elasticity equations in exterior domains

J. Math. Anal. Appl. 359 (2009) 464–481

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Journal of Mathematical Analysis and Applications www.elsevier.com/locate/jmaa

Large time behavior of anisotropic electromagnetic/elasticity equations in exterior domains Cleverson R. da Luz a , G. Perla Menzala b,a,∗ a b

Institute of Mathematics, Federal University of Rio de Janeiro, PO Box 68530, Rio de Janeiro, RJ, Brazil National Laboratory of Scientific Computation (LNCC/MCT), Av. Getulio Vargas 333, Petrópolis, RJ, 25651-070, Brazil

a r t i c l e

i n f o

a b s t r a c t

Article history: Received 20 March 2009 Available online 6 June 2009 Submitted by Steven G. Krantz Keywords: Anisotropic Maxwell equations Anisotropic elasticity equations Coupled system in exterior domains Large time behavior as t → +∞

We find uniform rates of decay of the total energy of the coupled system of anisotropic electromagnetic/elasticity model in exterior domains provided mild dissipative effects are present. The decay of the total energy is of polynomial type. The conclusions of this paper improve previous results on the subject. © 2009 Elsevier Inc. All rights reserved.

1. Introduction This paper is devoted to investigate the dynamics of coupled models of hyperbolic systems in an exterior domain. More precisely, we consider the coupled system of anisotropic elastic waves with anisotropic Maxwell equations. The main motivation to study such coupled systems arise from recent technological uses of “smart materials”, crystal optics and in general nanotechnology. In the study of dynamical systems is very important to know qualitative and quantitative properties of the solutions which can be use to obtain new composite materials. This article presents conclusive results on the asymptotic behavior of the coupled system of anisotropic electromagnetic/elasticity model in exterior domains. We find uniform rates of decay (of polynomial type) of the total energy as t → +∞. In our case, the anisotropic Maxwell equations are very special: The energy in general does not propagate along the normals to the fronts but along rays which are distinct from the normals (see [12]). Thus, in these kind of mediums the so-called permittivity and permeability are no more scalar-valued functions but 3 × 3 symmetric matrices. As a consequence, the anisotropic Maxwell equations cannot be reduce (in general) to a second order vector-wave equation for which a large amount of results are available. Let us describe the model we want to study: Let Ω ⊆ R3 be an exterior domain (unbounded with bounded complement) with compact smooth boundary ∂Ω . We consider the coupled system

  3  ∂u ∂ + ut + γ curl E = 0, utt − A i j (x) ∂ xi ∂xj i , j =1

*

Corresponding author at: National Laboratory of Scientific Computation (LNCC/MCT), Av. Getulio Vargas 333, Petrópolis, RJ, 25651-070, Brazil. E-mail addresses: [email protected] (C.R. da Luz), [email protected] (G.P. Menzala).

0022-247X/$ – see front matter doi:10.1016/j.jmaa.2009.05.060

©

2009 Elsevier Inc. All rights reserved.

(1.1)

C.R. da Luz, G.P. Menzala / J. Math. Anal. Appl. 359 (2009) 464–481

465

 (x) E t − curl H + σ E − γ curl ut = 0,

(1.2)

μ(x) H t + curl E = 0,

(1.3)

μ(x) H = 0

(1.4)

div





in Ω × (0, +∞). Here x = (x1 , x2 , x3 ) belongs to Ω and t is the time variable. In the above system we denote by u = (u 1 , u 2 , u 3 ) the displacement vector, E = ( E 1 , E 2 , E 3 ) the dielectric induction (or electric displacement), H = ( H 1 , H 2 , H 3 ) the magnetic induction (or magnetic flux density),  (x) and μ(x) denote the electric permittivity and magnetic permeability, respectively. They are 3 × 3 symmetric matrices and uniformly positive definite almost everywhere for x in Ω . The constants σ > 0 and γ are the coefficients of the electric conductivity and coupling constant, respectively. We complement system (1.1)–(1.4) with initial conditions

(u , ut , E , H )|t =0 = (u 0 , u 1 , E 0 , H 0 ) in Ω

(1.5)

and boundary conditions

u = 0,

η × E = 0 on ∂Ω × (0, +∞)

(1.6)

where η = η(x) denotes the unit normal vector at x ∈ ∂Ω pointing the exterior of Ω . Here × is the usual vector product. The total energy associated with system (1.1)–(1.6) is given by

  3 2  ∂u ∂u L(t ) = ut (t ) + A i j (x) (t ) · (t ) +  (x) E (t ) · E (t ) + μ(x) H (t ) · H (t ) dx. 2 ∂xj ∂ xi 1

Ω

Formally, under reasonable assumptions on  (x),

dL dt

(1.7)

i , j =1

 (t ) = −

ut (t ) 2 dx − σ

Ω



μ(x) and A i j (x) we can prove that

E (t ) 2 dx.

Ω

Therefore L(t ) is non-increasing for any t  0. Our main interest in the article is to find uniform rates of decay of L(t ) as t → +∞. As we will see in Section 4 (Theorem 4.1), L(t ) decays uniformly at the rate (1 + t )−1 . Furthermore, it we choose more regular data, several terms in higher order energy do decay at faster rates (Theorem 4.2). There is a large literature concerning the decay of hyperbolic problems in exterior domains. In a pioneer work, C.S. Morawetz [15] considered the scalar wave equation utt − u = 0 in exterior domains with initial data of compact support in R3 and proved the decay of the local energy at rate t −1 as t → +∞. Using Huygens’ principle she obtained the exponential decay. The “obstacle” was assume to be star-shaped. Later on, G. Dassios [2] obtained similar results for the system of elastic waves. B.V. Kapitonov [9] studied the same problem for general hyperbolic systems which included the Maxwell equations in exterior domains with Silver–Muller boundary conditions. E. Zuazua [20] considered the semilinear wave equation with localized damping in unbounded domains. More recently, M. Nakao [19] considered the scalar wave equation

utt − u + a(x)ut = 0

(1.8)

in an exterior domain with Dirichlet boundary conditions. The term a(x)ut was acting as a “localized” dissipation in part of the boundary and “near” infinity. In [17] was obtained a polynomial rate of decay of the total energy as t → +∞. Further contributions of the same author were given in [18,16]. Later on, R. Ikehata [7,8] also obtained uniform rates of decay of the solutions of (1.8) with a(x) ≡ 1 in exterior domains. In [6], M.V. Ferreira and G. Perla Menzala considered the coupled isotropic Maxwell equations with the system of elastic waves proving that the “second order” energy

  3 2  2 2 ∂ ut ∂ ut L1 (t ) = utt (t ) + Ai j (t ) · (t ) +  E t (t ) + μ H t (t ) dx 2 ∂xj ∂ xi 1

Ω

i , j =1

decays uniformly as t → +∞. We can observe that the  final results of [6]  (see 3 Theorems 3.1 and 3.2) say that all terms of L(t ) (in the isotropic case) given by (1.7) except Ω μ| H (t )|2 dx and Ω i , j =1 A i j ∂∂xu (t ) · ∂∂xu (t ) dx decay uniformly as j

i

t → +∞. The results presented in this article improve the ones in [6] considering model (1.1)–(1.6) in an appropriate function spaces. Inspired in the work of M. Nakao [19] and R. Ikehata [8] we adapted their methods to our more complicated situation. We do not require to assume geometrical conditions on the obstacle. In forthcoming work [14] we apply the results presented in this article considering the system of anisotropic electromagnetic/elasticity with a nonlinear damping in exterior domains.

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2. Notations and assumptions Let Ω ⊆ R3 be an open set which is an exterior domain (unbounded with bounded complement). The boundary of Ω is assume to be compact and smooth. We denote by M the set of 3 × 3 matrices which are symmetric, uniformly positive definite and whose entries belong to L ∞ (Ω). Thus, if α = α (x) belongs to M, there exists a number α0 > 0 such that

α0 | v |2  v α (x) v T for any v in R3 almost everywhere in x ∈ Ω . Here, v T = we will always assume the following conditions: (H1): The matrices  and μ belong to M. (H2): There exists a positive constant β such that 3 



A i j (x) v j · v i  β

i , j =1

 v1  v2 v3

whenever v = ( v 1

v2

v 3 ) , | v |2 =

3  | v i |2

3

j =1

v 2j . From now on,

(2.1)

i =1 ij

for any vectors v i ∈ R3 , i = 1, 2, 3. The entries C kl (x) are given by ij

C kl (x) = (1 − δil δ jk )aikjl (x) + δik δ jl ailjk (x) where

 δlk =

1 if l = k, 0 if l = k

and aikjl (x) are the Cartesian components of the elastic tensor with the symmetric properties

ai jkl (x) = a jikl (x) = akli j (x)

a.e. in Ω.

(2.2)

1,∞

All ai jkl belong to W (Ω), γ is real coupling constant and σ > 0. Observe that the symmetric assumptions (2.2) imply that the transpose of A i j (x) is A ji (x). In the simplest case, when the medium is isotropic then the constants ai jkl are given by

ai jkl = λ1 δi j δkl + λ2 (δik δ jl + δil δ jk ) where λ1 and λ2 are Lame’s constants. The operator

L=

  3  ∂ ∂ A i j (x) ∂ xi ∂xj

(2.3)

i , j =1

reduces to λ2  + (λ1 + λ2 )∇ div and condition (2.1) for A i j (x) holds with β = λ2 . Observe that (H1) implies that  and μ are invertible a.e. in Ω . In fact, since the belong to M then their eigenvalues are positive. Consequently, the determinant of each one of them is also positive. Hence,  (x) and μ(x) are invertible. We can easily prove that the entries of  −1 and μ−1 belong to L ∞ (Ω). Given α in M we consider the weighted space



2

L (Ω; α ) =





z(x) = z1 (x), z2 (x), z3 (x) in such a way that

 T

z(x)α (x) z (x) dx =

3  



αi, j (x)zi (x)z j (x) dx < +∞

i , j =1 Ω

Ω

with the following inner product: If v, u are in L 2 (Ω; α ) then

( v , u ) L 2 (Ω;α ) =

3  



and norm

v (x)α (x)u T (x) dx

αi, j (x) v i (x)u j (x) dx =

i , j =1 Ω

Ω



v 2L 2 (Ω;α ) =

v (x)α (x) v T (x) dx. Ω

If α ≡ the identity matrix, we denote [ L 2 (Ω)]3 = [ L 2 (Ω; I )]3 . For any α ∈ M it is clear that L 2 (Ω; α ) ∼ = [ L 2 (Ω)]3 and 2 3 the norms · L 2 (Ω;α ) and · (the usual norm in [ L (Ω)] ) are equivalent. All notations used in this article are standard and follow the ones given in [3]. From now on we will denote by C a positive constant which may be of different value from line to line. This is just to simplify notations.

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467

3. Well-posedness In this section we briefly describe the function spaces where we will consider the solution {u , E , H } of problem (1.1)–(1.6). We rewrite system (1.1)–(1.3) as

d dt

U (t ) = ( A + B )U (t ),

U (0) = U 0

where U (t ) = (u (t ), ut (t ), E (t ), H (t )), U 0 = (u 0 , u 1 , E 0 , H 0 ),



A (u , ut , E , H ) = ut , Lu − u − γ curl E , γ  −1 curl ut +  −1 curl H , −μ−1 curl E and



B (u , ut , E , H ) = 0, u − ut , −σ  −1 E , 0





where L is given as in (2.3). Thus, it is natural to consider the linear space

3

3 × L 2 (Ω) × L 2 (Ω;  ) × L 2 (Ω; μ)

X = H 01 (Ω)

with the inner product

v , w X =

  3 

Ω

i , j =1

   ∂ v1 ∂ w1 A i j (x) (x) · (x) + v 1 (x) · w 1 (x) dx + v 2 (x) · w 2 (x) dx ∂xj ∂ xi Ω

+ ( v 3 , w 3 ) L 2 (Ω; ) + ( v 4 , w 4 ) L 2 (Ω;μ) for any v = ( v 1 , v 2 , v 3 , v 4 ), w = ( w 1 , w 2 , w 3 , w 4 ) in X . Next, we consider the unbounded linear operator A : D ( A ) ⊆ X → X , with domain

3

D ( A ) = H 2 (Ω) ∩ H 01 (Ω) given by

3 × H 01 (Ω) × H 0 (curl; Ω) × H (curl; Ω)



A w = w 2 , L w 1 − w 1 − γ curl w 3 , γ  −1 curl w 2 +  −1 curl w 4 , −μ−1 curl w 3 for any w = ( w 1 , w 2 , w 3 , w 4 ) ∈ D ( A ). In (3.1) we denote by



3

H (curl; Ω) = v in L 2 (Ω) with inner product



v , u H (curl;Ω) =



(3.1)



3 

such that curl v belongs to L 2 (Ω)



v (x) · u (x) + curl v (x) · curl u (x) dx

Ω

and



H 0 (curl; Ω) = w in H (curl; Ω) such that η × w |∂Ω = 0



η = η(x) is the unit normal vector at x ∈ ∂Ω pointing the exterior of Ω . The definition of this subspace requires ψ ¯ 3 into [C 1 (∂Ω)]3 extends by continuity some explanation: It is well known [3] that the map w −→ η × w |∂Ω from [C 01 (Ω)] − 1/ 2 3 (∂Ω)] . Therefore η × w |∂Ω makes sense for any w in H (curl; Ω) by to a continuous linear map from H (curl; Ω) into [ H ˜ w ) = η × w with ψ˜ the extension map of ψ . It can be verify that H 0 (curl; Ω) is a closed subspace of H (curl; Ω) denoting ψ( where

(see [4]) and the following property





v (x) · curl u (x) dx = Ω

curl v (x) · u (x) dx

(3.2)

Ω

holds provided v belongs to H 0 (curl; Ω) and u belongs to H (curl; Ω). Conversely, if (3.2) is valid for any u in H (curl; Ω) and some v in H (curl; Ω) then v ∈ H 0 (curl; Ω) (see [3, Chapter IX]). The linear operator B : X → X is given by



B w = 0, w 1 − w 2 , −σ  −1 w 3 , 0



(3.3)

for any w = ( w 1 , w 2 , w 3 , w 4 ) in X . Theorem 3.1. Let Ω ⊆ R3 be as in Section 2 and (H1) and (H2) hold. If (u 0 , u 1 , E 0 , H 0 ) belongs to D ( A ) ∩ Y , then system (1.1)–(1.6) has a unique (strong) solution

    (u , ut , E , H ) which belongs to C [0, +∞); D ( A ) ∩ Y ∩ C 1 [0, +∞); Y

where Y = {( w 1 , w 2 , w 3 , w 4 ) in X such that div(μ w 4 ) = 0 in Ω}. Furthermore, if (u 0 , u 1 , E 0 , H 0 ) ∈ Y then system (1.1)–(1.6) has a unique weak solution (u , ut , E , H ) ∈ C ([0, +∞); Y ).

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Proof. We will prove that A is skew-adjoint. First, we claim that D ( A ) ⊆ D ( A ∗ ) and A ∗ v = − A v for any v in D ( A ). Here A ∗ denotes the adjoint of A. In fact, given v in X to prove that v belongs to D ( A ∗ ) it is sufficient to prove the existence of some g in X such that

A w, v X = w, g X

for any w in D ( A ).

Using the definition of A clearly we have

  3 

     3  ∂ w2 ∂ v1 ∂ ∂ w1 A w, v X = Ai j + w2 · v1 + Ai j · · v 2 dx − [ w 1 · v 2 + γ curl w 3 · v 2 ] dx ∂xj ∂ xi ∂ xi ∂xj i , j =1 Ω i , j =1 Ω  −1     −1  −1 + γ  curl w 2 , v 3 L 2 (Ω; ) +  curl w 4 , v 3 L 2 (Ω; ) − μ curl w 3 , v 4 L 2 (Ω;μ) . (3.4) Since v 1 ∈ [ H 2 (Ω) ∩ H 01 (Ω)]3 and w 2 ∈ [ H 01 (Ω)]3 then, using the divergence theorem we deduce 3    i , j =1 Ω

Ai j

      3   3  ∂ ∂ v1 ∂ w2 ∂ v1 ∂ ∂ v1 · · w 2 dx = + Ai j Ai j · w 2 dx ∂xj ∂ xi ∂ xi ∂xj ∂ xi ∂xj j =1 Ω i =1

=

3   3   j =1 ∂Ω i =1

 ∂ v1 Ai j · w 2 η i d Γ = 0. ∂xj

(3.5)

Using the fact that the transpose of A i j is A ji we deduce from (3.5) that

  3  Ω i , j =1

      3 ∂ w2 ∂ v1 ∂ ∂ v1 Ai j dx = − Ai j · · w 2 dx = − L v 1 · w 2 dx. ∂xj ∂ xi ∂ xi ∂xj Ω i , j =1

(3.6)

Ω

Similarly

       3 3  ∂ ∂ w1 ∂ w1 ∂ v2 · v 2 dx = − · Ai j Ai j dx. ∂ xi ∂xj ∂xj ∂ xi

Ω i , j =1

Using (3.2) we have



w 3 ,  −1 curl v 2

(3.7)

Ω i , j =1



 = L 2 (Ω; )

 w 3 · curl v 2 dx =

Ω

curl w 3 · v 2 dx. Ω

Therefore



−γ



curl w 3 · v 2 dx = w 3 , −γ  −1 curl v 2

 L 2 (Ω; )

.

Ω

Similarly



γ  −1 curl w 2 , v 3  and



 −1 curl w 4 , v 3



μ−1 curl w 3 , v 4



 =γ L 2 (Ω; )

w 2 · curl v 3 dx, Ω

  = w 4 , μ−1 curl v 3 L 2 (Ω;μ) L 2 (Ω; )

 L 2 (Ω;μ)

  = w 3 ,  −1 curl v 4 L 2 (Ω; ) .

Substitution of the above identities into (3.4) shows that

A w, v X = w, g X

for any w ∈ D ( A )

where g = (− v 2 , − L v 1 + v 1 + γ curl v 3 , −γ  −1 curl v 2 −  −1 curl v 4 , μ−1 curl v 3 ) which proves our claim. Next, we want to prove that D ( A ∗ ) ⊆ D ( A ) and A ∗ v = − A v for any v = ( v 1 , v 2 , v 3 , v 4 ) in D ( A ∗ ). Let v ∈ D ( A ∗ ). Then, there exists g = ( g 1 , g 2 , g 3 , g 4 ) in X such that

A w, v X = w, g X

for any w in D ( A ).

Let w 4 ∈ [C0∞ (Ω)]3 then w



 −1 curl w 4 , v 3



L 2 (Ω; )

= (0, 0, 0, w 4 ) ∈ D ( A ). Substitution in (3.8) give us

= ( w 4 , g 4 ) L 2 (Ω;μ) .

(3.8)

C.R. da Luz, G.P. Menzala / J. Math. Anal. Appl. 359 (2009) 464–481

469

Consequently,

g 4 = μ−1 curl v 3

3

in L 2 (Ω)

and

v 3 ∈ H (curl; Ω).

Next, choosing w 1 , w 2 and w 3 in [C0∞ (Ω)]3 we consider w = ( w 1 , 0, 0, 0), w = (0, w 2 , 0, 0) and w = (0, 0, w 3 , 0) to follow the same procedure as above to find that g 1 = − v 2 in [ L 2 (Ω)]3 , g 2 = − L v 1 + v 1 + γ curl v 3 in [ L 2 (Ω)]3 and g 3 = −γ  −1 curl v 2 −  −1 curl v 4 in [ L 2 (Ω)]3 . Thus, we obtain g = ( g 1 , g 2 , g 3 , g 4 ) and we easily verify that A ∗ v = − A v. To conclude the proof of our claim remains to prove that v 3 belongs to H 0 (curl; Ω). Rewriting A ∗ v = − A v in their components, using (3.6) and (3.7) we obtain the identity

 −γ Ω



 curl w 3 · v 2 dx + γ Ω



=γ

curl w 2 · v 3 dx + Ω



w 2 · curl v 3 dx − γ Ω



curl w 4 · v 3 dx = Ω

w 4 · curl v 3 dx



w 3 · curl v 4 dx + Ω

for any w in D ( A ). Thus, since v 2 , w 2 belong to

curl w 3 · v 4 dx Ω



w 3 · curl v 2 dx − Ω



 curl w 4 · v 3 dx −

[ H 01 (Ω)]3

w 4 · curl v 3 dx

(3.9)

Ω

and w 3 ∈ H 0 (curl; Ω) we deduce from (3.9) and (3.2) that

for any w 4 ∈ H (curl; Ω).

Ω

Thus, v 3 ∈ H 0 (curl; Ω) which completes the proof of our claim. By Stone’s Theorem A is the infinitesimal generator of a group of unitary operators. The linear operator B given by (3.3) is bounded. In fact, if w = ( w 1 , w 2 , w 3 , w 4 ) belongs to X and denoting by w 3 = ( w 3,1 , w 3,2 , w 3,3 ) we have

 2

B w 2X =  0, w 1 − w 2 , −σ  −1 w 3 , 0  X = w 1 − w 2 2 +

 2 w 1 2 + 2 w 2 2 + σ 2

3 





T  −σ  −1 w 3  −σ  −1 w 3 dx

Ω

i−, j1 w 3,i w 3, j dx  C w 2X

i , j =1 Ω

for some positive constant C . Here we used that the entries (denoted by i−j 1 ) of  −1 belong to L ∞ (Ω) for any i , j = 1, 2, 3. It follows that the operator A + B with domain D ( A + B ) = D ( A ) is the infinitesimal generator of a semigroup of bounded operators. In order to conclude the proof of Theorem 3.1 we need to verify (1.4). We apply the divergence to Eq. (1.3) to obtain

div









μ(x) H t (t ) + div curl E (t ) = 0 in Ω × (0, +∞)

in the distributional sense. Therefore div(μ(x) H (t )) = div(μ(x) H 0 ) a.e. in Ω and t  0. Thus, in order to have (1.4) for any t  0 is enough to choose initial data in the subspace Y which is the assumption in Theorem 3.1. 2 4. Asymptotic behavior In this section we study the asymptotic behavior of the solution (u , ut , E , H ) of problem (1.1)–(1.6). Theorem 4.1. Let (u , ut , E , H ) be the global weak solution of problem (1.1)–(1.6) obtained in Theorem 3.1. Suppose that μ H 0 = curl ψ0 for some ψ0 ∈ H 0 (curl; Ω). Then,

L(t )  C I 0 (1 + t )−1 for any t  0 for some positive constant C . Here L(t ) denotes the total energy given by (1.7) and I 0 = L(0) + u 0 2 + ψ0 2 . Proof. Clearly, if we prove the above result for (u 0 , u 1 , E 0 , H 0 ) ∈ D ( A ) ∩ Y then by a standard limiting process the same decay still holds for the weak solution of problem (1.1)–(1.6) for any (u 0 , u 1 , E 0 , H 0 ) ∈ Y . Let us take the inner product in [ L 2 (Ω)]3 of Eq. (1.1) with ut (t ), (1.2) with E (t ) and (1.3) with H (t ). Addition of all the identities give us





utt (t ), ut (t ) +

  3  Ω i , j =1

Ai j

   ∂ ut ∂u (t ) · (t ) dx + E t (t ), E (t ) L 2 (Ω; ) ∂xj ∂ xi

  2 2  + H t (t ), H (t ) L 2 (Ω;μ) + ut (t ) + σ  E (t ) = 0 

(4.1)

470

C.R. da Luz, G.P. Menzala / J. Math. Anal. Appl. 359 (2009) 464–481

where ( , ) and · denote the inner product and norm respectively in [ L 2 (Ω)]3 . Observe that we obtained (4.1) using (3.2). Identity (4.1) implies

d dt

  2 2 L(t ) + ut (t ) + σ  E (t ) = 0.

(4.2)

Therefore

t L(t ) +

  u s (s)2 ds + σ

0

t

   E (s)2 ds = L(0).

(4.3)

0

Multiplying identity (4.2) by (1 + t ) and integrate over [0, t ] give us (after integration by parts)

t

 2 (1 + s)u s (s) ds + σ

(1 + t )L(t ) + 0

t

t

 2 (1 + s) E (s) ds = L(0) +

L(s) ds.

0

0

Therefore in order to obtain the conclusion of Theorem 4.1 is sufficient to prove that there exists a positive constant C 1 such that

+∞ L(s) ds  C 1 I 0 .

(4.4)

0

Consider the auxiliary vector-valued functions

t W (t ) =

t E (s) ds

F (t ) =

and

0

H (s) ds. 0

Straightforward calculation shows that W and F satisfy the following initial–boundary-value problem

 W t − curl F + σ W − γ curl u =  E 0 − γ curl u 0 ,

(4.5)

μ F t + curl W = μ H 0

(4.6)

in Ω × (0, +∞), with initial conditions

W (0) = 0,

F (0) = 0 in Ω

and boundary condition

W × η = 0 on ∂Ω × (0, +∞). We differentiate Eq. (4.5) with respect to time and take the inner product in [ L 2 (Ω)]3 of the resulting identity with W (t ). Next, we take the inner product in [ L 2 (Ω)]3 of (4.6) with F t (t ). Adding the resulting identities we obtain





W tt (t ), W (t )

L 2 (Ω; )

2   σ d  + F t (t ), F t (t ) L 2 (Ω;μ) + W (t ) − γ





curl ut (t ) · W (t ) dx −

2 dt

Ω

[μ H 0 ] · F t (t ) dx = 0. Ω

(4.7) Let us rewrite (4.7) in the form

 2  2 2 σ d  −  W t (t ) L 2 (Ω; ) +  F t (t )L 2 (Ω;μ) + W (t ) dt   2 dt d d −γ curl u (t ) · W (t ) dx + γ curl u (t ) · W t (t ) dx − [μ H 0 ] · F (t ) dx = 0. d



W t (t ), W (t )

L 2 (Ω; )

dt

(4.8)

dt

Ω

Ω

Ω

Integration of identity (4.8) over [0, t ] give us

σ  2

2 W (t ) +

t

   H (s)22 L

t ds = (Ω;μ)

0

   E (s)22 L



0

L 2 (Ω; )

+γ

curl u (t ) · W (t ) dx Ω



t  −γ

curl u (s) · E (s) dx ds + 0 Ω

because W t (t ) = E (t ) and F t (t ) = H (t ).





ds − E (t ), W (t ) (Ω; )

[μ H 0 ] · F (t ) dx Ω

(4.9)

C.R. da Luz, G.P. Menzala / J. Math. Anal. Appl. 359 (2009) 464–481

Next we find estimates for the last four terms on the right-hand side of (4.9). By assumption

ψ0 ∈ H 0 (curl; Ω). Therefore





[μ H 0 ] · F (t ) dx =

μ H 0 = curl ψ0 for some

 curl ψ0 · F (t ) dx =

Ω

471

Ω

ψ0 · curl F (t ) dx

(4.10)

Ω

due to (3.2). Using (4.5) we know that

curl F =  W t + σ W − γ curl u −  E 0 + γ curl u 0 . Substitution into (4.10) give us



      [μ H 0 ] · F (t ) dx = E (t ), ψ0 L 2 (Ω, ) + σ W (t ), ψ0 − γ curl u (t ), ψ0 − ( E 0 , ψ0 ) L 2 (Ω, ) + γ (curl u 0 , ψ0 ). (4.11)

Ω

Let v = ( v 1 , v 2 , v 3 ) any element in [ H 1 (Ω)]3 . Then, due to assumption (2.1), the inequalities

3  3    ∂ v j 2 dx = 2

curl v  2 ∂x 2

i

i , j =1 Ω

i =1 Ω

   3  ∂ v 2 ∂v ∂v dx  2β −1 · A dx i j ∂x ∂ x j ∂ xi i

(4.12)

Ω i , j =1

hold. Using (4.12) and the equivalence of the norms · L 2 (Ω;α ) and · , we find the following estimate for the left-hand side of (4.11):

    3      ∂u ∂u [μ H 0 ] · F (t ) dx  C I 0 + C  E (t )2 + σ  W (t )2 + C Ai j (t ) · (t ) dx 8 ∂xj ∂ xi

(4.13)

Ω i , j =1

Ω

for some positive constant C . Now, for any δ > 0 we can estimate

 t    curl u (s) · E (s) dx ds − E (t ), W (t ) L 2 (Ω; ) + γ curl u (t ) · W (t ) dx − γ δ



−1 

0 Ω

Ω

 2  2 2 E (t ) L 2 (Ω; ) + δ  W (t ) L 2 (Ω; ) + |γ |δ −1 curl u (t )

  2 |γ | + |γ |δ  W (t ) +

t

2

  curl u (s)2 ds + |γ |

t

2

0

   E (s)2 ds.

(4.14)

0

The equivalence of the norms · L 2 (Ω; ) and · tell us that W (t ) 2L 2 (Ω; )  C W (t ) 2 for some C > 0. Thus, from (4.13) and (4.14) we deduce from (4.9) the estimate



σ 2



− C δ − |γ |δ −

σ  8

2 W (t ) +

t

   H (s)22

L (Ω;μ)

ds

0

t C

     E (s)2 ds + δ −1  E (t )22

L (Ω; )

+ |γ |δ



−1 

2 |γ | curl u (t ) +

0

+

t

2

  curl u (s)2 ds

0

|γ |

t

2

     E (s)2 ds + C I 0 + C  E (t )2 + C

  3  Ω i , j =1

0

 ∂u ∂u Ai j (t ) · (t ) dx. ∂xj ∂ xi

(4.15)

Choosing δ  σ4 (C + |γ |)−1 , using (4.3) together with (4.12) we deduce from (4.15) the estimate

σ  8

2 W (t ) +

t

   H (s)22

L (Ω;μ)

0

ds  C I 0 + C

t   3  0 Ω i , j =1

 ∂u ∂u Ai j (s) · (s) dx ds. ∂xj ∂ xi

(4.16)

Due to (4.3) and (4.16) in order to prove (4.4) remains only to estimate the last term on the right-hand side of (4.16). Let us take inner product in [ L 2 (Ω)]3 of Eq. (1.1) with u (t ) give us

472

C.R. da Luz, G.P. Menzala / J. Math. Anal. Appl. 359 (2009) 464–481

d dt

2   ut (t ), u (t ) − ut (t ) +

  3  Ω i , j =1

    ∂u ∂u 1 d u (t )2 + γ curl E (t ), u (t ) = 0. Ai j (t ) · (t ) dx + ∂xj ∂ xi 2 dt

(4.17)

Integrating identity (4.17) over [0, t ) and using (3.2) we obtain

1 2

 u (t )2 +

t   3 

Ai j

0 Ω i , j =1

t

1

2

= u 0 + 2

 ∂u ∂u (s) · (s) dx ds ∂xj ∂ xi

    u s (s)2 ds − ut (t ), u (t ) + (u 1 , u 0 ) − γ

0

t





E (s), curl u (s) ds.

(4.18)

0

Let us estimate the last three terms of the right-hand side of (4.18): For any δ > 0 and using (4.12) we deduce

t     E (s), curl u (s) ds − ut (t ), u (t ) − γ 0

 2 1  2  ut (t ) + u (t ) + (2δ)−1

t

4

   E (s)2 ds + δ

2

0

 2 1  2  ut (t ) + u (t ) + (2δ)−1

t

t

4

  curl u (s)2 ds

0

   E (s)2 ds + β −1 δ

0

t   3  0 Ω i , j =1

 ∂u ∂u Ai j (s) · (s) dx ds. ∂xj ∂ xi

Using the above estimate (with 0 < 2δ < β ) together with (4.18) we obtain

 u (t )2 + 1

1 4

2

t   3  0 Ω i , j =1

 t t 2  2    ∂u ∂u −1      E (s)2 ds. Ai j (s) · (s) dx ds  C I 0 + u s (s) ds + ut (t ) + (2δ) ∂xj ∂ xi 0

0

(4.19) The right-hand side of (4.19) is less than or equal to C I 0 for some positive constant C due to (4.3). Returning to (4.16) +∞

H (s) 2L 2 (Ω;μ) ds  C I 0 for some positive constant C . This proves (4.4) and concludes the proof of we conclude that 0 Theorem 4.1.

2

For truly strong solutions of problem (1.1)–(1.6) we can obtain better rates of decay. Theorem 4.2. Let (u , ut , E , H ) be the global strong solution of problem (1.1)–(1.6) obtained in Theorem 3.1. Suppose that curl ψ0 for some ψ0 ∈ H 0 (curl; Ω) and (u 0 + u 1 ) ∈ [ L 6/5 (Ω)]3 . Then,

(a)

  utt (t )2 +

  3  Ω i , j =1 −3

 C I 1 (1 + t ) (b)

  3  Ω i , j =1

(c)

Ai j

Ai j

μH0 =

  2  2  2  2  2 ∂ ut ∂ ut (t ) · (t ) dx +  E t (t ) +  H t (t ) + curl E (t ) + ut (t ) +  Lu (t ) ∂xj ∂ xi

for any t  0;

  2  2 ∂u ∂u (t ) · (t ) dx +  E (t ) + curl H (t )  C I 1 (1 + t )−2 for any t  0; ∂xj ∂ xi

    u (t )2 +  H (t )2  C I 1 (1 + t )−1 for any t  0

for some positive constant C and I 1 given by

I 1 = u 0 2[ H 2 (Ω)]3 + u 1 2[ H 1 (Ω)]3 + u 0 + u 1 2[ L 6/5 (Ω)]3 + E 0 2H (curl;Ω) + H 0 2H (curl;Ω) + ψ0 2 . Proof. We will assume more regular initial data, say (u 0 , u 1 , E 0 , H 0 ) ∈ D ( A 2 ) ∩ Y . If the proof is done in this case then by density of D ( A 2 ) in D ( A ) the result will be valid for initial data in D ( A ) ∩ Y .

C.R. da Luz, G.P. Menzala / J. Math. Anal. Appl. 359 (2009) 464–481

473

We take the derivative of Eqs. (1.1)–(1.4) with respect to t to obtain

uttt − Lut + utt + γ curl E t = 0,

(4.20)

 E tt − curl H t + σ E t − γ curl utt = 0,

(4.21)

μ H tt + curl E t = 0,

(4.22)

div(μ H t ) = 0

(4.23)

in Ω × (0, +∞). From the original system (1.1)–(1.4) we also obtain

⎧ ⎨ utt (0) = Lu 0 − u 1 − γ curl E 0 ≡ u 2 , E (0) =  −1 curl H 0 − σ  −1 E 0 + γ  −1 curl u 1 ≡ E 1 , ⎩ t H t (0) = −μ−1 curl E 0 ≡ H 1

in Ω . The boundary conditions are

u t = 0,

E t × η = 0 on ∂Ω × (0, +∞).

Let us take the inner product in [ L 2 (Ω)]3 of (4.20) with utt (t ), (4.21) with E t (t ) and (4.22) with H t (t ), using (3.2) and adding the corresponding identities we obtain



d dt

  utt (t )2 +

  3  Ω i , j =1

   2  2 ∂ ut ∂ ut Ai j (t ) · (t ) dx +  E t (t )L 2 (Ω; ) +  H t (t ) L 2 (Ω;μ) ∂xj ∂ xi

  2 2 + 2utt (t ) + 2σ  E t (t ) = 0.

(4.24)

We multiply identity (4.24) by (1 + t )3 and integrate over [0, t ). Integration by parts give us

 2 (1 + t )3 utt (t ) + (1 + t )3

  3 

Ai j

Ω i , j =1

 2 + (1 + t )3  H t (t )L 2 (Ω;μ) + 2

t

  2 ∂ ut ∂ ut (t ) · (t ) dx + (1 + t )3  E t (t ) L 2 (Ω; ) ∂xj ∂ xi

 2 (1 + s)3 u ss (s) ds + 2σ

0

  3 

= u 2 2 +

Ω i , j =1

t  (1 + s)2

+3 t +3

 2 (1 + s)3  E s (s) ds

0

 t  2 ∂ u1 ∂ u1 Ai j dx + E 1 2L 2 (Ω; ) + H 1 2L 2 (Ω;μ) + 3 (1 + s)2 u ss (s) ds · ∂xj ∂ xi 0

3   i , j =1

0 Ω

t

 t  2 ∂ us ∂ us Ai j (s) · (s) dx ds + 3 (1 + s)2  E s (s) L 2 (Ω; ) ds ∂xj ∂ xi 0

 2 (1 + s)2  H s (s) L 2 (Ω;μ) ds.

(4.25)

0

Next, let us take the inner product in [ L 2 (Ω)]3 of Eq. (1.1) with utt (t ), (4.21) with E (t ) and (1.3) with H t (t ), using (3.2) and adding the corresponding identities we obtain



d dt

1 2

=

   ut (t )2 + σ  E (t )2 + 2

  3  Ω i , j =1

  3  Ω i , j =1

  2  2    ∂ ut ∂u Ai j (t ) · (t ) dx + E t (t ), E (t ) L 2 (Ω; ) +utt (t ) +  H t (t ) L 2 (Ω;μ) ∂xj ∂ xi

  2 ∂ ut ∂ ut Ai j (t ) · (t ) dx +  E t (t ) L 2 (Ω, ) . ∂xj ∂ xi

(4.26)

Multiply identity (4.26) by (1 + t )2 . Integrate over [0, t ) and perform integration by parts to deduce

t 0



2 (1 + s) u ss (s) ds + 2

t

   2 2 σ 2 1 (1 + s)2  H s (s) L 2 (Ω;μ) ds + (1 + t )2 ut (t ) + (1 + t )2  E (t ) 2

0

2

474

C.R. da Luz, G.P. Menzala / J. Math. Anal. Appl. 359 (2009) 464–481

1

σ

2

2

= u 1 2 +

t

E 0 2 +

 2 (1 + s)u s (s) ds + σ

0 3  

t  (1 + s)2

+

i , j =1

0 Ω

− (1 + t )2

  3  Ω i , j =1

t  +2

(1 + s) t

+2

 2 (1 + s) E (s) ds

0

 t  2 ∂ us ∂ us Ai j (s) · (s) dx ds + (1 + s)2  E s (s) L 2 (Ω; ) ds ∂xj ∂ xi 0

    3  ∂ ut ∂u ∂ u0 ∂ u1 · Ai j (t ) · (t ) dx + Ai j dx ∂xj ∂ xi ∂xj ∂ xi Ω i , j =1

3   i , j =1

0 Ω

t

Ai j

   ∂ us ∂u (s) · (s) dx ds − (1 + t )2 E t (t ), E (t ) L 2 (Ω; ) + ( E 1 , E 0 ) L 2 (Ω; ) ∂xj ∂ xi

  (1 + s) E s (s), E (s) L 2 (Ω; ) ds.

(4.27)

0

We want to find estimates especially for the third and four terms on the right-hand side of identity (4.27). Clearly, the following inequalities are valid: For any δ > 0

   3  3   ∂ u ∂ u t ∂ ut ∂u 2 2 dx 1 + t ) A ( t ) · ( t ) dx C ( 1 + t ) ( t ) ( t ) −(  ij ∂ x ∂x ∂xj ∂ xi j i

(i)

Ω i , j =1

i , j =1 Ω

 C1δ

−1

3  3    ∂ u 2 3 (1 + t ) ∂ x (t ) dx + C 1 δ(1 + t ) j j =1 Ω

i =1 Ω

∂ ut 2 ∂ x (t ) dx i

for some positive constant C 1 .

−1    2   −(1 + t )2 E t (t ), E (t ) 2  δ (1 + t )2  E t (t )22 + δ(1 + t )2  E (t ) L 2 (Ω; ) . L (Ω; ) L (Ω; ) t   3   ∂ us ∂u (1 + s) Ai j (s) · (s) dx ds 2 ∂xj ∂ xi

(ii) (iii)

i , j =1

0 Ω

C

3 

t



i , j =1 0 Ω

∂ u ∂ u s (1 + s) (s) (s) dx ds ∂ x j ∂ x i

2 3 t  3 t    ∂ u 2 2 ∂ us  C1 (1 + s) (s) dx ds. ∂ x (s) dx ds + C 1 ∂x j

j =1 0 Ω

i

i =1 0 Ω

t t t  2  2     2 (1 + s) E s (s), E (s) L 2 (Ω; ) ds  (1 + s) E s (s) L 2 (Ω; ) ds + (1 + s) E (s) L 2 (Ω; ) ds.

(iv)

0

0

0

   3  1 σ ∂ u 0 ∂ u 1 2 2 ( E 1 , E 0 ) 2 · +

u

+

E

+ A dx  C I1 1 0 i j L (Ω; ) 2 2 ∂xj ∂ xi

(v)

Ω i , j =1

for some positive constant C . From (4.27), (i)–(v), the equivalence of the norms · L 2 (Ω; ) and · and assumption (2.1) we deduce

t

  2 2 σ (1 + s)2  H s (s) L 2 (Ω;μ) ds + (1 + t )2  E (t ) 2

0

t  2

 C I1 + C

(1 + s) 0 Ω

3   i , j =1

 t t   2 2 ∂ us ∂ us Ai j (s) · (s) dx ds + (1 + s)u s (s) ds + C (1 + s)2  E s (s) ds ∂xj ∂ xi 0

0

C.R. da Luz, G.P. Menzala / J. Math. Anal. Appl. 359 (2009) 464–481

t +C

  3 

 2 (1 + s) E (s) ds + β −1 C δ −1 (1 + t )

Ai j

Ω i , j =1

0

+ β −1 C δ(1 + t )3

  3 

Ai j

Ω i , j =1

475

 ∂u ∂u (t ) · (t ) dx ∂xj ∂ xi

  t   3  ∂ ut ∂u ∂ ut ∂u (t ) · (t ) dx + β −1 C Ai j (s) · (s) dx ds ∂xj ∂ xi ∂xj ∂ xi 0 Ω i , j =1

  2 2 + δ −1 (1 + t )2  E t (t ) L 2 (Ω; ) + C δ(1 + t )2  E (t ) .

(4.28)

Using Theorem 4.1 and (4.19) we get bounds for some terms in (4.28) to obtain

t

    2 2 σ (1 + s)2  H s (s) L 2 (Ω;μ) ds + − C δ (1 + t )2  E (t ) 2

0 3  

t  2

 C I1 + C

(1 + s)

i , j =1

0 Ω

+ β −1 C δ(1 + t )3

  3  Ω i , j =1

 t  2 ∂ us ∂ us Ai j (s) · (s) dx ds + C (1 + s)2  E s (s) ds ∂xj ∂ xi 0

  2 ∂ ut ∂ ut Ai j (t ) · (t ) dx + δ −1 (1 + t )2  E t (t ) L 2 (Ω; ) . ∂xj ∂ xi

(4.29)

We need to get estimates on the terms for the right-hand side of (4.29). We take the inner product in [ L 2 (Ω)]3 of Eq. (4.20) with ut (t ) to obtain

1 d 2 dt

 ut (t )2 +

  3 

Ai j

Ω i , j =1

    2 ∂ ut d ∂ ut (t ) · (t ) dx = − utt (t ) · ut (t ) dx + utt (t ) − γ E t (t ) · curl ut (t ) dx. ∂xj ∂ xi dt Ω

Ω

(4.30)

We integrate (4.30) over [0, t ] to obtain for any δ > 0 the estimate

1 2

 ut (t )2 +

t   3 

Ai j

0 Ω i , j =1

 ∂ us ∂ us (s) · (s) dx ds ∂xj ∂ xi

 2 1  2  C I 1 + utt (t ) + ut (t ) +

t

4

  u ss (s)2 ds + γ δ −1

0

t

   E s (s)2 ds + γ δ

0

t

  curl u s (s)2 ds.

(4.31)

0

Integration of (4.24) over [0, t ] give us

  utt (t )2 +

  3 

Ai j

Ω i , j =1

t +2

  2  2 ∂ ut ∂ ut (t ) · (t ) dx +  E t (t ) L 2 (Ω; ) +  H t (t )L 2 (Ω;μ) ∂xj ∂ xi

  u ss (s)2 ds + 2σ

0

t

   E s (s)2 ds  C I 1 .

(4.32)

0

Using (4.12) and combining (4.32) with (4.31) it follows that

t   3 

Ai j

0 Ω i , j =1

 ∂ us ∂ us (s) · (s) dx ds  C I 1 . ∂xj ∂ xi

(4.33)

Integrating (4.26) over [0, t ] and estimating the right-hand side we obtain for any δ > 0

t

  u ss (s)2 ds +

0

t

   H s (s)22

L (Ω;μ)

ds +

1  2

2

ut (t ) +

σ  2

2

E (t )

0

1

2

= u 1 + 2

σ 2

t 2

E0 +

   E s (s)22 L

0

ds + (Ω; )

t   3  0 Ω i , j =1

Ai j

   ∂ us ∂ us (s) · (s) dx ds − E t (t ), E (t ) L 2 (Ω; ) ∂xj ∂ xi

476

C.R. da Luz, G.P. Menzala / J. Math. Anal. Appl. 359 (2009) 464–481

+ ( E 1 , E 0 ) L 2 (Ω; ) −

  3  Ω i , j =1

t

   E s (s)2 ds +

 C I1 + C

    3  ∂ ut ∂u ∂ u0 ∂ u1 Ai j (t ) · (t ) dx + Ai j dx · ∂xj ∂ xi ∂xj ∂ xi Ω i , j =1

  3 

t

Ai j

0 Ω i , j =1

0

  2 ∂ us ∂ us (s) · (s) dx ds + δ −1  E t (t )L 2 (Ω; ) ∂xj ∂ xi

3  3    ∂ u 2  2 dx + C   + C δ E (t ) + C ( t ) ∂x j j =1 Ω

i =1 Ω

∂ ut 2 ∂ x (t ) dx.

(4.34)

i

Choosing δ > 0 small enough, using (4.32), (4.33), (4.3) and (2.1) we deduce from (4.34)

t

   H s (s)22

L (Ω;μ)

ds  C I 1

(4.35)

0

for some positive constant C . Next, we multiply identity (4.24) by (1 + t ) and integrate by parts and use (4.32), (4.33) and (4.35) to obtain

 2 (1 + t )utt (t ) + (1 + t )

  3  Ω i , j =1

t +2

 2 (1 + s)u ss (s) ds + 2σ

  2  2 ∂ ut ∂ ut Ai j (t ) · (t ) dx + (1 + t ) E t (t ) L 2 (Ω; ) + (1 + t ) H t (t ) L 2 (Ω;μ) ∂xj ∂ xi

t

0

 2 (1 + s) E s (s) ds  C I 1

(4.36)

0

for some positive constant C . Multiplying (4.24) by (1 + t )2 and integrating by parts over [0, t ] to obtain

 2 (1 + t )2 utt (t ) + (1 + t )2

  3 

Ai j

Ω i , j =1



2

+ (1 + t )

2 H t (t ) L 2 (Ω;μ) + 2

t

  2 ∂ ut ∂ ut (t ) · (t ) dx + (1 + t )2  E t (t ) L 2 (Ω; ) ∂xj ∂ xi 

2 (1 + s) u ss (s) ds + 2σ 2

0

= u 2 2 +

  3  Ω i , j =1

t +2

+2

 2 (1 + s)2  E s (s) ds

0

 ∂ u1 ∂ u1 Ai j dx + E 1 2L 2 (Ω; ) + H 1 2L 2 (Ω;μ) · ∂xj ∂ xi

 2 (1 + s)u ss (s) ds + 2

0

t

t

t  (1 + s)

i , j =1

0 Ω

 2 (1 + s) H s (s) L 2 (Ω;μ) ds + 2

0

3  

t

 ∂ us ∂ us Ai j (s) · (s) dx ds ∂xj ∂ xi

 2 (1 + s) E s (s) L 2 (Ω; ) ds.

(4.37)

0

Multiplying (4.26) by (1 + t ), integrating by parts over [0, t ], using (2.1), the equivalence of the norms · L 2 (Ω;α ) and ·

and estimating the right-hand side of the identity, we deduce for any δ > 0

t

 2  2 σ (1 + s) H s (s) L 2 (Ω;μ) ds + (1 + t ) E (t ) 2

0

 C I1 +

1

t

2 0

  u s (s)2 ds + C

t 0

   E (s)2 ds + C

t  (1 + s) 0 Ω

3   i , j =1

Ai j

 ∂ us ∂ us (s) · (s) dx ds ∂xj ∂ xi

C.R. da Luz, G.P. Menzala / J. Math. Anal. Appl. 359 (2009) 464–481

t +C

 2 (1 + s) E s (s) ds + C δ −1

Ai j

Ω i , j =1

0

+C

  3 

t   3 

Ai j

0 Ω i , j =1

477

    3  ∂u ∂ ut ∂u ∂ ut (t ) · (t ) dx + C δ(1 + t )2 Ai j (t ) · (t ) dx ∂xj ∂ xi ∂xj ∂ xi Ω i , j =1

  2  2 ∂u ∂u (s) · (s) dx ds + δ −1 (1 + t ) E t (t )L 2 (Ω; ) + C δ(1 + t ) E (t ) . ∂xj ∂ xi

Using (4.3), (4.36), (4.19) and choosing δ > 0 sufficiently small we obtain

t

 2  2 σ (1 + s) H s (s) L 2 (Ω;μ) ds + (1 + t ) E (t ) 4

0

1

2

 C I 1 + (1 + t ) 4

  3  Ω i , j =1

  t  3   ∂ ut ∂ us ∂ ut ∂ us Ai j (t ) · (t ) dx + C (1 + s) Ai j (s) · (s) dx ds. ∂xj ∂ xi ∂xj ∂ xi

(4.38)

i , j =1

0 Ω

Multiplying (4.30) by (1 + t ) and integrating by parts we obtain the estimate

1 2

 2 (1 + t )ut (t ) +

t  (1 + s)

 C I1 + C

 ∂ us ∂ us (s) · (s) dx ds ∂xj ∂ xi

Ai j

i , j =1

0 Ω

t

3  

      u s (s)2 ds + (1 + t )utt (t )2 + 1 (1 + t )ut (t )2 + C

t

4

0

0

t

+ γ δ −1

 2 (1 + s)u ss (s) ds

 2 (1 + s) E s (s) ds + γ δ

0

t

 2 (1 + s)curl u s (s) ds.

(4.39)

0

Using (4.3), (4.36), (4.12) and choosing δ > 0 sufficiently small we obtain from (4.39) 3  

t  (1 + s)

 ∂ us ∂ us (s) · (s) dx ds  C I 1 ∂xj ∂ xi

Ai j

i , j =1

0 Ω

(4.40)

for some positive constant C . Using (4.40), (4.38) and (4.36) we obtain an estimate for the right-hand side of (4.37):



2 1 (1 + t ) utt (t ) + (1 + t )2 2

2



2

+ (1 + t )

  3  Ω i , j =1

2 H t (t ) L 2 (Ω;μ) + 2

t

  2 ∂ ut ∂ ut Ai j (t ) · (t ) dx + (1 + t )2  E t (t )L 2 (Ω; ) ∂xj ∂ xi 

2 (1 + s) u ss (s) ds + 2σ 2

0

t

 2 (1 + s)2  E s (s) ds  C I 1 .

(4.41)

0

Multiplying (4.30) by (1 + t )2 , integrating over [0, t ] then integration by parts and estimating the right-hand side we obtain for any δ > 0

1 2



2 (1 + t ) ut (t ) + 2

t  (1 + s)2 0 Ω

t

1

= u 1 2 + 2

3  

Ai j

i , j =1

 2 (1 + s)u s (s) ds − (1 + t )2

0

t  +2



 utt (t ) · ut (t ) dx + Ω

t (1 + s)u ss (s) · u s (s) dx ds +

0 Ω

 ∂ us ∂ us (s) · (s) dx ds ∂xj ∂ xi

0

u 2 · u 1 dx Ω



2 (1 + s) u ss (s) ds − γ 2

t  (1 + s)2 E s (s) · curl u s (s) dx ds 0 Ω

478

C.R. da Luz, G.P. Menzala / J. Math. Anal. Appl. 359 (2009) 464–481

t  C I1 + C

   2 2 1 2 (1 + s)u s (s) ds + (1 + t )2 utt (t ) + (1 + t )2 ut (t ) + C

t

4

0

t

+ γ δ −1

 2 (1 + s)2 u ss (s) ds

0

 2 (1 + s)2  E s (s) ds + γ δ

t

0

 2 (1 + s)2 curl u s (s) ds.

(4.42)

0

Using Theorem 4.1, (4.41), (4.12) and choosing δ > 0 sufficiently small we deduce from (4.42)

t  (1 + s)2

3  

Ai j

i , j =1

0 Ω

 ∂ us ∂ us (s) · (s) dx ds  C I 1 . ∂xj ∂ xi

(4.43)

Next, we use estimates (4.41), (4.43) and choosing δ > 0 sufficiently small we obtain from (4.29)

t

  2 2 σ 1 (1 + s)2  H s (s) L 2 (Ω;μ) ds + (1 + t )2  E (t )  C I 1 + (1 + t )3 4

4

0

  3 

Ai j

Ω i , j =1

 ∂ ut ∂ ut (t ) · (t ) dx. ∂xj ∂ xi

(4.44)

Now we use the information obtained in (4.44), (4.43) and (4.41) into (4.25) to deduce



2 1 (1 + t ) utt (t ) + (1 + t )3 3

t +2

4

  3  Ω i , j =1



2 (1 + s) u ss (s) ds + 2σ 3

0

t

   2 2 ∂ ut ∂ ut Ai j (t ) · (t ) dx + (1 + t )3  E t (t )L 2 (Ω; ) + (1 + t )3  H t (t )L 2 (Ω;μ) ∂xj ∂ xi

  2 2 3σ (1 + s)3  E s (s) ds + (1 + t )2  E (t )  C I 1 . 4

(4.45)

0

Finally, we want to obtain an estimate for u (t ) in the norm [ L 2 (Ω)]3 . Multiplying (4.17) by (1 + t ) and integrating by parts we obtain the estimate

1 2

 2 (1 + t )u (t ) +

(1 + s)

t

  u (s)2 ds + C

0

+ γ δ −1

t

Ai j

i , j =1

0 Ω

 C I1 + C

3  

t 

t

 ∂u ∂u (s) · (s) dx ds ∂xj ∂ xi

 2  2 1  2 (1 + s)u s (s) ds + (1 + t )ut (t ) + (1 + t )u (t ) 4

0

 2 (1 + s) E (s) ds + γ δ

0

t

 2 (1 + s)curl u (s) ds.

(4.46)

0

Using Theorem 4.1, (4.12) and choosing δ > 0 sufficiently small we obtain from (4.46)

1 4

 2 1 (1 + t )u (t ) +

t  (1 + s)

2

3   i , j =1

0 Ω

 t  2 ∂u ∂u Ai j (s) · (s) dx ds  C I 1 + C u (s) ds. ∂xj ∂ xi

(4.47)

0

Let us define

t V (t ) =

u (s) ds 0

then V solves of the following system:

V tt − L V + V t + γ curl W = u 0 + u 1 V (0) = 0, V t (0) = u 0 V =0

in Ω × (0, +∞),

in Ω,

(4.48) (4.49)

on ∂Ω × (0, +∞).

(4.50)

We take the inner product in [ L (Ω)] of Eq. (4.48) with V t (t ) to obtain 2

3

C.R. da Luz, G.P. Menzala / J. Math. Anal. Appl. 359 (2009) 464–481

  3 

1 d 2 dt

=

  V t (t )2 + 1 d

2 dt

d



Ω i , j =1

479

   2 ∂V ∂V   Ai j (t ) · (t ) dx + V t (t ) + γ curl W (t ) · V t (t ) dx ∂xj ∂ xi Ω

(u 0 + u 1 ) · V (t ) dx.

dt

(4.51)

Ω

Integrating (4.51) over [0, t ] and substitution (4.6) give us

1 2

  V t (t )2 + 1 2

  3 

Ai j

Ω i , j =1

t

1

= u 0 2 + γ



2

 t  2 ∂V ∂V (t ) · (t ) dx +  V s (s) ds ∂xj ∂ xi 0



F s (s), V s (s)

t  ds − γ L 2 (Ω;μ)

0

 C I 1 + γ δ −1

t

   H (s)22 L

ds + C δ (Ω;μ)

2

Ω i , j =1

 (u 0 + u 1 ) · V (t ) dx Ω

   V s (s)2 ds − γ

0

Using Theorem 4.1, (3.2), the assumption obtain

  3 

ds

 [μ H 0 ] · V (s) dx ds +

0 Ω

t

0

1

d



 [μ H 0 ] · V (t ) dx + Ω

(u 0 + u 1 ) · V (t ) dx. Ω

μ H 0 = curl ψ0 for some ψ0 ∈ H 0 (curl; Ω) and choosing δ sufficiently small we

  t   ∂V ∂V 1   V s (s)2 ds  C I 1 − γ ψ0 · curl V (t ) dx + (u 0 + u 1 ) · V (t ) dx. Ai j (t ) · (t ) dx + ∂xj ∂ xi 2 0

Ω

Ω

Thus from (4.12), we conclude that

1 4

  3  Ω i , j =1

  t  ∂V 1  ∂V  V s (s)2 ds  C I 1 + (u 0 + u 1 ) · V (t ) dx. Ai j (t ) · (t ) dx + ∂xj ∂ xi 2 0

(4.52)

Ω

We use the well-known Sobolev inequality when n = 3 (see [1])

v 2L 6 (Ω)  C ∇ v 2 . Thus, from (4.52) we obtain

1 4

  3  Ω i , j =1

Ai j

 t   2 ∂V ∂V 1   V s (s)2 ds  C I 1 + δ −1 u 0 + u 1 2 6/5   (t ) · (t ) dx + 3 + δ V (t ) [ L 6 (Ω)]3 [ L (Ω)] ∂xj ∂ xi 2 0

 2  C 1 I 1 + δ −1 u 0 + u 1 2[ L 6/5 (Ω)]3 + C 1 δ ∇ V (t ) .

Using assumption (2.1), the fact that V t (t ) = u (t ) and choosing δ sufficiently small we obtain

t

  u (s)2 ds  C I 1 .

0

Returning to (4.47) we conclude that

 2 (1 + t )u (t ) +

t  (1 + s) 0 Ω

3   i , j =1

Ai j

 ∂u ∂u (s) · (s) dx ds  C I 1 ∂xj ∂ xi

(4.53)

for some positive constant C . To conclude the proof of Theorem 4.1 remains only to show that

(1 + t )2

  3  Ω i , j =1

Ai j

  2 ∂u ∂u (t ) · (t ) dx + (1 + t )3 ut (t )  C I 1 , ∂xj ∂ xi

∀t > 0.

480

C.R. da Luz, G.P. Menzala / J. Math. Anal. Appl. 359 (2009) 464–481

Taking the inner product of (1.1) with ut (t ) we obtain

  3 

1 d 2 dt

 ut (t )2 + 1 d

2 dt

Ai j

Ω i , j =1

   2 ∂u ∂u (t ) · (t ) dx + ut (t ) + γ curl E (t ) · ut (t ) dx = 0. ∂xj ∂ xi

(4.54)

Ω

Multiplying (4.54) by (1 + t ) , integrating by parts over [0, t ], using Theorem 4.1, (4.53) and Eq. (1.3) we estimate the right-hand side to obtain 2

1 2



2 1 (1 + t ) ut (t ) + (1 + t )2 2

2

t

  3  Ω i , j =1

 t  2 ∂u ∂u Ai j (t ) · (t ) dx + (1 + s)2 u s (s) ds ∂xj ∂ xi 0

 (1 + s)2 curl E (s) · u s (s) dx ds

 C I1 − γ 0 Ω

 C1 I1 + γ δ

−1

t



2

(1 + s)

2 H s (s) L 2 (Ω;μ) ds + C 1 δ

0

t

 2 (1 + s)2 u s (s) ds.

0

Using (4.44) and (4.45) we conclude

1 2

2

(1 + t )

  3  Ω i , j =1

 t  2 ∂u ∂u 1 Ai j (t ) · (t ) dx + (1 + s)2 u s (s) ds  C I 1 . ∂xj ∂ xi 2

(4.55)

0

Multiplying (4.30) by (1 + t ) and integrating by parts we obtain the estimate 3

1 2



2 (1 + t ) ut (t ) + 3

t  (1 + s)3 0 Ω

t  C I1 + C

3  

Ai j

i , j =1

 ∂ us ∂ us (s) · (s) dx ds ∂xj ∂ xi



  2 2 1 2 (1 + s) u s (s) ds + (1 + t )3 utt (t ) + (1 + t )3 ut (t ) + C 2

4

0

+ γ δ −1

t

t

 2 (1 + s)3 u ss (s) ds

0

 2 (1 + s)3  E s (s) ds + γ δ

0

t

 2 (1 + s)3 curl u s (s) ds.

0

Using (4.55), (4.45) and (4.12) we deduce

 2 (1 + t )3 ut (t )  C I 1 .

(4.56)

All conclusions of Theorem 4.2 follow from estimates (4.45), (4.53), (4.55), (4.56), Theorem 4.1 and Eqs. (1.1)–(1.3).

2

6/ 5 3 2 Remark 4.1. The  assumption (u 0 + u 1 ) ∈ [ L (Ω)] in Theorem 4.2 used to prove the L -decay of u (t ) can be replace by the condition Ω |x|2 |u 0 + u 1 |2 dx to prove the same type of result. This new assumption would allow us to use Hardy’s inequality as a crucial step.

5. Summary section Motivated by the interaction of two distinct physical phenomenons (mechanical and electromagnetic) outside a compact body, we considered the linearly coupled system (1.1)–(1.6). After proving well-posedness of the model we obtain new results on the asymptotic behavior of the total energy using multiplicative techniques. Uniform rates of decay of polynomial type are obtained. Some other terms belonging to higher order energies also decay polynomially, some of them at higher rates. Due to the anisotropic case we are treating, we can not reduce (in general) the Maxwell equations to a second order vector wave equation for which a large amount of results are available. The decay of the L 2 -norm of the displacement vector u (x, t ) is not part of the total energy but still is an important physical information especially in unbounded domains. In Theorem 4.2 we obtained a uniform rate of decay of such norm. This result is also new. The main difficulties we found were due to the fact that both equations in the model are anisotropic. Besides, a boundary condition on the magnetic field H is not present in the model. This paper improves results given in [6]. Most of the known results on the subject are of local type [2,9] or in bounded domains [5,10,11,13]. Future directions of research would be: Some applications of this

C.R. da Luz, G.P. Menzala / J. Math. Anal. Appl. 359 (2009) 464–481

481

article would be to consider nonlinear perturbations of type f (ut , ∇ u , curl E , curl H ) and study the behavior of the energy as t → +∞ for small initial data. Also, the possibility to study model (1.1)–(1.6) considering only one global dissipation or two localized dissipations could be interesting. We could consider thermal effects and study the asymptotic behavior of the corresponding total energy as t → +∞. There are no results available in the literature for those kind of questions. Acknowledgments The second author (G.P.M.) was partially supported by a Research Grant of CNPq (Proc. 306282/2003-8) and Project Universal (Proc. 474296/2008-3) from the Brazilian Government (MCT, Brazil). He would like to express his gratitude for such important support.

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