P, we have
      T λ ( u , v ) ( t ) λ m 0 σ ( T ) G 1 ( σ ( s ) , s ) f ( s , A 2 u , A 1 u , A 0 u , A 2 v , A 1 v , A 0 v ) Δ s m ( M 3 + ε 1 ) i = 0 2 | A i u | + i = 0 2 | A i v | 0 σ ( T ) G 1 ( σ ( s ) , s ) q 1 ( s ) Δ s M 3 m 2 M R 2 0 σ ( T ) G 1 ( σ ( s ) , s ) q 1 ( s ) Δ s = R 2 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-59/MediaObjects/13661_2011_Article_104_Equap_HTML.gif
      Consequently, for all (u, v) ∈ Ω2P, we have
      | | T ( u , v ) | | | | T λ ( u , v ) | | | | ( u , v ) | | . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-59/MediaObjects/13661_2011_Article_104_Equ23_HTML.gif
      (3.4)
      Finally, from μ > N3/g, we can choose ε2 > 0 such that μg > N3 + ε2. then, there exists l 2 > m M R http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-59/MediaObjects/13661_2011_Article_104_IEq48_HTML.gif such that for any i = 1 6 | φ i | < l 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-59/MediaObjects/13661_2011_Article_104_IEq49_HTML.gif and tI,
      g ( t , φ 1 , , φ 6 ) N 3 + ε 2 μ q 3 ( t ) i = 1 6 | φ i | . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-59/MediaObjects/13661_2011_Article_104_Equaq_HTML.gif
      Take
      R 3 = m M - 1 l 2 > R 1 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-59/MediaObjects/13661_2011_Article_104_Equar_HTML.gif
      For all (u, v) ∈ Ω3P, where Ω3 = {(u, v) ∈ Q; ||(u, v)|| < R3}, from (3.3), we have
      i = 0 2 A i u ( t ) + i = 0 2 A i v ( t ) m M R 3 = l 2 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-59/MediaObjects/13661_2011_Article_104_Equ24_HTML.gif
      (3.5)
      Thus, for all (u, v) ∈ Ω3P, we have
      T μ ( u , v ) ( t ) μ m 0 σ ( T ) G 2 ( σ ( s ) , s ) g ( s , A 2 u , A 1 u , A 0 u , A 2 v , A 1 v , A 0 v ) Δ s m ( N 3 + ε 2 ) i = 0 2 | A i u | + i = 0 2 | A i v | 0 σ ( T ) G 2 ( σ ( s ) , s ) q 3 ( s ) Δ s N 3 m 2 M R 3 0 σ ( T ) G 2 ( σ ( s ) , s ) q 3 ( s ) Δ s = R 3 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-59/MediaObjects/13661_2011_Article_104_Equas_HTML.gif
      Consequently, for all (u, v) ∈ Ω3P, we have
      | | T ( u , v ) | | | | T μ ( u , v ) | | | | ( u , v ) | | . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-59/MediaObjects/13661_2011_Article_104_Equ25_HTML.gif
      (3.6)

      From (3.2), (3.4), and (ii) of Lemma 2.5, it follows that system (2.1) has one positive solution (u1, v1) ∈ P with R2 ≤ ||(u1, v1)|| ≤ R1. Therefore, from Lemma 2.1, it follows that system (1.1) has one positive solution (x1, y1). In the same way, from (3.2), (3.6), and (i) of Lemma 2.5, it follows that system (2.1) has one positive solution (u2, v2) ∈ P with R1 ≤ ||(u2, v2)|| ≤ R3. Therefore, from Lemma 2.1, it follows that system (1.1) has one positive solution (x2, y2). Above all, system (1.1) has at least two positive solutions. This completes the proof.

      Theorem 3.2. Assume that (H1) holds. Suppose further that

      (H3) there exist a constant R0 > 0, and two functions w i (t) ∈ C rd (I, R+) (i = 1, 2) satisfying 0 < 0 σ ( T ) G i ( σ ( s ) , s ) w i ( s ) Δ s < + http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-59/MediaObjects/13661_2011_Article_104_IEq50_HTML.gif such that
      f ( t , φ 1 , , φ 6 ) R 0 w 1 ( t ) , t I , i = 1 6 | φ i | > R 0 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-59/MediaObjects/13661_2011_Article_104_Equ26_HTML.gif
      (3.7)
      or
      g ( t , φ 1 , , φ 6 ) R 0 w 2 ( t ) , t I , i = 1 6 | φ i | > R 0 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-59/MediaObjects/13661_2011_Article_104_Equ27_HTML.gif
      (3.8)
      Then system (1.1) has at least two positive solutions for each λ ( M 5 , M 6 F 0 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-59/MediaObjects/13661_2011_Article_104_IEq51_HTML.gif and μ ( N 5 , N 6 G 0 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-59/MediaObjects/13661_2011_Article_104_IEq52_HTML.gif, where
      M 5 = m 2 M 0 σ ( T ) G 1 ( σ ( s ) , s ) w 1 ( s ) Δ s - 1 , N 5 = m 2 M 0 σ ( T ) G 2 ( σ ( s ) , s ) w 2 ( s ) Δ s - 1 , M 6 = O 1 M N 0 σ ( T ) G 1 ( σ ( s ) , s ) q 2 ( s ) Δ s - 1 , N 6 = o 2 M N 0 σ ( T ) G 2 ( σ ( s ) , s ) q 4 ( s ) Δ s - 1 , F α = max { f 0 , f } < , G α = max { g 0 , g } < . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-59/MediaObjects/13661_2011_Article_104_Equat_HTML.gif

      Proof. We only prove the case in which (3.7) holds. The other case in which (3.8) holds can be proved similarly.

      Take
      R 1 = m M - 1 R 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-59/MediaObjects/13661_2011_Article_104_Equau_HTML.gif
      and let Ω 4 = { ( u , v ) Q ; | | ( u , v ) | | < R 1 } http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-59/MediaObjects/13661_2011_Article_104_IEq53_HTML.gif. For any tI, (u, v) ∈ ∂Ω4P, it follows from λ > M5 and (H3) that
      T λ ( u , v ) ( t ) = λ 0 σ ( T ) H 1 ( t , s ) f ( s , A 2 u , A 1 u , A 0 u , A 2 v , A 1 v , A 0 v ) Δ s M 5 m 0 σ ( T ) G 1 ( σ ( s ) , s ) f ( s , A 2