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 that U1∈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,and gx=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, where α depends either on p- or p+,
(iii)(25)∑k=1nβUk-βUk-122≤CT,u0.
Proof. (i) From Lemma 5, 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-1Uk dx+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=1nminUk1,pxp+,Uk1,pxp-≤τC1∑k=1nUkL1Ω+∫Ωψ∗βu0dx.We set (35)Ukpxα=minUk1,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 all u,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. If 1<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-∇vpx dx≤∫Ω∇upx-2∇u-∇vpx-2∇v.∇u-vpx/2×∇u+∇v2-pxpx/2 dx.Then, by Holder’s inequality we get (43)δp+/2∫Ω∇u-∇vpx dx≤∇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-vdxand J=∫Ω∇u+∇vpxdx,and α and ξ be such that (45)Iα/2=maxIp-/2,Ip+/2
and (46)Jξ=maxJ2-p-/2,J2-p+/2.Then, we get (47)δp+/2∫Ω∇u-∇vpx dx≤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. Assume p(x)≥2. Then, for all u,v∈W01,p(x)(Ω), we have(49)12p-ρpx∇u-v≤-Δpxu+Δpxv,u-v.
Proof. As p(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′for 1<px<2,p′-for px≥2, for all x∈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. Let ω1 and ω2 be two solutions of (14).
First case: suppose that 1<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=maxρpxω1-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 that p(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′x dx+τp+CMp+/p′-p-∫Ωω1-ω2px dx.
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 Theorem 6. 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)signξfx,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 Theorem 7, 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=minUk1,pxp+,Uk1,pxp-,Ukrxα′=minUkrxr+,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).