231 S.ChallalELASTICBEHAVIOROFATWO-DIMENSIONALLATTICEBEAM Rend.Sem.Mat.Univ.Pol.TorinoVol.57,4(1999)

S. Challal

ELASTIC BEHAVIOR OF A TWO-DIMENSIONAL LATTICE BEAM

Abstract.

We study the linearized elasticity system and the dependence of the displace- ment on a small parameterεcharacterizing the length and the size of the period of the constitutive elements (bars or layers) of the structure. We show that, when ε→0, the structure becomes equivalent to a beam governed by the Bernoulli law.

1. Introduction

In this paper, we study the asymptotic behavior of linear elasticity equations in a two-dimensional domain perforated with holes periodically distributed in one direction. The size of the period is of the order of a small parameterεwhich is also the order of the thickness of the domain. This models many structures used in engineering such as lattice beams. We letεgo to zero and look for laws governing the structure. The limit structure is governed by the Bernoulli law (see [4]).

The main difficulties are to construct an extension operator of the displacement and the fact that the thickness and the period are in the same line of order. We overcome the last difficulty by using techniques developed by D. Caillerie [1] for thin elastic and periodic plates.

In section 2, we define the problem. In section 3, we give an a priori estimate and built an extension operator. Section 4 is devoted to a formal asymptotic study. In the last section we prove the convergence of our initial problem to the homogenized problem obtained in section 4.

2. Statement of the problem

The structure considered here is a two-dimensional periodic truss. The repeated element is named the basic cell. It may be simple (a single pattern in the basic cell) or complex (sev- eral patterns constituting the basic cell) (see Figure 1). To describe such family of structures, we denote by Y the representative cell.

Y =(0,L)×(−K/2,K/2), L,K>0.

The part of Y occupied by the material is denoted by Y^∗, the “hole” T = Y \Y^∗ does not intersect the boundary∂Y . We assume that∂Y^∗is Lipschitz continuous. We consider structures for which the number n₀of elementary cells is large and the inverseεof n₀will be taken as a small parameter. Our structure is then composed of identical cells which are homothetic in the ratioεto the basic cell Y (see Figure 2). We denote by L the length of the structure and we set:

_ε = (0,L)×(−εK/2, εK/2), Y_ε=εY, Y_ε^∗=εY^∗

^∗_ε : the part ofεoccupied by the material= ∪ⁿ_i=0⁰⁻¹τ_(xi,0)Y_ε^∗ τ_(xi,0) : translation of vector(x_i,0), x_i =iεL 0≤i≤n₀ T_ε : the set of holes.

231

We also use the notations:

0_ε¹ = [0,L]× {εK/2}: the upper boundary of the lattice beam^∗_ε 0_ε² = [0,L]× {−εK/2}: the lower boundary of the lattice beam^∗_ε 0₀^ε = {0} ×(−εK/2, εK/2) (resp.0^ε_L= {L} ×(−εK/2, εK/2)):

the left(resp. the right)boundary of the lattice beam^∗_ε. The current point inεis denoted by x =(x₁,x₂).

We assume the material to be anisotropic and satisfying the equations of linearized elastic- ity:

(1)















∂_jeσ_{i j}^ε + ef_i = 0 i=1,2 in^∗_ε eσ_{i j}^ε = ea^ε_{i j kh}ε_kh(eu_ε) i,j=1,2

eσ_{i j}^εn_j = F_iε^k i=1,2 on0_ε^k k=1,2 eσ_{i j}^εn_j = 0 i=1,2 on∂T_ε

eu_εi = 0 i=1,2 on0₀^ε∪0^ε_L

where n is the unit normal directed towards the exterior of^∗_ε,euε =(euε₁,euε₂)is the displace- ment,(eσ_{i j}^ε)is the stress tensor and ε_{i j}(euε)= 1

2

∂euε_i

∂x_j +∂eu_εj

∂x_i

is the linearized strain tensor.

The elasticity coefficientsea^ε_{i j kh}are defined by:

ea_{i j kh}^ε (x)= 1

ε²a_{i j kh} x₁ ε ,x₂

ε

where a_{i j kh}(y)(i,j,k,h=1,2)are bounded functions defined for y∈ =

∪_n∈(τ_{(n L,0)}(Y^∗)), Y₁-periodic (i.e. periodic in y₁∈Y₁=(0,L)) and satisfy:

(2)









i) a_{i j kh}(y)=a_{j ikh}(y)=a_{khi j}(y) for a.e. y∈Y^∗

ii) ∃m>0 such that∀τ=(τ_{i j})1≤i,j≤2∈⁴ τ_{i j}=τ_{j i}i,j=1,2

mτ_{i j}τ_{i j} ≤a_{j ikh}(y)τ_{i j}τ_kh for a.e. y∈Y^∗

iii) ∃M such that M=sup_y∈Y^∗{a_{j ikh}(y),i,j,k,h=1,2}.

We also set: (

a^ε_{i j kh}(x) = a_{i j kh}(^x_ε¹,x₂) a^ε_{i j kh}(x) = ¹

ε²a_{i j kh}^ε (x).

The structure is submitted to body forces ef = (ef₁, ef₂)and the upper and lower boundaries to forces F_ε¹=(F₁¹

ε ,F₂¹)and F_ε²=(F₁²

ε ,F₂²). There is no applied surface force on the boundary of the holes and the lattice beam is supposed to be clamped on0₀^ε∪0^ε_L. We assume

ef(x₁,x₂)= f(x₁,x₂/ε)∈[L²(ε)]², F_ε¹,F_ε²∈[L²(0,L)]². A weak formulation of problem (1) is

(3)

( Findeu_ε=(eu_ε1,eu_ε2)∈V_εsuch that:

R

^∗_εea^ε_{i j kh}ε_{i j}(euε)ε_kh(v)=R

^∗_ε ef+R

0_ε¹F_ε¹v+R

0²_εF_ε²v ∀v∈Vε

where Vε= {v∈[H¹(^∗_ε)]²/ v=0 on0^ε₀∪0^ε_L}is a Hilbert space provided with the usual norm of [H¹(^∗_ε)]².

By Lax-Milgram’s theorem and Korn’s inequality, we have a unique solution to problem (3).

REMARK1. The elastic modulus depend onε. This shows that the structure must be more rigid since it is more thin. The difference between the longitudinal and the transverse forces comes from the fact that the structure is more rigid under traction than under flexion.

We are interested in the dependence ofeu_ε onε(the length of the thickness of the lattice beam isεK and the period in the x₁direction isεL). For this, we first dilate our domain in the x₂direction, then we construct an extension operator to get an estimate for the displacement in a fixed domain.

3. A priori estimate

Let us introduce the following new variables:

y₁=x₁, y₂= x₂ ε .

Under this change of variables the set^∗_ε is expanded to ^∗_ε and0₀^ε(resp. 0^ε_L,0^k_ε, k= 1,2) becomes0₀(resp. 0_L,0^k, k= 1,2). We set for any functioneϕdefined on^∗_ε: ϕ(y₁,y₂)= e

ϕ(y₁, εy₂).

Then (1) can be written:

(4)























∂₁σ_i1^ε +¹_ε∂₂σ_i2^ε + f_i = 0 in _ε^∗ i=1,2

σ_{i j}^ε = a^ε_{i j 1k}^∂u_∂y^εk

1 +¹_εa^ε_{i j 2k}^∂u_∂y^εk

2

(y₁,y₂) i,j=1,2

σ_{1 j}^εn_j = ^F

k

ε1 on0^k k=1,2

σ_{2 j}^εn_j = F₂^k on0^k k=1,2

σ_{i j}^εn_j = 0 on∂ _ε i=1,2

u_εi = 0 on0₀∪0_L i=1,2

where∂ ε=∂ _ε^∗\0¹∪0²∪0₀∪0_L.

Let us takeeu_εas a test function in (3) and use (2) ii), we obtain:

(5) m

ε² Z

^∗_ε

ε_{i j}(eu_ε)ε_{i j}(eu_ε)≤ Z

^∗_ε e feu_ε+

Z

0_ε¹ F_ε¹eu_ε+

Z

0_ε² F_ε²eu_ε.

Letuˆε=(ˆu_ε1,uˆ_ε2)be the field defined on _ε^∗by:

(6) uˆ_ε1(y₁,y₂)= 1

εeu_ε1(y₁, εy₂), uˆ_ε2(y₁,y₂)=eu_ε2(y₁, εy₂).

Sinceeuε∈Vε, thenuˆε∈Hεwhere Hεis the Hilbert space defined by:

Hε= {v∈[H¹( _ε^∗)]²/ v=0 on0₀∪0_L} It is easy to see that we have:

ε₁₁(uˆε)(y₁,y₂)= ¹_εε₁₁(euε)(x₁,x₂), ε₂₂(uˆε)(y₁,y₂)=ε.ε₂₂(euε)(x₁,x₂), ε₁₂(uˆ_ε)(y₁,y₂)=ε₁₂(eu_ε)(x₁,x₂) with(x₁,x₂)=(y₁, εy₂).

So we deduce from (5)

(7) mεR

∗ε(ε₁₁(ˆuε))²+ ¹

ε⁴(ε₂₂(uˆε))²+ ²

ε²(ε₁₂(uˆε))²

≤R

∗

εε[εf₁uˆ_ε1+ f₂uˆ_ε2]+R

0¹F₁¹uˆ_ε1+F₂¹uˆ_ε2+R

0²F₁²uˆ_ε1+F₂²uˆ_ε2. Let us now prove the following lemma:

LEMMA1. Let H be the Hilbert space defined by: H= {v∈[H¹(Y)]²/ v=0 on0₀∪ 0_L}. There exists an extension operator Pε∈(Hε,H)such that:

(8) Z

Y

ε_{i j}(Pεv)ε_{i j}(Pεv)≤c Z

∗ ε

(ε₁₁(v))²+ 1

ε⁴(ε₂₂(v))²+ 2

ε²(ε₁₂(v))²∀v∈ Hε where c is a constant independent ofε.

Proof. It is done in two steps:

1^st step: it is a lemma due to Conca [3].

LEMMA2. There exists an extension operator S∈ ([H¹(Y^∗)]²,[H¹(Y)]²)and a con- stant c such that:

(9)

Z

Y

ε_{i j}(Sw)ε_{i j}(Sw)≤c Z

Y^∗

ε_{i j}(w)ε_{i j}(w) ∀w∈[H¹(Y^∗)]².

2^nd step: from definition of _ε^∗we have: _ε^∗= ∪ⁿ⁻¹_i=0τ_(xi,0)(ϕ(Y^∗))whereϕis the change of variable defined by:ϕ:(y₁,y₂)7−→(εy₁,y₂).

First letv∈[H¹(ϕ(Y^∗))]². Thenvoϕ ∈[H¹(Y^∗)]²and the functionwdefined byw = (v₁oϕ,1

εv₂oϕ)∈[H¹(Y^∗)]². From Lemma 2, Sw∈[H¹(Y)]²and satisfies (9). Set eSv=((Sw)₁oϕ⁻¹, ε(Sw)₂oϕ⁻¹).

We haveeSv∈[H¹(ϕ(Y))]²and the following inequality:

(10) Z

ϕ(Y)

ε_{i j}(eSv)ε_{i j}(eSv)≤c Z

ϕ(Y^∗)

(ε₁₁(v))²+ 1

ε⁴(ε₂₂(v))²+ 2

ε²(ε₁₂(v))² indeed we have

R

ϕ(Y)ε_{i j}(eSv)ε_{i j}(eSv) = εR

Y 1

ε²(ε₁₁(Sw))²+ε²(ε₂₂(Sw))²+2(ε₁₂(Sw))²

≤ ¹_εR

Yε_{i j}(Sw)ε_{i j}(Sw) (sinceε <1)

≤ ^c_εR

Y^∗ε_{i j}(w)ε_{i j}(w) (by(9))

and Z

Y^∗

ε_{i j}(w)ε_{i j}(w)= 1 ε Z

ϕ(Y^∗)

ε²(ε₁₁(v))²+ 1

ε²(ε₂₂(v))²+2(ε₁₂(v))² then (10) holds.

Next, ifv∈[H¹( ^∗_ε))]², we define the extension operator P_εby:

Pεv|_ϕ(Y) = eS(v|_ϕ(Y∗)) P_εv|_{τ (xi,0)(ϕ(Y))} = eS(v_|

τ (xi,0)(ϕ(Y∗))oτ (x_i,0))oτ (−x_i,0) i=1, . . . ,n₀−1 and we verify the inequality (8).

We have also the following inequalities:

LEMMA3. There exists a constant c independent ofεsuch that:∀v∈Hε, we have:

i) R

Y(ε₁₁(Pεv))² ≤ cR

∗

ε(ε₁₁(v))²+ ¹

ε⁴(ε₂₂(v))²+ ²

ε²(ε₁₂(v))² ii) R

Y(ε₂₂(Pεv))² ≤ cε⁴R

∗

ε(ε₁₁(v))²+ ¹

ε⁴(ε₂₂(v))²+ ²

ε²(ε₁₂(v))² iii) R

Y(ε₁₂(Pεv))² ≤ cε²R

∗

ε(ε₁₁(v))²+ ¹

ε⁴(ε₂₂(v))²+ ²

ε²(ε₁₂(v))². Proof. Using the same notations as in the proof of Lemma 2, we have forv∈H_ε:

R

ϕ(Y)(ε₁₁(eSv))² = εR

Y 1

ε²(ε₁₁(Sw))²

≤ ¹_εR

Yε_{i j}(Sw)ε_{i j}(Sw)

≤ ^c_εR

Y^∗ε_{i j}(w)ε_{i j}(w)

≤ cR

ϕ(Y^∗)(ε₁₁(v))²+ ¹

ε⁴(ε₂₂(v))²+ ²

ε²(ε₁₂(v))² then we deduce i). To prove ii) and iii) one can see that we have:

R

ϕ(Y)(ε₂₂(eSv))² = εR

Yε²(ε₂₂(Sw))²

≤ ε³R

Yε_{i j}(Sw)ε_{i j}(Sw)

≤ cε³R

Y^∗ε_{i j}(w)ε_{i j}(w)

≤ cε⁴R

ϕ(Y^∗)(ε₁₁(v))²+ ¹

ε⁴(ε₂₂(v))²+ ²

ε²(ε₁₂(v))²

and R

ϕ(Y)(ε₁₂(eSv))² = εR

Y(ε₁₂(Sw))²

≤ εR

Yε_{i j}(Sw)ε_{i j}(Sw)

≤ cεR

Y^∗ε_{i j}(w)ε_{i j}(w)

≤ cε²R

ϕ(Y^∗)(ε₁₁(v))²+ ¹

ε⁴(ε₂₂(v))²+ ²

ε²(ε₁₂(v))².

COROLLARY1. Letuˆεdefined by (6). Then we have:

i) |Pεuˆ_ε|_H ≤ c/ε, ii) |ε₁₁(P_εuˆ_ε)|_L2(Y) ≤ c/ε iii) |ε₂₂(P_εuˆ_ε)|_L2(Y) ≤ cε, iv) |ε₁₂(P_εuˆ_ε)|_L2(Y) ≤ c where c is a constant independent ofε.

Proof. From (7), Korn’s inequality and Lemma 1, we get:

|Pεuˆε|²_H ≤ _ε^c

|f|_L2(Y)|Pεuˆε|_L2(Y) + |F¹|_L2(0¹)|γ (Pεuˆε)|_L2(0¹) + |F²|_L2(0²)|γ (Pεuˆε)|_L2(0²)

where^{0 0}denotes the extension by 0 in Y\ ^∗_εandγthe trace operator.

Now by Poincar´e’s inequality and the continuity ofγ, we get i).

ii), iii) and iv) are consequences of lemma 3 and the following inequality:

(11)

Z

∗ε

(ε₁₁(uˆ_ε))²+ 1

ε⁴(ε₂₂(uˆ_ε))²+ 2

ε²(ε₁₂(ˆu_ε))²≤ c

ε|P_εuˆ_ε|_H ≤ c ε².

Now, we deduce estimates on the stress tensor defined by:

ˆ

σ_{i j}^ε =εσ_{i j}^ε =εa^ε_{i j 1k}∂u_εk

∂y₁ +1

εa^ε_{i j 2k}∂u_εk

∂y₂

(y₁,y₂) (y₁,y₂)∈ _ε^∗

which can be written ˆ σ_{i j}^ε =ε

εa^ε_{i j 11}ε₁₁(uˆε)+2a^ε_{i j 12}ε₁₂(ˆuε)+1

εa^ε_{i j 22}ε₂₂(uˆε) .

Then by (11) we have

| ˆσ_{i j}^ε|_L2(

ε∗)≤c

|ε₁₁(uˆε)|_L2(

ε∗)+1

ε|ε₁₂(ˆuε)|_L2(

∗ε)+ 1

ε²|ε₂₂(uˆε)|_L2(

ε∗) ≤c/ε

and

(12) | ˆσ_{i j}^ε|_L2(Y)≤c/ε.

As a consequence of estimates of Corollary 1 i), (12) and the fact that the spaces H¹(Y)and L²(Y)are reflexif, we obtain:

THEOREM1. There existsuˆ^∗ ∈[H¹(Y)]²andσˆ^∗∈[L²(Y)]⁴such that we have, up to a subsequence,

ε.(Pε(uˆε)) *uˆ^∗ in [H¹(Y)]² (13)

ε.(σˆ^ε) *σˆ^∗ in [L²(Y)]⁴.

In order to identifyuˆ^∗andσˆ^∗, we first do a formal study in the following section which allows us to get the homogenized problem. In Section 5, we justify this result.

4. A formal asymptotic study

In this section, we use a formal method to give the asymptotic behavior of the structure. For this, we look for uεandσ^εin the form:

u_εi(x₁,y₂) = ¹_εu⁻¹_i (x₁)+u⁰_i(x₁,^x_ε¹,y₂)+εu¹_i(x₁,^x_ε¹,y₂)+

ε²u²_i(x₁,^x_ε¹,y₂)+. . . σ_{i j}^ε(x₁,y₂) = ¹

ε³σ_{i j}⁻³(x₁)+ ¹

ε²σ_{i j}⁻²(x₁,^x_ε¹,y₂)+

1

εσ_{i j}⁻¹(x₁,^x_ε¹,y₂)+. . . , i,j=1,2

where the functions u^m_i (x₁,y₁,y₂),σ_{i j}^m(x₁,y₁,y₂)are Y₁-periodic (y₁∈ Y₁= (0,L))(y₁= x₁

ε).

We take back these expansions in the equations of (4) and we identify the terms of the same line of order ofε. We obtain:

(14)













∂σ_i1⁻³

∂y₁ +^∂σ

−3 i2

∂y₂ = 0

∂σ_i1^m

∂x₁ +^∂σ

m i1

∂y₁ +^∂σ

m+1 i2

∂y₂ = 0 for m6=0

∂σ_i1⁰

∂x₁ +^∂σ

1 i1

∂y₁ +^∂σ

1 i2

∂y₂ + f_i = 0 for m=0

(15)







σ_{1 j}^mn_j =0 for m6= −1, σ_{1 j}⁻¹n_j =F₁^k σ_{2 j}^mn_j =0 for m6=0, σ_{2 j}⁻¹n_j =F₂^k

σ_{i j}^mn_j =0 ∀m on∂Y_int^∗ , ∀x₁∈(0,L) i=1,2

on0^k k=1,2

where∂Y_int^∗ denotes the interior boundary of Y^∗and

(16) σ_{i j}^m=a_{i j 1k}∂u^m+2_k

∂x₁ +a_{i j hk}∂u^m+3_k

∂y_h ∀m.

Let us define:







N^m(x₁)= _|Y¹

1| R

Y^∗σ₁₁^m(x₁,y₁,y₂)d y₁d y₂=σ^m₁₁ : the normal force T^m(x₁)= _|Y¹

1| R

Y^∗σ₁₂^m(x₁,y₁,y₂)d y₁d y₂=σ^m₁₂ : the transverse shearing force M^m(x₁)= _|Y¹

1| R

Y^∗y₂σ₁₁^m(x₁,y₁,y₂)d y₁d y₂ : the bending couple.

Integrating the equilibrium equations (14) on Y^∗and using the boundary conditions (15), we

obtain 









∂σ^m_i1

∂x₁ = 0 ∀m6= −2,−1,0

∂σ⁻²₁₁

∂x1 +F₁¹+F₁² = 0 and ^∂σ

−2 11

∂x1 =0

∂σ⁻¹₁₁

∂x₁ = 0 and ^∂σ

−1 11

∂x₁ +F₂¹+F₂²=0.

Integrating the equation∂σ₁₁⁻²

∂x₁ +

∂σ_{1 j}⁻¹

∂y_j =0 on Y^∗after multiplying by y₂, we get

∂

∂x₁ Z

Y^∗ y₂σ₁₁⁻²

− Z

Y^∗

σ₁₂⁻¹+ |Y₁|K

2(F₁¹−F₁²)=0.

Then we deduce the equilibrium equation of the homogenized structure:

(17)







 d N⁻²

d x₁ (x₁) = −(F₁¹+F₁²)

d T⁻¹

d x₁ (x₁) = −(F₂¹+F₂²)

d M⁻²

d x1 (x₁)−T⁻¹(x₁) = −^K₂(F₁¹−F₁²).

Now, we are looking for the constitutive law of the structure.

First let us consider the problem satisfied byσ_{i j}⁻³. We have:

(18)







∂σ_{i j}⁻³

∂y_j =0 in Y^∗, σ_{i j}⁻³n_j=0 on∂Y_int^∗ ∪0¹∪0² σ_{i j}⁻³ Y₁−periodic, σ_{i j}⁻³=a_{i j 1k}^∂u

−1

∂xk₁ +a_{i j hk}^∂u

0

∂y_hk.

In these equations the variable x₁appears like a parameter. If we suppose the function u⁻¹ well-known, then we can write problem (18) as:

(19)











∂

∂yj

a_{i j hk}^∂u

0 k

∂yh

= −_∂y^∂

j

a_{i j 1k}^∂u

−1 k

∂x₁

in Y^∗ a_{i j hk}^∂u

0 k

∂y_hn_j= −a_{i j 1k}^∂u

−1 k

∂x₁ n_j on∂Y_int^∗ ∪0¹∪0² u⁰_k Y₁-periodic.

A weak formulation associated to (19) is:

(20)





Find u⁰∈ (Y^∗)such that : R

Y^∗a_{i j hk}^∂u

0 k

∂y_h

∂ψi

∂y_j = −R

Y^∗a_{i j 1k}^∂u

−1 k

∂x₁

∂ψi

∂y_j ∀ψ∈ (Y^∗) where (Y^∗)is the Hilbert space defined by:

(Y^∗) = {ψ ∈ H_loc¹ ( ),Y₁-periodic such that Z

Y^∗

ψ = 0} and equiped with the norm kψk = ε_{i j}(ψ )ε_{i j}(ψ )1/2

.

Applying Lax-Milgram’s theorem, we conclude the existence and uniqueness of a solution u⁰of (19).

The linearity of problem (20) allows us to introduce the following functions (21)





Findχ^α1∈ (Y^∗)such that : R

Y^∗a_{i j hk}^∂χ

α1 k

∂y_h

∂ψi

∂y_j = −R

Y^∗a_{i j 1α}^∂ψ_∂yⁱ

j ∀ψ ∈ (Y^∗)

and since u⁻¹depends only on x₁, we can write for a function ˇ

u⁰(x₁): u⁰_k=χ_k^α1∂u⁻¹_α

∂x₁ + ˇu⁰_k(x₁).

From definition (21) ofχ^α1, we can verify thatχ_k²¹=(m(y₂)−y₂)δ_k1with m(y₂)= 1

|Y^∗| Z

Y^∗

y₂. So u⁰can be written:

u⁰_k=χ_k^11∂u

−1 1

∂x1 +(m(y₂)−y₂)δ_k1^∂u

−1 2

∂x1 + ˇu⁰_k(x₁)andσ_{i j}⁻³= a_{i j 11}+a_{i j kh}^∂χ

11 k

∂yh

∂u⁻¹₁

∂x1 . Then integrating on Y^∗, we get:

σ⁻³_{i j} =c_{i j}∂u⁻¹₁

∂x₁ with c_{i j}= 1

|Y^∗| Z

Y^∗

a_{i j 11}+a_{i j kh}∂χ_k¹¹

∂y_h .

Takingψ = (y₂−m(y₂),0) (resp. ψ = (0,y₂−m(y₂))) in (21) forα = 1, we obtain c₁₂=c₂₁=0 (resp. c₂₂=0). So u⁻¹₁ is a solution of the problem:

(22)



 d d x1 c₁₁^du

−1 1

d x1

= 0 in(0,L) u⁻¹₁ (0) = u⁻¹₁ (L)=0.

LEMMA4. We have : c₁₁>0.

Proof. Letψ=(y₁,0)then a_11kh∂ψ_k

∂y_h =a₁₁₁₁and c₁₁= 1

|Y₁| Z

Y^∗ a_11kh ∂

∂y_h(ψ_k+χ_k¹¹).

Now takeχ¹¹as a test function in (21) forα=1, we get:

Z

Y^∗ a_{i j kh} ∂

∂y_h(ψ_k+χ_k¹¹)∂χ_i¹¹

∂y_j =0.

Then

c₁₁ = 1

|Y₁| Z

Y^∗ a_11kh ∂

∂y_h(ψ_k+χ_k¹¹)+ 1

|Y₁| Z

Y^∗ a_{i j kh} ∂

∂y_h(ψ_k+χ_k¹¹)∂χ_i¹¹

∂y_j

= 1

|Y₁| Z

Y^∗ a_{i j kh} ∂

∂y_h(ψ_k+χ_k¹¹) ∂

∂y_j(ψ_i+χ_i¹¹).

By (2) ii), we have: c₁₁≥ m

|Y₁|kψ+χ¹¹k.

Ifkψ+χ¹¹k = 0 thenψ +χ¹¹ = (a+by₁,a−by₂)with a,b∈ . Soχ¹¹−(a,a− by₂) = (b−1)(y₁,0). If b = 1, χ¹¹ = (a,a− y₂) and satisfy

Z

Y^∗

a_{i j kh}∂χ_k¹¹

∂y_h

∂ϕ_i

∂y_j =

− Z

Y^∗

a_{i j 22}∂ϕ_i

∂y_j ∀ϕ ∈ (Y^∗)which contradicts (21). So b 6= 1. Butχ¹¹−(a,a−by₂) is Y₁-periodic and(b−1)(y₁,0)is not Y₁-periodic. The lemma follows.

From this lemma and (22), we deduce that u⁻¹₁ =0 andσ⁻¹=0. So u⁰becomes equal to:

(23) u⁰₁=(m(y₂)−y₂)∂u⁻¹₂

∂x₁ + ˇu⁰₁(x₁) and u⁰₂= ˇu⁰₂(x₁).

To get more information, we compute now functions u¹andσ⁻². Using (14), (15) and (16), let us consider the problem:

(24)







∂σ_{i j}⁻²

∂y_j =0 in Y^∗, σ_{i j}⁻²n_j=0 on∂Y_int^∗ ∪0¹∪0² σ_{i j}⁻² Y₁−periodic, σ_{i j}⁻²=a_{i j 1k}^∂u

0 k

∂x1 +a_{i j hk}^∂u

1 k

∂yh. A weak formulation of (24) is:





Find u¹∈ (Y^∗)such that : R

Y^∗a_{i j hk}^∂u

1 k

∂y_h

∂ψi

∂y_j = −R

Y^∗a_{i j 1k}^∂u

0 k

∂x₁

∂ψi

∂y_j ∀ψ∈ (Y^∗).

Noting the linearity of this problem, let us considerχ¹²be the unique solution of : (25)





Findχ¹²∈ (Y^∗)such that : R

Y^∗a_{i j hk}^∂χ

12 k

∂yh

∂ψ_i

∂yj = −R

Y^∗a_{i j 11}(m(y₂)−y₂)^∂ψ_∂yⁱ

j ∀ψ∈ (Y^∗).

Then we can write:

u¹_k=χ_k¹¹∂uˇ⁰₁

∂x₁ +(m(y₂)−y₂)δ_k1∂uˇ⁰₂

∂x₁ +χ_k¹²∂²u⁻¹₂

∂x₁² + ˇu¹_k(x₁) and

σ_{i j}⁻²=

a_{i j 11}+a_{i j kh}∂χ_k¹¹

∂y_h ∂uˇ⁰₁

∂x₁+

a_{i j 11}(m(y₂)−y₂)+a_{i j kh}∂χ_k¹²

∂y_h ∂²u⁻¹₂

∂x₁² . Integrating on Y^∗, we obtain the constitutive law of the structure

(26) N⁻²(x₁)=s¹¹∂uˇ⁰₁

∂x₁+s¹²∂²u⁻¹₂

∂x₁² , M⁻²(x₁)=s²¹∂uˇ⁰₁

∂x₁ +s²²∂²u⁻¹₂

∂x₁²

where(s^αν)(α, ν=1,2)are given by:

(27)



















s¹¹ = _|Y¹

1| R

Y^∗

a_11kh^∂χ

11 k

∂y_h +a₁₁₁₁ ,

s¹² = _|Y¹

1| R

Y^∗

a_11kh^∂χ

12 k

∂y_h +a₁₁₁₁(m(y₂)−y₂) s²¹ = _|Y¹

1| R

Y^∗y₂ a_11kh^∂χ

11 k

∂y_h +a₁₁₁₁ ,

s²² = _|Y¹

1| R

Y^∗y₂ a_11kh^∂χ

12 k

∂yh +a₁₁₁₁(m(y₂)−y₂) .

Taking into account the boundary conditions of uε on 0₀∪0_L and (23), the homogenized problem is finally given by (17), (26), (27) and the boundary conditions:

(28) u⁻¹₂ (0)=u⁻¹₂ (L)=0, ∂u⁻¹₂

∂x₁ (0)= ∂u⁻¹₂

∂x₁ (L)=0, uˇ⁰₁(0)= ˇu⁰₁(L)=0.

The remainder of this section is devoted to prove the existence and uniqueness of a solution to the homogenized problem. It suffices for this to verify that the matrix S = (s^µν)_µ,ν=1,2is invertible. First, we introduce the following matrix: