Open Access

General decay for Kirchhoff plates with a boundary condition of memory type

Boundary Value Problems20122012:129

DOI: 10.1186/1687-2770-2012-129

Received: 30 July 2012

Accepted: 25 October 2012

Published: 7 November 2012

Abstract

In this paper we consider Kirchhoff plates 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.

MSC:35B40, 74K20, 35L70.

Keywords

Kirchhoff plates general decay rate memory term relaxation function

1 Introduction

We consider the following Kirchhoff plates with a memory condition at the boundary:
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-129/MediaObjects/13661_2012_Article_220_Equ1_HTML.gif
(1.1)
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-129/MediaObjects/13661_2012_Article_220_Equ2_HTML.gif
(1.2)
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-129/MediaObjects/13661_2012_Article_220_Equ3_HTML.gif
(1.3)
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-129/MediaObjects/13661_2012_Article_220_Equ4_HTML.gif
(1.4)
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-129/MediaObjects/13661_2012_Article_220_Equ5_HTML.gif
(1.5)
where a C 1 ( Ω ¯ ) and Ω is an open bounded set of R 2 with a regular boundary Γ. We divide the boundary into two parts:
Γ = Γ 0 Γ 1 with  Γ ¯ 0 Γ ¯ 1 = ;  and  Γ 0 .
Let us denote by ν = ( ν 1 , ν 2 ) the external unit normal to Γ, and let us denote by η = ( ν 2 , ν 1 ) the unit tangent positively oriented on Γ. We are denoting by B 1 , B 2 the following differential operators:
B 1 u = Δ u + ( 1 μ ) B 1 u , B 2 u = Δ u ν + ( 1 μ ) B 2 u ,
where B 1 and B 2 are given by
B 1 u = 2 ν 1 ν 2 u x y ν 1 2 u y y ν 2 2 u x x , B 2 u = η [ ( ν 1 2 ν 2 2 ) u x y + ν 1 ν 2 ( u y y u x x ) ] ,

and the constant μ, 0 < μ < 1 2 , represents Poisson’s ratio.

In (1.1), u denotes the position of the plate. The integral equations (1.3) and (1.4) describe the memory effects which can be caused, for example, by the interaction with another viscoelastic element. The relaxation functions g 1 , g 2 C 1 ( 0 , ) are positive and nondecreasing. This system models the small transversal vibrations of a thin plate whose Poisson coefficient is equal to μ. We assume that there exists x 0 R 2 such that
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-129/MediaObjects/13661_2012_Article_220_Equ6_HTML.gif
(1.6)
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-129/MediaObjects/13661_2012_Article_220_Equ7_HTML.gif
(1.7)

If we denote the compactness of Γ 1 by m ( x ) = x x 0 , the condition (1.7) implies that there exists a small positive constant δ 0 such that 0 < δ 0 m ( x ) ν ( x ) , x Γ 1 .

The uniform stabilization of Kirchhoff plates with linear or nonlinear boundary feedback was investigated by several authors; see, for example, [13] among others. The uniform decay for plates with memory was studied in [46] and the references therein. There exists a large body of literature regarding viscoelastic problems with the memory term acting in the domain or at the boundary (see [712]). Rivera and Racke [13] investigated the decay results for magneto-thermo-elastic system. Santos et al. [14] studied the asymptotic behavior of the solutions of a nonlinear wave equation of Kirchhoff type with a boundary condition of memory type. Cavalcanti and Guesmia [15] proved the general decay rates of solutions to a nonlinear wave equation with a boundary condition of memory type. Park and Kang [16] studied the exponential decay for the multi-valued hyperbolic differential inclusion with a boundary condition of memory type. Kafini [17] showed the decay results for viscoelastic diffusion equations in the absence of instantaneous elasticity. They proved that the energy decays uniformly exponentially or algebraically at the same rate as the relaxation functions. In the present work, we generalize the earlier decay results of the solution of (1.1)-(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 [18], Santos and Soufyane [19], and Mustafa and Messaoudi [20] proved the general decay for the wave equation, von Karman plate system, and Timoshenko system with viscoelastic boundary conditions, respectively.

The paper is organized as follows. In Section 2 we present some notations and material needed for our work. In Section 3 we prove the general decay of the solutions to the Kirchhoff plates with a memory condition at the boundary.

2 Preliminaries

In this section, we present some material needed in the proof of our main result. We use the standard Lebesgue and Sobolev spaces with their usual scalar products and norms. Define the following space:
W = { w H 2 ( Ω ) ; w = w ν = 0  on  Γ 0 } .
First, we shall use Eqs. (1.3) and (1.4) to estimate the values B 1 and B 2 on Γ 1 . Denoting by
( g v ) ( t ) = 0 t g ( t s ) v ( s ) d s ,
the convolution product operator and differentiating Eqs. (1.3) and (1.4), we arrive at the following Volterra equations:
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-129/MediaObjects/13661_2012_Article_220_Equf_HTML.gif
Applying the Volterra inverse operator, we get
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-129/MediaObjects/13661_2012_Article_220_Equg_HTML.gif
where the resolvent kernels of g i g i ( 0 ) satisfy
k i + 1 g i ( 0 ) g i k i = 1 g i ( 0 ) g i , i = 1 , 2 .
Denoting by τ i = 1 g i ( 0 ) , for i = 1 , 2 , we obtain
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-129/MediaObjects/13661_2012_Article_220_Equ8_HTML.gif
(2.1)
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-129/MediaObjects/13661_2012_Article_220_Equ9_HTML.gif
(2.2)

Therefore, we use (2.1) and (2.2) instead of the boundary conditions (1.3) and (1.4).

Let us define the bilinear form a ( , ) as follows:
a ( u , v ) = Ω { u x x v x x + u y y v y y + μ ( u x x v y y + u y y v x x ) + 2 ( 1 μ ) u x y v x y } d x d y .
(2.3)

We state the following lemma which will be useful in what follows.

Lemma 2.1 ([2])

Let u and v be functions in H 4 ( Ω ) W . Then we have
Ω ( Δ 2 u ) v d x = a ( u , v ) + Γ 1 { ( B 2 u ) v ( B 1 u ) v ν } d Γ .
(2.4)
Let us denote
( g v ) ( t ) : = 0 t g ( t s ) | v ( t ) v ( s ) | 2 d s .

The following lemma states an important property of the convolution operator.

Lemma 2.2 For g , v C 1 ( [ 0 , ) : R ) , we have
( g v ) v t = 1 2 g ( t ) | v ( t ) | 2 + 1 2 g v 1 2 d d t [ g v ( 0 t g ( s ) d s ) | v | 2 ] .

The proof of this lemma follows by differentiating the term g v .

We formulate the following assumption:

(A1) Let a C 1 ( Ω ¯ ) satisfy a ( x ) a 0 > 0 in Ω for some a 0 .

Let us introduce the energy function
E ( t ) = 1 2 [ Ω | u t | 2 d x + a ( u , u ) + τ 1 Γ 1 ( k 1 ( t ) | u | 2 k 1 u ) d Γ + τ 2 Γ 1 ( k 2 ( t ) | u ν | 2 k 2 u ν ) d Γ ] .

In these conditions, we are able to prove the existence of a strong solution.

Theorem 2.1 Let k i C 2 ( R + ) be such that
k i , k i , k i 0 ( i = 1 , 2 ) .
If ( u 0 , u 1 ) ( H 4 ( Ω ) W ) × W satisfy the compatibility condition
B 2 u 0 τ 1 u 1 = 0 , B 1 u 0 + τ 2 u 1 ν on Γ 1 ,
(2.5)
then there is only one solution u of the system (1.1)-(1.5) satisfying
u L ( 0 , T : H 4 ( Ω ) W ) , u t L ( 0 , T : W ) , u t t L ( 0 , T : L 2 ( Ω ) ) .

Proof See Park and Kang [16]. □

3 General decay

In this section, we show that the solution of the system (1.1)-(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:
  1. (H)
    k i : R + R + is a twice differentiable function such that
    k i ( 0 ) > 0 , lim t k i ( t ) = 0 , k i ( t ) 0 ,
     
and there exists a nonincreasing continuous function ξ i : R + R + satisfying
k i ( t ) ξ i ( t ) k i ( t ) , i = 1 , 2 , t 0 .
(3.1)

The following identity will be used later.

Lemma 3.1 ([2])

For every v H 4 ( Ω ) and for every μ R , we have
Ω ( m v ) Δ 2 v d x = a ( v , v ) + 1 2 Γ m ν [ v x x 2 + v y y 2 + 2 μ v x x v y y + 2 ( 1 μ ) v x y 2 ] d Γ + Γ [ ( B 2 v ) m v ( B 1 v ) ν ( m v ) ] d Γ .

Our point of departure will be to establish some inequalities for the strong solution of the system (1.1)-(1.5).

Lemma 3.2 The energy functional E satisfies, along the solution of (1.1)-(1.5), the estimate
E ( t ) Ω a ( x ) | u t | 2 d x τ 1 2 Γ 1 ( | u t | 2 k 1 2 ( t ) | u 0 | 2 k 1 ( t ) | u | 2 + k 1 u ) d Γ τ 2 2 Γ 1 ( | u t ν | 2 k 2 2 ( t ) | u 0 ν | 2 k 2 ( t ) | u ν | 2 + k 2 u ν ) d Γ .
(3.2)
Proof Multiplying Eq. (1.1) by u t , integrating over Ω, and using Lemma 2.1, we get
1 2 d d t { Ω | u t | 2 d x + a ( u , u ) } = Γ 1 ( B 2 u ) u t d Γ + Γ 1 ( B 1 u ) u t ν d Γ Ω a ( x ) | u t | 2 d x .

Substituting the boundary terms by (2.1) and (2.2) and using Lemma 2.2 and the Young inequality, our conclusion follows. □

Let us consider the following binary operator:
( k φ ) ( t ) : = 0 t k ( t s ) ( φ ( t ) φ ( s ) ) d s .
Then applying the Holder inequality for 0 α 1 , we have
| ( k φ ) ( t ) | 2 [ 0 t | k ( s ) | 2 ( 1 α ) d s ] ( | k | 2 α φ ) ( t ) .
(3.3)
Let us define the functional
ψ ( t ) = Ω ( m u ) u t d x .

The following lemma plays an important role in the construction of the desired functional.

Lemma 3.3 Suppose that the initial data ( u 0 , u 1 ) ( H 4 ( Ω ) W ) × W and satisfies the compatibility condition (2.5). Then the solution of the system (1.1)-(1.5) satisfies
d d t ψ ( t ) 1 2 Γ 1 m ν | u t | 2 d Γ 1 2 Ω | u t | 2 d x ( 1 C 0 2 ϵ λ 0 2 ) a ( u , u ) + 1 2 Ω a ( x ) | u t | 2 d x + 2 τ 1 2 ϵ Γ 1 { | u t | 2 + k 1 2 ( t ) | u | 2 + k 1 2 ( t ) | u 0 | 2 + | k 1 u | 2 } d Γ + 2 τ 2 2 ϵ Γ 1 { | u t ν | 2 + k 2 2 ( t ) | u ν | 2 + k 2 2 ( t ) | u 0 ν | 2 + | k 2 u ν | 2 } d Γ ( 1 2 ϵ λ 0 2 δ 0 ) Γ 1 m ν [ u x x 2 + u y y 2 + 2 μ u x x u y y + 2 ( 1 μ ) u x y 2 ] d Γ .
Proof Differentiating ψ, using Eq. (1.1), and taking v = u in Lemma 3.1, we get
d d t ψ ( t ) = Ω ( m u t ) u t d x + Ω ( m u ) u t t d x = 1 2 Γ 1 m ν | u t | 2 d Γ Ω | u t | 2 d x a ( u , u ) Γ [ ( B 2 u ) ( m u ) ( B 1 u ) ν ( m u ) ] d Γ 1 2 Γ m ν [ u x x 2 + u y y 2 + 2 μ u x x u y y + 2 ( 1 μ ) u x y 2 ] d Γ Ω a ( x ) u t ( m u ) d x .
(3.4)
Let us next examine the integrals over Γ 0 in (3.4). Since u = u ν = 0 on Γ 0 , we have B 1 u = B 2 u = 0 on Γ 0 and
ν ( m u ) = ( m ν ) Δ u , u x x 2 + u y y 2 + 2 μ u x x u y y + 2 ( 1 μ ) u x y 2 = ( Δ u ) 2 on  Γ 0
(3.5)
since
u x x u y y ( u x y ) 2 = 0 on  Γ 0 .
Therefore, from (3.4) and (3.5), we have
d d t ψ ( t ) 1 2 Γ 1 m ν | u t | 2 d Γ Ω | u t | 2 d x a ( u , u ) + 1 2 Γ 0 m ν ( Δ u ) 2 d Γ 1 2 Γ 1 m ν [ u x x 2 + u y y 2 + 2 μ u x x u y y + 2 ( 1 μ ) u x y 2 ] d Γ Γ 1 ( B 2 u ) ( m u ) d Γ + Γ 1 ( B 1 u ) ν ( m u ) d Γ Ω a ( x ) u t ( m u ) d x .
(3.6)
Using the Young inequality, we have
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-129/MediaObjects/13661_2012_Article_220_Equ19_HTML.gif
(3.7)
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-129/MediaObjects/13661_2012_Article_220_Equ20_HTML.gif
(3.8)
where ϵ is a positive constant. Since the bilinear form a ( u , u ) is strictly coercive on W, using the trace theory, we obtain
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-129/MediaObjects/13661_2012_Article_220_Equ21_HTML.gif
(3.9)
where λ 0 is a constant depending on Ω and μ. Further, one has
| Ω a ( x ) u t ( m u ) d x | 1 2 Ω a ( x ) | u t | 2 d x + C 0 2 a ( u , u ) ,
(3.10)
where Ω | u | 2 d x C 0 a ( u , u ) with some constant C 0 > 0 . Substituting the inequalities (3.7)-(3.10) into (3.6) and taking into account the fact that m ν 0 on Γ 0 , we have
d d t ψ ( t ) 1 2 Γ 1 m ν | u t | 2 d Γ 1 2 Ω | u t | 2 d x ( 1 C 0 2 ϵ λ 0 2 ) a ( u , u ) + 1 2 ϵ Γ 1 | B 1 u | 2 + | B 2 u | 2 d Γ ( 1 2 ϵ λ 0 2 δ 0 ) Γ 1 m ν [ u x x 2 + u y y 2 + 2 μ u x x u y y + 2 ( 1 μ ) u x y 2 ] d Γ + 1 2 Ω a ( x ) | u t | 2 d x .
Since the boundary conditions (2.1) and (2.2) can be written as
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-129/MediaObjects/13661_2012_Article_220_Equv_HTML.gif

our conclusion follows. □

Let us introduce the Lyapunov functional
L ( t ) = N E ( t ) + ψ ( t ) ,

with N > 0 . Now we are in a position to show the main result of this paper.

Theorem 3.1 Let ( u 0 , u 1 ) W × L 2 ( Ω ) . Suppose the resolvent kernels k 1 , k 2 satisfy the conditions (H) and (3.1). Then there exist constants ω , C > 0 such that for some t 0 large enough, the solution of (1.1)-(1.5) satisfies
E ( t ) C E ( 0 ) e ω 0 t ξ ( s ) d s , t t 0 , if u 0 = u 0 ν = 0 on Γ 1 .
(3.11)
Otherwise,
E ( t ) C ( E ( 0 ) + 0 t k 0 ( s ) e ω t 0 s ξ ( τ ) d τ d s ) e ω 0 t ξ ( s ) d s
(3.12)
for all t t 0 , where
ξ ( t ) = min { ξ 1 ( t ) , ξ 2 ( t ) }
and
k 0 ( t ) = Γ 1 k 1 2 ( t ) | u 0 | 2 d Γ + Γ 1 k 2 2 ( t ) | u 0 ν | 2 d Γ .
Proof Applying the inequality (3.3) with α = 1 2 in Lemma 3.3 and from Lemma 3.2, we obtain
L ( t ) ( N 1 2 ) Ω a ( x ) | u t | 2 d x τ 1 N 2 Γ 1 { | u t | 2 k 1 2 ( t ) | u 0 | 2 k 1 ( t ) | u | 2 + k 1 u } d Γ τ 2 N 2 Γ 1 { | u t ν | 2 k 2 2 ( t ) | u 0 ν | 2 k 2 ( t ) | u ν | 2 + k 2 u ν } d Γ ( 1 C 0 2 ϵ λ 0 2 ) a ( u , u ) + 1 2 Γ 1 m ν | u t | 2 d Γ + 2 τ 1 2 ϵ Γ 1 { | u t | 2 + k 1 2 ( t ) | u | 2 + k 1 2 ( t ) | u 0 | 2 k 1 ( 0 ) k 1 u } d Γ + 2 τ 2 2 ϵ Γ 1 { | u t ν | 2 + k 2 2 ( t ) | u ν | 2 + k 2 2 ( t ) | u 0 ν | 2 k 2 ( 0 ) k 2 u ν } d Γ 1 2 Ω | u t | 2 d x ( 1 2 ϵ λ 0 2 δ 0 ) × Γ 1 m ν [ u x x 2 + u y y 2 + 2 μ u x x u y y + 2 ( 1 μ ) u x y 2 ] d Γ .
(3.13)
Then, choosing N large enough and 0 < ϵ < min { 2 C 0 λ 0 , δ 0 λ 0 } , we obtain
L ( t ) c 0 E ( t ) + c Γ 1 k 1 2 ( t ) | u 0 | 2 d Γ + c Γ 1 k 2 2 ( t ) | u 0 ν | 2 d Γ c Γ 1 k 1 u d Γ c Γ 1 k 2 u ν d Γ , t t 0 .
On the other hand, we can choose N even larger so that
L ( t ) E ( t ) .
(3.14)
If ξ ( t ) = min { ξ 1 ( t ) , ξ 2 ( t ) } , t 0 , then using (3.1) and (3.2), we have
ξ ( t ) L ( t ) c 0 ξ ( t ) E ( t ) + c ξ ( t ) Γ 1 k 1 2 ( t ) | u 0 | 2 d Γ + c ξ ( t ) Γ 1 k 2 2 ( t ) | u 0 ν | 2 d Γ c ξ 1 ( t ) Γ 1 k 1 u d Γ c ξ 2 ( t ) Γ 1 k 2 u ν d Γ c 0 ξ ( t ) E ( t ) + c ξ ( t ) Γ 1 k 1 2 ( t ) | u 0 | 2 d Γ + c ξ ( t ) Γ 1 k 2 2 ( t ) | u 0 ν | 2 d Γ + c Γ 1 k 1 u d Γ + c Γ 1 k 2 u ν d Γ c 0 ξ ( t ) E ( t ) + c ξ ( t ) Γ 1 k 1 2 ( t ) | u 0 | 2 d Γ + c ξ ( t ) Γ 1 k 2 2 ( t ) | u 0 ν | 2 d Γ c E ( t ) , t t 0 ,
which gives
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-129/MediaObjects/13661_2012_Article_220_Equab_HTML.gif
Using the fact that ξ is a nonincreasing continuous function as ξ 1 and ξ 2 are nonincreasing, and so ξ is differentiable, with ξ ( t ) 0 , for a.e. t, then we infer that
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-129/MediaObjects/13661_2012_Article_220_Equac_HTML.gif
Since using (3.14),
F = ξ L + c E E ,
(3.15)
we obtain for some positive constant ω,
F ( t ) ω ξ ( t ) F ( t ) + c Γ 1 k 1 2 ( t ) | u 0 | 2 d Γ + c Γ 1 k 2 2 ( t ) | u 0 ν | 2 d Γ , t t 0 .
(3.16)
Case 1: If u 0 = u 0 ν = 0 on Γ 1 , then (3.16) reduces to
F ( t ) ω ξ ( t ) F ( t ) , t t 0 .
A simple integration over ( t 0 , t ) yields
F ( t ) F ( t 0 ) e ω t 0 t ξ ( s ) d s , t t 0 .
By using (3.2) and (3.15), we then obtain for some positive constant C
E ( t ) C E ( 0 ) e ω 0 t ξ ( s ) d s , t t 0 .

Thus, the estimate (3.11) is proved.

Case 2: If ( u 0 , u 0 ν ) ( 0 , 0 ) on Γ 1 , then (3.16) gives
F ( t ) ω ξ ( t ) F ( t ) + c k 0 ( t ) , t t 0 ,
(3.17)
where
k 0 ( t ) = Γ 1 k 1 2 ( t ) | u 0 | 2 d Γ + Γ 1 k 2 2 ( t ) | u 0 ν | 2 d Γ .
In this case, we introduce
G ( t ) : = F ( t ) c e ω t 0 t ξ ( s ) d s t 0 t k 0 ( s ) e ω t 0 s ξ ( τ ) d τ d s .
(3.18)
A simple differentiation of G, using (3.17), leads to
G ( t ) = F ( t ) + ω ξ ( t ) c e ω t 0 t ξ ( s ) d s t 0 t k 0 ( s ) e ω t 0 s ξ ( τ ) d τ d s c k 0 ( t ) ω ξ ( t ) G ( t ) , t t 0 .
Again, a simple integration over ( t 0 , t ) yields
G ( t ) G ( t 0 ) e ω t 0 t ξ ( s ) d s , t t 0 ,
which implies, for all t t 0 ,
F ( t ) ( F ( t 0 ) + c t 0 t k 0 ( s ) e ω t 0 s ξ ( τ ) d τ d s ) e ω t 0 t ξ ( s ) d s .
By using (3.15), we deduce
E ( t ) C ( E ( 0 ) + 0 t k 0 ( s ) e ω t 0 s ξ ( τ ) d τ d s ) e ω t 0 t ξ ( s ) d s , t t 0 .

Consequently, by the boundedness of ξ, (3.12) is established. □

Remark 3.1 Estimates (3.11) and (3.12) are also true for t [ 0 , t 0 ] by virtue of continuity and boundedness of E ( t ) and ξ.

Remark 3.2 Note that the exponential and polynomial decay estimates are only particular cases of (3.11) and (3.12). More precisely, we have exponential decay for ξ 1 ( t ) c 1 and ξ 2 ( t ) c 2 and polynomial decay for ξ 1 ( t ) = c 1 ( 1 + t ) 1 and ξ 2 ( t ) c 2 , where c 1 and c 2 are positive constants.

Example 3.1 As in [20], we give some examples to illustrate the energy decay rates given by (3.11).
  1. (1)
    If k 1 ( t ) = k 2 ( t ) = a e b ( 1 + t ) p , 0 < p 1 , then for i = 1 , 2 , k i ( t ) ξ ( t ) k i ( t ) , where ξ ( t ) = b p ( 1 + t ) p 1 . For suitably chosen positive constants a and b, k i satisfies (H) and (3.11) gives
    E ( t ) c e ω b ( 1 + t ) p .
     
  2. (2)
    If k 1 ( t ) = a 1 ( 1 + t ) q , q > 0 , and k 2 ( t ) = a 2 e b ( 1 + t ) p , 0 < p 1 , then for i = 1 , 2 , k i ( t ) ξ ( t ) k i ( t ) , where ξ ( t ) = q ( 1 + t ) 1 . Then
    E ( t ) c ( 1 + t ) ω q .
     
The above two examples are included in the following more general one.
  1. (3)
    For any nonincreasing functions k i ( t ) , i = 1 , 2 , which satisfy (H), ξ i = k k are also nonincreasing differentiable functions, and c ξ 1 ( t ) ξ 2 ( t ) , for some 0 < c 1 , (3.11) gives
    E ( t ) c [ k 1 ( t ) ] ω .
     

Author’s contributions

The work was realized by the author.

Declarations

Acknowledgements

The author thanks the anonymous referee for a careful review. This work was supported by the Dong-A University research fund.

Authors’ Affiliations

(1)
Department of Mathematics, Dong-A University

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© Kang; licensee Springer. 2012

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