## 5 Main Results
### Illustration of Lemma 1
\right),\quad t\in[0,1], \tag{138}\]  
which, along with (73), yields  
\[\bar{z}_{2,1}(t)\equiv 0,\quad t\in[0,1]. \tag{139}\]Furthermore, due to Equations (72), (131), (134), and (135),  
\[\bar{\lambda}_{u2,1}(t)=-\frac{d\bar{z}_{2,0}(t)}{dt}-\bar{\lambda}_{v20}(t), \quad t\in[0,1]. \tag{140}\]  
To complete the construction of the functions \(z_{2}^{1}(t,\varepsilon)\) and \(\lambda_{u2}^{1}(t,\varepsilon)\) (see Equation (102)), we need to obtain \(z_{2,1}^{0}(\xi)\), \(\lambda_{u2,1}^{0}(\xi)\), \(\xi\geq 0\) and \(z_{2,1}^{f}(\eta)\), \(\lambda_{u2,1}^{f}(\eta)\), \(\eta\leq 0\). Due to the data of Example (131), the functions \(f_{x}(\xi)\) and \(f_{y}(\xi)\) (see Equation (85)) are identically zero. This circumstance, along with the equality \(z_{2,1}(0)=0\) and Equations (87)-(90), yields  
\[z_{2,1}^{0}(\xi)\equiv 0,\quad\lambda_{u2,1}^{0}(\xi)\equiv 0,\quad\xi\geq 0. \tag{141}\]  
Furthermore, using Equations (100), (131), and (140), and with the fact that \(\bar{\lambda}_{v20}(1)=0\), we obtain  
\[z_{2,1}^{f}(\eta)=-\sqrt{2}\frac{d\bar{z}_{2,0}(t_{f})}{dt}\exp\left(\frac{ \eta}{\sqrt{2}}\right),\quad\lambda_{u2,1}^{f}(\eta)=\frac{d\bar{z}_{2,0}(t_{ f})}{dt}\exp\left(\frac{\eta}{\sqrt{2}}\right),\quad\eta\leq 0. \tag{142}\]  
Thus, using the Equations (134), (135), and (137)-(142), the functions of Equation (102) become  
\[z_{1}^{1}(t,\varepsilon)=z_{1,0}(t),\quad t\in[0.1],\] \[z_{2}^{1}(t,\varepsilon)=z_{2,0}(t)+\exp\left(-\frac{t}{ \varepsilon\sqrt{2}}\right)-\varepsilon\sqrt{2}\frac{d\bar{z}_{2,0}(t_{f})}{ dt}\exp\left(\frac{t-1}{\varepsilon\sqrt{2}}\right),\quad t\in[0,1],\] \[\lambda_{u1}^{1}(t,\varepsilon)=\bar{\lambda}_{u1,0}(t),\quad t \in[0.1],\] \[\lambda_{u2}^{1}(t,\varepsilon)=\frac{1}{\sqrt{2}}\exp\left(- \frac{t}{\varepsilon\sqrt{2}}\right)\] \[+\varepsilon\bigg{(}-\frac{d\bar{z}_{2,0}(t)}{dt}-\bar{\lambda}_{ v,20}(t)+\frac{d\bar{z}_{2,0}(t_{f})}{dt}\exp\left(\frac{t-1}{\varepsilon\sqrt{2}} \right)\bigg{)},\quad t\in[0,1],\] \[\lambda_{v}^{1}(t,\varepsilon)=\left( \begin{array}{c}\lambda_ {v1}^{1}(t,\varepsilon)\\ \lambda_{v2}^{1}(t,\varepsilon)\end{array} \right)\] \[=\left( \begin{array}{c}\bar{\lambda}_{v,10}(t)\\ \bar{\lambda}_{v,20}(t)+\varepsilon\sqrt{2}\exp\left(-\frac{t}{\varepsilon \sqrt{2}}\right)\end{array} \right),\quad t\in[0,1],\] \[\lambda_{u,v}^{1}(t,\varepsilon)=\left( \begin{array}{c}\lambda_ {u,v1}^{1}(t,\varepsilon)\\ \lambda_{u,v2}^{1}(t,\varepsilon)\end{array} \right)=\left( \begin{array}{c} \bar{\lambda}_{u,v10}(t)\\ \bar{\lambda}_{u,v20}(t)\end{array} \right),\quad t\in[0,1]. \tag{143}\]  
Consider the following values:  
\[\Delta z_{1}(\varepsilon)\stackrel{{\triangle}}{{=}}\max_{t\in[0,1]}\big{|}z_{1}(t,\varepsilon)-z_{1}^{1}(t,\varepsilon)\big{|},\quad\Delta z_{2 }(\varepsilon)\stackrel{{\triangle}}{{=}}\max_{t\in[0,1]}\big{|}z_ {2}(t,\varepsilon)-z_{2}^{1}(t,\varepsilon)\big{|},\] \[\Delta\lambda_{u1}(\varepsilon)\stackrel{{\triangle}}{{= }}\max_{t\in[0,1]}\big{|}\lambda_{u1}(t,\varepsilon)-\lambda_{u1}^{1}(t, \varepsilon)\big{|},\quad\Delta\lambda_{u2}(\varepsilon)\stackrel{{ \triangle}}{{=}}\max_{t\in[0,1]}\big{|}\lambda_{u2}(t,\varepsilon)-\lambda_{ u2}^{1}(t,\varepsilon)\big{|}, \tag{144}\] \[\Delta\lambda_{v2}(\varepsilon)\stackrel{{\triangle}}{{= }}\big{|}\lambda_{v2}(t,\varepsilon)-\lambda_{v2}^{1}(t,\varepsilon)\big{|}, \quad\Delta\lambda_{u,v1}(\varepsilon)\stackrel{{\triangle}}{{=}} \big{|}\lambda_{u,v1}(t,\varepsilon)-\lambda_{u,v1}^{1}(t,\varepsilon)\big{|},\]  
where \(z_{1}(t,\varepsilon)\), \(z_{2}(t,\varepsilon)\), \(\lambda_{u1}(t,\varepsilon)\), \(\lambda_{u2}(t,\varepsilon)\), \(\lambda_{v2}(t,\varepsilon)\), \(\lambda_{u,v1}(t,\varepsilon)\) are the corresponding components of the solution to the boundary-value Problems (26) and (27) subject to Data (131). **Remark 12**. : _Note that, in (144), the components \(\lambda_{v1}(t,\varepsilon)\) and \(\lambda_{u,v2}(t,\varepsilon)\) of the solution to Problems (26) and (27) are absent.