## 4 Non-Selfadjoint Operator Pencil Generated by the Harvester System
However, to keep the paper self-contained, we outline the proof of the next statement. **Theorem 2**. : _(1) Under the assumptions in Equations (24), (25) and (74), Equation (72) has four linearly independent solutions with the following asymptotic approximations as \(|\lambda|\to\infty\):_  
\[\psi_{1}(\lambda,x)=\sqrt[4]{\frac{K(x)}{\rho(x)}}\ \ \exp\big{(}i\lambda\eta(x) \big{)}\Big{(}1+O\big{(}|\lambda|^{-1}\big{)}\Big{)}, \tag{75}\]  
\[\psi_{2}(\lambda,x)=\sqrt[4]{\frac{EI(x)}{R\rho(x)}}\ \ \exp\big{(}i\lambda \widehat{\eta}(x)\big{)}\Big{(}1+O\big{(}|\lambda|^{-1}\big{)}\Big{)}, \tag{76}\]  
\[\psi_{3}(\lambda,x)=\sqrt[4]{\frac{K(x)}{\rho(x)}}\ \ \exp\big{(}-i\lambda\eta(x) \big{)}\Big{(}1+O\big{(}|\lambda|^{-1}\big{)}\Big{)}, \tag{77}\]  
\[\psi_{4}(\lambda,x)=\sqrt[4]{\frac{EI(x)}{R\rho(x)}}\ \ \exp\big{(}-i\lambda \widehat{\eta}(x)\big{)}\Big{(}1+O\big{(}|\lambda|^{-1}\big{)}\Big{)}, \tag{78}\]  
_where_  
\[\eta(x)=\int_{0}^{x}\sqrt{\frac{\rho(\tau)}{K(\tau)}}d\tau,\qquad\widehat{ \eta}(x)=\int_{0}^{x}\sqrt{\frac{R\rho(\tau)}{EI(\tau)}}d\tau. \tag{79}\]  
_All estimates \(O\big{(}|\lambda|^{-1}\big{)}\) in (75)-(78) are uniform with respect to \(x\in[0,L]\). The functions \(\psi_{j}(\lambda,x)\), \(j=1,2,3,4\) are analytic functions of \(\lambda\)._  
_(2) Each function \(\psi_{j}(\lambda,x)\) has four continuous derivatives with respect to \(x\in[0,L]\) and these derivatives can be approximated as follows when \(|\lambda|\to\infty\):_  
\[ \begin{gathered}\bigg{(}\frac{d}{dx}\bigg{)}^{n}\psi_{j}(\lambda,x)=(i\lambda)^{n}\big{(}\gamma_{j}(x)\big{)}^{n}\big{|}(\gamma_{j}(x))\big{|} ^{-1/4}\exp\Bigg{(}i\lambda\int_{0}^{x}\gamma_{j}(\tau)d\tau\Bigg{)}\Big{(}1+ O\big{(}|\lambda|^{-1}\big{)}\Big{)},\\ n=1,2,3,4,\qquad j=1,2,3,4,\end{gathered}  \tag{80}\]  
_where_  
\[\gamma_{1,3}(x)=\pm\sqrt{\frac{\rho(x)}{K(x)}}\qquad\text{and}\qquad\gamma_{ 2,4}(x)=\pm\sqrt{\frac{S\rho(x)}{EI(x)}}. \tag{81}\]  
_All estimates \(O\big{(}|\lambda|^{-1}\big{)}\) in Equation (80) are uniform with respect to \(x\in[0,L]\)._  
Proof. : First, we consider the equation obtained from Equation (72) in which the right-hand side has been replaced with zero. For the latter equation, we prove that it has four linearly independent solutions with the approximations given precisely by Equations (75)-(78). Secondly, we refer to the standard methods of asymptotic analysis (see [PERSON] [36], [PERSON] [40], [PERSON] [37]) to prove that the lower order terms do not destroy the asymptotic behavior of the solutions given by Equations (75)-(78). Thus, we consider the equation  
\[y^{\prime\prime\prime\prime}+\lambda^{2}\big{(}A(x)+B(x)\big{)}y^{\prime\prime}+ \lambda^{4}A(x)B(x)=0, \tag{82}\]  
where \(B(x)=S\rho(x)/EI(x)\) and \(A(x)=\rho(x)/K(x)\). Let us reduce this equation to the first order linear system, i.e., we consider the linearization of Equation (82) with respect to \(\lambda\) (see [PERSON] [34] and [PERSON] [36]). To this end, we make the substitutions  
\[y(x)=y_{1}(x),\qquad y_{1}^{\prime}(x)=\lambda y_{2}(x),\qquad y_{2}^{\prime} (x)=\lambda y_{3}(x),\qquad y_{3}^{\prime}(x)=\lambda y_{4}(x). \tag{83}\]  
Equation (82) can be rewritten in the following form:  
\[\gamma^{\prime}(x)=\lambda\mathcal{A}(x)Y(x), \tag{84}\]  
where  
\[\mathcal{A}(x)=\left[ \begin{array}{cccc}0&1&0&0\\ 0&0&1&0\\ 0&0&0&1\\ -A(x)B(x)&0&-\big{(}A(x)+B(x)\big{)}&0\end{array} \right],\qquad Y(x)=\left[ \begin{array}{c}y_{1}(x)\\ y_{2}(x)\\ y_{3}(x)\\ y_{4}(x)\end{array} \right]. \tag{85}\]  
In the next step, we reduce the system in Equation (84) to the asymptotically diagonal form. To this end, we have to calculate the eigenvalues and the eigenvectors of the matrix \(\mathcal{A}(x)\).