\documentclass[reqno]{amsart}
\usepackage{hyperref}

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2017 (2017), No. 102, pp. 1--13.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu}
\thanks{\copyright 2017 Texas State University.}
\vspace{8mm}}

\begin{document}
\title[\hfilneg EJDE-2017/102\hfil Uniqueness of traveling wave solutions]
{Uniqueness of traveling wave solutions for non-monotone cellular neural
networks with distributed delays}

\author[H.-L. Zhou, Z. Yu \hfil EJDE-2017/102\hfilneg]
{Hui-Ling Zhou, Zhixian Yu}

\address{Hui-Ling Zhou \newline
College of Science,
University of Shanghai for Science and Technology,
Shanghai 200093, China}
\email{hlzhou@163.com}

\address{Zhixian Yu (corresponding author) \newline
College of Science,
University of Shanghai for Science and Technology,
Shanghai 200093, China}
\email{zxyu0902@163.com}


\thanks{Submitted March 1, 2017. Published April 11, 2017.}
\subjclass[2010]{35C07, 92D25, 35B35}
\keywords{ Cellular neural network; uniqueness; non-monotone;
\hfill\break\indent asymptotic behavior; distributed delays}

\begin{abstract}
 In this article,  we study the uniqueness of traveling wave solutions
 for non-monotone cellular neural networks with distributed delays.
 First we establish a priori asymptotic behavior of the traveling wave 
 solutions at infinity. Then, based on Ikehara's theorem, we prove the 
 uniqueness of the solution $\psi(n-ct)$ with $c\leq c_*$, where $c_*<0$
 is the critical wave speed.
\end{abstract}

\maketitle
\numberwithin{equation}{section}
\newtheorem{theorem}{Theorem}[section]
\newtheorem{lemma}[theorem]{Lemma}
\newtheorem{proposition}[theorem]{Proposition}
\newtheorem{remark}[theorem]{Remark}
\allowdisplaybreaks

\section{Introduction}

  In this article, we study the uniqueness of traveling wave solution
for the non-monotone cellular neural networks with distributed delays
\begin{equation} \label{1.1}
\begin{aligned}
x'_n(t)&=-x_n(t)+\sum_{i=1}^m\int_0^\tau a_iJ_i(y)f(x_{n-i}(t-y))dy\\
&\quad +\alpha\int_0^\tau J_{m+1}(y)f(x_n(t-y))dy
+\sum_{j=1}^l\beta_j\int_0^\tau J_{m+1+j}(y)f(x_{n+j}(t-y))dy,
\end{aligned}
\end{equation}
where the constants $n\in Z$, $m,l\in N$, $\tau\geq 0$, and the varible
 $t\in \mathbb{R}$.
We use the following assumptions:
\begin{itemize}
\item[(H0)]  (i) $\alpha>0$, $a_1>0$, $a_i\geq 0$ ($i=2,\ldots,m$),
$\beta_1>0$ and $\beta_j\geq0$ $(j=2,\ldots,l)$.
 $a=\sum_{i=1}^ma_i$ and $\beta=\sum_{j=1}^l\beta_j$.
\\
 (ii) $J_i:[0,\tau]\to (0,+\infty)$ is
the piecewise continuous function satisfying
$\int_0^{\tau}J_i(y)dy=1$, where $0<\tau<\infty$.

\item[(H1)] $f\in C([0,b],[0,\frac{b}{a+\alpha+\beta}])$,
$f(0)=0$, $\alpha f'(0)\geq 1$ and there exists $K>0$ with $K\leq b$ such that
\[
(a+\alpha+\beta)f(K)=K,\quad |f(u)-f(v)|\leq f'(0)|u-v|\quad
\text{for }u,v\in[0,b].
\]

\item[(H2)] $(a+\alpha+\beta)f(u)>u$ for $u\in(0,K)$ and
$(a+\alpha+\beta)f(u)<u$ for $u\in(K,b]$.

\item[(H3)] There exist $\sigma>0$, $\delta>0$ and $M>0$ such that
\begin{equation*}
f(u)\geq f'(0)u-Mu^{1+\sigma} \quad\text{for }u\in[0,\delta].
\end{equation*}
\end{itemize}

A traveling wave solution \eqref{1.1} with speed $c$  is a nonnegative bounded
solution of the form $u_n(t) = \psi(n -ct)$ satisfying $\psi(-\infty)=0$
and $\liminf_{\xi\to +\infty}\psi(\xi)>0$. Substituting
$u_n(t) = \psi(n -ct)$ in \eqref{1.1}, we have the
wave profile equation
\begin{equation} \label{1.2}
\begin{aligned}
-c\psi'(\xi)
&=-\psi(\xi)+\sum_{i=1}^m\int_0^\tau a_iJ_i(y)f(\psi(\xi-i+cy))dy \\
&\quad +\alpha\int_0^\tau J_{m+1}(y)f(\psi(\xi+cy))dy \\
&\quad +\sum_{j=1}^l\beta_j\int_0^\tau J_{m+1+j}(y)f(\psi(\xi+j+cy))dy.
\end{aligned}
\end{equation}

When the output function $f$ is monotone, the existence of traveling wave solutions
for many versions of CNNs \eqref{1.1} with delays or without delays has been
widely investigated. see for example
\cite{Hsu-Li-Yang,Hsu-Lin-Shen,Hsu-Yang1,Hsu-Yang2,Hsu-Yang,Li-Huang,
Li-Li,Liu-Weng-Xu,Weng-Wu,Wu-Hsu-12,Wu-Hsu-14,Yu-Mei,Yu-Yuan-T,Yu-Yuan-Hsu-Jiang}.
The existence of entire solutions for \eqref{1.1} has been investigated by
 Wu and Hsu \cite{Wu-Hsu-12,Wu-Hsu-14}.
Letting $J_i=\delta(y-\tau_i)$, $i=1,\dots,m+l+1$, \eqref{1.1} reduces to the
multiple discrete delays equation
\begin{equation} \label{0.2}
\begin{aligned}
w_n'(t)&=-w_n(t)+\sum_{i=1}^ma_if(w_{n-i}(t-\tau_i))
 +\alpha f(w_n(t-\tau_{m+1})) \\
&\quad +\sum_{j=1}^l\beta_jf(w_{n+j}(t-\tau_{m+1+j})).
\end{aligned}
\end{equation}
Yu and Mei \cite{Yu-Mei} investigated uniqueness and stability of traveling
wave solutions for \eqref{0.2} with the monotone output function. In \cite{Yu-Mei}
the authors used the technique in \cite{Carr-Chmaj} to study uniqueness
of travelling wave soluitons for \eqref{0.2} with discrete delays.
We will extend this method to \eqref{1.1} with distributed delays.

For the non-monotone output function $f$, Yu et al.\ \cite{Yu-Yuan-Hsu-Peng}
only established the existence of non-critical traveling wave solutions.
 Yu and Zhao \cite{Yu-Zhao} further established the existence of the spreading
speed, its coincidence with the minimal wave speed and the existence of
critical waves for the non-monotone DCNNs \eqref{1.1}. We summarize the
existence of traveling wave solutions of \eqref{1.1} with the non-monotone output
function in \cite{Yu-Yuan-Hsu-Peng,Yu-Zhao} as follows.

\begin{proposition}\label{prop1.1}
Assume that {\rm (H0)-(H3)} hold. Then there exists $c^*<0$
(which is given in Lemma \ref{lem2.1}) such that for any $c\leq c^*$,
\eqref{1.1} admits a non-negative traveling wave solution
$\psi(n-ct)$ with the wave
speed $c^*<0$ and satisfying
\begin{equation}\label{1.4}
\psi(-\infty)=0\quad \text{and}\quad
0<\liminf_{\xi\to +\infty}\psi(\xi)\leq\limsup_{\xi\to +\infty}\psi(\xi)\leq b.
\end{equation}
\end{proposition}

The uniqueness of monotone travelling wave solutions for various evolution systems
 has been established; see for example
 \cite{Aguerrea,Berestyski-Nirenberg,Chen,Chen-Fu-Guo,Ma-Wu,
Ma-Zou} and the references therein.
 The proof of uniqueness strongly relies on the monotonicity of travelling waves.
It seems very difficult to extend the techniques in those literatures to
the non-monotone evolution systems because the wave profile may lose
the monotonicity and the study of the corresponding uniqueness is very limited,
see, e.g., \cite{Diekmann-Kaper,Fang-Zhao,Fang-Wei-Zhao1}.
Recently, the authors in \cite{Yu-12,Yu-Mei-C13} extend the technique
in \cite{Carr-Chmaj} to non-monotone lattice equations with discrete delays.
In this article, we extend the technique in \cite{Carr-Chmaj} to non-monotone
CNNs  with distributed delays.

The rest of this article is organized as follows.
Section 2 is devoted to studying the asymptotic behavior of the traveling wave
solutions.
In Section 3, we prove the uniqueness of the solution.

\section{Asymptotic behavior of traveling wave solutions}

In this section, we consider the asymptotic behavior at negative infinity
of any traveling wave solutions of \eqref{1.1}.
The characteristic equation of \eqref{1.2} at 0 is
\begin{equation} \label{2.1}
\begin{aligned}
\Delta(c,\lambda)
&=-c\lambda+1-f'(0)\Big[\sum_{i=1}^m a_i\int_0^\tau J_i(y)e^{\lambda(-i+cy)}dy\\
&\quad +\alpha\int_0^\tau J_{m+1}(y)e^{\lambda cy}dy
 +\sum_{j=1}^l\beta_j\int_0^\tau J_{m+1+j}(y)e^{\lambda(j+cy)}dy\Big].
\end{aligned}
\end{equation}

\begin{lemma}[{\cite[Lemma 2.1]{Yu-Yuan-Hsu-Jiang}}] \label{lem2.1}
Assume that {\rm(H0)} and $\alpha f'(0)\geq1$ hold. Then there
exist a unique pair of $c_*<0$ and $\lambda_*>0$ such that
\begin{itemize}
\item[(i)] $\Delta(c_*,\lambda_*)=0$,
$\frac{\partial\Delta(c,\lambda)}{\partial\lambda}|_{c=c_*,\lambda=\lambda_*}=0$;

\item[(ii)] For any $c>c_*$ and $\lambda\in[0,+\infty)$,
$\triangle(c,\lambda)<0$;

\item[(iii)] For any $c<c_*$, $\triangle(c,\lambda)=0$ has two positive
roots $\lambda_2\geq\lambda_1>0$. Moreover, if $c<c_*$,
$\triangle(c,\lambda)>0$ for any $\lambda\in (\lambda_1,\lambda_2)$;
if $c=c_*$, then $\lambda_1=\lambda_2=\lambda_*$.
\end{itemize}
\end{lemma}

Now we give a different version of Ikehara's Theorem, which can
 be found in \cite{Carr-Chmaj}.

\begin{proposition}\label{prop2.1}
Let $F(\lambda):=\int_0^{+\infty}u(x)e^{-\lambda x}dx$, where $u(x)$
is a positive decreasing function. Assume $F(\lambda)$ can be
written as
\[
F(\lambda)=\frac{h(\lambda)}{(\lambda+\mu)^{k+1}},
\]
where $k>-1$ and $h(\lambda)$ is analytic in the strip $-\mu\leq
\Re\lambda<0$. Then
\begin{align*}
\lim_{x\to  +\infty }\frac{u(x)}{x^k e^{-\mu x}}=\frac{h(-\mu)}{\Gamma(\mu+1)}.
\end{align*}
\end{proposition}

\begin{remark}\label{remark2.1} \rm
Changing the variable $t=-x$, and modifying the proof for Ikehara's
Theorem given in \cite{Ellison}, we can show the following version of
Proposition \ref{prop2.1}.
Let $F(\lambda):=\int_{-\infty}^0 u(t)e^{-\lambda t}dt$, where $u(t)$ is
a positive increasing function. Assume $F(\lambda)$ can be written as
\begin{align*}
F(\lambda)=\frac{h(\lambda)}{(\mu-\lambda)^{k+1}},
\end{align*}
where $k>-1$ and $h(\lambda)$ is analytic in the strip
$\mu-\epsilon< \Re\lambda\leq\mu$ for some $0<\epsilon<\mu$. Then
\begin{align*}
\lim_{x\to  -\infty }\frac{u(x)}{|x|^k e^{\mu x}}=\frac{h(\mu)}{\Gamma(\mu+1)}.
\end{align*}
\end{remark}

To apply Ikehara's Theorem, we need to assure that  traveling wave solutions 
are positive.

\begin{lemma}\label{lem2.2}
Assume that {\rm (H0)--(H3)} hold and let $\psi(n-ct)$ be a non-negative
 traveling wave of \eqref{1.1} with $c\leq c_*$ satisfying
\eqref{1.4}. Then $\psi(\xi)>0$ for $\xi\in\mathbb{R}$.
\end{lemma}

\begin{proof}
 Assume that there exists $\xi_0$ such that $\psi(\xi_0)=0$. Without loss
of generality, we may assume $\xi_0$  is the left-most point. According to
$\psi(\xi)\geq0$ for $\xi\in\mathbb{R}$, we can easily see that $\psi(\xi)$
attains the minimum at $\xi_0$ and $\psi'(\xi_0)=0$.
According to (H0) and (H1), it follows from \eqref{1.2} that
$$
\int_0^\tau J_{m+1}(y)f(\psi(\xi_0+cy))dy=0,
$$
which implies that $f(\psi(\xi_0+cy))=0$ for any $y\in[0,\tau]$.
Thus, choosing some sufficiently small number $y_0>0$, we can obtain
$\psi(\xi_0+cy_0)=0$ according to the continuity of $\psi(\xi)$ and $c<0$.
This contradicts to the choice of $\xi_0$, and completes the proof.
\end{proof}

\begin{lemma}\label{lem2.3}
Assume that {\rm (H1)--(H3)} hold and let $\psi(n-ct)$ be any non-negative
traveling wave of \eqref{1.1} with $c\leq c_*$ and satisfy
\eqref{1.4}. Then there exists a positive number $\rho>0$ such that
$\psi(\xi)=O(e^{\rho\xi})$ as $\xi\to  -\infty$.
\end{lemma}

\begin{proof}
Since $f'(0)(a+\alpha+\beta)>1$, there exists
$\epsilon_0 >0$ such that
\begin{align*}
A:=(1-\epsilon_0)f'(0)(a+\alpha+\beta)-1>0.
\end{align*}
For such $\epsilon_0 >0$, there exist $\delta_1>0$ such that
$f(u)\geq(1-\epsilon_0)f'(0)u$ for any $u\in[0,\delta_1]$. Since
$\psi(-\infty)=0$, there exists $M>0$ and $\forall\xi\leq -M$ such
that $\psi(\xi)<\delta_1 $. Integrating \eqref{1.2} from $\eta$ to $\xi $
with $\xi\leq-l-M$, it follows that
\begin{equation} \label{2.2}
\begin{aligned}
&-c[\psi(\xi)-\psi(\eta)] \\
&=-\int_\eta^\xi \psi(x)dx+\sum_{i=1}^m
a_i\int_\eta^\xi \int_0^\tau J_i(y)f(\psi(x-i+cy))\,dy\,dx \\
&\quad+\alpha\int_\eta^\xi \int_0^\tau
J_{m+1}(y)f(\psi(x+cy))\,dy\,dx \\
&\quad +\sum_{j=1}^l
\beta_j\int_\eta^\xi \int_0^\tau J_{m+1+j}(y)f(\psi(x+j+cy))\,dy\,dx \\
&\geq -\int_\eta^\xi
\psi(x)dx+f'(0)(1-\epsilon_0)\Big[\sum_{i=1}^m a_i\int_\eta^\xi
\int_0^\tau J_i(y)\psi(x-i+cy)\,dy\,dx   \\
&\quad +\alpha\int_\eta^\xi
\int_0^\tau J_{m+1}(y)\psi(x+cy)\,dy\,dx \\
&\quad +\sum_{j=1}^l
\beta_j\int_\eta^\xi \int_0^\tau J_{m+1+j}(y)\psi(x+j+cy)\,dy\,dx\Big] \\
&=A\int_\eta^\xi
\psi(x)dx+f'(0)(1-\epsilon_0)\Big[\alpha\int_\eta^\xi \int_0^\tau
J_{m+1}(y)(\psi(x+cy)-\psi(x))\,dy\,dx \\
&\quad +\sum_{i=1}^m
a_i\int_\eta^\xi \int_0^\tau
J_i(y)(\psi(x-i+cy)-\psi(x))\,dy\,dx  \\
&\quad +\sum_{j=1}^l
\beta_j\int_\eta^\xi \int_0^\tau J_{m+1+j}(y)(\psi(x+j+cy)-\psi(x))\,dy\,dx\Big].
\end{aligned}
\end{equation}
Since $\psi(x)$ is differentiable, we have
\begin{align*}
\int_\eta^\xi(\psi(x-i+cy)-\psi(x))dx
&=\int_\eta^\xi\int_0^{-i+cy}\psi'(x+s)dsdx\\
&=\int_0^{-i+cy}(\psi(\xi+s)-\psi(\eta+s))ds.
\end{align*}
Similarly,
\begin{gather*}
\int_\eta^\xi(\psi(x+cy)-\psi(x))dx=\int_0^{cy}(\psi(\xi+s)-\psi(\eta+s))ds,\\
\int_\eta^\xi(\psi(x+j+cy)-\psi(x))dx=\int_0^{j+cy}(\psi(\xi+s)-\psi(\eta+s))ds.
\end{gather*}
Letting $\eta\to  -\infty$ in \eqref{2.2}, we obtain
\begin{equation} \label{2.3}
\begin{aligned}
&A\int_{-\infty}^\xi\psi(x)dx \\
&\leq-c\psi(\xi)-f'(0)(1-\epsilon_0)\Big[\sum_{i=1}^m
a_i\int_0^\tau\int_0^{-i+cy}J_i(y)\psi(\xi+s)\,ds\,dy\\
&\quad +\alpha\int_0^\tau\int_0^{cy}J_{m+1}(y)\psi(\xi+s)\,ds\,dy\\
&\quad +\sum_{j=1}^l
\beta_j\int_0^\tau\int_0^{j+cy}J_{m+1+j}(y)\psi(\xi+s)\,ds\,dy\Big].
\end{aligned}
\end{equation}
From \eqref{2.3}, we know that $\int_{-\infty}^\xi\psi(x)dx<{+\infty}$.
Letting $\Phi(\xi)=\int_{-\infty}^\xi\psi(x)dx $ and integrating
\eqref{2.3} from ${-\infty}$ to $\xi$, we have
\begin{equation} \label{2.4}
\begin{aligned}
&A\int_{-\infty}^\xi\Phi(x)dx \\
&\leq-c\Phi(\xi)-f'(0)(1-\epsilon_0)\Big[\sum_{i=1}^ma_i
 \int_0^\tau\int_0^{-i+cy}J_{i}(y)\Phi(\xi+s)\,ds\,dy\\
&\quad +\alpha\int_{0}^\tau\int_{0}^{cy}J_{m+1}(y)\Phi(\xi+s)\,ds\,dy \\
&\quad +\sum_{j=1}^l\beta_j\int_0^\tau\int_0^{j+cy}J_{m+1+j}(y)\Phi(\xi+s)\,ds\,dy
 \Big]\\
&\leq\varrho\Phi(\xi+\kappa)
\end{aligned}
\end{equation}
for some $\kappa>0$ and $\varrho>0$ according to the monotonicity of
$\Phi(\xi)$, Letting $\varpi>0$ such that $\varrho<A\varpi$, and for
$\xi\leq-l-M$, it follows that
\begin{equation} \label{2.5}
\Phi(\xi-\varpi)\leq\frac{1}{\varpi}\int_{\xi-\varpi}^\xi\Phi(x)dx
\leq\frac{1}{\varpi}\int_{-\infty}^\xi\Phi(x)dx
\leq\frac{\varrho}{A\varpi}\Phi(\xi+\kappa).
\end{equation}
Define $h(\xi)=\Phi(\xi)e^{-\rho\xi}$, where
$\rho=\frac{1}{\rho+\varpi}\ln\frac{A\varpi}{\varrho}>0$. Hence,
\begin{align*}
h(\xi-\varpi)=\Phi(\xi-\varpi)e^{-\rho(\xi-\varpi)}
\leq\frac{\varrho}{A\varpi}e^{\rho(\kappa+\varpi)}h(\xi+\kappa)=h(\xi+\kappa),
\end{align*}
which implies $h$ is bounded. Therefore, $\Phi(\xi)=O(e^{\rho\xi})$
when $\xi\to {-\infty}$. Integrating \eqref{1.2} from $-\infty$ to
$\xi$, it follows from (H2) that
\begin{equation} \label{2.6}
\begin{aligned}
-c\psi(\xi)
&=\sum_{i=1}^ma_i\int_{0}^\tau\int_{-\infty}^\xi J_i(y)f(\psi(x-i+cy))\,dx\,dy\\
&\quad +\alpha\int_0^\tau\int_{-\infty}^\xi J_{m+1}(y)f(\psi(x+cy))\,dx\,dy\\
&\quad +\sum_{j=1}^l\beta_j\int_0^\tau\int_{-\infty}^\xi J_{m+1+j}(y)f(\psi(x+j+cy))
 \,dx\,dy-\Phi(\xi)\\
&\leq f'(0)\sum_{i=1}^ma_i\int_{0}^\tau\int_{-\infty}^\xi J_i(y)\psi(x-i+cy)\,dx\,dy\\
&\quad +\alpha f'(0)\int_0^\tau\int_{-\infty}^\xi J_{m+1}(y)\psi(x+cy)\,dx\,dy\\
&\quad +f'(0)\sum_{j=1}^l\beta_j\int_0^\tau\int_{-\infty}^\xi J_{m+1+j}(y)
 \psi(x+j+cy)\,dx\,dy-\Phi(\xi)\\
&=-\Phi(\xi)+f'(0)\sum_{i=1}^ma_i\int_{0}^\tau J_i(y)\Phi(\xi-i+cy)dy\\
&\quad +\alpha f'(0)\int_0^\tau J_{m+1}(y)\Phi(\xi+cy)dy\\
&\quad +f'(0)\sum_{j=1}^l\beta_j\int_0^\tau J_{m+1+j}(y)\Phi(\xi+j+cy)dy.
\end{aligned}
\end{equation}
Thus, we have $\psi(\xi)=O(e^{\rho\xi})$ when $\xi\to -\infty$.
With the help of Ikehara's theorem, we obtain the asymptotic
behavior of traveling wave solutions at $-\infty$.
\end{proof}

\begin{proposition}\label{prop2.2}
Assume that {\rm (H1)--(H3)} hold and let $\psi(n-ct)$ be any non-negative
traveling wave of \eqref{1.1} with the wave speed $c\leq c_*$ and satisfy
\eqref{1.4}. Then
\begin{equation}\label{2.7}
\lim_{\xi\to  -\infty}\frac{\psi(\xi)}{e^{\lambda_1 \xi}}\,
\text{ exists for } c< c_*,\quad
\lim_{\xi\to -\infty}\frac{\psi(\xi)}{|\xi|e^{\lambda_* \xi}}\, \text{ exists for }
 c= c_*.
\end{equation}
\end{proposition}

\begin{proof}
According to Lemma \ref{lem2.3}, we define the two-sided
Laplace transform of $\psi$ for $0<\Re\lambda<\rho$,
\begin{align*}
L(\lambda)\equiv\int_{-\infty}^{+\infty}\psi(x)e^{-\lambda x}dx.
\end{align*}
We claim that $L(\lambda)$ is analytic for $0<\Re\lambda<\lambda_1$ and
has a singularity at $\lambda=\lambda_1$. Note that
\begin{align*}
&-c\psi'(\xi)+\psi(\xi)-f'(0)\sum_{i=1}^ma_i\int_0^\tau J_i(y)\psi(\xi-i+cy)dy\\
&-\alpha f'(0)\int_0^\tau  J_{m+1}(y)\psi(\xi+cy)dy\\
&-f'(0)\sum_{j=1}^l\beta_j\int_0^\tau J_i{m+1+j}\psi(\xi+j+cy)dy\\
&=\sum_{i=1}^ma_i\int_0^\tau
 J_i(y)[f(\psi(\xi-i+cy))-f'(0)\psi(\xi-i+cy)]dy\\
&\quad +\alpha \int_0^\tau
 J_{m+1}(y)[f(\psi(\xi+cy))-f'(0)\psi(\xi-i+cy)]dy\\
&\quad +\sum_{j=1}^l\beta_j\int_0^\tau
 J_{m+1+j}[f(\psi(\xi+j+cy))-f'(0)\psi(\xi-i+cy)]dy\\
&=:Q(\psi)(\xi).
\end{align*}
Multiplying the two sides of the above equality by $e^{-\lambda\xi}$
and integrating $\xi$ on $\mathbb{R}$, we obtain
\begin{align}\label{2.8}
\Delta(c,\lambda)L(\lambda)=\int_{-\infty}^{+\infty}e^{-\lambda
x}Q(\psi)(x)dx.
\end{align}
We know that the left-hand side of \eqref{2.8} is analytic for
$0<\Re\lambda<\rho$. According to (H3), for any $\overline{u}>0$,
there exists $\overline{d}>0$ such that
$f(u)\geq f'(0)u-\overline{d}u^{\sigma+1}$, for all
$u\in[0,\overline{u}]$, where
$\overline{d}:=\max\{d,\gamma^{-(\sigma+1)}
\max_{u\in[\gamma,\overline{u}]}\{f'(0)u-f(u)\}\}$. Thus,
\begin{equation} \label{2.800}
\begin{aligned}
&-\overline{d}\Big[\sum_{i=1}^ma_i\int_0^\tau
J_i(y)\psi^{\sigma+1}(\xi-i+cy)dy+\alpha\int_0^\tau
J_{m+1}(y)\psi^{\sigma+1}(\xi+cy)dy \\
&+\sum_{j=1}^l\beta_j\int_0^\tau
J_{m+1+j}\psi^{\sigma+1}(\xi+j+cy)dy\Big] \\
&\leq Q(\psi)(\xi)\leq0.
\end{aligned}
\end{equation}
Choose $\upsilon>0$ such that $\frac{\upsilon}{\sigma}<\rho$. Then
for any $\Re\lambda\in(0,\rho+\upsilon)$, we have
% \label{2.9}
\begin{align*}
&\big|\int_{-\infty}^{+\infty}e^{-\lambda x}Q(\psi)(x)dx\big|\\
&\leq \overline{d}\int_{-\infty}^{+\infty}e^{-\lambda \xi}
 \Big[\sum_{i=1}^ma_i\int_0^\tau  J_i(y)\psi^{\sigma+1}(\xi-i+cy)dy \\
&\quad +\alpha\int_0^\tau
 J_{m+1}(y)\psi^{\sigma+1}(\xi+cy)dy\\
&\quad +\sum_{j=1}^l\beta_j\int_0^\tau
 J_{m+1+j}\psi^{\sigma+1}(\xi+j+cy)dy\Big]d\xi\\
&=\overline{d}[\sum_{i=1}^ma_i\int_0^\tau
 J_i(y)e^{\lambda(-i+cy)}dy+\alpha\int_0^\tau J_{m+1}(y)e^{\lambda cy}dy\\
&\quad +\sum_{j=1}^l\beta_j\int_0^\tau
 J_{m+1+j}e^{\lambda(j+cy)}dy]\int_{-\infty}^{+\infty}e^{-\lambda x}
 \psi^{\sigma+1}(x)dx\\
&\leq \overline{d}\Big[\sum_{i=1}^ma_i\int_0^\tau
 J_i(y)e^{\lambda(-i+cy)}dy+\alpha\int_0^\tau J_{m+1}(y)e^{\lambda cy}dy\\
&\quad +\sum_{j=1}^l\beta_j\int_0^\tau
 J_{m+1+j}e^{\lambda(j+cy)}dy\Big]L(\lambda-\upsilon)
 \Big(\sup_{\xi\in\mathbb{R}}\psi(\xi)e^{\frac{-\upsilon\xi}{\sigma}}
 \Big)^\sigma<+\infty.
\end{align*}
We use properties of Laplace transform \cite[p. 58]{Widder}. Since
$\psi>0$ according to Lemma \ref{lem2.2}, there exists a real number
 $D$ such that $L(\lambda)$ is analytic for $0<\Re\lambda<D$ and $L(\lambda)$
has a singularity at
$\lambda=D$. Thus, when $c\leq c_*$, $L(\lambda)$ is analytic for
$\Re\lambda\in(0,\lambda_1)$ and $L(\lambda)$ has a singularity at
$\lambda=\lambda_1$.

According to \eqref{2.8}, we have
\[
F(\lambda):=\int_{-\infty}^0\psi(x)e^{-\lambda x}dx
=\frac{\int_{-\infty}^{+\infty}e^{-\lambda x}Q(\psi)(x)dx}{\Delta(c,\lambda)}
-\int_0^{+\infty}\psi(x)e^{-\lambda x}dx.
\]
Define $H(\lambda)=F(\lambda)(\lambda_1-\lambda)^{k+1}$, where $k=0$
if $c<c_*$ and $k=1$ if $c=c_*$.

We claim that $H(\lambda)$ is analytic in the strip
$S:=\{\lambda\in C|0<\Re\lambda\leq\lambda_1\}$. Indeed,
 define
\[
G(\lambda)=\frac{\int_{-\infty}^{+\infty}e^{-\lambda x}Q(\psi)(x)dx}
{\Delta(c,\lambda)/(\lambda_1-\lambda)^{k+1}}
=L(\lambda)(\lambda_1-\lambda)^{k+1}.
\]
It is easily seen that $G(\lambda)$ is analytic in the strip $\{\lambda\in
C|0<\Re\lambda<\lambda_1\}$.

To prove that $G(\lambda)$ is analytic for $\Re\lambda=\lambda_1$, we only need
to prove that  $\Delta(c,\lambda)=0$ does not have any zero with
$\Re\lambda=\lambda_1$ other than
$\lambda=\lambda_1$. Indeed, letting
$\lambda=\lambda_1+i\widetilde{\lambda}$, we have
\begin{equation} \label{2.10}
\begin{aligned}
0&=-c\widetilde{\lambda}+1-f'(0)\Big[\sum_{k=1}^ma_k\int_0^\tau
J_k(y)e^{\lambda_1(-k+cy)}\cos(-k+cy)\widetilde{\lambda}dy  \\
&\quad +\alpha\int_0^\tau J_{m+1}(y)e^{\lambda_1cy}\cos cy\widetilde{\lambda}dy \\
&\quad +\sum_{j=1}^l\beta_j\int_0^\tau
J_{m+1+j}(y)e^{\lambda_1(j+cy)}\cos(j+cy)\widetilde{\lambda}dy\Big]
\end{aligned}
\end{equation}
and
\begin{equation} \label{2.11}
\begin{aligned}
0&=-c\widetilde{\lambda}-f'(0)\Big[\sum_{k=1}^ma_k\int_0^\tau
 J_k(y)e^{\lambda_1(-k+cy)}\sin(-k+cy)\widetilde{\lambda}dy \\
&\quad +\alpha\int_0^\tau
 J_{m+1}(y)e^{\lambda_1cy}\sin cy\widetilde{\lambda}dy \\
&\quad +\sum_{j=1}^l\beta_j\int_0^\tau
J_{m+1+j}(y)e^{\lambda_1(j+cy)}\sin(j+cy)\widetilde{\lambda}dy\Big].
\end{aligned}
\end{equation}
It follows from \eqref{2.10} and \eqref{2.11} that $\widetilde{\lambda}=0$.

According to the above argument, $G(\lambda)$
is analytic in $S$,  and $H(\lambda)$ is also analytic in $S$. Moreover,
 we claim that $H(\lambda_1)>0$. Indeed, notice that
$H(\lambda_1)=G(\lambda_1)$. On the other hand,
$\int_{-\infty}^{+\infty}e^{-\lambda_1 x}Q(\psi)(x)dx<0$ according to
\eqref{2.800} and
$\lim_{\lambda\to \lambda_1^-}\Delta(c,\lambda)/(\lambda_1-\lambda)^{k+1}<0$
according to Lemma \ref{lem2.1}.

Since $\psi(\xi)$ may be non-monotone, Ikehara's Theorem could be directly used.
 Thus, we need to make a function transformation, i.e.,
 $\widehat{\psi}(\xi)=\psi(\xi)e^{p\xi}$, where $p=\frac{-1}{c}>0$.
It follows from \eqref{1.2} that
\begin{align*}
\widehat{\psi}'(\xi)
&= \frac{-1}{c}\Big[\sum_{i=1}^m\int_0^\tau
a_iJ_i(y)f(\psi(\xi-i+cy))dy +\alpha\int_0^\tau
J_{m+1}(y)f(\psi(\xi+cy))dy\\
&\quad +\sum_{j=1}^l\beta_j\int_0^\tau
J_{m+1+j}(y)f(\psi(\xi+j+cy))dy\Big]e^{p\xi}>0.
\end{align*}
Therefore, $\widehat{\psi}(\xi)$ is increasing and $\widehat{\psi}(\xi)>0$.
Now we apply the Ikehara's Theorem to $\widehat{\psi}(\xi)$. Let
\[
\widehat{F}(\lambda):=\int_{-\infty}^0\widehat{\psi}(x)e^{-\lambda x}dx
=F(\lambda-p).
\]
Then
\[
\widehat{F}(\lambda)=\frac{\widehat{H}(\lambda)}{((\lambda_1+p)-\lambda)^{k+1}},
\]
where $\widehat{H}(\lambda)=H(\lambda-p)$ is analytic for
$p<\Re\lambda\leq\lambda_1+p$ and $\widehat{H}(\lambda_1+p)=H(\lambda_1)>0$.
According to Remark \ref{remark2.1}, the limits exists and
\[
\lim_{\xi\to -\infty}\frac{\widehat{\psi}(\xi)}{|\xi|^ke^{(\lambda_1+p)\xi}}
=\lim_{\xi\to -\infty}\frac{\psi(\xi)}{|\xi|^ke^{\lambda_1\xi}},
\]
i.e.,
\[
\lim_{\xi\to  -\infty}\frac{\psi(\xi)}{e^{\lambda_1 \xi}}\,
\text{ exists for } c< c_*,\quad
 \lim_{\xi\to -\infty}\frac{\psi(\xi)}{|\xi|e^{\lambda_* \xi}} \text{ exists for }
 c= c_*.
\]
This completes the proof.
\end{proof}

\section{Uniqueness of traveling wave solutions}

In this section, we show the following unique result of traveling wave solutions of
\eqref{1.1}.

\begin{theorem} \label{thm3.1}
Assume that {\rm (H1)--(H3)} hold.
Let $\psi(n-ct)$ be a traveling wave of \eqref{1.1}
with the wave speed $c\leq c_*$, which is given in Proposition \ref{prop1.1}.
 If $\phi(n-ct)$ is any non-negative traveling wave of \eqref{1.1} with the
same wave speed $c$ satisfying \eqref{1.4}, then $\phi$ is a translation
of $\psi$; more precisely, there exists $\bar{\xi}\in\mathbb{R}$ such that
$\phi(n-ct)=\psi(n-ct+\bar{\xi})$.
\end{theorem}

\begin{proof}
 From Proposition \ref{prop2.2}, there exist two positive numbers
$\vartheta_1$ and $\vartheta_2$ such that
\[
\lim_{\xi\to -\infty}\frac{\phi(\xi)}{|\xi|^k e^{\lambda_1 \xi}}=\vartheta_1, \quad
\lim_{\xi\to -\infty}\frac{\psi(\xi)}{|\xi|^k e^{\lambda_1 \xi}}=\vartheta_2
\]
where $k=0$ for $c<c_*$, and $k=1$ for $c=c_*$. For $\epsilon>0$, define
\begin{equation} \label{3.1}
\omega(\xi):=\frac{\phi(\xi)-\psi(\xi+\overline{\xi})}{e^{\lambda_1\xi}}
\text{ for }c<c_*,\quad
\omega_\epsilon(\xi):=\frac{\phi(\xi)-\psi(\xi+\overline{\xi})}{(\epsilon|\xi|+1)
e^{\lambda_*\xi}} \text{ for } c=c_*,
\end{equation}
where
$\overline{\xi}=\frac{1}{\lambda_1}\ln\frac{\vartheta_1}{\vartheta_2}$.
Then $\omega(\pm\infty)=0$ and $\omega_\epsilon(\pm\infty)=0$.

First, we consider $c<c_*$. Since $\omega(\pm\infty)=0$,
$\sup_{\xi\in \mathbb{R}}\{\omega(\xi)\}$ and
$\inf_{\xi\in \mathbb{R}}\{\omega(\xi)\}$ are finite. Without loss of generality,
 we assume $\sup_{\xi\in \mathbb{R}}\{\omega(\xi)\}\geq |\inf_{\xi\in
\mathbb{R}}\{\omega(\xi)\}|$. If $\omega(\xi)\not\equiv0$, there exists
$\xi_0$ such that
\[
\omega(\xi_0)=\max_{\xi\in \mathbb{R}}{\omega(\xi)}
=\sup_{\xi\in \mathbb{R}}{\omega(\xi)}>0 ,\quad \omega'(\xi_0)=0.
\]
We claim that for all $i,j\in\mathbb{Z}$, we have
\[
\omega(\xi_0-i+cy)=\omega(\xi_0+cy)=\omega(\xi_0+j+cy)=\omega(\xi_0)
\]
for $y\in[0,\tau]$. Suppose on the contrary that one of three
inequalities $\omega(\xi_0-i+cy)<\omega(\xi_0)$,
$\omega(\xi_0+cy)<\omega(\xi_0)$ and
$\omega(\xi_0+j+cy)<\omega(\xi_0)$ for some $i_0$, $j_0$ must
hold. According to \eqref{1.2}, \eqref{3.1} and (H2), we obtain
%\label{3.2}
\begin{align*}
0&=c\omega'(\xi_0)\\
&=-c\lambda_1\omega(\xi_0)+\omega(\xi_0) \\
&\quad -e^{-\lambda_1\xi_0}\sum_{i=1}^ma_i\int_0^\tau
 J_i(y)[f(\phi(\xi_0-i+cy))-f(\psi(\xi_0+\overline{\xi}-i+cy))]dy\\
&\quad -e^{-\lambda_1\xi_0}\alpha\int_0^\tau
 J_{m+1}(y)[f(\phi(\xi_0+cy))-f(\psi(\xi_0+\overline{\xi}+cy))]dy\\
&\quad -e^{-\lambda_1\xi_0}\sum_{j=1}^l\beta_j\int_0^\tau
 J_{m+1+j}(y)[f(\phi(\xi_0+j+cy))-f(\psi(\xi_0+\overline{\xi}+j+cy))]dy\\
&>-c\lambda_1\omega(\xi_0)+\omega(\xi_0)-f'(0)\omega(\xi_0)
 \Big[\sum_{i=1}^ma_i\int_0^\tau J_i(y)e^{\lambda_1(-i+cy)}dy\\
&\quad +\alpha\int_0^\tau J_{m+1}(y)e^{\lambda_1cy}dy+\sum_{j=1}^l\beta_j
 \int_0^\tau J_{m+1+j}(y)e^{\lambda_1(cy+j)}dy\Big]\\
&=-\omega(\xi_0)\triangle(c,\lambda_1)=0,
\end{align*}
which is a contradiction. Thus, $\omega(\xi_0+cy_0)=\omega(\xi_0)$ also holds
for $y_0\in (0,\tau)$. Again by bootstrapping,
$\omega(\xi_0+kcy_0)=\omega(\xi_0)$ for all $k\in \mathbb{Z}$ and
$\omega(+\infty)=0$. Therefore, we have
$\phi(\xi)\equiv\psi(\xi+\overline{\xi})$ for $\xi\in \mathbb{R}$, which
contradicts to $\omega(\xi)\not\equiv0$.

Next, we consider $c=c_*$. Assume $\sup_{\xi\in
\mathbb{R}}\{\omega(\xi)\}\geq |\inf_{\xi\in \mathbb{R}}\{\omega(\xi)\}|$. If
$\omega_\epsilon(\xi)\neq0$, there exists $\xi_0^\epsilon$ such that
\begin{align*}
\omega_\epsilon(\xi_0^\epsilon)=\max_{\xi\in
\mathbb{R}}\{\omega_\epsilon(\xi)\}=\sup_{\xi\in \mathbb{R}}\{\omega_\epsilon(\xi)\}>0
,\quad \omega_\epsilon'(\xi_0^\epsilon)=0.
\end{align*}

Now we divide this part into three cases:
\smallskip

\noindent\textbf{Case 1:} Suppose that $\xi_0^\epsilon\to +\infty$ as
$\epsilon\to 0$.  It follows from \eqref{3.1} and \eqref{1.2} that
\begin{align*}
&c_*[\phi'(\xi_0^\epsilon)-\psi'(\xi_0^\epsilon+\overline{\xi})] \\
&= \phi(\xi_0^\epsilon)-\sum_{i=1}^ma_i\int_0^\tau
J_i(y)[f(\phi(\xi_0^\epsilon-i+c_*y))
 -f(\psi(\xi_0^\epsilon+\overline{\xi}-i+c_*y))]dy\\
&\quad -\alpha\int_0^\tau
J_{m+1}(y)[f(\phi(\xi_0^\epsilon+c_*y))
 -f(\psi(\xi_0^\epsilon+\overline{\xi}+c_*y))]dy
 -\psi(\xi_0^\epsilon+\overline{\xi})\\
&\quad -\sum_{j=1}^l\beta_j\int_0^\tau J_{m+1+j}(y)
f(\phi(\xi_0^\epsilon+j+c_*y))
 -f(\psi(\xi_0^\epsilon+\overline{\xi}+j+c_*y))]dy.
\end{align*}
We claim that for all $i,j\in\mathbb{Z}$,
\begin{align*}
\omega_\epsilon(\xi_0^\epsilon-i+c_*y)=\omega_\epsilon(\xi_0^\epsilon+c_*y)=\omega_\epsilon(\xi_0^\epsilon+j+c_*y)=\omega_\epsilon(\xi_0^\epsilon)
\end{align*}
for $y\in[0,\tau]$. Suppose for the contrary that one of three
inequalities
$\omega_\epsilon(\xi_0^\epsilon-i+c_*y)<\omega_\epsilon(\xi_0^\epsilon)$,
$\omega_\epsilon(\xi_0^\epsilon+c_*y)<\omega_\epsilon(\xi_0^\epsilon)$
and
$\omega_\epsilon(\xi_0^\epsilon+j+c_*y)<\omega_\epsilon(\xi_0^\epsilon)$
for some $i_0$ and $j_0$ must hold. Choose $\epsilon>0$ sufficiently small such
that $\xi_0^\epsilon-m+c_*\tau>0$. Thus,
% \label{3.3}
\begin{align*}
&-c_*\omega_\epsilon(\xi_0^\epsilon)\epsilon
 -c_*\lambda_*\omega_\epsilon(\xi_0^\epsilon)(\epsilon\xi_0^\epsilon+1)\\
&\leq-\omega_\epsilon(\xi_0^\epsilon)(\epsilon|\xi_0^\epsilon|+1) \\
&\quad +f'(0)\sum_{i=1}^ma_i\int_0^\tau
J_i(y)e^{\lambda_*(-i+c_*y)}[\epsilon|\xi_0^\epsilon-i+c_*y|+1]
 \omega_\epsilon(\xi_0^\epsilon-i+c_*y)dy\\
&\quad +\alpha f'(0)\int_0^\tau
 J_{m+1}(y)e^{\lambda_*c_*y}[\epsilon|\xi_0^\epsilon+c_*y|+1]
 \omega_\epsilon(\xi_0^\epsilon+c_*y)dy\\
&\quad +f'(0)\sum_{j=1}^l\beta_j\int_0^\tau
J_{m+1+j}(y)e^{\lambda_*(j+c_*y)}[\epsilon|\xi_0^\epsilon+j+c_*y|+1]
 \omega_\epsilon(\xi_0^\epsilon+j+c_*y)dy\\
&<-\omega_\epsilon(\xi_0^\epsilon)(\epsilon|\xi_0^\epsilon|+1)
 +f'(0)\sum_{i=1}^ma_i\int_0^\tau
 J_i(y)e^{\lambda_*(-i+c_*y)}[\epsilon|\xi_0^\epsilon-i+c_*y|+1]\omega_\epsilon(\xi_0^\epsilon)dy\\
&\quad +\alpha f'(0)\int_0^\tau
 J_{m+1}(y)e^{\lambda_*c_*y}[\epsilon|\xi_0^\epsilon+c_*y|+1]\omega_\epsilon(\xi_0^\epsilon)dy\\
&\quad +f'(0)\sum_{j=1}^l\beta_j\int_0^\tau
J_{m+1+j}(y)e^{\lambda_*(j+c_*y)}[\epsilon|\xi_0^\epsilon+j+c_*y|+1]
 \omega_\epsilon(\xi_0^\epsilon)dy,
\end{align*}
it follows that
\begin{equation} \label{3.4}
\begin{aligned}
&-c_*\omega_\epsilon(\xi_0^\epsilon)\epsilon+\omega_\epsilon(\xi_0^\epsilon)(\epsilon\xi_0^\epsilon+1)\Delta(c_*;\lambda_*)\\
&<f'(0)\sum_{i=1}^ma_i\int_0^\tau
J_i(y)e^{\lambda_*(-i+c_*y)}\epsilon(-i+c_*y)\omega_\epsilon(\xi_0^\epsilon)dy\\
&\quad +\alpha f'(0)\int_0^\tau J_{m+1}(y)e^{\lambda_*(c_*y)}\epsilon
c_*y\omega_\epsilon(\xi_0^\epsilon)dy\\
&\quad +f'(0)\sum_{j=1}^l\beta_j\int_0^\tau
J_{m+1+j}(y)e^{\lambda_*(j+c_*y)} [\epsilon(j+c_*y)]
\omega_\epsilon(\xi_0^\epsilon)dy.
\end{aligned}
\end{equation}
This  contradicts
$\frac{\partial\Delta(c,\lambda)}{\partial\lambda}|_{c=c_*,\lambda=\lambda_*}=0$.
Repeating the arguments, we have
$\omega_\epsilon(\xi_0^\epsilon)=\omega_\epsilon(\xi_0^\epsilon+kc_*y_0)$
for all $k\in Z$ and some $y_0\in(0,\tau)$.
It follows from that $\omega_\epsilon(+\infty)=0$, we can obtain
$\phi(\xi)\equiv\psi(\xi+\overline{\xi})$ for $\xi\in \mathbb{R}$, which
contradicts $\omega_\epsilon(\xi)\not\equiv0$.

Similar to the process in \cite{Carr-Chmaj},
$\phi(\xi)\equiv\psi(\xi+\overline{\xi})$ for $\xi\in \mathbb{R}$ still holds for
\smallskip

\noindent\textbf{Case 2:}
Suppose $\xi_0^\epsilon\to  -\infty$ as
$\epsilon\to  0$ and {\bf Case 3:} Suppose ${\xi_0^\epsilon}$ is bounded.
This completes the proof.
\end{proof}


\subsection*{Acknowledgements}
Zhi-Xian Yu was supported by the National Natural Science Foundation of China,
by the Shanghai Leading Academic Discipline Project(No. XTKX2012),
by the Innovation Program of Shanghai Municipal Education Commission (No.14YZ096),
 and by the Hujiang Foundation of China (B14005).



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\end{document}
