Appendix: Proof of Theorem 1
Consider the following Lyapunov–Krasovskii functional candidate:
$$ V\bigl(t,y(t)\bigr)=\sum^5_{i=1}V_{i} \bigl(t,y(t)\bigr), $$
(15)
where
$$\begin{aligned} V_{1}\bigl(t,y(t)\bigr)&= \biggl[ y(t)-D\int_{t-\sigma}^t y(s)\,\mathrm{d}s \biggr]^T{P} \biggl[ y(t)-D\int_{t-\sigma}^t y(s)\,\mathrm{d}s \biggr], \\ V_{2}\bigl(t,y(t)\bigr)& = \int_{t-{\tau_1}}^t {y}(s)^T{Q_1}y(s)\,\mathrm{d}s +\int_{t-{\tau_2}}^t {y}(s)^T{Q_2}y(s)\,\mathrm{d}s \\ &\quad +\int_{t-\tau(t)}^t {y}(s)^T{Q_3}y(s)\, \mathrm{d}s +\int_{t-\tau(t)}^t {g}\bigl(y(s) \bigr)^T{Q_4}g\bigl(y(s)\bigr)\,\mathrm{d}s, \\ V_{3}\bigl(t,y(t)\bigr)&={\tau_2}\int _{-{\tau_2}}^0 \int_{t+ \theta}^t \bigl({y}(s)^T{Q_5}y(s)+\dot{y}(s)^T{Q_6} \dot{y}(s) \bigr)\,\mathrm{d}s\,\mathrm{d}\theta \\ &\quad +({\tau_2}-{\tau_1})\int_{-{\tau_2}}^{-\tau_1} \int_{t+\theta}^t \bigl({y}(s)^T{Q_7}y(s)+ \dot{y}(s)^T{Q_8}\dot{y}(s) \bigr)\,\mathrm{d}s\,\mathrm{d} \theta, \\ V_{4}\bigl(t,y(t)\bigr)&=\int_{t - \tau_2 }^t {\int_\theta^t {\int_\beta^t {\dot{y}{{(s)}^T} {Q_9}\dot{y}(s)\,\mathrm{d}s\,\mathrm{d} \beta\,\mathrm{d}\theta}}} \\ &\quad +\int_{t-{\tau_2}}^{t-\tau_1}{\int_\theta^t {\int_\beta^t {\dot{y}{{(s)}^T}Q_{10} \dot{y}(s)\,\mathrm{d}s\,\mathrm{d}\beta\,\mathrm{d}\theta}}} \\ &\quad +\sigma\int_{-\sigma}^0 \int_{t+\theta}^t \bigl({y}(s)^T{Q_{11}}y(s)+\dot{y}^TQ_{12} \dot{y}(s) \bigr)\,\mathrm{d}s\,\mathrm{d}\theta\\ &\quad + \int_{t-\sigma}^t {y}(s)^T{Q_{13}}y(s)\,\mathrm{d}s, \\ V_{5}\bigl(t,y(t)\bigr)&=\sum_{j=1}^n r_{1j}\int_0^\infty{k_j}( \theta)\int_{t-\theta}^t g_j^2 \bigl({y_j}(s)\bigr)\,\mathrm{d}s\,\mathrm{d}\theta \end{aligned}$$
with \(P=\operatorname{diag}\{p_{1}, p_{2}, \ldots, p_{n} \}\).
Calculating the upper right derivative of V(t,y(t)) along the solution of (2) at the continuous interval t∈[t
k−1,t
k
),k∈Z
+, we get that
$$ D^+V\bigl(t,y(t)\bigr)=\sum^5_{i=1}D^+V_{i} \bigl(t,y(t)\bigr), $$
(16)
where
$$\begin{aligned} D^+V_{1}\bigl(t,y(t)\bigr) &= 2\sum ^n_{i=1}p_{i} \biggl({y_i}(t)-d_i \int_{t-\sigma }^t {y_i}(s)\,\mathrm{d}s \biggr)\frac{\mathrm{d}}{\mathrm{d}t} \biggl[{y_i}(t)-d_i\int _{t-\sigma}^t {y_i}(s)\,\mathrm{d}s \biggr], \end{aligned}$$
(17)
$$\begin{aligned} D^+V_{2}\bigl(t,y(t)\bigr) &= \sum^3_{i=1}{y}(t)^T{Q_i}y(t)-{y}(t-{ \tau_1})^T{Q_1}y(t-{\tau_1}) \\ &\quad -{y}(t-{\tau_2})^T{Q_2}y(t- { \tau_2})-\bigl(1-\dot{\tau} (t)\bigr)y\bigl(t-\tau(t) \bigr)^T \\ &\quad \times {Q_3}y\bigl(t-\tau(t)\bigr) \\ &\quad + {g}\bigl(y(t)\bigr)^T{Q_4}g\bigl(y(t)\bigr)-\bigl(1- \dot{\tau} (t)\bigr){g}\bigl(y\bigl(t-\tau(t)\bigr)\bigr)^T \\ &\quad \times {Q_4}g \bigl(y\bigl(t-\tau(t)\bigr)\bigr), \end{aligned}$$
(18)
$$\begin{aligned} D^+V_{3}\bigl(t,y(t)\bigr)& = \tau_2^2 \bigl(y(t)^T{Q_5}y(t)+{\dot{y}}(t)^T{Q_6} \dot{y}(t) \bigr) \\ &\quad -{\tau_2}\int_{t-{\tau_2}}^t \bigl({y(s)^T{Q_5}y(s)}+ {{\dot{y}}(s)^T{Q_6} \dot{y}(s)} \bigr) \,\mathrm{d}s \\ &\quad + {({\tau_2}-{\tau_1})^2} \bigl({y}(t)^T{Q_7}y(t)+{\dot{y}}(t)^T{Q_8} \dot{y}(t) \bigr) \\ &\quad -({\tau_2}-{\tau_1})\int_{t-{\tau_2}}^{t-{\tau_1}} \bigl({{y}(s)^T{Q_7}y(s)}+ {{\dot{y}}(s)^T{Q_8} \dot{y}(s)} \bigr)\,\mathrm{d}s, \end{aligned}$$
(19)
$$\begin{aligned} {D^+ V_4}\bigl(t,y(t)\bigr) &= \frac{1}{2}{\tau_2 ^2}\dot{y}{(t)^T} {Q_9}\dot{y}(t) - \int _{t - \tau_2 }^t {\int_\theta^t {\dot{y}{{(s)}^T} {Q_9}\dot{y}(s)} } \,\mathrm{d}s\,\mathrm{d} \theta \\ &\quad + \frac{1}{2}{({\tau_2}-{\tau_1})^2} \dot{y}{(t)^T}Q_{10}\dot{y}(t)- \int_{t - \tau_2 }^{t-{\tau_1}} {\int_\theta^t {\dot{y}{{(s)}^T}Q_{10} \dot{y}(s)} } \,\mathrm{d}s\,\mathrm{d}\theta \\ &\quad + {\sigma^2} \bigl({y}(t)^T{Q_{11}}y(t)+ \dot{y}(t)^TQ_{12}\dot{y}(t) \bigr) \\ &\quad +{y}(t)^TQ_{13}y(t)-{y}(t- \sigma)^TQ_{13}y(t-\sigma) \\ &\quad -\sigma\int_{t-\sigma}^t \bigl({{y}(s)^T{Q_{11}}y(s)}+{ \dot{y}(s)^TQ_{12}\dot{y}(s)} \bigr) \,\mathrm{d}s, \end{aligned}$$
(20)
$$\begin{aligned} D^+V_{5}\bigl(t,y(t)\bigr)&=\sum_{j=1}^n {r_{1j}\int_0^\infty {{k_j}(\theta)g_j^2\bigl({y_j}(t )\bigr)} } \,\mathrm{d}\theta \\ &\quad -\sum_{j= 1}^n {r_{1j}\int_0^\infty {{k_j}(\theta)g_j^2\bigl({y_j}(t- \theta)\bigr)} } \,\mathrm{d}\theta. \end{aligned}$$
(21)
Using (5) gives
$$\begin{aligned} D^+V_{1}\bigl(t,y(t)\bigr) &= 2\sum^n_{i=1}p_{i} \biggl({y_i}(t)-d_i\int_{t-\sigma }^t {y_i}(s)\mathrm{d}s \biggr) \Biggl\{-{d_i} {y_i}(t)+\sum_{j=1}^n {{a_{ij}} {g_j}\bigl({y_j}(t)\bigr)} \\ &\quad +\sum_{j=1}^n {{b_{ij}} {g_j}\bigl({y_j}\bigl(t-\tau(t)\bigr)\bigr)} \\ &\quad +\bigwedge _{j=1}^n {\alpha_{ij}}\int _{-\infty}^t \, \mathrm{d}s \\ &\quad -\bigwedge_{j=1}^n {\alpha_{ij}} \int_{-\infty}^t {{k_j}(t-s){f_j} \bigl(x_j^*\bigr)} \,\mathrm{d}s \\ &\quad -\bigvee_{j=1}^n {\beta_{ij}}\int_{-\infty}^t {{k_j}(t-s){f_j}\bigl(x_j^*\bigr)\, \mathrm{d}s} \\ &\quad +\bigvee_{j=1}^n {\beta_{ij}} \int_{-\infty}^t {{k_j}(t-s){f_j} \bigl({y_j}(s)+x_j^*\bigr)\,\mathrm{d}s} \Biggr\}. \end{aligned}$$
(22)
Based on Lemma 3, we obtain the following inequalities:
$$\begin{aligned} &\Biggl| \bigwedge_{j=1}^n \alpha_{ij} \int_{-\infty}^t {k_j}(t-s){f_j} \bigl({y_j}(s)+x_j^*\bigr) \mathrm{d}s-\bigwedge _{j=1}^n \alpha_{ij}\int _{-\infty}^t {k_j}(t-s){f_j} \bigl(x_j^*\bigr)\, \mathrm{d}s \Biggr| \\ &\quad \le\sum_{j=1}^n | \alpha_{ij} |\times\biggl|\int_{-\infty }^t {k_j}(t- s){f_j}\bigl({y_j}(s)+x_j^* \bigr) \,\mathrm{d}s-\int_{-\infty}^t {k_j}(t-s){f_j}\bigl(x_j^*\bigr)\, \mathrm{d}s \biggr| \\ &\quad =\sum_{j=1}^n | \alpha_{ij} |\times\biggl|\int_{-\infty}^t {k_j}(t- s){g_j}\bigl({y_j}(s)\bigr)\, \mathrm{d}s\biggr|, \\ &\Biggl| \bigvee_{j=1}^n \beta_{ij} \int_{-\infty}^t {k_j}(t-s){f_j} \bigl({y_j}(s)+x_j^*\bigr)\, \mathrm{d}s-\bigvee _{j=1}^n \beta_{ij}\int _{-\infty}^t {k_j}(t-s){f_j} \bigl(x_j^*\bigr) \,\mathrm{d}s \Biggr| \\ &\quad \le\sum_{j=1}^n | \beta_{ij} |\times\biggl|\int_{-\infty }^t {k_j}(t- s){f_j}\bigl({y_j}(s)+x_j^* \bigr)\, \mathrm{d}s-\int_{-\infty}^t {k_j}(t-s){f_j}\bigl(x_j^*\bigr)\, \mathrm{d}s \biggr| \\ &\quad =\sum_{j=1}^n | \beta_{ij} |\times\biggl|\int_{-\infty}^t {k_j}(t- s){g_j}\bigl({y_j}(s)\bigr)\, \mathrm{d}s\biggr|. \end{aligned}$$
By applying Lemmas 1 and 4 we get the following inequalities for any positive diagonal matrices R
2,R
4:
$$\begin{aligned} &2\sum^n_{i=1}p_{im}y_i(t) \Biggl( \bigwedge_{j=1}^n \alpha_{ij} \int_{-\infty}^t {k_j}(t-s){f_j} \bigl({y_j}(s)+x_j^*\bigr) \mathrm{d}s \\ &\qquad -\bigwedge _{j=1}^n \alpha_{ij}\int _{-\infty}^t {k_j}(t-s){f_j} \bigl(x_j^*\bigr)\, \mathrm{d}s \\ & \qquad + \bigvee _{j=1}^n \beta_{ij}\int _{-\infty}^t {k_j}(t-s){f_j} \bigl({y_j}(s)+x_j^*\bigr) \,\mathrm{d}s-\bigvee _{j=1}^n \beta_{ij}\int _{-\infty}^t {k_j}(t-s){f_j} \bigl(x_j^*\bigr)\, \mathrm{d}s \Biggr) \\ &\quad \le2\sum^n_{i=1}p_{im}\bigl|{y_i}(t)\bigr| \Biggl( \Biggl| \bigwedge_{j=1}^n \alpha _{ij}\int_{-\infty}^t {k_j}(t-s){f_j} \bigl({y_j}(s)+x_j^*\bigr) \,\mathrm{d}s \\ &\qquad -\bigwedge_{j=1}^n \alpha_{ij} \int_{-\infty}^t {k_j}(t-s){f_j} \bigl(x_j^*\bigr) \,\mathrm{d}s \Biggr| + \Biggl|\bigvee_{j=1}^n \beta_{ij}\int_{-\infty}^t {k_j}(t-s){f_j}\bigl({y_j}(s)+x_j^* \bigr)\, \mathrm{d}s \\ &\qquad -\bigvee_{j=1}^n \beta_{ij}\int_{-\infty}^t {k_j}(t-s){f_j}\bigl(x_j^*\bigr) \,\mathrm{d}s\Biggr | \Biggr) \\ &\quad \le2\bigl|y(t)\bigr|^TP_m \bigl(|\alpha|+| \beta|\bigr) \biggl|\int_{-\infty}^t {k}(t- s){g}\bigl({y}(s)\bigr) \,\mathrm{d}s\biggr| \\ &\quad \le\bigl|y(t)\bigr|^TP_m \bigl(|\alpha|+| \beta|\bigr)R_2^{-1} \bigl(|\alpha|+ |\beta|\bigr)P_m\bigl|y(t)\bigr| \\ &\qquad +\biggl|\int_{-\infty}^t {k}(t- s){g} \bigl({y}(s)\bigr) \,\mathrm{d}s\biggr|^T R_2\biggl |\int_{-\infty}^t {k}(t- s){g}\bigl({y}(s)\bigr) \,\mathrm{d}s\biggr| \\ &\quad \le ny(t)^TP_m\varUpsilon R_2^{-1} \varUpsilon P_my(t)+ \zeta(t)^T\varpi_7^TR_2 \varpi_7\zeta(t), \end{aligned}$$
(23)
$$\begin{aligned} &{-}2\sum^n_{i=1}p_{im}d_i \int_{t-\sigma}^t {y_i}(s)\,\mathrm{d}s \Biggl( \bigwedge_{j=1}^n \alpha_{ij}\int _{-\infty}^t {k_j}(t-s){f_j} \bigl({y_j}(s)+x_j^*\bigr)\, \mathrm{d}s \\ &\qquad -\bigwedge _{j=1}^n \alpha_{ij}\int _{-\infty}^t {k_j}(t-s){f_j} \bigl(x_j^*\bigr)\, \mathrm{d}s + \bigvee _{j=1}^n \beta_{ij}\int _{-\infty}^t {k_j}(t-s){f_j} \bigl({y_j}(s)+x_j^*\bigr)\, \mathrm{d}s \\ &\qquad -\bigvee _{j=1}^n \beta_{ij}\int _{-\infty}^t {k_j}(t-s){f_j} \bigl(x_j^*\bigr)\, \mathrm{d}s \Biggr) \\ &\quad \le n \biggl(\int_{t-\sigma}^t {y}(s) \mathrm{d}s \biggr)^TP_mD\varUpsilon R_4^{-1} \varUpsilon DP_m\int_{t-\sigma}^t {y}(s) \mathrm{d}s+ \zeta(t)^T\varpi_7^TR_4 \varpi_7\zeta(t). \end{aligned}$$
(24)
When 0<τ(t)<τ
2, according to Lemma 2, we derive
$$\begin{aligned} &{-}{\tau_2}\int_{t-{\tau_2}}^t {y(s)^T{Q_5}y(s)}\,\mathrm{d}s \\ &\quad =-{\tau_2}\int_{t-\tau(t)}^t {y(s)^T{Q_5}y(s)}\,\mathrm{d}s-{\tau_2}\int _{t-{\tau_2}}^{t-\tau(t)} {y(s)^T{Q_5}y(s)}\, \mathrm{d}s \\ &\quad \leq\xi(t)^T \biggl[-\frac{\tau_2}{\tau(t)}\varpi_9^T{Q_5} \varpi_9 -\frac{\tau_2}{\tau_2-\tau(t)}\varpi_{11}^T{Q_5} \varpi_{11} \biggr]\xi(t) \\ &\quad =\xi(t)^T \biggl[-\varpi_9^T{Q_5} \varpi_9-\frac{{\tau_2}-\tau(t)}{\tau (t)}\varpi_9^T{Q_5} \varpi_9 \\ &\qquad -\varpi_{11}^T{Q_5} \varpi_{11}-\frac{\tau(t)}{\tau_2-\tau(t)}\varpi_{11}^T{Q_5} \varpi_{11} \biggr]\xi(t). \end{aligned}$$
(25)
Based on the reciprocal convex technique of [11, 22], from (7) we have
$$\xi(t)^T \begin{bmatrix} \sqrt{\frac{{\tau_2}-\tau(t)}{\tau(t)}}\varpi_9\\ -\sqrt{\frac{\tau(t)}{{\tau_2}-\tau(t)}}\varpi_{11} \end{bmatrix} ^T \begin{bmatrix} Q_5& X_9\\ *& Q_5 \end{bmatrix} \begin{bmatrix} \sqrt{\frac{{\tau_2}-\tau(t)}{\tau(t)}}\varpi_9\\ -\sqrt{\frac{\tau(t)}{{\tau_2}-\tau(t)}}\varpi_{11} \end{bmatrix} \xi(t)\ge0, $$
which implies
$$\begin{aligned} &\xi(t)^T \biggl[\frac{{\tau_2}-\tau(t)}{\tau(t)}\varpi_9^T{Q_5} \varpi_9 +\frac{\tau(t)}{\tau_2-\tau(t)}\varpi_{11}^T{Q_5} \varpi_{11} \biggr]\xi(t) \\ &\quad \geq\xi(t)^T \bigl(\varpi_9^TX_9 \varpi_{11}+\varpi_{11}^TX_9^T \varpi_9 \bigr)\xi(t). \end{aligned}$$
(26)
Then, we can get from (25) and (26) that
$$\begin{aligned} &{-}{\tau_2}\int_{t-{\tau_2}}^t {y(s)^T{Q_5}y(s)}\,\mathrm{d}s \\ &\quad \leq\xi(t)^T \bigl(-\varpi_9^T{Q_5} \varpi_9-\varpi_{11}^T{Q_5}\varpi _{11}-\varpi_9^TX_9 \varpi_{11} -\varpi_{11}^TX_9^T \varpi_9 \bigr)\xi(t) \\ &\quad =-\xi(t)^T \begin{bmatrix} \varpi_9\\ \varpi_{11} \end{bmatrix} ^T \begin{bmatrix} Q_5& X_9\\ * &Q_5 \end{bmatrix} \begin{bmatrix} \varpi_9\\ \varpi_{11} \end{bmatrix} \xi(t). \end{aligned}$$
(27)
Note that when τ(t)=0 or τ(t)=τ
2, we have ϖ
9
ξ(t)=0 or ϖ
11
ξ(t)=0, respectively. Thus, inequality (27) still holds.
On the other hand, from Lemma 2 we have
$$\begin{aligned} &\tau_2\int_{t-{\tau_2}}^t { \dot{y}(s)^TQ_6\dot{y}(s)} \,\mathrm{d}s \\ &\quad =\tau_2\int_{t-\tau(t)}^t { \dot{y}(s)^TQ_6\dot{y}(s)} \,\mathrm{d}s+\tau _2\int_{t-{\tau_2}}^{t-\tau(t)} {\dot{y}(s)^TQ_6 \dot{y}(s)} \,\mathrm{d}s \\ &\quad = \bigl[\tau(t)+\bigl({\tau_2}-\tau(t)\bigr) \bigr]\int _{t-\tau(t)}^t {\dot{y}(s)^TQ_6 \dot{y}(s)}\, \mathrm{d}s \\ &\qquad +\bigl[\bigl({\tau_2}-\tau(t)\bigr)+\tau(t) \bigr]\int _{t-{\tau_2}}^{t-\tau(t)} {\dot{y}(s)^TQ_6 \dot{y}(s)} \,\mathrm{d}s \\ &\quad \ge\biggl\{1+\frac{{{\tau_2}-\tau(t)}}{\tau_2} \biggr\}\times \tau(t)\int_{t-\tau(t)}^t {\dot{y}(s)^TQ_6\dot{y}(s)} \,\mathrm{d}s \\ &\qquad + \biggl\{1+\frac{{\tau(t)}}{\tau_2} \biggr\} \times\bigl({\tau _2}-\tau (t)\bigr)\int_{t-{\tau_2}}^{t-\tau(t)} {\dot{y}(s)^TQ_6 \dot{y}(s)}\, \mathrm{d}s \\ &\quad \ge\biggl\{1+\frac{{{\tau_2}-\tau(t)}}{\tau_2} \biggr\} \biggl (\int_{t-\tau (t)}^t \dot{y}(s)\,\mathrm{d}s \biggr)^TQ_6 \biggl(\int _{t-\tau(t)}^t \dot{y}(s)\,\mathrm{d}s \biggr) \\ &\qquad + \biggl\{1+\frac{{\tau(t)}}{\tau_2} \biggr\} \biggl(\int _{t-{\tau _2}}^{t-\tau(t)} \dot{y}(s)\,\mathrm{d}s \biggr)^TQ_6 \biggl(\int _{t-{\tau _2}}^{t-\tau(t)} \dot{y}(s)\,\mathrm{d}s \biggr) \\ &\quad = \biggl\{1+\frac{{{\tau_2}-\tau(t)}}{\tau_2} \biggr\} \bigl [y(t)-y\bigl(t-\tau(t)\bigr) \bigr]^TQ_6 \bigl[y(t)-y\bigl(t-\tau(t)\bigr) \bigr] \\ &\qquad + \biggl\{1+\frac{{\tau(t)}}{\tau_2} \biggr\} \bigl[y\bigl (t-\tau(t)\bigr)-y(t- \tau_2) \bigr]^TQ_6 \bigl[y\bigl(t-\tau(t) \bigr)-y(t-\tau_2) \bigr]. \end{aligned}$$
(28)
Similarly, we get that
$$\begin{aligned} &\tau_{12}\int_{t-{\tau_2}}^{t-{\tau_1}} { \dot{y}(s)^TQ_8\dot{y}(s)}\, \mathrm{d}s \\ &\quad \ge\biggl\{1+\frac{{{\tau_2}-\tau(t)}}{{ {\tau_2-\tau_1 } }} \biggr\} \bigl[y(t-\tau_1)-y \bigl(t-\tau(t)\bigr) \bigr]^TQ_8 \bigl[y(t- \tau_1)-y\bigl(t-\tau(t)\bigr) \bigr] \\ &\qquad + \biggl\{1+\frac{{\tau(t)-\tau_1}}{{ {\tau_2-\tau _1 } }} \biggr\} \bigl[y\bigl(t-\tau(t) \bigr) -y(t-\tau_2) \bigr]^TQ_8 \bigl[y \bigl(t-\tau(t)\bigr)-y(t-\tau_2) \bigr]. \end{aligned}$$
(29)
When τ
1<τ(t)<τ
2, from Lemma 2 we get
$$\begin{aligned} &{-}({\tau_2}-{\tau_1})\int_{t-{\tau_2}}^{t-{\tau_1}} {{y}(s)^T{Q_7}y(s)}\,\mathrm{d}s \\ &\quad =-({\tau_2}-{\tau_1})\int_{t-\tau(t)}^{t-{\tau_1}} {{y}(s)^T{Q_7}y(s)}\,\mathrm{d}s-({\tau_2}-{ \tau_1})\int_{t-{\tau _2}}^{t-\tau(t)} {{y}(s)^T{Q_7}y(s)}\,\mathrm{d}s \\ &\quad \leq\xi(t)^T \biggl[-\frac{{\tau_2}-{\tau_1}}{\tau(t)-{\tau _1}}\varpi_{10}^T{Q_7} \varpi_{10} -\frac{{\tau_2}-{\tau_1}}{\tau_2-\tau(t)}\varpi_{11}^T{Q_7} \varpi_{11} \biggr]\xi(t) \\ &\quad =\xi(t)^T \biggl[-\varpi_{10}^T{Q_7} \varpi_{10}-\frac{{\tau_2}-\tau (t)}{\tau(t)-{\tau_1}}\varpi_{10}^T{Q_7} \varpi_{10} \\ &\qquad -\varpi_{11}^T{Q_7} \varpi_{11}-\frac{\tau (t)-{\tau_1}}{\tau_2-\tau(t)}\varpi_{11}^T{Q_7} \varpi_{11} \biggr]\xi(t). \end{aligned}$$
(30)
In addition, from (8) we have
$$\xi(t)^T \begin{bmatrix} \sqrt{\frac{{\tau_2}-\tau(t)}{\tau(t)-\tau _1}}\varpi _{10}\\ -\sqrt{\frac{\tau(t)-\tau_1}{{\tau_2}-\tau(t)}}\varpi_{11} \end{bmatrix} ^T \begin{bmatrix} Q_7& X_{10}\\ *&Q_7 \end{bmatrix} \begin{bmatrix} \sqrt{\frac{{\tau_2}-\tau(t)}{\tau(t)-\tau _1}}\varpi _{10}\\ -\sqrt{\frac{\tau(t)-\tau_1}{{\tau_2}-\tau(t)}}\varpi_{11} \end{bmatrix} \xi(t)\ge0, $$
which implies
$$\begin{aligned} &\xi(t)^T \biggl[\frac{{\tau_2}-\tau(t)}{\tau(t)-{\tau_1}}\varpi _{10}^T{Q_7} \varpi_{10} +\frac{\tau(t)-{\tau_1}}{\tau_2-\tau(t)}\varpi_{11}^T{Q_7} \varpi_{11} \biggr]\xi(t) \\ &\quad \geq\xi(t)^T \bigl(\varpi_{10}^TX_{10} \varpi_{11}+\varpi_{11}^TX_{10}^T \varpi_{10} \bigr)\xi(t). \end{aligned}$$
(31)
Then, we can get from (30) and (31) that
$$\begin{aligned} & {-}({\tau_2}-{\tau_1})\int_{t-{\tau_2}}^{t-{\tau_1}} {{y}(s)^T{Q_7}y(s)}\,\mathrm{d}s \\ &\quad \leq\xi(t)^T \bigl(-\varpi_{10}^T{Q_7} \varpi_{10}-\varpi_{11}^T{Q_7} \varpi_{11}-\varpi_{10}^TX_{10} \varpi_{11} -\varpi_{11}^TX_{10}^T \varpi_{10} \bigr)\xi(t) \\ &\quad =-\xi(t)^T \begin{bmatrix} \varpi_{10}\\ \varpi_{11} \end{bmatrix} ^T \begin{bmatrix} Q_7& X_{10}\\ *&Q_7 \end{bmatrix} \begin{bmatrix} \varpi_{10}\\ \varpi_{11} \end{bmatrix} \xi(t). \end{aligned}$$
(32)
Note that when τ(t)=τ
1 or τ(t)=τ
2, we have ϖ
10
ξ(t)=0 or ϖ
11
ξ(t)=0, respectively. Thus, inequality (32) still holds.
It is easy to verify that
$$\begin{aligned} &\int_{t - \tau_2 }^t {\int_\theta^t {\dot{y}{{(s)}^T} {Q_9}\dot{y}(s)} } \,\mathrm{d}s\,\mathrm{d} \theta\\ &\quad = \int_{t - \tau_2 }^t { ( {\tau_2 - t + s} )} \dot{y}{(s)^T} {Q_9}\dot{y}(s)\,\mathrm{d}s \\ &\quad =\int_{t - \tau(t)}^t { ( {\tau_2 - t + s} )} \dot{y}{(s)^T} {Q_9}\dot{y}(s)\,\mathrm{d}s+\int _{t - \tau_2 }^{t - \tau(t)} { ( {\tau_2 - t + s} )} \dot{y}{(s)^T} {Q_9}\dot{y}(s)\,\mathrm{d}s. \end{aligned}$$
Considering the case 0<τ(t)≤τ
2, based on Lemmas 1 and 2, we get the following inequalities for any matrices X
1,X
2 of appropriate dimensions
$$\begin{aligned} &{-} \int_{t - \tau(t)}^t { ( {\tau_2 - t + s} )} \dot{y}{(s)^T} {Q_9}\dot{y}(s)\,\mathrm{d}s \\ &\quad = - \int_{t - \tau(t)}^t { \bigl[ {\tau(t) - t + s} \bigr]} \dot{y}{(s)^T} {Q_9}\dot{y}(s)\,\mathrm{d}s - \int _{t - \tau(t)}^t { \bigl[ {\tau_2- \tau(t)} \bigr]} \dot{y}{(s)^T} {Q_9}\dot{y}(s)\,\mathrm{d}s \\ &\quad \le\int_{t - \tau(t)}^t { \bigl[ {\tau(t) - t + s} \bigr]} \bigl\{ {\xi{(t)^T} {\mathcal{X}_1}Q_9^{ - 1} \mathcal{X}_1^T\xi(t) + 2\xi{(t)^T} { \mathcal{X}_1}\dot{y}(s)} \bigr\}\,\mathrm{d}s \\ &\qquad - \frac{{\tau_2 - \tau(t)}}{{\tau(t)}}{ \biggl( {\int_{t - \tau (t)}^t { \dot{y}(s)\,\mathrm{d}s} } \biggr)^T} {Q_9} \biggl( {\int _{t - \tau(t)}^t {\dot{y}(s)\,\mathrm{d}s} } \biggr) \\ &\quad \le\frac{1}{2}\tau{(t)^2}\xi{(t)^T} { \mathcal{X}_1}Q_9^{ - 1}\mathcal {X}_1^T\xi(t)+ 2\xi{(t)^T} { \mathcal{X}_1} \biggl[ {\tau(t)y(t) - \int_{t - \tau(t)}^t {y(s)\,\mathrm{d}s} } \biggr] \\ &\qquad - \frac{{\tau_2 - \tau(t)}}{{\tau_2 }} \bigl[ {y{(t)} - y\bigl(t-\tau(t)\bigr) } \bigr]^T{Q_9} \bigl[ {y(t) - y\bigl(t-\tau(t)\bigr) } \bigr]. \end{aligned}$$
(33)
It is easy to see that (33) holds for any t with τ(t)=0. Again by Lemma 1, the following inequalities hold for any matrices X
3,X
4 of appropriate dimensions:
$$\begin{aligned} &{-} \int_{t - \tau_2 }^{t - \tau(t)} { ( {\tau_2 - t + s} )} \dot{y}{(s)^T} {Q_9}\dot{y}(s)\,\mathrm{d}s \\ &\quad \le- \int_{t - \tau_2 }^{t - \tau(t)} {(\tau_2 } - t + s) \bigl\{ {\xi{(t)^T} {\mathcal{X}_2}Q_9^{ - 1}{ \mathcal{X}_2}\xi(t) + 2\xi{(t)^T} {\mathcal{X}_2} \dot{y}(s)} \bigr\}\,\mathrm{d}s \\ &\quad = \frac{1}{2}{ \bigl[ {\tau_2 - \tau(t)} \bigr]^2}\xi{(t)^T} {\mathcal{X}_2}Q_9^{ - 1}{ \mathcal{X}_2}\xi(t) \\ &\qquad + 2\xi{(t)^T} {\mathcal{X} _2} \biggl\{ { \bigl[ { \tau_2 - \tau(t)} \bigr]{y\bigl(t-\tau(t)\bigr) } - \int _{t - \tau_2 }^{t - \tau(t)} {y(s)\,\mathrm{d}s} } \biggr\}. \end{aligned}$$
(34)
Furthermore, we have
$$\begin{aligned} &\int_{t-\tau_2}^{t-\tau_1} {\int_\theta^t {\dot{y}{{(s)}^T} {Q_{10}}\dot{y}(s)} } \,\mathrm{d}s\,\mathrm{d} \theta\\ &\quad = \int_{t - \tau_2 }^{t-\tau_1} { ( {\tau_2 - t + s} )} \dot{y}{(s)^T} {Q_{10}}\dot{y}(s)\,\mathrm{d}s \\ &\quad =\int_{t - \tau(t)}^{t-\tau_1} { ( {\tau_2 - t + s} )} \dot{y}{(s)^T} {Q_{10}}\dot{y}(s)\,\mathrm{d}s+\int _{t - \tau_2 }^{t - \tau(t)} { ( {\tau_2 - t + s} )} \dot{y}{(s)^T} {Q_{10}}\dot{y}(s)\,\mathrm{d}s. \end{aligned}$$
When τ
1<τ(t)≤τ
2, based on Lemmas 1 and 2, the following inequalities hold for any matrices X
5,X
6 of appropriate dimensions
$$\begin{aligned} &{-} \int_{t - \tau(t)}^{t-\tau_1} { ( {\tau_2 - t + s} )} \dot{y}{(s)^T} {Q_{10}}\dot{y}(s)\,\mathrm{d}s \\ &\quad = - \int_{t - \tau(t)}^{t-\tau_1} { \bigl[ {\tau(t) - t + s} \bigr]} \dot{y}{(s)^T} {Q_{10}}\dot{y}(s)\,\mathrm{d}s - \int _{t - \tau(t)}^{t-\tau _1} { \bigl[ {\tau_2 - \tau(t)} \bigr]} \dot{y}{(s)^T} {Q_{10}}\dot{y}(s)\,\mathrm{d}s \\ &\quad \le\int_{t - \tau(t)}^{t-\tau_1} { \bigl[ {\tau(t) - t + s} \bigr]} \bigl\{ {\xi{(t)^T} {\mathcal{X}_3}Q_{10}^{ - 1} \mathcal{X}_3^T\xi(t) + 2\xi{(t)^T} { \mathcal{X}_3}\dot{y}(s)} \bigr\}\,\mathrm{d}s \\ &\qquad - \frac{{\tau_2 - \tau(t)}}{{\tau(t)-\tau_1}}{ \biggl( {\int_{t - \tau (t)}^{t-\tau_1} { \dot{y}(s)\,\mathrm{d}s} } \biggr)^T} {Q_{10}} \biggl( {\int _{t - \tau(t)}^{t-\tau_1} {\dot{y}(s)\,\mathrm{d}s} } \biggr) \\ &\quad \le\frac{1}{2} \bigl[\tau(t)-\tau_1 \bigr]^2 \xi{(t)^T} {\mathcal{X}_3}Q_{10}^{ - 1} \mathcal{X}_3^T\xi(t)+ 2\xi{(t)^T} {\mathcal {X}_3} \biggl[ {\tau(t)y(t) - \int_{t - \tau(t)}^{t-\tau_1} {y(s)\,\mathrm{d}s} } \biggr] \\ &\qquad - \frac{{\tau_2 - \tau(t)}}{\tau_2-\tau_1} \bigl[ {y{{(t-\tau_1)}} - y \bigl(t-\tau(t)\bigr) } \bigr]^T{Q_{10}} \bigl[ {y(t- \tau_1) - y\bigl(t-\tau(t)\bigr) } \bigr]. \end{aligned}$$
(35)
It is easy to see that (35) holds for any t with τ(t)=0. Again by Lemma 1, the following inequalities hold for any matrices X
7,X
8 of appropriate dimensions
$$\begin{aligned} &{-} \int_{t - \tau_2 }^{t - \tau(t)} { ( {\tau_2 - t + s} )} \dot{y}{(s)^T} {Q_{10}}\dot{y}(s)\,\mathrm{d}s \\ &\quad \le- \int_{t - \tau_2 }^{t - \tau(t)} {(\tau_2 } - t + s) \bigl\{ {\xi{(t)^T} {\mathcal{X}_4}Q_{10}^{ - 1}{ \mathcal{X}_4}\xi(t) + 2\xi{(t)^T} {\mathcal{X}_4} \dot{y}(s)} \bigr\}\,\mathrm{d}s \\ &\quad = \frac{1}{2}{ \bigl[ {\tau_2 - \tau(t)} \bigr]^2}\xi{(t)^T} {\mathcal{X}_4}Q_{10}^{ - 1}{ \mathcal{X}_4}\xi(t) \\ &\qquad + 2\xi{(t)^T} {\mathcal{X} _4} \biggl\{ { \bigl[ { \tau_2 - \tau(t)} \bigr]{y\bigl(t-\tau(t)\bigr) } - \int _{t - \tau_2 }^{t - \tau(t)} {y(s)\,\mathrm{d}s} } \biggr\}. \end{aligned}$$
(36)
Using Lemma 2 again, we obtain that
$$\begin{aligned} &\sigma\int_{t-\sigma}^t {{y}(s)^TQ_{11}y(s)}\, \mathrm{d}s \ge{ \biggl( {\int_{t-\sigma}^t {y(s)}\, \mathrm{d}s} \biggr)^T}Q_{11} \biggl( {\int _{t- \sigma}^t {y(s)} \,\mathrm{d}s} \biggr), \\ & \begin{aligned}[t] \sigma\int_{t-\sigma}^t {\dot{y}(s)^TQ_{12} \dot{y}(s)} \,\mathrm{d}s &\ge{ \biggl( {\int_{t-\sigma}^t {\dot{y}(s)} \,\mathrm{d}s} \biggr)^T}Q_{12} \biggl( {\int _{t- \sigma}^t {\dot{y}(s)}\, \mathrm{d}s} \biggr) \\ & ={ \bigl[ y(t)-y(t-\sigma) \bigr]^T}Q_{12} \bigl[ y(t)-y(t-\sigma) \bigr]. \end{aligned} \end{aligned}$$
(37)
By the Cauchy–Schwarz inequality and Eq. (3), the following inequality holds:
$$\begin{aligned} &\sum_{j=1}^n {r_{1j}\int _0^\infty{{k_j}(\theta)g_j^2 \bigl({y_j}(t )\bigr)} } \,\mathrm{d}\theta-\sum _{j= 1}^n {r_{1j}\int_0^\infty {{k_j}(\theta)g_j^2\bigl({y_j}(t- \theta)\bigr)} } \,\mathrm{d}\theta \\ &\quad = {g}\bigl(y(t)\bigr)^TR_1g\bigl(y(t)\bigr)-\sum _{j=1}^n {r_{1j}\int _0^\infty{{k_j}(\theta)\,\mathrm{d} \theta} \int_0^\infty{{k_j}(\theta )g_j^2\bigl({y_j}(t- \theta)\bigr)} }\, \mathrm{d}\theta \\ &\quad \le{g}\bigl(y(t)\bigr)^TR_1g\bigl(y(t)\bigr)-\sum _{j=1}^n {r_{1j}{{ \biggl( {\int _0^\infty{{k_j}(\theta){g_j} \bigl({y_j}(t-\theta)\bigr)\,\mathrm{d}\theta} } \biggr)}^2}} \\ &\quad = {g}\bigl(y(t)\bigr)^TR_1g\bigl(y(t)\bigr)- \biggl( { \int_{-\infty}^t {k(s)g\bigl(y(s)\bigr)\,\mathrm{d}s} } \biggr)R_1 \biggl( {\int_{-\infty}^t {k(s)g\bigl(y(s)\bigr)\,\mathrm{d}s} } \biggr). \end{aligned}$$
(38)
Moreover, based on (H2), the following matrix inequalities hold for any positive diagonal matrices \(U=\operatorname{diag}\{ u_{1},u_{2},\dots, u_{n}\}\) and \(W=\operatorname{diag}\{ w_{1},w_{2},\dots, w_{n}\}\):
$$\begin{aligned} 0& \le\sum_{i=1}^n {-{u_{i}} \bigl( {g_i^2\bigl({y_i}(t) \bigr)-l_i^2y_i^2(t)} \bigr)} \\ &=\xi(t)^T \bigl(-\varpi_{5}^T{U} \varpi_{5}+\varpi_1^TL{U}L \varpi_1 \bigr)\xi(t), \end{aligned}$$
(39)
$$\begin{aligned} 0& \le\sum_{i=1}^n {-{w_{i}} \bigl( {g_i^2\bigl({y_i}\bigl(t-\tau(t) \bigr)\bigr)-l_i^2y_i^2\bigl(t- \tau(t)\bigr)} \bigr)} \\ &=\xi(t)^T \bigl(-\varpi_6^T{W} \varpi_6+{\varpi_2^T}L{W}L \varpi_2 \bigr)\xi(t). \end{aligned}$$
(40)
Again by utilizing Lemmas 1 and 4 we get the following inequality with any positive diagonal matrices R
3 and \(S=\operatorname{diag}\{ s_{1}, s_{2},\dots, s_{n}\}\):
$$\begin{aligned} 0 &= 2\sum_{i=1}^n \dot{y}_{i}(t)^Ts_{i} \Biggl[-\dot{y}_{i}(t)-{d_i} {y_i}(t-\sigma)+ \sum_{j=1}^n a_{ij}{g_j} \bigl({y_j}(t)\bigr) \\ &\quad +\sum_{j=1}^n b_{ij}{g_j}\bigl({y_j}\bigl(t-\tau(t)\bigr) \bigr) \\ &\quad + \bigwedge_{j=1}^n \alpha_{ij}\int_{-\infty}^t {k_j}(t- s){f_j}\bigl({y_j}(s)+x_j^* \bigr) \,\mathrm{d}s- \bigwedge_{j= 1}^n \alpha_{ij}\int_{-\infty}^t {k_j}(t-s){f_j}\bigl(x_j^*\bigr)\, \mathrm{d}s \\ &\quad +\bigvee_{j=1}^n \beta_{ij}\int_{-\infty}^t {k_j}(t-s){f_j}\bigl({y_j}(s)+x_j^* \bigr)\,\mathrm{d}s -\bigvee_{j=1}^n \beta _{ij}\int_{-\infty}^t {k_j}(t-s){f_j} \bigl(x_j^*\bigr)\,\mathrm{d}s \Biggr] \\ &\le2\xi(t)^T \bigl\{{\varpi_8^T} {S} (- \varpi_8-D\varpi_{13}+{A}\varpi_{5}+{B} \varpi_{6} ) \\ &\quad +n\varpi_8^T{S}\varUpsilon R_3^{-1}\varUpsilon{S}\varpi_8+ \varpi _7^T{R_3}\varpi_7 \bigr\}\xi(t). \end{aligned}$$
(41)
Substituting (17)–(41) into (16), we derive
$$ {\rm D}^+{V}\bigl(t,y(t)\bigr)\leq\xi(t)^T (\varLambda+ \varLambda_R+\varLambda_\tau) \xi(t),\quad t \in[{t_{k- 1}},{t_k}),k \in{\mathbf{Z}_+}, $$
(42)
where
$$\begin{aligned} \varLambda_R&=n\varpi_1^TP\varUpsilon R_2^{-1}\varUpsilon P\varpi_1+n{\varpi _8^T} {S}\varUpsilon R_3^{-1} \varUpsilon{S}\varpi_8+n\varpi_{12}^TPD\varUpsilon R_4^{-1}\varUpsilon DP\varpi_{12}, \\ \varLambda_\tau&=\frac{1}{2}\tau{(t)^2} { \mathcal{X}_1}Q_9^{ - 1}\mathcal {X}_1^T+ \tau(t)\operatorname{sym} \bigl[( \mathcal{X}_1 +\mathcal{X}_3)\varpi_1 \bigr]+ \frac{1}{2}{ \bigl[ {\tau_2 - \tau(t)} \bigr]^2} {\mathcal{X}_2}Q_9^{ - 1}{ \mathcal{X}_2} \\ &\quad - \frac{{\tau_2 - \tau(t)}}{\tau_2}(\varpi_1-\varpi_2)^T(Q_6+Q_9) (\varpi_1-\varpi_2) \\ &\quad + \bigl[ {\tau_2 - \tau(t)} \bigr]\operatorname{sym} \bigl[({\mathcal{X} _2}+{\mathcal{X} _4})\varpi_2 \bigr] \\ &\quad + \frac{1}{2} \bigl[\tau(t)-\tau_1 \bigr]^2{ \mathcal{X}_3}Q_{10}^{-1}\mathcal{X}_3^T- \frac{{\tau_2-\tau(t)}}{\tau_2-\tau_1} (\varpi_3-\varpi_2)^T(Q_8+Q_{10}) (\varpi_3-\varpi_2) \\ &\quad + \frac{1}{2}{ \bigl[ {\tau_2 - \tau(t)} \bigr]^2} {\mathcal{X}_4}Q_{10}^{ - 1}{ \mathcal{X}_4} \\ &\quad - (\varpi_2-\varpi_4)^T \biggl[\frac{{\tau(t)}}{\tau_2}Q_6+\frac{{\tau (t)-\tau_1}}{{ {\tau_2-\tau_1 } }}Q_8 \biggr](\varpi_2-\varpi_4). \end{aligned}$$
Since the coefficient
$$\frac{1}{2} \bigl({\mathcal{X}_1}Q_9^{ - 1} \mathcal{X}_1^T+{\mathcal{X}_2}Q_9^{ - 1}{ \mathcal{X}_2} +{\mathcal{X}_3}Q_{10}^{ - 1} \mathcal{X}_3^T+{\mathcal{X}_4}Q_{10}^{ - 1}{ \mathcal{X}_4} \bigr) $$
of τ(t)2 in Λ
τ
is nonnegative definite, based on Lemma 6, we have that Λ+Λ
R
+Λ
τ
<0 if and only if the following two inequalities hold simultaneously:
$$\begin{aligned} &\varLambda+\varLambda_R+\frac{1}{2} {\tau_2^2} {\mathcal{X}_1}Q_9^{ - 1}\mathcal{X}_1^T+ \tau_2\operatorname{sym} \bigl[(\mathcal{X}_1+\mathcal {X}_3)\varpi_1 \bigr] \\ &\quad + \frac{1}{2} [\tau_2-\tau_1 ]^2{\mathcal{X}_3}Q_{10}^{-1} \mathcal{X}_3^T - (\varpi_2- \varpi_4)^T(Q_6+Q_8) (\varpi _2-\varpi_4)<0, \end{aligned}$$
(43)
$$\begin{aligned} & \varLambda+\varLambda_R+\frac{1}{2} {\tau_1^2} {\mathcal{X}_1}Q_9^{ - 1}\mathcal{X}_1^T+ \tau_1\operatorname{sym} \bigl[(\mathcal{X}_1+\mathcal {X}_3)\varpi_1 \bigr] \\ &\quad + \frac{1}{2}{ [ {\tau_2-\tau_1} ]^2} \bigl({\mathcal{X}_2}Q_9^{-1}{ \mathcal{X}_2}+{\mathcal{X}_4}Q_{10}^{ - 1}{ \mathcal{X}_4} \bigr) \\ &\quad - \frac{\tau_2-\tau_1}{\tau_2}(\varpi_1-\varpi_2)^T(Q_6+Q_9) (\varpi_1-\varpi_2) + ( {\tau_2- \tau_1} )\operatorname{sym} \bigl[({\mathcal{X} _2}+{ \mathcal{X} _4})\varpi_2 \bigr] \\ &\quad -(\varpi_3-\varpi_2)^T(Q_8+Q_{10}) (\varpi_3-\varpi_2)-\frac{\tau_1}{\tau_2} ( \varpi_2-\varpi_4)^TQ_6(\varpi _2-\varpi_4)<0. \end{aligned}$$
(44)
From the well-known Schur complement, we deduce that inequalities (43) and (44) are equivalent to inequalities (9) and (10), respectively. Therefore, if inequalities (9) and (10) hold, then from (42) we derive that
$$\mathrm{D}^+{V}\bigl(t,y(t)\bigr)<0\quad \forall t\in[t_{k - 1},t_k), \ k \in\mathbf{Z}_+. $$
When t=t
k
, k∈Z
+, from condition (H5) we have
$$ V\bigl(t_k,y(t_k)\bigr)={V} \bigl({t_k^-},y\bigl({t_k^-}\bigr)\bigr)+ y \bigl({t_k^-}\bigr)^T \bigl[{(I - {\varGamma _k})^T}P(I - {\varGamma_k})-P \bigr]y \bigl({t_k^-}\bigr). $$
(45)
On the other hand, it follows from (6) that
$$ \begin{pmatrix} I & 0 \\ 0 & P^{ - 1} \end{pmatrix} \begin{pmatrix} P & (I - \varGamma_k)P \\ *& P \\ \end{pmatrix} \begin{pmatrix} I & 0 \\ 0 & P^{ - 1} \end{pmatrix} \ge0, $$
that is,
$$ \begin{pmatrix} P & I - {\varGamma_k} \\ * & P^{ - 1} \end{pmatrix} \ge0. $$
From the Schur complement we have
$$ P - {(I - {\varGamma_k})^T}P(I - { \varGamma_k}) \ge0. $$
(46)
By combining (45) with (46) we obtain
$${V}\bigl({t_k},y({t_k})\bigr)\le{V} \bigl({t_k^-},y\bigl({t_k^-}\bigr)\bigr),\quad k \in{ \mathbf{Z}_ + }. $$
Therefore, system (5) is asymptotically stable, which implies that the equilibrium point of model (2) is globally asymptotically stable. This ends the proof of Theorem 1.
