## 4 Non-Selfadjoint Operator Pencil Generated by the Harvester System
Let us fix the choice of \(\mathbf{T}_{1}(x)\) by the following conditions:  
\[\big{(}\mathbf{T}_{1}\big{)}_{j\!j}=0\qquad\text{and}\qquad\big{(}\mathbf{T}_ {1}\big{)}_{j\!k}=\big{(}\lambda_{j}(x)-\lambda_{k}(x)\big{)}^{-1}\bigg{(} \mathbf{T}_{0}^{-1}(x)\frac{d\mathbf{T}_{0}(x)}{dx}\bigg{)}_{j\!k},\ j,k=1,2,3,4, \tag{98}\]  
where \(\big{\{}\lambda_{j}(x)\big{\}}_{j=1}^{4}\) are given in Equations (87). Note, since \(A(x)\
eq B(x)\), the matrix \(\mathbf{T}_{1}(x)\) from Equation (98) is well-defined. It can be verified by a straightforward calculation that \(\mathcal{B}(x)\), defined by Equation (97), is diagonal if \(\mathbf{T}_{1}(x)\) is defined by Equation (98). We use the notation \(\mathbf{\Lambda}_{1}(x)\) for the matrix \(\mathcal{B}(x)\) with the aforementioned choice of \(\mathbf{T}_{1}(x)\). With this choice of \(T_{1}\), Equation (95) for \(W\) can be modified to the following equation:  
\[W^{\prime}(x)=\Big{[}\lambda\mathbf{\Lambda}_{0}(x)+\mathbf{\Lambda}_{1}(x)+O\big{(} \lambda^{-1}\big{)}\Big{]}W(x),\qquad|\lambda|\to\infty, \tag{99}\]  
where the estimate \(O\big{(}\lambda^{-1}\big{)}\) is uniform with respect to \(x\in[0,L]\) and \(\mathbf{\Lambda}_{1}(x)\) is a diagonal matrix whose diagonal elements are equal to:  
\[\mu_{j}(x)=-\bigg{[}\mathbf{T}_{0}^{-1}(x)\frac{d\mathbf{T}_{0}(x)}{dx}\bigg{]} _{jj},\qquad j=1,2,3,4. \tag{100}\]  
Using standard methods of asymptotic analysis (such as in [PERSON] [36], [PERSON] [40], [PERSON] [37]), we obtain that the solutions \(W_{j}(\lambda,x)\) of the system in Equation (99) are asymptotically close to the solutions \(W_{j}^{0}(\lambda,x)\) of the following system:  
\[W^{\prime}(x)=\Big{[}\lambda\mathbf{\Lambda}_{0}(x)+\mathbf{\Lambda}_{1}(x)\Big{]}W(x). \tag{101}\]  
(Explicit formulas for \(W_{j}^{0}(\lambda,x),j=1,2,3,4\), are given in Equation (104) below.) Namely, the following relations are valid:  
\[W_{j}(\lambda,x)=W_{j}^{0}(\lambda,x)\Big{(}1+O\big{(}\lambda^{-1}\big{)} \Big{)},\qquad j=1,2,3,4,\qquad|\lambda|\to\infty, \tag{102}\]  
and the estimates \(O\big{(}\lambda^{-1}\big{)}\) are uniform with respect to \(x\in[0,L]\). To complete the proof, we have to calculate the diagonal elements of the matrix \(\mathbf{T}_{0}^{-1}(x)\frac{d\mathbf{T}_{0}(x)}{dx}\). Straightforward calculations yield the following results:  
\[\mu_{1}(x)=\mu_{3}(x)=-\frac{A^{\prime}(x)}{4A(x)},\qquad\mu_{2}(x)=\mu_{4}(x) =-\frac{B^{\prime}(x)}{4B(x)}. \tag{103}\]  
Substituting Equation (103) into the system in Equation (101) and taking into account Equation (90), we immediately obtain the following expression for the \(k\)-th component of the four-vector \(W_{j}^{0}(\lambda,x)\):  
\[\big{[}W_{j}^{0}(\lambda,x)\big{]}_{k}= \begin{cases}0,&k\
eq j,\\ \big{(}A(x)\big{)}^{-1/4}\exp\bigg{\{}\lambda\int_{0}^{x}\lambda_{j}(t)dt \bigg{\}},&k=j,j=1,3;\end{cases}  \tag{104}\]  
\[\big{[}W_{j}^{0}(\lambda,x)\big{]}_{k}= \begin{cases}0,&k\
eq j,\\ \big{(}B(x)\big{)}^{-1/4}\exp\bigg{\{}\lambda\int_{0}^{x}\lambda_{j}(t)dt \bigg{\}}\big{(}1+O\big{(}|\lambda|^{-1}\big{)}\Big{)},&k=j,j=2,4;\end{cases}  \tag{105}\]  
Using Equations (102), (104) and (105), we immediately obtain the asymptotic approximations for the solutions of Equation (99). Since \(Y_{j}(\lambda,x)=\mathbf{T}_{0}(x)Z(\lambda,x)\) and \(Z(\lambda,x)\) is given by Equation (94), we obtain  
\[Y_{j}(\lambda,x)=\mathbf{T}_{0}(x)\Big{[}\mathbb{I}+\lambda^{-1}\mathbf{T}_{1 }(x)\Big{]}W_{j}(\lambda,x),\qquad j=1,2,3,4.\]  
Taking into account the explicit Equation (89) for \(\mathbf{T}_{0}(x)\), we obtain that the first components of the vectors \(Y_{j}(\lambda,x),\;j=1,2,3,4\) are given exactly by the formulas from the second line of (105). Statement 1 of Theorem 2 is completely shown. The results of Statement 2 are almost straightforward consequences of Equations (83) and the explicit form of the matrix \(\mathbf{T}_{0}(x)\). Asymptotics of the Spectrum of the Operator \(\mathbb{L}\).