I'm trying to prove the local stability of a nonlinear system and got the following inequality.
$
\|x(t)\|\leq c_1\|x(t_0)\|\exp(-c_2(t-t_0))+c_3\epsilon_m\cdots
$(i)
where $c_1, c_2, c_3$ are positive constants.
Now, since the approximation error has upper bound $\epsilon_m=\sup_{x \in \mathcal{X}}\| \epsilon(x)\|$ only within a compact set $\mathcal{X}$ of feasible system states, to complete the proof, I have to prove that $x$ always lies in $\mathcal{X}$.
But I have no idea where to start or what assumption is needed to justify the stability.
Off the top of my head, the assumption $x_0 \in \mathcal{X}, c_3<1$ seems to work cuz with this assumption, $x$ starts inside the set and exponentially converges to a smaller ball than $\epsilon_m$.
Please let me know your thought
Here's the background for your information.
The system is $\dot{x}=Ax+B(u+f_u(x))$ where $f_u$ is an unknown disturbance. By the control input $u=-\hat{f}_u(x)$, the closed-loop dynamics becomes $\dot{x}=Ax+B(u+\epsilon)$ with the estimation error $\epsilon=f_u(x)-\hat{f}_u(x)$. (I've got (i) by applying the Comparison lemma to this system.)
$f_u(x), \hat{f}_u(x)$ has the same Lipschitz constant, so does $\epsilon$
$\mathcal{X}$ is where $\epsilon$ is bounded, which means $\|f_u(x^\prime)-\hat{f}_u(x) \|< L_u \|x^\prime - x\|$ with $L_u$ being the Lipshitz constant of $f_u$