Stability of a Timoshenko System with Localized Kelvin–Voigt Dissipation

Abstract

We consider the Timoshenko beam with localized Kelvin–Voigt dissipation distributed over two components: one of them with constitutive law of the type $C^1$, and the other with discontinuous law. The third component is simply elastic, where the viscosity is not effective. Our main result is that the decay depends on the position of the components. We will show that the system is exponentially stable if and only if the component with discontinuous constitutive law is not in the center of the beam. When the discontinuous component is in the middle, the solution decays polynomially.

1 Introduction

We consider a Timoshenko beam configured in the interval $] 0, \ell [$, and divided into three components: an elastic part configured over the interval $I_E$, without any dissipative mechanism, and two viscous components, one of them configured over $I_C$ has a $C ^ 1$ constitutive law, the other viscous component over $I_D$ with discontinuous constitutive law. These components can be distributed over any of the intervals $I_1 =] 0, \ell _0 [$, $I_2 =] \ell _0, \, \ell _1 [$, $I_3 =] \ \ell _1, \ell [$. Denoting by $\widetilde{I}=I_1\cup I_2\cup I_3$, we consider

$$\begin{aligned} \rho _1\,\varphi _{tt} - S_x= & {} 0 \text { in }\widetilde{I}\times (0,+\infty ), \end{aligned}$$

(1.1)

$$\begin{aligned} \rho _2\,\psi _{tt} -M_x+S= & {} 0 \text { in }\widetilde{I}\times (0,+\infty ), \end{aligned}$$

(1.2)

with initial conditions

$$\begin{aligned} \varphi (x,0)=\varphi _0(x), \,\, \varphi _t(x,0)=\varphi _1(x), \,\, \psi (x,0)=\psi _0(x), \,\, \psi _t(x,0)=\psi _1(x) \quad \text {in }(0,\ell ), \end{aligned}$$

(1.3)

and Dirichlet boundary conditions:

$$\begin{aligned} \varphi (0,t)=\varphi (\ell ,t)=\psi (0,t)= \psi (\ell ,t)=0\qquad \text {in }(0,\,+\infty ) \end{aligned}$$

(1.4)

Here, S and M are given, respectively, by:

$$\begin{aligned} S=\kappa (\varphi _{x}+\psi )+\widetilde{\kappa }(\varphi _{xt}+\psi _t),\qquad M=b\psi _x+\tilde{b}\psi _{xt} \end{aligned}$$

(1.5)

where $\rho _1$, $\rho _2$, $\kappa$, and b positive constants for simplicity. To see more details of the model, we refer to [15]. The functions $\tilde{\kappa }$ and $\tilde{b}$ are non negative, where $\tilde{\kappa }=\kappa _0(x)+\kappa _1(x)$, $\tilde{b}=b_0(x)+b_1(x)$. Here $\kappa _0, b_0\in C^1(I_D)$ are discontinuous functions of the first kind over $]0,\,\ell [$, vanishing outside of $I_D$ and positive inside $I_D$. Instead, $\kappa _1(x)$ and $b_1(x)$, are $C^1$ functions vanishing outside of $I_C$ and positive inside $I_C$.

Finally, we consider the transmission conditions,

$$\begin{aligned} \varphi (\ell _i^-)=\varphi (\ell _i^+),\quad \psi (\ell _i^-)=\psi (\ell _i^+),\quad S(\ell _i^-)=S(\ell _i^+),\quad M(\ell _i^-)=M(\ell _i^+). \end{aligned}$$

(1.6)

for $i=0,1$. Note that condition (1.6) implies $S,M\in H^1(0,\ell )$. If we have more points of discontinuity, the set $\widetilde{I}$ have to be modified.

To get the uniform rate of decay, we consider the following hypotheses (to be used in Lemma 3.3)

$$\begin{aligned} |b_1'(x)|^2\le c|b_1(x)|,\qquad |\kappa _1'(x)|^2\le c|\kappa _1(x)|,\quad \forall x\in \overline{I_C} \end{aligned}$$

(1.7)

Additionally, we assume that there exists positive constants $C_1$ and $C_2$ such that

$$\begin{aligned} C_1\kappa _1(x)\le b_1(x)\le C_2\kappa _1(x) \end{aligned}$$

(1.8)

As a typical example of a function $\widetilde{\kappa }(x)$, ($\widetilde{b}(x)$ is similar) is given in the following graphics

Fig. 1

The discontinuous component $I_D$ is in the center of the beam

Fig. 2

Here the continuous component $I_C$ is in the center of the beam

Fig. 3

The elastic component $I_C$ is in the center of the beam

In the case of Fig. 1 we have not exponential stability, and in case of Figs. 2 and 3 the system is exponentially stable.

In [11], the authors consider the transmission problem of Timoshenko beam composed by N components, each of them being either purely elastic (E), or a Kelvin–Voigt viscoelastic material (discontinuous constitutive law V), or an elastic material inserted with a frictional damping mechanism (F). The authors prove that the Timoshenko model is exponentially stable if and only if all the elastic components are connected with one component with frictional damping. Otherwise, there is no exponential stability, but a polynomial decay of the energy as $1/t^{2}$. On the other hand, Liu and Liu in [8] and Cheng et al. [3], proved that the wave equation with localized Kelvin–Voigt viscoelastic damping (with discontinuous constitutive law) is not exponentially stable. In [1] was proved that the corresponding semigroup decays polynomially to zero. On the other hand, Liu and Rao in [9] proved that when the localized viscoelastic damping has a $C^1$-constitutive law, then the corresponding semigroup is exponentially stable. Therefore, for localized viscoelastic damping, the regularity of the constitutive law is important and completely changes the asymptotic properties.

In this work we consider the two types of localized viscoelastic damping (continuous and discontinuous constitutive law) and we prove that the exponential stability depends on the order of the viscoelastic components of the beam. That is, we will show that the semigroup is exponentially stable if and only if the discontinuous component is not in the center of the beam. Furthermore, in case of lack of exponential stability, we show that the semigroup decays polynomially to zero.

The remainder part of this paper is organized as follows. In Sect. 2 we show the well-posedness of the model. In Sect. 3 we show the the exponential stability provided the discontinuous component is not in the center of the beam, and the polynomial stability, in case of the discontinuous component is in the center. Finally, in Sect. 4 we show the lack of exponential stability.

2 The Semigroup Approach

The energy of the system is given by:

$$\begin{aligned} E(t)=\frac{1}{2}\int _0^\ell \left( \rho _1|\varphi _t|^2 +\rho _2|\psi _t|^2+b|\psi _x|^2+\kappa |\varphi _x+\psi |^2\right) \,dx \end{aligned}$$

(2.1)

Multiplying Eq. (1.1) by $\varphi _t$ and Eq. (1.2) by $\psi _t$, summing up the product result we arrive to

$$\begin{aligned} \frac{d}{dt}E(t)=-\int _0^{\ell }\tilde{b}|\psi _{xt}|^2\,dx-\int _0^{\ell }\tilde{\kappa }|\varphi _{xt}+\psi _t|^2\,dx\le 0. \end{aligned}$$

(2.2)

We denote by $\mathcal {H}$ the phase space given by:

$$\begin{aligned} \mathcal {H}=H_0^1(0,\ell )\times L^2(0,\,\ell )\times H_0^1(0,\,\ell )\times L^2(0,\,\ell ) \end{aligned}$$

For $U=(\varphi ,\,\Phi ,\,\psi ,\,\Psi )^t$ we define

$$\begin{aligned} \Vert U\Vert _{\mathcal {H}}^2=\int _0^{\ell }\rho _1|\Phi (s)|^2 +\rho _2|\Psi (s)|^2+b|\psi _x(s)|^2+\kappa |\varphi _x(s)+\psi (s)|^2\,ds \end{aligned}$$

Taking $U=(\varphi ,\psi ,\varphi _t,\psi _t)^\top$, system (1.1)–(1.2) can be written as

$$\begin{aligned} U_t=AU \end{aligned}$$

with $\mathcal {A}:D(\mathcal {A})\subset \mathcal {H}\rightarrow \mathcal {H}$ is the linear operator defined by:

$$\begin{aligned} \mathcal {A}U =\left( \Phi \; ,\; \dfrac{1}{\rho _1}S_x\; ,\; \Psi \; ,\; \dfrac{1}{\rho _2}(M_x-S)\right) ^\top \end{aligned}$$

where M and S are given in (1.5). The domain is given by:

$$\begin{aligned} D(\mathcal {A})=\{U\in \mathcal {H}:\Phi ,\,\Psi \in H_0^1(0,\,\ell ),\,\varphi ,\,\psi \in H^2(I_E),\,S,\,M \in H^1(0,\ell )\} \end{aligned}$$

Note the operator $\mathcal {A}$ is dissipative,

$$\begin{aligned} \text{ Re }\,\langle \mathcal {A}U,\,U\rangle =-\int _0^{\ell }\tilde{b}|\Psi _{x}|^2\,dx-\int _0^{\ell }\tilde{\kappa }|\Phi _{x}+\Psi |^2\,dx\le 0 \end{aligned}$$

So we have the following result.

Theorem 2.1

The operator $\mathcal {A}$ defined by (2) is the infinitesimal generator of a contractions semigroup S(t) over the space $\mathcal {H}$.

Proof

It is no difficult to show that $0\in \rho (\mathcal {A})$. Hence, to use Theorem 1.2.4 in [10] to show the the desired result, we only need to prove that the domain D(A) is dense. But this follows by using Theorem 4.6 Chapter 1 of [13], this because $\mathcal {H}$ is reflexive and $\mathcal {A}$ is a dissipative operator; thus, it is deduced that $D(\mathcal {A})$ is dense. We conclude that $\mathcal {A}$ is the infinitesimal generator of a contractions $C_0$-semigroup (see [4]). $\square$

3 The Asymptotic Behavior

The main tool we use is the characterizations of the exponential and polynomial stabilization due to Prüss [14]–Huang [6]–Gearhart [5] and Borichev and Tomilov [2], respectively.

Theorem 3.1

Let S(t) be a contraction $C_{0}$-semigroup, generated by $\mathcal{A}$ over a Hilbert space $\mathcal{H}$. Then, in Prüss [14] is established that there exists $C, \gamma >0$ verifying

$$\begin{aligned} \Vert S(t)\Vert \le Ce^{-\gamma t}\quad \Leftrightarrow \quad i\,\mathbb {R}\subset \varrho (\mathcal{A}) \, \text{ and } \, \Vert (i\,\lambda \,I - \mathcal{A})^{-1}\Vert _{\mathcal{L}(\mathcal{H})} \leqslant M, \;\; \forall \,\lambda \in \mathbb {R}. \end{aligned}$$

(3.1)

To polynomial stability, Borichev and Tomilov [2] result establish that there exists $C>0$ such that

$$\begin{aligned} \Vert \mathcal{S}(t)\mathcal{A}^{-1}\Vert \leqslant \frac{C}{t^{1/\alpha }}\;\; \Leftrightarrow \;\; i\mathbb {R}\subset \varrho (\mathcal{A}) \, \text{ and } \,\Vert (i\lambda \,I - \mathcal{A})^{-1}\Vert \leqslant M |\lambda |^{\alpha },\;\; \forall \lambda \in \mathbb {R} \end{aligned}$$

(3.2)

Hence, to show the uniform rate of decay we use the resolvent equation, given by:

$$\begin{aligned} i\lambda U-\mathcal {A}U=F \end{aligned}$$

(3.3)

Taking $U=(\varphi ,\,\Phi ,\,\psi ,\,\Psi )^t$ and $F=(f_1,\,f_2,\,f_3,\,f_4)^t$ we can rewrite (3.3) as

$$\begin{aligned} i\lambda \varphi -\Phi&=f_1 \end{aligned}$$

(3.4)

$$\begin{aligned} i\rho _1\lambda \Phi -S_x&=\rho _1f_2 \end{aligned}$$

(3.5)

$$\begin{aligned} i\lambda \psi -\Psi&=f_3 \end{aligned}$$

(3.6)

$$\begin{aligned} i\rho _2\lambda \Psi -M_x+S&=\rho _2f_4 \end{aligned}$$

(3.7)

Lemma 3.1

$i\mathbb {R}\subseteq \rho (\mathcal {A})$

Proof

Since $0\in \rho (A)$, the set

$$\begin{aligned} \mathcal {N}=\{s\in \mathbb {R}^+:]-is,\,is[\subset \rho (\mathcal {A})\} \end{aligned}$$

is not empty. Let us denote by $\sigma =\sup \mathcal {N}$. If $\sigma =\infty$ we have that $i\mathbb {R}\subset \rho (\mathcal {A})$, hence there is nothing to prove. So, let us suppose that $\sigma <\infty$, we will arrive to a contradiction. This implies that $i\mathbb {R}\not \subseteq \rho (\mathcal {A})$. Then, exists a sequence $\{\lambda _n\}\subseteq \mathbb {R}$ such that $\lambda _n\rightarrow \sigma <+\infty$ and

$$\begin{aligned} \Vert (i\lambda _nI-\mathcal {A})^{-1}\Vert _{\mathcal {L}(\mathcal {H})}\rightarrow +\infty \end{aligned}$$

Hence, exists a sequence $\{f_n\}\subseteq \mathcal {H}$ with $\Vert f_n\Vert _{\mathcal {H}}=1$ and $\Vert (i\lambda _nI-\mathcal {A})^{-1}f_n\Vert _{\mathcal {H}}\rightarrow \infty$. Denoting by:

$$\begin{aligned} \tilde{U}_n=(i\lambda _nI-\mathcal {A})^{-1}f_n,\quad U_n=\dfrac{\tilde{U}_n}{\Vert \tilde{U}_n\Vert }\,\quad F_n=\dfrac{f_n}{\Vert \tilde{U}_n\Vert } \end{aligned}$$

we get:

$$\begin{aligned} i\lambda _nU_n-\mathcal {A}U_n=F_n\rightarrow 0 \end{aligned}$$

(3.8)

Note that $\Vert \mathcal {A}U_n\Vert \le C$. Therefore $U_n$ is bounded in $D(\mathcal {A})$. This implies in particular that $\Psi _n$ and $\Phi _n$ are bounded in $H^1(0,\,\ell )$ and also $\psi$ and $\varphi$ are bounded in $H^2(I_E)$; therefore, there exists a subsequence (we still denote in the same way) such that:

$$\begin{aligned}&\Phi _n\rightarrow \Phi ,\qquad \Psi _n\rightarrow \Psi ,\qquad \text { strong in } L^2(0,\,\ell ) \end{aligned}$$

(3.9)

$$\begin{aligned}&\varphi _{n,x}+\psi _n\rightarrow \varphi _{x}+\psi ,\qquad \psi _{n,x}\rightarrow \psi ,\qquad \text { strong in } L^2(I_E) \end{aligned}$$

(3.10)

Taking inner product to (3.8)

$$\begin{aligned} i\lambda _n\Vert U_n\Vert ^2-\langle \mathcal {A}U_n,U_n\rangle =\langle F_n,\,U_n\rangle \rightarrow 0 \end{aligned}$$

and taking real part:

$$\begin{aligned} -\text{ Re }\,\langle \mathcal {A}U_n,U_n\rangle =\int _0^{\ell } (\tilde{b}|\Psi _x^n|^2+\tilde{\kappa } |\Phi _x^n+\Psi ^n|^2)\,dx\rightarrow 0 \end{aligned}$$

(3.11)

That implies:

$$\begin{aligned} \Phi _{n,x}+\Psi _n\rightarrow 0,\qquad \Psi _{n,x}\rightarrow 0\qquad \text { strong in } \in L^2(I_C\cup I_D) \end{aligned}$$

Therefore we have

$$\begin{aligned} \varphi _{n,x}+\psi _n\rightarrow 0,\qquad \psi _{n,x}\rightarrow 0\qquad \text { strong in } \in L^2(I_C\cup I_D) \end{aligned}$$

(3.12)

From (3.9), (3.10) and (3.12), we get that $U_n\rightarrow U$ strongly in $\mathcal {H}$. Since $\mathcal {A}$ is closed, we conclude that U satisfies:

$$\begin{aligned} i\sigma U-\mathcal {A}U= 0 \end{aligned}$$

Moreover, using (3.12) into (3.5)–(3.7) we conclude that $\Phi =\Psi =0$, so relations (3.4)–(3.6) implies that $\varphi =\psi =0$, hence $U\equiv 0$ over $I_C\cup I_D$. Since $I_E=[\alpha ,\,\beta ]$ is linked to $I_C$ or $I_D$ on $\alpha$ or $\beta$, we get that $U(\alpha )=0$ or $U(\beta )=0$. So we have that over $]\alpha ,\,\beta [$ it verifies that:

$$\begin{aligned} -\sigma ^2\varphi +\kappa (\varphi _x+\psi )=0,\qquad -\sigma ^2\psi +b\psi _{xx}+\kappa (\varphi _x+\psi )=0 \end{aligned}$$

with

$$\begin{aligned} \varphi (\alpha )=\psi (\alpha )=\varphi _x(\alpha )=\psi _x(\alpha )=0 \end{aligned}$$

It is a second order initial value problem verifying $\varphi =\psi =0$ over $]\alpha ,\,\beta [$. From where it follows that $U\equiv 0$ on $\mathcal {H}$, which is a contradiction. This finish the proof.$\square$

A key result that we are going to use in this work, is given by the following:

Lemma 3.2

For $g\in H^1(a,\,b)$:

$$\begin{aligned} \int _a^b|g|^2\,dx\le C\left| \int _a^b g\,dx\right| ^2+\int _a^b|g_x|^2\,dx \end{aligned}$$

Proof

In fact, for any $a\le x<y\le b$ we have

$$\begin{aligned} g(y)-g(x)=\int _x^yg_x\; ds \quad \Rightarrow \quad (b-a)g(y)-\int _a^bg(x)\;dx =\int _a^b\int _x^yg_x\; ds\;dx, \end{aligned}$$

therefore, taking absolute value

$$\begin{aligned} (b-a)|g(y)|\le \left| \int _a^bg(x)\;dx \right| +(b-a)\int _a^b|g_x|\;dx, \end{aligned}$$

Since $(b-a)\int _a^b|g_x|\;dx\le (b-a)^{3/2}\left( \int _a^b|g_x|^2\;dx\right) ^{1/2}$, squaring and integrating once more over [a, b] our conclusion follows. .$\square$

The dissipativity of the operator $\mathcal {A}$ implies that

$$\begin{aligned}&\int _{I_C}\kappa _1|\Phi _{x}+\Psi |^2+b_1|\Psi _{x}|^2\,dx+\int _{I_D}\kappa _0|\Phi _{x}+\Psi |^2+b_0|\Psi _{x}|^2\,dx\nonumber \\&\quad =\text{ Re }\,(U,F)_{\mathcal {H}}\le \Vert U\Vert \Vert F\Vert \end{aligned}$$

(3.13)

Lemma 3.3

Let us suppose that condition (1.7)–(1.8) holds, then any solution of (3.4)–(3.7) satisfies

$$\begin{aligned} \int _{I_C} \kappa _1|\lambda \Phi |^2+b_1|\lambda \Psi |^2\,dx\le C_{\varepsilon }\Vert U\Vert \Vert F\Vert +C_{\varepsilon }\Vert F\Vert ^2+\varepsilon \Vert U\Vert ^2 \end{aligned}$$

Proof

The resolvent system over $I_C$ is written as:

$$\begin{aligned} i\lambda \rho _1\Phi -[\kappa (\varphi _{x}+\psi )]_x-[\kappa _1(\Phi _{x}+\Psi )]_x&=\rho _1f_2,\quad \text{ in }\quad I_C \end{aligned}$$

(3.14)

$$\begin{aligned} i\lambda \rho _2\Psi -(b\psi _x)_x-(b_1\Psi _x)_x+\kappa (\varphi _{x}+\psi )+\kappa _1(\Phi _{x}+\Psi )&=\rho _2f_4 ,\quad \text{ in }\quad I_C \end{aligned}$$

(3.15)

Multiplying (3.14) by $\overline{i\lambda \kappa _1\Phi }$ and integrating over $I_C=[a,b]$

$$\begin{aligned} \int _a^b\rho _1\kappa _1|\lambda \Phi |^2\,dx&=\int _a^b[\kappa (\varphi _{x}+\psi )+\kappa _1(\Phi _{x}+\Psi )]_x\overline{i\lambda \kappa _1\Phi }\,dx\nonumber \\&\quad +\int _a^b\rho _1f_2\overline{i\lambda \kappa _1\Phi }\,dx\nonumber \\&=\mathfrak {G}+\mathfrak {G}_0+\int _a^b\rho _1f_2\overline{i\lambda \kappa _1\Phi }\,dx \end{aligned}$$

(3.16)

where $\mathfrak {G}=\int _a^b[\kappa _1(\Phi _{x}+\Psi )]i\lambda \overline{(\kappa _1'\Phi +\kappa _1\Phi _x)}\,dx$ and $\mathfrak {G}_0=\int _a^b[\kappa (\varphi _{x}+\psi )]i\lambda \overline{(\kappa _1'\Phi +\kappa _1\Phi _x)}\,dx$. Estimating $\mathfrak {G}$ (the estimation of $\mathfrak {G}_0$ is similar after using Eqs. (3.4) and (3.6))

$$\begin{aligned} \mathfrak {G}= & {} \int _a^b[\kappa _1(\Phi _{x}+\Psi )]i\lambda \overline{(\kappa _1'\Phi +\kappa _1(\Phi _x+\Psi ))}\,dx\\&-\int _a^b[\kappa _1(\Phi _{x}+\Psi )]i\lambda \overline{(\kappa _1\Psi )}\,dx \end{aligned}$$

Taking the real part of the above relation and using (3.13), we get:

$$\begin{aligned} \text{ Re }\,\mathfrak {G}&=\text{ Re }\,\int _a^b[\kappa _1(\Phi _{x}+\Psi )]i\lambda \overline{(\kappa _1'\Phi )}\,dx -\text{ Re }\,\int _a^b[\kappa _1(\Phi _{x}+\Psi )]i\lambda \overline{(\kappa _1\Psi )}\,dx\nonumber \\&\le \epsilon \Vert \lambda \Phi \Vert ^2+ \epsilon \Vert \lambda \Psi \Vert ^2+C_{\epsilon }\Vert U\Vert \Vert F\Vert \end{aligned}$$

(3.17)

Similarly, using (3.4), (3.6) and (3.13), we get:

$$\begin{aligned}&\text{ Re }\,\int _a^b[\kappa (\varphi _{x}+\psi )]i\lambda \overline{(\kappa _1'\Phi +\kappa _1\Phi _x)}\,dx\nonumber \\&\quad =\text{ Re }\,\int _a^b[\kappa (\Phi _{x}+\Psi )+\kappa (f_{1,x}+f_3)]\overline{(\kappa _1'\Phi +\kappa _1\Phi _x)}\,dx\nonumber \\&\quad \le \epsilon \int _a^b\Vert \Phi \Vert ^2+ \Vert \Psi \Vert ^2\;dx+C_{\epsilon }\Vert U\Vert \Vert F\Vert +C_{\epsilon }\Vert F\Vert ^2 \end{aligned}$$

(3.18)

Thus, substitution of (3.17) and (3.18) into (3.16) yields

$$\begin{aligned} \int _a^b\kappa _1|\lambda \Phi |^2\,dx\le \epsilon \Vert \lambda \Phi \Vert ^2+ \epsilon \Vert \lambda \Psi \Vert ^2+C_{\epsilon }\Vert U\Vert \Vert F\Vert +C_{\epsilon }\Vert F\Vert ^2 \end{aligned}$$

(3.19)

for $|\lambda |>1$. Multiplying (3.15) by $\overline{i\lambda b_1\Psi }$ and using the same above procedure, we get

$$\begin{aligned} \int _a^b\rho _2b_1|\lambda \Psi |^2\,dx&\le \epsilon \Vert \lambda \Psi \Vert ^2+C_{\epsilon }\Vert U\Vert \Vert F\Vert +C_{\epsilon }\Vert F\Vert ^2 \end{aligned}$$

From the last two inequalities, our conclusion follows. $\square$

Let us introduce the following notations

$$\begin{aligned} \mathcal {E}_{\varphi }= & {} \frac{(\kappa q \rho _1)'}{2}|\Phi |^2+\frac{q'}{2}|S|^2, \qquad \mathcal {I}_{\varphi } =\frac{\kappa q \rho _1}{2}|\Phi |^2+\frac{q}{2}|S|^2 \end{aligned}$$

(3.20)

$$\begin{aligned} \mathcal {E}_{\psi }= & {} \frac{(b q\rho _2)'}{2}|\Psi |^2+\frac{q'}{2}|M|^2, \qquad \mathcal {I}_{\psi } =\frac{b q \rho _2}{2}|\Phi |^2+\frac{q}{2}|M|^2 \end{aligned}$$

(3.21)

$$\begin{aligned} \mathcal {E}= & {} \mathcal {E}_{\varphi }+ \mathcal {E}_{\psi }, \qquad \mathcal {I}=\mathcal {I}_{\varphi }+ \mathcal {I}_{\psi } \end{aligned}$$

(3.22)

and

$$\begin{aligned} \mathcal {L}=\int _a^b \mathcal {E}(s)\,ds-\int _a^b\rho _1q\Phi \overline{\psi }\,dx+\int _a^b\kappa q S\bar{M}\,dx \end{aligned}$$

(3.23)

Taking $q(x)=\dfrac{\mathrm {e}^{nx}-\mathrm {e}^{na}}{n}$ we have $q'(x)=e^{nx}\gg q(x)$, for n large. Note that

$$\begin{aligned} \left| \int _a^b\rho _1q\Phi \overline{\psi }\,dx\right| \le \frac{1}{n} \int _a^b\rho _1q'\left| \Phi \overline{\psi }\right| \,dx\le \frac{c}{n}\int _a^b \mathcal {E}(s)\,ds \end{aligned}$$

similarly we have

$$\begin{aligned} \left| \int _a^b\kappa q S\bar{M}\,dx\right| \le \frac{c}{n}\int _a^b \mathcal {E}(s)\,ds \end{aligned}$$

Hence, for n large enough we have

$$\begin{aligned} C_0\int _a^b \mathcal {E}\,dx\le \mathcal {L} \le C_1 \int _a^b \mathcal {E}\,dx \end{aligned}$$

(3.24)

Remark 3.1

Recalling the definition of S and M we get

$$\begin{aligned} \int _a^b |S|^2\,dx&\le c\int _a^b \kappa |\varphi _x+\psi |^2\,dx+\int _a^b |\widetilde{\kappa }(\Phi _x+\Psi ) |^2\,dx \end{aligned}$$

Using the dissipative properties

$$\begin{aligned} \int _a^b |S|^2\,dx\le c\int _a^b |\varphi _x+\psi |^2\,dx+c\Vert U\Vert \Vert F\Vert \end{aligned}$$

Similarly

$$\begin{aligned} \int _a^b |M|^2\,dx\le c\int _a^b |\psi _x|^2\,dx+c\Vert U\Vert \Vert F\Vert \end{aligned}$$

from where it follows that

$$\begin{aligned}&\int _a^b |\Phi |^2+|\varphi _x+\psi |^2+ |\Psi |^2+|\psi _x|^2\,dx \le \int _a^b \mathcal {E}\,dx+c\Vert U\Vert \Vert F\Vert \\&\int _a^b \mathcal {E}\,dx\le c\int _a^b |\Phi |^2+|\varphi _x+\psi |^2+ |\Psi |^2+|\psi _x|^2\,dx +c\Vert U\Vert \Vert F\Vert \end{aligned}$$

Lemma 3.4

Over $[a,b]\subset I_C\cup I_E$ we have

$$\begin{aligned} \left| \mathcal {L}-\mathcal {I}(s)\Big |_a^b\right| \le C_{\varepsilon }\Vert U\Vert \Vert F\Vert +C_{\varepsilon }\Vert F\Vert ^2+\varepsilon \Vert U\Vert ^2 \end{aligned}$$

Instead over $I_D=[a,b]$

$$\begin{aligned} \left| \mathcal {L}-\mathcal {I}(s)\Big |_a^b\right| \le \varepsilon \Vert U\Vert ^2+C_{\varepsilon }|\lambda |^2\Vert U\Vert \Vert F\Vert ^2+\Vert F\Vert ^2 \end{aligned}$$

Proof

Multiplying (3.5) by $q\bar{S}$ and (3.7) by $q\bar{M}$ we have

$$\begin{aligned} i\lambda \rho _1\Phi q\bar{S}-S_xq\bar{S}&=\rho _1f_2q\bar{S},\\ i\lambda \rho _2\Psi q\bar{M}-M_xq\bar{M} +q S\bar{M}&= \rho _2f_4q\bar{M}. \end{aligned}$$

The above equations implies

$$\begin{aligned} -\frac{\rho _1\kappa q}{2} \frac{d}{dx}|\Phi |^2-\frac{q}{2}\frac{d}{dx}|S|^2= & {} \rho _1f_2q\bar{S}+\rho _1q\Phi \kappa (\overline{f_{1,x}+f_{3}})-i\lambda \rho _1 q\Phi (\kappa \overline{\psi }\\&+\tilde{\kappa } \overline{(\Phi _{x}+\Psi )})\\&-\frac{\rho _2bq}{2} \frac{d}{dx}|\Psi |^2 -\frac{1}{2}q\frac{d}{dx}|M|^2 +q S\bar{M}= \rho _2f_4q\bar{M}\\&+\rho _2q\Psi b\overline{f_{3,x}}-i\lambda \rho _2\Psi q[\overline{\tilde{b}\Psi _{x}}] \end{aligned}$$

Summing up the two equations we get

$$\begin{aligned} -\frac{d}{dx}\mathcal {I}(x)+\mathcal {E}(x)=R_3+\rho _1\kappa q\Phi \overline{\Psi }-q S\bar{M}\underbrace{-i\lambda \rho _1\tilde{\kappa }q\Phi \overline{(\Phi _{x}+\Psi )}-i\lambda \rho _2\tilde{b}q\Psi \overline{\Psi _{x}}}_{:=J(x)}, \end{aligned}$$

where $R_3=\rho _1f_2q\bar{S}+\rho _1q\Phi \kappa (\overline{f_{1,x}+f_{3}})+ \rho _2f_4q\bar{M}+\rho _2q\Psi b\overline{f_{3,x}}$. Note that when $[a,b]\subset I_C\cup I_E$, from Lemma 3.3 we get

$$\begin{aligned} \left| \int _a^b J(x)\,dx\right| \le C_{\varepsilon }\Vert U\Vert \Vert F\Vert +C_{\varepsilon }\Vert F\Vert ^2+\varepsilon \Vert U\Vert ^2 \end{aligned}$$

(3.25)

Over $I_D$ we get

$$\begin{aligned} \left| \int _{I_D} J(x)\,dx\right| \le \varepsilon \Vert U\Vert ^2+C_{\varepsilon }|\lambda |^2\Vert U\Vert \Vert F\Vert ^2+\Vert F\Vert ^2 \end{aligned}$$

(3.26)

for $\lambda$ large enough. After an integration using the above inequalities our conclusion follows. $\square$

Now, we are in condition to establish our main result.

Theorem 3.2

The system is exponentially stable if the viscous discontinuous part $I_D$ is not in the center of the beam.

Proof

Since $I_D$ is not in the middle then $0\in I_D$ or $\ell \in I_D$; hence, because of the boundary conditions, Poincaré inequality is valid for $\Phi$ and $\Psi$. So we have

$$\begin{aligned} \int _{I_D}|\Psi |^2\,dx\le C_p\int _{I_D}|\Psi _x|^2\,dx\le C\Vert U\Vert \Vert F\Vert \end{aligned}$$

(3.27)

Using that $i\lambda \psi =\Psi +f_3$ we get

$$\begin{aligned} \int _{\ell _1}^{\ell }|\psi _x|^2+|\Psi |^2\,dx\le C\Vert U\Vert \Vert F\Vert +C\Vert F\Vert ^2 \end{aligned}$$

Using Poincare’s and the triangular inequality, we get

$$\begin{aligned} \int _{I_D}|\Phi |^2\,dx\le c\int _{I_D}|\Phi _x|^2\,dx\le C\int _{I_D}\kappa |\Phi _{x}+\Psi |^2+|\Psi |^2\,dx\le C\Vert U\Vert \Vert F\Vert +C\Vert F\Vert ^2 \end{aligned}$$

So we have

$$\begin{aligned} \int _{I_D}|\Phi |^2+|\psi _x|^2+|\varphi _{x}+\psi |^2+|\Psi |^2\,dx\le C\Vert U\Vert \Vert F\Vert +C\Vert F\Vert ^2 \end{aligned}$$

Integrating (3.5) and (3.7) over $[a,b]\subset I_C$, we get

$$\begin{aligned} i\lambda \rho _1\int _{a}^{b} \Phi \,dx-S(b^-)+S(a^+)&= \int _{a}^{b} \rho _1f_2\,dx \end{aligned}$$

(3.28)

$$\begin{aligned} i\lambda \rho _2\int _{a}^{b}\Psi \,dx-M(b^-)+M(a^+)&= \int _{a}^{b}\rho _2 f_4\,dx \end{aligned}$$

(3.29)

From Lemma 3.4 we get

$$\begin{aligned} \left| \int _{a}^{b} \Phi \,dx\right| +\left| \int _{a}^{b} \Psi \,dx\right|&\le \frac{C}{|\lambda |}\Vert U\Vert ^{1/2}\Vert F\Vert ^{1/2}+ \frac{C}{|\lambda |}\Vert U\Vert + \frac{C}{|\lambda |}\Vert F\Vert \end{aligned}$$

(3.30)

$$\begin{aligned} \int _{a}^{b}|\Psi |^2\,dx\le & {} c\left| \int _{a}^{b} \Psi \,dx\right| ^2+C\int _{a}^{b}b_1|\Psi _x|^2\,dx\le C\Vert U\Vert \Vert F\Vert \\&+ \frac{C}{|\lambda |^2}\Vert U\Vert ^2+ \frac{C}{|\lambda |^2}\Vert F\Vert ^2 \end{aligned}$$

Using (3.4), (3.6) and (3.13), we get

$$\begin{aligned} \int _{a}^b|\Phi |^2+|\psi _x|^2+|\varphi _{x}+\psi |^2+|\Psi |^2\,dx\le C\Vert U\Vert \Vert F\Vert + \frac{C}{|\lambda |^2}\Vert U\Vert ^2+C\Vert F\Vert ^2 \end{aligned}$$

From Lemma 3.4

$$\begin{aligned} \mathcal {I}(a) \le C\Vert U\Vert \Vert F\Vert + \varepsilon \Vert U\Vert ^2+ C\Vert F\Vert ^2 \end{aligned}$$

(3.31)

Since $I_D$ is not in the center, then $\overline{I_C}\cup \overline{I_E}=[0,\ell _2]$ or $I_C\cup I_E=[\ell _0,\ell ]$. Let us assume the later case. Using the observability Lemma 3.4 once more

$$\begin{aligned} \int _{\ell _0 }^a|\Phi |^2+|\psi _x|^2+|\varphi _{x}+\psi |^2+|\Psi |^2\,dx\le & {} c\mathcal {I}(a) +C\Vert U\Vert \Vert F\Vert \nonumber \\&+ \frac{C}{|\lambda |^2}\Vert U\Vert ^2+C\Vert F\Vert ^2\nonumber \\\le & {} C\Vert U\Vert \Vert F\Vert + \varepsilon \Vert U\Vert ^2+ C\Vert F\Vert ^2\nonumber \\ \end{aligned}$$

(3.32)

for $\lambda$ large. Using the observability over the interval $[a,\ell ]$, we get

$$\begin{aligned} \int _a^{\ell }|\Phi |^2+|\psi _x|^2+|\varphi _{x}+\psi |^2+|\Psi |^2\,dx\le & {} C\Vert U\Vert \Vert F\Vert + \varepsilon \Vert U\Vert ^2+ C\Vert F\Vert ^2\nonumber \\ \end{aligned}$$

(3.33)

From (3.27), (3.32) and (3.33) we get

$$\begin{aligned} \Vert U\Vert ^2&=\int _0^{\ell }|\Phi |^2+|\psi _x|^2+|\varphi _{x}+\psi |^2+|\Psi |^2\,dx\nonumber \\&\le C\Vert U\Vert \Vert F\Vert + \varepsilon \Vert U\Vert ^2+ C\Vert F\Vert ^2 \end{aligned}$$

(3.34)

from where we get that $\Vert U\Vert \le C\Vert F\Vert$. So our conclusion follows. $\square$

Finally, we finish this section showing the polynomial decay when the discontinuous viscous part is in the center of the beam. We use the result given in [2].

Theorem 3.3

Suppose that the viscoelastic discontinuous part $V_D$ is in the center of the beam. Then, the energy of the system decays polynomially, and:

$$\begin{aligned} \Vert S(t)U_0\Vert \le Ct^{-1}\Vert U_0\Vert _{D(\mathcal {A})} \end{aligned}$$

(3.35)

Proof

Denoting $V_D=[\ell _0,\,\ell _1]$. Using (3.28) and (3.29) for $a=\ell _0$ and $b=\ell _1$ we have:

$$\begin{aligned} \int _{I_D}|\Phi |^2+|\psi _x|^2+|\varphi _{x}+\psi |^2+|\Psi |^2\,dx\le C\Vert U\Vert \Vert F\Vert + \frac{C}{|\lambda |^2}\Vert U\Vert ^2+C\Vert F\Vert ^2 \end{aligned}$$

(3.36)

Using the same procedure as in Theorem 3.2, we get

$$\begin{aligned} \int _{I_C}|\Phi |^2+|\psi _x|^2+|\varphi _{x}+\psi |^2+|\Psi |^2\,dx\le C\Vert U\Vert \Vert F\Vert + \frac{C}{|\lambda |^2}\Vert U\Vert ^2+C\Vert F\Vert ^2 \end{aligned}$$

(3.37)

Let us suppose that $\ell _1\in \overline{I_E}$. Using Lemma 3.4 over $I_D=]\ell _0,\ell _1[$, we have

$$\begin{aligned} \mathcal {I}(\ell _1^+)\le \int _{I_D}|\Phi |^2+|\psi _x|^2+|\varphi _{x}+\psi |^2+|\Psi |^2\,dx + \varepsilon \Vert U\Vert ^2+ C_{\varepsilon }|\lambda |^2\Vert F\Vert ^2 \end{aligned}$$

(3.38)

Since $S(\ell _1^-)=S(\ell _1^+)$ and $M(\ell _1^-)=M(\ell _1^+)$ we have

$$\begin{aligned} \int _{I_E}|\Phi |^2+|\psi _x|^2+|\varphi _{x}+\psi |^2+|\Psi |^2\,dx\le & {} \mathcal {I}(\ell _1^-)+C\Vert U\Vert _{\mathcal {H}}\Vert F\Vert _{\mathcal {H}}\nonumber \\\le & {} \varepsilon \Vert U\Vert ^2+ C_{\varepsilon }|\lambda |^2\Vert F\Vert ^2 \end{aligned}$$

(3.39)

From the above inequality we get

$$\begin{aligned} \Vert U\Vert ^2 \le C_\varepsilon |\lambda |^2\Vert F\Vert ^2+\varepsilon \Vert U\Vert ^2 \end{aligned}$$

and the polynomial decay is a consequence of the Borichev-Tomilov theorem. $\square$

4 Lack of Exponential Stability

Our starting point is the boundary estimate of the Timoshenko system.

$$\begin{aligned} \rho _1{\varphi }_{tt} - \kappa ({\varphi }_{x}+{\psi })_x= & {} 0\nonumber \\ \rho _2{\psi }_{tt} -b{\psi }_{xx}+\kappa ({\varphi }_{x}+{\psi })= & {} 0 \end{aligned}$$

(4.1)

over some interval [a, b], then we have

Theorem 4.1

Let us suppose that the solution of system (4.1) $(\varphi ,\varphi _t,\psi ,\psi _t)$ is bounded in $C(0,T;[H^1(a,b)\times L^2(a,b)]^2)$. Then we have that

$$\begin{aligned} \int _0^t\rho _1|\varphi _t(\alpha ,\tau )|^2+\rho _2|\psi _t(\alpha ,\tau )|^2+\kappa |\varphi _x(\alpha ,\tau )|^2+b|\psi _x(\alpha ,\tau )|^2d\tau \le C_E \end{aligned}$$

for any $\alpha \in [a,b]$.

Proof

The proof is well known now, we develop here only the main ideas for completeness. Multiplying (4.1)$_1$ by $q\varphi _x$ and (4.1)$_2$ by $q\psi _x$ to get

$$\begin{aligned} \frac{d}{dt}\left( \rho _1{\varphi }_{t}q\varphi _x\right) -q\rho _1{\varphi }_{t}\varphi _{tx} - q\kappa {\varphi }_{xx}\varphi _x= & {} -\kappa {\psi }_xq\varphi _x\\ \frac{d}{dt}\left( \rho _2{\psi }_{t}q\psi _x\right) -q\rho _2{\psi }_{t}\psi _{tx} -qb{\psi }_{xx}\psi _x= & {} -\kappa {\varphi }_{x}q\psi _x-\kappa {\psi }q\psi _x \end{aligned}$$

Summing up the above inequalities we get

$$\begin{aligned}&-\frac{q}{2}\frac{d}{dx}\left( \rho _1|\varphi _{t}|^2 +\kappa |{\varphi }_{x}|^2+\rho _2|\psi _{t}|^2 +b{\psi }_{x}|^2\right) =-\kappa {\psi }q\psi _x\\&\quad -q\frac{d}{dt}\left( \rho _1{\varphi }_{t}\varphi _x+\rho _2{\psi }_{t}\psi _x\right) \end{aligned}$$

Integrating over $[\alpha ,\beta ]\times [0,t]$, with $\beta \in [a,b]$, and taking $q=x-\beta$, we get:

$$\begin{aligned} \left| \int _0^t\int _\alpha ^\beta q\frac{d}{dt}\left( \rho _1{\varphi }_{t}\varphi _x+\rho _2{\psi }_{t}\psi _x\right) \;dxd\tau \right| =\left| \int _\alpha ^\beta \left. \left( \rho _1{\varphi }_{t}\varphi _x+\rho _2{\psi }_{t}\psi _x\right) \right| _{t=0}^{\tau =t}\;dx\right| \le C_E, \end{aligned}$$

the last inequality is a consequence of the hypotheses, where $C_E$ is a positive constant that depends on the initial data. So, our conclusion follows.

Here we consider that $I_D$ is in the middle of the beam. Our main tool is the following theorem due to [12].

Theorem 4.2

Let H be a Hilbert space and $H_0$ a closed subspace of H. Let S(t) be a contractions semigroup over H and $S_0(t)$ an unitary group over $H_0$. If the difference $S(t)-S_0(t)$ is a compact operator from $H_0$ over H, then S(t) is not exponentially stable. $\square$

Theorem 4.3

The semigroup S(t) is not exponentially stable when the viscous discontinuous part is in the center of the beam.

Proof

Let be the spaces:

$$\begin{aligned} \mathbb {L}_0= & {} \{f\in L^2(0,\,\ell ):f\Big |_{[\ell _0,\,\ell ]}=0\},\quad V_0=H_0^1(0,\,\ell )\cap \mathbb {L}_0,\\ H_0= & {} V_0\times \mathbb {L}_0 \times V_0\times \mathbb {L}_0 \end{aligned}$$

Let us consider the model over $[0,\,\ell _0]$:

$$\begin{aligned} \rho _1\tilde{\varphi }_{tt} - \kappa (\tilde{\varphi }_{x}+\tilde{\psi })_x= & {} 0\nonumber \\ \rho _2\tilde{\psi }_{tt} -b\tilde{\psi }_{xx}+\kappa (\tilde{\varphi }_{x}+\tilde{\psi })= & {} 0 \nonumber \\ \tilde{\varphi }(0,t)=\tilde{\varphi }(\ell _0,t)=\tilde{\psi }(0,t)=\tilde{\psi }(\ell _0,t)= & {} 0 \end{aligned}$$

(4.2)

Let $S_0$ be the semigroup over $H_0$ (null extensions on $[\ell _0,\ell ]$) associated to (4.2). So we have

$$\begin{aligned} \Vert S_0(t)U_0\Vert ^2=\Vert U_0\Vert ^2,\quad \forall U_0\in H_0 \end{aligned}$$

(4.3)

Now, we are going to prove that $S(t)-{S}_0(t):H_0\rightarrow H$ is a compact operator, where

$$\begin{aligned} S(t)U_0^{m}=(\varphi ^m,\,\varphi _t^m,\,\psi ^m,\,\psi _t^m)\in H,\qquad S_0(t)U_0^{m}=(\tilde{\varphi }^m,\,\tilde{\varphi }_t^m,\,\tilde{\psi }^m,\,\tilde{\psi }_t^m)\in H_0 \end{aligned}$$

Let be: $v^m:=\varphi ^m-\tilde{\varphi }^m,\qquad w^m:=\psi ^m-\tilde{\psi }^m$. By definition we have

$$\begin{aligned} v^m(x,t)= {\left\{ \begin{array}{ll} \varphi ^m-\tilde{\varphi }^m,\text { if } x\in [0,\,\ell _0]\\ \varphi ^m\quad \quad ,\text { if } x\notin [0,\,\ell _0] \end{array}\right. };\quad w^m(x,t)= {\left\{ \begin{array}{ll} \psi ^m-\tilde{\psi }^m,\text { if } x\in [0,\,\ell _0]\\ \psi ^m\quad ,\text { if } x\notin [0,\,\ell _0] \end{array}\right. } \end{aligned}$$

Moreover, v and w verifies

$$\begin{aligned} \rho _1v_{tt} - \kappa (v_{x}+w)_x-\tilde{\kappa }(v_{xt}+w_t)_x&=0 \end{aligned}$$

(4.4)

$$\begin{aligned} \rho _2w_{tt} -bw_{xx}-\tilde{b}w_{xxt}+\kappa (v_{x}+w) +\tilde{\kappa }(v_{xt}+w_t)&=0 \end{aligned}$$

(4.5)

Multiplying (4.4) by $v_t$, (4.5) by $w_t$, and integrating over $[0,\,\ell ]$, we obtain:

$$\begin{aligned}&\int _0^{\ell }\left( \rho _1|v_t|^2 +\rho _2|w_t|^2+b|w_x|^2+\kappa |v_x+w|^2\right) \,dx=\nonumber \\&\quad \kappa v_xv_t\Bigg |_0^{\ell _{0}}+b w_xw_t\Bigg |_0^{\ell _{0}}-\int _{\ell _0}^\ell \tilde{\kappa }|v_{xt}+w_t|^2 +\tilde{b}|w_{xt}|^2\,dx \end{aligned}$$

(4.6)

Using the boundary conditions we get

$$\begin{aligned} \kappa v_xv_t\Bigg |_0^{\ell _{0}}+b w_xw_t\Bigg |_0^{\ell _{0}}=-\kappa \tilde{\varphi }_x(\ell _0^-,t)\varphi _t(\ell _0^-,t)-b\tilde{\psi }_x(\ell _0^-,t)\psi _t(\ell _0^-,t) \end{aligned}$$

Denoting by $\mathfrak {U}^m(t)=[S(t)-S_0(t)]U_0^m=(v^m,v_t^m,w^m,w_t^m)$, integrating (4.6) over $[0,\,t]$, recalling the definition of the norm of the phase space $\mathcal {H}$ we get

$$\begin{aligned}&\int _0^t\Vert \mathfrak {U}^m(t)\Vert _{\mathcal {H}}^2\,dt+\int _0^t\int _{\ell _0}^{\ell }\tilde{\kappa }|v_{xt}^m+w_t^m|^2 +\tilde{b}|w_{xt}^m|^2\,dx\,dt\nonumber \\&\quad =-\int _0^t(\kappa \tilde{\varphi }_x^m({\ell }_0^-,t)\varphi _t^m(\ell _0^-,t)+b\tilde{\psi }_x(\ell _0^-,t)\psi _t(\ell _0^-,t))\,dt \end{aligned}$$

(4.7)

using Theorem 4.1 we have that $\tilde{\varphi }_x^m(\ell _0^-,t)$ and $\tilde{\psi }_x^m(\ell _0^-,t)$ are bounded. So there exists a subsequence, we still denote in the same way, such that

$$\begin{aligned} \tilde{\varphi }_x^m(\ell _0^-,t) \rightarrow \tilde{\varphi }_x(\ell _0^-,t)\text { weak in } L^2(0,\,T),\, \tilde{\psi }_x^m(\ell _0^-,t) \rightarrow \tilde{\psi }_x(\ell _0^-,t)\text { weak in } L^2(0,\,T) \end{aligned}$$

We only need to prove that

$$\begin{aligned} \left( {\varphi }_x^m(\ell _0^-,t)\;,\;{\psi }_x^m(\ell _0^-,t)\right) \rightarrow \left( {\varphi }_x (\ell _0^-,t)\;,\;{\psi }_x(\ell _0^-,t)\right) \text { strong in } L^2(0,\,T)\times L^2(0,\,T) \end{aligned}$$

(4.8)

which implies the norm convergence in (4.7). To do that we use (3.13) and (1.1)–(1.2) to get

$$\begin{aligned} \varphi _t^m, \psi _t^m\in L^2(0,T;H^1(I_D)),\quad \varphi _{tt}^m, \psi _{tt}^m\in L^2(0,T;H^{-1}(I_D)) \end{aligned}$$

Since $H^1\subset H^{1-\delta }\subset H^{-1}$ where the first inclusion is a compact embedding, the compactness Theorem of Lions-Aubin (see [7]) implies that there exists a subsequence (we still denote in the same way) such that

$$\begin{aligned} (\varphi _t^m,\psi _t^m) \rightarrow (\varphi _t,\psi _t)\text { strong in } L^2(0,\,T;H^{1-\delta }(I_D)\times H^{1-\delta }(I_D)). \end{aligned}$$

Using that the embedding $H^{1-\delta }({I_D})\subset C(\overline{{I_D}})$ is compact, we have:

$$\begin{aligned} (\varphi _t^m,\psi _t^m) \rightarrow (\varphi _t,\psi _t)\text { strong in } L^2(0,\,T;C({V_D})\times C({V_D})) \end{aligned}$$

This implies (4.8). Hence inequality (4.7) implies the convergence in norm of $\mathfrak {U}^m$. So, $S(t)-\tilde{S}_0(t)$ is a compact operator. Then our conclusion follows. $\square$

In summary, we can state the following theorem:

Theorem 4.4

The Timoshenko system is exponentially stable if and only if the viscoelastic discontinuous part is not in the middle of the beam. Otherwise, the system only has polynomial rate of decay.