## 3 Algorithm Description
\[S_{(\zeta at_{\rm s})_{v_{2}}}(t)=\frac{n^{t-\zeta_{2}}}{\Big{|}n^{t+\frac{t_{ \rm v}}{t}-1}-n^{a}\Big{|}}+\frac{n^{t-\zeta_{v_{2}}}_{v}}{\Big{|}n^{t_{\rm v}+ \frac{t_{\rm v}}{t}-1}-n^{a_{\rm v}}_{\rm v}\Big{|}}+q_{0} \tag{13}\]  
Now, taking any value of \(S_{v_{1}}\) such that \(S_{v_{1}}<S_{f}\) and \(q_{0}>S_{v_{1}}\), then \(\zeta_{1}=t_{\rm s}-log_{n}\Big{[}\Big{(}\Big{|}S_{f}-S_{v_{1}}\Big{|}\Big{)} \big{(}\big{|}n^{t_{\rm s}}-n^{a}\big{|}\big{)}\Big{]}\) and \(\zeta_{v_{1}}=t_{\rm v}-log_{n_{\rm v}}\Big{[}\big{(}|q_{0}-S_{v_{1}}|\big{)} \Big{(}\Big{|}n^{t_{\rm v}}_{\rm v}-n^{a}_{\rm v}\Big{|}\Big{)}\Big{]}\), it obtains the following trajectory function (14) at \(i=1\):  
\[S_{(\zeta at_{\rm s})_{v_{1}}}(t)=\frac{n^{t-\zeta_{1}}}{\Big{|}n^{t+\frac{t_{ \rm v}}{t}-1}-n^{a}\Big{|}}-\frac{n^{t-\zeta_{v_{1}}}_{v}}{\Big{|}n^{t+\frac{t_ {\rm v}}{t}-1}-n^{a_{\rm v}}_{\rm v}\Big{|}}+q_{0}. \tag{14}\]  
Additionally, making:  
\[-\frac{n^{t-\zeta_{v_{3}}}_{v}}{\Big{|}n^{t_{\rm v}}_{\rm v}-n^{a_{\rm v}}_{ \rm v}\Big{|}}+q_{0}=S_{v_{3}}=\theta_{v}-\frac{n^{t_{\rm v}-\zeta_{1}}}{\Big{|} n^{t_{\rm v}+\frac{t_{\rm v}}{t_{\rm v}}-1}-n^{a}\Big{|}} \tag{15}\]  
Using \(S_{v_{3}}\) in (15) to calculate \(\zeta_{3}\) and \(\zeta_{v_{3}}\) using formula (10), then, the following trajectory function (16) at \(i=3\) is  
\[S_{(\zeta at_{\rm s})_{v_{3}}}(t)=\frac{n^{t-\zeta_{3}}}{\Big{|}n^{t+\frac{t_{ \rm v}}{t}-1}-n^{a}\Big{|}}-\frac{n^{t-\zeta_{v_{3}}}_{v}}{\Big{|}n^{t_{\rm v}+ \frac{t_{\rm v}}{t}-1}-n^{a_{\rm v}}_{\rm v}\Big{|}}+q_{0}. \tag{16}\]  
Then, for the 4 th iteration (\(i=4\)), the following sequence (17) is obtained:  
\[\frac{n^{t_{\rm v}-\zeta_{v_{4}}}_{v}}{\Big{|}n^{t_{\rm v}}_{\rm v}-n^{a_{\rm v }}_{\rm v}\Big{|}}+q_{0}=S_{v_{4}}=\theta_{v}-\frac{n^{t_{\rm v}-\zeta_{2}}}{ \Big{|}n^{t_{\rm v}+\frac{t_{\rm v}}{t}-1}-n^{a}\Big{|}}\;. \tag{17}\]  
With \(\zeta_{4}=t_{\rm s}-log_{n}\Big{[}\Big{(}\Big{|}S_{f}-S_{v_{4}}\Big{|}\Big{)} \big{(}\big{|}n^{t_{\rm s}}-n^{a}\big{|}\big{)}\Big{]}\) and \(\zeta_{v_{4}}=t_{\rm v}-log_{n_{\rm v}}\Big{[}\big{(}|S_{v_{4}}-q_{0}|\big{)} \Big{(}\Big{|}n^{t_{\rm v}}_{\rm v}-n^{a}_{\rm v}\big{|}\Big{)}\Big{]}\) by using formula (10). Additionally, for the 5 th iteration, then:  
\[-\frac{n_{v}^{t_{v}-\zeta_{v5}}}{\left|n_{v}^{t_{v}}-n_{v}^{a_{v}}\right|}+q_{0}= S_{v5}=\theta_{v}-\frac{n^{t_{v}-\zeta_{3}}}{\left|n^{t_{v}+\frac{t_{v}}{t_{v}}-1}-n^{ a}\right|}\;. \tag{18}\]  
With \(\zeta_{5}=t_{s}-log_{n}\Big{[}\Big{(}\left|S_{f}-S_{v5}\right|\Big{)}\big{(} \left|n^{t_{s}}-n^{a}\right|\big{)}\Big{]}\) and \(\zeta_{v5}=t_{v}-log_{n_{v}}\Big{[}\big{(}\left|q_{0}-S_{v5}\right|\big{)} \Big{(}\left|n^{t_{v}}_{v}-n^{a}_{v}\right|\Big{)}\Big{]}\) by using formula (10). Repeating this process \(k\) times, it has:  
\[\frac{n_{v}^{t_{v}-\zeta_{v5}}}{\left|n_{v}^{t_{v}}-n_{v}^{a_{v}}\right|}+q_{0 }=S_{v5}=\theta_{v}-\frac{n^{t_{v}-\zeta_{k-2}}}{\left|n^{t_{v}+\frac{t_{v}}{ t_{v}}-1}-n^{a}\right|}\;. \tag{19}\]  
Using formula (10), \(\zeta_{k}=t_{s}-log_{n}\Big{[}\Big{(}\left|S_{f}-S_{v5}\right|\Big{)}\big{(} \left|n^{t_{s}}-n^{a}\right|\big{)}\Big{]}\) and (20) is obtained representing the trajectory in in the \(k\)-iteration:  
\[S_{(\zeta_{at_{s}})_{v_{k}}}(t)=\frac{n^{t-\zeta_{k}}}{\left|n^{t+\frac{t}{t}- 1}-n^{a}\right|}+\frac{n_{v}^{t-\zeta_{v5}}}{\left|n_{v}^{t_{v}+\frac{t_{v}}{ t_{v}}-1}-n^{a_{v}}\right|}+q_{0}, \tag{20}\]  
Such that:  
\[ \begin{split} S_{(\zeta_{at_{s}})_{v_{k}}}(t_{v})&= \frac{n^{t_{v}-\zeta_{k}}}{\left|n^{t_{v}+\frac{t_{v}}{t_{v}}-1}-n^{a}\right|} +\frac{n_{v}^{t_{v}-\zeta_{k}}}{\left|n_{v}^{t_{v}-\frac{t_{v}}{t_{v}}-1}-n^{ a}\right|}+q_{0}=\frac{n^{t_{v}-\zeta_{k}}}{\left|n^{t_{v}+\frac{t_{v}}{t_{v}}-1}-n^{ a}\right|}+\big{(}S_{v5}\big{)}\\ &=\frac{n^{t_{v}-\zeta_{k}}}{\left|n^{t_{v}+\frac{t_{v}}{t_{v}}-1} -n^{a}\right|}+\left(\theta_{v}-\frac{n^{t_{v}-\zeta_{k-2}}}{\left|n^{t_{v}+ \frac{t_{v}}{t_{v}}-1}-n^{a}\right|}\right),\end{split}