ANISOTROPIC DISCRETE

. In this paper, we study the existence and multiplicity of nontrivial solutions for an anisotropic discrete nonlinear problem with variable exponent. The analysis makes use of variational methods and critical point theory.

In the case when p(t) = p (a constant), the problem has been studied by A.R. El Amrouss and O. Hammouti in [7]. They obtained the existence of at least two nontrivial solutions, by using the critical point theory and Mountain Pass Theorem. Here, we will generalize this result.
We want to notice that problem (P) could be regarded as a discrete analogue of the variable exponent anisotropic problem where Ω ⊂ R N , N ≥ 3, is a bounded domain with smooth boundary, f ∈ C Ω × R, R is a given function that satisfy certain properties, p i (x) are continuous functions on Ω, with p i (x) ≥ 2 for (i, x) ∈ [1, N ] Z × Ω.
In the previous decades, the nonlinear difference equations have been intensively used for the mathematical modelling of various problems in different disciplines of science, such as mechanical engineering, statistics, computing, ecology, optimal control, neural network, electrical circuit analysis, population dynamics, economics, biology and other fields; (see, for example [2,17]). In this context, anisotropic discrete nonlinear problems seem to have attracted a great deal of attention due to its usefulness of modelling some more complicated phenomenon such us fluid dynamics and nonlinear elasticity. We refer the reader to [1,5,6,8,11,12] and references therein, where they can find the detailed background as well as many different approaches and techniques applied in the related area.
As is well known, critical point theory, variational methods and also monotonicity methods are powerful tools to investigate the existence and multiplicity of solutions of various problems, see the monographs [4,9,10,[15][16][17].
In this paper, we shall study the existence and multiplicity of nontrivial solutions of (P), via min-max methods and Mountain Pass Theorem. Let To state our main results, we consider the following conditions: with and It is easy to see that γ N > 0 and we will see later that γ N is finite.
Clearly, we have After a simple calculation, we get Then F satisfies the conditions (H 1 ), (H 2 ) and (H 3 ), but f does not satisfy the conditions of the article [12].
By the expression of f we have The main results in this paper are the following theorems. The rest of this paper is organized as follows. Section 2 contains some preliminary lemmas. The main results will be proved in Section 3.

Preliminary lemmas
The vector space E N defined in (2) is an N -dimensional Hilbert space with the inner product while the corresponding norm is given by We list also some inequalities that will be used later.
Lemma 2.1 ( [12]). For every u ∈ E N , we have: Remark 2.2. From (A 6 ), it is easy to see that γ N defined in (1) is finite. Let u ∈ E N , we consider the functional as follows It is easy to see that Φ ∈ C 1 (E N , R) and its derivative Φ ′ (u) at u ∈ E N is given by for any v ∈ E N . By the summation by parts formula, Φ ′ can be written as for any v ∈ E N . Thus, if u ∈ E N is a critical point of Φ, then u is a solution of (P). Now, we consider the truncated problem For u ∈ E N , we denote by u + = max(u, 0) and u − = max(−u, 0) the positive and negative parts of u. It is clear to see that u Lemma 2.3. All solutions of (P + ) (resp. (P − )) are non-negative (resp. non positive) solutions of (P).
It is easy to see that ∆u Let u be a solution of (P + ), or equivalently let u be a critical point of Φ + . Taking we have Therefore, we deduce that On the other hand

Proof of Theorem 1.3
To apply the Mountain Pass Theorem, we will do separate studies of the (P S) condition compactness of Φ ± and its geometry.
Lemma 3.1. Assume that (H 1 ) holds; then the functional Φ + satisfies the (P S) condition.
Proof. Let (u n ) ⊂ E N be a (P S) sequence for the functional Φ + , i.e., |Φ + (u n )| ≤ C and Φ ′ + (u n ) −→ 0 as n → ∞, where C is a constant. Let us show that (u n ) is bounded in E N . Since u n = u + n − u − n , we will prove that (u + n ) and (u − n ) are bounded. Suppose that (u − n ) is unbounded. Then there exists an integer n 0 > 0 such that ∥u − n ∥ ≥ N + 1 for n ≥ n 0 .
Since ∆u + Using the above inequality and (A 2 ), we obtain for any n ≥ n 0 This implies that Since Φ ′ + (u n ) −→ 0 as n → ∞, then for any ε > 0, there exists an integer n 1 with n 1 ≥ n 0 such that ∥Φ ′ + (u n )∥ < ε, ∀n ≥ n 1 . Combining the preceding inequality and (5), we get ∥u − n ∥ p − −1 ≤ (ε + 1) N p − −2 2 for any n ≥ n 1 . Which means that (u − n ) is bounded. Thus we obtain a contradiction. Now, we will prove that (u + n ) is bounded. According to (H 1 ), there exists R > 0 such that On the other hand, by continuity of Thus, we deduce that According to (A 5 ), (A 8 ) and (7), we obtain N t=1 If ∥u + n ∥ = 0 for any integer n ≥ 0, then (u + n ) is bounded. Otherwise, by (A 6 ) we have So, we deduce that If (u + n ) is unbounded, up to a subsequence we may assume that ∥u + n ∥ → ∞. Then in view of (6) and the fact that (u − n ) is bounded, we get what is a contradiction, hence (u + n ) is bounded. It follows that (u n ) is bounded. □ Lemma 3.2. Assume that (H 3 ) holds; then there exist r > 0 and α > 0 such that Φ + (u) ≥ α, for all u ∈ E N with ∥u∥ = r.
Similarly, using Φ − , we show that there furthermore exists a non-positive solution.
Now, by contradiction we prove that Φ is anti-coercive. Let K ∈ R and (u n ) ⊂ E N such that ∥u n ∥ −→ ∞ and Φ(u n ) ≥ K. Putting v n = u n ||u n || , one has ∥v n ∥ = 1. Since dimE N < ∞, there exists v ∈ E N such that ∥v n − v∥ −→ 0, as n → ∞ and ∥v∥ = 1.
For t ∈ Θ, |u n (t)| −→ ∞. Using (1), we have From the condition (H 2 ), we deduce that The sequence (u n (t)) is bounded for any t ∈ [1, N ] Z ⧹Θ and F is continuous. Hence, there exists a constant M ∈ R such that Therefore, we get This a contradiction. Hence Φ is anti-coercive on E N . So, we can choose e large enough to ensure that Φ(e) < 0, and that any (P S) sequence (u n ) is bounded. In view of the fact that the dimension of E N is finite, we see that Φ satisfies the (P S) condition. Since Φ(0) = 0, then all the conditions of Lemma 2.5 are satisfied. Thus Φ possesses a critical value c ≥ α = 1 2p + N p + −2 2 ρ p + > 0, where c = inf h∈Γ max s∈[0,1] Φ(h(s)), and Γ = {h ∈ C([0, 1] , E N )/ h(0) = 0, h(1) = e}. Let u 1 ∈ E N such that Φ(u 1 ) = c. Clearly, u 1 is a nontrivial solution of the problem (P). On the other hand, since Φ is bounded from above and anti-coercive, then there is a maximum point of Φ at some u 2 ∈ E N i.e., Φ(u 2 ) = sup u∈E N Φ(u). Using (9), we obtain Φ(u 2 ) = sup u∈E N Φ(u) ≥ sup u∈∂Bρ Φ(u) > 0. Hence u 2 is a nontrivial solution of the problem (P).
Consequently, the problem (P) has at least two nontrivial solutions. □