Multiple positive doubly periodic solutions for a singular semipositone telegraph equation with a parameter

  • Fanglei Wang1Email author and

    Affiliated with

    • Yukun An2

      Affiliated with

      Boundary Value Problems20132013:7

      DOI: 10.1186/1687-2770-2013-7

      Received: 26 July 2012

      Accepted: 29 December 2012

      Published: 16 January 2013

      Abstract

      In this paper, we study the multiplicity of positive doubly periodic solutions for a singular semipositone telegraph equation. The proof is based on a well-known fixed point theorem in a cone.

      MSC:34B15, 34B18.

      Keywords

      semipositone telegraph equation doubly periodic solution singular cone fixed point theorem

      1 Introduction

      Recently, the existence and multiplicity of positive periodic solutions for a scalar singular equation or singular systems have been studied by using some fixed point theorems; see [19]. In [10], the authors show that the method of lower and upper solutions is also one of common techniques to study the singular problem. In addition, the authors [11] use the continuation type existence principle to investigate the following singular periodic problem:
      ( | u | p 2 u ) + h ( u ) u = g ( u ) + c ( t ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-7/MediaObjects/13661_2012_Article_267_Equa_HTML.gif
      More recently, using a weak force condition, Wang [12] has built some existence results for the following periodic boundary value problem:
      { u t t u x x + c 1 u t + a 11 ( t , x ) u + a 12 ( t , x ) v = f 1 ( t , x , u , v ) + χ 1 ( t , x ) , v t t v x x + c 2 v t + a 21 ( t , x ) u + a 22 ( t , x ) v = f 2 ( t , x , u , v ) + χ 2 ( t , x ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-7/MediaObjects/13661_2012_Article_267_Equb_HTML.gif

      The proof is based on Schauder’s fixed point theorem. For other results concerning the existence and multiplicity of positive doubly periodic solutions for a single regular telegraph equation or regular telegraph system, see, for example, the papers [1317] and the references therein. In these references, the nonlinearities are nonnegative.

      On the other hand, the authors [18] study the semipositone telegraph system
      { u t t u x x + c 1 u t + a 1 ( t , x ) u = b 1 ( t , x ) f ( t , x , u , v ) , v t t v x x + c 2 v t + a 2 ( t , x ) v = b 2 ( t , x ) g ( t , x , u , v ) , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-7/MediaObjects/13661_2012_Article_267_Equc_HTML.gif

      where the nonlinearities f, g may change sign. In addition, there are many authors who have studied the semipositone equations; see [19, 20].

      Inspired by the above references, we are concerned with the multiplicity of positive doubly periodic solutions for a general singular semipositone telegraph equation
      { u t t u x x + c u t + a ( t , x ) u = λ f ( t , x , u ) , u ( t + 2 π , x ) = u ( t , x + 2 π ) = u ( t , x ) , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-7/MediaObjects/13661_2012_Article_267_Equ1_HTML.gif
      (1)
      where c > 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-7/MediaObjects/13661_2012_Article_267_IEq1_HTML.gif is a constant, λ > 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-7/MediaObjects/13661_2012_Article_267_IEq2_HTML.gif is a positive parameter, a ( t , x ) C ( R × R , R ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-7/MediaObjects/13661_2012_Article_267_IEq3_HTML.gif, f ( t , x , u ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-7/MediaObjects/13661_2012_Article_267_IEq4_HTML.gif may change sign and is singular at u = 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-7/MediaObjects/13661_2012_Article_267_IEq5_HTML.gif, namely,
      lim u 0 + f ( t , x , u ) = + . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-7/MediaObjects/13661_2012_Article_267_Equd_HTML.gif

      The main method used here is the following fixed-point theorem of a cone mapping.

      Lemma 1.1 [21]

      Let E be a Banach space, and K E http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-7/MediaObjects/13661_2012_Article_267_IEq6_HTML.gif be a cone in E. Assume Ω 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-7/MediaObjects/13661_2012_Article_267_IEq7_HTML.gif, Ω 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-7/MediaObjects/13661_2012_Article_267_IEq8_HTML.gif are open subsets of E with 0 Ω 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-7/MediaObjects/13661_2012_Article_267_IEq9_HTML.gif, Ω ¯ 1 Ω 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-7/MediaObjects/13661_2012_Article_267_IEq10_HTML.gif, and let T : K ( Ω ¯ 2 Ω 1 ) K http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-7/MediaObjects/13661_2012_Article_267_IEq11_HTML.gif be a completely continuous operator such that either
      1. (i)

        T u u http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-7/MediaObjects/13661_2012_Article_267_IEq12_HTML.gif, u K Ω 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-7/MediaObjects/13661_2012_Article_267_IEq13_HTML.gif and T u u http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-7/MediaObjects/13661_2012_Article_267_IEq14_HTML.gif, u K Ω 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-7/MediaObjects/13661_2012_Article_267_IEq15_HTML.gif; or

         
      2. (ii)

        T u u http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-7/MediaObjects/13661_2012_Article_267_IEq14_HTML.gif, u K Ω 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-7/MediaObjects/13661_2012_Article_267_IEq13_HTML.gif and T u u http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-7/MediaObjects/13661_2012_Article_267_IEq12_HTML.gif, u K Ω 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-7/MediaObjects/13661_2012_Article_267_IEq15_HTML.gif.

         

      Then T has a fixed point in K ( Ω ¯ 2 Ω 1 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-7/MediaObjects/13661_2012_Article_267_IEq16_HTML.gif.

      The paper is organized as follows. In Section 2, some preliminaries are given. In Section 3, we give the main result.

      2 Preliminaries

      Let 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-7/MediaObjects/13661_2012_Article_267_IEq17_HTML.gif be the torus defined as
      2 = ( R / 2 π Z ) × ( R / 2 π Z ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-7/MediaObjects/13661_2012_Article_267_Eque_HTML.gif
      Doubly 2π-periodic functions will be identified to be functions defined on 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-7/MediaObjects/13661_2012_Article_267_IEq17_HTML.gif. We use the notations
      L p ( 2 ) , C ( 2 ) , C α ( 2 ) , D ( 2 ) = C ( 2 ) , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-7/MediaObjects/13661_2012_Article_267_Equf_HTML.gif

      to denote the spaces of doubly periodic functions with the indicated degree of regularity. The space D ( 2 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-7/MediaObjects/13661_2012_Article_267_IEq18_HTML.gif denotes the space of distributions on 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-7/MediaObjects/13661_2012_Article_267_IEq17_HTML.gif.

      By a doubly periodic solution of Eq. (1) we mean that a u L 1 ( 2 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-7/MediaObjects/13661_2012_Article_267_IEq19_HTML.gif satisfies Eq. (1) in the distribution sense, i.e.,
      2 u ( φ t t φ x x c φ t + a ( t , x ) φ ) d t d x = λ 2 f ( t , x , u ) φ d t d x . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-7/MediaObjects/13661_2012_Article_267_Equg_HTML.gif
      First, we consider the linear equation
      u t t u x x + c u t ξ u = h ( t , x ) , in  D ( 2 ) , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-7/MediaObjects/13661_2012_Article_267_Equ2_HTML.gif
      (2)

      where c > 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-7/MediaObjects/13661_2012_Article_267_IEq1_HTML.gif, μ R http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-7/MediaObjects/13661_2012_Article_267_IEq20_HTML.gif, and h ( t , x ) L 1 ( 2 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-7/MediaObjects/13661_2012_Article_267_IEq21_HTML.gif.

      Let £ ξ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-7/MediaObjects/13661_2012_Article_267_IEq22_HTML.gif be the differential operator
      £ ξ u = u t t u x x + c u t ξ u , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-7/MediaObjects/13661_2012_Article_267_Equh_HTML.gif
      acting on functions on 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-7/MediaObjects/13661_2012_Article_267_IEq17_HTML.gif. Following the discussion in [14], we know that if ξ < 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-7/MediaObjects/13661_2012_Article_267_IEq23_HTML.gif, £ ξ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-7/MediaObjects/13661_2012_Article_267_IEq22_HTML.gif has the resolvent R ξ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-7/MediaObjects/13661_2012_Article_267_IEq24_HTML.gif,
      R ξ : L 1 ( 2 ) C ( 2 ) , h i ( t , x ) u i ( t , x ) , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-7/MediaObjects/13661_2012_Article_267_Equi_HTML.gif

      where u ( t , x ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-7/MediaObjects/13661_2012_Article_267_IEq25_HTML.gif is the unique solution of Eq. (2), and the restriction of R ξ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-7/MediaObjects/13661_2012_Article_267_IEq24_HTML.gif on L p ( 2 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-7/MediaObjects/13661_2012_Article_267_IEq26_HTML.gif ( 1 < p < http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-7/MediaObjects/13661_2012_Article_267_IEq27_HTML.gif) or C ( 2 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-7/MediaObjects/13661_2012_Article_267_IEq28_HTML.gif is compact. In particular, R ξ : C ( 2 ) C ( 2 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-7/MediaObjects/13661_2012_Article_267_IEq29_HTML.gif is a completely continuous operator.

      For ξ = c 2 / 4 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-7/MediaObjects/13661_2012_Article_267_IEq30_HTML.gif, the Green function G ( t , x ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-7/MediaObjects/13661_2012_Article_267_IEq31_HTML.gif of the differential operator £ ξ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-7/MediaObjects/13661_2012_Article_267_IEq22_HTML.gif is explicitly expressed; see Lemma 5.2 in [14]. From the definition of G ( t , x ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-7/MediaObjects/13661_2012_Article_267_IEq31_HTML.gif, we have
      http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-7/MediaObjects/13661_2012_Article_267_Equj_HTML.gif

      For convenience, we assume the following condition holds throughout this paper:

      (H1) a ( t , x ) C ( 2 , R ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-7/MediaObjects/13661_2012_Article_267_IEq32_HTML.gif, 0 a ( t , x ) c 2 4 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-7/MediaObjects/13661_2012_Article_267_IEq33_HTML.gif on 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-7/MediaObjects/13661_2012_Article_267_IEq17_HTML.gif, and 2 a ( t , x ) d t d x > 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-7/MediaObjects/13661_2012_Article_267_IEq34_HTML.gif.

      Finally, if −ξ is replaced by a ( t , x ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-7/MediaObjects/13661_2012_Article_267_IEq35_HTML.gif in Eq. (2), the author [13] has proved the following unique existence and positive estimate result.

      Lemma 2.1 Let h ( t , x ) L 1 ( 2 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-7/MediaObjects/13661_2012_Article_267_IEq21_HTML.gif. Then Eq. (2) has a unique solution u ( t , x ) = P [ h ( t , x ) ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-7/MediaObjects/13661_2012_Article_267_IEq36_HTML.gif, P : L 1 ( 2 ) C ( 2 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-7/MediaObjects/13661_2012_Article_267_IEq37_HTML.gif is a linear bounded operator with the following properties:
      1. (i)

        P : C ( 2 ) C ( 2 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-7/MediaObjects/13661_2012_Article_267_IEq38_HTML.gif is a completely continuous operator;

         
      2. (ii)
        If h ( t , x ) > 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-7/MediaObjects/13661_2012_Article_267_IEq39_HTML.gif, a.e ( t , x ) 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-7/MediaObjects/13661_2012_Article_267_IEq40_HTML.gif, P [ h ( t , x ) ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-7/MediaObjects/13661_2012_Article_267_IEq41_HTML.gif has the positive estimate
        G ̲ h L 1 P [ h ( t , x ) ] G ¯ G ̲ a L 1 h L 1 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-7/MediaObjects/13661_2012_Article_267_Equ3_HTML.gif
        (3)
         

      3 Main result

      Theorem 3.1 Assume (H1) holds. In addition, if f ( t , x , u ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-7/MediaObjects/13661_2012_Article_267_IEq4_HTML.gif satisfies

      (H2) lim u 0 + f ( t , x , u ) = + http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-7/MediaObjects/13661_2012_Article_267_IEq42_HTML.gif, uniformly ( t , x ) 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-7/MediaObjects/13661_2012_Article_267_IEq43_HTML.gif,

      (H3) f : 2 × ( 0 , + ) ( , + ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-7/MediaObjects/13661_2012_Article_267_IEq44_HTML.gif is continuous,

      (H4) there exists a nonnegative function h ( t , x ) C ( 2 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-7/MediaObjects/13661_2012_Article_267_IEq45_HTML.gif such that
      f ( t , x , u ) + h ( t , x ) 0 , ( t , x ) 2 , u > 0 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-7/MediaObjects/13661_2012_Article_267_Equk_HTML.gif

      (H5) 2 F ( t , x ) d t d x = + http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-7/MediaObjects/13661_2012_Article_267_IEq46_HTML.gif, where the limit function F ( t , x ) = lim inf u + f ( t , x , u ) u http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-7/MediaObjects/13661_2012_Article_267_IEq47_HTML.gif,

      then Eq. (1) has at least two positive doubly periodic solutions for sufficiently small λ.

      C ( 2 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-7/MediaObjects/13661_2012_Article_267_IEq28_HTML.gif is a Banach space with the norm u = max ( t , x ) 2 | u ( t , x ) | http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-7/MediaObjects/13661_2012_Article_267_IEq48_HTML.gif. Define a cone K C ( 2 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-7/MediaObjects/13661_2012_Article_267_IEq49_HTML.gif by
      K = { u C ( 2 ) : u 0 , u ( t , x ) δ u } , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-7/MediaObjects/13661_2012_Article_267_Equl_HTML.gif

      where δ = G ̲ 2 a L 1 G ¯ ( 0 , 1 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-7/MediaObjects/13661_2012_Article_267_IEq50_HTML.gif. Let K r = { u K : u = r } http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-7/MediaObjects/13661_2012_Article_267_IEq51_HTML.gif, [ u ] + = max { u , 0 } http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-7/MediaObjects/13661_2012_Article_267_IEq52_HTML.gif. By Lemma 2.1, it is easy to obtain the following lemmas.

      Lemma 3.2 If h ( t , x ) C ( 2 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-7/MediaObjects/13661_2012_Article_267_IEq45_HTML.gif is a nonnegative function, the linear boundary value problem
      { u t t u x x + c u t + a ( t , x ) u = λ h ( t , x ) , u ( t + 2 π , x ) = u ( t , x + 2 π ) = u ( t , x ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-7/MediaObjects/13661_2012_Article_267_Equm_HTML.gif
      has a unique solution ω ( t , x ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-7/MediaObjects/13661_2012_Article_267_IEq53_HTML.gif. The function ω ( t , x ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-7/MediaObjects/13661_2012_Article_267_IEq53_HTML.gif satisfies the estimates
      λ G ̲ h L 1 ω ( t , x ) = λ P ( h ( t , x ) ) λ G ¯ G ̲ a L 1 h L 1 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-7/MediaObjects/13661_2012_Article_267_Equn_HTML.gif
      Lemma 3.3 If the boundary value problem
      { u t t u x x + c u t + a ( t , x ) u = λ [ f ( t , x , [ u ( t , x ) ω ( t , x ) ] + ) + h ( t , x ) ] , u ( t + 2 π , x ) = u ( t , x + 2 π ) = u ( t , x ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-7/MediaObjects/13661_2012_Article_267_Equo_HTML.gif

      has a solution u ˜ ( t , x ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-7/MediaObjects/13661_2012_Article_267_IEq54_HTML.gif with u ˜ > λ G ¯ 2 G ̲ 3 a L 1 2 h L 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-7/MediaObjects/13661_2012_Article_267_IEq55_HTML.gif, then u ( t , x ) = u ˜ ( t , x ) ω ( t , x ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-7/MediaObjects/13661_2012_Article_267_IEq56_HTML.gif is a positive doubly periodic solution of Eq. (1).

      Proof of Theorem 3.1 Step 1. Define the operator T as follows:
      ( T u ) ( t , x ) = λ P [ f ( t , x , [ u ( t , x ) ω ( t , x ) ] + ) + h ( t , x ) ] . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-7/MediaObjects/13661_2012_Article_267_Equp_HTML.gif

      We obtain the conclusion that T ( K { u K : [ u ( t , x ) ω ( t , x ) ] + = 0 } ) K http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-7/MediaObjects/13661_2012_Article_267_IEq57_HTML.gif, and T : K { u K : [ u ( t , x ) ω ( t , x ) ] + = 0 } K http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-7/MediaObjects/13661_2012_Article_267_IEq58_HTML.gif is completely continuous.

      For any u K { u K : [ u ( t , x ) ω ( t , x ) ] + = 0 } http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-7/MediaObjects/13661_2012_Article_267_IEq59_HTML.gif, then [ u ( t , x ) ω ( t , x ) ] + > 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-7/MediaObjects/13661_2012_Article_267_IEq60_HTML.gif, and T is defined. On the other hand, for u K { u K : [ u ( t , x ) ω ( t , x ) ] + = 0 } http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-7/MediaObjects/13661_2012_Article_267_IEq59_HTML.gif, the complete continuity is obvious by Lemma 2.1. And we can have
      ( T u ) ( t , x ) = λ P [ f ( t , x , [ u ( t , x ) ω ( t , x ) ] + ) + h ( t , x ) ] λ G ̲ f ( t , x , [ u ( t , x ) ω ( t , x ) ] + ) + h ( t , x ) L 1 G ̲ G ̲ a L 1 G ¯ T ( u ) δ T u . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-7/MediaObjects/13661_2012_Article_267_Equq_HTML.gif

      Thus, T ( K { u K : u ( t , x ) ω ( t , x ) } ) K http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-7/MediaObjects/13661_2012_Article_267_IEq61_HTML.gif.

      Now we prove that the operator T has one fixed point u ˜ K http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-7/MediaObjects/13661_2012_Article_267_IEq62_HTML.gif and u ˜ > λ G ¯ 2 G ̲ 3 a L 1 2 h L 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-7/MediaObjects/13661_2012_Article_267_IEq55_HTML.gif for all sufficiently small λ.

      Since 2 F ( t , x ) d t d x = + http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-7/MediaObjects/13661_2012_Article_267_IEq46_HTML.gif, there exists r 1 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-7/MediaObjects/13661_2012_Article_267_IEq63_HTML.gif such that
      2 f ( t , x , u ) u d t d x 1 δ , u δ r 1 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-7/MediaObjects/13661_2012_Article_267_Equr_HTML.gif
      Furthermore, we have 2 f ( t , x , δ r 1 ) d t d x r 1 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-7/MediaObjects/13661_2012_Article_267_IEq64_HTML.gif. It follows that
      http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-7/MediaObjects/13661_2012_Article_267_Equs_HTML.gif
      Let Φ ( t , x ) = max { f ( t , x , u ) : δ 2 r 1 u r 1 } + h ( t , x ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-7/MediaObjects/13661_2012_Article_267_IEq65_HTML.gif. Then Φ L 1 ( 2 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-7/MediaObjects/13661_2012_Article_267_IEq66_HTML.gif and 2 Φ ( t , x ) d t d x > 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-7/MediaObjects/13661_2012_Article_267_IEq67_HTML.gif. Set
      λ = min { δ 2 2 G ̲ h L 1 , 2 G ̲ a L 1 G ¯ Φ L 1 } . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-7/MediaObjects/13661_2012_Article_267_Equt_HTML.gif
      For any u K r 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-7/MediaObjects/13661_2012_Article_267_IEq68_HTML.gif and 0 < λ < λ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-7/MediaObjects/13661_2012_Article_267_IEq69_HTML.gif, we can verify that
      u ( t , x ) ω ( t , x ) δ u ω ( t , x ) = δ r 1 ω ( t , x ) δ r 1 λ G ¯ G ̲ a L 1 h L 1 δ r 1 δ r 1 2 = δ r 1 2 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-7/MediaObjects/13661_2012_Article_267_Equu_HTML.gif
      Then we have
      T u = λ P [ f ( t , x , [ u ( t , x ) ω ( t , x ) ] + ) + h ( t , x ) ] λ G ¯ G ̲ a L 1 f ( t , x , [ u ( t , x ) ω ( t , x ) ] + ) + h ( t , x ) L 1 λ G ¯ G ̲ a L 1 Φ ( t , x ) L 1 < 2 r 1 = u . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-7/MediaObjects/13661_2012_Article_267_Equv_HTML.gif
      On the other hand,
      lim inf u + f ( t , x , u ω ( t , x ) ) u = lim inf u + f ( t , x , u ) u = F ( t , x ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-7/MediaObjects/13661_2012_Article_267_Equw_HTML.gif
      By the Fatou lemma, one has
      http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-7/MediaObjects/13661_2012_Article_267_Equx_HTML.gif
      Hence, there exists a positive number r 2 > δ r 2 > r 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-7/MediaObjects/13661_2012_Article_267_IEq70_HTML.gif such that
      2 f ( t , x , u ω ( t , x ) ) + h ( t , x ) u d t d x λ 1 δ 1 G ̲ 1 ( 4 π 2 ) 1 , u δ r 2 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-7/MediaObjects/13661_2012_Article_267_Equy_HTML.gif
      Hence, we have
      2 f ( t , x , u ω ( t , x ) ) + h ( t , x ) d t d x λ 1 G ̲ 1 ( 4 π 2 ) 1 r 2 , u δ r 2 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-7/MediaObjects/13661_2012_Article_267_Equz_HTML.gif
      For any u K r 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-7/MediaObjects/13661_2012_Article_267_IEq71_HTML.gif, we have δ r 2 = δ u u ( t , x ) u = r 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-7/MediaObjects/13661_2012_Article_267_IEq72_HTML.gif. On the other hand, since 0 < λ < λ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-7/MediaObjects/13661_2012_Article_267_IEq69_HTML.gif, we can get
      u ( t , x ) ω ( t , x ) δ r 2 ω ( t , x ) δ r 2 δ λ G ¯ G ̲ a L 1 δ r 2 δ > 0 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-7/MediaObjects/13661_2012_Article_267_Equaa_HTML.gif
      From above, we can have
      T u λ P [ f ( t , x , [ u ( t , x ) ω ( t , x ) ] + ) + h ( t , x ) ] λ G ̲ f ( t , x , [ u ( t , x ) ω ( t , x ) ] + ) + h ( t , x ) L 1 λ G ̲ 4 π 2 λ 1 G ̲ 1 ( 4 π 2 ) 1 r 2 = r 2 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-7/MediaObjects/13661_2012_Article_267_Equab_HTML.gif
      Therefore, by Lemma 1.1, the operator T has a fixed point u ˜ ( t , x ) K http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-7/MediaObjects/13661_2012_Article_267_IEq73_HTML.gif and
      http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-7/MediaObjects/13661_2012_Article_267_Equac_HTML.gif

      So, Eq. (1) has a positive solution u ˆ ( t , x ) = u ˜ ( t , x ) ω ( t , x ) δ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-7/MediaObjects/13661_2012_Article_267_IEq74_HTML.gif.

      Step 2. By conditions (H2) and (H3), it is clear to obtain that
      u 0 = inf { u K : f ( t , x , u ) 0 , ( t , x ) 2 } > 0 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-7/MediaObjects/13661_2012_Article_267_Equad_HTML.gif
      Let r 4 = min { δ 2 , δ u 0 2 } http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-7/MediaObjects/13661_2012_Article_267_IEq75_HTML.gif. For any u ( 0 , r 4 ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-7/MediaObjects/13661_2012_Article_267_IEq76_HTML.gif, we have f ( t , x , u ) > 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-7/MediaObjects/13661_2012_Article_267_IEq77_HTML.gif. Then define the operator A as follows:
      ( A u ) ( t , x ) = λ P ˆ [ f ( t , x , u ( t , x ) ) ] . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-7/MediaObjects/13661_2012_Article_267_Equae_HTML.gif

      It is easy to prove that A ( K { u C ( 2 ) : 0 < u < r 4 } ) K http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-7/MediaObjects/13661_2012_Article_267_IEq78_HTML.gif, and A : K { u C ( 2 ) : 0 < u < r 4 } K http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-7/MediaObjects/13661_2012_Article_267_IEq79_HTML.gif is completely continuous.

      And for any ρ > 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-7/MediaObjects/13661_2012_Article_267_IEq80_HTML.gif, define
      M ( ρ ) = max { f ( t , x , u ) : u R + , δ ρ u ρ , ( t , x ) 2 } > 0 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-7/MediaObjects/13661_2012_Article_267_Equaf_HTML.gif
      Furthermore, for any u K r 4 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-7/MediaObjects/13661_2012_Article_267_IEq81_HTML.gif, we have
      A u = λ P ˆ [ f ( t , x , u ( t , x ) ) ] λ G ¯ G ̲ a L 1 f ( t , x , u ( t , x ) ) L 1 λ G ¯ G ̲ a L 1 M ( r 4 ) 4 π 2 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-7/MediaObjects/13661_2012_Article_267_Equag_HTML.gif
      Thus, from the above inequality, there exists λ ¯ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-7/MediaObjects/13661_2012_Article_267_IEq82_HTML.gif such that
      A u < u , for  u K r 4 , 0 < λ < λ ¯ . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-7/MediaObjects/13661_2012_Article_267_Equah_HTML.gif
      Since lim u 0 + f ( t , x , u ) = http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-7/MediaObjects/13661_2012_Article_267_IEq83_HTML.gif, then there is 0 < r 3 < r 4 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-7/MediaObjects/13661_2012_Article_267_IEq84_HTML.gif such that
      f ( t , x , u ) μ u , for  u R +  with  0 < u r 3 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-7/MediaObjects/13661_2012_Article_267_Equai_HTML.gif
      where μ satisfies λ G ̲ μ δ > 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-7/MediaObjects/13661_2012_Article_267_IEq85_HTML.gif. For any u K r 3 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-7/MediaObjects/13661_2012_Article_267_IEq86_HTML.gif, then we have
      f ( t , x , u ) μ u ( t , x ) , for  ( t , x ) 2 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-7/MediaObjects/13661_2012_Article_267_Equaj_HTML.gif
      By Lemma 2.1, it is clear to obtain that
      A u = λ P ˆ [ f ( t , x , u ( t , x ) ) ] λ G ̲ f ( t , x , u ( t , x ) ) L 1 λ G ̲ μ δ r 3 > r 3 = u . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-7/MediaObjects/13661_2012_Article_267_Equak_HTML.gif

      Therefore, by Lemma 1.1, A has a fixed point in u ¯ ( t , x ) K http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-7/MediaObjects/13661_2012_Article_267_IEq87_HTML.gif and u ¯ r 4 δ 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-7/MediaObjects/13661_2012_Article_267_IEq88_HTML.gif, which is another positive periodic solution of Eq. (1).

      Finally, from Step 1 and Step 2, Eq. (1) has two positive doubly periodic solutions u ˆ ( t , x ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-7/MediaObjects/13661_2012_Article_267_IEq89_HTML.gif and u ¯ ( t , x ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-7/MediaObjects/13661_2012_Article_267_IEq90_HTML.gif for sufficiently small λ. □

      Example

      Consider the following problem:
      { u t t u x x + 2 u t + sin 2 ( t + x ) u = λ [ 1 u + min { u 2 , u | 1 t π | | 1 x π | } 10 ] , u ( t + 2 π , x ) = u ( t , x + 2 π ) = u ( t , x ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-7/MediaObjects/13661_2012_Article_267_Equal_HTML.gif

      It is clear that f ( t , x , u ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-7/MediaObjects/13661_2012_Article_267_IEq4_HTML.gif satisfies the conditions (H1)-(H5).

      Declarations

      Acknowledgements

      The authors would like to thank the referees for valuable comments and suggestions for improving this paper.

      Authors’ Affiliations

      (1)
      College of Science, Hohai University
      (2)
      Department of Mathematics, Nanjing University of Aeronautics and Astronautics

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      © Wang and An; licensee Springer. 2013

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