Open Access Research

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

Li Yin1, Yunrui Guo2, Guizhen Zhi1 and Qihu Zhang1*

Author Affiliations

1 Department of Mathematics and Information Science, Zhengzhou University of Light Industry, Zhengzhou, Henan 450002, China

2 Department of Mathematics, Henan Institute of Science and Technology, Xinxiang, Henan 453003, China

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Boundary Value Problems 2011, 2011:42  doi:10.1186/1687-2770-2011-42


The electronic version of this article is the complete one and can be found online at: http://www.boundaryvalueproblems.com/content/2011/1/42


Received:1 July 2011
Accepted:1 November 2011
Published:1 November 2011

© 2011 Yin et al; licensee Springer.

This is an Open Access article distributed under the terms of the Creative Commons Attribution License (http://creativecommons.org/licenses/by/2.0), which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

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

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M1','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M1">View MathML</a>

(1)

where u: [0, T] → ℝN, with the following impulsive boundary conditions

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M2','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M2">View MathML</a>

(2)

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M3','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M3">View MathML</a>

(3)

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M4','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M4">View MathML</a>

(4)

where p C ([0, T], ℝ) and p(r) > 1, -Δp(r) u:= -(w(r) |u'|p(r)-2 u'(r))' is called weighted p(r)-Laplacian; 0 < r1 < r2 < ⋯ < rk < T; Ai, Bi C(ℝN × ℝN, ℝN); 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 ∈ ℝN, vj will denote the j-th component of v; the inner product in ℝN will be denoted by 〈·,·〉; |·| will denote the absolute value and the Euclidean norm on ℝN. Denote J = [0, T], J' = [0, T]\{r0, r1,..., rk+1}, J0 = [r0, r1], Ji = (ri, ri+1], i = 1, ..., k, where r0 = 0, rk+1 = T. Denote <a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M5','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M5">View MathML</a> the interior of Ji, i = 0, 1,..., k. Let PC(J, ℝN) = {x: J → ℝN | x C(Ji, ℝN), i = 0, 1,..., k, and <a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M6','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M6">View MathML</a> exists for i = 1,..., k}; w PC(J, ℝ) satisfies 0 < w(r), ∀r J', and <a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M7','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M7">View MathML</a>; <a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M8','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M8">View MathML</a>, <a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M9','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M9">View MathML</a> and <a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M10','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M10">View MathML</a> exists for i = 0, 1,..., k}. For any x = (x1,..., xN) ∈ PC(J, ℝN), denote |xi|0 = suprJ' |xi(r)|. Obviously, PC(J, ℝN) is a Banach space with the norm <a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M11','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M11">View MathML</a>, PC1(J, ℝN) is a Banach space with the norm <a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M12','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M12">View MathML</a>. In the following, PC(J, ℝN) and PC1(J, ℝN) will be simply denoted by PC and PC1, respectively. Let L1 = L1(J, ℝN) with the norm <a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M13','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M13">View MathML</a>, ∀x L1, where <a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M14','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M14">View MathML</a>. We will denote

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M15','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M15">View MathML</a>

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M16','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M16">View MathML</a>

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 [1-4]). Many results have been obtained on this problems, for example [1-25]. 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 -Δp; for example, if Ω ⊂ ℝN is a bounded domain, the Rayleigh quotient

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M17','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M17">View MathML</a>

is zero in general, and only under some special conditions λp(·) > 0 (see [8,17-19]), but the property of λp > 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 [26-28] 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 [29-37]). 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 -Δp, 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

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M18','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M18">View MathML</a>

(5)

where Φp(x) = |x|p-2 x, 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, Ii 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 [24-26,38,40].

Let N ≥ 1, the function f: J × ℝN × ℝN → ℝN is assumed to be Caratheodory; by this, we mean:

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

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

(iii) for each R > 0, there is a αR L1 (J, ℝ), such that, for almost every t J and every (x, y) ∈ ℝN × ℝN with |x| ≤ R, |y| ≤ R, one has

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M19','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M19">View MathML</a>

We say a function u: J → ℝN is a solution of (1) if u PC1 with w(·) |u'|p(·)-2 u'(·) absolutely continuous on every <a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M5','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M5">View MathML</a>, 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) ⊂ X Y 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: X1 Y1 and the continuous linear projectors P: X X1 and Q: Y Y1 with KerQ = ImL and ImP = X1.

Let Ω be an open bounded subset of X with Ω ∩ D(L) ≠ ∅. Operator <a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M20','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M20">View MathML</a> 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

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M21','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M21">View MathML</a>

It is easy to see that the operator <a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M22','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M22">View MathML</a>, M = (L + ΛP)-1 (S + ΛP) is well defined, and

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M23','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M23">View MathML</a>

If M is continuous and compact, then S is called L-compact, and the Leray-Schauder degree of IX - M (where IX is the identity mapping of X) is well defined in Ω, and we will denote it by dLS (IX - M, Ω, 0). This number is independent of the choice of P, Q and Λ (up to a sign) and we can define

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M24','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M24">View MathML</a>

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) = dLS (IX - M, Ω, 0).

There are many papers about coincidence degree and its applications (see [40-43]).

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

(ii) (Homotopy invariant property). If <a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M25','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M25">View MathML</a> is continuous, L-compact and Lx H(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 Lx Sx for each x ∈ ∂Ω. If Sε is such that supx∈∂Ω||Sεx||Y is sufficiently small, then Lx Sx + Sεx for all x ∈ ∂Ω and d(L - S - Sε, Ω) = d(L - S, Ω).

For any (r, x) ∈ (J × ℝN), 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

(i) For any r ∈ [0, T], φp(r)(·) is strictly monotone, i.e.,

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M26','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M26">View MathML</a>

(ii) There exists a function η: [0, +∞) → [0, +∞), η(s) → +∞ as s → +∞, such that

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M27','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M27">View MathML</a>

It is well known that φp(r)(·) is a homeomorphism from ℝN to ℝN for any fixed r J. Denote

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M28','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M28">View MathML</a>

It is clear that <a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M29','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M29">View MathML</a> is continuous and sends bounded sets to bounded sets, and <a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M30','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M30">View MathML</a> where <a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M31','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M31">View MathML</a>. Let X = {(x1, x2) | x1 PC, x2 PC} 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

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M32','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M32">View MathML</a>

where y1, y2 L1, zm ∈ ℝN, m = 1,..., 2(k + 1), then X and Y are Banach spaces.

Define L: D(L) ⊂ X Y and S: X Y as the following

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M33','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M33">View MathML</a>

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M34','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M34">View MathML</a>

where

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M35','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M35">View MathML</a>

(6)

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

Since

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M36','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M36">View MathML</a>

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

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M37','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M37">View MathML</a>

at the same time the projectors P: X X and Q: Y Y satisfy

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M38','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M38">View MathML</a>

Since ImQ is isomorphic to KerL, there exists an isomorphism Λ: KerL ImQ. It is easy to see that L |D(L)∩KerP : D(L) ∩ KerP ImL is invertible. We denote the inverse of that mapping by Kp, then Kp : ImL D(L) ∩ KerP as

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M39','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M39">View MathML</a>

then

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M40','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M40">View MathML</a>

Proposition 2.5 (i) Kp(·) is continuous;

(ii) Kp (I - Q)S is continuous and compact.

Proof. (i) It is easy to see that Kp(·) is continuous. Moreover, the operator <a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M41','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M41">View MathML</a> sends equi-integrable set of L1 to relatively compact set of PC.

(ii) It is easy to see that Kp(I - Q)Sx X, ∀x X. Since <a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M42','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M42">View MathML</a> and f is Caratheodory, it is easy to check that S is a continuous operator from X to Y, and the operators (x1, x2) → φq(r) ((w(r))-1 x2) and (x1, x2) → f (r, x1, φq(r) ((w(r))-1 x2)) both send bounded sets of X to equi-integrable set of L1. Obviously, Ai, Bi and QS are compact continuous. Since f is Caratheodory, by using the Ascoli-Arzela theorem, we can show that the operator <a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M43','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M43">View MathML</a> is continuous and compact. This completes the proof.

Denote

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M44','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M44">View MathML</a>

where Ai, Bi are defined in (6), i = 1,..., k.

Consider

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M45','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M45">View MathML</a>

Define <a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M46','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M46">View MathML</a> as M(·, ·) = (L + ΛP)-1 (S(·, ·) + ΛP), then

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M47','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M47">View MathML</a>

Since (I - Q)S(·, 0) = 0 and Kp (0) = 0, we have

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M48','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M48">View MathML</a>

It is easy to see that all the solutions of Lx = S(x, 0) belong to KerL, then

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M49','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M49">View MathML</a>

Notice that P |KerL = IKerL, then

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M50','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M50">View MathML</a>

Proposition 2.6 (continuation theorem) (see [40]). Suppose that L is a Fredholm operator of index zero and S is L-compact on <a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M51','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M51">View MathML</a>, where Ω is an open bounded subset of X. If the following conditions are satisfied,

(i) for each λ ∈ (0, 1), every solution x of

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M52','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M52">View MathML</a>

is such that x ∉ ∂Ω;

(ii) QS(x, 0) ≠ 0 for x ∈ ∂Ω ∩ KerL and dB-1 QS(·,0), Ω ∩ KerL, 0) ≠ 0, then the operator equation Lx = S(x, 1) has one solution lying in <a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M51','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M51">View MathML</a>.

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 dB) 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

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M53','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M53">View MathML</a>

(7)

where J' = [0, T]\{r0, r1, ..., rk+1}, ai, bi ∈ ℝN; g L1.

If u is a solution of (7), then we have

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M54','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M54">View MathML</a>

(8)

Denote ρ0 = w(0)φp(0) (u'(0)). Obviously, ρ0 is dependent on g, ai, bi. Define F: L1 PC as

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M55','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M55">View MathML</a>

By (8), we have

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M56','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M56">View MathML</a>

(9)

If a ≠ 0, then the boundary condition <a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M57','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M57">View MathML</a> implies that

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M58','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M58">View MathML</a>

The boundary condition <a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M59','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M59">View MathML</a> implies that

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M60','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M60">View MathML</a>

Denote H = L1 × ℝ2kN with the norm

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M61','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M61">View MathML</a>

then H is a Banach space. For fixed h H, we denote

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M62','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M62">View MathML</a>

Lemma 2.7 The mapping Θh(·) has the following properties

(i) For any fixed h H, the equation

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M63','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M63">View MathML</a>

(10)

has a unique solution ρ(h) ∈ ℝN.

(ii) The mapping ρ: H → ℝN, defined in (i), is continuous and sends bounded sets to bounded sets. Moreover, <a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M64','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M64">View MathML</a>, where h = (g, ai, bi) ∈ H, <a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M65','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M65">View MathML</a>, the notation p# means <a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M66','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M66">View MathML</a>.

Proof. (i) From Proposition 2.4, it is immediate that

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M67','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M67">View MathML</a>

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

Let <a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M68','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M68">View MathML</a>. Since <a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M69','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M69">View MathML</a> and F(g) ∈ PC, if |ρ| > R0, it is easy to see that there exists a j0 such that, the j0-th component <a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M70','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M70">View MathML</a> of ρ satisfies

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M71','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M71">View MathML</a>

(11)

Obviously,

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M72','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M72">View MathML</a>

then

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M73','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M73">View MathML</a>

(12)

and

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M74','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M74">View MathML</a>

(13)

By (11) and (12), the j0-th component of <a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M75','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M75">View MathML</a> keeps the same sign of <a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M86','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M86">View MathML</a> on J and

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M76','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M76">View MathML</a>

(14)

Combining (13) and (14), the j0-th component <a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M77','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M77">View MathML</a> of <a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M78','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M78">View MathML</a> satisfies

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M79','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M79">View MathML</a>

From the definition <a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M80','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M80">View MathML</a>, we have <a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M81','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M81">View MathML</a>, then <a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M82','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M82">View MathML</a>, and

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M83','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M83">View MathML</a>

Without loss of generality, we may assume that <a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M218','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M218">View MathML</a>, then we have

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M84','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M84">View MathML</a>

Therefore, the j0-th component of <a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M85','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M85">View MathML</a> keeps the same sign of <a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M86','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M86">View MathML</a>. Since the j0-th component of <a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M87','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M87">View MathML</a> keeps the same sign of <a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M86','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M86">View MathML</a>, a, b, c, d ∈ [0, +∞) and ad + bc > 0, we can easily see that the j0-th component of Θh(ρ) keeps the same sign of <a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M86','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M86">View MathML</a>, and thus

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M88','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M88">View MathML</a>

Let us consider the equation

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M89','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M89">View MathML</a>

(15)

According to the above discussion, all the solutions of (15) belong to b(R0 + 1) = {x ∈ ℝN| |x| < R0 + 1}. So, we have

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M90','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M90">View MathML</a>

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

In this way, we define a mapping ρ(h): H → ℝN, which satisfies

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M91','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M91">View MathML</a>

(ii) By the proof of (i), we also obtain ρ sends bounded set to bounded set, and

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M92','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M92">View MathML</a>

It only remains to prove the continuity of ρ. Let {un} is a convergent sequence in H and un u, as n → +∞. Since {ρ(un)} is a bounded sequence, it contains a convergent subsequence <a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M93','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M93">View MathML</a> satisfies <a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M94','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M94">View MathML</a> as j → +∞. Since Θh(ρ) consists of continuous functions, and

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M95','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M95">View MathML</a>

Letting j → +∞, we have

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M96','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M96">View MathML</a>

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

This completes the proof.

If a = 0, the boundary condition <a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M97','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M97">View MathML</a> implies that

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M98','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M98">View MathML</a>

Since ad + bc > 0, we have c > 0. Thus,

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M99','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M99">View MathML</a>

the boundary condition <a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M100','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M100">View MathML</a> implies that

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M101','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M101">View MathML</a>

Denote G: H → ℝN as

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M102','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M102">View MathML</a>

It is easy to see that

Lemma 2.8 The function G(·) is continuous and sends bounded sets to bounded sets. Moreover, <a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M103','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M103">View MathML</a>, where <a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M104','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M104">View MathML</a>, the notation p* means <a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M105','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M105">View MathML</a>.

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 Ci 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

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M106','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M106">View MathML</a>

(16)

has no solution on ∂Ω.

(20) (0, 0) ∈ Ω.

Then, problem (1)-(4) has a solution u satisfies <a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M107','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M107">View MathML</a>, where v = w(r)φp(r)(u'(r)), ∀r J'.

Proof. Let us consider the following operator equation

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M108','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M108">View MathML</a>

(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 <a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M109','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M109">View MathML</a>, ∀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

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M110','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M110">View MathML</a>

The problem (??) is a usual differential equation. Hence,

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M111','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M111">View MathML</a>

where c1, c2 ∈ ℝN are constants. The boundary value condition of (??) holds,

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M112','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M112">View MathML</a>

Since (ad + bc) > 0, we have

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M113','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M113">View MathML</a>

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

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M114','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M114">View MathML</a>

(18)

Now, it is clear that the following problem

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M115','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M115">View MathML</a>

(19)

is equivalent to problem (1)-(4), and (18) tells us that problem (19) will have a solution if we can show that

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M116','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M116">View MathML</a>

Since by hypothesis (20), this last degree

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M117','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M117">View MathML</a>

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

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M118','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M118">View MathML</a>

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, ℝ), and 1<β - β + < p -;

(30) <a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M119','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M119">View MathML</a>when |u| + |v| is large enough, where <a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M120','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M120">View MathML</a>;

(40) <a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M121','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M121">View MathML</a>when |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

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M122','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M122">View MathML</a>

(20)

For any λ ∈ (0, 1], x = (x1, x2) = (u, v) is a solution of (20) if and only if <a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M123','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M123">View MathML</a> and u(r) is a solution of the following

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M124','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M124">View MathML</a>

(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 (un, λn) of solutions for (21), such that ||un||1 > 1 and ||un||1 → +∞ when n → +∞, λn ∈ (0, 1]. Since (un, λn) are solutions of (21), we have

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M125','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M125">View MathML</a>

for any r J', where <a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M126','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M126">View MathML</a> and

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M127','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M127">View MathML</a>

By computation, we have

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M128','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M128">View MathML</a>

(22)

Denote

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M129','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M129">View MathML</a>

We claim that

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M130','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M130">View MathML</a>

(23)

If it is false, without loss of generality, we may assume that

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M131','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M131">View MathML</a>

then for any n = 1, 2, ..., there is a jn ∈ {1, ..., N} such that the jn-th component <a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M132','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M132">View MathML</a> of ρn satisfies

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M133','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M133">View MathML</a>

Thus, when n is large enough, the jn-th component <a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M134','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M134">View MathML</a> of Γn(r) keeps the same sign as <a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M132','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M132">View MathML</a> and satisfies

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M135','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M135">View MathML</a>

When n is large enough, we can conclude that the jn-th component <a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M136','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M136">View MathML</a> of F{φq(r)n(r)]} (T) keeps the same sign as <a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M132','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M132">View MathML</a> and satisfies

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M137','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M137">View MathML</a>

(24)

Since

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M138','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M138">View MathML</a>

from (22) and (24), we can see that <a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M139','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M139">View MathML</a> keeps the same sign as <a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M132','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M132">View MathML</a>, when n is large enough.

But the boundary value conditions (4) mean that

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M140','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M140">View MathML</a>

It is a contradiction. Thus (23) is valid. Therefore,

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M141','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M141">View MathML</a>

It means that

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M142','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M142">View MathML</a>

(25)

From (22), (23) and (25), for any r J, we have

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M143','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M143">View MathML</a>

then

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M144','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M144">View MathML</a>

(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

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M145','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M145">View MathML</a>

we have

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M146','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M146">View MathML</a>

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) <a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M147','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M147">View MathML</a>when |u| + |v| is large enough, where 0 < θ < 1;

(40) <a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M148','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M148">View MathML</a>when |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

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M149','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M149">View MathML</a>

(27)

If (x1, x2) is a solution of (27) when λ = 0, then (x1, x2) is a solution of the following usual equation

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M150','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M150">View MathML</a>

Then, we have

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M151','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M151">View MathML</a>

For any λ ∈ (0, 1], x = (x1, x2) = (u, v) is a solution of (27) if and only if <a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M152','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M152">View MathML</a> and u(r) is a solution of the following

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M153','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M153">View MathML</a>

(28)

We only need to prove that all the solutions of (28) are uniformly bounded for λ ∈ (0, 1].

In fact, if it is false, we can find a sequence (un, λn) of solutions for (28), such that ||un||1 > 1 and ||un||1 → +∞ when n → +∞. Since (un, λn) are solutions of (28), we have

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M154','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M154">View MathML</a>

where <a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M155','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M155">View MathML</a>.

From conditions (20) and (40), we have

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M156','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M156">View MathML</a>

Thus,

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M157','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M157">View MathML</a>

(29)

Denote

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M158','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M158">View MathML</a>

By solving un(r), we have

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M159','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M159">View MathML</a>

(30)

where <a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M160','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M160">View MathML</a>.

From condition (30), we have

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M161','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M161">View MathML</a>

The boundary value condition implies

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M162','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M162">View MathML</a>

(31)

From (31) and conditions (20), (30) and (40), we have

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M163','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M163">View MathML</a>

(32)

From (30) and (32), we have

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M164','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M164">View MathML</a>

(33)

From (29) and (33), we can conclude that {||un||1} is uniformly bounded for λ ∈ (0, 1]. This completes the proof.

Now, let us consider the following mixed type boundary value condition

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M165','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M165">View MathML</a>

(34)

Theorem 3.4 Assume that the following conditions hold

(10) <a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M166','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M166">View MathML</a>, for r J uniformly, where β(r) ∈ C(J, ℝ), and 1 < β - β +< p-;

(20) <a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M167','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M167">View MathML</a>when |u| + |v| is large enough, where <a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M168','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M168">View MathML</a>;

(30) <a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M169','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M169">View MathML</a>when |u| + |v| is large enough, where 0 ≤ ε < β + - 1.

Then, problem (1) with (2), (3) and (34) has at least one solution.

Proof. It is similar to the proof of Theorem 3.2 and Theorem 3.3, we omit it.

Denote

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M170','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M170">View MathML</a>

where f* (r, u, v) is Caratheodory.

Let us consider

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M171','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M171">View MathML</a>

(35)

Theorem 3.5 Under the conditions of Theorem 3.2, Theorem 3.3 or Theorem 3.4, then (35) with (2), (3) and (4) or (34) has at least a solution when δ is small enough.

Proof We only need to prove the existence of solutions under the conditions of Theorem 3.2, the rest is similar. If δ = 0, the proof of Theorem 3.2 means that all the solutions of (35) with (2), (3) and (4) are bounded and belong to U(R0) = {u PC1| ||u||1 < R0}. Define Sδ : X Y as

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M172','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M172">View MathML</a>

where <a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M173','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M173">View MathML</a>, <a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M174','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M174">View MathML</a>.

Since f* (r, u, v) is a Caratheodory function, we have ||Sδx - S0x||Y → 0 as δ → 0, for <a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M175','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M175">View MathML</a> uniformly. According to Proposition 2.3, we get the existence of solutions.

In the following, we will consider the existence of nonnegative solutions. For any x = (x1, ..., xN) ∈ ℝN, the notation x ≥ 0 means xi ≥ 0 for any i = 1, ..., N.

Theorem 3.6 We assume

(i) f(r, u, v) ≤ 0, ∀(r, u, v) ∈ J × ℝN × ℝN;

(ii) for any i = 1, ..., k, Bi(u, v) ≤ 0,∀(u, v) ∈ ℝN × ℝN;

(iii) for any i = 1, ..., k, j = 1, ..., N, <a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M176','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M176">View MathML</a>∀(u, v) ∈ ℝN × ℝN.

Then, the solution u in Theorem 3.2, Theorem 3.3 or Theorem 3.4 is nonnegative.

Proof We only need to prove that the solution u in Theorem 3.2 is nonnegative, and the rest is similar. Denote

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M177','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M177">View MathML</a>

where

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M178','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M178">View MathML</a>

Similar to (8) and (9), we have

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M179','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M179">View MathML</a>

(36)

where

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M180','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M180">View MathML</a>

and ρ is the solution (unique) of

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M181','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M181">View MathML</a>

(37)

Denote

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M182','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M182">View MathML</a>

From (i), (ii) and (36), we can see that Φ(r) is decreasing, namely

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M183','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M183">View MathML</a>

(38)

We claim that

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M184','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M184">View MathML</a>

(39)

If it is false, then there exists some j0 ∈ {1, ..., N}, such that the j0-th component <a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M86','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M86">View MathML</a> of ρ satisfies

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M185','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M185">View MathML</a>

(40)

Combining (i), (ii), (iii) and (40), we can see that the j0-th component <a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M186','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M186">View MathML</a> of D is negative. It is a contradiction to (37). Thus, (39) is valid. So, we have

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M187','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M187">View MathML</a>

We claim that

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M188','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M188">View MathML</a>

(41)

If it is false. Then, there exists some j1 ∈ {1,..., N}, such that the j1-th component <a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M189','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M189">View MathML</a> of Φ(T) satisfies

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M190','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M190">View MathML</a>

(42)

From (38) and (42), we have

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M191','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M191">View MathML</a>

Combining (i), (ii), (iii) and (42), we can see that the j1-th component <a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M192','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M192">View MathML</a> of D is positive. It is a contradiction to (37). Thus, (41) is valid.

If c > 0. We have

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M193','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M193">View MathML</a>

Since Φ(r) is decreasing, Φ(0) = ρ ≥ 0 and Φ(T) ≤ 0, for any j = 1, ..., N, there exists ξj J such that

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M194','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M194">View MathML</a>

Combining condition (iii), we can conclude that uj(r) is increasing on [0, ξj], and uj(r) is decreasing on (ξj, T]. Notice that u(0) ≥ 0 and u(T) ≥ 0, then we have u(r) ≥ 0,∀r ∈ [0, T].

If c = 0, boundary condition (4) means that Φ(T) = 0. Since Φ(r) is decreasing, we get that Φ(r) ≥ 0. Combining condition (iii), we can conclude that u(r) is increasing on J, namely u(t2) - u(t1) ≥ 0,∀t2, t1 J, t2 > t1. Notice that u(0) ≥ 0, then we have u(r) ≥ 0,∀t J. This completes the proof.

Corollary 3.7 We assume

(i) f(r, u, v) ≤ 0,∀(r, u, v) ∈ J × ℝN × ℝN with u ≥ 0;

(ii) for any i = 1, ..., k, Bi(u, v) ≤ 0,∀(u, v) ∈ ℝN × ℝN with u ≥ 0;

(iii) for any i = 1, ..., k, j = 1, ..., N, <a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M195','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M195">View MathML</a>,∀(u, v) ∈ ℝN × ℝN with u ≥ 0.

Then, we have

(10) Under the conditions of Theorem 3.2 or Theorem 3.3, (1)-(4) has a nonnegative solution.

(20) Under the conditions of Theorem 3.4, (1) with (2), (3) and (34) has a nonnegative solution.

Proof We only need to prove that (1)-(4) has a nonnegative solution under the conditions of Theorem 3.2, and the rest is similar. Define

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M196','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M196">View MathML</a>

where

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M197','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M197">View MathML</a>

Denote

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M198','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M198">View MathML</a>

then <a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M199','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M199">View MathML</a> satisfies Caratheodory condition, and <a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M200','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M200">View MathML</a> for any (r, u, v) ∈ J × ℝN × ℝN.

For any i = 1, ..., k, we denote

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M201','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M201">View MathML</a>

then <a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M202','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M202">View MathML</a> and <a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M203','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M203">View MathML</a> are continuous and satisfy

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M204','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M204">View MathML</a>

Obviously, we have

(20)' <a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M205','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M205">View MathML</a>, for r J uniformly, where β(r) ∈ C(J, ℝ), and1 < β - β +< p- ;

(30)' <a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M206','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M206">View MathML</a> when |u| + |v| is large enough, where <a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M207','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M207">View MathML</a>;

(40)' <a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M208','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M208">View MathML</a> when |u| + |v| is large enough, where 0 ≤ ε < β + - 1.

Let us consider

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M209','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M209">View MathML</a>

(43)

From Theorem 3.2 and Theorem 3.6, we can see that (43) has a nonnegative solution u. Since u ≥ 0, we have ϕ(u) = u, and then

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M210','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M210">View MathML</a>

Thus, u is a nonnegative solution of (1)-(4). This completes the proof.

4 Examples

Example 4.1. Consider the following problem

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M211','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M211">View MathML</a>

where p(r) = 5 + cos 3r, <a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M212','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M212">View MathML</a>, 0 ≤ g(r) ∈ L1, e0, e1 ∈ ℝN, w(r) = 3 + sin r, σ is a nonnegative parameter.

Obviously, <a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M213','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M213">View MathML</a> is Caratheodory, q(r) ≤ 3.5 < 4 ≤ p (r) ≤ 6, then the conditions of Theorem 3.5 are satisfied, then (P1) has a solution when δ > 0 is small enough. Moreover, when σ = 0, the conditions of Corollary 3.7 are satisfied, then (P1) has a nonnegative solution.

Example 4.2. Consider the following problem

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M214','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M214">View MathML</a>

where p(r) = 5 + cos 3r, <a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M215','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M215">View MathML</a>, 0 ≤ g(r) ∈ L1, e0, e1 ∈ ℝN, w(r) = 3+sin r, σ is a nonnegative parameter.

Obviously, <a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M216','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/42/mathml/M216">View MathML</a> is Caratheodory, 1 < q(r) < 2 < 4 ≤ p (r) ≤ 6, then conditions of Theorem 3.5 are satisfied, then (P2) has a solution when δ > 0 is small enough. Moreover, when σ = 0, the conditions of Corollary 3.7 are satisfied, and (P2) has a nonnegative solution.

Competing interests

The authors declare that they have no competing interests.

Authors' contributions

All authors typed, read and approved the final manuscript.

Acknowledgements

The authors would like to thank the referees very much for their helpful comments and suggestions. This study was partly supported by the National Science Foundation of China (10701066 & 10926075 & 10971087) and China Postdoctoral Science Foundation funded project (20090460969) and the Natural Science Foundation of Henan Education Committee (2008-755-65).

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