Existence and uniqueness of solutions for Steklov problem with variable exponent

In this article, we give some results on the existence and uniqueness of solutions concerned a class of elliptic problems involving p(x)−Laplacian with Steklov boundary condition. We give also some su cient conditions to assure the existence of a positive solution.

Problems like (1.1) have been extensively considered in the literature in the recent years. In [11], the authors have studied the case f (x, u) = λ|u| p(x)−2 u.
When f (x, u) is independent of the second variable u, the uniqueness for (1.1) holds easily, accurately for the case p is constant. When f (x, u) depends on the second variable and it is nonincreasing in u, the uniqueness for (1.1) can be reached (see [2] for example).
Our purpose is to give results on the existence of solutions under the typical growth condition in which we will deal with the existence of constant sign solutions, in fact, since the non-homogeneity of the p(x)−Laplacian. A distinguish feature is that we consider the variable exponent problem when the nonlinear term f neither satisfy sub p − growth condition, nor satisfy sup p + growth condition, for instance, when the nonlinearity satises the following typical growth condition On the other side, we prove the uniqueness for (1.1) which seems rarely studied for those problems. So, we state our main results as form as three theorems.
Then (1.1) has a nontrivial solution.
Theorem 1.2. Suppose that the following assumption hold: Under the condition (H 2 ), the problem (1.1) has a unique solution which is nontrivial. Theorem 1.3. Under the assumptions (H 0 ) and (H 3 ), the problem (1.1) has a nontrivial solution in then such solution is unique.
The rest of this paper is organized as follows. In Section 2, we recall some preliminaries on variable exponent spaces. In Section 3, by the Galerkin method and other approaches, we shall establish the results of existence and uniqueness of a solution for problem (1.1).

Preliminaries
For the integrity of the paper, we recall some basic facts which will be used later. For more detail we refer to [13,14,19]. For p ∈ C + (Ω), we designate the variable exponent Lebesgue space by is a continuous and bounded operator.
As in the constant exponent case, the generalized Lebesgue-Sobolev space W 1,p(x) (Ω) is dened as With such norms, L p(x) (Ω) and W 1,p(x) (Ω) are separable, reexive and uniformly convex Banach spaces.
Then the following statements are equivalent each other: Let a : ∂Ω → R be a measurable. Dene the weighted variable exponent Lebesgue space by Recall the following embedding theorem.
Then, there exists a compact embedding Let us recall the following interesting results: In view of the assumption (H 1 ), we have that then φ + is coercive and even, so its minimizer can be achieved at certain point v ≥ 0.
On other hand, denote In virtue of (H 0 ) and (H 1 ) we have that Accordingly , for u > 1, it is easy to see that There exists Ω l such that where Ω = k l=1 Ω l . Thus, for u < 1, we obtain is strictly positive in a neighborhood of zero. Thereby, we may nd positive constants α > 0 and β > 0 such that Now, we are ready to claim that φ + veries (P S), the Palais Smale condition. In fact, let (u n ) be a sequence in Because the operator u → ∂Ω F + (x, u) dσ x is weakly continuous and its derivative is compact, then we only prove that (u n ) is bounded in W 1,p(x) (Ω). Otherwise, we may nd a subsequence still denoted by (u n ) such that u n → ∞ when n → ∞. From the coercivity, φ + (u n ) → ∞, which is a contradiction with the fact that φ + (u n ) is bounded. By regarding Proposition 2.3, it follows that φ + has a critical point ω which is nontrivial (because φ + (ω) > 0 = φ + (0)). What means that ω is a nontrivial solution for the following problem Next, we check that We conclude that ω ≤ v a.e in Ω. Meanwhile, we have Hence ω − = 0 a.e in Ω, so ω ≥ 0.
Then ω is a solution of (1.1) and the proof is achieved. Then has unique L ∞ solution u 1 which is nonegative (see [12]and [20]). Denote A standard argument shows that ψ ∈ C 1 (W 1,p(x) (Ω), R), since p − > 1 and f is bounded. When u > 1 we have ψ(u) ≥ 1 p + u p − − K 1 u , with K 1 is a positive constant. Then ψ is coercive, and since it is sequentially weakly lower continuous, we conclude that ψ has a global minimizer u ∈ W 1,p(x) (Ω) s.t ψ ( u) = 0. Thereby, u veries Taking u − as test function and keeping in mind that f (x, u) = f (x, 0), for u < 0, then we have which implies that u − = 0 and then u ≥ 0.
Meanwhile, f (x, u) ≤ m, according to comparison principle see shao gao deng..., we have u ≤ u 1 . Accordingly, and then u is a solution of (1.1), which is nontrivial because f (x, 0) ≡ 0. b) Uniqueness: Let recall the following formulas: where x.y is the inner product in R N . Let u and v two solutions of (1.1), viewing the last inequalities, we have 0 ≤ [u>v] ∇u| p(x)−2 ∇u − |∇v| p(x)−2 ∇v ∇u − ∇v dx + [u>v] u| Thus, ∇u(x) = ∇v(x) and u( Similarly, we prove v ≤ u a.e x ∈ Ω, hence, u = v. Remark 2.1. If we suppose that f (x, u)u ≥ 0, so the problem has no solution which is = 0.
Dene V n = span{e 1 , ..., e n }. It is known that V n and R N are isomorphic and for η ∈ R N , we have an

Dene the function
Our method consists in considering a class of auxiliary problems, Now, by applying the Lemma 2.1, let check the existence of solution u n for the problem (2.6).
For u > 1, by (H 3 ), it yields This shows that there is R > 1 such that A n u, u ≥ 0 if u = R, and then (2.6) has a solution u n ∈ V n such that u n ≤ R, ∀n ∈ N.
From this point, passing to a subsequence if necessary, then u n u in W 1,p(x) (Ω), u n → u a.e. inΩ.
By using the fact of the continuity of the Nemytskii operator from L r( Since W 1,p(x) (Ω) → L r(x) (Ω) compactly, tending n → ∞, so we infer that Thus, u is a solution of problem (1.1). Now let u and v be two solutions of problem (1.1), we have which implies that (u − v) = 0, and the proof will be completed.
Remark 2.2. Under the hypothesis (H 3 ). If we suppose that there exists a 0 > 0 and δ > 0 such that where q 0 ∈ C(Ω) with q 0 < p − . Then we get that the Energy functional associated to (1.1) φ is coercive and has a global minimizer u 1 which is non trivial. Indeed, x v 0 ∈ X \ {0} and t > 0 is small enough, so from (H 3 ) we have that because q 0 < p − . By using the Mountain Pass Theorem and the fact that φ is coercive, we construct a continuous curve From this point of view, we can show the existence of an other critical point u 2 ∈ X of φ such that u 2 = u 1 and u 2 is nontrivial.