## 4 Non-Selfadjoint Operator Pencil Generated by the Harvester System
For the eigenvalues, we have the following equation:  
\[\det\big{(}\lambda I-\mathcal{A}(x)\big{)}=\lambda^{4}+\lambda^{2}\big{(}A(x) +B(x)\big{)}+A(x)B(x)=0. \tag{86}\]  
Due to Equation (74), Equation (86) has four different roots given by the formulas  
\[\lambda_{1,3}(x)=\pm i\sqrt{A(x)},\qquad\lambda_{2,4}(x)=\pm i\sqrt{B(x)}. \tag{87}\]  
The eigenvector \(F_{j}(x)\), corresponding to the eigenvalue \(\lambda_{j}(x)\), has the form  
\[F_{j}(x)=\Big{[}1,\ \ \lambda_{j}(x),\ \ \lambda_{j}^{2}(x),\ \ \lambda_{j}^{3}(x) \Big{]}^{\mathrm{T}},\qquad j=1,2,3,4. \tag{88}\]  
Let \(\mathbf{T}_{0}(x)\) be the following Vandermonde matrix:  
\[ \begin{split}\mathbf{T}_{0}(x)&=\left[\begin{array}{ ccccc}1&1&1&1\\ \lambda_{1}(x)&\lambda_{2}(x)&\lambda_{3}(x)&\lambda_{4}(x)\\ \lambda_{1}^{2}(x)&\lambda_{2}^{2}(x)&\lambda_{3}^{2}(x)&\lambda_{4}^{2}(x)\\ \lambda_{1}^{3}(x)&\lambda_{2}^{3}(x)&\lambda_{3}^{3}(x)&\lambda_{4}^{3}(x) \end{array}\right]\\ &=\left[\begin{array}{cccc}1&1&1&1\\ iA^{1/2}(x)&-iA^{1/2}(x)&iB^{1/2}(x)&-iB^{1/2}(x)\\ -A(x)&-A(x)&-B(x)&-B(x)\\ -iA^{3/2}(x)&iA^{3/2}(x)&-iB^{3/2}(x)&iB^{3/2}(x)\end{array}\right].\end{split}  \tag{89}\]  
The matrix \(\mathbf{T}_{0}(x)\) reduces \(\mathcal{A}(x)\) to the diagonal form, i.e.,  
\[\mathbf{T}_{0}^{-1}(x)\mathcal{A}(x)\mathbf{T}_{0}(x)\equiv\Lambda_{0}(x)= \left[ \begin{array}{cccc}i\sqrt{A(x)}&0&0&0\\ 0&-i\sqrt{A(x)}&0&0\\ 0&0&i\sqrt{B(x)}&0\\ 0&0&0&-i\sqrt{B(x)}\end{array} \right]. \tag{90}\]One can show by a straightforward calculation that  
\[\det\mathbf{T}_{0}(x)=-4\sqrt{A(x)B(x)}\big{(}A(x)-B(x)\big{)}^{2}. \tag{91}\]  
Since due to Equation (74), \(A(x)\
eq B(x)\), there exists \(\mathbf{T}_{0}^{-1}(x)\). Let us look for a solution of Equation (84) in the form  
\[Y(x)=\mathbf{T}_{0}(x)Z(x).\]  
Rewriting Equation (84) with respect to \(Z\), we have  
\[\mathbf{T}_{0}(x)Z^{\prime}(x)=\lambda\mathcal{A}(x)\mathbf{T}_{0}(x)Z(x)- \frac{d\mathbf{T}_{0}(x)}{dx}Z(x). \tag{92}\]  
Applying matrix \(\mathbf{T}_{0}^{-1}(x)\) from the left, we obtain from Equation (92) the following equation for \(Z(x)\):  
\[Z^{\prime}(x)=\lambda\Lambda_{0}(x)Z(x)-\mathbf{T}_{0}^{-1}(x)\frac{d\mathbf{ T}_{0}(x)}{dx}Z(x) \tag{93}\]  
with \(\Lambda_{0}(x)\) being introduced in Equation (90). It is convenient to look for \(Z\) in the following form [36]:  
\[Z(x)=\Big{(}\mathbf{I}+\lambda^{-1}\mathbf{T}_{1}(x)\Big{)}W(x), \tag{94}\]  
where \(4\times 4\) matrix \(\mathbf{T}_{1}(x)\) is chosen below. Equation (93) can be rewritten for \(W\) as  
\[ \begin{split} W^{\prime}(x)&=\Big{(}\mathbf{I}+ \lambda^{-1}\mathbf{T}_{1}(x)\Big{)}^{-1}\bigg{(}\lambda\Lambda_{0}(x)- \mathbf{T}_{0}^{-1}(x)\frac{d\mathbf{T}_{0}(x)}{dx}\bigg{)}\Big{(}\mathbf{I}+ \lambda^{-1}\mathbf{T}_{1}(x)\Big{)}W(x)\\ &-\lambda^{-1}\Big{(}\mathbf{I}+\lambda^{-1}\mathbf{T}_{1}(x) \Big{)}\frac{d\mathbf{T}_{1}(x)}{dx}W(x).\end{split}  \tag{95}\]  
If we assume that \(\mathbf{T}_{1}(x)\) and its first derivative exist and are bounded as functions of \(x\in[-L,0]\), then for large \(\,\mid\!\lambda\!\mid\,\), the matrix of the system in Equation (95) has the following asymptotical form:  
\[ \begin{split}&\Big{(}\mathbf{I}-\lambda^{-1}\mathbf{T}_{1}(x)+ O\big{(}\lambda^{-2}\big{)}\Big{)}\bigg{(}\lambda\Lambda_{0}(x)-\mathbf{T}_{0}^{-1}(x) \frac{d\mathbf{T}_{0}(x)}{dx}\bigg{)}\Big{(}\mathbf{I}+\lambda^{-1}\mathbf{T }_{1}(x)\Big{)}+O\big{(}\lambda^{-1}\big{)}\\ =&\lambda\Lambda_{0}(x)+\Big{[}\Lambda_{0}(x), \mathbf{T}_{1}(x)\Big{]}-\mathbf{T}_{0}^{-1}(x)\frac{d\mathbf{T}_{0}(x)}{dx}+ O\big{(}\lambda^{-1}\big{)},\end{split}  \tag{96}\]  
where \(\Big{[}\Lambda_{0}(x),\mathbf{T}_{1}(x)\Big{]}\) is the commutator of \(\Lambda_{0}(x)\) and \(\mathbf{T}_{1}(x)\). Now, we are in a position to choose the matrix \(\mathbf{T}_{1}(x)\). We do this in such a way that the matrix \(\mathcal{B}(x)\) defined by  
\[\mathcal{B}(x)=\Big{[}\Lambda_{0}(x),\mathbf{T}_{1}(x)\Big{]}-\mathbf{T}_{0}^{ -1}(x)\frac{d\mathbf{T}_{0}(x)}{dx} \tag{97}\]  
is a diagonal matrix.