## 4 Non-Selfadjoint Operator Pencil Generated by the Harvester System
\begin{array}{c}C_{1}(\lambda)\\ C_{2}(\lambda)\end{array} \right]=\mathfrak{R}_{r}(\lambda)\mathfrak{R}_{l}^{-1}( \lambda)\left[ \begin{array}{c}C_{1}(\lambda)\\ C_{2}(\lambda)\end{array} \right]. \tag{151}\]  
This equation has nontrivial solutions if and only if  
\[\det\left(\mathfrak{R}_{r}(\lambda)\mathfrak{R}_{l}^{-1}(\lambda)-I\right)=0, \tag{152}\]  
where \(I\) is the identity matrix. In the sequel, we need the following result. **Lemma 2**. : _The set of roots of Equation (152) is located in a strip parallel to the real axis._  
Proof. : Notice Equation (152) has the same set of roots as the following equation:  
\[\det\left(\mathbb{D}(\lambda)-\mathbb{E}(\lambda)\mathfrak{R}_{l}^{-1}( \lambda)\mathbb{E}(\lambda)\right)=0, \tag{153}\]  
where \(\mathbb{E}(\lambda)\) is defined in Equation (117). Using the contradiction argument, assume that there exists a sequence of roots of Equation (153) denoted by \(\left\{\lambda_{m}\right\}_{m=1}^{\infty}\) such that \(x_{m}=\Re\lambda_{m}\to\infty\) and \(y_{m}=\Im\lambda_{m}\to\infty\) as \(m\to\infty\). Rewriting Equation (153) for \(\lambda=\lambda_{m}\), we obtain (see Equation (119)):  
\[ \begin{split}&\left[d_{11}(\lambda_{m})-e^{2}(\lambda_{m}) \Big{(}1+\mathcal{O}\Big{(}\frac{1}{\lambda_{m}}\Big{)}\Big{)}\right]\left[d_ {22}(\lambda_{m})-\widehat{e}^{2}(\lambda_{m})\Big{(}1+\mathcal{O}\Big{(} \frac{1}{\lambda_{m}}\Big{)}\Big{)}\right]-\\ &\left[d_{12}(\lambda_{m})-\mathcal{O}\Big{(}\frac{1}{\lambda_{m }}\Big{)}e(\lambda_{m})\widehat{e}(\lambda_{m})\right]\left[d_{21}(\lambda_{m })+2\Big{(}\frac{SK(0)}{EI(0)}\Big{)}^{1/4}\Big{(}1+\mathcal{O}\Big{(}\frac{1} {\lambda_{m}}\Big{)}\Big{)}e(\lambda_{m})\widehat{e}(\lambda_{m})\right]=0. \end{split}  \tag{154}\]  
It can be readily checked that, when \(y_{m}\to\infty\), we have \(e\big{(}\lambda_{m}\big{)}\to 0\) and \(\widehat{e}\big{(}\lambda_{m}\big{)}\to 0\). Hence, Equation (154) can be written in the form  
\[d_{11}\big{(}\lambda_{m}\big{)}d_{22}\big{(}\lambda_{m}\big{)}-d_{12}\big{(} \lambda_{m}\big{)}d_{21}\big{(}\lambda_{m}\big{)}=o(1),\]  
which means that the left-hand side approaches its limit:  
\[\lim_{m\to\infty}d_{11}\big{(}\lambda_{m}\big{)}d_{22}\big{(}\lambda_{m}\big{)} =\frac{\alpha-\sqrt{K(L)\rho(L)}}{\alpha+\sqrt{K(L)\rho(L)}}\
eq 0,\]  
while the right-hand side approaches \(0\) as \(m\to\infty\). This contradiction means that the set of all roots that belongs to the upper half-plane can be located in a strip parallel to the real axis. Using a similar argument, we can show that the set of all roots that belongs to the lower half-plane can be located in a strip parallel to the real axis as well. In the following, we consider Equation (153) in a strip on the complex plane parallel to the real axis. Let us multiply matrices \(\mathbb{D}(\lambda)\) of (148) and \(\mathbb{E}^{-1}(\lambda)\mathfrak{R}_{l}(\lambda)\mathbb{E}^{-1}(\lambda)\). As the result, we reduce the spectral equation (Equation (153)) to the form  
\[\det\left[ \begin{array}{cc}\left(d_{11}(\lambda)\big{(}e(\lambda)\big{)}^{ -2}-1\right)\Big{(}1+\mathcal{O}\big{(}\lambda^{-1}\big{)}\Big{)}&\mathcal{O} \big{(}\lambda^{-1}\big{(}e(\lambda)\widehat{e}(\lambda)\big{)}^{-1}\\ \left(d_{21}\big{(}\lambda\big{)}-2\Big{(}\frac{SK(0)}{EI(0)}\Big{)}^{1/4}d_{ 21}\big{(}\lambda\big{)}\Big{)}\Big{(}e(\lambda)\widehat{e}(\lambda)\Big{)} ^{-1}\Big{(}1+\mathcal{O}\big{(}\lambda^{-1}\big{)}\Big{)}&-\left(d_{22}( \lambda)\big{(}\widehat{e}(\lambda)\big{)}^{-2}+1\right)\Big{(}1+\mathcal{O} \big{(}\lambda^{-1}\big{)}\Big{)}\end{array} \right]=0.