Boundary Homogenization and Reduction of Dimension in a Kirchhoff–Love Plate

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SIAM J. MATH. ANAL. c© 2008 Society for Industrial and Applied MathematicsVol. 39, No. 6, pp. 1764–1787

BOUNDARY HOMOGENIZATION AND REDUCTION OFDIMENSION IN A KIRCHHOFF–LOVE PLATE∗

DOMINIQUE BLANCHARD† , ANTONIO GAUDIELLO‡ , AND TARAS A. MEL’NYK§

Abstract. We investigate the asymptotic behavior, as ε tends to 0+, of the transverse displace-ment of a Kirchhoff–Love plate composed of two domains Ω+

ε ∪ Ω−ε ⊂ R

2 depending on ε in thefollowing way. The set Ω+

ε is a union of fine teeth, having small cross section of size ε and constantheight, ε-periodically distributed on the upper side of a horizontal thin strip with vanishing height hε,as ε tends to 0+. The structure is clamped on the top of the teeth, with a free boundary elsewhere,and subjected to a transverse load. As ε tends to 0+, we obtain a “continuum” bending model ofrods in the limit domain of the comb, while the limit displacement is independent of the verticalvariable in the rescaled (with respect to hε) strip. We show that the displacement in the strip isequal to the displacement on the base of the teeth if hε � ε4. However, if the strip is thin enough(i.e., hε � ε4), we show that microscopic oscillations of the displacement in the strip, between thebasis of the teeth, may produce a limit average field different from that on the base of the teeth; i.e.,a discontinuity in the transmission condition may appear in the limit model.

Key words. Kirchhoff–Love plate, rough boundary, thick junctions, homogenization, dimensionreduction

AMS subject classifications. 74K20, 35B27

DOI. 10.1137/070685919

1. Introduction. Consider a bounded three-dimensional (3D) plate with smallthickness and with middle surface Ω+

ε ∪ Ω−ε ⊂ R2 like a comb, where ε is a small

positive parameter. Precisely, Ω+ε is a set of fine teeth, with small cross section of size

ε and constant height, ε-periodically distributed on the upper side of a horizontal thinstrip Ω−

ε having a small height hε (see Figure 1). The structure is clamped on the topof the teeth, with a free boundary elsewhere, and subjected to a transverse load (seeRemark 2.3 for other compatible boundary conditions). The transverse displacementis assumed to satisfy the Kirchhoff–Love equation.

The problem under investigation here pertains to the field of stationary problemsposed on a domain which has a so-called rough boundary or highly oscillating boundary.Boundary-value problems involving rough boundaries appear in many fields of physicsand engineering sciences such as the scattering of acoustic waves on small periodicobstacles, the free vibrations of strongly nonhomogeneous elastic bodies, and thebehavior of fluids over rough walls or of coupled fluid-solid periodic structures. Thereader can find a detailed bibliography about this subject in [2], [4], and [16]. Moregenerally, for the study of thin structures and multistructures, we refer the readerto the following monographs: [7], [9], [12], [13], [14], [17], [19], [20], [22], and thereferences therein.

∗Received by the editors March 21, 2007; accepted for publication (in revised form) July 25, 2007;published electronically February 15, 2008.

http://www.siam.org/journals/sima/39-6/68591.html†Universite de Rouen, UMR 6085, F-76821 Mont Saint Aignan Cedex, France; and Laboratoire

d’Analyse Numerique, Universite P. et M. Curie, Case Courrier 187, 75252 Paris Cedex 05, France([email protected]).

‡DAEIMI, Universita degli Studi di Cassino, via G. Di Biasio 43, 03043 Cassino (FR), Italy([email protected]).

§Kyiv Nat. Taras Shevchenko Univ., Volodymyrs’ka Str. 64, 01033 Kyiv, Ukraine ([email protected]).

1764


OSCILLATING KIRCHHOFF–LOVE PLATES 1765

Fig. 1. The middle surface of our 3D plate.

A physical example for the specific model and structure considered here appearsin MEMS1 acceleration sensors (see [15]) where the acceleration component in thedirection of the motion corresponds to a transverse load applied over the thin stripand the deformation of the teeth is then proportional to the acceleration itself. Thesesensors are widely used in automotive applications (antilock braking system (ABS),electronic stability program (ESP), airbags, etc.). Our geometrical structure can alsomodel a thin elastic grid used to filter a liquid flow in which case the transverseexternal forces are due to the action of the fluid on the grid (turn Figure 1 upsidedown).

It is often impossible to approach these problems directly with numerical methods,because the rough boundary of a comb requires a large number of mesh points intheir neighborhood. Thus, the computational cost associated with such a problemgrows rapidly when the scale of periodicity gets smaller. Moreover, it can occurthat the required discretization step becomes too small for the machine precision.Then, the goal is to construct accurate and numerically implementable asymptoticapproximations.

The aim of this paper is to mark a first step in this direction by finding, via anasymptotic analysis as ε tends to zero, a handier limit model which approximatesthe Kirchhoff–Love equation in the comb. In our case, the asymptotic analysis isstill more complicated in consequence of the competition between the parameter ofperiodicity ε and the parameter of thickness hε which vanishes as ε tends to zero.

According to an idea introduced by Ciarlet and Destuynder (see [8]), the termsin the thin strip Ω−

ε are rescaled in a fixed domain Ω−. Moreover, in the following,Ω+ denotes the “limit domain” of the teeth (see Figure 2). In the limit process,under suitable convergence assumptions on the rescaled loads, we obtain, in Ω+, a

1Microelectromechanical systems.


1766 D. BLANCHARD, A. GAUDIELLO, AND T. A. MEL’NYK

Fig. 2. The limit middle surface.

continuum bending model of rods subjected to a force f depending on the limit ofthe transverse loads on the teeth. The rods are clamped on the upper side Γ andsubjected on the lower side Σ to applied forces but without applied momentum. Theforces on Σ depend on the limit density g of the transverse loads on the thin strip Ω−

ε

and on the measure of the cross section of the reference tooth.

The limit solution is independent of the vertical variable in Ω− and meets aDirichlet transmission condition between Ω+ and Ω− if hε � ε4, or if hε � ε4 and ifg is negligible. However, if the strip is thin enough, i.e., hε � ε4, and the transverseloads on the thin strip are strong enough, a discontinuity in the Dirichlet transmissioncondition appears. Roughly speaking, this means that microscopic oscillations of thedisplacement in the strip, between the basis of the teeth of Ω+

ε , produce a limit averagefield different from that on the base of the teeth.

Now, we give a short review on problems involving rough boundaries with twosmall geometrical parameters in competition. As far as we know, these problems havebeen recently treated in [3], [5], and [6]. Precisely, in [3] and [6] the authors studiedthe asymptotic behavior of a monotone nonlinear second-order Neumann problem,with growth p − 1 (p ∈ ]1,+∞[), in a multidomain of RN (N ≥ 2) composed of a“forest” of cylinders with fixed height and small cross section of size ε, distributedwith ε-periodicity upon an asymptotically flat part of thickness hε. They proved thathε = εp is a critical size for the thickness of the thin domain. In [5], the homogenizationprocess for the junction of a periodic family of elastic rods with a thin elastic plate isstudied in the setting of the linearized elasticity. Here different critical sizes appearleading to various limit models: bending-bending model for the rods and the two-dimensional (2D) plate, rigid-bending, or microscopic effects. In [11] the authorsconsidered a thin multidomain of RN (N ≥ 2) consisting (e.g., in a 3D setting) of onlyone vertical rod upon a horizontal disk. In this thin multidomain they introduced abulk energy density of the kind W (D2U), where W is a convex function with growthp ∈ ]1,+∞[. By assuming that the two volumes tend to zero with the same rate,under suitable boundary conditions, they showed that the limit problem is uncoupled



if 1 < p ≤ N−12 , “partially” coupled if N−1

2 < p ≤ N − 1, and coupled if N − 1 < p.The main difficulty in analyzing the problem considered in the present paper is

twofold and it is essentially due to the fourth order of the Kirchhoff–Love model.First, deriving a priori estimates on the displacement for this operator and for anoscillating domain after rescaling is more intricate than for second-order problems.Second, as far as the homogenization process is concerned, the use of the method ofoscillating test functions, introduced by Tartar in [21], is also more complicated thanfor second-order problems since here we have to take into account also the oscillationsof the second derivatives of the solution.

Our paper is organized as follows. In section 2 the problem and the main resultsare stated. In section 3, by making use of the results proved in [3], some a priorinorm-estimates are obtained. In section 4, these estimates provide some convergenceresults in L2-norm, in the weak topology of L2, or in the setting of the two-scaleconvergence method, proposed by Nguetseng in [18] and developed by Allaire in [1].For convenience of the reader, in this section we also recall the definition and themain properties of the two-scale convergence. Finally, in section 5, the limit problemis derived by making use of the method of oscillating test functions.

2. Statement of the problem and main results. Let ω = ]a, b[ with 0 <a < b < 1, c, d ∈ ]0,+∞[, and let {ε}, {hε} ⊂ ]0, 1[ be two sequences converging tozero. For every ε, consider the 3D plate with small thickness t > 0 and with middlesurface Ω+

ε ∪ Ω−ε ⊂ R2 having the shape of a comb (see Figure 1), where

Ω+ε =

⋃{k∈N: εb+εk<c}

(εω + εk) × [0, d[

is a set of fine teeth of small cross section εω and constant height d, ε-periodicallydistributed on the upper basis of the thin strip:

Ω−ε = ]0, c[ × ] − hε, 0[,

which has a vanishing height hε and constant basis. Moreover, denote by Γε the topof the teeth of the middle surface:

Γε =⋃

{k∈N: εb+εk<c}(εω + εk) × {d}.

When the plate is clamped on Γε×] − t2 ,

t2 [, with a free boundary on (∂(Ω+

ε ∪Ω−ε )\

Γε) × ] − t2 ,

t2 [, and it is subjected to a transverse load, the Kirchhoff–Love equation

satisfied by the transverse displacement Uε of the middle surface Ω+ε ∪Ω−

ε is given by(see pages 205–207 in [10])

(2.1)

⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩

Et3

12(1 − μ2)Δ2Uε = Fε in Ω+

ε ∪ Ω−ε ,

Uε = ∂nUε = 0 on Γε,

ΔUε + (1 − μ)(2n1n2∂

2x1x2

Uε − n22∂

2x1Uε − n2

1∂2x2Uε

)= 0

on ∂(Ω+ε ∪ Ω−

ε ) \ Γε,

∂nΔUε + (1 − μ)∂τ[n1n2

(∂2x2Uε − ∂2

x1Uε

)+ (n2

1 − n22)∂

2x1x2

Uε

]= 0

on ∂(Ω+ε ∪ Ω−

ε ) \ Γε,



where Fε ∈ L2(Ω+ ∪ Ω−ε ) represents the transverse load, Ω+ = ]0, c[ × ]0, d[ is the

“limit domain” of the comb, n = (n1, n2) and τ denote the exterior unit normal andthe unit tangent to Ω+

ε ∪Ω−ε , respectively, μ ∈ ]0, 1

2 [ is the Poisson ratio, and E > 0 isthe Young modulus of the plate. In the following, M will denote the flexural rigiditymodulus of the plate, i.e.,

(2.2) M =Et3

12(1 − μ2).

We work on the weak formulation of problem (2.1) (see pages 205–207 in [10]):

(2.3)⎧⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎩

Uε ∈ H2(Ω+ε ∪ Ω−

ε ), Uε = ∂nUε = 0 on Γε,

M

∫Ω+

ε ∪Ω−ε

ΔUεΔV + (1 − μ)(2∂2

x1x2Uε ∂

2x1x2

V − ∂2x1Uε ∂

2x2V − ∂2

x2Uε ∂

2x1V)dx

=

∫Ω+

ε ∪Ω−ε

FεV dx ∀V ∈ H2(Ωε ∪ Ω−ε ) : V = ∂nV = 0 on Γε.

The goal of our paper is to study the asymptotic behavior of problem (2.3) as εtends to zero. To this aim, by following an idea of Ciarlet and Destuynder (see [8]),problem (2.3) can be reformulated on a domain independent of hε by using the maps

(x1, x2) ∈ Ω− = ]0, c[ × ] − 1, 0[ −→ (x1, hεx2) ∈ Ω−ε .

Namely, by setting

{fε(x) = Fε(x) a.e. x ∈ Ω+,fε(x) = Fε(x1, hεx2) a.e. x ∈ Ω−,

(2.4) {uε(x) = Uε(x) a.e. x ∈ Ω+

ε ,uε(x) = Uε(x1, hεx2) a.e. x ∈ Ω−,

and Ωε = Ω+ε ∪ Ω−, it turns out that uε belongs to the following space:

Vε ={v ∈ H1(Ωε) : v+ ∈ H2(Ω+

ε ), v− ∈ H2(Ω−),

v = 0, Dv = 0 on Γε, ∂x1v+ = ∂x1v

− on Σ \ ∂Ωε,

hε∂x2v+ = ∂x2v

− on Σ \ ∂Ωε} ,

where v+ = v|Ω

+ε

, v− = v|Ω− , Σ = ]0, c[ × {0} (note that, since v = 0 on Γε, the

boundary condition ∂nv = 0 on Γε is equivalent to Dv = 0 on Γε). Moreover, uε isthe unique solution of the following problem:



(2.5)⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩

uε ∈ Vε,

M

∫Ω+

ε

ΔuεΔv + (1 − μ)(2∂2

x1x2uε ∂

2x1x2

v − ∂2x1uε ∂

2x2v − ∂2

x2uε ∂

2x1v)dx

+Mhε

∫Ω−

(∂2x1uε +

1

h2ε

∂2x2uε

)(∂2x1v +

1

h2ε

∂2x2v

)dx

+M(1 − μ)hε

∫Ω−

21

hε∂2x1x2

uε1

hε∂2x1x2

v − ∂2x1uε

1

h2ε

∂2x2v − 1

h2ε

∂2x2uε ∂

2x1v dx

=

∫Ω+

ε

fεvdx + hε

∫Ω−

fεvdx ∀v ∈ Vε.

The study of the asymptotic behavior of problem (2.5) will be performed underthe following assumption:

(2.6)

{fε|Ω+

→ f strongly in L2(Ω+),

hεfε|Ω− → g strongly in L2(Ω−),

as ε tends to zero. Moreover, the following spaces will be involved:

(2.7)W 2(Ω+) =

{v ∈ L2(Ω+) : ∂x2v ∈ L2(Ω+), ∂2

x2v ∈ L2(Ω+), v = ∂x2v = 0 on Γ

},

where Γ = ]0, c[ × {d} and

H2per(]0, 1[) =

{v ∈ H2(]0, 1[) : v(0) = v(1), v′(0) = v′(1)

}with v′ denoting the first derivative of v. Note that H2

per(]0, 1[) is the closure ofC∞

per([0, 1]) with respect to the H2(]0, 1[)-norm, where C∞per([0, 1]) is the set of func-

tions in C∞(R) which are 1-periodic.In the following, v denotes the zero-extension to Ω+ of any function v defined in

a subset of Ω+, and

(2.8) |ω| = b− a.

We will show that the limit problem depends on

(2.9) limε→0

ε4

hε= l ∈ [0,+∞[

and∫ 0

−1g(x1, x2)dx2. Precisely, the following main result will be proved.

Theorem 2.1. Let uε be the unique solution of problem (2.5). Let W 2(Ω+) bethe space defined in (2.7). Assume (2.6) and (2.9). Then,

uε ⇀ |ω|u weakly in W 2(Ω+),

∂2x1uε ⇀ −μ|ω|∂2

x2u weakly in L2(Ω+),

˜∂2x1x2

uε ⇀ 0 weakly in L2(Ω+),



as ε → 0, where u is the unique solution of the following problem:

(2.10)

⎧⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎩

u ∈ W 2(Ω+),

|ω|Et3

12

∫Ω+

∂2x2u ∂2

x2v dx = |ω|

∫Ω+

fv dx

+

∫ c

0

(∫ 0

−1

g(x1, x2) dx2

)v(x1, 0) dx1 ∀v ∈ W 2(Ω+)

with |ω| defined in (2.8), μ ∈ ]0, 12 [ the Poisson ratio, E > 0 the Young modulus, t

denoting the small thickness of the 3D plate (see problem (2.1)), and f and g givenby (2.6). Moreover,

‖∂x2uε‖L2(Ω−) ≤ Ch34ε ,

∥∥∂2x1x2

uε

∥∥L2(Ω−)

≤ Ch12ε ,

∥∥∂2x2uε

∥∥L2(Ω−)

≤ Ch32ε

for every ε, where C is a positive constant independent of ε, and

(2.11) uε ⇀ u|Σ +

∫ 1

0

v0 dy1 weakly in L2(Ω−),

as ε → 0, where v0 = 0 if l = 0 in (2.9), while if l ∈ ]0,+∞[, v0 (= v0(x1, y1)) is theunique solution of the following problem:

(2.12)

⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩

v0 ∈ L2(]0, c[, H2per(]0, 1[)),

v0(x1, y1) = 0 in ]0, c[ × ω,

Et3

12

1

l

∫]0,c[ × ]0,1[

∂2y1v0(x1, y1)∂

2y1ϕ(x1, y1)dx1dy1

=

∫]0,c[ × ]0,1[

(∫ 0

−1

g(x1, x2)dx2

)ϕ (x1, y1) dx1dy1

∀ϕ ∈ L2(]0, c[, H2per(]0, 1[)) : ϕ (x1, y1) = 0 in ]0, c[ × sω

with u|Σ denoting the function in L2(Ω−) independent of x2 and equal, on Σ, to thetrace of the solution u of (2.10). Furthermore, the convergence of the energies holds:

limε→0

{Et3

12(1 − μ2)

∫Ω+

ε

|Δuε|2 + 2(1 − μ)(|∂2

x1x2uε|2 − ∂2

x1uε ∂

2x2uε

)dx

+Et3

12(1 − μ2)hε

∫Ω−

[ ∣∣∣∣∂2x1uε +

1

h2ε

∂2x2uε

∣∣∣∣2

+2(1 − μ)

(∣∣∣∣ 1

hε∂2x1x2

uε

∣∣∣∣2 − 1

h2ε

∂2x2uε ∂

2x1uε

)]dx

}

=Et3

12

(|ω|∫

Ω+

|∂2x2u|2dx +

1

l

∫]0,c[ × ]0,1[

∣∣∂2y1v0(x1, y1)

∣∣2 dx1dy1

),

where ∞ · 0 means 0.



Proof. Theorem 2.1 is an immediate consequence of Proposition 4.3, Corollary4.4, and Propositions 5.1 and 5.2 (see sections 4 and 5). For the sake of clarity, wedetail the proof of (2.11) and (2.12). If l = 0 in (2.9), convergence (4.6), Corollary4.4, and (ii) of Proposition 4.2 provide that

uε ⇀

∫ 1

0

u0(·, y1) dy1 = u|Σ weakly in L2(Ω−).

If l > 0, by setting v0(x1, y1) = u0(x1, y1) − u|Σ(x1, 0) in ]0, c[ × ω, convergence(4.6), equality (4.10), and (ii) of Proposition 4.2 provide that

uε ⇀

∫ 1

0

u0(·, y1) dy1 = u|Σ +

∫ 1

0

v0(·, y1)dy1 weakly in L2(Ω−),

and, by virtue of (5.9) (since ∂2y1u0 = ∂2

y1v0), v0 is the unique solution of the following

problem:

⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩

v0 ∈ L2(]0, c[, H2per(]0, 1[)),

v0(x1, y1) = 0 in ]0, c[ × ω,

M1 − μ2

l

∫]0,c[ × ]0,1[

∂2y1v0(x1, y1)∂

2y1ϕ(x1, y1)dx1dy1

=

∫]0,c[ × ]0,1[

(∫ 0

−1

g(x1, x2)dx2

)ϕ (x1, y1) dx1dy1

∀ϕ ∈ L2(]0, c[, H2per(]0, 1[)) : ϕ (x1, y1) = 0 in ]0, c[ × ω,

which coincides with problem (2.12) by virtue of definition (2.2). Note that, in prob-

lem (2.12), v0 = 0 if∫ 0

−1g(x1, x2)dx2 = 0 a.e. in ]0, c[.

The convergences of the energies are obtained by passing to the limit, as ε tendsto zero, in (2.5) with v = uε and by making use of assumption (2.6), the convergencesof {uε}ε, and the equation satisfied by u and v0.

Remark 2.2. Problems (2.10) and (2.12) are the weak formulation of the followingproblems:

(2.13)

⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩

Et3

12

∂4u

∂x42

= f in Ω+,

u =∂u

∂x2= 0 on Γ,

∂2u

∂x22

= 0 on Σ,

∂3u

∂x32

=12

|ω|Et3

∫ 0

−1

g(x1, x2) dx2 on Σ,



and for a.e. x1 ∈ ]0, c[

(2.14)

⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩

Et3

12

1

l

∂4v0

∂y41

(x1, y1) =

∫ 0

−1

g(x1, x2)dx2 for y1 ∈ ]0, a[,

Et3

12

1

l

∂4v0

∂y41

(x1, y1) =

∫ 0

−1

g(x1, x2)dx2 for y1 ∈ ]b, 1[,

v0(x1, a) =∂v0

∂y1(x1, a) = v0(x1, b) =

∂v0

∂y1(x1, b) = 0,

v0(x1, 0) = v0(x1, 1),

∂v0

∂y1(x1, 0) =

∂v0

∂y1(x1, 1),

∂2v0

∂y21

(x1, 0) =∂2v0

∂y21

(x1, 1),

∂3v0

∂y31

(x1, 0) =∂3v0

∂y31

(x1, 1),

v0(x1, y1) = 0 for y1 ∈ ω = ]a, b[,

respectively.The solution of problem (2.14) can be explicitly computed by solving a linear

system of 8 equations with 8 unknowns. Then, for a.e. x1 ∈ ]0, c[, it results that

(2.15) v0(x1, y1) =

⎧⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎩

l

2Et3(a− y1)

2(1 − b + y1)2

∫ 0

−1

g(x1, x2)dx2 for y1 ∈ [0, a[,

0 for y1 ∈ ω = [a, b],

l

2Et3(1 + a− y1)

2(b− y1)2

∫ 0

−1

g(x1, x2)dx2 for y1 ∈ ]b, 1],

and consequently

(2.16)∫ 1

0

v0(x1, y1) dy1

=l

2Et3

{1

30+

a

6+

a2

3+

a3

3+

a4

6+

a5

30− b

6− 2ab

3− a2b− 2a3b

3− a4b

6

+b2

3+ ab2 + a2b2 +

a3b2

3− b3

3− 2ab3

3− a2b3

3+

b4

6+

ab4

6− b5

30

}∫ 0

−1

g(x1, x2)dx2.

In the limit domain Ω+ of the comb, we obtain a continuum bending model ofrods subjected to a force f , clamped on the upper side Γ, and subjected on the lowerside Σ to applied forces but without applied momentum. The forces on Σ depend on



the limit density g of the transverse loads on the thin strip Ω−ε and on the measure

of the cross section ω of the reference tooth. The force f depends on the limit of thetransverse loads on the teeth.

The limit solution meets a Dirichlet transmission condition between Ω+ and therescaled strip Ω− if hε � ε4, or if hε � ε4 and

∫ 0

−1g(x1, x2)dx2 = 0 a.e. in ]0, c[.

However, if the strip is thin enough and the transverse loads on the thin strip are

strong enough, i.e., hε � ε4 and∫ 0

−1g(x1, x2)dx2 = 0 in a subset of ]0, c[ with positive

measure, a discontinuity in the Dirichlet transmission condition appears. Roughlyspeaking, this means that microscopic oscillations of the displacement in the strip,between the basis of the teeth of Ω+

ε , produce a limit average field different from that

on the base of the teeth. We point out that (2.15) provides that∫ 1

0v0(x1, y1) dy1 = 0

in a subset of ]0, c[ with positive measure if and only if∫ 0

−1g(x1, x2)dx2 = 0 in the same

subset. Consequently, by taking into account the definition of g in (2.6), for obtainingthe additional term in (2.11) when hε � ε4, it is necessary that the transverse loadsin the thin strip Ω−

ε are strong enough to avoid that limε→0(hε

∫Ω−

ε|Fε|2dx) = 0. For

instance, if Fε = ε−4α in Ω−ε , the additional term in the displacement of the strip

intervenes when α = 1 and it is given by formula (2.16) with g = 1; it does not appearwhen α < 1.

As regards the Laplacian, in [6] the authors proved that hε � ε2 is a critical sizefor the thickness of the thin strip. In particular, if hε � ε2, they gave an example inwhich g = 0 and the sequence of the solutions is not even bounded in L1(Ω−). In ourpaper, as regards the case hε � ε4, we think that a deterministic limit model mayhardly be expected, but we have no example to validate it.

Remark 2.3. Let us give a few comments about the boundary condition on∂(Ω+

ε ∪ Ω−ε ). Indeed, it is physically reasonable to assume that the plate is clamped

on a part of its boundary. An alternative to the condition Uε = ∂nUε = 0 on Γε couldbe to impose Uε = ∂nUε = 0 on ]0, c[ × {−hε} (i.e., on the lower lateral surfaceof the 3D plate modeled by the 2D plate Ω−

ε ). A similar and easier analysis thanthat developed below shows that, in this case, the limit problem (2.13) (also when

limε→0ε4

hε= +∞) is replaced by the following one:

⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩

Et3

12

∂4u

∂x42

= f in Ω+,

∂2u

∂x22

=∂3u

∂x32

= 0 on Γ,

u =∂u

∂x2= 0 on Σ.

Moreover, it results that (see Remark 3.3)

uε → 0 strongly in H1(Ω−),

and in this case there are no oscillations in the strip.

Concerning the original problem (2.3), the result below immediately follows fromTheorem 2.1.

Corollary 2.4. Let Uε be the solution of problem (2.3) under the assumptionsof Theorem 2.1 with {fε}ε defined by (2.4).



Then, it results that

Uε ⇀ |ω|u weakly in W 2(Ω+),

∂2x1Uε ⇀ −μ|ω|∂2

x2u weakly in L2(Ω+),

˜∂2x1x2

Uε ⇀ 0 weakly in L2(Ω+),

limε→0

1

|Ω−ε |

∫Ω−

ε

Uε dx =1

c

∫ c

0

(u|Σ +

∫ 1

0

v0dy1

)dx1,

limε→0

1

|Ω−ε |α

∫Ω−

ε

∂x2Uεdx= lim

ε→0

1

|Ω−ε |β

∫Ω−

ε

∂2x1x2

Uεdx

= limε→0

1

|Ω−ε |β

∫Ω−

ε

∂2x2Uεdx = 0 ∀α <

3

4,∀β <

1

2,

where u is the weak solution of problem (2.13), u|Σ denotes the trace of u on Σ, andv0 = 0 if l = 0 in (2.9), while if l ∈ ]0,+∞[, v0 (= v0(x1, y1)) is the solution ofproblem (2.14). Furthermore, the energies converge in the sense that

limε→0

(Et3

12(1 − μ2)

∫Ω+

ε⋃

Ω−ε

|ΔUε|2 + 2(1 − μ)(|∂2

x1x2Uε|2 − ∂2

x1Uε ∂

2x2Uε

)dx

)

=Et3

12

(|ω|∫

Ω+

|∂2x2u|2dx +

1

l

∫]0,c[ × ]0,1[

∣∣∂2y1v0(x1, y1)

∣∣2 dx1dy1

),

where ∞ · 0 means 0.

3. A priori norm-estimates. Define

D2(v) =

⎛⎝ ∂2x1v ∂2

x1x2v

∂2x1x2

v ∂2x2v

⎞⎠ , v ∈ H2(Ω+ε );

D2ε(v) =

⎛⎜⎜⎜⎝∂2x1v

1

hε∂2x1x2

v

1

hε∂2x1x2

v1

h2ε

∂2x2v

⎞⎟⎟⎟⎠ , v ∈ H2(Ω−),

for every ε. This section is devoted to proving the following a priori norm-estimates.Proposition 3.1. Let uε be the solution of problem (2.5). Assume (2.9) and

(2.6). Then, there exists a positive constant C, independent of ε, such that

(3.1) ‖uε‖H2(Ω+ε ) ≤ C,

(3.2)∥∥∥h 1

2ε D2

ε(uε)∥∥∥

(L2(Ω−))4≤ C

for every ε.



To prove Proposition 3.1, the following result is required.

Lemma 3.2. There exists a positive constant C, independent of ε, such that

(3.3)

‖v‖2L2(Ω−) ≤ C

(‖v‖2

L2(Σ\∂Ωε)+ ε2 ‖∂x1v‖

2L2(Ω−) + ‖∂x2v‖

2L2(Ω−)

)∀v ∈ H1(Ω−);

(3.4) ‖v‖2L2(Σ\∂Ωε)

≤ C(‖v‖2

L2(Ω+ε ) + ‖∂x2v‖

2L2(Ω+

ε )

)∀v ∈ H1(Ω+

ε );

(3.5)

‖v‖2H2(Ω+

ε ) ≤ C∥∥D2v

∥∥2

(L2(Ω+ε ))

4 ∀v ∈{v ∈ H2(Ω+

ε ) : v = 0, Dv = 0 on Γε

}for every ε.

Proof. The proof of inequality (3.3) is performed in the proof of Proposition 3.3in [3]. Standard arguments using Poincare inequality and Sobolev trace theorem leadto inequalities (3.4) and (3.5). Obviously, boundary conditions v = 0 and Dv = 0 onΓε intervene for obtaining estimate (3.5). In the case of other boundary conditions,see Remark 3.3.

Proof of Proposition 3.1. In the following, C denotes any positive constant inde-pendent of ε.

By choosing v = uε in (2.5), it results that

M

∫Ω+

ε

∣∣∂2x1uε

∣∣2 +∣∣∂2

x2uε

∣∣2 + 2μ∂2x1uε ∂

2x2uε + 2(1 − μ)

∣∣∂2x1x2

uε

∣∣2 dx+Mhε

∫Ω−

∣∣∂2x1uε

∣∣2 +

∣∣∣∣ 1

h2ε

∂2x2uε

∣∣∣∣2 + 2μ∂2x1uε

1

h2ε

∂2x2uε + 2(1 − μ)

∣∣∣∣ 1

hε∂2x1x2

uε

∣∣∣∣2 dx=

∫Ω+

ε

fεuεdx + hε

∫Ω−

fεuεdx

for every ε. Consequently, by taking into account that −α2 −β2 ≤ 2αβ, for α, β ∈ R,and by making use of assumption (2.6), one obtains that∫

Ω+ε

∣∣∂2x1uε

∣∣2 +∣∣∂2

x2uε

∣∣2 − μ∣∣∂2

x1uε

∣∣2 − μ∣∣∂2

x2uε

∣∣2 + 2(1 − μ)∣∣∂2

x1x2uε

∣∣2 dx+hε

∫Ω−

∣∣∣∣∂2x1uε

∣∣∣∣2 +

∣∣∣∣ 1

h2ε

∂2x2uε

∣∣∣∣2 −μ

∣∣∣∣∂2x1uε

∣∣∣∣2 −μ

∣∣∣∣ 1

h2ε

∂2x2uε

∣∣∣∣2 + 2(1−μ)

∣∣∣∣ 1

hε∂2x1x2

uε

∣∣∣∣2dx≤ C

(‖uε‖L2(Ω+

ε ) + ‖uε‖L2(Ω−)

)for every ε, that is,

(3.6)∥∥D2uε

∥∥2

(L2(Ω+ε ))

4 + hε

∥∥D2εuε

∥∥2

(L2(Ω−))4≤ C

(‖uε‖L2(Ω+

ε ) + ‖uε‖L2(Ω−)

)for every ε.



On the other hand, by applying (3.3) three times and by recalling that ∂x2u−ε =

hε∂x2u+ε on Σ \ ∂Ωε, one obtains that

‖uε‖2L2(Ω−) ≤ C

(‖uε‖2

L2(Σ\∂Ωε)+ ε2 ‖∂x1uε‖2

L2(Ω−) + ‖∂x2uε‖2L2(Ω−)

)≤ C ‖uε‖2

L2(Σ\∂Ωε)

+Cε2(‖∂x1

uε‖2L2(Σ\∂Ωε)

+ ε2∥∥∂2

x1uε

∥∥2

L2(Ω−)+∥∥∂2

x1x2uε

∥∥2

L2(Ω−)

)+C

(∥∥∂x2u−ε

∥∥2

L2(Σ\∂Ωε)+ ε2

∥∥∂2x1x2

uε

∥∥2

L2(Ω−)+∥∥∂2

x2uε

∥∥2

L2(Ω−)

)= C

(‖uε‖2

L2(Σ\∂Ωε)+ ε2 ‖∂x1uε‖2

L2(Σ\∂Ωε)+∥∥hε∂x2u

+ε

∥∥2

L2(Σ\∂Ωε)

)+C

(ε4∥∥∂2

x1uε

∥∥2

L2(Ω−)+ ε2

∥∥∂2x1x2

uε

∥∥2

L2(Ω−)+∥∥∂2

x2uε

∥∥2

L2(Ω−)

)for every ε, from which, by taking into account (3.4) and that ε < 1 and hε < 1, itfollows that

(3.7)

‖uε‖2L2(Ω−)

≤ C(‖uε‖2

H2(Ω+ε ) + ε4

∥∥∂2x1uε

∥∥2

L2(Ω−)+∥∥∂2

x1x2uε

∥∥2

L2(Ω−)+∥∥∂2

x2uε

∥∥2

L2(Ω−)

)

= C

[‖uε‖2

H2(Ω+ε )

+hε

(ε4

hε

∥∥∂2x1uε

∥∥2

L2(Ω−)+ hε

∥∥∥∥ 1

hε∂2x1x2

uε

∥∥∥∥2

L2(Ω−)

+ h3ε

∥∥∥∥ 1

h2ε

∂2x2uε

∥∥∥∥2

L2(Ω−)

)]

≤ C

[‖uε‖2

H2(Ω+ε )

+hε

(ε4

hε

∥∥∂2x1uε

∥∥2

L2(Ω−)+

∥∥∥∥ 1

hε∂2x1x2

uε

∥∥∥∥2

L2(Ω−)

+

∥∥∥∥ 1

h2ε

∂2x2uε

∥∥∥∥2

L2(Ω−)

)]for every ε.

By combining (3.6) with (3.7), making use of (3.5), and assuming that the limit(2.9) is finite, one has that

‖uε‖2H2(Ω+

ε ) + hε

∥∥D2εuε

∥∥2

(L2(Ω−))4≤ C

(‖uε‖2

H2(Ω+ε ) + hε

∥∥D2εuε

∥∥2

(L2(Ω−))4

) 12

for every ε, which provides estimates (3.1) and (3.2).Remark 3.3. Let us remark that in the case of the alternative boundary conditions

mentioned in Remark 2.3, deriving of estimates (3.1) and (3.2) is easier (and they

hold true also when limε→0ε4

hε= +∞). Indeed, they follow easily from (3.6) by using



∂x2u−ε = hε∂x2

u+ε on Σ \ ∂Ωε and the Poincare inequality in Ω− (which is licit since

one has that uε = ∂x2uε = 0 on ]0, c[ × {−1}). We point out that now inequality(3.2) provides that uε → 0 strongly in H1(Ω−).

Corollary 3.4. Let uε be the solution of problem (2.5). Assume (2.9) and (2.6).Then, there exists a positive constant C, independent of ε, such that∥∥ε2∂2

x1uε

∥∥L2(Ω−)

≤ C,(3.8) ∥∥∥∥∥ 1

h12ε

∂2x1x2

uε

∥∥∥∥∥L2(Ω−)

≤ C,(3.9)

∥∥∥∥∥ 1

h32ε

∂2x2uε

∥∥∥∥∥L2(Ω−)

≤ C,(3.10)

∥∥∥∥∥ 1

h34ε

∂x2uε

∥∥∥∥∥L2(Ω−)

≤ C,(3.11)

‖ε∂x1uε‖L2(Ω−) ≤ C,(3.12)

‖uε‖L2(Ω−) ≤ C(3.13)

for every ε.

Proof. Estimates (3.8), (3.9), and (3.10) follow immediately from estimate (3.2).By combining estimate (3.2) with inequalities (3.3) and (3.4) and by recalling that∂x2u

−ε = hε∂x2u

+ε on Σ \ ∂Ωε, it is easy to obtain (3.11), (3.12), and (3.13).

4. Convergence results. For convenience of the reader, we recall the definitionand the main properties of the two-scale convergence. We refer the reader to [1] and[18] for the proofs.

Definition 4.1. Let O ⊂ RN and Y = ]0, 1[N . A sequence {vε}ε ⊂ L2(O)is said to two-scale converge to a limit v ∈ L2(O × Y ) if, for any function φ inC∞

0

(O, C∞

per(Y )), it results that

limε→0

∫Ovε(x)φ

(x,

x

ε

)dx =

∫O×Y

v(x, y)φ(x, y) dxdy.

Proposition 4.2. Let O ⊂ RNand Y = ]0, 1[N .

(i) Let {vε}ε ⊂ L2(O) be a sequence converging to v strongly in L2(O). Then,{vε}ε two-scale converges to the same limit v.

(ii) Let {vε}ε ⊂ L2(O) be a bounded sequence, two-scale converging to v ∈ L2(O×Y ). Then, {vε} converges to

∫Yv(·, y)dy weakly in L2(O).

(iii) Let {vε}ε be a bounded sequence in L2(O). Then, there exist a subsequenceof {ε}, still denoted by {ε}, and a function v ∈ L2(O × Y ) such that {vε}ε two-scaleconverges to v.

(iv) Let {vε}ε ⊂ W 1,2(O) be a sequence such that {vε}ε and {εDvε}ε are boundedin L2(O) and (L2(O))N , respectively. Then, there exist a subsequence of {ε}, still



denoted by {ε}, and a function v ∈ L2(O,W 1,2per(Y )) such that {vε}ε and {εDvε}ε

two-scale converge to v and Dyv, respectively.The a priori norm-estimates of the solution uε of problem (2.5), obtained in section

3, provide the following convergence result.Proposition 4.3. Let uε be the solution of problem (2.5). Let W 2(Ω+) be the

space defined in (2.7). Assume (2.6) and (2.9). Then,

(4.1) ‖∂x2uε‖L2(Ω−) ≤ Ch

34ε ,

(4.2)∥∥∂2

x1x2uε

∥∥L2(Ω−)

≤ Ch12ε ,

∥∥∂2x2uε

∥∥L2(Ω−)

≤ Ch32ε

for every ε, where C is a positive constant independent of ε. Moreover, there ex-ist a subsequence of {ε}, still denoted by {ε}, u ∈ W 2(Ω+), η, ζ ∈ L2(Ω+), u0(=u0(x1, y1)) ∈ L2(]0, c[, H2

per(]0, 1[)), and ξ(= ξ((x1, x2), (y1, y2))) ∈ L2(Ω−×]0, 1[2)such that

(4.3) uε ⇀ |ω|u weakly in W 2(Ω+),

(4.4) ∂2x1uε ⇀ η weakly in L2(Ω+),

(4.5) ˜∂2x1x2

uε ⇀ ζ weakly in L2(Ω+),

(4.6) {u−ε }ε two-scale converges to u0,

(4.7) {ε∂x1u−ε }ε two-scale converges to ∂y1u0,

(4.8) {ε2∂2x1u−ε }ε two-scale converges to ∂2

y1u0,

(4.9)

{1

h32ε

∂2x2u−ε

}ε

two-scale converges to ξ,

as ε → 0, and

(4.10) u0(x1, y1) = u|Σ(x1, 0) in ]0, c[ × ω.

Proof. Estimates (4.1) and (4.2) follow from estimates (3.11), (3.9), and (3.10).Convergences (4.3), (4.4), and (4.5) are a consequence of estimate (3.1). Estimates(3.13), (3.12), (3.8), (4.1), and (4.2) provide convergences (4.6), (4.7), and (4.8) withu0 ∈ L2(Ω−, H2

per(]0, 1[)). Moreover, u0 is independent of x2, too. In fact, it resultsthat

0 = limε→0

∫Ω−

∂x2uεϕ(x,

x1

ε

)dx = − lim

ε→0

∫Ω−

uε∂x2ϕ(x,

x1

ε

)dx

= −∫

Ω−×]0,1[

u0(x, y1)∂x2ϕ (x, y1) dxdy1 ∀ϕ ∈ C∞

0

(Ω− × ]0, 1[

).

Convergence (4.9) springs from estimate (3.10).



Finally, statement (4.10) can be obtained by arguing as in the proof of Propo-sition 6.4 in [6]. More precisely, let χω be the ]0, 1[-periodic extension to R of thecharacteristic function of ω = ]a, b[ with respect to ]0, 1[. Then, by proceeding as inthe proof of (6.7) in [6], i.e., by using mainly (4.1) and (4.6), one obtains that

(4.11){u−ε (·, 0)χω

( ·ε

)}ε

two-scale converges to u0χω,

as ε → 0. On the other hand, by proceeding as in the proof of (6.13) in [6], i.e., byusing mainly (3.1), one proves that

(4.12) {u+ε (·, 0)}ε two-scale converges to u(·, 0)χω,

as ε → 0. Finally, in order to obtain (4.10) it is enough to pass to the two-scale limitin

u−ε (x1, 0)χω

(x1

ε

)= u+

ε (x1, 0) for a.e. x1 ∈ ]0, c[,

as ε → 0, and to make use of (4.11) and (4.12).If l = 0 in (2.9), then u0 can be completely identified in terms of u.Corollary 4.4. Let uε be the solution of problem (2.5). Assume (2.9), with

l = 0, and (2.6). Let u ∈ W 2(Ω+) and u0 ∈ L2(]0, c[, H2per(]0, 1[)) be satisfying

Proposition 4.3. Then,

(4.13) u0(x1, y1) = u|Σ(x1, 0) in ]0, c[ × ]0, 1[.

Proof. Assumption (2.9) with l = 0 and estimate (3.2) ensure that

ε2∂2x1uε → 0 strongly in L2(Ω−),

as ε → 0. Consequently, by virtue of (4.8), it results that

(4.14) ∂2y1u0 = 0 in ]0, c[ × ]0, 1[.

By combining (4.10) with (4.14), one obtains (4.13).

5. The limit problem. The following proposition is devoted to identify thelimit problem in Ω+.

Proposition 5.1. Let uε be the solution of problem (2.5). Assume (2.9) and(2.6). Let u ∈ W 2(Ω+) and η, ζ ∈ L2(Ω+) satisfy Proposition 4.3. Then,

(5.1) η = −μ|ω|∂2x2u a.e. in Ω+,

(5.2) ζ = 0 a.e. in Ω+,

and u ∈ W 2(Ω+) is the unique solution of

(5.3)

M |ω|(1 − μ2)

∫Ω+

∂2x2u∂2

x2v dx

= |ω|∫

Ω+

fv dx +

∫ c

0

(∫ 0

−1

g(x1, x2) dx2

)v(x1, 0) dx1 ∀v ∈ W 2(Ω+),



where μ ∈ ]0, 12 [ is the Poisson ratio, M > 0 represents the flexural rigidity modulus

of the plate (see problem (2.5)), and f , g ∈ L2(Ω−) are given by (2.6).

Proof. At first, claim (5.1) will be proved. To this aim, choose v(x1, x2) = ε2ψ1(x1

ε )ϕ(x1, x2) as test function in (2.5), where ψ1 is the 1-periodic function defined by

ψ1(y1) = 12y1(y1 − 1) in [0, 1] and ϕ ∈ C∞

0 (Ω+) (point out that v ∈ C∞(Ω+ε )⋂C0(Ω

+)⊂ Vε). Then, it results that

(5.4)

M

∫Ω+

ε

Δuε

(ϕ + 2εψ′

1

(x1

ε

)∂x1

ϕ + ε2ψ1

(x1

ε

)∂2x1ϕ + ε2ψ1

(x1

ε

)∂2x2ϕ)dx

+M(1 − μ)

∫Ω+

ε

2∂2x1x2

uε

(εψ′

1

(x1

ε

)∂x2

ϕ + ε2ψ1

(x1

ε

)∂2x1x2

ϕ)dx

−M(1 − μ)

∫Ω+

ε

∂2x1uεε

2ψ1

(x1

ε

)∂2x2ϕdx

−M(1 − μ)

∫Ω+

ε

∂2x2uε

(ϕ + 2εψ′

1

(x1

ε

)∂x1ϕ + ε2ψ1

(x1

ε

)∂2x1ϕ)dx

=

∫Ω+

ε

fεε2ψ1

(x1

ε

)ϕdx

for every ε. By passing to the limit, as ε → 0, in (5.4) and by making use of (4.3),(4.4), (4.5), and (2.6), it is easy to see that∫

Ω+

ηϕ + μ|ω|∂2x2uϕdx = 0 ∀ϕ ∈ C∞

0 (Ω+),

which provides (5.1).

In the next step, it will be proved that the function ζ ∈ L2(Ω+) is independent ofx2. To this aim, choose v(x1, x2) = εψ2(

x1

ε )ϕ(x1, x2) as test function in (2.5), whereψ2 is the 1-periodic function defined by ψ2(y1) = −y1 + 1

2 in [0, 1[ and ϕ ∈ C∞0 (Ω+)

(point out that v ∈ C∞(Ω+ε ) and supp v ⊂ Ω+, consequently v ∈ Vε). Then, it results

that

(5.5)

M

∫Ω+

ε

Δuε

(−2∂x1ϕ + εψ2

(x1

ε

)∂2x1ϕ + εψ2

(x1

ε

)∂2x2ϕ)dx

+M(1 − μ)

∫Ω+

ε

2∂2x1x2

uε

(−∂x2ϕ + εψ2

(x1

ε

)∂2x1x2

ϕ)dx

−M(1 − μ)

∫Ω+

ε

∂2x1uεεψ2

(x1

ε

)∂2x2ϕdx

−M(1 − μ)

∫Ω+

ε

∂2x2uε

(−2∂x1

ϕ + εψ2

(x1

ε

)∂2x1ϕ)dx

=

∫Ω+

ε

fεεψ2

(x1

ε

)ϕdx



for every ε. By passing to the limit, as ε → 0, in (5.5) and by making use of (4.3),(4.4), (4.5), (2.6), and (5.1), it is easy to see that∫

Ω+

2μ|ω|∂2x2u∂x1ϕ− 2μ|ω|∂2

x2u∂x1ϕ− 2(1 − μ)ζ∂x2ϕdx = 0 ∀ϕ ∈ C∞

0 (Ω+),

that is, ∫Ω+

ζ∂x2ϕdx = 0 ∀ϕ ∈ C∞0 (Ω+),

which provides that ζ is independent of x2.In the third step, claim (5.2) will be proved. To this aim, choose

v(x1, x2) =

⎧⎪⎪⎨⎪⎪⎩εψ2

(x1

ε

)φ(x2)ϕ(x1) in Ω+

ε ,

εψ2

(x1

ε

)ϕ(x1) in Ω−

as test function in (2.5), where ψ2 is defined as above, φ ∈ C∞([0, d]) is such thatφ = 1 in [0, d

4 ], φ = 0 in [3d4 , d], and ϕ ∈ C∞0 (]0, c[) (it is evident that v ∈ Vε). Then,

it results that

(5.6)

M

∫Ω+

ε

Δuε

(−2φ∂x1

ϕ + εψ2

(x1

ε

)φ∂2

x1ϕ + εψ2

(x1

ε

)ϕ∂2

x2φ)dx

+M(1 − μ)

∫Ω+

ε

2∂2x1x2

uε

(−ϕ∂x2φ + εψ2

(x1

ε

)∂x1ϕ∂x2φ

)dx

−M(1 − μ)

∫Ω+

ε

∂2x1uεεψ2

(x1

ε

)ϕ∂2

x2φdx

−M(1 − μ)

∫Ω+

ε

∂2x2uε

(−2φ∂x1ϕ + εψ2

(x1

ε

)φ∂2

x1ϕ)dx

+Mh12ε

∫Ω−

h12ε ∂

2x1uε

(−2∂x1ϕ + εψ2

(x1

ε

)∂2x1ϕ)dx

+Mh12ε

∫Ω−

μ1

h32ε

∂2x2uε

(−2∂x1

ϕ + εψ2

(x1

ε

)∂2x1ϕ)dx

=

∫Ω+

ε

fεεψ2

(x1

ε

)φϕdx +

∫Ω−

hεfεεψ2

(x1

ε

)ϕdx

for every ε. By passing to the limit, as ε → 0, in (5.6) and by making use of (4.3),(4.4), (4.5), (5.1), (3.2), and (2.6), it is easy to see that∫

Ω+

ζϕ∂x2φdx = 0 ∀ϕ ∈ C∞0 (]0, c[),

from which, by recalling the assumptions on φ and that ζ is independent of x2, itfollows that ∫ c

0

ζ(x1)ϕ(x1) dx = 0 ∀ϕ ∈ C∞0 (]0, c[),

that is, (5.2).



Now, the limit problem satisfied by u will be identified. To this aim, choose

v(x1, x2) =

⎧⎨⎩ϕ(x1, x2) in Ω+

ε ,

ϕ(x1, 0) + hεx2 (∂x2ϕ) (x1, 0) in Ω−

as test function in (2.5), where ϕ ∈ C∞(Ω+) and ϕ = 0, Dϕ = 0 on Γ. Then, itresults that

(5.7)

M

∫Ω+

ε

ΔuεΔϕ + (1 − μ)(2∂2

x1x2uε ∂

2x1x2

ϕ− ∂2x1uε ∂

2x2ϕ− ∂2

x2uε ∂

2x1ϕ)dx

+Mh12ε

∫Ω−

(h

12ε ∂

2x1uε +

1

h32ε

∂2x2uε

)((∂2x1ϕ)(x1, 0) + hεx2

(∂3x21x2

ϕ)

(x1, 0))dx

+M(1 − μ)

∫Ω−

2∂2x1x2

uε

(∂2x2x1

ϕ)(x1, 0) dx

−M(1 − μ)h12ε

∫Ω−

1

h32ε

∂2x2uε

((∂2x1ϕ)(x1, 0) + hεx2

(∂3x21x2

ϕ)

(x1, 0))dx

=

∫Ω+

ε

fεϕdx + hε

∫Ω−

fε (ϕ(x1, 0) + hεx2 (∂x2ϕ) (x′, 0)) dx

for every ε. By passing to the limit, as ε → 0, in (5.7), making use of (4.3), (4.4),(4.5), (5.1), (5.2), (3.2), and (2.6), and recalling that

χΩ+ε⇀ |ω| weakly in L2(Ω+),

it is easy to see that

(5.8)

M |ω|(1 − μ2)

∫Ω+

∂2x2u∂2

x2ϕdx = |ω|

∫Ω+

fϕ dx

+

∫ c

0

(∫ 0

−1

g(x1, x2) dx2

)ϕ(x1, 0) dx1 ∀ϕ ∈ C∞(Ω+) : ϕ = 0, Dϕ = 0 on Γ,

which, by density arguments, provides that u ∈ W 2(Ω+) is the unique solution of(5.3).

Now, the limit problem in Ω− will be identified. Point out that the two-scaleconvergence method will be explicitly used in Proposition 5.2, while it is not necessaryin Proposition 5.1 for identifying the limit problem in Ω+.

Proposition 5.2. Let uε be the solution of problem (2.5). Assume (2.9), with



l ∈ ]0,+∞[, and (2.6). Let u0 ∈ L2(]0, c[, H2per(]0, 1[)) satisfy Proposition 4.3. Then,

(5.9)

M1 − μ2

l

∫]0,c[ × ]0,1[

∂2y1u0(x1, y1)∂

2y1ϕ(x1, y1)dx1dy1

=

∫]0,c[ × ]0,1[

(∫ 0

−1

g(x1, x2)dx2

)ϕ (x1, y1) dx1dy1

∀ϕ ∈ L2(]0, c[, H2per(]0, 1[)) : ϕ (x1, y1) = 0 in ]0, c[ × ω,

where μ ∈ ]0, 12 [ is the Poisson ratio, M > 0 represents the flexural rigidity modulus

of the plate (see problem (2.5)), and g ∈ L2(Ω−) is given by (2.6).Proof. In the following, ε takes values in a subsequence satisfying Proposition 4.3.The proof of (5.9) will be performed in two steps.At first, it will be proved that

(5.10)∫ 0

−1

∫ 1

0

ξ((x1, x2), (y1, y2))dx2dy2 = −μl−12 ∂2

y1u0(x1, y1) a.e. in ]0, c[ × (]0, 1[),

where ξ ∈ L2(Ω− × ]0, 1[2) satisfies Proposition 4.3. To this aim, choose

v(x1, x2) =

{0 in Ω+

ε ,

h32ε x2

2ϕ(x1,

x1

ε

)in Ω−

as test function in (2.5), where ϕ (= ϕ(x1, y1)) ∈ C∞0 (]0, c[ × (]0, 1[)) (point out that

ϕ(x1, ·) admits an intrinsic 1-periodic extension on R). Then, it results that

(5.11)

Mhε

∫Ω−

∂2x1uε∂

2x1

(h

32ε x

22ϕ(x1,

x1

ε

))dx

+Mhε

∫Ω−

∂2x1uε

1

h2ε

∂2x2

(h

32ε x

22ϕ(x1,

x1

ε

))dx

+Mhε

∫Ω−

1

h2ε

∂2x2uε ∂2

x1

(h

32ε x

22ϕ(x1,

x1

ε

))dx

+Mhε

∫Ω−

1

h2ε

∂2x2uε

1

h2ε

∂2x2

(h

32ε x

22ϕ(x1,

x1

ε

))dx

+M(1 − μ)hε

∫Ω−

21

hε∂2x1x2

uε1

hε∂2x1x2

(h

32ε x

22ϕ(x1,

x1

ε

))−M(1 − μ)hε

∫Ω−

∂2x1uε

1

h2ε

∂2x2

(h

32ε x

22ϕ(x1,

x1

ε

))−M(1 − μ)hε

∫Ω−

1

h2ε

∂2x2uε ∂

2x1

(h

32ε x

22ϕ(x1,

x1

ε

))dx

= hε

∫Ω−

fεh32ε x

22ϕ(x1,

x1

ε

)dx



for every ε.Now, pass to the limit, as ε → 0, in each term of (5.11), after having simplified the

second term with the sixth one, as well as the third with the seventh. At first, Note

that the assumption l = 0 in (2.9) gives limε→0h2ε

ε = limε→0h2ε

ε2 = limε→0h

12ε

ε = 0 and

limε→0h

12ε

ε2 = l−12 .

From (2.9), with l ∈ ]0,+∞[, and estimate (3.2) it follows that

(5.12)

limε→0

(hε

∫Ω−

∂2x1uε∂

2x1

(h

32ε x

22ϕ(x1,

x1

ε

))dx

)

= limε→0

∫Ω−

h12ε ∂

2x1uεx

22

(h2ε∂

2x1ϕ + 2

h2ε

ε∂2x1y1

ϕ +h2ε

ε2∂2y1ϕ

)(x1,

x1

ε

)dx = 0.

From (2.9), with l ∈ ]0,+∞[, and (4.8) it follows that

(5.13)

limε→0

∫Ω−

hε∂2x1uε

1

h2ε

∂2x2

(h

32ε x

22ϕ(x1,

x1

ε

))dx = lim

ε→0

∫Ω−

h12ε ∂

2x1uε2ϕ

(x1,

x1

ε

)dx

= limε→0

∫Ω−

2h

12ε

ε2ε2∂2

x1uεϕ

(x1,

x1

ε

)dx = 2l−

12

∫]0,c[ × ]0,1[

∂2y1u0(x1, y1)ϕ(x1, y1)dx1dy1.

From (2.9), with l ∈ ]0,+∞[, and the second estimate in (4.2) (which involveslimε→0 ‖∂2

x2uε‖L2(Ω−) = 0) it follows that

(5.14)

limε→0

∫Ω−

hε1

h2ε

∂2x2uε ∂2

x1

(h

32ε x

22ϕ(x1,

x1

ε

))dx

= limε→0

∫Ω−

∂2x2uεx

22

(h

12ε ∂

2x1ϕ + 2

h12ε

ε∂2x1y1

ϕ +h

12ε

ε2∂2y1ϕ

)(x1,

x1

ε

)dx = 0.

From (4.9) it follows that

(5.15)

limε→0

∫Ω−

hε1

h2ε

∂2x2uε

1

h2ε

∂2x2

(h

32ε x

22ϕ(x1,

x1

ε

))dx= lim

ε→0

∫Ω−

1

h32ε

∂2x2uε2ϕ

(x1,

x1

ε

)dx

= 2

∫Ω−×]0,1[2

ξ((x1, x2), (y1, y2))ϕ (x1, y1) d(x1, x2)d(y1, y2).

From (2.9), with l ∈ ]0,+∞[, and the first estimate in (4.2) it follows that

(5.16)

limε→0

∫Ω−

hε21

hε∂2x1x2

uε1

hε∂2x1x2

(h

32ε x

22ϕ(x1,

x1

ε

))dx

= limε→0

∫Ω−

4∂2x1x2

uε x2

(h

12ε ∂x1ϕ +

h12ε

ε∂y1ϕ

)(x1,

x1

ε

)dx = 0.


(5.17) limε→0

∫Ω−

hεfεh32ε x

22ϕ(x1,

x1

ε

)dx = 0.



Then, by passing to the limit, as ε → 0, in (5.11) and by making use of (5.12)–(5.17), one obtains that

2l−12

∫]0,c[ × ]0,1[

∂2y1u0(x1, y1)ϕ(x1, y1)dx1dy1

+2

∫Ω−×]0,1[2

ξ((x1, x2), (y1, y2))ϕ (x1, y1) d(x1, x2)d(y1, y2)

−(1 − μ)2l−12

∫]0,c[ × ]0,1[

∂2y1u0(x1, y1)ϕ(x1, y1)dx1dy1 = 0 ∀ϕ ∈ C∞

0 (]0, c[ × (]0, 1[)),

that is, (5.10).Now, to prove (5.9), choose v(x1, x2) = ϕ(x1,

x1

ε ) as test function in (2.5), whereϕ (= ϕ (x1, y1)) ∈ C∞ ([0, c], C∞

per ([0, 1]))

such that ϕ (x1, y1) = 0 in [0, c]× ω. Then,it results that

(5.18)

Mhε

∫Ω−

∂2x1uε∂

2x1

(ϕ(x1,

x1

ε

))dx+Mhε

∫Ω−

1

h2ε

∂2x2uε ∂2

x1

(ϕ(x1,

x1

ε

))dx

−M(1 − μ)hε

∫Ω−

1

h2ε

∂2x2uε ∂

2x1

(ϕ(x1,

x1

ε

))dx = hε

∫Ω−

fεϕ(x1,

x1

ε

)dx

for every ε.Pass to the limit, as ε → 0, in each term of (5.18).From (2.9), with l ∈ ]0,+∞[, and (4.8) it follows that

(5.19)

limε→0

∫Ω−

hε∂2x1uε∂

2x1

(ϕ(x1,

x1

ε

))dx

= limε→0

∫Ω−

h12ε

ε2ε2∂2

x1uε

(h

12ε ∂

2x1ϕ + 2

h12ε

ε∂2x1y1

ϕ +h

12ε

ε2∂2y1ϕ

)(x1,

x1

ε

)dx

=1

l

∫]0,c[ × ]0,1[

∂2y1u0(x1, y1)∂

2y1ϕ(x1, y1)dx1dy1.

From (2.9), with l ∈ ]0,+∞[, (4.9), and (5.10) it follows that

(5.20)

limε→0

∫Ω−

hε1

h2ε

∂2x2uε ∂2

x1

(ϕ(x1,

x1

ε

))dx

= limε→0

∫Ω−

1

h32ε

∂2x2uε

(h

12ε ∂

2x1ϕ + 2

h12ε

ε∂2x1y1

ϕ +h

12ε

ε2∂2y1ϕ

)(x1,

x1

ε

)dx

=

∫Ω−×]0,1[2

ξ((x1, x2), (y1, y2))1

l12

∂2y1ϕ (x1, y1) d(x1, x2)d(y1, y2)

= −μ

l

∫]0,c[ × ]0,1[

∂2y1u0(x1, y1)∂

2y1ϕ(x1, y1)dx1dy1.




(5.21) limε→0

∫Ω−

hεfεϕ(x1,

x1

ε

)dx =

∫Ω−×]0,1[

g(x1, x2)ϕ (x1, y1) d(x1, x2)dy1.

Then, by passing to the limit, as ε → 0, in (5.18) and by making use of (5.19)–(5.21),one obtains that

M1 − μ2

l

∫]0,c[ × ]0,1[

∂2y1u0(x1, y1)∂

2y1ϕ(x1, y1)dx1dy1

=

∫]0,c[ × ]0,1[

(∫ 0

−1

g(x1, x2)dx2

)ϕ (x1, y1) dx1dy1

∀ϕ ∈ C∞ ([0, c], C∞per ([0, 1])

)such that ϕ (x1, y1) = 0 in [0, c] × ω,

which provides (5.9) by density arguments.

Acknowledgment. The authors wish to thank Professor Giovanni Busatto forthe helpful discussions they had on MEMS.

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Boundary Homogenization and Reduction of Dimension in a Kirchhoff–Love Plate

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