# Entropy Solution for Doubly Nonlinear Elliptic Anisotropic Problems with Robin Boundary Conditions

Authors: I. Ibrango; S. Ouaro
Journal: International Journal of Differential Equations
 (2015)

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## Abstract

We study in this paper nonlinear anisotropic problems with Robin boundary conditions. We prove, by using the technic of monotone operators in Banach spaces, the existence of a sequence of weak solutions of approximation problems associated with the anisotropic Robin boundary value problem. For the existence and uniqueness of entropy
solutions, we prove that the sequence of weak solutions converges to a measurable function which is the entropy solution of the anisotropic Robin boundary value problem.


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## Body

## 1. Introduction

The aim of this paper is to study the following nonlinear anisotropic elliptic Robin boundary value problem:(1)-∑i=1N∂∂xiaix,∂u∂xi+bu=finΩ,∑i=1Naix,∂u∂xiηi=-γuon∂Ω,where Ω is an open bounded domain of RN (N≥3) with smooth boundary and meas(Ω)>0, f∈L1(Ω), η=(η1,…,ηN) is the unit outward normal on ∂Ω, and γ(u)=|u|r(x)-2u.All papers on problems like (1) considered particular cases of function b. Indeed, in [1], Bonzi et al. studied the following problems:(2)-∑i=1N∂∂xiaix,∂u∂xi+upMx-2u=finΩ,∑i=1Naix,∂u∂xiηi=-urx-2uon∂Ω,where f∈L1(Ω). The authors use minimization technics used in [2] or [3] (see also [4, 5]) to prove the existence and uniqueness of entropy solution.The Robin type boundary conditions in the variable exponents setting are new and interesting problems and were for the first time studied by Boureanu and Radulescu in [3]. The main difficulty for the study of problem in [3] was the definition of an admissible space of solutions. The authors defined the appropriate space and obtained its properties which permit them to use minimization method to prove the existence of weak solutions to the following problem:(3)-∑i=1N∂∂xiaix,∂u∂xi+bxupMx-2u=fx,uinΩ,u≥0inΩ,∑i=1Naix,∂u∂xiηi=gx,uon∂Ω.Since we consider L1-data f instead of the function f(x,u) considered in [3], the suitable notion of solution is the entropy solution introduced by Bénilan et al. in [6] (see also [7]). With the Robin type boundary conditions, the values of the solutions at the boundary must be precise and the notion of solutions considered must include the boundary condition. In this paper, as the function b is more general, it is not possible to use minimization technic to get the existence of solution. Therefore, we used the technic of monotone operators in Banach spaces (see [8]) to get the existence of entropy solutions of (1).For presenting our main result, we first have to describe the data involved in our problem. LetΩ be a bounded domain in RN (N≥3) with smooth boundary domain ∂Ω and p→(·)=(p1(·),…,pN(·)) such that, for any i=1,…,N, pi(·):Ω¯→[2;N) is a continuous function with(4)1<pi-≔ess⁡infx∈Ω⁡pix≤ess⁡supx∈Ω⁡pix≔pi+<+∞.For any i=1,…,N, let ai:Ω×R→R be a Carathéodory function satisfying the following:(i)
There exists a positive constantC1 such that(5)aix,ξ≤C1jix+ξpix-1,for almost every x∈Ω and for every ξ∈R, where ji is a nonnegative function lying in Lpi′(·)(Ω), with (1/pix)+(1/pi′x)=1.(ii)
Forξ,η∈R with ξ≠η and for almost every x∈Ω, there exists a positive constant C2 such that(6)aix,ξ-aix,ηξ-η≥C2ξ-ηpixifξ-η≥1,C2ξ-ηpi-ifξ-η<1.(iii)
There exists a positive constantC3 such that(7)aix,ξ·ξ≥C3ξpix,for ξ∈R and for almost every x∈Ω.The hypotheses on ai are classical in the study of nonlinear problems (see [1, 3]).The functionb is such that(8)b:R⟶Ris continuous, surjective, nondecreasingwithb0=0.Throughout this paper, for any i=1,…,N, we assume that(9)p¯N-1Np¯-1<pi-<p¯N-1N-p¯,pi+-pi--1pi-<p¯-Np¯N-1,∑i=1N1pi->1,where N/p¯=∑i=1N‍(1/pi-).We put for allx∈Ω, (10)pMx≔max⁡p1x,…,pNx,pmx≔min⁡p1x,…,pNxand for all x∈∂Ω, (11)p∂x=N-1pxN-pxifpx<N,+∞ifpx≥N.We make the following assumption:(12)r∈CΩ¯with1<r-≤r+<minx∈∂Ω⁡p1∂x,…,pN∂x.Note that the function γ is continuous, defined on R with γ(t)t≥0 for all t in R and γ(0)=0.A prototype example that is covered by our assumptions is the following anisotropicp→(·)-harmonic system:(13)-∑i=1N∂∂xi∂u∂xipix-2∂u∂xi=f,which, in the particular case when pi=p for any i=1,…,N, is the p-Laplace equation.The rest of the paper is organized as follows. In Section2, we present some preliminary results. In Section 3, we study the existence and uniqueness of entropy solution.


## 2. Preliminaries

We recall in this section some definitions and basic properties of anisotropic Lebesgue and Sobolev spaces with variable exponent. Set(14)C+Ω¯=p∈CΩ¯such thatminx∈Ω¯⁡px>1.For any p∈C+(Ω¯), the variable exponent Lebesgue space is defined by (15)Lp·Ω≔u:Ω⟶R,measurable  such  that∫Ωupxdx<∞, endowed with the so-called Luxemburg norm (16)up·≔inf⁡λ>0such  that∫Ωuxλpxdx≤1.The p(·)-modular of the Lp(·)(Ω) space is the mapping ρp(·):Lp(·)(Ω)→R defined by (17)ρp·u≔∫Ωupxdx.For any u∈Lp(·)(Ω), the following inequality (see [9, 10]) will be used later:(18)min⁡up·p-;up·p+≤ρp·u≤max⁡up·p-;up·p+.For any u∈Lp(·)(Ω) and v∈Lq(·)(Ω) with (1/px)+(1/qx)=1 in Ω, we have the following Hölder type inequality:(19)∫Ωuvdx≤1p-+1q-up·vq·.If Ω is bounded and p,q∈C+(Ω¯) such that p(x)≤q(x) for any x∈Ω, then the embedding Lq(·)(Ω)↪Lp(·)(Ω) is continuous (see [11, Theorem  2.8]).Herein, we need the following anisotropic Sobolev space with variable exponent:(20)W1,p→·Ω≔u∈LpM·Ωsuch  that∂u∂xi∈Lpi·Ω,i=1,…,N.W1,p→(·)(Ω) is a separable and reflexive Banach space (see [2]) under the norm (21)up→·=upM·+∑i=1N∂u∂xipi·.We need the following embedding and trace results.Theorem 1 (see [9, Corollary  2.1]).
LetΩ⊂RN(N≥3) be a bounded open set and for all i=1,…,N, pi∈L∞(Ω), pi(x)≥1 a.e. in Ω. Then, for any q∈L∞(Ω) with q(x)≥1 a.e. in Ω such that (22)ess⁡infx∈Ω⁡pMx-qx>0,one has the compact embedding(23)W1,p→·Ω↪Lq·Ω.Theorem 2 (see [3, Theorem  6]).
LetΩ⊂RN(N≥2) be a bounded open set with smooth boundary and let p→(·)∈(C+(Ω¯))N, r∈C(Ω¯) satisfy the condition(24)1≤rx<min⁡p1∂x,…,pN∂x,∀x∈∂Ω.Then, there is a compact boundary trace embedding (25)W1,p→·Ω↪Lr·∂Ω.We introduce the numbers(26)q=Np¯-1N-1,q∗=Np¯-1N-p¯=NqN-q.The following result is due to Troisi (see [12]).Theorem 3.
Letp1,…,pN∈[1,+∞); g∈W1,(p1,…,pN)(Ω) and (27)q=p¯∗ifp¯∗<N,q∈1,+∞ifp¯∗≥N.Then, there exists a constant C4>0 depending on N,p1,…,pN if p¯<N and also on q and meas(Ω) if p¯≥N such that(28)gLqΩ≤C4∏i=1NgLpMΩ+∂g∂xiLpiΩ1/N.In this paper, we will use the Marcinkiewicz spaceMq(Ω) (1<q<+∞) as the set of measurable functions g:Ω→R for which the distribution function(29)λgk=meas⁡x∈Ω:gx>k,k≥0satisfies an estimate of the form(30)λgk≤Ck-q,for some finite constantC>0.We will use the following pseudonorm in Mq(Ω):(31)gMqΩ≔inf⁡C>0:λgk≤Ck-q,∀k>0.For any k>0, the truncation function Tk is defined by(32)Tks=max⁡-k;min⁡k;s.It is clear that limk→+∞⁡Tk(s)=s and Tks=mins;k.In order to simplify the notation, for anyv∈W1,p→(·)(Ω), we use v instead of v|∂Ω for the trace of v on ∂Ω.SetT1,p→(·)(Ω) as the set of the measurable functions u:Ω→R such that, for any k>0, Tk(u)∈W1,p→(·)(Ω). We define the space Ttr1,p→(·)(Ω) as the set of functions u∈T1,p→(·)(Ω) such that there exists a sequence (un)n⊂W1,p→(·)(Ω) satisfying(33)un⟶ua.e.  inΩ,∂Tkun∂xi⟶∂Tku∂xiinL1Ω,∀k>0,there  exists  a  measurable  functionvon∂Ωsuch  thatun⟶va.e.  on∂Ω.We need the following lemma proved in [13].Lemma 4.
Letg be a nonnegative function in W1,p→(·)(Ω). Assume p¯<N and there exists a constant C>0 such that(34)∫ΩTkgpM-dx+∑i=1N∫g≤k∂g∂xipi-dx≤C1+k,∀k>0.Then, there exists a constant D, depending on C, such that(35)gMq∗Ω≤D,where q∗=N(p¯-1)/(N-p¯).


## 3. Entropy Solutions

The notion of entropy solutions to problem (1) where the data f belongs to L1(Ω) is the following.Definition 5.
A measurable functionu∈Ttr1,p→(·)(Ω) is an entropy solution of problem (1) if b(u)∈L1(Ω), γ(u)∈L1(∂Ω), and, for every k>0,(36)∑i=1N∫Ωaix,∂u∂xi∂∂xiTku-φdx+∫ΩbuTku-φdx+∫∂ΩγuTku-φdσ≤∫ΩfTku-φdx,for every φ∈W1,p→(·)(Ω)∩L∞(Ω).The existence result is the following theorem.Theorem 6.
Assume that (4)–(12) hold. Then, problem (1) admits at least one entropy solution.Proof. The proof is done in three steps.Step 1 (the approximate problem).
We define the reflexive space(37)E=W1,p→·Ω×LpM·∂Ω.Let X0 be the subspace of E defined by (38)X0=u,v∈E:v=τu,where τ(u) is the trace of u∈Ttr1,p(·)(Ω) in the usual sense, since u∈W1,p→(·)(Ω). In the sequel, we will identify an element (u,v)∈X0 with its representative u∈W1,p→(·)(Ω).
For anyn∈N and ε>0, we consider the sequence of approximate problems:(39)∑i=1N∫Ωaix,∂un∂xi∂v∂xidx+ε∫ΩunpMx-2unvdx+∫ΩTnbunvdx+∫∂ΩTnγunvdσ=∫Ωfnvdx,where fn=Tn(f), and we define an operator An by (40)Anu,v=Au,v+∫ΩTnbuvdx+∫∂ΩTnγuvdσ∀u,v∈X0,where (41)Au,v=∑i=1N∫Ωaix,∂u∂xi∂v∂xidx+ε∫ΩupMx-2uvdx.Note that(42)fn∞≤f1meas⁡Ω,fn1=∫Ωfndx≤∫Ωfdx=f1,fn→n→+∞n→+∞finL1Ω,a.e.  inΩ.
Assertion  1 (the operator
A
is of type M). (i) The operator A is monotone. Indeed, for u,v∈W1,p→(·)(Ω), we have(43)Au-Av,u-v=Au,u-v+Av,v-u=∫Ω∑i=1Naix,∂u∂xi∂u-v∂xidx+ε∫ΩupMx-2uu-vdx+∫Ω∑i=1Naix,∂v∂xi∂v-u∂xidx+ε∫ΩvpMx-2vv-udx=∫Ω∑i=1Naix,∂u∂xi-aix,∂v∂xi∂u∂xi-∂v∂xidx+ε∫ΩupMx-2u-vpMx-2vu-vdx.Therefore,(44)Au-Av,u-v≥0,since for i=1,…,N, for almost every x∈Ω, ai(x,·) and t↦|t|pM(x)-2t are monotone.
(ii) The operatorA is hemicontinuous. Indeed, for every u,v in W1,p→(·)(Ω), let (45)φ:t∈R⟼φt=Au+tv,vand let t,t0∈R be such that t→t0. We have w=u+tv→w0=u+t0v in W1,p→(·)(Ω).
Using the Hölder type inequality, there existsi0∈{1,…,N} such that (46)φt-φt0=Au+tv,v-Au+t0v,v≤∑i=1N∫Ωaix,∂w∂xi-aix,∂w0∂xi∂v∂xidx+ε∫ΩwpMx-2w-w0pMx-2w0vdx≤N1pi0-+1pi0′-ai0x,∂w∂xi0-ai0x,∂w0∂xi0pi0′·∂v∂xi0pi0·+ε1pM-+1pM′-wpMx-2w-w0pMx-2w0pM′·vpM·.Let us denote ψi0(x,w)=ai0(x,∂w/∂xi0).
Using assumption (5) and [11, Theorems  4.1 and 4.2] we have ψi0(x,w)→ψi0(x,w0) in Lpi0′(·)(Ω). Then, we deduce that φ is continuous; namely, the operator A is hemicontinuous.
Since the operatorA ismonotone andhemicontinuous, then according to Lemma  2.1 in [8], A is of type M.
Assertion  2 (the operator
A
n
is of type M). Indeed, let (uk)k∈N be a sequence in X0 such that(47)uk⇀uweakly  inX0,Anuk⇀χweakly  inX0′,limsupk→+∞⁡Anuk,uk≤χ,u.Since (48)Tnbuu≥0,Tnγuu≥0,by Fatou’s lemma, we obtain that(49)liminfk→+∞⁡∫ΩTnbukukdx+∫∂ΩTnγukukdσ≥∫ΩTnbuudx+∫∂ΩTnγuudσand thanks to the Lebesgue dominated convergence theorem, we have(50)limk→+∞⁡∫ΩTnbukvdx+∫∂ΩTnγukvdσ=∫ΩTnbuvdx+∫∂ΩTnγuvdσ,for all v in X0. Consequently, (51)Tnbuk+Tnγuk⇀Tnbu+Tnγuweakly  inX0′.Therefore, we deduce that (52)Auk⇀χ-Tnbu+Tnγuweakly  inX0′.As in Assertion  1, we prove that the operator A is of type M, so we have (53)Au=χ-Tnbu+Tnγu.Thus, it follows that (54)Anu=χ.Hence, An is of type M.
Assertion  3 (the operator
A
n
is coercive). Indeed, since (55)Tnbuu+Tnγuu≥0,then(56)Anu,u≥Au,u.According to (7), we have(57)Au,u≥C3∫Ω∑i=1N∂u∂xipixdx+ε∫ΩupMxdx.Denote (58)I=i∈1,…,N:∂u∂xipi·≤1,J=i∈1,…,N:∂u∂xipi·>1.We have (59)∫Ω∑i=1N∂u∂xipixdx=∑i∈I∫Ω∂u∂xipixdx+∑i∈J∫Ω∂u∂xipixdx≥∑i∈I∂u∂xipi·pi++∑i∈J∂u∂xipi·pi-≥∑i∈J∂u∂xipi·pi-≥∑i∈J∂u∂xipi·pm-≥∑i=1N∂u∂xipi·pm--∑i∈I∂u∂xipi·pm-≥∑i=1N∂u∂xipi·pm--N.Using the convexity of the application t∈R+↦tpm-, pm->1, we obtain (60)∫Ω∑i=1N∂u∂xipixdx≥1Npm--1∑i=1N∂u∂xipi·pm--N.Then,(61)Anu,u≥C3Npm--1∑i=1N∂u∂xipi·pm-+ε∫ΩupMxdx-C3N.
(i) Assume|u|pM(·)>1. Then, (18) gives ∫Ω‍|u|pM(x)dx≥|u|pM(·)pm-.
So, combining (56) and (61) we get (62)Anu,u≥C∑i=1N∂u∂xipi·pm-+upM·pm--C3N≥C2pm--1up→·pm--C3N,whereC=min⁡C3Npm--1;ε.
(ii) Assume|u|pM(·)≤1. Then, combining (56) and (61) we get (63)Anu,u≥C∑i=1N∂u∂xipi·pm-+upM·pm--1-C3N+ε∫ΩupMxdx≥C2pm--1up→·pm--1-C3N,whereC=min⁡C3Npm--1;1.Consequently, since pm->1, the operator An is coercive.
Besides, the operatorAn is bounded and hemicontinuous.
Then, for anyFn=Tn(f)∈E′⊂X0′, we can deduce the existence of a function un∈X0 such that (64)Anun,v=Fn,v,∀v∈X0.Namely, un is a weak solution of problem (39).We are now going to prove that these approximated solutionsun tend, as n goes to infinity, to a measurable function u which is an entropy solution of problem (1). To start with, we establish some a priori estimates.Step 2 (a priori estimates).
Assume (4)–(12) and let un be a solution of problem (39). We have the following results.Lemma 7.
There exists a constantC5>0 such that(65)∫ΩTkunpM-dx+∑i=1N∫un≤k∂un∂xipi-dx≤C5k+1.Proof.
Let us takeTk(un) as a test function in (39). Since (66)∫ΩTnbunTkundx+ε∫ΩunpMx-2unTkun+∫∂ΩTnγunTkundσ≥0,using relation (7), we obtain(67)C3∑i=1N∫un≤k∂un∂xipixdx≤kf1.Then, we have (68)∑i=1N∫un≤k∂un∂xipi-dx=∑i=1N∫un≤k;∂un/∂xi>1∂un∂xipi-dx+∑i=1N∫un≤k;∂un/∂xi≤1∂un∂xipi-dx≤∑i=1N∫un≤k∂un∂xipixdx+N·meas⁡Ω≤kC3f1+N·meas⁡Ω.Moreover, we have (69)∫ΩTkunpM-dx=∫Tkun≤1TkunpM-dx+∫Tkun>1TkunpM-dx≤meas⁡Ω+∫Tkun>1kpM-dx≤meas⁡Ω1+kpM-.Therefore, we get (70)∫ΩTkunpM-dx+∑i=1N∫un≤k∂un∂xipi-dx≤meas⁡Ω1+N+kpM-+kC3f1≤C51+k,where C5=max{meas⁡(Ω)(1+N+kpM-);(1/C3)f1}.Lemma 8.
For anyk>0, there exist two constants C7>0 and C8>0 such that (i)
unMq∗(Ω)≤C7,(ii)
∂un/∂xiMpi-q/p¯(Ω)≤C8, ∀i=1,…,N.Proof.
(i) This is a consequence of Lemmas4 and 7.
(ii)(a)
Letα≥1. For any k≥1, we have (71)λ∂un/∂xiα=meas⁡x∈Ω:∂un∂xi>α=meas⁡∂un∂xi>α;un≤k+meas⁡∂un∂xi>α;un>k≤∫∂un/∂xi>α;un≤kdx+λunk≤∫un≤k1α∂un∂xipi-dx+λunk≤α-pi-C′k+Ck-q∗.Then, there exists a positive constant C6 such that(72)λ∂un/∂xiα≤C6kα-pi-+k-q∗.Let us consider the function (73)g:1,+∞⟶R,t⟼gt=tαpi-+t-q∗.We have g′(t)=0 for t=(q∗αpi-)1/(q∗+1). Thus, if we take k=(q∗αpi-)1/(q∗+1)≥1 in (72) we get (74)λ∂un/∂xiα≤C6kq∗+1q∗1αpi-≤C6′α-q∗/q∗+1pi-≤C6′α-pi-q/p¯,∀α≥1,whereC6′is a positive constant.(b)
If0≤α<1, we have (75)λ∂un/∂xiα=meas⁡∂un∂xi>α≤meas⁡Ω≤meas⁡Ωα-pi-q/p¯.Then (76)λ∂un/∂xiα≤C6′+meas⁡Ωα-pi-q/p¯,∀α≥0.Therefore, we deduce that there exists a positive constant C8 such that (77)∂un∂xiMpi-q/p¯Ω≤C8,∀i=1,…,N.Step 3 (existence of entropy solution).
Using Lemma8, we have the following useful lemma (see [13]).Lemma 9.
Fori=1,…,N, as n→+∞, one has(78)aix,∂un∂xi⟶aix,∂u∂xiinL1Ωa.e.x∈Ω.In order to pass to the limit in relation (39), we also need the following convergence results which can be proved as in [7] (see also [1, 13]).Proposition 10.
Assume (4)–(12). If un∈W1,p→(·)(Ω) is a weak solution of (Pn) then the sequence (un)n∈N∗ is Cauchy in measure. In particular, there exist a measurable function u and a subsequence still denoted by un such that un→u in measure.Proposition 11.
Assume (4)–(12). If un∈W1,p→(·)(Ω) is a weak solution of (Pn) then (i)
there existss>1 such that un→u a.e. in Ω and moreover un⇀u in W1,s(Ω),(ii)
for alli=1,…,N, ∂un/∂xi converges strongly in L1(Ω). Moreover ai(x,∂un/∂xi) converges to ai(x,∂u/∂xi) in L1(Ω) strongly and in Lpi′(·)(Ω) weakly for all i=1,…,N,(iii)
un converges to some measurable function v a.e. in ∂Ω.We can now pass to the limit in relation (39).Letφ∈W1,p→(·)(Ω)∩L∞(Ω) and choosing Tk(un-φ) as a test function in (39), we get(79)∑i=1N∫Ωaix,∂un∂xi∂∂xiTkun-φdx+∫ΩTnbunTkun-φdx+ε∫ΩunpMx-2unTkun-φdx+∫∂ΩTnγunTkun-φdσ=∫ΩfnTkun-φdx.For the right-hand side of (79), we have(80)∫ΩfnxTkun-φdx⟶∫ΩfxTku-φdx,since fn converges strongly to f in L1(Ω) and Tk(un-φ) converges weakly-∗ to Tk(u-φ) in L∞(Ω) and a.e. in Ω.For the first term of (79), we have (see [13])(81)liminfn→+∞⁡∑i=1N∫Ωaix,∂un∂xi∂∂xiTkun-φdx≥∑i=1N∫Ωaix,∂u∂xi∂∂xiTku-φdx.We now focus our attention on the second term of (79). We have(82)TnbunTkun-φ⟶buTku-φa.e.x∈Ω,TnbunTkun-φ≤kbun.Now we show that |b(un)|≤f1/meas(Ω). Indeed, let us denote by (83)Hδs=min⁡s+δ;1,sign0+⁡s=1ifs>0,0ifs≤0.If θ is a maximal monotone operator defined on R, we denote by θ0 the main section of θ; that is, (84)θ0s=minimal  absolute  value  ofθsifθs≠∅,+∞ifs,+∞∩Dθ=∅,-∞if-∞,s∩Dθ=∅.Remark that as δ goes to 0, Hδ(s) goes to sign0+(s).We takeφ=Hδ(un-M) as a test function in (39) for the weak solution un and M>0 (a constant to be chosen later) to get(85)∑i=1N∫Ωaix,∂un∂xi∂∂xiHδun-Mdx+ε∫ΩunpMx-2unHδun-Mdx+∫ΩTnbunHδun-Mdx+∫∂ΩTnγunHδun-Mdσ=∫ΩfnHδun-Mdx.We have (86)∑i=1N∫Ωaix,∂un∂xi∂∂xiHδun-Mdx=1δ∑i=1N∫un-M+/δ<1aix,∂un∂xi∂∂xiun-M+dx=1δ∑i=1N∫0<un-M<δaix,∂un∂xi∂∂xiundx≥0according  to7,∫ΩunpMx-2unHδun-Mdx=∫un-M+/δ<1unpMx-2unun-M+δdx+∫un-M+/δ≥1unpMx-2undx≥1δ∫M<un<M+δunpMx-2unun-Mdx≥0.Note also that since the function γ is nondecreasing with γ(0)=0, we have (87)∫∂ΩTnγunHδun-Mdσ=∫∂Ω∩un-M+/δ<1TnγunHδun-Mdσ+∫∂Ω∩un-M+/δ≥1TnγunHδun-Mdσ=∫∂Ω∩un-M+/δ<1Tnγunun-M+δdσ+∫∂Ω∩un-M+/δ≥1Tnγundσ=1δ∫∂Ω∩0<un-M<δTnγunun-Mdσ+∫∂Ω∩un-M≥δTnγundσ≥0.Then, (85) gives (88)∫ΩTnbunHδun-Mdx≤∫ΩfnHδun-Mdx,which is equivalent to saying (89)∫ΩTnbun-TnbMHδun-Mdx≤∫Ωfn-TnbMHδun-Mdx.We now let δ go to 0 in the above inequality to obtain(90)∫ΩTnbun-TnbM+dx≤∫Ωfn-TnbMsign0+⁡un-Mdx.Choosing M=b0-1(fn∞) in the above inequality (since b is surjective), we obtain(91)∫ΩTnbun-Tnfn∞+dx≤∫Ωfn-Tnfn∞sign0+⁡un-b0-1fn∞dx.For any n>f1/meas(Ω), we have (92)∫Ωfn-Tnfn∞sign0+⁡un-b0-1fn∞dx=∫Ωfn-fn∞sign0+⁡un-b0-1fn∞dx≤0.Then, (91) gives (93)∫ΩTnbun-fn∞+dx≤0,∀n>f1meas⁡Ω.Hence, for all n>f1/meas(Ω), we have (Tn(b(un))-fn∞)+=0 a.e. in Ω, which implies that(94)Tnbun≤fn∞∀n>f1meas⁡Ω.Let us remark that as un is a weak solution of (39), then (-un) is a weak solution to the following problem:(95)
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,where a~i(x,ξ)=-ai(x,-ξ), b~(s)=-b(-s), and f~n=-fn.According to (94) we deduce that (96)Tn-bun≤fn∞∀n>f1meas⁡⁡Ω.Therefore(97)Tnbun≥-fn∞∀n>f1meas⁡⁡Ω.It follows from (94) and (97) that, for all n>f1/meas⁡⁡(Ω), Tnbun≤fn∞ which implies (98)bun≤fn∞≤f1meas⁡⁡Ωa.e.   inΩ.We can now use the Lebesgue dominated convergence theorem to get(99)limn→+∞⁡∫ΩTnbunTkun-φdx=∫ΩbuTku-φdx.By using again the Lebesgue dominated convergence theorem, we obtain(100)∫∂ΩTnγunTkun-φdσ⟶∫∂ΩγuTku-φdσ.For the third term of (79), let us prove that (101)liminfn→+∞⁡ε∫ΩunpMx-2unTkun-φdx≥0forε⟶0.We have (102)∫ΩunpMx-2unTkun-φdx=∫ΩunpMx-2un-φpMx-2φTkun-φdx+∫ΩφpMx-2φTkun-φdx.Since the quantity (|un|pM(x)-2un-|φ|pM(x)-2φ)Tk(un-φ) is nonnegative and since for all x in Ω, the application ξ↦|ξ|pM(x)-2ξ is continuous, we have (103)unpMx-2un-φpMx-2φTkun-φ⟶upMx-2u-φpMx-2φTku-φa.e.  inΩand by Fatou’s lemma, it follows that(104)liminfn→+∞⁡∫ΩunpMx-2un-φpMx-2φTkun-φdx≥∫ΩupMx-2u-φpMx-2φTku-φdx.We have (105)∫ΩφpMx-2φdx=∫ΩφpMx-1φdx≤∫Ωφ∞pMx-1dx≤∫φ∞≤1φ∞pMx-1dx+∫φ∞>1φ∞pMx-1dx≤meas⁡⁡Ω+φ∞pM+-1meas⁡⁡Ω<+∞.Hence, |φ|pM(x)-2φ∈L1(Ω).SinceTk(un-φ) converges weakly-∗ to Tk(u-φ) in L∞(Ω) and |φ|pM(x)-2φ∈L1(Ω), it follows that(106)limn→+∞⁡∫ΩφpMx-2φTkun-φdx=∫ΩφpMx-2φTku-φdx.By adding (104) and (106), we get(107)liminfn→+∞⁡∫ΩunpMx-2unTkun-φdx≥∫ΩupMx-2uTku-φdx.Since (108)∫ΩupMx-2uTku-φdx≤k∫ΩupMx-1dx<+∞,thus we get(109)liminfn→+∞⁡ε∫ΩunpMx-2unTkun-φdx≥0forε⟶0.Combining (80), (81), (99), (100), and (109) we obtain(110)∑i=1N∫Ωaix,∂u∂xi∂∂xiTku-φdx+∫ΩbuTku-φdx+∫∂ΩγuTku-φdσ≤∫ΩfTku-φdx.Then, u is an entropy solution of problem (1). That completes the proof of Theorem 6.We now state the uniqueness result of entropy solution.Theorem 12.
Assume that (4)–(12) hold true and let u be an entropy solution of (1). Then, u is unique.Proof. The proof is done in two steps.Step 1 (a priori estimates).
We consider the following.Lemma 13.
Assume (4)–(12) and f∈L1(Ω). Let u be an entropy solution of (1). Then(111)∑i=1N∫u≤k∂u∂xipixdx≤kC3f1and there exists a positive constant C9 such that(112)bu1≤C9·meas⁡Ω+f1.Proof.
Let us takeφ=0 in the entropy inequality (36).(i)
By the fact that∫Ω‍b(u)Tk(u)dx+∫∂Ω‍γ(u)Tk(u)dσ≥0, using (7), we get (111).(ii)
Also, using the fact that∑i=1N‍∫Ω‍ai(x,∂u/∂xi)(∂/∂xi)Tk(u)dx+∫∂Ω‍γ(u)Tk(u)dσ≥0, relation (36) gives(113)∫ΩbuTkudx≤∫ΩfxTkudx.By (113), we deduce that (114)∫u≤kbuTkudx+∫u>kbuTkudx≤kf1which imply that (115)∫u>kbuTkudx≤kf1or (116)∫u>kbudx+∫u<-k-budx≤f1.Therefore, (117)∫u>kbudx≤f1.So, we obtain (118)∫Ωbudx=∫u≤kbudx+∫u>kbudx≤∫u≤kbudx+f1.Since the function b is nondecreasing, then (119)∫u≤kbudx≤max⁡bk;b-k·meas⁡Ω.Consequently, there exists a constant C9=max{b(k);|b(-k)|} such that (120)bu1≤C9·meas⁡Ω+f1.Lemma 14.
Assume (4)–(12) and let f∈L1(Ω). If u is an entropy solution of (1), then there exists a constant D which depends on f and Ω such that(121)meas⁡u>k≤Dmin⁡bk,b-k,∀k>0and a constant D′>0 which depends on f and Ω such that(122)meas⁡∂u∂xi>k≤D′k1/pM-′,∀k≥1.Proof.
(i) For anyk>0, relation (112) gives (123)∫u>kmin⁡bk,b-kdx≤∫u>kbudx≤C9·meas⁡Ω+f1.Therefore, (124)min⁡bk,b-k·meas⁡u>k≤C9·meas⁡Ω+f1=D;that is, (125)meas⁡u>k≤Dmin⁡bk,b-k.
(ii) See [7] for the poof of (122).Lemma 15.
Assume (4)–(12) and let f∈L1(Ω). If u is an entropy solution of (1), then(126)limh→+∞⁡∫Ωfχu>h-tdx=0,where h>0 and t>0.Proof.
Since the functionb is surjective, according to Lemma 14-(121), we have (127)limh→+∞⁡meas⁡u>h-t=0and as f∈L1(Ω), it follows by using the Lebesgue dominated convergence theorem that (128)limh→+∞⁡∫Ωfχu>h-tdx=0.Lemma 16.
Assume (4)–(12) and let f∈L1(Ω). If u is an entropy solution of (1), then there exists a positive constant K such that(129)ρpi′·∂u∂xipix-1χF≤K,∀i=1,…,N,where F=h<u≤h+k, h>0, k>0.Proof.
Letφ=Th(u) as test function in the entropy inequality (36). We get (130)∑i=1N∫Ωaix,∂u∂xi∂∂xiTku-Thudx+∫ΩbuTku-Thudx+∫∂ΩγuTku-Thudσ≤∫ΩfTku-Thudx.Thus, (131)∑i=1N∫h<u≤h+kaix,∂u∂xi∂u∂xidx≤kf1and using (7), we have (132)∫F∂u∂xipixdx≤kC3f1,∀i=1,…,N.Consequently, (133)ρpi′·∂u∂xipix-1χF≤K,∀i=1,…,N.Step 2 (uniqueness of entropy solution).
Leth>0 and u,v be two entropy solutions of (1). We write the entropy inequality corresponding to the solution u, with Th(v) as test function, and to the solution v, with Th(u) as test function to get(134)∑i=1N∫Ωaix,∂u∂xi∂∂xiTku-Thvdx+∫ΩbuTku-Thvdx+∫∂ΩγuTku-Thvdσ≤∫ΩfTku-Thvdx,∑i=1N∫Ωaix,∂v∂xi∂∂xiTkv-Thudx+∫ΩbvTkv-Thudx+∫∂ΩγvTkv-Thudσ≤∫ΩfTkv-Thudx.Upon addition, we get(135)∑i=1N∫Ωaix,∂u∂xi∂∂xiTku-Thvdx+∑i=1N∫Ωaix,∂v∂xi∂∂xiTkv-Thudx+∫ΩbuTku-Thvdx+∫ΩbvTkv-Thudx+∫∂ΩγuTku-Thvdσ+∫∂ΩγvTkv-Thudσ≤∫ΩfTku-Thv+Tkv-Thudx.Define the following sets: (136)E1≔u-v≤k;v≤h;E2≔E1∩u≤h,E3≔E1∩u>h.We start with the first integral in (135). We have (137)∑i=1N∫u-Thv≤kaix,∂u∂xi∂∂xiTku-Thvdx=∑i=1N∫u-Thv≤k∩v≤haix,∂u∂xi∂∂xiTku-Thvdx+∑i=1N∫u-Thv≤k∩v>haix,∂u∂xi∂∂xiTku-Thvdx=∑i=1N∫u-v≤k∩v≤haix,∂u∂xi∂u-v∂xidx+∑i=1N∫u-hsign⁡v≤k∩v>haix,∂u∂xi∂u∂xidx≥∑i=1N∫E1aix,∂u∂xi∂∂xiu-vdx≥∑i=1N∫E2aix,∂u∂xi∂∂xiu-vdx+∑i=1N∫E3aix,∂u∂xi∂∂xiu-vdx.Then, we obtain(138)∑i=1N∫u-Thv≤kaix,∂u∂xi∂∂xiTku-Thvdx≥∑i=1N∫E2aix,∂u∂xi∂∂xiu-vdx-∑i=1N∫E3aix,∂u∂xi∂v∂xidx.According to (5) and the Hölder type inequality, we have (139)∑i=1N∫E3aix,∂u∂xi∂v∂xidx≤C1∑i=1N∫E3jix+∂u∂xipix-1∂v∂xidx≤C1∑i=1Njipi′·+∂u∂xipix-1pi′·,h<u≤h+k∂v∂xipi·,h-k<v≤h,where (140)∂u∂xipix-1pi′·,h<u≤h+k=∂u∂xipix-1Lpi′·h<u≤h+k.Thanks to relation (18) and Lemma 16, the quantity (|ji|pi′(·)+||∂u/∂xi|pi(x)-1|pi′(·),h<u≤h+k) is finite for all i=1,…,N.
According to Lemma15, the quantity |∂v/∂xi|pi(·),h-k<v≤h converges to zero as h goes to infinity. Consequently, the last integral of (138) converges to zero as h goes to infinity. Then,(141)∑i=1N∫u-Thv≤kaix,∂u∂xi∂∂xiTku-Thvdx≥Ih+∑i=1N∫E2aix,∂u∂xi∂∂xiu-vdx,with limh→+∞Ih=0.
We may adopt the same procedure to treat the second term in (135) to obtain(142)∑i=1N∫v-Thu≤kaix,∂v∂xi∂∂xiTkv-Thudx≥Jh-∑i=1N∫E2aix,∂v∂xi∂∂xiu-vdx,with limh→+∞⁡Jh=0.
For the other terms in the left-hand side of (135), we denote (143)Kh=∫ΩbuTku-Thvdx+∫ΩbvTkv-Thudx,Lh=∫∂ΩγuTku-Thvdσ+∫∂ΩγvTkv-Thudσ.We have (144)buTku-Thv⟶buTku-va.e.  inΩsinceh⟶+∞,buTku-Thv≤kbu∈L1Ω.Then, by the Lebesgue dominated convergence theorem, we obtain (145)limh→+∞⁡∫ΩbuTku-Thvdx=∫ΩbuTku-vdx,limh→+∞⁡∫ΩbvTkv-Thudx=∫ΩbvTkv-udx.Then,(146)limh→+∞⁡Kh=∫Ωbu-bvTku-vdx.In the same way, we get(147)limh→+∞⁡Lh=∫∂Ωγu-γvTku-vdσ.Now, consider the right-hand side of inequality (135). We have (148)limh→+∞⁡fTku-Thv+Tkv-Thu=0a.e.  inΩ,fxTku-Thv+Tkv-Thu≤2kf∈L1Ω.By the Lebesgue dominated convergence theorem, we obtain(149)limh→+∞⁡∫ΩfxTku-Thv+Tkv-Thudx=0.After passing to the limit as h goes to +∞ in (135), we get(150)∑i=1N∫u-v≤kaix,∂u∂xi-aix,∂v∂xi∂∂xiu-vdx+∫Ωbu-bvTku-vdx+∫∂Ωγu-γvTku-vdσ≤0.Since b, γ, and ai(x,·) are monotone, then (151)∫Ωbu-bvTku-vdx=0,(152)∫∂Ωγu-γvTku-vdσ=0,(153)∫u-v≤k∑i=1Naix,∂u∂xi-aix,∂v∂xi∂∂xiu-vdx=0.According to (6), we deduce from (153) that (154)u-v=ca.e.x∈Ω,wherecis  a  real  constant.For x fixed in ∂Ω, s↦|s|p(x)-2s is nondecreasing and vanishes at 0. Then, (155)γux-γvxTkux-vx≥0,∀x∈∂Ω,∀k>0.Now, by inequality above and (152), we deduce that for all k∈N∗ there exists Ck⊂∂Ω with meas(Ck)=0 such that, for all x∈∂Ω∖Ck, (156)γux-γvxTkux-vx=0.Therefore,(157)γux-γvxux-vx=0,∀x∈∂Ω∖⋃k∈N∗Ck.As p->1, the following relation is true for any ξ,η∈R, ξ≠η (cf. [14]): (158)ξpx-2ξ-ηpx-2ηξ-η>0.From inequality above and (157), we get (159)u-v=0a.e.  on∂Ω.Finally as (160)u-v=ca.e.inΩ,u-v=0a.e.  on∂Ω,it follows that (161)u=va.e.  inΩ.That completes the proof of Theorem 12.



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*Source: 919608-2015-11-24.xml*