STRONGLY NONLINEAR VARIATIONAL PARABOLIC EQUATIONS WITH p(x)-GROWTH∗

2016-11-24ElhoussineAZROULBadrLAHMIDepartmentofMathematicsFacultyofSciencesDharElMahrazUniversitySidiMohamedBenAbdellahLaboratoryLAMABox1796Atlases30000Moroccomailazroulelhoussinegmailcomlahmibadrgmailcom

Acta Mathematica Scientia(English Series) 2016年5期

Elhoussine AZROULBadr LAHMI Department of Mathematics,Faculty of Sciences Dhar El Mahraz,University Sidi Mohamed Ben Abdellah,Laboratory LAMA,P.O.Box 1796 Atlas F`es 30000,Morocco E-mail:azroulelhoussine@gmail.com;lahmi.badr@gmail.com

Ahmed YOUSSFI†University Sidi Mohamed Ben Abdellah,National School of Applied Sciences,Laboratory of Engineering,Systems and Applications,P.O.Box 72-Principale,F`es 30000,Morocco E-mail:ahmed.youssfi@gmail.com

STRONGLY NONLINEAR VARIATIONAL PARABOLIC EQUATIONS WITH p(x)-GROWTH∗

We study the Dirichlet problem associated to strongly nonlinear parabolic equations involving p(x)structure inQ).We prove the existence of weak solutions by applying Galerkin’s approximation method.

nonlinear parabolic equations;existence;variable exponents;weak solution 2010 MR Subject Classification35K55;46E30;35D30

1　Introduction

Let Ω be an open bounded subset of ℝN,N≥2,with a Lipschitz boundary denoted by∂Ω. Fixing a final time T>o,we denote by Q the cylinder Ω×]o,T[and Γ=∂Ω×]o,T[its lateral surface.We consider the following strongly nonlinear parabolic initial-boundary problem

where Au=−div a(x,t,u,∇u)+ao(x,t,u,∇u)is an operator which mapsits dual W−1,xLp′(x)(Q).The functions a and aosatisfy Leray-Lions-like conditions with respect to the variable exponent p(x)which is only dependent on the space variable x.

The nonlinear term g(x,t,s,ξ)satisfies

and the sign condition

Spaces with variable exponent are relevant in the study of non-Newtonian fluids.The underlying integral energy appearing in the modeling of the so called electrorheological fluids (see for instance[16,17])isRΩ|∇u|p(x)dx.Accordingly,this naturally leads to study these fluids in space W1,p(x)o(Ω).Such energies occur also in Image Restoration[4,14]and Elasticity[23].

A great deal of work was developed around elliptic problems with p(x)-growth that is,involving the space

Problems with p(x)-growth can be regarded as a kind of nonstandard growth problems(see [15]).Parabolic problems with variable exponent appear in many applications,for instance [4,14].The study of such problems raises many mathematical difficulties.Note that(1.1)has a more complicated nonlinearity than in the classical case p(x)≡p because it is nonhomogeneous. Furthermore,an integral version of the Poincar´e inequality is no longer valid in general.

Given the assumptions we have made,we think that the space(Q)is a reasonable framework in which we can discuss our problem where p(x)is only dependent on the space variable x.This space is a particular case of the general framework of the inhomogeneous Musielak-Orlicz spacesLM(Q)built from the Musielak N-function M(x,ξ)=|ξ|p(x)which is related to the actual growth of the functions a and ao.For the Orlicz framework,one can see [7].

Recently,Fu-Pan in[9]proved an existence result of weak solutions for the problem(1.1) with g≡o in the spaceQ).In this work,we are mainly concerned with the existence of weak solutions for the strongly nonlinear problem(1.1)when uo∈L2(Ω),f∈W−1,xLp′(x)(Q) and the perturbation g has a critical p(x)-growth with respect to∇u restricted by the integration variable exponent of the underlying energy space W1,xoLp(x)(Q).

The paper is organized as follows:in Section 2 we recall some notations and properties of Lebesgue and Sobolev spaces with variable exponent.In addition,we give the definition of(Q)and it’s dual.Section 3 contains results that are usable subsequently. Section 4 contains basic assumptions and the main result,while Section 5 is devoted to the proof of main result.

2　Preliminaries

We recall some definitions and basic properties of variable exponent Lebesgue and Sobolev spaces Lp(x)(Ω),W1,p(x)(Ω)and the spaces W1,xLp(x)(Q)(see e.g.[5,6,1o,12]for more details).Let Ω be an open bounded subset of ℝN,N≥2,with a Lipschitz boundary denoted by∂Ω.Let p:Ω→[1,+∞)be a measurable real-valued function.Define

The Lebesgue space with variable exponent Lp(·)(Ω)consists of all measurable functions u:Ω→ℝ such that the modular ρp(x)(u):=RΩ|u(x)|p(x)dx is finite.It is a Banach space with respect to the so-called Luxemburg norm

If 1

Indeed,for the left-hand side inequality,we can write

Thus,by the definition of the norm

For the other inequality,we have to discuss two cases.

1)If ρp(·)(u)≤1 then the inequality holds true.

2)If ρp(·)(u)>1,we can write

Norm and modular convergence are equivalent and we have

It is a Banach space under the normWe also defineAn interesting feature of variable exponent Sobolev space is that smooth functions are not dense in it without additional assumptions on the exponent p(·).Thus,if the exponent p(·)satisfies 1≤p−≤p(x)≤p+<+∞and the log-H¨older continuity condition,i.e.there is a constant C>o such that for every x,ythenis dense in Lp(·)(Ω)(see[19,24]).If(2.2)is fulfilled,the spaces Lp(x)(Ω)and W1,p(·)(Ω)are reflexive and separable([5,1o,12])and so isThe dual space of(Ω),denoted W−1,p′(·)(Ω),is equipped with the norm

where the infinimum is taken on all possible decompositions v=vo−div F with vo∈Lp′(x)(Q) and F=(v1,···,vN)∈(Lp′(x)(Q))N.We recall the Poincar´e inequality(see[11,Theorem 4.3] and[19,Theorem 3]).

Lemma 2.1Let Ω be a bounded open subset of ℝN.Let p∈satisfy 1≤p−≤p(x)≤p+<+∞and the log-H¨older continuity condition(2.4).Then,there exists a constant C>o depending only on Ω and the function p such that

Note that the nonstandard version of the Poincar´e type inequality of integral form do not hold(see[5]).

Lemma 2.2(see[8])Suppose that 1≤p(x)<∞.Let{vn}nbe bounded in Lp(x)(Ω).If vn→v a.e.in Ω,then vn⇀v weakly in Lp(x)(Ω).

Extending a variable exponent p :for all(x,t)∈Q,we may also consider the generalized Lebesgue space which shares the same type of properties as Lp(x)(Ω):

We define the space W1,xLp(x)(Q)as the following:

where∇u stands for the gradient of u with respect to the space variable x.Thus defined, W1,xLp(x)(Q)is a Banach space with the norm

where the infinimum is taken on all possible decompositioni=o,1,···,N.Moreover,we have the following continuous embeddings

3　Auxilliary Results

Lemma 3.1Consider a sequence(wn)∈W−1,xLp′(x)(Q)+ L1(Q)such that wn⇀w

where(hn)is bounded in W−1,xLp′(x)(Q)and(gn)is bounded in L1(Q).Then,up to a subsequence,wn→w strongly in L1(Q)and a.e.in Q.

ProofObserve that(wn)is bounded in L1(o,T;W−1,1(Ω)),as shown by the following continuous embeddings

On the other hand,the sequencethe space L1(o,T;W−1,1(Ω))as well.Thus,by[2o,Lemma 4],one has for all n∈ℕ,‖τhwn−→o as h→o.Let oo such that

By the compactness of the embeddingrelatively compact in L1(Ω)and also in W−1,1(Ω).Thus,by[2o,Theorem 1],we can deduce that(wn)is relatively compact in L1(o,T;W−1,1(Ω)).Since the embeddingis continuous,by[9,Lemma 2.13]we have(wn)is relatively compact in L1(Q).Thus,up to a subsequence,wn→w strongly in L1(Q)and a.e.in Q.

Lemma 3.2Assume that(2.2)and(2.4)hold true.If wthen there exists a sequence

Lemma 3.3Let Ω be a bounded open subset of ℝN.Assume that(2.2)and(2.4)hold truethen u∈C([o,T],L2(Ω)).

Thus,we can writ e

implying that(vi)is a Cauchy sequence in C(ℝ,L2(Ω))and therefore converges uniformly on ℝ to a function v∈C(ℝ,L2(Ω)).Since vi→for almost every t∈[o,T].Thus,u∈C([o,T],L2(Ω)).

Lemma 3.4Assume that(2.2)and(2.4)hold true.Let F:ℝ→ℝ be uniformly Lipschitzian with F(o)=o.If w∈(Q)+L1(Q), then F(w)∈Moreover,if the set D of discontinuity points of F′is finite,then

ProofConsider firstly the caseLemma 3.2,there exists a sequence(wn)insuch that wn→w strongly in

Thus,for a subsequence,one has wn→w a.e.in Q.Consequently,F(wn)→F(w)a.e. in Q.On the other hand,there exists a positive constant M such thatWe deduce that F(wn)remains bounded inSo that by Lemma 3.1 we can deduce that for a subsequence,F(wn)→v a.e.in Q.Thus,F(w)=v a.e.Q which

The case where F is general can be handled similarly without essential difficulty by taking convolutions with mollifier ρnin ℝ.We haveuniformly in every compact.Then,by the first case we obtain

4　Basic Assumptions and Main Result

Let p(·)be a variable exponent satisfying(2.2)and(2.4).Denote Q=Ω×(o,T),whereΩ is a bounded open subset of ℝNand T>o.We assume that a(x,t,s,ξ)and ao(x,t,s,ξ): Q×ℝ×ℝN→ℝNare Carath´eodory functions(i.e.,continuous with respect to(t,s,ξ)∈ℝ×ℝNfor a.e.x∈Ω and measurable with respect to x for all(t,s,ξ)∈(o,T)×ℝ×ℝN)which satisfies, for a.e.(x,t)∈Q,for every s∈ℝ and ξ,ξ′∈ℝNwith ξ/=ξ′,the following assumptions

where C(x,t),Co(x,t)In the sequel,without loss of generality,we can assume that C=Coand α=γ=1.Conditions(4.1)–(4.3)are the nonstandard parabolic version of the so-called Leray-Lions conditions.The lower order term g:Q×ℝ×ℝN→ℝ is a Carath´eodory function(i.e.,continuous with respect to(t,s,ξ)∈ℝ×ℝNfor a.e.x∈Ω and measurable with respect to x for all(t,s,ξ)∈(o,T)×ℝ×ℝN)satisfying

for a.e.(x,t)∈Q and all(s,ξ)∈ℝ×ℝNwhere θ∈L1(Q)and b:ℝ+→ℝ+is a continuous and increasing function.It should be noted here,that there is no growth restriction on the perturbation g as a function of the unknown.The sign condition(4.7)is needed to provide a priori estimates for approximate solutions for variational equations.

In what follows,〈,〉denotes the duality pairing betweenQτ=Ω×(o,τ)for τ∈(o,T]and χEstands for the characteristic function of a measurable subset E of Ω.Our main result is the following.

Theorem 4.1Let f∈W−1,xLp′(x)(Q)and uo∈L2(Ω).Assume that(4.1),(4.2), (4.3),(4.4),(4.5),(4.6)and(4.7)are fulfilled.Then,there exists at least one weak solution u of problem(1.1)in the following sense:u∈such that g(·,·,u,∇u)∈L1(Q),g(·,·,u,∇u)u∈L1(Q),u(x,o)=uo(x)for a.e.x∈Ω and for all τ∈(o,T]

Observe that by Lemma 3.3,the test function φ belongs to C([o,T];L2(Ω))implying that the second term in(4.8)is well defined.For the remaining terms,it’s easy to check that they are well defined too.

5　Proof of Main Result

We divide the proof into six steps.

Step 1Galerkin solutions

We choose a sequence of functions(Ω)orthonormalwith respect to the Hilbert space L2(Ω)such that(Ω)with s large enough such as s>

Define Wk=C1([o,T],Vk).with respect to the norm

We also note that there exists a sequence

Definition 5.1A function un∈Wnis called Galerkin solution of(1.1)if un(o)=uon

for all v∈Wn.

for i=1,···,n.Note that the function yn(t,·)is continuous because a,aoand gnare both Carath´eodory functions.We obtain the following system of ordinary differential equations

where(F)=Rfωdx and(v)=Ruωdx,i=1,···,n.Multiplying the first equation

niΩniniΩoniby d(t)and using(4.4),(4.5)and(4.7)one has yn(t,d)d≥o.Then,Young’s inequality yields

By virtue of Gronwall’s lemma one has

Step 2A priori estimates

Let unbe a Galerkin solution of(1.1)and let τ∈[o,T].Choosing unas test function in (5.1),we get

for every τ∈[o,T].By(4.4),(4.5)and(4.7)we can write

Which implies that

From now on,we denote by c various positive constant not depending on n which may vary from line to line.By virtue of(2.3)and Young’s inequality,we deduce from(5.4)

Thus,the sequence(un)nis uniformly bounded inThanks to(5.3)and(4.2),we

This allows us to assert that(Aun)is bounded in

Furthermore,(5.2)implies

Going back to(5.2),we obtain

Let m>o be fixed.By(4.6)and(5.1o)we can write

Hence,the sequence (g(x,t,un,∇un))nis uniformly bounded in L1(Q).Thus,

By using Lemma 3.1,we conclude that there exists a subsequence of(un),still denoted by (un),such that un→u strongly in L1(Q).Collecting the above results,we conclude that there exists a subsequence of(un),a measurable functions uwhich belongs respectively to

Integrating(5.1)between o and T we get

for all v∈Wn.In fact,formulation(5.14)remains valid,by density argument,for everysuch that‖v−such that

Let us use hjas test in(5.14),we can writeis bounded in W,we can easily deduce thatdxdt tends to o as j→∞.The passage to the limit in the remaining integrals in(5.14)is an easy task.

Step 3Almost everywhere convergence of the gradients

Given a real number k>o,we define the truncation function Tkby Tk(s)=max(−k, min(s,k))for every s in ℝ.We shall prove that∇un→∇u a.e.in Q.To this end,we prove that for all k>o,

Thanks to(5.12)and Lemma 3.4,we have Tk(un)Convergence(5.13)implies, by using Lebesgue’s dominated convergence theorem,that

Moreover,(5.9)implies that

It’s easy to see that(Tk(vj))µ∈i=o,1,···,N which implies that(Tk(vj))µ→Tk(u) strongly inhas the following properties(see e.g.[21]for similar properties)

Henceforward,we will denote by ε(n,µ,i,s)all quantities such that

this will be the order in which the parameters we use will tend to infinity,that is,first n,then j,µ,i and finally s.Similarly,we write only ε(n)to mean that the limit is made only on the specified parameter n.

We have fn→f strongly in Wweakly inas j→∞andasµ→∞,then we can write

On the other hand,using(5.8),(5.15)and(5.19)we get

Let Gk(s)=s−Tk(s).We can write

which implies

On the other hand,the fact that

For what concerns J3,an integration by parts gives

Observe that φ′>o andthat one has

Since Tk(vj)→Tk(u)for σ∗(L∞,L1)as j→∞,we get

Combining all these estimates,we get

Observe that∇Tk(un)⇀∇Tk(u)weakly in(Lp(x)(Q))Nand by(4.1)and Lebesgue’s dominated convergence theorem

On other hand,by using(4.1),(5.18)and again Lebesgue’s dominated convergence theorem, we get

It is not hard to check that(4.4)implies a(x,t,s,o)=o.Therefore,

We now estimate I3,writing

Letting n→∞,by vertue of the weak convergence of a(x,t,un,∇un)the convergence ofwhich is zero,

we obtain

Letting j→∞,(5.18)enables us to get

Passing to the limit asµ→+∞,one has

Finally,letting s→+∞we get

We turn now to estimate I4.We have as above,letting first n,then j andµgo to infinity

Using(4.4)and(4.6),the third term of the left-hand of(5.2o)can be estimated by

Since θ(x,t)∈L1(Q),we get

The second term of(5.24)reads as

As above,we can write

Combining(5.2o),(5.21),(5.22),(5.23)and(5.25)we get

so that by(5.17),we obtain

Then,it follows

Each integral of the right-hand side is of the form ε(n,j,s),therefore

On the other hand,for r≤s we have Z

which implies,by passing to the limit superior over n with(4.3),that

Then,we let successivelyµ→∞,i→∞and s→∞to obtain

as n→∞.Hence,we can deduce as in the elliptic case[3,Lemma 5],that

We also deduce that for a subsequence denoted(un)

Since r and k are arbitrary,we conclude that there exists a subsequence of(un)also denoted (un)such that

Therefore,we have

Using(5.8)and Lemma 2.2,we deduce that

Step 4Equi-integrability of the nonlinearities

We shall prove that

We have g(x,t,un,∇un)→g(x,t,u,∇u)a.e.in Q.Let E be a subset of Q and let m>o.We can write

By(4.4),the growth condition(4.6)and(5.1o),we obtain

For ε>o there exists m=m(ε)such thatsuch that b(m)θ(x,t)dxdt

We claim that

Indeed,going back to(5.26)and(5.27),we can write

It follows by passing to the limit superior over n,

Then,we pass to the limit as jZ,µ,i and s tend successively to∞to obtain

On the other hand,using Fatou’s lemma we get

Since a(x,t,Tm(un),∇Tm(un))·∇Tm(un)≥o,we conclude that

as n→∞.Therefore,by Vitali’s theorem there exists δ2>o such that

whenever meas(E)<δ2.So that for meas(E)

Applying Vitali’s theorem,we obtain

In addition,by Fatou’s Lemma,we get

Step 5u∈C([o,T],L2(Ω))

Let τ∈(o,T].Choosing un−ωiµ,jas test function in(5.14),we get

Then,by using Fatou’s lemma and Lebesgue’s dominated convergence theorem we can pass to the limit superior respectively over n,j,µand k,to obtain

The positiveness of the last integral is due to the fact that

implying

Thus,the sequence(un)is a Cauchy sequence in C([o,T];L2(Ω)).Furthermore,thanks to(5.13) the limit of unin L1(Q)is u,then un→u in C([o,T];L2(Ω)).Finally,

Step 6Passage to the limit

Let τ∈(o,T]be arbitrary and let φ∈As un⇀u in L2(Q), we can write using(5.32)

Furthermore,‖φi‖∞≤const‖v‖∞.Using φiχ(o,τ)as a test function in(5.33),we obtain

We shall pass to the limit as i→∞in each term of(5.35).Observe first that in view of Lemma 3.3,one has φi→v in C([o,T],L2(Ω)).So that we get

Using(5.34)and the fact that the sequence(φi)is uniformly bounded,we can easily pass to the limit in the remaining terms in(5.35)to obtain

We now prove the energy equality following the line of[7].Let k>o.By(5.12)and Lemma 3.4,we know that Tk(un)is an admissible test function in(5.14).Taking it so,it yields

We have|Sk(un(τ))|≤k|un(τ)|and un(τ)→u(τ)in L2(Ω)as n→∞,then by using(5.28), (5.3o),(5.31)and Lebesgue theorem,we can pass to the limit as n→∞to obtain

The proof of Theorem 4.1 is achieved.

AcknowledgementsThe authors would like to thank warmly the two referees for their valuable comments and suggestions that improved the earlier version of the manuscript.

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∗April 24,2015;revised July 21,2015.†

Ahmed YOUSSFI.

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Acta Mathematica Scientia(English Series)

2016年5期