# Mean Square Consensus of Nonlinear Multi-Agent Systems under Markovian Impulsive Attacks
## 3 Main Results
Therefore, the restrictive condition \(\gamma<0\) in (16) is required._  
_In the following, we further assume that the Markovian chain obeyed by the impulsive sequence \(\{t_{k}\}_{k=1}^{+\infty}\) is irreducible and that it adopts a stationary distribution \(\varepsilon=(\varepsilon_{1},\varepsilon_{2},\ldots,\varepsilon_{p})\)._  
**Theorem 2**. : _Under Assumptions 1 and 2, then the consensus of system (1) with controller (2) under Markovian impulsive attacks can be reached almost surely (a.s.), if there exist symmetric positive definite matrix \(P\) and feedback gain \(K\) such that_  
\[\sum_{i\in\Gamma}\lambda_{i}\varepsilon_{i}\ln(\mathbb{E}[\eta_{i}^{2}])+ \gamma<0. \tag{17}\]  
_where \(\gamma=\rho\beta\); \(\rho=\frac{\lambda_{\max}(P)}{\lambda_{\min}(P)}\); and \(\beta=\lambda_{\max}(\Pi)\), where \(\Pi=H^{\mathbb{T}}+H+\kappa\bar{B}^{\mathbb{T}}+\kappa\bar{B}\), and \(\varepsilon_{i},i\in\Gamma\) is the element of the stationary distribution adopted by the impulsive sequence._Proof. : From (11), we get that for \(\forall t\geq t_{0}\)  
\[ \begin{split} V(t,e(t))&\leq\prod_{l=1}^{k}\eta_{l}^{2 }e^{\gamma(t-t_{0})}V(t_{0},e(t_{0}))\\ &=\prod_{k=1}^{\bar{N}(t,t_{0})}\eta_{k}^{2}e^{\gamma(t-t_{0})}V( t_{0},e(t_{0})).\end{split}  \tag{18}\]  
Note that the impulsive interval \(\tau(k)\sim Exp(\lambda_{i}),i\in\Gamma\); then using the Ergodic theorem [29], the strong law of large number, and the properties of the homogeneous Markovian chain, for \(i\in\Gamma\), we can obtain that  
\[ \begin{split}\lim_{t\to+\infty}\frac{\bar{N}_{i}(t,t_{0})}{t-t_{0} }&=\lim_{t\to+\infty}\frac{\bar{N}_{i}(t,t_{0})}{T_{i}(t,t_{0})} \frac{T_{i}(t,t_{0})}{t-t_{0}}\\ &=\lim_{t\to+\infty}\frac{\bar{N}_{i}(t,t_{0})}{\sum_{k=1}^{\bar {N}_{i}(t,t_{0})}\tau_{k}}\bar{\varepsilon}_{i}\\ &=\lambda_{i}\bar{\varepsilon}_{i},\quad a.s.\end{split}  \tag{19}\]  
Then, it follows from (17) and (19) that  
\[ \begin{split}\mathbb{E}[V(t,e(t))]\\ &\leq\prod_{k=1}^{\bar{N}(t,t_{0})}\mathbb{E}[\eta_{k}^{2}]e^{ \gamma(t-t_{0})}\mathbb{E}[V(t_{0},e(t_{0}))]\\ &=\prod_{i=1}^{p}(\mathbb{E}[\eta_{i}^{2}])^{\bar{N}_{i}(t_{0})} e^{\gamma(t-t_{0})}\mathbb{E}[V(t_{0},e(t_{0}))]\\ &=e^{(\sum_{i\in\Gamma}\lambda_{i}\bar{\varepsilon}_{i}\ln( \mathbb{E}[\eta_{i}^{2}])+\gamma)(t-t_{0})}\mathbb{E}[V(t_{0},e(t_{0}))]\\ &=\bar{\omega}_{1}(\mathbb{E}[V(t_{0},e(t_{0}))],t-t_{0})\quad a.s.\end{split}  \tag{20}\]  
where the function \(\bar{\omega}_{1}(r,s):=e^{(\sum_{i\in\Gamma}\lambda_{i}\bar{\varepsilon}_{i}\ln (\mathbb{E}[\eta_{i}^{2}])+\gamma)s}r\). Define \(\bar{\omega}_{2}(r,s):=\frac{\bar{\omega}_{1}(rs)}{\epsilon}\) for an arbitrarily small \(\epsilon\in(0,1)\). By employing the [PERSON]'s inequality [[30], p. 111, (18.1)] to (20), we have for \(t\geq t_{0}\)  
\[ \begin{split}\mathcal{P}\{V(t,e(t))&\leq\bar{ \omega}_{2}(\mathbb{E}[V(t_{0},e(t_{0}))],t-t_{0})\}\\ &\geq 1-\frac{\mathbb{E}[V(t,e(t))]}{\bar{\omega}_{2}(\mathbb{E}[V(t_{ 0},e(t_{0}))],t-t_{0})}\geq 1-\epsilon.\end{split}  \tag{21}\]  
From (20), we know that \(\lim_{t\to+\infty}\mathbb{E}[V(t,e(t))]=0\). Hence, the consensus of system (1) can be achieved almost surely, and the proof of Theorem 2 is completed. **Remark 5**. : _Compared with the parameters in Theorem 1, from (17), one can observe that the condition \(\gamma<\min_{j\in\Gamma}\{v\}\) is removed by employing the properties of stationary distribution of Markovian chain._