Mathematical justification of the obstacle problem in the case of piezoelectric plate

2.1. Constitutive equations of piezoelectricity

A piezoelectric material is described by three tensors: the fourth order symmetric positive definite stiffness tensor C = ( C i j k l ) , the third order piezoelectric tensor P = ( P k i j ) , and the second order symmetric positive definite dielectric tensor d = ( d i j ) . The set of data ( C , P , d ) can be represented by a symmetric matrix (see [6]) C 11 C 12 C 13 C 14 C 15 C 16 P 11 P 21 P 31 C 21 C 22 C 23 C 24 C 25 C 26 P 12 P 22 P 32 C 31 C 32 C 33 C 34 C 35 C 36 P 13 P 23 P 33 C 41 C 42 C 43 C 44 C 45 C 46 P 14 P 24 P 34 C 51 C 52 C 53 C 54 C 55 C 56 P 15 P 25 P 35 C 61 C 62 C 63 C 64 C 65 C 66 P 16 P 26 P 36 P 11 P 12 P 13 P 14 P 15 P 16 d 11 d 12 d 13 P 21 P 22 P 23 P 24 P 25 P 26 d 21 d 22 d 23 P 31 P 32 P 33 P 34 P 35 P 36 d 31 d 32 d 33 , where we used the abbreviation 11 → 1 , 22 → 2 , 33 → 3 , 23 = 32 → 4 , 13 = 31 → 5 , 12 = 21 → 6 . For example, the component C 1111 is denoted by C 11 , C 1131 by C 15 , the component P 123 by P 14 , P 233 by P 23 , etc.

There exists a positive constant c such that the tensors C, P, d satisfy the following relations:

There exists positive c such that, for every second order 3 × 3 symmetric tensor M = ( M i j ) and for every vector of dimension 3, θ = ( θ k ) we have (2.1) C i j k l = C i j l k = C k l i j , C i j k l M k l M i j ⩾ c ∑ i , j = 1 3 M i j 2 , d k l = d l k , d k l θ k θ l ⩾ c ∑ k = 1 3 θ k 2 , P m k l = P m l k . The constitutive laws that describe, in the linear case, the behavior of static piezoelectricity couple the second order symmetric stress tensor σ = ( σ k l ) and the displacement vector D = ( D k ) to the second order symmetric strain tensor e = ( e k l ) and the electric vector field E = ( E k ) through the relations (for a complete description we refer to [7] and [3]): (2.2) σ i j = C i j k l e k l − P k i j E k , D k = P k i j e i j + d k l E l . Let us recall that the dielectric tensor d is related to the electric susceptibility tensor d ′ = ( d k l ′ ) and to the electric permittivity of the free space d 0 by d = d 0 I identity + d ′ . The polarization vector X = ( X k ) is given component-wise by X k = P k i j e i j + d k l ′ E l , so that the electric displacement D is related to the polarization vector by D = d 0 E + X .

2.2. Equilibrium equations of piezoelectricity

Now let us write, in the linear case, the equilibrium equations of a piezoelectric body in the reference configuration Ω ‾ where Ω is a connected open subset of R 3 with smooth boundary. When subjected to the action of mechanical forces with density f = ( f k ) : Ω ‾ → R 3 and electric charge of density ρ e , the body undergoes a mechanical displacement field u = ( u k ) : Ω ‾ → R 3 and an electric potential φ : Ω ‾ → R , which solve the system of partial differential equations (2.3) div σ ( u , φ ) = − f , div D ( u , φ ) = ρ e , here div stands for the divergence operator of a tensor or of a vector. In the static linearized case considered in this paper the strain tensor and the electric field depend upon the unknowns by the relations e = 1 2 ( ∇ u + ( ∇ u ) T ) , E = − ∇ φ , or component-wise by e i j = 1 2 ( ∂ i u j + ∂ j u i ) , E k = − ∂ k φ .

2.3. Boundary conditions

The equilibrium equations have to be supplemented by conditions posed on the boundary ∂ Ω . Let ∂ Ω = Γ 0 ∪ Γ ˆ be a partition of the boundary with area Γ 0 > 0 . On Γ 0 the body is clamped and is subjected to null electric potential, and on Γ ˆ the body is subjected to applied surface force with density g = ( g k ) : Γ ˆ → R 3 and is subjected to electric charge with density ϑ : Γ ˆ → R 3 . Then the boundary conditions read: (2.4) u = 0 on Γ 0 , φ = 0 on Γ 0 , σ ( u , φ ) ν = g on Γ ˆ , D ( u , φ ) · ν = ϑ on Γ ˆ , where ν is the unit normal vector to ∂ Ω . The contact condition will be introduced later in Section 3.2. Another partition of the boundary with more general condition is considered in Section 6.

Let us recall that problem (2.3)–(2.4) has a unique weak solution in appropriate space [14].

3.1. Three-dimensional problem

We consider a family of plates with reference configuration Ω ‾ ε made of piezoelectric material where three characteristics tensors C, P and d are independent of ε. We denote by σ ε , e ε , D ε , E ε , respectively, the stress tensor, the strain tensor, the electric displacement field and the electric vector field and by ( u ε , φ ε ) the mechanical displacement field and the electric potential. The constitutive equations posed in Ω ε are: (3.1) σ ε ( u ε , φ ε ) = C e ε ( u ε ) − P E ε ( φ ε ) , D ε ( u ε , φ ε ) = P e ε ( u ε ) + d E ε ( φ ε ) . Let x ε = ( x k ε ) be a generic point in Ω ‾ ε , with x α ε ∈ ω and x 3 ε ∈ ( − ε , ε ) and let ∂ i ε = ∂ ∂ x i ε . Then the linear strain tensor e ε ( u ε ) is defined by e ε ( u ε ) = 1 2 ( ∇ ε u ε + ( ∇ ε u ε ) T ) or component-wise by e i j ε ( u ε ) = 1 2 ( ∂ i ε u j ε + ∂ j ε u i ε ) , and the electric field vector E ε ( φ ε ) is given by E k ε ( φ ε ) = − ∂ k ε φ ε . For a plate subjected to applied body forces with density f ε and to electric charge with density ρ e ε the equilibrium problem posed in Ω ε reads: (3.2) div ε σ ε ( u ε , φ ε ) = − f ε , div ε D ε ( u ε , φ ε ) = ρ e ε .

We consider that the body

is fixed on the whole lateral surface Γ 0 ε , this condition can be relaxed as indicated in Section 6,

3.2. The boundary conditions for a body in contact with a plane

We focus now on the unilateral contact with an horizontal plane set at level − ε . For any point x − ε = ( x 1 ε , x 2 ε , − ε ) on the lower face Γ − ε the non-penetrability condition reads ( O x − ε + u ε ( x − ε ) ) · ν ⩾ 0 , in other words (3.3) ∀ x − ε ∈ Γ − ε , − ε + u 3 ε ( x − ε ) ⩾ − ε ⟹ u 3 ε ( x − ε ) ⩾ 0 . The so-called Signorini conditions which give the full description of the unilaterality, are classically obtained by adding the following constraints to the non-penetrability condition:

No tensile forces but only compressive forces are exerted on the boundary by the obstacle;

Let us note that in the case of a partial clamping on a part of the lateral surface, we can keep the same boundary conditions by restricting the lateral surface to Γ 0 ε = γ 0 × [ − ε , ε ] with γ 0 = ∂ ω . In Section 6 we use the “new definition of upper face” that includes the remaining part of the lateral surface Γ + ε ∪ ( γ ∖ γ 0 ) × [ − ε , ε ] .

3.3. The variational inequality in Ω ε

The natural functional framework for problem (3.2), (3.5) is the product vector space K ε ( Ω ε ) × Q ε ( Ω ε ) where K ε ( Ω ε ) is the convex set (3.6) K ε ( Ω ε ) = { v ε ∈ H 1 ( Ω ε ) , v ε = 0 on Γ 0 ε , v 3 ε ⩾ 0 on Γ − ε } and Q ε ( Ω ε ) is the vector space (3.7) Q ε ( Ω ε ) = { ψ ε ∈ H 1 ( Ω ε ) , ψ ε = 0 on Γ 0 ε } . Hence the weak solution ( u ε , φ ε ) to problem (3.2), (3.5), is given by the following variational inequality: (3.8) Find ( u ε , φ ε ) ∈ K ε ( Ω ε ) × Q ε ( Ω ε ) such that: ∫ Ω ε C i j k l e k l ε ( u ε ) e i j ε ( v ε − u ε ) d x ε − ∫ Ω ε d i j ∂ j ε φ ε ∂ i ε ( ψ ε − φ ε ) d x ε + ∫ Ω ε P k i j ∂ k ε φ ε e i j ε ( v ε − u ε ) d x ε + ∫ Ω ε P k i j ∂ k ε ( ψ ε − φ ε ) e i j ε ( u ε ) d x ε ⩾ ∫ Ω ε f ε · ( v ε − u ε ) d x ε + ∫ Γ + ε g ε · ( v ε − u ε ) d a ε − ∫ Ω ε ρ e ε ( ψ ε − φ ε ) d x ε + ∫ Γ + ε ϑ ε ( ψ ε − φ ε ) d a ε ∀ ( v ε , ψ ε ) ∈ K ε ( Ω ε ) × Q ε ( Ω ε ) , where d a ε is the area element of the boundary ∂ Ω ε . Based on classical arguments ([10]).

For any fixed ε > 0 , the weak solution associated to problem (3.2), (3.5) is given by the unique solution (3.8) (for a proof see [10]).

Let us now introduce the scaling procedure in order to establish the convergence theorem.

4.1. Scalings of the unknowns and test functions

We now change the domain Ω ε having middle surface ω and thickness 2 ε into a fixed domain Ω with the same middle surface and fixed thickness 2 via the simple geometrical transformation defined as follows: let x ε = ( x k ε ) be a generic point in Ω ‾ ε and the corresponding point x = ( x k ) in Ω ‾ with x α ε = x α ∈ ω and x 3 ε = ε x 3 ∈ ( − ε , ε ) . This induces ∂ α ε = ∂ ∂ x α ε = ∂ ∂ x α and ∂ 3 ε = ∂ ∂ x 3 ε = 1 ε ∂ ∂ x 3 . By analogy, the boundary of the domain Ω is divided into three parts: ∂ Ω = Γ − ∪ Γ + ∪ Γ 0 , Γ − = ω × { − 1 } , Γ + = ω × { 1 } , Γ 0 = γ × [ − 1 , 1 ] .

We follow [12] and to any functions defined on Ω ‾ ε we associate scaled functions defined on Ω ‾ as: (4.1) u α ε = ε 2 u α ( ε ) , u 3 ε = ε u 3 ( ε ) , v α ε = ε 2 v α , v 3 ε = ε v 3 , φ ε = ε 2 φ ( ε ) , ψ ε = ε 2 ψ .

Along with the scaling procedure, we let e = ( e i j ) to denote the scaled linearized strain tensor, the components of which are (4.2) e α β ε ( v ε ) = ε 2 e α β ( v ) , e α 3 ε ( v ε ) = ε e α 3 ( v ) , e 33 ε ( v ε ) = e 33 ( v ) , where e i j ( v ) = 1 2 ( ∂ j v i + ∂ i v j ) .

4.2. Assumptions on the data

In order to obtain a non-trivial limit problem by asymptotic analysis, it is essential, as it has been shown in [12] to scale the data in accordance with the scalings of the unknowns. More precisely we assume that there exist functions f ∈ L 2 ( Ω ) , g ∈ L 2 ( Γ + ) , ρ e ∈ L 2 ( Ω ) , ϑ ∈ L 2 ( Γ + ) independent of ε such that (4.3) f α ε = ε 2 f α , f 3 ε = ε 3 f 3 ∀ x ∈ Ω , g α ε = ε 3 g α , g 3 ε = ε 4 g 3 ∀ x ∈ Γ + , ρ e ε = ε 2 ρ e ∀ x ∈ Ω , ϑ ε = ε 2 ϑ ∀ x ∈ Γ + .

4.3. The contact condition posed in the fixed domain

After the scaling process, the non-penetrability condition holds on Γ − and reads (4.4) v 3 ( x 1 , x 2 , − 1 ) ⩾ 0 ∀ ( x 1 , x 2 ) ∈ ω , and the corresponding functional space is K ( Ω ) = { v ∈ H 1 ( Ω ) , v = 0 on Γ 0 , v 3 ⩾ 0 on Γ − } .

4.4. The equilibrium problem in the fixed domain Ω = ω × ( − 1 , 1 )

Let us note Q ( Ω ) the vector space associated to Q ε ( Ω ε ) : Q ( Ω ) = { ψ ∈ H 1 ( Ω ) , ψ = 0 on Γ 0 } . Replacing ( u ε , φ ε ) by their scaled values ( u ( ε ) , φ ( ε ) ) and ( v ε , ψ ε ) by ( v , ψ ) given in (4.1) in the problem (3.8), yields the following problem posed over the fixed domain Ω: (4.5) Find ( u ( ε ) , φ ( ε ) ) ∈ K ( Ω ) × Q ( Ω ) such that: ε 5 ∫ Ω C α β σ τ e σ τ ( u ( ε ) ) e α β ( v − u ( ε ) ) d x − ε 5 ∫ Ω d α β ∂ β φ ( ε ) ∂ α ( ψ − φ ( ε ) ) d x + ε 5 ∫ Ω P γ α β ∂ γ φ ( ε ) e α β ( v − u ( ε ) ) d x + ε 5 ∫ Ω P γ α β ∂ γ ( ψ − φ ( ε ) ) e α β ( u ( ε ) ) d x + 2 ε 4 ∫ Ω C α β σ 3 ( e σ 3 ( u ( ε ) ) e α β ( v − u ( ε ) ) + e α β ( u ( ε ) ) e σ 3 ( v − u ( ε ) ) ) d x − ε 4 ∫ Ω d α 3 ( ∂ 3 φ ( ε ) ∂ α ( ψ − φ ( ε ) ) + ∂ α φ ( ε ) ∂ 3 ( ψ − φ ( ε ) ) ) d x + 2 ε 4 ∫ Ω P γ α 3 ∂ γ φ ( ε ) e α 3 ( v − u ( ε ) ) d x + ε 4 ∫ Ω P 3 α β ∂ 3 φ ( ε ) e α β ( v − u ( ε ) ) d x + 2 ε 4 ∫ Ω P γ α 3 ( ∂ γ ( ψ − φ ( ε ) ) e α 3 ( u ( ε ) ) + ∂ 3 ( ψ − φ ( ε ) ) e α β ( u ( ε ) ) ) d x + 4 ε 3 ∫ Ω C α 3 σ 3 e σ 3 ( u ( ε ) ) e α 3 ( v − u ( ε ) ) d x + ε 3 ∫ Ω C α β 33 ( e 33 ( u ( ε ) ) e α β ( v − u ( ε ) ) + e α β ( u ( ε ) ) e 33 ( v − u ( ε ) ) ) d x + 2 ε 3 ∫ Ω P 3 α 3 ( ∂ 3 φ ( ε ) e α 3 ( v − u ( ε ) ) + ∂ 3 ( ψ − φ ( ε ) ) e α 3 ( u ( ε ) ) ) d x + ε 3 ∫ Ω P γ 33 ( ∂ γ φ ( ε ) e 33 ( v − u ( ε ) ) + ∂ γ ( ψ − φ ( ε ) ) e 33 ( u ( ε ) ) ) d x − ε 3 ∫ Ω d 33 ∂ 3 φ ( ε ) ∂ 3 ( ψ − φ ( ε ) ) d x + 2 ε 2 ∫ Ω C α 333 ( e 33 ( u ( ε ) ) e α 3 ( v − u ( ε ) ) + e α 3 ( u ( ε ) ) e 33 ( v − u ( ε ) ) ) d x + ε 2 ∫ Ω P 333 ( ∂ 3 φ ( ε ) e 33 ( v − u ( ε ) ) + ∂ 3 ( ψ − φ ( ε ) ) e 33 ( u ( ε ) ) ) d x + ε ∫ Ω C 3333 e 33 ( u ( ε ) ) e 33 ( v − u ( ε ) ) d x ⩾ ε 5 ∫ Ω ( f · ( v − u ( ε ) ) − ρ e ( ψ − φ ( ε ) ) ) d x + ε 5 ∫ Γ + ( g · ( v − u ( ε ) ) + ϑ ( ψ − φ ( ε ) ) ) d a ∀ ( v , ψ ) ∈ K ( Ω ) × Q ( Ω ) , where d a is the area element of the boundary ∂ Ω .

Theorem 4.1.

For all fixed ε > 0 problem (4.5) has a unique weak solution.

Proof.

As pointed out before we use classical arguments ([10]) to prove the existence and uniqueness of the weak solution to the variational inequality problem (4.5). □

5. Two-dimensional limit scaled solution in the plate

The aim of this section is to establish that when ε tends to zero, the sequence { ( u ( ε ) , φ ( ε ) ) } ε converges to a limit ( u , φ ) which solves a two-dimensional obstacle problem. An important preliminary point here is the following lemma, which is a new version of Korn’s inequality.

Lemma 5.1.

For all v ∈ H 1 ( Ω ) , the mapping v → { ∑ i j | e i j ( v ) | 0 , Ω 2 } 1 2 is a norm over the set K ( Ω ) , which is equivalent to the norm ‖ · ‖ 1 , Ω .

Proof.

The proof follows from the fact that the set K ( Ω ) is a closed subset of the vector space { v ∈ H 1 ( Ω ) , v = 0 on Γ 0 } . □

5.1. Variational solution

We recall the definition of the functional spaces K ( Ω ) , Q ( Ω ) K ( Ω ) = { v ∈ H 1 ( Ω ) , v = 0 on Γ 0 , v 3 ⩾ 0 on Γ − } , Q ( Ω ) = { ψ ∈ H 1 ( Ω ) , ψ = 0 on Γ 0 } . We are now in a position to prove the main result of the paper.

Theorem 5.1.

Assume f ∈ L 2 ( Ω ) , ρ e ∈ L 2 ( Ω ) , g ∈ L 2 ( Γ + ) , ϑ ∈ L 2 ( Γ + ) . Then

As ε tends to 0, the family { ( u ( ε ) , φ ( ε ) ) } ε > 0 converges strongly in the set K ( Ω ) × Q ( Ω ) to a limit { ( u , φ ) } .

The limit u is a Kirchhoff–Love displacement field, namely there exists ζ = ( ζ H , ζ 3 ) ∈ V H ( ω ) × K 3 ( ω ) , such that u α = ζ α − x 3 ∂ α ζ 3 , u 3 = ζ 3 , with the definition of the bi-dimensional functional spaces V H ( ω ) = { η H = ( η α ) ∈ H 1 ( ω ) , η H = 0 on γ } , K 3 ( ω ) = { η 3 ∈ H 2 ( ω ) , η 3 = ∂ ν η 3 = 0 on γ , η 3 ⩾ 0 in ω } .

The limit electric potential φ belongs to the bi-dimensional functional space Q ( ω ) = { ψ ∈ H 1 ( Ω ) , ψ = 0 on γ } , and the limit solution ( ζ , φ ) = ( ζ H , ζ 3 , φ ) solves the following coupled problem:

Find ( ζ H , ζ 3 , φ ) ∈ V H ( ω ) × K 3 ( ω ) × Q ( ω ) such that: (5.1) 2 3 ∫ ω C ˜ α β σ τ ∂ σ τ ζ 3 ∂ α β ( η 3 − ζ 3 ) d ω ⩾ ∫ ω ( p 3 ( η 3 − ζ 3 ) − s α ∂ α ( η 3 − ζ 3 ) ) d ω ∀ η 3 ∈ K 3 ( ω ) , 2 ∫ ω n ˜ α β ( ζ H , φ ) e α β ( η H ) d ω + 2 ∫ ω D ˜ γ ( ζ H , φ ) ∂ γ ψ d ω = ∫ ω ( p α η α − ( ρ ˜ e − ϑ ψ ) ) d ω ∀ ( η H , ψ ) ∈ V H ( ω ) × Q ( ω ) , where the new constitutive bi-dimensional law is given by n ˜ α β ( ζ H , φ ) = C ˜ α β σ τ e σ τ ( ζ H ) + P ˜ γ α β ∂ γ φ , D ˜ γ ( ζ H , φ ) = P ˜ γ α β e α β ( ζ H ) − d ˜ γ β ∂ β φ and the mechanical forces and electric charges are given by p i ( x 1 , x 2 ) : = ∫ − 1 1 f i d x 3 + g i , s α ( x 1 , x 2 ) : = ∫ − 1 1 x 3 f α d x 3 + g α , ρ ˜ e ( x 1 , x 2 ) : = ∫ − 1 1 ρ e d x 3 , where the new bi-dimensional elasticity tensor C ˜ , piezoelectricity tensor P ˜ and dielectric tensor d ˜ are given in (5.13) .

Proof.

The proof is broken into five steps. In the first one, we introduce a new scaled strain tensor R ( ε ) and with some boundness result we establish that the sequence { ( u ( ε ) ) , φ ( ε ) } converges weakly to a limit ( u , φ ) , where u is a Kirchhoff–Love field. The second one deals with technical results about the components of this strain tensor. In the third step we show that the convergence of the family { ( u ( ε ) ) , φ ( ε ) } towards ( u , φ ) is strong. The fourth step completes the proof by deducing the limit de-coupled transverse and membrane variational problems. Finally in the last step, we show that the membrane problem is of a piezoelectric form. Step I.

Let us introduce the following scaled symmetric tensor R ( ε ) = ( R i j ( ε ) ) ∈ L 2 ( Ω ) and the scaled vector τ ( ε ) = ( τ i ( ε ) ) ∈ L 2 ( Ω ) by R α β ( ε ) ( v ) = e α β ( v ) , R α 3 ( ε ) ( v ) = 1 ε e α 3 ( v ) , R 33 ( ε ) ( v ) = 1 ε 2 e 33 ( v ) , τ α ( ε ) ( ψ ) = ∂ α ψ , τ 3 ( ε ) ( ψ ) = 1 ε ∂ 3 ψ . We introduce R ( ε ) ( u ( ε ) ) and τ ( ε ) ( φ ( ε ) ) in the variational inequality (4.5) and get (5.2) ∫ Ω C α β σ τ R σ τ ( ε ) ( u ( ε ) ) e α β ( v − u ( ε ) ) d x + 2 ∫ Ω C α β σ 3 R σ 3 ( ε ) ( u ( ε ) ) e α β ( v − u ( ε ) ) d x + ∫ Ω C α β 33 R 33 ( ε ) ( u ( ε ) ) e α β ( v − u ( ε ) ) d x − ∫ Ω d α β τ β ( ε ) ( φ ( ε ) ) ∂ α ( ψ − φ ( ε ) ) d x − ∫ Ω d α 3 τ 3 ( ε ) ( φ ( ε ) ) ∂ α ( ψ − φ ( ε ) ) d x + ∫ Ω P γ α β τ γ ( ε ) ( φ ( ε ) ) e α β ( v − u ( ε ) ) d x + ∫ Ω P γ α β ∂ γ ( ψ − φ ( ε ) ) R α β ( ε ) ( u ( ε ) ) d x + ∫ Ω P 3 α β τ 3 ( ε ) ( φ ( ε ) ) e α β ( v − u ( ε ) ) d x + 2 ∫ Ω P γ α 3 ∂ γ ( ψ − φ ( ε ) ) R α 3 ( ε ) ( u ( ε ) ) d x + ∫ Ω P γ 33 ∂ γ ( ψ − φ ( ε ) ) R 33 ( ε ) ( u ( ε ) ) d x + 2 ε ∫ Ω C α 3 σ τ R σ τ ( u ( ε ) ) ( ε ) e α 3 ( v − u ( ε ) ) d x + 4 ε ∫ Ω C α 3 σ 3 R σ 3 ( ε ) ( u ( ε ) ) e α 3 ( v − u ( ε ) ) d x + 2 ε ∫ Ω C α 333 R 33 ( ε ) ( u ( ε ) ) e α 3 ( v − u ( ε ) ) d x − 1 ε ∫ Ω d 3 α τ α ( ε ) ( φ ( ε ) ) ∂ 3 ( ψ − φ ( ε ) ) d x − 1 ε ∫ Ω d 33 τ 3 ( ε ) ( φ ( ε ) ) ∂ 3 ( ψ − φ ( ε ) ) d x + 2 ε ∫ Ω P γ α 3 τ γ ( ε ) ( φ ( ε ) ) e α 3 ( v − u ( ε ) ) d x + 2 ε ∫ Ω P 3 α 3 τ 3 ( ε ) ( φ ( ε ) ) e α 3 ( v − u ( ε ) ) d x + 1 ε ∫ Ω P 3 α β ∂ 3 ( ψ − φ ( ε ) ) R α β ( ε ) ( u ( ε ) ) d x + 2 ε ∫ Ω P 3 α 3 ∂ 3 ( ψ − φ ( ε ) ) R α 3 ( ε ) ( u ( ε ) ) d x + 1 ε ∫ Ω P 333 ∂ 3 ( ψ − φ ( ε ) ) R 33 ( ε ) ( u ( ε ) ) d x + 1 ε 2 ∫ Ω C 33 σ τ R σ τ ( ε ) ( u ( ε ) ) e 33 ( v − u ( ε ) ) d x + 2 ε 2 ∫ Ω C 33 σ 3 R σ 3 ( ε ) ( u ( ε ) ) e 33 ( v − u ( ε ) ) d x + 1 ε 2 ∫ Ω C 3333 R 33 ( ε ) ( u ( ε ) ) e 33 ( v − u ( ε ) ) d x + 1 ε 2 ∫ Ω P γ 33 τ γ ( ε ) ( φ ( ε ) ) e 33 ( v − u ( ε ) ) d x + 1 ε 2 ∫ Ω P 333 τ 3 ( ε ) ( φ ( ε ) ) e 33 ( v − u ( ε ) ) d x ⩾ ∫ Ω f · ( v − u ( ε ) ) d x + ∫ Γ + g · ( v − u ( ε ) ) d a − ∫ Ω ρ e ( ψ − φ ( ε ) ) d x + ∫ Γ + ϑ ( ψ − φ ( ε ) ) d a , and next by introducing the new tensors R ( ε ) ( v − u ( ε ) ) and τ ( ε ) ( ψ − φ ( ε ) ) into inequality (5.2) we obtain, (5.3) ∫ Ω C i j k l R k l ( ε ) ( u ( ε ) ) R i j ( ε ) ( v − u ( ε ) ) d x − ∫ Ω d i j τ j ( ε ) ( φ ( ε ) ) τ i ( ε ) ( ψ − φ ( ε ) ) d x + ∫ Ω P k i j τ k ( ε ) ( φ ( ε ) ) R i j ( ε ) ( v − u ( ε ) ) d x + ∫ Ω P k i j τ k ( ε ) ( ψ − φ ( ε ) ) R i j ( ε ) ( u ( ε ) ) d x ⩾ ∫ Ω f · ( v − u ( ε ) ) d x + ∫ Γ + g · ( v − u ( ε ) ) d a − ∫ Ω ρ e ( ψ − φ ( ε ) ) d x + ∫ Γ + ϑ ( ψ − φ ( ε ) ) d a . In inequality (5.3) we take v = 0 and ψ = φ ( ε ) and get the inequality − ∫ Ω C i j k l R k l ( ε ) ( u ( ε ) ) R i j ( ε ) ( u ( ε ) ) d x − ∫ Ω P k i j τ k ( ε ) ( φ ( ε ) ) R i j ( ε ) ( u ( ε ) ) d x ⩾ − ∫ Ω f · u ( ε ) d x − ∫ Γ + g · u ( ε ) d a . Next we let v = u ( ε ) and ψ = 2 φ ( ε ) in inequality (5.3), and we get: − ∫ Ω d i j τ j ( ε ) ( φ ( ε ) ) τ i ( ε ) ( φ ( ε ) ) d x + ∫ Ω P k i j τ k ( ε ) ( φ ( ε ) ) R i j ( ε ) ( u ( ε ) ) d x ⩾ − ∫ Ω ρ e φ ( ε ) d x + ∫ Γ + ϑ φ ( ε ) d a . Adding the previous two inequalities yields, ∫ Ω C i j k l R k l ( ε ) ( u ( ε ) ) R i j ( ε ) ( u ( ε ) ) d x + ∫ Ω d i j τ j ( ε ) ( φ ( ε ) ) τ i ( ε ) ( φ ( ε ) ) d x ⩽ ∫ Ω [ f · u ( ε ) + ρ e φ ( ε ) ] d x + ∫ Γ + [ g · u ( ε ) − ϑ φ ( ε ) ] d a . From this inequality, using the coerciveness properties of tensors C and d, we get | R ( ε ) ( u ( ε ) ) | 0 , Ω 2 + | τ ( ε ) ( φ ( ε ) ) | 0 , Ω 2 ⩽ c ( ‖ u ( ε ) ‖ 0 , Ω + ‖ φ ( ε ) ‖ 0 , Ω ) ⩽ c ( ‖ u ( ε ) ‖ 1 , Ω + ‖ φ ( ε ) ‖ 1 , Ω ) . We recall Korn inequality and Poincare inequality: there exists c > 0 such that ‖ u ( ε ) ‖ 1 , Ω 2 ⩽ c | e ( u ( ε ) ) | 0 , Ω 2 , ‖ φ ( ε ) ‖ 1 , Ω 2 ⩽ c | τ ( ε ) ( φ ( ε ) ) | 0 , Ω 2 . Therefore for ε ⩽ 1 , there exist c > 0 : ‖ u ( ε ) ‖ 1 , Ω 2 + ‖ φ ( ε ) ‖ 1 , Ω 2 ⩽ c ( | e ( u ( ε ) ) | 0 , Ω 2 + ‖ τ ( ε ) ( φ ( ε ) ) ‖ 0 , Ω 2 ) ⩽ c ( | R ( ε ) ( u ( ε ) ) | 0 , Ω 2 + | τ ( ε ) ( φ ( ε ) ) | 0 , Ω 2 ) ⩽ c ( ‖ u ( ε ) ‖ 1 , Ω + ‖ φ ( ε ) ‖ 1 , Ω ) , by ( a + b ) 2 ⩽ 2 ( a 2 + b 2 ) , these inequalities imply that the norms ‖ u ( ε ) ‖ 1 , Ω , ‖ φ ( ε ) ‖ 1 , Ω , | R ( ε ) ( u ( ε ) ) | 0 , Ω , | τ ( ε ) ( φ ( ε ) ) | 0 , Ω are uniformly bounded. Therefore there exist u ∈ H 1 ( Ω ) , φ ∈ H 1 ( Ω ) and R ∈ L 2 ( Ω ) , τ ∈ L 2 ( Ω ) such that we have the weak convergences: u ( ε ) ⇀ u in H 1 ( Ω ) , φ ( ε ) ⇀ φ in H 1 ( Ω ) , R ( ε ) ( u ( ε ) ) ⇀ R ( u ) in L 2 ( Ω ) , τ ( ε ) ( φ ( ε ) ) ⇀ τ ( φ ) in L 2 ( Ω ) . Moreover, from the definition of R ( ε ) and τ ( ε ) , we have the bounds: | e α 3 ( u ( ε ) ) | 0 , Ω ⩽ c ε , | e 33 ( u ( ε ) ) | 0 , Ω ⩽ c ε 2 , | ∂ 3 φ ( ε ) | 0 , Ω ⩽ c ε . Hence e i 3 ( u ( ε ) ) ⇀ 0 in L 2 ( Ω ) , ∂ 3 φ ( ε ) ⇀ 0 in L 2 ( Ω ) , thus | e i 3 ( u ) | 0 , Ω ⩽ lim inf ε → 0 | e i 3 ( u ( ε ) ) | 0 , Ω = 0 and | ∂ 3 φ | 0 , Ω ⩽ lim inf ε → 0 | ∂ 3 φ ( ε ) | 0 , Ω = 0 . From the one hand, since e i 3 ( u ) = 0 , we deduce that there exist a bi-dimensional field ζ = ( ζ i ) such that ζ α ∈ H 1 ( ω ) and ζ 3 ∈ H 2 ( ω ) and u is a Kirchhoff–Love displacement field u α = ζ α − x 3 ∂ α ζ 3 , u 3 = ζ 3 . From the other hand, since ∂ 3 φ = 0 , the limit φ is independent of x 3 .

The following lemma will be used several times in the next step. Lemma 5.2.

Let the operator A ( ε ) : u ( ε ) → H 1 ( Ω ) which satisfies the weak convergence A ( ε ) ( u ( ε ) ) ⇀ A ( u ) ∈ L 2 ( Ω ) . If u ( ε ) ∈ K ( Ω ) solves the variational inequality (5.4) ∫ Ω A ( ε ) ( u ( ε ) ) · ∂ 3 ( v − u ( ε ) ) ⩾ ε ∫ Ω f · ( v − u ( ε ) ) ∀ v ∈ K ( Ω ) then A ( u ) = 0 .

Proof.

In (5.4) we first let v = 0 and then v = 2 u ( ε ) . This yields ∫ Ω A ( ε ) ( u ( ε ) ) · ∂ 3 u ( ε ) = ε ∫ Ω f · u ( ε ) . And (5.4) now reads ∫ Ω A ( ε ) ( u ( ε ) ) · ∂ 3 v ⩾ ε ∫ Ω f · v ∀ v ∈ K ( Ω ) which in turn implies ∫ Ω A ( u ) · ∂ 3 v ⩾ 0 ∀ v ∈ K ( Ω ) . We consider the 2 cases of the in-plane components and the vertical component of the displacement fields.

Since v α belongs to a vector space, the previous inequality becomes an equality ∫ Ω A α ( u ) ∂ 3 v α = 0 ∀ v α ∈ H 1 ( Ω ) and following [8] we get A α ( u ) = 0 .

For the transverse component we consider z , t ∈ D ( Ω ) , z > 0 and select v 3 ⩾ 0 under the form: v 3 ( x ) = z ( x 1 , x 2 ) + max − 1 < s < 1 | t ( x 1 , x 2 , s ) | + ∫ − 1 x 3 t ( x 1 , x 2 , s ) d s . A direct computation ∫ Ω A 3 ( u ) ∂ 3 v 3 = ∫ Ω A 3 ( u ) t ( x 1 , x 2 , x 3 ) = 0 for all t ∈ D ( Ω ) implies A 3 ( u ) . □

Step II.

Before we establish the strong convergence, let us compute τ 3 ( φ ) and R i 3 ( u ) .

The weak convergences established in Step I imply: R α β ( ε ) ( u ( ε ) ) ⇀ R α β ( u ) = e α β ( u ) in L 2 ( Ω ) , τ α ( ε ) φ ( ε ) = ∂ α φ ( ε ) ⇀ τ α ( φ ) = ∂ α φ in L 2 ( Ω ) . We let v = u ( ε ) in (5.2), and multiply the inequality by ε, we obtain: ∫ Ω ( 2 P 3 α 3 R α 3 ( ε ) ( u ( ε ) ) + P 333 R 33 ( ε ) ( u ( ε ) ) − d 33 τ 3 ( ε ) ( φ ( ε ) ) − d 3 γ ∂ γ φ ( ε ) + P 3 α β e α β ( ε ) ( u ( ε ) ) ) ∂ 3 ( ψ − φ ( ε ) ) d x ⩾ ε ∫ Ω d α β τ β ( ε ) ( φ ( ε ) ) ∂ α ( ψ − φ ( ε ) ) d x + ε ∫ Ω d α 3 τ 3 ( ε ) ( φ ( ε ) ) ∂ α ( ψ − φ ( ε ) ) d x − ε ∫ Ω P γ α β ∂ γ ( ψ − φ ( ε ) ) R α β ( ε ) ( u ( ε ) ) d x − 2 ε ∫ Ω P γ α 3 ∂ γ ( ψ − φ ( ε ) ) R α 3 ( ε ) ( u ( ε ) ) d x − ε ∫ Ω P γ 33 ∂ γ ( ψ − φ ( ε ) ) R 33 ( ε ) ( u ( ε ) ) d x − ε ∫ Ω ρ e ( ψ − φ ( ε ) ) d x + ε ∫ Γ + ϑ ( ψ − φ ( ε ) ) d a ∀ ψ ∈ Q ( Ω ) . The weak convergences and Lemma 5.2 yields (5.5) 2 P 3 α 3 R α 3 ( u ) + P 333 R 33 ( u ) − d 33 τ 3 ( φ ) = d 3 γ ∂ γ φ − P 3 α β e α β ( u ) . Similarly in (5.2), we let ψ = φ ( ε ) , v 3 = u 3 ( ε ) , we multiply by ε and get: ∫ Ω ( 2 C α 3 σ τ R σ τ ( u ( ε ) ) ( ε ) + 4 C α 3 σ 3 R σ 3 ( ε ) ( u ( ε ) ) + 2 C α 333 R 33 ( ε ) ( u ( ε ) ) ) ∂ 3 ( v α − u α ( ε ) ) d x + 2 ∫ Ω ( P γ α 3 τ γ ( ε ) ( φ ( ε ) ) + P 3 α 3 τ 3 ( ε ) ( φ ( ε ) ) ) ∂ 3 ( v α − u α ( ε ) ) d x ⩾ − ε ∫ Ω C α β σ τ R σ τ ( ε ) ( u ( ε ) ) e α β ( v − u ( ε ) ) d x − 2 ε ∫ Ω C α β σ 3 R σ 3 ( ε ) ( u ( ε ) ) e α β ( v − u ( ε ) ) d x − ε ∫ Ω C α β 33 R 33 ( ε ) ( u ( ε ) ) e α β ( v − u ( ε ) ) d x − ε ∫ Ω P γ α β τ γ ( ε ) ( φ ( ε ) ) e α β ( v − u ( ε ) ) d x − ε ∫ Ω P 3 α β τ 3 ( ε ) ( φ ( ε ) ) e α β ( v − u ( ε ) ) d x + ε ∫ Ω f α ( v α − u α ( ε ) ) d x + ε ∫ Γ + g α ( v α − u α ( ε ) ) d a ∀ v α ∈ H 1 ( Ω ) , v α = 0 on Γ 0 , from this, we obtain: 2 C α 3 σ 3 R σ 3 ( u ) + C α 333 R 33 ( u ) + P 3 α 3 τ 3 ( φ ) = − P γ α 3 ∂ γ φ − C α 3 α β e α β ( u ) . Finally, in (5.2), we let ψ = φ ( ε ) , v α = u α ( ε ) , multiply by ε 2 , we have: ∫ Ω ( C 33 σ τ R σ τ ( ε ) ( u ( ε ) ) + 2 C 33 σ 3 R σ 3 ( ε ) ( u ( ε ) ) + C 3333 R 33 ( ε ) ( u ( ε ) ) ) ∂ 3 ( v 3 − u 3 ( ε ) ) d x + ∫ Ω ( P γ 33 τ γ ( ε ) ( φ ( ε ) ) + P 333 τ 3 ( ε ) ( φ ( ε ) ) ) ∂ 3 ( v 3 − u 3 ( ε ) ) d x ⩾ − ε ∫ Ω C α 3 σ τ R σ τ ( u ( ε ) ) ( ε ) ∂ α ( v 3 − u 3 ( ε ) ) d x − 2 ε ∫ Ω C α 3 σ 3 R σ 3 ( ε ) ( u ( ε ) ) ∂ α ( v 3 − u 3 ( ε ) ) d x − ε ∫ Ω C α 333 R 33 ( ε ) ( u ( ε ) ) ∂ α ( v 3 − u 3 ( ε ) ) d x − ε ∫ Ω P γ α 3 τ γ ( ε ) ( φ ( ε ) ) ∂ α ( v 3 − u 3 ( ε ) ) d x − ε ∫ Ω P 3 α 3 τ 3 ( ε ) ( φ ( ε ) ) ∂ α ( v 3 − u 3 ( ε ) ) d x + ε 2 ∫ Ω f 3 ( v 3 − u 3 ( ε ) ) d x + ε 2 ∫ Γ + g 3 ( v 3 − u 3 ( ε ) ) d a ∀ v 3 ∈ H 1 ( Ω ) , v 3 = 0 on Γ 0 , v 3 ⩾ 0 on Γ − , 2 C 33 α 3 R α 3 ( u ) + C 3333 R 33 ( u ) + P 333 τ 3 ( φ ) = − P γ 33 ∂ γ φ − C 33 α β e α β ( u ) . Thus R i 3 ( u ) satisfy the following linear system (5.6) 2 C i 3 α 3 R α 3 ( u ) + C i 333 R 33 ( u ) + P 3 i 3 τ 3 ( φ ) = − P γ i 3 ∂ γ φ − C i 3 α β e α β ( u ) . We get τ 3 ( φ ) from (5.5) and substitute it in (5.6), so the three unknowns R i 3 ( u ) solve the linear system of three equations: (5.7) 2 C i 3 α 3 ′ R α 3 ( u ) + C i 333 ′ R 33 ( u ) = − P γ i 3 ′ ∂ γ φ − C i 3 α β ′ e α β ( u ) , where C i j k l ′ = C i j k l + P 3 i j P 3 k l d 33 and P k i j ′ = P k i j − P 3 i j d 3 k d 33 . To show that this system has a unique solution we first note that since d 33 > 0 , the symmetry and positivity of tensor C yield the symmetry and positivity of tensor C ′ , more precisely: C i j k l ′ A k l A i j = C i j k l A k l A i j + P 3 i j P 3 k l d 33 A k l A i j ⩾ c A k l A i j ∀ A i j ∈ R , A i j = A j i . Next we note that system (5.7) can be written as 2 C i 3 α 3 ′ x α + C i 333 ′ x 3 = f i , the determinant of this linear system is 2 C 1313 ′ 2 C 1323 ′ C 1333 ′ 2 C 2313 ′ 2 C 2323 ′ C 2333 ′ 2 C 3313 ′ 2 C 3323 ′ C 3333 ′ = 4 C 1313 ′ C 1323 ′ C 1333 ′ C 2313 ′ C 2323 ′ C 2333 ′ C 3313 ′ C 3323 ′ C 3333 ′ . With A k α = 0 we get C i j k l ′ A k l A i j = C i 3 j 3 ′ A i 3 A j 3 > 0 , therefore system (5.7) has a unique solution, which reads: (5.8) R α 3 ( u ) = − 1 2 △ ( △ γ α ∂ γ φ + △ ζ η α e ζ η ( u ) ) , R 33 ( u ) = − 1 △ ( △ γ 3 ∂ γ φ + △ ζ η 3 e ζ η ( u ) ) , τ 3 ( φ ) = − 1 d 33 [ ( 1 △ P 3 k 3 △ γ ρ k − P 3 γ ρ ) e γ ρ ( u ) + ( 1 △ P 3 k 3 △ γ k + d 3 γ ) ∂ γ φ ] , where △ = ϵ p q r C 13 p 3 ′ C 23 q 3 ′ C 33 r 3 ′ , △ γ 1 = ϵ p q r P γ p 3 ′ C 23 q 3 ′ C 33 r 3 ′ , △ γ 2 = ϵ p q r C 13 p 3 ′ P γ q 3 ′ C 33 r 3 ′ , △ γ 3 = ϵ p q r C 13 p 3 ′ C 23 q 3 ′ P γ r 3 ′ , △ ζ η 1 = ϵ p q r C 3 p ζ η ′ C 23 q 3 ′ C 33 r 3 ′ , △ ζ η 2 = ϵ p q r C 13 p 3 ′ C 3 q ζ η ′ C 33 r 3 ′ , △ ζ η 3 = ϵ p q r C 13 p 3 ′ C 23 q 3 ′ C 3 r ζ η ′ , and ϵ is the Levi-Civitta symbol ϵ i j k = 1 , ( i , j , k ) is even permutation of ( 1 , 2 , 3 ) , − 1 , ( i , j , k ) is odd permutation of ( 1 , 2 , 3 ) .

Step III.

We now show that the whole family { u ( ε ) , φ ( ε ) } ε converges strongly.

Let us introduce the notation ∫ Ω C A : A d x = ∫ Ω C i j k l A k l A i j d x for all second order symmetric tensor A and let ∫ Ω d v : v d x = ∫ Ω d i j v i v j d x for all vector v ∈ R 3 , c | R ( ε ) ( u ( ε ) ) − R ( u ) | 0 , Ω 2 ⩽ ∫ Ω C ( R ( ε ) ( u ( ε ) ) − R ( u ) ) : ( R ( ε ) ( u ( ε ) ) − R ( u ) ) d x ⩽ ∫ Ω C R ( u ) : ( R ( u ) − 2 R ( ε ) ( u ( ε ) ) ) d x + ∫ Ω C R ( ε ) ( u ( ε ) ) : R ( ε ) ( u ( ε ) ) d x . Since we have already established the weak convergences R ( ε ) ⇀ R in L 2 ( Ω ) as ε → 0 , then lim ε → 0 c | R ( ε ) ( u ( ε ) ) − R ( u ) | 0 , Ω 2 ⩽ − ∫ Ω C R ( u ) : R ( u ) d x + lim ε → 0 ∫ Ω C R ( ε ) ( u ( ε ) ) : R ( ε ) ( u ( ε ) ) d x . In (5.3), let ψ = φ ( ε ) , v = u , and pass to the limit, we get: ∫ Ω C R ( u ) : R ( u ) d x − lim ε → 0 ∫ Ω C R ( ε ) ( u ( ε ) ) : R ( ε ) ( u ( ε ) ) d x ⩾ 0 . Then we have lim ε → 0 | R ( ε ) ( u ( ε ) ) − R ( u ) | 0 , Ω 2 ⩽ 0 , which combined with the previous inequality as ε → 0 , so we get the strong convergence.

By the definition of R ( ε ) , we get: | e ( u ( ε ) ) − e ( u ) | 0 , Ω 2 ⩽ ∑ α , β | R α β ( ε ) ( u ( ε ) ) − R α β ( u ) | 0 , Ω 2 + 2 ε 2 ∑ α | R α 3 ( ε ) ( u ( ε ) ) − R α 3 ( u ) | 0 , Ω 2 + ε 4 | R 33 ( ε ) ( u ( ε ) ) − R 33 ( u ) | 0 , Ω 2 , which implies that the sequence { e ( u ( ε ) ) } ε converges strongly in L 2 ( Ω ) to e ( u ) . Therefore by Korn’s inequality the sequence u ( ε ) converges strongly in H 1 ( Ω ) to u.

Similarly, c | τ ( ε ) ( φ ( ε ) ) − τ ( φ ) | 0 , Ω 2 ⩽ ∫ Ω d ( τ ( ε ) ( φ ( ε ) ) − τ ( φ ) ) : ( τ ( ε ) ( φ ( ε ) ) − τ ( φ ) ) d x ⩽ ∫ Ω d τ ( φ ) : ( τ ( φ ) − 2 τ ( ε ) ( φ ( ε ) ) ) d x + ∫ Ω d τ ( ε ) ( φ ( ε ) ) : τ ( ε ) ( φ ( ε ) ) d x . Since we have already established the weak convergence τ ( ε ) ( φ ( ε ) ) ⇀ τ ( φ ) in L 2 ( Ω ) as ε → 0 , we get lim ε → 0 c | τ ( ε ) ( φ ( ε ) ) − τ ( φ ) | 0 , Ω 2 ⩽ − ∫ Ω d τ ( φ ) : τ ( φ ) d x + lim ε → 0 ∫ Ω d τ ( ε ) ( φ ( ε ) ) : τ ( ε ) ( φ ( ε ) ) d x . We let v = u ( ε ) in (5.3), which yields: (5.9) − ∫ Ω d i j τ j ( ε ) ( φ ( ε ) ) τ i ( ε ) ( ψ − φ ( ε ) ) d x + ∫ Ω P k i j τ k ( ε ) ( ψ − φ ( ε ) ) R i j ( ε ) ( u ( ε ) ) d x ⩾ − ∫ Ω ρ e ( ψ − φ ( ε ) ) d x + ∫ Γ + ϑ ( ψ − φ ( ε ) ) d a ∀ ψ ∈ Q ( Ω ) . Using the fact that ψ belongs to a vector space this inequality can be rewritten, tacking ψ = φ ( ε ) , as the equality: ∫ Ω d i j τ j ( ε ) ( φ ( ε ) ) τ i ( ε ) ( φ ( ε ) ) d x = ∫ Ω P k i j τ k ( ε ) ( φ ( ε ) ) R i j ( ε ) ( u ( ε ) ) d x − ∫ Ω ρ e φ ( ε ) d x + ∫ Γ + ϑ φ ( ε ) d a , then as ε → 0 , − ∫ Ω d τ ( φ ) : τ ( φ ) d x + lim ε → 0 ∫ Ω d τ ( ε ) ( φ ( ε ) ) : τ ( ε ) ( φ ( ε ) ) d x = − ∫ Ω d τ ( φ ) : τ ( φ ) d x + lim ε → 0 ( ∫ Ω P k i j τ k ( ε ) ( φ ( ε ) ) R i j ( ε ) ( u ( ε ) ) d x + ∫ Ω ρ e φ ( ε ) d x − ∫ Γ + ϑ φ ( ε ) d x ) = 0 , so | τ ( ε ) ( φ ( ε ) ) − τ ( φ ) | 0 , Ω 2 ⩽ 0 , we get the strong convergence, the sequence { τ ( φ ( ε ) ) } ε converges strongly in L 2 ( Ω ) to τ ( φ ) , then the sequence of φ ( ε ) converges strongly to φ in H 1 ( Ω ) .

Step IV.

For any Kirchhoff–Love vector field v = ( ζ H , ζ 3 ) ∈ V H ( ω ) × K 3 ( ω ) we have e α β ( v ) = e α β ( η H ) − x 3 ∂ α β η 3 , e i 3 ( v ) = 0 . We pass to the limit in (5.3) and get for all vector field v = ( ζ H , ζ 3 ) ∈ V H ( ω ) × K 3 ( ω ) the variational inequality: ∫ Ω C α β σ τ ( e σ τ ( ζ H ) − x 3 ∂ σ τ ζ 3 ) ( e α β ( η H − ζ H ) − x 3 ∂ α β ( η 3 − ζ 3 ) ) d x + 2 ∫ Ω C α β σ 3 [ − 1 2 △ ( △ γ σ ∂ γ φ + △ σ τ σ ( e σ τ ( ζ H ) − x 3 ∂ σ τ ζ 3 ) ) ] × ( e α β ( η H − ζ H ) − x 3 ∂ α β ( η 3 − ζ 3 ) ) d x + ∫ Ω C α β 33 [ − 1 △ ( △ γ 3 ∂ γ φ + △ σ τ 3 ( e σ τ ( ζ H ) − x 3 ∂ σ τ ζ 3 ) ) ] × ( e α β ( η H − ζ H ) − x 3 ∂ α β ( η 3 − ζ 3 ) ) d x − ∫ Ω d α β ∂ β φ ∂ α ( ψ − φ ) d x − ∫ Ω d α 3 [ − 1 d 33 [ ( 1 △ P 3 k 3 △ σ τ k − P 3 σ τ ) ( e σ τ ( ζ H ) − x 3 ∂ σ τ ζ 3 ) + ( 1 △ P 3 k 3 △ γ k + d 3 γ ) ∂ γ φ ] ] ∂ α ( ψ − φ ) d x + ∫ Ω P γ α β ∂ γ φ ( e α β ( η H − ζ H ) − x 3 ∂ α β ( η 3 − ζ 3 ) ) d x + ∫ Ω P γ α β ∂ γ ( ψ − φ ) ( e α β ( ζ H ) − x 3 ∂ α β ζ 3 ) d x + ∫ Ω P 3 α β [ − 1 d 33 ( 1 △ P 3 k 3 △ σ τ k − P 3 σ τ ) ( e σ τ ( ζ H ) − x 3 ∂ σ τ ζ 3 ) × ( e α β ( η H − ζ H ) − x 3 ∂ α β ( η 3 − ζ 3 ) ) ] d x + 2 ∫ Ω P γ σ 3 ∂ γ ( ψ − φ ) [ − 1 2 △ ( △ γ σ ∂ γ φ + △ σ τ σ ( e σ τ ( ζ H ) − x 3 ∂ σ τ ζ 3 ) ) ] d x + ∫ Ω P γ 33 ∂ γ ( ψ − φ ) [ − 1 △ ( △ γ 3 ∂ γ φ + △ σ τ 3 ( e σ τ ( ζ H ) − x 3 ∂ σ τ ζ 3 ) ) ] d x ⩾ ∫ Ω [ f α ( η α − ζ α − x 3 ∂ α ( η 3 − ζ 3 ) ) + f 3 ( η 3 − ζ 3 ) ] d x + ∫ Γ + [ g α ( η α − ζ α − x 3 ∂ α ( η 3 − ζ 3 ) ) + g 3 ( η 3 − ζ 3 ) ] d a − ∫ Ω ρ e ( ψ − φ ) d x + ∫ Γ + ϑ ( ψ − φ ) d a . Then we get 2 ∫ ω C ˜ α β σ τ e σ τ ( ζ H ) e α β ( η H − ζ H ) d ω + 2 3 ∫ ω C ˜ α β σ τ ∂ σ τ ζ 3 ∂ α β ( η 3 − ζ 3 ) d ω + 2 ∫ ω P ˜ γ α β ∂ γ φ e α β ( η H − ζ H ) d ω + 2 ∫ ω P ˜ γ α β ♠ ∂ γ ( ψ − φ ) e α β ( ζ H ) d ω − 2 ∫ ω d ˜ γ β ∂ β φ ∂ γ ( ψ − φ ) d ω ⩾ ∫ ω p H · ( η H − ζ H ) d ω − ∫ ω ( ρ ˜ e − ϑ ) ( ψ − φ ) d ω + ∫ ω p 3 ( η 3 − ζ 3 ) d ω − ∫ ω s α ∂ α ( η 3 − ζ 3 ) d ω . This inequality can be decoupled as: (5.10) 2 3 ∫ ω C ˜ α β σ τ ∂ σ τ ζ 3 ∂ α β ( η 3 − ζ 3 ) d ω ⩾ ∫ ω p 3 ( η 3 − ζ 3 ) d ω − ∫ ω s α ∂ α ( η 3 − ζ 3 ) d ω ∀ η 3 ∈ K 3 ( ω ) , 2 ∫ ω n ˜ α β ( ζ H , φ ) e α β ( η H ) d ω + 2 ∫ ω D ˜ γ ( ζ H , φ ) ∂ γ ψ d ω = ∫ ω p H · η H d ω − ∫ ω ( ρ ˜ e − ϑ ) ψ d ω ∀ ( η H , ψ ) ∈ V H ( ω ) × Q ( ω ) , where the constitutive law is n ˜ α β ( ζ H , φ ) = C ˜ α β σ τ e σ τ ( ζ H ) + P ˜ γ α β ∂ γ φ , D ˜ γ ( ζ H , φ ) = P ˜ γ α β ♠ e α β ( ζ H ) − d ˜ γ β ∂ β φ and the mechanical forces and electric charges are given by p i : = ∫ − 1 1 f i d x 3 + g i , s α : = ∫ − 1 1 x 3 f α d x 3 + g α , ρ ˜ e : = ∫ − 1 1 ρ e d x 3 , where the new characteristics of the piezoelectric plate are (5.11) C ˜ α β σ τ = C α β σ τ ′ − 1 △ C α β k 3 ′ △ σ τ k , C i j k l ′ = C i j k l + P 3 i j P 3 k l d 33 , P ˜ γ α β = P γ α β ′ − 1 △ C α β k 3 ′ △ γ k , P ˜ γ α β ♠ = P γ α β ′ − 1 △ P γ k 3 ′ △ α β k , P k i j ′ = P k i j − P 3 i j d 3 k d 33 , d ˜ γ β = d γ β ′ + 1 △ P γ k 3 ′ △ β k , d i j ′ = d i j − d i 3 d 3 j d 33 , and (5.12) △ = ϵ p q r C 13 p 3 ′ C 23 q 3 ′ C 33 r 3 ′ , △ γ 1 = ϵ p q r P γ p 3 ′ C 23 q 3 ′ C 33 r 3 ′ , △ γ 2 = ϵ p q r C 13 p 3 ′ P γ q 3 ′ C 33 r 3 ′ , △ γ 3 = ϵ p q r C 13 p 3 ′ C 23 q 3 ′ P γ r 3 ′ , △ ζ η 1 = ϵ p q r C 3 p ζ η ′ C 23 q 3 ′ C 33 r 3 ′ , △ ζ η 2 = ϵ p q r C 13 p 3 ′ C 3 q ζ η ′ C 33 r 3 ′ , △ ζ η 3 = ϵ p q r C 13 p 3 ′ C 23 q 3 ′ C 3 r ζ η ′ , with the Levi-Civitta symbol ϵ i j k = + 1 , ( i , j , k ) is an even permutation of ( 1 , 2 , 3 ) , − 1 , ( i , j , k ) is an odd permutation of ( 1 , 2 , 3 ) . We note that the constitutive law which relates D ˜ γ and n ˜ α β to ( ζ H , φ ) has not the symmetric property of piezoelectricity. The next step will tackle this incoherence.

Step V.

According to the definition given in (5.12), we have P ˜ γ α β = P γ α β ′ − 1 △ C α β k 3 ′ △ γ k , C α β k 3 ′ △ γ k = C α β 13 ′ △ γ 1 + C α β 23 ′ △ γ 2 + C α β 33 ′ △ γ 3 C α β k 3 ′ △ γ k = C α β 13 ′ P γ p 3 ′ C 23 q 3 ′ C 33 r 3 ′ + C α β 23 ′ C 13 p 3 ′ P γ q 3 ′ C 33 r 3 ′ + C α β 33 ′ C 13 p 3 ′ C 23 q 3 ′ P γ r 3 ′ C α β k 3 ′ △ γ k = P γ 13 ′ ( C α β 13 ′ C 2323 ′ C 3333 ′ − C α β 13 ′ C 2333 ′ C 3323 ′ + C α β 23 ′ C 1333 ′ C 3323 ′ − C α β 23 ′ C 1323 ′ C 3333 ′ C α β k 3 ′ △ γ k = + C α β 33 ′ C 1323 ′ C 2333 ′ − C α β 33 ′ C 1333 ′ C 2323 ′ ) C α β k 3 ′ △ γ k = + P γ 23 ′ ( C α β 13 ′ C 2333 ′ C 3313 ′ − C α β 13 ′ C 2313 ′ C 3333 ′ + C α β 23 ′ C 1313 ′ C 3333 ′ − C α β 23 ′ C 1333 ′ C 3313 ′ C α β k 3 ′ △ γ k = + C α β 33 ′ C 1333 ′ C 2313 ′ − C α β 33 ′ C 1313 ′ C 2333 ′ ) C α β k 3 ′ △ γ k = + P γ 33 ′ ( C α β 13 ′ C 2313 ′ C 3323 ′ − C α β 13 ′ C 2323 ′ C 3313 ′ + C α β 23 ′ C 1313 ′ C 3333 ′ − C α β 23 ′ C 1333 ′ C 3313 ′ C α β k 3 ′ △ γ k = + C α β 33 ′ C 1313 ′ C 2323 ′ − C α β 33 ′ C 1323 ′ C 2313 ′ ) C α β k 3 ′ △ γ k = ϵ p q r ( P γ 13 ′ C α β p 3 ′ C 23 q 3 ′ C 33 r 3 ′ + P γ 23 ′ C 13 p 3 ′ C α β q 3 ′ C 33 r 3 ′ + P γ 33 ′ C 13 p 3 ′ C 23 q 3 ′ C α β r 3 ′ ) , and similarly P ˜ γ α β ♠ = P γ α β ′ − 1 △ P γ k 3 ′ △ α β k = P γ α β ′ − 1 △ P γ 13 ′ △ α β 1 − 1 △ P γ 23 ′ △ α β 2 − 1 △ P γ 33 ′ △ α β 3 = P γ α β ′ − 1 △ ϵ p q r ( P γ 13 ′ C 3 p α β ′ C 23 q 3 ′ C 33 r 3 ′ + P γ 23 ′ C 13 p 3 ′ C 3 q α β ′ C 33 r 3 ′ + P γ 33 ′ C 13 p 3 ′ C 23 q 3 ′ C 3 r α β ′ ) . By the properties of the determinant and symmetry (2.1), we get P ˜ γ α β = P ˜ γ α β ♠ . So finally the new bi-dimensional piezoelectric tensors are given by (5.13) C ˜ α β σ τ = C α β σ τ ′ − 1 △ C α β k 3 ′ △ σ τ k , C i j k l ′ = C i j k l + P 3 i j P 3 k l d 33 , P ˜ γ α β = P γ α β ′ − 1 △ C α β k 3 ′ △ γ k , P k i j ′ = P k i j − P 3 i j d 3 k d 33 , d ˜ γ β = d γ β ′ + 1 △ P γ k 3 ′ △ β k , d i j ′ = d i j − d i 3 d 3 j d 33 , and △, △ γ k , △ ζ η k are given in (5.12).

Hence we get the desired variational inequalities. □

5.2. Strong formulation