In this section, we present some lemmas and definitions to prove our main results in Section 3.

**Definition 2.1** Let \overline{x} be the unique positive equilibrium of model (1.2). If there exist constants \lambda >0, \tilde{K}>0 and T>0 such that every solution x(t;\phi ) to the initial value problem (1.2) and (1.4) always satisfies

|x(t;\phi )-\overline{x}|\le \tilde{K}{e}^{-\lambda t}\phantom{\rule{1em}{0ex}}\text{for all}tT,

then \overline{x} is said to be globally exponentially stable.

**Definition 2.2** All solutions are uniformly permanent if there exist positive constants *m* and *M* such that for any solution x(t), we have

m<\underset{t\to \mathrm{\infty}}{lim\hspace{0.17em}inf}x(t)\le \underset{t\to \mathrm{\infty}}{lim\hspace{0.17em}sup}x(t)<M.

**Lemma 2.1** *There exists a unique positive global solution of model* (1.2) *and* (1.4) *on the interval* [0,+\mathrm{\infty}).

*Proof* Because of \phi \in {C}_{+}, we have {x}_{t}(\phi )\in {C}_{+} by using Theorem 5.2.1 in [20]. Set x(t)=x(t;\phi ) for t\in [0,\eta (\phi )). From the variation of constants formula and initial condition \phi (0)>0, we obtain

x(t)=\phi (0){e}^{-\alpha t}+\beta {\int}_{0}^{t}{e}^{-\alpha (t-s)}{x}^{\gamma}(s-\tau ){e}^{-\delta x(s-\tau )}\phantom{\rule{0.2em}{0ex}}ds>0

for all t\in [0,\eta (\phi )).

It remains to show \eta (\phi )=+\mathrm{\infty}.

For the sake of contradiction, assume that \eta (\phi ) is bounded. Note the fact that {sup}_{x\in [0,+\mathrm{\infty})}{x}^{\gamma}{e}^{-x}={\gamma}^{\gamma}{e}^{-\gamma}, we obtain from (1.2) that

\begin{array}{rcl}{x}^{\prime}(t)& =& \frac{\beta}{{\delta}^{\gamma}}{(\delta x(t-\tau ))}^{\gamma}{e}^{-\delta x(t-\tau )}-\alpha x(t)\\ \le & \frac{\beta {\gamma}^{\gamma}}{{\delta}^{\gamma}{e}^{\gamma}}-\alpha x(t),\phantom{\rule{1em}{0ex}}t\ge 0.\end{array}

(2.1)

This leads to

x(t)\le \phi (0){e}^{-\alpha t}+\frac{\beta {\gamma}^{\gamma}}{{\delta}^{\gamma}{e}^{\gamma}\alpha}(1-{e}^{-\alpha t}),\phantom{\rule{1em}{0ex}}t\ge 0,

which excludes the possibility that {lim}_{t\to \eta {(\phi )}^{-}}x(t)=\mathrm{\infty}. Hence it violates Theorem 2.3.1 in [21]. So we obtain the existence of the unique global positive solution of (1.2) and (1.4) on [0,+\mathrm{\infty}). Therefore Lemma 2.1 is proved. □

**Lemma 2.2**
*Suppose that there exists a positive constant*
K\in (\kappa ,\tilde{\kappa})
*such that*

\frac{\beta {\gamma}^{\gamma}{\delta}^{1-\gamma}}{K{e}^{\gamma}}<\alpha <\frac{\beta {\delta}^{1-\gamma}}{{e}^{\kappa}},

(2.2)

*then solutions of* (1.2) *and* (1.4) *are uniformly permanent with*

m=\frac{\kappa}{\delta},\phantom{\rule{2em}{0ex}}M=\frac{K}{\delta}.

*Proof* Let x(t)=x(t;\phi ). By Lemma 2.1, x(t)>0 for t\ge 0. From Theorem 1.6.1 in [22] and (2.1), we get that the solution x(t) to the initial value problem (1.2) and (1.4) is not greater than the solution to the initial value problem

{y}^{\prime}(t)=\frac{\beta {\gamma}^{\gamma}}{{\delta}^{\gamma}{e}^{\gamma}}-\alpha y(t),\phantom{\rule{1em}{0ex}}t\ge 0,\phantom{\rule{2em}{0ex}}y(0)=x(0).

In view of (2.2) and

y(t)=x(0){e}^{-\alpha t}+\frac{\beta {\gamma}^{\gamma}}{{\delta}^{\gamma}{e}^{\gamma}\alpha}(1-{e}^{-\alpha t}),

we have

\underset{t\to \mathrm{\infty}}{lim\hspace{0.17em}sup}x(t)\le \underset{t\to \mathrm{\infty}}{lim\hspace{0.17em}sup}y(t)=\frac{\beta {\gamma}^{\gamma}}{{\delta}^{\gamma}{e}^{\gamma}\alpha}<\frac{K}{\delta}.

We next show that l={lim\hspace{0.17em}inf}_{t\to \mathrm{\infty}}x(t)>0. Otherwise, we assume that l=0. For each t\ge 0, we define

m(t)=max\left\{\xi \right|\xi \le t,x(\xi )=\underset{0\le s\le t}{min}x(s)\}.

It follows from l=0 that m(t)\to \mathrm{\infty} as t\to \mathrm{\infty} and that

\underset{t\to \mathrm{\infty}}{lim}x(m(t))=0.

From the definition of m(t), we know that {x}^{\prime}(m(t))\le 0 or

\alpha x(m(t))\ge \beta {x}^{\gamma}(m(t)-\tau ){e}^{-\delta x(m(t)-\tau )}.

(2.3)

Then

\begin{array}{rl}0& =\underset{t\to \mathrm{\infty}}{lim}\alpha x(m(t))\\ \ge \underset{t\to \mathrm{\infty}}{lim}\beta {x}^{\gamma}(m(t)-\tau ){e}^{-\delta x(m(t)-\tau )}\\ \ge \underset{t\to \mathrm{\infty}}{lim}\beta {x}^{\gamma}(m(t)-\tau ){e}^{-K}\\ \ge 0,\end{array}

which implies that

\underset{t\to \mathrm{\infty}}{lim}x(m(t)-\tau )=0.

This yields that there exists {t}_{1}>0 such that

0<\delta x(m(t)-\tau )<\kappa ,\phantom{\rule{1em}{0ex}}t\ge {t}_{1}.

(2.4)

In view of (2.3), (2.4) and \kappa ,\gamma \in (0,1), for t\ge {t}_{1}, we have

\begin{array}{rcl}\alpha & \ge & \beta {\delta}^{1-\gamma}\underset{t\to \mathrm{\infty}}{lim}\frac{{(\delta x(m(t)-\tau ))}^{\gamma}}{\delta x(m(t))}{e}^{-\delta x(m(t)-\tau )}\\ \ge & \beta {\delta}^{1-\gamma}\underset{t\to \mathrm{\infty}}{lim}\frac{\delta x(m(t)-\tau )}{\delta x(m(t))}{e}^{-\kappa}\\ \ge & \frac{\beta {\delta}^{1-\gamma}}{{e}^{\kappa}},\end{array}

which obviously contradicts with (2.2). Thus, we have proved that l>0.

Finally, we prove that l>\frac{\kappa}{\delta}. Again, by way of contradiction, we assume that l\le \frac{\kappa}{\delta}. By the fluctuation lemma [[22], Lemma A.1], there exists a sequence {\{{t}_{k}\}}_{k\ge 1} such that

{t}_{k}\to \mathrm{\infty},\phantom{\rule{2em}{0ex}}x({t}_{k})\to \underset{t\to \mathrm{\infty}}{lim\hspace{0.17em}inf}x(t),\phantom{\rule{2em}{0ex}}{x}^{\prime}({t}_{k})\to 0\phantom{\rule{1em}{0ex}}\text{as}k\to \mathrm{\infty}.

Since \{{x}_{{t}_{k}}\} is bounded and equicontinuous, by the Ascoli-Arzelá theorem, there exists a subsequence, still denoted by itself for simplicity of notation, such that

{x}_{{t}_{k}}\to \tilde{\phi}\phantom{\rule{1em}{0ex}}\text{for some}\tilde{\phi}\in {C}_{+}.

Moreover,

l=\tilde{\phi}(0)\le \tilde{\phi}(\theta )\le \frac{K}{\delta}<\frac{\tilde{\kappa}}{\delta}\phantom{\rule{1em}{0ex}}\text{for}\theta \in [-\tau ,0).

Then recall that \delta l\le \kappa <\gamma <1, and it follows from

{x}^{\prime}({t}_{k})=\beta {x}^{\gamma}({t}_{k}-\tau ){e}^{-\delta x({t}_{k}-\tau )}-\alpha x({t}_{k}),

that (taking limits)

\begin{array}{rcl}0& =& \frac{\beta}{{\delta}^{\gamma}}{(\delta \tilde{\phi}(-\tau ))}^{\gamma}{e}^{-\delta \tilde{\phi}(-\tau )}-\alpha l\\ \ge & \frac{\beta}{{\delta}^{\gamma}}{(\delta l)}^{\gamma}{e}^{-\delta l}-\alpha l\\ \ge & l(\beta {\delta}^{1-\gamma}{(\delta l)}^{\gamma -1}{e}^{-\kappa}-\alpha )\\ >& l(\frac{\beta {\delta}^{1-\gamma}}{{e}^{\kappa}}-\alpha ),\end{array}

which contradicts with (2.2). This proves that l>\frac{\kappa}{\delta}. Hence the proof of Lemma 2.2 is completed. □

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