Open Access

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

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 , https://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] → 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 , https://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 , https://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 , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-42/MediaObjects/13661_2011_Article_87_Equ4_HTML.gif
(4)

where p C ([0, T], ) 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( N × N , 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 N , v j will denote the j-th component of v; the inner product in N will be denoted by 〈·,·〉; |·| will denote the absolute value and the Euclidean norm on 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 https://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, N ) = {x: J N | x C(J i , N ), i = 0, 1,..., k, and lim r r i + x ( r ) https://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}; w PC(J, ) satisfies 0 < w(r), r J', and ( w ( ) ) - 1 p ( ) - 1 L 1 ( 0 , T ) https://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 ) https://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 ) https://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 ) https://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, N ), denote |x i |0 = suprJ'|x i (r)|. Obviously, PC(J, N ) is a Banach space with the norm x 0 = ( i = 1 N x i 0 2 ) 1 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-42/MediaObjects/13661_2011_Article_87_IEq7_HTML.gif, PC1(J, N ) is a Banach space with the norm x 1 = x 0 + ( w ( r ) ) 1 p ( r ) - 1 x 0 https://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, N ) and PC1(J, N ) will be simply denoted by PC and PC1, respectively. Let L1 = L1(J, N ) with the norm x L 1 = ( i = 1 N x i L 1 2 ) 1 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-42/MediaObjects/13661_2011_Article_87_IEq9_HTML.gif, x L1, where x i L 1 = 0 T x i ( r ) d r https://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 ) , https://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 ) . https://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 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 https://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 , https://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, s L [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, f C ([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 × N × N N is assumed to be Caratheodory; by this, we mean:
  1. (i)

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

     
  2. (ii)

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

     
  3. (iii)
    for each R > 0, there is a α R L 1 (J, ), such that, for almost every t J and every (x, y) N × N with |x| ≤ R, |y| ≤ R, one has
    f ( t , x , y ) α R ( t ) . https://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 N is a solution of (1) if u PC1 with w(·) |u'|p(·)-2u'(·) absolutely continuous on every J i o https://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 = X1 coKerL and Y = Y1 ImL. 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 https://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 . https://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 https://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 * . https://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 ) . https://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 https://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 × 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 ; https://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 . https://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 N to N for any fixed r J. Denote
φ p ( r ) - 1 ( x ) = x 2 - p ( r ) p ( r ) - 1 x , for x N \ { 0 } , φ p ( r ) - 1 ( 0 ) = 0 . https://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 ( ) https://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 ) ( ) https://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 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-42/MediaObjects/13661_2011_Article_87_IEq16_HTML.gif. Let X = {(x1, x2) | x1 PC, x2 PC} with the norm ||(x1, x2)|| X = || x1||0 + ||x2||0, Y = L1 × L1 × 2(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 , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-42/MediaObjects/13661_2011_Article_87_Equk_HTML.gif

where y1, y2 L1, z m 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 ) , https://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 ) , https://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 . https://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 } , https://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 , https://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 ) . https://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 , https://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 ) ) ) ) . https://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 https://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)Sx X, x X. Since ( w ( r ) ) - 1 p ( r ) - 1 L 1 https://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 https://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 ) , https://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 ) . https://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 https://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 . https://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 ) . https://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 ) . https://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 ) . https://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 ¯ https://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 ) https://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 ¯ https://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 , https://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 N ; g L1.

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 . https://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 . https://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 . https://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 https://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 . https://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 https://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 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-42/MediaObjects/13661_2011_Article_87_Equab_HTML.gif
Denote H = L1 × 2kNwith the norm
h H = g L 1 + i = 1 k a i + i = 1 k b i , h = ( g , a i , b i ) H , https://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 h H, 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 ) . https://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 h H, the equation
    Θ h ( ρ ) = 0 https://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) N .
  1. (ii)

    The mapping ρ: H 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 ] https://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 https://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 https://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 , https://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 ] https://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 ) https://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 https://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 . https://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 ] , https://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 ] , https://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 ] . https://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 ) https://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 https://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 . https://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 ) https://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 ) https://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 . https://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 ( ) https://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 https://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 https://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 . https://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 https://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 . https://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 ) https://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 https://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 ) https://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 https://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 https://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 . https://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 ] . https://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 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 . https://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 N , which satisfies
Θ h ( ρ ( h ) ) = 0 . https://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 . https://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 ) } https://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 * https://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 , https://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 , https://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 https://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 . https://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 , https://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 https://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 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-42/MediaObjects/13661_2011_Article_87_Equar_HTML.gif
Denote G: H 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 ) . https://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 ] https://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 https://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 https://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 ) https://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 ¯ https://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)), r J'.

Proof. Let us consider the following operator equation
L x = S ( x , λ ) . https://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 ) ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-42/MediaObjects/13661_2011_Article_87_IEq48_HTML.gif, r J'.

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 . https://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 , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-42/MediaObjects/13661_2011_Article_87_Equau_HTML.gif
where c1, c2 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 . https://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 , https://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 ] . https://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 ) https://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 . https://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 , https://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 , ) . https://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 r J uniformly, where β(r) C(J, ), and 1<β - β+ < p-;

(30) i = 1 k A i ( u , v ) C 1 ( u + v ) θ https://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 https://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 https://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 ) . https://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 ) https://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 . https://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 , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-42/MediaObjects/13661_2011_Article_87_Equba_HTML.gif
for any r J', where ρ n = w ( 0 ) φ p ( 0 ) ( u n ( 0 ) ) https://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 ) . https://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 . https://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 . https://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 . https://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 , , https://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 https://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 , https://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 ) https://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 https://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 , https://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 ) https://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 https://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 . https://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 , , https://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 ) https://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 https://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 , https://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 . https://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 . https://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 r J, 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 , https://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 . https://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 , https://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 ) . https://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) = {x X | || 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 r J uniformly, where 0 ≤ ε min(1, p- - 1);

(30) i = 1 k A i ( u , v ) C 1 ( u + v ) θ https://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 https://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 ) . https://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 . https://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