## 3 Operator Reformulation of the Problem
\begin{split}\mathcal{D}\big{(}\mathbb{L}^{*}\big{)}=\bigg{\{}& \Psi\in\mathcal{H}:\psi_{0},\psi_{2}\in H^{2}(0,L);\;\psi_{1},\psi_{3}\in H^{1 }(0,L),\;\psi_{4}\in\mathbb{C};\;\psi_{0}(0)=\psi_{2}(0)=0,\\ &\psi_{1}(0)=\psi_{3}(0)=0,\;\;\psi_{0}(L)-\psi_{2}^{\prime}(L)=0, \;\;EI(L)\psi_{0}^{\prime}(L)=-\frac{\theta}{\mathbb{C}}\psi_{4}\bigg{\}}.\end{split}  \tag{26}\]  
Proof. : We have to derive the matrix differential expression and the boundary conditions in such a way that, for any \(\mathbb{F}\in\mathcal{D}\big{(}\mathbb{L}\big{)}\) and \(\mathbb{G}\in\mathcal{D}\big{(}\mathbb{L}^{*}\big{)}\), the following relation holds:  
\[\Big{(}\mathbb{L}\mathbb{F},\mathbb{G}\Big{)}_{\mathcal{H}}=\Big{(}\mathbb{F },\mathbb{L}^{*}\mathbb{G}\Big{)}_{\mathcal{H}}. \tag{27}\]  
Let \(\mathbb{F}=\Big{(}\varphi_{0}(x),\varphi_{1}(x),\varphi_{2}(x),\varphi_{3}(x),\varphi_{4}\Big{)}^{T}\) and \(\mathbb{G}=\Big{(}g_{0}(x),g_{1}(x),g_{2}(x),g_{3}(x),g_{4}\Big{)}^{T}\). We have for any \(\mathbb{F}\in\mathcal{H}\):  
\[\Big{(}\mathbb{L}\mathbb{F}\Big{)}(x)=-i\left[ \begin{array}{c}\varphi_{1}( x)\\ \frac{1}{S\varphi(x)}\bigg{(}\Big{(}EI(x)\varphi_{0}^{\prime}(x)\Big{)}^{ \prime}-K(x)\varphi_{0}(x)\bigg{)}+\frac{K(x)}{S\rho(x)}\varphi_{2}^{\prime}(x )\\ \varphi_{3}(x)\\ -\frac{1}{\rho(x)}\Big{(}K(x)\varphi_{0}(x)\Big{)}^{\prime}+\frac{1}{\rho(x) }\Big{(}K(x)\varphi_{2}^{\prime}(x)\Big{)}^{\prime}\\ -\frac{\theta}{\mathbb{C}}\varphi_{1}(L)-\frac{1}{\mathbb{C}R}\varphi_{4} \end{array} \right]\equiv\left[ \begin{array}{c}\widetilde{\varphi_{0}}(x)\\ \widetilde{\varphi_{1}}(x)\\ \widetilde{\varphi_{2}}(x)\\ \widetilde{\varphi_{3}}(x)\\ \widetilde{\varphi_{4}}\end{array} \right]. \tag{28}\]  
Evaluating the inner product induced by norm (Equation (19)) yields  
\[ \begin{split}\Big{(}\mathbb{L}\mathbb{F},\mathbb{G}\Big{)}_{ \mathcal{H}}=&-i\left(\left(\begin{array}{c}\widetilde{\varphi_{0}}(x)\\ \widetilde{\varphi_{1}}(x)\\ \widetilde{\varphi_{2}}(x)\\ \widetilde{\varphi_{3}}(x)\\ \widetilde{\varphi_{4}}\end{array}\right),\left(\begin{array}{c}g_{0}(x)\\ g_{1}(x)\\ g_{2}(x)\\ g_{3}(x)\\ g_{4}\end{array}\right)\right)_{\mathcal{H}}=-\frac{i}{2}\left\{\int_{0}^{L} \rho(\xi)\widetilde{\varphi_{3}}(\xi)\overline{\widetilde{g_{3}}}(\xi)d \xi+\right.\\ &\left.\int_{0}^{L}S\rho(\xi)\widetilde{\varphi_{1}}(\xi)\overline{ \widetilde{g_{1}}}(\xi)d\xi+\int_{0}^{L}EI(\xi)\widetilde{\varphi_{0}^{\prime }}(\xi)\overline{\widetilde{g_{0}^{\prime}}}(\xi)d\xi+\right.\\ &\left.\int_{0}^{L}K(\xi)\Big{(}\widetilde{\varphi_{0}}(\xi)- \widetilde{\varphi_{2}^{\prime}}(\xi)\Big{)}\Big{(}\overline{\widetilde{g_{0} }}(\xi)-\overline{\widetilde{g_{2}^{\prime}}}(\xi)\Big{)}d\xi+\widetilde{ \varphi_{4}}\overline{\widetilde{g_{4}}}\right\}\equiv-\frac{i}{2}\sum_{j=1}^ {5}I_{j}.\end{split}  \tag{29}\]  
Now, we modify the integrals \(I_{j}\), \(j=1,...,5\). Evaluating the first integral, we have  
\[ \begin{split} I_{1}&=\int_{0}^{L}\rho(\xi)\bigg{[}- \frac{1}{\rho(\xi)}\Big{(}K(\xi)\varphi_{0}(\xi)\Big{)}^{\prime}+\frac{1}{\rho( \xi)}\Big{(}K(\xi)\varphi_{2}^{\prime}(\xi)\Big{)}^{\prime}\bigg{]}\overline{ \widetilde{g}}_{3}(\xi)d\xi\\ &=-K(\xi)\varphi_{0}(\xi)\overline{\widetilde{g}}_{3}(\xi)\Big{|} _{0}^{L}+K(\xi)\varphi_{2}^{\prime}(\xi)\overline{\widetilde{g}}_{3}(\xi) \Big{|}_{0}^{L}+\int_{0}^{L}\Big{(}\varphi_{0}(\xi)-\varphi_{2}^{\prime}(\xi) \Big{)}\overline{\widetilde{g^{\prime}}}_{3}(\xi)d\xi\\ &=-K(L)\Big{(}\varphi_{0}(L)-\varphi_{2}^{\prime}(L)\Big{)} \overline{\widetilde{g^{\prime}}}_{3}(L)+K(0)\Big{(}\varphi_{0}(0)-\varphi_{2} ^{\prime}(0)\Big{)}\overline{\widetilde{g}}_{3}(0)+\\ &\int_{0}^{L}K(\xi)\Big{(}\varphi_{0}(\xi)-\varphi_{2}^{\prime}(\xi) \Big{)}\overline{\widetilde{g}}_{3}(\xi)d\xi.\end{split}