Existence of solutions and nonnegative solutions for a class of p ( t ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq1_HTML.gif-Laplacian differential systems with multipoint and integral boundary value conditions

  • Guizhen Zhi1,

    Affiliated with

    • Yunrui Guo2,

      Affiliated with

      • Yan Wang1 and

        Affiliated with

        • Qihu Zhang1Email author

          Affiliated with

          Boundary Value Problems20132013:106

          DOI: 10.1186/1687-2770-2013-106

          Received: 12 November 2011

          Accepted: 12 April 2013

          Published: 26 April 2013

          Abstract

          This paper explores the existence of solutions for a class of p ( t ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq1_HTML.gif-Laplacian differential systems with multipoint and integral boundary value conditions via Leray-Schauder’s degree. Moreover, the existence of nonnegative solutions is discussed.

          MSC:34B10.

          Keywords

          p ( t ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq1_HTML.gif-Laplacian Leray-Schauder degree fixed point

          1 Introduction

          In this paper, we consider the existence of solutions for the following system:
          ( P ) { p 1 ( t ) u = δ 1 f 1 ( t , u , u , v , v ) , t ( 0 , 1 ) , p 2 ( t ) v = δ 2 f 2 ( t , u , u , v , v ) , t ( 0 , 1 ) , u ( 1 ) = i = 1 m 2 α i u ( ξ i ) + e 1 , lim t 1 | u | p 1 ( t ) 2 u ( t ) = 0 1 k ( t ) | u | p 1 ( t ) 2 u ( t ) d t + e 2 , v ( 0 ) k 1 v ( 0 ) = 0 1 e ( t ) v ( t ) d t , v ( 1 ) + k 2 v ( 1 ) = i = 1 m 2 β i v ( η i ) , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_Equa_HTML.gif

          where p l C ( [ 0 , 1 ] , R ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq2_HTML.gif, p l ( t ) > 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq3_HTML.gif ( l = 1 , 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq4_HTML.gif); p ( t ) γ : = ( | γ | p ( t ) 2 γ ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq5_HTML.gif is called p ( t ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq1_HTML.gif-Laplacian; 0 < ξ 1 < < ξ m 2 < 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq6_HTML.gif, 0 < η 1 < < η m 2 < 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq7_HTML.gif; α i 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq8_HTML.gif, β i 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq9_HTML.gif ( i = 1 , , m 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq10_HTML.gif) and 0 < i = 1 m 2 α i < 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq11_HTML.gif, 0 < i = 1 m 2 β i < 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq12_HTML.gif; k ( t ) , e ( t ) L 1 ( 0 , 1 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq13_HTML.gif, they are both nonnegative, σ 1 = 0 1 k ( t ) d t ( 0 , 1 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq14_HTML.gif, σ 2 = 0 1 e ( t ) d t ( 0 , 1 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq15_HTML.gif; e 1 , e 2 R N http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq16_HTML.gif; k 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq17_HTML.gif and k 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq18_HTML.gif are nonnegative constants; δ 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq19_HTML.gif and δ 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq20_HTML.gif are positive parameters.

          The study of differential equations and variational problems with variable exponent growth conditions has attracted more and more attention in recent years. Many results have been obtained on these problems, for example, [116]. We refer to [3, 12, 16] for the applied background of these problems. If p ( t ) p http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq21_HTML.gif (a constant), p ( t ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq22_HTML.gif becomes the well-known p-Laplacian. If p ( t ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq1_HTML.gif is a general function, p ( t ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq22_HTML.gif represents a non-homogeneity and possesses more nonlinearity, thus p ( t ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq22_HTML.gif is more complicated than p http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq23_HTML.gif (see [7]).

          In recent years, because of the wide mathematical and physical background (see [1719]), the existence of positive solutions for the p-Laplacian equation group has received extensive attention. Especially, when p = 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq24_HTML.gif, the existence of positive solutions for the equation group boundary value problems has been obtained (see [2025]). On the integral boundary value problems, we refer to [2630]. But as for the p ( t ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq1_HTML.gif-Laplacian equation group, there are few papers dealing with the existence of solutions, especially the existence of solutions for the systems with multipoint and integral boundary value problems. Therefore, when p ( t ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq1_HTML.gif is a general function, this paper mainly investigates the existence of solutions for a class of p ( t ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq1_HTML.gif-Laplacian differential systems with multipoint and integral boundary value conditions. Moreover, we discuss the existence of nonnegative solutions.

          Let N 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq25_HTML.gif and J = [ 0 , 1 ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq26_HTML.gif, the function f l = ( f l 1 , , f l N ) : J × R N × R N × R N × R N R N http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq27_HTML.gif, ( l = 1 , 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq28_HTML.gif) is assumed to be Carathéodory, by which we mean:
          1. (i)

            For almost every t J http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq29_HTML.gif, the function f l ( t , , , , ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq30_HTML.gif is continuous;

             
          2. (ii)

            For each ( x , y , z , w ) R N × R N × R N × R N http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq31_HTML.gif, the function f l ( , x , y , z , w ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq32_HTML.gif is measurable on J;

             
          3. (iii)
            For each R > 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq33_HTML.gif, there are β R , ρ R L 1 ( J , R ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq34_HTML.gif such that, for almost every t J http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq29_HTML.gif and every ( x , y , z , w ) R N × R N × R N × R N http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq35_HTML.gif with | x | R http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq36_HTML.gif, | y | R http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq37_HTML.gif, | z | R http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq38_HTML.gif, | w | R http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq39_HTML.gif, one has
            | f 1 ( t , x , y , z , w ) | β R ( t ) , | f 2 ( t , x , y , z , w ) | ρ R ( t ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_Equb_HTML.gif
             
          Throughout the paper, we denote
          http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_Equc_HTML.gif

          The inner product in R N http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq40_HTML.gif will be denoted by , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq41_HTML.gif, | | http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq42_HTML.gif will denote the absolute value and the Euclidean norm on R N http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq43_HTML.gif. For N 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq25_HTML.gif, we set C = C ( J , R N ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq44_HTML.gif, C 1 = { γ C γ C } http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq45_HTML.gif; W = { ( u , v ) u , v C 1 } http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq46_HTML.gif. For any γ ( t ) = ( γ 1 ( t ) , , γ N ( t ) ) C http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq47_HTML.gif, we denote | γ i | 0 = max t [ 0 , 1 ] | γ i ( t ) | http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq48_HTML.gif, γ 0 = ( i = 1 N | γ i | 0 2 ) 1 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq49_HTML.gif and γ 1 = γ 0 + γ 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq50_HTML.gif. For any ( u , v ) W http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq51_HTML.gif, we denote ( u , v ) = u 1 + v 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq52_HTML.gif. Spaces C, C 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq53_HTML.gif and W will be equipped with the norm 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq54_HTML.gif, 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq55_HTML.gif and http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq56_HTML.gif, respectively. Then ( C , 0 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq57_HTML.gif, ( C 1 , 1 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq58_HTML.gif and ( W , ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq59_HTML.gif are Banach spaces. Denote L 1 = L 1 ( J , R N ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq60_HTML.gif with the norm γ L 1 = [ i = 1 N ( 0 1 | γ i | d t ) 2 ] 1 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq61_HTML.gif.

          We say a function ( u , v ) : J R N http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq62_HTML.gif is a solution of (P) if ( u , v ) W http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq51_HTML.gif satisfies the differential equation in (P) a.e. on J and the boundary value conditions.

          In this paper, we always use C i http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq63_HTML.gif to denote positive constants if this does not lead to confusion. Denote
          b = inf t J b ( t ) , b + = sup t J b ( t ) for any  b C ( J , R ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_Equd_HTML.gif
          We say f l http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq64_HTML.gif ( l = 1 , 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq28_HTML.gif) satisfies a sub- ( p l 1 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq65_HTML.gif growth condition if f l http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq64_HTML.gif satisfies
          lim | x | + | y | + | z | + | w | + f l ( t , x , y , z , w ) ( | x | + | y | + | z | + | w | ) q l ( t ) 1 = 0 for  t J  uniformly , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_Eque_HTML.gif

          where q l ( t ) C ( J , R ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq66_HTML.gif, and 1 < q l q l + < p l http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq67_HTML.gif. We say f l http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq64_HTML.gif satisfies a general growth condition if f l http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq64_HTML.gif does not satisfy a sub- ( p l 1 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq65_HTML.gif growth condition.

          We will discuss the existence of solutions for (P) in the following two cases:
          1. (i)

            f l http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq64_HTML.gif satisfies a sub- ( p l 1 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq65_HTML.gif growth condition for l = 1 , 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq4_HTML.gif;

             
          2. (ii)

            f l http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq64_HTML.gif satisfies a general growth condition for l = 1 , 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq4_HTML.gif.

             

          This paper is organized as follows. In Section 2, we do some preparation. In Section 3, we discuss the existence of solutions of (P). Finally, in Section 4, we discuss the existence of nonnegative solutions for (P).

          2 Preliminary

          For any ( t , x ) J × R N http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq68_HTML.gif, denote φ p l ( t , x ) = | x | p l ( t ) 2 x http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq69_HTML.gif ( l = 1 , 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq28_HTML.gif). Obviously, φ p l http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq70_HTML.gif has the following properties.

          Lemma 2.1 (see [5])

          φ p l http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq70_HTML.gif is a continuous function and satisfies the following:
          1. (i)
            For any t [ 0 , 1 ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq71_HTML.gif, φ p l ( t , ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq72_HTML.gif is strictly monotone, that is,
            φ p l ( t , x 1 ) φ p l ( t , x 2 ) , x 1 x 2 > 0 for any x 1 , x 2 R N , x 1 x 2 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_Equf_HTML.gif
             
          2. (ii)
            There exists a function β l : [ 0 , + ) [ 0 , + ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq73_HTML.gif, β l ( s ) + http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq74_HTML.gif as s + http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq75_HTML.gif, such that
            φ p l ( t , x ) , x β l ( | x | ) | x | for all x R N . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_Equg_HTML.gif
             
          It is well known that φ p l ( t , ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq72_HTML.gif is a homeomorphism from R N http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq40_HTML.gif to R N http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq43_HTML.gif for any fixed t [ 0 , 1 ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq71_HTML.gif. For any t J http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq29_HTML.gif, denote by φ p l 1 ( t , ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq76_HTML.gif the inverse operator of φ p l ( t , ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq72_HTML.gif, then
          φ p l 1 ( t , x ) = | x | 2 p l ( t ) p l ( t ) 1 x , for  x R N { 0 } , φ p l 1 ( t , 0 ) = 0 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_Equh_HTML.gif

          It is clear that φ p l 1 ( t , ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq76_HTML.gif is continuous and sends bounded sets into bounded sets.

          Let us now consider the following problem:
          ( φ p 1 ( t , u ( t ) ) ) = g 1 ( t ) , t ( 0 , 1 ) , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_Equ1_HTML.gif
          (1)
          with the boundary value condition
          u ( 1 ) = i = 1 m 2 α i u ( ξ i ) + e 1 , lim t 1 | u | p 1 ( t ) 2 u ( t ) = 0 1 k ( t ) | u | p 1 ( t ) 2 u ( t ) d t + e 2 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_Equ2_HTML.gif
          (2)
          where g 1 L 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq77_HTML.gif. If u is a solution of (1) with (2), by integrating (1) from 0 to t, we find that
          φ p 1 ( t , u ( t ) ) = φ p 1 ( 0 , u ( 0 ) ) 0 t g 1 ( s ) d s . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_Equ3_HTML.gif
          (3)
          Denote a 1 = φ p 1 ( 0 , u ( 0 ) ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq78_HTML.gif. It is easy to see that a 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq79_HTML.gif is dependent on g 1 ( ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq80_HTML.gif. Define operator F : L 1 C http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq81_HTML.gif as
          F ( g 1 ) ( t ) = 0 t g 1 ( s ) d s , t J , g 1 L 1 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_Equi_HTML.gif
          From (3), we have
          u ( t ) = φ p 1 1 [ t , a 1 F ( g 1 ) ] . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_Equ4_HTML.gif
          (4)
          By integrating (4) from 0 to t, we find that
          u ( t ) = u ( 0 ) + F { φ p 1 1 [ t , a 1 F ( g 1 ) ] } ( t ) , t J . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_Equj_HTML.gif
          From (2), we have
          a 1 = 0 1 g 1 ( t ) d t 0 1 k ( t ) 0 t g 1 ( s ) d s d t + e 2 1 σ 1 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_Equk_HTML.gif
          and
          u ( 0 ) = i = 1 m 2 α i 0 ξ i φ p 1 1 [ t , a 1 F ( g 1 ) ( t ) ] d t 0 1 φ p 1 1 [ t , a 1 F ( g 1 ) ( t ) ] d t + e 1 1 i = 1 m 2 α i . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_Equl_HTML.gif
          For fixed h 1 L 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq82_HTML.gif, we define a 1 : L 1 R N http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq83_HTML.gif as
          a 1 ( h 1 ) = 0 1 h 1 ( t ) d t 0 1 k ( t ) 0 t h 1 ( s ) d s d t + e 2 1 σ 1 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_Equ5_HTML.gif
          (5)

          It is easy to obtain the following lemma.

          Lemma 2.2 a 1 : L 1 R N http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq84_HTML.gif is continuous and sends bounded sets of L 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq85_HTML.gif to bounded sets of R N http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq40_HTML.gif. Moreover,
          | a 1 ( h 1 ) | 2 N 1 σ 1 ( h 1 L 1 + | e 2 | ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_Equ6_HTML.gif
          (6)

          It is clear that a 1 ( ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq86_HTML.gif is a compact continuous mapping.

          Let us now consider another problem
          ( φ p 2 ( t , v ( t ) ) ) = g 2 ( t ) , t ( 0 , 1 ) , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_Equ7_HTML.gif
          (7)
          with the boundary value condition
          v ( 0 ) k 1 v ( 0 ) = 0 1 e ( t ) v ( t ) d t , v ( 1 ) + k 2 v ( 1 ) = i = 1 m 2 β i v ( η i ) , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_Equ8_HTML.gif
          (8)
          where g 2 L 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq87_HTML.gif. Similar to the discussion of the solutions of (1) with (2), we have
          v ( t ) = φ p 2 1 [ t , a 2 F ( g 2 ) ] , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_Equm_HTML.gif
          and
          v ( t ) = v ( 0 ) + F { φ p 2 1 [ t , a 2 F ( g 2 ) ] } ( t ) , t J , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_Equn_HTML.gif

          where a 2 : = φ p 2 ( 0 , v ( 0 ) ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq88_HTML.gif, F ( g 2 ) ( t ) = 0 t g 2 ( s ) d s http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq89_HTML.gif for any t J http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq29_HTML.gif.

          From v ( 0 ) k 1 v ( 0 ) = 0 1 e ( t ) v ( t ) d t http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq90_HTML.gif, we have
          v ( 0 ) = k 1 φ p 2 1 ( 0 , a 2 ) + 0 1 e ( t ) 0 t φ p 2 1 [ s , a 2 F ( g 2 ) ( s ) ] d s d t 1 σ 2 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_Equ9_HTML.gif
          (9)
          From v ( 1 ) + k 2 v ( 1 ) = i = 1 m 2 β i v ( η i ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq91_HTML.gif, we have
          v ( 0 ) = i = 1 m 2 β i 0 η i φ p 2 1 [ t , a 2 F ( g 2 ) ( t ) ] d t 0 1 φ p 2 1 [ t , a 2 F ( g 2 ) ( t ) ] d t 1 i = 1 m 2 β i k 2 φ p 2 1 [ 1 , a 2 F ( g 2 ) ( 1 ) ] 1 i = 1 m 2 β i . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_Equ10_HTML.gif
          (10)
          From (9) and (10), we have
          http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_Equo_HTML.gif
          For fixed h 2 C http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq92_HTML.gif, we denote
          Λ h 2 ( a 2 ) = k 1 φ p 2 1 ( 0 , a 2 ) + 0 1 e ( t ) 0 t φ p 2 1 [ s , a 2 h 2 ( s ) ] d s d t 1 σ 2 i = 1 m 2 β i 0 η i φ p 2 1 [ t , a 2 h 2 ( t ) ] d t 0 1 φ p 2 1 [ t , a 2 h 2 ( t ) ] d t 1 i = 1 m 2 β i + k 2 φ p 2 1 [ 1 , a 2 h 2 ( 1 ) ] 1 i = 1 m 2 β i . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_Equp_HTML.gif
          Lemma 2.3 The function Λ h 2 ( ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq93_HTML.gif has the following properties:
          1. (i)
            For any fixed h 2 C http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq92_HTML.gif, the equation
            Λ h 2 ( a 2 ) = 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_Equ11_HTML.gif
            (11)
             
          has a unique solution a 2 ˜ ( h 2 ) R N http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq94_HTML.gif.
          1. (ii)
            The function a 2 ˜ : C R N http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq95_HTML.gif, defined in (i), is continuous and sends bounded sets to bounded sets. Moreover,
            | a 2 ˜ ( h 2 ) | 3 N h 2 0 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_Equq_HTML.gif
             
          Proof (i) It is easy to see that
          Λ h 2 ( a 2 ) = k 1 φ p 2 1 ( 0 , a 2 ) + 0 1 e ( t ) 0 t φ p 2 1 [ s , a 2 h 2 ( s ) ] d s d t 1 σ 2 + i = 1 m 2 β i η i 1 φ p 2 1 [ t , a 2 h 2 ( t ) ] d t + k 2 φ p 2 1 [ 1 , a 2 h 2 ( 1 ) ] 1 i = 1 m 2 β i + 0 1 φ p 2 1 [ t , a 2 h 2 ( t ) ] d t . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_Equr_HTML.gif
          From Lemma 2.1, it is immediate that
          Λ h 2 ( x ) Λ h 2 ( y ) , x y > 0 for  x , y R N  with  x y , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_Equs_HTML.gif

          and hence, if (11) has a solution, then it is unique.

          Let t 0 = 3 N h 2 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq96_HTML.gif. Suppose | a 2 | > t 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq97_HTML.gif. Since h 2 C http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq92_HTML.gif, it is easy to see that there exists an i { 1 , , N } http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq98_HTML.gif such that the i th component a 2 i http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq99_HTML.gif of a 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq100_HTML.gif satisfies
          | a 2 i | | a 2 | N > 3 h 2 0 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_Equt_HTML.gif
          Thus ( a 2 i h 2 i ( t ) ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq101_HTML.gif keeps sign on J and
          | a 2 i h 2 i ( t ) | | a 2 i | h 2 0 2 | a 2 | 3 N > 2 h 2 0 , t J . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_Equu_HTML.gif
          Obviously, | a 2 h 2 ( t ) | 4 | a 2 | 3 2 N | a 2 i h 2 i ( t ) | http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq102_HTML.gif, then
          | a 2 h 2 ( t ) | 2 p 2 ( t ) p 2 ( t ) 1 | a 2 i h 2 i ( t ) | > 1 2 N | a 2 i h 2 i ( t ) | 1 p 2 ( t ) 1 > 1 2 N [ 2 h 2 0 ] 1 p 2 ( ζ ) 1 , ζ J , t J . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_Equv_HTML.gif
          Thus the i th component Λ h 2 i ( a 2 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq103_HTML.gif of Λ h 2 ( a 2 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq104_HTML.gif is nonzero and keeps sign, and then we have
          Λ h 2 ( a 2 ) 0 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_Equw_HTML.gif
          Let us consider the equation
          λ Λ h 2 ( a 2 ) + ( 1 λ ) a 2 = 0 , λ [ 0 , 1 ] . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_Equ12_HTML.gif
          (12)
          It is easy to see that all the solutions of (12) belong to b ( t 0 + 1 ) = { x R N | x | < t 0 + 1 } http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq105_HTML.gif. So, we have
          d B [ Λ h 2 ( a 2 ) , b ( t 0 + 1 ) , 0 ] = d B [ I , b ( t 0 + 1 ) , 0 ] 0 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_Equx_HTML.gif

          which implies the existence of solutions of Λ h 2 ( a 2 ) = 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq106_HTML.gif.

          In this way, we define a function a 2 ˜ ( h 2 ) : C [ 0 , 1 ] R N http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq107_HTML.gif, which satisfies
          Λ h 2 ( a 2 ˜ ( h 2 ) ) = 0 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_Equy_HTML.gif
          1. (ii)
            By the proof of (i), we also obtain that a 2 ˜ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq108_HTML.gif sends bounded sets to bounded sets, and
            | a 2 ˜ ( h 2 ) | 3 N h 2 0 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_Equz_HTML.gif
             

          It only remains to prove the continuity of a 2 ˜ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq108_HTML.gif. Let { v n } http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq109_HTML.gif be a convergent sequence in C and v n v http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq110_HTML.gif as n + http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq111_HTML.gif. Since { a 2 ˜ ( v n ) } http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq112_HTML.gif is a bounded sequence, then it contains a convergent subsequence { a 2 ˜ ( v n j ) } http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq113_HTML.gif. Let a 2 ˜ ( v n j ) a 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq114_HTML.gif as j + http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq115_HTML.gif. Since Λ v n j ( a 2 ˜ ( v n j ) ) = 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq116_HTML.gif, letting j + http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq115_HTML.gif, we have Λ v ( a 0 ) = 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq117_HTML.gif. From (i), we get a 0 = a 2 ˜ ( v ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq118_HTML.gif, it means that a 2 ˜ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq108_HTML.gif is continuous. The proof is completed. □

          Now, we define the operator a 2 : L 1 R N http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq119_HTML.gif as
          a 2 ( v ) = a 2 ˜ ( F ( v ) ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_Equ13_HTML.gif
          (13)

          It is clear that a 2 ( ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq120_HTML.gif is continuous and sends bounded sets of L 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq85_HTML.gif into bounded sets of R N http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq40_HTML.gif, and hence it is a compact continuous mapping.

          If u is a solution of (1) with (2), we have
          u ( t ) = u ( 0 ) + F { φ p 1 1 [ t , a 1 F ( g 1 ) ] } ( t ) , t J , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_Equaa_HTML.gif
          and
          u ( 0 ) = i = 1 m 2 α i 0 ξ i φ p 1 1 [ t , a 1 F ( g 1 ) ( t ) ] d t 0 1 φ p 1 1 [ t , a 1 F ( g 1 ) ( t ) ] d t + e 1 1 i = 1 m 2 α i . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_Equab_HTML.gif
          If u is a solution of (7) with (8), we have
          v ( t ) = v ( 0 ) + F { φ p 2 1 [ t , a 2 F ( g 2 ) ] } ( t ) , t J , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_Equac_HTML.gif
          and
          v ( 0 ) = k 1 φ p 2 1 ( 0 , a 2 ) + 0 1 e ( t ) 0 t φ p 2 1 [ s , a 2 F ( g 2 ) ( s ) ] d s d t 1 σ 2 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_Equad_HTML.gif
          We denote
          http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_Equae_HTML.gif

          Lemma 2.4 The operators K l http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq121_HTML.gif ( l = 1 , 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq28_HTML.gif) are continuous and send equi-integrable sets in L 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq85_HTML.gif to relatively compact sets in C 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq122_HTML.gif.

          Proof We only prove that the operator K 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq123_HTML.gif is continuous and sends equi-integrable sets in L 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq85_HTML.gif to relatively compact sets in C 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq53_HTML.gif, the rest is similar.

          It is easy to check that K 1 ( h 1 ) ( t ) C 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq124_HTML.gif for all h 1 L 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq82_HTML.gif. Since
          http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_Equaf_HTML.gif

          it is easy to check that K 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq123_HTML.gif is a continuous operator from L 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq85_HTML.gif to C 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq53_HTML.gif.

          Let now U be an equi-integrable set in L 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq85_HTML.gif, then there exists ρ L 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq125_HTML.gif such that
          | u ( t ) | ρ ( t ) a.e. in  J  for any  u L 1 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_Equag_HTML.gif

          We want to show that K 1 ( U ) ¯ C 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq126_HTML.gif is a compact set.

          Let { u n } http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq127_HTML.gif be a sequence in K 1 ( U ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq128_HTML.gif, then there exists a sequence { h n } U http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq129_HTML.gif such that u n = K 1 ( h n ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq130_HTML.gif. For any t 1 , t 2 J http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq131_HTML.gif, we have
          | F ( h n ) ( t 1 ) F ( h n ) ( t 2 ) | = | 0 t 1 h n ( t ) d t 0 t 2 h n ( t ) d t | = | t 1 t 2 h n ( t ) d t | | t 1 t 2 ρ ( t ) d t | . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_Equah_HTML.gif

          Hence the sequence { F ( h n ) } http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq132_HTML.gif is uniformly bounded and equicontinuous. By the Ascoli-Arzela theorem, there exists a subsequence of { F ( h n ) } http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq132_HTML.gif (which we still denote by { F ( h n ) } http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq132_HTML.gif) convergent in C. According to the bounded continuous of the operator a 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq79_HTML.gif, we can choose a subsequence of { a 1 ( h n ) F ( h n ) } http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq133_HTML.gif (which we still denote by { a 1 ( h n ) F ( h n ) } http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq133_HTML.gif) which is convergent in C, then φ p 1 ( t , K 1 ( h n ) ( t ) ) = a 1 ( h n ) F ( h n ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq134_HTML.gif is convergent in C.

          From the definition of K 1 ( h n ) ( t ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq135_HTML.gif and the continuity of φ p 1 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq136_HTML.gif, we can see that K 1 ( h n ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq137_HTML.gif is convergent in C. Thus, { u n } http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq127_HTML.gif is convergent in C 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq53_HTML.gif. This completes the proof. □

          Let us define P 1 , P 2 : C 1 C 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq138_HTML.gif as
          P 1 ( h 1 ) = i = 1 m 2 α i K 1 ( h 1 ) ( ξ i ) K 1 ( h 1 ) ( 1 ) + e 1 1 i = 1 m 2 α i , P 2 ( h 2 ) = k 1 φ p 2 1 ( 0 , a 2 ( h 2 ) ) + 0 1 e ( t ) K 2 ( h 2 ) ( t ) d t 1 σ 2 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_Equai_HTML.gif

          It is easy to see that P 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq139_HTML.gif and P 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq140_HTML.gif are both compact continuous.

          We denote N f l ( u , v ) : [ 0 , 1 ] × C 1 L 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq141_HTML.gif ( l = 1 , 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq28_HTML.gif) the Nemytskii operator associated to f l http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq64_HTML.gif defined by
          N f l ( u , v ) ( t ) = f l ( t , u ( t ) , u ( t ) , v ( t ) , v ( t ) ) a.e. on  J . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_Equaj_HTML.gif
          Lemma 2.5 ( u , v ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq142_HTML.gif is a solution of (P) if and only if ( u , v ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq142_HTML.gif is a solution of the following abstract equation:
          ( S ) { u = P 1 ( δ 1 N f 1 ( u , v ) ) + K 1 ( δ 1 N f 1 ( u , v ) ) , v = P 2 ( δ 2 N f 2 ( u , v ) ) + K 2 ( δ 2 N f 2 ( u , v ) ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_Equak_HTML.gif

          Proof If ( u , v ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq142_HTML.gif is a solution to (P), according to the proof before Lemma 2.5, it is easy to obtain that ( u , v ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq142_HTML.gif is a solution to (S).

          Conversely, if ( u , v ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq142_HTML.gif is a solution to (S), then
          u ( 1 ) = P 1 ( δ 1 N f 1 ( u , v ) ) + K 1 ( δ 1 N f 1 ( u , v ) ) ( 1 ) = i = 1 m 2 α i K 1 ( δ 1 N f 1 ( u , v ) ) ( ξ i ) K 1 ( δ 1 N f 1 ( u , v ) ) ( 1 ) + e 1 1 i = 1 m 2 α i + K 1 ( δ 1 N f 1 ( u , v ) ) ( 1 ) = i = 1 m 2 α i K 1 ( δ 1 N f 1 ( u , v ) ) ( ξ i ) i = 1 m 2 α i K 1 ( δ 1 N f 1 ( u , v ) ) ( 1 ) + e 1 1 i = 1 m 2 α i = i = 1 m 2 α i [ u ( ξ i ) u ( 0 ) ] i = 1 m 2 α i [ u ( 1 ) u ( 0 ) ] + e 1 1 i = 1 m 2 α i = i = 1 m 2 α i u ( ξ i ) i = 1 m 2 α i u ( 1 ) + e 1 1 i = 1 m 2 α i , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_Equal_HTML.gif
          which implies
          u ( 1 ) = i = 1 m 2 α i u ( ξ i ) + e 1 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_Equam_HTML.gif
          It follows from (S) that
          φ p 1 ( t , u ( t ) ) = a 1 ( δ 1 N f 1 ( u , v ) ) F ( δ 1 N f 1 ( u , v ) ) ( t ) , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_Equan_HTML.gif
          then
          φ p 1 ( 1 , u ( 1 ) ) = a 1 ( δ 1 N f 1 ( u , v ) ) F ( δ 1 N f 1 ( u , v ) ) ( 1 ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_Equao_HTML.gif
          By the condition of the mapping a 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq79_HTML.gif, we have
          φ p 1 ( 1 , u ( 1 ) ) = 0 1 δ 1 N f 1 ( u , v ) ( t ) d t 0 1 k ( t ) 0 t δ 1 N f 1 ( u , v ) ( s ) d s d t + e 2 1 σ 1 0 1 δ 1 N f 1 ( u , v ) ( t ) d t = σ 1 0 1 δ 1 N f 1 ( u , v ) ( t ) d t 0 1 k ( t ) 0 t δ 1 N f 1 ( u , v ) ( s ) d s d t + e 2 1 σ 1 = σ 1 [ a 1 φ p 1 ( 1 , u ( 1 ) ) ] 0 1 k ( t ) [ a 1 φ p 1 ( t , u ( t ) ) ] d t + e 2 1 σ 1 = σ 1 φ p 1 ( 1 , u ( 1 ) ) + 0 1 k ( t ) φ p 1 ( t , u ( t ) ) d t + e 2 1 σ 1 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_Equap_HTML.gif
          and then
          φ p 1 ( 1 , u ( 1 ) ) = 0 1 k ( t ) φ p 1 ( t , u ( t ) ) d t + e 2 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_Equaq_HTML.gif
          From (S), we have
          v ( t ) = φ p 2 1 [ t , a 2 F ( δ 2 N f 2 ( u , v ) ) ] , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_Equar_HTML.gif
          and
          v ( 0 ) = P 2 ( δ 2 N f 2 ( u , v ) ) = k 1 φ p 2 1 ( 0 , a 2 ) + 0 1 e ( t ) K 2 ( δ 2 N f 2 ( u , v ) ) ( t ) d t 1 σ 2 = k 1 φ p 2 1 ( 0 , a 2 ) + 0 1 e ( t ) v ( t ) d t σ 2 v ( 0 ) 1 σ 2 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_Equas_HTML.gif
          then
          v ( 0 ) = k 1 φ p 2 1 ( 0 , a 2 ) + 0 1 e ( t ) v ( t ) d t . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_Equat_HTML.gif
          Thus
          v ( 0 ) k 1 v ( 0 ) = k 1 φ p 2 1 ( 0 , a 2 ) + 0 1 e ( t ) v ( t ) d t k 1 φ p 2 1 ( 0 , a 2 ) = 0 1 e ( t ) v ( t ) d t . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_Equau_HTML.gif
          From (S), we have
          v ( 1 ) = P 2 ( δ 2 N f 2 ( u , v ) ) + K 2 ( δ 2 N f 2 ( u , v ) ) ( 1 ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_Equav_HTML.gif
          By the condition of the mapping a 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq100_HTML.gif, we have
          v ( 1 ) = P 2 ( δ 2 N f 2 ( u , v ) ) + K 2 ( δ 2 N f 2 ( u , v ) ) ( 1 ) = i = 1 m 2 β i η i 1 φ p 2 1 [ t , a 2 F ( δ 2 N f 2 ( u , v ) ) ( t ) ] d t + k 2 φ p 2 1 [ 1 , a 2 F ( δ 2 N f 2 ( u , v ) ) ( 1 ) ] 1 i = 1 m 2 β i = i = 1 m 2 β i [ v ( 1 ) v ( η i ) ] + k 2 φ p 2 1 [ 1 , a 2 F ( δ 2 N f 2 ( u , v ) ) ( 1 ) ] 1 i = 1 m 2 β i , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_Equaw_HTML.gif
          which implies that
          v ( 1 ) = i = 1 m 2 β i v ( η i ) k 2 φ p 2 1 [ 1 , a 2 F ( δ 2 N f 2 ( u , v ) ) ( 1 ) ] . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_Equax_HTML.gif
          Since v ( 1 ) = φ p 2 1 [ 1 , a 2 F ( δ 2 N f 2 ( u , v ) ) ( 1 ) ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq143_HTML.gif, then we have
          v ( 1 ) + k 2 v ( 1 ) = i = 1 m 2 β i v ( η i ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_Equay_HTML.gif
          Moreover, from (S), it is easy to obtain
          ( φ p 1 ( t , u ( t ) ) ) = δ 1 N f 1 ( u , v ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_Equaz_HTML.gif
          and
          ( φ p 2 ( t , v ( t ) ) ) = δ 2 N f 2 ( u , v ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_Equba_HTML.gif

          Hence ( u , v ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq142_HTML.gif is a solution of (P).

          This completes the proof. □

          3 Existence of solutions

          In this section, we apply Leray-Schauder’s degree to deal with the existence of solutions for (P), when f l http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq64_HTML.gif satisfies a sub- ( p l 1 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq65_HTML.gif growth condition or a general growth condition ( l = 1 , 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq28_HTML.gif).

          We denote (S) as
          ( u , v ) = A ( u , v ) = ( Ψ f 1 ( u , v ) , Φ f 2 ( u , v ) ) , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_Equbb_HTML.gif
          where
          http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_Equbc_HTML.gif

          Theorem 3.1 If f l http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq64_HTML.gif satisfies a sub- ( p l 1 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq65_HTML.gif growth condition, then the problem (P) has at least one solution for any fixed parameter δ l http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq144_HTML.gif ( l = 1 , 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq28_HTML.gif).

          Proof Denote
          A λ ( u , v ) = ( Ψ λ f 1 ( u , v ) , Φ λ f 2 ( u , v ) ) , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_Equbd_HTML.gif
          where
          http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_Eqube_HTML.gif
          According to Lemma 2.5, we know that (P) has the same solution of
          ( u , v ) = A λ ( u , v ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_Equ14_HTML.gif
          (14)

          when λ = 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq145_HTML.gif.

          It is easy to see that the operators P 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq139_HTML.gif and P 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq140_HTML.gif are compact continuous. According to Lemma 2.2, Lemma 2.3 and Lemma 2.4, we can see that Ψ λ f 1 ( u , v ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq146_HTML.gif and Φ λ f 2 ( u , v ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq147_HTML.gif are compact continuous from C 1 × [ 0 , 1 ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq148_HTML.gif to C 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq53_HTML.gif, thus A λ ( u , v ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq149_HTML.gif is compact continuous from W × [ 0 , 1 ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq150_HTML.gif to W.

          We claim that all the solutions of (14) are uniformly bounded for λ [ 0 , 1 ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq151_HTML.gif. In fact, if it is false, we can find a sequence of solutions { ( ( u n , v n ) , λ n ) } http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq152_HTML.gif for (14) such that ( u n , v n ) + http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq153_HTML.gif as n + http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq154_HTML.gif.

          From Lemma 2.2, we have
          | a 1 ( λ n δ 1 N f 1 ( u n , v n ) ) | C 1 ( N f 1 ( u n , v n ) L 1 + | e 2 | ) C 2 ( 1 + ( u n , v n ) ) q 1 + 1 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_Equbf_HTML.gif
          which together with the sub- ( p 1 1 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq155_HTML.gif growth condition of f 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq156_HTML.gif implies that
          http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_Equ15_HTML.gif
          (15)
          From (14), we have
          | u n ( t ) | p 1 ( t ) 2 u n ( t ) = a 1 ( λ n δ 1 N f 1 ( u n , v n ) ) F ( λ n δ 1 N f 1 ( u n , v n ) ) , t J , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_Equbg_HTML.gif
          then
          | u n ( t ) | p 1 ( t ) 1 | a 1 ( λ n δ 1 N f 1 ( u n , v n ) ) | + | F ( λ n δ 1 N f 1 ( u n , v n ) ) | C 4 ( 1 + ( u n , v n ) ) q 1 + 1 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_Equbh_HTML.gif
          Denote α 1 = q 1 + 1 p 1 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq157_HTML.gif. From the above inequality we have
          u n ( t ) 0 C 5 ( 1 + ( u n , v n ) ) α 1 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_Equ16_HTML.gif
          (16)
          It follows from (14) and (15) that
          | u n ( 0 ) | C 6 ( 1 + ( u n , v n ) ) α 1 , where  α 1 = q 1 + 1 p 1 1 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_Equbi_HTML.gif
          For any j = 1 , , N http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq158_HTML.gif, we have
          | u n j ( t ) | = | u n j ( 0 ) + 0 t ( u n j ) ( r ) d r | | u n j ( 0 ) | + | 0 t ( u n j ) ( r ) d r | [ C 7 + C 5 ] ( 1 + ( u n , v n ) ) α 1 C 8 ( 1 + ( u n , v n ) ) α 1 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_Equbj_HTML.gif
          which implies that
          | u n j | 0 C 9 ( 1 + ( u n , v n ) ) α 1 , j = 1 , , N ; n = 1 , 2 , . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_Equbk_HTML.gif
          Thus
          u n 0 C 10 ( 1 + ( u n , v n ) ) α 1 , n = 1 , 2 , . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_Equ17_HTML.gif
          (17)

          It follows from (16) and (17) that u n 1 C 11 ( 1 + ( u n , v n ) ) α 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq159_HTML.gif.

          Similarly, we have v n 1 C 12 ( 1 + ( u n , v n ) ) α 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq160_HTML.gif, where α 2 = q 2 + 1 p 2 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq161_HTML.gif.

          Thus, { ( u n , v n ) } http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq162_HTML.gif is bounded.

          Thus, we can choose a large enough R 0 > 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq163_HTML.gif such that all the solutions of (14) belong to B ( R 0 ) = { ( u , v ) W ( u n , v n ) < R 0 } http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq164_HTML.gif. Therefore, the Leray-Schauder degree d L S [ I A λ ( u , v ) , B ( R 0 ) , 0 ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq165_HTML.gif is well defined for each λ [ 0 , 1 ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq151_HTML.gif, and
          d L S [ I A 1 ( u , v ) , B ( R 0 ) , 0 ] = d L S [ I A 0 ( u , v ) , B ( R 0 ) , 0 ] . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_Equbl_HTML.gif
          Denote
          u 0 = i = 1 m 2 α i 0 ξ i φ p 1 1 [ t , a 1 ( 0 ) ] d t 0 1 φ p 1 1 [ t , a 1 ( 0 ) ] d t + e 1 1 i = 1 m 2 α i + 0 r φ p 1 1 [ t , a 1 ( 0 ) ] d t , v 0 = k 1 φ p 2 1 [ 0 , a 2 ( 0 ) ] + 0 1 e ( t ) 0 t φ p 2 1 [ r , a 2 ( 0 ) ] d r d t 1 σ 2 + 0 r φ p 2 1 [ t , a 2 ( 0 ) ] d t , } http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_Equ18_HTML.gif
          (18)

          where a 1 ( 0 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq166_HTML.gif and a 2 ( 0 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq167_HTML.gif are defined in (5) and (13), then ( u 0 , v 0 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq168_HTML.gif is the unique solution of ( u , v ) = A 0 ( u , v ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq169_HTML.gif.

          It is easy to see that ( u , v ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq142_HTML.gif is a solution of ( u , v ) = A 0 ( u , v ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq169_HTML.gif if and only if ( u , v ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq142_HTML.gif is a solution of the following system:
          p 1 ( t ) u = 0 , t ( 0 , 1 ) , p 2 ( t ) v = 0 , t ( 0 , 1 ) , u ( 1 ) = i = 1 m 2 α i u ( ξ i ) + e 1 , lim t 1 | u | p 1 ( t ) 2 u ( t ) = 0 1 k ( t ) | u | p 1 ( t ) 2 u ( t ) d t + e 2 , v ( 0 ) k 1 v ( 0 ) = 0 1 e ( t ) v ( t ) d t , v ( 1 ) + k 2 v ( 1 ) = i = 1 m 2 β i v ( η i ) . } http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_Equ19_HTML.gif
          (19)
          Obviously, (19) possesses a unique solution ( u 0 , v 0 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq168_HTML.gif. Note that ( u 0 , v 0 ) B ( R 0 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq170_HTML.gif, we have
          d L S [ I A 1 ( u , v ) , B ( R 0 ) , 0 ] = d L S [ I A 0 ( u , v ) , B ( R 0 ) , 0 ] 0 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_Equbm_HTML.gif

          Therefore (P) has at least one solution. This completes the proof. □

          In the following, we investigate the existence of solutions for (P) when f l http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq64_HTML.gif satisfies a general growth condition.

          Denote
          http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_Equbn_HTML.gif

          Assume the following.

          (A1) Let a positive constant ε be such that ( u 0 , v 0 ) Ω ε http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq171_HTML.gif, | P 1 ( 0 ) | < θ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq172_HTML.gif, | P 2 ( 0 ) | < θ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq173_HTML.gif and | a 1 ( 0 ) | < min t J ( θ 3 ) p 1 ( t ) 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq174_HTML.gif, | a 2 ( 0 ) | < min t J ( θ 2 ) p 2 ( t ) 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq175_HTML.gif, where ( u 0 , v 0 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq168_HTML.gif is defined in (18), a 1 ( ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq86_HTML.gif and a 2 ( ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq120_HTML.gif are defined in (5) and (13), respectively.

          It is easy to see that Ω ε http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq176_HTML.gif is an open bounded domain in W. We have the following theorem.

          Theorem 3.2 Assume that (A1) is satisfied. If positive parameters δ 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq19_HTML.gif and δ 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq20_HTML.gif are small enough, then the problem (P) has at least one solution on Ω ε ¯ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq177_HTML.gif.

          Proof Similarly, we denote A λ ( u , v ) = ( Ψ λ f 1 ( u , v ) , Φ λ f 2 ( u , v ) ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq178_HTML.gif. By Lemma 2.5, ( u , v ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq142_HTML.gif is a solution of
          { p 1 ( t ) u = λ δ 1 f 1 ( t , u , u , v , v ) , t ( 0 , 1 ) , p 2 ( t ) v = λ δ 2 f 2 ( t , u , u , v , v ) , t ( 0 , 1 ) , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_Equbo_HTML.gif
          with (2) and (8) if and only if ( u , v ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq142_HTML.gif is a solution of the following abstract equation:
          ( u , v ) = A λ ( u , v ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_Equ20_HTML.gif
          (20)

          From the proof of Theorem 3.1, we can see that A λ ( u , v ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq149_HTML.gif is compact continuous from W × [ 0 , 1 ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq150_HTML.gif to W. According to Leray-Schauder’s degree theory, we only need to prove that

          (1) ( u , v ) = A λ ( u , v ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq179_HTML.gif has no solution on Ω ε http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq180_HTML.gif for any λ [ 0 , 1 ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq151_HTML.gif,

          (2) d L S [ I A 0 ( u , v ) , Ω ε , 0 ] 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq181_HTML.gif,

          then we can conclude that the system (P) has a solution on Ω ε ¯ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq182_HTML.gif.

          (1) If there exists a λ [ 0 , 1 ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq151_HTML.gif and ( u , v ) Ω ε http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq183_HTML.gif is a solution of (20), then ( u , v ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq142_HTML.gif and λ satisfy
          u ( t ) = φ p 1 1 [ t , a 1 F ( λ δ 1 N f 1 ( u , v ) ) ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_Equbp_HTML.gif
          and
          v ( t ) = φ p 2 1 [ t , a 2 F ( λ δ 2 N f 2 ( u , v ) ) ] . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_Equbq_HTML.gif
          Since ( u , v ) Ω ε http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq183_HTML.gif, there exists an i such that | u i | 0 + | ( u i ) | 0 = ε http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq184_HTML.gif or | v i | 0 + | ( v i ) | 0 = ε http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq185_HTML.gif.
          1. (i)

            If | u i | 0 + | ( u i ) | 0 = ε http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq184_HTML.gif.

             

          ( i http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq186_HTML.gif) Suppose that | u i | 0 > 2 θ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq187_HTML.gif, then | ( u i ) | 0 < ε 2 θ = θ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq188_HTML.gif. On the other hand, for any t , t J http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq189_HTML.gif, we have

          | u i ( t ) u i ( t ) | = | t t ( u i ) ( r ) d r | 0 1 | ( u i ) ( r ) | d r < θ . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_Equbr_HTML.gif
          This implies that | u i ( t ) | > θ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq190_HTML.gif for each t J http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq29_HTML.gif.
          Note that ( u , v ) Ω ε ¯ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq191_HTML.gif, then | f 1 ( t , u , u , v , v ) | β N ε ( t ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq192_HTML.gif, holding | F ( N f 1 ) | 0 1 β N ε ( t ) d t http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq193_HTML.gif. Since P 1 ( ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq194_HTML.gif is continuous, when 0 < δ 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq195_HTML.gif is small enough, from (A1), we have
          | u ( 0 ) | = | P 1 ( λ δ 1 N f 1 ( u , v ) ) | < θ . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_Equbs_HTML.gif

          It is a contradiction to | u i ( t ) | > θ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq196_HTML.gif for each t J http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq29_HTML.gif.

          ( ii http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq197_HTML.gif) Suppose that | u i | 0 2 θ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq198_HTML.gif, then θ | ( u i ) | 0 ε http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq199_HTML.gif. This implies that | ( u i ) ( t 2 ) | θ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq200_HTML.gif for some t 2 J http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq201_HTML.gif, and we can find
          θ | ( u i ) ( t 2 ) | | ( u ) ( t 2 ) | = | φ p 1 1 [ t 2 , a 1 F ( λ δ 1 N f 1 ( u , v ) ) ( t 2 ) ] | . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_Equ21_HTML.gif
          (21)
          Since ( u , v ) Ω ε ¯ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq191_HTML.gif and f 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq156_HTML.gif is Carathéodory, it is easy to see that
          | f 1 ( t , u , u , v , v ) | β N ε ( t ) , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_Equbt_HTML.gif
          thus
          | δ 1 F ( N f 1 ) | δ 1 0 1 β N ε ( t ) d t . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_Equbu_HTML.gif
          From Lemma 2.2, a 1 ( ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq86_HTML.gif is continuous, then we have
          | a 1 ( λ δ 1 N f 1 ) | | a 1 ( 0 ) | as  δ 1 0 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_Equbv_HTML.gif
          When 0 < δ 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq195_HTML.gif is small enough, from (A1) and (21), we can conclude that
          θ | φ p 1 1 [ t , a 1 F ( λ δ 1 N f 1 ( u , v ) ) ( t ) ] | < θ 3 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_Equbw_HTML.gif
          It is a contradiction. Thus | u i | 0 + | ( u i ) | 0 ε http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq202_HTML.gif.
          1. (ii)

            If | v i | 0 + | ( v i ) | 0 = ε http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq203_HTML.gif. Similar to the proof of (i), we get a contradiction. Thus | v i | 0 + | ( v i ) | 0 ε http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq204_HTML.gif.

             

          Summarizing this argument, for each λ [ 0 , 1 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq205_HTML.gif, ( u , v ) = A λ ( u , v ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq179_HTML.gif has no solution on Ω ε http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq180_HTML.gif when positive parameters δ 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq19_HTML.gif and δ 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq20_HTML.gif are small enough.

          (2) Since ( u 0 , v 0 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq168_HTML.gif (where ( u 0 , v 0 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq168_HTML.gif is defined in (18)) is the unique solution of ( u , v ) = A 0 ( u , v ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq169_HTML.gif, and (A1) holds ( u 0 , v 0 ) Ω ε http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq171_HTML.gif, we can see that the Leray-Schauder degree
          d L S [ I A 0 ( u , v ) , Ω ε , 0 ] 0 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_Equbx_HTML.gif

          This completes the proof. □

          As applications of Theorem 3.2, we have the following.

          Corollary 3.3 Assume that f l ( t , u , u , v , v ) = μ l ( t ) | u | m l ( t ) 2 u ( t ) + γ l ( t ) | u | n l ( t ) 2 u ( t ) + μ ˜ l ( t ) | v | m ˜ l ( t ) 2 v ( t ) + γ ˜ l ( t ) | v | n ˜ l ( t ) 2 v ( t ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq206_HTML.gif, where l = 1 , 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq4_HTML.gif; m l , n l , m ˜ l , n ˜ l , μ l , γ l , μ ˜ l , γ ˜ l C ( J , R ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq207_HTML.gif satisfy max t J p l ( t ) < m l , n l , m ˜ l , n ˜ l http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq208_HTML.gif, t J http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq209_HTML.gif. If | e 1 | http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq210_HTML.gif and | e 2 | http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq211_HTML.gif are small enough, then the problem (P) possesses at least one solution.

          Proof It is easy to have
          | f l ( t , u , u , v , v ) | | μ l ( t ) | | u | m l ( t ) 1 + | γ l ( t ) | | u | n l ( t ) 1 + | μ ˜ l ( t ) | | v | m ˜ l ( t ) 1 + | γ ˜ l ( t ) | | v | n ˜ l ( t ) 1 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_Equby_HTML.gif
          From μ l , γ l , μ ˜ l , γ ˜ l C ( J , R ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq212_HTML.gif and the definition of Ω ε http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq176_HTML.gif, we have
          | f l ( t , u , u , v , v ) | C 13 ε m l ( t ) 1 + C 14 ε n l ( t ) 1 + C 15 ε m ˜ l ( t ) 1 + C 16 ε n ˜ l ( t ) 1 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_Equbz_HTML.gif
          Since max t J p l ( t ) < m l , n l , m ˜ l , n ˜ l http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq208_HTML.gif, then there exists a small enough ε such that
          | f l ( t , u , u , v , v ) | 1 σ 1 4 N ( θ 3 ) p l ( t ) 1 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_Equca_HTML.gif
          From Lemma 2.2 and the small enough | e 2 | http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq211_HTML.gif, we have
          | a 1 ( δ 1 f 1 ) | 2 N 1 σ 1 ( δ 1 f 1 L 1 + | e 2 | ) < ( θ 3 ) p 1 ( t ) 1 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_Equcb_HTML.gif

          then | a 1 ( 0 ) | < min t J ( θ 3 ) p 1 ( t ) 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq213_HTML.gif is valid.

          Similarly, we have | a 2 ( 0 ) | < min t J ( θ 2 ) p 2 ( t ) 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq175_HTML.gif.

          Obviously, it follows from | a 1 ( 0 ) | < min t J ( θ 3 ) p 1 ( t ) 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq214_HTML.gif, | a 2 ( 0 ) | < min t J ( θ 2 ) p 2 ( t ) 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq215_HTML.gif and the small enough | e 1 | http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq210_HTML.gif that ( u 0 , v 0 ) Ω ε http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq171_HTML.gif, | P 1 ( 0 ) | < θ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq172_HTML.gif, and | P 2 ( 0 ) | < θ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq173_HTML.gif.

          Thus, the conditions of (A1) are satisfied, then the problem (P) possesses at least one solution. □

          Corollary 3.4 Assume that f l ( t , u , u , v , v ) = μ l ( t ) | u | m l ( t ) 2 u ( t ) + γ l ( t ) | u | n l ( t ) 2 u ( t ) + μ ˜ l ( t ) | v | m ˜ l ( t ) 2 v ( t ) + γ ˜ l ( t ) | v | n ˜ l ( t ) 2 v ( t ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq206_HTML.gif, where l = 1 , 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq4_HTML.gif; m l , n l , m ˜ l , n ˜ l , μ l , γ l , μ ˜ l , γ ˜ l C ( J , R ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq216_HTML.gif satisfy min t J p l ( t ) m l , n l , m ˜ l , n ˜ l max t J p l ( t ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq217_HTML.gif. If | e 1 | http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq210_HTML.gif, | e 2 | http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq211_HTML.gif and δ l http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq144_HTML.gif are small enough, then the problem (P) possesses at least one solution.

          Proof From Lemma 2.2, we have
          | a 1 ( δ 1 f 1 ) | 2 N 1 σ 1 ( δ 1 f 1 L 1 + | e 2 | ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_Equcc_HTML.gif
          Since a 1 ( δ 1 f 1 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq218_HTML.gif is dependent on the small enough δ 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq219_HTML.gif and | e 2 | http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq211_HTML.gif, then it follows from the continuity of a 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq79_HTML.gif that | a 1 ( 0 ) | http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq220_HTML.gif is small enough, which implies that
          | a 1 ( 0 ) | < min t J ( θ 3 ) p 1 ( t ) 1 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_Equcd_HTML.gif

          Similarly, we have | a 2 ( 0 ) | < min t J ( θ 2 ) p 2 ( t ) 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq175_HTML.gif.

          From | a 1 ( 0 ) | < min t J ( θ 3 ) p 1 ( t ) 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq213_HTML.gif, | a 2 ( 0 ) | < min t J ( θ 2 ) p 2 ( t ) 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq221_HTML.gif and the small enough | e 1 | http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq210_HTML.gif and | e 2 | http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq211_HTML.gif, it is easy to have that ( u 0 , v 0 ) Ω ε http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq171_HTML.gif, | P 1 ( 0 ) | < θ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq172_HTML.gif, and | P 2 ( 0 ) | < θ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq173_HTML.gif.

          Thus, the conditions of (A1) are satisfied, then the problem (P) possesses at least one solution. □

          We denote
          http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_Equce_HTML.gif

          Assume the following.

          (A2) Let positive constants ε 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq222_HTML.gif and ε 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq223_HTML.gif be such that ( u 0 , v 0 ) Ω ε 1 , ε 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq224_HTML.gif, | P 1 ( 0 ) | < θ 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq225_HTML.gif, | P 2 ( 0 ) | < θ 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq226_HTML.gif and | a 1 ( 0 ) | < min t J ( θ 1 3 ) p 1 ( t ) 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq227_HTML.gif, | a 2 ( 0 ) | < min t J ( θ 2 2 ) p 2 ( t ) 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq228_HTML.gif, where ( u 0 , v 0 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq168_HTML.gif is defined in (18), a 1 ( ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq86_HTML.gif and a 2 ( ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq120_HTML.gif are defined in (5) and (13), respectively.

          It is easy to see that Ω ε 1 , ε 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq229_HTML.gif is an open bounded domain in W. We have the following.

          Corollary 3.5 Assume that
          http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_Equcf_HTML.gif

          where ϵ, ϵ ˜ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq230_HTML.gif are positive constants; m , n , m ˜ , n ˜ , ϱ , ϱ ˜ , μ , γ , μ ˜ , γ ˜ , ϰ , ϰ ˜ C ( J , R ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq231_HTML.gif satisfy 1 < m , n , m ˜ , n ˜ < min t J p 1 ( t ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq232_HTML.gif, and max t J p 2 ( t ) < ϱ , ϱ ˜ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq233_HTML.gif, t J http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq209_HTML.gif. Then the problem (P) possesses at least one solution.

          Proof Similar to the proof of Theorem 3.2, we only need to prove that (A2) is satisfied, then we can conclude that the problem (P) possesses at least one solution.

          From μ , γ , μ ˜ , γ ˜ C ( J , R ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq234_HTML.gif and the definition of Ω ε 1 , ε 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq229_HTML.gif, it is easy to have that
          | f 1 ( t , u , u , v , v ) | C 17 ε 1 m ( t ) 1 + C 18 ε 1 n ( t ) 1 + C 19 ε 2 m ˜ ( t ) 1 + C 20 ε 2 n ˜ ( t ) 1 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_Equcg_HTML.gif
          where we suppose ε 2 < 1 < ε 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq235_HTML.gif. Since 1 < m , n , m ˜ , n ˜ < min t J p 1 ( t ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq232_HTML.gif, then there exists a big enough ε 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq222_HTML.gif such that
          | f 1 ( t , u , u , v , v ) | 1 σ 1 3 N ( θ 1 3 ) p 1 ( t ) 1 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_Equch_HTML.gif
          From Lemma 2.2, we have
          | a 1 ( δ 1 f 1 ) | 2 N 1 σ 1 ( δ 1 f 1 L 1 + | e 2 | ) < ( θ 1 3 ) p 1 ( t ) 1 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_Equci_HTML.gif

          then | a 1 ( 0 ) | < min t J ( θ 1 3 ) p 1 ( t ) 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq227_HTML.gif is valid.

          From ϰ , ϰ ˜ C ( J , R ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq236_HTML.gif and the definition of Ω ε 1 , ε 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq229_HTML.gif, we have
          | f 2 ( t , u , u , v , v ) | C 21 ε 1 ϵ ε 2 ϱ ( t ) 1 + C 22 ε 1 ϵ ˜ ε 2 ϱ ˜ ( t ) 1 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_Equcj_HTML.gif
          Since max t J p 2 ( t ) < ϱ , ϱ ˜ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq237_HTML.gif, then there exists a ε 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq223_HTML.gif such that ε 2 < ( C 23 C 21 ε 1 ϵ + C 22 ε 1 ϵ ˜ ) 1 ϱ p 2 + http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq238_HTML.gif (where ϱ = min { ϱ , ϱ ˜ } http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq239_HTML.gif), which implies that
          | f 2 ( t , u , u , v , v ) | 1 4 N ( θ 2 2 ) p 2 ( t ) 1 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_Equck_HTML.gif
          From Lemma 2.3, we have
          | a 2 ( δ 2 f 2 ) | 3 N δ 2 f 2 0 < ( θ 2 2 ) p 2 ( t ) 1 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_Equcl_HTML.gif

          then | a 2 ( 0 ) | < min t J ( θ 2 2 ) p 2 ( t ) 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq228_HTML.gif is valid.

          Obviously, it follows from | a 1 ( 0 ) | < min t J ( θ 1 3 ) p 1 ( t ) 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq227_HTML.gif and | a 2 ( 0 ) | < min t J ( θ 2 2 ) p 2 ( t ) 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq240_HTML.gif that ( u 0 , v 0 ) Ω ε 1 , ε 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq241_HTML.gif, | P 1 ( 0 ) | < θ 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq225_HTML.gif, and | P 2 ( 0 ) | < θ 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq226_HTML.gif.

          Thus, the conditions of (A2) are satisfied, then the problem (P) possesses at least one solution. □

          Corollary 3.6 Assume that
          http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_Equcm_HTML.gif

          where ϵ, ϵ ˜ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq230_HTML.gif are positive constants; ϱ , ϱ ˜ , m , n , m ˜ , n ˜ , ϰ , ϰ ˜ , μ , γ , μ ˜ , γ ˜ C ( J , R ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq242_HTML.gif satisfy max t J p 1 ( t ) < ϱ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq243_HTML.gif, ϱ ˜ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq244_HTML.gif, and 1 < m , n , m ˜ , n ˜ < min t J p 2 ( t ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq245_HTML.gif, t J http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq209_HTML.gif. If | e 1 | http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq210_HTML.gif and | e 2 | http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq211_HTML.gif are small enough, then the problem (P) possesses at least one solution.

          Proof Similar to the proof of Corollary 3.5, we conclude that (A2) is satisfied. Then the problem (P) possesses at least one solution. □

          4 Existence of nonnegative solutions

          In the following, we deal with the existence of nonnegative solutions of (P). For any x = ( x 1 , , x N ) R N http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq246_HTML.gif, the notation x 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq247_HTML.gif ( x > 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq248_HTML.gif) means x j 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq249_HTML.gif ( x j > 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq250_HTML.gif) for any j = 1 , , N http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq158_HTML.gif. For any x , y R N http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq251_HTML.gif, the notation x y http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq252_HTML.gif means x y 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq253_HTML.gif, the notation x > y http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq254_HTML.gif means x y > 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq255_HTML.gif.

          Theorem 4.1 We assume that

          (10) δ 1 f 1 ( t , x , y , z , w ) 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq256_HTML.gif, ( t , x , y , z , w ) J × R N × R N × R N × R N http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq257_HTML.gif;

          (20) δ 2 f 2 ( t , x , y , z , w ) 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq258_HTML.gif, ( t , x , y , z , w ) J × R N × R N × R N × R N http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq257_HTML.gif;

          (30) e 1 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq259_HTML.gif;

          (40) e 2 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq260_HTML.gif.

          Then every solution of (P) is nonnegative.

          Proof (i) We shall show that u ( t ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq261_HTML.gif is nonnegative.

          If ( u , v ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq262_HTML.gif is a solution of (P), from Lemma 2.5, we have
          φ p 1 ( t , u ( t ) ) = a 1 ( δ 1 N f 1 ( u , v ) ) 0 t δ 1 f 1 ( s , u , u , v , v ) d s , t J , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_Equcn_HTML.gif
          which together with (5), (10) and (40) implies that
          http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_Equco_HTML.gif

          Thus u ( t ) 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq263_HTML.gif for any t J http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq29_HTML.gif. Holding u ( t ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq264_HTML.gif is decreasing, namely u ( t 1 ) u ( t 2 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq265_HTML.gif for any t 1 , t 2 J http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq131_HTML.gif with t 1 < t 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq266_HTML.gif.

          According to the boundary value condition (2) and condition (30), we have
          u ( 1 ) = i = 1 m 2 α i u ( ξ i ) + e 1 i = 1 m 2 α i u ( 1 ) + e 1 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_Equcp_HTML.gif
          then
          u ( 1 ) e 1 1 i = 1 m 2 α i 0 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_Equcq_HTML.gif
          Thus u ( t ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq261_HTML.gif is nonnegative.
          1. (ii)

            We shall show that v ( t ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq267_HTML.gif is nonnegative.

             
          If ( u , v ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq262_HTML.gif is a solution of (P), From Lemma 2.5, we have
          v ( t ) = v ( 0 ) + F { φ p 2 1 [ t , a 2 F ( δ 2 N f 2 ( u , v ) ) ] } ( t ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_Equcr_HTML.gif
          We claim that a 2 ( δ 2 N f 2 ( u , v ) ) 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq268_HTML.gif. If it is false, then there exists some j { 1 , , N } http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq269_HTML.gif such that a 2 j ( δ 2 N f 2 ( u , v ) ) < 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq270_HTML.gif, which together with condition (20) implies that
          [ a 2 ( δ 2 N f 2 ( u , v ) ) F ( δ 2 N f 2 ( u , v ) ) ( t ) ] j < 0 , t J . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_Equ22_HTML.gif
          (22)
          Similar to the proof of Lemma 2.3, the boundary value condition (8) implies
          0 = k 1 φ p 2 1 ( 0 , a 2 ) + 0 1 e ( t ) 0 t φ p 2 1 [ s , a 2 F ( δ 2 N f 2 ( u , v ) ) ( s ) ] d s d t 1 σ 2 + i = 1 m 2 β i η i 1 φ p 2 1 [ t , a 2 F ( δ 2 N f 2 ( u , v ) ) ( t ) ] d t + k 2 φ p 2 1 [ 1 , a 2 F ( δ 2 N f 2 ( u , v ) ) ( 1 ) ] 1 i = 1 m 2 β i + 0 1 φ p 2 1 [ t , a 2 F ( δ 2 N f 2 ( u , v ) ) ( t ) ] d t . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_Equ23_HTML.gif
          (23)

          From (22) and a 2 j ( δ 2 N f 2 ( u , v ) ) < 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq270_HTML.gif, we get a contradiction to (23).

          Thus a 2 ( δ 2 N f 2 ( u , v ) ) 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq268_HTML.gif.

          We claim that
          a 2 ( δ 2 N f 2 ( u , v ) ) F ( δ 2 N f 2 ( u , v ) ) ( 1 ) 0 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_Equ24_HTML.gif
          (24)
          If it is false, then there exists some j { 1 , , N } http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq269_HTML.gif such that
          [ a 2 ( δ 2 N f 2 ( u , v ) ) F ( δ 2 N f 2 ( u , v ) ) ( 1 ) ] j > 0 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_Equcs_HTML.gif
          which together with condition (20) implies
          [ a 2 ( δ 2 N f 2 ( u , v ) ) F ( δ 2 N f 2 ( u , v ) ) ( t ) ] j > 0 , t J . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_Equ25_HTML.gif
          (25)

          From (25) and a 2 ( δ 2 N f 2 ( u , v ) ) 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq268_HTML.gif, we get a contradiction to (23). Thus (24) is valid.

          Denote
          Γ ( t ) = a 2 ( δ 2 N f 2 ( u , v ) ) F ( δ 2 N f 2 ( u , v ) ) ( t ) , t J . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_Equct_HTML.gif
          Obviously, Γ ( 0 ) = a 2 ( δ 2 N f 2 ( u , v ) ) 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq271_HTML.gif, Γ ( 1 ) 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq272_HTML.gif, and Γ ( t ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq273_HTML.gif is decreasing, i.e., Γ ( t ) Γ ( t ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq274_HTML.gif for any t , t J http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq275_HTML.gif with t t http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq276_HTML.gif. For any j = 1 , , N http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq158_HTML.gif, there exist ζ j J http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq277_HTML.gif such that
          Γ j ( t ) 0 , t [ 0 , ζ j ] and Γ j ( t ) 0 , t [ ζ j , T ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_Equcu_HTML.gif
          We can conclude that v j ( t ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq278_HTML.gif is increasing on [ 0 , ζ j ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq279_HTML.gif, and v j ( t ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq278_HTML.gif is decreasing on [ ζ j , T ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq280_HTML.gif. Thus
          min { v j ( 0 ) , v j ( 1 ) } = inf t I v j ( t ) , j = 1 , , N . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_Equcv_HTML.gif
          For any fixed j { 1 , , N } http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq269_HTML.gif, if
          v j ( 0 ) = inf t I v j ( t ) , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_Equcw_HTML.gif
          which together with (8) implies that
          v j ( 0 ) = 0 1 e ( t ) v j ( t ) d t + k 1 ( v ) j ( 0 ) 0 1 e ( t ) v j ( 0 ) d t + k 1 ( v ) j ( 0 ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_Equ26_HTML.gif
          (26)
          From a 2 ( δ 2 N f 2 ( u , v ) ) 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_IEq268_HTML.gif, we have
          ( v ) j ( 0 ) = ( φ p 2 1 [ 0 , a 2 ] ) j 0 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_Equ27_HTML.gif
          (27)
          It follows from (26) and (27) that
          v j ( 0 ) k 1 ( v ) j ( 0 ) 1 σ 2 0 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_Equcx_HTML.gif
          If
          v j ( 1 ) = inf t I v j ( t ) , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_Equ28_HTML.gif
          (28)
          from (8) and (28), we have
          v j ( 1 ) = i = 1