Polynomial stability of a piezoelectric beam with magnetic effect and a boundary dissipation of the fractional derivative type

Abstract

This work studies the asymptotic behavior of a waves coupled system with a boundary dissipation of the fractional derivative type. We prove well-posedness and polynomial stability based on the semigroup approach, the energy method, and the result of stability.

1. Introduction

In this work, we first consider the piezoelectric coupled system given by

(1.1)\begin{eqnarray} \left\lbrace \begin{array}{l} \rho u_{tt} - \chi u_{xx} + \gamma \beta v_{xx} + \mu_{1}(t) u_{t} = 0,\qquad x\in [0,\,L],\quad t \geq 0, \\ \mu v_{tt} - \beta v_{xx} + \gamma \beta u_{xx} = 0,\qquad x\in [0,L],\quad t \geq 0 \\ u(x, 0) = u_{0}(x),\qquad u_{t}(x, 0) = u_{1}(x) \end{array} \right. \end{eqnarray}

where $u=u(x,t)$ represents the longitudinal displacement of the center line, $v=v(x, t)$ is the total load of the electric displacement along the transverse direction at each point, ρ denotes the mass density per unit volume, χ is the elastic stiffness, γ is the piezoelectric coefficient, µ is the magnetic permeability, and β is the water resistance coefficient of the beam and the prescribed voltage on electrodes of the beam. It is a well-known fact that a piezoelectric beam is an elastic beam, which has electrodes at its top and bottom surfaces, and in order to prevent fringing effects, it is insulated at the edges, as well as connected to an external electric circuit. The characteristics mentioned before are the simplest structures on which to study the interaction between the electrical and mechanical energy in these systems, see [10] for more information. Experimentally, it is observed that the magnetic effects are minor in the overall dynamics in polarized ceramics, and therefore these effects are ignored in the piezoelectric beam models. The piezoelectric beam models have a unique characteristic of converting mechanical energy into electrical and magnetic energy, and vice versa and, hence, are a competitive choice for many tasks in industry, particularly those involving control of structures. In fact, piezoelectric materials have been employed in civil, industrial, automotive, aeronautic, and space structures. The coupled system (1.1) has been studied from different points of views. Ramos et al. [15], for example, inserted a mechanical dissipative term $\delta u_{t}$ into Equation (1.1)₁ for χ > 0 and an appropriate boundary condition to prove that the energy decays exponentially. Later, Ramos et al. [14] showed that under different boundary conditions, the system is exponentially stable, regardless of any relationship between system parameters, and also showed that the exponential stability is equivalent to an exact observability at the boundary. The topic has been thoroughly explained in [13, 17, 18] and supported by the multiple references therein.

On the other hand, fractional calculus is a mathematical tool which studies integral and differential operators of arbitrary order, not necessarily of integer order. For the past three decades or so, a number of scientists have shown a growing interest on the study of fractional calculus. Several engineering fields, applied sciences, and mathematical physics have benefited from this ascending wave of growing applications in this area. Moreover, space sciences, fluid mechanics, porous media flows, and viscoelastic and biological processes are areas in which fractional differential equations have become a favoured tool to seek new paths.

Fractional derivatives have been used in the model frequency–dependent damping behaviour of many viscoelastic materials, as well as in modelling many chemical processes. A diversity of problems from the scientific applied areas, including analysis of viscoelastic materials, heat conduction in materials with memory, electrodynamics with memory, signal processing, among others, can be modelled with fractional differential calculus. Models involving fractional derivatives represent an important number of natural phenomena in a more realistic way than models involving classical derivatives. For more information, we would like to direct the reader to [9, 11, 16] and the references therein.

The objective of this paper is to study polynomial stabilization in the case of $\mu_{1}(t) \equiv 0$ in Equation (1.1) by means of a boundary viscoelastic damper, the action of which is to cause a feedback force opposite the direction of the fractional derivative of the position of a limit point. By using an energy method, it is shown that the energy of the system decays polynomially. The method is motivated by the fact that the input–output relationship as generalized diffusion equation can be expressed in terms of fractional derivative operator and following the ideas given by [1, 3, 11, 12]. In order to achieve results of polynomial stability, the dissipation given in the domain was eliminated and placed on the border. The imposed dissipation is of a fractional derivative type. Thus, we have focused on the following system:

(1.2)\begin{eqnarray} \left\lbrace \begin{array}{l} \rho u_{tt} - \chi u_{xx} + \gamma \beta v_{xx} = 0,\qquad x\in [0,L],\ t \geq 0, \\ \mu v_{tt} - \beta v_{xx} + \gamma \beta u_{xx} = 0,\qquad x\in [0, L],\ t \geq 0, \\ u(x,0) = u_{0}(x),\qquad u_{t}(x,0) = u_{1}(x), \\ v(x,0) = v_{0}(x),\qquad v_{t}(x,0) = v_{1}(x), \\ u(0,t) = 0, \qquad\mbox{for}\ t\geq 0, \\ v(0,t) = 0, \qquad\mbox{for}\ t\geq 0, \\ \chi u_{x}(L,t) - \gamma \beta v_{x}(L,t)=-\partial_{t}^{\alpha,\eta}u(L,t), \qquad\mbox{for}\ t\geq 0, \\ \beta v_{x}(L,t) - \gamma \beta u_{x}(L,t) = -\partial_{t}^{\alpha,\eta}v(L,t), \qquad\mbox{for}\ t\geq 0, \end{array} \right. \end{eqnarray}

where $\ 0 \lt \alpha \lt 1,$ $\eta \geq 0 $ and $u=u(x,t),$ $v=v(x,t)$ are real-valued functions.

Lemma 1.1. For every solution (u, v) of the system (1.2), the total energy ${\cal E}_{1}: \mathbb{R}^{+}\rightarrow \mathbb{R}^{+}$ is given at time t by

(1.3)\begin{eqnarray} \frac{\rm d}{{\rm d}t}{\cal E}_{1}(t) = -\partial_{t}^{\alpha, \eta}u(L,t) - \partial_{t}^{\alpha, \eta}v(L,t)\ \partial_{t}v(L,t), \end{eqnarray}

along with

(1.4)\begin{eqnarray} {\cal E}_{1}(t) = \frac{1}{2}\left[\rho\|u_{t}\|_{L^{2}(0,L)}^{2} + \mu\|v_{t}\|_{L^{2}(0,L)}^{2}\right] + \int_{0}^{L}X^{\rm T}AX\, {\rm d}x \end{eqnarray}

and the positive definite quadratic form

(1.5)\begin{eqnarray} X^{\rm T}AX = \left( \begin{array}{cc} u_{x} & v_{x} \end{array} \right)\left( \begin{array}{cc} \chi & - \gamma \beta \\ - \gamma \beta & \beta \\ \end{array} \right)\left( \begin{array}{c} u_{x} \\ v_{x} \\ \end{array} \right) = \chi u_{x}^{2} + \beta v_{x}^{2} - 2\gamma \beta u_{x} v_{x} \geq 0. \end{eqnarray}

Proof. Straightforward calculations.

Remark 1.2. We have seen that $\frac{{\rm d}{\cal E}_{1}(t)}\,{{\rm d}t} \neq 0,$ which follows from Equation (1.3), and it is not energy conserving. Moreover, due to the negativity of Equation (1.3), the system (1.2) is a dissipative kind of energy.

The interest in studying the fractional differential equations arose as a separate topic about 40 years ago. It has been of special interest to prove the existence of solutions of Cauchy-type problems involving the Caputo fractional derivatives. Several articles published during the last decade have developed engineering applications in viscoelastic damping and structural mechanics (see [5, 11]). A systematic and rigorous study of problems of this kind involving fractional differential equations can be found in recent literature; for a brief review, refer to [7, 8]. Although numerous theoretical applications of fractional calculus operators have been found during an extended period time, many mathematicians and applied researchers have tried to model real processes using fractional differential equations. However, achieving results has been complex due in part to the fact that many of the properties of the ordinary derivative do not translate analogously to the case of the fractional derivative operator $D^{\alpha},$ such as, for example, a clear geometric or physical meaning, product rules, and chain rules.

In this work, we have obtained polynomial stability for a piezoelectric beam with a boundary dissipation of the fractional derivative type. In order to obtain the partial result, we used a similar approach to Mbodje and Montseny [12]. This relationship help us achieve our main result in the following:

Theorem 1.3. The semigroup $\{{\cal S}_{\cal A}(t)\}_{t\geq 0}$ is polynomially stable and

\begin{eqnarray*} \|{\cal S}_{{\cal A}}(t)U_{0}\|_{{\cal H}} \leq \frac{1}{t^{1/2(1 - \alpha)}}\|U_{0}\|_{{\cal D}({\cal A})}. \end{eqnarray*}

This paper is organized as follows. In $\S$ 2, we reformulate the model (1.2) into an augmented system, which couples the system (1.1) with a suitable diffusion equation. In $\S$ 3, we establish the well-posedness of the system (2.10). In $\S$ 4, we show the polynomial stability of the corresponding semigroup.

Throughout this paper, C is a generic constant, not necessarily the same at each occasion. It will change from line to line and depends on the indicated quantities.

2. Augmented Model

This section is concerned with the reformulation of the model (1.2) with boundary conditions (1.2) $_{5,\,6,\,7,\,8}$ into an augmented system. To fulfill this goal, we will use a method based on the input–output relationship in diffusion equations which realize the fractional derivative operator. This tool in fractional calculus allows integrals and derivatives of any positive order. This can be considered a branch of mathematical physics which deals with integro-differential equations, where integrals are of a convolution type and exhibit weakly singular kernels of power law type. There are multiple ways in which fractional derivatives and integrals can be defined. These are not all equivalent to each other, but each of them has its own advantages and disadvantages.

Let $0 \lt \alpha \lt 1.$ We define the Caputo fractional integral of order α > 0 as

(2.1)\begin{eqnarray} J^{\alpha}f(t) = \frac{1}{\Gamma(\alpha)}\int_{0}^{t}(t - \tau)^{\alpha - 1} f(\tau)\, {\rm d}\tau, \end{eqnarray}

where Γ is the gamma function, $f \in L^{1}(0,t),$ and t > 0.

The Caputo fractional derivative operator of order α > 0 is defined by

(2.2)\begin{equation} D_{t}^{\alpha}f(t) = I^{1 - \alpha}D_{t}f(t) := \frac{1}{\Gamma(1 - \alpha)}\int_{0}^{t}(t - \tau)^{-\alpha}\, \frac{{\rm d}f}{{\rm d}\tau}(\tau)\, {\rm d}\tau, \end{equation}

with $f \in W^{1,1}(0,t)$ and t > 0. If the function f(t) represents the strained history within a viscoelastic material whose relaxation function is $[\Gamma(1 - \alpha)t^{\alpha}]^{-1}$, then the material will experience at any time t a total stress given the expression $D^{\alpha}f(t).$ Also, note that D ^α is a left inverse of $J^{\alpha},$ but in general, it is not a right inverse. More precisely, we have

\begin{eqnarray*} D_{t}^{\alpha}I^{\alpha}f = f, \qquad I^{\alpha}D_{t}^{\alpha}f(t) = f(t) - f(0). \end{eqnarray*}

For the proof of the above mentioned equalities and more properties of fractional calculus, we suggest the reader to refer [16].

We consider slightly different versions of (2.1) and (2.2). For this work, we have focused on [6], where Choi and MacCamy establish the following definition of fractional integro-differential operators with exponential weight.

Let $0 \lt \alpha \lt 1, \eta \ge 0.$ The exponential fractional integral of order α is defined by

\begin{equation*} I^{\alpha, \eta}f(t) = \frac{1}{\Gamma(\alpha)}\int_{0}^{t}{\rm e}^{-\eta(t - \tau)} (t - \tau)^{\alpha - 1}f(\tau)\, {\rm d}\tau, \end{equation*}

with $f \in L^{1}(0,t)$ and t > 0. The exponential fractional derivative operator of order α is given by

(2.3)\begin{equation} D_{t}^{\alpha, \eta}f(t) = I^{1 - \alpha,\eta}D_{t}f(t) := \frac{1}{\Gamma(1 - \alpha)}\int_{0}^{t}{\rm e}^{-\eta(t - \tau)}(t - \tau)^{-\alpha}\,\frac{{\rm d}f}{{\rm d}\tau}(\tau)\, {\rm d}\tau, \end{equation}

where $f \in W^{1, 1}(0,t)$ and t > 0. Note that the two operators (2.2) and (2.3) differ just by their kernels. $D^{\alpha,\eta}$ is merely Caputo’s fractional derivative operator with the exception of its exponential factor.

Theorem 2.1. ([11])

Let µ be the function

(2.4)\begin{eqnarray} \mu(\xi) = |\xi|^{(2\alpha - 1)/2},\quad \xi\in\mathbb{R},\quad 0\lt\alpha\lt1. \end{eqnarray}

Then the relation between the Input U and the Output O of the following system

(2.5)\begin{eqnarray} & & \varphi_{t}(\xi,t) + \xi^{2}\varphi(\xi,t) - U(t)\mu(\xi) = 0,\quad \xi\in\mathbb{R},\quad t\gt0,\\ & & \varphi(\xi,0) = 0 \nonumber \end{eqnarray}

(2.6)\begin{eqnarray}& & O = \mathfrak{C} \int_{\mathbb{R}}\mu(\xi)\varphi(\xi,t)\, {\rm d}\xi \end{eqnarray}

is given by $O = I^{1 - \alpha,\eta}U = D^{\alpha,\eta}U,$ where $\mathfrak{C} = \pi^{-1}\,\sin(\alpha \pi)$ and

\begin{eqnarray*} [I^{\alpha,\eta}f](t) = {\rm e}^{-\eta t}\int_{0}^{t}\frac{(t - s)^{\alpha - 1}}{\Gamma(\alpha)}\,{\rm e}^{\eta s} f(s)\, {\rm d}s. \end{eqnarray*}

Proof. Multiplying Equation (2.5) by ${\rm e}^{t\xi^{2}},$ integrating over $[0,t]$, and performing straightforward calculations, we have

(2.7)\begin{eqnarray} \varphi(\xi,t) = \int_{0}^{t}{\rm e}^{-\xi^{2}(t - s)}U(s)\mu(\xi)\, {\rm d}\xi. \end{eqnarray}

Moreover, replacing Equation (2.4) into Equation (2.7), we get

(2.8)\begin{eqnarray} \varphi(\xi,t) = \int_{0}^{t}|\xi|^{(2\alpha - 1)/2}\,{\rm e}^{-\xi^{2}(t - s)}U(s)\, {\rm d}\xi. \end{eqnarray}

To continue the process of making a clear relation between output and input, we turn Equation (2.8) into Equation (2.6) having Equation (2.9) as result:

(2.9)\begin{eqnarray} O = \mathfrak{C} \int_{0}^{t}\left(2\int_{0}^{\infty}|\xi|^{(2\alpha - 1)/2}\,{\rm e}^{-\xi^{2}(t - s)}\, {\rm d}\xi\right)U(s)\, {\rm d}s. \end{eqnarray}

With the change of variable $m = \xi^{2}(t - s)$ in the above equality, we are left with

\begin{eqnarray*} O & = & \mathfrak{C} \int_{0}^{t}(t - s)^{-\alpha}\left(\int_{0}^{\infty}m^{\alpha - 1}\,{\rm e}^{-m}\, {\rm d}m\right)U(s)\, {\rm d}s = \mathfrak{C} \int_{0}^{t}(t - s)^{-\alpha}\Gamma(\alpha)U(s)\, {\rm d}s \\ & = & \frac{1}{\Gamma(1 - \alpha)}\int_{0}^{t}(t - s)^{-\alpha}U(s)\, {\rm d}s = \left[I^{1 - \alpha}U\right](t) \end{eqnarray*}

and considering that $ \mathfrak{C} = \frac{1}{\Gamma(\alpha)\Gamma(1 - \alpha)}.$

Lemma 2.2. If $ \lambda \in D = \{\lambda\in \mathbb{C}: {\rm Re}\lambda + \eta \gt0\} \cup \{\lambda\in \mathbb{C}: {\rm Im}\lambda \neq 0\}. $ Then

\begin{eqnarray*} \int_{\mathbb{R}}\frac{\mu^{2}(\xi)}{\xi^{2} + \eta + \lambda}\, {\rm d}\xi = \frac{\pi}{\sin(\alpha \pi)}(\eta + \lambda)^{\alpha - 1}. \end{eqnarray*}

Proof. See [11].

Remark 2.3. In what remains of this paper, we will write $\mathfrak{C} = \pi^{-1}\,\sin(\alpha \pi).$

Now, we reformulate system (1.2) using Theorem 2.1. Therefore, the system is included into the augmented model

(2.10)\begin{eqnarray} \left\lbrace \begin{array}{l} \rho u_{tt} - \chi u_{xx} + \gamma \beta v_{xx} = 0,\qquad x\in [0,L],\ t \geq 0, \\ \mu v_{tt} - \beta v_{xx} + \gamma \beta u_{xx} = 0,\qquad x\in [0, L],\ t \geq 0, \\ \partial_{t}\varphi(\xi, t) + \left(\xi^{2} + \eta\right) \varphi(\xi, t) - u_{t}(L, t) \mu(\xi) = 0, \\ \partial_{t}\phi(\xi, t) + \left(\xi^{2} + \eta\right) \phi(\xi, t) - v_{t}(L, t) \mu(\xi) = 0, \\ u(x, 0) = u_{0}(x),\qquad u_{t}(x, 0) = u_{1}(x), \\ v(x, 0) = v_{0}(x),\qquad v_{t}(x, 0) = v_{1}(x), \\ u(0, t) = 0, \qquad\mbox{for}\ t\geq 0, \\ v(0, t) = 0, \qquad\mbox{for}\ t\geq 0, \\ \chi u_{x}(L, t) - \gamma \beta v_{x}(L, t)=- \mathfrak{C} \int_{\mathbb{R}}\mu(\xi) \varphi(\xi, t)\, {\rm d}\xi, \qquad\mbox{for}\ t\geq 0, \\ \beta v_{x}(L, t) - \gamma \beta u_{x}(L, t) = - \mathfrak{C} \int_{\mathbb{R}}\mu(\xi) \phi(\xi, t)\, {\rm d}\xi, \qquad\mbox{for}\ t\geq 0. \end{array} \right. \end{eqnarray}

Remark 2.4. The relation between the boundary conditions (1.2) and (2.10) are given by

\begin{eqnarray*} \chi u_{x}(L, t) - \gamma \beta v_{x}(L, t) = - \mathfrak{C} \int_{\mathbb{R}}\mu(\xi) \varphi(\xi, t)\, {\rm d}\xi = -\partial_{t}^{\alpha, \eta}u(L, t) \\ \beta v_{x}(L, t) - \gamma \beta u_{x}(L, t) = - \mathfrak{C} \int_{\mathbb{R}}\mu(\xi) \phi(\xi, t)\, {\rm d}\xi = -\partial_{t}^{\alpha, \eta}v(L, t). \end{eqnarray*}

Lemma 2.5. Let $(u, v, \varphi, \phi)$ be a solution of the system (2.10). Then, the energy functional is defined by

(2.11)\begin{align} {\cal E}(t) & = {\cal E}_{1}(t) + {\cal E}_{2}(t) \nonumber \\ & = \frac{1}{2}\left[\|u_{t}\|_{L^{2}(0, L)}^{2} + \|v_{t}\|_{L^{2}(0, L)}^{2} + \int_{0}^{L}X^{\rm T} A X\,{\rm d}x + \mathfrak{C} \|\varphi\|_{L^{2}(\mathbb{R})}^{2} + \mathfrak{C} \|\phi\|_{L^{2}(\mathbb{R})}^{2}\right], \end{align}

where ${\cal E}_{1}(t)$ and ${\cal E}_{2}(t)$ are the energies associated with the waves coupled systems and diffusion equations, respectively, and $q(u,\ v)$ is the positive definite quadratic form given in Equation (1.5), satisfying

(2.12)\begin{align} \frac{\rm d}{{\rm d}t}{\cal E}(t) = - \mathfrak{C}\int_{\mathbb{R}}\left(\xi^{2} + \eta\right)\varphi^{2}(\xi, t)\, {\rm d}\xi - \mathfrak{C}\int_{\mathbb{R}}\left(\xi^{2} + \eta\right)\phi^{2}(\xi, t)\, {\rm d}\xi \leq 0. \end{align}

The following theorems are fundamental to the proof of our results:

Theorem 2.6. ([2])

Let ${\cal A}$ be the generator of a uniformly bounded C₀-semigroup $\{{\cal S}(t)\}_{t\geq 0}$ on a Hilbert space $\cal H$. If:

1. i) ${\cal A}$ does not have eigenvalues on $\mathbb{R}.$

2. ii) the intersection of the spectrum $\sigma({\cal A})$ with $i\mathbb{R}$ is at most a countable set.

Then the semigroup $\{{\cal S}(t)\}_{t\geq 0}$ is asymptotically stable, that is,

\begin{eqnarray*} \|{\cal S}(t)z\|_{{\cal H}}\rightarrow 0\qquad\mbox{as}\ t\rightarrow \infty,\ \ \mbox{for any}\ z\in{\cal H}. \end{eqnarray*}

Theorem 2.7. ([4])

Let ${\cal A}$ be the generator of a uniformly bounded C₀-semigroup $\{{\cal S}(t)\}_{t\geq 0}$ on a Hilbert space $\cal H.$ If

\begin{eqnarray*} i\mathbb{R}\subseteq\varrho({\cal A})\quad\mbox{and}\quad \sup_{|\beta|\geq 1}\frac{1}{\beta^{\ell}}\|(i\,\beta I - {\cal A})^{-1}\|_{{\cal L}({\cal H})} \lt M \end{eqnarray*}

for some $\ell$, then there exist c such that

\begin{eqnarray*} \|S(t) U_{0}\|^{2} \leq \frac{c}{t^{\frac{2}{\ell}}}\|U_{0}\|_{{\cal D}({\cal A})}^{2}. \end{eqnarray*}

3. Setting of the Semigroup

In this section, we use results from the semigroup theory of linear operators in order to obtain an existence theorem of the system (2.10). Moreover, afterwards we will use the following standard $L^{2}(0,L)$ space, where the scalar product and the norm are denoted by

\begin{equation*} \langle\varphi, \psi\rangle_{L^{2}(0, L)} = \int_{0}^{L}\varphi \overline{\psi} \, {\rm d}x,\qquad \|\psi\|_{L^{2}(0, L)}^{2} = \int_{0}^{L}|\psi|^{2}\, {\rm d}x. \end{equation*}

In a similar way, let $L^{2}(\mathbb{R})$ be the Hilbert space of all measurable square integrable functions on the real line with the inner product

\begin{eqnarray*} \langle\varphi, \psi\rangle_{L^{2}(\mathbb{R})} = \int_{\mathbb{R}}\varphi \overline{\psi} \, {\rm d}\xi, \qquad \varphi, \psi\in L^{2}(\mathbb{R}). \end{eqnarray*}

We define $\displaystyle \mathbb{H}(0, L) = \{\Psi\in H^{1}(0, L):\ \Psi(0)=0\}.$ Then

(3.1)\begin{eqnarray} \mathcal{H}= \left[L^{2}(0, L)\right]^{2}\times \left[\mathbb{H}(0, L)\right]^{2} \times \left[L^{2}(\mathbb{R})\right]^{2}, \end{eqnarray}

equipped with the inner product given by

(3.2)\begin{eqnarray} & & \langle U, \widetilde{U}\rangle_{{\cal H}} = \rho\int_{0}^{L}w \overline{\widetilde{w}}\, {\rm d}x + \mu\int_{0}^{L}z \overline{\widetilde{z}}\, {\rm d}x + \chi\int_{0}^{L}u_{x} \overline{\widetilde{u}}_{x}\, {\rm d}x + \beta\int_{0}^{L}v_{x} \overline{\widetilde{v}}_{x}\, {\rm d}x \nonumber \\& & - \gamma \beta\int_{0}^{L}u_{x} \overline{\widetilde{v}}_{x}\, {\rm d}x - \gamma \beta\int_{0}^{L}v_{x} \overline{\widetilde{u}}_{x}\, {\rm d}x + \mathfrak{C} \int_{\mathbb{R}}\varphi \overline{\widetilde{\varphi}}\, {\rm d}x + \mathfrak{C} \int_{\mathbb{R}}\phi \overline{\widetilde{\phi}}\, {\rm d}x, \end{eqnarray}

where $ U=(u, v, w, z, \varphi, \phi)^{\rm T}\ $ and $\widetilde{U} =(\widetilde{u}, \widetilde{v}, \widetilde{w}, \widetilde{z}, \widetilde{\varphi}, \widetilde{\phi})^{\rm T}.$ The norm is given by

\begin{eqnarray*} \|U\|_{\cal H}^{2} = \rho \|w\|_{L^{2}(0, L)}^{2} + \mu \|z\|_{L^{2}(0, L)}^{2} + \int_{0}^{L}X^{\rm T} A X\, {\rm d}x + \mathfrak{C} \|\varphi\|_{L^{2}({\mathbb{R}})}^{2} + \mathfrak{C} \|\phi\|_{L^{2}({\mathbb{R}})}^{2}. \end{eqnarray*}

In order to advance, we will transform the initial boundary value problem (2.10) to an abstract problem in the Hilbert space $\mathcal{H}.$ To achieve this, we will introduce the functions $u_{t} = w,$ and $v_{t} = z$ and rewrite the system (2.10) as the following initial value problem:

(3.3)\begin{eqnarray} \frac{\rm d}{{\rm d}t}U(t) = {\cal A}U(t),\quad U(0) = U_{0},\quad \forall\; t \gt 0, \end{eqnarray}

where $U=(u, v, w, z, \varphi, \phi)^{\rm T}\ $ and $\ U_{0}=(u_{0}, v_{0}, w_{1}, z_{1}, \varphi_{0}, \phi_{0})^{\rm T}, $ and the operator $\,\mathcal{A}:\mathcal{D}(\mathcal{A})\subset \mathcal{H}\rightarrow \mathcal{H}$ is given by

(3.4)\begin{equation} \mathcal{A}\left( \begin{array}{c} u \\ v\\ w \\ z \\ \varphi \\ \phi \end{array} \right) =\left( \begin{array}{c} w\\ z \\ \frac{1}{\rho}\left[\chi u_{xx} - \gamma \beta v_{xx}\right] \\ \frac{1}{\mu}\left[\beta v_{xx} - \gamma \beta u_{xx}\right] \\ -(\xi^{2} + \eta) \varphi + w(L, t) \mu(\xi) \\ -(\xi^{2} + \eta) \phi + z(L, t) \mu(\xi) \end{array} \right). \end{equation}

with domain

\begin{align*} &\mathcal{D}({\cal A})\nonumber \\ &\!\!= \left\{ U \in {\cal H} \left| \begin{array}{cl} &\!\!\!\!\!\!\! u, v\in H^{2}(0, L)\cap \mathbb{H}(0, L), w, z\in \mathbb{H}(0, L) \\ &\!\!\!\!\!\!\! - (\xi^{2} + \eta) \varphi + w(L, t) \mu(\xi) \in L^{2}({\mathbb{R}})\\ &\!\!\!\!\!\!\!- (\xi^{2} + \eta) \phi + z(L, t) \mu(\xi) \in L^{2}({\mathbb{R}}) \\ &\!\!\!\!\!\!\! \chi u_{x}(L, t) - \gamma \beta v_{x}(L, t) + \mathfrak{C} \int_{-\infty}^{\infty}\mu(\xi) \varphi(\xi, t)\, {\rm d}\xi = 0, \quad |\xi| \varphi \in L^{2}(-\infty, \infty) \\ &\!\!\!\!\!\!\! \chi v_{x}(L, t) - \gamma \beta u_{x}(L, t) + \mathfrak{C} \int_{-\infty}^{\infty}\mu(\xi) \phi(\xi, t)\, {\rm d}\xi = 0, \quad |\xi| \phi \in L^{2}(-\infty, \infty) \end{array} \right. \right\}.\end{align*}

Proposition. 3.1. The operator ${\cal A}$ is the infinitesimal generator of the contraction semigroup $\{{\cal S}_{{\cal A}}(t)\}_{t\geq 0}.$

Proof. We will show that ${\cal A}$ is a dissipative operator and $0\in \varrho({\cal A})$ the resolvent set of ${\cal A}.$ Then the conclusion will follow Lumer–Phillips’ theorem. If $U = (u, v, w, z, \varphi, \phi)^{\rm T}\in {\cal D}({\cal A}),$ then from Equations (3.3) and (1.4) and using the energy ${\cal E}(t)$ definition, we get

(3.5)\begin{align} {\rm Re}\langle{\cal A}U, U\rangle_{\cal H} = -\mathfrak{C}\int_{\mathbb{R}}\left(\xi^{2} + \eta\right) |\varphi(\xi, t)|^{2}\, {\rm d}\xi - \mathfrak{C}\int_{\mathbb{R}}\left(\xi^{2} + \eta\right) |\phi(\xi, t)|^{2}\, {\rm d}\xi \leq 0. \end{align}

In fact, let $U = (u, v, w, z, \varphi, \phi) \in {\cal D}({\cal A}),$ and using Equation (2.10) $_{9,10}$, we have

\begin{align*} \langle {\cal A}U, U\rangle_{\cal H} & = \chi\int_{0}^{L}u_{xx} \overline{w}\, {\rm d}x - \gamma \beta\int_{0}^{L}v_{xx} \overline{w}\, {\rm d}x + \beta\int_{0}^{L}v_{xx} \overline{z}\, {\rm d}x - \gamma \beta\int_{0}^{L}u_{xx} \overline{z}\, {\rm d}x \\ & + \ \chi\int_{0}^{L}w_{x} \overline{u}_{x}\,{\rm d}x + \beta\int_{0}^{L}z_{x} \overline{v}_{x}\,{\rm d}x - \gamma \beta\int_{0}^{L}w_{x} \overline{v}_{x}\, {\rm d}x - \gamma \beta\int_{0}^{L}z_{x} \overline{u}_{x}\, {\rm d}x \\ & - \mathfrak{C}\int_{\mathbb{R}}\left(\xi^{2} + \eta\right)|\varphi|^{2}\,{\rm d}\xi + \mathfrak{C} w(L,\,t)\int_{\mathbb{R}}\mu(\xi) \overline{\varphi}\, {\rm d}x \\ & - \mathfrak{C}\int_{\mathbb{R}}\left(\xi^{2} + \eta\right)|\phi|^{2}\, {\rm d}\xi + \mathfrak{C} z(L, t) \int_{\mathbb{R}}\mu(\xi) \overline{\phi}\, {\rm d}x. \end{align*}

Subsequently, if we integrate by parts, we have

\begin{align*} \displaystyle \int_{0}^{L}u_{xx} \overline{w}\, {\rm d}x &= \displaystyle \overline{w}(L, t) u_{x}(L, t) - \int_{0}^{L}u_{x} \overline{w}_{x}\, {\rm d}x, \\ \displaystyle \int_{0}^{L}v_{xx} \overline{w}\, {\rm d+}x &= \overline{w}(L, t) v_{x}(L, t) - \int_{0}^{L}v_{x} \overline{w}_{x}\, {\rm d}x \\ \displaystyle \int_{0}^{L}v_{xx} \overline{z}\, {\rm dx} &= \overline{z}(L, t) v_{x}(L, t) - \int_{0}^{L}v_{x} \overline{z}_{x}\, {\rm dx}, \\ \quad\displaystyle \int_{0}^{L}u_{xx} \overline{z}\,{\rm d}x &= \overline{z}(L, t) u_{x}(L, t) - \int_{0}^{L}u_{x} \overline{z}_{x}\, {\rm d}x. \end{align*}

Replacing these integrals in $\langle {\cal A}U,\ U\rangle_{\cal H}$, we obtain

\begin{align*} \langle {\cal A}U, U\rangle_{\cal H} & = \displaystyle \chi \overline{w}(L, t) u_{x}(L, t) - \chi\int_{0}^{L}u_{x} \overline{w}_{x}\, {\rm d}x - \gamma \beta \overline{w}(L, t) v_{x}(L, t) + \gamma \beta\int_{0}^{L}v_{x} \overline{w}_{x}\, {\rm d}x \\ & + \displaystyle \beta \overline{z}(L, t) v_{x}(L, t) - \beta\int_{0}^{L}v_{x} \overline{z}_{x}\, {\rm d}x - \gamma \beta \overline{z}(L, t) u_{x}(L, t) + \gamma \beta\int_{0}^{L}u_{x} \overline{z}_{x}\, {\rm d}x \\ & + \displaystyle \chi\int_{0}^{L}w_{x} \overline{u}_{x}\, {\rm d}x + \beta\int_{0}^{L}z_{x} \overline{v}_{x}\, {\rm d}x - \gamma \beta\int_{0}^{L}w_{x} \overline{v}_{x}\, {\rm d}x - \gamma \beta\int_{0}^{L}z_{x} \overline{u}_{x}\, {\rm d}x \\ & - \displaystyle \mathfrak{C}\int_{\mathbb{R}}\left(\xi^{2} + \eta\right)|\varphi|^{2}\, {\rm d}\xi + \mathfrak{C} w(L, t)\int_{\mathbb{R}}\mu(\xi) \overline{\varphi}\, {\rm d}x \\ & - \displaystyle \mathfrak{C}\int_{\mathbb{R}}\left(\xi^{2} + \eta\right)|\phi|^{2}\, {\rm d}\xi + \mathfrak{C} z(L,\,t)\int_{\mathbb{R}}\mu(\xi)\overline{\phi}\, {\rm d}x. \end{align*}

Grouping and simplifying the before mentioned, we are left with

\begin{align*} \langle {\cal A}U, U\rangle_{\cal H} & = - \displaystyle \int_{0}^{L}(u_{x} \overline{w} - \overline{u_{x} \overline{w}})\,{\rm d}x - \beta\int_{0}^{L}(v_{x} \overline{z} - \overline{v_{x} \overline{z}})\,{\rm d}x + \gamma \beta\int_{0}^{L}(\overline{w}_{x} v_{x} - \overline{\overline{w}_{x} v_{x}})\,{\rm d}x\\ & + \displaystyle \gamma \beta\int_{0}^{L}(\overline{z}_{x} u_{x} - \overline{\overline{z}_{x} u_{x}})\,{\rm d}x + [\chi u_{x}(L, t) - \gamma \beta v_{x}(L, t)] \overline{w}(L, t) \\ & + [\beta v_{x}(L, t) - \gamma \beta u_{x}(L, t)] \overline{z}(L, t) +\displaystyle \mathfrak{C} w(L, t) \int_{\mathbb{R}}\mu(\xi) \overline{\varphi}(\xi, t)\, {\rm d}x\\ & + \displaystyle \mathfrak{C} z(L,t) \int_{\mathbb{R}}\mu(\xi)\overline{\phi}(\xi, t)\, {\rm d}x - \displaystyle \mathfrak{C }\int_{\mathbb{R}}\left(\xi^{2} + \eta\right)|\varphi|^{2}\, {\rm d}\xi - \mathfrak{C}\int_{\mathbb{R}}\left(\xi^{2} + \eta\right)|\phi|^{2}\, {\rm d}\xi. \end{align*}

Considering the boundary conditions (2.10) $_{9,\,10}$ and replacing them in the above equality, we have as a result

\begin{align*} \langle {\cal A}U, U\rangle_{\cal H} & = \displaystyle - 2 i \chi {\rm Im}\int_{0}^{L}u_{x} \overline{w}\,{\rm d}x - 2 i \beta\,{\rm Im}\int_{0}^{L}v_{x} \overline{z}\,{\rm d}x \\ & + \displaystyle 2 i \gamma \beta\,{\rm Im}\int_{0}^{L}\overline{w}_{x} v_{x}\,{\rm d}x + 2 i \gamma \beta {\rm Im}\int_{0}^{L}\overline{z}_{x} u_{x}\,{\rm d}x \\ & - \displaystyle \left[\mathfrak{C} \overline{w}(L, t) \int_{\mathbb{R}}\mu(\xi) \varphi(\xi, t)\, {\rm d}\xi - \overline{\mathfrak{C} \overline{w}(L, t) \int_{\mathbb{R}}\mu(\xi) \varphi(\xi, t)}\, {\rm d}\xi \right] \\ & - \displaystyle \left[\mathfrak{C} \overline{z}(L, t) \int_{\mathbb{R}}\mu(\xi) \phi(\xi, t)\, {\rm d}\xi - \overline{\mathfrak{C} \overline{z}(L, t) \int_{\mathbb{R}}\mu(\xi) \phi(\xi, t)}\, {\rm d}\xi \right] \\ & - \displaystyle \mathfrak{C}\int_{\mathbb{R}}\left(\xi^{2} + \eta\right)|\varphi|^{2}\, {\rm d}\xi - \mathfrak{C}\int_{\mathbb{R}}\left(\xi^{2} + \eta\right)|\phi|^{2}\, {\rm d}\xi. \end{align*}

Consequently,

\begin{align*} \langle {\cal A}U, U\rangle_{\cal H} & = \displaystyle - 2 i \chi {\rm Im}\int_{0}^{L}u_{x} \overline{w}\,{\rm d}x - 2 i \beta Im\int_{0}^{L}v_{x} \overline{z}\,{\rm d}x + 2 i \gamma \beta {\rm Im}\int_{0}^{L}\overline{w}_{x} v_{x}\,{\rm d}x \\ &+ 2 i \gamma \beta Im\int_{0}^{L}\overline{z}_{x} u_{x}\,{\rm d}x - 2 i {\rm Im}\left[\mathfrak{C} \overline{w}(L, t) \int_{\mathbb{R}}\mu(\xi) \varphi(\xi, t)\, {\rm d}\xi\right] \\ & - 2 i {\rm Im}\left[\mathfrak{C} \overline{z}(L, t) \int_{\mathbb{R}}\mu(\xi) \phi(\xi, t)\, {\rm d}\xi\right] -\ \mathfrak{C} \int_{\mathbb{R}}\left(\xi^{2} + \eta\right)|\varphi|^{2}\, {\rm d}\xi \\ &\quad- \mathfrak{C}\int_{\mathbb{R}}\left(\xi^{2} + \eta\right)|\phi|^{2}\, {\rm d}\xi. \end{align*}

Taking the real part in the equality mentioned above, it follows properly that ${\cal A}$ is a dissipative operator and the statement (3.5) holds true.

Remark 3.2. From Equation (2.10) $_{9,\,10}$, we have

\begin{eqnarray*} \chi u_{x}(L, t) - \gamma \beta v_{x}(L, t)=- \mathfrak{C} \int_{\mathbb{R}}\mu(\xi) \varphi(\xi, t)\, {\rm d}\xi \equiv \varphi_1\\ - \gamma \beta u_{x}(L, t) + \beta v_{x}(L, t) = - \mathfrak{C} \int_{\mathbb{R}}\mu(\xi) \phi(\xi, t)\, {\rm d}\xi \equiv \varphi_2, \end{eqnarray*}

and equivalently $ A X_x = \varphi,$ where $\varphi= ( \varphi_1, \varphi_2)^{\rm T}$ and

(3.6)\begin{eqnarray} A = \left( \begin{array}{cc} \chi & - \gamma \beta \\ - \gamma \beta & \beta \\ \end{array} \right), \qquad X = \left( \begin{array}{c} u\\ v\\ \end{array} \right) \end{eqnarray}

If $\ \mathfrak{J} = \det(A) = \chi \beta - \gamma^{2} \beta^{2} \neq 0\ $, then

(3.7)\begin{eqnarray} A^{-1} = \left( \begin{array}{cc} \chi & -\gamma \beta \\ - \gamma \beta & \beta \\ \end{array} \right)^{-1} = \frac{1}{\mathfrak{J}}\left( \begin{array}{cc} \beta & \gamma \beta \\ \gamma \beta & \chi \\ \end{array} \right). \end{eqnarray}

and X _x is given by

(3.8)\begin{eqnarray} \left( \begin{array}{c} u_{x} \\ v_{x}\\ \end{array} \right) = \frac{1}{\mathfrak{J}}\left( \begin{array}{cc} \beta & \gamma \beta \\ \gamma \beta & \chi \\ \end{array} \right)\left( \begin{array}{c} \varphi_1 \\ \varphi_2\\ \end{array} \right) = \frac{1}{\mathfrak{J}}\left( \begin{array}{c} \beta \varphi_1 + \gamma \beta \varphi_2 \\ \gamma \beta \varphi_1+ \chi \varphi_2\\ \end{array} \right). \nonumber \end{eqnarray}

We will prove that the operator $\lambda I - {\cal A}$ is surjective for λ > 0. For this, let $F =(f_1, f_{2}, f_{3}, f_{4}, f_{5}, f_{6})^{T}\in {\cal H},$ we seek $ U=(u, v, w, z, \varphi, \phi)^{\rm T}\in {\cal D}({\cal A}) $ such that $\lambda I - {\cal A}U=F,$ that is,

(3.8)\begin{eqnarray} \left\lbrace \begin{array}{l} \lambda u - w = f_{1} \qquad \mbox{in}\ \mathbb{H}(0,\,L) \\ \lambda v - z = f_{2} \qquad \mbox{in}\ \mathbb{H}(0,\,L) \\ \rho \lambda w - \chi u_{xx} + \gamma \beta v_{xx} = \rho f_{3} \qquad \mbox{in}\ L^{2}(0,\,L) \\ \mu \lambda z - \beta v_{xx} + \gamma \beta u_{xx} = \mu f_{4} \qquad \mbox{in}\ L^{2}(0, L) \\ \lambda \varphi + (\xi^{2} + \eta) \varphi - w(L, t) \mu(\xi) = f_{5}\qquad \mbox{in}\ L^{2}({\mathbb{R}}) \\ \lambda \phi + (\xi^{2} + \eta) \phi - z(L, t) \mu(\xi) = f_{6}\qquad \mbox{in}\ L^{2}({\mathbb{R}}). \end{array} \right. \end{eqnarray}

From Equation (3.8) $_{5,\,6}$, we make clear that

(3.9)\begin{eqnarray} \left\lbrace \begin{array}{l} \varphi(\xi, t) = \displaystyle \frac{f_{5}(\xi) + w(L, t) \mu(\xi)}{\xi^{2} + \eta + \lambda} \\ \phi(\xi, t) = \displaystyle \frac{f_{6}(\xi) + z(L, t) \mu(\xi)}{\xi^{2} + \eta + \lambda} \end{array} \right. \end{eqnarray}

Note that, from Equation (3.8) $_{1,\,2}$, we have that $w, z \in \mathbb{H}(0, L)$, and by replacing w and z into Equation (3.8) $_{3,\,4}$, respectively, it yields that

(3.10)\begin{eqnarray} \left\lbrace \begin{array}{l} \rho \lambda^{2} u - \chi u_{xx} + \gamma \beta v_{xx} = \rho f_{3} + \rho \lambda f_{1} \\ \mu \lambda^{2} v - \beta v_{xx} + \gamma \beta u_{xx} = \mu f_{4} + \mu \lambda f_{2} \end{array} \right. \end{eqnarray}

To solve Equation (3.10), we need to find $u, v\in H^{2}(0, L)\cap \mathbb{H}(0,L)$ such that

(3.11)\begin{eqnarray} \left\lbrace \begin{array}{l} \displaystyle \int_{0}^{L}\left[\rho \lambda^{2} u - \chi u_{xx} + \gamma \beta v_{xx}\right]\zeta\, {\rm d}x = \int_{0}^{L}\left[\rho f_{3} + \rho \lambda f_{1}\right]\zeta\, {\rm d}x \\ \displaystyle \int_{0}^{L}\left[\mu \lambda^{2} v - \beta v_{xx} + \gamma \beta u_{xx}\right]\aleph\, {\rm d}x = \int_{0}^{L}\left[\mu f_{4} + \mu \lambda f_{2}\right]\aleph\, {\rm d}x \end{array} \right. \end{eqnarray}

for all $\zeta, \aleph \in \mathbb{H}(0,L).$ Integrating by parts Equation (3.11)₁, we have

\begin{eqnarray*} & & \rho \lambda^{2}\int_{0}^{L}u \zeta\, {\rm d}x + \chi\int_{0}^{L}u_{x} \zeta_{x}\, {\rm d}x - \gamma \beta\int_{0}^{L}v_{x} \zeta_{x}\, {\rm d}x \\ & & - [ \chi u_{x}(L, t) - \gamma \beta v_{x}(L, t)] \zeta(L, t) = \int_{0}^{L}\left[\rho f_{3} + \rho \lambda f_{1}\right]\zeta\, {\rm d}x. \end{eqnarray*}

Now, using the same idea with Equation (3.11)₂, we obtain

\begin{eqnarray*} & & \mu \lambda^{2}\int_{0}^{L}v \aleph\, {\rm d}x + \beta\int_{0}^{L}v_{x}\,\aleph_{x}\, {\rm d}x - \gamma \beta\int_{0}^{L}u_{x} \aleph_{x}\, {\rm d}x -\ [ \beta v_{x}(L, t) - \gamma \beta u_{x}(L, t)] \aleph(L, t) \\ & & = \int_{0}^{L}\left[\mu f_{4} + \mu \lambda f_{2}\right]\aleph\, {\rm d}x. \end{eqnarray*}

Moreover, using Equation (2.10) $_{9,\,10}$, we have

\begin{eqnarray*} & & \rho \lambda^{2}\int_{0}^{L}u \zeta\, {\rm d}x + \chi\int_{0}^{L}u_{x} \zeta_{x}\, {\rm d}x - \gamma \beta\int_{0}^{L}v_{x} \zeta_{x}\, {\rm d}x +\left[\mathfrak{C} \int_{\mathbb{R}}\mu(\xi) \varphi(\xi, t)\, {\rm d}\xi\right]\zeta(L, t) \\ & & = \int_{0}^{L}\left[\rho f_{3} + \rho \lambda f_{1}\right]\zeta\, {\rm d}x \end{eqnarray*}

and

\begin{eqnarray*} & & \mu \lambda^{2}\int_{0}^{L}v \aleph\, {\rm d}x + \beta\int_{0}^{L}v_{x} \aleph_{x}\, {\rm d}x - \gamma \beta\int_{0}^{L}u_{x} \aleph_{x}\, {\rm d}x + \left[\mathfrak{C} \int_{\mathbb{R}}\mu(\xi) \phi(\xi, t)\, {\rm d}\xi\right]\aleph(L, t) \\ & & = \int_{0}^{L}\left[\mu f_{4} + \mu \lambda f_{2}\right]\aleph\, {\rm d}x. \end{eqnarray*}

From Equation (3.9) $_{1, 2}$, we have, respectively,

(3.12)\begin{eqnarray} & & \rho \lambda^{2}\int_{0}^{L}u \zeta\, {\rm d}x + \chi\int_{0}^{L}u_{x} \zeta_{x}\, {\rm d}x - \gamma \beta\int_{0}^{L}v_{x} \zeta_{x}\, {\rm d}x +\left[\mathfrak{C} \int_{\mathbb{R}}\frac{\mu(\xi) f_{5}(\xi)\, {\rm d}\xi}{\xi^{2} + \eta + \lambda} \right]\zeta(L, t) \nonumber \\ & & + \left[\mathfrak{C} \int_{\mathbb{R}}\frac{\mu^{2}(\xi)\, {\rm d}\xi}{\xi^{2} + \eta + \lambda} \right]w(L, t) \zeta(L, t) = \int_{0}^{L}\left[\rho f_{3} + \rho \lambda f_{1}\right]\zeta\, {\rm d}x \end{eqnarray}

and

(3.13)\begin{eqnarray} & & \mu \lambda^{2}\int_{0}^{L}v \aleph\, {\rm d}x + \beta\int_{0}^{L}v_{x} \aleph_{x}\, {\rm d}x - \gamma \beta\int_{0}^{L}u_{x} \aleph_{x}\, {\rm d}x + \left[\mathfrak{C} \int_{\mathbb{R}}\frac{\mu(\xi) f_{6}(\xi)\, {\rm d}\xi}{\xi^{2} + \eta + \lambda} \right]\aleph(L,t) \nonumber \\ & & + \left[\mathfrak{C} \int_{\mathbb{R}}\frac{\mu^{2}(\xi)\, {\rm d}\xi}{\xi^{2} + \eta + \lambda}\right]z(L, t) \aleph(L, t) = \int_{0}^{L}\left[\mu f_{4} - \mu \lambda f_{2}\right]\aleph\, {\rm d}x. \end{eqnarray}

Since $w(L, t) = \lambda u(L, t) - f_{1}(L) $ and $ z(L, t) = \lambda v(L, t) - f_{2}(L), $ replacing these functions into Equations (3.12) and (3.13), respectively, we obtain

(3.14)\begin{eqnarray} & & \rho \lambda^{2}\int_{0}^{L}u \zeta\, {\rm d}x + \chi\int_{0}^{L}u_{x} \zeta_{x}\, {\rm d}x - \gamma \beta\int_{0}^{L}v_{x} \zeta_{x}\, {\rm d}x \nonumber \\ & + & \left[\mathfrak{C} \int_{\mathbb{R}}\frac{\mu(\xi)\,f_{5}(\xi)\, {\rm d}\xi}{\xi^{2} + \eta + \lambda} \right]\zeta(L, t) + \left[\mathfrak{C} \int_{\mathbb{R}}\frac{\mu^{2}(\xi)\, {\rm d}\xi}{\xi^{2} + \eta+ \lambda} \right]\lambda u(L, t) \zeta(L, t) \nonumber \\& = & \int_{0}^{L}\left[\rho f_{3} + \rho \lambda f_{1}\right]\zeta\, {\rm d}x + \left[\mathfrak{C} \int_{\mathbb{R}}\frac{\mu^{2}(\xi)\, {\rm d}\xi}{\xi^{2} + \eta + \lambda} \right]f_{1}(L) \zeta(L, t) \end{eqnarray}

and

(3.15)\begin{eqnarray} & & \mu \lambda^{2}\int_{0}^{L}v \aleph\, {\rm d}x + \beta\int_{0}^{L}v_{x} \aleph_{x}\, {\rm d}x - \gamma \beta\int_{0}^{L}u_{x} \aleph_{x}\, {\rm d}x \nonumber \\ & & +\left[\mathfrak{C} \int_{\mathbb{R}}\frac{\mu(\xi) f_{6}(\xi)\, {\rm d}\xi}{\xi^{2} + \eta + \lambda} \right]\aleph(L, t) + \left[\mathfrak{C} \int_{\mathbb{R}}\frac{\mu^{2}(\xi)\, {\rm d}\xi}{\xi^{2} + \eta + \lambda}\right]\lambda v(L,\,t) \aleph(L, t) \nonumber \\ & & = \int_{0}^{L}\left[\mu f_{4} - \mu \lambda f_{2}\right]\aleph\, {\rm d}x + \left[\mathfrak{C} \int_{\mathbb{R}}\frac{\mu^{2}(\xi)\, {\rm d}\xi}{\xi^{2} + \eta + \lambda}\right]f_{2}(L) \aleph(L, t). \end{eqnarray}

The system in Equations (3.14)–(3.15) represents the equivalent to the problem

(3.16)\begin{eqnarray} \mathfrak{a}\left(\left(u, v), (\zeta, \aleph\right)\right) = {\cal L}(\zeta, \chi), \end{eqnarray}

where the bilinear form continuous and coercive $\mathfrak{a} \left[\mathbb{H}^{1}(0, L)\times \mathbb{H}^{1}(0,L)\right]^{2} \rightarrow\mathbb{R}$ and the continuous linear form ${\cal L}: \mathbb{H}^{1}(0,L)\times \mathbb{H}^{1}(0,L) \rightarrow\mathbb{R}$ are defined by

(3.17)\begin{eqnarray} \mathfrak{a}\left(\left(u, v), (\zeta, \chi\right)\right) & = & \rho \lambda^{2}\int_{0}^{L}u \zeta\, {\rm d}x + \chi\int_{0}^{L}u_{x} \zeta_{x}\, {\rm d}x - \gamma \beta\int_{0}^{L}v_{x} \zeta_{x}\, {\rm d}x \nonumber \\ & + & \mu \lambda^{2}\int_{0}^{L}v \aleph\, {\rm d}x + \beta\int_{0}^{L}v_{x} \aleph_{x}\, {\rm d}x - \gamma\,\beta\int_{0}^{L}u_{x} \aleph_{x}\, {\rm d}x \nonumber \\ & + & \left[\mathfrak{C} \int_{\mathbb{R}}\frac{\mu(\xi) f_{5}(\xi)\, {\rm d}\xi}{\xi^{2} + \eta + \lambda} \right]\zeta(L, t) + \left[\mathfrak{C} \int_{\mathbb{R}}\frac{\mu^{2}(\xi)\, {\rm d}\xi}{\xi^{2} + \eta + \lambda} \right]\lambda u(L, t) \zeta(L, t) \nonumber \\& + & \left[\mathfrak{C} \int_{\mathbb{R}}\frac{\mu(\xi) f_{6}(\xi)\, {\rm d}\xi}{\xi^{2} + \eta + \lambda} \right]\aleph(L, t) + \left[\mathfrak{C} \int_{\mathbb{R}}\frac{\mu^{2}(\xi)\, {\rm d}\xi}{\xi^{2} + \eta + \lambda}\right]\lambda v(L, t) \aleph(L, t) \nonumber \end{eqnarray}

and

\begin{eqnarray} {\cal L}(\zeta, \chi) & = & \int_{0}^{L}\left[\rho f_{3} + \rho \lambda f_{1}\right]\zeta\, {\rm d}x + \left[\mathfrak{C} \int_{\mathbb{R}}\frac{\mu^{2}(\xi)\, {\rm d}\xi}{\xi^{2} + \eta + \lambda} \right]f_{1}(L) \zeta(L, t) \nonumber \\ & + & \int_{0}^{L}\left[\mu f_{4} - \mu \lambda f_{2}\right]\aleph\, {\rm d}x + \left[\mathfrak{C} \int_{\mathbb{R}}\frac{\mu^{2}(\xi)\, {\rm d}\xi}{\xi^{2} + \eta + \lambda}\right]f_{2}(L) \aleph(L, t). \nonumber \end{eqnarray}

By applying Lax–Milgram theorem, it follows that for all $(\zeta, \chi)\in \mathbb{H}^{1}(0, L)\times \mathbb{H}^{1}(0, L)$, the problem (3.16) admits a unique solution $(u, v)\in \mathbb{H}^{1}(0, L)\times \mathbb{H}^{1}(0, L).$ Using elliptic regularity, stated in Equations (3.14)–(3.15), it follows that $(u, v)\in \mathbb{H}^{1}(0, L)\times \mathbb{H}^{1}(0, L).$ Therefore, the operator $\lambda\,I - {\cal A}$ is surjective for any λ > 0.

Now, applying the Hille–Yosida theorem, we have the following result

Theorem 3.3. (a) If $U_{0}\in {\cal H},$ then the system (3.3) has a unique mild solution

(3.17)\begin{eqnarray} U\in C\left(\mathbb{R}_{0}^{+},\ {\cal H}\right). \end{eqnarray}

(b) If $U_{0}\in {\cal D}({\cal A}),$ then the system (3.3) has a unique strong solution

(3.18)\begin{eqnarray} U\in C\left(\mathbb{R}_{0}^{+}, {\cal D}({\cal A})\right)\cap C^{1}\left(\mathbb{R}_{0}^{+}, {\cal H}\right). \end{eqnarray}

4. Asymptotic behavior

In this section, we study the asymptotic behavior for the solution of our problem. First, we prove that the semigroup $\{{\cal S}_{{\cal A}}(t)\}_{t\geq 0}$ is asymptotically stable. To show this, we apply the Theorem 2.6 of Arend–Batty [2]. Next, using Theorem 2.7 of Borichev–Tomilov, we show that $\{{\cal S}_{\cal A}(t)\}_{t\geq 0}$ is polynomially stable.

To achieve our first result, we need the following lemmas:

Lemma 4.1. λ = 0 is not an eigenvalue of ${\cal A}.$

Proof. By applying Equation (3.4), we have as a result $U = (u, v, w, z, \varphi, \phi)^{\rm T}\in \ker({\cal A})$ if and only if ${\cal A}U = 0,$ that is,

(4.1)\begin{eqnarray} \left\lbrace \begin{array}{l} w = 0 \\ z = 0 \\ \chi u_{xx} - \gamma \beta v_{xx} = 0 \\ \beta v_{xx} - \gamma \beta u_{xx} = 0 \\ (\xi^{2} + \eta) \varphi - w(L, t) \mu(\xi) = 0 \\ (\xi^{2} + \eta) \phi - z(L, t) \mu(\xi) = 0. \end{array} \right. \end{eqnarray}

From Equation (4.1) $_{1, 2, 5, 6}$, we have $w = z = \varphi = \phi = 0.$ Multiplying Equation (4.1) $_{3, 4}$ by u and v, respectively, and integrating over $x\in (0,\,L)$ and using the definition of $\mathbb{H}(0, L)$, it follows that

(4.2)\begin{eqnarray} \begin{cases} -\displaystyle \chi\int_{0}^{L}u_{x}^{2}\, {\rm d}x + \gamma \beta\int_{0}^{L}u_{x} v_{x}\, {\rm d}x + [\chi u_{x}(L, t) - \gamma \beta v_{x}(L, t)] u(L, t) = 0 \\ - \displaystyle \beta\int_{0}^{L}v_{x}^{2}\, {\rm d}x + \gamma \beta\int_{0}^{L}u_{x} v_{x}\, {\rm d}x + [\beta v_{x}(L, t) - \gamma \beta u_{x}(L, t)] v(L, t) = 0. \end{cases} \end{eqnarray}

Using the boundary conditions (2.10) $_{9, 10}$ and that $\varphi = \phi = 0$, we obtain

(4.3)\begin{eqnarray} \begin{cases} \displaystyle \chi\int_{0}^{L}u_{x}^{2}\, {\rm d}x - \gamma \beta\int_{0}^{L}u_{x} v_{x}\, {\rm d}x = 0 \\ \displaystyle \beta\int_{0}^{L}v_{x}^{2}\, {\rm d}x - \gamma \beta\int_{0}^{L}u_{x} v_{x}\, {\rm d}x = 0. \end{cases} \end{eqnarray}

Adding Equation (4.3)₁ and Equation (4.3)₂, we obtain

\begin{eqnarray*} \chi\int_{0}^{L}u_{x}^{2}\, {\rm d}x + \beta\int_{0}^{L}v_{x}^{2}\, {\rm d}x - 2 \gamma \beta\int_{0}^{L}u_{x} v_{x}\, {\rm d}x = 0. \end{eqnarray*}

We write this equality as

(4.4)\begin{eqnarray} \int_{0}^{L}X^{\rm T} A X\, {\rm d}x = 0 \end{eqnarray}

where $X^{\rm T} A X$ is the positive definite quadratic form given by Equation (1.5).

From Equation (4.4), we have found that $u_{x} = v_{x} = 0,$ which involves $u(x, t) = C,$ where C is a constant, but $u(0, t) = 0,$ then $u \equiv 0.$ Similarly, we obtain $v \equiv 0.$ Then we conclude that $(u, v, w, z, \varphi, \phi)^{\rm T}=0$ and the Lemma follows.

Lemma 4.2. It verifies that

(4.5)\begin{eqnarray}\sigma({\cal A}) \cap i \mathbb{R} = \sigma({\cal A}) \cap \{i \lambda, \lambda\in\mathbb{R}, \lambda\neq 0\} = \emptyset. \end{eqnarray}

Proof. We will prove the equality by contradiction. In order to do that, we suppose there exists $\lambda\in\mathbb{R},$ λ ≠ 0, and $U\neq 0,$ such that $\left(i \lambda - {\cal A}\right)U = 0.$ Then we have

(4.6)\begin{eqnarray} \left\lbrace \begin{array}{l} i \lambda u - w = 0, \qquad w(L, t) = i \lambda u(L, t) \\ i \lambda v - z = 0, \qquad z(L, t) = i \lambda v(L, t) \\ i \rho \lambda w - \chi u_{xx} + \gamma \beta v_{xx} = 0 \\ i \mu \lambda z - \beta v_{xx} + \gamma \beta u_{xx} = 0 \\ i \lambda \varphi + (\xi^{2} + \eta) \varphi - w(L, t) \mu(\xi) = 0 \\ i \lambda \phi + (\xi^{2} + \eta) \phi - z(L, t) \mu(\xi) = 0. \end{array} \right. \end{eqnarray}

According to Equation (3.5), we imply $\varphi(\xi,\,t) = 0$ and $\phi(\xi,\,t) = 0,$ and by replacing these in Equation (4.6) $_{5, 6}$, we obtain $ w(L, t) = z(L, t) = 0;$ hence, from Equation (4.6) $_{1, 2}$ it is implied that $u(L, t) = v(L, t) = 0.$ Moreover, by revisiting Equation (2.10) $_{9, 10} $ it follows that $u_{x}(L, t) = v_{x}(L, t) = 0.$

Now, replacing Equation (4.6) $_{1, 2}$ onto (4.6) $_{3, 4}$, we obtain

(4.7)\begin{eqnarray} \left\lbrace \begin{array}{l} \chi u_{xx} - \gamma \beta v_{xx} = - \rho \lambda^{2} u \\ - \gamma \beta u_{xx} + \beta v_{xx} = - \mu \lambda^{2} v. \end{array} \right. \end{eqnarray}

To clarify the conclusion, we rewrite Equation (4.7) as the initial value problem:

(4.8)\begin{eqnarray} \begin{cases} A X_{xx} = B X \\ X(1, t) = 0,\\ X_{x}(1, t) = 0, \end{cases} \quad \iff \quad \begin{cases} X_{xx} = A^{-1}B X \\ X(1, t) = 0,\\ X_{x}(1, t) = 0, \end{cases} \end{eqnarray}

where $A, X$ are given by Equation (3.6) and A ⁻¹ by Equation (3.7) for $\mathfrak{J} = \chi \beta - \gamma^{2} \beta^{2} \not = 0,$ and

(4.9)\begin{equation} B = \left( \begin{array}{cc} - \rho \lambda^{2} & 0 \\ 0 & - \mu \lambda^{2} \\ \end{array} \right).\\ \end{equation}

Using the Picard theorem for ordinary differential equations, the system in Equation (4.8) has a unique solution $(u, v) = 0$; of these and Equation (4.6) $_{1, 2}$, we have w = 0 and z = 0. Therefore, $(u, v, w, z, \varphi, \phi) = 0$, and we conclude that ${\cal A}$ does not have eigenvalues on $\mathbb{R}.$

In the Theorem 2.6, the condition (ii) holds if we show that $\sigma({\cal A})\cap \{i \mathbb{R}\}$ is at most a countable set.

In fact, we will show that the operator $i \lambda I - {\cal A}$ is surjective for λ ≠ 0. For this purpose, and given $F =(f_{1}, f_{2}, f_{3}, f_{4}, f_{5}, f_{6})^{\rm T}\in {\cal H},$ and $U=(u, v, w, z, \varphi, \phi)^{\rm T}\in {\cal D}({\cal A})$ such that $i \lambda I - {\cal A}U=F,$ that is,

(4.10)\begin{eqnarray} \left\lbrace \begin{array}{l} i \lambda u - w = f_{1} \\ i \lambda v - z = f_{2} \\ i \rho \lambda w - \chi u_{xx} + \gamma \beta v_{xx} = \rho f_{3} \\ i \mu \lambda z - \beta v_{xx} + \gamma \beta u_{xx} = \mu f_{4} \\ i \lambda \varphi + (\xi^{2} + \eta) \varphi - w(L, t) \mu(\xi) = f_{5} \\ i \lambda \phi + (\xi^{2} + \eta) \phi - z(L, t) \mu(\xi) = f_{6}. \end{array} \right. \end{eqnarray}

with the initial condition

(4.11)\begin{eqnarray} \begin{cases} \displaystyle \chi u_{x}(L, t) - \gamma \beta v_{x}(L, t) = - \mathfrak{C} \int_{\mathbb{R}}\mu(\xi) \varphi(\xi, t)\, {\rm d}\xi \\ \displaystyle \beta v_{x}(L, t) - \gamma \beta u_{x}(\mu,t) = - \mathfrak{C} \int_{\mathbb{R}}\mu(\xi) \phi(\xi, t)\, {\rm d}\xi \end{cases} \end{eqnarray}

From Equation (4.10) $_{5, 6}$, we have

(4.12)\begin{eqnarray} \begin{cases} \varphi(\xi, t) = \displaystyle \frac{f_{5}(\xi) + w(L, t) \mu(\xi)}{\xi^{2}+ \eta + i \lambda} \\ \phi(\xi, t) = \displaystyle \frac{f_{6}(\xi) + z(L , t) \mu(\xi)}{\xi^{2} + \eta + i \lambda} \end{cases} \end{eqnarray}

Note that from Equation (4.10) $_{1, 2}$, it follws that $w, z \in \mathbb{H}(0, L).$ Now, replacing Equation (4.10) $_{1, 2}$ into Equation (4.10) $_{3, 4}$ we get, respectively,

(4.13)\begin{eqnarray} \begin{cases} - \rho \lambda^{2} u - \chi u_{xx} + \gamma \beta v_{xx} = \rho f_{3} + i \rho \lambda f_{1} \\ - \mu \lambda^{2} v - \beta v_{xx} + \gamma \beta u_{xx} = \mu f_{4} + i \mu \lambda f_{2}. \end{cases} \end{eqnarray}

Solving system (4.13) is equivalent to finding $u, v\in H^{2}(0, L)\cap \mathbb{H}(0, L)$ such that

(4.14)\begin{eqnarray} \begin{cases} \displaystyle \int_{0}^{L}\left[- \rho \lambda^{2} u - \chi u_{xx} + \gamma \beta v_{xx}\right]\zeta\, {\rm d}x = \int_{0}^{L}\left[\rho f_{3} + i \rho \lambda f_{1}\right]\zeta\, {\rm d}x \\ \displaystyle \int_{0}^{L}\left[- \mu \lambda^{2} v - \beta v_{xx} + \gamma \beta u_{xx}\right]\aleph\, {\rm d}x = \int_{0}^{L}\left[\mu f_{4} + i \mu \lambda f_{2}\right]\aleph\, {\rm d}x, \end{cases} \end{eqnarray}

for all $V = (\zeta, \aleph) \in \mathbb{H}(0, L)\times \mathbb{H}(0, L).$ Integrating Equation (4.14)₁ by parts

\begin{eqnarray*} & & \int_{0}^{L}- \rho \lambda^{2} u \zeta\, {\rm d}x + \chi\int_{0}^{L}u_{x} \zeta_{x}\, {\rm d}x - \gamma \beta\int_{0}^{L}v_{x} \zeta_{x}\, {\rm d}x \nonumber \\ & & [-\chi u_{x}(L, t) + \gamma \beta v_{x}(L, t)] \zeta(L, t) = \int_{0}^{L}\left[\rho f_{3} + i \rho \lambda f_{1}\right]\zeta\, {\rm d}x. \end{eqnarray*}

If we replace Equation (2.10)₉ in the above equality, we will get

\begin{eqnarray*} & & \int_{0}^{L}- \rho \lambda^{2} u \zeta\, {\rm d}x + \chi\int_{0}^{L}u_{x} \zeta_{x}\, {\rm d}x - \gamma \beta\int_{0}^{L}v_{x} \zeta_{x}\, {\rm d}x \nonumber \\ & & +\ \left[\mathfrak{C} \int_{\mathbb{R}}\mu(\xi) \varphi(\xi, t)\, {\rm d}\xi\right]\zeta(L, t) = \int_{0}^{L}\left[\rho f_{3} + i \rho \lambda f_{1}\right]\zeta\, {\rm d}x. \end{eqnarray*}

Now, using Equation (4.12)₁, it follows that

\begin{align} &\int_{0}^{L}- \rho \lambda^{2} u \zeta\, {\rm d}x + \chi\int_{0}^{L}u_{x} \zeta_{x}\, {\rm d}x - \gamma \beta\int_{0}^{L}v_{x} \zeta_{x}\, {\rm d}x \nonumber +\ \left[\mathfrak{C} \int_{\mathbb{R}}\frac{\mu^{2}(\xi)}{\xi^{2} + \eta + i \lambda}\, {\rm d}\xi\right]w(L, t) \zeta(L, t) \end{align}

(4.15)\begin{align}& =\int_{0}^{L}\left[\rho f_{3} + i \rho \lambda f_{1}\right]\zeta\, {\rm d}x - \left[\mathfrak{C} \int_{\mathbb{R}}\frac{\mu(\xi) f_{5}(\xi)}{\xi^{2} + \eta + i \lambda}\, {\rm d}\xi\right]\zeta(L, t). \end{align}

By replacing Equation (4.10)₁ onto Equation (4.15), we have

(4.16)\begin{align} &\int_{0}^{L}- \rho \lambda^{2} u \zeta\, {\rm d}x + \chi\int_{0}^{L}u_{x} \zeta_{x}\, {\rm d}x - \gamma \beta\int_{0}^{L}v_{x} \zeta_{x}\, {\rm d}x \nonumber \\&\quad +\ i\left[\mathfrak{C} \int_{\mathbb{R}}\frac{\mu^{2}(\xi)}{\xi^{2} + \eta + i \lambda}\, {\rm d}\xi\right]\lambda u(L, t) \zeta(L, t) \nonumber \\ &= \int_{0}^{L}\left[\rho f_{3} + i \rho \lambda f_{1}\right]\zeta\, {\rm d}x - \left[\mathfrak{C} \int_{\mathbb{R}}\frac{\mu(\xi) f_{5}(\xi)}{\xi^{2} + \eta + i\,\lambda}\, {\rm d}\xi\right]\zeta(L, t) \nonumber \\&\quad + \left[\mathfrak{C} \int_{\mathbb{R}}\frac{\mu^{2}(\xi)}{\xi^{2} + \eta + i \lambda}\, {\rm d}\xi\right]f_{1}(L) \zeta(L, t).\nonumber \\ \end{align}

Similarly to what we find in Equation (4.16), we obtain, starting from Equation (4.14)₂ that

(4.17)\begin{align} &\int_{0}^{L}- \mu \lambda^{2} v \aleph\, {\rm d}x + \beta\int_{0}^{L}v_{x} \aleph_{x}\, {\rm d}x - \gamma\,\beta\int_{0}^{L}u_{x} \aleph_{x}\, {\rm d}x \nonumber \\&\quad+\ i\left[\mathfrak{C} \int_{\mathbb{R}}\frac{\mu^{2}(\xi)}{\xi^{2} + \eta + i\,\lambda}\, {\rm d}\xi\right]\lambda v(L, t) \aleph(L, t) \nonumber \\ &= \int_{0}^{L}\left[\mu f_{4} + i \mu \lambda f_{2}\right]\aleph\, {\rm d}x - \left[\mathfrak{C} \int_{\mathbb{R}}\frac{\mu(\xi) f_{6}(\xi)}{\xi^{2} + \eta + i \lambda}\, {\rm d}\xi\right]\aleph(L, t) \nonumber \\&\quad + \left[\mathfrak{C} \int_{\mathbb{R}}\frac{\mu^{2}(\xi)}{\xi^{2} + \eta + i \lambda}\, {\rm d}\xi\right]f_{2}(L)\aleph(L,\,t). \nonumber \\ \end{align}

The system (4.16)–(4.17) is equivalent to the problem

(4.18)\begin{eqnarray}- \langle{\cal L}_{\lambda}U, V\rangle_{\left[\mathbb{H}(0, L)\right]^{2}} + \langle U, V\rangle_{\left[\mathbb{H}(0, L)\right]^{2}} = \Phi(V), \end{eqnarray}

with

\begin{align*} & \langle{\cal L}_{\lambda}U, V\rangle_{\left[\mathbb{H}(0, L)\right]^{2}} = - \lambda^{2}\int_{0}^{L}\left[\rho u \zeta + \mu v \aleph\right]\,{\rm d}x \\ & - i\left[\mathfrak{C} \int_{\mathbb{R}}\frac{\mu^{2}(\xi)\, {\rm d}\xi}{\xi^{2} + \eta + i \lambda}\right]\lambda u(L, t) \zeta(L, t) - {\rm i}\left[\mathfrak{C} \int_{\mathbb{R}}\frac{\mu^{2}(\xi)\, {\rm d}\xi}{\xi^{2} + \eta + i \lambda}\right]\lambda v(L, t) \aleph(L,\,t) \end{align*}

and in addition to

\begin{eqnarray*} \langle U, V\rangle_{\left[\mathbb{H}(0, L)\right]^{2}} = \chi\int_{0}^{L}u_{x}\,\zeta_{x}\, {\rm d}x + \beta\int_{0}^{L}v_{x} \aleph_{x}\, {\rm d}x - \gamma \beta\int_{0}^{L}\left[v_{x} \zeta_{x} + u_{x}\,\aleph_{x}\right]\,{\rm d}x. \end{eqnarray*}

Using that $L^{2}(0, L) \stackrel{c}{\hookrightarrow} H^{-1}(0, L)$ and $\mathbb{H}(0, L)\stackrel{c}{\hookrightarrow} L^{2}(0, L)$, it follows that the operator ${\cal L}_{\lambda}$ is compact from $L^{2}(0, L)$ into $L^{2}(0, L).$ By using Fredholm alternative, to prove the existence of solution, Equation (4.18) is reduce to show that 1 is not a eigenvalue of ${\cal L}_{\lambda}.$ In fact, by contradiction, if 1 is an eigenvalue, then there exists $U\neq 0,$ such that

(4.19)\begin{eqnarray}\langle{\cal L}_{\lambda}U, V\rangle_{\mathbb{H}(0, L)} = \langle U, V\rangle_{\mathbb{H}(0, L)},\quad \forall\; V\in \mathbb{H}(0, L). \end{eqnarray}

In particular, for U = V, we obtain:

(4.20)\begin{align} & - \lambda^{2}\int_{0}^{L}\left[\rho |u|^{2} + \mu |v|^{2}\right]\,{\rm d}x - i\left[\mathfrak{C} \int_{\mathbb{R}}\frac{\mu^{2}(\xi)\, {\rm d}\xi}{\xi^{2} + \eta + i \lambda}\right]\lambda\,|u(L, t)|^{2} \nonumber\\ &\quad- {\rm i}\left[\mathfrak{C} \int_{\mathbb{R}}\frac{\mu^{2}(\xi)\, {\rm d}\xi}{\xi^{2} + \eta + i \lambda}\right]\lambda |v(L, t)|^{2} \nonumber \\ & = \chi\int_{0}^{L}|u_{x}|^{2}\, {\rm d}x + \beta\int_{0}^{L}|v_{x}|^{2}\, {\rm d}x - \gamma \beta\int_{0}^{L}\left[|u_{x}|^{2} + |v_{x}|^{2}\right]\,{\rm d}x. \end{align}

Thus, by the above equation, the imaginary terms are equal to zero, then we should have

(4.21)\begin{eqnarray}u(L, t) = 0,\qquad \mbox{and} \ v(L,\,t) = 0. \end{eqnarray}

Consequently, in Equation (4.20), we have

(4.22)\begin{eqnarray} - \lambda^{2}\int_{0}^{L}\left[\rho |u|^{2} + \mu |v|^{2}\right]\,{\rm d}x = \int_{0}^{L}X^{\rm T} A X\, {\rm d}x. \end{eqnarray}

Furthermore, from Equation (4.19), integrating by parts and performing straightforward calculations, we deduce that $ u_{x}(L, t) = 0 \qquad \mbox{and}\ v_{x}(L, t) = 0;$ and, from this premise Equation (4.14), we sustain

(4.23)\begin{eqnarray} \begin{cases} - \rho \lambda^{2} u - \chi u_{xx} + \gamma \beta v_{xx} = 0 \\ - \mu \lambda^{2} v - \beta v_{xx} + \gamma \beta u_{xx} = 0. \end{cases} \end{eqnarray}

Now, we rewrite Equation (4.23) as the initial value problem

(4.24)\begin{eqnarray} \begin{cases} X_{xx} = A^{-1} B X \\ X(L, t) = 0,\qquad X_{x}(L, t) = 0, \end{cases} \end{eqnarray}

where A, X, and A ⁻¹ are defined in Remark 3.2, and B has been given by Equation (4.9).

Due to the Picard theorem for ordinary differential equation, we arrive at Equation (4.24), which has a unique solution u = 0 and v = 0. Moreover, from Equation (4.6) equality, we obtain w = 0 and z = 0. Therefore $U=(u, v, w, z, \varphi, \phi)=0.$

Lemma 4.3. If $\eta \neq 0,$ we have that $\ 0\in\varrho({\cal A}).$

Proof. We have to prove that given $F=(f_{1}, f_{2}, f_{3}, f_{4}, f_{5}, f_{6})\in {\cal H}$, there exists a unique $U=(u, v, w, z, \varphi, \phi)$ in ${\cal D}({\cal A}),$ such that $-{\cal A}U=F$, namely

(4.25)\begin{eqnarray} \left\lbrace \begin{array}{l} - w = f_{1} \ \\ - z = f_{2} \ \\ - \chi u_{xx} + \gamma \beta v = \rho f_{3} \\ - \beta v_{xx} + \gamma \beta u = \mu f_{4} \\ (\xi^{2} + \eta) \varphi - w(L, t) \mu(\xi) = f_{5} \\ (\xi^{2} + \eta) \phi - z(L, t) \mu(\xi) = f_{6} \end{array}, \right. \end{eqnarray}

with the initial condition

\begin{eqnarray*} \left\lbrace \begin{array}{l} \displaystyle \chi u_{x}(L, t) - \gamma \beta v_{x}(L, t) = - \mathfrak{C} \int_{\mathbb{R}}\mu(\xi) \varphi(\xi, t)\, {\rm d}\xi \\ \displaystyle \beta v_{x}(L, t) - \gamma \beta u_{x}(L, t) = - \mathfrak{C} \int_{\mathbb{R}}\mu(\xi) \phi(\xi, t)\, {\rm d}\xi. \end{array} \right. \end{eqnarray*}

To achieve this, we use Equations (4.16) and (4.17) with $\lambda = 0,$ and it follows that

(4.26)\begin{eqnarray} & & \chi\int_{0}^{L}u_{x} \zeta_{x}\, {\rm d}x - \gamma \beta\int_{0}^{L}v_{x} \zeta_{x}\, {\rm d}x = \rho\int_{0}^{L}f_{3} \zeta\, {\rm d}x - \left[\mathfrak{C} \int_{\mathbb{R}}\frac{\mu(\xi) f_{5}(\xi)}{\xi^{2} + \eta + i \lambda}\, {\rm d}\xi\right]\zeta(L, t) \nonumber \\& & + \left[\mathfrak{C} \int_{\mathbb{R}}\frac{\mu^{2}(\xi)}{\xi^{2} + \eta + i \lambda}\, {\rm d}\xi\right]f_{1}(L) \zeta(L,\,t) \end{eqnarray}

and

(4.27)\begin{eqnarray} & & \beta\int_{0}^{L}v_{x} \aleph_{x}\, {\rm d}x - \gamma \beta\int_{0}^{L}u_{x} \aleph_{x}\, {\rm d}x = \mu\int_{0}^{L}f_{4} \aleph\, {\rm d}x - \left[\mathfrak{C} \int_{\mathbb{R}}\frac{\mu(\xi) f_{6}(\xi)}{\xi^{2} + \eta + i \lambda}\, {\rm d}\xi\right]\aleph(L, t) \nonumber \\& & + \left[\mathfrak{C} \int_{\mathbb{R}}\frac{\mu^{2}(\xi)}{\xi^{2} + \eta + i \lambda}\, {\rm d}\xi\right]f_{2}(L) \aleph(L, t). \end{eqnarray}

The system in Equations (4.26) and (4.27) is equivalent to the problem

(4.28)\begin{eqnarray}\mathfrak{a}_{\eta}\left((u, v), (\zeta, \aleph)\right) = {\cal L}_{\eta}(\zeta, \aleph), \end{eqnarray}

where the bilinear form is continuous and coercive $\mathfrak{a}_{\eta}: \left[\mathbb{H}(0,\,L)\times \mathbb{H}(0, L)\right]^{2}\rightarrow\mathbb{R},$ and the continuous linear ${\cal L}_{\eta}: \left[\mathbb{H}(0, L)\right]^{2}\rightarrow\mathbb{R}$ are defined by

\begin{align} \mathfrak{a}_{\eta}\left((u, v), (\zeta, \chi)\right) = & \chi\int_{0}^{L}u_{x} \zeta_{x}\, {\rm d}x + \beta\int_{0}^{L}u_{x} \aleph_{x}\, {\rm d}x \nonumber - \gamma \beta\int_{0}^{L}v_{x} \zeta_{x}\, {\rm d}x - \gamma \beta\int_{0}^{L}u_{x} \aleph_{x}\, {\rm d}x \end{align}

and

\begin{align*} {\cal L}(\zeta, \aleph) &= \int_{0}^{L}f_{2} \zeta\, {\rm d}x + \int_{0}^{L}f_{4} \aleph\, {\rm d}x \nonumber -\left[\mathfrak{C} \int_{\mathbb{R}}\frac{\mu(\xi) f_{5}(\xi)}{\xi^{2} + \eta}\, {\rm d}\xi\right]\zeta(L, t) \nonumber\\ &\quad- \left[\mathfrak{C} \int_{\mathbb{R}}\frac{\mu(\xi) f_{6}(\xi)}{\xi^{2} + \eta}\, {\rm d}\xi\right]\aleph(L, t) + \left[\mathfrak{C} \int_{\mathbb{R}}\frac{\mu^{2}(\xi)}{\xi^{2} + \eta}\, {\rm d}\xi\right]f_{1}(L) \zeta(L, t)\nonumber \\ &\quad + \left[\mathfrak{C} \int_{\mathbb{R}}\frac{\mu^{2}(\xi)}{\xi^{2} + \eta}\, {\rm d}\xi\right]f_{2}(L) \aleph(L, t). \end{align*}

Applying the Lax–Milgran theorem, we have that for all $(\zeta, \aleph)\in \mathbb{H}(0, L)\times \mathbb{H}(0, L)$, the problem (4.28) admits a unique solution $(u, v)\in \mathbb{H}(0, L)\times \mathbb{H}(0, L).$ Then by using elliptic regularity, stated in Equations (4.26) and (4.27), it is deduced that $(u, v) \in H^{2}(0, L)\times H^{2}(0, L).$ Therefore, the operator ${\cal A}$ is surjective.

Lemma 4.4. Let ${\cal A}$ be defined by Equation (3.4), then

(4.29)\begin{equation} \mathcal{A}^{*}\left( \begin{array}{c} u \\ v\\ w\\ z\\ \varphi \\ \phi \end{array} \right) =\left( \begin{array}{c} - w\\ - z\\ \chi u_{xx} - \gamma \beta v_{xx} \\ \beta v_{xx} - \gamma \beta u_{xx} \\ - (\xi^{2} + \eta) \varphi - w(L, t) \mu(\xi) \\ - (\xi^{2} + \eta) \phi - z(L, t) \mu(\xi) \end{array} \right). \end{equation}

with the domain

\begin{align*} &\mathcal{D}({\cal A}^*) \\ &= \left\{ U \in {\cal H} \left| \begin{array}{cl} &\!\!\!\!\!\! u, v \in H^{2}(0, L)\cap \mathbb{H}(0,L), w, z\in \mathbb{H}(0, L) \\ &\!\!\!\!\!\! -(\xi^{2} + \eta)\varphi - w(L, t) \mu(\xi) \in L^{2}(\mathbb{R}) \\ &\!\!\!\!\!\! - (\xi^{2} + \eta) \varphi - z(L, t) \mu(\xi) \in L^{2}(\mathbb{R}) \\ & \!\!\!\!\!\!\chi u_{x}(L, t) - \gamma \beta v_{x}(L, t) + \mathfrak{C} \int_{-\infty}^{\infty}\mu(\xi) \varphi(\xi, t)\, {\rm d}\xi = 0, \qquad |\xi| \varphi \in L^{2}(\mathbb{R}) \\ &\!\!\!\!\!\! \chi v_{x}(L, t) - \gamma \beta u_{x}(L, t) + \mathfrak{C} \int_{-\infty}^{\infty}\mu(\xi) \phi(\xi, t)\, {\rm d}\xi = 0, \qquad |\xi|\,\phi \in L^{2}(\mathbb{R}) \end{array} \right. \right\} \end{align*}

Proof. It is a straightforward verification to show that $\ \langle {\cal A}G, G\rangle = \langle G, {\cal A}^{*}G\rangle.\ $ In fact, let $G = (u, v, w, z, \varphi, \phi)$ and $\widetilde{G} = (\widetilde{u}, \widetilde{v}, \widetilde{w}, \widetilde{z}, \widetilde{\varphi}, \widetilde{\phi}).$ Then

\begin{eqnarray*} \langle {\cal A}U, U\rangle_{\cal H} & = & \chi\int_{0}^{L}u_{xx} \widetilde{w}\, {\rm d}x - \gamma \beta\int_{0}^{L}v_{xx} \widetilde{w}\, {\rm d}x + \beta\int_{0}^{L}v_{xx} \widetilde{z}\, {\rm d}x - \gamma \beta\int_{0}^{L}u_{xx} \widetilde{z}\, {\rm d}x \\ & + & \chi\int_{0}^{L}w_{x} \widetilde{u}_{x}\, {\rm d}x + \beta\int_{0}^{L}z_{x} \widetilde{v}_{x}\, {\rm d}x - \gamma \beta\int_{0}^{L}w_{x} \widetilde{v}_{x}\, {\rm d}x - \gamma \beta\int_{0}^{L}z_{x} \widetilde{u}_{x}\, {\rm d}x \\ & - & \mathfrak{C}\int_{\mathbb{R}}\left(\xi^{2} + \eta\right)\varphi \widetilde{\varphi}\, {\rm d}\xi + w(L, t) \mathfrak{C}\int_{\mathbb{R}}\mu(\xi) \widetilde{\varphi}\, {\rm d}x \\ & - & \mathfrak{C}\int_{\mathbb{R}}\left(\xi^{2} + \eta\right)\phi \widetilde{\phi}\, {\rm d}\xi + z(L, t) \mathfrak{C}\int_{\mathbb{R}}\mu(\xi) \widetilde{\phi}\, {\rm d}x. \end{eqnarray*}

If we integrate by parts the first eight terms, grouping and replacing on before the equality, we obtain

\begin{eqnarray*} \langle {\cal A}U, U\rangle_{\cal H} & = & - \chi\int_{0}^{L}\widetilde{u}_{xx} w\, {\rm d}x + \gamma \beta\int_{0}^{L}\widetilde{v}_{xx} w\, {\rm d}x - \beta\int_{0}^{L}\widetilde{v}_{xx} z\, {\rm d}x + \gamma \beta\int_{0}^{L}\widetilde{u}_{xx} z\, {\rm d}x \nonumber \\ & - & \chi\int_{0}^{L}u_{x} \widetilde{w}_{x}\, {\rm d}x - \beta\int_{0}^{L}v_{x} \widetilde{z}_{x}\, {\rm d}x + \gamma \beta\int_{0}^{L}u_{x} \widetilde{z}_{x}\, {\rm d}x + \gamma \beta\int_{0}^{L}v_{x} \widetilde{w}_{x}\, {\rm d}x \nonumber \\ & + & \left[\chi u_{x}(L, t) - \gamma \beta v_{x}(L, t)\right]\widetilde{w}(L, t) + \left[\beta v_{x}(L, t) - \gamma \beta u_{x}(L, t)\right]\widetilde{z}(L, t) \\ & + & \left[\chi \widetilde{u}_{x}(L, t) - \gamma \beta \widetilde{v}_{x}(L, t)\right]w(L, t) + \left[\beta \widetilde{v}_{x}(L, t) - \gamma \beta \widetilde{u}_{x}(L, t)\right]z(L, t) \\ & - & \mathfrak{C}\int_{\mathbb{R}}\left(\xi^{2} + \eta\right)\varphi \widetilde{\varphi}\, {\rm d}\xi + \mathfrak{C} w(L, t)\int_{\mathbb{R}}\mu(\xi) \widetilde{\varphi}\, {\rm d}x \\ & - & \mathfrak{C}\int_{\mathbb{R}}\left(\xi^{2} + \eta\right)\phi \widetilde{\phi}\, {\rm d}\xi + \mathfrak{C} z(L,\,t) \int_{\mathbb{R}}\mu(\xi) \widetilde{\phi}\, {\rm d}x. \end{eqnarray*}

Revisiting the boundary conditions (2.10) $_{9, 10},$ we have

\begin{align*} \langle {\cal A}U, U\rangle_{\cal H} & = - \chi\int_{0}^{L}\widetilde{u}_{xx} w\, {\rm d}x + \gamma \beta\int_{0}^{L}\widetilde{v}_{xx} w\, {\rm d}x - \beta\int_{0}^{L}\widetilde{v}_{xx} z\, {\rm d}x + \gamma \beta\int_{0}^{L}\widetilde{u}_{xx} z\, {\rm d}x \nonumber \\ & \quad - \chi\int_{0}^{L}u_{x} \widetilde{w}_{x}\, {\rm d}x - \beta\int_{0}^{L}v_{x} \widetilde{z}_{x}\, {\rm d}x + \gamma \beta\int_{0}^{L}u_{x} \widetilde{z}_{x}\, {\rm d}x + \gamma \beta\int_{0}^{L}v_{x} \widetilde{w}_{x}\, {\rm d}x \nonumber \\ & \quad - \widetilde{w}(L, t)\ \mathfrak{C}\int_{\mathbb{R}}\mu(\xi) \varphi\, {\rm d}x - \widetilde{z}(L, t) \mathfrak{C}\int_{\mathbb{R}}\mu(\xi) \phi\, {\rm d}x \nonumber \\ & \quad - \mathfrak{C}\int_{\mathbb{R}}\left(\xi^{2} + \eta\right)\varphi \widetilde{\varphi}\, {\rm d}\xi - \mathfrak{C}\int_{\mathbb{R}}\left(\xi^{2} + \eta\right)\phi \widetilde{\phi}\, {\rm d}\xi \nonumber \\ & = \rho\int_{0}^{L}\frac{1}{\rho}\left[\chi \widetilde{u}_{xx}\, {\rm d}x - \gamma \beta \widetilde{v}_{xx}\right](-w)\, {\rm d}x + \mu\int_{0}^{L}\frac{1}{\mu}\left[\beta \widetilde{v}_{xx}\, {\rm d}x - \gamma \beta \widetilde{u}_{xx}\right](-z)\, {\rm d}x \nonumber \\ & \quad + \chi\int_{0}^{L}u_{x} (-\widetilde{w}_{x})\, {\rm d}x + \beta\int_{0}^{L}v_{x} (-\widetilde{z}_{x})\, {\rm d}x - \gamma \beta\int_{0}^{L}u_{x} (-\widetilde{z}_{x})\, {\rm d}x \nonumber \\ & \quad - \gamma \beta\int_{0}^{L}v_{x} (-\widetilde{w}_{x})\, {\rm d}x + \mathfrak{C}\int_{\mathbb{R}}\varphi\left[-\left(\xi^{2} + \eta\right)\widetilde{\varphi} + (-\widetilde{w}(L, t)) \mu(\xi)\right]\,{\rm d}\xi \nonumber \\ & \quad + \mathfrak{C}\int_{\mathbb{R}}\phi\left[-\left(\xi^{2} + \eta\right)\widetilde{\varphi} + (-\widetilde{z}(L, t)) \mu(\xi)\right]\,{\rm d}\xi = \langle G, {\cal A}^{*}G\rangle, \end{align*}

where ${\cal A}^{*}$ is given in Equation (4.29). Lemma 4.4 follows.

Remark 4.5. An operator may be injective and bounded below, but not necesarily invertible, for example, the unilateral shift on $l^{2}(\mathbb{N}).$ This shift operator is isometry and bounded inferiorly, but it is not invertible since it is not surjective. The set of λ for which $\lambda\,I - {\cal A}$ is injective but does not have a dense range is known as the residual spectrum of ${\cal A}$ and is denoted by $\sigma_{r}({\cal A}).$

\begin{eqnarray*} \sigma_{r}({\cal A}) & = & \left\{\lambda\in\mathbb{C} : \lambda I - {\cal A} \rm{has\ an\ inverse}, \rm{bounded\ or\ unbounded}, \right. \\ & & \left. \ \rm{with\ domain\ not\ as \ dense\ in}\ {\cal H} \right\} \end{eqnarray*}

Theorem 4.6. The set of residual spectrum verifies that $\sigma_{r}({\cal A}) = \emptyset.$

Proof. We consider that the eigenvalues of ${\cal A}$ are symmetric on the real axis; hence, we will show that $\sigma_{r}({\cal A}) = \sigma_{p}({\cal A}^{*}).$

In fact, let $\lambda\in\mathbb{C}$ such that ${\cal A}^{*}U = \lambda\,U$ and $0\neq U=(u,\,v,\,w,\,z,\,\varphi,\,\phi)\in \mathcal{D}({\cal A}^{*}).$ We recall the definition (4.29) and therefore we have

(4.30)\begin{eqnarray} \left\lbrace \begin{array}{l} \lambda u + w = 0 \\ \lambda v + z = 0 \\ \rho \lambda w - \left[\chi u_{xx} - \gamma \beta v_{xx}\right] = 0 \\ \mu \lambda z - \left[\beta v_{xx} - \gamma \beta u_{xx}\right] = 0 \\ \lambda \varphi + (\xi^{2} + \eta) \varphi + w(L, t)\,\mu(\xi) = 0 \\ \lambda \phi + (\xi^{2} + \eta) \phi + z(L, t) \mu(\xi) = 0 \end{array} \right. \end{eqnarray}

Replacing Equation (4.30) $_{1, 2}$ into (4.30) $_{3, 4}$, we obtain

(4.31)\begin{eqnarray} \left\lbrace \begin{array}{l} - \rho \lambda^{2} u - \left[\chi u_{xx} - \gamma \beta v_{xx}\right] = 0 \\ - \mu \lambda^{2} v - \left[\beta v_{xx} - \gamma \beta u_{xx}\right] = 0 \end{array} \right. \end{eqnarray}

with the boundary condition

(4.32)\begin{eqnarray} \left\lbrace \begin{array}{l} u(0, t) = 0,\qquad v(0, t) = 0 \\ u_{x}(L, t) = - \lambda (\lambda + \eta)^{\alpha - 1} u(L, t), \qquad v_{x}(L, t) = - \lambda (\lambda + \eta)^{\alpha - 1} v(L, t). \end{array} \right. \end{eqnarray}

The system (4.31) and (4.32) matches with the eigenvalue problem of ${\cal A}.$ We conclude that ${\cal A}^{*}$ has the same eigenvalues as that of ${\cal A}$ and the theorem follows.

Considering the previous lemmas and utilizing Theorem 2.6, we have proved the following:

Theorem 4.7. The $C_0-$semigroup of contraction ${\cal S}_{{\cal A}}(t)_{t\geq 0}$ is asymptotically stable, which means

\begin{eqnarray*} \|{\cal S}_{{\cal A}}(t)U_0\|_{{\cal H}}\rightarrow 0\quad\mbox{as}\ t\rightarrow \infty,\quad\mbox{for}\ U_0\in{\cal H}. \end{eqnarray*}

Therefore, we are able to prove the second main theorem of this section.

Theorem 4.8. Let η ≠ 0. The semigroup ${\cal S}_{{\cal A}}(t)_{t\geq 0}$ is polynomially stable and

(4.33)\begin{eqnarray} \|{\cal S}_{{\cal A}}(t)U_{0}\|_{{\cal H}} \leq \frac{1}{t^{1/2 (1 - \alpha)}} \|U_{0}\|_{{\cal D}({\cal A})}. \end{eqnarray}

Proof. We will analyze the resolvent equation $(i \lambda I - {\cal A})U = F,$ with $\lambda\in\mathbb{R},$ which in length is

(4.34)\begin{eqnarray} \left\lbrace \begin{array}{l} i \lambda u - w = f_{1} \\ i \lambda v - z = f_{2} \\ i \rho \lambda w - \chi u_{xx} + \gamma \beta v_{xx} = \rho f_{3} \\ i \mu \lambda z - \beta v_{xx} + \gamma \beta u_{xx} = \mu f_{4} \\ (\xi^{2} + \eta + i \lambda) \varphi - w(1, t) \mu(\xi) = f_{5} \\ (\xi^{2} + \eta + i \lambda) \phi - z(1, t) \mu(\xi) = f_{6} \end{array} \right. \end{eqnarray}

where $F = (f_{1}, f_{2}, f_{3}, f_{4}, f_{5}, f_{6}).$ Revisiting Equation (3.5), we have

\begin{eqnarray*} \left|{\rm Re}\langle {\cal A}U,\ U\rangle_{{\cal H}}\right| \leq \|U\|_{{\cal H}}\ \|F\|_{{\cal H}}, \end{eqnarray*}

that is,

(4.35)\begin{eqnarray} \left\lbrace \begin{array}{l} \displaystyle \mathfrak{C}\int_{\mathbb{R}}\left(\xi^{2} + \eta\right)|\varphi|^{2}\, {\rm d}\xi \leq \|U\|_{{\cal H}} \|F\|_{{\cal H}} \\ \\ \displaystyle \mathfrak{C}\int_{\mathbb{R}}\left(\xi^{2} + \eta\right)|\phi|^{2}\, {\rm d}\xi \leq \|U\|_{{\cal H}} \|F\|_{{\cal H}}. \end{array} \right. \end{eqnarray}

Furthermore, using $\left||a| - |b| \right| \leq \left|a - b\right|$ into Equation (4.34) $_{1,\,2}$ we have, respectively,

\begin{eqnarray*} \left\lbrace \begin{array}{l} \left||\lambda||u(L, t)| - |f_{1}(L)| \right| \leq \left| i \lambda u(L, t) - f_{1}(L) \right| = \left|w(L, t)\right| \\ \\ \left| |\lambda| |v(L, t)| - |f_{2}(L)| \right| \leq \left| i \lambda v(L, t) - f_{2}(L) \right| = \left|z(L, t)\right|, \end{array} \right. \end{eqnarray*}

then

\begin{eqnarray*} \left\lbrace \begin{array}{l} \left| |\lambda| |u(L, t)| - |f_{1}(L)| \right|^{2} \leq \left|w(L, t)\right|^{2} \\ \\ \left| |\lambda| |v(L, t)| - |f_{2}(L)| \right|^{2} \leq \left|z(L, t)\right|^{2}. \end{array} \right. \end{eqnarray*}

Thus,

\begin{eqnarray} \left\lbrace \begin{array}{l} |\lambda|^{2} |u(L, t)|^{2} \leq C |f_{1}(L)|^{2} + C \left|w(L, t)\right|^{2} \nonumber \\ \\ |\lambda|^{2} |v(L, t)|^{2} \leq C |f_{2}(L)|^{2} + C \left|z(L, t)\right|^{2}. \nonumber \end{array} \right. \end{eqnarray}

Subsequently, from Equation (2.10) $_{9, 10}$ and by using the Cauchy–Schwartz inequality, we have

\begin{eqnarray*} |\chi u_{x}(L, t) - \gamma \beta v_{x}(L, t)|^{2} & = & \left| \mathfrak{C} \int_{\mathbb{R}}\mu(\xi) \varphi(\xi, t)\, {\rm d}\xi\right|^{2} \nonumber \\ & \leq & \left|\mathfrak{C}\right|^{2} \left|\int_{\mathbb{R}}(\xi^{2} + \eta)^{-1/2} \mu(\xi) (\xi^{2} + \eta)^{1/2} \varphi(\xi, t)\, {\rm d}\xi\right|^{2} \nonumber \\ & \leq & \left(\int_{\mathbb{R}}(\xi^{2} + \eta)^{-1}\left|\mu(\xi)\right|^{2} d\xi\right)\left|\mathfrak{C}\right|^{2}\int_{\mathbb{R}}(\xi^{2} + \eta)\left|\varphi(\xi, t)\right|^{2}\, {\rm d}\xi \nonumber \\ & \leq & C\|U\|_{{\cal H}} \|F\|_{{\cal H}} \end{eqnarray*}

as well as

\begin{eqnarray*} |\beta v_{x}(L, t) - \gamma \beta u_{x}(L, t)|^{2} & = & \left| \mathfrak{C} \int_{\mathbb{R}}\mu(\xi) \phi(\xi, t)\, {\rm d}\xi\right|^{2} \nonumber \\ & \leq & \left|\mathfrak{C}\right|^{2} \left|\int_{\mathbb{R}}(\xi^{2} + \eta)^{-1/2} \mu(\xi) (\xi^{2} + \eta)^{1/2} \phi(\xi, t)\, {\rm d}\xi\right|^{2} \nonumber \\ & \leq & \left(\int_{\mathbb{R}}(\xi^{2} + \eta)^{-1}\left|\mu(\xi)\right|^{2} \,{\rm d}\xi\right)\left|\mathfrak{C}\right|^{2}\int_{\mathbb{R}}(\xi^{2} + \eta)\left|\phi(\xi, t)\right|^{2}\,{\rm d}\xi \nonumber \\ & \leq & C \|U\|_{{\cal H}} \|F\|_{{\cal H}}. \end{eqnarray*}

Ultimately, by considering Remark 3.2, we obtain

\begin{eqnarray} \left\lbrace \begin{array}{l} |u_{x}(L, t)|^{2} \leq C \|U\|_{{\cal H}} \|F\|_{{\cal H}} \nonumber \\ \\ |v_{x}(L, t)|^{2} \leq C \|U\|_{{\cal H}} \|F\|_{{\cal H}}. \nonumber \end{array} \right. \end{eqnarray}

Moreover, from Equation (4.34) $_{5, 6}$, we can derive, respectively, that

(4.36)\begin{eqnarray} \left\lbrace \begin{array}{l} w(L, t) \mu(\xi) = (\xi^{2} + \eta + i \lambda) \varphi - f_{5}(\xi) \\ \\ z(L, t) \mu(\xi) = (\xi^{2} + \eta + i \lambda) \phi - f_{6}(\xi). \end{array} \right. \end{eqnarray}

An by multiplying Equation (4.36)₁ by $(\xi^{2} + \eta + i \lambda)^{-1} \mu(\xi)$, we have

\begin{eqnarray*} (\xi^{2} + \eta + i \lambda)^{-1} \mu^{2}(\xi) w(L, t) = \mu(\xi) \varphi - (\xi^{2} + \eta + i \lambda)^{-1} \mu(\xi) f_{5}(\xi). \end{eqnarray*}

Then by applying absolute values and integrating them over $\xi\in \mathbb{R}$, we have the following result:

\begin{eqnarray*} & & \left(\int_{\mathbb{R}}|(\xi^{2} + \eta + i \lambda)|^{-1} |\mu(\xi)|^{2}\, {\rm d}\xi\right)|w(L, t)| \\ & \leq & \int_{\mathbb{R}}|\mu(\xi)| |\varphi|\, {\rm d}\xi + \int_{\mathbb{R}}|(\xi^{2} + \eta + i \lambda)|^{-1} |\mu(\xi)| |f_{5}(\xi)|\, {\rm d}\xi \\ & \leq & \int_{\mathbb{R}}(\xi^{2} + \eta)^{-1/2} |\mu(\xi)| (\xi^{2} + \eta)^{1/2} |\varphi|\, {\rm d}\xi + \int_{\mathbb{R}}|(\xi^{2} + \eta + i \lambda)|^{-1} |\mu(\xi)| |f_{5}(\xi)|\, {\rm d}\xi. \end{eqnarray*}

Once again, by using the Cauchy–Schwartz inequality and having straightforward estimates in front, it follows that

\begin{eqnarray*} \left(\int_{\mathbb{R}}\frac{|\mu(\xi)|^{2}\, {\rm d}\xi}{(\xi^{2} + \eta + |\lambda|)} \right)|w(L,\,t)| & \leq & \left(\int_{\mathbb{R}}(\xi^{2} + \eta)^{-1}|\mu(\xi)|^{2}\, {\rm d}\xi\right)^{1/2} \left(\int_{\mathbb{R}}(\xi^{2} + \eta) |\varphi|^{2}\, {\rm d}\xi\right)^{1/2} \\ & + & \left(\int_{\mathbb{R}}\frac{|\mu(\xi)|^{2}\, {\rm d}\xi}{(\xi^{2} + \eta + |\lambda|)^{2}}\right)^{1/2}\left(\int_{\mathbb{R}}|f_{5}|^{2}\, {\rm d}\xi\right)^{1/2}. \end{eqnarray*}

Also by applying power square on both sides of the inequality, as well as using $2 a b \leq a^{2} + b^{2}$, we are able to obtain

\begin{eqnarray} & & \left(\int_{\mathbb{R}}\frac{|\mu(\xi)|^{2}\, {\rm d}\xi}{(\xi^{2} + \eta + |\lambda|)}\right)^{2}|w(L, t)|^{2} \leq 2\left(\int_{\mathbb{R}}\frac{|\mu(\xi)|^{2}\, {\rm d}\xi}{(\xi^{2} + \eta)}\right) \left(\int_{\mathbb{R}}(\xi^{2} + \eta) |\varphi|^{2}\, {\rm d}\xi\right) \nonumber \\ & & + 2\left(\int_{\mathbb{R}}\frac{|\mu(\xi)|^{2}\, {\rm d}\xi}{(\xi^{2} + \eta + |\lambda|)^{2}}\right)\left(\int_{\mathbb{R}}|f_{5}(\xi)|^{2}\, {\rm d}\xi\right). \nonumber \end{eqnarray}

Hence, by using Equation (4.35) $_{1},$ we get

\begin{eqnarray*} & &\left(\int_{\mathbb{R}}\frac{|\mu(\xi)|^{2}}{|\lambda| + \xi^{2} + \eta}\, {\rm d}\xi\right)^{2}|w(L, t)|^{2} \leq \nonumber \\ & & C\left(\int_{\mathbb{R}}\frac{|\mu(\xi)|^{2}}{(\xi^{2} + \eta)}\, {\rm d}\xi\right)\|U\|_{{\cal H}}\ \|F\|_{{\cal H}} + C\left(\int_{\mathbb{R}} \frac{|\mu(\xi)|^{2}}{(|\lambda| + \xi^{2} + \eta)^{2}}\, {\rm d}\xi\right)\|F\|_{{\cal H}}^{2}. \nonumber \end{eqnarray*}

Applying Lemma 2.2, it follows that

(4.37)\begin{eqnarray}& & |w(L, t)|^{2} \leq C |\lambda|^{2 - 2 \alpha} \|U\|_{{\cal H}} \|F\|_{{\cal H}} + C \|F\|_{{\cal H}}^{2}. \end{eqnarray}

In a similar way, we extrapolate the term (4.36) $_{2},$ resulting in

\begin{eqnarray} & & |z(L, t)|^{2} \leq C |\lambda|^{2 - 2 \alpha} \|U\|_{{\cal H}} \|F\|_{{\cal H}} + C \|F\|_{{\cal H}}^{2}.\nonumber \end{eqnarray}

Consequently, we get

\begin{eqnarray} |u_{x}(L, t)|^{2} + |v_{x}(L, t)|^{2} \leq C\left(\frac{1}{|\lambda|} + 1\right)\|U\|_{{\cal H}} \|F\|_{{\cal H}} + C \frac{1}{|\lambda|^{2}} \|F\|_{{\cal H}}^{2}. \nonumber \end{eqnarray}

To conclude the verification of the theorem, we need the following lemma:

Lemma 4.9. Let $q\in H^{1}(0, L).$ Therefore, we have

\begin{align} \nonumber \int_{0}^{L}q_{x}\left[\rho |w|^{2} + \chi |u_{x}|^{2}\right]\,{\rm d}x = \rho q |w|^{2}\left/_{0}^{L}\right. + \chi q |u_{x}|^{2}\left/_{0}^{L}\right. + R_{1} \end{align}

where $\displaystyle R_{1} = 2 \rho\int_{0}^{L}q {\rm Re}(f_{3} \overline{u}_{x})\, {\rm d}x + 2 \rho\int_{0}^{L}q {\rm Re}(\overline{f}_{1x} w)\, {\rm d}x $ and also

\begin{align} \nonumber \int_{0}^{L}q_{x}\left[\mu |z|^{2} + \beta |v_{x}|^{2}\right]\,{\rm d}x = \mu q |z|^{2}\left/_{0}^{L}\right. + \beta q |v_{x}|^{2}\left/_{0}^{L}\right. + R_{2} \end{align}

along with $\displaystyle R_{2} = 2 \mu\int_{0}^{L}q {\rm Re}(f_{4} \overline{v}_{x})\, {\rm d}x + 2 \mu\int_{0}^{L}q {\rm Re}(\overline{f}_{2x} z)\, {\rm d}x. $

Note that for $R_{i},$ $(i = 1, 2)$, we have

(4.38)\begin{eqnarray} |R_{i}| \leq C \|U\|_{{\cal H}} \|F\|_{{\cal H}}. \end{eqnarray}

Proof. To demonstrate Lemma 4.9, we will multiply Equation (4.34)₃ by $q \overline{u}_{x}$ and integrate over $(0, L)$, and we have that

\begin{eqnarray*} - \rho\int_{0}^{L}q (\overline{i \lambda u_{x}}) w\, {\rm d}x - \chi\int_{0}^{L}q u_{xx} \overline{u}_{x}\, {\rm d}x + \gamma \beta\int_{0}^{L}q v_{xx} \overline{u}_{x}\, {\rm d}x = \rho\int_{0}^{L}q f_{3} \overline{u}_{x}\, {\rm d}x. \end{eqnarray*}

From Equation (4.34)₁, we have $\overline{i \lambda u_{x}} = \overline{w}_{x} + \overline{f}_{1x}.$ Then

\begin{eqnarray*} - \rho\int_{0}^{L}q (\overline{w}_{x} + \overline{f}_{1x}) w\, {\rm d}x - \chi\int_{0}^{L}q \overline{u}_{x} u_{xx}\, {\rm d}x + \gamma \beta\int_{0}^{L}q \overline{u}_{x} v_{xx}\, {\rm d}x = \rho\int_{0}^{L}q f_{3} \overline{u}_{x}\, {\rm d}x. \end{eqnarray*}

Thus,

\begin{align*} & - \rho\int_{0}^{L}q w \overline{w}_{x}\, {\rm d}x - \chi\int_{0}^{L}q \overline{u}_{x} u_{xx}\, {\rm d}x + \gamma \beta\int_{0}^{L}q \overline{u}_{x} v_{xx}\, {\rm d}x \nonumber \\ & = \rho\int_{0}^{L}q f_{3} \overline{u}_{x}\, {\rm d}x + \rho\int_{0}^{L}q \overline{f}_{1x} w\, {\rm d}x. \end{align*}

Taking the real part of the aforementioned equality, it follows that

\begin{align*} & - \rho\int_{0}^{L}q {\rm Re}(w \overline{w}_{x})\, {\rm d}x - \chi\int_{0}^{L}q {\rm Re}(\overline{u}_{x} u_{xx})\, {\rm d}x + \gamma \beta\int_{0}^{L}q {\rm Re}(\overline{u}_{x} v_{xx})\, {\rm d}x \\ & = \rho\int_{0}^{L}q {\rm Re}(f_{3} \overline{u}_{x})\, {\rm d}x + \rho\int_{0}^{L}q {\rm Re}(\overline{f}_{1x} w)\, {\rm d}x. \end{align*}

Using the precedent of ${\rm Re}(\Phi \overline{\Phi}_{x}) = \frac{1}{2} \frac{\rm d}{{\rm d}x}|\Phi|^{2},$ we have

\begin{eqnarray*} & &- \frac{1}{2} \rho\int_{0}^{L}q \frac{\rm d}{{\rm d}x}(|w|^{2})\, {\rm d}x - \frac{1}{2} \chi\int_{0}^{L}q \frac{\rm d}{{\rm d}x}(|u_{x}|^{2})\, {\rm d}x + \gamma \beta\int_{0}^{L}q {\rm Re}(\overline{u}_{x} v_{xx})\, {\rm d}x \\ & = & \rho\int_{0}^{L}q {\rm Re}(f_{3} \overline{u}_{x})\, {\rm d}x + \rho\int_{0}^{L}q {\rm Re}(\overline{f}_{1x} w)\, {\rm d}x. \end{eqnarray*}

By integrating by parts the first two integrals, we get

\begin{eqnarray*} & & \rho\int_{0}^{L}q_{x} |w|^{2}\, {\rm d}x + \chi\int_{0}^{L}q_{x}|u_{x}|^{2}\, {\rm d}x + 2 \gamma \beta\int_{0}^{L}q {\rm Re}(\overline{u}_{x} v_{xx})\, {\rm d}x \nonumber \\ & = & 2 \rho\int_{0}^{L}q {\rm Re}(f_{3} \overline{u}_{x})\, {\rm d}x + 2 \rho\int_{0}^{L}q {\rm Re}(\overline{f}_{1x} w)\, {\rm d}x + \rho q |w|^{2}\left/_{0}^{L}\right. + \chi q |u_{x}|^{2}\left/_{0}^{L}\right. \end{eqnarray*}

It continues as

(4.39)\begin{equation} \int_{0}^{L}q_{x}\left[\rho |w|^{2} + \chi |u_{x}|^{2}\right]\,{\rm d}x = \rho q |w|^{2}\left/_{0}^{L}\right. + \chi q |u_{x}|^{2}\left/_{0}^{L}\right. + 2 \gamma \beta\int_{0}^{L}q {\rm Re}(\overline{u}_{x} v_{xx})\, {\rm d}x + R_{1}, \end{equation}

where $R_{1} = 2 \rho\int_{0}^{L}q {\rm Re}(f_{3} \overline{u}_{x})\, {\rm d}x + 2 \rho\int_{0}^{L}q {\rm Re}(\overline{f}_{1x} w)\, {\rm d}x.$

Therefore, performing similar calculations onto Equation (4.39), we obtain

(4.40)\begin{equation} \int_{0}^{L}q_{x}\left[\mu |z|^{2} + \beta |v_{x}|^{2}\right]\,{\rm d}x = \mu q |z|^{2}\left/_{0}^{L}\right. + \beta q |v_{x}|^{2}\left/_{0}^{L}\right. + 2 \gamma \beta\int_{0}^{L}q {\rm Re}(u_{xx} \overline{v}_{x})\, {\rm d}x + R_{2} \end{equation}

with $ R_{2} = 2 \mu\int_{0}^{L}q {\rm Re}(f_{4} \overline{v}_{x})\, {\rm d}x + 2 \mu\int_{0}^{L}q {\rm Re}(\overline{f}_{2x} z)\, {\rm d}x. $

The lemma is corroborated.

Now, we continue with the proof of Theorem 4.8. Taking $q(x) = x$ in Lemma 4.9, we get

(4.41)\begin{equation} \int_{0}^{L}\left[\rho |w|^{2} + \chi |u_{x}|^{2}\right]\,{\rm d}x = \rho L |w(L, t)|^{2} + \chi L |u_{x}(L, t)|^{2} +\ 2 \gamma \beta\int_{0}^{L}x {\rm Re}(\overline{u}_{x} v_{xx})\, {\rm d}x + R_{1}, \end{equation}

where

\begin{equation*} R_{1} = 2\int_{0}^{L}x {\rm Re}(f_{3} \overline{u}_{x})\, {\rm d}x + 2\int_{0}^{L}x {\rm Re}(\overline{f}_{1x} w)\, {\rm d}x, \end{equation*}

and

(4.42)\begin{equation} \int_{0}^{L}\left[\mu |z|^{2} + \beta |v_{x}|^{2}\right]\,{\rm d}x = \mu L |z(L, t)|^{2} + \beta L |v_{x}(L, t)|^{2} + 2 \gamma \beta\int_{0}^{L}x {\rm Re}(u_{xx} \overline{v}_{x})\, {\rm d}x + R_{2} \end{equation}

with

\begin{equation*} R_{2} = 2\int_{0}^{L}x {\rm Re}(f_{4} \overline{v}_{x})\, {\rm d}x + 2\int_{0}^{L}x {\rm Re}(\overline{f}_{2x} z)\, {\rm d}x. \end{equation*}

Moreover, integrating by parts and bounding the following integral,

(4.43)\begin{eqnarray} & &2 \gamma \beta\int_{0}^{L}x {\rm Re}(\overline{u}_{x} v_{xx})\, {\rm d}x + 2 \gamma \beta\int_{0}^{L}x {\rm Re}(u_{xx} \overline{v}_{x})\, {\rm d}x \nonumber \\ & = & 2 \gamma \beta L \overline{u}_{x}(L, t) v_{x}(L, t) - 2 \gamma \beta\int_{0}^{L} {\rm Re}(\overline{u}_{x} v_{x})\, {\rm d}x - 2 \gamma \beta {\rm Re}\int_{0}^{L}x \left(\overline{u}_{xx} v_{x} - \overline{\overline{u}_{xx} v_{x}}\right)\,{\rm d}x \nonumber \\ & = & 2 \gamma \beta L \overline{u}_{x}(L, t) v_{x}(L, t) - 2 \gamma \beta\int_{0}^{L} {\rm Re}(\overline{u}_{x} v_{x})\, {\rm d}x \nonumber - 4 \gamma \beta {\rm Re}\int_{0}^{L}x \left(i Im(u_{x} \overline{v})\right)\,{\rm d}x \nonumber \\ & = & 2 \gamma \beta L \overline{u}_{x}(L, t) v_{x}(L, t) - 2 \gamma \beta \int_{0}^{L} {\rm Re}(\overline{u}_{x} v_{x})\, {\rm d}x \nonumber \\ & \leq & \gamma \beta L |u_{x}(L, t)|^{2} + \gamma \beta L |v_{x}(L, t)|^{2} +\ \gamma \beta\int_{0}^{L}|u_{x}|^{2}\, {\rm d}x + \gamma \beta\int_{0}^{L}|v_{x}|^{2}\, {\rm d}x \end{eqnarray}

By adding Equation (4.41) with Equation (4.42) and using Equations (4.43) and (4.38), we have that

(4.44)\begin{align} & \int_{0}^{L}\left[\rho |w|^{2} + \mu |z|^{2} + (\chi - \gamma \beta) |u_{x}|^{2} + \beta (1 - \gamma) |v_{x}|^{2}\right]\,{\rm d}x \nonumber \\ \leq &\ \rho L |w(L, t)|^{2} + \mu L |z(L, t)|^{2} + (\chi + \gamma \beta) L |u_{x}(L, t)|^{2} + \beta(1 + \gamma) L |v_{x}(L, t)|^{2}\nonumber \\ & + C \|U\|_{{\cal H}}\|F\|_{{\cal H}}, \end{align}

where $\chi \geq \gamma \beta$ and $1\geq \gamma.$ Furthermore, from Equation (4.35), we are able to confirm

(4.45)\begin{eqnarray} \begin{cases} \mathfrak{C}\int_{\mathbb{R}}|\varphi|^{2}\, {\rm d}\xi \leq \mathfrak{C}\int_{\mathbb{R}}\left(\xi^{2} + \eta\right)|\varphi|^{2}\, {\rm d}\xi \leq C \|U\|_{{\cal H}} \|F\|_{{\cal H}} \\ \\ \mathfrak{C}\int_{\mathbb{R}}|\phi|^{2}\, {\rm d}\xi \leq \mathfrak{C}\int_{\mathbb{R}}\left(\xi^{2} + \eta\right)|\phi|^{2}\, {\rm d}\xi \leq C \|U\|_{{\cal H}}\|F\|_{{\cal H}}. \end{cases} \end{eqnarray}

Then by adding Equations (4.44) and (4.45), it follows that

(4.46)\begin{align} & \|U\|_{\cal H}^{2} \leq C\int_{0}^{L}|u|^{2}\, {\rm d}x + C\int_{0}^{L}|v|^{2}\, {\rm d}x \nonumber \\ & + C\left|\lambda\right|^{2 \alpha - 2}\|U\|_{{\cal H}} \|F\|_{{\cal H}} + C \|U\|_{{\cal H}} \|F\|_{{\cal H}} + C \|F\|_{{\cal H}}^{2}. \end{align}

From Equation (4.34) $_{1, 2}$, we also have

(4.47)\begin{eqnarray}|u|^{2} & \leq & \frac{|w|^{2} + 2 |w| |f_{1}| + |f_{1}|^{2}}{|\lambda|^{2}} \leq \frac{2 |w|^{2} + 2 |f_{1}|^{2}}{|\lambda|^{2}} \end{eqnarray}

and

(4.48)\begin{eqnarray}|v|^{2} \leq \frac{2 |z|^{2} + 2 |f_{2}|^{2}}{|\lambda|^{2}} \qquad \mbox{for} \ \lambda\neq 0. \end{eqnarray}

Replacing Equations (4.47) and (4.48) onto Equation (4.46), we get

(4.49)\begin{eqnarray}\|U\|_{\cal H}^{2} & \leq & C |\lambda|^{2 - 2 \alpha} \|U\|_{{\cal H}} \|F\|_{{\cal H}} + C \|U\|_{{\cal H}} \|F\|_{{\cal H}} + C \|F\|_{{\cal H}}^{2} + \frac{C}{|\lambda|^{2}} \|U\|_{{\cal H}}^{2} \nonumber \\ & & + \frac{C}{|\lambda|^{2}} \|F\|_{{\cal H}}^{2} + \frac{C}{|\lambda|^{2}} \|U\|_{{\cal H}} \|F\|_{{\cal H}},\qquad \mbox{for}\ \lambda \neq 0. \end{eqnarray}

If $|\lambda|\gt1$, we have

\begin{eqnarray*} \|U\|_{{\cal H}}^{2} \leq |\lambda|^{4 (1 - \alpha)} \|F\|_{{\cal H}}^{2} \Longleftrightarrow \|U\|_{{\cal H}} \leq |\lambda|^{2 (1 - \alpha)} \|F\|_{{\cal H}}. \end{eqnarray*}

Consequently, it follows that

\begin{eqnarray*} \frac{1}{|\lambda|^{2 (1 - \alpha)}} \|(i \lambda I - {\cal A})^{-1}\|_{{\cal L}({\cal H})} \leq C,\qquad \forall\ \lambda\in\mathbb{R}, \end{eqnarray*}

for a positive constant $C.$ The conclusion is obtained by applying Theorem 2.7.