## 3 Main Results
\] then we obtain \[V_{0}(t)=\frac{1}{2}\tilde{\sigma}^{T}(t)\Omega\tilde{\xi}(t)\text{, so }\frac{\lambda_{min}(\Omega)}{2}\tilde{\xi}^{T}(t)\tilde{\xi}(t)\leq V_{0}(t) \leq\frac{\lambda_{max}(\Omega)}{2}\tilde{\xi}^{T}(t)\tilde{\xi}(t)\text{.}\] Thus, we have \[\dot{V}_{0}(t)\leq-\frac{2\lambda_{min}(\Psi)}{\lambda_{max}(\Omega)}V_{0}(t)\] Let \(\rho_{1}=\frac{2\lambda_{min}(\Psi)}{\lambda_{max}(\Omega)}\), then \(\rho_{1}>0\), and \(\dot{V}_{0}(t)\leq-\rho_{1}V_{0}(t)\text{,}t\in[t_{k}^{i},t_{k+1}^{i})\). Then we can obtain \[V_{0}(t)\leq e^{-\rho_{1}(t-t_{n}^{i})}V_{0}(t_{n}^{i})\text{,}t\in[t_{n}^{i} \text{,}t_{s_{1}+1})\] \[V_{0}(t_{n}^{i})\leq e^{-\rho_{1}(t_{n}^{i}-t_{n-1}^{i})}V_{0}(t_{n-1}^{i})\] Thus, we can obtain \[V_{0}(t) \leq e^{-\rho_{1}(t-t_{n}^{i})}V_{0}(t_{n}^{i})\] \[\leq e^{-\rho_{1}(t-t_{n-1}^{i})}V_{0}(t_{n-1}^{i})\] \[\leq\cdots\] \[\leq e^{-\rho_{1}(t-t_{n})}V_{0}(t_{s_{1}})\] When the system is under a consecutive asynchronous attack, namely, \(t\in[t_{s_{1}+1},t_{s_{1}+1})\), the communication topology is switching; thus, we choose the Lyapunov function as follows: \[V_{\tilde{\sigma}(t)}(t)=\frac{1}{2}\tilde{\sigma}^{T}(t)( \begin{bmatrix}Q_{ \tilde{\sigma}(t)}&0\\ 0&Q_{\tilde{\sigma}(t)}\end{bmatrix} \otimes I_{n})\tilde{\sigma}^{T}(t)\]  
Thus, we can obtain  
\[V_{m}(t) \leq e^{\rho_{2}(t-t_{p}^{i})}V_{m}(t_{p}^{i})\] \[\leq e^{\rho_{2}(t-t_{p-1}^{i})}V_{m}(t_{p-1}^{i})\] \[\leq\cdots\] \[\leq e^{\rho_{2}(t-t_{\gamma_{1}+m})}V_{m}(t_{s_{l}+m})\]  
When \(t\in[t_{s_{l+1}-1},t_{s_{l+1}})\), we can obtain  
\[V_{\delta(t)}(t)\leq e^{\rho_{2}(t-t_{\gamma_{l+1}-1})}V_{\delta(t_{s_{l+1}-1}) }(t_{s_{l+1}-1})\]  
According to condition (2), one has  
\[V_{\delta(t_{s_{l+1}-1}^{+})}(t_{s_{l+1}-1}^{+})\leq\mu_{2}V_{\delta(t_{s_{l+1} -1}^{-})}(t_{s_{l+1}-1}^{-})\]  
Because \(V_{\delta(t)}(t)\) is a right-continuous function, then one has  
\[V_{\delta(t_{s_{l+1}-1})}(t_{s_{l+1}-1})\leq\mu_{2}V_{\delta(t_{s_{l+1}-1}^{- })}(t_{s_{l+1}-1}^{-})\leq\mu_{2}e^{\rho_{2}(t_{s_{l+1}-1}-t_{s_{l+1}-2})}V_{ \delta(t_{s_{l+1}-2})}(t_{s_{l+1}-2})\]  
So we have \(V_{\delta(t)}(t)\leq\mu_{2}e^{\rho_{2}(t-t_{\gamma_{l+1}-2})}V_{\delta(t_{s_{l +1}-2})}(t_{s_{l+1}-2})\). Repeating the above step, we can obtain  
\[V_{\delta(t)}(t)\leq\mu_{2}^{N(t_{l}+1,t_{l+1})-1}e^{\rho_{2}(t-t_{\gamma_{l+1 }})}V_{\delta(t_{s_{l+1}})}(t_{s_{l+1}})\]  
Similarly, according to condition (2), we can obtain  
\[V_{\delta(t_{s_{l+1}}^{+})}(t_{s_{l+1}}^{+})\leq\mu_{3}V_{\delta(t_{s_{l+1}}^{ -})}(t_{s_{l+1}}^{-})\]  
\[V_{\delta(t_{s_{l+1}}^{+})}(t_{s_{l+1}}^{+})\leq\mu_{1}V_{\delta(t_{s_{l+1}}^{ -})}(t_{s_{l+1}}^{-})\]  
When \(t\in[t_{s_{l}},t_{s_{l}+1})\), there is no DoS attack; we have  
\[\leq e^{\rho_{2}(t-t_{p-1}^{i})}V_{m}(t_{p-1}^{i})\]0 \[\leq e^{\rho_{2}(t-t_{p-1}^{i})}V_{m}(t_{p-1}^{i})\]1  
Thus, when \(t\in[t_{s_{l+1}-1},t_{s_{l+1}})\), we have  
\[\leq e^{\rho_{2}(t-t_{p-1}^{i})}V_{m}(t_{p-1}^{i})\]2  
Moreover, we can obtain  
\[\leq e^{\rho_{2}(t-t_{p-1}^{i})}V_{m}(t_{p-1}^{i})\]3 \[\leq e^{\rho_{2}(t-t_{p-1}^{i})}V_{m}(t_{p-1}^{i})\]4 \[\leq e^{\rho_{2}(t-t_{p-1}^{i})}V_{m}(t_{p-1}^{i})\]5  
Let \(\mu=max\{\mu_{1},\mu_{2},\mu_{3}\}\), then we can obtain  
\[\leq e^{\rho_{2}(t-t_{p-1}^{i})}V_{m}(t_{p-1}^{i})\]6  
Furthermore, it is noted that \(T_{l}\leq T_{r},f_{l}\leq f\) and \(N(t_{s_{l}+1},t_{s_{l+1}})=T_{l}f_{l}\leq Tf,\tau_{l}\geq\tau_{\text{D}}\); then we have  
\[\leq e^{\rho_{2}(t-t_{p-1}^{i})}V_{m}(t_{p-1}^{i})\]7Let \(\omega=\mu^{Tf+1}e^{\rho_{2}T-\rho_{1}\tau_{0}}\); it is clear that \(\omega<1\) according to condition (3). Similar to the above analysis, one has  
\[\leq e^{\rho_{2}(t-t_{p-1}^{i})}V_{m}(t_{p-1}^{i})\]8  
Further, it is clear that \(\forall t\in[t_{\varepsilon_{1}},t_{\varepsilon_{1+1}})\),  
\[\leq e^{\rho_{2}(t-t_{p-1}^{i})}V_{m}(t_{p-1}^{i})\]9  
Meanwhile, we can obtain  
\[\leq\cdots\]0  
where \(\
u_{1}=min\{\lambda_{min}(Q_{m}),\lambda_{min}(\Omega)\}\),\(\
u_{2}=min\{\lambda_{max}(Q_{m}),\lambda_{max}(\Omega)\}\).