# Exponential Decay to Thermoelastic Systems over Noncylindrical Domains

Authors: Luci Harue Fatori; Michelle Klaiber
Journal: International Journal of Differential Equations
 (2011)

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## Abstract

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.


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## Body

## 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 [5–7].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, [8–10] 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 domainsQ̂⊂ℝ2 of the form (1.1)Q̂={(x,t)∈R2;x=K(t)y,y∈(-1,1),t∈(0,T)},
with lateral boundary (1.2)Σ̂=⋃0<t<T{{-K(t)×{t}}∪{K(t)×{t}}},
where K:[0,T]→ℝ+ is a given C2 function. Then our problem is(1.3)utt-uxx+αθx=0inQ̂,(1.4)θt-kθxx+βuxt=0inQ̂,
with initial conditions(1.5)u(x,0)=u0(x),ut(x,0)=u1(x),θ(x,0)=θ0(x),-K(0)<x<K(0),
and boundary conditions(1.6)u(-K(t),t)=u(K(t),t)=0,θx(-K(t),t)=θ(K(t),t)=0,0<t<T,
where α, β, and k are positive real constants.The functionK(t) and the constants α,β, and k satisfy the following conditions.(H1)  K∈C2([0,T],ℝ+) and (1.7)K0=min⁡0≤t≤TK(t)>0.(H2)  There exists a positive constant K1 such that (1.8)1-(K′(t)y)2>K1.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.
LetΩt and Ω0 be the intervals (-K(t),K(t)), 0<t<T, and (-K(0),K(0)), respectively. Then, given u0,θ0∈H01(Ω0)∩H2(Ω0) and u1∈H01(Ω0), there exist unique functions
(1.9)u:Q̂⟶R,θ:Q̂⟶R
satisfying the following conditions:
(1.10)u∈L∞(0,T;H01(Ωt)∩H2(Ωt)),ut∈L∞(0,T;H01(Ωt)),utt∈L∞(0,T;L2(Ωt)),θ∈L2(0,T,H2(Ωt)),θt∈L2(0,T;H01(Ωt)),
which are solutions of (1.3)–(1.6) in Q̂.


## 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 Q̂. 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 (2.1)0<δ0≤K′(t)≤δ1<1,t≥0.(H4)  There exists a positive constant δ2 such that (2.2)0<K(t)K′(t)≤δ2,t≥0.Let us introduce the energy functional(2.3)E(t)=E(t;u,θ)=12∫-K(t)K(t)(|ut|2+|ux|2+αβ|θ|2)dx.Our main result is the following.Theorem 2.1.
Under the hypotheses (H1)–(H4), there exist positive constantsC̃ and γ such that
(2.4)E(t;u,θ)≤C̃E(0;u,θ)e-γt.The proof of Theorem2.1 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
(2.5)ddtE(t;u,θ)≤-kαβ∫-K(t)K(t)|θx|2dx-C0[|ux(K(t),t)|2+|ux(-K(t),t)|2],
where C0=(δ0/2)(1-δ12)>0.Proof.
From hypothesisu(K(t),t)=0=u(-K(t),t) it follows that
(2.6)ut(K(t),t)=-K′(t)ux(K(t),t)eut(-K(t),t)=K′(t)ux(-K(t),t).
Multiplying (1.3) by ut, integrating in the variable x, and from (2.6) we obtain
(2.7)∫-K(t)K(t)uttutdx=12[ddt∫-K(t)K(t)|ut|2dx-K′(t)|ut(K(t),t)|2-K′(t)|ut(-K(t),t)|2]=12[ddt∫-K(t)K(t)|ut|2dx-K′(t)3|ux(K(t),t)|2-K′(t)3|ux(-K(t),t)|2].
Now, applying integration by parts and using (2.6) it follows that
(2.8)∫-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)uxutxdx=K′(t)2|ux(K(t),t)|2+K′(t)2|ux(-K(t),t)|2+12ddt∫-K(t)K(t)|ux|2dx.
Thus, from inequalities (2.7) and (2.8) we have
(2.9)12ddt∫-K(t)K(t)(|ut|2+|ux|2)dx=12K′(t)3[|ux(K(t),t)|2+|ux(-K(t),t)|2]-K′(t)2[|ux(K(t),t)|2+|ux(-K(t),t)|2]-α∫-K(t)K(t)θxutdx.
Multiplying (1.4) by θ and integrating in the variable x and using (2.6) we obtain
(2.10)12ddt∫-K(t)K(t)|θ|2dx=-k∫-K(t)K(t)|θx|2dx+β∫-K(t)K(t)utθxdx.
Multiplying (2.10) by α/β and summing with (2.9) it follows that
(2.11)12ddt∫-K(t)K(t)[|ut|2+|ux|2+αβ|θ|2]dx=-kαβ∫-K(t)K(t)|θx|2dx+K′(t)32[|ux(K(t),t)|2+|ux(-K(t),t)|2]-K′(t)2[|ux(K(t),t)|2+|ux(-K(t),t)|2].
Thus, following the hypothesis (H3),(2.12)12ddt∫-K(t)K(t)[|ut|2+|ux|2+αβ|θ|2]dx≤-kαβ∫-K(t)K(t)|θx|2dx-K′(t)2(1-δ12)[|ux(K(t),t)|2+|ux(-K(t),t)|2]≤-kαβ∫-K(t)K(t)|θx|2dx-δ02(1-δ12)[|ux(K(t),t)|2+|ux(-K(t),t)|2],
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 Lemma2.2, one gets
(2.13)ddt∫-K(t)K(t)utudx≤∫-K(t)K(t)|ut|2dx-12∫-K(t)K(t)|ux|2dx+Cpα22∫-K(t)K(t)|θx|2dx,
where Cp is Poincare's constant.Proof.
From the outline conditionu(-K(t),t)=u(K(t),t)=0   follows that
(2.14)ddt∫-K(t)K(t)utudx=∫-K(t)K(t)|ut|2dx+∫-K(t)K(t)uuttdx.
Replacingutt=uxx-αθx  in the derivative above we get
(2.15)ddt∫-K(t)K(t)utudx=∫-K(t)K(t)|ut|2dx-∫-K(t)K(t)|ux|2dx+α∫-K(t)K(t)uxθdx.
Applying Cauchy-Schwartz's inequality, Young's inequality, and Poincare's inequality in (2.15) we have
(2.16)ddt∫-K(t)K(t)utudx≤∫-K(t)K(t)|ut|2dx-12∫-K(t)K(t)|ux|2dx+Cpα22∫-K(t)K(t)|θx|2dx.
Therefore our conclusion follows.To estimate the term∫-K(t)K(t)|ut|2dx of the energy we introduce the function q=∫-K(t)xθds. By these conditions we have the following lemma.Lemma 2.4.
With the same hypothesis of Lemma2.2, there are positive constants C1 and C2 such that
(2.17)ddt∫-K(t)K(t)utqdx≤C1∫-K(t)K(t)|θx|2dx+β32∫-K(t)K(t)|ux|2dx-β4∫-K(t)K(t)|ut|2dx+ɛK(t)|ux(K(t),t)|2+C2[|ux(K(t),t)|2+|ux(-K(t),t)|2],
where C1=((Cp/2δ0)+αCp+(k2/2β)+(Cp/2)+(8/β)) and C2=δ2(1+2βδ1+δ13).Proof.
Calculate the derivative(2.18)ddt∫-K(t)K(t)utqdx=∫-K(t)K(t)∂∂t(utq)dx+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).
From (1.3) and recording that q=∫-K(t)xθds, we get
(2.19)I1=:ux(K(t),t)q(K(t),t)-∫-K(t)K(t)uxqxdx+∫-K(t)K(t)αθ[ddx∫-K(t)xθds]dx+∫-K(t)K(t)utdx[∫-K(t)xθtds]+K′(t)ut(K(t),t)q(K(t),t).
Asqx=θ and qt=∫-K(t)xθtds we obtain
(2.20)I1=ux(K(t),t)∫-K(t)K(t)θdx-∫-K(t)K(t)uxθdx+α∫-K(t)K(t)|θ|2dx+∫-K(t)K(t)utqtdx+K′(t)ut(K(t),t)∫-K(t)K(t)θdx.
Now, integrating  (1.4)  from  -K(t)  to  x, multiplying by ut, and after integrating from  -K(t)  to  K(t), it  follows that
(2.21)I2=:∫-K(t)K(t)utqtdx=k∫-K(t)K(t)θxutdx-β∫-K(t)K(t)|ut|2dx+βut(-K(t),t)∫-K(t)K(t)utdx.
Replacing(I2) in (I1) and from (2.6) we get
(2.22)I1=ddt∫-K(t)K(t)utqdx=ux(K(t),t)∫-K(t)K(t)θdx-∫-K(t)K(t)uxθdx+α∫-K(t)K(t)|θ|2dx+k∫-K(t)K(t)θxutdx-β∫-K(t)K(t)|ut|2dx+βK′(t)ux(-K(t),t)∫-K(t)K(t)utdx-[K′(t)]2ux(K(t),t)∫-K(t)K(t)θdx.
Estimating some terms of (2.22) we obtain
(2.23)ddt∫-K(t)K(t)utqdx≤α∫-K(t)K(t)|θ|2dx+K′(t)K(t)|ux(K(t),t)|2+Cp2K′(t)∫-K(t)K(t)|θx|2dx-∫-K(t)K(t)uxθdx-β4∫-K(t)K(t)|ut|2dx+(k22β+Cp2)∫-K(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.
Applying Poincare's inequality in the first term of the previous inequality, using the hypothesis (H3), and grouping the common terms, we obtain(2.24)ddt∫-K(t)K(t)utqdx≤(αCp+Cp2δ0+k22β+Cp2)∫-K(t)K(t)|θx|2dx-∫-K(t)K(t)uxθdx-β4∫-K(t)K(t)|ut|2dx+K(t)K′(t)(1+2βδ1+δ13)[|ux(-K(t),t)|2+|ux(K(t),t)|2].
From hypothesis (H4) we have(2.25)ddt∫-K(t)K(t)utqdx≤C1∫-K(t)K(t)|θx|2dx+β32∫-K(t)K(t)|ux|2dx-β4∫-K(t)K(t)|ut|2dx+C2[|ux(K(t),t)|2+|ux(-K(t),t)|2].
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 Theorem2.1.Proof of Theorem2.1.
Consider the functional(2.26)F(t)=∫-K(t)K(t)(β8utu+utq)dx.
From Lemmas 2.3 and 2.4 we obtain
(2.27)ddtF(t)≤-β32∫-K(t)K(t)|ux|2dx-β8∫-K(t)K(t)|ut|2dx+(C1+α2βCp16)∫-K(t)K(t)|θx|2dx+C2[|ux(K(t),t)|2+|ux(-K(t),t)|2].
Finally we introduce the functional(2.28)L(t)=F(t)+NE(t),
where N∈ℕ will be chosen later.
From Lemma2.2 and from (2.27) it follows that
(2.29)ddtL(t)≤-(Nk-αC1β-αβ2Cp16)αβ∫-K(t)K(t)|θx|2dx-(Nδ02-C2)[|ux(K(t),t)|2+|ux(-K(t),t)|2]-β32∫-K(t)K(t)|ux|2dx-β8∫-K(t)K(t)|ut|2dx.
TakingN sufficiently large we find that there is a positive constant C3 such that
(2.30)ddtL(t)≤-C3[∫-K(t)K(t)|ux|2dx+∫-K(t)K(t)|ut|2dx+αβ∫-K(t)K(t)|θx|2dx].
Observe thatℒ(t) and E(t) are equivalents, that is, there exists positive constant C4 satisfying
(2.31)N2E(t;u,θ)≤L(t)≤C4E(t;u,θ).
Therefore,(2.32)ddtL(t)≤-C3C4L(t).
Now, from equivalence (2.31) it follows that
(2.33)E(t)≤C̃E(0)e-γt,
where C̃=2C4/N and γ=C3/C4. The proof is now complete.



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