## 4 Stability Analysis
Then, substituting (41) and (42) into (40) yields  
\[\dot{V}_{i,j}\leq-\lambda_{i,j}c_{1,j}\xi_{i,j}^{2}+c_{3,j}\big{|}\xi_{i,j} \big{|} \tag{43}\]  
From (43), it follows that \(\dot{V}_{i,j}\) is negative when \(\big{|}\xi_{i,j}\big{|}\leq c_{3,j}/\lambda_{i,j}c_{1,j}\) and subsequently that  
\[\big{|}\xi_{i,j}\big{|}\leq\xi_{i,j}^{*}=\frac{c_{3,j}}{\lambda_{i,j}c_{1,j}} \tag{44}\]  
which implies  
\[\big{|}e_{i,j}(t)\big{|}\leq c_{4,j}=1-\frac{1}{\xi_{i,j}^{*2}}<1 \tag{45}\]As a result, the control signal \(\alpha_{i,j}\) is bounded. Moreover, we also can conclude the boundedness of \(x_{i,j+1}\) by noting (24). Finally, the time derivative of \(\alpha_{i,j}\) is  
\[\dot{\alpha}_{i,j}=-\lambda_{i,j}\dot{\xi}_{i,j} \tag{46}\]  
where  
\[ \begin{array}{l}\left|\dot{\xi}_{i,j}\right|\leq\frac{(1+\dot{c}_{i,j}^{2})}{ k_{i}(1-\dot{c}_{i,j}^{2})^{2}}(\left|f_{i,j}(\vec{x}_{i,j})\right|+\left|g_{i,j}( \vec{x}_{i,j})(k_{i}e_{i,j+1}+\sigma(t)x_{i,j+1}^{0}+\alpha_{i,j})\right|\\ \left.+\left|\dot{\sigma}(t)x_{i,j}^{0}\right|+\left|\dot{k}_{i}e_{i,j}\right| +b_{i,j}(\vec{x}_{i,j}))\right.\end{array}  \tag{47}\]  
Noting (45) and using the same analysis as (42), it also easy to conclude the boundedness of \(\dot{\xi}_{i,j}\) and hence \(\dot{\alpha}_{i,j}\). Step \(n\): Consider the following Lyapunov functions  
\[V_{i,n}=\frac{1}{2}\log\frac{1}{1-e_{i,n}^{2}} \tag{48}\]  
Let \(\xi_{i,n}=\frac{1}{1-e_{i,n}^{2}}\). Similar as the former steps, we can have  
\[\dot{V}_{i,n}\leq-\lambda_{i,n}c_{1,n}\xi_{i,n}^{2}+c_{3,n}|\xi_{i,n}| \tag{49}\]  
where \(c_{1,n}\) and \(c_{3,n}\) are some unknown positive constants. It follows from (49) that \(\dot{V}_{i,n}\) is negative when \(|\xi_{i,n}|\leq c_{3,n}/\lambda_{i,n}c_{1,n}\) and subsequently that  
\[|\xi_{i,n}|\leq\dot{\xi}_{i,n}^{*}=\frac{c_{3,n}}{\lambda_{i,n}c_{1,n}} \tag{50}\]  
which implies  
\[|e_{i,n}(t)|\leq c_{4,n}=1-\frac{1}{\dot{\xi}_{i,n}^{*2}}<1 \tag{51}\]  
As a result, the control signal \(\alpha_{i,j}\) is bounded. Moreover, we also can conclude the boundedness of \(u_{i}\). Notice that (36), (45) and (51) imply that \(e_{i}(t)\in\Omega^{\prime}_{e}\), for \(\forall t\in[0,t_{\max}),i=1,2,\ldots,N\), where the set \(\Omega^{\prime}_{e}\) is nonempty and compact, defined as  
\[\Omega^{\prime}_{e}=[-c_{4,1},c_{4,1}]\times[-c_{4,2},c_{4,2}]\cdots\times[- c_{4,n},c_{4,n}]\]  
Owing to (36), (45) and (51) it is straightforward to verify that \(\Omega^{\prime}_{e}\subset\Omega_{e}\). Therefore, assuming \(t_{\max}<+\infty\) dictates the existence of a time instant \(t^{\prime}\in[0,t_{\max})\), such that \(e_{i}(t^{\prime})\
otin\Omega^{\prime}_{e}\), which is a clear contradiction. Therefore, \(t_{\max}=+\infty\). Hence, all closed-loop signals remain bounded and moreover \(e_{i}(t)\in\Omega^{\prime}_{e}\subset\Omega_{e},for\ \forall t\geq 0\). Furthermore, from (36) we conclude that  
\[|e_{i,1}(t)|\leq c_{4,1}<1 \tag{52}\]  
Then, for all \(t\geq 0\). In view of Lemma 2 and (52), we have  
\[ \begin{array}{l}\lim_{t\rightarrow+\infty}|y_{i}-y_{d}|=\lim_{t \rightarrow+\infty}|q_{i,1}-y_{d}+y_{i}-q_{i,1}|\\ \leq\lim_{t\rightarrow+\infty}|q_{i,1}-y_{d}|+\lim_{t\rightarrow+\infty}|y_{ i}-q_{i,1}|\\ \leq\lim_{t\rightarrow+\infty}|k_{i}e_{i,1}(t)|\\ \leq k_{i}\end{array}  \tag{53}\]  
This completes the proof. \[f_{2}(X_{i})=\left[ \begin{array}{c}l_{\beta}\beta_{i}+l_{p}p_{i}+l_{q}q_{i}+l_{ r}r_{i}+(l_{\beta\alpha}\beta_{i}+l_{ra}r_{i})\Delta\alpha_{i}-i_{1}q_{i}r_{i}\\ m_{\pm}\Delta\alpha_{i}+m_{q}q_{i}+i_{2}p_{i}r_{i}-m_{\hat{\alpha}}(\beta_{0}/ V_{i})(\cos\theta_{i}\cos\phi_{i}-\cos\theta_{0})\\ n_{\beta}\beta_{i}+n_{r}r_{i}+n_{p}p_{i}+n_{p}p_{i}\Delta\alpha_{i}-i_{3}p_{i} q_{i}+n_{q}q_{i}\end{array}