Q:

Question about stability of a nonlinear dynamical system

Let $A(x) \in \mathbb{R}^{n\times n}$ with $x(t) \in \mathbb{R}^{n\times 1}$, $B(x) \in \mathbb{R}^{n \times m}$ $m \leq n$ and $F \in \mathbb{R^{m\times 1}}$ I have the following dynamical nonlinear system: 
$$ A(x) \ddot{x} = B(x) F$$
Let $k(x) \in \mathbb{R}^+$ be such that $-k(x)\mathbb{I} + A(x)$ has negative eigenvalues. 
The above system can be written:
$$ \left(-k(x)\mathbb{I} + A(x)\right) \ddot{x} - \dot{x} + k(x) \ddot{x} + \dot{x} - B(x) F = 0$$   My question regards the following observation:
The dynamical system 
$$ \left( -k(x) \mathbb{I} + A(x)\right) \ddot{x} - \dot{x} = 0$$ is asymptotically stable ($\dot{x},\ddot{x} \to 0$), because the system matrix is Hurwitz.
If also a control $F = F(x,\dot{x})$ is found such that 
$$ k(x) \ddot{x} + \dot{x} - B(x) F(x,\dot{x}) = 0$$ is stable ($\dot{x} \to 0$ and $\ddot{x} \to 0$). 
Can I somehow prove that the above system is stable with $F(x,\dot{x})$ control law? 

A:

Consider a simple (linear) counterexample, take $A(x)=a> 0,k(x)=k\in \mathbb{R}^+,B(x)=b\neq 0,F(x,\dot{x})=f\dot{x}$, all scalars, but further assume no signs. It follows from $\ddot{x}=(-k+a)^{-1}\dot{x}$ that for stability we want $k>a$. Then over to the next equation, we have
\begin{equation}
\ddot{x} = \frac{bf-1}{k}\dot{x},
\end{equation}
so for stability we want $(bf-1)/k<0$. Since $k\in \mathbb{R}^+$ we can for example take $f:=\frac{1}{b}(1-\varepsilon)$, $1\gg\varepsilon>0$. Then finally consider your system $\ddot{x}=(bf/a) \dot{x}$, plug in $f$ and get 
\begin{equation}
\ddot{x} = \frac{1-\varepsilon}{a}\dot{x},
\end{equation}
which is definitely not stable.
So finding an $F(x,\dot{x})$ which stabilizes your second equation is not sufficient. Which intuitively makes sense since both equations contain $(x,\dot{x},\ddot{x})$, so approaching them separately is rather difficult.

