It is critical to discuss the persistence and extinction for an ecological population system. In this section, we will mainly investigate the persistence in mean and extinction of system (2).
Theorem 5.1
-
(1)
If
\(r<0.5\sigma_{11}^{2}\), populations
x
and
y
of system (2) will go extinct a.s., namely, \(\lim_{t\rightarrow+\infty}x(t)=\lim_{t\rightarrow +\infty}y(t)=0\), a.s.
-
(2)
If
\(r>0.5\sigma_{11}^{2}\), population
x
is weakly persistent a.s., namely, \(x^{*}>0\).
-
(a)
If
\(r>0.5\sigma_{11}^{2}\)
and
\(\frac{a_{21}(r-0.5\sigma_{11}^{2})}{(1-\tau')a_{11}}-(d+0.5\sigma_{21}^{2})<0\), population
y
of system (2) will go extinct a.s., namely, \(\lim_{t\rightarrow+\infty}y(t)=0\), a.s.
-
(b)
If
\(r>0.5\sigma_{11}^{2}\)
and
\(\frac{a_{21}(r-0.5\sigma_{11}^{2})}{(1-\tau')a_{11}}-(d+0.5\sigma_{21}^{2})>0\), then
\(\langle y(t)\rangle^{*}\leq\frac{\frac{a_{21}(r-0.5\sigma _{11}^{2})}{(1-\tau')a_{11}}-(d+0.5\sigma_{21}^{2})}{a_{22}}\), a.s.
-
(3)
If
\(r=0.5\sigma_{11}^{2}\), then the population
x
is nonpersistent in the mean a.s., namely, \(\langle x\rangle^{*}=0\).
Proof
Applying Itô’s formula to system (2), we have
$$\begin{aligned} \begin{aligned} d\ln x ={}& \bigl[r-a_{11}x-a_{12}x \bigl(t-\tau_{1}(t) \bigr)-a_{13}y \bigl(t- \tau_{2}(t) \bigr) \bigr]\,dt \\ &-0.5 \bigl[\sigma_{11}^{2}+\sigma_{12}^{2}x^{2}+ \sigma_{13}^{2}x^{2} \bigl(t-\tau _{1}(t) \bigr)+\sigma_{14}^{2}y^{2} \bigl(t- \tau_{2}(t) \bigr) \bigr]\,dt \\ &+\sigma_{11}\,dB_{11}(t)+\sigma_{12}x\,dB_{12}(t)+ \sigma_{13}x \bigl(t-\tau _{1}(t) \bigr) \,dB_{13}(t)\\&+ \sigma_{14}y \bigl(t-\tau_{2}(t) \bigr)\,dB_{14}(t),\end{aligned} \\ \begin{aligned} d\ln y ={}& \bigl[-d+a_{21}x \bigl(t-\tau_{3}(t) \bigr)-a_{22}y(t)-a_{23}y \bigl(t-\tau_{4}(t) \bigr) \bigr]\, dt \\ &-0.5 \bigl[\sigma_{21}^{2}+\sigma_{23}^{2}y^{2}+ \sigma_{22}^{2}x^{2} \bigl(t-\tau _{3}(t) \bigr)+\sigma_{24}^{2}y^{2} \bigl(t- \tau_{4}(t) \bigr) \bigr]\,dt \\ &+\sigma_{21}\,dB_{21}(t)+\sigma_{23}y\,dB_{23}(t)+ \sigma_{22}x \bigl(t-\tau _{3}(t) \bigr) \,dB_{22}(t)\\&+ \sigma_{24}y \bigl(t-\tau_{4}(t) \bigr)\,dB_{24}(t). \end{aligned} \end{aligned}$$
Integrating both sides from 0 to t, we get
$$\begin{aligned}& \begin{aligned}[b] \ln x(t)/x(0) ={}& \int_{0}^{t}r-a_{11}x(s)-a_{12}x \bigl(s-\tau_{1}(s) \bigr)-a_{13}y \bigl(s-\tau _{2}(s) \bigr)\,ds \\ &-0.5 \int_{0}^{t}\sigma_{11}^{2}+ \sigma_{12}^{2}x^{2}(s)+\sigma_{13}^{2}x^{2} \bigl(s-\tau _{1}(s) \bigr)+\sigma_{14}^{2}y^{2} \bigl(s-\tau_{2}(s) \bigr)\,ds \\ &+ \int_{0}^{t}\sigma_{11}\, dB_{11}(s)+ \int_{0}^{t}\sigma_{12}x(s) \,dB_{12}(s)+ \int_{0}^{t}\sigma_{13}x \bigl(s-\tau _{1}(s) \bigr)\,dB_{13}(s)\\ &+ \int_{0}^{t}\sigma_{14}y \bigl(s- \tau_{2}(s) \bigr)\,dB_{14}(s), \end{aligned} \end{aligned}$$
(11)
$$\begin{aligned}& \begin{aligned}[b] \ln y(t)/y(0) ={}& \int_{0}^{t}-d+a_{21}x \bigl(s-\tau _{3}(s) \bigr)-a_{22}y(s)-a_{23}y \bigl(s- \tau_{4}(s) \bigr)\,ds \\ &-0.5 \int_{0}^{t}\sigma_{21}^{2}+ \sigma_{23}^{2}y^{2}(s)+\sigma_{22}^{2}x^{2} \bigl(s-\tau _{3}(s) \bigr)+\sigma_{24}^{2}y^{2} \bigl(s-\tau_{4}(s) \bigr)\,ds \\ &+ \int_{0}^{t}\sigma_{21}\, dB_{21}(s)+ \int_{0}^{t}\sigma_{23}y(s) \,dB_{23}(s)+ \int_{0}^{t}\sigma_{22}x \bigl(s-\tau _{3}(s) \bigr)\,dB_{22}(s)\\ &+ \int_{0}^{t}\sigma_{24}y \bigl(s- \tau_{4}(s) \bigr)\,dB_{24}(s), \end{aligned} \end{aligned}$$
(12)
where
$$\begin{aligned}& N_{1}(t)= \int_{0}^{t} \sigma_{11} \,dB_{11}(s),\qquad N_{2}(t)= \int_{0}^{t}\sigma _{12}x(s) \,dB_{12}(s), \\& N_{3}(t)= \int_{0}^{t}\sigma_{13}{x \bigl(s- \tau_{1}(s) \bigr)}\,dB_{13}(s),\qquad N_{4}(t)= \int _{0}^{t}\sigma_{14}{y \bigl(s- \tau_{2}(s) \bigr)}\,dB_{14}(s), \\& N_{5}(t)= \int_{0}^{t} \sigma_{21} \,dB_{21}(s),\qquad N_{6}(t)= \int_{0}^{t}\sigma _{22}x \bigl(s- \tau_{3}(s) \bigr)\,dB_{22}(s), \\& N_{7}(t)= \int_{0}^{t}\sigma_{23}y(s) \,dB_{23}(s),\qquad N_{8}(t)= \int_{0}^{t}\sigma _{24}y \bigl(s- \tau_{4}(s) \bigr)\,dB_{24}(s). \end{aligned}$$
Noting that \(N_{i}(t)\) is a local martingale, we have
$$ \begin{gathered} \bigl\langle N_{2}(t), N_{2}(t) \bigr\rangle = \int_{0}^{t} \sigma_{21}^{2}x^{2}(s) \,ds, \qquad\bigl\langle N_{3}(t), N_{3}(t) \bigr\rangle = \int_{0}^{t}\sigma_{23}^{2}x^{2} \bigl(s-\tau_{1}(s) \bigr)\, ds, \\ \bigl\langle N_{4}(t), N_{4}(t) \bigr\rangle = \int_{0}^{t}\sigma_{14}^{2}{y^{2} \bigl(s-\tau_{2}(s) \bigr)}\, ds,\qquad \bigl\langle N_{6}(t), N_{6}(t) \bigr\rangle = \int_{0}^{t}\sigma_{22}^{2}{x^{2} \bigl(s-\tau _{3}(s) \bigr)}\,ds, \\ \bigl\langle N_{7}(t), N_{7}(t) \bigr\rangle = \int_{0}^{t} \sigma_{23}^{2}y^{2}(s) \,ds, \qquad\bigl\langle N_{8}(t), N_{8}(t) \bigr\rangle = \int_{0}^{t}\sigma_{24}^{2}y^{2} \bigl(s-\tau_{4}(s) \bigr)\,ds. \end{gathered} $$
By the exponential martingale inequality (7) (choose \(T=k\), \(\alpha=1\), \(\beta=2\ln k\)), we have
$$ P \biggl\{ \sup_{0\leq t\leq\gamma k} \biggl[N_{i}(t)- \frac{1}{2} \bigl\langle N_{i}(t), N_{i}(t) \bigr\rangle \biggr]>2\ln k \biggr\} \leq1/k^{2},\quad i=2,3,4,6,7,8. $$
By virtue of the Borel-Cantelli lemma, for almost all \(\omega\in\Omega\), there exists a \(k_{0}(\omega)\) such that, for all \(k\geq k_{0}(\omega)\),
$$ N_{i}(t)\leq\frac{1}{2} \bigl\langle N_{i}(t), N_{i}(t) \bigr\rangle +2\ln k,\quad 0\leq t\leq k, k\geq k_{0}( \omega). $$
Substituting above inequalities into (11) and (12), we have
$$\begin{aligned}& \begin{aligned}[b] \ln x(t)/x(0) \leq{}& \int_{0}^{t}r-0.5\sigma_{11}^{2}-a_{11}x(s)-a_{12}x \bigl(s-\tau _{1}(s) \bigr)-a_{13}y \bigl(s- \tau_{2}(s) \bigr)\,ds\\ &+N_{1}(t)+6\ln k \\ \leq{}& \int_{0}^{t}r-0.5\sigma_{11}^{2} \,ds+N_{1}(t)+6\ln k, \end{aligned} \end{aligned}$$
(13)
$$\begin{aligned}& \begin{aligned}[b] \ln y(t)/y(0) \leq{}& \int_{0}^{t}-d-0.5\sigma_{21}^{2}+a_{21}x \bigl(s-\tau _{3}(s) \bigr)-a_{22}y(s)-a_{23}y \bigl(s-\tau_{4}(s) \bigr)\,ds\\ &+N_{5}(t)+6\ln k \\ \leq{}& \int_{0}^{t}-d-0.5\sigma_{21}^{2}+a_{21}x \bigl(s-\tau_{3}(s) \bigr)\,ds+N_{5}(t)+6\ln k, \end{aligned} \end{aligned}$$
(14)
for all \(0\leq t\leq k\), \(k\geq k_{0}\). When \(0\leq k-1\leq t\leq k\), by (13)
$$ \frac{\ln x(t)/x(0)}{t}\leq r-0.5\sigma_{11}^{2}+ \frac{N_{1}(t)}{t}+\frac {6\ln k}{k-1}. $$
(1) If \(r-0.5\sigma_{11}^{2}<0\), have \(\lim_{t\rightarrow+\infty} x(t)=0\). In combination with (14) we easily obtain \(\lim_{t\rightarrow+\infty} y(t)=0\) a.s.
(2) If \(r-0.5\sigma_{11}^{2}>0\), suppose this assertion is not true, that is to say \(P(S)>0\), where \(S=\{x^{*}=0\}\). Then for \(\omega\in S \), we have \(\lim_{t\rightarrow+\infty}x(t,\omega)=0\), hence,
$$ \begin{aligned}[b] \lim_{t\rightarrow+\infty} \bigl\langle x(t,\omega) \bigr\rangle &=\lim_{t\rightarrow+\infty} \bigl\langle x^{2}(t,\omega) \bigr\rangle =\lim_{t\rightarrow +\infty} \bigl\langle x \bigl(t-\tau_{1}(t),\omega \bigr) \bigr\rangle =\lim _{t\rightarrow+\infty} \bigl\langle x^{2} \bigl(t- \tau_{1}(t),\omega \bigr) \bigr\rangle \\ &=\lim_{t\rightarrow+\infty} \bigl\langle y \bigl(t-\tau_{2}(t), \omega \bigr) \bigr\rangle =\lim_{t\rightarrow+\infty} \bigl\langle y^{2} \bigl(t-\tau_{2}(t),\omega \bigr) \bigr\rangle =0. \end{aligned} $$
(15)
Then, according to the law of large numbers for local martingales, \(\frac{N_{i}(t)}{t}=0\). By virtue of (12) and (15) we have
$$ \biggl[\frac{\ln x(t)/x(0)}{t} \biggr]^{*}=r-0.5\sigma_{11}^{2}>0. $$
(16)
By Theorem 4.1 we obtain
$$ \biggl[\frac{\ln x(t)/x(0)}{t} \biggr]^{*}\leq \biggl[\frac{\ln x(t)/x(0)}{\ln t} \biggr]^{*} \biggl[ \frac {\ln t}{t} \biggr]^{*}\leq0. $$
Obviously, this is a contradiction, namely \(x^{*}>0\).
For all \(0\leq t\leq k\), \(k\geq k_{0}\). Then, \(\forall\epsilon>0\), there exists a \(T>0\), when \(0< k-1\leq T\leq t\leq k\) and \([\frac{6\ln k}{t}]^{*}<\epsilon\).
It follows from (13) that
$$ \frac{\ln x(t)/x(0)}{t} \leq r-0.5\sigma_{11}^{2}+ \epsilon-a_{11} \bigl\langle x(t) \bigr\rangle +\frac {N_{1}(t)}{t},\quad t \geq T. $$
(17)
If \(r-0.5\sigma_{11}^{2}>0\), we have \(\langle x(t)\rangle^{*}\leq\frac {r-0.5\sigma_{11}^{2}}{a_{11}}\).
In the same way, by (14)
$$ \begin{aligned} \frac{\ln y(t)/y(0)}{t} \leq{}&{-}d-0.5\sigma_{21}^{2}+t^{-1} \int _{0}^{t}a_{21}x \bigl(s- \tau_{3}(s) \bigr)\,ds-a_{22} \bigl\langle y(t) \bigr\rangle + \frac {N_{5}(t)}{t}+\frac{6\ln k}{t} \\ \leq{}&{-}d-0.5\sigma_{21}^{2}+\frac{a_{21}}{t(1-\tau')} \biggl[ \int_{-\tau (0)}^{0}x(s)\,ds+ \int_{0}^{t}x(s)\,ds \biggr]-a_{22} \bigl\langle y(t) \bigr\rangle \\ &+\frac {N_{5}(t)}{t}+\frac{6\ln k}{t}, \end{aligned} $$
for all \(0\leq t\leq k\), \(k\geq k_{0}\). \(\forall\epsilon>0\), there exists a \(T>0\), when \(0< k-1\leq T\leq t\leq k\), \([\frac{6\ln k}{t}]^{*}<\epsilon /2\) and \(\frac{a_{21}}{t(1-\tau')}\int_{-\tau(0)}^{0}x(s)\,ds<\epsilon/2\).
We have
$$ \begin{aligned} \frac{\ln y(t)/y(0)}{t} &\leq\frac{a_{21}(r-0.5\sigma_{11}^{2})}{a_{11}(1-\tau')}- \bigl(d+0.5\sigma _{21}^{2} \bigr)-a_{22} \bigl\langle y(t) \bigr\rangle +\frac{N_{5}(t)}{t}. \end{aligned} $$
(a) If \(\frac{a_{21}(r-0.5\sigma_{11}^{2})}{a_{11}(1-\tau')}-(d+0.5\sigma _{21}^{2})<0\), we have \(\lim_{t\rightarrow+\infty} y(t)=0\), a.s. (b) If \(\frac{a_{21}(r-0.5\sigma_{11}^{2})}{a_{11}(1-\tau')}- (d+0.5\sigma _{21}^{2})>0\), we have \(\langle y(t)\rangle^{*}\leq\frac{\frac{a_{21}(r-0.5\sigma _{11}^{2})}{(1-\tau')a_{11}}-(d+0.5\sigma_{21}^{2})}{a_{22}}\), a.s.
(3) If \(r-0.5\sigma_{11}^{2}=0\), by (17) we have \(\langle x(t)\rangle^{*}\leq\frac{\epsilon}{a_{11}}\). Since ϵ is arbitrary, we have \(\langle x(t)\rangle^{*}=0\), a.s.
The proof of Theorem 5.1 is completed. □
Remark 5.1
The definition of weak persistence is not a very appropriate one for stochastic models. Many authors have introduced some more appropriate definitions of permanence for stochastic population models, for example, stochastic persistence in probability (see [21, 22]) or a new definition of stochastic permanence (see [23]). We will continue to study them in future work.