Rates of decay to non homogeneous Timoshenko model with ...

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J. Differential Equations ••• (••••) •••–•••www.elsevier.com/locate/jde

Rates of decay to non homogeneous Timoshenko model

with tip body ✩

Jaime E. Muñoz Rivera a,∗, Andrés I. Ávila b

a Laboratório Nacional de Computação Científica, Petrópolis, RJ, Brazilb Departamento de Ingeniería Matemática, CMCC, Universidad de La Frontera, Temuco, Chile

Received 30 October 2013; revised 30 December 2014

Abstract

We consider the uniform stabilization of a hybrid elastic model consisting of a Timoshenko beam and a tip load at the free end of the beam. Our main result proves that the semigroup eAt associated to this model is not exponentially stable. Moreover, we prove that the semigroup decays polynomially to zero as t−1/2. When the damping mechanism is effective only on the boundary of the rotational angle, the solution also decays polynomially as t−1/2 provided the wave speeds are equal. Otherwise it decays as t−1/4 for any initial data taken in D(A).© 2015 Elsevier Inc. All rights reserved.

MSC: 35L20; 35P05

Keywords: Uniform stabilization; Polynomial decay; Timoshenko’s beam; Tip body; Semigroup theory; Essential radius

1. Introduction

Understanding and measuring the vibration of beams has been a standard problem in Struc-tural Engineering. Since the seminal book of Timoshenko [30], different elastic bars are modeled as an infinite large system of particles. Indeed, obtaining a uniform stabilization is one of the most required properties of the system. Among different beam models, Timoshenko model has

✩ This work was supported by FONDECYT 1080439 and MEC 80100007.* Corresponding author.

E-mail address: [email protected] (J.E. Muñoz Rivera).

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2 J.E. Muñoz Rivera, A.I. Ávila / J. Differential Equations ••• (••••) •••–•••

Fig. 1. Timoshenko beam with tip body and parameters.

proved to be simple and it considers enough complexity to include shear deformation. We are interested in the stabilization of the Timoshenko beam with one clamped end and the other end rigidly attached to a tip body modeling a sealed container with a granular material, for example sand. This granular material provides damping to the motion of the system by internal friction. The physical setting of the problem is shown in Fig. 1.

Consider a bar of length L and let ϕ be the deflection from equilibrium and ψ the slope of the deflection curve. The Timoshenko model can be written for (x, t) ∈]0, L[×]0, T [ as

ρ1(x)ϕtt = Sx, ρ2(x)ψtt = Mx − S, (1.1)

where

S = κ(x)(ϕx + ψ), M = b(x)ψx, (1.2)

with ρ1 = ρA, ρ2 = ρI , κ = KAG, b = EI . Here S stands for the shear force, M for the bend-ing moment, ρ denotes the density, A the cross-sectional area, I is the area moment of inertia, K is the shear coefficient for measuring the stiffness of materials (K < 1), E and G are elastic constants. We also consider the initial conditions

ϕ(x,0) = ϕ0(x), ϕt (x,0) = ϕ1(x), ψ(x,0) = ψ0(x), ψt (x,0) = ψ1(x), (1.3)

and Dirichlet boundary condition at the clamped left end x = 0

ϕ(0, t) = ψ(0, t) = 0. (1.4)

For the right end, we assume that the container is rigidly attached at x = L with mass m and a center of mass O ′ located at distance d from the beam end.

We assume that the damping effect of the internal granular material can be represented by damping coefficients k0 and k1 for ϕ and ψ respectively. The force balance at the end x = L is

mϕtt (L, t) + k0ϕt (L, t) + κ(L)(ϕx(L, t) + ψ(L, t)

)︸︷︷︸ = 0. (1.5)

=S(L,t)


J.E. Muñoz Rivera, A.I. Ávila / J. Differential Equations ••• (••••) •••–••• 3

The first two terms are the contribution of the inertia of the tip body. Indeed, the precise vertical component of the inertial term of the container is m(ϕ + d sin θ)tt where θ is the angle between the

−−−−→OO ′ segment and the x-axis. But in the linear approximation we neglect d sin θ . The second

term represents the damping that the granular material provides, which is assumed to be propor-tional to the velocity with damping coefficient k0. The last two terms represent the shear force S(x, t). Similarly for ψ , we obtain the second right end boundary condition

Imψtt (L, t) + k1ψt(L, t) + b(L)ψx(L, t)︸︷︷︸=M(L,t)

= 0, (1.6)

where Im is the inertial moment of the tip and M(x, t) is the bending moment.Several authors have studied the exponential stability of the system. In the seminal paper

[13], Kim and Renardy used two right end control by time derivatives of ϕ and ψ . See also [36]. Bassam et al. [6] study the case of partial dissipation by introducing only one boundary control effective over the rotation-angle equation. Other example of partial dissipation was given in [2], where the authors introduce boundary controls in both sides of the beam. Recently, Han and Xu [11] extended the above result for time delay boundary feedbacks. For memory dissipation see [5,8,17,20,23,24]. For other types of dissipations, see [12,19,21,22,25,35,37].

To get a better understanding of the Timoshenko model, some authors have studied: spectral asymptotic formulas [1,29,32], distributed port Hamiltonian [18], and computing eigenvalues and eigenvectors [31]. Another way for studying this problem has been numerical approaches. First, finite element methods were used for computing eigenvalues and eigenvectors [38] and improvements were made in [34]. The thermo-viscoelastic tip body system was also solved by finite elements [4]. Next, separation of variables were also used in [27]. Li et al. [15,16] used reduction order for obtaining a difference scheme. On the other hand, an interesting and recent application of nanotubes as micro-mass sensors for tip masses was studied in [28].

In case of tip body, Andrews and Shillor [3] showed that damping is critical for the exponen-tial stabilization of the energy. We also quote the work [10], where the author claims to prove exponential stability to system (1.1)–(1.6) provided condition Z [10, p. 395] is valid. The point is that such condition assumes that the solution verifies a convenient inequality that is not possible to check and maybe is not correct in general.

We consider the asymptotic behavior of the hybrid Timoshenko model. Our main result shows that system (1.1)–(1.6) is not exponentially stable. Therefore it is interesting to search for others types of decay. In this framework, we prove that the semigroup S(t) = eAt associated to system (1.1)–(1.6), defined over H, decays polynomially as t− 1

2 , that is,

∥∥S(t)U0∥∥H ≤ ct−1/2‖U0‖D(A),

where A stands for the infinitesimal generator of S(t) and D(A) stands for the domain of A. Using standard arguments the above inequality can be extended to

∥∥S(t)U0∥∥H ≤ ct−k/2‖U0‖D(Ak).

Indeed, the more regularity the initial data, the faster the energy decays. We also consider the case of partial dissipation effective only on the rotational angle ψ . To give a fully characterization of the polynomial decay we introduce the number



χ := k

ρ1− b

ρ2. (1.7)

Where χ denotes the difference of the speed of propagation of the waves equations in (1.1). We prove that if χ = 0 then the solution decays polynomially as t− 1

2 for any initial data in D(A) in general we have ∥∥S(t)U0

∥∥H ≤ ct−k/2‖U0‖D(Ak).

Otherwise, if χ �= 0 then the solution decays as∥∥S(t)U0∥∥H ≤ ct−k/4‖U0‖D(Ak).

Here we use the Borichev and Tomilov’s result [7] (see Theorem 3.1 stated below).Finally, let us denote by ω(S), ωσ (A) and ωess(S) the type of the semigroup S(t), the spec-

trum upper bound of A, and the essential type of the semigroup S(t) respectively. Note that

ω(S) = max{ωσ (A),ωess(S)

}. (1.8)

To show the lack of exponential stability, that is ω(S) = 0, we use the Weyl’s Theorem [26, Theorem XIII.14], [33]: If the difference of two operators is compact, then the essential spec-trum radii are the same. Using Weyl’s Theorem we are able to show that ωess(S) = 0. Since ωσ (A) ≤ 0 for any dissipative operator, using (1.8) we get the result.

This paper is organized as follows. In Section 2, we show the well possedness of the model. In Section 3, we show the polynomial decay of the semigroup associated to the Timoshenko’s model with tip. In Section 4, we consider the case of partial dissipation. Finally, in Section 5 we prove the lack of the exponential stability.

2. The semigroup approach

To define the semigroup problem associated to (1.1)–(1.4), we consider the right end boundary conditions

ϕt (L, t) = u(t), ψt (L, t) = v(t) (2.1)

where u and v solve the system

mut(t) + k0u(L, t) + κ(L)(ϕx(L, t) + ψ(L, t)

) = 0, (2.2)

Imvt (t) + k1v(t) + b(L)ψx(L, t) = 0, (2.3)

with the initial conditions

u(0) = ϕ1(L), v(0) = ψ1(L). (2.4)

Let us denote by U := (ϕ, ϕt , ψ, ψt, u, v)′, then U satisfies the following Cauchy problem

Ut =AU, U(t = 0) = U0, (2.5)



where U0 := (ϕ0, ϕ1, ψ0, ψ1, u0, v0)′ and A : D(A) ⊆ H → H is given by

A =

⎛⎜⎜⎜⎜⎜⎜⎜⎝

0 I 0 0 0 01ρ1

[κ/∂x]x 0 κ/ρ1∂x 0 0 00 0 0 I 0 0

−κ/ρ2∂x 0 1ρ2

[b∂x]x − κ/ρ2 0 0 0

− 1m

γ1 0 − 1m

γ2 0 − k0m

I 0

0 0 − 1Im

γ3 0 0 − k1Im

I

⎞⎟⎟⎟⎟⎟⎟⎟⎠

, (2.6)

where γi is a trace operator for i = 1, 2, 3 given by

γ1(ϕ) = k(L)ϕx(L), γ2(ψ) = k(L)ψ(L), γ3(ψ) = b(L)ψx(L).

By H we denote the phase space associated to system (1.1), that is,

H := H 1∗ (0,L) × L2(0,L) × H 1∗ (0,L) × L2(0,L) × C2, (2.7)

where H 1∗ (0, L) = {w ∈ H 1(0, L) : w(0) = 0}. Therefore, the operator domain will be

D(A) := {U ∈H;AU ∈H,Φ(L) = u,Ψ (L) = v

}, (2.8)

and the inner product in H is defined as

(U1,U2)H :=L∫

0

[ρ1Φ1Φ2 + ρ2Ψ1Ψ 2 + bψ1,xψ2,x + κ(ϕ1,x + ψ1)(ϕ2,x + ψ2)

]dx

+ mu1(L)u2(L) + Imv1(L)v2(L), (2.9)

where Uj = (ϕj , Φj , ψj , Ψj , uj , vj )′ ∈H, j = 1, 2. The corresponding norm is

‖U‖2H = ∥∥(ϕ,Φ,ψ,Ψ,u, v)′

∥∥2H ≡

L∫0

[ρ1|Φ|2 + ρ2|Ψ |2 + b|ψx |2 + κ|ϕx + ψ |2]dx

+ m∣∣u(L)

∣∣2 + Im

∣∣v(L)∣∣2

. (2.10)

It is not difficult to see that D(A) is dense in H and A is dissipative, that is, a straightforward calculations show that

Re(AU,U)H = −k0|u|2 − k1|v|2 ≤ 0. (2.11)

Now, consider the resolvent equation

iλU −AU = F,



where U = (ϕ, Φ, ψ, Ψ, u, v)′ ∈ H and F = (f1, f2, f3, f4, f5, f6)′ ∈ H. Taking inner product

and its real part in the above equation and using identity (2.11), we get

k0|u|2 + k1|v|2 = Re (U,F )H. (2.12)

Rewriting the resolvent equation in terms of the components, we get

λϕ − Φ = f1, λρ1Φ − [κ(ϕx + ψ)

]x

= f2, (2.13)

λψ − Ψ = f3, λρ2Ψ − [bψx]x + κ(ϕx + ψ) = f4, (2.14)

mλu + k0u + κ(L)(ϕx + ψ)(L) = f5, Imλv + k1v + b(L)ψx(L) = f6, (2.15)

ϕ(0) = 0, ψ(0) = 0. (2.16)

Theorem 2.1. The operator A is the infinitesimal generator of a C0 semigroup of contractions.

Proof. We will show that 0 ∈ ρ(A). In fact, taking λ = 0 in (2.13)–(2.15) we get

−[κ(ϕx + ψ)

]x

= f2 ∈ L2(0, l), −[bψx]x + κ(ϕx + ψ) = f4 ∈ L2(0, l),

κ(L)(ϕx(L) + ψ(L)

) = −k0f1(L) + f5 ∈ C, b(L)ψx(L) = −k1f3(L) + f6 ∈ C,

ϕ(0, t) = ψ(0, t) = 0.

With −Φ = f1, −Ψ = f3. Standard computations show that the above problem is well posed implying that the operator A is a bijection between D(A) and the space H. Since A is closed, by the Closed Graph Theorem, we conclude that 0 ∈ ρ(A). Since A is dissipative, we have that A is the infinitesimal generator of a C0 semigroup. See [14]. �3. Polynomial rate of decay

Our result on the polynomial stability is based on the recent work of Borichev and Tomilov [7]. Here, we rephrase their Theorem 4.2 in a compact form.

Theorem 3.1. Let S(t) be a bounded C0-semigroup on a Hilbert space H with generator A such that iR ⊂ ρ(A). Then

1

|η|α∥∥(iηI −A)−1

∥∥ ≤ C, ∀η ∈R ⇔ ∥∥S(t)A−1∥∥ ≤ c

t1/α.

To consider the non homogeneous model, let us introduce the following notations:

Iϕ(β) = ρ1q1κ∣∣Φ(β)

∣∣2 + q1∣∣κ(

ϕx(β) + ψ(β))∣∣2

, (3.1)

Iψ(β) = ρ2q2b∣∣Ψ (β)

∣∣2 + q2∣∣bψx(β)

∣∣2, (3.2)

N 2 =L∫

0

ρ1|Φ|2 + ρ2|Ψ |2 + b|ψx |2 + κ|ϕx + ψ |2dx. (3.3)



Lemma 3.2. Let us suppose that ϕ, Φ , ψ , Ψ are strong solutions to the system (2.13)–(2.15). Then there exists a positive constant c such that

N 2 ≤ c(Iφ(β) + Iψ(β) + ‖F‖2) and Iφ(β) + Iψ(β) ≤ c

(N 2 + ‖F‖2),

for β = 0, L.

Proof. We will prove the inequalities for β = L. The case β = 0 is similar. Let λ ∈ iR and q1 ∈ C2(0, L) such that q1(0) = 0. Multiplying the second equation in (2.13) by q1S, where S = κ(ϕx + ψ) we get

L∫0

λρ1Φq1S −L∫

0

Sxq1S ds =L∫

0

f2q1S ds. (3.4)

Using (2.14), the first term can be decomposed as

L∫0

λρ1Φq1S ds = −L∫

0

ρ1Φq1κλϕx −L∫

0

ρ1Φq1κλψ ds

= −L∫

0

ρ1Φq1κΦx −L∫

0

ρ1q1κΦΨ ds −L∫

0

ρ1q1κΦ[f3,x + f4]ds

︸︷︷︸:=R1

.

Therefore, taking the real part we get

Re

L∫0

λρ1Φq1S ds = −1

2

L∫0

ρ1q1κd

dx|Φ|2 ds − Re

L∫0

ρ1q1κΦΨ ds + R1,

where R1 contains terms with f3 and f4 and satisfies

|R1| ≤ cN‖F‖.

Using (3.4), it follows that

Iϕ(t) − 1

2

L∫0

([ρ1q1κ]x |Φ|2 + q1,x

∣∣κ(ϕx + ψ)∣∣2)

dx − Re

L∫0

ρ1q1κΦΨ dx = R2, (3.5)

where R2 contains the term f2 and

|R2| ≤ cN‖F‖.



Similarly, let us take q2 ∈ C2(0, L) such that q2(0) = 0 and multiplying the second equation in (2.14) by q2bψx we get

L∫0

λρ2Ψ q2bψx dx

︸︷︷︸J1

−L∫

0

[bψx]xq2bψx dx

︸︷︷︸:=J2

+L∫

0

κ(ϕx + ψ)q2bψx dx

︸︷︷︸J3

=L∫

0

f4q2bψx dx.

Consider the first term and using the first equation of (2.14) with ψx ,

ReJ1 = −Re

L∫0

ρ2Ψ q2b(Ψx + f3,x) ds

= −1

2Re

L∫0

ρ2q2bd

dx|Ψ |2 ds − Re

L∫0

ρ2Ψ q2bf3,x ds.

Now, integrating by parts the third term and using (2.13) and (2.14), we get

J3 = −L∫

0

[κ(ϕx + ψ)

]xq2bψ dx −

L∫0

κ(ϕx + ψ)[q2b]xψ dx + P(L)

=L∫

0

ρ1Φq2b(Ψ + f3) dx + 1

λ

L∫0

κ(ϕx + ψ)[q2b]x(Ψ + f3) dx + R3 + P(L)

=L∫

0

ρ1q2bΦΨ dx + P(L) + R4,

where P(L) = q2(L)bψ(L)k(L)(ϕx + ψ)(L) and R4 is such that

|R4| ≤ cN‖F‖ + c

|λ|N2.

Summing J1, J2 and J3 it follows from the last two equalities that

Iψ(L) −L∫

0

([ρ2q2b]x |Ψ |2 + q2,x |bψx |2)dx + Re

L∫0

ρ1q2bΦΨ dx

= −Re(P(L) + R4

). (3.6)



Then, we choose q1 and q2 such that

ρ1q2b =x∫

0

ens ds, ρ1q1κ =x∫

0

ens ds.

Adding up the inequalities (3.5) and (3.6), we conclude that

L∫0

[ρ1q1κ]x |Φ|2 + q1,x

∣∣κ(ϕx + ψ)∣∣2 + [ρ2q2b]x |Ψ |2 + q2,x |bψx |2 ds = Iψ(L) + Iφ(L) + R5,

where R5 is such that

|R5| ≤ cN‖F‖ + c

|λ|N2.

Denote by q3 := ρ2q2b = ρ2ρ1

∫ x

0 ens ds. Therefore, we must estimate the derivative of qi(x) =γi(x)

∫ x

0 ens ds, where γ satisfies γ0 ≤ γ (x) ≤ γ1 for 0 < γ0 < γ1. For n large enough, we get

qx = enxγ + γ ′(x)

x∫0

ens ds = nenxγ + γ ′[enx − 1]n

≥ 1

n

([nγ0 − γ ′︸︷︷︸

> 3n4 γ0

]enx + γ ′) ≥ 1

n

(3

4nγ0e

nx − 2‖γ ′‖∞)

≥ γ0

2exn.

This implies that there exists a constant c2 > 0 such that

L∫0

[ρ1q1κ]x |Φ|2 + q1,x

∣∣κ(ϕx + ψ)∣∣2 + [ρ2q2b]x |Ψ |2 + q2,x |bψx |2 ds ≥ c2

nN 2.

By the definition of N 2

N 2 ≤ c0

(Iψ + Iφ +N‖F‖ + 1

|λ|N2)

,

where c0 does not depend on n. Also, we have

Iψ + Iφ ≤ c0(N 2 +N‖F‖).

Taking λ large enough, our conclusion follows. �Theorem 3.3. The solution of the Timoshenko system with tip body decays polynomially as

∥∥U(t)∥∥ ≤ 1√

t‖U0‖D(A).



Proof. First we show that iR ⊂ �(A). In fact, since A is a closed operator and D(A) has com-pact embedding on the phase space H, we conclude that the spectrum σ(A) is discrete. Therefore we show that there are no imaginary eigenvalues. By contradiction, let us suppose that there ex-ists an imaginary eigenvalue then iλ ∈ σ(A). Then there exists U �= 0 satisfying

AU = iλU.

Taking inner product in H with U and using (2.11), we get that

u = v = 0, Ψ (L) = Φ(L) = 0.

This implies that

ψ(L) = φ(L) = φx(L) = ψx(L) = 0.

We consider system (2.13)–(2.15) with F = 0 as an initial value problem with null data, therefore φ = ψ = 0, which implies that U = 0. But this is a contradiction with our assumption, then we have that iR ⊂ �(A).

Finally, using (2.11) and Lemma 3.2 we conclude for λ large

‖U‖2 ≤ cIφ + cIψ + c‖F‖2

≤ c|λ|2|u|2 + c|λ|2|v|2 + c‖F‖2

≤ c|λ|2‖U‖‖F‖ + c‖F‖2.

For a constant C > 0, we conclude that

1

|λ|2 ‖U‖ ≤ C.

Finally, using Theorem 3.1 our result follows. �4. Partial dissipation

In applications it is important to reduce dissipative mechanism. Therefore, in this section we assume that dissipative properties of the system are produced only by the bending moment of the beam (see [2]) therefore model (1.1)–(1.4) changes to

ρ1ϕtt − κ(ϕx + ψ)x = 0, ρ2ψtt − bψxx + κ(ϕx + ψ) = 0, (4.1)

with the boundary conditions at the right end x = L

S(L, t) + αϕ(L, t) = 0 (4.2)

M(L, t) + Imψtt (L, t) + k1ψt(L, t) = 0 (4.3)



and for the left end x = 0

ϕ(0, t) = 0 (4.4)

M(0, t) − Imψtt (0, t) − k2ψt(0, t) = 0 (4.5)

with initial conditions given by (1.3). Here we assume that all coefficients are constant. Note that system (4.1)–(4.5) has no dissipative mechanism effective over the transversal deflection, even then we show that the corresponding semigroup also decays polynomially. Our starting point is to show the well possedness of system (4.1)–(4.5). To do that let us denote by U :=(ϕ, ϕt , ψ, ψt, v, w)′, then U satisfies the following Cauchy problem

Ut = AU, U(t = 0) = U0, (4.6)

where U0 := (ϕ0, ϕ1, ψ0, ψ1, u0, v0, w0)′ and A : D(A) ⊆ H → H is given by

A=

⎛⎜⎜⎜⎜⎜⎜⎜⎝

0 I 0 0 0 0κρ1

∂xx 0 κ/ρ1∂x 0 0 00 0 0 I 0 0

−κ/ρ2∂x 0 bρ2

∂xx − κ/ρ2 0 0 0

0 0 − 1Im

γ3 0 − k1Im

I 0

0 0 1Im

γ4 0 0 k2Im

I

⎞⎟⎟⎟⎟⎟⎟⎟⎠

, (4.7)

where γ3(ψ) = bψx(L), and γ4(ψ) = bψx(0). By H we denote the space

H := H 1∗ (0,L) × L2(0,L) × H 1(0,L) × L2(0,L) × C2, (4.8)

which is a Hilbert space with the norm,

‖U‖2H = ∥∥(ϕ,Φ,ψ,Ψ,v,w)′

∥∥2H ≡

L∫0

[ρ1|Φ|2 + ρ2|Ψ |2 + b|ψx |2 + κ|ϕx + ψ |2]dx

+ α|ϕ(L)|2 + Im|v|2 + Im|w|2.

Therefore the operator domain for A will be

D(A) := {U ∈ H; AU ∈H, S(L) + αϕ(L) = 0,Ψ (L) = v,Ψ (0) = w

}. (4.9)

Following the computations as in Section 2, we get

Re(AU,U

)H = −k1|v|2 − k2|w|2 ≤ 0 (4.10)

and the resolvent equation is given by



λϕ − Φ = f1, λρ1Φ − k(ϕx + ψ)x = f2 (4.11)

λψ − Ψ = f3, λρ2Ψ − bψxx + κ(ϕx + ψ) = f4 (4.12)

Imλv + k1v + bψx(L) = f5, Imλw + k2w + bψx(0) = f6 (4.13)

S(L) − αϕ(L) = 0, ϕ(0) = 0. (4.14)

Now, condition (2.12) reads

k1|v|2 + k2|w|2 = Re(U,F )H, (4.15)

where F = (f1, f2, f3, f4, f5, f6). Therefore we have

Theorem 4.1. The operator A is the infinitesimal generator of a C0 semigroup of contractions.

Proof. As in Theorem 2.1 we prove that there exists only one solution to system (4.1)–(4.2) for λ = 0 we have −Φ = f1, −Ψ = f3 and

−κ(ϕx + ψ)x = f2, −bψxx + κ(ϕx + ψ) = f4, (4.16)

with the following boundary conditions

ϕ(0) = S(L) + αϕ(L) = 0, bψx(0) = f6 + k2f3(0), bψx(L) = f5 + k1f3(L).

The above problem is well posed. In fact, let us denote by m(x) the function, m(x) = Ax2 +Bx, where A and B are such that

bmx(0) = bψx(0), bmx(L) = bψx(L).

Denote by ψ = ψ − m(x), the above system can be written as

−[κ(ϕx + ψ

)]x

= f, −bψxx + κ(ϕx + ψ

) = g, (4.17)

ϕ(0) = S(L) + αϕ(L) = 0, bψx(0) = 0, bψx(L) = 0, (4.18)

with f = f2 − κmx and g = f4 + bmxx − κm. To show the well possedness of the problem, we consider the Hilbert space V = H 1

0 (0, L) × H 1(0, L) and the bilinear form

a(ϕ1, ϕ2,ψ1,ψ2) =

L∫0

κ(ϕ1

x + ψ1)(ϕ2x + ψ2

) + bψ1xψ2

x dx + αϕ1(L)ϕ2(L).

Note that a is continuous and symmetric. To check the coercivity, let us consider the identity

−κϕxx = κψx − κ(ϕxx + ψx)



and multiply by ϕ to get

αϕ(L)2 + κϕ(L)ψ(L) + κ

L∫0

|ϕx |2 dx = −κ

L∫0

ψxϕ dx + κ

L∫0

(ϕx + ψ)ϕx dx + αϕ(L)2.

Using Poincaré’s inequality, there exists a constant c > 0 such that

L∫0

|ϕx |2 dx ≤ c

L∫0

|ψx |2 dx + c

L∫0

|ϕx + ψ |2 dx + cε

∣∣ϕ(L)∣∣2 + ε

∣∣ψ(L)∣∣2

. (4.19)

On the other hand, note that

L∫0

|ψ |2 dx =L∫

0

|ϕx + ψ |2 dx −L∫

0

|ϕx |2 dx − 2Re

L∫0

ϕxψ dx

≤L∫

0

|ϕx + ψ |2 dx + c

L∫0

|ϕx |2 dx + 1

2

L∫0

|ψ |2 dx.

It follows that

L∫0

|ψ |2 dx ≤ c

L∫0

|ϕx + ψ |2 dx + c

L∫0

|ϕx |2 dx.

Using (4.19) and Poincaré’s inequality, we get

L∫0

|ψ |2 + |ϕ|2 dx ≤ c

L∫0

|ϕx + ψ |2 dx + c

L∫0

|ψx |2 dx + cε |ϕ(L)|2 + ε∣∣ψ(L)

∣∣2.

Since

∣∣ψ(L)∣∣2 ≤ c

L∫0

|ψ |2 + |ψx |2 dx,

for ε small enough we get

L∫0

|ψ |2 + |ϕ|2 dx ≤ c

L∫0

|ϕx + ψ |2 dx + c

L∫0

|ψx |2 dx + cε

∣∣ϕ(L)∣∣2

.

Therefore a is coercive. From Lax–Milgram Lemma, the well possedness follows. In particu-lar 0 ∈ ρ(A) hence A generates a C0 semigroup of contractions. �



Now we are in condition to show the main result of this section

Theorem 4.2. The semigroup defined by Timoshenko system (4.1)–(4.2) decays polynomially as

∥∥eAtU0∥∥ ≤ c

t1/2‖U0‖D(A),

provided χ = 0, where χ is given in (1.7). Otherwise

∥∥eAtU0∥∥ ≤ c

t1/4‖U0‖D(A).

Proof. To prove that iR ⊂ �(A) we proceed as in Theorem 3.3. Let us suppose that there is an imaginary eigenvalue iλU − AU = 0. Using (4.10), we have that v = w = 0. From relations (4.12), (4.13) and recalling that F = 0, we have that ψ satisfies

ψxx + α2ψ = −κ

bϕx, where bα2 = λ2ρ2 − κ (4.20)

with boundary conditions

ψ(0) = ψx(0) = ψ(L) = ψx(L) = 0.

Multiplying Eq. (4.20) by ψx , integrating over ]0, L[, and taking the real part, it follows that

Re

L∫0

ϕxψx dx = 0. (4.21)

Multiplying Eq. (4.12) by ψ , integrating over ]0, L[, and taking the real part, we find that

Re

L∫0

ΦΨ dx = 0 ⇒ Re

L∫0

ϕψ dx = 0. (4.22)

Multiplying Eq. (4.20) by ϕ, integrating over ]0, L[, taking the real part, and using (4.21)and (4.22), we get ϕ(L) = 0. Therefore, because of the boundary condition (4.18) we get that ϕx(L) = 0. From Lemma 3.2 applied to β = L, we conclude that ϕ = ψ = 0. Thus U = 0, which is a contradiction. Therefore there are no imaginary eigenvalues. Finally, we will show that the resolvent operator is uniformly bounded over the imaginary axis. Multiplying Eq. (2.13) by ψand (2.14) by ψx we get

iρ1λΦψ − κϕxxψ − kψxψ = f2ψ, (4.23)

iρ2λΨ ψx − bψxxψx + κϕxψx + κψψx = f4ψx. (4.24)



Integrating by parts, taking the real part, and using (4.2) we get

Re

L∫0

iρ1λΦψ dx + κRe

L∫0

ϕxψx dx + ReAψ = Re

L∫0

f2ψ dx (4.25)

Re

L∫0

κϕxψx dx + Bψ = Re

L∫0

f4ψx dx, (4.26)

where

Aψ = κ

2

∣∣ψ(L)∣∣2 + κ

2

∣∣ψ(0)∣∣2 + κϕx(0)ψ(0) + αReϕ(L)ψ(L),

Bψ = ρ2∣∣Ψ (L)

∣∣2 − ρ2∣∣Ψ (0)

∣∣2 + b∣∣ψx(L)

∣∣2 − b∣∣ψx(0)

∣∣2 + κ∣∣ψ(L)

∣∣2 − κ∣∣ψ(0)

∣∣2.

Because of the boundary conditions of ψ and relation (4.15), we get

|Bψ | ≤ c|λ|2‖U‖‖F‖. (4.27)

Using Lemma 3.2 we get

|Aψ | ≤ α∣∣ϕ(L)ψ(L)

∣∣ + ∣∣κϕx(0)ψ(0)∣∣ + c‖U‖‖F‖

≤ α∣∣ϕ(L)ψ(L)

∣∣ + c‖U‖∣∣ψ(0)∣∣ + c‖U‖‖F‖

for λ large enough. Therefore, from (4.26) we have that

∣∣∣∣∣Re

L∫0

ϕxψx dx

∣∣∣∣∣ ≤ c|λ|2‖U‖‖F‖ (4.28)

for λ large enough. Using the above inequality into (4.25), we get that

∣∣∣∣∣Re

L∫0

ΦΨ dx

∣∣∣∣∣ ≤ c|λ|2‖U‖‖F‖ + c‖U‖∣∣ψ(0)∣∣ + α

∣∣ϕ(L)ψ(L)∣∣, (4.29)

which implies that

∣∣∣∣∣Re

L∫0

φψ dx

∣∣∣∣∣ ≤ c‖U‖‖F‖ + α

|λ|2∣∣ϕ(L)ψ(L)

∣∣ + c

|λ|2 ‖U‖∣∣ψ(0)∣∣ + c‖F‖2. (4.30)



Multiplying Eq. (2.13) by ψ and taking real part, we get

−Re

L∫0

ΦΨ dx + k

ρ1Re

L∫0

ϕxψx dx = R6. (4.31)

Multiplying Eq. (2.14) by ϕ and taking real part, we get

−Re

L∫0

Ψ Φ dx + b

ρ2Re

L∫0

ψxϕx dx + κ

ρ2

∣∣ϕ(L)∣∣2 + κ

ρ2Re

L∫0

ψϕ dx = R7 (4.32)

where Ri for i = 6, 7 is such that

|Ri | ≤ c‖U‖‖F‖

Taking the difference of (4.31) and (4.32) us using that k/ρ1 = b/ρ2 we get that

∣∣ϕ(L)∣∣2 ≤ c‖U‖‖F‖ + α

|λ|2∣∣ϕ(L)ψ(L)

∣∣ + c

|λ|2 ‖U‖∣∣ψ(0)∣∣ + c‖F‖2,

which implies that

∣∣Φ(L)∣∣2 ≤ c|λ|2‖U‖‖F‖ + c‖U‖∣∣ψ(0)

∣∣ + c|λ|2‖F‖2.

From (4.14) we also have that

∣∣ϕx(L)∣∣2 ≤ c|λ|2‖U‖‖F‖ + c‖U‖∣∣ψ(0)

∣∣ + c|λ|2‖F‖2.

Then using Lemma 3.2 for β = L we get

L∫0

|Ψ |2 + |ψx |2 + |Φ|2 + |ϕx |2 ds ≤ cIψ(L) + cIφ(L) + c‖U‖‖F‖.

Thus,

L∫0

|Ψ |2 + |ψx |2 + |Φ|2 + |ϕx |2 ds ≤ c|λ|2‖U‖‖F‖

and for λ large enough

‖U‖2 ≤ c|λ|4‖F‖2.



Using Theorem 3.1, the first part of this theorem follows. In case of χ �= 0, from identity (4.32)and inequalities (4.28)–(4.30) we get

∣∣ϕ(L)∣∣2 ≤ c|λ|2‖U‖‖F‖ + c‖F‖2 + α

∣∣ϕ(L)ψ(L)∣∣ + c‖U‖∣∣ψ(0)

∣∣,which implies that

∣∣Φ(L)∣∣2 ≤ c|λ|4‖U‖‖F‖ + c|λ|2‖F‖2 + c‖U‖∣∣Ψ (0)

∣∣.From (4.14) we also have that

∣∣ϕx(L)∣∣2 ≤ c|λ|4‖U‖‖F‖ + c|λ|2‖F‖2 + c‖U‖∣∣Ψ (0)

∣∣.For λ large. Then using Lemma 3.2 we get

L∫0

|Ψ |2 + |ψx |2 + |Φ|2 + |ϕx |2 ds ≤ cIψ(L) + cIφ(L) + c‖U‖‖F‖.

Using the same above reasoning we get

L∫0

|Ψ |2 + |ψx |2 + |Φ|2 + |ϕx |2 ds ≤ c|λ|4‖U‖‖F‖

and for λ large enough

‖U‖2 ≤ c|λ|8‖F‖2.

Our result follows from this inequality and Theorem 3.1. �5. The lack of exponential stability

In this section we prove the lack of exponential stability of the Timoshenko system (1.1)–(1.6). The proof of the lack of exponential stability to system (4.1)–(4.5) is similar. Let us recall the definition of the type of a semigroup eAt ,

ω(A) := limt→∞

ln(‖eAt‖)t

= inft>0

ln(‖eAt‖)t

.

Note that ω(A) = 0 implies ‖eAt‖ = 1, therefore to prove the lack of exponential stability, it is enough to show that ω(A) = 0. Note that the spectral radius of the semigroup Rσ (eAt ) is given by

Rσ

(eAt

) = eω(A)t .



It is well known that the radius of the essential spectrum is invariant by compact perturbations

ress(S) = ress(S + K),

where K is a compact operator, ress(S) is the essential spectrum radius of S (see [9]) given by

ress(S) = max{|λ|;λ ∈ σess(S)

} = inf{R > 0;λ ∈ σ(S), |λ| > R ⇒ λ ∈ σd(S)

},

and σd(S) is the set of isolated eigenvalues of S with finite multiplicity. To prove the lack of exponential stability the semigroup, we use the formula


}.

For dissipative systems, we have that ωσ (A) ≤ 0. We will show that ωess(S) = 0, which implies that ω(S) = 0.

Let us denote by S the semigroup defined by system (1.1)–(1.6). Note that solving for ϕ(L)

and ψ(L), the boundary conditions (1.5)–(1.6) can be written as

ϕt (L, t) = e− κ0m

tϕ1(L) + 1

m

t∫0

e− κ0m

(t−s)S(L, s) ds,

ψt (L, t) = e− κ1

Imtψ1(L) + 1

Im

t∫0

e− κ1

Im(t−s)

M(L, s) ds.

Denote by

D1(t) = e− κ0m

tϕ1(L), K1(t) = 1

m

t∫0

e− κ0m

(t−s)[ϕx(L, s) + ψ(L, s)

]ds, (5.1)

D2(t) = e− κ1

Imtψ1(L), K2(t) = 1

Im

t∫0

e− κ1

Im(t−s)

ψx(L, s) ds. (5.2)

Let us introduce a conservative system of Timoshenko’s type

ρ1(x)ϕtt = Sx, ρ2(x)ψtt = Mx − k(x)(ϕx + ψ

), (5.3)

where S = k(x)(ϕx + ψ) and M = b(x)ψx . The initial condition is given by

ϕ(x,0) = ϕ0(x), ϕt (x,0) = ϕ1(x), ψ(x,0) = ψ0(x), ψt (x,0) = ψ1 (5.4)

with the Dirichlet boundary conditions

ϕ(0, t) = ψ(0, t) = 0, S(L, t) + mϕtt (L, t) = M(L, t) + Imψtt (L, t) = 0. (5.5)



Denoting by E(t) the associated energy of system (5.3)–(5.5)

E(t) = 1

2

L∫0

ρ1|ϕt |2 + ρ2∣∣ψt

∣∣2 + b∣∣Ψ ∣∣2 + k

∣∣ϕx + ψ∣∣2

dx + m

2

∣∣ϕt (L, t)∣∣2 + Im

2

∣∣ψt (L, t)∣∣2

.

We have that ddt

E(t) = 0 ⇒ E(t) = E(0). Let us denote by S0 the semigroup defined by sys-tem (5.3)–(5.5) generated by A with k1 = k2 = 0. Note that ‖S0(t)U0‖ = ‖U0‖. Similarly, the boundary condition (5.5) can be rewritten as

ϕ(0, t) = ψ(0, t) = 0. (5.6)

Solving (5.5) we get

ϕt (L, t) = ϕt (L,0) + 1

m

t∫0

S(L, s) ds, ψt (L, t) = ψt (L,0) + 1

Im

t∫0

M(L, s) ds. (5.7)

Denote by

D3 = ϕ1(L), K3(t) = 1

m

t∫0

[ϕx(L, s) + ψ(L, s)

]ds, (5.8)

D4 = ψ1(L), K4(t) = 1

Im

t∫0

ψx(L, s) ds. (5.9)

Using the above notations, we have that U = ϕ − ϕ and V = ψ − ψ satisfy,

ρ1(x)Utt = [κ(x)(Ux + V )

]x, ρ2(x)Vtt = [

b(x)Vx

]x

− k(x)(Ux + V ), (5.10)

with initial conditions

U(x,0) = 0, Ut (x,0) = 0, V (x,0) = 0, Vt (x,0) = 0 (5.11)

and boundary conditions

U(0, t) = V (0, t) = 0, Ut (L, t) = G1(t), Vt (L, t) = G2(t), (5.12)

where

G1(t) = D1(t) − D3 + K1(t) − K4(t), (5.13)

G2(t) = D2(t) − D4 + K2(t) − K4(t). (5.14)



Let us introduce the following notation,

E(Y,Z, t) =L∫

0

[ρ1|Yt |2 + ρ2|Zt |2 + b|Zx |2 + κ|Yx + Z|2]dx,

IY,Z(t) = ∣∣Yt (L, t)∣∣2 + ∣∣Yx(L, t) + Z(L, t)

∣∣2,

JZ(t) = ∣∣Zt(L, t)∣∣2 + ∣∣Zx(L, t)

∣∣2.

As in Lemma 3.2, we have an estimate for the following evolution system.

Lemma 5.1. Let us assume that there exists a weak solution to system

ρ1Ytt − κ(Yx + Z)x = 0, ρ2Ztt − bZxx + k(Yx + Z) = 0, (5.15)

with finite first order energy E(t) in L1. Then, we have that there exists positive constant such that

T∫0

IY,Z(t) + JY (t) dt ≤ C

T∫0

E(s) ds + CE(T ) + CE(0).

Moreover, for T large enough we get that there exist positive constant such that the reverse inequality also holds, that is

T∫0

E(s) ds ≤ C

T∫0

IY,Z(t) + JZ(t) dt + CE(T ) + CE(0).

Proof. Multiply the first equation in (5.10) by q(ϕx + ψ) and the second equation in (5.10)by qψx . Use the same approach as in Lemma 3.2 to prove our conclusion. �Lemma 5.2. Let the operators Di , Ki be considered as functions

Di :H → L2(0, T ), U0 �→ Di(t), Ki :H → L2(0, T ), U0 �→ Ki(t),

where Di(t) and Ki(t) are given by (5.1)–(5.2) and (5.8)–(5.9), ϕ and ψ are solutions of system (5.10), and ϕ and ψ are solutions of system (5.3)–(5.5). Then, they are compact operators.

Proof. Because of the energy identity and Lemma 5.1, we get that

t �→ S(L, t), t �→ M(L, t)

are bounded in L2(0, T ). This implies that the functions t �→ Ki(U0, t) map bounded sets of H in bounded set of H 1(0, T ). Therefore t �→ Ki(U0, t) is a compact application from H to L2(0, T ). �



Theorem 5.3. The Timoshenko system with tip body is not exponentially stable.

Proof. To prove the lack of exponential stability, we show that the difference of the semigroups S − S0 is a compact operator, where S is the semigroup associated to model (1.1)–(1.6) and S0by (5.3)–(5.5). Therefore, the spectral essential radiuses of S and S0 are equal. That is

ωess(S) = ωess(S0).

Since S0 is unitary, then ωess(S0) = 0. Denoting by ω(S) the type of semigroup, ωσ (A) the spec-tral upper bound of the resolvent set σ(A), and by ωess(S) the essential type of the semigroup eAt , we have that


} = 0.

Therefore there is no exponential stability for S .In fact, to show that S − S0 is compact, let us multiply the first and second equation in (5.10)

by Ut and Vt respectively, to get

d

dtE(U,V, t) = S(L, t)(D1 + K1) + M(L, t)(D2 + K2).

Integrating over time, we have that

E(U,V, t) =t∫

0

S(L, s)(G1) + M(L, s)(G2) ds.

From the first inequality of Lemma 5.1, we get that

T∫0

IU,V (t) + JU(t) dt ≤4∑

i=1

T∫0

|Di |2 + |Ki |2 ds.

It follows that

T∫0

E(U,V, s) ds ≤ C

4∑i=1

T∫0

|Di |2 + |Ki |2 ds.

Note that Di(U0) and Ki(U0) are compact sets in L2(0, T ) if U0 is a bounded set. Therefore, for any bounded sequence of initial data Uμ

0 in the phase space H there exists a subsequence (we still denote in the same way) such that DiU

μ0 and KiU

μ0 converges strongly in L2(0, T ). By the

linearity, this implies that Uμ, V μ also converges strongly in H and (S − S0)(Uμ0 ) converges

strongly. This implies that S − S0 is a compact operator. This fact completes the proof. �Acknowledgments

The authors would like to thank the anonymous reviewers for their valuable comments and suggestions to improve the quality of the paper.



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