Now, we give our main results.
Theorem 3.1
Let
\((x_{1}(t), x_{2}(t),u(t))^{T}\)
be a solution of system (1.2). If condition (H1) or (H3) holds, then the unique positive equilibrium of system (1.2) is globally asymptotically stable, that is,
$$\lim_{t\rightarrow\infty}x_{i}(t)=x_{i}^{*},\quad i=1,2,\qquad \lim_{t\rightarrow\infty}u(t)=u^{*}. $$
Proof
Define a Lyapunov function as follows:
$$V_{1}(t)=\sum_{i=1}^{2} \eta_{i}\biggl(x_{i}-x_{i}^{*}-x_{i}^{*} \ln\frac {x_{i}}{x_{i}^{*}}\biggr)+\frac{c\eta_{2}}{2d}\bigl(u-u^{*} \bigr)^{2}, $$
where \(\eta_{1}=1\), and \(\eta_{2}\) is a positive constant to be determined.
Note that system (1.2) can be rewritten as
$$ \begin{gathered} \dot{x}_{1}(t) = x_{1}(t) \bigl[-a_{11}\bigl(x_{1}(t)-x_{1}^{*} \bigr)-a_{12}\bigl(x_{2}(t-\tau_{1})-x_{2}^{*} \bigr) \bigr], \\ \dot{x}_{2}(t) = x_{2}(t) \bigl[a_{21} \bigl(x_{1}(t-\tau _{2})-x_{1}^{*} \bigr)-a_{22}\bigl(x_{2}(t)-x_{2}^{*}\bigr)-c \bigl(u(t)-u^{*}\bigr) \bigr], \\ \dot{u}(t) = -e\bigl(u(t)-u^{*}\bigr)+d\bigl(x_{2}(t)-x_{2}^{*} \bigr). \end{gathered} $$
(3.1)
Calculating the derivative of \(V_{1}(t)\) along the solution \((x_{1}(t), x_{2}(t),u(t))^{T}\) of system (1.2), we have
$$ \begin{aligned}[b] \dot{V}_{1}(t)&= \bigl(x_{1}(t)-x_{1}^{*} \bigr) \bigl[-a_{11}\bigl(x_{1}(t)-x_{1}^{*} \bigr)-a_{12}\bigl(x_{2}(t-\tau _{1})-x_{2}^{*} \bigr)\bigr] \\ &\quad {}+\eta_{2}\bigl(x_{2}(t)-x_{2}^{*}\bigr) \bigl[a_{21}\bigl(x_{1}(t-\tau _{2})-x_{1}^{*} \bigr)-a_{22}\bigl(x_{2}(t)-x_{2}^{*}\bigr)-c \bigl(u(t)-u^{*}\bigr) \bigr] \\ &\quad{}+\frac{c\eta_{2}}{d}\bigl(u(t)-u^{*}\bigr) \bigl[-e \bigl(u(t)-u^{*}\bigr)+d\bigl(x_{2}(t)-x_{2}^{*}\bigr) \bigr] \\ &\leq-a_{11}\bigl(x_{1}(t)-x_{1}^{*} \bigr)^{2}-a_{22}\eta_{2}\bigl(x_{2}(t)-x_{2}^{*} \bigr)^{2} \\ &\quad{}-a_{12}\bigl(x_{1}(t)-x_{1}^{*}\bigr) \bigl(x_{2}(t-\tau_{1})-x_{2}^{*}\bigr)+\eta _{2}a_{21}\bigl(x_{2}(t)-x_{2}^{*}\bigr) \bigl(x_{1}(t-\tau_{2})-x_{1}^{*}\bigr). \end{aligned} $$
(3.2)
By the inequality \(ab\leq\frac{\theta}{2}a^{2}+\frac{1}{2\theta}b^{2}\), \(\theta>0\), from (3.2) it follows that
$$ \begin{aligned}[b] \dot{V}_{1}(t)&\leq-a_{11} \bigl(x_{1}(t)-x_{1}^{*}\bigr)^{2}-a_{22} \eta _{2}\bigl(x_{2}(t)-x_{2}^{*}\bigr)^{2} \\ &\quad{}+a_{12} \biggl[\frac{1}{2\theta_{1}}\bigl(x_{1}(t)-x_{1}^{*} \bigr)^{2}+\frac{\theta _{1}}{2}\bigl(x_{2}(t- \tau_{1})-x_{2}^{*}\bigr)^{2} \biggr] \\ &\quad{}+\eta_{2}a_{21} \biggl[\frac{1}{2\theta_{2}} \bigl(x_{1}(t-\tau _{2})-x_{1}^{*} \bigr)^{2}+\frac{\theta_{2}}{2}\bigl(x_{2}(t)-x_{2}^{*} \bigr)^{2} \biggr] \\ &=-a_{11}\bigl(x_{1}(t)-x_{1}^{*} \bigr)^{2}-a_{22}\eta_{2}\bigl(x_{2}(t)-x_{2}^{*} \bigr)^{2} \\ &\quad{}+\frac{a_{12}}{2\theta_{1}}\bigl(x_{1}(t)-x_{1}^{*} \bigr)^{2}+\frac{a_{12}\theta _{1}}{2}\bigl(x_{2}(t- \tau_{1})-x_{2}^{*}\bigr)^{2} \\ &\quad{}+\frac{\eta_{2}a_{21}}{2\theta_{2}}\bigl(x_{1}(t-\tau_{2})-x_{1}^{*} \bigr)^{2}+\frac {\eta_{2}a_{21}\theta_{2}}{2}\bigl(x_{2}(t)-x_{2}^{*} \bigr)^{2}. \end{aligned} $$
(3.3)
Let
$$V_{2}(t)=\frac{\eta_{2}a_{21}}{2\theta_{2}} \int_{t-\tau _{2}}^{t}\bigl(x_{1}(s)-x_{1}^{*} \bigr)^{2}\,ds+\frac{a_{12}\theta_{1}}{2} \int_{t-\tau _{1}}^{t}\bigl(x_{2}(s)-x_{2}^{*} \bigr)^{2}\,ds. $$
Calculating the derivative of \(V_{2}(t)\), we obtain
$$ \begin{aligned}[b] \dot{V}_{2}(t)&=\frac{\eta_{2}a_{21}}{2\theta_{2}} \bigl(x_{1}(t)-x_{1}^{*}\bigr)^{2}- \frac{\eta _{2}a_{21}}{2\theta_{2}}\bigl(x_{1}(t-\tau_{2})-x_{1}^{*} \bigr)^{2} \\ &\quad{}+\frac{a_{12}\theta_{1}}{2}\bigl(x_{2}(t)-x_{2}^{*} \bigr)^{2}-\frac{a_{12}\theta _{1}}{2}\bigl(x_{2}(t- \tau_{1})-x_{2}^{*}\bigr)^{2}. \end{aligned} $$
(3.4)
Define \(V(t)=V_{1}(t)+V_{2}(t)\). It follows from (3.3) and (3.4) that
$$ \begin{aligned}[b] \dot{V}(t) &\leq- \biggl(a_{11}-\frac{a_{12}}{2\theta_{1}}- \frac{\eta _{2}a_{21}}{2\theta_{2}} \biggr) \bigl(x_{1}(t)-x_{1}^{*} \bigr)^{2}\\ &\quad {}- \biggl(a_{22}\eta _{2}- \frac{a_{12}\theta_{1}}{2}-\frac{\eta_{2}a_{21}\theta_{2}}{2} \biggr) \bigl(x_{2}(t)-x_{2}^{*} \bigr)^{2} . \end{aligned} $$
(3.5)
Denote \(\delta_{1}=a_{11}-\frac{a_{12}}{2\theta_{1}}-\frac{\eta _{2}a_{21}}{2\theta_{2}}\) and \(\delta_{2}=a_{22}\eta_{2}-\frac{a_{12}\theta_{1}}{2}-\frac{\eta _{2}a_{21}\theta_{2}}{2}\).
Then taking \(\eta_{2}=\frac{a_{12}}{a_{21}}\) and \(\theta_{1}=\theta_{2}=\frac{a_{11}a_{22}+a_{12}a_{21}}{2a_{11}a_{21}}\), we have
$$ \delta_{1}=\frac{a_{11}(a_{11}a_{22}-a_{12}a_{21})}{a_{12}a_{21}+a_{11}a_{22}}, \qquad\delta_{2}= \frac{a_{12}(a_{11}a_{22}-a_{12}a_{21})}{2a_{11}a_{21}}. $$
(3.6)
Then, (H1) or (H3) shows that \(\delta_{i}>0\), \(i=1,2\). It is easy to see that
$$ \dot{V}(t) \leq-\delta_{1}\bigl(x_{1}(t)-x_{1}^{*} \bigr)^{2}-\delta_{2}\bigl(x_{2}(t)-x_{2}^{*} \bigr)^{2} . $$
(3.7)
Therefore, \(V(t)\) is nonincreasing. By Lemma 2.2, \(\dot{x}_{i}(t)\), \(i=1,2\), are bounded. So \(\vert x_{i}(t)-x_{i}^{*} \vert \), \(i=1,2\), are uniformly continuous on \([0, +\infty)\). Integrating both sides of (3.7) on the interval \([T,t)\), we have
$$V(t) +\delta_{1} \int_{T}^{t}\bigl(x_{1}(s)-x_{1}^{*} \bigr)^{2}\,ds+\delta_{2} \int _{T}^{t}\bigl(x_{2}(s)-x_{2}^{*} \bigr)^{2}\,ds< V(T) . $$
It follows from Lemma 2.2 and the initial condition \(\phi_{i}\) that \(x_{i}(t)\), \(i=1,2\), are bounded for \(t\in R\), that is, there exists \(M>0\) such that \(0< x_{i}(t)< M\), \(i=1,2\), \(t\in R\). Obviously, \(V_{1}(T)\) is bounded, and
$$ \begin{aligned}[b] V_{2}(t)&=\frac{\eta_{2}a_{21}}{2\theta_{2}} \int_{t-\tau _{2}}^{t}\bigl(x_{1}(s)-x_{1}^{*} \bigr)^{2}\,ds+\frac{a_{12}\theta_{1}}{2} \int_{t-\tau _{1}}^{t}\bigl(x_{2}(s)-x_{2}^{*} \bigr)^{2}\,ds \\ &\leq\frac{\eta_{2}a_{21}}{2\theta_{2}}\tau_{2}\bigl(M+x_{1}^{*} \bigr)^{2}+\frac {a_{12}\theta_{1}}{2}\tau_{1}\bigl(M+x_{2}^{*} \bigr)^{2}< +\infty. \end{aligned} $$
(3.8)
Therefore
$$\int_{T}^{t}\bigl(x_{i}(s)-x_{i}^{*} \bigr)^{2}\,ds< \frac{V(T)}{\delta_{i}}< +\infty,\quad i=1,2. $$
From this inequality it follows that \((x_{i}(s)-x_{i}^{*})^{2}\in L^{1}[0,+\infty)\), \(i=1,2\). By Barbalat’s lemma (see [30]) we conclude that
$$\lim_{t\rightarrow \infty}\bigl(x_{i}(t)-x_{i}^{*} \bigr)^{2}=0,\quad i=1,2, $$
and therefore
$$\lim_{t\rightarrow\infty}x_{i}(t)=x_{i}^{*},\quad i=1,2. $$
By the third equation of system (1.2) we have
$$\lim_{t\rightarrow\infty}u(t)=u^{*}. $$
This completes the proof of Theorem 3.1. □
Corollary 3.1
Let
\((x_{1}(t), x_{2}(t),u(t))^{T}\)
be a solution of system (1.2). If condition (H5) holds, then the unique positive equilibrium of system (1.2) is globally asymptotically stable, that is,
$$\lim_{t\rightarrow\infty}x_{i}(t)=x_{i}^{*},\quad i=1,2,\qquad \lim_{t\rightarrow\infty}u(t)=u^{*}. $$
Proof
(H5) implies (H3), so system (1.2) has a positive globally asymptotically stable equilibrium. □
Remark 3.1
If (H1) holds, then systems (1.1) and (1.2) are globally stable. Theorem 3.1 implies that the feedback control keeps the property of stability of system (1.2) but only changes the position of the unique positive equilibrium. That is, feedback control of system (1.2) leads to the number of the prey population increased (\(x_{1}^{*}=\frac {e(r_{1}a_{22}-r_{2}a_{12})+r_{1}cd}{e(a_{11}a_{22}+a_{12}{a_{21})+cda_{11}}}>\bar {x}_{1}=\frac{r_{1}a_{22}-r_{2}a_{12}}{a_{11}a_{22}+a_{12}{a_{21}}}\)) and the number of the predator population decreased (\(x_{2}^{*}= \frac {e(r_{2}a_{11}+{r_{1}}a_{21})}{e(a_{11}a_{22}+a_{12}{a_{21}})+cda_{11}}<\bar {x}_{2}=\frac{r_{2}a_{11}+r_{1}a_{21}}{a_{11}a_{22}+a_{12}{a_{21}}}\)).
Remark 3.2
If (H2) holds, system (1.1) is extinct. (H5) implies (H2), and system (1.2) has a unique positive equilibrium, which is globally asymptotically stable. Theorem 3.1 implies that the proper feedback control on predator species can change extinct prey species to be permanent.
Theorem 3.2
Assume that (H4) holds. Let
\((x_{1}(t), x_{2}(t),u(t))^{T}\)
be a solution of system (1.2). Then
$$\lim_{t\rightarrow \infty}x_{1}(t)=0,\qquad \int_{0}^{+\infty} x_{1}(s)\,ds< +\infty. $$
Proof
Condition (H4) implies that
$$ \frac{r_{1}}{r_{2}}< \frac{a_{12}}{a_{22}},\qquad \frac{c}{e}< \frac{a_{12}r_{2}-a_{22}r_{1}}{dr_{1}}. $$
(3.9)
By (3.9) we can choose positive constants \(\alpha>0\), \(\beta>0\), \(\gamma>0\) such that
$$\frac{r_{1}}{r_{2}}< \frac{\beta}{\alpha}< \frac{a_{12}}{a_{22}} $$
and
$$\frac{c}{e}< \frac{\gamma}{\beta}< \frac{\alpha a_{12}- \beta a_{22}}{d\beta}< \frac{a_{12}r_{2}-a_{22}r_{1}}{dr_{1}}. $$
Thus there exists \(\delta>0\) such that
$$ -\beta{r_{2}}+\alpha{r_{1}}< -\delta< 0,\qquad \beta a_{22}+\gamma d-\alpha a_{12}< 0,\qquad\beta c-\gamma e< 0. $$
(3.10)
Consider a Lyapunov functional of the form
$$V(t)=x_{1}^{\alpha}(t)x_{2}^{-\beta}(t)\exp \biggl[\gamma u(t)-\beta a_{21} \int_{t-\tau_{2}}^{t}x_{1}(s)\,ds-\alpha a_{12} \int_{t-\tau_{1}}^{t}x_{2}(s)\,ds \biggr]. $$
Calculating the derivative of V along the solution of system (1.2), we have
$$ \begin{aligned}[b]\dot{V}(t)&=V(t) \bigl[+\alpha \bigl(r_{1}-a_{11}x_{1}(t)-a_{12}x_{2}(t- \tau _{1}) \bigr) \\ &\quad{}-\beta \bigl(r_{2}+a_{21}x_{1}(t- \tau_{2})-a_{22}x_{2}(t)-cu(t) \bigr) \\ &\quad{}+\gamma \bigl(-eu(t)+dx_{2}(t) \bigr) \\ &\quad{}-\beta a_{21}x_{1}(t)+\beta a_{21}x_{1}(t- \tau_{2})-\alpha a_{12}x_{2}(t)+\alpha a_{12}x_{2}(t-\tau_{1}) \bigr] \\ &= V(t) \bigl[-\beta r_{2}+\alpha r_{1}+(-a_{11} \alpha-\beta a_{21})x_{1}(t) \\ &\quad{}+(\beta a_{22}+\gamma d-\alpha a_{12})x_{2}(t)+( \beta c-\gamma e)u(t) \bigr]. \end{aligned} $$
(3.11)
From inequalities (3.10) and (3.11) we obtain
$$\dot{V}(t)\leq-\delta V(t). $$
Integrating this inequality from 0 to t, we have
$$ V(t)\leq V(0)\exp(-\delta t). $$
(3.12)
By similar arguments as in the proof of Theorem 3.1, there exists \(M>0\) such that \(0< x_{i}(t)< M\), \(i=1,2\), \(t\in R\). So
$$ \begin{aligned}[b]V(0)&=x_{1}^{\alpha}(0)x_{2}^{-\beta}(0) \exp \biggl[\gamma u(0)-\beta a_{21} \int_{-\tau_{2}}^{0}x_{1}(s)\,ds-\alpha a_{12} \int_{-\tau_{1}}^{0}x_{2}(s)\,ds \biggr] \\ &\leq x_{1}^{\alpha}(0)x_{2}^{-\beta}(0)\exp \bigl[\gamma u(0)+\beta a_{21}M\tau+\alpha a_{12}M\tau \bigr] \\ &\leq +\infty. \end{aligned} $$
(3.13)
On the other hand,
$$ \begin{aligned}[b]V(t)&\geq x_{1}^{\alpha}(t)x_{2}^{-\beta}(t) \exp \biggl[-\beta a_{21} \int_{t-\tau_{2}}^{t}x_{1}(s)\,ds-\alpha a_{12} \int_{t-\tau_{1}}^{t}x_{2}(s)\,ds \biggr] \\ &\geq x_{1}^{\alpha}(t)x_{2}^{-\beta}(t)\exp [- \beta a_{21}M\tau-\alpha a_{12}M\tau ]. \end{aligned} $$
(3.14)
Combining inequalities (3.12), (3.13), and (3.14), we have
$$ x_{1}^{\alpha}(t) \leq x_{2}^{\beta}(t) \exp(\beta a_{21}M\tau+\alpha a_{12}M\tau) V(0)\exp(-\delta t). $$
(3.15)
Inequality (3.15) means that
$$x_{1}(t) \leq\lambda\exp\biggl(-\frac{\delta}{\alpha}t\biggr), $$
where \(\lambda= [M^{\beta}\exp(\beta a_{21}M\tau+\alpha a_{12}M\tau) V(0) ]^{\frac{1}{\alpha}}\). Hence, we obtain that
$$\lim_{t\rightarrow \infty}x_{1}(t)=0,\qquad \int_{0}^{+\infty} x_{1}(s)\,ds< +\infty. $$
This completes the proof of Theorem 3.2. □
Theorem 3.3
If condition (H4) holds, then the equilibrium
\((x_{1}^{**}, x_{2}^{**}, u^{**})=(0,\frac{er_{2}}{e a_{22}+cd}, \frac{dr_{2}}{e a_{22}+cd})\)
of system (1.2) is globally asymptotically stable, that is,
$$\lim_{t\rightarrow \infty}x_{1}(t)=0,\qquad\lim _{t\rightarrow \infty}x_{2}(t)=x_{2}^{**},\qquad \lim_{t\rightarrow \infty}u(t)=u^{**}. $$
Proof
Let \((x_{1}(t), x_{2}(t),u(t))^{T}\) be a solution of system (1.2). It follows from Theorem 3.2 that \(\lim_{t\rightarrow\infty}x_{1}(t)=0\), and it is easy to obtain that system (1.2) has an equilibrium \((x_{1}^{**}, x_{2}^{**}, u^{**})=(0,\frac{er_{2}}{e a_{22}+cd}, \frac{dr_{2}}{e a_{22}+cd})\). Therefore, we only need to verify
$$\lim_{t\rightarrow \infty}x_{2}(t)=x_{2}^{**}, \qquad\lim_{t\rightarrow \infty}u(t)=u^{**}. $$
As before, we define a Lyapunov function as follows:
$$ V_{1}(t)= x_{2}-x_{2}^{**}-x_{2}^{**} \ln\frac{x_{2}}{x_{2}^{**}}+\frac{c}{2d}\bigl(u-u^{**} \bigr)^{2}. $$
(3.16)
Calculating the derivative of \(V_{1}(t)\) along the solution of system (1.2), we obtain that
$$ \begin{aligned}[b] V'_{1}(t)&= \bigl(x_{2}(t)-x_{2}^{**}\bigr) \bigl[a_{21}x_{1}(t-\tau _{2})-a_{22} \bigl(x_{2}(t)-x_{2}^{**}\bigr)-c \bigl(u(t)-u^{**}\bigr) \bigr] \\ &\quad{}+ \frac{c}{d}\bigl(u(t)-u^{**}\bigr) \bigl[-e \bigl(u(t)-u^{**}\bigr)+d\bigl(x_{2}(t)-x_{2}^{**} \bigr) \bigr] \\ &\leq -a_{22}\bigl(x_{2}(t)-x_{2}^{**} \bigr)^{2}+a_{21}\bigl(M+x_{2}^{**} \bigr)x_{1}(t-\tau _{2})-\frac{ce}{d} \bigl(u(t)-u^{**}\bigr)^{2}. \end{aligned} $$
(3.17)
Let
$$V_{2}(t)=a_{21}\bigl(M+x_{2}^{**}\bigr) \int_{t-\tau_{2}}^{t}x_{1}(s)\,ds. $$
Calculating the derivative of \(V_{2}(t)\) along the solution of system (1.2), we have
$$ V'_{2}(t)\leq a_{21} \bigl(M+x_{2}^{**}\bigr)x_{1}(t)-a_{21} \bigl(M+x_{2}^{**}\bigr)x_{1}(t- \tau_{2}). $$
(3.18)
Define
$$V(t)=V_{1}(t)+V_{2}(t). $$
It follows from (3.17) and (3.18) that
$$ V'(t)\leq -a_{22}\bigl(x_{2}(t)-x_{2}^{**} \bigr)^{2}+a_{21}\bigl(M+x_{2}^{**} \bigr)x_{1}(t)-\frac {ce}{d}\bigl(u(t)-u^{**} \bigr)^{2}. $$
(3.19)
Integrating both sides of (3.19) on the interval \([T, t)\), we obtain
$$V(t)+ a_{22} \int_{T}^{t}\bigl(x_{2}(s)-x_{2}^{**} \bigr)^{2}\,ds+\frac{ce}{d} \int_{T}^{t}\bigl(u(s)-u^{**} \bigr)^{2}\,ds \leq V(T)+a_{21}\bigl(M+x_{2}^{**} \bigr) \int_{T}^{t}x_{1}(s)\,ds. $$
Obviously, \(V_{1}(t)\) is bounded, and
$$V_{2}(t)=a_{21}\bigl(M+x_{2}^{**}\bigr) \int_{t-\tau_{2}}^{t}x_{1}(s)\,ds\leq a_{21}\bigl(M+x_{2}^{**}\bigr)\tau M< +\infty. $$
By (H5) it follows Theorem 3.2 that \(\int_{0}^{+\infty} x_{1}(s)\,ds<+\infty\), and so \(\int_{T}^{t} x_{1}(s)\,ds<+\infty\). Hence, we have
$$\int_{T}^{t}\bigl(x_{2}(s)-x_{2}^{**} \bigr)^{2}\,ds< +\infty,\qquad \int _{T}^{t}\bigl(u(s)-u^{**} \bigr)^{2}\,ds< +\infty . $$
Similarly to the analysis of the proof of Theorem 3.1, by Barbalat’s lemma (see [30]) we have
$$\lim_{t\rightarrow\infty}\bigl(x_{2}(t)-x_{2}^{**} \bigr)^{2}=0,\qquad \lim_{t\rightarrow\infty}\bigl(u(t)-u^{**} \bigr)^{2}=0, $$
that is,
$$\lim_{t\rightarrow\infty}x_{2}(t)=x_{2}^{**}, \qquad \lim_{t\rightarrow\infty}u(t)=u^{**}. $$
This completes the proof of Theorem 3.3. □
Remark 3.3
By comparative analysis of (H3) and (H4) note that when the feedback control repression \(\frac{cd}{e}\) remains small, feedback control on predator species has no influence on the extinction of system (1.2).