1 Introduction

In this work, we consider the Timoshenko system with past history given by

$$\begin{aligned} \begin{array}{lll} \rho w_{tt}(x,t)-\kappa \left[ \varphi (x,t)+w_{x}(x,t)\right] _{x} &{} =0, &{} \mathtt {\ in\ }\left( 0,L\right) \times \mathbb {R}_{+}, \\ I_{\rho }\varphi _{tt}(x,t)-\varphi _{xx}(x,t)+\int _{0}^{\infty }g(s)\varphi _{xx}(x,t-s)ds &{} &{} \\ \quad +\kappa \left[ \varphi +w_{x}\right] (x,t) &{} =0, &{} \mathtt {\ in}\left( 0,L\right) \times \mathbb {R}_{+}, \end{array} \end{aligned}$$
(1)

subject to the following initial conditions

$$\begin{aligned} w(x,0)=w_{0}(x),\quad w_{t}(x,0)=w_{1}(x),\quad \varphi (x,0)=\varphi _{0}(x),\quad \varphi _{t}(x,0)=\varphi _{1}(x), \end{aligned}$$
(2)

and the boundary conditions

$$\begin{aligned} w(x,0)=\varphi (x,0)&=0,&\mathtt { in }\quad (0,L), \end{aligned}$$
(3)
$$\begin{aligned} Mw_{tt}(L,t)-\kappa \left[ \varphi +w_{x}\right] (L,t)=0,&\mathtt {\ in} \quad \mathbb {R}_{+}, \end{aligned}$$
(4)
$$\begin{aligned} J\varphi _{tt}(L,t)-\varphi _{x}(L,t)+\int _{0}^{\infty }g(s)\varphi _{x}(L,t-s)ds=0,&\mathtt {\ in }\quad \mathbb {R}_{+}. \end{aligned}$$
(5)

In (1), t,  x denote, respectively, the time variable and the space variable along the beam of length L in its equilibrium configuration, \(S=k(w_{x}+\varphi )\) and \(M=\varphi _{x}\) denote the shear force and the bending moment, respectively. \(w=w(x,t)\) is the transversal displacement, and \(\varphi =\varphi (x,t)\) is the rotation angle of the filament, \(\rho =\tilde{\rho }A,\) \(I_{\rho }=\tilde{\rho }I\), \( \kappa =KAG,\) \(b=EI\), where \(\tilde{\rho }\) denotes the density, A is the cross-sectional area, I is the area moment of inertia, E is the modulus of elasticity, K is the shear factor, and G is the shear modulus.

The Timoshenko system is usually considered as describing the transverse vibration of a beam and ignoring damping effects of any nature. There are several publications concerning the stabilization of the Timoshenko system with different kinds of damping. We cite, for example, [1, 7, 13, 15, 16, 19, 20, 27, 41, 43].

Soufyane and Wehbe [40] proved that if the wave propagation speeds are equals (i.e., \(\frac{k_{1}}{\rho _{1}}=\frac{\rho _{2}}{k_{2}}\)) the Timoshenko system with one internal distributed dissipation is exponentially stable; otherwise, only the strong stability holds. Rivera and Racke [32] improved the previous results and showed an exponential decay of the solution of the system when the coefficient of the feedback admits an indefinite sign.

Kim and Renardy [21] considered the system of Timoshenko beam together with two boundary controls and proved exponential decay for the total energy, using multiplicative techniques.

Wehbe and Youssef [43] improved the results of [8], where an optimal polynomial energy decay rate and nonuniform stability of the Timoshenko system with only one dissipation on the boundary has been established.

On the other hand, many authors have treated the system of the Timoshenko beam with nonlinear dissipation acting on boundary or in the domain (see [3, 9, 12, 20, 26, 31]).

The Timoshenko model with dissipation memory has been investigated by many authors in recent years, and many results concerning existence and asymptotic behavior have been established (see [7, 17, 18, 25, 27]).

In the case of the Timoshenko model with boundary dissipation over the shear force effective in only one side, we can cite [4] where authors proved that to get the exponential stability of the related contraction semigroup, the equality of the wave propagation is not enough, it is necessary additional conditions over the coefficient of the system.

In [33], the authors showed that the semigroup associated with the Timoshenko model with tip body decays to zero as \(t^{\frac{1}{2}}\) when the damping mechanism is effective only on the boundary of rotational angle. When wave speeds are equal, they showed that the solution also decays polynomially as \(t^{\frac{1}{2}}\). Otherwise, it decays as \(t^{\frac{1}{4}}\) for any initial data taken in D(A).

Raposo et al. [39] proved an exponential decay of the solution for Timoshenko beam system with linear frictional dissipative terms.

Alabau-Boussouira [3] studied the asymptotic behavior for the Timoshenko system with a nonlinear damping and proved a general semi-explicit formula for the decay rate of the energy in the case of the same wave speeds.

The purpose of our paper is to show the well-posedness of Timoshenko model with past history and to give an energy decay estimates of the solutions to the problem (1)–(5). The influence of viscoelastic dampings on the rate of decay of the solutions is examined.

This paper is organized as follows. In Sect. 2, some preliminary results are introduced. In Sect. 3, we show the well-posedness of Timoshenko model with past history by using the method of semigroup approach and the Lumer–Phillips theorem. In the last section, we prove that the semigroup associated with our model of Timoshenko beam decays polynomially to zero when the wave speeds are not equals.

2 Preliminaries

In this section, we introduce some assumptions and some functional spaces. We establish some useful inequalities, which will be used for the remaining of the present paper.

Assumptions on the kernel

We suppose that the kernel g(t) is a \(C^{1}(\mathbb {R}_{+};\mathbb {R}_{+})\) is nonincreasing function satisfying

H\(_{1})\) \(g(0)>0\) and

$$\begin{aligned} 1-\int _{0}^{\infty }g(s){\mathrm{d}}s=1-g_{\infty }=l>0 \end{aligned}$$
(6)

In addition, we assume that

$$\begin{aligned} \exists k_{0},k_{1}>0:-k_{0}g(t)\le g^{\prime }(t)\le -k_{1}g(t) \end{aligned}$$
(7)

In order to prove the well-posedness of (1)–(5) by using semigroups theory, we introduce some functional spaces.

Let us define the energy space \(\mathcal {H}\) by

$$\begin{aligned} \mathcal {H}=H_{0}^{1}(0,L)\times L^{2}(0,L)\times H_{0}^{1}(0,L)\times L^{2}(0,L)\times \mathbb {C} \times \mathbb {C} \times L_{g}(0,L) \end{aligned}$$

where \(L_{g}(I)\) is the weighted Sobolev space defined by

$$\begin{aligned} L_{g}(I)=\left\{ v:\mathbb {R}_{+}\rightarrow H_{0}^{1}\left( 0,L\right) ,\quad \int _{0}^{L}\int _{0}^{\infty }g(s)v_{x}^{2}{\mathrm{d}}s{\mathrm{d}}x<+\infty \right\} . \end{aligned}$$

The space \(L_{g}(I)\) is endowed with the inner product

$$\begin{aligned} \left\langle v,\overline{v}\right\rangle _{g}=\int _{0}^{\infty }g(s)\left\langle v_{x}(s),\overline{v}_{x}(s)\right\rangle {\mathrm{d}}s. \end{aligned}$$

The energy space \(\mathcal {H}\) is equipped with the inner product defined by

$$\begin{aligned} \left\langle U,\widehat{U}\right\rangle _{\mathcal {H}}= & {} \int _{0}^{L}\left[ \tilde{b}v_{3x}w_{3x}+\rho v_{2}w_{2}+I_{\rho }v_{4}w_{4}\right] {\mathrm{d}}x \\&+\kappa \int _{0}^{L}\left( v_{1x}+v_{3}\right) \left( w_{1x}+w_{3}\right) {\mathrm{d}}x \\&+Mv_{5}w_{5}+Jv_{6}w_{6}+\left\langle v_{7},w_{7}\right\rangle _{L_{g}\left( I\right) }, \end{aligned}$$

where \(U=(v_{1},v_{2},v_{3},v_{4},v_{5},v_{6},v_{7})^{T}\), \(\widehat{U} =(w_{1},w_{2},w_{3},w_{4},w_{5},w_{6},w_{7})^{T}\in \mathcal {H}\). Then, the corresponding norm is

$$\begin{aligned} \left\| U\right\| _{\mathcal {H}}^{2}= & {} \tilde{b}\left\| v_{3x}\right\| _{2}^{2}+\rho \left\| v_{2}\right\| _{2}^{2}+I_{\rho }\left\| v_{4}\right\| _{2}^{2} \\&+\kappa \left\| \partial _{x}v_{1}+v_{3}\right\| _{2}^{2}+M\left| v_{5}\right| ^{2}+J\left| v_{6}\right| ^{2}+\left\| v_{7}\right\| _{*}^{2}. \end{aligned}$$

where

$$\begin{aligned} \tilde{b}=1-\int _{0}^{+\infty }g(s){\mathrm{d}}s, \end{aligned}$$

\(\left\| .\right\| _{2}^{2}\) is the usual norm in \(L^{2}(0,L),\) and

$$\begin{aligned} \left\| v_{7}\right\| _{*}^{2}=\int _{0}^{L}\int _{0}^{\infty }g(s)v_{7x}^{2}{\mathrm{d}}s{\mathrm{d}}x. \end{aligned}$$

3 Well-Posedness

In this section, we will establish the well-posedness of system (1)–(5) by the semigroup approach. To this aim, as in Dafermos [14], we introduce the following auxiliary change of variable

$$\begin{aligned} \eta (x,t,s)=\varphi (x,t)-\varphi (x,t-s),\quad \mathtt { for }\quad (x,t,s)\in (0,L)\times \mathbb {R}_{+}\times \mathbb {R}_{+}. \end{aligned}$$
(8)

This function satisfies the initial conditions

$$\begin{aligned} \eta (0,t,s)=0,\mathtt { in }\quad \mathbb {R}_{+}\times \mathbb {R}_{+},\quad \eta (x,t,0)=0,\quad \mathtt { in }\quad \left[ 0,L\right] \times \mathbb {R}_{+}, \end{aligned}$$

and the equation

$$\begin{aligned} \eta _{t}+\eta _{s}-\varphi _{t}=0\quad \mathtt { in }\quad (0,L)\times \mathbb {R}_{+}\times \mathbb {R}_{+}. \end{aligned}$$
(9)

So, we can rewrite (1)-(5) in the following way:

$$\begin{aligned} \rho w_{tt}(x,t)-\kappa \left[ \varphi (x,t)+w_{x}(x,t)\right] _{x}= & {} 0, \end{aligned}$$
(10)
$$\begin{aligned} I_{\rho }\varphi _{tt}(x,t)-\tilde{b}\varphi _{xx}(x,t)+\int _{0}^{\infty }g(s)\eta _{xx}(x,t,s){\mathrm{d}}s+\kappa \left[ \varphi +w_{x}\right] (x,t)= & {} 0, \end{aligned}$$
(11)
$$\begin{aligned} Mw_{tt}(L,t)-\kappa \left[ \varphi +w_{x}\right] (L,t)= & {} 0, \end{aligned}$$
(12)
$$\begin{aligned} J\varphi _{tt}(L,t)-\tilde{b}\varphi _{x}(L,t)+\int _{0}^{\infty }g(s)\eta _{x}(L,t,s){\mathrm{d}}s= & {} 0.\mathtt {\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ } \end{aligned}$$
(13)

Let

$$\begin{aligned} \eta _{0}(x,s)=\eta (x,0,s)=\varphi _{0}(x,0)-\varphi _{0}(x,s),\quad \mathtt {for}\quad (x,s)\in (0,L)\times \mathbb {R} _{+}. \end{aligned}$$

To define the semigroup approach associated with (10)–(13), we consider the following condition of the right end contour of beam

$$\begin{aligned} w_{t}(L,t)=\mathbf {u}(t),\quad \varphi _{t}(L,t)=\mathbf {v}(t),\quad \mathtt {for}\quad t>0, \end{aligned}$$

where \(\mathbf {u}\) and \(\mathbf {v}\) solve the system

$$\begin{aligned} M\mathbf {u}_{t}(L,t)-\kappa \left[ \varphi +w_{x}\right] (L,t)= & {} 0,\quad \mathtt { on }\quad \mathbb {R}_{+}, \\ J\mathbf {v}_{t}(L,t)-\tilde{b}\varphi _{x}(L,t)+\int _{0}^{\infty }g(s)\eta _{x}(L,t,s){\mathrm{d}}s= & {} 0,\quad \mathtt { on }\quad \mathbb {R}_{+}, \end{aligned}$$

under the initial condition

$$\begin{aligned} \mathbf {u}(0)=w_{1}(L),\quad \mathbf {v}(0)=\varphi _{1}(L),\quad \mathtt {for}\quad t>0. \end{aligned}$$

Now, for \(U=(v_{1},v_{2},v_{3},v_{4},v_{5},v_{6},v_{7})^{T}\) and \( U_{0}=\left( w_{0},w_{1},\varphi _{0},\varphi _{1},\mathbf {u}_{0},\mathbf {v} _{0},\eta _{0}\right) ^{T},\) where

$$\begin{aligned} \mathbf {u}(t)=v_{2}(L,t),\quad \mathbf {v}(t)=v_{4}(L,t),\quad \mathtt {for}\quad t>0, \end{aligned}$$

the system is equivalent to the abstract linear order Cauchy problem

$$\begin{aligned} \left\{ \begin{array}{c} U_{t}=\mathcal {A}U \\ U(0)=U_{0} \end{array} \right. , \end{aligned}$$
(14)

where \(\mathcal {A}\) is the linear operator defined by

$$\begin{aligned} \mathcal {A}=\left( \begin{array}{ccccccc} 0 &{} I_{d} &{} 0 &{} 0 &{} 0 &{} 0 &{} 0 \\ \frac{\kappa }{\rho }\partial _{xx} &{} \frac{\kappa }{\rho }\partial _{x} &{} 0 &{} 0 &{} 0 &{} 0 &{} 0 \\ 0 &{} 0 &{} 0 &{} I_{d} &{} 0 &{} 0 &{} 0 \\ -\frac{\kappa }{I_{\rho }}\partial _{x} &{} 0 &{} \frac{\tilde{b}}{I_{\rho }} \partial _{xx}-\frac{\kappa }{I_{\rho }}I_{d} &{} 0 &{} 0 &{} 0 &{} -\frac{EI}{ I_{\rho }}\int _{0}^{\infty }g(s)\partial _{xx}{\mathrm{d}}s \\ \frac{k}{M}T_{1} &{} 0 &{} \frac{k}{M}T_{2} &{} 0 &{} 0 &{} 0 &{} 0 \\ 0 &{} 0 &{} \frac{b}{J}T_{3} &{} 0 &{} 0 &{} 0 &{} -\frac{b}{J}\int _{0}^{\infty }g(s)T_{4}{\mathrm{d}}s \\ 0 &{} 0 &{} I_{d} &{} 0 &{} 0 &{} 0 &{} -\partial _{s} \end{array} ,\right) , \end{aligned}$$

\(I_{d}\) denotes the identity operator and \(T_{i}\), \(i=1,2,3,4\) are the trace operators given by

$$\begin{aligned} \begin{array}{ll} T_{1}(w)=w_{x}(L,t), &{} T_{2}(\varphi )=\varphi (L,t), \\ T_{3}(\varphi )=\tilde{b}\varphi _{x}(L,t), &{} T_{4}(\eta )=\eta _{x}(L,t). \end{array} \end{aligned}$$
$$\begin{aligned} D(\mathcal {A})=\left\{ \begin{array}{c} V=(v_{1},v_{2},v_{3},v_{4},v_{5},v_{6},v_{7})^{T}\in \mathcal {H,}\quad \mathcal {A}V\in \mathcal {H,} \\ v_{7}(x,t,0)=0,\quad \mathbf {u}(t)=v_{2}(L,t),\quad \mathbf {v}(t)=v_{4}(L,t). \end{array} \right\} , \end{aligned}$$

It is not difficult to see that \(\mathcal {H}\) is a Hilbert space and that \(D\left( \mathcal {A}\right) \) is dense in \(\mathcal {H}\).

The energy of system (14) is given by

$$\begin{aligned} \mathcal {E}(t)=\frac{1}{2}\left\| U(t)\right\| _{\mathcal {H}}^{2}. \end{aligned}$$
(15)

Lemma 1

Let \(U=(v_{1},v_{2},v_{3},v_{4},v_{5},v_{6},v_{7})^{T}\) be a regular solution of problem (14). Then, the functional energy defined in (15) satisfies

$$\begin{aligned} \frac{\mathcal {E}(t)}{{\mathrm{d}}t}=\frac{1}{2}\int _{0}^{L}\int _{0}^{\infty }g^{\prime }(s)\mid v_{7x}\mid ^{2}{\mathrm{d}}s{\mathrm{d}}x\le 0,\quad \forall t\ge 0. \end{aligned}$$
(16)

From Lemma 1, we deduce that the system (14) is dissipative in the sense that its energy is a nonincreasing function with respect to the time variable t.

Now, we state and prove a result of existence, uniqueness and regularity of the solution.

Theorem 1

The operator \(\mathcal {A}\) defined in (14) is the infinitesimal generator of a \(C_{0}\)-semigroup S(t) of contraction in \(\mathcal {H}\). Thus, for any initial data \(U_{0} \in \mathcal {H}\), problem (14) has a unique weak solution \(U \in C\left( [0, \infty [, \mathcal {H} \right) \). Moreover, if \(U_{0} \in D\left( \mathcal {A}\right) \), then U is a strong solution of (14), that is, \( U \in C\left( [0, \infty [,D\left( \mathcal {A}\right) \right) \cap C^{1}\left( [0, \infty [, \mathcal {H} \right) \).

Proof

To prove Theorem 1, we use the semigroup approach and the Lumer–Phillips theorem. For this purpose, we show firstly that the operator \(\mathcal {A}\) is dissipative in the phase space \(\mathcal {H}\). Indeed, for any \(U(t)\in D(\mathcal {A})\),

$$\begin{aligned} \langle \mathcal {A}U,U\rangle _{\mathcal {H}}= & {} \frac{1}{2} \int _{0}^{L}\int _{0}^{\infty }g^{\prime }(s)\left| v_{7x}\right| ^{2}{\mathrm{d}}s{\mathrm{d}}x \nonumber \\\le & {} -\frac{\kappa }{2}\int _{0}^{L}\int _{0}^{\infty }g(s)\left| v_{7x}\right| ^{2}{\mathrm{d}}s{\mathrm{d}}x\le 0. \end{aligned}$$
(17)

which implies that the operator \(\mathcal {A}\) is a dissipative.

Now, for any \(F=\left( f_{1},f_{2},f_{3},f_{4},f_{5},f_{6},f_{7}\right) ^{T}\in \mathcal {H}\), we show the existence of \( U=(v_{1},v_{2},v_{3},v_{4},v_{5},v_{6},v_{7})^{T}\in D\left( \mathcal {A} \right) \), unique solution of the following equation

$$\begin{aligned} \lambda U-\mathcal {A}U=F. \end{aligned}$$
(18)

Equivalently, one must consider the system given by

$$\begin{aligned} \lambda v_{1}-v_{2}= & {} f_{1}, \end{aligned}$$
(19)
$$\begin{aligned} \lambda \rho v_{2}-\kappa \left[ v_{1x}+v_{3}\right] _{x}= & {} \rho f_{2}, \end{aligned}$$
(20)
$$\begin{aligned} \lambda v_{3}-v_{4}= & {} f_{3}, \end{aligned}$$
(21)
$$\begin{aligned} \lambda I_{\rho }v_{4}-\left( \tilde{b}v_{3xx}-\int _{0}^{\infty }g(s)v_{7xx}{\mathrm{d}}s\right) +\kappa \left[ v_{1x}+v_{3}\right]= & {} I_{\rho }f_{4}, \end{aligned}$$
(22)
$$\begin{aligned} M\lambda v_{5}-\kappa \left[ v_{1x}+v_{3}\right]= & {} Mf_{5}, \end{aligned}$$
(23)
$$\begin{aligned} J\lambda v_{6}-\left( \tilde{b}v_{3x}-\int _{0}^{\infty }g(s)v_{7x}{\mathrm{d}}s\right)= & {} Jf_{6}, \end{aligned}$$
(24)
$$\begin{aligned} \lambda v_{7}+v_{7s}-v_{4}= & {} f_{7}. \end{aligned}$$
(25)

We will show that \(0\in \rho (\mathcal {A})\). In fact, taking \(\lambda =0\) in (19-25), equivalently, we have the following system

$$\begin{aligned} -v_{2}= & {} f_{1}, \end{aligned}$$
(26)
$$\begin{aligned} -\kappa \left[ v_{1x}+v_{3}\right] _{x}= & {} \rho f_{2}, \end{aligned}$$
(27)
$$\begin{aligned} -v_{4}= & {} f_{3}, \end{aligned}$$
(28)
$$\begin{aligned} -\left( \tilde{b}v_{3xx}-\int _{0}^{\infty }g(s)v_{7xx}{\mathrm{d}}s\right) +\kappa \left[ v_{1x}+v_{3}\right]= & {} I_{\rho }f_{4}, \end{aligned}$$
(29)
$$\begin{aligned} -\kappa \left[ v_{1x}+v_{3}\right]= & {} Mf_{5}, \end{aligned}$$
(30)
$$\begin{aligned} -\left( \tilde{b}v_{3x}-\int _{0}^{\infty }g(s)v_{7x}{\mathrm{d}}s\right)= & {} Jf_{6}, \end{aligned}$$
(31)
$$\begin{aligned} v_{7s}-v_{4}= & {} f_{7}. \end{aligned}$$
(32)

From (26) and (28), we have

$$\begin{aligned} -v_{1}=f_{1} \end{aligned}$$
(33)

and

$$\begin{aligned} -v_{4}=f_{3}. \end{aligned}$$
(34)

The equation (32) implies

$$\begin{aligned} v_{7s}-v_{4}= & {} f_{7} \end{aligned}$$

that is

$$\begin{aligned} v_{7s}=f_{7}-f_{3}. \end{aligned}$$
(35)

Integrating (35) between 0 and s, we get

$$\begin{aligned} v_{7}(s)=\int \limits _{0}^{s}(f_{7}-f_{3})(\tau ){\mathrm{d}}\tau . \end{aligned}$$

Next, for the unique solvability of solution \(\left( v_{1},v_{3}\right) \) to the equations (27), (29) (30) and (31), we define the following bilinear form

$$\begin{aligned} a(\left( v_{1},v_{3}\right) ,\left( v_{2},v_{4}\right) )=b(w_{2},w_{4}), \end{aligned}$$
(36)

where

$$\begin{aligned} a(\left( v_{1},v_{3}\right) ,\left( w_{2},w_{4}\right) )= & {} \kappa \int _{0}^{L}\left[ v_{3}+v_{1x}\right] \left[ w_{4}+w_{2x}\right] {\mathrm{d}}x +\tilde{b}\int _{0}^{L}v_{3}w_{4x}{\mathrm{d}}x, \end{aligned}$$

and

$$\begin{aligned} b(w_{2},w_{4})= & {} \int _{0}^{L}f_{2}w_{2}{\mathrm{d}}x-f_{5}w_{5}+\int _{0}^{L}f_{4}w_{4}{\mathrm{d}}x-f_{6}w_{4} \\&+\int _{0}^{L}\int _{0}^{\infty }g(s)\int _{0}^{s}(f_{7}-f_{3})(\tau )w_{4x}{\mathrm{d}}\tau {\mathrm{d}}s{\mathrm{d}}x. \end{aligned}$$

It is easy to show that a is a bilinear continuous coercive form, and b is Linear continuous form. The conclusion follows from the Lax–Milgram. Thus, \(0\in \rho (\mathcal {A})\). By the resolvent identity, for small \(\lambda >0,\) we have \(R(\lambda I- \mathcal {A})=\mathcal {H}\) ( Theorem 1.2.4 in [24]), and therefore by Lumer–Phillips ([24] Theorem 4.3), we deduce that \(\mathcal {A}\) generates a \(C_{0}\)-semigroup of contraction \(\mathcal {S}(t)\) in \(\mathcal {H}\). The proof is thus complete.

4 Polynomial Rate of Decay

In this section, we want to prove Theorem 2, namely the polynomial stability of system (14) by using the frequency domain method (see [11]).

Theorem 2

Let \(\mathcal {S}(t)\) be a bounded \(C_{0}\)-semigroup on a Hilbert space \(\mathcal {H}\) with generator \(\mathcal {A}\) such that \(i\mathbb {R}\subset \rho (\mathcal {A}).\) Then,

$$\begin{aligned} \frac{1}{\left| \eta \right| ^{\alpha }}\left\| \left( i\eta I-\mathcal {A}\right) ^{-1}\right\| \le C,\quad \forall \eta \in \mathbb {R}\mathcal {\Leftrightarrow }\left\| S(t)\mathcal {A}^{-1}\right\| \le \frac{C}{t^{\frac{1}{\alpha }}}. \end{aligned}$$
(37)

Let introduce the following notations

$$\begin{aligned} \mathcal {L}^{2}=\int _{0}^{L}\left[ \rho v_{2}^{2}+I_{\rho }v_{4}^{2}+\left( \tilde{b}v_{3x}-\int _{0}^{\infty }g(s)\left| v_{7x}\right| \right) ^{2}+\kappa \left[ v_{3}+v_{1x}\right] ^{2}\right] {\mathrm{d}}x, \end{aligned}$$
$$\begin{aligned} \Pi _{1}(B)= & {} -\rho q\left| v_{2}(B)\right| ^{2}-q\kappa \left| v_{1x}(B)+v_{3}(B)\right| ^{2}, \\ \Pi _{2}(B)= & {} -I_{\rho }q\left| \varphi _{t}\right| ^{2}(B)-q\left| \tilde{b}\varphi _{x}(B)-\int _{0}^{\infty }g(s)\eta _{x}(B){\mathrm{d}}s\right| ^{2},\mathtt {\ }B=0,L. \end{aligned}$$

Lemma 2

Let us assume that \(\left( v_{1},v_{2},v_{3},v_{4}\right) \) is strong solution to the system (18), then there exists a positive constant c such that

$$\begin{aligned} \mathcal {L}^{2}\le c\left( \Pi _{1}(B)+\Pi _{2}(B)+\left\| F\right\| _{\mathcal {H}}^{2}\right. \end{aligned}$$

and

$$\begin{aligned} \Pi _{1}(B)+\Pi _{2}(B)\le c\left( \mathcal {L}^{2}+\left\| F\right\| _{\mathcal {H}}^{2}\right) . \end{aligned}$$

Proof

We will prove the inequalities for \(B=L.\) The case \(B=0\) is similar. Let \( \lambda \in i \mathbb {R} \) and \(q\in C^{2}(0,L)\) such that \(q(0)=0\) and

$$\begin{aligned} q_{x}\ge \tilde{c}, \end{aligned}$$
(38)

where \(\tilde{c}\) is a positive constant.

In order to get \(\left\| v_{3}+v_{1x}\right\| ^{2}\) and \(\left\| v_{2}\right\| ^{2}\), we multiply the Eq. (20) by \(q \overline{\left[ v_{3}+v_{1x}\right] },\) to obtain

$$\begin{aligned} \rho \int _{0}^{L}\lambda v_{2}q\overline{\left[ v_{3}+v_{1x}\right] } {\mathrm{d}}x-\kappa \int _{0}^{L}\left[ v_{3}+v_{1x}\right] _{x}q\overline{\left[ v_{3}+v_{1x}\right] }{\mathrm{d}}x=\rho \int _{0}^{L}f_{2}q\overline{\left[ v_{3}+v_{1x} \right] }{\mathrm{d}}x. \end{aligned}$$
(39)

Using (19) and (21), we have

$$\begin{aligned} \overline{\lambda v_{1x}}= & {} \overline{\left( v_{2x}+f_{1x}\right) }, \\ \overline{\lambda v_{3}}= & {} \overline{\left( v_{4}+f_{3}\right) }, \end{aligned}$$

Then, the first term in the left hand side gives

$$\begin{aligned} I= & {} \rho \int _{0}^{L}\lambda v_{2}q\overline{\left[ v_{3}+v_{1x}\right] } {\mathrm{d}}x=-\rho \int _{0}^{L}v_{2}q\overline{\left( v_{2x}+f_{1x}\right) }{\mathrm{d}}x-\rho \int _{0}^{L}v_{2}q\overline{\left( v_{2}+f_{3}\right) }{\mathrm{d}}x \\= & {} -\rho \int _{0}^{L}v_{2}q\overline{v_{2x}}{\mathrm{d}}x-\rho \int _{0}^{L}v_{2}q \overline{v_{4}}{\mathrm{d}}x-\rho \int _{0}^{L}v_{2}q_{1}\overline{\left( f_{3}+f_{1x}\right) }{\mathrm{d}}x \end{aligned}$$

Now, integrating by parts, we obtain

$$\begin{aligned} \rho \int _{0}^{L}\lambda v_{2}q\overline{\left[ v_{3}+v_{1x}\right] }{\mathrm{d}}x= & {} -\left[ \frac{\rho }{2}q\left| v_{2}\right| ^{2}\right] _{0}^{L}+ \frac{\rho }{2}\int _{0}^{L}q_{x}\left| v_{2}\right| ^{2}{\mathrm{d}}x\\&-\rho \int _{0}^{L}v_{2}q\overline{v_{4}}{\mathrm{d}}x-\rho \int _{0}^{L}v_{2}q\overline{\left( f_{3}+f_{1x}\right) }{\mathrm{d}}x. \end{aligned}$$

Therefore, taking the real part, we get

$$\begin{aligned}&\mathcal {R}e\int _{0}^{L}\rho \lambda v_{2}q\overline{\left[ v_{3}+v_{1x} \right] }{\mathrm{d}}x\\&\quad =-\frac{\rho }{2}\int _{0}^{L}q\frac{{\mathrm{d}}}{{\mathrm{d}}x}\left| v_{2}\right| ^{2}{\mathrm{d}}x-\mathcal {R}e\int _{0}^{L}\rho v_{2}q\overline{v_{4}} {\mathrm{d}}x-\rho \int _{0}^{L}v_{2}q\overline{\left( f_{3} +f_{1x}\right) }{\mathrm{d}}x.\\&\mathcal {R}e\int _{0}^{L}\rho \lambda v_{2}q\overline{\left[ v_{3}+v_{1x} \right] }{\mathrm{d}}x\\&\quad =-\frac{\rho }{2}\left[ q\left| v_{2}\right| ^{2}\right] _{0}^{L}+\frac{\rho }{2}\int _{0}^{L}q_{x}\left| v_{2}\right| ^{2}{\mathrm{d}}x- \mathcal {R}e\int _{0}^{L}\rho v_{2}q\overline{v_{4}}{\mathrm{d}}x+R_{1}, \end{aligned}$$

where \(\mathcal {R}e\) denotes the real part and \(R_{1}\) contains terms with \(f_{3}\) and \(f_{1}\) and satisfies

$$\begin{aligned} \left| R_{1}\right| \le c\left\| U\right\| _{\mathcal {H} }\left\| F\right\| _{\mathcal {H}}. \end{aligned}$$

Using (39), it follows that

$$\begin{aligned}&-\left[ \frac{1}{2}\left[ \rho q\left| v_{2}\right| ^{2}+\kappa q\left| v_{3}+v_{1x}\right| ^{2}\right] _{0}^{L}\right] \nonumber \\&\qquad +\frac{1}{2}\int _{0}^{L}\left( \rho q_{x}\left| v_{2}\right| ^{2}+\kappa q_{1x}\left| v_{3}+v_{1x}\right| ^{2}\right) {\mathrm{d}}x \nonumber \\&\qquad -\mathcal {R}e\int _{0}^{L}\rho v_{2}q\overline{v_{4}}{\mathrm{d}}x \nonumber \\&\quad =\rho \int _{0}^{L}f_{2}q\overline{\left[ v_{3}+v_{1x}\right] }{\mathrm{d}}x+R_{1} \nonumber \\&\quad =R_{2}, \end{aligned}$$
(40)

where \(R_{2}\) can be estimated by

$$\begin{aligned} \left| R_{2}\right| \le c\left\| U\right\| _{\mathcal {H} }\left\| F\right\| _{\mathcal {H}}. \end{aligned}$$

Similarly, in order to get \(\left\| \tilde{b}v_{3x}-\int _{0}^{\infty }g(s)v_{7x}{\mathrm{d}}s\right\| ^{2}\) and \(\left\| v_{4}\right\| ^{2}\), we multiply the Eq. (22) by \(q\left( \overline{\tilde{b} v_{3x}-\int _{0}^{\infty }g(s)v_{7x}{\mathrm{d}}s}\right) \) to obtain

$$\begin{aligned}&\int _{0}^{L}\lambda I_{\rho }v_{4}q\left( \overline{\tilde{b} v_{3x}-\int _{0}^{\infty }g(s)v_{7x}{\mathrm{d}}s}\right) {\mathrm{d}}x \nonumber \\&-\int _{0}^{L}q\left( \overline{\tilde{b}v_{3x}-\int _{0}^{\infty }g(s)v_{7x}{\mathrm{d}}s}\right) _{x}\left( \overline{\tilde{b}v_{3x}-\int _{0}^{\infty }g(s)v_{7x}{\mathrm{d}}s}\right) {\mathrm{d}}x \nonumber \\&+\int _{0}^{L}\kappa \left[ v_{3}+v_{1x}\right] q\left( \overline{\tilde{b} v_{3x}-\int _{0}^{\infty }g(s)v_{7x}{\mathrm{d}}s}\right) {\mathrm{d}}x \nonumber \\= & {} I_{\rho }\int _{0}^{L}f_{4}q\left( \overline{\tilde{b}v_{3x}-\int _{0}^{ \infty }g(s)v_{7x}{\mathrm{d}}s}\right) {\mathrm{d}}x. \end{aligned}$$
(41)

Using (21) and (25), we have

$$\begin{aligned} \overline{\lambda v_{3x}}= & {} \overline{\left( v_{4x}+f_{3x}\right) },\nonumber \\ \overline{\lambda v_{7x}}= & {} \overline{f_{7}+v_{4}-v_{7s}}, \end{aligned}$$
(42)

Now, taking the real part of the first term in the left hand of (41), we arrive at

$$\begin{aligned} J_{1}= & {} \mathcal {R}e\int _{0}^{L}\lambda I_{\rho }v_{4}q\left( \overline{ \tilde{b}v_{3x}-\int _{0}^{\infty }g(s)v_{7x}{\mathrm{d}}s}\right) {\mathrm{d}}x \\= & {} -\mathcal {R}e\int _{0}^{L}I_{\rho }\tilde{b}qv_{4}\overline{\lambda v_{3x}} {\mathrm{d}}x+\mathcal {R}e\int _{0}^{L}I_{\rho }v_{4}q\int _{0}^{\infty }g(s)\quad \overline{\lambda v_{7x}}{\mathrm{d}}s{\mathrm{d}}x. \end{aligned}$$

Then, using (42), we deduce

$$\begin{aligned} J_{1}= & {} -\mathcal {R}e\int _{0}^{L}I_{\rho }v_{4}q\overline{\left( v_{4x}+f_{3x}\right) }{\mathrm{d}}x+\mathcal {R}e\int _{0}^{L}I_{\rho }v_{4}q\int _{0}^{\infty }g(s)\overline{\left( f_{7,x}+v_{4x}-v_{7xs}\right) } {\mathrm{d}}s{\mathrm{d}}x \\= & {} -\mathcal {R}e\int _{0}^{L}I_{\rho }v_{4}q\overline{v_{4x}}{\mathrm{d}}x+\mathcal {R} e\int _{0}^{L}I_{\rho }v_{4}q\int _{0}^{\infty }g(s)\overline{v_{4x}}{\mathrm{d}}s{\mathrm{d}}x\\&- \mathcal {R}e\int _{0}^{L}\int _{0}^{\infty }I_{\rho }v_{4}qg(s)\overline{ v_{7xs}}{\mathrm{d}}s{\mathrm{d}}x \\&+\mathcal {R}e\int _{0}^{L}\int _{0}^{\infty }I_{\rho }v_{4}qg(s)\overline{ f_{7x}}{\mathrm{d}}s{\mathrm{d}}x-\mathcal {R}e\int _{0}^{L}I_{\rho }v_{4}q\overline{f_{3x}}{\mathrm{d}}x \\= & {} -\frac{1}{2}\mathcal {R}e\int _{0}^{L}I_{\rho }q\frac{{\mathrm{d}}}{{\mathrm{d}}x}\left| v_{4}\right| ^{2}{\mathrm{d}}x+\mathcal {R}e\int _{0}^{L}I_{\rho }q\int _{0}^{\infty }g(s)v_{4}\overline{v_{4x}}{\mathrm{d}}s{\mathrm{d}}x \\&-\mathcal {R}e\int _{0}^{L}I_{\rho }q\int _{0}^{\infty }g(s)v_{4}\overline{ v_{7xs}}{\mathrm{d}}s{\mathrm{d}}x-\mathcal {R}e\int _{0}^{L}I_{\rho }v_{4}q\overline{f_{3x}}{\mathrm{d}}x \\&+\mathcal {R}e\int _{0}^{L}I_{\rho }q\int _{0}^{\infty }g(s)v_{4}\overline{ f_{7x}}{\mathrm{d}}s{\mathrm{d}}x. \\= & {} -\frac{1}{2}\left[ I_{\rho }q\left| v_{4}\right| ^{2}\right] _{0}^{L}+\frac{1}{2}\mathcal {R}e\int _{0}^{L}I_{\rho }q_{x}\left| v_{4}\right| ^{2}{\mathrm{d}}x \\&+\mathcal {R}e\int _{0}^{L}I_{\rho }q\int _{0}^{\infty }g(s)v_{4}\overline{ v_{4x}}{\mathrm{d}}s{\mathrm{d}}x-\mathcal {R}e\int _{0}^{L}I_{\rho }q\int _{0}^{\infty }g(s)v_{4} \overline{v_{7xs}}{\mathrm{d}}s{\mathrm{d}}x \\&-\mathcal {R}e\int _{0}^{L}I_{\rho }qv_{4}\overline{f_{3x}}{\mathrm{d}}x+\mathcal {R} e\int _{0}^{L}I_{\rho }q\int _{0}^{\infty }g(s)v_{4}\overline{f_{7x}}{\mathrm{d}}s{\mathrm{d}}x. \end{aligned}$$

Integrating by parts the second term of the left hand side of (41), we get

$$\begin{aligned} J_{2}=-\frac{1}{2}\left( q\left| \tilde{b}v_{3x}-\int _{0}^{\infty }g(s)v_{7x}{\mathrm{d}}s\right| ^{2}\right) _{0}^{L}+\frac{1}{2}\int _{0}^{L}q_{x} \left| \tilde{b}v_{3x}-\int _{0}^{\infty }g(s)v_{7x}{\mathrm{d}}s\right| ^{2}{\mathrm{d}}x. \end{aligned}$$

Combining Eqs. (20)–(21), we obtain

$$\begin{aligned} -\kappa \left( v_{3}+v_{1x}\right) _{x}= & {} \rho f_{2}-\lambda \rho v_{2}, \end{aligned}$$
(43)
$$\begin{aligned} \overline{v_{3}}= & {} \overline{\frac{1}{\lambda }\left( v_{4}+f_{3}\right) }, \end{aligned}$$
(44)
$$\begin{aligned} \overline{v_{7}}= & {} \overline{\frac{1}{\lambda }\left( f_{7}+v_{4}-v_{7s}\right) }, \end{aligned}$$
(45)

Now, integrating by parts the third term of the left hand side of (41) and using (20), we deduce

$$\begin{aligned} J_{3}= & {} \int _{0}^{L}\kappa \left[ v_{3}+v_{1x}\right] q\left( \overline{ \tilde{b}v_{3x}-\int _{0}^{\infty }g(s)v_{7x}{\mathrm{d}}s}\right) {\mathrm{d}}x \end{aligned}$$
(46)
$$\begin{aligned}= & {} \left[ \kappa \left( v_{3}+v_{1x}\right) q\left( \overline{\tilde{b} v_{3}-\int _{0}^{\infty }g(s)v_{7}{\mathrm{d}}s}\right) \right] _{0}^{L} \end{aligned}$$
(47)
$$\begin{aligned}&-\int _{0}^{L}\kappa \left( v_{3}+v_{1x}\right) _{x}q\left( \overline{ \tilde{b}v_{3}-\int _{0}^{\infty }g(s)v_{7}{\mathrm{d}}s}\right) {\mathrm{d}}x \end{aligned}$$
(48)
$$\begin{aligned}&-\int _{0}^{L}\kappa \left( v_{3}+v_{1x}\right) q_{x}\left( \overline{ \tilde{b}v_{3}-\int _{0}^{\infty }g(s)v_{7}{\mathrm{d}}s}\right) {\mathrm{d}}x \end{aligned}$$
(49)
$$\begin{aligned}= & {} \left[ \kappa \left( v_{3}+v_{1x}\right) q\left( \overline{\tilde{b} v_{3}-\int _{0}^{\infty }g(s)v_{7}{\mathrm{d}}s}\right) \right] _{0}^{L}+J_{3}^{\prime }+J_{3}^{\prime \prime }, \end{aligned}$$
(50)

where

$$\begin{aligned} J_{3}^{\prime }=-\int _{0}^{L}\kappa \left( v_{3}+v_{1x}\right) _{x}\tilde{b}q \overline{v_{3}}{\mathrm{d}}x-\int _{0}^{L}\kappa \left( v_{3}+v_{1x}\right) \tilde{b} q_{x}\overline{v_{3}}{\mathrm{d}}x, \end{aligned}$$
(51)

and

$$\begin{aligned} J_{3}^{\prime \prime }=\int _{0}^{L}\kappa \left( v_{3}+v_{1x}\right) _{x}q\int _{0}^{\infty }g(s)\overline{v_{7}}{\mathrm{d}}s{\mathrm{d}}x+\int _{0}^{L}\kappa \left( v_{3}+v_{1x}\right) q_{x}\int _{0}^{\infty }g(s)\overline{v_{7}}{\mathrm{d}}s{\mathrm{d}}x. \end{aligned}$$
(52)

Substituting (43) and (44) in (51), we get

$$\begin{aligned} J_{3}^{\prime }= & {} -\int _{0}^{L}\kappa \left( v_{3}+v_{1x}\right) _{x}\tilde{ b}q\overline{v_{3}}{\mathrm{d}}x-\int _{0}^{L}\kappa \left( v_{3}+v_{1x}\right) \tilde{b} q_{x}\overline{v_{3}}{\mathrm{d}}x \nonumber \\= & {} \int _{0}^{L}\left( \rho f_{2}-\lambda \rho v_{2}\right) \tilde{b}q \overline{\frac{1}{\left| \lambda \right| }\left( v_{4}+f_{3}\right) }{\mathrm{d}}x-\int _{0}^{L}\kappa \left( v_{3}+v_{1x}\right) \tilde{b}q_{x}\overline{ \frac{1}{\left| \lambda \right| }\left( v_{4}+f_{3}\right) }{\mathrm{d}}x \nonumber \\= & {} \int _{0}^{L}\rho v_{2}\tilde{b}q\overline{v_{4}}{\mathrm{d}}x-\frac{1}{\left| \lambda \right| }\int _{0}^{L}\rho f_{2}\tilde{b}q\overline{\left( v_{4}+f_{3}\right) }{\mathrm{d}}x \nonumber \\&+\frac{1}{\left| \lambda \right| }\int _{0}^{L}\kappa \left( v_{3}+v_{1x}\right) \tilde{b}q_{x}\overline{\left( v_{4}+f_{3}\right) } {\mathrm{d}}x+\int _{0}^{L}\rho v_{2}\tilde{b}q\overline{f_{3}}{\mathrm{d}}x. \end{aligned}$$
(53)

Similarly, using (43) and (45), \(J_{3}^{\prime \prime }\) becomes

$$\begin{aligned} J_{3}^{\prime \prime }= & {} \int _{0}^{L}\kappa \left( v_{3}+v_{1x}\right) _{x}q\int _{0}^{\infty }g(s)\overline{v_{7}}{\mathrm{d}}s{\mathrm{d}}x+\int _{0}^{L}\kappa \left( v_{3}+v_{1x}\right) q_{x}\int _{0}^{\infty }g(s)\overline{v_{7}}{\mathrm{d}}s{\mathrm{d}}x \nonumber \\= & {} \int _{0}^{L}\left( \lambda \rho v_{2}-\rho f_{2}\right) q\int _{0}^{\infty }g(s)\overline{\frac{1}{\left| \lambda \right| }\left( f_{7}+v_{4}-v_{7s}\right) }{\mathrm{d}}s{\mathrm{d}}x \nonumber \\&+\int _{0}^{L}\kappa \left( v_{3}+v_{1x}\right) q_{x}\int _{0}^{\infty }g(s) \overline{\frac{1}{\left| \lambda \right| }\left( f_{7}+v_{4}-v_{7s}\right) }{\mathrm{d}}s{\mathrm{d}}x \nonumber \\= & {} -\int _{0}^{L}\rho v_{2}q\int _{0}^{\infty }g(s)\overline{\left( f_{7}+v_{4}-v_{7s}\right) }{\mathrm{d}}s{\mathrm{d}}x \nonumber \\&+\frac{1}{\left| \lambda \right| }\int _{0}^{L}\rho f_{2}q\int _{0}^{\infty }g(s)\overline{\left( f_{7}+v_{4}-v_{7s}\right) }{\mathrm{d}}s{\mathrm{d}}x \nonumber \\&-\frac{1}{\left| \lambda \right| }\int _{0}^{L}\kappa \left( v_{3}+v_{1x}\right) q_{x}\int _{0}^{\infty }g(s)\overline{\left( f_{7}+v_{4}-v_{7s}\right) }{\mathrm{d}}s{\mathrm{d}}x. \end{aligned}$$
(54)

Injecting (53) and (54) in (50), we get

$$\begin{aligned} J_{3}= & {} \int _{0}^{L}\rho v_{2}\tilde{b}q\overline{v_{4}}{\mathrm{d}}x-\int _{0}^{L}\rho v_{2}q\int _{0}^{\infty }g(s)\overline{v_{4}}{\mathrm{d}}s{\mathrm{d}}x \\&+\int _{0}^{L}\rho v_{2}q\int _{0}^{\infty }g(s)\overline{\eta _{s}} {\mathrm{d}}s{\mathrm{d}}x+P(L)+R_{4}, \end{aligned}$$

where

$$\begin{aligned} P(L)=\left[ \kappa \left( v_{3}+v_{1x}\right) q\left( \overline{\tilde{b} v_{3}+\int _{0}^{\infty }g(s)v_{7}{\mathrm{d}}s}\right) \right] _{0}^{L}, \end{aligned}$$

and

$$\begin{aligned} R_{4}= & {} -\frac{1}{\left| \lambda \right| }\int _{0}^{L}f_{2}\tilde{b} q\overline{\left( v_{4}+f_{3}\right) }{\mathrm{d}}x+\frac{1}{\left| \lambda \right| }\int _{0}^{L}f_{2}q\int _{0}^{\infty }g(s)\overline{\left( f_{7}+v_{4}-v_{7s}\right) }{\mathrm{d}}s{\mathrm{d}}x \\&+\frac{1}{\left| \lambda \right| }\int _{0}^{L}\kappa \left( v_{3}+v_{1x}\right) q_{x}\overline{\left( v_{4}+f_{3}\right) }{\mathrm{d}}x \\&-\frac{1}{\left| \lambda \right| }\int _{0}^{L}\kappa \left( v_{3}+v_{1x}\right) q_{x}\int _{0}^{\infty }g(s)\overline{\left( f_{7}+v_{4}-v_{7s}\right) }{\mathrm{d}}s{\mathrm{d}}x. \\&+\int _{0}^{L}\rho v_{2}q\overline{f_{3}}{\mathrm{d}}x-\int _{0}^{L}\rho v_{2}q\int _{0}^{\infty }g(s)\overline{f_{7}}{\mathrm{d}}s{\mathrm{d}}x. \end{aligned}$$

Making use Young’s inequality, we obtain

$$\begin{aligned} \frac{1}{\left| \lambda \right| }\int _{0}^{L}\kappa \left( v_{3}+v_{1x}\right) q_{x}\overline{v_{4}}{\mathrm{d}}x\le \frac{c}{\left| \lambda \right| }\left\| U\right\| _{\mathcal {H}}^{2} \end{aligned}$$
(55)

and

$$\begin{aligned}&\frac{1}{\left| \lambda \right| }\int _{0}^{L}\kappa \left( v_{3}+v_{1x}\right) q_{x}\int _{0}^{\infty }g(s)\overline{\left( f_{7}+v_{4}-v_{7s}\right) }{\mathrm{d}}s{\mathrm{d}}x\nonumber \\&\quad \le \frac{c}{\left| \lambda \right| }\left\| U\right\| _{\mathcal {H}}\left\| F\right\| _{ \mathcal {H}} +\frac{c}{\left| \lambda \right| }\left\| U\right\| _{ \mathcal {H}}^{2}. \end{aligned}$$
(56)

Combining (55) and (56), we arrive at

$$\begin{aligned} \left| R_{4}\right| \le \frac{c}{\left| \lambda \right| } \left\| U\right\| _{\mathcal {H}}\left\| F\right\| _{\mathcal {H}}+ \frac{c}{\left| \lambda \right| }\left\| U\right\| _{\mathcal {H }}^{2}. \end{aligned}$$

Summing the estimates of \(J_{1},\) \(J_{2}\) and \(J_{3}\), we obtain

$$\begin{aligned}&-\left[ \frac{I_{\rho }}{2}q\left| v_{4}\right| ^{2}+\frac{q}{2} \left| \tilde{b}v_{3x}-\int _{0}^{\infty }g(s)v_{7x}{\mathrm{d}}s\right| ^{2} \right] _{0}^{L} \nonumber \\&+\frac{I_{\rho }}{2}\mathcal {R}e\int _{0}^{L}q_{x}\left| v_{4}\right| ^{2}{\mathrm{d}}x+\frac{1}{2}\int _{0}^{L}q_{x}\left| \tilde{b} v_{3x}-\int _{0}^{\infty }g(s)v_{7x}{\mathrm{d}}s\right| ^{2}{\mathrm{d}}x \nonumber \\&-\mathcal {R}e\int _{0}^{L}I_{\rho }v_{4}q\overline{f_{3x}}{\mathrm{d}}x +\mathcal {R} e\int _{0}^{L}I_{\rho }v_{4}q\int _{0}^{\infty }g^{\prime }(s)\overline{v_{7x}} {\mathrm{d}}s{\mathrm{d}}x\nonumber \\&+\mathcal {R}e\int _{0}^{L}\left( \int _{0}^{\infty }g(s){\mathrm{d}}s\right) I_{\rho }v_{4}q\overline{v_{4x}}{\mathrm{d}}x +\mathcal {R}e\int _{0}^{L}\rho v_{2}\tilde{b}q\overline{v_{4}}{\mathrm{d}}x\nonumber \\&-\mathcal {R} e\int _{0}^{L}\left( \int _{0}^{\infty }g(s){\mathrm{d}}s\right) \rho v_{2}q\overline{ v_{4}}{\mathrm{d}}x+\mathcal {R}e\int _{0}^{L}I_{\rho }q\int _{0}^{\infty }g(s)v_{4} \overline{f_{7x}}{\mathrm{d}}s{\mathrm{d}}x \nonumber \\&-\mathcal {R}e\int _{0}^{L}\rho v_{2}q\int _{0}^{\infty }g^{\prime }(s) \overline{v}{\mathrm{d}}s{\mathrm{d}}x \nonumber \\= & {} -\mathcal {R}e\left[ R_{4}+P(L)\right] +\mathcal {R}e\int _{0}^{L}I_{\rho }v_{4}q\overline{f_{3x}}{\mathrm{d}}x-\mathcal {R}e\int _{0}^{L}I_{\rho }q\int _{0}^{\infty }g(s)v_{4}\overline{f_{7x}}{\mathrm{d}}s{\mathrm{d}}x. \nonumber \\ \end{aligned}$$
(57)

The right hand side of (57) can be easily estimated by

$$\begin{aligned}&-\mathcal {R}e\left[ R_{4}+P(L)\right] +\mathcal {R}e\int _{0}^{L}I_{\rho }v_{4}q\overline{f_{3x}}{\mathrm{d}}x -\mathcal {R}e\int _{0}^{L}I_{\rho }q\int _{0}^{\infty }g(s)v_{4}\overline{ f_{7x}}{\mathrm{d}}s{\mathrm{d}}x \\&\quad \le \frac{c}{\left| \lambda \right| }\left\| U\right\| _{ \mathcal {H}}\left\| F\right\| _{\mathcal {H}}+\frac{c}{\left| \lambda \right| }\left\| U\right\| _{\mathcal {H}}^{2}. \end{aligned}$$

Adding up the equalities, (17) and (57), we conclude that

$$\begin{aligned}&\frac{I_{\rho }}{2}\int _{0}^{L}q_{x}\left| v_{4}\right| ^{2}{\mathrm{d}}x+ \frac{1}{2}\int _{0}^{L}q_{x}\left| \tilde{b}v_{3x}-\int _{0}^{\infty }g(s)v_{7x}{\mathrm{d}}s\right| ^{2}{\mathrm{d}}x \nonumber \\&\qquad +\frac{\rho }{2}\int _{0}^{L}q_{x}\left| v_{2}\right| ^{2}{\mathrm{d}}x+\frac{ \kappa }{2}\int _{0}^{L}q_{x}\left| v_{3}+v_{1x}\right| ^{2}{\mathrm{d}}x \nonumber \\&\quad =\Pi _{1}(L)+\Pi _{2}(L)+R_{5}, \end{aligned}$$
(58)

where \(R_{5}\) is estimated as follows:

$$\begin{aligned} \left| R_{5}\right| \le \left| R_{4}\right| \le \left\| U\right\| _{\mathcal {H}}\left\| F\right\| _{\mathcal {H}}+\frac{c}{ \left| \lambda \right| }\left\| U\right\| _{\mathcal {H}}^{2}. \end{aligned}$$

Taking the real part of (58) and using (38), we get

$$\begin{aligned}&\frac{I_{\rho }}{2}\int _{0}^{L}q_{x}\left| v_{4}\right| ^{2}{\mathrm{d}}x+ \frac{1}{2}\int _{0}^{L}q_{x}\left| \tilde{b}v_{3x}-\int _{0}^{\infty }g(s)v_{7x}{\mathrm{d}}s\right| ^{2}{\mathrm{d}}x \\&+\frac{\rho }{2}\int _{0}^{L}q_{x}\left| v_{2}\right| ^{2}{\mathrm{d}}x+\frac{ \kappa }{2}\int _{0}^{L}q_{x}\left| v_{3}+v_{1x}\right| ^{2}{\mathrm{d}}x\ge \tilde{C}\mathcal {L}^{2}. \end{aligned}$$

Setting

$$\begin{aligned} \mathcal {L}^{2}=\int _{0}^{L}\left[ \rho v_{2}^{2}+I_{\rho }v_{4}^{2}+\left( \tilde{b}v_{3x}-\int _{0}^{\infty }g(s)v_{7x}{\mathrm{d}}s\right) ^{2}+\kappa \left[ v_{3}+v_{1x}\right] ^{2}\right] {\mathrm{d}}x, \end{aligned}$$

then

$$\begin{aligned}&\left[ \frac{I_{\rho }}{2}\int _{0}^{L}\left| v_{4}\right| ^{2}{\mathrm{d}}x+ \frac{1}{2}\int _{0}^{L}\left| \tilde{b}v_{3x}-\int _{0}^{\infty }g(s)v_{7x}{\mathrm{d}}s\right| ^{2}{\mathrm{d}}x\right. \\&\left. +\frac{\rho }{2}\int _{0}^{L}q_{x}\left| v_{2}\right| ^{2}{\mathrm{d}}x+ \frac{\kappa }{2}\int _{0}^{L}q_{x}\left| v_{3}+v_{1x}\right| ^{2}{\mathrm{d}}x \right] \\\le & {} \tilde{c}_{0}\left( \Pi _{1}(L)+\Pi _{2}(L)+c\left\| U\right\| _{\mathcal {H}}\left\| F\right\| _{\mathcal {H}}+\frac{c}{\left| \lambda \right| }\left\| U\right\| _{\mathcal {H}}^{2}\right) , \end{aligned}$$

also, we have

$$\begin{aligned} \Pi _{1}(L)+\Pi _{2}(L)\le \tilde{c}_{0}\left( \left\| U\right\| _{\mathcal { H}}^{2}+\left\| U\right\| _{\mathcal {H}}\left\| F\right\| _{ \mathcal {H}}\right) , \end{aligned}$$

Taking \(\lambda \) large enough, our conclusion follows.

Theorem 3

The solution of the Timoshenko system with past history decays polynomially as

$$\begin{aligned} \left\| U(t)\right\| _{\mathcal {H}} \le \frac{1}{\sqrt{t}}\left\| U_{0}\right\| _{D(\mathcal {A})}. \end{aligned}$$

Proof

First we show that \(i \mathbb {R} \subset \) \(\rho (\mathcal {A})\). In fact, since \(\mathcal {A}\) is a closed operator and \(D(\mathcal {A})\) has compact embedding on the phase space \( \mathcal {H}\) , we conclude that the spectrum \(\sigma (\mathcal {A})\) is discrete. Therefore we show that there are no imaginary eigenvalues. By contradiction, let us suppose that there exits an imaginary eigenvalue, then \( i\lambda \in \sigma (\mathcal {A})\). Then, there exists \(U\ne 0\) satisfying

$$\begin{aligned} \mathcal {A}U\mathcal {=}i\lambda U. \end{aligned}$$
(59)

Taking the inner product of (59) with U and using (40), we get

$$\begin{aligned} \mathcal {R}e\left( i\lambda \left\| U(t)\right\| _{\mathcal {H} }^{2}\right) =\mathcal {R}e\left( \mathcal {A}U,U\right) _{\mathcal {H}}=0. \end{aligned}$$

It follows that w, \(\varphi \), satisfies

$$\begin{aligned} \left\{ \begin{array}{l} -\kappa \left[ \varphi _{x}+w_{xx}\right] +\rho \lambda ^{2}w=0 \\ -EI\left( \tilde{b}\varphi _{xx}+\int _{0}^{\infty }g(s)\eta _{xx}(x,s){\mathrm{d}}s\right) +\kappa \left[ \varphi +w_{x}\right] +I_{\rho }\lambda ^{2}\varphi =0 \\ w(L)=\varphi (L)=w_{x}(L)=\varphi _{x}(L)=0. \end{array} \right. \end{aligned}$$
(60)

We can write the first three equation in (60) as follows:

$$\begin{aligned} \left\{ \begin{array}{l} X^{\prime }=MX,\mathtt {\ in\ }(0,L) \\ X(L)=0. \end{array} \right. \end{aligned}$$

with \(X=\left( w,w_{x},\varphi ,\varphi _{x}\right) ^{T}\) and

$$\begin{aligned} M=\left( \begin{array}{cccc} 0 &{} I_{d} &{} 0 &{} 0 \\ -\frac{\rho }{\kappa }\lambda ^{2} &{} 0 &{} I_{d} &{} 0 \\ 0 &{} 0 &{} I_{d} &{} 0 \\ 0 &{} -\frac{\kappa }{EI\tilde{b}} &{} 0 &{} -\frac{\kappa +I_{\rho }\lambda ^{2}}{ EI\tilde{b}} \end{array} \right) , \end{aligned}$$

Using ordinary differential equation theory, we deduce that system (60 ) has a unique trivial solution \(X=0\) in (0, L). Which implies that \(w=c\), \(\varphi =0,\) in (0, L). Because \(w(0)=0,\) we conclude that \(w=0\) in (0, L). Then, \(U=0,\) which is a contradiction.

Therefore, \(i \mathbb {R} \subset \rho (\mathcal {A}).\)

Finally, using (40) and Lemma 2, we conclude for \(\lambda \) large

$$\begin{aligned} \left\| U\right\| _{\mathcal {H}}^{2}\le & {} c\Pi _{1}(B)+c\Pi _{2}(B)+c\left\| F\right\| _{\mathcal {H}}^{2} \\\le & {} c\left| \lambda \right| ^{2}\left| v_{5}\right| ^{2}+c\left| \lambda \right| ^{2}\left| v_{6}\right| ^{2}+c\left\| F\right\| _{\mathcal {H}}^{2} \\\le & {} c\left| \lambda \right| ^{2}\left\| U\right\| _{ \mathcal {H}}\left\| F\right\| _{\mathcal {H}}+c\left\| F\right\| _{\mathcal {H}}^{2}. \end{aligned}$$

For a constant \(C>0\), we conclude that

$$\begin{aligned} \frac{1}{\left| \lambda \right| ^{2}}\left\| U\right\| _{ \mathcal {H}}\le C. \end{aligned}$$

Finally, using Theorem 2 our result holds true.