## 7 The Harvester Model with Constant Physical Parameters. The Left Reflection Matrix
 \tag{171}\]At this moment, it is convenient to introduce new notations:  
\[A=\frac{\rho}{2K},\qquad B=\frac{S\rho}{2 EI},\qquad C=\frac{\rho}{EI}, \tag{172}\]  
and represent the roots of Equation (171) in the form  
\[\Big{(}z^{2}\Big{)}_{1,2}=-\lambda^{2}\big{(}A+B\big{)}\pm\sqrt{\lambda^{4}(A- B)^{2}+\lambda^{2}C}. \tag{173}\]  
For definiteness, we assume \(A>B\). Then, the following asymptotic approximations are valid for the roots of Equation (171):  
\[ \begin{split}\Big{(}z^{2}\Big{)}_{1,2}&=-\lambda^ {2}\big{(}A+B\big{)}\pm\lambda^{2}(A-B)\bigg{(}1+\frac{C}{\lambda^{2}(A-B)^{2} }\bigg{)}^{1/2}\\ &=-\lambda^{2}\big{(}A+B\big{)}\pm\lambda^{2}(A-B)\bigg{(}1+\frac {C}{2\lambda^{2}(A-B)^{2}}+O\bigg{(}\frac{1}{\lambda^{4}}\bigg{)}\bigg{)}. \end{split}  \tag{174}\]  
Rewriting Equation (174) in terms of the physical parameters of the structure, we get  
\[\Big{(}z^{2}\Big{)}_{1}=-\lambda^{2}\ \ \frac{\rho}{K}-\frac{1}{EI/K-S}+O \bigg{(}\frac{1}{\lambda^{2}}\bigg{)}, \tag{175}\]  
\[\Big{(}z^{2}\Big{)}_{2}=-\lambda^{2}\ \ \frac{S\rho}{EI}+\frac{1}{EI/K-S}+O \bigg{(}\frac{1}{\lambda^{2}}\bigg{)}. \tag{176}\]  
Let \(\Lambda(\lambda)\) and \(\widetilde{\Lambda}(\lambda)\) be defined by  
\[\Lambda(\lambda) =-\lambda^{2}\big{(}A+B\big{)}-\sqrt{\lambda^{4}(A-B)^{2}+ \lambda^{2}C}, \tag{177}\] \[\widetilde{\Lambda}(\lambda) =-\lambda^{2}\big{(}A+B\big{)}+\sqrt{\lambda^{4}(A-B)^{2}+ \lambda^{2}C}. \tag{178}\]  
The general solution of Equation (169) can be given as the following linear combination of the exponentials:  
\[ \begin{split}\Omega(\lambda,x)&=C_{1}(\lambda) \exp\bigg{(}\sqrt{\Lambda(\lambda)}x\bigg{)}+C_{2}(\lambda)\exp\bigg{(}\sqrt{ \widetilde{\Lambda}(\lambda)}x\bigg{)}\\ &+D_{1}(\lambda)\exp\bigg{(}-\sqrt{\Lambda(\lambda)}x\bigg{)}+D _{2}(\lambda)\exp\bigg{(}-\sqrt{\widetilde{\Lambda}(\lambda)}x\bigg{)},\end{split}  \tag{179}\]  
where \(C_{j}(\lambda)\) and \(D_{j}(\lambda)\), \(j=1,2\), are arbitrary functions of \(\lambda\). Now, we establish two constraints on the coefficients in order to satisfy the left-end boundary conditions in Equation (163). First, we rewrite these conditions \(\Big{(}\varphi(\lambda,0)=w(\lambda,0)=0\Big{)}\) in terms of \(\varphi\). To this end, we use Equation (166) and have  
\[w(\lambda,x)=\frac{K}{\rho\lambda^{2}}\Big{(}\varphi^{\prime}(\lambda,x)-w^{ \prime\prime}(\lambda,x)\Big{)}\]  
and also from Equation (167) we have  
\[\varphi(\lambda,x)-w^{\prime}(\lambda,x)=\frac{S\rho}{K}\lambda^{2}\varphi( \lambda,x)+\frac{EI}{K}\varphi^{\prime\prime}(\lambda,x).\]Differentiating this equation once and substituting the result for \(\Big{(}\varphi^{\prime}-w^{\prime\prime}\Big{)}\) into the formula for \(w\), we obtain  
\[w(\lambda,x)=\frac{K}{\rho\lambda^{2}}\bigg{(}\frac{S\rho}{K}\lambda^{2}\varphi ^{\prime}(\lambda,x)+\frac{EI}{K}\varphi^{\prime\prime\prime\prime}(\lambda,x) \bigg{)}=S\varphi^{\prime}(\lambda,x)+\frac{EI}{\lambda^{2}\rho}\varphi^{ \prime\prime\prime\prime}(\lambda,x). \tag{180}\]  
Therefore, the second boundary condition at the left end becomes:  
\[\lambda^{2}S\rho\varphi^{\prime}(\lambda,0)+EI\varphi^{\prime\prime\prime}( \lambda,0)=0. \tag{181}\]  
Now, use the general solution of Equation (169) in the form of Equation (179) and obtain that, for the solution to satisfy the left-end boundary conditions, the coefficients must satisfy the following linear system:  
\[ \begin{cases}C_{1}(\lambda)+D_{1}(\lambda)+C_{2}(\lambda)+D_{2}(\lambda)=0.\\ \lambda^{2}S\rho\Big{(}\sqrt{\Lambda(\lambda)}\;C_{1}(\lambda)-\sqrt{\Lambda( \lambda)}\;D_{1}(\lambda)+\sqrt{\widetilde{\Lambda}(\lambda)}\;C_{2}(\lambda) -\sqrt{\widetilde{\Lambda}(\lambda)}\;D_{1}(\lambda)\Big{)}\\ +EI\Big{(}\Lambda^{3/2}(\lambda)\;C_{1}(\lambda)-\Lambda^{3/2}(\lambda)\;D_{1 }(\lambda)+\widetilde{\Lambda}^{3/2}(\lambda)\;C_{2}(\lambda)-\widetilde{ \Lambda}^{3/2}(\lambda)\;D_{2}(\lambda)\Big{)}=0.\end{cases}  \tag{182}\]  
The second equation can be modified as  
\[