# Exponential stabilisation of a star-shaped network of fluid-conveying pipes
induced via vertex control

Xiao Xuan Feng1,3 ,  Gen Qi Xu2 and  Mahyar Mahinzaeim1,∗

###### Abstract.

In this paper we study the equation governing the small vibrations of a
stretched fluid-conveying pipe in a star-shaped network. We examine the
question exponential stabilisability of such a system by means of feedback
control torques applied at the inner vertex of the system. An approach to the
study of the stabilisability problem for the closed-loop system is presented
using the spectral approach previously introduced by the authors for star
graphs of beams. When the tension in the pipes is very large compared to
fluid-flow velocity, we establish a positive result that gives exponential
decay of the energy of solutions of the closed-loop system.

MSC2020: 37L15, 93D23, 37C10, 34B45, 35P10, 47B06
Keywords: pipe conveying fluid, pipe network, spectral problem on a metric
star graph, spectral analysis, Riesz basis property, exponential stability
1 Research Center for Complex Systems, Aalen University, Germany.
2 Department of Mathematics, Tianjin University, China.
3 School of Mathematics and Information Science, Hebei Normal University of
Science and Technology, China.
Emails<EMAIL_ADDRESS><EMAIL_ADDRESS><EMAIL_ADDRESS>
∗ Corresponding author.

##  1\. Introduction and system description

In recent years one very active area of mathematical systems theory has been
the investigation of control, or control related, problems in mechanics and
mathematical physics on networks or metric graphs. Much of the research has
focused on the problems of controllability and stabilisability of vibrating
elastic (string, beam, and plate) networks by insertion of control action into
the boundary and vertex conditions, and there are already adequate texts on
these subjects, e.g. [4, 9, 2, 17]. The reason for investigating such systems
is not difficult to explain: remarkable results are obtained in networked
elastic systems when control is exercised. For example, elastic systems that
are unstable can be stabilised by boundary or vertex control when a network
setup is considered, and vice versa. A conclusive answer as to whether or not
all vibrating elastic networks can be stabilised by such control action cannot
be expected, of course, and must be considered individually for each problem.
This is specially true when it comes to the treatment of flow-induced
oscillations problems on networks and is typical of the particular kind of
stabilisability problem we are interested in here.

In a recent paper [10], the authors have studied (among other things) the
stabilisation problem for a single vibrating fluid-conveying pipe, represented
by the partial differential equation

$\frac{\partial^{4}{w}\left(s,t\right)}{\partial
s^{4}}-(\gamma-\eta^{2})\,\frac{\partial^{2}{w}\left(s,t\right)}{\partial
s^{2}}+2\beta\eta\frac{\partial^{2}{w}\left(s,t\right)}{\partial s\partial
t}+\frac{\partial^{2}{w}\left(s,t\right)}{\partial t^{2}}=0$

which together with boundary conditions

$\displaystyle{w}\left(0,t\right)=\left.\frac{\partial^{2}{w}\left(s,t\right)}{\partial
s^{2}}\right|_{s=0}$ $\displaystyle=0,$
$\displaystyle\left.\left(\frac{\partial^{2}{w}\left(s,t\right)}{\partial
s^{2}}+\kappa\frac{\partial^{2}{w}\left(s,t\right)}{\partial s\partial
t}\right)\right|_{s=1}$ $\displaystyle=0,$
$\displaystyle\left.\left(\frac{\partial^{3}{w}\left(s,t\right)}{\partial
s^{3}}-(\gamma-\eta^{2})\,\frac{\partial{w}\left(s,t\right)}{\partial
s}\right)\right|_{s=1}$ $\displaystyle=0$

and given initial conditions ${w}\left(s,0\right)$,
$\left.\left(\partial{w}/\partial t\right)\left(s,t\right)\right|_{t=0}$ forms
what we called in [10] a closed-loop system. Here ${w}\left(s,t\right)$ for
$0\leq s\leq 1$, $t\geq 0$, represents the deflection of a thin horizontal
pipe of unit length, subjected to a “nonfollowing” external tensile force
proportional to a parameter $\gamma>0$ and carrying the stationary flow of an
ideal incompressible fluid with velocity $\eta\geq 0$, pinned at $s=0$, and
subjected at $s=1$ to a torque feedback control proportional to $\kappa\geq
0$. The parameter $\beta\in\left(0,1\right)$ depends only on the pipe and
fluid densities. It was shown that the energy of solutions of the closed-loop
system decreases exponentially fast as $t\rightarrow\infty$ via the control
for $\kappa>0$ as long as $\gamma>\eta^{2}$, the latter restriction meaning
that the tension in the pipe is much bigger than the fluid-flow velocity.

In a different paper [11], vertex feedback stabilisability for a star-network
setup of Beck’s Problem with damped, pinned-elastic ends is studied (refer to
[3, 19] for a physical description of Beck’s Problem and its variants). In its
essential characteristics – partial differential equations, boundary and
vertex conditions – this system is similar to an interpretation as a star-
network setup of the above closed-loop system with $\eta=0$ (i.e. when there
is no flow). The purpose of the paper is to show that our approach in [11],
suitably modified, can accomplish the same end of studying feedback
stabilisability for a star-network setup of the closed-loop system when
$\eta>0$.

(We note that recently Aissa et al. [1] and Khemmoudj [8] have considered an
alternative approach to the question of feedback stabilisability in a single
pipe conveying fluid, which assumes that the controls are achieved via time-
varying boundary feedbacks. For a review of early work on pipe problems, see
[13, 14].)

###  1.1. System description

Let us now precisely describe the system we study in this paper. Referring to
Fig. 1 we consider an equilateral $3$-edge metric star graph
$\mathbf{G}\coloneqq\left\\{\mathbf{V},\mathbf{E}\right\\}$, where
$\mathbf{V}$ and $\mathbf{E}$ are the sets of vertices
$\left\\{0\right\\}\cup\left\\{a_{k}\right\\}_{k=1}^{3}$ and edges
$\left\\{e_{k}\right\\}_{k=1}^{3}$, respectively, such that each edge $e_{k}$
connecting the inner vertex at the origin $0$ to the outer vertex $a_{k}$ is
of unit length and is identified with the interval $0\leq s_{k}\leq 1$. The
value $s_{k}=1$ corresponds to the outer vertices, and $s_{k}=0$ corresponds
to the inner vertex.

Figure 1. A star-shaped network with vertex control.

Let ${w}_{k}\left(s_{k},t\right)$ be the deflection of the edge $e_{k}$ for
$0\leq s_{k}\leq 1$ at time $t\geq 0$ from the equilibrium position of the
network which we identify with $\mathbf{G}$. Consider the partial differential
equation

$\frac{\partial^{4}{w}_{k}\left(s_{k},t\right)}{\partial
s_{k}^{4}}-(\gamma-\eta^{2})\,\frac{\partial^{2}{w}_{k}\left(s_{k},t\right)}{\partial
s_{k}^{2}}+2\beta\eta\frac{\partial^{2}{w}_{k}\left(s_{k},t\right)}{\partial
s_{k}\partial t}+\frac{\partial^{2}{w}_{k}\left(s_{k},t\right)}{\partial
t^{2}}=0,\quad k=1,2,3.$ (1.1)

As initial conditions we require that

${w}_{k}\left(s_{k},0\right)=g_{k}\left(s_{k}\right),\quad\left.\frac{\partial{w}_{k}\left(s_{k},t\right)}{\partial
t}\right|_{t=0}=h_{k}\left(s_{k}\right),\quad s_{k}\in e_{k},\quad k=1,2,3,$
(1.2)

where the known functions $g_{k}$, $h_{k}$ are smooth. For boundary and vertex
conditions we proceed as in [11] (to which we refer the reader if an
explanation of their physical meaning is desired). The pinned boundary
conditions at the outer vertices $a_{k}$ imply

${w}_{k}\left(1,t\right)=\frac{\partial^{2}{w}_{k}\left(s_{k},t\right)}{\partial
s_{k}^{2}}\Big{|}_{s_{k}=1}=0,\quad k=1,2,3.$ (1.3)

Further, the deflections ${w}_{k}\left(s_{k},t\right)$ are continuous at the
inner vertex so that

${w}_{j}\left(0,t\right)={w}_{k}\left(0,t\right),\quad j,k=1,2,3.$ (1.4)

The remaining connectivity conditions at the inner vertex have to do with
control torques applied to a system, and also have to do with systems which in
the absence of control may be called “energy-conservative systems”. Thus we
have

$\displaystyle\left.\left(\frac{\partial^{2}{w}_{k}\left(s_{k},t\right)}{\partial
s_{k}^{2}}-\alpha\frac{\partial{w}_{k}\left(s_{k},t\right)}{\partial
s_{k}}-\kappa\frac{\partial^{2}{w}_{k}\left(s_{k},t\right)}{\partial
s_{k}\partial t}\right)\right|_{s_{k}=0}$ $\displaystyle=0,\quad k=1,2,3,$
(1.5)
$\displaystyle\sum_{k=1}^{3}\left.\left[\frac{\partial^{3}{w}_{k}\left(s_{k},t\right)}{\partial
s_{k}^{3}}-(\gamma-\eta^{2})\,\frac{\partial{w}_{k}\left(s_{k},t\right)}{\partial
s_{k}}+\beta\eta\frac{\partial{w}_{k}\left(s_{k},t\right)}{\partial
t}\right]\right|_{s_{k}=0}$ $\displaystyle=0,$ (1.6)

where the feedback control parameters $\alpha,\kappa\geq 0$, conditioned on
the restriction $\gamma>\eta^{2}$, should be chosen so as to exponentially
stabilise the resulting closed-loop system (1.1)–(1.6) in an appropriate
sense.

We point out that one important difference between these conditions and those
in [11] is the “addition” of extra terms
$\beta\eta\left.\left(\partial{w}_{k}/\partial
t\right)\left(s_{k},t\right)\right|_{s_{k}=0}$ to the condition (1.6), which,
as we have noted above, states that the system is energy-conservative when
there is no control, $\kappa=0$. In fact, it may be verified readily that a
variational derivation (conservative Hamilton’s principle) of (1.1) will lead
to the vertex condition (1.6). Then (1.6) also has the physical interpretation
as the force balance condition at the inner vertex just as in [11].

Although there are several approaches to verification of exponential
stabilisability, a direct spectral approach is the natural one to follow. In
particular, this approach is used in [11, 10] (see also the many other papers
by the second author which appeared over the past years) and should be of
considerable interest to the engineer wishing to use only spectral information
– location and asymptotics of eigenvalues, typically – for the boundary-
eigenvalue problem associated with the closed-loop system to study its
stability, as in the case for lumped parameter systems.

For the spectral approach to work for a semigroup formulation of the closed-
loop system, in principle, we must show that it has a system operator which
has a compact resolvent and whose root vectors (eigen- and associated vectors)
form a Riesz basis for the underlying Hilbert state space. Then the familiar
Spectrum Determined Growth Assumption (definition at the end of Section 2) is
satisfied and one is justified to use spectral information as a criterion for
exponential stabilisability alone, as it is explained in detail in [11, 10]
and elsewhere

The organisation of the paper follows. In Section 2 an operator formalism in
an appropriate Hilbert state space is given so that it will be possible to
study well-posedness of the closed-loop system in the setting of strongly
continuous semigroups of bounded linear operators (abbreviated henceforth to
$C_{0}$-semigroups). Section 3 deals with a complete spectral analysis
(existence, location, multiplicity and asymptotics for eigenvalues) of the
closed-loop system operator. The completeness, minimality, and Riesz basis
properties of its root vectors are investigated in Section 4, where the
satisfaction of the Spectrum Determined Growth Assumption is verified. We
conclude in Section 5 with a positive result for the exponential stability
property of the closed-loop system.

##  2\. Operator formulation and well-posedness

We begin this section with a summary of the restrictions on the parameters
specified in the Introduction:

$\alpha,\kappa\geq 0,\quad\beta\in\left(0,1\right),\quad\eta\geq
0,\quad\gamma>\eta^{2}.$

We shall henceforth put $s$ in place of the variable $0\leq s_{k}\leq 1$,
$k=1,2,3$, and set

${v}_{k}\left(s,t\right)=\frac{\partial{w}_{k}\left(s,t\right)}{\partial
t},\quad
x_{k}\left(s,t\right)=\left({w}_{k}\left(s,t\right),{v}_{k}\left(s,t\right)\right),\quad
k=1,2,3,$

and

${x}\left(s,t\right)=\left({x}_{1}\left(s,t\right),{x}_{2}\left(s,t\right),{x}_{3}\left(s,t\right)\right).$

We then introduce the space
$H^{2}_{*}\left(0,1\right)\coloneqq\left\\{w\in{H}^{2}\left(0,1\right)~{}\middle|~{}w\left(1\right)=0\right\\}$,
where ${H}^{r}\left(0,1\right)$, $r\in\mathbb{N}_{0}$, denotes the usual
Sobolev space of order $r$ related to ${L}^{2}\left(0,1\right)$. It follows
that $H^{2}_{*}\left(0,1\right)$, equipped with the inner product

$\left(w,\widetilde{w}\right)=\int^{1}_{0}w^{\prime\prime}\left(s\right)\overline{\widetilde{w}^{\prime\prime}\left(s\right)}\,ds+(\gamma-\eta^{2})\int^{1}_{0}w^{\prime}\left(s\right)\overline{\widetilde{w}^{\prime}\left(s\right)}\,ds+\alpha
w^{\prime}\left(0\right)\overline{\widetilde{w}^{\prime}\left(0\right)},$

is a Hilbert space.

Let now ${L}^{2}\left(\mathbf{G}\right)$ be the metric space of vector-valued
functions $v=\left(v_{1},v_{2},v_{3}\right)$ for which
$v_{k}\in{L}^{2}\left(0,1\right)$, $k=1,2,3$. We similarly define the space
$H^{2}_{*}\left(\mathbf{G}\right)$ of vector-valued functions
$w=\left(w_{1},w_{2},w_{3}\right)$ for which $w_{k}\in
H^{2}_{*}\left(0,1\right)$, $k=1,2,3$, and
$w_{j}\left(0\right)=w_{k}\left(0\right)$, $j,k=1,2,3$. In the Hilbert space
$\mathscr{X}=H^{2}_{*}\left(\mathbf{G}\right)\times
L^{2}\left(\mathbf{G}\right)$, i.e.,

$\mathscr{X}\coloneqq\left\\{x=\left\\{x_{k}\right\\}^{3}_{k=1}~{}\middle|~{}\begin{gathered}x_{k}=\left(w_{k},v_{k}\right)\in
H^{2}_{*}\left(0,1\right)\times L^{2}\left(0,1\right),\\\
w_{j}\left(0\right)=w_{k}\left(0\right),\quad
j,k=1,2,3\end{gathered}\right\\}$

with the inner product (inducing an energy-motivated norm on $\mathscr{X}$)

$\displaystyle\left(x,\widetilde{x}\right)_{\mathscr{X}}\coloneqq\left(w,\widetilde{w}\right)_{H^{2}_{*}\left(\mathbf{G}\right)}+\left(v,\widetilde{v}\right)_{L^{2}\left(\mathbf{G}\right)},$

where

$\displaystyle\left(w,\widetilde{w}\right)_{H^{2}_{*}\left(\mathbf{G}\right)}=\sum_{k=1}^{3}\left[\int^{1}_{0}w_{k}^{\prime\prime}\left(s\right)\overline{\widetilde{w}_{k}^{\prime\prime}\left(s\right)}\,ds+(\gamma-\eta^{2})\int^{1}_{0}w_{k}^{\prime}\left(s\right)\overline{\widetilde{w}_{k}^{\prime}\left(s\right)}\,ds+\alpha
w_{k}^{\prime}\left(0\right)\overline{\widetilde{w}_{k}^{\prime}\left(0\right)}\right],$
$\displaystyle\left(v,\widetilde{v}\right)_{L^{2}\left(\mathbf{G}\right)}=\sum_{k=1}^{3}\int^{1}_{0}v_{k}\left(s\right)\overline{\widetilde{v}_{k}\left(s\right)}\,ds,$

we define the operators $\mathcal{A}$, $\mathcal{B}$ on the domains

$\displaystyle\mathscr{D}\left(\mathcal{A}\right)$
$\displaystyle=\left\\{x=\left\\{x_{k}\right\\}^{3}_{k=1}\in\mathscr{X}~{}\middle|~{}\begin{gathered}x_{k}=\left(w_{k},v_{k}\right)\in(H^{4}\left(0,1\right)\cap
H^{2}_{*}\left(0,1\right))\times H^{2}_{*}\left(0,1\right),\\\
w^{\prime\prime}_{k}\left(1\right)=0,\quad
w^{\prime\prime}_{k}\left(0\right)-\alpha w^{\prime}_{k}\left(0\right)-\kappa
v^{\prime}_{k}\left(0\right)=0,\quad k=1,2,3,\\\
\sum_{k=1}^{3}\,\bigl{[}w^{\left(3\right)}_{k}\left(0\right)-(\gamma-\eta^{2})\,w_{k}^{\prime}\left(0\right)+\beta\eta
v_{k}\left(0\right)\bigr{]}=0\end{gathered}\right\\},$ (2.4)
$\displaystyle\mathscr{D}\left(\mathcal{B}\right)$
$\displaystyle=\left\\{x=\left\\{x_{k}\right\\}^{3}_{k=1}\in\mathscr{X}~{}\middle|~{}x_{k}=\left(w_{k},v_{k}\right)\in
H^{2}_{*}\left(0,1\right))\times H^{1}_{*}\left(0,1\right),\quad
k=1,2,3\right\\}$ (2.5)

by

$\displaystyle\mathcal{A}x$
$\displaystyle\coloneqq\bigl{\\{}\bigl{(}v_{k},-w_{k}^{\left(4\right)}+(\gamma-\eta^{2})\,w^{\prime\prime}_{k}\bigr{)}\bigr{\\}}_{k=1}^{3},$
(2.6) $\displaystyle\mathcal{B}x$
$\displaystyle\coloneqq\bigl{\\{}\bigl{(}0,-2\beta\eta
v_{k}^{\prime}\bigr{)}\bigr{\\}}_{k=1}^{3},$ (2.7)

respectively. Clearly
$\mathscr{D}\left(\mathcal{A}\right)\subset\mathscr{D}\left(\mathcal{B}\right)$,
and one may verify quite readily that $\mathcal{B}$ is relatively compact with
respect to $\mathcal{A}$ (in the sense of [7, Section IV.1.3]). The closed-
loop system may be formulated abstractly thus:

$\left\\{\begin{split}\dot{{x}}\left(t\right)&=\mathcal{T}{x}\left(t\right),\quad\mathcal{T}\coloneqq\mathcal{A}+\mathcal{B},\quad\mathscr{D}\left(\mathcal{T}\right)=\mathscr{D}\left(\mathcal{A}\right),\\\
{{x}}\left(0\right)&=x_{0},\end{split}\right.$ (2.8)

where

${x}\left(t\right)=\left\\{\left({w}_{k}\left(\,\cdot\,,t\right),{v}_{k}\left(\,\cdot\,,t\right)\right)\right\\}^{3}_{k=1},\quad
x_{0}=\left\\{\left(g_{k},h_{k}\right)\right\\}^{3}_{k=1}.$

Substitute ${x}\left(t\right)=x\exp\left(\lambda t\right)$, $x\in\mathscr{X}$,
in (2.8) and note for later reference that

$\mathcal{T}x=\lambda x,\quad
x\in\mathscr{D}\left(\mathcal{A}\right),\quad\lambda\in\mathbb{C},$ (2.9)

which is the spectral problem for the closed-loop system with spectral
parameter $\lambda$. (Refer to standard textbooks for the standard definitions
from functional analysis of the spectral theory of linear operators in Hilbert
space.)

Our central well-posedness result is the following:

######  Theorem 2.1.

The closed-loop system is well posed in the sense that (2.8) has a unique
solution
${x}\in{C}^{1}\left(\left(0,\infty\right);\mathscr{X}\right)\cap{C}\left(\left[0,\infty\right);\mathscr{D}\left(\mathcal{A}\right)\right)$
given by

${x}\left(t\right)=\mathbb{S}\left(t\right)x_{0},\quad
x_{0}\in\mathscr{D}\left(\mathcal{A}\right),$

where $\mathbb{S}\left(t\right)$ is a contractive $C_{0}$-semigroup on
$\mathscr{X}$ with infinitesimal generator $\mathcal{T}$.

We will prove the theorem with the help of the following lemma.

######  Lemma 2.1.

The following statements hold:

  1. cmrmn(1)

$\mathcal{T}$ is closed and densely defined.

  2. cmrmn(2)

$0\in\varrho\left(\mathcal{T}\right)$ (the resolvent set of $\mathcal{T}$) and
$\mathcal{T}^{-1}$ is compact.

  3. cmrmn(3)

$\mathcal{T}$ is maximal dissipative for $\kappa>0$ and skewadjoint for
$\kappa=0$.

###### Proof.

To prove statement 2 it will be shown, first of all, that $\mathcal{T}$ is
injective and surjective. To this end for $\lambda=0$ we consider the solution
$x$ of (2.9). Then we clearly have $v_{k}=0$, $k=1,2,3$, and the
$w_{k}=w_{k}\left(\lambda,s\right)$ satisfy the boundary-eigenvalue problem

$\left\\{\begin{aligned}
w^{\left(4\right)}_{k}-(\gamma-\eta^{2})\,w^{\prime\prime}_{k}&=0,&k&=1,2,3,\\\
w_{k}\left(1\right)=w^{\prime\prime}_{k}\left(1\right)&=0,&k&=1,2,3,\\\
w_{j}\left(0\right)&=w_{k}\left(0\right),&j,k&=1,2,3,\\\
w^{\prime\prime}_{k}\left(0\right)-\alpha
w^{\prime}_{k}\left(0\right)&=0,&k&=1,2,3,\\\
\sum^{3}_{k=1}\,\bigl{[}w^{\left(3\right)}_{k}\left(0\right)-(\gamma-\eta^{2})\,w^{\prime}_{k}\left(0\right)\bigr{]}&=0.&&&\end{aligned}\right.$
(2.10)

Multiply the differential equations in (2.10) by the conjugate of $w_{k}$,
$k=1,2,3$, and integrate from $0$ to $1$. Integrating by parts, making use of
the boundary and vertex conditions
$w_{k}\left(1\right)=w^{\prime\prime}_{k}\left(1\right)=0$,
$w^{\prime\prime}_{k}\left(0\right)-\alpha w^{\prime}_{k}\left(0\right)=0$,
$k=1,2,3$, we get

$\displaystyle 0$
$\displaystyle=\int_{0}^{1}\left|w_{k}^{\prime\prime}\left(s\right)\right|^{2}ds+(\gamma-\eta^{2})\int_{0}^{1}\left|w_{k}^{\prime}\left(s\right)\right|^{2}ds+\alpha\left|w_{k}^{\prime}\left(0\right)\right|^{2}$
$\displaystyle\qquad-\bigl{[}w^{\left(3\right)}_{k}\left(0\right)-(\gamma-\eta^{2})\,w^{\prime}_{k}\left(0\right)\bigr{]}\,\overline{w_{k}\left(0\right)},\quad
k=1,2,3.$

Subsequent summation over $k=1,2,3$, using the vertex conditions
$w_{j}\left(0\right)=w_{k}\left(0\right)$, $j,k=1,2,3$, and
$\sum^{3}_{k=1}\,\bigl{[}w^{\left(3\right)}_{k}\left(0\right)-(\gamma-\eta^{2})\,w^{\prime}_{k}\left(0\right)\bigr{]}=0$,
yields

$0=\sum_{k=1}^{3}\left[\int^{1}_{0}\left|w_{k}^{\prime\prime}\left(s\right)\right|^{2}ds+(\gamma-\eta^{2})\int^{1}_{0}\left|w_{k}^{\prime}\left(s\right)\right|^{2}ds+\alpha\left|w_{k}^{\prime}\left(0\right)\right|^{2}\right]=\left\|w\right\|^{2}_{{H}_{*}^{2}\left(\mathbf{G}\right)}.$

The fact that $v_{k}=0$, $k=1,2,3$, implies that
$\left\|v\right\|_{{L}^{2}\left(\mathbf{G}\right)}=0$ and hence that
$\left\|x\right\|_{\mathscr{X}}=0$, or equivalently, that $x=0$. It follows
from this that $\ker\mathcal{T}=0$ and so $\mathcal{T}$ is injective. Thus,
$0$ is not an eigenvalue.

For the surjectivity, we proceed as follows. Let
$\widetilde{x}\in\mathscr{X}$, $x\in\mathscr{D}\left(\mathcal{A}\right)$
(arbitrary) and consider the equation

$\mathcal{T}x=\widetilde{x},$ (2.11)

or equivalently in coordinates,

$\left\\{\begin{aligned} v_{k}&=\widetilde{w}_{k},&k&=1,2,3,\\\
-w^{\left(4\right)}_{k}+(\gamma-\eta^{2})\,w^{\prime\prime}_{k}-2\beta\eta
v_{k}^{\prime}&=\widetilde{v}_{k},&k&=1,2,3,\\\
w_{k}\left(1\right)=w^{\prime\prime}_{k}\left(1\right)&=0,&k&=1,2,3,\\\
w_{j}\left(0\right)&=w_{k}\left(0\right),&j,k&=1,2,3,\\\
w^{\prime\prime}_{k}\left(0\right)-\alpha w^{\prime}_{k}\left(0\right)-\kappa
v^{\prime}_{k}\left(0\right)&=0,&k&=1,2,3,\\\
\sum^{3}_{k=1}\,\bigl{[}w^{\left(3\right)}_{k}\left(0\right)-(\gamma-\eta^{2})\,w_{k}^{\prime}\left(0\right)+\beta\eta
v_{k}\left(0\right)\bigr{]}&=0,&&&\end{aligned}\right.$ (2.12)

Write the differential equations in (2.12) in weak form as

$\int_{0}^{1}\biggl{[}-w^{\left(4\right)}_{k}\left(s\right)+(\gamma-\eta^{2})\,w^{\prime\prime}_{k}\left(s\right)-2\beta\eta
v_{k}^{\prime}\left(s\right)\biggr{]}\,\overline{\phi_{k}\left(s\right)}\,ds=\int_{0}^{1}\widetilde{v}_{k}\left(s\right)\overline{\phi_{k}\left(s\right)}\,ds,\quad
k=1,2,3,$

for the $\phi_{k}$ in an appropriate class of test functions satisfying the
conditions $\phi_{k}\left(1\right)=0$, $k=1,2,3$, and
$\phi_{j}\left(0\right)=\phi_{k}\left(0\right)$, $j,k=1,2,3$. Again,

$\displaystyle\sum_{k=1}^{3}\int_{0}^{1}\left(\widetilde{v}_{k}\left(s\right)+2\beta\eta\widetilde{w}_{k}^{\prime}\left(s\right)\right)\overline{\phi_{k}\left(s\right)}\,ds+\beta\eta\sum_{k=1}^{3}\widetilde{w}_{k}\left(0\right)\overline{\phi_{k}\left(0\right)}+\kappa\sum_{k=1}^{3}\widetilde{w}_{k}^{\prime}\left(0\right)\overline{\phi_{k}^{\prime}\left(0\right)}$
$\displaystyle\hskip
43.36464pt=-\sum_{k=1}^{3}\left[\int_{0}^{1}w_{k}^{\prime\prime}\left(s\right)\overline{\phi_{k}^{\prime\prime}\left(s\right)}\,ds+(\gamma-\eta^{2})\int_{0}^{1}w_{k}^{\prime}\left(s\right)\overline{\phi_{k}^{\prime}\left(s\right)}\,ds+\alpha
w_{k}^{\prime}\left(0\right)\overline{\phi_{k}^{\prime}\left(0\right)}\right]$

employing integration by parts and using the boundary and vertex conditions,
and where we have taken into account that $v_{k}=\widetilde{w}_{k}$,
$k=1,2,3$. We define the bilinear form $\left<\,\cdot\,,\,\cdot\,\right>$ on
${H}_{*}^{2}\left(\mathbf{G}\right)$ by

$\left<u,\widetilde{u}\right>\coloneqq\sum_{k=1}^{3}\left[\int_{0}^{1}u_{k}^{\prime\prime}\left(s\right)\overline{\widetilde{u}_{k}^{\prime\prime}\left(s\right)}\,ds+(\gamma-\eta^{2})\int_{0}^{1}u_{k}^{\prime}\left(s\right)\overline{\widetilde{u}_{k}^{\prime}\left(s\right)}\,ds+\alpha
u_{k}^{\prime}\left(0\right)\overline{\widetilde{u}_{k}^{\prime}\left(0\right)}\right]$

for any $u,\widetilde{u}\in{H}_{*}^{2}\left(\mathbf{G}\right)$. Clearly from
Schwarz’s inequality,

$\left|\left<u,\widetilde{u}\right>\right|\leq\left\|u\right\|_{{H}_{*}^{2}\left(\mathbf{G}\right)}\left\|\widetilde{u}\right\|_{{H}_{*}^{2}\left(\mathbf{G}\right)},\quad
u,\widetilde{u}\in{H}_{*}^{2}\left(\mathbf{G}\right).$

Moreover,
$\left<u,u\right>=\left\|u\right\|_{{H}_{*}^{2}\left(\mathbf{G}\right)}$, and
so $\left<\,\cdot\,,\,\cdot\,\right>$ is coercive. Now define

$f\left(\phi\right)\coloneqq\sum_{k=1}^{3}\int_{0}^{1}\left(\widetilde{v}_{k}\left(s\right)+2\beta\eta\widetilde{w}_{k}^{\prime}\left(s\right)\right)\overline{\phi_{k}\left(s\right)}\,ds+\beta\eta\sum_{k=1}^{3}\widetilde{w}_{k}\left(0\right)\overline{\phi_{k}\left(0\right)}+\kappa\sum_{k=1}^{3}\widetilde{w}_{k}^{\prime}\left(0\right)\overline{\phi_{k}^{\prime}\left(0\right)},$

a bounded conjugate linear functional of $\phi$ in
${H}_{*}^{2}\left(\mathbf{G}\right)$ for fixed
$\widetilde{x}=\left\\{\left(\widetilde{w}_{k},\widetilde{v}_{k}\right)\right\\}^{3}_{k=1}$.
Then there exists a unique
$w=\left\\{w_{k}\right\\}^{3}_{k=1}\in{H}_{*}^{2}\left(\mathbf{G}\right)$ such
that $\left<w,\phi\right>+f\left(\phi\right)=0$ (by the Lax–Milgram theorem)
and therefore that $w_{k}=w_{k}\left(s\right)$, $k=1,2,3$, satisfies the
differential equations in (2.12). Since
$x=\left\\{\left(w_{k},v_{k}\right)\right\\}^{3}_{k=1}=\left\\{\left(w_{k},\widetilde{w}_{k}\right)\right\\}^{3}_{k=1}$,
we have
$\left\\{\left(w_{k},\widetilde{w}_{k}\right)\right\\}^{3}_{k=1}\in\mathscr{D}\left(\mathcal{A}\right)$
and (2.11) is satisfied for any given
$\widetilde{x}=\left\\{\left(\widetilde{w}_{k},\widetilde{v}_{k}\right)\right\\}^{3}_{k=1}\in\mathscr{X}$.
This proves surjectivity, and hence bijectivity of $\mathcal{T}$. The closed
graph theorem now shows that the inverse $\mathcal{T}^{-1}$ of $\mathcal{T}$
is closed and bounded, so it remains only to show that $\mathcal{T}^{-1}$ is
compact. But this is clear, for
$\mathscr{D}\left(\mathcal{A}\right)\subset\mathscr{X}$ and the spaces
$\mathscr{D}\left(\mathcal{A}\right)$, $\mathscr{X}$ are closed subspaces of
${H}^{4}\left(\mathbf{G}\right)\times{H}^{2}\left(\mathbf{G}\right)$,
${H}^{2}\left(\mathbf{G}\right)\times{L}^{2}\left(\mathbf{G}\right)$,
respectively, and hence by the Sobolev embedding theorem,
$\mathscr{D}\left(\mathcal{A}\right)$ is compactly embedded in $\mathscr{X}$.

In order to establish statement 3, we compute for any
$x\in\mathscr{D}\left(\mathcal{A}\right)$,

$\Re\left(\mathcal{T}x,x\right)_{\mathscr{X}}=\left(\mathcal{T}x,x\right)_{\mathscr{X}}+\left(x,\mathcal{T}x\right)_{\mathscr{X}}=-\kappa\sum_{k=1}^{3}\left|v_{k}^{\prime}\left(0\right)\right|^{2},$
(2.13)

after the usual integrations by parts and using the boundary and vertex
conditions. The previous arguments show that $\mathcal{T}$ is maximal
dissipative for $\kappa>0$. Indeed, since
$0\in\varrho\left(\mathcal{T}\right)$, there exists $\lambda>0$ in
$\varrho\left(\mathcal{T}\right)$ and
$\operatorname{Ran}\left(\lambda\mathcal{I}-\mathcal{T}\right)=\mathscr{X}$
(because $\varrho\left(\mathcal{T}\right)$ is open); hence the maximality of
$\mathcal{T}$. It is skewsymmetric when $\kappa=0$ for then we have
$\left(\mathcal{T}x,x\right)_{\mathscr{X}}+\left(x,\mathcal{T}x\right)_{\mathscr{X}}=0$
from (2.13). The resulting skewadjointness of $\mathcal{T}$ for $\kappa=0$ can
be obtained by the same standard arguments as in the proof of [11, Lemma 3.2].

Finally, statements 2 and 3 imply statement 1. ∎

###### Proof of Theorem 2.1.

Since, by Lemma 2.1, $\mathcal{T}$ is a closed, densely defined, maximal
dissipative operator in $\mathscr{X}$, the proof follows immediately from the
Lumer–Phillips theorem, see [15, Theorem 1.4.3]. ∎

In preparation for what follows, we recall that for any $C_{0}$-semigroup
$\mathbb{S}\left(t\right)$ on $\mathscr{X}$, there exist constants $M$,
$\varpi$ such that $\left\|\mathbb{S}\left(t\right)\right\|_{\mathscr{X}}\leq
Me^{\varpi t}$, $t\geq 0$. The semigroup is said to be exponentially stable if
$\varpi<0$. In this case, solutions of the closed-loop system are such that
$\left\|\mathbb{S}\left(t\right)x_{0}\right\|_{\mathscr{X}}\leq Me^{\varpi
t}\left\|x_{0}\right\|_{\mathscr{X}}$ and so, as $t\rightarrow\infty$,
$\left\|{x}\left(t\right)\right\|_{\mathscr{X}}\rightarrow 0$ exponentially
(recall that $\left\|{x}\left(t\right)\right\|^{2}_{\mathscr{X}}$ is a measure
of the energy of the closed-loop system at a given time $t$). The goal of the
paper is to prove the exponential stability of the closed-loop system for
which the growth bound $\varpi_{0}$ satisfies
$\varpi_{0}=\sup\left\\{\Re\left(\lambda\right)~{}\middle|~{}\lambda\in\sigma\left(\mathcal{T}\right)\right\\}$,
$\sigma\left(\mathcal{T}\right)$ being the spectrum of $\mathcal{T}$, and
hence the Spectrum Determined Growth Assumption will be satisfied.

##  3\. Spectral analysis

We return to the spectral problem (2.9) and analyse the spectrum of the
closed-loop system operator $\mathcal{T}$ in detail. It follows from Lemma 2.1
that $\mathcal{T}$ has compact resolvent and
$0\in\varrho\left(\mathcal{T}\right)$. It is known then (see, e.g., [5,
Corollary XI.8.4]) that the spectrum $\sigma\left(\mathcal{T}\right)$ is a
purely discrete set, consisting only of eigenvalues of finite algebraic
multiplicity which accumulate only at infinity. Moreover, because
$\mathcal{T}$ is maximal dissipative, for any
$\lambda\in\sigma\left(\mathcal{T}\right)$ we have
$\Re\left(\lambda\right)\leq 0$.

Further information about the location of the eigenvalues are obtained in the
next result.

######  Theorem 3.1.

The spectrum of $\mathcal{T}\coloneqq\mathcal{A}+\mathcal{B}$ as defined by
(2.4)–(2.7) is symmetric with respect to the real axis of the complex plane,
the eigenvalues of $\mathcal{T}$ being confined to the open left half-plane
when $\kappa>0$.

###### Proof.

Since $\mathcal{T}$ is a real operator we obtain the first assertion about
symmetry of the spectrum. Indeed, conjugation of (2.9) shows that
$\overline{x}$ satisfies its conjugate spectral problem and is the eigenvector
of $\mathcal{T}$ which corresponds to the eigenvalue $\overline{\lambda}$.
Next, we prove the second assertion. To do this, we take the inner product of
(2.9) with the corresponding $x$ to obtain for the real parts of the resulting
expression, taking into account (2.13),

$\Re\left(\lambda\right)=\frac{\Re\left(\mathcal{T}x,x\right)_{\mathscr{X}}}{\left\|x\right\|_{\mathscr{X}}^{2}}\leq
0.$

We must show that if $\kappa>0$ then $\Re\left(\lambda\right)<0$. Let
$\lambda$ be an eigenvalue with $\Re\left(\lambda\right)=0$ and let
$x=\left\\{\left(w_{k},v_{k}\right)\right\\}_{k=1}^{3}$ be the corresponding
eigenvector. Then, because $v_{k}=\lambda w_{k}$, $k=1,2,3$, replacing the
$v_{k}^{\prime}\left(0\right)$ in (2.13) by $\lambda
w_{k}^{\prime}\left(0\right)$ we arrive at

$\Re\left(\mathcal{T}x,x\right)_{\mathscr{X}}=-\kappa\left|\lambda\right|^{2}\sum_{k=1}^{3}\left|w_{k}^{\prime}\left(0\right)\right|^{2}=0$

and

$\sum_{k=1}^{3}\left|w_{k}^{\prime}\left(0\right)\right|^{2}=0$

since $\kappa>0$, $\lambda\neq 0$. So $w_{k}^{\prime}\left(0\right)=0$,
$k=1,2,3$, and the $w_{k}=w_{k}\left(\lambda,s\right)$ satisfy the boundary-
eigenvalue problem

$\left\\{\begin{aligned}
w^{\left(4\right)}_{k}-(\gamma-\eta^{2})\,w^{\prime\prime}_{k}+2\lambda\beta\eta
w^{\prime}_{k}&=-\lambda^{2}w_{k},&k&=1,2,3,\\\
w_{k}\left(1\right)=w^{\prime\prime}_{k}\left(1\right)&=0,&k&=1,2,3,\\\
w_{j}\left(0\right)&=w_{k}\left(0\right),&j,k&=1,2,3,\\\
w^{\prime}_{k}\left(0\right)=w^{\prime\prime}_{k}\left(0\right)&=0,&k&=1,2,3,\\\
\sum^{3}_{k=1}\,\bigl{(}w^{\left(3\right)}_{k}\left(0\right)+\lambda\beta\eta
w_{k}\left(0\right)\bigr{)}&=0.&&&\end{aligned}\right.$ (3.1)

Setting $\lambda=i\mu$, $\mu\in\mathbb{R}$, consider the boundary-eigenvalue
problem

$\left\\{\begin{aligned}
\varphi^{\left(4\right)}-(\gamma-\eta^{2})\,\varphi^{\prime\prime}+2i\beta\eta\mu\varphi^{\prime}&=\mu^{2}\varphi,&&\\\
\varphi\left(1\right)=\varphi^{\prime\prime}\left(1\right)&=0,&&\\\
\varphi^{\prime}\left(0\right)=\varphi^{\prime\prime}\left(0\right)&=0.&&\end{aligned}\right.$
(3.2)

Let $\varphi=\varphi\left(\lambda,s\right)\neq 0$ be a solution of (3.2). Then
solutions of (3.1) are of the form
$w_{k}\left(\lambda,s\right)=c_{k}\varphi\left(\lambda,s\right)$, $k=1,2,3$,
where $c_{k}$ are arbitrary constants. The vertex condition
$w_{j}\left(0\right)=w_{k}\left(0\right)$, $j,k=1,2,3$, together with
$\sum^{3}_{k=1}\,\bigl{(}w^{\left(3\right)}_{k}\left(0\right)+i\beta\eta\mu
w_{k}\left(0\right)\bigr{)}=0$ implies that $c_{j}=c_{k}\equiv c$,
$j,k=1,2,3$, and hence that

$3c\,\bigl{(}\varphi^{\left(3\right)}\left(0\right)+i\beta\eta\mu\varphi\left(0\right)\bigr{)}=0.$

Using
$\varphi^{\left(3\right)}\left(0\right)+i\beta\eta\mu\varphi\left(0\right)\neq
0$ gives $c=0$. This shows that (3.1) has only zero solutions. Thus
$\mathcal{T}$ has no eigenvalues on the imaginary axis,
$\Re\left(\lambda\right)<0$, if $\kappa>0$. ∎

###  3.1. Eigenvalues and eigenvectors

We have shown that studying the spectrum of $\mathcal{T}$ reduces to studying
its discrete spectrum, consisting only of eigenvalues. To pursue this further,
we will show in the next theorem how to determine these eigenvalues and the
corresponding eigenvectors.

######  Theorem 3.2.

Let $\varphi=\varphi\left(\lambda,s\right)$ be a nonzero solution of the
differential equation

$\varphi^{\left(4\right)}-(\gamma-\eta^{2})\,\varphi^{\prime\prime}+2\lambda\beta\eta\varphi^{\prime}=-\lambda^{2}\varphi$
(3.3)

for $\lambda\in\mathbb{C}$ satisfying the boundary conditions

$\displaystyle\varphi\left(1\right)=\varphi^{\prime\prime}\left(1\right)$
$\displaystyle=0,$ (3.4)
$\displaystyle\varphi^{\prime\prime}\left(0\right)-\left(\alpha+\lambda\kappa\right)\varphi^{\prime}\left(0\right)$
$\displaystyle=0.$ (3.5)

Define by

$D_{1}\left(\lambda\right)\coloneqq\varphi^{\left(3\right)}\left(\lambda,0\right)-(\gamma-\eta^{2})\,\varphi^{\prime}\left(\lambda,0\right)+\lambda\beta\eta\varphi\left(\lambda,0\right),\quad
D_{2}\left(\lambda\right)\coloneqq\varphi\left(\lambda,0\right)$

the corresponding characteristic functions. The eigenvalues of $\mathcal{T}$
are the solutions of $D_{1}\left(\lambda\right)=0$ and
$D_{2}\left(\lambda\right)=0$,

$\sigma\left(\mathcal{T}\right)=\left\\{\lambda\in\mathbb{C}~{}\middle|~{}D_{1}\left(\lambda\right)=0\right\\}\cup\left\\{\lambda\in\mathbb{C}~{}\middle|~{}D_{2}\left(\lambda\right)=0\right\\}.$

If $D_{1}\left(\lambda\right)=0$ and $D_{2}\left(\lambda\right)\neq 0$, then
an eigenvector $x=x\left(\lambda\right)$ corresponding to the eigenvalue
$\lambda$ is given by

$x\left(\lambda\right)=\bigl{\\{}\left(\varphi\left(\lambda,\,\cdot\,\right),\lambda\varphi\left(\lambda,\,\cdot\,\right)\right);\left(\varphi\left(\lambda,\,\cdot\,\right),\lambda\varphi\left(\lambda,\,\cdot\,\right)\right);\left(\varphi\left(\lambda,\,\cdot\,\right),\lambda\varphi\left(\lambda,\,\cdot\,\right)\right)\bigr{\\}}.$
(3.6)

If $D_{1}\left(\lambda\right)\neq 0$ and $D_{2}\left(\lambda\right)=0$, then
there are two linearly independent eigenvectors
$x_{1}=x_{1}\left(\lambda\right)$, $x_{2}=x_{2}\left(\lambda\right)$ for the
eigenvalue $\lambda$ (i.e., $\lambda$ has geometric multiplicity two) given by

$\displaystyle x_{1}\left(\lambda\right)$
$\displaystyle=\bigl{\\{}\left(\varphi\left(\lambda,\,\cdot\,\right),\lambda\varphi\left(\lambda,\,\cdot\,\right)\right);-\frac{1}{2}\left(\varphi\left(\lambda,\,\cdot\,\right),\lambda\varphi\left(\lambda,\,\cdot\,\right)\right);-\frac{1}{2}\left(\varphi\left(\lambda,\,\cdot\,\right),\lambda\varphi\left(\lambda,\,\cdot\,\right)\right)\bigr{\\}},$
(3.7) $\displaystyle x_{2}\left(\lambda\right)$
$\displaystyle=\bigl{\\{}\left(0,0\right);\left(\varphi\left(\lambda,\,\cdot\,\right),\lambda\varphi\left(\lambda,\,\cdot\,\right)\right);-\left(\varphi\left(\lambda,\,\cdot\,\right),\lambda\varphi\left(\lambda,\,\cdot\,\right)\right)\bigr{\\}},$
(3.8)

respectively.

###### Proof.

Let us first note from (2.9) that if $\lambda\in\mathbb{C}$ is an eigenvalue
of $\mathcal{T}$ with corresponding eigenvector $x$, then the
$w_{k}=w_{k}\left(\lambda,s\right)$ are nonzero solutions of the boundary-
eigenvalue problem

$\left\\{\begin{aligned}
w^{\left(4\right)}_{k}-(\gamma-\eta^{2})\,w^{\prime\prime}_{k}+2\lambda\beta\eta
w^{\prime}_{k}&=-\lambda^{2}w_{k},&k&=1,2,3,\\\
w_{k}\left(1\right)=w^{\prime\prime}_{k}\left(1\right)&=0,&k&=1,2,3,\\\
w_{j}\left(0\right)&=w_{k}\left(0\right),&j,k&=1,2,3,\\\
w^{\prime\prime}_{k}\left(0\right)-\left(\alpha+\lambda\kappa\right)w^{\prime}_{k}\left(0\right)&=0,&k&=1,2,3,\\\
\sum^{3}_{k=1}\,\bigl{[}w^{\left(3\right)}_{k}\left(0\right)-(\gamma-\eta^{2})\,w^{\prime}_{k}\left(0\right)+\lambda\beta\eta
w_{k}\left(0\right)\bigr{]}&=0.&&&\end{aligned}\right.$ (3.9)

Comparing (3.3)–(3.5) with (3.9) we observe that
$w_{k}\left(\lambda,s\right)=c_{k}\varphi\left(\lambda,s\right)$, $k=1,2,3$.
Substitution in $w_{j}\left(0\right)=w_{k}\left(0\right)$, $j,k=1,2,3$, and
$\sum^{3}_{k=1}\,\bigl{[}w^{\left(3\right)}_{k}\left(0\right)-(\gamma-\eta^{2})\,w^{\prime}_{k}\left(0\right)+\lambda\beta\eta
w_{k}\left(0\right)\bigr{]}=0$ gives
$c_{j}\varphi\left(\lambda,0\right)=c_{k}\varphi\left(\lambda,0\right)$,
$j,k=1,2,3$, and

$\sum^{3}_{k=1}c_{k}\,\bigl{[}\varphi^{\left(3\right)}\left(\lambda,0\right)-(\gamma-\eta^{2})\,\varphi^{\prime}\left(\lambda,0\right)+\lambda\beta\eta\varphi\left(\lambda,0\right)\bigr{]}=0.$
(3.10)

Thus two cases arise.

  1. Case 1. $D_{2}\left(\lambda\right)\neq 0$. Here $c_{j}=c_{k}\equiv c\neq 0$, $j,k=1,2,3$, since $\varphi\left(\lambda,0\right)\neq 0$, and from (3.10) we have therefore

$\varphi^{\left(3\right)}\left(\lambda,0\right)-(\gamma-\eta^{2})\,\varphi^{\prime}\left(\lambda,0\right)+\lambda\beta\eta\varphi\left(\lambda,0\right)=0.$
(3.11)

Then $\lambda$ is an eigenvalue if and only if the boundary-eigenvalue problem
(3.3)–(3.5), (3.11) has a nonzero solution, in which case $\lambda$ is a zero
of $D_{1}$ and (3.6) follows since

$x\left(\lambda\right)=\left\\{c_{k}\left(\varphi\left(\lambda,\,\cdot\,\right),\lambda\varphi\left(\lambda,\,\cdot\,\right)\right)\right\\}^{3}_{k=1}$
(3.12)

with the $c_{k}\equiv c$.

  2. Case 2. $D_{2}\left(\lambda\right)=0$. In this case we consider (3.3)–(3.5) subject to the additional boundary condition

$\varphi\left(\lambda,0\right)=0.$ (3.13)

Therefore, $\lambda$ is an eigenvalue if and only if the boundary-eigenvalue
problem (3.3)–(3.5), (3.13) admits a nonzero solution. Using (3.13) in (3.10),
we obtain

$\sum^{3}_{k=1}c_{k}\,\bigl{[}\varphi^{\left(3\right)}\left(\lambda,0\right)-(\gamma-\eta^{2})\,\varphi^{\prime}\left(\lambda,0\right)\bigr{]}=\bigl{[}\varphi^{\left(3\right)}\left(\lambda,0\right)-(\gamma-\eta^{2})\,\varphi^{\prime}\left(\lambda,0\right)\bigr{]}\,\sum^{3}_{k=1}c_{k}=0.$

Let

$\varphi^{\left(3\right)}\left(\lambda,0\right)-(\gamma-\eta^{2})\,\varphi^{\prime}\left(\lambda,0\right)=0.$
(3.14)

It is easy to see that any solution (3.3)–(3.5), (3.13), (3.14) must be the
zero solution. Consequently, $\sum^{3}_{k=1}c_{k}=0$ and
$\varphi^{\left(3\right)}\left(\lambda,0\right)-(\gamma-\eta^{2})\,\varphi^{\prime}\left(\lambda,0\right)\neq
0$, and it follows that associated with $\lambda$ there exist two linearly
independent eigenvectors $x_{1}\left(\lambda\right)$ and
$x_{2}\left(\lambda\right)$ given by (3.7) and (3.8), respectively. With this
choice of eigenvectors $x_{1}\left(\lambda\right)$,
$x_{2}\left(\lambda\right)$ we have in fact $x_{1}\left(\lambda\right)\perp
x_{2}\left(\lambda\right)$. The theorem is proven. ∎

###  3.2. Asymptotics of eigenvalues

Our aim in this subsection is to discuss eigenvalue asymptotics. The approach
we take here is based on the “asymptotic spectral problem” given in [10] for
the single pipe case. By Theorem 3.1, we need only consider eigenvalues in the
left half-plane $\Re\left(\lambda\right)\leq 0$. Moreover, those eigenvalues
with nonzero imaginary part occur in conjugate pairs $\lambda$,
$\overline{\lambda}$, so that we may restrict attention to
$\frac{\pi}{2}\leq\arg\left(\lambda\right)\leq\pi$. As usual, we use the
standard substitution $\lambda=i\rho^{2}$ and take
$0\leq\arg\left(\rho\right)\leq\frac{\pi}{4}$. Define the sector ${S}$ in the
complex plane by

${S}\coloneqq\left\\{\rho\in\mathbb{C}~{}\middle|~{}0\leq\arg\left(\rho\right)\leq\frac{\pi}{4}\right\\}.$

For ${S}$ the four roots of $-1$ can be ordered so that

$\Re\left(-\rho\right)\leq\Re\left(i\rho\right)\leq\Re\left(-i\rho\right)\leq\Re\left(\rho\right),\quad\rho\in{S},$

and from elementary considerations we have

$\Re\left(-\rho\right)=\left|\rho\right|\cos\left(\arg\left(\rho\right)\right)\leq-\frac{\sqrt{2}}{2}\left|\rho\right|<0$

and

$\Re\left(i\rho\right)=\left|\rho\right|\cos\left(\arg\left(\rho\right)+\frac{\pi}{2}\right)=-\left|\rho\right|\sin\left(\arg\left(\rho\right)\right)\leq
0.$

Using this, it follows that for $\rho\in{S}$ we have as
$\left|\rho\right|\rightarrow\infty$

$\left|e^{i\rho}\right|\leq
1,\quad\left|e^{-\rho}\right|=\mathcal{O}\,(e^{-b\left|\rho\right|})\rightarrow
0,$

for some constant $b>0$. We shall use these observations henceforth without
explicit mention.

The underlying idea in the proof of the next theorem is that it is enough to
base it on asymptotic properties in $\left|\lambda\right|$ of the solutions to
(3.3), according to Theorem 3.2. We are here concerned with the case
$\lambda=i\rho^{2}$, and (3.3) becomes therefore

$\varphi^{\left(4\right)}-(\gamma-\eta^{2})\,\varphi^{\prime\prime}+2i\beta\eta\rho^{2}\varphi^{\prime}=\rho^{4}\varphi.$
(3.15)

For convenience, we state a result from [10].

######  Lemma 3.1.

In the sector ${S}$, there exists a fundamental system
$\left\\{\varphi_{r}\left(\rho,\,\cdot\,\right)\right\\}^{4}_{r=1}$ of the
differential equation (3.15) which has the following asymptotic expressions
for large $\left|\rho\right|$:

$\varphi^{\left(m\right)}_{r}\left(\rho,s\right)=\left(i^{r}\rho\right)^{m}e^{i^{r}\rho
s}\left(1+\Phi_{r}\left(s\right)+\dfrac{i^{r}\Phi_{r1}\left(s\right)+m\Phi^{\prime}_{r}\left(s\right)}{i^{r}\rho}+\mathcal{O}\,(\rho^{-2})\right),\quad
r=1,2,3,4,$

for $m=0,1,2,3$, $0\leq s\leq 1$, where

$\Phi_{r}\left(s\right)=-1+e^{\left(-1\right)^{r+1}\frac{i\beta\eta}{2}s},\quad\Phi_{r1}\left(s\right)=\frac{\left(-i\right)^{r}}{4}\left(\frac{\beta^{2}\eta^{2}}{2}+\gamma-\eta^{2}\right)se^{\left(-1\right)^{r+1}\frac{i\beta\eta}{2}s},\quad
r=1,2,3,4.$

Let us note from the lemma that, in ${S}$, since $\omega_{r}=i^{r}$ and
$\omega^{2}_{r}=\left(-1\right)^{r}$, with $\omega_{1}=-\omega_{3}=i$ and
$\omega_{2}=-\omega_{4}=-1$

$\varphi^{\left(m\right)}_{r}\left(\rho,s\right)=\left(\rho\omega_{r}\right)^{m}e^{\rho\omega_{r}s}\left(1+\Phi_{r}\left(s\right)+\frac{\omega_{r}\Phi_{r1}\left(s\right)+m\Phi^{\prime}_{r}\left(s\right)}{\rho\omega_{r}}+\mathcal{O}\,(\rho^{-2})\right),\quad
r=1,2,3,4,$

which we shall use subsequently.

We know that linear combinations of the form
$\varphi\left(\rho,s\right)=\sum_{r=1}^{4}a_{r}\varphi_{r}\left(\rho,s\right)$
(the constants $a_{r}$ possibly depending on the spectral parameter) satisfy
the differential equation (3.15). It is immediately obvious from Theorem 3.2
that $D_{1}\left(\lambda\right)=0$ is equivalent to the condition that the
system of equations

$\left\\{\begin{split}\sum^{4}_{r=1}a_{r}\varphi_{r}\left(\rho,1\right)&=0,\\\\[1.00006pt]
\sum^{4}_{r=1}a_{r}\varphi^{\prime\prime}_{r}\left(\rho,1\right)&=0,\\\\[1.00006pt]
\sum^{4}_{r=1}a_{r}\,\bigl{[}\varphi^{\prime\prime}_{r}\left(\rho,0\right)-(\alpha+i\kappa\rho^{2})\,\varphi_{r}\left(\rho,0\right)\bigr{]}&=0,\\\\[1.00006pt]
\sum^{4}_{r=1}a_{r}\,\bigl{[}\varphi_{r}^{\left(3\right)}\left(\rho,0\right)-(\gamma-\eta^{2})\,\varphi_{r}^{\prime}\left(\rho,0\right)+i\beta\eta\rho^{2}\varphi_{r}\left(\rho,0\right)\bigr{]}&=0\end{split}\right.$
(3.16)

should have nonzero solutions. Similarly, $D_{2}\left(\lambda\right)=0$ in
Theorem 3.2 is equivalent to the condition that

$\left\\{\begin{split}\sum^{4}_{r=1}a_{r}\varphi_{r}\left(\rho,1\right)&=0,\\\\[1.00006pt]
\sum^{4}_{r=1}a_{r}\varphi^{\prime\prime}_{r}\left(\rho,1\right)&=0,\\\\[1.00006pt]
\sum^{4}_{r=1}a_{r}\,\bigl{[}\varphi^{\prime\prime}_{r}\left(\rho,0\right)-(\alpha+i\kappa\rho^{2})\,\varphi_{r}\left(\rho,0\right)\bigr{]}&=0,\\\\[1.00006pt]
\sum^{4}_{r=1}a_{r}\varphi_{r}\left(\rho,0\right)&=0\end{split}\right.$ (3.17)

should have nonzero solutions. Let

$\begin{split}\Delta_{1}\left(\lambda\right)&\coloneqq\left|\begin{matrix}\varphi_{1}\left(\rho,1\right)\\\\[1.00006pt]
\varphi_{1}^{\prime\prime}\left(\rho,1\right)\\\\[1.00006pt]
\varphi_{1}^{\prime\prime}\left(\rho,0\right)-(\alpha+i\kappa\rho^{2})\,\varphi_{1}^{\prime}\left(\rho,0\right)\\\\[1.00006pt]
\varphi^{\left(3\right)}_{1}\left(\rho,0\right)-(\gamma-\eta^{2})\,\varphi^{\prime}_{1}\left(\rho,0\right)+i\beta\eta\rho^{2}\varphi_{1}\left(\rho,0\right)\end{matrix}\right.\\\\[1.99997pt]
&\hskip 65.04034pt\begin{matrix}\varphi_{2}\left(\rho,1\right)\\\\[1.00006pt]
\varphi_{2}^{\prime\prime}\left(\rho,1\right)\\\\[1.00006pt]
\varphi_{2}^{\prime\prime}\left(\rho,0\right)-(\alpha+i\kappa\rho^{2})\,\varphi_{2}^{\prime}\left(\rho,0\right)\\\\[1.00006pt]
\varphi^{\left(3\right)}_{2}\left(\rho,0\right)-(\gamma-\eta^{2})\,\varphi^{\prime}_{2}\left(\rho,0\right)+i\beta\eta\rho^{2}\varphi_{2}\left(\rho,0\right)\end{matrix}\\\\[1.99997pt]
&\hskip 65.04034pt\begin{matrix}\varphi_{3}\left(\rho,1\right)\\\\[1.00006pt]
\varphi_{3}^{\prime\prime}\left(\rho,1\right)\\\\[1.00006pt]
\varphi_{3}^{\prime\prime}\left(\rho,0\right)-(\alpha+i\kappa\rho^{2})\,\varphi_{3}^{\prime}\left(\rho,0\right)\\\\[1.00006pt]
\varphi^{\left(3\right)}_{3}\left(\rho,0\right)-(\gamma-\eta^{2})\,\varphi^{\prime}_{3}\left(\rho,0\right)+i\beta\eta\rho^{2}\varphi_{3}\left(\rho,0\right)\end{matrix}\\\\[1.99997pt]
&\left.\hskip
130.08731pt\begin{matrix}\varphi_{4}\left(\rho,1\right)\\\\[1.00006pt]
\varphi_{4}^{\prime\prime}\left(\rho,1\right)\\\\[1.00006pt]
\varphi_{4}^{\prime\prime}\left(\rho,0\right)-(\alpha+i\kappa\rho^{2})\,\varphi_{4}^{\prime}\left(\rho,0\right)\\\\[1.00006pt]
\varphi^{\left(3\right)}_{4}\left(\rho,0\right)-(\gamma-\eta^{2})\,\varphi^{\prime}_{4}\left(\rho,0\right)+i\beta\eta\rho^{2}\varphi_{4}\left(\rho,0\right)\end{matrix}\right|\end{split}$

and

$\begin{split}\Delta_{2}\left(\lambda\right)&\coloneqq\left|\begin{matrix}\varphi_{1}\left(\rho,1\right)&\varphi_{2}\left(\rho,1\right)\\\\[1.00006pt]
\varphi_{1}^{\prime\prime}\left(\rho,1\right)&\varphi_{2}^{\prime\prime}\left(\rho,1\right)\\\\[1.00006pt]
\varphi_{1}^{\prime\prime}\left(\rho,0\right)-(\alpha+i\kappa\rho^{2})\,\varphi_{1}^{\prime}\left(\rho,0\right)&\varphi_{2}^{\prime\prime}\left(\rho,0\right)-(\alpha+i\kappa\rho^{2})\,\varphi_{2}^{\prime}\left(\rho,0\right)\\\\[1.00006pt]
\varphi_{1}\left(\rho,0\right)&\varphi_{2}\left(\rho,0\right)\end{matrix}\right.\\\\[1.99997pt]
&\quad\left.\hskip
86.72267pt\begin{matrix}\varphi_{3}\left(\rho,1\right)&\varphi_{4}\left(\rho,1\right)\\\\[1.00006pt]
\varphi_{3}^{\prime\prime}\left(\rho,1\right)&\varphi_{4}^{\prime\prime}\left(\rho,1\right)\\\\[1.00006pt]
\varphi_{3}^{\prime\prime}\left(\rho,0\right)-(\alpha+i\kappa\rho^{2})\,\varphi_{3}^{\prime}\left(\rho,0\right)&\varphi_{4}^{\prime\prime}\left(\rho,0\right)-(\alpha+i\kappa\rho^{2})\,\varphi_{4}^{\prime}\left(\rho,0\right)\\\\[1.00006pt]
\varphi_{3}\left(\rho,0\right)&\varphi_{4}\left(\rho,0\right)\end{matrix}\right|\end{split}$

be the characteristic determinants associated with (3.16) and (3.17),
respectively. Clearly the zeros of the characteristic functions $D_{1}$,
$D_{2}$ coincide (counted with multiplicities) with those of $\Delta_{1}$,
$\Delta_{2}$, respectively. So there are again two cases.

  1. Case 1. Asymptotic zeros of $\Delta_{1}$. It is a straightforward calculation to show that

$\displaystyle\varphi_{r}\left(\rho,1\right)$
$\displaystyle=e^{\rho\omega_{r}}e^{-\frac{i\beta\eta}{2}\omega^{2}_{r}}$
$\displaystyle\quad\times\left[1+\frac{1}{4\rho\omega_{r}}\left(\frac{\beta^{2}\eta^{2}}{2}+\gamma-\eta^{2}\right)+\mathcal{O}\,(\rho^{-2})\right],$
$\displaystyle\varphi^{\prime\prime}_{r}\left(\rho,1\right)$
$\displaystyle=\left(\rho\omega_{r}\right)^{2}e^{\rho\omega_{r}}e^{-\frac{i\beta\eta}{2}\omega^{2}_{r}}\,\biggl{[}1+\frac{1}{4\rho\omega_{r}}\,\biggl{(}\frac{\beta^{2}\eta^{2}}{2}+\gamma-\eta^{2}$
$\displaystyle\quad-4i\beta\eta\omega^{2}_{r}\biggr{)}+\mathcal{O}\,(\rho^{-2})\biggr{]},$
$\displaystyle\varphi^{\prime\prime}_{r}\left(\rho,0\right)-(\alpha+i\kappa\rho^{2})\,\varphi^{\prime}_{r}\left(\rho,0\right)$
$\displaystyle=i\kappa\rho^{2}\left(\rho\omega_{r}\right)\left(-1+\frac{i\beta\eta}{2\rho}\omega_{r}+\frac{\omega_{r}}{i\kappa\rho}+\mathcal{O}\,(\rho^{-2})\right),$
$\displaystyle\varphi^{\left(3\right)}_{r}\left(\rho,0\right)-(\gamma-\eta^{2})\,\varphi^{\prime}_{r}\left(\rho,0\right)+i\beta\eta\rho^{2}\varphi_{r}\left(\rho,0\right)$
$\displaystyle=\left(\rho\omega_{r}\right)^{3}\left(1-\frac{3i\beta\eta}{2\rho}\omega_{r}+\frac{i\beta\eta}{\rho}\omega_{r}+\mathcal{O}\,(\rho^{-2})\right)$

for $r=1,2,3,4$. Substituting these expressions in the equation
$\Delta_{1}\left(\lambda\right)=0$ and performing elementary computations, we
obtain that for $\left|\rho\right|$ large,

$\begin{split}&\mathopen{\bBigg@{6.5}}|\begin{matrix}e^{i\rho}e^{\frac{i\beta\eta}{2}}\left[1+\frac{1}{4i\rho}\left(\frac{\beta^{2}\eta^{2}}{2}+\gamma-\eta^{2}\right)+\mathcal{O}\,(\rho^{-2})\right]\\\\[1.00006pt]
-e^{i\rho}e^{\frac{i\beta\eta}{2}}\left[1+\frac{1}{4i\rho}\left(\frac{\beta^{2}\eta^{2}}{2}+\gamma-\eta^{2}+4i\beta\eta\right)+\mathcal{O}\,(\rho^{-2})\right]\\\\[1.00006pt]
-\kappa\left(-1-\frac{\beta\eta}{2\rho}+\frac{1}{\kappa\rho}+\mathcal{O}\,(\rho^{-2})\right)\\\\[1.00006pt]
-i\left(1+\frac{\beta\eta}{2\rho}+\mathcal{O}\,(\rho^{-2})\right)\end{matrix}\\\\[1.99997pt]
&\hskip 43.36464pt\begin{matrix}0\\\\[1.00006pt] 0\\\\[1.00006pt]
-i\kappa\left(-1-\frac{i\beta\eta}{2\rho}-\frac{1}{i\kappa\rho}+\mathcal{O}\,(\rho^{-2})\right)\\\\[1.00006pt]
-\left(1+\frac{i\beta\eta}{2\rho}+\mathcal{O}\,(\rho^{-2})\right)\\\\[8.99994pt]
e^{-i\rho}e^{\frac{i\beta\eta}{2}}\left[1-\frac{1}{4i\rho}\left(\frac{\beta^{2}\eta^{2}}{2}+\gamma-\eta^{2}\right)+\mathcal{O}\,(\rho^{-2})\right]\\\\[1.00006pt]
-e^{-i\rho}e^{\frac{i\beta\eta}{2}}\left[1-\frac{1}{4i\rho}\left(\frac{\beta^{2}\eta^{2}}{2}+\gamma-\eta^{2}+4i\beta\eta\right)+\mathcal{O}\,(\rho^{-2})\right]\\\\[1.00006pt]
\kappa\left(-1+\frac{\beta\eta}{2\rho}-\frac{1}{\kappa\rho}+\mathcal{O}\,(\rho^{-2})\right)\\\\[1.00006pt]
i\left(1-\frac{\beta\eta}{2\rho}+\mathcal{O}\,(\rho^{-2})\right)\end{matrix}\\\\[1.99997pt]
&\hskip
86.72267pt\begin{matrix}e^{-\frac{i\beta\eta}{2}}\left[1+\frac{1}{4\rho}\left(\frac{\beta^{2}\eta^{2}}{2}+\gamma-\eta^{2}\right)+\mathcal{O}\,(\rho^{-2})\right]\\\\[1.00006pt]
e^{-\frac{i\beta\eta}{2}}\left[1+\frac{1}{4\rho}\left(\frac{\beta^{2}\eta^{2}}{2}+\gamma-\eta^{2}-4i\beta\eta\right)+\mathcal{O}\,(\rho^{-2})\right]\\\\[1.00006pt]
0\\\\[1.00006pt]
0\end{matrix}\mathclose{\bBigg@{6.5}}|+\mathcal{O}\,(e^{-b\left|\rho\right|})=0,\end{split}$

where we have used that $\omega_{1}^{2}=\omega_{3}^{2}=-1$,
$\omega_{2}^{2}=\omega_{4}^{2}=1$ and
$e^{\rho\omega_{2}}=e^{-\rho\omega_{4}}=e^{-\rho}$,
$e^{\rho\omega_{1}}=e^{-\rho\omega_{3}}=e^{-i\rho}$ in ${S}$. So
$\Delta_{1}\left(\lambda\right)$ has an asymptotic representation of the form

$\displaystyle\Delta_{1}\left(\lambda\right)$
$\displaystyle=-8\cos\rho+\frac{\cos\rho}{\rho}\left[-\beta^{2}\eta^{2}-2\,(\gamma-\eta^{2})+\frac{4i}{\kappa}\right]$
$\displaystyle\qquad+\frac{i\sin\rho}{\rho}\left[i\beta^{2}\eta^{2}+2i\,(\gamma-\eta^{2})-\frac{4}{\kappa}\right]+\mathcal{O}\,(\rho^{-2})+\mathcal{O}\,(e^{-b\left|\rho\right|}).$

Let the sequence $\left\\{\lambda_{n}\right\\}$ represent the roots of the
equation $\frac{1}{2}\Delta_{1}\left(\lambda\right)=0$ and set
$\rho_{n}=\left(n+\frac{1}{2}\right)\pi+z_{n}$. Since

$\displaystyle\cos\rho_{n}$
$\displaystyle=\cos\left(n+\frac{1}{2}\right)\pi\cos
z_{n}-\sin\left(n+\frac{1}{2}\right)\pi\sin z_{n}=-\left(-1\right)^{n}\sin
z_{n},$ $\displaystyle\sin\rho_{n}$
$\displaystyle=\sin\left(n+\frac{1}{2}\right)\pi\cos
z_{n}+\cos\left(n+\frac{1}{2}\right)\pi\sin z_{n}=\left(-1\right)^{n}\cos
z_{n},$

it follows that the $z_{n}$ satisfy

$\sin z_{n}=\frac{\cos
z_{n}}{4\rho_{n}}\left(\frac{\beta^{2}\eta^{2}}{2}+\gamma-\eta^{2}+\frac{2i}{\kappa}\right)+\mathcal{O}\,(\rho^{-2}_{n})+\mathcal{O}\,(e^{-b\left|\rho_{n}\right|}).$

Thus, for large $n$,

$z_{n}=\frac{\frac{\beta^{2}\eta^{2}}{2}+\gamma-\eta^{2}+\frac{2i}{\kappa}}{4\left(n+\frac{1}{2}\right)\pi}+\mathcal{O}\,(n^{-2}).$

We write $\rho_{n}=\tau_{n}+z_{n}$ where
$\tau_{n}=\left(n+\frac{1}{2}\right)\pi$ to obtain, taking into account
$\lambda_{n}=i\rho_{n}^{2}=i\left(\tau_{n}+z_{n}\right)^{2}$,

$\lambda_{n}=-\frac{1}{\kappa}+i\left(\tau_{n}^{2}+\frac{\frac{\beta^{2}\eta^{2}}{2}+\gamma-\eta^{2}}{2}\right)+\mathcal{O}\,(\tau_{n}^{-1}),\quad\tau_{n}=\left(n+\frac{1}{2}\right)\pi.$

  2. Case 2. Asymptotic zeros of $\Delta_{2}$. In this case,

$\displaystyle\varphi_{r}\left(\rho,1\right)$
$\displaystyle=e^{\rho\omega_{r}}e^{-\frac{i\beta\eta}{2}\omega^{2}_{r}}$
$\displaystyle\quad\times\left[1+\frac{1}{4\rho\omega_{r}}\left(\frac{\beta^{2}\eta^{2}}{2}+\gamma-\eta^{2}\right)+\mathcal{O}\,(\rho^{-2})\right],$
$\displaystyle\varphi^{\prime\prime}_{r}\left(\rho,1\right)$
$\displaystyle=\left(\rho\omega_{r}\right)^{2}e^{\rho\omega_{r}}e^{-\frac{i\beta\eta}{2}\omega^{2}_{r}}\,\biggl{[}1+\frac{1}{4\rho\omega_{r}}\,\biggl{(}\frac{\beta^{2}\eta^{2}}{2}+\gamma-\eta^{2}$
$\displaystyle\quad-4i\beta\eta\omega^{2}_{r}\biggr{)}+\mathcal{O}\,(\rho^{-2})\biggr{]},$
$\displaystyle\varphi^{\prime\prime}_{r}\left(\rho,0\right)-(\alpha+i\kappa\rho^{2})\,\varphi^{\prime}_{r}\left(\rho,0\right)$
$\displaystyle=i\kappa\rho^{2}\left(\rho\omega_{r}\right)\left(-1+\frac{i\beta\eta}{2\rho}\omega_{r}+\frac{\omega_{r}}{i\kappa\rho}+\mathcal{O}\,(\rho^{-2})\right),$
$\displaystyle\varphi_{r}\left(\rho,0\right)$
$\displaystyle=1+\mathcal{O}\,(\rho^{-2})$

for $r=1,2,3,4$ and hence, using arguments analogous to those given in Case 1
we compute

$\Delta_{2}\left(\lambda\right)=4\sqrt{2}\cos\left(\rho+\frac{\pi}{4}\right)+\frac{\cos\rho}{\rho}\left[\beta^{2}\eta^{2}+2\,(\gamma-\eta^{2})\right]+\frac{i\sin\rho}{\rho}\left(\frac{8}{\kappa}\right)+\mathcal{O}\,(\rho^{-2})+\mathcal{O}\,(e^{-b\left|\rho\right|}).$

Let the sequence $\left\\{\lambda_{n}\right\\}$ be the roots of the equation
$\Delta_{2}\left(\lambda\right)=0$. Set
$\rho_{n}=\left(n+\frac{1}{4}\right)\pi+z_{n}$. Since

$\displaystyle\cos\rho_{n}$
$\displaystyle=\cos\left(n+\frac{1}{4}\right)\pi\cos
z_{n}-\sin\left(n+\frac{1}{4}\right)\pi\sin
z_{n}=\frac{\sqrt{2}}{2}\left(-1\right)^{n}\cos
z_{n}-\frac{\sqrt{2}}{2}\left(-1\right)^{n}\sin z_{n},$
$\displaystyle\sin\rho_{n}$
$\displaystyle=\sin\left(n+\frac{1}{4}\right)\pi\cos
z_{n}+\cos\left(n+\frac{1}{4}\right)\pi\sin
z_{n}=\frac{\sqrt{2}}{2}\left(-1\right)^{n}\cos
z_{n}+\frac{\sqrt{2}}{2}\left(-1\right)^{n}\sin z_{n},$

proceeding as before, we have that the $z_{n}$ in this case satisfy

$\sin z_{n}=\frac{\cos
z_{n}}{4\rho_{n}}\left(\frac{\beta^{2}\eta^{2}}{2}+\gamma-\eta^{2}+\frac{4i}{\kappa}\right)+\mathcal{O}\,(\rho^{-2}_{n})+\mathcal{O}\,(e^{-b\left|\rho_{n}\right|}).$

Therefore, for $n$ large,

$z_{n}=\frac{\frac{\beta^{2}\eta^{2}}{2}+\gamma-\eta^{2}+\frac{4i}{\kappa}}{4\left(n+\frac{1}{4}\right)\pi}+\mathcal{O}\,(n^{-2}),$

and a similar calculation to the above shows that

$\lambda_{n}=-\frac{2}{\kappa}+i\left(\tau_{n}^{2}+\frac{\frac{\beta^{2}\eta^{2}}{2}+\gamma-\eta^{2}}{2}\right)+\mathcal{O}\,(\tau_{n}^{-1}),\quad\tau_{n}=\left(n+\frac{1}{4}\right)\pi.$

We collect the foregoing results together into the following theorem.

######  Theorem 3.3.

Let $\kappa>0$. The spectrum of $\mathcal{T}$ consists of two branches of a
discrete set of eigenvalues,
$\sigma\left(\mathcal{T}\right)=\sigma^{\left(1\right)}\left(\mathcal{T}\right)\cup\sigma^{\left(2\right)}\left(\mathcal{T}\right)$,
where asymptotically, for large $n$,

$\displaystyle\sigma^{\left(1\right)}\left(\mathcal{T}\right)$
$\displaystyle=\left\\{\lambda\in\mathbb{C}~{}\middle|~{}\Delta_{1}\left(\lambda\right)=0\right\\}=\\{\lambda^{\left(1\right)}_{n},\overline{\lambda^{\left(1\right)}_{n}}{\\}}^{\infty}_{n=0},$
$\displaystyle\sigma^{\left(2\right)}\left(\mathcal{T}\right)$
$\displaystyle=\left\\{\lambda\in\mathbb{C}~{}\middle|~{}\Delta_{2}\left(\lambda\right)=0\right\\}=\\{\lambda^{\left(2\right)}_{n},\overline{\lambda^{\left(2\right)}_{n}}{\\}}^{\infty}_{n=0},$

with the sequences enumerated properly (in the sense of [11, Remark 4.2]).
Both branches are confined to a vertical strip in the open left half-plane
such that, as $n\rightarrow\infty$,

$|\Re\,(\lambda^{(j)}_{n})|\leq
C<\infty,\quad\Im\,(\lambda^{(j)}_{n})\rightarrow\infty,\quad j=1,2,$

some constant $C$. The eigenvalues belonging to
$\sigma_{1}\left(\mathcal{T}\right)$ have asymptotic representations

$\lambda^{\left(1\right)}_{n}=-\frac{1}{\kappa}+i\left((\tau^{\left(1\right)}_{n}{)}^{2}+\frac{\frac{\beta^{2}\eta^{2}}{2}+\gamma-\eta^{2}}{2}\right)+\mathcal{O}\,\bigl{(}(\tau^{\left(1\right)}_{n}{)}^{-1}\bigr{)},\quad\tau^{\left(1\right)}_{n}=\left(n+\frac{1}{2}\right)\pi,\quad
n\rightarrow\infty,$

and those belonging to $\sigma_{2}\left(\mathcal{T}\right)$ have asymptotic
representations

$\lambda^{\left(2\right)}_{n}=-\frac{2}{\kappa}+i\left((\tau^{\left(2\right)}_{n}{)}^{2}+\frac{\frac{\beta^{2}\eta^{2}}{2}+\gamma-\eta^{2}}{2}\right)+\mathcal{O}\,\bigl{(}(\tau^{\left(2\right)}_{n}{)}^{-1}\bigr{)},\quad\tau^{\left(2\right)}_{n}=\left(n+\frac{1}{4}\right)\pi,\quad
n\rightarrow\infty.$

###  3.3. Multiplicity of Eigenvalues

For the study of eigenvalue multiplicities, we quote the following useful
result from [16, Corollary 4.2.2].

######  Lemma 3.2.

Let $\mathcal{A}$ be an operator on a Hilbert space $\mathscr{X}$, and let
$\lambda$ be an eigenvalue of $\mathcal{A}$ with corresponding eigenvector
$x$, i.e. $\left(\lambda\mathcal{I}-\mathcal{A}\right)x=0$, $x\neq 0$. If
there is a nonzero element
$z\in\ker\,(\overline{\lambda}\mathcal{I}-\mathcal{A}^{*})$ such that the
inner product $\left(x,z\right)\neq 0$, then $\lambda$ is a simple eigenvalue
if $\operatorname{dim}\ker\left(\lambda\mathcal{I}-\mathcal{A}\right)=1$.

In order to use Lemma 3.2, we need to consider the spectral problem for the
adjoint operator for $\mathcal{T}$, $\mathcal{T}^{*}$. First we prove a
proposition.

######  Proposition 3.1.

The adjoint operator $\mathcal{T}^{*}$ is defined on the domain

$\mathscr{D}\left(\mathcal{T}^{*}\right)=\left\\{z=\left\\{z_{k}\right\\}^{3}_{k=1}\in\mathscr{X}~{}\middle|~{}\begin{gathered}z_{k}=\left(\widetilde{w}_{k},\widetilde{v}_{k}\right)\in({H}^{4}\left(0,1\right)\cap{H}_{*}^{2}\left(0,1\right))\times{H}_{*}^{2}\left(0,1\right),\\\
\widetilde{w}^{\prime\prime}_{k}\left(1\right)=0,\quad\widetilde{w}^{\prime\prime}_{k}\left(0\right)-\alpha\widetilde{w}^{\prime}_{k}\left(0\right)+\kappa\widetilde{v}^{\prime}_{k}\left(0\right)=0,\quad
k=1,2,3,\\\
\sum_{k=1}^{3}\,\bigl{[}\widetilde{w}^{\left(3\right)}_{k}\left(0\right)-(\gamma-\eta^{2})\,\widetilde{w}_{k}^{\prime}\left(0\right)+\beta\eta\widetilde{v}_{k}\left(0\right)\bigr{]}=0\end{gathered}\right\\},$
(3.18)

by

$\mathcal{T}z\coloneqq\bigl{\\{}\bigl{(}-\widetilde{v}_{k},\widetilde{w}_{k}^{\left(4\right)}-(\gamma-\eta^{2})\,\widetilde{w}^{\prime\prime}_{k}+2\beta\eta\widetilde{v}_{k}^{\prime}\bigr{)}\bigr{\\}}_{k=1}^{3}.$
(3.19)

###### Proof.

The proof is a formal calculation. First note that
$x\in\mathscr{D}\left(\mathcal{T}\right)\left(=\mathscr{D}\left(\mathcal{A}\right)\right)$
implies $v_{k}\left(1\right)=0$, $k=1,2,3$, and
$v_{j}\left(0\right)=v_{k}\left(0\right)$, $j,k=1,2,3$. Using this, it follows
from integration by parts that if $x\in\mathscr{D}\left(\mathcal{T}\right)$,
then for any
$z=\left\\{\left(\widetilde{w}_{k},\widetilde{v}_{k}\right)\right\\}^{3}_{k=1}\in{H}^{4}\left(\mathbf{G}\right)\times{H}^{2}\left(\mathbf{G}\right)\left(\supset\mathscr{D}\left(\mathcal{T}^{*}\right)\right)$,

$\displaystyle\left(\mathcal{T}x,z\right)_{\mathscr{X}}$
$\displaystyle=-\sum_{k=1}^{3}\left[\int^{1}_{0}w^{\prime\prime}_{k}\left(s\right)\overline{\widetilde{v}^{\prime\prime}_{k}\left(s\right)}ds+(\gamma-\eta^{2})\int^{1}_{0}w^{\prime}_{k}\left(s\right)\overline{\widetilde{v}^{\prime}_{k}\left(s\right)}ds+\alpha
w^{\prime}_{k}\left(0\right)\overline{\widetilde{v}^{\prime}_{k}\left(0\right)}\right]$
$\displaystyle\qquad+\sum_{k=1}^{3}\int^{1}_{0}v_{k}\left(s\right)\left[\overline{\widetilde{w}^{\left(4\right)}_{k}\left(s\right)}-(\gamma-\eta^{2})\,\overline{\widetilde{w}^{\prime\prime}_{k}\left(s\right)}+2\beta\eta\overline{\widetilde{v}^{\prime}_{k}\left(s\right)}\right]ds$
$\displaystyle\qquad+\sum_{k=1}^{3}v_{k}^{\prime}\left(1\right)\overline{\widetilde{w}_{k}^{\prime\prime}\left(1\right)}-\sum_{k=1}^{3}\biggl{[}w^{\left(3\right)}_{k}\left(1\right)-(\gamma-\eta^{2})\,w^{\prime}_{k}\left(1\right)+2\beta\eta
v_{k}\left(1\right)\biggr{]}\,\overline{\widetilde{v}_{k}\left(1\right)}$
$\displaystyle\qquad-\sum_{k=1}^{3}v_{k}^{\prime}\left(0\right)\Bigl{(}\overline{\widetilde{w}_{k}^{\prime\prime}\left(0\right)}-\alpha\overline{\widetilde{w}_{k}^{\prime}\left(0\right)}+\kappa\widetilde{v}^{\prime}_{k}\left(0\right)\Bigr{)}$
$\displaystyle\qquad+\sum_{k=1}^{3}v_{k}\left(0\right)\left[\overline{\widetilde{w}^{\left(3\right)}_{k}\left(0\right)}-(\gamma-\eta^{2})\,\overline{\widetilde{w}_{k}^{\prime}\left(0\right)}+\beta\eta\widetilde{v}^{\prime}_{k}\left(0\right)\right]$
$\displaystyle\qquad-\sum_{k=1}^{3}\Bigl{(}w^{\prime\prime}_{k}\left(0\right)-\alpha
w^{\prime}_{k}\left(0\right)-\kappa
v^{\prime}_{k}\left(0\right)\Bigr{)}\,\overline{\widetilde{v}^{\prime}_{k}\left(0\right)}$
$\displaystyle\qquad+\sum_{k=1}^{3}\biggl{[}w^{\left(3\right)}_{k}\left(0\right)-(\gamma-\eta^{2})\,w^{\prime}_{k}\left(0\right)+\beta\eta
v_{k}\left(0\right)\biggr{]}\,\overline{\widetilde{v}_{k}\left(0\right)}$

Let $\widetilde{v}_{k}\left(1\right)=0$, $k=1,2,3$, and let
$\widetilde{v}_{j}\left(0\right)=\widetilde{v}_{k}\left(0\right)$,
$j,k=1,2,3$. Further, let
$\widetilde{w}_{k}\left(1\right)=\widetilde{w}^{\prime\prime}_{k}\left(1\right)=0$,
$\widetilde{w}^{\prime\prime}_{k}\left(0\right)-\alpha\widetilde{w}^{\prime}_{k}\left(0\right)+\kappa\widetilde{v}^{\prime}_{k}\left(0\right)=0$,
$k=1,2,3$, $\widetilde{w}_{j}\left(0\right)=\widetilde{w}_{k}\left(0\right)$,
$j,k=1,2,3$, and
$\sum^{3}_{k=1}\,\bigl{[}\widetilde{w}^{\left(3\right)}_{k}\left(0\right)-(\gamma-\eta^{2})\,\widetilde{w}_{k}^{\prime}\left(0\right)+\beta\eta\widetilde{v}_{k}\left(0\right)\bigr{]}=0$.
Then we have
$\left\\{\left(\widetilde{w}_{k},\widetilde{v}_{k}\right)\right\\}^{3}_{k=1}\in\mathscr{D}\left(\mathcal{T}^{*}\right)$
and

$\displaystyle\left(\mathcal{T}x,z\right)_{\mathscr{X}}$
$\displaystyle=\left(x,\mathcal{T}^{*}z\right)_{\mathscr{X}}$
$\displaystyle=-\sum_{k=1}^{3}\left[\int^{1}_{0}w^{\prime\prime}_{k}\left(s\right)\overline{\widetilde{v}^{\prime\prime}_{k}\left(s\right)}ds+(\gamma-\eta^{2})\int^{1}_{0}w^{\prime}_{k}\left(s\right)\overline{\widetilde{v}^{\prime}_{k}\left(s\right)}ds+\alpha
w^{\prime}_{k}\left(0\right)\overline{\widetilde{v}^{\prime}_{k}\left(0\right)}\right]$
$\displaystyle\qquad+\sum_{k=1}^{3}\int^{1}_{0}v_{k}\left(s\right)\left[\overline{\widetilde{w}^{\left(4\right)}_{k}\left(s\right)}-(\gamma-\eta^{2})\,\overline{\widetilde{w}^{\prime\prime}_{k}\left(s\right)}+2\beta\eta\overline{\widetilde{v}^{\prime}_{k}\left(s\right)}\right]ds$

where $\mathscr{D}\left(\mathcal{T}^{*}\right)$, $\mathcal{T}^{*}$ are as in
the lemma, completing the proof. ∎

Let us now consider the spectral problem for $\mathcal{T}^{*}$ defined by
(3.18), (3.19),

$\mathcal{T}^{*}z=\mu z,\quad
z\in\mathscr{D}\left(\mathcal{T}^{*}\right),\quad\mu\in\mathbb{C},$ (3.20)

which in coordinates is given by

$\left\\{\begin{aligned}
\widetilde{w}^{\left(4\right)}_{k}-(\gamma-\eta^{2})\,\widetilde{w}^{\prime\prime}_{k}-2\mu\beta\eta\widetilde{w}_{k}^{\prime}&=-\mu^{2}\widetilde{w}_{k},&k&=1,2,3,\\\
\widetilde{w}_{k}\left(1\right)=\widetilde{w}^{\prime\prime}_{k}\left(1\right)&=0,&k&=1,2,3,\\\
\widetilde{w}_{j}\left(0\right)&=\widetilde{w}_{k}\left(0\right),&j,k&=1,2,3,\\\
\widetilde{w}^{\prime\prime}_{k}\left(0\right)-\left(\alpha+\mu\kappa\right)\widetilde{w}^{\prime}_{k}\left(0\right)&=0,&k&=1,2,3,\\\
\sum^{3}_{k=1}\,\bigl{[}\widetilde{w}^{\left(3\right)}_{k}\left(0\right)-(\gamma-\eta^{2})\,\widetilde{w}_{k}^{\prime}\left(0\right)+\mu\beta\eta\widetilde{w}_{k}\left(0\right)\bigr{]}&=0,&&&\end{aligned}\right.$
(3.21)

where we have taken into account that $v_{k}=-\mu\widetilde{w}_{k}$,
$k=1,2,3$. Proceeding formally as in the proof of Theorem 3.2 we let
$\psi=\psi\left(\mu,s\right)$ be a nonzero solution of the differential
equation

$\psi^{\left(4\right)}-(\gamma-\eta^{2})\,\psi^{\prime\prime}-2\mu\beta\eta\psi_{k}^{\prime}=-\mu^{2}\psi$
(3.22)

satisfying the boundary conditions

$\displaystyle\psi\left(1\right)=\psi^{\prime\prime}\left(1\right)$
$\displaystyle=0,$ (3.23)
$\displaystyle\psi^{\prime\prime}\left(0\right)-\left(\alpha+\mu\kappa\right)\psi^{\prime}\left(0\right)$
$\displaystyle=0.$ (3.24)

Obviously,
$\psi\,(\overline{\mu},\,\cdot\,)=\overline{\psi\,(\mu,\,\cdot\,)}$. It is
then clear that solutions $w_{k}=w_{k}\left(\mu,s\right)$ of (3.21) are of the
form $w_{k}\left(\mu,s\right)=b_{k}\psi\left(\mu,s\right)$, $k=1,2,3$, where
$b_{k}$ are arbitrary constants, and therefore we have in the case
$\psi\left(\mu,0\right)\neq 0$ that $\mu$ is an eigenvalue if and only if
(3.22)–(3.24) with the supplementary requirement

$\psi^{\left(3\right)}\left(\mu,0\right)-(\gamma-\eta^{2})\,\psi^{\prime}\left(\mu,0\right)-\mu\beta\eta\psi\left(\mu,0\right)=0$

has a nonzero solution. The corresponding eigenvector $z=z\left(\mu\right)$ is
given by

$z\left(\mu\right)=\left\\{b_{k}\left(\psi\left(\mu,\,\cdot\,\right),-\mu\psi\left(\mu,\,\cdot\,\right)\right)\right\\}^{3}_{k=1}$
(3.25)

with the $b_{k}\equiv b\neq 0$.

In the case

$\psi\left(\mu,0\right)=0,$

again continuing as in the proof of Theorem 3.2, we have
$\sum^{3}_{k=1}b_{k}=0$ and it follows that there are two linearly independent
eigenvectors for the eigenvalue $\mu$, $z_{1}=z_{1}\left(\mu\right)$,
$z_{2}=z_{2}\left(\mu\right)$, given by

$\displaystyle z_{1}\left(\mu\right)$
$\displaystyle=\bigl{\\{}\left(\psi\left(\mu,\,\cdot\,\right),-\mu\psi\left(\mu,\,\cdot\,\right)\right);-\frac{1}{2}\left(\psi\left(\mu,\,\cdot\,\right),-\mu\psi\left(\mu,\,\cdot\,\right)\right);-\frac{1}{2}\left(\psi\left(\mu,\,\cdot\,\right),-\mu\psi\left(\mu,\,\cdot\,\right)\right)\bigr{\\}},$
(3.26) $\displaystyle z_{2}\left(\mu\right)$
$\displaystyle=\bigl{\\{}\left(0,0\right);\left(\psi\left(\mu,\,\cdot\,\right),-\mu\psi\left(\mu,\,\cdot\,\right)\right);-\left(\psi\left(\mu,\,\cdot\,\right),-\mu\psi\left(\mu,\,\cdot\,\right)\right)\bigr{\\}},$
(3.27)

respectively.

We can now turn to the verification of the condition
$\left(x,z\right)_{\mathscr{X}}\neq 0$ in Lemma 3.2. To do this note first
that, since $\mathscr{X}$ is a Hilbert space,
$\sigma\left(\mathcal{T}^{*}\right)=\overline{\sigma\left(\mathcal{T}\right)}$.
Let $\lambda\in\sigma\left(\mathcal{T}\right)$,
$\mu\in\sigma\left(\mathcal{T}^{*}\right)$, and let
$x\left(\lambda\right)=\left\\{\left(w_{k}\left(\lambda,\,\cdot\,\right),v_{k}\left(\lambda,\,\cdot\,\right)\right)\right\\}_{k=1}^{3}$
and
$z\left(\mu\right)=\left\\{\left(\widetilde{w}_{k}\left(\mu,\,\cdot\,\right),\widetilde{v}_{k}\left(\mu,\,\cdot\,\right)\right)\right\\}_{k=1}^{3}$
be the eigenvectors corresponding to $\lambda$ and $\mu$, respectively. We
take the inner product of (2.9) with $z\left(\mu\right)$ and obtain, taking
into account (3.20),

$\lambda\left(x\left(\lambda\right),z\left(\mu\right)\right)_{\mathscr{X}}=\left(\mathcal{T}x\left(\lambda\right),z\left(\mu\right)\right)_{\mathscr{X}}=\left(x\left(\lambda\right),\mathcal{T}^{*}z\left(\mu\right)\right)_{\mathscr{X}}=\overline{\mu}\left(x\left(\lambda\right),z\left(\mu\right)\right)_{\mathscr{X}}.$

Consequently, for $\mu\neq\overline{\lambda}$,
$\left(x\left(\lambda\right),z\left(\mu\right)\right)_{\mathscr{X}}=0$. If we
set $\mu=\overline{\lambda}$,

$(x\left(\lambda\right),z\,(\overline{\lambda}){)}_{\mathscr{X}}=B\left(\lambda\right)\sum_{k=1}^{3}c_{k}\overline{b_{k}}$
(3.28)

with

$\displaystyle B\left(\lambda\right)$
$\displaystyle\coloneqq\int^{1}_{0}\varphi^{\prime\prime}\left(\lambda,s\right)\psi^{\prime\prime}\left(\lambda,s\right)ds+(\gamma-\eta^{2})\int^{1}_{0}\varphi^{\prime}\left(\lambda,s\right)\psi^{\prime}\left(\lambda,s\right)ds$
$\displaystyle\qquad+\alpha\varphi^{\prime}\left(\lambda,0\right)\psi^{\prime}\left(\lambda,0\right)-\lambda^{2}\int_{0}^{1}\varphi\left(\lambda,s\right)\psi\left(\lambda,s\right)ds,$

which is easily verified on using the fact that, by (3.12) and (3.25),
$w_{k}\left(\lambda,\,\cdot\,\right)=c_{k}\varphi\left(\lambda,\,\cdot\,\right)$,
$v_{k}\left(\lambda,\,\cdot\,\right)=c_{k}\lambda\varphi\left(\lambda,\,\cdot\,\right)$,
$\widetilde{w}_{k}\left(\lambda,\,\cdot\,\right)=b_{k}\psi\,(\overline{\lambda},\,\cdot\,)$,
$\widetilde{v}_{k}\left(\lambda,\,\cdot\,\right)=-b_{k}\overline{\lambda}\psi\,(\overline{\lambda},\,\cdot\,)$,
$k=1,2,3$, and that
$\overline{\psi\,(\overline{\lambda},\,\cdot\,)}=\psi\,(\lambda,\,\cdot\,)$.
Now integration by parts in the first and second integrals in
$B\left(\lambda\right)$ yields

$\displaystyle B\left(\lambda\right)$
$\displaystyle=-\lambda^{2}\int_{0}^{1}\varphi\left(\lambda,s\right)\psi\left(\lambda,s\right)ds+\int_{0}^{1}\biggl{[}\varphi^{\left(4\right)}\left(\lambda,s\right)-(\gamma-\eta^{2})\,\varphi^{\prime\prime}\left(\lambda,s\right)\biggr{]}\,\psi\left(\lambda,s\right)ds$
$\displaystyle\qquad+\varphi^{\prime\prime}\left(\lambda,1\right)\psi^{\prime}\left(\lambda,1\right)-\left[\varphi^{\left(3\right)}\left(\lambda,1\right)-(\gamma-\eta^{2})\,\varphi^{\prime}\left(\lambda,1\right)\right]\psi\left(\lambda,1\right)$
$\displaystyle\qquad-\left(\varphi^{\prime\prime}\left(\lambda,0\right)-\alpha\varphi^{\prime}\left(\lambda,0\right)\right)\psi^{\prime}\left(\lambda,0\right)+\left[\varphi^{\left(3\right)}\left(\lambda,0\right)-(\gamma-\eta^{2})\,\varphi^{\prime}\left(\lambda,0\right)\right]\psi\left(\lambda,0\right).$

Inserting the differential equation (3.3) and using that
$\varphi^{\prime\prime}\left(1\right)=0$,
$\varphi^{\prime\prime}\left(0\right)-\left(\alpha+\lambda\kappa\right)\varphi\left(0\right)=0$,
$\psi\left(1\right)=0$, we can write

$\displaystyle B\left(\lambda\right)$
$\displaystyle=-2\lambda^{2}\int_{0}^{1}\varphi\left(\lambda,s\right)\psi\left(\lambda,s\right)ds-2\lambda\beta\eta\int^{1}_{0}\varphi^{\prime}\left(\lambda,s\right)\psi\left(\lambda,s\right)ds-\lambda\kappa\varphi^{\prime}\left(\lambda,0\right)\psi^{\prime}\left(\lambda,0\right)$
$\displaystyle\qquad+\left[\varphi^{\left(3\right)}\left(\lambda,0\right)-(\gamma-\eta^{2})\,\varphi^{\prime}\left(\lambda,0\right)+\lambda\beta\eta\varphi\left(\lambda,0\right)\right]\psi\left(\lambda,0\right)-\lambda\beta\eta\varphi\left(\lambda,0\right)\psi\left(\lambda,0\right)$
$\displaystyle=-2\lambda^{2}\int_{0}^{1}\varphi\left(\lambda,s\right)\psi\left(\lambda,s\right)ds+2\lambda\beta\eta\int^{1}_{0}\psi^{\prime}\left(\lambda,s\right)\varphi\left(\lambda,s\right)ds-\lambda\kappa\varphi^{\prime}\left(\lambda,0\right)\psi^{\prime}\left(\lambda,0\right)$
$\displaystyle\qquad+\left[\psi^{\left(3\right)}\left(\lambda,0\right)-(\gamma-\eta^{2})\,\psi^{\prime}\left(\lambda,0\right)-\lambda\beta\eta\psi\left(\lambda,0\right)\right]\varphi\left(\lambda,0\right)+\lambda\beta\eta\psi\left(\lambda,0\right)\varphi\left(\lambda,0\right).$

Note that

$\displaystyle
2\lambda\beta\eta\int^{1}_{0}\varphi^{\prime}\left(\lambda,s\right)\psi\left(\lambda,s\right)ds+\lambda\beta\eta\varphi\left(\lambda,0\right)\psi\left(\lambda,0\right)$
$\displaystyle\hskip
43.36464pt=\lambda\beta\eta\int^{1}_{0}\left(\varphi^{\prime}\left(\lambda,s\right)\psi\left(\lambda,s\right)-\varphi\left(\lambda,s\right)\psi^{\prime}\left(\lambda,s\right)\right)ds.$

Hence,

$\displaystyle B\left(\lambda\right)$
$\displaystyle=-2\lambda^{2}\int_{0}^{1}\varphi\left(\lambda,s\right)\psi\left(\lambda,s\right)ds-\lambda\beta\eta\int^{1}_{0}\left(\varphi^{\prime}\left(\lambda,s\right)\psi\left(\lambda,s\right)-\varphi\left(\lambda,s\right)\psi^{\prime}\left(\lambda,s\right)\right)ds$
$\displaystyle\qquad-\lambda\kappa\varphi^{\prime}\left(\lambda,0\right)\psi^{\prime}\left(\lambda,0\right)+\left[\varphi^{\left(3\right)}\left(\lambda,0\right)-(\gamma-\eta^{2})\,\varphi^{\prime}\left(\lambda,0\right)+\lambda\beta\eta\varphi\left(\lambda,0\right)\right]\psi\left(\lambda,0\right)$
$\displaystyle=-2\lambda^{2}\int_{0}^{1}\varphi\left(\lambda,s\right)\psi\left(\lambda,s\right)ds-\lambda\beta\eta\int^{1}_{0}\left(\varphi^{\prime}\left(\lambda,s\right)\psi\left(\lambda,s\right)-\varphi\left(\lambda,s\right)\psi^{\prime}\left(\lambda,s\right)\right)ds$
$\displaystyle\qquad-\lambda\kappa\varphi^{\prime}\left(\lambda,0\right)\psi^{\prime}\left(\lambda,0\right)+\Delta_{1}\left(\lambda\right)\Delta_{2}\left(\lambda\right)$

Therefore, since $\lambda\in\sigma\left(\mathcal{T}\right)$,

$\displaystyle B\left(\lambda\right)$
$\displaystyle=-\lambda\biggl{[}2\lambda\int_{0}^{1}\varphi\left(\lambda,s\right)\psi\left(\lambda,s\right)ds+\beta\eta\int^{1}_{0}\left(\varphi^{\prime}\left(\lambda,s\right)\psi\left(\lambda,s\right)-\varphi\left(\lambda,s\right)\psi^{\prime}\left(\lambda,s\right)\right)ds$
$\displaystyle\qquad+\kappa\varphi^{\prime}\left(\lambda,0\right)\psi^{\prime}\left(\lambda,0\right)\biggr{]}.$

Clearly if $\lambda$ belongs to the branch
$\sigma_{1}\left(\mathcal{T}\right)$ of the spectrum
$\sigma\left(\mathcal{T}\right)$, then from (3.28) we have

$(x\left(\lambda\right),z\,(\overline{\lambda}){)}_{\mathscr{X}}=3bcB\left(\lambda\right)\neq
0,$

as $bc\neq 0$. If $\lambda$ belongs to the branch
$\sigma_{2}\left(\mathcal{T}\right)$ of $\sigma\left(\mathcal{T}\right)$ we
have

$(x_{1}\left(\lambda\right),z_{1}\,(\overline{\lambda}){)}_{\mathscr{X}}=\frac{1}{2}B\left(\lambda\right)\neq
0,\quad(x_{2}\left(\lambda\right),z_{2}\,(\overline{\lambda}){)}_{\mathscr{X}}=2B\left(\lambda\right)\neq
0$

where the eigenvectors $x_{j}\left(\lambda\right)$,
$z_{j}\,(\overline{\lambda})$, $j=1,2$, have the forms given by (3.7), (3.26)
and (3.8), (3.27), respectively. The calculations above imply that associated
with an eigenvector $x$ corresponding to an eigenvalue
$\lambda\in\sigma_{1}\left(\mathcal{T}\right)$ there is a nontrivial element
$z\in\ker\left(\overline{\lambda}\mathcal{I}-\mathcal{T}^{*}\right)$ such that
$\left(x,z\right)_{\mathscr{X}}\neq 0$. Then each
$\lambda\in\sigma_{1}\left(\mathcal{T}\right)$ is simple since, by Theorem
3.2, it has a one-dimensional geometric eigenspace,
$\dim\ker\left(\lambda\mathcal{I}-\mathcal{T}\right)=1$. Similarly, it also
follows from the calculations above that associated with an eigenvector $x$
corresponding to an eigenvalue $\lambda\in\sigma_{2}\left(\mathcal{T}\right)$
there is a nontrivial element
$z\in\ker\left(\overline{\lambda}\mathcal{I}-\mathcal{T}^{*}\right)$ such that
$\left(x,z\right)_{\mathscr{X}}\neq 0$. Hence, we obtain that each
$\lambda\in\sigma_{2}\left(\mathcal{T}\right)$ is semisimple and has
multiplicity $2$ since
$\dim\ker\left(\lambda\mathcal{I}-\mathcal{T}\right)=2$. We have proven the
following theorem.

######  Theorem 3.4.

All eigenvalues of $\mathcal{T}$ are semisimple, i.e. there are no second-
order root vectors (associated vectors) corresponding to any eigenvalue
$\lambda\in\sigma\left(\mathcal{T}\right)$. Each eigenvalue belonging to
$\sigma_{1}\left(\mathcal{T}\right)$ has multiplicity $1$, and each eigenvalue
belonging to $\sigma_{2}\left(\mathcal{T}\right)$ has multiplicity $2$.

##  4\. Completeness, minimality, and Riesz basis properties of eigenvectors

We begin by collecting three familiar operator results from the literature
which we require in the sequel. The first is due to [6, Theorem V.8.1], the
second to [12, Lemma 2.4], and the third to [18, Theorem 1.1].

######  Lemma 4.1.

Let $\mathcal{K}$ be a compact skewadjoint operator on a Hilbert space
$\mathscr{X}$ with $\ker\mathcal{K}=\left\\{0\right\\}$, and let $\mathcal{S}$
be a real operator on $\mathscr{X}$ which has finite rank. Let

$\mathcal{A}=\mathcal{K}+\kappa\mathcal{S},\quad\kappa\geq 0.$

Then the root vectors of the operator $\mathcal{A}$ are complete in
$\mathscr{X}$.

######  Lemma 4.2.

Let $\mathcal{A}$ be a compact operator on $\mathscr{X}$ and
$\ker\mathcal{A}=\left\\{0\right\\}$. Then the root vectors of $\mathcal{A}$
are minimal in $\mathscr{X}$.

######  Lemma 4.3.

Let $\mathscr{X}$ be a separable Hilbert space and let $\mathcal{A}$ be the
infinitesimal generator of a $C_{0}$-semigroup $\mathbb{U}\left(t\right)$ on
$\mathscr{X}$. Suppose that the following conditions hold:

  1. cmrmn(i)

$\sigma\left(\mathcal{A}\right)=\sigma^{(1)}\left(\mathcal{A}\right)\cup\sigma^{(2)}\left(\mathcal{A}\right)$
where the branch
$\sigma^{(2)}\left(\mathcal{A}\right)=\\{\lambda_{k}{\\}}^{\infty}_{k=1}$,
consisting entirely of isolated eigenvalues of finite algebraic multiplicity;

  2. cmrmn(ii)

$\sup_{k\geq 1}m_{a}\left(\lambda_{k}\right)<\infty$ with
$m_{a}\left(\lambda_{k}\right)\coloneqq\dim
E\left(\lambda_{k},\mathcal{A}\right)\mathscr{X}$, the
$E\left(\lambda_{k},\mathcal{A}\right)$ being the Riesz projections associated
with the eigenvalues $\lambda_{k}$; and

  3. cmrmn(iii)

there is a real constant $\nu$ such that

$\sup\,\\{\Re\left(\lambda\right)~{}|~{}\lambda\in\sigma^{(1)}\left(\mathcal{A}\right)\\}\leq\nu\leq\inf\,\\{\Re\left(\lambda\right)~{}|~{}\lambda\in\sigma^{(2)}\left(\mathcal{A}\right)\\},$

and

$\inf_{k\neq j}\left|\lambda_{k}-\lambda_{j}\right|>0.$ (4.1)

Then the following statements hold:

  1. cmrmn(1)

There exist two $\mathbb{U}\left(t\right)$-invariant closed subspaces
$\mathscr{X}_{1}$ and $\mathscr{X}_{2}$ with the properties

    1. cmrmn(i)

$\sigma\left(\mathcal{A}|_{\mathscr{X}_{1}}\right)=\sigma^{(1)}\left(\mathcal{A}\right)$
and
$\sigma\left(\mathcal{A}|_{\mathscr{X}_{2}}\right)=\sigma^{(2)}\left(\mathcal{A}\right)$;
and

    2. cmrmn(ii)

$\left\\{E\left(\lambda_{k},\mathcal{A}\right)\mathscr{X}_{2}\right\\}^{\infty}_{k=1}$
forms a Riesz basis of subspaces for $\mathscr{X}_{2}$, and

$\mathscr{X}=\overline{{\mathscr{X}_{1}\oplus\mathscr{X}_{2}}}.$

  2. cmrmn(2)

If $\sup_{k\geq
1}\left\|E\left(\lambda_{k},\mathcal{A}\right)\right\|<\infty$, then

$\mathscr{D}\left(\mathcal{A}\right)\subset\mathscr{X}_{1}\oplus\mathscr{X}_{2}\subset\mathscr{X}.$

  3. cmrmn(3)

$\mathscr{X}$ can be decomposed into the topological direct sum

$\mathscr{X}=\mathscr{X}_{1}\oplus\mathscr{X}_{2}$

if and only if $\sup_{n\geq
1}\,\bigl{\|}\sum\limits^{n}_{k=1}E\left(\lambda_{k},\mathcal{A}\right)\bigr{\|}<\infty$.

###  4.1. Completeness and minimality

We shall now show that the root vectors of $\mathcal{T}$, all of which shown
in Theorem 3.4 to be eigenvectors, are minimal complete in $\mathscr{X}$. The
result can be obtained by application of a similar method to the one in [11]
to construct the inverse operator $\mathcal{T}^{-1}$, which we know by Lemma
2.1 to exist and be compact, and subsequent application of Lemmas 4.1 and 4.2.
To this end, we need first of all the following theorem.

######  Theorem 4.1.

Let $\mathcal{T}_{0}$ be the skewadjoint part of $\mathcal{T}$, i.e.,
$\mathcal{T}\coloneqq\mathcal{A}+\mathcal{B}$ as defined by (2.4)–(2.7) with
$\kappa=0$. Then $\mathcal{T}^{-1}=\mathcal{T}_{0}^{-1}+\kappa\mathcal{S}$,
where $\mathcal{S}$ is a real operator on $\mathscr{X}$ which is of finite
rank.

###### Proof.

Consider the problem (2.12). If we integrate the differential equation twice
from $0$ to $1$, making use of the boundary conditions
$w_{k}\left(1\right)=w_{k}^{\prime\prime}\left(1\right)=0$, $k=1,2,3$, we
arrive at

$\begin{split}&w_{k}^{\prime\prime}\left(s\right)-(\gamma-\eta^{2})\,w_{k}\left(s\right)+\left[w_{k}^{\left(3\right)}\left(0\right)-(\gamma-\eta^{2})\,w_{k}^{\prime}\left(0\right)+\beta\eta
v_{k}\left(0\right)\right]\\\ &\hskip
173.44534pt\times\left(1-s\right)=-\widetilde{V}_{k}\left(s\right)-\widetilde{W}_{k}\left(s\right),\quad
k=1,2,3,\end{split}$ (4.2)

with the integral terms

$\widetilde{V}_{k}\left(s\right)=-\int^{1}_{s}dt\int^{t}_{0}\widetilde{v}_{k}\left(r\right)dr,\quad\widetilde{W}_{k}\left(s\right)=-2\beta\eta\int^{1}_{s}dt\int^{t}_{0}\widetilde{w}^{\prime}_{k}\left(r\right)dr.$

For brevity, let us set
$\widetilde{\left(V;W\right)}_{k}\left(\,\cdot\,\right)\coloneqq\widetilde{V}_{k}\left(\,\cdot\,\right)+\widetilde{W}_{k}\left(\,\cdot\,\right)$
in this proof. The solutions of (4.2) then take the form

$\begin{split}w_{k}\left(s\right)&=c_{k}\sinh\sqrt{\gamma-\eta^{2}}\left(1-s\right)+\left[w_{k}^{\left(3\right)}\left(0\right)-(\gamma-\eta^{2})\,w_{k}^{\prime}\left(0\right)+\beta\eta
v_{k}\left(0\right)\right]\\\
&\qquad\times\frac{1}{\sqrt{\gamma-\eta^{2}}}\int^{1}_{s}\left(1-r\right)\sinh\sqrt{\gamma-\eta^{2}}\left(s-r\right)dr\\\
&\qquad+\frac{1}{\sqrt{\gamma-\eta^{2}}}\int^{1}_{s}\sinh\sqrt{\gamma-\eta^{2}}\left(s-r\right)\widetilde{\left(V;W\right)}_{k}\left(r\right)dr,\quad
k=1,2,3,\end{split}$ (4.3)

with arbitrary constant $c_{k}$. Thus the following equations are obtained

$\begin{split}w_{k}\left(0\right)&=c_{k}\sinh\sqrt{\gamma-\eta^{2}}-\left[w_{k}^{\left(3\right)}\left(0\right)-(\gamma-\eta^{2})\,w_{k}^{\prime}\left(0\right)+\beta\eta
v_{k}\left(0\right)\right]\\\
&\qquad\times\frac{1}{\sqrt{\gamma-\eta^{2}}}\int^{1}_{0}\left(1-r\right)\sinh\sqrt{\gamma-\eta^{2}}\,r\,dr\\\
&\qquad-\frac{1}{\sqrt{\gamma-\eta^{2}}}\int^{1}_{0}\sinh\sqrt{\gamma-\eta^{2}}\,r\,\widetilde{\left(V;W\right)}_{k}\left(r\right)dr,\\\
w_{k}^{\prime}\left(0\right)&=-c_{k}\sqrt{\gamma-\eta^{2}}\cosh\sqrt{\gamma-\eta^{2}}+\left[w_{k}^{\left(3\right)}\left(0\right)-(\gamma-\eta^{2})\,w_{k}^{\prime}\left(0\right)+\beta\eta
v_{k}\left(0\right)\right]\\\
&\qquad\times\int^{1}_{0}\left(1-r\right)\cosh\sqrt{\gamma-\eta^{2}}\,r\,dr+\int^{1}_{0}\cosh\sqrt{\gamma-\eta^{2}}\,r\,\widetilde{\left(V;W\right)}_{k}\left(r\right)dr,\\\
w_{k}^{\prime\prime}\left(0\right)&=c_{k}\,(\gamma-\eta^{2})\sinh\sqrt{\gamma-\eta^{2}}-\left[w_{k}^{\left(3\right)}\left(0\right)-(\gamma-\eta^{2})\,w_{k}^{\prime}\left(0\right)+\beta\eta
v_{k}\left(0\right)\right]\\\
&\qquad\times\left[\sqrt{\gamma-\eta^{2}}\int^{1}_{0}\left(1-r\right)\sinh\sqrt{\gamma-\eta^{2}}\,r\,dr+1\right]\\\
&\qquad-\sqrt{\gamma-\eta^{2}}\int^{1}_{0}\sinh\sqrt{\gamma-\eta^{2}}\,r\,\widetilde{\left(V;W\right)}_{k}\left(r\right)dr-\widetilde{\left(V;W\right)}_{k}\left(0\right)\end{split}$

for $k=1,2,3$. If we sum each of these three equations over $k=1,2,3$, using
the vertex condition
$\sum^{3}_{k=1}\,\bigl{[}w^{\left(3\right)}_{k}\left(0\right)-(\gamma-\eta^{2})\,w_{k}^{\prime}\left(0\right)+\beta\eta
v_{k}\left(0\right)\bigr{]}=0$, we obtain

$\left\\{\begin{split}\sum_{k=1}^{3}w_{k}\left(0\right)&=\sinh\sqrt{\gamma-\eta^{2}}\,\sum_{k=1}^{3}c_{k}-\frac{1}{\sqrt{\gamma-\eta^{2}}}\int^{1}_{0}\sinh\sqrt{\gamma-\eta^{2}}\,r\sum_{k=1}^{3}\widetilde{\left(V;W\right)}_{k}\left(r\right)dr\\\
\sum_{k=1}^{3}w_{k}^{\prime}\left(0\right)&=-\sqrt{\gamma-\eta^{2}}\cosh\sqrt{\gamma-\eta^{2}}\sum_{k=1}^{3}c_{k}+\int^{1}_{0}\cosh\sqrt{\gamma-\eta^{2}}\,r\sum_{k=1}^{3}\widetilde{\left(V;W\right)}_{k}\left(r\right)dr\\\
\sum_{k=1}^{3}w_{k}^{\prime\prime}\left(0\right)&=(\gamma-\eta^{2})\,\sinh\sqrt{\gamma-\eta^{2}}\sum_{k=1}^{3}c_{k}-\sqrt{\gamma-\eta^{2}}\int^{1}_{0}\sinh\sqrt{\gamma-\eta^{2}}\,r\sum_{k=1}^{3}\widetilde{\left(V;W\right)}_{k}\left(r\right)dr\\\
&\qquad-\sum_{k=1}^{3}\widetilde{\left(V;W\right)}_{k}\left(0\right).\end{split}\right.$
(4.4)

Using the vertex condition $w^{\prime\prime}_{k}\left(0\right)-\alpha
w^{\prime}_{k}\left(0\right)-\kappa v^{\prime}_{k}\left(0\right)=0$,
$k=1,2,3$, taking into account that the $v_{k}=\widetilde{w}_{k}$, and summing
over $k=1,2,3$ we find, again using
$\sum^{3}_{k=1}\,\bigl{[}w^{\left(3\right)}_{k}\left(0\right)-(\gamma-\eta^{2})\,w_{k}^{\prime}\left(0\right)+\beta\eta
v_{k}\left(0\right)\bigr{]}=0$, that

$\displaystyle\sum_{k=1}^{3}c_{k}$
$\displaystyle=\frac{\sum_{k=1}^{3}\left[\widetilde{\left(V;W\right)}_{k}\left(0\right)+\int^{1}_{0}\bigl{(}\sqrt{\gamma-\eta^{2}}\sinh\sqrt{\gamma-\eta^{2}}\,r+\alpha\cosh\sqrt{\gamma-\eta^{2}}\,r\bigr{)}\,\widetilde{\left(V;W\right)}_{k}\left(r\right)dr\right]}{(\gamma-\eta^{2})\,\sinh\sqrt{\gamma-\eta^{2}}+\alpha\sqrt{\gamma-\eta^{2}}\cosh\sqrt{\gamma-\eta^{2}}}$
$\displaystyle\qquad+\frac{\kappa\sum_{k=1}^{3}\widetilde{w}_{k}^{\prime}\left(0\right)}{(\gamma-\eta^{2})\,\sinh\sqrt{\gamma-\eta^{2}}+\alpha\sqrt{\gamma-\eta^{2}}\cosh\sqrt{\gamma-\eta^{2}}}$
$\displaystyle\coloneqq b\left(\widetilde{w},\widetilde{v}\right)+\kappa
H\left(\widetilde{w}\right),$

where

$H\left(\widetilde{w}\right)\coloneqq\frac{\sum_{k=1}^{3}\widetilde{w}_{k}^{\prime}\left(0\right)}{(\gamma-\eta^{2})\,\sinh\sqrt{\gamma-\eta^{2}}+\alpha\sqrt{\gamma-\eta^{2}}\cosh\sqrt{\gamma-\eta^{2}}}.$

Clearly the first equation in (4.4) can be written as

$\sum_{k=1}^{3}w_{k}\left(0\right)=\left(b\left(\widetilde{w},\widetilde{v}\right)+\kappa
H\left(\widetilde{w}\right)\right)\sinh\sqrt{\gamma-\eta^{2}}-\frac{1}{\sqrt{\gamma-\eta^{2}}}\int^{1}_{0}\sinh\sqrt{\gamma-\eta^{2}}\,r\sum_{k=1}^{3}\widetilde{\left(V;W\right)}_{k}\left(r\right)dr.$

Since $w_{j}\left(0\right)=w_{k}\left(0\right)\equiv w\left(0\right)$,
$j,k=1,2,3$, it follows that

$\displaystyle w\left(0\right)$
$\displaystyle=\frac{1}{3}\left[\left(b\left(\widetilde{w},\widetilde{v}\right)+\kappa
H\left(\widetilde{w}\right)\right)\sinh\sqrt{\gamma-\eta^{2}}-\frac{1}{\sqrt{\gamma-\eta^{2}}}\int^{1}_{0}\sinh\sqrt{\gamma-\eta^{2}}r\sum_{k=1}^{3}\widetilde{\left(V;W\right)}_{k}\left(r\right)dr\right]$
$\displaystyle\coloneqq c\left(\widetilde{w},\widetilde{v}\right)+\kappa
G\left(\widetilde{w}\right)$

where

$G\left(\tilde{w}\right)\coloneqq\frac{H\left(\widetilde{w}\right)}{3}\sinh\sqrt{\gamma-\eta^{2}}.$

Thus we have

$\sum_{k=1}^{3}c_{k}=b\left(\widetilde{w},\widetilde{v}\right)+\kappa
H\left(\widetilde{w}\right),\quad
w\left(0\right)=c\left(\widetilde{w},\widetilde{v}\right)+\kappa
G\left(\widetilde{w}\right),$

and hence the following algebraic equations for $c_{k}$,
$w^{\left(3\right)}_{k}\left(0\right)-(\gamma-\eta^{2})\,w_{k}^{\prime}\left(0\right)+\beta\eta
v_{k}\left(0\right)$:

$\left\\{\begin{split}&c_{k}\sinh\sqrt{\gamma-\eta^{2}}-\frac{w_{k}^{\left(3\right)}\left(0\right)-(\gamma-\eta^{2})\,w_{k}^{\prime}\left(0\right)+\beta\eta
v_{k}\left(0\right)}{\sqrt{\gamma-\eta^{2}}}\int^{1}_{0}\left(1-r\right)\sinh\sqrt{\gamma-\eta^{2}}\,r\,dr\\\
&\hskip
86.72267pt=c\left(\widetilde{w},\widetilde{v}\right)+\frac{1}{\sqrt{\gamma-\eta^{2}}}\int^{1}_{0}\sinh\sqrt{\gamma-\eta^{2}}\,r\,\widetilde{\left(V;W\right)}_{k}\left(r\right)dr,\\\
&c_{k}\left(\sinh\sqrt{\gamma-\eta^{2}}+\frac{\alpha\cosh\sqrt{\gamma-\eta^{2}}}{\sqrt{\gamma-\eta^{2}}}\right)-\frac{w_{k}^{\left(3\right)}\left(0\right)-(\gamma-\eta^{2})\,w_{k}^{\prime}\left(0\right)+\beta\eta
v_{k}\left(0\right)}{\sqrt{\gamma-\eta^{2}}}\\\ &\hskip
43.36464pt\times\left[\int^{1}_{0}\left(1-r\right)\left(\sinh\sqrt{\gamma-\eta^{2}}\,r+\frac{\alpha\cosh\sqrt{\gamma-\eta^{2}}\,r}{\sqrt{\gamma-\eta^{2}}}\right)dr+1\right]\\\
&\hskip
86.72267pt=\frac{1}{\sqrt{\gamma-\eta^{2}}}\int^{1}_{0}\left(\sinh\sqrt{\gamma-\eta^{2}}\,r+\frac{\alpha\cosh\sqrt{\gamma-\eta^{2}}\,r}{\sqrt{\gamma-\eta^{2}}}\right)\widetilde{\left(V;W\right)}_{k}\left(r\right)dr\\\
&\hskip
86.72267pt\qquad+\frac{\widetilde{\left(V;W\right)}_{k}\left(0\right)+\kappa\widetilde{w}_{k}^{\prime}\left(0\right)}{\gamma-\eta^{2}}\end{split}\right.$
(4.5)

for $k=1,2,3$. We observe that the determinant of the coefficient matrix
formed by (4.5) does not vanish,

$\left|\begin{matrix}\sinh\sqrt{\gamma-\eta^{2}}&-\int^{1}_{0}\left(1-r\right)\sinh\sqrt{\gamma-\eta^{2}}\,r\,dr\\\
\alpha\cosh\sqrt{\gamma-\eta^{2}}&-\alpha\int^{1}_{0}\left(1-r\right)\cosh\sqrt{\gamma-\eta^{2}}\,r\,dr-1\\\
\end{matrix}\right|\neq 0,$

whence there are unique solution pairs $c_{k}$,
$w^{\left(3\right)}_{k}\left(0\right)-(\gamma-\eta^{2})\,w_{k}^{\prime}\left(0\right)+\beta\eta
v_{k}\left(0\right)$ of the form

$\displaystyle c_{k}$
$\displaystyle=c_{k}\left(\widetilde{w},\widetilde{v}\right)+\kappa
d_{k}\left(\widetilde{w}\right),$ $\displaystyle k$ $\displaystyle=1,2,3,$
$\displaystyle
w_{k}^{\left(3\right)}\left(0\right)-(\gamma-\eta^{2})\,w_{k}^{\prime}\left(0\right)+\beta\eta
v_{k}\left(0\right)$
$\displaystyle=M_{k}\left(\widetilde{w},\widetilde{v}\right)+\kappa
N_{k}\left(\widetilde{w}\right),$ $\displaystyle k$ $\displaystyle=1,2,3.$

Inserting these into (4.3), we find

$\displaystyle w_{k}\left(s\right)$
$\displaystyle=c_{k}\left(\widetilde{w},\widetilde{v}\right)\sinh\sqrt{\gamma-\eta^{2}}\left(1-s\right)+\frac{M_{k}\left(\widetilde{w},\widetilde{v}\right)}{\sqrt{\gamma-\eta^{2}}}\int^{1}_{s}\left(1-r\right)\sinh\sqrt{\gamma-\eta^{2}}\left(s-r\right)dr$
$\displaystyle\qquad+\frac{1}{\sqrt{\gamma-\eta^{2}}}\int^{1}_{s}\sinh\sqrt{\gamma-\eta^{2}}\left(s-r\right)\widetilde{\left(V;W\right)}_{k}\left(r\right)dr$
$\displaystyle\qquad+\kappa\left[d_{k}\left(\widetilde{w}\right)\sinh\sqrt{\gamma-\eta^{2}}\left(1-s\right)+\frac{N_{k}\left(\widetilde{w}\right)}{\sqrt{\gamma-\eta^{2}}}\int^{1}_{s}\left(1-r\right)\sinh\sqrt{\gamma-\eta^{2}}\left(s-r\right)dr\right]$
$\displaystyle\coloneqq
w_{k}\left(s,\widetilde{w},\widetilde{v}\right)+\kappa\left(d_{k}\left(\widetilde{w}\right)\phi_{1}\left(s\right)+N_{k}\left(\widetilde{w}\right)\phi_{2}\left(s\right)\right)$

where $\phi_{1}\left(s\right)=\sinh\sqrt{\gamma-\eta^{2}}\left(1-s\right)$,
$\phi_{2}\left(s\right)=\frac{1}{\sqrt{\gamma-\eta^{2}}}\int^{1}_{s}\left(1-r\right)\sinh\sqrt{\gamma-\eta^{2}}\left(s-r\right)dr$.
Therefore we have

$\displaystyle\mathcal{T}^{-1}\left\\{\left(\widetilde{w}_{k},\widetilde{v}_{k}\right)\right\\}_{k=1}^{3}$
$\displaystyle=\left\\{\left(w_{k},v_{k}\right)\right\\}_{k=1}^{3}$
$\displaystyle=\left\\{\left(w_{k}\left(s,\widetilde{w},\widetilde{v}\right),v_{k}\right)\right\\}_{k=1}^{3}+\kappa\left\\{d_{k}\left(\widetilde{w}\right)\left(\phi_{1},0\right)+N_{k}\left(\widetilde{w}\right)\left(\phi_{2},0\right)\right\\}_{k=1}^{3}$
$\displaystyle=\mathcal{T}_{0}^{-1}\left\\{\left(\widetilde{w}_{k},\widetilde{v}_{k}\right)\right\\}_{k=1}^{3}+\kappa\mathcal{S}\left\\{\left(\widetilde{w}_{k},\widetilde{v}_{k}\right)\right\\}_{k=1}^{3},$

taking into account that $v_{k}=\widetilde{w}_{k}$, $k=1,2,3$, where
$\mathcal{T}^{-1}_{0}$ is a compact skewadjoint operator and $\mathcal{S}$ is
a bounded operator of rank $2$ on $\mathscr{X}$. In particular if
$\mathcal{T}x=y$, $y\in\mathscr{X}$,
$x\in\mathscr{D}\left(\mathcal{A}\right)$, it is easily seen that

$\Re\left(\mathcal{T}x,x\right)_{\mathscr{X}}=\Re\,(\mathcal{T}^{-1}y,y{)}_{\mathscr{X}}=\Re\,((\mathcal{T}_{0}^{-1}+\kappa\mathcal{S})\,y,y{)}_{\mathscr{X}}=\kappa\Re\,(\mathcal{S}y,y{)}_{\mathscr{X}}=\kappa\,(\mathcal{S}y,y{)}_{\mathscr{X}}.$

This completes the proof of the theorem. ∎

Combined with Lemmas 4.1 and 4.2, if we identify $\mathcal{A}$ with
$\mathcal{T}^{-1}$ (recall Lemma 2.1) and note that the geometric eigenspaces
of $\mathcal{A}$ and $\mathcal{T}$ corresponding, respectively, to eigenvalues
$\lambda$ and $\lambda^{-1}$ coincide, Theorem 4.1 leads to the following
result.

######  Theorem 4.2.

The eigenvectors of $\mathcal{T}$ are minimal complete in $\mathscr{X}$.

###  4.2. Riesz basis property

There remains the verification of the Riesz basis property of the
eigenvectors. We use the result of [18, Theorem 1.1] to prove the following
theorem.

######  Theorem 4.3.

There exists a sequence of eigenvectors corresponding to a properly enumerated
sequence of eigenvalues of $\mathcal{T}$ which is a Riesz basis for
$\mathscr{X}$.

###### Proof.

We identify the operator $\mathcal{A}$ in [18, Theorem 1.1] with $\mathcal{T}$
and begin by noting that, by Theorem 3.4, for any
$\lambda\in\sigma\left(\mathcal{T}\right)$,
$E\left(\lambda,T\right)\mathscr{X}$ is equal to the geometric eigenspace
$\ker\left(\lambda\mathcal{I}-\mathcal{T}\right)$ and so $\dim
E\left(\lambda,T\right)\mathscr{X}=1$ or $\dim
E\left(\lambda,T\right)\mathscr{X}=2$. Denote by $\left\\{\lambda_{\pm
n}\right\\}^{\infty}_{n=1}$ the properly enumerated sequence of eigenvalues of
$\mathcal{T}$, i.e. indexed in such a way that
$\lambda_{n}=\overline{\lambda_{-n}}$ whenever
$\Im\left({\lambda}_{n}\right)\neq 0$. Let us take
$\sigma^{\left(2\right)}\left(\mathcal{T}\right)=\sigma\left(\mathcal{T}\right)=\left\\{\lambda_{\pm
n}\right\\}^{\infty}_{n=1}$ and
$\sigma^{\left(1\right)}\left(\mathcal{T}\right)=\left\\{-\infty\right\\}$. It
now follows from Theorem 3.3 that all conditions of Lemma 4.3 are satisfied.
In particular, by virtue of Theorem 3.3, the eigenvalues are simple for large
enough $n$, and thus their sequence is interpolating (because
$|\Re\,(\lambda_{\pm n})|<\infty$ and the separation condition (4.1) holds).
It also follows from Theorem 4.2 that

$\overline{\operatorname{span}\left\\{E\left(\lambda,\mathcal{T}\right)\mathscr{X}~{}\middle|~{}\lambda\in\sigma^{\left(2\right)}\left(\mathcal{T}\right)\right\\}}=\mathscr{X}_{2}=\mathscr{X}.$

Hence, combining these results, we have by Lemma 4.3 that there exists a
sequence of eigenvectors of $\mathcal{T}$ corresponding to
$\left\\{\lambda_{\pm n}\right\\}^{\infty}_{n=1}$ which is a Riesz basis for
$\mathscr{X}$. ∎

As a corollary of the theorem we obtain the following result.

######  Theorem 4.4.

$\mathcal{T}$ satisfies the Spectrum Determined Growth Assumption.

##  5\. Exponential stability result

We now have all the ingredients to apply the spectral approach to obtain the
final result of the paper.

######  Theorem 5.1.

If $\kappa>0$, then there exist constants $M,\varepsilon>0$ such that

$\left\|\mathbb{S}\left(t\right)x_{0}\right\|_{\mathscr{X}}\leq
Me^{-\varepsilon t}\left\|x_{0}\right\|_{\mathscr{X}},\quad
x_{0}\in\mathscr{D}\left(\mathcal{A}\right),\quad t\geq 0,$

where $\mathbb{S}\left(t\right)$ is the $C_{0}$-semigroup of contractions
generated by $\mathcal{T}\coloneqq\mathcal{A}+\mathcal{B}$ defined by
(2.4)–(2.7); and consequently we have for the solutions of the closed-loop
system (2.8), as $t\rightarrow\infty$,
$\left\|\bm{x}\left(t\right)\right\|_{\mathscr{X}}\rightarrow 0$
exponentially.

###### Proof.

Since, by Theorem 4.4, $\mathcal{T}$ satisfies the Spectrum Determined Growth
Assumption, we have
$\varpi_{0}=\sup\left\\{\Re\left(\lambda\right)~{}\middle|~{}\lambda\in\sigma\left(\mathcal{T}\right)\right\\}$.
It follows from Theorem 3.1 that
$i\mathbb{R}\subset\varrho\left(\mathcal{T}\right)$, and according to Theorem
3.3 the two branches $\sigma^{\left(1\right)}\left(\mathcal{T}\right)$,
$\sigma^{\left(2\right)}\left(\mathcal{T}\right)$ have asymptotes
$\Re\left(\lambda\right)\sim-\frac{1}{\kappa}$,
$\Re\left(\lambda\right)\sim-\frac{2}{\kappa}$, respectively. Thus
$\sup\left\\{\Re\left(\lambda\right)~{}\middle|~{}\lambda\in\sigma\left(\mathcal{T}\right)\right\\}<0$
and hence
$\sup\left\\{\Re\left(\lambda\right)~{}\middle|~{}\lambda\in\sigma\left(\mathcal{T}\right)\right\\}\leq-\varepsilon<0$
for some $\varepsilon>0$. With this the proof is complete. ∎

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