## 5 Main Results
### Illustration of Lemma 1  
To illustrate this lemma, we should construct the functions in (102) subject to Data (131). By \(\bar{\lambda}_{{\rm v}1,0}(t)\) and \(\bar{\lambda}_{{\rm v}2,0}(t)\), we denote the upper and lower components of the vector \(\bar{\lambda}_{{\rm v}0}(t)\). Similarly, by \(\bar{\lambda}_{{\rm u},{\rm v}10}(t)\) and \(\bar{\lambda}_{{\rm u},{\rm v}20}(t)\), we denote the upper and lower components of the vector \(\bar{\lambda}_{{\rm u},{\rm v}0}(t)\). Thus,  
\[\bar{\lambda}_{{\rm v}0}(t)=\left( \begin{array}{c}\bar{\lambda}_{{\rm v}10} (t)\\ \bar{\lambda}_{{\rm v}20}(t)\end{array} \right)\!\!\!,\quad\bar{\lambda}_{{\rm u },{\rm v}0}(t)=\left( \begin{array}{c}\bar{\lambda}_{{\rm u},{\rm v}10}(t) \\ \bar{\lambda}_{{\rm u},{\rm v}20}(t)\end{array} \right)\!\!\!. \tag{132}\]  
Using these notations and Data (131), we can rewrite Problems (39) and (37) as \[\frac{d\bar{z}_{1,0}(t)}{dt}=-\bar{\lambda}_{\text{u},10}(t)+\bar{ \lambda}_{\text{v},20}(t),\quad z_{1,0}(0)=\sqrt{2},\] \[\frac{d\bar{\lambda}_{\text{u},10}(t)}{dt}=-2\bar{z}_{1,0}(t), \quad\bar{\lambda}_{\text{u},10}(1)=0,\] \[\frac{d\bar{\lambda}_{\text{v},10}(t)}{dt}=0,\quad\bar{\lambda}_ {\text{v},10}(1)=0,\] \[\frac{d\bar{\lambda}_{\text{v},20}(t)}{dt}=-2\bar{\lambda}_{\text {u},\text{v},20}(t),\quad\bar{\lambda}_{\text{v},20}(1)=0,\] \[\frac{d\bar{\lambda}_{\text{u},\text{v},10}(t)}{dt}=\bar{\lambda }_{\text{u},10}(t)-2\bar{\lambda}_{\text{v},10}(t)+2\bar{\lambda}_{\text{v},2 0}(t),\quad\bar{\lambda}_{\text{u},\text{v},10}(0)=0,\] \[\frac{d\bar{\lambda}_{\text{u},\text{v},20}(t)}{dt}=-\bar{ \lambda}_{\text{u},10}(t)+2\bar{\lambda}_{\text{v},10}(t)-2\bar{\lambda}_{ \text{v},20}(t),\quad\bar{\lambda}_{\text{u},\text{v},20}(0)=0. \tag{133}\]  
Problem (133) yields a unique solution in the time interval \([0,1]\)  
\[\bar{z}_{1,0}(t)=\big{(}C_{1}+C_{2}(t+\sqrt{2})\big{)}\exp(\sqrt{ 2}t)+\big{(}C_{3}+C_{4}(t-\sqrt{2})\big{)}\exp(-\sqrt{2}t),\] \[\bar{\lambda}_{\text{u},10}(t)=-\big{(}C_{1}\sqrt{2}+C_{2}(\sqrt {2}t+1)\big{)}\exp(\sqrt{2}t)+\big{(}C_{3}\sqrt{2}+C_{4}(\sqrt{2}t-1)\big{)} \exp(-\sqrt{2}t),\] \[\bar{\lambda}_{\text{v},10}(t)=0,\] \[\bar{\lambda}_{\text{u},\text{v},10}(t)=\Big{(}C_{1}+C_{2}(t+2 \sqrt{2})\big{)}\exp(\sqrt{2}t)+\big{(}C_{3}+C_{4}(t-2\sqrt{2})\big{)}\exp(- \sqrt{2}t),\] \[\frac{d\bar{\lambda}_{\text{u},10}(t)}{dt}=-2\bar{z}_{1,0}(t), \quad\bar{\lambda}_{\text{u},10}(1)=0,\]0  
where  
\[\frac{d\bar{\lambda}_{\text{u},10}(t)}{dt}=-2\bar{z}_{1,0}(t), \quad\bar{\lambda}_{\text{u},10}(1)=0,\]1  
Using Equations (38), (131), and (134), we obtain for \(t\in[0,1]\)  
\[\frac{d\bar{\lambda}_{\text{u},10}(t)}{dt}=-2\bar{z}_{1,0}(t), \quad\bar{\lambda}_{\text{u},10}(1)=0,\]2  
yielding  
\[\frac{d\bar{\lambda}_{\text{u},10}(t)}{dt}=-2\bar{z}_{1,0}(t), \quad\bar{\lambda}_{\text{u},10}(1)=0,\]3  
Using Equations (60), (65), (131), and (136), we directly obtain  
\[\frac{d\bar{\lambda}_{\text{u},10}(t)}{dt}=-2\bar{z}_{1,0}(t), \quad\bar{\lambda}_{\text{u},10}(1)=0,\]4 \[z_{1,1}^{0}(\bar{\xi})\equiv 0,\quad\lambda_{\text{u},1}^{0}( \bar{\xi})\equiv 0,\quad\lambda_{\text{u},\text{v},1}^{0}(\bar{\xi})\equiv\left( \begin{array}{cc}0\\ 0\end{array} \right),\quad\bar{\xi}\geq 0,\] \[\lambda_{\text{v},1}^{0}(\bar{\xi})=\left( \begin{array}{cc}0\\ \sqrt{2}\exp\Bigl{(}-\frac{\bar{\xi}}{\sqrt{2}}\Bigr{)}\end{array} \right), \quad\bar{\xi}\geq 0. \tag{137}\]  
Taking into account that \(z_{1,1}^{0}(\bar{\xi})\equiv 0\) and \(\xi\geq 0\), and solving the boundary-value Problems (74) and (71), we immediately have  
\[\bar{z}_{1,1}(t)\equiv 0,\quad\bar{\lambda}_{\text{u},1}(t)\equiv 0,\quad\bar{\lambda}_{\text{v},1}(t)\equiv\left( \begin{array}{cc}0\\ 0\end{array} \right),\quad\bar{\lambda}_{\text{u},\text{v},1}(t)\equiv\left( \begin{array}{cc}0\\ 0\end{array}