## 4 Stability Analysis  
**Lemma 1**.: ([50])_. Consider the system \(\dot{x}=f(x)\). For the given positive scalar constants, \(c_{1}\), \(c_{2}\), \(0<\gamma<1\), and \(\delta\), if the continuous function \(V(x)\) satisfies \(\dot{V}(x)\leq-c_{1}V(x)-c_{2}V^{\gamma}(x)+\delta\), then the trajectory of the system \(\dot{x}=f(x)\) is finite time stable and convergence time is bounded as follows:_  
\[\begin{array}{c}T_{s}\leq\max[t_{0}+\frac{1}{vc_{1}(1-\gamma)}\ln\frac{vc_{1}V^{1 -\gamma}(t_{0})+c_{2}}{c_{2}},\\ t_{0}+\frac{1}{c_{1}(1-\gamma)}\ln\frac{c_{1}V^{1-\gamma}(t_{0})+vc_{2}}{vc_{2}}]. \end{array} \tag{81}\]  
**Theorem 5**.: _If the sliding mode surfaces are defined as (23), (24), (43)-(46), (66), and (67), then systems (7)-(12) are controlled by virtual controls (33), (57), (58), (74) and control inputs (34), (59), (60), (75), and the unknown disturbances are estimated by the nonlinear disturbance observer provided in (78). Furthermore, the equilibrium points of (7)-(12) are asymptotically stable, and the trajectory of the closed-loop signals is bounded in finite time as follows:_  
\[\begin{array}{c}\lim\limits_{t\to T_{\rm s}}V(\zeta)\leq\min\biggl{\{} \frac{\Delta}{(1-v)A_{\rm 1 min}},\Bigl{(}\frac{\Delta}{(1-v)A_{\rm 2 min}} \Bigr{)}^{1/2}\biggr{\}}\\ T_{\rm s}\leq\max[t_{0}+\frac{2}{vA_{\rm 1 min}}\ln\frac{vA_{\rm 1 min}V^{1/2}(t_{0})+A_{ \rm 2 min}}{A_{\rm 2 min}}\,\mu_{0}+\frac{2}{A_{\rm 1 min}}\ln\frac{A_{\rm 1 min}V^{1/2}(t_{0})+vA_{ \rm 2 min}}{vA_{\rm 2 min}}]\end{array} \tag{82}\]  
if the parameter conditions guarantee positive definiteness of \(Q_{k}\), i.e.,  
\[k_{k1}>\frac{2\epsilon_{k}^{3}}{c_{k}}+\frac{c_{k}}{8}+\epsilon_{k},k_{k2}>2 \epsilon_{k}k_{k1}. \tag{83}\]  
**Proof.** By defining the Lyapunov function as \(V=\sum\limits_{k=1}^{3}V_{k}\) using the results from (39  
given that \(V\geq\frac{\Delta}{(1-v)\Lambda_{\rm 1 min}}\), we obtain the following expression:  
\[\dot{V}\leq-v\Lambda_{\rm 1 min}V-\Lambda_{\rm 2 min}V^{1/2}. \tag{90}\]  
From Lemma 1, the boundedness of \(V(\zeta)\) decreases argument \(\zeta\)  
\[\zeta\in\bigg{\{}V(\zeta)\leq\frac{\Delta}{(1-v)\Lambda_{\rm 1 min}}\bigg{\}}. \tag{91}\]  
Thus, the convergence time \(T_{s}\) is given as follows:  
\[T_{s}\leq t_{0}+\frac{2}{v\Lambda_{\rm 1 min}}\ln\frac{v\Lambda_{\rm 1 min}V^{1/2}(t _{0})+\Lambda_{\rm 2 min}}{\Lambda_{\rm 2 min}}. \tag{92}\]  
Next, given that \(V\geq\frac{\Delta}{(1-v)\Lambda_{\rm 2 min}}\), we obtain the following expression:  
\[\dot{V}\leq-\Lambda_{\rm 1 min}V-v\Lambda_{\rm 2 min}V^{1/2}. \tag{93}\]  
This leads to the following:  
\[\zeta\in\bigg{\{}V^{1/2}(\zeta)\leq\frac{\Delta}{(1-v)\Lambda_{\rm 2 min}} \bigg{\}}, \tag{94}\]  
\[T_{s}\leq t_{0}+\frac{2}{\Lambda_{\rm 1 min}}\ln\frac{\Lambda_{\rm 1 min}V^{1/2}( t_{0})+v\Lambda_{\rm 2 min}}{v\Lambda_{\rm 2 min}}. \tag{95}\]  
Therefore, \(V(\zeta)\) is a continuously decreasing strong Lyapunov function, and the equilibrium point \(\zeta=0\) is reached in finite time from every initial condition. This also guarantees that the equilibrium point \(s=0\) is reached in finite time from every initial condition. This concludes the proof. \(\square\)  
**Theorem 6**.: _The finite time convergence of the compensating dynamics given in (25), (47), and (68) is also guaranteed by the finite time convergence of \(\zeta_{ij}\). Thus, the finite time boundedness of \(\xi_{ij}\) is demonstrated by defining the following expression:_  
\[V_{\xi}=\sum_{j=1}^{2}\frac{1}{2}\xi_{\rm jj}^{2}+\sum_{j=1}^{2}\frac{1}{2} \xi_{\rm jj}^{2}+\sum_{j=1}^{2}\frac{1}{2}\xi_{\rm 5j}^{2}+\sum_{j=1}^{2} \frac{1}{2}\xi_{\rm 5j}^{2}. \tag{96}\]  
The convergence time \(T_{\xi r}\) is then obtained as  
\[T_{\xi r}\leq\frac{4V_{\xi}^{1/4}(0)}{\lambda\Xi_{\xi}}. \tag{97}\]