Energy decay for Maxwell's equations with Ohm's law in partially cubic domains

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Energy decay for Maxwell’s equations with Ohm’s law in partially cubic domains

Kim Dang Phung

To cite this version:

Kim Dang Phung. Energy decay for Maxwell’s equations with Ohm’s law in partially cubic domains.

Communications on Pure and Applied Analysis, AIMS American Institute of Mathematical Sciences,

2013, 12 (5), pp.2229-2266. �10.3934/cpaa.2013.12.2229�. �hal-00845761�

Energy decay for Maxwell’s equations with Ohm’s law in partially cubic domains

Kim Dang Phung

Université d’Orléans, Laboratoire MAPMO, CNRS UMR 7349, Fédération Denis Poisson, FR CNRS 2964,

Bâtiment de Mathématiques

B.P. 6759, 45067 Orléans Cedex 2, France.

E-mail: kim_dang_phung@yahoo.fr

Abstract .- We prove a polynomial energy decay for the Maxwell’s equations with Ohm’s law in partially cubic domains with trapped rays. We extend the results of polynomial decay for the scalar damped wave equation in partially rectangular or cubic domain. Our approach have some similitude with the construction of re‡ected gaussian beams.

Keywords .- Maxwell’s equations; decay estimates; trapped ray.

1 Introduction and main result

The problems dealing with Maxwell’s equations with nonzero conductivity are not only theoretical interesting but also very important in many industrial applications (see e.g. [3], [7], [8], [14]). This paper is concerned with the energy decay of Maxwell’s equations with Ohm’s law in a bounded cylinder A R ³ with trapped rays. Precisely, let > 0 and D be an open simply connected bounded set in R ² with C ² boundary @D. Consider A = D 2(0; ) with boundary @A = 0 0 [ 0 1 where 0 0 = D 2f0; g and 0 1 = @D 2 ( 0 ; ). The domain A is occupied by an electromagnetic medium of constant electric permittivity " o and constant magnetic permeability _o . For the sake of simplicity, we assume from now that " o _o = 1.

Let E and H denote the electric and magnetic …elds respectively. De…ne the energy by E (t) = 1

2 Z

A

" o j E (x; t) j ² + _o j H (x; t) j ²

dx . (1.1)

The Maxwell’s equations with Ohm’s law are described by 8 >

> >

> <

> >

> >

:

" o @ t E 0 curl H + E = 0 in A 2 [0; + 1 ) _o @ t H + curl E = 0 in A 2 [0; + 1 ) div ( _o H) = 0 in A 2 [0; + 1 ) E 2 = H 1 = 0 on @A 2 [0; + 1 ) (E; H) ( 1 ; 0) = (E o ; H o ) in A .

(1.2)

Here, (E o ; H o ) is the initial data in the energy space L ² (A) ⁶ and denotes the outward unit normal

vector to @A. The conductivity is such that 2 L ¹ (A) and 0. It is well-known that when

the conductivity is identically null, then the above system is conservative and when is bounded from below by a positive constant, then an exponential energy decay rate holds for the Maxwell’s equations with Ohm’s law in the energy space. The situation becomes more delicate if the conductivity is locally active, i.e., when the following constraint holds

= 1 _j ! a where ! A; a 2 L ¹ (A) and a constant > 0 .

Here and hereafter, we denote by 1 _j1 the characteristic function of a set in the place where 1 stays.

From the works of Bardos, Lebeau and Rauch [2] for the scalar wave operator, there is a geometric condition on the location of ! which implies an exponential decay rate for the Maxwell’s equations with Ohm’s law in the energy space. In our paper, ! is a non-empty connected open subset ! of A as follows. Consider

! =

x ⁰ 2 D; inf

y

⁰

2 @D j x ⁰ 0 y ⁰ j < r o

2 ( 0 ; ) for some small r o 2 (0; =2) .

Such case of ! does not satisfy the geometric condition of the works of Bardos, Lebeau and Rauch [2]

and we do not hope an exponential decay rate for the the Maxwell’s equations with Ohm’s law in the energy space.

Our geometry presents parallel trapped rays and can be compared to the one in [12] or in [4],[10]

for the two dimensional case. It generalizes the cube (see [8]) and therefore explicit and analytical results are harder to obtain. Our main result is as follows.

Theorem 1 .- If = 1 _j ! a where a 2 L ¹ (A) and a constant > 0, then there exist c > 0 and D > 0 such that for any t 0

E (t) c t ^D

E (0) + k (curl E o ; curl H o ) k ² L

²

(A)

⁶

for every solution of the system (1.2) of Maxwell’s equations with Ohm’s law with initial data (E o ; H o ) 2 L ² (A) ⁶ such that 8

> >

<

> >

:

(curl E o ; curl H o ) 2 L ² (A) ⁶ , div H o = 0 in A E o 2 = H o 1 = 0 on @A

div 0

1 _j A n ! E o 1

= 0 in A n! .

In literature, the exponential energy decay rate of a linear dissipative system can be deduced from an observability estimate with (1.1). Precisely, in order to get an exponential decay rate in the energy space we should have the following observability inequality

9 C; T c > 0 8 0 Z

A j (E; H) ( 1 ; ) j ² dx C Z +T

c

Z

A

j E j ² dxdt or simply, in virtue of a semigroup property,

9C; T c > 0 Z

A

j(E o ; H o )j ² dx C Z T

c

0

Z

A

jEj ² dxdt

for any initial data (E o ; H o ) in the energy space L ² (A) ⁶ . We can also look for establishing the

above observability inequality for any initial data in the energy space intersecting suitable invariant

subspaces. Such estimate is established in [11] under the geometric control condition of Bardos, Lebeau

and Rauch [2] for the scalar wave operator on (A; !) and when the conductivity has the property

that (x) constant > 0 for all x 2 ! and (x) = 0 for all x 2 A n! . Now, our main result

gives a polynomial energy decay with regular initial data when the conductivity is only active on a

neighborhood of the lateral boundary. Our proof is based on a new kind of observation inequality (see

(4.1) below) which can also be seen as an interpolation estimate. It relies with the construction of a

particular solution for the operator i@ s + h 0

1 0 @ ² _t 1

inspired by the gaussian beam techniques (see [13] and [15]).

The plan of the paper is as follows. In the next section, we recall the known results about the Maxwell’s equations with Ohm’s law that will be used. Section 3 contains the proof of our main result, while Section 4 is concerned with the interpolation estimate. In Section 5, we present an interpolation estimate with weight functions, while Section 6 includes its proof. Finally, two appendix are added dealing with inequalities involving Fourier analysis and the Poisson summation formula.

2 The Maxwell’s equations with Ohm’s law

We begin to recall some well-known results concerning the Maxwell’s equations with Ohm’s law: well- posedness, energy identity, standard orthogonal decomposition and asymptotic behaviour in time of the energy of the electromagnetic …eld.

2.1 Well-posedness of the problem

Let us introduce the spaces

V = L ² (A) ³ 2 n

G 2 L ² (A) ³ ; div G = 0; G 1 _j @A = 0 o

, (2.1.1)

W = n

(F; G) 2 L ² (A) ⁶ ; curl F 2 L ² (A) ³ ; F 2 _j @A = 0;

div G = 0; G 1 _j @A = 0; curl G 2 L ² (A) ³ o

. (2.1.2)

It is well-known that if (E o ; H o ) 2 V , there is a unique weak solution (E; H) 2 C ⁰ ([0; + 1 ) ; V ).

Further, if (E o ; H o ) 2 W , there is a unique strong solution (E; H) 2 C ⁰ ([0; + 1 ) ; W ) \ C ¹ ([0; + 1 ) ; V ).

We can easily check that the energy E , de…ned by (1.1), is a continuous positive non-increasing real function on [0; + 1 ). Further for any initial data (E o ; H o ) 2 W and any t 2 > t 1 0,

E (t 2 ) 0 E (t 1 ) + Z t

2

t

1

Z

A

(x) j E (x; t) j ² dxdt = 0 (2.1.3) and

E 1 (t 2 ) 0 E 1 (t 1 ) + Z t

2

t

1

Z

A

(x) j @ t E (x; t) j ² dxdt = 0 , (2.1.4) where

E 1 (t) = 1 2

Z

A

" o j@ t E (x; t)j ² + _o j@ t H (x; t)j ²

dx . (2.1.5)

2.2 Orthogonal decomposition

Both E and _o H can be described, by means of the scalar and vector potentials p and A with the Coulomb gauge, in a unique way as follows.

Proposition 2.1 -. For any (E o ; H o ) 2 W , there is a unique (p; A) 2 C ¹ 0

[0; + 1 ) ; H ₀ ¹ (A) 1 2 C ²

[0; + 1 ) ; H ¹ (A) ³

such that the solution (E; H ) of the Maxwell’s equations with Ohm’s law (1.2)

satis…es

E = 0r p 0 @ t A

_o H = curl A (2.2.1)

8 <

:

" o _o @ _t ² A + curl curl A = _o (0" o @ t rp + E) in A 2 [0; +1) div A = 0 in A 2 [0; +1) A 2 = 0 on @A 2 [0; +1)

(2.2.2) and we have the following relations

kEk ² L

²

(A)

³

= krpk ² L

²

(A)

³

+ k@ t Ak ² L

²

(A)

³

, (2.2.3) k " o @ t r p k L

²

(A)

³

k E k L

²

(A)

³

, (2.2.4) 9 c > 0 k A k ² L

²

(A)

³

c k curl A k ² L

²

(A)

³

. (2.2.5) Further, since curl H 2 L ² (A) ³ , curl curl A 2 L ² (A) ³ and div A 2 H ₀ ¹ (A).

The proof is essentially given in [11, p. 121] from a Hodge decomposition and is omitted here.

Now, the vector …eld A has the nice property of free divergence and satis…es a second order vector wave equation with homogeneous boundary condition A 2 = div A = 0 and with a second member in C ¹

[0; + 1 ) ; L ² (A) ³

bounded by 2 _o k E k L

²

(A)

³

.

2.3 Invariant subspaces, asymptotic behavior and exponential energy decay

Let ! + be a non-empty connected open subset of A with a Lipschitz boundary @! + . In this subsection we suppose that

= 1 _j !

+

a where a 2 L ¹ (A) and a constant > 0 .

Denote ! ₀ = A n ! + and suppose that its boundary @! ₀ is Lipschitz and has no more than two connected components D ₁ ; D ₂ .

We recall that the range of the curl, curl H ¹ (! ₀ ) ³ , is closed in L ² (! ₀ ) ³ (see [6, p. 257] or [5, p.

54]) and

curl H ¹ (! ₀ ) ³ = (

U 2 L ² (! ₀ ) ³ ; div U = 0 in ! ₀ ; Z

D

_i

U 1 = 0 for i 2 f 1; 2 g )

. (2.3.1)

Its orthogonal space for the 0

L ² (! ₀ ) 1 3

norm is curl H ¹ (! ₀ ) ³ ?

= n

V 2 L ² (! ₀ ) ³ ; curl V = 0 in ! ₀ ; V 2 = 0 on @! ₀ o

. (2.3.2)

Let us introduce S =

curl H ¹ (! ₀ ) ³ \ L ² (A) ³

2 L ² (A) ³ . The space W \ S is stable for

the system of Maxwell’s equations with Ohm’s law, which can be seen by multiplying by V 2

curl H ¹ (! ₀ ) ³ ?

the equation " o @ t E 0 curl H + E = 0. Then, we can add the following well- posedness result. If (E o ; H o ) 2 W \ S , there is a unique solution (E; H) 2 C ⁰ ([0; + 1 ) ; W \ S ) \ C ¹ ([0; + 1 ) ; V \ S ).

It has been proved (see [11, p. 124]) that if ! ₀ is a non-empty connected open set then lim

t ! + 1 E (t) = 0 for any initial data (E o ; H o ) 2 V \ S . Further, the following result (see [11, p. 124]) plays a key role.

Proposition 2.2 -. If ! ₀ is a non-empty connected open set, then there exists c > 0 such that for all initial data (E o ; H o ) 2 W \ S of the system (1.2) of Maxwell’s equations with Ohm’s law, we have

8t 0 E (t) cE 1 (t) . (2.3.3)

Remark 2.3 -. The estimate (2.3.3) is still true under the assumption @! + \ @A 6= ; and without adding the hypothesis saying that ! ₀ is a connected set. Indeed, the proof given in [11, p. 127] can be divided into two steps. In the …rst step, we begin to establish the existence of c > 0 such that

kr p k ² L

²

(!

+

)

³

+ k @ t A k ² L

²

(A)

³

+ k H k ² L

²

(A)

³

c

E 1 (t) + p E (t) p

E 1 (t)

(2.3.4) for any (E o ; H o ) 2 W \ S . Here, we used a standard compactness-uniqueness argument for H , (2.2.5) of Proposition 2.1 for @ t A, and for r p from the fact that (x) constant > 0 for all x 2 ! + and (2.1.5). Till now, we did not utilize that ! ₀ is a connected set. The second step (see [11, p. 128]) did consist to prove that

krpk ² _L

²

_(!

₀

₎

³

c

krpk ² _L

²

_(!

₊

₎

³

+ k@ t Ak ² _L

²

_(A)

³

. (2.3.5)

Finally, we concluded by virtue of (2.2.3) of Proposition 2.1. This last estimate becomes easier to obtain under the assumption @! + \ @A 6 = ; and without adding the hypothesis saying that ! ₀ is a connected set. Indeed, since (E o ; H o ) 2 W \ S and 0 1p = div E, p 2 H ₀ ¹ (A) solves the following

elliptic system 8

<

:

1p = 0 in ! ₀ p _{j @!}

₀

2 H ¹⁼² (@! ₀ ) p = 0 on @! + \ @A .

(2.3.6) Thus, by the elliptic regularity, the trace theorem and the Poincaré inequality, we have the following estimate

krpk L

²

(!

₀

)

³

c 1 kpk H

¹⁼²

(@!

₀

) c 2 kpk H

¹⁼²

(@!

+

) c 3 krpk L

²

(!

+

)

³

(2.3.7) for suitable constants c 1 ; c 2 ; c 3 > 0. Hence, combining (2.3.4) and (2.3.7) with (2.2.5), (2.3.3) follows if @! + \ @A 6 = ; and without adding the hypothesis saying that ! ₀ is a connected set.

The exponential energy decay rate for the Maxwell’s equations with Ohm’s law in the energy space is as follows.

Proposition 2.4 -. Assume that

(i) = 1 _j !

+

a where a 2 L ¹ (A) and a constant > 0;

(ii) @! + \ @A 6 = ; or ! ₀ is a non-empty connected open set;

(iii) at any point in ! ₀ and any direction, the generalized ray of the scalar wave operator @ _t ² 0 1 is uniquely de…ned.

Let # be a subset of A such that any generalized ray of the scalar wave operator @ _t ² 0 1 meets #.

If # \ A ! + , then there exist c > 0 and C > 0 such that for any t 0, we have

E (t) ce ^{0 Ct} E (0) , (2.3.8)

for every solution of the system (1.2) of Maxwell’s equations with Ohm’s law with initial data (E o ; H o ) 2

V \ S .

The proof of Proposition 2.4 is done in [11, p. 129] when ! ₀ is a non-empty connected open set.

Here, we only recall the key points of the proof. From the geometric control condition, the following estimate holds without using the fact that ! ₀ is a non-empty connected open set.

9 C; T c > 0 8 0 E 1 () C Z T

c

+

Z

A

(x) j @ t E (x; t) j ² + (x) j E (x; t) j ²

dxdt (2.3.9) for any initial data (E o ; H o ) 2 W \ S . Next by (2.3.3), we deduced that

9 C; T c > 0 8 0 E () + E ¹ () C Z T

c

+

Z

A

(x) j @ t E (x; t) j ² + (x) j E (x; t) j ² dxdt .

(2.3.10) Finally, we concluded by virtue of a semigroup property with (2.1.3) and (2.1.4). Notice that no assumption div (E ( 1 ; )) = 0 in A is set up. This is important because the free divergence of the electric …eld is not preserved by the Maxwell’s equations with Ohm’s law. Now, thanks to Remark 2.3, the proof works as well when @! + \ @A 6 = ; and without adding the hypothesis saying that ! ₀ is a connected set

3 Proposition 3.1 and proof of Theorem 1

Let us consider the solution U of the following system 8 >

> >

> >

> <

> >

> >

> >

:

@ _t ² U + curl curl U = 0 in A 2 R div U = 0 in A 2 R U 2 = 0 on @A 2 R (U (1; 0) ; @ t U (1; 0)) = 0

U ⁰ ; U ¹ 1

in A , 0 U ⁰ ; U ¹ 1

2 X , 0 U ¹ ; curl curl U ⁰ 1

2 X ,

(3.1)

where

X = n

(F; G) 2 L ² (A) ⁶ ; curl F 2 L ² (A) ³ ; F 2 _j @A = 0; div F = 0; div G = 0 o

. (3.2) It is well-known that the above system is well-posed with a unique solution U so that (U ( 1 ; t) ; @ t U ( 1 ; t)) and 0

@ t U ( 1 ; t) ; @ _t ² U ( 1 ; t) 1

belong to X for any t 2 R. Let us de…ne the following two conservations of energies.

G (U; 0) = G (U; t) Z

A

j @ t U (x; t) j ² + j curl U (x; t) j ²

dx , (3.3)

G (@ t U; 0) = G (@ t U; t) Z

A

j curl curl U (x; t) j ² + j curl @ t U (x; t) j ²

dx . (3.4)

Further, for such solution U , the following two inequalities hold by standard compactness-uniqueness argument and classical embedding (see [1] and [5, p. 50]).

G (U; t) cG (@ t U; t) , (3.5)

k U ( 1 ; t) k ² H

¹

(A)

³

c k curl U ( 1 ; t) k ² L

²

(A)

³

, (3.6) for some c > 0 and any t 2 R.

Let # r

o

(@D) be an r o -neighborhood of the boundary of D, that is

# r

o

(@D) =

x ⁰ 2 D; inf

y

⁰

2 @D jx ⁰ 0 y ⁰ j < r o

for some small r o 2 (0; =2) . (3.7)

Recall that = 1 _j ! a where

! = # r

o

(@D) 2 (0; ) . (3.8)

Consider

! o = (D n # r

o

(@D) ) 2 ( 0 2r o ; 0 r o ) . (3.9) Proposition 3.1 -. For any T o > 0, there exist c; D > 0 such that for any h > 0, the solution U of (3.1) satis…es

Z T

o

0

Z

!

o

j @ t U (x; t) j ² dxdt c 1 h ^D

Z c

_hD¹

0 c

_hD¹

Z

!

j @ t U (x; t) j ² + j U (x; t) j ²

dxdt + h G (@ t U; 0) . (3.10)

We shall leave the proof of Proposition 3.1 till later (see Section 4). Now we turn to prove Theorem 1.

Remark that (E; H ) ( 1 ; 0) 2 W \S and therefore from the previous section for any t 0, (E; H) ( 1 ; t) 2 W \ S .

Step 1 .- Denote (E 1 ; H 1 ) the electromagnetic …eld of the following Maxwell’s equations with Ohm’s

law 8

> >

> >

<

> >

> >

:

" o @ t E 1 0 curl H 1 + 0

+ 1 _j !

o

1 E 1 = 0 in A 2 [0; +1) _o @ t H 1 + curl E 1 = 0 in A 2 [0; + 1 ) div ( _o H 1 ) = 0 in A 2 [0; + 1 ) E 1 2 = H 1 1 = 0 on @A 2 [0; + 1 ) (E 1 ; H 1 ) ( 1 ; T h ) = (E; H) ( 1 ; T h ) in A .

(3.11)

Notice that (E; H ) ( 1 ; T h ) 2 W \S 0

V \ S +1

_j!o

1

. Therefore by Proposition 2.4, there exist c; C > 0 (independent of T h ) such that for any t 0 we have

Z

A

" o j E 1 (x; t + T h ) j ² + _o j H 1 (x; t + T h ) j ²

dx ce ^{0 Ct} E (T h ) . (3.12) On the other hand, let (E 2 ; H 2 ) = (E 1 ; H 1 ) 0 (E; H). Then it solves

8 >

> >

> <

> >

> >

:

" o @ t E 2 0 curl H 2 + 0

+ 1 _j !

o

1 E 2 = 0 1 _j !

o

E in A 2 [0; + 1 ) _o @ t H 2 + curl E 2 = 0 in A 2 [0; + 1 ) div ( _o H 2 ) = 0 in A 2 [0; + 1 ) E 2 2 = H 2 1 = 0 on @A 2 [0; +1) (E 2 (1; T h ) ; H 2 (1; T h )) = (0; 0) in A ,

(3.13)

and by a standard energy method, we get that for any t 0 Z

A

" o j E 2 (x; t + T h ) j ² + _o j H 2 (x; t + T h ) j ² dx t

" o

Z t+T

h

T

h

Z

!

o

j E (x:s) j ² dxds . (3.14) Now we are able to bound the quantity E (T h ) = E (t + T h ) + R t+T

h

T

h

R

A (x) j E (x; s) j ² dxds as follows.

By using (1.1), (2.1.3), (3.12) and (3.14), we deduce that E (T h ) 2ce ^{0 Ct} E (T h ) + 2t

" o

Z t+T

h

T

h

Z

!

o

jE (x:s)j ² dxds + Z t+T

h

T

h

Z

A

(x) jE (x; s)j ² dxds (3.15) which implies by taking t large enough, the existence of constants C; T c > 1 such that

E (T h ) C

Z T

c

+T

h

T

h

Z

A

(x) j E (x; t) j ² dx + Z

!

o

j E (x; t) j ² dx

dt . (3.16)

Step 2 .- Recall the existence of the vector potential A from Proposition 2.1 and let U be the

solution of 8

> >

<

> >

:

@ _t ² U + curl curl U = 0 in A 2 R div U = 0 in A 2 R U 2 = 0 on @A 2 R (U; @ t U ) ( 1 ; 0) = (A; @ t A) ( 1 ; 0) in A ,

(3.17)

then by a standard energy method, for any T 1 > 0 Z T

1

0

Z

A

j @ t (U 0 A) j ² + j curl (U 0 A) j ²

dxdt T ₁ ² Z T

1

0

Z

A

_o j0 " o @ t r p + E j ² dxdt (3.18) which implies from (2.2.4) of Proposition 2.1 that

Z T

1

0

Z

A

j @ t (U 0 A) j ² + j curl (U 0 A) j ²

dxdt 4 _o T ₁ ² Z T

1

0

Z

A j E j ² dxdt . (3.19) Now we are able to bound the quantity E (0) = E (T h ) + R T

h

0

R

A (x) jE (x; s)j ² dxds as follows.

Since E = 0rp + @ t (U 0 A) 0 @ t U , we deduce by (2.1.3) and (3.16) that E (0) C

Z T

c

+T

h

T

h

Z

A

j E j ² dx + Z

!

o

j0r p + @ t (U 0 A) 0 @ t U j ² dx

dt + Z T

h

0

Z

A

j E j ² dxdt C 0

1 + T _h ² 1 Z ^T

^c

^+T

^h

0

Z

A

j E j ² + jr p j ²

dxdt + C

Z T

c

+T

h

T

h

Z

!

o

j @ t U j ² dxdt ,

(3.20) where in the last line, we used (3.19). Here and hereafter, C will be used to denote a generic constant, not necessarily the same in any two places.

Step 3 .- Now we …x T h = c _h ¹

D

where c and D are given by Proposition 3.1. Taking T o = T c in Proposition 3.1, we obtain that for any h > 0,

Z T

c

0

Z

!

o

j @ t U j ² dxdt T h

Z T

h

0 T

h

Z

!

j @ t U j ² + j U j ²

dxdt + h G (@ t U; 0) . (3.21) By a translation in time and (3.4), the above inequality implies that

Z T

c

+T

h

T

h

Z

!

o

j @ t U j ² dxdt T h

Z 2T

h

0

Z

!

j @ t U j ² + j U j ²

dxdt + h G (@ t U; 0) . (3.22) But from (3.19),

Z 2T

h

0

Z

! j @ t U j ² dxdt = Z 2T

h

0

Z

! j0 E 0 r p + @ t (U 0 A) j ² dxdt CT _h ²

Z 2T

h

0

Z

A

j E j ² + jr p j ² dxdt

(3.23)

and Z 2T

h

0

Z

! j U j ² dxdt 2 Z 2T

h

0

Z

A j U 0 A j ² dxdt + 2 Z 2T

h

0

Z

! j A j ² dxdt C

Z 2T

h

0

Z

A j curl (U 0 A) j ² dxdt + 2 Z 2T

h

0

Z

! j A j ² dxdt CT _h ²

Z 2T

h

0

Z

A

jEj ² dxdt + 2 Z 2T

h

0

Z

!

jAj ² dxdt .

(3.24)

Therefore, plugging (3.23) and (3.24) into (3.22), we get Z T

c

+T

h

T

h

Z

!

o

j @ t U j ² dxdt CT _h ³ Z 2T

h

0

Z

A

j E j ² + jr p j ² dx +

Z

! j A j ² dx

dt + h G (@ t U; 0) . (3.25) Finally, combining (3.20) and (3.25), we have

E (0) CT _h ³ Z 2T

h

0

Z

A

j E j ² + jr p j ² dx +

Z

! j A j ² dx

dt + h G (@ t U; 0) . (3.26)

In conclusion, we proved that there exist c; D > 0 such that for any h > 0, the solution (E; H) of (1.2) with initial data in W \ S satis…es

E (0) c 1 h ^D

Z c

_hD¹

0

Z

A

j E j ² + jr p j ² dx +

Z

! j A j ² dx

dt + h G (@ t U; 0) . (3.27)

Step 4 .- By formula (2.1.3) and since G 0

@ t U; mc _h ¹

D

1 = G (@ t U; 0) for any m, this last inequality becomes

N E (0) 0 P

m=0;::;N 0 1

Z mc

_hD¹

0

Z

A

j E j ² dxdt

= P

m=0;::;N 0 1 E 0 mc _h ¹

D

1

c _h ¹

D

P

m=0;::;N 0 1

Z (m+1)c

_hD¹

mc

_hD¹

Z

A

j E j ² + jr p j ² dx +

Z

! j A j ² dx

dt + N h G (@ t U; 0) ,

(3.28)

for any N > 1. We choose N 2 0

c _h ¹

D

; 1 + c _h ¹

D

3 . Therefore, there exist c; D > 0 such that for any h > 0,

E (0) c Z c (

h¹

)

^D

0

Z

A

jEj ² + jrpj ² dx +

Z

!

jAj ² dx

dt + hG (@ t U; 0) . (3.29) On the other hand, since (U; @ t U ) (1; 0) = (A; @ t A) (1; 0),

G (@ t U; 0) = k curl E ( 1 ; 0) k ² L

²

(A)

³

+ k _o curl H ( 1 ; 0) k ² L

²

(A)

³

= ² _o k @ t H ( 1 ; 0) k ² L

²

(A)

³

+ k (@ t E + _o E) ( 1 ; 0) k ² L

²

(A)

³

c (E 1 (0) + E (0)) c k(E o ; H o )k ² _{D( M )}

(3.30)

where M is the m-accretive operator in V \ S with domain D ( M ) = W \ S , de…ned as follows.

k (F; G) k ² _V = " o k F k ² L

²

(A)

³

+ _o k G k ² L

²

(A)

³

, M =

1

"

o

0 _" ¹

_o

curl

1

_o

curl 0

!

. (3.31)

Therefore, combining (3.29) and (3.30), we get the existence of constants c; D > 0 such that for any h > 0,

E (0) c Z c (

h¹

)

^D

0

Z

A

j E j ² + jr p j ² dx +

Z

! j A j ² dx

dt + ch k (E o ; H o ) k ² D( M ) . (3.32) Similarly, we get the existence of constants c; D > 0 such that for any 0 and h > 0,

E () c

Z +c (

h¹

)

^D

Z

A

jEj ² + jrpj ² dx +

Z

!

jAj ² dx

dt + h k(E o ; H o )k ² D( M ) . (3.33)

Step 5 .- Denote (T (t)) _{t 0} the unique semigroup of contractions generated by 0M. First, suppose that (E o ; H o ) 2 D 0

M ³ 1

and let us de…ne the functional of energy E 2 (t) = 1

2 Z

A

" o

C C@ ² _t E (x; t) C C ² + _o C C@ ² _t H (x; t) C C ²

dx (3.34)

which satis…es

E 2 (t 2 ) 0 E 2 (t 1 ) + Z t

2

t

1

Z

A

(x) C C@ _t ² E (x; t) C C ² dxdt = 0 . (3.35)

Let X o = 0M ² (E o ; H o ), then (T (t)) _{t 0} X o = 0

@ _t ² E; @ _t ² H 1

, kT () X o k ² _V = 2E 2 () and kX o k ² _{D( M )} c k(E o ; H o )k ² D( M

³

) . Further, by uniqueness of the orthogonal decomposition in (2.1.1) of Proposition 2.1, (3.33) implies that for any (E o ; H o ) 2 D 0

M ³ 1 E 2 () c

Z +c (

h¹

)

^D

Z

A

C C@ ² _t E C C ² + C C@ _t ² r p C C ² dx +

Z

!

C C@ _t ² A C C ² dx

dt + ch k (E o ; H o ) k ² D( M

³

) . (3.36) Since by Proposition 2.2, E () cE 1 () and in a similar way E 1 () cE 2 () for some c > 0, taking account of the …rst line of (2.2.1) and (2.2.4), (3.36) becomes

E () + E 1 () + E 2 () c

Z +c (

¹h

)

^D

Z

A

j E j ² + j @ t E j ² + C C@ ² _t E C C ²

dxdt + h k (E o ; H o ) k ² D( M

³

) . (3.37) Step 6 .- Denote

H () = E 2 () + E 1 () + E () k (E o ; H o ) k ² D( M

³

)

. (3.38)

By (2.1.3), (2.1.4) and (3.35), the function H is a continuous positive decreasing real function on [0; +1), bounded by one. Taking h = ¹ ₂ H () in (3.37), we get the existence of constants c; D > 0 such that for any 0,

E () + E 1 () + E 2 () c

Z +c (

_H¹()

)

^D

Z

A

jEj ² + j@ t Ej ² + C C@ _t ² E C C ²

dxdt , (3.39) that is, using (2.1.3), (2.1.4) and (3.35),

H () c

H () 0 H

c

H () D

+

8 0 . (3.40)

From [12, p. 122, Lemma B], we deduce that there exist C; D > 0 such that for any t > 0 E (t) + E 1 (t) + E 2 () C

t ^D k (E o ; H o ) k ² D( M

³

) , (3.41) that is

kT (t) Y o k ² D( M

²

) C

t ^D k Y o k ² D( M

³

) 8 Y o 2 D 0 M ³ 1

. (3.42)

Since M is an m-accretive operator in V \ S with dense domain, one can restrict it to D 0 M ² 1

in a way that its restriction operator is m-accretive. Thus the following two properties holds.

8 Z o 2 D 0 M ² 1

9 !Y o 2 D 0 M ³ 1

Y o + M Y o = Z o ; (3.43)

kY o k D( M

²

) kY o + MY o k D( M

²

) 8Y o 2 D 0 M ³ 1

. (3.44)

Consequently,

kT (t) Z o k ² D( M ) = kT (t) (Y o + M Y o ) k ² D( M ) by (3.43) C kT (t) Y o k ² D( M

²

)

t ^C

^D

k Y o k ² D( M

³

) by (3.42) _t ^C

^D

k Y o k ² D( M

²

) + k Y o + M Y o k ² D( M

²

)

_t ^C

^D

k Y o + M Y o k ² D( M

²

) by (3.44) _t ^C

^D

k Z o k ² D( M

²

) by (3.43)

(3.45)

for suitable positive constant C > 0 which is not necessarily the same in any two places.

Now, suppose that (E o ; H o ) 2 D (M). Since M is an m-accretive operator in V \ S with dense domain, one can restrict it to D (M) in a way that its restriction operator is m-accretive. Thus the following two properties holds.

8 (E o ; H o ) 2 D ( M ) 9 !Z o 2 D 0 M ² 1

Z o + M Z o = (E o ; H o ) ; (3.46) k Z o k D( M ) k Z o + M Z o k D( M ) 8 Z o 2 D 0

M ² 1

. (3.47)

We conclude that

2 E (t) = kT (t) (E o ; H o ) k ² _V = kT (t) (Z o + M Z o ) k ² _V by (3.46) C kT (t) Z o k ² D( M )

_t ^C

^D

k Z o k ² D( M

²

) by (3.45) _t ^C

^D

k Z o k ² D( M ) + k Z o + M Z o k ² D( M )

_t ^C

^D

k Z o + M Z o k ² D( M ) by (3.47) _t ^C

^D

k(E o ; H o )k ² D( M ) by (3.46) _t ^C

^D

(E (0) + E 1 (0))

(3.48)

for suitable positive constant C > 0 which is not necessarily the same in any two places.

4 Proposition 4.1 and proof of propositon 3.1

Recall that ! o = (D n # r

o

(@D) ) 2 ( 0 2r o ; 0 r o ). In this Section, we establish an interpolation estimate in L ² norm in order to prove Proposition 3.1.

Proposition 4.1 -. There exist c; D > 0 such that for any T o > 0, and h 2 (0; 1], the solution U of (3.1) satis…es

e ^{0 cT}

^o2

Z T

o

0

Z

!

o

j U (x; t) j ² dxdt c 1 h ^D

Z c

_hD¹

0 c

_hD¹

Z

!

j @ t U (x; t) j ² + j U (x; t) j ²

dxdt + ch G (U; 0) . (4.1)

We shall leave the proof of Proposition 4.1 till later (see Section 5). Now we turn to prove Propo- sition 3.1 as follows.

Let 2 C ₀ ¹ (0; 5T o =3) be such that = 1 in (T o =3; 4T o =3). Then, Z 4T

o

=3

T

o

=3

Z

!

o

j @ t U (x; t) j ² dxdt

Z 5T

o

=3 0

Z

!

o

j (t) @ t U (x; t)j ² dxdt

Z 5T

o

=3 0

Z

!

o

C C2 (t) ⁰ (t) @ t U (x; t) + ² (t) @ _t ² U (x; t) C C j U (x; t) j dxdt

^{p C} _h Z 5T

o

=3

0

Z

!

o

j U (x; t) j ² dxdt + p

h G (@ t U; 0) 8 h > 0 ^{p C} _h

"

c _h ¹

D

Z c

_hD¹

0 c

_hD¹

Z

!

j @ t U (x; t) j ² + j U (x; t) j ²

dxdt + ch G (U; 0)

# + p

h G (@ t U; 0)

(4.2)

where in the last line, we used Proposition 4.1. Therefore, it implies thanks to (3.5) that for any T o > 0, there exist c; D > 0 such that for any h 2 (0; 1],

Z 4T

o

=3 T

o

=3

Z

!

o

j @ t U (x; t) j ² dxdt c 1 h ^D

Z c

_hD¹

0 c

_hD¹

Z

!

j @ t U (x; t) j ² + j U (x; t) j ²

dxdt + ch G (@ t U; 0) . (4.3)

By a translation in time and (3.3), we obtain (3.10) for any h 2 (0; 1]. Since Z T

o

0

Z

!

o

j@ t U (x; t)j ² dxdt ChG (@ t U; 0) 8h > 1 , (4.4) we get the desired estimate of Proposition 3.1.

5 Proposition 5.1 and proof of Proposition 4.1

Notice that C ₀ ¹ (B (x o ; r o =2)) C ₀ ¹ (A) for any x o 2 ! o , where B (x o ; r) denotes the ball of center x o

and radius r.

Let ` 2 C ¹ 0 R ³ 1

be such that 0 ` (x) 1, ` = 1 in R ³ n ! , ` (x) ` o > 0 for any x 2 ! o ,

` = @ ` = 0 on 0 1 and both r ` and 1` have support in !.

The proof of Proposition 4.1 comes from the following result.

Proposition 5.1 -. There exist c; E; > 0 such that for any x o 2 ! o and h 2 (0; 1], 1, the solution U of (3.1) satis…es

Z

A 2R

(x) e ⁰

¹²

(

¹h

j x 0 x

o

j

²

+t

²

) ` (x) j U (x; t) j ² dxdt c ^{p 1} G (U; 0) + ce ⁰

^ch1

_h

^E

G (U; 0) + c _h

^E

k (U; @ t U) k ² _L

2

! 2

0 c

^E_h

;c

_h^E

6

, where 2 C ₀ ¹ (B (x o ; r o =2)), 0 1.

We shall leave the proof of Proposition 5.1 till later (see Section 6). Now we turn to prove Propo- sition 4.1.

Let h 2 (0; h o ] where h o = min

1; (r o =8) ²

. We begin by covering ! o with a …nite collection of balls B

x ⁱ _o ; 2 p h

for i 2 I with x ⁱ _o 2 ! o and where I is a countable set such that the number of elements of I is _h ^c p

^o

h for some constant c o > 0 independent of h. Then, for each x ⁱ _o , we introduce _x

ⁱ_o

2 C ₀ ¹ 0

B 0

x ⁱ _o ; r o =2 11

C ₀ ¹ (A) be such that 0 _x

ⁱ_o

1 and _x

ⁱ_o

= 1 on B 0

x ⁱ _o ; r o =4 1 B

x ⁱ _o ; 2 p h

. Consequently, for any T o > 0,

e ⁰

¹²

^T

^o2

Z T

o

0

Z

!

o

jU (x; t)j ² dxdt _` ¹

_o

Z T

o

0

Z

!

o

e ⁰

¹²

^t

²

` (x) jU (x; t)j <sup>2</sup> dxdt <sub>`</sub> ¹

_o

e ² P

i 2 I

Z T

o

0

Z

B ( ^x

ⁱo

;2 p h ) _x

i

o

(x) e ⁰

¹²

1

h

j ^{x 0 x}

^io

j

²

^+t

²

` (x) j U (x; t) j <sup>2</sup> dxdt <sub>`</sub> ¹

_o

e ² P

i 2 I

Z

A 2R

_x

i

o

(x) e ⁰

¹²

1

h

j ^{x 0 x}

^io

j

²

^+t

²

` (x) j U (x; t) j ² dxdt .

(5.1) By virtue of Proposition 5.1, R

A 2R _x

i

o

(x) e ⁰

¹²

1

h

j ^{x 0 x}

^io

j

²

^+t

²

` (x) j U (x; t) j ² dxdt is bounded indepen- dently of x ⁱ _o and it implies that for some constants c; E; > 0,

e ⁰

¹²

^T

^o2

Z T

o

0

Z

!

o

jU (x; t)j ² dxdt c _h ^{p 1} _h

p 1

G (U; 0) + ce ⁰

^ch1

_h

^E

G (U; 0) + c _h

^E

k(U; @ t U )k ² _L

2

! 2

0 c

^E_h

;c

^E_h

6

.

(5.2)

Finally, we choose 1 such that = 0 h

o

h

1 ⁵

in order that ¹

h p h

p 1

Ch for some C > 0. Then there exist c; D > 0 such that for any h 2 (0; h o ],

e ⁰

¹²

^T

^o2

Z T

o

0

Z

!

o

j U (x; t) j ² dxdt c 1 h ^D

Z c

_hD¹

0 c

_hD¹

Z

!

j @ t U (x; t) j ² + j U (x; t) j ²

dxdt + ch G (U; 0) . (5.3) Since e ⁰

¹²

^T

^o2

R T

o

0

R

!

o

j U j ² dxdt Ch G (U; 0) for any h h o , the proof of Proposition 4.1 is complete.

6 Proof of Proposition 5.1

Let h 2 (0; 1], x o 2 ! o , 2 C ₀ ¹ (B (x o ; r o =2)) be such that 0 1. Let us introduce

a o (x; t) = e ⁰

¹⁴

(

h¹

j x 0 x

o

j

²

+t

²

) and ' (x; t) = (x) a o (x; t) . (6.1) By the Fourier inversion formula,

Z

A 2R

(x) e ⁰

¹⁴

(

h¹

j x 0 x

o

j

²

+t

²

) ` (x) j U (x; t) j ² dxdt

= Z

A 2R

' (x; t) a o (x; t) ` (x) jU (x; t)j ² dxdt

= Z

A 2R

1 (2)

⁴

Z

R

⁴

e ^i(x+t) d 'U (; ) dd

1 a o (x; t) ` (x) U (x; t) dxdt

= Z

A 2R 1 (2)

⁴

Z

R

³

Z

j j <

e ^i(x+t) d 'U (; ) dd

!

1 a o (x; t) ` (x) U (x; t) dxdt +

Z

A 2R 1 (2)

⁴

Z

R

³

Z

j j

e ^i(x+t) d 'U (; ) dd

!

1 a o (x; t) ` (x) U (x; t) dxdt

(6.2)

for any 1. Here we recall that F b (; ) =

Z

R

⁴

e ^{0 i(x+t)} F (x; t) dxdt and F (x; t) = 1 (2) ⁴

Z

R

⁴

e ^i(x+t) F b (; ) dd (6.3) when F and F b belong to L ¹ 0

R ⁴ 1 ³

. On the other hand, from (A1) of Appendix A, (3.6), (3.3) and the fact that each component of U solves the wave equation, we have that for any U solution of (3.1),

C C C C C Z

A 2R 1 (2)

⁴

Z

R

³

Z

j j

e ^i(x+t) d 'U (; ) dd

!

1 a o (x; t) ` (x) U (x; t) dxdt C C C C C c ^{p 1} G (U; 0) .

(6.4)

It remains to study the following quantity Z

A 2R

1 (2) ⁴

Z

R

³

Z

j j <

e ^{i(x+t )} d 'U (; ) dd

!

1 a o (x; t) ` (x) U (x; t) dxdt . (6.5) We claim that there exists c; E; > 0 such that for any x o 2 ! o and h 2 (0; 1], 1, we have

C C C C C Z

A 2R 1 (2)

⁴

Z

R

³

Z

j j <

e ^i(x+t) d 'U (; ) dd

!

1 a o (x; t) ` (x) U (x; t) dxdt C C C C C c p ¹

G (U; 0) + ce ⁰

^ch1

_h

^E

G (U; 0) + c _h

^E

k (U; @ t U ) k ² _L

2

! 2

0 c

_h^E

;c

^E_h

6

,

(6.6)

which implies Proposition 5.1 using (6.2), (6.4).

The proof of our claim is divided into many subsections. In the next subsection, we shall introduce suitable sequences of Fourier integral operators. First, we add a new variable s 2 [0; L]. Next, we construct a particular solution of the equation (6.1.10) below for (x; t; s) 2 R ⁴ 2 [0; L] with good properties on 0 0 .

6.1 Fourier integral operators

Let ' 2 C ₀ ¹ (A) and f = f (x; t) 2 L ¹ 0 R; L ² 0

R ³ 11

be such that 'f c 2 L ¹ 0 R ⁴ 1

. Let h 2 (0; 1], L 1, 1 and (x o ; _o3 ) 2 ! o 2 (2Z + 1). Denote x = (x 1 ; x 2 ; x 3 ) and x o = (x o1 ; x o2 ; x o3 ). First, let us introduce for any (x; t; s) 2 R ⁴ 2 [0; L] and n 2 Z,

(A (x o ; _o3 ; n) f ) (x; t; s)

= ₍₂₎ ¹

4

Z

R

²

Z

_o3

+1

_o3

0 1

Z

j j <

e ^i(x

¹

¹

^+x

²

²

^+t) e ⁱ

h ( 0 1)

ⁿ

x

3

+2n

^o3

jo3j

i

₃

e ^{0 i} ( j j

²

0

²

) ^hs 'f c (; ) a

x 1 0 x o1 0 2 ₁ hs; x 2 0 x o2 0 2 ₂ hs; ( 0 1) ⁿ x 3 + 2n _j

^o3

o3

j 0 x o3 0 2 ₃ hs; t + 2 hs; s dd

(6.1.1) where = ( ₁ ; ₂ ; ₃ ) 2 R ² 2 [ _o3 0 1; _o3 + 1],

a (x; t; s) = 1

(is + 1) ³⁼² e ⁰

^{4h1 jxj}

2 is+1

! 1

p 0ihs + 1 e ⁰

^{14 t}

2 0ihs+1

. (6.1.2)

Next, let us introduce for any (x; t; s) 2 R ⁴ 2 [0; L], (A (x o ; _o3 ) f ) (x; t; s) =

2P+1 X

n= 0 2Q

( 0 1) ⁿ ( A (x o ; _o3 ; n) f ) (x; t; s) , (6.1.3)

(B (x o ; _o3 ) f ) (x; t; s) =

2P+1 X

n= 0 2Q

( A (x o ; _o3 ; n) f ) (x; t; s) , (6.1.4) where (P; Q) 2 N ² is the …rst couple of integer numbers satisfying

8 <

:

P ₄ ¹ p

( j _o3 j + 2) (L ² + 1) + 2 ( j _o3 j + 1) L , Q 4 ¹

p ( j _o3 j + 2) (L ² + 1) + 2 0 r o

. (6.1.5)

We check after a lengthy but straightforward calculation that for any (x; t; s) 2 R ⁴ 2 [0; L], 0 i@ s + h 0

1 0 @ ² _t 11

(A (x o ; _o3 ) f ) (x; t; s) = 0 , 0 i@ s + h 0

1 0 @ _t ² 11

(B (x o ; _o3 ) f ) (x; t; s) = 0 , (6.1.6) and that for any (x 1 ; x 2 ; t; s) 2 R ³ 2 [0; L],

8 >

> >

<

> >

> :

(A (x o ; _o3 ) f)

x 1 ; x 2 ; _j

^o3

o3

j ; t; s

= 0 , (A (x o ; _o3 ) f )

x 1 ; x 2 ; 0 _j

^o3_o3

_j ; t; s

= (A (x o ; _o3 ; 02Q) f )

x 1 ; x 2 ; 0 _j

^o3_o3

_j ; t; s + (A (x o ; _o3 ; 2P + 1) f )

x 1 ; x 2 ; 0 _j

^o3_o3

_j ; t; s ,

(6.1.7)

8 >

> >

<

> >

> :

@ x

3

(B (x o ; _o3 ) f)

x 1 ; x 2 ; _j

^o3

o3

j ; t; s

= 0 ,

@ x

3

(B (x o ; _o3 ) f )

x 1 ; x 2 ; 0 _j

^o3_o3

_j ; t; s

= @ x

3

(A (x o ; _o3 ; 02Q) f )

x 1 ; x 2 ; 0 _j

^o3_o3

_j ; t; s 0@ x

3

(A (x o ; _o3 ; 2P + 1) f )

x 1 ; x 2 ; 0 _j

^o3_o3

_j ; t; s

.

(6.1.8)

Let f j = f j (x; t) 2 L ¹ 0 R; L ² 0

R ³ 11

be such that 'f d j 2 L ¹ 0 R ⁴ 1

for any j 2 f 1; 2; 3 g . Let us introduce

F = 0

@ f 1

f 2

f 3

1

A and V (x o ; _o3 ) F = 0

@

A (x o ; _o3 ) f 1

A (x o ; _o3 ) f 2

B (x o ; _o3 ) f 3

1

A (6.1.9)

then 0

i@ s + h 0

1 0 @ _t ² 11 (V (x o ; _o3 ) F ) (x; t; s) = 0 8 (x; t; s) 2 R ⁴ 2 [0; L] . (6.1.10) On another hand, let U be the solution of (3.1). Denote

U = 0

@ u 1

u 2

u 3

1 A then

8 <

:

8j 2 f1; 2; 3g @ _t ² u j 0 1u j = 0 in A 2 R u 1 = u 2 = 0 on 0 0 2 R

@ x

3

u 3 = 0 on 0 0 2 R

(6.1.11)

because div U = 0 and U 2 = 0. Further, by (3.6) and (3.3),

9 c > 0 k u j ( 1 ; t) k ² H

¹

(A) + k @ t u j ( 1 ; t) k ² L

²

(A) c G (U; 0) 8 j 2 f 1; 2; 3 g . (6.1.12)

By multiplying the equation (6.1.10) by ` (x) U (x; t) and integrating by parts over A 2 [ 0 T; T ] 2 [0; L], we have that for all (x o ; _o3 ) 2 ! o 2 (2Z + 1) and all h 2 (0; 1], L 1, T > 0,

0 = 0 i Z

A

Z T 0 T

(V (x o ; _o3 ) F) ( 1 ; 1 ; 0) 1 `U dxdt +i

Z

A

Z T 0 T

(V (x o ; _o3 ) F) (1; 1; L) 1 `U dxdt 0h

Z

0

0

Z T 0 T

( Z L 0

A (x o ; _o3 ) f 1 ds

!

`@ u 1 + Z L

0

A (x o ; _o3 ) f 2 ds

!

`@ u 2

) ddt +h

Z

0

0

Z T 0 T

Z L 0

@ (B (x o ; _o3 ) f 3 ) ds

!

`u 3 ddt 0 h

Z Z

0

0

\ @!

Z T 0 T

Z L 0

B (x o ; _o3 ) f 3 ds

!

@ `u 3 ddt 0h R

A

" Z L 0

@ t (V (x o ; _o3 ) F) ds

! 1 `U 0

Z L 0

(V (x o ; _o3 ) F ) ds

! 1 `@ t U

# T 0 T

dx +h

Z

!

Z T 0 T

Z L 0

V (x o ; _o3 ) F ds

!

1 [2 ( r ` 1 r ) U + 1`U ] dxdt I 1 + I 2 + I 3 + I 4 + I 5 + I 6 + I 7 .

(6.1.13)

The di¤erent terms of the last equality will be estimated separately. The quantity I ¹ will allow us to recover (6.5). The dispersion property for the one dimensional Schrödinger operator will be used for making I 2 small for large L. We treat I 3 (resp. I 4 ) by applying the formula (6.1.7) (resp. (6.1.8)).

The quantity I 5 and I 7 will correspond to a term localized in !. Finally, an appropriate choice of T will bound I 6 and give the desired inequality (6.9.2) below.

6.2 Estimate for I ¹ (the term at s = 0 )

We estimate I 1 = 0i Z

A

Z T 0 T

(V (x o ; _o3 ) F ) (x; t; 0) 1 ` (x) U (x; t) dxdt as follows.

Lemma 6.1 .- There exists c > 0 such that for any (x o ; _o3 ) 2 ! o 2 (2Z + 1) and h 2 (0; 1], 1, T > 0, we have

C C C C C I 1 + i

Z

A 2R 1 (2)

⁴

Z

R

²

Z

_o3

+1

_o3

0 1

Z

j j <

e ^{i(x+t )} 'F c (; ) dd

!

1 a (x 0 x o ; t; 0) ` (x) U (x; t) dxdt C C C C C c

e ⁰

^ch1

+ e ⁰

^Tc2

p G (U; 0)

Z

R

²

Z

_o3

+1

_o3

0 1

Z

j j <

C C

C c 'F (; ) C C C dd

! .

(6.2.1) Proof .- We start with the third component of V (x o ; _o3 ) F . First, from (6.1.1) and (6.1.4) whenever s = 0,

(B (x o ; _o3 ) f ) (x; t; 0)

=

2P+1 P

n= 0 2Q

h a

x 1 0 x o1 ; x 2 0 x o2 ; ( 0 1) ⁿ x 3 + 2n _j

^o3

o3

j 0 x o3 ; t; 0

1 (2)

⁴

Z

R

²

Z

_o3

+1

_o3

0 1

Z

j j <

e ^i(x

¹

¹

^+x

²

²

^{+t )} e ⁱ

h ( 0 1)

ⁿ

x

3

+2n

^o3

jo3j

i

₃

'f c (; ) dd

#

= a (x 0 x o ; t; 0) ₍₂₎ ¹

4

Z

R

²

Z

_o3

+1

_o3

0 1

Z

j j <

e ^i(x+t) 'f c (; ) dd

+ P

n 2f0 2Q; 111 ;2P+1 gnf 0 g

h a

x 1 0 x o1 ; x 2 0 x o2 ; (01) ⁿ x 3 + 2n _j

^o3

o3

j 0 x o3 ; t; 0

1 (2)

⁴

Z

R

²

Z

_o3

+1

_o3

0 1

Z

j j <

e ^i(x

¹

¹

^+x

²

²

^+t) e ⁱ

h ( 0 1)

ⁿ

x

3

+2n

^o3

jo3j

i

₃

'f c (; ) dd

# . (6.2.2) Next, we estimate the discrete sum over f0 2Q; 1 1 1 ; 2P + 1 g nf 0 g . By (6.1.2),

C C C C C

P

n 2f0 2Q; 111 ;2P +1 gnf 0 g

h a

x 1 0 x o1 ; x 2 0 x o2 ; (01) ⁿ x 3 + 2n _j

^o3

o3

j 0 x o3 ; t; 0

1 (2)

⁴

Z

R

²

Z

_o3

+1

_o3

0 1

Z

j j <

e ^i(x

¹

¹

^+x

²

²

^+t) e ⁱ

h ( 0 1)

ⁿ

x

3

+2n

^o3

jo3j

i

₃

'f c (; ) dd #CC C C C ₍₂₎ ¹

⁴

Z

R

²

Z

_o3

+1

_o3

0 1

Z

j j <

C C

C c 'f (; ) C C C dd e ⁰

^t42

e ⁰

^{j(x10xo1;x20xo2)j}

2

4h

P

n 2Znf 0 g

e ⁰

(01)nx3+2no3 jo3j0xo3

2

4h

.

(6.2.3) Remark that for any n 2 Z nf0g , x o3 2 [ 0 2r o ; 0 r o ] and x 3 2 [0; ],

0

(01) ⁿ x 3 + 2n _o3

j _o3 j 0 x o3

2

04 ² (jnj 0 1) ² 0 r _o ² . (6.2.4) Indeed,

2 j n j = C C C 2n _j

^o3

o3

j C C C

C C C ( 0 1) ⁿ x 3 + 2n _j

^o3

o3

j 0 x o3

C C

C + j ( 0 1) ⁿ x 3 0 x o3 j C C C ( 0 1) ⁿ x 3 + 2n _j

^o3

o3

j 0 x o3

C C

C + 2 0 r o .

(6.2.5)

Therefore, for some c > 0,

e ⁰

^{j(x10xo1;x20xo2)j}

2

4h

X

n 2Znf 0 g

e ⁰

(01)nx3+2no3 jo3j0xo3

2

4h

ce ⁰

^ch1

. (6.2.6)

Now, we deduce from (6.2.2), (6.2.3) and (6.2.6) that C C

C C C 0 i

Z

A

Z T 0 T

(B (x o ; _o3 ) f) (x; t; 0) ` (x) u 3 (x; t) dxdt

+i Z

A

Z T 0 T

1 (2)

⁴

Z

R

²

Z

_o3

+1

_o3

0 1

Z

j j <

e ^i(x+t) 'f c (; ) dd

!

a (x 0 x o ; t; 0) ` (x) u 3 (x; t) dxdt C C C C C

= C C C C C 0i

Z

A

Z T 0 T

P

n 2f0 2Q; 111 ;2P+1 gnf 0 g

h a

x 1 0 x o1 ; x 2 0 x o2 ; (01) ⁿ x 3 + 2n _j

^o3

o3

j 0 x o3 ; t; 0

1 (2)

⁴

Z

R

²

Z

_o3

+1

_o3

0 1

Z

j j <

e ^i(x

¹

¹

^+x

²

²

^+t) e ⁱ

h

( 0 1)

ⁿ

x

3

+2n

_j^o3_o3j

i

₃

'f c (; ) dd

#!

` (x) u 3 (x; t) dxdt C C C C C ₍₂₎ ^c

⁴

e ⁰

^ch1

Z

A

Z T 0 T

e ⁰

^t42

j ` (x) u 3 (x; t) j dxdt Z

R

²

Z

_o3

+1

_o3

0 1

Z

j j <

C C

C c 'f (; ) C C C dd

!

ce ⁰

^ch1

p G (U; 0)

Z

R

²

Z

_o3

+1

_o3

0 1

Z

j j <

C C

C c 'f (; ) C C C dd

!

(6.2.7) where in the last line we have used, from Cauchy-Schwarz inequality and conservation of energy (3.3), the fact that the solution U has the following property

Z T 0 T

e ⁰

^t42

Z

A

j` (x) u 3 (x; t)j dxdt c p jAj

Z ₁

01

e ⁰

^t42

dt p G (U; 0) c p

G (U; 0) .

(6.2.8)

Here and hereafter, c will be used to denote a generic constant, not necessarily the same in any two places. On the other hand, by using Cauchy-Schwarz inequality and conservation of energy, we have

C C C C C i

Z

A

Z

Rn ( 0 T;T ) 1 (2)

⁴

Z

R

²

Z

_o3

+1

_o3

0 1

Z

j j <

e ^i(x+t) 'f c (; ) dd

!

e ⁰

¹⁴

(

h¹

j x 0 x

o

j

²

+t

²

) ` (x) u 3 (x; t) dxdt C C C C C ce ⁰

¹⁸

^T

²

p

G (U; 0) Z

R

²

Z

_o3

+1

_o3

0 1

Z

j j <

C C

C c 'f (; ) C C C dd

! .

(6.2.9) Now, we cut the integral on time into two parts to obtain

Z

A 2R 1 (2)

⁴

Z

R

²

Z

_o3

+1

_o3

0 1

Z

j j <

e ^i(x+t) 'f c (; ) dd

!

a (x 0 x o ; t; 0) ` (x) u 3 (x; t) dxdt

= Z

A

Z T 0 T

1 (2)

⁴

Z

R

²

Z

_o3

+1

_o3

0 1

Z

j j <

e ^i(x+t) 'f c (; ) dd

!

a (x 0 x o ; t; 0) ` (x) u 3 (x; t) dxdt +

Z

A

Z

Rn ( 0 T;T) 1 (2)

⁴

Z

R

²

Z

_o3

+1

_o3

0 1

Z

j j <

e ^i(x+t) 'f c (; ) dd

!

e ⁰

¹⁴

(

¹h

j x 0 x

o

j

²

+t

²

) ` (x) u 3 (x; t) dxdt . (6.2.10) We conclude from (6.2.7), (6.2.9) and (6.2.10) that

C C C C C 0 i

Z

A

Z T 0 T

(B (x o ; _o3 ) f ) (x; t; 0) ` (x) u 3 (x; t) dxdt

+i Z

A 2R 1 (2)

⁴

Z

R

²

Z

_o3

+1

_o3

0 1

Z

j j <

e ^{i(x+t )} 'f c (; ) dd

!

a (x 0 x o ; t; 0) ` (x) u 3 (x; t) dxdt C C C C C c

e ⁰

^ch1

+ e ⁰

¹⁸

^T

²

p G (U; 0)

Z

R

²

Z

_o3

+1

_o3

0 1

Z

j j <

C C

C c 'f (; ) C C C dd

! .

(6.2.11)