Existence of solutions for weighted p(r)-Laplacian impulsive system mixed type boundary value problems

  • Li Yin1,

    Affiliated with

    • Yunrui Guo2,

      Affiliated with

      • Guizhen Zhi1 and

        Affiliated with

        • Qihu Zhang1Email author

          Affiliated with

          Boundary Value Problems20112011:42

          DOI: 10.1186/1687-2770-2011-42

          Received: 1 July 2011

          Accepted: 1 November 2011

          Published: 1 November 2011

          Abstract

          This paper investigates the existence of solutions for weighted p(r)-Laplacian impulsive system mixed type boundary value problems. The proof of our main result is based upon Gaines and Mawhin's coincidence degree theory. Moreover, we obtain the existence of nonnegative solutions.

          Keywords

          Weighted p(r)-Laplacian impulsive system coincidence degree

          1 Introduction

          In this paper, we mainly consider the existence of solutions for the weighted p(r)-Laplacian system
          - ( w ( r ) u p ( r ) - 2 u ( r ) ) + f ( r , u ( r ) , ( w ( r ) ) 1 p ( r ) - 1 u ( r ) ) = 0 , r ( 0 , T ) , r r i , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-42/MediaObjects/13661_2011_Article_87_Equ1_HTML.gif
          (1)
          where u: [0, T] → ℝ N , with the following impulsive boundary conditions
          lim r r i + u ( r ) - lim r r i - u ( r ) = A i ( lim r r i - u ( r ) , lim r r i - ( w ( r ) ) 1 p ( r ) - 1 u ( r ) ) , i = 1 , , k , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-42/MediaObjects/13661_2011_Article_87_Equ2_HTML.gif
          (2)
          lim r r i + w ( r ) u p ( r ) - 2 u ( r ) - lim r r i - w ( r ) u p ( r ) - 2 u ( r ) = B i ( lim r r i - u ( r ) , lim r r i - ( w ( r ) ) 1 p ( r ) - 1 u ( r ) ) , i = 1 , , k , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-42/MediaObjects/13661_2011_Article_87_Equ3_HTML.gif
          (3)
          a u ( 0 ) - b lim r 0 + ( w ( r ) ) 1 p ( r ) - 1 u ( r ) = 0 , and c u ( T ) + d lim r T - w ( r ) u p ( r ) - 2 u ( r ) = 0 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-42/MediaObjects/13661_2011_Article_87_Equ4_HTML.gif
          (4)

          where pC ([0, T], ℝ) and p(r) > 1, -Δp(r)u:= -(w(r) |u'|p(r)-2u'(r))' is called weighted p(r)-Laplacian; 0 < r1 < r2 < ⋯ < r k < T; A i , B i C(ℝ N × ℝ N , ℝ N ); a, b, c, d ∈ [0, +∞), ad + bc > 0.

          Throughout the paper, o(1) means functions which uniformly convergent to 0 (as n → +∞); for any v ∈ ℝ N , v j will denote the j-th component of v; the inner product in ℝ N will be denoted by 〈·,·〉; |·| will denote the absolute value and the Euclidean norm on ℝ N . Denote J = [0, T], J' = [0, T]\{r0, r1,..., rk+1}, J0 = [r0, r1], J i = (r i , ri+1], i = 1, ..., k, where r0 = 0, rk+1= T. Denote J i o http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-42/MediaObjects/13661_2011_Article_87_IEq1_HTML.gif the interior of J i , i = 0, 1,..., k. Let PC(J, ℝ N ) = {x: J → ℝ N | xC(J i , ℝ N ), i = 0, 1,..., k, and lim r r i + x ( r ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-42/MediaObjects/13661_2011_Article_87_IEq2_HTML.gif exists for i = 1,..., k}; wPC(J, ℝ) satisfies 0 < w(r), ∀rJ', and ( w ( ) ) - 1 p ( ) - 1 L 1 ( 0 , T ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-42/MediaObjects/13661_2011_Article_87_IEq3_HTML.gif; P C 1 ( J , N ) = { x P C ( J , N ) x C ( J i o , N ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-42/MediaObjects/13661_2011_Article_87_IEq4_HTML.gif, lim r r i + w ( r ) x p ( r ) - 2 x ( r ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-42/MediaObjects/13661_2011_Article_87_IEq5_HTML.gif and lim r r i + 1 - w ( r ) x p ( r ) - 2 x ( r ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-42/MediaObjects/13661_2011_Article_87_IEq6_HTML.gif exists for i = 0, 1,..., k}. For any x = (x1,..., x N ) ∈ PC(J, ℝ N ), denote |x i |0 = suprJ'|x i (r)|. Obviously, PC(J, ℝ N ) is a Banach space with the norm x 0 = ( i = 1 N x i 0 2 ) 1 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-42/MediaObjects/13661_2011_Article_87_IEq7_HTML.gif, PC1(J, ℝ N ) is a Banach space with the norm x 1 = x 0 + ( w ( r ) ) 1 p ( r ) - 1 x 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-42/MediaObjects/13661_2011_Article_87_IEq8_HTML.gif. In the following, PC(J, ℝ N ) and PC1(J, ℝ N ) will be simply denoted by PC and PC1, respectively. Let L1 = L1(J, ℝ N ) with the norm x L 1 = ( i = 1 N x i L 1 2 ) 1 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-42/MediaObjects/13661_2011_Article_87_IEq9_HTML.gif, ∀xL1, where x i L 1 = 0 T x i ( r ) d r http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-42/MediaObjects/13661_2011_Article_87_IEq10_HTML.gif. We will denote
          u ( r i + ) = lim r r i + u ( r ) , u ( r i - ) = lim r r i - u ( r ) , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-42/MediaObjects/13661_2011_Article_87_Equa_HTML.gif
          w ( 0 ) u p ( 0 ) - 2 u ( 0 ) = lim r 0 + w ( r ) u p ( r ) - 2 u ( r ) , w ( T ) u p ( T ) - 2 u ( T ) = lim r T - w ( r ) u p ( r ) - 2 u ( r ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-42/MediaObjects/13661_2011_Article_87_Equb_HTML.gif
          The study of differential equations and variational problems with nonstandard p(r)-growth conditions is a new and interesting topic. It arises from nonlinear elasticity theory, electro-rheological fluids, image processing, etc. (see [14]). Many results have been obtained on this problems, for example [125]. If p(r) ≡ p (a constant), (1) is the well-known p-Laplacian system. If p(r) is a general function, -Δp(r)represents a nonhomogeneity and possesses more nonlinearity, thus -Δp(r)is more complicated than -Δ p ; for example, if Ω ⊂ ℝ N is a bounded domain, the Rayleigh quotient
          λ p ( ) = inf u W 0 1 , p ( x ) ( Ω ) \ { 0 } Ω 1 p ( x ) u p ( x ) d x Ω 1 p ( x ) u p ( x ) d x http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-42/MediaObjects/13661_2011_Article_87_Equc_HTML.gif

          is zero in general, and only under some special conditions λp(·)> 0 (see [8, 1719]), but the property of λ p > 0 is very important in the study of p-Laplacian problems.

          Impulsive differential equations have been studied extensively in recent years. Such equations arise in many applications such as spacecraft control, impact mechanics, chemical engineering and inspection process in operations research (see [2628] and the references therein). It is interesting to note that p(r)-Laplacian impulsive boundary problems are about comparatively new applications like ecological competition, respiratory dynamics and vaccination strategies. On the Laplacian impulsive differential equation boundary value problems, there are many results (see [2937]). There are many methods to deal with this problem, e.g., subsupersolution method, fixed point theorem, monotone iterative method and coincidence degree. Because of the nonlinearity of -Δ p , results on the existence of solutions for p-Laplacian impulsive differential equation boundary value problems are rare (see [38, 39]). On the Laplacian (p(x) ≡ 2) impulsive differential equations mixed type boundary value problems, we refer to [30, 32, 34].

          In [39], Tian and Ge have studied nonlinear IBVP
          - ( ρ ( t ) Φ p ( x ( t ) ) ) + s ( t ) Φ p ( x ( t ) ) = f ( t , x ( t ) ) , t t i , a . e . t [ a , b ] , lim t t i + ρ ( t ) Φ p ( x ( t ) ) - lim t t i - ρ ( t ) Φ p ( x ( t ) ) = I i ( x ( t i ) ) , i = 1 , , l , α x ( a ) - β x ( a ) = σ 1 , γ x ( b ) + σ x ( b ) = σ 2 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-42/MediaObjects/13661_2011_Article_87_Equ5_HTML.gif
          (5)

          where Φ p (x) = |x|p-2x, p > 1, ρ, sL [a, b] with essin f[a, b]ρ > 0, and essin f[a,b]s > 0, 0 < ρ(a), p(b) <∞, σ1 ≤ 0, σ2 ≥ 0, α, β, γ, σ > 0, a = t0 < t1 < ⋯ < tl < tl+1 = b, I i C([0, +∞), [0, ∞)), i = 1,..., l, fC ([a, b] × [0, +∞), [0, ∞)), f(·, 0) is nontrivial. By using variational methods, the existence of at least two positive solutions was obtained.

          In [24, 25], the present author investigates the existence of solutions of p(r)-Laplacian impulsive differential equation (1-3) with periodic-like or multi-point boundary value conditions.

          In this paper, we consider the existence of solutions for the weighted p(r)-Laplacian impulsive differential system mixed type boundary value condition problems, when p(r) is a general function. The proof of our main result is based upon Gaines and Mawhin's coincidence degree theory. Since the nonlinear term f in (5) is independent on the first-order derivative, and the impulsive conditions are simpler than (2), our main results partly generalized the results of [30, 32, 34, 39]. Since the mixed type boundary value problems are different from periodic-like or multi-point boundary value conditions, and this paper gives two kinds of mixed type boundary value conditions (linear and nonlinear), our discussions are different from [24, 25] and have more difficulties. Moreover, we obtain the existence of nonnegative solutions. This paper was motivated by [2426, 38, 40].

          Let N ≥ 1, the function f: J × ℝ N × ℝ N → ℝ N is assumed to be Caratheodory; by this, we mean:
          1. (i)

            for almost every tJ, the function f(t, ·, ·) is continuous;

             
          2. (ii)

            for each (x, y) ∈ ℝ N × ℝ N , the function f(·, x, y) is measurable on J;

             
          3. (iii)
            for each R > 0, there is a α R L 1 (J, ℝ), such that, for almost every tJ and every (x, y) ∈ ℝ N × ℝ N with |x| ≤ R, |y| ≤ R, one has
            f ( t , x , y ) α R ( t ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-42/MediaObjects/13661_2011_Article_87_Equd_HTML.gif
             

          We say a function u: J → ℝ N is a solution of (1) if uPC1 with w(·) |u'|p(·)-2u'(·) absolutely continuous on every J i o http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-42/MediaObjects/13661_2011_Article_87_IEq1_HTML.gif, i = 0, 1,..., k, which satisfies (1) a.e. on J.

          This paper is divided into three sections; in the second section, we present some preliminary. Finally, in the third section, we give the existence of solutions and nonnegative solutions of system (1)-(4).

          2 Preliminary

          Let X and Y be two Banach spaces and L: D(L) ⊂ XY be a linear operator, where D(L) denotes the domain of L. L will be a Fredholm operator of index 0, i.e., ImL is closed in Y and the linear spaces KerL and coImL have the same dimension which is finite. We define X1 = KerL and Y1 = coImL, so we have the decompositions X = X1coKerL and Y = Y1ImL. Now, we have the linear isomorphism Λ: X1Y1 and the continuous linear projectors P: XX1 and Q: YY1 with KerQ = ImL and ImP = X1.

          Let Ω be an open bounded subset of X with Ω ∩ D(L) ≠ ∅. Operator S : Ω ¯ Y http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-42/MediaObjects/13661_2011_Article_87_IEq11_HTML.gif be a continuous operator. In order to define the coincidence degree of (L, S) in Ω, as in [40, 41], denoted by d(L - S, Ω), we assume that
          L x S x for all  x Ω . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-42/MediaObjects/13661_2011_Article_87_Eque_HTML.gif
          It is easy to see that the operator M : Ω ¯ X http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-42/MediaObjects/13661_2011_Article_87_IEq12_HTML.gif, M = (L + ΛP)-1 (S + ΛP) is well defined, and
          L x * = S x *  if and only if  x * = M x * . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-42/MediaObjects/13661_2011_Article_87_Equf_HTML.gif
          If M is continuous and compact, then S is called L-compact, and the Leray-Schauder degree of I X - M (where I X is the identity mapping of X) is well defined in Ω, and we will denote it by d LS (I X - M, Ω, 0). This number is independent of the choice of P, Q and Λ (up to a sign) and we can define
          d ( L - S , Ω ) : = d L S ( I X - M , Ω , 0 ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-42/MediaObjects/13661_2011_Article_87_Equg_HTML.gif

          Definition 2.1. (see [40, 41]) The coincidence degree of (L, S) in Ω, denoted by d(L - S, Ω), is defined as d(L - S, Ω) = d LS (I X - M, Ω, 0).

          There are many papers about coincidence degree and its applications (see [4043]).

          Proposition 2.2. (see [40]) (i) (Existence property). If d(L - S, Ω) ≠ 0, then there exists x ∈ Ω such that Lx = Sx.
          1. (ii)

            (Homotopy invariant property). If H : Ω ¯ × [ 0 , 1 ] Y http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-42/MediaObjects/13661_2011_Article_87_IEq13_HTML.gif is continuous, L-compact and LxH(x, λ) for all x ∈ ∂Ω and λ ∈ [0, 1], then d(L - H (·, λ), Ω) is independent of λ.

             

          The effect of small perturbations is negligible, as is proved in the next Proposition (see [41] Theorem III.3, page 24).

          Proposition 2.3. Assume that LxSx for each x ∈ ∂Ω. If S ε is such that supx∈∂Ω||S ε x|| Y is sufficiently small, then LxSx + S ε x for all x ∈ ∂Ω and d(L - S - S ε , Ω) = d(L - S, Ω).

          For any (r, x) ∈ (J × ℝ N ), denote φp(r)(x) = |x|p(r)-2x. Obviously, φ has the following properties

          Proposition 2.4 (see [41]) φ is a continuous function and satisfies
          1. (i)
            For any r ∈ [0, T], φ p(r)(·) is strictly monotone, i.e.,
            φ p ( r ) ( x 1 ) - φ p ( r ) ( x 2 ) , x 1 - x 2 > 0 , x 1 , x 2 N , x 1 x 2 ; http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-42/MediaObjects/13661_2011_Article_87_Equh_HTML.gif
             
          2. (ii)
            There exists a function η: [0, +∞) → [0, +∞), η(s) → +∞ as s → +∞, such that
            φ p ( r ) ( x ) , x η ( x ) x , for all  x N . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-42/MediaObjects/13661_2011_Article_87_Equi_HTML.gif
             
          It is well known that φp(r)(·) is a homeomorphism from ℝ N to ℝ N for any fixed rJ. Denote
          φ p ( r ) - 1 ( x ) = x 2 - p ( r ) p ( r ) - 1 x , for x N \ { 0 } , φ p ( r ) - 1 ( 0 ) = 0 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-42/MediaObjects/13661_2011_Article_87_Equj_HTML.gif
          It is clear that φ p ( r ) - 1 ( ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-42/MediaObjects/13661_2011_Article_87_IEq14_HTML.gif is continuous and sends bounded sets to bounded sets, and φ p ( r ) - 1 ( ) = φ q ( r ) ( ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-42/MediaObjects/13661_2011_Article_87_IEq15_HTML.gif where 1 p ( r ) + 1 q ( r ) 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-42/MediaObjects/13661_2011_Article_87_IEq16_HTML.gif. Let X = {(x1, x2) | x1PC, x2PC} with the norm ||(x1, x2)|| X = || x1||0 + ||x2||0, Y = L1 × L1 × ℝ2(k + 1)N, and we define the norm on Y as
          ( y 1 , y 2 , z 1 , , z 2 ( k + 1 ) ) Y = y 1 L 1 + y 2 L 1 + m = 1 2 ( k + 1 ) z m , ( y 1 , y 2 , z 1 , , z 2 ( k + 1 ) ) Y , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-42/MediaObjects/13661_2011_Article_87_Equk_HTML.gif

          where y1, y2L1, z m ∈ ℝ N , m = 1,..., 2(k + 1), then X and Y are Banach spaces.

          Define L: D(L) ⊂ XY and S: XY as the following
          L x = ( x 1 , x 2 , Δ x 1 ( r i ) , Δ x 2 ( r i ) , 0 , 0 ) , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-42/MediaObjects/13661_2011_Article_87_Equl_HTML.gif
          S x = φ q ( r ) x 2 w ( r ) , f ( r , x 1 , φ q ( r ) ( x 2 ) ) , A i , B i , a x 1 ( 0 ) - b φ q ( 0 ) ( x 2 ( 0 ) ) , c x 1 ( T ) + d x 2 ( T ) , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-42/MediaObjects/13661_2011_Article_87_Equm_HTML.gif
          where
          Δ x j ( r i ) = x j ( r i + ) - x j ( r i - ) , j = 1 , 2 , i = 1 , , k ; A i = A i ( x 1 ( r i - ) , φ q ( r i ) ( x 2 ( r i - ) ) ) , B i = B i ( x 1 ( r i - ) , φ q ( r i ) ( x 2 ( r i - ) ) ) , i = 1 , , k . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-42/MediaObjects/13661_2011_Article_87_Equ6_HTML.gif
          (6)

          Obviously, the problem (1)-(4) can be written as Lx = Sx, where L: XY is a linear operator, S: XY is a nonlinear operator, and X and Y are Banach spaces.

          Since
          I m L = { ( y 1 , y 2 , a i , b i , 0 , 0 ) y 1 , y 2 L 1 , a i , b i N , i = 1 , , k } , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-42/MediaObjects/13661_2011_Article_87_Equn_HTML.gif
          we have dimKerL = dim(Y/ImL) = 2N < +∞ is even and ImL is closed in Y, then L is a Fredholm operator of index zero. Define
          P : X X , ( x 1 , x 2 ) ( x 1 ( 0 ) , x 2 ( 0 ) ) , Q : Y Y , ( y 1 , y 2 , a i , b i , h 1 , h 2 ) ( 0 , 0 , 0 , 0 , h 1 , h 2 ) , y 1 , y 2 L 1 , a i , b i , h 1 , h 2 N , i = 1 , , k , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-42/MediaObjects/13661_2011_Article_87_Equo_HTML.gif
          at the same time the projectors P: XX and Q: YY satisfy
          d i m ( I m P ) = d i m ( K e r L ) = d i m ( Y I m L ) = d i m ( I m Q ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-42/MediaObjects/13661_2011_Article_87_Equp_HTML.gif
          Since ImQ is isomorphic to KerL, there exists an isomorphism Λ: KerLImQ. It is easy to see that L |D(L)∩KerP: D(L) ∩ KerPImL is invertible. We denote the inverse of that mapping by K p , then K p : ImLD(L) ∩ KerP as
          K p z = 0 t y 1 ( r ) d r + r i < t a i , 0 t y 2 ( r ) d r + r i < t b i , z = ( y 1 , y 2 , a i , b i , 0 , 0 ) I m L , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-42/MediaObjects/13661_2011_Article_87_Equq_HTML.gif
          then
          K p ( I Q ) S x = ( 0 t φ q ( r ) ( ( w ( r ) ) 1 x 2 ) d r + r i < t A i ( x 1 ( r i ) , φ q ( r i ) ( x 2 ( r i ) ) ) , 0 t f ( r , x 1 , φ q ( r ) ( x 2 ) ) d r + r i < t B i ( x 1 ( r i ) , φ q ( r i ) ( x 2 ( r i ) ) ) ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-42/MediaObjects/13661_2011_Article_87_Equr_HTML.gif
          Proposition 2.5 (i) K p (·) is continuous;
          1. (ii)

            K p (I - Q)S is continuous and compact.

             
          Proof. (i) It is easy to see that K p (·) is continuous. Moreover, the operator Ψ ( y ) = 0 t y ( r ) d r http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-42/MediaObjects/13661_2011_Article_87_IEq17_HTML.gif sends equi-integrable set of L1 to relatively compact set of PC.
          1. (ii)

            It is easy to see that K p (I - Q)SxX, ∀xX. Since ( w ( r ) ) - 1 p ( r ) - 1 L 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-42/MediaObjects/13661_2011_Article_87_IEq18_HTML.gif and f is Caratheodory, it is easy to check that S is a continuous operator from X to Y, and the operators (x 1, x 2) → φ q(r)((w(r))-1 x 2) and (x 1, x 2) → f (r, x 1, φ q(r)((w(r))-1 x 2)) both send bounded sets of X to equi-integrable set of L 1. Obviously, A i , B i and QS are compact continuous. Since f is Caratheodory, by using the Ascoli-Arzela theorem, we can show that the operator K p ( I - Q ) S : Ω ¯ X http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-42/MediaObjects/13661_2011_Article_87_IEq19_HTML.gif is continuous and compact. This completes the proof.

             
          Denote
          S ( x , λ ) = λ φ q ( r ) x 2 w ( r ) , λ p ( r ) f ( r , x 1 , φ q ( r ) ( x 2 ) ) , λ 2 A i , λ p ( r i ) B i , a x 1 ( 0 ) - b φ q ( 0 ) ( x 2 ( 0 ) ) , c x 1 ( T ) + d x 2 ( T ) , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-42/MediaObjects/13661_2011_Article_87_Equs_HTML.gif

          where A i , B i are defined in (6), i = 1,..., k.

          Consider
          L x = S ( x , λ ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-42/MediaObjects/13661_2011_Article_87_Equt_HTML.gif
          Define M ( , ) : Ω ¯ × [ 0 , 1 ] X http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-42/MediaObjects/13661_2011_Article_87_IEq20_HTML.gif as M(·, ·) = (L + ΛP)-1 (S(·, ·) + ΛP), then
          M ( , λ ) = ( L + Λ P ) - 1 ( S ( , λ ) + Λ P ) = ( K p + Λ - 1 ) ( ( I - Q ) S ( , λ ) + Q S ( , λ ) + Λ P ) = K p ( I - Q ) S ( , λ ) + Λ - 1 ( Q S ( , λ ) + Λ P ) = K p ( I - Q ) S ( , λ ) + Λ - 1 Q S ( , λ ) + P . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-42/MediaObjects/13661_2011_Article_87_Equu_HTML.gif
          Since (I - Q)S(·, 0) = 0 and K p (0) = 0, we have
          d ( L - S ( , 0 ) , Ω ) = d L S ( I X - M ( , 0 ) , Ω , 0 ) = d L S ( I X - Λ - 1 Q S ( , 0 ) - P , Ω , 0 ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-42/MediaObjects/13661_2011_Article_87_Equv_HTML.gif
          It is easy to see that all the solutions of Lx = S(x, 0) belong to KerL, then
          d L S ( I X - Λ - 1 Q S ( , 0 ) - P , Ω , 0 ) = d B ( I K e r L - Λ - 1 Q S ( , 0 ) - P K e r L , Ω K e r L , 0 ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-42/MediaObjects/13661_2011_Article_87_Equw_HTML.gif
          Notice that P | KerL = I KerL , then
          d ( L - S ( , 0 ) , Ω ) = d L S ( I X - M ( , 0 ) , Ω , 0 ) = d B ( Λ - 1 Q S ( , 0 ) , Ω K e r L , 0 ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-42/MediaObjects/13661_2011_Article_87_Equx_HTML.gif
          Proposition 2.6 (continuation theorem) (see [40]). Suppose that L is a Fredholm operator of index zero and S is L-compact on Ω ¯ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-42/MediaObjects/13661_2011_Article_87_IEq21_HTML.gif, where Ω is an open bounded subset of X. If the following conditions are satisfied,
          1. (i)
            for each λ ∈ (0, 1), every solution x of
            L x = S ( x , λ ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-42/MediaObjects/13661_2011_Article_87_Equy_HTML.gif
             
          is such that x ∉ ∂Ω;
          1. (ii)

            QS(x, 0) ≠ 0 for x ∈ ∂Ω ∩ KerL and d B -1 QS(·,0), Ω ∩ KerL, 0) ≠ 0, then the operator equation Lx = S(x, 1) has one solution lying in Ω ¯ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-42/MediaObjects/13661_2011_Article_87_IEq21_HTML.gif.

             

          The importance of the above result is that it gives sufficient conditions for being able to calculate the coincidence degree as the Brouwer degree (denoted with d B ) of a related finite dimensional mapping. It is known that the degree of finite dimensional mappings is easier to calculate. The idea of the proof is the use of the homotopy of the problem Lx = S(x, 1) with the finite dimensional one Lx = S(x, 0).

          Let us now consider the following simple impulsive problem
          ( w ( r ) φ p ( r ) ( u ( r ) ) ) = g ( r ) , r J , lim r r i + u ( r ) - lim r r i - u ( r ) = a i , i = 1 , , k , lim r r i + w ( r ) φ p ( r ) ( u ( r ) ) - lim r r i - w ( r ) φ p ( r ) ( u ( r ) ) = b i , i = 1 , , k , a u ( 0 ) - b lim r 0 + ( w ( r ) ) 1 p ( r ) - 1 u ( r ) = 0 , and c u ( T ) + d lim r T - w ( r ) u p ( r ) - 2 u ( r ) = 0 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-42/MediaObjects/13661_2011_Article_87_Equ7_HTML.gif
          (7)

          where J' = [0, T]\{r0, r1, ..., rk+1}, a i , b i ∈ ℝ N ; gL1.

          If u is a solution of (7), then we have
          w ( r ) φ p ( r ) ( u ( r ) ) = w ( 0 ) φ p ( 0 ) ( u ( 0 ) ) + r i < r b i + 0 r g ( t ) d t , r J . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-42/MediaObjects/13661_2011_Article_87_Equ8_HTML.gif
          (8)
          Denote ρ0 = w(0)φp(0)(u'(0)). Obviously, ρ0 is dependent on g, a i , b i . Define F: L1PC as
          F ( g ) ( r ) = 0 r g ( t ) d t , r J , g L 1 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-42/MediaObjects/13661_2011_Article_87_Equz_HTML.gif
          By (8), we have
          u ( r ) = u ( 0 ) + r i < r a i + F φ q ( r ) ( w ( r ) ) - 1 ρ 0 + r i < r b i + F ( g ) ( r ) ( r ) , r J . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-42/MediaObjects/13661_2011_Article_87_Equ9_HTML.gif
          (9)
          If a ≠ 0, then the boundary condition a u ( 0 ) - b lim r 0 + ( w ( r ) ) 1 p ( r ) - 1 u ( r ) = 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-42/MediaObjects/13661_2011_Article_87_IEq22_HTML.gif implies that
          u ( r ) = b a φ q ( 0 ) ( ρ 0 ) + r i < r a i + F φ q ( r ) ( w ( r ) ) - 1 ρ 0 + r i < r b i + F ( g ) ( r ) ( r ) , r J . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-42/MediaObjects/13661_2011_Article_87_Equaa_HTML.gif
          The boundary condition c u ( T ) + d lim r T - w ( r ) u p ( r ) - 2 u ( r ) = 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-42/MediaObjects/13661_2011_Article_87_IEq23_HTML.gif implies that
          c b a φ q ( 0 ) ( ρ 0 ) + c i = 1 k a i + c F φ q ( r ) ( w ( r ) ) - 1 ρ 0 + r i < r b i + F ( g ) ( r ) ( T ) + d ρ 0 + i = 1 k b i + F ( g ) ( T ) = 0 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-42/MediaObjects/13661_2011_Article_87_Equab_HTML.gif
          Denote H = L1 × ℝ2kNwith the norm
          h H = g L 1 + i = 1 k a i + i = 1 k b i , h = ( g , a i , b i ) H , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-42/MediaObjects/13661_2011_Article_87_Equac_HTML.gif
          then H is a Banach space. For fixed hH, we denote
          Θ h ( ρ ) = c b a φ q ( 0 ) ( ρ ) + c i = 1 k a i + c F φ q ( r ) ( w ( r ) ) - 1 ρ + r i < r b i + F ( g ) ( r ) ( T ) + d ρ + i = 1 k b i + F ( g ) ( T ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-42/MediaObjects/13661_2011_Article_87_Equad_HTML.gif
          Lemma 2.7 The mapping Θ h (·) has the following properties
          1. (i)
            For any fixed hH, the equation
            Θ h ( ρ ) = 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-42/MediaObjects/13661_2011_Article_87_Equ10_HTML.gif
            (10)
             
          has a unique solution ρ(h) ∈ ℝ N .
          1. (ii)

            The mapping ρ: H → ℝ N , defined in (i), is continuous and sends bounded sets to bounded sets. Moreover, ρ ( h ) 3 N [ ( 2 N E + 1 E i = 1 k a i ) p # - 1 + i = 1 k b i + g L 1 ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-42/MediaObjects/13661_2011_Article_87_IEq24_HTML.gif, where h = (g, a i , b i ) ∈ H, E = 0 T ( w ( r ) ) - 1 p ( r ) - 1 d r http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-42/MediaObjects/13661_2011_Article_87_IEq25_HTML.gif, the notation p # means C p # - 1 = C p + - 1 , C > 1 C p - - 1 , C 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-42/MediaObjects/13661_2011_Article_87_IEq26_HTML.gif.

             
          Proof. (i) From Proposition 2.4, it is immediate that
          Θ h ( ρ 1 ) - Θ h ( ρ 2 ) , ρ 1 - ρ 2 > 0 , f o r ρ 1 ρ 2 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-42/MediaObjects/13661_2011_Article_87_Equae_HTML.gif

          and hence, if (10) has a solution, then it is unique.

          Let R 0 = 3 N [ ( 2 N E + 1 E i = 1 k a i ) p # - 1 + i = 1 k b i + g L 1 ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-42/MediaObjects/13661_2011_Article_87_IEq27_HTML.gif. Since ( w ( r ) ) - 1 p ( r ) - 1 L 1 ( 0 , T ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-42/MediaObjects/13661_2011_Article_87_IEq28_HTML.gif and F(g) ∈ PC, if |ρ| > R0, it is easy to see that there exists a j0 such that, the j0-th component ρ j 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-42/MediaObjects/13661_2011_Article_87_IEq29_HTML.gif of ρ satisfies
          ρ j 0 1 N ρ . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-42/MediaObjects/13661_2011_Article_87_Equ11_HTML.gif
          (11)
          Obviously,
          | r i < r b i + F ( g ) ( r ) | r i < r b i + | F ( g ) ( r ) | i = 1 k | b i | + g L 1 , r 0, T ] , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-42/MediaObjects/13661_2011_Article_87_Equaf_HTML.gif
          then
          r i < r b i + F ( g ) ( r ) i = 1 k b i + g L 1 R 0 3 N < ρ 3 N , r [ 0 , T ] , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-42/MediaObjects/13661_2011_Article_87_Equ12_HTML.gif
          (12)
          and
          ρ + r i < r b i + F ( g ) ( r ) ρ + r i < r b i + F ( g ) ( r ) < 4 ρ 3 , r [ 0 , T ] . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-42/MediaObjects/13661_2011_Article_87_Equ13_HTML.gif
          (13)
          By (11) and (12), the j0-th component of ρ + r i < r b i + F ( g ) ( r ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-42/MediaObjects/13661_2011_Article_87_IEq30_HTML.gif keeps the same sign of ρ j 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-42/MediaObjects/13661_2011_Article_87_IEq31_HTML.gif on J and
          ρ j 0 + r i < r b i j 0 + F ( g ) j 0 ( r ) ρ j 0 - r i < r b i j 0 + F ( g ) j 0 ( r ) > 2 ρ 3 N , r J . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-42/MediaObjects/13661_2011_Article_87_Equ14_HTML.gif
          (14)
          Combining (13) and (14), the j0-th component φ q ( r ) j 0 ( w ( r ) ) - 1 ρ + r i < r b i + F ( g ) ( r ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-42/MediaObjects/13661_2011_Article_87_IEq32_HTML.gif of φ q ( r ) ( w ( r ) ) - 1 ρ + r i < r b i + F ( g ) ( r ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-42/MediaObjects/13661_2011_Article_87_IEq33_HTML.gif satisfies
          φ q ( r ) j 0 ( ρ + r i < r b i + F ( g ) ( r ) ) = ρ + r i < r b i + F ( g ) ( r ) q ( r ) - 2 ρ j 0 + r i < r b i j 0 + F ( g ) j 0 ( r ) > 2 3 N ρ + r i < r b i + F ( g ) ( r ) q ( r ) - 2 ρ > 1 2 N ρ + r i < r b i + F ( g ) ( r ) q ( r ) - 1 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-42/MediaObjects/13661_2011_Article_87_Equag_HTML.gif
          From the definition φ q ( r ) ( ) = φ p ( r ) - 1 ( ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-42/MediaObjects/13661_2011_Article_87_IEq34_HTML.gif, we have 1 p ( r ) + 1 q ( r ) 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-42/MediaObjects/13661_2011_Article_87_IEq35_HTML.gif, then q ( r ) - 1 = 1 p ( r ) - 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-42/MediaObjects/13661_2011_Article_87_IEq36_HTML.gif, and
          φ q ( r ) j 0 ρ + r i < r b i + F ( g ) ( r ) > 1 2 N ρ + r i < r b i + F ( g ) ( r ) 1 p ( r ) - 1 1 2 N ρ - r i < r b i + F ( g ) ( r ) 1 p ( r ) - 1 1 2 N 2 N E + 1 E i = 1 k a i p # - 1 1 p ( r ) - 1 1 2 N 2 N E + 1 E i = 1 k a i = E + 1 E i = 1 k a i . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-42/MediaObjects/13661_2011_Article_87_Equah_HTML.gif
          Without loss of generality, we may assume that ρ j 0 > 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-42/MediaObjects/13661_2011_Article_87_IEq37_HTML.gif, then we have
          F j 0 φ q ( r ) ( w ( r ) ) - 1 ρ + r i < r b i + F ( g ) ( r ) ( T ) > ( E + 1 ) i = 1 k a i i = 1 k a i j 0 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-42/MediaObjects/13661_2011_Article_87_Equai_HTML.gif
          Therefore, the j0-th component of i = 1 k a i + F { φ q ( r ) [ ( w ( r ) ) - 1 ( ρ + r i < r b i + F ( g ) ( r ) ) ] } ( T ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-42/MediaObjects/13661_2011_Article_87_IEq38_HTML.gif keeps the same sign of ρ j 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-42/MediaObjects/13661_2011_Article_87_IEq31_HTML.gif. Since the j0-th component of ρ + i = 1 k b i + F ( g ) ( T ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-42/MediaObjects/13661_2011_Article_87_IEq39_HTML.gif keeps the same sign of ρ j 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-42/MediaObjects/13661_2011_Article_87_IEq31_HTML.gif, a, b, c, d ∈ [0, +∞) and ad + bc > 0, we can easily see that the j0-th component of Θ h (ρ) keeps the same sign of ρ j 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-42/MediaObjects/13661_2011_Article_87_IEq31_HTML.gif, and thus
          Θ h ( ρ ) 0 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-42/MediaObjects/13661_2011_Article_87_Equaj_HTML.gif
          Let us consider the equation
          λ Θ h ( ρ ) + ( 1 - λ ) ρ = 0 , λ [ 0 , 1 ] . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-42/MediaObjects/13661_2011_Article_87_Equ15_HTML.gif
          (15)
          According to the above discussion, all the solutions of (15) belong to b(R0 + 1) = {x ∈ ℝ N | |x| < R0 + 1}. So, we have
          d B [ Θ h ( ρ ) , b ( R 0 + 1 ) , 0 ] = d B [ I , b ( R 0 + 1 ) , 0 ] 0 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-42/MediaObjects/13661_2011_Article_87_Equak_HTML.gif

          It means the existence of solutions of Θ h (ρ) = 0.

          In this way, we define a mapping ρ(h): H → ℝ N , which satisfies
          Θ h ( ρ ( h ) ) = 0 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-42/MediaObjects/13661_2011_Article_87_Equal_HTML.gif
          1. (ii)
            By the proof of (i), we also obtain ρ sends bounded set to bounded set, and
            ρ ( h ) 3 N 2 N E + 1 E i = 1 k a i p # - 1 + i = 1 k b i + g L 1 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-42/MediaObjects/13661_2011_Article_87_Equam_HTML.gif
             
          It only remains to prove the continuity of ρ. Let {u n } is a convergent sequence in H and u n u, as n → +∞. Since {ρ(u n )} is a bounded sequence, it contains a convergent subsequence { ρ ( u n j ) } http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-42/MediaObjects/13661_2011_Article_87_IEq40_HTML.gif satisfies ρ ( u n j ) ρ * http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-42/MediaObjects/13661_2011_Article_87_IEq41_HTML.gif as j → +∞. Since Θ h (ρ) consists of continuous functions, and
          Θ u n j ( ρ ( u n j ) ) = 0 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-42/MediaObjects/13661_2011_Article_87_Equan_HTML.gif
          Letting j → +∞, we have
          Θ u ( ρ * ) = 0 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-42/MediaObjects/13661_2011_Article_87_Equao_HTML.gif

          from (i) we get ρ* = ρ(u), it means that ρ is continuous.

          This completes the proof.

          If a = 0, the boundary condition a u ( 0 ) - b lim r 0 + ( w ( r ) ) 1 p ( r ) - 1 u ( r ) = 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-42/MediaObjects/13661_2011_Article_87_IEq42_HTML.gif implies that
          lim r 0 + ( w ( r ) ) 1 p ( r ) - 1 u ( r ) = 0 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-42/MediaObjects/13661_2011_Article_87_Equap_HTML.gif
          Since ad + bc > 0, we have c > 0. Thus,
          u ( r ) = u ( 0 ) + r i < r a i + F φ q ( r ) ( w ( r ) ) - 1 r i < r b i + F ( g ) ( r ) ( r ) , r J , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-42/MediaObjects/13661_2011_Article_87_Equaq_HTML.gif
          the boundary condition c u ( T ) + d lim r T - w ( r ) u p ( r ) - 2 u ( r ) = 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-42/MediaObjects/13661_2011_Article_87_IEq43_HTML.gif implies that
          u ( 0 ) + i = 1 k a i + F φ q ( r ) ( w ( r ) ) - 1 r i < r b i + F ( g ) ( r ) ( T ) + d c i = 1 k b i + F ( g ) ( T ) = 0 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-42/MediaObjects/13661_2011_Article_87_Equar_HTML.gif
          Denote G: H → ℝ N as
          G ( h ) = - i = 1 k a i - F φ q ( r ) ( w ( r ) ) - 1 r i < r b i + F ( g ) ( r ) ( T ) - d c i = 1 k b i + F ( g ) ( T ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-42/MediaObjects/13661_2011_Article_87_Equas_HTML.gif

          It is easy to see that

          Lemma 2.8 The function G(·) is continuous and sends bounded sets to bounded sets. Moreover, G ( h ) 3 N ( c + d ) c [ i = 1 k a i + E ( i = 1 k b i + g L 1 ) 1 p * - 1 + i = 1 k b i + g L 1 ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-42/MediaObjects/13661_2011_Article_87_IEq44_HTML.gif, where E = 0 T ( w ( r ) ) - 1 p ( r ) - 1 d r http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-42/MediaObjects/13661_2011_Article_87_IEq45_HTML.gif, the notation p* means C 1 p * - 1 = C 1 p + - 1 , C 1 C 1 p - - 1 , C > 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-42/MediaObjects/13661_2011_Article_87_IEq46_HTML.gif.

          3 Main results and proofs

          In this section, we will apply coincidence degree to deal with the existence of solutions for (1)-(4). In the following, we always use C and C i to denote positive constants, if it cannot lead to confusion.

          Theorem 3.1 Assume that Ω is an open bounded set in X such that the following conditions hold.

          (10) For each λ ∈ (0, 1) the problem
          L x = S ( x , λ ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-42/MediaObjects/13661_2011_Article_87_Equ16_HTML.gif
          (16)

          has no solution on ∂Ω.

          (20) (0, 0) ∈ Ω.

          Then, problem (1)-(4) has a solution u satisfies ( u , v ) Ω ¯ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-42/MediaObjects/13661_2011_Article_87_IEq47_HTML.gif, where v = w(r)φp(r)(u'(r)), ∀rJ'.

          Proof. Let us consider the following operator equation
          L x = S ( x , λ ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-42/MediaObjects/13661_2011_Article_87_Equ17_HTML.gif
          (17)

          It is easy to see that x = (x1, x2) is a solution of Lx = S(x, 1) if and only if x1(r) is a solution of (1)-(4) and x 2 ( r ) = w ( r ) φ p ( r ) ( x 1 ( r ) ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-42/MediaObjects/13661_2011_Article_87_IEq48_HTML.gif, ∀rJ'.

          According to Proposition 2.5, we can conclude that S(·, ·) is L-compact from X × [0, 1] to Y. We assume that for λ = 1, (16) does not have a solution on ∂Ω, otherwise we complete the proof. Now from hypothesis (10), it follows that (16) has no solutions for (x, λ) ∈ ∂Ω × (0, 1]. For λ = 0, (17) is equivalent to Lx = S(x, 0), namely the following usual problem
          x 1 = 0 , r ( 0 , T ) , x 2 = 0 , r ( 0 , T ) , a x 1 ( 0 ) - b φ q ( 0 ) ( x 2 ( 0 ) ) = 0 , c x 1 ( T ) + d x 2 ( T ) = 0 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-42/MediaObjects/13661_2011_Article_87_Equat_HTML.gif
          The problem (??) is a usual differential equation. Hence,
          x 1 c 1 , x 2 c 2 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-42/MediaObjects/13661_2011_Article_87_Equau_HTML.gif
          where c1, c2 ∈ ℝ N are constants. The boundary value condition of (??) holds,
          a c 1 - b φ q ( 0 ) ( c 2 ) = 0 , c c 1 + d c 2 = 0 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-42/MediaObjects/13661_2011_Article_87_Equav_HTML.gif
          Since (ad + bc) > 0, we have
          c 1 = 0 , c 2 = 0 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-42/MediaObjects/13661_2011_Article_87_Equaw_HTML.gif
          which together with hypothesis (20), implies that (0, 0)∈ Ω. Thus, we have proved that (16) has no solution on ∂Ω × [0, 1]. It means that the coincidence degree d[L - S(·, λ), Ω] is well defined for each λ ∈ [0, 1]. From the homotopy invariant property of that degree, we have
          d [ L - S ( , 1 ) , Ω ] = d [ L - S ( , 0 ) , Ω ] . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-42/MediaObjects/13661_2011_Article_87_Equ18_HTML.gif
          (18)
          Now, it is clear that the following problem
          L x = S ( x , 1 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-42/MediaObjects/13661_2011_Article_87_Equ19_HTML.gif
          (19)
          is equivalent to problem (1)-(4), and (18) tells us that problem (19) will have a solution if we can show that
          d [ L - S ( , 0 ) , Ω ] 0 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-42/MediaObjects/13661_2011_Article_87_Equax_HTML.gif
          Since by hypothesis (20), this last degree
          d [ L - S ( , 0 ) , Ω ] = d B [ ω * , Ω 2 N , 0 ] 0 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-42/MediaObjects/13661_2011_Article_87_Equay_HTML.gif

          where ω*(c1, c2) = (ac1 - q(0)(c2), cc1 + dc2). This completes the proof.

          Our next theorem is a consequence of Theorem 3.1. Denote
          z - = min r J z ( r ) , z + = max r J z ( r ) , for z C ( J , ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-42/MediaObjects/13661_2011_Article_87_Equaz_HTML.gif

          Theorem 3.2 Assume that the following conditions hold

          (10) a > 0;

          (20) lim|u| + |v| → +∞(f(r, u, v)/(|u| + |v|)β(r) -1) = 0, for rJ uniformly, where β(r) ∈ C(J, ℝ), and 1<β - β+ < p-;

          (30) i = 1 k A i ( u , v ) C 1 ( u + v ) θ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-42/MediaObjects/13661_2011_Article_87_IEq49_HTML.gifwhen |u| + |v| is large enough, where 0 < θ < p - - 1 p + - 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-42/MediaObjects/13661_2011_Article_87_IEq50_HTML.gif;

          (40) i = 1 k B i ( u , v ) C 2 ( u + v ) ε http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-42/MediaObjects/13661_2011_Article_87_IEq51_HTML.gifwhen |u| + |v| is large enough, where 0 ≤ ε < β+ - 1.

          Then, problem (1)-(4) has at least one solution.

          Proof. Now, we consider the following operator equation
          L x = S ( x , λ ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-42/MediaObjects/13661_2011_Article_87_Equ20_HTML.gif
          (20)
          For any λ ∈ (0, 1], x = (x1, x2) = (u, v) is a solution of (20) if and only if v ( r ) = 1 λ p ( r ) - 1 w ( r ) φ p ( r ) ( u ( r ) ) ( r J ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-42/MediaObjects/13661_2011_Article_87_IEq52_HTML.gif and u(r) is a solution of the following
          1 λ p ( r ) - 1 w ( r ) φ p ( r ) ( u ( r ) ) ) = λ p ( r ) f ( r , u , 1 λ ( w ( r ) ) 1 p ( r ) - 1 u ) r ( 0 , T ) , r r i , lim r r i + u ( r ) - lim r r i - u ( r ) = λ 2 A i ( lim r r i - u ( r ) , 1 λ lim r r i - ( w ( r ) ) 1 p ( r ) - 1 u ( r ) ) , i = 1 , , k , lim r r i + 1 λ p ( r ) - 1 w ( r ) φ p ( r ) ( u ( r ) ) - lim r r i - 1 λ p ( r ) - 1 w ( r ) φ p ( r ) ( u ( r ) ) = λ p ( r i ) B i ( lim r r i - u ( r ) , 1 λ lim r r i - ( w ( r ) ) 1 p ( r ) - 1 u ( r ) ) , i = 1 , , k , a u ( 0 ) - b 1 λ ( w ( 0 ) ) 1 p ( 0 ) - 1 u ( 0 ) = 0 , and c u ( T ) + d 1 λ p ( T ) - 1 w ( T ) u p ( T ) - 2 u ( T ) = 0 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-42/MediaObjects/13661_2011_Article_87_Equ21_HTML.gif
          (21)
          We claim that all the solutions of (21) are uniformly bounded for λ ∈ (0, 1]. In fact, if it is false, we can find a sequence (u n , λ n ) of solutions for (21), such that ||u n ||1 > 1 and ||u n ||1 → +∞ when n → +∞, λ n ∈ (0, 1]. Since (u n , λ n ) are solutions of (21), we have
          w ( r ) u n p ( r ) - 2 u n ( r ) = λ n p ( r ) - 1 1 λ n p ( 0 ) - 1 ρ n + r i < r λ n p ( r i ) B i + 0 r λ n p ( t ) f t , u n , 1 λ n ( w ( t ) ) 1 p ( t ) - 1 u n d t , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-42/MediaObjects/13661_2011_Article_87_Equba_HTML.gif
          for any rJ', where ρ n = w ( 0 ) φ p ( 0 ) ( u n ( 0 ) ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-42/MediaObjects/13661_2011_Article_87_IEq53_HTML.gif and
          A i = A i lim r r i - u n ( r ) , 1 λ n lim r r i - ( w ( r ) ) 1 p ( r ) - 1 u n ( r ) , B i = B i lim r r i - u n ( r ) , 1 λ n lim r r i - ( w ( r ) ) 1 p ( r ) - 1 u n ( r ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-42/MediaObjects/13661_2011_Article_87_Equbb_HTML.gif
          By computation, we have
          r i < r λ n 2 A i C 1 λ n 2 u n 0 + 1 λ n ( w ( r ) ) 1 p ( r ) - 1 u n 0 θ λ n C 3 u n 1 θ , r i < r λ n p ( r i ) B i C 2 λ n p ( r i ) u n 0 + 1 λ n ( w ( r ) ) 1 p ( r ) - 1 u n 0 β + - 1 C 4 u n 1 β + - 1 , 0 T λ n p ( r ) f r , u n , 1 λ n ( w ( r ) ) 1 p ( r ) - 1 u n d r C 5 λ n p - u n 0 + 1 λ n ( w ( r ) ) 1 p ( r ) - 1 u n 0 β + - 1 C 5 u n 1 β + - 1 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-42/MediaObjects/13661_2011_Article_87_Equ22_HTML.gif
          (22)
          Denote
          Γ n ( r ) = 1 λ n p ( 0 ) - 1 ρ n + r i < r λ n p ( r i ) B i + 0 r λ n p ( t ) f t , u n , 1 λ n ( w ( t ) ) 1 p ( t ) - 1 u n d t , r J . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-42/MediaObjects/13661_2011_Article_87_Equbc_HTML.gif
          We claim that
          1 λ n p ( 0 ) - 1 ρ n 3 N C 6 u n 1 θ * ( p + - 1 ) + u n 1 β + - 1 , n = 1 , 2 , , where  θ * θ , p - - 1 p + - 1 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-42/MediaObjects/13661_2011_Article_87_Equ23_HTML.gif
          (23)
          If it is false, without loss of generality, we may assume that
          1 λ n p ( 0 ) - 1 ρ n > 3 N ( C 4 + C 5 ) u n 1 θ * ( p + - 1 ) + u n 1 β + - 1 , n = 1 , 2 , , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-42/MediaObjects/13661_2011_Article_87_Equbd_HTML.gif
          then for any n = 1, 2, ..., there is a j n ∈ {1, ..., N} such that the j n -th component ρ n j n http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-42/MediaObjects/13661_2011_Article_87_IEq54_HTML.gif of ρ n satisfies
          1 λ n p ( 0 ) - 1 ρ n j n > 3 ( C 4 + C 5 ) u n 1 θ * ( p + - 1 ) + u n 1 β + - 1 , n = 1 , 2 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-42/MediaObjects/13661_2011_Article_87_Eqube_HTML.gif
          Thus, when n is large enough, the j n -th component Γ n j n ( r ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-42/MediaObjects/13661_2011_Article_87_IEq55_HTML.gif of Γ n (r) keeps the same sign as ρ n j n http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-42/MediaObjects/13661_2011_Article_87_IEq54_HTML.gif and satisfies
          Γ n j n ( r ) > ( C 4 + C 5 ) u n 1 θ * ( p + - 1 ) + u n 1 β + - 1 , r J , n = 1 , 2 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-42/MediaObjects/13661_2011_Article_87_Equbf_HTML.gif
          When n is large enough, we can conclude that the j n -th component F j n { φ q ( r ) [ Γ n ( r ) ] } ( T ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-42/MediaObjects/13661_2011_Article_87_IEq56_HTML.gif of F{φq(r) n (r)]} (T) keeps the same sign as ρ n j n http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-42/MediaObjects/13661_2011_Article_87_IEq54_HTML.gif and satisfies
          F j n { φ q ( r ) [ Γ n ( r ) ] } > C 7 u n 1 θ * , r J . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-42/MediaObjects/13661_2011_Article_87_Equ24_HTML.gif
          (24)
          Since
          u n ( r ) = u n ( 0 ) + r i < r λ n 2 A i + λ n F { φ q ( r ) ( w ( r ) ) - 1 Γ n ( r ) } = b a φ q ( 0 ) 1 λ n p ( 0 ) - 1 ρ n + r i < r λ n 2 A i + λ n F { φ q ( r ) [ ( w ( r ) ) - 1 Γ n ( r ) ] } , r J , n = 1 , 2 , , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-42/MediaObjects/13661_2011_Article_87_Equbg_HTML.gif

          from (22) and (24), we can see that u n j n ( r ) ( r J ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-42/MediaObjects/13661_2011_Article_87_IEq57_HTML.gif keeps the same sign as ρ n j n http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-42/MediaObjects/13661_2011_Article_87_IEq54_HTML.gif, when n is large enough.

          But the boundary value conditions (4) mean that
          c u n ( T ) + d 1 λ n p ( T ) - 1 lim r T - w ( r ) u n p ( r ) - 2 u n ( r ) = c u n ( T ) + d Γ n ( T ) = 0 , n = 1 , 2 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-42/MediaObjects/13661_2011_Article_87_Equbh_HTML.gif
          It is a contradiction. Thus (23) is valid. Therefore,
          w ( r ) u n ( r ) p ( r ) - 1 C 7 u n 1 θ * ( p + - 1 ) + u n 1 β + - 1 , r J . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-42/MediaObjects/13661_2011_Article_87_Equbi_HTML.gif
          It means that
          ( w ( r ) ) 1 p ( r ) - 1 u n 0 o ( 1 ) u n 1 ,  where  o ( 1 )  tends to  0  uniformly as  n . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-42/MediaObjects/13661_2011_Article_87_Equ25_HTML.gif
          (25)
          From (22), (23) and (25), for any rJ, we have
          u n ( r ) = u n ( 0 ) + 0 r u n ( t ) d t + r i < r λ n 2 A i u n ( 0 ) + 0 r u n ( t ) d t + r i < r λ n 2 A i u n ( 0 ) + 0 r ( w ( t ) ) - 1 p ( t ) - 1 ( w ( t ) ) 1 p ( t ) - 1 u n ( t ) d t + C 3 u n 1 θ b a 1 λ n ( w ( 0 ) ) 1 p ( 0 ) - 1 u n ( 0 ) + E ( w ( r ) ) 1 p ( r ) - 1 u n ( r ) 0 + C 3 u n 1 θ o ( 1 ) u n 1 , where  o ( 1 )  tends to  0  uniformly as  n , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-42/MediaObjects/13661_2011_Article_87_Equbj_HTML.gif
          then
          u n 0 o ( 1 ) u n 1 , where  o ( 1 )  tends to  0  uniformly as  n . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-42/MediaObjects/13661_2011_Article_87_Equ26_HTML.gif
          (26)

          From (25) and (26), we get that all the solutions of (20) are uniformly bounded for any λ ∈ (0, 1].

          When λ = 0, if (x1, x2) is a solution of (20), then (x1, x2) is a solution of the following usual equation
          x 1 = 0 , r ( 0 , T ) , x 2 = 0 , r ( 0 , T ) , a x 1 ( 0 ) - b φ q ( 0 ) ( x 2 ( 0 ) ) = 0 , c x 1 ( T ) + d x 2 ( T ) = 0 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-42/MediaObjects/13661_2011_Article_87_Equbk_HTML.gif
          we have
          ( x 1 , x 2 ) = ( 0 , 0 ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-42/MediaObjects/13661_2011_Article_87_Equbl_HTML.gif

          Thus, there exists a large enough R0 > 0 such that all the solutions of (20) belong to B(R0) = {xX | || x || X < R0}. Thus, (20) has no solution on ∂B (R0). From theorem 3.1, we obtain that (1)-(4) has at least one solution. This completes the proof.

          Theorem 3.3 Assume that the following conditions hold

          (10) a = 0;

          (20) lim|u| + |v| → +∞f(r, u, v)/(|u| + |v|)ε = 0 for rJ uniformly, where 0 ≤ ε min(1, p- - 1);

          (30) i = 1 k A i ( u , v ) C 1 ( u + v ) θ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-42/MediaObjects/13661_2011_Article_87_IEq58_HTML.gifwhen |u| + |v| is large enough, where 0 < θ < 1;

          (40) i = 1 k B i ( u , v ) C 2 ( u + v ) ε http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-42/MediaObjects/13661_2011_Article_87_IEq59_HTML.gifwhen |u| + |v| is large enough, where 0 ≤ ε < min(1, p- - 1).

          Then, problem (1)-(4) has at least one solution.

          Proof Now, we consider the following operator equation
          L x = S ( x , λ ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-42/MediaObjects/13661_2011_Article_87_Equ27_HTML.gif
          (27)
          If (x1, x2) is a solution of (27) when λ = 0, then (x1, x2) is a solution of the following usual equation
          x 1 = 0 , r ( 0 , T ) , x 2 = 0 , r ( 0 , T ) , a x 1 ( 0 ) - b φ q ( 0 ) ( x 2 ( 0 ) ) = 0 , c x 1 ( T ) + d x 2 ( T ) = 0 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-42/MediaObjects/13661_2011_Article_87_Equbm_HTML.gif
          Then, we have
          ( x 1 , x 2 ) = ( 0 , 0 ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-42/MediaObjects/13661_2011_Article_87_Equbn_HTML.gif
          For any λ ∈ (0, 1], x = (x1, x2) = (u, v) is a solution of (27) if and only if v ( r ) = 1 λ p ( r ) - 1 w ( r ) φ p ( r ) ( u ( r ) ) ( r J ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-42/MediaObjects/13661_2011_Article_87_IEq60_HTML.gif and u(r) is a solution of the following
          ( 1 λ p ( r ) - 1 w ( r ) φ p ( r ) ( u ( r ) ) ) = λ p ( r ) f ( r , u , 1 λ ( w ( r ) ) 1 p ( r ) - 1 u ) , r ( 0 , T ) , r r i , lim r