OUARO Stanislasand OUEDRAOGO Arouna

Universit′e de Ouagadougou,Unit′e de Formation et de Recherche en Sciences Exactes et Appliqu′ees,D′epartement de Math′ematiques B.P.7021 Ouagadougou 03, Burkina Faso.

Received 29 January 2013;Accepted 6 January 2014

L1Existenceand Uniquenessof EntropySolutions to Nonlinear Multivalued Elliptic Equations with Homogeneous Neumann Boundary Condition and Variable Exponent

OUARO Stanislas?and OUEDRAOGO Arouna

Universit′e de Ouagadougou,Unit′e de Formation et de Recherche en Sciences Exactes et Appliqu′ees,D′epartement de Math′ematiques B.P.7021 Ouagadougou 03, Burkina Faso.

Received 29 January 2013;Accepted 6 January 2014

.In this work,we study the following nonlinear homogeneous Neumann boundary value problem β(u)?diva(x,?u)?f in ?,a(x,?u)·η=0 on??,where ?is a smooth bounded open domain in RN,N≥3 with smooth boundary?? and η the outer unit normal vector on??.We prove the existence and uniqueness of an entropy solution for L1-data f.The functional setting involves Lebesgue and Sobolev spaces with variable exponent.

Elliptic equation;variable exponent;entropy solution;L1-data;Neumann boundary condition.

1 Introduction

The paper is motivated by phenomena which are described by the homogeneous Neumann boundary value problem of the form

where η is the unit outward normal vector on??,? is a smooth bounded open domain in RN,N≥3,β=?j is a maximal monotone graph in R2with dom(β)bounded on R and 0∈β(0),f∈L1(?)and a is a Leray-Lions operator which involves variable exponents.

Note that j is a nonnegative,convex and l.s.c.function on R and,?j is the subdifferential of j.We set

Recall that a Leray-Lions operator which involves variable exponents is a Carath′eodory function a(x,ξ):?×RN?→RN(i.e.a(x,ξ)is continuous in ξ for a.e.x∈?and measurable in x for every ξ∈RN)such that:

?There exists a positive constant C1such that

for almost every x∈? and for every ξ∈RNwhere j is a nonnegative function in Lp′(.)(?), with 1/p(x)+1/p′(x)=1.

?The following inequalities hold

for almost every x∈? and for every ξ,η∈RN,with ξ/=η,and

for almost every x∈?,C＞0 and for every ξ∈RN.

In this paper,we make the following assumption on the variable exponent:

where p?:=essinfx∈?p(x)and p+:=esssupx∈?p(x).

As the exponent p(·)appearing in(1.2)and(1.4)depends on the variable x,the functional setting for the study of problem(1.1)involves Lebesgue and Sobolev spaces with variable exponents Lp(·)(?)and W1,p(·)(?).In the next section,we will make a brief presentation of the variable exponent spaces.

Many results are known as regards to elliptic problems in the variational setting for Dirichlet or Dirichlet-Neumann problems(cf.[1-9]).

Problem(1.1)can be viewed as an extension of the following

where ? is a smooth bounded open domain in RN,N≥3 and η the outer unit normal vector on??.b:R→R is a continuous,nondecreasing function,surjective such that b(0)=0,f∈L1(?)and a is a Lerray-Lions operator which involves variable exponents.

Problem(1.6)was studiedby Bonzi,Nyanquini and Ouaro(cf.[2])where they proved the existence and uniqueness of an entropy solution.An equivalent notion of solution iscalled renormalized solution.The concept of renormalized solutions was introduced by Diperna and Lions in[10].This notionwas thenadapted tothe studyofvarious problems of PDEs.In[9],Wittbold and Zimmermann adapted this notion of solution to a new and interesting problem of the form

They proved for F locally Lipschitz continuous,β maximal monotone mapping with 0∈β(0),f∈L1(?)and a continuous exponent p(·),the existence and uniqueness of a renormalized solution for problem(1.7).

In this work,we consider an homogeneous Neumann boundary condition instead of the Dirichlet boundary condition considered in[9].One of the main difficulty which appears in this case is the famous Poincar′e inequality which d?sn’t apply and even the Poincar′e-Wirtinger inequality also.An other difficulty is that,since we assume that the domain of β is bounded,it appears in the definition of the solution,a bounded Radon diffuse measure in order to take into account the border of the domain.The techniques used in this work are close to those used in[5,7].

We denote byMb(?)the space of bounded Radon measure in ?,equipped with its standard norm||·||Mb(?).Given ν∈Mb(?),we say that ν is diffuse with respect to the capacity W1,p(·)(?)(p(·)?capacity for short)if ν(E)=0 for every set E such that Capp(·)(E,?)=0,where the Sobolev p(·)-capacity of E is defined by

with

The remaining part of the paper is the following:in Section 2,we introduce some notations and functional spaces.In Section 3,we prove the existence of entropy solution to the problem(1.1)and in Section 4,we prove the uniqueness of entropy solution.

2 Assumptions and preliminary

As the exponent p(·)appearing in(1.2)and(1.4)depends on the variable x,we must work with Lebesgue and Sobolev spaces with variable exponents.

We define the Lebesgue space with variable exponent Lp(·)(?)as the set of all measurable function u:??→R for which the convex modular

is finite.If the exponent is bounded,i.e.,if p+＜+∞,then the expression

defines a norm in Lp(·)(?),called the Luxembourg norm.The space(Lp(·)(?),|·|p(·))is a separable Banach space.Moreover,if 1＜p?≤p+＜+∞,then Lp(·)(?)is uniformly convex,hence reflexive,and its dual space is isomorphic to Lp′(·)(?),where

Finally,we have the H¨older type inequality:

for all u∈Lp(·)(?)and v∈Lp′(·)(?).

Now,let

which is a Banach space equipped with the following norm

The space(W1,p(·)(?),||u||1,p(·))is a separable and reflexive Banach space.For the interested reader,more details about Lebesgue and Sobolev spaces with variable exponent can be found in[11](see also[12]).

An important role in manipulating the generalized Lebesgue and Sobolev spaces is played by the modular ρp(·)of the space Lp(·)(?).We have the following result(cf.[13]):

Lemma 2.1.If un,u∈Lp(·)(?)and p+＜+∞,then the following properties hold:

For a measurable function u:?→R,we introduce the function

Then we have the following lemma(see[14,15]).

Lemma 2.2.If u∈W1,p(·)(?),then the following properties hold:

For any given l,k＞0,we define the function hlby hl(r)=min?(l+1?|r|)+,1?and the truncation function Tk:R→R by Tk(s)=max{?k,min(k,s)}.

For any l0,we consider the function h0=hl0defined by

Let γ be a maximal monotone operator defined on R.We recall the definition of the main section γ0of γ:

We write for any u:?→R and k≥0,{|u|≤k(＜k,＞k,≥k,=k)}for the set{x∈?/|u(x)|≤k(＜k,＞k,≥k,=k)}.

Before introducing the notion of entropy solution for the problem(1.1),we define the following spaces which are similar to that introduced in[16,17].We note

As in[17],we can prove that for u∈T1,p(·)(?),there exists a unique measurable function w:??→R such that?Tk(u)=wχ{|w|＜k}?k＞0.This function w will be denoted?u.

WedefineT1,p(·)

H(?)(see[2])as thesetoffunctions u∈T1,p(·)(?)suchthat thereexists a sequence(uδ)δ＞0∈W1,p(·)(?)satisfying the following conditions:

The symbolHin thenotationis related tothe fact that weconsiderherehomogeneous Neumann boundary condition.

To end this section,we give some useful convergence results.

Lemma 2.3.Letbe a sequence of maximal monotone graphs such that βn→β in the sense of graphs(i.e.for(x,y)∈β,there existssuch that xn→x and yn→y).We consider two sequencesand.We suppose that:is bounded in.Then z∈dom(β).Proof.For the proof of Lemma 2.3,we need the“biting lemma of Chacon”.Let us recall it.

Lemma 2.4.(The“biting lemma of Chacon”)[19].Let ??RNbe a bounded open domain and(fn)n≥1a bounded sequence in L1(?).Then there exists f∈L1(?),a sequence(fnk)k≥1and a sequence of measurable sets(Ej)j≥1,Ej??,?j∈N?with Ej+1?Ejand limj→+∞|Ej|=0,such that for any j∈N?,fnk?f in L1(?Ej).

Since the sequence(wn)n≥1is boundedin L1(?),usingthe“biting lemma ofChacon”, thereexistsw∈L1(?),asubsequence(wnk)k≥1and asequenceofmeasurable sets(Ej)j∈N?in ? such that?j∈N?,Ej+1?Ej,limj→+∞|Ej|=0 and?j∈N?,wnk?w in L1(?Ej).Since znk?z in L1(?)and so in L1(?Ej),?j∈N and βnk→β in the sense of graphs,we have w∈β(z)a.e.in ?Ej.Thus z∈dom(β)a.e.in ?Ej.Finally,we obtain z∈dom(β)a.e.in?.

Lemma 2.5.(cf.[13],Theorem 1.4)Let u,un∈Lp(·)(?),n=1,2,···.Then the following statements are equivalent to each other:

Lemma 2.6.(Lebesgue generalized convergence theorem)Let(fn)n∈Nbe a sequence of measurable functions and f a measurable function such that fn→f a.e.in ?.Let(gn)n∈N?L1(?)such that for all n∈N,|fn|≤gna.e.in ? and gn→g in L1(?).Then

3 Statement of the main results

We introduce the following concepts of solution for problem(1.1).Definition 3.1.A solution of(1.1)is a couple(u,b)∈T1,p(·)H(?)×L1(?),such that

and

Remark 3.1.If(u,b)isasolutionoftheproblem(1.1)then,itsatisfythefollowing entropic formulation

for any ξ∈W1,p(·)(?)∩L∞(?)such that ξ∈dom(β)LN?a.e.in ?.

Our main result is the following.

Theorem 3.1.Assume that(1.2)-(1.5)hold true and f∈L1(?),there exists a unique entropy solution to problem(1.1).Moreover

Proof.The proof of Theorem 3.1 is divided into several steps.

3.1 Regularized problem

For every ?＞0,we consider the Yosida regularisation β?of β given by

and we set

According to Proposition 2.11 in[18],we have

Note thatβ?is a nondecreasing and Lipschitz-continu?ou?s function.We also define the functionfor any x∈?.Thenf??＞0is a sequence of bounded functions which converges strongly toand such that

Lemma 3.1.The Yosida regularisation β?is a surjective operator.

Proof.Since dom(β)?[m,M],then?r∈R,J?(r)=?I+?β??1(r)∈[m,M].Consequently

and

As β?is a maximal monotone graph,thanks to([18],Corollaire 2.3),we conclude that β?is surjective.

Now,we consider the approximated problem

where the notation u?is used for any solution of the problem(3.4).

Definition 3.2.A weak solution of(3.4)is a measurable function u?∈W1,p(·)(?)such thatand

We have the following result according to[2].

Theorem 3.2.Assume that(1.2)-(1.5)hold and f?∈L∞(?).Then,there exists a unique weak solution u?to problem(3.4).

3.2 A priori estimates

Lemma 3.2.Assume that(1.2)-(1.5)hold and f∈L1(?).Let u?be a weak solution of(3.4). Then,for all k＞0,Z

and

where C is a positive constant.

Proof.Taking ?=Tk(u?)as a test function in(3.5),we get

Using(1.4)and the fact that β?is nondecreasing and β?(0)=0,we deduce from(3.8)that

Then,

Now,using(1.4),we have a(x,?Tk(u?))·?Tk(u?)≥C1|?Tk(u?)|p(x)≥0.Then,we get from(3.8)that

The proof is complete.

Lemma 3.3.The sequence(β?(u?))?＞0is uniformly bounded in L1(?). Proof.According to(3.7),we get which gives,if we let k→0,Z

The proof is complete.

Lemma 3.4.The sequence(β?(Tk(u?)))?＞0is uniformly bounded in L1(?). Proof.We have for all k＞0,

Then,according to Lemma 3.3,(β?(Tk(u?)))?＞0is uniformly bounded in L1(?).

Lemma 3.5.Assume that(1.2)-(1.5)hold true and f∈L1(?).Let u?be a weak solution of(3.4), then

and

where C is a positive constant.

Proof.We start by the proof of(3.9).By Lemma 3.3,we have

which implies thatZ

Since β?is nondecreasing and β?(0)=0,we have

and

Then,we deduce from(3.11)that

i.e.

Now,we prove(3.10).For k,λ≥0,set

According to(3.9),we have

Using the fact that the function λ→Φ(k,λ)is nonincreasing,we get for k＞0 and λ＞0 that

Observe that since

we obtain

Note that

By the inequalities above,thanks to(3.6),we obtain

Combining(3.13)and(3.14),we obtain

Coming back to(3.12)and using(3.15)we arrive at

In particular,

Setting λ=kp?in(3.16)gives(3.10).

3.3 Convergence results

Lemma 3.6.(i)For all k＞0,Tk(u?)→Tk(u)in Lp?and a.e.in ?,as ?→0.

(ii)There exists a measurable function u:?→R such that u∈dom(β)a.e.in ? and u?→u in measure and a.e.in ?,as ?→0.Proof.For k＞0,the sequence(?Tk(u?))?＞0is bounded in Lp(·)(?);hence the sequence (Tk(u?))?＞0is bounded in W1,p(·)(?).Then,up to a subsequence we can assume that for any k＞0,(Tk(u?))?＞0converges weakly to σkin W1,p(·)(?)and so(Tk(u?))?＞0converges strongly to σkin Lp?.Let s＞0 and define

where k＞0 is to be fixed.We note that

and hence

Let ?＞0.Using Lemma 3.5,we choose k=k(?)such that

for all n,m≥n0(s,?).

Finally,from(3.17)-(3.19),we obtain

Hence,the sequence(u?)?＞0is a Cauchy sequence in measure.We can extract a subsequence such that u?→u a.e.in ?.As for k＞0,Tkis continuous,then Tk(u?)→Tk(u) a.e.in ? and σk=Tk(u)a.e.in ?.Finally,using Lemma 2.3 we deduce that for all k＞0,Tk(u)∈dom(β)a.e.in ? and as dom(β)is bounded,we deduce that u∈dom(β) a.e.in ?.

Lemma 3.7.For all k＞0,

Step 1:We prove that for every function h∈W1,+∞(?),h≥0 with a compact support (supp(h)?[?l,l]?R),

Let us take ?=h(u?)(Tk(u?)?Tk(u)),k＞0 as a test function in(3.5).We have

For any r＞0,sufficiently small,we consider

For any k＞0,Tk(ur)∈W1,p(·)(?).Furthermore,we have

Note that

Since β?is nondecreasing we have

We deduce that h(u?)β???Tk(u?)?Tk(ur)?∈L1(?). Since

Then,by the Lebesgue dominated convergence theorem,we get

Ash(u?)a(x,?u?)=h(u?)a(x,?Tl(u?))isuniformly boundedinLp′(·)(?)N(by assumption(1.2)and relation(3.6))and?(Tk(ur)?Tk(u))?0 as r→0,then

Recall that m+r≤ur≤M?r and by Lemma 3.6(part(ii)),u∈dom(β)?[m,M].Then

Thus,for the third term of I2,we have

where C(h,l,?,‖f‖1)is a constant depending on h,l,? and‖f‖1.Then,we get

Therefore,combining(3.24)-(3.26)we obtain

Now,let us see that

Indeed,

since 0∈β(0),m+r≤0≤M?r and Tkis nondecreasing.It follows that

Then,by the Lebesgue generalized convergence theorem,we get

Passing to the limit in(3.22)as ?→0 and combining(3.27)-(3.29),we obtain(3.21).Step 2:We prove that

Let us take for l＞0,?=T1(u??Tl(u?))as a test function in(3.5).We have

We have

Then

If u?＜?l or u?＞l,then u?andhave the same sign.We conclude that the second term of the left-hand side of(3.31)is nonnegative,i.e.

The first term of(3.31)is written as follows

As in step 1,we show that

We also have T1(u?Tl(u))→0 a.e.in ? as l→+∞and? ?fT1(u?Tl(u))??≤|f|∈L1(?). Then,using the Lebesgue dominated convergence theorem,we obtain

We deduce that

Passing to the limit as ?→0,to the limit as l→+∞in(3.31)and using(3.32)-(3.34),we deduce(3.30).

Step 3:We prove that for every k＞0,

For ν＞k,we have

Since ν＞k,on the set{|u?|≤k},it follows that hν(u?)=1 and we get

The second term of the right-hand side of(3.36)can be written as

Using the Lebesgue dominated convergence theorem,we deduce that

The sequence(a(x,?Tν+1(u?)))?＞0is bounded inLp′(·)(?)N,then it converges weakly in?Lp′(·)(?))Nto Γν+1.By the Lebesgue dominated convergence theorem,we find

i.e.

Considering the third term of the right-hand side of(3.36),we have

Using the result of step 2,we obtain

Applying(3.21)with h replaced by hν,ν＞k in(3.36)and using(3.37)-(3.39),it follows that

Therefore,(3.35)follows.

Step 4:Now,we prove by standard monotonicity arguments that for all k＞0,Φk= a(x,?Tk(u))a.e.in ?.Let ?∈D(?)and λ∈R?.Using(3.35),(1.3)and Lemma 3.6,we get

Dividing the first and the last integrals of(3.40)by λ＞0 and by λ＜0 respectively,passing to the limit with λ→0 it follows that

This means that

and so Φk=a(x,?Tk(u))inD′(?)for all k＞0.Hence Φk=a(x,?Tk(u))a.e.in ? and we have

(ii)From(3.35)and(1.3),we deduce that for all k＞0

Now,set

g?(x)→0 strongly in L1(?)as ?→0.Up to a subsequence,g?(x)→0 a.e.in ?,which means that there exists ω?? such that meas(ω)=0 and g?(x)→0 in ?ω.Let x∈?ω.Using assumptions(1.4)and(1.2),it follows that the sequence??Tk(u?(x))??＞0is bounded in RNand so we can extract a subsequence which converges to some θ in RN.

Passing to the limit in the expression of g?(x),it follows that

and it yields θ=?Tk(u),?x∈?ω.

(iii)The continuity of a(x,ξ)with respect to ξ∈RNgives us

Therefore

and as

it follows by using the Lebesgue dominated convergence theorem that

which means that

We have:

?h?is a sequence of measurable functions,h is a measurable function and according to (ii),h?→h a.e.in ?.

?Using(iii),we have(g?)?＞0?L1(?),g?→g a.e.in ?,g?→g in L1(?)and using(1.4), we have|h?|≤Cg?.

Then,by Lemma 2.6,we have

Remark 3.2.By Lemma 3.6 and Lemma 3.7-(iv),we deduce that u∈T1,p(·)H(?).

The following lemma is useful for the next.

Lemma 3.8.For any h∈C1c(R)and ?∈W1,p(·)(?)∩L∞(?), Proof.For any h∈C1c(R)and ?∈W1,p(·)(?)∩L∞(?),we have

Set

Then,by the Lebesgue dominated convergence theorem,we get that lim?→0ρp(·)(ψ?1)=0. Hence,

For the term ψ?2we consider

for some l＞0 such that supp(h)?[?l,l].Set

We have Θ?2(x)→0 a.e.x∈? as ?→0 and

Since?Tl(u?)→?Tl(u)strongly in?Lp(·)(?)?N,we get

which is equivalent to say

Then|?Tl(u?)??Tl(u)|p(·)→0 strongly in L1(?).By the Lebesgue generalized convergence theorem,one has

Hence,

Set

We have Θ?3(x)→0 a.e.x∈? as ?→0 and

withsome l＞0 suchthat supp(h)?[?l,l].Then,by the Lebesguedominatedconvergence theorem,we get

Hence,

Thanks to(3.42)-(3.44),we get

and the lemma is proved.

Now,we want to pass to the limit in the first integral of(3.5).Since,for any k＞0, (hk(u?)β?(u?))?＞0is bounded in L1(?),there exists zk∈Mb(?),such that

Moreover,for any ?∈W1,p(·)(?)∩L∞(?),we have

which implies that zk∈Mp(·)b(?)and,for any k≤l,

Let us consider the Radon measure z defined by

for any ?∈W1,p(·)(?)∩L∞(?).Indeed,let k0＞0 be such that supp(h)?[?k0,k0],

Moreover,we have

Lemma 3.9.The Radon-Nikodym decomposition of the measure z given by(3.45)with respect toLN,

satisfies the following properties

Proof.Since,for any ?＞0,z?∈?j?(u?),we have

Then,for any h∈Cc(R),h≥0 and k＞0 such that supp(h)?[?k,k],we have

In addition,for any 0＜?＜??,we have

and,integrating over ? yields

As ?→0,we get by using Fatou’s Lemma

we have

So,

From the inequality above,we have

Using the Radon-Nikodym decomposition of z we have z=bLN+μwithμ⊥LN,b∈L1(?),then comparing the regular part and the singular part of(3.47),for any h∈Cc(R), we obtain

and

From(3.48)we get

so that b∈?j(u)LN?a.e in ?.As to(3.49),this implies that for any t∈dom(j),

and

In particular,this implies that

and so that

and the proof of the Lemma 3.9 is finished.

To finish the proof of Theorem 3.1,we consider ?∈W1,p(·)(?)∩L∞(?)and h∈C1c(R). Then,we take h(u?)? as test function in(3.5).We get

By the Lebesguedominated convergence theorem,we have for the term in the right hand side of(3.52),

The first term of(3.52)can be written as

for some l0＞0 so that,by Lemma 3.7-(i)and Lemma 3.9,we have

Thanks to the convergence of Lemma 3.9 and Lemma 3.7-(i)we have from(3.52)

Letting ? goes to 0 in(3.52),we obtain

In(3.53),we take h∈C1c(R)such that[m,M]?supp(h)?[?l,l]and h(s)=1 forall s∈[?l,l]. As u∈dom(β),then h(u)=1 and it yields that(u,b)is a solution of the problem(1.1). Now,let us prove the uniqueness of the solution for the problem(1.1).Suppose that (u1,w1),(u2,w2)are two solutions of the problem(1.1).For u1,we choose ξ=u2as test function in(3.2)to get

Similarly we get for u2Adding these two last inequalities yields

For any k＞0,from(3.54)it yields

From(3.55),it follows that there exists a constant c such that u1?u2=c a.e.in ?.At last, let us see that w1=w2a.e.in ? and ν1=ν2.Indeed for any ?∈D(?),taking ? as test function in(3.1)for the solutions(u1,w1)and(u1,w2),after substraction,we get

Hence

Therefore

Since the Radon-Nikodym decomposition of a measure is unique,we get

To end the proof of Theorem 3.1,we prove(3.3).We take ?=T1(u??Tn(u?))as test function in(3.1)to get

Since

we have from equality(3.56),

Thanks to(3.34),we have

Using assumption(1.3),it follows if we let ?→0 and n→+∞r(nóng)espectively in(3.57),

Therefore,we get(3.3).

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10.4208/jpde.v27.n1.1 March 2014

?Corresponding author.Email addresses:souaro@univ-ouaga.bf,ouaro@yahoo.fr(S.Ouaro), arounaoued2002@yahoo.fr(A.Ouedraogo)

AMS Subject Classifications:35K55,35D05

Chinese Library Classifications:O175.26