## 4 Stability Analysis  
**Theorem 1**. : _The errors of the system (11) with the control law (36) are uniformly ultimately bounded, if the parameter \(k<1\) and the time constant of the filter \(T<2\)._  
Proof of Theorem 1.: Assume there exists a positive definite matrix \(\mathbf{P}\) such that \(A_{2}^{\mathsf{T}}\mathbf{P}+\mathbf{P}A_{2}=-I\). Construct a Lyapunov function  
\[V=\frac{1}{2}\epsilon_{1}^{2}+\frac{1}{2}s^{2}+\frac{1}{2}\epsilon_{2d}^{2}+ \tilde{\mathbf{e}}^{\mathsf{T}}\mathbf{P}\tilde{\mathbf{e}}. \tag{37}\]  
Substituting (36) into (35) yields  
\[\dot{V}=e_{1}e_{2}-k_{1}\epsilon_{1}^{2}+s\Big{(}\dot{f}-f-ks-D^{1-\beta} \epsilon sgn(s)\Big{)}+\tilde{x}_{2d}\dot{\xi}_{2d}+2\tilde{\mathbf{e}}^{\mathsf{ T}}\mathbf{P}\tilde{\mathbf{e}}. \tag{38}\]  
A proper selection of the gain \(\mathbf{L}\) ensures that the error dynamics of the LESO is stable. Thus, it is reasonable to assume that \(\dot{f}\approx f_{total}\). Then, rewrite (38) as  
\[\dot{V} =e_{1}e_{2}-k_{1}\epsilon_{1}^{2}-s\Big{(}ks+D^{1-\beta}\epsilon sgn (s)\Big{)}+\tilde{x}_{2d}\dot{\xi}_{2d}+2\tilde{\mathbf{e}}^{\mathsf{T}}\mathbf{P} \tilde{\mathbf{e}} \tag{39}\] \[=e_{1}e_{2}-k_{1}\epsilon_{1}^{2}-ks^{2}-sD^{1-\beta}\epsilon sgn (s)+\tilde{x}_{2d}\dot{\xi}_{2d}+2\tilde{\mathbf{e}}^{\mathsf{T}}\mathbf{P}\tilde\mathbf{{e}}\] \[=e_{1}e_{2}-k_{1}\epsilon_{1}^{2}+(1-k)s^{2}-s^{2}+\dot{s}\dot{s}+ \tilde{x}_{2d}\dot{\xi}_{2d}+2\tilde{\mathbf{e}}^{\mathsf{T}}\mathbf{P}\tilde{\mathbf{e}}.\]  
Let \(\mathbf{e}_{12}= \begin{bmatrix}e_{1}&e_{2}&D^{\alpha}e_{2}\end{bmatrix} ^{\mathsf{T}}\), and choose a symmetric matrix \(\mathbf{Q}\)  
\[\mathbf{Q}= \begin{bmatrix}k_{1}+(k-1)\lambda_{1}^{2}&-0.5&(k-1)\lambda_{1}\\ -0.5&0&0\\ (k-1)\lambda_{1}&0&k-1\end{bmatrix}  \tag{40}\]  
to make  
\[\mathbf{e}_{12}^{\mathsf{T}}\mathbf{Q}\mathbf{e}_{12} =-e_{1}e_{2}+k_{1}\epsilon_{1}^{2}-(1-k)(\lambda_{1}\epsilon_{1 }+D^{\alpha}\epsilon_{2})^{2} \tag{41}\] \[=-e_{1}e_{2}+k_{1}\epsilon_{1}^{2}-(1-k)s^{2}\]hold. Substituting (41) into (39) yields  
\[\dot{V}=-e_{12}^{\rm T}\mathbf{Q}e_{12}-s^{2}+s\dot{s}+\tilde{x}_{2d}\dot{x}_{2d}+2 \tilde{\mathbf{e}}^{\rm T}\mathbf{P}\dot{\mathbf{e}}. \tag{42}\]  
Calculating the determinant of \(\mathbf{Q}\) gives  
\[|\mathbf{Q}|=-0.25(k-1). \tag{43}\]  
It can be seen that \(k\) should satisfy \(k<1\) to guarantee that \(|\mathbf{Q}|>0\) holds. Thus \(-e_{12}^{\rm T}\mathbf{Q}e_{12}<0\) holds and there exists a positive constant \(\mu\) such that the inequality  
\[\dot{V}\leq-\mu e_{1}^{2}-s^{2}+s\dot{s}+\tilde{x}_{2d}\dot{\tilde{x}}_{2d}+2 \tilde{\mathbf{e}}^{\rm T}\mathbf{P}\dot{\mathbf{e}} \tag{44}\]  
is satisfied. Assume \(\dot{s}\) is bounded and \(|\dot{s}|\leq S_{M}\), where \(S_{M}\) is a positive constant. According to [PERSON]'s inequality, it is easy to obtain  
\[ \begin{split}\tilde{x}_{2d}\dot{\tilde{x}}_{2d}&= -\frac{1}{T}\tilde{x}_{2d}^{2}+\tilde{x}_{2d}B\\ &\leq-\frac{1}{T}\tilde{x}_{2d}^{2}+|\tilde{x}_{2d}B|\\ &\leq-\frac{1}{T}\tilde{x}_{2d}^{2}+|\tilde{x}_{2d}|B_{M}\\ &\leq-\frac{1}{T}\tilde{x}_{2d}^{2}+\frac{1}{2}\tilde{x}_{2d}^{2} +\frac{1}{2}B_{M}^{2}\end{split}  \tag{45}\]  
and  
\[ \begin{split} s\dot{s}&\leq\frac{1}{2}s^{2}+\frac{1 }{2}s^{2}\\ &\leq\frac{1}{2}s^{2}+\frac{1}{2}S_{M}^{2}.\end{split}  \tag{46}\]  
According to (16), we have  
\[ \begin{split} 2\tilde{\mathbf{e}}^{\rm T}\mathbf{P}\dot{\mathbf{e}}& =2\tilde{\mathbf{e}}^{\rm T}\mathbf{P}(A_{2}\tilde{\mathbf{e}}+B_{2}h)\\ &=-\tilde{\mathbf{e}}^{\rm T}\tilde{\mathbf{e}}+2\tilde{\mathbf{e}}^{\rm T} \mathbf{P}B_{2}h\\ &\leq-\tilde{\mathbf{e}}^{\rm T}\tilde{\mathbf{e}}+2M_{1}M_{2}||\mathbf{P} \mathbf{B}_{2}||.\end{split}  \tag{47}\]  
Substituting (45)-(47) into (44) yields  
\[\dot{V}\leq-\mu e_{1}^{2}-\frac{1}{2}s^{2}-\left(\frac{1}{T}-\frac{1}{2} \right)\tilde{x}_{2d}^{2}-\tilde{\mathbf{e}}^{\rm T}\tilde{\mathbf{e}}+\frac{1}{2}S_{M }^{2}+\frac{1}{2}B_{M}^{2}+2M_{1}M_{2}||\mathbf{P}\mathbf{B}_{2}||. \tag{48}\]  
Assume \(\frac{1}{T}-\frac{1}{2}>0\).