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International Journal of Differential Equations
Volume 2011, Article ID 403264, 10 pages
Research Article

Exponential Decay to Thermoelastic Systems over Noncylindrical Domains

Department of Mathematics, State University of Londrina, Campus Universitario, 86051-990 Londrina, PR, Brazil

Received 13 May 2011; Accepted 4 July 2011

Academic Editor: Alberto Cabada

Copyright © 2011 Luci Harue Fatori and Michelle Klaiber. 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.


This paper is concerned with linear thermoelastic systems defined in domains with moving boundary. The uniform rate of decay of the energy associated is proved.

1. Introduction

In the study of asymptotic behavior for thermoelastic systems, a pioneering work is the one by Dafermos [1] concerned with the classical linear thermoelasticity for inhomogeneous and anisotropic materials, where the existence of a unique global solution and asymptotic stability of the system were proved. The existence of solution and asymptotic behavior to thermoelastic systems has been investigated extensively in the literature. For example, Muñoz Rivera [2] showed that the energy of the linear thermoelastic system (on cylindrical domain) decays to zero exponentially as 𝑡. In [3], Burns et al. proved the energy decay for a linear thermoelastic bar. The asymptotic behaviour of a semigroup of the thermoelasticity was established in [4]. Concerning nonlinear thermoelasticity we can cite [57].

In the last two decades, several well-known evolution partial differential equations were extended to domains with moving boundary, which is also called noncylindrical problems. See, for instance, [810] and the references therein. In this work we studied the linear thermoelastic system in a noncylindrical domain with Dirichlet boundary conditions. This problem was early considered by Caldas et al. [11], which concluded that the energy associated to the system decreases inversely proportional to the growth of the functions that describes the noncylindrical domain. However they did not establish a rate of decay. The goal in the present work is to provide a uniform rate of decay for this noncylindrical problem.

Let us consider noncylindrical domains 𝑄2 of the form 𝑄=(𝑥,𝑡)2,;𝑥=𝐾(𝑡)𝑦,𝑦(1,1),𝑡(0,𝑇)(1.1) with lateral boundary Σ=0<𝑡<𝑇{{𝐾(𝑡)×{𝑡}}{𝐾(𝑡)×{𝑡}}},(1.2) where 𝐾[0,𝑇]+ is a given 𝐶2 function. Then our problem is𝑢𝑡𝑡𝑢𝑥𝑥+𝛼𝜃𝑥𝜃=0in𝑄,(1.3)𝑡𝑘𝜃𝑥𝑥+𝛽𝑢𝑥𝑡=0in𝑄,(1.4) with initial conditions𝑢(𝑥,0)=𝑢0(𝑥),𝑢𝑡(𝑥,0)=𝑢1(𝑥),𝜃(𝑥,0)=𝜃0(𝑥),𝐾(0)<𝑥<𝐾(0),(1.5) and boundary conditions𝑢(𝐾(𝑡),𝑡)=𝑢(𝐾(𝑡),𝑡)=0,𝜃𝑥(𝐾(𝑡),𝑡)=𝜃(𝐾(𝑡),𝑡)=0,0<𝑡<𝑇,(1.6) where 𝛼, 𝛽, and 𝑘 are positive real constants.

The function 𝐾(𝑡) and the constants 𝛼,𝛽, and 𝑘 satisfy the following conditions.

(H1) 𝐾𝐶2([0,𝑇],+) and 𝐾0=min0𝑡𝑇𝐾(𝑡)>0.(1.7)

(H2) There exists a positive constant 𝐾1 such that 𝐾1(𝑡)𝑦2>𝐾1.(1.8)

Problem (1.3)–(1.6) is slightly different from the one of [11] with respect to condition (1.6). Indeed, they assumed that 𝜃(𝐾(𝑡),𝑡)=𝜃(𝐾(𝑡),𝑡)=0, for all 𝑡[0,𝑇]. Because of this mixed boundary condition in (1.6), we are able to construct a suitable Liapunov functional to derive decay rates of the energy. This is sufficient to provide a uniform rate of decay for this noncylindrical problem.

The existence and uniqueness of global solutions are derived by the arguments of [11] step by step, that is, to prove that the result of existence and uniqueness is based on transforming the system (1.3)–(1.6) into another initial boundary-value problem defined over a cylindrical domain whose sections are not time-dependent. This is done using a suitable change of variable. Then to show the existence and uniqueness for this equivalent system using Galerkin Methods and the existence result on noncylindrical domains will follows using the inverse of the transformation.

Therefore, we have the following result.

Theorem 1.1. Let Ω𝑡 and Ω0 be the intervals (𝐾(𝑡),𝐾(𝑡)), 0<𝑡<𝑇, and (𝐾(0),𝐾(0)), respectively. Then, given 𝑢0,𝜃0𝐻10(Ω0)𝐻2(Ω0) and 𝑢1𝐻10(Ω0), there exist unique functions 𝑢𝑄,𝜃𝑄(1.9) satisfying the following conditions: 𝑢𝐿0,𝑇;𝐻10Ω𝑡𝐻2Ω𝑡,𝑢𝑡𝐿0,𝑇;𝐻10Ω𝑡,𝑢𝑡𝑡𝐿0,𝑇;𝐿2Ω𝑡,𝜃𝐿20,𝑇,𝐻2Ω𝑡,𝜃𝑡𝐿20,𝑇;𝐻10Ω𝑡,(1.10) which are solutions of (1.3)–(1.6) in 𝑄.

2. Energy Decay

In [11] the authors proved that the energy associated with (1.3)–(1.6) decays at the rate 1/[𝐾(𝑡)]𝛾1 with 𝛾1>0; that is, the energy is decreasing inverserly proportional to the increase of sections of 𝑄. We make a slightly difference from the one of [11] with respect to the hypotheses about 𝐾; we are able to construct a suitable Liapunov functional to derive decay rates of the energy. This is done with the thermal dissipation only. More specifically, in this section we prove that the energy associated with (1.3)–(1.6) decays exponentially. Instead considering an auxiliary problem, we work directly on the original problem (1.3)-(1.4) in its noncylindrical domain.

In order to decay rates of the energy let us suppose the following hypotheses.

(H3) There exist positive constants 𝛿0 and 𝛿1 such that 0<𝛿0𝐾(𝑡)𝛿1<1,𝑡0.(2.1)

(H4) There exists a positive constant 𝛿2 such that 0<𝐾(𝑡)𝐾(𝑡)𝛿2,𝑡0.(2.2)

Let us introduce the energy functional 1𝐸(𝑡)=𝐸(𝑡;𝑢,𝜃)=2𝐾(𝑡)𝐾(𝑡)||𝑢𝑡||2+||𝑢𝑥||2+𝛼𝛽||𝜃||2𝑑𝑥.(2.3)

Our main result is the following.

Theorem 2.1. Under the hypotheses (H1)–(H4), there exist positive constants 𝐶 and 𝛾 such that 𝐸(𝑡;𝑢,𝜃)𝐶𝐸(0;𝑢,𝜃)𝑒𝛾𝑡.(2.4)

The proof of Theorem 2.1 is given by using multipliers techniques. The notations and function spaces used here are standard and can be found, for instance, in the book by Lions [8].

Lemma 2.2. Let (𝑢,𝜃) be solution of (1.3)–(1.5) given by Theorem 1.1; then one obtains 𝑑𝑑𝑡𝐸(𝑡;𝑢,𝜃)𝑘𝛼𝛽𝐾(𝑡)𝐾(𝑡)||𝜃𝑥||2𝑑𝑥𝐶0||𝑢𝑥||(𝐾(𝑡),𝑡)2+||𝑢𝑥||(𝐾(𝑡),𝑡)2,(2.5) where 𝐶0=(𝛿0/2)(1𝛿21)>0.

Proof. From hypothesis 𝑢(𝐾(𝑡),𝑡)=0=𝑢(𝐾(𝑡),𝑡) it follows that 𝑢𝑡(𝐾(𝑡),𝑡)=𝐾(𝑡)𝑢𝑥(𝐾(𝑡),𝑡)𝑒𝑢𝑡(𝐾(𝑡),𝑡)=𝐾(𝑡)𝑢𝑥(𝐾(𝑡),𝑡).(2.6)
Multiplying (1.3) by 𝑢𝑡, integrating in the variable 𝑥, and from (2.6) we obtain 𝐾(𝑡)𝐾(𝑡)𝑢𝑡𝑡𝑢𝑡1𝑑𝑥=2𝑑𝑑𝑡𝐾(𝑡)𝐾(𝑡)||𝑢𝑡||2𝑑𝑥𝐾||𝑢(𝑡)𝑡||(𝐾(𝑡),𝑡)2𝐾||𝑢(𝑡)𝑡||(𝐾(𝑡),𝑡)2=12𝑑𝑑𝑡𝐾(𝑡)𝐾(𝑡)||𝑢𝑡||2𝑑𝑥𝐾(𝑡)3||𝑢𝑥||(𝐾(𝑡),𝑡)2𝐾(𝑡)3||𝑢𝑥||(𝐾(𝑡),𝑡)2.(2.7)
Now, applying integration by parts and using (2.6) it follows that 𝐾(𝑡)𝐾(𝑡)𝑢𝑥𝑥𝑢𝑡𝑑𝑥=𝑢𝑥(𝐾(𝑡),𝑡)𝑢𝑡(𝐾(𝑡),𝑡)+𝑢𝑥(𝐾(𝑡),𝑡)𝑢𝑡(𝐾(𝑡),𝑡)+𝐾(𝑡)𝐾(𝑡)𝑢𝑥𝑢𝑡𝑥=𝐾𝑑𝑥(𝑡)2||𝑢𝑥||(𝐾(𝑡),𝑡)2+𝐾(𝑡)2||𝑢𝑥||(𝐾(𝑡),𝑡)2+12𝑑𝑑𝑡𝐾(𝑡)𝐾(𝑡)||𝑢𝑥||2𝑑𝑥.(2.8) Thus, from inequalities (2.7) and (2.8) we have 12𝑑𝑑𝑡𝐾(𝑡)𝐾(𝑡)||𝑢𝑡||2+||𝑢𝑥||21𝑑𝑥=2𝐾(𝑡)3||𝑢𝑥||(𝐾(𝑡),𝑡)2+||𝑢𝑥||(𝐾(𝑡),𝑡)2𝐾(𝑡)2||𝑢𝑥||(𝐾(𝑡),𝑡)2+||𝑢𝑥||(𝐾(𝑡),𝑡)2𝛼𝐾(𝑡)𝐾(𝑡)𝜃𝑥𝑢𝑡𝑑𝑥.(2.9)
Multiplying (1.4) by 𝜃 and integrating in the variable 𝑥 and using (2.6) we obtain 12𝑑𝑑𝑡𝐾(𝑡)𝐾(𝑡)||𝜃||2𝑑𝑥=𝑘𝐾(𝑡)𝐾(𝑡)||𝜃𝑥||2𝑑𝑥+𝛽𝐾(𝑡)𝐾(𝑡)𝑢𝑡𝜃𝑥𝑑𝑥.(2.10)
Multiplying (2.10) by 𝛼/𝛽 and summing with (2.9) it follows that 12𝑑𝑑𝑡𝐾(𝑡)𝐾(𝑡)||𝑢𝑡||2+||𝑢𝑥||2+𝛼𝛽||𝜃||2𝑑𝑥=𝑘𝛼𝛽𝐾(𝑡)𝐾(𝑡)||𝜃𝑥||2+𝐾𝑑𝑥(𝑡)32||𝑢𝑥||(𝐾(𝑡),𝑡)2+||𝑢𝑥||(𝐾(𝑡),𝑡)2𝐾(𝑡)2||𝑢𝑥||(𝐾(𝑡),𝑡)2+||𝑢𝑥||(𝐾(𝑡),𝑡)2.(2.11)
Thus, following the hypothesis (H3), 12𝑑𝑑𝑡𝐾(𝑡)𝐾(𝑡)||𝑢𝑡||2+||𝑢𝑥||2+𝛼𝛽||𝜃||2𝑑𝑥𝑘𝛼𝛽𝐾(𝑡)𝐾(𝑡)||𝜃𝑥||2𝐾𝑑𝑥(𝑡)21𝛿21||𝑢𝑥||(𝐾(𝑡),𝑡)2+||𝑢𝑥||(𝐾(𝑡),𝑡)2𝑘𝛼𝛽𝐾(𝑡)𝐾(𝑡)||𝜃𝑥||2𝛿𝑑𝑥021𝛿21||𝑢𝑥||(𝐾(𝑡),𝑡)2+||𝑢𝑥||(𝐾(𝑡),𝑡)2,(2.12) which concludes the demonstration.

To estimate the term 𝐾(𝑡)𝐾(𝑡)|𝑢𝑥|2𝑑𝑥 of the energy we use the following lemma.

Lemma 2.3. With the same hypothesis of Lemma 2.2, one gets 𝑑𝑑𝑡𝐾(𝑡)𝐾(𝑡)𝑢𝑡𝑢𝑑𝑥𝐾(𝑡)𝐾(𝑡)||𝑢𝑡||21𝑑𝑥2𝐾(𝑡)𝐾(𝑡)||𝑢𝑥||2𝐶𝑑𝑥+p𝛼22𝐾(𝑡)𝐾(𝑡)||𝜃𝑥||2𝑑𝑥,(2.13) where 𝐶𝑝 is Poincare's constant.

Proof. From the outline condition 𝑢(𝐾(𝑡),𝑡)=𝑢(𝐾(𝑡),𝑡)=0 follows that 𝑑𝑑𝑡𝐾(𝑡)𝐾(𝑡)𝑢𝑡𝑢𝑑𝑥=𝐾(𝑡)𝐾(𝑡)||𝑢𝑡||2𝑑𝑥+𝐾(𝑡)𝐾(𝑡)𝑢𝑢𝑡𝑡𝑑𝑥.(2.14)
Replacing 𝑢𝑡𝑡=𝑢𝑥𝑥𝛼𝜃𝑥 in the derivative above we get 𝑑𝑑𝑡𝐾(𝑡)𝐾(𝑡)𝑢𝑡𝑢𝑑𝑥=𝐾(𝑡)𝐾(𝑡)||𝑢𝑡||2𝑑𝑥𝐾(𝑡)𝐾(𝑡)||𝑢𝑥||2𝑑𝑥+𝛼𝐾(𝑡)𝐾(𝑡)𝑢𝑥𝜃𝑑𝑥.(2.15)
Applying Cauchy-Schwartz's inequality, Young's inequality, and Poincare's inequality in (2.15) we have 𝑑𝑑𝑡𝐾(𝑡)𝐾(𝑡)𝑢𝑡𝑢𝑑𝑥𝐾(𝑡)𝐾(𝑡)||𝑢𝑡||21𝑑𝑥2𝐾(𝑡)𝐾(𝑡)||𝑢𝑥||2𝐶𝑑𝑥+𝑝𝛼22𝐾(𝑡)𝐾(𝑡)||𝜃𝑥||2𝑑𝑥.(2.16) Therefore our conclusion follows.

To estimate the term 𝐾(𝑡)𝐾(𝑡)|𝑢𝑡|2𝑑𝑥 of the energy we introduce the function 𝑞=𝑥𝐾(𝑡)𝜃𝑑𝑠. By these conditions we have the following lemma.

Lemma 2.4. With the same hypothesis of Lemma 2.2, there are positive constants 𝐶1 and 𝐶2 such that 𝑑𝑑𝑡𝐾(𝑡)𝐾(𝑡)𝑢𝑡𝑞𝑑𝑥𝐶1𝐾(𝑡)𝐾(𝑡)||𝜃𝑥||2𝛽𝑑𝑥+32𝐾(𝑡)𝐾(𝑡)||𝑢𝑥||2𝛽𝑑𝑥4𝐾(𝑡)𝐾(𝑡)||𝑢𝑡||2||𝑢𝑑𝑥+𝜀𝐾(𝑡)𝑥||(𝐾(𝑡),𝑡)2+𝐶2||𝑢𝑥||(𝐾(𝑡),𝑡)2+||𝑢𝑥||(𝐾(𝑡),𝑡)2,(2.17) where 𝐶1=((𝐶𝑝/2𝛿0)+𝛼𝐶𝑝+(𝑘2/2𝛽)+(𝐶𝑝/2)+(8/𝛽)) and 𝐶2=𝛿2(1+2𝛽𝛿1+𝛿31).

Proof. Calculate the derivative 𝑑𝑑𝑡𝐾(𝑡)𝐾(𝑡)𝑢𝑡𝑞𝑑𝑥=𝐾(𝑡)𝐾(𝑡)𝜕𝑢𝜕𝑡𝑡𝑞𝑑𝑥+𝐾(𝑡)𝑢𝑡(𝐾(𝑡),𝑡)𝑞(𝐾(𝑡),𝑡)+𝐾(𝑡)𝑢𝑡=(𝐾(𝑡),𝑡)𝑞(𝐾(𝑡),𝑡)𝐾(𝑡)𝐾(𝑡)𝑢𝑡𝑡𝑞𝑑𝑥+𝐾(𝑡)𝐾(𝑡)𝑢𝑡𝑞𝑡𝑑𝑥+𝐾(𝑡)𝑢𝑡(𝐾(𝑡),𝑡)𝑞(𝐾(𝑡),𝑡).(2.18)
From (1.3) and recording that 𝑞=𝑥𝐾(𝑡)𝜃𝑑𝑠, we get 𝐼1=𝑢𝑥(𝐾(𝑡),𝑡)𝑞(𝐾(𝑡),𝑡)𝐾(𝑡)𝐾(𝑡)𝑢𝑥𝑞𝑥𝑑𝑥+𝐾(𝑡)𝐾(𝑡)𝑑𝛼𝜃𝑑𝑥𝑥𝐾(𝑡)+𝜃𝑑𝑠𝑑𝑥𝐾(𝑡)𝐾(𝑡)𝑢𝑡𝑑𝑥𝑥𝐾(𝑡)𝜃𝑡𝑑𝑠+𝐾(𝑡)𝑢𝑡(𝐾(𝑡),𝑡)𝑞(𝐾(𝑡),𝑡).(2.19)
As 𝑞𝑥=𝜃 and 𝑞𝑡=𝑥𝐾(𝑡)𝜃𝑡𝑑𝑠 we obtain 𝐼1=𝑢𝑥(𝐾(𝑡),𝑡)𝐾(𝑡)𝐾(𝑡)𝜃𝑑𝑥𝐾(𝑡)𝐾(𝑡)𝑢𝑥𝜃𝑑𝑥+𝛼𝐾(𝑡)𝐾(𝑡)||𝜃||2+𝑑𝑥𝐾(𝑡)𝐾(𝑡)𝑢𝑡𝑞𝑡𝑑𝑥+𝐾(𝑡)𝑢𝑡(𝐾(𝑡),𝑡)𝐾(𝑡)𝐾(𝑡)𝜃𝑑𝑥.(2.20)
Now, integrating (1.4) from 𝐾(𝑡) to 𝑥, multiplying by 𝑢𝑡, and after integrating from 𝐾(𝑡) to 𝐾(𝑡), it follows that 𝐼2=𝐾(𝑡)𝐾(𝑡)𝑢𝑡𝑞𝑡𝑑𝑥=𝑘𝐾(𝑡)𝐾(𝑡)𝜃𝑥𝑢𝑡𝑑𝑥𝛽𝐾(𝑡)𝐾(𝑡)||𝑢𝑡||2𝑑𝑥+𝛽𝑢𝑡(𝐾(𝑡),𝑡)𝐾(𝑡)𝐾(𝑡)𝑢𝑡𝑑𝑥.(2.21)
Replacing (𝐼2) in (𝐼1) and from (2.6) we get 𝐼1=𝑑𝑑𝑡𝐾(𝑡)𝐾(𝑡)𝑢𝑡𝑞𝑑𝑥=𝑢𝑥(𝐾(𝑡),𝑡)𝐾(𝑡)𝐾(𝑡)𝜃𝑑𝑥𝐾(𝑡)𝐾(𝑡)𝑢𝑥𝜃𝑑𝑥+𝛼𝐾(𝑡)𝐾(𝑡)||𝜃||2𝑑𝑥+𝑘𝐾(𝑡)𝐾(𝑡)𝜃𝑥𝑢𝑡𝑑𝑥𝛽𝐾(𝑡)𝐾(𝑡)||𝑢𝑡||2𝑑𝑥+𝛽𝐾(𝑡)𝑢𝑥(𝐾(𝑡),𝑡)𝐾(𝑡)𝐾(𝑡)𝑢𝑡𝐾𝑑𝑥(𝑡)2𝑢𝑥(𝐾(𝑡),𝑡)𝐾(𝑡)𝐾(𝑡)𝜃𝑑𝑥.(2.22)
Estimating some terms of (2.22) we obtain 𝑑𝑑𝑡𝐾(𝑡)𝐾(𝑡)𝑢𝑡𝑞𝑑𝑥𝛼𝐾(𝑡)𝐾(𝑡)||𝜃||2𝑑𝑥+𝐾||𝑢(𝑡)𝐾(𝑡)𝑥||(𝐾(𝑡),𝑡)2+𝐶𝑝2𝐾(𝑡)𝐾(𝑡)𝐾(𝑡)||𝜃𝑥||2𝑑𝑥𝐾(𝑡)𝐾(𝑡)𝑢𝑥𝛽𝜃𝑑𝑥4𝐾(𝑡)𝐾(𝑡)||𝑢𝑡||2𝑘𝑑𝑥+2+𝐶2𝛽𝑝2𝐾(𝑡)𝐾(𝑡)||𝜃𝑥||2𝑑𝑥+2𝛽𝐾(𝑡)𝐾(𝑡)2||𝑢𝑥||(𝐾(𝑡),𝑡)2+𝐾(𝑡)𝐾(𝑡)4||𝑢𝑥||(𝐾(𝑡),𝑡)2.(2.23)
Applying Poincare's inequality in the first term of the previous inequality, using the hypothesis (H3), and grouping the common terms, we obtain 𝑑𝑑𝑡𝐾(𝑡)𝐾(𝑡)𝑢𝑡𝑞𝑑𝑥𝛼𝐶𝑝+𝐶𝑝2𝛿0+𝑘2+𝐶2𝛽𝑝2𝐾(𝑡)𝐾(𝑡)||𝜃𝑥||2𝑑𝑥𝐾(𝑡)𝐾(𝑡)𝑢𝑥𝛽𝜃𝑑𝑥4𝐾(𝑡)𝐾(𝑡)||𝑢𝑡||2𝑑𝑥+𝐾(𝑡)𝐾(𝑡)1+2𝛽𝛿1+𝛿31||𝑢𝑥||(𝐾(𝑡),𝑡)2+||𝑢𝑥||(𝐾(𝑡),𝑡)2.(2.24)
From hypothesis (H4) we have 𝑑𝑑𝑡𝐾(𝑡)𝐾(𝑡)𝑢𝑡𝑞𝑑𝑥𝐶1𝐾(𝑡)𝐾(𝑡)||𝜃𝑥||2𝛽𝑑𝑥+32𝐾(𝑡)𝐾(𝑡)||𝑢𝑥||2𝛽𝑑𝑥4𝐾(𝑡)𝐾(𝑡)||𝑢𝑡||2𝑑𝑥+𝐶2||𝑢𝑥||(𝐾(𝑡),𝑡)2+||𝑢𝑥||(𝐾(𝑡),𝑡)2.(2.25) where 𝐶1 and 𝐶2 are positive constants. This concludes the demonstration of the lemma.

Now we use the above auxiliary lemmas to conclude the proof of Theorem 2.1.

Proof of Theorem 2.1. Consider the functional (𝑡)=𝐾(𝑡)𝐾(𝑡)𝛽8𝑢𝑡𝑢+𝑢𝑡𝑞𝑑𝑥.(2.26) From Lemmas 2.3 and 2.4 we obtain 𝑑𝛽𝑑𝑡(𝑡)32𝐾(𝑡)𝐾(𝑡)||𝑢𝑥||2𝛽𝑑𝑥8𝐾(𝑡)𝐾(𝑡)||𝑢𝑡||2+𝐶𝑑𝑥1+𝛼2𝛽𝐶𝑝16𝐾(𝑡)𝐾(𝑡)||𝜃𝑥||2𝑑𝑥+𝐶2||𝑢𝑥||(𝐾(𝑡),𝑡)2+||𝑢𝑥||(𝐾(𝑡),𝑡)2.(2.27)
Finally we introduce the functional (𝑡)=(𝑡)+𝑁𝐸(𝑡),(2.28) where 𝑁 will be chosen later.
From Lemma 2.2 and from (2.27) it follows that 𝑑𝑑𝑡(𝑡)𝑁𝑘𝛼𝐶1𝛽𝛼𝛽2𝐶𝑝𝛼16𝛽𝐾(𝑡)𝐾(𝑡)||𝜃𝑥||2𝑑𝑥𝑁𝛿02𝐶2||𝑢𝑥||(𝐾(𝑡),𝑡)2+||𝑢𝑥||(𝐾(𝑡),𝑡)2𝛽32𝐾(𝑡)𝐾(𝑡)||𝑢𝑥||2𝛽𝑑𝑥8𝐾(𝑡)𝐾(𝑡)||𝑢𝑡||2𝑑𝑥.(2.29)
Taking 𝑁 sufficiently large we find that there is a positive constant 𝐶3 such that 𝑑𝑑𝑡(𝑡)𝐶3𝐾(𝑡)𝐾(𝑡)||𝑢𝑥||2𝑑𝑥+𝐾(𝑡)𝐾(𝑡)||𝑢𝑡||2𝛼𝑑𝑥+𝛽𝐾(𝑡)𝐾(𝑡)||𝜃𝑥||2𝑑𝑥.(2.30)
Observe that (𝑡) and 𝐸(𝑡) are equivalents, that is, there exists positive constant 𝐶4 satisfying 𝑁2𝐸(𝑡;𝑢,𝜃)(𝑡)𝐶4𝐸(𝑡;𝑢,𝜃).(2.31)
Therefore, 𝑑𝐶𝑑𝑡(𝑡)3𝐶4𝐿(𝑡).(2.32)
Now, from equivalence (2.31) it follows that 𝐸(𝑡)𝐶𝐸(0)𝑒𝛾𝑡,(2.33) where 𝐶=2𝐶4/𝑁 and 𝛾=𝐶3/𝐶4. The proof is now complete.


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