Hindawi Publishing Corporation
Abstract and Applied Analysis
Volume 2013, Article ID 682061, 8 pages
http://dx.doi.org/10.1155/2013/682061
Research Article
General Decay for the Degenerate Equation with
a Memory Condition at the Boundary
Su-Young Shin and Jum-Ran Kang
Department of Mathematics, Dong-A University, Saha-Ku, Busan 604-714, Republic of Korea
Correspondence should be addressed to Jum-Ran Kang; pointegg@hanmail.net
Received 21 December 2012; Accepted 5 March 2013
Academic Editor: Abdelaziz Rhandi
Copyright © 2013 S.-Y. Shin and J.-R. Kang. This is an open access article distributed under the Creative Commons Attribution
License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly
cited.
We consider a degenerate equation with a memory condition at the boundary. For a wider class of relaxation functions, we establish
a more general decay result, from which the usual exponential and polynomial decay rates are only special cases.
1. Introduction
The main purpose of this paper is to investigate the asymptotic behavior of the solutions of the degenerate equation with
a memory condition at the boundary
functions 𝑔𝑖 (𝑖 = 1, 2) are positive and nondecreasing, the
function 𝑓 ∈ 𝐶1 (R) and
B1 𝑢 = Δ𝑢 + (1 − 𝜇) 𝐵1 𝑢,
B2 𝑢 =
𝜕𝐵 𝑢
𝜕Δ𝑢
+ (1 − 𝜇) 2 ,
𝜕]
𝜕𝜏
𝐵1 𝑢 = 2]1 ]2 𝑢𝑥𝑦 − ]21 𝑢𝑦𝑦 − ]22 𝑢𝑥𝑥 ,
𝐾 (𝑥) 𝑢󸀠󸀠 + Δ2 𝑢 + 𝑓 (𝑢) = 0 in 𝑄 = Ω × (0, ∞) ,
𝜕𝑢
𝑢=
=0
𝜕]
(1)
(6)
on Γ0 × (0, ∞) ,
𝑡
(2)
−𝑢 + ∫ 𝑔1 (𝑡 − 𝑠) B2 𝑢 (𝑠) 𝑑𝑠 = 0
on Γ1 × (0, ∞) ,
(3)
𝑡
𝜕𝑢
+ ∫ 𝑔2 (𝑡 − 𝑠) B1 𝑢 (𝑠) 𝑑𝑠 = 0
𝜕]
0
on Γ1 × (0, ∞) ,
(4)
0
𝑢 (0, 𝑥) = 𝑢0 (𝑥) ,
𝐵2 𝑢 = (]21 − ]22 ) 𝑢𝑥𝑦 + ]1 ]2 (𝑢𝑦𝑦 − 𝑢𝑥𝑥 ) ,
𝑢󸀠 (0, 𝑥) = 𝑢1 (𝑥) in Ω,
(5)
where Ω is a bounded domain of R𝑛 with a smooth boundary
Γ and let us assume that Γ, can be divided into two nonnull
parts Γ = Γ0 ∪ Γ1 and Γ0 ∩ Γ1 = 0 and 𝐾 ∈ 𝐶1 (Ω) and
𝐾(𝑥) ≥ 0 for all 𝑥 ∈ Ω which satisfies some appropriate
conditions. ] is the unit outward normal to Γ and 𝜏 = (−]2 , ]1 )
the corresponding unit tangent vector. Here, the relaxation
and the constant 𝜇, 0 < 𝜇 < 1/2, represents Poisson’s ratio.
From the physical point of view, we know that the memory effect described in integral equations (3) and (4) can be
caused by the interaction with another viscoelastic element.
In fact, the boundary conditions (3) and (4) mean that Ω is
composed of a material which is clamped in a rigid body in
Γ0 and is clamped in a body with viscoelastic properties in
the complementary part of its boundary named Γ1 . Problems
related to (1)–(5) are interesting not only from the point of
view of PDE general theory, but also due to its applications in
mechanics.
The existence of global solutions and exponential decay to
the degenerate equation with 𝜕Ω = Γ0 has been investigated
by several authors. See Cavalcanti et al. [1] and Menezes
et al. [2]. For instance, when 𝐾(𝑥) is equal to 1, (1) describes
the transverse deflection 𝑢(𝑥, 𝑡) of beams. There exists a
large body of literature regarding viscoelastic problems with
the memory term acting in the domain or at the boundary
2
Abstract and Applied Analysis
(see [3–17]). Santos et al. [18] studied the asymptotic behavior
of the solutions of a nonlinear wave equation of Kirchhoff
type with boundary condition of memory type. Cavalcanti
et al. [19] proved the uniform decay rates of solutions to a
degenerate system with a memory condition at the boundary.
Santos and Junior [20] investigated the stability of solutions
for Kirchhoff plate equations with a boundary memory
condition. Rivera et al. [21] showed the asymptotic behavior
to a von Karman plate with boundary memory conditions.
Park and Kang [22] studied the exponential decay for the
Kirchhoff plate equations with nonlinear dissipation and
boundary memory condition. They proved that the energy
decays uniformly exponentially or algebraically with the same
rate of decay as the relaxation functions. In the present work,
we generalize the earlier decay results of the solution of
(1)–(5). More precisely, we show that the energy decays at
the rate similar to the relaxation functions, which are not
necessarily decaying like polynomial or exponential functions. In fact, our result allows a larger class of relaxation
functions. Recently, Messaoudi and Soufyane [23], Mustafa
and Messaoudi [24], and Santos and Soufyane [25] proved
the general decay for the wave equation, Timoshenko system,
and von Karman plate system with viscoelastic boundary
conditions, respectively.
The organization of this paper is as follows. In Section 2,
we present some notations and material needed for our work
and state the existence result to system (1)–(5). In Section 3,
we prove the general decay of the solutions to the degenerate
equation with a memory condition at the boundary.
We assume that there exists 𝑥0 ∈ R𝑛 such that
Γ0 = {𝑥 ∈ Γ : ] (𝑥) ⋅ (𝑥 − 𝑥0 ) ≤ 0} ,
(11)
Γ1 = {𝑥 ∈ Γ : ] (𝑥) ⋅ (𝑥 − 𝑥0 ) > 0} .
(12)
If we denote the compactness of Γ1 by 𝑚(𝑥) = 𝑥 − 𝑥0 , condition (12) implies that there exists a small positive constant 𝛿0
such that 0 < 𝛿0 ≤ 𝑚(𝑥) ⋅ ](𝑥), for all 𝑥 ∈ Γ1 .
Next, we will use (3) and (4) to estimate the values B1 and
B2 on Γ1 . Denoting by
𝑡
(𝑔 ∗ 𝜑) (𝑡) = ∫ 𝑔 (𝑡 − 𝑠) 𝜑 (𝑠) 𝑑𝑠
0
(13)
the convolution product operator and differentiating (3) and
(4), we arrive at the following Volterra equations:
B2 𝑢 +
1
1
𝑔1󸀠 ∗ B2 𝑢 =
𝑢󸀠 ,
𝑔1 (0)
𝑔1 (0)
1
1 𝜕𝑢󸀠
𝑔2󸀠 ∗ B1 𝑢 = −
B1 𝑢 +
.
𝑔2 (0)
𝑔2 (0) 𝜕]
(14)
Applying the Volterra inverse operator, we get
B2 𝑢 =
1
{𝑢󸀠 + 𝑘1 ∗ 𝑢󸀠 } ,
𝑔1 (0)
𝜕𝑢󸀠
1
𝜕𝑢󸀠
B1 𝑢 = −
{
+ 𝑘2 ∗
},
𝑔2 (0) 𝜕]
𝜕]
(15)
where the resolvent kernels satisfy
2. Preliminaries
In this section, we introduce some notations and establish the
existence of solutions of the problem (1)–(5).
Note that, because of condition (2), the solution of system
(1)–(5) must belong to the following space:
𝜕V
= 0 on Γ0 } .
𝑊 = {V ∈ 𝐻 (Ω) : V =
𝜕]
2
(7)
𝑘𝑖 +
1
1
𝑔󸀠 ∗ 𝑘𝑖 = −
𝑔󸀠 ,
𝑔𝑖 (0) 𝑖
𝑔𝑖 (0) 𝑖
𝑎 (𝑢, V) = ∫ {𝑢𝑥𝑥 V𝑥𝑥 + 𝑢𝑦𝑦 V𝑦𝑦 + 𝜇 (𝑢𝑥𝑥 V𝑦𝑦 + 𝑢𝑦𝑦 V𝑥𝑥 )
Ω
(8)
+ 2 (1 − 𝜇) 𝑢𝑥𝑦 V𝑥𝑦 } 𝑑𝑥 𝑑𝑦.
(16)
Denoting that 𝜏1 = 1/𝑔1 (0) and 𝜏2 = 1/𝑔2 (0), we have
B2 𝑢 = 𝜏1 {𝑢󸀠 + 𝑘1 (0) 𝑢 − 𝑘1 (𝑡) 𝑢0 + 𝑘1󸀠 ∗ 𝑢} ,
B1 𝑢 = −𝜏2 {
Let us define the bilinear form 𝑎(⋅, ⋅) as follows:
∀𝑖 = 1, 2.
𝜕𝑢
𝜕𝑢
𝜕𝑢󸀠
𝜕𝑢
+ 𝑘2 (0)
− 𝑘2 (𝑡) 0 + 𝑘2󸀠 ∗
}.
𝜕]
𝜕]
𝜕]
𝜕]
(17)
Therefore, we use (17) instead of the boundary conditions (3)
and (4).
Let us denote that
𝑡
2
Since Γ0 ≠ 0, we know that 𝑎(𝑢, 𝑢) is equivalent to the 𝐻 (Ω)
norm on 𝑊; that is,
𝑐0 ‖𝑢‖2𝐻2 (Ω) ≤ 𝑎 (𝑢, 𝑢) ≤ 𝐶0 ‖𝑢‖2𝐻2 (Ω) ,
(9)
and here and in the sequel, we denote by 𝑐0 and 𝐶0 generic
positive constants. Simple calculation, based on the integration by parts formula, yields
(Δ2 𝑢, V) = 𝑎 (𝑢, V) + (B2 𝑢, V)Γ − (B1 𝑢,
𝜕V
) .
𝜕] Γ
(𝑔 ⬦ V) (𝑡) := ∫ 𝑔 (𝑡 − 𝑠) |V(𝑡) − V (𝑠)|2 𝑑𝑠.
0
The following lemma states an important property of the
convolution operator.
Lemma 1. For 𝑔, V ∈ 𝐶1 ([0, ∞) : R), one has
1
1
1 𝑑
(𝑔 ∗ V) V󸀠 = − 𝑔 (𝑡) |V (𝑡)|2 + 𝑔󸀠 ⬦ V −
2
2
2 𝑑𝑡
𝑡
(10)
(18)
× [𝑔 ⬦ V − (∫ 𝑔 (𝑠) 𝑑𝑠) |V|2 ] .
0
(19)
Abstract and Applied Analysis
3
The proof of this lemma follows by differentiating the
term 𝑔 ⬦ V.
Lemma 2 (see [26]). Suppose that 𝑓 ∈ 𝐿2 (Ω), 𝑔 ∈ 𝐻1/2 (Γ1 )
and ℎ ∈ 𝐻3/2 (Γ1 ); then, any solution of
𝑎 (V, 𝑤) = ∫ 𝑓𝑤 𝑑𝑥 + ∫ 𝑔𝑤 𝑑Γ + ∫ ℎ
Ω
Γ1
Γ1
𝜕𝑤
𝑑Γ,
𝜕]
∀𝑤 ∈ 𝑊,
3. General Decay
In this section, we show that the solution of system (1)–
(5) may have a general decay not necessarily of exponential
or polynomial type. For this we consider that the resolvent
kernels satisfy the following hypothesis.
(H) 𝑘𝑖 : R+ → R+ is twice differentiable function such
that
(20)
𝑘𝑖 (0) > 0,
4
satisfies V ∈ 𝐻 (Ω) and also
Δ V = 𝑓,
B1 V = ℎ,
B2 V = 𝑔
(21)
∀𝑠 ∈ R.
(22)
𝑧
(30)
𝐹 (𝑧) = ∫ 𝑓 (𝑠) 𝑑𝑠,
0
∀𝑠 ∈ R,
(23)
󵄨
󵄨
󵄨󵄨
󵄨 󵄨𝜌−1 󵄨
𝜌−1
󵄨󵄨𝑓 (𝑥) − 𝑓 (𝑦)󵄨󵄨󵄨 ≤ 𝑐 (1 + |𝑥| + 󵄨󵄨󵄨𝑦󵄨󵄨󵄨 ) 󵄨󵄨󵄨𝑥 − 𝑦󵄨󵄨󵄨 ,
∀𝑥, 𝑦 ∈ R
(24)
∞
(A2) 𝐾 ∈ 𝐶 (Ω);
∩ 𝐿 (Ω) with 𝐾(𝑥) ≥ 0, for all
𝑥 ∈ Ω, and satisfy the following condition
∇𝐾 ⋅ 𝑚 ≥ 0 in Ω.
+
1
2
2
2
+ V𝑦𝑦
+ 2𝜇V𝑥𝑥 V𝑦𝑦 + 2 (1 − 𝜇) V𝑥𝑦
] 𝑑Γ
∫ 𝑚 ⋅ ] [V𝑥𝑥
2 Γ
+ ∫ [(B2 V) 𝑚 ⋅ ∇V − (B1 V)
for some 𝑐 > 0 and 𝜌 ≥ 1 such that (𝑛 − 2)𝜌 ≤ 𝑛.
𝐻02 (Ω)
∫ (𝑚 ⋅ ∇V) Δ2 V𝑑𝑥
= 𝑎 (V, V)
for some 𝜂 > 0 with the following growth condition:
Γ
𝐸 (𝑡) =
𝑘𝑖 , −𝑘𝑖󸀠 , 𝑘𝑖󸀠󸀠 ≥ 0
(𝑖 = 1, 2) .
(26)
If 𝑢0 ∈ 𝑊 ∩ 𝐻4 (Ω), 𝑢1 ∈ 𝑊, satisfying the compatibility
condition
B2 𝑢0 = 𝜏1 𝑢1
𝑜𝑛 Γ1 ,
(27)
then there is only one solution 𝑢 of the system (1)–(5) satisfying
𝑢 ∈ 𝐿∞ (0, 𝑇 : 𝑊 ∩ 𝐻4 (Ω)) ,
𝑢󸀠 ∈ 𝐿∞ (0, 𝑇 : 𝑊) ,
2
𝑢 ∈ 𝐿 (0, 𝑇 : 𝐿 (Ω)) .
(28)
(31)
1
1
󵄨 󵄨2
∫ 𝐾 (𝑥) 󵄨󵄨󵄨󵄨𝑢󸀠 󵄨󵄨󵄨󵄨 𝑑𝑥 + 𝑎 (𝑢, 𝑢)
2 Ω
2
+ ∫ 𝐹 (𝑢) 𝑑𝑥 +
Ω
(25)
Theorem 3 (see [27]). Consider assumptions (A1)-(A2) and let
𝑘𝑖 ∈ 𝐶2 (R+ ) be such that
𝜕
(𝑚 ⋅ ∇V)] 𝑑Γ.
𝜕]
Let us introduce the energy function
+
The well-posedness of system (1)–(5) is given by the
following theorem.
∞
𝑖 = 1, 2, ∀𝑡 ≥ 0.
Ω
Additionally, we suppose that 𝑓 is superlinear; that is,
󸀠󸀠
(29)
Lemma 4 (see [26]). For every V ∈ 𝐻4 (Ω) and for every 𝜇 ∈
R, one has
𝑓 (𝑠) 𝑠 ≥ 0,
𝜕𝑢
B1 𝑢0 = −𝜏2 1 ,
𝜕]
𝑘𝑖󸀠 (𝑡) ≤ 0,
The following identity will be used later.
(A1) Let 𝑓 ∈ 𝐶1 (R) satisfy
1
𝑘𝑖󸀠󸀠 (𝑡) ≥ −𝜉𝑖 (𝑡) 𝑘𝑖󸀠 (𝑡) ,
𝑜𝑛 Γ1 .
We formulate the following assumptions.
𝑓 (𝑠) 𝑠 ≥ (2 + 𝜂) 𝐹 (𝑠) ,
(𝑡) = 0,
and there exists a nonincreasing continuous function 𝜉𝑖 :
R+ → R+ satisfying
𝜕V
V=
= 0 𝑜𝑛 Γ0 ,
𝜕]
2
lim 𝑘
𝑡→∞ 𝑖
𝜏1
∫ (𝑘 (𝑡) |𝑢|2 − 𝑘1󸀠 ⬦ 𝑢) 𝑑Γ
2 Γ1 1
󵄨󵄨 𝜕𝑢 󵄨󵄨2
𝜏2
𝜕𝑢
󵄨 󵄨
) 𝑑Γ.
∫ (𝑘2 (𝑡) 󵄨󵄨󵄨 󵄨󵄨󵄨 − 𝑘2󸀠 ⬦
󵄨󵄨 𝜕] 󵄨󵄨
2 Γ1
𝜕]
(32)
Now, we establish some inequalities for the strong solution of
system (1)–(5).
Lemma 5. The energy functional 𝐸 satisfies, along the solution
of (1)–(5), the estimate
𝐸󸀠 (𝑡) ≤ −
−
𝜏1
󵄨 󵄨2
󵄨 󵄨2
∫ (󵄨󵄨󵄨𝑢󸀠 󵄨󵄨󵄨 −𝑘12 (𝑡) 󵄨󵄨󵄨𝑢0 󵄨󵄨󵄨 −𝑘1󸀠 (𝑡) |𝑢|2 +𝑘1󸀠󸀠 ⬦ 𝑢) 𝑑Γ
2 Γ1 󵄨 󵄨
󵄨󵄨 󸀠 󵄨󵄨2
󵄨󵄨 𝜕𝑢 󵄨󵄨2
𝜏2
󵄨 𝜕𝑢 󵄨󵄨
󵄨󵄨 − 𝑘22 (𝑡) 󵄨󵄨󵄨󵄨 0 󵄨󵄨󵄨󵄨
∫ (󵄨󵄨󵄨󵄨
󵄨󵄨 𝜕] 󵄨󵄨
2 Γ1 󵄨󵄨 𝜕] 󵄨󵄨󵄨
󵄨󵄨 𝜕𝑢 󵄨󵄨2
𝜕𝑢
󵄨 󵄨
− 𝑘2󸀠 (𝑡) 󵄨󵄨󵄨 󵄨󵄨󵄨 + 𝑘2󸀠󸀠 ⬦
) 𝑑Γ.
󵄨󵄨 𝜕] 󵄨󵄨
𝜕]
(33)
4
Abstract and Applied Analysis
Proof. Multiplying (1) by 𝑢󸀠 , integrating over Ω, and using
(10), we get
1 𝑑
󵄨 󵄨2
{∫ 𝐾󵄨󵄨󵄨𝑢󸀠 󵄨󵄨󵄨 𝑑𝑥 + 𝑎 (𝑢, 𝑢) + 2 ∫ 𝐹 (𝑢) 𝑑𝑥}
2 𝑑𝑡 Ω 󵄨 󵄨
Ω
= − ∫ (B2 𝑢) 𝑢󸀠 𝑑Γ + ∫ (B1 𝑢)
Γ1
Γ1
𝜕𝑢󸀠
𝑑Γ.
𝜕]
Proof. Differentiating 𝜓 using (1) and Lemma 4, we get
𝑛
𝜓󸀠 (𝑡) = ∫ [𝑚 ⋅ ∇𝑢󸀠 + ( − 𝜃) 𝑢󸀠 ] 𝐾𝑢󸀠 𝑑𝑥
2
Ω
𝑛
+ ∫ [𝑚 ⋅ ∇𝑢 + ( − 𝜃) 𝑢] 𝐾𝑢󸀠󸀠 𝑑𝑥
2
Ω
(34)
=
Substituting the boundary terms by (17) and using Lemma 1
and the Young inequality, our conclusion follows.
−
𝑡
0
−
(35)
𝑡
0
𝑛
𝜓 (𝑡) = ∫ [𝑚 ⋅ ∇𝑢 + ( − 𝜃) 𝑢] 𝐾𝑢󸀠 𝑑𝑥.
2
Ω
(37)
The following lemma plays an important role in the construction of the desired functional.
Lemma 6. Suppose that the initial data (𝑢0 , 𝑢1 ) ∈ (𝐻4 (Ω) ∩
𝑊) × 𝑊, satisfying the compatibility condition (27). Then, the
solution of system (1)–(5) satisfies
(39)
Let us next examine the integrals over Γ0 in (39). Since 𝑢 =
𝜕𝑢/𝜕] = 0 on Γ0 , we have 𝐵1 𝑢 = 𝐵2 𝑢 = 0 on Γ0 and
𝜕
(𝑚 ⋅ ∇𝑢) = (𝑚 ⋅ ]) Δ𝑢,
𝜕]
2
2
2
+ 𝑢𝑦𝑦
+ 2𝜇𝑢𝑥𝑥 𝑢𝑦𝑦 + 2 (1 − 𝜇) 𝑢𝑥𝑦
= (Δ𝑢)2
𝑢𝑥𝑥
on Γ0 ,
(40)
since
on Γ0 .
(41)
Therefore, from (39) and (40), we have
𝑛
󵄨 󵄨2
− 𝜃 ∫ 𝐾󵄨󵄨󵄨󵄨𝑢󸀠 󵄨󵄨󵄨󵄨 𝑑𝑥 − (1 + − 𝜃 − 𝜖𝜆 0 ) 𝑎 (𝑢, 𝑢)
2
Ω
𝜓󸀠 (𝑡) ≤
𝑛𝜂
2𝜏 2
− 2𝜃 − 𝜂𝜃) ∫ 𝐹 (𝑢) 𝑑𝑥 + 1
2
𝜖
Ω
1
󵄨󵄨 󸀠 󵄨󵄨2
󵄨󵄨 𝜕𝑢 󵄨󵄨2
󵄨󵄨 𝜕𝑢 󵄨󵄨2
2𝜏2 2
󵄨 𝜕𝑢 󵄨󵄨
󵄨󵄨 + 𝑘22 (𝑡) 󵄨󵄨󵄨󵄨 󵄨󵄨󵄨󵄨 + 𝑘22 (𝑡) 󵄨󵄨󵄨󵄨 0 󵄨󵄨󵄨󵄨
∫ {󵄨󵄨󵄨󵄨
󵄨󵄨 𝜕] 󵄨󵄨
󵄨󵄨 𝜕] 󵄨󵄨
𝜖 Γ1 󵄨󵄨 𝜕] 󵄨󵄨󵄨
󵄨󵄨
𝜕𝑢 󵄨󵄨󵄨󵄨2
󵄨
+ 󵄨󵄨󵄨𝑘2󸀠 ∘
󵄨 } 𝑑Γ
󵄨󵄨
𝜕] 󵄨󵄨󵄨
1
𝑛
󵄨 󵄨2
∫ ∇𝐾 ⋅ 𝑚󵄨󵄨󵄨󵄨𝑢󸀠 󵄨󵄨󵄨󵄨 𝑑𝑥 − (1 + − 𝜃) 𝑎 (𝑢, 𝑢)
2 Ω
2
𝑛
+ 𝑛 ∫ 𝐹 (𝑢) 𝑑𝑥 − ( − 𝜃) ∫ 𝑓 (𝑢) 𝑢𝑑𝑥
2
Ω
Ω
+
1
∫ 𝑚 ⋅ ](Δ𝑢)2 𝑑Γ
2 Γ0
−
1
2
2
2
+𝑢𝑦𝑦
+2𝜇𝑢𝑥𝑥 𝑢𝑦𝑦 +2 (1−𝜇) 𝑢𝑥𝑦
] 𝑑Γ
∫ 𝑚 ⋅ ] [𝑢𝑥𝑥
2 Γ1
𝑛
− ∫ (B2 𝑢) [(𝑚 ⋅ ∇𝑢) + ( − 𝜃) 𝑢] 𝑑Γ
2
Γ1
1 𝜖𝜆
2
2
− ( − 0 ) ∫ 𝑚 ⋅ ] [𝑢𝑥𝑥
+ 𝑢𝑦𝑦
+ 2𝜇𝑢𝑥𝑥 𝑢𝑦𝑦
2
𝛿0
Γ1
+ 2 (1 −
1
󵄨 󵄨2
󵄨 󵄨2
∫ 𝑚 ⋅ ]𝐾󵄨󵄨󵄨󵄨𝑢󸀠 󵄨󵄨󵄨󵄨 𝑑Γ − 𝜃 ∫ 𝐾󵄨󵄨󵄨󵄨𝑢󸀠 󵄨󵄨󵄨󵄨 𝑑𝑥
2 Γ1
Ω
−
󵄨 󵄨2
󵄨2
󵄨 󵄨2 󵄨
× ∫ {󵄨󵄨󵄨󵄨𝑢󸀠 󵄨󵄨󵄨󵄨 + 𝑘12 (𝑡) |𝑢|2 + 𝑘12 (𝑡) 󵄨󵄨󵄨𝑢0 󵄨󵄨󵄨 + 󵄨󵄨󵄨󵄨𝑘1󸀠 ∘ 𝑢󵄨󵄨󵄨󵄨 } 𝑑Γ
Γ
+
Γ
𝜕
𝑛
𝜕𝑢
(𝑚 ⋅ ∇𝑢) + ( − 𝜃) ] 𝑑Γ.
𝜕]
2
𝜕]
2
𝑢𝑥𝑥 𝑢𝑦𝑦 − 𝑢𝑥𝑦
=0
1
󵄨 󵄨2
𝜓󸀠 (𝑡) ≤ ∫ 𝑚 ⋅ ]𝐾󵄨󵄨󵄨󵄨𝑢󸀠 󵄨󵄨󵄨󵄨 𝑑Γ
2 Γ1
−(
+ ∫ (B1 𝑢) [
(36)
Let us define the functional
1
2
2
2
+𝑢𝑦𝑦
+2𝜇𝑢𝑥𝑥 𝑢𝑦𝑦 +2 (1−𝜇) 𝑢𝑥𝑦
] 𝑑Γ
∫ 𝑚 ⋅ ] [𝑢𝑥𝑥
2 Γ
𝑛
− ∫ (B2 𝑢) [(𝑚 ⋅ ∇𝑢) + ( − 𝜃) 𝑢] 𝑑Γ
2
Γ
Then applying the Holder inequality for 0 ≤ 𝛼 ≤ 1 we have
|(𝑘 ∘ 𝑢) (𝑡)|2 ≤ [∫ |𝑘 (𝑠)|2(1−𝛼) 𝑑𝑠] (|𝑘|2𝛼 ⬦ 𝑢) (𝑡) .
1
𝑛
󵄨 󵄨2
∫ ∇𝐾 ⋅ 𝑚󵄨󵄨󵄨󵄨𝑢󸀠 󵄨󵄨󵄨󵄨 𝑑𝑥 − (1 + − 𝜃) 𝑎 (𝑢, 𝑢)
2 Ω
2
𝑛
+ 𝑛 ∫ 𝐹 (𝑢) 𝑑𝑥 − ( − 𝜃) ∫ 𝑓 (𝑢) 𝑢𝑑𝑥
2
Ω
Ω
Let us consider the following binary operator:
(𝑘 ∘ 𝑢) (𝑡) := ∫ 𝑘 (𝑡 − 𝑠) (𝑢 (𝑡) − 𝑢 (𝑠)) 𝑑𝑠.
1
󵄨 󵄨2
󵄨 󵄨2
∫ 𝑚 ⋅ ]𝐾󵄨󵄨󵄨󵄨𝑢󸀠 󵄨󵄨󵄨󵄨 𝑑Γ − 𝜃 ∫ 𝐾󵄨󵄨󵄨󵄨𝑢󸀠 󵄨󵄨󵄨󵄨 𝑑𝑥
2 Γ1
Ω
+ ∫ (B1 𝑢) [
2
𝜇) 𝑢𝑥𝑦
] 𝑑Γ.
Γ1
(38)
𝑛
𝜕
𝜕𝑢
(𝑚 ⋅ ∇𝑢) + ( − 𝜃) ] 𝑑Γ.
𝜕]
2
𝜕]
(42)
Abstract and Applied Analysis
5
Using the Young inequality, we get
B1 𝑢 = −𝜏2 {
󵄨󵄨
󵄨󵄨
𝑛
󵄨󵄨
󵄨
󵄨󵄨∫ (B2 𝑢) [(𝑚 ⋅ ∇𝑢) + ( − 𝜃) 𝑢] 𝑑Γ󵄨󵄨󵄨
󵄨󵄨 Γ
󵄨󵄨
2
󵄨 1
󵄨
≤
𝜕𝑢
𝜕𝑢
𝜕𝑢󸀠
𝜕𝑢
+ 𝑘2 (𝑡)
− 𝑘2 (𝑡) 0 − 𝑘2󸀠 ∘
},
𝜕]
𝜕]
𝜕]
𝜕]
(47)
2
1
𝑛
󵄨
󵄨2
∫ 󵄨󵄨󵄨B2 𝑢󵄨󵄨󵄨 𝑑Γ + 𝜖 ∫ (|𝑚 ⋅ ∇𝑢|2 + ( − 𝜃) |𝑢|2 ) 𝑑Γ,
2𝜖 Γ1
2
Γ1
our conclusion follows.
Let us introduce the Lyapunov functional
(43)
󵄨󵄨
󵄨󵄨
𝜕
𝑛
𝜕𝑢
󵄨󵄨
󵄨
󵄨󵄨∫ (B1 𝑢) [ (𝑚 ⋅ ∇𝑢) + ( − 𝜃) ] 𝑑Γ󵄨󵄨󵄨
󵄨󵄨 Γ
󵄨󵄨
𝜕]
2
𝜕]
󵄨 1
󵄨
≤
1
󵄨
󵄨2
∫ 󵄨󵄨󵄨B1 𝑢󵄨󵄨󵄨 𝑑Γ
2𝜖 Γ1
L (𝑡) = 𝑁𝐸 (𝑡) + 𝜓 (𝑡) ,
with 𝑁 > 0. Now, we are in a position to show the main result
of this paper.
(44)
󵄨󵄨 𝜕
󵄨󵄨2
2 󵄨󵄨 𝜕𝑢 󵄨󵄨2
𝑛
󵄨
󵄨 󵄨
󵄨
+ 𝜖 ∫ (󵄨󵄨󵄨 (𝑚 ⋅ ∇𝑢)󵄨󵄨󵄨 + ( − 𝜃) 󵄨󵄨󵄨 󵄨󵄨󵄨 ) 𝑑Γ,
󵄨󵄨 𝜕] 󵄨󵄨
󵄨󵄨
2
Γ1 󵄨󵄨 𝜕]
where 𝜖 is a positive constant. Since the bilinear form 𝑎(𝑢, 𝑢)
is strictly coercive on 𝑊, using the trace theory, we obtain
2
𝑛
∫ (|𝑚 ⋅ ∇𝑢|2 + ( − 𝜃) |𝑢|2 ) 𝑑Γ
2
Γ1
Theorem 7. Let (𝑢0 , 𝑢1 ) ∈ 𝑊 × 𝐿2 (Ω). Suppose that the
resolvent kernels 𝑘1 , 𝑘2 satisfy the condition (H). Then, there
exist constants 𝜔, 𝐶 > 0 such that, for some 𝑡0 large enough,
the solution of (1)–(5) satisfies
𝑡
𝐸 (𝑡) ≤ 𝐶𝐸 (0) 𝑒−𝜔 ∫0 𝜉(𝑠)𝑑𝑠 ,
󵄨󵄨 𝜕
2 󵄨󵄨 𝜕𝑢
𝑛
󵄨
󵄨 󵄨
󵄨
+ ∫ (󵄨󵄨󵄨 (𝑚 ⋅ ∇𝑢)󵄨󵄨󵄨 + ( − 𝜃) 󵄨󵄨󵄨 󵄨󵄨󵄨 ) 𝑑Γ
󵄨󵄨 𝜕] 󵄨󵄨
󵄨󵄨
2
Γ1 󵄨󵄨 𝜕]
󵄨󵄨2
𝑡
𝜕𝑢0
= 0 𝑜𝑛 Γ1 .
𝜕]
(49)
𝑡
𝑑𝑠) 𝑒−𝜔 ∫0 𝜉(𝑠)𝑑𝑠 , (50)
for all 𝑡 ≥ 𝑡0 , where
Γ1
(45)
where 𝜆 0 is a constant depending on Ω, 𝜇, 𝜃 and 𝑛. Substituting inequalities (43)–(45) into (42) and taking into account
that 𝑚 ⋅ ] ≤ 0 on Γ0 , as well as (23) and (25), we have
1
󵄨 󵄨2
󵄨 󵄨2
∫ 𝑚 ⋅ ]𝐾󵄨󵄨󵄨󵄨𝑢󸀠 󵄨󵄨󵄨󵄨 𝑑Γ − 𝜃 ∫ 𝐾󵄨󵄨󵄨󵄨𝑢󸀠 󵄨󵄨󵄨󵄨 𝑑𝑥
2 Γ1
Ω
+
0
0
2
2
2
× ∫ 𝑚 ⋅ ] [𝑢𝑥𝑥
+ 𝑢𝑦𝑦
+ 2𝜇𝑢𝑥𝑥 𝑢𝑦𝑦 + 2 (1 − 𝜇) 𝑢𝑥𝑦
] 𝑑Γ,
−(
𝑠
𝜔 ∫𝑡 𝜉(𝜏)𝑑𝜏
𝐸 (𝑡) ≤ 𝐶 (𝐸 (0) + ∫ 𝑘0 (𝑠) 𝑒
𝜆
≤ 𝜆 0 𝑎 (𝑢, 𝑢) + 0
𝛿0
− (1 +
∀𝑡 ≥ 𝑡0 , if 𝑢0 =
Otherwise,
󵄨󵄨2
𝜓󸀠 (𝑡) ≤
(48)
𝜉 (𝑡) = min {𝜉1 (𝑡) , 𝜉2 (𝑡)} ,
󵄨󵄨 𝜕𝑢 󵄨󵄨2
󵄨 󵄨2
󵄨
󵄨
𝑘0 (𝑡) = ∫ 𝑘12 (𝑡) 󵄨󵄨󵄨𝑢0 󵄨󵄨󵄨 𝑑Γ + ∫ 𝑘22 (𝑡) 󵄨󵄨󵄨 0 󵄨󵄨󵄨 𝑑Γ.
󵄨󵄨 𝜕] 󵄨󵄨
Γ1
Γ1
Proof. Applying inequality (36) with 𝛼 = 1/2 in Lemma 6 and
from Lemma 5, we obtain
󵄨 󵄨2
L󸀠 (𝑡) ≤ − ∫ 𝜃𝐾󵄨󵄨󵄨󵄨𝑢󸀠 󵄨󵄨󵄨󵄨 𝑑𝑥
Ω
𝑛
− 𝜃 − 𝜖𝜆 0 ) 𝑎 (𝑢, 𝑢)
2
𝑛𝜂
− 2𝜃 − 𝜂𝜃) ∫ 𝐹 (𝑢) 𝑑𝑥
2
Ω
1
󵄨
󵄨2 󵄨
󵄨2
∫ (󵄨󵄨B 𝑢󵄨󵄨 + 󵄨󵄨B 𝑢󵄨󵄨 ) 𝑑Γ
2𝜖 Γ1 󵄨 1 󵄨 󵄨 2 󵄨
−
𝜏1 𝑁
󵄨 󵄨2
󵄨 󵄨2
∫ {󵄨󵄨󵄨󵄨𝑢󸀠 󵄨󵄨󵄨󵄨 −𝑘12 (𝑡) 󵄨󵄨󵄨𝑢0 󵄨󵄨󵄨 −𝑘1󸀠 (𝑡) |𝑢|2 +𝑘1󸀠󸀠 ⬦𝑢} 𝑑Γ
2 Γ1
−
󵄨󵄨 󸀠 󵄨󵄨2
󵄨󵄨 𝜕𝑢 󵄨󵄨2
󵄨󵄨 𝜕𝑢 󵄨󵄨2
𝜏2 𝑁
󵄨 𝜕𝑢 󵄨󵄨
󵄨󵄨 − 𝑘22 (𝑡) 󵄨󵄨󵄨󵄨 0 󵄨󵄨󵄨󵄨 − 𝑘2󸀠 (𝑡) 󵄨󵄨󵄨󵄨 󵄨󵄨󵄨󵄨
∫ {󵄨󵄨󵄨󵄨
󵄨󵄨 𝜕] 󵄨󵄨
󵄨󵄨 𝜕] 󵄨󵄨
2 Γ1 󵄨󵄨 𝜕] 󵄨󵄨󵄨
+ 𝑘2󸀠󸀠 ⬦
1 𝜖𝜆
2
2
+ 𝑢𝑦𝑦
+ 2𝜇𝑢𝑥𝑥 𝑢𝑦𝑦
− ( − 0 ) ∫ 𝑚 ⋅ ] [𝑢𝑥𝑥
2
𝛿0
Γ1
− (1 +
2
+ 2 (1 − 𝜇) 𝑢𝑥𝑦
] 𝑑Γ.
(46)
Since the boundary conditions (17) can be written as
B2 𝑢 = 𝜏1 {𝑢󸀠 + 𝑘1 (𝑡) 𝑢 − 𝑘1 (𝑡) 𝑢0 − 𝑘1󸀠 ∘ 𝑢} ,
(51)
𝑛𝜂
𝑛
− 𝜃 − 𝜖𝜆 0 ) 𝑎 (𝑢, 𝑢) − ( − 2𝜃 − 𝜂𝜃)
2
2
× ∫ 𝐹 (𝑢) 𝑑𝑥+
Ω
𝜕𝑢
} 𝑑Γ
𝜕]
2𝜏1 2
󵄨 󵄨2
󵄨 󵄨2
∫ {󵄨󵄨󵄨𝑢󸀠 󵄨󵄨󵄨 +𝑘2 (𝑡) |𝑢|2+𝑘12 (𝑡) 󵄨󵄨󵄨𝑢0 󵄨󵄨󵄨
𝜖 Γ1 󵄨 󵄨 1
− 𝑘1 (0) 𝑘1󸀠 ⬦ 𝑢} 𝑑Γ
6
Abstract and Applied Analysis
󵄨󵄨 󸀠 󵄨󵄨2
󵄨󵄨 𝜕𝑢 󵄨󵄨2
󵄨󵄨 𝜕𝑢 󵄨󵄨2
2𝜏2 2
󵄨 𝜕𝑢 󵄨󵄨
󵄨󵄨 + 𝑘22 (𝑡) 󵄨󵄨󵄨󵄨 󵄨󵄨󵄨󵄨 + 𝑘22 (𝑡) 󵄨󵄨󵄨󵄨 0 󵄨󵄨󵄨󵄨
+
∫ {󵄨󵄨󵄨󵄨
󵄨󵄨 𝜕] 󵄨󵄨
󵄨󵄨 𝜕] 󵄨󵄨
𝜖 Γ1 󵄨󵄨 𝜕] 󵄨󵄨󵄨
󵄨 󵄨2
≤ −𝑐0 𝜉 (𝑡) 𝐸 (𝑡) + 𝑐𝜉 (𝑡) ∫ 𝑘12 (𝑡) 󵄨󵄨󵄨𝑢0 󵄨󵄨󵄨 𝑑Γ
Γ1
󵄨󵄨 𝜕𝑢 󵄨󵄨2
󵄨
󵄨
+ 𝑐𝜉 (𝑡) ∫ 𝑘22 (𝑡) 󵄨󵄨󵄨 0 󵄨󵄨󵄨 𝑑Γ − 𝑐𝐸󸀠 (𝑡) ,
󵄨󵄨 𝜕] 󵄨󵄨
Γ1
𝜕𝑢
− 𝑘2 (0) 𝑘2󸀠 ⬦
} 𝑑Γ
𝜕]
+
(56)
1 𝜖𝜆
1
󵄨 󵄨2
∫ 𝑚 ⋅ ]𝐾󵄨󵄨󵄨󵄨𝑢󸀠 󵄨󵄨󵄨󵄨 𝑑Γ − ( − 0 )
2 Γ1
2
𝛿0
which gives
2
2
2
× ∫ 𝑚 ⋅ ] [𝑢𝑥𝑥
+𝑢𝑦𝑦
+2𝜇𝑢𝑥𝑥 𝑢𝑦𝑦 +2 (1 − 𝜇) 𝑢𝑥𝑦
] 𝑑Γ.
Γ1
󵄨 󵄨2
𝜉 (𝑡) L󸀠 (𝑡) + 𝑐𝐸󸀠 (𝑡) ≤ −𝑐0 𝜉 (𝑡) 𝐸 (𝑡) + 𝑐𝜉 (𝑡) ∫ 𝑘12 (𝑡) 󵄨󵄨󵄨𝑢0 󵄨󵄨󵄨 𝑑Γ
Γ1
󵄨󵄨 𝜕𝑢 󵄨󵄨2
󵄨
󵄨
+ 𝑐𝜉 (𝑡) ∫ 𝑘22 (𝑡) 󵄨󵄨󵄨 0 󵄨󵄨󵄨 𝑑Γ,
󵄨󵄨 𝜕] 󵄨󵄨
Γ1
(52)
We take 𝜃 and 𝜖 so small such that
𝑛𝜂
− 2𝜃 − 𝜂𝜃 > 0,
2
1+
𝑛
− 𝜃 − 𝜖𝜆 0 > 0,
2
1 𝜖𝜆 0
> 0.
−
2
𝛿0
(53)
L (𝑡) ≤ −𝑐0 𝐸 (𝑡) + 𝑐 ∫
Γ1
𝑘12
Γ1
Γ1
(57)
󸀠
(𝜉L + 𝑐𝐸) (𝑡) ≤ 𝜉 (𝑡) L󸀠 (𝑡) + 𝑐𝐸󸀠 (𝑡)
󵄨 󵄨2
≤ −𝑐0 𝜉 (𝑡) 𝐸 (𝑡) + 𝑐𝜉 (𝑡) ∫ 𝑘12 (𝑡) 󵄨󵄨󵄨𝑢0 󵄨󵄨󵄨 𝑑Γ
Γ1
󵄨󵄨 𝜕𝑢 󵄨󵄨2
󵄨
󵄨
+ 𝑐𝜉 (𝑡) ∫ 𝑘22 (𝑡) 󵄨󵄨󵄨 0 󵄨󵄨󵄨 𝑑Γ,
󵄨󵄨
󵄨
𝜕]
Γ1
󵄨
󵄨󵄨 𝜕𝑢 󵄨󵄨2
󵄨
󵄨 󵄨2
󵄨
(𝑡) 󵄨󵄨󵄨𝑢0 󵄨󵄨󵄨 𝑑Γ + 𝑐 ∫ 𝑘22 (𝑡) 󵄨󵄨󵄨 0 󵄨󵄨󵄨 𝑑Γ
Γ1
󵄨󵄨 𝜕] 󵄨󵄨
− 𝑐 ∫ 𝑘1󸀠 ⬦ 𝑢𝑑Γ − 𝑐 ∫ 𝑘2󸀠 ⬦
∀𝑡 ≥ 𝑡0 .
Using the fact that 𝜉 is a nonincreasing continuous function
as 𝜉1 and 𝜉2 are nonincreasing, and so 𝜉 is differentiable, with
𝜉󸀠 (𝑡) ≤ 0, for a.e. 𝑡, then we infer that
Since 𝐾 ∈ 𝐿∞ (Ω) and then choosing 𝑁 large enough, we
obtain
󸀠
∀𝑡 ≥ 𝑡0 ,
𝜕𝑢
𝑑Γ,
𝜕]
∀𝑡 ≥ 𝑡0 .
(58)
Since using (55),
∀𝑡 ≥ 𝑡0 .
(54)
𝐹 = 𝜉L + 𝑐𝐸 ∼ 𝐸,
(59)
we obtain, for some positive constant 𝜔,
On the other hand, we can choose 𝑁 even larger so that
L (𝑡) ∼ 𝐸 (𝑡) .
󵄨 󵄨2
𝐹󸀠 (𝑡) ≤ −𝜔𝜉 (𝑡) 𝐹 (𝑡) + 𝑐 ∫ 𝑘12 (𝑡) 󵄨󵄨󵄨𝑢0 󵄨󵄨󵄨 𝑑Γ
Γ1
(55)
+ 𝑐∫
If 𝜉(𝑡) = min{𝜉1 (𝑡), 𝜉2 (𝑡)}, 𝑡 ≥ 𝑡0 , then, using (30) and (33),
we have
Γ1
𝑘22
󵄨󵄨 𝜕𝑢 󵄨󵄨2
󵄨
󵄨
(𝑡) 󵄨󵄨󵄨 0 󵄨󵄨󵄨 𝑑Γ,
󵄨󵄨 𝜕] 󵄨󵄨
(60)
∀𝑡 ≥ 𝑡0 .
Case 1. If 𝑢0 = 𝜕𝑢0 /𝜕] = 0 on Γ1 , then (60) reduces to
󵄨 󵄨2
𝜉 (𝑡) L󸀠 (𝑡) ≤ −𝑐0 𝜉 (𝑡) 𝐸 (𝑡) + 𝑐𝜉 (𝑡) ∫ 𝑘12 (𝑡) 󵄨󵄨󵄨𝑢0 󵄨󵄨󵄨 𝑑Γ
Γ1
− 𝑐𝜉1 (𝑡) ∫
Γ1
Γ1
𝑘2󸀠
𝜕𝑢
⬦
𝑑Γ
𝜕]
󵄨 󵄨2
≤ −𝑐0 𝜉 (𝑡) 𝐸 (𝑡) + 𝑐𝜉 (𝑡) ∫ 𝑘12 (𝑡) 󵄨󵄨󵄨𝑢0 󵄨󵄨󵄨 𝑑Γ
Γ1
󵄨󵄨 𝜕𝑢 󵄨󵄨2
󵄨
󵄨
+ 𝑐𝜉 (𝑡) ∫ 𝑘22 (𝑡) 󵄨󵄨󵄨 0 󵄨󵄨󵄨 𝑑Γ
󵄨󵄨 𝜕] 󵄨󵄨
Γ1
Γ1
Γ1
𝑡
−𝜔 ∫𝑡 𝜉(𝑠)𝑑𝑠
⬦ 𝑢𝑑Γ − 𝑐𝜉2 (𝑡) ∫
+ 𝑐 ∫ 𝑘1󸀠󸀠 ⬦ 𝑢𝑑Γ + 𝑐 ∫ 𝑘2󸀠󸀠 ⬦
∀𝑡 ≥ 𝑡0 .
(61)
A simple integration over (𝑡0 , 𝑡) yields
󵄨󵄨 𝜕𝑢 󵄨󵄨2
󵄨
󵄨
+ 𝑐𝜉 (𝑡) ∫ 𝑘22 (𝑡) 󵄨󵄨󵄨 0 󵄨󵄨󵄨 𝑑Γ
󵄨󵄨
󵄨
𝜕]
Γ1
󵄨
𝑘1󸀠
𝐹󸀠 (𝑡) ≤ −𝜔𝜉 (𝑡) 𝐹 (𝑡) ,
𝐹 (𝑡) ≤ 𝐹 (𝑡0 ) 𝑒
0
,
∀𝑡 ≥ 𝑡0 .
(62)
By using (33) and (59), we then obtain for some positive
constant 𝐶
𝑡
𝐸 (𝑡) ≤ 𝐶𝐸 (0) 𝑒−𝜔 ∫0 𝜉(𝑠)𝑑𝑠 ,
∀𝑡 ≥ 𝑡0 .
(63)
Thus, estimate (49) is proved.
𝜕𝑢
𝑑Γ
𝜕]
Case 2. If (𝑢0 , (𝜕𝑢0 /𝜕])) ≠ (0, 0) on Γ1 , then (60) gives
𝐹󸀠 (𝑡) ≤ −𝜔𝜉 (𝑡) 𝐹 (𝑡) + 𝑐𝑘0 (𝑡) ,
∀𝑡 ≥ 𝑡0 ,
(64)
Abstract and Applied Analysis
7
where
󵄨󵄨 𝜕𝑢 󵄨󵄨2
󵄨
󵄨 󵄨2
󵄨
𝑘0 (𝑡) = ∫ 𝑘12 (𝑡) 󵄨󵄨󵄨𝑢0 󵄨󵄨󵄨 𝑑Γ + ∫ 𝑘22 (𝑡) 󵄨󵄨󵄨 0 󵄨󵄨󵄨 𝑑Γ.
󵄨󵄨 𝜕] 󵄨󵄨
Γ1
Γ1
(65)
(3) For any nonincreasing functions 𝑘𝑖 (𝑡), 𝑖 = 1, 2,
which satisfy (H), 𝜉𝑖 = −𝑘󸀠 /𝑘 are also nonincreasing
differentiable functions, and 𝑐𝜉1 (𝑡) ≤ 𝜉2 (𝑡), for some
0 < 𝑐 ≤ 1, and (49) gives
In this case, we introduce
𝜔
𝑡
𝑡
−𝜔 ∫𝑡 𝜉(𝑠)𝑑𝑠
𝐺 (𝑡) := 𝐹 (𝑡) − 𝑐𝑒
𝑠
𝜔 ∫𝑡 𝜉(𝜏)𝑑𝜏
∫ 𝑘0 (𝑠) 𝑒
0
𝐸 (𝑡) ≤ 𝑐[𝑘1 (𝑡)] .
0
𝑡0
𝑑𝑠.
(66)
A simple differentiation of 𝐺, using (64), leads to
𝑡
0
𝑠
𝜔 ∫𝑡 𝜉(𝜏)𝑑𝜏
× ∫ 𝑘0 (𝑠) 𝑒
𝑑𝑠 − 𝑐𝑘0 (𝑡)
0
𝑡0
(67)
References
∀𝑡 ≥ 𝑡0 .
≤ −𝜔𝜉 (𝑡) 𝐺 (𝑡) ,
Again, a simple integration over (𝑡0 , 𝑡) yields
𝑡
−𝜔 ∫𝑡 𝜉(𝑠)𝑑𝑠
𝐺 (𝑡) ≤ 𝐺 (𝑡0 ) 𝑒
0
,
∀𝑡 ≥ 𝑡0 ,
(68)
which implies, for all 𝑡 ≥ 𝑡0 ,
𝑡
𝑠
𝜔 ∫𝑡 𝜉(𝜏)𝑑𝜏
𝐹 (𝑡) ≤ (𝐹 (𝑡0 ) + 𝑐 ∫ 𝑘0 (𝑠) 𝑒
0
𝑡0
𝑡
−𝜔 ∫𝑡 𝜉(𝑠)𝑑𝑠
𝑑𝑠) 𝑒
0
. (69)
By using (59), we deduce that
𝑡
𝑠
𝜔 ∫𝑡 𝜉(𝜏)𝑑𝜏
𝐸 (𝑡) ≤ 𝐶 (𝐸 (0) + ∫ 𝑘0 (𝑠) 𝑒
0
0
𝑡
−𝜔 ∫𝑡 𝜉(𝑠)𝑑𝑠
𝑑𝑠) 𝑒
0
,
(70)
∀𝑡 ≥ 𝑡0 .
Consequently, by the boundedness of 𝜉, (50) is established.
Remark 8. Note that the exponential and polynomial decay
estimates are only particular cases of (49) and (50). More
precisely, we have exponential decay for 𝜉1 (𝑡) ≡ 𝑐1 and 𝜉2 (𝑡) ≡
𝑐2 and polynomial decay for 𝜉1 (𝑡) = 𝑐1 (1 + 𝑡)−1 and 𝜉2 (𝑡) ≡ 𝑐2 ,
where 𝑐1 and 𝑐2 are positive constants.
Example 9. As in [24], we give some examples to illustrate the
energy decay rates given by (49).
𝑝
(1) If 𝑘1 (𝑡) = 𝑘2 (𝑡) = 𝑎𝑒−𝑏(1+𝑡) , 0 < 𝑝 ≤ 1, then, for 𝑖 =
1, 2, 𝑘𝑖󸀠󸀠 (𝑡) ≥ −𝜉(𝑡)𝑘𝑖󸀠 (𝑡), where 𝜉(𝑡) = 𝑏𝑝(1 + 𝑡)𝑝−1 . For
suitably chosen positive constants 𝑎 and 𝑏, 𝑘𝑖 satisfies
(H) and (49) gives
𝑝
𝐸 (𝑡) ≤ 𝑐𝑒−𝜔𝑏(1+𝑡) .
(71)
(2) If 𝑘1 (𝑡) = 𝑎1 /(1 + 𝑡)𝑞 , 𝑞 > 0, and 𝑘2 (𝑡) =
𝑝
𝑎2 𝑒−𝑏(1+𝑡) , 0 < 𝑝 ≤ 1, then, for 𝑖 = 1, 2, 𝑘𝑖󸀠󸀠 (𝑡) ≥
−𝜉(𝑡)𝑘𝑖󸀠 (𝑡), where 𝜉(𝑡) = 𝑞(1 + 𝑡)−1 . Then
𝐸 (𝑡) ≤
𝑐
.
(1 + 𝑡)𝜔𝑞
Acknowledgment
This research was supported by Basic Science Research Program through the National Research Foundation of Korea
(NRF) funded by the Ministry of Education, Science and
Technology (Grant no. 2012R1A1A3011630).
𝑡
−𝜔 ∫𝑡 𝜉(𝑠)𝑑𝑠
𝐺󸀠 (𝑡) = 𝐹󸀠 (𝑡) + 𝜔𝜉 (𝑡) 𝑐𝑒
(73)
(72)
The aforementioned two examples are included in the
following more general one.
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