## 3 Main Results
### Adaptive Rules for Parameters
\Big{]}\mathbf{\mathcal{T}}_{0i} \widetilde{\mathbf{X}}_{ij_{k}}^{t}\] \[=-2k_{p}^{\varphi}\big{(}v_{t}^{i}\big{)}^{2}+v_{t}^{i}\sum_{k=1} ^{L}\pi_{t}^{j_{k}}\]  
where \(k_{p}^{\varphi}=\sum_{k=1}^{L}k_{ij_{k}}^{2}\), \(\pi_{t}^{i}=-\bigg{[}k_{ij_{k}}^{1}&k_{ij_{k}}^{2}\dot{\psi}_{t}^{j_{k}}\Big{]} \mathbf{\mathcal{T}}_{0i}\widetilde{\mathbf{X}}_{ij_{k}}^{t}+cos\Big{(}\psi_{r}^{j_{k}} \Big{)}a_{t}^{j_{k}}+k_{ij_{k}}^{2}v_{t}^{j_{k}}cos\Big{(}\psi_{r}^{j_{k}}\Big{)}\). Using \(\mathbf{\mathcal{T}}_{0i}\widetilde{\mathbf{X}}_{ij_{k}}^{t}=\mathbf{T}\Big{(}\psi_{r}^{j_{ k}}\Big{)}\mathbf{\mathcal{T}}_{0i}\widetilde{\mathbf{X}}_{ij_{k}}^{t}\) and defining \(\mathbf{\mathcal{T}}_{0i}\widetilde{\mathbf{X}}_{ij_{k}}^{t}=\Big{[}x_{ij_{k}}^{t}(t),y_{ ij_{k}}^{t}(t)\Big{]}^{\mathsf{T}}\), we have  
\[\pi_{t}^{j_{k}}=-sin\Big{(}\psi_{r}^{j_{k}}\Big{)}a_{j_{k}}(t)-cos \Big{(}\psi_{r}^{j_{k}}\Big{)}b_{j_{k}}(t) \tag{30}\]where  
\[\left\{ \begin{array}{l}a_{i_{k}}(t)=k_{i_{lk}}^{1}y_{i_{lk}}^{e}(t)-k_{i_{lk}}^{2 }x_{i_{lk}}^{e}(t)\dot{\psi}_{t}^{i_{k}}\\ b_{j}(t)=-a_{i_{lk}}^{h_{k}}-k_{i_{lk}}^{2}\dot{v}_{t}^{i_{k}}\\ k_{i_{lk}}^{1}x_{i_{lk}}^{e}(t)+k_{i_{lk}}^{2}y_{i_{lk}}^{e}(t)\dot{\psi}_{t}^{ i_{k}}\end{array} \right.\]  
Firstly, we give the following in Equation (31):  
\[ \begin{array}{l}\left|-sin\left(\psi_{r}^{i_{lk}}\right)a_{j_{k}}(t)-cos \left(\psi_{r}^{i_{lk}}\right)b_{j_{k}}(t)\right|\\ \leq\sqrt{\left(a_{j_{k}}(t)\right)^{2}+\left(b_{j_{k}}(t)\right)^{2}}\end{array}  \tag{31}\]  
Assuming \(x=\left|sin\left(\psi_{r}^{i_{lk}}\right)\right|\) yields the function \(f_{x}=\left|a_{j_{k}}(t)\right|x+b_{j_{k}}(t)\sqrt{1-x^{2}},0\leq x\leq 1\). Then, taking the derivative of \(f_{x}\), we have:  
\[\dot{f}_{x}=\left|a_{j_{k}}(t)\right|-\frac{b_{j_{k}}(t)x}{\sqrt{1-x^{2}}} \tag{32}\]  
Referring to (32), \(f_{x}\) is maximum when \(x\sqrt{\left(a_{j_{k}}(t)\right)^{2}+\left(b_{j_{k}}(t)\right)^{2}}=\left|a_{ j_{k}}(t)\right|\). Defining \(\sigma_{j_{k}}=\alpha_{j_{k}}+\alpha_{j_{k}}\beta_{j_{k}}\left|\omega_{j_{k}}\right|\), we have the following inequation based on (30) and Theorem 1:  
\[ \begin{array}{l}\left|\pi_{t}^{i_{lk}}\right|\leq\sqrt{\left(a_{j_{k}}(t) \right)^{2}+\left(b_{j_{k}}(t)\right)^{2}}\\ \leq\sqrt{2\left(\sigma_{j_{k}}\right)^{2}+\left|k_{i_{lk}}^{2}v_{j_{k}}+a_{j_ {k}}\right|^{2}+2\left|k_{i_{lk}}^{2}v_{j_{k}}+a_{j_{k}}\right|\sigma_{j_{k}} }\end{array}  \tag{33}\]  
Consequently, \(\left|\pi_{t}^{i_{lk}}\right|\leq 2k_{r}^{v}v_{m}\) holds and we can obtain \(v_{t}^{i}\leq v_{m}\) with \(\pi_{t}^{i_{lk}}>0\) or \(v_{t}^{i}\geq-v_{m}\) with \(\pi_{t}^{i_{lk}}>0\). Based on the equality \(a_{i_{lk}}(t_{k+1})=\pi_{t_{k}}^{j_{lk}}-k_{i_{lk}}^{2}\dot{v}_{t_{k}}^{i}\) and the inequality \(v_{t_{k}}^{i}\geq v_{-}\geq 0\), we have \(v_{t_{k+1}}^{i}\geq-v_{m}\) with \(\left|v_{t_{k}}^{i}\right|\leq v_{m}\) and \(\pi_{t_{k}}^{j_{lk}}\leq 0\), or \(v_{t_{k+1}}^{i}\leq v_{m}\) with \(\left|v_{t_{k}}^{i}\right|\leq v_{m}\) and \(\pi_{t_{k}}^{i_{lk}}\geq 0\). Thus, the velocity constraint can be satisfied with the condition \(v_{t_{k}}^{i}\geq v_{-}\geq 0\), the challenge is how to make sure that \(\left(\widetilde{X}_{i_{lk}}^{t}\right)^{\rm T}\widetilde{X}_{i_{lk}}^{t}\) is decreasing. Given that \(v_{t_{k+1}}^{i}\) decreases if \(\widetilde{X}_{i_{lk}}^{t}\) has a positive value, there must be \(a_{i_{lk}}(t_{k+1})\leq 0\) and \(v_{t_{k+1}}^{i}=v_{-}<v_{t_{k+1}}^{i_{k}}\). The boundedness analysis of the velocity is finished. As for the angular rate, we have:  
\[ \begin{array}{l}\dot{\psi}_{t}^{i}=-\frac{\sum_{k=1}^{L}\left[-\right.\ \ k_{i_{lk}}^{2}\dot{\psi}_{t}^{i_{lk}}\left.k_{i_{lk}}^{1}\right]\mathbf{\mathcal{ F}}_{\alpha}\widetilde{X}_{i_{lk}}^{i_{lk}}}{v_{t}^{i}}\\ -\sum_{k=1}^{L}\frac{k_{i_{lk}}^{2}\dot{v}_{t}^{i_{lk}}sin\left(\dot{v}_{t}^{i_ {lk}}\right)+sin\left(\dot{v}_{t}^{i_{lk}}\right)\dot{\pi}_{t}^{i_{k}}}{v_{t} ^{i}}-\dot{\psi}_{t}^{i_{k}}\end{array}  \tag{34}\]  
Define \(\dot{\psi}_{t}^{i}=\sum_{k=1}^{L}\theta_{t}^{i_{lk}}\) with  
\[