## 3 Global Stability Analysis
### Stability Analysis for the First Tank
\tag{29}\]  
The main properties of \(f_{v1}\) are:  
\[(Pi)f_{v1}\ is\ continuous\ with\ respect\ to\ x_{1}\, \tag{30}\]  
\[(Pii)\ f_{v1}=0\ for\ x_{1}\ \in\ \Big{[}x_{1}^{l},x_{1}^{u}\Big{]}, \tag{31}\]  
\[(Piii)\ f_{v1}\
eq 0\ for\ x_{1}\ \
otin\ \Big{[}x_{1}^{l},x_{1}^{u}\Big{]}, \tag{32}\]  
\[P(iv)\ sgn(f_{v1})=sgn(x_{1})\
eq 0\ for\ x_{1}\ \
otin\ \Big{[}x_{1}^{l},x_{1}^{u} \Big{]}, \tag{33}\]  
\[(Pv)\ sgn(f_{v1})=sgn(g_{x1})=sgn(g_{x1}-\delta_{sin})\
eq 0\ for\ x_{1}\ \
otin\ \Big{[}x_{1}^{l},x_{1}^{u}\Big{]}. \tag{34}\]  
A dead-zone Lyapunov function that satisfies \(dV_{1}/dx_{1}=f_{v1}\) (Equation (17)) is:  
\[V_{1}=\frac{1}{2}f_{v1}^{2}. \tag{35}\]  
The main properties of \(V_{1}\) are:  
\[V_{1}>0\ for\ \ \ x_{1}\ \
otin\ \Big{[}x_{1}^{l},x_{1}^{u}\Big{]}, \tag{36}\]  
\[V_{1}=0\ for\ x_{1}\ \in\ \Big{[}x_{1}^{l},x_{1}^{u}\Big{]}. \tag{37}\]  
\(\square\)  
**Task 3. ** Arrangement of the \(dV_{1}/dt\) expression in terms of a non-positive function of \(f_{v1}=dV_{1}/dx_{1}\). From properties (31), (34), it follows that the term \(f_{v1}(g_{x1}-\delta_{sin})\) appearing in Equation (18) satisfies:  
\[f_{v1}(g_{x1}-\delta_{sin})=|f_{v1}||(g_{x1}-\delta_{sin})|\geq 0. \tag{38}\]  
From the definition of \(g_{x1}\) (19) and definitions of \(x_{1}^{l},\ x_{1}^{u}\) (24), (25), it follows that:  
\[[g_{x1}-\delta_{sin}]\geq|g_{t1}|, \tag{39}\]  
\[g_{t1}=\left\{ \begin{array}{c}g_{x1}-max\{\delta_{sin}\}\ \ for\ \ x_{1}>x_{1}^{u}\\ 0\ for\ \ x_{1}\in\ \Big{[}x_{1}^{l},\ x_{1}^{u}\Big{]}\\ g_{x1}-min\{\delta_{sin}\}\ \ for\ \ x_{1}<x_{1}^{l}\end{array} \right.. \tag{40}\]  
From Equations (38) and (39), it follows that:  
\[f_{v1}(g_{x1}-\delta_{sin})\geq|f_{v1}||(g_{t1})=f_{v1}g_{t1}\geq 0. \tag{41}\]From the property (21), definition of \(g_{t1}\) (40), and definition of \(f_{v1}\) (29), it follows that:  
\[|g_{t1}|\geq|f_{v1}|. \tag{42}\]  
Therefore, Equation (41) yields:  
\[f_{v1}(g_{x1}-\delta_{sin})\geq f_{v1}^{2}\geq 0. \tag{43}\]  
Using this in Equation (18) yields:  
\[\frac{dV_{1}}{dt}\leq-Df_{v1}^{2}. \tag{44}\]  
Accounting for the definition of \(V_{1}\) (35), we have:  
\[\frac{dV_{1}}{dt}\leq-2 DV_{1}\leq 0.\]  
\(\square\)  
**Task 4. ** Integration of the \(dV_{1}/dt\) expression. From the above expression, it follows that:  
\[V_{1}\leq V_{1t0}e^{-2D(t-t_{o})}. \tag{45}\]  
That is, \(V_{1}\)(35) converges exponentially to zero, and consequently \(f_{v1}\) converges exponentially to zero. Further, accounting for the definition of \(f_{v1}\) (29), it follows that \(x_{1}\) converges asymptotically to \(\Omega_{x1}=\left\{x_{1}:x_{1}^{l}\leq x_{1}\leq x_{1}^{u}\right\}\). From this convergence result and the definition of \(x_{1}\) (8), it follows that \(S_{1}\) converges asymptotically to \(\Omega_{51}\); where,  
\[\Omega_{51}=\left[S_{1}^{l}\ S_{1}^{u}\right], \tag{46}\]  
\[S_{1}^{l}=x_{1}^{l}+S_{1}^{eq},\ S_{1}^{u}=x_{1}^{u}+S_{1}^{eq}. \tag{47}\]  
\(S_{1}^{eq}\) is provided by Equation (5), the bounds \(x_{1}^{l}\), \(x_{1}^{u}\) are defined in Equations (24) and (25). This completes the proof of Ti. From Equation (44) and properties (31) and (32), it follows that \(\frac{dV_{1}}{dt}<0\ for\ x_{1}\ \
ot\in\ \Omega_{x1}\) and \(\frac{dV_{1}}{dt}\leq 0\ for\ x_{1}\in\ \Omega_{x1}\). Therefore, the set \(\Omega_{x1}\) is invariant, and consequently, \(\Omega_{51}\) is invariant. This completes the proof of Ti. \(\square\)  
**Remark 1**. : _The property \(|g_{t}|\geq|f_{v1}|\) (42) is crucial for proving the asymptotic convergence of \(x_{1}\)._  
**Remark 2**. : _The invariance of the set \(\Omega_{51}\) stated in Theorem 1 implies that once the state \(S_{1}\) is inside the convergence region \(\Omega_{51}\), that is, \(S_{1}\in\Omega_{51}\), it remains inside afterwards._