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 [Reference Samko, Kilbas and Marichev16].

We consider slightly different versions of (2.1) and (2.2). For this work, we have focused on [Reference Choi and Maccamy6], 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. ([Reference Mbodje11])

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*}

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. ([Reference Arendt and Batty2])

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. ([Reference Borichev and Tomilov4])

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*}

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 [Reference Arendt and Batty2]. 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:

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 $U=(u, v, w, z, \varphi, \phi)=0.$

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*}

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.