\documentclass[reqno]{amsart}
\usepackage{hyperref}

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2015 (2015), No. 152, pp. 1--21.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
\newline ftp ejde.math.txstate.edu}
\thanks{\copyright 2015 Texas State University - San Marcos.}
\vspace{9mm}}

\begin{document}
\title[\hfilneg EJDE-2015/152\hfil Traveling wavefronts]
{Traveling wavefronts in nonlocal diffusive predator-prey
 system with Holling type II functional response}

\author[S. Li, P. Weng \hfil EJDE-2015/152\hfilneg]
{Shuang Li, Peixuan Weng}

\address{Shuang Li \newline
School of Mathematics, South China Normal University,
Guangzhou 510631,  China}
\email{lemonls@163.com}

\address{Peixuan Weng (corresponding author) \newline
School of Mathematics, South China Normal University,
Guangzhou 510631,  China}
\email{wengpx@scnu.edu.cn}

\thanks{Submitted June 24, 2013. Published June 10, 2015.}
\subjclass[2010]{45K05, 35K57, 92D25}
\keywords{Traveling wavefront; nonlocal diffusive predator-prey system;
\hfill\break\indent functional response of Holling type II; upper-lower solutions}

\begin{abstract}
 This article concerns the existence of traveling wavefronts for a nonlocal
 diffusive predator-prey system with functional response of Holling type II.
 We first establish the existence principle for  the  system with a general
 functional response by using a fixed point theorem and upper-lower solution
 technique. We apply this result  to a predator-prey model with Holling type II
 functional response. We deduce the existence of traveling wavefronts that
 connect the zero equilibrium and the positive equilibrium.
\end{abstract}

\maketitle
\numberwithin{equation}{section}
\newtheorem{theorem}{Theorem}[section]
\newtheorem{lemma}[theorem]{Lemma}
\newtheorem{remark}[theorem]{Remark}
\newtheorem{definition}[theorem]{Definition}
\allowdisplaybreaks

\section{Introduction}

We consider the reaction system based on  the predator-prey
interaction model with nonlocal diffusion
\begin{equation}\label{eq1.1}
\begin{gathered}
\frac{\partial u(x,t)}{\partial t}=d_1[(J_1*u)(x,t)-u(x,t)]
 + h_1(u(x,t))-f(u(x,t))v(x,t),\\
\frac{\partial v(x,t)}{\partial t}=d_2[(J_2*v)(x,t)-v(x,t)]
 +h_2(v(x,t))+\rho f(u(x,t))v(x,t),
\end{gathered}
\end{equation}
where  $u$ and $v$ are the densities of the prey and
 the predator, respectively; $d_1>0$, $d_2>0$ are the diffusion coefficients;
 $J_i(x) (i=1,2)$ are the kernel functions describing the spatial migration
probability of  individuals and is given by
 $$
(J_1*u)(x,t)=\int_{\mathbb{R}}J_1(x-y)u(y,t)dy,\quad
(J_2*v)(x,t)=\int_{\mathbb{R}}J_2(x-y)v(y,t)dy;
$$
 $f(u)$ is the predator response function;
$h_1(u)$ is the growth function of prey which is a positive function
within the maximal carrying capacity of the prey,
and $h_2(v)$ is the growth function of the predator; 
 $\rho\in (0, 1)$ is the transmission coefficient.
If the predator only depends on the prey given in \eqref{eq1.1},
then $h_2(v)$ is a negative function. Otherwise, it may be positive.


A classical mathematical model for  describing the spatial-temporal pattern
for prey-predator (abbreviated as P-P) species is
\begin{equation}\label{eq1.2}
\begin{gathered}
u_{t}= D_1u_{xx}+\alpha u(\beta-u)-f(u)v,\\
v_{t}= D_2v_{xx}-dv +\rho f(u)v,
\end{gathered}
\end{equation}
where in \eqref{eq1.2}, the predator has only the  $u$-species as its
food resource. Assuming that the predator has resource other  than
the $u$-species, and obeys the logistic dynamical growth rule  without
the $u$-species, then the P-P model is
\begin{equation}\label{eq1.3}
\begin{gathered}
u_{t}= D_1u_{xx}+\alpha u(\beta-u)-f(u)v,\\
v_{t}= D_2v_{xx}+\gamma v(\delta-v)+\rho f(u)v.
\end{gathered}
\end{equation}
Both \eqref{eq1.2} and \eqref{eq1.3} used the Laplacian operator
$\Delta :=\frac{\partial^ {2}}{\partial x^{2}}$  to model the diffusion of species,
which is a local operator which suggests  that  the population at the
 location $x$   can only be influenced by the variation of  the population near
the  location $x$.   In \eqref{eq1.1}, at time $t$,  the total individuals
of $u$-species and $v$-species moving from the whole space  into the location
$x$ are expressed as  $\int^{\infty}_{-\infty}J_1(x-y)[u(t,y)-u(t,x)]dy$ and
$\int^{\infty}_{-\infty}J_2(x-y)[v(t,y)-v(t,x)]dy$, respectively.
Generally speaking,  one may call \eqref{eq1.1} a system with  nonlocal diffusion,
and correspondingly,  call \eqref{eq1.2} and \eqref{eq1.3} as systems with
local diffusion.
In recent years,  models with  nonlocal diffusion
 have  attracted much  attention, e.g., see
\cite{B1,B2,B3,B4,Chen,Linguo,p,pll,YW,zll}.
Similar to the study of traveling wave solutions of  reaction-reaction systems
with local diffusion (e.g., see \cite{vvv} and \cite{wz}), the  traveling wave
solutions for the nonlocal reaction-diffusion systems are important in describing
the phenomena in physical process, biological process, and
other fields (e.g., see \cite{B2,p,pll,yy}).


Pioneering  works on the existence of traveling wave solutions  connecting two
steady states (point-to-point orbit) for diffusive
predators-prey systems \eqref{eq1.2} are found in
Dunbar \cite{D1,D2,D3}  with $f(u)=uv$ ($D_1=0$ and
$D_1\neq 0$) and $f(u)=\frac{u}{1+Eu}$ ($D_1=0$) by using the shooting method,
 which is based on a variant of Wazewski's theorem \cite{D1,D2,D3} and LaSalle's
invariant principle. Following Dunbar's ideas,  \eqref{eq1.2} with
$f(u)=\frac{u}{1+Eu}$ ($D_1\neq 0$) and $f(u)=\frac{u^2}{1+Eu^2}$ ($D_1=0$)
for point-to-point orbits were proved  by Huang, Lu \& Ruan \cite{HLR} and
 Li \& Wu \cite{LW}, respectively.
Also  Yang \& Yang \cite{HYYY} considered a model with a more general form.
 Lin, Weng \& Wu \cite{LWW,W} considered a P-P model with Sigmoidal response
function and simplified Dunbar's method by constructing a
bounded set $\mathbb{W}$ to replace the unbounded Wazewski set in \cite{D1,D2,D3}.
 See also  Huang \cite{Huang} for further  development.


Combining  fixed point theory  with the method of upper-lower solutions is
 effective in obtaining the existence of solutions for mixed quasi-monotonic
reaction-diffusion systems.  For P-P system \eqref{eq1.3},
$\alpha u(\beta-u)-f(u)v$ is non-increasing on $v$ and
$\gamma v(\delta-v)+\rho f(u)v$ is nondecreasing on $u$.
Thus,  \eqref{eq1.3} is a mixed quasi-monotonic system (same for \eqref{eq1.2}).
In this article, we  obtain  traveling wavefronts for the  P-P
system \eqref{eq1.1} only by using the upper-lower solution technique
and fixed point theorem.

This article is organized as follows.
In Section 2, some preliminaries are done and an existence theorem
of traveling wavefronts connecting two steady states for \eqref{eq1.1}
is derived  briefly   using fixed point theorem. As applications of
the existence theorem, we need to find a pair of upper-lower solutions
 for the wave profile system with given functions $h_1$, $h_2$ and $f$.
The main contribution  of this article is the construction and verification
of upper-lower solutions for the wave profile systems with  logistic
growth and Holling type II functional response. These works are done in Section 3.

\section{An existence theorem of traveling wavefronts}

We adopt the usual notation for the standard
ordering in $\mathbb{R}^{2}$.  Let $|\cdot|$ denote the Euclidean norm
in $\mathbb{R}^{2}$ and $\|\cdot\|$ denote the supermum norm in space
$C(\mathbb{R},\mathbb{R}^{2})$. Let $\mathbf{0}=(0,0)$.


We make the following assumptions on $h_1(u)$, $h_2(v)$ and $f(u)$.
\begin{itemize}
\item[(H1)] There exist two positive numbers $u_0$,$v_0$
such that $h_1(u_0)-f(u_0)v_0=0$, $h_2(v_0)+\rho
f(u_0)v_0=0$, and $f(0)=h_1(0)=h_2(0)=0$;

\item[(H2)] $f$, $h_1$ and $h_2$ are Lipschitz continuous functions
on any compact interval;

\item[(H3)] $f$ is nondecreasing on $[0,+\infty)$.
\end{itemize}


\begin{remark}\label{R2.0} \rm
(H1) guarantees that the system \eqref{eq1.1} has a trivial
steady state $(0,0)$ and a positive steady state $(u_0,v_0)$.
On the other hand, (H1) and (H3) imply that $f(u)\geq 0$ for $u\geq0$.
\end{remark}

We first consider systems of the form
\begin{equation}\label{eq2.1}
\begin{gathered}
\frac{\partial u(x,t)}{\partial t}
 =d_1\int_{\mathbb{R}}J_1(x-y)[u(y,t)-u(x,t)]dy+f_1(u(x,t),v(x,t)),\\
\frac{\partial v(x,t)}{\partial t}
 =d_2\int_{\mathbb{R}}J_2(x-y)[v(y,t)-v(x,t)]dy+f_2(u(x,t),v(x,t)).
\end{gathered}
\end{equation}
Here we assume that $J_i$ and $f_i$ $(i=1,2)$,  satisfy
\begin{itemize}
\item[(J1)] $\int_{\mathbb{R}}J_i(x)dx=1$, $J_i(x)\geq 0$,
$J_i(x)=J_i(-x)$ for $x\in \mathbb{R}$; for $\nu\in (0, +\infty]$,
$\int_{\mathbb{R}}J_{i}(x)e^{\nu x}dx<\infty$;

\item[(F1)] $f_i\in C(\mathbb{R},\mathbb{R})$ ($i=1,2$),
$f_1(0,0)=f_2(0,0)=0$, $f_1(u_0,v_0)=f_2(u_0,v_0)=0$;
 there exist two positive constants $L_1>0$, $L_2>0$
such that
\begin{gather*}
|f_1(\varphi_1,\psi_1)-f_1(\varphi_2,\psi_2)|\leq
L_1\|\Phi-\Psi\|, \\
|f_2(\varphi_1,\psi_1)-f_2(\varphi_2,\psi_2)|\leq
L_2\|\Phi-\Psi\|,
\end{gather*}
for any $\Phi=(\varphi_1,\psi_1)$,
$\Psi=(\varphi_2,\psi_2)\in C(\mathbb{R},\mathbb{R}^{2})$ with
$0\leq\Phi(t), \Psi(t)\leq \mathbf{K}$, and $\mathbf{K}=(K_1,K_2)\geq (u_0,v_0)$
is some vector in $\mathbb{R}^{2}$ which will be given later;

\item[(F2)] $f_1(u,v)$ is non-increasing on $v$, and $f_2(u,v)$ is
non-decreasing on $u$, where $(u,v)\in [\mathbf{0},\mathbf{K}]$.
\end{itemize}

A traveling wave solution of \eqref{eq2.1} is a  solution
with the form $(u(t,x),w(t,x))=(\varphi(x+ct),\psi(x+ct))$, where
$(\varphi,\psi)\in C^{1}(\mathbb{R},\mathbf{K})$ is the wave profile
which propagates at a constant velocity $c>0$.
Substituting $(u(t,x),v(t,x))=(\varphi(x+ct),\psi(x+ct))$ into
\eqref{eq2.1} and replacing $x+ct$ by $t$, then we obtain
\begin{equation}\label{eq2.3}
\begin{gathered}
c\varphi'(t)=d_1(J_1*\varphi)(t)-d_1\varphi(t)+f_1(\varphi(t),\psi(t)),\\
c\psi'(t)=d_2(J_2*\psi)(t)-d_2\psi(t)+f_2(\varphi(t),\psi(t)),
\end{gathered}
\end{equation}
where $(J_1*\varphi)(t)=\int_{\mathbb{R}}J_1(t-s)\varphi(s)ds$ and
$(J_2*\psi)(t)=\int_{\mathbb{R}}J_2(t-s)\psi(s)ds$.
Recalling the physical and biological motivation  (see e.g., \cite{B2,f,m}),
we also require that the traveling solution $(\varphi,\psi)$
satisfies the asymptotic boundary  conditions
\begin{equation} \label{eq2.2}
\lim_{t\to-\infty}(\varphi(t),\psi(t))=(0,0),\quad
\lim_{t\to+\infty}(\varphi(t),\psi(t))=(u_0,v_0).
\end{equation}
We call a solution of \eqref{eq2.3} satisfying \eqref{eq2.2} as a traveling
wavefront of \eqref{eq2.1}.

 Let $\beta>0$ be large.
For any $(\varphi,\psi)\in
C_{[0,\mathbf{K}]}:=C(\mathbb{R},[\mathbf{0},\mathbf{K}])$, define the operator
$H=(H_1,H_2):C_{[0,\mathbf{K}]}\to
C(\mathbb{R},\mathbb{R}^{2})$ by
\begin{equation}\label{eq2.4}
\begin{gathered}
H_1(\varphi,\psi)(t)=\beta_1\varphi(t)+d_1(J_1*\varphi)(t)
-d_1\varphi(t)+f_1(\varphi(t),\psi(t)),\\
H_2(\varphi,\psi)(t)=\beta_2\psi(t)+d_2(J_2*\psi)(t)
-d_2\psi(t)+f_2(\varphi(t),\psi(t)).\\
\end{gathered}
\end{equation}
Then \eqref{eq2.3} can be written as
\begin{equation}\label{eq2.5}
\begin{gathered}
c\varphi'(t)=-\beta_1\varphi(t)+H_1(\varphi,\psi)(t),\\
c\psi'(t)=-\beta_2\psi(t)+H_2(\varphi,\psi)(t).
\end{gathered}
\end{equation}
By \eqref{eq2.5}, we further define
$F=(F_1,F_2):C_{[0,\mathbf{K}]}\to C(\mathbb{R},\mathbb{R}^{2})$ by
\begin{equation}\label{eq2.6}
\begin{gathered}
F_1(\varphi,\psi)(t)=\frac{1}{c}e^{-\frac{\beta_1}{c}t}
\int^{t}_{-\infty}e^{\frac{\beta_1}{c}s}
H_1(\varphi,\psi)(s)ds,\\
F_2(\varphi,\psi)(t)=\frac{1}{c}e^{-\frac{\beta_2}{c}t}
\int^{t}_{-\infty}e^{\frac{\beta_2}{c}s}
H_2(\varphi,\psi)(s)ds.
\end{gathered}
\end{equation}
 Thus, a fixed point of $F$ is a solution of \eqref{eq2.5}, which is
a traveling wavefront of \eqref{eq2.1}, and vice verse.
Therefore, in what follows, we shall search for the fixed point of $F$
connecting  $(0,0)$ and  $(u_0,v_0)$.

 We introduce an exponential decay norm as follows.
 Let $\mu>0$ such that
$\mu<\min\{\frac{\beta_1}{c},\frac{\beta_2}{c}\}$. Define
$$
B_{\mu}(\mathbb{R},\mathbb{R}^{2})
=\big\{u(t)\in
C(\mathbb{R},\mathbb{R}^{2}):\sup_{t\in \mathbb{R}}|u(t)|e^{-\mu|t|}<\infty\big\},
$$
and  $|u(t)|_{\mu}=\sup_{t\in \mathbb{R}}|u(t)|e^{-\mu|t|}$ for
$u\in B_{\mu}(\mathbb{R},\mathbb{R}^{2})$.
Then  $(B_{\mu}(\mathbb{R},\mathbb{R}^{2}),|\cdot|_{\mu})$ is a Banach space.

We give a definition of upper and lower solutions of \eqref{eq2.3} as follows.

\begin{definition}\label{D2.1} \rm
 A pair of continuous functions
$\overline{\Phi}(t)=(\overline{\varphi}(t),\overline{\psi}(t))$
and
$\underline{\Phi}(t)=(\underline{\varphi}(t),\underline{\psi}(t))\in
C_{[0,\mathbf{K}]}$ is called a pair of upper-lower solutions
 of \eqref{eq2.3}, respectively, if
$\overline{\Phi}'(t)$ and $\underline{\Phi}{'}(t)$ exist for
$t\in \mathbb{R}\setminus \mathbb{T}$, which are bounded and satisfy
\begin{gather*}
c\overline{\varphi}'(t)\geq d_1(J_1*\overline{\varphi})(t)-d_1\overline{\varphi}(t)
+f_1(\overline{\varphi}(t),\underline{\psi}(t)),\quad
t\in \mathbb{R}\setminus \mathbb{T}, \\
c\overline{\psi}'(t)\geq  d_2(J_2*\overline{\psi})(t)-d_2\overline{\psi}(t)
+f_2(\overline{\varphi}(t),\overline{\psi}(t)), \quad
t\in \mathbb{R}\setminus \mathbb{T};
\end{gather*}
\begin{gather*}
c\underline{\varphi}'(t)\leq d_1(J_1*\underline{\varphi})(t)
 -d_1\underline{\varphi}(t)
+ f_1(\underline{\varphi}(t),\overline{\psi}(t)),\quad
 t\in \mathbb{R}\setminus \mathbb{T}, \\
c\underline{\psi}'(t)\leq  d_2(J_2*\underline{\psi})(t)-d_2\underline{\psi}(t)
 +f_2(\underline{\varphi}(t),\underline{\psi}(t)), \quad
t\in \mathbb{R}\setminus \mathbb{T}.
\end{gather*}
Here $\mathbb{T}$=$(t_1,t_2,\dots ,t_{n})$ is
a  set of finite points with $t_1<t_2<\dots <t_{n}$.
\end{definition}


In this article, we assume that a pair of upper-lower  solutions
$\overline{\Phi}(t)=(\overline{\varphi}(t),\overline{\psi}(t))$ and
 $\underline{\Phi}(t)=(\underline{\varphi}(t),\underline{\psi}(t))$
of \eqref{eq2.3} satisfies
\begin{itemize}
\item[(P1)] $(0,0)\leq(\underline{\varphi},\underline{\psi})(t)
\leq(\overline{\varphi},\overline{\psi})(t)\leq(K_1,K_2),\
t\in\mathbb{R}$;

\item[(P2)] $\lim\limits_{t\to -\infty}(\overline{\varphi},\overline{\psi})(t)=(0,0)$,
$\lim\limits_{t\to+\infty}(\underline{\varphi},\underline{\psi})(t)
=\lim\limits_{t\to+\infty}(\overline{\varphi},\overline{\psi})(t)
=(u_0,v_0)$.
\end{itemize}

Now we state an existence theorem for traveling wave solution of \eqref{eq2.1}.
A similar proof  can be found in  \cite{hw, YW,zll}, and  we omit its proof  here.

\begin{theorem} \label{T2.1}
Assume that {\rm (J1), (F1), (F2)} hold. If \eqref{eq2.3} has a pair of upper-lower
 solutions $\overline{\Psi}=(\overline{\varphi},\overline{\psi})$
and
$\underline{\Psi}=(\underline{\varphi},\underline{\psi})$
satisfying {\rm (P1)--(P2)}. Then \eqref{eq2.3} has a  solution
satisfying \eqref{eq2.2}. That is, \eqref{eq2.1} has a traveling wavefront
satisfying \eqref{eq2.2}.
\end{theorem}


\begin{remark}\label{R2.1} \rm
Let $f_1(u,v)=h_1(u) -f(u)v$, $f_2(u,v)=h_2(v)+\rho f(u)v$,
where $h_i$ ($i=1,2$) and $f$ satisfy (H1)--(H3). It is obvious that
$f_1$ and $f_2$ satisfy (F1) and (F2). Therefore, the conclusion in Theorem
 \ref{T2.1} holds for system \eqref{eq1.1} if $J_i$ ($i=1,2$) satisfy (J1).
\end{remark}

\section{Prey-predator system with Holling type II response}

We consider the  prey-predator system with Holling type II response,
\begin{equation}\label{eq3.1}
\begin{gathered}
\frac{\partial u(x,t)}{\partial t}=d_1(J_1*u)(x,t)-d_1u(x,t)
 +\alpha u(x,t)(\beta-u(x,t))-\frac{u(x,t)v(x,t)}{1+u(x,t)},\\
\frac{\partial v(x,t)}{\partial t}=d_2(J_2*v)(x,t)-d_2v(x,t)
+\gamma v(x,t)(\delta-v(x,t))+\rho\frac{u(x,t)v(x,t)}{1+u(x,t)}.
\end{gathered}
\end{equation}
in which $x\in \mathbb{R}$, $t\geq 0$, and all parameters are positive,
and $J_{i}$ ($i=1,2$) satisfy  (J1).

Obviously, \eqref{eq3.1} has three trivial equilibria as follows:
$$
E_0=(0,0),\quad E_1=(\beta,0),\quad E_2=(0,\delta).
$$
 Furthermore, we assume that \eqref{eq3.1} has a unique positive equilibrium
$E_{3}=(u_0,v_0)$ satisfying
\begin{equation}\label{eq3.2}
\alpha\beta-\alpha u_0-\frac{v_0}{1+u_0}=0,\quad
\gamma\delta-\gamma v_0+\frac{\rho u_0}{1+u_0}=0.
\end{equation}
It is clear that $u_0<\beta$, $v_0>\delta$   and $\gamma v_0>\frac{\rho u_0}{1+u_0}$.

Let us consider the  algebraic system
\begin{equation}\label{eq3.2*}
\alpha\beta-\alpha u -\frac{v}{1+u}=0,\quad
\gamma\delta-\gamma v+\frac{\rho u}{1+u}=0,
\end{equation}
which is equivalent to
$$
(\alpha\beta-\alpha u)(1+u)=v,\quad (\gamma\delta-\gamma v)(1+u)+\rho u=0.
$$
Substituting the first equation into the second equation, we obtain
$$
[\gamma\delta(1+u)-\gamma (\alpha\beta-\alpha u)(1+u)^2]+\rho u=0.
$$
That is,
$$
Q(u):=\alpha u^3+(2\alpha-\alpha \beta)u^2-(2\alpha\beta-\alpha
-\delta-\frac{\rho}{\gamma})u+(\delta-\alpha\beta)=0.
$$
Note $Q(0)=\delta-\alpha\beta$, $Q(-\infty)=-\infty$ and
$Q(\infty)=\infty$, and $Q(u)=0$ is a cubic algebraic equation.
If $\delta-\alpha\beta<0$, then $Q(u)=0$ has a positive real root $u=u_0$. Since
$$
Q(\beta)=\alpha\beta^3+2\alpha \beta^2-\alpha\beta^3-2\alpha \beta^2
+\alpha\beta+\delta\beta+\frac{\rho}{\gamma}\beta+\delta-\alpha\beta
=\delta\beta+\frac{\rho}{\gamma}\beta+\delta>0,
$$
we have $u_0<\beta$. Let $v_0=(\alpha\beta-\alpha u_0)(1+u_0)$, then
$v_0>0$ and $(u_0,v_0)$ is a positive solution of \eqref{eq3.2*}.
Therefore, the assumption \eqref{eq3.2} is feasible.

Assume $c>0$. Let $u(x,t)=\varphi(x+ct)$, $v(x,t)=\psi(x+ct)$,
and denote the traveling wave coordinate $x+ct$ still by $t$, then the
corresponding wave profile system is
\begin{equation}\label{eq3.3}
\begin{gathered}
c\varphi'(t)=d_1(J_1*\varphi)(t)-d_1\varphi(t)
 \alpha \varphi(t)(\beta-\varphi(t))-\frac{\varphi(t)\psi(t)}{1+\varphi(t)},\\
c\psi'(t)= d_2(J_2*\psi )(t)-d_2\psi(t)+\gamma \psi(t)(\delta-\psi(t))
 +\rho\frac{\varphi(t)\psi(t)}{1+\varphi(t)}.
\end{gathered}
\end{equation}

We first do some preliminaries. For $\lambda\in \mathbb{R}$ and $c>0$, define
\begin{gather*}
\Delta_1(\lambda,c)=d_1\int_{\mathbb{R}}J_1(y)(e^{\lambda y}-1)dy
 -c\lambda+\alpha\beta,\\
\Delta_2(\lambda,c)=d_2\int_{\mathbb{R}}J_2(y)(e^{\lambda y}-1)dy
 -c\lambda+\delta_2.
\end{gather*}
Here $\delta_2$ is some constant which will be given in the following subsections.
By (J1) and some direct calculations, we have the following lemma.

\begin{lemma}\label{L3.1}
The following conclusions hold.

(i) There is a $c_1^*>0$, such that for any given $c>c_1^{*}$,
$\Delta_1(\lambda,c)$ has two distinct real roots
$\lambda_1(c)$ and $\lambda_2(c)$ satisfying $0<\lambda_1(c)<\lambda_2(c)$
and
$$
\Delta_1(\lambda,c) \begin{cases}
>0,& \text{either }0 < \lambda < \lambda_1(c) \text{ or } \lambda > \lambda_2(c),\\
<0,& \lambda_1(c) < \lambda < \lambda_2(c).
\end{cases}
$$
(ii) There is a $c_2^*\geq 0$, such that for any given $c>c_2^{*}$,
$\Delta_2(\lambda,c)$ has two distinct real roots
$\lambda_{3}(c)$ and $\lambda_{4}(c)(>0)$ satisfying
$\lambda_{3}(c)<\lambda_{4}(c)$ and
$$
\Delta_2(\lambda,c)\begin{cases}
>0,& \text{either }\lambda < \lambda_{3}(c) \text{ or } \lambda > \lambda_{4}(c),\\
<0,& \lambda_{3}(c) < \lambda < \lambda_{4}(c).\\
\end{cases}
$$
\end{lemma}

\begin{remark}\label{R3.a} \rm
If $\delta_2>0$, then $c_2^*>0$ and $\lambda_{3}(c)>0$. If $\delta_2= 0$,
then $c_2^*=0$ and $\lambda_{3}(c)= 0$. If $\delta_2< 0$, then
$c_2^*=0$ and $\lambda_{3}(c)<0$.
\end{remark}

\subsection{Traveling waves connecting $E_0$ and $E_3$}

In this subsection, we are interested in the solution of \eqref{eq3.3}
with asymptotic boundary conditions
\begin{equation} \label{eq3.4}
\lim_{t\to-\infty}(\varphi(t),\psi(t))=(0,0),\quad
\lim_{t\to+\infty}(\varphi(t),\psi(t))=(u_0,v_0).
\end{equation}
Here, we choose $K_1=\beta$,  $K_2=\delta+\frac{\rho}{\gamma}\beta$,
then we have $K_1>u_0$ and $K_2>v_0$.

To construct  lower-upper solutions of  \eqref{eq3.3},  we need the following lemma.

\begin{lemma}\label{L3.2}
Suppose that
\begin{equation}\label{3.5}
\alpha u_0 \geq \frac{(3+2\sqrt{2})v_0}{1+u_0},\quad
\gamma v_0 \geq \frac{4(\sqrt{2}-1)\rho u_0}{1+u_0}
\end{equation}
hold, then there exist $\varepsilon_1\in(0,(\sqrt{2}-1)u_0]$
and $\varepsilon_2\in(0,\frac{v_0}{2}]$ such that
\begin{equation} \label{3.6}
\begin{gathered}
-\alpha\varepsilon_1^{2}+(2\sqrt{2}-2)\alpha u_0\varepsilon_1+
\frac{u_0v_0}{1+u_0}-\frac{2(u_0-\varepsilon_1)v_0}
{1+(u_0-\varepsilon_1)} > \varepsilon_0,\\
-\gamma\varepsilon_2^{2}+\gamma v_0\varepsilon_2
-\frac{\rho v_0\varepsilon_1}
{(1+u_0)(1+u_0-\varepsilon_1)} > \varepsilon_0,
\end{gathered}
\end{equation}
where  $\varepsilon_0>0$ is some constant. Particularly, one can choose
$\varepsilon_1=(\sqrt{2}-1)u_0$ and  $\varepsilon_2=v_0/2$.
\end{lemma}

\begin{proof}
The proof of the first inequality is similar to that of \cite[Lemma 4.2]{hw}.
We only prove the second inequality. Let
\begin{align*}
g_1(\varepsilon_2)=-\gamma\varepsilon_2^{2}+\gamma v_0
\varepsilon_2,\quad
g_2(\varepsilon_1)=\frac{\rho v_0\varepsilon_1}{(1+u_0)(1+u_0-\varepsilon_1)}.
\end{align*}
Note that $g_1(0)=0$,
$\max{g_1(\varepsilon_2)}=g_1(\frac{v_0}{2})=\frac{1}{4}\gamma v_0^2$.
If $\varepsilon_1\in(0,(\sqrt{2}-1)u_0]$, then
$$
g_2(\varepsilon_1)=\frac{\rho v_0\varepsilon_1}{(1+u_0)(1+u_0-\varepsilon_1)}
\leq \frac{(\sqrt{2}-1)\rho v_0u_0}{(1+u_0)[1+(2-\sqrt{2})u_0]}
< \frac{(\sqrt{2}-1)\rho v_0u_0}{1+u_0}.
$$
If $\gamma v_0 \geq \frac{4(\sqrt{2}-1)\rho u_0}{1+u_0}$, then for any
$\varepsilon_1\in (0,(\sqrt{2}-1)u_0]$, there exist
$\varepsilon_2^*\in(0,v_0/2]$ such that
$g_1(\varepsilon_2)> g_2(\varepsilon_1)$ for
$\varepsilon_2^*<\varepsilon_2\leq v_0/2$.
The proof is complete.
\end{proof}

\begin{remark}\label{R3.2} \rm
If we assume that
\begin{equation}\label{3.7}
\alpha \beta>\frac{(4+2\sqrt{2})v_0}{1+u_0}, \quad
\gamma\delta>\frac{(4\sqrt{2}-5)\rho u_0}{1+u_0},
\end{equation}
then  from \eqref{3.7}  we derive that
\begin{equation}\label{3.7*}
\begin{gathered}
\alpha u_0=\alpha\beta-\frac{v_0}{1+u_0}>\frac{(3+2\sqrt{2})v_0}{1+u_0},\\
\alpha u_0=\alpha\beta-\frac{v_0}{1+u_0}>\alpha\beta-\frac{\alpha\beta}{4+2\sqrt{2}}
=\frac{(2+\sqrt{2})\alpha\beta}{4},\\
\gamma v_0= \gamma\delta+\frac{\rho u_0}{1+u_0}\geq \frac{4(\sqrt{2}-1)\rho u_0}{1+u_0},\\
\gamma\delta=\gamma v_0-\frac{\rho u_0}{1+u_0}\geq\gamma v_0-\frac{\gamma v_0}{4(\sqrt{2}-1)}>\frac{1}{3}\gamma v_0.
\end{gathered}
\end{equation}
That is, \eqref{3.5} holds.
\end{remark}


In this subsection, let $\delta_2=\gamma\delta$, then we have from Remark
\ref{R3.a} that $c_2^*>0$ and $\lambda_3(c)>0$. We  denote $\lambda_{i}(c)$
by $\lambda_{i}$ in what follows, where $i=1,2,3,4$.
 Let $c>c^{*}:=\max\{c_1^*,c_2^*\}$,  $q>1$ and
\begin{equation} \label{3.8}
\eta\in\big(1,\min\{\frac{\lambda_2}{\lambda_1},\frac{\lambda_{4}}{\lambda_{3}},
\frac{\lambda_1+\lambda_{3}}{\lambda_1},
\frac{\lambda_1+\lambda_{3}}{\lambda_{3}},2\}\big).
\end{equation}
Define two  continuous functions
$$
L_1(t):=e^{\lambda_1t}-qe^{\eta\lambda_1t},\quad
L_2(t):=e^{\lambda_{3}t}-qe^{\eta\lambda_{3}t}.
$$
Let $L_i(t)=0,i=1,2$, we obtain
$$
\bar{t}_2= -\frac{1}{(\eta-1)\lambda_1}\ln q <0,\quad
\bar{t}_4=-\frac{1}{(\eta-1)\lambda_{3}}\ln q <0.
$$
Let $\varepsilon_1, \varepsilon_2$ be defined as in Lemma \ref{L3.2}.
For any given $q>1$,  we can choose small
$\lambda>0$ such that $t_2<\bar{t}_2,t_4<\bar{t}_4$ and
$$
u_0-\varepsilon_1e^{-\lambda t_2}=L_1(t_2), \quad
v_0-\varepsilon_2e^{-\lambda t_{4}}=L_2(t_{4}).
$$
Using the above constants, define the continuous vector functions:
\begin{gather*}
\overline{\varphi}(t)=\begin{cases}
e^{\lambda_1 t}, &t \leq t_1,\\
\min\{K_1,u_0+u_0e^{-\lambda t}\},&t \geq t_1,
\end{cases}
\quad
\underline{\varphi}(t)= \begin{cases}
e^{\lambda_1 t}-qe^{\eta\lambda_1 t}, & t \leq t_2,\\
u_0-\varepsilon_1e^{-\lambda t},& t \geq t_2,
\end{cases}
\\
\overline{\psi}(t)=\begin{cases}
e^{\lambda_{3} t}+qe^{\eta\lambda_{3} t}, &t \leq t_{3},\\
\min\{K_2,v_0+v_0e^{-\lambda t}\}, & t \geq t_{3},
\end{cases}
\quad
\underline{\psi}(t)=\begin{cases}
e^{\lambda_{3} t}-qe^{\eta\lambda_{3} t}, & t \leq t_{4},\\
v_0-\varepsilon_2e^{-\lambda t}, &t \geq t_{4},
\end{cases}
\end{gather*}
where  $\lambda>0$ is small and will be determined  later.
We can see
$\overline{\varphi}(t),\overline{\psi}(t),
\underline{\varphi}(t),\underline{\psi}(t)$
satisfy (P1)--(P2) in Section 2. Moreover, if $q>1$ is large enough,
then it is obvious that
$$
t_3<0\quad\text{and}\quad t_1 \geq \max \{t_2,t_{3},t_{4}\}.
$$


\begin{lemma}\label{L3.3}
Assume \eqref{3.7} holds. If $q>1$ is large  and $\lambda>0$ is small, then
$(\overline{\varphi}(t),\overline{\psi}(t))$ and
$(\underline{\varphi}(t),\underline{\psi}(t))$ is a pair of  upper solution
and  lower solution of \eqref{eq3.3}.
\end{lemma}

\begin{proof}
Firstly, we note the following facts: for $t\in \mathbb{R}$,
\begin{equation}\label{ul}
\begin{gathered}
\overline{\varphi}(t)\leq e^{\lambda_1 t}, \quad
 \overline{\varphi}(t)\leq u_0+u_0 e^{-\lambda t}, \quad
 \overline{\varphi}(t)\leq K_1,\\
\overline{\psi}(t)\leq e^{\lambda_3 t}+qe^{\eta\lambda_3 t}\quad
 \overline{\psi}(t)\leq  v_0+v_0e^{-\lambda t},\quad
 \overline{\psi}(t)\leq K_2,\\
\underline{\varphi}(t)\geq e^{\lambda_1 t}-qe^{\eta\lambda_1 t},\quad
 \underline{\varphi}(t)\geq u_0-\varepsilon_1e^{-\lambda t}, \quad
 \underline{\varphi}(t)\geq 0,\\
\underline{\psi}(t)\geq e^{\lambda_3 t}-qe^{\eta\lambda_3 t},\quad
 \underline{\psi}(t)\geq v_0-\varepsilon_2e^{-\lambda t}, \quad
 \underline{\psi}(t)\geq 0.
\end{gathered}
\end{equation}
These inequalities will be used in the following arguments without
extra explanations.

 Now we consider  $\overline{\varphi}(t)$.
If $t<t_1$, $\overline{\varphi}(t)=e^{\lambda_1t}$ and
$\underline{\psi}(t)\geq0$, then
\begin{align*}
&d_1(J_1*\overline{\varphi})(t)-d_1\overline{\varphi}(t)
 -c\overline{\varphi'}(t)+\alpha \overline{\varphi}(t)(\beta-\overline{\varphi}(t))
 -\frac{\overline{\varphi}(t)\underline{\psi}(t)}{1+\overline{\varphi}(t)}\\
&\leq e^{\lambda_1t}\Delta_1(\lambda_1,c)=0.
\end{align*}
If $t\geq t_1$, we have $\overline{\varphi}(t)=K_1=\beta$, then the result
is clear. Otherwise, $\overline{\varphi}(t)=u_0+u_0e^{-\lambda t}$,
$\underline{\psi}(t)\geq v_0-\varepsilon_2e^{-\lambda t}$ implies that
\begin{align*}
& d_1(J_1*\overline{\varphi})(t)-d_1\overline{\varphi}(t)-c\overline{\varphi'}(t)
 +\alpha \overline{\varphi}(t)(\beta-\overline{\varphi}(t))
 -\frac{\overline{\varphi}(t)\underline{\psi}(t)}{1+\overline{\varphi}(t)}\\
&\leq  d_1\int_{R}J_1(y-t)(u_0+u_0e^{-\lambda y})dy
 -d_1(u_0+u_0e^{-\lambda t})+c\lambda u_0e^{-\lambda t}\\
&\quad +\alpha (u_0+u_0e^{-\lambda t})[\beta-(u_0+u_0e^{-\lambda t})]
 -\frac{(u_0+u_0e^{-\lambda t}) \underline{\psi}(t)}{1+u_0+u_0e^{-\lambda t}}\\
&=u_0e^{-\lambda t}\Delta_1(-\lambda,c)+\alpha\beta u_0-\alpha u_0^{2}-2\alpha u_0^{2}e^{-\lambda t}-\alpha u_0^{2}e^{-2\lambda t}-\frac{(u_0+u_0e^{-\lambda t}) \underline{\psi}(t)}{1+u_0+u_0e^{-\lambda t}}\\
&= u_0e^{-\lambda t}[\Delta_1(-\lambda,c)-2(2-\sqrt{2})\alpha u_0]\\
&\quad -u_0[(2\sqrt{2}-2)\alpha u_0e^{-\lambda t}+\alpha u_0e^{-2\lambda t}+
\frac{(1+e^{-\lambda t}) \underline{\psi}(t)}{1+u_0+u_0e^{-\lambda t}}
 -\frac{v_0}{1+u_0}].
\end{align*}
Note from the second inequality in \eqref{3.7*}  that
$\Delta_1(0,c)-2(2-\sqrt{2})\alpha u_0=\alpha\beta-2(2-\sqrt{2})\alpha u_0<0$,
 and thus $\Delta_1(-\lambda,c)-2(2-\sqrt{2})\alpha u_0<0$ if $\lambda>0$ is small.
Then there exists a constant $\lambda_1^{*}$ such that
$\Delta_1(-\lambda,c)-2(2-\sqrt{2})\alpha u_0<0$ for any
$\lambda\in(0,\lambda_1^{*})$.

 On the other hand, let
\begin{align*}
I_1(\lambda,t)
&:=(2\sqrt{2}-2)\alpha u_0e^{-\lambda t}+\alpha u_0e^{-2\lambda t}
 + \frac{(1+e^{-\lambda t}) \underline{\psi}(t)}{1+u_0+u_0e^{-\lambda t}}
 -\frac{v_0}{1+u_0}\\
&\geq (2\sqrt{2}-2)\alpha u_0e^{-\lambda t}+\alpha u_0e^{-2\lambda t}
 +\frac{(1+e^{-\lambda t}) (v_0-\varepsilon_2e^{-\lambda t})}
 {1+u_0+u_0e^{-\lambda t}}
 -\frac{v_0}{1+u_0}.
\end{align*}
Let $t\geq t_1$ and $x:=e^{-\lambda t}\in (0,\infty)$. Then
\begin{align*}
I_1(\lambda,t)
&\geq (2\sqrt{2}-2)\alpha u_0x+\alpha u_0x^2
+\frac{(1+x) (v_0-\varepsilon_2x)}{1+u_0+u_0x}-\frac{v_0}{1+u_0}\\
&=\Big((2\sqrt{2}-2)\alpha u_0x(1+u_0+u_0x)+\alpha u_0x^2(1+u_0+u_0x)\\
&\quad +(1+x) (v_0-\varepsilon_2x)
 -\frac{v_0}{1+u_0}(1+u_0+u_0x)\Big)
  \big/ \big(1+u_0+u_0x\big)\\
&=\frac{x\hat{g}(x)}{1+u_0+u_0x}.
\end{align*}
Here
\begin{gather*}
\hat{g}(x)=\hat{a}x^2+\hat{b}x+\hat{c},\quad \hat{a}:=\alpha u_0,\\
 \hat{b}:=\alpha u_0(1+u_0)-\varepsilon_2+(2\sqrt{2}-2)\alpha u_0^2,\\
\hat{c}:=\frac{v_0}{1+u_0}-\varepsilon_2+(2\sqrt{2}-2)\alpha u_0(1+u_0).
\end{gather*}
Let $\varepsilon_2=v_0/2$. Note that we have from \eqref{3.7},
\begin{gather*}
\begin{aligned}
\hat{b}&= \alpha u_0(1+u_0)-\varepsilon_2+(2\sqrt{2}-2)\alpha u_0^2\\
&\geq \alpha u_0(1+u_0)-\frac{v_0}{2}\\
&= [\alpha\beta-\frac{v_0}{1+u_0}](1+u_0)-\frac{v_0}{2}\\
&=\alpha\beta(1+u_0)-\frac{3v_0}{2}\\
&>(4+\sqrt{2})v_0-\frac{3v_0}{2}\geq  0,
\end{aligned}
\\
\begin{aligned}
 \hat{g}(0)&= \frac{v_0}{1+u_0}-\varepsilon_2+(2\sqrt{2}-2)\alpha u_0(1+u_0),\\
&\geq \frac{v_0}{1+u_0}-\frac{v_0}{2}+(2\sqrt{2}-2)(3+2\sqrt{2})v_0>0,
\end{aligned}
\\
 \hat{g}'(x)= 2\hat{a}x+\hat{b}>0\text{ for }x>0.
\end{gather*}
Therefore, $\hat{f}(x):=x\hat{g}(x)$ is increasing on $x\in (0,\infty)$.
Since $\hat{f}(0)=0$, we have
$\hat{f}(x)>0$ for $x\in (0,\infty)$. That is $I_1(\lambda,t)>0$
for $t\geq t_1$. Thus $\overline{\varphi}(t)$ satisfies
the definition of upper solution.

We now consider $\overline{\psi}(t)$. If $t\leq t_{3}$, then
$\overline{\psi}(t)=e^{\lambda_{3}t}+qe^{\eta\lambda_{3}t}$,
$\overline{\varphi}(t)\leq e^{\lambda_1t}$. Noting
$\eta\lambda_3\leq \lambda_1+\lambda_3$ and $t_3<0$, we have
\begin{equation}\label{lt3}
\begin{aligned}
&d_2(J_2*\overline{\psi})(t)-d_2\overline{\psi}(t)-c\overline{\psi'}(t)
 +\gamma \overline{\psi}(t)(\delta-\overline{\psi}(t))
 +\frac{\rho\overline{\varphi}(t)\overline{\psi}(t)}{1+\overline{\varphi}(t)}\\
&\leq
d_2\int_{R}J_2(y-t)(e^{\lambda_{3}y}+qe^{\eta\lambda_{3}y})dy-d_2(e^{\lambda_{3}t}+
qe^{\eta\lambda_{3}t})-c(\lambda_{3}e^{\lambda_{3}t}
 +q\eta\lambda_{3} e^{\eta\lambda_{3}t})\\
&\quad +\gamma(e^{\lambda_{3}t}+qe^{\eta\lambda_{3}t})(\delta-e^{\lambda_{3}t}
 -qe^{\eta\lambda_{3}t})
 +\frac{\rho\overline{\varphi}(t)(e^{\lambda_{3}t}
 +qe^{\eta\lambda_{3}t})}{1+\overline{\varphi}(t)}\\
&= e^{\lambda_{3}t}\Delta_2(\lambda_{3},c)
 +qe^{\eta\lambda_{3}t}\Delta_2(\eta\lambda_{3},c)
 -\gamma(e^{\lambda_{3}t}+qe^{\eta\lambda_{3}t})^{2}
 +\frac{\rho\overline{\varphi}(t)
(e^{\lambda_{3}t}+qe^{\eta\lambda_{3}t})}{1+\overline{\varphi}(t)}\\
&\leq qe^{\eta\lambda_{3}t}\Delta_2(\eta\lambda_{3},c)
 -\gamma(e^{\lambda_{3}t}+qe^{\eta\lambda_{3}t})^{2}
 +\frac{\rho e^{\lambda_1t}(e^{\lambda_{3}t}
 +qe^{\eta\lambda_{3}t})}{1+e^{\lambda_1t}}\\
&\leq qe^{\eta\lambda_{3}t}\Delta_2(\eta\lambda_{3},c)
 +\rho e^{(\lambda_1+\lambda_{3})t}+q\rho e^{\lambda_1t+\eta\lambda_{3}t}\\
&\leq e^{\eta\lambda_{3}t}(q\Delta_2(\eta\lambda_{3},c)+\rho
 +q\rho e^{\lambda_1t}).
\end{aligned}
\end{equation}
Note $\Delta_2(\eta\lambda_{3},c)<0$. Let $q>1$ be large enough,
then $-t_{3}>0$ is also large enough such that
\[
q\Delta_2(\eta\lambda_{3},c)+\rho+q\rho e^{\lambda_1t_{3}}
=q[\Delta_2(\eta\lambda_{3},c)+\rho e^{\lambda_1t_{3}}]+\rho<0,
\]
which leads to $q\Delta_2(\eta\lambda_{3},c)+\rho+q\rho e^{\lambda_1t}<0$
for $t\leq t_{3}$.

If $t\geq t_{3}$ and  $\overline{\psi}(t)=K_2$, then
from the definition of $K_1$, $K_2$ we have
\begin{equation} \label{gt3}
\begin{aligned}
&d_2(J_2*\overline{\psi})(t)-d_2\overline{\psi}(t)-c\overline{\psi'}(t)
+\gamma \overline{\psi}(t)(\delta-\overline{\psi}(t))
+\frac{\rho\overline{\varphi}(t)\overline{\psi}(t)}{1+\overline{\varphi}(t)}\\
&\leq \gamma K_2(\delta-K_2)+\rho K_1K_2=0.
\end{aligned}
\end{equation}
Otherwise, $\overline{\psi}(t)=v_0+v_0e^{-\lambda t}$,
$\overline{\varphi}(t)\leq u_0+u_0e^{-\lambda t}$
implies that
\begin{align*}
&d_2(J_2*\overline{\psi})(t)-d_2\overline{\psi}(t)-c\overline{\psi'}(t)
 +\gamma \overline{\psi}(t)(\delta-\overline{\psi}(t))
 + \frac{\rho\overline{\varphi}(t)\overline{\psi}(t)}{1+\overline{\varphi}(t)}\\
&\leq d_2\int_{R}J_2(y-t)(v_0+v_0e^{-\lambda y})dy-d_2(v_0+v_0e^{-\lambda t})
 +c\lambda v_0e^{-\lambda t}\\
&\quad +\gamma(v_0+v_0e^{-\lambda t})(\delta-v_0-v_0e^{-\lambda t})
+\frac{\rho\overline{\varphi}(t)(v_0+v_0e^{-\lambda t})}{1+\overline{\varphi}(t)}\\
&\leq v_0e^{-\lambda t}\Delta_2(-\lambda,c)+\gamma\delta v_0
 -\gamma(v_0+v_0e^{-\lambda t})^{2}
 +\frac{\rho(u_0+u_0e^{-\lambda t})(v_0+v_0e^{-\lambda t})}{1+u_0
 +u_0e^{-\lambda t}}\\
&= v_0e^{-\lambda t}\Delta_2(-\lambda,c)-2\gamma v_0^{2}
 e^{-\lambda t}-\gamma v_0^{2}e^{-2\lambda t}-\frac{\rho u_0v_0}{1+u_0}
 +\frac{\rho u_0v_0(1+e^{-\lambda t})^{2}}{1+u_0+u_0e^{-\lambda t}}\\
&= v_0e^{-\lambda t}\big[\Delta_2(-\lambda,c)-\gamma v_0
 +\frac{\frac{1}{2}\rho u_0}{1+u_0+u_0e^{-\lambda t}}\big]\\
&\quad -v_0\big[\gamma v_0e^{-\lambda t}+\gamma v_0e^{-2\lambda t}
 +\frac{\rho u_0}{1+u_0}-\frac{\rho u_0(1+\frac{3}{2}e^{-\lambda t}
 +e^{-2\lambda t})}{1+u_0+u_0e^{-\lambda t}}\big].
\end{align*}
Note  that $\Delta_2(0,c)-\gamma v_0
 +\frac{\frac{1}{2}\rho u_0}{1+u_0+u_0e^{-\lambda t}}\leq\gamma\delta-\gamma v_0
 + \frac{\frac{1}{2}\rho u_0}{1+u_0}<0$.
Thus  there exists a constant $\lambda_2^{*}$ such that
$\Delta_2(-\lambda,c)-\gamma v_0+\frac{\frac{1}{2}\rho u_0}{1+u_0
 +u_0e^{-\lambda t}}<0$ for any $\lambda\in(0,\lambda_2^{*})$.

On the other hand,  from \eqref{3.7*} we have
\begin{align*}
 I_2(\lambda,t):
&=\gamma v_0e^{-\lambda t}+\gamma v_0e^{-2\lambda t}
 +\frac{\rho u_0}{1+u_0}-\frac{\rho u_0(1+\frac{3}{2}e^{-\lambda t}
 +e^{-2\lambda t})}{1+u_0+u_0e^{-\lambda t}}\\
 &\geq  \gamma v_0e^{-\lambda t}+\gamma v_0e^{-2\lambda t}
 +\frac{\rho u_0}{1+u_0}-\frac{\rho u_0(1+\frac{3}{2}e^{-\lambda t}
 +e^{-2\lambda t})}{1+u_0}\\
 &= \gamma v_0e^{-\lambda t}+\gamma v_0e^{-\lambda t}
 -\frac{\rho u_0}{1+u_0}(\frac{3}{2}e^{-\lambda t}+e^{-2\lambda t})\\
&= e^{-\lambda t}\big[\gamma v_0-\frac{3}{2}\frac{\rho u_0}{1+u_0}
 +(\gamma v_0-\frac{\rho u_0}{1+u_0})e^{-\lambda t}\big]>0,
\end{align*}
uniformly for $t>t_{3}$. Therefore, $\overline{\psi}(t)$ satisfies the
definition of  upper solution.

 Now we consider $\underline{\varphi}(t)$.
If $t\leq t_2$, then $\underline{\varphi}(t)=e^{\lambda_1t}-qe^{\eta\lambda_1t}$,
$0\leq \overline{\psi}(t)\leq e^{\lambda_{3}t}+qe^{\eta\lambda_{3}t}$. 
Thus we have
\begin{equation} \label{lt2}
\begin{aligned}
&d_1(J_1*\underline{\varphi})(t)-d_1\underline{\varphi}(t)
 -c\underline{\varphi'}(t)+\alpha \underline{\varphi}(t)
 (\beta-\underline{\varphi}(t))-
\frac{\underline{\varphi}(t)\overline{\psi}(t)}{1+\underline{\varphi}(t)}\\
&\geq d_1\int_{R}J_1(y-t)(e^{\lambda_1y}-qe^{\eta\lambda_1y})dy
 -d_1(e^{\lambda_1t}-qe^{\eta\lambda_1t})\\
&\quad-c(\lambda_1e^{\lambda_1t}-q\eta\lambda_1e^{\eta\lambda_1t})
 +\alpha(e^{\lambda_1t}-qe^{\eta\lambda_1t})[\beta-e^{\lambda_1t}
 +qe^{\eta\lambda_1t}]\\
&\quad -\frac{(e^{\lambda_1t}-qe^{\eta\lambda_1t})\overline{\psi}(t)}
 {1+e^{\lambda_1t}-qe^{\eta\lambda_1t}}\\
&= e^{\lambda_1t}\Delta_1(\lambda_1,c)
 -qe^{\eta\lambda_1t}\Delta_1(\eta\lambda_1,c)
  -\alpha(e^{\lambda_1t}-qe^{\eta\lambda_1t})^{2} \\
&\quad  -\frac{(e^{\lambda_1t}-qe^{\eta\lambda_1t})\overline{\psi}(t)}
 {1+e^{\lambda_1t}-qe^{\eta\lambda_1t}}\\
&\geq -qe^{\eta\lambda_1t}\Delta_1(\eta\lambda_1,c)
 -\alpha(e^{\lambda_1t}-qe^{\eta\lambda_1t})^{2}
 -\frac{(e^{\lambda_1t}-qe^{\eta\lambda_1t})(e^{\lambda_{3}t}
 +qe^{\eta\lambda_{3}t})}{1+e^{\lambda_1t}-qe^{\eta\lambda_1t}}\\
&\geq -qe^{\eta\lambda_1t}\Delta_1(\eta\lambda_1,c)-\alpha e^{2\lambda_1t}
 -e^{\lambda_1t}(e^{\lambda_{3}t}+qe^{\eta\lambda_{3}t})\\
&\geq -e^{\eta\lambda_1t}[q\Delta_1(\eta\lambda_1,c)+\alpha+1
 +qe^{(\lambda_1+\eta\lambda_{3}-\eta\lambda_1)t}].
\end{aligned}
\end{equation}
Note that $\Delta_1(\eta\lambda_1,c)<0$ by Lemma \ref{L3.1}, and
from \eqref{3.8} that $\lambda_1+\eta\lambda_{3}-\eta\lambda_1>0$.
Let $q>1$ be large enough, then $-t_2>0$ is also enough such that
\begin{align*}
&q\Delta_1(\eta\lambda_1,c)+\alpha+1+qe^{(\lambda_1+\eta\lambda_{3}
 -\eta\lambda_1)t_2}\\
&= q[\Delta_1(\eta\lambda_1,c)+e^{(\lambda_1+\eta\lambda_{3}-\eta\lambda_1)t_2}]
+(\alpha+1)<0,
\end{align*}
which leads to
$$
-e^{\eta\lambda_1t}[q\Delta_1(\eta\lambda_1,c)+\alpha+1+
qe^{(\lambda_1+\eta\lambda_{3}-\eta\lambda_1)t}]\geq0\quad\text{for  } t\leq t_2.
$$

If $t\geq t_2$, $\underline{\varphi}(t)=u_0-\varepsilon_1e^{-\lambda t}$, 
$0\leq\overline{\psi}(t)\leq v_0+v_0e^{-\lambda t}$, then
\begin{align*}
&d_1(J_1*\underline{\varphi})(t)-d_1\underline{\varphi}(t)
 -c\underline{\varphi'}(t)+\alpha \underline{\varphi}(t)
 (\beta-\underline{\varphi}(t))
 -\frac{\underline{\varphi}(t)\overline{\psi}(t)}{1+\underline{\varphi}(t)}\\
&\geq d_1\int_{R}J_1(y-t)(u_0-\varepsilon_1e^{-\lambda y})dy
 -d_1(u_0-\varepsilon_1e^{-\lambda t})- c\lambda \varepsilon_1e^{-\lambda t}\\
&\quad +\alpha(u_0-\varepsilon_1e^{-\lambda t})[\beta-(u_0-\varepsilon_1
 e^{-\lambda t})]-\frac{(u_0-\varepsilon_1e^{-\lambda t})
 \overline{\psi}(t)}{1+u_0-\varepsilon_1e^{-\lambda t}}\\
&=-\varepsilon_1e^{-\lambda t}\Delta_1(-\lambda,c)+\alpha\beta u_0
 -\alpha(u_0-\varepsilon_1e^{-\lambda t})^{2}
 -\frac{(u_0-\varepsilon_1e^{-\lambda t})\overline{\psi}(t)}
 {1+u_0-\varepsilon_1e^{-\lambda t}}\\
&= \varepsilon_1e^{-\lambda t}[-\Delta_1(-\lambda,c)+(4-2\sqrt{2})\alpha u_0]
 +(2\sqrt{2}-2)\alpha u_0\varepsilon_1e^{-\lambda t}
 -\alpha\varepsilon_1^{2}e^{-2\lambda t}\\
&\quad +\frac{u_0v_0}{1+u_0}-\frac{(u_0-\varepsilon_1e^{-\lambda t})
 \overline{\psi}(t)}{1+u_0-\varepsilon_1e^{-\lambda t}}.
\end{align*}
Note that $-\Delta_1(0,c)+(4-2\sqrt{2})\alpha u_0=-\alpha\beta+(4-2\sqrt{2})\alpha u_0>0$ 
by the second inequality in \eqref{3.7*}. We can choose $\lambda_{3}^{*}>0$ 
such that $-\Delta_1(-\lambda,c)+(4-2\sqrt{2})\alpha u_0>0$ for 
$\lambda\in(0,\lambda_{3}^{*})$.

Let
\[
\begin{aligned}
&I_{3}(\lambda,t)\\
&:=(2\sqrt{2}-2)\alpha u_0\varepsilon_1e^{-\lambda t}
-\alpha\varepsilon_1^{2}e^{-2\lambda t}+\frac{u_0v_0}{1+u_0}-
\frac{(u_0-\varepsilon_1e^{-\lambda t})\overline{\psi}(t)}{1+u_0-
\varepsilon_1e^{-\lambda t}}\\
&\geq (2\sqrt{2}-2)\alpha u_0\varepsilon_1e^{-\lambda t}
-\alpha\varepsilon_1^{2}e^{-2\lambda t}+\frac{u_0v_0}{1+u_0}-
\frac{(u_0-\varepsilon_1e^{-\lambda t})(v_0+v_0e^{-\lambda t})}{1+u_0-
\varepsilon_1e^{-\lambda t}}.
\end{aligned}
\]
By Lemma \ref{L3.2}, we have
$$
I_{3}(\lambda,0)\geq(2\sqrt{2}-2)\alpha u_0\varepsilon_1-\alpha\varepsilon_1^{2}+
\frac{u_0v_0}{1+u_0}-\frac{2v_0(u_0-\varepsilon_1)}{1+u_0-\varepsilon_1}
>\varepsilon_0>0.
$$
Choose $\xi_1>1$ satisfying
\begin{equation}\label{3.10}
(2\sqrt{2}-2)\alpha u_0\varepsilon_1\xi-\alpha(\varepsilon_1\xi)^{2}+
\frac{u_0v_0}{1+u_0}-\frac{(v_0+v_0\xi)(u_0-\varepsilon_1\xi)}
{1+u_0-\varepsilon_1\xi}>\frac{\varepsilon_0}{2}>0
\end{equation}
for $\xi\in[1,\xi_1]$.
Let  $\lambda_{3}^{*}>0$ be small enough, such that
$e^{-\lambda^*_3 t_2}\leq \xi_1$.
For given $\lambda\in(0,\lambda_{3}^{*})$, $t\in [t_2,0]$, 
the above relations leads to  $e^{-\lambda t}\in [1,\xi_1]$. 
Therefore,  $I_{3}(\lambda,t)>0$ for $t\in [t_2,0]$.

Let $t>0$ and  $x:=e^{-\lambda t}\in (0,1)$, 
$\bar{a}:=(2\sqrt{2}-2)\alpha u_0$, $\bar{b}:=\frac{u_0v_0}{1+u_0}$. Then
\begin{align*}
&(2\sqrt{2}-2)\alpha u_0\varepsilon_1e^{-\lambda t}
-\alpha\varepsilon_1^{2}e^{-2\lambda t}+\frac{u_0v_0}{1+u_0}-
\frac{(u_0-\varepsilon_1e^{-\lambda t})(v_0+v_0e^{-\lambda t})}{1+u_0-
\varepsilon_1e^{-\lambda t}}\\
&= \bar{a}\varepsilon_1x
-\alpha\varepsilon_1^{2}x^2+\bar{b}-
\frac{(u_0-\varepsilon_1x)(v_0+v_0x)}{1+u_0-
\varepsilon_1x}\\
&= \frac{1}{1+u_0-
\varepsilon_1x}[\bar{a}\varepsilon_1(1+u_0)x-\bar{a}\varepsilon_1^2x^2
 -\alpha(1+u_0)\varepsilon_1^2x^2+\alpha \varepsilon_1^3x^3+\bar{b}(1+u_0)\\
&\quad-\bar{b}\varepsilon_1 x-u_0v_0-u_0v_0x+\varepsilon_1v_0x+\varepsilon_1v_0x^2]
\\
&= \Big(\bar{a}\varepsilon_1(1+u_0)x-\bar{a}\varepsilon_1^2x^2
 -\alpha(1+u_0)\varepsilon_1^2x^2+\alpha \varepsilon_1^3x^3
 -\bar{b}\varepsilon_1 x
 -u_0v_0x+\varepsilon_1v_0x\\
&\quad+\varepsilon_1v_0x^2\Big)
\big/(1+u_0- \varepsilon_1x)
\\
&= \frac{\tilde{f}(x)}{1+u_0-\varepsilon_1x}.
\end{align*}
Here $\tilde{f}(x):=x\tilde{g}(x)=x\{\tilde{a}x^2+\tilde{b}x+\tilde{c}\}$ and
\begin{gather*}
\tilde{a}:=\alpha \varepsilon_1^3,\quad 
\tilde{b}:=\varepsilon_1v_0-\bar{a}\varepsilon_1^2-\alpha(1+u_0)\varepsilon_1^2
=\varepsilon_1[v_0-\bar{a}\varepsilon_1-\alpha(1+u_0)\varepsilon_1],\\
\tilde{c}:=\bar{a}\varepsilon_1(1+u_0)-\bar{b}\varepsilon_1-u_0v_0
 +\varepsilon_1v_0.
\end{gather*}
Let $\varepsilon_1=(\sqrt{2}-1)u_0$, then we have from the assumption  
$\alpha u_0\geq \frac{(3+2\sqrt{2})v_0}{1+u_0}$  that
\begin{align*}
\tilde{c}
&= (2\sqrt{2}-2)(\sqrt{2}-1)\alpha u_0^2(1+u_0)-u_0v_0
 +\frac{\varepsilon_1v_0}{1+u_0}\\
&\geq  2(3-2\sqrt{2})(3+2\sqrt{2})u_0v_0-u_0v_0+\frac{\varepsilon_1v_0}{1+u_0}\\
&= 2u_0v_0-u_0v_0+\frac{\varepsilon_1v_0}{1+u_0}=u_0v_0
 +\frac{\varepsilon_1v_0}{1+u_0}>0.
\end{align*}
There are three cases.
\smallskip

\noindent\textbf{Case 1:} 
If $\tilde{b}\geq 0$, then a similar discussion 
to $I_1(\lambda,t)$ yields to $\tilde{g}(x)\geq 0$, and then 
$I_3(\lambda,t)\geq 0$ for $\lambda\in \in(0,\lambda_{3}^{*})$ and  $t>0$.
\smallskip

\noindent\textbf{Case 2:}
 If $\tilde{b}< 0$ and $\tilde{b}^2-4\tilde{a}\tilde{c}\leq 0$, then the equation 
$\tilde{g}(x)=0$ has either no real roots or a double real root.
Therefore, $g(x)\geq 0$ for $x\in (0,\infty)$, which leads to 
$I_3(\lambda,t)\geq 0$ for $\lambda \in(0,\lambda_{3}^{*})$ and  $t>0$.
\smallskip

\noindent\textbf{Case 3:} 
If $\tilde{b}< 0$ and $\tilde{b}^2-4\tilde{a}\tilde{c}> 0$, then the 
equation $\tilde{g}(x)=0$ has two real roots: 
$x_{1,2}=\frac{1}{2\tilde {a}}(-\tilde{b}\pm\sqrt{\tilde{b}^2-4\tilde{a}\tilde{c}})$. 
Consider the small root 
$x_1=\frac{1}{2\tilde {a}}(-\tilde{b}-\sqrt{\tilde{b}^2-4\tilde{a}\tilde{c}})$. 
We derive from $\alpha u_0\geq \frac{(3+2\sqrt{2})v_0}{1+u_0}$  that
\begin{gather*}
\alpha u_0(1+u_0)\geq (3+2\sqrt{2})v_0>(\sqrt{2}+1)v_0=\frac{v_0}{\sqrt{2}-1},\\
2(\sqrt{2}-1)^2\alpha u_0+(\sqrt{2}-1)\alpha u_0(1+u_0)-v_0
 >2(\sqrt{2}-1)^2\alpha u_0^2,\\
-\tilde{b}=\bar{a}\varepsilon_1+\alpha(1+u_0)\varepsilon_1-v_0
>2\alpha \varepsilon_1^2=2\tilde{a},\\
-\tilde{b}-2\tilde{a}>0.
\end{gather*}
Let $x_1<1$, which is equivalent to
$$
-\tilde{b}-2\tilde{a}<\sqrt{\tilde{b}^2-4\tilde{a}\tilde{c}}
\; \Leftrightarrow \; 
(\tilde{b}+2\tilde{a})^2<\tilde{b}^2-4\tilde{a}\tilde{c}\;
\Leftrightarrow \; \tilde{a}+\tilde{b}+\tilde{c}<0.
$$
It is  derived that
\begin{align*}
\tilde{a}+\tilde{b}+\tilde{c}
&=(\sqrt{2}-1)^3\alpha u_0^3-[2(\sqrt{2}-1)\alpha u_0
 +\alpha (1+u_0)](\sqrt{2}-1)^2 u_0^2\\
&\quad +[2(\sqrt{2}-1)\alpha u_0-\frac{u_0v_0}{1+u_0}+2v_0](\sqrt{2}-1) u_0-u_0v_0\\
&=-(\sqrt{2}-1)^3\alpha u_0^3-(\sqrt{2}-1)^2\alpha u_0^2(1+u_0)
 +2(\sqrt{2}-1)^2\alpha u_0^2\\
&\quad -\frac{(\sqrt{2}-1)u_0^2v_0}{1+u_0}+2(\sqrt{2}-1)u_0v_0-u_0v_0<0.
\end{align*}
Therefore,  $x_1<1$. Since we already known that $\tilde{f}(1)=\tilde{g}(1)>0$ 
(see \eqref{3.10}) and $\tilde{g}(0)=\tilde{c}>0$, by the graph of $\tilde{g}$ 
we know that $x_2<1$, which leads to 
$0>\tilde{b}+2\tilde{a}>\sqrt{\tilde{b}^2-4\tilde{a}\tilde{c}}$. 
This is a contradiction, and thus Case 3 is impossible. 
Now, we conclude  $\tilde{g}(x)>0$ for $x\in (0,1)$. 
That is, $I_{3}(\lambda,t)>0$ for $t>0$.
Summarizing the above discussion, we know that $\underline{\varphi}(t)$ 
satisfies the definition of  lower solution.


 Now we consider $\underline{\psi}(t)$.
If $t\leq t_{4}$, then $\underline{\psi}(t)=e^{\lambda_{3}t}-qe^{\eta\lambda_{3}t}$, 
$\underline{\varphi}(t)\geq0$. Noting $\eta\leq 2$, we have
\begin{align*}
&d_2(J_2*\underline{\psi})(t)-d_2\underline{\psi}(t)
 -c\underline{\psi'}(t)+\gamma \underline{\psi}(t)(\delta-\underline{\psi}(t))
 +\frac{\rho\underline{\varphi}(t)\underline{\psi}(t)}{1+\underline{\varphi}(t)}\\
&\geq d_2\int_{R}J_2(y-t)(e^{\lambda_{3}y}-qe^{\eta\lambda_{3}y})dy
 -d_2(e^{\lambda_{3}t}-qe^{\eta\lambda_{3}t})
 +c(\lambda_{3}e^{\lambda_{3} t}-\eta\lambda_{3}qe^{\eta\lambda_{3}t})\\
&\quad +\gamma(e^{\lambda_{3}t}-qe^{\eta\lambda_{3}t})
 (\delta-e^{\lambda_{3}t}+qe^{\eta\lambda_{3}t})
\\
&= -qe^{\eta\lambda_{3}t}\Delta_2(\eta\lambda_{3},c)-
\gamma(e^{\lambda_{3}t}-qe^{\eta\lambda_{3}t})^{2}\\
&\geq -qe^{\eta\lambda_{3}t}\Delta_2(\eta\lambda_{3},c)-\gamma e^{2\lambda_{3}t}\\
&\geq e^{\eta\lambda_{3}t}(-q\Delta_2(\eta\lambda_{3},c)-\gamma)\geq0
\end{align*}
for $q>1$ large enough.

Note that the function $y=\frac{x}{1+x}$ is nondecreasing for 
$x\in (-1,\infty)$. If 
$\underline{\varphi}(t)\geq u_0-\varepsilon_1e^{-\lambda t}>-1$ for $t\geq t_4$, 
then
$$
\frac{\underline{\varphi}(t)}{1+\underline{\varphi}(t)}
\geq \frac{ u_0-\varepsilon_1e^{-\lambda t}}{1+ u_0-\varepsilon_1e^{-\lambda t}}.
$$
In fact, if $\lambda>0$ is small enough, one can have 
$(\sqrt{2}-1)e^{-\lambda  t_4}<\frac{u_0+1}{u_0}$, which leads to
 $$
u_0-(\sqrt{2}-1)u_0e^{-\lambda t_4}>-1 \;\Rightarrow\; u_0-\varepsilon_1
e^{-\lambda t}>-1\quad\text{for }\varepsilon_1<(\sqrt{2}-1)u_0, t\geq t_4.
$$
Now, if $t\geq t_{4}$, then $\underline{\psi}(t)=v_0-\varepsilon_2e^{-\lambda t}$ 
and  $\underline{\varphi}(t)\geq u_0-\varepsilon_1e^{-\lambda t}>-1$ 
(assuming $\lambda>0$ small). Thus
\begin{align*}
& d_2(J_2*\underline{\psi})(t)-d_2\underline{\psi}(t)-c\underline{\psi'}(t)
 +\gamma \underline{\psi}(t)(\delta-\underline{\psi}(t))
 + \frac{\rho\underline{\varphi}(t)\underline{\psi}(t)}{1+\underline{\varphi}(t)}\\
&\geq d_2\int_{R}J_2(y-t)(v_0-\varepsilon_2e^{-\lambda y})dy
 -d_2(v_0-\varepsilon_2e^{-\lambda t})-c\lambda\varepsilon_2 e^{-\lambda t}\\
&\quad +\gamma(v_0-\varepsilon_2e^{-\lambda t})(\delta-v_0
 +\varepsilon_2e^{-\lambda t})+\frac{\rho\underline{\varphi}(t)
 (v_0-\varepsilon_2e^{-\lambda t})}{1+\underline{\varphi}(t)}\\
&\geq -\varepsilon_2e^{-\lambda t}\Delta_2(-\lambda,c)
 +\gamma\delta v_0-\gamma(v_0-\varepsilon_2e^{-\lambda t})^{2}
 +\frac{\rho \underline{\varphi}(t)(v_0-\varepsilon_2
 e^{-\lambda t})}{1+u_0-\varepsilon_1e^{-\lambda t}}\\
&=-\varepsilon_2e^{-\lambda t}\Delta_2(-\lambda,c)+\gamma\delta v_0-
\gamma v_0^{2}+2\varepsilon_2\gamma v_0e^{-\lambda t}
 -\gamma\varepsilon_2^{2}e^{-2\lambda t} \\
&\quad +\frac{\rho\underline{\varphi}(t)(v_0
 -\varepsilon_2e^{-\lambda t})}{1+u_0-\varepsilon_1e^{-\lambda t}}\\
&\geq \varepsilon_2e^{-\lambda t}[\Delta_2(-\lambda,c)
 +\gamma v_0-\frac{\frac{m-1}{m}\rho(u_0-\varepsilon_1
 e^{-\lambda t})}{1+u_0-\varepsilon_1e^{-\lambda t}}]
 +\varepsilon_2\gamma v_0e^{-\lambda t}-
\gamma\varepsilon_2^{2}e^{-2\lambda t}\\
&\quad -\frac{\rho u_0v_0}{1+u_0}
 +\frac{\rho v_0(u_0-\varepsilon_1e^{-\lambda t})}{1+u_0-\varepsilon_1e^{-\lambda t}}
 -\frac{\frac{1}{m}\rho\varepsilon_2e^{-\lambda t}(u_0-\varepsilon_1
 e^{-\lambda t})}{1+u_0-\varepsilon_1e^{-\lambda t}}.
\end{align*}
Here $m$ is some positive integer. Note
$$
\Delta_2(0,c)+\gamma v_0-\frac{\frac{m-1}{m}\rho(u_0-\varepsilon_1
e^{-\lambda t})}{1+u_0-\varepsilon_1e^{-\lambda t}}
\geq -\gamma\delta+\gamma v_0-\frac{\frac{m-1}{m}\rho u_0}{1+u_0}>0,
$$ 
we can choose $\lambda_{4}^{*}>0$ such that 
\[
\Delta_2(-\lambda,c)+\gamma v_0
-\frac{\frac{m-1}{m}\rho(u_0-\varepsilon_1e^{-\lambda t})}
{1+u_0-\varepsilon_1e^{-\lambda t}}>0
\]
 for $\lambda\in(0,\lambda_{4}^{*})$.

Let
\begin{align*}
I_{4}(\lambda,t)
&:=\varepsilon_2\gamma v_0e^{-\lambda t}
 -\gamma\varepsilon_2^{2}e^{-2\lambda t}
 -\frac{\rho u_0v_0}{1+u_0}+\frac{\rho v_0(u_0-\varepsilon_2e^{-\lambda t})}
 {1+u_0-\varepsilon_1e^{-\lambda t}}\\
&\quad -\frac{\frac{1}{m}\rho\varepsilon_2e^{-\lambda t}(u_0-\varepsilon_1
 e^{-\lambda t})}{1+u_0-\varepsilon_1e^{-\lambda t}}\\
&= \varepsilon_2\gamma v_0e^{-\lambda t}-\gamma\varepsilon_2^{2}
 e^{-2\lambda t}-\frac{\rho v_0\varepsilon_1e^{-\lambda t}}{(1+u_0)
 (1+u_0-\varepsilon_1e^{-\lambda t})} \\
&\quad -\frac{\frac{1}{m}\rho\varepsilon_2e^{-\lambda t}(u_0
 -\varepsilon_1e^{-\lambda t})}{1+u_0-\varepsilon_1e^{-\lambda t}}.
\end{align*}
Obviously, we have $I_{4}(\lambda,\infty)=0$. By Lemma \ref{L3.2}, we obtain
$$
I_{4}(\lambda,0)=\varepsilon_2\gamma v_0-\gamma\varepsilon_2^{2}
-\frac{\rho v_0\varepsilon_1}{(1+u_0)(1+u_0-\varepsilon_1)}
-\frac{\frac{1}{m}\rho\varepsilon_2(u_0-\varepsilon_1)}{1+u_0-\varepsilon_1}
>\frac{\varepsilon_0}{2}>0
$$
if $m$ is large.
Similar to the argument for $I_3(\lambda,t)$, we can also obtain that 
$I_{4}(\lambda,t)>0$ uniformly for $t\geq t_{4}$. We omit the details here. 
Therefore, $\underline{\psi}(t)$ satisfies the definition of lower solution.

Taking $\lambda\in(0,\min_{1\leq i\leq 4}\{\lambda_{i}^{*}\})$, we see 
that $(\overline{\varphi}(t),\overline{\psi}(t))$ and 
$(\underline{\varphi}(t),\underline{\psi}(t))$ is a pair of upper-lower 
solutions of \eqref{eq3.3}.
\end{proof}


From the above statements, the following result is obvious. 
Note that the following asymptotic behaviors of $(\varphi(t),\psi(t))$ 
can be obtained from the forms of lower solutions and sandwich method.

\begin{theorem}\label{T3.1} 
Assume \eqref{3.7} holds and $c>c^{*}=\max\{c_1^*,c_2^*\}$. 
Then \eqref{eq3.1} has a traveling wave solution $(\varphi,\psi)$  with speed
$c$ connecting $E_0=(0,0)$ and $E_{3}=(u_0,v_0)$. Furthermore, 
\begin{equation}\label{last}
\lim_{t\to-\infty}\varphi(t)e^{-\lambda_1t}
=\lim_{t\to-\infty}\psi(t)e^{-\lambda_{3}t}=1.
\end{equation}
\end{theorem}


 \subsection{The particular case  $\delta=0$}

 If $\delta=0$, then \eqref{eq3.1} reduces to
\begin{equation}\label{eq3.21}
\begin{gathered}
\begin{aligned}
\frac{\partial u(x,t)}{\partial t}
&=d_1(J_1*u)(x,t)-d_1u(x,t)+\alpha u(x,t)(\beta-u(x,t))\\
&\quad -\frac{u(x,t)v(x,t)}{1+u(x,t)},
\end{aligned} \\
\frac{\partial v(x,t)}{\partial t}=d_2(J_2*v)(x,t)-d_2v(x,t)
-\gamma v^{2}(x,t)+\rho\frac{u(x,t)v(x,t)}{1+u(x,t)}.
\end{gathered}
\end{equation}
This is a P-P system with a mortality rate function  $\gamma v^{2}$ 
for the predator, which depends  on the prey as its unique resource.
 \eqref{eq3.21} has three equilibria
$$
{E}_0=(0,0),\quad {E}_1=(\beta,0),\quad 
{E}_2=(u_0,v_0),
$$
where the positive equilibrium ${E}_2=(u_0,v_0)$ satisfying
\begin{equation}\label{3.22}
\alpha\beta-\alpha u_0-\frac{v_0}{1+u_0}=0,\quad 
\gamma v_0-\frac{\rho u_0}{1+u_0}=0.
\end{equation}
It is clear that $u_0<\beta$.

Assume that $c>0$, let $u(x,t)=\varphi(x+ct)$, $v(x,t)=\psi(x+ct)$, 
and denote the traveling wave coordinate $x+ct$ still by $t$. 
Then we can obtain a wave profile system \eqref{eq3.3} with $\delta=0$. 
Similar to  $\S$3.1, consider the traveling waves connecting $E_0$ and $E_2$.
Here, we can choose ${K}_1=\beta$,  ${K}_2=\frac{\rho\beta}{\gamma}$.

Let us consider Lemma \ref{L3.2}. Assume that
\begin{equation}\label{3.71}
\alpha \beta>\frac{(4+2\sqrt{2})v_0}{1+u_0}
\end{equation}
which yields to
\begin{equation}\label{3.51}
\begin{gathered}
\alpha u_0=\alpha\beta-\frac{v_0}{1+u_0} \geq \frac{(3+2\sqrt{2})v_0}{1+u_0},\\
\alpha u_0=\alpha\beta-\frac{v_0}{1+u_0}>\alpha\beta-\frac{\alpha\beta}{4+2\sqrt{2}}
=\frac{(2+\sqrt{2})\alpha\beta}{4}.
\end{gathered}
\end{equation}
It is easy to see, that $\delta$ has no effect on the selection of $\varepsilon_1$,
thus under the first inequality of \eqref{3.51}, we still have 
$\varepsilon_1\in(0,(\sqrt{2}-1)u_0]$ such that the first inequality 
in \eqref{3.6} holds. Since $\delta=0$, we have 
$\gamma v_0=\frac{\rho u_0}{1+u_0}$ (thus the second inequality of \eqref{3.5} 
is false), and we need to re-choose a $\varepsilon_2\in(0,\frac{v_0}{2}]$ 
and to find an inequality similar to the second one of \eqref{3.6}.
Let
\begin{gather*}
g_1(\varepsilon_2)= -\gamma\varepsilon_2^{2}+2\gamma v_0
\varepsilon_2,\\
g_2(\varepsilon_2)= \frac{\rho u_0v_0}{1+u_0}
+\frac{\rho u_0\varepsilon_2}{50(1+u_0)}
-\frac{\rho(u_0-\varepsilon_1)(v_0-\varepsilon_2)}{1+u_0-\varepsilon_1}.
\end{gather*}
Note that
\begin{gather*}
\max\{g_1(\varepsilon_2)\}=g_1(v_0)=\gamma v_0 ^{2},\quad 
g_1(\frac{v_0}{2})=\frac{3}{4}\gamma v_0^{2},\\
g_1(0)=0,\quad 
g_2(0)=\frac{\rho u_0v_0}{1+u_0}
-\frac{\rho(u_0-\varepsilon_1)v_0}{1+u_0-\varepsilon_1}>0.
\end{gather*}
If $\varepsilon_1\in(0,(\sqrt{2}-1)u_0]$, then
\begin{align*}
g_2(\frac{v_0}{2})
&=\frac{\rho u_0v_0}{1+u_0}+\frac{\rho u_0v_0}{100(1+u_0)}
-\frac{\frac{1}{2}\rho(u_0-\varepsilon_1)v_0
}{1+u_0-\varepsilon_1}\\
&\leq \frac{\rho u_0v_0}{1+u_0}+\frac{\rho u_0v_0}{100(1+u_0)}
-\frac{\frac{2-\sqrt{2}}{2}\rho u_0v_0
}{1+(2-\sqrt{2})u_0}\\
&\leq \frac{\rho u_0v_0}{1+u_0}+\frac{\rho u_0v_0}{100(1+u_0)}
-\frac{\frac{2-\sqrt{2}}{2}\rho u_0v_0
}{1+u_0}\\
&= \frac{\sqrt{2}}{2}\frac{\rho u_0v_0}{1+u_0}+
\frac{\rho u_0v_0}{100(1+u_0)}\\
&= (\frac{\sqrt{2}}{2}+\frac{1}{100})\gamma v_0^{2}
<g_1(\frac{v_0}{2}),
\end{align*}
 then there exist $\varepsilon_2^{*}\in (0,v_0/2]$ such that 
$g_1(\varepsilon_2^{*})=g_2(\varepsilon_2^{*})$ and 
\begin{equation}\label{epsilon2}
 g_1(\varepsilon_2)> g_2(\varepsilon_2)\quad \text{for }
 \varepsilon_2^{*}<\varepsilon_2\leq \frac{v_0}{2}.
\end{equation}
In particular, one can choose $ \varepsilon_2=v_0/2$.

In this subsection, let $\delta_2=0$, and then  $\lambda_3=\lambda_{3}(c)=0$ 
for $c>c_2^*=0$. Correspondingly,
 $c>c^*:= c^*_1$. We  denote $\lambda_{i}(c)$ by $\lambda_{i}$, where $i=1,2,3,4$.
 Assume that $\lambda_4>\lambda_1$. Choose $\zeta>0$ with 
$\lambda_1>\lambda_{4}-\zeta>0$, $q>1$ and
\begin{equation} \label{3.23}
\eta\in\big(1,\min\{\frac{\lambda_2}{\lambda_1},\frac{\lambda_1+\lambda_{4}-\zeta}
{\lambda_1},2\}\big).
\end{equation}
Define continuous functions
\begin{gather*}
\overline{\varphi}(t)=\begin{cases}
e^{\lambda_1 t}, &t \leq t_1,\\
\min\{{K}_1,u_0+u_0e^{-\lambda t}\},& t \geq t_1,
\end{cases}
\quad
\underline{\varphi}(t)=\begin{cases}
e^{\lambda_1 t}-qe^{\eta\lambda_1 t}, & t \leq t_2,\\
u_0-\varepsilon_1e^{-\lambda t}, & t \geq t_2,
\end{cases}\\
\overline{\psi}(t)=\begin{cases}
e^{\lambda_{4} t}+qe^{(\lambda_{4}-\zeta)t}, & t \leq t_{3},\\
\min\{{K}_2,v_0+v_0e^{-\lambda t}\}, &t \geq t_{3},
\end{cases}
\quad
\underline{\psi}(t)=\begin{cases}
0, &t \leq t_{4},\\
v_0-\varepsilon_2e^{-\lambda t}, & t \geq t_{4},
\end{cases}
\end{gather*}
where $q>1$ is large enough and $\lambda>0$ is small which will be defined later.
We can see that
$\overline{\varphi}(t)$, $\overline{\psi}(t)$, $\underline{\varphi}(t)$,
$\underline{\psi}(t)$  satisfy (P1)--(P2) in Section 2.
 Moreover, if $q>1$ is large enough, then it is clear that
$$
t_2<0,\quad t_{3}<0,\quad t_4<0,\quad t_1 \geq \max \{t_2,t_{3}\}.
$$

\begin{lemma}\label{L3.5}
Assume that \eqref{3.71} holds and  $\lambda_4>\lambda_1$, $q>1$ is large 
enough and $\lambda>0$ is small. Then
$(\overline{\varphi}(t),\overline{\psi}(t))$ is an upper solution and
$(\underline{\varphi}(t),\underline{\psi}(t))$ is a lower solution of \eqref{eq3.21}.
\end{lemma}

\begin{proof} 
 We omit the argument for  $\overline{\varphi}(t)$ since it is similar
to the corresponding argument in Lemma \ref{L3.3}.

  We now consider $\overline{\psi}(t)$.
If $t\leq t_{3}$, then 
$\overline{\psi}(t)=e^{\lambda_{4}t}+qe^{(\lambda_{4}-\zeta)t}$,  
$\overline{\varphi}(t)\leq e^{\lambda_1t}$. Using 
$\lambda_4$ and $\lambda_4-\zeta$ to replace $\lambda_3$ and 
$\eta\lambda_3$ in \eqref{lt3} respectively, we obtain
\begin{align*}
& d_2(J_2*\overline{\psi})(t)-d_2\overline{\psi}(t)-c\overline{\psi'}(t)
 -\gamma \overline{\psi}^{2}(t)
 + \frac{\rho\overline{\varphi}(t)\overline{\psi}(t)}{1+\overline{\varphi}(t)}\\
&\leq qe^{(\lambda_{4}-\zeta)t}\Delta_2(\lambda_{4}-\zeta,c)
-\gamma (e^{\lambda_{4}t}+qe^{(\lambda_{4}-\zeta)t})^2
+\frac{\rho e^{\lambda_1t}\cdot (e^{\lambda_{4}t}+qe^{(\lambda_{4}-\zeta)t})}
{1+e^{\lambda_1t}}\\
&\leq e^{(\lambda_{4}-\zeta)t}[q\Delta_2(\lambda_{4}-\zeta,c)
 -\gamma q^2e^{(\lambda_{4}-\zeta)t}+\rho+\rho qe^{\lambda_1t}].
\end{align*}
Note $\Delta_2(\lambda_{4}-\zeta,c)<0$. Let $q>1$ be large enough, 
then $\rho-\gamma q<0$, and thus we have from $\lambda_1>\lambda_{4}-\zeta>0$ 
and $t_3<0$ that
\begin{align*}
& q\Delta_2(\lambda_{4}-\zeta,c)-\gamma q^2e^{(\lambda_{4}-\zeta)t_3}+
\rho qe^{\lambda_1t_3}+\rho\\
&\leq q[\Delta_2(\lambda_{4}-\zeta,c)+e^{\lambda_1t_3}(\rho-\gamma q)]+\rho<0,
\end{align*}
which leads to
$$
e^{(\lambda_{4}-\zeta)t}[q\Delta_2(\lambda_{4}-\zeta,c)
-\gamma q^2e^{(\lambda_{4}-\zeta)t}+\rho+\rho qe^{\lambda_1t}]<0
\quad \text{for } t\leq t_{3}.
$$

If $t\geq t_{3}$ and $\overline{\psi}(t)=K_2$, then  the proof is similar 
to \eqref{gt3}. Otherwise, $\overline{\psi}(t)=v_0+v_0e^{-\lambda t}$, 
$\overline{\varphi}(t)\leq u_0+u_0e^{-\lambda t}$
implies 
\begin{align*}
&d_2(J_2*\overline{\psi})(t)-d_2\overline{\psi}(t)-c\overline{\psi'}(t)
-\gamma \overline{\psi}^{2}(t)+
\frac{\rho\overline{\varphi}(t)\overline{\psi}(t)}{1+\overline{\varphi}(t)}\\
&\leq v_0e^{-\lambda t}\Delta_2(-\lambda,c)-
\gamma(v_0+v_0e^{-\lambda t})^{2}+
\frac{\rho(u_0+u_0e^{-\lambda t})(v_0+v_0e^{-\lambda t})}{1+u_0+u_0e^{-\lambda t}}\\
&= v_0e^{-\lambda t}\Delta_2(-\lambda,c)-
\frac{\rho u_0v_0(1+e^{-\lambda t})^{2}}{1+u_0}
+ \frac{\rho u_0v_0(1+e^{-\lambda t})^{2}}{1+u_0+u_0e^{-\lambda t}}
\\
&= v_0e^{-\lambda t}\Delta_2(-\lambda,c)
 -\frac{\rho u_0^{2}v_0e^{-\lambda t}(1+e^{-\lambda t})^{2}}
 {(1+u_0)(1+u_0+u_0e^{-\lambda t})}\\
&\leq  v_0e^{-\lambda t}[\Delta_2(-\lambda,c)
 -\frac{\rho u_0^{2}}{(1+u_0)(1+u_0+u_0e^{-\lambda t})}].
\end{align*}
Note that for $t\geq t_3$, 
\begin{align*}
\Delta_2(0,c)-\frac{\rho u_0^{2}}{(1+u_0)(1+u_0+u_0e^{-\lambda t})}
&=-\frac{\rho u_0^{2}}{(1+u_0)(1+u_0+u_0e^{-\lambda t})}\\
&\leq -\frac{\rho u_0^{2}}{(1+u_0)(1+u_0+u_0e^{-\lambda t_3})}<0,
\end{align*}
and thus  there exists a constant $\lambda_2^{*}$ such that
$\Delta_2(-\lambda,c)-\frac{\rho u_0^{2}}{(1+u_0)(1+u_0+u_0e^{-\lambda t})}<0$ 
for any $\lambda\in(0,\lambda_2^{*})$.
Therefore, $\overline{\psi}(t)$ satisfies the definition of  upper solution.


 If $t\leq t_2$, then 
$0\leq \underline{\varphi}(t)=e^{\lambda_1t}-qe^{\eta\lambda_1t}\leq e^{\lambda_1t}$, 
$0\leq\overline{\psi}(t)\leq e^{\lambda_{4}t}+qe^{(\lambda_{4}-\zeta)t}$. 
Using $\lambda_4-\zeta$ to replace $\eta\lambda_3$ in \eqref{lt2},   
we have from \eqref{3.23} that
\begin{align*}
& d_1(J_1*\underline{\varphi})(t)-d_1\underline{\varphi}(t)
 -c\underline{\varphi'}(t)+\alpha \underline{\varphi}(t)
 (\beta-\underline{\varphi}(t))-
\frac{\underline{\varphi}(t)\overline{\psi}(t)}{1+\underline{\varphi}(t)}\\
&\geq -e^{\eta\lambda_1t}[q{\Delta}_1(\eta\lambda_1,c)+(\alpha+1) 
+q e^{(\lambda_1+\lambda_{4}-\zeta-\eta\lambda_1)t}].
\end{align*}
The coming argument for $t\leq t_2$, and a further  discussion for $t\geq t_2$
are  the same as the corresponding argument in Lemma \ref{L3.3}, so we omit it.
Then we know that $\underline{\varphi}(t)$ satisfies the definition of  
lower solution.

 If $t\leq t_{4}$, $\underline{\psi}(t)=0$, $\underline{\varphi}(t)\geq0$, 
the following inequality is clear:
$$
d_2(J_2*\underline{\psi})(t)-d_2\underline{\psi}(t)
-c\underline{\psi'}(t)-\gamma \underline{\psi}^{2}(t)
+ \frac{\rho\underline{\varphi}(t)\underline{\psi}(t)}{1+\underline{\varphi}(t)}
\geq 0.
$$
If $t\geq t_{4}$, then $\underline{\psi}(t)=v_0-\varepsilon_2e^{-\lambda t}$, 
$\underline{\varphi}(t)\geq u_0-\varepsilon_1e^{-\lambda t}>-1$ 
($\lambda>0$ small), and thus
\begin{align*}
& d_2(J_2*\underline{\psi})(t)-d_2\underline{\psi}(t)-c\underline{\psi'}(t)
-\gamma \underline{\psi}^{2}(t)
+ \frac{\rho\underline{\varphi}(t)\underline{\psi}(t)}{1+\underline{\varphi}(t)}\\
&\geq -\varepsilon_2e^{-\lambda t}\Delta_2(-\lambda,c)
 -\gamma(v_0-\varepsilon_2e^{-\lambda t})^{2}
 +\frac{\rho(u_0-\varepsilon_1e^{-\lambda t})(v_0-\varepsilon_2
 e^{-\lambda t})}{1+u_0-\varepsilon_1e^{-\lambda t}}\\
&= \varepsilon_2e^{-\lambda t}[-{\Delta}_2(-\lambda,c)+\frac{\rho u_0}{50(1+u_0)}]
+2\varepsilon_2\gamma v_0e^{-\lambda t}-\gamma\varepsilon_2^{2}e^{-2\lambda t}
 -\frac{\rho u_0v_0}{1+u_0}\\
&\quad -\frac{\rho u_0\varepsilon_2e^{-\lambda t}}{50(1+u_0)}
 +\frac{\rho(u_0-\varepsilon_1e^{-\lambda t})(v_0-\varepsilon_2
 e^{-\lambda t})}{1+u_0-\varepsilon_1e^{-\lambda t}}.
\end{align*}
Note $-{\Delta}_2(0,c)+\frac{\rho u_0}{50(1+u_0)}=\frac{\rho u_0}{50(1+u_0)}>0$,
thus  there exists a constant $\lambda_{4}^{*}$ such that 
$-{\Delta}_2(-\lambda,c)+\frac{\rho u_0}{50(1+u_0)}>0$ for any
$\lambda\in(0,\lambda_{4}^{*})$.

Let
\begin{align*}
&I_{4}(\lambda,t)\\
&:=2\varepsilon_2\gamma v_0e^{-\lambda t}-\gamma\varepsilon_2^{2}
 e^{-2\lambda t}-\frac{\rho u_0v_0}{1+u_0}
 -\frac{\rho u_0\varepsilon_2e^{-\lambda t}}{50(1+u_0)}
 +\frac{\rho(u_0-\varepsilon_1e^{-\lambda t})(v_0-\varepsilon_2
 e^{-\lambda t})}{1+u_0-\varepsilon_1e^{-\lambda t}}\\
&=\frac{99}{50}\varepsilon_2\gamma v_0e^{-\lambda t}
 -\gamma\varepsilon_2^{2}e^{-2\lambda t}-\frac{\rho u_0v_0}{1+u_0}
 +\frac{\rho(u_0-\varepsilon_1e^{-\lambda t})(v_0
 -\varepsilon_2e^{-\lambda t})}{1+u_0-\varepsilon_1e^{-\lambda t}}.
\end{align*}
Obviously, we have $I_{4}(\lambda,\infty)=0$ and
\begin{align*}
I_{4}(\lambda,0)
&=g_1(\varepsilon_2)+g_2(\varepsilon_2)\\
&=2\varepsilon_2\gamma v_0-\gamma\varepsilon_2^{2}
 -\frac{\rho u_0v_0}{1+u_0}-\frac{\rho u_0\varepsilon_2}{50(1+u_0)}
+\frac{\rho(u_0-\varepsilon_1)(v_0-\varepsilon_2)}{1+u_0-\varepsilon_1}.
\end{align*}
In view of \eqref{epsilon2}, there exists $\varepsilon_2\in(0,\frac{v_0}{2}]$ 
such that $I_{4}(\lambda,0)>\varepsilon_0>0$ where $\varepsilon_0>0$ 
is a constant. Similar to the argument in \eqref{3.10}, we can also obtain 
$I_{4}(\lambda,t)>0$ uniformly for $t\in [ t_{4},0]$ and 
$\lambda\in(0,\lambda_{4}^{*})$.

Now we show that $I_4(\lambda,t)>0$ uniformly for $t>0$. 
Let $x=e^{-\lambda t}$ and $\kappa=\frac{99}{50}$. Then
\begin{align*}
I_{4}(\lambda,t)
&=\kappa\varepsilon_2\gamma v_0x-\gamma\varepsilon_2^{2}x^2
 -\frac{\rho u_0v_0}{1+u_0}
 +\frac{\rho(u_0-\varepsilon_1x)(v_0-\varepsilon_2x)}{1+u_0-\varepsilon_1x}\\
&=\frac{1}{1+u_0-\varepsilon_1x}\{[\kappa\varepsilon_2\gamma v_0x
 -\gamma\varepsilon_2^{2}x^2-\frac{\rho u_0v_0}{1+u_0}]
 [1+u_0-\varepsilon_1x] \\
&\quad +\rho(u_0-\varepsilon_1x)(v_0-\varepsilon_2x)\}\\
&=\frac{1}{1+u_0-\varepsilon_1x}\{x[\breve{a}x^2+\breve{b}x
 +\breve{c}]\}\\
&=:\frac{1}{1+u_0-\varepsilon_1x}\{x\breve{g}(x)\},
\end{align*}
where
\begin{gather*}
\breve{a}:=\gamma \varepsilon_1\varepsilon_2^2,\quad 
\breve{b}:=-\kappa \varepsilon_1\varepsilon_2\gamma v_0
 -\varepsilon_2^2\gamma(1+u_0)+\rho \varepsilon_1\varepsilon_2,\\
\breve{c}:=\kappa \varepsilon_2\gamma v_0(1+u_0)-\rho\varepsilon_1v_0
-\rho \varepsilon_2u_0+\frac{\rho\varepsilon_1u_0v_0}{1+u_0}.
\end{gather*}

Let $\varepsilon_1=(\sqrt{2}-1)u_0$ and $\varepsilon_2=\frac{v_0}{2}$. 
Then from $\frac{\rho u_0}{1+u_0}=\gamma v_0$  we have
\begin{gather*}
\begin{aligned}
\breve{c}
&=\kappa \varepsilon_2\gamma v_0(1+u_0)-\rho\varepsilon_1v_0
 -\rho \varepsilon_2u_0+\frac{\rho\varepsilon_1u_0v_0}{1+u_0}\\
&=\gamma v_0^2[\frac{\kappa}{2}-(\sqrt{2}-1)-\frac{1}{2}]
 +\gamma u_0v_0^2[\frac{\kappa}{2}-(\sqrt{2}-1)-\frac{1}{2}+(\sqrt{2}-1)]>0,
\end{aligned}\\
\begin{aligned}
\breve{b}&=-\kappa \varepsilon_1\varepsilon_2\gamma v_0-\varepsilon_2^2\gamma(1+u_0)
 +\rho \varepsilon_1\varepsilon_2\\
&=\gamma v_0^2[-\frac{1}{4}+\frac{\sqrt{2}-1}{2}]
 +\gamma u_0v_0^2[-\frac{\kappa(\sqrt{2}-1)}{2}-\frac{1}{4}
 +\frac{\sqrt{2}-1}{2}]<0,\\
\end{aligned}
\\
\begin{aligned}
&\breve{a}+\breve{b}+\breve{c}\\
&=\gamma v_0^2[-\frac{1}{4}+\frac{\sqrt{2}-1}{2}+\frac{\kappa}{2}-(\sqrt{2}-1)
-\frac{1}{2}]\\
&\quad +\gamma u_0v_0^2[\frac{\sqrt{2}-1}{4}-\frac{\kappa(\sqrt{2}-1)}{2}
-\frac{1}{4}+\frac{\sqrt{2}-1}{2}+\frac{\kappa}{2}-\frac{1}{2}]\\
&=\gamma v_0^2[-\frac{3}{4}-\frac{\sqrt{2}-1}{2}+\frac{\kappa}{2}]
+\gamma u_0v_0^2[-\frac{3}{4}+\frac{3(\sqrt{2}-1)}{4}
-\frac{\kappa(\sqrt{2}-1)}{2}+\frac{\kappa}{2}]>0.
\end{aligned}
\end{gather*}
Since $\breve{a}+\breve{b}+\breve{c}>0$ is equivalent to $x_1>1$, where
 $x_1$ is the smallest real root of $\breve{g}(x)=0$, similar to proving 
procedure of Lemma \ref{L3.3}, we know that $I_{4}(\lambda,t)>0$ uniformly 
for $t>0$ and $\lambda\in(0,\lambda_{4}^{*})$.
 Therefore, $\underline{\psi}(t)$ satisfies the definition of  lower solution.

 Let $\lambda^*=\min_{i=1,2,3,4}\{\lambda_i^*\}$. Summarizing the above discussion,
we know that $(\overline{\varphi}(t),\overline{\psi}(t))$ and
$(\underline{\varphi}(t),\underline{\psi}(t))$ satisfy the definition of
 upper-lower solutions.
\end{proof}


From the above statements, the following result is obvious.

\begin{theorem}\label{T3.3}  
Assume \eqref{3.71} holds, $c>c^{*}:=c_1^*$ and $\lambda_4>\lambda_1$. 
Then \eqref{eq3.21} has a traveling wave solution with speed
$c$ connecting ${E}_0=(0,0)$ and ${E}_2=(u_0,v_0)$.
 \end{theorem}


\subsection*{Concluding remarks}

Using the upper-lower solution method, we proved  the existence 
of traveling wavefronts connecting  $E_0$ and $E_3$ for \eqref{eq1.1} 
with $h_1=\alpha u(\beta-u),h_2=\gamma v(\delta -v)$ and $f$  
of Holling type II, with either $\delta>0$ or $\delta=0$.

Assume that $\delta>0$. Note that the predator in  system \eqref{eq3.1} has 
logistic growth, which implies that it has
resources other than the prey. Therefore, it is possible to consider the 
immigration of the prey into the residence area of predator,
 and further a co-existence steady state may be reached  as $x+ct\to\infty$.   
Similar arguments show the existence  of traveling waves  connecting $E_2$ 
and $E_3$.

We did not discuss the existence of traveling wavefronts connecting $E_1$ 
and $E_3$,  which raised in the case of mild invasions.
Furthermore, we don't know if  traveling wave solutions can be studied using 
the  upper-lower solution technique under one of the following situations:
 \begin{itemize}
 \item[(i)] $h_1=\alpha u(\beta-u),h_2=-dv$ and  $f=\frac{u}{1+u}$ (Holling type  II);
 \item[(ii)] $h_1=\alpha u(\beta-u),h_2=\gamma v(\delta -v)$ and $f=u$ (Holling type  I).
 \end{itemize}
 These problems are still open, and we leave it for future investigation.

\subsection*{Acknowledgments}
Research  is supported by the NSF of China (11171120), and
by the Doctoral Program of Higher Education of China (20094407110001).


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\end{document}