CUBO, A Mathematical Journal Vol.22, No¯ 01, (85–124). April 2020 http: // doi. org/ 10. 4067/ S0719-06462020000100085 Nonlinear elliptic p(u)− Laplacian problem with Fourier boundary condition Stanislas Ouaro 1 and Noufou Sawadogo 2 LAboratoire de Mathématiques et Informatique (LA.M.I) 1 UFR. Sciences Exactes et Appliquées Université Joseph KI ZERBO 03 BP 7021 Ouaga 03, Ouagadougou, Burkina Faso 2 UFR. Sciences et Techniques, Université Nazi Boni, 01 BP 1091 Bobo 01, Bobo-Dioulasso, Burkina Faso ouaro@yahoo.fr, noufousawadogo858@yahoo.fr ABSTRACT We study a nonlinear elliptic p(u)− Laplacian problem with Fourier boundary con- ditions and L1− data. The existence and uniqueness results of entropy solutions are established. RESUMEN Estudiamos un problema p(u)−Laplaciano eĺıptico nolineal con condiciones de borde Fourier y datos L1. Se establecen resultados de existencia y unicidad de soluciones de entroṕıa. Keywords and Phrases: variable exponent, p(u)−Laplacian, Young measure, Fourier boundary condition, entropy solution. 2010 AMS Mathematics Subject Classification: 35J60, 35D05, 76A05. http://doi.org/10.4067/S0719-06462020000100085 86 Stanislas Ouaro & Noufou Sawadogo CUBO 22, 1 (2020) 1 Introduction In this paper, we consider the following nonlinear Fourier boundary value problem { b(u) − diva(x, u, ∇u) = f in Ω a(x, u, ∇u).η + λu = g on ∂Ω, (1.1) where Ω ⊆ RN (N ≥ 3) is a bounded open domain with Lipschitz boundary ∂Ω, η is the outer unit normal vector on ∂Ω and λ > 0. The operator diva(x, u, ∇u) is called p(u)-Laplacian. It is more complicated than p(x)-Laplacian in the term of nonlinearity. A prototype of this operator is div ( |∇u|p(u)−2.∇u ) . The variable exponent p depend both on the space variable x and on the unknown solution u. The problem (1.1) is a generalization of the following nonlinear problem { b(u) − diva(x, ∇u) = f in Ω a(x, ∇u).η + λu = g on ∂Ω, studied in [15] by Nyanquini and Ouaro. The authors used an auxiliary result due to Le (see [16], Theorem 3.1) to prove the existence of the weak solution when f ∈ L∞(Ω), g ∈ L∞(∂Ω) and by approximation methods they obtained the entropy solution when f ∈ L1(Ω), g ∈ L1(∂Ω). In the present paper, as the function (x, z, η) 7→ a(x, z, η) is more general than (x, η) 7→ a(x, η), it is difficult to use the sub-supersolution method, as in [16], to get the existence of the weak solution. Therefore, we use the technic of pseudo-monotone operators in Banach spaces, some a priori estimates and the convergence in term of Young measure to obtain the existence of entropy solutions of problem (1.1). Indeed, we define an approximation problem, and we prove that this problem has a solution un which converges to u, an entropy solution of problem (1.1). In this paper, we consider the following basic assumptions on the data for the study of the problem (1.1). (A1) f and g are some functions such as f ∈ L 1(Ω), g ∈ L1(∂Ω) and g 6≡ 0. (A2) b is nondecreasing surjective and continuous function defined on R such that b(0) = 0. Problem (1.1) is adapted into a generalized Leray-Lions framework under the assumption that a : Ω × (R × RN) → RN is a Carathéodory function with: (A3) a(x, z, 0) = 0 for all z ∈ R, and a.e. x ∈ Ω; (A4) ( a(x, z, ξ) − a(x, z, η) ) .(ξ − η) > 0 for all ξ, η ∈ RN, ξ 6= η, as well as the growth and the coercivity assumptions with variable exponent (A5) ∣ ∣a(x, z, ξ) ∣ ∣ p ′(x,z) ≤ C1 ( |ξ|p(x,z) + M(x) ) and CUBO 22, 1 (2020) Nonlinear elliptic p(u)− Laplacian problem with Fourier . . . 87 (A6) a(x, z, ξ).ξ ≥ 1 C2 |ξ|p(x,z). Here, C1 and C2 are positive constants and M is a positive function such that M ∈ L 1(Ω). p : Ω×R → [p−, p+] is a Carathéodory function, 1 < p− ≤ p+ < +∞ and p ′(x, z) = p(x, z) p(x, z) − 1 is the conjugate exponent of p(x, z), with p− := ess inf (x,z)∈Ω×R p(x, z) and p+ := ess sup (x,z)∈Ω×R p(x, z). The study of p(u)-Laplacian problem was recently developped by Andreianov et al. (see [2]). These authors established the partial existence and uniqueness result to the weak solution in the cases of homogeneous Dirichlet boundary condition. The interest of the study of this kind of problem is due to the fact that they can model various phenomena which arise in the study of elastic mechanic (see [6]), electrorheological fluids (see [20]) or image restoration (see [9]). In this paper, we study the existence of the weak solution for approximation problem and we also establish the existence and uniqueness results of the entropy solution when the data are in L1. In this work, we use the Sobolev embedding results and the convergence in term of Young measure (see [10, 12]). Here, we consider a Fourier boundary condition which bring some difficulties to treat the term at the boundary. We were inspired by the work of Ouaro and Tchousso (see [15]), where the authors defined for the first time a new space by taking into account the boundary. For the next part of the paper (section 2), we introduce some preliminary results. In section 3, we study the existence and uniqueness of entropy solution when the data are in L1. 2 Preliminary • We will use the so-called truncation function Tk(s) := { s if |s| ≤ k ksign0(s) if |s| > k , where sign0(s) :=      1 if s > 0 0 if s = 0 −1 if s < 0. The truncation function possesses the following properties. Tk(−s) = −Tk(s), |Tk(s)| = min{|s|, k}, lim k→+∞ Tk(s) = s and lim k→0 1 k Tk(s) = sign0(s). 88 Stanislas Ouaro & Noufou Sawadogo CUBO 22, 1 (2020) We also need to truncate vector valued-function with the help of the mapping hm : R N −→ RN, hm(λ) =    λ, if |λ| ≤ m m λ |λ| if |λ| > m, where m > 0. For a Lebesgue measurable set A ⊂ Ω, χA denotes its characteristic function and meas(A) denotes its Lebesgue measure. Let u : Ω → R be a function and k ∈ R, we write {|u| ≤ k} or [|u| ≤ k] for the set {x ∈ Ω : |u(x)| ≤ k}, (respectively, ≥, =, <, >). The function a(., ., .) appearing in (1.1) satisfies a generalized Leray-Lions assumptions given in Introduction. View that, a(., ., .) satisfies (A5) and (A6), we must work in Lebesgue and Sobolev spaces with variable exponent, that depend on x and on u(x). For the study of problem (1.1), we need the Sobolev spaces W 1,π(.)(Ω), where π(.) = p(., u(.)). Definition 1. Let π : Ω −→ [1, +∞) be a measurable function for π(.) = p(., u(.)). •Lπ(.)(Ω) is the space of all measurable function f : Ω −→ R such that the modular ρπ(.)(f) := ∫ Ω |f|π(x)dx < +∞. If p+ is finite, this space is equipped with the Luxembourg norm ||f||Lπ(.)(Ω) := inf { λ > 0; ρπ(.) ( f λ ) ≤ 1 } . In the sequel, we will use the same notation Lπ(.)(Ω) for the space (Lπ(.)(Ω))N of vector-valued functions. •W 1,π(.)(Ω) is the space of all functions f ∈ Lπ(.)(Ω) such that the gradient of f (taken in the sense of distributions) belongs to Lπ(.)(Ω). The space W 1,π(.)(Ω) is equipped with the norm ||u||W 1,π(.)(Ω) := ||u||Lπ(.)(Ω) + ||∇u||Lπ(.)(Ω). When 1 < p− ≤ π(.) ≤ p+ < +∞, all the above spaces are separable and reflexive Banach spaces. We denote πn(x) := p(x, un(x)). Proposition 1. (See [1], Proposition 2.3) For all measurable function π : Ω → [p−, p+], the following properties hold. i) Lπ(.)(Ω) and W 1,π(.)(Ω) are separable and reflexive Banach spaces. ii) Lπ ′(.)(Ω) can be identified with the dual space of Lπ(.)(Ω), and the following Hölder type inequality holds: ∀f ∈ Lπ(.)(Ω), g ∈ Lπ ′(.)(Ω), ∣ ∣ ∣ ∣ ∫ Ω fgdx ∣ ∣ ∣ ∣ ≤ 2||f||Lπ(.)(Ω)||g||Lπ′(.)(Ω). CUBO 22, 1 (2020) Nonlinear elliptic p(u)− Laplacian problem with Fourier . . . 89 iii) One has ρπ(.)(f) = 1 if and only if ||f||Lπ(.)(Ω) = 1; further, if ρπ(.)(f) ≤ 1, then ||f|| p+ Lπ(.)(Ω) ≤ ρπ(.)(f) ≤ ||f|| p− Lπ(.)(Ω) ; if ρπ(.)(f) ≥ 1, then ||f|| p− Lπ(.)(Ω) ≤ ρπ(.)(f) ≤ ||f|| p+ Lπ(.)(Ω) . In particular, if (fn)n∈N is a sequence in L π(.)(Ω), then ||fn||Lπ(.)(Ω) tends to zero (resp., to infinity) if and only if ρπ(.)(fn) tends to zero (resp., to infinity), as n → +∞. For a measurable function f ∈ W 1,π(.)(Ω) we introduce the following notation: ρ1,π(.)(f) = ∫ Ω |f|π(.)dx + ∫ Ω |∇f|π(.)dx. Replacing p(x) by π(x) in [8]-Proposition 2.2, we obtain the following result that is fundamental in this paper. Proposition 2. ( See [23, 24] ) If f ∈ W 1,π(.)(Ω), the following properties hold: i) ||f||W 1,π(.)(Ω) > 1 ⇒ ||f|| p− W 1,π(.)(Ω) < ρ1,π(.)(f) < ||f|| p+ W 1,π(.)(Ω) ; ii) ||f||W 1,π(.)(Ω) < 1 ⇒ ||f|| p+ W 1,π(.)(Ω) < ρ1,π(.)(f) < ||f|| p− W 1,π(.)(Ω) ; iii) ||f||W 1,π(.)(Ω) < 1 (respectively = 1; > 1) ⇔ ρ1,π(.)(f) < 1 (respectively = 1; > 1). The following lemma prove that the space W 1,π(.)(Ω) is stable by truncation. Lemma 2.1. If u ∈ W 1,π(.)(Ω) then Tk(u) ∈ W 1,π(.)(Ω). Now, we give the embedding results. Proposition 3. (See [1], Proposition 2.4) Assume that π : Ω → [p−, p+] has a representative which can be extended into a continuous function up to the boundary ∂Ω and satisfying the log-Hölder continuity assumption: ∃L > 0, ∀x, y ∈ Ω, x 6= y, − ( log |x − y| ) |π(x) − π(y)| ≤ L. (2.1) i) Then, D(Ω) is dense in W 1,π(.)(Ω). ii) W 1,π(.)(Ω) is embedded into Lπ ∗(.)(Ω), where π∗(.) is the Sobolev embedding exponent defined as in (2.2) below. If q is a measurable variable exponent such that ess inf x∈Ω (π∗(.) − q(.)) > 0, then the embedding of W 1,π(.)(Ω) into Lq(.)(Ω) is compact. 90 Stanislas Ouaro & Noufou Sawadogo CUBO 22, 1 (2020) For a given π(.), a function taking values in [p−, p+], π ∗(.) denotes the optimal Sobolev embedding defined for any x ∈ Ω by π ∗(x) =       Nπ(x) N − π(x) if π(x) < N any real value if π(x) = N +∞ if π(x) > N. (2.2) Put π∂(x) := ( π(x) )∂ :=    (N − 1)π(x) N − π(x) if π(x) < N +∞ if π(x) ≥ N. (2.3) Proposition 4. (See [18], Proposition 2.3 ) Let π(.) ∈ C(Ω) and p− > 1. If q(x) ∈ C(∂Ω) satisfies the condition: 1 ≤ q(x) < π∂(x), ∀x ∈ ∂Ω, then, there is a compact embedding W 1,π(.)(Ω) →֒ Lq(.)(∂Ω). In particular there is compact embedding W 1,π(.)(Ω) →֒ Lπ(.)(∂Ω). TYoung measures and nonlinear weak-* convergence. Throughout the paper, we denote by δc the Dirac measure on R d (d ∈ N), concentrated at the point c ∈ Rd. In the following theorem, we gather the results of Ball [7], Pedregal [19] and Hungerbühler [13] which will be needed for our purposes (we limit the statement to the case of a bounded domain Ω). Let us underline that the results of (ii),(iii), expressed in terms of the convergence in measure, are very convenient for the applications that we have in mind. Theorem 2.1. (i) Let Ω ⊂ RN, N ∈ N, and a sequence (vn)n∈N of R d -valued functions, d ∈ N , such that (vn)n∈N is equi-integrable on Ω. Then, there exists a subsequence (nk)k∈N and a parametrized family (νx)x∈Ω of probability measures on R d (d ∈ N) , weakly measurable in x with respect to the Lebesgue measure in Ω, such that for all Carathéodory function F : Ω × Rd → Rt, t ∈ N, we have lim k→+∞ ∫ Ω F(x, vnk )dx = ∫ Ω ∫ Rd F(x, λ)dνx(λ)dx, (2.4) CUBO 22, 1 (2020) Nonlinear elliptic p(u)− Laplacian problem with Fourier . . . 91 whenever the sequence (F(., vn(.)))n∈N is equi-integrable on Ω. In particular, v(x) := ∫ Rd λdνx(λ) (2.5) is the weak limit of the sequence (vnk )k∈N in L 1(Ω). The family (νx)x∈Ω is called the Young measure generated by the subsequence (vnk )k∈N. (ii) If Ω is of finite measure, and (νx)x∈Ω is the Young measure generated by a sequence (vn)n∈N, then νx = δv(x) for a.e. x ∈ Ω ⇔ vn converges in measure in Ω to v as n → +∞. (iii) If Ω is of finite measure, (un)n∈N generates a Dirac Young measure (δu(x))x∈Ω on R d1, and (vn)n∈N generates a Young measure (νx)x∈Ω on R d2, then the sequence (un, vn)n∈N generates the Young measure (δu(x) ⊗ νx)x∈Ω on R d1+d2. Whenever a sequence (vn)n∈N generates a Young measure (νx)x∈Ω, following the terminology of [11] we will say that (vn)n∈N nonlinear weak-* converges, and (νx)x∈Ω is the nonlinear weak-* limit of the sequence (vn)n∈N. In the case where (vn)n∈N possesses a nonlinear weak-* convergent subsequence, we will say that it is nonlinear weak-* compact. ([1], Theorem 2.10(i)) It means that any equi-integrable sequence of measurable functions is nonlinear weak-* compact on Ω. Lemma 2.2. (See [1], Theorem 3.11 and [2] Step 2 of proof of Theorem 2.6). Assume that (un)n∈N converges a.e. on Ω to some function u, then |p(x, un(x)) − p(x, u(x))| converges in measure to 0 on Ω, and for all bounded subset K of RN, sup ξ∈K |a(x, un(x), ξ) − a(x, u(x), ξ)| converges in measure to 0 on Ω. (2.6) For the sequel, we assume that p(., .) is log Hölder continuous uniformly on Ω × [−M, M] and p− > N. We recall some notations. For any u ∈ W 1,π(.)(Ω), we denote by τ(u) the trace of u on ∂Ω in the usual sense. We will identify at boundary u and τ(u). Set T 1,π(.)(Ω) = { u : Ω → R, measurable such that Tk(u) ∈ W 1,π(.)(Ω), for any k > 0 } . 3 Entropy solution In this part, we study the existence and uniqueness of the entropy solution to the problem (1.1). We give some notations. 92 Stanislas Ouaro & Noufou Sawadogo CUBO 22, 1 (2020) We define T 1,π(.) tr (Ω) as the set of the functions u ∈ T 1,π(.)(Ω) such that there exists a sequence (un)n∈N ⊂ W 1,p+(Ω) satisfying the following conditions: (C1) un → u a.e. in Ω. (C2) ∇Tk(un) → ∇Tk(u) in L 1(Ω). (C3) There exists a measurable function v on ∂Ω, such that un → v a.e. on ∂Ω. The function v is the trace of u in the generalized sense as introduced in [4, 5]. In the sequel the trace of u ∈ T 1,π(.) tr (Ω) on ∂Ω will be denoted tr(u). If u ∈ W 1,π(.)(Ω), tr(u) coincides with τ(u) in the usual sense. Moreover, for u ∈ T 1,π(.) tr (Ω) and for all k > 0, tr(Tk(u)) = Tk(tr(u)) and if ϕ ∈ W 1,π(.)(Ω) then u − ϕ ∈ T 1,π(.) tr (Ω) and tr(u − ϕ) = tr(u) − tr(ϕ). As in [1]-Proposition 3.5, we give the following result. Proposition 5. Let u ∈ T 1,π(.)(Ω). There exists a unique measurable function w : Ω → RN such that ∇Tk(u) = wχ{|u| 0. The function w is denoted by ∇u. Moreover, if u ∈ W 1,π(.)(Ω) then w ∈ Lπ(.)(Ω) and w = ∇u in the usual sense. Remark 3.1. The space T 1,π(.) tr (Ω) in our context will be a subset of T 1,π(.)(Ω) consisting to the function can be approximated by function of W 1,p+(Ω). Since the weak solution of approximated problem (3.2) belongs to W 1,p+(Ω). Now, we introduce the notion of entropy solution due to Ouaro and al. [14, Definition 3.1]. Definition 2. A measurable function u : Ω → R for π(.) = p(., u(.)) is called entropy solution of the problem (1.1) if u ∈ T 1,π(.) tr (Ω), b(u) ∈ L 1(Ω), u ∈ L1(∂Ω) and for all k > 0, ∫ Ω b(u)Tk(u − ϕ)dx + ∫ Ω a(x, u, ∇u).∇Tk(u − ϕ)dx + λ ∫ ∂Ω uTk(u − ϕ)dσ ≤ ∫ Ω fTk(u − ϕ)dx + ∫ ∂Ω gTk(u − ϕ)dσ, (3.1) where ϕ ∈ W 1,π(.)(Ω) ∩ L∞(Ω). The following theorem gives existence result of entropy solution. Theorem 3.2. Assume that (A3) − (A6) hold and f ∈ L 1(Ω), g ∈ L1(∂Ω). Then, there exists at least one entropy solution to the problem (1.1). The proof of Theorem 3.2 is done in two parts. CUBO 22, 1 (2020) Nonlinear elliptic p(u)− Laplacian problem with Fourier . . . 93 Part 1: The approximate problem. Let fn = Tn(f) and gn = Tn(g). Then, fn ∈ L ∞(Ω) and gn ∈ L ∞(∂Ω). Moreover, (fn)n∈N strongly converges to f in L1(Ω) and (gn)n∈N strongly converges to g in L 1(∂Ω) such that ||fn||L1(Ω) ≤ ||f||L1(Ω) and ||gn||L1(∂Ω) ≤ ||g||L1(∂Ω). We consider the following problem      Tn(b(un)) − diva(x, un, ∇un) − ε△p+un + ε|un| p+−2un = fn in Ω ( a(x, un, ∇un) + ε|∇un| p+−2∇un ) .η + λTn(un) = gn on ∂Ω, (3.2) where −△p+un := − N ∑ i=1 ∂ ∂xi (∣ ∣ ∣ ∣ ∂un ∂xi ∣ ∣ ∣ ∣ p+−2 ∂un ∂xi ) . In this part, we show that the problem (3.2) admits at least one weak solution un, for all ε > 0. We define the following reflexive space E = W 1,p+(Ω) × Lp+(∂Ω). Let X0 = {(u, v) ∈ E : v = τ(u)}. In the sequel, we will identify an element (u, v) ∈ X0 with its representative u ∈ W 1,p+(Ω) (since W 1,p+(Ω) →֒→֒ Lp+(∂Ω)). Theorem 3.3. There exists at least one weak solution un for the problem (3.2) in the sense that un ∈ X0 and for all v ∈ X0, ∫ Ω Tn(b(un))vdx + ∫ Ω a(x, un, ∇un)∇vdx + ∫ ∂Ω λTn(un)vdσ + ε ∫ Ω [ |un| p+−2unv + |∇un| p+−2∇un∇v ] dx = ∫ Ω fnvdx + ∫ ∂Ω gnvdσ. (3.3) To prove the Theorem 3.3, we need the following result. Lemma 3.1. (See [22], Corollary 2.2). If an operator A is of type (M), bounded and coercive on a separable Banach space to its dual, then A is surjective. We define the operator An by Anu = Au + Bnu, 94 Stanislas Ouaro & Noufou Sawadogo CUBO 22, 1 (2020) where < Au, v >= ∫ Ω a(x, u, ∇u)∇vdx and < Bnu, v >= ∫ Ω Tn(b(u))vdx + λ ∫ ∂Ω Tn(u)vdσ + ε ∫ Ω [ |u|p+−2uv + |∇u|p+−2∇u∇v ] dx, with u, v ∈ X0. Proof of the Theorem 3.3. The proof is organized in three Steps. Step 1: An is bounded. By using Hölder type inequality and (A5) with constant exponent p+, we deduce that A is bounded. Moreover, Bn is bounded. Indeed, let u ∈ F , where F is a bounded subset of X0. As b is onto, we have < Bnu, v > = ∫ Ω Tn(b(u))vdx + λ ∫ ∂Ω Tn(u)vdσ + ε ∫ Ω [ |u|p+−2uv + |∇u|p+−2∇u∇v ] dx ≤ ∫ Ω |b(u)||v|dx + λ ∫ ∂Ω |u||v|dσ + ε ∫ Ω [ |u|p+−1|v| + |∇u|p+−1|∇v| ] dx ≤ C(λ) ( ||v||L1(Ω) + ||v||L1(∂Ω) ) + ε [ ||u|| p+ (p+) ′ L p+(Ω) ||v||Lp+(Ω) + ||∇u|| p+ (p+) ′ L p+ (Ω) ||∇v||Lp+ (Ω) ] ≤ C(λ) ( ||v||L1(Ω) + ||v||L1(∂Ω) ) + C(ε)||v||W 1,p+ (Ω). Therefore, An is bounded. We recall the following notion: Definition 3. An operator A : V → V ′ is type of (M) if: un ⇀ u in V A(un) ⇀ χ in V ′ lim sup n→∞ < A(un), un >≤< χ, u >        ⇒ χ = A(u). Step 2: An is pseudo-monotone. Let (uk)k∈N be a sequence in X0 such that       uk ⇀ u in X0 Anuk ⇀ χ in X ′ 0 lim sup k→∞ < Anuk, uk >=< χ, u > . We will prove that χ = Anu. As Tn(b(uk))uk ≥ 0 and λTn(uk)uk ≥ 0, by Fatou’s Lemma, we deduce that lim inf k→∞ ( ∫ Ω Tn(b(uk))ukdx + λ ∫ ∂Ω Tn(uk)ukdσ ) ≥ ∫ Ω Tn(b(u))udx + λ ∫ ∂Ω Tn(u)udσ. CUBO 22, 1 (2020) Nonlinear elliptic p(u)− Laplacian problem with Fourier . . . 95 One the other hand, thanks to the Lebesgue dominated convergence Theorem, we have lim k→∞ ( ∫ Ω Tn(b(uk))vdx + λ ∫ ∂Ω Tn(uk)vdσ + ε ∫ Ω [ |uk| p+−2ukv + |∇uk| p+−2∇uk∇v ] dx ) = ∫ Ω Tn(b(u))vdx + λ ∫ ∂Ω Tn(u)vdσ + ε ∫ Ω [ |u|p+−2uv + |∇u|p+−2∇u∇v ] dx, for any v ∈ X0. Therefore, for k large enough, Tn(b(uk))+λTn(uk)+ε [ |uk| p+−2uk+|∇uk| p+−2∇uk ] ⇀ Tn(b(u))+λTn(u)+ε [ |u|p+−2u+|∇u|p+−2∇u ] in X′0. Thus, Auk ⇀ χ − ( Tn(b(u)) + λTn(u) + ε[|u| p+−2u + |∇u|p+−2∇u ] ) in X′0, as k → +∞. Now, we are going to prove that A is of type (M). Let us set a1(u, v, w) = ∫ Ω a(x, u, ∇v)∇wdx. Then, w 7→ a1(u, v, w) is continuous on W 1,p+(Ω), thus a1(u, v, w) = 〈 A(u, v), w 〉 , A(u, v) ∈ (W 1,p+ (Ω))′, and verify A(u, u) = Au, where Au := −diva(x, u, ∇u). Let us prove that A is of type of Calculus of variation. • As A(u, .) is bounded, we prove that v 7→ A(u, v) is hemi-continuous from W 1,p+(Ω) → (W 1,p+ (Ω))′. Since a(x, u, ∇(v1 + tv2)) ⇀ a(x, u, ∇v1) in L p ′ +(Ω) as t → 0 and u, v1, v2 ∈ W 1,p+(Ω) then, a1(u, v1 + tv2, w) → a1(u, v1, w) as t → 0. In the same manner we prove that u 7→ A(u, v) is hemi-continuous from W 1,p+(Ω) → (W 1,p+ (Ω))′. Moreover, for all u, v ∈ W 1,p+(Ω), we have < A(u, u) − A(u, v), u − v > = < A(u, u), u − v > − < A(u, v), u − v > = a1(u, u, u − v) − a1(u, v, u − v) = ∫ Ω a(x, u, ∇u)∇(u − v)dx − ∫ Ω a(x, u, ∇v)∇(u − v)dx = ∫ Ω ( a(x, u, ∇u) − a(x, u, ∇v) ) ∇(u − v)dx ≥ 0. 96 Stanislas Ouaro & Noufou Sawadogo CUBO 22, 1 (2020) • Let us suppose that uk ⇀ u in W 1,p+(Ω) and < A(uk, uk) − A(uk, u), uk − u >→ 0. We prove that ∀v ∈ W 1,p+(Ω), A(uk, v) ⇀ A(u, v) in (W 1,p+(Ω))′. Let’s set ∫ Ω Fkdx = 〈 A(uk, uk) − A(uk, u), uk − u 〉 , then Fk → 0. As uk ⇀ u, we have a(x, uk, ∇v) ⇀ a(x, u, ∇v) in L p ′ +(Ω) (see [17], Lemma 2.2 with m = 1). Therefore, A(uk, v) ⇀ A(u, v) in (W 1,p+(Ω))′. • Now, we suppose that uk ⇀ u in W 1,p+(Ω) and A(uk, v) ⇀ Θ in (W 1,p+(Ω))′. We prove that 〈 A(uk, v), uk 〉 → 〈 Θ, u 〉 . Then, by using ([17], Lemma 2.1), we obtain that a(x, uk, ∇v) → a(x, u, ∇v) in L p ′ +(Ω) and thus, a1(uk, v, uk) → a1(u, v, u). Therefore, < A(uk, v), uk >= a1(uk, v, uk) →< A(u, v), u > and Θ = A(u, v). Hence, A is of type of Calculus of variation. Finally, by using ([17], Proposition 2.6 and Proposition 2.5), we prove that A is of type (M). As the operator A is of type (M), so we have immediately Au = χ − ( Tn(b(u)) + λTn(u) + ε [ |u|p+−2u + |∇u|p+−2∇u ] ) . Therefore, we deduce that Anu = χ. Step 3: An is coercive. < Anu, u > = ∫ Ω a(x, u, ∇u).∇udx + ∫ Ω Tn(b(u))udx + λ ∫ ∂Ω Tn(u)udx + ε ∫ Ω [ |u|p+ + |∇u|p+ ] dx ≥ ε ∫ Ω [ |u|p+ + |∇u|p+ ] dx ≥ ε||u|| p+ W 1,p+ (Ω) . We deduce that < Anu, u > ||u|| W 1,p+ (Ω) → +∞ as ||u|| W 1,p+ (Ω) → +∞. CUBO 22, 1 (2020) Nonlinear elliptic p(u)− Laplacian problem with Fourier . . . 97 Hence, An is coercive. Then, according to Lemma 3.1, An is surjective. Thus, for any Fn =< Tn(f), Tn(g) >⊂ E ′ ⊂ X′0, there exists at least one solution un ∈ X0 of the problem < Anun, v >=< Fn, v > for all v ∈ X0. Therefore, un is a weak solution of the problem (3.2). This ends the proof of Theorem 3.3. Remark 3.4. If un is a weak solution of the problem (3.2), then un ∈ W 1,πn(.)(Ω), since W 1,p+(Ω) →֒ W 1,πn(.)(Ω) continuously. Moreover, a(x, un, ∇un) satisfies (A3) − (A6) with variable exponent πn(x) := p(x, un(x)). Part 2: A priori estimates and convergence results. This part is done in three steps, we make a priori estimates, some convergence results and other based on the Young measure and nonlinear weak−∗ convergence. Step 1: A priori estimates Lemma 3.2. Suppose that (A3) − (A6) hold with variable exponent πn(.) and fn ∈ L ∞(Ω), gn ∈ L∞(∂Ω). Let un be a weak solution of (3.2). Then, for all k > 0, ∫ Ω ∣ ∣∇Tk(un) ∣ ∣ πn(.) dx ≤ C2k ( ||f||L1(Ω) + ||g||L1(∂Ω) ) , (3.4) ∫ Ω ∣ ∣Tn(b(un)) ∣ ∣dx ≤ ||f||L1(Ω) + ||g||L1(∂Ω), (3.5) ∫ ∂Ω ∣ ∣Tn(un) ∣ ∣dx ≤ 1 λ ( ||f||L1(Ω) + ||g||L1(∂Ω) ) . (3.6) Proof of Lemma 3.2. By taking v = Tk(un) in the weak formulation (3.3), we obtain ∫ Ω Tn(b(un))Tk(un)dx + ∫ Ω a(x, un, ∇un).∇Tk(un)dx + ∫ ∂Ω λTn(un)Tk(un)dσ + ε ∫ Ω [ |un| p+−2unTk(un) + |∇un| p+−2∇un∇Tk(un) ] dx = ∫ Ω fnTk(un)dx + ∫ ∂Ω gnTk(un)dσ. (3.7) Since all the terms of the left hand side of (3.7) are nonnegative, we deduce that ∫ Ω a(x, un, ∇un).∇Tk(un)dx ≤ ∫ Ω fnTk(un)dx + ∫ ∂Ω gnTk(un)dσ. (3.8) 98 Stanislas Ouaro & Noufou Sawadogo CUBO 22, 1 (2020) By using (A6) and (3.8), we get ∫ Ω ∣ ∣∇Tk(un) ∣ ∣ πn(.) dx ≤ C2 ( ∫ Ω fnTk(un)dx + ∫ ∂Ω gnTk(un)dσ ) ≤ C2k ( ||f||L1(Ω) + ||g||L1(∂Ω) ) . From (3.7), we deduce that ∫ Ω Tn(b(un))Tk(un)dx ≤ ∫ Ω fnTk(un)dx + ∫ ∂Ω gnTk(un)dσ ≤ k ( ||f||L1(Ω) + ||g||L1(∂Ω) ) (3.9) and λ ∫ ∂Ω Tn(un)Tk(un)dx ≤ ∫ Ω fnTk(un)dx + ∫ ∂Ω gnTk(un)dσ ≤ k ( ||f||L1(Ω) + ||g||L1(∂Ω) ) . (3.10) Dividing (3.9) and (3.10) by k and letting k goes to 0, we obtain ∫ Ω Tn(b(un))sign0(un)dx ≤ ||f||L1(Ω) + ||g||L1(∂Ω) and λ ∫ ∂Ω Tn(un)sign0(un)dx ≤ ||f||L1(Ω) + ||g||L1(∂Ω). Hence, ∫ Ω ∣ ∣Tn(b(un)) ∣ ∣dx ≤ ||f||L1(Ω) + ||g||L1(∂Ω) and ∫ ∂Ω ∣ ∣Tn(un) ∣ ∣dx ≤ 1 λ ( ||f||L1(Ω) + ||g||L1(∂Ω) ) . Lemma 3.3. Assume that (A3)-(A6) hold. If un is a weak solution of the problem (3.2), fn ∈ L∞(Ω) and gn ∈ L ∞(∂Ω), then for all k > 0 ∫ Ω ∣ ∣∇Tk(un) ∣ ∣ p− dx ≤ C ( ||f||L1(Ω), ||g||L1(∂Ω), meas(Ω) ) (k + 1) (3.11) and ∫ ∂Ω ∣ ∣Tk(un) ∣ ∣dσ ≤ 1 λ ( ||f||L1(Ω) + ||g||L1(∂Ω) ) , (3.12) for all n ≥ k > 0. CUBO 22, 1 (2020) Nonlinear elliptic p(u)− Laplacian problem with Fourier . . . 99 Proof of Lemma 3.3. Firstly, we prove (3.11). We know that ∫ Ω ∣ ∣∇Tk(un) ∣ ∣ πn(.) dx ≤ C2k ( ||f||L1(Ω) + ||g||L1(∂Ω) ) . (3.13) Let us note that ∫ Ω ∣ ∣∇Tk(un) ∣ ∣ p− dx = ∫ {|∇Tk(un)|>1} ∣ ∣∇Tk(un) ∣ ∣ p− dx + ∫ {|∇Tk(un)|≤1} ∣ ∣∇Tk(un) ∣ ∣ p− dx ≤ ∫ {|∇Tk(un)|>1} ∣ ∣∇Tk(un) ∣ ∣ p− dx + meas(Ω) ≤ ∫ Ω ∣ ∣∇Tk(un) ∣ ∣ πn(.) dx + meas(Ω). (3.14) By using (3.13) and (3.14), we get ∫ Ω ∣ ∣∇Tk(un) ∣ ∣ p− dx ≤ max ( C2 ( ||f||L1(Ω) + ||g||L1(∂Ω) ) , meas(Ω) ) (k + 1) := C ( ||f||L1(Ω), ||g||L1(∂Ω), meas(Ω) ) (k + 1). (3.15) Now, from the formula (3.6), we obtain ||Tn(un)||L1(∂Ω) ≤ 1 λ ( ||f||L1(Ω) + ||g||L1(∂Ω) ) and as |Tk(un)| ≤ |Tn(un)| for all n ≥ k > 0, one deduces that ∫ ∂Ω |Tk(un)|dσ ≤ 1 λ ( ||f||L1(Ω) + ||g||L1(∂Ω) ) . Lemma 3.4. For any k > 0, we have ||Tk(un)||W 1,πn(.)(Ω) ≤ 1 + C ( k, f, g, p−, p+, meas(Ω) ) and for all k ≥ 1, meas ( {|un| > k} ) ≤ C min ( b(k), |b(−k)| ). Proof of Lemma 3.4. By using (3.4), we have ∫ Ω ∣ ∣∇Tk(un) ∣ ∣ πn(.) dx ≤ C2k ( ||f||L1(Ω) + ||g||L1(∂Ω) ) . (3.16) We also have ∫ Ω |Tk(un)| πn(.)dx = ∫ {|un|≤k} |Tk(un)| πn(.)dx + ∫ {|un|>k} |Tk(un)| πn(.)dx. Furthermore, ∫ {|un|>k} |Tk(un)| πn(.)dx = ∫ {|un|>k} k πn(.)dx ≤ { kp+meas(Ω) if k ≥ 1 meas(Ω) if k < 1 100 Stanislas Ouaro & Noufou Sawadogo CUBO 22, 1 (2020) and ∫ {|un|≤k} |Tk(un)| πn(.)dx ≤ ∫ {|un|≤k} kπn(.)dx ≤ { kp+meas(Ω) if k ≥ 1 meas(Ω) if k < 1. This allow us to write ∫ Ω |Tk(un)| πn(.)dx ≤ 2(1 + kp+)meas(Ω). (3.17) Hence, adding (3.16) and (3.17) one gets ρ1,πn(.)(Tk(un)) ≤ C2k ( ||f||L1(Ω) + ||g||L1(∂Ω) ) + 2(1 + kp+)meas(Ω). For ||Tk(un)||W 1,πn(.)(Ω) ≥ 1, we have according to Proposition 2 that ||Tk(un)|| p− W 1,πn(.)(Ω) ≤ ρ1,πn(.)(Tk(un)) ≤ [ C2k ( ||f||L1(Ω) + ||g||L1(∂Ω) ) + 2(1 + kp+)meas(Ω) ] , which implies that ||Tk(un)||W 1,πn(.)(Ω) ≤ [ C2k ( ||f||L1(Ω) + ||g||L1(∂Ω) ) + 2(1 + kp+)meas(Ω) ] 1 p− := C(k, f, g, p+, p−, meas(Ω)). Thus, ||Tk(un)||W 1,πn(.)(Ω) < 1 + C(k, f, g, p+, p−, meas(Ω)). Moreover, from (3.5), we have ∫ ∂Ω ∣ ∣Tn(b(un)) ∣ ∣dx ≤ ||f||L1(Ω) + ||g||L1(∂Ω). We deduce that the sequence (Tn(b(un)))n∈N∗ is uniformly bounded in L 1(Ω). Thus, (b(un))n∈N∗ is uniformly bounded in L1(Ω). So, there exists a positive constant C such that ∫ Ω |b(un)|dx ≤ C. Furthermore, for all k ≥ 1, we have ∫ {|un|>k} |b(un)|dx ≤ ∫ Ω |b(un)|dx ≤ C. As b is continuous, nondecreasing and surjective, we infer ∫ {|un|>k} min ( b(k), |b(−k)| ) dx ≤ ∫ {|un|>k} |b(un)|dx ≤ C. Therefore, meas ( {|un| > k} ) ≤ C min ( b(k), |b(−k)| ), ∀k ≥ 1. CUBO 22, 1 (2020) Nonlinear elliptic p(u)− Laplacian problem with Fourier . . . 101 Then, the proof of Lemma 3.4 is complete. From the Lemma 3.4, we deduce that for any k > 0, the sequence ( Tk(un) ) n∈N is uniformly bounded in W 1,πn(.)(Ω) and also in W 1,p−(Ω). Then, up to a subsequence still denoted Tk(un), we can assume that for any k > 0, Tk(un) weakly converges to sk in W 1,p−(Ω) and also Tk(un) strongly converges to sk in L p−(Ω). By using the above a priori estimates, we obtain the following convergence results . Step 2: The convergence results The proof of the following proposition use the Lemma 3.4. Proposition 6. Assume that (A3) − (A6) hold and let un be a weak solution of the problem (3.2), then the sequence (un)n∈N is Cauchy in measure. In particular, there exists a measurable function u and a subsequence still denoted un such that un → u in measure, as n → +∞. As (un)n∈N is a Cauchy sequence in measure, so (up to a subsequence) it converges almost everywhere to some measurable function u. As for any k > 0, Tk is continuous; then Tk(un) → Tk(u) a.e. x ∈ Ω, so sk = Tk(u). Therefore, Tk(un) ⇀ Tk(u) in W 1,p−(Ω) and by compact embedding Theorem, we have Tk(un) → Tk(u) in L p−(Ω) (respectively in Lp−(∂Ω)) and a.e. in Ω (respectively a.e. on ∂Ω). Lemma 3.5. un converges a.e. on ∂Ω to some function v. Proof of Lemma 3.5 Since Tk(un) ⇀ Tk(u) in W 1,p−(Ω) and W 1,p−(Ω) →֒ Lp−(∂Ω) (compact embedding), then Tk(un) → Tk(u) in L p−(∂Ω) and a.e. on ∂Ω. Therefore, Tk(un) → Tk(u) in L 1(∂Ω) and a.e. in ∂Ω. We deduce that there exists E ⊂ ∂Ω such that Tk(un) → Tk(u) on ∂Ω \ E with µ(E) = 0, where µ is area measure on ∂Ω. For every k > 0, let Ek = {x ∈ ∂Ω such that |Tk(u)| < k} and F = ∂Ω \ ⋃ k>0 Ek. By using Fatou’s Lemma, we have ∫ ∂Ω ∣ ∣Tk(u) ∣ ∣dσ ≤ lim inf n→+∞ ∫ ∂Ω |Tk(un) ∣ ∣dσ ≤ ||f||L1(Ω) + ||g||L1(∂Ω) λ . (3.18) 102 Stanislas Ouaro & Noufou Sawadogo CUBO 22, 1 (2020) Now, we use (3.18) to get µ(F) = 1 k ∫ F ∣ ∣Tk(u) ∣ ∣dσ ≤ 1 k ∫ ∂Ω ∣ ∣Tk(u) ∣ ∣dσ ≤ ||f||L1(Ω) + ||g||L1(∂Ω) kλ . We obtain µ(F) = 0, as k goes to ∞. Let’ s now define on ∂Ω the function v by v(x) = Tk(u(x)), x ∈ Ek. We take x ∈ ∂Ω r (E ∪ F), then there exists k > 0 such that x ∈ Ek and we have un(x) − v(x) = ( un(x) − Tk(un(x)) ) + ( Tk(un(x)) − Tk(u(x)) ) . Since x ∈ Ek, we have |Tk(u(x))| < k and so |Tk(un(x))| < k, from which we deduce that |un(x)| < k. Therefore, un(x) − v(x) = Tk(un(x)) − Tk(u(x)) → 0, as n → +∞. This means that un converges to v a.e. on ∂Ω, but for all x ∈ Ek, Tk(u(x)) = u(x). Thus, v = u a.e. on ∂Ω. Therefore, un → u a.e. on ∂Ω. The following assertions are based on the Young measure and nonlinear weak −∗ convergence re- sults (see [7, 19, 13]). Step 3: The convergence in term of Young measure Assertion 1 The sequence (∇Tk(un))n∈N converges to a Young measure ν k x(λ) on R N in the sense of the non- linear weak-* convergence and ∇Tk(u) = ∫ RN λdν k x(λ). (3.19) Proof. Using Lemma 3.3, ∇Tk(un) is uniformly bounded in L p−(Ω), so, equi-integrable on Ω. Moreover, ∇Tk(un) weakly converges to ∇Tk(u) in L p−(Ω). Therefore, using the representation of weakly convergence sequences in L1(Ω) in terms of Young measures (see Theorem 2.1 and formula (2.5)), we can write ∇Tk(u) = ∫ RN λdνkx(λ) � Assertion 2. |λ|π(.) is integrable with respect to the measure νkx(λ)dx on R N × Ω, moreover, Tk(u) ∈ W 1,π(.)(Ω). Proof. We know that p(., un(.)) → p(., u(.)) in measure on Ω. Now, using Theorem 2.1 (ii), CUBO 22, 1 (2020) Nonlinear elliptic p(u)− Laplacian problem with Fourier . . . 103 (iii) (p(., un(.)), ∇Tk(un))n∈N converges on R × R N to Young measure µkx = δπ(x) ⊗ ν k x. Thus, we can apply the weak convergence properties (2.4) to the Carathéodory function Fm(x, λ0, λ) ∈ Ω × (R × R N) 7→ |hm(λ)| λ0 with m ∈ N, where hm is defined in the preliminaries. Then, we obtain ∫ Ω×RN |hm(λ)| π(x)dνkx(λ)dx = ∫ Ω×(R×RN ) |hm(λ)| λ0 dµkx(λ0, λ)dx = ∫ Ω ∫ R×RN Fm(x, λ0, λ)dµ k x(λ0, λ)dx = lim n→+∞ ∫ Ω Fm(x, p(x, un(x)), ∇Tk(un(x)))dx = lim n→+∞ ∫ Ω |hm(∇Tk(un))| p(.,un(.))dx ≤ lim n→+∞ ∫ Ω |∇Tk(un)| p(.,un(.))dx ≤ C2k ( ||f||L1(Ω) + ||g||L1(∂Ω) ) (using (3.4)). hm(λ) → λ, as m → +∞ and m 7→ hm(λ) is increasing. Then, using Lebesgue convergence Theorem , we deduce from last inequality that ∫ Ω×RN |λ|π(x)dνkx(λ)dx ≤ C2k ( ||f||L1(Ω) + ||g||L1(∂Ω) ) . Hence, |λ|π(.) is integrable with respect to the measure νkx(λ)dx on R N × Ω. From (3.19), the last inequality and Jensen inequality, we get ∫ Ω |∇Tk(u)| π(x)dx = ∫ Ω ∣ ∣ ∣ ∣ ∫ RN λdνkx(λ) ∣ ∣ ∣ ∣ π(x) dx ≤ ∫ Ω×RN |λ|π(x)dνkxdx < ∞. Thus, ∇Tk(u) ∈ L π(.)(Ω). Moreover, ∫ Ω |Tk(u)| π(.) dx ≤ max ( k p+, k p− ) meas(Ω). Hence, Tk(u) ∈ Lπ(.)(Ω) and we conclude that Tk(u) ∈ W 1,π(.)(Ω). � Assertion 3. i) The sequence ( Φkn ) n∈N defined by Φkn := a(x, un, ∇Tk(un)) is equi-integrable on Ω. ii) The sequence ( Φkn ) n∈N weakly converges to Φk in L1(Ω) and we have Φk(x) = ∫ RN a(x, u, λ)dνkx(λ). (3.20) Proof. i) Using the growth assumption (A5) with variable exponent p(., un(.)) and relation (3.4), we deduce that (Φkn) is bounded in L π ′ n(.)(Ω), so, Lπ ′ n(.)− equi-integrable on Ω. Moreover, as π′n(.) > 1, we obtain |a(x, un, ∇Tk(un))| ≤ 1 + |a(x, un, ∇Tk(un))| π ′ n (.) . 104 Stanislas Ouaro & Noufou Sawadogo CUBO 22, 1 (2020) Thus, for all subset E ⊂ Ω, we have ∫ E |a(x, un, ∇Tk(un))|dx ≤ meas(E) + ∫ E |a(x, un, ∇Tk(un))| π ′ n(.)dx. Therefore, for meas(E) small enough, (Φkn) is equi-integrable on Ω. ii) Set Φ̃kn = a(x, u(x), ∇vn) with ∇vn = ∇Tk(un).χSn where Sn = { x ∈ Ω, |π(x) − πn(x)| < 1 2 } . Applying (A5) with variable exponent π(.) on a(x, u(x), ∇vn), we have for all subset E ⊂ Ω, ∫ E |a(x, u(x), ∇vn)|dx ≤ C ∫ E ( 1 + M(x) + |∇vn| π(.)−1 ) dx ≤ C ∫ E ( 1 + M(x))dx + ∫ E∩Sn |∇Tk(un)| π(.)−1dx. The first term of the right hand side of the last inequality is small for meas(E) small enough. For x ∈ Sn, π(x) < πn(x) + 1 2 , thus ∫ E∩Sn |∇Tk(un)| π(.)−1 dx ≤ ∫ E∩Sn ( 1 + |∇Tk(un)| πn(.)− 1 2 ) dx and ∫ Ω |∇Tk(un)| (πn(.)− 1 2 )(2πn(.)) ′ dx = ∫ Ω |∇Tk(un)| πn(.)dx < ∞, which is equivalent to saying |∇Tk(un)| πn(.)− 1 2 ∈ L(2πn(.)) ′ (Ω). Now, using Hölder type inequality, ∫ E∩Sn |∇Tk(un)| π(.)−1dx ≤ ∫ E ( 1 + |∇Tk(un)| πn(.)− 1 2 ) dx ≤ meas(E) + 2 ∣ ∣ ∣ ∣∇Tk(un) ∣ ∣ ∣ ∣ Lπn(.)(Ω) ||χE||L2πn(.)(Ω). (3.21) According to Proposition 1, ||χE||L2πn(.)(Ω) ≤ max { ( ρ2πn(.) ( χE )) 1 2p− , ( ρ2πn(.)(χE) ) 1 2p+ } = max { ( meas(E) ) 1 2p− , ( meas(E) ) 1 2p+ } . The right-hand side of (3.21) is uniformly small for meas(E) small, and the equi-integrability of Φ̃kn follows. Therefore, (up to a subsequence) Φ̃ k n weakly converges in L 1(Ω) to Φ̃k, as n → +∞. Now, we prove that Φ̃k = Φk; more precisely, we show that Φ̃kn − Φ k n strongly converges in L 1(Ω) to 0. Let β > 0, by (3.4), ∫ Ω |∇Tk(un)| πn(.)dx is uniformly bounded, which implies that ∫ Ω |∇Tk(un)|dx is finite, since ∫ Ω |∇Tk(un)|dx ≤ ∫ Ω (1 + |∇Tk(un)| πn(x))dx. By Chebyschev Inequality, we have meas({|∇Tk(un)| > L}) ≤ ∫ Ω |∇Tk(un)|dx L . CUBO 22, 1 (2020) Nonlinear elliptic p(u)− Laplacian problem with Fourier . . . 105 Therefore, sup n∈N meas({|∇Tk(un)| > L}) tends to 0 for L large enough. Since Φ̃ k n − Φ k n is equi- integrable, there exists δ = δ(β) such that for all A ⊂ Ω, meas(A) < δ and ∫ A |Φ̃kn − Φ k n|dx < β 4 . Therefore, if we choose L large enough, we get ∫ Ω |∇Tk(un)|dx L < δ, so meas({|∇Tk(un)| > L}) < δ. Hence, ∫ {|∇Tk(un)|>L} |Φ̃kn − Φ k n|dx < β 4 . By Lemma 2.2, we also have meas ({ x ∈ Ω; sup λ∈K |a(x, un(x), λ) − a(x, u(x), λ)| ≥ σ }) −→ 0, as n → +∞. Thus, by the above equi-integrability, for all σ > 0, there exists n0 = n0(σ, L) ∈ N such that for all n ≥ n0, ∫ { x∈Ω; sup|λ|≤L |a(x,un(x),λ)−a(x,u(x),λ)|≥σ } |Φ̃ k n − Φ k n|dx < β 4 . Using the definition of Φkn and Φ̃ k n, we have Φkn − Φ̃ k n = a(x, un(x), ∇Tk(un)) − a(x, u(x), ∇Tk(un)) on Sn. Now, we reason on Sn,L,σ := { x ∈ Ω; sup |λ|≤L |a(x, un(x), λ) − a(x, u(x), λ)| < σ, |∇Tk(un)| ≤ L } . We get ∫ Sn,L,σ |Φ̃kn − Φ k n|dx ≤ ∫ Sn,L,σ sup |λ|≤L |a(x, un(x), λ) − a(x, u(x), λ)|dx ≤ σmeas(Ω). We observe that ∫ Sn |Φ̃kn − Φ k n|dx = ∫ Sn∩Sn,L,σ |Φ̃kn − Φ k n|dx + ∫ Sn\Sn,L,σ |Φ̃kn − Φ k n|dx and Sn \ Sn,L,σ ⊂ { x ∈ Ω; sup |λ|≤L |a(x, un(x), λ) − a(x, u(x), λ)| ≥ σ } ∪ { |∇Tk(un)| > L } . Consequently, by choosing σ = σ(β) < β 4meas(Ω) , we get ∫ Sn |Φ̃kn − Φn|dx < β 4 + β 4 + β 4 = 3β 4 , 106 Stanislas Ouaro & Noufou Sawadogo CUBO 22, 1 (2020) for all n ≥ n0(σ, L). By Lemma 2.2, we also have meas({x ∈ Ω, |π(x) − πn(x)| ≥ 1 2 }) → 0 for n large enough; which means that meas(Ω \ Sn) converges to 0 for n large enough. Thus, ∫ Ω\Sn |Φ̃kn − Φ k n|dx = ∫ Ω\Sn |Φkn|dx ≤ β 4 . Therefore, for all β > 0 there exists n0 = n0(β) such that for all n ≥ n0, ∫ Ω |Φ̃kn − Φ k n|dx ≤ β. Hence, Φ̃kn − Φ k n strongly converges to 0 in L 1(Ω). We prove that Φk(x) = ∫ RN a(x, u(x), λ)dνkx(λ) a.e. x ∈ Ω and Φ k ∈ Lπ ′(.)(Ω). Notice that lim n→+∞ ∫ Ω |∇Tk(un)|(1 − χSn)dx = lim n→+∞ ∫ Ω\Sn |∇Tk(un)|dx = 0, since (∇Tk(un))n∈N is equi-integrable and meas(Ω \ Sn) converges to 0 for n large enough. Therefore, (∇Tk(un))n∈N and ∇Tk(un)χSn converge to the same Young measure ν k x(λ). Moreover, by applying Theorem 2.1 i) to the Carathéodory function F(x, (λ0, λ)) := a(x, λ0, λ), we infer that Φ̃(x) = Φ(x) = ∫ RN a(x, u(x), λ)dνkx(λ) a.e. x ∈ Ω. Using (A5), it follows that |a(x, u(x), λ)| π ′(.) ≤ C(M(x) + |λ|π(.)). Thus, with Jensen Inequality, it follows that ∫ Ω |Φk(x)|π ′(.)dx = ∫ Ω ∣ ∣ ∣ ∣ ∫ RN a(x, u(x), λ)dνkx(λ) ∣ ∣ ∣ ∣ π ′(.) dx ≤ ∫ Ω×RN ∣ ∣a(x, u(x), λ)|π ′(.)dνkx(λ)dx ≤ C ∫ Ω×RN ( M(x) + |λ|π(.) ) dνkx(λ)dx < ∞. Hence, Φk ∈ Lπ ′(.)(Ω). � Assertion 4 (a) For all k′ > k > 0, we have Φk = Φk ′ χ{|u| 0, ∫ Ω Φk.∇Tk(u)dx ≥ ∫ Ω×RN a(x, u(x), λ).λdνkx(λ)dx. (3.22) (c) The “div-curl” inequality holds: ∫ Ω×RN ( a(x, u(x), λ) − a(x, u(x), ∇Tk(u(x)) ) (λ − ∇Tk(u(x)))dν k x(λ)dx ≤ 0. (3.23) CUBO 22, 1 (2020) Nonlinear elliptic p(u)− Laplacian problem with Fourier . . . 107 (d) For all k > 0, Φk = a(x, u(x), ∇Tk(u)) for a.e. x ∈ Ω and ∇Tk(un) converges to ∇Tk(u) in measure on Ω, as n → +∞. Proof. (a) Let k′ > k > 0 and gkn := a(x, un, ∇Tk′(un))χ[|u| k > 0, then, we get hkn := a(x, un, ∇Tk′(un))χ[|un| 0. Moreover, un → u a.e. on Ω, then χ[|un| 0, we have ∫ Ω S ′ M (un)a(x, un, ∇TM (un)).∇ϕdx = ∫ {|∇ϕ|≤L} S ′ M (un)Φ M n .∇ϕdx + ∫ {|∇ϕ|>L} S′M (un)Φ M n .∇ϕdx. (3.30) For the first term of the right-hand side of (3.30), we have ∫ {|∇ϕ|≤L} S′M (un)Φ M n .∇ϕdx → ∫ {|∇ϕ|≤L} S′M (u)Φ M .∇ϕdx, as n → +∞. (3.31) Thanks ΦMn ⇀ Φ M in L1(Ω) and ∇ϕS′M (un)χ{|∇ϕ|≤L} → ∗ ∇ϕS′M (u)χ{|∇ϕ|≤L} in L ∞(Ω). Fur- thermore, the second term of the right hand-side of (3.30) converges to zero for L large enough, uniformly in n. Indeed, using Hölder type inequality and the fact that Lp+(Ω) →֒ Lπn(.)(Ω), we get ∣ ∣ ∣ ∣ ∫ {|∇ϕ|>L} ΦMn ∇ϕS ′ M (un)dx ∣ ∣ ∣ ∣ ≤ C||S′M ||L∞(R)||Φ M n ||Lπ′n(.)(Ω)||∇ϕχ{|∇ϕ|>L}||Lπn(.)(Ω) ≤ C ( p−, ||S ′ M ||L∞(R), meas(Ω) ) ||ΦMn ||Lπ′n(.)(Ω)||∇ϕ||L p+ (Ω)meas ( {|∇ϕ| > L} ) . From (A5), (3.4) and Proposition 2, we obtain ||ΦnM||Lπ′n(.)(Ω) < C. CUBO 22, 1 (2020) Nonlinear elliptic p(u)− Laplacian problem with Fourier . . . 109 Moreover, ϕ ∈ C∞(Ω) and C∞(Ω) is dense in the space W 1,p+(Ω). Then, by Proposition 2 and the fact that lim L→+∞ meas({|∇ϕ| > L}) = 0, we get meas ( {|∇ϕ| > L} ) ||ΦMn ||Lπ′n(.)(Ω)||∇ϕ||L p+ (Ω) → 0, as L → +∞. Hence, the second term of the right hand-side of (3.30) converges to zero, as L tends to infinity. Thus, as n → +∞ and L → +∞ in (3.30), we deduce (3.29). Let us consider the third term of left hand-side of (3.24), we obtain ∫ Ω |S′′M (un)|a(x, un, ∇TM(un)).∇TM (un)ϕdx ≤ C ∫ {|un| 0 fixed, Tk(u) ∈ W 1,π(.)(Ω) and the exponent π(.) verify (2.1). Therefore, C∞(Ω) is dense in W 1,π(.)(Ω), so, we replace ϕ by Tk(u). Now, for M > k, thanks to (a), we replace Φ M .∇Tk(u) by Φk.∇Tk(u) in (3.33). S′M converges a.e. to 1 on R, as M → +∞, then using the monotone convergence theorem in the first term of left hand-side of (3.33) and dominated convergence theorem in the other term of (3.33), we get ∫ Ω [ b(u)Tk(u) + Φ k .∇Tk(u)]dx + λ ∫ ∂Ω uTk(u)dσ = ∫ Ω fTk(u)dx + ∫ ∂Ω gTk(u)dσ. (3.34) 110 Stanislas Ouaro & Noufou Sawadogo CUBO 22, 1 (2020) The relation (3.7) is equivalent to ∫ Ω Tn(b(un))Tk(un)dx + ∫ Ω a(x, un, ∇Tk(un)).∇Tk(un)dx + ∫ ∂Ω λTn(un)Tk(un)dσ + ε ∫ Ω [ |un| p+−2unTk(un) + |∇un| p+−2∇un∇Tk(un) ] dx = ∫ Ω fnTk(un)dx + ∫ ∂Ω gnTk(un)dσ. (3.35) The sequences ( Tn(b(un))Tk(un) ) n∈N , ( Tn(un)Tk(un) ) n∈N are nonnegative and converge a.e. in Ω to b(u)Tk(u) and a.e. on ∂Ω to uTk(u). By Fatou’s Lemma, we have lim inf n→+∞ ∫ Ω Tn(b(un))Tk(un)dx ≥ ∫ Ω b(u)Tk(u)dx (3.36) and λ lim inf n→+∞ ∫ ∂Ω Tn(un)Tk(un)dx ≥ λ ∫ ∂Ω uTk(u)dσ. (3.37) Now, we consider the right hand side of (3.35). We have |fnTk(un)| ≤ k|f| ∈ L 1(Ω), fnTk(un) → fTk(u) a.e. in Ω and |gnTk(un)| ≤ k|g| ∈ L 1(∂Ω), gnTk(un) → gTk(u) a.e. on ∂Ω. Thus, by Lebesgue dominated convergence Theorem ∫ Ω fnTk(un)dx → ∫ Ω fTk(u)dx, as n → +∞ (3.38) and ∫ ∂Ω gnTk(un)dσ → ∫ ∂Ω gTk(u)dσ, as n → +∞. (3.39) Combining (3.36),(3.37), (3.38), (3.39) and using (3.35), we get lim inf n→+∞ ( ∫ Ω fnTk(un)dx + ∫ ∂Ω gnTk(un)dσ ) − ( ∫ Ω b(u)Tk(u)dx + λ ∫ ∂Ω uTk(u)dσ ) ≥ lim inf n→+∞ ( ∫ Ω fnTk(un)dx + ∫ ∂Ω gnTk(un)dσ − ∫ Ω Tn(b(un))Tk(un)dx − λ ∫ ∂Ω Tn(un)Tk(un)dσ ) , which is equivalent to ∫ Ω fTk(u)dx + ∫ ∂Ω gTk(u)dσ − ( ∫ Ω b(u)Tk(u)dx + λ ∫ ∂Ω uTk(u)dσ ) ≥ lim inf n→+∞ ∫ Ω a(x, un, ∇Tk(un))∇Tk(un)dx + ε ∫ Ω [ |un| p+−2unTk(un) + |∇un| p+−2∇un∇Tk(un) ] dx ≥ lim inf n→+∞ ∫ Ω a(x, un, ∇Tk(un)∇Tk(un)dx. Thus, by using the relation (3.34), we obtain ∫ Ω Φk∇Tk(u)dx ≥ lim inf n→+∞ ∫ Ω a(x, un, ∇Tk(un))∇Tk(un)dx. (3.40) CUBO 22, 1 (2020) Nonlinear elliptic p(u)− Laplacian problem with Fourier . . . 111 (c) From [1]-Lemma 2.1, m 7→ a(x, un, hm(∇Tk(un))).hm(∇Tk(un)) is increasing and converges to a(x, un, ∇Tk(un)).∇Tk(un) for m large enough. Thus, we deduce that a(x, un, hm(∇Tk(un))).hm(∇Tk(un)) ≤ a(x, un, ∇Tk(un)).∇Tk(un) = Φ k n.∇Tk(un). Therefore, using (b) and Theorem 2.1, we get ∫ Ω Φk.∇Tk(u)dx ≥ lim inf n→+∞ ∫ Ω Φkn.∇Tk(un)dx ≥ lim n→+∞ ∫ Ω a(x, un, hm(∇Tk(un))).hm(∇Tk(un))dx = ∫ Ω×RN a(x, u, hm(λ)).hm(λ)dν k x(λ)dx. (3.41) Using Lebesgue convergence Theorem in (3.41), we get for m large enough ∫ Ω Φk.∇Tk(u)dx ≥ ∫ Ω×RN a(x, u, λ).λdνkx(λ)dx. (3.42) We have ∫ Ω×RN (a(x, u(x), λ) − a(x, u(x), ∇Tk(u(x)))(λ − ∇Tk(u(x)))dν k x(λ)dx = ∫ Ω×RN a(x, u(x), λ).λdνkx(λ)dx − ∫ Ω×RN a(x, u(x), λ).∇Tk(u(x))dν k x(λ)dx − ∫ Ω×RN a(x, u(x), ∇Tk(u(x))).λdν k x(λ)dx + ∫ Ω×RN a(x, u(x), ∇Tk(u(x))).∇Tk(u(x))dν k x(λ)dx = ∫ Ω×RN a(x, u(x), λ).λdνkx(λ)dx − ∫ Ω ( ∫ RN a(x, u(x), λ)dνkx(λ) ) ∇Tk(u(x))dx − ∫ Ω a(x, u(x), ∇Tk(u(x))). ( ∫ RN λdνkx ) dx + ∫ Ω a(x, u(x), ∇Tk(u(x))).∇Tk(u(x)) ( ∫ RN dνkx ) dx = ∫ Ω×RN a(x, u(x), λ).λdνkx(λ)dx − ∫ Ω Φk.∇Tk(u(x))dx ≤ 0. We pass from the first equality to the second equality by using Fubini-Tonelli Theorem and from the second inequality to the third one by using (3.19), (3.20) and the fact that νx is probability measures on RN. Finally (3.42) give us the desired inequality. (d) Using (3.23) and the strict monotonicity assumption (A4), we deduce that ( a(x, u(x), λ) − a(x, u(x), ∇Tk(u(x)) )( λ − ∇Tk(u(x)) ) = 0 a.e. x ∈ Ω, λ ∈ RN. Thus, λ = ∇Tk(u(x)) a.e. x ∈ Ω with respect to the measure ν k x on R N. Therefore, the measure νkx reduces to the Dirac measure δ∇Tk(u(x)). Using (3.20), we obtain Φk = ∫ RN a(x, u(x), λ)dνkx(λ) = a(x, u(x), ∇Tk(u(x))) a.e. x ∈ Ω. 112 Stanislas Ouaro & Noufou Sawadogo CUBO 22, 1 (2020) Now, by using Theorem 2.1 (ii), we deduce that ∇Tk(un) converges in measure to ∇Tk(u). � Lemma 3.6. u is an entropy solution of (1.1). Proof of the Lemma 3.6. Let un be a weak solution of the problem (3.2). Then, by Assertion 4−(d), (∇Tk(un))n∈N converges to ∇Tk(u) in measure, thus (up to a subsequence still denoted (∇Tk(un))n∈N), (∇Tk(un))n∈N converges to ∇Tk(u) a.e. Ω. Moreover, we deduce from Lemma 3.4 that ∇Tk(un) is uniformly bounded in Lp−(Ω), so, p−−equi-integrable on Ω. Then, by using Vitali’s Theorem ∇Tk(un) → ∇Tk(u) in L p−(Ω), which implies that ∇Tk(un) → ∇Tk(u) in L 1(Ω). Furthermore, thanks to Assertion 2, u ∈ T 1,π(.)(Ω) and it follows from Lemma 3.5 that un → u a.e on ∂Ω. Therefore, u ∈ T 1,π(.) tr (Ω). Now, using Lemma 3.2, the fact that Tn(b(un)) → b(u) a.e. in Ω and un → u a.e. on ∂Ω, it follows from Fatou’s Lemma that ∫ Ω |b(u)| ≤ lim inf n→+∞ ∫ Ω ∣ ∣Tn(b(un)) ∣ ∣dx ≤ ||f||L1(Ω) + ||g||L1(∂Ω) and ∫ ∂Ω |u| ≤ lim inf n→+∞ ∫ ∂Ω ∣ ∣Tn(un) ∣ ∣dx ≤ 1 λ ( ||f||L1(Ω) + ||g||L1(∂Ω) ) . Hence, b(u) ∈ L1(Ω) and u ∈ L1(∂Ω). Let ϕ ∈ C∞(Ω), then we can choose Tk(un − ϕ) as a test function in (3.3) (C ∞(Ω) is dense in the space W 1,p+(Ω) and Tk(un − ϕ) ∈ L ∞(∂Ω)) to get ∫ Ω Tn(b(un))Tk(un − ϕ)dx + ∫ Ω a(x, un, ∇un).∇Tk(un − ϕ)dx + ∫ ∂Ω λTn(un)Tk(un − ϕ)dσ + ε ∫ Ω [ |∇un| p+−2∇un∇Tk(un − ϕ) + |un| p+−2unTk(un − ϕ) ] dx = ∫ Ω fnTk(un − ϕ)dx + ∫ ∂Ω gnTk(un − ϕ)dσ. (3.43) For the first term of the left hand side of (3.43), we have ∫ Ω Tn(b(un))Tk(un − ϕ)dx = ∫ Ω [ Tn(b(un)) − Tn(b(ϕ)) ] Tk(un − ϕ)dx + ∫ Ω Tn(b(ϕ))Tk(un − ϕ)dx. By Fatou’s Lemma, we infer lim inf n→+∞ ∫ Ω Tn(b(un))Tk(un − ϕ)dx ≥ ∫ Ω b(u)Tk(u − ϕ)dx, (3.44) CUBO 22, 1 (2020) Nonlinear elliptic p(u)− Laplacian problem with Fourier . . . 113 since, [ Tn(b(un)) − Tn(b(ϕ)) ] Tk(un − ϕ) → ( b(u − b(ϕ) ) Tk(u − ϕ) a.e. with [ Tn(b(un)) − Tn(b(ϕ)) ] Tk(un − ϕ) ≥ 0 and Tn(b(ϕ))Tk(un − ϕ) → b(ϕ)Tk(u − ϕ) in L 1(Ω). In the same manner lim inf n→+∞ λ ∫ ∂Ω Tn(un)Tk(un − ϕ)dσ ≥ λ ∫ ∂Ω uTk(u − ϕ)dσ. (3.45) For the fourth term of the left hand side of (3.43), we prove that lim n→+∞ ε ∫ Ω [ |∇un| p+−2∇un∇Tk(un − ϕ) + |un| p+−2unTk(un − ϕ) ] dx ≥ 0 as ε → 0. (3.46) Setting l = k + ||ϕ||L∞(Ω) we have, ε ∫ Ω |∇un| p+−2∇un∇Tk(un − ϕ)dx = ε ∫ {|un−ϕ|1} |∇ϕ|πn(.)dx ≤ meas(E) + ∫ E |∇ϕ|p+ dx, since |∇ϕ|p+ , M ∈ L1(Ω) and |∇Tl(un)| πn(.) is equi-integrable ( using density argument for C∞(Ω) and (3.4) ) . Then, we obtain lim meas(E)→0 ∫ E a(x, un, ∇Tl(un))∇ϕχ{|un−ϕ| N and Ω is a bounded open domain with Lipschitz boundary ∂Ω. Therefore, the inequality (3.59) holds true for ϕ ∈ W 1,π(.)(Ω) ∩ L∞(Ω). Hence, u is an entropy solution of (1.1). � Now, we state the uniqueness result of entropy solution. This result uses the same arguments as [2]-Theorem 2.8. Theorem 3.5. Assume that b is strictly increasing. Assume that a = a(x, z, η) satisfies (A3)−(A6) and M constant. Moreover, a satisfies: for all bounded subset K of R × RN, there exists a constant C(K) such that a.e. x ∈ Ω, for all (z, η), (z̃, η) ∈ K, |a(x, z, η) − a(x, z̃, η)| ≤ C(K)|z − z̃|. (3.60) Finally, suppose the following regularity property: for all f ∈ L∞(Ω) and g ∈ L∞(∂Ω) there exists an entropy solution of (1.1), which is Lipchitz continuous on Ω. (3.61) Then, for all f ∈ L1(Ω) and g ∈ L1(∂Ω) the problem (1.1) admits a unique entropy solution. CUBO 22, 1 (2020) Nonlinear elliptic p(u)− Laplacian problem with Fourier . . . 117 Remark 3.6. As in [2, Theorem 2.8], the condition (3.61) goes back to idea of [3]. Moreover, in the Theorem 3.5 the relation (3.60) is used to obtain the inequality (3.69) below. Proof. The proof of this theorem is done in two steps. Step 1. A priori estimates. Lemma 3.7. If v is an entropy solution of (1.1), there exists a positive constant C such that ρp(.,v(.)) ( |∇v|χF ) ≤ Ck, where F = {h − k < |v| < h}, h > k > 0. Proof. Let ϕ = Th−k(v) as test function in the entropy inequality (3.1), we get ∫ Ω a(x, v, ∇v).∇Tk(v − Th−k(v))dx + ∫ Ω b(v)Tk(v − Th−k(v))dx + λ ∫ ∂Ω vTk(v − Th−k(v))dσ ≤ ∫ Ω fTk(v − Th−k(v))dx + ∫ ∂Ω gTk(v − Th−k(v))dσ. Thus, ∫ {h−k<|v| h} ) ≤ ||f||L1(Ω) + ||g||L1(∂Ω) min ( b(h), |b(−h)| ) , ∀h ≥ 1. Proof. Let us take ϕ = 0 and k = h in entropy inequality (3.1). Since ∫ Ω a(x, u, ∇u).∇Th(u)dx + λ ∫ ∂Ω uTh(u)dσ ≥ 0, the relation (3.1) gives ∫ Ω b(u)Th(u)dx ≤ ∫ Ω fTh(u)dx + ∫ ∂Ω gTh(u)dσ. 118 Stanislas Ouaro & Noufou Sawadogo CUBO 22, 1 (2020) Then, ∫ {|u|≤h} b(u)Th(u)dx + ∫ {|u|>h} b(u)Th(u)dx ≤ h ( ||f||L1(Ω) + ||g||L1(∂Ω) ) , or ∫ {|u|>h} b(u)Th(u) h dx = ∫ {u>h} b(u)dx + ∫ {u<−h} −b(u)dx ≤ ( ||f||L1(Ω) + ||g||L1(∂Ω) ) . Therefore, ∫ {|u|>h} |b(u)|dx ≤ ||f||L1(Ω) + ||g||L1(∂Ω). Since b is nondecreasing, we deduce ∫ {|u|>h} min(b(h), |b(−h)|)dx ≤ ∫ {|u|>h} |b(u)| ≤ ||f||L1(Ω) + ||g||L1(∂Ω), ∀h ≥ 1. So, meas ( {|u| > h} ) ≤ ||f||L1(Ω) + ||g||L1(∂Ω) min ( b(h), |b(−h)| ) , ∀h ≥ 1. � Step 2. Uniqueness. The existence has already been proved. Now, we show the uniqueness. For more details see [2]- Proof of Theorem 2.8. Let u be a Lipschitz continuous entropy solution of (1.1) with f ∈ L∞(Ω), g ∈ L∞(∂Ω) and v be an entropy solution, with f̂ ∈ L1(Ω), ĝ ∈ L1(∂Ω). Since Ω is open bounded domain with smooth boundary ∂Ω, the space of Lipschitz functions C0,1(Ω) and W 1,∞(Ω) are homeomorphic and they can be identified. Therefore, u belongs to W 1,∞(Ω). Thus, for all h > 0, we can write the entropy inequality corresponding to the solution u, with Th(v) as a test function and to the solution v, with Th(u) as a test function. For all k > 0, we get      ∫ Ω a(x, u, ∇u).∇Tk(u − Th(v))dx + ∫ Ω b(u)Tk(u − Th(v))dx +λ ∫ ∂Ω uTk(u − Th(v))dσ ≤ ∫ Ω fTk(u − Th(v))dx + ∫ ∂Ω gTk(u − Th(v))dσ (3.62) and      ∫ Ω a(x, v, ∇v).∇Tk(v − Th(u))dx + ∫ Ω b(v)Tk(v − Th(u))dx +λ ∫ ∂Ω vTk(v − Th(u))dσ ≤ ∫ Ω f̂Tk(v − Th(u))dx + ∫ ∂Ω ĝTk(v − Th(u))dσ. (3.63) CUBO 22, 1 (2020) Nonlinear elliptic p(u)− Laplacian problem with Fourier . . . 119 Adding (3.62) and (3.63) we obtain                           ∫ Ω a(x, u, ∇u).∇Tk(u − Th(v))dx + ∫ Ω a(x, v, ∇v).∇Tk(v − Th(u))dx + ∫ Ω b(u)Tk(u − Th(v))dx + ∫ Ω b(v)Tk(v − Th(u))dx +λ ∫ ∂Ω uTk(u − Th(v))dσ + λ ∫ ∂Ω vTk(v − Th(u))dσ ≤ ∫ Ω [ fTk(u − Th(v)) + f̂Tk(v − Th(u)) ] dx + ∫ ∂Ω [ gTk(u − Th(v)) + ĝTk(v − Th(u)) ] dσ. (3.64) Set A = {0 < |u − v| < k, |v| ≤ h}; B = A ∩ {|u| ≤ h}; C = A ∩ {|u| > h} and A′ = {0 < |v − u| < k, |u| ≤ h}; B′ = A′ ∩ {|v| ≤ h}; C′ = A′ ∩ {|v| > h}. We start with the first integral in (3.64). We have ∫ {0<|u−Th(v)|h} a(x, u, ∇u).∇Tk(u − Th(v))dx = ∫ {0<|u−v|h} a(x, u, ∇u).∇udx ≥ ∫ A a(x, u, ∇u)∇(u − v)dx = ∫ B a(x, u, ∇u)∇(u − v)dx + ∫ C a(x, u, ∇u)∇(u − v)dx. Then, we get        ∫ {0<|u−Th(v)|