As indicated by the Remark 2.1, overfishing will leads to the extinction of both species, hence, from now on, we make the following assumption:

$$\begin{array}{}{\displaystyle {b}_{i}^{L}{e}^{-{r}_{i}^{M}{\tau}_{i}}>{q}_{i}^{M}{E}_{i}^{M},\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}i=1,2.}\end{array}$$(3.1)

Before stating the main results of this section, we introduce a set of conditions

$$\begin{array}{}{\displaystyle \frac{{b}_{1}^{L}{e}^{-{r}_{1}^{M}{\tau}_{1}}-{q}_{1}^{M}{E}_{1}^{M}}{{b}_{2}^{M}{e}^{-{r}_{2}^{L}{\tau}_{2}}-{q}_{2}^{L}{E}_{2}^{L}}>max\{\frac{{a}_{12}^{M}}{{a}_{22}^{L}},\frac{{a}_{11}^{M}}{{a}_{21}^{L}},\frac{{d}_{1}^{M}}{{d}_{2}^{L}}\}}\end{array}$$(3.2)

$$\begin{array}{}{\displaystyle \frac{{b}_{1}^{L}{e}^{-{r}_{1}^{M}{\tau}_{1}}-{q}_{1}^{M}{E}_{1}^{M}}{{b}_{2}^{M}{e}^{-{r}_{2}^{L}{\tau}_{2}}-{q}_{2}^{L}{E}_{2}^{L}}>max\{\frac{{a}_{12}^{M}}{{a}_{22}^{L}},\frac{{a}_{11}^{M}+{d}_{1}^{M}{M}_{2}}{{a}_{21}^{L}}\}}\end{array}$$(3.3)

$$\begin{array}{}{\displaystyle \frac{{b}_{1}^{L}{e}^{-{r}_{1}^{M}{\tau}_{1}}-{q}_{1}^{M}{E}_{1}^{M}}{{b}_{2}^{M}{e}^{-{r}_{2}^{L}{\tau}_{2}}-{q}_{2}^{L}{E}_{2}^{L}}>max\{\frac{{a}_{12}^{M}+{d}_{1}^{M}{M}_{1}}{{a}_{22}^{L}},\frac{{a}_{11}^{M}}{{a}_{21}^{L}}\}}\end{array}$$(3.4)

$$\begin{array}{}{\displaystyle \frac{{b}_{2}^{L}{e}^{-{r}_{2}^{M}{\tau}_{2}}-{q}_{2}^{M}{E}_{2}^{M}}{{b}_{1}^{M}{e}^{-{r}_{1}^{L}{\tau}_{1}}-{q}_{1}^{L}{E}_{1}^{L}}>max\{\frac{{a}_{21}^{M}}{{a}_{11}^{L}},\frac{{a}_{22}^{M}}{{a}_{12}^{L}},\frac{{d}_{2}^{M}}{{d}_{1}^{L}}\}}\end{array}$$(3.5)

$$\begin{array}{}{\displaystyle \frac{{b}_{2}^{L}{e}^{-{r}_{2}^{M}{\tau}_{2}}-{q}_{2}^{M}{E}_{2}^{M}}{{b}_{1}^{M}{e}^{-{r}_{1}^{L}{\tau}_{1}}-{q}_{1}^{L}{E}_{1}^{L}}>max\{\frac{{a}_{21}^{M}}{{a}_{11}^{L}},\frac{{a}_{22}^{M}+{d}_{2}^{M}{M}_{1}}{{a}_{12}^{L}}\}}\end{array}$$(3.6)

$$\begin{array}{}{\displaystyle \frac{{b}_{2}^{L}{e}^{-{r}_{2}^{M}{\tau}_{2}}-{q}_{2}^{M}{E}_{2}^{M}}{{b}_{1}^{M}{e}^{-{r}_{1}^{L}{\tau}_{1}}-{q}_{1}^{L}{E}_{1}^{L}}>max\{\frac{{a}_{21}^{M}+{d}_{2}^{M}{M}_{2}}{{a}_{11}^{L}},\frac{{a}_{22}^{M}}{{a}_{12}^{L}}\}}\end{array}$$(3.7)

where *M*_{i}, *i* = 1, 2 are defined by (2.2).

Before we begin to prove the main results, we need several Lemmas again.

#### Lemma 3.1

*Let* (*x*_{1}(*t*), *y*_{1}(*t*), *x*_{2}(*t*), *y*_{2}(*t*))^{T} *be any solution of system (1.15) with initial conditions (1.16) and (1.17). Assume that (3.2) or (3.3) or (3.4) holds*, *then there exists a α* > 0 *such that x*_{1}(*t*) ≥ *α for all t* ≥ 0.

#### Proof

We first show that the conclusion of Lemma 3.1 holds under the assumption (3.2). It follows from Lemma 2.3 that $\begin{array}{}{\displaystyle \underset{t\to +\mathrm{\infty}}{lim\u2006sup}{x}_{2}(t)\le \frac{{b}_{2}^{M}{e}^{-{r}_{2}^{L}{\tau}_{1}}-{q}_{2}^{L}{E}_{2}^{L}}{{a}_{22}^{L}}.}\end{array}$ Given $\begin{array}{}{\displaystyle \epsilon =\frac{1}{2}(\frac{{b}_{1}^{L}{e}^{-{r}_{1}^{M}{\tau}_{1}}-{q}_{1}^{M}{E}_{1}^{M}}{{a}_{12}^{M}}-\frac{{b}_{2}^{M}{e}^{-{r}_{2}^{L}{\tau}_{2}}-{q}_{2}^{M}{E}_{2}^{M}}{{a}_{22}^{L}}),}\end{array}$ there exists a *T* ≥ 0 such that for all *t* ≥ *T*

$$\begin{array}{}{\displaystyle {x}_{2}(t)\le \frac{{b}_{2}^{M}{e}^{-{r}_{2}^{L}{\tau}_{1}}-{q}_{2}^{L}{E}_{2}^{L}}{{a}_{22}^{L}}+\epsilon =\frac{1}{2}(\frac{{b}_{1}^{L}{e}^{-{r}_{1}^{M}{\tau}_{1}}-{q}_{1}^{M}{E}_{1}^{M}}{{a}_{12}^{M}}+\frac{{b}_{2}^{M}{e}^{-{r}_{2}^{L}{\tau}_{2}}-{q}_{2}^{M}{E}_{2}^{M}}{{a}_{22}^{L}}).}\end{array}$$

So, for all *t* ≥ *T*, from the first equation of system (1.15), it follows that

$$\begin{array}{}\begin{array}{rll}\dot{{x}_{1}}(t)& \ge & {b}_{1}^{L}{e}^{-{r}_{1}^{M}{\tau}_{1}}{x}_{1}(t-{\tau}_{1})-{a}_{11}^{M}{x}_{1}^{2}(t)-{a}_{12}^{M}{x}_{1}(t){x}_{2}(t)\\ & & -{d}_{1}^{M}{x}_{1}^{2}(t){x}_{2}(t)-{q}_{1}^{M}{E}_{1}^{M}{x}_{1}(t)\\ & \ge & {b}_{1}^{L}{e}^{-{r}_{1}^{M}{\tau}_{1}}{x}_{1}(t-{\tau}_{1})-({a}_{11}^{M}+\frac{{d}_{1}^{M}}{2}(\frac{{b}_{1}^{L}{e}^{-{r}_{1}^{M}{\tau}_{1}}-{q}_{1}^{M}{E}_{1}^{M}}{{a}_{12}^{M}}+\frac{{b}_{2}^{M}{e}^{-{r}_{2}^{L}{\tau}_{2}}-{q}_{2}^{M}{E}_{2}^{M}}{{a}_{22}^{L}})){x}_{1}^{2}(t)\\ & & -(\frac{1}{2}({b}_{1}^{L}{e}^{-{r}_{1}^{M}{\tau}_{1}}-{q}_{1}^{M}{E}_{1}^{M}+{a}_{12}^{M}\frac{{b}_{2}^{M}{e}^{-{r}_{2}^{L}{\tau}_{2}}-{q}_{2}^{M}{E}_{2}^{M}}{{a}_{22}^{L}})+{q}_{1}^{M}{E}_{1}^{M}){x}_{1}(t)\\ & \stackrel{\mathrm{d}\mathrm{e}\mathrm{f}}{=}& A{x}_{1}(t-{\tau}_{1})-B{x}_{1}(t)-C{x}_{1}^{2}(t).\end{array}\end{array}$$

Let *u*(*t*) be a solution of the following equation

$$\begin{array}{}{\displaystyle \dot{u}(t)=Au(t-{\tau}_{1})-Bu(t)-C{u}^{2}(t),}\end{array}$$

with *u*(*T* + *τ*_{1}) = *x*_{1}(*T* + *τ*_{1}). It follows from condition (3.1) that

$$\begin{array}{}{\displaystyle A-B=\frac{1}{2}({b}_{1}^{L}{e}^{-{r}_{1}^{M}{\tau}_{1}}-{q}_{1}^{M}{E}_{1}^{M}-{a}_{12}^{M}\frac{{b}_{2}^{M}{e}^{-{r}_{2}^{L}{\tau}_{2}}-{q}_{2}^{M}{E}_{2}^{M}}{{a}_{22}^{L}})>0.}\end{array}$$

From Lemma 2.2

$$\begin{array}{}{\displaystyle \underset{t\to +\mathrm{\infty}}{lim}u(t)=\frac{A-B}{C}={\alpha}_{1}>0.}\end{array}$$

Therefore, we obtain

$$\begin{array}{}{\displaystyle {\underset{\_}{x}}_{1}=\underset{t\to +\mathrm{\infty}}{lim\u2006inf}{x}_{1}(t)\ge {\alpha}_{1}>0.}\end{array}$$

Given $\begin{array}{}{\displaystyle \epsilon =\frac{1}{2}{\alpha}_{1},}\end{array}$ there exists a *T*_{1} ≥ *T* such that

$$\begin{array}{}{\displaystyle {x}_{1}(t)\ge {\underset{\_}{x}}_{1}-\frac{{\alpha}_{1}}{2}\ge {\alpha}_{1}-\frac{1}{2}{\alpha}_{1}=\frac{{\alpha}_{1}}{2},\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}t\ge {T}_{1}.}\end{array}$$

Let *α*_{2} = min{*x*_{1}(*t*) : 0 ≤ *t* ≤ *T*_{1}} > 0 and $\begin{array}{}{\displaystyle \alpha =min\{\frac{{\alpha}_{1}}{2},{\alpha}_{2}\}>0.}\end{array}$ It follows that *x*_{1}(*t*) ≥ *α* > 0 for all *t* ≥ 0.

Noting that above proof only use the fact

$$\begin{array}{}{\displaystyle \frac{{b}_{1}^{L}{e}^{-{r}_{1}^{M}{\tau}_{1}}-{q}_{1}^{M}{E}_{1}^{M}}{{a}_{12}^{M}}>\frac{{b}_{2}^{M}{e}^{-{r}_{2}^{L}{\tau}_{2}}-{q}_{2}^{M}{E}_{2}^{M}}{{a}_{22}^{L}}.}\end{array}$$

Condition (3.3) and (3.4) all implies this inequality holds, hence, under the assumption of (3.2) or (3.3) or (3.4), the conclusion of Lemma 3.1 holds. This ends the proof of Lemma 3.1.

Our main results are the following Theorems.

#### Theorem 3.1

*Assume that (3.2) holds. Then*

$$\begin{array}{}{\displaystyle {m}_{1}\le \underset{t\to +\mathrm{\infty}}{lim\u2006inf}{x}_{1}(t)\le \underset{t\to +\mathrm{\infty}}{lim\u2006sup}{x}_{1}(t)\le {M}_{1}.}\\ {\displaystyle {n}_{1}\le \underset{t\to +\mathrm{\infty}}{lim\u2006inf}{y}_{1}(t)\le \underset{t\to +\mathrm{\infty}}{lim\u2006sup}{y}_{1}(t)\le {N}_{1}.}\\ {\displaystyle \underset{t\to +\mathrm{\infty}}{lim}{x}_{2}(t)=0,\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\underset{t\to +\mathrm{\infty}}{lim}{y}_{2}(t)=0,}\end{array}$$

*where*

$$\begin{array}{}{\displaystyle {m}_{1}=\frac{{b}_{1}^{L}{e}^{-{r}_{1}^{M}{\tau}_{1}}-{q}_{1}^{M}{E}_{1}^{M}}{{a}_{11}^{M}},\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}{n}_{1}=\frac{{b}_{1}^{L}{m}_{1}}{{r}_{1}^{M}}(1-{e}^{-{r}_{1}^{L}{\tau}_{1}}).}\end{array}$$

Noting that system (1.9) is the special case of system (1.15), (*q*_{i}(*t*) ≡ 0, *E*_{i}(*t*) ≡ 0, *i* = 1, 2) Then as a direct corollary of Theorem 3.1, we have

#### Corollary 3.1

*Assume that in system (1.9)*

$$\begin{array}{}{\displaystyle \frac{{b}_{1}^{L}{e}^{-{r}_{1}^{M}{\tau}_{1}}}{{b}_{2}^{M}{e}^{-{r}_{2}^{L}{\tau}_{2}}}>max\{\frac{{a}_{12}^{M}}{{a}_{22}^{L}},\frac{{a}_{11}^{M}}{{a}_{21}^{L}},\frac{{d}_{1}^{M}}{{d}_{2}^{L}}\}}\end{array}$$

*hold. Then*

$$\begin{array}{}{\displaystyle {m}_{1}\le \underset{t\to +\mathrm{\infty}}{lim\u2006inf}{x}_{1}(t)\le \underset{t\to +\mathrm{\infty}}{lim\u2006sup}{x}_{1}(t)\le {M}_{1}.}\\ {n}_{1}\le \underset{t\to +\mathrm{\infty}}{lim\u2006inf}{y}_{1}(t)\le \underset{t\to +\mathrm{\infty}}{lim\u2006sup}{y}_{1}(t)\le {N}_{1}.\\ \underset{t\to +\mathrm{\infty}}{lim}{x}_{2}(t)=0,\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\underset{t\to +\mathrm{\infty}}{lim}{y}_{2}(t)=0,\end{array}$$

*where*

$$\begin{array}{}{\displaystyle {m}_{1}=\frac{{b}_{1}^{L}{e}^{-{r}_{1}^{M}{\tau}_{1}}}{{a}_{11}^{M}},\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}{n}_{1}=\frac{{b}_{1}^{L}{m}_{1}}{{r}_{1}^{M}}(1-{e}^{-{r}_{1}^{L}{\tau}_{1}}).}\end{array}$$

As a direct corollary of Theorem 3.2 and 3.3, we have

#### Corollary 3.2

*Assume that in system (1.9)*

$$\begin{array}{}{\displaystyle \frac{{b}_{1}^{L}{e}^{-{r}_{1}^{M}{\tau}_{1}}}{{b}_{2}^{M}{e}^{-{r}_{2}^{L}{\tau}_{2}}}>max\{\frac{{a}_{12}^{M}}{{a}_{22}^{L}},\frac{{a}_{11}^{M}+{d}_{1}^{M}{M}_{11}}{{a}_{21}^{L}}\}}\end{array}$$

*hold*, *where* $\begin{array}{}{\displaystyle {M}_{11}=\frac{{b}_{1}^{M}{e}^{-{r}_{1}^{L}{\tau}_{1}}}{{a}_{11}^{L}}}\end{array}$. *Then the conclusion of Corollary 3.1 holds*.

#### Corollary 3.3

*Assume that in system (1.9)*

$$\begin{array}{}{\displaystyle \frac{{b}_{1}^{L}{e}^{-{r}_{1}^{M}{\tau}_{1}}}{{b}_{2}^{M}{e}^{-{r}_{2}^{L}{\tau}_{2}}}>max\{\frac{{a}_{12}^{M}+{d}_{1}^{M}{M}_{22}}{{a}_{22}^{L}},\frac{{a}_{11}^{M}}{{a}_{21}^{L}}\}}\end{array}$$(3.8)

*hold*, *where* $\begin{array}{}{\displaystyle {M}_{22}=\frac{{b}_{2}^{M}{e}^{-{r}_{2}^{L}{\tau}_{1}}}{{a}_{22}^{L}}}\end{array}$. *Then the conclusion of Corollary 3.1 holds*.

It follows from Lemma 2.1 and 2.3 that *x*_{i}(*t*), *i* = 1, 2 are bounded and positive for all *t* ≥ 0. Let $\begin{array}{}{\displaystyle {\underset{\_}{x}}_{1}=\underset{t\to +\mathrm{\infty}}{lim\u2006inf}{x}_{1}(t)\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{and}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{\overline{x}}_{2}=\underset{t\to +\mathrm{\infty}}{lim\u2006sup}{x}_{2}(t)}\end{array}$. For above *ε*_{1} > 0, it follows from Lemma 2.3 that

$$\begin{array}{}{\displaystyle {\underset{\_}{x}}_{1}<{M}_{1}+{\epsilon}_{1},\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}{\overline{x}}_{2}<{M}_{2}+{\epsilon}_{1}.}\end{array}$$(3.9)

From Lemma 3.1 we know that *x*_{1} ≥ *α* > 0. Obviously, *x*_{2} ≥ 0. To prove $\begin{array}{}{\displaystyle \underset{t\to +\mathrm{\infty}}{lim}}\end{array}$ *x*_{2}(*t*) = 0, it suffices to show that *x*_{2} = 0. In order to get a contradiction, we suppose that *x*_{2} > 0. According to the Fluctuation lemma (Lemma 2.4), there exist sequences *γ*_{n} → +∞, *σ*_{n} → +∞ such that $\begin{array}{}{\displaystyle {x}_{1}^{{}^{\prime}}({\gamma}_{n})\to 0,{x}_{2}^{{}^{\prime}}({\sigma}_{n})}\end{array}$ → 0, *x*_{1}(*γ*_{n}) → *x*_{1} and *x*_{2}(*σ*_{n}) → *x*_{2} as *n* → +∞. It follows from the first equation of system (1.15) that

$$\begin{array}{}\begin{array}{rll}\dot{{x}_{1}}({\gamma}_{n})& =& {b}_{1}({\gamma}_{n}-{\tau}_{1}){e}^{-{\int}_{{\gamma}_{n}-{\tau}_{1}}^{{\gamma}_{n}}{r}_{1}(s)ds}{x}_{1}({\gamma}_{n}-{\tau}_{1})-{a}_{11}({\gamma}_{n}){x}_{1}^{2}({\gamma}_{n})\\ & & -{a}_{12}({\gamma}_{n}){x}_{1}({\gamma}_{n}){x}_{2}({\gamma}_{n})-{d}_{1}({\gamma}_{n}){x}_{1}^{2}({\gamma}_{n}){x}_{2}({\gamma}_{n})-{q}_{1}({\gamma}_{n}){E}_{1}({\gamma}_{n}){x}_{1}({\gamma}_{n})\\ & \ge & {b}_{1}^{L}{e}^{-{r}_{1}^{M}{\tau}_{1}}\underset{t\ge {\gamma}_{n}-{\tau}_{1}}{inf}{x}_{1}(t)-{a}_{11}^{M}{x}_{1}^{2}({\gamma}_{n})\\ & & -{a}_{12}^{M}{x}_{1}({\gamma}_{n})\underset{t\ge {\gamma}_{n}}{sup}{x}_{2}(t)-{d}_{1}^{M}{x}_{1}^{2}({\gamma}_{n})\underset{t\ge {\gamma}_{n}}{sup}{x}_{2}(t)-{q}_{1}^{M}{E}_{1}^{M}{x}_{1}({\gamma}_{n}).\end{array}\end{array}$$(3.10)

By taking the limit of the above inequality as *n* → +∞, we obtain the inequality

$$\begin{array}{}{\displaystyle {b}_{1}^{L}{e}^{-{r}_{1}^{M}{\tau}_{1}}-{q}_{1}^{M}{E}_{1}^{M}\le {a}_{11}^{M}{\underset{\_}{x}}_{1}+{a}_{12}^{M}{\overline{x}}_{2}+{d}_{1}^{M}{\underset{\_}{x}}_{1}{\overline{x}}_{2}.}\end{array}$$(3.11)

From the third equation of system (1.15), by a similar argument as above, we obtain

$$\begin{array}{}{\displaystyle {b}_{2}^{M}{e}^{-{r}_{2}^{L}{\tau}_{2}}-{q}_{2}^{L}{E}_{2}^{L}\ge {a}_{21}^{L}{\underset{\_}{x}}_{1}+{a}_{22}^{L}{\overline{x}}_{2}+{d}_{2}^{L}{\underset{\_}{x}}_{1}{\overline{x}}_{2}.}\end{array}$$(3.12)

(3.11) is equivalent to

$$\begin{array}{}{\displaystyle 1\le \frac{{a}_{11}^{M}}{{b}_{1}^{L}{e}^{-{r}_{1}^{M}{\tau}_{1}}-{q}_{1}^{M}{E}_{1}^{M}}{\underset{\_}{x}}_{1}+\frac{{a}_{12}^{M}}{{b}_{1}^{L}{e}^{-{r}_{1}^{M}{\tau}_{1}}-{q}_{1}^{M}{E}_{1}^{M}}{\overline{x}}_{2}+\frac{{d}_{1}^{M}}{{b}_{1}^{L}{e}^{-{r}_{1}^{M}{\tau}_{1}}-{q}_{1}^{M}{E}_{1}^{M}}{\underset{\_}{x}}_{1}{\overline{x}}_{2}.}\end{array}$$(3.13)

(3.12) is equivalent to

$$\begin{array}{}{\displaystyle 1\ge \frac{{a}_{21}^{L}}{{b}_{2}^{M}{e}^{-{r}_{2}^{L}{\tau}_{2}}-{q}_{2}^{L}{E}_{2}^{L}}{\underset{\_}{x}}_{1}+\frac{{a}_{22}^{L}}{{b}_{2}^{M}{e}^{-{r}_{2}^{L}{\tau}_{2}}-{q}_{2}^{L}{E}_{2}^{L}}{\overline{x}}_{2}+\frac{{d}_{2}^{L}}{{b}_{2}^{M}{e}^{-{r}_{2}^{L}{\tau}_{2}}-{q}_{2}^{L}{E}_{2}^{L}}{\underset{\_}{x}}_{1}{\overline{x}}_{2}.}\end{array}$$(3.14)

(3.13) together with (3.13) leads to

$$\begin{array}{}{\displaystyle {A}_{1}{\underset{\_}{x}}_{1}+{A}_{2}{\overline{x}}_{2}+{A}_{3}{\underset{\_}{x}}_{1}{\overline{x}}_{2}\ge 0,}\end{array}$$(3.15)

where

$$\begin{array}{}{\displaystyle {A}_{1}=\frac{{a}_{11}^{M}}{{b}_{1}^{L}{e}^{-{r}_{1}^{M}{\tau}_{1}}-{q}_{1}^{M}{E}_{1}^{M}}-\frac{{a}_{21}^{L}}{{b}_{2}^{M}{e}^{-{r}_{2}^{L}{\tau}_{2}}-{q}_{2}^{L}{E}_{2}^{L}},}\\ {A}_{2}={\displaystyle \frac{{a}_{12}^{M}}{{b}_{1}^{L}{e}^{-{r}_{1}^{M}{\tau}_{1}}-{q}_{1}^{M}{E}_{1}^{M}}-\frac{{a}_{22}^{L}}{{b}_{2}^{M}{e}^{-{r}_{2}^{L}{\tau}_{2}}-{q}_{2}^{L}{E}_{2}^{L}},}\\ {A}_{3}={\displaystyle \frac{{d}_{1}^{M}}{{b}_{1}^{L}{e}^{-{r}_{1}^{M}{\tau}_{1}}-{q}_{1}^{M}{E}_{1}^{M}}-\frac{{d}_{2}^{L}}{{b}_{2}^{M}{e}^{-{r}_{2}^{L}{\tau}_{2}}-{q}_{2}^{L}{E}_{2}^{L}}.}\end{array}$$

It follows from (3.2) that *A*_{i} < 0, *i* = 1, 2, 3, this together with the fact *x*_{1} > 0, *x*_{2} > 0 leads to

$$\begin{array}{}{\displaystyle {A}_{1}{\underset{\_}{x}}_{1}+{A}_{2}{\overline{x}}_{2}+{A}_{3}{\underset{\_}{x}}_{1}{\overline{x}}_{2}<0,}\end{array}$$(3.16)

which is contradiction with (3.15). Then we obtain $\begin{array}{}{\displaystyle \underset{t\to +\mathrm{\infty}}{lim}}\end{array}$ *x*_{2}(*t*) = 0. Since

$$\begin{array}{}{\displaystyle {y}_{2}(t)=\underset{t-{\tau}_{2}}{\overset{t}{\int}}{b}_{2}(s){x}_{2}(s){e}^{{\int}_{t}^{s}{r}_{2}(u)du}ds,}\end{array}$$

it immediately follows that

$$\begin{array}{}{\displaystyle \underset{t\to +\mathrm{\infty}}{lim}{y}_{2}(t)=0.}\end{array}$$

Above analysis shows that for 0 < *ε* < $\begin{array}{}{\displaystyle \frac{({b}_{1}^{L}{e}^{-{r}_{1}^{M}{\tau}_{1}}-{q}_{1}^{M}{E}_{1}^{M})}{{a}_{12}^{M}}}\end{array}$, there exists a *T*_{1} > 0, such that for all *t* ≥ *T*_{1}, *y*_{2}(*t*) < *ε*. Lemma 2.3 had showed that

$$\begin{array}{}{\displaystyle \underset{t\to +\mathrm{\infty}}{lim\u2006sup}{x}_{1}(t)\le {M}_{1},\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\underset{t\to +\mathrm{\infty}}{lim\u2006sup}{y}_{1}(t)\le {N}_{1}.}\end{array}$$

To end the proof of Theorem 3.1, it’s enough to show that

$$\begin{array}{}{\displaystyle \underset{t\to +\mathrm{\infty}}{lim\u2006inf}{x}_{1}(t)\ge {m}_{1},\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\underset{t\to +\mathrm{\infty}}{lim\u2006inf}{y}_{1}(t)\ge {n}_{1}.}\end{array}$$

For *t* ≥ *T*_{1} + *τ*, from the first equation of system (1.15), we have

$$\begin{array}{}{\displaystyle {\dot{x}}_{1}(t)\ge {b}_{1}^{L}{e}^{-{r}_{1}^{M}{\tau}_{1}}{x}_{1}(t-{\tau}_{1})-({a}_{11}^{M}+{d}_{1}^{M}\epsilon ){x}_{1}^{2}(t)-({a}_{12}^{M}\epsilon +{q}_{1}^{M}{E}_{1}^{M}){x}_{1}(t).}\end{array}$$(3.17)

Let *u*(*t*) be the solution of the equation

$$\begin{array}{}{\displaystyle \dot{u}={b}_{1}^{L}{e}^{-{r}_{1}^{M}{\tau}_{1}}{u}_{1}(t-{\tau}_{1})-({a}_{11}^{M}+{d}_{1}^{M}\epsilon ){u}_{1}^{2}(t)-({a}_{12}^{M}\epsilon +{q}_{1}^{M}{E}_{1}^{M})u(t)}\end{array}$$

with *u*(*T*_{1} + *τ*) = *x*_{1}(*T*_{1} + *τ*). It follows from Lemma 2.2 that

$$\begin{array}{}{\displaystyle \underset{t\to +\mathrm{\infty}}{lim}u(t)=\frac{{b}_{1}^{L}{e}^{-{r}_{1}^{M}{\tau}_{1}}-({a}_{12}^{M}\epsilon +{q}_{1}^{M}{E}_{1}^{M})}{{a}_{11}^{M}+{d}_{1}^{M}\epsilon}.}\end{array}$$

Therefore, we have

$$\begin{array}{}{\displaystyle \underset{t\to +\mathrm{\infty}}{lim\u2006inf}{x}_{1}(t)\ge \frac{{b}_{1}^{L}{e}^{-{r}_{1}^{M}{\tau}_{1}}-({a}_{12}^{M}\epsilon +{q}_{1}^{M}{E}_{1}^{M})}{{a}_{11}^{M}+{d}_{1}^{M}\epsilon}.}\end{array}$$

Setting *ε* → 0, it follows that

$$\begin{array}{}{\displaystyle \underset{t\to +\mathrm{\infty}}{lim\u2006inf}{x}_{1}(t)\ge \frac{{b}_{1}^{L}{e}^{-{r}_{1}^{M}{\tau}_{1}}-{q}_{1}^{M}{E}_{1}^{M}}{{a}_{11}^{M}}{\stackrel{\mathrm{d}\mathrm{e}\mathrm{f}}{m}}_{1}.}\end{array}$$

Noting that

$$\begin{array}{}{\displaystyle {y}_{1}(t)=\underset{t-{\tau}_{2}}{\overset{t}{\int}}{b}_{1}(s){x}_{1}(s){e}^{{\int}_{t}^{s}{r}_{1}(u)du}ds.}\end{array}$$

From this, one could easily obtain

$$\begin{array}{}{\displaystyle \underset{t\to +\mathrm{\infty}}{lim\u2006inf}{y}_{1}(t)\ge \frac{{b}_{1}^{L}{m}_{1}}{{r}_{1}^{M}}(1-{e}^{-{r}_{1}^{L}{\tau}_{1}}).}\end{array}$$

The proof of Theorem 3.1 is completed.

Let (*x*_{1}(*t*), *y*_{1}(*t*), *x*_{2}(*t*), *y*_{2}(*t*))^{T} be any solution of system (1.15) with initial conditions (1.16) and (1.17). It follows from (3.3) that there exists a *ε*_{2} > 0 enough small, such that

$$\begin{array}{}{\displaystyle \frac{{b}_{1}^{L}{e}^{-{r}_{1}^{M}{\tau}_{1}}-{q}_{1}^{M}{E}_{1}^{M}}{{b}_{2}^{M}{e}^{-{r}_{2}^{L}{\tau}_{2}}-{q}_{2}^{L}{E}_{2}^{L}}>max\{\frac{{a}_{12}^{M}}{{a}_{22}^{L}},\frac{{a}_{11}^{M}+{d}_{1}^{M}({M}_{2}+{\epsilon}_{2})}{{a}_{21}^{L}}\}.}\end{array}$$(3.18)

Let *x*_{1}and *x*_{2} be defined as that of Lemma 3.3. For above *ε*_{2} > 0, it follows from Lemma 2.3 that

$$\begin{array}{}{\displaystyle {\underset{\_}{x}}_{1}<{M}_{1}+\epsilon ,\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}{\overline{x}}_{2}<{M}_{2}+{\epsilon}_{2}.}\end{array}$$(3.19)

From Lemma 3.1 we know that *x*_{1} ≥ *α* > 0. Obviously, *x*_{2} ≥ 0. To prove $\begin{array}{}{\displaystyle \underset{t\to +\mathrm{\infty}}{lim}}\end{array}$ *x*_{2}(*t*) = 0, it suffices to show that *x*_{2} = 0. In order to get a contradiction, we suppose that *x*_{2} > 0. Already, by using the Fluctuation lemma, we had established the inequalities (3.11) and (3.12). Now, from (3.11) and (3.19), we have

$$\begin{array}{}{\displaystyle {b}_{1}^{L}{e}^{-{r}_{1}^{M}{\tau}_{1}}-{q}_{1}^{M}{E}_{1}^{M}\le ({a}_{11}^{M}+{d}_{1}^{M}({M}_{2}+\epsilon )){\underset{\_}{x}}_{1}+{a}_{12}^{M}{\overline{x}}_{2},}\end{array}$$(3.20)

which is equivalent to

$$\begin{array}{}{\displaystyle 1\le \frac{{a}_{11}^{M}+{d}_{1}^{M}({M}_{2}+\epsilon )}{{b}_{1}^{L}{e}^{-{r}_{1}^{M}{\tau}_{1}}-{q}_{1}^{M}{E}_{1}^{M}}{\underset{\_}{x}}_{1}+\frac{{a}_{12}^{M}}{{b}_{1}^{L}{e}^{-{r}_{1}^{M}{\tau}_{1}}-{q}_{1}^{M}{E}_{1}^{M}}{\overline{x}}_{2}.}\end{array}$$(3.21)

Also, it follows from (3.12) that

$$\begin{array}{}{\displaystyle 1\ge \frac{{a}_{21}^{L}}{{b}_{2}^{M}{e}^{-{r}_{2}^{L}{\tau}_{2}}-{q}_{2}^{L}{E}_{2}^{L}}{\underset{\_}{x}}_{1}+\frac{{a}_{22}^{L}}{{b}_{2}^{M}{e}^{-{r}_{2}^{L}{\tau}_{2}}-{q}_{2}^{L}{E}_{2}^{L}}{\overline{x}}_{2}.}\end{array}$$(3.22)

(3.21) combine with (3.22) leads to

$$\begin{array}{}{\displaystyle {B}_{1}{\underset{\_}{x}}_{1}+{B}_{2}{\overline{x}}_{2}\ge 0,}\end{array}$$(3.23)

where

$$\begin{array}{}{\displaystyle {B}_{1}=\frac{{a}_{11}^{M}+{d}_{1}^{M}({M}_{2}+\epsilon )}{{b}_{1}^{L}{e}^{-{r}_{1}^{M}{\tau}_{1}}-{q}_{1}^{M}{E}_{1}^{M}}-\frac{{a}_{21}^{L}}{{b}_{2}^{M}{e}^{-{r}_{2}^{L}{\tau}_{2}}-{q}_{2}^{L}{E}_{2}^{L}}.}\end{array}$$

$$\begin{array}{}{\displaystyle {B}_{2}=\frac{{a}_{12}^{M}}{{b}_{1}^{L}{e}^{-{r}_{1}^{M}{\tau}_{1}}-{q}_{1}^{M}{E}_{1}^{M}}-\frac{{a}_{22}^{L}}{{b}_{2}^{M}{e}^{-{r}_{2}^{L}{\tau}_{2}}-{q}_{2}^{L}{E}_{2}^{L}}.}\end{array}$$

Condition (3.18) implies that *B*_{i} < 0, *i* = 1, 2. This together with the fact *x*_{1} > 0, *x*_{2} > 0 leads to

$$\begin{array}{}{\displaystyle {B}_{1}{\underset{\_}{x}}_{1}+{B}_{2}{\overline{x}}_{2}<0,}\end{array}$$(3.24)

which is contradiction with (3.23). Then we obtain $\begin{array}{}{\displaystyle \underset{t\to +\mathrm{\infty}}{lim}}\end{array}$ *x*_{2}(*t*) = 0. The rest of the proof is similar to that of the proof of Theorem 3.1, and we omit the detail here.

Let (*x*_{1}(*t*), *y*_{1}(*t*), *x*_{2}(*t*), *y*_{2}(*t*))^{T} be any solution of system (1.15) with initial conditions (1.16) and (1.17). It follows from (3.4) that there exists a *ε*_{3} > 0 enough small, such that

$$\begin{array}{}{\displaystyle \frac{{b}_{1}^{L}{e}^{-{r}_{1}^{M}{\tau}_{1}}-{q}_{1}^{M}{E}_{1}^{M}}{{b}_{2}^{M}{e}^{-{r}_{2}^{L}{\tau}_{2}}-{q}_{2}^{L}{E}_{2}^{L}}>max\{\frac{{a}_{12}^{M}+{d}_{1}^{M}({M}_{1}+{\epsilon}_{3})}{{a}_{22}^{L}},\frac{{a}_{11}^{M}}{{a}_{21}^{L}}\}.}\end{array}$$(3.25)

Let *x*_{1} and *x*_{2} be defined as that of Lemma 3.3. For above *ε*_{3} > 0, it follows from Lemma 2.3 that

$$\begin{array}{}{\displaystyle {\underset{\_}{x}}_{1}<{M}_{1}+{\epsilon}_{3},\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}{\overline{x}}_{2}<{M}_{2}+{\epsilon}_{3}.}\end{array}$$(3.26)

From Lemma 3.1 we know that *x*_{1} ≥ *α* > 0. Obviously, *x*_{2} ≥ 0. To prove $\begin{array}{}{\displaystyle \underset{t\to +\mathrm{\infty}}{lim}}\end{array}$ *x*_{2}(*t*) = 0, it suffices to show that *x*_{2} = 0. In order to get a contradiction, we suppose that *x*_{2} > 0. Already, by using the Fluctuation lemma, we had established the inequalities (3.11) and (3.12). Now, from (3.11) and (3.26), we have

$$\begin{array}{}{\displaystyle {b}_{1}^{L}{e}^{-{r}_{1}^{M}{\tau}_{1}}-{q}_{1}^{M}{E}_{1}^{M}\le {a}_{11}^{M}{\underset{\_}{x}}_{1}+({a}_{12}^{M}+{d}_{1}^{M}({M}_{1}+{\epsilon}_{3})){\overline{x}}_{2},}\end{array}$$(3.27)

which is equivalent to

$$\begin{array}{}{\displaystyle 1\le \frac{{a}_{11}^{M}}{{b}_{1}^{L}{e}^{-{r}_{1}^{M}{\tau}_{1}}-{q}_{1}^{M}{E}_{1}^{M}}{\underset{\_}{x}}_{1}+\frac{{a}_{12}^{M}+{d}_{1}^{M}({M}_{1}+{\epsilon}_{3})}{{b}_{1}^{L}{e}^{-{r}_{1}^{M}{\tau}_{1}}-{q}_{1}^{M}{E}_{1}^{M}}{\overline{x}}_{2}.}\end{array}$$(3.28)

Also, it follows from (3.12) that

$$\begin{array}{}{\displaystyle 1\ge \frac{{a}_{21}^{L}}{{b}_{2}^{M}{e}^{-{r}_{2}^{L}{\tau}_{2}}-{q}_{2}^{L}{E}_{2}^{L}}{\underset{\_}{x}}_{1}+\frac{{a}_{22}^{L}}{{b}_{2}^{M}{e}^{-{r}_{2}^{L}{\tau}_{2}}-{q}_{2}^{L}{E}_{2}^{L}}{\overline{x}}_{2}.}\end{array}$$(3.29)

(3.28) combine with (3.29) leads to

$$\begin{array}{}{\displaystyle {C}_{1}{\underset{\_}{x}}_{1}+{C}_{2}{\overline{x}}_{2}\ge 0,}\end{array}$$(3.30)

where

$$\begin{array}{}{\displaystyle {C}_{1}=\frac{{a}_{11}^{M}}{{b}_{1}^{L}{e}^{-{r}_{1}^{M}{\tau}_{1}}-{q}_{1}^{M}{E}_{1}^{M}}-\frac{{a}_{21}^{L}}{{b}_{2}^{M}{e}^{-{r}_{2}^{L}{\tau}_{2}}-{q}_{2}^{L}{E}_{2}^{L}}.}\\ {C}_{2}={\displaystyle \frac{{a}_{12}^{M}+{d}_{1}^{M}({M}_{1}+{\epsilon}_{3})}{{b}_{1}^{L}{e}^{-{r}_{1}^{M}{\tau}_{1}}-{q}_{1}^{M}{E}_{1}^{M}}-\frac{{a}_{22}^{L}}{{b}_{2}^{M}{e}^{-{r}_{2}^{L}{\tau}_{2}}-{q}_{2}^{L}{E}_{2}^{L}}.}\end{array}$$

Condition (3.25) implies that *C*_{i} < 0, *i* = 1, 2. This together with the fact *x*_{1} > 0, *x*_{2} > 0 leads to

$$\begin{array}{}{\displaystyle {C}_{1}{\underset{\_}{x}}_{1}+{C}_{2}{\overline{x}}_{2}<0,}\end{array}$$(3.31)

which is contradiction with (3.30). Then we obtain $\begin{array}{}{\displaystyle \underset{t\to +\mathrm{\infty}}{lim}}\end{array}$ *x*_{2}(*t*) = 0. The rest of the proof is similar to that of the proof of Theorem 3.1, and we omit the detail here.

Concerned with the extinction of the first species, we have the following result.

#### Theorem 3.4

*Assume that (3.5) or (3.6) or (3.7) hold. Then*

$$\begin{array}{}{\displaystyle {m}_{2}\le \underset{t\to +\mathrm{\infty}}{lim\u2006inf}{x}_{2}(t)\le \underset{t\to +\mathrm{\infty}}{lim\u2006sup}{x}_{2}(t)\le {M}_{2}.}\\ {n}_{2}\le \underset{t\to +\mathrm{\infty}}{lim\u2006inf}{y}_{2}(t)\le \underset{t\to +\mathrm{\infty}}{lim\u2006sup}{y}_{2}(t)\le {N}_{2}.\\ \underset{t\to +\mathrm{\infty}}{lim}{x}_{1}(t)=0,\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\underset{t\to +\mathrm{\infty}}{lim}{y}_{1}(t)=0,\end{array}$$

*where*

$$\begin{array}{}{\displaystyle {m}_{2}=\frac{{b}_{2}^{L}{e}^{-{r}_{2}^{M}{\tau}_{2}}-{q}_{2}^{M}{E}_{2}^{M}}{{a}_{22}^{M}},\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}{n}_{2}=\frac{{b}_{2}^{L}{m}_{2}}{{r}_{2}^{M}}(1-{e}^{-{r}_{2}^{L}{\tau}_{2}}).}\end{array}$$

Since the proof of Theorem 3.4 is similar to that of Theorems 3.1-3.3, we omit the detail here.

As a direct corollary of Theorem 2.4, we have

#### Corollary 3.4

*Assume that in system (1.9)*, *one of the following three inequalities holds*.

$$\begin{array}{c}{\displaystyle \frac{{b}_{2}^{L}{e}^{-{r}_{2}^{M}{\tau}_{2}}}{{b}_{1}^{M}{e}^{-{r}_{1}^{L}{\tau}_{1}}}>max\{\frac{{a}_{21}^{M}}{{a}_{11}^{L}},\frac{{a}_{22}^{M}}{{a}_{12}^{L}},\frac{{d}_{2}^{M}}{{d}_{1}^{L}}\},}\\ {\displaystyle \frac{{b}_{2}^{L}{e}^{-{r}_{2}^{M}{\tau}_{2}}}{{b}_{1}^{M}{e}^{-{r}_{1}^{L}{\tau}_{1}}}>max\{\frac{{a}_{21}^{M}}{{a}_{11}^{L}},\frac{{a}_{22}^{M}+{d}_{2}^{M}{M}_{22}}{{a}_{12}^{L}}\},}\\ {\displaystyle \frac{{b}_{1}{2}^{L}{e}^{-{r}_{2}^{M}{\tau}_{2}}}{{b}_{1}^{M}{e}^{-{r}_{1}^{L}{\tau}_{1}}}>max\{\frac{{a}_{21}^{M}+{d}_{2}^{M}{M}_{11}}{{a}_{11}^{L}},\frac{{a}_{22}^{M}}{{a}_{12}^{L}}\},}\end{array}$$

*where* $\begin{array}{}{\displaystyle {M}_{ii}=\frac{{b}_{i}^{M}{e}^{-{r}_{i}^{L}{\tau}_{i}}}{{a}_{ii}^{L}}}\end{array}$, *i* = 1, 2. *Then*

$$\begin{array}{}{\displaystyle {m}_{2}\le \underset{t\to +\mathrm{\infty}}{lim\u2006inf}{x}_{2}(t)\le \underset{t\to +\mathrm{\infty}}{lim\u2006sup}{x}_{2}(t)\le {M}_{2}.}\\ {n}_{2}\le \underset{t\to +\mathrm{\infty}}{lim\u2006inf}{y}_{2}(t)\le \underset{t\to +\mathrm{\infty}}{lim\u2006sup}{y}_{2}(t)\le {N}_{2}.\\ \underset{t\to +\mathrm{\infty}}{lim}{x}_{1}(t)=0,\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\underset{t\to +\mathrm{\infty}}{lim}{y}_{1}(t)=0,\end{array}$$

*where*

$$\begin{array}{}{\displaystyle {m}_{2}=\frac{{b}_{2}^{L}{e}^{-{r}_{2}^{M}{\tau}_{2}}}{{a}_{22}^{M}},\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}{n}_{2}=\frac{{b}_{2}^{L}{m}_{2}}{{r}_{2}^{M}}(1-{e}^{-{r}_{2}^{L}{\tau}_{2}}).}\end{array}$$

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