In this section, we shall show that system (1.2) has a set of eigenvectors which form a Riesz basis in Hilbert space $\mathcal{\mathscr{H}}$.
The key idea is to apply some basic result of Pavlov [13], which provides a useful way to verify the Riesz basis property for the eigenvectors of a linear operator in Hilbert space.

Since the matrix $\left(\begin{array}{cc}\hfill 0\hfill & \hfill -c\left(x\right)\hfill \\ \hfill -c\left(x\right)\hfill & \hfill 0\hfill \end{array}\right)$ is symmetric with distinct eigenvalues ${\lambda}_{i}={(-1)}^{i+1}c(x)$, $i=1,2$, introducing the change of variables

$\left(\begin{array}{c}\hfill p\hfill \\ \hfill v\hfill \end{array}\right)=\frac{-1}{2}\left(\begin{array}{cc}\hfill 1\hfill & \hfill -1\hfill \\ \hfill -1\hfill & \hfill -1\hfill \end{array}\right)\left(\begin{array}{c}\hfill u\hfill \\ \hfill w\hfill \end{array}\right),\left(\begin{array}{c}\hfill f\hfill \\ \hfill g\hfill \end{array}\right)=\frac{-1}{2}\left(\begin{array}{cc}\hfill 1\hfill & \hfill -1\hfill \\ \hfill -1\hfill & \hfill -1\hfill \end{array}\right)\left(\begin{array}{c}\hfill h\hfill \\ \hfill k\hfill \end{array}\right),\alpha :=\frac{1-{\alpha}_{1}}{1+{\alpha}_{1}}\mathit{\hspace{1em}}\text{and}\mathit{\hspace{1em}}\beta :=\frac{1-{\beta}_{1}}{1+{\beta}_{1}},$

our system becomes

$\{\begin{array}{cc}\hfill \frac{d}{dt}\left(\begin{array}{c}\hfill p(x,t)\hfill \\ \hfill v(x,t)\hfill \end{array}\right)+\left(\begin{array}{cc}\hfill c(x)\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill -c(x)\hfill \end{array}\right)\frac{d}{dx}\left(\begin{array}{c}\hfill p(x,t)\hfill \\ \hfill v(x,t)\hfill \end{array}\right)& =\left(\begin{array}{c}\hfill f(x,t)\hfill \\ \hfill g(x,t)\hfill \end{array}\right)\hfill \\ \hfill p(0,t)& =\alpha v(0,t),\hfill \\ \hfill p(L,t)& =\beta v(L,t).\hfill \end{array}$(4.1)

To turn system (1.2) into a framework of semigroups, we
introduce the underlying state Hilbert space $\mathcal{\mathscr{H}}:={L}^{2}[0,L]\times {L}^{2}[0,L]$.
System (4.1) is then written as an evolutionary equation in
$\mathcal{\mathscr{H}}$,

$\frac{d}{dt}W(t)=\mathcal{\mathcal{A}}W(t),t>0,$

with $W(t)=\left(\begin{array}{c}\hfill p(\cdot ,t)\hfill \\ \hfill v(\cdot ,t)\hfill \end{array}\right)$ and $\mathcal{\mathcal{A}}:\mathcal{\mathcal{D}}(\mathcal{\mathcal{A}})\subset \mathcal{\mathscr{H}}\to \mathcal{\mathscr{H}}$ defined by

$\mathcal{\mathcal{A}}:=C(x)\frac{d}{dx}:=\left(\begin{array}{cc}\hfill c(x)\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill -c(x)\hfill \end{array}\right)\frac{d}{dx}$(4.2)

with domain

$\mathcal{\mathcal{D}}(\mathcal{\mathcal{A}})=\{\left(\begin{array}{c}\hfill p(x)\hfill \\ \hfill v(x)\hfill \end{array}\right)\in {H}^{1}[0,L]\times {H}^{1}[0,L]:p(0)=\alpha v(0),p(L)=\beta v(L)\}.$(4.3)

In this part, our goal is to give some important properties about a linear hyperbolic system to verify the assumptions of Theorem 3.1 in several steps.
We shall assume without loss of generality that $p(x):=p(x,t)$, $v(x):=v(x,t)$, $f(x):=f(x,t)$ and $g(x):=g(x,t)$.
In the following, we establish some preliminary results about the operator $\mathcal{\mathcal{A}}$.

#### Proposition 4.1.

*Let $\mathcal{A}$ be the operator given in (4.2) and (4.3).
Then the following assertions are true:*

(i)

$\mathcal{\mathcal{D}}(\mathcal{\mathcal{A}})$
* is dense on *
$\mathcal{\mathscr{H}}$.

(ii)

*The canonical injection *
*i*
* from *
$\mathcal{\mathcal{D}}(\mathcal{\mathcal{A}})$
* into *
$\mathcal{\mathscr{H}}$
* is compact.*

#### Proof.

(i) It is seen that $\mathcal{\mathcal{D}}(\mathcal{\mathcal{A}})$ is a closed subset of the Sobolev space ${H}^{1}[0,L]\times {H}^{1}[0,L]$ which is dense in $\mathcal{\mathscr{H}}$.
Then assertion (i) follows immediately.

(ii) On the other hand, we infer from the Sobolev theorem
[5, Theorem VIII.7, p. 129] that the injection from ${H}^{1}[0,L]$ into ${L}^{2}[0,L]$ is compact.
So the injection from $\mathcal{\mathcal{D}}(\mathcal{\mathcal{A}})$ into $\mathcal{\mathscr{H}}$ is too.
∎

#### Theorem 4.1.

*The following statements hold:*

(i)

*The operator *
$\mathcal{\mathcal{A}}$
* generates a *
${C}_{0}$
*-semigroup on *
$\mathcal{\mathscr{H}}$.

(ii)

*The resolvent expression *
$R(\lambda ,\mathcal{\mathcal{A}})$
* of *
$\mathcal{\mathcal{A}}$
* can be represented as*

$R(\lambda ,\mathcal{\mathcal{A}})\left(\begin{array}{c}\hfill f\hfill \\ \hfill g\hfill \end{array}\right)(x)=M(x,0,\lambda )\left(\begin{array}{c}\hfill \alpha v(0)\hfill \\ \hfill v(0)\hfill \end{array}\right)-{\int}_{0}^{x}M(x,s,\lambda ){C}^{-1}(s)\left(\begin{array}{c}\hfill f(s)\hfill \\ \hfill g(s)\hfill \end{array}\right)ds\mathit{\hspace{1em}}\mathit{\text{for all}}\left(\begin{array}{c}\hfill f\hfill \\ \hfill g\hfill \end{array}\right)\in H,$

*where
*

$M(x,s,\lambda )=\left(\begin{array}{cc}\hfill {e}_{1}(x,s,\lambda )\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill {e}_{2}(x,s,\lambda )\hfill \end{array}\right),{e}_{i}(x,s,\lambda )={e}^{{(-1)}^{i+1}\lambda {\int}_{s}^{x}{c}^{-1}(r)dr},i=1,2,$

*and*

$\{\begin{array}{cc}\hfill v(0)& =-\frac{1}{\mathrm{\Delta}(\lambda )}\left[{\int}_{0}^{L}\u3008M(L,s,\lambda )\left(\begin{array}{cc}\hfill -{c}^{-1}(s)\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill {c}^{-1}(s)\hfill \end{array}\right)\left(\begin{array}{c}\hfill f(s)\hfill \\ \hfill g(s)\hfill \end{array}\right),\left(\begin{array}{c}\hfill 1\hfill \\ \hfill -\beta \hfill \end{array}\right)\u3009ds\right],\hfill \\ \hfill \mathrm{\Delta}(\lambda )& =\alpha {e}_{1}(L,0,\lambda )-\beta {e}_{2}(L,0,\lambda ).\hfill \end{array}$

#### Proof.

(i) Since the entries of the matrix $C(x)$ are real-valued ${C}^{1}$-functions in *x*, $\mathcal{\mathcal{A}}$ generates an evolution operator.

(ii) First, to get the resolvent expression of the operator $\mathcal{\mathcal{A}}$, we consider the inhomogeneous equation expressed as

$(\lambda -\mathcal{\mathcal{A}})U=F,$(4.4)

where $\lambda \in \u2102$,
$U:=\left(\begin{array}{c}\hfill p\left(x\right)\hfill \\ \hfill v\left(x\right)\hfill \end{array}\right)\in \mathcal{\mathcal{D}}(\mathcal{\mathcal{A}})$ and
$F:=\left(\begin{array}{c}\hfill f\left(x\right)\hfill \\ \hfill g\left(x\right)\hfill \end{array}\right)$ is arbitrary in $\mathcal{\mathscr{H}}$.

An elementary calculation by the variation of constants reveals that the general solution of equation (4.4) is given by

$R(\lambda ,\mathcal{\mathcal{A}})\left(\begin{array}{c}\hfill f(x)\hfill \\ \hfill g(x)\hfill \end{array}\right)=M(x,0,\lambda )\left(\begin{array}{c}\hfill \alpha v(0)\hfill \\ \hfill v(0)\hfill \end{array}\right)-{\int}_{0}^{x}M(x,s,\lambda ){C}^{-1}(s)\left(\begin{array}{c}\hfill f(s)\hfill \\ \hfill g(s)\hfill \end{array}\right)ds,$

where we designate by $M(x,s,\lambda )$ the matrix

$M(x,s,\lambda )=\left(\begin{array}{cc}\hfill {e}_{1}(x,s,\lambda )\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill {e}_{2}(x,s,\lambda )\hfill \end{array}\right),{e}_{i}(x,s,\lambda )={e}^{{(-1)}^{i+1}\lambda {\int}_{s}^{x}{c}^{-1}(r)dr},i=1,2.$

As a preparation for attacking the expression of $v(0)$, it is sufficient to solve
$p(L)=\beta v(L)$
by recalling the resolvent of $\mathcal{\mathcal{A}}$.
So we get

$[\alpha {e}_{1}(L,0,\lambda )-\beta {e}_{2}(L,0,\lambda )]v(0)=-\left[{\int}_{0}^{L}-{e}_{1}(L,s,\lambda ){c}^{-1}(s)f(s)-\beta {e}_{2}(L,s,\lambda ){c}^{-1}(s)g(s)ds\right].$

Denote by $\mathrm{\Delta}(\lambda )$ the expression

$\mathrm{\Delta}(\lambda )=\alpha {e}_{1}(L,0,\lambda )-\beta {e}_{2}(L,0,\lambda ),$

which is an entire function on λ.
Using the above equality with this notation, we get

$\mathrm{\Delta}(\lambda )v(0)=-{\int}_{0}^{L}\u3008M(L,s,\lambda )\left(\begin{array}{cc}\hfill -{c}^{-1}(s)\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill {c}^{-1}(s)\hfill \end{array}\right)\left(\begin{array}{c}\hfill f(s)\hfill \\ \hfill g(s)\hfill \end{array}\right),\left(\begin{array}{c}\hfill 1\hfill \\ \hfill -\beta \hfill \end{array}\right)\u3009ds.$

A short computation shows that a complex number λ belongs to the resolvent set of $\mathcal{\mathcal{A}}$ if and only if $\mathrm{\Delta}(\lambda )\ne 0$; in this case,

$v(0)=-\frac{1}{\mathrm{\Delta}(\lambda )}\left[{\int}_{0}^{L}\u3008M(L,s,\lambda )\left(\begin{array}{cc}\hfill -{c}^{-1}(s)\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill {c}^{-1}(s)\hfill \end{array}\right)\left(\begin{array}{c}\hfill f(s)\hfill \\ \hfill g(s)\hfill \end{array}\right),\left(\begin{array}{c}\hfill 1\hfill \\ \hfill -\beta \hfill \end{array}\right)\u3009ds\right],$

which completes the proof.
∎

#### Theorem 4.2.

$\mathcal{\mathcal{A}}$
* is a discrete operator in $\mathcal{H}$; in other words, for any $\lambda \mathrm{\in}\rho \mathit{}\mathrm{(}\mathcal{A}\mathrm{)}$, $R\mathit{}\mathrm{(}\lambda \mathrm{,}\mathcal{A}\mathrm{)}$ is compact on $\mathcal{H}$.
Therefore, the spectrum $\sigma \mathit{}\mathrm{(}\mathcal{A}\mathrm{)}$ consists only of isolated eigenvalues.
*

#### Proof.

It follows from Proposition 4.1 together with the fact $\rho (\mathcal{\mathcal{A}})\ne \mathrm{\varnothing}$, that $\mathcal{\mathcal{A}}$ has a compact resolvent on $\mathcal{\mathscr{H}}$.
Thus its spectrum consists only of isolated eigenvalues, which are exactly the zeros set of $\mathrm{\Delta}(\lambda )$.
∎

For each $\lambda \in \sigma (\mathcal{\mathcal{A}})$, all eigenvectors
associated with λ can be represented as

$\left(\begin{array}{c}\hfill p(x,\lambda )\hfill \\ \hfill v(x,\lambda )\hfill \end{array}\right)=\left(\begin{array}{c}\hfill \alpha {e}_{1}(x,0,\lambda )v(0)\hfill \\ \hfill {e}_{2}(x,0,\lambda )v(0)\hfill \end{array}\right)\mathit{\hspace{1em}}\text{for all non-zero}v(0)\text{satisfying}\mathrm{\Delta}(\lambda )v(0)=0.$

The next theorem gives information about the distribution of the eigenvalues of the operator $\mathcal{\mathcal{A}}$.

#### Theorem 4.3.

*Each eigenvalue of $\mathcal{A}$ is simple and given by*

${\lambda}_{n}=\frac{1}{2{c}_{L}}\mathrm{log}\left|\frac{(1-{\beta}_{1})(1+{\alpha}_{1})}{(1-{\alpha}_{1})(1+{\beta}_{1})}\right|+i\frac{n\pi}{{c}_{L}},n\in \mathbb{Z}.$

#### Proof.

An elementary calculation reveals that the zeros set of $\mathrm{\Delta}(\lambda )$ consists of an infinite number of complex eigenvalues of the operator $\mathcal{\mathcal{A}}$, which are formulated by the form

${\lambda}_{n}=\frac{1}{2{c}_{L}}\mathrm{log}\left|\frac{\beta}{\alpha}\right|+i\frac{n\pi}{{c}_{L}},n\in \mathbb{Z},$

where $\alpha ,\beta $ are defined as above and ${c}_{L}$ is defined by

${c}_{L}:={\int}_{0}^{L}{c}^{-1}(r)dr.$

Obviously, we observe from the above formula that each eigenvalue ${\lambda}_{n}$ is simple.
∎

The corresponding eigenvectors associated to ${\lambda}_{n}$ are

$\left(\begin{array}{c}\hfill p(x,{\lambda}_{n})\hfill \\ \hfill v(x,{\lambda}_{n})\hfill \end{array}\right)=\left(\begin{array}{c}\hfill \alpha {e}_{1}(x,0,{\lambda}_{n})v(0)\hfill \\ \hfill {e}_{2}(x,0,{\lambda}_{n})v(0)\hfill \end{array}\right)=\left(\begin{array}{c}\hfill \alpha {e}^{{c}_{x}[\frac{1}{2{c}_{L}}\mathrm{log}|\frac{\beta}{\alpha}|+i\frac{n\pi}{{c}_{L}}]}v(0)\hfill \\ \hfill {e}^{-{c}_{x}[\frac{1}{2{c}_{L}}\mathrm{log}|\frac{\beta}{\alpha}|+i\frac{n\pi}{{c}_{L}}]}v(0)\hfill \end{array}\right),$

where ${c}_{x}:={\int}_{0}^{x}{c}^{-1}(r)dr$.

Therefore, the eigenvectors associated to ${\lambda}_{n}$ have the asymptotic form

$\left(\begin{array}{c}\hfill p(x,{\lambda}_{n})\hfill \\ \hfill v(x,{\lambda}_{n})\hfill \end{array}\right)=\left(\begin{array}{cc}\hfill \alpha {e}^{\frac{{c}_{x}}{2{c}_{L}}\mathrm{log}|\frac{\beta}{\alpha}|}v(0)\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill {e}^{-\frac{{c}_{x}}{2{c}_{L}}\mathrm{log}|\frac{\beta}{\alpha}|}v(0)\hfill \end{array}\right)\left(\begin{array}{c}\hfill {e}^{i\frac{{c}_{x}}{{c}_{L}}n\pi}\hfill \\ \hfill {e}^{-i\frac{{c}_{x}}{{c}_{L}}n\pi}\hfill \end{array}\right).$

It is easily seen that

$H(x):=\left(\begin{array}{cc}\hfill \alpha {e}^{\frac{{c}_{x}}{2{c}_{L}}\mathrm{log}|\frac{\beta}{\alpha}|}v(0)\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill {e}^{-\frac{{c}_{x}}{2{c}_{L}}\mathrm{log}|\frac{\beta}{\alpha}|}v(0)\hfill \end{array}\right)$

is invertible in *x* and $|H(x)|\ne 0$.

The motivation for our studies is based on proving the Riesz basis property of $\{{e}^{{\lambda}_{n}t}\}$ in ${L}^{2}[0,2{c}_{L}]$.
To obtain such property, explicit estimates for sine-type functions turn out to be good.

#### Lemma 4.1.

$\{{e}^{{\lambda}_{n}t}\}$
* forms a Riesz basis for ${L}^{\mathrm{2}}\mathit{}\mathrm{[}\mathrm{0}\mathrm{,}\mathrm{2}\mathit{}{c}_{L}\mathrm{]}$.*

#### Proof.

It should be noted that our results are aimed to provide a description for the Riesz basis property of $\{{e}^{i\mathrm{Im}{\lambda}_{n}t}\}$ since the real part of the eigenvalues $\{{\lambda}_{n}\}$ is independent of *n*.

To claim this, let $g(\mu )$ be defined by
$g(\mu ):=\mathrm{sin}({c}_{L}\mu )$.
The zeros set $\{{\mu}_{n}\}$ of $g(\mu )$ is related by

${\mu}_{n}=\frac{n\pi}{{c}_{L}}.$

Obviously, $g(\mu )$ is uniformly bounded on the real axis, and hence *g* belongs to the Cartwright class.
Thus its indicator diagram is an interval.
Furthermore, it is easy to show that $g(\mu )$ is of exponential type with type ${c}_{L}=\frac{T}{2}$.
In fact, let $z:=x+iy\in \u2102$.
Then

$|g(z)|=\sqrt{{\mathrm{sin}}^{2}({c}_{L}x){\mathrm{cosh}}^{2}({c}_{L}y)+{\mathrm{cos}}^{2}({c}_{L}x){\mathrm{sinh}}^{2}({c}_{L}y)}\le |\mathrm{cosh}({c}_{L}y)|\le {e}^{{c}_{L}|y|}\le {e}^{{c}_{L}|z|}.$

Moreover, from the last expression, we have

${C}_{1}{e}^{{c}_{L}|y|}\le |g(x+iy)|\le {C}_{2}{e}^{{c}_{L}|y|},$

where $0<{C}_{1}<\frac{1}{2}$, ${C}_{2}\ge 1$ and all $y\in \mathbb{R}$ whenever *y* is sufficiently large.
In what follows, item (ii) of Definition 2.2 is satisfied.

Moreover, it is seen that the zeros $\{{\mu}_{n}\}$ are separated.
Indeed, let $n\in \mathbb{Z}$.
Then we have

$|{\mu}_{n+1}-{\mu}_{n}|=\left|\frac{(n+1)\pi}{{c}_{L}}-\frac{n\pi}{{c}_{L}}\right|=\frac{\pi}{{c}_{L}}.$

As a consequence of the above sentences, we get the Riesz basis property by Theorem 2.1 for $\{{e}^{i\frac{n\pi}{{c}_{L}}t}\}$ in ${L}^{2}[0,2{c}_{L}]$ and hence for $\{{e}^{{\lambda}_{n}t}\}$ in ${L}^{2}[0,2{c}_{L}]$.
∎

#### Proposition 4.2.

*Let $\mathcal{A}$ be the generator of a ${C}_{\mathrm{0}}$-semigroup in a Hilbert space $\mathcal{H}$.
Then the spectrum of its adjoint is given by
$\sigma \mathit{}\mathrm{(}{\mathcal{A}}^{\mathrm{*}}\mathrm{)}\mathrm{=}\mathrm{\{}\overline{\lambda}\mathrm{:}\lambda \mathrm{\in}\sigma \mathit{}\mathrm{(}\mathcal{A}\mathrm{)}\mathrm{\}}$.*

#### Proof.

Since $\mathcal{\mathscr{H}}$ is a Hilbert space and $T(t)$ is the generator of a ${C}_{0}$-semigroup $\mathcal{\mathcal{A}}$, then ${T}^{*}(t)$ is also a ${C}_{0}$-semigroup, and its generator is ${\mathcal{\mathcal{A}}}^{*}$.
Moreover, its spectrum consists of an isolated eigenvalue $\overline{\lambda}$, where $\lambda \in \sigma (\mathcal{\mathcal{A}})$.
∎

#### Theorem 4.4.

*The resolvent of the operator $\mathcal{A}$ belongs to the Carleman class ${C}_{\mathrm{1}\mathrm{+}\epsilon}$ for every $\epsilon \mathrm{>}\mathrm{0}$.*

#### Proof.

A short computation from the expression of the eigenvalues ${\lambda}_{n}$ of the operator $\mathcal{\mathcal{A}}$, which is given in Theorem 4.3, shows that

$|{\lambda}_{n}|=\frac{n\pi}{|{c}_{L}|}\sqrt{1+\frac{1}{4{n}^{2}{\pi}^{2}}{\mathrm{log}}^{2}\left|\frac{\beta}{\alpha}\right|}.$

Therefore,

$|{\lambda}_{n}|{\sim}_{+\mathrm{\infty}}\frac{n\pi}{|{c}_{L}|}.$

Obviously, $0\in \rho (\mathcal{\mathcal{A}})$, so let $K:={\mathcal{\mathcal{A}}}^{-1}$, and let ${s}_{n}({\mathcal{\mathcal{A}}}^{-1})$ be the eigenvalue of the operator $\sqrt{{({\mathcal{\mathcal{A}}}^{-1})}^{*}{\mathcal{\mathcal{A}}}^{-1}}$.
It is clear that

$\sum _{n\in \mathbb{Z}}\frac{1}{{|{\lambda}_{n}|}^{1+\epsilon}}<\mathrm{\infty}\mathit{\hspace{1em}}\text{for every}\epsilon >0.$

Then ${\mathcal{\mathcal{A}}}^{-1}$ belongs to the Carleman class ${C}_{1+\epsilon}$ for every $\epsilon >0$.
∎

As a consequence of Theorem 2.2, we have the following corollary.

#### Corollary 4.1.

*The resolvent of $\mathcal{A}$ can be represented as two entire functions*

$R(\lambda ,\mathcal{\mathcal{A}})=\frac{F(\lambda ,\mathcal{\mathcal{A}})}{\mathrm{\Delta}(\lambda )}.$

#### Proof.

Let $\lambda \in \rho (\mathcal{\mathcal{A}})$.
Then we have

$R(\lambda ,\mathcal{\mathcal{A}})={(\lambda -\mathcal{\mathcal{A}})}^{-1}={({[\lambda {\mathcal{\mathcal{A}}}^{-1}-I]}^{-1}\mathcal{\mathcal{A}})}^{-1}={\mathcal{\mathcal{A}}}^{-1}{(\lambda {\mathcal{\mathcal{A}}}^{-1}-I)}^{-1}.$

Denoting $K={\mathcal{\mathcal{A}}}^{-1}$, then

$R(\lambda ,\mathcal{\mathcal{A}})=K{(\lambda K-I)}^{-1}=K{R}_{\lambda}(K).$(4.5)

Since the resolvent of $\mathcal{\mathcal{A}}$ belongs to the class ${C}_{1+\epsilon}$ for $\epsilon >0$, then we infer from Theorem 2.2 with equation (4.5) that the resolvent of $\mathcal{\mathcal{A}}$ can be represented under the form
$\frac{F(\lambda ,\mathcal{\mathcal{A}})}{\mathrm{\Delta}(\lambda )}$,
where $F(\lambda ,\mathcal{\mathcal{A}})$ and $\mathrm{\Delta}(\lambda )$ are two entire functions.
∎

Now we are in the position to provide the denseness of the eigenvectors of the operator $\mathcal{\mathcal{A}}$.

#### Theorem 4.5.

*The system of the eigenvectors of the operator $\mathcal{A}$ is dense in $\mathcal{H}$.*

#### Proof.

From Theorem 4.4, the resolvent of $\mathcal{\mathcal{A}}$ belongs to the Carleman class ${C}_{1+\epsilon}$ for $\epsilon >0$
since $R(\lambda ,\mathcal{\mathcal{A}})$ is an entire function of λ and the orders of both entire functions of $F(\lambda ,\mathcal{\mathcal{A}})$ and $\mathrm{\Delta}(\lambda )$ are less than or equal to 1.
That is, there is a $\epsilon >0$ such that

$\parallel R(\lambda ,\mathcal{\mathcal{A}})\parallel =O({e}^{{|\lambda |}^{1+\epsilon}})\mathit{\hspace{1em}}\text{as}|\lambda |\to \mathrm{\infty}.$

As $\mathcal{\mathcal{A}}$ generates an analytic semigroup, it follows that $\parallel R(\lambda ,\mathcal{\mathcal{A}})\parallel $ is uniformly bounded in both real and imaginary axis.
We may assume without loss of generality that $\parallel R(\lambda ,\mathcal{\mathcal{A}})\parallel $ is uniformly bounded in the right complex plane, particularly on the imaginary axis.

Set

${S}_{j}=\{\lambda \in \u2102:\mathrm{Arg}{\gamma}_{j}\le \mathrm{Arg}\lambda \le \mathrm{Arg}{\gamma}_{j+1}\},j=1,2.$

By assumption, $R(\lambda ,\mathcal{\mathcal{A}})$ is bounded on the boundary ${S}_{j}$, and
$\parallel R(\lambda ,\mathcal{\mathcal{A}})\parallel =O({e}^{{|\lambda |}^{1+\epsilon}})$
for all $\lambda \in {S}_{j}$,
where $\epsilon >0$ is chosen so that $1+\epsilon <2$.
Applying the Phragmen–Lindelof theorem to $R(\lambda ,\mathcal{\mathcal{A}})$ in each ${S}_{j}$ (see [17, Theorem 10, p. 80]), we known that $R(\lambda ,\mathcal{\mathcal{A}})$ is uniformly bounded in ${S}_{j}$ and so in the whole complex plane.
Therefore, all conditions of Theorem 2.3 are satisfied with $p=1+\epsilon $ for $\epsilon >0$, $m=3$ and
${\gamma}_{2}=\{\lambda :\mathrm{Arg}\lambda =\pi \}$.
This result in the completeness of the generalized eigenvectors of $\mathcal{\mathcal{A}}$ on $\mathcal{\mathscr{H}}$.
∎

We summarize all the results proved in this section by mentioning the following result.

#### Theorem 4.6.

*The eigenvectors associated to the operator $\mathcal{A}$ form a Riesz basis in $\mathcal{H}$.*

#### Proof.

From the previous results presented in this part, we infer that all the hypotheses of Theorem 3.1 are fulfilled, which makes us conclude that the system of eigenvectors associated to $\mathcal{\mathcal{A}}$ forms a Riesz basis in the state Hilbert space $\mathcal{\mathscr{H}}$.
∎

As a consequence of the spectral mapping theorem for the ${C}_{0}$-semigroup generated by $\mathcal{\mathcal{A}}$, we have the following result.

#### Corollary 4.2.

*Our system satisfies the spectrum-determined growth assumption, i.e.,
$S\mathit{}\mathrm{(}\mathcal{A}\mathrm{)}\mathrm{=}w\mathit{}\mathrm{(}\mathcal{A}\mathrm{)}$.*

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