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ScienceDirect

Ann. I. H. Poincaré – AN 32 (2015) 159–182

www.elsevier.com/locate/anihpc

Permeability through a perforated domain for the incompressible 2D Euler equations

V. Bonnaillie-Noël

a

, C. Lacave

b,

, N. Masmoudi

c

aIRMAR - UMR6625, ENS Rennes, Univ. Rennes 1, CNRS, UEB, av Robert Schuman, 35170 Bruz, France

bUniversité Paris-Diderot (Paris 7), Institut de Mathématiques de Jussieu - Paris Rive Gauche, UMR 7586 - CNRS, Bâtiment Sophie Germain, Case 7012, 75205 Paris Cedex 13, France

cCourant Institute, 251 Mercer St., New York, NY 10012, USA

Received 17 June 2013; received in revised form 26 October 2013; accepted 12 November 2013 Available online 15 November 2013

Abstract

We investigate the influence of a perforated domain on the 2D Euler equations. Small inclusions of sizeεare uniformly dis- tributed on the unit segment or a rectangle, and the fluid fills the exterior. These inclusions are at least separated by a distanceεα and we prove that forαsmall enough (namely, less than 2 in the case of the segment, and less than 1 in the case of the square), the limit behavior of the ideal fluid does not feel the effect of the perforated domain at leading order whenε→0.

©2013 Elsevier Masson SAS. All rights reserved.

1. Presentation

The homogenization of the Stokes operator and of the incompressible Navier–Stokes equations in a porous medium is by now a very classical problem[29,31,2,26]. Recently, more attention was given to the homogenization of other fluid models such as the compressible Navier–Stokes system[10,24], the acoustic system[11,4]and the incompress- ible Euler system[27,21,16].

The goal of this paper is to study the effect of small inclusions of sizeεon the behavior of an ideal fluid governed by the 2D Euler system. One can expect that for very small holes which are well separated, the effect of the inclusions disappears at the limit. This is in the spirit of[8,3]where critical sizes of the holes were studied.

1.1. The perforated domain

LetKbe a smooth simply-connected compact set ofR2, which is the shape of the inclusions. More precisely, we assume that∂Kis aC1,1Jordan curve. Without loss of generality, we assume that 0∈K(−1,1)2. Letα >0 and

* Corresponding author.

E-mail addresses:bonnaillie@math.cnrs.fr(V. Bonnaillie-Noël),lacave@math.jussieu.fr(C. Lacave),masmoudi@cims.nyu.edu (N. Masmoudi).

URLs:http://www.ens-rennes.fr/math/people/virginie.bonnaillie(V. Bonnaillie-Noël),http://www.math.jussieu.fr/~lacave/(C. Lacave), http://www.math.nyu.edu/faculty/masmoudi/(N. Masmoudi).

0294-1449/$ – see front matter ©2013 Elsevier Masson SAS. All rights reserved.

http://dx.doi.org/10.1016/j.anihpc.2013.11.002

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Fig. 1. Geometrical settings.

μ∈ [0,1]be two parameters which represent how the inclusions fill the square[0,1]2. Fori1,j1 andε >0, we define

zε,αi,j :=

ε+2(i−1) ε+εα

, ε+2(j−1)

ε+εα

=(ε, ε)+2 ε+εα

(i−1, j−1), (1.1)

the centers of the inclusions of sizeε:

Kε,αi,j :=zε,αi,j +εK. (1.2)

The geometrical setting is represented inFig. 1(in the case whereK=B(0,1)).

LetNε,α= [2(ε1++εαα)](where[x]denotes the integer part ofx) be the number of inclusions, of sizeεand separated by 2εα, that we can distribute on the unit segment[0,1](seeFig. 1(a)). In vertical axis, we assume that there are [(Nε,α)μ]inclusions of sizeεat distance 2εα(withμ∈ [0,1]). For shorter, we denote byn1the number of inclusions along the horizontal axis andn2those on the vertical axis:

n1:=Nε,α and n2:=

(Nε,α)μ .

We denote byRε,α,μthe rectangle containing all the inclusions:

Rε,α,μ= 0,2

ε+εα

n1−2εα

× 0,2

ε+εα

n2−2εα

. (1.3)

Then the total number of inclusions inRε,α,μequals n1n2(Nε,α)1+μ

1+2εα 2(ε+εα)

1+μ

1

+εα)1+μ, (1.4)

as soon asεis small enough.

We notice that ifμ=0, thenn2=1 and there are just inclusions along a line. Ifμ=1, then there are as many inclusions in both directions and in this case the rectangleRε,α,μis almost the square[0,1]2.

We defineΩε,α,μthe domain Ωε=Ωε,α,μ:=R2\

n

1

i=1 n2

j=1

Kε,αi,j

. (1.5)

Since the parametersαandμare fixed and we are interested in the limitε→0, the indicesα,μwill often be omitted in the notation for shorter.

1.2. The Euler equations

Letuε=uε(t, x)=(uε1(t, x), uε2(t, x))be the velocity of an incompressible, ideal flow inΩε. The evolution is governed by the Euler equations

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⎧⎪

⎪⎪

⎪⎨

⎪⎪

⎪⎪

tuε+uε· ∇uε= −∇pε in(0,)×Ωε, divuε=0 in[0,∞)×Ωε, uε·n=0 in[0,∞)×∂Ωε,

|xlim|→∞uε(t, x)=0 fort∈ [0,∞), uε(0, x)=uε0(x) inΩε.

(1.6)

Letωεbe the vorticity defined by ωε:=curluε=1uε22uε1. The velocity and the vorticity satisfy

⎧⎪

⎪⎨

⎪⎪

divuε=0 in[0,∞)×Ωε, curluε=ωε in[0,∞)×Ωε, uε·n=0 in[0,∞)×∂Ωε,

|xlim|→∞uε(t, x)=0 fort∈ [0,∞).

(1.7)

The initial velocity in(1.6)has to verify:

divuε0=0 inΩε, lim

|x|→∞uε0(x)=0, uε0·n=0 on∂Ωε. (1.8)

As our domain depends onε, it is standard to give the initial data in terms of an initial vorticity independent ofε.

Physically, it is relevant to consider the following setting: we assume that the fluid is steadyuε0≡0 at timet <0 (thenωε(0,·)≡0 anduε0has zero circulation around each inclusion) and at timet=0 we add, by an exterior force, a vorticityω0. More precisely, letω0Cc(R2), then we infer that there exists a unique vector fielduε0verifying(1.8) which has zero circulation around each inclusion and whose curl is: curluε0=ωε0:=ω0|Ωε (see e.g.[18,20]).

Then, the vorticity allows us to give an initial condition independent ofε, but the main advantage of the vorticity for the 2D Euler equations comes from the nature of the equations governing the vorticity:

⎧⎪

⎪⎪

⎪⎪

⎪⎪

⎪⎪

⎪⎪

⎪⎨

⎪⎪

⎪⎪

⎪⎪

⎪⎪

⎪⎪

⎪⎪

tωε+uε· ∇ωε=0 in(0,)×Ωε, divuε=0 in[0,∞)×Ωε, uε·n=0 in[0,∞)×∂Ωε, curluε=ωε in[0,∞)×Ωε,

|xlim|→∞uε(t, x)=0 fort∈ [0,∞),

Kε,αi,j

uε(0, s)·τds=0 for alli, j, ωε(0,·)=ω0 inΩε.

(1.9)

We can show that the two systems(1.6)and(1.9)are equivalent, but we obtain more properties from the second system because it is a transport equation. Thanks to this structure, forω0Cc(R2), Kikuchi establishes in[18]that there exists a unique global strong solutionuεof(1.6), such thatωεbelongs to L(R+;L1∩Lε)). Actually, for a strong solutionuε, the transport nature of(1.9)implies that:

• the Lpnorm of the vorticity is conserved for anyp∈ [1,∞]:

ωε(t,·)

Lpε)=ωε0

Lpε)ω0Lp(R2),t0,∀p∈ [1,+∞]; (1.10)

• the total mass of the vorticity is conserved:

Ωε

ωε(t, x)dx=

Ωε

ω0(x)dx; (1.11)

• at any timet0, the vorticity is compactly supported (but the size of the support can grow);

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• the circulation ofuεaround each inclusion is conserved (Kelvin’s theorem):

Kε,αi,j

uε(t, s)·τds=0, ∀t0,∀i, j. (1.12)

1.3. Issue and former results

For(α, μ)(0,)× [0,1), the complementary of Ωε converges, in the Hausdorff sense, to the unit segment [0,1] × {0}and for(α, μ)(0,)× {1}, it converges to the unit square[0,1]2. Indeed, we check easily that

dH n

2

j=1 n1

i=1

Kε,αi,j,Rε,α,μ

=max

sup

x

i,jKε,αi,j

d(x,Rε,α,μ), sup

x∈Rε,α,μ

d

x,

i,j

Kε,αi,j

max

0,√ 2

ε+εα ,

and

dH

Rε,α,1,[0,1]2

max

0,2(ε+εα)

forμ=1, dH

Rε,α,μ,[0,1] × {0}

max

2(ε+εα)1μ,2(ε+εα)

forμ∈ [0,1).

The issue of this article is to determine the limit of (uε, ωε)whenεtends to zero, for different values of αandμ, and to compare the limit with the solution in the full plane, or in the exterior of a segment, or in the exterior of a square.

The well-posedness of the Euler equations in the full plane is well known since McGrath[25]. In the exterior of a sharp domain, let us mention that the existence of a global weak-solution to the Euler equations in the exterior of the segment, such thatω0∈L(R+;L1∩L(R2\([0,1] × {0}))), is established in[19]. Such a result is recently extended to the exterior of any connected compact set in[12], for example outside the unit square.

Physically, we can preview that we do not feel the presence of the inclusions for smallα (i.e.εα ε) and for small μ, whereas it should appear a wall for μ∈ [0,1)and α large, and the unit square for μ=1 and α large.

Moreover, we can think that the criticalαshould be a decreasing function in terms ofμ.

The study of the Euler equations in the exterior of one small obstacle was initiated by Iftimie, Lopes Filho and Nussenzveig Lopes in [16]. In that paper, the authors consider only one obstacle which shrinks homotetically to a point, and indeed, if the initial circulation is zero, then their result reads as the solution (uε, ωε)converges to the solution in the full plane. Later Lopes Filho has treated in[22]the case of several obstacles in a bounded domain when one of them shrinks to a point. The final result is the same: if initially the circulation is zero, we do not feel the presence of the point at the limit. Finally the last generalization can be found in[20]where an infinite number of obstacles is considered. We quote here the theorem in the case where all the initial circulations are equal to zero:

Theorem 1.1.Letω0Cc(R2). Let us also fixR0>0such thatsuppω0B(0, R0). For any sequences{zki}i=1···nkB(0, R0)nk, there exist a subsequence, again denotedk, and a sequenceεk∈R+ tending to zero such that the solutions (uk, ωk)of(1.9)in

Ωk:=R2\ n

k

i=1

B zki, εk

,

with initial vorticityω0|Ωk and initial circulations0around the balls, verify (a) ukustrongly inLploc(R+×R2)for anyp∈ [1,2);

(b) ωk ωweakinL(R+;Lq(R2))for anyq∈ [1,∞];

(c) the limit pair(u, ω)is the unique solution of the Euler equations in the full plane, with initial vorticityω0. In that theorem, we have extended uk andωk by zero in k)c. Therefore, we could consider zεi,j as in our configuration (see the first subsection), however there is no control onεk in terms of the distance between the points.

The size of the ball can be very small compare to this distance (i.e.α1), and the goal of this article is to get this control. Let us mention that all the works cited before[12,16,19,20,22]consider also non-zero initial circulations, and

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in particular around small obstacles[16,20,22]the authors find a reminiscent term which appears from the vanishing obstacles. Removing the assumption of zero initial circulations in the present work could be the subject of a future research.

Before stating our result, we also mention a work with the opposite result. The third author and Lions have treated in[21]a case which is close to our configuration with(α, μ)=(1,1). We write “close” because that article considers bounded domains[0,1]2\(n2

j=1

n1

i=1Kε,αi,j)and the initial condition is not exactly as ours. Nevertheless, in the spirit of homogenization and two scale convergence, the authors prove that the limit solution is not the Euler solution in the unit square but rather a two-scale system that describes the limit behavior. In particular the limit solution depends on the shape of the obstacles.

1.4. Result

As we can expect, our main result reads asfor anyμthere exists a criticalαc(μ),such that for anyαless than αc(μ),the perforated domain is perfectly permeable, i.e. the presence of the inclusions does not perturb the behavior of a perfect fluid. More precisely:

Theorem 1.2.LetΩεdefined in(1.1)–(1.5), then for allμ∈ [0,1], we define αc(μ)=2−μ.

Letω0be a smooth function compactly supported inR2(0, αc(μ))and any sequenceε→0, then the solutions (uε, ωε)of(1.9)inΩεwith initial vorticityω0|Ωε and initial circulations0around the inclusions, verify:

(a) uεustrongly inL2loc(R+×R2);

(b) ωε ωweakinL(R+×R2);

(c) the limit pair(u, ω)is the unique global solution to the Euler equations in the full planeR2, with initial vortic- ityω0.

Again, in the previous theorem and in all the sequel, we extend(uε, ωε)by zero inside the inclusions.

We note that the functionμαc(μ) is continuous, decreasing, positive, such that αc(0)=2 and αc(1)=1.

Finally, this result does not depend on the shapeKof the inclusions. This result can be extended to the case where the inclusions have not the same shape (seeRemark 2.3).

Even if zero circulation are treated in the previous theorem, the goal here is to investigate the effect of the ratio distance/size of the inclusions, an important parameter not controlled in[20]. Such a question is investigated by the first author on some elliptic problems such that the Laplace and Navier equations in[7,5,6], and we emphasize that the Euler equations are linked to such a problem. Indeed we already have a good control of the Lpnorm of the vorticity, and the velocity can be deduced from the vorticity by a kernel of the type∇ 1.

We note that a possible extension can be made considering a less regularω0, belonging to the space L1∩Lp(R2) for somep >2.

An important future work will be to prove that this αc(μ) is indeed critical, in the sense that we note a non- negligible effect from the inclusions ifααc(μ). In fact, the result in[21]is already a first hint that it is indeed the case at least in the caseμ=1 and for any type of obstacles.

Our result should be compared with critical values obtained with other equations. The study of the behavior of a flow through a porous medium has a long story in the homogenization framework. The most common set- ting is to consider a bounded domain Ω containing many tiny solid obstacles, distributed in each direction. For the Stokes equations with Dirichlet boundary condition, Cioranescu and Murat considered the case where the ratio Rε:=(size of the inclusions)/(distance)ise1/ε2/ε, and they obtained in[8]that the limit equation contains an addi- tional term due to the holes. Concerning Stokes and Navier–Stokes, Allaire extensively treated the previous problem, for e.g. in[3]he showed that ifRεe1/ε2(the rate of Cioranescu–Murat), the limit is the Stokes system (hence we do not feel the presence of the inclusions). IfRεe1/ε2/ε, we get the Darcy law (which was well known in the case where the ratio isε/ε, see references in[2]and also[13]for different ratios). And ifRε=e1/ε2/ε, we get the Brinkman type law. Therefore, the above study has treated every case for the viscous problem, and we note that

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the critical ratee1/ε2is very small compared toε/ε, which is the rate obtained in the case of the square (μ=1, α=αc(1)=1). Even if the Euler equations are related to the Laplace equation via the stream functions, the inclu- sions can generate different effects: there is no effect in some regimes for the ideal flow whereas there is for the Stokes equation.

However, an important question is to understand what is the role of the viscosity in the determination of the critical rate (see[27]for more motivation). For a modified Euler equations, Mikeli´c and Paoli[27]and Lions and Masmoudi [21]consider a bounded domain perforated in both direction where the rate isε/ε, and the limit homogenized system takes into account of the inclusions. Our result is complementary of these articles.

There are also many works concerning inclusions distributed on the unit segment (through grids, sieves or porous walls, we refer e.g. to Conca and Sepúlveda[9]and Sánchez-Palenlencia[30]). In this setting, the study of the Stokes and Navier–Stokes system is performed by Allaire[3], where he obtained the similar result than before, except that the critical rate ise1/ε/ε, which is naturally bigger thane1/ε2/ε, but which stays to be very small compared to our rate:ε/ε2(μ=0,α=αc(0)=2).

1.5. Plan of the paper

Thanks to the transport nature of the equation governing the vorticity, we will deduce easily from(1.10)the point (b) ofTheorem 1.2 from the Banach–Alaoglu theorem. In the sequel, we keep the notation ε even if we extract a subsequence. Indeed, as the limit pair is unique, we will be able to conclude that the limit is the same for any subsequence, so for the full sequence.

The difficulty is to prove (a), i.e. thatuε=uε[ωε]converges touwith u(x):=KR2[ω](x)= 1

R2

(xy)

|xy|2ω(y)dy, ∀x∈R2. (1.13)

This formula is the well-known Biot–Savart law in the full plane, i.e. which gives the unique vector field inR2which is divergence free, tending to zero at infinity, and whose curl isω. We need a strong convergence for the velocity in order to pass to the limit in the vorticity equation, and to conclude that the limit pair is well a weak solution of the Euler equations in the full plane. By uniqueness of weak solution (see[17]), it will end the proof of (c) andTheorem 1.2.

The main idea is to introduce an explicit modification ofKR2[ωε], denoted byvε[ωε], in order to have a tangent vector field inΩε whose curl isωεplus a small error term. In Section2, we recall the explicit formula of the Biot–

Savart law in the exterior of one obstacle K, thanks to the Riemann mapping which sendsKc toB(0,1)c, and we present a construction of this modification, based on some cut-off functions around each inclusion.

Then we will write the decomposition:

uεu = uε

ωε

vε ωε

+ vε

ωε

KR2

ωε + KR2

ωεω

=: rε ωε

wε ωε

+KR2

ωεω

. (1.14)

The central part of this article will be Section3: at timet fixed, we will look for the critical value ofα(in terms ofμ), below which the convergence ofwεto zero in L2(R2)holds. This will follow from a careful study of the explicit formula. Next, we will simply note thatrε is the Leray projection ofwε. As this projector is orthogonal in L2, this will give the convergence ofrεto zero in L2(R2). Thanks to these two convergences, we will prove in Section4the main theorem.

In the sequel,Cwill denote a constant independent of the underlying parameter (which will often beε), the value of which can possibly change from a line to another.

2. Explicit formula of the correction

InΩε, we note thatuε (solving(1.7)and having zero circulation around each inclusion) andKR2[ωε](see(1.13)) are divergence free, with the same curl and the same limit at infinity. Moreover, they have the same circulations because we compute by the Stokes formula that

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Kε,αi,j

KR2

ωε

(s)·τds=

Kε,αi,j

ωε(x)dx=0.

The only differences are thatKR2[ωε]is not tangent to∂Kε,αi,j, and that we do not have an explicit formula ofuε in terms ofωε. The goal of this section is to correct this lack of tangency.

In this section, we fix the timet, i.e. we considerf as a function depending only onx∈R2, belonging in L1∩ L(R2)whose support is bounded.

2.1. The Biot–Savart law in an exterior domain

In the full plane, we know that there is a unique vector fieldusatisfying inR2: divu=0, curlu=f, lim

|x|→∞u(x)=0, which is given by the standard Biot–Savart formula:

u(x)=KR2[f](x):= 1 2π

R2

(xy)

|xy|2f (y)dy= 1 2π∇

R2

ln|xy|f (y)dy, ∀x∈R2. (2.1) It is also well known (see e.g.[23]) that there is a universal constantCsuch that

KR2[f]

L(R2) 1 2π

R2

|f (y)|

|xy|dyL(R2)Cf1/2L1(R2)f1/2L(R2), (2.2) and iff is compactly supported, we have the following behavior at infinity:

KR2[f](x)=

R2f (y)dy 2π

x

|x|2+O 1

|x|2

.

We note here that consideringu0=KR2[ω0] ∈L2(R2)is too restrictive because it would imply that ω0=0.

In the exterior of a unit disk in dimension 2, we have again an explicit formula for the Biot–Savart law: there exists a unique vector fieldu[f]solving inR2\B(0,1):

divu[f] =0, curlu[f] =f, lim

|x|→∞u[f](x)=0, u[f] ·n|∂B(0,1)=0,

∂B(0,1)

u[f](s)·τds=0.

This vector fieldu[f]is given explicitly by:

u[f](x)= 1 2π

B(0,1)c

xy

|xy|2xy

|xy|2

f (y)dy+

B(0,1)cf (y)dy 2π

x

|x|2

= 1 2π∇

B(0,1)c

ln|xy||x|

|xy| f (y)dy,

with the notationz=z/|z|2 (coming from the image method in order to have a tangent vector field). As we have mentioned in the introduction, solving the elliptic equation(1.7)is equivalent to solving ψ=f, whereψis con- stant on the boundary (here, the boundary has only one connected component) and settingu:= ∇ψ. Hence, the previous Biot–Savart law comes from the explicit formula of the Green’s function inB(0,1)c. Another advantage of the dimension two is that we can extend this formula to the exterior of any simply-connected compact setK: thanks to the complex analysis (identifyingR2andC) and the fact that holomorphic function is a good change of variable for the Laplace problem. By the Riemann mapping theorem, there exists a unique biholomorphismT mappingKcto

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B(0,1)cand verifyingT()= ∞andT()∈R+. The last condition reads in the Laurent decomposition ofT at infinity:

T(z)=βz+γ+Oz→∞

1 z

, withβ∈R+. Then, we will use several times that

T(z)=βz+h(z), (2.3)

whereh is a holomorphic function satisfying at infinityh(z)=O(1)andh(z)=O(1/|z|2). Of course we have a similar behavior forT1.

In the sequel, we will need a kind of mean value theorem in a non-convex domain given by the following lemma:

Lemma 2.1.We assume thatKis a compact set such that∂Kis aC1,1Jordan curve. There existsCsuch that T(x)T(y)C|xy|,(x, y)

Kc2

, T1(x)T1(y)C|xy|,(x, y)

B(0,1)c2

. Proof. As long as the boundary isC1,α, we can extend the definition ofT andDT continuously up the boundary due to Kellogg–Warschawski theorem (see[28, Theorem 3.6]). Hence, by the behavior at infinity (see(2.3)), we infer that DT is uniformly bounded onKc. The same argument gives also thatT1is bounded onB(0,1)c.

By the connectivity ofKc, we know that for anyx, yKc, there exists a smooth pathγinKcjoiningxandy, and we have

T(x)T(y)= 1

0

DT γ (t)

γ(t)dt

DTL(γ ).

Therefore, it is sufficient to prove that there existsa1 such thatKc isa-quasiconvex, that is, for all pointsx, y there exists a rectifiable pathγ joiningx, yand satisfying

(γ )a|xy|.

We note easily thatB(0,1)cis π2-quasiconvex which ends the proof forT1.

ConcerningT, we remark thatKc cannot be quasiconvex if∂Khas a double point or a cusp. Conversely, if∂K is aC1Jordan curve, it is rather classical to show that there existsa1 such thatKcisa-quasiconvex. We refer to Hakobyan and Herron[15]for recent development about quasiconvexity. This kind of problem is although extensively studied in complex analysis, and Ahlfors shows in[1]the following equivalence in dimension two:

∂Kis a quasidisk ⇔ Kcis quasiconvex

where it is known that a Jordan curve, piecewiseC1, is a quasidisk iff∂Khas no cusp (see e.g.[14]). 2 Next, with the definitions(1.1)–(1.2), we setTi,jε,αas

Ti,jε,α(z)=T

zzi,jε,α ε

, (2.4)

the unique biholomorphism which maps(Kε,αi,j)c toB(0,1)cand satisfiesTi,jε,α()= ∞and(Ti,jε,α)()∈R+. Let us note that

Ti,jε,α1

(z)=εT1(z)+zε,αi,j. (2.5)

From these formulas andLemma 2.1, we will often use the following Lipschitz estimates:

Ti,jε,α

LipC

ε and Ti,jε,α1

LipCε. (2.6)

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Then we infer that there exists a unique vector fielduεi,j[f]solving inR2\Kε,αi,j: divuεi,j[f] =0, curluεi,j[f] =f, lim

|x|→∞uεi,j[f](x)=0, uεi,j[f] ·n|Kε,αi,j =0,

Kε,αi,j

uεi,j[f](s)·τds=0,

which is given explicitly by:

uεi,j[f](x)= 1 2π∇

(Kε,αi,j)c

ln|Ti,jε,α(x)Ti,jε,α(y)||Ti,jε,α(x)|

|Ti,jε,α(x)Ti,jε,α(y)| f (y)dy

= 1 2π∇

(Kε,αi,j)c

lnε|Ti,jε,α(x)Ti,jε,α(y)| |Ti,jε,α(x)|

β|Ti,jε,α(x)Ti,jε,α(y)| f (y)dy. (2.7) For more details and literature on this problem, we refer to[16, Section 2].

A useful estimate for the next section is:

Lemma 2.2.There existC1,C2,C3,C4four positive numbers such that for allε >0,α >0,μ∈ [0,1],i∈ {1, . . . , n1}, j∈ {1, . . . , n2},r >0, we have

Ti,jε,α

∂B zi,jε,α, r

Ki,jε,αc

B

0, C1

r ε

\B

0, C2

r ε

and

Ti,jε,α1

∂B(0, r+1)

B

zε,αi,j, εC3(r+1)

\B

zε,αi,j, εC4(r+1) .

Proof. With the definitions ofKε,αi,j andTi,jε,α (see(1.2)and(2.4)) we have to prove that there existC1,C2,C3,C4

such that for allεandrwe have:

T

∂B

0,r ε

Kc

B

0, C1

r ε

\B

0, C2

r ε

and T1

∂B(0, r+1)

B

0, C3(r+1)

\B

0, C4(r+1) .

The second point is obvious, becauseT1is continuous fromB(0,1)ctoKcand asT1(z)/z→1/β as|z| → ∞, we can infer thatz→ |T1(z)|/|z|has an upper and lower positive bounds. Indeed, we have assumed that there is a small neighborhood of zero insideK.

Actually, the first point is the same. Indeed, we are looking forC1, C2such that for alls >0 we have T

∂B(0, s)Kc

B(0, C1s)\B(0, C2s).

So, the conclusion comes with the same argument applied to z→ |T(z)|/|z| whereT is continuous fromKc to B(0,1)c. 2

Remark 2.3.Even if the shape ofK(via the bounds on the conformal map) is taking into account in the convergence proof, the limit vector field is independent on. Our analysis can be extended to the case where the inclusions have different shapesKk if we have uniform bounds (sup|βk|<∞, infβk>0, suphkL, and uniform constantsC1, C2,C3,C4inLemma 2.2). For example, this is the case when we consider a finite number of shapesK1,K2, . . . ,Kp (smooth simply-connected compact subsets of[−1,1]2).

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2.2. Definition and properties ofvε[f]

A similar modification was introduced in[20]in the case of a finite number of balls, whose centers are fixed and whose radii tend to zero. Our case is more difficult because the centers change, the shape of the inclusion is more general than a ball and the number of inclusions tends to infinity. The idea is to definevεsuch that it is equal to(2.7) in a neighborhood ofKε,αi,j and to(2.1)far away.

For this, let us define some cut-off functionsϕi,jε . LetϕC(R)be a positive non-increasing function such that ϕ(s)=

1 ifs1/2, 0 ifs1.

We define the cut-off functionϕi,jε onΩεby ϕi,jε (x)=ϕ

1

εαx−zε,αi,j

ε . This function isCalmost everywhere and satisfies

0ϕi,jε 1, ϕi,jε (x)=

1 ifxzε,αi,jε+ε2α, 0 ifxzε,αi,jε+εα,

and we recall thatKε,αi,j ⊂ {x∈R2, xzε,αi,jε}. From the definition, we note that ∇ϕi,jε

Lε) C

εα, (2.8)

meas

supp∇ϕi,jε

=4

ε+εα2

−4

ε+εα 2

2

=4εα+1+3ε4 εα

ε+εα

. (2.9)

Concerning the support ofϕi,jε we have meas

suppϕi,jε

=4

ε+εα2

ε2meas(K),

so meas(suppϕi,jε )=O(εα+εα))ifK= [−1,1]2, and meas(suppϕi,jε )=O(ε+ε2)if not. In any case, we have

meas

suppϕi,jε

4

ε+εα2

8

ε2+ε

. (2.10)

Moreover, we note easily that all the supports are disjoints, i.e. for all α >0, μ(0,1],(i, k)∈ {1, . . . , n1}2 and (j, l)∈ {1, . . . , n2}2, we have

ϕi,jε ϕk,lε ≡0 iff (i, j )=(k, l). (2.11)

Now, we can simply define our correction as:

vε[f] := ∇ψε[f], (2.12)

with

ψε[f](x):= 1 2π

1−

n2

j=1 n1

i=1

ϕi,jε (x)

Ωε

ln|xy|f (y)dy

+ 1 2π

n2

j=1 n1

i=1

ϕi,jε (x)

Ωε

lnε|Ti,jε,α(x)Ti,jε,α(y)||Ti,jε,α(x)| β|Ti,jε,α(x)Ti,jε,α(y)| f (y)dy

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= 1 2π

Ωε

ln|xy|f (y)dy

− 1 2π

n2

j=1 n1

i=1

ϕi,jε (x)

Ωε

ln β|xy|

ε|Ti,jε,α(x)Ti,jε,α(y)|f (y)dy + 1

n2

j=1 n1

i=1

ϕi,jε (x)

Ωε

ln |Ti,jε,α(x)|

|Ti,jε,α(x)Ti,jε,α(y)|f (y)dy.

From this definition and the previous subsection, it appears obvious that:

divvε[f] =0 inΩε, lim

|x|→∞vε[f](x)=0, vε[f] ·n|∂Ωε=0,

Kε,αi,j

vε[f](s)·τds=0, ∀i, j. (2.13)

We can also note that the curl ofvε[f]is equal tof inΩεplus some terms localized on the support of∇ϕεi,j. In this article, we do not need to estimate precisely this quantity, so we do not write its expression.

3. Convergence at fixed time

As we have said in the introduction, we want to decomposeuεuas in(1.14)and to pass to limit in each terms.

In this section, we fixed the time, i.e. we considerf as a function in Lc (R2). Then, we introduceuε[f]such that:

divuε[f] =0 inΩε, curluε[f] =f inΩε, lim

|x|→∞uε[f](x)=0, uε[f] ·n|∂Ωε=0,

Kε,αi,j

uε[f](s)·τds=0, ∀i, j, (3.1)

andvε[f]the correction ofKR2[f1Ωε], i.e.vε[f]given by(2.12).

LetM0>0 be fixed, the goal here is to prove the convergence of wε[f] :=KR2[f1Ωε] −vε[f] and rε[f] :=uε[f] −vε[f] to zero uniformly inf verifying

fL1L(R2)M0,

where we have extended by zerovε[f]anduε[f]inside the inclusions.

3.1. Convergence ofwε[f] =KR2[f1Ωε] −vε[f] First, in the inclusions, we prove that

Proposition 3.1.For allp∈ [1,∞), wε[f]

Lp(R2\Ωε)→0 asε→0, ∀(α, μ)(0,)× [0,1)∪(0,1)× {1}, (3.2) uniformly inf verifying

fL1L(R2)M0.

Proof. Indeed, we haveKR2[f1Ωε] −vε[f] =KR2[f1Ωε]onR2\Ωεand by(2.2)we write that KR2[f1Ωε]

Lp(R2\Ωε)KR2[f1Ωε]

L(R2)

meas

R2\Ωε1/p

CM0

meas

R2\Ωε1/p

.

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