Dynamic friction (Lugre) Model

$$\dot \zeta = \dot q_{1} - \frac{\vert\dot q_{1}\vert}{g(q_{1})}\zeta$$ (4.12)

$$\zeta =$$ (4.13)

$$F = \sigma_{0}\zeta + \sigma_{1}\dot \zeta + \sigma_{2}\dot q_{1} + F_{g}$$ (4.14)

$$g(\dot q_{1}) = F_{c} + (F_{s}-F{c})e^{(\frac {\dot q_{1}}{V_{s}})^2}$$ (4.15)

$$\dot q_{1} = \dot x_{1}=$$ (4.16)

$$F_{f} = \sigma_{0}z + \sigma_{1}\dot z + \sigma_{2}\dot x$$ (4.17)

$$\dot z = \dot x - \frac{\vert\dot x\vert}{g(x)}Z$$ (4.18)

$$\sigma_{0}g(\dot x) = F_{c}+(F_{s}-F_{c})e^{(-(\frac{\dot x}{V_{s}}^2))}$$ (4.19)

assume $x_{3} = z$

$$\dot x_{1} = x_{2}$$ (4.20)

$$\dot x_{2} = u - [ \sigma_{0} x_{3} + \sigma_{1}\dot x_{3} + \sigma_{2} x_{2}]$$ (4.21)

$$\dot x_{2} = u - [ \sigma_{0} x_{3} + \sigma_{1}x_{2}- \frac {\sigma_{1}\vert x_{2}\vert x_{3}}{g(x_{2})}+\sigma_{2} x_{2}]$$ (4.22)

now analyze

$$f_{a} = \frac{\vert x_{2}\vert}{g(x_{2})}x_{3}$$ (4.23)

$x_{3}$ will be Lipschitz as here we can see that $\Vert f(x) - f(y)\Vert \leq k \Vert X-Y\Vert$ so the system as a whole will be Lipschitz.