Let \(E(t)= ((e_{1}(t))^{T}, (e_{2}(t))^{T},\ldots , (e_{N}(t))^{T} )^{T}\), \(d=2L+\frac{1}{1-\rho }\), \(\sigma_{\psi }=t_{\psi }-t_{ \psi -1}\) be the impulsive intervals, *I* be an identity matrix with appropriate dimension, \(\lambda_{\max }\) be the largest eigenvalues of \(dI+(U(t)\otimes B)(U(t)\otimes B)^{T}\), \(\mathcal{G}(t,\xi (t))=\{ \eta \mid t-\xi (t)< t_{\eta }< t, \eta =1,2,\ldots \}\), \(\underline{ \mathcal{G}}(t,\xi (t))\) and \(\overline{\mathcal{G}}(t,\xi (t))\) be the minimum and maximum values of \(\mathcal{G}(t,\xi (t))\), \(\delta (t)=(1+ \vartheta (t))^{2}\), from the definition of \(\vartheta (t)\), we have \(\delta (t)=1\) for \(t\neq t_{\psi }\).

### Theorem 1

*Suppose that Assumptions*
1
*and*
2
*hold*. *If there exists a constant*
\(\gamma >0\)
*such that*

$$\begin{aligned} \ln \delta (t_{\psi })+\gamma +\lambda_{\max } \sigma_{\psi }< 0,\quad \psi =1,2,\ldots , \end{aligned}$$

(4)

*holds*, *then network* (2) *can achieve synchronization*.

### Proof

Consider the following Lyapunov function:

$$\begin{aligned} V(t)&= \sum_{k=1}^{N}e_{k}^{T}(t)e_{k}(t)+ \frac{\delta (t)}{1-\rho } \Biggl( \int_{t-\xi (t)}^{t_{\underline{\mathcal{G}}(t,\xi (t))}^{-}} \sum_{k=1}^{N}e_{k}^{T}( \theta )e_{k}(\theta )\,d\theta \\ &\quad {}+\sum_{q=\underline{\mathcal{G}}(t,\xi (t))}^{ \overline{\mathcal{G}}(t,\xi (t))-1} \int_{t_{q}^{+}}^{t_{q+1}^{-}} \sum_{k=1}^{N}e_{k}^{T}( \theta )e_{k}(\theta )\,d\theta + \int_{t_{\overline{\mathcal{G}}(t,\xi (t))}^{+}}^{t}\sum_{k=1}^{N}e _{k}^{T}(\theta )e_{k}(\theta )\,d\theta \Biggr) \end{aligned}$$

for \(t\in (t_{\psi -1},t_{\psi }]\), \(\psi =1,2,\ldots \) .

When \(t\in (t_{\psi -1},t_{\psi })\),

$$\begin{aligned} V(t)&= \sum_{k=1}^{N}e_{k}^{T}(t)e_{k}(t)+ \frac{1}{1-\rho } \Biggl( \int_{t-\xi (t)}^{t_{\underline{\mathcal{G}}(t,\xi (t))}^{-}}\sum_{k=1} ^{N}e_{k}^{T}(\theta )e_{k}(\theta )\,d\theta \\ &\quad {}+\sum_{q=\underline{\mathcal{G}}(t,\xi (t))}^{ \overline{\mathcal{G}}(t,\xi (t))-1} \int_{t_{q}^{+}}^{t_{q+1}^{-}} \sum_{k=1}^{N}e_{k}^{T}( \theta )e_{k}(\theta )\,d\theta + \int_{t_{\overline{\mathcal{G}}(t,\xi (t))}^{+}}^{t}\sum_{k=1}^{N}e _{k}^{T}(\theta )e_{k}(\theta )\,d\theta \Biggr), \end{aligned}$$

and the derivative of \(V(t)\) is

$$\begin{aligned} \dot{V}(t)&=2\sum_{k=1}^{N}e_{k}^{T}(t) \dot{e}_{k}(t)+\frac{1}{1- \rho }\sum_{k=1}^{N}e_{k}^{T}(t)e_{k}(t) \\ &\quad {}-\frac{(1-\dot{\xi }(t))}{1-\rho }\sum_{k=1}^{N}e_{k}^{T} \bigl(t-\xi (t)\bigr)e _{k}\bigl(t-\xi (t)\bigr) \\ & = 2\sum_{k=1}^{N}e_{k}^{T}(t) \hat{g}\bigl(e_{k}(t)\bigr)+2\sum_{k=1}^{N} \sum_{j=1}^{N}\mu_{kj}(t)e_{k}^{T}(t)Be_{j} \bigl(t-\xi (t)\bigr) \\ &\quad {}+\frac{1}{1-\rho }\sum_{k=1}^{N}e_{k}^{T}(t)e_{k}(t)- \frac{(1- \dot{\xi }(t))}{1-\rho }\sum_{k=1}^{N}e_{k}^{T} \bigl(t-\xi (t)\bigr)e_{k}\bigl(t- \xi (t)\bigr). \end{aligned}$$

According to Assumptions 1 and 2,

$$\begin{aligned} V(t)&\leq 2LE^{T}(t)E(t)+E^{T}(t) \bigl(U(t)\otimes B\bigr) \bigl(U(t)\otimes B\bigr)^{T}E(t) \\ &\quad {}+E^{T}\bigl(t-\xi (t)\bigr)E\bigl(t-\xi (t)\bigr)+ \frac{1}{1-\rho }E^{T}(t)E(t) \\ &\quad {}-\frac{(1-\dot{\xi }(t))}{1-\rho }E^{T}\bigl(t-\xi (t)\bigr)E\bigl(t-\xi (t)\bigr) \\ &= E^{T}(t) \bigl(dI+\bigl(U(t)\otimes B\bigr) \bigl(U(t)\otimes B \bigr)^{T} \bigr)E(t) \\ &\quad {}-\frac{(\rho -\dot{\xi }(t))}{1-\rho }E^{T}\bigl(t-\xi (t)\bigr)E\bigl(t-\xi (t)\bigr) \\ &\leq \lambda_{\max } E^{T}(t)E(t)\leq \lambda_{\max } V(t), \end{aligned}$$

which gives

$$\begin{aligned} V(t)\leq V(t_{\psi -1})\exp \bigl(\lambda_{\max }(t-t_{\psi -1}) \bigr),\quad t \in (t_{\psi -1},t_{\psi }). \end{aligned}$$

(5)

When \(t=t_{\psi }\),

$$\begin{aligned} V\bigl(t_{\psi }^{+}\bigr) &= \sum _{k=1}^{N}e_{k}^{T} \bigl(t_{\psi }^{+}\bigr)e_{k}\bigl(t_{ \psi }^{+} \bigr)+\frac{\delta (t_{\psi }^{+})}{1-\rho } \Biggl( \int_{t_{\psi }^{+}-\xi (t_{\psi }^{+})}^{t_{\underline{\mathcal{G}}(t, \xi (t))}^{-}}\sum_{k=1}^{N}e_{k}^{T}( \theta )e_{k}(\theta )\,d\theta \\ &\quad {}+\sum_{q=\underline{\mathcal{G}}(t,\xi (t))}^{ \overline{\mathcal{G}}(t,\xi (t))-1} \int_{t_{q}^{+}}^{t_{q+1}^{-}} \sum_{k=1}^{N}e_{k}^{T}( \theta )e_{k}(\theta )\,d\theta + \int_{t_{\overline{\mathcal{G}}(t,\xi (t))}^{+}}^{t_{\psi }^{+}}\sum_{k=1}^{N}e_{k}^{T}( \theta )e_{k}(\theta )\,d\theta \Biggr) \\ &= \bigl(1+\vartheta (t_{\psi })\bigr)^{2}\sum _{k=1}^{N}e_{k}^{T} \bigl(t_{\psi }^{-}\bigr)e _{k}\bigl(t_{\psi }^{-} \bigr)+\frac{\delta (t_{\psi }^{-})}{1-\rho } \Biggl( \int_{t_{\psi }^{-}-\xi (t_{\psi }^{-})}^{t_{\underline{\mathcal{G}}(t, \xi (t))}^{-}}\sum_{k=1}^{N}e_{k}^{T}( \theta )e_{k}(\theta )\,d\theta \\ &\quad {}+\sum_{q=\underline{\mathcal{G}}(t,\xi (t))}^{ \overline{\mathcal{G}}(t,\xi (t))-1} \int_{t_{q}^{+}}^{t_{q+1}^{-}} \sum_{k=1}^{N}e_{k}^{T}( \theta )e_{k}(\theta )\,d\theta + \int_{t_{\overline{\mathcal{G}}(t,\xi (t))}^{+}}^{t_{\psi }^{-}}\sum_{k=1}^{N}e_{k}^{T}( \theta )e_{k}(\theta )\,d\theta \Biggr) \\ &= \delta (t_{\psi })V\bigl(t_{\psi }^{-}\bigr). \end{aligned}$$

(6)

When \(\psi =1\), from inequalities (5) and (6),

$$\begin{aligned} &V\bigl(t_{1}^{-}\bigr) \leq V(t_{0})\exp ( \lambda_{\max }\sigma_{1}), \\ &V\bigl(t_{1}^{+}\bigr) \leq \delta (t_{1})V \bigl(t_{1}^{-}\bigr) \leq \delta (t_{1})V(t _{0})\exp (\lambda_{\max } \sigma_{1}). \end{aligned}$$

When \(\psi =2\),

$$\begin{aligned}& V\bigl(t_{2}^{-}\bigr) \leq V\bigl(t_{1}^{+} \bigr)\exp (\lambda_{\max }\sigma_{2}) \\& \hphantom{V(t_{2}^{-})}\leq \delta (t_{1})V(t_{0})\exp \bigl( \lambda_{\max } (\sigma_{1}+ \sigma_{2}) \bigr), \\& V\bigl(t_{2}^{+}\bigr) \leq \delta (t_{2})V \bigl(t_{2}^{-}\bigr) \\& \hphantom{V(t_{2}^{+})}\leq \delta (t_{2})\delta (t_{1})V(t_{0}) \exp \bigl(\lambda_{\max } (\sigma_{1}+\sigma_{2}) \bigr) \\& \hphantom{V(t_{2}^{+})}=V(t_{0})\prod_{\pi =1}^{2} \delta (t_{\pi })\exp (\lambda_{\max } \sigma_{\pi }). \end{aligned}$$

By mathematical induction, for any positive integer *ψ*,

$$\begin{aligned} V\bigl(t_{\psi }^{+}\bigr)\leq V(t_{0})\prod _{\pi =1}^{\psi }\delta (t_{\pi }) \exp ( \lambda_{\max }\sigma_{\pi }). \end{aligned}$$

If condition (4) holds,

$$\begin{aligned} \delta (t_{\pi })\exp (\lambda_{\max }\sigma_{\pi })\leq \exp (- \gamma ),\quad \pi =1,2,\ldots , \end{aligned}$$

and

$$\begin{aligned} V\bigl(t_{\psi }^{+}\bigr)\leq V(t_{0})\exp (-\psi \gamma ), \end{aligned}$$

which implies \(V(t_{\psi }^{+})\rightarrow 0\) as \(\psi \rightarrow \infty \).

Then, for \(t\in (t_{\psi },t_{\psi +1}]\),

$$\begin{aligned} V(t)\leq V\bigl(t_{\psi }^{+}\bigr)\exp \bigl( \lambda_{\max }(t-t_{\psi })\bigr), \end{aligned}$$

which implies \(V(t)\rightarrow 0\) as \(t\rightarrow \infty \), i.e., the synchronization is achieved. This completes the proof. □

### Remark 2

It is clear that \(V(t)\) is the quadratic sum of synchronization errors, i.e., \(V(t)\geq 0\) and \(V(t)=0\) if and only if \(e_{k}(t)=0\), \(k=1,2,\ldots ,N\). In particular, if the set \(\mathcal{G}(t,\xi (t))\) is an empty set, the Lyapunov function is

$$\begin{aligned} V(t)= &\sum_{k=1}^{N}e_{k}^{T}(t)e_{k}(t)+ \frac{\delta (t)}{1-\rho } \int_{t-\xi (t)}^{t}\sum_{k=1}^{N}e_{k}^{T}( \theta )e_{k}(\theta )\,d\theta ; \end{aligned}$$

if the set \(\mathcal{G}(t,\xi (t))\) only has an element *ϱ*, \(\underline{\mathcal{G}}(t,\xi (t))=\overline{\mathcal{G}}(t,\xi (t))= \varrho \),

$$\begin{aligned} V(t)&= \sum_{k=1}^{N}e_{k}^{T}(t)e_{k}(t)+ \frac{\delta (t)}{1-\rho } \Biggl( \int_{t-\xi (t)}^{t_{\varrho }^{-}}\sum_{k=1}^{N}e_{k}^{T}( \theta )e_{k}(\theta )\,d\theta+ \int_{t_{\varrho }^{+}}^{t}\sum_{k=1}^{N}e_{k}^{T}( \theta )e_{k}( \theta )\,d\theta \Biggr). \end{aligned}$$

### Remark 3

From condition (4), it is easy to see the relationship among system parameters, impulsive gains and impulsive intervals. Clearly, for any given network, we can calculate the needed impulsive gains and intervals such that condition (4) holds. As we know, different networks usually have different system parameters. That is, impulsive controllers with fixed impulsive gains and intervals are not always valid for different networks. In the following, we introduce adaptive strategy to design unified impulsive controllers.

### Theorem 2

*Suppose that Assumptions*
1
*and*
2
*hold*. *If there exists a constant*
\(\gamma >0\)
*such that*

$$\begin{aligned} \ln \delta (t_{\psi })+\gamma +\widehat{M}(t_{\psi }) \sigma_{\psi }< 0,\quad \psi =1,2,\ldots , \end{aligned}$$

(7)

*holds*, *where*
\(\dot{\widehat{M}}(t)=\varepsilon \sum_{k=1}^{N}e_{k} ^{T}(t)e_{k}(t)\)
*and*
\(\varepsilon >0\)
*is the adaptive gain*, *then network* (2) *can achieve synchronization*.

### Proof

Consider the following Lyapunov function:

$$\begin{aligned} V(t)&= \sum_{k=1}^{N}e_{k}^{T}(t)e_{k}(t)+ \frac{\delta (t)}{1-\rho } \Biggl( \int_{t-\xi (t)}^{t_{\underline{\mathcal{G}}(t,\xi (t))}^{-}} \sum_{k=1}^{N}e_{k}^{T}( \theta )e_{k}(\theta )\,d\theta \\ &\quad {}+\sum_{q=\underline{\mathcal{G}}(t,\xi (t))}^{ \overline{\mathcal{G}}(t,\xi (t))-1} \int_{t_{q}^{+}}^{t_{q+1}^{-}} \sum_{k=1}^{N}e_{k}^{T}( \theta )e_{k}(\theta )\,d\theta + \int_{t_{\overline{\mathcal{G}}(t,\xi (t))}^{+}}^{t}\sum_{k=1}^{N}e _{k}^{T}(\theta )e_{k}(\theta )\,d\theta \Biggr) \\ &\quad {}+\frac{\delta (t)}{2\varepsilon }\bigl(\widehat{M}(t)-\lambda_{\max } \bigr)^{2} \end{aligned}$$

for \(t\in (t_{\psi -1},t_{\psi }],\psi =1,2,\ldots \) .

When \(t\in (t_{\psi -1},t_{\psi })\),

$$\begin{aligned} V(t)= &\sum_{k=1}^{N}e_{k}^{T}(t)e_{k}(t)+ \frac{1}{1-\rho } \Biggl( \int_{t-\xi (t)}^{t_{\underline{\mathcal{G}}(t,\xi (t))}^{-}}\sum_{k=1} ^{N}e_{k}^{T}(\theta )e_{k}(\theta )\,d\theta \\ &+\sum_{q=\underline{\mathcal{G}}(t,\xi (t))}^{ \overline{\mathcal{G}}(t,\xi (t))-1} \int_{t_{q}^{+}}^{t_{q+1}^{-}} \sum_{k=1}^{N}e_{k}^{T}( \theta )e_{k}(\theta )\,d\theta + \int_{t_{\overline{\mathcal{G}}(t,\xi (t))}^{+}}^{t}\sum_{k=1}^{N}e _{k}^{T}(\theta )e_{k}(\theta )\,d\theta \Biggr) \\ &+\frac{1}{2\varepsilon }\bigl(\widehat{M}(t)-\lambda_{\max } \bigr)^{2}, \end{aligned}$$

and the derivative of \(V(t)\) is

$$\begin{aligned} \dot{V}(t)&= 2\sum_{k=1}^{N}e_{k}^{T}(t) \hat{g}\bigl(e_{k}(t)\bigr)+2\sum_{k=1} ^{N}\sum_{j=1}^{N} \mu_{kj}(t)e_{k}^{T}(t)Be_{j}\bigl(t-\xi (t)\bigr) \\ &\quad {}+\frac{1}{1-\rho }\sum_{k=1}^{N}e_{k}^{T}(t)e_{k}(t)- \frac{(1- \dot{\xi }(t))}{1-\rho }\sum_{k=1}^{N}e_{k}^{T} \bigl(t-\xi (t)\bigr)e_{k}\bigl(t- \xi (t)\bigr) \\ &\quad {}+\bigl(\widehat{M}(t)-\lambda_{\max }\bigr)\sum _{k=1}^{N}e_{k}^{T}(t)e_{k}(t) \\ &\leq \widehat{M}(t)V(t)\leq \widehat{M}(t_{\psi })V(t), \end{aligned}$$

which gives

$$ V(t)\leq V(t_{\psi -1})\exp \bigl(\widehat{M}(t_{\psi }) (t-t_{\psi -1}) \bigr),\quad t\in (t_{\psi -1},t_{\psi }). $$

When \(t=t_{\psi }\),

$$\begin{aligned} V\bigl(t_{\psi }^{+}\bigr)&= \sum_{k=1}^{N}e_{k}^{T} \bigl(t_{\psi }^{+}\bigr)e_{k}\bigl(t_{ \psi }^{+} \bigr)+\frac{\delta (t_{\psi }^{+})}{1-\rho } \Biggl( \int_{t_{\psi }^{+}-\xi (t_{\psi }^{+})}^{t_{\underline{\mathcal{G}}(t, \xi (t))}^{-}}\sum_{k=1}^{N}e_{k}^{T}( \theta )e_{k}(\theta )\,d\theta \\ &\quad {}+\sum_{q=\underline{\mathcal{G}}(t,\xi (t))}^{ \overline{\mathcal{G}}(t,\xi (t))-1} \int_{t_{q}^{+}}^{t_{q+1}^{-}} \sum_{k=1}^{N}e_{k}^{T}( \theta )e_{k}(\theta )\,d\theta + \int_{t_{\overline{\mathcal{G}}(t,\xi (t))}^{+}}^{t_{\psi }^{+}}\sum_{k=1}^{N}e_{k}^{T}( \theta )e_{k}(\theta )\,d\theta \Biggr) \\ &\quad {}+\frac{\delta (t_{\psi }^{+})}{2\varepsilon }\bigl(\widehat{M}\bigl(t_{\psi } ^{+} \bigr)-\lambda_{\max }\bigr)^{2} \\ &= \delta (t_{\psi })V\bigl(t_{\psi }^{-}\bigr). \end{aligned}$$

Thus, similar to the proof of Theorem 1, the proof can be completed. □

### Remark 4

Generally, when the impulsive interval \(\sigma_{ \psi }\) is fixed, for any given *γ*, we can choose

$$\begin{aligned} &-\exp \biggl(-\frac{\gamma +\widehat{M}(t_{\psi })\sigma_{\psi }}{2} \biggr)-1+\varrho \leq \vartheta (t_{\psi })\leq \exp \biggl(- \frac{\gamma +\widehat{M}(t_{\psi })\sigma_{\psi }}{2} \biggr)-1-\varrho , \end{aligned}$$

such that condition (7) holds, where *ϱ* is an arbitrary small positive constant.

### Remark 5

From condition (7), when \(\vartheta (t_{ \psi })\) and *γ* are fixed, we can estimate the control instants \(t_{\psi }\) through finding the maximum value of

$$ t_{\psi }< t_{\psi -1}- \bigl(\delta (t_{\psi })+\gamma \bigr) \widehat{M} ^{-1}(t_{\psi }), $$

with \(t_{0}=0\), \(\psi =1,2,\ldots \) .