Consider the following linear system
$$\dot{x}=\sum\limits_{i=1}^{m}{{{\alpha }_{i}}}\left( t \right)\cdot {{A}_{i}}\cdot x \quad (1)
$$
where, $x\in {{\mathbb{R}}^{n}}$ represents the state vector, $\sum\limits_{i=1}^{m}{{{\alpha }_{i}}}\left( t \right)=1$, ${{\alpha }_{i}}\left( t \right)$are functions and ${{\alpha }_{i}}\left( t \right)>0$, ${{A}_{i}}\in {{\mathbb{R}}^{n\times n}}$ are constant matrices, $i=1,\cdots ,m$.
Once system (1) is stable , then the Euclidean norm of vector $x$ is bounded. Now, the questions are as the following: 1) what’s the upper bound for $\left\| x \right\|$ when system (1) is stable; 2) what’s the stability condition for system (1)? (Lyapunov function may be an appropriate and conventional method to derive the conditions for the stability of system (1).)
2 Answers
This is a more general case than the one of a switched dynamical system: it is obtained as the case in which $\alpha_i(t)$ are constant along discrete time steps of length $h$, and at each of these time steps one of the $\alpha_i(t)$ is one and all the other ones are zero).
As you can read a few lines above on the same Wikipedia page, that special case already is an undecidable problem, and even approximation algorithms are NP-hard.
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$\begingroup$ Thank you for the answer. In fact, an upper bound of stable system (1) is my concern. Maybe, it is difficult to obtain solution for system (1). Nevertheless, an appropriate upper bound of the state vector norm for system (1) may be not a difficult problem. $\endgroup$ Commented Aug 31, 2014 at 0:12
I realize I came across this question late, but decided to answer in case it's of any help. Thinking the original dynamics as a linear time varying system $\dot{x}(t) = A(t) x(t)$ with $A(t) = \sum_{i=1}^{m}\alpha_{i}(t)A_{i}$, the state can be solved as $x(t) = \Phi\left(t,0\right)x_{0}$, where $x_{0}$ is the initial state, and $\Phi\left(t,\tau\right)$ is the state transition matrix. It is well known that if $A(t)A(\tau) = A(\tau)A(t)$ for all $t, \tau$, then the state transition matrix $\Phi(t,\tau) = \exp\left(\int_{\tau}^{t}A(\sigma)d\sigma\right)$.
If the OP's matrices $A_{i}$ pairwise commute, then indeed $A(t)A(\tau) = A(\tau)A(t)$, and the state transition matrix $\Phi(t,0) = \prod_{i=1}^{m}\exp\left(A_{i}\beta_{i}(t)\right) = \exp\left(\sum_{i=1}^{m}A_{i}\beta_{i}(t)\right)$, where $\beta_{i}(t) = \int_{0}^{t}\alpha_{i}(\sigma)d\sigma > 0$. The stability condition then boils down to the norm of the state transition matrix being bounded: $\parallel \Phi(t,0)\parallel \leq \gamma$ (uniform stability), $\parallel \Phi(t,0)\parallel \leq \gamma \exp\left(-\delta t\right)$, for some $\gamma, \delta > 0$ (uniform asymptotic stability).
