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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 ¯ ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-129/MediaObjects/13661_2012_Article_220_IEq1_HTML.gif and Ω is an open bounded set of R 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-129/MediaObjects/13661_2012_Article_220_IEq2_HTML.gif 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 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-129/MediaObjects/13661_2012_Article_220_Equa_HTML.gif
Let us denote by ν = ( ν 1 , ν 2 ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-129/MediaObjects/13661_2012_Article_220_IEq3_HTML.gif the external unit normal to Γ, and let us denote by η = ( ν 2 , ν 1 ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-129/MediaObjects/13661_2012_Article_220_IEq4_HTML.gif the unit tangent positively oriented on Γ. We are denoting by B 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-129/MediaObjects/13661_2012_Article_220_IEq5_HTML.gif, B 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-129/MediaObjects/13661_2012_Article_220_IEq6_HTML.gif 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 , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-129/MediaObjects/13661_2012_Article_220_Equb_HTML.gif
where B 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-129/MediaObjects/13661_2012_Article_220_IEq7_HTML.gif and B 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-129/MediaObjects/13661_2012_Article_220_IEq8_HTML.gif 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 ) ] , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-129/MediaObjects/13661_2012_Article_220_Equc_HTML.gif

and the constant μ, 0 < μ < 1 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-129/MediaObjects/13661_2012_Article_220_IEq9_HTML.gif, 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 , ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-129/MediaObjects/13661_2012_Article_220_IEq10_HTML.gif 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 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-129/MediaObjects/13661_2012_Article_220_IEq11_HTML.gif 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 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-129/MediaObjects/13661_2012_Article_220_IEq12_HTML.gif by m ( x ) = x x 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-129/MediaObjects/13661_2012_Article_220_IEq13_HTML.gif, the condition (1.7) implies that there exists a small positive constant δ 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-129/MediaObjects/13661_2012_Article_220_IEq14_HTML.gif such that 0 < δ 0 m ( x ) ν ( x ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-129/MediaObjects/13661_2012_Article_220_IEq15_HTML.gif, x Γ 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-129/MediaObjects/13661_2012_Article_220_IEq16_HTML.gif.

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 } . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-129/MediaObjects/13661_2012_Article_220_Equd_HTML.gif
First, we shall use Eqs. (1.3) and (1.4) to estimate the values B 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-129/MediaObjects/13661_2012_Article_220_IEq5_HTML.gif and B 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-129/MediaObjects/13661_2012_Article_220_IEq6_HTML.gif on Γ 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-129/MediaObjects/13661_2012_Article_220_IEq12_HTML.gif. Denoting by
( g v ) ( t ) = 0 t g ( t s ) v ( s ) d s , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-129/MediaObjects/13661_2012_Article_220_Eque_HTML.gif
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 ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-129/MediaObjects/13661_2012_Article_220_IEq17_HTML.gif satisfy
k i + 1 g i ( 0 ) g i k i = 1 g i ( 0 ) g i , i = 1 , 2 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-129/MediaObjects/13661_2012_Article_220_Equh_HTML.gif
Denoting by τ i = 1 g i ( 0 ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-129/MediaObjects/13661_2012_Article_220_IEq18_HTML.gif, for i = 1 , 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-129/MediaObjects/13661_2012_Article_220_IEq19_HTML.gif, 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 ( , ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-129/MediaObjects/13661_2012_Article_220_IEq20_HTML.gif 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 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-129/MediaObjects/13661_2012_Article_220_Equ10_HTML.gif
(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 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-129/MediaObjects/13661_2012_Article_220_IEq21_HTML.gif. 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 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-129/MediaObjects/13661_2012_Article_220_Equ11_HTML.gif
(2.4)
Let us denote
( g v ) ( t ) : = 0 t g ( t s ) | v ( t ) v ( s ) | 2 d s . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-129/MediaObjects/13661_2012_Article_220_Equi_HTML.gif

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

Lemma 2.2 For g , v C 1 ( [ 0 , ) : R ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-129/MediaObjects/13661_2012_Article_220_IEq22_HTML.gif, 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 ] . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-129/MediaObjects/13661_2012_Article_220_Equj_HTML.gif

The proof of this lemma follows by differentiating the term g v https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-129/MediaObjects/13661_2012_Article_220_IEq23_HTML.gif.

We formulate the following assumption:

(A1) Let a C 1 ( Ω ¯ ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-129/MediaObjects/13661_2012_Article_220_IEq1_HTML.gif satisfy a ( x ) a 0 > 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-129/MediaObjects/13661_2012_Article_220_IEq24_HTML.gif in Ω for some a 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-129/MediaObjects/13661_2012_Article_220_IEq25_HTML.gif.

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 ] . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-129/MediaObjects/13661_2012_Article_220_Equk_HTML.gif

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

Theorem 2.1 Let k i C 2 ( R + ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-129/MediaObjects/13661_2012_Article_220_IEq26_HTML.gif be such that
k i , k i , k i 0 ( i = 1 , 2 ) . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-129/MediaObjects/13661_2012_Article_220_Equl_HTML.gif
If ( u 0 , u 1 ) ( H 4 ( Ω ) W ) × W https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-129/MediaObjects/13661_2012_Article_220_IEq27_HTML.gif 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 , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-129/MediaObjects/13661_2012_Article_220_Equ12_HTML.gif
(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 ) ) . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-129/MediaObjects/13661_2012_Article_220_Equm_HTML.gif

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 + https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-129/MediaObjects/13661_2012_Article_220_IEq28_HTML.gif is a twice differentiable function such that
    k i ( 0 ) > 0 , lim t k i ( t ) = 0 , k i ( t ) 0 , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-129/MediaObjects/13661_2012_Article_220_Equn_HTML.gif
     
and there exists a nonincreasing continuous function ξ i : R + R + https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-129/MediaObjects/13661_2012_Article_220_IEq29_HTML.gif satisfying
k i ( t ) ξ i ( t ) k i ( t ) , i = 1 , 2 , t 0 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-129/MediaObjects/13661_2012_Article_220_Equ13_HTML.gif
(3.1)

The following identity will be used later.

Lemma 3.1 ([2])

For every v H 4 ( Ω ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-129/MediaObjects/13661_2012_Article_220_IEq30_HTML.gif and for every μ R https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-129/MediaObjects/13661_2012_Article_220_IEq31_HTML.gif, 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 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-129/MediaObjects/13661_2012_Article_220_Equo_HTML.gif

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 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-129/MediaObjects/13661_2012_Article_220_Equ14_HTML.gif
(3.2)
Proof Multiplying Eq. (1.1) by u t https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-129/MediaObjects/13661_2012_Article_220_IEq32_HTML.gif, 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 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-129/MediaObjects/13661_2012_Article_220_Equp_HTML.gif

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 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-129/MediaObjects/13661_2012_Article_220_Equq_HTML.gif
Then applying the Holder inequality for 0 α 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-129/MediaObjects/13661_2012_Article_220_IEq33_HTML.gif, we have
| ( k φ ) ( t ) | 2 [ 0 t | k ( s ) | 2 ( 1 α ) d s ] ( | k | 2 α φ ) ( t ) . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-129/MediaObjects/13661_2012_Article_220_Equ15_HTML.gif
(3.3)
Let us define the functional
ψ ( t ) = Ω ( m u ) u t d x . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-129/MediaObjects/13661_2012_Article_220_Equr_HTML.gif

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 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-129/MediaObjects/13661_2012_Article_220_IEq34_HTML.gif 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 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-129/MediaObjects/13661_2012_Article_220_Equs_HTML.gif
Proof Differentiating ψ, using Eq. (1.1), and taking v = u https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-129/MediaObjects/13661_2012_Article_220_IEq35_HTML.gif 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 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-129/MediaObjects/13661_2012_Article_220_Equ16_HTML.gif
(3.4)
Let us next examine the integrals over Γ 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-129/MediaObjects/13661_2012_Article_220_IEq36_HTML.gif in (3.4). Since u = u ν = 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-129/MediaObjects/13661_2012_Article_220_IEq37_HTML.gif on Γ 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-129/MediaObjects/13661_2012_Article_220_IEq36_HTML.gif, we have B 1 u = B 2 u = 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-129/MediaObjects/13661_2012_Article_220_IEq38_HTML.gif on Γ 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-129/MediaObjects/13661_2012_Article_220_IEq36_HTML.gif 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 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-129/MediaObjects/13661_2012_Article_220_Equ17_HTML.gif
(3.5)
since
u x x u y y ( u x y ) 2 = 0 on  Γ 0 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-129/MediaObjects/13661_2012_Article_220_Equt_HTML.gif
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 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-129/MediaObjects/13661_2012_Article_220_Equ18_HTML.gif
(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 ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-129/MediaObjects/13661_2012_Article_220_IEq39_HTML.gif 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 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-129/MediaObjects/13661_2012_Article_220_IEq40_HTML.gif 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 ) , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-129/MediaObjects/13661_2012_Article_220_Equ22_HTML.gif
(3.10)
where Ω | u | 2 d x C 0 a ( u , u ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-129/MediaObjects/13661_2012_Article_220_IEq41_HTML.gif with some constant C 0 > 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-129/MediaObjects/13661_2012_Article_220_IEq42_HTML.gif. Substituting the inequalities (3.7)-(3.10) into (3.6) and taking into account the fact that m ν 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-129/MediaObjects/13661_2012_Article_220_IEq43_HTML.gif on Γ 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-129/MediaObjects/13661_2012_Article_220_IEq36_HTML.gif, 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 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-129/MediaObjects/13661_2012_Article_220_Equu_HTML.gif
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 ) , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-129/MediaObjects/13661_2012_Article_220_Equw_HTML.gif

with N > 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-129/MediaObjects/13661_2012_Article_220_IEq44_HTML.gif. 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 ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-129/MediaObjects/13661_2012_Article_220_IEq45_HTML.gif. Suppose the resolvent kernels k 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-129/MediaObjects/13661_2012_Article_220_IEq46_HTML.gif, k 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-129/MediaObjects/13661_2012_Article_220_IEq47_HTML.gif satisfy the conditions (H) and (3.1). Then there exist constants ω , C > 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-129/MediaObjects/13661_2012_Article_220_IEq48_HTML.gif such that for some t 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-129/MediaObjects/13661_2012_Article_220_IEq49_HTML.gif 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 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-129/MediaObjects/13661_2012_Article_220_Equ23_HTML.gif
(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 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-129/MediaObjects/13661_2012_Article_220_Equ24_HTML.gif
(3.12)
for all t t 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-129/MediaObjects/13661_2012_Article_220_IEq50_HTML.gif, where
ξ ( t ) = min { ξ 1 ( t ) , ξ 2 ( t ) } https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-129/MediaObjects/13661_2012_Article_220_Equx_HTML.gif
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 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-129/MediaObjects/13661_2012_Article_220_Equy_HTML.gif
Proof Applying the inequality (3.3) with α = 1 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-129/MediaObjects/13661_2012_Article_220_IEq51_HTML.gif 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 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-129/MediaObjects/13661_2012_Article_220_Equ25_HTML.gif
(3.13)
Then, choosing N large enough and 0 < ϵ < min { 2 C 0 λ 0 , δ 0 λ 0 } https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-129/MediaObjects/13661_2012_Article_220_IEq52_HTML.gif, 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 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-129/MediaObjects/13661_2012_Article_220_Equz_HTML.gif
On the other hand, we can choose N even larger so that
L ( t ) E ( t ) . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-129/MediaObjects/13661_2012_Article_220_Equ26_HTML.gif
(3.14)
If ξ ( t ) = min { ξ 1 ( t ) , ξ 2 ( t ) } https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-129/MediaObjects/13661_2012_Article_220_IEq53_HTML.gif, t 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-129/MediaObjects/13661_2012_Article_220_IEq54_HTML.gif, 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 , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-129/MediaObjects/13661_2012_Article_220_Equaa_HTML.gif
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 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-129/MediaObjects/13661_2012_Article_220_IEq55_HTML.gif and ξ 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-129/MediaObjects/13661_2012_Article_220_IEq56_HTML.gif are nonincreasing, and so ξ is differentiable, with ξ ( t ) 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-129/MediaObjects/13661_2012_Article_220_IEq57_HTML.gif, 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 , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-129/MediaObjects/13661_2012_Article_220_Equ27_HTML.gif
(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 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-129/MediaObjects/13661_2012_Article_220_Equ28_HTML.gif
(3.16)
Case 1: If u 0 = u 0 ν = 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-129/MediaObjects/13661_2012_Article_220_IEq58_HTML.gif on Γ 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-129/MediaObjects/13661_2012_Article_220_IEq12_HTML.gif, then (3.16) reduces to
F ( t ) ω ξ ( t ) F ( t ) , t t 0 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-129/MediaObjects/13661_2012_Article_220_Equad_HTML.gif
A simple integration over ( t 0 , t ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-129/MediaObjects/13661_2012_Article_220_IEq59_HTML.gif yields
F ( t ) F ( t 0 ) e ω t 0 t ξ ( s ) d s , t t 0 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-129/MediaObjects/13661_2012_Article_220_Equae_HTML.gif
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 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-129/MediaObjects/13661_2012_Article_220_Equaf_HTML.gif

Thus, the estimate (3.11) is proved.

Case 2: If ( u 0 , u 0 ν ) ( 0 , 0 ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-129/MediaObjects/13661_2012_Article_220_IEq60_HTML.gif on Γ 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-129/MediaObjects/13661_2012_Article_220_IEq12_HTML.gif, then (3.16) gives
F ( t ) ω ξ ( t ) F ( t ) + c k 0 ( t ) , t t 0 , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-129/MediaObjects/13661_2012_Article_220_Equ29_HTML.gif
(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 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-129/MediaObjects/13661_2012_Article_220_Equag_HTML.gif
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 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-129/MediaObjects/13661_2012_Article_220_Equ30_HTML.gif
(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 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-129/MediaObjects/13661_2012_Article_220_Equah_HTML.gif
Again, a simple integration over ( t 0 , t ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-129/MediaObjects/13661_2012_Article_220_IEq59_HTML.gif yields
G ( t ) G ( t 0 ) e ω t 0 t ξ ( s ) d s , t t 0 , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-129/MediaObjects/13661_2012_Article_220_Equai_HTML.gif
which implies, for all t t 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-129/MediaObjects/13661_2012_Article_220_IEq50_HTML.gif,
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 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-129/MediaObjects/13661_2012_Article_220_Equaj_HTML.gif
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 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-129/MediaObjects/13661_2012_Article_220_Equak_HTML.gif

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 ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-129/MediaObjects/13661_2012_Article_220_IEq61_HTML.gif by virtue of continuity and boundedness of E ( t ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-129/MediaObjects/13661_2012_Article_220_IEq62_HTML.gif 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 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-129/MediaObjects/13661_2012_Article_220_IEq63_HTML.gif and ξ 2 ( t ) c 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-129/MediaObjects/13661_2012_Article_220_IEq64_HTML.gif and polynomial decay for ξ 1 ( t ) = c 1 ( 1 + t ) 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-129/MediaObjects/13661_2012_Article_220_IEq65_HTML.gif and ξ 2 ( t ) c 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-129/MediaObjects/13661_2012_Article_220_IEq64_HTML.gif, where c 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-129/MediaObjects/13661_2012_Article_220_IEq66_HTML.gif and c 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-129/MediaObjects/13661_2012_Article_220_IEq67_HTML.gif 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 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-129/MediaObjects/13661_2012_Article_220_IEq68_HTML.gif, 0 < p 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-129/MediaObjects/13661_2012_Article_220_IEq69_HTML.gif, then for i = 1 , 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-129/MediaObjects/13661_2012_Article_220_IEq70_HTML.gif, k i ( t ) ξ ( t ) k i ( t ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-129/MediaObjects/13661_2012_Article_220_IEq71_HTML.gif, where ξ ( t ) = b p ( 1 + t ) p 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-129/MediaObjects/13661_2012_Article_220_IEq72_HTML.gif. For suitably chosen positive constants a and b, k i https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-129/MediaObjects/13661_2012_Article_220_IEq73_HTML.gif satisfies (H) and (3.11) gives
    E ( t ) c e ω b ( 1 + t ) p . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-129/MediaObjects/13661_2012_Article_220_Equal_HTML.gif
     
  2. (2)
    If k 1 ( t ) = a 1 ( 1 + t ) q https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-129/MediaObjects/13661_2012_Article_220_IEq74_HTML.gif, q > 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-129/MediaObjects/13661_2012_Article_220_IEq75_HTML.gif, and k 2 ( t ) = a 2 e b ( 1 + t ) p https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-129/MediaObjects/13661_2012_Article_220_IEq76_HTML.gif, 0 < p 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-129/MediaObjects/13661_2012_Article_220_IEq69_HTML.gif, then for i = 1 , 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-129/MediaObjects/13661_2012_Article_220_IEq77_HTML.gif, k i ( t ) ξ ( t ) k i ( t ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-129/MediaObjects/13661_2012_Article_220_IEq78_HTML.gif, where ξ ( t ) = q ( 1 + t ) 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-129/MediaObjects/13661_2012_Article_220_IEq79_HTML.gif. Then
    E ( t ) c ( 1 + t ) ω q . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-129/MediaObjects/13661_2012_Article_220_Equam_HTML.gif
     
The above two examples are included in the following more general one.
  1. (3)
    For any nonincreasing functions k i ( t ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-129/MediaObjects/13661_2012_Article_220_IEq80_HTML.gif, i = 1 , 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-129/MediaObjects/13661_2012_Article_220_IEq77_HTML.gif, which satisfy (H), ξ i = k k https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-129/MediaObjects/13661_2012_Article_220_IEq81_HTML.gif are also nonincreasing differentiable functions, and c ξ 1 ( t ) ξ 2 ( t ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-129/MediaObjects/13661_2012_Article_220_IEq82_HTML.gif, for some 0 < c 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-129/MediaObjects/13661_2012_Article_220_IEq83_HTML.gif, (3.11) gives
    E ( t ) c [ k 1 ( t ) ] ω . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-129/MediaObjects/13661_2012_Article_220_Equan_HTML.gif
     

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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