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

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.