Traveling Waves in a Stage-Structured Predator–Prey Model with Holling Type Functional Response

Abstract

This article is mainly concerned with a predator–prey model with Holling type functional response, where the stage structure and harvesting are considered. Firstly, we discuss the local stability of various equilibria by using the characteristic equation technique. Then, by constructing upper and lower solutions, we prove the existence of traveling waves connecting two steady states.

1 Introduction

The dynamics between species are very complex. Since the predator–prey model with diffusion exists widely in the natural environment, a variety of models have been studied. A fundamental predator–prey model with diffusion is given by

$$\begin{aligned} {\left\{ \begin{array}{ll} \frac{\partial u}{\partial t}=D_1\frac{\partial ^2u}{\partial x^2}+\alpha u(\beta -u)-f(u)v,\\ \frac{\partial v}{\partial t}=D_2\frac{\partial ^2v}{\partial x^2}-dv+\rho f(u)v. \end{array}\right. } \end{aligned}$$

(1)

There have been a lot of works on system (1). In [1,2,3], by using the shooting method, Dunbar established the existence of various kinds of traveling waves for system (1) with Holling type-I and Holling type-II functional responses. Following Dunbar’s ideas, traveling waves of (1) with different kinds of f(u) were studied in [4,5,6,7,8,9,10,11,12].

As we all know, species at different stages may behave differently, so it seems necessary to investigate the predator–prey models with stage structure. Recently, Zhang et al. [13], Ge and He [14], Zhang and Xu [15], and Ge et al. [16] studied the properties of traveling wave solutions of various kinds of predator–prey models with stage structure.

On the other hand, for economic reasons, human needs to exploit biological resources and harvest some biological species. Therefore, it is necessary to study the suitable population model with harvesting. Recently, Hong and Weng [17], Lv et al. [18], Hong and Weng [19], and Xia et al. [20] concerned the traveling waves of various kinds of predator–prey models, where the harvesting and stage structure were considered.

In this paper, we will concentrate on the following predator–prey system with Holling type functional response

$$\begin{aligned} {\left\{ \begin{array}{ll} \frac{\partial u_1}{\partial t}=D_1\frac{\partial ^2u_1}{\partial x^2}+au_2(x,t)-d_1u_1(x,t)-a_{11}u_1^2(x,t)\\ \quad \quad \quad -ae^{-d_1\tau }\int _{-\infty }^{+\infty }G(\tau ,x-y)u_2(y,t-\tau )\hbox {d}y,\\ \frac{\partial u_2}{\partial t}=D_2\frac{\partial ^2u_2}{\partial x^2}+ae^{-d_1\tau }\int _{-\infty }^{+\infty }G(\tau ,x-y)u_2(y,t-\tau )\hbox {d}y -(d_2+q_2e_2)u_2(x,t)\\ \quad \quad \quad -a_{22}u_2^2(x,t)-\frac{a_{23}u_2^p(x,t)v(x,t)}{1+mu_2^p(x,t)},\\ \frac{\partial v}{\partial t}=D_3\frac{\partial ^2v}{\partial x^2}+\left( a_1-bv(x,t)\right) v(x,t)-q_3e_3v(x,t)+\frac{a_{32}u_2^p(x,t)v(x,t)}{1+mu_2^p(x,t)}, \end{array}\right. } \end{aligned}$$

(2)

where

$$\begin{aligned} G(\tau ,y)=\frac{1}{\sqrt{4\pi D_1\tau }}e^{-\frac{y^2}{4D_1\tau }},\\p\in {\mathbb {N}},\quad p\ge 2. \end{aligned}$$

We make the following assumptions for model (2).

• (A1) All parameters are positive. $u_1(x,t),u_2(x,t),v(x,t)$ represent the densities of the immature and mature prey–predator species at time t and location x, respectively; $D_i(i=1,2,3)$ are the diffusion coefficients.

• (A2) a is the birth rate of immature prey population; $d_i(i=1,2)$ are the death rate of immature and mature prey population, respectively; $a_{ii}(i=1,2)$ are the intra-specific competition rate of immature and mature prey population, respectively; $a_1$ and $\frac{b}{a_1}$ are the growth rate and environmental carrying capacity of predator population, respectively; $q_i(i=2,3)$ are the catch ability coefficient of the mature prey and predator species, respectively; $e_i(i=2,3)$ are the harvesting effort of the mature prey and predator species, respectively.

• (A3) The term $ae^{-d_1\tau }\int _{-\infty }^{+\infty }G(\tau ,x-y)u_2(y,t-\tau )\hbox {d}y$ stands for the number of prey population which leave immature individuals to mature individuals at time t and location x. $\frac{u_2^p(x,t)v(x,t)}{1+mu_2^p(x,t)}$ represents the Holling type functional response.

This article is organized as follows. We analyze the stability of the equilibria first in Sect. 2. In Sect. 3, by using Schauder’s fixed point theorem, an existence theorem of traveling waves connecting two steady states is derived. The main contribution of this article is the construction and verification of upper–lower solutions. These works are done in Sect. 4.

2 Local Stability of Equilibria

Note that $u_1(x,t)$ is independent of the last two equations of system (2); for simplicity , we denote $u_2(x,t)$ by u(x, t) to obtain the following system

$$\begin{aligned} {\left\{ \begin{array}{ll} \frac{\partial u}{\partial t}=D_2\frac{\partial ^2u}{\partial x^2}+ae^{-d_1\tau }\int _{-\infty }^{+\infty }G(\tau ,x-y)u(y,t-\tau )\hbox {d}y -(d_2+q_2e_2)u(x,t)\\ \quad \quad \;\;-a_{22}u^2(x,t)-\frac{a_{23}u^p(x,t)v(x,t)}{1+mu^p(x,t)},\\ \frac{\partial v}{\partial t}=D_3\frac{\partial ^2v}{\partial x^2}+\left( a_1-bv(x,t)\right) v(x,t)-q_3e_3v(x,t)+\frac{a_{32}u^p(x,t)v(x,t)}{1+mu^p(x,t)}. \end{array}\right. } \end{aligned}$$

(3)

Let

$$\begin{aligned} \vartheta _1=ae^{-d_1\tau }-d_2-q_2e_2,\quad \vartheta _2=a_1-q_3e_3. \end{aligned}$$

It is easy to know that system (3) has the following equilibria

$$\begin{aligned} E_0(0,0),\quad E_1\left( \frac{\vartheta _1}{a_{22}},0\right) ,\quad E_2\left( 0,\frac{\vartheta _2}{b}\right) . \end{aligned}$$

We assume that

$$\begin{aligned} \vartheta _1>0,\quad \vartheta _2>0 \end{aligned}$$

due to the background of our system. The linearized system of (3) at any equilibrium $(u^*,v^*)$ is

$$\begin{aligned} {\left\{ \begin{array}{ll} \frac{\partial u}{\partial t}=D_2\frac{\partial ^2u}{\partial x^2}+ae^{-d_1\tau }\int _{-\infty }^{+\infty } \frac{1}{\sqrt{4\pi D_1\tau }}e^{-\frac{(x-y)^{2}}{4D_1\tau }}u(y,t-\tau )\hbox {d}y -(d_2+q_2e_2)u(x,t)\\ \quad \quad \;\;-2a_{22}u^{*}u(x,t)-\frac{pa_{23}(u^{*})^{p-1}v^{*}}{(1+m(u^{*})^p)^2}u(x,t)-\frac{a_{23}(u^{*})^p}{1+m(u^{*})^p}v(x,t),\\ \frac{\partial v}{\partial t}=D_3\frac{\partial ^2v}{\partial x^2}+(a_1-2bv^{*})v(x,t)-q_3e_3v(x,t) +\frac{pa_{32}(u^{*})^{p-1}v^{*}}{(1+m(u^{*})^p)^2}u(x,t)\\ \quad \quad \;\;+\frac{a_{32}(u^{*})^p}{1+m(u^{*})^p}v(x,t). \end{array}\right. } \end{aligned}$$

(4)

Let $\lambda$ be a complex number and $\sigma$ be a real number. We know that Eq. (4) admits non-trivial solutions with the form

$$\begin{aligned} \left( \begin{aligned}&u(x,t)\\ {}&v(x,t)\end{aligned}\right) =\left( \begin{aligned}&c_1\\ {}&c_2\end{aligned}\right) e^{\lambda t+i\sigma x} \end{aligned}$$

if and only if

$$\begin{aligned} \left| \begin{aligned}&\chi _1(\lambda ,\sigma ,u^*,v^*)+\frac{pa_{23}(u^{*})^{p-1}v^{*}}{(1+m(u^{*})^p)^2}\quad \quad \quad \frac{a_{23}(u^{*})^p}{1+m(u^{*})^p}\\&\quad \quad \quad -\frac{pa_{32}(u^{*})^{p-1}v^{*}}{(1+m(u^{*})^p)^2}\quad \quad \quad \chi _2(\lambda ,\sigma ,u^*,v^*)-\frac{a_{32}(u^{*})^p}{1+m(u^{*})^p}\end{aligned}\right| =0, \end{aligned}$$

(5)

where

$$\begin{aligned} \chi _1(\lambda ,\sigma ,u^*,v^*)=\lambda +D_2\sigma ^2-ae^{-d_1\tau -\lambda \tau }e^{-D_1\sigma ^2\tau }+d_2+q_2e_2+2a_{22}u^{*},\\\chi _2(\lambda ,\sigma ,u^*,v^*)=\lambda +D_3\sigma ^2-a_1+2bv^{*}+q_3e_3. \end{aligned}$$

(5) is equivalent to

$$\begin{aligned} \begin{aligned}&\left[ \chi _1(\lambda ,\sigma ,u^*,v^*)+\frac{pa_{23}(u^{*})^{p-1}v^{*}}{(1+m(u^{*})^p)^2}\right] \left[ \chi _2(\lambda ,\sigma ,u^*,v^*)-\frac{a_{32}(u^{*})^p}{1+m(u^{*})^p}\right] \\&\quad +\frac{pa_{23}a_{32}(u^{*})^{2p-1}v^*}{(1+m(u^{*})^p)^3}=0. \end{aligned} \end{aligned}$$

(6)

Theorem 1

$E_0(0,0)$ is an unstable equilibrium.

Proof

Letting $(u^*,v^*)=(0,0)$ in (6), we obtain

$$\begin{aligned} (\lambda +D_2\sigma ^2-ae^{-d_1\tau -\lambda \tau }e^{-D_1\sigma ^2\tau }+d_2+q_2e_2)(\lambda +D_3\sigma ^2-a_1+q_3e_3)=0. \end{aligned}$$

Thus, either

$$\begin{aligned} \lambda +D_2\sigma ^2-ae^{-d_1\tau -\lambda \tau }e^{-D_1\sigma ^2\tau }+d_2+q_2e_2=0, \end{aligned}$$

(7)

or

$$\begin{aligned} \lambda +D_3\sigma ^2-a_1+q_3e_3=0. \end{aligned}$$

(8)

Let

$$\begin{aligned} f(\lambda ,\sigma )=-D_2\sigma ^2+ae^{-d_1\tau -\lambda \tau }e^{-D_1\sigma ^2\tau }-d_2-q_2e_2. \end{aligned}$$

Then,

$$\begin{aligned} f(0,0)=ae^{-d_1\tau }-d_2-q_2e_2=\vartheta _1>0, \end{aligned}$$

Therefore, there exists a $\sigma _*>0$ such that $f(0,\sigma _*)>0$. Since $f(\infty ,\sigma _*)<0$, the equation $\lambda =f(\lambda ,\sigma _*)$ has at least one root $\lambda _*>0$. And Eq. (7) with $\sigma =\sigma _*$ has at least one root $\lambda _*>0$. That is, $E_0(0,0)$ is an unstable equilibrium. $\square$

Theorem 2

$E_1\left( \nicefrac {\vartheta _1}{a_{22}},0\right)$ is an unstable equilibrium.

Proof

Letting $(u^*,v^*)=\left( \nicefrac {\vartheta _1}{a_{22}},0\right)$ in (6), we obtain that either

$$\begin{aligned} \lambda +D_2\sigma ^2-ae^{-d_1\tau -\lambda \tau }e^{-D_1\sigma ^2\tau }+d_2+q_2e_2+2\vartheta _1=0, \end{aligned}$$

(9)

or

$$\begin{aligned} \lambda +D_3\sigma ^2-a_1+q_3e_3-\frac{a_{32}\vartheta _1^p}{a_{22}^p+m\vartheta _1^p}=0. \end{aligned}$$

(10)

It is easy to see that Eq. (10) has at least one root $(\lambda _*,\sigma _*)$ with $\lambda _*>0$.

$E_1\left( \nicefrac {\vartheta _1}{a_{22}},0\right)$ is an unstable equilibrium. $\square$

Theorem 3

$E_2\left( 0,\nicefrac {\vartheta _2}{b}\right)$ is an unstable equilibrium.

Proof

Letting $(u^*,v^*)=\left( 0,\nicefrac {\vartheta _2}{b}\right)$ in (6), it follows that either

$$\begin{aligned} \lambda +D_2\sigma ^2-ae^{-d_1\tau -\lambda \tau }e^{-D_1\sigma ^2\tau }+d_2+q_2e_2=0, \end{aligned}$$

(11)

or

$$\begin{aligned} \lambda +D_3\sigma ^2+\vartheta _2=0. \end{aligned}$$

(12)

From the proof of Theorem 1, we know that Eq. (11) has at least one root $(\lambda _*,\sigma _*)$ with $\lambda _*>0$. $E_2\left( 0,\nicefrac {\vartheta _2}{b}\right)$ is an unstable equilibrium. $\square$

Now, we consider the stability of the positive equilibrium $E_3(u^+,v^+)$, which is given by the following system

$$\begin{aligned} {\left\{ \begin{array}{ll} \vartheta _1-a_{22}u-\frac{a_{23}u^{p-1}v}{1+mu^p}=0,\\ \vartheta _2-bv+\frac{a_{32}u^p}{1+mu^p}=0. \end{array}\right. } \end{aligned}$$

The above system can be rewritten as

$$\begin{aligned} h(u)= & {} A_0u^{2p+1}+A_1u^{2p}+A_2u^{2p-1}+A_3u^{p+1}\nonumber \\&+A_4u^p+A_5u^{p-1}+A_6u+A_7=0, \end{aligned}$$

(13)

where

$$\begin{aligned} A_0= & {} bm^2a_{22},\quad A_1=-\,bm^2\vartheta _1,\quad A_2=ma_{23}\vartheta _2+a_{23}a_{32},\\ A_3= & {} 2bma_{22},\quad A_4=-\,2bm\vartheta _1,\quad A_5=a_{23}\vartheta _2,\quad A_6=ba_{22},\quad A_7=-b\vartheta _1. \end{aligned}$$

Noting $A_0>0$ and $A_7<0$, thus Eq. (13) possesses at least one root $u^+>0$.

Theorem 4

If $a_{22}\ge \frac{a_{23}\left( m(u^{+})^p+1-p\right) (u^+)^{p-2}v^+}{(1+m(u^{+})^p)^2}$ holds, then $E_3(u^+,v^+)$ is a locally

asymptotically stable equilibrium.

Proof

Denote

$$\begin{aligned}&\gamma _1=\frac{a_{32}}{1+m(u^{+})^p},\quad \gamma _2=\frac{pa_{23}}{(1+m(u^{+})^p)^2},\\&\begin{aligned}A(\mu _1,\omega _1,\sigma _1)=&\mu _1+D_2\sigma _1^2-ae^{-d_1\tau -\mu _1\tau }e^{-D_1\sigma _1^2\tau }\cos (\omega _1\tau )+d_2+q_2e_2\\&+2a_{22}u^++\gamma _2(u^+)^{p-1}v^+,\end{aligned}\\&B(\mu _1,\omega _1,\sigma _1)=ae^{-d_1\tau -\mu _1\tau }e^{-D_1\sigma _1^2\tau }\sin (\omega _1\tau )+\omega _1. \end{aligned}$$

Letting $(u^*,v^*)=(u^+,v^+)$ in (6), from $\vartheta _2-bv^++\gamma _1(u^{+})^p=0$, we have

$$\begin{aligned} \begin{aligned} \lambda&=-\frac{\gamma _1\gamma _2(u^{+})^{2p-1}v^+}{\lambda +D_2\sigma ^2-ae^{-d_1\tau -\lambda \tau }e^{-D_1\sigma ^2\tau }+d_2+q_2e_2+2a_{22}u^++\gamma _2(u^+)^{p-1}v^+}\\&\quad -\,(D_3\sigma ^2-a_1+2bv^++q_3e_3-\gamma _1(u^{+})^p)\\&=-\frac{\gamma _1\gamma _2(u^{+})^{2p-1}v^+}{\lambda +D_2\sigma ^2-ae^{-d_1\tau -\lambda \tau }e^{-D_1\sigma ^2\tau }+d_2+q_2e_2+2a_{22}u^++\gamma _2(u^+)^{p-1}v^+}\\&\quad -\,(D_3\sigma ^2+bv^+). \end{aligned} \end{aligned}$$

(14)

Suppose that there is a $(\mu _1+i\omega _1,\sigma _1)$ satisfying (14) with $\mu _1\ge 0$, and note that

$$\begin{aligned} \vartheta _1-a_{22}u^+-\frac{a_{23}(u^+)^{p-1}v^+}{1+m(u^{+})^p}=0. \end{aligned}$$

The direct computation gives us

$$\begin{aligned} \begin{aligned} 0&\le \mu _1=-\frac{A(\mu _1,\omega _1,\sigma _1)\gamma _1\gamma _2(u^{+})^{2p-1}v^+}{A(\mu _1,\omega _1,\sigma _1)^2+B(\mu _1,\omega _1,\sigma _1)^2}-(D_3\sigma _1^2+bv^+)\\&\le -\frac{-ae^{-d_1\tau -\mu _1\tau }e^{-D_1\sigma _1^2\tau }\cos (\omega _1\tau )+d_2+q_2e_2+2a_{22}u^++\gamma _2(u^+)^{p-1}v^+}{A(\mu _1,\omega _1,\sigma _1)^2+B(\mu _1,\omega _1,\sigma _1)^2}\\&\quad \times \gamma _1\gamma _2(u^{+})^{2p-1}v^+-(D_3\sigma _1^2+bv^+)\\&=-\frac{ae^{-d_1\tau }-ae^{-d_1\tau -\mu _1\tau }e^{-D_1\sigma _1^2\tau }\cos (\omega _1\tau )+a_{22}u^++\gamma _2(u^+)^{p-1}v^+-\frac{a_{23}(u^+)^{p-1}v^+}{1+m(u^{+})^p}}{A(\mu _1,\omega _1,\sigma _1)^2+B(\mu _1,\omega _1,\sigma _1)^2}\\&\quad \times \gamma _1\gamma _2(u^{+})^{2p-1}v^+-(D_3\sigma _1^2+bv^+)\\&\le -\gamma _1\gamma _2(u^{+})^{2p-1}v^+\times \frac{a_{22}u^++\gamma _2(u^+)^{p-1}v^+-\frac{a_{23}(u^+)^{p-1}v^+}{1+m(u^{+})^p}}{A(\mu _1,\omega _1,\sigma _1)^2+B(\mu _1,\omega _1,\sigma _1)^2}-(D_3\sigma _1^2+bv^+)\\&=-\gamma _1\gamma _2(u^{+})^{2p-1}v^+\times \frac{a_{22}u^+-\frac{a_{23}\left( m(u^{+})^p+1-p\right) (u^+)^{p-1}v^+}{(1+m(u^{+})^p)^2}}{A(\mu _1,\omega _1,\sigma _1)^2+B(\mu _1,\omega _1,\sigma _1)^2}-(D_3\sigma _1^2+bv^+)<0 \end{aligned} \end{aligned}$$

as $a_{22}\ge \frac{a_{23}\left( m(u^{+})^p+1-p\right) (u^+)^{p-2}v^+}{(1+m(u^{+})^p)^2}$, a contradiction. Thus, if $(\lambda ,\sigma )=(\mu +i\omega ,\sigma )$ satisfies (14), we must have $\mu <0$. $E_3(u^+,v^+)$ is a locally asymptotically stable equilibrium. $\square$

3 Existence of Traveling Waves

Substituting $(u(x,t),v(x,t))=(\phi (x+ct),\psi (x+ct))$ into (3), for simplicity, denoting the traveling wave coordinate $x+ct$ by t, it follows that

$$\begin{aligned} {\left\{ \begin{array}{ll} D_2\phi ''(t)-c\phi '(t)+f_2(\phi ,\psi )(t)=0,\\ D_3\psi ''(t)-c\psi '(t)+f_3(\phi ,\psi )(t)=0, \end{array}\right. } \end{aligned}$$

(15)

where

$$\begin{aligned}\begin{aligned}f_2(\phi ,\psi )(t)&= ae^{-d_1\tau }\int _{-\infty }^{+\infty }\frac{1}{\sqrt{4\pi D_1\tau }}e^{-\frac{y^{2}}{4D_1\tau }}\phi (t-y-c\tau )\hbox {d}y -(d_2+q_2e_2)\phi (t)\\&\quad -\,a_{22}\phi ^2(t)-\frac{a_{23}\phi ^p(t)\psi (t)}{1+m\phi ^p(t)},\end{aligned}\\f_3(\phi ,\psi )(t)=(a_1-b\psi (t))\psi (t)-q_3e_3\psi (t)+\frac{a_{32}\phi ^p(t)\psi (t)}{1+m\phi ^p(t)}. \end{aligned}$$

We will look for the non-trivial and nonnegative solutions of system (15) which satisfy the following asymptotic boundary conditions

$$\begin{aligned} \lim _{t\rightarrow -\infty }(\phi (t),\psi (t))=(0,0),\quad \lim _{t\rightarrow +\infty }(\phi (t),\psi (t))=(u^+,v^+). \end{aligned}$$

(16)

Let

$$\begin{aligned} M_2\ge \max \left\{ u^+,\frac{ae^{-d_1\tau }-d_2-q_2e_2}{a_{22}}\right\} ,\quad M_3\ge \max \left\{ v^+,\frac{a_1-q_3e_3+a_{32}M_2^p}{b}\right\} ,\nonumber \\ \end{aligned}$$

(17)

and

$$\begin{aligned} C_{[0,M]}({\mathbb {R}},{\mathbb {R}}^2):=\left\{ (\phi ,\psi )\in C({\mathbb {R}},{\mathbb {R}}^2)| 0\le \phi (t)\le M_2,0\le \psi (t)\le M_3\quad \text{ for }\,t\in {\mathbb {R}}\right\} . \end{aligned}$$

Choosing

$$\begin{aligned} \beta _2\ge d_2+q_2e_2+2a_{22}M_2+pa_{23}M_2^{p-1}M_3,\quad \beta _3\ge 2bM_3+q_3e_3-a_1, \end{aligned}$$

(18)

we define two operators $H=(H_2,H_3)$ and $F=(F_2,F_3)$ from $C_{[0,M]}({\mathbb {R}},{\mathbb {R}}^2)$ to $C({\mathbb {R}},{\mathbb {R}}^2)$ by

$$\begin{aligned}&H_2(\phi ,\psi )(t)=f_2(\phi ,\psi )(t)+\beta _2\phi (t),\quad H_3(\phi ,\psi )(t)=f_3(\phi ,\psi )(t)+\beta _3\psi (t),\\&\begin{aligned}&F_2(\phi ,\psi )(t)\\&=\frac{1}{D_2(\lambda _{22}-\lambda _{21})}\left[ \int _{-\infty }^t e^{\lambda _{21}(t-s)} H_2(\phi ,\psi )(s)\hbox {d}s+ \int _{t}^{+\infty } e^{\lambda _{22}(t-s)} H_2(\phi ,\psi )(s)\hbox {d}s\right] ,\end{aligned}\\&\begin{aligned}&F_3(\phi ,\psi )(t)\\&=\frac{1}{D_3(\lambda _{32}-\lambda _{31})}\left[ \int _{-\infty }^t e^{\lambda _{31}(t-s)} H_3(\phi ,\psi )(s)\hbox {d}s+ \int _{t}^{+\infty } e^{\lambda _{32}(t-s)} H_3(\phi ,\psi )(s)\hbox {d}s\right] , \end{aligned} \end{aligned}$$

where

$$\begin{aligned} \lambda _{21}= & {} \frac{c-\sqrt{c^2+4\beta _2D_2}}{2D_2}<0,\quad \lambda _{22}=\frac{c+\sqrt{c^2+4\beta _2D_2}}{2D_2}>0,\\ \lambda _{31}= & {} \frac{c-\sqrt{c^2+4\beta _3D_3}}{2D_3}<0,\quad \lambda _{32}=\frac{c+\sqrt{c^2+4\beta _3D_3}}{2D_3}>0. \end{aligned}$$

For $\mu \in (0,\min \{-\lambda _{21},\lambda _{22},-\lambda _{31},\lambda _{32}\})$, define

$$\begin{aligned} B_{\mu }({\mathbb {R}},{\mathbb {R}}^2)=\left\{ (\phi ,\psi )\in C({\mathbb {R}},{\mathbb {R}}^2)|\sup _{t\in {\mathbb {R}}}|(\phi ,\psi )(t)|e^{-\mu |t|}<\infty \right\} \end{aligned}$$

and

$$\begin{aligned} |(\phi ,\psi )|_\mu =\sup _{t\in {\mathbb {R}}}|(\phi ,\psi )(t)|e^{-\mu |t|}. \end{aligned}$$

Then, it is easy to check that $(B_{\mu }({\mathbb {R}},{\mathbb {R}}^2),|\cdot |_\mu )$ is a Banach space.

The operators $H_i(i=2,3)$ and $F_i(i=2,3)$ admit the following properties.

Lemma 1

For sufficiently large $\beta _2,\beta _3$ satisfying (18), we have

$$\begin{aligned} H_2(\phi _1,\psi _1)(t)\ge H_2(\phi _2,\psi _1)(t),\quad H_2(\phi _1,\psi _1)(t)\le H_2(\phi _1,\psi _2)(t),\\H_3(\phi _1,\psi _1)(t)\ge H_3(\phi _2,\psi _1)(t),\quad H_3(\phi _1,\psi _1)(t)\ge H_3(\phi _1,\psi _2)(t), \end{aligned}$$

for $t\in {\mathbb {R}}$ with $0\le \phi _2(t)\le \phi _1(t)\le M_2, 0\le \psi _2(t)\le \psi _1(t)\le M_3$.

Proof

Let

$$\begin{aligned} f(\phi )=\frac{\phi ^p}{1+m\phi ^p}. \end{aligned}$$

Then,

$$\begin{aligned} f'(\phi )=\frac{p\phi ^{p-1}}{(1+m\phi ^p)^2}. \end{aligned}$$

$f(\phi )$ is increasing on $[0,+\infty )$. For $0\le \phi _2\le \phi _1\le M_2$, according to Lagrange mean value theorem, we have

$$\begin{aligned} f(\phi _1)-f(\phi _2)=f'(\xi )(\phi _1-\phi _2)=\frac{p\xi ^{p-1}}{(1+m\xi ^p)^2}(\phi _1-\phi _2), \end{aligned}$$

where $\phi _2\le \xi \le \phi _1$. Thus,

$$\begin{aligned} 0\le f(\phi _1)-f(\phi _2)\le pM_2^{p-1}(\phi _1-\phi _2). \end{aligned}$$

From the definition of $H=(H_2,H_3)$ and note (18), we have

$$\begin{aligned}&\begin{aligned}&H_2(\phi _1,\psi _1)(t)- H_2(\phi _2,\psi _1)(t)\\&=f_2(\phi _1,\psi _1)(t)+\beta _2\phi _1(t)-f_2(\phi _2,\psi _1)(t)-\beta _2\phi _2(t)\\&=ae^{-d_1\tau }\int _{-\infty }^{+\infty }\frac{1}{\sqrt{4\pi D_1\tau }}e^{-\frac{y^{2}}{4D_1\tau }}(\phi _1(t-y-c\tau )-\phi _2(t-y-c\tau ))\hbox {d}y\\&\quad -\,(d_2+q_2e_2)(\phi _1(t)-\phi _2(t))-a_{22}\left( \phi _1^2(t)-\phi _2^2(t)\right) \\&\quad -\,a_{23}\left( \frac{\phi _1^p(t)}{1+m\phi _1^p(t)}-\frac{\phi _2^p(t)}{1+m\phi _2^p(t)}\right) \psi _1(t) +\beta _2\left( \phi _1(t)-\phi _2(t)\right) \\&\ge -(d_2+q_2e_2)(\phi _1(t)-\phi _2(t))-a_{22}\left( \phi _1(t)+\phi _2(t)\right) (\phi _1(t)-\phi _2(t))\\&\quad -\,pa_{23}M_2^{p-1}(\phi _1(t)-\phi _2(t))\psi _1(t)+\beta _2(\phi _1(t)-\phi _2(t))\\&\ge \left( -d_2-q_2e_2-2a_{22}M_2-pa_{23}M_2^{p-1}M_3+\beta _2\right) (\phi _1(t)-\phi _2(t))\ge 0, \end{aligned}\\&\begin{aligned}&H_2(\phi _1,\psi _1)(t)- H_2(\phi _1,\psi _2)(t)\\&=f_2(\phi _1,\psi _1)(t)-f_2(\phi _1,\psi _2)(t)\\&=-\frac{a_{23}\phi _1^p(t)(\psi _1(t)-\psi _2(t))}{1+m\phi _1^p(t)}\le 0, \end{aligned}\\&\begin{aligned}&H_3(\phi _1,\psi _1)(t)- H_3(\phi _2,\psi _1)(t)\\&=f_3(\phi _1,\psi _1)(t)-f_3(\phi _2,\psi _1)(t)\\&=a_{32}\left( \frac{\phi _1^p(t)}{1+m\phi _1^p(t)}-\frac{\phi _2^p(t)}{1+m\phi _2^p(t)}\right) \psi _1(t)\ge 0, \end{aligned}\\&\begin{aligned}&H_3(\phi _1,\psi _1)(t)- H_3(\phi _1,\psi _2)(t)\\&=f_3(\phi _1,\psi _1)(t)+\beta _3\psi _1(t)-f_3(\phi _1,\psi _2)(t)-\beta _3\psi _2(t)\\&=(a_1-b\psi _1(t))\psi _1(t)-q_3e_3\psi _1(t)+\frac{a_{32}\phi _1^p(t)\psi _1(t)}{1+m\phi _1^p(t)} \end{aligned}\\&\begin{aligned}&\quad -\,(a_1-b\psi _2(t))\psi _2(t)+q_3e_3\psi _2(t)-\frac{a_{32}\phi _1^p(t)\psi _2(t)}{1+m\phi _1^p(t)}\\&\quad +\beta _3(\psi _1(t)-\psi _2(t))\\&\ge (a_1-q_3e_3)(\psi _1(t)-\psi _2(t))-b(\psi _1(t)+\psi _2(t))(\psi _1(t)-\psi _2(t))\\&\quad +\beta _3(\psi _1(t)-\psi _2(t))\\&\ge (a_1-q_3e_3-2bM_3+\beta _3)(\psi _1(t)-\psi _2(t))\ge 0. \end{aligned} \end{aligned}$$

$\square$

Lemma 2

For sufficiently large $\beta _2,\beta _3$ satisfying (18), we have

$$\begin{aligned} F_2(\phi _1,\psi _1)(t)\ge & {} F_2(\phi _2,\psi _1)(t),\quad F_2(\phi _1,\psi _1)(t)\le F_2(\phi _1,\psi _2)(t),\\ F_3(\phi _1,\psi _1)(t)\ge & {} F_3(\phi _2,\psi _1)(t),\quad F_3(\phi _1,\psi _1)(t)\ge F_3(\phi _1,\psi _2)(t), \end{aligned}$$

for $t\in {\mathbb {R}}$ with $0\le \phi _2(t)\le \phi _1(t)\le M_2, 0\le \psi _2(t)\le \psi _1(t)\le M_3$.

Proof

From the definition of $F_i(i=2,3)$, we can see that $F_i(i=2,3)$ enjoys the same properties as $H_i(i=2,3)$ stated in Lemma 1. $\square$

Lemma 3

$F=(F_2,F_3)$ is continuous with respect to the norm $|\cdot |_\mu$ in $B_{\mu }({\mathbb {R}},{\mathbb {R}}^2)$.

Proof

Note that

$$\begin{aligned} \begin{aligned}&\int _{-\infty }^{+\infty }G(\tau ,y)e^{\mu |y+c\tau |}\hbox {d}y\le \int _{-\infty }^{+\infty }\frac{1}{\sqrt{4\pi D_1\tau }}e^{-\frac{y^{2}}{4D_1\tau }}e^{\mu (|y|+c\tau )}\hbox {d}y\\&\quad =\int _{-\infty }^{+\infty }\frac{1}{\sqrt{4\pi D_1\tau }}e^{-\frac{(|y|-2D_1\mu \tau )^{2}}{4D_1\tau }}e^{(D_1\mu ^2+c\mu )\tau }\hbox {d}y=e^{(D_1\mu ^2+c\mu )\tau }. \end{aligned} \end{aligned}$$

If $\varPhi =(\phi _1,\psi _1), \varPsi =(\phi _2,\psi _2)\in B_{\mu }({\mathbb {R}},{\mathbb {R}}^2)$, we have

$$\begin{aligned} \begin{aligned}&|H_2(\phi _1,\psi _1)(t)-H_2(\phi _2,\psi _2)(t)|e^{-\mu |t|}\\&\le |f_2(\phi _1,\psi _1)(t)-f_2(\phi _2,\psi _2)(t)|e^{-\mu |t|} +\beta _2|\phi _1(t)-\phi _2(t)|e^{-\mu |t|}\\&\le \left| ae^{-d_1\tau }\int _{-\infty }^{+\infty }G(\tau ,y)(\phi _1(t-y-c\tau )-\phi _2(t-y-c\tau ))\hbox {d}y\right. \\&\quad -\,(d_2+q_2e_2)(\phi _1(t)-\phi _2(t))-a_{22}(\phi _1^2(t)-\phi _2^2(t))\\&\quad \left. -\,a_{23}\left( \frac{\phi _1^p(t)\psi _1(t)}{1+m\phi _1^p(t)}-\frac{\phi _2^p(t)\psi _2(t)}{1+m\phi _2^p(t)}\right) \right| e^{-\mu |t|} +\beta _2|\varPhi -\varPsi |_\mu \\&\le \left\{ ae^{-d_1\tau }\int _{-\infty }^{+\infty }G(\tau ,y)|\phi _1(t-y-c\tau )-\phi _2(t-y-c\tau )|\hbox {d}y\right. \\&\quad +(d_2+q_2e_2)|\phi _1(t)-\phi _2(t)|+a_{22}|\phi _1(t)+\phi _2(t)||\phi _1(t)-\phi _2(t)|\\&\quad \left. +\,a_{23}\left| \frac{\phi _1^p(t)\psi _1(t)}{1+m\phi _1^p(t)}-\frac{\phi _2^p(t)\psi _2(t)}{1+m\phi _2^p(t)}\right| \right\} e^{-\mu |t|} +\beta _2|\varPhi -\varPsi |_\mu \\&\le ae^{-d_1\tau }\int _{-\infty }^{+\infty }G(\tau ,y)e^{\mu |y+c\tau |}\hbox {d}y|\varPhi -\varPsi |_\mu +(d_2+q_2e_2)|\varPhi -\varPsi |_\mu \\&\quad +2a_{22}M_2|\varPhi -\varPsi |_\mu +a_{23}\left( M_2^p+pM_2^{p-1}M_3\right) |\varPhi -\varPsi |_\mu +\beta _2|\varPhi -\varPsi |_\mu \\&\le \kappa _2|\varPhi -\varPsi |_\mu , \end{aligned} \end{aligned}$$

where

$$\begin{aligned} \kappa _2=ae^{-d_1\tau }e^{(D_1\mu ^2+c\mu )\tau }+d_2+q_2e_2+2a_{22}M_2+a_{23}\left( M_2^p+pM_2^{p-1}M_3\right) +\beta _2. \end{aligned}$$

Therefore, $H_2:B_{\mu }({\mathbb {R}},{\mathbb {R}}^2)\rightarrow B_{\mu }({\mathbb {R}},{\mathbb {R}})$ is continuous with respect to the norm $|\cdot |_\mu$. Using the same method, we can prove that

$$\begin{aligned} |H_3(\phi _1,\psi _1)(t)-H_3(\phi _2,\psi _2)(t)|e^{-\mu |t|}\le \kappa _3|\varPhi -\varPsi |_\mu , \end{aligned}$$

where

$$\begin{aligned} \kappa _3=a_1+q_3e_3+2bM_3+a_{32}\left( M_2^p+pM_2^{p-1}M_3\right) +\beta _3. \end{aligned}$$

And thus, $H_3:B_{\mu }({\mathbb {R}},{\mathbb {R}}^2)\rightarrow B_{\mu }({\mathbb {R}},{\mathbb {R}})$ is continuous with respect to the norm $|\cdot |_\mu$.

The proof of $F_2$ and $F_3$ which are continuous with respect to the norm $|\cdot |_\mu$ can be found in [19], and we omit it here. The proof is complete. $\square$

Definition 1

A pair of continuous functions ${\overline{\rho }}(t)=({\overline{\phi }}(t),{\overline{\psi }}(t))$ and ${\underline{\rho }}(t)=({\underline{\phi }}(t),$ ${\underline{\psi }}(t))$ are called a pair of upper–lower solutions of (15), if there exists a set ${\mathcal {S}}=\{s_i\in {\mathbb {R}},i=1,2,\ldots ,n\}$ with finite points such that ${\overline{\rho }}(t)$ and ${\underline{\rho }}(t)$ are twice continuously differentiable on ${\mathbb {R}}\backslash {\mathcal {S}}$ and satisfy

$$\begin{aligned} {\left\{ \begin{array}{ll} D_2{\overline{\phi }}''(t)-c{\overline{\phi }}'(t)+f_2({\overline{\phi }},{\underline{\psi }})(t)\le 0,\\ D_3{\overline{\psi }}''(t)-c{\overline{\psi }}'(t)+f_3({\overline{\phi }},{\overline{\psi }})(t)\le 0, \end{array}\right. } \end{aligned}$$

and

$$\begin{aligned} {\left\{ \begin{array}{ll} D_2{\underline{\phi }}''(t)-c{\underline{\phi }}'(t)+f_2({\underline{\phi }},{\overline{\psi }})(t)\ge 0,\\ D_3{\underline{\psi }}''(t)-c{\underline{\psi }}'(t)+f_3({\underline{\phi }},{\underline{\psi }})(t)\ge 0 \end{array}\right. } \end{aligned}$$

for $t\in {\mathbb {R}}\backslash {\mathcal {S}}$.

Now we assume that a pair of upper–lower solutions $({\overline{\phi }}(t),{\overline{\psi }}(t))$ and $({\underline{\phi }}(t),{\underline{\psi }}(t))$ are given such that

• (P1) $(0,0)\le ({\underline{\phi }}(t),{\underline{\psi }}(t))\le ({\overline{\phi }}(t),{\overline{\psi }}(t))\le (M_2,M_3)$.

• (P2) $\lim _{t\rightarrow -\infty }({\overline{\phi }}(t),{\overline{\psi }}(t))=(0,0),$

  $\lim _{t\rightarrow +\infty }({\underline{\phi }}(t),{\underline{\psi }}(t)) =\lim _{t\rightarrow +\infty }({\overline{\phi }}(t),{\overline{\psi }}(t))=(u^+,v^+)$.

• (P3) ${\overline{\phi }}'(t+)\le {\overline{\phi }}'(t-)$ and ${\underline{\phi }}'(t+)\ge {\underline{\phi }}'(t-)$ for $t\in {\mathbb {R}}$.

Define a profile set $\varGamma$ as follows:

$$\begin{aligned} \varGamma =\left\{ (\phi ,\psi )\in C({\mathbb {R}},{\mathbb {R}}^2)| ({\underline{\phi }}(t),{\underline{\psi }}(t))\le (\phi (t),\psi (t))\le ({\overline{\phi }}(t),{\overline{\psi }}(t))\; \text{ for }\; t\in {\mathbb {R}}\right\} . \end{aligned}$$

We have the following results. And the proofs can be found in [19], and we omit it here.

Lemma 4

$F(\varGamma )\subset \varGamma$.

Lemma 5

$F:\varGamma \rightarrow \varGamma$ is compact with respect to the decay norm $|\cdot |_\mu$.

Theorem 5

Suppose that there are a pair of upper–lower solutions ${\overline{\rho }}(t)=({\overline{\phi }}(t),{\overline{\psi }}(t))$ and ${\underline{\rho }}(t)=({\underline{\phi }}(t),{\underline{\psi }}(t))$ for (15) satisfying (P1)-(P3). Then, system (15) has a traveling wave solution.

4 Construction of Upper–Lower Solutions

Firstly, we have the following lemma.

Lemma 6

Assume that

$$\begin{aligned} a_{22}\ge \frac{(3+2\sqrt{2})a_{23}(u^+)^{p-2}M_3}{1+m(u^{+})^p},\quad bv^+\ge \frac{4a_{32}(u^{+})^p}{1+m(u^{+})^p} \end{aligned}$$

(19)

hold. Then, there exist $\varepsilon _1\in \left( 0,(\sqrt{2}-1)u^+\right)$ and $\varepsilon _2\in \left( 0,\nicefrac {v^+}{2}\right)$ such that

$$\begin{aligned} {\left\{ \begin{array}{ll} (2\sqrt{2}-2)a_{22}u^+\varepsilon _1-a_{22}\varepsilon _1^2+\frac{a_{23}(u^{+})^pv^+}{1+m(u^{+})^p} -\frac{a_{23}(u^{+}-\varepsilon _1)^pM_3}{1+m(u^{+}-\varepsilon _1)^p}>\varepsilon _0,\\ bv^+\varepsilon _2-b\varepsilon _2^2-\frac{a_{32}(u^{+})^pv^+}{1+m(u^{+})^p}>\varepsilon _0, \end{array}\right. } \end{aligned}$$

(20)

where $\varepsilon _0>0$ is a constant.

Proof

Let

$$\begin{aligned} g_1(\varepsilon _1)= & {} (2\sqrt{2}-2)a_{22}u^+\varepsilon _1-a_{22}\varepsilon _1^2,\\ g_2(\varepsilon _1)= & {} -\frac{a_{23}(u^{+})^pv^+}{1+m(u^{+})^p} +\frac{a_{23}(u^{+}-\varepsilon _1)^pM_3}{1+m(u^{+}-\varepsilon _1)^p},\\ g_3(\varepsilon _2)= & {} bv^+\varepsilon _2-b\varepsilon _2^2. \end{aligned}$$

Obviously,

$$\begin{aligned} g_1(0)= & {} g_1\left( (2\sqrt{2}-2)u^+\right) =0,\\ \max \{g_1(\varepsilon _1)\}= & {} g_1\left( (\sqrt{2}-1)u^+\right) =(3-2\sqrt{2})a_{22}(u^{+})^2,\\ g_2(\varepsilon _1)\le & {} \frac{a_{23}(u^{+})^pM_3}{1+m(u^{+})^p},\\ \max \{g_3(\varepsilon _2)\}= & {} g_3\left( \frac{v^+}{2}\right) =\frac{1}{4}b(v^{+})^2. \end{aligned}$$

If (19) hold, then there exist $\varepsilon _i^*(i=1,2)$ such that

$$\begin{aligned} 0<\varepsilon _1^*<(\sqrt{2}-1)u^+,\quad 0<\varepsilon _2^*<\frac{v^+}{2}, \end{aligned}$$

and

$$\begin{aligned} g_1(\varepsilon _1)\ge g_2(\varepsilon _1)\quad \text{ for }\;\varepsilon _1^*\le \varepsilon _1<(\sqrt{2}-1)u^+,\\g_3(\varepsilon _2)\ge \frac{a_{32}(u^{+})^pv^+}{1+m(u^{+})^p} \quad \text{ for }\;\varepsilon _2^*\le \varepsilon _2<\frac{v^+}{2}. \end{aligned}$$

$\square$

The following assumption will be imposed throughout this section.

$$\begin{aligned} {\left\{ \begin{array}{ll} \vartheta _1>\frac{(4+2\sqrt{2})a_{23}(u^+)^{p-1}M_3}{1+m(u^{+})^p},\\ \vartheta _2>\max \left\{ 2^{p+1}-p-3,4(\sqrt{2})^p-1\right\} \times \frac{a_{32}(u^{+})^p}{1+m(u^{+})^p}. \end{array}\right. } \end{aligned}$$

(21)

Remark 1

If $\vartheta _1>\frac{(4+2\sqrt{2})a_{23}(u^+)^{p-1}M_3}{1+m(u^{+})^p}$, then we know that

$$\begin{aligned} a_{22}u^+= & {} \vartheta _1-\frac{a_{23}(u^+)^{p-1}v^+}{1+m(u^{+})^p}>\frac{(3+2\sqrt{2})a_{23}(u^+)^{p-1}M_3}{1+m(u^{+})^p},\\ a_{22}u^+= & {} \vartheta _1-\frac{a_{23}(u^+)^{p-1}v^+}{1+m(u^{+})^p}>\vartheta _1-\frac{\vartheta _1}{4+2\sqrt{2}}=\frac{(2+\sqrt{2})\vartheta _1}{4}. \end{aligned}$$

On the other hand, since $a_{22}\ge \frac{(3+2\sqrt{2})a_{23}(u^+)^{p-2}M_3}{1+m(u^{+})^p}$, we have

$$\begin{aligned} a_{22}>\frac{a_{23}(u^+)^{p-2}v^+}{1+m(u^{+})^p}>\frac{a_{23}\left( m(u^{+})^p+1-p\right) (u^+)^{p-2}v^+}{\left( 1+m(u^{+})^p\right) ^2}. \end{aligned}$$

Thus, by Theorem 4, we know that $(u^+,v^+)$ is locally asymptotic stable. Since

$$\begin{aligned} \vartheta _2>\max \left\{ 2^{p+1}-p-3,4(\sqrt{2})^p-1\right\} \times \frac{a_{32}(u^{+})^p}{1+m(u^{+})^p}, \end{aligned}$$

we have that

$$\begin{aligned}bv^+=\vartheta _2+\frac{a_{32}(u^{+})^p}{1+m(u^{+})^p}>\max \left\{ 2^{p+1}-p-2,4(\sqrt{2})^p\right\} \times \frac{a_{32}(u^{+})^p}{1+m(u^{+})^p}.\end{aligned}$$

That is, (21) implies (19).

It is not difficult to verify

$$\begin{aligned} \int _{-\infty }^{+\infty }G(\tau ,y)e^{-\lambda (y+c\tau )}\hbox {d}y =\int _{-\infty }^{+\infty }\frac{1}{\sqrt{4\pi D_1\tau }}e^{-\frac{y^{2}}{4D_1\tau }}e^{-\lambda (y+c\tau )}\hbox {d}y =e^{(D_1\lambda ^2-c\lambda )\tau }.\nonumber \\ \end{aligned}$$

(22)

In order to construct upper–lower solutions for (15), we consider the following functions

$$\begin{aligned} {\left\{ \begin{array}{ll} \varDelta _1(\lambda ,c):=D_2\lambda ^2-c\lambda +ae^{-d_1\tau }e^{(D_1\lambda ^2-c\lambda )\tau }-d_2-q_2e_2,\\ \varDelta _2(\lambda ,c):=D_3\lambda ^2-c\lambda +a_1-q_3e_3. \end{array}\right. } \end{aligned}$$

(23)

By a direct calculation, we have

$$\begin{aligned} \varDelta _1(\lambda ,0)= & {} D_2\lambda ^2+ae^{-d_1\tau }e^{D_1\lambda ^2\tau }-d_2-q_2e_2>ae^{-d_1\tau }-d_2\\&-q_2e_2>0\quad \text{ for }\;\lambda \ge 0,\\ \frac{\partial \varDelta _1(\lambda ,c)}{\partial c}= & {} -\lambda -\lambda \tau ae^{-d_1\tau }e^{(D_1\lambda ^2-c\lambda )\tau }\le 0\quad \text{ for } \text{ any } \text{ given }\;\lambda>0,\\ \varDelta _1(0,c)= & {} ae^{-d_1\tau }-d_2-q_2e_2>0,\\ \varDelta _1(+\infty ,c)= & {} +\infty \quad \text{ for } \text{ all }\;c>0,\\ \frac{\partial ^2\varDelta _1(\lambda ,c)}{\partial \lambda ^2}= & {} 2D_2+2D_1\tau ae^{-d_1\tau }e^{(D_1\lambda ^2-c\lambda )\tau }\\&+(2D_1\lambda -c)^2\tau ^2ae^{-d_1\tau }e^{(D_1\lambda ^2-c\lambda )\tau }>0. \end{aligned}$$

Let

$$\begin{aligned} c_1=\inf _{\lambda >0}\frac{D_2\lambda ^2+ae^{-d_1\tau }e^{(D_1\lambda ^2-c\lambda )\tau }-d_2-q_2e_2}{\lambda } \end{aligned}$$

and

$$\begin{aligned} c_2=2\sqrt{D_3(a_1-q_3e_3)}. \end{aligned}$$

From the above observation, we conclude that $\varDelta _1(\lambda ,c)=0$ has two distinct positive roots $\lambda _1(c)$ and $\lambda _2(c)$ for $c>c_1$. And $\varDelta _2(\lambda ,c)=0$ has two distinct positive roots $\lambda _3(c)$ and $\lambda _4(c)$ for $c>c_2$.

Let $c>c^*:=\max \{c_1,c_2\}$ be fixed, $q>1, \lambda _i=\lambda _i(c) (i=1,2,3,4)$,

$$\begin{aligned} \eta \in \left( 1,\min \left\{ 2,\frac{\lambda _2}{\lambda _1},\frac{\lambda _4}{\lambda _3}, \frac{p\lambda _1+\lambda _3}{\lambda _1},\frac{p\lambda _1+\lambda _3}{\lambda _3}\right\} \right) . \end{aligned}$$

(24)

Denote

$$\begin{aligned} L_1(t):=e^{\lambda _1t}-qe^{\eta \lambda _1t},\quad L_2(t):=e^{\lambda _3t}-qe^{\eta \lambda _3t}. \end{aligned}$$

Then, $L_1({\bar{t}}_3)=0, L_2({\bar{t}}_4)=0$ with ${\bar{t}}_3=-\frac{\ln q}{(\eta -1)\lambda _1}, {\bar{t}}_4=-\frac{\ln q}{(\eta -1)\lambda _3}$. Let $\varepsilon _1,\varepsilon _2$ be defined as in Lemma 6. For sufficiently small $\lambda >0$, we choose $q>1$ large enough such that

$$\begin{aligned} u^+-\varepsilon _1e^{-\lambda t_3}=L_1(t_3)=e^{\lambda _1t_3}-qe^{\eta \lambda _1t_3},\quad v^+-\varepsilon _2e^{-\lambda t_4}=L_2(t_4)=e^{\lambda _3t_4}-qe^{\eta \lambda _3t_4}, \end{aligned}$$

respectively. Here,

$$\begin{aligned} t_3\le {\bar{t}}_3<0,\quad t_4\le {\bar{t}}_4 <0, \end{aligned}$$

and $|t_i|(i=3,4)$ are large enough. Now we define the following functions

$$\begin{aligned} {\overline{\phi }}(t)= & {} {\left\{ \begin{array}{ll} e^{\lambda _1t},&{}t\le t_1,\\ \min \{M_2,u^++u^+e^{-\lambda t}\},&{}t>t_1, \end{array}\right. }\\ {\overline{\psi }}(t)= & {} {\left\{ \begin{array}{ll} e^{\lambda _3t}+qe^{\eta \lambda _3t},&{}t\le t_2,\\ \min \{M_3,v^++v^+e^{-\lambda t}\},&{}t>t_2, \end{array}\right. }\\ {\underline{\phi }}(t)= & {} {\left\{ \begin{array}{ll} e^{\lambda _1t}-qe^{\eta \lambda _1t},&{}t\le t_3,\\ u^+-\varepsilon _1e^{-\lambda t},&{}t>t_3, \end{array}\right. }\quad {\underline{\psi }}(t)= {\left\{ \begin{array}{ll} e^{\lambda _3t}-qe^{\eta \lambda _3t},&{}t\le t_4,\\ v^+-\varepsilon _2e^{-\lambda t},&{}t>t_4. \end{array}\right. } \end{aligned}$$

We can verify that $({\overline{\phi }}(t),{\overline{\psi }}(t))$ and $({\underline{\phi }}(t),{\underline{\psi }}(t))$ satisfy (P1)–(P3). Furthermore, if $q>1$ is large enough, then it is easy to know that

$$\begin{aligned} t_1\ge \max \{t_2,t_3,t_4\}. \end{aligned}$$

Lemma 7

Assume that (21) holds and $q>1$ is large enough. Then $({\overline{\phi }}(t),{\overline{\psi }}(t))$ is an upper solution of (15).

Proof

For $({\overline{\phi }}(t),{\overline{\psi }}(t))\in C({\mathbb {R}},{\mathbb {R}}^2)$, if $t\le t_1$, then ${\overline{\phi }}(t)=e^{\lambda _1t}$. If $t-y-c\tau \le t_1$, then ${\overline{\phi }}(t-y-c\tau )=e^{\lambda _1(t-y-c\tau )}$; if $t-y-c\tau >t_1$, then ${\overline{\phi }}(t-y-c\tau )=\min \{M_2,u^++u^+e^{-\lambda (t-y-c\tau )}\}\le e^{\lambda _1(t-y-c\tau )}$. Thus,

$$\begin{aligned}\begin{aligned}&D_2{\overline{\phi }}''(t)-c{\overline{\phi }}'(t) +ae^{-d_1\tau }\int _{-\infty }^{+\infty }\frac{1}{\sqrt{4\pi D_1\tau }}e^{-\frac{y^{2}}{4D_1\tau }}{\overline{\phi }}(t-y-c\tau )\hbox {d}y -(d_2+q_2e_2){\overline{\phi }}(t)\\&\quad -\,a_{22}{\overline{\phi }}^2(t) -\frac{a_{23}{\overline{\phi }}^p(t){\underline{\psi }}(t)}{1+m{\overline{\phi }}^p(t)}\\&\le D_2\lambda _1^2e^{\lambda _1t}-c\lambda _1e^{\lambda _1t} +ae^{-d_1\tau }\int _{-\infty }^{+\infty }\frac{1}{\sqrt{4\pi D_1\tau }}e^{-\frac{y^{2}}{4D_1\tau }}e^{\lambda _1(t-y-c\tau )}\hbox {d}y -(d_2+q_2e_2)e^{\lambda _1t}\\&= e^{\lambda _1t}[D_2\lambda _1^2-c\lambda _1+ae^{-d_1\tau }e^{(D_1\lambda _1^2-c\lambda _1)\tau }-d_2-q_2e_2]\\&=e^{\lambda _1t}\varDelta _1(\lambda _1,c)=0. \end{aligned} \end{aligned}$$

When $t>t_1$, if ${\overline{\phi }}(t)=M_2$, note that ${\overline{\phi }}(t-y-c\tau )\le M_2$. We have from the definition of $M_2$ that

$$\begin{aligned} \begin{aligned}&D_2{\overline{\phi }}''(t)-c{\overline{\phi }}'(t) +ae^{-d_1\tau }\int _{-\infty }^{+\infty }\frac{1}{\sqrt{4\pi D_1\tau }}e^{-\frac{y^{2}}{4D_1\tau }}{\overline{\phi }}(t-y-c\tau )\hbox {d}y -(d_2+q_2e_2){\overline{\phi }}(t)\\&\quad -\,a_{22}{\overline{\phi }}^2(t) -\frac{a_{23}{\overline{\phi }}^p(t){\underline{\psi }}(t)}{1+m{\overline{\phi }}^p(t)}\\&\le ae^{-d_1\tau }M_2-(d_2+q_2e_2)M_2-a_{22}M_2^2\le 0. \end{aligned} \end{aligned}$$

Otherwise, ${\overline{\phi }}(t)=u^++u^+e^{-\lambda t}, {\underline{\psi }}(t)=v^+-\varepsilon _2e^{-\lambda t}$. If $t-y-c\tau \le t_1$, then ${\overline{\phi }}(t-y-c\tau )=e^{\lambda _1(t-y-c\tau )}\le u^++u^+e^{-\lambda (t-y-c\tau )}$; if $t-y-c\tau >t_1$, then ${\overline{\phi }}(t-y-c\tau )=u^++u^+e^{-\lambda (t-y-c\tau )}$. Therefore, we have

$$\begin{aligned}&\begin{aligned}&D_2{\overline{\phi }}''(t)-c{\overline{\phi }}'(t) +ae^{-d_1\tau }\int _{-\infty }^{+\infty }\frac{1}{\sqrt{4\pi D_1\tau }}e^{-\frac{y^{2}}{4D_1\tau }}{\overline{\phi }}(t-y-c\tau )\hbox {d}y\\&\quad -(d_2+q_2e_2){\overline{\phi }}(t)\\&\quad -\,a_{22}{\overline{\phi }}^2(t) -\frac{a_{23}{\overline{\phi }}^p(t){\underline{\psi }}(t)}{1+m{\overline{\phi }}^p(t)}\\&\le D_2\lambda ^2u^+e^{-\lambda t}+c\lambda u^+e^{-\lambda t}\\&\quad +ae^{-d_1\tau }\int _{-\infty }^{+\infty }\frac{1}{\sqrt{4\pi D_1\tau }}e^{-\frac{y^{2}}{4D_1\tau }}\left( u^++u^+e^{-\lambda (t-y-c\tau )}\right) \hbox {d}y\\&\quad -\,(d_2+q_2e_2)\left( u^++u^+e^{-\lambda t}\right) -a_{22}\left( u^++u^+e^{-\lambda t}\right) ^2\\&\quad -\,\frac{a_{23}\left( u^++u^+e^{-\lambda t}\right) ^p(v^+-\varepsilon _2e^{-\lambda t})}{1+m\left( u^++u^+e^{-\lambda t}\right) ^p}\end{aligned}\\&\begin{aligned}&=D_2\lambda ^2u^+e^{-\lambda t}+c\lambda u^+e^{-\lambda t}+ae^{-d_1\tau }u^++ae^{-d_1\tau }u^+e^{-\lambda t}e^{(D_1\lambda ^2+c\lambda )\tau }\\&\quad -\,(d_2+q_2e_2)u^+ -(d_2+q_2e_2)u^+e^{-\lambda t}-a_{22}(u^{+})^2-2a_{22}(u^{+})^2e^{-\lambda t}\\&\quad -\,a_{22}(u^{+})^2e^{-2\lambda t} -\frac{a_{23}\left( u^++u^+e^{-\lambda t}\right) ^p(v^+-\varepsilon _2e^{-\lambda t})}{1+m\left( u^++u^+e^{-\lambda t}\right) ^p}\\&=u^+e^{-\lambda t} \varDelta _1(-\lambda ,c)-2a_{22}(u^{+})^2e^{-\lambda t}-a_{22}(u^{+})^2e^{-2\lambda t}+\frac{a_{23}(u^{+})^pv^+}{1+m(u^{+})^p}\\&\quad -\,\frac{a_{23}\left( u^++u^+e^{-\lambda t}\right) ^p(v^+-\varepsilon _2e^{-\lambda t})}{1+m\left( u^++u^+e^{-\lambda t}\right) ^p}\\&\le u^+e^{-\lambda t}\left[ \varDelta _1(-\lambda ,c)-\frac{3}{2}a_{22}u^{+}\right] -u^+\left[ \frac{1}{2}a_{22}u^{+}e^{-\lambda t}-\frac{a_{23}(u^{+})^{p-1}v^+}{1+m(u^{+})^p}\right. \\&\quad \left. +\frac{a_{23}(u^+)^{p-1}\left( 1+e^{-\lambda t}\right) ^p(v^+-\varepsilon _2e^{-\lambda t})}{1+m(u^{+})^p\left( 1+e^{-\lambda t}\right) ^p}\right] . \end{aligned} \end{aligned}$$

Since $a_{22}\ge \frac{(3+2\sqrt{2})a_{23}(u^+)^{p-2}M_3}{1+m(u^{+})^p}$, we have

$$\begin{aligned} \begin{aligned}&\varDelta _1(0,c)-\frac{3}{2}a_{22}u^{+} =\vartheta _1-\frac{3}{2}a_{22}u^{+} =\frac{a_{23}(u^{+})^{p-1}v^+}{1+m(u^{+})^p}-\frac{1}{2}a_{22}u^{+}\\&\le \frac{a_{23}(u^{+})^{p-1}v^+}{1+m(u^{+})^p}-\frac{(3+2\sqrt{2})a_{23}(u^+)^{p-1}M_3}{2(1+m(u^{+})^p)}<0, \end{aligned} \end{aligned}$$

and there exists $\lambda _1^*$ such that $\varDelta _1(-\lambda ,c)-\frac{3}{2}a_{22}u^{+}<0$ for $\lambda \in (0,\lambda _1^*)$. On the other hand, using $\varepsilon _2\in \left( 0,\nicefrac {v^+}{2}\right)$, we obtain

$$\begin{aligned} \begin{aligned} I_1(\lambda ,t)&:=\frac{1}{2}a_{22}u^{+}e^{-\lambda t}-\frac{a_{23}(u^{+})^{p-1}v^+}{1+m(u^{+})^p} +\frac{a_{23}(u^+)^{p-1}\left( 1+e^{-\lambda t}\right) ^p(v^+-\varepsilon _2e^{-\lambda t})}{1+m(u^{+})^p\left( 1+e^{-\lambda t}\right) ^p}\\&\ge \frac{1}{2}a_{22}u^{+}e^{-\lambda t}-\frac{a_{23}(u^{+})^{p-1}v^+}{1+m(u^{+})^p} +\frac{a_{23}(u^+)^{p-1}(v^+-\varepsilon _2e^{-\lambda t})}{1+m(u^{+})^p}\\&=\frac{1}{2}a_{22}u^{+}e^{-\lambda t}-\frac{a_{23}(u^+)^{p-1}\varepsilon _2e^{-\lambda t}}{1+m(u^{+})^p}\\&=u^{+}e^{-\lambda t}\left[ \frac{1}{2}a_{22}-\frac{a_{23}\varepsilon _2(u^+)^{p-2}}{1+m(u^{+})^p}\right] \\&>u^{+}e^{-\lambda t}\left[ \frac{1}{2}a_{22}-\frac{a_{23}(u^+)^{p-2}v^+}{2(1+m(u^{+})^p)}\right] >0. \end{aligned} \end{aligned}$$

Hence, we can get

$$\begin{aligned} \begin{aligned}&D_2{\overline{\phi }}''(t)-c{\overline{\phi }}'(t) +ae^{-d_1\tau }\int _{-\infty }^{+\infty }\frac{1}{\sqrt{4\pi D_1\tau }}e^{-\frac{y^{2}}{4D_1\tau }}{\overline{\phi }}(t-y-c\tau )\hbox {d}y-(d_2+q_2e_2){\overline{\phi }}(t)\\&-a_{22}{\overline{\phi }}^2(t) -\frac{a_{23}{\overline{\phi }}^p(t){\underline{\psi }}(t)}{1+m{\overline{\phi }}^p(t)}\le 0 \end{aligned} \end{aligned}$$

for any $\lambda \in (0,\lambda _1^*)$.

If $t\le t_2$, then ${\overline{\psi }}(t)=e^{\lambda _3t}+qe^{\eta \lambda _3t}, {\overline{\phi }}(t)=e^{\lambda _1t}$. We have

$$\begin{aligned} \begin{aligned}&D_3{\overline{\psi }}''(t)-c{\overline{\psi }}'(t)+(a_1-b{\overline{\psi }}(t)){\overline{\psi }}(t)-q_3e_3{\overline{\psi }}(t) +\frac{a_{32}{\overline{\phi }}^p(t){\overline{\psi }}(t)}{1+m{\overline{\phi }}^p(t)}\\&\le D_3(e^{\lambda _3t}+qe^{\eta \lambda _3t})''-c(e^{\lambda _3t}+qe^{\eta \lambda _3t})' +\left( a_1-b(e^{\lambda _3t}+qe^{\eta \lambda _3t})\right) (e^{\lambda _3t}+qe^{\eta \lambda _3t})\\&\quad -\,q_3e_3(e^{\lambda _3t}+qe^{\eta \lambda _3t}) +\frac{a_{32}e^{p\lambda _1t}(e^{\lambda _3t}+qe^{\eta \lambda _3t})}{1+me^{p\lambda _1t}}\\&= qe^{\eta \lambda _3t}\varDelta _2(\eta \lambda _3,c)-b(e^{\lambda _3t}+qe^{\eta \lambda _3t})^2+ \frac{a_{32}e^{p\lambda _1t}(e^{\lambda _3t}+qe^{\eta \lambda _3t})}{1+me^{p\lambda _1t}}\\&\le qe^{\eta \lambda _3t}\varDelta _2(\eta \lambda _3,c) +a_{32}e^{(p\lambda _1+\lambda _3)t}+a_{32}qe^{(p\lambda _1+\eta \lambda _3)t}\\&\le e^{\eta \lambda _3t}\left[ q\varDelta _2(\eta \lambda _3,c) +a_{32}+a_{32}qe^{p\lambda _1t}\right] . \end{aligned} \end{aligned}$$

Note $\varDelta _2(\eta \lambda _3,c)<0$ by (24). Let $q>1$ be large enough, then $-t_2>0$ is also large enough such that

$$\begin{aligned} q\varDelta _2(\eta \lambda _3,c) +a_{32}+a_{32}qe^{p\lambda _1t_2} =q\left[ \varDelta _2(\eta \lambda _3,c)+a_{32}e^{p\lambda _1t_2}\right] +a_{32}<0, \end{aligned}$$

which leads to

$$\begin{aligned} q\varDelta _2(\eta \lambda _3,c)+a_{32}+a_{32}qe^{p\lambda _1t}<0\quad \text{ for }\quad t\le t_2. \end{aligned}$$

When $t>t_2$, if ${\overline{\psi }}(t)=M_3$, note that ${\overline{\phi }}(t)\le M_2$, using $M_3\ge \frac{a_1-q_3e_3+a_{32}M_2^p}{b}$, we have

$$\begin{aligned} \begin{aligned}&D_3{\overline{\psi }}''(t)-c{\overline{\psi }}'(t)+(a_1-b{\overline{\psi }}(t)){\overline{\psi }}(t)-q_3e_3{\overline{\psi }}(t) +\frac{a_{32}{\overline{\phi }}^p(t){\overline{\psi }}(t)}{1+m{\overline{\phi }}^p(t)}\\&\quad =(a_1-bM_3)M_3-q_3e_3M_3+\frac{a_{32}{\overline{\phi }}^p(t)M_3}{1+m{\overline{\phi }}^p(t)}\\&\quad \le M_3(a_1-bM_3-q_3e_3+a_{32}M_2^p)\le 0. \end{aligned} \end{aligned}$$

Otherwise, ${\overline{\psi }}(t)=v^++v^+e^{-\lambda t}, {\overline{\phi }}(t)\le u^++u^+e^{-\lambda t}$, then

$$\begin{aligned}&\begin{aligned}&D_3{\overline{\psi }}''(t)-c{\overline{\psi }}'(t)+(a_1-b{\overline{\psi }}(t)){\overline{\psi }}(t)-q_3e_3{\overline{\psi }}(t) +\frac{a_{32}{\overline{\phi }}^p(t){\overline{\psi }}(t)}{1+m{\overline{\phi }}^p(t)}\\&\quad \le D_3\lambda ^2v^+e^{-\lambda t}+c\lambda v^+ e^{-\lambda t}+\left( a_1-b(v^++v^+e^{-\lambda t})\right) (v^++v^+e^{-\lambda t})\\&\qquad -\,q_3e_3(v^++v^+e^{-\lambda t}) +\frac{a_{32}(u^++u^+e^{-\lambda t})^p(v^++v^+e^{-\lambda t})}{1+m(u^++u^+e^{-\lambda t})^p}\\&\quad = D_3\lambda ^2v^+e^{-\lambda t}+c\lambda v^+ e^{-\lambda t}+a_1v^+e^{-\lambda t}-2b(v^{+})^2 e^{-\lambda t}-b(v^{+})^2e^{-2\lambda t}\\&\qquad -\,q_3e_3v^+e^{-\lambda t} +\frac{a_{32}(u^{+})^pv^+(1+e^{-\lambda t})^{p+1}}{1+m(u^{+})^p(1+e^{-\lambda t})^p}-\frac{a_{32}(u^{+})^pv^+}{1+m(u^{+})^p}\end{aligned}\\&\begin{aligned}&\le v^+e^{-\lambda t}[\varDelta _2(-\lambda ,c)-bv^+] -v^+\left[ bv^{+} e^{-\lambda t}+bv^{+} e^{-2\lambda t}+\frac{a_{32}(u^{+})^p}{1+m(u^{+})^p}\right. \\&\qquad \left. -\frac{a_{32}(u^{+})^p(1+e^{-\lambda t})^{p+1}}{1+m(u^{+})^p}\right] . \end{aligned} \end{aligned}$$

Note that $\varDelta _2(0,c)-bv^+=\vartheta _2-bv^+=-\frac{a_{32}(u^{+})^p}{1+m(u^{+})^p}<0$; thus, there exists $\lambda _2^*$ such that $\varDelta _2(-\lambda ,c)-bv^+<0$ for $\lambda \in (0,\lambda _2^*)$. On the other hand, let

$$\begin{aligned} I_2(\lambda ,t) :=bv^{+} e^{-\lambda t}+bv^{+} e^{-2\lambda t}+\frac{a_{32}(u^{+})^p}{1+m(u^{+})^p}-\frac{a_{32}(u^{+})^p(1+e^{-\lambda t})^{p+1}}{1+m(u^{+})^p}. \end{aligned}$$

Note that for $p\in {\mathbb {N}}, p\ge 2$, we have $2^{p+1}-p-2>2^p-\nicefrac {1}{2}$. By (21) and Remark 1, we have

$$\begin{aligned} I_2(\lambda ,0)=2bv^{+}-\frac{(2^{p+1}-1)a_{32}(u^{+})^p}{1+m(u^{+})^p}>0. \end{aligned}$$

Thus, we can choose $\delta _0>0$ such that

$$\begin{aligned} bv^{+} \delta +bv^{+} \delta ^2+\frac{a_{32}(u^{+})^p}{1+m(u^{+})^p}-\frac{a_{32}(u^{+})^p(1+\delta )^{p+1}}{1+m(u^{+})^p}>0 \end{aligned}$$

for $\delta \in [1,1+\delta _0]$. Let $\delta ^*=1+\delta _0$.

If $t\in (t_2,0]$, noting $e^{-\lambda t}$ is decreasing on $(t_2,0]$, we can choose $\lambda _3^*>0$ small enough such that $e^{-\lambda _3^*t_2}=1+\delta _0=\delta ^*$. Thus, we have $1\le e^{-\lambda t}<\delta ^*$ for $t\in (t_2,0]$ and $\lambda \in (0,\lambda _3^*)$. Therefore, $I_2(\lambda ,t)>0$ for $t\in (t_2,0]$.

If $t>0$, noting $2^{p+1}-p-2>p+1$, we have

$$\begin{aligned} \begin{aligned}I_2(\lambda ,t)&=bv^{+} e^{-\lambda t}+bv^{+} e^{-2\lambda t}-\frac{a_{32}(u^{+})^p}{1+m(u^{+})^p}\left( (1+e^{-\lambda t})^{p+1}-1\right) \\&=e^{-\lambda t}\left[ bv^{+}+bv^{+} e^{-\lambda t}-\frac{a_{32}(u^{+})^p}{1+m(u^{+})^p}\left( \frac{(1+e^{-\lambda t})^{p+1}-1}{e^{-\lambda t}}\right) \right] \\&>e^{-\lambda t}\left\{ bv^{+}+bv^{+} e^{-\lambda t}-\frac{a_{32}(u^{+})^p}{1+m(u^{+})^p}\right. \\&\quad \left. \times \left[ p+1+\left( \left( {\begin{array}{c}p+1\\ 2\end{array}}\right) +\left( {\begin{array}{c}p+1\\ 3\end{array}}\right) +\cdots +\left( {\begin{array}{c}p+1\\ p+1\end{array}}\right) \right) e^{-\lambda t}\right] \right\} \\&=e^{-\lambda t}\left\{ bv^{+}-\frac{(p+1)a_{32}(u^{+})^p}{1+m(u^{+})^p} +\left[ bv^{+}-\frac{(2^{p+1}-p-2)a_{32}(u^{+})^p}{1+m(u^{+})^p}\right] e^{-\lambda t} \right\} \\&>0.\end{aligned} \end{aligned}$$

Summarizing the above discussions, we have

$$\begin{aligned} \begin{aligned} D_3{\overline{\psi }}''(t)-c{\overline{\psi }}'(t)+(a_1-b{\overline{\psi }}(t)){\overline{\psi }}(t)-q_3e_3{\overline{\psi }}(t) +\frac{a_{32}{\overline{\phi }}^2(t){\overline{\psi }}(t)}{1+m{\overline{\phi }}^2(t)}\le 0 \end{aligned} \end{aligned}$$

for any $\lambda \in (0,\min \{\lambda _2^*,\lambda _3^*\})$. Thus, $({\overline{\phi }}(t),{\overline{\psi }}(t))$ is an upper solution of (15). $\square$

Lemma 8

Assume that (21) holds and $q>1$ is large enough. Then, $({\underline{\phi }}(t),{\underline{\psi }}(t))$ is a lower solution of (15).

Proof

For $({\underline{\phi }}(t),{\underline{\psi }}(t))\in C({\mathbb {R}},{\mathbb {R}}^2)$, if $t\le t_3$, then ${\underline{\phi }}(t)=e^{\lambda _1t}-qe^{\eta \lambda _1t}, {\overline{\psi }}(t)\le e^{\lambda _3t}+qe^{\eta \lambda _3t}$. If $t-y-c\tau \le t_3$, then ${\underline{\phi }}(t-y-c\tau )=e^{\lambda _1(t-y-c\tau )}-qe^{\eta \lambda _1(t-y-c\tau )}$; if $t-y-c\tau >t_3$, then ${\underline{\phi }}(t-y-c\tau )=u^+-\varepsilon _1e^{-\lambda (t-y-c\tau )}\ge e^{\lambda _1(t-y-c\tau )}-qe^{\eta \lambda _1(t-y-c\tau )}$. Therefore, we have

$$\begin{aligned} \begin{aligned}&D_2{\underline{\phi }}''(t)-c{\underline{\phi }}'(t) +ae^{-d_1\tau }\int _{-\infty }^{+\infty }\frac{1}{\sqrt{4\pi D_1\tau }}e^{-\frac{y^{2}}{4D_1\tau }}{\underline{\phi }}(t-y-c\tau )\hbox {d}y\\&\qquad -(d_2+q_2e_2){\underline{\phi }}(t)\\&\qquad -\,a_{22}{\underline{\phi }}^2(t) -\frac{a_{23}{\underline{\phi }}^p(t){\overline{\psi }}(t)}{1+m{\underline{\phi }}^p(t)}\\&\quad \ge D_2\left( e^{\lambda _1t}-qe^{\eta \lambda _1t}\right) ''-c\left( e^{\lambda _1t}-qe^{\eta \lambda _1t}\right) '\\&\qquad +ae^{-d_1\tau }\int _{-\infty }^{+\infty }\frac{1}{\sqrt{4\pi D_1\tau }}e^{-\frac{y^{2}}{4D_1\tau }}\left( e^{\lambda _1(t-y-c\tau )}-qe^{\eta \lambda _1(t-y-c\tau )}\right) \hbox {d}y\\&\qquad -\,(d_2+q_2e_2)\left( e^{\lambda _1t}-qe^{\eta \lambda _1t}\right) -a_{22}\left( e^{\lambda _1t}-qe^{\eta \lambda _1t}\right) ^2\\&\qquad -\,\frac{a_{23}\left( e^{\lambda _1t}-qe^{\eta \lambda _1t}\right) ^p(e^{\lambda _3t}+qe^{\eta \lambda _3t})}{1+m\left( e^{\lambda _1t}-qe^{\eta \lambda _1t}\right) ^p}\\&\quad =-qe^{\eta \lambda _1t}\varDelta _1(\eta \lambda _1,c)-a_{22}\left( e^{\lambda _1t}-qe^{\eta \lambda _1t}\right) ^2\\&\qquad -\frac{a_{23}\left( e^{\lambda _1t}-qe^{\eta \lambda _1t}\right) ^p(e^{\lambda _3t}+qe^{\eta \lambda _3t})}{1+m\left( e^{\lambda _1t}-qe^{\eta \lambda _1t}\right) ^p}\\&\quad \ge -qe^{\eta \lambda _1t}\varDelta _1(\eta \lambda _1,c)-a_{22}e^{2\lambda _1 t}-a_{23}e^{p\lambda _1t}(e^{\lambda _3t}+qe^{\eta \lambda _3t})\\&\quad \ge -e^{\eta \lambda _1t}\left[ q\varDelta _1(\eta \lambda _1,c)+a_{22}+a_{23}+a_{23}qe^{(p\lambda _1+\eta \lambda _3-\eta \lambda _1)t}\right] . \end{aligned} \end{aligned}$$

Note from (24) that $\varDelta _1(\eta \lambda _1,c)<0$ and $p\lambda _1+\eta \lambda _3-\eta \lambda _1>0$. Let $q>1$ be large enough, then $-t_3>0$ is also large enough such that

$$\begin{aligned} \begin{aligned}&q\varDelta _1(\eta \lambda _1,c)+a_{22}+a_{23}+a_{23}qe^{(p\lambda _1+\eta \lambda _3-\eta \lambda _1)t_3}\\&\quad =q\left[ \varDelta _1(\eta \lambda _1,c)+a_{23}e^{(p\lambda _1+\eta \lambda _3-\eta \lambda _1)t_3}\right] +a_{22}+a_{23}<0, \end{aligned} \end{aligned}$$

which leads to

$$\begin{aligned} q\varDelta _1(\eta \lambda _1,c)+a_{22}+a_{23}+a_{23}qe^{(p\lambda _1+\eta \lambda _3-\eta \lambda _1)t} <0\quad \text{ for }\,t\le t_3. \end{aligned}$$

Therefore,

$$\begin{aligned} -e^{\eta \lambda _1t}\left[ q\varDelta _1(\eta \lambda _1,c)+a_{22}+a_{23}+a_{23}qe^{(p\lambda _1+\eta \lambda _3-\eta \lambda _1)t}\right] \ge 0. \end{aligned}$$

If $t>t_3$, then ${\underline{\phi }}(t)=u^+-\varepsilon _1e^{-\lambda t}, {\overline{\psi }}(t)\le v^++v^+e^{-\lambda t}$. If $t-y-c\tau \le t_3$, then ${\underline{\phi }}(t-y-c\tau )=e^{\lambda _1(t-y-c\tau )}-qe^{\eta \lambda _1(t-y-c\tau )}\ge u^+-\varepsilon _1e^{-\lambda (t-y-c\tau )}$; if $t-y-c\tau >t_3$, then ${\underline{\phi }}(t-y-c\tau )=u^+-\varepsilon _1e^{-\lambda (t-y-c\tau )}$. Thus,

$$\begin{aligned} \begin{aligned}&D_2{\underline{\phi }}''(t)-c{\underline{\phi }}'(t) +ae^{-d_1\tau }\int _{-\infty }^{+\infty }\frac{1}{\sqrt{4\pi D_1\tau }}e^{-\frac{y^{2}}{4D_1\tau }}{\underline{\phi }}(t-y-c\tau )\hbox {d}y\\&\qquad -(d_2+q_2e_2){\underline{\phi }}(t)\\&\qquad -\,a_{22}{\underline{\phi }}^2(t) -\frac{a_{23}{\underline{\phi }}^p(t){\overline{\psi }}(t)}{1+m{\underline{\phi }}^p(t)}\\&\quad \ge -D_2\lambda ^2\varepsilon _1e^{-\lambda t}-c\lambda \varepsilon _1e^{-\lambda t}\\&\qquad +ae^{-d_1\tau } \int _{-\infty }^{+\infty }\frac{1}{\sqrt{4\pi D_1\tau }}e^{-\frac{y^{2}}{4D_1\tau }}\left( u^+-\varepsilon _1e^{-\lambda (t-y-c\tau )}\right) \hbox {d}y\\&\qquad -\,(d_2+q_2e_2)\left( u^+-\varepsilon _1e^{-\lambda t}\right) -a_{22}\left( u^+-\varepsilon _1e^{-\lambda t}\right) ^2 -\frac{a_{23}\left( u^+-\varepsilon _1e^{-\lambda t}\right) ^p{\overline{\psi }}(t)}{1+m\left( u^+-\varepsilon _1e^{-\lambda t}\right) ^p}\\&\quad =-D_2\lambda ^2\varepsilon _1e^{-\lambda t}-c\lambda \varepsilon _1e^{-\lambda t}+ae^{-d_1\tau }\left( u^+-\varepsilon _1e^{-\lambda t}e^{(D_1\lambda ^2+c\lambda )\tau }\right) \\&\qquad -\,(d_2+q_2e_2)\left( u^+-\varepsilon _1e^{-\lambda t}\right) -a_{22}\left( u^+-\varepsilon _1e^{-\lambda t}\right) ^2 -\frac{a_{23}\left( u^+-\varepsilon _1e^{-\lambda t}\right) ^p{\overline{\psi }}(t)}{1+m\left( u^+-\varepsilon _1e^{-\lambda t}\right) ^p}\\&\quad =-\varepsilon _1e^{-\lambda t}\varDelta _1(-\lambda ,c)+2a_{22}u^+\varepsilon _1e^{-\lambda t} -a_{22}\varepsilon _1^2e^{-2\lambda t}+\frac{a_{23}(u^{+})^pv^+}{1+m(u^{+})^p}\\&\qquad -\,\frac{a_{23}\left( u^+-\varepsilon _1e^{-\lambda t}\right) ^p{\overline{\psi }}(t)}{1+m\left( u^+-\varepsilon _1e^{-\lambda t}\right) ^p}\\&\quad =\varepsilon _1e^{-\lambda t}\left[ -\varDelta _1(-\lambda ,c)+(4-2\sqrt{2})a_{22}u^+\right] +(2\sqrt{2}-2)a_{22}u^+\varepsilon _1e^{-\lambda t} -a_{22}\varepsilon _1^2e^{-2\lambda t}\\&\qquad +\frac{a_{23}(u^{+})^pv^+}{1+m(u^{+})^p} -\frac{a_{23}\left( u^+-\varepsilon _1e^{-\lambda t}\right) ^p{\overline{\psi }}(t)}{1+m\left( u^+-\varepsilon _1e^{-\lambda t}\right) ^p}. \end{aligned} \end{aligned}$$

Note that $-\varDelta _1(0,c)+(4-2\sqrt{2})a_{22}u^+=-\vartheta _1+(4-2\sqrt{2})a_{22}u^+>0$ by (21) and Remark 1. We can choose $\lambda _4^*>0$ such that $-\varDelta _1(-\lambda ,c)+(4-2\sqrt{2})a_{22}u^+>0$ for $\lambda \in (0,\lambda _4^*)$. Let

$$\begin{aligned} \begin{aligned}&I_3(\lambda ,t)\\&:=(2\sqrt{2}-2)a_{22}u^+\varepsilon _1e^{-\lambda t} -a_{22}\varepsilon _1^2e^{-2\lambda t}+\frac{a_{23}(u^{+})^pv^+}{1+m(u^{+})^p} -\frac{a_{23}\left( u^+-\varepsilon _1e^{-\lambda t}\right) ^p{\overline{\psi }}(t)}{1+m\left( u^+-\varepsilon _1e^{-\lambda t}\right) ^p}. \end{aligned} \end{aligned}$$

By Lemma 6, we can choose $\delta _1>0$ such that

$$\begin{aligned} (2\sqrt{2}-2)a_{22}u^+\delta -a_{22}\delta ^2+\frac{a_{23}(u^{+})^pv^+}{1+m(u^{+})^p} -\frac{a_{23}\left( u^+-\delta \right) ^pM_3}{1+m\left( u^+-\delta \right) ^p}>\frac{\varepsilon _0}{2}>0 \end{aligned}$$

for $\delta \in [\varepsilon _1,\varepsilon _1+\delta _1]$. Let $\delta ^{**}=\varepsilon _1+\delta _1$.

If $t\in (t_3,0]$, noting that $\varepsilon _1e^{-\lambda t}$ is decreasing on $(t_3,0]$, we can choose $\lambda _5^*>0$ small enough such that $\varepsilon _1e^{-\lambda _5^*t_3}=\varepsilon _1+\delta _1=\delta ^{**}$. Thus, we have $\varepsilon _1\le \varepsilon _1e^{-\lambda t}<\delta ^{**}$ for $t\in (t_3,0]$ and $\lambda \in (0,\lambda _5^*)$. Therefore,

$$\begin{aligned} \begin{aligned}&I_3(\lambda ,t)\\&\quad \ge (2\sqrt{2}-2)a_{22}u^+\varepsilon _1e^{-\lambda t} -a_{22}\varepsilon _1^2e^{-2\lambda t}+\frac{a_{23}(u^{+})^pv^+}{1+m(u^{+})^p}\\&\qquad -\frac{a_{23}\left( u^+-\varepsilon _1e^{-\lambda t}\right) ^pM_3}{1+m\left( u^+-\varepsilon _1e^{-\lambda t}\right) ^p}\\&\quad>\frac{\varepsilon _0}{2}>0. \end{aligned} \end{aligned}$$

(25)

If $t>0$, then we have

$$\begin{aligned} \begin{aligned} I_3(\lambda ,t)&\ge (2\sqrt{2}-2)a_{22}u^+\varepsilon _1e^{-\lambda t} -a_{22}\varepsilon _1^2e^{-2\lambda t}+\frac{a_{23}(u^{+})^pv^+}{1+m(u^{+})^p}\\&\quad -\,\frac{a_{23}\left( u^+-\varepsilon _1e^{-\lambda t}\right) ^p(v^++v^+e^{-\lambda t})}{1+m\left( u^+-\varepsilon _1e^{-\lambda t}\right) ^p}\\&\ge (2\sqrt{2}-2)a_{22}u^+\varepsilon _1e^{-\lambda t} -a_{22}\varepsilon _1^2e^{-\lambda t}+\frac{a_{23}(u^{+})^pv^+}{1+m(u^{+})^p}\\&\quad -\,\frac{a_{23}(u^+)^p(v^++v^+e^{-\lambda t})}{1+m(u^+)^p}\\&=(2\sqrt{2}-2)a_{22}u^+\varepsilon _1e^{-\lambda t} -a_{22}\varepsilon _1^2e^{-\lambda t}-\frac{a_{23}(u^+)^pv^+e^{-\lambda t}}{1+m(u^+)^p}\\&=e^{-\lambda t}\left[ (2\sqrt{2}-2)a_{22}u^+\varepsilon _1 -a_{22}\varepsilon _1^2-\frac{a_{23}(u^+)^pv^+}{1+m(u^+)^p}\right] . \end{aligned} \end{aligned}$$

Since $\max \left\{ (2\sqrt{2}-2)a_{22}u^+\varepsilon _1 -a_{22}\varepsilon _1^2\right\} =(3-2\sqrt{2})a_{22}(u^{+})^2$ and

$$\begin{aligned} a_{22}\ge \frac{(3+2\sqrt{2})a_{23}(u^+)^{p-2}M_3}{1+m(u^{+})^p}, \end{aligned}$$

we know that there exists $\varepsilon _1^{**}(0<\varepsilon _1^{**}<(\sqrt{2}-1)u^+)$ such that

$$\begin{aligned} (2\sqrt{2}-2)a_{22}u^+\varepsilon _1 -a_{22}\varepsilon _1^2-\frac{a_{23}(u^+)^pv^+}{1+m(u^+)^p}>0 \quad \text{ for }\; \varepsilon _1\in \left( \varepsilon _1^{**},(\sqrt{2}-1)u^+\right) . \end{aligned}$$

Taking $\varepsilon _1'=\max \{\varepsilon _1^{*},\varepsilon _1^{**}\}$, we obtain $I_3(\lambda ,t)\ge 0$ for $\varepsilon _1\in \left( \varepsilon _1',(\sqrt{2}-1)u^+\right)$.

Summarizing the above discussions, we have

$$\begin{aligned} \begin{aligned}&D_2{\underline{\phi }}''(t)-c{\underline{\phi }}'(t) +ae^{-d_1\tau }\int _{-\infty }^{+\infty }\frac{1}{\sqrt{4\pi D_1\tau }}e^{-\frac{y^{2}}{4D_1\tau }}{\underline{\phi }}(t-y-c\tau )\hbox {d}y -(d_2+q_2e_2){\underline{\phi }}(t)\\&\quad -a_{22}{\underline{\phi }}^2(t) -\frac{a_{23}{\underline{\phi }}^p(t){\overline{\psi }}(t)}{1+m{\underline{\phi }}^p(t)}\ge 0 \end{aligned} \end{aligned}$$

for any $\lambda \in (0,\min \{\lambda _4^*,\lambda _5^*\})$.

For ${\underline{\psi }}(t)$, if $t\le t_4$, then ${\underline{\psi }}(t)=e^{\lambda _3t}-qe^{\eta \lambda _3t}$. Therefore, we obtain

$$\begin{aligned} \begin{aligned}&D_3{\underline{\psi }}''(t)-c{\underline{\psi }}'(t) +(a_1-b{\underline{\psi }}(t)){\underline{\psi }}(t)-q_3e_3{\underline{\psi }}(t) +\frac{a_{32}{\underline{\phi }}^p(t){\underline{\psi }}(t)}{1+m{\underline{\phi }}^p(t)}\\&\quad \ge D_3\left( e^{\lambda _3t}-qe^{\eta \lambda _3t}\right) ''-c\left( e^{\lambda _3t}-qe^{\eta \lambda _3t}\right) '\\&\qquad +(a_1-b(e^{\lambda _3t}-qe^{\eta \lambda _3t}))(e^{\lambda _3t}-qe^{\eta \lambda _3t})-q_3e_3(e^{\lambda _3t}-qe^{\eta \lambda _3t})\\&\quad \ge -qe^{\eta \lambda _3t}\varDelta _2(\eta \lambda _3,c)-be^{2\lambda _3 t}\\&\quad \ge e^{\eta \lambda _3t}\left[ -q\varDelta _2(\eta \lambda _3,c)-b\right] \ge 0\quad \text{ for } \text{ sufficiently } \text{ large }\;q. \end{aligned} \end{aligned}$$

If $t>t_4$, then ${\underline{\psi }}(t)=v^+-\varepsilon _2e^{-\lambda t}, {\underline{\phi }}(t)\ge u^+-\varepsilon _1e^{-\lambda t}$, we have

$$\begin{aligned} \begin{aligned}&D_3{\underline{\psi }}''(t)-c{\underline{\psi }}'(t) +(a_1-b{\underline{\psi }}(t)){\underline{\psi }}(t)-q_3e_3{\underline{\psi }}(t) +\frac{a_{32}{\underline{\phi }}^p(t){\underline{\psi }}(t)}{1+m{\underline{\phi }}^p(t)}\\&\quad \ge -D_3\lambda ^2\varepsilon _2e^{-\lambda t}-c\lambda \varepsilon _2e^{-\lambda t} +\left( a_1-b(v^+-\varepsilon _2e^{-\lambda t})\right) (v^+-\varepsilon _2e^{-\lambda t})\\&\qquad -\,q_3e_3(v^+-\varepsilon _2e^{-\lambda t}) +\frac{a_{32}(u^+-\varepsilon _1e^{-\lambda t})^p(v^+-\varepsilon _2e^{-\lambda t})}{1+m(u^+-\varepsilon _1e^{-\lambda t})^p}\\&\quad =-D_3\lambda ^2\varepsilon _2e^{-\lambda t}-c\lambda \varepsilon _2e^{-\lambda t}+a_1(v^+-\varepsilon _2e^{-\lambda t})-bv^{+2}+2bv^+\varepsilon _2e^{-\lambda t}\\&\qquad -\,b\varepsilon _2^2e^{-2\lambda t}-q_3e_3(v^+-\varepsilon _2e^{-\lambda t}) +\frac{a_{32}(u^+-\varepsilon _1e^{-\lambda t})^p(v^+-\varepsilon _2e^{-\lambda t})}{1+m(u^+-\varepsilon _1e^{-\lambda t})^p}\\&\quad =-\varepsilon _2e^{-\lambda t}\varDelta _2(-\lambda ,c)+2bv^+\varepsilon _2e^{-\lambda t}-b\varepsilon _2^2e^{-2\lambda t} -\frac{a_{32}(u^{+})^pv^+}{1+m(u^{+})^p}\\&\qquad +\frac{a_{32}(u^+-\varepsilon _1e^{-\lambda t})^p(v^+-\varepsilon _2e^{-\lambda t})}{1+m(u^+-\varepsilon _1e^{-\lambda t})^p}\\&\quad \ge \varepsilon _2e^{-\lambda t}[-\varDelta _2(-\lambda ,c)+bv^+]+bv^+\varepsilon _2e^{-\lambda t}-b\varepsilon _2^2e^{-2\lambda t} -\frac{a_{32}(u^{+})^pv^+}{1+m(u^{+})^p}\\&\qquad +\frac{a_{32}(u^+-\varepsilon _1e^{-\lambda t})^p(v^+-\varepsilon _2e^{-\lambda t})}{1+m(u^+)^p}. \end{aligned} \end{aligned}$$

Since $-\varDelta _2(0,c)+bv^+=-\vartheta _2+bv^+=\frac{a_{32}(u^{+})^p}{1+m(u^{+})^p}>0$. We can choose $\lambda _6^*>0$ such that $-\varDelta _2(-\lambda ,c)+bv^+>0$ for $\lambda \in (0,\lambda _6^*)$. Let

$$\begin{aligned} I_4(\lambda ,t) :=bv^+\varepsilon _2e^{-\lambda t}-b\varepsilon _2^2e^{-2\lambda t} -\frac{a_{32}(u^{+})^pv^+}{1+m(u^{+})^p} +\frac{a_{32}(u^+-\varepsilon _1e^{-\lambda t})^p(v^+-\varepsilon _2e^{-\lambda t})}{1+m(u^+)^p}. \end{aligned}$$

By Lemma 6, we can choose $\delta _2>0$ such that

$$\begin{aligned} bv^+\delta -b\delta ^2 -\frac{a_{32}(u^{+})^pv^+}{1+m(u^{+})^p}>\frac{\varepsilon _0}{2}>0 \end{aligned}$$

for $\delta \in [\varepsilon _2,\varepsilon _2+\delta _2]$. Let $\delta ^{***}=\varepsilon _2+\delta _2$.

If $t\in (t_4,0]$, noting that $\varepsilon _2e^{-\lambda t}$ is decreasing on $(t_4,0]$, we can choose $\lambda _7^*>0$ small enough such that $\varepsilon _2e^{-\lambda _7^*t_4}=\varepsilon _2+\delta _2=\delta ^{***}$. Thus, we have $\varepsilon _2\le \varepsilon _2e^{-\lambda t}<\delta ^{***}$ for $t\in (t_4,0]$ and $\lambda \in (0,\lambda _7^*)$. Therefore,

$$\begin{aligned} I_4(\lambda ,t) \ge bv^+\varepsilon _2e^{-\lambda t}-b\varepsilon _2^2e^{-2\lambda t} -\frac{a_{32}(u^{+})^pv^+}{1+m(u^{+})^p}>\frac{\varepsilon _0}{2}>0. \end{aligned}$$

(26)

If $t>0$, using $\varepsilon _1\in \left( 0,(\sqrt{2}-1)u^+\right)$ and $\varepsilon _2\in \left( 0,\nicefrac {v^+}{2}\right)$, we get

$$\begin{aligned} \begin{aligned}&I_4(\lambda ,t)\\&\quad \ge bv^+\varepsilon _2e^{-\lambda t}-b\varepsilon _2^2e^{-\lambda t} -\frac{a_{32}(u^{+})^pv^+}{1+m(u^{+})^p} +\frac{a_{32}(u^+-\varepsilon _1e^{-\lambda t})^p(v^+-\varepsilon _2e^{-\lambda t})}{1+m(u^+)^p}\\&\quad \ge bv^+\varepsilon _2e^{-\lambda t}-\frac{1}{2}bv^+\varepsilon _2e^{-\lambda t} -\frac{a_{32}(u^{+})^pv^+}{1+m(u^{+})^p} +\frac{a_{32}(u^+-\varepsilon _1e^{-\lambda t})^p(v^+-\varepsilon _2e^{-\lambda t})}{1+m(u^+)^p}\\&\quad =e^{-\lambda t}\left[ \frac{1}{2}bv^+\varepsilon _2 +\frac{a_{32}}{1+m(u^{+})^p}\left( (u^+-\varepsilon _1e^{-\lambda t})^p(v^+-\varepsilon _2e^{-\lambda t})-(u^{+})^pv^+\right) \right] .\end{aligned} \end{aligned}$$

If p is odd, we have

$$\begin{aligned} \begin{aligned}&(u^+-\varepsilon _1e^{-\lambda t})^p(v^+-\varepsilon _2e^{-\lambda t})-(u^{+})^pv^+\\&\quad \ge -(u^{+})^p\varepsilon _2e^{-\lambda t}-C_p^1(u^{+})^{p-1}v^+\varepsilon _1e^{-\lambda t} -C_p^2(u^{+})^{p-2}\varepsilon _1^2\varepsilon _2e^{-\lambda t}\\&\qquad -\,C_p^3(u^{+})^{p-3}v^+\varepsilon _1^3e^{-\lambda t} -\cdots -C_p^pv^+\varepsilon _1^pe^{-\lambda t}\\&\quad \ge -(u^{+})^pv^+e^{-\lambda t}\left[ \frac{1}{2}+C_p^1(\sqrt{2}-1)+\frac{1}{2}C_p^2(\sqrt{2}-1)^2\right. \\&\qquad \left. +C_p^3(\sqrt{2}-1)^3+\cdots +C_p^p(\sqrt{2}-1)^p\right] \\&\quad \ge -(\sqrt{2})^p(u^{+})^pv^+e^{-\lambda t}. \end{aligned} \end{aligned}$$

If p is even, we have

$$\begin{aligned} \begin{aligned}&(u^+-\varepsilon _1e^{-\lambda t})^p(v^+-\varepsilon _2e^{-\lambda t})-(u^{+})^pv^+\\&\quad \ge -(u^{+})^p\varepsilon _2e^{-\lambda t}-C_p^1(u^{+})^{p-1}v^+\varepsilon _1e^{-\lambda t} -C_p^2(u^{+})^{p-2}\varepsilon _1^2\varepsilon _2e^{-\lambda t}\\&\qquad -\,C_p^3(u^{+})^{p-3}v^+\varepsilon _1^3e^{-\lambda t} -\cdots -C_p^p\varepsilon _1^p\varepsilon _2e^{-\lambda t}\\&\quad \ge -(u^{+})^pv^+e^{-\lambda t}\left[ \frac{1}{2}+C_p^1(\sqrt{2}-1)+\frac{1}{2}C_p^2(\sqrt{2}-1)^2\right. \\&\quad \left. +C_p^3(\sqrt{2}-1)^3+\cdots +\frac{1}{2}C_p^p(\sqrt{2}-1)^p\right] \\&\quad \ge -(\sqrt{2})^p(u^{+})^pv^+e^{-\lambda t}. \end{aligned} \end{aligned}$$

Therefore,

$$\begin{aligned} \begin{aligned} I_4(\lambda ,t)&\ge e^{-\lambda t}\left[ \frac{1}{2}bv^+\varepsilon _2 -\frac{(\sqrt{2})^pa_{32}(u^{+})^pv^+}{1+m(u^{+})^p}\right] . \end{aligned} \end{aligned}$$

On the other hand, if $\varepsilon _2=\nicefrac {v^+}{2}$, then

$$\begin{aligned} \frac{1}{4}b(v^+)^2 -\frac{(\sqrt{2})^pa_{32}(u^{+})^pv^+}{1+m(u^{+})^p}>0. \end{aligned}$$

Thus, there exists $\varepsilon _2^{**}\left( 0<\varepsilon _2^{**}<\nicefrac {v^+}{2}\right)$ such that

$$\begin{aligned} \frac{1}{2}bv^+\varepsilon _2 -\frac{(\sqrt{2})^pa_{32}(u^{+})^pv^+}{1+m(u^{+})^p}>0\quad \text{ for }\; \varepsilon _2\in \left( \varepsilon _2^{**},\frac{v^+}{2}\right) . \end{aligned}$$

Taking $\varepsilon '_2=\max \{\varepsilon _2^{*},\varepsilon _2^{**}\}$, we obtain $I_4(\lambda ,t)\ge 0$ for $\varepsilon _2\in \left( \varepsilon '_2,\nicefrac {v^+}{2}\right)$.

Summarizing the above discussions, we have

$$\begin{aligned} D_3{\underline{\psi }}''(t)-c{\underline{\psi }}'(t) +(a_1-b{\underline{\psi }}(t)){\underline{\psi }}(t)-q_3e_3{\underline{\psi }}(t) +\frac{a_{32}{\underline{\phi }}^p(t){\underline{\psi }}(t)}{1+m{\underline{\phi }}^p(t)}\ge 0 \end{aligned}$$

for $\lambda \in (0,\min \{\lambda _6^*,\lambda _7^*\})$. Thus, $({\underline{\phi }}(t),{\underline{\psi }}(t))$ is a lower solution of (15). $\square$

Now we obtain and state the main result in this paper.

Theorem 6

Assume that

$$\begin{aligned} {\left\{ \begin{array}{ll} \vartheta _1>\max \left\{ a_{22}u^+,\frac{(4+2\sqrt{2})a_{23}(u^+)^{p-1}M_3}{1+m(u^{+})^p}\right\} ,\\ \vartheta _2>\max \left\{ 2^{p+1}-p-3,4(\sqrt{2})^p-1\right\} \times \frac{a_{32}(u^{+})^p}{1+m(u^{+})^p} \end{array}\right. } \end{aligned}$$

(27)

hold. Then, for every $c>c^*$, system (3) has a traveling wave solution $(\phi (t),\psi (t))$ connecting two equilibria $E_0(0,0)$ and $E_3(u^+,v^+)$. Furthermore,

$$\begin{aligned} \lim _{t\rightarrow -\infty }\phi (t)e^{-\lambda _1 t}=\lim _{t\rightarrow -\infty }\psi (t)e^{-\lambda _3 t}=1. \end{aligned}$$

(28)

Proof

The existence result can be obtained by Theorem 5 and Lemmas 7–8. And now we prove (28). Because

$$\begin{aligned} {\underline{\phi }}(t)\le \phi (t)\le {\overline{\phi }}(t),\quad {\underline{\psi }}(t)\le \psi (t)\le {\overline{\psi }}(t), \end{aligned}$$

we have

$$\begin{aligned}&e^{\lambda _1t}-qe^{\eta \lambda _1t}\le \phi (t)\le e^{\lambda _1t}\quad \text{ for }\;t<t_3,\\&e^{\lambda _3t}-qe^{\eta \lambda _3t}\le \psi (t)\le e^{\lambda _3t}+qe^{\eta \lambda _3t}\quad \text{ for }\;t<\min \{t_2,t_4\}, \end{aligned}$$

which leads to

$$\begin{aligned}&1-qe^{(\eta -1)\lambda _1t}\le \phi (t)e^{-\lambda _1 t}\le 1\quad \text{ for }\;t<t_3,\\&1-qe^{(\eta -1)\lambda _3t}\le \psi (t)e^{-\lambda _3 t}\le 1+qe^{(\eta -1)\lambda _3t}\quad \text{ for }\;t<\min \{t_2,t_4\}. \end{aligned}$$

Therefore, (28) is proved. $\square$