DISCRETE AND CONTINUOUS
DYNAMICAL SYSTEMS
Volume 20, Number 4, April 2008
Website: http://aimSciences.org
pp. 1057–1093
BOUNDARY STABILIZATION FOR THE WAVE EQUATION
IN A BOUNDED CYLINDRICAL DOMAIN
Kim Dang Phung
Yangtze Center of Mathematics, Sichuan University
Chengdu 610064, China
(Communicated by Irena Lasiecka)
Abstract. We provide a polynomial decay rate for the energy of the wave
equation with a dissipative boundary condition in a cylindrical trapped domain. A new kind of interpolation estimate for the wave equation with mixed
Dirichlet-Neumann boundary condition is established from a construction based
on a Fourier integral operator involving a good choice of weight functions.
1. Introduction and main results. In this paper we study the behavior of the
energy for the wave equation in a bounded cylindrical domain Ω in R3 with a
boundary damping.
Let D be an open connected bounded set in R2 with C 2 boundary ∂D and let
Ω = D × (−1, 1) ⊂ R3 with boundary ∂Ω = Γ0 ∪ Γ1 where Γ0 = D × {−1, 1} and
Γ1 = ∂D × (−1, 1). Let mD ∈ C (Γ1 ), mN ∈ C (Γ1 ) be two functions bounded
above by some positive constants. Denote
©
W ≡ (w0 , w1 ) ∈ H 1 (Ω) × L2 (Ω) ; (w1 , ∆w0 ) ∈ H 1 (Ω) × L2 (Ω) , ∂ν w0 |Γ0 = 0,
©
2
1
= (w0 , w1 ) ∈ H (Ω) × H (Ω) ; ∂ν w0 |Γ0
mD ∂ν w0 + mN w1 |Γ1 = 0}
ª
= 0, mD ∂ν w0 + mN w1 |Γ1 = 0 .
We consider the following wave equation

∂t2 w − ∆w = 0



 m (x) ∂ w + m (x) ∂ w = 0
D
ν
N
t

∂
w=0
ν



(w (·, 0) , ∂t w (·, 0)) = (w0 , w1 ) ∈ W
in Ω × R+
on Γ1 × R+
on Γ0 × R+
(1.1)
in Ω ,
where ν is the outward normal vector to ∂Ω and ∂ν is the normal derivative. We
define by E (w, t) the energy of the solution w = w (x, t),
Z ³
´
2
2
E (w, t) ≡
|∂t w (x, t)| + |∇w (x, t)| dx
Ω
Z tZ
(1.2)
2
= E (w, 0) − 2
α (x) |∂ν w (x, θ)| dσdθ ,
0
Γ1
2000 Mathematics Subject Classification. Primary: 35L05, 35S30; Secondary: 49J20.
Key words and phrases. Wave equation, Mixed Dirichlet-Neumann boundary condition, Polynomial energy decay rate.
This work is supported by the NCET of China under grant NCET-04-0882, the NSF of China
under grant 10525105 and 10771149, and by China Postdoctoral Science Foundation.
1057
1058
KIM DANG PHUNG
mD
where α = m
.
N
Then the main result of this paper is the following
Theorem 1.1. There exist C > 0 and δ > 0 such that for any t > 0, the solution
w of (1.1) satisfies
E (w, t) ≤
C
(E (w, 0) + E (∂t w, 0)) .
tδ
Boundary stabilization for the wave equation in a bounded domain have been discussed by very many mathematicians, notably, [1], [5], [4], [6], [7], [9], [10], [11], [8],
[12], [13], to mention a few. In comparison with the existing literature:
1. the damping region chosen here does not satisfy a “geometric optics condition”. Indeed, there exists a ray which bounces infinitely on Γ0 without meeting Γ1 . Consequently, we can not expect a uniform and exponential decay
rate of the energy;
2. the polynomial decay rate established here improves the logarithmic one, available without any assumption on the dynamics of the rays, which can be deduced using Carleman estimates. Notice that with our particular geometry,
we only need to pay attention to the trapped rays parallel to the x3 −axis.
Let us now give some ideas on the proof. It is by now well-known that the strong
stabilization (i.e., a uniform and exponential decay rate of the energy) of the damped
wave equation is equivalent to an observability estimate for the wave equation. On
the other hand, the logarithmic decay rate of the energy for the wave, on a smooth
compact Riemannian manifold with boundary ∂M , damped with the boundary
condition ∂ν w + α (x) ∂t w |∂M = 0 where α ∈ C ∞ , with respect to a stronger norm
of the initial data is deduced in [11] from an interpolation inequality which traduces
a logarithmic type of dependence (it is also worthwhile to mention the works of [2]
and [3] who established a decay rate (logarithmic and polynomial) for regular initial
data from precise estimate on the resolvent). Our approach will consist to get a
Hölder type of dependence for the wave equation with mixed Dirichlet-Neumann
boundary condition.
Let {µj , `j }j≥1 be the couple satisfying the spectral problem

in Ω

 −∆`j = µj `j
`j = 0
on Γ1

 ∂ ` =0
on Γ .
ν j
0
The corresponding normalized eigenfunctions, i.e., k`j kL2 (Ω) = 1 and eigenvalues
0 < µ1 ≤ µ2 ≤ µ3 ≤ · · ·, take the form
`j (x1 , x2 , x3 ) = φm (x1 , x2 ) cn cos
³ nπ
´
³ nπ ´2
(x3 + 1) , µj = λm +
,
2
2
(1.3)
where n ∈ N, cn > 0 and {λm }m≥1 , 0 < λ1 < λ2 ≤ λ3 ≤ · · ·, are the eigenvalues
of −∆ on D with Dirichlet boundary condition with the corresponding normalized
eigenfunctions {φm }m≥1 .
Denote
©
ª
V = (u0 , u1 ) ∈ H 1 (Ω) × L2 (Ω) ; u0 |Γ1 = 0 ,
BOUNDARY STABILIZATION FOR THE WAVE EQUATION
1059
and let u = u (x, t) be the solution of the wave equation





∂t2 u − ∆u = 0
u=0
in Ω × R
on Γ1 × R




∂ν u = 0
on Γ0 × R
(u (·, 0) , ∂t u (·, 0)) = (u0 , u1 ) ∈ V
Suppose that u0 =
P
j≥1
2
ku0 kH 2 (Ω) '
b0j `j and u1 =
X
P
j≥1
(1.4)
in Ω .
b1j `j are such that
X ¯ ¯2
¯ ¯2
2
µj ¯b1j ¯ < +∞ ,
µ2j ¯b0j ¯ < +∞ and ku1 kH 1 (Ω) '
j≥1
j≥1
then it can be checked that
u (·, t) =
X
j≥1
"
#
1
¢
¡
¢
¡
b
√
√
j
b0j cos t µj + √ sin t µj
`j ,
µj
(1.5)
¡
¢
and (u, ∂t u) ∈ C 1 (R; V ) ∩ C R; H 2 (Ω) × H 1 (Ω) . Let define by E (u, t) the energy
of the solution u = u (x, t),
Z ³
´
2
2
E (u, t) ≡
|∂t u (x, t)| + |∇u (x, t)| dx = E (u, 0) .
(1.6)
Ω
The Hölder type of dependence for the wave equation with mixed Dirichlet-Neumann
boundary condition that we will prove is the following
Theorem 1.2. There exist C > 0 and δ > 0 such that for any h ∈ (0, 1] and initial
data (u0 , u1 ) ∈ H 2 (Ω) × H 1 (Ω), such that ∂ν u0 |Γ0 = u0 |Γ1 = u1 |Γ1 = 0, the
solution u of (1.5) satisfies
Z
Z
E (u, 0) ≤ C
Γ1
1
C( h
)
1/δ
2
|∂ν u (x, t)| dσdt + hE (∂t u, 0) .
0
The first step in the proof of Theorem 1.2 consists in checking that we have the
following observability estimate (see Theorem 2.1 in section 2): there exist C > 0
and T > 0 such that any solution u of (1.4) satisfies
ÃZ Z
!
Z Z
T
E (u, 0) ≤ C
T
2
|∂ν u (x, t)| dσdt +
Γ1
0
2
|∂t u (x, t)| dxdt
ωo
,
0
where ωo = D × (−1/4, 1/4). Then, in the next step of the proof, it remains to
bound the last term in the above estimate and more precisely to ³check that a´Hölder
¡ ¢1/δ
type of dependence between ∂t u in ωo × (0, T ) and ∂ν u in Γ1 × 0, C h1
holds
(see (4.8) in section 4). To this end, we argue locally around any point xo ∈ ωo . Let
us denote B (xo , r) the ball of center xo and radius r. For any xo ∈ ωo , we introduce
χ = χxo (x) ∈ C0∞ (B (xo , ro )) be such that 0 ≤ χ ≤ 1 and χ = 1 on B (xo , ro /2),
for some small ro ∈ (0, 1/8). Now, we have the following four remarks.
1060
KIM DANG PHUNG
Remark
1. If there
¡
¢ exist two constants c > 0, γ > 1 and F = F (x, t, s) ∈
C ∞ R4 × (0, +∞) such that for any L ≥ 1 and h ∈ (0, 1],
¡
¢¢
¡
i∂s + h ∆ − ∂t2 F = 0
in R4 × (0, L)



Z
³
´


γ


on Γ0 × (0, L)
|∂ν F (·, t, ·)| dt = O e−c/h



R




0 6= F = O (1)
on Γ1 × R × (0, L)


lim F = lim ∂t F = 0
on R3 × (0, L)

t→±∞
t→±∞




1 2
1 2
1 d2


on Ω × R
F (·, ·, s = 0) = −χe− 2 t ∂t2 u + χ 2 e− 2 t u



2
dt

µ
¶
Z



1


|F (·, t, s = L)| dt = O
on R3 ,
γ
L
R
(1.7)
then by multiplying the first equality of (1.7) by u and integrating by parts over
Ω × R × [0, L], we have
Z
Z
2
−t2 /2
i
χ (x) e
|∂t u (x, t)| dxdt = i
F (x, t, 0) u (x, t) dxdt
Ω×R
Ω×R
Ã
!
Z
Z
Z
L
=i
F (x, t, L) u (x, t) dxdt + h
Ω×R
ÃZ
Z
!
L
−h
∂ν F (x, t, s) ds u (x, t) dσdt
Γ0 ×R
0
F (x, t, s) ds ∂ν u (x, t) dσdt
Γ1 ×R
0
and an interpolation estimate between ∂t u in ωo and ∂ν u in Γ1 may be expected by
choosing L > 1/h large enough and h adequately. Our construction of an admissible
F relies on Fourier integral operators and it requires some changes with the above
F but the idea still holds.
¡
¢
Remark 2. It is easy to find a solution a = a (x, t, s) ∈ C ∞ R4 × (0, +∞) solving
the first equality of (1.7) and satisfying the fourth and sixth equality of (1.7).
Indeed, it is sufficient to take the product of four, mono-dimensional, solutions of
the Schrödinger equation i∂s ± h∂ 2 and to use the natural dispersive property of
the linear Schrödinger equation. Let us define, for any (x, t, s) = (x1 , x2 , x3 , t, s) ∈
R4 × (0, +∞),


Ã
!
x2
|(x1 ,x2 )|2
1
1
t2
3
− 14 ihs+1
− 4h
is+1
e
e
e− 4 −ihs+1




√
√
a (x, t, s) =
.
(1.8)
ihs + 1
is + 1
−ihs + 1
We get by a simple computation the following identities
¡
¡
¢¢
i∂s + h ∆ − ∂t2 a (x, t, s) = 0 ∀ (x, t, s) ∈ R4 × (0, +∞) ,
1
−
|(x1 ,x2 )|2
1
2
2
4
x3
(hs)2 +1
1
e
− t4 (hs)12 +1
− 4h
s2 +1 e
|a (x, t, s)| = ¡√
.
e
µ
¶
¢1/2 q
3/2
s2 + 1
2
(hs) + 1
(1.9)
Let us denote
ao (x, t) = a (x − xo , t, 0) and ϕ (x, t) = χ (x) ao (x, t) .
(1.10)
BOUNDARY STABILIZATION FOR THE WAVE EQUATION
1061
Remark 3. In order to get the fifth equality of (1.7), we will construct two Fourier
2
integral operators: one corresponding to f1 = −χe−t /2 ∂t2 u and the second one for
¡
¡ ¢¢
2
2
d
−t /2
f2 = 12 χ dt
u. Let us define for any f = f (x, t) ∈ L∞ R; L2 R3 such that
2e
¡ ¢
c ∈ L1 R4 ,
ϕf
Z
2
2
1
(Af ) (x, t, s) =
ei(xξ+tτ ) e−i(|ξ| −τ )hs
4
(2π) R4
c (ξ, τ ) dξdτ
× a (x − xo − 2ξhs, t + 2τ hs, s) ϕf
c is the Fourier transform of ϕf ,
where ϕf
Z
c
ϕf (ξ, τ ) ≡
e−i(xξ+tτ ) ϕ (x, t) f (x, t) dxdt .
R4
Then, we get by a simple computation the following identities
¢¢
¡
¡
i∂s + h ∆ − ∂t2 (Af ) (x, t, s) = 0 ∀ (x, t, s) ∈ R4 × (0, +∞) ,
2
1 2
(Af ) (x, t, s = 0) = χ (x) |ao (x, 0)| e− 2 t f (x, t) .
Remark 4. Now, the second and fourth equalities of (1.7) require a microlocal
partition of unity. Indeed, the second equality of (1.7) holds if we restrict the
integral over τ ∈ R to a bounded interval τ ∈ (−R, R), for some R ≥ 1. Concerning
the fourth equality of (1.7), we decompose the integral over ξ3 ∈ R, ξ3 being the
third component of ξ ∈ R3 and the dual variable of x3 , into a infinite sum of integrals
as follows
Z
Z ξo3 +1
X
(· · ·) dξ3 =
(· · ·) dξ3 .
R
ξo3 ∈(2Z+1)
ξo3 −1
We arrive at the construction of (B (xo , ξo3 , n = 0) f ) (x, t, s) (see (3.2) in section 3).
Thus, for each (xo , ξo3 ) ∈ ωo × (2Z + 1), we will be able to construct a “reflected”
operator in order that the normal derivative at x3 = 1 vanishes. Our construction
of the “reflected” operators in order to get the desired boundary condition is inspired by the construction of the gaussian beams in a convex bounded domain by
superposing different reflected gaussian beams [14]. More precisely, we arrive at the
study of the operator (BP,Q (xo , ξo3 ) f ) (x, t, s) (see (3.1) in section 3) and we need
to check the computations in remark 1 with BP,Q (xo , ξo3 ) f instead of F .
The work is organized as follows. In section 2, we establish the observability
estimate. Section 3 is devoted to the construction of the Fourier integral operator
and the “reflected” operators. In section 4 and section 5, we prove Theorem 1.2
and Theorem 1.1 respectively. Finally, two appendices gather the proofs of some
useful inequalities and technical results used in the paper.
2. Observability estimate. This section is devoted to get some inequalities by
the multiplier method (see e.g., [10]). Our first result is an observability estimate.
Theorem 2.1. There exist C > 0 and T > 0 such that any solution u of (1.4)
satisfies
Z TZ
Z TZ
2
2
E (u, 0) ≤ C
|∂ν u (x, t)| dσdt + C
|∂t u (x, t)| dxdt .
0
Γ1
0
ωo
The next two results establish the regularity of the normal derivative. The arguments are well-known [10]. For completeness, we shortly give the proofs.
1062
KIM DANG PHUNG
Proposition 1. There exists c > 0 such that for any g ∈ C 1 (R) ∩ L1 (R) such that
g 0 ∈ L1 (R), any solution u of (1.4) satisfies
Z Z
Z
2
g (t) |∂ν u (x, t)| dσdt ≤ cE (u, 0) (|g (t)| + |g 0 (t)|) dt .
R
Γ1
R
¡
¢
2
2
Proposition
2.
There
exists
c
>
0
such
that
for
any
T
>
0
and
v
∈
C
R,
L
(Ω)
∩
¡
¢
¡
¢
1
1
2
C R, H (Ω) ∩ C R, H (Ω) satisfying
 2
in Ω × R

 ∂t v − ∆v = 0
∂ν v = 0
on Γ0 × R
(2.1)


E (v, 0) = 0
,
the following estimate holds
Z TZ
Z
2
|∂ν v (x, t)| dσdt +
0
Γ1
Z
T
Z
Z
2
T
Z
2
|∂t v (x, t)| dσdt −
0
Γ1
|∇tan v (x, t)| dσdt
0
Γ1
T
≤ cE (v, T ) + c
E (v, t) dt
0
2
2
2
where |∇v| = |∂ν v| + |∇tan v| on Γ1 × R.
The above three results are based on the following well-known identity (see [10]).
3
1
1
Lemma¡ 2.2. Let ¢H : Ω →
¡ R 1be a ¢vector¡ field 2of class
¢ C and g = g (t) ∈ C (R),
2
2
1
Φ ∈ C R, L (Ω) ∩ C R, H (Ω) ∩ C R, H (Ω) . Let −∞ < T1 < T2 < +∞.
Then we have the following identity
Z T2 Z
¡
¢
2g (t) H (x) · ∇Φ (x, t) ∂t2 Φ (x, t) − ∆Φ (x, t) dxdt
T1
Z
Ω
T2 Z
+
2g (t) H (x) · ∇Φ (x, t) ∂ν Φ (x, t) dσdt
T1
Z T2
Z
∂Ω
+
T1
·
∂Ω
Z
= 2g (t)
Z
T2
Z
Ω
T1
Z T2
Z
T1
T2
Z
Ω
+
T1
³
´
2
2
g (t) divH (x) |∂t Φ (x, t)| − |∇Φ (x, t)| dxdt
Ω
+
T1
¸T 2
H (x) · ∇Φ (x, t) ∂t Φ (x, t) dx
2g 0 (t) H (x) · ∇Φ (x, t) ∂t Φ (x, t) dxdt
−
Z
³
´
2
2
g (t) H (x) · ν (x) |∂t Φ (x, t)| − |∇Φ (x, t)| dσdt
Ω
2g (t) ∂xj Hi (x) ∂xi Φ (x, t) ∂xj Φ (x, t) dxdt .
Proof. We multiply ∂t2 Φ (x, t) − ∆Φ (x, t) by 2g (t) H (x) · ∇Φ (x, t) and integrate by
parts.
Proof. In order to prove Proposition 1, we begin to apply Lemma 2.2 to g ∈ C 1 (R)∩
L1 (R) such that g 0 ∈ L1 (R), H (x) = (ν1 (x) , ν2 (x) , 0) and Φ = u. Then, we
conclude using the conservation of energy (1.6) and a standard density argument.
BOUNDARY STABILIZATION FOR THE WAVE EQUATION
1063
Proof. Proposition 2 is a direct consequence of Lemma 2.2 applied to g = 1, T1 = 0,
T2 = T , H (x) = (ν1 (x) , ν2 (x) , 0) and Φ = v.
Proof. Let us now give the proof of Theorem 2.1. Let ψ1 = ψ1 (x3 ),ψ2 = ψ2 (x3 ) ∈
C ∞ (R) be such that ψ1 (x3 ) = 1 for x3 ≤ −1/8, 0 ≤ ψ1 ≤ 1, ψ1 (x3 ) = 0 for
x3 ≥ 1/8, and ψ1 + ψ2 = 1. Let u
e = ψ1 (x3 ) u. It solves
 2
∂t u
e − ∆e
u = −2∂x3 ψ1 ∂x3 u − ∂x23 ψ1 u
in Ω × R





u
e=0
on Γ1 × R
(2.2)

∂ν u
e=0
on D × {−1} × R




u
e = ∂ν u
e=0
on D × {1} × R .
Now, we apply Lemma 2.2 to g = 1, H (x) = (x − xo ) for some xo ∈ R3 and Φ = u
e,
it gives
Z T2 Z ³
Z T2 Z ³
´
´
2
2
2
2
|∂t u
e| + |∇e
u| dxdt +
2 |∂t u
e| − |∇e
u| dxdt
T1
Ω
T1
· Z
¸T2
+ 2
(x − xo ) · ∇e
u∂t u
edx
Z
T2
Z
Ω
T1
¡
¢
2 (x − xo ) · ∇e
u ∂t2 u
e − ∆e
u dxdt
=
T1
Z
Ω
T2
Ω
(2.3)
Z
2
+
(x − xo ) · ν (x) |∂ν u
e| dσdt
T1
T2
Z
Γ1
Z
³
´
2
2
(x − xo ) · ν (x) |∂t u
e| − |∇e
u| dσdt .
+
D×{−1}
T1
On the other hand, multiplying (2.2) by u
e and integrating by parts, we also have
Z T2 Z ³
´
2
2
|∂t u
e| − |∇e
u| dxdt
T1
Ω
·Z
=
¸T2
u
e∂t u
edx
Ω
Z
T2
Z
−
T1
T1
Ω
¡
¢
u
e ∂t2 u
e − ∆e
u dxdt .
(2.4)
0
Consequently, denoting ∇ u
e = (∂x1 u
e, ∂x2 u
e), thanks to (2.4), (2.3) becomes
Z T2 Z ³
´
2
2
|∂t u
e| + |∇e
u| dxdt
T1
Ω
·Z
= −2
¸T2
(e
u + (x − xo ) · ∇e
u) ∂t u
edx
Z
Ω
T1
Z
T2
+2
Z
T1
Ω
T2 Z
+
¡
¢
(e
u + (x − xo ) · ∇e
u) ∂t2 u
e − ∆e
u dxdt
2
(x − xo ) · ν (x) |∂ν u
e| dσdt
T1
Z T2
Γ1
Z
+
T1
D×{−1}
³
´
2
2
(x − xo ) · ν (x) |∂t u
e| − |∇0 u
e| dσdt .
(2.5)
1064
KIM DANG PHUNG
On the other hand, multiplying (2.2) by 2∂x3 u
e and integrating by parts, we also
have
·Z
¸T2
Z T2 Z
³
´
2
2
|∂t u
e| − |∇0 u
e| dσdt + 2
∂x3 u
e∂t u
edx
T1
D×{−1}
Ω
T1
T2
·Z
T1
¸T2
d
2
=−
|∂t u
e| dxdt + 2
∂x 3 u
e∂t u
edx
T1
Ω dx3
Ω
T1
Z T2 Z
Z T2 Z
d
d
2
2
−
|∂x3 u
e| dxdt +
|∇0 u
e| dxdt
dx
dx
3
3
T1
Ω
T1
Ω
Z T2 Z
¢
¡ 2
=
e − ∆e
u dxdt .
2∂x3 u
e ∂t u
Z
Z
(2.6)
Ω
Consequently, from (2.5) and (2.6), noticing that the support of ∂t2 u
e − ∆e
u is in
D × [−1/8, 1/8] × R, we find
Z T2
Z T2 Z ³
´
2
2
E (e
u, t) dt =
|∂t u
e| + |∇e
u| dxdt
T1
T1
Ω
≤ cE (e
u, T2 ) + cE (e
u, T1 )
Z T2 Z
Z
³
´
2
2
+c
|∂x3 u| + |u| dxdt + c
T1
D×(−1/8,1/8)
T2
(2.7)
Z
2
|∂ν u
e| dσdt .
T1
Γ1
By same arguments applied to ve = ψ2 (x3 ) u, we get that
Z T2
Z T2 Z ³
´
2
2
E (e
v , t) dt =
|∂t ve| + |∇e
v | dxdt
T1
T1
Ω
≤ cE (e
v , T2 ) + cE (e
v , T1 )
Z T2 Z
Z
³
´
2
2
+c
|∂x3 u| + |u| dxdt + c
T1
D×(−1/8,1/8)
From (2.7), (2.8) and
Z T2
Z
E (u, t) dt ≤
T1
Z
T2
T2
Z
T1
³
´
2
2
|∂t u| + |∇u| dxdt ,
D×(−1/8,1/8)
we finally obtain that
Z T2
Z
E (u, t) dt ≤ cE (u, T2 ) + cE (u, T1 ) + c
T2
Γ1
E (e
v , t) dt
T1
+
Z
2
|∂ν ve| dσdt .
T2
E (e
u, t) dt +
Z
(2.8)
Z
T1
T1
T1
T2
Z
³
+c
T2
Z
2
|∂ν u| dσdt
T1
Γ1
2
2
2
2
2
|∂t u| + |∇u| + |u|
T1
(2.9)
´
dxdt .
D×(−1/8,1/8)
Using localized energy estimates, we also have that
Z T2 Z
2
|∇u| dxdt
T1
D×(−1/8,1/8)
Z
T2
Z
³
≤ cE (u, T2 ) + cE (u, T1 ) + c
|∂t u| + |u|
T1
D×(−1/4,1/4)
(2.10)
´
dxdt .
BOUNDARY STABILIZATION FOR THE WAVE EQUATION
1065
From (2.9) and (2.10), we find by choosing (T2 − T1 ) > 0 large enough and using
conservation of energy (1.6) that there exist c > 0 and T > 0 such that
Z T
Z TZ
Z TZ
Z TZ
2
2
2
E (u, t) dt ≤ c
|∂ν u| dσdt + c
|∂t u| dxdt + c
|u| dxdt .
0
0
Γ1
0
ωo
0
ωo
Finally, using a standard uniqueness-compactness argument, we can eliminate the
last term in the right hand side of the above estimate in order to get the desired
observability inequality.
¡
¡ ¢¢
3. Fourier integral operators. Let f = f (x, t) ∈ L∞ R; L2 R3 be such that
¡ ¢
c ∈ L1 R4 , (xo , ξo3 ) ∈ ωo × (2Z + 1) and (P, Q) ∈ N2 . Let us introduce, for any
ϕf
(x, t, s) ∈ R4 × (0, +∞),
(BP,Q (xo , ξo3 ) f ) (x, t, s) =
2P
+1
X
(B (xo , ξo3 , n) f ) (x, t, s) ,
(3.1)
n=−2Q
where
=
(B (xo , ξo3 , n) f ) (x, t, s)
Z Z ξo3 +1 Z
1
4
(2π)
R2
ξo3 −1
h
i
ξ
i (−1)n x3 +2n |ξo3 | ξ3
ei(x1 ξ1 +x2 ξ2 +tτ ) e
o3
|τ |<R
2
2
c (ξ, τ )
× e−i(|ξ| −τ )hs ϕf
³
× a x1 − xo1 − 2ξ1 hs, x2 − xo2 − 2ξ2 hs,
n
(−1) x3 + 2n
(3.2)
¶
ξo3
− xo3 − 2ξ3 hs, t + 2τ hs, s dξdτ .
|ξo3 |
Recall that we have introduced in section 1, h ∈ (0, 1], R ≥ 1, ϕ (x, t) = χ (x) ao (x, t)
where χ = χxo (x) ∈ C0∞ (B (xo , ro )) is such that 0 ≤ χ ≤ 1 and χ = 1 on
B (xo , ro /2), for some small ro ∈ (0, 1/8), xo = (xo1 , xo2 , xo3 ), ξ = (ξ1 , ξ2 , ξ3 ) ∈ R3 ,
x = (x1 , x2 , x3 ) ∈ R3 , and ao (x, t) = a (x − xo , t, 0), (see (1.8 for the definition of
a (x, t, s)).
We check after a lengthy but straightforward calculation that for any (x, t, s) ∈
R4 × (0, +∞),
¡
¡
¢¢
i∂s + h ∆ − ∂t2 (BP,Q (xo , ξo3 ) f ) (x, t, s) = 0 ,
(3.3)
and also that for any (x1 , x2 , t, s) ∈ D × R × (0, +∞),


 ∂x3 (BP,Q (xo , ξo3 ) f ) (Λ+ ) = 0
∂x3 (BP,Q (xo , ξo3 ) f ) (Λ− ) = ∂x3 (B (xo , ξo3 , −2Q) f ) (Λ− )


+ ∂x3 (B (xo , ξo3 , 2P + 1) f ) (Λ− ) ,
³
´
³
´
ξo3
where Λ+ = x1 , x2 , |ξξo3
,
t,
s
and
Λ
=
x
,
x
,
−
,
t,
s
.
−
1
2
|
|ξ
|
o3
o3
Clearly, we also have the following equalities
lim (BP,Q (xo , ξo3 ) f ) (x, t, s) = lim ∂t (BP,Q (xo , ξo3 ) f ) (x, t, s) = 0 ,
t→±∞
t→±∞
(3.4)
(3.5)
1066
KIM DANG PHUNG
(BP,Q (xo , ξo3 ) f ) (x, t, 0)
¶
2P
+1 · µ
X
ξo3
n
=
a x1 − xo1 , x2 − xo2 , (−1) x3 + 2n
− xo3 , t, 0
|ξo3 |
n=−2Q
Z Z ξo3 +1 Z
1
×
ei(x1 ξ1 +x2 ξ2 +tτ )
4
(2π) R2 ξo3 −1 |τ |<R
¸
h
i
ξ
i (−1)n x3 +2n |ξo3 | ξ3 c
o3
×e
ϕf (ξ, τ ) dξdτ .
(3.6)
On another hand, by multiplying (3.3) by u (x, t), solution of (1.5), and integrating
by parts over Ω×R×[0, L], using (3.5), we have that for all (xo , ξo3 ) ∈ ωo ×(2Z + 1),
for all (P, Q) ∈ N2 and all h ∈ (0, 1], L ≥ 1,
Z
0=−i
BP,Q (xo , ξo3 ) f (x, t, 0) u (x, t) dxdt
Ω×R
Z
+i
BP,Q (xo , ξo3 ) f (x, t, L) u (x, t) dxdt
Ω×R
ÃZ
!
Z
L
(3.7)
+h
∂ν BP,Q (xo , ξo3 ) f (x, t, s) ds u (x, t) dσdt
Γ0 ×R
Z
0
ÃZ
!
L
−h
BP,Q (xo , ξo3 ) f (x, t, s) ds ∂ν u (x, t) dσdt
Γ1 ×R
0
≡ I1 + I2 + I3 + I4 .
Our goal consists in estimating separately the four integrals in (3.7), i.e., the four
terms I1 , I2 , I3 and I4 in order to get the following result.
¡
¡ ¢¢
Proposition 3. There exists c > 0 such that for any f ∈ L∞ R; L2 R3 such
¡ ¢
c ∈ L1 R4 , (xo , ξo3 ) ∈ ωo × (2Z + 1) and h ∈ (0, 1], L ≥ 1, ε ∈ (0, 1),
that ϕf
β ∈ (0, 1/2], R ≥ 1, we have
¯Z
¯
Ã
!
Z Z ξo3 +1 Z
¯
¯
1
¯
¯
i(xξ+tτ ) c
e
ϕf
(ξ,
τ
)
dξdτ
a
(x,
t)
u
(x,
t)
dxdt
¯
¯
o
¯ Ω×R (2π)4 R2 ξo3 −1 |τ |<R
¯
Z
Z
Z
¯
¯
ξo3 +1
³
´
p
√
1
¯c
¯
≤ c 1 + hL + L2 e− ch2β
E (u, 0)
¯ϕf (ξ, τ )¯ dξdτ
1
hL
1+
+c √
hL
ÃZ
+ c εh
Z
Z
p
Z
E (∂t u, 0)
ξo3 −1
|τ |<R
R2
µZ
ξo3 −1
R2
|τ |<R
¶1/2
¯
¯2
¯c
¯
ϕf
(ξ,
τ
)
dξ
dξ
dξ3 dτ
¯
¯
1 2
Z
!1/2
2
Γ1
ξo3 +1
|t|≤2Rhε+
Z
ξo3 −1
1
hβ
|∂ν u (x, t)| dσdt
(hε+1)
¯
¯
¯c
¯
¯ϕf (ξ, τ )¯ dξdτ
×
R2
ξo3 +1 Z
|τ |<R
!1/2
¶ ÃZ Z
L
2
|∂ν u (x, t)| dσdt
+c
hL + ln
ε
Γ1 |t|≤2RhL+ 1β (hL+1)
h
µZ ¯
¶1/2
Z ξo3 +1 Z
¯2
¯c
¯
×
dξ3 dτ
¯ϕf (ξ, τ )¯ dξ1 dξ2
µ
√
ξo3 −1
|τ |<R
R2
BOUNDARY STABILIZATION FOR THE WAVE EQUATION
1067
with u being the solution of (1.5).
In the next five subsections, we estimate Ij for j = 1, 2, 3, 4 and we prove Proposition 3.
3.1. Estimate
for I1 (the term at s = 0). We estimate the integral given by
R
I1 = −i Ω×R (BP,Q (xo , ξo3 ) f ) (x, t, 0) u (x, t) dxdt as follows.
Lemma 3.1. There exists c > 0 such that for any (xo , ξo3 ) ∈ ωo × (2Z + 1),
(P, Q) ∈ N2 and h ∈ (0, 1], R ≥ 1, we have
¯
¯
¯ I1
¯
Ã
!
Z
Z Z ξo3 +1 Z
¯
1
¯
i(xξ+tτ ) c
+i
e
ϕf
(ξ,
τ
)
dξdτ
a
(x,
t)
u
(x,
t)
dxdt
¯
o
4
¯
(2π) R2 ξo3 −1 |τ |<R
Ω×R
Z
Z
Z
¯
¯
ξ
+1
o3
1 p
¯
¯c
≤ ce− ch
E (u, 0)
¯ϕf (ξ, τ )¯ dξdτ .
R2
ξo3 −1
|τ |<R
Proof. Since from (3.6)
(BP,Q (xo , ξo3 ) f ) (x, t, 0)
Z Z ξo3 +1 Z
1
c (ξ, τ ) dξdτ
= ao (x, t)
ei(xξ+tτ ) ϕf
4
(2π) R2 ξo3 −1 |τ |<R
¶
· µ
X
ξo3
n
− xo3 , t, 0
+
a x1 − xo1 , x2 − xo2 , (−1) x3 + 2n
|ξo3 |
n={−2Q,···,2P +1}\{0}
×
Z
1
4
(2π)
Z
ξo3 +1
Z
e
R2
ξo3 −1
i(x1 ξ1 +x2 ξ2 +tτ )
h
i
ξ
i (−1)n x3 +2n |ξo3 | ξ3
e
o3
#
c (ξ, τ ) dξdτ
ϕf
|τ |<R
(3.1.1)
we only need to compute the discrete sum over {−2Q, · · ·, 2P + 1} \{0} . It comes
from (1.9) with s = 0,
¯
¯
¶
· µ
X
¯
ξo3
n
¯
−
x
,
t,
0
a
x
−
x
,
x
−
x
,
(−1)
x
+
2n
1
o1
2
o2
3
o3
¯
|ξo3 |
¯n={−2Q,···,2P +1}\{0}
#¯
Z Z ξo3 +1 Z
i
h
¯
ξo3
n
1
ξ
i
(−1)
x
+2n
¯
3 c
3
i(x1 ξ1 +x2 ξ2 +tτ )
|ξo3 |
×
ϕf
(ξ,
τ
)
dξdτ
e
e
¯
4
¯
2
(2π) R ξo3 −1 |τ |<R
Z Z ξo3 +1 Z
¯
¯
1
¯c
¯
≤
ϕf
(ξ,
τ
)
¯
¯ dξdτ
4
(2π) R2 ξo3 −1 |τ |<R
¯ µ
¶¯
X
¯
¯
¯a x1 − xo1 , x2 − xo2 , (−1)n x3 + 2n ξo3 − xo3 , t, 0 ¯
×
¯
¯
|ξo3 |
n={−2Q,···,2P +1}\{0}
Z Z ξo3 +1 Z
¯
¯
1
¯c
¯
ϕf
(ξ,
τ
)
≤
¯
¯ dξdτ
4
(2π) R2 ξo3 −1 |τ |<R
× e
−
|(x1 −xo1 ,x2 −xo2 )|2
4
e
2
− t4
X
µ
e
−
ξo3
−xo3
|ξo3 |
4h
(−1)n x3 +2n
¶2
.
n={−2Q,···,2P +1}\{0}
(3.1.2)
1068
KIM DANG PHUNG
But, there exists c > 0 such that for all (P, Q) ∈ N2 , (xo3 , ξo3 ) ∈ [1/4, 1/4]×(2Z + 1)
and x3 ∈ (−1, 1),
µ
X
−
ξo3
−xo3
|ξo3 |
4h
(−1)n x3 +2n
e
¶2
n={−2Q,···,2P +1}\{0}
³
´2
|ξ |
1
2n+ ξo3 [(−1)n x3 −xo3 ]
− 4h
X
=
e
o3
1
≤ ce− ch
(3.1.3)
n={−2Q,···,2P +1}\{0}
and notice that u is solution of (1.4) and satisfies using Cauchy-Schwarz and (1.6),
Z
p Z
t2
t2
e− 4 |u (x, t)| dxdt ≤ |Ω| e− 4 ku (·, t)kL2 (Ω) dt
Ω×R
R
≤ c k(u0 , u1 )kH 1 (Ω)×L2 (Ω) .
(3.1.4)
Finally, we get the desired estimate from (3.1.1 )-(3.1.2)-(3.1.3)-(3.1.4).
3.2. Estimate
for I2 (the term at s = L). We estimate the integral given by
R
I2 = i Ω×R BP,Q (xo , ξo3 ) f (x, t, L) u (x, t) dxdt as follows.
Lemma 3.2. There exists c > 0 such that for any (xo , ξo3 ) ∈ ωo × (2Z + 1),
(P, Q) ∈ N2 and h ∈ (0, 1], L ≥ 1, R ≥ 1, we have
µ
|I2 | ≤ c
1
1 + hL
√
hL
¶
Z
p
ξo3 +1
µZ
Z
E (∂t u, 0)
ξo3 −1
|τ |<R
R2
¶1/2
¯
¯2
¯c
¯
dξ3 dτ .
¯ϕf (ξ, τ )¯ dξ1 dξ2
Proof. First, we have the following uniform bound with respect to (P, Q).
¯ Z
¯
¯
¯
¯i
BP,Q (xo , ξo3 ) f (x, t, L) u (x, t) dxdt¯¯
¯
¯ Ω×R
¯
¯ 2P +1 Z
¯
¯ X
¯
¯
= ¯i
(B (xo , ξo3 , n) f ) (x, t, L) u (x, t) dxdt¯¯
¯ n=−2Q Ω×R
¯
¯
¯
Z
X¯
¯
¯
≤
(B (xo , ξo3 , n) f ) (x, t, L) u (x, t) dxdt¯¯ .
¯
n∈Z
(3.2.1)
Ω×R
Now, our goal will consist in estimating this last quantity. Recall that from (3.2)
=
(B (xo , ξo3 , n) f ) (x, t, L)
Z Z ξo3 +1 Z
1
4
(2π)
R2
ξo3 −1
i(x1 ξ1 +x2 ξ2 +tτ )
e
|τ |<R
e
h
i
ξ
i (−1)n x3 +2n |ξo3 | ξ3
o3
·
¸
|(x1 −xo1 −2ξ1 hL,x2 −xo2 −2ξ2 hL)|2
1
− 14
ihL+1
× e
e
ihL + 1

µ
¶2 
ξ
·
¸
(−1)n x3 +2n o3 −xo3 −2ξ3 hL
2
|ξ
|
1
1 (t+2τ hL)
o3
1
 √ 1
iL+1
e− 4h
e− 4 −ihL+1 dξdτ
× √
iL + 1
−ihL + 1
−i(|ξ|2 −τ 2 )hL
c (ξ, τ )
ϕf
(3.2.2)
BOUNDARY STABILIZATION FOR THE WAVE EQUATION
1069
and that from (1.5),
"
#
X b0j ¡ √
√ ¢
√ ¢
b1j ¡ it√µj
it µj
−it µj
−it µj
u (x, t) =
e
+e
+ √
e
−e
`j (x)
2
2i µj
j≥1
Ã
!
Ã
!
X b0j
X b0j
√
√
b1j
b1j
it µj
+ √
− √
=
`j (x) e
+
`j (x) e−it µj .
2
2i µj
2
2i µj
j≥1
j≥1
(3.2.3)
Therefore, from (3.2.2)-(3.2.3) and thanks to (1.8) with s = L, we write
Z
(B (xo , ξo3 , n) f ) (x, t, L) u (x, t) dxdt
Ω×R
=
1
1
1
√
ihL + 1 iL + 1 (2π)4
Z Z ξo3 +1 Z
XZ
×
`j (x)
j≥1
R2
Ω
ξo3 −1
i
h
ξ
i (−1)n x3 +2n |ξo3 | ξ3
ei(x1 ξ1 +x2 ξ2 ) e
o3
|τ |<R
2
2
c (ξ, τ )
× e−i(|ξ| −τ )hL ϕf

¶2 
µ
ξ
·
¸
(−1)n x3 +2n o3 −xo3 −2ξ3 hL
2
|ξ
|
|(x
−x
−2ξ
hL,x
−x
−2ξ
hL)|
1
1
o3
1
o1
1
2
o2
2
e− 4h
 dx
ihL+1
iL+1
× e− 4
µZ
·
¸
(t+2τ hL)2
1
− 14 −ihL+1
√
×
e
e
−ihL + 1
!
Ã
!
# !
"ÃR
0
1
√
√
b
b1j
b
b0j
j
j
eit µj +
e−it µj dt dξdτ .
+ √
− √
×
2
2i µj
2
2i µj
itτ
On another hand, a straightforward computation yields
Z
√
ei(τ ±
ZR
=
µj )t
√
ei(τ ±
2
1 (t+2τ hL)
−ihL+1
e− 4
dt
µj )(t−2τ hL)
e− 4 −ihL+1 dt
R
√
i(τ ± µj )(−2τ hL)
Z
t2
1
√
µj )t
ei(τ ±
=e
1
t2
e− 4 −ihL+1 dt
R
√
i(τ ± µj )(−2τ hL)
=e
2
√
√ √
2 π −ihL + 1 e−(−ihL+1)(τ ± µj )
where we have used the following formula
Z
−itς
e
R
e
− z2 t2
√
1 2
2π
dt = √ e− 2z ς ,
z
Re z ≥ 0 and z 6= 0 .
1070
KIM DANG PHUNG
Consequently,
Z
(B (xo , ξo3 , n) f ) (x, t, L) u (x, t) dxdt
Ω×R
=
1
1
1
√
ihL + 1 iL + 1 (2π)4
Z Z ξo3 +1 Z
XZ
×
`j (x)
R2
Ω
j≥1
ξo3 −1
h
i
ξ
i (−1)n x3 +2n |ξo3 | ξ3
ei(x1 ξ1 +x2 ξ2 ) e
o3
|τ |<R
2
2
c (ξ, τ )
× e−i(|ξ| −τ )hL ϕf

µ
¶2 
ξ
µ
¶
(−1)n x3 +2n o3 −xo3 −2ξ3 hL
2
|ξo3 |
1
1 |(x1 −xo1 −2ξ1 hL,x2 −xo2 −2ξ2 hL)|
 dx
e− 4h
ihL+1
iL+1
× e− 4
"Ã
!
2
√
√
b0j
b1j
√
×
+ √
ei(τ + µj )(−2τ hL) 2 πe−(−ihL+1)(τ + µj )
2
2i µj
Ã
!
#
2
√
√
b0j
b1j
√
+
− √
ei(τ − µj )(−2τ hL) 2 πe−(−ihL+1)(τ − µj ) dξdτ .
2
2i µj
Thus,
¯Z
¯
¯
¯
Ω×R
¯
¯
(B (xo , ξo3 , n) f ) (x, t, L) u (x, t) dxdt¯¯
1
≤q
2
(hL) + 1
¡√
1
L2 + 1
¢1/2
√
Z
2 π X
4
(2π)
j≥1
Z
|`j (x)|
Ω

¶
µ
|(x1 −xo1 −2ξ1 hL,x2 −xo2 −2ξ2 hL)|2
1
−1
(hL)2 +1
e− 4h
× e 4
µ
Z
R2
ξo3 +1
ξo3 −1
(−1)n x3 +2n
Z
¯
¯
¯c
¯
¯ϕf (ξ, τ )¯
|τ |<R
¶2
ξo3
−xo3 −2ξ3 hL
|ξo3 |
L2 +1

 dx
¯ ¯
¯!
ï
¯ b0
b1j ¯¯ ¯¯ b0j
b1j ¯¯
¯ j
× ¯ + √ ¯ + ¯ − √ ¯ dξdτ .
¯2
2i µj ¯ ¯ 2
2i µj ¯
Then, there exists c > 0 such that
¯
X ¯¯Z
¯
¯
¯
(B
(x
,
ξ
,
n)
f
)
(x,
t,
L)
u
(x,
t)
dxdt
o o3
¯
¯
Ω×R
n∈Z
¯
¯!
Ã
³
√ ´ X ¯ 0 ¯ ¯¯ b1j ¯¯
1
1
¯
¯
bj + ¯ √ ¯
≤ cq
¡√
¢1/2 1 + hL
¯ µj ¯
2
L2 + 1
j≥1
(hL) + 1
Z Z ξo3 +1 Z
¯
¯
2
¯c
¯ − 1 (hL) |(ξ1 ,ξ2 )|2
dξdτ
×
¯ϕf (ξ, τ )¯ e 2 (hL)2 +1
R2
ξo3 −1
(3.2.4)
|τ |<R
by using the following inequalities
X − 1 ³2n+ |ξo3 | [(−1)n x3 −xo3 −2ξ3 hL]´2 1
p
ξo3
L2 +1 ≤ 1 +
e 4h
πh (L2 + 1) ,
n∈Z
e
− 14
|(x1 −xo1 −2ξ1 hL,x2 −xo2 −2ξ2 hL)|2
(hL)2 +1
≤e
− 12
(hL)2
|(ξ1 ,ξ2 )|2
(hL)2 +1
2
1 |(x1 −xo1 ,x2 −xo2 )|
(hL)2 +1
e4
,
BOUNDARY STABILIZATION FOR THE WAVE EQUATION
1071
p
p
R
and Ω |`j (x)| dx ≤
|Ω| k`j kL2 (Ω) =
|Ω|. On another hand, using CauchySchwarz inequality, we have the following inequalities
Z Z ξo3 +1 Z
¯
¯
2
¯c
¯ − 1 (hL) |(ξ1 ,ξ2 )|2
dξdτ
¯ϕf (ξ, τ )¯ e 2 (hL)2 +1
R2
ξo3 −1
µZ
|τ |<R
(hL)2
− (hL)2 +1 |(ξ1 ,ξ2 )|2
e
≤
R2
Z
ξo3 +1
Z
µZ
×
ξo3 −1
|τ |<R
R2
¶1/2
dξ1 dξ2
¶1/2
¯
¯2
¯c
¯
dξ3 dτ
¯ϕf (ξ, τ )¯ dξ1 dξ2
(3.2.5)
v Ã
!Z
u
µZ ¯
¶1/2
2
¯2
ξo3 +1 Z
u
(hL) + 1
¯c
¯
t
≤ π
ϕf
(ξ,
τ
)
dξ
dξ
dξ3 dτ ,
¯
¯
1 2
2
(hL)
ξo3 −1
|τ |<R
R2
¯
¯! v
¯
¯! v
Ã
Ã
u
¯
¯ b1 ¯ 2 u
uX
X ¯ ¯ ¯¯ b1j ¯¯
X ¯¯ 1 ¯2
¯
¯
¯
¯ u
j
0
2
¯b0j ¯ + ¯ √ ¯ ≤ t
¯
¯
¯
¯
t
µj bj + ¯ √ ¯
¯ µj ¯
¯ µj ¯
¯ µj ¯
j≥1
j≥1
j≥1
v
uX ¯ ¯2
√
¯1¯
u
¯ ¯ ,
≤ 2 k(∆u0 , ∇u1 )kL2 (Ω)×L2 (Ω) t
¯ µj ¯
(3.2.6)
j≥1
and by using the Weyl formula, we know that for some c > 0,
¯ ¯
X ¯ 1 ¯2
XX
XX
1
1
¯ ¯ =
³
´2 ≤ c
2 < +∞ .
¯ µj ¯
2
(m + n2 )
j≥1
m≥1 n≥0 λm + (nπ/2)
m≥1 n≥0
(3.2.7)
As a result of the above (3.2.4)-(3.2.5)-(3.2.6 )-(3.2.7), we finally obtain the existence
of a constant c > 0 such that
¯
¯
X ¯Z
¯
¯
(B (xo , ξo3 , n) f ) (x, t, L) u (x, t) dxdt¯¯
¯
n∈Z
≤ cq
Ω×R
1
1
³
1+
√
´
hL k(∆u0 , ∇u1 )kL2 (Ω)×L2 (Ω)
¡√
¢1/2
2
L2 + 1
(hL) + 1
vÃ
!
u
µZ ¯
¶1/2
¯2
u (hL)2 + 1 Z ξo3 +1 Z
¯c
¯
t
×
dξ3 dτ ,
¯ϕf (ξ, τ )¯ dξ1 dξ2
2
(hL)
ξo3 −1
|τ |<R
R2
from which thanks to (3.2.1) the desired estimate easily follows.
3.3. Estimate
for I3 (the term on Γi0 ). We estimate the integral given by I3 =
hR
R
L
h Γ0 ×R 0 ∂ν BP,Q (xo , ξo3 ) f (x, t, s) ds u (x, t) dσdt as follows.
Lemma 3.3. There exists c > 0 such that for any ξo3 (2Z + 1) and L ≥ 1, there
exists (P, Q) ∈ N2 such that for any xo ∈ ωo and h ∈ (0, 1], R ≥ 1, we have
Z Z ξo3 +1 Z
¯
¯
1 p
¯c
¯
2 − 8h
|I3 | ≤ cL e
E (u, 0)
¯ϕf (ξ, τ )¯ dξdτ .
R2
ξo3 −1
|τ |<R
1072
KIM DANG PHUNG
Proof. Recall that on Γ0 = D × {−1, 1},
∂¯ ν (BP,Q (xo , ξo3 ) f ) (x, t, s)
¯ ∂ (B
(x , ξ ) f ) (x1 , x2 , 1, t, s)
= ¯¯ x3 P,Q o o3
−∂x3 (BP,Q (xo , ξo3 ) f ) (x1 , x2 , −1, t, s)
for x3 = 1
for x3 = −1 .
Therefore, using (3.4), we get that on Γ0 × R × (0, +∞),
|∂ν (BP,Q (xo , ξo3 ) f ) (x, t, s)| ≤ |∂x3 (B (xo , ξo3 , −2Q) f ) (x1 , x2 , −1, t, s)|
+ |∂x3 (B (xo , ξo3 , 2P + 1) f ) (x1 , x2 , −1, t, s)|
+ |∂x3 (B (xo , ξo3 , −2Q) f ) (x1 , x2 , 1, t, s)|
+ |∂x3 (B (xo , ξo3 , 2P + 1) f ) (x1 , x2 , 1, t, s)| .
Since
∂x3 (B (xo , ξo3 , n) f ) (x, t, s)
Z ξo3 +1 Z
h
i
n Z
ξo3
n
(−1)
i(x1 ξ1 +x2 ξ2 +tτ ) i (−1) x3 +2n |ξo3 | ξ3
=
e
e
4
(2π) R2 ξo3 −1 |τ |<R
³
2
2
c (ξ, τ ) (iξ3 a + ∂x a) x1 − xo1 − 2ξ1 hs, x2 − xo2 − 2ξ2 hs,
× e−i(|ξ| −τ )hs ϕf
3
¶
ξo3
n
(−1) x3 + 2n
− xo3 − 2ξ3 hs, t + 2τ hs, s dξdτ ,
|ξo3 |
we obtain that
|∂x3 (B (xo , ξo3 , n) f ) (x, t, s)|
Z Z ξo3 +1 Z
¯
¯
1
¯c
¯
≤
ϕf
(ξ,
τ
)
¯
¯
4
(2π) R2 ξo3 −1 |τ |<R
³
× (|ξ3 | |a| + |∂x3 a|) x1 − xo1 − 2ξ1 hs, x2 − xo2 − 2ξ2 hs,
¶
ξo3
n
(−1) x3 + 2n
− xo3 − 2ξ3 hs, t + 2τ hs, s dξdτ .
|ξo3 |
But from (1.8)-(1.9),
(|ξ3 | |a| + |∂x3 a|) (x, t, s) ≤
µ
¶
2
x2
1
1
−t
− 3 1
|ξ3 | +
e 4h s2 +1 e 4 (hs)2 +1 ,
h
therefore
³
(|ξ3 | |a| + |∂x3 a|) x1 − xo1 − 2ξ1 hs, x2 − xo2 − 2ξ2 hs,
¶
ξo3
n
(−1) x3 + 2n
− xo3 − 2ξ3 hs, t + 2τ hs, s
|ξo3 |
³
´2
¶
µ
ξ
1
1
− (−1)n x3 +2n |ξo3 | −xo3 −2ξ3 hs
−(t+2τ hs)2
1
4h(s2 +1)
4((hs)2 +1)
o3
e
≤ |ξ3 | +
e
.
h
Now, our goal will consist in choosing (P, Q) ∈ N2 large enough such that for any
s ∈ (0, L), h ∈ (0, 1], x3 ∈ {±1}, xo3 ∈ (−1/4, 1/4), ξ3 ∈ (ξo3 − 1, ξo3 + 1), we have
BOUNDARY STABILIZATION FOR THE WAVE EQUATION
¡
¢
(|ξ3 | h + 1) s2 + 1 ≤
µ
n
(−1) x3 + 2n
¶2
ξo3
− xo3 − 2ξ3 hs
|ξo3 |
1073
∀n ∈ {−2Q, 2P + 1}
in order to get
´2
³
µ
¶
µ
¶
ξ
1
1 1
− (−1)n x3 +2n |ξo3 | −xo3 −2ξ3 hs
1
1
4h(s2 +1)
o3
|ξ3 | +
e
≤ |ξ3 | +
e−(|ξ3 |+ h ) 4
h
h
1 1
1
≤ 8e−(|ξ3 |+ h ) 8 ≤ 8e− 8h .
When n = −2Q, then for ξ3 ∈ (ξo3 − 1, ξo3 + 1),
¯
¯
¯
¯
¯(−1)n x3 + 2n ξo3 − xo3 − 2ξ3 hs¯
¯
¯
|ξo3 |
½
|4Q + 2 |ξ3 | hs − (x3 − xo3 )|
for ξo3 ∈ (2Z + 1) , ξo3 > 0
=
|4Q + 2 |ξ3 | hs + (x3 − xo3 )|
for ξo3 ∈ (2Z + 1) , ξo3 < 0 ,
and the desired estimate will hold if
p
(|ξo3 | + 2) (L2 + 1) ≤ 4Q − |x3 − xo3 | .
But for any x3 ∈ {±1} and xo3 ∈ (−1/4, 1/4), we have |x3 − xo3 | ≤ 5/4. Consequently, we can choose Q being the first integer satisfying
´
1 ³p
Q≥
(|ξo3 | + 2) (L2 + 1) + 2 .
4
When n = 2P + 1, then for ξ3 ∈ (ξo3 − 1, ξo3 + 1),
¯
¯
¯
¯
¯(−1)n x3 + 2n ξo3 − xo3 − 2ξ3 hs¯
¯
¯
|ξo3 |
½
|4P − 2 |ξ3 | hs + 2 − x3 − xo3 |
=
|4P − 2 |ξ3 | hs + 2 + x3 + xo3 |
for ξo3 ∈ (2Z + 1) , ξo3 > 0
for ξo3 ∈ (2Z + 1) , ξo3 < 0 ,
and the desired estimate will hold if
p
(|ξo3 | + 2) (L2 + 1) + 2 (|ξo3 | + 1) L ≤ 4P + 2 − |x3 + xo3 | .
But for any x3 ∈ {±1} and xo3 ∈ (−1/4, 1/4), we have 3/4 ≤ 2 − |x3 − xo3 |.
Consequently, we can choose P being the first integer satisfying
P ≥
´
1 ³p
(|ξo3 | + 2) (L2 + 1) + 2 (|ξo3 | + 1) L .
4
With the above choice of (P, Q) ∈ N2 , for any s ∈ (0, L), h ∈ (0, 1], x3 ∈ {±1},
xo3 ∈ (−1/4, 1/4), ξ3 ∈ (ξo3 − 1, ξo3 + 1), we have ∀n ∈ {−2Q, 2P + 1}
|∂x3 (B (xo , ξo3 , n) f ) (x, t, s)|
¯
¯
1
R R ξo3 +1 R
−(t+2τ hs)2
1
¯c
¯ − 8h
8
4((hs)2 +1)
≤ (2π)
ϕf
(ξ,
τ
)
e
dξdτ .
¯
¯ e
4
2
R
ξo3 −1 |τ |<R
Consequently,
1074
KIM DANG PHUNG
¯ Z
¯
"Z
#
¯
¯
L
¯
¯
∂ν (BP,Q (xo , ξo3 ) f ) (x, t, s) ds u (x, t) dσdt¯
¯h
¯ Γ0 ×R 0
¯
"Z
#
Z
L
≤h
|∂ν (BP,Q (xo , ξo3 ) f ) (x, t, s)| ds |u (x, t)| dσdt
Γ0 ×R
≤ 4h e
0
Z
8
1
− 8h
4
(2π)
Z
Z
R2
ξo3 +1
ξo3 −1
Z
L
e
|u (x, t)| dσ
×
Γ0 ×R
≤ 4h e
Z
Z
8
1
− 8h
4
(2π)
LZ
Z
Z
R2
¯
¯
¯c
¯
¯ϕf (ξ, τ )¯ dξ
|τ |<R
−(t+2τ hs)2
0
ξo3 +1
ξo3 −1
Z
1
4((hs)2 +1)
¯
¯
¯
¯c
¯ϕf (ξ, τ )¯ dξ
|τ |<R
2
×
|u (x, t − 2τ hs)| dσdτ e
0
dtdτ ds
− t4
1
(hs)2 +1
dtds
Γ0 ×R
which implies by Cauchy-Schwarz inequality and (1.6), the existence of a constant
c > 0 such that
¯ Z
¯
"Z
#
¯
¯
L
¯
¯
∂ν (BP,Q (xo , ξo3 ) f ) (x, t, s) ds u (x, t) dσdt¯
¯h
¯ Γ0 ×R 0
¯
Z Z ξo3 +1 Z
¯
¯
1
¯c
¯
≤ c e− 8h
¯ϕf (ξ, τ )¯ dξdτ k(u0 , u1 )kH 1 (Ω)×L2 (Ω)
Z
L
R2
Z
ξo3 −1
|τ |<R
2
− t4 (hs)12 +1
×
dtds
e
0
R
Z
Z
ξo3 +1
1
≤ c e− 8h
R2
Z
ξo3 −1
|τ |<R
¯
¯
¢
¡ √
¯c
¯
¯ϕf (ξ, τ )¯ dξdτ k(u0 , u1 )kH 1 (Ω)×L2 (Ω) 3 πL2 .
This completes the proof.
3.4. Estimate
for I4 (the term on Γ1 ).
i We estimate the integral given by I4 =
hR
R
L
−h Γ1 ×R 0 (BP,Q (xo , ξo3 ) f ) (x, t, s) ds ∂ν u (x, t) dσdt as follows.
Lemma 3.4. There exists c > 0 such that for any (xo , ξo3 ) ∈ ωo × (2Z + 1),
(P, Q) ∈ N2 and h ∈ (0, 1], L ≥ 1, ε ∈ (0, 1), β ∈ (0, 1/2], R ≥ 1, we have
!1/2
µ
¶ ÃZ Z
√
L
2
|I4 | ≤ c
hL + ln
|∂ν u (x, t)| dσdt
ε
Γ1 |t|≤2RhL+ 1β (hL+1)
h
µZ ¯
¶1/2
Z ξo3 +1 Z
¯2
¯c
¯
×
dξ3 dτ
¯ϕf (ξ, τ )¯ dξ1 dξ2
ξo3 −1
ÃZ
|τ |<R
R2
!1/2
Z
2
|∂ν u (x, t)| dσdt
+ c εh
Z
Z
Γ1
ξo3 +1
|t|≤2Rhε+
Z
ξo3 −1
(hε+1)
¯
¯
¯c
¯
¯ϕf (ξ, τ )¯ dξdτ
×
R2
1
hβ
|τ |<R
Z
p
√
− 12β
ch
+ c hL e
E (u, 0)
Z
R2
ξo3 +1
ξo3 −1
Z
|τ |<R
¯
¯
¯c
¯
¯ϕf (ξ, τ )¯ dξdτ .
BOUNDARY STABILIZATION FOR THE WAVE EQUATION
1075
Proof. First, we have a uniform bound with respect to (P, Q). Indeed,
¯Z
¯
"Z
#
¯
¯
L
¯
¯
(BP,Q (xo , ξo3 ) f ) (x, t, s) ds ∂ν u (x, t) dσdt¯
¯
¯ Γ1 ×R 0
¯
¯
¯


¯Z
¯
Z L 2P
+1
X
¯
¯
¯


=¯
(B (xo , ξo3 , n) f ) (x, t, s) ds ∂ν u (x, t) dσdt¯¯
¯ Γ1 ×R 0 n=−2Q
¯
"Z
#
Z
LX
≤
|(B (xo , ξo3 , n) f ) (x, t, s)| ds |∂ν u (x, t)| dσdt .
Γ1 ×R
Next, we estimate
0
n∈Z
P
|(B (xo , ξo3 , n) f ) (x, t, s)| as follows. From (1.8)-(1.9),
n∈Z
X
|(B (xo , ξo3 , n) f ) (x, t, s)|
n∈Z
≤
Z
1
(2π)
4
× ¡√
Z
R2
ξo3 +1
ξo3 −1
1
s2 + 1
×
n∈Z
≤
1
(2π)
4
× ¡√
e
Z
R2
s2 + 1
1
2
(−1)n x3 +2n
ξo3 +1
ξo3 −1
1
|τ |<R
¶3/2 e
− 14
|(x1 −xo1 −2ξ1 hs,x2 −xo2 −2ξ2 hs)|2
(hs)2 +1
− 14
e
(t+2τ hs)2
(hs)2 +1
(hs) + 1
1
− 4h
Z
¯
¯
¯c
¯
¯ϕf (ξ, τ )¯
¢1/2 µq
µ
X
Z
ξo3
−xo3 −2ξ3 hs
|ξo3 |
s2 +1
Z
¶2
dξdτ
2
¯
¯
2
o1 ,x2 −xo2 )|
¯c
¯ − 1 (hs) |(ξ1 ,ξ2 )|2 14 |(x1 −x(hs)
2 +1
e
¯ϕf (ξ, τ )¯ e 2 (hs)2 +1
|τ |<R
1
¢1/2 µq
2
³
´
(t+2τ hs)2
p
− 14 (hs)2 +1
dξdτ ,
¶3/2 4 + c h (s2 + 1) e
(hs) + 1
where we used the following two inequalities
X
e
³
´2
|ξ |
1
− 4h
2n+ ξo3 [(−1)n x3 −xo3 −2ξ3 hs] s21+1
o3
≤1+
p
πh (s2 + 1) ,
n∈Z
e
− 41
|(x1 −xo1 −2ξ1 hs,x2 −xo2 −2ξ2 hs)|2
(hs)2 +1
≤e
− 12
(hs)2
|(ξ1 ,ξ2 )|2
(hs)2 +1
2
1 |(x1 −xo1 ,x2 −xo2 )|
(hs)2 +1
e4
Consequently, there exists c > 0 such that
¯Z
¯
"Z
#
¯
¯
L
¯
¯
(BP,Q (xo , ξo3 ) f ) (x, t, s) ds ∂ν u (x, t) dσdt¯
¯
¯ Γ1 ×R 0
¯
.
1076
KIM DANG PHUNG
Z
L
Z
Z
ξo3 +1
Z
¯
¯
2
¯c
¯ − 1 (hs) |(ξ1 ,ξ2 )|2
dξdτ
¯ϕf (ξ, τ )¯ e 2 (hs)2 +1
|τ |<R

≤c
R2
0

ξo3 −1

³
´
p
1
1

2 + 1) 
×  ¡√
1
+
πh
(s

µ
¶
¢
3/2

 s2 + 1 1/2 q
2
(hs) + 1
Z
(t+2τ hs)2
−1
|∂ν u (x, t)| e 4 (hs)2 +1 dσdtds .
×
Γ1 ×R
On the other hand, from Cauchy-Schwarz inequality, we know that
Z
(t+2τ hs)2
1
−
4
(hs)2 +1 |∂ u (x, t)| dσdt
e
ν
Γ1 ×R
µZ
≤
e
−
(t+2τ hs)2
1
4
(hs)2 +1
Γ1 ×R
µZ
×
e
(t+2τ hs)2
1
4
(hs)2 +1
¶1/2
dσdt
(3.4.1)
Γ1 ×R
µZ
≤
e
Γ1 ×R
Ã
×
−
¶1/2
2
|∂ν u (x, t)| dσdt
−
(t+2τ hs)2
1
4
(hs)2 +1
r
|Γ1 |
³
´
2
¶1/2
2
|∂ν u (x, t)| dσdt
!1/2
.
4π (hs) + 1
Now, we split the integral over t ∈ R into two parts, in order to get
Z
(t+2τ hs)2
1
−
2
4
(hs)2 +1 |∂ u (x, t)| dσdt
e
ν
Γ ×R
Z1 Z
2
≤
|∂ν u (x, t)| dσdt
√
Γ1
|t+2τ hs|≤
Z
Z
+
Z
Γ1
Z
1
hβ
√
(hs)2 +1
1
hβ
e
−
(t+2τ hs)2
1
4
(hs)2 +1
(hs)2 +1<|t+2τ hs|
2
|∂ν u (x, t)| dσdt
(3.4.2)
2
≤
Γ1
|t|≤|2τ hs|+ 1β
h
1
Z
|∂ν u (x, t)| dσdt
(hs+1)
−
+ e− 8h2β
e
(t+2τ hs)2
1
8
(hs)2 +1
2
|∂ν u (x, t)| dσdt
Γ1 ×R
for any β ∈ (0, 1/2]. Then, we bound the last term on the right hand side by using
Proposition 1 which says that there exists c > 0 such that
Z
(t+2τ hs)2
1
−
2
8
(hs)2 +1 |∂ u (x, t)| dσdt
e
ν
Γ1 ×R
Z
2
1
−t
2
≤ c k(u0 , u1 )kH 1 (Ω)×L2 (Ω) e 16 (hs)2 +1 dt
(3.4.3)
R
r
³
´
2
2
≤ c 16π (hs) + 1 k(u0 , u1 )kH 1 (Ω)×L2 (Ω) .
BOUNDARY STABILIZATION FOR THE WAVE EQUATION
1077
From (3.4.1)-(3.4.2)-(3.4.3), we deduce that there exists c > 0 such that
Z
−
e
(t+2τ hs)2
1
4
(hs)2 +1
|∂ν u (x, t)| dσdt
Γ1 ×R
ÃZ
!1/2 µq
Z
2
|∂ν u (x, t)| dσdt
≤c
Γ1
−
+c e
1
hβ
|t|≤|2τ hs|+
1
ch2β
q
2
¶1/2
(hs) + 1
(hs+1)
2
(hs) + 1 k(u0 , u1 )kH 1 (Ω)×L2 (Ω) .
Therefore,
¯Z
¯
"Z
#
¯
¯
L
¯
¯
(BP,Q (xo , ξo3 ) f ) (x, t, s) ds ∂ν u (x, t) dσdt¯
¯
¯ Γ1 ×R 0
¯
Z L Z Z ξo3 +1 Z
¯
¯
2
¯c
¯ − 1 (hs) |(ξ1 ,ξ2 )|2
dξdτ
≤c
¯ϕf (ξ, τ )¯ e 2 (hs)2 +1
R2
0

ξo3 −1
|τ |<R

1

×  ¡√
 s2 + 1¢1/2 µq
Ã
 Z
×


1
2
(hs) + 1
³
´
p

¶3/2 1 + πh (s2 + 1) 

!1/2 µq
Z
2
Γ1
(3.4.4)
|t|≤|2τ hs|+
1
hβ
|∂ν u (x, t)| dσdt
2
¶1/2
(hs) + 1
(hs+1)
¾
2
(hs) + 1 k(u0 , u1 )kH 1 (Ω)×L2 (Ω) ds .
q
1
+e− ch2β
But,
Z
Z
R2
Z
ξo3 +1
ξo3 −1
ξo3 +1
Z
Z
¯
¯
2
¯c
¯ − 1 (hs) |(ξ1 ,ξ2 )|2
dξdτ
¯ϕf (ξ, τ )¯ e 2 (hs)2 +1
|τ |<R
µZ
≤
ξo3 −1
µZ
(hs)2
− (hs)2 +1 |(ξ1 ,ξ2 )|2
×
e
R2
Z
R2
|τ |<R
¶1/2
¯
¯2
¯c
¯
¯ϕf (ξ, τ )¯ dξ1 dξ2
ξo3 +1
≤
ξo3 −1
|τ |<R
dξ1 dξ2
dξ3 dτ
v Ã
!
u
¶1/2
2
¯
¯2
u
(hs)
+
1
¯c
¯
,
dξ3 dτ tπ
¯ϕf (ξ, τ )¯ dξ1 dξ2
2
(hs)
R2
µZ
Z
¶1/2
1078
KIM DANG PHUNG
and due to singularity at s = 0, we will cut the integral over s ∈ (0, L) into two
parts, (0, L) = (0, ε) ∪ (ε, L) for some ε ∈ (0, 1), in order to get from (3.4.4)
¯Z
¯
"Z
#
¯
¯
L
¯
¯
(BP,Q (xo , ξo3 ) f ) (x, t, s) ds ∂ν u (x, t) dσdt¯
¯
¯ Γ1 ×R 0
¯
Z Z ξo3 +1 Z
¯
¯
¯c
¯
≤c
¯ϕf (ξ, τ )¯ dξdτ
R2
ξo3 −1
ÃZ
|τ |<R
!1/2
Z
2
×
Γ1
|t|≤2Rhε+

Z
ε
 ¡√
1
hβ
|∂ν u (x, t)| dσdt
(hε+1)
1
³
1
´
√

1 + hs  ds
¢1/2 q
2
s2 + 1
(hs) + 1
µZ ¯
¶1/2
Z ξo3 +1 Z
¯2
¯c
¯
+c
dξ3 dτ
¯ϕf (ξ, τ )¯ dξ1 dξ2
×
0
ξo3 −1
ÃZ
R2
|τ |<R
!1/2
Z
2
×
Γ1
Z
L
×
ε
|t|≤2RhL+
1
hβ
|∂ν u (x, t)| dσdt
(hL+1)

µ
¶
 1 + 1 ¡√ 1 ¢ q
1/2
hs
s2 + 1
1
+ c e− ch2β k(u0 , u1 )kH 1 (Ω)×L2 (Ω)
Z
L
0

1

 ¡√
 s2 + 1¢1/2 µq
2
(hs) + 1
Z Z ξo3 +1 Z

×

³
√ ´
1 + hs  ds
1
R2
³
1
2
ξo3 −1
1+
¶1/2
(hs) + 1
¯
¯
¯c
¯
¯ϕf (ξ, τ )¯ dξdτ
|τ |<R

√
´

hs  ds .

Now we use the following inequalities
Z
ε
0
Z
ε
¡√
L
¡√
1
s2 + 1
1
s2 + 1
Z L
ε
³
1
¢1/2 q
1+
2
√
(hs) + 1
³
1
¢1/2 q
2
1+
√
(hs) + 1
1
1
q
hs ¡√s2 + 1¢1/2
1
2
´
hs ds ≤ c ε ,
√
´
L
hs ds ≤ c √ ,
h
1 L
ds ≤ c ln ,
h
ε
(hs) + 1

√
Z L
√
1
1
1
L
 ¡√

q
hs ds ≤ c √ ,
¢
1/2
hs
2
h
ε
s2 + 1
(hs) + 1

BOUNDARY STABILIZATION FOR THE WAVE EQUATION
1079
where c > 0 is independent on h ∈ (0, 1], L ≥ 1 and ε ∈ (0, 1) to conclude that
¯Z
¯
"Z
#
¯
¯
L
¯
¯
(BP,Q (xo , ξo3 ) f ) (x, t, s) ds ∂ν u (x, t) dσdt¯
¯
¯ Γ1 ×R 0
¯
Z Z ξo3 +1 Z
¯
¯
¯c
¯
≤cε
¯ϕf (ξ, τ )¯ dξdτ
R2
ξo3 −1
ÃZ
|τ |<R
!1/2
Z
2
|∂ν u (x, t)| dσdt
×
Γ1
|t|≤2Rhε+
Ã√
1
hβ
!Z
1 L
L
+ c √ + ln
h
ε
h
ÃZ Z
×
Γ1
|t|≤2RhL+
(hε+1)
µZ
Z
ξo3 +1
ξo3 −1
R2
|τ |<R
¶1/2
¯2
¯
¯
¯c
dξ3 dτ
¯ϕf (ξ, τ )¯ dξ1 dξ2
!1/2
2
|∂ν u (x, t)| dσdt
1
hβ
(hL+1)
√
Z Z ξo3 +1 Z
¯
¯
L − 12β
¯c
¯
ch
k(u0 , u1 )kH 1 (Ω)×L2 (Ω)
+c√ e
¯ϕf (ξ, τ )¯ dξdτ .
2
h
R
ξo3 −1
|τ |<R
3.5. Proof of Proposition 3. From (1.6), Lemma 3.1, Lemma 3.2, Lemma 3.3,
Lemma 3.4, we deduce that there exists c > 0 such that for any (xo , ξo3 ) ∈ ωo ×
(2Z + 1) and h ∈ (0, 1], L ≥ 1, ε ∈ (0, 1), β ∈ (0, 1/2], R ≥ 1,
¯
¯Z
!
Ã
Z Z ξo3 +1 Z
¯
¯
1
¯
¯
i(xξ+tτ ) c
e
ϕf (ξ, τ ) dξdτ ao (x, t) u (x, t) dxdt¯
¯
4
¯
¯ Ω×R (2π) R2 ξo3 −1 |τ |<R
Z
Z
Z
¯
¯
ξo3 +1
1 p
¯c
¯
E (u, 0)
≤ c e− ch
¯ϕf (ξ, τ )¯ dξdτ
µ
+c
¶
1
hL
1+
√
hL
ξo3 −1
Z
p
Z
µ
√
L
+c
hL + ln
ε
µZ
Z ξo3 +1 Z
×
ÃZ
ξo3 −1
Z
ξo3 +1 Z
R2
|τ |<R
ξo3 −1
Z
¯
¯
¯c
¯
¯ϕf (ξ, τ )¯ dξdτ
|τ |<R
!1/2
2
Γ1
R2
|t|≤2RhL+
1
hβ
|∂ν u (x, t)| dσdt
(hL+1)
¶1/2
¯
¯2
¯c
¯
dξ3 dτ
¯ϕf (ξ, τ )¯ dξ1 dξ2
!1/2
Z
2
Γ1
ξo3 +1
|t|≤2Rhε+
Z
×
R2
R2
|τ |<R
¶1/2
¯
¯2
¯c
¯
dξ3 dτ
¯ϕf (ξ, τ )¯ dξ1 dξ2
Z
+ c εh
Z
µZ
Z
E (u, 0)
¶ ÃZ
ξo3 −1
|τ |<R
ξo3 +1
E (∂t u, 0)
p
1
+ cL2 e− 8h
R2
ξo3 −1
1
hβ
|∂ν u (x, t)| dσdt
(hε+1)
¯
¯
¯c
¯
¯ϕf (ξ, τ )¯ dξdτ
|τ |<R
Z
p
√
1
E (u, 0)
+ c hL e− ch2β
R2
Z
ξo3 +1
ξo3 −1
Z
¯
¯
¯c
¯
¯ϕf (ξ, τ )¯ dξdτ ,
|τ |<R
and the conclusion of Proposition 3 follows.
1080
KIM DANG PHUNG
4. Interpolation estimate. This section is devoted to prove Theorem 1.2.
We consider initial data (u0 , u1 ) ∈ H 6 (Ω)×H 5 (Ω) with the adequate
compatibility
¡
¢
boundary conditions in order that the solution of (1.5) u ∈ C 4 R; H 2 (Ω) . The
desired result will follow by density arguments. ¡
¢
By classical arguments, we can extend u ∈ C 4 R; H 2 (Ω) by u = u (x, t) outside
D, ∂D being of class C 2 in order to get u = u a.e. in Ω × R and
2
2
k∂tm u (·, ·, x3 , t)kH 2 (R2 ) ≤ co k∂tm u (·, ·, x3 , t)kH 2 (D) ∀ |m| ≤ 4
°
°2
°
°2
0
° m m0
°
°
°
∀ |m| ≤ 4, |m0 | ≤ 2
°∂t ∂x3 u (·, ·, x3 , t)° 2 2 ≤ co °∂tm ∂xm3 u (·, ·, x3 , t)° 2
L (R )
L (D)
(4.1)
for a.e. (x3 , t) ∈ (−1, 1) × R, where co > 0 only depends on D.
On the other hand, by integrating by parts on the time variable, we get
Z
2
2
2
χ (x) |ao (x, 0)| e−t |∂t u| dxdt
Ω×R
Z
2
2
=−
χ (x) |ao (x, 0)| e−t ∂t2 uudxdt
Z Ω×R
¢
2
2¡
2
+
χ (x) |ao (x, 0)| −1 + 2t2 e−t |u| dxdt ,
(4.2)
Ω×R
which implies, using (1.10), that
Z
Z
2
2 −t2
2
2
2
χ (x) |ao (x, 0)| e
|∂t u| dxdt +
χ (x) |ao (x, 0)| e−t |u| dxdt
Ω×R
Ω×R
Z
1 2
2
=−
χ (x) |ao (x, t)| e− 2 t ∂t2 uudxdt
Ω×R
Z
2
2
2
+2
χ (x) |ao (x, 0)| t2 e−t |u| dxdt
¯Z Ω×R
¯
¯
¯
− 12 t2 2
¯
≤¯
∂t u (x, t) ao (x, t) u (x, t) dxdt¯¯
ϕ (x, t) e
Ω×R
¯
¯Z
¯
¯
¯
+ 4¯
ϕ (x, t) u (x, t) ao (x, t) u (x, t) dxdt¯¯ .
Ω×R
1 2
From the Fourier inversion formula, when f = e− 2 t ∂t2 u or f = u, (see (A1) in
¡ ¢
c ∈ L1 R4 )
appendix A for ϕf
Z
ϕ (x, t) f (x, t) ao (x, t) u (x, t) dxdt
Ω×R
!
Ã
Z
Z
1
i(xξ+tτ ) c
e
ϕf (ξ, τ ) dξdτ ao udxdt
=
4
(2π) R4
Ω×R
Ã
!
(4.3)
Z
Z Z
1
i(xξ+tτ ) c
=
e
ϕf (ξ, τ ) dξdτ ao udxdt
4
(2π) R3 R≤|τ |
Ω×R
Ã
!
Z
Z Z
1
i(xξ+tτ ) c
+
e
ϕf (ξ, τ ) dξdτ ao udxdt .
4
(2π) R3 |τ |<R
Ω×R
BOUNDARY STABILIZATION FOR THE WAVE EQUATION
1081
1 2
But from (A4) in appendix A, when f = e− 2 t ∂t2 u or f = u, there exists c > 0,
independent on u, such that for any xo ∈ ωo and h ∈ (0, 1], R ≥ 1,
¯Z
¯
Ã
!
Z Z
¯
¯
1
¯
c (ξ, τ ) dξdτ ao udxdt¯¯
ei(xξ+tτ ) ϕf
¯
4
¯ Ω×R (2π) R3 R≤|τ |
¯
r
p
1p
≤c
E (u, 0) E (∂t u, 0) .
R
(4.4)
Now, by summing over ξo3 ∈ (2Z + 1) in the inequality of Proposition 3, we obtain,
1 2
using (A2)-(A3) in appendix A, when f = e− 2 t ∂t2 u or f = u, that there exists
c > 0, independent on u, such that for any xo ∈ ωo and h ∈ (0, 1], L ≥ 1, ε ∈ (0, 1),
β ∈ (0, 1/2], R ≥ 1,
¯Z
¯
Ã
!
Z Z
¯
¯
1
¯
c (ξ, τ ) dξdτ ao (x, t) u (x, t) dxdt¯¯
ei(xξ+tτ ) ϕf
¯
4
¯ Ω×R (2π) R3 |τ |<R
¯
¯
¯
!
Ã
¯
¯
Z Z ξo3 +1 Z
Z
¯
¯ X
1
i(xξ+tτ ) c
¯
e
ϕf (ξ, τ ) dξdτ ao udxdt¯¯
=¯
4
¯
¯ξo3 ∈(2Z+1) Ω×R (2π) R2 ξo3 −1 |τ |<R
¯Z
¯
!
Ã
Z Z ξo3 +1 Z
¯
¯
X
1
¯
¯
i(xξ+tτ ) c
e
ϕf
(ξ,
τ
)
dξdτ
a
udxdt
≤
¯
¯
o
¯ Ω×R (2π)4 R2 ξo3 −1 |τ |<R
¯
ξo3 ∈(2Z+1)
µ
¶
´
³
p
√
1
R3 p
E (u, 0)
E (∂t u, 0)
≤ c 1 + hL + L2 e− ch2β
h
¶
µ
µ 2 1/4
¶
1
1 p
R h p
+ c√
1+
E (∂t u, 0)
E (∂t u, 0)
hL
h
hL
ÃZ Z
!1/2 µ
¶
R3 p
2
+ c εh
|∂ν u (x, t)| dσdt
E (∂t u, 0)
h
Γ1 |t|≤2Rhε+ 1β (hε+1)
h
Ã
!1/2
µ
¶ Z Z
√
L
2
+c
|∂ν u (x, t)| dσdt
hL + ln
ε
Γ1 |t|≤2RhL+ 1β (hL+1)
h
µ 2 1/4
¶
R h p
×
E (∂t u, 0) .
h
(4.5)
Finally, from (4.2)-(4.3)-(4.4)-(4.5), we get that there exists c > 0 such that for any
xo ∈ ωo and h ∈ (0, 1], L ≥ 1, ε ∈ (0, 1), β ∈ (0, 1/2], R ≥ 1, the solution u of
(1.4) with initial data (u0 , u1 ) ∈ H 2 (Ω) × H 1 (Ω), ∂ν u0 |Γ0 = u0 |Γ1 = u1 |Γ1 = 0,
1082
KIM DANG PHUNG
satisfies
Z
2
2
2
χ (x) |ao (x, 0)| e−t |∂t u (x, t)| dxdt
Ω×R
r
p
1p
E (u, 0) E (∂t u, 0)
R
³
´ R3
p
p
√
1
+ c 1 + hL + L2
e− ch2β
E (u, 0) E (∂t u, 0)
h
µ
¶ 2 1/4
1
1
R h
+ c√
1+
E (∂t u, 0)
hL
h
hL
(4.6)
ÃZ Z
!1/2
3p
R
2
+ c εh
E (∂t u, 0)
|∂ν u (x, t)| dσdt
h
Γ1 |t|≤2Rhε+ 1β (hε+1)
h
¶
µ
√
L R2 h1/4 p
+c
hL + ln
E (∂t u, 0)
ε
h
ÃZ Z
!1/2
2
×
|∂ν u (x, t)| dσdt
.
≤c
Γ1
|t|≤2RhL+
1
hβ
(hL+1)
We assume that the number of sets covering D is a finite number and that
³ the
´
√
number of interval Ij such that |Ij | = h covering [−1/4, 1/4] is of order O √1h
N
√
£ j j+1 ¤
√
S
since [−1/4, 1/4] ⊂
h < ho ≡
4N , 4N for any integer N > 1. Taking
j=−(N +1)
³ ´
¡ j ro ¢
ro
1
,
we
can
cover
where xjo ∈ ωo
ω
o by a number of order O √h of balls B xo , 2
2
in order that there exist a constant
³ ´co > 0 only dependent on Ω and a bounded set
J ⊂ N satisfying card (J) = O √1h , such that we have the following inequalities
Z
2
|∂t u (x, t)| dxdt
ωo ×(0,T )
Z
≤e
T2
2
2
e−t |∂t u (x, t)| dxdt
ωo ×(0,T )
≤e
T
2
co
XZ
j∈J
2
≤ eT co
XZ
j∈J
D×Ij ×R
Ω×R
¯ ¡
¯2
¢
2
χxjo (x) ¯a x − xjo , 0, 0 ∂t u (x, t)¯ e−t dxdt
(4.7)
¯ ¡
¯2
¢
2
χxjo (x) ¯a x − xjo , 0, 0 ∂t u (x, t)¯ e−t dxdt .
From (4.6) and (4.7), we deduce that there exists c > 0 such that for any h ∈ (0, ho ),
L ≥ 1, ε ∈ (0, 1), β ∈ (0, 1/2], R ≥ 1, the solution u of (1.4) with initial data
BOUNDARY STABILIZATION FOR THE WAVE EQUATION
1083
(u0 , u1 ) ∈ H 2 (Ω) × H 1 (Ω), ∂ν u0 |Γ0 = u0 |Γ1 = u1 |Γ1 = 0, satisfies
√ Z
2
h
|∂t u (x, t)| dxdt
r
ωo ×(0,T )
p
1p
E (u, 0) E (∂t u, 0)
R
³
´ R3
p
p
√
1
+ c 1 + hL + L2
e− ch2β
E (u, 0) E (∂t u, 0)
h
¶ 2 1/4
µ
1
R h
1
+ c√
E (∂t u, 0)
1+
hL
h
hL
ÃZ Z
!1/2
R3 p
2
+ c εh
E (∂t u, 0)
|∂ν u (x, t)| dσdt
h
Γ1 |t|≤2Rhε+ 1β (hε+1)
h
¶
µ
√
L R2 h1/4 p
hL + ln
E (∂t u, 0)
+c
ε
h
ÃZ Z
!1/2
2
×
|∂ν u (x, t)| dσdt
.
≤c
Γ1
|t|≤2RhL+
1
hβ
(hL+1)
1/2
Now, we take R = h12 , L = hh14 and ε,β adequately in order to get the existence
of c > 0 such that for any h ∈ (0, ho ), the solution u of (1.4) with initial data
(u0 , u1 ) ∈ H 2 (Ω) × H 1 (Ω), ∂ν u0 |Γ0 = u0 |Γ1 = u1 |Γ1 = 0, satisfies
Z
2
|∂t u (x, t)| dxdt
ωo ×(0,T )
p
√ p
≤ c h E (u, 0) E (∂t u, 0) + chE (∂t u, 0)
ÃZ Z
!1/2
h1/2 p
2
+ c 10 E (∂t u, 0)
|∂ν u (x, t)| dσdt
1/2
h
Γ1 |t|≤c h 15
(4.8)
h
which gives from Theorem 2.1 and Cauchy-Schwarz inequality that there exists
c > 0 such that for any h ∈ (0, ho ), the solution u of (1.4) with initial data (u0 , u1 ) ∈
H 2 (Ω) × H 1 (Ω), ∂ν u0 |Γ0 = u0 |Γ1 = u1 |Γ1 = 0, satisfies
Z Z
1
2
E (u, 0) ≤ c 18
|∂ν u (x, t)| dσdt + chE (∂t u, 0) .
1/2
h
Γ1 |t|≤c h 15
h
By conservation of energy (1.6), this last inequality becomes, for any N > 1,
Z Z
X
1
2
N E (u, 0) ≤ c 18
|∂ν u (x, t)| dσdt + chN E (∂t u, 0)
1/2
h
Γ1 t∈(n,n+1)2c h 15
n=0,..,N −1
h
which gives by choosing N adequately the existence of a constant c > 0 such that for
any h ∈ (0, ho ), the solution u of (1.4) with initial data (u0 , u1 ) ∈ H 2 (Ω) × H 1 (Ω),
∂ν u0 |Γ0 = u0 |Γ1 = u1 |Γ1 = 0, satisfies
Z
Z
E (u, 0) ≤ c
Γ1
1/2
c hh33
2
|∂ν u (x, t)| dσdt + chE (∂t u, 0) .
0
Since the above estimate is also true for any h ≥ ho , this completes the proof of
Theorem 1.2.
1084
KIM DANG PHUNG
5. Proof of Theorem 1.1. This section is devoted to prove the polynomial decay
rate for the boundary damped wave equation.
First, recall
2,¢ there¡ exists c >¢ 0 such that for any T > 0
¡ that from
¢ Proposition
¡
and v ∈ C 2 R, L2 (Ω) ∩ C 1 R, H 1 (Ω) ∩ C R, H 2 (Ω) satisfying (2.1), we have
Z
T
Z
Z
T
2
Z
Z
2
|∂ν v (x, t)| dσdt ≤
0
Γ1
0
T
E (v, t) dt
|∇tan v (x, t)| dσdt + cE (v, T ) + c
Γ1
0
(5.1)
where |∇v| = |∂ν v| + |∇tan v| on Γ1 × R. Also, the following energy estimates
hold for any T > 0,
2
2
2
"Z
#1/2 "Z
Z
T
E (v, T ) ≤ 2
#1/2
Z
2
|∂t v (x, t)| dσdt
0
Z
T
2
"Z
T
|∂ν v (x, t)| dσdt
Γ1
T
0
#1/2 "Z
Z
T
2
E (v, t) dt ≤ 2T
#1/2
Z
2
|∂t v (x, t)| dσdt
0
0
,
Γ1
|∂ν v (x, t)| dσdt
Γ1
0
.
Γ1
(5.2)
On the other hand, by standard trace and interpolation theorems, there exists c > 0,
only dependent on Ω, such that for any T > 0,
Z
Z
T
2
|∇tan v (x, t)| dσdt
0
Γ1
Z
T
≤c
(E (v, t))
0
Ã
≤c
T3
4h3
Z
T
0
1/4
³
´3/4
2
k∇v (·, t)kH 1 (Ω)
dt
3h
E (v, t) dt +
4T
Z
T
0
(5.3)
!
2
k∇v (·, t)kH 1 (Ω)
dt
∀h ∈ (0, 1] .
¡ ¢1/δ
where
From (5.1)-(5.2)-(5.3), Cauchy-Schwarz inequality and choosing T = C h1
0
C > 0 and δ > 0 are given by Theorem 1.2, we deduce
that
there
exist
c
>
0 and
¡
¢
¡
¢
2
2
1
1
γ>
0
such
that
for
any
h
∈
(0,
1],
any
solution
v
∈
C
R,
L
(Ω)
∩C
R,
H
(Ω)
∩
¡
¢
2
C R, H (Ω) of (2.1) satisfies
Z
Z
Γ1
0
1
C( h
)
1/δ
1
|∂ν v (x, t)| dσdt ≤ c γ
h
2
Z
Z
1
C( h
)
0
h
+c
T
Γ1
Z
0
0
1/δ
2
|∂t v (x, t)| dσdt
0
T
2
k∇v (·, t)kH 1 (Ω)
(5.4)
.
Consequently, from Theorem 1.2, (5.4) and a standard decomposition of the solution
of the wave equation,¡we obtain¢ that there
exist c¢ > 0 ¡and γ > 0 ¢such that for any
¡
h ∈ (0, 1] and z ∈ C 2 R, L2 (Ω) ∩ C 1 R, H 1 (Ω) ∩ C R, H 2 (Ω) satisfying
(
∂t2 z − ∆z = 0
in Ω × R
∂ν z = 0
on Γ0 × R ,
BOUNDARY STABILIZATION FOR THE WAVE EQUATION
1085
the following estimate holds
E (z, 0) ≤ c
1
hγ
Z
Z
Γ1
Z
h
+c
T
c h1γ
³
2
|∂t z (x, t)| + |∂ν z (x, t)|
2
´
dσdt
0
T
2
k∇z (·, t)kH 1 (Ω) + hE (∂t z, 0) ,
0
¡ ¢1/δ
.
for any h ∈ (0, 1], where T = C h1
Applying this last inequality to the solution w of (1.1), it gives the existence of
c > 0 and γ > 0 such that for any h ∈ (0, 1], the solution w of (1.1) satisfies
1
E (w, 0) ≤ c γ
h
Z
Z
Γ1
c h1γ
2
|∂t w (x, t)| dσdt + ch (E (w, 0) + E (∂t w, 0)) .
0
By the energy identity (1.2), this last inequality becomes, for any N > 1,
µ
¶
Z Z
X
1
N
2
|∂t w (x, t)| dσdt
N E w, c γ ≤ c γ
h
h
Γ1 t∈(n,n+1)c h1γ
n=0,..,N −1
+ chN (E (w, 0) + E (∂t w, 0))
which gives by choosing N adequately that there exist C > 0 and δ > 0 such that
for any h ∈ (0, 1], the solution w of (1.1) satisfies
Z
1/δ
1
C( h
)
E (w, 0) ≤ C
Z
2
|∂t w (x, t)| dσdt + Ch (E (w, 0) + E (∂t w, 0)) .
0
Γ1
We conclude with Proposition 4 in appendix B.
Appendix A. ¡Here,
¢ we also denote by F (g) = gb the Fourier transform of an
element g ∈ L1 R4 . The goal of this appendix is to prove, with the notations of
the above sections, the following four inequalities,
Z
¯
¯
¯c
¯
¯ϕu (ξ, τ )¯ dξdτ +
Z
R4
Z
R4
Z
R3
¯ ³
¯
´
1 2
¯
¯
¯F ϕe− 2 t ∂t2 u (ξ, τ )¯ dξdτ < +∞ ,
Z
¯
¯
¯c
¯
¯ϕu (ξ, τ )¯ dξdτ +
R3
|τ |<R
Z
|τ |<R
¯ ³
¯
´
1 2
¯
¯
¯F ϕe− 2 t ∂t2 u (ξ, τ )¯ dξdτ
R3 p
E (∂t u, 0) ,
≤c
h
µZ
Z Z
R
|τ |<R
Z Z
R2
≤c
|τ |<R
(A2)
¶1/2
¯2
¯
¯
¯c
dξ3 dτ
¯ϕu (ξ, τ )¯ dξ1 dξ2
µZ
+
R
(A1)
R2
¶1/2
¯2
¯ ³
´
¯
¯
− 12 t2 2
dξ3 dτ
∂t u (ξ, τ )¯ dξ1 dξ2
¯F ϕe
R2 h1/4 p
E (∂t u, 0) ,
h
(A3)
1086
KIM DANG PHUNG
¯Z
¯
Ã
!
Z Z
¯
¯
1
¯
c (ξ, τ ) dξdτ ao (x, t) u (x, t) dxdt¯¯
ei(xξ+tτ ) ϕu
¯
4
¯ Ω×R (2π) R3 R≤|τ |
¯
¯Z
!
Ã
Z Z
¯
³
´
1
¯
i(xξ+tτ )
− 21 t2 2
+¯
e
F
ϕe
∂
u
(ξ, τ ) dξdτ
t
¯ Ω×R (2π)4 R3 R≤|τ |
(A4)
¯
¯
× ao (x, t) u (x, t) dxdt¯
r
p
1p
≤c
E (u, 0) E (∂t u, 0) ,
R
where c > 0 only depends on Ω.
To this end, we begin to compute
Z
2
|F ((1 − ∆) (ϕu)) (ξ, τ )| dξ
R3
¯Z
¶ ¯2
µZ
¯
¯
−ixξ
−itτ
¯
=
e
e
(1 − ∆) (ϕu) (x, t) dt dx¯¯ dξ
¯
R3
R3
R
¶¯2
Z ¯µZ
¯
¯
−itτ
¯
¯ dx
=
(x,
t)
dt
e
(1
−
∆)
(ϕu)
¯
¯
R3
R
¶¯2
Z ¯µZ
2
¯
¯
− t4
¯
|(1 − ∆) (χ (x) ao (x, 0) u (x, t))| dt ¯¯ dx
≤
e
¯
R3
R
¶ µZ
¶¯
Z ¯µZ
2
2
¯
¯
2
− t4
− t4
¯
dt
e
|(1 − ∆) (χ (x) ao (x, 0) u (x, t))| dt ¯¯ dx
≤
e
¯
R
R3
R
2
√ Z
2
− t4
≤ 4π
e
|(1 − ∆) (χ (x) ao (x, 0) u (x, t))| dtdx
4
R
Z
Z
t2
2
≤c
e− 4 ku (·, ·, x3 , t)kH 2 (R2 ) dtdx3
(−1,1) R
Z
¯2
t2 ¯
+c
e− 4 ¯∂x23 u (x, t)¯ dtdx
R2 ×(−1,1)×R
Z
t2
c
2
+√
e− 4 |∂x3 u (x, t)| dtdx
h R2 ×(−1,1)×R
Z
t2
c
2
+
e− 4 |u (x, t)| dtdx .
h R2 ×(−1,1)×R
Z
Consequently,
Z
c
2
|F ((1 − ∆) (ϕu)) (ξ, τ )| dξ ≤ cE (∂t u, 0) + E (u, 0) < +∞ ,
h
R3
(A5)
and similarly,
Z
¯
¯ ¡
¢
¡
¢
¡
¢
¯F (1 − ∆) ∂t2 (ϕu) (ξ, τ )¯2 dξ ≤ cE ∂t3 u, 0 + c E ∂t2 u, 0 < +∞ , (A6)
h
R3
Z ¯ ³
¯
³
´´
¢
¡
¢ c ¡
1 2
¯
¯2
¯F (1 − ∆) ϕe− 2 t ∂t2 u (ξ, τ )¯ dξ ≤ cE ∂t3 u, 0 + E ∂t2 u, 0
h
(A7)
R3
< +∞ ,
BOUNDARY STABILIZATION FOR THE WAVE EQUATION
1087
Z
¯2
¯ ³
´´
³
¡
¢ c ¡
¢
1 2
¯
¯
¯F (1 − ∆) ∂t2 ϕe− 2 t ∂t2 u (ξ, τ )¯ dξ ≤ cE ∂t5 u, 0 + E ∂t4 u, 0
h
(A8)
R3
< +∞ .
On another hand, we observe that
³
´
³
´
³
´
1 2
1 2
1 2
2
F ϕe− 2 t ∂t2 u = − |τ | F ϕe− 2 t u + 3iτ F ϕte− 2 t u
à Ã
!
µ ¶2 !
(A9)
3
3
− 21 t2
+F ϕ − +
t
u ,
e
2
2
because
3t2
1 2
ϕe− 2 t ∂t2 u = χ (x) ao (x, 0) e− 4 ∂t2 u
· ³
µ ³
¶
¸
´
2
2 ´
d
d2 ³ − 3t2 ´
2
− 3t4
− 3t4
4
= χ (x) ao (x, 0) ∂t e
e
u − 2∂t
u + 2 e
u .
dt
dt
Proof of (A1) .- From (A2), in order to check that the Fourier transform of ϕu
¡ ¢
1 2
and the one of ϕe− 2 t ∂t2 u are in L1 R4 , it is enough to prove that
Z Z
Z Z
¯
¯
¯
¯ ³
´
1 2
¯c
¯
¯
¯
¯ϕu (ξ, τ )¯ dξdτ +
¯F ϕe− 2 t ∂t2 u (ξ, τ )¯ dξdτ < +∞ .
R3
R3
R≤|τ |
− 21 t2
But, for g = u or g = e
Z
R≤|τ |
∂t2 u, we have
Z
Z
Z
|c
ϕg (ξ, τ )| dξdτ =
R3

Z
≤
R≤|τ |
1 
2 
|τ |
Z
≤π
R3
R≤|τ |
R≤|τ |
1/2
Z
1
³
R3
1 + |ξ|
µZ
1
2
2
|τ |
R≤|τ |
R3
2

´2 dξ 
³
|τ |
2
|τ |
2
µZ
R3
³
1 + |ξ|
2
1 + |ξ|
2
´
´ |c
ϕg (ξ, τ )| dξdτ
¯ ¡
¯
¢
¯F (1 − ∆) ∂t2 (ϕg) (ξ, τ )¯2 dξ
¯ ¡
¯
¢
¯F (1 − ∆) ∂t2 (ϕg) (ξ, τ )¯2 dξ
¶1/2
¶1/2
dτ ,
which gives the desired result using (A6) and (A8).
Proof of (A2) .- We have
³
´
2
Z Z
Z Z
¯
¯
¯
¯
1
+
|ξ|
¯c
¯
c (ξ, τ )¯¯ dξdτ
³
´ ¯¯ϕu
¯ϕu (ξ, τ )¯ dξdτ =
2
R3 |τ |<R
R3 |τ |<R 1 + |ξ|

1/2
¶1/2
µZ
Z
Z
1


2
≤
dτ
|F ((1 − ∆) (ϕu)) (ξ, τ )| dξ

³
´2 dξ 
|τ |<R
R3 1 + |ξ|2
R3
µZ
¶1/2
Z
2
2
|F ((1 − ∆) (ϕu)) (ξ, τ )| dξ
≤π
dτ
|τ |<R
R3
R
≤ c E (∂t u, 0) ,
h
using (A5). In a similar way, using (A9) with R ≥ 1, we obtain
ÃZ
!
Z Z
¯ ³
¯
´
c
¯
¯
2
− 12 t2 2
∂t u (ξ, τ )¯ dξdτ ≤
E (∂t u, 0) .
|τ | dτ
¯F ϕe
h
R3 |τ |<R
|τ |<R
dτ
1088
KIM DANG PHUNG
Proof of (A3) .- First, we compute, using Parseval identity,
Z
µZ
¶1/2
¯
¯2
¯c
¯
dξ3 dτ
¯ϕu (ξ, τ )¯ dξ1 dξ2
|τ |<R
R2
Z
³Z ¯¯Z
¯
e−i(x1 ξ1 +x2 ξ2 )
¯
Z
|ξ3 |≤2R
Z
=
|ξ3 |≤2R
R2
|τ |<R
·Z
e
−i(x3 ξ3 +tτ )
R2
Z
R2
!1/2
¯2
¸
¯
ϕ (x, t) u (x, t) dx3 dt dx1 dx2 ¯¯ dξ1 dξ2
dξ3 dτ
ÃZ
Z
=
2π
|ξ3 |≤2R
ÃZ
≤ cR2
|τ |<R
¯Z
¯
¯
¯
2
e
−i(x3 ξ3 +tτ )
R2
R
!1/2
¯2
¯
ϕ (x, t) u (x, t) dx3 dt¯¯ dx1 dx2
dξ3 dτ
¯Z
¯
!1/2
Z
¯
¯
2
2
t2
¯
¯
2
e− 4 |u (x, t)| dx3 dt
e−x3 /(ch) e−t /4 dx3 dt¯ dx1 dx2
¯
¯
R2 ¯ (−1,1)×R
R2
ÃZ
!1/2
t2
≤ cR2 h1/4
2
e− 4 |u (x, t)| dxdt
.
R2 ×(−1,1)×R
Now, we obtain
µZ
Z Z
R
R2
|τ |<R
Z
Z
¶1/2
¯
¯2
¯c
¯
ϕu
(ξ,
τ
)
dξ
dξ
dξ3 dτ
¯
¯
1 2
µZ
=
|ξ3 |≤2R
R2
|τ |<R
Z
Z
+
2R<|ξ3 |
Ã
×
Z
2π
R
|τ |<R
¯Z
¯
¯
¯
2
R2
¶1/2
¯
¯2
¯c
¯
(ξ,
τ
)
ϕu
dξ
dξ
dξ3 dτ
¯
¯
1 2
1
|ξ32 − τ 2 |
!1/2
¯2
¯
¡ 2
¢ −i(x3 ξ3 +tτ )
2
ξ3 − τ e
ϕ (x, t) u (x, t) dx3 dt¯¯ dx1 dx2
dξ3 dτ
!1/2
ÃZ
2 1/4
≤ cR h
e
2
− t4
2
|u (x, t)| dxdt
R2 ×(−1,1)×R
 Ã
!1/2
Z
t2
1
2
e− c |u (x, t)| dxdt
+ ch1/4 
h
R2 ×(−1,1)×R
ÃZ
+
e
R2 ×(−1,1)×R
2
− t4
!1/2 
¯
¯¡ 2
¢
2
¯ ∂t − ∂x2 u (x, t)¯ dxdt

3
where we have used the following inequalities
¯¡ 2
¯
¯¢
¯
¡
¢
¯ ∂t − ∂x2 ϕ (x, t)¯ ≤ c χ (x) + |∂x3 χ (x)| + ¯∂x2 χ (x)¯ e−(x3 −xo3 )2 /(ch) e−t2 /c ,
3
3
h
BOUNDARY STABILIZATION FOR THE WAVE EQUATION
1089
¯
¯Z
¯
¯
¡ 2
¢ −i(x3 ξ3 +tτ )
2
¯
¯
u
(x,
t)
dx
dt
ξ
−
τ
e
ϕ
(x,
t)
3 ¯
¯ 2 3
R
¯Z
¯
¯
¯
¡
¢
= ¯¯
e−i(x3 ξ3 +tτ ) ∂t2 − ∂x23 [ϕ (x, t) u (x, t)] dx3 dt¯¯
2
ZR
Z
¯
¯¡ 2
¯
¯¡
¢
¢
¯ ∂t − ∂x2 ϕ (x, t)¯ |u (x, t)| dx3 dt +
≤
ϕ (x, t) ¯ ∂t2 − ∂x23 u (x, t)¯ dx3 dt
3
R2
≤
c
h
R2
ÃZ
!1/2 µZ
t2
2
e− c |u (x, t)| dx3 dt
e−
R2
(−1,1)×R
"Z
e
+
#1/2 ·Z
¯¡ 2
¯2
¢
2
¯ ∂t − ∂x u (x, t)¯ dx3 dt
3
2
− t4
(−1,1)×R
 Ã
!1/2
Z
t2
1
2
≤ ch1/4 
e− c |u (x, t)| dx3 dt
h
(−1,1)×R
ÃZ
+
e
2
− t4
(−1,1)×R
Z
2R<|ξ3 |
Z
|τ |<R
¶1/2
t2
e− c dx3 dt
(x3 −xo3 )2
ch
−
e
(x3 −xo3 )2
4h
e
2
− t4
R2
¸1/2
dx3 dt
!1/2 
¯¡ 2
¯
¢
2
¯ ∂t − ∂x2 u (x, t)¯ dx3 dt
 ,
3
1
dτ dξ3 ≤
|ξ32 − τ 2 |
ÃZ
!Ã Z
!
4
1
dξ3 ≤ c .
dτ
3 2R<|ξ3 | ξ32
|τ |<R
Proof of (A4) .- Introduce
Ã
!
Z
Z Z
1
H (g) =
ei(xξ+tτ ) ϕ
cg (ξ, τ ) dξdτ ao u (x, t) dxdt .
4
(2π) R3 R≤|τ |
Ω×R
By an integration by parts,
Ã
·Z
¸ !
Z
Z
1
1 itτ
−iθτ
H (g) = −
∂t (ao u (x, t))
e
e
ϕg (x, θ) dθ dτ dxdt .
2π R≤|τ | iτ
Ω×R
R
It follows using Cauchy-Schwarz inequality and Parseval identity that
Ã
!1/2
Z
h 1 Z
1
|H (g)| ≤
|∂t (ao u (x, t))|
dτ
2
2π
Ω×R
R≤|τ | τ

ÃZ ¯Z
¯2 !1/2
¯
¯
¯ e−iθτ ϕg (x, θ) dθ¯ dτ
 dxdt
×
¯
¯
R
R
Z
≤
|∂t (ao u (x, t))|
Ω×R

!1/2 µ Z
¶1/2
1
2
 dxdt
dτ
2π |ϕg (x, θ)| dθ
2
R≤|τ | τ
R
Ã
!
r
Z
1
2
≤
|∂t (ao u (x, t))| √
kϕg (x, ·)kL2 (R) dxdt
2π R
Ω×R
r
Z
1
1
≤√
k∂t (ao u) (·, t)kL2 (Ω) dt kϕgkL2 (Ω×R) .
π R R

1
×
2π
ÃZ
1090
KIM DANG PHUNG
Since we have the following estimates
°
°
1 2
°
°
kϕukL2 (Ω×R) + °ϕe− 2 t ∂t2 u° 2
µZ
L (Ω×R)
Z
1 2
e− 2 t
≤
q
≤
R
√
Ω
¶1/2 µZ
¶1/2
Z
¯ 2
¯2
2
− 21 t2
¯
¯
|u (x, t)| dxdt
+
e
∂t u (x, t) dxdt
R
³ p
´
p
2π c E (u, 0) + E (∂t u, 0) ,
Ω
Z
R
k∂t (ao u) (·, t)kL2 (Ω) dt
¶1/2
¶1/2
Z µZ
2
2
dt
dt +
|ao ∂t u (x, t)| dx
|∂t ao u (x, t)| dx
Z µZ
≤
R
Z
Ω
R
µZ
|t| − 1 t2
e 4
R 2
p
≤ c E (u, 0) ,
Ω
¶1/2
µZ
¶1/2
Z
2
2
− 41 t2
|u (x, t)| dx
dt + e
|∂t u (x, t)| dx
dt
≤
Ω
R
Ω
we conclude that
r
¯ ³ 1 2
´¯
p
1p
¯
¯
−2t 2
E (u, 0) E (∂t u, 0) .
|H (u)| + ¯H e
∂t u ¯ ≤ c
R
That completes the proof of (A4).
Appendix B.
Proposition 4. Let δ > 0. The following three assertions are equivalent.
(i): There exists C > 0 such that the solution w of (1.1) satisfies
C
(E (w, 0) + E (∂t w, 0)) ∀t > 0 .
tδ
(ii): There exists C > 0 such that the solution w of (1.1) satisfies
E (w, t) ≤
Z
³
C
E (w, 0) ≤ C
E(w,0)+E(∂t w,0)
E(w,0)
´1/δ
Z
0
2
α (x) |∂ν w (x, t)| dσdt .
Γ1
(iii): There exists C > 0 such that for any h ∈ (0, 1], the solution w of (1.1)
satisfies
Z C ( h1 )1/δ Z
2
E (w, 0) ≤ C
α (x) |∂ν w (x, t)| dσdt + Ch (E (w, 0) + E (∂t w, 0)) .
0
Γ1
Proof. (i) ⇒ (ii). Suppose that
E (w, T ) ≤
C
(E (w, 0) + E (∂t w, 0))
Tδ
∀T > 0 .
Therefore from (1.2)
Z
T
Z
2
E (w, 0) = E (w, T ) + 2
α (x) |∂ν w (x, t)| dσdt
0
Γ1
C
≤ δ (E (w, 0) + E (∂t w, 0)) + 2
T
Z
T
Z
2
α (x) |∂ν w (x, t)| dσdt .
0
Γ1
BOUNDARY STABILIZATION FOR THE WAVE EQUATION
By choosing
1091
µ
¶1/δ
E (w, 0) + E (∂t w, 0)
T = 2C
E (w, 0)
we get
³
´
E(w,0)+E(∂t w,0) 1/δ
2C
E(w,0)
Z
E (w, 0) ≤ 4
Z
2
α (x) |∂ν w (x, t)| dσdt .
0
Γ1
(ii) ⇒ (i). Conversely, suppose that
Z
³
c
E (w, 0) ≤ 2c
E(w,0)+E(∂t w,0)
E(w,0)
´1/δ
Z
2
α (x) |∂ν w (x, t)| dσdt
Γ1
0
for some c > 1, then using a translation in time, for any s ≥ 0
E (w, s)
≤ 2c
(E (w, 0) + E (∂t w, 0))
Denoting G (s) =
Ã
µ
G s+c
³
´
(E(w,0)+E(∂t w,0)) 1/δ
s+c
E(w,s)
Z
s
which gives
¶1/δ
³
1
G(s)
´1/δ
"
Ã
≤ G (s) ≤ c G (s) − G s + c
µ
G s+c
If c1 s ≤ c
α (x) |∂ν w (x, t)|
dσdt .
E (w, 0) + E (∂t w, 0)
we have
!
Ã
Let us introduce c1 =
2
Γ1
E(w,s)
E(w,0)+E(∂t w,0) ,
1
G (s)
Z
¡ 1+c ¢1/δ
c
1
G (s)
¶1/δ !
≤
µ
1
G (s)
¶1/δ !#
c
G (s) .
1+c
− 1 > 0. We distinguish two cases.
³
, then G (s) ≤
c 1
c1 s
´δ
and
µ
¶δ
c 1
G ((1 + c1 ) s) ≤
.
c1 s
µ
³
´1/δ
³
´1/δ ¶
1
1
If c1 s > c G(s)
, then G ((1 + c1 ) s) ≤ G s + c G(s)
and
G ((1 + c1 ) s) ≤
c
G (s) .
1+c
³
´δ
Let us denote d (s) = cc1 1s .
Then by induction, we obtain ∀s > 0, ∀n ∈ N, n ≥ 1,
·
µ
¶
c
s
G ((1 + c1 ) s) ≤ max d (s) ,
d
, · · ·,
1+c
(1 + c1 )
µ
¶n µ
¶ µ
¶n+1 µ
¶#
c
s
c
s
,
d
,
G
.
n
n
1+c
1+c
(1 + c1 )
(1 + c1 )
Now, remark that with the above choice of c1 ,
µ
¶
c
s
d
= d (s)
1+c
(1 + c1 )
∀s > 0 .
1092
KIM DANG PHUNG
Consequently,
Ã
µ
G ((1 + c1 ) s) ≤ max d (s) ,
Ã
µ
≤ max d (s) ,
c
1+c
c
1+c
and we conclude that
E (w, s)
= G (s) ≤ d
E (w, 0) + E (∂t w, 0)
µ
¶n+1
µ
G
¶n+1 !
s
n
(1 + c1 )
¶!
∀n ≥ 1
∀n ≥ 1 ,
s
1 + c1
¶
µ
=
c (1 + c1 )
c1
¶δ
1
sδ
∀s > 0 .
(ii) ⇒ (iii). Suppose that
Z
³
C
E (w, 0) ≤ C
E(w,0)+E(∂t w,0)
E(w,0)
´1/δ
0
If
E(w,0)+E(∂t w,0)
E(w,0)
2
α (x) |∂ν w (x, t)| dσdt .
Γ1
≤ h1 , then
Z
1/δ
1
C( h
)
E (w, 0) ≤ C
0
If
Z
E(w,0)+E(∂t w,0)
E(w,0)
Z
2
α (x) |∂ν w (x, t)| dσdt .
Γ1
> h1 , then
E (w, 0) ≤ h (E (w, 0) + E (∂t w, 0)) .
Consequently, for any h > 0,
Z C ( h1 )1/δ Z
2
E (w, 0) ≤ C
α (x) |∂ν w (x, t)| dσdt + h (E (w, 0) + E (∂t w, 0)) .
0
Γ1
(iii) ⇒ (ii). Conversely, suppose that for there exists C > 0 such that for any
h ∈ (0, 1],
Z C ( h1 )1/δ Z
2
E (w, 0) ≤ C
α (x) |∂ν w (x, t)| dσdt + Ch (E (w, 0) + E (∂t w, 0)) .
0
Γ1
Then, we deduce that there exists C > 0 such that for any h > 0,
Z C ( h1 )1/δ Z
2
E (w, 0) ≤ C
α (x) |∂ν w (x, t)| dσdt + h (E (w, 0) + E (∂t w, 0)) .
0
Γ1
We conclude by choosing h =
E(w,0)
1
2 E(w,0)+E(∂t w,0) .
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Received January 2007; revised October 2007.
E-mail address: kim dang phung@yahoo.fr