Electronic Journal of Differential Equations, Vol. 2014 (2014), No. 202, pp. 1–18.
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
ftp ejde.math.txstate.edu
GENERAL BOUNDARY STABILIZATION RESULT OF
MEMORY-TYPE THERMOELASTICITY WITH SECOND SOUND
FAIROUZ BOULANOUAR, SALAH DRABLA
Abstract. In this article we consider an n-dimensional system of visco-thermoelasticity with second sound, where a viscoelastic dissipation is acting on a
part of the boundary. We prove an explicit general decay rate result without
imposing u0 = 0 as in [17]. This allows a larger class of relaxation functions and
initial data, hence, generalizes some previous results existing in the literature.
1. Introduction
The Classical Fourier law of heat conduction expresses that the heat flux within
a medium is proportional to the local temperature gradient in the system. A well
known consequence of this law is that heat perturbations propagates with an infinite
speed. Experiments showed that heat conduction in some dielectric crystals at low
temperatures is free of this paradox and disturbances, which are almost entirely
thermal, propagate in a finite speed. This phenomenon in dielectric crystals is
called second sound (see [6]). To overcome this physical paradox, Maxwell [7],
Cattaneo [3] adopted a non-classical heat flux Maxwell-Cattaneo law to get rid of
this unphysical results. This Maxwell-Cattaneo relation contains an extra inertial
term with respect to the Fourier law
τ0 qt + q + κ∇θ = 0,
where q = q(x, t) ∈ Rn is the heat flux vector, τ0 is the relaxation time and κ is the
heat conductivity. The conservation of energy equation introduces the hyperbolic
equation, which describes heat propagation with finite speed.
Result concerning existence, blow up, and asymptotic behavior of smooth, as
well as weak solutions in thermoelasticity with second sound have been established
over the past two decades by many mathematicians. Tarabek [20] treated problems
related to
utt − a(ux , θ, q)uxx + b(ux , θ, q)θx = α1 (ux , θ)qqx
θt + g(ux , θ, q)qx + d(ux , θ, q)utx = α2 (ux , θ)qqt
τ (ux , θ)qt + q + k(ux , θ)θx = 0,
2000 Mathematics Subject Classification. 35B37, 35L55, 74D05, 93D15, 93D20.
Key words and phrases. Thermoelasticity with second sound; viscoelastic damping;
decay; relaxation function; boundary stabilization; convexity.
c
2014
Texas State University - San Marcos.
Submitted August 1, 2014. Published September 30, 2014.
1
(1.1)
2
F. BOULANOUAR, S. DRABLA
EJDE-2014/202
in both bounded and unbounded situations and established global existence results
for small initial data. He also showed that these “ classical” solutions tend to
equilibrium as t tends to infinity; however, no rate of decay has been discussed. In
his work, Tarabek used the usual energy argument and exploited some relations
from the second law of thermodynamics to overcome the difficulty arising from the
lack of Poincaré’s inequality in the unbounded domains. Racke [18] discussed (1.1)
and established exponential decay results for several linear and nonlinear initial
boundary value problems. In particular he studied (1.1), with α1 = α2 = 0, and
for a rigidly clamped medium with temperature hold constant on the boundary. i.e
u(t, 0) = u(t, 1) = 0,
θ(t, 0) = θ(t, 1) = θ̄,
t ≥ 0,
and showed that, for small enough initial data and classical solutions decay exponentially to the equilibrium state. Messaoudi and Said-Houari [9] extended the
decay result of [18] to the case when α1 6= 0, α2 6= 0.
For the multi-dimensional case (n = 2, 3), Racke [19] established an existence
result for the following n-dimensional problem
utt − µ∆u − (µ + λ)∇(div u) + β∇θ = 0
θt + γ div q + δ div ut = 0
ut (., 0) = u1 ,
u=θ=0
in Ω × (0, +∞)
in Ω × (0, +∞)
τ qt + q + κ∇θ = 0
u(., 0) = u0 ,
in Ω × (0, +∞)
θ(., 0) = θ0 ,
q(., 0) = q0
(1.2)
in Ω
on Γ × [0, +∞),
where Ω is a bounded domain of Rn , with a smooth boundary Γ, u = u(x, t),
q = q(x, t) ∈ Rn , and µ, λ, β, γ, δ, τ, κ, are positive constants, where µ, λ are Lame
moduli and τ is the relaxation time, a small parameter compared to the others.
In particular if τ = 0, (1.2) reduces to the system of classical thermoelasticity, in
which the heat flux is given by Fourier’s law instead of Cattaneo’s law. He also
proved, under the conditions rotu = rotq = 0, an exponential decay result for (1.2).
This result applies automatically to the radially symmetric solution, since it is only
a special case.
Messaoudi [8] considered (1.2), in the presence of a source term, and proved a
blow up result for solutions with negative initial energy. This result was extended
later to certain solutions with positive energy by Messaoudi and Said-Houari [10].
In this article, we are concerned with the system
utt − µ∆u − (µ + λ)∇(div u) + β∇θ = 0
cθt + κ div q + β div ut = 0
τ0 qt + q + κ∇θ = 0
u(., 0) = u0 ,
ut (., 0) = u1 ,
u=0
Z
u(x, t) = −
t
in Ω × (0, +∞)
in Ω × (0, +∞)
in Ω × (0, +∞)
θ(., 0) = θ0 ,
q(., 0) = q0
in Ω
on Γ0 × [0, +∞)
(1.3)
∂u
+ (µ + λ)(div u)ν)(s)ds on Γ1 × [0, +∞)
∂ν
θ = 0 on Γ × [0, +∞),
g(t − s)(µ
0
which models the transverse vibration of a thin elastic body, taking in account
the heat conduction given by Cattaneo’s law. Here, Ω is a bounded domain of Rn
(n ≥ 2) with a smooth boundary Γ, such that {Γ0 , Γ1 } is a partition of Γ, ν is the
EJDE-2014/202
BOUNDARY STABILIZATION
3
outward normal to Γ, u = u(x, t) ∈ Rn is the displacement vector, q = q(x, t) ∈ Rn
is the heat flux vector, θ = θ(x, t) is the difference temperature and the relaxation
function g is a positive differentiable function. The coefficients c, κ, β, µ, λ, τ0 are
positive constants, where µ, λ are Lame moduli and τ0 is the relaxation time, a
small parameter compared to the others. The boundary condition on Γ1 is the
nonlocal boundary condition responsible for the memory effect.
Messaoudi and Al-Shehri treated system (1.3) of thermoelasticity with second
sound in [13] subject to boundary condition of memory type. If k is the resolvent
kernel of −g 0 /g(0), they showed in [12] that the energy decays at the same rate
as of (−k 0 ), while in [13], when (−k 0 ) decays exponentially, the energy decays at a
polynomial rate. Recently, Mustafa [17] treated system (1.3) for k satisfying
k(0) > 0,
lim k(t) = 0,
t→∞
00
0
k (t) ≥ H(−k (t)),
k 0 (t) ≤ 0,
∀t > 0,
(1.4)
(1.5)
where H is a positive function, which is linear or strictly increasing, strictly convex
of class C 2 on (0, r], r < 1, and H(0) = 0 and proved for u0 = 0 on Γ1 , an explicit
energy decay formula which is not necessarily of exponential or polynomial-type
decay.
Our aim in this work is to investigate (1.3) for resolvent kernels satisfying (1.4)
and (1.5), when u0 6= 0 on Γ1 is taken into account. The proof is based on the
multiplier method and makes use of some estimates of [17] with the necessary
modification needed to obtain our result. The paper is organized as follows. In
section 2, we present some notations and material needed for our work. In section
3, we establish some technical lemmas and state our main theorem, while the proof
of our main result will be given in section 4.
2. Notation and transformation
In this section we introduce some notation and prove some lemmas. To establish
our result, we shall make the following assumption:
(A1) The partition Γ0 and Γ1 are closed, disjoint, with meas(Γ0 ) > 0 and satisfying
Γ1 = {x ∈ Γ : m(x).ν ≥ δ > 0},
Γ0 = {x ∈ Γ : m(x).ν ≤ 0},
where m(x) = x − x0 , for some x0 ∈ Rn .
Similarly to [11, 12, 14], applying Volterra’s inverse operator, the boundary condition
Z t
∂u
+ (µ + λ)(div u)ν (s)ds, on Γ1 × [0, +∞), (2.1)
u(x, t) = −
g(t − s) µ
∂ν
0
can be transformed into
µ
∂u
1
+ (µ + λ)(div u)ν = −
(ut + k ∗ ut ),
∂ν
g(0)
on Γ1 × [0, +∞),
where * denotes the convolution product
Z t
(ϕ ∗ ψ)(t) =
ϕ(t − s)ψ(s) ds,
0
4
F. BOULANOUAR, S. DRABLA
EJDE-2014/202
and k is the resolvent kernel of (−g 0 /g(0)) which satisfies
1
1 0
k+
(g 0 ∗ k) = −
g.
g(0)
g(0)
Taking η = 1/g(0), we arrive at
∂u
+ (µ + λ)(div u)ν = −η(ut + k(0)u − k(t)u0 + k 0 ∗ u), on Γ1 × [0, +∞). (2.2)
∂ν
Then, we will use the boundary relation (2.2) instead of the fourth equation in
(1.3). Let us define
Z t
ϕ(t − s)|ψ(t) − ψ(s)|2 ds,
(ϕ ◦ ψ)(t) =
0
Z t
(ϕ♦ψ)(t) =
ϕ(t − s)(ψ(t) − ψ(s))ds.
µ
0
By using Hôlder’s inequality, we have
Z t
2
|(ϕ♦ψ)(t)| ≤
|ϕ(s)|ds (|ϕ| ◦ ψ)(t).
(2.3)
0
Lemma 2.1 ([15]). If ϕ, ψ ∈ C 1 (R+ ), then
1
1
1 d
(ϕ ∗ ψ)ψt = − ϕ(t)|ψ(t)|2 + ϕ0 ◦ ψ −
ϕ◦ψ−
2
2
2 dt
Z
t
ϕ(s)ds |ψ(t)|2 . (2.4)
0
Let us define
V = {v ∈ (H 1 (Ω)) : v = 0 on Γ0 }.
The well-posedness of system (1.3) is presented in the following theorem, which can
be proved, using the Galerkin method as in [1, 4] and the reference therein.
Theorem 2.2. Let k ∈ W 2,1 (R+ ) ∩ W 1,∞ (R+ ), u0 ∈ ((H 2 (Ω)) ∩ V )n , θ0 ∈ H01 (Ω),
q0 ∈ (H 1 (Ω))n , and u1 ∈ V n , with
∂u0
+ ηu0 = 0 on Γ1 .
∂ν
Then there exists a unique strong solution u of system (1.3), such that
(2.5)
u ∈ C(R+ ; (H 2 (Ω) ∩ V )n ) ∩ C 1 (R+ ; V n ),
θ ∈ C(R+ ; H01 (Ω)) ∩ C 1 (R+ ; L2 (Ω)),
q ∈ C(R+ ; (H 1 (Ω))n ) ∩ C 1 (R+ ; (L2 (Ω))n ).
3. Decay of solutions
In this section we discuss the asymptotic behavior of the solutions of system
(1.3) when the resolvent kernel k satisfies the assumption
(B1) k : R+ → R+ is a C 2 function such that
k(0) > 0,
lim k(t) = 0,
t→∞
k 0 (t) ≤ 0,
and there exists a positive function H ∈ C 1 (R+ ), with H(0) = 0, and H is
linear or strictly increasing and strictly convex C 2 function on (0, r], r < 1,
such that
k 00 (t) ≥ H(−k 0 (t)), ∀t > 0.
EJDE-2014/202
BOUNDARY STABILIZATION
5
It is a routine procedure to define the first-order energy of system (1.3) by (see
Lemma 3.2 below).
Z
2
1
E1 (t) =
|ut | + µ|∇u|2 + (µ + λ)(div u)2 + cθ2 + τ0 q 2 dx
2 Ω
Z
Z
(3.1)
η
η
(k 0 ◦ u)(t)dΓ1 +
k(t)|u|2 dΓ1 .
−
2 Γ1
2 Γ1
Now, we differentiate (1.3), with respect to t, to obtain
uttt − µ∆ut − (µ + λ)∇(div ut ) + β∇θt = 0
cθtt + κ div qt + β div utt = 0
τ0 qtt + qt + κ∇θt = 0
in Ω × (0, +∞)
in Ω × (0, +∞)
(3.2)
in Ω × (0, +∞)
and the boundary condition (2.2) to obtain
µ
∂ut
+ (µ + λ)(div ut )ν = −η(utt + k(0)ut + k 0 ∗ ut ),
∂ν
on Γ1 × R+ .
(3.3)
Consequently, similar computations yield the second-order energy of system (1.3):
Z
1
|utt |2 + µ|∇ut |2 + (µ + λ)(div ut )2 + cθt2 + τ0 qt2 dx
E2 (t) =
2 Ω
Z
Z
η
η
−
(k 0 ◦ ut )(t)dΓ1 +
k(t)|ut |2 dΓ1 .
2 Γ1
2 Γ1
Theorem 3.1. Given (u0 , u1 , θ0 , q0 ) ∈ (H 2 (Ω) ∩ V )n × V n × H01 (Ω) × (H 1 (Ω))n ,
we assume that (A1) and (B1) hold. Then there exist positive constants c1 , c2 , k1 ,
k2 , k3 , ε0 and t1 such that:
(I) In the special case H(t) = ctp , where 1 ≤ p < 3/2, the solution of (1.3)
satisfies
R
R
Z
1
c1 + c2 t [k(s)
Z ∞
|u0 |2 dΓ1 ]2p−1 ds 2p−1
η
t1
Γ1
E1 (t) ≤
|u0 |2 dΓ1
−
k 2 (s)ds
t
2 Γ1
t
(3.4)
for all t ≥ t1 .
(II) In the general case, the solution of (1.3) satisfies
R
Rt
k2 + k3 (
|u0 |2 dΓ1 ) t1 H0 (k(s))ds Γ1
−1
E1 (t) ≤ k1 H1
t
(3.5)
Z
Z ∞
η
2
2
−
|u0 | dΓ1
k (s)ds for all t ≥ t1 ,
2 Γ1
t
where
H1 (t) = tH00 (ε0 t),
H0 (t) = H(D(t)),
provided that D is a positive C 1 function, with D(0) = 0, for which H0 is strictly
increasing and strictly convex C 2 function on (0, r] and
Z +∞
−k 0 (s)
ds < +∞.
(3.6)
−1 00
H0 (k (s))
0
6
F. BOULANOUAR, S. DRABLA
EJDE-2014/202
Remarks.
R ∞ 1. If u0 = 0 on Γ1 , we then obtain the results in Mustafa [17].
2. If 0 H0 (k(s))ds < +∞, then (3.5) reduces to
Z
Z ∞
c
η
E1 (t) ≤ k1 H1−1 ( ) −
|u0 |2 dΓ1
k 2 (s)ds,
t
2 Γ1
t
which clearly shows that limt→∞ E1 (t) = 0, in this case.
3. The usual decay rate estimate, already proved for k satisfying k 00 ≥ d(−k 0 )p ,
1 ≤ p < 3/2, is a special case of our result. We will provide a “simpler” proof for
this special case.
4. The condition k 00 ≥ d(−k 0 )p , 1 ≤ p < 3/2 assumes −k 0 (t) ≤ ωe−dt when p = 1
1
and −k 0 (t) ≤ ω/t p−1 when 1 < p < 3/2. Our result allows resolvent kernels whose
derivatives are not necessarily of exponential or polynomial decay. For instance, if
√
k 0 (t) = − exp(− t),
then k 00 (t) = H(−k 0 (t)) where, for t ∈ (0, r], r < 1,
H(t) =
t
.
2 ln(1/t)
Since
H 0 (t) =
1 + ln(1/t)
2[ln(1/t)]2
and H 00 (t) =
ln(1/t) + 2
,
2t[ln(1/t)]3
the function H satisfies hypothesis (B1) on the interval (0, r] for any 0 < r < 1.
Also, by taking D(t) = tα , (3.6) is satisfied for any α > 1. Therefore, an explicit
rate of decay can be obtained by Theorem 3.1. The function H0 (t) = H(tα ) has
derivative
αtα−1 [1 + ln(1/tα )]
H00 (t) =
.
2[ln(1/tα )]2
Then, we do some direct calculations and use (3.5) to deduce that
Rt
R
k2 + k3
|u0 |2 dΓ1 t1 H0 (k(s))ds 1/(2α)
Γ1
E1 (t) ≤ k1
t
Z
Z ∞
η
−
|u0 |2 dΓ1
k 2 (s)ds, ∀t ≥ t1 ,
2 Γ1
t
for any α > 1, where H0 (k(s)) =
decays at the following rate
E1 (t) ≤ k1
k2 + k3 (
η
−
2
If
R∞
0
(k(s))α
2 ln(1/(k(s))α ) .
R
Γ1
|u0 |2 dΓ1 )
t
Z
2
|u0 | dΓ1
Γ1
Therefore, taking α → 1, the energy
Z
Rt
t1
H(k(s))ds 1/2
(3.7)
∞
2
k (s)ds,
∀t ≥ t1 .
t
H(k(s))ds < +∞, then equation (3.7) reduces to
Z
Z ∞
C
η
2
E1 (t) ≤ 1 −
|u0 | dΓ1
k 2 (s)ds,
2 Γ1
t2
t
∀t ≥ t1 .
5. The well-known Jensen’s inequality will be of essential use in establishing our
main result. If F is a convex function on [a, b], f : Ω → [a, b] and h are integrable
EJDE-2014/202
BOUNDARY STABILIZATION
7
R
functions on Ω, h(x) ≥ 0, and Ω h(x)dx = k > 0, then Jensen’s inequality states
that
h1 Z
i 1Z
F
f (x)h(x)dx ≤
F [f (x)]h(x)dx.
k Ω
k Ω
6. As in [17] and since limt→∞ k(t) = 0, limt→+∞ (−k 0 (t)) cannot be equal to a
positive number, and so it is natural to assume that limt→+∞ (−k 0 (t)) = 0, in the
same way, we deduce that limt→+∞ k 00 (t) = 0. Hence there is t1 > 0 large enough
such that k 0 (t1 ) < 0 and
max{k(t), −k 0 (t), k 00 (t)} < min{r, H(r), H0 (r)},
∀t ≥ t1 .
(3.8)
As k 0 is nondecreasing, k 0 (0) < 0 and k 0 (t1 ) < 0, then k 0 (t) < 0 for any t ∈ [0, t1 ]
and
0 < −k 0 (t1 ) ≤ −k 0 (t) ≤ −k 0 (0), ∀t ∈ [0, t1 ].
Therefore, since H is a positive continuous function, we have
a ≤ H(−k 0 (t)) ≤ b,
∀t ∈ [0, t1 ],
for some positive constants a and b. Consequently, for all t ∈ [0, t1 ],
a
a
k 00 (t) ≥ H(−k 0 (t)) ≥ a = 0 k 0 (0) ≥ 0 k 0 (t)
k (0)
k (0)
which gives
k 00 (t) ≥ d(−k 0 (t)),
∀t ∈ [0, t1 ],
(3.9)
for some positive constant d.
Lemma 3.2. Under the assumptions of Theorem 3.1, the energies of the solution
of (1.3) satisfy
Z
Z
Z
η
η
E10 (t) ≤ −
|q|2 dx −
|ut |2 dΓ1 + k 0 (t)
|u|2 dΓ1
2 Γ1
2
Ω
Γ1
Z
Z
(3.10)
η
η
−
(k 00 ◦ u)(t)dΓ1 + k 2 (t)
|u0 |2 dΓ1 .
2 Γ1
2
Γ1
Z
E20 (t) ≤ −
|qt |2 dx ≤ 0.
(3.11)
Ω
Proof. Multiplying (1.3)1 by ut , (1.3)2 by θ, and (1.3)3 by q and integrating over Ω,
using integration by parts, the boundary conditions (2.2) and (2.4), one can easily
find that
Z
Z
Z
η
E10 (t) = −
|q|2 dx − η
|ut |2 dΓ1 + k 0 (t)
|u|2 dΓ1
2
Ω
Γ1
Γ1
Z
Z
η
00
−
(k ◦ u)(t)dΓ1 + η
k(t)(u0 .ut )dΓ1
2 Γ1
Γ1
Young’s inequality then yields
Z
Z
Z
1
1
k(t)(u0 .ut )dΓ1 ≤
|ut |2 dΓ1 + k 2 (t)
|u0 |2 dΓ1 ,
2 Γ1
2
Γ1
Γ1
and consequently, we obtain (3.10) for strong solutions. Estimate (3.11) is established in a similar way using (3.2) and (3.3).
8
F. BOULANOUAR, S. DRABLA
EJDE-2014/202
Remark 3.3. If u0 = 0 on Γ1 , then E10 (t) ≤ 0, hence E1 (t) ≤ E1 (0). If u0 =
6 0 on
Γ1 , then
Z
Z t
η
2
E1 (t) ≤ E1 (0) +
|u0 | dΓ1
k 2 (s)ds ≤ A,
(3.12)
2 Γ1
0
for some A > 0.
Lemma 3.4. Under the assumptions (A1) and (B1), the solution of (1.3) satisfies:
for any > 0,
Z
d
ut .[M + (n − 1)u]dx
dt Ω
Z
Z
Z
Z
µ+λ
≤−
|ut |2 dx − µ
(div u)2 dx + C
|∇u|2 dx −
|∇θ|2 dx
2
Ω
Ω
Ω
Ω
Z
Z
Z
1
µδ
2
2
|∇u| dΓ1 − (µ + λ)δ
(div u) dΓ1 + C(1 + )
|ut |2 dΓ1 (3.13)
−
2 Γ1
Γ1
Γ1
Z
Z
C
2
2
|u| dΓ1 −
+ Ck (t)
(k 0 ◦ u)(t)dΓ1
Γ1
Γ1
Z
Z
1 2
2
+
|u| dΓ1 + C(1 + )k (t)
|u0 |2 dΓ1 ,
Γ1
Γ1
where
M = (M1 , M2 , ..., Mn )T ,
such that Mi = 2m.∇ui ,
and C is a “generic” positive constant independent of .
For a proof of the above lemma, see [13, 17].
4. Proof of main results
In this section we prove our main result.
Proof of Theorem 3.1. Taking E(t) = E1 (t) + E2 (t), we define
Z
L(t) = N E(t) +
ut .[M + (n − 1)u]dx.
(4.1)
Ω
From (3.10), (3.11), and (3.13), we obtain
Z
Z
Z
Z
Nη
N
L0 (t) ≤ −N
|q|2 dx − N
|qt |2 dx −
|ut |2 dΓ1 − η
(k 00 ◦ u)(t)dΓ1
2
2
Ω
Ω
Γ1
Γ1
Z
Z
Z
µ+λ
2
2
2
−
|ut | dx − µ
|∇u| dx −
(div u) dx
2
Ω
Ω
Ω
Z
Z
Z
µδ
−
|∇u|2 dΓ1 − (µ + λ)δ
(div u)2 dΓ1 + C
|ut |2 dΓ1
2 Γ1
Γ1
Γ1
Z
Z
Z
Z
C
C
+ k 2 (t)
|u|2 dΓ1 −
(k 0 ◦ u)(t)dΓ1 + |u|2 dΓ1 + C
|∇θ|2 dx
Γ1
Γ1
Γ1
Ω
Z
Z
1 2
η 2
2
+ C(1 + )k (t)
|u0 | dΓ1 + N k (t)
|u0 |2 dΓ1 .
2
Γ1
Γ1
By using (1.3)3 and
Z
Γ1
|u|2 dΓ1 ≤ c0
Z
Ω
|∇u|2 dx
(4.2)
EJDE-2014/202
and
BOUNDARY STABILIZATION
|∇θ|2 dx, for some positive constants c0 and cp , we arrive at
Z
Z
Z
L0 (t) ≤ −(N − C1 )
|q|2 dx − (N − C1 )
|qt |2 dx −
|ut |2 dx
Ω
Ω
Ω
Z
Z
C 2
2
|∇u|2 dx
−
k(t)|u| dΓ1 − µ − c0 − k (t) − c0 k(t)
Ω
Γ1
Z
Z
µ+λ
N
2
−
(div u) dx −
η−C
|ut |2 dΓ1
(4.3)
2
2
Ω
Γ1
Z
Z
−C
(k 0 ◦ u)(t)dΓ1 −
|θ|2 dx
Γ1
Ω
Z
N
1 η + C(1 + ) k 2 (t)
|u0 |2 dΓ1 .
+
2
Γ1
R
Ω
|θ|2 dx ≤ cp
9
R
Ω
At this point, we choose our constants carefully. We first, fix so small that
c0 = 12 µ and pick N large enough so that L ∼ E,
a1 =
N
η−C ≥0
2
and a2 = N − C1 > 0.
Thus, (4.3) simplifies to
Z
µ C
Z
0
2
2
|∇u|2 dx
L (t) ≤ −
|ut | dx −
− k (t) − c0 k(t)
2
Ω
Ω
Z
Z
Z
Z
µ+λ
−
(div u)2 dx −
k(t)|u|2 dΓ1 − a2
|q|2 dx −
|θ|2 dx
2
Ω
Γ1
Ω
Ω
Z
Z
0
2
2
−C
(k ◦ u)(t)dΓ1 + Ck (t)
|u0 | dΓ1 .
Γ1
Γ1
Using the fact that limt→+∞ k(t) = 0, we obtain
Z
Z
(k 0 ◦ u)(t)dΓ1 ,
|u0 |2 dΓ1 − c
L0 (t) ≤ −mE1 (t) + Ck 2 (t)
∀t ≥ t1 ,
(4.4)
Γ1
Γ1
for some t1 large enough and some positive constants m and c.
Now we use (3.9), and (3.10) to conclude that, for any t ≥ t1 ,
Z t1
Z
0
−
k (s)
|u(t) − u(t − s)|2 dΓ1 ds
0
1
≤
d
Γ1
Z
t1
k 00 (s)
Z
|u(t) − u(t − s)|2 dΓ1 ds
Z
η
≤ −c[E10 (t) − k 2 (t)
|u0 |2 dΓ1 ].
2
Γ1
0
(4.5)
Γ1
Next we take F (t) = L(t) + cE1 (t), which is clearly equivalent to E(t), and use
(4.4) and (4.5), to obtain: for all t ≥ t1 with some new positive constant C > 0,
Z
Z
Z t
F 0 (t) ≤ −mE1 (t)+Ck 2 (t)
|u0 |2 dΓ1 −c
k 0 (s)
|u(t)−u(t−s)|2 dΓ1 ds. (4.6)
Γ1
t1
Γ1
Similarly to [17] we consider two cases:
(I) H(t) = ctp and 1 ≤ p < 3/2: If 1 < p < 3/2, one can easily show that
R +∞
[−k 0 (s)]1−δ0 ds < +∞ for any δ0 < 2 − p. Using this fact, (3.10), (3.12) and
0
10
F. BOULANOUAR, S. DRABLA
EJDE-2014/202
(4.2) and choosing t1 even larger if needed, we deduce that, for all t ≥ t1
Z
Z t
η(t) =
[−k 0 (s)]1−δ0
|u(t) − u(t − s)|2 dΓ1 ds
t1
Γ1
Z
t
0
≤2
1−δ0
Z
[−k (s)]
t1
(|u(t)|2 + |u(t − s)|2 ) dΓ1 ds
(4.7)
Γ1
+∞
Z
[−k 0 (s)]1−δ0 ds < 1.
≤ cA
0
Then, Jensen’s inequality, (3.10), hypothesis (B1), and (4.7) lead to
Z t
Z
0
−
k (s)
|u(t) − u(t − s)|2 dΓ1 ds
t1
t
Z
Γ1
[−k 0 (s)]δ0 [−k 0 (s)]1−δ0
=
t1
Z t
=
Z
|u(t) − u(t − s)|2 dΓ1 ds
Γ1
δ0
[−k 0 (s)]p−1+δ0 ( p−1+δ0 ) [−k 0 (s)]1−δ0
Z
|u(t) − u(t − s)|2 dΓ1 ds
Γ1
t1
Z
δ0
h 1 Z t
i p−1+δ
0
≤ η(t)
[−k 0 (s)](p−1+δ0 ) [−k 0 (s)]1−δ0
|u(t) − u(t − s)|2 dΓ1 ds
η(t) t1
Γ1
Z
δ0
hZ t
i p−1+δ
0
0
p
2
≤
[−k (s)]
|u(t) − u(t − s)| dΓ1 ds
t1
Γ1
Z
hZ t
≤c
k 00 (s)
t1
δ0
i p−1+δ
0
|u(t) − u(t − s)|2 dΓ1 ds
Γ1
Z
δ0
h
i p−1+δ
η 2 0
0
≤ c − E1 (t) + k (t)
|u0 |2 dΓ1
.
2
Γ1
Then, particularly for δ0 = 1/2, we find that (4.6) becomes
Z
Z
1
i 2p−1
h
η
|u0 |2 dΓ1 )
+ Ck 2 (t)
|u0 |2 dΓ1 ,
F 0 (t) ≤ −mE1 (t) + c − E10 (t) + k 2 (t)(
2
Γ1
Γ1
for all t ≥ t1 . However,
Z
Z ∞
0
η
2
F (t) +
|u0 | dΓ1
k 2 (s)ds
2 Γ1
t
≤ F 0 (t)
Z
1
h
Z
i 2p−1
η
|u0 |2 dΓ1
+ Ck 2 (t)
|u0 |2 dΓ1 ,
≤ −mE1 (t) + c − E10 (t) + k 2 (t)
2
Γ1
Γ1
hence, for all t ≥ t1 , we have
Z
Z ∞
0
η
F (t) +
|u0 |2 dΓ1
k 2 (s)ds
2 Γ1
t
Z
h
Z ∞
i
η
≤ −m E1 (t) +
k 2 (s)ds
|u0 |2 dΓ1
2 Γ1
t
Z
Z
1
h
i 2p−1
η
+ c − E10 (t) + k 2 (t)
|u0 |2 dΓ1
+ Ck 2 (t)
|u0 |2 dΓ1
2
Γ1
Γ1
EJDE-2014/202
+m
BOUNDARY STABILIZATION
η
2
Z
|u0 |2 dΓ1
Γ1
Z
∞
k 2 (s)ds.
(4.8)
t
Using hypothesis (B1), for all t ≥ t1 , we have
Z ∞
Z ∞
Z ∞
k 2 (s)ds ≤ k(t)
k(s)ds ≤ k(t)
k(s)ds, and k 2 (t) ≤ c0 k(t).
t
11
t
(4.9)
0
Then (4.8) becomes, for some positive constant C
Z
Z ∞
0
η
|u0 |2 dΓ1
F (t) +
k 2 (s)ds
2 Γ1
t
Z
Z ∞
i
h
η
2
|u0 | dΓ1
k 2 (s)ds
≤ −m E1 (t) +
2 Γ1
t
Z
1
Z
i 2p−1
h
η
2
2
0
|u0 | dΓ1
+ Ck(t)
|u0 |2 dΓ1 .
+ c − E1 (t) + k (t)
2
Γ1
Γ1
2p−2
R
R∞
Now we multiply by E1 (t) + η2 ( Γ1 |u0 |2 dΓ1 ) t k 2 (s)ds
, using that E10 (t) ≤
R
η 2
2
2 k (t)( Γ1 |u0 | dΓ1 ) we obtain
Z
h
Z ∞
ih
η
2
F (t) +
|u0 | dΓ1
k 2 (s)ds E1 (t)
2 Γ1
t
Z
Z ∞
i2p−2 0
η
+
|u0 |2 dΓ1
k 2 (s)ds
2 Γ1
t
Z
Z ∞
0
η
2
|u0 | dΓ1 )
k 2 (s)ds
≤ F (t) + (
2 Γ1
t
Z
h
Z ∞
i2p−2
η
2
× E1 (t) +
|u0 | dΓ1
k 2 (s)ds
2 Γ1
t
Z
Z ∞
i2p−1
h
η
|u0 |2 dΓ1
k 2 (s)ds
≤ −m E1 (t) +
2 Γ1
t
Z
Z ∞
i2p−2
h
η
|u0 |2 dΓ1
k 2 (s)ds
+ c E1 (t) +
2 Γ1
t
Z
1
h
i 2p−1
η
× − E10 (t) + k 2 (t)(
|u0 |2 dΓ1 )
2
Γ1
Z
Z
Z ∞
i2p−2
h
ih
η
|u0 |2 dΓ1 )
k 2 (s)ds
.
+ C k(t)
|u0 |2 dΓ1 E1 (t) + (
2 Γ1
t
Γ1
2p−1
Then, applying Young’s inequality, with σ = 2p − 1, and σ 0 = 2p−2
, for
Z
Z
Z
1
∞
h
i 2p−1
h
i2p−2
η
η
− E10 (t) + k 2 (t)
|u0 |2 dΓ1
E1 (t) + (
|u0 |2 dΓ1 )
k 2 (s)ds
2
2 Γ1
Γ1
t
and
Z
ih
Z ∞
i2p−2
η
|u0 |2 dΓ1 E1 (t) +
|u0 |2 dΓ1
k 2 (s)ds
2 Γ1
Γ1
t
Z
h
k(t)
we obtain, for C > 0,
Z
h
Z ∞
i
η
F (t) +
|u0 |2 dΓ1
k 2 (s)ds
2 Γ1
t
12
F. BOULANOUAR, S. DRABLA
EJDE-2014/202
Z
Z ∞
h
i2p−2 0
η
× E1 (t) + (
k 2 (s)ds
|u0 |2 dΓ1 )
2 Γ1
t
Z
h
Z ∞
i2p−1
η
≤ −m E1 (t) +
|u0 |2 dΓ1
k 2 (s)ds
2 Γ1
t
Z ∞
i2p−1
h ηZ
|u0 |2 dΓ1
k 2 (s)ds
+ 2ε E1 t +
2 Γ1
t
Z
h
Z
i
η
+ Cε − E10 (t) + k 2 (t)
|u0 |2 dΓ1 + C[k(t)
|u0 |2 dΓ1 ]2p−1 .
2
Γ1
Γ1
Consequently, for 2ε < m, we obtain
Z
Z
h
Z ∞
i2p−1
h
i2p−1
η
0
0
2
2
F0 (t) ≤ −m E1 (t)+
|u0 | dΓ1
k (s)ds
+C k(t)
|u0 |2 dΓ1
,
2 Γ1
t
Γ1
where m0 is some positive constant and
Z
Z ∞
i
h
η
|u0 |2 dΓ1
k 2 (s)ds
F0 (t) = F (t) +
2 Γ1
t
Z ∞
Z
h
i2p−2
η
2
|u0 | dΓ1 )
× E1 (t) + (
k 2 (s)ds
2 Γ1
t
Z
Z ∞
i
h
η
2
|u0 | dΓ1 )
k 2 (s)ds .
+ Cε E1 (t) + (
2 Γ1
t
Also, it is easy to show that Rthis inequality is true for p = 1. Once again, we use
the fact that E10 (t) ≤ η2 k 2 (t)( Γ1 |u0 |2 dΓ1 ) to deduce that
Z
Z ∞
h
i2p−1 0
η
2
t E1 (t) + (
|u0 | dΓ1 )
k 2 (s)ds
2 Γ1
t
Z
h
Z ∞
i2p−1
η
≤ E1 (t) +
|u0 |2 dΓ1
k 2 (s)ds
2 Γ1
t
Z
i2p−1
Ch
1 0
|u0 |2 dΓ1
≤ − 0 F0 (t) + 0 k(t)
,
m
m
Γ1
for all t ≥ t1 . A simple integration over (t1 , t) yields
Z
h
Z ∞
i2p−1
η
t E1 (t) +
|u0 |2 dΓ1
k 2 (s)ds
2 Γ1
t
Z th
Z
i2p−1
1
C
≤ 0 F0 (t1 ) + 0
k(s)
|u0 |2 dΓ1
ds
m
m t1
Γ1
Z
Z
∞
i2p−1
h
η
+ t1 E1 (t1 ) +
|u0 |2 dΓ1
k 2 (s)ds
2 Γ1
t1
Z th
Z
i2p−1
C
≤ c1 + 0
k(s)
|u0 |2 dΓ1
,
m t1
Γ1
hence
h
E1 (t) +
η
2
Z
Γ1
|u0 |2 dΓ1
Z
∞
i2p−1
k 2 (s)ds
t
Z
Z th
i2p−1
c1
C
≤
k(s)
|u0 |2 dΓ1
ds.
+ 0
t
m t t1
Γ1
EJDE-2014/202
BOUNDARY STABILIZATION
13
Therefore,
E1 (t) ≤
c1 + c2
η
−
2
Z
Rt
t1
[k(s)
R
|u0 |2 dΓ1
Γ1
Γ1
1
|u0 |2 dΓ1 ]2p−1 ds 2p−1
t
Z
∞
k 2 (s)ds.
t
(II) The general case: As in [17], we define
Z t
Z
−k 0 (s)
I(t) =
|u(t) − u(t − s)|2 dΓ1 ds
−1 00
t1 H0 (k (s)) Γ1
where H0 is such that (3.6) is satisfied. As in (4.7), we find that for all t ≥ t1 , I(t)
satisfies
I(t) < 1.
(4.10)
We also assume, without loss of generality that I(t) ≥ β0 > 0, for all t ≥ t1 ;
otherwise (4.6) yields an explicit decay. In addition, we define λ(t) by
Z
Z t
−k 0 (s)
|u(t) − u(t − s)|2 dΓ1 ds,
λ(t) =
k 00 (s) −1 00
H
(k
(s))
Γ1
t1
0
and infer from (B1) and the properties of H0 and D that
−k 0 (s)
−k 0 (s)
−k 0 (s)
≤
=
≤ k0 ,
D−1 (−k 0 (s))
H0−1 (k 00 (s))
H0−1 (H(−k 0 (s)))
for some positive constantk0 . Then, using (3.8), (3.10), and (3.12), one can easily
see that for all t ≥ t1 , λ(t) satisfies
Z t
Z
|u(t) − u(t − s)|2 dΓ1 ds
λ(t) ≤ k0
k 00 (s)
Γ1
t1
Z
t
≤ cA
k 00 (s)ds ≤ cA(−k 0 (t1 ))
(4.11)
t1
< min{r, H(r), H0 (r)},
for t1 even larger (if needed). In addition, we can easily see that
Z
h
i
η 2
0
λ(t) ≤ −c E1 (t) − k (t)
|u0 |2 dΓ1 , ∀t ≥ t1 .
2
Γ1
(4.12)
Since H0 is strictly convex on (0, r], and H0 (0) = 0, we have
H0 (θx) ≤ θH0 (x),
provided 0 ≤ θ ≤ 1, and x ∈ (0, r]. Then using this fact, (4.10) , (4.11), and
Jensen’s inequality leads to
Z t
Z
1
−k 0 (s)
λ(t) =
I(t)H0 [H0−1 (k 00 (s))] −1 00
|u(t) − u(t − s)|2 dΓ1 ds
I(t) t1
H0 (k (s)) Γ1
Z t
Z
−k 0 (s)
1
−1 00
≥
H0 [I(t)H0 (k (s))] −1 00
|u(t) − u(t − s)|2 dΓ1 ds
I(t) t1
H0 (k (s)) Γ1
Z
1 Z t
−k 0 (s)
≥ H0
I(t)H0−1 (k 00 (s)) −1 00
|u(t) − u(t − s)|2 dΓ1 ds
I(t) t1
H0 (k (s)) Γ1
Z
Z t
= H0 −
k 0 (s)
|u(t) − u(t − s)|2 dΓ1 ds , ∀t ≥ t1 .
t1
Γ1
14
F. BOULANOUAR, S. DRABLA
EJDE-2014/202
This implies
Z
t
−
k 0 (s)
Z
t1
|u(t) − u(t − s)|2 dΓ1 ds ≤ H0−1 (λ(t)),
∀t ≥ t1 ,
Γ1
and (4.6) becomes
F 0 (t) ≤ −mE1 (t) + Ck 2 (t)
Z
|u0 |2 dΓ1 + cH0−1 (λ(t)),
∀t ≥ t1 .
(4.13)
Γ1
Now, using that H00 > 0, H000 > 0, ε0 < r, c0 > 0, and
Z
η
|u0 |2 dΓ1 ),
E10 (t) ≤ k 2 (t)(
2
Γ1
we find that the functional
R
R∞ 2
2
E1 (t) + η (
2 Γ1 |u0 | dΓ1 ) t k (s)ds
0
R
R
F (t)
F1 (t) = H0 ε0
∞
E1 (0) + η2 ( Γ1 |u0 |2 dΓ1 ) 0 k 2 (s)ds
Z
Z ∞
η
2
+ c0 E1 (t) + (
|u0 | dΓ1 )
k 2 (s)ds
2 Γ1
t
satisfies
F10 (t)
R
E10 (t) − η2 ( Γ1 |u0 |2 dΓ1 )k 2 (t)
R
R∞
= ε0
η
E1 (0) + 2 ( Γ1 |u0 |2 dΓ1 ) 0 k 2 (s)ds
R
R∞ 2
2
E1 (t) + η (
2 Γ1 |u0 | dΓ1 ) t k (s)ds
00
R
R
× H0 ε0
F (t)
∞
E1 (0) + η2 ( Γ1 |u0 |2 dΓ1 ) 0 k 2 (s)ds
R
R∞ 2
2
E1 (t) + η (
2 Γ1 |u0 | dΓ1 ) t k (s)ds
0
R
R
+ H0 ε0
F 0 (t)
∞
E1 (0) + η2 ( Γ1 |u0 |2 dΓ1 ) 0 k 2 (s)ds
Z
i
h
η 2
0
|u0 |2 dΓ1 )
+ c0 E1 (t) − k (t)(
2
Γ1
R∞ 2
R
2
E1 (t) + η (
|u
0 | dΓ1 ) t k (s)ds
2 Γ1
0
R
R∞
≤ H0 ε0
F 0 (t)
η
2
2
E1 (0) + 2 ( Γ1 |u0 | dΓ1 ) 0 k (s)ds
Z
i
h
η 2
0
|u0 |2 dΓ1 ) , ∀t ≥ t1 .
+ c0 E1 (t) − k (t)(
2
Γ1
Hence, using (4.13), for all t ≥ t1 , we obtain
F10 (t)
≤ −mE1 (t)H00 ε0
E1 (t) + η2 (
R
Γ1
η
2 ( Γ1
R
|u0 |2 dΓ1 )
R∞
Rt∞
k 2 (s)ds E1 (0) +
|u0 |2 dΓ1 ) 0 k 2 (s)ds
R
R∞ 2
2
E1 (t) + η (
2 Γ1 |u0 | dΓ1 ) t k (s)ds
2
2
0
R
R
+C
|u0 | dΓ1 k (t)H0 ε0
∞
E1 (0) + η2 ( Γ1 |u0 |2 dΓ1 ) 0 k 2 (s)ds
Γ1
R
R∞ 2
2
E1 (t) + η (
2 Γ1 |u0 | dΓ1 ) t k (s)ds
−1
0
R
R
+ cH0 (λ(t))H0 ε0
∞
E1 (0) + η2 ( Γ1 |u0 |2 dΓ1 ) 0 k 2 (s)ds
Z
h
i
η
+ c0 E10 (t) − k 2 (t)(
|u0 |2 dΓ1 ) .
2
Γ1
Z
(4.14)
EJDE-2014/202
BOUNDARY STABILIZATION
15
Let H0∗ be the convex conjugate of H0 in the sense of Young (see [2, p. 61-64]),
then
H0∗ (s) = s(H00 )−1 (s) − H0 [(H00 )−1 (s)], if s ∈ (0, H00 (r)],
(4.15)
and H0∗ satisfies the Young inequality
AB ≤ H0∗ (A) + H0 (B),
if A ∈ (0, H00 (r)], B ∈ (0, r],
(4.16)
with
A=
H00
ε0
E1 (t) + η2 (
R
E1 (0) + η2 (
Γ1
R
|u0 |2 dΓ1 )
|u0 |2 dΓ1 )
Γ1
R∞
Rt∞
0
k 2 (s)ds k 2 (s)ds
,
B = H0−1 (λ(t)) .
Using (4.11), (4.14), and (4.16), we arrive at
R
R∞ 2
2
E1 (t) + η (
|u
|
dΓ
)
k (s)ds 0
1
2
Γ
R 1
Rt∞
F10 (t) ≤ −mE1 (t)H00 ε0
η
E1 (0) + 2 ( Γ1 |u0 |2 dΓ1 ) 0 k 2 (s)ds
R
R∞ 2
Z
2
E1 (t) + η (
2 Γ1 |u0 | dΓ1 ) t k (s)ds
2
2
0
R
R
+ C(
|u0 | dΓ1 )k (t)H0 ε0
∞
Γ1
E1 (0) + η2 Γ1 |u0 |2 dΓ1 0 k 2 (s)ds
R
R∞ 2
2
E1 (t) + η (
2 Γ1 |u0 | dΓ1 ) t k (s)ds
∗
0
R
R
+ cλ(t) + cH0 H0 ε0
∞
E1 (0) + η2 ( Γ1 |u0 |2 dΓ1 ) 0 k 2 (s)ds
Z
i
h
η
|u0 |2 dΓ1 .
+ c0 E10 (t) − k 2 (t)
2
Γ1
Then using (4.12) and (4.15), we obtain that for all t ≥ t1 ,
R∞ 2
R
2
E1 (t) + η (
2 Γ1 |u0 | dΓ1 ) t k (s)ds
0
0
R∞
R
F1 (t) ≤ −mE1 (t)H0 ε0
η
E1 (0) + 2 ( Γ1 |u0 |2 dΓ1 ) 0 k 2 (s)ds
R
R
Z
E1 (t) + η2 Γ |u0 |2 dΓ1 t∞ k 2 (s)ds 1
R
R
|u0 |2 dΓ1 H00 ε0
+ Ck 2 (t)
∞
η
Γ1
E1 (0) + 2 Γ1 |u0 |2 dΓ1 0 k 2 (s)ds
Z
h
i
η
|u0 |2 dΓ1
− c E10 (t) − k 2 (t)
2
Γ1
R
R
∞
η
E1 (t) + 2 Γ1 |u0 |2 dΓ1 t k 2 (s)ds
R
R
+ c(ε0
)
∞
E1 (0) + η2 Γ1 |u0 |2 dΓ1 0 k 2 (s)ds
R
R∞ 2
2
E1 (t) + η (
2 Γ1 |u0 | dΓ1 ) t k (s)ds
0
R
R
× H0 ε 0
∞
E1 (0) + η2 ( Γ1 |u0 |2 dΓ1 ) 0 k 2 (s)ds
Z
h
i
η
+ c0 E10 (t) − k 2 (t)(
|u0 |2 dΓ1 ) .
2
Γ1
Consequently, with a suitable choice of c0 , we obtain, for all t ≥ t1 ,
R
R∞ 2
2
E1 (t) + η (
2 Γ1 |u0 | dΓ1 ) t k (s)ds
0
R
R∞
F1 (t) ≤ −m
η
E1 (0) + 2 ( Γ1 |u0 |2 dΓ1 ) 0 k 2 (s)ds
R
R∞ 2
2
E1 (t) + η (
2 Γ1 |u0 | dΓ1 ) t k (s)ds
0
R
R
× H0 ε 0
∞
E1 (0) + η2 ( Γ1 |u0 |2 dΓ1 ) 0 k 2 (s)ds
16
F. BOULANOUAR, S. DRABLA
EJDE-2014/202
Z ∞
|u0 |2 dΓ1
k 2 (s)ds
2 Γ1
t
R
R∞ 2
2
E1 (t) + η (
2 Γ1 |u0 | dΓ1 ) t k (s)ds
0
R
R
× H0 ε 0
∞
E1 (0) + η2 ( Γ1 |u0 |2 dΓ1 ) 0 k 2 (s)ds
R
R
∞
Z
E1 (t) + η2 Γ1 |u0 |2 dΓ1 t k 2 (s)ds
2
2
0
R
R∞
+ C(k (t)
|u0 | dΓ1 )H0 (ε0
)
E1 (0) + η2 ( Γ1 |u0 |2 dΓ1 ) 0 k 2 (s)ds
Γ1
R
R∞ 2
2
E1 (t) + η (
2 Γ1 |u0 | dΓ1 ) t k (s)ds
R
R∞
+ c ε0
η
E1 (0) + 2 ( Γ1 |u0 |2 dΓ1 ) 0 k 2 (s)ds
R
R∞ 2
2
E1 (t) + η (
2 Γ1 |u0 | dΓ1 ) t k (s)ds
0
R
R∞
× H0 ε 0
.
(4.17)
η
E1 (0) + 2 ( Γ1 |u0 |2 dΓ1 ) 0 k 2 (s)ds
+m
η Z
Using (4.9), for some positive constant C, (4.17) becomes
R
R∞ 2
2
E1 (t) + η (
2 Γ1 |u0 | dΓ1 ) t k (s)ds
0
R
R∞
F1 (t) ≤ −m
η
2
2
E1 (0) + 2 ( Γ1 |u0 | dΓ1 ) 0 k (s)ds
R
R∞ 2
2
E1 (t) + η (
|u
|
dΓ
)
k (s)ds 0
1
2
Γ
R 1
Rt∞
× H00 ε0
η
E1 (0) + 2 ( Γ1 |u0 |2 dΓ1 ) 0 k 2 (s)ds
R∞ 2
R
Z
2
E1 (t) + η (
2 Γ1 |u0 | dΓ1 ) t k (s)ds
2
0
R
R
|u0 | dΓ1 k(t)H0 ε0
+C
∞
E1 (0) + η2 ( Γ1 |u0 |2 dΓ1 ) 0 k 2 (s)ds
Γ1
R∞ 2
R
2
E1 (t) + η (
2 Γ1 |u0 | dΓ1 ) t k (s)ds
R
R
+ c ε0
∞
E1 (0) + η2 ( Γ1 |u0 |2 dΓ1 ) 0 k 2 (s)ds
R∞ 2
R
2
E1 (t) + η (
2 Γ1 |u0 | dΓ1 ) t k (s)ds
0
R
R∞
.
× H0 ε0
η
2
2
E1 (0) + 2 ( Γ1 |u0 | dΓ1 ) 0 k (s)ds
(4.18)
In a similar way, using (4.15) and (4.16), we find that, for t1 even larger (if needed),
R
R∞
h
E1 (t) + η2 ( Γ1 |u0 |2 dΓ1 ) t k 2 (s)ds i
0
R
R∞
k(t) H0 (ε0
)
E1 (0) + η2 ( Γ1 |u0 |2 dΓ1 ) 0 k 2 (s)ds
R
R∞ 2
2
E1 (t) + η (
2 Γ1 |u0 | dΓ1 ) t k (s)ds
R
R
≤ ε0
∞
E1 (0) + η2 ( Γ1 |u0 |2 dΓ1 ) 0 k 2 (s)ds
R
R∞ 2
2
E1 (t) + η (
2 Γ1 |u0 | dΓ1 ) t k (s)ds
0
R
R
× H0 ε 0
∞
E1 (0) + η2 ( Γ1 |u0 |2 dΓ1 ) 0 k 2 (s)ds
+ H0 (k(t)),
∀t ≥ t1 .
So, with a suitable choice of ε0 , (4.18) becomes
R
R∞ 2
2
E1 (t) + η (
2 Γ1 |u0 | dΓ1 ) t k (s)ds
0
R
R∞
F1 (t) ≤ −`
η
E1 (0) + 2 ( Γ1 |u0 |2 dΓ1 ) 0 k 2 (s)ds
R
R∞ 2
2
E1 (t) + η (
2 Γ1 |u0 | dΓ1 ) t k (s)ds
0
R
R
× H0 ε0
∞
E1 (0) + η2 ( Γ1 |u0 |2 dΓ1 ) 0 k 2 (s)ds
EJDE-2014/202
BOUNDARY STABILIZATION
Z
+ C(
17
|u0 |2 dΓ1 )H0 (k(t)),
Γ1
for all t ≥ t1 .
Hence,
F10 (t)
R
R∞ 2
2
E1 (t) + η (
2 Γ1 |u0 | dΓ1 ) t k (s)ds
R
R
≤ −`H1
∞
E1 (0) + η2 ( Γ1 |u0 |2 dΓ1 ) 0 k 2 (s)ds
Z
+C
|u0 |2 dΓ1 H0 (k(t)), ∀t ≥ t1 ,
(4.19)
Γ1
where H1 (t) = tH00 (ε0 t).
Since H10 (t) = H00 (ε0 t) + ε0 tH000 (ε0 t), then using the strict convexity of H0 on
0
(0, r], we find that
1 (t) > 0, on (0, 1]. Thus, taking in account that
R H12(t), H
η 2
0
E1 (t) ≤ 2 k (t) Γ1 |u0 | dΓ1 and (4.19), for all t ≥ t1 , we have
R
R∞ 2
2
i0
h
E1 (t) + η (
2 Γ1 |u0 | dΓ1 ) t k (s)ds
R
R∞
tH1
η
E1 (0) + 2 ( Γ1 |u0 |2 dΓ1 ) 0 k 2 (s)ds
R
R∞ 2
2
E1 (t) + η (
2 Γ1 |u0 | dΓ1 ) t k (s)ds
R
R
≤ H1
∞
E1 (0) + η2 ( Γ1 |u0 |2 dΓ1 ) 0 k 2 (s)ds
Z
C
1
|u0 |2 dΓ1 H0 (k(t)).
≤ − F10 (t) +
`
`
Γ1
A simple integration over (t1 , t) yields
R
R∞ 2
2
E1 (t) + η (
2 Γ1 |u0 | dΓ1 ) t k (s)ds
R∞
R
tH1
η
E1 (0) + 2 ( Γ1 |u0 |2 dΓ1 ) 0 k 2 (s)ds
R
R
∞
E1 (t1 ) + η2 Γ1 |u0 |2 dΓ1 t1 k 2 (s)ds
R∞
R
≤ t 1 H1 (
)
E1 (0) + η2 ( Γ1 |u0 |2 dΓ1 ) 0 k 2 (s)ds
Z
Z t
C
1
2
|u0 | dΓ1
H0 (k(s))ds.
+ F1 (t1 ) +
`
`
Γ1
t1
This gives, for some positive constant k2 and for all t ≥ t1 ,
R
R∞ 2
Z
2
E1 (t) + η (
Z t
k
C
2
2 Γ1 |u0 | dΓ1 ) t k (s)ds
2
R
R∞
H1
+
|u0 | dΓ1
≤
H0 (k(s))ds.
t `t Γ1
E1 (0) + η2 ( Γ1 |u0 |2 dΓ1 ) 0 k 2 (s)ds
t1
Therefore, for some positive constants k1 and k3 , we obtain
R
Rt
k2 + k3 (
|u0 |2 dΓ1 ) t1 H0 (k(s))ds Γ1
−1
E1 (t) ≤ k1 H1
t
Z
Z ∞
η
|u0 |2 dΓ1
k 2 (s)ds.
−
2 Γ1
t
This completes the proof.
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Fairouz Boulanouar
Department of Mathematics, Faculty of Sciences, University Farhat Abbas of Setif1,
Setif 19000, Algeria
E-mail address: boulanoir b@yahoo.com
Salah Drabla
Department of Mathematics, Faculty of Sciences, University Farhat Abbas of Setif1,
Setif 19000, Algeria
E-mail address: drabla s@univ-setif.dz