## 3 Design of a New FSMC
### Structure FSMC
\[\dot{S}(\mathbf{x})= \Bigl{(}\hat{f}(\mathbf{x},\varphi_{f}^{*})-\hat{f}(\mathbf{x}, \varphi_{f})\Bigr{)}+\Bigl{(}\hat{g}_{1}(\mathbf{x},\varphi_{g}^{*})-\hat{g}_ {1}(\mathbf{x},\varphi_{g})-\varepsilon_{0}\lambda^{T}(\mathbf{x})\lambda( \mathbf{x})\Bigr{)}\,u_{s}(t) \tag{24}\] \[-\rho_{ad}\operatorname{sat}(S(\mathbf{x})/\Phi)+\Omega(\mathbf{ x},\varphi_{f},\varphi_{g})\]  
Substituting \(\hat{g}_{1}(\mathbf{x},\varphi_{g})=\varphi_{g}{}^{T}\lambda(\mathbf{x})\), \(\hat{f}(\mathbf{x},\varphi_{f})=\varphi_{g}{}^{T}\lambda(\mathbf{x})\) and their optimal functions into (24), the following is obtained. \[\dot{S}(\mathbf{x})=\psi_{f}^{T}\lambda(\mathbf{x})+\psi_{g}^{T}\,\lambda( \mathbf{x})u_{s}(t)-\rho_{ad}\operatorname{sat}(S(\mathbf{x})/\Phi)+\Omega( \mathbf{x},\varphi_{f},\varphi_{g}) \tag{25}\]  
Then, choose a Lyaponov function as below. \[V_{1}=\frac{1}{2}S(\mathbf{x})^{2}+\frac{1}{2}\psi_{f}{}^{T}\psi_{f}+\frac{1}{ 2}\psi_{g}{}^{T}\psi_{g} \tag{26}\]  
The time derivative of the above function is determined as follows. \[\dot{V}_{1}=S(\mathbf{x})\dot{S}(\mathbf{x})+\psi_{f}{}^{T}\dot{ \psi}_{f}+\psi_{g}{}^{T}\dot{\psi}_{g} \tag{27}\] \[=\psi_{f}{}^{T}\Bigl{(}S(\mathbf{x})\lambda(\mathbf{x})+\dot{\psi}_ {f}\Bigr{)}+\psi_{g}{}^{T}\Bigl{(}S(\mathbf{x})\lambda(\mathbf{x})\,u_{s}(t)+ \dot{\psi}_{g}\Bigr{)}\] \[\rho_{ad}=\max\Bigl{(}\eta_{1}\frac{\Omega_{0}\Phi}{|S(\mathbf{x})|+\varepsilon},\,\eta_{2}\Bigl{(}\varepsilon+\exp\Bigl{(}\eta_{3}S(\mathbf{x})\mathrm{sgn} (\hat{S}(\mathbf{x})\Bigr{)}\Bigr{)}\Bigr{)}\quad if\quad|S(\mathbf{x})|< \Phi\,, \tag{22}\]0  
Differentiating (27) with respect to time, the followings equations are achieved. \[\rho_{ad}=\max\Bigl{(}\eta_{1}\frac{\Omega_{0}\Phi}{|S(\mathbf{x})|+\varepsilon},\,\eta_{2}\Bigl{(}\varepsilon+\exp\Bigl{(}\eta_{3}S(\mathbf{x})\mathrm{sgn} (\hat{S}(\mathbf{x})\Bigr{)}\Bigr{)}\Bigr{)}\quad if\quad|S(\mathbf{x})|< \Phi\,, \tag{22}\]1\[\rho_{ad}=\max\Bigl{(}\eta_{1}\frac{\Omega_{0}\Phi}{|S(\mathbf{x})|+\varepsilon},\,\eta_{2}\Bigl{(}\varepsilon+\exp\Bigl{(}\eta_{3}S(\mathbf{x})\mathrm{sgn} (\hat{S}(\mathbf{x})\Bigr{)}\Bigr{)}\Bigr{)}\quad if\quad|S(\mathbf{x})|< \Phi\,, \tag{22}\]2  
In case \(\dot{g}_{1}(\mathbf{x},\varphi_{g})\
eq 0\), then \(\varepsilon_{0}=0\), so \(\dot{\psi}_{g}=-\dot{\varphi}_{g}\). With reference to (28) and (29), as well as the update laws (20), the following result is then obtained. \[\rho_{ad}=\max\Bigl{(}\eta_{1}\frac{\Omega_{0}\Phi}{|S(\mathbf{x})|+\varepsilon},\,\eta_{2}\Bigl{(}\varepsilon+\exp\Bigl{(}\eta_{3}S(\mathbf{x})\mathrm{sgn} (\hat{S}(\mathbf{x})\Bigr{)}\Bigr{)}\Bigr{)}\quad if\quad|S(\mathbf{x})|< \Phi\,, \tag{22}\]3  
In the case of \(\dot{g}_{1}(\mathbf{x},\varphi_{g})=0\), without losing generality, if \(\Delta t\) in (30) is chosen to be the sampling time (meaning \(\Delta t=\varepsilon_{0}\)), then (29) becomes as below. \[\rho_{ad}=\max\Bigl{(}\eta_{1}\frac{\Omega_{0}\Phi}{|S(\mathbf{x})|+\varepsilon},\,\eta_{2}\Bigl{(}\varepsilon+\exp\Bigl{(}\eta_{3}S(\mathbf{x})\mathrm{sgn} (\hat{S}(\mathbf{x})\Bigr{)}\Bigr{)}\Bigr{)}\quad if\quad|S(\mathbf{x})|< \Phi\,, \tag{22}\]4  
Therefore, from (27), (28), and (31), the following equation is obtained. \[ \begin{array}{rl}\dot{V}_{1}=&\psi_{f}{}^{T}\Big{(}S(\mathbf{x})\lambda( \mathbf{x})-\dot{\varphi}_{f}\Big{)}+\psi_{g}{}^{T}\Big{(}S(\mathbf{x})\lambda (\mathbf{x})u_{s}(t)\,-\,\dot{\varphi}_{g}-\Delta\lambda(\mathbf{x})\Big{)} \\ &\qquad\qquad-\rho_{ad}\,S(\mathbf{x})\,\mathrm{sat}(S(\mathbf{x})/\Phi)+S( \mathbf{x})\Omega(\mathbf{x},\varphi_{f},\varphi_{g}).\end{array}  \tag{32}\]  
In fact, there are two cases related to (32), as follows. \[ \begin{array}{rl}\dot{V}_{1}=-|S(\mathbf{x})|\rho_{ad}+S(\mathbf{x})\Omega (\mathbf{x},\varphi_{f},\varphi_{g})\leq|S(\mathbf{x})|(-\rho_{ad}+\Omega_{0 }),\;\mathrm{if}|S(\mathbf{x})/\Phi|\geq 1\\ \dot{V}_{1}=-\frac{S^{2}(\mathbf{x})}{\Phi}\rho_{ad}+S(\mathbf{x})\Omega( \mathbf{x},\varphi_{f},\varphi_{g})\leq|S(\mathbf{x})|\Big{(}\Omega_{0}-\frac {|S(\mathbf{x})|}{\Phi}\rho_{ad}\Big{)},\;\mathrm{if}|S(\mathbf{x})/\Phi|<1 \end{array}