Positive periodic solutions for a second-order functional differential equation

  • Yongxiang Li1Email author and

    Affiliated with

    • Qiang Li1

      Affiliated with

      Boundary Value Problems20122012:140

      DOI: 10.1186/1687-2770-2012-140

      Received: 12 June 2012

      Accepted: 12 November 2012

      Published: 27 November 2012

      Abstract

      In this paper, the existence results of positive ω-periodic solutions are obtained for the second-order functional differential equation

      u ¨ ( t ) = f ( t , u ( t ) , u ˙ ( t τ 1 ( t ) ) , , u ˙ ( t τ n ( t ) ) ) , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-140/MediaObjects/13661_2012_Article_224_Equa_HTML.gif

      where f : R × [ 0 , ) × R n R http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-140/MediaObjects/13661_2012_Article_224_IEq1_HTML.gif is a continuous function which is ω-periodic in t, τ i C ( R , [ 0 , ) ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-140/MediaObjects/13661_2012_Article_224_IEq2_HTML.gif is a ω-periodic function, i = 1 , 2 , , n http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-140/MediaObjects/13661_2012_Article_224_IEq3_HTML.gif. Our discussion is based on the fixed point index theory in cones.

      MSC:34C25, 47H10.

      Keywords

      functional differential equation positive periodic solution cone fixed point index

      1 Introduction

      In this paper, we discuss the existence of positive ω-periodic solutions of the second-order functional differential equation with the delay terms of first-order derivative in nonlinearity,
      u ¨ ( t ) = f ( t , u ( t ) , u ˙ ( t τ 1 ( t ) ) , , u ˙ ( t τ n ( t ) ) ) , t R , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-140/MediaObjects/13661_2012_Article_224_Equ1_HTML.gif
      (1)

      where f : R × [ 0 , ) × R n R http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-140/MediaObjects/13661_2012_Article_224_IEq4_HTML.gif is a continuous function which is ω-periodic in t and τ i C ( R , [ 0 , ) ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-140/MediaObjects/13661_2012_Article_224_IEq5_HTML.gif is a ω-periodic delay function, i = 1 , 2 , , n http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-140/MediaObjects/13661_2012_Article_224_IEq6_HTML.gif.

      For the second-order differential equation without delay and the first-order derivative term in nonlinearity,
      u ¨ ( t ) = f ( t , u ( t ) ) , t R , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-140/MediaObjects/13661_2012_Article_224_Equ2_HTML.gif
      (2)

      the existence problems of periodic solutions have attracted many authors’ attention and concern. Many theorems and methods of nonlinear functional analysis have been applied to research the periodic problems of Equation (2), such as the upper and lower solutions method and monotone iterative technique [14], the continuation method of topological degree [57], variational method and critical point theory [810], the theory of the fixed point index in cones [1116], etc.

      In recent years, the existence of periodic solutions for the second-order delayed differential equations have also been researched by many authors; see [1724] and the references therein. In some practice models, only positive periodic solutions are significant. In [20, 21, 23], the authors obtained the existence of positive periodic solutions for some delayed second-order differential equations as a special form of the following equation:
      u ¨ ( t ) + b ( t ) u ˙ ( t ) + a ( t ) u ( t ) = f ( t , u ( t τ 1 ( t ) ) , , u ( t τ n ( t ) ) ) , t R , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-140/MediaObjects/13661_2012_Article_224_Equ3_HTML.gif
      (3)
      by using Krasnoselskii’s fixed point theorem of cone mapping or the theory of the fixed point index in cones. In these works, the positivity of Green’s function of the corresponding linear second-order periodic problems plays an important role. The positivity guarantees that the integral operators of the second-order periodic problems are cone-preserving in the cone
      P = { u C [ 0 , ω ] u ( t ) σ u , t [ 0 , ω ] } http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-140/MediaObjects/13661_2012_Article_224_Equ4_HTML.gif
      (4)

      in the Banach space C [ 0 , ω ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-140/MediaObjects/13661_2012_Article_224_IEq7_HTML.gif, where σ > 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-140/MediaObjects/13661_2012_Article_224_IEq8_HTML.gif is a constant. Hence, the fixed point theorems of cone mapping can be applied to periodic problems of the second-order delay equation (3) as well as Equation (2) (for Equation (2), see [1116]). However, few people consider the existence of positive periodic solutions of Equation (1). Since the nonlinearity of Equation (1) explicitly contains the delayed first-order derivative term, the corresponding integral operator has no definition on the cone P. Thus, the argument methods used in [20, 21, 23] are not applicable to Equation (1).

      The purpose of this paper is to discuss the existence of positive periodic solutions of Equation (1). We will use a different method to treat Equation (1). Our main results will be given in Section 3. Some preliminaries to discuss Equation (1) are presented in Section 2.

      2 Preliminaries

      Let C ω ( R ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-140/MediaObjects/13661_2012_Article_224_IEq9_HTML.gif denote the Banach space of all continuous ω-periodic function u ( t ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-140/MediaObjects/13661_2012_Article_224_IEq10_HTML.gif with the norm u C = max 0 t ω | u ( t ) | http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-140/MediaObjects/13661_2012_Article_224_IEq11_HTML.gif. Let C ω 1 ( R ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-140/MediaObjects/13661_2012_Article_224_IEq12_HTML.gif be the Banach space of all continuous differentiable ω-periodic function u ( t ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-140/MediaObjects/13661_2012_Article_224_IEq10_HTML.gif with the norm
      u C 1 = u C + u ˙ C . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-140/MediaObjects/13661_2012_Article_224_Equb_HTML.gif

      Generally, C ω n ( R ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-140/MediaObjects/13661_2012_Article_224_IEq13_HTML.gif denotes the n th-order continuous differentiable ω-periodic function space for n N http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-140/MediaObjects/13661_2012_Article_224_IEq14_HTML.gif. Let C ω + ( R ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-140/MediaObjects/13661_2012_Article_224_IEq15_HTML.gif be the cone of all nonnegative functions in C ω ( R ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-140/MediaObjects/13661_2012_Article_224_IEq16_HTML.gif.

      Let M ( 0 , π 2 ω 2 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-140/MediaObjects/13661_2012_Article_224_IEq17_HTML.gif be a constant. For h C ω ( R ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-140/MediaObjects/13661_2012_Article_224_IEq18_HTML.gif, we consider the linear second-order differential equation
      u ¨ ( t ) + M u ( t ) = h ( t ) , t R . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-140/MediaObjects/13661_2012_Article_224_Equ5_HTML.gif
      (5)
      The ω-periodic solutions of Equation (5) are closely related to the linear second-order boundary value problem
      { u ¨ ( t ) + M u ( t ) = 0 , 0 t ω , u ( 0 ) u ( ω ) = 0 , u ˙ ( 0 ) u ˙ ( ω ) = 1 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-140/MediaObjects/13661_2012_Article_224_Equ6_HTML.gif
      (6)
      see [14]. It is easy to see that problem (6) has a unique solution which is explicitly given by
      U ( t ) = cos β ( t ω 2 ) 2 β sin β ω 2 , 0 t ω , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-140/MediaObjects/13661_2012_Article_224_Equ7_HTML.gif
      (7)

      where β = M http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-140/MediaObjects/13661_2012_Article_224_IEq19_HTML.gif. By [[14], Lemma 1], we have

      Lemma 2.1 Let M ( 0 , π 2 ω 2 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-140/MediaObjects/13661_2012_Article_224_IEq17_HTML.gif. Then, for every h C ω ( R ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-140/MediaObjects/13661_2012_Article_224_IEq20_HTML.gif, the linear equation (5) has a unique ω-periodic solution u ( t ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-140/MediaObjects/13661_2012_Article_224_IEq10_HTML.gif which is given by
      u ( t ) = t ω t U ( t s ) h ( s ) d s : = S h ( t ) , t R . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-140/MediaObjects/13661_2012_Article_224_Equ8_HTML.gif
      (8)

      Moreover, S : C ω ( R ) C ω 1 ( R ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-140/MediaObjects/13661_2012_Article_224_IEq21_HTML.gif is a completely continuous linear operator.

      Since U ( t ) > 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-140/MediaObjects/13661_2012_Article_224_IEq22_HTML.gif, for every t [ 0 , ω ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-140/MediaObjects/13661_2012_Article_224_IEq23_HTML.gif, by (8), if h C ω + ( R ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-140/MediaObjects/13661_2012_Article_224_IEq24_HTML.gif and h ( t ) 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-140/MediaObjects/13661_2012_Article_224_IEq25_HTML.gif, then the ω-periodic solution of Equation (5) u ( t ) > 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-140/MediaObjects/13661_2012_Article_224_IEq26_HTML.gif for every t R http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-140/MediaObjects/13661_2012_Article_224_IEq27_HTML.gif, and we term it the positive ω-periodic solution. Let
      U ¯ = max 0 t ω U ( t ) = 1 2 β sin β ω 2 , U ̲ = min 0 t ω U ( t ) = cos β ω 2 2 β sin β ω 2 , U ¯ 1 = max 0 t ω | U ˙ ( t ) | = max 0 t ω | sin β ( t ω 2 ) | 2 sin β ω 2 = 1 2 , σ = U ̲ U ¯ = cos β ω 2 , C 0 = U ¯ 1 U ̲ = β tan β ω 2 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-140/MediaObjects/13661_2012_Article_224_Equ9_HTML.gif
      (9)
      Define a set K in C ω 1 ( R ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-140/MediaObjects/13661_2012_Article_224_IEq12_HTML.gif by
      K = { u C ω 1 ( R ) u ( t ) σ u C , | u ˙ ( τ ) | C 0 u ( t ) , τ , t R } . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-140/MediaObjects/13661_2012_Article_224_Equ10_HTML.gif
      (10)

      It is easy to verify that K is a closed convex cone in C ω 1 ( R ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-140/MediaObjects/13661_2012_Article_224_IEq12_HTML.gif.

      Lemma 2.2 Let M ( 0 , π 2 ω 2 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-140/MediaObjects/13661_2012_Article_224_IEq17_HTML.gif. Then, for every h C ω + ( R ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-140/MediaObjects/13661_2012_Article_224_IEq28_HTML.gif, the positive ω-periodic solution of Equation (5) u = S h K http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-140/MediaObjects/13661_2012_Article_224_IEq29_HTML.gif. Namely, S ( C ω + ( R ) ) K http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-140/MediaObjects/13661_2012_Article_224_IEq30_HTML.gif.

      Proof Let h C ω + ( R ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-140/MediaObjects/13661_2012_Article_224_IEq28_HTML.gif, u = S h http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-140/MediaObjects/13661_2012_Article_224_IEq31_HTML.gif. For every t R http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-140/MediaObjects/13661_2012_Article_224_IEq27_HTML.gif, from (8) it follows that
      u ( t ) = t ω t U ( t s ) h ( s ) d s U ¯ t ω t h ( s ) d s = U ¯ 0 ω h ( s ) d s , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-140/MediaObjects/13661_2012_Article_224_Equc_HTML.gif
      and therefore,
      u C U ¯ 0 ω h ( s ) d s . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-140/MediaObjects/13661_2012_Article_224_Equd_HTML.gif
      Using (8), we obtain that
      u ( t ) = t ω t U ( t s ) h ( s ) d s U ̲ t ω t h ( s ) d s = U ̲ 0 ω h ( s ) d s σ u C . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-140/MediaObjects/13661_2012_Article_224_Eque_HTML.gif
      For every τ R http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-140/MediaObjects/13661_2012_Article_224_IEq32_HTML.gif, since
      u ˙ ( τ ) = τ ω τ U ˙ ( τ s ) h ( s ) d s , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-140/MediaObjects/13661_2012_Article_224_Equf_HTML.gif
      we have
      | u ˙ ( τ ) | τ ω τ | U ˙ ( τ s ) | h ( s ) d s U ¯ 1 τ ω τ h ( s ) d s = U ¯ 1 0 ω h ( s ) d s = C 0 U ̲ 0 ω h ( s ) d s C 0 u ( t ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-140/MediaObjects/13661_2012_Article_224_Equg_HTML.gif

      Hence, u K http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-140/MediaObjects/13661_2012_Article_224_IEq33_HTML.gif. □

      Now we consider the nonlinear delay equation (1). Hereafter, we assume that the nonlinearity f satisfies the condition

      (F0) There exists M ( 0 , π 2 ω 2 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-140/MediaObjects/13661_2012_Article_224_IEq17_HTML.gif such that
      f ( t , x , y 1 , , y n ) + M x 0 , x 0 , t R , ( y 1 , , y n ) R n . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-140/MediaObjects/13661_2012_Article_224_Equh_HTML.gif
      Let f 1 ( t , x , y 1 , , y n ) = f ( t , x , y 1 , , y n ) + M x http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-140/MediaObjects/13661_2012_Article_224_IEq34_HTML.gif, then f 1 ( t , x , y 1 , , y n ) 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-140/MediaObjects/13661_2012_Article_224_IEq35_HTML.gif for x 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-140/MediaObjects/13661_2012_Article_224_IEq36_HTML.gif, t R http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-140/MediaObjects/13661_2012_Article_224_IEq27_HTML.gif, ( y 1 , , y n ) R n http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-140/MediaObjects/13661_2012_Article_224_IEq37_HTML.gif, and Equation (1) is rewritten to
      u ¨ ( t ) + M u ( t ) = f 1 ( t , u ( t ) , u ˙ ( t τ 1 ( t ) ) , , u ˙ ( t τ n ( t ) ) ) , t R . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-140/MediaObjects/13661_2012_Article_224_Equ11_HTML.gif
      (11)
      For every u K http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-140/MediaObjects/13661_2012_Article_224_IEq33_HTML.gif, set
      F ( u ) ( t ) : = f 1 ( t , u ( t ) , u ˙ ( t τ 1 ( t ) ) , , u ˙ ( t τ n ( t ) ) ) , t R . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-140/MediaObjects/13661_2012_Article_224_Equ12_HTML.gif
      (12)
      Then F : K C ω + ( R ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-140/MediaObjects/13661_2012_Article_224_IEq38_HTML.gif is continuous. We define an integral operator A : K C ω 1 ( R ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-140/MediaObjects/13661_2012_Article_224_IEq39_HTML.gif by
      A u ( t ) = t ω t U ( t s ) F ( u ) ( s ) d s = ( S F ) ( t ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-140/MediaObjects/13661_2012_Article_224_Equ13_HTML.gif
      (13)

      By the definition of the operator S, the positive ω-periodic solution of Equation (1) is equivalent to the nontrivial fixed point of A. From assumption (F0), Lemma 2.1 and Lemma 2.2, we easily see that

      Lemma 2.3 A ( K ) K http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-140/MediaObjects/13661_2012_Article_224_IEq40_HTML.gif and A : K K http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-140/MediaObjects/13661_2012_Article_224_IEq41_HTML.gif is completely continuous.

      We will find the non-zero fixed point of A by using the fixed point index theory in cones. We recall some concepts and conclusions on the fixed point index in [25, 26]. Let E be a Banach space and K E http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-140/MediaObjects/13661_2012_Article_224_IEq42_HTML.gif be a closed convex cone in E. Assume Ω is a bounded open subset of E with the boundary Ω, and K Ω http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-140/MediaObjects/13661_2012_Article_224_IEq43_HTML.gif. Let A : K Ω ¯ K http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-140/MediaObjects/13661_2012_Article_224_IEq44_HTML.gif be a completely continuous mapping. If A u u http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-140/MediaObjects/13661_2012_Article_224_IEq45_HTML.gif for any u K Ω http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-140/MediaObjects/13661_2012_Article_224_IEq46_HTML.gif, then the fixed point index i ( A , K Ω , K ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-140/MediaObjects/13661_2012_Article_224_IEq47_HTML.gif has a definition. One important fact is that if i ( A , K Ω , K ) 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-140/MediaObjects/13661_2012_Article_224_IEq48_HTML.gif, then A has a fixed point in K Ω http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-140/MediaObjects/13661_2012_Article_224_IEq49_HTML.gif. The following two lemmas are needed in our argument.

      Lemma 2.4 ([26])

      Let Ω be a bounded open subset of E with θ Ω http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-140/MediaObjects/13661_2012_Article_224_IEq50_HTML.gif and A : K Ω ¯ K http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-140/MediaObjects/13661_2012_Article_224_IEq51_HTML.gif be a completely continuous mapping. If λ A u u http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-140/MediaObjects/13661_2012_Article_224_IEq52_HTML.gif for every u K Ω http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-140/MediaObjects/13661_2012_Article_224_IEq46_HTML.gif and 0 < λ 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-140/MediaObjects/13661_2012_Article_224_IEq53_HTML.gif, then i ( A , K Ω , K ) = 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-140/MediaObjects/13661_2012_Article_224_IEq54_HTML.gif.

      Lemma 2.5 ([26])

      Let Ω be a bounded open subset of E and A : K Ω ¯ K http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-140/MediaObjects/13661_2012_Article_224_IEq55_HTML.gif be a completely continuous mapping. If there exists an e K { θ } http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-140/MediaObjects/13661_2012_Article_224_IEq56_HTML.gif such that u A u μ e http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-140/MediaObjects/13661_2012_Article_224_IEq57_HTML.gif for every u K Ω http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-140/MediaObjects/13661_2012_Article_224_IEq58_HTML.gif and μ 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-140/MediaObjects/13661_2012_Article_224_IEq59_HTML.gif, then i ( A , K Ω , K ) = 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-140/MediaObjects/13661_2012_Article_224_IEq60_HTML.gif.

      In the next section, we will use Lemma 2.4 and Lemma 2.5 to discuss the existence of positive ω-periodic solutions of Equation (1).

      3 Main results

      We consider the existence of positive ω-periodic solutions of the functional differential equation (1). Let f C ( R × [ 0 , ) × R n ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-140/MediaObjects/13661_2012_Article_224_IEq61_HTML.gif satisfy assumption (F0) and f ( t , x , y 1 , , y n ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-140/MediaObjects/13661_2012_Article_224_IEq62_HTML.gif be ω-periodic in t. Let C 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-140/MediaObjects/13661_2012_Article_224_IEq63_HTML.gif be the constant defined by (9) and I = [ 0 , ω ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-140/MediaObjects/13661_2012_Article_224_IEq64_HTML.gif. For convenience, we introduce the notations
      f 0 = lim inf x 0 + min t I , | y i | C 0 | x | , i = 1 , , n ( f ( t , x , y 1 , , y n ) / x ) , f 0 = lim sup x 0 + max t I , | y i | C 0 | x | , i = 1 , , n ( f ( t , x , y 1 , , y n ) / x ) , f = lim inf x + min t I , | y i | C 0 | x | , i = 1 , , n ( f ( t , x , y 1 , , y n ) / x ) , f = lim sup x + max t I , | y i | C 0 | x | , i = 1 , , n ( f ( t , x , y 1 , , y n ) / x ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-140/MediaObjects/13661_2012_Article_224_Equi_HTML.gif

      Our main results are as follows.

      Theorem 3.1 Let f C ( R × [ 0 , ) × R n ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-140/MediaObjects/13661_2012_Article_224_IEq65_HTML.gif and f ( t , x , y 1 , , y n ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-140/MediaObjects/13661_2012_Article_224_IEq66_HTML.gif be ω-periodic in t, τ 1 , , τ n C ω + ( R ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-140/MediaObjects/13661_2012_Article_224_IEq67_HTML.gif. If f satisfies assumption (F0) and the condition

      (F1) f 0 < 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-140/MediaObjects/13661_2012_Article_224_IEq68_HTML.gif, f > 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-140/MediaObjects/13661_2012_Article_224_IEq69_HTML.gif,

      then Equation (1) has at least one positive ω-periodic solution.

      Theorem 3.2 Let f C ( R × [ 0 , ) × R n ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-140/MediaObjects/13661_2012_Article_224_IEq65_HTML.gif and f ( t , x , y 1 , , y n ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-140/MediaObjects/13661_2012_Article_224_IEq66_HTML.gif be ω-periodic in t, τ 1 , , τ n C ω + ( R ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-140/MediaObjects/13661_2012_Article_224_IEq67_HTML.gif. If f satisfies assumption (F0) and the conditions

      (F2) f 0 > 0 , f < 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-140/MediaObjects/13661_2012_Article_224_IEq70_HTML.gif,

      then Equation (1) has at least one positive ω-periodic solution.

      In Theorem 3.1, the condition (F1) allows f ( t , x , y 1 , , y n ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-140/MediaObjects/13661_2012_Article_224_IEq71_HTML.gif to be superlinear growth on x and y 1 , , y n http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-140/MediaObjects/13661_2012_Article_224_IEq72_HTML.gif. For example,
      f ( t , x , y 1 , , y n ) = x 2 + y 1 2 + + y n 2 1 4 π 2 ω 2 ( 2 + sin 2 π t ω ) x http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-140/MediaObjects/13661_2012_Article_224_Equj_HTML.gif

      satisfies (F0) with M = 3 4 π 2 ω 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-140/MediaObjects/13661_2012_Article_224_IEq73_HTML.gif and (F1) with f 0 = 1 4 π 2 ω 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-140/MediaObjects/13661_2012_Article_224_IEq74_HTML.gif and f = + http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-140/MediaObjects/13661_2012_Article_224_IEq75_HTML.gif.

      In Theorem 3.2, the condition (F2) allows f ( t , x , y 1 , , y n ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-140/MediaObjects/13661_2012_Article_224_IEq71_HTML.gif to be sublinear growth on x and y 1 , , y n http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-140/MediaObjects/13661_2012_Article_224_IEq72_HTML.gif. For example,
      f ( t , x , y 1 , , y n ) = x + | y 1 | + + | y n | 1 4 π 2 ω 2 ( 2 + sin 2 π t ω ) x http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-140/MediaObjects/13661_2012_Article_224_Equk_HTML.gif

      satisfies (F0) with M = 3 4 π 2 ω 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-140/MediaObjects/13661_2012_Article_224_IEq73_HTML.gif and (F2) with f 0 = + http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-140/MediaObjects/13661_2012_Article_224_IEq76_HTML.gif and f = 1 4 π 2 ω 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-140/MediaObjects/13661_2012_Article_224_IEq77_HTML.gif.

      Proof of Theorem 3.1 Choose the working space E = C ω 1 ( R ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-140/MediaObjects/13661_2012_Article_224_IEq78_HTML.gif. Let K C ω 1 ( R ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-140/MediaObjects/13661_2012_Article_224_IEq79_HTML.gif be the closed convex cone in C ω 1 ( R ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-140/MediaObjects/13661_2012_Article_224_IEq12_HTML.gif defined by (10) and A : K K http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-140/MediaObjects/13661_2012_Article_224_IEq41_HTML.gif be the operator defined by (13). Then the positive ω-periodic solution of Equation (1) is equivalent to the nontrivial fixed point of A. Let 0 < r < R < + http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-140/MediaObjects/13661_2012_Article_224_IEq80_HTML.gif and set
      Ω 1 = { u C ω 1 ( R ) u C 1 < r } , Ω 2 = { u C ω 1 ( R ) u C 1 < R } . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-140/MediaObjects/13661_2012_Article_224_Equ14_HTML.gif
      (14)

      We show that the operator A has a fixed point in K ( Ω 2 Ω ¯ 1 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-140/MediaObjects/13661_2012_Article_224_IEq81_HTML.gif when r is small enough and R is large enough.

      By f 0 < 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-140/MediaObjects/13661_2012_Article_224_IEq68_HTML.gif and the definition of f 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-140/MediaObjects/13661_2012_Article_224_IEq82_HTML.gif, there exist ε ( 0 , M ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-140/MediaObjects/13661_2012_Article_224_IEq83_HTML.gif and δ > 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-140/MediaObjects/13661_2012_Article_224_IEq84_HTML.gif such that
      f ( t , x , y 1 , , y n ) ε x , t I , 0 x δ , | y i | C 0 x , i = 1 , , n . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-140/MediaObjects/13661_2012_Article_224_Equ15_HTML.gif
      (15)
      Let r ( 0 , δ ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-140/MediaObjects/13661_2012_Article_224_IEq85_HTML.gif. We now prove that A satisfies the condition of Lemma 2.4 in K Ω 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-140/MediaObjects/13661_2012_Article_224_IEq86_HTML.gif, namely λ A u u http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-140/MediaObjects/13661_2012_Article_224_IEq52_HTML.gif for every u K Ω 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-140/MediaObjects/13661_2012_Article_224_IEq87_HTML.gif and 0 < λ 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-140/MediaObjects/13661_2012_Article_224_IEq88_HTML.gif. In fact, if there exist u 0 K Ω 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-140/MediaObjects/13661_2012_Article_224_IEq89_HTML.gif and 0 < λ 0 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-140/MediaObjects/13661_2012_Article_224_IEq90_HTML.gif such that λ 0 A u 0 = u 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-140/MediaObjects/13661_2012_Article_224_IEq91_HTML.gif, then by the definition of A and Lemma 2.1, u 0 C ω 2 ( R ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-140/MediaObjects/13661_2012_Article_224_IEq92_HTML.gif satisfies the delay differential equation
      u ¨ 0 ( t ) + M u 0 ( t ) = λ 0 f 1 ( t , u 0 ( t ) , u ˙ 0 ( t τ 1 ( t ) ) , , u ˙ 0 ( t τ n ( t ) ) ) , t R . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-140/MediaObjects/13661_2012_Article_224_Equ16_HTML.gif
      (16)
      Since u 0 K Ω 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-140/MediaObjects/13661_2012_Article_224_IEq89_HTML.gif, by the definitions of K and Ω 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-140/MediaObjects/13661_2012_Article_224_IEq93_HTML.gif, we have
      0 σ u 0 C u 0 ( t ) u 0 C u 0 C 1 = r < δ , | u ˙ 0 ( t τ i ( t ) ) | C 0 u 0 ( t ) , i = 1 , , n , t R . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-140/MediaObjects/13661_2012_Article_224_Equ17_HTML.gif
      (17)
      Hence, from (15) it follows that
      f ( t , u 0 ( t ) , u ˙ 0 ( t τ 1 ( t ) ) , , u ˙ 0 ( t τ n ( t ) ) ) ε u 0 ( t ) , t R . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-140/MediaObjects/13661_2012_Article_224_Equl_HTML.gif
      By this, (16) and the definition of f 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-140/MediaObjects/13661_2012_Article_224_IEq94_HTML.gif, we have
      u ¨ 0 ( t ) + M u 0 ( t ) λ 0 ( M u 0 ( t ) ε u 0 ( t ) ) ( M ε ) u 0 ( t ) , t R . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-140/MediaObjects/13661_2012_Article_224_Equm_HTML.gif
      Integrating both sides of this inequality from 0 to ω and using the periodicity of u 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-140/MediaObjects/13661_2012_Article_224_IEq95_HTML.gif, we obtain that
      M 0 ω u 0 ( t ) d t ( M ε ) 0 ω u 0 ( t ) d t . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-140/MediaObjects/13661_2012_Article_224_Equn_HTML.gif
      Since 0 ω u 0 ( t ) d t ω σ u 0 C > 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-140/MediaObjects/13661_2012_Article_224_IEq96_HTML.gif, it follows that M M ε http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-140/MediaObjects/13661_2012_Article_224_IEq97_HTML.gif, which is a contradiction. Hence, A satisfies the condition of Lemma 2.4 in K Ω 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-140/MediaObjects/13661_2012_Article_224_IEq98_HTML.gif. By Lemma 2.4, we have
      i ( A , K Ω 1 , K ) = 1 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-140/MediaObjects/13661_2012_Article_224_Equ18_HTML.gif
      (18)
      On the other hand, since f > 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-140/MediaObjects/13661_2012_Article_224_IEq69_HTML.gif, by the definition of f http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-140/MediaObjects/13661_2012_Article_224_IEq99_HTML.gif, there exist ε 1 > 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-140/MediaObjects/13661_2012_Article_224_IEq100_HTML.gif and H > 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-140/MediaObjects/13661_2012_Article_224_IEq101_HTML.gif such that
      f ( t , x , y 0 , , y n ) ε 1 x , t I , x H , | y i | C 0 x , i = 1 , , n . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-140/MediaObjects/13661_2012_Article_224_Equ19_HTML.gif
      (19)
      Choose R > max { 1 + C 0 σ H , δ } http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-140/MediaObjects/13661_2012_Article_224_IEq102_HTML.gif and let e ( t ) 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-140/MediaObjects/13661_2012_Article_224_IEq103_HTML.gif. Clearly, e K { θ } http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-140/MediaObjects/13661_2012_Article_224_IEq104_HTML.gif. We show that A satisfies the condition of Lemma 2.5 in K Ω 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-140/MediaObjects/13661_2012_Article_224_IEq105_HTML.gif, namely u A u μ v http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-140/MediaObjects/13661_2012_Article_224_IEq106_HTML.gif for every u K Ω 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-140/MediaObjects/13661_2012_Article_224_IEq107_HTML.gif and μ 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-140/MediaObjects/13661_2012_Article_224_IEq59_HTML.gif. In fact, if there exist u 1 K Ω 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-140/MediaObjects/13661_2012_Article_224_IEq108_HTML.gif and μ 1 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-140/MediaObjects/13661_2012_Article_224_IEq109_HTML.gif such that u 1 A u 1 = μ 1 e http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-140/MediaObjects/13661_2012_Article_224_IEq110_HTML.gif, since u 1 μ 1 e = A u 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-140/MediaObjects/13661_2012_Article_224_IEq111_HTML.gif, by the definition of A and Lemma 2.1, u 1 C ω 2 ( R ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-140/MediaObjects/13661_2012_Article_224_IEq112_HTML.gif satisfies the differential equation
      u ¨ 1 ( t ) + M ( u 1 ( t ) μ 1 ) = f 1 ( t , u 1 ( t ) , u ˙ 1 ( t τ 1 ( t ) ) , , u ˙ 1 ( t τ n ( t ) ) ) , t R . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-140/MediaObjects/13661_2012_Article_224_Equ20_HTML.gif
      (20)
      Since u 1 K Ω 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-140/MediaObjects/13661_2012_Article_224_IEq108_HTML.gif, by the definition of K, we have
      u 1 ( t ) σ u 1 C , | u ˙ 1 ( τ ) | C 0 u 1 ( t ) , τ , t I . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-140/MediaObjects/13661_2012_Article_224_Equ21_HTML.gif
      (21)
      By the latter inequality of (21), we have that u ˙ 1 C C 0 u 1 C http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-140/MediaObjects/13661_2012_Article_224_IEq113_HTML.gif. This implies that u 1 C 1 = u 1 C + u ˙ 1 C ( 1 + C 0 ) u 1 C http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-140/MediaObjects/13661_2012_Article_224_IEq114_HTML.gif. Consequently,
      u 1 C 1 1 + C 0 u 1 C 1 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-140/MediaObjects/13661_2012_Article_224_Equ22_HTML.gif
      (22)
      By (22) and the former inequality of (21), we have
      u 1 ( t ) σ u 1 C σ 1 + C 0 u 1 C 1 = σ R 1 + C 0 > H , t I . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-140/MediaObjects/13661_2012_Article_224_Equo_HTML.gif
      From this, the latter inequality of (21) and (19), it follows that
      f ( t , u 1 ( t ) , u ˙ 1 ( t τ 1 ( t ) ) , , u ˙ 1 ( t τ n ( t ) ) ) ε 1 u 1 ( t ) , t I . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-140/MediaObjects/13661_2012_Article_224_Equp_HTML.gif
      By this inequality, (20) and the definition of f 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-140/MediaObjects/13661_2012_Article_224_IEq94_HTML.gif, we have
      u 1 ( n ) ( t ) + M ( u 1 ( t ) μ 1 ) ( M + ε 1 ) u 1 ( t ) , t I . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-140/MediaObjects/13661_2012_Article_224_Equq_HTML.gif
      Integrating this inequality on I and using the periodicity of u 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-140/MediaObjects/13661_2012_Article_224_IEq115_HTML.gif, we get that
      M 0 ω u 1 ( t ) d t ω M μ 1 ( M + ε 1 ) 0 ω u 1 ( t ) d t . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-140/MediaObjects/13661_2012_Article_224_Equr_HTML.gif
      Since 0 ω u 1 ( t ) d t ω σ u 1 C > 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-140/MediaObjects/13661_2012_Article_224_IEq116_HTML.gif, from this inequality it follows that M M + ε 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-140/MediaObjects/13661_2012_Article_224_IEq117_HTML.gif, which is a contradiction. This means that A satisfies the condition of Lemma 2.5 in K Ω 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-140/MediaObjects/13661_2012_Article_224_IEq105_HTML.gif. By Lemma 2.5,
      i ( A , K Ω 2 , K ) = 0 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-140/MediaObjects/13661_2012_Article_224_Equ23_HTML.gif
      (23)
      Now, by the additivity of fixed point index, (18) and (23), we have
      i ( A , K ( Ω 2 Ω ¯ 1 ) , K ) = i ( A , K Ω 2 , K ) i ( A , K Ω 1 , K ) = 1 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-140/MediaObjects/13661_2012_Article_224_Equs_HTML.gif

      Hence, A has a fixed-point in K ( Ω 2 Ω ¯ 1 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-140/MediaObjects/13661_2012_Article_224_IEq118_HTML.gif, which is a positive ω-periodic solution of Equation (1). □

      Proof of Theorem 3.2 Let Ω 1 , Ω 2 C ω 1 ( R ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-140/MediaObjects/13661_2012_Article_224_IEq119_HTML.gif be defined by (14). We prove that the operator A defined by (13) has a fixed point in K ( Ω 2 Ω ¯ 1 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-140/MediaObjects/13661_2012_Article_224_IEq81_HTML.gif if r is small enough and R is large enough.

      By f 0 > 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-140/MediaObjects/13661_2012_Article_224_IEq120_HTML.gif and the definition of f 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-140/MediaObjects/13661_2012_Article_224_IEq121_HTML.gif, there exist ε > 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-140/MediaObjects/13661_2012_Article_224_IEq122_HTML.gif and δ > 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-140/MediaObjects/13661_2012_Article_224_IEq84_HTML.gif such that
      f ( t , x , y 1 , , y n ) ε x , t I , 0 < x δ , | y i | C 0 x , i = 1 , , n . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-140/MediaObjects/13661_2012_Article_224_Equ24_HTML.gif
      (24)
      Let r ( 0 , δ ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-140/MediaObjects/13661_2012_Article_224_IEq85_HTML.gif and e ( t ) 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-140/MediaObjects/13661_2012_Article_224_IEq103_HTML.gif. We prove that A satisfies the condition of Lemma 2.5 in K Ω 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-140/MediaObjects/13661_2012_Article_224_IEq86_HTML.gif, namely u A u μ e http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-140/MediaObjects/13661_2012_Article_224_IEq123_HTML.gif for every u K Ω 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-140/MediaObjects/13661_2012_Article_224_IEq87_HTML.gif and μ 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-140/MediaObjects/13661_2012_Article_224_IEq59_HTML.gif. In fact, if there exist u 0 K Ω 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-140/MediaObjects/13661_2012_Article_224_IEq89_HTML.gif and μ 0 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-140/MediaObjects/13661_2012_Article_224_IEq124_HTML.gif such that u 0 A u 0 = μ 0 e http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-140/MediaObjects/13661_2012_Article_224_IEq125_HTML.gif, since u 0 μ 0 e = A u 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-140/MediaObjects/13661_2012_Article_224_IEq126_HTML.gif, by the definition of A and Lemma 2.1, u 0 C ω 2 ( R ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-140/MediaObjects/13661_2012_Article_224_IEq92_HTML.gif satisfies the delay differential equation
      u ¨ 0 ( t ) + M ( u 0 ( t ) μ 0 ) = f 1 ( t , u 0 ( t ) , u ˙ 0 ( t τ 1 ( t ) ) , , u ˙ 0 ( t τ n ( t ) ) ) , t R . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-140/MediaObjects/13661_2012_Article_224_Equ25_HTML.gif
      (25)
      Since u 0 K Ω 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-140/MediaObjects/13661_2012_Article_224_IEq89_HTML.gif, by the definitions of K and Ω 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-140/MediaObjects/13661_2012_Article_224_IEq93_HTML.gif, u 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-140/MediaObjects/13661_2012_Article_224_IEq95_HTML.gif satisfies (17). From (17) and (24) it follows that
      f 1 ( t , u 0 ( t ) , u ˙ 0 ( t τ 1 ( t ) ) , , u ˙ 0 ( t τ n ( t ) ) ) ε u 0 ( t ) , t R . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-140/MediaObjects/13661_2012_Article_224_Equt_HTML.gif
      By this, (25) and the definition of f 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-140/MediaObjects/13661_2012_Article_224_IEq94_HTML.gif, we have
      u 0 ( t ) + M u 0 ( t ) = f 1 ( t , u 0 ( t ) , u ˙ 0 ( t τ 1 ( t ) ) , , u ˙ 0 ( t τ n ( t ) ) ) + M μ 0 ( M + ε ) u 0 ( t ) , t R . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-140/MediaObjects/13661_2012_Article_224_Equu_HTML.gif
      Integrating this inequality on [ 0 , ω ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-140/MediaObjects/13661_2012_Article_224_IEq127_HTML.gif and using the periodicity of u 0 ( t ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-140/MediaObjects/13661_2012_Article_224_IEq128_HTML.gif, we obtain that
      M 0 ω u 0 ( t ) d t ( M + ε ) 0 ω u 0 ( t ) d t . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-140/MediaObjects/13661_2012_Article_224_Equv_HTML.gif
      Since 0 ω u 0 ( t ) d t ω σ u 0 C > 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-140/MediaObjects/13661_2012_Article_224_IEq96_HTML.gif, from this inequality it follows that M M + ε http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-140/MediaObjects/13661_2012_Article_224_IEq129_HTML.gif, which is a contradiction. Hence, A satisfies the condition of Lemma 2.5 in K Ω 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-140/MediaObjects/13661_2012_Article_224_IEq86_HTML.gif. By Lemma 2.5, we have
      i ( A , K Ω 1 , K ) = 0 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-140/MediaObjects/13661_2012_Article_224_Equ26_HTML.gif
      (26)
      Since f < 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-140/MediaObjects/13661_2012_Article_224_IEq130_HTML.gif, by the definition of f http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-140/MediaObjects/13661_2012_Article_224_IEq131_HTML.gif, there exist ε 1 ( 0 , M ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-140/MediaObjects/13661_2012_Article_224_IEq132_HTML.gif and H > 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-140/MediaObjects/13661_2012_Article_224_IEq101_HTML.gif such that
      f ( t , x , y 1 , , y n ) ε 1 x , t I , x H , | y i | C 0 x , i = 1 , , n . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-140/MediaObjects/13661_2012_Article_224_Equ27_HTML.gif
      (27)
      Choosing R > max { 1 + C 0 σ H , δ } http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-140/MediaObjects/13661_2012_Article_224_IEq102_HTML.gif, we show that A satisfies the condition of Lemma 2.4 in K Ω 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-140/MediaObjects/13661_2012_Article_224_IEq105_HTML.gif, namely λ A u u http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-140/MediaObjects/13661_2012_Article_224_IEq133_HTML.gif for every u K Ω 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-140/MediaObjects/13661_2012_Article_224_IEq107_HTML.gif and 0 < λ 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-140/MediaObjects/13661_2012_Article_224_IEq134_HTML.gif. In fact, if there exist u 1 K Ω 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-140/MediaObjects/13661_2012_Article_224_IEq108_HTML.gif and 0 < λ 1 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-140/MediaObjects/13661_2012_Article_224_IEq135_HTML.gif such that λ 1 A u 1 = u 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-140/MediaObjects/13661_2012_Article_224_IEq136_HTML.gif, then by the definition of A and Lemma 2.1, u 1 C ω 2 ( R ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-140/MediaObjects/13661_2012_Article_224_IEq137_HTML.gif satisfies the differential equation
      u ¨ 1 ( t ) + M u 1 ( t ) = λ 1 f 1 ( t , u 1 ( t ) , u ˙ 1 ( t τ 1 ( t ) ) , , u ˙ 1 ( t τ n ( t ) ) ) , t R . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-140/MediaObjects/13661_2012_Article_224_Equ28_HTML.gif
      (28)
      Since u 1 K Ω 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-140/MediaObjects/13661_2012_Article_224_IEq108_HTML.gif, by the definition of K, u 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-140/MediaObjects/13661_2012_Article_224_IEq115_HTML.gif satisfies (21). From the second inequality of (21), it follows that (22) holds. By (22) and the first inequality of (21), we have
      u 1 ( t ) σ u 1 C σ ( 1 + C 0 ) u 1 C 1 = σ R ( 1 + C 0 ) > H , t R . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-140/MediaObjects/13661_2012_Article_224_Equw_HTML.gif
      From this, the second inequality of (21) and (27), it follows that
      f ( t , u 1 ( t ) , u ˙ 1 ( t τ 1 ( t ) ) , , u ˙ 1 ( t τ n ( t ) ) ) ε 1 u 1 ( t ) , t R . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-140/MediaObjects/13661_2012_Article_224_Equx_HTML.gif
      By this and (28), we have
      u ¨ 1 ( t ) + M u 1 ( t ) λ 1 ( M u 1 ( t ) ε 1 u 1 ( t ) ) ( M ε 1 ) u 1 ( t ) , t R . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-140/MediaObjects/13661_2012_Article_224_Equy_HTML.gif
      Integrating this inequality on [ 0 , ω ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-140/MediaObjects/13661_2012_Article_224_IEq127_HTML.gif and using the periodicity of u 1 ( t ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-140/MediaObjects/13661_2012_Article_224_IEq138_HTML.gif, we obtain that
      M 0 ω u 1 ( t ) d t ( M ε 1 ) 0 ω u 1 ( t ) d t . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-140/MediaObjects/13661_2012_Article_224_Equz_HTML.gif
      Since 0 ω u 1 ( t ) d t ω σ u 1 C > 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-140/MediaObjects/13661_2012_Article_224_IEq139_HTML.gif, from this inequality it follows that M M ε 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-140/MediaObjects/13661_2012_Article_224_IEq140_HTML.gif, which is a contradiction. This means that A satisfies the condition of Lemma 2.4 in K Ω 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-140/MediaObjects/13661_2012_Article_224_IEq105_HTML.gif. By Lemma 2.4,
      i ( A , K Ω 2 , K ) = 1 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-140/MediaObjects/13661_2012_Article_224_Equ29_HTML.gif
      (29)
      Now, from (26) and (29), it follows that
      i ( A , K ( Ω 2 Ω ¯ 1 ) , K ) = i ( A , K Ω 2 , K ) i ( A , K Ω 1 , K ) = 1 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-140/MediaObjects/13661_2012_Article_224_Equaa_HTML.gif

      Hence, A has a fixed-point in K ( Ω 2 Ω ¯ 1 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-140/MediaObjects/13661_2012_Article_224_IEq118_HTML.gif, which is a positive ω-periodic solution of Equation (1). □

      Example 1 Consider the following second-order differential equation with delay:
      u ¨ ( t ) = a 1 ( t ) u ( t ) + a 2 ( t ) u 2 ( t ) + a 3 ( t ) u ˙ 2 ( t ω / 2 ) , t R , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-140/MediaObjects/13661_2012_Article_224_Equ30_HTML.gif
      (30)
      where a i ( t ) C ω ( R ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-140/MediaObjects/13661_2012_Article_224_IEq141_HTML.gif, i = 1 , 2 , 3 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-140/MediaObjects/13661_2012_Article_224_IEq142_HTML.gif. If π 2 ω 2 < a 1 ( t ) < 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-140/MediaObjects/13661_2012_Article_224_IEq143_HTML.gif and a 2 ( t ) , a 3 ( t ) > 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-140/MediaObjects/13661_2012_Article_224_IEq144_HTML.gif for t [ 0 , ω ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-140/MediaObjects/13661_2012_Article_224_IEq23_HTML.gif, we can verify that
      f ( t , x , y ) = a 1 ( t ) x + a 2 ( t ) x 2 + a 3 ( t ) y 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-140/MediaObjects/13661_2012_Article_224_Equab_HTML.gif

      satisfies the conditions (F0) and (F1) for n = 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-140/MediaObjects/13661_2012_Article_224_IEq145_HTML.gif. By Theorem 3.1, the delay equation (30) has at least one positive ω-periodic solution.

      Example 2 Consider the functional differential equation
      u ¨ ( t ) = c 1 ( t ) u ( t ) + c 2 ( t ) u 2 ( t ) 3 + c 3 ( t ) u ˙ 2 ( t τ ( t ) ) 3 , t R , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-140/MediaObjects/13661_2012_Article_224_Equ31_HTML.gif
      (31)
      where c i ( t ) C ω ( R ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-140/MediaObjects/13661_2012_Article_224_IEq146_HTML.gif, i = 1 , 2 , 3 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-140/MediaObjects/13661_2012_Article_224_IEq142_HTML.gif, and τ C ω + ( R ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-140/MediaObjects/13661_2012_Article_224_IEq147_HTML.gif. If π 2 ω 2 < c 1 ( t ) < 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-140/MediaObjects/13661_2012_Article_224_IEq148_HTML.gif and c 2 ( t ) , c 3 ( t ) > 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-140/MediaObjects/13661_2012_Article_224_IEq149_HTML.gif for t [ 0 , ω ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-140/MediaObjects/13661_2012_Article_224_IEq23_HTML.gif. We easily see that
      f ( t , x , y ) = c 1 ( t ) x + c 2 ( t ) | x | 2 / 3 + c 3 ( t ) | y | 2 / 3 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-140/MediaObjects/13661_2012_Article_224_Equac_HTML.gif

      satisfies the conditions (F0) and (F2) for n = 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-140/MediaObjects/13661_2012_Article_224_IEq145_HTML.gif. By Theorem 3.2, the functional differential equation (31) has a positive ω-periodic solution.

      Declarations

      Acknowledgements

      The research was supported by the NNSFs of China (11261053, 11061031).

      Authors’ Affiliations

      (1)
      Department of Mathematics, Northwest Normal University

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