In this section, we establish a permanence result for system (1.4). First, let us state several lemmas which will be useful in proving the main results.
Now let us consider the first order difference equation
$$ y(n+1)=Ay(n)+B, \quad n=1,2,\ldots, $$
(2.1)
where A and B are positive constants.
Lemma 2.1
[12]
Assume that
\(|A|<1\), for any initial value
\(y(0)\), there exists a unique solution
\(y(n)\)
of (2.1) which can be expressed as follows:
$$ y(n)=A^{n}\bigl(y(0)-y^{\ast}\bigr)+y^{\ast}, $$
(2.2)
where
\(y^{\ast}=B/(1-A)\). Thus, for any solution
\(y(n)\)
of system (2.1),
$$ \lim_{n\rightarrow+\infty} y(n)=y^{\ast}. $$
(2.3)
Lemma 2.2
[12]
Let
\(n\in N^{+}_{n_{0}}=\{n_{0}, n_{0}+1,\ldots,n_{0}+l,\ldots\}\), \(r\geq0\). For any fixed
n, \(g(n,r)\)
is a nondecreasing function with respect to
r, and for
\(n\geq n_{0}\), the following inequalities hold:
$$ y(n+1)\leq g\bigl(n,y(n)\bigr), \qquad u(n+1)\geq g\bigl(n,u(n) \bigr). $$
(2.4)
If
\(y(n_{0})\leq u(n_{0})\), then
\(y(n)\leq u(n)\)
for all
\(n\geq n_{0}\).
Now let us consider the following single species discrete model:
$$ N(n+1)=N(n)\exp\bigl(a(n)-b(n)N(n)\bigr), $$
(2.5)
where \(a(n)\) and \(b(n)\) are strictly positive sequences of real numbers defined for \(n\in N=\{0,1,2,\ldots\}\) and \(0< a^{l}\leq a^{u}\), \(0< b^{l}\leq b^{u}\). We have the following lemma.
Lemma 2.3
Any solution of system (2.5) with initial condition
\(N(0)>0\)
satisfies
$$ m\leq \liminf_{n\rightarrow+\infty}N(n)\leq \limsup_{n\rightarrow+\infty}N(n) \leq M, $$
(2.6)
where
$$ M=\frac{1}{b^{l}}\exp\bigl(a^{u}-1\bigr), \qquad m= \frac{a^{l}}{b^{u}}\exp\bigl(a^{l}-b^{u}M\bigr). $$
(2.7)
Lemma 2.4
[13]
Let
\(x(n)\)
and
\(b(n)\)
be nonnegative sequences defined on
N
and
\(c\geq0\)
is a constant. If
$$ x(n)\leq c+ \sum^{n-1}_{s=0}b(s)x(s), \quad \textit{for } n\in N, $$
(2.8)
then
$$ x(n)\leq c \prod^{n-1}_{s=0}\bigl[1+b(s) \bigr],\quad \textit{for } n\in N. $$
(2.9)
Lemma 2.5
[14]
Assume that
\(A>0\)
and
\(y(0)>0\), and further suppose that
$$y(n+1)\leq Ay(n)+B(n), \quad n=1,2,\ldots. $$
Then, for any integer
\(k\leq n\),
$$y(n)\leq A^{k}y(n-k)+ \sum^{k-1}_{i=0}A^{i}B(n-i-1). $$
Proposition 2.6
Assume that
$$ -r^{l}_{2}+\frac{\alpha^{u}_{21}}{a^{l}}>0 $$
(2.10)
holds, then
$$\limsup_{n\rightarrow+\infty}x_{i}(n)\leq M_{i},\qquad \limsup_{n\rightarrow+\infty}u_{i}(n)\leq W_{i}, \quad i=1,2, $$
where
$$\begin{aligned}& M_{1}= \frac{1}{d^{l}_{1}}\exp\biggl\{ r^{u}_{1}+ \frac{\alpha^{u}_{12}}{b^{l}}-1\biggr\} , \\& M_{2}=\exp\biggl\{ 2\biggl(-r^{l}_{2}+ \frac{\alpha^{u}_{21}}{a^{l}}\biggr)\biggr\} , \\& W_{i}= \frac{r^{u}_{i}M_{i}}{\eta^{l}_{i}}\quad (i=1,2). \end{aligned}$$
Proof
Let \((x_{1}(n), x_{2}(n),u_{1}(n),u_{2}(n))\) be any positive solution of system (1.4), it follows from the first equation of system (1.4) that
$$x_{1}(n+1)\leq x_{1}(n)\exp \biggl\{ r_{1}(n)-d_{1}(n)x_{1}(n)+ \frac {\alpha_{12}(n)}{ b(n)} \biggr\} . $$
By applying Lemmas 2.2 and 2.3, we have
$$ \limsup_{n\rightarrow+\infty}x_{1}(n)\leq \frac {1}{d^{l}_{1}}\exp \biggl\{ r^{u}_{1}+ \frac{\alpha^{u}_{12}}{b^{l}}-1\biggr\} \stackrel{ \mathrm {def}}{=}M_{1}. $$
(2.11)
Let \(x_{2}(n)=\exp\{v(n)\}\), then
$$\begin{aligned} v(n+1) \leq&v(n)+ \biggl(-r^{l}_{2}+ \frac{\alpha ^{u}_{21}}{a^{l}} \biggr) \\ =& \sum^{n}_{s=0}c(s)v(s)+ \biggl(-r^{l}_{2}+ \frac {\alpha^{u}_{21}}{a^{l}} \biggr), \end{aligned}$$
(2.12)
where
$$c(s)=\left \{ \textstyle\begin{array}{l@{\quad}l} 0,&0\leq s \leq n-1, \\ 1,&s=n. \end{array}\displaystyle \right . $$
Condition (2.10) shows that Lemma 2.4 could be applied to (2.12), it immediately follows that
$$v(n+1)\leq2\biggl(-r^{l}_{2}+ \frac{\alpha^{u}_{21}}{a^{l}}\biggr). $$
This is
$$ \limsup_{n\rightarrow+\infty}x_{2}(n)\leq\exp \biggl\{ 2 \biggl(-r^{l}_{2}+ \frac{\alpha^{u}_{21}}{a^{l}}\biggr) \biggr\} \stackrel{\mathrm {def}}{=}M_{2}. $$
(2.13)
For any small enough positive constant ε, it follows from (2.11) and (2.13) that there exists a large enough \(N_{0}\) such that
$$ x_{i}(n)\leq M_{i\varepsilon}, \quad \forall n\geq N_{0}, i=1,2. $$
(2.14)
From the third and fourth equations of the system (1.4) and (2.14), we can obtain
$$\begin{aligned}& u_{1}(n+1) \leq \bigl(1-\eta^{l}_{1} \bigr)u_{1}(n)+q^{u}_{1}M_{1\varepsilon}, \\& u_{2}(n+1) \leq \bigl(1-\eta^{l}_{2} \bigr)u_{1}(n)+q^{u}_{2}M_{2\varepsilon}. \end{aligned}$$
By applying Lemmas 2.1 and 2.2, it immediately follows that
$$\begin{aligned}& \limsup_{n\rightarrow+\infty}u_{1}(n) \leq \frac{q^{u}_{1}M_{1\varepsilon}}{\eta^{l}_{1}}, \\& \limsup_{n\rightarrow+\infty}u_{2}(n) \leq \frac{q^{u}_{2}M_{2\varepsilon}}{\eta^{l}_{2}}. \end{aligned}$$
Setting \(\varepsilon\rightarrow0\) in the above inequalities leads to
$$\begin{aligned}& \limsup_{n\rightarrow+\infty}u_{1}(n) \leq \frac{q^{u}_{1}M_{1}}{\eta^{l}_{1}} \stackrel{\mathrm{def}}{=}W_{1}, \\& \limsup_{n\rightarrow+\infty}u_{2}(n) \leq \frac{q^{u}_{2}M_{2}}{\eta^{l}_{2}} \stackrel{\mathrm{def}}{=}W_{2}. \end{aligned}$$
This completes the proof of Proposition 2.6. □
Theorem 2.7
In addition to (2.10), assume further that
$$ -r^{u}_{2}+ \frac{\alpha^{l}_{21}m_{1}}{1+a^{u}m_{1}}-e^{u}_{1}W_{2}>0, $$
(2.15)
then the system (1.4) is permanent.
Proof
By applying Proposition 2.6, it is easy to see that, to end the proof of Theorem 2.7, it is enough to show that under the conditions of Theorem 2.7,
$$\liminf_{n\rightarrow+\infty}x_{i}(n)\geq m_{i},\qquad \liminf_{n\rightarrow+\infty}u_{i}(n)\geq w_{i}, \quad i=1,2. $$
From Proposition 2.6, we know that for the above ε, there exists a \(N_{1}>N_{0}\) such that
$$ x_{i}(n)\leq M_{i\varepsilon}, \qquad u_{i}(n)\leq W_{i\varepsilon}, \quad i=1,2 \mbox{ for all } n>N_{1}. $$
(2.16)
From the first equation of system (1.4) and (2.16), we have
$$\begin{aligned} x_{1}(n+1) \geq& x_{1}(n)\exp\bigl\{ r^{l}_{1}-d^{u}_{1} M_{1\varepsilon }-e^{u}_{1}W_{1\varepsilon}\bigr\} \\ \geq& x_{1}(n)\exp\bigl\{ -d^{u}_{1} M_{1\varepsilon}-e^{u}_{1}W_{1\varepsilon}\bigr\} \\ \stackrel{\mathrm{def}}{=}& x_{1}(n)\exp\{D_{\varepsilon}\} \end{aligned}$$
for all \(n>N_{1}\), where \(D_{\varepsilon}=-d^{u}_{1} M_{1\varepsilon }-e^{u}_{1}W_{1\varepsilon}\). So for \(n\geq k\) we have
$$x_{1}(n-k)\leq x_{1}(n)\exp\{-D_{\varepsilon}k\}. $$
From the third equation of system (1.4), we have
$$\begin{aligned} u_{1}(n+1) \leq&\bigl(1-\eta^{l}_{1}\bigr) u_{1}(n)+ q^{u}_{1}x_{1}(n) \\ \stackrel{\mathrm{def}}{=}& Au_{1}(n)+B(n), \end{aligned}$$
where \(A=1-\eta^{l}_{1}\), \(B=q^{u}_{1}x_{1}(n)\). Then Lemma 2.5 implies that, for any \(n\geq k\),
$$\begin{aligned} u_{1}(n) \leq& A^{k}u_{1}(n-k)+ \sum ^{k-1}_{i=0}A^{i}B(n-i-1) \\ =&A^{k}u_{1}(n-k)+ \sum^{k-1}_{i=0}A^{i}q^{u}_{1}x_{1}(n-i-1) \\ \leq& A^{k}u_{1}(n-k)+q^{u}_{1}x_{1}(n) \sum^{k-1}_{i=0}A^{i}\exp\bigl\{ -D_{\varepsilon}(i+1)\bigr\} . \end{aligned}$$
Note that
$$0\leq A^{k}u_{1}(n-k)\leq A^{k}W_{1\varepsilon} \rightarrow0,\quad k\rightarrow+\infty. $$
We can choose \(N_{2}=\max\{N_{1}, \frac{\ln P_{1}}{\ln{A}} \}+1\), where \(P_{1}= \frac{r^{u}_{1}}{e^{u}_{1}W_{1\varepsilon}}\). As \(n>N_{2}\), we have \(r^{l}_{1}-e^{u}_{1}A^{N_{2}}W_{1\varepsilon}>0\), then we get
$$\begin{aligned} u_{1}(n) \leq& A^{N_{2}}u_{1}(n-N_{2})+q^{u}_{1}x_{1}(n) \sum^{N_{2}-1}_{i=0}A^{i}\exp\bigl\{ -D_{\varepsilon}(i+1)\bigr\} \\ \leq& A^{N_{2}}W_{1\varepsilon}+q^{u}_{1}x_{1}(n) \sum^{N_{2}-1}_{i=0}A^{i}\exp\bigl\{ -D_{\varepsilon}(i+1)\bigr\} \\ \stackrel{\mathrm{def}}{=}&A^{N_{2}}W_{1\varepsilon}+G_{\varepsilon}x_{1}(n), \end{aligned}$$
where \(G_{\varepsilon}=q^{u}_{1} \sum^{N_{2}-1}_{i=0}A^{i}\exp\{ -D_{\varepsilon}(i+1)\}\).
Considering the first equation of system (1.4), we have
$$\begin{aligned} x_{1}(n+1) \geq& x_{1}(n)\exp\bigl\{ r^{l}_{1}-d^{u}_{1}x_{1}(n)-e^{u}_{1}u_{1}(n) \bigr\} \\ \geq& x_{1}(n)\exp\bigl\{ r^{l}_{1}-e^{u}_{1} \bigl(A^{N_{2}}W_{1\varepsilon }+G_{\varepsilon}x_{1}(n) \bigr)- d^{u}_{1}x_{1}(n)\bigr\} \\ =& x_{1}(n)\exp\bigl\{ r^{l}_{1}-e^{u}_{1} A^{N_{2}}W_{1\varepsilon}- \bigl(e^{u}_{1}G_{\varepsilon}+d^{u}_{1} \bigr)x_{1}(n)\bigr\} \\ \stackrel{\mathrm{def}}{=}&x_{1}(n)\exp\bigl\{ E_{1\varepsilon}- E_{2\varepsilon}x_{1}(n)\bigr\} , \end{aligned}$$
where \(E_{1\varepsilon}=r^{l}_{1}-e^{u}_{1} A^{N_{2}}W_{1\varepsilon}\), \(E_{2\varepsilon}=e^{u}_{1}G_{\varepsilon}+d^{u}_{1}\).
By applying Lemmas 2.2 and 2.3, it immediately follows that
$$ \liminf_{n\rightarrow+\infty}x_{1}(n)\geq \frac{E_{1\varepsilon}}{E_{2\varepsilon}} \exp \{E_{1\varepsilon}-E_{2\varepsilon}M_{1}\}. $$
(2.17)
Setting \(\varepsilon\rightarrow0\) in (2.17) leads to
$$ \liminf_{n\rightarrow+\infty}x_{1}(n)\geq \frac{E_{1}}{E_{2}} \exp \{E_{1}-E_{2}M_{1}\}\stackrel{\mathrm{def}}{=}m_{1}. $$
(2.18)
Then we assume that \(\varepsilon<\frac{1}{2}m_{1}\), from (2.18) we know that there exists a large enough \(N_{2}>N_{1}\) such that
$$ x_{1}(n)\geq m_{1\varepsilon}, \quad \forall n\geq T_{2}. $$
(2.19)
From the second equation of system (1.4), (2.16), and (2.19), we have
$$\begin{aligned} x_{2}(n+1) =&x_{2}(n)\exp \biggl\{ -r_{2}(n)+ \frac{\alpha_{21}(n)}{a(n)}- \frac{\alpha_{21}(n)}{a(n)} \\ &{}\times \biggl( \frac{1+b(n)x_{2}(n)}{ 1+a(n)x_{1}(n)+b(n)x_{2}(n)} \biggr) -e_{2}(n)u_{2}(n) \biggr\} \\ \geq& x_{2}(n)\exp \biggl\{ -r_{2}(n)+ \frac{\alpha_{21}(n)}{a(n)}- \frac{\alpha_{21}(n)}{a(n) (1+a(n)m_{1\varepsilon} )} \\ &{} -e_{2}(n)W_{2\varepsilon}- \frac{\alpha_{21}(n)b(n)}{ a(n) (1+a(n)m_{1\varepsilon} )}x_{2}(n) \biggr\} \\ =& x_{2}(n)\exp \biggl\{ -r_{2}(n)+ \frac{\alpha _{21}(n)m_{1\varepsilon}}{ 1+a(n)m_{1\varepsilon}}-e_{2}(n)W_{2\varepsilon} \\ &{}- \frac{\alpha_{21}(n)b(n)}{ a(n) (1+a(n)m_{1\varepsilon} )}x_{2}(n) \biggr\} \end{aligned}$$
(2.20)
for all \(n>N_{2}\).
By applying Lemmas 2.2 and 2.3, it immediately follows that
$$\begin{aligned} \liminf_{n\rightarrow+\infty}x_{2}(n) \geq& \frac{ [ ( \frac{\alpha ^{l}_{21}m_{1\varepsilon}}{1+a^{u}m_{1\varepsilon}} )-r^{u}_{2}-e^{u}_{2}W_{2\varepsilon} ] (a^{l}(1+a^{l}m_{1\varepsilon}) )}{\alpha^{u}_{21}b^{u}} \\ &{}\times \exp \biggl\{ \frac{\alpha^{l}_{21}m_{1\varepsilon}}{ 1+a^{u}m_{1\varepsilon}}-r^{u}_{2}-e^{u}_{2}W_{2\varepsilon}- \frac{\alpha^{u}_{21}b^{u}}{ a^{l}(1+a^{l}m_{1\varepsilon})}M_{2} \biggr\} . \end{aligned}$$
Setting \(\varepsilon\rightarrow0\), we have
$$\begin{aligned} \liminf_{n\rightarrow+\infty}x_{2}(n) \geq& \frac{ [ ( \frac{\alpha^{l}_{21}m_{1}}{1+a^{u}m_{1}} )-r^{u}_{2}-e^{u}_{2}W_{2} ] (a^{l}(1+a^{l}m_{1}))}{\alpha^{u}_{21}b^{u}} \\ &{}\times \exp \biggl\{ \frac{\alpha^{l}_{21}m_{1}}{ 1+a^{u}m_{1}}-r^{u}_{2}-e^{u}_{2}W_{2}- \frac{\alpha^{u}_{21}b^{u}}{ a^{l}(1+a^{l}m_{1})}M_{2} \biggr\} \stackrel{\mathrm{def}}{=}m_{2}. \end{aligned}$$
(2.21)
Without loss of generality, we may assume that \(\varepsilon<(1/2)\min\{ m_{1},m_{2}\}\). It follows from (2.18) and (2.21) that there exists a large enough \(N_{3}>N_{2}\), such that
$$ x_{i}\geq m_{i\varepsilon}, \quad i=1,2, \forall n\geq N_{3}. $$
(2.22)
From the third and fourth equations of the system (1.4) and (2.22), we have
$$\begin{aligned}& u_{1}(n+1) \geq \bigl(1-\eta^{u}_{1} \bigr)u_{1}(n)+q^{l}_{1}m_{1\varepsilon}, \\& u_{2}(n+1) \geq \bigl(1-\eta^{u}_{2} \bigr)u_{1}(n)+q^{l}_{2}m_{2\varepsilon}. \end{aligned}$$
By applying Lemmas 2.1 and 2.2, it immediately follows that
$$\begin{aligned}& \liminf_{n\rightarrow+\infty}u_{1}(n) \geq \frac{q^{l}_{1}m_{1\varepsilon}}{\eta^{u}_{1}}, \\& \liminf_{n\rightarrow+\infty}u_{2}(n) \geq \frac{q^{l}_{2}m_{2\varepsilon}}{\eta^{u}_{2}}. \end{aligned}$$
Setting \(\varepsilon\rightarrow0\) in the above inequalities leads to
$$\begin{aligned}& \liminf_{n\rightarrow+\infty}u_{1}(n) \geq \frac{q^{l}_{1}m_{1}}{\eta^{u}_{1}} \stackrel{\mathrm{def}}{=}w_{1}, \\& \liminf_{n\rightarrow+\infty}u_{2}(n) \geq \frac{q^{l}_{2}m_{2}}{\eta^{u}_{2}} \stackrel{\mathrm{def}}{=}w_{2}. \end{aligned}$$
This completes the proof. □
Theorem 2.8
Assume that the inequality
$$ r^{l}_{2}> \frac{\alpha^{u}_{21}}{a^{l}} $$
(2.23)
holds. Let
\((x_{1}(n), x_{2}(n),u_{1}(n),u_{2}(n))\)
be any positive solution of system (1.4), then
\(x_{2}(n)\rightarrow0\), \(u_{2}(n)\rightarrow0\)
as
\(n\rightarrow+\infty\).
Proof
Equation (2.23) is equivalent to the following inequality:
$$ -r^{l}_{2}+ \frac{\alpha^{u}_{21}}{a^{l}}< 0. $$
(2.24)
From (2.24), there exists a \(\delta>0\) such that
$$ -r^{l}_{2}+ \frac{\alpha^{u}_{21}}{a^{l}}< -\delta< 0. $$
(2.25)
Let \((x_{1}(n), x_{2}(n),u_{1}(n),u_{2}(n))\) be any positive solution of system (1.4). For any \(q\in N\), according to the second equation of system (1.4), we obtain
$$\begin{aligned} \ln \frac{x_{2}(q+1)}{x_{2}(q)} =&-r_{2}(q)+ \frac{\alpha_{21}(q)x_{1}(q)}{ 1+a(q)x_{1}(q)+b(q)x_{2}(q)}-e_{2}(q)u_{2}(q) \\ \leq&r_{2}(q)+ \frac{\alpha_{21}(q)x_{1}(q)}{ 1+a(q)x_{1}(q)+b(q)x_{2}(q)} \\ \leq& -r^{l}_{2}+ \frac{\alpha^{u}_{21}}{a^{l}}< -\delta< 0. \end{aligned}$$
Summating both sides of the above inequalities from 0 to \(n-1\), we obtain
$$ \ln \frac{x_{2}(n)}{x_{2}(0)}< -\delta n, $$
(2.26)
then
$$ x_{2}(n)< x_{2}(0)\exp\{-\delta n\}. $$
(2.27)
From (2.27), \(x_{2}(n)\rightarrow0\) as \(n\rightarrow+\infty\).
Further, consider the fourth equation of system (1.4). Applying Lemma 2.3 in [15], we easily obtain \(u_{2}(n)\rightarrow0\) as \(n\rightarrow+\infty\). This completes the proof of Theorem 2.8. □
Theorem 2.9
Assume that (2.23) holds, and further assume that
$$ \begin{aligned} &\eta^{l}_{1}>e^{u}_{1}, \\ &\min \biggl[d^{l}_{1}, \frac{2}{M_{1}}-d^{u}_{1} \biggr]>q^{u}_{1}, \end{aligned} $$
(2.28)
then, for any two positive solutions
\((x_{1}(n),x_{2}(n),u_{1}(n),u_{2}(n))\)
and
\((x^{\ast}_{1}(n),x^{\ast}_{2}(n), u^{\ast}_{1}(n),u^{\ast}_{2}(n)) \)
of the system, we have
$$\lim_{n\rightarrow+\infty}\bigl(x_{1}(n)-x^{\ast}_{1}(n) \bigr)=0,\qquad \lim_{n\rightarrow+\infty}\bigl(u_{1}(n)-u^{\ast}_{1}(n) \bigr)=0. $$
Proof
By conditions (2.28), there exist a positive constant ε and δ such that
$$ \begin{aligned} &\eta^{l}_{1}-e^{u}_{1}> \delta, \\ &\min \biggl[d^{l}_{1}, \frac{2}{M_{1\varepsilon}}-d^{u}_{1} \biggr]-q^{u}_{1}>\delta. \end{aligned} $$
(2.29)
From Theorems 2.7 and 2.8, for the above ε, there exists \(N_{4}>N_{3}\) such that
$$m_{i\varepsilon} \leq x_{1}(n), \qquad x^{\ast}_{1}(n) \leq M_{i\varepsilon},\qquad x_{2}(n)\leq\varepsilon. $$
Using the mean value theorem, one has
$$ \ln x_{1}(n)-\ln x^{\ast}_{1}(n)= \frac{1}{\theta(n)}\bigl(x_{1}(n)- x^{\ast}_{1}(n) \bigr), $$
(2.30)
where \(\theta(n)\) is between \(x_{1}(n)\) and \(x^{\ast}_{1}(n)\).
Now we define
$$V_{1}(n)=\bigl\vert \ln x_{1}(n)-\ln x^{\ast}_{1}(n)\bigr\vert ,\qquad V_{2}(n)=\bigl\vert u_{1}(n)-u^{\ast}_{1}(n)\bigr\vert . $$
From the first equation of the system, we have
$$\begin{aligned} \Delta V_{1}(n) =&\bigl\vert \ln x_{1}(n+1)-\ln x^{\ast}_{1}(n+1)\bigr\vert -\bigl\vert \ln x_{1}(n)-\ln x^{\ast}_{1}(n)\bigr\vert \\ \leq& \bigl\vert \ln x_{1}(n)-\ln x^{\ast}_{1}(n)-d_{1}(n) \bigl(x_{1}(n)-x^{\ast}_{1}(n)\bigr) \bigr\vert \\ &{}- \bigl\vert \ln x_{1}(n)-\ln x^{\ast}_{1}(n)\bigr\vert +\alpha_{12}(n) \biggl\vert \frac{x_{2}(n)}{1+a(n)x_{1}(n)+b(n)x_{2}(n)} \\ &{}- \frac{x^{\ast}_{2}(n)}{1+a(n)x^{\ast}_{1}(n)+b(n)x^{\ast}_{2}(n)}\biggr\vert + e_{1}(n)\bigl\vert u_{1}(n)-u^{\ast}_{1}(n)\bigr\vert \\ \leq& -\biggl( \frac{1}{\theta(n)}- \biggl\vert \frac{1}{\theta(n)}-d_{1}(n) \biggr\vert \biggr)\bigl\vert x_{1}(n)-x^{\ast}_{1}(n) \bigr\vert \\ &{}+e^{u}_{1}\bigl\vert u_{1}(n)-u^{\ast}_{1}(n) \bigr\vert +\alpha^{u}_{12} \frac{x_{2}(n)+x^{\ast}_{2}(n)}{1+a^{l}m_{1\varepsilon} +b^{l}\varepsilon}. \end{aligned}$$
(2.31)
From the third equation of the system, we have
$$\begin{aligned} \Delta V_{2}(n) =&\bigl\vert u_{1}(n+1)-u^{\ast}_{1}(n+1) \bigr\vert -\bigl\vert u_{1}(n)-u^{\ast}_{1}(n) \bigr\vert \\ \leq& -\eta^{l}_{1}\bigl\vert u_{1}(n)-u^{\ast}_{1}(n) \bigr\vert +q^{u}_{1}\bigl\vert x_{1}(n)-x^{\ast}_{1}(n) \bigr\vert . \end{aligned}$$
(2.32)
Now we define a Lyapunov function as follows:
$$V(n)=V_{1}(n)+V_{2}(n). $$
From (2.31) and (2.32), we have
$$\begin{aligned} \Delta V(n) \leq&-\biggl\{ \min\biggl[d^{l}_{1}, \frac{2}{M_{1\varepsilon}}-d^{u}_{1}\biggr] -q^{u}_{1} \biggr\} \bigl\vert x_{1}(n)-x^{\ast}_{1}(n)\bigr\vert -\bigl\{ \eta^{l}_{1}-e^{u}_{1} \bigr\} \\ &{}\times\bigl\vert u_{1}(n)-u^{\ast}_{1}(n) \bigr\vert +\alpha^{u}_{12} \frac{x_{2}(n)+x^{\ast}_{2}(n)}{1+a^{l}m_{1\varepsilon} +b^{l}\varepsilon} \\ \leq& -\delta\bigl(\bigl\vert x_{1}(n)-x^{\ast}_{1}(n) \bigr\vert +\bigl\vert u_{1}(n)-u^{\ast}_{1}(n) \bigr\vert \bigr)+\alpha^{u}_{12} \frac{x_{2}(n)+x^{\ast}_{2}(n)}{1+a^{l}m_{1\varepsilon} +b^{l}\varepsilon}. \end{aligned}$$
Summating both sides of the above inequalities from \(N_{4}\) to n, we have
$$\begin{aligned} \sum^{n}_{p=N_{4}}\bigl(V(p+1)-V(p)\bigr) \leq& -\delta \sum^{n}_{p=N_{4}} \bigl(\bigl\vert x_{1}(n)-x^{\ast}_{1}(n)\bigr\vert +\bigl\vert u_{1}(n)-u^{\ast}_{1}(n)\bigr\vert \bigr) \\ &{} + \frac{\alpha^{u}_{12}}{1+a^{l}m_{1\varepsilon} +b^{l}\varepsilon} \sum^{n}_{p=N_{4}} \bigl(x_{2}(n)+x^{\ast}_{2}(n)\bigr). \end{aligned}$$
Hence
$$\begin{aligned}& V(n+1)+\delta \sum^{n}_{p=N_{4}} \bigl(\bigl\vert x_{1}(n)-x^{\ast}_{1}(n)\bigr\vert +\bigl\vert u_{1}(n)-u^{\ast}_{1}(n)\bigr\vert \bigr) \\& \quad \leq V(N_{4}) + \frac{\alpha^{u}_{12}}{1+a^{l}m_{1\varepsilon} +b^{l}\varepsilon} \sum ^{n}_{p=N_{4}}\bigl(x_{2}(n)+x^{\ast}_{2}(n) \bigr). \end{aligned}$$
From Theorem 2.8, we have \(\sum^{+\infty }_{n=0}x_{2}(n)<+\infty\) and \(\sum^{+\infty}_{n=0}x^{\ast}_{2}(n)<+\infty\). We notice that \(V(N_{4})\) is bounded. So from the above inequalities, we have
$$\sum^{n}_{p=N_{4}} \bigl(\bigl\vert x_{1}(n)-x^{\ast}_{1}(n)\bigr\vert +\bigl\vert u_{1}(n)-u^{\ast}_{1}(n)\bigr\vert \bigr)\leq+ \infty. $$
Therefore
$$\sum^{+\infty}_{p=N_{4}} \bigl(\bigl\vert x_{1}(n)-x^{\ast}_{1}(n)\bigr\vert +\bigl\vert u_{1}(n)-u^{\ast}_{1}(n)\bigr\vert \bigr)\leq+ \infty. $$
This means that
$$\lim_{n\rightarrow+\infty} \bigl(\bigl\vert x_{1}(n)-x^{\ast}_{1}(n) \bigr\vert +\bigl\vert u_{1}(n)-u^{\ast}_{1}(n) \bigr\vert \bigr)=0. $$
Consequently
$$\begin{aligned}& \lim_{n\rightarrow+\infty}\bigl(x_{1}(n)-x^{\ast}_{1}(n) \bigr)=0, \\& \lim_{n\rightarrow+\infty}\bigl(u_{1}(n)-u^{\ast}_{1}(n) \bigr)=0. \end{aligned}$$
This completes the proof of Theorem 2.9. □