First in this section we will give some notations as follows:
$$\begin{aligned}& u^{*}= \frac{r_{1}^{u}}{D_{1}^{l}}\exp\bigl\{ r_{1}^{u} \tau_{1}\bigr\} , \qquad u_{*}=\frac{r^{l}_{1}B^{l}-C_{1}^{u}}{B^{l}D_{1}^{u}}\exp \biggl\{ \biggl( \frac {r^{l}_{1}B^{l}-C_{1}^{u}}{B^{l}}-D_{1}^{u} u^{*} \biggr)\tau_{1} \biggr\} , \\& v^{*}=\frac{A^{l}r_{2}^{u}+C_{2}^{u}}{A^{l}D_{2}^{l}}\exp \biggl\{ {\frac {A^{l}r_{2}^{u}+C_{2}^{u}}{A^{l}}\tau_{2}} \biggr\} , \qquad v_{*}=\frac {r_{2}^{l}}{D_{2}^{u}}\exp \bigl\{ \bigl(r_{2}^{l}-D_{2}^{u}v^{*} \bigr)\tau_{2} \bigr\} . \end{aligned}$$
Theorem 3.1
Assume that (C1) and (C2) hold, suppose further that
-
(C3)
\(C^{u}_{1}< B^{l}r^{l}_{1}\),
then any positive solution
\((u(t),v(t))^{T}\)
of system (6) satisfies
$$\begin{aligned}& u_{*}\leq \liminf_{t\rightarrow+\infty} u(t)\leq \limsup_{t\rightarrow+\infty} u(t)\leq u^{*}, \\& v_{*}\leq \liminf_{t\rightarrow+\infty} v(t)\leq \limsup_{t\rightarrow+\infty} v(t)\leq v^{*}. \end{aligned}$$
Proof
Let \((u(t),v(t))^{T}\) be any solution of system (6), then from the first equation of (6),
$$u'(t)\leq u(t) \bigl[r^{u}_{1}-D^{l}_{1}u(t- \tau_{1}) \bigr]. $$
By the first conclusion of Lemma 2.3, we have
$$ \limsup_{t\rightarrow+\infty} u(t)\leq \frac{r_{1}^{u}}{D_{1}^{l}}\exp\bigl\{ r_{1}^{u}\tau_{1}\bigr\} =u^{*}. $$
(19)
Similarly, by the second equation of system (6),
$$v'(t)\leq v(t) \biggl[r^{u}_{2}-D^{l}_{2}v(t- \tau_{2})+\frac{C^{u}_{2}u(t-\tau _{3})}{1+A^{l}u(t-\tau_{3})} \biggr]\leq v(t) \biggl[ \biggl(r^{u}_{2}+\frac {C^{u}_{2}}{A^{l}}\biggr)-D^{l}_{2}v(t- \tau_{2}) \biggr] , $$
which leads to
$$ \limsup_{t\rightarrow+\infty} v(t)\leq\frac {A^{l}r_{2}^{u}+C_{2}^{u}}{A^{l}D_{2}^{l}}\exp \biggl\{ { \frac{A^{l}r_{2}^{u}+C_{2}^{u}}{A^{l}}\tau _{2}} \biggr\} =v^{*}. $$
(20)
Therefore, there exists a \(T_{1}>0\), such that \(v(t)\leq v^{*}\) when \(t>T_{1}\).
Thus, when \(t>T_{1}+\tau\), from the first equation of system (6) again we have
$$u'(t)\geq u(t) \biggl[r^{l}_{1}-D^{u}_{1}u(t- \tau_{1})-\frac {C^{u}_{1}v^{*}}{1+B^{l}v^{*}} \biggr]\geq u(t) \biggl[ \biggl(r^{l}_{1}-\frac {C^{u}_{1}}{B^{l}}\biggr)-D^{u}_{1}u(t- \tau_{1}) \biggr]. $$
Then by the second conclusion of Lemma 2.3, we have
$$ \liminf_{t\rightarrow+\infty} u(t)\geq\frac {r^{l}_{1}B^{l}-C_{1}^{u}}{B^{l}D_{1}^{u}}\exp \biggl\{ \biggl( \frac {r^{l}_{1}B^{l}-C_{1}^{u}}{B^{l}}-D_{1}^{u} u^{*} \biggr)\tau_{1} \biggr\} =u_{*}. $$
(21)
On the other hand, from the second equation of system (6) again, we can get
$$v(t)\geq v(t) \bigl[r^{l}_{2}-D^{u}_{2}v(t- \tau_{2}) \bigr]. $$
Then, by the second conclusion of Lemma 2.3 again, we have
$$ \liminf_{t\rightarrow+\infty} v(t)\geq\frac{r_{2}^{l}}{D_{2}^{u}}\exp \bigl\{ \bigl(r_{2}^{l}-D_{2}^{u}v^{*}\bigr) \tau_{2} \bigr\} =v_{*}. $$
(22)
Thus, we complete the proof of this theorem by combining (19), (20), (21), and (22). □
Theorem 3.2
Assume that (C1)-(C3) hold, then any positive solution
\((x(t),y(t))^{T}\)
of system (4) satisfies
$$\begin{aligned}& H^{l}_{1}u_{*}\leq \liminf_{t\rightarrow+\infty} x(t)\leq \limsup_{t\rightarrow+\infty} x(t)\leq H^{u}_{1}u^{*}, \\& H^{l}_{2}v_{*}\leq \liminf_{t\rightarrow+\infty} y(t)\leq \limsup_{t\rightarrow+\infty} y(t)\leq H^{u}_{2}v^{*}. \end{aligned}$$
Proof
Since \((x(t),y(t))^{T}\) is a solution of system (4), then by the second conclusion of Lemma 2.2,
$$ \bigl(u(t),v(t)\bigr)^{T}= \biggl( \frac{x(t)}{H_{1}(t)}, \frac{y(t)}{H_{2}(t)} \biggr)^{T} $$
(23)
is a solution of system (6).
Then it follows from Theorem 3.1 that
$$\begin{aligned}& u_{*}\leq \liminf_{t\rightarrow+\infty} \frac {x(t)}{H_{1}(t)}\leq \liminf _{t\rightarrow+\infty} u(t)\leq \limsup_{t\rightarrow+\infty} u(t)=\limsup _{t\rightarrow+\infty} \frac{x(t)}{H_{1}(t)}\leq u^{*}, \end{aligned}$$
(24)
$$\begin{aligned}& v_{*}\leq \liminf_{t\rightarrow+\infty} \frac {y(t)}{H_{2}(t)}\leq \liminf _{t\rightarrow+\infty} v(t)\leq \limsup_{t\rightarrow+\infty} v(t)=\limsup _{t\rightarrow+\infty} \frac{y(t)}{H_{2}(t)}\leq v^{*}, \end{aligned}$$
(25)
which implies that
$$\begin{aligned}& H^{l}_{1}u_{*}\leq \liminf_{t\rightarrow+\infty} x(t)\leq \limsup_{t\rightarrow+\infty} x(t)\leq H^{u}_{1}u^{*}, \end{aligned}$$
(26)
$$\begin{aligned}& H^{l}_{2}v_{*}\leq \liminf_{t\rightarrow+\infty} y(t)\leq \limsup_{t\rightarrow+\infty} y(t)\leq H^{u}_{2}v^{*}. \end{aligned}$$
(27)
This completes the proof of this theorem. □
Remark 3.1
Suppose that (C1)-(C3) hold, then system (6) is permanent.
Before we discuss the existence and uniformly asymptotically stability of a unique almost periodic solution of system (4), we will introduce the following notations first:
$$\begin{aligned}& \alpha_{1}=K_{4}\tau_{2}+K_{5} \tau_{4}; \qquad \beta_{1}=D^{l}_{1}u_{*}- \sqrt {K_{1}}-(K_{2}+K_{3})\tau_{1}-(K_{1}+K_{3}) \tau_{3}; \\& \alpha_{2}=D^{l}_{2}v_{*}+P-(\tau_{2}+ \tau_{4})L_{3}-L_{4}\tau_{2}-L_{5} \tau_{4}; \qquad \beta _{2}=L_{1} \tau_{1}+L_{2}\tau_{3}, \end{aligned}$$
where
$$\begin{aligned}& K_{1}= \biggl[\frac{A^{u}C^{u}_{1}u^{*}v^{*}}{M^{2}N} \biggr]^{2},\qquad K_{2}=\bigl(D^{u}_{1}u^{*}\bigr)^{2} ; \\& K_{3}=\frac{A^{u}C^{u}_{1}D^{u}_{1}(u^{*})^{2}v^{*}}{M^{2}N},\qquad K_{4}=\frac {C^{u}_{2}D^{u}_{2}u^{*}v^{*}}{M^{2}N}, \qquad K_{5}=\frac{B^{u}(C^{u}_{2}u^{*})^{2}v^{*}}{M^{3}N^{3}} ; \\& L_{1}=\frac{C^{u}_{1}D^{u}_{1}u^{*}v^{*}}{MN^{2}}, \qquad L_{2}=\frac {A^{u}(C^{u}_{1}v^{*})^{2}u^{*}}{M^{3}N^{3}}, \qquad L_{3}=\frac{B^{u}C^{u}_{2}D^{u}_{2}u^{*}(v^{*})^{2}}{MN^{2}} ; \\& L_{4}=\bigl(D^{u}_{2}v^{*}\bigr)^{2}, \qquad L_{5}= \biggl[\frac {B^{u}C^{u}_{2}u^{*}v^{*}}{MN^{2}} \biggr]^{2}, \qquad M_{1}=\frac {C^{u}_{2}u^{*}}{M^{2}N},\qquad M_{2}=\frac{C^{u}_{1}v^{*}}{MN^{2}} ; \\& M=1+A^{l}u_{*}, \qquad N=1+B^{l}v_{*},\qquad P= \frac {B^{l}C^{l}_{2}u_{*}v_{*}}{(1+A^{u}u^{*})(1+B^{u}v^{*})^{2}} . \end{aligned}$$
Theorem 3.3
Assume that (C1)-(C3) hold, further assume that
-
(C4)
positive
\(\lambda_{1}\)
and
\(\lambda_{2}\)
exist, such that
\(\frac{\alpha_{1}}{\beta_{1}}<\frac{\lambda_{1}}{\lambda _{2}}<\frac{\alpha_{2}}{\beta_{2}}\),
then system (4) admits a unique almost periodic solution which is uniformly asymptotically stable.
Proof
At first, we prove that system (6) has a unique uniformly asymptotically stable almost periodic solution.
In order to achieve this aim, we take a transformation
$$ u(t)=e^{x_{1}(t)}, \qquad v(t)=e^{y_{1}(t)},\quad t\in R^{+}. $$
(28)
Then system (6) is transformed into following system:
$$ \left \{ \textstyle\begin{array}{l} x'_{1}(t)=r_{1}(t)-D_{1}(t)e^{x_{1}(t-\tau_{1})}-\frac {C_{1}(t)e^{y_{1}(t-\tau_{4})}}{(1+A(t)e^{x_{1}(t-\tau _{3})})(1+B(t)e^{y_{1}(t-\tau_{4})})}, \\ y'_{1}(t)=r_{2}(t)-D_{2}(t)e^{y_{1}(t-\tau_{2})}+\frac {C_{2}(t)e^{x_{1}(t-\tau_{3})}}{(1+A(t)e^{x_{1}(t-\tau _{3})})(1+B(t)e^{y_{1}(t-\tau_{4})})}. \end{array}\displaystyle \right . $$
(29)
Suppose that \(U_{1}(t)= (x_{1}(t),y_{1}(t) )^{T}\) and \(U_{2}(t)= (x_{2}(t),y_{2}(t) )^{T}\) are any two solutions of system (29), then the product system of (29) reads
$$ \left \{ \textstyle\begin{array}{l} x'_{1}(t)=r_{1}(t)-D_{1}(t)e^{x_{1}(t-\tau_{1})}-\frac {C_{1}(t)e^{y_{1}(t-\tau_{4})}}{(1+A(t)e^{x_{1}(t-\tau _{3})})(1+B(t)e^{y_{1}(t-\tau_{4})})}, \\ y'_{1}(t)=r_{2}(t)-D_{2}(t)e^{y_{1}(t-\tau_{2})}+\frac {C_{2}(t)e^{x_{1}(t-\tau_{3})}}{(1+A(t)e^{x_{1}(t-\tau _{3})})(1+B(t)e^{y_{1}(t-\tau_{4})})}, \\ x'_{2}(t)=r_{1}(t)-D_{1}(t)e^{x_{2}(t-\tau_{1})}-\frac {C_{1}(t)e^{y_{2}(t-\tau_{4})}}{(1+A(t)e^{x_{2}(t-\tau _{3})})(1+B(t)e^{y_{2}(t-\tau_{4})})}, \\ y'_{2}(t)=r_{2}(t)-D_{2}(t)e^{y_{2}(t-\tau_{2})}+\frac {C_{2}(t)e^{x_{2}(t-\tau_{3})}}{(1+A(t)e^{x_{2}(t-\tau _{3})})(1+B(t)e^{y_{2}(t-\tau_{4})})}. \end{array}\displaystyle \right . $$
(30)
Denote
$$S^{*}= \bigl\{ \phi=(x_{1t},y_{1t})^{T}\in C\bigl([- \tau,0],R^{2}\bigr)|\ln u_{*}\leq x_{1t}\leq \ln u^{*},\ln v_{*}\leq y_{1t}\leq \ln v^{*}, t\in R^{+} \bigr\} . $$
For any \(\Phi=(\phi_{1},\phi_{2})^{T}=(x_{1t},y_{1t})^{T}\in S^{*}\), we can choose \(\Psi=(\psi_{1},\psi_{2})^{T}=(x_{2t},y_{2t})^{T}\in S^{*}\) such that
$$ \bigl\vert \Phi(0)-\Psi(0)\bigr\vert _{0}=\bigl\vert \phi_{1}(0)-\psi_{1}(0)\bigr\vert +\bigl\vert \phi_{1}(0)-\psi _{1}(0)\bigr\vert >0. $$
(31)
Consider a Lyapunov functional \(V(t)=V(t,\Phi,\Psi )=V(t,(x_{1t},y_{1t})^{T},(x_{2t},y_{2t})^{T})\) defined on \(R^{+}\times S^{*}\times S^{*}\) as follows:
$$ V(t)=V_{1}(t)+V_{2}(t)+V_{3}(t)+V_{4}(t), $$
(32)
where
$$\begin{aligned}& V_{1}(t)=\lambda_{1}\bigl\vert x_{1}(t)-x_{2}(t) \bigr\vert +\lambda_{2}\bigl\vert y_{1}(t)-y_{2}(t) \bigr\vert ; \end{aligned}$$
(33)
$$\begin{aligned}& V_{2}(t) =\lambda_{1}K_{1} \int^{-\tau_{3}}_{-2\tau_{3}} \int ^{t}_{t+s}\bigl\vert x_{1}(r)-x_{2}(r) \bigr\vert \,dr\,ds \\& \hphantom{V_{2}(t) ={}}{}+\lambda_{1}K_{2} \int^{-\tau_{1}}_{-2\tau_{1}} \int ^{t}_{t+s}\bigl\vert x_{1}(r)-x_{2}(r) \bigr\vert \,dr\,ds \\& \hphantom{V_{2}(t) ={}}{} +\lambda_{1}K_{3} \int^{-\tau_{1}}_{-\tau_{1}-\tau_{3}} \int ^{t}_{t+s}\bigl\vert x_{1}(r)-x_{2}(r) \bigr\vert \,dr\,ds \\& \hphantom{V_{2}(t) ={}}{} +\lambda_{1}K_{3} \int^{-\tau_{3}}_{-\tau_{1}-\tau_{3}} \int ^{t}_{t+s}\bigl\vert x_{1}(r)-x_{2}(r) \bigr\vert \,dr\,ds \\& \hphantom{V_{2}(t) ={}}{} +\lambda_{2}K_{4} \int^{-\tau_{3}}_{-\tau_{2}-\tau_{3}} \int ^{t}_{t+s}\bigl\vert x_{1}(r)-x_{2}(r) \bigr\vert \,dr\,ds \\& \hphantom{V_{2}(t) ={}}{} +\lambda_{2}K_{5} \int^{-\tau_{3}}_{-\tau_{3}-\tau_{4}} \int ^{t}_{t+s}\bigl\vert x_{1}(r)-x_{2}(r) \bigr\vert \,dr\,ds; \end{aligned}$$
(34)
$$\begin{aligned}& V_{3}(t) =\lambda_{1}L_{1} \int^{-\tau_{4}}_{-\tau_{1}-\tau_{4}} \int ^{t}_{t+s}\bigl\vert y_{1}(r)-y_{2}(r) \bigr\vert \,dr\,ds \\& \hphantom{V_{3}(t) ={}}{}+\lambda_{1}L_{2} \int^{-\tau_{4}}_{-\tau_{3}-\tau_{4}} \int ^{t}_{t+s}\bigl\vert y_{1}(r)-y_{2}(r) \bigr\vert \,dr\,ds \\& \hphantom{V_{3}(t) ={}}{} +\lambda_{2}L_{3} \int^{-\tau_{2}}_{-\tau_{2}-\tau_{4}} \int ^{t}_{t+s}\bigl\vert y_{1}(r)-y_{2}(r) \bigr\vert \,dr\,ds \\& \hphantom{V_{3}(t) ={}}{} +\lambda_{2}L_{3} \int^{-\tau_{4}}_{-\tau_{2}-\tau_{4}} \int ^{t}_{t+s}\bigl\vert y_{1}(r)-y_{2}(r) \bigr\vert \,dr\,ds \\& \hphantom{V_{3}(t) ={}}{} +\lambda_{2}L_{4} \int^{-\tau_{2}}_{-2\tau_{2}} \int ^{t}_{t+s}\bigl\vert y_{1}(r)-y_{2}(r) \bigr\vert \,dr\,ds \\& \hphantom{V_{3}(t) ={}}{} +\lambda_{2}L_{5} \int^{-\tau_{4}}_{-2\tau_{4}} \int^{t}_{t+s}\bigl\vert y_{1}(r)-y_{2}(r) \bigr\vert \,dr\,ds; \end{aligned}$$
(35)
$$\begin{aligned}& V_{4}(t)=\lambda_{2}M_{1} \int^{t}_{t-\tau_{3}}\bigl\vert x_{1}(s)-x_{2}(s) \bigr\vert \,ds+\lambda _{1}M_{2} \int^{t}_{t-\tau_{4}}\bigl\vert y_{1}(s)-y_{2}(s) \bigr\vert \,ds. \end{aligned}$$
(36)
According to the definitions of \(S^{*}\) and \(V(t)=V(t,\Phi,\Psi)\), there is some positive constant M large enough such that
$$ V(t)=V(t,\Phi,\Psi)\leq M. $$
(37)
By the structure of \(V(t)\), it is easy to see that
$$ V(t)\geq V_{1}(t)\geq\min\{\lambda_{1}, \lambda_{2}\} \bigl(\bigl\vert x_{1}(t)-x_{2}(t) \bigr\vert +\bigl\vert y_{1}(t)-y_{2}(t)\bigr\vert \bigr)=\underline{\lambda}\bigl\vert \Phi (0)-\Psi(0)\bigr\vert _{0}>0, $$
(38)
where \(\underline{\lambda}=\min\{\lambda_{1},\lambda_{2}\}>0\).
Moreover, by the integrative inequality and the absolute-value inequality properties we have
$$\begin{aligned} V(t) \leq&\frac{1}{2} \bigl\{ 2(\lambda_{1}+ \lambda_{2})+\lambda_{1}\tau ^{2}(K_{1}+K_{2}+2K_{3}+L_{1}+L_{2}) \\ &{}+\lambda_{2}\tau^{2}(K_{4}+K_{5}+2L_{3}+L_{4}+L_{5})+ \tau(\lambda_{2} M_{1}+\lambda _{1} M_{2}) \bigr\} \\ &{}\times \sup_{s\in[-\tau,0]} \bigl[\bigl\vert x_{1t}(s)-x_{2t}(s) \bigr\vert +\bigl\vert y_{1t}(s)-y_{2t}(s)\bigr\vert \bigr] \\ =&\bar{\lambda}\|\Phi-\Psi\|, \end{aligned}$$
(39)
where
$$\begin{aligned} \bar{\lambda} :=&\frac{1}{2} \bigl\{ 2(\lambda_{1}+ \lambda_{2})+\lambda _{1}\tau^{2}(K_{1}+K_{2}+2K_{3}+L_{1}+L_{2}) \\ &{}+\lambda_{2}\tau^{2}(K_{4}+K_{5}+2L_{3}+L_{4}+L_{5})+ \tau(\lambda_{2} M_{1}+\lambda _{1} M_{2}) \bigr\} . \end{aligned}$$
Let \(u,v\in C(R^{+},R^{+})\), choose \(u=\underline{\lambda}s\), \(v=\bar {\lambda}s\), then the first condition of Lemma 2.4 is satisfied.
For \(\forall \bar{\Phi}=(x_{1t},y_{1t})^{T},\bar{\Psi }=(x_{2t},y_{2t})^{T},\hat{\Phi}=(x^{*}_{1t},y^{*}_{1t})^{T},\hat{\Psi }=(x^{*}_{2t},y^{*}_{2t})^{T}\in S^{*}\), then from the previous definitions of \(V_{i}(t)\), \(i=1,2,3,4\), it is easy to calculate that
$$\begin{aligned}& \bigl\vert V(t,\bar{\Phi},\bar{\Psi})- V(t,\hat{\Phi},\hat{\Psi })\bigr\vert \\& \quad = \lambda_{1} \bigl\vert \bigl\vert x_{1}(t)-x_{2}(t) \bigr\vert -\bigl\vert x^{*}_{1}(t)-x^{*}_{2}(t)\bigr\vert \vert +\lambda _{2} \vert \bigl\vert y_{1}(t)-y_{2}(t) \bigr\vert -\bigl\vert y^{*}_{1}(t)-y^{*}_{2}(t)\bigr\vert \bigr\vert \\& \qquad {} + \lambda_{1}K_{1} \int^{-\tau_{3}}_{-2\tau_{3}} \int^{t}_{t+s} \bigl\vert \bigl\vert x_{1}(r)-x_{2}(r)\bigr\vert -\bigl\vert x^{*}_{1}(r)-x^{*}_{2}(r)\bigr\vert \bigr\vert \,dr\,ds \\& \qquad {} + \lambda_{1}K_{2} \int^{-\tau_{1}}_{-2\tau_{1}} \int^{t}_{t+s} \bigl\vert \bigl\vert x_{1}(r)-x_{2}(r)\bigr\vert -\bigl\vert x^{*}_{1}(r)-x^{*}_{2}(r)\bigr\vert \bigr\vert \,dr\,ds \\& \qquad {} + \lambda_{1}K_{3} \int^{-\tau_{1}}_{-\tau_{1}-\tau_{3}} \int^{t}_{t+s} \bigl\vert \bigl\vert x_{1}(r)-x_{2}(r)\bigr\vert -\bigl\vert x^{*}_{1}(r)-x^{*}_{2}(r)\bigr\vert \bigr\vert \,dr\,ds \\& \qquad {} + \lambda_{1}K_{3} \int^{-\tau_{3}}_{-\tau_{1}-\tau_{3}} \int^{t}_{t+s} \bigl\vert \bigl\vert x_{1}(r)-x_{2}(r)\bigr\vert -\bigl\vert x^{*}_{1}(r)-x^{*}_{2}(r)\bigr\vert \bigr\vert \,dr\,ds \\& \qquad {} + \lambda_{2}K_{4} \int^{-\tau_{3}}_{-\tau_{2}-\tau_{3}} \int^{t}_{t+s} \bigl\vert \bigl\vert x_{1}(r)-x_{2}(r)\bigr\vert -\bigl\vert x^{*}_{1}(r)-x^{*}_{2}(r)\bigr\vert \bigr\vert \,dr\,ds \\& \qquad {} + \lambda_{2}K_{5} \int^{-\tau_{3}}_{-\tau_{3}-\tau_{4}} \int^{t}_{t+s} \bigl\vert \bigl\vert x_{1}(r)-x_{2}(r)\bigr\vert -\bigl\vert x^{*}_{1}(r)-x^{*}_{2}(r)\bigr\vert \bigr\vert \,dr\,ds \\& \qquad {} +\lambda_{1} L_{1} \int^{-\tau_{4}}_{-\tau_{1}-\tau_{4}} \int^{t}_{t+s} \bigl\vert \bigl\vert y_{1}(r)-y_{2}(r)\bigr\vert -\bigl\vert y^{*}_{1}(r)-y^{*}_{2}(r)\bigr\vert \bigr\vert \,dr\,ds \\ & \qquad {} +\lambda_{1} L_{2} \int^{-\tau_{4}}_{-\tau_{3}-\tau_{4}} \int^{t}_{t+s} \bigl\vert \bigl\vert y_{1}(r)-y_{2}(r)\bigr\vert -\bigl\vert y^{*}_{1}(r)-y^{*}_{2}(r)\bigr\vert \bigr\vert \,dr\,ds \\ & \qquad {} +\lambda_{2} L_{3} \int^{-\tau_{2}}_{-\tau_{2}-\tau_{4}} \int^{t}_{t+s} \bigl\vert \bigl\vert y_{1}(r)-y_{2}(r)\bigr\vert -\bigl\vert y^{*}_{1}(r)-y^{*}_{2}(r)\bigr\vert \bigr\vert \,dr\,ds \\ & \qquad {} +\lambda_{2} L_{3} \int^{-\tau_{4}}_{-\tau_{2}-\tau_{4}} \int^{t}_{t+s} \bigl\vert \bigl\vert y_{1}(r)-y_{2}(r)\bigr\vert -\bigl\vert y^{*}_{1}(r)-y^{*}_{2}(r)\bigr\vert \bigr\vert \,dr\,ds \\ & \qquad {} +\lambda_{2} L_{4} \int^{-\tau_{2}}_{-2\tau_{2}} \int^{t}_{t+s} \bigl\vert \bigl\vert y_{1}(r)-y_{2}(r)\bigr\vert -\bigl\vert y^{*}_{1}(r)-y^{*}_{2}(r)\bigr\vert \bigr\vert \,dr\,ds \\ & \qquad {} +\lambda_{2} L_{5} \int^{-\tau_{4}}_{-2\tau_{4}} \int^{t}_{t+s} \bigl\vert \bigl\vert y_{1}(r)-y_{2}(r)\bigr\vert -\bigl\vert y^{*}_{1}(r)-y^{*}_{2}(r)\bigr\vert \bigr\vert \,dr\,ds \\ & \qquad {} +\lambda_{2} M_{1} \int^{t}_{t-\tau_{3}} \bigl\vert \bigl\vert x_{1}(s)-x_{2}(s)\bigr\vert -\bigl\vert x^{*}_{1}(s)-x^{*}_{2}(s)\bigr\vert \bigr\vert \,ds \\ & \qquad {} +\lambda_{1} M_{2} \int^{t}_{t-\tau_{4}} \bigl\vert \bigl\vert y_{1}(s)-y_{2}(s)\bigr\vert -\bigl\vert y^{*}_{1}(s)-y^{*}_{2}(s)\bigr\vert \bigr\vert \,ds. \end{aligned}$$
Then it follows from inequality (19) in Lemma 2.5 that
$$\begin{aligned}& \bigl\vert V(t,\bar{\Phi},\bar{\Psi})- V(t,\hat{\Phi},\hat{\Psi })\bigr\vert \\& \quad \leq \lambda_{1} \bigl\vert \bigl\vert x_{1}(t)-x^{*}_{1}(t) \bigr\vert +\bigl\vert x_{2}(t)-x^{*}_{2}(t)\bigr\vert \bigr\vert +\lambda_{2} \bigl\vert \bigl\vert y_{1}(t)-y^{*}_{1}(t) \bigr\vert +\bigl\vert y_{2}(t)-y^{*}_{2}(t)\bigr\vert \bigr\vert \\& \qquad {}+ \lambda_{1}K_{1} \int^{-\tau_{3}}_{-2\tau_{3}} \int^{t}_{t+s}\bigl\vert \bigl\vert x_{1}(r)-x^{*}_{1}(r)\bigr\vert +\bigl\vert x_{2}(r)-x^{*}_{2}(r)\bigr\vert \bigr\vert \,dr\,ds \\& \qquad {}+ \lambda_{1}K_{2} \int^{-\tau_{1}}_{-2\tau_{1}} \int^{t}_{t+s} \bigl\vert \bigl\vert x_{1}(r)-x^{*}_{1}(r)\bigr\vert +\bigl\vert x_{2}(r)-x^{*}_{2}(r)\bigr\vert \bigr\vert \,dr\,ds \\& \qquad {}+ \lambda_{1}K_{3} \int^{-\tau_{1}}_{-\tau_{1}-\tau_{3}} \int ^{t}_{t+s}\bigl\vert \bigl\vert x_{1}(r)-x^{*}_{1}(r)\bigr\vert +\bigl\vert x_{2}(r)-x^{*}_{2}(r)\bigr\vert \bigr\vert \,dr\,ds \\& \qquad {}+ \lambda_{1}K_{3} \int^{-\tau_{3}}_{-\tau_{1}-\tau_{3}} \int ^{t}_{t+s}\bigl\vert \bigl\vert x_{1}(r)-x^{*}_{1}(r)\bigr\vert +\bigl\vert x_{2}(r)-x^{*}_{2}(r)\bigr\vert \bigr\vert \,dr\,ds \\& \qquad {}+ \lambda_{2}K_{4} \int^{-\tau_{3}}_{-\tau_{2}-\tau_{3}} \int^{t}_{t+s} \bigl\vert \bigl\vert x_{1}(r)-x^{*}_{1}(r)\bigr\vert +\bigl\vert x_{2}(r)-x^{*}_{2}(r)\bigr\vert \bigr\vert \,dr\,ds \\& \qquad {}+ \lambda_{2}K_{5} \int^{-\tau_{3}}_{-\tau_{3}-\tau_{4}} \int^{t}_{t+s} \bigl\vert \bigl\vert x_{1}(r)-x^{*}_{1}(r)\bigr\vert +\bigl\vert x_{2}(r)-x^{*}_{2}(r)\bigr\vert \bigr\vert \,dr\,ds \\& \qquad {}+ \lambda_{1} L_{1} \int^{-\tau_{4}}_{-\tau_{1}-\tau_{4}} \int^{t}_{t+s} \bigl\vert \bigl\vert y_{1}(r)-y^{*}_{1}(r)\bigr\vert +\bigl\vert y_{2}(r)-y^{*}_{2}(r)\bigr\vert \bigr\vert \,dr\,ds \\& \qquad {} +\lambda_{1} L_{2} \int^{-\tau_{4}}_{-\tau_{3}-\tau_{4}} \int^{t}_{t+s} \bigl\vert \bigl\vert y_{1}(r)-y^{*}_{1}(r)\bigr\vert +\bigl\vert y_{2}(r)-y^{*}_{2}(r)\bigr\vert \bigr\vert \,dr\,ds \\& \qquad {} +\lambda_{2} L_{3} \int^{-\tau_{2}}_{-\tau_{2}-\tau_{4}} \int^{t}_{t+s} \bigl\vert \bigl\vert y_{1}(r)-y^{*}_{1}(r)\bigr\vert +\bigl\vert y_{2}(r)-y^{*}_{2}(r)\bigr\vert \bigr\vert \,dr\,ds \\& \qquad {} +\lambda_{2} L_{3} \int^{-\tau_{4}}_{-\tau_{2}-\tau_{4}} \int^{t}_{t+s} \bigl\vert \bigl\vert y_{1}(r)-y^{*}_{1}(r)\bigr\vert +\bigl\vert y_{2}(r)-y^{*}_{2}(r)\bigr\vert \bigr\vert \,dr\,ds \\& \qquad {} +\lambda_{2} L_{4} \int^{-\tau_{2}}_{-2\tau_{2}} \int^{t}_{t+s} \bigl\vert \bigl\vert y_{1}(r)-y^{*}_{1}(r)\bigr\vert +\bigl\vert y_{2}(r)-y^{*}_{2}(r)\bigr\vert \bigr\vert \,dr\,ds \\& \qquad {} +\lambda_{2} L_{5} \int^{-\tau_{4}}_{-2\tau_{4}} \int^{t}_{t+s} \bigl\vert \bigl\vert y_{1}(r)-y^{*}_{1}(r)\bigr\vert +\bigl\vert y_{2}(r)-y^{*}_{2}(r)\bigr\vert \bigr\vert \,dr\,ds \\& \qquad {} +\lambda_{2} M_{1} \int^{t}_{t-\tau_{3}} \bigl\vert \bigl\vert x_{1}(s)-x^{*}_{1}(s)\bigr\vert +\bigl\vert x_{2}(s)-x^{*}_{2}(s)\bigr\vert \bigr\vert \,ds \\& \qquad {} +\lambda_{1} M_{2} \int^{t}_{t-\tau_{4}}\bigl\vert \bigl\vert y_{1}(s)-y^{*}_{1}(s)\bigr\vert +\bigl\vert y_{2}(s)-y^{*}_{2}(s)\bigr\vert \bigr\vert \,ds, \end{aligned}$$
which yields
$$\begin{aligned}& \bigl\vert V(t,\bar{\Phi},\bar{\Psi})- V(t,\hat{\Phi},\hat{\Psi })\bigr\vert \\& \quad \leq \bar{\lambda} \sum_{i=1}^{2} \sup_{s\in[-\tau ,0]} \bigl[\bigl\vert x_{it}(s)-x^{*}_{it}(s) \bigr\vert +\bigl\vert y_{it}(s)-y^{*}_{it}(s)\bigr\vert \bigr] \\& \quad = \bar{\lambda} \bigl(\Vert \bar{\Phi}-\hat{\Phi} \Vert +\Vert \bar{ \Psi }-\hat{\Psi} \Vert \bigr). \end{aligned}$$
(40)
This means the second condition of Lemma 2.4 is satisfied.
Finally, we will prove the last condition of Lemma 2.4.
In fact, calculating the right derivative \(D^{+}V_{1}(t)\) of \(V_{1}(t)\) along the solutions of system (30) we have
$$ D^{+}V_{1}(t)= \bigl(x'_{1}(t)-x'_{2}(t) \bigr)\operatorname{sgn}\bigl(x_{1}(t)-x_{2}(t)\bigr)+ \bigl(y'_{1}(t)-y'_{2}(t) \bigr) \operatorname{sgn}\bigl(y_{1}(t)-y_{2}(t)\bigr). $$
(41)
It follows from the mean-value theorem and the product system (30) that
$$\begin{aligned}& x'_{1}(t)-x'_{2}(t) \\& \quad = D_{1}(t)e^{x_{2}(t-\tau_{1})}-D_{1}(t)e^{x_{1}(t-\tau_{1})}+ \frac {C_{1}(t)e^{y_{2}(t-\tau_{4})}}{(1+A(t)e^{x_{2}(t-\tau _{3})})(1+B(t)e^{y_{2}(t-\tau_{4})})} \\& \qquad {}- \frac{C_{1}(t)e^{y_{1}(t-\tau_{4})}}{(1+A(t)e^{x_{1}(t-\tau _{3})})(1+B(t)e^{y_{1}(t-\tau_{4})})}. \end{aligned}$$
Then
$$\begin{aligned}& x'_{1}(t)-x'_{2}(t) \\& \quad = -D_{1}(t)e^{\theta_{1}(t)} \bigl[x_{1}(t- \tau_{1})-x_{2}(t-\tau_{1}) \bigr] \\& \qquad {}+ \frac{A(t)C_{1}(t)e^{\theta_{2}(t)+\theta_{3}(t)}}{(1+A(t)e^{\theta _{2}(t)})^{2}(1+B(t)e^{\theta_{3}(t)})} \bigl[x_{1}(t-\tau_{3})-x_{2}(t- \tau _{3}) \bigr] \\& \qquad {}- \frac{C_{1}(t)e^{\theta_{3}(t)}}{(1+A(t)e^{\theta _{2}(t)})(1+B(t)e^{\theta_{3}(t)})^{2}} \bigl[y_{1}(t-\tau_{4})-y_{2}(t- \tau _{4}) \bigr], \end{aligned}$$
(42)
where \(\theta_{1}(t)\) lie between \(x_{1}(t-\tau_{1})\) and \(x_{2}(t-\tau_{1})\), \(\theta_{2}(t)\) lie between \(x_{1}(t-\tau_{3})\) and \(x_{2}(t-\tau_{3})\), \(\theta_{3}(t)\) lie between \(y_{1}(t-\tau_{4})\) and \(y_{2}(t-\tau_{4})\).
That is,
$$\begin{aligned}& x'_{1}(t)-x'_{2}(t) \\& \quad = -D_{1}(t)e^{\theta_{1}(t)} \bigl[x_{1}(t)-x_{2}(t) \bigr]+D_{1}(t)e^{\theta _{1}(t)} \int^{t}_{t-\tau_{1}} \bigl[x'_{1}(s)-x'_{2}(s) \bigr]\,ds \\& \qquad {}+ \frac{A(t)C_{1}(t)e^{\theta_{2}(t)+\theta_{3}(t)}}{(1+A(t)e^{\theta _{2}(t)})^{2}(1+B(t)e^{\theta_{3}(t)})} \bigl[x_{1}(t)-x_{2}(t) \bigr] \\ & \qquad {}- \frac{A(t)C_{1}(t)e^{\theta_{2}(t)+\theta_{3}(t)}}{(1+A(t)e^{\theta _{2}(t)})^{2}(1+B(t)e^{\theta_{3}(t)})} \int^{t}_{t-\tau_{3}} \bigl[x'_{1}(s)-x'_{2}(s) \bigr]\,ds \\ & \qquad {}- \frac{C_{1}(t)e^{\theta_{3}(t)}}{(1+A(t)e^{\theta _{2}(t)})(1+B(t)e^{\theta_{3}(t)})^{2}} \bigl[y_{1}(t-\tau_{4})-y_{2}(t- \tau _{4}) \bigr]. \end{aligned}$$
(43)
Integrating both sides of (42) on the interval \([t-\tau_{1},t]\), we have
$$\begin{aligned} \int^{t}_{t-\tau_{1}} \bigl[x'_{1}(s)-x'_{2}(s) \bigr]\,ds =&- \int^{t}_{t-\tau _{1}}D_{1}(s)e^{\theta_{1}(s)} \bigl[x_{1}(s-\tau_{1})-x_{2}(s- \tau_{1}) \bigr]\,ds \\ & {}+ \int^{t}_{t-\tau_{1}}\frac{A(s)C_{1}(s)e^{\theta_{2}(s)+\theta _{3}(s)}}{(1+A(s)e^{\theta_{2}(s)})^{2}(1+B(s)e^{\theta_{3}(s)})} \bigl[x_{1}(s-\tau_{3})-x_{2}(s-\tau_{3}) \bigr]\,ds \\ &{}- \int^{t}_{t-\tau_{1}}\frac{C_{1}(s)e^{\theta_{3}(s)}}{(1+A(s)e^{\theta _{2}(s)})(1+B(s)e^{\theta_{3}(s)})^{2}} \bigl[y_{1}(s-\tau_{4})-y_{2}(s-\tau _{4}) \bigr]\,ds \\ =& - \int^{-\tau_{1}}_{-2\tau_{1}}D_{1}(t+s+ \tau_{1})e^{\theta_{1}(t+s+\tau _{1})} \bigl[x_{1}(t+s)-x_{2}(t+s) \bigr]\,ds \\ & {}+ \int^{-\tau_{3}}_{-\tau_{1}-\tau_{3}}\frac{A(t+s+\tau_{3})C_{1}(t+s+\tau _{3})e^{\theta_{2}(t+s+\tau_{3})+\theta_{3}(t+s+\tau_{3})}}{(1+A(t+s+\tau _{3})e^{\theta_{2}(s)})^{2}(1+B(t+s+\tau_{3})e^{\theta_{3}(t+s+\tau _{3})})} \\ & {}\times \bigl[x_{1}(t+s)-x_{2}(t+s) \bigr]\,ds \\ & {}- \int^{-\tau_{4}}_{-\tau_{1}-\tau_{4}} \frac{C_{1}(t+s+\tau_{4})e^{\theta _{3}(t+s+\tau_{4})}}{(1+A(t+s+\tau_{4})e^{\theta_{2}(t+s+\tau _{4})})(1+B(t+s+\tau_{4})e^{\theta_{3}(t+s+\tau_{4})})^{2}} \\ & {}\times \bigl[y_{1}(t+s)-y_{2}(t+s) \bigr]\,ds, \end{aligned}$$
which implies that
$$\begin{aligned} \biggl\vert \int^{t}_{t-\tau_{1}} \bigl[x'_{1}(s)-x'_{2}(s) \bigr]\,ds\biggr\vert \leq& \sqrt{K_{2}} \int^{-\tau_{1}}_{-2\tau_{1}} \bigl\vert x_{1}(t+s)-x_{2}(t+s) \bigr\vert \,ds \\ &{}+\sqrt{K_{1}} \int^{-\tau_{3}}_{-\tau_{1}-\tau_{3}} \bigl\vert x_{1}(t+s)-x_{2}(t+s) \bigr\vert \,ds \\ &{}+M_{2} \int^{-\tau_{4}}_{-\tau_{1}-\tau_{4}} \bigl\vert y_{1}(t+s)-y_{2}(t+s) \bigr\vert \,ds. \end{aligned}$$
(44)
Similarly, we can get
$$\begin{aligned} \biggl\vert \int^{t}_{t-\tau_{3}} \bigl[x'_{1}(s)-x'_{2}(s) \bigr]\,ds \biggr\vert \leq& \sqrt{K_{2}} \int^{-\tau_{1}}_{-\tau_{1}-\tau_{3}} \bigl\vert x_{1}(t+s)-x_{2}(t+s) \bigr\vert \,ds \\ &{}+\sqrt{K_{1}} \int^{-\tau_{3}}_{-2\tau_{3}} \bigl\vert x_{1}(t+s)-x_{2}(t+s) \bigr\vert \,ds \\ &{}+M_{2} \int^{-\tau_{4}}_{-\tau_{3}-\tau _{4}} \bigl\vert y_{1}(t+s)-y_{2}(t+s) \bigr\vert \,ds. \end{aligned}$$
(45)
Thus,
$$\begin{aligned}& \bigl(x'_{1}(t)-x'_{2}(t) \bigr)\operatorname{sgn}\bigl(x_{1}(t)-x_{2}(t)\bigr) \\ & \quad \leq -\bigl(D^{l}_{1}u_{*}-\sqrt{K_{1}} \bigr)\bigl\vert x_{1}(t)-x_{2}(t)\bigr\vert \\& \qquad {}+K_{3} \int^{-\tau _{3}}_{-\tau_{1}-\tau_{3}} \bigl\vert x_{1}(t+s)-x_{2}(t+s) \bigr\vert \,ds \\& \qquad {}+ K_{2} \int^{-\tau_{1}}_{-2\tau_{1}} \bigl\vert x_{1}(t+s)-x_{2}(t+s) \bigr\vert \,ds+L_{1} \int^{-\tau_{4}}_{-\tau_{1}-\tau_{4}}\bigl\vert y_{1}(t+s)-y_{2}(t+s) \bigr\vert \,ds \\& \qquad {}+ K_{1} \int^{-\tau_{3}}_{-2\tau_{3}} \bigl\vert x_{1}(t+s)-x_{2}(t+s) \bigr\vert \,ds+K_{3} \int^{-\tau_{1}}_{-\tau_{1}-\tau_{3}} \bigl\vert x_{1}(t+s)-x_{2}(t+s) \bigr\vert \,ds \\& \qquad {}+ L_{2} \int^{-\tau_{4}}_{-\tau_{3}-\tau_{4}}\bigl\vert y_{1}(t+s)-y_{2}(t+s) \bigr\vert \,ds+M_{2} \bigl\vert y_{1}(t- \tau_{4})-y_{2}(t-\tau_{4})\bigr\vert . \end{aligned}$$
(46)
By the same argument, from the second and the fourth equation of system (30) we can derive that
$$\begin{aligned}& y'_{1}(t)-y'_{2}(t) \\& \quad = D_{2}(t)e^{y_{2}(t-\tau_{2})}-D_{2}(t)e^{y_{1}(t-\tau_{2})}+ \frac {C_{2}(t)e^{x_{1}(t-\tau_{3})}}{(1+A(t)e^{x_{1}(t-\tau _{3})})(1+B(t)e^{y_{1}(t-\tau_{4})})} \\& \qquad {}- \frac{C_{2}(t)e^{x_{2}(t-\tau_{3})}}{(1+A(t)e^{x_{2}(t-\tau _{3})})(1+B(t)e^{y_{2}(t-\tau_{4})})} \\& \quad = -D_{2}(t)e^{\theta_{4}(t)} \bigl[y_{1}(t- \tau_{2})-y_{2}(t-\tau_{2}) \bigr] \\& \qquad {}+ \frac{C_{2}(t)e^{\theta_{5}(t)}}{(1+A(t)e^{\theta _{5}(t)})^{2}(1+B(t)e^{\theta_{6}(t)})} \bigl[x_{1}(t-\tau_{3})-x_{2}(t- \tau _{3}) \bigr] \\& \qquad {}- \frac{B(t)C_{2}(t)e^{\theta_{5}(t)+\theta_{6}(t)}}{(1+A(t)e^{\theta _{5}(t)})(1+B(t)e^{\theta_{6}(t)})^{2}} \bigl[y_{1}(t-\tau_{4})-y_{2}(t- \tau _{4}) \bigr], \end{aligned}$$
(47)
where \(\theta_{4}(t)\) lie between \(y_{1}(t-\tau_{2})\) and \(y_{2}(t-\tau_{2})\), \(\theta_{5}(t)\) lie between \(x_{1}(t-\tau_{3})\) and \(x_{2}(t-\tau_{3})\), \(\theta_{6}(t)\) lie between \(y_{1}(t-\tau_{4})\) and \(y_{2}(t-\tau_{4})\).
Integrating both sides of (47) on the interval \([t-\tau_{2},t]\), we have
$$\begin{aligned}& \int^{t}_{t-\tau_{2}} \bigl[y'_{1}(s)-y'_{2}(s) \bigr]\,ds \\& \quad = \int^{t}_{t-\tau_{2}} D_{2}(s)e^{\theta_{4}(s)} \bigl[y_{1}(s-\tau _{2})-y_{2}(s- \tau_{2}) \bigr]\,ds \\& \qquad {}+ \int^{t}_{t-\tau_{2}} \frac{C_{2}(s)e^{\theta_{5}(s)}}{(1+A(s)e^{\theta _{5}(s)})^{2}(1+B(s)e^{\theta_{6}(s)})} \bigl[x_{1}(s-\tau_{3})-x_{2}(s-\tau _{3}) \bigr]\,ds \\& \qquad {}- \int^{t}_{t-\tau_{2}} \frac{B(s)C_{2}(s)e^{\theta_{5}(s)+\theta _{5}(s)}}{(1+A(s)e^{\theta_{5}(s)})(1+B(s)e^{\theta_{6}(s)})^{2}} \bigl[y_{1}(s-\tau_{4})-y_{2}(s- \tau_{4}) \bigr]\,ds \\& \quad = - \int^{-\tau_{2}}_{-2\tau_{2}}D_{2}(t+s+ \tau_{2})e^{\theta_{4}(t+s+\tau _{2})} \bigl[y_{1}(t+s)-y_{2}(t+s) \bigr]\,ds \\& \qquad {}+ \int^{-\tau_{3}}_{-\tau_{2}-\tau_{3}}\frac{C_{2}(t+s+\tau_{3})e^{\theta _{5}(t+s+\tau_{3})} [x_{1}(t+s)-x_{2}(t+s) ]}{(1+A(t+s+\tau_{3}) e^{\theta_{5}(t+s+\tau_{3})})^{2} (1+B(t+s+\tau_{3}) e^{\theta_{6}(t+s+\tau _{3})})}\,ds \\& \qquad {}- \int^{-\tau_{4}}_{-\tau_{2}-\tau_{4}} \frac{B(t+s+\tau _{4})C_{2}(t+s+\tau_{4})e^{\theta_{5}(t+s+\tau_{4})+\theta_{5} (t+s+\tau_{4})}}{(1+A(t+s+\tau_{4})e^{\theta_{5}(t+s+\tau _{4})})(1+B(s)e^{\theta_{6}(t+s+\tau_{4})})^{2}} \\& \qquad {}\times \bigl[y_{1}(t+s)-y_{2}(t+s) \bigr]\,ds, \end{aligned}$$
(48)
which leads to
$$\begin{aligned} \begin{aligned}[b] &\biggl\vert \int^{t}_{t-\tau_{2}}\bigl[y'_{1}(s)-y'_{2}(s) \bigr]\,ds\biggr\vert \\ &\quad \leq \sqrt{L_{4}} \int^{-\tau_{2}}_{-2\tau_{2}} \bigl\vert y_{1}(t+s)-y_{2}(t+s) \bigr\vert \,ds +\sqrt{L_{5}} \int^{-\tau_{4}}_{-\tau_{2}-\tau_{4}} \bigl\vert y_{1}(t+s)-y_{2}(t+s) \bigr\vert \,ds \\ &\qquad {}+ M_{1} \int^{-\tau_{3}}_{-\tau_{2}-\tau_{3}}\bigl\vert x_{1}(t+s)-x_{2}(t+s) \bigr\vert \,ds. \end{aligned} \end{aligned}$$
(49)
By the same argument, we can obtain
$$\begin{aligned}& \biggl\vert \int^{t}_{t-\tau_{4}} \bigl[y'_{1}(s)-y'_{2}(s) \bigr]\,ds \biggr\vert \\& \quad \leq \sqrt{L_{4}} \int^{-\tau_{2}}_{-\tau_{2}-\tau_{4}} \bigl\vert y_{1}(t+s)-y_{2}(t+s) \bigr\vert \,ds+\sqrt{L_{5}} \int^{-\tau_{4}}_{-2\tau_{4}} \bigl\vert y_{1}(t+s)-y_{2}(t+s) \bigr\vert \,ds \\& \qquad {}+ M_{1} \int^{-\tau_{3}}_{-\tau_{3}-\tau_{4}} \bigl\vert x_{1}(t+s)-x_{2}(t+s) \bigr\vert \,ds. \end{aligned}$$
(50)
On the other hand, it follows from (47) that
$$\begin{aligned}& y'_{1}(t)-y'_{2}(t) \\ & \quad =D_{2}(t)e^{\theta_{4}(t)} \int^{t}_{t-\tau_{2}} \bigl[y_{1}(s)-y_{2}(s) \bigr]\,ds \\ & \qquad {}- D_{2}(t)e^{\theta_{4}(t)} \bigl[y_{1}(t)-y_{2}(t) \bigr]-\frac {B(t)C_{2}(t)e^{\theta_{5}(t)+\theta_{6}(t)}}{(1+A(t)e^{\theta _{5}(t)})(1+B(t)e^{\theta_{6}(t)})^{2}} \bigl[y_{1}(t)-y_{2}(t) \bigr] \\ & \qquad {}+ \frac{B(t)C_{2}(t)e^{\theta_{5}(t)+\theta_{6}(t)}}{(1+A(t)e^{\theta _{5}(t)})(1+B(t)e^{\theta_{6}(t)})^{2}} \int^{t}_{t-\tau_{4}} \bigl[y_{1}(s)-y_{2}(s) \bigr]\,ds \\ & \qquad {}- \frac{C_{2}(t)e^{\theta_{5}(t)}}{(1+A(t)e^{\theta _{5}(t)})^{2}(1+B(t)e^{\theta_{6}(t)})} \bigl[x_{1}(t-\tau_{3})-x_{2}(t- \tau _{3}) \bigr]. \end{aligned}$$
(51)
Thus, combined (51) with (49)-(50), one can deduce that
$$\begin{aligned}& \bigl(y'_{1}(t)-y'_{2}(t) \bigr)\operatorname{sgn}\bigl(y_{1}(t)-y_{2}(t)\bigr) \\ & \quad \leq -\bigl(D^{l}_{2}v_{*}+P\bigr)\bigl\vert y_{1}(t)-y_{2}(t)\bigr\vert +L_{4} \int^{-\tau _{2}}_{-2\tau_{2}} \bigl\vert y_{1}(t+s)-y_{2}(t+s) \bigr\vert \,ds \\ & \qquad {}+ L_{3} \int^{-\tau_{4}}_{-\tau_{2}-\tau_{4}} \bigl\vert y_{1}(t+s)-y_{2}(t+s) \bigr\vert \,ds+K_{4} \int^{-\tau_{3}}_{-\tau_{2}-\tau_{3}}\bigl\vert x_{1}(t+s)-x_{2}(t+s) \bigr\vert \,ds \\ & \qquad {}+ L_{3} \int^{-\tau_{2}}_{-\tau_{2}-\tau_{4}} \bigl\vert y_{1}(t+s)-y_{2}(t+s) \bigr\vert \,ds+L_{5} \int^{-\tau_{4}}_{-2\tau_{4}} \bigl\vert y_{1}(t+s)-y_{2}(t+s) \bigr\vert \,ds \\ & \qquad {}+ K_{5} \int^{-\tau_{3}}_{-\tau_{3}-\tau_{4}} \bigl\vert x_{1}(t+s)-x_{2}(t+s) \bigr\vert \,ds+M_{1} \bigl\vert x_{1}(t- \tau_{3})-x_{2}(t-\tau_{3})\bigr\vert . \end{aligned}$$
(52)
Therefore,
$$\begin{aligned} D^{+}V_{1}(t) \leq&-\lambda_{1}\bigl(D^{l}_{1}u_{*}- \sqrt{K_{1}}\bigr)\bigl\vert x_{1}(t)-x_{2}(t) \bigr\vert -\lambda_{2}\bigl(D^{l}_{2}v_{*}+P\bigr) \bigl\vert y_{1}(t)-y_{2}(t)\bigr\vert \\ &{}+ \lambda_{1}K_{1} \int^{-\tau_{3}}_{-2\tau_{3}} \bigl\vert x_{1}(t+s)-x_{2}(t+s) \bigr\vert \,ds+\lambda_{1}K_{2} \int^{-\tau_{1}}_{-2\tau_{1}} \bigl\vert x_{1}(t+s)-x_{2}(t+s) \bigr\vert \,ds \\ &{}+ \lambda_{1}K_{3} \int^{-\tau_{1}}_{-\tau_{1}-\tau_{3}} \bigl\vert x_{1}(t+s)-x_{2}(t+s) \bigr\vert \,ds+\lambda_{1}K_{3} \int^{-\tau_{3}}_{-\tau_{1}-\tau _{3}} \bigl\vert x_{1}(t+s)-x_{2}(t+s) \bigr\vert \,ds \\ &{}+ \lambda_{2}K_{4} \int^{-\tau_{3}}_{-\tau_{2}-\tau_{3}} \bigl\vert x_{1}(t+s)-x_{2}(t+s) \bigr\vert \,ds+\lambda_{2}K_{5} \int^{-\tau_{3}}_{-\tau_{3}-\tau _{4}} \bigl\vert x_{1}(t+s)-x_{2}(t+s) \bigr\vert \,ds \\ &{}+ \lambda_{1}L_{1} \int^{-\tau_{4}}_{-\tau_{1}-\tau_{4}} \bigl\vert y_{1}(t+s)-y_{2}(t+s) \bigr\vert \,ds+\lambda_{1}L_{2} \int^{-\tau_{4}}_{-\tau_{3}-\tau _{4}}\bigl\vert y_{1}(t+s)-y_{2}(t+s) \bigr\vert \,ds \\ &{}+ \lambda_{2}L_{3} \int^{-\tau_{2}}_{-\tau_{2}-\tau_{4}} \bigl\vert y_{1}(t+s)-y_{2}(t+s) \bigr\vert \,ds+\lambda_{2}L_{3} \int^{-\tau_{4}}_{-\tau_{2}-\tau _{4}} \bigl\vert y_{1}(t+s)-y_{2}(t+s) \bigr\vert \,ds \\ &{}+ \lambda_{2}L_{4} \int^{-\tau_{2}}_{-2\tau_{2}} \bigl\vert y_{1}(t+s)-y_{2}(t+s) \bigr\vert \,ds+\lambda_{2}L_{5} \int^{-\tau_{4}}_{-2\tau_{4}} \bigl\vert y_{1}(t+s)-y_{2}(t+s) \bigr\vert \,ds \\ &{}+ \lambda_{2}M_{1} \bigl\vert x_{1}(t- \tau_{3})-x_{2}(t-\tau_{3})\bigr\vert + \lambda_{1}M_{2} \bigl\vert y_{1}(t- \tau_{4})-y_{2}(t-\tau_{4})\bigr\vert . \end{aligned}$$
(53)
On the other hand, according to the structure of \(V_{i}(t)\), \(i=2,3,4\), we can easily calculate the right derivatives \(D^{+}V_{2}(t)\), \(D^{+}V_{3}(t)\), and \(D^{+}V_{4}(t)\) along the solution of system (30) as follows:
$$\begin{aligned}& D^{+}V_{2}(t)= (\lambda_{1}\tau_{3}K_{1}+ \lambda_{1}\tau_{1}K_{2}+\lambda_{1} \tau _{3}K_{3}+\lambda_{1}\tau_{1}K_{3}+ \lambda_{2}\tau_{2}K_{4}+\lambda_{2}\tau _{4}K_{5})\times\bigl\vert x_{1}(t)-x_{2}(t) \bigr\vert \\& \hphantom{D^{+}V_{2}(t)={}}{}- \lambda_{1}K_{1} \int^{-\tau_{3}}_{-2\tau _{3}}\bigl\vert x_{1}(t+s)-x_{2}(t+s) \bigr\vert \,ds-\lambda_{1}K_{2} \int^{-\tau_{1}}_{-2\tau _{1}}\bigl\vert x_{1}(t+s)-x_{2}(t+s) \bigr\vert \,ds \\& \hphantom{D^{+}V_{2}(t)={}}{}- \lambda_{1}K_{3} \int^{-\tau_{1}}_{-\tau_{1}-\tau _{3}}\bigl\vert x_{1}(t+s)-x_{2}(t+s) \bigr\vert \,ds-\lambda_{1}K_{3} \int^{-\tau_{3}}_{-\tau_{1}-\tau _{3}}\bigl\vert x_{1}(t+s)-x_{2}(t+s) \bigr\vert \,ds \\& \hphantom{D^{+}V_{2}(t)={}}{}- \lambda_{2}K_{4} \int^{-\tau_{3}}_{-\tau_{2}-\tau _{3}}\bigl\vert x_{1}(t+s)-x_{2}(t+s) \bigr\vert \,ds \\& \hphantom{D^{+}V_{2}(t)={}}{}-\lambda_{2}K_{5} \int^{-\tau_{3}}_{-\tau_{3}-\tau _{4}}\bigl\vert x_{1}(t+s)-x_{2}(t+s) \bigr\vert \,ds; \end{aligned}$$
(54)
$$\begin{aligned}& D^{+}V_{3}(t)= (\lambda_{1}\tau_{1}L_{1}+ \lambda_{1}\tau_{3}L_{2}+\lambda_{2}\tau _{4}L_{3}+\lambda_{2}\tau_{2}L_{3}+ \lambda_{2}\tau_{2}L_{4}+\lambda_{2}\tau _{4}L_{5})\times\bigl\vert y_{1}(t)-y_{2}(t) \bigr\vert \\& \hphantom{D^{+}V_{3}(t)={}}{}- \lambda_{1}L_{1} \int^{-\tau_{4}}_{-\tau_{1}-\tau _{4}}\bigl\vert y_{1}(t+s)-y_{2}(t+s) \bigr\vert \,ds-\lambda_{1}L_{2} \int^{-\tau_{4}}_{-\tau_{3}-\tau _{4}}\bigl\vert y_{1}(t+s)-y_{2}(t+s) \bigr\vert \,ds \\& \hphantom{D^{+}V_{3}(t)={}}{}- \lambda_{2}L_{3} \int^{-\tau_{2}}_{-\tau_{2}-\tau _{4}}\bigl\vert y_{1}(t+s)-y_{2}(t+s) \bigr\vert \,ds-\lambda_{2}L_{3} \int^{-\tau_{4}}_{-\tau_{2}-\tau _{4}}\bigl\vert y_{1}(t+s)-y_{2}(t+s) \bigr\vert \,ds \\& \hphantom{D^{+}V_{3}(t)={}}{}- \lambda_{2}L_{4} \int^{-\tau_{2}}_{-2\tau _{2}}\bigl\vert y_{1}(t+s)-y_{2}(t+s) \bigr\vert \,ds\,ds \\& \hphantom{D^{+}V_{3}(t)={}}{}-\lambda_{2}L_{5} \int^{-\tau_{4}}_{-2\tau _{4}}\bigl\vert y_{1}(t+s)-y_{2}(t+s) \bigr\vert \,ds; \end{aligned}$$
(55)
$$\begin{aligned}& D^{+}V_{4}(t) =\lambda_{2}M_{1}\bigl[\bigl\vert x_{1}(t)-x_{2}(t)\bigr\vert -\bigl\vert x_{1}(t-\tau _{3})-x_{2}(t-\tau_{3}) \bigr\vert \bigr] \\& \hphantom{D^{+}V_{4}(t) ={}}{}+\lambda_{1}M_{2}\bigl[\bigl\vert y_{1}(t)-y_{2}(t)\bigr\vert -\bigl\vert y_{1}(t-\tau_{4})-y_{2}(t-\tau _{4}) \bigr\vert \bigr]. \end{aligned}$$
(56)
From (53)-(56), by a simple reduction we can easily give the following estimations for the right derivatives of \(V(t)\):
$$\begin{aligned} D^{+}V(t) \leq&-\bigl\{ \lambda_{1}\bigl(D^{l}_{1}u_{*}- \sqrt{K_{1}}\bigr)-(\lambda _{1}\tau_{3}K_{1}+ \lambda_{1}\tau_{1}K_{2}+\lambda_{1} \tau_{3}K_{3}+\lambda_{1}\tau _{1}K_{3} \\ &{} +\lambda_{2}\tau_{2}K_{4}+ \lambda_{2}\tau_{4}K_{5}) \bigr\} \times \bigl\vert x_{1}(t)-x_{2}(t)\bigr\vert -\bigl\{ \lambda_{2}\bigl(D^{l}_{2}v_{*}+P\bigr)-(\lambda _{1}\tau_{1}L_{1} \\ &{}+\lambda_{1}\tau_{3}L_{2}+ \lambda_{2}\tau_{4}L_{3}+\lambda _{2} \tau_{2}L_{3}+\lambda_{2}\tau_{2}L_{4}+ \lambda_{2}\tau_{4}L_{5})\bigr\} \times\bigl\vert y_{1}(t)-y_{2}(t)\bigr\vert \\ =&-(\lambda_{1}\beta_{1}-\lambda_{2} \alpha_{1}) \bigl\vert x_{1}(t)-x_{2}(t) \bigr\vert -(\lambda_{2}\alpha_{2}-\lambda_{1} \beta_{2})\bigl\vert y_{1}(t)-y_{2}(t)\bigr\vert . \end{aligned}$$
(57)
If we denote \(\delta=\min \{\beta_{1}-\frac{\lambda _{2}}{\lambda_{1}}\alpha_{1}, \alpha_{2}-\frac{\lambda_{1}}{\lambda_{2}}\beta _{2} \}\), then \(\delta>0\) by the condition (C4).
Thus, it follows from equalities (37) and (38) that
$$\begin{aligned} D^{+}V(t) =&-\biggl(\beta_{1}-\frac{\lambda_{2}}{\lambda_{1}}\alpha_{1} \biggr)\lambda_{1} \bigl\vert x_{1}(t)-x_{2}(t) \bigr\vert \\ &{}-\biggl(\alpha_{2}-\frac{\lambda_{1}}{\lambda _{2}}\beta_{2} \biggr)\lambda_{2}\bigl\vert y_{1}(t)-y_{2}(t) \bigr\vert \\ \leq&-\delta\bigl[\lambda_{1}\bigl\vert x_{1}(t)-x_{2}(t) \bigr\vert +\lambda_{2} \bigl\vert y_{1}(t)-y_{2}(t) \bigr\vert \bigr] \\ =&-\delta V_{1}(t)=-\delta\frac{V_{1}(t)}{V(t)}V(t) \\ \leq&-\frac{\underline{\lambda} \vert \Phi(0)-\Psi(0)\vert _{0}}{M}\delta V(t). \end{aligned}$$
(58)
Let \(\gamma= \frac{\underline{\lambda}|\Phi (0)-\Psi(0)|_{0}}{M}\delta\), then \(\gamma>0\) and
$$ D^{+}V(t)\leq-\gamma V(t), \quad t\in R^{+}. $$
(59)
This means the last condition of Lemma 2.4 is satisfied. By Lemma 2.4, the system admits a unique uniformly asymptotically stable almost periodic solution \((x(t),y(t))^{T}\).
Thus, by the transformation (28), we can conclude that system (6) admits a unique uniformly asymptotically stable almost periodic solution \((u(t),v(t))^{T}=(e^{x(t)},e^{y(t)})^{T}\).
Finally, we will explain that system (4) has a unique uniformly asymptotically stable almost periodic solution.
In fact, from Lemma 2.2, we know that
$$\bigl(x(t),y(t)\bigr)^{T}= \biggl(\prod_{0< t_{k}< t}(1+h_{1k})u(t), \prod_{0< t_{k}< t}(1+h_{2k})v(t) \biggr)^{T} $$
is a solution of system (4). Since the condition (C2) holds, similar to the proofs of Lemma 31 and Theorem 79 in [28], we can prove that \((x(t),y(t))^{T}\) is almost periodic. Therefore, \((x(t),y(t))^{T}\) is the unique uniformly asymptotically stable almost periodic solution of system (4) because of the uniqueness and the uniformly asymptotical stability of \((u(t),v(t))^{T}\).
This completes the proof of this theorem. □