Asymptotic behaviour of thin ferroelectric materials N. Aïssa -K. Hamdache R.I. N January 2004

Transcription

1 ECOLE POLYTECHNIQUE CENTRE DE MATHÉMATIQUES APPLIQUÉES UMR CNRS PALAISEAU CEDEX (FRANCE). Tél: Fax: Asymptotic behaviour of thin ferroelectric materials N. Aïssa -K. Hamdache R.I. N January 2004

2 Asymptotic behaviour of thin ferroelectric materials Naïma Aïssa 1,2 and Kamel Hamdache 1 (1) Centre de Mathématiques Appliquées, CNRS & Ecole Polytechnique Palaiseau Cedex, France (2) Faculté de Mathématiques. USTHB BP 32, El Alia, Bab Ezzouar, Alger, Algérie aissa@cmapx.polytechnique.fr, hamdache@cmapx.polytechnique.fr January 9, 2004 Abstract We discuss the behaviour of a model of ferroelectric material represented by a thin cylinder with small thickness ν > 0. This model is described by a couple of Maxwell equations satisfied by the electromagnetic field (H, E) and electric polarization field P. We give a complete description of the limit model as ν 0 in the linear case when (E, H) and P satisfy a Silver- Müller type boundary condition. When the potential is nonlinear and P satisfies the boundary condition P n = 0 we prove the strong convergence of the polarization field which allows to give the description of the behaviour of the nonlinear problem when the thickness ν tends to 0. We observe that the behaviour is very sensitive to the choice of the boundary conditions satisfied by the different fields. Key words. Maxwell equations, Ferroelectric material, Thin cylinders AMS subject classifications. 35L10,35K05. 1 Introduction We shall discuss the model equations of ferroelectric materials introduced by Greenberg et al. see [8] and also [7]. The characteristic feature of ferroelectric cristal is the appearance of a spontaneous electric dipole. It can be reversed, with no net change in magnitude, by an applied electric field. The current density j of the ferroelectric domain Ω is driven by the difference between the electric equilibrium field Ê(P ) and the electric field E where P is the spontaneous electric polarization). If one denotes by m the internal magnetic field then the model equations introduced in [8] takes the form in R + Ω ɛ( t P + θ j) = curl m µ( t m + θα m) = curl P t j + θα j = γθ(ê(p ) E). (1) This set of equations is completed by initial conditions P (0) = P 0, m(0) = m 0, j(0) = j 0 and boundary conditions which will be discussed later. Eliminating the variables j and m we get the following Maxwell equation satisfied by P t 2 P + (ɛµ) 1 curl 2 P + a t P = γθ(ê(p ) E) (2) 1

3 where a = θα. We set curl 2 P = curl (curl P ). The parameters ɛ > 0 and µ > 0 are the permititivity and the magnetic permeabilty of the vaccum and the others ones are some physical constants. The equilibrium electric field is given by Ê(P ) = φ ( P 2 )P where φ is a two wells potential satisfying some hypotheses given later. The electric displacement D is linked to the electric and polarization fields E and P by the law D = ɛ(e + P ). Hence the electromagnetic field (H, E) satisfies in R + Ω the Maxwell equations µ t H curl E = 0, ɛ t (E + P ) + curl H + σe = 0 (3) where σ > 0 is the constant conductivity of Ω. The initial contitions are E(0) = E 0, H(0) = H 0. The boundary conditions satisfied by E and P take an important place in the characterization of the thin limit behaviour of the model. If m satisfies the boundary condition m n = 0 on Ω where n is the unit outward normal to Ω, one deduces by using (1) that P satisfies the boundary condition curl P n = 0. (4) This condition was proposed in [8]. If more generally m satisfies a Silver-Müller type boundary condition like n m + ρ n (P n) = 0 with ρ 0 is a function defined on Ω then,we obtain directly from (1) the boundary condition for P curl P n + ρµ n (( t + a)p n) = 0. (5) Formally, we have the Green formula Ω curl 2 P t P dx = 1 d 2 dt ( Ω curl P 2 dx+ + µa Ω ρ P n 2 dα) + µ Ω ρ tp n 2 dα This boundary condition will be used in our work only for the linear case when φ is given by φ( P 2 ) = k P 2 /2. For the nonlinear case, we will use the boundary condition (6) P n = 0 (7) instead of (5). It can be deduced from the boundary condition curl m n = θj n. The reason we use (7) is that we can prove the H 1 -regularity of the polarization field P which allow to control the nonlinear term Ê(P ). We refer to the work of [1] where the boundary condition (5) is considered. For the linear as well as the nonlinear case we use for the electric field E, the Silver-Müller boundary condition H n + β n (E n) = 0, P n = 0 (8) where β 0 is some function defined on Ω. We have formally the Green formula (curl H E curl E H)dx = H n Edα = β E n 2 dα. (9) Ω Ω The equilibrium electric field Ê(P ) is given by Ê(P ) = φ ( P 2 )P where φ : R R is a C 2 potential function, defined in [8], such that φ(0) = 0, r 2 0 > 0 with φ(r 2 0) < 0 is the location of the unique minimum of φ(r 2 ) and φ(r 2 ) > 0 for r 2 r 2 1 and φ satisfies the following hypotheses φ(s) C 0 s for s +, φ (s) C 1, sφ (2) (s) C 2 for s 0. (10) where C 0, C 1 > 0 and C 2 > 0 are constants depending only of φ. It follows that there exists C > 0 depending only of C 1 and C 2 such that Ω 2

4 Using (11) we get for all A, B R 3. (sφ (2) (s 2 )) C for s 0. (11) Aφ ( A 2 ) Bφ ( B 2 ) C A B. (12) Let us precise the models we shall discuss. To simplify the presentation we equate to 1 all the parameters appearing in the model except a > 0 and σ to measure the size of the dissipation process. Let ν > 0 be fixed representing the thickness of the cylinder Ω ν = Ω (0, ν) with cross section Ω R 2 assumed to be an open bounded and regular domain. We denote by n the outward unit normal to Ω ν. The generic point x Ω ν is denoted by x = ( x, x 3 ) where x = (x 1, x 2 ) Ω and 0 < x 3 < ν. The electromagnetic field (H ν, E ν ) satisfies in R + Ω ν the problem coupled to the polarization equation t H ν curl E ν = 0, t (E ν + P ν ) + curl H ν + σe ν = 0 H ν (0) = H 0, E ν (0) = E 0 in Ω ν H ν n + β ν (x 3 )n (E ν n) = 0 in R + Ω ν (13) 2 t P ν + curl 2 P ν + a t P ν + φ ( P ν 2 )P ν = E ν in R + Ω ν P ν (0) = P 0, t P ν (0) = P 1 in Ω ν P ν n = 0 in R + Ω ν. For the linear case we have φ ( P 2 ) = k where k > 0 is a constant. The polarization P satisfies the equation 2 t P ν + curl 2 P ν + a t P ν + kp ν = E ν in R + Ω ν (14) P ν (0) = P 0, t P ν (0) = P 1 in Ω ν (15) curl P ν n + ρ ν (x 3 )n (( t + a)p ν n) = 0 in R + Ω ν where β ν and ρ ν are two functions depending of the variable x 3 and the thickness parameter ν. Let O be an open, bounded and regular domain of R 3 or R 2. We denote by L 2 (O) the Lebesgues space (L 2 (O)) 3 equipped with the usual norm denoted by and the scalar product ( ; ). The norm of the Sobolev space H 1 (O) is denoted by H 1. Let H(curl, O) be the usual Hilbert space used in the theory of Maxwell equations equipped with the norm H. We also use the Banach space L p (R + ; L 2 (O)) for p 1, p 2 and the Hilbert space L 2 (R + ; L 2 (O)) with norms denoted respectively by p and. The existence and regularity of solutions (H ν, E ν, P ν ) to problem (13)-(14) as well as to (13)- (15) has been proved in [2] and [9] in the nonlinear case with the boundary conditions E ν n = 0 and either curl P ν n = 0 or P ν n = 0. The Silver-Muller boundary condition satisfied by E ν and H ν is usual in the existence theory of solutions to Maxwell equations. Following the lines of the proof given in [2] we way prove the same results in the case of Silver-Müller type boundary conditions used in our problem (13)-(15). We have for the linear problem (13)-(15) the result, Theorem 1.1 (The linear case.) Let ρ ν, β ν L (0, ν) be fixed such that ρ ν (x 3 ) 0 and β ν (x 3 ) 0 a.e. We assume that the initial data are such that 3

5 H 0, E 0, P 1 L 2 (Ω ν ), P 0 H(curl, Ω ν ) P 0 n L 2 ( Ω ν ). Then, there exists a unique weak solution (H ν, E ν, P ν ) to problem (13)-(15) such that H ν,e ν L (R + ; L 2 (Ω ν )) and P ν L (R + ; H(curl, Ω ν )). The tangential traces H ν n, E ν n, t P ν n belong to L 2 (R + ; L 2 ( Ω ν )) and P ν n L (R + ; L 2 ( Ω ν )). Moreover for all t 0, we have the energy inequality (16) where the energy at the time t is defined by E ν (t) + 2 t 0 (a tp ν (s) 2 + σ E ν (s) 2 + β ν E ν n 2 + ρ ν t P ν (s) n 2 )ds E ν 0 (17) and the initial energy E ν 0 E ν (t) = t P ν (t) 2 + k P ν (t) 2 + curl P ν (t) 2 + a ρ ν P ν (t) n 2 + H ν (t) 2 + E ν (t) 2 (18) is given by E ν 0 = P k P curl P a ρ ν P 0 n 2 + H E 0 2. (19) For the nonlinear problem (13)-(14) we have Theorem 1.2 (The nonlinear case) We assume that φ satisfies hypotheses (10) and the initial data are such that H 0, E 0, P 1 L 2 (Ω ν ), P 0 H(curl, Ω ν ) (20) Then, there exists a unique weak solution (H ν, E ν, P ν ) to problem (13)-(14) such that H ν,e ν L (R + ; L 2 (Ω ν )) and P ν L (R + ; H(curl, Ω ν )). The tangential traces H ν n, E ν n belong to L 2 (R + ; L 2 ( Ω ν )). Moreover for all t 0, we have the energy inequality E ν (t) + 2 t 0 (a t P ν (s) 2 + σ E ν (s) 2 + β ν E ν n 2 )ds E ν 0 (21) where the energy at the time t is defined by E ν (t) = t P ν (t) 2 + φ( P ν (t) 2 )dx + curl P ν (t) 2 + H ν (t) 2 + E ν (t) 2 (22) Ω ν and the initial energy E0 ν is given by E0 ν = P φ( P 0 2 )dx + curl P H E 0 2. Ω ν (23) Using hypotheses (10)- (11) satisfied by φ, we get for all t 0 P ν (t) 2 C( φ( P ν (t) 2 )dx + Ω ν ) (24) Ω ν where C > 0 depends only of φ and, Ω ν = ν Ω is the Lebesgue measure of Ω ν. Finally we have the regularity result 4

6 Lemma 1.1 (Regularity) Let (H ν, E ν, P ν ) be a weak solution of either problem (13)-(14) or problem (13)-(15). If in both cases (H 0, E 0, P 0, P 1 ) satisfies H 0, E 0, P 1, curl P 0 H(curl, Ω ν ), P 0 n L 2 ( Ω ν ), (25) and moreover for the linear case we have P 1 n L 2 ( Ω ν ) then t H ν, t E ν and 2 t P ν L (R + ; L 2 (Ω ν )). Further, H ν, E ν belong to L (R + ; H(curl, Ω ν )) and curl P ν, t P ν L (R + ; H(curl, Ω ν )). The following regularity result is devoted to the nonlinear problem (13)-(14). Proposition 1.1 (H 1 -regularity) We assume that the open and bounded domain Ω is convex, the data (H 0, E 0, P 0 ) are independent of the variable x 3 satisfying (20) with div H 0, div E 0, div P 0, div P 1 are in L 2 ( Ω). Then, for all fixed T > 0 there exists C T > 0 such that for all ν > 0 the solution (H ν, E ν, P ν ) of problem (13)-(14) satisties for all T > 0, the uniform bound P ν W 1, (0,T ;H 1 (Ω ν )) + div H ν 2 + div E ν 2 νc T. (26) Proof. Let (H ν, E ν, P ν ) associated with the initial data satisfying the hypotheses stated in the proposition. The initial energy E0 ν of the system satisfies E ν 0 = O(ν). (27) Using the estimate P ν (t) 2 C( φ( P ν (t) 2 )dx + Ω ν ) and the energy inequality we deduce Ω ν that P ν (t) 2 + curl P ν (t) 2 = O(ν), H ν (t) 2 + E ν (t) 2 = O(ν). To prove the proposition, we follow the lines of the proof given in [2] and [9] and we use a classical result of the regularity of Maxwell fields [3]. For ν fixed, let (H ɛ, E ɛ, P ɛ ) be the solution of the problem (13)-(14) when we replace φ ( P 2 )P by φ ( ρ ɛ P 2 )(ρ ɛ P ) where ρ ɛ is a nonnegative regularizing sequence with unit mass in R 3. We shall prove the estimate (26) for P ɛ (indeed P ν,ɛ ) then we deduce the whished result by letting ɛ 0. Of course (H ɛ, E ɛ, P ɛ ) satisfies the same energy estimate as (H ν, E ν, P ν ). Moreover the initial energy estimate satisfies also E0 ν = O(ν) and we have ρ ɛ P ɛ (t) 2 = O(ν) uniformly with respect ɛ. Let us consider the compatibility equations associated with the regularized problem of (13)-(14). Setting u ɛ = div H ɛ, v ɛ = div E ɛ and w ɛ = div P ɛ we get t u ɛ = 0, t (v ɛ + w ɛ ) + σv ɛ = 0 2 t w ɛ + a t w ɛ v ɛ = F ɛ (28) with F ɛ = div (φ ( ρ ɛ P ɛ 2 )(ρ ɛ P ɛ )) = φ ( ρ ɛ P ɛ 2 )ρ ɛ div P ɛ + 2φ (2) ( ρ ɛ P ɛ 2 ) k,j (ρ ɛ P ɛ k )(ρ ɛ P ɛ j )(ρ ɛ k P ɛ ). Setting W ɛ = (v ɛ, w ɛ, t w ɛ ) we get the equation where W 0 = ( div E 0, div P 0, div P 1 ) and C = t W ɛ + CW ɛ = S ɛ (P ɛ ), W ɛ (0) = W 0 (29) σ a, S ɛ (P ɛ ) = 0 0 F ɛ (30) Notice that we have W 0 = O(ν). Using the hypotheses (10)-(11), there exists C > 0 which is independent of ν and ɛ such that S ɛ (P ɛ (t)) C P ɛ (t) (31) 5

7 for all t 0. Consequently W ɛ satisfies the estimate W ɛ (t) 2 e δt ( W C t 0 P ɛ (s) 2 ds) (32) where the constant δ > 0 and C > 0 depend only of φ (but not of ν and ɛ). Recall that we have W 0 2 = ν W 0 2 L 2 ( b Ω). Since Ων is a cylinder with convex cross section, then using the estimate given in lemma 2.17 of [3] we get P ɛ (t) 2 curl P ɛ (t) 2 + div P ɛ (t) 2 (33) for all t 0. Let T > 0 be fixed. For t [0, T ], we deduce (by using w ɛ (t) 2 W ɛ (t) 2 ) the estimate t W ɛ (t) 2 νc T + Ce δt ( curl P ɛ (s) 2 + W ɛ (s) 2 )ds (34) 0 where C T > 0 is independent of ν and ɛ. Finally the estimate of curl P ɛ 2 νc implies the inequality W ɛ (t) 2 νc T + C T t 0 W ɛ (s) 2 ds and finally we obtain P ɛ (t) 2 + div P ɛ (t) 2 + div t P ɛ (t) 2 νc T. (35) Since P ɛ P ν strongly in L (R + ; L 2 (Ω ν )) as ɛ 0 see [2], [9] then, using the convexity of the L 2 -norm we deduce that (E ν, P ν ) satisfies the estimate P ν (t) 2 + div E ν (t) 2 + div t P ν (t) 2 νc T and P ν (t) 2 = O(ν) on [0, T ]. This concludes the proof of the proposition. The content of this work is the following. In the next section we introduce the change of variable and the new problems (linear and nonlinear) setted in the fixed cylinder Ω 1 with thickness 1 which is denoted Ω in the sequel. We prove uniform bounds with respect to the parameter ν for the scaled solution (h ν, e ν, p ν ) of the new problems. In section 3, we discuss the behaviour of the solution of the linear problem and section 5, we give the behaviour of the solutions of the nonlinear problem. 2 Uniform bounds for the scaled solutions In the sequel Ω denotes the cylinder with thickness ν = 1. Let (u 1, u 2, u 3 ) be the canonical basis of R 3. The generic point x of Ω ν is denoted by x = ( x, x 3 ) Ω (0, ν) and x = (x 1, x 2 ). For a vector valued function we set f = ( f, f 3 ) with f = (f 1, f 2 ) and f 3 = f u 3. The partial derivative of a function g with respect to the variable x j is denoted j g. The curl operator of a vector valued function f in R 3 or in R 2 is defined by curl f = ( 2 f 3 3 f 2, 3 f 1 1 f 3, 1 f 2 2 f 1 ) or ĉurl f = 1 f 2 2 f 1 = div (f u 3 ) and the curl of a scalar function is defined by Curl (f u 3 ) = ( 2 f 3, 1 f 3, 0). The div operator of a vector valued function in R 3 or in R 2 is defined by div f = 1 f f f 3 or by div f = 1 f f 2. Finally the gradient operator is written as = + x3 u 3. Notice that curl f is a vector of R 3 and curl f u 3 = ĉurl f. Moreover we have f n R 3 and (f n) u 3 = f n = f 1 n 2 f 2 n 1 R. Since this work concerns the reduction of the dimension of problems (13)-(15) and (13)-(14) then we assume that the initial data H 0, E 0, P 0, P 1 are independent of the variable x 3. Hence, for the linear problem we assume that H 0 = H 0 ( x), E 0 = E 0 ( x), P 0 = P 0 ( x), P 1 = P 1 ( x) P 0, curl P 0, P 1, H 0, E 0 H(curl, Ω) P 0 n, P 1 n L 2 ( Ω) 6 (36)

8 Notice for example that we have curl P 0 = Curl P3 0 + (ĉurl P 0 )u 3. The generic point x of Ω is denoted by x = ( x, z) with 0 < z < 1. We set Q = R + Ω, Q = R + Ω and for T > 0 fixed we set Ω T = Ω (0, T ) and Ω T = Ω (0, T ). We introduce the change of variable z = x 3 /ν. For a vector valued function F ν ( x, x 3 ) defined in Ω ν we introduce the scaled function f ν ( x, z) defined in Ω associated with F ν by setting F ν ( x, x 3 ) = f ν ( x, x 3 /ν). Notice that we have F ν = f ν and 3 F ν = 1 ν zf ν. More generally we get curl F ν = 1 ν z(f ν u 3 ) + Curl (f ν u 3 ) + (ĉurl f ν )u 3 div F ν = div f ν + 1 ν (37) zf3 ν. In the sequel we use the notations Notice that θ ν ( 1 f ν 2 2 f ν 1 )u 3 = 0. curl ν f ν = (θ ν, 1 f ν 2 2 f ν 1 ) θ ν = ( 2 f ν 3 1 ν zf ν 2, 1 ν zf ν 1 1 f ν 3 ) div ν f ν = div f ν + 1 ν zf ν 3. Let (H ν, E ν, P ν ) be the global solution of either problem (13)-(15) or (13)-(14) associated with the initial data (H 0, E 0, P 0, P 1 ) satisfying the hypotheses (20)-(36). Let h ν (t, x, z) = H ν (t, x, νz), e ν (t, x, z) = E ν (t, x, νz) and p ν (t, x, z) = P ν (t, x, νz) be the scaled solution defined in R + Ω and associated with (H ν, E ν, P ν ). (38) 2.1 The linear problem The scaled solution (h ν, e ν, p ν ) satisfies in R + Ω the problem t h ν curl ν e ν = 0 in R + Ω t (e ν + p ν ) + curl ν h ν + σe ν = 0 in R + Ω coupled to the polarization equation The energy inequality becomes h ν (0) = H 0 ( x), e ν (0) = E 0 ( x) in Ω h ν n + β ν n (e ν n) = 0 in R + Ω 2 t p ν + curl 2 νp ν + a t p ν + k p ν = e ν in R + Ω p ν (0) = P 0 ( x), t p ν (0) = P 1 ( x) in Ω curl ν p ν n + ρ ν n (( t + a)p ν n) = 0 in R + Ω. (39) (40) E ν (t) + t 0 (a t p ν (s) 2 + σ e ν (s) 2 + β ν e ν n 2 + ρ ν t p ν (s) n 2 )ds E ν 0 (41) where the energy at the time t is given by E ν (t) = t p ν (t) 2 + k p ν (t) 2 + curl ν p ν (t) 2 + a (42) ρ ν p ν (t) n 2 + h ν (t) 2 + e ν (t) 2 and the initial energy E0 ν, by using hypotheses (20)-(36), becomes 7

9 E ν 0 = P k P ĉurl P Curl P a ρ ν P 0 n 2 + H E 0 2. Notice that the initial energy E ν 0 is uniformly bounded with respect to the parameter ν if ρ ν is bounded uniformly in L (0, 1). (43) We assume that the functions β ν and ρ ν are given by β ν (z) = β 0 if 0 < z < 1, β ν (0) = νβ 0 > 0, β ν (1) = νβ 1 > 0 ρ ν (z) = ρ 0 if 0 < z < 1, ρ ν (0) = νρ 0 > 0, ρ ν (1) = νρ 1 > 0 (44) Then, the energy inequality implies the following uniform bounds Lemma 2.1 Under hypotheses (20)-(36) and (44) there exists C > 0 which is independent of ν such that the solution (h ν, e ν, p ν ) of problem (39)-(40) satisfies the nestimates h ν 2 + e ν 2 + p ν 2 + e ν 2 + t p ν 2 C (45) t p ν p ν 1 1 p ν θ ν 2 C moreover we have (e ν u 3 ) z=0,1 2 L 2 (R + ;L 2 ( b Ω)) + êν n 2 L 2 (R + ;L 2 ( b Ω (0,1))) C (h ν u 3 ) z=0,1 2 L 2 (R + ;L 2 ( b Ω)) Cν, ĥν n 2 L 2 (R + ;L 2 ( b Ω (0,1))) C. (46) and (p ν u 3 ) z=0,1 2 L (R + ;L 2 ( b Ω)) + pν n 2 L (R + ;L 2 ( b Ω (0,1))) C ( t p ν u 3 ) z=0,1 2 L 2 (R + ;L 2 ( b Ω)) + t p ν n 2 L 2 (R + ;L 2 ( b Ω (0,1))) C (47) Notice that from the boundary condition satisfied by p ν and the previous estimates we deduce that (curl ν p ν u 3 ) z=0,1 2 L 2 loc (R+ ;L 2 ( Ω)) b Cν. (48) We assume that the initial data are as in lemma 1.1. We get Lemma 2.2 Under hypotheses (25) the solution has a time regularity property (see theorem 1) which implies the following estimates t h ν 2 + t e ν 2 + t 2 p ν 2 C curl ν h ν 2 + curl ν e ν 2 + curl 2 (49) νp ν 2 + curl ν t p ν 2 C where C is independent of ν. 2.2 The nonlinear problem The scaled solution (h ν, e ν, p ν ) satisfies in R + Ω the problem (39) coupled to the polarization equation The energy inequality becomes 2 t p ν + curl 2 νp ν + a t p ν + φ ( p ν 2 )p ν = e ν in R + Ω p ν (0) = P 0 ( x), t p ν (0) = P 1 ( x) in Ω p ν n = 0 in R + Ω. 8 (50)

10 E ν (t) + 2 t 0 (a t p ν (s) 2 + σ e ν (s) 2 + β ν e ν n 2 E ν 0 (51) where the energy at the time t is given by E ν (t) = t p ν (t) 2 + φ( p ν (t) 2 )dx + curl ν p ν (t) 2 + h ν (t) 2 + e ν (t) 2 (52) Ω and the initial energy E0 ν, by using hypotheses (20)-(36), becomes E0 ν = P φ( P 0 2 )dx + ĉurl P Curl P H E 0 2. (53) Ω Notice that the initial energy E ν 0 is uniformly bounded if β ν is bounded uniformly in L (0, 1). From the energy inequality we deduce the following uniform bounds Lemma 2.3 If β ν is given as in (44) then, under hypotheses (20)-(36), there exists C > 0 which is independent of ν such that the solution (h ν, e ν, p ν ) of problem (39)-(50) satisfies the estimates h ν 2 + e ν 2 + p ν 2 + e ν 2 + t p ν 2 C (54) t p ν p ν 1 1 p ν θ ν 2 C moreover we have the bounds (e ν u 3 ) z=0,1 2 L 2 (R + ;L 2 ( b Ω)) + êν n 2 L 2 (R + ;L 2 ( b Ω (0,1))) C (h ν u 3 ) z=0,1 2 L 2 (R + ;L 2 ( b Ω)) Cν, ĥν n 2 L 2 (R + ;L 2 ( b Ω (0,1))) C. (55) Assuming the initial data to be more regular as in lemma 1.1 then we have Lemma 2.4 Under hypotheses (25) the solution of (39)-(50) satisfies estimates t h ν 2 + t e ν t p ν 2 C where C is independent of ν. curl ν h ν 2 + curl ν e ν 2 + curl 2 νp ν 2 + curl ν t p ν 2 C (56) 2.3 Weak convergences For a subsequence still denoted (h ν, e ν, p ν ), where (h ν, e ν, p ν ) is either the solution of the linear problem (39)-(40) or the nonlinear problem (39)-(50), the following convergences hold. (h ν, e ν, p ν ) (h, e, p) in L (R + ; L 2 (Ω)) weakly, (e ν, t p ν ) (e, t p) in L 2 (R + ; L 2 (Ω)) weakly, (57) t p ν t p in L (R + ; L 2 (Ω)) weakly The estimate in L (R + ; L 2 (Ω)) of the curl ν of the solution gives the weak- convergences ĉurl p ν ĉurl p, θν θ ĉurl ĥν ĉurl ĥ, ĉurl êν ĉurl ê (58) where θ L (R + ; L 2 (Ω)) is some vector valued function. Since we have ĉurl ĥ, ĉurl ê, ĉurl p L (R + ; L 2 (Ω)) we deduce that the traces ĥ n, ê n and p n are well defined in L (R + ; H 1/2 ( Ω (0, 1)). Hence the weak- convergence of that traces holds at least in L (R + ; H 1/2 ( Ω (0, 1)). 9

11 Now let us consider the bounds of the traces of e ν, h ν and p ν. First of all, using estimate (49) we deduce that h ν, e ν, p ν and t p ν are uniformly bounded in W 1, (R + ; L 2 (Ω)). Hence, h, e, p and t p belong to W 1, (R + ; L 2 (Ω)). Finally we get Lemma 2.5 (Initial data) The traces of h, e, p and t p at t = 0 make sense in L 2 (Ω) and we have Next, from (46) we get the convergences h(0) = H 0, e(0) = E 0, p(0) = P 0, t p(0) = P 1 (59) Lemma 2.6 (Convergence of the traces) The traces of the solutions (h ν, e ν ) of either (39)- (40) or (39)-(50) satisfy (e ν u 3 ) z=0,1 A 0,1 in L 2 (R + ; L 2 ( Ω)) weak (h ν u 3 ) z=0,1 0 in L 2 (R + ; L 2 (60) ( Ω)) strong. The tangential traces of the polarization p ν for the linear problem (39)-(40) satisfy (p ν u 3 ) z=0,1 B 0,1 in L (R + ; L 2 ( Ω)) weak, ( t p ν u 3 ) z=0,1 t B 0,1 in L 2 (R + ; L 2 ( Ω)) weak (curl ν p ν u 3 ) z=0,1 0 in L 2 loc (R+ ; L 2 ( Ω)) strong (61) where A 0, A 1, B 0 and B 1 are some vector valued functions belonging to the spaces defined by the weak convergences. Before we pass to the limit in problem (39)-(40) and (39)-(50) let us give the remark Remark 2.1 Assume that the function β ν and ρ ν are independent of ν say β ν = β > 0 and ρ ν = ρ > 0 then the energy inequality (41) and (51) imply, in particular, the following bounds (p ν u 3 ) z=0,1 2 L (R + ;L 2 ( b Ω)) + ( tp ν u 3 ) z=0,1 2 L 2 (R + ;L 2 ( b Ω)) Cν (e ν u 3 ) z=0,1 2 L (R + ;L 2 ( b Ω)) Cν. (62) and then A 0,1 = B 0,1 = 0. We shall use for the solution (h ν, e ν, p ν ) of (39)-(40) as well as for (39)-(50) the following convergence results. We dedote by v ν one of the fields h ν, e ν, p ν or. : rot ν p ν. Proposition 2.1 Let v ν be a uniformly bounded sequence in L (R + ; H(curl ν, Ω) such that the tangential trace v ν n is uniformly bounded in L p (R + ; L 2 ( Ω)) with p = or p = 2. Then there exists a subsequence such that v ν u, ĉurl vν ĉurl v weakly- in L (R + ; L 2 (Ω)), v ν n v n = v 1 n 2 v 2 n 1 weakly in L p (R + ; H 1/2 ( Ω (0, 1)) satisfying the properties v is independent of z, v L (R + ; H( rot, Ω)) (v ν u 3 ) z=0,1 (v u 3 ) z=0,1 weakly in L p (R + ; L 2 ( Ω)) (63) 1 0 (vν n) u 3 dz (v n) u 3 weakly in L p (R + ; L 2 ( Ω)) where (v n) u 3 = v 1 n 2 v 2 n 1 (which is independent of z). 10

12 We use the following Green formula Q curl νv ν ϕdxdt = Q vν curl ν ϕdxdt R + Ω (0,1) b (vν n) ϕdαdzdt 1 ν bq (vν u 3 ) z=1 ϕ z=1 d xdt + 1 ν bq (vν u 3 ) z=0 ϕ z=0 d xdt (64) for all test function ϕ D(R + Ω [0, 1]). Choosing ϕ = (νϕ 1, νϕ 2, 0) with ϕ j D(R + Ω (0, 1)) for j = 1, 2. We have curl ν ϕ = z (ϕ u 3 ) + νcurl ϕ 3 + ν(ĉurl ϕ)u 3. Passing to the limit in the Green formula we obtain Q ( v 1 z ϕ 2 + v 2 z ϕ 1 )dxdt = 0 and then z v = 0 in the sense of distributions. Next let V j L p (R + ; L 2 ( Q)) be the weak limit of (v ν u 3 ) z=j for j = 1, 2. Next, we choose in the Green formula ϕ = (νzϕ 1 (t, x), νzϕ 2 (t, x), 0) and pass to the limit we get b Q ( v 1 ϕ 2 + v 2 ϕ 1 )d xdt b Q V 1 ϕd xdt = 0 which implies that V 1 = u u 3. Finally with the test function ϕ = (ν(1 z)ϕ 1 (t, x), ν(1 z)ϕ 2 (t, x), 0) we get b Q (v 1 ϕ 2 v 2 ϕ 1 )d xdt bq V 0 ϕd xdt = 0 and then V 0 = v u 3. Let g L p (R + ; L 2 (R + Ω (0, 1)) be the weak limit of the sequence (v ν n) u 3. We choose in the Green formula ϕ = (0, 0, ϕ 3 (t, x)) with ϕ 3 D(R + Ω). We have curl ν ϕ = Curl ϕ 3. Then passing to the limit in the Green formula we get b Q ( 1 v 2 2 v 1 )ϕ 3 d xdt = b Q (v 1 2 ϕ 3 v 2 1 ϕ 3 )d xdt R + b Ω 1 0 gdzϕ 3dαdt. Since we have v L (R + ; H(ĉurl, Ω) then we deduce that 1 0 gdz = v 1n 2 v 2 n 1. Remark 2.2 Notice (ĥ, ê, p) is independent of z. For vν = e ν we have e u 3 = A 0 = A 1, for v ν = h ν we have h u 3 = 0 and then ĥ = 0. For vν = p ν we have p u 3 = B 0 = B 1. Moreover we have curl ν p ν, curl ν (curl ν p ν ) L (R + ; L 2 (Ω)) and (curl ν p ν n) z=0,1 = O(ν). Let θ be the weak- limit of θ ν defined by curl ν p ν = θ ν + ĉurl pν u 3 then θ is independent of z and θ = 0. Proposition 2.2 Let v ν be a uniformly bounded sequence in L (R + ; H(curl ν, Ω)) satisfying v ν n = 0 on R + Ω. Then θ satisfies 1 0 θdz = 1 0 Curl v 3dz in R + Ω and 1 0 v 3dz = 0 on R + Ω. Moreover we have v = 0 Proof. Since we have curl ν v ν = θ ν + ĉurl vν u 3, the Green formula gives Q θν ϕdxdt + Q ( 1v ν 2 2 v ν 1 )ϕ 3 dxdt = Q vν 1 ( 2 ϕ 3 1 ν zϕ 2 )dxdt + Q vν 2 ( 1 ν zϕ 1 1 ϕ 3 )dxdt + Q vν 3 ( 1 ϕ 2 2 ϕ 1 )dxdt (65) for all test function ϕ D(R + Ω (0, 1]). Passing to the limit we get with ϕ = ( ϕ(t, x), 0) we get Q θ ϕdxdt = Q v 3( 1 ϕ 2 2 ϕ 1 )dxdt (66) integrating by parts we get Q θ ϕdxdt = Q Curl v 3 ϕdxdt R + Ω (0,1) b v 3(n 1 ϕ 2 n 2 ϕ 1 )dαdzdt. It follows that 1 0 Curl v 3dz = 1 0 θdz which belongs to L (R + ; L 2 ( Ω)). The trace of v 3 on Ω makes sense in L (R + ; H 1/2 ( Ω)) and we have 1 0 v 3dz = 0 on R + Ω. Next writting rot ν v ν = 1 ν z(v ν u 3 ) + Curl v3 ν + ĉurl vν u 3 and using the Green formulation with ϕ = νψ we get O(ν) = Q ( vν 1 z ψ 2 + v2 ν z ψ 1 )dxdt + O(ν). Passing to the limit we get Q ( v 1 z ψ 2 + v 2 z ψ 1 )dxdt = 0. It follows that v is independent of the variable z and using the previous proposition and the boundary condition (v ν n) z=0,1 = 0 we deduce that v = 0. 11

13 3 Convergence of the linear problem We shall prove the following result. Theorem 3.1 Let (H 0, E 0, P 0, P 1 ) be initial data which are independent of the variable z and satisfying the hypotheses (36). We assume that the functions β ν, ρ ν are as in lemma 2.1. Let (h ν, e ν, p ν ) be the solution of problem (39)-(40) associated with the initial data (H 0, E 0, P 0, P 1 ). Then there exists a subsequence still denoted (h ν, e ν, p ν ) converging weakly- in L (R + ; L 2 (Ω)) to (h, e, p) which is independent of the variable z and ĥ = 0. The weak star limit (h, e, p) is the solution in R + Ω of the problem t h 3 ĉurl ê = 0, t(ê + p) + Curl h 3 + (σ + β 1 + β 0 )ê = 0 ( t 2 + a t + k) p + Curl ĉurl p + (ρ 1 + ρ 0 )( t + a)p = ê ê(0) = Ê0, p(0) = P 0, t p(0) = p 1, h 3 (0) = H3 0 in Ω, (67) h 3 = β(e 1 n 2 e 2 n 1 ), ĉurl p = ρ( t + a)(p 1 n 2 p 2 n 1 ) on R + Ω and the system of o.d.e t (e 3 + p 3 ) + σe 3 = 0, ( 2 t + a t + k)p 3 = e 3 e 3 (0) = E 0 3, p 3 (0) = P 0 3, t p 3 (0) = P 1 3 in Ω. (68) Moreover, we have h 3 L (R + ; H 1 (Ω)), ê L (R + ; H(ĉurl, Ω)), p L (R + ; L 2 ( Ω)) and ĉurl p L ; H + ( Ω)). Let (h ν, e ν, p ν ) be the solution associated with (H 0, E 0, P 0, P 1 ) satisfying the hypotheses of the theorem and let (h, e, p) be the weak- limit of a subsequence. Applying proposition 2.1 to h ν, e ν and p ν we deduce that (h, e, p) satisfies the conclusions of the proposition. The weak formulations associated with problem (39)-(40) take the forms Q hν t ϕdxdt Q eν curl ν ϕdxdt + R + Ω (0,1) b eν n ϕdαdzdt + 1 ν R + Ω b(eν u 3 ) z=1 ϕ z=1 d xdt 1 ν R + Ω b(eν u 3 ) z=0 ϕ z=0 d xdt = (69) Ω H0 ϕ(0)dx and, Q (eν + p ν ) t φdxdt + Q hν curl ν φdxdt +β R + b Ω (0,1) eν n φ ndαdzdt +β 1 R + b Ω (eν u 3 ) z=1 (φ u 3 ) z=1 d xdt β 0 R + b Ω (eν u 3 ) z=0 (φ u 3 ) z=0 d xdt + σ Q eν φdxdt = Ω (E0 + P 0 )φ(0)dx. for all regular test functions ϕ, φ defined on Q. Here we used the boundary condition h ν n + β ν n (e ν n) = 0. The polarization field p ν satisfies Q pν ( t 2 a t + k)ψdxdt + Q curl νp ν curl ν ψdxdt Q eν ψdxdt +ρ R + Ω (0,1) b ( t + a)p ν n ψ ndαdzdt +ρ 1 R + Ω b(( t + a)p ν u 3 ) z=1 (ψ u 3 ) z=1 d xdt (71) ρ 0 R + b Ω (( t + a)p ν u 3 ) z=0 (ψ u 3 ) z=0 d xdt = Ω (P 1 ψ(0) P 0 ( t ψ(0) aψ(0))dx. (70) 12

14 for all test function ψ defined in Q. Here we used the boundary condition curl ν p ν b + ρ ν n (( t + a)p ν n) = 0. We have the convergences R + Ω hν n ν φdαdt β R + Ω (0,1) b e n φ ndαdzdt +β 1 R + Ω b A 1 φ z=1 u 3 d xdt β 0 R + Ω b A (72) 0 φ z=0 u 3 d xdt appearing in the weak formulation (70) and R + Ω curl νp ν n ν ψdαdt ρ R + b Ω (0,1) ( t + a)p ν n ψ ndαdzdt + R + b Ω (ρ 1( t + a)b 1 ψ z=1 u 3 ρ 0 ( t + a)b 0 ψ z=0 u 3 )d xdt. (73) appearing in the third weak formulation (71). The compatibilty conditions for problem (39)- (40) (obtained by using div ν (curl ν ) = 0) can be written as t ( div ĥν + 1 ν zh ν 3) = 0 t ( div (ê ν + p ν ) + 1 ν z(e ν 3 + p ν 3)) + σ( div ê ν + 1 ν zh ν 3) = 0 (74) ( t 2 + a t + k)( div p ν + 1 ν zp ν 3) ( div ê ν + 1 ν ze ν 3) = 0. passing to the limit in the sense of distributions we get the result Lemma 3.1 The functions h 3, e 3 and p 3 are independent of the variable z Let us consider in (69) and (70) test functions of the form ϕ = ( ϕ, ϕ 3 ) and φ = ( φ, φ 3 ) respectively which are independent of the variable z. One observes that curl ν ϕ = ( 2 ϕ 3, 1 ϕ 3, 1 ϕ 2 2 ϕ 1 ). Hence, we obtain Q hν t ϕdxdt Q eν (Curl ϕ 3 + ĉurl ϕu 3)dxdt + R + Ω (0,1) b eν n ϕdαdzdt+ and 1 ν R + Ω b(eν u 3 ) z=1 ϕ z=1 d xdt 1 ν Ω b H 0 ϕ(0)d x R + b Ω (eν u 3 ) z=0 ϕ z=0 d xdt = Q (eν + p ν ) t φdxdt + Q hν (Curl φ 3 + ĉurl φu 3 )dxdt +β R + b Ω (0,1) eν n φ ndαdzdt R + b Ω (β 1(e ν u 3 ) z=1 φ z=1 u 3 + β 0 (e ν u 3 ) z=0 φ z=0 u 3 )d xdt +σ Q eν φdxdt = Ω (E0 + P 0 ) φ(0)d x. Recalling that ĥ = 0 then we choose ϕ = (0, 0, ϕ 3) in (75) we get Q hν 3 t ϕ 3 dxdt Q êν Curl ϕ 3 dxdt + R + b Ω (0,1) (n 2e ν 1 n 1 e ν 2)ϕ 3 dαdzdt = b Ω H 0 3 ϕ 3 (0)d x and passing to the limit we get b Q h 3 t ϕ 3 d xdt b Q ê Curl ϕ 3 d xdt + R + b Ω (n 2e 1 n 1 e 2 )ϕ 3 dαdt = b Ω H 0 3 ϕ 3 (0)d x. Hence, the magnetic field h = ( 0, h 3 ) satisfies in the sense of distributions the problem t h 3 ĉurl ê = 0 in R+ Ω h 3 (0) = H 0 3 in Ω, Next, we pass to the limit in (76) with test function φ which are independent of z and use the result A 0 = A 1 = e u 3. We get (75) (76) (77) 13

15 Q b(e + p) t φd xdt + Q b h 3 ĉurl φd xdt +β R + Ω b e n φ ndαdt +(β 1 + β 0 ) R + Ω b e u 3 φ u 3 d xdt + σ Q b e φd xdt = b Ω (E 0 + P 0 )φ(0)d x. (78) Integrating by parts we get (notice that h 3 L (R + ; H 1 ( Ω))) t (ê + p) + Curl h 3 + (σ + β 1 β 0 )ê = 0 t (e 3 + p 3 ) + σe 3 = 0 h 3 + β(e 1 n 2 e 2 n 1 n) = 0 e(0) + p(0) = E 0 + P 0 in Ω Let us consider the weak formulation (71) associated with the polarization equation (40). Let us recall the notation used curl ν p ν = θ ν + ĉurl pν u 3 where θ ν = 1 ν z(p ν u 3 ) + Curl p ν 3. Using proposition 2.1 and lemma 2.1 we deduce that p u 3 = B 0 = B 1 and we have that p 3 is independent of z). Passing to the limit we get bq p ( 2 t a t + k)ψdxdt + Q θ Curl ψ 3d xdzdt+ bq (ĉurl p)(ĉurl ψ)dxdt Q e ψd xdt +ρ R + Ω b( t + a)p n ψ ndαdt ρ 1 R + Ω b( (80) t + a)b 1 ψ u 3 d xdt +ρ 0 R + b Ω ( t + a)b 0 ψ u 3 d xdt = b Ω (P 1 ψ(0) P 0 ( t ψ(0) aψ(0))dx. In proposition 2.1 and remark 2.1 we have shown that θ = 0. Integrating by parts in (80), we get in R + Ω ( t 2 + a t + k) p + Curl ĉurl p + (ρ 1 + ρ 0 )( t + a) p = ê ĉurl p = ρ(( t + a)(p 1 n 2 p 2 n 1 )) on R + Ω (81) p(0) = P 0, t p(0) = P 1 in Ω (79) and ( 2 t + a t + k)p 3 = e 3 in R + Ω p 3 (0) = P3 0, t p 3 (0) = P3 1 in Ω. Hence the main theorem is proved. (82) 4 Convergence of the nonlinear problem In the sequel we assume that Ω is a bounded, regular and convex domain of R 2 and the initial data H 0, E 0, P 0 and P 1 which are independent of the variable x 3 satisfy the hypothesis H 0 = (0, 0, H3 0 ), E 0 = (Ê0, E3), 0 P 0 = (0, 0, P3 0 ), P 1 = (0, 0, P3 1 ) H3 0, P3 0, P3 1 H 1 ( Ω), div (83) E 0 L 2 ( Ω) We shall prove the following result 14

16 Theorem 4.1 Under hypotheses (83)-(44), there exists a subsequence (h ν, e ν, p ν ) of solutions to the nonlinear problem (39)-(50) such that h ν (0, h 3 ), e ν e, p ν (0, p 3 ) weakly- in L (R + ; L 2 ( Ω)) and (h 3, e, p 3 ) is independent of the variable z. Moreover p ν p strongly in L 2 (0, T ; L 2 ( Ω)), div h ν z h 3 and div e ν div e weakly- in L (R + ; L 2 ( Ω)). Further, (h 3, e, p 3 ) satisfies in R + Ω the limit problem t h 3 ĉurl ê = 0, t ê + Curl h 3 + (σ β 1 + β 0 )ê = 0, t (e 3 + p 3 ) + σe 3 = 0 2 t p 3 + a t p 3 p 3 + φ ( p 3 2 )p 3 = e 3 (84) h 3 (0) = H 0 3, ê(0) = Ê0, e 3 (0) = E 0 3, p 3 (0) = P 0 3, t p 3 (0) = P 1 3 p 3 = 0, h 3 = β(n 1 e 2 n 2 e 1 ) on R + Ω. The proof of the theorem is essentially proved in section 3, at the end of the proof of theorem 3.1 and by the strong convergence result given in proposition 1.1. Let us precise tha convergences we obtain in this case. Let (h ν, e ν, p ν ) be the associated scalled solution of (39)-(50). Proposition 1.1 implies the following uniform bound in L (0, T, L 2 (Ω)), ν p ν 2 + ν t p ν 2 + div ν h ν 2 + div ν e ν 2 C T. (85) combining this bound, the results of propositions 2.1 and 2.2 and the energy inequality associated with problem (39)-(50) we get Lemma 4.1 For a subsequence (still denoted (h ν, e ν, p ν )) we have (h ν, e ν, p ν ) (h, e, p) weakly- in L (R + ; L 2 (Ω)) where (h, e, p) is independent of the variable z. Moreover p ν p, φ ( p ν 2 )p ν φ ( p 2 )p in L (0, T ; L 2 (Ω)) strong p ν p, div ĥ ν 0, div ê ν div ê in L (0, T ; L 2 (86) (Ω)) weakly for all T > 0. Furthermore, we have, ĥ = 0, p = 0, θ = Curl p 3 and p 3 = 0 on R + Ω. (87) Let us consider problem (39)-(50) with the boundary condition p ν n = 0 on R + Ω. The convergence for the Maxwell equation follows the lines of the proof given in theorem 2.1. The weak formulation of the polarization equation becomes Q pν ( t 2 a t )ψdxdt + Q curl νp ν curl ν ψdxdt Q eν ψdxdt = = Q φ ( p ν (t) 2 )p ν ψdxdt Ω (P 1 ψ(0) P 0 (88) ( t ψ(0) aψ(0))dx. for all test function ψ satisfying the boundary condition ψ n = 0 on R + Ω. Since p = 0 we use test functions of the form ψ = (0, 0, ψ 3 ) which are independent of the variable z. Then passing to the limit by using lemma 4.1 we get the result stated in theorem 4.1. Remark 4.1 If we consider problem (39)-(50) with the Silver-Müller boundary condition curl ν p ν n+ρ ν n (p ν n) = 0 on R + Ω then all the results obtained for problem (39)-(40) remain true. The main point we have to prove is that φ ( p ν 2 )p ν φ ( p 2 )p weakly star in L (R + ; L 2 ( Ω)). Of course the weak- convergence follows from the bounds given by the energy inequality but, to 15

17 identify the weak- limit the energy inequality is not sufficient to do that. In [1], it is proved an uniform bound (with respect to ν) in L (0, T ; H 1/2 (Ω)) for p ν which allows to pass to the limit in the nonlinear term and to identify the zero thickness limit of the problem. References [1] N. Aïssa. Regularity of vector fields and behaviour of thin ferroelectric materials with nonlinear potential. In preparation. [2] H. Ammari and K. Hamdache. Global existence and regularity of solutions to a system of nonlinear Maxwell equations. J. Math. Anal. Appl., 286, 51 63, (2003) [3] C. Amrouche, C. Bernardi, M. Dauge and V. Girault Vector potentials in three nonsmoth domains, Math. Meth. Apl. Sci. Vol. 21, (1998) [4] F. Chaput. Matériaux céramiques et ferroélectriques. Fascicule de l Ecole Polytechnique, (1997) [5] R. Dautray and J.L. Lions. Analyse mathématique et calcul numérique pour les sciences et les techniques. Masson. Paris, T.2, Chp. IX. (1985) [6] L. Eyraud. Diélectriques solides anisotropes et ferroélectricité. Gauthier-Villard. Paris, (1967) [7] M. Fabrizio and A. Morro. Models of electromagnetic materials. Math. and Comp. Modelling,34, , (2001) [8] J.M. Greenberg, R.C. Mac Camy and C.V. Coffman. On the long time behavior of ferrolectric systems. Physica D,134, , (1999) [9] K. Hamdache and I. Ngningone. Time harmonic solutions to a model of ferroelectric materials. Preprint CMAP, N. 497, (2002) [10] H. Sachse. Les ferroélectriques. Dunod. Paris, (1958) [11] L. Tartar. On the characterization of traces of a Sobolev spaces used in Maxwell s equations. Proceeding. Bordeaux. A.Y Leroux Edt., November 6-7 (1998) 16