COMMUNICATIONS ON
PURE AND APPLIED ANALYSIS
Volume 12, Number 5, September 2013
doi:10.3934/cpaa.2013.12.2229
pp. 2229–2266
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
(Communicated by Hong-Ming Yin)
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 reflected gaussian beams.
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 Ω ⊂ R3 with trapped rays. Precisely, let ρ > 0 and D be an open simply
connected bounded set in R2 with C 2 boundary ∂D. Consider Ω = D × (−ρ, ρ)
with boundary ∂Ω = Γ0 ∪ Γ1 where Γ0 = D × {−ρ, ρ} and Γ1 = ∂D × (−ρ, ρ). The
domain Ω is occupied by an electromagnetic medium of constant electric permittivity εo and constant magnetic permeability µo . For the sake of simplicity, we assume
from now that εo µo = 1.
Let E and H denote the electric and magnetic fields respectively. Define the
energy by
Z 1
2
2
E (t) =
εo |E (x, t)| + µo |H (x, t)| dx .
(1.1)
2 Ω
The Maxwell’s equations with Ohm’s law are described by

in Ω × [0, +∞)

 εo ∂t E − curl H + σE = 0


µo ∂t H + curl E = 0
in Ω × [0, +∞)

div (µo H) = 0
in Ω × [0, +∞)
(1.2)


E
×
ν
=
H
·
ν
=
0
on
∂Ω
×
[0,
+∞)



(E, H) (·, 0) = (Eo , Ho )
in Ω .
6
Here, (Eo , Ho ) is the initial data in the energy space L2 (Ω) and ν denotes the
outward unit normal vector to ∂Ω. The conductivity σ is such that σ ∈ L∞ (Ω)
and σ ≥ 0. It is well-known that when the conductivity is identically null, then
the above system is conservative and when σ is bounded from below by a positive
constant, then an exponential energy decay rate holds for the Maxwell’s equations
2000 Mathematics Subject Classification. Primary: 35B40, 35Q61; Secondary: 93D15.
Key words and phrases. Maxwell’s equations, decay estimates, trapped ray.
2229
2230
KIM DANG PHUNG
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
σ = 1|ω a where ω ⊂ Ω, ω 6= Ω, a ∈ L∞ (Ω) and a ≥ constant > 0.
Here and hereafter, we denote by 1|· the characteristic function of a set in the place
where · stays. From the works of Bardos, Lebeau and Rauch [2] for the scalar
wave operator, there is a geometric condition on the location of ω which implies an
exponential decay rate for the Maxwell’s equations with Ohm’s law in the energy
space. In our paper, ω is a non-empty connected open subset of Ω as follows.
Consider
ω = x0 ∈ D; 0inf |x0 − y 0 | < ro × (−ρ, ρ) for some small ro ∈ (0, ρ/2) .
y ∈∂D
Such case of ω 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
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.
Main Theorem. If σ = 1|ω a where a ∈ L∞ (Ω) and a ≥ constant > 0, then there
exist c > 0 and γ > 0 such that for any t ≥ 0
c 2
E (t) ≤ γ E (0) + k(curl Eo , curl Ho )kL2 (Ω)6
t
for every solution of the system (1.2) of Maxwell’s equations with Ohm’s law with
6
initial data (Eo , Ho ) ∈ L2 (Ω) such that

6
2

 (curl Eo , curl Ho ) ∈ L (Ω) ,

div Ho = 0
in Ω
E
×
ν
=
H
·
ν
=
0
on ∂Ω

o
o


div 1|Ω\ω Eo = 0
in Ω \ω .
In literature, the exponential energy decay rate of a linear dissipative system can
be deduced from an observability estimate. Precisely, in order to get an exponential
decay rate in the energy space we should have the following observability inequality
Z
Z ζ+Tc Z
2
2
∃C, Tc > 0 ∀ζ ≥ 0
|(E, H) (x, ζ)| dx ≤ C
σ |E| dxdt
Ω
ζ
or simply, in virtue of a semigroup property,
Z
Z
2
∃C, Tc > 0
|(Eo , Ho )| dx ≤ C
Ω
0
Tc
Ω
Z
2
σ |E| dxdt
Ω
6
for any initial data (Eo , Ho ) in the energy space L2 (Ω) . 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 (Ω, ωBLR ) and when the conductivity has the property
that σ (x) ≥ constant > 0 for all x ∈ ωBLR and σ (x) = 0 for all x ∈ Ω \ωBLR .
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 Proposition 4 below) which
ENERGY DECAY FOR MAXWELL’S EQUATIONS WITH OHM’S LAW
2231
can also be seen as an interpolation estimate. It relies
with the construction of a
particular solution for the operator i∂s + h ∆ − ∂t2 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 wellknown 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 field.
2.1. Well-posedness of the problem. Let us introduce the spaces
n
o
3
3
V = L2 (Ω) × G ∈ L2 (Ω) ; div G = 0, G · ν|∂Ω = 0 ,
n
6
3
W = (F, G) ∈ L2 (Ω) ; curl F ∈ L2 (Ω) , F × ν|∂Ω = 0,
div G = 0, G · ν|∂Ω = 0, curl G ∈ L2 (Ω)
3
o
.
It is well-known that if (Eo , Ho ) ∈ V, there is a unique weak solution (E, H) ∈
C 0 ([0, +∞) , V). Further, if (Eo , Ho ) ∈ W, there is a unique strong solution
(E, H) ∈ C 0 ([0, +∞) , W) ∩ C 1 ([0, +∞) , V). We can easily check that the energy E, defined by (1.1), is a continuous positive non-increasing real function on
[0, +∞). Further for any initial data (Eo , Ho ) ∈ W and any t2 > t1 ≥ 0,
Z t2 Z
2
σ (x) |E (x, t)| dxdt = 0
(2.1.1)
E (t2 ) − E (t1 ) +
Ω
t1
and
Z
t2
Z
E1 (t2 ) − E1 (t1 ) +
where
E1 (t) =
1
2
Z 2
σ (x) |∂t E (x, t)| dxdt = 0 ,
t1
(2.1.2)
Ω
2
2
εo |∂t E (x, t)| + µo |∂t H (x, t)|
dx .
(2.1.3)
Ω
2.2. Orthogonal decomposition. Both E and µo 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 1. For any (Eo , Ho ) ∈ W, there is a unique
3
(p, A) ∈ C 1 [0, +∞) , H01 (Ω) × C 2 [0, +∞) , H 1 (Ω)
such that the solution (E, H) of the Maxwell’s equations with Ohm’s law (1.2) satisfies
E = −∇p − ∂t A
(2.2.1)
µo H = curl A

in Ω × [0, +∞)
 εo µo ∂t2 A + curl curl A = µo (−εo ∂t ∇p + σE)
div A = 0
in Ω × [0, +∞)
(2.2.2)

A×ν =0
on ∂Ω × [0, +∞)
2232
KIM DANG PHUNG
and we have the following relations
2
2
2
kEkL2 (Ω)3 = k∇pkL2 (Ω)3 + k∂t AkL2 (Ω)3 ,
(2.2.3)
kεo ∂t ∇pkL2 (Ω)3 ≤ kσEkL2 (Ω)3 ,
(2.2.4)
2
∃c > 0
2
kAkL2 (Ω)3 ≤ c kcurl AkL2 (Ω)3 .
3
(2.2.5)
3
Further, since curl H ∈ L2 (Ω) , curl curl A ∈ L2 (Ω) and div A ∈ H01 (Ω).
The proof is essentially given in [11, p. 121] from a Hodge decomposition and is
omitted here. Now, the vector field A has the nice property of free divergence and
satisfies a second order vector wave equation (2.2.2) with homogeneous
boundary
condition A × ν = div A = 0 and with a second member in C 1 [0, +∞) , L2 (Ω)
3
bounded by 2µo kσEkL2 (Ω)3 .
2.3. Invariant subspaces, asymptotic behavior and exponential energy decay. Let ω+ be a non-empty connected open subset of Ω with a Lipschitz boundary
∂ω+ . In this subsection we suppose that
σ = 1|ω+ a where a ∈ L∞ (Ω) and a ≥ constant > 0 .
Denote ω− = Ω \ω+ and suppose that its boundary ∂ω− is Lipschitz and has no
more than two connected components γ1 , γ2 .
3
3
We recall that the range of the curl, curl H 1 (ω− ) , is closed in L2 (ω− ) (see [6,
p. 257] or [5, p. 54]) and
Z
3
3
curl H 1 (ω− ) = U ∈ L2 (ω− ) ; div U = 0 in ω− ,
U · ν = 0 for i ∈ {1, 2} .
γi
3
Its orthogonal space for the L2 (ω− ) norm is
⊥ n
o
3
3
curl H 1 (ω− )
= V ∈ L2 (ω− ) ; curl V = 0 in ω− , V × ν = 0 on ∂ω− .
3
3
3
Let us introduce Sσ = curl H 1 (ω− ) ∩ L2 (Ω) × L2 (Ω) . The space W ∩ Sσ
is stable for the system of Maxwell’s equations with Ohm’s law, which can be seen
⊥
3
by multiplying by V ∈ curl H 1 (ω− )
the equation εo ∂t E − curl H + σE = 0.
Then, we can add the following well-posedness result. If (Eo , Ho ) ∈ W ∩ Sσ , there
is a unique solution (E, H) ∈ C 0 ([0, +∞) , W ∩ Sσ ) ∩ C 1 ([0, +∞) , V ∩ Sσ ).
It has been proved (see [11, p. 124]) that if ω− is a non-empty connected open set
then lim E (t) = 0 for any initial data (Eo , Ho ) ∈ V ∩ Sσ . Further, the following
t→+∞
result (see [11, p. 124]) plays a key role.
Proposition 2. If ω− is a non-empty connected open set, then there exists c > 0
such that for all initial data (Eo , Ho ) ∈ W ∩ Sσ of the system (1.2) of Maxwell’s
equations with Ohm’s law, we have
∀t ≥ 0
E (t) ≤ cE1 (t) .
(2.3.1)
Remark 1. The estimate (2.3.1) is still true under the assumption ∂ω+ ∩ ∂Ω 6= ∅
and without adding the hypothesis saying that ω− is a connected set.
ENERGY DECAY FOR MAXWELL’S EQUATIONS WITH OHM’S LAW
2233
Indeed, the proof given in [11, p. 127] can be divided into two steps. In the first
step, we begin to establish the existence of c > 0 such that
p
p
2
2
2
k∇pkL2 (ω+ )3 + k∂t AkL2 (Ω)3 + kHkL2 (Ω)3 ≤ c E1 (t) + E (t) E1 (t)
(2.3.2)
for any (Eo , Ho ) ∈ W ∩ Sσ . Here, we used a standard compactness-uniqueness
argument for H, (2.2.5) of Proposition 1 for ∂t A, and for ∇p from the fact that
σ (x) ≥ constant > 0 for all x ∈ ω+ and (2.1.3). Till now, we did not utilize that
ω− is a connected set. The second step (see [11, p. 128]) did consist to prove that
2
2
2
k∇pkL2 (ω− )3 ≤ c k∇pkL2 (ω+ )3 + k∂t AkL2 (Ω)3 .
Finally, we concluded by virtue of (2.2.3) of Proposition 1. This last estimate
becomes easier to obtain under the assumption ∂ω+ ∩ ∂Ω 6= ∅ and without adding
the hypothesis saying that ω− is a connected set. Indeed, since (Eo , Ho ) ∈ W ∩ Sσ
and −∆p = div E, p ∈ H01 (Ω) solves the following elliptic system

∆p = 0 in ω−

p|∂ω− ∈ H 1/2 (∂ω− )

p = 0 on ∂ω+ ∩ ∂Ω .
Thus, by the elliptic regularity, the trace theorem and the Poincaré inequality, we
have the following estimate
k∇pkL2 (ω− )3 ≤ c1 kpkH 1/2 (∂ω− ) ≤ c2 kpkH 1/2 (∂ω+ ) ≤ c3 k∇pkL2 (ω+ )3
(2.3.3)
for suitable constants c1 , c2 , c3 > 0. Hence, combining (2.3.2) and (2.3.3) with
(2.2.5), (2.3.1) follows if ∂ω+ ∩ ∂Ω 6= ∅ and without adding the hypothesis saying
that ω− 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 3. Assume that
(i) σ = 1|ω+ a where a ∈ L∞ (Ω) and a ≥ constant > 0;
(ii) ∂ω+ ∩ ∂Ω 6= ∅ or ω− is a non-empty connected open set;
(iii) at any point in ω− and any direction, the generalized ray of the scalar wave
operator ∂t2 − ∆ is uniquely defined.
Let ϑ be a subset of Ω such that any generalized ray of the scalar wave operator
∂t2 − ∆ meets ϑ. If ϑ ∩ Ω ⊂ ω+ , then there exist c > 0 and β > 0 such that for any
t ≥ 0, we have
E (t) ≤ ce−βt E (0)
for every solution of the system (1.2) of Maxwell’s equations with Ohm’s law with
initial data (Eo , Ho ) ∈ V ∩ Sσ .
The proof of Proposition 3 is done in [11, p. 129] when ω− 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 ω− is a non-empty connected open set.
Z Tc +ζ Z 2
2
∃C, Tc > 0 ∀ζ ≥ 0 E1 (ζ) ≤ C
σ |∂t E| + σ |E| dxdt
ζ
Ω
for any initial data (Eo , Ho ) ∈ W ∩ Sσ . Next by (2.3.1), we deduced that
Z Tc +ζ Z 2
2
∃C, Tc > 0 ∀ζ ≥ 0 E (ζ) + E1 (ζ) ≤ C
σ |∂t E| + σ |E| dxdt .
ζ
Ω
2234
KIM DANG PHUNG
Finally, we concluded by virtue of a semigroup property with (2.1.1) and (2.1.2).
Notice that no assumption div E (·, ζ) = 0 in Ω is set up. This is important because
the free divergence of the electric field is not preserved by the Maxwell’s equations
with Ohm’s law. Now, thanks to Remark 1, the proof works as well when ∂ω+ ∩∂Ω 6=
∅ and without adding the hypothesis saying that ω− is a connected set
3. Proposition 4 and proof of Main Theorem.
U of the following system

∂t2 U + curl curl U = 0




div U = 0



U × ν = 0
U 0, U 1

 (U (·, 0) , ∂t U (·, 0)) =

0

U , U 1 ∈ X ,



1
U , curl curl U 0 ∈ X ,
Let us consider the solution
in Ω × R
in Ω × R
on ∂Ω × R
in Ω ,
(3.1)
where
n
o
6
3
X = (F, G) ∈ L2 (Ω) ; curl F ∈ L2 (Ω) , F × ν|∂Ω = 0, div F = 0, div G = 0 .
It is well-known that the above system is well-posed
with a unique solution U so
that (U (·, t) , ∂t U (·, t)) and ∂t U (·, t) , ∂t2 U (·, t) belong to X for any t ∈ R. Let
us define the following two conservations of energies.
Z 2
2
G (U, 0) = G (U, t) ≡
|∂t U (x, t)| + |curl U (x, t)| dx ,
(3.2)
Ω
G (∂t U, 0) = G (∂t U, t) ≡
Z 2
|curl curl U (x, t)| + |curl ∂t U (x, t)|
2
dx .
(3.3)
Ω
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) ,
2
(3.4)
2
kU (·, t)kH 1 (Ω)3 ≤ c kcurl U (·, t)kL2 (Ω)3 ,
(3.5)
for some c > 0 and any t ∈ R.
Let ϑro (∂D) be an ro -neighborhood of the boundary of D, that is
ϑro (∂D) = x0 ∈ D; 0inf |x0 − y 0 | < ro for some small ro ∈ (0, ρ/2) .
y ∈∂D
Recall that σ = 1|ω a where ω = ϑro (∂D) × (−ρ, ρ). Consider
ωo = (D \ϑro (∂D) ) × (ρ − 2ro , ρ − ro ) .
(3.6)
Proposition 4. For any To > 0, there exist c, γ > 0 such that for any h > 0, the
solution U of (3.1) satisfies
Z To Z
2
|∂t U (x, t)| dxdt
0
ωo
Z c h1γ Z (3.7)
1
2
2
|∂t U (x, t)| + |U (x, t)| dxdt + hG (∂t U, 0) .
≤c γ
h −c h1γ ω
We shall leave the proof of Proposition 4 till later (see Section 4). Now we turn
to prove our Main Theorem.
ENERGY DECAY FOR MAXWELL’S EQUATIONS WITH OHM’S LAW
2235
Proof of Main Theorem. Remark that (E, H) (·, 0) ∈ W ∩ Sσ and therefore from
the previous section for any t ≥ 0, (E, H) (·, t) ∈ W ∩ Sσ .
Step 1 .- Denote (E1 , H1 ) the electromagnetic field of the following Maxwell’s
equations with Ohm’s law

εo ∂t E1 − curl H1 + σ + 1|ωo E1 = 0
in Ω × [0, +∞)




µ
∂
H
+
curl
E
=
0
in
Ω × [0, +∞)

o t 1
1
div (µo H1 ) = 0
in Ω × [0, +∞)


E1 × ν = H1 · ν = 0
on ∂Ω × [0, +∞)



(E1 , H1 ) (·, Th ) = (E, H) (·, Th )
in Ω .
Notice that (E, H) (·, Th ) ∈ W ∩ Sσ ⊂ V ∩ Sσ+1|ωo . Therefore by Proposition 3,
there exist c, β > 0 (independent of Th ) such that for any t ≥ 0 we have
Z 2
2
εo |E1 (x, t + Th )| + µo |H1 (x, t + Th )| dx ≤ ce−βt E (Th ) .
(3.8)
Ω
On the other hand, let (E2 , H2 ) = (E1 , H1 ) − (E, H). Then it solves

in Ω × [0, +∞)
εo ∂t E2 − curl H2 + σ + 1|ωo E2 = −1|ωo E




µo ∂t H2 + curl E2 = 0
in Ω × [0, +∞)

div (µo H2 ) = 0
in Ω × [0, +∞)


E2 × ν = H2 · ν = 0
on ∂Ω × [0, +∞)



(E2 (·, Th ) , H2 (·, Th )) = (0, 0)
in Ω ,
and by a standard energy method, we get that for any t ≥ 0
Z t+Th Z
Z t
2
2
2
|E (x.s)| dxds .
εo |E2 (x, t + Th )| + µo |H2 (x, t + Th )| dx ≤
ε o Th
ωo
Ω
(3.9)
R t+Th R
2
Now we are able to bound E (Th ) = E (t + Th ) + Th
σ (x) |E (x, s)| dxds as
Ω
follows. By using (1.1), (2.1.1), (3.8) and (3.9), we deduce that
E (Th ) ≤ 2ce−βt E (Th )
Z
Z
Z t+Th Z
2t t+Th
2
2
+
|E (x.s)| dxds +
σ (x) |E (x, s)| dxds
ε o Th
ωo
Th
Ω
which implies by taking t large enough, the existence of constants C, Tc > 1 such
that
Z Tc +Th Z
Z
2
2
E (Th ) ≤ C
σ |E| dx +
|E| dx dt .
(3.10)
Th
Ω
ωo
Step 2 .- Recall the existence of the vector potential A from Proposition 1 and
let U be the solution of

∂t2 U + curl curl U = 0
in Ω × R



div U = 0
in Ω × R
U ×ν =0
on ∂Ω × R



(U, ∂t U ) (·, 0) = (A, ∂t A) (·, 0)
in Ω ,
then by a standard energy method, for any T1 > 0
Z T1 Z 2
2
|∂t (U − A)| + |curl (U − A)| dxdt
0
ZΩ T1 Z
2
2
≤ µo T1
|εo ∂t ∇p − σE| dxdt
0
Ω
2236
KIM DANG PHUNG
which implies from (2.2.4) of Proposition 1 that
Z
Z T1 Z 2
2
|∂t (U − A)| + |curl (U − A)| dxdt ≤ 4µo T12
0
T1
Z
2
|σE| dxdt .
0
Ω
Ω
(3.11)
Z
Th
Z
2
Now we are able to bound E (0) = E (Th )+
σ (x) |E (x, t)| dxdt as follows.
0
Ω
Since E = −∇p + ∂t (U − A) − ∂t U , we deduce by (2.1.1) and (3.10) that
E (0)
Z Th Z
2
≤
σ |E| dxdt
Z
Z
Z
2
2
+C
σ |E| dx +
|−∇p + ∂t (U − A) − ∂t U | dx dt
Th
ωo
Z Ω Z Z Tc +Th Z
Tc +Th
2
2
2
2
≤ C 1 + Th
σ |E| + |∇p| dxdt + C
|∂t U | dxdt ,
0
Ω
Tc +Th
0
Ω
Th
ωo
(3.12)
where in the last line, we used (3.11). Here and hereafter, C will be used to denote
a generic constant, not necessarily the same in any two places.
Step 3 .- Now we fix Th = c h1γ where c and γ are given by Proposition 4. Taking
To = Tc in Proposition 4, we obtain that for any h > 0,
Z Tc Z
Z Th Z 2
2
2
|∂t U | dxdt ≤ Th
|∂t U | + |U | dxdt + hG (∂t U, 0) .
0
−Th
ωo
ω
By a translation in time and (3.3), the above inequality implies that
Z Tc +Th Z
Z 2Th Z 2
2
2
|∂t U | dxdt ≤ Th
|∂t U | + |U | dxdt + hG (∂t U, 0) . (3.13)
Th
ωo
0
But from (3.11),
Z 2Th Z
2
|∂t U | dxdt
0
Z
ω
2Th
Z
2
|−E − ∇p + ∂t (U − A)| dxdt
0
Z ω2Th Z 2
2
≤ CTh2
σ |E| + |∇p| dxdt
=
ω
0
(3.14)
Ω
and
Z
2Th
Z
2
|U | dxdt
0
2Th
Z
Z
ω
2Th
Z
2
≤2
Z
2
|U − A| dxdt + 2
Z0
2Th
ZΩ
0
2
≤C
Zω
|A| dxdt
2Th Z
2
|A| dxdt
|curl (U − A)| dxdt + 2
0Z
ΩZ
ω
Z 2Th Z 0
2Th
2
2
2
σ |E| dxdt + 2
|A| dxdt .
≤ CTh
0
Ω
0
ω
(3.15)
Therefore, plugging (3.14) and (3.15) into (3.13), we get
Z Tc +Th Z
Z 2Th Z Z
2
2
2
2
3
|∂t U | dxdt ≤ CTh
σ |E| + |∇p| dx +
|A| dx dt
Th
ωo
0
Ω
ω
+hG (∂t U, 0) .
(3.16)
ENERGY DECAY FOR MAXWELL’S EQUATIONS WITH OHM’S LAW
2237
Finally, combining (3.12) and (3.16), we have
Z 2Th Z Z
2
2
2
3
E (0) ≤ CTh
σ |E| + |∇p| dx +
|A| dx dt + hG (∂t U, 0) .
0
Ω
ω
In conclusion, we proved that there exist c, γ > 0 such that for any h > 0, the
solution (E, H) of (1.2) with initial data in W ∩ Sσ satisfies
Z c h1γ Z Z
1
2
2
2
σ |E| + |∇p| dx +
E (0) ≤ c γ
|A| dx dt + hG (∂t U, 0) .
h 0
Ω
ω
Step 4 .- By formula (2.1.1) and since G ∂t U, mc h1γ = G (∂t U, 0) for any m, this
last inequality becomes
Z mc h1γ Z
X
X
1
2
σ |E| dxdt =
E mc γ
N E (0) −
h
Ω
m=0,..,N −1
m=0,..,N −1 0
1 Z
Z
Z
(m+1)c
γ
X
h
1
2
2
2
≤c γ
σ |E| + |∇p| dx +
|A| dx dt
h
mc h1γ
Ω
ω
m=0,..,N −1
+N hG (∂t U, 0) ,
for any N > 1. We choose N ∈ c h1γ , 1 + c h1γ . Therefore, there exist c, γ > 0 such
that for any h > 0,
Z
Z c( h1 )γ Z 2
2
2
σ |E| + |∇p| dx +
|A| dx dt + hG (∂t U, 0) . (3.17)
E (0) ≤ c
0
ω
Ω
On the other hand, since (U, ∂t U ) (·, 0) = (A, ∂t A) (·, 0),
G (∂t U, 0)
2
2
= kcurl E (·, 0)kL2 (Ω)3 + kµo curl H (·, 0)kL2 (Ω)3
2
2
= µ2o k∂t H (·, 0)kL2 (Ω)3 + k(∂t E + µo σE) (·, 0)kL2 (Ω)3
2
≤ c (E1 (0) + E (0)) ≤ c k(Eo , Ho )kD(M)
(3.18)
where M is the m-accretive operator in V ∩ Sσ with domain D (M) = W ∩ Sσ ,
defined as follows.
2
2
2
k(F, G)kV = εo kF kL2 (Ω)3 + µo kGkL2 (Ω)3 ,


1
1
σ
− curl


εo
M =  1εo
 .
curl
0
µo
Therefore, combining (3.17) and (3.18), we get the existence of constants c, γ > 0
such that for any h > 0,
Z c( h1 )γ Z Z
2
2
2
2
E (0) ≤ c
σ |E| + |∇p| dx +
|A| dx dt + ch k(Eo , Ho )kD(M) .
0
Ω
ω
Similarly, we get the existence of constants c, γ > 0 such that for any ζ ≥ 0 and
h > 0,
Z ζ+c( h1 )γ Z Z
2
2
2
2
E (ζ) ≤ c
σ |E| + |∇p| dx +
|A| dx dt + h k(Eo , Ho )kD(M) .
ζ
Ω
ω
(3.19)
2238
KIM DANG PHUNG
Step 5 .- Denote (T (t))t≥0 the unique semigroup of contractions generated by
−M. First, suppose that (Eo , Ho ) ∈ D M3 and let us define the functional of
energy
Z 2 2
1
E2 (t) =
εo ∂t2 E (x, t) + µo ∂t2 H (x, t) dx
2 Ω
which satisfies
Z t2 Z
2
(3.20)
σ (x) ∂t2 E (x, t) dxdt = 0 .
E2 (t2 ) − E2 (t1 ) +
t1
Ω
2
Let Xo = −M (Eo , Ho ), then (T (t))t≥0 Xo = ∂t2 E, ∂t2 H , kT (ζ) Xo kV = 2E2 (ζ)
2
2
and kXo kD(M) ≤ c k(Eo , Ho )kD(M3 ) . Further, by uniqueness of the orthogonal
decomposition
in (2.2.1) of Proposition 1, (3.19) implies that for any (Eo , Ho ) ∈
D M3
Z ζ+c( h1 )γ Z Z
2 2 2 2 2 2
E2 (ζ) ≤ c
σ ∂t E + ∂t ∇p
dx +
∂t A dx dt
(3.21)
ζ
Ω
ω
2
+ch k(Eo , Ho )kD(M3 ) .
2
Since by Proposition 2, E (ζ) ≤ cE1 (ζ) and in a similar way E1 (ζ) ≤ cE2 (ζ) for
some c > 0, taking account of the first line of (2.2.1) and (2.2.4), (3.21) becomes
Z ζ+c( h1 )γ Z 2 2
2
σ |E| + σ |∂t E| + σ ∂t2 E dxdt
E (ζ) + E1 (ζ) + E2 (ζ) ≤ c
ζ
Ω
2
+h k(Eo , Ho )kD(M3 ) .
(3.22)
Step 6 .- Denote
H (ζ) =
E2 (ζ) + E1 (ζ) + E (ζ)
2
k(Eo , Ho )kD(M3 )
.
By (2.1.1), (2.1.2) and (3.20), the function H is a continuous positive decreasing
real function on [0, +∞), bounded by one. Taking h = 12 H (ζ) in (3.22), we get the
existence of constants c, γ > 0 such that for any ζ ≥ 0,
γ
1
Z ζ+c( H(ζ)
) Z 2 2
2
E (ζ) + E1 (ζ) + E2 (ζ) ≤ c
σ |E| + σ |∂t E| + σ ∂t2 E dxdt ,
ζ
that is,
Ω
γ
c
+ζ
∀ζ ≥ 0 .
H (ζ)
From [12, p. 122, Lemma B], we deduce that there exist C, γ > 0 such that for any
t>0
C
2
E (t) + E1 (t) + E2 (ζ) ≤ γ k(Eo , Ho )kD(M3 ) ,
t
that is
C
2
2
kT (t) Yo kD(M2 ) ≤ γ kYo kD(M3 ) ∀Yo ∈ D M3 .
(3.23)
t
Since M is an m-accretive operator in V ∩Sσ with dense domain, one can restrict
it to D M2 in a way that its restriction operator is m-accretive. Thus the following
two properties holds.
∀Zo ∈ D M2
∃!Yo ∈ D M3
Yo + MYo = Zo ;
(3.24)
3
kYo kD(M2 ) ≤ kYo + MYo kD(M2 ) ∀Yo ∈ D M .
(3.25)
H (ζ) ≤ c H (ζ) − H
ENERGY DECAY FOR MAXWELL’S EQUATIONS WITH OHM’S LAW
2239
Consequently,
2
2
kT (t) Zo kD(M)
= kT (t) (Yo + MYo )kD(M) by (3.24)
2
≤ C kT (t) Yo kD(M2 )
C
2
≤ γ kYo kD(M3 ) by (3.23)
t C
2
2
≤ γ kYo kD(M2 ) + kYo + MYo kD(M2 )
t
C
2
≤ γ kYo + MYo kD(M2 ) by (3.25)
t
C
2
≤ γ kZo kD(M2 ) by (3.24)
t
(3.26)
for suitable positive constant C > 0 which is not necessarily the same in any two
places.
Now, suppose that (Eo , Ho ) ∈ D (M). Since M is an m-accretive operator in
V ∩ Sσ 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.
∀ (Eo , Ho ) ∈ D (M) ∃!Zo ∈ D M2
Zo + MZo = (Eo , Ho ) ;
(3.27)
2
kZo kD(M) ≤ kZo + MZo kD(M) ∀Zo ∈ D M .
(3.28)
We conclude that
2
2E (t) = kT (t) (Eo , Ho )kV
2
= kT (t) (Zo + MZo )kV by (3.27)
2
≤ C kT (t) Zo kD(M)
C
2
≤ γ kZo kD(M2 ) by (3.26)
t 2
2
≤ tCγ kZo kD(M) + kZo + MZo kD(M)
C
2
≤ γ kZo + MZo kD(M) by (3.28)
t
C
2
≤ γ k(Eo , Ho )kD(M) by (3.27)
t
C
≤ γ (E (0) + E1 (0))
t
for suitable positive constant C > 0 which is not necessarily the same in any two
places.
4. Proposition 5 and proof of Proposition 4. Recall that ωo is defined by
(3.6). In this Section, we establish an interpolation estimate in L2 norm in order to
prove Proposition 4.
Proposition 5. There exist c, γ > 0 such that for any To > 0, and h ∈ (0, 1], the
solution U of (3.1) satisfies
2
e−cTo
Z
To
Z
2
|U (x, t)| dxdt
0
ωo
Z c h1γ Z 1
2
2
|∂
U
(x,
t)|
+
|U
(x,
t)|
dxdt
t
hγ −c h1γ ω
+chG (U, 0) .
≤c
We shall leave the proof of Proposition 5 till later (see Section 5). Now we turn
to prove Proposition 4 as follows.
2240
KIM DANG PHUNG
Proof of Proposition 4. Let ψ ∈ C0∞ (0, 5To /3) be such that ψ = 1 in (To /3, 4To /3).
Then,
Z 4To /3 Z
2
|∂t U (x, t)| dxdt
ZTo5T/3o /3 Zωo
2
≤
|ψ (t) ∂t U (x, t)| dxdt
Z0 5To /3 Zωo
2ψ (t) ψ 0 (t) ∂t U (x, t) + ψ 2 (t) ∂t2 U (x, t) |U (x, t)| dxdt
≤
0
o
Z
Z 5Toω/3
√
C
2
√
≤
|U (x, t)| dxdt + hG (∂t U, 0) ∀h > 0
h "0
#
Z cωh1oγ Z √
1
C
2
2
|∂t U | + |U | dxdt + chG (U, 0) + hG (∂t U, 0)
≤√ c γ
h h −c h1γ ω
where in the last line, we used Proposition 5. Therefore, it implies thanks to (3.4)
that for any To > 0, there exist c, γ > 0 such that for any h ∈ (0, 1],
Z 4To /3 Z
Z c h1γ Z 1
2
2
2
|∂t U | dxdt ≤ c γ
|∂t U | + |U | dxdt + chG (∂t U, 0) .
h −c h1γ ω
To /3
ωo
By a translation in time and (3.3), we obtain (3.7) for any h ∈ (0, 1]. Since
Z To Z
2
|∂t U | dxdt ≤ ChG (∂t U, 0) ∀h > 1 ,
0
ωo
we get the desired estimate of Proposition 4.
5. Proposition 6 and proof of Proposition 5. Notice that the following inclusion holds C0∞ (B (xo , ro /2)) ⊂ C0∞ (Ω) for any xo ∈ ωo , where B (xo , r) denotes the
ball of center xo and
radius r.
Let ` ∈ C ∞ R3 be such that 0 ≤ ` (x) ≤ 1, ` = 1 in R3 \ω , ` (x) ≥ `o > 0 for
any x ∈ ωo , ` = ∂ν ` = 0 on Γ1 and both ∇` and ∆` have support in ω.
The proof of Proposition 5 comes from the following result.
Proposition 6. There exist c, δ, η > 0 such that for any xo ∈ ωo and h ∈ (0, 1],
λ ≥ 1, the solution U of (3.1) satisfies
Z
2
2
1 1
2
χ (x) e− 2 ( h |x−xo | +t ) ` (x) |U (x, t)| dxdt
Ω×R
δ
1
λδ
1 λ
2
≤ c √ G (U, 0) + ce− ch η G (U, 0) + c η k(U, ∂t U )k 2 λδ λδ 6 ,
L ω× −c hη ,c hη
h
h
λ
where χ ∈ C0∞ (B (xo , ro /2)), 0 ≤ χ ≤ 1.
We shall leave the proof of Proposition 6 till later (see Section 6). Now we turn
to prove Proposition 5.
2
Proof of Proposition 5. Let h ∈ (0, ho ] where ho = min 1, (ro /8) . We begin by
√ covering ωo with a finite collection of balls B xio , 2 h for i ∈ I with xio ∈ ωo
and where I is a countable set such that the number of elements of I is hc√oh
for some constant co > 0 independent of h. Then, for each xio , we introduce
ENERGY DECAY FOR MAXWELL’S EQUATIONS WITH OHM’S LAW
2241
χxio ∈ C0∞ B xio , ro /2 ⊂ C0∞ (Ω) be such that 0 ≤ χxio ≤ 1 and χxio = 1 on
√ B xio , ro /4 ⊃ B xio , 2 h . Consequently, for any To > 0,
1 Z To Z
− To2
2
e 2
|U (x, t)| dxdt
ωo
Z To Z0
2
− 12 t2
1
e
≤ `o
` (x) |U (x, t)| dxdt
0
ω
o
Z
Z
2
1
1 P To
− 21 h
|x−xio | +t2 ` (x) |U (x, t)|2 dxdt
i
(x)
e
≤ e2
χ
xo
√
`o i∈I 0
B (xio ,2 h)
Z
2
1 P
− 1 1 x−xio | +t2
2
` (x) |U (x, t)| dxdt .
≤ e2
χxio (x) e 2 h |
`o i∈I Ω×R
2
R
− 1 1 x−xio | +t2
2
` (x) |U (x, t)| dxdt has
By virtue of Proposition 6, Ω×R χxio (x) e 2 h |
the property to be bounded independently of xio and it implies that for some constants c, δ, η > 0,
1
2
e− 2 To
1
≤c √
h h
Z
To
Z
2
|U (x, t)| dxdt
ωo
0
δ
1
λδ
1 λ
√ G (U, 0) + ce− ch η G (U, 0) + c η k(U, ∂t U )k2 2 λδ λδ 6 .
L ω× −c hη ,c hη
h
h
λ
5
1 √1
≤ Ch for some
Finally, we choose λ ≥ 1 such that λ = hho in order that h√
h λ
C > 0. Then there exist c, γ > 0 such that for any h ∈ (0, ho ],
− 12 To2
Z
e
0
To
Z
1
|U | dxdt ≤ c γ
h
ωo
2
Z
c h1γ
−c h1γ
Z 2
|∂t U | + |U |
2
dxdt + chG (U, 0) .
ω
1 2 RT R
2
Since e− 2 To 0 o ωo |U | dxdt ≤ ChG (U, 0) for any h ≥ ho , the proof of Proposition
5 is complete.
6. Proof of Proposition 6. We will denote
Z
e−i(xξ+tτ ) F (x, t) dxdt
Fb (ξ, τ ) =
R4
and recall the Fourier inversion formula
Z
1
F (x, t) =
ei(xξ+tτ ) Fb (ξ, τ ) dξdτ
4
(2π) R4
3
when F and Fb belong to L1 R4 . Let h ∈ (0, 1], xo ∈ ωo , χ ∈ C0∞ (B (xo , ro /2))
be such that 0 ≤ χ ≤ 1. Let us introduce
2
ao (x, t) = e− 4 ( h |x−xo |
1
1
+t2 )
and ϕ (x, t) = χ (x) ao (x, t) .
2242
KIM DANG PHUNG
By the Fourier inversion formula,
Z
2
χ (x) e− 4 ( h |x−xo |
1
1
+t2 )
2
` (x) |U (x, t)| dxdt
ZΩ×R
2
ϕ (x, t) ao (x, t) ` (x) |U (x, t)| dxdt
!
Z
Z
1
i(xξ+tτ ) d
e
=
ϕU (ξ, τ ) dξdτ · ao (x, t) ` (x) U (x, t) dxdt
4
(2π) R4
Ω×R
!
Z Z
Z
1
i(xξ+tτ ) d
e
ϕU (ξ, τ ) dξdτ · ao `U dxdt
=
4
(2π) R3 |τ |<λ
Ω×R
!
Z Z
Z
1
d (ξ, τ ) dξdτ · ao `U dxdt
ei(xξ+tτ ) ϕU
+
4
(2π) R3 |τ |≥λ
Ω×R
=
Ω×R
(6.1)
for any λ ≥ 1. On the other hand, from (A1) of Appendix A, (3.5), (3.2) and the
fact that each component of U solves the wave equation, we have that for any U
solution of (3.1),
Z
!
Z Z
1
d (ξ, τ ) dξdτ · ao (x, t) ` (x) U (x, t) dxdt
ei(xξ+tτ ) ϕU
4
Ω×R (2π) R3 |τ |≥λ
1
≤ c √ G (U, 0) .
λ
(6.2)
It remains to study the following quantity
Z
Ω×R
Z
1
4
(2π)
!
Z
e
R3
i(xξ+tτ ) d
ϕU (ξ, τ ) dξdτ
· ao (x, t) ` (x) U (x, t) dxdt . (6.3)
|τ |<λ
We claim that there exist c, δ, η > 0 such that for any xo ∈ ωo and h ∈ (0, 1], λ ≥ 1,
we have
Z
!
Z Z
1
d (ξ, τ ) dξdτ · ao (x, t) ` (x) U (x, t) dxdt
ei(xξ+tτ ) ϕU
4
Ω×R (2π) R3 |τ |<λ
δ
δ
1
λ
1 λ
2
≤ c √ G (U, 0) + ce− ch η G (U, 0) + c η k(U, ∂t U )k 2 λδ λδ 6 ,
L ω× −c hη ,c hη
h
h
λ
(6.4)
which implies Proposition 6 using (6.1) and (6.2).
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 ∈ [0, L]. Next, we construct a particular solution of the equation
(6.1.9) below for (x, t, s) ∈ R4 × [0, L] with good properties on Γ0 .
6.1. Fourier integral operators. Let ϕ ∈ C0∞ (Ω) and consider f = f (x, t) ∈
c ∈ L1 R4 . Let h ∈ (0, 1], L ≥ 1, λ ≥ 1 and
L∞ R; L2 R3 be such that ϕf
(xo , ξo3 ) ∈ ωo × (2Z + 1). Denote x = (x1 , x2 , x3 ) and xo = (xo1 , xo2 , xo3 ). First,
ENERGY DECAY FOR MAXWELL’S EQUATIONS WITH OHM’S LAW
2243
let us introduce for any (x, t, s) ∈ R4 × [0, L] and n ∈ Z,
=
(A (xo , ξo3 , n) f ) (x, t, s)
Z Z ξo3 +1 Z
1
i
h
ξo3
n
i(x1 ξ1 +x2 ξ2 +tτ ) i (−1) x3 +2n |ξo3 | ρ ξ3 −i(|ξ|2 −τ 2 )hs
e
e
4
(2π) R2 ξo3 −1 |τ |<λ
c (ξ, τ ) a (x1 − xo1 − 2ξ1 hs, x2 − xo2 − 2ξ2 hs
×ϕf
ξo3
n
ρ − xo3 − 2ξ3 hs, t + 2τ hs, s dξdτ
, (−1) x3 + 2n
|ξo3 |
e
(6.1.1)
where ξ = (ξ1 , ξ2 , ξ3 ) ∈ R2 × [ξo3 − 1, ξo3 + 1],
!
2
1 |x|
t2
1
1
− 4h is+1
− 14 −ihs+1
√
e
e
.
a (x, t, s) =
3/2
−ihs + 1
(is + 1)
(6.1.2)
Next, let us introduce for any (x, t, s) ∈ R4 × [0, L],
(A (xo , ξo3 ) f ) (x, t, s) =
2P
+1
X
n
(−1) (A (xo , ξo3 , n) f ) (x, t, s) ,
(6.1.3)
n=−2Q
(B (xo , ξo3 ) f ) (x, t, s) =
2P
+1
X
(A (xo , ξo3 , n) f ) (x, t, s) ,
(6.1.4)
n=−2Q
where (P, Q) ∈ N2 is the first couple of integer numbers satisfying

1 p

 P ≥
(|ξo3 | + 2) (L2 + 1) + 2 (|ξo3 | + 1) L ,
4ρ 1 p

 Q≥
(|ξo3 | + 2) (L2 + 1) + 2ρ − ro .
4ρ
(6.1.5)
We check after a lengthy but straightforward calculation that for any (x, t, s) ∈
R4 × [0, L],
i∂s + h ∆ − ∂t2 (A (xo , ξo3 ) f ) (x, t, s) = 0 ,
(6.1.6)
i∂s + h ∆ − ∂t2 (B (xo , ξo3 ) f ) (x, t, s) = 0 ,
o3
and that for any (x1 , x2 , t, s) ∈ R3 × [0, L], (A (xo , ξo3 ) f ) x1 , x2 , |ξξo3
| ρ, t, s = 0
and
ξo3
ρ, t, s
(A (xo , ξo3 ) f ) x1 , x2 , −
|ξo3 |
ξo3
= (A (xo , ξo3 , −2Q) f ) x1 , x2 , −
ρ, t, s
(6.1.7)
|ξo3 |
ξo3
+ (A (xo , ξo3 , 2P + 1) f ) x1 , x2 , −
ρ, t, s ,
|ξo3 |
ρ,
t,
s
= 0 , and
∂x3 (B (xo , ξo3 ) f ) x1 , x2 , |ξξo3
o3 |
ξo3
∂x3 (B (xo , ξo3 ) f ) x1 , x2 , −
ρ, t, s
|ξo3 |
ξo3
= ∂x3 (A (xo , ξo3 , −2Q) f ) x1 , x2 , −
ρ, t, s
(6.1.8)
|ξo3 |
ξo3
−∂x3 (A (xo , ξo3 , 2P + 1) f ) x1 , x2 , −
ρ, t, s .
|ξo3 |
2244
KIM DANG PHUNG
dj ∈ L1 R4 for any j ∈
Let fj = fj (x, t) ∈ L∞ R; L2 R3 be such that ϕf
{1, 2, 3}. Let us introduce




f1
A (xo , ξo3 ) f1
F =  f2  and V (xo , ξo3 ) F =  A (xo , ξo3 ) f2 
f3
B (xo , ξo3 ) f3
then by (6.1.6)
i∂s + h ∆ − ∂t2
(V (xo , ξo3 ) F ) (x, t, s) = 0 ∀ (x, t, s) ∈ R4 × [0, L] .
On another hand, let U be


u1
U =  u2  then
u3
the solution of (3.1). Denote

 ∀j ∈ {1, 2, 3} ∂t2 uj − ∆uj = 0
u1 = u2 = 0

∂x3 u3 = 0
(6.1.9)
in Ω × R
on Γ0 × R
on Γ0 × R
because div U = 0 and U × ν = 0. Further, by (3.5) and (3.2),
2
2
kuj (·, t)kH 1 (Ω) + k∂t uj (·, t)kL2 (Ω) ≤ cG (U, 0) ,
(6.1.10)
for some c > 0 and any (t, j) ∈ R × {1, 2, 3}.
By multiplying the equation (6.1.9) by ` (x) U (x, t) and integrating by parts over
Ω × [−T, T ] × [0, L], we have that for all (xo , ξo3 ) ∈ ωo × (2Z + 1) and all h ∈ (0, 1],
L ≥ 1, T > 0,
Z Z T
Z Z T
−i
(V (xo , ξo3 ) F ) (·, ·, 0) · `U dxdt+i
(V (xo , ξo3 ) F ) (·, ·, L) · `U dxdt
Ω −T
Ω −T
!
!
)
Z Z ( Z
Z
T
L
−h
L
A (xo , ξo3 ) f1 ds `∂ν u1 +
Γ0 −T
Z
Z
A (xo , ξo3 ) f2 ds `∂ν u2
0
T
!
L
Z
+h
dσdt
0
∂ν (B (xo , ξo3 ) f3 ) ds `u3 dσdt
−T
Γ0
Z
0
Z
T
Z
−h
!
L
B (xo , ξo3 ) f3 ds ∂ν `u3 dσdt
Γ0 ∩∂ω
Z " Z
−h
−T
0
!
L
∂t (V (xo , ξo3 ) F ) ds
Ω
· `U −
0
Z Z
T
Z
L
(V (xo , ξo3 ) F ) ds
· `∂t U
dx
−T
!
V (xo , ξo3 ) F ds
−T
#T
!
L
0
+h
ω
Z
· [2 (∇` · ∇) U + ∆`U ] dxdt
0
≡I1 + I2 + I3 + I4 + I5 + I6 + I7 = 0 .
(6.1.11)
The different terms of the last equality will be estimated separately. The quantity
I1 will allow us to recover (6.3). The dispersion property for the one dimensional
Schrödinger operator will be used for making I2 small for large L. We treat I3
(resp. I4 ) by applying the formula (6.1.7) (resp. (6.1.8)). The quantity I5 and I7
will correspond to a term localized in ω. Finally, an appropriate choice of T will
bound I6 and give the desired inequality (6.9.1) below.
6.2. Estimate for I1 (the term at s = 0). We estimate the quantity I1 =
R RT
−i Ω −T (V (xo , ξo3 ) F ) (x, t, 0) · ` (x) U (x, t) dxdt as follows.
ENERGY DECAY FOR MAXWELL’S EQUATIONS WITH OHM’S LAW
2245
Lemma 6.1. There exists c > 0 such that for any (xo , ξo3 ) ∈ ωo × (2Z + 1) and
h ∈ (0, 1], λ ≥ 1, T > 0, we have
!
Z
Z Z ξo3 +1 Z
1
c (ξ, τ ) dξdτ
ei(xξ+tτ ) ϕF
I1 + i
4
(2π) R2 ξo3 −1 |τ |<λ
Ω×R
·a (x − xo , t, 0) ` (x) U (x, t) dxdt|
(6.2.1)
!
Z Z ξo3 +1 Z
1
p
2
T
c
≤ c e− ch + e− c
G (U, 0)
ϕF (ξ, τ ) dξdτ .
R2
ξo3 −1
|τ |<λ
Proof. We start with the third component of V (xo , ξo3 ) F . First, from (6.1.1) and
(6.1.4) whenever s = 0,
(B (xo , ξo3 ) f ) (x, t, 0)
2P
+1 X
ξo3
n
ρ − xo3 , t, 0
=
a x1 − xo1 , x2 − xo2 , (−1) x3 + 2n
|ξo3 |
n=−2Q
#
Z Z ξo3 +1 Z
i
h
ξo3
n
1
ρ
ξ
i
(−1)
x
+2n
3
3
|ξo3 |
c dξdτ
×
ei(x1 ξ1 +x2 ξ2 +tτ ) e
ϕf
4
(2π) R2 ξo3 −1 |τ |<λ
Z Z ξo3 +1 Z
1
c dξdτ
ei(xξ+tτ ) ϕf
= a (x − xo , t, 0)
4
(2π) Rh2 ξo3 −1 |τ |<λ
P
n
o3
+
a x1 − xo1 , x2 − xo2 , (−1) x3 + 2n |ξξo3
| ρ − xo3 , t, 0
n∈{−2Q,···,2P +1}\{0}
#
Z Z ξo3 +1 Z
h
i
ξo3
n
1
i
(−1)
x
+2n
ρ
ξ
3
3 c
|ξo3 |
×
ei(x1 ξ1 +x2 ξ2 +tτ ) e
ϕf dξdτ .
4
(2π) R2 ξo3 −1 |τ |<λ
(6.2.2)
Next, we estimate the discrete sum over {−2Q, · · ·, 2P + 1} \{0} . By (6.1.2),
X
ξo3
n
ρ
−
x
,
t,
0
a
x
−
x
,
x
−
x
,
(−1)
x
+
2n
o3
1
o1
2
o2
3
|ξo3 |
n∈{−2Q,···,2P +1}\{0}
#
Z Z ξo3 +1 Z
i
h
ξo3
n
1
i(x1 ξ1 +x2 ξ2 +tτ ) i (−1) x3 +2n |ξo3 | ρ ξ3 c
e
e
ϕf (ξ, τ ) dξdτ 4
2
(2π) R ξo3 −1 |τ |<λ
Z Z ξo3 +1 Z
1
c
≤
ϕf (ξ, τ ) dξdτ
4
2
(2π) R ξo3 −1 |τ |<λ
2
ξ
(−1)n x3 +2n o3 ρ−xo3
X
2
|ξo3 |
|(x −x ,x −x )|2
−
− t4 − 1 o14h2 o2
4h
×e
e
e
.
n∈Z\{0}
(6.2.3)
Remark that for any n ∈ Z \{0} , xo3 ∈ [ρ − 2ro , ρ − ro ] and x3 ∈ [−ρ, ρ],
2
ξo3
n
2
− (−1) x3 + 2n
ρ − xo3
≤ −4ρ2 (|n| − 1) − ro2 ,
|ξo3 |
because
ξo3 2ρ |n| = 2n
ρ
|ξo3 |
ξo3
n
n
≤ (−1) x3 + 2n
ρ − xo3 + |(−1) x3 − xo3 |
|ξo3 |
ξo3
n
≤ (−1) x3 + 2n
ρ − xo3 + 2ρ − ro .
|ξo3 |
2246
KIM DANG PHUNG
Therefore, for some c > 0,
e
−
|(x1 −xo1 ,x2 −xo2 )|2
4h
X
e
−
ξo3
ρ−xo3
|ξo3 |
4h
(−1)n x3 +2n
2
1
≤ ce− ch .
(6.2.4)
n∈Z\{0}
Now, we deduce from (6.2.2), (6.2.3) and (6.2.4) that
Z Z
T
(B (xo , ξo3 ) f ) (x, t, 0) ` (x) u3 (x, t) dxdt
−i
Ω −T
!
Z Z T
Z Z ξo3 +1 Z
1
c (ξ, τ ) dξdτ
+i
ei(xξ+tτ ) ϕf
4
(2π) R2 ξo3 −1 |τ |<λ
Ω −T
t, 0) ` (x) u3 (x, t) dxdt|
×a (x − xo ,
Z Z T
X

= −i
[a (x1 − xo1 , x2 − xo2
Ω −T
n∈{−2Q,···,2P +1}\{0}
ξo3
n
, (−1) x3 + 2n
ρ − xo3 , t, 0
|ξo3 |
#!
Z Z ξo3 +1 Z
h
i
ξo3
n
1
i(x1 ξ1 +x2 ξ2 +tτ ) i (−1) x3 +2n |ξo3 | ρ ξ3 c
×
e
e
ϕf dξdτ
4
(2π) R2 ξo3 −1 |τ |<λ
× ` (x) u3 (x, t) dxdt|
!
Z Z T
Z Z ξo3 +1 Z
t2
c
1
− ch
c dξdτ
≤
e− 4 |` (x) u3 (x, t)| dxdt
ϕf
4e
(2π)
R2 ξo3 −1
Ω −T
|τ |<λ
!
Z Z ξo3 +1 Z
1 p
c
− ch
G (U, 0)
≤ ce
ϕf (ξ, τ ) dξdτ
R2
ξo3 −1
|τ |<λ
(6.2.5)
where in the last line we have used, from Cauchy-Schwarz inequality and conservation of energy (3.2), the fact that the solution U has the following property
Z
T
2
e
−T
− t4
Z
|` (x) u3 (x, t)| dxdt
Ω
Z ∞
2
p
p
− t4
≤ c |Ω|
e
dt
G (U, 0)
−∞
p
≤ c G (U, 0) .
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
Z Z
!
Z Z ξo3 +1 Z
1
i(xξ+tτ ) c
e
ϕf (ξ, τ ) dξdτ
i
Ω R\(−T,T ) (2π)4 R2 ξo3 −1 |τ |<λ
2
2
1 1
×e− 4 ( h |x−xo | +t ) ` (x) u3 (x, t) dxdt
!
Z Z ξo3 +1 Z
p
c
− 81 T 2
≤ ce
G (U, 0)
ϕf (ξ, τ ) dξdτ .
R2
ξo3 −1
|τ |<λ
(6.2.6)
ENERGY DECAY FOR MAXWELL’S EQUATIONS WITH OHM’S LAW
2247
Now, we cut the integral on time into two parts to obtain
!
Z
Z Z ξo3 +1 Z
1
i(xξ+tτ ) c
e
ϕf (ξ, τ ) dξdτ
4
(2π) R2 ξo3 −1 |τ |<λ
Ω×R
×a (x − xo , t, 0) ` (x) u3 (x, t) dxdt
!
Z Z T
Z Z ξo3 +1 Z
1
i(xξ+tτ ) c
=
e
ϕf (ξ, τ ) dξdτ
4
(2π) R2 ξo3 −1 |τ |<λ
Ω −T
×a (x − xo , t, 0) ` (x) u3 (x, t) dxdt
!
Z Z
Z Z ξo3 +1 Z
1
i(xξ+tτ ) c
e
ϕf (ξ, τ ) dξdτ
+
4
(2π) R2 ξo3 −1 |τ |<λ
Ω R\(−T,T )
2
2
1 1
×e− 4 ( h |x−xo | +t ) ` (x) u (x, t) dxdt .
(6.2.7)
3
We conclude from (6.2.5), (6.2.6) and (6.2.7) that
Z Z
T
(B (xo , ξo3 ) f ) (x, t, 0) ` (x) u3 (x, t) dxdt
−i
Ω −T
!
Z Z ξo3 +1 Z
Z
1
i(xξ+tτ ) c
e
ϕf (ξ, τ ) dξdτ
+i
4
(2π) R2 ξo3 −1 |τ |<λ
Ω×R
×a (x − xo , t, 0) ` (x) u3 (x, t) dxdt|
!
Z Z ξo3 +1 Z
p
1
1 2
c (ξ, τ ) dξdτ .
G (U, 0)
≤ c e− ch + e− 8 T
ϕf
R2
ξo3 −1
|τ |<λ
Similarly, using (6.1.3),
Z Z
T
(A (xo , ξo3 ) f ) (x, t, 0) ` (x) uj (x, t) dxdt
−i
Ω −T
!
Z Z ξo3 +1 Z
R
i(xξ+tτ ) c
1
+i Ω×R (2π)4
e
ϕf (ξ, τ ) dξdτ
R2
ξo3 −1
|τ |<λ
×a (x − xo , t, 0) ` (x) uj (x, t) dxdt|
Z Z
1
p
1 2
G (U, 0)
≤ c e− ch + e− 8 T
R2
ξo3 +1
ξo3 −1
Z
|τ |<λ
c
ϕf (ξ, τ ) dξdτ
!
.
This completes the proof.
6.3. Estimate for I2 (the term at s = L). We estimate the quantity I2 =
Z Z T
i
(V (xo , ξo3 ) F ) (x, t, L) · ` (x) U (x, t) dxdt as follows.
Ω
−T
Lemma 6.2. There exists c > 0 such that for any (xo , ξo3 ) ∈ ωo × (2Z + 1) and
h ∈ (0, 1], L ≥ 1, λ ≥ 1, T > 0, we have
!
Z Z ξo3 +1 Z
p
1
1
c
− ch
|I2 | ≤ c √ + e
G (U, 0)
ϕF (ξ, τ ) dξdτ . (6.3.1)
L
R2 ξo3 −1
|τ |<λ
Proof. We start with the third component of V (xo , ξo3 ) F . First,
X
|(B (xo , ξo3 ) f ) (x, t, L)| ≤
|(A (xo , ξo3 , n) f ) (x, t, L)|
n∈Z\{−2Q,···,2P
+1}
X
+
(A (xo , ξo3 , n) f ) (x, t, L) .
n∈Z
(6.3.2)
2248
KIM DANG PHUNG
Next, by (6.1.1) and (6.1.2),
X
|(A (xo , ξo3 , n) f ) (x, t, L)|
n∈Z\{−2Q,···,2P +1}

Z
1
≤
(2π)
4
R2
×
√
× √
ξo3 +1
Z
Z
L2 + 1
e
1
− 4h
|(x1 −xo1 −2ξ1 hL,x2 −xo2 −2ξ2
L2 +1
1
− 4h
X
+1
2
e
1/2 
1/2 e
− 41
(t+2τ hL)2
(hL)2 +1
(hL) + 1
hL)|2

1
L2
1


c
ϕf (ξ, τ )  q

|τ |<λ
ξo3 −1
1





2
ξ
(−1)n x3 +2n o3 ρ−xo3 −2ξ3 hL
|ξo3 |
L2 +1

 dξdτ .
n∈Z\{−2Q,···,2P +1}
(6.3.3)
When ξ3 ∈ (ξo3 − 1, ξo3 + 1) with ξo3 ∈ (2Z + 1),
√
L2 + 1
≤ 4P ρ − 2 (|ξo3 | + 1) L from our choice of P
≤ 4P ρ − 2 |ξ3 | hL + 2ρ − |x3 | − |xo3 | because |x3 | + |xo3 | ≤ 2ρ − ro
ξo3
n
≤ 4P ρ − 2 |ξ3 | hL + 2ρ +
[− (−1) x3 − xo3 ] ∀n ∈ Z
|ξo3 |
thus
X
e
1
− 4h
2
ξ
(−1)n x3 +2n o3 ρ−xo3 −2ξ3 hL
|ξo3 |
L2 +1
n≥2P +2
=
X
e
1
− 4h
ξo3
−(−1)n x3 −xo3 ]
|ξo3 | [
L2 +1
2
2nρ+4P ρ−2|ξ3 |hL+2ρ+
n≥1
2
ξo3
−(−1)n x3 −xo3 ]
|ξo3 | [
2
L +1
≤
X
e
1
− 4h
(2nρ)2
L2 +1
e
1
− 4h
n≥1
≤e
1
− 4h
X
n≥1
e
1
−h
(nρ)2
L2 +1
(6.3.4)
4P ρ−2|ξ3 |hL+2ρ+
≤e
1
− 4h
!
√ √ √ 2
π h L +1
.
2
ρ
Also,
√
L2 + 1
≤ 4Qρ − 2ρ + ro from our choice of Q
≤ 4Qρ + 2 |ξ3 | hL − |x3 | − |xo3 | because |x3 | + |xo3 | ≤ 2ρ − ro
ξo3
n
≤ 4Qρ + 2 |ξ3 | hL −
[(−1) x3 − xo3 ] ∀n ∈ Z
|ξo3 |
ENERGY DECAY FOR MAXWELL’S EQUATIONS WITH OHM’S LAW
2249
thus
X
e
n≤−2Q−1 ≤
X
e
1
− 4h
1
− 4h
(−1)n x3 +2n
ξo3
ρ−xo3 −2ξ3 hL
|ξo3 |
L2 +1
2
2
ξ
2nρ+4Qρ+2|ξ3 |hL− o3 [(−1)n x3 −xo3 ]
|ξo3 |
L2 +1
n≥1
≤
X
e
2
1 (2nρ)
− 4h
L2 +1
e
1
− 4h
n≥1
≤e
1
− 4h
X
e
1
−h
(nρ)2
L2 +1
(6.3.5)
2
ξ
4Qρ+2|ξ3 |hL− o3 [(−1)n x3 −xo3 ]
|ξo3 |
L2 +1
≤e
1
− 4h
n≥1
!
√ √ √ 2
π h L +1
.
2
ρ
Therefore, from (6.3.3), (6.3.4) and (6.3.5), we get that
X
|(A (xo , ξo3 , n) f ) (x, t, L)|
n∈Z\{−2Q,···,2P +1}
1
1 √ 1
√
(2π)4 L2 +1 ( L2 +1)1/2
≤
√
√ √
1
π h L2 +1
e− 4h
ρ


Z
ξo3 +1
Z
Z

c

ϕf (ξ, τ )  q

|τ |<λ
×
R2
ξo3 −1
1
2
1/2 e
− 41
(t+2τ hL)2
(hL)2 +1
(hL) + 1


 dξdτ .

(6.3.6)
Now, by (6.1.1) and (6.1.2),
X
(A (xo , ξo3 , n) f ) (x, t, L)
n∈Z

≤
Z
1
4
(2π)
×
√
Z
R2
ξo3 +1
Z
ξo3 −1
1
e
L2

1
− 4h

c

ϕf (ξ, τ )  q

|τ |<λ
1
2
(hL) + 1
hL)|2
1/2 e
− 14
(t+2τ hL)2
(hL)2 +1




|(x1 −xo1 −2ξ1 hL,x2 −xo2 −2ξ2
L2 +1
+1

2 
ξ
i
h
(−1)n x3 +2n o3 ρ−xo3 −2ξ3 hL
ξo3
n
X
|ξ
|
o3
1
i (−1) x3 +2n |ξ | ρ ξ3 − 1
dξdτ


4h
iL+1
o3
√
×
e
e
iL + 1
n∈Z


≤√
c
L2 + 1
Z
R2
Z
ξo3 +1
ξo3 −1

(t+2τ hL)2 
1
− 41 (hL)2 +1 
c

e
ϕf (ξ, τ )  q
 dξdτ
1/2


|τ |<λ
2
(hL) + 1
Z
(6.3.7)
2250
KIM DANG PHUNG
because from Appendix B with z =
4h
ρ2
(iL + 1) we know that
2 ξ
h
i
(−1)n x3 +2n o3 ρ−xo3 −2ξ3 hL
X
ξo3
n
|ξ
|
o3
i (−1) x3 +2n |ξ | ρ ξ3 − 1
√ 1
4h
iL+1
o3
e
e
iL + 1
n∈Z
√ √
2 h
π
ρ
≤
+√
.
ρ
2
h
Finally, (6.3.2), (6.3.6) and (6.3.7) imply that
|(B
(xo , ξo3 ) f ) (x, t, L)|

1
1
− 1
+ √
≤ c √
1/2 e 4h 
2
L2 + 1
L +1

Z
Z
ξo3 +1
×
R2
ξo3 −1

Z

c

ϕf (ξ, τ )  q

|τ |<λ
1
2
1/2 e
− 41
(t+2τ hL)2
(hL)2 +1
(hL) + 1


 dξdτ

and we conclude that
Z Z
T
(B (xo , ξo3 ) f ) (x, t, L) ` (x) u3 (x, t) dxdt
i
Ω −T


Z Z ξo3 +1 Z
1
1
1
c
− 4h

≤ c √
e
+ √
ϕf
(ξ,
τ
)
dξ
1/2
2
2
L2 + 1
R
ξo3 −1
|τ |<λ
L +1


T
Z
×
−T
≤c
1


 q

2
1/2 e
− 14
(t+2τ hL)2
(hL)2 +1
(6.3.8)
Z

|` (x) u3 (x, t)| dxdtdτ

 Ω
(hL) + 1
!
Z Z ξo3 +1 Z
p
1
1
−
c (ξ, τ ) dξdτ
√ + e 4h
G (U, 0)
ϕf
L
R2 ξo3 −1
|τ |<λ
where in the last line we have used the fact that the solution U has the following
property, from Cauchy-Schwarz inequality and conservation of energy (3.2),
Z
T
e
−T
− 41
(t+2τ hL)2
(hL)2 +1
Z
|`u3 | dxdt
Ω
Z
∞
p
e
dt
G (U, 0)
−∞q
p
p
√
2
≤ c |Ω| 2 π (hL) + 1
G (U, 0) .
p
≤ c |Ω|
− 14
t2
(hL)2 +1
Similarly,
Z Z
T
(A (xo , ξo3 ) f ) (x, t, L) ` (x) uj (x, t) dxdt
i
Ω −T
!
Z Z ξo3 +1 Z
p
1
1
−
c (ξ, τ ) dξdτ ,
≤ c √ + e 4h
G (U, 0)
ϕf
L
R2 ξo3 −1
|τ |<λ
(6.3.9)
ENERGY DECAY FOR MAXWELL’S EQUATIONS WITH OHM’S LAW
2251
using the estimate
|(A (xo , ξo3 ) f ) (x, t, L)|
X
≤
|(A (xo , ξo3 , n) f ) (x, t, L)|
n∈Z\{−2Q,···,2P
+1}
X
n
(−1) (A (xo , ξo3 , n) f ) (x, t, L)
+
n∈Z
and
2 ξ
h
i
(−1)n x3 +2n o3 ρ−xo3 −2ξ3 hL
X
ξo3
n
|ξo3 |
1
1
n i (−1) x3 +2n |ξo3 | ρ ξ3 − 4h
√
iL+1
(−1)
e
e
iL + 1
√ √ n∈Z
2 h
ρ
π
≤
+√
ρ
2
h
deduced from Appendix B with z =
complete the proof.
4h
ρ2
(iL + 1). The estimate (6.3.8) and (6.3.9)
6.4. Estimate for I3 (the boundary term with A). We estimate the quantity
Z
I3
Z
T
Z
= −h
Γ0
Z
−T
Z T
0
Z
!
L
L
+h
A (xo , ξo3 ) f1 (x, t, s) ds ` (x) ∂ν u1 (x, t) dσdt
!
A (xo , ξo3 ) f2 (x, t, s) ds ` (x) ∂ν u2 (x, t) dσdt
Γ0
−T
0
as follows.
Lemma 6.3. There exists c > 0 such that for any (xo , ξo3 ) ∈ ωo × (2Z + 1) and
h ∈ (0, 1], L ≥ 1, λ ≥ 1, T > 0, we have
1
− 4h
|I3 | ≤ chLe
p
Z
Z
ξo3 +1
Z
G (U, 0)
R2
ξo3 −1
|τ |<λ
c
ϕF (ξ, τ ) dξdτ
!
.
(6.4.1)
Proof. First, by (6.1.7), we deduce that
Z
(A (xo , ξo3 ) f ) (x, t, s) ` (x) ∂x3 uj (x, t) dσ Z Γ0
ξo3
≤
(A (xo , ξo3 , −2Q) f ) x1 , x2 , − |ξo3 | ρ, t, s Γ0
ξo3
× `∂x3 uj x1 , x2 , −
ρ, t dσ
|ξo3 |
Z ξo3
+
(A (xo , ξo3 , 2P + 1) f ) x1 , x2 , − |ξo3 | ρ, t, s Γ0
ξo3
× `∂x3 uj x1 , x2 , −
ρ, t dσ .
|ξo3 |
(6.4.2)
2252
KIM DANG PHUNG
Next, recall that
|(A (xo , ξo3 , n) f ) (x, t, s)|

≤
Z
1
(2π)
4
√
×
× √
Z
R2
ξo3 +1

Z
ξo3 −1

c

ϕf (ξ, τ )  q

|τ |<λ
|(x1 −xo1 −2ξ1 hs,x2 −xo2 −2ξ2
1
− 1
s2 +1
e 4h
s2 + 1

1
s2 + 1
1/2 e
1
− 4h
1
2
1/2 e
− 41
(t+2τ hs)2
(hs)2 +1
(hs) + 1
hs)|2
2
ξ
2nρ−2|ξ3 |hs+ o3 [(−1)n x3 −xo3 ]
|ξo3 |
s2 +1





 dξdτ .
(6.4.3)
Here, for any s ∈ [0, L], h ∈ (0, 1], x3 ∈ [−ρ, ρ], xo3 ∈ [ρ − 2ro , ρ − ro ], ξ3 ∈
(ξo3 − 1, ξo3 + 1), ξo3 ∈ (2Z + 1), we have chosen (P, Q) ∈ N2 (only depending on
(ξo3 , L) see (6.1.5)) such that
s2 + 1 ≤
2nρ − 2 |ξ3 | hs +
2
ξo3
n
[(−1) x3 − xo3 ]
|ξo3 |
when n ∈ {−2Q, 2P + 1} .
(6.4.4)
Indeed, for any x3 ∈ [−ρ, ρ] and xo3 ∈ [ρ − 2ro , ρ − ro ],
√
√
s2 + 1 ≤ L2 + 1 ≤ 4P ρ − 2 (|ξo3 | + 1) L from our choice of P
≤ 4P
ρ − 2 |ξ3 | hs + 2ρ − |x3 + xo3 | because
ro ≤ 2ρ − |x3 + xo3 |
ξ
o3
≤ 4P ρ − 2 |ξ3 | hs + 2ρ +
[−x3 − xo3 ]
|ξo3 |
and
√
√
≤ L2 + 1 ≤ 4Qρ − 2ρ + ro from our choice of Q
≤ 4Qρ
+ 2 |ξ3 | hs − |x3 − xo3 | because
|x3 − xo3 | ≤ 2ρ − ro
ξ
o3
[x3 − xo3 ] .
≤ −4Qρ − 2 |ξ3 | hs +
|ξo3 |
s2 + 1
So (6.4.4) implies that
e
1
− 4h
2
ξ
2nρ−2|ξ3 |hs+ o3 [(−1)n x3 −xo3 ]
|ξo3 |
s2 +1
1
≤ e− 4h
when n ∈ {−2Q, 2P + 1} .
(6.4.5)
Therefore, from (6.4.3) and (6.4.5), for any s ∈ [0, L], h ∈ (0, 1], x3 ∈ [−ρ, ρ],
xo3 ∈ [ρ − 2ro , ρ − ro ], ξ3 ∈ (ξo3 − 1, ξo3 + 1), and n ∈ {−2Q, 2P + 1},
|(A (xo , ξo3 , n) f ) (x, t, s)|
1
1
1
− 1
√
≤
√
1/2 e 4h
4
2
s +1
(2π)
s2 + 1

Z
Z
ξo3 +1 Z
×
R2
ξo3 −1

c

ϕf (ξ, τ )  q

|τ |<λ

1
2
(hs) + 1
1/2 e
− 41
(t+2τ hs)2
(hs)2 +1


 dξdτ .

(6.4.6)
ENERGY DECAY FOR MAXWELL’S EQUATIONS WITH OHM’S LAW
On the other hand, by Cauchy-Schwarz inequality
Z
Z T
(t+2τ hs)2
− 41 (hs)2 +1
e
|` (x) ∂x3 uj (x, t)| dσ dt
−T
Γ0
!
1/2 Z
Z
Z
T
≤c
e
− 41
(t+2τ hs)2
(hs)2 +1
T
e
dt
−T
√
≤c 2 π
−T
1/2
q
2
(hs) + 1
Z
T
(t+2τ hs)2
(hs)2 +1
!1/2
2
|`∂x3 uj | dσdt
Γ0
Z
− 14
e
−T
− 41
2253
(t+2τ hs)2
(hs)2 +1
!1/2
2
|`∂x3 uj | dσdt
.
Γ0
(6.4.7)
Next, by multiplying the equation ∂t2 uj − ∆uj = 0 by g`2 ∇uj · W where g (t) =
−1
(t+2τ hs)2
e 4 (hs)2 +1 and W = W (x) is a smooth vector field such that W = ν on Γ0 (see
[9, p. 29]), we get, after integrations by parts and by Cauchy-Schwarz inequality,
observing that `uj = 0 on ∂Ω for any j ∈ {1, 2} and (6.1.10),
Z Z
2
g (t) |` (x) ∂x3 uj (x, t)| dσdt
R
Z Z Γ0
d 2
2
2
(6.4.8)
|uj | + |∇uj | + |∂t uj | dxdt
≤c
g + g dt
Ω
qR
2
≤ c (hs) + 1G (U, 0) .
Therefore, (6.4.7) and (6.4.8) imply that
Z
Z T
q
(t+2τ hs)2
p
−1
2
|` (x) ∂x3 uj (x, t)| dσ dt ≤ c (hs) + 1 G (U, 0) . (6.4.9)
e 4 (hs)2 +1
−T
Γ0
We conclude from (6.4.2) and (6.4.6) that
Z Z
!
Z L
T
A (xo , ξo3 ) f (x, t, s) ds ` (x) ∂ν uj (x, t) dσdt
h
Γ0 −T
0
Z Z ξo3 +1 Z
h −1
c
4h
≤
ϕf (ξ, τ ) dξdτ
4e
2
(2π)
R
ξo3 −1
|τ |<λ
Z L
Z T
Z
(t+2τ hs)2
1
1
− 14 (hs)2 +1
×
e
|`∂x3 uj | dσdtds
√
3/2 q
1/2
0
−T
Γ0
s2 + 1
2
(hs) + 1
!
Z Z ξo3 +1 Z
1 p
c (ξ, τ ) dξdτ ,
≤ chLe− 4h G (U, 0)
ϕf
R2
ξo3 −1
|τ |<λ
where in the last inequality, we used (6.4.9). This completes the proof.
6.5. Estimatefor I4 (the boundary termwith ∂ν B). We estimate the quantity
R RT RL
I4 = h Γ0 −T 0 ∂ν B (xo , ξo3 ) f3 (x, t, s) ds ` (x) u3 (x, t) dσdt as follows.
Lemma 6.4. There exists c > 0 such that for any (xo , ξo3 ) ∈ ωo × (2Z + 1) and
h ∈ (0, 1], L ≥ 1, λ ≥ 1, T > 0, we have
!
Z Z ξo3 +1 Z
1 p
c
− 4h
(6.5.1)
|I4 | ≤ chLe
G (U, 0)
ϕF (ξ, τ ) dξdτ .
R2
ξo3 −1
|τ |<λ
2254
KIM DANG PHUNG
Proof. First, by (6.1.8), we deduce that
Z
(∂
B
(x
,
ξ
)
f
)
(x,
t,
s)
`
(x)
u
(x,
t)
dσ
ν
o o3
3
Z Γ0
∂x3 (A (xo , ξo3 , −2Q) f ) x1 , x2 , − ξo3 ρ, t, s ≤
|ξo3 |
Γ0
ξo3
× `u3 x1 , x2 , −
ρ, t dσ
|ξ
|
o3
Z ξo3
+
∂x3 (A (xo , ξo3 , 2P + 1) f ) x1 , x2 , − |ξo3 | ρ, t, s Γ0 ξo3
× `u3 x1 , x2 , −
ρ, t dσ .
|ξo3 |
(6.5.2)
Next, recall that by (6.1.1) and (6.1.2),
|∂x3 (A (xo , ξo3 , n) f ) (x, t, s)|

≤
Z
1
4
(2π)
R2
Z

ξo3 +1 Z
ξo3 −1

c

ϕf (ξ, τ )  q

|τ |<λ
1
2
1/2 e
− 41
(t+2τ hs)2
(hs)2 +1
(hs) + 1
hs)|2




|(x1 −xo1 −2ξ1 hs,x2 −xo2 −2ξ2
1
− 1
s2 +1
e 4h
+1

2 
ξ
2nρ−2|ξ3 |hs+ o3 [(−1)n x3 −xo3 ]
|ξo3 |
1
1
1
− 4h
 dξdτ .
s2 +1
× √
1/2 |ξ3 | + √ e
2
h
s +1
(6.5.3)
Here, for any s ∈ [0, L], h ∈ (0, 1], x3 ∈ [−ρ, ρ], xo3 ∈ [ρ − 2ro , ρ − ro ], ξ3 ∈
(ξo3 − 1, ξo3 + 1), ξo3 ∈ (2Z + 1), we have chosen (P, Q) ∈ N2 (only depending on
(ξo3 , L) see (6.1.5)) such that
2
ξo3
n
2
(h |ξ3 | + 1) s + 1 ≤ 2nρ − 2 |ξ3 | hs +
[(−1) x3 − xo3 ]
(6.5.4)
|ξo3 |
×
√
s2
when n ∈ {−2Q, 2P + 1}. Indeed, for any x3 ∈ [−ρ, ρ] and xo3 ∈ [ρ − 2ro , ρ − ro ]
p
2
p(h |ξ3 | + 1) (s + 1)
2
≤ (|ξo3 | + 2) (L + 1) ≤ 4P ρ − 2 (|ξo3 | + 1) L
≤ 4P
ρ − 2 |ξ3 | hs + 2ρ − |x3 + xo3 | because
ro ≤ 2ρ − |x3 + xo3 |
ξ
o3
≤ 4P ρ − 2 |ξ3 | hs + 2ρ +
[−x3 − xo3 ]
|ξo3 |
and
p
2
p(h |ξ3 | + 1) (s + 1)
2
≤ (|ξo3 | + 2) (L + 1) ≤ 4Qρ − 2ρ + ro
≤ 4Qρ + 2 |ξ3 | hs − |x3 − xo3 | because
|x3 − xo3 | ≤ 2ρ − ro
ξ
o3
≤ −4Qρ − 2 |ξ3 | hs +
[x3 − xo3 ] .
|ξo3 |
So (6.5.4) implies that when n ∈ {−2Q, 2P + 1}
1
− 1
|ξ3 | + √ e 8h
h
2
ξ
2nρ−2|ξ3 |hs+ o3 [(−1)n x3 −xo3 ]
|ξo3 |
s2 +1
1
≤ 16e− 16h .
(6.5.5)
ENERGY DECAY FOR MAXWELL’S EQUATIONS WITH OHM’S LAW
2255
Therefore, from (6.5.3) and (6.5.5), for any s ∈ [0, L], h ∈ (0, 1], x3 ∈ [−ρ, ρ],
xo3 ∈ [ρ − 2ro , ρ − ro ], ξ3 ∈ (ξo3 − 1, ξo3 + 1), and n ∈ {−2Q, 2P + 1},
|∂x3 (A (xo , ξo3 , n) f ) (x, t, s)|
1
−
1
1
√
≤
√
1/2 ce ch
4
s2 + 1
(2π)
s2 + 1

1
Z
Z
ξo3 +1
×
R2
ξo3 −1

Z

c

ϕf (ξ, τ )  q

|τ |<λ
1
2
1/2 e
− 41
(t+2τ hs)2
(hs)2 +1
(hs) + 1


 dξdτ .

(6.5.6)
On the other hand, by Cauchy-Schwarz inequality, a trace theorem and conservation
of energy (3.2), we have
Z
Z T
(t+2τ hs)2
− 14 (hs)2 +1
e
|` (x) u3 (x, t)| dσ dt
−T
Γ0
!
!1/2
1/2 Z
Z
Z
T
≤c
− 41
e
(t+2τ hs)2
(hs)2 +1
T
dt
e
−T
− 41
−T
(t+2τ hs)2
(hs)2 +1
2
|u3 (x, t)| dσdt
Γ0
!1/2
1/2 Z T
q
(t+2τ hs)2
√
− 41 (hs)2 +1
2
2
≤ c 2 π (hs) + 1
ku3 (·, t)kH 1 (Ω) dt
e
−T
q
p
2
≤c
(hs) + 1
G (U, 0) .
(6.5.7)
We conclude from (6.5.2) and (6.5.6) that
Z Z
!
Z L
T
∂ν (B (xo , ξo3 ) f ) (x, t, s) ds ` (x) u3 (x, t) dσdt
h
Γ0 −T
0
Z Z ξo3 +1 Z
h
1
c
− ch
≤
ϕf (ξ, τ ) dξdτ
4 ce
2
(2π)
R
ξo3 −1
|τ |<λ
Z T
Z
Z L
(t+2τ hs)2
1
1
− 41 (hs)2 +1
×
e
|`u3 | dσdtds
1/2
√
3/2 q
0
−T
Γ0
s2 + 1
2
(hs) + 1
!
Z Z ξo3 +1 Z
1 p
c
− ch
≤ chLe
G (U, 0)
ϕf (ξ, τ ) dξdτ .
R2
ξo3 −1
|τ |<λ
where in the last inequality, we used (6.5.7). This completes the proof.
6.6. Estimate for I5 (the
boundary term on Γ0 ∩∂ω). We estimate the quantiR
R T R L
B
(xo , ξo3 ) f3 (x, t, s) ds ∂ν ` (x) u3 (x, t) dσdt as follows.
Γ0 ∩∂ω −T
0
ty I5 = −h
Lemma 6.5. There exists c > 0 such that for any (xo , ξo3 ) ∈ ωo × (2Z + 1) and
h ∈ (0, 1], L ≥ 1, λ ≥ 1, T > 0, we have
√ |I5 | ≤ ch 1 + hL k(u3 , ∂t u3 )kL2 (ω×(−1−T,T +1))2
!
Z Z ξo3 +1 Z
(6.6.1)
c
×
ϕF (ξ, τ ) dξdτ .
R2
ξo3 −1
|τ |<λ
2256
KIM DANG PHUNG
Proof. Since
P
|(B (xo , ξo3 ) f ) (x, t, s)| ≤
|(A (xo , ξo3 , n) f ) (x, t, s)|
n∈Z
Z Z ξo3 +1 Z
1
c
≤
ϕf
(ξ,
τ
)
4
(2π) R2 ξo3 −1 |τ |<λ
!
2
(t+2τ hs)2
1 |(x1 −xo1 −2ξ1 hs,x2 −xo2 −2ξ2 hs)|
1
− 14 (hs)2 +1
− 4h
1
2
s +1
1/2 e
√
e
× √
2
(hs)2 +1
 s +1

2
ξo3
ρ−xo3 −2ξ3 hs
|
X − 1 (−1)n x3 +2n |ξo3
1
 dξdτ
s2 +1
× √
e 4h
1/2
s2 + 1
n∈Z
and
X
e
1
− 4h
√ √ p
π
h s2 + 1
≤2+
ρ
2
ξ
(−1)n x3 +2n o3 ρ−xo3 −2ξ3 hs
|ξo3 |
s2 +1
n∈Z
2
(see Appendix B with z = 4h
ρ2 s + 1 ), we have
Z
!
Z T Z L
B (xo , ξo3 ) f ds ∂ν ` (x) u3 (x, t) dσdt
h
Γ0 ∩∂ω −T
0
√ √ √
π
Z Z ξo3 +1 Z
2+1
Z L
2
+
h
s
ρ
1
1
c
√
≤h
ϕf (ξ, τ )
√
1/2
4
s2 + 1
(2π) R2 ξo3 −1 |τ |<λ
0
s2 + 1 

Z
Z
1


×  q

2
T
e
1/2
− 14
(t+2τ hs)2
(hs)2 +1
−T
(hs) + 1
Z ξo3 +1 Z
Z


|∂ν `u3 (·, t)| dσdt dsdξdτ

Γ0 ∩∂ω
c
ϕf (ξ, τ ) dξdτ
2
R ξo3 −1
|τ |<λ
√ √
Z ∞
1/2
Z L 1 + h s2 + 1
t2
1
− 21 (hs)
2 +1
×
dt
ds
e
1/2
√
3/2 q
0
−∞
s2 + 1
2
(hs) + 1
!
1/2
Z Z
≤ ch
T
2
×
|∂ν `u3 | dσdt
−T
.
Γ0 ∩∂ω
Now, we shall treat the term
R
T
−T
(6.6.2)
1/2
|∂ν `u3 | dσdt
as follows. Let W =
Γ0 ∩∂ω
2
R
W (x) be a smooth vector field such that W = ν on Γ0 (see [9, p. 29]). Since
i
h
2
2
2
= 2u (∇u · W ) (∇` · W ) + u2 ∇ (∇` · W ) · W
div W u2 (∇` · W )
2
+u2 (∇` · W ) div W ,
we have the following trace theorem
Z
Z
Z
2
2
2
2
|∂ν `u| dσ ≤ c |u| dx + c |∇u| (∇` · W ) dx .
Γ0
ω
(6.6.3)
ω
2
Next, by multiplying the equation ∂t2 u3 − ∆u3 = 0 by u3 (∇` · W ) g where g ∈
C0∞ (−1 − T, T + 1) and g = 1 in (−T, T ), we get, after integrations by parts and
ENERGY DECAY FOR MAXWELL’S EQUATIONS WITH OHM’S LAW
2257
by Cauchy-Schwarz inequality, observing that ∂ν u3 ∂ν ` = 0 on ∂Ω,
Z Z T
Z Z T +1 2
2
2
2
|∇u3 | (∇` · W ) dxdt ≤ c
|u3 | + |∂t u3 | dxdt .
(6.6.4)
Ω
−T
ω
−1−T
Therefore, combining (6.6.3), (6.6.4) and (6.6.2), we conclude that
Z
!
Z T Z L
B (xo , ξo3 ) f ds ∂ν ` (x) u3 (x, t) dσdt
h
Γ0 ∩∂ω −T
0
√ ≤ ch 1 + hL k(u3 , ∂t u3 )kL2 (ω×(−1−T,T +1))2
Z Z ξo3 +1 Z
c
×
ϕf (ξ, τ ) dξdτ .
R2
ξo3 −1
|τ |<λ
This completes the proof.
6.7. Estimate for I6 (the term at t = ±T ). We estimate the quantity
Z " Z
I6
= −h
· ` (x) U (x, t)
∂t (V (xo , ξo3 ) F ) (x, t, s) ds
Ω
#t=T
!
L
0
Z " Z
+h
(V (xo , ξo3 ) F ) (x, t, s) ds
Ω
t=−T
#t=T
!
L
dx
· ` (x) ∂t U (x, t)
0
dx
t=−T
as follows.
Lemma 6.6. There exists c > 0 such that for any (xo , ξo3 ) ∈ ωo × (2Z + 1) and
h ∈ (0, 1], L ≥ 1, λ ≥ 1, we have
!
Z Z ξo3 +1 Z
1 p
c
−h
|I6 | ≤ chLλe
G (U, 0)
(6.7.1)
ϕF (ξ, τ ) dξdτ
R2
ξo3 −1
|τ |<λ
when
T =4
λhL √
1
√ + hL + √
2
h
.
Proof. Since
|∂t (A (xo , ξo3 , n) f ) (x, t, s)|
Z Z ξo3 +1 Z
1
c
≤
ϕf
(ξ,
τ
)
4
2
(2π)
R
ξ
−1
|τ
|<λ
o3
|(x1 −xo1 −2ξ1 hs,x2 −xo2 −2ξ2 hs)|2
1
− 1
s2 +1
e 4h
× √
 s2 + 1
× √
1
s2 + 1
1/2 e
1
− 4h
2
ξ
(−1)n x3 +2n o3 ρ−xo3 −2ξ3 hs
|ξo3 |
s2 +1


× |τ | +




(t+2τ hs)2 
1 |t + 2τ hs|  
1
− 41 (hs)2 +1 

q
e
 q
 dξdτ ,
1/2
2


2
2
(hs) + 1
(hs) + 1
2258
KIM DANG PHUNG
we have
|(A (xo , ξo3 ) f ) (x, ±T, s)| + |∂t (A (xo , ξo3 ) f ) (x, ±T, s)|
+ |(B (xo , ξo3 ) f ) (x, ±T, s)| + |∂t (B (xo , ξo3 ) f ) (x, ±T, s)|
Z Z ξo3 +1 Z
1
c
≤c
ϕf (ξ, τ ) √ 2
2
s +1
ξo3 −1
|τ |<λ
R

2 
ξ
(−1)n x3 +2n o3 ρ−xo3 −2ξ3 hs
|ξo3 |
X
1
1
−

s2 +1
e 4h
× √
1/2
2
s +1
n∈Z


!
(±T +2τ hs)2
1
|±T
+
2τ
hs|
− 14 (hs)2 +1
1
 √
1/2 e
× 1 + λ + q
dξdτ
2
2
(hs)2 +1
(hs) + 1
√ √
Z Z ξo3+1 Z
1 (±T+2τ hs)2
1
1+ h s2 +1
c − 8 (hs)2+1
dξdτ .
≤c √
ϕf e
1/2 λ
3/2 q
2
ξo3 −1 |τ |<λ
R
s2 +1
2
(hs) +1
(6.7.2)
On the other hand,
e
− 81
(±T +2τ hs)2
(hs)2 +1
Now, when T = 4
implies that
2
≤e
λhL
√
2
− T16
1
(hL)2 +1
2
1 (2λhL)
e 8 (hL)2 +1 ∀s ∈ [0, L] , |τ | < λ .
√
2
2
+ hL + √1h , then 12 (λhL) + h1 (hL) + 1 ≤
−T
2
2
1 (2λhL)
1
(6.7.3)
2
T
16
which
1
e 16 (hL)2 +1 e 8 (hL)2 +1 ≤ e− h .
(6.7.4)
In conclusion, combining (6.7.2), (6.7.3), (6.7.4) and conservation of energy (3.2),
we get
!
#T
Z " Z L
−h
∂t (V (xo , ξo3 ) F ) (·, t, ·) ds · `U (·, t)
dx
Ω
0
−T
!
#T
Z "
Z L
−h
−
(V (xo , ξo3 ) F ) (·, t, ·) ds · `∂t U (·, t)
dx
Ω
0
−T
!
Z Z ξo3 +1 Z
1 p
c
≤ chLλe− h G (U, 0)
ϕF (ξ, τ ) dξdτ .
R2
ξo3 −1
|τ |<λ
This completes the proof.
6.8. Estimate for I7 (the internal term in ω). We estimate the quantity
!
Z Z T Z L
I7 = h
V (xo , ξo3 ) F (x, t, s) ds
ω
−T
0
× [2 (∇` (x) · ∇) U (x, t) + ∆` (x) U (x, t)] dxdt
as follows.
Lemma 6.7. There exists c > 0 such that for any (xo , ξo3 ) ∈ ωo × (2Z + 1) and
h ∈ (0, 1], L ≥ 1, λ ≥ 1, T > 0, we have
√ |I7 | ≤ ch 1 + hL k(U, ∂t U )kL2 (ω×(−1−T,T +1))6
Z Z ξo3 +1 Z
(6.8.1)
c
×
ϕF (ξ, τ ) dξdτ .
R2
ξo3 −1
|τ |<λ
ENERGY DECAY FOR MAXWELL’S EQUATIONS WITH OHM’S LAW
2259
Proof. We start with the third component of V (xo , ξo3 ) F . Since
|(B (xo , ξo3 ) f ) (x, t, s)| ≤
X
|(A (xo , ξo3 , n) f ) (x, t, s)|
n∈Z
≤
Z
1
Z
ξo3 +1
Z
c
ϕf (ξ, τ )
4
(2π) R2 ξo3 −1 |τ |<λ
!
2
(t+2τ hs)2
1 |(x1 −xo1 −2ξ1 hs,x2 −xo2 −2ξ2 hs)|
1
− 14 (hs)2 +1
− 4h
1
2 +1
s
√
1/2 e
× √
e
2
(hs)2 +1
 s +1
2 
ξo3
ρ−xo3 −2ξ3 hs
|
X − 1 (−1)n x3 +2n |ξo3
1
4h

 dξdτ
s2 +1
e
×
√
1/2
2
s +1
n∈Z
and
X
e
1
− 4h
2
ξ
(−1)n x3 +2n o3 ρ−xo3 −2ξ3 hs
|ξo3 |
s2 +1
≤2+
n∈Z
(see Appendix B with z =
4h
ρ2
√ √ p
π
h s2 + 1
ρ
s2 + 1 ), we have
|(B (xo , ξo3 ) f ) (x, t, s)|
Z Z ξo3 +1 Z
(t+2τ hs)2
1
1
− 41 (hs)2 +1
c
dξdτ
≤
ϕf (ξ, τ ) q
1/2 e
4
(2π) R2 ξo3 −1 |τ |<λ
2
(hs) + 1
√π √ √
2
1
2+ ρ
h s +1
×√
.
√
2 +1 1/2
2
s
(
)
s +1
Consequently, we deduce from the above that
Z Z
T
h
ω −T
Z Z T
≤ ch
L
Z
0
Z
L
(B (xo , ξo3 ) f ) (x, t, s) ds [2∇`∇u3 + ∆`u3 ] (x, t) dxdt
!
!
|(B (xo , ξo3 ) f ) (x, t, s)| ds (|∇`| |∇u3 | + |u3 |) (x, t) dxdt
√ √ √
π
Z Z ξo3 +1 Z
Z L
2+1
h
s
2
+
ρ
1
1
c
√
≤ ch
ϕf (ξ, τ )
√
1/2
4
2+1
s
(2π) R2 ξo3 −1 |τ |<λ
0
s2 + 1
Z T
Z
(t+2τ hs)2
1
− 41 (hs)2 +1
×
(|∇`| |∇u3 | + |u3 |) dxdtdsdξdτ
q
1/2 e
ω
−T
−T
0
2
(hs) + 1
ω
2260
KIM DANG PHUNG
which implies using Cauchy-Schwarz inequality
Z Z
!
Z L
T
(B (xo , ξo3 ) f ) (x, t, s) ds [2∇`∇u3 + ∆`u3 ] (x, t) dxdt
h
ω −T
0
Z Z ξo3 +1 Z
1
c
≤ ch
ϕf (ξ, τ ) dξdτ
4
(2π) R2 ξo3 −1 |τ |<λ
√ √
Z ∞
1/2
Z L 1 + h s2 + 1
t2
1
− 21 (hs)
2 +1
e
ds
×
dt
1/2
√
3/2 q
0
−∞
s2 + 1
2
(hs) + 1
!1/2
Z Z T 2
2
2
|∇`| |∇u3 | + |u3 | dxdt
×
ω
−T
!1/2
Z Z T 2
2
2
≤ ch 1 + hL
|∇`| |∇u3 | + |u3 | dxdt
ω −T
Z Z ξo3 +1 Z
c
×
ϕf (ξ, τ ) dξdτ .
√
R2
ξo3 −1
|τ |<λ
Similarly, for any j ∈ {1, 2},
Z Z
T
h
ω −T
(A (xo , ξo3 ) f ) (x, t, s) ds [2∇`∇uj + ∆`uj ] (x, t) dxdt
0
!1/2
Z
Z
T √ 2
2
2
|∇`| |∇uj | + |uj | dxdt
≤ ch 1 + hL
ω −T
Z Z ξo3 +1 Z
c
×
ϕf (ξ, τ ) dξdτ .
R2
Z
ξo3 −1
!
L
|τ |<λ
Now, we shall bound the term
R R T
ω
−T
2
2
2
|∇`| |∇uj | + |uj |
dxdt
1/2
for any j ∈
{1, 2, 3} by the quantity k(U, ∂t U )kL2 (ω×(−1−T,T +1))6 . By multiplying the equation
2
∂t2 uj − ∆uj = 0 by uj |∇`| g where g ∈ C0∞ (−1 − T, T + 1) and g = 1 in (−T, T ),
we get, after integrations by parts and by Cauchy-Schwarz inequality, observing
that uj ∂ν uj |∇`| = 0 on ∂Ω,
Z Z
T
2
Z Z
2
T +1
|∇uj | |∇`| dxdt ≤ c
Ω
−T
ω
2
|uj | + |∂t uj |
2
dxdt ,
−1−T
for any j ∈ {1, 2, 3}. This completes the proof.
6.9. Key inequality. From now,
T =4
λhL √
1
√ + hL + √
2
h
.
By (6.1.11), (6.2.1), (6.3.1), (6.4.1), (6.5.1), (6.6.1), (6.7.1) and (6.8.1), there exists
c > 0 such that for any (xo , ξo3 ) ∈ ωo × (2Z + 1) and h ∈ (0, 1], L ≥ 1, λ ≥ 1, we
ENERGY DECAY FOR MAXWELL’S EQUATIONS WITH OHM’S LAW
2261
have
Z
!
Z Z ξo3 +1 Z
1
c (ξ, τ ) dξdτ
ei(xξ+tτ ) ϕF
Ω×R (2π)4 R2 ξo3 −1 |τ |<λ
·a (x − xo , t, 0) ` (x) U (x, t) dxdt|
Z Z ξo3 +1 Z
p
1
1
c
− ch
≤ c (1 + hLλ) e
G (U, 0)
+√
ϕF dξdτ
L
|τ |<λ
R2 ξo3 −1
√ +ch 1 + hL k(U, ∂t U )kL2 (ω×(−1−T,T +1))6
Z Z ξo3 +1 Z
c
×
ϕF dξdτ .
R2
ξo3 −1
(6.9.1)
|τ |<λ
By summing over ξo3 ∈ (2Z + 1), it implies that
Z
!
Z Z
1
i(xξ+tτ ) c
e
ϕF (ξ, τ ) dξdτ
Ω×R (2π)4 R3 |τ |<λ
·a(x − xo , t, 0) ` (x) U (x, t) dxdt|
Z Z
p
1
1
c
− ch
+√
G (U, 0)
≤ c (1 + hLλ) e
ϕF dξdτ
3
L
R
|τ |<λ
Z Z
√
c
+ch 1 + hL k(U, ∂t U )kL2 (ω×(−1−T,T +1))6
ϕF dξdτ .
R3
(6.9.2)
|τ |<λ
On the other hand, from (A2) of Appendix A,
Z Z
√
1 p
c
2
G (U, 0)
ϕF (ξ, τ ) dξdτ ≤ c λ λ +
h
R3 |τ |<λ
(6.9.3)
whenever F = (f1 , f2 , f3 ) with fj = uj . Therefore, by (6.9.2) with (6.9.3), we obtain
that
Z
!
Z Z
1
i(xξ+tτ ) d
e
ϕU (ξ, τ ) dξdτ
Ω×R (2π)4 R3 |τ |<λ
·a(x − xo , t, 0) ` (x) U (x, t) dxdt|
√
1
1
≤ c (1 + hLλ) e− ch + √
λ λ2 + h1 G (U, 0)
L
√ √ 1 p
2
+ch 1 + hL k(U, ∂t U )kL2 (ω×(−1−T,T +1))6
λ λ +
G (U, 0)
h
1 √
1
1
1 √
≤ c√
λ λ2 +
G (U, 0) + ce− ch λ (1 + hLλ) λ2 +
G (U, 0)
h
h
L
√ 2 √
2
+ch 1 + hL
Lλ 1 + hλ2 k(U, ∂t U )kL2 (ω×(−1−T,T +1))6 ,
where in√the last inequality we used Cauchy-Schwarz. Next, we choose L ≥ 1 so
2
that √1L λ λ2 + h1 = √1λ i.e., L = λ2 λ2 + h1 . Then there exist c, δ, η > 0 such
that for any xo ∈ ωo and h ∈ (0, 1], λ ≥ 1, we have
Z
!
Z Z
1
i(xξ+tτ ) d
e
ϕU (ξ, τ ) dξdτ
Ω×R (2π)4 R3 |τ |<λ
·a (x − xo , t, 0) ` (x) U (x, t) dxdt|
δ
1
λδ
1 λ
2
≤ c √ G (U, 0) + ce− ch η G (U, 0) + c η k(U, ∂t U )kL2 (ω×(−1−T,T +1))6 ,
h
h
λ
2262
KIM DANG PHUNG
and further
1+T =1+4
λhL √
1
√ + hL + √
2
h
≤c
λδ
.
hη
This implies our claim (6.4) since ao (x, t) = a (x − xo , t, 0). This completes the
proof of Proposition 6.
Appendix A. The goal of this Appendix A is to prove the two following inequalities
(A1) and (A2) below.
Lemma A. Let
2
1
1
ao (x, t) = e− c1 h |x−xo | e− c2 t
2
1
2
1
2
ϕ (x, t) = φ (x) e− c3 h |x−xo | e− c4 t
for some c1 , c2 , c3 , c4 > 0 and φ ∈ C0∞ (Ω). Let ` ∈ C ∞ R3 be such that 0 ≤
` (x) ≤ 1. There
exists c > 0 such that for any h ∈ (0, 1], λ ≥ 1 and any u ∈
C 1 R, H 1 (Ω) ∩ C 2 R, L2 (Ω) satisfying
and
∂t2 u − ∆u = 0
in Ω × R ,
and
ku (·, t)kL2 (Ω) ≤ R0 ,
k∂t u (·, t)kL2 (Ω) + k∇u (·, t)kL2 (Ω) ≤ R1 ,
we have
Z
!
Z Z
1
i(xξ+tτ )
e
ϕu
c
(ξ,
τ
)
dξdτ
a
(x,
t)
`
(x)
u
(x,
t)
dxdt
o
Ω×R (2π)4 R3 |τ |≥λ
r
1
≤c
R0 (R0 + R1 )
λ
(A1)
and
Z Z
√ 1
√
1
|ϕu
c (ξ, τ )| dξdτ ≤ c λ λ2 +
R0 + c λ √ R1 .
(A2)
h
h
R3 |τ |<λ
Proof of A1. Introduce
Z
R (f ) =
Ω×R
Z
1
(2π)
4
!
Z
e
R3
i(xξ+tτ ) c
ϕf (ξ, τ ) dξdτ
ao (x, t) ` (x) u (x, t) dxdt .
|τ |≥λ
Thus,
Z
|R (f )| = Ω×R
×∂t
Z
= ` (x) ∂t (ao u (x, t))
Z
!
Z
1
1 itτ
−iθτ
e
e
(ϕf ) (x, θ) dθ dτ dxdt .
2π |τ |≥λ iτ
R
Ω×R
×
ao (x, t) ` (x) u (x, t)
!
Z Z
1 i(xξ+tτ ) c
1
e
ϕf (ξ, τ ) dξdτ dxdt
4
iτ
3
(2π) R |τ |≥λ
ENERGY DECAY FOR MAXWELL’S EQUATIONS WITH OHM’S LAW
2263
It follows using Cauchy-Schwarz inequality and Parseval identity that
Z
|R (f )| ≤
|` (x) ∂t (ao u (x, t))|
Ω×R


"Z
#1/2 "Z Z
2 #1/2
1
1
e−iθτ (ϕf ) (x, θ) dθ dτ
 dxdt
dτ
×
2π |τ |≥λ τ 2
R
R
Z
≤
|∂t (ao u (x, t))|
Ω×R


"Z
#1/2 Z
1/2
1
1
2
 dxdt
×
dτ
2π |(ϕf ) (x, θ)| dθ
2π |τ |≥λ τ 2
R
!
r
Z
1
2
≤
|∂t (ao u (x, t))| √
k(ϕf ) (x, ·)kL2 (R) dxdt
2π λ
Ω×R
r Z
1
1
k∂t (ao u) (·, t)kL2 (Ω) dt kϕf kL2 (Ω×R) .
≤√
π λ R
R
It remains to estimate R k∂t (ao u) (·, t)kL2 (Ω) dt. We have
Z
k∂t (ao u) (·, t)kL2 (Ω) dt
1/2
1/2
ZR Z
Z Z
2
2
≤
|∂t ao u (x, t)| dx
dt +
|ao ∂t u (x, t)| dx
dt
R
Ω
R
Ω
Z
1/2 Z
1/2 !
Z
2 |t|
2
2
− c1 t2
≤
e 2
|u (x, t)| dx
+
|∂t u (x, t)| dx
dt
c2
R
Ω
Ω
≤ c (R0 + R1 ) ,
where in the last line we used
1/2
Z
Z
t2
2
|u (x, t)| dxdt
≤ cR0 .
e− c
Ω
R
We conclude that
r
|R (u)| ≤ c
1
R0 (R0 + R1 ) .
λ
That completes the proof of (A1).
Proof of A2. By Cauchy-Schwarz inequality,
Z Z
c
ϕf (ξ, τ ) dξdτ
3
R
|τ
|<λ
Z Z
1
2 c
=
1
+
|ξ|
ϕf
(ξ,
τ
)
dξdτ
2
R3 |τ |<λ 1 + |ξ|

1/2
Z Z
Z
2 1/2
1


\
≤
dξ
(1
−
∆)
(ϕf
)
(ξ,
τ
)
dτ
dξ

2 
3
3
2
|τ |<λ
R
R
1 + |ξ|
"Z Z
#1/2
2
√
2
\
≤π λ
.
(1 − ∆) (ϕf ) (ξ, τ ) dξdτ
R3
|τ |<λ
2264
KIM DANG PHUNG
Since
(1 − ∆) (ϕu)
= ϕu − ϕ∆u + ∆ϕu + 2∇ϕ∇u
= −ϕ∂t2 u + ϕu + ∆ϕu + 2∇ϕ∇u
= −ϕ∂t2 u + ϕu + ∆ϕu + 2∇ϕ∇u
= −∂t2 (ϕu) + ∂t2 ϕu + 2∂t ϕ∂t u + ϕu + ∆ϕu + 2∇ϕ∇u
2
2
and ∂\
c we get when f = u, using Parseval identity
t (ϕu) = −τ ϕu,
Z Z
|ϕu
c (ξ, τ )| dξdτ
R3 |τ |<λ
√ √
≤ c λλ2 kϕukL2 (Ω×R) + c λ ∂t2 ϕuL2 (Ω×R) + k∂t ϕ∂t ukL2 (Ω×R)
√ +c λ kϕukL2 (Ω×R) + k∆ϕukL2 (Ω×R) + k∇ϕ∇ukL2 (Ω×R) .
2
1
1
2
1
On the other hand, remark that ∂xj ϕ (x, t) = hj/2
φj (x) e− c3 h |x−xo | e− c4 t for some
2
1
1 2
φj ∈ C0∞ (Ω) and ∂t2 ϕ (x, t) + |∂t ϕ (x, t)| ≤ cφ (x) e− c3 h |x−xo | e− c t . It implies
that
Z Z
√ 2
√
1
1
|ϕu
c (ξ, τ )| dξdτ ≤ c λλ R0 + c λ
R0 + √ R1 .
h
h
R3 |τ |<λ
We conclude that there exists c > 0 such that for any h ∈ (0, 1] and λ ≥ 1,
Z Z
√ 1
√
1
R0 + c λ √ R1 .
|ϕu
c (ξ, τ )| dξdτ ≤ c λ λ2 +
h
h
R3 |τ |<λ
That completes the proof of (A2).
Appendix B. The goal of this Appendix B is to prove the two following inequalities.
Lemma B. For any x, y, C, R ∈ R, any z ∈ C, Re z > 0,
√
1 X
2
π
n − z1 (2n+C(−1)n +R)2 inx i(−1)n y (−1) e
+√
,
e e
√
≤
z
2
Re z
n∈Z
√
1 X 1
n
2
n π
2
e− z (2n+C(−1) +R) einx ei(−1) y ≤
+√
.
√
z
2
Re z
n∈Z
Proof. First we recall the Poisson summation formula. Let u ∈ C 2 (R, C) be such
that for any k ∈ {0, 1, 2}, the functions x 7−→ 1 + x2 u(k) (x) are bounded on R.
Then for any x ∈ R,
Z
X
X
u (x + n) =
u
b (2πn) e2πinx where u
b (2πn) =
u (t) e−2πint dt .
n∈Z
R
n∈Z
−2πiBx
Next, by choosing u (x) = v (x) e
for some B ∈ R and v ∈ C 2 (R, C) such
that for any k ∈ {0, 1, 2}, the functions x 7−→ 1 + x2 v (k) (x) are bounded on R,
we obtain that for any x, B ∈ R,
X
X
vb (2π (n + B)) e2πinx =
v (x + n) e−2πiB(x+n) ,
n∈Z
n∈Z
R
−2πi(n+B)t
where vb (2π (n + B)) = R v (t) e
dt. Now, for any z ∈ C such that Re z >
√
2
1
− z2 x2
0, we take v (x) = e
in order that vb (2π (n + B)) = √2π
e− 2z (2π(n+B)) . Thus,
z
ENERGY DECAY FOR MAXWELL’S EQUATIONS WITH OHM’S LAW
2265
after simple changes, the following formula holds for any x, B ∈ R, any z ∈ C, Re z >
0, a > 0,
√ X
2
z
1 X − a (n+B)2 i2nx
π
√
e z
e− a (x+πn) e−i2B(x+πn) .
e
= √
z
a
n∈Z
n∈Z
Finally, we deduce that for any x, y, C, R ∈ R, any z ∈ C, Re z > 0,
1 X
n − z1 (2n+C(−1)n +R)2 inx i(−1)n y (−1) e
e e
√
z
n∈Z
1 X 1
1 X − 1 (4n+2−C+R)2 i(2n+1)x −iy − z (4n+C+R)2 i2nx iy
= √
e
e
e −√
e z
e
e z
z
n∈Z
n∈Z
2
2
1 X − 4z2 (n+ C+R
1 X − 4z2 (n+ 2−C+R
) ei2nx 4 ) ei2nx − e−iy eix √
4
= eiy √
e
e
z
z
n∈Z
√ n∈Z
X
2
C+R
z
π
= eiy
e− 42 (x+πn) e−i 2 (x+πn)
4
n∈Z
√ X
2
2−C+R
z
π
(x+πn)
−
−i
(x+πn)
2
e
−e−iy
e 42
4
√ X n∈Z
√
2
Re z 2 x
π
π
2
≤
e− 42 π ( π +n) ≤
+√
,
2
2
Re z
n∈Z
and similarly
1 X 1
− z (2n+C(−1)n +R)2 inx i(−1)n y e
e e
√
z
n∈Z
X
X 1
2
2
1
1
iy 1
e− z (4n+C+R) ei2nx + e−iy eix √
e− z (4n+2−C+R) ei2nx = e √
z
z
n∈Z
n∈Z
√
2
π
≤
+√
.
2
Re z
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KIM DANG PHUNG
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Received January 2012; revised June 2012.
E-mail address: kim dang phung@yahoo.fr