Extinction
In this section, we investigate the conditions for the extinction of the two microorganisms of system (2) under the stochastic white noise disturbance.
Lemma 3.1
Let
\((S(t), x_{1}(t), x_{2}(t),c(t))\)
be a solution of system (2) with initial value
\((S(0), x_{1}(0),x_{2}(0),c(0))\in R ^{4}_{+}\). Then
$$\begin{aligned}& \lim_{t\rightarrow +\infty }\frac{\int_{0}^{t}\frac{\sigma_{i} S( \theta)}{a_{i}+x_{i}(\theta)}\,\mathrm{d}B_{i}(\theta)}{t}=0, \\& \lim_{t\rightarrow +\infty } \frac{\int_{0}^{t}\sigma_{i} S(\theta) \,\mathrm{d}B_{i}(\theta)}{t}=0,\quad i=1,2, \textit{a.s.} \end{aligned}$$
Proof
Let \(Z(t)=\int_{0}^{t}\frac{\sigma_{i} S(\theta)}{a_{i}+x_{i}(\theta)}\,\mathrm{d}B_{i}(\theta)\) and \(\xi >2\). From Lemma 2.2 and Burkholder-Davis-Gundy inequality (see [52]) we have
$$\begin{aligned} E \Bigl[ \sup_{0\leq \theta \leq t} \bigl\vert Z( \theta) \bigr\vert ^{\xi } \Bigr] \leq& C _{\xi }E \biggl[ \int_{0}^{t}\frac{\sigma_{i}^{2} S^{2}(\theta)}{(a_{i}+x _{i}(\theta))^{2}}\,\mathrm{d}\theta \biggr] ^{\frac{\xi }{2}} \\ \leq& C_{\xi }t^{\frac{\xi }{2}}E \biggl[ \sup_{0\leq \theta \leq t} \frac{ \sigma_{i}^{\xi } S^{\xi }(\theta)}{(a_{i}+x_{i}(\theta))^{\xi }} \biggr] \\ \leq& M_{\xi }C_{\xi }t^{\frac{\xi }{2}}, \end{aligned}$$
where \(M_{\xi }= ( \frac{S_{0}\sigma_{i}}{a_{i}} ) ^{\xi }\). Let ε be an arbitrary positive constant. Then we can observe that
$$\begin{aligned} \mathbb{P} \Bigl\{ \omega:\sup_{k\delta \leq t\leq (k+1)\delta } \bigl\vert Z(t) \bigr\vert ^{ \xi }>(k\delta)^{1+\varepsilon +\frac{\xi }{2}} \Bigr\} \leq& \frac{E ( \vert Z((k+1)\delta)\vert ^{\xi } ) }{(k\delta)^{1+\varepsilon +\frac{ \xi }{2}}} \\ \leq& \frac{M_{\xi }C_{\xi }[(k+1)\delta ]^{\frac{\xi }{2}}}{(k \delta)^{1+\varepsilon +\frac{\xi }{2}}} \\ \leq& \frac{2^{\frac{\xi }{2}}M_{\xi }C_{\xi }}{(k\delta)^{1+\varepsilon }}. \end{aligned}$$
By the Borel-Cantelli lemma and Doob’s martingale inequality (see [52]), for almost all \(\omega \in \Omega \), we have that
$$\begin{aligned} \sup_{k\delta \leq t\leq (k+1)\delta } \bigl\vert Z(t) \bigr\vert ^{\xi }\leq (k\delta)^{1+ \varepsilon +\frac{\xi }{2}} \end{aligned}$$
(6)
for all but finitely many k. Thus, there exists a positive \(k_{0}(\omega)\) such that, for almost all \(\omega \in \Omega \), (6) holds when \(k\geq k_{0}(\omega)\). Hence, if \(k\geq k _{0}(\omega)\) and \(k\delta \leq t\leq (k+1)\delta \), then, for almost all \(\omega \in \Omega \),
$$\frac{\ln \vert Z(t)\vert ^{\xi }}{\ln t}\leq \frac{ ( 1+\varepsilon +\frac{ \xi }{2} ) \ln (k\delta)}{\ln (k\delta)}=1+\varepsilon +\frac{ \xi }{2}. $$
Thus we have
$$\limsup_{t\rightarrow +\infty }\frac{\ln \vert Z(t)\vert }{\ln t}\leq \frac{1+ \varepsilon +\frac{\xi }{2}}{\xi }. $$
Letting \(\varepsilon \rightarrow 0\), we obtain
$$\limsup_{t\rightarrow +\infty }\frac{\ln \vert Z(t)\vert }{\ln t}\leq \frac{1}{2}+ \frac{1}{\xi }. $$
Then, for an arbitrary small positive constant ϵ
\((\epsilon <\frac{1}{2}-\frac{1}{\xi })\), there exist a constant \(T(\omega)\) and a set \(\Omega_{\epsilon }\) such that \(\mathbb{P}(\Omega_{\epsilon }) \geq 1-\epsilon \) and, for \(t\geq T(\omega)\), \(\omega \in \Omega_{\epsilon }\),
$$\ln \bigl\vert Z(t) \bigr\vert \leq \biggl( \frac{1}{2}+ \frac{1}{\xi }+\epsilon \biggr) \ln t. $$
Therefore,
$$\limsup_{t\rightarrow +\infty }\frac{Z(t)}{t}\leq \limsup _{t\rightarrow +\infty }\frac{t^{\frac{1}{2}+\frac{1}{\xi }+ \epsilon }}{t}=0. $$
Note that
$$\liminf_{t\rightarrow +\infty }\frac{\vert Z(t)\vert }{t}\geq 0. $$
Then we have
$$\lim_{t\rightarrow +\infty }\frac{\vert Z(t)\vert }{t}=0\quad \mbox{a.s.} $$
that is,
$$\lim_{t\rightarrow +\infty }\frac{Z(t)}{t}=\lim_{t\rightarrow +\infty } \frac{\int_{0}^{t}\frac{\sigma_{i} S(\theta)}{a_{i}+x_{i}(\theta)} \,\mathrm{d}B_{i}(\theta)}{t}=0 \quad \mbox{a.s.} $$
Similarly, we can obtain
$$\lim_{t\rightarrow +\infty }\frac{\int_{0}^{t}\sigma_{i} S(\theta) \,\mathrm{d}B_{i}(\theta)}{t}=0,\quad i=1,2, \mbox{a.s.} $$
This completes the proof of Lemma 3.1. □
Define
$$\begin{aligned}& \mathcal{R}_{1}^{*}=\frac{\mu_{1} S_{0}}{a_{1}(Q+\frac{r_{1}u }{h \tau })}-\frac{\sigma_{1}^{2} S_{0}^{2}}{2a_{1}^{2}(Q+\frac{r_{1}u }{h \tau })}= \mathcal{R}_{1}-\frac{\sigma_{1}^{2} S_{0}^{2}}{2a_{1}^{2}(Q+\frac{r _{1}u }{h\tau })}, \\& \mathcal{R}_{2}^{*}=\frac{\mu_{2} S_{0}}{a_{2}(Q+\frac{r_{2}u }{h \tau })}-\frac{\sigma_{2}^{2} S_{0}^{2}}{2a_{2}^{2}(Q+\frac{r_{2}u }{h \tau })}= \mathcal{R}_{2}-\frac{\sigma_{2}^{2} S_{0}^{2}}{2a_{2}^{2}(Q+\frac{r _{2}u }{h\tau })}, \end{aligned}$$
where \(\mathcal{R}_{1}\), \(\mathcal{R}_{2}\) are the thresholds of the deterministic system (1) given in (5).
Theorem 3.1
Let
\((S(t), x_{1}(t), x_{2}(t),c(t))\)
be the solution of system (2) with initial value
\((S(0), x_{1}(0),x_{2}(0),c(0))\in R ^{4}_{+}\). If (i) \(\sigma_{i}>\frac{\mu_{i}}{\sqrt{2 ( Q+\frac{r _{i}u }{h\tau } ) }}\)
for
\(i=1,2\)
or (ii) \(\mathcal{R}_{i}^{*}<1\)
and
\(\sigma_{i}\leq \sqrt{\frac{a_{i}\mu_{i}}{S_{0}}}\)
for
\(i=1,2\), then the two microorganisms of system (2) go to extinction almost surely, that is, \(\lim_{t\rightarrow +\infty }x_{i}(t)=0\) (\(i=1,2\)) a.s.; moreover, \(\lim_{t\rightarrow +\infty }S(t)=S_{0}\)
a.s. and
\(\lim_{t\rightarrow +\infty }c(t)=c^{*}(t)\)
for
\(t\in (n\tau,(n+1) \tau ]\)
and
\(n\in Z^{+}\).
Proof
Applying Itô’s formula to system (2) leads to
$$\begin{aligned} \mathrm{d}\ln x_{i}(t) =& \biggl( \frac{\mu_{i} S(t)}{a_{i}+x_{i}(t)}-Q-r _{i}c(t)-\frac{\sigma_{i}^{2}S^{2}(t)}{2(a_{i}+x_{i}(t))^{2}} \biggr) \, \mathrm{d}t \\ &{}+\frac{\sigma_{i} S(t)}{a_{i}+x_{i}(t)} \,\mathrm{d}B_{i}(t),\quad i=1,2. \end{aligned}$$
(7)
Case (i). Integrating both sides of (7) from 0 to t results in
$$\begin{aligned} \ln x_{i}(t) =&-\frac{\sigma_{i}^{2}}{2} \int_{0}^{t} \biggl( \frac{\mu _{i} }{\sigma_{i}^{2}}- \frac{ S(t)}{a_{i}+x_{i}(t)} \biggr) ^{2} \,\mathrm{d}t-Qt-r_{i} \int_{0}^{t}c(\theta)\,\mathrm{d}\theta + \frac{\mu _{i}^{2}}{2\sigma_{i}^{2}}t+M_{i}(t)+\ln x_{i}(0) \\ \leq &-Qt-r_{i} \int_{0}^{t}c(\theta)\,\mathrm{d}\theta + \frac{\mu_{i} ^{2}}{2\sigma_{i}^{2}}t+M_{i}(t)+\ln x_{i}(0), \end{aligned}$$
(8)
where \(M_{i}(t)=\int_{0}^{t}\frac{\sigma_{i} S(\theta)}{a_{i}+x_{i}( \theta)}\,\mathrm{d}B_{i}(\theta)\), \(i=1,2\). Dividing both sides of (8) by t, we observe that
$$\begin{aligned} \frac{\ln x_{i}(t)}{t}\leq - \biggl( Q+r_{i} \bigl\langle c(t) \bigr\rangle -\frac{ \mu_{i}^{2}}{2\sigma_{i}^{2}} \biggr) +\frac{M_{i}(t)}{t}+ \frac{\ln x _{i}(0)}{t}. \end{aligned}$$
(9)
The process \(M_{i}(t)\) (\(i = 1, 2\)) is a local continuous martingale with \(M_{i}(0)=0\), and from Lemma 3.1 we have
$$\lim_{t\rightarrow +\infty }\frac{M_{i}(t)}{t}=0, \quad i=1,2, \mbox{a.s.} $$
Since \(\sigma_{i}>\frac{\mu_{i}}{\sqrt{2 ( Q+\frac{r_{i}u }{h \tau } ) }}\) for \(i=1,2\), we have \(- ( Q+r_{i}\langle c(t) \rangle -\frac{\mu_{i}^{2}}{2\sigma_{i}^{2}} ) <0\).
Taking the limit superior of both sides of (9), we can observe that
$$\begin{aligned} \limsup_{t\rightarrow +\infty }\frac{\ln x_{i}(t)}{t}\leq - \biggl( Q+r _{i} \bigl\langle c(t) \bigr\rangle -\frac{\mu_{i}^{2}}{2\sigma_{i}^{2}} \biggr) < 0 \quad \mbox{a.s.}, \end{aligned}$$
which implies \(\lim_{t\rightarrow +\infty }x_{i}(t)=0\), \(i=1,2\), a.s.
Case (ii). Integrating both sides of (7) from 0 to t and dividing by t lead to
$$\begin{aligned} \frac{\ln x_{i}(t)}{t} =&\frac{1}{t} \int_{0}^{t} \biggl( \frac{\mu_{i} S( \theta)}{a_{i}+x_{i}(\theta)}-Q-r_{i}c( \theta)-\frac{\sigma_{i} ^{2}S^{2}(\theta)}{2(a_{i}+x_{i}(\theta))^{2}} \biggr) \,\mathrm{d} \theta +\frac{M_{i}(t)}{t}+ \frac{\ln x_{i}(0)}{t} \\ \leq & \biggl( \frac{\mu_{i} S_{0}}{a_{i}}- \bigl(Q+r_{i} \bigl\langle c(t) \bigr\rangle \bigr)-\frac{\sigma_{i}^{2}S_{0}^{2}}{2a_{i}^{2}} \biggr) + \frac{M_{i}(t)}{t}+ \frac{\ln x_{i}(0)}{t} \\ =& \bigl(Q+r_{i} \bigl\langle c(t) \bigr\rangle \bigr) \biggl( \frac{\mu_{i} S_{0}}{a_{i}(Q+r _{i}\langle c(t)\rangle)}-\frac{\sigma_{i}^{2}S_{0}^{2}}{2a_{i}^{2}(Q+r _{i}\langle c(t)\rangle)}-1 \biggr) \\ &{} +\frac{M_{i}(t)}{t}+ \frac{\ln x _{i}(0)}{t}. \end{aligned}$$
(10)
Taking the limit superior of both sides of (10) yields
$$\begin{aligned} \limsup_{t\rightarrow +\infty }\frac{\ln x_{i}(t)}{t}\leq (Q+r_{i} \overline{c}) \bigl(\mathcal{R}_{i}^{*}-1 \bigr)< 0 \quad \mbox{a.s.}, \end{aligned}$$
which means \(\lim_{t\rightarrow +\infty }x_{i}(t)=0\) a.s.
Without loss of generality, we may assume that \(0< x_{i}(t)<\varepsilon _{i}\) (\(i=1,2\)) for an arbitrarily small positive quantity \(\varepsilon_{i}\) and all \(t\geq 0\). By the first equation of system (2) we have
$$\begin{aligned} \frac{\mathrm{d}S(t)}{\mathrm{d}t}\geq Q \bigl(S_{0}-S(t) \bigr)- \biggl( \frac{u _{1} \varepsilon_{1}}{\delta_{1}a_{1}}+ \frac{u_{2} \varepsilon_{2}}{ \delta_{2}a_{2}}+\frac{\sigma_{1} \varepsilon_{1}}{\delta_{1}a_{1}} \bigl\vert \dot{B}_{1}(t) \bigr\vert +\frac{\sigma_{2} \varepsilon_{2}}{\delta_{2}a _{2}} \bigl\vert \dot{B}_{2}(t) \bigr\vert \biggr) S(t). \end{aligned}$$
(11)
As \(\varepsilon_{1}\rightarrow 0\) and \(\varepsilon_{2}\rightarrow 0\), taking the limit inferior of both sides of (11) gives
$$\begin{aligned} \liminf_{t\rightarrow +\infty }S(t)\geq S_{0} \quad \mbox{a.s.} \end{aligned}$$
(12)
By the proof of Lemma 2.2 we have
$$\lim_{t\rightarrow +\infty }S(t)\leq S_{0}+\varepsilon_{1}+ \varepsilon_{2}\quad \mbox{a.s.} $$
Then, letting \(\varepsilon_{1}\rightarrow 0\) and \(\varepsilon_{2} \rightarrow 0\), we have
$$\begin{aligned} \limsup_{t\rightarrow +\infty }S(t)\leq S_{0} \quad \mbox{a.s.} \end{aligned}$$
(13)
From (12) and (13) we have
$$\begin{aligned} \lim_{t\rightarrow +\infty }S(t)=S_{0}\quad \mbox{a.s.} \end{aligned}$$
From (3) and Lemma 2.1 we can observe that
$$\lim_{t\rightarrow +\infty }c(t)=c^{*}(t) $$
for \(t\in (n\tau,(n+1)\tau ]\) and \(n\in Z^{+}\). □
Remark 3.1
Theorem 3.1 shows that the two microorganisms will die out if the white noise disturbance is large or \(\mathcal{R}_{i}^{*}<1\) and the white noise disturbance is not too large. Note that the expression of \(\mathcal{R}_{i}^{*}\) is a difference compared with two thresholds of system (1), \(\mathcal{R}_{i}\). This implies that the conditions for \(x_{i}(t)\) to go to extinction in the deterministic system (1) are stronger than in the corresponding stochastic model (2).
Persistence in mean
Theorem 3.2
Let
\((S(t), x_{1}(t), x_{2}(t),c(t))\)
be a solution of system (2) with initial value
\((S(0), x_{1}(0),x_{2}(0),c(0))\in R ^{4}_{+}\).
-
(i)
If
\(\mathcal{R}_{1}^{*}>1\), \(\mathcal{R}_{2}^{*}<1\), and
\(\sigma_{2}\leq \sqrt{\frac{a_{2}\mu_{2}}{S_{0}}}\), then the microorganism
\(x_{2}\)
dies out, and the microorganism
\(x_{1}\)
is persistent in mean; moreover, \(x_{1}\)
satisfies
$$\begin{aligned} \liminf_{t\rightarrow +\infty } \bigl\langle x_{1}(t) \bigr\rangle \geq \frac{a _{1}\delta_{1} Q(Q+r_{1}\overline{c})}{(\mu_{1}+\delta_{1} Q)(Q+r_{1}c ^{*}(0))} \bigl(\mathcal{R}_{1}^{*}-1 \bigr) \quad \textit{a.s.} \end{aligned}$$
-
(ii)
If
\(\mathcal{R}_{2}^{*}>1\), \(\mathcal{R}_{1}^{*}<1\), and
\(\sigma_{1}\leq \sqrt{\frac{a_{1}\mu_{1}}{S_{0}}}\), then the microorganism
\(x_{1}\)
dies out, and the microorganism
\(x_{2}\)
is persistent in mean; moreover, \(x_{2}\)
satisfies
$$\begin{aligned} \liminf_{t\rightarrow +\infty } \bigl\langle x_{2}(t) \bigr\rangle \geq \frac{a _{2}\delta_{2} Q(Q+r_{2}\overline{c})}{(\mu_{2}+\delta_{2} Q)(Q+r_{2}c ^{*}(0))} \bigl(\mathcal{R}_{2}^{*}-1 \bigr) \quad \textit{a.s.} \end{aligned}$$
-
(iii)
If
\(\mathcal{R}_{1}^{*}>1\)
and
\(\mathcal{R}_{2}^{*}>1\), then the two microorganisms
\(x_{1}\)
and
\(x_{2}\)
are persistent in mean; moreover, \(x_{1}\)
and
\(x_{2}\)
satisfy
$$\begin{aligned} \liminf_{t\rightarrow +\infty } \bigl\langle x_{1}(t)+x_{2}(t) \bigr\rangle \geq \frac{1}{\Delta_{\max }}\sum_{i=1}^{2}a_{i}(Q+r_{i} \overline{c}) \bigl( \mathcal{R}_{i}^{*}-1 \bigr) \quad \textit{a.s.}, \end{aligned}$$
where
$$\begin{aligned} \Delta_{\max } =&\max \biggl\{ \bigl(Q+r_{1}c^{*}(0) \bigr) \biggl( \frac{\mu_{1}+\mu _{2}}{\delta_{1}Q}+1 \biggr), \bigl(Q+r_{2}c^{*}(0) \bigr) \biggl( \frac{\mu_{1}+\mu _{2}}{\delta_{2}Q}+1 \biggr) \biggr\} . \end{aligned}$$
Proof
Case (i). By Theorem 3.1, since \(\mathcal{R}_{2}^{*}<1\) and \(\sigma_{2}\leq \sqrt{\frac{a_{2}\mu_{2}}{S_{0}}}\), we have \(\lim_{t\rightarrow +\infty }x_{2}(t)=0\) a.s. Since \(\mathcal{R}_{1}^{*}>1\), we have that, for ε small enough such that \(0< x_{2}(t)<\varepsilon \) for all t large enough,
$$\frac{\mu_{1} ( S_{0}-(\frac{Q+r_{2}c^{*}(0)}{\delta_{3}Q}\varepsilon)) }{a_{1}(Q+r\overline{c})}-\frac{\sigma_{1}^{2} S_{0}^{2}}{2a _{1}^{2}(Q+r_{1}\overline{c})}>1\quad \mbox{a.s.} $$
Integrating both sides of system (2) from 0 to t and dividing by t yield
$$\begin{aligned} \Theta (t) \triangleq &\frac{S(t)-S(0)}{t}+\frac{1}{\delta_{1}}\frac{x _{1}(t)-x_{1}(0)}{t}+ \frac{1}{\delta_{2}}\frac{x_{2}(t)-x_{2}(0)}{t} \\ \geq &QS_{0}-Q \bigl\langle S(t) \bigr\rangle - \biggl( \frac{Q+r_{1}c^{*}(0)}{ \delta_{1}} \biggr) \bigl\langle x_{1}(t) \bigr\rangle - \biggl( \frac{Q+r_{2}c^{*}(0)}{ \delta_{2}} \biggr) \bigl\langle x_{2}(t) \bigr\rangle \\ \geq &QS_{0}-Q \bigl\langle S(t) \bigr\rangle - \biggl( \frac{Q+r_{1}c^{*}(0)}{ \delta_{1}} \biggr) \bigl\langle x_{1}(t) \bigr\rangle - \biggl( \frac{Q+r_{2}c^{*}(0)}{ \delta_{2}} \biggr) \varepsilon. \end{aligned}$$
Then we get
$$\begin{aligned} \bigl\langle S(t) \bigr\rangle \geq \biggl( S_{0}- \biggl( \frac{Q+r_{2}c^{*}(0)}{ \delta_{2}Q} \biggr) \varepsilon \biggr) - \biggl( \frac{Q+r_{1}c^{*}(0)}{ \delta_{1}Q} \biggr) \bigl\langle x_{1}(t) \bigr\rangle - \frac{\Theta (t)}{Q}. \end{aligned}$$
(14)
Applying Itô’s formula to system (2) leads to
$$\begin{aligned}& \mathrm{d} \bigl( a_{1}\ln x_{1}(t)+x_{1}(t) \bigr) \\& \quad = \biggl( \mu_{1}S(t)-a_{1} \bigl(Q+r_{1}c(t) \bigr)- \bigl(Q+r_{1}c(t) \bigr)x_{1}(t)-\frac{a _{1}\sigma_{1}^{2}S^{2}(t)}{2(a_{1}+x_{1}(t))^{2}} \biggr) \,\mathrm{d}t+ \sigma_{1} S(t)\,\mathrm{d}B_{1}(t) \\& \quad \geq \biggl( \mu_{1}S(t)-a_{1} \bigl(Q+r_{1}c(t) \bigr)- \bigl(Q+r_{1}c^{*}(0) \bigr)x_{1}(t)- \frac{ \sigma_{1}^{2}S_{0}^{2}}{2a_{1}} \biggr) \,\mathrm{d}t+\sigma_{1} S(t) \, \mathrm{d}B_{1}(t). \end{aligned}$$
(15)
Integrating on both sides of (15) from 0 to t and dividing by t yield
$$\begin{aligned} &\frac{a_{1} ( \ln x_{1}(t)-\ln x_{1}(0)) }{t}+\frac{x_{1}(t)-x _{1}(0)}{t} \\ &\quad \geq \mu_{1} \bigl\langle S(t) \bigr\rangle -a_{1} \bigl(Q+r_{1} \bigl\langle c(t) \bigr\rangle \bigr)- \bigl(Q+r _{1}c^{*}(0) \bigr) \bigl\langle x_{1}(t) \bigr\rangle -\frac{\sigma_{1}^{2}S_{0}^{2}}{2a _{1}}+\frac{M_{1}(t)}{t} \\ &\quad \geq \mu_{1} \biggl( S_{0}- \biggl( \frac{Q+r_{2}c^{*}(0)}{\delta_{2}Q} \biggr) \varepsilon \biggr) -a_{1} \bigl(Q+r_{1} \bigl\langle c(t) \bigr\rangle \bigr)-\frac{\sigma _{1}^{2}S_{0}^{2}}{2a_{1}} \\ &\qquad {} - \biggl( \frac{\mu_{1}(Q+r_{1}c^{*}(0))}{\delta_{1}Q}+ \bigl(Q+r_{1}c ^{*}(0) \bigr) \biggr) \bigl\langle x_{1}(t) \bigr\rangle - \frac{\mu_{1}\Theta (t)}{Q}+ \frac{M _{1}(t)}{t} \\ &\quad =a_{1} \bigl(Q+r_{1} \bigl\langle c(t) \bigr\rangle \bigr) \biggl( \frac{\mu_{1}(S_{0}-\frac{Q+r _{2}c^{*}(0)}{\delta_{2}Q}\varepsilon)}{a_{1}(Q+r_{1}\langle c(t) \rangle)}-\frac{\sigma_{1}^{2}S_{0}^{2}}{2a_{1}(Q+r_{1}\langle c(t) \rangle)}-1 \biggr) \\ &\qquad {} - \biggl( \frac{\mu_{1}(Q+r_{1}c^{*}(0))}{\delta_{1}Q}+ \bigl(Q+r_{1}c ^{*}(0) \bigr) \biggr) \bigl\langle x_{1}(t) \bigr\rangle - \frac{\mu_{1}\Theta (t)}{Q}+ \frac{M _{1}(t)}{t}, \end{aligned}$$
(16)
where \(M_{1}(t)=\int_{0}^{t}\sigma_{1} S(\theta)\,\mathrm{d}B_{1}( \theta)\). Inequality (16) can be rewritten as
$$\begin{aligned} \bigl\langle x_{1}(t) \bigr\rangle \geq & \frac{1}{\Delta } \biggl[ a_{1} \bigl(Q+r_{1} \bigl\langle c(t) \bigr\rangle \bigr) \biggl( \frac{\mu_{1}(S_{0}-\frac{Q+r_{2}c^{*}(0)}{ \delta_{2}Q}\varepsilon)}{a_{1}(Q+r_{1}\langle c(t)\rangle)}-\frac{ \sigma_{1}^{2}S_{0}^{2}}{2a_{1}^{2}(Q+r_{1}\langle c(t)\rangle)}-1 \biggr) \\ &{} -\frac{\mu_{1}\Theta (t)}{Q}+\frac{M_{1}(t)}{t}- \biggl( \frac{a _{1} ( \ln x_{1}(t)-\ln x_{1}(0)) }{t}+ \frac{x_{1}(t)-x_{1}(0)}{t} \biggr) \biggr] \\ \geq & \textstyle\begin{cases} \frac{1}{\Delta } [ a_{1}(Q+r_{1}\langle c(t)\rangle) ( \frac{ \mu_{1}(S_{0}-\frac{Q+r_{2}c^{*}(0)}{\delta_{2}Q}\varepsilon)}{a_{1}(Q+r _{1}\langle c(t)\rangle)}-\frac{\sigma_{1}^{2}S_{0}^{2}}{2a_{1}^{2}(Q+r _{1}\langle c(t)\rangle)}-1) \\ \quad {} -\frac{\mu_{1}\Theta (t)}{Q}+\frac{M_{1}(t)}{t}+\frac{a _{1}\ln x_{1}(0)}{t}-\frac{x_{1}(t)-x_{1}(0)}{t} ] , \quad 0< x_{1}(t)< 1; \\ \frac{1}{\Delta } [ a_{1}(Q+r_{1}\langle c(t)\rangle) ( \frac{ \mu_{1}(S_{0}-\frac{Q+r_{2}c^{*}(0)}{\delta_{2}Q}\varepsilon)}{a_{1}(Q+r _{1}\langle c(t)\rangle)}-\frac{\sigma_{1}^{2}S_{0}^{2}}{2a_{1}^{2}(Q+r _{1}\langle c(t)\rangle)}-1) \\ \quad {} -\frac{\mu_{1}\Theta (t)}{Q}+\frac{M_{1}(t)}{t}-\frac{a _{1} ( \ln x_{1}(t)-\ln x_{1}(0)) }{t}- \frac{x_{1}(t)-x_{1}(0)}{t} ] ,\quad 1\leq x_{1}(t), \end{cases}\displaystyle \end{aligned}$$
(17)
where \(\Delta =\frac{(Q+r_{1}c^{*}(0))(\mu_{1}+\delta_{1} Q)}{\delta _{1} Q}\).
By Lemma 3.1 we have \(\lim_{t\rightarrow +\infty }\frac{M _{1}(t)}{t}=0\) a.s. According to Lemma 2.2, we can find that \(x_{1}(t)\leq \delta_{1} S_{0}\). Thus we have \(\lim_{t\rightarrow +\infty }\frac{x_{1}(t)}{t}=0\) and \(\lim_{t\rightarrow +\infty }\frac{\ln x_{1}(t)}{t}=0 \) a.s. as \(x_{1}(t)\geq 1\) and \(\lim_{t\rightarrow +\infty }\Theta (t)=0\) a.s. Taking the limit inferior of both sides of (17) results in
$$\begin{aligned} \liminf_{t\rightarrow +\infty } \bigl\langle x_{1}(t) \bigr\rangle \geq & \frac{a _{1}(Q+r_{1}\overline{c})}{\Delta } \biggl[ \frac{\mu_{1} S_{0}}{a_{1}(Q+r _{1}\overline{c})}-\frac{\sigma_{1}^{2}S_{0}^{2}}{2a_{1}^{2}(Q+r_{1} \overline{c})}-1 \biggr] \\ =&\frac{a_{1}\delta_{1} Q(Q+r_{1}\overline{c})}{(\mu_{1}+\delta_{1} Q)(Q+r _{1}c^{*}(0))} \bigl(\mathcal{R}_{1}^{*}-1 \bigr)>0. \end{aligned}$$
Similarly, we can prove Case (ii), and we omit it here.
Case (iii). Note that
$$\begin{aligned} \bigl\langle S(t) \bigr\rangle =S_{0}- \frac{Q+r_{2}\langle c(t)\rangle }{\delta _{2}Q} \bigl\langle x_{2}(t) \bigr\rangle - \frac{Q+r_{1}\langle c(t)\rangle }{ \delta_{1}Q} \bigl\langle x_{1}(t) \bigr\rangle - \frac{\Theta (t)}{Q}. \end{aligned}$$
(18)
Define
$$V(t)=\ln \bigl[ x_{1}^{a_{1}}(t)x_{2}^{a_{2}}(t) \bigr] + \bigl[ x_{1}(t)+x _{2}(t) \bigr] . $$
Note that \(V(t)\) is bounded. Then we have
$$\begin{aligned} D^{+}V(t) =& \Biggl[ (\mu_{1}+ \mu_{2})S(t)-\sum_{i=1}^{2} \bigl(Q+r_{i}c(t) \bigr) \bigl(a _{i}+x_{i}(t) \bigr)-\sum_{i=1}^{2}\frac{a_{i}\sigma_{i}^{2}S^{2}(t)}{2(a _{i}+x_{i}(t))^{2}} \Biggr] \,\mathrm{d}t \\ &+\sum_{i=1}^{2}\sigma_{i} S(t) \,\mathrm{d}B_{i}(t) \\ \geq & \Biggl[ (\mu_{1}+\mu_{2})S(t)-\sum _{i=1}^{2}a_{i} \bigl(Q+r_{i}c(t) \bigr)- \sum_{i=1}^{2}x_{i}(t) \bigl(Q+r_{i} c^{*}(0) \bigr)-\sum _{i=1}^{2}\frac{\sigma _{i}^{2}S_{0}^{2}}{2a_{i}} \Biggr] \,\mathrm{d}t \\ &+\sum_{i=1}^{2}\sigma_{i} S(t) \,\mathrm{d}B_{i}(t). \end{aligned}$$
(19)
Integrating both sides of (19) from 0 to t and dividing by t yield
$$\begin{aligned}& \frac{V(t)}{t}-\frac{V(0)}{t} \\& \quad \geq ( \mu_{1}+\mu_{2}) \bigl\langle S(t) \bigr\rangle -\sum_{i=1}^{2}a_{i} \bigl(Q+r_{i} \bigl\langle c(t) \bigr\rangle \bigr)- \sum _{i=1} ^{2} \bigl\langle x_{i}(t) \bigr\rangle \bigl(Q+r_{i}c^{*}(0) \bigr) \\& \qquad {}-\sum_{i=1}^{2} \frac{\sigma_{i}^{2}S_{0}^{2}(t)}{2a_{i}}+\sum_{i=1} ^{2} \frac{M_{i}}{t} \\& \quad =(\mu_{1}+\mu_{2})S_{0}-\sum _{i=1}^{2} \bigl(Q+r_{i} \bigl\langle c(t) \bigr\rangle \bigr)a _{i}-\sum_{i=1}^{2} \frac{\sigma_{i}^{2}S_{0}}{2a_{i}} \\& \qquad {}-\sum_{i=1}^{2} \biggl[ \frac{(\mu_{1}+\mu_{2})(Q+r_{i}\langle c(t) \rangle )}{\delta_{i}Q}+ \bigl(Q+r_{i}c^{*}(0) \bigr) \biggr] \bigl\langle x_{i}(t) \bigr\rangle \\& \qquad {} -\frac{\mu_{1}+\mu_{2}}{Q}\Theta (t)+\sum_{i=1}^{2} \frac{M_{i}}{t} \\& \quad \geq (\mu_{1}+\mu_{2})S_{0}-\sum _{i=1}^{2} \bigl(Q+r_{i} \bigl\langle c(t) \bigr\rangle \bigr)a_{i}-\sum_{i=1}^{2} \frac{\sigma_{i}^{2}S_{0}}{2a_{i}}- \Delta_{\max } \bigl[ \bigl\langle x_{1}(t) \bigr\rangle + \bigl\langle x_{2}(t) \bigr\rangle \bigr] \\& \qquad {}-\frac{\mu_{1}+\mu_{2}}{Q}\Theta (t)+\sum_{i=1}^{2} \frac{M_{i}}{t}, \end{aligned}$$
(20)
where \(M_{i}(t)=\int_{0}^{t}\sigma_{i} S(\theta)\,\mathrm{d}B_{i}( \theta)\). Inequality (20) can be rewritten as
$$\begin{aligned} \bigl\langle x_{1}(t) \bigr\rangle + \bigl\langle x_{2}(t) \bigr\rangle \geq &\frac{1}{ \Delta_{\max }} \Biggl[ ( \mu_{1}+\mu_{2})S_{0}-\sum _{i=1}^{2} \bigl(Q+r_{i} \bigl\langle c(t) \bigr\rangle \bigr)a_{i}-\sum_{i=1}^{2} \frac{\sigma_{i}^{2}S_{0}}{2a _{i}} \\ &{} -\frac{\mu_{1}+\mu_{2}}{Q}\Theta (t)-\frac{V(t)}{t}+ \frac{V(0)}{t}+ \sum_{i=1}^{2} \frac{M_{i}}{t} \Biggr] . \end{aligned}$$
(21)
Since \(0< S\leq S_{0}\), we have
$$\limsup_{t\rightarrow +\infty }\frac{\langle M_{i}(t),M_{i}(t) \rangle_{t}}{t}\leq \sigma^{2}S_{0}^{2}< \infty\quad \mbox{a.s.} $$
By Lemma 3.1 we observe that \(\lim_{t\rightarrow + \infty }\frac{M_{i}(t)}{t}=0\) a.s. for \(i=1,2\). According to Lemma 2.2, we have \(\lim_{t\rightarrow +\infty }\Theta (t)=0\) and \(\lim_{t\rightarrow +\infty }\frac{V(t)}{t}=0\).
Taking the limit inferior of both sides of (21) yields
$$\begin{aligned} \liminf_{t\rightarrow +\infty } \bigl\langle x_{1}(t)+x_{2}(t) \bigr\rangle \geq \frac{1}{\Delta_{\max }}\sum_{i=1}^{2}a_{i}(Q+r_{i} \overline{c}) \bigl( \mathcal{R}_{i}^{*}-1 \bigr)>0 \quad \mbox{a.s.} \end{aligned}$$
This completes the proof of Theorem 3.2. □
Remark 3.2
Theorem 3.2 shows that the two microorganisms will be persistent if the white noise disturbances are small enough such that \(\mathcal{R}_{i}^{*}>1\); conversely, if the white noise disturbances are large enough, then the two microorganisms will go to extinction. This implies that the stochastic disturbance may cause the populations to die out.