Now, we will discuss extinction and persistence of system (1.3). For any positive solution \((x(t),u(t))\) of system (1.3) we first introduce some useful definitions.
Definition 4.1
System (1.3) is said to be extinction almost surely, if
$$\lim_{t\rightarrow\infty}x(t)=0\quad \mbox{and}\quad \lim _{t\rightarrow\infty }u(t)=0 \quad \mbox{a.s.}; $$
nonpersistence in the mean, if
$$\lim_{t\rightarrow\infty}\bigl\langle x(t)\bigr\rangle =0\quad \mbox{and} \quad \lim_{t\rightarrow\infty}\bigl\langle u(t)\bigr\rangle =0 \quad \mbox{a.s.}; $$
uniform persistence in the mean, if there are positive constants m and M such that
$$m\le\langle x\rangle_{\ast}\le\langle x\rangle^{\ast}\le M\quad \mbox{and} \quad m\le\langle u\rangle_{\ast}\le\langle u \rangle^{\ast}\le M\quad \mbox{a.s.} $$
For convenience and simplicity in the following discussion, we denote \(b(t)=r(t)0.5\sigma^{2}(t)\) and \((x(t),u(t))=(x(t,0,x_{0},u_{0}),u(t,0,x_{0},u_{0}))\) for any \((x_{0},u_{0})\in R_{+}^{2}\). Applying Itô’s formula to \(\ln x(t)\), we have
$$ \mathrm{d}\ln x(t)=\bigl(b(t)a(t)x(t)c(t)u(t)\bigr)\,\mathrm{d}t+\sigma (t)\, \mathrm{d}B(t). $$
(4.1)
Then we have
$$ \ln x(t)=\ln x_{0}+ \int_{0}^{t}\bigl(b(s)a(s)x(s)c(s)u(s)\bigr)\, \mathrm {d}s+M(t), $$
(4.2)
where \(M(t)=\int_{0}^{t}\sigma(s)\,\mathrm{d}B(s)\). By the second equation of system (1.3) we have
$$ u(t)u_{0}= \int_{0}^{t}e(s)u(s)\,\mathrm{d}s+ \int_{0}^{t}f(s)x(s)\, \mathrm {d}s . $$
(4.3)
Note that \(M(t)\) is a local martingale. Making use of the strong law of large numbers for local martingales (see Mao [27]), we have
$$ \lim_{t\rightarrow\infty}\frac{M(t)}{t}=0\quad \mbox{a.s.} $$
(4.4)
We denote \(\Omega_{0}=\{\lim_{t\rightarrow\infty}M(t)/t=0\}\), obviously, \(\mathbb{P}(\Omega_{0})=1\).
Theorem 4.1
If (H_{2}) holds and
\(\langle b\rangle^{\ast}<0\), then system (1.3) will go to extinction almost surely.
Proof
For any \(\omega\in\Omega_{0}\), from (4.2) we have
$$ \frac{\ln x(t,\omega)}{t}\le\frac{\ln x_{0}}{t}+\frac{1}{t} \int _{0}^{t}b(s)\,\mathrm{d}s+\frac{M(t,\omega)}{t}. $$
(4.5)
Making use of (4.4) we obtain
$$\limsup_{t\rightarrow\infty}\frac{\ln x(t,\omega)}{t}\le\langle b \rangle^{\ast}. $$
That is to say, \(\lim_{t\rightarrow\infty}x(t,\omega)=0\) for \(\langle b\rangle^{\ast}<0\). Now, we will prove \(\lim_{t\rightarrow\infty }u(t,\omega )=0\). Since \(\lim_{t\rightarrow\infty}x(t,\omega)=0\), then for any \(\alpha_{0}>0\), there is a positive constant \(T_{0}\) such that
$$\bigl\vert x(t,\omega)\bigr\vert < \frac{\alpha_{0}}{ f_{u}} \quad \mbox{for all }t \ge T_{0}. $$
Consequently, from (4.3) we have
$$\frac{\mathrm{d}u(t,\omega)}{\mathrm{d}t}\lee(t)u(t,\omega )+\alpha_{0}\quad \mbox{for all }t\ge T_{0}. $$
We consider the comparison equation
$$ \frac{\mathrm{d}v(t)}{\mathrm{d}t}= e(t)v(t)+\alpha_{0}. $$
(4.6)
By (H_{2}) and Lemma 2.3 with \(m(t)\equiv0\) and \(v(0)=0\), we see for any positive constant ε that there are constants \(\delta=\delta(\varepsilon)\) and \(T_{1}=T_{1}(u(T_{0}))>T_{0}\) such that when \(\alpha_{0}<\delta\), we have
$$\bigl\vert v(t)\bigr\vert < \varepsilon \quad \mbox{for all }t\ge T_{1}, $$
where \(v(t)\) is the solution of system (4.6) with initial condition \(v(T_{0})=u(T_{0},\omega)\). Therefore, by the comparison theorem, we obtain
$$u(t,\omega)< \varepsilon \quad \mbox{for all }t\ge T_{1}. $$
Since ε is arbitrary, we have \(\lim_{t\rightarrow\infty }u(t,\omega)=0\). This complete the proof of the theorem, for \(\mathbb {P}(\Omega_{0})=1\). □
Remark 4.1
If \(c(t)\equiv0\), we can obtain system (3.8). In Theorem 7 in [22], the authors obtained the extinction of system (3.8) under the same conditions with Theorem 4.1. Hence, if \(\langle b\rangle^{\ast}<0\), the feedback control cannot change the extinction of the species x.
Theorem 4.2
Suppose
\(\langle b\rangle^{\ast}=0\), we have

(i)
if
\(\langle a\rangle_{\ast}>0\), \(\langle c\rangle_{\ast}>0\), and (H_{2}) hold, then
\(\liminf_{t\rightarrow\infty}x(t)=0\)
and
\(\liminf_{t\rightarrow\infty}u(t)=0\)
a.s.;

(ii)
if
\(a_{l}, e_{l}>0\), then system (1.3) will be nonpersistent in the mean a.s.
Proof
(i) First of all, we will prove \(\liminf_{t\rightarrow\infty}x(t,\omega)=0\) for all \(\omega\in\Omega_{0}\). Otherwise, there is a positive constant \(\varepsilon_{0}\) such that
$$\liminf_{t\rightarrow\infty}x(t,\omega_{0})>\varepsilon_{0} \quad \mbox{for some }\omega_{0}\in \Omega_{0}. $$
Hence, by \(\langle a\rangle_{\ast}>0\) and \(\langle b\rangle^{\ast}=0\), for any positive constant \(\varepsilon<\varepsilon_{0}\) there is a positive constant \(T_{0}\) such that
$$ x(t,\omega_{0})\ge\varepsilon \quad \mbox{and} \quad \frac{M(t,\omega _{0})}{t}< \frac{\langle a\rangle_{\ast}}{8}\varepsilon\quad \mbox{for all }t\ge T_{0}. $$
(4.7)
And there is a positive constant \(T_{1}=T_{1}(\varepsilon)>T_{0}\) such that
$$ \frac{1}{t} \int_{T_{0}}^{t} \bigl(b(s)\varepsilon a(s) \bigr)\, \mathrm {d}s< \frac {\langle a\rangle_{\ast}}{2}\varepsilon\quad \mbox{for all }t\ge T_{1}. $$
(4.8)
Then from (4.2), (4.7), and (4.8) we have
$$\begin{aligned} \begin{aligned} \ln x(t,\omega_{0})\ln x(T_{0},\omega_{0})& \le \int_{T_{0}}^{t} \bigl(b(s)a(s)\varepsilon \bigr)\, \mathrm{d}s+M(t,\omega_{0})M(T_{0},\omega _{0}) \\ &\le\frac{\langle a\rangle_{\ast}}{4}t\quad \mbox{for all }t\ge T_{1}. \end{aligned} \end{aligned}$$
Consequently, we have
$$x(t,\omega_{0})\le x(T_{0},\omega_{0})\exp \biggl\{ \frac{\langle a\rangle_{\ast}}{4}t\biggr\} . $$
Letting \(t\rightarrow\infty\) we have \(\limsup_{t\rightarrow\infty }x(t,\omega_{0})\le0\), which is a contradiction. Therefore,
$$\liminf_{t\rightarrow\infty}x(t,\omega)=0 \quad \mbox{for all }\omega\in \Omega_{0}. $$
Now, we will prove \(\liminf_{t\rightarrow\infty}u(t,\omega)=0\) for all \(\omega\in\Omega_{0}\). Otherwise, there is a \(\eta_{0}>0\) such that
$$\liminf_{t\rightarrow\infty}u(t,\omega_{0})\ge2\eta_{0} \quad \mbox{for some }\omega_{0}\in \Omega_{0}. $$
Consequently, we see that there is a positive constant \(T_{0}\) such that
$$u(t,\omega_{0})\ge\eta_{0}\quad \mbox{for all }t\ge T_{0}. $$
From (4.2) we can obtain
$$\ln x(t,\omega_{0})\ln x(T_{0},\omega_{0})\le \int_{T_{0}}^{t} \bigl(b(s)\eta _{0}c(s) \bigr)\,\mathrm{d}s+M(t,\omega_{0})M(T_{0}, \omega_{0}) $$
for all \(t\ge T_{0}\). Dividing the two side of above equation by t and letting \(t\rightarrow\infty\), we can get
$$\limsup_{t\rightarrow\infty}\frac{\ln x(t,\omega_{0})}{t}\le\langle b \rangle^{\ast}\eta _{0}\langle c\rangle_{\ast}< 0\quad \mbox{for }\langle c\rangle_{\ast}>0. $$
This leads to \(\lim_{t\rightarrow\infty}x(t,\omega_{0})=0\). By the proof of Theorem 4.1 we can obtain \(\lim_{t\rightarrow\infty}u(t, \omega_{0})=0\). This is a contradiction. Therefore, the proof of (i) is completed.
(ii) \(\langle b\rangle^{\ast}=0\) and (4.4) imply that, for any \(\varepsilon>0\) and \(\omega\in\Omega_{0}\), there is a positive constant \(T_{0}\) such that
$$\int_{0}^{t} b(s)\,\mathrm{d}s\le\frac{\varepsilon t}{2} \quad \mbox{and}\quad M(t,\omega)\le\frac{\varepsilon t}{2}\quad \mbox{for all }t\ge T_{0}. $$
Then it follows from (4.2) that
$$ \ln x(t,\omega)\le\ln x_{0}+\varepsilon t a_{l} \int_{0}^{t}x(s,\omega )\,\mathrm{d}s. $$
Let \(h(t)=\int_{0}^{t}x(s,\omega)\,\mathrm{d}s\), then we deduce that
$$e^{ a_{l}h(t)}\,\mathrm{d}h(t)\le x_{0}e^{\varepsilon t}\, \mathrm{d}t. $$
Integrating this inequality from \(T>T_{0}\) to t results in
$$e^{ a_{l}h(t)}\le e^{ a_{l}h(T)}+\frac{x_{0} a_{l}}{\varepsilon }\bigl(e^{\varepsilon t} e^{\varepsilon T}\bigr). $$
It follows that
$$h(t)\le\frac{1}{ a_{l}}\ln\biggl(e^{ a_{l}h(T)}+\frac{x_{0} a_{l}}{\varepsilon } \bigl(e^{\varepsilon t} e^{\varepsilon T}\bigr)\biggr). $$
Using L’Hospital’s rule we get
$$\limsup_{t\rightarrow\infty}\frac{h(t)}{t}\le\frac{\varepsilon}{ a_{l}}. $$
Since ε is arbitrary and \(x(t,\omega)>0\) (\(t>0\)), we can obtain \(\lim_{t\rightarrow\infty}\langle x(t,\omega)\rangle=0\).
Now, we will prove \(\lim_{t\rightarrow\infty}\langle u(t,\omega )\rangle=0\). Dividing both sides of equation (4.3) by t, we get
$$\begin{aligned} \frac{1}{t} \int_{0}^{t}u(s,\omega)\,\mathrm{d}s \le& \frac{ f_{u}}{ e_{l}t} \int _{0}^{t}x(s,\omega)\,\mathrm{d}s \frac{u(t,\omega)}{ e_{l}t}+\frac{u_{0}}{ e_{l}t} \\ \le&\frac{ f_{u}}{ e_{l}t} \int_{0}^{t}x(s,\omega)\,\mathrm{d}s+ \frac{u_{0}}{ e_{l}t}. \end{aligned}$$
From \(\lim_{t\rightarrow\infty}\langle x(t,\omega)\rangle=0\), letting \(t\rightarrow\infty\) we obtain \(\lim_{t\rightarrow\infty}\langle u(t,\omega )\rangle=0\). Since \(\mathbb{P}(\Omega_{0})=1\), this completes the proof of the theorem. □
Theorem 4.3
If
\(e_{l}>0\)
and
\(\langle b\rangle^{\ast}>0\), then species
x
will be weakly persistent in the mean a.s., i.e. \(\langle x\rangle ^{\ast}>0\)
a.s.
Proof
We claim that \(\Omega_{0}\subset\{\langle x\rangle ^{\ast}>0\}\). If the claim is not true, then \(\{\langle x\rangle^{\ast}=0\}\cap \Omega_{0}\neq\varnothing\). By the proof of (ii) in Theorem 4.2, if \(e_{l}>0\), we have \(\langle u(t,\omega)\rangle^{\ast}=0\) for any \(\omega\in\{ \langle x\rangle^{\ast}=0\}\cap\Omega_{0}\). It is easy to see that
$$ \bigl\langle a(t)x(t,\omega)\bigr\rangle ^{\ast}\le a_{u}\bigl\langle x(t,\omega )\bigr\rangle ^{\ast}=0 \quad \mbox{and}\quad \bigl\langle c(t)u(t,\omega)\bigr\rangle ^{\ast}\le c_{u}\bigl\langle u(t,\omega)\bigr\rangle ^{\ast}=0. $$
(4.9)
From (4.2) we get
$$\begin{aligned} \frac{\ln x(t,\omega)}{t} =&\frac{\ln x_{0}}{t}+\frac{1}{t} \int _{0}^{t}b(s)\,\mathrm{d}s\frac{1}{t} \int_{0}^{t}a(s)x(s,\omega)\, \mathrm {d}s \\ &{} \frac{1}{t} \int _{0}^{t}c(s)u(s,\omega)\,\mathrm{d}s+ \frac{M(t,\omega)}{t}. \end{aligned}$$
Combining this equation with (4.4) and (4.9) we have
$$ \limsup_{t\rightarrow\infty}\frac{\ln x(t,\omega)}{t}=\langle b\rangle^{\ast}. $$
Hence, there are a positive constant \(T_{0}\) and a time sequence \(\{t_{n}\} \) with \(t_{n}\ge T_{0}\) and \(t_{n+1}t_{n}\ge1\) for all \(n\ge1\) such that
$$ \frac{\ln x(t_{n},\omega)}{t_{n}}>\frac{\langle b\rangle^{\ast}}{2} \quad \mbox{and} \quad \frac{M(t,\omega)}{t}< \frac{\langle b\rangle ^{\ast}}{8}\quad \mbox{for all }t>T_{0}. $$
(4.10)
Let \(\bar{b}=\sup_{t\ge0}\{b(s)\}\). For any positive constant \(\Delta t<\min\{1,\langle b\rangle^{\ast}t_{1}/(8\bar{b})\}\) from (4.2) we have
$$\begin{aligned} \ln x(t_{n},\omega)\ln x(t,\omega) \le& \int_{t}^{t_{n}}b(s)\,\mathrm{d} s+M(t_{n}, \omega)M(t,\omega) \\ \le&\bar{b}\Delta t+\frac{\langle b\rangle^{\ast}}{4} t_{n}\quad \mbox{for all }t \in[t_{n}\Delta t,t_{n}]. \end{aligned}$$
Combining with (4.10) we obtain
$$\begin{aligned} \ln x(t,\omega) \ge&\ln x(t_{n},\omega)\bar{b}\Delta t \frac {\langle b\rangle^{\ast}}{4} t_{n} \\ \ge&\frac{\langle b\rangle^{\ast}}{4}t_{n}\bar{b}\Delta t \\ \ge&\frac{\langle b\rangle^{\ast}}{8}t_{n}\quad \mbox{for all }t\in [t_{n}\Delta t,t_{n}]\mbox{ and }n\ge1. \end{aligned}$$
Consequently,
$$\begin{aligned} \frac{1}{t_{n}} \int_{t_{1}}^{t_{n}} x(s,\omega)\,\mathrm{d}s \ge& \frac {1}{t_{n}}\sum_{t_{1}< t_{m}\le t_{n}} \int_{t_{m}\Delta t}^{t_{m}}x(s,\omega )\,\mathrm{d}s \\ \ge&\frac{1}{t_{n}}\sum_{t_{1}< t_{m}\le t_{n}}\Delta t\exp\biggl\{ \frac{\langle b\rangle^{\ast}}{8}t_{m}\biggr\} \\ \ge&\frac{\Delta t}{t_{n}}\exp\biggl\{ \frac{\langle b\rangle^{\ast}}{8}t_{n}\biggr\} . \end{aligned}$$
Since \(\langle b\rangle^{\ast}>0\), \(\lim_{n\rightarrow\infty}\frac {1}{t_{n}}\int_{t_{1}}^{t_{n}} x(s,\omega)\,\mathrm{d}s=+\infty\), which contradicts with \(\omega\in\{\langle x\rangle^{\ast}=0\}\cap\Omega_{0}\). Therefore, \(\Omega_{0}\subset\{\langle x\rangle^{\ast}>0\}\), i.e.
\(\langle x\rangle^{\ast}>0\) a.s. □
Remark 4.2
In Theorem 9 in [22], the authors studied the weakly persistent in the mean of system (3.8) with the conditions \(a_{l}>0\) and \(\langle b\rangle^{\ast}>0\). Obviously, from Theorem 4.3 we can obtain the same result with [22] only under the condition \(\langle b\rangle^{\ast}>0\). Therefore, the result in [22] is improved by Theorem 4.3.
Remark 4.3
In this theorem, due to shortage of the analysis techniques on the stochastic model, the weakly persistent in the mean of u case has not been studied. But we can see that the feedback control does not affect the persistence property of the species x under the conditions in this theorem.
Theorem 4.4
Assume
\(a_{l}>0\), \(e_{l}>0\), \(f_{l}>0\), and
\(\langle b\rangle_{\ast}>0\). Then system (1.3) will be uniform permanent in the mean a.s. Moreover,
$$\underline{x}\le\langle x\rangle_{\ast}\le\langle x\rangle^{\ast}\le \bar{x}\quad \textit{and}\quad \underline{u}\le\langle u\rangle_{\ast}\le\langle u\rangle^{\ast}\le \bar{u} \quad \textit{a.s.}, $$
where
\(\underline{x}=\langle b\rangle_{\ast}e_{l}/( a_{u} e_{l}+ c_{u} f_{u})\), \(\underline{u}=f_{l} e_{l}\langle b\rangle_{\ast}/e_{u}( a_{u} e_{l}+ c_{u} f_{u})\), \(\bar{x}=(\langle b\rangle^{\ast}e_{u}( a_{u} e_{l}+ c_{u} f_{u})c_{l}f_{l} e_{l}\langle b\rangle_{\ast})/ a_{l}e_{u}( a_{u} e_{l}+ c_{u} f_{u})\), and
\(\bar{u}=f_{u}(\langle b\rangle^{\ast}e_{u}( a_{u} e_{l}+ c_{u} f_{u})c_{l}f_{l} e_{l}\langle b\rangle_{\ast})/a_{l}e_{l}e_{u}( a_{u} e_{l}+ c_{u} f_{u})\).
Proof
From equation (4.3) we have
$$ \int_{0}^{t}e(s)u(s)\,\mathrm{d}s= \int_{0}^{t}f(s)x(s)\,\mathrm{d}su(t)+u_{0} \le f_{u} \int_{0}^{t}x(s)\,\mathrm{d}s+u_{0}. $$
Consequently, we have
$$ \int_{0}^{t}u(s)\,\mathrm{d}s\le\frac{f_{u}}{ e_{l}} \int_{0}^{t}x(s)\, \mathrm {d}s+ \frac {u_{0}}{e_{l}}. $$
(4.11)
For any \(\varepsilon>0\) and \(\omega\in\Omega_{0}\), there is a T such that
$$\bigl\langle b(t)\bigr\rangle >\langle b\rangle_{\ast}\frac{\varepsilon }{2} \quad \mbox{and}\quad \frac{M(t)}{t}>\frac{\varepsilon }{2}\quad \mbox{for all }t>T. $$
Substituting these inequalities and (4.11) into equation (4.2) we get
$$\ln x(t)\ln x_{0}\ge\nu t\biggl( a_{u}+ \frac{ c_{u} f_{u}}{ e_{l}}\biggr) \int _{0}^{t}x(s)\,\mathrm{d} s\frac{ c_{u}u_{0}}{ e_{l}} \quad \mbox{for all }t\ge T, $$
where \(\nu=\langle b\rangle_{\ast}\varepsilon\). Let \(g(t)=\int _{0}^{t}x(s)\,\mathrm{d}s\), then we have
$$\ln\frac{\mathrm{d}g(t)}{\mathrm{d}t}\ln x_{0}\ge\nu t\frac{ c_{u}u_{0}}{ e_{l}}\biggl( a_{u}+\frac{ c_{u} f_{u}}{ e_{l}}\biggr)g(t). $$
Consequently,
$$\exp\biggl\{ \biggl( a_{u}+\frac{ c_{u} f_{u}}{ e_{l}}\biggr)g(t)\biggr\} \frac{\mathrm {d}g(t)}{\mathrm{d}t}\ge x_{0}\exp\biggl\{ \nu t\frac{ c_{u}u_{0}}{ e_{l}}\biggr\} . $$
Integrating this inequality from T to t we have
$$\exp\biggl\{ \biggl( a_{u}+\frac{ c_{u} f_{u}}{ e_{l}}\biggr)g(t)\biggr\} \ge \exp\biggl\{ \biggl( a_{u}+\frac{ c_{u} f_{u}}{ e_{l}}\biggr)g(T)\biggr\} + \frac{x_{0}( a_{u}+\frac{ c_{u} f_{u}}{ e_{l}})}{\nu\exp\{\frac{ c_{u}u_{0}}{ e_{l}}\}}\bigl(\exp\{\nu t\}\exp\{\nu T\}\bigr). $$
Taking the logarithm of both sides yields
$$g(t)\ge\biggl( a_{u}+\frac{ c_{u} f_{u}}{ e_{l}}\biggr)^{1}\ln \biggl(\exp\biggl\{ \biggl( a_{u}+\frac { c_{u} f_{u}}{ e_{l}}\biggr)g(T)\biggr\} + \frac{x_{0}( a_{u}+\frac{ c_{u} f_{u}}{ e_{l}})}{\nu\exp\{\frac{ c_{u}u_{0}}{ e_{l}}\}}\bigl(\exp\{\nu t\}\exp\{\nu T\}\bigr) \biggr). $$
That is to say,
$$\begin{aligned} \langle x\rangle_{\ast} \ge&\liminf_{t\rightarrow\infty}\biggl[\biggl( a_{u}+\frac { c_{u} f_{u}}{ e_{l}}\biggr)t\biggr]^{1} \\ &{}\times\ln\biggl(\exp \biggl\{ \biggl( a_{u}+\frac{ c_{u} f_{u}}{ e_{l}}\biggr)g(T)\biggr\} + \frac{x_{0}( a_{u}+\frac{ c_{u} f_{u}}{ e_{l}})}{\nu\exp\{\frac{ c_{u}u_{0}}{ e_{l}}\}}\bigl(\exp\{\nu t\}\exp\{\nu T\}\bigr) \biggr). \end{aligned}$$
Using L’Hospital’s rule, we can obtain
$$\langle x\rangle_{\ast}\ge\frac{\nu}{ a_{u}+\frac{ c_{u} f_{u}}{ e_{l}}}=\frac{\nu e_{l}}{ a_{u} e_{l}+ c_{u} f_{u}}. $$
Since ε is arbitrary, we obtain
$$\langle x\rangle_{\ast}\ge\frac{\langle b\rangle_{\ast}e_{l}}{ a_{u} e_{l}+ c_{u} f_{u}}:=\underline{x}\quad \mbox{for all }\omega\in\Omega_{0}. $$
Now, we will prove \(\langle u\rangle_{\ast}\) also has a lower bound. From the above proof, we can see for any \(\varepsilon>0\) and \(\omega \in\Omega_{0}\) that there is a positive constant T such that
$$\int_{0}^{t}x(s)\,\mathrm{d}s\ge(\underline{x} \varepsilon)t\quad \mbox{for all }t\ge T. $$
Substituting the above inequality into (4.3), we have
$$ u(t)\ge e_{u} \int_{0}^{t}u(s)\,\mathrm{d}s+ f_{l}( \underline{x}\varepsilon )t\quad \mbox{for all }t\ge T. $$
Let \(h(t)=\int_{0}^{t}u(s)\,\mathrm{d}s\), then we have
$$ \frac{\mathrm{d}h(t)}{\mathrm{d}t}\ge e_{u}h(t)+ f_{l}(\underline {x} \varepsilon )t\quad \mbox{for all }t\ge T. $$
Consider the following comparison equation:
$$\frac{\mathrm{d}y(t)}{\mathrm{d}t}= e_{u}y(t)+ f_{l}(\underline {x} \varepsilon)t $$
with initial value \(y(T)=h(T)\). By the wellknown variationofconstants formula, we have
$$\begin{aligned} y(t) =& f_{l}(\underline{x}\varepsilon) \int_{T}^{t}s\exp\bigl\{  e_{u}(ts)\bigr\} \,\mathrm{d} s+h(T)\exp\bigl\{  e_{u}(tT)\bigr\} \\ =&\frac{ f_{l}}{ e_{u}}(\underline{x}\varepsilon) \biggl(\biggl(t \frac{1}{ e_{u}}\biggr)\biggl(T\frac{1}{ e_{u}}\biggr)\exp\bigl\{  e_{u}(tT)\bigr\} \biggr) +h(T)\exp\bigl\{  e_{u}(tT)\bigr\} . \end{aligned}$$
By the comparison theorem, we have
$$\liminf_{t\rightarrow\infty}\frac{h(t)}{t}\ge\liminf_{t\rightarrow\infty} \frac{y(t)}{t}=\frac{ f_{l}}{ e_{u}}(\underline{x}\varepsilon). $$
Since ε is arbitrary, we obtain
$$ \langle u\rangle_{\ast}=\liminf_{t\rightarrow\infty} \frac {h(t)}{t}\ge\frac{ f_{l}}{e_{u}}\underline{x}:=\underline{u}. $$
(4.12)
In the following, we will prove the upper bound of \(\langle x\rangle ^{\ast}\) and \(\langle u\rangle^{\ast}\). From (4.4) and (4.12), for any \(\varepsilon>0\) and \(\omega \in\Omega_{0}\) there exists a positive constant \(T_{0}\) such that
$$ \int_{0}^{t}b(s)\,\mathrm{d}s\le\biggl(\langle b \rangle^{\ast}+\frac {c_{l}\varepsilon}{3}\biggr)t,\qquad \int_{0}^{t}u(s)\,\mathrm{d}s\ge\biggl( \underline{u}\frac{\varepsilon }{3}\biggr)t \quad \mbox{and}\quad M(t)\le \frac{c_{l}\varepsilon}{3}t $$
(4.13)
for all \(t\ge T_{0}\). Substituting (4.13) into equation (4.2) we have
$$\begin{aligned} \ln x(t)\ln x_{0} \le&\biggl(\langle b\rangle^{\ast}+ \frac{c_{l}\varepsilon }{3}\biggr) t a_{l} \int_{0}^{t}x(s)\,\mathrm{d}sc_{l} \biggl(\underline{u}\frac{\varepsilon }{3}\biggr)t+\frac{c_{l}\varepsilon}{3}t \\ =&\bigl(\langle b\rangle^{\ast}c_{l}(\underline{u} \varepsilon)\bigr)ta_{l} \int _{0}^{t}x(s)\,\mathrm{d}s\quad \mbox{for all }t \ge T_{0}. \end{aligned}$$
Let \(k(t)=\int_{0}^{t}x(s)\,\mathrm{d}s\), then we have
$$\ln\frac{\mathrm{d}k(t)}{\mathrm{d}t}\ln x_{0}\le\rho ta_{l}k(t) \quad \mbox{for all }t\ge T_{0}, $$
where \(\rho=\langle b\rangle^{\ast}c_{l}(\underline{u}\varepsilon)\). Consequently,
$$\exp\bigl\{ a_{l}k(t)\bigr\} \frac{\mathrm{d}k(t)}{\mathrm{d}t}\le x_{0} \exp\{ \rho t\}\quad \mbox{for all }t\ge T_{0}. $$
Integrating this inequality from \(T_{0}\) to t we have
$$\exp\bigl\{ a_{l}k(t)\bigr\} \le\exp\bigl\{ a_{l}k(T_{0}) \bigr\} +\frac{a_{l}x_{0}}{\rho}\bigl(\exp\{ \rho t\}\exp\{\rho T_{0}\} \bigr). $$
Taking the logarithm of both sides yields
$$k(t)\le\frac{1}{a_{l}}\ln\biggl\{ \exp\bigl\{ a_{l}k(T_{0}) \bigr\} +\frac{a_{l}x_{0}}{\rho }\bigl(\exp\{\rho t\}\exp\{\rho T_{0}\}\bigr) \biggr\} . $$
That is to say,
$$\langle x\rangle^{\ast}\le\limsup_{t\rightarrow\infty} \frac {1}{a_{l}t}\ln\biggl\{ \exp\bigl\{ a_{l}k(T_{0})\bigr\} +\frac{a_{l}x_{0}}{\rho}\bigl(\exp\{\rho t\}\exp\{\rho T_{0}\}\bigr)\biggr\} . $$
Using L’Hospital’s rule, we can obtain \(\langle x\rangle^{\ast}\le\rho/a_{l}\). Since ε is arbitrary, we obtain
$$ \langle x\rangle^{\ast}\le\frac{\langle b\rangle^{\ast}c_{l}\underline {u}}{a_{l}}:=\bar{x}. $$
(4.14)
Rewriting equation (4.3) we have
$$\begin{aligned} \int_{0}^{t}u(s)\,\mathrm{d}s \le& \frac{f_{u}}{e_{l}} \int_{0}^{t}x(s)\, \mathrm {d}s\frac {u(t)}{e_{l}}+ \frac{u_{0}}{e_{l}} \\ \le&\frac{f_{u}}{e_{l}} \int_{0}^{t}x(s)\,\mathrm{d}s+\frac{u_{0}}{e_{l}}. \end{aligned}$$
Combining this inequality with equation (4.14), we have \(\langle u\rangle^{\ast}\le f_{u}\bar{x}/e_{l}:=\bar{u}\). This completes the proof. □
Remark 4.4
From Theorems 4.14.4, we can find that the feedback control is harmless to the permanence of the species under the randomized environment.