## 2 Main Results
### Controller Design and Stability Analysis
 \tag{37}\]By applying [PERSON]'s inequality, the following inequalities hold:  
\[-\frac{k_{2}z_{2}\zeta}{k_{a2}^{2}-z_{2}^{2}}g\rho \leq\frac{k_{2}}{k_{a2}^{2}-z_{2}^{2}}|z_{2}||\zeta||gp| \tag{38}\] \[\leq-\frac{k_{2}z_{2}^{2}G}{2(k_{a2}^{2}-z_{2}^{2})}-\frac{k_{2} \zeta^{2}G^{2}}{2(k_{a2}^{2}-z_{2}^{2})\bar{G}}\] \[\Delta u\zeta \leq\frac{1}{2k_{3}}\Delta u^{2}+\frac{k_{3}}{2}\zeta^{2}\]  
From (37) and (38), the time derivative of \(V\) is expressed as  
\[\dot{V} \leq-\frac{k_{1}z_{1}^{2}}{k_{a1}^{2}-z_{1}^{2}}+\frac{k_{2}z_{2} ^{2}}{k_{a2}^{2}-z_{2}^{2}}\bar{G}-\frac{k_{2}z_{2}^{2}}{2(k_{a2}^{2}-z_{2}^{2 })}\bar{G}-\frac{k_{2}\zeta^{2}\bar{G}^{2}}{2(k_{a2}^{2}-z_{2}^{2})\bar{G}}- \frac{\hat{\theta}z_{2}^{2}}{2a^{2}(k_{a2}^{2}-z_{2}^{2})^{2}}\Psi^{T}\Psi\bar {G} \tag{39}\] \[-\frac{z_{2}}{k_{a2}^{2}-z_{2}^{2}}\delta tanh(\frac{z_{2}}{ \lambda(k_{a2}^{2}-z_{2}^{2})})\bar{G}+\frac{z_{2}^{2}}{2a^{2}(k_{a2}^{2}-z_{2 }^{2})^{2}}\bar{G}\theta\Psi^{T}\Psi+\frac{|z_{2}|}{k_{a2}^{2}-z_{2}^{2}}\bar{ G}\delta+\frac{a^{2}}{2}\] \[+\bar{G}\delta\sigma_{2}\delta-\bar{G}\frac{z_{2}}{k_{a2}^{2}-z_{ 2}^{2}}tanh(\frac{z_{2}}{\lambda(k_{a2}^{2}-z_{2}^{2})})+\bar{G}\delta\sigma_{ 1}\hat{\theta}-\bar{G}\frac{z_{2}^{2}}{2a^{2}(k_{a2}^{2}-z_{2}^{2})^{2}}\Psi^ {T}\Psi\] \[-k_{3}\zeta^{2}-\frac{k_{4}}{k_{a2}^{2}-z_{2}^{2}}\zeta^{2}+\frac {1}{2k_{3}}\Delta u^{2}+\frac{k_{3}}{2}\zeta^{2}\] \[=-\frac{k_{1}z_{1}^{2}}{k_{a1}^{2}-z_{1}^{2}}+\frac{k_{2}z_{2}^{2 }}{(k_{a2}^{2}-z_{2}^{2})}\bar{G}-\frac{\zeta^{2}}{k_{a2}^{2}-z_{2}^{2}}(k_{4} +\frac{k_{2}\bar{G}^{2}}{2\bar{G}})+\bar{G}\delta\sigma_{2}\hat{\delta}+\bar{ G}\delta\sigma_{1}\hat{\theta}+\frac{a^{2}}{2}\] \[-\frac{k_{3}}{2}\zeta^{2}+\frac{1}{2k_{3}}\Delta u^{2}+\bar{G} \delta\left[\frac{|z_{2}|}{k_{a2}^{2}-z_{2}^{2}}-\frac{z_{2}}{k_{a2}^{2}-z_{ 2}^{2}}tanh(\frac{z_{2}}{\lambda(k_{a2}^{2}-z_{2}^{2})})\right]\] \[\leq-\frac{k_{1}z_{1}^{2}}{k_{a1}^{2}-z_{1}^{2}}-\frac{k_{2}z_{2} ^{2}}{2(k_{a2}^{2}-z_{2}^{2})}(-\bar{G})-\frac{k_{3}}{2}\zeta^{2}+\bar{G} \delta\sigma_{2}\hat{\theta}+\bar{G}\theta\sigma_{1}\hat{\theta}+\frac{a^{2}} {2}+\frac{\Delta u^{2}}{2k_{3}}+0.2785\lambda\bar{G}\delta\]  
Since \(\bar{\delta}=\delta-\hat{\delta}\) and \(\bar{\theta}=\theta-\hat{\theta}\), the following inequalities hold:  
\[ \begin{split}\bar{\delta}\hat{\delta}&=\bar{\delta} (\delta-\bar{\delta})\leq-\frac{1}{2}\bar{\delta}^{2}+\frac{1}{2}\delta^{2}\\ \bar{\theta}\hat{\theta}&=\bar{\theta}(\theta-\bar{ \theta})\leq-\frac{1}{2}\bar{\theta}^{2}+\frac{1}{2}\theta^{2}\end{split}  \tag{40}\]  
Considering (40) and Lemma 1, we have  
\[ \begin{split}\dot{V}&\leq-\frac{k_{1}z_{1}^{2}}{k_{a1 }^{2}-z_{1}^{2}}-\frac{k_{2}z_{2}^{2}}{2(k_{a2}^{2}-z_{2}^{2})}(-\bar{G})-\frac{ \bar{G}\sigma_{2}\bar{\delta}^{2}}{2}-\frac{\bar{G}\sigma_{1}\bar{\theta}^{2}}{ 2}-\frac{k_{3}}{2}\zeta^{2}\\ &+\frac{a^{2}}{2}+\frac{\Delta u^{2}}{2k_{3}}+0.2785\lambda\bar{G} \delta+\frac{\bar{G}\sigma_{2}}{2}\delta^{2}+\frac{\bar{G}\sigma_{1}}{2}\theta ^{2}\\ &\leq-k_{1}\ln\frac{k_{a1}^{2}}{k_{a1}^{2}-z_{1}^{2}}-\frac{-k_{2} \bar{G}}{2}\frac{k_{a2}^{2}}{k_{a2}^{2}-z_{2}^{2}}-\frac{\bar{G}\sigma_{2}\bar{ \delta}^{2}}{2}-\frac{\bar{G}\sigma_{1}\bar{\theta}^{2}}{2}-\frac{k_{3}}{2} \zeta^{2}\\ &+\frac{a^{2}}{2}+\frac{\Delta u^{2}}{2k_{3}}+0.2785\lambda\bar{G} \delta+\frac{\bar{G}\sigma_{2}}{2}\delta^{2}+\frac{\bar{G}\sigma_{1}}{2}\theta ^{2}\end{split}  \tag{41}\]  
It should be noted that the state \(\zeta\) of the anti-saturation compensator is affected by the value of \(|\Delta u|\). If \(|\Delta u|\) tends to infinity, the control input will be infinite when the system (5) tracks the reference trajectory which will cause the saturation compensator to cease to be effective; the system will not track the desired value. Based on this analysis, \(|\Delta u|\) should be bounded. **Remark 4**. : _In this paper, an anti-saturation compensator (33) is used to avoid the problem of input saturation. When the control input required by the system is greater than the actuator performance, \(\Delta u\
eq 0\).