## 4 Non-Selfadjoint Operator Pencil Generated by the Harvester System
\begin{split}& C_{1}(\lambda)\bigg{(}1+\mathcal{O}\bigg{(}\frac{1}{ \lambda}\bigg{)}\bigg{)}+C_{2}(\lambda)\bigg{(}\frac{SK(0)}{EI(0)}\bigg{)}^{1/ 4}\bigg{(}1+\mathcal{O}\bigg{(}\frac{1}{\lambda}\bigg{)}\bigg{)}=\\ -& D_{1}(\lambda)\bigg{(}1+\mathcal{O}\bigg{(}\frac{1 }{\lambda}\bigg{)}\bigg{)}-D_{2}(\lambda)\bigg{(}\frac{SK(0)}{EI(0)}\bigg{)}^{1/ 4}\bigg{(}1+\mathcal{O}\bigg{(}\frac{1}{\lambda}\bigg{)}\bigg{)},\end{split}  \tag{114}\]  
\[ \begin{split}& C_{1}(\lambda)\Big{(}SK(0)-EI(0)\Big{)}\bigg{(}1+ \mathcal{O}\bigg{(}\frac{1}{\lambda}\bigg{)}\bigg{)}+C_{2}(\lambda)\mathcal{O }\bigg{(}\frac{1}{\lambda}\bigg{)}=\\ & D_{1}(\lambda)\Big{(}SK(0)-EI(0)\Big{)}\bigg{(}1+\mathcal{O} \bigg{(}\frac{1}{\lambda}\bigg{)}\bigg{)}+D_{2}(\lambda)\mathcal{O}\bigg{(} \frac{1}{\lambda}\bigg{)}.\end{split}  \tag{115}\]  
This system can be written in a matrix form as  
\[ \begin{split}&\left[\begin{array}{cc}1+\mathcal{O}\big{(} \lambda^{-1}\big{)}&O\big{(}\lambda^{-1}\big{)}\\ 1+\mathcal{O}\big{(}\lambda^{-1}\big{)}&\Big{(}\frac{EI(0)}{SK(0)}\Big{)}^{1 /4}\Big{(}1+\mathcal{O}\big{(}\lambda^{-1}\big{)}\Big{)}\end{array}\right] \left[\begin{array}{c}C_{1}(\lambda)\\ C_{2}(\lambda)\end{array}\right]=\\ &\left[\begin{array}{cc}1+\mathcal{O}\big{(}\lambda^{-1}\big{)}&O\big{(} \lambda^{-1}\big{)}\\ -\bigg{(}1+\mathcal{O}\big{(}\lambda^{-1}\big{)}&-\Big{(}\frac{EI(0)}{SK(0)} \Big{)}^{1/4}\Big{(}1+\mathcal{O}\big{(}\lambda^{-1}\big{)}\Big{)}\end{array} \right]\left[\begin{array}{c}D_{1}(\lambda)\\ D_{2}(\lambda)\end{array}\right].\end{split}  \tag{116}\]  
Taking the inverse of the left-hand side matrix and applying it to the right-hand side, we finally obtain the desired result (Equation (108)). Using the boundary conditions at the right end of the beam, we derive the representation for the right-reflection matrix \(\mathfrak{R}_{r}(\lambda)\). We need the notations  
\[e(\lambda)=\exp\big{(}i\lambda\eta(L)\big{)},\quad\widehat{e}(\lambda)=\exp \big{(}i\lambda\widehat{\eta}(L)\big{)},\quad\mathbb{E}(\lambda)=diag\{e( \lambda),\;\widehat{e}(\lambda)\}, \tag{117}\]  
with \(\eta(L)\) and \(\widehat{\eta}(L)\) being given in Equation (79). The following result holds. **Theorem 4**. : _Let_  
\[\Theta(\lambda)=\frac{K^{\prime}(L)}{K(L)}-\frac{i\lambda\theta^{2}}{EI(L)C \Big{(}\frac{1}{\mathcal{C}\mathcal{R}}+i\lambda\Big{)}}. \tag{118}\]  
_Assume that the parameter \(\alpha\) satisfies the following condition:_  
\[|\alpha|\
eq\sqrt{\rho(L)K(L)}. \tag{119}\]  
_For the solution \(\Psi(\lambda,x)\) of Equation (106) to satisfy the right-end boundary conditions in Equations (67) and (69), the coefficients \(C_{j}(\lambda)\), \(D_{j}(\lambda)\), \(j=1,2\), must be related by means of the right-reflection matrix \(\mathfrak{R}_{r}(\lambda)\):_  
\[\left[ \begin{array}{c}C_{1}(\lambda)\\ C_{2}(\lambda)\end{array} \right]=\mathfrak{R}_{r}(\lambda)\left[ \begin{array}[] {c}D_{1}(\lambda)\\ D_{2}(\lambda)\end{array} \right], \tag{120}\]  
_with \(\mathfrak{R}_{r}(\lambda)\) being given by_  
\[\mathfrak{R}_{r}(\lambda)=\mathbb{E}^{-1}(\lambda)\left[ \begin{array}{cc}d_{ 11}(\lambda)&d_{12}(\lambda)\\ d_{21}(\lambda)&d_{22}(\lambda)\end{array} \right]\mathbb{E}^{-1}(\lambda). \tag{121}\]_The following asymptotic approximations are valid for the entries \(d_{jk}(\lambda),\ j=1,2,\ k=1,2,\) as \(|\lambda|\to\infty\):_  
\[ \begin{split} d_{11}(\lambda)&=\frac{\alpha-\sqrt{K( L)\rho(L)}}{\alpha+\sqrt{K(L)\rho(L)}}\bigg{(}1+O\bigg{(}\frac{1}{\lambda} \bigg{)}\bigg{)},\qquad d_{12}(\lambda)=O\bigg{(}\frac{1}{\lambda}\bigg{)},\\ d_{21}(\lambda)&=\frac{2\sqrt[4]{E(L)K(L)/S}}{ \alpha/\sqrt{\rho(L)}+\sqrt{K(L)}}\bigg{(}1+O\bigg{(}\frac{1}{\lambda}\bigg{)} \bigg{)},\qquad d_{22}(\lambda)=1+O\bigg{(}\frac{1}{\lambda}\bigg{)}.\end{split}  \tag{122}\]  
Proof. : Let us represent the condition in Equation (67) in the asymptotical form:  
\[