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      Lipschitz Continuity and Explicit Form of Solution in a Class of Free Boundary Problem with Neumann Boundary Condition

      2023-04-16 18:51:16SAADIAbderachid

      SAADI Abderachid

      Department of Mathematics,University of Msila,Algeria.

      Laboratory of Nonlinear Partial Differential Equations and History of Mathematics,ENS,Kouba,Algeria.

      Abstract. We consider a class of free boundary problems with Neumann boundary conditions.We would like to give certain results with regularity of solutions(mainly the local interior and boundary Lipschitz continuity).We will also show an explicit form of solution under well-specified conditions.

      Key Words: Lipschitz continuity;free boundary;Neumann boundary condition.

      1 Introduction

      In this article,we are interested in studying a free boundary problem with Neumann boundary conditions,whose weak formulation is as follows

      where Ω be aC1bounded domain of Rn,and Γ2,Γ3=(Γ3,i)1≤i≤Nare relatively open connected subsets of?Ω and Γ1=?Ω(Γ2∪Γ3),x=(x′,xn),e=(0,...,0,1),a(x)=(aij(x))is annbynmatrix satisfying these conditions below for Λ,λ>0

      In addition,H(x)is aC1()vector function not necessarily depending on the matrixa(x),satisfying these conditions follow for the positive constantsandc

      For a.e.x∈Γ3,the functionβ(x,·)is a nonnegative,nondecreasing function satisfying

      Andφis a nonnegative Lipschitz continuous function.

      This problem describes of some free boundaries,such as the aluminum electrolysis problem [1],the problem of lubrication [2],and the dam problem with leaky boundary conditions[3-6].The separation in the dam problem is between the part of the domain where the water is and the rest.The separation in the case of the aluminum electrolysis problem is between the liquid and solid region.The separation in the lubrication problem is between the nil-pressure zone and the other region.

      In[7],Carrillo and Chipot showed the existence of solution,the lower semi-continuity of the function representing the free boundary and the uniqueness of the solution,which they called”S3-connected solution”for the Dirichlet condition on Γ1∪Γ2.The existence proved in[3]in the casea(x)=InandH(x)=a(x)e.

      In[8],Chipot considered the problem in the case whereH(x)=h(x)·ewithe=(1,0)andhx1≥0 inD′(Ω).He proved that the free boundary is a graph of a continuous functionx1=?(x2).Challal and Lyaghfouri[9]proved the same result under weaker assumptions.This result is generalized in the article published by Saadi[10].

      In[11],Lyaghfouri generalized the result shown by Carrillo and Chipot in[7]to the case

      In[6],Chipot and Lyaghfouri showed that ifH(x)=a(x)·ethen,

      Moreover,fora(x)=I2then,?is a Lipschitz continuous function,and theS3-connected solution is unique.

      For more results in the case of a nonlinear operator,see for example[9,11-15].

      This work is divided into four parts.In the second section,we highlighted the problem and suggested some preliminary results that have been proven before.In the third section,these results are needed to give an explicit form ofχ.In the fourth section,we finally show the local interior and boundary Lipschitz continuity.We did this by using the methods from[16,17].

      2 Statement of the problem and preliminary results

      This paragraph is meant to show the weak formulation of the problem that needs to be solved.The reader is asked to look at reference[6]to learn more about the origin of the problem.The authors gave a well-detailed description on the dam problem.They arrived at the following strong formulation:

      whereAis a subset of Ω such that?Athe boundary ofAis partitioned into four parts:S1=Γ1,the free boundaryS2=?A∩Ω,S3=Γ3,andS4?Γ2.

      Now,we consider the following generalized problem:

      Assuming that the free boundary ofAis smooth enough,then for a smooth test functionξwe can write

      Using the divergence formula,we get

      hence,

      Let’s say thatξ ≥0 on Γ2,and taking into accountS4?Γ2,we get

      Prolonginguby 0 outsideA,and calling it alwaysu,gives us

      When we replaceχAby a functionχ∈L∞(Ω)that makesχ=1 ifu>0,we get the weak formulation of the problem

      For the existence of a solution of (P),we refer for example to [6].The steps are the same,but we replace the functiona(x)·ewith the functionH(x).The following results were established in[4,5,18]:

      Proposition 2.1.We have the following in D′(Ω):

      Proposition 2.2.We also have

      Following[4,5,19],we consider the differential equation

      whereh∈πxn(Ω),w∈πx′(Ω∩{xn=h}).This system has a maximal solutionX(·,w,h)defined on(α-(w,h),α+(w,h)),and continuous on the open set

      Moreover,we have

      We will denote in the sequelX(t,w,h),α-(w,h),α+(w,h)byX(t,w),α-(w),α+(w).

      Definition 2.1.For any h∈πxn(Ω),we define the set as:

      and the mappings

      where

      is the curvilinear abscissa of the point X(t,w)in the curve X(·,w).

      Note that

      Now,let us look at a few properties:

      Proposition 2.3.α- and α+are C1functionsa.e.on πxn(Ω∩[xn=h]).

      Proof.Leth∈πxn(Ω)andw0∈πxn(Ω∩{xn=h}).Then,following the steps of the proof of Proposition 2.1 in[4],since Γ isC1there existsη >0 small enough,and aC1functionσ=(σk)(w)such that one of the following is true for everywinBη(w0)

      Assume thati=n,i.e.Xn(α+(w),w)=σ(X1(α+(w),w),···,Xn-1(α+(w),w)),this mean thatα+(w)satisfies

      So,the implicit function theorem shows the existence ofδ∈(0,η)and a unique functionf∈C1(Bδ(w0),R)such thatf(w0)=α+(w0)and

      SinceF(α+(w),w)=0,we know thatα+(w)=f(w)andα+∈C1(Bδ(w0)).The same proof holds forα-.

      Now,we set:Yh(t,w)=det(J Th) andZh(t,w)=det(J Sh).The next proposition is an extension of the Proposition 2.2 in[4]:

      Proposition 2.4.We have

      We will denote respectively by:Th,Yhthe functionsTh?,and denote bythe functions(u?Th?)and(χ?Th?).The following proposition confirms thatχis decreasing:

      Proposition 2.5.Let(u,χ)be a solution of(P),then for each h∈πxn(Ω)we have

      Using the same arguments from the proof of Theorem 3.1 in[4],we have the following proposition:

      Proposition 2.6.Let(u,χ)be a solution of(P).and X0=Th(t0,w0)∈Th(Dh).

      So,we can use the following function to describe the free boundary:

      Definition 2.2.For each h∈πh(Ω),we define the functionΦh on πx′(Ω∩{xn=h})

      The proof of the next proposition is analogous to that of Proposition 3.1 in[4].

      Proposition 2.7.Φh is lower semi continuous at each w∈πx(Ω∩{xn=h}).Moreover,

      3 Explicit form of χ

      First,we have the following lemma which is established in[6]:

      Lemma 3.1.Let(u,χ)be a solution of(P).Ch a connected component of {τ<Φh(w)} such that:∩Γ3=?.We set: Zτ0={πw(Ch)×(τ0,+∞)∩Sh(Dh).Then,we have

      Next,we can talk about the following theorem:

      Theorem 3.1.Let(u,χ)be a solution of(P),andbe a point ofΩ.We denote by Br(w0,τ0)a ball of center(w0,τ0)and radius r contained in Sh(Dh),and set

      If=0in Br(w0,τ0),then we have

      1.=0in Cr={(w,τ)∈Sh(Dh):|w-w0|<r,τ >τ0}∪Br(w0,τ0),

      Proof.1.According to case 2 of Proposition 2.6,we have=0 inCr.

      2.Applying Lemma 3.1 with domains of typeZτ0,we obtain

      So,χ=0 a.e.inTh(Zτ0).

      Thus,it leads to the conclusion that for all domainsZτ0?Cr.Hence,χ=0 a.e.inTh(Cr).

      3.Sinceu=0 inTh(Cr)then,div(χH(X))=0 inD(Th(Cr)).

      which leads to

      On the other hand

      We deduce from Proposition 2.3 thatx′(x′,σ(x′)) is a parametrization of Γ3∩?Th(Cr),then we can write:

      But we have(x′,σ(x′))=Th(α+(w),w))=g(w)andgis aC1function such that

      From(2.9),we obtain

      Then,

      Hence,Yh(α+(w),w)=(-1)n+1H·ν(α+(w),w)detJ g(α+(w),w).We deduce that

      Using the change of variablegin(3.5),we can show that

      It follows from(3.4)and(3.6)that

      4 Lipschitz continuity of u

      First,the local interior Lipschitz continuity is given by the following theorem:

      Theorem 4.1.Let(u,χ)be a solution of(P).Then,

      To prove this theorem,we refer to[9]whenH(x)=h(x)e,and to[18]for more general situation.

      Now,we have the following theorem that proves the Lipschitz continuity up to the boundary:

      Theorem 4.2.Assume thatΓ3is of class C1,1,and let(u,χ)be a solution of(P).Then,u∈

      We will need two lemmas to prove this theorem.

      Lemma 4.1.It is enough to establish the result whenΓ3?[xn=0].

      Proof.Letx0∈Γ3,then there exists a neighbourhoodVofx0in Rn,and aC1,1bijectionψ:such thatandψ(Q0)=V∩Γ3,where

      Now,set fory∈Q-

      On the other hand,we have

      Lemma 4.2.Let’s sayΓ3?[xn=0]is a constant function and let∈Γ3and R>0be such that?Ω.Then,

      where C is a positive constant,depends only to λ,Λ,M and R.

      Proof.Letx0∈Γ3andR >0 be such that(x0)?Ω.Letz0=(x′,x0n+R),and let ΩR=B2R(z0)∩Ω.We consider the functionv(x)=ψ(d1(x))defined forx∈ΩR,whered1(x)=|x-z0|-Rand

      Taking into account thatψ′≥0,ψ′′≤0,we get the formula below by using the same arguments as in[17]:

      On the other hand,on Γ3we have

      Now,we have

      Ifψ(R)≥maxΩR u,then by(4.5)we have

      Using(4.3)and(4.4),we get

      Adding(4.7)to(4.8),we obtain

      Sinceβ(x,·)is non decreasing,andξ=0 on ΓR∩{u<v},we get

      Then,(4.9)can be written

      From(1.1),we deduce that

      Letting0 in(4.10),we get-div(a(x)?)(u-v)+≤0 inD′(ΩR).

      So,from(4.5),(4.6)we get(u-v)+≤0 in ΩR.This lead tou≤vin ΩR.Sincev(x0)=ψ(0)andψ(R)=0,we deduce that for allx∈ΩR

      The lemma is true because

      Proof of Theorem4.2.We will follow the steps that were used to prove the Theorem 3.1 in[17].Letx0∈Q0andR>0 such thatB-(x0,3R)?Ω.We shall prove that?uis bounded inB-(x0,r)by a constantCdepending only onλ,Λ,M,bandR.We have two cases:

      1)B-(x0,2R)?{u>0}

      Since-div(a(x)?u)=divH(x)inD′(O),whereOis a neighbourhood ofwe deduce from Corollary 8.36 in[20]and the remark that follows,that

      In particular,?u≤Cfor allx∈B-(x0,2R).

      2)?xf∈B-(x0,2R)∩{u=0}

      Letx∈B-(x0,R) such thatu(x)>0,andr=dist(x,{u=0}).We haver ≤|x-xf|<2R,Br(x)∩Ω?{u>0}and((x)∩Ω)?(B-(x0,3R)∩Ω).We also distinguish two cases:

      a)Br(x)∩Q0=?:

      In this case,from the Lemma 3.2 in[9],we haveu(x)≤cr,wherecdepends only onλ,Λ,,candR.The functionis defined inBO(1),and we have

      wherear(y)=a(x+r(x)y),Hr(y)=H(x+r(x)y).

      Applying Theorems 8.17 and 8.18 in [20] to the equality (4.11),we get for a positive constantp>nand another positive constantC1depending only onλ,Λ,b,p

      But we have

      Then,?uis uniformly bounded in

      b)?y1∈Br(x)∩Q0:

      There are two cases:

      b1)(x)?Ω:

      By Lemma 4.2,for all(x) there exists a positive constantCdepending onlyλ,Λ,MandRsuch that

      Then,by arguing exactly ofa)we can prove that?uis uniformly bounded in(x).

      Moreover,ursatisfies

      Applying Corollary 8.36 of[20],and the remark that follow,we get

      In particular,|?ur|is uniformly bounded.Hence,|?u|is uniformly bounded in

      Acknowledgement

      I am immensely grateful toPr Abdeslam Lyaghfourifor his efforts that have greatly contributed to the production of this work.

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