## 5 Numerical Experiments
 \tag{25}\]  
From the initial conditions (22), we have  
\[p_{i}(t_{f})=y_{0,i}^{f},\quad\dot{p}_{i}(t_{f})=\dot{y}_{0,i}^{f},\quad i=\{ 1,2\}. \tag{26}\]  
The solution of Cauchy problem (25) and (26) in reverse time has the form  
\[p_{1}(t)=\xi_{f}(t-t_{f})+x_{0,1}^{f},\quad p_{2}(t)=(x_{0,1}^{f}+x_{0,3}^{f} \xi_{f})(t-t_{f})+x_{0,2}^{f}.\]Substituting \(t=t_{0}\) in these formulas we get the initial conditions on \(p_{i}\) and \(\dot{p}_{i}\) which allows us to find the initial conditions on \(\dot{y}_{i}\) and \(\dot{\overline{y}}_{i}\), using (24) and (25). With (22) we have the full set of initial conditions for (23). Solving the Cauchy problem (23) with the initial conditions we can receive the solution of point-to-point problem in terms of flat output. Then, (19) give us the explicit representations for the coordinates of admissible trajectories and the corresponding controls. For brevity, we give the control representations only  
\[u_{0,1}(t)=-\frac{1}{T^{3}}\bigg{(}-\xi_{0}(t-t_{f})(3t-2t_{0}-t_ {f})T+(t-t_{0})(-\xi_{f}(3t-t_{0}-2t_{f})T-\] \[-6(t-t_{f})(x_{0,1}^{0}-x_{0,1}^{f})\bigg{)},\] \[u_{0,2}(t)=\frac{1}{m_{1}^{2}}\bigg{(}m_{1}\Big{(}2(-t_{f}+ \Delta t_{3})m_{2}+(8x_{0,1}^{f}+4x_{0,1}^{0})\Delta t_{3}t_{f}+\] \[+m_{7}-12 tm_{4}+2t_{f}m_{3}+(-8x_{0,1}^{f}-4x_{0,1}^{0})\Delta t_ {3}t_{f}+\xi_{0}Tm_{5}+\xi_{f}Tm_{6}\Big{)}-\] \[-2m_{8}\Big{(}\xi_{0}\Delta t_{2}(2t_{f}-\Delta t_{3})m_{9}- \Delta t_{1}(-2\Delta t_{2}m_{10}-m_{11})\Big{)}\bigg{)},\]  
where \(\xi_{0}=-8\), \(\xi_{f}=-4\) and  
\[T=t_{f}-t_{0},\quad n_{1}=3t-2t_{0}-t_{f},\quad n_{2}=2t_{0}+t_{f }+3x_{0,3}^{f},\] \[n_{3}=3t-t_{0}-2t_{f},\quad n_{4}=x_{0,1}^{0}-x_{0,1}^{f},\] \[\Delta t_{1}=t-t_{0},\quad\Delta t_{2}=t-t_{f},\quad\Delta t_{3}= t_{0}+t_{f},\] \[p_{1}(t)=\xi_{f}(t-t_{f})+x_{0,1}^{f},\quad p_{2}(t)=(x_{0,1}^{f}+x_{0,3}^{f} \xi_{f})(t-t_{f})+x_{0,2}^{f}.\]0 \[p_{1}(t)=\xi_{f}(t-t_{f})+x_{0,1}^{f},\quad p_{2}(t)=(x_{0,1}^{f}+x_{0,3}^{f} \xi_{f})(t-t_{f})+x_{0,2}^{f}.\]1 \[p_{1}(t)=\xi_{f}(t-t_{f})+x_{0,1}^{f},\quad p_{2}(t)=(x_{0,1}^{f}+x_{0,3}^{f} \xi_{f})(t-t_{f})+x_{0,2}^{f}.\]2 \[p_{1}(t)=\xi_{f}(t-t_{f})+x_{0,1}^{f},\quad p_{2}(t)=(x_{0,1}^{f}+x_{0,3}^{f} \xi_{f})(t-t_{f})+x_{0,2}^{f}.\]3 \[p_{1}(t)=\xi_{f}(t-t_{f})+x_{0,1}^{f},\quad p_{2}(t)=(x_{0,1}^{f}+x_{0,3}^{f} \xi_{f})(t-t_{f})+x_{0,2}^{f}.\]4 \[p_{1}(t)=\xi_{f}(t-t_{f})+x_{0,1}^{f},\quad p_{2}(t)=(x_{0,1}^{f}+x_{0,3}^{f} \xi_{f})(t-t_{f})+x_{0,2}^{f}.\]5 \[p_{1}(t)=\xi_{f}(t-t_{f})+x_{0,1}^{f},\quad p_{2}(t)=(x_{0,1}^{f}+x_{0,3}^{f} \xi_{f})(t-t_{f})+x_{0,2}^{f}.\]6 \[p_{1}(t)=\xi_{f}(t-t_{f})+x_{0,1}^{f},\quad p_{2}(t)=(x_{0,1}^{f}+x_{0,3}^{f} \xi_{f})(t-t_{f})+x_{0,2}^{f}.\]7 \[p_{1}(t)=\xi_{f}(t-t_{f})+x_{0,1}^{f},\quad p_{2}(t)=(x_{0,1}^{f}+x_{0,3}^{f} \xi_{f})(t-t_{f})+x_{0,2}^{f}.\]8 \[p_{1}(t)=\xi_{f}(t-t_{f})+x_{0,1}^{f},\quad p_{2}(t)=(x_{0,1}^{f}+x_{0,3}^{f} \xi_{f})(t-t_{f})+x_{0,2}^{f}.\]9 \[u_{0,1}(t)=-\frac{1}{T^{3}}\bigg{(}-\xi_{0}(t-t_{f})(3t-2t_{0}-t_ {f})T+(t-t_{0})(-\xi_{f}(3t-t_{0}-2t_{f})T-\]0 \[u_{0,1}(t)=-\frac{1}{T^{3}}\bigg{(}-\xi_{0}(t-t_{f})(3t-2t_{0}-t_ {f})T+(t-t_{0})(-\xi_{f}(3t-t_{0}-2t_{f})T-\]1 \[u_{0,1}(t)=-\frac{1}{T^{3}}\bigg{(}-\xi_{0}(t-t_{f})(3t-2t_{0}-t_ {f})T+(t-t_{0})(-\xi_{f}(3t-t_{0}-2t_{f})T-\]2  
The Figure 0(a) shows the results of numerical simulation of closed-loop systems along \(u_{0}(t)\) and \(u_{1}(t,\varepsilon)=u_{0}(t)+\varepsilon u_{1}(t)\) for \(\varepsilon=0.1\), \(k_{u}=2\). The initial approximations \(\alpha^{(0)}=1\), \(c_{i,l}^{(0)}=0\) were used for the \(u_{1}(t)\) calculation. Achievement of a given calculation accuracy for \(M^{a}\) occurs in 14 iterations and does not depend on the parameter \(\varepsilon\). The Figure 0(b) shows the trajectories of systems for \(\varepsilon=1\) along four controls, namely, \(u_{0}(t)\), \(u_{1}(t,\varepsilon)\) and \(u_{\omega}(t,\varepsilon)=u_{1}(t,\varepsilon)+\varepsilon\omega(t)\), here the correction \(\omega\) was calculated for \(k_{\omega}=1\), and \(\hat{u}(t)\).