Two constant sign solutions for a nonhomogeneous Neumann boundary value problem

. We consider a nonlinear Neumann problem with a nonhomogeneous elliptic diﬀerential operator. With some natural conditions for its structure and some general assumptions on the growth of the reaction term we prove that the problem has two nontrivial solutions of constant sign. In the proof we use variational methods with truncation and minimization techniques.


Introduction
Let Ω ⊆ R N be a bounded domain with a C 1,α boundary ∂Ω, where α ∈ (0, 1] is a positive constant. In this paper we are looking for smooth solutions to the following Neumann problem    −div a(∇u(z)) = f (z, u(z)) a.e. in Ω, ∂u ∂n a = 0 on ∂Ω, (1.1) where ∂u ∂na = (a(∇u(z)), n(z)) R N with n(·) = (n 1 (·), . . . , n N (·)) the outward unit normal vector on ∂Ω. On the continuous map a = (a i ) N i=1 : R N → R N we impose certain conditions (see Section 3) to obtain a p-Laplacian type operator, which unifies several important differential operators. Similar conditions are studied widely in literature (see Damascelli [2], Montenegro [6], Motreanu-Papageorgiou [7]), as they allow us to apply the regularity results of Lieberman [5]. The reaction term f : Ω × R → R is a Carathéodory function. We assume that f (z, ·) has a positive and negative z-dependant zero and we are interested in the existence of constant sign positive and negative solutions of Problem (1.1), imposing some growth conditions on f (z, ·) only near zero, without any control in ±∞. [48]

Liliana Klimczak
The rest of this paper is organized as follows. In Section 2 we provide mathematical preliminaries and recall the main mathematical tools which will be employed in this paper. In Section 3 we formulate the assumptions on maps a and f and provide a basic example of the reaction term f . Next we prove the existence theorem using variational and truncation methods.

Mathematical background
In this paper we will denote by (·, ·) R N the scalar product in R N . Also · denotes the norm in Sobolev space W 1,p (Ω). We will assume that 1 < p < ∞.
In the analysis of Problem (1.1) we will use the positive cone ≥ 0 for all z ∈Ω and ∂u ∂n a = 0 on ∂Ω and its interior given by Below we present main mathematical tools which will be needed in the proofs of our results.
Definition 2.1 Let φ : X ⊇ M → R be a functional on a subset M of the Banach space X. We say that φ is • weakly sequentially lower semicontinuous on M iff for each u ∈ M and each sequence {u n } n ⊆ M such that u n → u weakly in X, we have • weakly coercive iff lim Theorem 2.2 (25.D in Zeidler [9]) Suppose that the functional φ : X ⊇ M → R has the following three properties: ii. φ is weakly sequentially lower semicontinuous, iii. φ is weakly coercive.
Then φ has a minimum on M . with some constants δ > 0, c 0 > 0. We define H(ξ) = ξ 0 h(t) dt. By W 1,H (Ω) we denote the class of functions which are weakly differentiable in the set Ω with Let α ∈ (0, 1], Λ, Λ 1 , M 0 > 0 be positive constants and let Ω ⊆ R N be a bounded domain with C 1,α boundary.  Let Ω ⊆ R N be a domain. Suppose that A ∈ C 1 (R + ) is such that function s → sA(s) is strictly increasing in R + and sA(s) → 0 as s → 0 + . Let B ∈ L ∞ loc (Ω × R + × R N ) satisfy the following condition holds, i.e. if u is a classical distribution solution of (2.2) with u(z 0 ) = 0 at some point z 0 ∈ Ω, then u ≡ 0 in Ω. By classical distribution solution we mean a function u ∈ C 1 (Ω), which satisfies (2.2) in the distribution sense.
To deal with the boundary condition in Problem (1.1), we introduce the following function space framework, due to Casas-Fernández [1]: If Ω has a Lipschitz boundary ∂Ω, we have that the space We denote the space of traces on ∂Ω by W 1/p ,p (∂Ω), endowed with the usual norm, and denote the trace of u ∈ W 1,p (Ω) on ∂Ω by γ 0 (u). Let us also consider the space . We denote the dual space of T p (∂Ω) by T −p (∂Ω) and the duality brackets by ·, · T . We have Also there exists a unique linear continuous map From this result one can obtain the following Green's formula.

Problem setting
In this section we formulate our assumptions on the continuous map a and the Carathéodory reaction term f in Problem (1.1).

Two constant sign solutions for a nonhomogeneous Neumann boundary value problem [51]
Assumption H(a) 1 The function a : Assumption H(a) 2 There exist some constants δ, c 0 , c 1 , c 2 , c 3 > 0, q ∈ (1, p) and there exists a function Assumption H(a) 3 For all y, ξ ∈ R N such that y = 0 we have Assumption H(a) 4 There exists some constants µ ∈ (1, q] and τ ∈ (1, p] such that the map t → G 0 (t Then G is strictly convex, G(0) = 0 and ∇G(y) = a(y) for y ∈ R N \{0}. Moreover, for all y ∈ R N G(y) ≤ (a(y), y) R N . (3.5) Proof. As a 0 > 0, G 0 is strictly increasing. Thus the function t → G 0 (t 1 τ ) is strictly inceasing on (0, ∞). So G 0 is also strictly convex, because from H(a) 4 we have for any λ ∈ (0, 1), s, t > 0, s = t Here we have used the fact, that for τ > 1 the function t → t τ is strictly convex on (0, ∞). To show that G is also strictly convex, let λ ∈ (0, 1). Then by the definition of the norm and the properties of G 0 (strict monotonicity and strict convexity) we get [52]

Liliana Klimczak
To obtain the gradient of G, for i = 1, . . . , N and y ∈ R N \{0} we compute 4 ). Next we show that inequality (3.5) holds. For y = 0 the inequality is true. Let y ∈ R N \{0}, b ∈ R N . It follows from convexity of G that As ∇G(y) = a(y) and G(0) = 0, for b = 0 we get (c) for all y ∈ R N we have Proof. (a) Strict monotonicity of a is equivalent to the strict convexity of G (see Zeidler [9], Proposition 25.10). As a is monotone and continuous, it is also maximal monotone (Zeidler [9]). (b) and (c) The proof is similar to the proof of Lemma 2.1 in Damascelli [2]. We use the fact that Proof. To prove the first inequality we observe, that from (3.7) we have Thus We prove the second inequality using the Cauchy-Schwarz inequality, Lemma 3.2 and Proposition 3.1: (see (3.5), (3.6)).

Liliana Klimczak
Our assumptions on the Carathéodory map f : Ω × R → R are the following.

Existence of two constant sign solutions
In this section we state the main result of this paper.