Help! Lyapunov proof for calculated torque control with friction term for robot I want to prove asymptotic stability for a Calculated torque control with friction compensation. I was told to find "an already proved" system but I have had no luck while searching for books and papers. If anyone know of a book where I could finbd some info about it or have candidate Lyapunov function, that would help a lot. What I was trying to do was to use other Lyapunov function which I have already used for PD control and tirque control but still don't know how to bound the friction term.
system's dynamic model:
\begin{equation}\label{eq:sis2}
\mathbf{M}(\mathbf{q})\ddot{\mathbf{q}}+\mathbf{C}(\mathbf{q},\dot{\mathbf{q}})\dot{\mathbf{q}}+\mathbf{g}(\mathbf{q})=\boldsymbol{\tau}
\end{equation}
Control law:
\begin{equation}\label{eq:tau2}
\boldsymbol{\tau}=\mathbf{M}(\mathbf{q})[\mathbf{\ddot{q}}_{d}
+K_{v}\Delta \mathbf{\dot{q}}+K_{p}\Delta \mathbf{q}]+ \mathbf{C}(\mathbf{\dot{q}},\mathbf{q})\mathbf{\dot{q}}+\mathbf{g}(\mathbf{q})+\mathbf{B}\dot{\mathbf{q}}
\end{equation}
where $K_{p}$, $K_{v}$ $\in \mathbb{R}^{n\times n}$ are the proportional and derivative gains matrices; respectively. These gains are symmetric and positive definite. The terms $\Delta \mathbf{q} = \mathbf{q}_{d}-\mathbf{q}$, $\Delta \mathbf{\dot{\mathbf{q}}} = \mathbf{\dot{q}}_{d}-\mathbf{\dot{q}}$ are the angular position and velocity errors, where $\mathbf{q}_{d}$, $\mathbf{\dot{q}}_{d}$ and $\mathbf{\ddot{q}}_{d}$ are the desired angular position, velocity and acceleration, respecticely. $\mathbf{B}$ is a friction gains symmetric and positive definite martrix. 
This is the closed-loop system:
\begin{equation}\label{eq:es3}
\frac{d}{dt}\begin{bmatrix}
\Delta \mathbf{q}\\
\Delta \mathbf{\dot{q}}
\end{bmatrix}=\begin{bmatrix}
\Delta \mathbf{\dot{q}}\\
 -K_{p}\Delta \mathbf{q} - K_{v}\Delta \mathbf{\dot{q}}-\mathbf{M}^{-1}(\mathbf{q})\mathbf{B}\dot{\mathbf{q}}
\end{bmatrix}=\begin{bmatrix}
0&I\\
-K_{p}&-K_{v}-\mathbf{M}^{-1}(\mathbf{q})\mathbf{B}
\end{bmatrix}\begin{bmatrix}
\Delta \mathbf{q}\\
\Delta \mathbf{\dot{q}}
\end{bmatrix}
\end{equation}
Here I present a pair of candidate functions I was testing:
\begin{equation}\label{eq:L}
 V(\Delta \mathbf{q},\mathbf{q})=\frac{1}{2}\dot{\mathbf{q}}^{T}\mathbf{M}(\mathbf{q})\dot{\mathbf{q}}+\frac{1}{2}\Delta \mathbf{q}^{T}K_{p}\Delta \mathbf{q}
\end{equation}
and
\begin{equation}\label{eq:Lya}
\begin{aligned}
V=&\frac{1}{2}\begin{bmatrix}
\Delta \mathbf{q}\\
\Delta \mathbf{\dot{q}}
\end{bmatrix}^{T}\begin{bmatrix}
K_{p}+\epsilon K_{v}&\epsilon I\\
\epsilon I&I
\end{bmatrix}\begin{bmatrix}
\Delta \mathbf{q}\\
\Delta \mathbf{\dot{q}}
\end{bmatrix}\\
=&\frac{1}{2}[\Delta \mathbf{q} + \epsilon \Delta \mathbf{\dot{q}}]^{T}[\Delta \mathbf{q} + \epsilon \Delta \mathbf{\dot{q}}]+\frac{1}{2}\Delta \mathbf{q}^{T}[K_{p}+\epsilon K_{v}-\epsilon^{2}I]\Delta \mathbf{q}
\end{aligned}
\end{equation}
where $\epsilon$ satisfices $K_{v}-\epsilon I>0$ and $K_{p}+\epsilon K_{v}-\epsilon^{2}I>0$.
I'll appreciate if anyone could guide me here.
 A: I hope this answers your question , but you don't need the term $B \dot{q}$ it is making the system difficult to study and also your system will never converge to the desired reference , to demonstrate this substitute by $\tau$ in the equation and the system will be reduced to :
$$
 \Delta\ddot{q} + K_v \Delta \dot{q} +K_p \Delta q +M^{-1}B \dot{q}=0
$$
if you want you system to converge to the desired reference then you need to ensure that  $\Delta q=\Delta\dot{q}=0$ is an equilibrium point of the system i.e. $\Delta \ddot{q}=0 \implies M^{-1}B \dot{q}=0$ , so unless $\dot{q}=0$ , $B$ must be zero , so the system should actually be :
$$\Delta\ddot{q} +K_v \Delta \dot{q} +K_p \Delta q =0$$
so a suitable lyapanuv function would be :
$$
V=\frac{1}{2} \Delta\dot{q}^T\Delta\dot{q}+\frac{1}{2} \Delta q^T K_p\Delta q \implies \dot{V}=-\Delta\dot{q}^TK_v\Delta\dot{q} \leq 0
$$
if you want to keep the term $B\dot{q}$ you have to add the term $-B\dot{q}_d$ , but it is a redundancy as you already have the term $K_v \Delta \dot{q}$ .
