## 3 Operator Reformulation of the Problem
\begin{split}&\Big{(}\mathbb{F},\mathbb{L}^{*}\mathbb{G}\Big{)}_{ \mathcal{H}}=\frac{i}{2}\Bigg{\{}\int_{0}^{L}\bigg{[}\rho(\xi)\varphi_{3}(\xi) \overline{\overline{\dot{s}_{3}}}(\xi)+S\rho(\xi)\varphi_{1}(\xi)\overline{ \dot{\overline{s}_{1}}}(\xi)+EI(\xi)\varphi_{0}^{\prime}(\xi)\overline{ \overline{\dot{s}_{0}^{\prime}}}(\xi)\\ &+K(\xi)\Big{(}\varphi_{0}(\xi)-\varphi_{2}^{\prime}(\xi)\Big{)} \Big{(}\overline{\dot{\overline{s}_{0}}}(\xi)-\overline{\overline{\dot{s}_{2}^ {\prime}}}(\xi)\Big{)}\bigg{]}d\xi+\theta\overline{s_{1}}(L)\varphi_{4}-\frac {1}{CR}\overline{\dot{s}_{4}}\varphi_{4}\Bigg{\}}\equiv\frac{i}{2}\sum_{j=1}^{ 5}I_{j}^{*},\end{split}  \tag{35}\]  
where \(I_{1}^{*}=\int_{0}^{L}\rho(\xi)\varphi_{3}(\xi)\overline{\dot{\overline{s}_{3 }}}(\xi)d\xi\), \(I_{2}^{*}=\int_{0}^{L}S\rho(\xi)\varphi_{1}(\xi)\overline{\dot{\overline{s}_{1 }}}(\xi)d\xi\), \(I_{3}^{*}=\int_{0}^{L}EI(\xi)\varphi_{0}^{\prime}(\xi)\overline{\dot{ \overline{s}_{0}^{\prime}}}(\xi)d\xi\), \(I_{4}^{*}=\int_{0}^{L}K(\xi)\Big{(}\varphi_{0}(\xi)-\varphi_{2}^{\prime}(\xi) \Big{)}\Big{(}\overline{\dot{\overline{s}_{0}}}(\xi)-\overline{\dot{\overline {s}_{2}}}(\xi)\Big{)}d\xi\), and \(I_{5}^{*}=\theta\overline{s_{1}}(L)\varphi_{4}-\frac{1}{CR}\overline{\dot{s}_{ 4}}\varphi_{4}\). Evaluating the first integral, we have  
\[ \begin{split} I_{1}^{*}&=\int_{0}^{L}\rho(\xi) \varphi_{3}(\xi)\bigg{(}-\frac{1}{\rho(x)}\Big{(}K(\xi)\overline{\dot{s}_{0}}( \xi)\Big{)}^{\prime}+\frac{1}{\rho(\xi)}\Big{(}K(\xi)\overline{\dot{s}_{2}^{ \prime}}(\xi)\Big{)}^{\prime}\bigg{)}d\xi\\ &=-K(\xi)\overline{\dot{s}_{0}}(\xi)\varphi_{3}(\xi)\bigg{|}_{0}^ {L}+K(\xi)\overline{\dot{s}_{2}^{\prime}}(\xi)\varphi_{3}(\xi)\bigg{|}_{0}^{L }+\int_{0}^{L}K(\xi)\Big{(}\overline{\dot{s}_{0}}(\xi)-\overline{\dot{s}_{2}^ {\prime}}(\xi)\Big{)}\varphi_{3}^{\prime}(\xi)d\xi\\ &+K(L)\varphi_{3}(L)\Big{(}\overline{\dot{s}_{2}^{\prime}}(L)- \overline{\dot{s}_{0}}(L)\Big{)}+K(0)\varphi_{3}(0)\overline{\dot{s}_{0}(0)}- K(0)\varphi_{3}(0)\overline{\dot{s}_{2}(0)}.\end{split} \]  
Since \(\mathbb{G}\in\mathcal{D}(\mathbb{L}^{*})\), i.e., \(g_{0}(L)-g_{2}^{\prime}(L)=0\) and \(\mathbb{F}\in\mathcal{D}(\mathbb{L})\), i.e., \(\varphi_{3}(0)=0\), we obtain that  
\[I_{1}^{*}=\int_{0}^{L}K(\xi)\Big{(}\overline{\dot{s}_{0}}(\xi)-\overline{ \dot{s}_{2}^{\prime}}(\xi)\Big{)}\varphi_{3}(\xi)d\xi.\] (36a) Evaluating the second integral from Equation ( 35 ), we get \[ \begin{split} I_{2}^{*}&=\int_{0}^{L}\bigg{[} \Big{(}EI(\xi)\overline{\dot{s}_{0}}^{\prime}(\xi)\Big{)}^{\prime}-K(\xi) \overline{\dot{s}_{0}}(\xi)+K(\xi)\overline{\dot{s}_{2}^{\prime}}(\xi)\bigg{]} \varphi_{1}(\xi)d\xi\\ &=EI(L)\overline{\dot{s}_{0}}^{\prime}(L)\varphi_{1}(L)-\int_{0}^ {L}EI(\xi)\overline{\dot{s}_{0}}^{\prime}(\xi)\varphi_{1}^{\prime}(\xi)d\xi- \int_{0}^{L}K(\xi)\Big{(}\overline{\dot{s}_{0}}(\xi)-\overline{\dot{s}_{2}^ {\prime}}(\xi)\Big{)}\varphi_{1}(\xi)d\xi.\end{split} \] (36b) For integrals \(I_{3}^{*}\), \(I_{4}^{*}\) and \(I_{5}^{*}\), we obtain the following expressions: \[I_{3}^{*}=\int_{0}^{L}EI(\xi)\overline{\dot{s}_{1}^{\prime}}(\xi)\varphi_{0}^{ \prime}(\xi)d\xi,\] (36c) \[I_{4}^{*}=\int_{0}^{L}K(\xi)\Big{(}\varphi_{0}(\xi)-\varphi_{2}^{\prime}(\xi) \Big{)}\Big{(}\overline{\dot{s}_{1}}(\xi)-\overline{\dot{s}_{3}^{\prime}}(\xi) \Big{)}d\xi,\] (36d) \[I_{5}^{*}=\theta\overline{\dot{s}_{1}}(L)\varphi_{4}-\frac{1}{CR} \overline{\dot{s}_{4}}\varphi_{4}.\] (36e) Collecting Equations ( 36a )-( 36e ), we obtain from Equation ( 34 )  
Comparing Equation (33) with Equation (37) yields the result. **Theorem 1**. : _The operator \(\mathbb{L}\) has the following properties._  
* \(\mathbb{L}\) _is an unbounded closed non-selfadjoint operator in_ \(\mathcal{H}\)_._
* _The inverse operator_ \(\mathbb{L}^{-1}\) _exists and is a compact operator in_ \(\mathcal{H}\)_. Therefore,_ \(\mathbb{L}\) _has purely discrete spectrum of normal eigenvalues (see_ _Gohberg and [PERSON]_ _[_33_]__)._  
Proof.