# Boundary Stabilization of Memory Type for the Porous-Thermo-Elasticity System

Authors: Abdelaziz Soufyane; Mounir Afilal; Mama Chacha
Journal: Abstract and Applied Analysis
 (2009)

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## Abstract

We consider the one-dimensional viscoelastic Porous-Thermo-Elastic system. We establish a general decay results. The usual exponential and polynomial decay rates are only special cases.


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## Body

## 1. Introduction

An increasing interest has been developed in recent
years to determine the decay behavior of the solutions of several elasticity
problems. It is known that combining the elasticity equations with thermal
effects provokes stability of solutions in the one-dimensional case [1]. Several results concerning
the exponential or the polynomial decay of solutions for the thermoelastic
systems were obtained by [2–6].A sample model describing the one-dimensional
porous-thermo-elasticity, which was developed in [7, 8], is given by the following system:(1.1)ρutt=μuxx+bvx−βθx,in(0,L)×ℝ+,Jvtt=αvxx−bux−ξv+mθ−τvt,in(0,L)×ℝ+,cθt=κθxx−βuxt−mvt,in(0,L)×ℝ+,where t denotes the
time variable, x is the space
variable,   the functions u
is the
displacement, v is the volume
fraction of the solid elastic material, and the function θ is the
temperature difference.  The coefficients ρ, μ, J, α, ξ, τ, c,   and κ are positive
constants. b is a constant
such that b2<μξ.Casas and Quintanilla [7] considered the above system
and used the semigroup theory and the method developed by Liu and Zheng
[4] to establish the
exponential decay of the solution under the boundary conditions of the
form(1.2)u(x,t)=vx(x,t)=θx(x,t)=0,x=0,L,t∈(0,∞).Soufyane [9] considered the following
system:
(1.3)utt=uxx+vx−θx,in(0,L)×ℝ+,vtt=vxx−ux−v+θ−∫0tg(t−s)vxx(s)ds,in(0,L)×ℝ+,θt=θxx−uxt−vt,in(0,L)×ℝ+,u(x,0)=u0(x),ut(x,0)=u1(x),v(x,0)=v0(x),vt(x,0)=v1(x),θ(x,0)=θ0(x),u(0,t)=u(L,t)=vx(0,t)=vx(L,t)=θ(0,t)=θ(L,t)=0,t≥0.He proved that the solution of
(1.3) decays exponentially if the function g decays
exponentially, and the solutions (1.3) decay polynomially if the function g decays
polynomially.Recently Pamplona et al. [10] considered the follwing
system:(1.4)ρutt=μuxx+bvx−βθx+γuxxt,in(0,π)×ℝ+,Jvtt=αvxx−bux−ξv+mθ,in(0,π)×ℝ+,cθt=κθxx−βuxt−mvt,in(0,π)×ℝ+,u(x,0)=u0(x),ut(x,0)=u1(x),v(x,0)=v0(x),vt(x,0)=v1(x),θ(x,0)=θ0(x),u(0,t)=u(π,t)=v(0,t)=v(π,t)=θx(0,t)=θx(π,t)=0,t≥0.They proved that the system is
not exponential stable, and they showed that the solution decays polynomially.In this paper we are concerned with the following
model:(1.5)ρutt=μuxx+bvx−βθx,in(0,L)×ℝ+,Jvtt=αvxx−bux−ξv+mθ,in(0,L)×ℝ+,cθt=κθxx−βuxt−mvt,in(0,L)×ℝ+,with the initial
conditions(1.6)u(x,0)=u0(x),ut(x,0)=u1(x),v(x,0)=v0(x),vt(x,0)=v1(x),θ(x,0)=θ0(x),and the boundary
conditions
(1.7)u(0,t)=v(0,t)=θ(0,t)=θ(L,t)=0,t≥0,(1.8)u(L,t)=−∫0tg1(t−s)[μux(L,s)+bv(L,s)]ds,(1.9)v(L,t)=−∫0tg2(t−s)αvx(L,s)ds. Our main interest concerns the
asymptotic behavior of the solution of the system above. That is, whether the
dissipation given by the boundary memory effect is strong enough to stabilize
the whole system. And what type of rate of decay may we expect (exponential
decay or polynomial decay?).  We obtain
an exponential decay or polynomial decay result under some conditions on gi(i=1,2). Our proof is
based on the multiplier techniques.This work is divided into four sections. In Section2
we introduce some notations and some material needed for our work. In Section 3
we state and prove the exponential decay of the solutions of our studied
system. Section 4 is devoted to the polynomial decay.


## 2. Preliminaries

In this section we introduce some notations and we
study the existence of regular and weak solutions to system (1.5)–(1.9). First,   we
will use   (1.8) and (1.9) to estimate the boundary terms ux(L,t) and vx(L,t).Defining the convolution product operator
by(2.1)(g∗φ)(t)=∫0tg(t−s)φ(s)ds,and differentiating equation (1.8)
we obtain(2.2)μux(L,t)+bv(L,t)+1g1(0)(g1′∗u)(L,t)=−1g1(0)ut(L,t)∀t≥0.Applying Volterra's inverse
operator, we get(2.3)μux(L,t)+bv(L,t)=−1g1(0)[ut(L,t)+(k1∗u)(L,t)],∀t≥0,where the resolvent kernel k1 satisfies(2.4)k1(t)+1g1(0)(g1′∗k1)(t)=−1g1(0)g1′(t).Denoting by η1=1/g1(0), we arrive at(2.5)μux(L,t)+bv(L,t)=−η1[ut(L,t)+k1(0)u(L,t)−k1(t)u(L,0)+(k1′∗u)(L,t)],∀t≥0.A similar procedure leads
to(2.6)αvx(L,t)=−η2[vt(L,t)+k2(0)v(L,t)−k2(t)v(L,0)+(k2′∗v)(L,t)],∀t≥0,where η2=1/g2(0).Reciprocally, taking, in a natural way, the initial
datau0(L)=v0(L)=0, the identities
(2.5) and (2.6) imply (1.8) and (1.9).Since we are interested in relaxation functions of
exponential or polynomial type and identities (2.5)-(2.6) involve the resolvent
kernels ki(i=1,2), we want to know
if ki has the same
properties. The following lemma answers this question. Let h be a relaxation
function and k its resolvent
kernel, that is,(2.7)k(t)−(k∗h)(t)=h(t).Lemma 2.1 (see [11]).
Ifh
is a positive
continuous function, then   k
is also
positive and continuous. Moreover,
(1) If there exist positive constantsc0  and  γ  with  c0<γ  such that
(2.8)h(t)≤c0e−γt, then the function  k  satisfies (2.9)k(t)≤c0(γ−ϵ)(γ−ϵ−c0)e−ϵt,   for all 0<ϵ<γ−c0.
(2) If(2.10)cp:=supt∈ℝ+∫0t(1+t)p(1+t−s)−p(1+s)−pds<+∞,   for a given p>1
and if there
exists a positive constant   c0  with  c0cp<1, for which (2.11)h(t)≤c0(1+t)−p,   then the function k  satisfies (2.12)k(t)≤c01−c0cp(1+t)−p.Based on Lemma2.1, we will use
(2.5)-(2.6) instead of (1.8)-(1.9). We then define(2.13)(goφ)(t):=∫0tg(t−s)|φ(t)−φ(s)|2ds,(g⊙φ)(t):=∫0tg(t−s)(φ(t)−φ(s))ds. By using Hölder's inequality for 0≤μ≤1, we have(2.14)|(g⊙φ)(t)|2≤(∫0t|g(s)|2(1−μ)ds)(|g|2μoφ)(t).Lemma 2.2 (see [12]).
Ifg,φ∈C1(ℝ+),  then (2.15)(g∗φ)φt=−12g(t)|φ(t)|2+12g′oφ−12ddt(goφ−(∫0tg(s)ds)|φ(t)|2).


## 3. Exponential Decay

In this section we study the asymptotic behavior of
the solutions of system (1.5)–(1.9), when the resolvent kernels ki(i=1,2) satisfy, for γi>0, the following conditions:(3.1)ki(0)>0,ki(t)≥0,ki′(t)≤0,ki′′(t)≥−γiki′(t).These assumptions imply that ki′ converges
exponentially to 0, that
is,(3.2)0≤−ki′(t)≤Ce−γit.We define the first-order energy
of system (1.5)–(1.9) by(3.3)E(t):=12∫0L[ρ|ut(x,t)|2+J|vt(x,t)|2+c|θ(x,t)|2+μ|ux(x,t)|2]dx+12∫0L[α|vx(x,t)|2+ξ|v(x,t)|2+2bux(x,t)v(x,t)]dx+η12(k1(t)u2(L,t)−(k1′ou)(L,t)+η22(k2(t)v2(L,t)−(k2′ov)(L,t).In the sequel we define by V1:={u∈H1(0,L):u(0)=0}. We are now ready to state our first result.Theorem 3.1.
Given((u0,u1),(v0,v1),θ0)∈((V1×L2(0,L))2×L2(0,L)), assume that (3.1) holds   with  (3.4)supt∈[t0,∞[(ki(t))smallenough.  Assume further that   b is a small number,
then the energy   E
satisfies the
following decay estimates:  (3.5)E(t)≤c1E(0)e−ωt,ifu0(L)=v0(L)=0.Otherwise,(3.6)E(t)≤c1E(0)e−ωt(1+∫0tk12(s)eωsds+∫0tk22(s)eωsds),  where   c1   and   ω   are positive
constants independent of the initial data.Proof.
The
main idea is to construct a Lyapunov functionalℒ(t) equivalent to E(t). To do this we use the multiplier techniques. The
proof of Theorem 3.1 will be achieved with the help of a sequence of lemmas.Lemma 3.2.
Under
the assumptions of Theorem3.1, the energy of the solution of (1.5)–(1.9) satisfies  (3.7)dEdt≤−κ∫0L|θx|2dx−η12ut2(L,t)+η12k12(t)|u0(L)|2+η12k1′(t)|u(L,t)|2−η12(k1′′ou)(L,t)−η22vt2(L,t)+η22k22(t)|v0(L)|2+η22k2′(t)|v(L,t)|2−η22(k2′′ov)(L,t).Proof.
Multiplying first
equation of (1.5) by ut, multiplying second equation of (1.5) by vt and third
equation of (1.5) by θ, and
integrating by parts over [0,L], we
obtain(3.8)d2dt∫0L(ρ|ut|2+μ|ux|2)dx=−b∫0Lutxvdx+β∫0Luxtθdx+[μux(L,t)+bv(L,t)]ut(L,t),d2dt∫0L(J|vt|2+α|vx|2+ξ|v|2)dx=−b∫0Lvtuxdx+m∫0Lθvtdx+αvx(L)vt(L),d2dt∫0Lc|θ|2dx=−κ∫0L|θx|2dx−m∫0Lθvtdx−β∫0Lθuxtdx.By a summation of these three
identities, we get(3.9)d2dt∫Ω(ρ|ut|2+μ|ux|2+J|vt|2+α|vx|2+ξ|v|2+c|θ|2+2buxv)dx≤−κ∫0L|θx|2dx+[μux(L,t)+bv(L,t)]ut(L,t)+αvx(L,t)vt(L,t),using (2.5), (2.6), and Lemma 2.2, we
obtain(3.10)dEdt≤−κ∫0L|θx|2dx−η12ut2(L,t)+η12k12(t)|u0(L)|2+η12k1′(t)|u(L,t)|2−η12(k1′′ou)(L,t)−η22vt2(L,t)+η22k22(t)|v0(L)|2+η22k2′(t)|v(L,t)|2−η22(k2′′ov)(L,t),which ends the proof of Lemma
3.2.Lemma 3.3.
Under
the assumptions of Theorem3.1, the energy of the solution of (1.5)–(1.9) satisfies  (3.11)ddt∫0L(2xux+(1−ε0)u)ρutdx+ddt∫0L(2xvx+(1−ε0)v)Jvtdx≤−ε0∫0Lρut2dx−(2−ε0)2μ∫0Lux2dx−2b(1−ε0)∫0Luxvdx−ε0∫0LJvt2dx−(2−ε0)2α∫0Lvx2dx+C∫0Lθx2dx+(1−ε0)μu(L)ux(L)+ρLut2(L)+μLux2(L)−(1−ε0)αv(0)vx(0),
where   ε0   is a small
positive number.Proof.
We multiply first equation of (1.5) by 2xux+(1−ε0)u to
obtain(3.12)ddt∫0L(2xux+(1−ε0)u)ρutdx=∫0L(2xutx+(1−ε0)ut)ρutdx+∫0L(2xux+(1−ε0)u)(μuxx+bvx−βθx)dx.Integrating by parts, we
get(3.13)ddt∫0L(2xux+(1−ε0)u)ρutdx=−ε0∫0Lρut2dx−(2−ε0)μ∫0Lux2dx+b∫0L2xuxvxdx−b(1−ε0)∫0Luxvdx+(1−ε0)μu(L,t)ux(L,t)+ρLut2(L,t)+μLux2(L,t)−β∫0L(2xux+(1−ε0)u)θxdx.Similarly, we multiply second
equation of (1.5)
by 2xvx+(1−ε0)v and integrate
over ]0,L[, using
integration by parts, to arrive at(3.14)ddt∫0L(2xvx+(1−ε0)v)Jvtdx=−ε0∫0LJvt2dx−(2−ε0)α∫0Lvx2dx+ε0∫0Lξv2dx+αLvt2(L,t)+μLvx2(L,t)−ξLv2(L,t)−(1−ε0)αv(L,t)vx(L,t)−∫0Lb(2xvx+(1−ε0)v)uxdx+m∫0L(2xvx+(1−ε0)v)θdx.Summing the above two identities
and using Poincare's and Young's inequalities and taking ε0 small, we deduce
that(3.15)ddt∫0L(2xux+(1−ε0)u)ρutdx+ddt∫0L(2xvx+(1−ε0)v)Jvtdx≤−ε0∫0Lρut2dx−(2−ε0)2μ∫0Lux2dx−2b(1−ε0)∫0Luxvdx−ε0∫0LJvt2dx−(2−ε0)2α∫0Lvx2dx+C∫0Lθx2dx+(1−ε0)μu(L,t)ux(L,t)+ρLut2(L,t)+μLux2(L,t)−αLvt2(L,t)+μLvx2(L,t)−ξLv2(L,t)+(1−ε0)αv(L,t)vx(L,t).The proof of Lemma 3.3 is
completed.Now, we introduce the Lyapunov functional. So, forN>0 large enough,
let(3.16)ℒ(t)=NE(t)+∫0L(2xux+(1−ε0)u)ρutdx+∫0L(2xvx+(1−ε0)v)Jvtdx.Applying Young's inequality and
Poincaré's inequality to the boundary terms, we have, for ε>0,(3.17)(1−ε0)μu(L,t)ux(L,t)+(1−ε0)αv(L,t)vx(L,t)≤ε[μu2(L,t)+αv2(L,t)]+Cε[μux2(L,t)+αvx2(L,t)]≤εC(∫0L|ux|2dx+∫0L|vx|2dx)+Cε[μux2(L,t)+αvx2(L,t)].By rewriting the boundary
conditions (2.5)-(2.6) as(3.18)μux(L,t)=−bv(L,t)−η1(ut+k1(t)u−k1(t)u0−k1′⊙u)(L,t),αvx(L,t)=−η2(vt+k2(t)v−k2(t)v0−k2′⊙v)(L,t),and combining all above
relations, using the fact that b is a small number,
the condition 17, taking N large enough, ε0 very small, and ε≪ε0, we obtain(3.19)dℒdt(t)≤−κN2∫0L|θx|2dx−Nη18ut2(L,t)+Nη1k12(t)|u0(L)|2−Nη12k1′′ou(L,t)−Nη28vt2(L,t)+Nη2k22(t)|v0(L)|2−Nη22k2′′ov(L,t)−ε0∫0Lρut2dx−(1−ε0)8μ∫0Lux2dx−ε0∫0LJvt2dx−(1−ε0)8α∫0Lvx2dx+Cε((−k1′⊙u)2+(−k2′⊙v)2)(L,t).Applying inequality (2.14)
with μ=1/2, using the trace formula we have, for some positive
constant c0, the following estimate:(3.20)ddtℒ(t)≤−c0E(t)+C[k12(t)|u0(L)|2+k22(t)|v0(L)|2].Also, by direct computations, it
is easy to check that, for N large, we
have(3.21)N2E(t)≤ℒ(t)≤2NE(t).Therefore (3.20)
becomes(3.22)ddtℒ(t)≤−ωℒ(t)+C[k12(t)|u0(L)|2+k22(t)|v0(L)|2],for some positive constants ω. At this point we
distinguish two cases.Case 1.
Ifu0(L)=v0(L)=0, then (3.22)
reduces(3.23)ddtℒ(t)≤−ωℒ(t),∀t≥0.A simple integration over (0,t) yields(3.24)ℒ(t)≤ℒ(0)e−ωt,∀t≥0.By using (3.21), estimate (3.5) is
proved.Case 2.
Ifu0(L)≠0 or v0(L)≠0, then (3.22)
gives(3.25)ddtℒ(t)≤−ωℒ(t)+C1k12(t)+C2k22(t),where(3.26)C1=C|u0(L)|2,C2=C|v0(L)|2.In this case we introduce the
following functional:(3.27)F(t):=ℒ(t)−C1e−ωt∫0tk12(t)eωsds−C2e−ωt∫0tk22(t)eωsds.A simple differentiation of F, using (3.25), leads to(3.28)dFdt(t)≤−ωF(t).Again a simple integration over (0,t) yields(3.29)F(t)≤F(0)e−ωt,∀t≥0.A combination of (3.21), (3.27), and
(3.29) then yields the estimate (3.6). This completes the proof of Theorem 3.1.


## 4. Polynomial Decay

In this section we study the asymptotic behavior of
the solutions of system (1.5)–(1.9) when the resolvent kernels ki(i=1,2) satisfy(4.1)ki(0)>0,ki(t)≥0,ki′(t)≤0,ki′′(t)≥γi(−ki′(t))1+qi,for (q1,q2)≠(0,0),0≤qi<1/2, and some
positive constants γi. These
assumptions imply that ki′ decays
polynomially to 0 if qi>0. That
is,(4.2)0≤−ki′(t)≤C(1+t)−1/qi.The following lemmas will play an
important role in the sequel.Lemma 4.1.
(see [13]) Let  p>1, 0≤r<1, and t≥0.   Then for  0<r,  (4.3)(|k′|oϕ(L,t))1+1/(1−r)(1+p)≤2(∥ϕ∥L∞(0,T,H1(0,L))2∫0t|k′(s)|rds)1/(1−r)(1+p)(|k′|1+1/(1+p)oϕ(L,t)), and for  r=0,  (4.4)(|k′|oϕ(L,t))1+1/(1+p)≤2(∫0t∥ϕ(·,s)∥H1(0,L)2ds+t∥ϕ(·,t)∥H1(0,L)2)1/(1+p)(|k′|1+1/(1+p)oϕ(L,t)), where(4.5)p=1q−1.Theorem 4.2.
Given  that((u0,v0),(u1,v1,θ0))∈(V2×(L2(0,L))3), assume that
b small number and (4.1)   hold. Then there exists a positive  constant  λ>0,
for which the
energy  E
satisfies, for
all t≥0,
the following
decay estimates: (4.6)E(t)≤λ(1+t)1/q,ifu0(L)=v0(L)=0on.
Otherwise,(4.7)E(t)≤λ(1+t)(1−r)/q[1+∫0t(k12(s)+k22(s))(1+s)(1−r)(p+1)ds],where(4.8)q=max{q1,q2},1p+1<r<pp+1,p=min{p1,p2}.Moreover, if(4.9)∫0∞(k12(s)+k22(s))(1+s)(1−r)(p+1)ds<+∞for some r, satisfying
(4.8), then (4.7) reduces to (4.6).Proof.
By using (4.1)
in (3.7), we easily see that(4.10)dEdt≤−κ∫0L|θx|2dx−η12ut2(L,t)+η12k12(t)|u0(L)|2+η12k1′(t)|u(L,t)|2−η1γ12(−k1′(t))1+q1ou(L,t)−η22vt2(L)+η22k22(t)|v0(L)|2+η22k2′(t)|v(L)|2−η2γ22(−k2′(t))1+q2ov(L,t).By defining the functional ℒ(t) as in (3.16), we
get(4.11)dℒdt(t)≤−κN2∫0L|θx|2dx−Nη18ut2(L,t)+Nη1k12(t)|u0(L)|2−Nη1γ12(−k1′(t))1+q1ou(L,t)−Nη28vt2(L,t)+Nη2k22(t)|v0(L)|2−Nη2γ22(−k2′(t))1+q2ov(L,t)−ε0∫0Lρut2dx−(1−ε0)8μ∫0Lux2dx−ε0∫0LJvt2dx−(1−ε0)8α∫0Lvx2dx+Cε((−k1′⊙u)2+(−k2′⊙v)2)(L,t).Applying inequality (2.14) for ki′ with μ=(p1+2)/2(p1+1)
if i=1 and μ=(p2+2)/2(p2+1) if i=2, we get(4.12)|k1′⊙u|2≤C(−k1′)1+q1ou,|k2′⊙v|2≤C(−k2′)1+q2ov.Using the  above inequalities and
taking N large, then for
some positive constant c1, we
obtain(4.13)ddtℒ(t)≤−c1[∫0L|θ|2dx+∫0Lρut2dx+∫0Lμux2dx+∫0LJvt2dx+∫0Lαvx2dx+(−k1′(t))1+q1ou(L,t)+(−k2′(t))1+q2ov(L,t)]+Cε(k12(t)|u0(L)|2+k22(t)|v0(L)|2)−Nη18ut2(L,t)−Nη28vt2(L,t).Now, we fix 0<r<1 such that 1/(p+1)<r<p/(p+1) and p=min{p1,p2}. From (4.1) we
get(4.14)∫0t|ki′(s)|rds≤C∫0∞(1+t)−r/qids<∞,i=1,2.[−k1′(t)ou(L)]1+1/(1−r)(p+1)≤C[(−k1′(t))1+1/(p1+1)ou(L)][−k2′(t)ov(L)]1+1/(1−r)(p+1)≤C[(−k2′(t))1+1/(p2+1)ov(L)].Consequently we
have(4.15)CE(t)1+1/(1−r)(p+1)≤∫0L[ρ|ut|2+J|vt|2+c|θ|2+μ|ux|2+[α|vx|2]dx+[(−k1′(t))1+q1ou(L)]+[(−k2′(t))1+q2ov(L)].Inserting (4.15) into (4.13) and
using (3.21), we deduce that(4.16)ddtℒ(t)≤−C(ℒ(t))1+1/(1−r)(p+1)+Cε(k12(t)|u0(L)|2+k22(t)|v0(L)|2).Here, we distinguish two cases.
Case 1  (u0(L)=v0(L)=0).
In this case
(4.16) reduces to(4.17)ddtℒ(t)≤−C(ℒ(t))1+1/(1−r)(p+1).A simple integration over (0,t) gives(4.18)ℒ(t)≤C(1+t)(1−r)(p+1),∀t≥0.As a consequence of (4.18) and the
fact that (1−r)(p+1)>1, we easily verify that(4.19)∫0∞ℒ(t)dt+supt≥0tℒ(t)<∞.Therefore, by using Lemma 4.1,
(3.3), and (3.21), we have(4.20)[|k1′|ou(L)]1+1/(1+p)≤C[|k1′|1+1/(1+p1)ou(L)][|k2′|ov(L)]1+1/(1+p)≤C[|k2′|1+1/(1+p2)ov(L)],which implies
that(4.21)[|k1′|1+1/(1+p1)ou(L)](p+1)/(p+2)≥C[|k1′|ou(L)][|k2′|1+1/(1+p2)ov(L)](p+1)/(p+2)≥C[|k2′|ov(L)].Consequently we
have(4.22)CE(t)1+1/(p+1)≤∫0Lρ|ut|2+J|vt|2+c|θ|2+μ|ux|2+[α|vx|2]dx+[(−k1′(t))1+q1ou(L)]+[(−k2′(t))1+q2ov(L)].Inserting (4.22) into (4.13), with u0(L)=v0(L)=0, we deduce
that(4.23)ddtℒ(t)≤−C(ℒ(t))1+1/(p+1).A simple integration then leads
to(4.24)ℒ(t)≤C(1+t)p+1=C(1+t)1/q,∀t≥0.where q=max{q1′q2}. By using (3.21),
estimate (4.6) is established.
Case 2  (u0(L)≠0 or v0(L)≠0).
In this case
(4.16) gives that(4.25)ddtℒ(t)≤−C(ℒ(t))1+1/(1−r)(p+1)+C1k12(t)+C2k22(t),where(4.26)C1=Cε|u0(L)|2,C2=Cε|v0(L)|2.We then introduce the following
functional H(t):=ℒ(t)−g(t), where(4.27)g(t):=C1(1+t)−(1−r)(p+1)∫0tk12(s)(1+s)(1−r)(p+1)ds+C2(1+t)−(1−r)(p+1)∫0tk22(s)(1+s)(1−r)(p+1)ds.By using (4.1), it is easy to show
that, for some t0>0, we have(4.28)g(t)1+1/(1−r)(p+1)=(1+t)−1−(1−r)(p+1)×[∫0t(C1k12(s)+C2k22(s))(1+s)(1−r)(p+1)ds]1+1/(1−r)(p+1)≥γ0(1+t)−1−(1−r)(p+1)×[∫0t(C1k12(s)+C2k22(s))(1+s)(1−r)(p+1)ds],∀t≥t0,where(4.29)γ0=[∫0t0(C1k12(s)+C2k22(s))(1+s)(1−r)(p+1)ds]1/(1−r)(p+1).Hence, simple calculations
give(4.30)g′(t)+Cg(t)1+1/(1−r)(p+1)≥C1k12(t)+C2k22(t),∀t≥t0.Thanks to (4.25)–(4.30), we
have(4.31)H′(t)≤−C[ℒ(t)1+1/(1−r)(p+1)−g(t)1+1/(1−r)(p+1)].By using the fact
that(4.32)ℒ(t)1+1/(1−r)(p+1)=[H(t)+g(t)]1+1/(1−r)(p+1)≥[H(t)1+1/(1−r)(p+1)+g(t)1+1/(1−r)(p+1)]estimate (4.31)
gives(4.33)H′(t)≤−CH(t)1+1/(1−r)(p+1),∀t≥t0.A simple integration of (4.33) over (t0,t) then leads
to(4.34)H(t)≤C(1+t)(1−r)(p+1),∀t≥t0.Therefore, using (4.27) we get, for
all, t≥t0,(4.35)ℒ(t)≤C(1+t)(1−r)(p+1)[1+∫0t(k12(s)+k22(s))(1+s)(1−r)(p+1)ds]. Again, recalling (3.21) and using the continuity and the boundedness of E, estimate
(4.7) is established.
If, in addition,(4.36)∫0∞(k12(s)+k22(s))(1+s)(1−r)(p+1)ds<+∞for
some r, satisfying (4.8) then (4.35) takes the
form(4.37)ℒ(t)≤C(1+t)(1−r)(p+1).By repeating (4.18)–(4.24), the
desired estimate is established.



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*Source: 280790-2009-06-02.xml*