# Semidiscretization for a Doubly Nonlinear Parabolic Equation Related to the p(x)-Laplacian

Authors: Hamid El Bahja; Abderrahmane El Hachimi; Ali Alami Idrissi
Journal: International Journal of Differential Equations
 (2019)

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## Abstract

This paper studies a time discretization for a doubly nonlinear parabolic equation related to the p(x)-Laplacian by using Euler-forward scheme. We investigate existence, uniqueness, and stability questions and prove existence of the global compact attractor.


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## Body

## 1. Introduction

The investigation of the asymptotic behavior for nonlinear parabolic equations involving the so-called p-Laplacian operator has been addressed by several authors in the last decades, in both bounded and unbounded domains, with constant or variable exponents (see [1–8]). One way to treat this question is to analyze the existence and structure (regularity and finite or infinite dimensionality of the attractor generated by the solutions of the governed equation (see [9]). The existence of the global attractor for the related semigroup acting on the natural weak energy space Lp(Ω) has been proved in [7, 10, 11].In this paper, our goal is to study the time discretization for a doubly nonlinear parabolic equation associated with the p (x)-Laplacian, where in addition to usual questions of existence, uniqueness, and stability of the solutions, we will be concerned with the existence of absorbing sets and the global attractor as well. The problem under consideration is of the form(1)∂βu∂t-Δpxu+fx,t,u=0inΩ×0,∞,u=0on∂Ω×0,∞,βut=0=βu0inΩ.where Δp(x)u=div⁡(∇up(x)-2∇u), p∈C(Ω¯) with 1<p(x)<+∞, β is a nonlinearity of porous media type, f is a nonlinearity of reaction type, and Ω is an open bounded set of RN with smooth boundary.Existence results and qualitative properties concerning the solutions of the continuous problem (1) and more general problems have been obtained by many authors in the last decade. We cote the papers [1–6, 12] and the references therein.Our motivation to study problem (1) is the fact that it is considered in particular as a model of an important class of non-Newtonian fluids which are well known as electrorheological fluids (see [13]).This paper is organized as follows: In Section2, we give some preliminaries and notation. In Section 3, we discretize problem (1) by using Euler-forward scheme and obtain existence, uniqueness, and stability results.Finally, in Section4 we show the existence of absorbing sets in W01,p(x)(Ω)∩L∞(Ω), which in turn ensures the existence of a compact global attractor.


## 2. Preliminaries

We begin with a review of some basic results that will be needed in the subsequent sections. The known results are stated without proofs. We shall however provide references where the proofs can be found.We first introduce the spaceLp(.)(Ω) and W1,p(.)(Ω) and state some of their properties.LetΩ stand for a regular open bounded set of RN and p:Ω→]1,+∞[ be a measurable bounded function as a variable exponent. Denote (2)p-≔ess⁡infx∈Ω⁡pxandp+≔sup⁡infx∈Ω⁡px.We define the variable exponent Lebesgue space Lp(x)(Ω) by (3)LpxΩ=u;uis  a  mesurable  real-valued  function,ρpxu=∫Ωuxpxdx<∞,endowed with the Luxembourg norm (4)upx=inf⁡λ>0,∫Ωuxλpx≤1.The following results can be found in [14–17].Lemma 1.
Letp:Ω→]1,+∞[ be a measurable function with 1<p-≤p(x)≤p+<∞. Then, we have (5)min⁡ρpxu1/p-,ρpxu1/p+≤upx≤max⁡ρpxu1/p-,ρpxu1/p+,for all u∈Lp(x)(Ω).Proposition 2.
The space(Lp(x)(Ω),.p(x)) is a separable, uniform convex Banach space, and its conjugate space is Lq(x)(Ω), where 1/p(x)+1/q(x)=1. Moreover, for any u∈Lp(x)(Ω) and v∈Lq(x)(Ω), we have(6)∫Ωuvdx≤1p-+1q-upxvqx.LetW1,p(x)(Ω) denote the space of measurable functions u such that u and the distributional derivative ∇u are in Lp(x)(Ω). The norm (7)u1,px=upx+∇upxmakes W1,p(x)(Ω) a Banach space.Let(8)C+Ω-=px:Ω-→p-,p+⊂1,∞;pis  a  continous  function,We say that p∈C+(Ω) satisfies the log-Hölder condition in Ω if (9)∀x,y∈Ω,x-y<1,px-py≤ωx-y,where ω satisfies (10)lim⁡supτ→0+⁡ωτln⁡1τ<∞.It is well known that if p(x) satisfies the log-Hölder condition (9), then the space C∞(Ω) is dense in W1,px(Ω). Moreover, we can define the Sobolev space with zero boundary values, W01,p(x)(Ω) as the completion of C0∞(Ω), with respect to the norm .W1,p(x)(Ω).Let us recall the following versions of Poincaré’s inequality.Lemma 3.
Ifp:Ω→]1,+∞[ is continuous in Ω¯, then there exists a constant C such that (11)upx≤C∇upx,for all u∈W01,p(x)(Ω) and thus u1,p(x) and ∇up(x) are equivalent norms in W01,p(x)(Ω).Let us next consider the modular version of Poincaré’s inequality.Lemma 4.
Letp(x) be an element of L∞(Ω) and let u∈W1,p(x)(Ω). There exists a constant C depending only on Ω such that (12)ρpxu≤Cρpx∇u.


## 3. The Semidiscretized Problem: Existence, Uniqueness, and Stability

Letβ be a continuous increasing function with β(0)=0. For t∈R, we set (13)ψt=∫0tβτdτ.We consider the following Euler-forward scheme associated with (1):(14)βUn-τΔpxUn+τfx,nτ,Un=βUn-1inΩ,Un=0on∂Ω,βU0=βu0inΩ,where Nτ=T, with T being a fixed positive real, and 1≤n≤N. We shall be concerned with the following two cases: u0∈L∞(Ω) or u0∈L2(Ω).


### 3.1. Case 1:u0∈L∞(Ω)

We assume the following hypotheses:( H 1 ) the function β is an increasing and continuous from R to R such that β(u)≤Cuα-1 for any u∈R with 1≤α<p-.( H 2 ) for ξ∈R, the map (x,t)↦f(x,t,ξ) is measurable and, a.e. in Ω×R+, ξ↦f(x,t,ξ) is continuous. Furthermore, we assume that there exists C1>0, such that, for a.e. (x,t)∈Ω×R+, we have sign⁡(ξ).f(x,t,ξ)≥-C1.( H 3 ) there exists C2>0, such that, for almost (x,t)∈Ω×R+, ξ↦f(x,t,ξ)+C2β(ξ) is increasing.Lemma 5.
Assume(H1) and (H2). Then, for all n∈{0,..,N}, we have Un∈L∞(Ω).Proof.
To show thatU1∈L∞(Ω), we can write (17) as(15)-τΔpxU1=βu0-βU1-τfx,τ,U1,U1∈W01,px.Then, by (H1), (H2), and Theorem 4.1 of [18], we can conclude that U1∈L∞(Ω). Then, by a simple induction, we deduce that Un∈L∞(Ω) for all n=0,..,N.Theorem 6.
Assume(H1), (H2), and (H3). For n=0,…,N, there exists a unique solution Un of (14) in W01,p(x)(Ω)∩L∞(Ω) provided that 0<τ<1/C2.Proof.
We can write (14) as (16)-τΔpxUn=βUn-1-βUn-τfx,nτ,UnUn∈W01,pxΩ.By using (H1), (H2), and applying Theorem 4.3 of [19] and Lemma 5, we deduce the existence of at least one solution Un∈W01,p(x)(Ω)∩L∞(Ω) for n=1,…,N.
Let us now prove the uniqueness. For simplicity, we set(17)ω=Un,f¯x,ω=fx,nτ,Un,andgx=βUn-1.Then, problem (14) reads(18)-τΔpxω+τf¯x,ω+βω=gx,ω∈W01,px.If ω1 and ω2 are two solutions of (14), then(19)-τΔpxω1+τΔpxω2+τf¯x,ω1-f¯x,ω2+βω1-βω2=0Multiplying (19) by ω1-ω2 and integrating over Ω give (20)-τΔpxω1+τΔpxω2,ω1-ω2+τ∫Ωf¯x,ω1-f¯x,ω2ω1-ω2dx+∫Ωβω1-βω2ω1-ω2dx=0,where .,. denotes the pairing between W1,px(Ω) and W-1,qx(Ω).
Then, applying(H3) yields(21)∫Ωf¯x,ω1-f¯x,ω2ω1-ω2dx≥-C2∫Ωβω1-βω2ω1-ω2dx.Now by using (21) and the monotonicity of the p(x)-Laplacian operator, (20) reduces to(22)1-τC2∫Ωβω1-βω2ω1-ω2dx≤0Then by (H1), we get ω1=ω2 for τ<1/C2.Theorem 7.
Assume(H1) and (H2). Then, there exists a constant C(T,u0)>0, depending on T,u0,β, and Ω, but not on N, such that, for all n=1,..,N,
(i)(23)Un∞≤CT,u0,
(ii)(24)∫Ωψ∗βUndx+τ∑k=1nUk1,pxα≤CT,u0,whereαdependseitheronp-orp+,
(iii)(25)∑k=1nβUk-βUk-122≤CT,u0.Proof.
(i) From Lemma5, we have Un∈L∞(Ω). Then, multiplying (14) by βUnkβ(Un) and integrating over Ω, we get(26)∫ΩβUnk+2dx-τ∫ΩΔpxUnβUnkβUndx+τ∫ΩβUnkβUnfx,nτ,Undx=∫ΩβUnkβUnβUn-1dx.Since β(0)=0 and β and -Δp(x) are monotone, then we have (27)-τ∫ΩΔpxUnβUnkβUndx≥0.Therefore, we obtain(28)βUnk+2k+2≤βUnk+2k+1βUn-1k+2+CτβUnk+2k+1.Hence,(29)βUnk+2≤βUn-1k+2+CτβUnk+2k+1.By simple induction, we get(30)βUnk+2≤βU0k+2+NCτ.Finally, as k→∞, we obtain (23).
(ii) In order to prove (24), we multiply (14) by Uk (with k instead of n). By using (H2), we get (31)∫ΩβUk-βUk-1Ukdx+τρpx∇Uk≤τC1Uk1.Thanks to the properties of the Legendre transformation, we get (32)∫Ωψ∗βUkdx-∫Ωψ∗βUk-1dx≤∫ΩβUk-βUk-1Ukdx.Then, we have(33)∫Ωψ∗βUkdx-∫Ωψ∗βUk-1dx+τρpx∇Uk≤τC1UkL1Ω.Finally, after summation of (33) from k=1 to n, we deduce that(34)∫Ωψ∗βUndx+τ∑k=1nmin⁡Uk1,pxp+,Uk1,pxp-≤τC1∑k=1nUkL1Ω+∫Ωψ∗βu0dx.We set (35)Ukpxα=min⁡Uk1,pxp+,Uk1,pxp-.Then, the continuity of β and the use of Lemma 5 allow us to conclude to the proof of point (24).
(iii) To prove point (25), we multiply the first equation of (17) by β(Uk). By using (H2), we get(36)∫ΩβUk-βUk-1βUkdx-τ∫ΩΔpxUkβUkdx≤C1τβUkL1Ω.
With the aid of the elementary identity,(37)2aa-b=a2-b2+a-b2,for any reals a and b, we get from (36) that(38)βUkL2Ω2-βUk-1L2Ω2+βUk-βUk-1L2Ω2≤CτβUkL1Ω.Now, we take the sum of (38) from k=1 to n to obtain(39)βUnL2Ω2+∑k=1nβUk-βUk-1L2Ω2≤CτβUkL1Ω+βu0L2Ω2.Thus, by (H1) and Lemma 5 we deduce (25).Lemma 8.
For allu,v∈W01,p(x)(Ω), there exists a positive constant α depending either on p+ or p- such that, for 1<p(x)<2, we have(40)ρpx∇u-v2/α≤C-Δpxu+Δpxv,u-v.Proof.
If1<p(x)<2, for any x∈Ω, then we have the following inequality for any ξ,η∈RN: (41)ξ+η2-pxξpx-2ξ-ηpx-2ηξ-η≥δξ-η2By setting ξ=∇u and η=∇v and integrating over Ω, we get (42)δp+/2∫Ω∇u-∇vpxdx≤∫Ω∇upx-2∇u-∇vpx-2∇v.∇u-vpx/2×∇u+∇v2-pxpx/2dx.Then, by Holder’s inequality we get (43)δp+/2∫Ω∇u-∇vpxdx≤∇upx-2∇u-∇vpx-2∇v.∇u-vpx/22/px×∇u+∇v2-pxpx/22/2-px.Let (44)I=∫Ω∇upx-2∇u-∇vpx-2∇v.∇u-vdxandJ=∫Ω∇u+∇vpxdx,and α and ξ be such that (45)Iα/2=max⁡Ip-/2,Ip+/2
and(46)Jξ=max⁡J2-p-/2,J2-p+/2.Then, we get (47)δp+/2∫Ω∇u-∇vpxdx≤2ξp+Iα/2Jξ.Therefore, we have (48)δp+/2∫Ω∇u-∇vpxdx≤2ξp+-Δpxu+Δpxv,u-vα/2ρpx∇u+ρpx∇vξ.Hence, by (24) of Theorem 7 we get the desired result.Lemma 9.
Assumep(x)≥2. Then, for all u,v∈W01,p(x)(Ω), we have(49)12p-ρpx∇u-v≤-Δpxu+Δpxv,u-v.Proof.
Asp(x)≥2, for any x∈Ω, then we have the following inequality for any ξ,η∈RN: (50)ξpx-2ξ-ηpx-2ηξ-η≥12p-ξ-ηpxBy setting ξ=∇u and η=∇v and integrating over Ω, we get (51)12p-∫Ω∇u-∇vpxdx≤∫Ω∇upx-2∇u-∇vpx-2∇v∇u-vdxHence(52)12p-ρpx∇u-v≤-Δpxu+Δpxv,u-v.We can also derive a uniqueness result for problem (17) if we replace (H3) by the following hypothesis:( H 4 ) for all M>0, there exists CM>0 such that, if ξ+ξ′≤M, then (53)ft,x,ξ-ft,x,ξ′θ≤CMβξ-βξ′ξ-ξ′,where (54)θ=σ′for1<px<2,p′-forpx≥2,for  allx∈Ω,with σ′ being a positive constant to be prescribed below.Proposition 10.
Assume(H1), (H2), and (H4). Then, problem (14) has a unique solution for all 0<τ<η, where η is a prescribed constant.Proof.
Letω1 and ω2 be two solutions of (14).
First case: suppose that1<p(x)<2, for all x∈Ω. Then, from (20) and by using Lemma 8 and Holder’s inequality, we get(55)τCρpx∇ω1-ω22/α+∫Ωβω1-βω2ω1-ω2dx≤τf¯x,ω1-f¯x,ω2L∞Ωω1-ω2L1Ω.Let λ be such that (56)ρpxω1-ω21/λ=max⁡ρpxω1-ω21/p-,ρpxω1-ω21/p+,and(57)σ=2λα,1σ+1σ′=1.Then, by (H1), (H2), and (H4), Lemma 4, and Young’s inequality, we get(58)τCρpx∇ω1-ω22/α≤1σ′-1∫Ωβω1-βω2ω1-ω2dx+C′τσρpx∇ω1-ω22/α.Therefore, for 0<τ<C/C′1/σ-1, we get ω1=ω2.
Second case: suppose thatp(x)≥2, for all x∈Ω. From (20) and by using Lemma 9 and Young’s inequality, we get(59)τ12p-ρpx∇ω1-ω2+∫Ωβω1-βω2ω1-ω2dx≤1p′-CM∫Ωf¯x,ω1-f¯x,ω2p′xdx+τp+CMp+/p′-p-∫Ωω1-ω2pxdx.
Then, by using(H1) and (H4) we get (60)τ12p-ρpx∇ω1-ω2≤1p′--1∫Ωβω1-βω2ω1-ω2dx+τp+CMp+/p′-p-∫Ωω1-ω2pxdx.Thus, from Lemma 4 we get (61)τ12p-ρpx∇ω1-ω2≤τp+CMp+/p′-p-ρpx∇ω1-ω2.Hence, when 0<τ<(1/2)p-p-/C.CMp+/p′-1/p+-1 we have ω1=ω2.


### 3.2. Case 2:u0∈L2(Ω)

Theorem 11.
Assume that(H1), (H2), and (H3) hold true. Then, for n=0,…,N, there exists a unique solution Un of (14) in W01,p(x)(Ω) provided that 0<τ<1/C where C is some positive constant.Proof.
The proofs of existence and uniqueness are the same as those of Theorem6. Therefore, we omit them.Now, we consider the following assumption:( H 5 ) for any ξ∈R, the map (x,t)↦f(x,t,ξ) is measurable and, a.e. in Ω×R+, ξ↦f(x,t,ξ) is continuous. Furthermore, we assume that there exist r∈C+(Ω) with r(x)>sup⁡(2,p(x)) and positive constants C5 and C6 such that (62)sign⁡ξfx,t,ξ≥C5ξrx-1-C6.Then, we have the following stability theorem.Theorem 12.
Assume that(H1) and (H5) are fulfilled. Then, there exists a constant C(T,u0)>0 such that, for all n=1,..,N,(63)∫Ωψ∗βUndx+τ∑k=1nUk1,pxα+Cτ∑k=1nUkqxα′≤CT,u0,(64)max1≤k≤n⁡∑k=1nβUk22+βUk-βUk-122≤CT,u0,where α and α′ are two constants each depending either on p+ or on p-.Proof.
Since the proof is nearly the same as that of Theorem7, we just sketch it.
The argument that allowed us to get (34), with (H5), allows also us to write(65)∫Ωψ∗βUndx+τ∑k=1Nρpx∇Uk+τ∑k=1NρrxUk≤τC6∑k=1NUkL1Ω+∫Ωψ∗βu0dx,By using Lemmas 4 and 5, (H1) and Young’s inequality, we get that for all η>0 there exists Cη(T,u0)>0 such that(66)∫Ωψ∗βUndx+τ∑k=1nρpx∇Uk+τ∑k=1nρrxUk≤ητ∑k=1Nρpx∇Uk+CηT,u0.Since ψ∗(β(u)) is positive then, for a suitable choice of η, we infer from (66) that(67)τ∑k=1nρpxUk≤C~ηT,u0.By taking α and α′ such that (68)Uk1,pxα=min⁡Uk1,pxp+,Uk1,pxp-,Ukrxα′=min⁡Ukrxr+,Ukrxq-and using (66) and (67), we deduce that (69)∫Ωψ∗βUndx+τ∑k=1nUk1,pxα+Cτ∑k=1nUkrxα′≤CT,u0.
As in (39), by using (H5), we get (70)βUnL2Ω2+∑k=1nβUk-βUk-1L2Ω2≤C1τβUkL1Ω+βu0L2Ω2.Hence, by (H1), (67), and Lemma 5, we deduce (64).


## 3.1. Case 1:u0∈L∞(Ω)

We assume the following hypotheses:( H 1 ) the function β is an increasing and continuous from R to R such that β(u)≤Cuα-1 for any u∈R with 1≤α<p-.( H 2 ) for ξ∈R, the map (x,t)↦f(x,t,ξ) is measurable and, a.e. in Ω×R+, ξ↦f(x,t,ξ) is continuous. Furthermore, we assume that there exists C1>0, such that, for a.e. (x,t)∈Ω×R+, we have sign⁡(ξ).f(x,t,ξ)≥-C1.( H 3 ) there exists C2>0, such that, for almost (x,t)∈Ω×R+, ξ↦f(x,t,ξ)+C2β(ξ) is increasing.Lemma 5.
Assume(H1) and (H2). Then, for all n∈{0,..,N}, we have Un∈L∞(Ω).Proof.
To show thatU1∈L∞(Ω), we can write (17) as(15)-τΔpxU1=βu0-βU1-τfx,τ,U1,U1∈W01,px.Then, by (H1), (H2), and Theorem 4.1 of [18], we can conclude that U1∈L∞(Ω). Then, by a simple induction, we deduce that Un∈L∞(Ω) for all n=0,..,N.Theorem 6.
Assume(H1), (H2), and (H3). For n=0,…,N, there exists a unique solution Un of (14) in W01,p(x)(Ω)∩L∞(Ω) provided that 0<τ<1/C2.Proof.
We can write (14) as (16)-τΔpxUn=βUn-1-βUn-τfx,nτ,UnUn∈W01,pxΩ.By using (H1), (H2), and applying Theorem 4.3 of [19] and Lemma 5, we deduce the existence of at least one solution Un∈W01,p(x)(Ω)∩L∞(Ω) for n=1,…,N.
Let us now prove the uniqueness. For simplicity, we set(17)ω=Un,f¯x,ω=fx,nτ,Un,andgx=βUn-1.Then, problem (14) reads(18)-τΔpxω+τf¯x,ω+βω=gx,ω∈W01,px.If ω1 and ω2 are two solutions of (14), then(19)-τΔpxω1+τΔpxω2+τf¯x,ω1-f¯x,ω2+βω1-βω2=0Multiplying (19) by ω1-ω2 and integrating over Ω give (20)-τΔpxω1+τΔpxω2,ω1-ω2+τ∫Ωf¯x,ω1-f¯x,ω2ω1-ω2dx+∫Ωβω1-βω2ω1-ω2dx=0,where .,. denotes the pairing between W1,px(Ω) and W-1,qx(Ω).
Then, applying(H3) yields(21)∫Ωf¯x,ω1-f¯x,ω2ω1-ω2dx≥-C2∫Ωβω1-βω2ω1-ω2dx.Now by using (21) and the monotonicity of the p(x)-Laplacian operator, (20) reduces to(22)1-τC2∫Ωβω1-βω2ω1-ω2dx≤0Then by (H1), we get ω1=ω2 for τ<1/C2.Theorem 7.
Assume(H1) and (H2). Then, there exists a constant C(T,u0)>0, depending on T,u0,β, and Ω, but not on N, such that, for all n=1,..,N,
(i)(23)Un∞≤CT,u0,
(ii)(24)∫Ωψ∗βUndx+τ∑k=1nUk1,pxα≤CT,u0,whereαdependseitheronp-orp+,
(iii)(25)∑k=1nβUk-βUk-122≤CT,u0.Proof.
(i) From Lemma5, we have Un∈L∞(Ω). Then, multiplying (14) by βUnkβ(Un) and integrating over Ω, we get(26)∫ΩβUnk+2dx-τ∫ΩΔpxUnβUnkβUndx+τ∫ΩβUnkβUnfx,nτ,Undx=∫ΩβUnkβUnβUn-1dx.Since β(0)=0 and β and -Δp(x) are monotone, then we have (27)-τ∫ΩΔpxUnβUnkβUndx≥0.Therefore, we obtain(28)βUnk+2k+2≤βUnk+2k+1βUn-1k+2+CτβUnk+2k+1.Hence,(29)βUnk+2≤βUn-1k+2+CτβUnk+2k+1.By simple induction, we get(30)βUnk+2≤βU0k+2+NCτ.Finally, as k→∞, we obtain (23).
(ii) In order to prove (24), we multiply (14) by Uk (with k instead of n). By using (H2), we get (31)∫ΩβUk-βUk-1Ukdx+τρpx∇Uk≤τC1Uk1.Thanks to the properties of the Legendre transformation, we get (32)∫Ωψ∗βUkdx-∫Ωψ∗βUk-1dx≤∫ΩβUk-βUk-1Ukdx.Then, we have(33)∫Ωψ∗βUkdx-∫Ωψ∗βUk-1dx+τρpx∇Uk≤τC1UkL1Ω.Finally, after summation of (33) from k=1 to n, we deduce that(34)∫Ωψ∗βUndx+τ∑k=1nmin⁡Uk1,pxp+,Uk1,pxp-≤τC1∑k=1nUkL1Ω+∫Ωψ∗βu0dx.We set (35)Ukpxα=min⁡Uk1,pxp+,Uk1,pxp-.Then, the continuity of β and the use of Lemma 5 allow us to conclude to the proof of point (24).
(iii) To prove point (25), we multiply the first equation of (17) by β(Uk). By using (H2), we get(36)∫ΩβUk-βUk-1βUkdx-τ∫ΩΔpxUkβUkdx≤C1τβUkL1Ω.
With the aid of the elementary identity,(37)2aa-b=a2-b2+a-b2,for any reals a and b, we get from (36) that(38)βUkL2Ω2-βUk-1L2Ω2+βUk-βUk-1L2Ω2≤CτβUkL1Ω.Now, we take the sum of (38) from k=1 to n to obtain(39)βUnL2Ω2+∑k=1nβUk-βUk-1L2Ω2≤CτβUkL1Ω+βu0L2Ω2.Thus, by (H1) and Lemma 5 we deduce (25).Lemma 8.
For allu,v∈W01,p(x)(Ω), there exists a positive constant α depending either on p+ or p- such that, for 1<p(x)<2, we have(40)ρpx∇u-v2/α≤C-Δpxu+Δpxv,u-v.Proof.
If1<p(x)<2, for any x∈Ω, then we have the following inequality for any ξ,η∈RN: (41)ξ+η2-pxξpx-2ξ-ηpx-2ηξ-η≥δξ-η2By setting ξ=∇u and η=∇v and integrating over Ω, we get (42)δp+/2∫Ω∇u-∇vpxdx≤∫Ω∇upx-2∇u-∇vpx-2∇v.∇u-vpx/2×∇u+∇v2-pxpx/2dx.Then, by Holder’s inequality we get (43)δp+/2∫Ω∇u-∇vpxdx≤∇upx-2∇u-∇vpx-2∇v.∇u-vpx/22/px×∇u+∇v2-pxpx/22/2-px.Let (44)I=∫Ω∇upx-2∇u-∇vpx-2∇v.∇u-vdxandJ=∫Ω∇u+∇vpxdx,and α and ξ be such that (45)Iα/2=max⁡Ip-/2,Ip+/2
and(46)Jξ=max⁡J2-p-/2,J2-p+/2.Then, we get (47)δp+/2∫Ω∇u-∇vpxdx≤2ξp+Iα/2Jξ.Therefore, we have (48)δp+/2∫Ω∇u-∇vpxdx≤2ξp+-Δpxu+Δpxv,u-vα/2ρpx∇u+ρpx∇vξ.Hence, by (24) of Theorem 7 we get the desired result.Lemma 9.
Assumep(x)≥2. Then, for all u,v∈W01,p(x)(Ω), we have(49)12p-ρpx∇u-v≤-Δpxu+Δpxv,u-v.Proof.
Asp(x)≥2, for any x∈Ω, then we have the following inequality for any ξ,η∈RN: (50)ξpx-2ξ-ηpx-2ηξ-η≥12p-ξ-ηpxBy setting ξ=∇u and η=∇v and integrating over Ω, we get (51)12p-∫Ω∇u-∇vpxdx≤∫Ω∇upx-2∇u-∇vpx-2∇v∇u-vdxHence(52)12p-ρpx∇u-v≤-Δpxu+Δpxv,u-v.We can also derive a uniqueness result for problem (17) if we replace (H3) by the following hypothesis:( H 4 ) for all M>0, there exists CM>0 such that, if ξ+ξ′≤M, then (53)ft,x,ξ-ft,x,ξ′θ≤CMβξ-βξ′ξ-ξ′,where (54)θ=σ′for1<px<2,p′-forpx≥2,for  allx∈Ω,with σ′ being a positive constant to be prescribed below.Proposition 10.
Assume(H1), (H2), and (H4). Then, problem (14) has a unique solution for all 0<τ<η, where η is a prescribed constant.Proof.
Letω1 and ω2 be two solutions of (14).
First case: suppose that1<p(x)<2, for all x∈Ω. Then, from (20) and by using Lemma 8 and Holder’s inequality, we get(55)τCρpx∇ω1-ω22/α+∫Ωβω1-βω2ω1-ω2dx≤τf¯x,ω1-f¯x,ω2L∞Ωω1-ω2L1Ω.Let λ be such that (56)ρpxω1-ω21/λ=max⁡ρpxω1-ω21/p-,ρpxω1-ω21/p+,and(57)σ=2λα,1σ+1σ′=1.Then, by (H1), (H2), and (H4), Lemma 4, and Young’s inequality, we get(58)τCρpx∇ω1-ω22/α≤1σ′-1∫Ωβω1-βω2ω1-ω2dx+C′τσρpx∇ω1-ω22/α.Therefore, for 0<τ<C/C′1/σ-1, we get ω1=ω2.
Second case: suppose thatp(x)≥2, for all x∈Ω. From (20) and by using Lemma 9 and Young’s inequality, we get(59)τ12p-ρpx∇ω1-ω2+∫Ωβω1-βω2ω1-ω2dx≤1p′-CM∫Ωf¯x,ω1-f¯x,ω2p′xdx+τp+CMp+/p′-p-∫Ωω1-ω2pxdx.
Then, by using(H1) and (H4) we get (60)τ12p-ρpx∇ω1-ω2≤1p′--1∫Ωβω1-βω2ω1-ω2dx+τp+CMp+/p′-p-∫Ωω1-ω2pxdx.Thus, from Lemma 4 we get (61)τ12p-ρpx∇ω1-ω2≤τp+CMp+/p′-p-ρpx∇ω1-ω2.Hence, when 0<τ<(1/2)p-p-/C.CMp+/p′-1/p+-1 we have ω1=ω2.


## 3.2. Case 2:u0∈L2(Ω)

Theorem 11.
Assume that(H1), (H2), and (H3) hold true. Then, for n=0,…,N, there exists a unique solution Un of (14) in W01,p(x)(Ω) provided that 0<τ<1/C where C is some positive constant.Proof.
The proofs of existence and uniqueness are the same as those of Theorem6. Therefore, we omit them.Now, we consider the following assumption:( H 5 ) for any ξ∈R, the map (x,t)↦f(x,t,ξ) is measurable and, a.e. in Ω×R+, ξ↦f(x,t,ξ) is continuous. Furthermore, we assume that there exist r∈C+(Ω) with r(x)>sup⁡(2,p(x)) and positive constants C5 and C6 such that (62)sign⁡ξfx,t,ξ≥C5ξrx-1-C6.Then, we have the following stability theorem.Theorem 12.
Assume that(H1) and (H5) are fulfilled. Then, there exists a constant C(T,u0)>0 such that, for all n=1,..,N,(63)∫Ωψ∗βUndx+τ∑k=1nUk1,pxα+Cτ∑k=1nUkqxα′≤CT,u0,(64)max1≤k≤n⁡∑k=1nβUk22+βUk-βUk-122≤CT,u0,where α and α′ are two constants each depending either on p+ or on p-.Proof.
Since the proof is nearly the same as that of Theorem7, we just sketch it.
The argument that allowed us to get (34), with (H5), allows also us to write(65)∫Ωψ∗βUndx+τ∑k=1Nρpx∇Uk+τ∑k=1NρrxUk≤τC6∑k=1NUkL1Ω+∫Ωψ∗βu0dx,By using Lemmas 4 and 5, (H1) and Young’s inequality, we get that for all η>0 there exists Cη(T,u0)>0 such that(66)∫Ωψ∗βUndx+τ∑k=1nρpx∇Uk+τ∑k=1nρrxUk≤ητ∑k=1Nρpx∇Uk+CηT,u0.Since ψ∗(β(u)) is positive then, for a suitable choice of η, we infer from (66) that(67)τ∑k=1nρpxUk≤C~ηT,u0.By taking α and α′ such that (68)Uk1,pxα=min⁡Uk1,pxp+,Uk1,pxp-,Ukrxα′=min⁡Ukrxr+,Ukrxq-and using (66) and (67), we deduce that (69)∫Ωψ∗βUndx+τ∑k=1nUk1,pxα+Cτ∑k=1nUkrxα′≤CT,u0.
As in (39), by using (H5), we get (70)βUnL2Ω2+∑k=1nβUk-βUk-1L2Ω2≤C1τβUkL1Ω+βu0L2Ω2.Hence, by (H1), (67), and Lemma 5, we deduce (64).


## 4. Absorbing Sets inW01,p(x)(Ω): Existence of the Attractor

In this section we consider the following problems: for all integern>0(71)βUn-τΔpxUn+τfx,nτ,Un=βUn-1inΩ,Un=0in∂Ω,with U0=u0 and τ fixed such that 0<τ<τ2 where τ2=min⁡(1,1/C2).We assume that(H1), (H2), and (H3) hold true in all the remaining section.The result of Theorem6 on the existence and uniqueness of the solution of (14) allows us to define a map Sτ on L∞(Ω)∩W1,p(x)(Ω) by setting (72)SτUn-1=Un.Since Sτ is continuous, we have (73)SτnU0=Un.The existence of an absorbing set in W01,p(x)(Ω)∩L∞(Ω) allows us to prove the existence of a global compact attractor (see [9]). This will be done next in Theorem 14.Proposition 13.
Ifτ satisfies τ<1/C2 then there is an absorbing set B in W01,p(x)(Ω)∩L∞(Ω). Namely, for any u0∈L∞(Ω) there exists n(τ) such that (74)UnL∞Ω+Un1,px≤C,∀n≥nτ.Proof.
We multiply (14) by Δn=(Un-Un-1). We obtain(75)βUn-βUn-1τ,Un-Un-1+∇Un,∇Δn+∫Ωfx,nτ,UnΔndx=0.Let us denote (76)Fβu=∫0ufx,nτ,ω+C2βω.By (H3), Fβ(u) is a convex function and hence satisfies the standard inequality (77)Fβ′uu-v≥Fβu-Fβv,Consequently, (78)fx,nτ,Un,Δn=fx,nτ,Un+C2βUn,Δn-C2βUn,Δn≥∫ΩFβUn-FβUn-1dx-C2βUn,Δn.Now, by using (H1), we get that ψ(u) is a convex function and hence we have (79)ψ′vu-v≥ψu-ψv.Thus, we obtain (80)∫ΩβUnUn-Un-1dx=∫ΩβUn-βUn-1Un-Un-1dx+∫ΩβUn-1Un-Un-1dx≤∫ΩβUn-βUn-1Un-Un-1dx+∫ΩψUn-ψUn-1dx.
The following inequality holds, for anya and b in RN: (81)1p+apx-1p-bpx≤apx-2a.a-b,
By settinga=∇Un and b=∇Un-1 and integrating over Ω, we get (82)1p+ρpxUn-1p-ρpxUn-1≤∫Ω∇Unpx-2∇Un∇Un-∇Un-1.Now, since τ<1/C2, then from (75), we deduce that (83)1p+ρpx∇Un+∫ΩFβUndx≤1p-ρpx∇Un-1+∫ΩFβUn-1dx+C2∫ΩψUn-ψUn-1dx.On the other hand, by writing (84)∫ΩFβudx=∫ΩFudx+C2∫Ωψudx,where F(u)=∫0uf(x,t,ω)dω, we have(85)1p+ρpx∇Un+∫ΩFUndx≤1p-ρpx∇Un-1+∫ΩFUn-1dx.
Denote the left hand side of (85) by yn. By using (H1) and relations (23) and (24) of Theorem 7 and taking Nτ=1, we deduce that there exists nτ such that (86)τ∑n=n0n0+Nyn≤a1foralln0≥nτThen, by applying the discrete version of the uniform Gronwall Lemma (see Lemma 7.5 of [11]) with hn=0, we obtain(87)1p+ρpx∇Un+∫ΩFUndx≤Cforalln≥nτ.Thus by Lemma 5, we deduce that(88)Un1,px≤Cforalln≥nτ,Therefore, from (88) and Theorem 7, we conclude to the desired relation (89)UnL∞Ω+Un1,px≤C,∀n≥nτ.Now we are able to state our result on the existence of a compact attractor.Theorem 14.
Suppose thatf(x,t,ξ)=f(x,ξ). Then, for u0∈L∞(Ω), the discretized problem (71) has an associated semigroup solution Sτ that maps L∞(Ω) into L∞(Ω)∩W01,p(x)(Ω). This semigroup has a compact attractor Aτ which is bounded in L∞(Ω)∩W01,p(x)(Ω).Proof.
The nonlinear mapSτ defines a semigroup from L∞(Ω) into L∞(Ω)∩W01,p(x)(Ω). By Proposition 13, the existence of an absorbing ball Bτ in L∞(Ω)∩W01,p(x)(Ω) is guaranteed.
We define theω-limit set of Bτ as (90)Aτ=ωBτ=⋂n≥0⋃m≥nSτmBτ¯.Then, by the results of Temam (see [9]), Aτ=ω(Bτ) is a compact attractor which attracts all bounded sets of L∞(Ω), which means that, for all u0∈L∞(Ω), we have (91)distAτ,Sτnu0↦0asn→+∞.



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