Existence results for a class of nonlocal problems involving p(x)-Laplacian

We study the existence of weak solutions for a p(x)-Kirchhoff problem. The main tool used is the variational method, more precisely, the Mountain Pass Theorem.


Introduction
In this paper, we consider the following problem: |∇u| p(x) dx div(|∇u| p(x)−2 ∇u) = λ(a(x)|u| q(x)−2 u + b(x)|u| r (x)−2 u) in , where ⊂ R N , N ≥ 3, is a bounded domain with smooth boundary ∂ , M : R + → R is a continuous function, λ is a positive number, p is Lipschitz continuous on , and q, r are continuous functions on with q − := inf x∈ q(x) > 1, r − := inf x∈ r (x) > 1, a(x), b(x) > 0 for x ∈ such that a ∈ L α(x) ( ), (x) . Hereafter, We will use the notations such as h − and h + , where The operator p(x) u := div(|∇u| p(x)−2 ∇u) is called the p(x)-Laplacian, and becomes p-Laplacian when p(x) ≡ p (a constant). The p(x)-Laplacian possesses more complicated properties than the p-Laplacian; for M. Allaoui (B) Department of Mathematics, Faculty of Sciences, University Mohamed I, Oujda, Morocco E-mail: allaoui19@hotmail.com example, it is inhomogeneous. The study of problems involving variable exponent growth conditions has a strong motivation due to the fact that they can model various phenomena which arise in the study of elastic mechanics [21] and image restoration [7]. The problem (1.1) is a generalization of a model introduced by Kirchhoff [18]. More precisely, Kirchhoff proposed a model given by the equation which extends the classical D'Alembert's wave equation, by considering the effects of the changes in the length of the strings during the vibrations. Lions [19] has proposed an abstract framework for the Kirchhoff-type equations. After the work by Lions [19], various equations of Kirchhoff-type have been studied extensively, see [2,5]. The study of Kirchhoff-type equations has already been extended to the case involving the p-Laplacian (for details, see [3,4,10,11]) and p(x)-Laplacian (see [9,12]). Motivated by the above papers and the results in [8,20], we consider (1.1) to study the existence of weak solutions.

Preliminary
For completeness, we first recall some facts on the variable exponent spaces L p(x) ( ) and W 1, p(x) ( ). For more details, see [13][14][15]. Suppose that is a bounded open domain of R N with smooth boundary ∂ and p ∈ C + ( ), where with the norm Define the variable exponent Sobolev space with the norm We denote by W are separable and uniformly convex Banach spaces.

Lemma 2.4 ([14]) Assume that is bounded and smooth.
• Let p be Lipschitz continuous and p + < N .
The Euler-Lagrange functional associated to (1.1) is for all u, v ∈ X , then we know that the weak solution of (1.1) corresponds to the critical point of the functional J λ . Hereafter, M(t) is supposed to verify the following assumptions: (M 0 ) There exists m 1 ≥ m 0 > 0 and μ ≥ ν > 1 such that An example of functions satisfying the assumptions (M 0 ) and (M 1 ): Throughout this paper, we assume the condition: For simplicity, we use C i , i = 1, 2..., to denote the general positive constants whose exact values may change from line to line.

Lemma 3.3
There exists e ∈ X with e > ρ (where ρ is given in Lemma 3.2) such that J λ (e) < 0.

Lemma 3.4 The functional J λ satisfies the Palais-Smale condition (PS).
Proof Suppose that (u n ) ⊂ X is a (PS) sequence; that is, (3.6) We prove that (u n ) is bounded in X . Arguing by contradiction we assume that, passing eventually to a subsequence, still denote by (u n ), u n → ∞ and u n > 1 for all n. By (3.6) and (M 0 ), (M 1 ), for n large enough, we have Dividing the above inequality by u n νp − , taking into account (2.1) holds true and passing to the limit as n → ∞, we obtain a contradiction. It follows that (u n ) is bounded in X . By the reflexivity of X , for a subsequence still denoted (u n ), we have u n u in X and a(x)|u n | q(x)−2 u n (u n − u) dx → 0, (3.8) and b(x)|u n | r (x)−2 u n (u n − u) dx → 0. (3.9) Since (u n ) is bounded in X , passing to a subsequence, if necessary, we may assume that If h 0 = 0 then (u n ) converges strongly to u = 0 in X and the proof is complete. If h 0 > 0 then since the function M is continuous, we obtain Thus, by (M 0 ), for sufficiently large n, we have (3.10) From (3.7), (3.8), (3.9) and (3.10), we deduce that A(u) := |∇u n | p(x)−2 ∇u n (∇u n − ∇u) dx → 0. According to the fact that A satisfies Condition (S + ) (see [17]), we have u n → u in X . This completes the proof.