Stabilization of a Quantum Equation under Boundary Connections with an Elastic Wave Equation

Abstract

The stability of coupled PDE systems is one of the most important topic because it covers realistic modeling of the most important physical phenomena. In fact, the stabilization of the energy of partial differential equations has been the main goal in solving many structural or microstructural dynamics problems. In this chapter, we investigate the stability of the Schrödinger-like quantum equation in interaction with the mechanical wave equation caused by the vibration of the Euler–Bernoulli beam, to effect stabilization, viscoelastic Kelvin-Voigt dampers are used through weak boundary connection. Firstly, we show that the system is well-posed via the semigroup approach. Then with spectral analysis, it is shown that the system operator of the closed-loop system is not of compact resolvent and the spectrum consists of three branches. Finally, the Riesz basis property and exponential stability of the system are concluded via comparison method in the Riesz basis approach.

1. Introduction

There are many coupled systems that have been addressed in the literature, and we can hint here that coupling may be through the association of PDEs with coefficients or via boundary conditions of PDEs. The coupling may be strong or weak as the characteristic is determined based on the results obtained after studying the stability or control. We can divide the coupled systems according to the coupling form. Firstly, the parabolic-hyperbolic coupled systems, such as heat wave system, that arise from the interaction of the fluid structure. See works [1, 2] where stability and control systems are analyzed. Secondly, we can refer heat-beam system through works [3, 4] where the researchers used an effective method for stabilization of the system. Thirdly, in the heat-Schrödinger system, the heat dynamic controller was applied for stabilization and Gevrey regularity property in the paper [5]. Finally, in the case of thermoelastic systems, the exponential stability and Riesz basis property of the coupled heat equation and elastic structure were discussed in reference [6]. The exponential stability of thermoplastic systems with microtemperature in reference [7], for the linear beam system coupled with thermal effect, we refer to the works [8, 9, 10, 11, 12]. For the nonlinear beam system with thermal effect, see reference [13].

From general result related to the previously mentioned research works, we can conclude that the heat equation plays the role of dynamic boundary feedback controller of the hyperbolic PDE. Also, for the interconnected system of Euler–Bernoulli beam and heat equation with boundary weak connections where the heat is the dynamic boundary controller to the whole system, which means that this subsystem can be presented as a controller for other subsystems.

Euler–Bernoulli beam equation with boundary energy dissipation is analyzed in the work [14], the problem is given as follows:

$$\left\{\begin{array}{ll}\rho y_{\mathit{tt}}+{\mathit{EIy}}_{\mathit{\text{xxxx}}}=0,& 0<x<1,\\ y\left(0,t\right)=y_x\left(0,t\right)=0& \\ -{\mathit{EIy}}_{\mathit{xxx}}\left(1,t\right)=-k_1^2{y}_t\left(1,t\right),& k_1\in \mathbb{R},\\ -{\mathit{EIy}}_{\mathit{xx}}\left(1,t\right)=k_2^2{y}_{\mathit{xt}}\left(1,t\right),& k_2\in \mathbb{R},\\ y\left(x,0\right)=y_0\left(x\right)y_t\left(x,0\right)=y_1\left(x\right),& 0\le x\le 1,\end{array}\right.$$E1

where $\rho$ denotes the mass density per unit length, $\mathit{EI}$ is the flexural rigidity coefficient. The authors extract some estimates of the resolvent operator on the imaginary axis by applying Huang’s¹ theorem to establish an exponential decay result.

For the asymptotic behavior of the wave equation, we introduce the following problem:

$$\left\{\begin{array}{ll}\frac{{\partial }^{2}w}{\partial t^2}-\mathrm{\Delta }w=0\text{in}& \mathrm{\Omega }\times \left(0,\infty \right),\\ w\left(x,t\right)=0\mathrm{on}& {\mathrm{\Gamma }}_0\times \left[0,\infty \right),\\ \frac{\partial w}{\partial \nu }+a\left(x\right)\frac{\partial w}{\partial t}=0\mathrm{on}& {\mathrm{\Gamma }}_1\times \left(0,\infty \right),\end{array}\right.$$E2

where $\nu$ is the unit normal of $\mathrm{\Gamma }$ pointing toward exterior of $\mathrm{\Omega }.$ The function $a\in C^1\left(\overline{{\mathrm{\Gamma }}_1}\right)$ with $a\left(x\right)\ge a_0>0$ on ${\mathrm{\Gamma }}_1.$ Problem (2) has been treated by Lagnese in [17], he used a multiplier method² and proved that the energy decay rate is obtained for solutions of wave type equations in a bounded region in ${\mathbb{R}}^n\left(n\ge 2\right)$ whose boundary consists partly of a nontrapping reflecting surface and partly of an energy absorbing surface. We can express this result, as follows:

$$E\left(t\right)\le f\left(t\right)E\left(0\right),t\ge 0,$$E3

with energy defined by

$$E\left(t\right)=\frac{1}{2}\left({‖w_t‖}_{L^2\left(\mathrm{\Omega }\right)}^2+{‖\nabla w‖}_{L^2\left(\mathrm{\Omega }\right)}^2\right).$$E4

The decay rate of solutions is a function $f\left(t\right)$ satisfying $f\left(t\right)\to 0$ as $t\to \infty .$ However, there are difficulties with some boundary condition problems, which makes the energy multiplier method ineffective in proving the exponential stability property.

Wazwaz [18], used the variational iteration method³ for the study of both linear and nonlinear Schrödinger equations, these problem is governed by the following equations:

$$\left\{\begin{array}{ll}{u}_t+{\mathit{iu}}_{\mathit{xx}}=0,& \\ u\left(x,0\right)=f\left(x\right),& i^2=-1\end{array}\right.$$E5

and

$$\left\{\begin{array}{l}{\mathit{iu}}_t+u_{\mathit{xx}}+\gamma {\left|u\right|}^{2r}u=0,r\ge 1,\\ u\left(x,0\right)=f\left(x\right),i^2=-1.\end{array}\right.$$E6

The variational iteration method was used to give rapid convergent successive approximations as well as to treat linear and non-linear problems in a uniform manner.

1.1 Statement of the problem

In this work, we consider stabilization for a Schrödinger equation through a boundary feedback dynamic controller interacted by an Euler–Bernoulli beam equation with Kelvin-Voigt damping⁴, the system is described by the following coupled partial differential equations:

$$\left\{\begin{array}{ll}{\partial }_t^{2}u+{\partial }_x^{4}u+\beta {\partial }_x^4{\partial }_{t}u=0,& 0<x<1,t>0,\\ {\partial }_{t}v+i{\partial }_x^{2}v=0,& 0<x<1,t>0,\end{array}\right.$$E7

boundary conditions are given by

$$\left\{\begin{array}{ll}u\left(1,t\right)={\partial }_{x}u\left(0,t\right)={\partial }_x^{2}u\left(1,t\right)=v\left(1,t\right)=0,& t\ge 0,\\ v\left(0,t\right)=\alpha {\partial }_{t}u\left(0,t\right),& t\ge 0,\\ \beta {\partial }_x^3{\partial }_{t}u\left(0,t\right)+{\partial }_x^{3}u\left(0,t\right)=-\mathit{\alpha i}{\partial }_{x}v\left(0,t\right),& t\ge 0,\end{array}\right.$$E8

the problem is associated with the following initial conditions:

$$u\left(x,0\right)=u_0\left(x\right),{\partial }_{t}u\left(x,0\right)=u_1\left(x\right),v\left(x,0\right)=v_0\left(x\right),0\le x\le 1.$$E9

1.2 Energy space

Initial condition (9) is in the following phase space:

$$\mathcal{H}=H_{\ast }^2\left(0,1\right)\times L^2\left(0,1\right)\times L^2\left(0,1\right),$$E10

where

$$H_{\ast }^2\left(0,1\right)=\left\{s|s\in H^2\left(0,1\right),{\partial }_{x}s\left(0\right)=s\left(1\right)=0\right\}.$$

1.3 Energies

The energy is the sum of the potential energy and the kinetic energy, given by

$$E\left(t\right)=\frac{1}{2}\left({‖u_t‖}_{L^2\left(0,1\right)}^2+{‖{\partial }_x^{2}u‖}_{L^2\left(0,1\right)}^2+{‖v‖}_{L^2\left(0,1\right)}^2\right).$$E11

Then, we have

$$\frac{d}{\mathit{dt}}E\left(t\right)=-\beta {‖{\partial }_x^2{\partial }_{t}u‖}_{L^2\left(0,1\right)}^2.$$E12

It is clear that $E\left(t\right)$ is nonincreasing with time.

1.4 Remark

1. The energy dissipation is related to the wave equation, that is, there are no explicit terms for a part of the Schrödinger subsystem.

2. We note that the weakness of the boundary connections for problems (7)–(9) lead to a complicated problem in stability analysis.

3. If we take the $\beta$ coefficient equal to zero in Eq. (12), the system becomes conservative.

1.5 Notations

1. ${⟨\cdot ,\cdot ⟩}_{L^2\left(0,1\right)}$ is the $L^2\left(0,1\right)-$inner product and ${‖\cdot ‖}_{L^2\left(0,1\right)}$ is the $L^2\left(0,1\right)-$norm.

2. The symbols $\Re \left(s\right)$ and$\Im \left(s\right)$ indicate the real part and the imaginary of a complex number $s.$

3. ${\left(s\right)}^T$ represents the transposed vector of $\left(s\right).$

2. Well-posedness

2.1 Setting of the semigroup

Setting $z={\left(u,{\partial }_{t}u=w,v\right)}^T$. Then, we introduce the norm in the Hilbert space $\mathcal{H}$ as follows:

$$\begin{array}{c}{‖z‖}_{\mathcal{H}}^2={‖u_t‖}_{L^2\left(0,1\right)}^2+{‖{\partial }_x^{2}u‖}_{L^2\left(0,1\right)}^2+{‖v‖}_{L^2\left(0,1\right)}^2\\ =2E\left(t\right),\end{array}$$E13

for $z_1,z_2\in \mathcal{H}$, the norm (13) is induced by the following inner product

$${⟨z_1,z_2⟩}_{L^2\left(0,1\right)}={⟨w_1,w_2⟩}_{L^2\left(0,1\right)}+{⟨{\partial }_x^2{u}_1,{\partial }_x^2{u}_2⟩}_{L^2\left(0,1\right)}+{⟨v_1,v_2⟩}_{L^2\left(0,1\right)}.$$E14

System (7) can be written as an abstract Cauchy problem in the phase space (10) as follows:

$$\left\{\begin{array}{l}\frac{d}{\mathit{dt}}z=\mathcal{A}z,t>0,\\ z\left(0\right)=z_0.\end{array}\right.$$E15

The solution at time $t>0$ to problem (15) can be written as:

$$z\left(t\right)=S\left(t\right)z_0=e^{t\mathcal{A}}{z}_0,$$

where the operator $\mathcal{A}:\mathcal{D}\left(\mathcal{A}\right)\subset \mathcal{H}\to \mathcal{H}$ is given by

$$\mathcal{A}z=\left(\begin{array}{c}w\\ -{\partial }_x^2\left({\partial }_x^{2}u+\beta {\partial }_x^{2}w\right)\\ -i{\partial }_x^{2}v\end{array}\right),$$E16

with domain

$$\mathcal{D}\left(\mathcal{A}\right)=\left\{z\in \mathcal{H},\mathcal{A}z\in \mathcal{H}\left|\begin{array}{c}{\partial }_x^{2}u+\beta {\partial }_x^{2}w\in H^2\left(0,1\right),\\ u\left(1\right)={\partial }_{x}u\left(0\right)={\partial }_x^{2}u\left(1\right)=v\left(1\right)=0,\\ v\left(0\right)=\mathit{\alpha w}\left(0\right),\\ \beta {\partial }_x^{3}w\left(0\right)+{\partial }_x^{3}u\left(0\right)=-\mathit{\alpha i}{\partial }_{x}v\left(0\right),\end{array}\right.\right\},$$E17

Theorem 1.1: Let $\mathcal{A}$ defined by (16). Then, ${\mathcal{A}}^{-1}$ exists and $\mathcal{A}$ generates a $C_0$-semigroup of contractions on $\mathcal{H}$.

Proof: We use the semigroup method, we shall show that:

1. The operator $\mathcal{A}$ is dissipative.

2. The operator $I_d-\mathcal{A}$ is onto ($I_d$ is the identity operator).

For the proof of (1). Firstly, we have $\mathcal{D}\left(\mathcal{A}\right)$ is dense in $\mathcal{H}$, that is,

$$\overline{\mathcal{D}\left(\mathcal{A}\right)}=\mathcal{H}.$$E18

Secondly, by applying the scalar product in the Hilbert space $\mathcal{H}$, we obtain

$$\begin{array}{l}{⟨\mathcal{A}z,z⟩}_{\mathcal{H}}={⟨{\partial }_x^{2}w,{\partial }_x^2\overline{u}⟩}_{L^2\left(0,1\right)}-{⟨{\partial }_x^2\left({\partial }_x^{2}u+\beta {\partial }_x^{2}w\right),\overline{w}⟩}_{L^2\left(0,1\right)}-{⟨i{\partial }_x^{2}v,\overline{v}⟩}_{L^2\left(0,1\right)}\\ ={⟨{\partial }_x^{2}w,{\partial }_x^2\overline{u}⟩}_{L^2\left(0,1\right)}+\left({\partial }_x^{3}u\left(0\right)+\beta {\partial }_x^{3}w\left(0\right)\right)\overline{w}\left(0\right)\\ +i{\partial }_{x}v\left(0\right)\overline{v}\left(0\right)+{⟨i{\partial }_{x}v,{\partial }_x\overline{v}⟩}_{L^2\left(0,1\right)}-{⟨{\partial }_x^{2}u+\beta {\partial }_x^{2}w,{\partial }_x^2\overline{w}⟩}_{L^2\left(0,1\right)}.\end{array}$$E19

By using boundary conditions (8), we get

$$\Re {⟨\mathcal{A}z,z⟩}_{\mathcal{H}}=-\beta {‖{\partial }_x^{2}w‖}_{L^2\left(0,1\right)}^2\le 0.$$E20

Then, the density property (18) and inequality (20) show that $\mathcal{A}$ is dissipative.

For the proof of (2), we shall solve the equation

$$\mathcal{A}z=F$$

for any $F={\left(f_1,f_2,f_3\right)}^T\in \mathcal{H}$, we can express the equation as follows:

$$\left\{\begin{array}{l}w=f_1,\\ {\partial }_x^2\left({\partial }_x^{2}u+\beta {\partial }_x^{2}w\right)=-f_2,\\ i{\partial }_x^{2}v=-f_3\end{array}\right.$$E21

By using the first equation of (21), we get

$$\left\{\begin{array}{l}{\partial }_x^{4}u=-f_2+\beta {\partial }_x^4{f}_1,\\ {\partial }_x^{2}v={\mathit{\text{if}}}_3.\end{array}\right.$$E22

We solve the following equation for the function $v$,

$$\left\{\begin{array}{l}{\partial }_x^{2}v={\mathit{\text{if}}}_3,\\ v\left(1\right)=0,v\left(0\right)=\alpha f_1\left(0\right),\end{array}\right.$$E23

to obtain

$$\left\{\begin{array}{l}v={\partial }_{x}v\left(0\right)x+i{\int }_0^x\left(x-y\right)f_3\left(y\right)\mathit{dy}+\alpha f_1\left(0\right),\\ {\partial }_{x}v\left(0\right)=-i{\int }_0^1\left(1-y\right)f_3\left(y\right)\mathit{dy}-\alpha f_1\left(0\right).\end{array}\right.$$E24

For $u$, we solve

$$\left\{\begin{array}{l}{\partial }_x^{4}u=-f_2+\beta {\partial }_x^4{f}_1,\\ u\left(1\right)={\partial }_{x}u\left(0\right)={\partial }_x^{2}u\left(1\right)=0,\\ \beta {\partial }_x^{3}w+{\partial }_x^{3}u\left(0\right)=-\mathit{i\alpha }{\partial }_{x}v\left(0\right),\end{array}\right.$$E25

to obtain

$$\left\{\begin{array}{l}u=-{\int }_0^x\left(1-x\right)g\left(y\right)\mathit{dy}-{\int }_x^1\left(1-y\right)g\left(y\right)\mathit{dy},\\ g\left(x\right)=\beta \left({\partial }_x^2{f}_1\left(1\right)-{\partial }_x^2{f}_1\left(x\right)\right)+{\int }_0^x\left(1-x\right)f_2\left(y\right)\mathit{dy}\\ +{\int }_x^1\left(1-y\right)f_2\left(y\right)\mathit{dy}+\mathit{i\alpha }{\partial }_{x}v\left(0\right)\left(1-x\right).\end{array}\right.$$E26

Eqs. (24) and (26) give a unique $z\in \mathcal{D}\left(\mathcal{A}\right)$ satisfying $\mathcal{A}z=F$.

It is easy to check that ${\mathcal{A}}^{-1}$ is bounded, that is,

$$0\in \rho \left(\mathcal{A}\right).$$

Therefore, the operator $\mathcal{A}$ generates a $C_0$-semigroup of contractions on $\mathcal{H}$ by the Lumer–Philips theorem [22].

3. Spectral analysis

We consider the following eigenvalue problem for the system operator $\mathcal{A}$. Let $\mathcal{A}z=\mathit{\lambda z}.$ Then, we have

$$\left\{\begin{array}{l}w=\mathit{\lambda u},\\ {\partial }_x^2\left({\partial }_x^{2}u+\beta {\partial }_x^{2}w\right)=-\mathit{\lambda w},\\ {\partial }_x^{2}v=\mathit{i\lambda v},\\ u\left(1\right)={\partial }_{x}u\left(0\right)={\partial }_x^{2}u\left(1\right)=v\left(1\right)=0,\\ \mathit{\alpha \lambda u}\left(0\right)=v\left(0\right),\\ \left(1+\mathit{\beta \lambda }\right){\partial }_x^{3}u\left(0\right)=-\mathit{i\alpha }{\partial }_{x}v\left(0\right).\end{array}\right.$$E27

The first and second equations of system (27) give the following system

$$\left\{\begin{array}{l}\left(1+\mathit{\beta \lambda }\right){\partial }_x^{4}u+{\lambda }^{2}u=0,\\ {\partial }_x^{2}v=\mathit{i\lambda v},\\ u\left(1\right)={\partial }_{x}u\left(0\right)={\partial }_x^{2}u\left(1\right)=v\left(1\right)=0,\\ \mathit{\alpha \lambda u}\left(0\right)=v\left(0\right),\\ \left(1+\mathit{\beta \lambda }\right){\partial }_x^{3}u\left(0\right)=-\mathit{i\alpha }{\partial }_{x}v\left(0\right).\end{array}\right.$$E28

Lemma

For any $\lambda \in {\sigma }_p\left(\mathcal{A}\right)$, it holds

$$\Re \left(\lambda \right)<0.$$E29

Proof: By Theorem 1.1, we have $\Re \left(\lambda \right)\le 0.$⁵ Letting $0\ne \lambda \in {\sigma }_p\left(\mathcal{A}\right)$ with $\Re \left(\lambda \right)=0$ and $z\in \mathcal{D}\left(\mathcal{A}\right)$ satisfying

$$\mathcal{A}z=\mathit{\lambda z}.$$E30

By using inequality 20, it follows that

$$0=\Re \left(\lambda \right){‖z‖}_{\mathcal{H}}^2=\Re {⟨\mathcal{A}z,z⟩}_{\mathcal{H}}=-\beta {‖{\partial }_x^{2}w‖}_{L^2\left(0,1\right)}^2.$$E31

From Eq. (31) and boundary conditions (28)₃, we have $w=0.$

From (27)₁ we have $u=0.$ Moreover, Eq. (30) gives

$$\left\{\begin{array}{l}{\partial }_x^{2}v=\mathit{i\lambda v},\\ v\left(0\right)=v\left(1\right)={\partial }_{x}v\left(0\right)=0.\end{array}\right.$$E32

It is easy to check that the above equation has only a trivial null solution $v=0.$ Hence, $z=0,$ and all the points that are located on the imaginary axis are not eigenvalues of $\mathcal{A}.$ Then the proof is completed.

Setting $\lambda ={\rho }^2$ in (28), when $1+{\mathit{\beta \rho }}^2\ne 0$, we obtain

$$\left\{\begin{array}{l}{\partial }_x^{4}u=\frac{-{\rho }^4}{1+{\mathit{\beta \rho }}^2}u,\\ {\partial }_x^{2}v=i{\rho }^{2}v,\\ u\left(1\right)={\partial }_{x}u\left(0\right)={\partial }_x^{2}u\left(1\right)=v\left(1\right)=0,\\ {\mathit{\alpha \rho }}^{2}u\left(0\right)=v\left(0\right),\\ \left(1+{\mathit{\beta \rho }}^2\right){\partial }_x^{3}u\left(0\right)=-\mathit{i\alpha }{\partial }_{x}v\left(0\right).\end{array}\right.$$E33

Let

$$a=\sqrt[4]{\frac{-{\lambda }^2}{1+\mathit{\beta \lambda }}}.$$

Then, the general solution of system (33) can be expressed as follows:

$$\begin{array}{l}u=c_{1}exp\left(\mathit{ax}\right)+c_{2}exp\left(-\mathit{ax}\right)+c_{3}exp\left(\mathit{iax}\right)+c_{4}exp\left(-\mathit{iax}\right),\\ v=d_{1}exp\left(\sqrt{i}\mathit{\rho x}\right)+d_{2}exp\left(-\sqrt{i}\mathit{\rho x}\right).\end{array}$$E34

By the boundary conditions of (33), we obtain that the constants $c_1,\cdots ,c_4$ and $d_1,d_2$ are not identical to zero if and only if $det\left(X\right)=0$, where

$$X=\left(\begin{array}{cccccc}{e}^a& e^{-a}& e^{\mathit{ia}}& e^{-\mathit{ia}}& 0& 0\\ a^2{e}^a& a^2{e}^{-a}& -a^2{e}^{\mathit{ia}}& -a^2{e}^{-\mathit{ia}}& 0& 0\\ a& -a& \mathit{ia}& -\mathit{ia}& 0& 0\\ 0& 0& 0& 0& e^{\sqrt{i}\rho }& -e^{\sqrt{i}\rho }\\ {\mathit{\alpha \rho }}^2& {\mathit{\alpha \rho }}^2& {\mathit{\alpha \rho }}^2& {\mathit{\alpha \rho }}^2& -1& -1\\ a^3& -a^3& -{\mathit{ia}}^3& {\mathit{ia}}^3& \frac{i\sqrt{i}\mathit{\alpha \rho }}{{\mathit{\beta \rho }}^2+1}& -\frac{i\sqrt{i}\mathit{\alpha \rho }}{{\mathit{\beta \rho }}^2+1}\end{array}\right),$$E35

by using boundary conditions (8), we get

$$c_2=-e^{2a}{c}_1,c_4=-e^{2\mathit{ia}}{c}_3,d_2=-e^{2\sqrt{i}\rho }{d}_1.$$

Then, the solution can be expressed by

$$u=c_1\left(e^{\mathit{ax}}-e^{a\left(2-x\right)}\right)+c_3\left(e^{\mathit{iax}}-e^{\mathit{ia}\left(2-x\right)}\right),v=d_1\left(e^{\sqrt{i}\mathit{\rho x}}-e^{\sqrt{i}\rho \left(2-x\right)}\right),$$

where $c_1,c_3,d_1$ are determined by the remaining three boundary conditions of (36) that $det\left(X\right)=0$ if and only if $det\left(\stackrel{\sim }{X}\right)=0$, where

$$\stackrel{\sim }{X}=\left(\begin{array}{ccc}1+e^{2a}& i+{\mathit{ie}}^{2\mathit{ia}}& 0\\ \left(1-e^{2a}\right){\mathit{\alpha \rho }}^2& \left(1-e^{2\mathit{ia}}\right){\mathit{\alpha \rho }}^2& -1+e^{2\sqrt{i}\rho }\\ a^3\left(1+e^{2a}\right)& -{\mathit{ia}}^3\left(1+e^{2\mathit{ia}}\right)& \frac{i\sqrt{i}\mathit{\alpha \rho }}{{\mathit{\beta \rho }}^2+1}\left(1+e^{2\sqrt{i}\rho }\right)\end{array}\right).$$E36

We recall the result of Lemma (29) and in light of this, we know that all eigenvalues have negative real parts. Thus, we only consider those $\lambda$ that lie in the second and third quadrants of the complex plane:

$$S≔\left\{\rho \in \mathbb{C}|\frac{\pi }{4}\le arg\rho \le \frac{3\pi }{4}\right\}.$$

Denote the region $S≔S_1\cup S_2\cup S_3$ such that

$$\begin{array}{l}{S}_1=\left\{\rho \in \mathbb{C}|\frac{\pi }{4}\le arg\rho \le \frac{3\pi }{8}\right\},\\ S_2=\left\{\rho \in \mathbb{C}|\frac{3\pi }{8}\le arg\rho \le \frac{5\pi }{8}\right\},\\ S_3=\left\{\rho \in \mathbb{C}|\frac{5\pi }{8}\le arg\rho \le \frac{3\pi }{4}\right\},\end{array}$$

the following theorem gives asymptotic distributions of the eigenvalues in $S_1,S_2$, and $S_3.$

Theorem 1.2: The eigenvalues of $\mathcal{A}$ have two families:

$${\sigma }_p\left(\mathcal{A}\right)=\left\{{\lambda }_{1n},n\in \mathbb{N}\right\}\cup \left\{{\lambda }_{2n}^{+},{\lambda }_{2n}^{-},n\in \mathbb{N}\right\},$$

where

$$\begin{array}{l}{\lambda }_{1n}={\mathit{\text{in}}}^2{\pi }^2+\frac{\sqrt{2}{\alpha }^2}{\sqrt[4]{\beta }}{e}^{\frac{5\mathit{i\pi }}{8}}\sqrt{\mathit{n\pi }}-\frac{{\alpha }^4}{\sqrt{\beta }}{e}^{\frac{\mathit{i\pi }}{4}}+O\left(n^{\frac{-1}{2}}\right),\\ {\lambda }_{2n}^{+}=-\beta {\left(\mathit{n\pi }-\frac{\pi }{2}\right)}^4+4\sqrt{\mathit{i\beta }}{\alpha }^2{\left(\mathit{n\pi }-\frac{\pi }{2}\right)}^2-2\sqrt{2i}{\alpha }^4\left(\mathit{n\pi }-\frac{\pi }{2}\right)\\ +\left(6i{\mathit{\pi \alpha }}^4-\frac{2\sqrt{i}{\alpha }^6}{\sqrt{\beta }}\right)+O\left(\frac{1}{n}\right),\\ {\lambda }_{2n}^{-}=-\frac{1}{\beta }-\frac{1}{{\beta }^3{\left(\mathit{n\pi }-\frac{\pi }{2}\right)}^4}+O\left(\frac{1}{n^8}\right).\end{array}$$E37

Therefore, we have

$$\Re \left({\lambda }_{1n}\right),\Re \left({\lambda }_{2n}^{+}\right)\to -\infty ,\Re \left({\lambda }_{2n}^{-}\right)\to -\frac{1}{\beta }\mathrm{as}n\to \infty .$$

Proof: When $\rho \in S_1,$ it has

$$\Re \left(\sqrt{i}\rho \right)=\mid \rho \mid cos\left(arg\left(\rho +\frac{\pi }{4}\right)\right)\le 0.$$

Since

$$a=\sqrt[4]{\frac{-{\lambda }^2}{1+\mathit{\beta \lambda }}}=\sqrt[4]{\frac{-{\rho }^4}{1+{\mathit{\beta \rho }}^2}}=\frac{\sqrt{\mathit{i\rho }}}{\sqrt[4]{\beta }}+O\left({\left|\rho \right|}^{-\frac{3}{2}}\right)\mathrm{as}\mid \rho \mid \to \infty .$$E38

Based on estimate (38), we can state that there is a positive constant ${\gamma }_1$ such that

$$\begin{array}{l}-\Re \left(a\right)=-\frac{\sqrt{\mid \rho \mid }}{\sqrt[4]{\beta }}cos\left(arg\left(\sqrt{\rho }+\frac{\pi }{4}\right)\right)\le -\frac{\sqrt{\mid \rho \mid }}{\sqrt[4]{\beta }}sin\left(\frac{\pi }{16}\right)<-{\gamma }_1\sqrt{\mid \rho \mid },\\ \Re \left(\mathit{ia}\right)=\frac{\sqrt{\mid \rho \mid }}{\sqrt[4]{\beta }}cos\left(arg\left(\sqrt{\rho }+\frac{3\pi }{4}\right)\right)\le -\frac{\sqrt{\mid \rho \mid }}{\sqrt[4]{\beta }}cos\left(\frac{\pi }{8}\right)<-{\gamma }_1\sqrt{\mid \rho \mid }.\end{array}$$

Therefore, we get the following estimates

$$\mid e^{-a}\mid =O\left(e^{-{\gamma }_1\sqrt{\mid \rho \mid }}\right),\mid e^{\mathit{ia}}\mid =O\left(e^{-{\gamma }_1\sqrt{\mid \rho \mid }}\right),\mid e^{\sqrt{i}\rho }\mid \le 1.$$E39

By multiplying some factors, we make each entry of the $det\left(\stackrel{\sim }{X}\right)$ be bounded as $\rho \to \infty$

$$\frac{1}{a^3{e}^{2a}}det\left(\stackrel{\sim }{X}\right)=\left|\begin{array}{ccc}1+e^{-2a}& i+{\mathit{ie}}^{2\mathit{ia}}& 0\\ \alpha e^{-2a}-\alpha & \alpha -\alpha e^{2\mathit{ia}}& -1+e^{2\sqrt{i}\rho }\\ 1+e^{-2a}& -i\left(1+e^{2\mathit{ia}}\right)& \frac{i\sqrt{i}{\mathit{\alpha \rho }}^3}{\left({\mathit{\beta \rho }}^2+1\right)a^3}\left(1+e^{2\sqrt{i}\rho }\right)\end{array}\right|.$$E40

By using the expression of $a$ and $\rho$, and the Taylor expansion, we obtain

$$\frac{i\sqrt{i}{\mathit{\alpha \rho }}^3}{\left({\mathit{\beta \rho }}^2+1\right)a^3}=\frac{\alpha }{\sqrt[4]{\beta }}\sqrt{\frac{1}{\rho }}+O\left({\left|\rho \right|}^{-\frac{5}{2}}\right).$$E41

By using Eqs. (41) and (39), we get

$$\begin{array}{l}\frac{1}{a^3{e}^{2a}}det\left(\stackrel{\sim }{X}\right)=\left|\begin{array}{ccc}1& i& 0\\ -\alpha & \alpha & -1+e^{2\sqrt{i}\rho }\\ 1& -i& \frac{\alpha }{\sqrt[4]{\beta }}\sqrt{\frac{1}{\rho }}\left(1+e^{2\sqrt{i}\rho }\right)\end{array}\right|+O\left({\left|\rho \right|}^{\frac{-5}{2}}\right)\\ =\left(\frac{\left(1+i\right){\alpha }^2}{\sqrt[4]{\beta }}\sqrt{\frac{1}{\rho }}-2i\right)+e^{2\sqrt{i}\rho }\left(\frac{\left(1+i\right){\alpha }^2}{\sqrt[4]{\beta }}\sqrt{\frac{1}{\rho }}+2i\right)\\ +O\left({\left|\rho \right|}^{\frac{-5}{2}}\right).\end{array}$$E42

From the previous equality, we can get $det\left(\stackrel{\sim }{X}\right)=0$ if and only if

$$e^{2\sqrt{i}\rho }=1-\frac{\left(1-i\right){\alpha }^2}{\sqrt[4]{\beta }}\sqrt{\frac{1}{\rho }}-\frac{i{\alpha }^4}{\sqrt{\beta }\rho }+O\left({\left|\rho \right|}^{\frac{-3}{2}}\right).$$E43

Suppose

$$2\sqrt{i}\rho =2\mathit{n\pi i}+O\left(n^{\frac{-1}{2}}\right),$$E44

where $n$ is a sufficiently large integer. Substituting Eq. (43) into Eq. (42), we arrive at

$$O\left(n^{\frac{-1}{2}}\right)=\frac{{\left(-1\right)}^{\frac{5}{8}}\sqrt{2}{\alpha }^2}{\sqrt[4]{\beta }\sqrt{\mathit{n\pi }}}-\frac{\sqrt{i}{\alpha }^4}{\mathit{n\pi }\sqrt{\beta }}+O\left(n^{\frac{-3}{2}}\right).$$E45

The roots of Eq. (42) have the following asymptotic expressions

$${\rho }_{1n}=\sqrt{i}\mathit{n\pi }+\frac{{\left(-1\right)}^{\frac{3}{8}}{\alpha }^2}{\sqrt[4]{\beta }\sqrt{2\mathit{n\pi }}}-\frac{{\alpha }^4}{2\mathit{n\pi }\sqrt{\beta }}+O\left(n^{\frac{-3}{2}}\right),n>N_1,$$E46

where $N_1$ is a sufficiently large positive integer. By $\lambda ={\rho }^2,$we have

$${\lambda }_{1n}={\mathit{\text{in}}}^2{\pi }^2+\frac{\sqrt{2}{\alpha }^2}{\sqrt[4]{\beta }}{e}^{\frac{\mathit{i\pi }}{4}}+O\left(n^{-\frac{1}{2}}\right).$$

By using the value of $a$ given by Eq. (38), we can obtain the expression of $a$ as follows:

$$a_{1n}=\frac{{\left(-1\right)}^{\frac{3}{8}}\sqrt{\mathit{\pi n}}}{\sqrt[4]{\beta }}+O\left(\frac{1}{n}\right).$$E47

Similarly, when $\rho \in S_2,$it is easier to verify that there exists a ${\gamma }_2>0$ such that

$$\left\{\begin{array}{l}\Re \left(\mathit{ia}\right)\le -{\gamma }_2\sqrt{\mid \rho \mid },\\ \Re \left(\sqrt{i}\rho \right)=\mid \rho \mid cos\left(arg\left(\rho +\frac{\pi }{4}\right)\right)\le \mid \rho \mid cos\left(\frac{5\pi }{8}\right).\end{array}\right.$$

Hence, we get the following estimations

$$\mid e^{\mathit{ia}}\mid =O\left(e^{-{\gamma }_2\sqrt{\mid \rho \mid }}\right),\mid e^{\sqrt{i}\rho }\mid =O\left(e^{-{\gamma }_2\mid \rho \mid }\right),$$

by using Eq. (38), we obtain

$$arg\left(a\right)=arg\left(\sqrt{\mathit{i\rho }}\right)\in \left(\frac{7\pi }{16},\frac{9\pi }{16}\right]\text{in}{S}_2.$$

Thus, the sign of $a$ is different under the two conditions:

$$arg\left(\rho \right)\in \left(\frac{7\pi }{16},\frac{\pi }{2}\right]\text{and}arg\left(\rho \right)\in \left(\frac{\pi }{2},\frac{9\pi }{16}\right].$$

Therefore, we conclude that

$$\begin{array}{c}\frac{1}{a^3{e}^a}det\left(\stackrel{\sim }{X}\right)=\left|\begin{array}{ccc}{e}^{-a}+e^a& i+{\mathit{ie}}^{2\mathit{ia}}& 0\\ e^{-a}\alpha -e^a\alpha & \alpha -\alpha e^{2\mathit{ia}}& -1+e^{2\sqrt{i}\rho }\\ e^{-a}+e^a& -i\left(1+e^{2\mathit{ia}}\right)& \frac{i\sqrt{i}{\mathit{\alpha \rho }}^3}{\left({\mathit{\beta \rho }}^2+1\right)a^3}\left(1+e^{2\sqrt{i}\rho }\right)\end{array}\right|\\ =\left|\begin{array}{ccc}{e}^{-a}+e^a& i& 0\\ e^{-a}\alpha -e^a\alpha & \alpha & -1\\ e^{-a}+e^a& -i& \frac{i\sqrt{i}{\mathit{\alpha \rho }}^3}{\left({\mathit{\beta \rho }}^2+1\right)a^3}\end{array}\right|+O\left(e^{-{\gamma }_2\sqrt{\mid \rho \mid }}\right)\\ =e^a\left(\frac{\sqrt{2}a{\alpha }^2}{\rho }-2i\right)-e^{-a}\left(\frac{\sqrt{2}\mathit{ia}{\alpha }^2}{\rho }+2i\right)+O\left(e^{-{\gamma }_2\sqrt{\mid \rho \mid }}\right).\end{array}$$

From the previous equality, it is seen that $det\left(\stackrel{\sim }{X}\right)=0$ if and only if

$$e^a\left(\frac{\sqrt{2}a{\alpha }^2}{\rho }-2i\right)-e^{-a}\left(\frac{\sqrt{2}\mathit{ia}{\alpha }^2}{\rho }+2i\right)+O\left(e^{-{\gamma }_2\sqrt{\mid \rho \mid }}\right)=0.$$E48

By using the expression of $a$ and $\rho$, we obtain

$$\rho =\sqrt{\beta }{a}^2-\frac{1}{2{\beta }^{\frac{3}{2}}{a}^2}+O\left(\frac{1}{{\left|a\right|}^4}\right),$$E49

which shows that $\mid a\mid ,\mid \rho \mid \to \infty$ at the same time. Now, substitute the value of $\rho$ given by (48) into equality (47), and we obtain

$$\begin{array}{l}{e}^a\left(-2i+\frac{\sqrt{2}{\alpha }^2}{\sqrt{\beta }a}+\frac{\sqrt{2}{\alpha }^2}{2{\beta }^{\frac{5}{2}}{a}^5}+O\left({\left|a\right|}^{-7}\right)\right)-e^{-a}\left(2i+\frac{i\sqrt{2}{\alpha }^2}{\sqrt{\beta }a}+\frac{i\sqrt{2}{\alpha }^2}{2{\beta }^{\frac{5}{2}}{a}^5}+O\left({\left|a\right|}^{-7}\right)\right)\\ +O\left(e^{-{\gamma }_2\mid a\mid }\right)=0.\end{array}$$

Letting $a=x+\mathit{iy}$, it is easily checked that $\overline{a}=x-\mathit{iy}$ also satisfies the same asymptotic equation above. Hence, we only need to analyze the asymptotic expression of $a$ located in the second quadrant. Given the value of $a$ given by (48), when $a$ is located on the second quadrant, $\Re \left(-a\right)\le 0$ and $\mid e^{-a}\mid \le 1$. Therefore,

$$e^{-2a}=-1+\frac{\left(1-i\right){\alpha }^2}{\sqrt{2\beta }a}-\frac{\left(1-i\right){\alpha }^4}{2a^2\beta }+\frac{\left(1-i\right){\alpha }^6}{2\sqrt{2}{a}^3{\beta }^{\frac{3}{2}}}+O\left(\frac{1}{a^4}\right),$$

and for the quadrant where $a$ is located, we have

$$\begin{array}{l}{a}_{2n}=i\left(\mathit{n\pi }-\frac{\pi }{2}\right)+\frac{\left(1+i\right){\alpha }^2}{\sqrt{2\beta }\left(\mathit{n\pi }-\frac{\pi }{2}\right)}-\frac{\left(1-i\right){\alpha }^4}{2\beta {\left(\mathit{n\pi }-\frac{\pi }{2}\right)}^2}\\ -\frac{\left(1+i\right){\alpha }^6}{2\sqrt{2}{\beta }^{\frac{3}{2}}{\left(\mathit{n\pi }-\frac{\pi }{2}\right)}^3}+O\left(\frac{1}{n^4}\right).\end{array}$$

Since $a=\sqrt[4]{\frac{{\lambda }^2}{1+\mathit{\beta \lambda }}}$ or ${\lambda }^2-\beta a^4\lambda -a^4=0,$ it has

$${\lambda }_{2n}^{\pm }=\frac{\beta a^4}{2}\left(1\pm \sqrt{1+\frac{4}{{\beta }^2{a}^4}}\right).$$

Using the Taylor expansion, we obtain the expressions of ${\lambda }_{2n}^{+}$ and ${\lambda }_{2n}^{-}$ given by (37). Moreover, by using $\lambda ={\rho }^2$, we have the asymptotic expressions of ${\rho }_{2n}^{+}$ and ${\rho }_{2n}^{-}$

$$\left\{\begin{array}{l}{\rho }_{2n}^{+}=i\sqrt{\beta }{\left(\mathit{n\pi }-\frac{\pi }{2}\right)}^2+2\sqrt{i}{\alpha }^2+O\left(n^{-1}\right),\\ {\rho }_{2n}^{-}=\frac{i}{\sqrt{\beta }}+\frac{i}{2{\beta }^{\frac{5}{2}}{\left(\mathit{n\pi }-\frac{\pi }{2}\right)}^4}+O\left(n^{-8}\right).\end{array}\right.$$E50

Similarly, in $S_3$, there exists ${\gamma }_3>0$ such that

$$\mid e^a\mid =O\left(e^{-{\gamma }_3\sqrt{\mid \rho \mid }}\right),\mid e^{\mathit{ia}}\mid =O\left(e^{-{\gamma }_3\sqrt{\mid \rho \mid }}\right),\mid e^{\sqrt{i}\rho }\mid =O\left(e^{-{\gamma }_3\mid \rho \mid }\right).$$

It is easy to check that there is no null point of $det\left(\stackrel{\sim }{X}\right)$, namely, there is no point spectrum in $S_3.$

According to the conclusion of Theorem 1.2, it is obvious that $-\frac{1}{\beta }$ is an accumulation point of the point spectrum of the operator $\mathcal{A}$. We thus have the following corollary.

Corollary

$${\sigma }_c\left(\mathcal{A}\right)=-\frac{1}{\beta }.$$E51

We next analyze the asymptotic expression of eigenfunctions of the operator $\mathcal{A}.$

Theorem 1.3: Let ${\sigma }_p\left(\mathcal{A}\right)=\left\{{\lambda }_{1n},n\in \mathbb{N}\right\}\cup \left\{{\lambda }_{2n}^{+},{\lambda }_{2n}^{-},n\in \mathbb{N}\right\}$ be the point spectrum of $\mathcal{A}$. Let ${\lambda }_{1n}={\rho }_{1n}^2,{\lambda }_{2n}^{+}={\left({\rho }_{2n}^{+}\right)}^2$ and ${\lambda }_{2n}^{-}={\left({\rho }_{2n}^{-}\right)}^2$ with ${\rho }_{1n},{\rho }_{2n}^{+}$ and ${\rho }_{2n}^{-}$ being given by Eqs. (45) and (49), respectively. Then, there are three families of approximated normalized eigenfunctions of $\mathcal{A}$

1. One family $\left\{z_{1n}=\left(u_{1n},\lambda u_{1n},v_{1n}\right),n\in \mathbb{N}\right\}$, where $z_{1n}$ is the eigenfunction of $\mathcal{A}$ corresponding to the eigenvalue ${\lambda }_{1n}$, has the following asymptotic expression:

   $$\left({\partial }_x^2{u}_{1n},\lambda u_{1n},v_{1n}\right)=\left(0,0,sin\left[a_n\left(1-x\right)\right]\right)+O_x\left(n^{\frac{-3}{4}}\right),$$E52

   where

   $$a_n=\mathit{n\pi }+\frac{{\left(-1\right)}^{\frac{1}{8}}{\alpha }^2}{\sqrt[4]{\beta }\sqrt{2\mathit{n\pi }}}+O\left(n^{-1}\right)$$E53

   and $O_x\left(n^{\frac{-3}{4}}\right)$ means that ${‖O_x\left(n^{\frac{-3}{4}}\right)‖}_{L^2\left(0,1\right)}=O\left(n^{\frac{-3}{4}}\right).$

2. The second family $\left\{z_{2n}^{+}=\left(u_{2n}^{+},\lambda u_{2n}^{+},v_{2n}^{+}\right),n\in \mathbb{N}\right\}$, where $z_{2n}^{+}$ is the eigenfunction of $\mathcal{A}$ corresponding to the eigenvalue ${\lambda }_{2n}^{+}$, has the following asymptotic expression:

   $$\left({\partial }_x^2{u}_{2n}^{+},\lambda u_{2n}^{+},v_{2n}^{+}\right)=\left(0,sin\left[\left(\mathit{n\pi }-\frac{\pi }{2}\right)\left(1-x\right)\right],0\right)+O_x\left(n^{-1}\right).$$E54

3. The third family $\left\{z_{2n}^{-}=\left(u_{2n}^{-},\lambda u_{2n}^{-},v_{2n}^{-}\right),n\in \mathbb{N}\right\}$, where $z_{2n}^{-}$ is the eigenfunction of $\mathcal{A}$ corresponding to the eigenvalue ${\lambda }_{2n}^{-}$, has the following asymptotic expression:

   $$\left({\partial }_x^2{u}_{2n}^{-},\lambda u_{2n}^{-},v_{2n}^{-}\right)=\left(sin\left[\left(\mathit{n\pi }-\frac{\pi }{2}\right)\left(1-x\right)\right]\mathrm{,0,0}\right)+O\left(n^{-1}\right).$$E55

The proof is limited to the first result declared in Theorem 1.3.

Proof: We look for $z_{1n}$ associated with ${\lambda }_{1n}$. From the expression ${\rho }_{1n}$ given by (45) and $a_{1n}$ given by (46) we have

$$\left\{\begin{array}{l}{e}^{-a_{1n}y}=e^{\frac{{\left(-1\right)}^{\frac{-11}{8}}\sqrt{\mathit{\pi x}}y}{\sqrt[4]{\beta }}+O\left(n^{-1}\right)},e^{{\mathit{ia}}_{1n}y}=e^{\frac{{\left(-1\right)}^{\frac{-7}{8}}\sqrt{\mathit{\pi x}}y}{\sqrt[4]{\beta }}+O\left(n^{-1}\right)},\\ e^{\pm \sqrt{i}{\rho }_{1n}\left(1-x\right)}=e^{\pm \mathit{in\pi }\left(1-x\right)+O\left(n^{-1}\right)},\end{array}\right.$$E56

and the following estimations:

$$\begin{array}{c}‖e^{-a_{1n}y}‖=O\left(n^{\frac{-1}{4}}\right),‖e^{{\mathit{ia}}_{1n}y}‖=O\left(n^{\frac{-1}{4}}\right);\\ ‖e^{\pm \sqrt{i}{\rho }_{1n}\left(1-x\right)}‖=O\left(1\right),\end{array}$$

where $y=x$ or $2-x\in \left[0,1\right].$ According to the matrix $\stackrel{\sim }{X}$ given by (36), for $\rho$ with (45) and $a_{1n}$ given by (46), we obtain

$$\begin{array}{c}{u}_1=\frac{1}{e^{2a}{e}^{\sqrt{i}\rho }}\left|\begin{array}{ccc}1+e^{2a}& i+{\mathit{ie}}^{2\mathit{ia}}& 0\\ \left(1-e^{2a}\right){\mathit{\alpha \rho }}^2& \left(1-e^{2\mathit{ia}}\right){\mathit{\alpha \rho }}^2& -1+e^{2\sqrt{i}\rho }\\ e^{\mathit{ax}}-e^{a\left(2-x\right)}& e^{\mathit{iax}}-e^{\mathit{ia}\left(2-x\right)}& 0\end{array}\right|\\ =\frac{e^{-\sqrt{i}\rho }-e^{\sqrt{i}}\rho }{{\rho }^2}\left|\begin{array}{cc}1+e^{-2a}& i+{\mathit{ie}}^{2\mathit{ia}}\\ e^{-a\left(2-x\right)}-e^{\mathit{ax}}& e^{\mathit{iax}}-e^{\mathit{ia}\left(2-x\right)}\end{array}\right|.\end{array}$$

By using estimates (39), we can write

$$\begin{array}{c}{u}_1=\frac{e^{-\sqrt{i}\rho }-e^{\sqrt{i}}\rho }{{\rho }^2}\left|\begin{array}{cc}1& i\\ e^{-a\left(2-x\right)}-e^{\mathit{ax}}& e^{\mathit{iax}}-e^{\mathit{ia}\left(2-x\right)}\end{array}\right|+O\left(e^{-{\gamma }_1\sqrt{\mid \rho \mid }}\right)\\ =\frac{1}{{\rho }^2}\left(e^{-\sqrt{i}\rho }-e^{\sqrt{i}\rho }\right)\left[\left(e^{\mathit{iax}}-e^{\mathit{ia}\left(2-x\right)}\right)-i\left(e^{-a\left(2-x\right)}-e^{-\mathit{ax}}\right)\right]+O\left(e^{-{\gamma }_1\sqrt{\mid \rho \mid }}\right).\end{array}$$

By the expression ${\rho }_{1n}$ given by (45), we can obtain

$$e^{-\sqrt{i}\rho }-e^{\sqrt{i}\rho }=-2isin\mathit{n\pi }+O\left(n^{\frac{-1}{2}}\right)=O\left(n^{\frac{-1}{2}}\right).$$

This together with estimates 77 gives, after a direct computation, that

$${\partial }_x^2{u}_1=\frac{a^2}{{\rho }^2}\left(e^{-\sqrt{i}\rho }-e^{\sqrt{i}\rho }\right)\left[\left(e^{\mathit{ia}\left(2-x\right)}-e^{\mathit{iax}}\right)-i\left(e^{-a\left(2-x\right)}-e^{-\mathit{ax}}\right)\right]+O\left(e^{-{\gamma }_1\sqrt{n}}\right)=O_x\left(n^{\frac{-7}{4}}\right),$$

and

$$\lambda u_1=\left(e^{-\sqrt{i}\rho }-e^{\sqrt{i}\rho }\right)\left[\left(e^{\mathit{iax}}-e^{\mathit{ia}\left(2-x\right)}\right)-i\left(e^{-a\left(2-x\right)}-e^{-\mathit{ax}}\right)\right]+O\left(e^{-{\gamma }_1\sqrt{n}}\right)=O_x\left(n^{\frac{-3}{4}}\right).$$

Here,

$$O_x\left(n^{\frac{-3}{4}}\right)\text{means\hspace{0.17em}that}{‖O_x\left(n^{\frac{-3}{4}}\right)‖}_{L^2\left(0,1\right)}=O\left(n^{\frac{-3}{4}}\right)\text{because}\parallel e^{-\mathit{ax}}\parallel =\parallel e^{\mathit{iax}}\parallel =O\left(n^{\frac{-1}{4}}\right).$$

Similarly, by using estimates (39) and (55), we have

$$\begin{array}{l}{v}_1=\frac{1}{e^{2a}{e}^{\sqrt{i}\rho }{\rho }^2}\left|\begin{array}{ccc}1+e^{2a}& i+{\mathit{ie}}^{2\mathit{ia}}& 0\\ \left(1-e^{2a}\right){\mathit{\alpha \rho }}^2& \left(1-e^{2\mathit{ia}}\right){\mathit{\alpha \rho }}^2& -1+e^{2\sqrt{i}\rho }\\ 0& 0& e^{\sqrt{i}\mathit{\rho x}}-e^{\sqrt{i}\rho \left(2-x\right)}\end{array}\right|\\ =-2\alpha \left(1+i\right)sin\left[a_n\left(1-x\right)\right]+O\left(e^{-{\gamma }_1\sqrt{n}}\right),\end{array}$$

where $a_n$ is given by (52). Let

$$z_{1n}=\frac{-1}{2\alpha \left(1+i\right)}{z}_1,$$

so, we obtain

$$\left({\partial }_x^2{u}_{1n},\lambda u_{1n},v_{1n}\right)=\left(0,0,sin\left[a_n\left(1-x\right)\right]\right)+O\left(n^{\frac{-3}{4}}\right).$$

The second and third results of Theorem 3 are obtained by the same procedure as before.

Corollary

$${\sigma }_r\ne \varnothing .$$E57

4. Riesz basis property

Lemma.(see [23])

Let ${\lambda }_n\in \mathbb{C},n=1,2,\cdots ,$be a sequence that satisfies ${sup}_n\mid \Im \left({\lambda }_n\right)\mid \le M$, where $M$ is a positive constant. Then the sine system $\left\{sin{\lambda }_{n}x,n\ge 1\right\}$ is a Riesz basis for $L^2\left(0,1\right)$ provided that the sequence ${\lambda }_n$ satisfies one of the following conditions:

$${sup}_n\mid \Re \left({\lambda }_n\right)-\mathit{n\pi }\mid <\frac{\pi }{4};$$

$${sup}_n\mid \Re \left({\lambda }_n\right)-\mathit{n\pi }+\frac{\pi }{2}\mid <\frac{\pi }{4}.$$

Lemma.(see [24])

Let $\mathcal{A}$ be a densely defined closed linear operator in a Hilbert space $\mathcal{H}$ with isolated eigenvalues ${\left\{{\lambda }_i\right\}}_{i=1}^{\infty }.$ Let ${\left\{{\varphi }_i\right\}}_{i=1}^{i=\infty }$ be a Riesz basis for $\mathcal{H}$. Suppose that there is an integer $N\ge 1$ and a sequence of generalized eigenvectors ${\left\{{\psi }_i\right\}}_{i=N}^{\infty }$ of $\mathcal{A}$ such that

$$\sum _{i=N}^{\infty }{‖{\psi }_i-{\varphi }_i‖}^2<\infty .$$

Then, there exists $M$ a number of generalized eigenvectors ${\left\{{\psi }_{i_0}\right\}}_{i=1}^M$ of $\mathcal{A}$ such that

$${\left\{{\psi }_{i_0}\right\}}_{i=1}^M\cup {\left\{{\psi }_i\right\}}_{i=M+1}^{\infty }$$

forms a Riesz basis for $\mathcal{H}$.

Theorem 1.4: The generalized eigenfunctions of $\mathcal{A}$ forms a Riesz basis for $\mathcal{H}$. As a result, all eigenvalues with large modules must be algebraically simple and, hence, the spectrum-determined growth condition holds for

$$e^{\mathcal{A}t}:\mathrm{\Phi }\left(\mathcal{A}\right)=S\left(\mathcal{A}\right)$$

where

$$\mathrm{\Phi }\left(\mathcal{A}\right)=inf\left\{\mathrm{\Phi }|\text{there\hspace{0.17em}exists\hspace{0.17em}an}M\text{such\hspace{0.17em}that}\parallel e^{\mathcal{A}t}\parallel \le {\mathit{Me}}^{\mathrm{\Phi }t}\right\},$$

and

$$S\left(\mathcal{A}\right)=sup\left\{\Re \left(\lambda \right)|\lambda \in \sigma \left(\mathcal{A}\right)\right\}.$$

Proof: By the bounded invertible mapping:

$$\mathbb{T}\left(u,w,v\right)=\left({\partial }_x^{2}u,w,v\right),$$

the space $\mathcal{H}$ is mapped onto

$$L^2\left(0,1\right)\times L^2\left(0,1\right)\times L^2\left(0,1\right).$$

The value of $a_n$ given by (52) satisfies

$$\underset{n}{sup}\mid \Im \left(a_n\right)\mid =sup\left|\frac{sin\frac{\pi }{8}{\alpha }^2}{\sqrt{2\mathit{n\pi }}\sqrt[4]{\beta }}\right|$$

is bounded and its real part satisfies

$$\underset{n}{sup}\left|\Re \left(a_n\right)-\mathit{n\pi }\right|=\underset{n}{sup}\left|\frac{cos\frac{\pi }{8}{\alpha }^2}{\sqrt{2\mathit{n\pi }\sqrt[4]{\beta }}}\right|\le \frac{\pi }{4}.$$

Then, it follows that the sequence

$$\left\{sin\left[a_n\left(1-x\right)\right],n=1,2,\cdots \right\},$$

forms a Riesz basis for $L^2\left(0,1\right)$. Similarly, the sequences

$$\left\{sin\left[\left(\mathit{n\pi }-\frac{\pi }{2}\right)\left(1-x\right)\right],n=1,2,\cdots \right\},$$

form a Riesz basis for $L^2\left(0,1\right).$

Let

$${\mathrm{\Psi }}_{1n}=\left(sin\left[a_n\left(1-x\right)\right]\mathrm{,0,0}\right),{\mathrm{\Psi }}_{2n}^{+}=\left(0,sin\left[\left(\mathit{n\pi }-\frac{\pi }{2}\right)\left(1-x\right)\right],0\right)$$

and

$${\mathrm{\Psi }}_{2n}^{-}=\left(0,0,sin\left[\left(\mathit{n\pi }-\frac{\pi }{2}\right)\left(1-x\right)\right]\right).$$

Then, the sequences

$${\left\{{\mathrm{\Psi }}_{1n}\right\}}_{n\ge 1}\cup {\left\{{\mathrm{\Psi }}_{2n}^{+}\right\}}_{n\ge 1}\cup {\left\{{\mathrm{\Psi }}_{2n}^{-}\right\}}_{n\ge 1}$$

forms a Riesz basis for the following space

$$L^2\left(0,1\right)\times L^2\left(0,1\right)\times L^2\left(0,1\right).$$

Therefore, by the expression of $z_{1n},z_{2n}^{+}$, and $z_{2n}^{-}$ given by (51), (53), and (54), respectively, this implies that there exists $N>0$ such that

$$\sum _{n\ge N}^{\infty }\left[{‖\mathbb{T}{z}_{1n}-{\mathrm{\Psi }}_{1n}‖}^2+{‖\mathbb{T}{z}_{2n}^{+}-{\mathrm{\Psi }}_{2n}^{+}‖}^2+{‖\mathbb{T}{z}_{2n}^{-}-{\mathrm{\Psi }}_{2n}^{-}‖}^2\right]\le \sum _{n\ge N}^{\infty }O\left(n^{-2}\right)<\infty .$$

This shows that there is a sequence of generalized eigenfunctions of $\mathcal{A}$, which forms a Riesz basis for $\mathcal{H}$, and all eigenvalues with large modulus must be algebraically simple.

5. Exponential stability

Theorem 1.5: The $C_0-$semigroup $S\left(t\right)$ generated by the operator $\mathcal{A}$ is exponentially stable, that is,

$$‖e^{\mathcal{A}t}‖\le {\mathit{Me}}^{\mathit{\omega t}},$$

where $M$ and $\omega$ are positive constants⁶.

Proof: By the asymptotic distribution of eigenvalues given by Theorem 1.2 and the continuous spectrum given by Eq. (50), in addition to the empty residual spectrum set given by Eq. (56), we conclude that $S\left(\mathcal{A}\right)=-\frac{1}{\beta }.$ The proof is completed by the spectrum-determined growth condition, which is similar to [24, 25, 26].

6. Conclusion

The main results of this work are similar to those mentioned in [27], the results are summarized as follows:

1. The system operator of the closed-loop system is not of compact resolvent and the spectrum consists of three branches.

2. By means of asymptotic analysis, the asymptotic expressions of eigenfunctions are obtained.

3. By the comparison method in the Riesz basis approach, exponential stability is obtained.

Conflict of interest

The authors declare no conflict of interest.