Stability of Timoshenko systems with past history - UFRJim.ufrj.br/~rivera/Art_Pub/TeseHugo.pdf ·...

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J. Math. Anal. Appl. ••• (••••) •••–•••www.elsevier.com/locate/jmaa

Stability of Timoshenko systems with past history ✩

Jaime E. Muñoz Rivera a,b, Hugo D. Fernández Sare b,∗

a Department of Research and Development, National Laboratory for Scientific Computation, Rua Getulio Vargas 333,Quitandinha CEP 25651-070, Petrópolis, Rio de Janeiro, Brazil

b Mathematics Institute, Federal University of Rio de Janeiro, Rio de Janeiro, Brazil

Received 9 October 2006

Submitted by M. Nakao

Abstract

We consider vibrating systems of Timoshenko type with past history acting only in one equation. We show that the dissipationgiven by the history term is strong enough to produce exponential stability if and only if the equations have the same wave speeds.Otherwise the corresponding system does not decay exponentially as time goes to infinity. In the case that the wave speeds of theequations are different, which is more realistic from the physical point of view, we show that the solution decays polynomially tozero, with rates that can be improved depending on the regularity of the initial data.© 2007 Elsevier Inc. All rights reserved.

Keywords: Timoshenko system; Exponential stability; Polynomial rate of decay

1. Introduction

We consider the following linear Timoshenko system with past history:

ρ1ϕtt − k(ϕx + ψ)x = 0 in (0,L) × (0,∞), (1.1)

ρ2ψtt − bψxx +∞∫

0

g(s)ψxx(x, t − s) ds + k(ϕx + ψ) = 0 in (0,L) × (0,∞), (1.2)

and initial conditions

ϕ(·,0) = ϕ0, ϕt (·,0) = ϕ1, ψ(·,0) = ψ0, ψt (·,0) = ψ1 in (0,L) (1.3)

with positive constants ρ1, ρ2, k, b.Here we are interested in the asymptotic behavior of the solutions. Our main tools are Prüss’ results on the ex-

ponential stability of semigroups, see [7,8]. In order to use these results, it is necessary to embed the problem into

✩ Supported by a CNPq-DLR grant and FAPERJ.* Corresponding author.

E-mail addresses: [email protected] (J.E. Muñoz Rivera), [email protected] (H.D. Fernández Sare).

Please cite this article in press as: J.E. Muñoz Rivera, H.D. Fernández Sare, Stability of Timoshenko systems with past history, J. Math. Anal.Appl. (2007), doi:10.1016/j.jmaa.2007.07.012

0022-247X/$ – see front matter © 2007 Elsevier Inc. All rights reserved.doi:10.1016/j.jmaa.2007.07.012


2 J.E. Muñoz Rivera, H.D. Fernández Sare / J. Math. Anal. Appl. ••• (••••) •••–•••

the context of semigroup, so some modifications in our original system (1.1)–(1.2) should be made. We introduce thenotation

ηt (x, s) := ψ(x, t) − ψ(x, t − s), (1.4)

then system (1.1)–(1.2) is rewritten as

ρ1ϕtt − k(ϕx + ψ)x = 0, (1.5)

ρ2ψtt −(

b −∞∫

0

g(s) ds

)ψxx −

∞∫0

g(s)ηtxx(x, s) ds + k(ϕx + ψ) = 0, (1.6)

ηt + ηs − ψt = 0, (1.7)

where the third equation is obtained differentiating (1.4) with respect to s. The initial conditions are given by

ϕ(·,0) = ϕ0, ϕt (·,0) = ϕ1, ψ(·,0) = ψ0, ψt (·,0) = ψ1 in (0,L), (1.8)

η0(·, s) = ψ0(·,0) − ψ0(·,−s) in (0,L) × (0,∞), (1.9)

which means that the history is considered as an initial value. We consider Dirichlet boundary conditions, but ourarguments can be used to prove similar results for other boundary conditions. Concerning the kernel g we considerthe following hypotheses:

g(t) > 0, ∃k0, k1, k2 > 0: −k0g(t) � g′(t) � −k1g(t),∣∣g′′(t)

∣∣� k2g(t), ∀t � 0, (1.10)

b := b −∞∫

0

g(s) ds > 0. (1.11)

Let us mention some energy decay results for dissipative Timoshenko systems. In [2], Kim and Renardy consid-ered a conservative Timoshenko system with two boundary feedbacks and they proved exponential stability for theenergy associated to the system. If the history term in (1.2) is replaced by the control function b(x)ψt , b > 0, thenSoufyane [9] proved exponential stability for the linearized system if and only if the waves speed of Eqs. (1.1), (1.2)are equal, that is,

ρ1

k= ρ2

b. (1.12)

Similar results are obtained by Rivera and Racke [5], where semigroups techniques are used. In [4] the sameauthors consider a dissipative Timoshenko system with a dissipation through a coupling to a heat equation, and theyshow exponential stability if and only if (1.12) holds.

In [1], Ammar Khodja et al. consider also a Timoshenko system with memory effect but considering null history,in that case the system is called a Volterra integro-differential system. For the Volterra problem they proved theexponential stability provided the wave speeds are equal. When the wave speeds are different, the authors consider aclass of kernels for which there is no exponential stability. No information is given concerning the decay in this case.

Introducing non-zero history on ψ makes the problem different from that considered in [1], so different techniqueshave to be used. The main result of this paper is to show that the system is exponentially stable if and only if the wavespeeds of Eqs. (1.1), (1.2) are equal, that is, if and only if (1.12) holds. Moreover, the class of kernel that we considerhere to prove the lack of exponential stability is larger than that considered in [1]. In particular our result implies thenon-exponential stability for singular kernels. When the identity (1.12) does not hold, which is more interesting fromthe physical point of view, we show that the first-order energy decays polynomially with rates that depend on theregularity of the initial data.

The paper is organized as follows. In Section 2 we establish the existence and uniqueness results to system (1.5)–(1.7). The exponential stability of the semigroup associated to this system is studied in Section 3. In Section 4 weshow the non-exponential stability of the semigroup. Finally, in Section 5 we show the polynomial decay when thewave speeds are different.



J.E. Muñoz Rivera, H.D. Fernández Sare / J. Math. Anal. Appl. ••• (••••) •••–••• 3

2. Existence and uniqueness

To facilitate our analysis we consider the following boundary conditions:

ϕ(0, t) = ϕ(L, t) = ψ(0, t) = ψ(L, t) = ηt (0, s) = ηt (L, s) = 0, s, t � 0. (2.1)

In view of (1.10), let L2g(R

+,H 10 ) be the Hilbert space of H 1

0 -valued functions on R+, endowed with the inner product

〈ϕ,ψ〉L2g(R+,H 1

0 ) =L∫

0

∞∫0

g(s)ϕx(s)ψx(s) ds dx.

Now, we use the Lumer–Phillips’ theorem (see [6]) to obtain existence and uniqueness results. Therefore, we formulateour problem as an abstract Cauchy problem.

We define U := (ϕ,ϕt ,ψ,ψt , ηt )′, so the system (1.5)–(1.7) is equivalent to

Ut = AU, U(0) = U0

where U0 := (ϕ0, ϕ1,ψ0,ψ1, η0)′ and A is given by

A :=

⎛⎜⎜⎜⎜⎜⎝0 I (·) 0 0 0

kρ1

∂2x (·) 0 k

ρ1∂x(·) 0 0

0 0 0 I (·) 0

− kρ2

∂x(·) 0 ( bρ2

∂2x − k

ρ2I )(·) 0 1

ρ2

∫∞0 g(s)∂2

x (·, s) ds

0 0 0 I (·) −∂s(·)

⎞⎟⎟⎟⎟⎟⎠ . (2.2)

Let

H := H 10 (0,L) × L2(0,L) × H 1

0 (0,L) × L2(0,L) × L2g

(R

+,H 10

).

It is not difficult to prove that H together with the norm

‖U‖2H = ∥∥(u1, u2, u3, u4, η

)∥∥2H (2.3)

= ρ1∥∥u2

∥∥2L2 + ρ2

∥∥u4∥∥2

L2 + b∥∥u3

x

∥∥2L2 + k

∥∥u1x + u3

∥∥2L2 + ‖η‖2

L2g(R+,H 1

0 )(2.4)

is a Hilbert space. The operator A has the following domain:

D(A) ={

U ∈H: u1, u3 ∈ H 2(0,L); u2, u4 ∈ H 10 (0,L),

∞∫0

g(s)ηtxx(x, s) ds ∈ L2(0,L), ηs ∈ L2

g

(R

+,H 10

), η(0) = 0

}.

Then the operator A, formally given by (2.2), is the infinitesimal generator of a contraction semigroup. In fact, notethat A is dissipative, because for any U ∈ D(A) we have

Re〈AU,U 〉H = 1

2

L∫0

∞∫0

g′(s)|ηx |2 ds dx

� −k1

2

L∫0

∞∫0

g(s)|ηx |2 ds dx � 0.

We also have Im(I − A) = H. Therefore, by the Lumer–Phillips’ theorem, it follows that A is the infinitesimalgenerator of a contraction semigroup.




Theorem 2.1. Assume that g satisfies (1.10)–(1.11) and that U0 ∈ D(A), then there exists a unique solution U =(ϕ,ϕt ,ψ,ψt , η) to system (1.5)–(1.7) with boundary conditions (2.1) satisfying

U ∈ C(R

+;D(A))∩ C1(

R+;H).

Moreover, if U0 ∈ D(An), then

U ∈ Cn−k(R

+;D(Ak))

, k = 0,1, . . . , n.

Remarks. For another boundary conditions, we denote A := ∂2x (·). We consider the following cases:

D(A) = H 2(0,L) ∩ H 10 (0,L),

D(A) ={

v ∈ H 2(0,L):

L∫0

v dx = 0

},

D(A) = {v ∈ H 2(0,L): v(0) = vx(L) = 0

},

D(A) = {v ∈ H 2(0,L): vx(0) = v(L) = 0

}.

We define H = L2(0,L) or L2∗(0,L), with

L2∗(0,L) :={

v ∈ L2(0,L):

L∫0

v dx = 0

}.

Then the semigroups formulation is made in the Hilbert spaces of type

H := D(A1/2)× H × D

(A1/2)× H × L2

g

(R

+,D(A1/2)). (2.5)

3. Exponential stability

First we consider the system (1.5)–(1.7) with boundary conditions (2.1) and the hypotheses over the kernel g

(1.10)–(1.11) hold. We shall demonstrate that the energy

E(t) = 1

2

L∫0

[ρ1ϕ

2t + ρ2ψ

2t + bψ2

x + k|ϕx + ψ |2 +∞∫

0

g(s)∣∣ηt

x

∣∣2 ds

]dx (3.1)

decays to zero exponentially as time goes to infinity provided condition (1.12) holds. We shall use Prüss’ results [3],which states that a semigroup eAt is exponentially stable if and only if the following conditions hold:

iR ⊂ �(A) (resolvent set) (3.2)

and

∃C > 0, ∀U ∈ D(A), ∀λ ∈ R:∥∥(iλI −A)−1

∥∥H � C. (3.3)

In fact, note that the resolvent equation (iλI −A)U = F is given by

iλu1 − u2 = f 1, (3.4)

iλρ1u2 − k

(u1

x + u3)x

= ρ1f2, (3.5)

iλu3 − u4 = f 3, (3.6)

iλρ2u4 − bu3

xx −∞∫

0

g(s)ηtxx(x, s) ds + k

(u1

x + u3)= ρ2f4, (3.7)

iλη + ηs − u4 = f 5, (3.8)




where

b0 :=∞∫

0

g(s) ds, b := b − b0 > 0. (3.9)

To prove condition (3.3) we will use a series of lemmas.

Lemma 3.1. Let us suppose that conditions (1.10) and (1.11) on g hold. Then there exists a positive constant C > 0,being independent of F ∈ H, such that

L∫0

∞∫0

g(s)|ηx |2 ds dx � C‖U‖H‖F‖H.

Proof. Multiplying (3.5) by u2 (in L2(0,L)) we get

iλρ2

L∫0

∣∣u2∣∣2 dx + k

L∫0

(u1

x + u3)u2x dx = ρ1

L∫0

f 2u2 dx

and, using Eq. (3.4),

iλρ2

L∫0

∣∣u2∣∣2 dx − iλk

L∫0

(u1

x + u3)u1x dx = ρ1

L∫0

f 2u2 dx + k

L∫0

(u1

x + u3)f 1x dx. (3.10)

On the other hand, multiplying Eq. (3.7) by u4 (in L2(0,L)) we get

iλρ2

L∫0

∣∣u4∣∣2 dx + b

L∫0

u3xu

4x dx +

L∫0

∞∫0

g(s)ηxu4x ds dx

︸︷︷︸:=I1

+ k

L∫0

(u1

x + u3)u4 dx

︸︷︷︸:=I2

= ρ2

L∫0

f 4u4 dx.

Substituting u4 given by (3.8), (3.6), into I1 and I2, respectively, we get

iλρ2

L∫0

∣∣u4∣∣2 dx − iλb

L∫0

∣∣u3x

∣∣2 dx − iλ

L∫0

∞∫0

g(s)|ηx |2 ds dx − iλk

L∫0

(u1

x + u3)u3 dx +L∫

0

∞∫0

g(s)ηxηxs ds

= ρ2

L∫0

f 4u4 dx + b

L∫0

f 3x u3

x dx + k

L∫0

(u1

x + u3)f 3 dx +L∫

0

∞∫0

g(s)ηxf 5x ds dx. (3.11)

Adding (3.10) and (3.11), using (1.10) and taking the real part our conclusion follows. �Lemma 3.2. With the same hypotheses as in Lemma 3.1 there exists C > 0 such that

ρ2

L∫0

∣∣u4∣∣2 dx � C‖U‖H‖F‖H + C‖U‖1/2

H ‖F‖1/2H(∥∥u3

x

∥∥L2 + ∥∥u1

x + u3∥∥

L2

).

Proof. Multiplying (3.7) by∫∞

g(s)η ds in L2(0,L) we get


0



iλρ2

L∫0

∞∫0

g(s)ηu4 ds dx

︸︷︷︸:=I3

+b

L∫0

∞∫0

g(s)ηxu3x ds dx +

L∫0

∣∣∣∣∣∞∫

0

g(s)ηx ds

∣∣∣∣∣2

dx + k

L∫0

∞∫0

g(s)(u1

x + u3)η ds dx

= ρ2

L∫0

∞∫0

g(s)ηf 4 ds dx.

From Lemma 3.1 we obtain

L∫0

∣∣∣∣∣∞∫

0

g(s)ηx ds

∣∣∣∣∣2

dx �∞∫

0

g(s) ds

L∫0

∞∫0

g(s)|ηx |2 ds dx � C‖U‖H‖F‖H.

Substituting η given by (3.8) into I3, using

Re

{ L∫0

∞∫0

g(s)ηsu4 ds dx

}� ρ2

2

L∫0

∣∣u4∣∣2 dx + C

L∫0

∞∫0

∣∣g′(s)∣∣|ηx |2 ds dx

and (1.10), our conclusion now immediately follows from Lemma 3.1. �To estimate u3 we introduce the multiplier w given by the solution of the Dirichlet problem

−wxx = u3x, w(0) = w(L) = 0. (3.12)

Note that w can be written as

w(x) = −x∫

0

u3(y) dy + x

L

L∫0

u3(y) dx ≡ G(u3)(x).

Under the above conditions we have

Lemma 3.3. With the same hypotheses as in Lemma 3.1, for any ε1 > 0 there exists Cε1 > 0 such that

b

L∫0

∣∣u3x

∣∣2 dx � Cε1‖U‖H‖F‖H + Cε1‖U‖1/2H ‖F‖1/2

H∥∥u1

x + u3∥∥

L2 + ε1ρ1∥∥u2

∥∥2L2 .

Proof. Multiplying (3.7) by u3 yields

iλρ2

L∫0

u4u3 dx

︸︷︷︸:=I4

+b

L∫0

∣∣u3x

∣∣2 dx +L∫

0

∞∫0

g(s)ηxu3x ds dx + k

L∫0

u1xu

3 dx + k

L∫0

∣∣u3∣∣2 dx = ρ2

L∫0

f 4u3 dx.

Substituting u3 given by (3.6) into I4 we get

b

L∫0

∣∣u3x

∣∣2 dx + k

L∫0

u1xu

3 dx + k

L∫0

∣∣u3∣∣2 dx

= ρ2

L∫ ∣∣u4∣∣2 dx −

L∫ ∞∫g(s)ηxu3

x ds dx + ρ2

L∫f 4u3 dx + ρ2

L∫u4f 3 dx. (3.13)


0 0 0 0 0



On the other hand, multiplying (3.5) by w we have

k

L∫0

u1xwx dx − k

L∫0

|wx |2 dx = ρ1

L∫0

u2[G(u4)+ G

(f 3)]

dx + ρ1

L∫0

f 2w dx. (3.14)

Since

L∫0

u1xwx dx = −

L∫0

u1xu

3 dx

we conclude from (3.13)–(3.14) that

b

L∫0

∣∣u3x

∣∣2 dx − k

( L∫0

|wx |2 dx −L∫

0

∣∣u3∣∣2 dx

)

= ρ2

L∫0

f 4u3 dx + ρ2

L∫0

u4f 3 dx + ρ1

L∫0

u2G(f 3)dx + ρ1

L∫0

f 2w dx + ρ2

L∫0

∣∣u4∣∣2 dx

−L∫

0

∞∫0

g(s)ηxu3x ds dx + ρ1

L∫0

u2G(u4)dx.

Note that, for any ε1 > 0 there exists Cε1 > 0 such that

Re

{ρ1

L∫0

u2G(u4)dx

}� ε1ρ1

∥∥u2∥∥2

L2 + Cε1ρ2∥∥u4

∥∥2L2 .

Finally, since

L∫0

|wx |2 dx �L∫

0

∣∣u3∣∣2 dx,

taking real part (and using Lemmas 3.2–3.1) our conclusion follows. �Our next step is to estimate the term ‖u1

x + u3‖2L2 . Here we shall use condition (1.12).

Lemma 3.4. With the same hypotheses as in Lemma 3.1 together with condition (1.12), for any ε2 > 0 there existsCε2 > 0 such that

k

L∫0

∣∣u1x + u3

∣∣2 dx � Cε2‖U‖H‖F‖H + Re

([bu3

x +∞∫

0

g(s)ηx ds

]u1

x

)x=L

x=0

+ (ε1 + ε2)∥∥u2

∥∥2L2 ,

where ε1 is given by the Lemma 3.3.

Proof. Multiplying (3.7) by u1x + u3 we have

iλρ2

L∫u4(u1

x + u3)dx −

([bu3

x +∞∫

g(s)ηx ds

]u1

x

)x=L

x=0

+ k

L∫ ∣∣u1x + u3

∣∣dx


0 0 0



+L∫

0

[bu3

x +∞∫

0

g(s)ηx ds

](u1

x + u3)xdx

︸︷︷︸:=I5

= ρ2

L∫0

f 4(u1x + u3

)dx.

Substituting (u1x + u3)x given by (3.5) into I5 we get

iλρ2

L∫0

u4u1x dx

︸︷︷︸:=I6

+ iλρ2

L∫0

u4u3 dx

︸︷︷︸:=I7

−([

bu3x +

∞∫0

g(s)ηx ds

]u1

x

)x=L

x=0

− iλbρ1

k

L∫0

u3xu

2 dx

−iλρ1

k

L∫0

∞∫0

g(s)ηxu2 ds dx

︸︷︷︸:=I8

−ρ1

k

L∫0

∞∫0

g(s)ηxf 2 ds dx

− bρ1

k

L∫0

u3xf

2 dx + k

L∫0

∣∣u1x + u3

∣∣2 dx = ρ2

L∫0

f 4(u1x + u3

)dx. (3.15)

Substituting u1 given by (3.4) and u4 given by (3.6) into I6 we obtain

I6 = −iλρ2

L∫0

u3u2x dx − ρ2

L∫0

u4f 1x dx + ρ2

L∫0

f 3u2x dx. (3.16)

Using (3.6) we get

I7 = −ρ2

L∫0

∣∣u4∣∣2 dx − ρ2

L∫0

u4f 3 dx. (3.17)

Finally, a substitution of η given by (3.8) yields

I8 = ρ1

k

L∫0

∞∫0

g(s)ηxsu2 ds dx − ρ1b0

k

L∫0

u4xu

2 dx − ρ1

k

L∫0

∞∫0

g(s)f 5x u2 ds dx.

From (3.6) we can rewrite I8 as

I8 = −ρ1

k

L∫0

∞∫0

g′(s)ηxu2 ds dx − iλρ1b0

k

L∫0

u3xu

2 dx + ρ1b0

k

L∫0

f 3x u2 dx − ρ1

k

L∫0

∞∫0

g(s)f 5x u2 ds dx. (3.18)

Using (3.16)–(3.18) in (3.15) we obtain

iλb

(ρ1

k− ρ2

b︸︷︷︸=0

) L∫0

u3u2x dx + k

L∫0

∣∣u1x + u3

∣∣2 dx

=([

bu3x +

∞∫0

g(s)ηx ds

]u1

x

)x=L

x=0

+ ρ2

L∫0

∣∣u4∣∣2 dx + ρ1

k

L∫0

∞∫0

g′(s)ηxu2 ds dx + ρ1b

k

L∫0

u3xf

2 dx

+ ρ1

k

L∫ ∞∫g(s)ηxf 2 ds dx + ρ2

L∫f 4(u1

x + u3)dx


0 0 0



+ ρ2

L∫0

u4f 3 dx + ρ2

L∫0

u4f 1x dx +

(ρ2 − ρ1b0

k

) L∫0

f 3x u2 dx + ρ1

k

L∫0

∞∫0

g(s)f 5x u2 ds dx.

Now, using (1.10) and the previous lemmas, our claim follows. �Noting that, when the boundary conditions are of mixed type, the boundary term in Lemma 3.4 is equal to zero. In

the case (2.1), this boundary term is not equal to zero. In the next lemma we make an estimation of the boundary term.

Lemma 3.5. Under the above notations, let us take q ∈ C1([0,L]) such that q(0) = −q(L) = 1, then there existC,Cq > 0 such that

(i)

−(

q(x)

2

∣∣∣∣∣bu3x +

∞∫0

g(s)ηx ds

∣∣∣∣∣2)x=L

x=0

� C‖U‖H‖F‖H + C‖U‖1/2H ‖F‖1/2

H∥∥u1

x + u3∥∥

L2

+ ε1Cρ1∥∥u2

∥∥L2 + Cq

∥∥u3x

∥∥L2

∥∥u1x + u3

∥∥L2,

and

(ii)

−(

q(x)

2|u1

x |2)x=L

x=0� C‖U‖H‖F‖H + Cq

(∥∥u1x + u3

∥∥2L2 + ρ1

∥∥u2∥∥2

L2

).

Proof. To prove (i), multiplying (3.7) by

q(x)

(bu3

x +∞∫

0

g(s)ηx ds

)

in L2(0,L) we have

iλρ2

L∫0

u4q(x)

(bu3

x +∞∫

0

g(s)ηx ds

)dx

︸︷︷︸:=I9

−L∫

0

(bu3

xx +∞∫

0

g(s)ηxx ds

)q(x)

(bu3

x +∞∫

0

g(s)ηx ds

)dx

︸︷︷︸:=I10

+ k

L∫0

q(x)(u1

x + u3)(bu3x +

∞∫0

g(s)ηx ds

)dx = ρ2

L∫0

f 4q(x)

(bu3

x +∞∫

0

g(s)ηx ds

)dx. (3.19)

From (3.6) and (3.8) we obtain

Re(I9) = ρ2b

2

L∫0

q ′(x)∣∣u4

∣∣2 dx + Re

{−bρ2

L∫0

q(x)u4f 3x dx − ρ2

L∫0

∞∫0

g′(s)q(x)u4ηx ds dx

− ρ2

L∫0

∞∫0

g(s)q(x)u4f 5x ds dx

}. (3.20)

Note that

Re(I10) = −(

q(x)

2

∣∣∣∣∣bu3x +

∞∫g(s)ηx ds

∣∣∣∣∣2)x=L

x=0

+ 1

2

L∫q ′(x)

∣∣∣∣∣bu3x +

∞∫g(s)ηx ds

∣∣∣∣∣2

dx. (3.21)


0 0 0



From (3.19)–(3.21), and using the previous lemmas we obtain (i). To get (ii), multiply (3.5) by q(x)u1x ,

iλρ1

L∫0

u2q(x)u1x dx

︸︷︷︸:=I11

−k

L∫0

u1xxq(x)u1

x dx − k

L∫0

u3xq(x)u1

x dx = ρ1

L∫0

f 2q(x)u1x dx.

Substituting u1 given by (3.4) into I11, using Lemma 3.3 and taking the real part our conclusion follows. �Lemma 3.6. There exists C > 0 such that

ρ1

L∫0

∣∣u2∣∣2 dx � C‖U‖H‖F‖H + 4k

∥∥u1x + u3

∥∥2L2 .

Proof. Multiplying Eq. (3.5) by u1 we get

iλρ1

L∫0

u2u1 dx

︸︷︷︸:=I12

+k

L∫0

(u1

x + u3)u1x dx = ρ1

L∫0

f 2u1 dx.

Substituting u1 given by (3.4) into I12 and taking real parts we get

ρ1

L∫0

∣∣u2∣∣2 dx � C‖U‖H‖F‖H + 2k

∥∥u1x + u3

∥∥2L2 + C

∥∥u3x

∥∥2L2 .

Using Lemma 3.3 for ε1 sufficiently small, our conclusion follows. �Now we are in the position to prove the main result of this section.

Theorem 3.7. Let us assume hypotheses (1.10) and (1.11) on g, suppose that initial data satisfies

ϕ0,ψ0 ∈ H 10 (0,L), η0 ∈ L2

g

(R

+,H 10

)and ϕ1,ψ1 ∈ L2(0,L)

and suppose that condition (1.12) holds. Then the energy E(t) decays exponentially to zero as time tends to infinity,that is, there exist positive constants C and α, being independent of the initial data, such that

E(t) � CE(0)e−αt , ∀t � 0.

Proof. We shall prove conditions (3.2) and (3.3) (see [7]). Let U = (u1, u2, u3, u4, η)′ and F = (f 1, f 2, f 3, f 4, f 5)′satisfy (3.4)–(3.9), then, from Lemma 3.1, we get

‖η‖2L2

g� C‖F‖H‖U‖H. (3.22)

From Lemma 3.2, for ε2 > 0, there exists C1 := C1(ε2) > 0 such that

ρ2∥∥u4

∥∥2L2 � C1‖F‖H‖U‖H + b

2

∥∥u3x

∥∥2L2 + ε2

2k∥∥u1

x + u3∥∥2

L2 . (3.23)

Also, from Lemma 3.3, we obtain

b∥∥u3

x

∥∥2L2 � Cε1‖F‖H‖U‖H + ε1ρ1

∥∥u2∥∥2

L2 + ε2

2k∥∥u1

x + u3∥∥2

L2 . (3.24)

Then, adding (3.23) and (3.24) we get




ρ2∥∥u4

∥∥2L2 + b

2

∥∥u3x

∥∥2L2 � C2‖F‖H‖U‖H + ε1ρ1

∥∥u2∥∥2

L2 + ε2k∥∥u1

x + u3∥∥2

L2 . (3.25)

On the other hand, from Lemmas 3.5 and 3.3, we obtain for each N > 0 and δ > 0,

−N

(q(x)

2

∣∣∣∣∣bu3x +

∞∫0

g(s)ηx ds

∣∣∣∣∣2)x=L

x=0

� CN‖F‖H‖U‖H + k

4

∥∥u1x + u3

∥∥2L2 + ε1CNρ1

∥∥u2∥∥2

L2 (3.26)

and

−δ

(q(x)

2

∣∣u1x

∣∣2)x=L

x=0� Cδ‖F‖H‖U‖H + k

4

∥∥u1x + u3

∥∥2L2 + δCqρ1

∥∥u2∥∥2

L2 . (3.27)

Adding (3.26), (3.27), using Lemma 3.4 and using

Re(z1z2) � σ |z1|2 + Cσ |z2|2, ∀z1, z2 ∈ C, ∀σ > 0,

we obtain, for any 0 < τ < 1, τ := τ(δ, ε1, ε2) > 0, that there exists Cτ > 0 such that

k

2

∥∥u1x + u3

∥∥2L2 � C4‖F‖H‖U‖H + τρ1

∥∥u2∥∥2

L2 . (3.28)

Finally, from Lemma 3.6, we have

2τρ1∥∥u2

∥∥2L2 � 2τC‖F‖H‖U‖H + 8τk

∥∥u1x + u3

∥∥2L2 . (3.29)

Adding (3.28) and (3.29) we conclude(1

2− 8τ

)k∥∥u1

x + u3∥∥2

L2 + τρ1∥∥u2

∥∥2L2 � C5‖F‖H‖U‖H. (3.30)

From (3.22), (3.25) and (3.30), we obtain for ε1, ε2 sufficiently small, that there exists C > 0 independent of λ

(and F , U ) such that

‖U‖2H � C‖F‖2

H, ∀U ∈ D(A).

This completes the proof. �Remark. For another boundary conditions, the elliptic problem (3.12) must change. For example, for

ϕ(0, t) = ϕ(L, t) = ψx(0, t) = ψx(L, t) = ηtx(0, s) = ηt

x(L, s) = 0,

w is given by the solution of the problem

−wxx = u3x, wx(0) = wx(L) = 0.

4. Non-exponential decay

Now we shall prove that condition (1.12) is also necessary for exponential stability in the case where the boundaryconditions are of mixed type. We use the following lemma.

Lemma 4.1. Let us suppose that g satisfies the conditions (1.10) and let us assume that

lims→0

√sg(s) = 0.

Then there exists C > 0 such that∣∣∣∣∣λ∞∫

0

g(s)e−iλs ds

∣∣∣∣∣� C,

uniformly in λ ∈ R.




Proof. Note that

∞∫0

g(s)e−iλsds =π/λ∫0

g(s)e−iλsds − 1

2

π/λ∫0

e−iλsg(s + π/λ)ds − 1

2

∞∫π/λ

e−iλs

[ s+π/λ∫s

g′(y) dy

]ds.

Then ∣∣∣∣∣π/λ∫0

g(s)e−iλs ds

∣∣∣∣∣�π/λ∫0

g(s) ds =π/λ∫0

√sg(s)√

sds.

Setting

ν(λ) = sups∈(0, π

λ)

√sg(s) → 0 as λ → ∞,

the above integral is less than or equal to

ν(λ)

π/λ∫0

ds√s

= 2√

πν(λ)√λ

.

The estimation of the second integral is similar. Concerning the last term, changing the order of integration, andmaking use of (1.10), we get∣∣∣∣∣

∞∫π/λ

e−iλs

[ s+π/λ∫s

g′(y) dy

]ds

∣∣∣∣∣�∞∫

π/λ

[ s∫s+π/λ

g′(y) dy

]ds = π

λg(π/λ),

which, multiplied by λ, tends to zero as λ → ∞. �Theorem 4.2. Let us suppose that (1.12) does not hold. Then the semigroup associated to system (1.5)–(1.7), withboundary conditions

ϕx(0, t) = ϕx(L, t) = ψ(0, t) = ψ(L, t) = ηt (0, s) = ηt (L, s) = 0, s, t � 0, (4.1)

is not exponentially stable.

Proof. From (2.5) let us consider the Hilbert space

H := H 1∗ (0,L) × L2∗(0,L) × H 10 (0,L) × L2(0,L) × L2

g

(R

+,H 10 (0,L)

).

Here the domain of the operator A is defined by

D(A) ={

U ∈H1: u1 ∈ H 2(0,L), u1x ∈ H 1

0 (0,L), u2 ∈ H 1∗ (0,L), u3 ∈ H 2(0,L),

u4 ∈ H 10 (0,L),

∞∫0

g(s)ηtxx(x, s) ds ∈ L2(0,L), ηs ∈ L2

g

(R

+,H 10

), η(0) = 0

}.

Now, from the previous analysis, we have that U = (ϕ,ϕt ,ψ,ψt , η)′ satisfies

d

dtU(t) = AU(t), U(0) = U0.

To show the lack of decay it is enough to show that the solution of

(iλnI −A)Un = Fn




satisfies

limn→∞‖Un‖H = ∞,

where

λ ≡ λn := nπ

δL(n ∈ N), δ :=

√ρ1

k.

As F we choose

F ≡ Fn := (0, f 2,0, f 4,0

)′,

where

f 2(x) := cos(δλx), f 4(x) := sin(δλx).

The solution U = (v1, v2, v3, v4, η)′ to (iλ −A)U = F , should satisfy

iλv1 − v2 = 0, (4.2)

iλv2 − k

ρ1v1xx − k

ρ1v3x = f 2, (4.3)

iλv3 − v4 = 0, (4.4)

iλv4 − b

ρ2v3xx + b0

ρ2v3xx − 1

ρ2

∞∫0


ρ2v1x + k

ρ2v3 = f 4, (4.5)

iλη + ηs − v4 = 0, (4.6)

where b0 := ∫∞0 g(s) ds. Eliminating v2, v4 we get

−λ2v1 − k

ρ1v1xx − k

ρ1v3x = f 2, (4.7)

−λ2v3 − b

ρ2v3xx + b0

ρ2v3xx − 1

ρ2

∞∫0


ρ2v1x + k

ρ2v3 = f 4, (4.8)

iλη + ηs − iλv3 = 0. (4.9)

This can be solved by

v1(x) = A cos(δλx), v3(x) = B sin(δλx), η(x, s) = ϕ(s) sin(δλx),

where A, B , ϕ(s) depend on λ and will be determined explicitly in the sequel. Note that this choose is just compatiblewith the boundary conditions. The system (4.7)–(4.9) is equivalent to

−λ2A + k

ρ1δ2λ2A − k

ρ1δλB = 1, (4.10)

−λ2B + b

ρ2δ2λ2B − b0

ρ2δ2λ2B + δ2λ2

ρ2

∞∫0

g(s)ϕ(s) ds − k

ρ2δλA + k

ρ2B = 1, (4.11)

iλϕ(s) + ϕ′(s) − iλB = 0. (4.12)

Solving (4.12) we get

ϕ(s) = Ce−iλs + B. (4.13)

Since η(0) = 0, then C = −B , and (4.13) becomes

ϕ(s) = B − Be−iλs . (4.14)




Then, from (4.14) we have∞∫

0

g(s)ϕ(s) ds =∞∫

0

g(s)[B − Be−iλs

]ds = Bb0 − B

∞∫0

g(s)e−iλs ds. (4.15)

Using (4.15), we have from (4.10)–(4.11) that A and B satisfy(k

ρ1δ2 − 1

)λ2A − k

ρ1δλB = 1, (4.16)

(b

ρ2δ2 − 1

)λ2B − δ2λ2

ρ2

( ∞∫0

g(s)e−iλs ds

)B − k

ρ2δλA + k

ρ2B = 1. (4.17)

Since kρ1

δ2 = 1, we conclude from (4.16) that

B = −√

ρ1

k· 1

λ. (4.18)

Substituting (4.18) into (4.17) we get

A = − 1

λ2− ρ2√

ρ1k

1

λ+ ρ1

k2

∞∫0

g(s)e−iλs ds + b

k

(ρ2

b− ρ1

k

).

Recalling that

v2 = iλv1 = iλA cos(δλx)

we get

v2(x) =(

− i

λ− iρ2√

ρ1k+ iρ1

k2λ

∞∫0

g(s)e−iλs ds + ib

k

(ρ2

b− ρ1

k

)λ

)cos(δλx).

Note that

∥∥v2∥∥2

L2∗=

L∫0

∣∣v2∣∣2 dx

= L

2

∣∣∣∣∣−1

λ− ρ2√

ρ1k+ ρ1

k2λ

∞∫0

g(s)e−iλs ds + b

k

(ρ2

b− ρ1

k

)λ

∣∣∣∣∣2

� −L

2

∣∣∣∣∣−1

λ− ρ2√

ρ1k+ ρ1

k2λ

∞∫0

g(s)e−iλs ds

∣∣∣∣∣2

︸︷︷︸bounded asλ→∞

+L

2

b2

k2

(ρ2

b− ρ1

k

)2

λ2

using Lemma 4.1. Therefore we have

limλ→∞‖Un‖2

H � limλ→∞

∥∥v2∥∥2

L2∗= ∞

which completes the proof. �Remark. The result also holds for the following boundary conditions:

ϕ(0, t) = ϕ(L, t) = ψx(0, t) = ψx(L, t) = ηtx(0, s) = ηt

x(L, s) = 0,

ϕx(0, t) = ϕ(L, t) = ψ(0, t) = ψx(L, t) = ηt (0, s) = ηtx(L, s) = 0,

ϕ(0, t) = ϕx(L, t) = ψx(0, t) = ψ(L, t) = ηtx(0, s) = ηt (L, s) = 0.




5. Polynomial decay

In this section we shall show the polynomial decay of the solutions of the system (1.5)–(1.7) with boundary condi-tions (2.1), when (1.12) does not hold. Let us introduce the second-order energy

E2(t) := E(ϕt ,ψt , ηt ).

Then from (3.1) and (1.10), results

d

dtE(t) � −k1

2

L∫0

( ∞∫0

g(s)∣∣ηt

x

∣∣2 ds

)dx, (5.1)

d

dtE2(t) � −k1

2

L∫0

( ∞∫0

g(s)∣∣ηt

xt

∣∣2 ds

)dx. (5.2)

We define w as the solution of the Dirichlet problem

−wxx = ψx, w(0) = w(L) = 0,

and we introduce the functional

F1(t) :=L∫

0

[ρ2ψtψ + ρ1ϕtw]dx.

Then the following lemma holds.

Lemma 5.1. For any ε1 > 0 there exists a positive constant Cε1 > 0 such that

d

dtF1(t) � −b1

2

L∫0

ψ2x dx + ε1

L∫0

ϕ2t dx + Cε1

L∫0

ψ2t dx + C

L∫0

( ∞∫0

g(s)∣∣ηt

x

∣∣2 ds

)dx. (5.3)

Proof. Multiplying Eq. (1.6) by ψ we get

d

dt

L∫0

ρ2ψtψ dx = ρ2

L∫0

ψ2t dx − b1

L∫0

ψ2x dx −

L∫0

( ∞∫0

g(s)ηtx ds

)ψx dx

− k

L∫0

ϕxψ dx − k

L∫0

ψ2 dx + ρ2

L∫0

ψ2t dx. (5.4)

Multiplying Eq. (1.5) by w we obtain

d

dt

L∫0

ρ1ϕtw dx = −k

L∫0

ϕψx dx + k

L∫0

w2x dx + ρ1

L∫0

ϕtwt dx. (5.5)

Equations (5.4) and (5.5) lead to

d

dtF1(t) = ρ2

L∫0

ψ2t dx − b1

L∫0

ψ2x dx − k

L∫0

ψ2 dx + k

L∫0

w2x dx

+ ρ1

L∫ϕtwt dx −

L∫ ( ∞∫g(s)ηt

x ds

)ψx dx.


0 0 0



Therefore, using

−L∫

0

( ∞∫0

g(s)ηtx ds

)ψx dx � δ

L∫0

ψ2x dx + Cδ

L∫0

( ∞∫0

g(s)∣∣ηt

x

∣∣2 ds

)dx

our conclusion follows. �Let us denote by K the functional

K(t) := −L∫

0

ρ2ψt

( ∞∫0

g(s)ηt (x, s) ds

)dx.

Let b0 := ∫∞0 g(s) ds. Using Eqs. (1.6)–(1.7) we get

d

dtK(t) = b1

L∫0

ψx

( ∞∫0

g(s)ηtx(t, s) ds

)dx +

L∫0

( ∞∫0

g(s)ηtx(x, s) ds

)2

dx − ρ2b0

L∫0

ψ2t dx

+ k

L∫0

(ϕx + ψ)

( ∞∫0

g(s)ηt (x, s) ds

)dx + ρ2

L∫0

ψt

( ∞∫0

g(s)ηs(x, s) ds

)dx.

SinceL∫

0

ψt

( ∞∫0

g(s)ηs(x, s) ds

)dx = −

L∫0

ψt

( ∞∫0

g′(s)ηs(x, s) ds

)dx

andL∫

0

( ∞∫0

g(s)ηtx(x, s) ds

)2

dx � b0

L∫0

( ∞∫0

g(s)∣∣ηt

x

∣∣2 ds

)dx, (5.6)

using Poincare’s inequality we conclude that, for any ε2 > 0 there exists Cε2 > 0 such that

d

dtK(t) � −ρ2b0

2

L∫0

ψ2t dx + ε2

L∫0

ψ2x dx + ε2

L∫0

ϕ2x dx + Cε2

L∫0

( ∞∫0

g(s)∣∣ηt

x

∣∣2 ds

)dx. (5.7)

Now, we define the functional E(t) as

E(t) := N(E(t) + E2(t)

)+ F1(t) + N2K(t), (5.8)

where N := N(ε1, ε2) > 0. Then, from (5.1)–(5.3) and (5.7) we get

d

dtE(t) � −NK1

2

L∫0

( ∞∫0

g(s)∣∣ηt

xt

∣∣2 ds

)dx −

(Nk1

2+ C + N2Cε2

) L∫0

( ∞∫0

g(s)∣∣ηt

x

∣∣2 ds

)dx

−(

b

2− N2ε2

) L∫0

ψ2x dx −

(N2ρ2b0

2− Cε1

) L∫0

ψ2t dx + ε1

L∫0

ϕ2t dx + N2ε2

L∫0

ϕ2x dx. (5.9)

Also, we define the functional F2(t) as

F2(t) := ρ2

L∫ψt(ϕx + ψ)dx + ρ1b

k

L∫ψxϕt dx + ρ1

k

L∫ ( ∞∫g(s)ηt

x(x, s) ds

)ϕt dx.


0 0 0 0



Under the above conditions we have

Lemma 5.2. For any ε3 > 0, there exists a constant Cε3 > 0 such that

d

dtF2(t) �

[(bψx +

∞∫0

g(s)ηtx(x, s) ds

)ϕx

]x=L

x=0

− k

L∫0

|ϕx + ψ |2 dx + ε3

L∫0

ϕ2t dx

+ ρ2

L∫0

ψ2t dx + Cε3

L∫0

( ∞∫0

g(s)∣∣ηt

x

∣∣2 1ds

)dx + Cε3

L∫0

( ∞∫0

g(s)∣∣ηt

xt

∣∣2 ds

)dx.

Proof. Multiplying Eq. (1.6) by (ϕx + ψ) and using (1.5) we get

d

dtF2(t) =

[(bψx +

∞∫0

g(s)ηtx(x, s) ds

)ϕx

]x=L

x=0

− k

L∫0

|ϕx + ψ |2 dx

+(

ρ1b

k− ρ2

) L∫0

ψxtϕt dx +L∫

0

( ∞∫0

g(s)ηtx(x, s) ds

)ϕt dx. (5.10)

On the other hand, from (1.7) we have

ψxt = ηtxt + ηt

xs .

Recalling that b0 = ∫∞0 g(s) ds, we deduce

b0

L∫0

ψxtϕt dx =L∫

0

( ∞∫0

g(s)ψxt ds

)ϕt dx

=L∫

0

( ∞∫0

g(s)[ηt

xt + ηtxs

]ds

)ϕt dx

=L∫

0

( ∞∫0

g(s)ηtxt ds

)ϕt dx +

L∫0

( ∞∫0

g(s)ηtxs ds

)ϕt dx

=L∫

0

( ∞∫0

g(s)ηtxt ds

)ϕt dx −

L∫0

( ∞∫0

g′(s)ηtx ds

)ϕt dx.

ThereforeL∫

0

ψxtϕt dx = 1

b0

L∫0

( ∞∫0

g(s)ηtxt ds

)ϕt dx − 1

b0

L∫0

( ∞∫0

g′(s)ηtx ds

)ϕt dx. (5.11)

Substituting (5.11) into (5.10) results

d

dtF2(t) =

[(bψx +

∞∫0

g(s)ηtx(x, s) ds

)ϕx

]x=L

x=0

− k

L∫0

|ϕx + ψ |2 dx

+ 1

b0

(ρ1b

k− ρ2

) L∫ ( ∞∫g(s)ηt

xt ds

)ϕt dx − 1

b0

(ρ1b

k− ρ2

) L∫ ( ∞∫g′(s)ηt

x ds

)ϕt dx


0 0 0 0



+L∫

0

( ∞∫0

g(s)ηtx(x, s) ds

)ϕt dx. (5.12)

Finally, using the hypothesis on g in (5.12), our conclusion follows. �Remark. When the boundary conditions are of mixed type, the boundary term in the last lemma is equal to zero. Forboundary conditions of Dirichlet type, we need to estimate this term. In order to deal with the boundary term we shallprove the following lemma.

Lemma 5.3. Let q ∈ C1([0,L]) satisfy q(0) = −q(L) = 2, and let us introduce the following functionals:

J1(t) := ρ2

L∫0

ψtq(x)

(bψx +

∞∫0

g(s)ηtx(x, s) ds

)dx,

J2(t) := ρ1

L∫0

ϕtq(x)ϕx dx.

Then there exist C1 > 0 and, for any ε > 0, a positive constant Cε > 0 such that

(i)

d

dtJ1(t) � −

[(bψx(L, t) +

∞∫0

g(s)ηtx(L, s) ds

)2

+(

bψx(0, t) +∞∫

0

g(s)ηtx(0, s) ds

)2]

+ C1

L∫0

ψ2t dx + ε

L∫0

ϕ2x dx + Cε

L∫0

ψ2x dx + Cε

L∫0

( ∞∫0

g(s)∣∣ηt

x

∣∣2 ds

)dx,

(ii)

d

dtJ2(t) � −k

[ϕ2

x(L, t) + ϕ2x(0, t)

]+ C1

L∫0

[ϕ2

t + ϕ2x + ψ2

x

]dx.

Proof. Using Eqs. (1.6)–(1.7) we get

d

dtJ1(t) = 1

2

[q(x)

(bψx +

∞∫0

g(s)ηtx ds

)2]x=L

x=0

− 1

2

L∫0

q ′(x)

(bψx +

∞∫0

g(s)ηtx(x, s) ds

)2

dx

− k

L∫0

q(x)(ϕx + ψ)

(bψx +

∞∫0

g(s)ηtx(x, s) ds

)dx − ρ2b

2

L∫0

q ′(x)ψ2t dx

+ ρ2

L∫0

ψtq(x)

( ∞∫0

g(s)[−ηxs + ψxt

]ds

)dx

= −[(

bψx(L, t) +∞∫

g(s)ηtx(L, s) ds

)2

+(

bψx(0, t) +∞∫

g(s)ηtx(0, s) ds

)2]


0 0



− 1

2

L∫0

q ′(x)

(bψx +

∞∫0

g(s)ηtx(x, s) ds

)2

dx

− kb

L∫0

q(x)ϕxψx dx − kb

2

L∫0

q ′(x)ψ2 dx − k

L∫0

q(x)ϕx

( ∞∫0

g(s)ηtx(x, s) ds

)dx

− k

L∫0

q(x)ψ

( ∞∫0

g(s)ηtx(x, s) ds

)dx − ρ2(b + 1)

2

L∫0

q ′(x)ψ2t dx

− ρ2

L∫0

q(x)ψt

( ∞∫0

g(s)ηts(x, s) ds

)dx.

Then, using (5.6) and the hypotheses (1.10) on g, conclusion (i) follows. To prove (ii) we use (1.5), that is

d

dtJ2(t) = −k

L∫0

q(x)ϕxxϕx dx + k

L∫0

q(x)ψxϕx dx + ρ1

L∫0

q(x)ϕtϕxt dx

= k

2

[q(x)ϕ2

x

]x=L

x=0 + k

L∫0

q(x)ψxϕx dx − ρ1

2

L∫0

q ′(x)ϕ2t dx. �

We define, for δ > 0 and N > 0, the following functional:

F3(t) := F2(t) + NJ1(t) + δJ2(t). (5.13)

From Lemmas 5.2 and 5.3 we conclude, observing

−k

2

L∫0

|ϕx + ψ |2 dx � −k

4

L∫0

ϕ2x dx + C

L∫0

ψ2x dx (C > 0),

that, for sufficiently small ε, ε3, δ > 0, large N and for any 0 < τ < 1, there exist Cτ > 0 and C2 > 0 such that

d

dtF3(t) � −k

2

L∫0

|ϕx + ψ |2 dx + C2τ

L∫0

ϕ2t dx

+ Cτ

L∫0

[ψ2

x + ψ2t +

∞∫0

g(s)∣∣ηt

x

∣∣2 ds +∞∫

0

g(s)∣∣ηt

xt

∣∣2 ds

]dx. (5.14)

Finally, we define the functional

F4(t) := −L∫

0

[ρ1ϕtϕ + ρ2ψtψ]dx. (5.15)

Using (1.5) and (1.6) we get

d

dtF4(t) � −ρ1

L∫0

ϕ2t dx − ρ2

L∫0

ψ2t dx + k

L∫0

|ϕx + ψ |2 dx + C

L∫0

[ψ2

x +∞∫

0

g(s)∣∣ηt

x

∣∣2 ds

]dx. (5.16)

Choosing τ small enough, we have




d

dt

{F3(t) + 2C2τ

ρ1F4(t)

}� −k

4

L∫0

|ϕx + ψ |2 dx − C2τ

L∫0

ϕ2t dx + Cτ

L∫0

[ψ2

t + ψ2x

]dx

+ Cτ

L∫0

[ ∞∫0

g(s)∣∣ηt

x

∣∣2 ds +∞∫

0

g(s)∣∣ηt

xt

∣∣2 ds

]dx. (5.17)

Now we are in the position to prove the polynomial rate of decay.

Theorem 5.4. Suppose that (1.10) holds and initial data verifies

ϕ0,ψ0 ∈ H 2 ∩ H 10 (0,L), η0 ∈ L2

g

(R

+,H 2 ∩ H 10

)and ϕ1,ψ1 ∈ H 1

0 (0,L).

Then the first-order energy E(t) decays polynomially to zero, that is, there exists a positive constant C, being inde-pendent of the initial data, such that

E(t) � C

t

(E(0) + E2(0)

).

Moreover, if U0 := (ϕ0, ϕ1,ψ0,ψ1, η)′ ∈ D(Ak), then∥∥T (t)U0∥∥H � Ck

tk

∥∥AkU0∥∥H. (5.18)

Proof. We define L(t) as

L(t) := E(t) + μ

{F3(t) + 2C2τ

ρ1F4(t)

}.

Choosing μ,ε1, ε2 small, N2,N large, and using the inequalities (5.9) and (5.17), we get

d

dtL(t) � −αE(t),

for some α > 0. Therefore

α

t∫0

E(s) ds �L(0) −L(t), ∀t � 0. (5.19)

On the other hand, it is not difficult to prove that there exists a constant β > 0 such that

L(0) −L(t) � β(E(0) + E2(0)

), ∀t � 0. (5.20)

From (5.19)–(5.20) we obtain

t∫0

E(s) ds � β

α

(E(0) + E2(0)

). (5.21)

Finally, since

d

dt

{tE(t)

}= E(t) + td

dtE(t) � E(t),

from (5.21) we get

E(t) � C

t

(E(0) + E2(0)

),

where C := βα

> 0. Finally, if U0 ∈ D(Ak), we use Prüss’ results [8] to obtain (5.18), which completes the proof. �



References

[1] F. Ammar Khodja, A. Benabdallah, J.E. Muñoz Rivera, R. Racke, Energy decay for Timoshenko systems of memory type, J. DifferentialEquations 194 (1) (2003) 82–115.

[2] J.U. Kim, Y. Renardy, Boundary control of the Timoshenko beam, SIAM J. Control Optim. 25 (6) (1987) 1417–1429.[3] Z. Liu, S. Zheng, Semigroups Associated with Dissipative Systems, Res. Notes Math., vol. 398, Chapman & Hall/CRC, Boca Raton, 1999.[4] J.E. Muñoz Rivera, R. Racke, Mildly dissipative nonlinear Timoshenko systems—Global existence and exponential stability, J. Math. Anal.

Appl. 276 (2002) 248–278.[5] J.E. Muñoz Rivera, R. Racke, Global stability for damped Timoshenko systems, Discrete Contin. Dyn. Syst. 9 (2003) 1625–1639.[6] A. Pazy, Semigroups of Linear Operators and Applications to Partial Differential Equations, Springer-Verlag, New York, 1983.[7] J. Prüss, On the spectrum of C0-semigroups, Trans. Amer. Math. Soc. 284 (2) (1984) 847–857.[8] J. Prüss, A. Bátkai, K. Engel, R. Schnaubelt, Polynomial stability of operator semigroups, Math. Nachr. 279 (2006) 1425–1440.[9] A. Soufyane, Stabilisation de la poutre de Timoshenko, C. R. Acad. Sci. Paris Sér. I 328 (1999) 731–734.


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