## 3 Operator Reformulation of the Problem
: The fact that \(\mathbb{L}\) is non-selfadjoint is due to two reasons: the matrix differential expressions for \(\mathbb{L}\) and \(\mathbb{L}^{*}\) are different and the domains \(\mathcal{D}(\mathbb{L})\) and \(\mathcal{D}(\mathbb{L}^{*})\) are different (see Equations (23) and (26)). Let us prove that \(\mathbb{L}^{-1}\) exists and is compact. To this end, we consider the equation  
\[\mathbb{L}\Psi=Y,\qquad Y\in\mathcal{H} \tag{38}\]  
and prove that this equation has a unique solution \(\Psi=\big{(}\psi_{0}(x),\psi_{1}(x),\psi_{2}(x),\psi_{3}(x),\psi_{4}\big{)}^{T} \in\mathcal{D}(\mathbb{L})\) for any \(Y\in\mathcal{H}\), \(Y\equiv\big{(}y_{0}(x),y_{1}(x),y_{2}(x),y_{3}(x),y_{4}\big{)}^{T}\). Rewriting Equation (38) component-wise, we obtain the following system:  
\[ \begin{split}(i)&\quad\psi_{1}(x)=iy_{0}(x),\\ (ii)&\quad\psi_{3}(x)=iy_{2}(x),\\ (iii)&\quad\big{(}EI(x)\psi_{0}^{\prime}(x)\big{)}^{ \prime}-K(x)\psi_{0}(x)+K(x)\psi_{2}^{\prime}(x)=iS\rho(x)y_{1},\\ (iv)&\quad-\big{(}K(x)\psi_{0}(x)\big{)}^{\prime}+ \big{(}K(x)\psi_{2}(x)\big{)}^{\prime}=i\rho(x)y_{3}(x),\\ (v)&\quad-\frac{\theta}{C}\psi_{1}(L)-\frac{1}{CR} \psi_{4}=iy_{4}.\end{split}  \tag{39}\]  
Integrating Equation (39) (iv) and using the condition \(\psi_{0}(L)-\psi_{2x}(L)=0\), we obtain  
\[K(x)\big{(}\psi_{0}(x)-\psi_{2}^{\prime}(x)\big{)}=i\int_{x}^{L}\rho(\tau)y_ {3}(\tau)d\tau. \tag{40}\]  
Substituting Equation (40) into Equation (39) (iii), we get  
\[\big{(}EI(x)\psi_{0}^{\prime}(x)\big{)}^{\prime}=K(x)\big{(}\psi_{0}(x)-\psi _{2}^{\prime}(x)\big{)}+iS\rho(x)y_{1}(x)=i\int_{x}^{L}\rho(\tau)y_{3}(\tau)d \tau+iS\rho(x)y_{1}(x).\]  
Integrating this equation and taking into account that \(E(L)\psi_{0}^{\prime}(L)=\theta\psi_{4}\), we obtain  
\[EI(x)\psi_{0}^{\prime}(x)=\theta\psi_{4}-i\int_{x}^{L}d\tau\int_{\tau}^{L}\rho( \eta)y_{3}(\eta)d\eta-iS\int_{x}^{L}\rho(\tau)y_{1}(\tau)d\tau. \tag{41}\]  
Using Equation (39) (i) we derive from Equation (39) (v) that  
\[\psi_{4}=-CR\bigg{(}iy_{4}-\frac{\theta}{C}\psi_{1}(L)\bigg{)}=-iR\Big{(}Cy_{4 }-\theta y_{0}(L)\Big{)}. \tag{42}\]  
Substituting this result into Equation (41), we obtain the following representation  
\[\psi_{0}^{\prime}(x)=-\frac{iR\theta}{EI(x)}\Big{(}Cy_{4}-\theta y_{0}(L) \Big{)}-\]  
\[\frac{i}{EI(x)}\int_{x}^{L}d\tau\int_{\tau}^{L}\rho(\eta)y_{3}(\eta)d\eta-\frac {iS}{EI(x)}\int_{x}^{L}\rho(\tau)y_{1}(\tau)d\tau,\]which upon integration yields the formula for \(\psi_{0}\), i.e.,  
\[\psi_{0}(x)= -iR\theta\Big{(}Cy_{4}-\theta y_{0}(L)\Big{)}\int_{0}^{x}\frac{d \tau}{EI(\tau)}-i\int_{0}^{x}\frac{d\tau}{EI(\tau)}\int_{\tau}^{L}d\eta\int_{ \eta}^{L}\rho(\xi)y_{3}(\xi)d\xi \tag{43}\] \[-iS\int_{0}^{x}\frac{d\tau}{EI(\tau)}\int_{\tau}^{L}\rho(\eta)y_{ 1}(\eta)d\eta.\]  
Integrating \(\psi_{2}^{\prime}\) from Equation (40), we derive that  
\[\psi_{2}(x)=\int_{0}^{x}\psi_{0}(\tau)d\tau-i\int_{0}^{x}\frac{d\tau}{K(\tau)} \int_{\tau}^{L}\rho(\eta)y_{3}(\eta)d\eta. \tag{44}\]  
Collecting together Equations (39) (i), (39) (ii), and (42)-(44), one can see that the following formula for the solution of Equation (38) is valid:  
\[\Psi(x)=\left[ \begin{array}{c}\left\{\begin{array}{c}-iR\theta\Big{(}Cy_{ 4}-\theta y_{0}(L)\Big{)}\int_{0}^{x}\frac{d\tau}{EI(\tau)}-i\int_{0}^{x}\frac {d\tau}{EI(\tau)}\int_{\tau}^{L}d\eta\int_{\eta}^{L}\rho(\xi)y_{3}(\xi)d\xi\\ -iS\int_{0}^{x}\frac{d\tau}{EI(\tau)}\int_{\tau}^{L}\rho(\eta)y_{1}(\eta)d\eta \end{array}\right.\\ \\ \left.\begin{array}{c}iy_{0}(x)\\ \int_{0}^{x}\psi_{0}(\tau)d\tau-i\int_{0}^{x}\frac{d\tau}{K(\tau)}\int_{\tau} ^{L}\rho(\eta)y_{3}(\eta)d\eta\\ -iR\Big{(}Cy_{4}-\theta y_{0}(L)\Big{)}\end{array}\right].\end{array} \right]. \tag{45}\]  
It is clear that \(\mathbb{L}^{-1}\) is defined on each \(Y\in\mathcal{H}\).