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Energy decay of solutions for a wave equation with a constant weak delay and a weak internal feedback

Abbes Benaissa

1

, Aissa Benguessoum

1

and Salim A. Messaoudi

2

1Laboratory of Analysis and Control of PDE’s, Djillali Liabes University, P. O. Box 89, Sidi Bel Abbes 22000, Algeria

2Department of Mathematics and Statistics, KFUPM, Dhahran 31261, Saudi Arabia

Received 2 September 2013, appeared 21 March 2014 Communicated by Paul Eloe

Abstract. In this paper, we consider the wave equation with a weak internal constant delay term:

u00(x,t)−xu(x,t) +µ1(t)u0(x,t) +µ2(t)u0(x,tτ) =0

in a bounded domain. Under appropriate conditions on µ1 and µ2, we prove global existence of solutions by the Faedo–Galerkin method and establish a decay rate estimate for the energy using the multiplier method.

Keywords:nonlinear wave equation, delay term, decay rate, multiplier method.

2010 Mathematics Subject Classification:35B40, 35L70.

1 Introduction

In this paper we investigate the decay properties of solutions for the initial boundary value problem for the linear wave equation of the form









u00(x,t)−xu(x,t) +µ1(t)u0(x,t) +µ2(t)u0(x,t−τ) =0 inΩ×]0,+[,

u(x,t) =0 onΓ×]0,+[,

u(x, 0) =u0(x), ut(x, 0) =u1(x) onΩ, ut(x,t−τ) = f0(x,t−τ) onΩ×]0,τ[,

(P)

where Ωis a bounded domain in IRn, n ∈ IN, with a smooth boundary∂Ω = Γ, τ > 0 is a time delay and the initial data(u0,u1,f0)belong to a suitable function space.

In absence of delay (µ2 = 0), the energy of problem (P) is exponentially decaying to zero provided that µ1is constant, see, for instance, [3,4, 7,8] and [12]. On the contrary, if µ1 = 0 and µ2 > 0 (a constant weight), that is, there exists only the internal delay, the system (P)

1Corresponding author. Email: benaissaabbes@yahoo.com

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becomes unstable (see, for instance, [5]). In recent years, the PDEs with time delay effects have become an active area of research since they arise in many practical problems (see, for example, [1, 19]). In [5], it was shown that a small delay in a boundary control could turn a well-behaved hyperbolic system into a wild one and, therefore, delay becomes a source of instability. To stabilize a hyperbolic system involving input delay terms, additional control terms will be necessary (see [13,15,20]). For instance, the authors of [13] studied the wave equation with a linear internal damping term with constant delay (τ = const in the problem (P) and determined suitable relations betweenµ1andµ2, for which the stability or alternatively instability takes place. More precisely, they showed that the energy is exponentially stable if µ2 < µ1 and they also found a sequence of delays for which the corresponding solution of (P) will be unstable ifµ2µ1. The main approach used in [13] is an observability inequality obtained with a Carleman estimate. The same results were obtained if both the damping and the delay are acting on the boundary. We also recall the result by Xu, Yung and Li [20], where the authors proved a result similar to the one in [13] for the one-space dimension by adopting the spectral analysis approach.

In [17], Nicaise, Pignotti and Valein extended the above result to higher space dimensions and established an exponential decay.

Our purpose in this paper is to give an energy decay estimate of the solution of problem (P) in the presence of a delay term with a weight depending on time. We use the Galerkin approximation scheme and the multiplier technique to prove our results.

2 Preliminaries and main results

First assume the following hypotheses:

(H1)µ1: IR+ →]0,+[is a non-increasing function of classC1(IR+)satisfying

µ01(t) µ1(t)

≤ M, (2.1)

(H2)µ2: IR+ → IR is a function of classC1(IR+), which is not necessarily positive or mono- tone, such that

µ2(t)βµ1(t), (2.2)

µ02(t)≤ Mµ˜ 1(t), (2.3) for some 0<β<1 and ˜M >0.

We now state a Lemma needed later.

Lemma 2.1(Martinez [10]). Let E: IR+→ IR+be a non increasing function andφ: IR+ → IR+an increasing C1function such that

φ(0) =0 and φ(t)→+ as t→+∞.

Assume that there existσ>−1andω>0such that Z +

S E1+σ(t)φ0(t)dt≤ 1

ωEσ(0)E(S), 0≤S<+∞. (2.4)

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Then

E(t) =0 ∀t≥ E(0)σ

ω|σ| , if −1<σ <0, (2.5) E(t)≤ E(0)

1+σ 1+ωσφ(t)

1σ

∀t≥0, if σ >0, (2.6)

E(t)≤ E(0)e1ωφ(t)∀t≥0, if σ=0. (2.7)

We introduce, as in [13], the new variable

z(x,ρ,t) =ut(x,t−τρ), x∈Ω, ρ∈ (0, 1), t>0. (2.8) Then, we have

τzt(x,ρ,t) +zρ(x,ρ,t) =0, inΩ×(0, 1)×(0,+). (2.9) Therefore, problem (P) takes the form:





















u00(x,t)−xu(x,t) +µ1(t)u0(x,t) +µ2(t)z(x, 1,t) =0, x∈Ω,t >0,

τzt(x,ρ,t) +zρ(x,ρ,t) =0, x∈Ω,ρ∈ (0, 1),t>0,

u(x,t) =0, x∈∂Ω,t>0,

z(x, 0,t) =u0(x,t), x∈,t >0, u(x, 0) =u0(x), ut(x, 0) =u1(x), x∈Ω,

z(x,ρ, 0) = f0(x,−τρ), x∈Ω,ρ∈ (0, 1).

(2.10)

Letξbe a positive constant such that

τβ<ξ <τ(2−β). (2.11)

We define the energy of the solution by:

E(t) = 1

2ku0(t)k22+1

2k∇xu(t)k22+ξ(t) 2

Z

Z 1

0 z2(x,ρ,t)dρdx, (2.12) where

ξ(t) =ξµ1(t). We have the following theorem.

Theorem 2.2. Let(u0,u1,f0)∈(H2()∩H01())×H01()×H01(Ω;H1(0, 1))satisfy the compat- ibility condition

f0, 0) =u1.

Assume that(H1)and(H2)hold. Then problem(P)admits a unique global weak solution

u∈ Lloc((−τ,∞); H2()∩H01()), u0 ∈ Lloc((−τ,∞); H01()), u00 ∈ Lloc((−τ,∞); L2()). Moreover, for some positive constants c, ω, we obtain the following decay property:

E(t)≤ cE(0)eω

Rt

0µ1(s)ds, ∀t ≥0. (2.13)

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Lemma 2.3. Let(u,z)be a solution to the problem(2.10). Then, the energy functional defined by(2.12) satisfies

E0(t)≤ −

µ1(t)− ξ(t)

2τ − |µ2(t)|

2

kut(x,t)k22ξ(t)

2τ − |µ2(t)|

2

kz(x, 1,t)k22

≤0. (2.14)

Proof. Multiplying the first equation in (2.10) byut(x,t), integrating overΩand using Green’s identity, we obtain:

1 2

∂tkut(x,t)k22+ 1 2

∂tk∇u(x,t)k22 +µ1(t)kut(x,t)k22+µ2(t)

Z

ut(x,t−τ)ut(x,t)dx=0.

(2.15)

We multiply the second equation in (2.10) byξ(t)zand integrate overΩ×(0, 1)to obtain:

ξ(t)τ Z

Z 1

0 zt(x,ρ,t)z(x,ρ,t)dρdx+ξ(t)

Z

Z 1

0 zρ(x,ρ,t)z(x,ρ,t))dρdx=0. (2.16) This yields

ξ(t)τ 2

d dt

Z

Z 1

0 z2(x,ρ,t)dρdx+ξ(t) 2

Z

Z 1

0

∂ρz2(x,ρ,t))dρdx=0, which gives

τ 2

d dt

ξ(t)

Z

Z 1

0 z2(x,ρ,t)dρdx

ξ0(t)

Z 1

0

Z

z2(x,ρ,t)dρdx

+ ξ(t) 2

Z

z2(x, 1,t)dx− ξ(t) 2

Z

u2t(x,t)dx=0.

Consequently, τ

2 d dt

ξ(t)

Z

Z 1

0 z2(x,ρ,t)dx

= τ 2ξ0(t)

Z

Z 1

0 z2(x,ρ,t)dρdx− ξ(t) 2

Z

z2(x, 1,t)dx+ ξ(t) 2

Z

u2t(x,t)dx.

(2.17)

Combination of (2.15) and (2.17) leads to 1

2

∂t

kut(x,t)k22+k∇u(x,t)k22+ξ(t)

Z

Z 1

0 z2(x,ρ,t)dρdx

=−µ1(t)kut(x,t)k22µ2(t)

Z

z(x, 1,t)ut(x,t)dx +1

2ξ0(t)

Z

Z 1

0 z2(x,ρ,t)dρdx−ξ(t) 2τ

Z

z2(x, 1,t)dx+ ξ(t)

2τ kut(x,t)k22.

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Recalling the definition ofE(t)in (2.12), we arrive at E0(t) = −

µ1(t) − ξ(t) 2τ

kut(x,t)k22µ2(t)

Z

z(x, 1,t)ut(x,t)dx +1

2ξ0(t)

Z

Z 1

0 z2(x,ρ,t)dρdx− ξ(t) 2τ

Z

z2(x, 1,t)dx.

≤ −

µ1(t)−ξ(t) 2τ

kut(x,t)k22µ2(t)

Z

z(x, 1,t)ut(x,t)dx

ξ(t) 2τ

Z

z2(x, 1,t)dx. (2.18)

Due to Young’s inequality, we have Z

z(x, 1,t)ut(x,t)dx≤ 1

2kut(x,t)k22+ 1

2kz(x, 1,t)k22. (2.19) Inserting (2.19) into (2.18), we obtain

E0(t)≤ −

µ1(t) − ξ(t)

2τ − |µ2(t)|

2

kut(x,t)k22ξ(t)

2τ − |µ2(t)|

2

kz(x, 1,t)k22

≤ −µ1(t) 1− ξ 2τ− β

2

!

kut(x,t)k22µ1(t) ξ 2τ− β

2

!

kz(x, 1,t)k22 ≤0. (2.20) This completes the proof of the lemma.

3 Global existence

Throughout this section we assumeu0 ∈ H2()∩H01()andu1∈ H01(),f0∈ L2(Ω;H1(0, 1)). We employ the Galerkin method to construct a global solution. Let T > 0 be fixed and denote byVk the space generated by{w1,w2, . . . ,wk}where the set{wk,k ∈ IN}is a basis of H2()∩H01().

Now, we define for 1≤ j≤ kthe sequenceφj(x,ρ)as follows:

φj(x, 0) =wj.

Then, we may extend φj(x, 0) by φj(x,ρ) over L2(×(0, 1))such that (φj)j form a basis of L2(Ω;H1(0, 1))and denote byZkthe space generated by{φ1,φ2, . . . ,φk}.

We construct approximate solutions(uk,zk),k =1, 2, 3, . . . , in the form uk(t) =

k j=1

gjk(t)wj, zk(t) =

k j=1

hjk(t)φj,

wheregjkandhjk(j=1, 2, . . . ,k)are determined by the following system of ordinary differen- tial equations:





(u00k(t),wj) + (∇xuk(t),∇xwj) +µ1(t)(u0k,wj) +µ2(t)(zk(., 1),wj) =0, 1≤ j≤ k,

zk(x, 0,t) =u0k(x,t),

(3.1)

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associated with the initial conditions uk(0) =u0k =

k j=1

(u0,wj)wj →u0in H2()∩H01()ask →+∞, (3.2)

u0k(0) =u1k =

k j=1

(u1,wj)wj →u1inH10()ask→+∞, (3.3) and

((τzkt+z,φj) =0,

1≤j≤k, (3.4)

zk(ρ, 0) =z0k =

k j=1

(f0,φj)φj → f0inL2(Ω;H1(0, 1))ask→+∞. (3.5) By virtue of the theory of ordinary differential equations, the system (3.1)–(3.5) has a unique local solution which is extended to a maximal interval [0,Tk[(with 0 < Tk ≤ +∞) by Zorn lemma. Note thatuk(t)is of classC2.

In the next step, we obtain a priori estimates for the solution of the system (3.1)–(3.5), so that it can be extended beyond[0,Tk[to obtain a solution defined for allt >0. Then, we utilize a standard compactness argument for the limiting procedure.

The first estimate.Since the sequencesu0k, u1kandz0k converge, then from (2.14) we can find a positive constantCindependent ofksuch that

Ek(t) +

Z t

0 a1(s)ku0k(s)k22ds+

Z t

0 a2(s)kzk(x, 1,s)k22ds≤ Ek(0)≤C, (3.6) where

Ek(t) = 1

2ku0k(t)k22+1

2k∇xuk(t)k22+ ξ(t) 2

Z

Z 1

0 z2k(x,ρ,t)dρdx, a1(t) =µ1(t) 1− ξ

2τ− β 2

!

anda2(t) =µ1(t) ξ 2τ− β

2

! . These estimates imply that the solution(uk,zk)exists globally in[0,+[.

Estimate (3.6) yields

(uk)is bounded inLloc(0,∞;H01()), (3.7) (u0k)is bounded inLloc(0,∞;L2()), (3.8) µ1(t)(u02k(t))is bounded inL1(×(0,T)), (3.9) µ1(t)(z2k(x,ρ,t))is bounded inLloc (0,∞;L1(×(0, 1))), (3.10) µ1(t)(z2k(x, 1,t))is bounded inL1(×(0,T)). (3.11) The second estimate.We first estimateu00k(0). Replacingwj byu00k(t)in (3.1) and takingt = 0, we obtain:

ku00k(0)k2≤ kxu0kk2+µ1(0)ku1kk2+|µ2(0)|kz0kk2

≤ kxu0k2+µ1(0)ku1k2+|µ2(0)|kz0k2

≤C.

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Differentiating (3.1) with respect tot, we get

(u000k (t) +xu0k(t) +µ1(t)u00k(t) +µ10(t)u0k(t) +µ2(t)z0k(1,t) +µ20(t)zk(1,t),wj) =0.

Multiplying byg00jk(t), summing overjfrom 1 tok, it follows that 1

2 d

dt ku00k(t)k22+k∇xu0k(t)k22+µ1(t)

Z

u002k(t)dx+µ10(t)

Z

u00k(t)u0k(t)dx +µ2(t)

Z

u00k(t)z0k(x, 1,t)dx+µ02(t)

Z

u00k(t)zk(x, 1,t)dx=0.

(3.12)

Differentiating (3.4) with respect tot, we get

τz00k(t) +

∂ρz0k,φj

=0.

Multiplying byh0jk(t), summing overjfrom 1 tok, it follows that τ

2 d

dtkz0k(t)k22+1 2

d

dρkz0k(t)k22=0. (3.13) Taking the sum of (3.12) and (3.13), we obtain that

1 2

d dt

ku00k(t)k22+k∇xu0k(t)k22+τ Z 1

0

kz0k(x,ρ,t)k2L2()

+µ1(t)

Z

u002k(t)dx+ 1 2

Z

|z0k(x, 1,t)|2dx

=−µ2(t)

Z

u00k(t)z0k(x, 1,t)dx−µ01(t)

Z

u00k(t)u0k(t)dx

µ02(t)

Z

u00k(t)zk(x, 1,t)dx+1

2ku00k(t)k22 Using(H1),(H2), Cauchy–Schwarz and Young’s inequalities, we obtain

1 2

d dt

ku00k(t)k22+k∇xu0k(t)k22+

Z 1

0 τkz0k(x,ρ,t)k2L2()

+µ1(t)

Z

u002k(t)dx+ 1 2

Z

|z0k(x, 1,t)|2dx

≤ |µ2(t)|ku00k(t)k2kz0k(x, 1,t)k2+|µ01(t)|ku00k(t)k2|ku0k(t)k2 +|µ02(t)|ku00k(t)k2kzk(x, 1,t)k2+ 1

2ku00k(t)k22

≤ |µ2(t)|2

2 ku00k(t)k22+1

2kz0k(x, 1,t)k22+ |µ10(t)|

4 ku00k(t)k22+|µ01(t)|ku0k(t)k22 +|µ20(t)|

4 ku00k(t)k22+|µ02(t)|kzk(x, 1,t)k22+1

2ku00k(t)k22

≤c0ku00k(t)k22+|µ01(t)|ku0kk22+|µ02(t)|kzk(x, 1,t)k22+1

2kz0k(x, 1,t)k22.

≤c0ku00k(t)k22+Mµ1(t)ku0kk22+Mµ˜ 1(t)kzk(x, 1,t)k22+1

2kz0k(x, 1,t)k22.

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Integrating the last inequality over(0,t)and using (3.6), we get

ku00k(t)k22+k∇xu0k(t)k22+τ Z 1

0

kz0k(x,ρ,t)k2L2()

ku00k(0)k22+k∇xu0k(0)k22+τ Z 1

0

kz0k(x,ρ, 0)k2L2()

+2M Z t

0 µ1(s)ku0k(s)k22ds +2 ˜M

Z t

0

µ1(s)kzk(x, 1,s)k22ds+2c0 Z t

0

ku00k(s)k22ds

≤C+C0 Z t

0

ku00k(s)k22+k∇xu0k(s)k22+τ Z 1

0

kz0k(x,ρ,s)k2L2()

ds.

Using Gronwall’s lemma, we deduce that ku00k(t)k22+k∇xu0k(t)k22+τ

Z 1

0

kz0k(x,ρ,t)k2L2()dρ≤CeC0T for allt∈IR+, therefore, we conclude that

(u00k)is bounded inLloc(0,∞;L2()), (3.14) (u0k)is bounded inLloc(0,∞;H01()), (3.15) (τz0k)is bounded inLloc(0,∞;L2(×(0, 1))). (3.16) Applying Dunford–Pettis’ theorem, we deduce from (3.7), (3.8), (3.9), (3.10), (3.11), (3.14), (3.15) and (3.16), replacing the sequenceuk with a subsequence if necessary, that

uk →u weak-star in Lloc (0,∞;H2()∩H01()), (3.17) u0k →u0 weak-star in Lloc(0,∞;H10()),

u00k →u00 weak-star in Lloc(0,∞;L2()), (3.18) u0kχ weak in L2(×(0,T);µ1(t)),

zk →z weak-star in Lloc(0,∞;H01(Ω;L2(0, 1)),

z0k →z0 weak-star in Lloc(0,∞;L2(×(0, 1))), (3.19) zk(x, 1,t)→ψ weak in L2(×(0,T),µ1(t))

for suitable functions

u∈ L(0,T;H2()∩H01()), z∈ L(0,T;L2(×(0, 1))), χ∈ L2(×(0,T);µ1(t)), ψ∈L2(×(0,T);µ1(t)), for allT≥0. We have to show thatuis a solution of (P).

From (3.15) we have that (u0k) is bounded in L(0,T;H01()). Then (u0k) is bounded in L2(0,T;H10()). Since(u00k)is bounded inL(0,T;L2()), then it is bounded inL2(0,T;L2()), too. Consequently,(u0k)is bounded inH1(Q).

Since the embeddingH1(Q),→ L2(Q)is compact, using the Aubin–Lions theorem [9], we can extract a subsequence(uς)of(uk)such that

u0ς→u0 strongly in L2(Q). (3.20) Therefore

u0ς→u0strongly and a.e. inQ. (3.21)

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Similarly we obtain

zς→zstrongly in L2(×(0, 1)×(0,T)) (3.22) and

zς→zstrongly and a.e. inΩ×(0, 1)×(0,T). (3.23) It follows at once from (3.17), (3.18), (3.19), (3.20) and (3.22) that for each fixedv∈L2(0,T;L2()) andw∈ L2(0,T;L2()×(0, 1))

Z T

0

Z

u00ςxuς+µ1(t)u0ς+µ2(t)zς v dx dt

Z T

0

Z

u00xu+µ1(t)u0+µ2(t)z

v dx dt,

(3.24)

Z T

0

Z 1

0

Z

τz0ς+ ∂ρzς

w dx dρdt→

Z T

0

Z 1

0

Z

τz0+ ∂ρz

w dx dρdt (3.25) asς→+. Thus the problem (P) admits a global weak solutionu.

Uniqueness. Let (u1,z1) and (u2,z2) be two solutions of problem (2.10). Then (w, ˜w) = (u1,z1)−(u2,z2)satisfies





















w00(x,t)−xw(x,t) +µ1(t)w0(x,t) +µ2(t)w˜(x, 1,t) =0, inΩ×]0,+[, τ0(x,ρ,t) +w˜ρ(x,ρ,t) =0, inΩ×]0, 1[×]0,+[

w(x,t) =0, onΩ×]0,+[

˜

w(x, 0,t) =w0(x,t), onΩ×]0,+[

w(x, 0) =0, w0(x, 0) =0, inΩ

˜

w(x,ρ, 0) =0, in×]0, 1[

(3.26)

Multiplying the first equation in (3.26) byw0, integrating overΩand using integration by parts, we get

1 2

d

dt(kw0k22+k∇xwk22) +µ1(t)kw0k22+µ2(t)(w˜(x, 1,t),w0) =0. (3.27) Multiplying the second equation in (3.26) by ˜w, integrating overΩ×(0, 1)and using integra- tion by parts, we get

τ 2

d

dtkw˜k22+1 2

d

dρkw˜k22=0. (3.28)

Then

τ 2

d dt

Z 1

0

kw˜k22dρ+ 1

2(kw˜(x, 1,t)k22− kw0k22) =0. (3.29) From (3.27), (3.29), using the Cauchy–Schwarz inequality we get

1 2

d dt

kw0k22+k∇xwk22+τ Z 1

0

kw˜k22

+µ1(t)kw0k22+1

2kw˜(x, 1,t)k22

=−µ2(t)(w˜(x, 1,t),w0) + 1 2kw0k22

≤ |µ2(t)|kw˜(x, 1,t)k2kw0k2+ 1 2kw0k22. Using Young’s inequality, we obtain

1 2

d dt

kw0k22+k∇xwk22+τ Z 1

0

kw˜k22

≤ckw0k22,

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wherecis a positive constant. Then integrating over(0,t), using Gronwall’s lemma, we con- clude that

kw0k22+k∇xwk22+τ Z 1

0

kw˜k22dρ=0.

Hence, uniqueness follows.

4 Asymptotic behavior

From now on, we denote byc various positive constants which may be different at different occurrences. We multiply the first equation of (2.10) byφ0Equ, whereφis a bounded function satisfying all the hypotheses of Lemma2.1. We obtain

0=

Z T

S Eqφ0 Z

u

u00∆u+µ1(t)u0+µ2(t)z(x, 1,t)

dx dt

=

Eqφ0 Z

uu0dx T

S

Z T

S

(qE0Eq1φ0+Eqφ00)

Z

uu0dx dt

−2 Z T

S Eqφ0 Z

u02dx dt+

Z T

S Eqφ0 Z

u02+|∇u|2 dxdt +

Z T

S

Eqφ0µ1(t)

Z

uu0dx dt+

Z T

S

Eqφ0µ2(t)

Z

uz(x, 1,t)dx dt.

Similarly, we multiply the second equation of (2.10) byEqφ0ξ(t)e2τρz(x,ρ,t)and get 0=

Z T

S Eqφ0 Z

Z 1

0 e2τρξ(t)z

τzt+zρ

dx dρdt

= 1

2Eqφ0ξ(t)τ Z

Z 1

0 e2τρz2dx dρ T

S

1 2

Z T

S

Z

Z 1

0 Eqφ0ξ(t)τe2τρ0z2dx dρdt +

Z T

S Eqφ0 Z

Z 1

0

ξ(t) 1

2

∂ρ e2τρz2

+τe2τρz2

dx dρdt

= 1

2Eqφ0ξ(t)τ Z

Z 1

0 e(t)ρz2dx dρ T

S

τ 2

Z T

S Eqφ0ξ(t)0

Z

Z 1

0 e2τρz2dx dρdt +1

2 Z T

S Eqφ0ξ(t)

Z

ez2(x, 1,t)−z2(x, 0,t)dx dt+

Z T

S Eqφ0ξ(t)τ Z 1

0

Z

e2τρz2dx dρdt.

Taking their sum, we obtain A

Z T

S Eq+1φ0dt≤ −

Eqφ0 Z

uu0dx T

S

+

Z T

S

(qE0Eq1φ0+Eqφ00)

Z

uu0dx dt +2

Z T

S Eqφ0 Z

u02dx dt−

Z T

S µ1(t)Eqφ0 Z

uu0dx dt

Z T

S µ2(t)Eqφ0 Z

uz(x, 1,t)dx dt− 1

2Eqφ0ξ(t)τ Z

Z 1

0 e2τρz2dx dρ T

S

+ 1 2

Z T

S Eqφ0ξ(t)0τ Z

Z 1

0

e2τρz2dx dρdt

1 2

Z T

S Eqφ0ξ(t)

Z

e(t)z2(x, 1,t)−z2(x, 0,t)dx dt,

(4.1)

(11)

where A = 2 min{1,e1}. Using the Cauchy–Schwarz and Poincaré’s inequalities and the definition ofEand assuming thatφ0 is a bounded non-negative function on IR+, we get

Eq(t)φ0 Z

uu0dx

≤cE(t)q+1. By recalling (2.14), we have

Z T

S

qE0Eq1φ0 Z

uu0dx

dt≤c Z T

S Eq(t)|E0(t)|dt≤c Z T

S Eq(t)(−E0(t))dt

≤cEq+1(S),

Z T

S Eqφ00 Z

uu0dx dt ≤c Z T

S Eq+1(t)(−φ00)≤cEq+1(S)

Z T

S

(−φ00)dt≤cEq+1(S),

and

Z T

S

Eqφ0 Z

u02dxdt≤c Z T

S

Eqφ0 1 µ1(t)

Z

µ1(t)u02dxdt

Z T

S Eq φ0

µ1(t) −E0

dt. (4.2)

Define

φ(t) =

Z t

0 µ1(s)ds. (4.3)

It is clear thatφis a non-decreasing function of classC1on IR+,φis bounded and

φ(t)→+ast →+∞. (4.4)

So, we deduce, from (4.2), that Z T

S Eqφ0 Z

u02dxdt≤c Z T

S Eq(−E0)dt≤cEq+1(S), (4.5) By the hypothesis(H1), Young’s and Poincaré’s inequality and (2.14), we have

Z T

S Eqφ0 Z

uu0dx dt

≤c Z T

S Eqφ0kuk2ku0k2dt

≤cε0 Z T

S Eqφ0kuk22dt+c(ε0)

Z T

S Eqφ0ku0k22dt

ε0c Z T

S Eqφ0k∇xuk22dt+c(ε0)

Z T

S Eqφ0ku0k22dt

ε0c Z T

S Eq+1φ0dt+cEq+1(S). Recalling thatξ0 ≤0 and the definition ofE, we have

Z T

S

(Eqξ(t))0τ Z

Z 1

0 e2τρz2dx dρdt≤

Z T

S

(Eq)0ξ(t)τ Z

Z 1

0 e2τρz2dx dρdt

≤c Z T

S Eq|E0|dt

≤c Z T

S Eq(−E0(t))dt

≤cEq+1(S),

(12)

Z T

S Eqξ(t)

Z

ez2(x, 1,t)dx dt≤ c Z T

S Eqξ(t)

Z

z2(x, 1,t)dx dt

≤ c Z T

S Eq −E0 dt

≤ cEq+1(S),

Z T

S Eqξ(t)

Z

z2(x, 0,t)dx dt=

Z T

S Eqξ(t)

Z

u02(x,t)dx dt

≤cEq+1(S),

where we have also used the Cauchy–Schwarz inequality. Combining these estimates and choosingε0 sufficiently small, we conclude from (4.1) that

Z T

S Eq+1φ0dt≤CEq+1(S). This ends the proof of Theorem2.2.

Acknowledgements

This work is partially supported by KFUPM under Grant #I N121029.

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