Uniform stabilization for a Timoshenko beam system with delays in fractional order internal dampings

Abstract

This work is concerned with a problem of a vibrating system of Timoshenko-type in a bounded one-dimensional domain under Dirichlet boundary conditions with two fractional time delays and two internal frictional dampings. Under a smallness condition on the fractional delay and by using a classical semigroup theory we prove existence and uniqueness of solutions. Furthermore, by a frequency domain approach we prove an exponential stability result.

1 Introduction

In this work, we deal with the existence and the internal stabilization of the Timoshenko beam system in bounded interval (0, L) in the presence of time fractional delays, that is,

where t denotes the time variable and x is the space variable in (0, L). Here, $w = w(x,t)$ is the transverse displacement of the beam and $\chi = \chi (x,t)$ is the rotation angle of the filament of the beam. The coefficients $\rho _1, \rho _2, K, b$ are, respectively, the density (the mass per unit length), the polar moment of inertia of a cross section, the shear modulus and Young’s modulus of elasticity times the moment of inertia of a cross section. The parameters $a_2$ and ${\tilde{a}}_2$ are positive constants and $a_1$ and ${\tilde{a}}_1$ are real constants. The Caputo’s fractional derivative $\partial _{t}^{\sigma , \kappa }$ of order $\sigma \in (0, 1)$ with respect to time variable t is defined by

$$\begin{aligned} \partial _{t}^{\sigma , \kappa }h(t)=\displaystyle \frac{1}{\Gamma (1-\sigma )} \displaystyle \int _{0}^{t} (t-s)^{-\sigma }e^{-\kappa (t-s)}\displaystyle \frac{dh}{ds}(s)\, ds,\ \kappa > 0 \end{aligned}$$

and $\tau _1, \tau _2 > 0$ are the time delays. We complement the system (P) with the following initial and boundary conditions

The initial data $(w_0, w_1,f_{0},\chi _{0},\chi _{1},g_{0})$ belongs to a suitable functional space.

The above problem can be regarded as a problem with a memory acting only on the time interval $(0, t-\tau )$.

During the last few years, an important amount of research has been devoted to the issue of the stabilization of Timoshenko system with various damping mechanisms. We mention some of these results.

In the case where no time delay is considered (i.e. $a_1 ={{{\tilde{a}}}}_{1} = 0$), Raposo et al. [24] proved the uniform decay of the solution for the following linear Timoshenko beam system with linear internal frictional dampings:

$$\begin{aligned} \left\{ \begin{array}{l} \rho _{1}w_{tt}-K(w_{x}+\chi )_{x}+w_{t}=0,\\ \rho _{2}\chi _{tt}-b\chi _{xx}+K(w_{x}+\chi )+\chi _{t}=0.\\ \end{array}\right. \end{aligned}$$

Messaoudi and Mustafa [19] (see also [21]) investigated the following Timoshenko beam system with nonlinear internal feedbacks:

$$\begin{aligned} \left\{ \begin{array}{l} \rho _{1}w_{tt}-K(w_{x}+\chi )_{x}+g_1(w_{t})=0,\\ \rho _{2}\chi _{tt}-b\chi _{xx}+K(w_{x}+\chi )+ g_2(\chi _{t})=0.\\ \end{array}\right. \end{aligned}$$

They established exponential and polynomial decay rate results. The case of a Timoshenko beam system with weakly nonlinear internal feedbacks was considered by Park and Kang in [21]. Shi and Feng [25] investigated the case of the uniform Timoshenko beam under two locally distributed feedbacks and proved that the vibration of the beam decays exponentially. They used the frequency domain approach combined with a multiplier method. Xu and Young [28] established an exponential stability of the uniform Timoshenko beam system by two pointwise control.

When time delay is present, the result concerning existence and asymptotic behavior of evolution systems have been established by many authors. They have shown that delays may destabilize a system which is uniformly asymptotically stable in the absence of delay, see [10] for more details.

Recenly, Guesmia and Soufiane [14] have considered the Timoshenko beam system with internal frictional dampings and discrete time delays, that is,

It is proved that the dissipation induced by the internal dampings is strong enough to stabilize (TM) in the presence of a small delays.

As mentioned before our internal fractional terms in (P) can be compared to internal terms with memory, the difference being that, for such a condition with memory, $\tau _i =0$. Indeed, it has been shown (see [1,2,3, 5, 7, 18] and [4]) that the fractional derivative $\partial _t^{\sigma , \kappa }$ forces the system to become dissipative and the solution to converge to the equilibrium state. In [7] Ammari et al., studied the stabilization for a class of evolution systems with fractional damping. They proved the polynomial stability of the system.

In [20], the authors consider the following wave equation with internal damping and time-distributed delay, that is,

$$\begin{aligned} u_{tt}-\Delta u+\mu _0 u_t+\displaystyle \int _{\tau _1}^{\tau _2}\mu (s)u_t(x, t-s)\, ds=0 \quad (x, t)\in \Omega \times (0, \infty ), \end{aligned}$$

where is an open bounded domain with a smooth boundary. They established an exponential decay result under the assumption

$$\begin{aligned} \mu _0\ge \displaystyle \int _{\tau _1}^{\tau _2}\mu (s)\, ds. \end{aligned}$$

Guesmia [13] considered the following Timoshenko system with infinite memory and distributed time delay, that is,

He proved an exponential decay result (by using suitable Lyapunov functionals) under the conditions

(A2) The function f is of class and satisfies, for some positive constant $\varpi$,

$$\begin{aligned} |f(s)|\le \varpi g(s) \hbox { and } |f'(s)|\le \varpi g(s). \end{aligned}$$

On the contrary to the discrete time delay models, which ignore the inherent memory effects, the delayed internal fractional term considered in this paper do take into account the past history of the solution. This is what makes the present case more realistic. In fact, the discrete case will be a special case which corresponds to the singular distribution kernel $t^{-\sigma } e^{-\kappa t}$ for $\sigma \rightarrow 1$ (at some time $\tau$).

Concerning the Timoshenko system with fractional delay terms $(P)-(P_c)$ considered in this work, as far as we know, introducing a fractional delay term in the internal feedback of Timoshenko system makes our problem different from those considered in the literature.

The paper is organized as follows. Section 2, briefly outlines the notation and also we reformulate the model $(P)-(P_c)$ into an augmented system, coupling the system (P) with a suitable diffusion equation. In Sect. 3, we establish the well-posedness of the system (9). Moreover, using a general criteria of Arendt-Batty and Lyubich-Vũ, we show the strong stability of our system. Finally, in the last Sect. 4, we prove that the Timoshenko beam System $(P)-(P_c)$ is uniformly exponentially stable.

2 Preliminaries

This section is concerned with the reformulation of the model $(P)-(P_c)$ into an augmented system. For that, we need the following claims.

Theorem 2.1

(see [18]) Let $\theta$ be the function:

$$\begin{aligned} \theta (\xi )=|\xi |^{(2\sigma -1)/2},\quad -\infty<\xi<+\infty ,\ 0<\sigma <1. \end{aligned}$$

(1)

Then the relationship between the ’input’ U and the ’output’ O of the system

$$\begin{aligned}&\partial _t\phi (\xi , t)+(\xi ^{2}+\kappa )\phi (\xi , t) -U(t)\theta (\xi )=0,\quad -\infty<\xi <+\infty ,\kappa> 0, t> 0,\end{aligned}$$

(2)

$$\begin{aligned}&\chi (\xi , 0)=0,\end{aligned}$$

(3)

$$\begin{aligned}&O(t)=(\pi )^{-1}\sin (\sigma \pi )\displaystyle \int _{-\infty }^{+\infty }\theta (\xi )\phi (\xi , t)\, d\xi \end{aligned}$$

(4)

is given by

$$\begin{aligned} O=I^{1-\sigma , \kappa }D U=D^{\sigma , \kappa }U, \end{aligned}$$

(5)

where

$$\begin{aligned} {[}I^{\sigma , \kappa }h](t)=\displaystyle \frac{1}{\Gamma (\sigma )}\displaystyle \int _{0}^{t} (t-s)^{\sigma -1}e^{-\kappa (t-s)}h(s)\, ds. \end{aligned}$$

Lemma 2.1

(see [1]) If then

$$\begin{aligned} \displaystyle \int _{-\infty }^{+\infty }\displaystyle \frac{\theta ^2(\xi )}{\mu +\kappa +\xi ^2}\, d\xi =\displaystyle \frac{\pi }{\sin \sigma \pi }(\mu +\kappa )^{\sigma -1}. \end{aligned}$$

We make the following hypotheses on the damping and the delay functions:

$$\begin{aligned} \left\{ \begin{array}{l} |a_1| \kappa _{1}^{\sigma _1-1}< a_2,\\ |{{{\tilde{a}}}}_1| \kappa _2^{\sigma _2-1}< {{{\tilde{a}}}}_2.\\ \end{array}\right. \end{aligned}$$

(6)

So, let us introduce the following new variables:

$$\begin{aligned} \left\{ \begin{array}{ll} z_{1}(x,p,t)=w_{t}(x,t-\tau _{1} p),&{}\quad \hbox { in } ]0,L[\times ]0, 1[\times ]0, +\infty [,\\ z_{2}(x,p,t)=\chi _{t}(x,t-\tau _{2} p), &{}\quad \hbox { in } ]0,L[\times ]0, 1[\times ]0, +\infty [.\\ \end{array}\right. \end{aligned}$$

(7)

Thus, we have

$$\begin{aligned} \left\{ \begin{array}{ll} \tau _{j}z_{jt}(x,p,t) + z_{jp}(x,p,t)=0, &{}\quad \hbox { in }]0,L[\times ]0, 1[\times ]0, +\infty [,\\ z_{1}(x,0,t)=w_t(x, t), z_{2}(x,0,t)=\chi _t(x, t), &{}\quad \hbox { in }]0,L[\times ]0, +\infty [.\\ \end{array}\right. \end{aligned}$$

(8)

Then, problem $(P)-(P_c)$ is equivalent to the following

$$\begin{aligned} \left\{ \begin{array}{l} \rho _{1}w_{tt}(x, t)-K(w_{x}+\chi )_{x}(x, t) + \zeta \displaystyle \int _{-\infty }^{+\infty }\theta _1(\xi )\phi _1(x,\xi , t)\, d\xi +a_{2}w_{t}(t)=0, \\ \rho _{2}\chi _{tt}(x, t)-b\chi _{xx}(x, t)+K(w_{x}+ \chi )(x, t) +{{{\tilde{\zeta }}}} \displaystyle \int _{-\infty }^{+\infty }\theta _2(\xi )\phi _2(x,\xi , t)\, d\xi +{{{\tilde{a}}}}_{2}\chi _{t}(t)=0,\\ \tau _{1}z_{1t}(x,p,t) + z_{1p}(x,p,t)=0, \\ \tau _{2}z_{2t}(x,p,t)+ z_{2p}(x,p,t)=0, \\ \partial _t\phi _{1}(x, \xi , t)+(\xi ^{2}+\kappa _1)\phi _{1}(\xi , t) -z_1(x, 1, t)\theta _1(\xi )=0, \\ \partial _t\phi _{2}(x, \xi , t)+(\xi ^{2}+\kappa _2)\phi _{2}(\xi , t) -z_2(x, 1, t)\theta _2(\xi )=0,\\ w(x,0)= w_{0}, \ \ w_{t}(x,0)= w_{1},\\ \chi (x,0)= \chi _{0}, \ \ \chi _{t}(x,0)= \chi _{1},\\ w_{t}(x, t-\tau _1)= f_{0}(x, t-\tau _1), \ \ \chi _{t}(x, t-\tau _2)= g_{0}(x, t-\tau _2),\\ \phi _{1}(x, \xi , 0)=0, \phi _{2}(x, \xi , 0)=0,\\ \end{array}\right. \end{aligned}$$

(9)

where $\zeta =(\pi )^{-1}\sin (\sigma _1\pi )a_{1}$ and ${{{\tilde{\zeta }}}} =(\pi )^{-1}\sin (\sigma _2\pi ){{{\tilde{a}}}}_{1}$.

Remark 2.1

$\zeta \int _{-\infty }^{+\infty }\theta _i(\xi )\phi _i(\xi , t)\, d\xi$ is an internal nonlocal force, where $\theta _i$ is a diffusive representation and $\phi _i$ is a diffusive state. We call (9)$_5$ and (9)$_6$ a diffusive realization associated with the Timoshenko system.

The energy of system (9) is given by

$$\begin{aligned} \begin{aligned} E(t)&=\displaystyle \frac{1}{2}\left( \rho _{1}\Vert w_t\Vert ^{2}_{L^2(0, L)}+\rho _{2}\Vert \chi _t\Vert ^{2}_{L^2(0, L)} +K\Vert (w_{x}+\chi )\Vert _{L^2(0, L)}^2+b \Vert \chi _{x}\Vert _{L^2(0, L)}^2 \right. \\&\quad +\nu _{1}\Vert z_{1}(x,p)\Vert _{L^2((0, L)\times (0, 1))}^2 +\nu _{2}\Vert z_{2}(x,p)\Vert _{L^2((0, L)\times (0, 1))}^2\\&\quad +|\zeta |\Vert \phi _{1}\Vert _{L^2((0, L)\times (-\infty , +\infty ))}^2\\&\quad \left. +|{{{\tilde{\zeta }}}}|\Vert \phi _{2}\Vert _{L^2((0, L)\times (-\infty , +\infty ))}^2\right) , \end{aligned} \end{aligned}$$

(10)

where $\nu _i$ are positive constants verifying

$$\begin{aligned} \left\{ \begin{array}{l} \tau _1 |\zeta | \left( \displaystyle \int _{-\infty }^{+\infty }\displaystyle \frac{\theta _1^2(\xi )}{\xi ^2+\kappa _1}\, d\xi \right)< \nu _1< \tau _1 \left( 2 a_{2} - |\zeta |\left( \displaystyle \int _{-\infty }^{+\infty }\displaystyle \frac{\theta _1^2(\xi )}{\xi ^2+\kappa _1}\, d\xi \right) \right) ,\\ \tau _2 |{{{\tilde{\zeta }}}}| \left( \displaystyle \int _{-\infty }^{+\infty }\displaystyle \frac{\theta _2^2(\xi )}{\xi ^2+\kappa _2}\, d\xi \right)< \nu _2 < \tau _2 \left( 2 {{{\tilde{a}}}}_{2} - |{{{\tilde{\zeta }}}}|\left( \displaystyle \int _{-\infty }^{+\infty }\displaystyle \frac{\theta _2^2(\xi )}{\xi ^2+\kappa _2}\, d\xi \right) \right) .\end{array} \right. \end{aligned}$$

(11)

Remark 2.2

Using Lemma 2.1, the condition (11) means that

$$\begin{aligned} \left\{ \begin{array}{l} \tau _1 |a_1|\kappa _1^{\sigma _1-1}<\nu _1<\tau _1(2a_2-|a_1|\kappa _1^{\sigma _1-1}),\\ \tau _2 |{{{\tilde{a}}}}_1|\kappa _2^{\sigma _2-1}<\nu _2<\tau _2(2{{{\tilde{a}}}}_2 -|{{{\tilde{a}}}}_1|\kappa _2^{\sigma _2-1}).\end{array} \right. \end{aligned}$$

Lemma 2.2

Let $(w, \phi _1,z_1, \chi ,\phi _2, z_2)$ be a regular solution of the system (9). Then there exists a positive constant C such that the energy functional E satisfies

$$\begin{aligned} \displaystyle \frac{dE(t)}{dt}\le & {} - C \left( \displaystyle \int _{0}^{L} \left( |w_{t}(x,t)|^2+ |z_{1}(x,1,t)|^2 \right) \, dx \right. \nonumber \\&\left. + \displaystyle \int _{0}^{L} \left( |\chi _{t}(x,t)|^{2}+ |z_{2}(x,1,t)|^{2} \right) \, dx\right) . \end{aligned}$$

(12)

Proof

Multiplying the first two equations in (9) by ${{{\overline{w}}}}_{t}$ and ${{{\overline{\chi }}}}_{t}$, respectively, and integration over (0, L), using integration by parts and the boundary conditions we obtain

$$\begin{aligned}&\displaystyle \frac{1}{2}\displaystyle \frac{d}{dt}\left\{ \rho _1\Vert w_{t}\Vert _{L^2(0, L)}^{2} +\rho _2\Vert \chi _{t}\Vert _{L^2(0, L)}^{2} +K\displaystyle \int _{0}^{L}|w_{x}+\chi |^2\,dx +b\displaystyle \int _{0}^{L}|\chi _{x}|^2\,dx\right\} \nonumber \\&\quad +\zeta \mathfrak {R}\displaystyle \int _{0}^{L}{{{\overline{w}}}}_{t} \displaystyle \int _{-\infty }^{+\infty }\theta _1(\xi )\phi _1(x,\xi , t)\, d\xi \, dx +a_{2} \displaystyle \int _{0}^{L}|w_{t}(t)|^{2}\, dx \nonumber \\&\quad +{\tilde{\zeta }}\mathfrak {R}\displaystyle \int _{0}^{L}{{{\overline{\chi }}}}_{t} \displaystyle \int _{-\infty }^{+\infty }\theta _2(\xi )\phi _2(x,\xi , t)\, d\xi \, dx +{\tilde{a}}_{2} \displaystyle \int _{0}^{L}|\chi _{t}(t)|^{2}\, dx=0.&\end{aligned}$$

(13)

A simple multiplication of (9)$_5$ and (9)$_6$ by $|\zeta | {{{\overline{\phi }}}}_1$ and $|{{{\tilde{\zeta }}}}| {{{\overline{\phi }}}}_2$, respectively, and integration over $(0, L) \times (-\infty , +\infty )$, yield

$$\begin{aligned}&\displaystyle \frac{1}{2}\displaystyle \frac{d}{dt}\left\{ |\zeta | \Vert \phi _1\Vert _{L^{2}((0, L)\times (-\infty , +\infty ))}^{2} +|{\tilde{\zeta }}|\Vert \phi _2\Vert _{L^{2}((0, L)\times (-\infty , +\infty ))}^{2}\right\} \nonumber \\&\quad +|\zeta | \displaystyle \int _{0}^{L}\displaystyle \int _{-\infty }^{+\infty }(\xi ^{2}+\kappa _1)| \phi _1(x,\xi , t)|^{2}\, d\xi \, dx\nonumber \\&\quad +|{\tilde{\zeta }}|\displaystyle \int _{0}^{L}\displaystyle \int _{-\infty }^{+\infty }(\xi ^{2}+\kappa _2)| \phi _2(x,\xi , t)|^{2}\, d\xi \, dx \nonumber \\&\quad -|\zeta |\mathfrak {R}\displaystyle \int _{0}^{L}z_{1}(x,1, t)\int _{-\infty }^{+\infty } \theta _1(\xi ){{{\overline{\phi }}}}_1(x, \xi , t)\, d\xi \, dx \nonumber \\&\quad -|{\tilde{\zeta }}|\mathfrak {R}\displaystyle \int _{0}^{L}z_{2}(x,1, t)\int _{-\infty }^{+\infty }\theta _2(\xi ){{{\overline{\phi }}}}_2(x, \xi , t)\,d\xi \, dx=0. \end{aligned}$$

(14)

Now, multiplying (9)$_3$ and (9)$_4$ by $\nu _1 {{{\overline{z}}}}_1$ and $\nu _2 {{{\overline{z}}}}_2$, respectively, and integration over $(0, L) \times (0, 1)$, we get

$$\begin{aligned}&\displaystyle \frac{1}{2}\displaystyle \frac{d}{dt}\left\{ \nu _1\Vert z_1\Vert _{L^{2}((0, L)\times (0, 1))}^{2}+\nu _2\Vert z_2\Vert _{L^{2}((0, L)\times (0, 1))}^{2}\right\} \nonumber \\&\quad + \displaystyle \frac{\nu _1\tau _1^{-1}}{2}\displaystyle \int _{0}^{L}\left( |z_{1}(x,1,t)|^2-|w_{t}(x,t)|^{2}\right) \nonumber \\&\quad + \displaystyle \frac{\nu _2\tau _2^{-1}}{2}\displaystyle \int _{0}^{L}\left( |z_{2}(x,1,t)|^2-|\chi _{t}(x,t)|^{2} \right) =0. \end{aligned}$$

(15)

By combining (13), (14) and (15), we obtain

$$\begin{aligned} E'(t)= & {} -a_{2} \Vert w_{t}\Vert ^{2}_{L^{2}}-{\tilde{a}}_{2} \Vert \chi _{t}\Vert ^{2}_{L^{2}} -|\zeta |\displaystyle \int _{0}^{L}\displaystyle \int _{-\infty }^{+\infty }(\xi ^{2}+\kappa _1)|\phi _1(x, \xi , t)|^{2}\, d\xi \, dx \nonumber \\&\quad -|{{{\tilde{\zeta }}}}|\displaystyle \int _{0}^{L}\displaystyle \int _{-\infty }^{+\infty }(\xi ^{2}+\kappa _2)|\phi _2(x, \xi , t)|^{2}\, d\xi \, dx \nonumber \\&\quad -\zeta \mathfrak {R}\displaystyle \int _{0}^{L}{{{\overline{w}}}}_{t}\displaystyle \int _{-\infty }^{+\infty }\theta _1(\xi )\phi _1(x,\xi , t)\, d\xi \, dx \nonumber \\&\quad -{{{\tilde{\zeta }}}} \mathfrak {R}\displaystyle \int _{0}^{L}{{{\overline{\chi }}}}_{t} \displaystyle \int _{-\infty }^{+\infty }\theta _2(\xi )\phi _2(x,\xi , t)\, d\xi \, dx \nonumber \\&\quad +|\zeta | \mathfrak {R}\displaystyle \int _{0}^{L} z_{1}(x,1, t)\int _{-\infty }^{+\infty } \theta _1(\xi ){{{\overline{\phi }}}}_1(x, \xi , t)\,d\xi \, dx \nonumber \\&\quad +|{{{\tilde{\zeta }}}}| \mathfrak {R}\displaystyle \int _{0}^{L} z_{2}(x,1, t) \int _{-\infty }^{+\infty }\theta _2(\xi ){{{\overline{\phi }}}}_2(x, \xi , t)\,d\xi \, dx \nonumber \\&\quad +\displaystyle \frac{\nu _1\tau _1^{-1}}{2} \displaystyle \int _{0}^{L} |w_{t}(x,t)|^2\, dx\, - \displaystyle \frac{\nu _1\tau _1^{-1}}{2} \displaystyle \int _{0}^{L}|z_{1}(x,1,t)|^2 \, dx \nonumber \\&\quad +\displaystyle \frac{\nu _2\tau _2^{-1}}{2} \displaystyle \int _{0}^{L} |\chi _{t}(x,t)|^2 \, dx\, - \displaystyle \frac{\nu _2\tau _2^{-1}}{2} \displaystyle \int _{0}^{L}|z_{2}(x,1,t)|^2 \, dx. \end{aligned}$$

(16)

Using Cauchy-Schwarz inequality, we get

$$\begin{aligned} \left| \int _{-\infty }^{+\infty }\theta _j(\xi ) \phi _j(x, \xi , t)\, d\xi \right| \le \left( \int _{-\infty }^{+\infty }\displaystyle \frac{\theta _j^2(\xi )}{\xi ^2+\kappa _j}\, d\xi \right) ^{\frac{1}{2}} \left( \int _{-\infty }^{+\infty }(\xi ^2+\kappa _j)|\phi _j(x, \xi , t)|^2\, d\xi \right) ^{\frac{1}{2}}. \end{aligned}$$

Then

$$\begin{aligned} \begin{aligned}&\left| \displaystyle \int _{0}^{L}z_{j}(x,1, t)\int _{-\infty }^{+\infty }\theta _j(\xi ){{{\overline{\phi }}}_j}(x, \xi , t)\, d\xi \, dx\right| \\&\quad \le \left( \displaystyle \int _{-\infty }^{+\infty }\displaystyle \frac{\theta _j^2(\xi )}{\xi ^2+\kappa _j}\, d\xi \right) ^{\frac{1}{2}} \Vert z_{j}(x,1, t)\Vert _{L^2(0, L)} \left( \displaystyle \int _{0}^{L}\int _{-\infty }^{+\infty }(\xi ^2+\kappa _j)|\phi _j(x, \xi , t)|^2\, dx\, d\xi \right) ^{\frac{1}{2}} \end{aligned} \end{aligned}$$

and

$$\begin{aligned} \left\{ \begin{aligned}&\left| \displaystyle \int _{0}^{L}{{{\overline{w}}}_{t}}(x, t)\int _{-\infty }^{ +\infty }\theta _1(\xi ) \phi _1(x, \xi , t)\, d\xi \, dx\right| \\&\quad \le \left( \displaystyle \int _{-\infty }^{+\infty }\displaystyle \frac{\theta _1^2(\xi )}{\xi ^2 +\kappa _1}\, d\xi \right) ^{\frac{1}{2}} \Vert w_{t}(x, t)\Vert _{L^2(0, L)} \left( \displaystyle \int _{0}^{L}\int _{-\infty }^{+\infty }(\xi ^2+\kappa _1)| \phi _1(x, \xi , t)|^2\, dx\, d\xi \right) ^{\frac{1}{2}}, \\&\left| \displaystyle \int _{0}^{L}{{{\overline{\chi }}}_{t}}(x, t)\int _{ -\infty }^{+\infty }\theta _2(\xi ) \phi _1(x, \xi , t)\, d\xi \, dx\right| \\&\quad \le \left( \displaystyle \int _{-\infty }^{+\infty }\displaystyle \frac{\theta _2^2(\xi )}{\xi ^2+\kappa _2}\, d\xi \right) ^{\frac{1}{2}} \Vert \chi _{t}(x, t)\Vert _{L^2(0, L)} \left( \displaystyle \int _{0}^{L}\int _{-\infty }^{+\infty }(\xi ^2+\kappa _2) |\phi _2(x, \xi , t)|^2\, dx\, d\xi \right) ^{\frac{1}{2}}.&\\\end{aligned} \right. \end{aligned}$$

Using Cauchy-Schwarz and Young’s inequalities, we obtain

$$\begin{aligned} \begin{aligned} E'(t)&\le \left( -a_{2}+\displaystyle \frac{|\zeta | I_1}{2}+ \displaystyle \frac{\nu _1\tau _1^{-1}}{2}\right) \displaystyle \int _{0}^{L}|w_{t}(x,t)|^2\, dx\,\\&\quad + \left( \displaystyle \frac{|\zeta | I_1}{2}-\displaystyle \frac{\nu _1\tau _1^{-1}}{2}\right) \displaystyle \int _{0}^{L} |z_{1}(x,1,t)|^2 \, dx\, \\&\quad +\left( -{\tilde{a}}_{2}+\displaystyle \frac{|{{{\tilde{\zeta }}}}| I_2}{2}+ \displaystyle \frac{\nu _2\tau _2^{-1}}{2}\right) \displaystyle \int _{0}^{L}|\chi _{t}(x,t)|^2\, dx\, \\&\quad + \left( \displaystyle \frac{|{{{\tilde{\zeta }}}}| I_2}{2}- \displaystyle \frac{\nu _2\tau _2^{-1}}{2}\right) \displaystyle \int _{0}^{L}|z_{2}(x,1,t)|^2 \, dx, \end{aligned} \end{aligned}$$

where $I_j=\displaystyle \int _{-\infty }^{+\infty }\displaystyle \frac{\theta _j^2(\xi )}{\xi ^2+\kappa _j}\, d\xi$, which implies

$$\begin{aligned} \begin{aligned} E'(t)&\le - C_1 \displaystyle \int _{0}^{L} \left( |w_{t}(x,t)|^2+ |z_{1}(x,1,t)|^2 \right) \, dx\, \\&\quad - C_2 \displaystyle \int _{0}^{L}\left( |\chi _{t}(x,t)|^2+ |z_{2}(x,1,t)|^2 \right) \, dx \end{aligned} \end{aligned}$$

with

$$\begin{aligned} \left\{ \begin{aligned} C_1&= \min \left\{ \left( a_{2}-\displaystyle \frac{|\zeta | I_1}{2}- \displaystyle \frac{\nu _1\tau _1^{-1}}{2}\right) ,\left( -\displaystyle \frac{|\zeta | I_1}{2}+ \displaystyle \frac{\nu _1\tau _1^{-1}}{2}\right) \right\} , \\ C_2&= \min \left\{ \left( {\tilde{a}}_{2}-\displaystyle \frac{|{\tilde{\zeta }}| I_2}{2}- \displaystyle \frac{\nu _2\tau _2^{-1}}{2}\right) ,\left( -\displaystyle \frac{|{\tilde{\zeta }}| I_2}{2}+ \displaystyle \frac{\nu _2\tau _2^{-1}}{2}\right) \right\} . \end{aligned} \right. \end{aligned}$$

Since $\nu _j$ is chosen satisfying assumption (11), the constant $C_j$ is positive. This completes the proof of the Lemma. $\square$

3 Well-posedness and strong stability

3.1 Well-posedness of the problem

In this part, we discuss the existence and uniqueness of solution of (9). We use the semigroup theory. Let us denote by U the vector-valued function

$$\begin{aligned} U=(w,{\widetilde{w}},z_{1},\phi _{1},\chi ,{\widetilde{\chi }},z_{2},\phi _{2})^{T}, \hbox { with } {\widetilde{w}}=w_t \hbox { and }{\widetilde{\chi }}=\chi _t. \end{aligned}$$

Then, system (9) is equivalent to the abstract linear first order Cauchy problem

$$\begin{aligned} U'={{{\mathcal {A}}}}U,\quad U(0)=U_0, \end{aligned}$$

(17)

where $U_0=(w_0, w_1, f_{0}(.,-.\tau ),0, \chi _{0},\chi _{1}, g_{0}(.,-.\tau ), 0)^{T}$ and the operator ${{{\mathcal {A}}}}$ is defined by

$$\begin{aligned} {{{\mathcal {A}}}}\begin{pmatrix}w\\ {\tilde{w}}\\ z_{1}\\ \phi _{1}\\ \chi \\ {\tilde{\chi }}\\ z_{2}\\ \phi _{2} \\ \end{pmatrix} = \begin{pmatrix}{\tilde{w}}\\ \frac{K}{\rho _{1}}(w_{x}+\chi )_{x}-\frac{\zeta }{\rho _{1}} \int _{-\infty }^{+\infty }\theta _1(\xi )\phi _1(x,\xi )\, d\xi -\frac{a_2}{\rho _{1}}{{{\tilde{w}}}}\\ \frac{-z_{1p}}{\tau _{1}}\\ -(\xi ^{2}+\kappa _1)\phi _{1}+z_1(x,1)\theta _1(\xi ) \\ {\tilde{\chi }} \\ \frac{b}{\rho _{2}}\chi _{xx}-\frac{K}{\rho }_{2}(w_{x}+\chi ) -\frac{{\tilde{\zeta }}}{\rho _{2}} \int _{-\infty }^{+\infty }\theta _2(\xi )\phi _2(x,\xi )\, d\xi -\frac{{{{\tilde{a}}}}_2}{\rho _{2}}{{{\tilde{\chi }}}}\\ \frac{-z_{2p}}{\tau _{2}} \\ -(\xi ^{2}+\kappa _2)\phi _{2}+z_2(x,1)\theta _2(\xi )\end{pmatrix} \end{aligned}$$

(18)

with domain

$$\begin{aligned} D({{{\mathcal {A}}}})=\left\{ \begin{array}{ll} (w, {\tilde{w}},z_{1},\phi _{1},\chi ,{\tilde{\chi }},z_{2},\phi _{2} )^{T}\quad \hbox { in } H: ({\tilde{w}},{\tilde{\chi }})=(z_{1},z_{2})(.,0) \hbox { on } (0,L), \\ -(\xi ^{2}+\kappa _1)\phi _{1}+z_1(x,1)\theta _1(\xi )\in L^2((0, L) \times (-\infty ,+\infty )), \\ -(\xi ^{2}+\kappa _2)\phi _{2}+z_2(x,1)\theta _2(\xi )\in L^2((0, L) \times (-\infty ,+\infty )), \\ |\xi |\phi _1\in L^{2}((0, L) \times (-\infty ,+\infty )),\\ |\xi |\phi _2\in L^{2}((0, L) \times (-\infty ,+\infty )).\\ \end{array}\right\} , \end{aligned}$$

(19)

where

$$\begin{aligned} \begin{aligned} {H}&=(H^{2}(0,L)\cap H^{1}_{0}(0,L))\times H^{1}_{0}(0,L)\times L^{2}(0,L, H^{1}(0,1))\times L^{2}((0, L) \times (-\infty ,+\infty ))\\&\quad \times (H^{2}(0,L)\cap H^{1}_{0}(0,L))\times H^{1}_{0}(0,L) \times L^{2}(0,L, H^{1}(0,1))\times L^{2}((0, L) \times (-\infty ,+\infty )) \end{aligned} \end{aligned}$$

and

$$\begin{aligned} H_{0}^{1}(0,L)=\{f\in H^{1}(0, L):f(0)=f(L)=0\}. \end{aligned}$$

Remark 3.1

The condition is imposed to insure that .

Now, let

$$\begin{aligned} {{{\mathcal {H}}}}:=(H^{1}_{0}(0,L) \times L^{2}(0,L) \times L^{2}((0,L) \times (0,1))\times L^{2}((0, L)\times (-\infty ,+\infty )))^2. \end{aligned}$$

(20)

${{{\mathcal {H}}}}$ is a Hilbert space endowed with the inner product defined, for $U=(w,{\widetilde{w}},z_{1},\phi _{1},\chi ,{\widetilde{\chi }},z_{2}, \phi _{2})^{T}$ and $U_*=(w_*,{{{\widetilde{w}}}}_*,z_{1*},\phi _{1*}, \chi _*,{{{\widetilde{\chi }}}}_*, z_{2*},\phi _{2*})^{T}\in {{{\mathcal {H}}}}$, by

$$\begin{aligned} \begin{aligned} \langle U, U_* \rangle _{{{\mathcal {H}}}}&=\displaystyle \int _{0}^{L} \{\rho _{1}{\widetilde{w}}\overline{{{{\widetilde{w}}}}_*} +\rho _{2}{\widetilde{\chi }} \overline{{{{\widetilde{\chi }}}}_*} +K(w_{x}+\chi )(\overline{w_{*x}}+\overline{\chi _*})+b \ \chi _{x}\overline{\chi _{*x}}\} \, dx\\&\quad +\nu _{1}\displaystyle \int _{0}^{L}\int _{0}^{1}z_{1}(x,p)\overline{z_{1*}} (x,p)dp\ dx+\nu _{2}\displaystyle \int _{0}^{L}\int _{0}^{1}z_{2}(x,p)\overline{z_{2*}}(x,p)dp\ dx\\&\quad +|\zeta |\displaystyle \int _{0}^{L}\int _{-\infty }^{+\infty } \phi _{1}\overline{\phi _{1*}} d\xi \, dx+|{{{\tilde{\zeta }}}}|\displaystyle \int _{0}^{L}\int _{-\infty }^{+\infty } \phi _{2}\overline{\phi _{2*}} d\xi \, dx. \end{aligned} \end{aligned}$$

Our first main result is the following.

Theorem 3.1

Let $U_{0} \in {{\mathcal {H}}}$, then there exists a unique solution $U\in C([0,+\infty ), {{\mathcal {H}}})$, of problem (17), Moreover if $U_{0} \in D({{{\mathcal {A}}}})$, then $U \in C([0,+\infty ),D({{{\mathcal {A}}}})) \cap C^{{1}}([0,+\infty ), {{\mathcal {H}}})$.

Proof

First, for any $U =(w,{\widetilde{w}},z_{1},\phi _{1},\chi ,{\widetilde{\chi }},z_{2},\phi _{2})^{T} \in \mathcal {D({{\mathcal {A}}})}$, we have

$$\begin{aligned} \begin{aligned} \mathfrak {R}\langle {{{\mathcal {A}}}} U, U \rangle _{{{\mathcal {H}}}}&= -a_2\displaystyle \int _{0}^{L}|{\widetilde{w}}|^2\, dx-{{{\tilde{a}}}}_2 \displaystyle \int _{0}^{L}|{\widetilde{\chi }}|^2\, dx\\&\quad -\zeta \mathfrak {R}\displaystyle \int _{0}^{L}\left( \displaystyle \int _{-\infty }^{+\infty }\theta _1(\xi ) \phi _1(x, \xi )\, d\xi \right) {\overline{{{\widetilde{w}}}}}\, dx\\&\quad -{{{\tilde{\zeta }}}}\mathfrak {R}\displaystyle \int _{0}^{L}\left( \displaystyle \int _{-\infty }^{ +\infty }\theta _2(\xi )\phi _2(x, \xi )\, d\xi \right) {\overline{{{\widetilde{\chi }}}}}\, dx\\&\quad - \frac{\nu _{1}}{\tau _1} \mathfrak {R}\displaystyle \int _{0}^{L}\int _{0}^{1} {{{\overline{z}}}_{1}(x,p)}z_{1p}(x,p) dp \ dx\\&\quad - \frac{\nu _{2}}{\tau _2}\mathfrak {R}\displaystyle \int _{0}^{L}\int _{0}^{1} {{{\overline{z}}}_{2}(x,p)} z_{2p}(x,p) dp \ dx\\&\quad +|\zeta | \mathfrak {R}\displaystyle \int _{0}^{L}\displaystyle \int _{-\infty }^{+\infty } [-(\xi ^{2} + \kappa _1)\phi _{1}(x, \xi )+z_1(x, 1)\theta _1(\xi )] { {{\overline{\phi }}}_{1}(x, \xi )}\ d\xi \, dx\\&\quad +|{{{\tilde{\zeta }}}}| \mathfrak {R}\displaystyle \int _{0}^{L}\displaystyle \int _{-\infty }^{ +\infty } [-(\xi ^{2} + \kappa _2)\phi _{2}(x, \xi )+z_2(x, 1) \theta _2(\xi )] {{{\overline{\phi }}}_{2}(x, \xi )}\ d\xi \, dx. \end{aligned} \end{aligned}$$

(21)

Using $z_1(x, 0)= {{{\tilde{w}}}}$ and $z_2(x, 0)= {{{\tilde{\chi }}}}$ note that

$$\begin{aligned} \begin{aligned}&\frac{\nu _{1}}{\tau _1}\mathfrak {R}\displaystyle \int _{0}^{L}\int _{0}^{1}{{{\overline{z}}}_{1} (x,p)}z_{1p}(x,p)\, dp \, dx + \frac{\nu _{2}}{\tau _2} \mathfrak {R}\displaystyle \int _{0}^{L}\int _{0}^{1} {{{\overline{z}}}_{2}(x,p)} z_{2p}(x,p)\, dp\, dx \\&\quad = \frac{\nu _{1}}{\tau _1} \displaystyle \int _{0}^{L}\int _{0}^{1} \frac{1}{2} \frac{\partial }{\partial p}|z_{1}(x,p)|^{2} dp \ dx + \frac{\nu _{2}}{\tau _2}\displaystyle \int _{0}^{L}\int _{0}^{1} \frac{1}{2} \frac{\partial }{\partial p} |z_{2}(x,p)|^{2}\, dp\, dx\\&\quad = \frac{\nu _{1}}{2\tau _1} \displaystyle \int _{0}^{L} \{|z_{1}(x,1)|^{2} - |{{{\tilde{w}}}}(x)|^{2}\} dx +\frac{\nu _{2}}{2\tau _2} \displaystyle \int _{0}^{L} \{|z_{2}(x,1)|^{2} - |{{{\tilde{\chi }}}}(x)|^{2}\} dx. \end{aligned} \end{aligned}$$

(22)

Then by applying Cauchy-Schwarz and young’s inequalities we obtain

$$\begin{aligned} \mathfrak {R}\langle {{{\mathcal {A}}}} U, U \rangle _{{{\mathcal {H}}}}&\le \left( -a_{2}+\displaystyle \frac{|\zeta | I_1}{2}+ \displaystyle \frac{\nu _1\tau _1^{-1}}{2}\right) \displaystyle \int _{0}^{L} |{\widetilde{w}}(x)|^{2}\, dx\nonumber \\&\quad + \left( \displaystyle \frac{|\zeta | I_1}{2}- \displaystyle \frac{\nu _1\tau _1^{-1}}{2}\right) \displaystyle \int _{0}^{L} |z_{1}(x,1)|^{2} \, dx \nonumber \\&\quad +\left( -{\tilde{a}}_{2}+\displaystyle \frac{|{{{\tilde{\zeta }}}}| I_2}{2}+ \displaystyle \frac{\nu _2\tau _2^{-1}}{2}\right) \displaystyle \int _{0}^{L} |{\widetilde{\chi }}(x)|^{2}\, dx\nonumber \\&\quad + \left( \displaystyle \frac{|{{{\tilde{\zeta }}}}| I_2}{2}- \displaystyle \frac{\nu _2\tau _2^{-1}}{2}\right) \displaystyle \int _{0}^{L}|z_{2}(x,1)|^{2} \, dx \nonumber \\&\le - C_1 \displaystyle \int _{0}^{L} \left( |{\widetilde{w}}(x)|^2+ |z_{1}(x,1)|^2 \right) \, dx\nonumber \\&\quad - C_2 \displaystyle \int _{0}^{L}\left( |{\widetilde{\chi }}(x)|^2+ |z_{2}(x,1)|^2 \right) \, dx\nonumber \\&\le 0, \end{aligned}$$

(23)

where $C_1$ and $C_2$ are positive constants and $I_j=\displaystyle \int _{-\infty }^{+\infty }\displaystyle \frac{\theta _j^2(\xi )}{\xi ^2+\kappa _j}\, d\xi$. Hence, ${{{\mathcal {A}}}}$ is a dissipative operator.

Next, we will prove that the operator $\mu I-{{{\mathcal {A}}}}$ is surjective for $\mu > 0$. For this purpose, let $F=(f_1, f_2, f_3, f_4, f_5, f_6, f_7, f_8)^{T}\in {{{\mathcal {H}}}}$, we seek $U=(w,{\widetilde{w}},z_{1},\phi _{1},\chi ,{\widetilde{\chi }},z_{2},\phi _{2})^{T}\in D({{{\mathcal {A}}}})$ satisfying $(\mu I-{{{\mathcal {A}}}})U=F$, that is,

$$\begin{aligned} \left\{ \begin{array}{l} \mu w-{\tilde{w}}=f_1, \\ \mu {\tilde{w}}-\frac{K}{\rho _1}(w_{x}+\chi )_{x} +\frac{\zeta }{\rho _{1}} \int _{-\infty }^{+\infty }\theta _1(\xi )\phi _1(x,\xi )\, d\xi +\frac{a_2}{\rho _{1}}{{{\tilde{w}}}}=f_2, \\ \mu z_1+\frac{z_{1p}}{\tau _1}=f_3, \\ \mu \phi _{1}+(\xi ^{2}+\kappa _1)\phi _{1}-z_1(x, 1)\theta _1(\xi )=f_{4}, \\ \mu \chi -{\tilde{\chi }}=f_5, \\ \mu {\tilde{\chi }}-\frac{b}{\rho _2}\chi _{xx}+\frac{K}{\rho _2}(w_{x}+\chi ) +\frac{{\tilde{\zeta }}}{\rho _{2}} \int _{-\infty }^{+\infty }\theta _2(\xi )\phi _2(x,\xi )\, d\xi +\frac{{\tilde{a}}_2}{\rho _{2}}{{{\tilde{\chi }}}}=f_6, \\ \mu z_2+\frac{z_{2p}}{\tau _2}=f_7, \\ \mu \phi _{2}+(\xi ^{2}+\kappa _2)\phi _{2}-z_2(x, 1)\theta _2(\xi )=f_{8}. \end{array} \right. \end{aligned}$$

(24)

From (24)$_4$ and (24)$_8$ we obtain

$$\begin{aligned} \left\{ \begin{aligned} \phi _1(x, \xi )&=\displaystyle \frac{f_4(x, \xi )+z_1(x, 1) \theta _1(\xi )}{\mu +\xi ^{2}+\kappa _1},\\ \phi _2(x, \xi )&=\displaystyle \frac{f_8(x, \xi )+z_2(x, 1)\theta _2(\xi )}{\mu +\xi ^{2}+\kappa _2} \end{aligned} \right. \end{aligned}$$

(25)

and by (24)$_1$ and (24)$_5$ we have

$$\begin{aligned} \left\{ \begin{aligned} {\tilde{w}}&=\mu w-f_1\in H_{0}^{1}(0, L), \\ {\tilde{\chi }}&=\mu \chi -f_5\in H_{0}^{1}(0, L). \end{aligned} \right. \end{aligned}$$

(26)

The thirth and seventh equations in (24) and that $(z_1,z_2)(x,0) =({\tilde{w}}, {\tilde{\chi }})(x)$ easily give

$$\begin{aligned} \left\{ \begin{aligned} z_1(x,p)&= \mu w(x) e^{-\mu p \tau _1} - f_1(x) e^{-\mu p \tau _1} + \tau _1 e^{-\mu p \tau _1} \displaystyle \int _{0}^{p}e^{\mu s \tau _1}f_3(x, s) \ ds, \\ z_2(x,p)&=\mu \chi (x) e^{-\mu p \tau _2} - f_5(x)e^{-\mu p \tau _2} + \tau _2 e^{-\mu p \tau _2} \displaystyle \int _{0}^{p}e^{\mu s \tau _2} f_7(x, s) \ ds. \end{aligned} \right. \end{aligned}$$

(27)

Puting

$$\begin{aligned} \left\{ \begin{aligned} (z_1)_0(x)&= -f_1(x) e^{-\mu \tau _1} + \tau _1 e^{-\mu \tau _1} \displaystyle \int _{0}^{1} e^{\mu \tau _1 s} f_3(x,s) \ ds,\\ (z_2)_0(x)&= -f_5(x) e^{-\mu \tau _2} + \tau _2 e^{-\mu \tau _2} \displaystyle \int _{0}^{1} e^{\mu \tau _2 s} f_7(x,s) \ ds. \end{aligned} \right. \end{aligned}$$

(28)

From (27) and (28) we have

$$\begin{aligned} \left\{ \begin{aligned} z_1(x,1)&= {\tilde{w}}(x) e^{-\mu \tau _1} + \tau _1 e^{-\mu \tau _1} \displaystyle \int _{0}^{1}e^{\mu s\tau _1 }f_3(x,s) \ ds =\mu w(x) e^{-\mu \tau _1} + (z_1)_0(x),\\ z_2(x,1)&= {\tilde{\chi }}(x) e^{-\mu \tau _2} + \tau _2 e^{-\mu \tau _2} \displaystyle \int _{0}^{1}e^{\mu s \tau _2 } f_7(x,s) \ ds =\mu \chi (x) e^{-\mu \tau _2} + (z_2)_0(x). \end{aligned} \right. \end{aligned}$$

(29)

By exploiting (25), (26) and (29), the second and sixth equations in (24) are equivalent to

$$\begin{aligned} \left\{ \begin{aligned}&\mu ^{2}w - \frac{K}{\rho _1}(w_{x}+\chi )_{x} + \mu w\left( \frac{a_1}{\rho _1} (\mu +\kappa _1)^{\sigma _1-1} e^{-\mu \tau _1}+\frac{a_2}{\rho _1}\right) =f_2 + \mu f_1 \\&\quad -\frac{a_1}{\rho _1}(\mu +\kappa _1)^{\sigma _1-1}(z_1)_0(x) +\frac{a_2}{\rho _1}f_1-\frac{\zeta }{\rho _{1}} \int _{-\infty }^{+\infty }\displaystyle \frac{\theta _1(\xi )}{\mu +\xi ^2+\kappa _1}f_4(x,\xi )\, d\xi \\&\mu ^{2}\chi - \frac{b}{\rho _2}\chi _{xx} + \frac{K}{\rho _2}(w_{x}+\chi ) +\mu \chi \left( \frac{{\tilde{a}}_1}{\rho _2} (\mu +\kappa _1)^{\sigma _2-1} \ e^{-\mu \tau _2}+\frac{{\tilde{a}}_2}{\rho _2}\right) =f_6 + \mu f_5 \\&\quad -\frac{{{{\tilde{a}}}}_1}{\rho _2}(\mu +\kappa _2)^{\sigma _2-1}(z_2)_0(x) +\frac{{\tilde{a}}_2}{\rho _2}f_5 -\frac{{\tilde{\zeta }}}{\rho _{2}} \int _{-\infty }^{+\infty }\displaystyle \frac{\theta _2(\xi )}{\mu +\xi ^2+\kappa _2}f_8(x,\xi )\, d\xi . \end{aligned} \right. \end{aligned}$$

(30)

So, we multiply the first and second equations in (30) by test functions ${{{\overline{w}}}_1}$ and ${{{\overline{w}}}_2}$, respectively, where $(w_{1}, w_{2})\in H_{0}^{1}\times H_{0}^{1}(0,L)$, and we integrate by parts with respect to x, obtaining the following variational formulation of (30):

$$\begin{aligned} a((w,\chi ),(w_1,w_2))= L(w_1,w_2),\quad \forall (w_1, w_2)\in H_{0}^{1}\times H_{0}^{1}(0,L), \end{aligned}$$

(31)

where

$$\begin{aligned} \begin{aligned}&a((w,\chi ),(w_1,w_2))\\&\quad =\displaystyle \int _{0}^{L} \left( \rho _1\mu ^{2}w {{{\overline{w}}}_{1}} + \rho _2\mu ^{2}\chi {{{\overline{w}}}_2}+ K(w_{x}+\chi )({{{\overline{w}}}_1}_x +{{{\overline{w}}}_2})+ b\chi _{x}{{{\overline{w}}}_2}_x\right) \ dx \\&\qquad +\displaystyle \int _{0}^{L}\left( \mu \left( a_1 (\mu +\kappa _1)^{\sigma _1-1} e^{-\mu \tau _1}+a_2\right) w {{{\overline{w}}}_{1}} +\mu \left( {\tilde{a}}_1 (\mu +\kappa _1)^{\sigma _2-1} \ e^{-\mu \tau _2}+{\tilde{a}}_2\right) \chi {{{\overline{w}}}_2}\right) \ dx \end{aligned} \end{aligned}$$

and

$$\begin{aligned} \begin{aligned} L(w_1,w_2)&= \displaystyle \int _{0}^{L}\left( \rho _1(f_2 + \mu f_1) -a_1(\mu +\kappa _1)^{\sigma _1-1}(z_1)_0(x)+a_2 f_1\right. \\&\quad \left. -\zeta \int _{-\infty }^{+\infty }\displaystyle \frac{\theta _1(\xi )}{\mu +\xi ^2+\kappa _1}f_4(x,\xi )\, d\xi \right) {{{\overline{w}}}_1} \ dx\\&\quad +\displaystyle \int _{0}^{L}\left( \rho _2(f_6 + \mu f_5) -{{{\tilde{a}}}}_1(\mu +\kappa _2)^{\sigma _2-1}(z_2)_0(x)+{\tilde{a}}_2 f_5 \right. \\&\quad \left. -{\tilde{\zeta }} \int _{-\infty }^{+\infty }\displaystyle \frac{\theta _2(\xi )}{\mu +\xi ^2+\kappa _2}f_8(x,\xi )\, d\xi \right) {{{\overline{w}}}_2} \ dx. \end{aligned} \end{aligned}$$

It is easy to verify that a is a sesquilinear, continuous and coercive form, and L is an antilinear and continuous form. So applying the Lax-Milgram theorem, we deduce for all $(w_1,w_2) \in H_{0}^{1}\times H_{0}^{1}(0, L)$ the problem (31) admits a unique solution $(w,\chi )\in H_{0}^{1}\times H_{0}^{1}(0, L)$. Using classical elliptic regularity, it follows from (31) that $(w,\chi ) \in (H^{2}(0, L) \cap H_{0}^{1}(0, L)) \times (H^{2}(0,L) \cap H_{0}^{1}(0, L))$. Therefore, the operator $\mu I- {{{\mathcal {A}}}}$ is surjective for any $\mu > 0$.

Since ${{{\mathcal {A}}}}$ is dissipative and $\mu I- {{{\mathcal {A}}}}$ is surjective, ${{{\mathcal {A}}}}$ is maximal monotone. Therefore, using Lummer-Phillips theorem (see [22]), we deduce that ${{{\mathcal {A}}}}$ is an infinitesimal generator of a linear contraction $C_0$-semigroup on ${{{\mathcal {H}}}}$. Consequently, the well-posedness results of Theorem 3.1 follow from the Hille–Yosida theorem (see [22]). $\square$

3.2 Strong stability

We now look for conditions to obtain the strong stability of the $C_0$-semigroup $(e^{t{{{\mathcal {A}}}}})_{t\ge 0}$. For this aim, we use the following general criteria of Arendt-Batty and Lyubich-Vũ in [8] and [17].

Theorem 3.2

[8]-[17] Let X be a reflexive Banach space and $(T(t))_{t\ge 0}$ be a $C_0-$semigroup generated by ${{{\mathcal {A}}}}$ on X. Assume that $(T(t))_{t\ge 0}$ is bounded and that no eigenvalues of ${{{\mathcal {A}}}}$ lie on the imaginary axis. If is countable, then $(T(t))_{t\ge 0}$ is stable.

Our aim is ensured by the following theorem.

Theorem 3.3

The $C_0$-semigroup $e^{t{{{\mathcal {A}}}}}$ is strongly stable in ${{{\mathcal {H}}}}$; i.e, for all $U_0\in {{{\mathcal {H}}}}$, the solution of (17) satisfies

$$\begin{aligned} \displaystyle \lim _{t\rightarrow \infty }\Vert e^{t{{{\mathcal {A}}}}}U_0\Vert _{{{\mathcal {H}}}}=0. \end{aligned}$$

To prove this result, we need some lemmas.

Lemma 3.1

${{{\mathcal {A}}}}$ does not have eigenvalues on .

Proof

We make a distinction between $i\mu = 0$ and $i\mu \not = 0$.

Case 1: By contradiction. We suppose that there is and $U\not =0$, such that ${{{\mathcal {A}}}}U=i\mu U$. Then

$$\begin{aligned} \left\{ \begin{array}{l} i\mu w-{\tilde{w}}=0, \\ i\mu {\tilde{w}}-\frac{K}{\rho _1}(w_{x}+\chi )_{x} +\frac{\zeta }{\rho _{1}} \int _{-\infty }^{+\infty }\theta _1(\xi )\phi _1(x,\xi )\, d\xi +\frac{a_2}{\rho _{1}}{{{\tilde{w}}}}=0, \\ i\mu z_1+\frac{z_{1p}}{\tau _1}=0, \\ i\mu \phi _{1}+(\xi ^{2}+\kappa _1)\phi _{1}-z_1(x, 1)\theta _1(\xi )=0, \\ i\mu \chi -{\tilde{\chi }}=0, \\ i\mu {\tilde{\chi }}-\frac{b}{\rho _2}\chi _{xx}+\frac{K}{\rho _2}(w_{x}+\chi ) +\frac{{\tilde{\zeta }}}{\rho _{2}} \int _{-\infty }^{+\infty }\theta _2(\xi )\phi _2(x,\xi )\, d\xi +\frac{{\tilde{a}}_2}{\rho _{2}}{{{\tilde{\chi }}}}=0, \\ i\mu z_2+\frac{z_{2p}}{\tau _2}=0, \\ i\mu \phi _{2}+(\xi ^{2}+\kappa _2)\phi _{2}-z_2(x, 1)\theta _2(\xi )=0. \end{array}\right. \end{aligned}$$

(32)

Now from (23) we have

$$\begin{aligned} {\tilde{w}}=0,\ {\tilde{\chi }}=0, \ z_j(x, 1)=0. \end{aligned}$$

(33)

Hence by (32)$_1$ and (32)$_5$ we obtain

$$\begin{aligned} w\equiv 0,\ \chi \equiv 0. \end{aligned}$$

(34)

Note that (32)$_3$ and (32)$_7$ give us $z_1= {\tilde{w}} e^{-i\mu p\tau _1}=0$ and $z_2= {\tilde{\chi }} e^{-i\mu p\tau _2}=0$. Then from (32)$_4$ and (32)$_8$, we deduce that $\phi _1, \phi _2\equiv 0$. Therefore $U\equiv 0$.

Case 2: Solving for $-{{{\mathcal {A}}}}U = 0$ leads to $U = 0$. Indeed if $\mu =0$, similarly as the case 1, we have ${\tilde{w}}=0,\ {\tilde{\chi }}=0, \ z_j(x, 1)=0, \phi _1=0, \phi _2=0$. Then, we deduce that

$$\begin{aligned} \left\{ \begin{array}{l} -\frac{K}{\rho _1}(w_{x}+\chi )_{x}=0, \\ -\frac{b}{\rho _2}\chi _{xx}=0, \\ w(0)=\chi (0)=w(L)=\chi (L)=0. \end{array}\right. \end{aligned}$$

(35)

Hence $w=\chi \equiv 0$. Then $U\equiv 0$. We obtain a contradiction.

Lemma 3.2

We have

where $\rho ({{{\mathcal {A}}}})$ is the resolvent set of the operator ${{{\mathcal {A}}}}$.

Proof

To prove this, we need the following generalization of the Lax-Milgram Lemma.

Lemma 3.3

(Lax–Milgram–Fredholm, see [11]) Let V and H be Hilbert spaces such that the embedding $V\subset H$ is compact and dense. Suppose that and are two bounded sesquilinear forms such that $a_{V}$ is V-coercive and is a continuous conjugate linear form. The equation

$$\begin{aligned} a_{H}(u, v)+a_{V}(u, v)=G(v), \qquad \forall v\in V \end{aligned}$$

has either a unique solution $u\in V$ for all $G\in V'$ or has a nontrivial solution for $G=0$.

We will prove that the operator $i\mu I-{{{\mathcal {A}}}}$ is surjective for $\mu \not =0$. For this purpose, given $F=(f_1, f_2, f_3, f_4, f_5, f_6, f_7, f_8)^{T}\in {{{\mathcal {H}}}}$, we seek $X=(w, {{{\tilde{w}}} }, z_1, \phi _1, \chi , {{{\tilde{\chi }}}}, z_2, \phi _2)^{T}\in D({{{\mathcal {A}}}})$ which is solution of $(i\mu I-{{{\mathcal {A}}}})X=F$, that is, the entries of X satisfy the system of equations

$$\begin{aligned} \left\{ \begin{array}{l} i\mu w-{\tilde{w}}=f_1, \\ i\mu {\tilde{w}}-\frac{K}{\rho _1}(w_{x}+\chi )_{x} +\frac{\zeta }{\rho _{1}} \int _{-\infty }^{+\infty }\theta _1(\xi )\phi _1(x,\xi )\, d\xi +\frac{a_2}{\rho _{1}}{{{\tilde{w}}}}=f_2, \\ i\mu z_1+\frac{z_{1p}}{\tau _1}=f_3, \\ i\mu \phi _{1}+(\xi ^{2}+\kappa _1)\phi _{1}-z_1(x, 1)\theta _1(\xi )=f_4, \\ i\mu \chi -{\tilde{\chi }}=f_5, \\ i\mu {\tilde{\chi }}-\frac{b}{\rho _2}\chi _{xx}+\frac{K}{\rho _2}(w_{x}+\chi ) +\frac{{\tilde{\zeta }}}{\rho _{2}} \int _{-\infty }^{+\infty }\theta _2(\xi )\phi _2(x,\xi )\, d\xi +\frac{{\tilde{a}}_2}{\rho _{2}}{{{\tilde{\chi }}}}=f_6, \\ i\mu z_2+\frac{z_{2p}}{\tau _2}=f_7, \\ i\mu \phi _{2}+(\xi ^{2}+\kappa _2)\phi _{2}-z_2(x, 1)\theta _2(\xi )=f_8. \end{array} \right. \end{aligned}$$

(36)

Substituting ${\tilde{w}}$ and ${\tilde{\chi }}$ in (36)$_1$ and (36)$_5$ into equations (36)$_2$ and (36)$_6$, we obtain

$$\begin{aligned} \left\{ \begin{aligned}&-\mu ^{2}w - \frac{K}{\rho _1}(w_{x}+\chi )_{x} + i\mu w\left( \frac{a_1}{\rho _1} (i\mu +\kappa _1)^{\sigma _1-1} e^{-i\mu \tau _1}+\frac{a_2}{\rho _1}\right) =f_2 + i\mu f_1 \\&\quad -\frac{a_1}{\rho _1}(i\mu +\kappa _1)^{\sigma _1-1}(z_1)_0(x) +\frac{a_2}{\rho _1}f_1-\frac{\zeta }{\rho _{1}} \int _{-\infty }^{ +\infty }\displaystyle \frac{\theta _1(\xi )}{i\mu +\xi ^2+\kappa _1}f_4(x,\xi )\, d\xi ,\\&-\mu ^{2}\chi - \frac{b}{\rho _2}\chi _{xx} + \frac{K}{\rho _2}(w_{x}+\chi ) +i\mu \chi \left( \frac{{\tilde{a}}_1}{\rho _2} (i\mu +\kappa _1)^{\sigma _2-1} \ e^{-i\mu \tau _2}+\frac{{\tilde{a}}_2}{\rho _2}\right) =f_6 + i\mu f_5 \\&\quad -\frac{{{{\tilde{a}}}}_1}{\rho _2}(i\mu +\kappa _2)^{\sigma _2-1} (z_2)_0(x)+\frac{{\tilde{a}}_2}{\rho _2}f_5 -\frac{{\tilde{\zeta }}}{\rho _{2}} \int _{-\infty }^{+\infty } \displaystyle \frac{\theta _2(\xi )}{i\mu +\xi ^2+\kappa _2}f_8(x,\xi )\, d\xi \end{aligned} \right. \end{aligned}$$

(37)

with $w(0)=w(L)=\chi (0)=\chi (L)=0$. Solving system (37) is equivalent to find $(w, \chi )\in (H^{2}\cap H_{0}^{1}(0, L))^2$ such that

$$\begin{aligned} \begin{aligned}&\displaystyle \int _{0}^{L} \left( -\rho _1\mu ^{2}w {{{\overline{w}}}_{1}} - \rho _2\mu ^{2}\chi {{{\overline{w}}}_2}+ K(w_{x}+\chi )({{{\overline{w}}}_1}_x +{{{\overline{w}}}_2})+ b\chi _{x}{{{\overline{w}}}_2}_x\right) \ dx\\&\qquad +\displaystyle \int _{0}^{L}\left( i\mu \left( a_1 (i\mu +\kappa _1)^{\sigma _1-1} e^{-i\mu \tau _1}+a_2\right) w {{{\overline{w}}}_{1}}\right. \\&\qquad \left. +\,i\mu \left( {\tilde{a}}_1 (i\mu +\kappa _1)^{\sigma _2-1} \ e^{-i\mu \tau _2}+{\tilde{a}}_2\right) \chi {{{\overline{w}}}_2}\right) \ dx\\&\quad =\displaystyle \int _{0}^{L}\left( \rho _1(f_2 + i\mu f_1) -a_1(i\mu +\kappa _1)^{\sigma _1-1}(z_1)_0(x)+a_2 f_1\right. \\&\qquad \left. -\zeta \int _{-\infty }^{+\infty }\displaystyle \frac{\theta _1(\xi )}{i\mu +\xi ^2+\kappa _1}f_4(x,\xi )\, d\xi \right) {{{\overline{w}}}_1} \ dx\\&\qquad +\displaystyle \int _{0}^{L}\left( \rho _2(f_6 + i\mu f_5) -{{{\tilde{a}}}}_1(i\mu +\kappa _2)^{\sigma _2-1}(z_2)_0(x)+{\tilde{a}}_2 f_5 \right. \\&\qquad \left. -{\tilde{\zeta }} \int _{-\infty }^{+\infty }\displaystyle \frac{\theta _2(\xi )}{i\mu +\xi ^2+\kappa _2}f_8(x,\xi )\, d\xi \right) {{{\overline{w}}}_2} \ dx \end{aligned} \end{aligned}$$

(38)

for all $W=(w_1, w_2)\in (H_{0}^{1}(0, L))^2$. The system (38) is equivalent to the problem

$$\begin{aligned} L_{\mu }(U, W)+a_{(H_{0}^{1}(0 , L))^2}(U, W)=l(W), \end{aligned}$$

(39)

where the sesquilinear forms , and the antilinear form are defined by

$$\begin{aligned} \begin{aligned} L_{\mu }(U, W)&=\displaystyle \int _{0}^{L} \left( -\rho _1\mu ^{2}w {{{\overline{w}}}_{1}} - \rho _2\mu ^{2}\chi {{{\overline{w}}}_2}\right) \, dx \\&\quad +\displaystyle \int _{0}^{L}\left( i\mu \left( a_1 (i\mu +\kappa _1)^{\sigma _1-1} e^{-i\mu \tau _1}+a_2\right) w {{{\overline{w}}}_{1}} \right. \\&\quad \left. +i\mu \left( {\tilde{a}}_1 (i\mu +\kappa _1)^{\sigma _2-1} \ e^{-i\mu \tau _2}+{\tilde{a}}_2\right) \chi {{{\overline{w}}}_2}\right) \ dx,\\ a_{(H_{0}^{1}(0 , L))^2}(U, W)&=\displaystyle \int _{0}^{L}\left( K(w_{x}+\chi )({{{\overline{w}}}_1}_x +{{{\overline{w}}}_2})+ b\chi _{x}{{{\overline{w}}}_2}_x\right) \ dx \end{aligned} \end{aligned}$$

and

$$\begin{aligned} \begin{aligned} l(W)&=\displaystyle \int _{0}^{L}\left( \rho _1(f_2 + i\mu f_1) -a_1(i\mu +\kappa _1 )^{\sigma _1-1}(z_1)_0(x)+a_2 f_1 \right. \\&\quad \left. -\zeta \int _{-\infty }^{+\infty } \displaystyle \frac{\theta _1(\xi )}{i\mu +\xi ^2+\kappa _1}f_4(x,\xi )\, d\xi \right) {{{\overline{w}}}_1} \ dx \\&\quad +\displaystyle \int _{0}^{L}\left( \rho _2(f_6 + i\mu f_5) -{{{\tilde{a}}}}_1 (i\mu +\kappa _2)^{\sigma _2-1}(z_2)_0(x)+{\tilde{a}}_2 f_5\right. \\&\quad \left. -{\tilde{\zeta }} \int _{-\infty }^{+\infty }\displaystyle \frac{\theta _2 (\xi )}{i\mu +\xi ^2+\kappa _2}f_8(x,\xi )\, d\xi \right) {{{\overline{w}}}_2} \ dx. \end{aligned} \end{aligned}$$

One can easily see that $L_{\mu }, a_{(H_{0}^{1}(0 , L))^2}$ and l are bounded. Furthermore

$$\begin{aligned} \mathfrak {R}a_{(H_{0}^{1}(0 , L))^2}(U, U)= b\Vert \chi _x\Vert _{L^2(0, L)}^2+K\Vert w_x+\chi \Vert _{L^2(0, L)}^{2}. \end{aligned}$$

Thus $a_{(H_{*}^{1}(0 , L))^2}$ is coercive. Consequently, by Lemma 3.3, proving the existence of U solution of (39) reduces to proving that (39) with $l\equiv 0$ has a notrivial solution. Indeed if there exists $U\not =0$, such that

$$\begin{aligned} L_{\mu }(U, W)+a_{(H_{0}^{1}(0 , L))^2}(U, W)=0\quad \forall W\in H_{0}^{1}(0 , L)\times H_{0}^{1}(0, L). \end{aligned}$$

(40)

Then $i\mu$ is an eigenvalue of ${{{\mathcal {A}}}}$. Therefore from Lemma 3.1 we deduce that $U=0$. $\square$

4 Exponential stability

The necessary and suficient conditions for the exponential stability of the $C_0$-semigroup of contractions on a Hilbert space were obtained by Gearhart [12] and Huang [15] independently, see also Pruss [23]. We will use the following result due to Gearhart.

Theorem 4.1

[23]-[15] Let $S(t)=e^{{{{\mathcal {A}}}}t}$ be a $C_0$-semigroup of contractions on Hilbert space ${{{\mathcal {H}}}}$. Then S(t) is exponentially stable if and only if

(41)

and

$$\begin{aligned} \overline{\displaystyle \lim _{|\beta |\rightarrow \infty }}\Vert (i\beta I -{{{\mathcal {A}}}})^{-1}\Vert _{{{{\mathcal {L}}}}({{{\mathcal {H}}}})}< \infty . \end{aligned}$$

(42)

Our main result is as follows.

Theorem 4.2

The semigroup ${S_{{{\mathcal {A}}}}(t)}_{t\ge 0}$ generated by ${{{\mathcal {A}}}}$ is exponentially stable.

Proof

We will study the resolvent equation , that is,

$$\begin{aligned} \left\{ \begin{array}{l} i\mu w-{\tilde{w}}=f_1, \\ i\mu {\tilde{w}}-\frac{K}{\rho _1}(w_{x}+\chi )_{x} +\frac{\zeta }{\rho _{1}} \int _{-\infty }^{+\infty }\theta _1(\xi )\phi _1(x,\xi )\, d\xi +\frac{a_2}{\rho _{1}}{{{\tilde{w}}}}=f_2, \\ i\mu z_1+\frac{z_{1p}}{\tau _1}=f_3, \\ i\mu \phi _{1}+(\xi ^{2}+\kappa _1)\phi _{1}-z_1(x, 1)\theta _1(\xi )=f_{4}, \\ i\mu \chi -{\tilde{\chi }}=f_5, \\ i\mu {\tilde{\chi }}-\frac{b}{\rho _2}\chi _{xx}+\frac{K}{\rho _2}(w_{x}+\chi ) +\frac{{\tilde{\zeta }}}{\rho _{2}} \int _{-\infty }^{+\infty }\theta _2(\xi )\phi _2(x,\xi )\, d\xi +\frac{{\tilde{a}}_2}{\rho _{2}}{{{\tilde{\chi }}}}=f_6, \\ i\mu z_2+\frac{z_{2p}}{\tau _2}=f_7, \\ i\mu \phi _{2}+(\xi ^{2}+\kappa _2)\phi _{2}-z_2(x, 1)\theta _2(\xi )=f_{8}, \end{array} \right. \end{aligned}$$

(43)

where $F=(f_1, f_2, f_3, f_4, f_5, f_6, f_7, f_8)^{T}$. Taking the inner product of (43) with U in ${{{\mathcal {H}}}}$ and then taking its real part yields

$$\begin{aligned} |Re\langle {{{\mathcal {A}}}}U, U\rangle _{{{\mathcal {H}}}}|\le \Vert U\Vert _{{{\mathcal {H}}}}\Vert F\Vert _{{{\mathcal {H}}}}. \end{aligned}$$

(44)

that is,

$$\begin{aligned} \displaystyle \int _{0}^L |{{{\tilde{w}}}}(x)|^2\, dx, \displaystyle \int _{0}^L |{{{\tilde{\chi }}}}(x)|^2\, dx,\quad \displaystyle \int _{0}^L |z_{j}(x,1)|^2 \, dx \le C\Vert U\Vert _{{{\mathcal {H}}}}\Vert F\Vert _{{{\mathcal {H}}}}. \end{aligned}$$

(45)

From (43)$_4$ and (43)$_8$ we have

$$\begin{aligned} \left\{ \begin{aligned} \phi _1&= \displaystyle \frac{z_1(x,1)\theta _1(\xi )+f_4}{i\mu +\xi ^{2}+\kappa _1},\\ \phi _2&= \displaystyle \frac{z_2(x,1)\theta _2(\xi )+f_8}{i\mu +\xi ^{2}+\kappa _2} \end{aligned} \right. \end{aligned}$$

(46)

and applying Young’s inequality we obtain

$$\begin{aligned}&\Vert \phi _1\Vert _{L^{2}((0, L)\times (-\infty , +\infty ))}\nonumber \\&\quad \le \left\| \displaystyle \frac{\theta _1(\xi )}{i\mu +\xi ^{2}+\kappa _1} \right\| _{L^{2}(-\infty , +\infty )} \Vert z_1(x,1)\Vert _{L^2(0, L)}\nonumber \\&\qquad +\left\| \displaystyle \frac{f_4}{i\mu +\xi ^{2}+\kappa _1} \right\| _{L^{2}((0, L)\times (-\infty , +\infty ))}\nonumber \\&\quad \le \left( 2 (1-\sigma _1)\displaystyle \frac{\pi }{\sin \sigma _1 \pi }(|\mu |+\kappa _1)^{\sigma _1-2}\right) ^{\frac{1}{2}} \Vert z_1(x,1)\Vert _{L^2(0, L)}\nonumber \\&\qquad +\displaystyle \frac{\sqrt{2}}{|\mu |+\kappa _1}\Vert f_4\Vert _{L^{2}((0, L)\times (-\infty , +\infty ))} \end{aligned}$$

(47)

and

$$\begin{aligned}&\Vert \xi \phi _1\Vert _{L^{2}((0, L)\times (-\infty , +\infty ))}\nonumber \\&\quad \le \left\| \displaystyle \frac{\xi \theta _1(\xi )}{i\mu +\xi ^{2}+\kappa _1}\right\| _{L^{2}(-\infty , +\infty )} \Vert z_1(x,1)\Vert _{L^2(0, L)}\nonumber \\&\qquad +\left\| \displaystyle \frac{\xi f_4}{i\mu +\xi ^{2}+\kappa _1}\right\| _{L^{2}((0, L)\times (-\infty , +\infty ))}\nonumber \\&\quad \le \left( 2 \sigma _1\displaystyle \frac{\pi }{\sin \sigma _1\pi }(|\mu |+\kappa _1)^{\sigma _1-1}\right) ^{\frac{1}{2}} \Vert z_1(x,1)\Vert _{L^2(0, L)}\nonumber \\&\qquad +\displaystyle \frac{\sqrt{2}}{\sqrt{|\mu |+\kappa _1}}\Vert f_4\Vert _{L^{2}((0, L)\times (-\infty , +\infty ))}. \end{aligned}$$

(48)

Let us introduce the following functions

$$\begin{aligned} {{{\mathcal {I}}}}_{w}(x)=|{{{\tilde{w}}}}(x)|^{2}+|{{{\tilde{\chi }}}}(x)|^{2} +K|w_x(x)+\chi (x)|^2+b|\chi _x(x)|^2 \end{aligned}$$

and

$$\begin{aligned} {{{\mathcal {E}}}}_{w}=\displaystyle \int _{0}^{L}{{{\mathcal {I}}}}_{w}(x)\, dx. \end{aligned}$$

Lemma 4.1

We have that

$$\begin{aligned} {{{\mathcal {E}}}}_{w}\le c \Vert F\Vert _{{{{\mathcal {H}}}}}^{2}+c' \Vert F\Vert _{{{{\mathcal {H}}}}}\Vert U\Vert _{{{{\mathcal {H}}}}}. \end{aligned}$$

(49)

for positive constants c and $c'$.

Proof

Multiplying equation (43)$_2$ by ${{{\overline{w}}}}$, integrating on (0, L), we get

$$\begin{aligned}&-\rho _1\displaystyle \int _{0}^{L}{{{\tilde{w}}}} (\overline{i\mu w })\, dx+K\displaystyle \int _{0}^{L} (w_x+\chi )\overline{w_x}\, dx \nonumber \\&\quad +\zeta \displaystyle \int _{0}^{L}{\bar{w}}\left( \displaystyle \int _{-\infty }^{+\infty }\theta _1(\xi )\phi _1(x,\xi )\, d\xi \right) \, dx + a_{2}\displaystyle \int _{0}^{L} {{{\tilde{w}}}}{\bar{w}}\, dx=\rho _1\displaystyle \int _{0}^{L}{\bar{w}}f_2 \, dx. \end{aligned}$$

(50)

From (43)$_1$, we get $i\mu w={{{\tilde{w}}}}+f_1$. Then

$$\begin{aligned}&-\rho _1\displaystyle \int _{0}^{L}|{{{\tilde{w}}}}|^2\, dx+K\displaystyle \int _{0}^{L} (w_x+\chi )\overline{w_x}\, dx \nonumber \\&\quad +\zeta \displaystyle \int _{0}^{L}{\bar{w}}\left( \displaystyle \int _{-\infty }^{+\infty }\theta _1(\xi )\phi _1(x,\xi )\, d\xi \right) \, dx + a_{2}\displaystyle \int _{0}^{L} {{{\tilde{w}}}}{\bar{w}}\, dx=\rho _1\displaystyle \int _{0}^{L}({\bar{w}}f_2 +{{{\tilde{w}}}}\bar{f_1}) \, dx. \end{aligned}$$

(51)

Multiplying equation (43)$_6$ by ${{{\overline{\chi }}}}$, integrating on (0, L), we get

$$\begin{aligned}&-\rho _2\displaystyle \int _{0}^{L}{{{\tilde{\chi }}}} (\overline{i\mu \chi })\, dx +b\displaystyle \int _{0}^{L}|\chi _x|^2\, dx+K\displaystyle \int _{0}^{L} (w_x+\chi ){\overline{\chi }}\, dx \nonumber \\&\quad +{{{\tilde{\zeta }}}} \displaystyle \int _{0}^{L}{\bar{\chi }}\left( \displaystyle \int _{-\infty }^{+\infty } \theta _2(\xi )\phi _2(x,\xi )\, d\xi \right) \, dx + {{{\tilde{a}}}}_{2}\displaystyle \int _{0}^{L} {{{\tilde{\chi }}}}{\bar{\chi }}\, dx =\rho _2\displaystyle \int _{0}^{L}{\bar{\chi }}f_6 \, dx. \end{aligned}$$

(52)

From (43)$_5$, we get $i\mu \chi ={{{\tilde{\chi }}}}+f_5$. Then

$$\begin{aligned}&-\rho _2\displaystyle \int _{0}^{L}|{{{\tilde{\chi }}}}|^2\, dx+b\displaystyle \int _{0}^{L}| \chi _x|^2\, dx+K\displaystyle \int _{0}^{L} (w_x+\chi ){\overline{\chi }}\, dx\nonumber \\&\quad +{{{\tilde{\zeta }}}} \displaystyle \int _{0}^{L}{\bar{\chi }}\left( \displaystyle \int _{-\infty }^{ +\infty }\theta _2(\xi )\phi _2(x,\xi )\, d\xi \right) \, dx+ {{{\tilde{a}}}}_{2}\displaystyle \int _{0}^{L} {{{\tilde{\chi }}}}{\bar{\chi }} \, dx=\rho _2\displaystyle \int _{0}^{L}({\bar{\chi }}f_6+{{{\tilde{\chi }}}}{\bar{f}}_5) \, dx. \end{aligned}$$

(53)

Combining (51) and (53), we get

$$\begin{aligned}&-\rho _1\displaystyle \int _{0}^{L}|{{{\tilde{w}}}}|^2\, dx-\rho _2\displaystyle \int _{0}^{L}|{{{\tilde{\chi }}}}|^2\, dx +K\displaystyle \int _{0}^{L} |w_x+\chi |^2\, dx+b\displaystyle \int _{0}^{L}|\chi _x|^2\, dx\nonumber \\&\quad +\zeta \displaystyle \int _{0}^{L}{\bar{w}}\left( \displaystyle \int _{-\infty }^{+\infty }\theta _1(\xi )\phi _1(x,\xi )\, d\xi \right) \, dx + {{{\tilde{\zeta }}}} \displaystyle \int _{0}^{L}{\bar{\chi }}\left( \displaystyle \int _{-\infty }^{+\infty }\theta _2(\xi )\phi _2(x,\xi )\, d\xi \right) \, dx\nonumber \\&\quad +a_{2}\displaystyle \int _{0}^{L} {{{\tilde{w}}}}{\bar{w}}\, dx+{{{\tilde{a}}}}_{2}\displaystyle \int _{0}^{L} {{{\tilde{\chi }}}} {\bar{\chi }}\, dx\nonumber \\&\quad =\rho _1\displaystyle \int _{0}^{L}({\bar{w}}f_2+{{{\tilde{w}}}}\bar{f_1}) \, dx +\rho _2\displaystyle \int _{0}^{L}({\bar{\chi }}f_6+{{{\tilde{\chi }}}}{\bar{f}}_5) \, dx. \end{aligned}$$

(54)

We can estimate

$$\begin{aligned} \begin{aligned}&\left| \displaystyle \int _{0}^{L}{\bar{w}}(\displaystyle \int _{-\infty }^{+\infty }\theta _1(\xi )\phi _1(x,\xi )\, d\xi )\, dx\right| \\&\quad \le \Vert w\Vert _{L^2(0, L)}\left( \displaystyle \int _{-\infty }^{+\infty }\displaystyle \frac{\theta _1^2(\xi )}{\xi ^2+\kappa _1}\, d\xi \right) ^{\frac{1}{2}} \left( \displaystyle \int _{0}^{L}\displaystyle \int _{-\infty }^{+\infty }(\xi ^2+\kappa _1)|\phi _1(x,\xi )|^2\, d\xi \, dx\right) ^{\frac{1}{2}}\\&\quad \le \displaystyle \frac{\varepsilon }{2} \left( \displaystyle \int _{-\infty }^{+\infty }\displaystyle \frac{\theta _1^2(\xi )}{\xi ^2+\kappa _1}\, d\xi \right) \Vert w\Vert _{L^2(0, L)}^2 +\displaystyle \frac{1}{2\varepsilon } \displaystyle \int _{0}^{L}\displaystyle \int _{-\infty }^{+\infty }(\xi ^2+\kappa )|\phi _1(x,\xi )|^2\, d\xi \, dx\\&\quad \le \displaystyle \frac{\varepsilon }{2} C \left( \displaystyle \int _{-\infty }^{+\infty }\displaystyle \frac{\theta _1^2(\xi )}{\xi ^2+\kappa _1}\, d\xi \right) \Vert w_x\Vert _{L^2(0, L)}^2 +\displaystyle \frac{1}{2\varepsilon } \displaystyle \int _{0}^{L}\displaystyle \int _{-\infty }^{+\infty }(\xi ^2+\kappa _1)|\phi _1(x,\xi )|^2\, d\xi \, dx.\\ \end{aligned} \end{aligned}$$

Moreover, we have

$$\begin{aligned} \begin{aligned} \left| \displaystyle \int _{0}^{L}{\bar{w}} {{{\tilde{w}}}}\, dx\right|&\le \Vert w\Vert _{L^2(0, L)}\Vert {{{\tilde{w}}}}\Vert _{L^2(0, L)}\\&\le \displaystyle \frac{\varepsilon }{2}C \Vert w_x\Vert _{L^2(0, L)}^2+\displaystyle \frac{1}{2\varepsilon } \Vert {{{\tilde{w}}}}\Vert _{L^2(0, L)}^2,\\ \left| \displaystyle \int _{0}^{L}{\bar{w}}f_2 \, dx\right|&\le \displaystyle \frac{\varepsilon }{2}C \Vert w_x\Vert _{L^2(0, L)}^2+\displaystyle \frac{1}{2\varepsilon } \Vert f_2 \Vert _{L^2(0, L)}^2,\\ \left| \displaystyle \int _{0}^{L} {{{\tilde{w}}}} \bar{f_1}\, dx\right|&\le \displaystyle \frac{\varepsilon }{2}\Vert {{{\tilde{w}}}}\Vert _{L^2(0, L)}^2+ \displaystyle \frac{1}{2\varepsilon }\Vert f_1\Vert _{L^2(0, L)}^2. \end{aligned} \end{aligned}$$

Similarly

$$\begin{aligned} \begin{aligned} \left| \displaystyle \int _{0}^{L}{\bar{\chi }}\left( \displaystyle \int _{-\infty }^{+\infty }\theta _1(\xi )\phi _2(x,\xi )\, d\xi \right) \, dx\right|&\le \displaystyle \frac{\varepsilon }{2} C \left( \displaystyle \int _{-\infty }^{+\infty }\displaystyle \frac{\theta _2^2(\xi )}{\xi ^2+\kappa _2}\, d\xi \right) \Vert \chi _x\Vert _{L^2(0, L)}^2 \\&\quad +\displaystyle \frac{1}{2\varepsilon } \displaystyle \int _{0}^{L}\displaystyle \int _{-\infty }^{+\infty }(\xi ^2+\kappa _2)|\phi _2(x,\xi )|^2\, d\xi \, dx\\ \left| \displaystyle \int _{0}^{L}{\bar{\chi }} {{{\tilde{\chi }}}}\, dx\right|&\le \Vert \chi \Vert _{L^2(0, L)}\Vert {{{\tilde{\chi }}}}\Vert _{L^2(0, L)}\\&\le \displaystyle \frac{\varepsilon }{2}C \Vert \chi _x\Vert _{L^2(0, L)}^2+\displaystyle \frac{1}{2\varepsilon } \Vert {{{\tilde{\chi }}}}\Vert _{L^2(0, L)}^2,\\ \left| \displaystyle \int _{0}^{L}{\bar{\chi }}f_6 \, dx\right|&\le \displaystyle \frac{\varepsilon }{2}C \Vert \chi _x\Vert _{L^2(0, L)}^2+\displaystyle \frac{1}{2\varepsilon } \Vert f_6 \Vert _{L^2(0, L)}^2,\\ \left| \displaystyle \int _{0}^{L} {{{\tilde{\chi }}}} \bar{f_5}\, dx\right|&\le \displaystyle \frac{\varepsilon }{2}\Vert {{{\tilde{\chi }}}}\Vert _{L^2(0, L)}^2+ \displaystyle \frac{1}{2\varepsilon }\Vert f_5\Vert _{L^2(0, L)}^2. \end{aligned} \end{aligned}$$

Choosing $\varepsilon$ small enough, we deduce (49) and the Lemma follows.

Now, it follows from the equations (43)$_3$ and (43)$_7$ that

$$\begin{aligned} \begin{aligned} z_1(x, \varrho )&=e^{-i\tau _1 \mu \varrho }z_1(x, 0) +\tau e^{-i\tau _1 \mu \varrho }\displaystyle \int _{0}^{\varrho }e^{-i\tau _1 \mu }f_4(x, r)\, dr\\&=e^{-i\tau _1 \mu \varrho }{{{\tilde{w}}}}(x) +\tau _1 e^{-i\tau _1 \mu \varrho }\displaystyle \int _{0}^{\varrho }e^{-i\tau _1 \mu }f_4(x, r)\, dr \end{aligned} \end{aligned}$$

and

$$\begin{aligned} z_2(x, \varrho )=e^{-i\tau _2 \mu \varrho }{{{\tilde{\chi }}}}(x) +\tau _2 e^{-i\tau _2 \mu \varrho }\displaystyle \int _{0}^{\varrho }e^{-i\tau _2 \mu }f_8(x, r)\, dr. \end{aligned}$$

Hence

$$\begin{aligned} \Vert z_1(x, \varrho )\Vert _{L^2((0, L)\times (0, 1))}\le \Vert {{{\tilde{w}}}}(x)\Vert _{L^2(0, L)}+\tau _1\Vert f_4(x, \varrho )\Vert _{L^2((0, L)\times (0, 1))} \end{aligned}$$

(55)

and

$$\begin{aligned} \Vert z_2(x, \varrho )\Vert _{L^2((0, L)\times (0, 1))}\le \Vert {{{\tilde{\chi }}}}(x)\Vert _{L^2(0, L)}+\tau _2\Vert f_8(x, \varrho )\Vert _{L^2((0, L)\times (0, 1))}. \end{aligned}$$

(56)

Finally, by using (47), (49), (55) and (56), we get

$$\begin{aligned} \Vert U\Vert _{{{{\mathcal {H}}}}}\le C \Vert F\Vert _{{{{\mathcal {H}}}}} \end{aligned}$$

for a positive constant C. It follows that

$$\begin{aligned} \Vert (i\mu I-{{{\mathcal {A}}}})^{-1}\Vert _{{{{\mathcal {H}}}}}\le C. \end{aligned}$$

The conclusion then follows by applying the Theorem 4.1.

Remark 4.1

We can extend the results of this paper to more general measure density instead of (1), that is $\theta _j$ is an even nonnegative measurable function such that

$$\begin{aligned} \displaystyle \int _{-\infty }^{\infty }\displaystyle \frac{\theta _j(\xi )^2}{1+\xi ^2}\, d\xi < \infty . \end{aligned}$$

(57)

$\square$