# Threshold dynamics of a nonlocal dispersal SIS epidemic model with free
boundaries ††thanks: The first author is supported by the Postgraduate
Research & Practice Innovation Program of Jiangsu Province (KYCX21-3188), the
second author is supported by the National Research Foundation of Korea(NRF)
grant funded by the Korea government (MSIT) (NRF-2022R1F1A1063068) and the
third author is supported by the National Natural Science Foundation of China
(Grant No. 12271470).

Yachun Tonga, Inkyung Ahnb and Zhigui Lina
a School of Mathematical Science, Yangzhou University, Yangzhou 225002, China
b Department of Mathematics, Korea University, Sejong 339-700, Republic of
Korea Corresponding author. Email<EMAIL_ADDRESS>(Z. Lin).

> Abstract. To study the influence of the moving front of the infected
> interval and the spatial movement of individuals on the spreading or
> vanishing of infectious disease, we consider a nonlocal SIS (susceptible-
> infected-susceptible) reaction-diffusion model with media coverage, hospital
> bed numbers and free boundaries. The principal eigenvalue of the integral
> operator is defined, and the impacts of the diffusion rate of infected
> individuals and interval length on the principal eigenvalue are analyzed.
> Furthermore, sufficient conditions for spreading and vanishing of the
> disease are derived. Our results show that large media coverage and hospital
> bed numbers are beneficial to the prevention and control of disease. The
> difference between the model with nonlocal diffusion and that with local
> diffusion is also discussed and nonlocal diffusion leads to more
> possibilities.
>
> MSC: 35K57, 92D30; secondary: 35R35.
>
> Keywords: SIS model; Free boundary; Nonlocal diffusion; Spreading and
> vanishing

##  1 Introduction

With the emergence and outbreak of COVID-19 [3, 30] in recent years,
infectious disease models have become one of the most popular research topics.
To study the spread and dynamics of COVID-19, most scholars use the SIR
(susceptible-infected-recovered) [3, 28], SEIR (susceptible-exposed-infected-
recovered) [30, 25] and SEAIR (susceptible-exposed-asymptomatic-infectious-
removed) [2, 46] models to describe the spread of COVID-19. Meanwhile, the
classical SIS model has received great attention in mathematical epidemiology.

Considering the impact of the spatial heterogeneity of the environment and the
movement of individuals on infectious diseases, Allen et al. in [1] proposed
and discussed an SIS reaction-diffusion system

$\left\\{\begin{array}[]{lll}S_{t}-d_{S}\Delta
S=-\frac{\beta(x)SI}{S+I}+\gamma(x)I,&\,t>0,\ x\in\Omega,\\\\[5.69054pt]
I_{t}-d_{I}\Delta I=\frac{\beta(x)SI}{S+I}-\gamma(x)I,&\,t>0,\
x\in\Omega,\\\\[5.69054pt] \frac{\partial S}{\partial\eta}=\frac{\partial
I}{\partial\eta}=0,&\,t>0,\ x\in\partial\Omega.\end{array}\right.$ (1.1)

Here, $\Omega\subset\mathbb{R}^{n}$ $(n\geq 1)$ is a bounded domain; $S(t,x)$
and $I(t,x)$ indicate the density of susceptible and infected individuals at
location $x$ and time $t$, respectively; $d_{S}$ and $d_{I}$ are positive
constants that account for the diffusion rate of susceptible and infected
individuals, respectively; and the positive bounded H$\ddot{o}$lder continuous
functions $\beta(x)$ and $\gamma(x)$ can be interpreted as rates of disease
transmission and recovery for $x\in\Omega$, respectively. The authors in [1]
mainly discussed the existence, uniqueness and stability of DFE (disease-free
equilibrium) and EE (endemic equilibrium) and used the basic reproduction
number ${\mathcal{R}}_{0}$ to characterize the risk of the region. Afterwards,
Peng and Liu [34] confirmed the conjecture proposed by Allen et al. in [1]
that a unique EE is globally asymptotically stable in some special cases.
Further results that the effect of individual movement (large or small) on the
existence and disappearance of disease were obtained in [33]. For more results
of the SIS reaction-diffusion model, one can see [24, 35, 42] and the
references therein.

It is easy to find that the above articles are devoted to the study of SIS
models on a fixed domain. In real life, the movement of species leads to
changes in biological habitats, and in mathematics, the free boundary can be
used to describe this phenomenon, such as the healing of wounds [10] and the
expansion of new species or invasive species [14, 6, 27, 38]. Free boundary
problems can also be used to describe the transmission of disease, such as the
SIRS model [7], SIS model [22], SIR model [23, 47] and references therein.

To explore the moving front of the infected individual, Wang and Guo [40]
introduced the free boundary and studied the dynamics of the following SIS
reaction-diffusion model:

$\left\\{\begin{array}[]{llllll}S_{t}-d\Delta S=\sigma-\mu
S-\beta(x)SI+\gamma(x)I,&t>0,\,x\in\mathbb{R},\\\\[5.69054pt] I_{t}-d\Delta
I=\beta(x)SI-\mu I-\gamma(x)I,&t>0,\,x\in(g(t),h(t)),\\\\[5.69054pt]
I(t,x)=0,&t>0,\,x\in\mathbb{R}\backslash(g(t),h(t)),\\\\[5.69054pt]
g^{\prime}(t)=-kI_{x}(t,g(t)),\ g(0)=-h_{0},&t\geq 0,\\\\[5.69054pt]
h^{\prime}(t)=-kI_{x}(t,h(t)),\ h(0)=h_{0},&t\geq 0,\\\\[5.69054pt]
S(0,x)=S_{0}(x),\ I(0,x)=I_{0}(x),&x\in\mathbb{R}.\end{array}\right.$ (1.2)

The basic reproduction number was given, and the spreading-vanishing dichotomy
was established. Some conditions for disease spreading or vanishing were
presented by investigating the effect of the diffusion rate $(d)$, initial
value ($I_{0}$) and expanding capability ($k$) on the asymptotic behavior of
the infected individuals.

It is widely known that random dispersal or local diffusion describes the
local behavior of the movements of organisms between adjacent spatial
locations [26]. Briefly, the classical Laplace diffusion operator is used to
describe that the movement of the infectious agent and infected population
only occurs between adjacent spatial positions [43]. However, Murray [32]
noted that a local or short-range diffusive flux proportional to the gradient
is not suitable to characterize some biological phenomena. In the real world,
the movements and interactions of some organisms occur at nonadjacent spatial
positions, and such dispersal is called nonlocal diffusion [13]. Nonlocal
diffusion can occur naturally through dispersal and migration or facilitated
by human activities. It can positively impact a population’s genetic diversity
and long-term viability, but it can also introduce disease or invasive species
into new territories. Local diffusion usually involves individuals migrating
short distances within a defined area, such as a habitat patch or a specific
population. Various factors, including random movement, resource competition,
and response to environmental conditions, can cause it. On the other hand,
nonlocal diffusion refers to the movement of individuals between different
regions or subpopulations. This migration typically involves individuals
moving long distances or crossing geographic barriers such as rivers or
mountain ranges.

Recently, nonlocal diffusion equations have attracted extensive attention and
have been used to characterize long-range dispersal in population ecology [6,
26]. In addition, scholars have extensively investigated infectious disease
models with nonlocal diffusion, such as the West Nile virus model [15], SIS
epidemic model [17, 44], and SIR reaction-diffusion model [45, 18]. For other
epidemic models with nonlocal diffusion, see references [41, 39, 9] and
references therein.

In addition, there are many factors that affect the spread of infectious
disease, such as the contact transmission rate and the recovery rate.
Educating the public about the disease through mass media (such as television,
radio, newspapers, billboards, internet, magazines, etc.), is one of the
important precautions. Therefore, media coverage can indirectly reduce the
contact rate between people and infectious diseases, thus reducing the contact
transmission rate of infectious diseases [36]. In general, the main factor
impacting the recovery rate is the availability of health care (such as the
number of physicians, nurses, hospital beds and isolation places). In fact,
health and medical institutions use the hospital bed-population ratio (HBPR)
(the number of hospital beds per 10000 people) as a method of reckoning
available resources to the public [31].

Taking into account nonlocal diffusion, media coverage and hospital bed
numbers, we consider the following nonlocal dispersal SIS epidemic model with
a free boundary:

$\left\\{\begin{array}[]{ll}S_{t}=d\mathcal{L}_{1}[S]+\sigma-\mu_{1}S-\beta(m(x),I,x)SI+\gamma(b(x),I,x)I,&t>0,\,x\in\mathbb{R},\\\\[5.69054pt]
I_{t}=d\mathcal{L}_{2}[I;g,h]-\mu_{2}I+\beta(m(x),I,x)SI-\gamma(b(x),I,x)I,&t>0,\,x\in(g(t),h(t)),\\\\[5.69054pt]
I(t,x)=0,&t\geq 0,\ x\in\mathbb{R}\backslash(g(t),h(t)),\\\\[5.69054pt]
h^{\prime}(t)=k\int_{g(t)}^{h(t)}\int_{h(t)}^{+\infty}J(x-y)I(t,x)dydx,&t>0,\\\\[5.69054pt]
g^{\prime}(t)=-k\int_{g(t)}^{h(t)}\int_{-\infty}^{g(t)}J(x-y)I(t,x)dydx,&t>0,\\\\[5.69054pt]
S(0,x)=S_{0}(x),\ g(0)=-h_{0},\ h(0)=h_{0},&x\in\mathbb{R},\\\\[5.69054pt]
I(0,x)=I_{0}(x),&x\in(-h_{0},h_{0}),\end{array}\right.$ (1.3)

where

$\mathcal{L}_{1}[S]=\int_{\mathbb{R}}J(x-y)S(t,y)dy-S(t,x),$
$\mathcal{L}_{2}[I;g,h]=\int_{g(t)}^{h(t)}J(x-y)I(t,y)dy-I(t,x),$

and $d,\,S(t,x)$ and $I(t,x)$ have the same epidemiological interpretation as
in (1.1). The constants $\sigma,\,\mu_{1}$ and $\mu_{2}$ are positive, where
$\sigma$ accounts for the environment carrying capability; the natural
mortality rate of the susceptible individuals is expressed by $\mu_{1}$, and
$\mu_{2}$ denotes the sum of the natural mortality and disease-caused death
rates of the infected individuals. The functions
$\beta(m(x),I,x),\,\gamma(b(x),I,x),\,m(x),\,b(x)$ are nonnegative, where
$m(x)$ represents the media coverage, and $b(x)$ stands for the number of
hospital beds. In this paper, we assume that

$(1)$ the contact infectious rate $\beta(m(x),I,x)$ is Lipschitz continuous
and monotonically decreasing in $m(x)$ and increasing in $I$;

$(2)$ the recovery rate $\gamma(b(x),I,x)$ is Lipschitz continuous and
increasing in $b(x)$ and monotonically decreasing in $I$;

$(3)$ $\beta_{I}(m(x),I,x)$ and $\gamma_{I}(b(x),I,x)$ are continuous and
bounded for $m(x)\in[0,\infty),I\in[0,\infty)$ and $x\in(-\infty,\infty)$.

For instance, Cui and Zhu [12] used the function $\beta(I)=\beta e^{mI}$ to
model the impact of media coverage on the transmission rate, and Shan and Zhu
[37] used the function
$\gamma(b,I,x)=\gamma_{0}+(\gamma_{1}-\gamma_{0})\frac{b}{b+I}$ to describe
the hospital resource impact factors.

Recalling that $S(t,x)$ denotes the density at point $x$ and time $t$, the
kernel function $J(x-y)$ is regarded as the probability distribution of
jumping from place $y$ to place $x$, then the integral operator
$\int_{\mathbb{R}}J(x-y)S(t,y)dy$ accounts for the rate at which the
individuals are gathering at point $x$ from all other places, and $-S(t,x)$ is
the rate at which the individuals are leaving at point $x$ to other places. In
addition, the infected individuals stay in the infected interval
$(g(t),h(t))$. We further suppose that the initial function $S_{0}(x)$
satisfies

$S_{0}(x)\in C(\mathbb{R})\cap L^{\infty}(\mathbb{R})\,\ \textrm{and}\,\
S_{0}(x)>0\,\ \textrm{in}\,\,\mathbb{R},$ (1.4)

and $I_{0}(x)$ satisfies

$I_{0}(x)\in C([-h_{0},h_{0}]),\,I_{0}(\pm h_{0})=0,\,I_{0}(x)>0\,\
\textrm{in}\,\,(-h_{0},h_{0}).$ (1.5)

For system (1.3), we assume that the kernel function
$J:\mathbb{R}\rightarrow\mathbb{R}$ is continuous and nonnegative and has the
properties

$\mathbf{(J)}:J\in C(\mathbb{R})\cap
L^{\infty}(\mathbb{R})\,{\rm\,is\,symmetric},\ J(0)>0,\
\int_{\mathbb{R}}J(x)dx=1.$

The free boundary conditions
$h^{\prime}(t)=k\int_{g(t)}^{h(t)}\int_{h(t)}^{+\infty}J(x-y)I(t,x)dydx$ and
$g^{\prime}(t)=-k\int_{g(t)}^{h(t)}\int_{-\infty}^{g(t)}J(x-y)I(t,x)dydx$ in
(1.3) imply that the expanding rate of the interval $(g(t),h(t))$ is
determined by the infected individuals and is proportional to the outward flux
of the infected individuals across the interval $(g(t),h(t))$ [5].

It is worth mentioning that there are links and differences between local
diffusion and nonlocal diffusion. Local diffusion, expressed by the Laplace
operator $\Delta u$ (the Laplace in $\mathbb{R}^{n}$, $n\geq 2$) or $u_{xx}$
(in one-dimensional space), is used to describe the influence between adjacent
positions, and nonlocal diffusion, expressed by the integral operator (is
given by $\int_{\mathbb{R}}J(x-y)u(t,y)dy-u(t,x)$), is used to describe long-
distance dispersal. However, the Laplace operator can be regarded as a local
approximation of a nonlocal diffusion operator. In fact, when $J(\cdot)$ is
symmetric and has compact supports, such as $J(x)=(1/\epsilon)K(x/\epsilon)$
with $0<\epsilon\ll 1$ and $K(x)$ is a general mollification function with
support $x\in[-1,1]$, we can transform nonlocal operators into local operators
by using the Taylor formula [29].

This article is organized as follows: the existence and uniqueness of the
global solution are given in Section 2. Section 3 is devoted to defining and
studying the properties of the principal eigenvalue. Section 4 gives some
sufficient conditions for the disease to spread or vanish. Finally, a brief
discussion is presented in Section 5.

##  2 Global existence and uniqueness

In this section, we assume that $h_{0}>0$, $S_{0}(x)$ and $I_{0}(x)$ satisfy
(1.4) and (1.5). For any given $T>0$, we first introduce the notations as
follows:

$\displaystyle\begin{array}[]{lll}&\mathbb{H}_{T}:=\\{h\in
C([0,T]):h(0)=h_{0},\,\ \inf\limits_{0\leq t_{1}<t_{2}\leq
T}\frac{h(t_{2})-h(t_{1})}{t_{2}-t_{1}}>0\\},\\\\[5.69054pt]
&\mathbb{G}_{T}:=\\{g\in C([0,T]):-g\in\mathbb{H}_{T}\\},\\\\[5.69054pt]
&D^{g,h}_{T}:=\\{(t,x)\in\mathbb{R}^{2}:0<t\leq T,\,\
g(t)<x<h(t)\\},\\\\[5.69054pt]
&D^{h_{0}}_{T}:=\\{(t,x)\in\mathbb{R}^{2}:0<t\leq T,\,\
-h_{0}<x<h_{0}\\},\\\\[5.69054pt]
&D^{\infty}_{T}:=\\{(t,x)\in\mathbb{R}^{2}:0<t\leq T,\,\
x\in\mathbb{R}\\},\\\\[5.69054pt] &X^{S_{0}}_{T}:=\\{\phi(t,x)\in
C(D^{\infty}_{T})\cap L^{\infty}(D^{\infty}_{T}):\phi(0,x)=S_{0}(x)\,\ {\rm
in}\,\ \mathbb{R},\,\ \phi(t,x)\geq 0\,\ {\rm in}\,\
D^{\infty}_{T}\\},\\\\[5.69054pt] &X^{I_{0}}_{T}:=\\{\psi(t,x)\in
C(D^{\infty}_{T}):\psi(0,x)=I_{0}(x)\,\ {\rm in}\,\ [-h_{0},h_{0}],\,\
\psi(t,x)\geq 0\,\ {\rm in}\,\ D^{g,h}_{T},\\\\[5.69054pt]
&\qquad\qquad\qquad\qquad\qquad\qquad\psi(t,x)=0\,\ {\rm for}\,\ t\in(0,T),\,\
x\in\mathbb{R}\backslash(g(t),h(t))\\}.\end{array}$

To prove the existence and uniqueness of the global solution of problem (1.3),
we first give the following result for problem (1.3) without a free boundary.

###### Lemma 2.1

For any given $T>0$ and $(g,h)\in\mathbb{H}_{T}\times\mathbb{G}_{T}$, the
problem

$\left\\{\begin{array}[]{ll}S_{t}=d\mathcal{L}_{1}[S]+\sigma-\mu_{1}S-\beta(m(x),I,x)SI+\gamma(b(x),I,x)I,&0<t\leq
T,\ x\in\mathbb{R},\\\\[5.69054pt]
I_{t}=d\mathcal{L}_{2}[I;g,h]-\mu_{2}I+\beta(m(x),I,x)SI-\gamma(b(x),I,x)I,&0<t\leq
T,\ x\in(g(t),h(t)),\\\\[5.69054pt] I(t,x)=0,&0\leq t\leq T,\
x\in\mathbb{R}\backslash(g(t),h(t)),\\\\[5.69054pt]
S(0,x)=S_{0}(x),&x\in\mathbb{R},\\\\[5.69054pt]
I(0,x)=I_{0}(x),&x\in(-h_{0},h_{0})\end{array}\right.$ (2.2)

admits a unique solution $(S_{g,h},I_{g,h})\in
C(\overline{D}_{T}^{\infty})\times C(\overline{D}_{T}^{g,h})$. Moreover,

$0<S_{g,h}(t,x)\leq A\qquad{\rm for}\,\ {\rm any}\,\,(t,x)\in D_{T}^{\infty},$
(2.3) $0<I_{g,h}(t,x)\leq A\qquad{\rm for}\,\ {\rm any}\,\,(t,x)\in
D_{T}^{g,h},$ (2.4)

where
$A=\max\\{\frac{\sigma}{\mu_{1}},\|S_{0}\|_{\infty}+\|I_{0}\|_{\infty}\\}$.

Proof. The main idea of this proof comes from [45]. We divide the proof into
three steps.

Step 1. The parameterized ODE problem.

For any given $x\in\mathbb{R}$, $s\in(0,T]$, denote

$\displaystyle
t_{x}=\left\\{\begin{array}[]{ll}t_{x}^{g},&x\in(g(s),-h_{0})\,\,\mbox{and}\,\,x=g(t_{x}^{g}),\\\\[5.69054pt]
0,&x\in[-h_{0},h_{0}],\\\\[5.69054pt]
t_{x}^{h},&x\in(h_{0},h(s))\,\,\mbox{and}\,\,x=h(t_{x}^{h}),\\\\[5.69054pt]
s,&x\in\mathbb{R}\backslash(g(s),h(s)).\end{array}\right.$

Clearly, $t_{x}>0$ for $x\in\mathbb{R}\backslash[-h_{0},h_{0}]$, $t_{x}<s$ for
$x\in(g(s),h(s))$. For any given $(\phi,\psi)\in X_{s}^{S_{0}}\times
X_{s}^{I_{0}}$, define

$A_{1}=\max\\{A,\,\frac{\sigma+\|\psi\|_{\infty}\sup\gamma}{\mu_{1}},\,\|\phi\|_{\infty}\\},\,\,\,\,A_{2}=\max\\{A,\,\frac{(d+A_{1}\sup\beta)\|\psi\|_{\infty}}{d+\mu_{2}}\\}.$

We discuss it in the following two cases:

Case 1: $x\in\mathbb{R}\backslash[-h_{0},h_{0}]$, $t\in[0,t_{x}]$.

Clearly, $I(t,x)=0$ for
$(t,x)\in[0,t_{x}]\times\mathbb{R}\backslash[-h_{0},h_{0}]$. Consider the ODE
problem

$\left\\{\begin{array}[]{lll}S_{t}=d\int_{\mathbb{R}}J(x-y)\phi(t,y)dy-
dS+\sigma-\mu_{1}S,&\,0<t\leq t_{x},\\\\[5.69054pt]
S(0,x)=S_{0}(x),&\,x\in\mathbb{R}\backslash[-h_{0},h_{0}].\end{array}\right.$
(2.6)

For any $S_{1},\,S_{2}\in[0,A_{1}]$,

$\begin{array}[]{llllll}&&|d\int_{\mathbb{R}}J(x-y)\phi(t,y)dy-
dS_{1}+\sigma-\mu_{1}S_{1}-d\int_{\mathbb{R}}J(x-y)\phi(t,y)dy+dS_{2}-\sigma+\mu_{1}S_{2}|\\\\[5.69054pt]
&=&(d+\mu_{1})|S_{1}-S_{2}|.\end{array}$

Therefore, $F:=d\int_{\mathbb{R}}J(x-y)\phi(t,y)dy-dS+\sigma-\mu_{1}S$ is
Lipschitz continuous in $S$ for $S\in[0,A_{1}]$. By the fundamental theory of
ODEs, problem (2.6) has a unique solution $S_{\phi}(t,x)$ defined in
$t\in[0,\widehat{t}_{x})$, and $S_{\phi}(t,x)$ is continuous in both $t$ and
$x$. To see that $t\rightarrow S(\cdot,x)$ can be uniquely extended to
$[0,t_{x}]$, we need to prove that if $S_{\phi}(t,x)$ is uniquely defined for
$t\in[0,\widehat{t}_{x}]$ with $\widehat{t}_{x}\in(0,t_{x}]$, then

$0\leq S_{\phi}(t,x)\leq
A_{1},\,\,\,\mbox{for}\,\,\,t\in[0,\widehat{t}_{x}]\,\,\mbox{and}\,\,x\in\mathbb{R}\backslash[-h_{0},h_{0}].$

Obviously,

$\begin{array}[]{llllll}&&d\int_{\mathbb{R}}J(x-y)\phi(t,y)dy-
dA_{1}+\sigma-\mu_{1}A_{1}\\\\[5.69054pt]
&\leq&d\|\phi\|_{\infty}-dA_{1}+\sigma-\mu_{1}A_{1}\\\\[5.69054pt]
&\leq&0,\end{array}$

and $\|S_{0}\|_{\infty}\leq A_{1}$. Thanks to the direct comparison argument,
one can derive $S_{\phi}(t,x)\leq A_{1}$ for $t\in[0,\widehat{{t}}_{x}]$ and
$x\in\mathbb{R}\backslash[-h_{0},h_{0}]$. We use similar method to prove that
$S_{\phi}(t,x)\geq 0$ for
$t\in[0,\widehat{t}_{x}],\,x\in\mathbb{R}\backslash[-h_{0},h_{0}]$.

Case 2: $x\in(g(s),h(s))$, $t\in[t_{x},s]$.

Define

$\widehat{S}_{\phi}(x)=\left\\{\begin{array}[]{ll}S_{0}(x),&x\in[-h_{0},h_{0}]\\\\[5.69054pt]
S_{\phi}(t_{x},x),&x\notin[-h_{0},h_{0}]\end{array}\right.\,\,\,\mbox{and}\,\,\,\
\widehat{I}(x)=\left\\{\begin{array}[]{ll}I_{0}(x),&x\in[-h_{0},h_{0}]\\\\[5.69054pt]
0,&x\notin[-h_{0},h_{0}].\end{array}\right.$

Consider the ODE problem

$\left\\{\begin{array}[]{lll}S_{t}=F_{1}(t,x,S,I),&\,t_{x}<t\leq
s,\\\\[5.69054pt] I_{t}=F_{2}(t,x,S,I),&\,t_{x}<t\leq s,\\\\[5.69054pt]
S(t_{x},x)=\widehat{S}_{\phi}(x),\,\,I(t_{x},x)=\widehat{I}(x),&\,x\in(g(s),h(s))\end{array}\right.$
(2.7)

with

$F_{1}=d\int_{\mathbb{R}}J(x-y)\phi(t,y)dy-
dS+\sigma-\mu_{1}S+\gamma(b,I,x)\psi-\beta(m,I,x)SI,$
$F_{2}=d\int_{g(t)}^{h(t)}J(x-y)\psi(t,y)dy-
dI-\mu_{2}I-\gamma(b,I,x)I+\beta(m,I,x)S\psi.$

For any $(S_{i},I_{i})\in[0,A_{1}]\times[0,A_{2}]$($i=1,\,2$), obviously,
$F_{i}(t,x,S,I)$ is Lipschitz continuous in $(S,I)$ for
$(S_{i},I_{i})\in[0,A_{1}]\times[0,A_{2}]$ by the continuity and monotonicity
of $\beta(m(x),I,x)$ and $\gamma(b(x),I,x)$, and it is uniformly continuous
for $x\in(g(s),h(s))$ and $t\in[t_{x},s]$. In addition, $F_{i}(t,x,S,I)$ is
continuous in all its variables in this range. Problem (2.7) has a unique
solution $(S_{\phi,\psi}(t,x),\,I_{\phi,\psi}(t,x))$ for $t\in[t_{x},s_{x})$,
and $(S_{\phi,\psi}(t,x),\,I_{\phi,\psi}(t,x))$ is continuous in both $t$ and
$x$ by the fundamental theorem of ODEs.

To show that $(S_{\phi,\psi}(t,x),\,I_{\phi,\psi}(t,x))$ can be uniquely
extended to $[t_{x},s]$, it suffices to prove that if
$(S_{\phi,\psi}(t,x),\,I_{\phi,\psi}(t,x))$ is uniquely defined for
$t\in[t_{x},\widehat{t}]$ with $\widehat{t}\in(t_{x},s]$, then

$0\leq S_{\phi,\psi}(t,x)\leq A_{1},\,\,0\leq I_{\phi,\psi}(t,x)\leq
A_{2}\,\,\,\,\mbox{for}\,\,\,t\in[t_{x},\widehat{t}].$ (2.8)

In fact, it is easy to see that

$\begin{array}[]{llllll}&&F_{1}(t,x,A_{1},A_{2})\\\\[5.69054pt]
&=&d\int_{\mathbb{R}}J(x-y)\phi(t,y)dy-
dA_{1}+\sigma-\mu_{1}A_{1}+\gamma(b,A_{2},x)\psi-\beta(m,A_{2},x)A_{1}A_{2}\\\\[5.69054pt]
&\leq&d\|\phi\|_{\infty}-dA_{1}+\sigma-\mu_{1}A_{1}+\gamma(b,A_{2},x)\|\psi\|-\beta(m,A_{2},x)A_{1}A_{2}\\\\[5.69054pt]
&<&d\|\phi\|_{\infty}-dA_{1}+\sigma-\mu_{1}A_{1}+\|\psi\|_{\infty}\sup\gamma\\\\[5.69054pt]
&\leq&0\end{array}$

and

$\begin{array}[]{llllll}&&F_{2}(t,x,A_{1},A_{2})\\\\[5.69054pt]
&=&d\int_{g(t)}^{h(t)}J(x-y)\psi(t,y)dy-
dA_{2}-\mu_{2}A_{2}-\gamma(b,A_{2},x)A_{2}+\beta(m,A_{2},x)A_{1}\psi\\\\[5.69054pt]
&\leq&d\int_{g(t)}^{h(t)}J(x-y)\psi(t,y)dy-
dA_{2}-\mu_{2}A_{2}+A_{1}\|\psi\|_{\infty}\sup\beta\\\\[5.69054pt]
&\leq&0.\end{array}$

Since $A_{1}\geq\|S_{0}\|_{\infty},\,A_{2}\geq\|I_{0}\|_{\infty}$, we have
$S_{\phi,\psi}(t,x)\leq A_{1}$ and $I_{\phi,\psi}(t,x)\leq A_{2}$ in
$t\in[t_{x},\widehat{t}]$ by the comparison argument. The left part of (2.8)
can be obtained similarly by using $F_{i}(t,x,0,0)\geq 0$ ($i=1,\,2$).

Step 2. A fixed point theorem.

For any $s\in(0,T)$, we note

$X_{s}^{S_{0}}:=\\{\phi|_{\overline{D}_{s}^{\infty}}:\phi\in
X_{T}^{S_{0}}\\},\,\,\,X_{s}^{I_{0}}:=\\{\psi|_{\overline{D}_{s}^{g,h}}:\psi\in
X_{T}^{I_{0}}\\}.$

Denote

$\displaystyle(\widehat{S}(t,x),\widehat{I}(t,x))=\left\\{\begin{array}[]{ll}(S_{\phi(t,x)},0),&x\in\mathbb{R}\backslash[-h_{0},h_{0}],\,\,t=[0,t_{x}],\\\\[5.69054pt]
(S_{\phi,\psi}(t,x),I_{\phi,\psi}(t,x)),&x\in(g(s),h(s)),\,\,t\in[t_{x},s],\end{array}\right.$

where $S_{\phi}(t,x),\,S_{\phi,\psi}(t,x)$ and $I_{\phi,\psi}(t,x))$ are given
in Step 1. By Step 1, for any $(\phi,\,\psi)$, we have a unique solution
$(\widehat{S},\widehat{I})$ for $t\in[0,s]$. It is easy to check that
$\widehat{S}(t,x)$ is continuous in $\overline{D}_{s}^{\infty}$, and
$\widehat{I}(t,x)$ is continuous in $\overline{D}_{s}^{g,h}$ due to the
continuous dependence of the ODE solution on the parameters. Therefore,
$(\widehat{S},\widehat{I})\in X_{s}^{S_{0}}\times X_{s}^{I_{0}}$. Note that
$X_{s}^{S_{0}}$ and $X_{s}^{I_{0}}$ are complete metric spaces, respectively,
with the norms

$d_{1}(\phi_{1},\phi_{2})=\|\phi_{1}-\phi_{2}\|_{C(\overline{D}_{s}^{\infty})},\,\,\,d_{2}(\psi_{1},\psi_{2})=\|\psi_{1}-\psi_{2}\|_{C(\overline{D}_{s}^{g,h})}.$

Hence, we find a mapping $\Gamma:X_{s}^{S_{0}}\times X_{s}^{I_{0}}\rightarrow
X_{s}^{S_{0}}\times X_{s}^{I_{0}}$ by
$\Gamma(\phi,\psi)=(\widehat{S},\widehat{I})$.

Setting

$M_{1}=\max\\{A,\,4\|S_{0}\|_{\infty},\,\frac{4\sigma}{\mu_{1}},\,\frac{4(\sigma+M_{2})}{\mu_{1}+d}\\},\,M_{2}=\max\\{A,\,2\|I_{0}\|_{\infty}\\}.$

Define

$X_{s}^{M_{1}}=\\{\phi|\,\phi\in
X_{s}^{S_{0}},\,\,\|\phi\|_{C(\overline{D}_{s}^{\infty})}\leq M_{1}\\},$
$X_{s}^{M_{2}}=\\{\psi|\,\psi\in
X_{s}^{I_{0}},\,\,\|\psi\|_{C(\overline{D}_{s}^{g,h})}\leq M_{2}\\}.$

Using the same arguments as Lemma 2.1 in [45], we can deduce that $\Gamma$ is
a contraction map and has a unique fixed point $(S^{*},\,I^{*})\in
X_{s}^{M_{1}}\times X_{s}^{M_{2}}$ for any $s\in(0,\widehat{s}]$ by the
contraction mapping theorem, where $\widetilde{s}$ relies on
$d,\,M_{1},\,\beta,\,\gamma$ and $M_{2}$.

To prove that $(S^{*},\,I^{*})$ is the unique solution (2.2) for $t\in[0,s]$
with $s\in(0,\widehat{s}]$, it suffices to discuss that any nonnegative
solution $(S,\,I)$ of (2.2) for $t\in[0,s]$ belongs to $X_{s}^{M_{1}}\times
X_{s}^{M_{2}}$.

We claim that

$S+I\leq
A\,\,\,\,\mbox{for}\,\,\,t\in[0,s]\,\,\,\,\mbox{and}\,\,\,x\in\mathbb{R},$
(2.10)

which implies that

$0\leq S(t,x)\leq A,\,\,\,\,(t,x)\in[0,s]\times\mathbb{R},$ $0\leq I(t,x)\leq
A,\,\,\,\,(t,x)\in[0,s]\times[g(t),h(t)].$

Consequently, we obtain that for any $s\in(0,\widehat{s}]$, (2.2) admits a
unique solution for $t\in[0,s]$. To complete the proof, it only needs to prove
that claim (2.10) is true.

Let $N=S+I$, then for $t\in[0,s]$ and $x\in(g(t),h(t))$,

$\begin{array}[]{lll}N_{t}&\leq&d\int_{\mathbb{R}}J(x-y)N(t,y)dy-
dN(t,x)-d(\int_{-\infty}^{g(t)}+\int_{h(t)}^{\infty})J(x-y)I(t,y)dy+\sigma-\mu_{1}N\\\\[5.69054pt]
&\leq&d\int_{\mathbb{R}}J(x-y)N(t,y)dy-dN(t,x)-\mu_{1}N+\sigma.\end{array}$

Since $\|S_{0}\|_{\infty}+\|I_{0}\|_{\infty}\leq A$, we have $N(t,x)\leq A$
for $t\in[0,s]$ and $x\in(g(t),h(t))$ by using the comparison principle.

While $t\in[0,s]$ and $x\in\mathbb{R}\backslash(g(t),h(t))$, then $I(t,x)=0$,
which implies

$N_{t}=d\int_{\mathbb{R}}J(x-y)N(t,y)dy-dN(t,x)+\sigma-\mu_{1}N.$

It is clear that $N\leq A$ for $t\in[0,s]$ and
$x\in\mathbb{R}\backslash(g(t),h(t))$. Next, we prove that (2.10) holds. We
argue by contradiction and suppose that
$\max\limits_{(t,x)\in[0,\,s]\times\mathbb{R}}N(t,x)>A$, there exists a point
$(t_{0},x_{0})\in[0,s]\times\mathbb{R}$ such that $\max N=N(t_{0},x_{0})>A$.
According to the above analysis, we can obtain that $x_{0}=g(t_{0})$ or
$x_{0}=h(t_{0})$. Without loss of generality, we assume that $x_{0}=g(t_{0})$.
Since $I(t_{0},g(t_{0}))=0$, $S(t_{0},x_{0})$ satisfies

$S_{t}(t_{0},x_{0})=d\int_{\mathbb{R}}J(x_{0}-y)S(t_{0},y)dy-
dS(t_{0},x_{0})+\sigma-\mu_{1}S(t_{0},x_{0}).$

Obviously, $S_{t}(t_{0},x_{0})\geq 0$ and $S(t_{0},x_{0})\leq A$, which
contradicts the assumption:
$\max\limits_{(t,x)\in[0,\,s]\times\mathbb{R}}N(t,x)>A$.

Step 3. Extension of the solution.

We now prove that the unique solution of (2.2) for $0<t\leq s$ can be extended
to $0<t\leq T$. In Step 2, $\widehat{s}$ depends only on $d,\,\gamma,\,\beta$,
and $A$. With the help of the iterative method, we obtain that problem (2.2)
has a unique solution for $t\in[0,T]$. We omit it here; see Step 3 of the
proof in Lemma 2.1 in [45] for more details.  $\Box$

###### Theorem 2.2

Assume that $\mathbf{(J)}$ holds. For any $S_{0}$ satisfying (1.4) and $I_{0}$
satisfying (1.5), problem (1.3) admits a unique positive solution
$(S(t,x),\,I(t,x);\,g(t),\,h(t))$ defined for all $t>0$.

Proof. We will prove this result by using Lemma 2.1 and the fixed point
theorem. For any given $T>0$ and
$(g^{*},h^{*})\in\mathbb{H}_{T}\times\mathbb{G}_{T}$, we know that (2.2) with
$(g,h)=(g^{*},h^{*})$ has a unique solution $(S^{*},I^{*})$. Define

$\left\\{\begin{array}[]{ll}\widetilde{g}=-h_{0}-k\int_{0}^{t}\int_{g^{*}(\tau)}^{h^{*}(\tau)}\int_{-\infty}^{g^{*}(\tau)}J(x-y)I^{*}(\tau,x)dydxd\tau,\\\\[5.69054pt]
\widetilde{h}=h_{0}+k\int_{0}^{t}\int_{g^{*}(\tau)}^{h^{*}(\tau)}\int_{h^{*}(\tau)}^{+\infty}J(x-y)I^{*}(\tau,x)dydxd\tau.\end{array}\right.$
(2.11)

In view of $\mathbf{(J)}$ and $J(0)>0$, there exist constants
$\epsilon_{0}\in(0,h_{0}/4)$ and $\delta_{0}>0$ such that

$J(x)\geq\delta_{0}\,\,\,\mbox{if}\,\,\,|x|\leq\epsilon_{0}.$

By virtue of the above inequality and proof of Theorem 2.1 in [5], there
exists

$T_{0}=T_{0}(k,A,h_{0},\epsilon_{0},I_{0},J)>0,$

such that, for any $T\in(0,T_{0}]$,

$\sup\limits_{0\leq t_{1}<t_{2}\leq
T}\frac{\widetilde{g}(t_{2})-\widetilde{g}(t_{1})}{t_{2}-t_{1}}\leq-k\eta_{1},\,\,\inf\limits_{0\leq
t_{1}<t_{2}\leq
T}\frac{\widetilde{h}(t_{2})-\widetilde{h}(t_{1})}{t_{2}-t_{1}}\geq
k\eta_{2},$ $\widetilde{h}(t)-\widetilde{g}(t)\leq
2h_{0}+\epsilon_{0}/4\,\,\,\mbox{for}\,\,t\in[0,T],$

where

$\eta_{1}=\frac{1}{4}\epsilon_{0}\delta_{0}e^{-(d+\mu_{2}+\sup\gamma)T_{0}}\int_{-h_{0}}^{-h_{0}+\frac{\epsilon_{0}}{4}}I_{0}(x)dx,\,\,\,\eta_{2}=\frac{1}{4}\epsilon_{0}\delta_{0}e^{-(d+\mu_{2}+\sup\gamma)T_{0}}\int_{h_{0}-\frac{\epsilon_{0}}{4}}^{h_{0}}I_{0}(x)dx.$

Let

$\displaystyle\begin{array}[]{lll}\Sigma_{T}:=\\{(g,h)\in\mathbb{H}_{T}^{h_{0}}\times\mathbb{G}_{T}^{h_{0}}:\sup\limits_{0\leq
t_{1}<t_{2}\leq
T}\frac{g(t_{2})-g(t_{1})}{t_{2}-t_{1}}\leq-k\eta_{1},\\\\[5.69054pt]
\qquad\qquad\inf\limits_{0\leq t_{1}<t_{2}\leq
T}\frac{h(t_{2})-h(t_{1})}{t_{2}-t_{1}}\geq k\eta_{2},\,h(t)-g(t)\leq
2h_{0}+\frac{\epsilon_{0}}{4}\,\,\,{\rm for}\,\ t\in[0,T]\\},\end{array}$

and define the mapping
$\mathcal{F}(g^{*},h^{*})=(\widetilde{g},\widetilde{h})$. Clearly, the above
analysis implies that

$\mathcal{F}(\Sigma_{T})\subset\Sigma_{T}\,\,\rm{for}\,\,T\in(0,T_{0}].$

In the following, similar to the proof of Theorem 1.1 in [45], we first prove
that $\mathcal{F}$ is a contraction mapping, and then $\mathcal{F}$ admits a
unique fixed point in $\Sigma_{T}$ by the contraction mapping theorem. Next,
we can derive that $(g,h)\in\Sigma_{T}$ holds for any solution $(S,I;g,h)$ of
(1.3) for $t\in[0,T]$, that is, $(S,I;g,h)$ is a unique solution of (1.3) for
$t\in[0,T]$. Finally, we can show that the solution $(S,I;g,h)$ of (1.3) is
uniquely extended to $t\in(0,+\infty)$. Here, we omit the proof here; see Step
3 in the proof of Theorem 1.1 in [45] for more details.

$\Box$

##  3 The eigenvalue problem

For any $-\infty<L_{1}<L_{2}<+\infty$ and $d>0$, denote

$\mathcal{L}_{\\{(L_{1},\,L_{2}),\,d\\}}[\phi](x)=d\int_{L_{1}}^{L_{2}}J(x-y)\phi(y)dy-d\phi(x).$

Now, we introduce some results on the principal eigenvalue of the linear
operator $\mathcal{L}_{\\{(L_{1},\,L_{2}),\,d\\}}+a(x):C([L_{1},L_{2}])\mapsto
C([L_{1},L_{2}])$ defined by

$(\mathcal{L}_{\\{(L_{1},\,L_{2}),\,d\\}}+a(x))[\phi](x)=d\int_{L_{1}}^{L_{2}}J(x-y)\phi(y)dy-d\phi(x)+a(x)\phi(x),$

where $a(x)=\frac{\sigma\beta(m(x),0,x)}{\mu_{1}}-\mu_{2}-\gamma(b(x),0,x)\in
C([L_{1},L_{2}])$ and $J$ satisfies $\mathbf{(J)}$. Furthermore, we assume

$\mathbf{(H)}:a(x)$ is Lipschitz continuous and achieves its maximum in
$[L_{1},L_{2}]$ at some point $x_{0}\in(L_{1},L_{2})$.

Define the generalized principal eigenvalue as

$\begin{array}[]{ll}&\lambda_{p}(\mathcal{L}_{\\{(L_{1},\,L_{2}),\,d\\}}+a(x))\\\
&:=\inf\\{\lambda\in\mathbb{R}\,|\,\exists\phi\in C([L_{1},L_{2}]),\,\phi>0\
{\rm\ s.\ t.}\
(\mathcal{L}_{\\{(L_{1},\,L_{2}),\,d\\}}+a(x))[\phi]\leq\lambda\phi{\rm\ in}\
(L_{1},L_{2})\\}.\end{array}$ (3.1)

Furthermore, we call it a principal eigenvalue if
$\lambda_{p}(\mathcal{L}_{\\{(L_{1},\,L_{2}),\,d\\}}+a(x))$ is an eigenvalue
of the operator $\mathcal{L}_{\\{(L_{1},\,L_{2}),\,d\\}}+a(x)$ with a
continuous and positive eigenfunction. Recalling that $a(x)$ is Lipschitz
continuous and achieves a global maximum in $(L_{1},L_{2})$, $a(x)$
automatically satisfies the condition
$\frac{1}{(\sup_{x\in(L_{1},\,L_{2})}a(x))-a(x)}\not\in L^{1}$. Therefore, it
follows from Theorem 1.1 or Theorem 1.2 in [11] that the generalized principal
eigenvalue $\lambda_{p}(\mathcal{L}_{\\{(L_{1},\,L_{2}),\,d\\}}+a(x))$ is a
principal eigenvalue.

In this section, we are interested in the properties of the generalized
principal eigenvalue
$\lambda_{p}(\mathcal{L}_{\\{(L_{1},\,L_{2}),\,d\\}}+a(x))$ with media
coverage $m(x)$ and hospital bed number $b(x)$ and asymptotic behavior of the
principal eigenvalue in large and small interval lengths $(L_{1},L_{2})$ or
diffusion rate $d$.

Before stating our primary result, we recall a useful proposition from [11].

###### Proposition 3.1

([11]) The following assertions hold:

$(i)$ Assume $(L_{1},L_{2})\subset(L_{3},L_{4})$. Then,

$\lambda_{p}(\mathcal{L}_{\\{(L_{1},\,L_{2}),\,d\\}}+a(x))\leq\lambda_{p}(\mathcal{L}_{\\{(L_{3},\,L_{4}),\,d\\}}+a(x)).$

$(ii)$ Fix $L_{1},L_{2}$ and suppose that $a_{1}(x)\leq a_{2}(x)$. Then,

$\lambda_{p}(\mathcal{L}_{\\{(L_{1},\,L_{2}),\,d\\}}+a_{1}(x))\leq\lambda_{p}(\mathcal{L}_{\\{(L_{1},\,L_{2}),\,d\\}}+a_{2}(x)).$

Moreover, if $a_{1}(x)+\delta<a_{2}(x)$ for some $\delta>0$, then

$\lambda_{p}(\mathcal{L}_{\\{(L_{1},\,L_{2}),\,d\\}}+a_{1}(x))<\lambda_{p}(\mathcal{L}_{\\{(L_{1},\,L_{2}),\,d\\}}+a_{2}(x)).$

$(iii)$ $\lambda_{p}(\mathcal{L}_{\\{(L_{1},\,L_{2}),\,d\\}}+a(x))$ is
Lipschitz continuous in $a(x)$. More precisely,

$|\lambda_{p}(\mathcal{L}_{\\{(L_{1},\,L_{2}),\,d\\}}+a_{1}(x))-\lambda_{p}(\mathcal{L}_{\\{(L_{1},\,L_{2}),\,d\\}}+a_{2}(x))|\leq\|a_{1}(x)-a_{2}(x)\|_{\infty}.$

Let us now analyze the impact of media coverage $m(x)$ and hospital bed number
$b(x)$ on the generalized principal eigenvalue. Obviously, the following
result holds by Proposition 3.1 $(ii)$.

###### Theorem 3.2

Suppose that $\mathbf{(J)}$ and $\mathbf{(H)}$ hold. Then,

$(i)$ $\lambda_{p}(\mathcal{L}_{\\{(L_{1},\,L_{2}),\,d\\}}+a(x))$ is strictly
monotone decreasing in $m(x)$.

$(ii)$ $\lambda_{p}(\mathcal{L}_{\\{(L_{1},\,L_{2}),\,d\\}}+a(x))$ is strictly
monotone decreasing in $b(x)$.

From now on, we discuss the effect of interval length on the principal
eigenvalue $\lambda_{p}(\mathcal{L}_{\\{(L_{1},L_{2}),\,d\\}}+a(x))$.

###### Theorem 3.3

Suppose that $\mathbf{(J)}$ and $\mathbf{(H)}$ hold; then, the following three
conclusions hold:

$(i)$ $\lambda_{p}(\mathcal{L}_{\\{(L_{1},\,L_{2}),\,d\\}}+a(x))$ is
continuous for $L_{1},L_{2}\in(-\infty,+\infty)$.

$(ii)$ $\lim\limits_{L_{1},\,L_{2}\to
0}\lambda_{p}(\mathcal{L}_{\\{(L_{1},\,L_{2}),\,d\\}}+a(x))=a(0)-d$.

$(iii)$
$\lim\limits_{-L_{1},L_{2}\to+\infty}\lambda_{p}(\mathcal{L}_{\\{(L_{1},\,L_{2}),\,d\\}}+a(x))=\sup\limits_{x\in\mathbb{R}}a(x)$.

Proof. The proof of $(i)$ is similar to Proposition 3.4 in [5], and we omit it
here.

$(ii)$ Due to the continuity of $a(x)$, for any given $\epsilon>0$, there
exists $h>0$ small enough such that

$|a(x)-a(0)|<\epsilon,\,\,x\in[-h,h].$

Since $\lambda_{p}(\mathcal{L}_{\\{(-h,\,h),\,d\\}}+a(x))$ is a principal
eigenvalue, there exists a positive function $\phi(x)\in C([-h,h])$ such that

$d\int_{-h}^{h}J(x-y)\phi(y)dy-d\phi(x)+a(x)\phi(x)=\lambda_{p}(\mathcal{L}_{\\{(-h,\,h),\,d\\}}+a(x))\phi(x),\,\,x\in[-h,h],$

which gives by integrating,

$\begin{array}[]{llllll}&&|\lambda_{p}(\mathcal{L}_{\\{(-h,\,h),\,d\\}}+a(x))-a(0)+d|\\\\[5.69054pt]
&=&|\frac{d\int_{-h}^{h}\int_{-h}^{h}J(x-y)\phi(y)\phi(x)dydx}{\int_{-h}^{h}\phi^{2}(x)dx}+\frac{\int_{-h}^{h}a(x)\phi^{2}(x)dx}{\int_{-h}^{h}\phi^{2}(x)dx}-a(0)|\\\\[5.69054pt]
&=&|\frac{d\int_{-h}^{h}\int_{-h}^{h}J(x-y)\phi(y)\phi(x)dydx}{\int_{-h}^{h}\phi^{2}(x)dx}+\frac{\int_{-h}^{h}(a(x)-a(0))\phi^{2}(x)dx}{\int_{-h}^{h}\phi^{2}(x)dx}|\\\\[5.69054pt]
&\leq&\frac{d\|J\|_{\infty}(\int_{-h}^{h}\phi(x)dx)^{2}}{\int_{-h}^{h}\phi^{2}(x)dx}+|\frac{\int_{-h}^{h}(a(x)-a(0))\phi^{2}(x)dx}{\int_{-h}^{h}\phi^{2}(x)dx}|\\\\[5.69054pt]
&\leq&2d\|J\|_{\infty}h+\epsilon\rightarrow\epsilon\,\,\,\mbox{as}\,\,\,h\rightarrow
0^{+}.\end{array}$

From the arbitrariness of $\epsilon$, we have

$|\lambda_{p}(\mathcal{L}_{\\{(-h,\,h),\,d\\}}+a(x))-a(0)+d|\rightarrow
0\,\,\,\mbox{as}\,\,\,h\rightarrow 0^{+},$

which together with the continuity of
$\lambda_{p}(\mathcal{L}_{\\{(L_{1},\,L_{2}),\,d\\}}+a(x))$ about
$L_{1},\,L_{2}$ give

$\lim\limits_{L_{1},\,L_{2}\to
0}\lambda_{p}(\mathcal{L}_{\\{(L_{1},\,L_{2}),\,d\\}}+a(x))=a(0)-d.$

$(iii)$ According to the monotonicity of
$\lambda_{p}(\mathcal{L}_{\\{(L_{1},\,L_{2}),\,d\\}}+a(x))$ with respect to
interval $(L_{1},L_{2})$ and function $a(x)$ yields

$\lambda_{p}(\mathcal{L}_{\\{(L_{1},\,L_{2}),\,d\\}}+a(x))\leq\lambda_{p}(\mathcal{L}_{\\{(L_{1},\,L_{2}),\,d\\}}+\sup\limits_{x\in\mathbb{R}}a(x))\leq\lambda_{p}(\mathcal{L}_{\\{(-\infty,\,+\infty),\,d\\}}+\sup\limits_{x\in\mathbb{R}}a(x)).$

Consider the following eigenvalue problem

$d\int_{-\infty}^{\infty}J(x-y)\phi(y)dy-d\phi(x)+\phi(x)\sup\limits_{x\in\mathbb{R}}a(x)=\lambda_{p}(\mathcal{L}_{\\{(-\infty,\,+\infty),\,d\\}}+\sup\limits_{x\in\mathbb{R}}a(x))\phi,\,\,x\in\mathbb{R}.$
(3.2)

It is easily seen that
$\lambda_{p}(\mathcal{L}_{\\{(-\infty,\,+\infty),\,d\\}}+\sup\limits_{x\in\mathbb{R}}a(x))=\sup\limits_{x\in\mathbb{R}}a(x)$.
So the principal eigenvalue
$\lambda_{p}(\mathcal{L}_{\\{(L_{1},\,L_{2}),\,d\\}}+a(x))\leq\sup\limits_{x\in\mathbb{R}}a(x)$.
Therefore,

$\limsup\limits_{-L_{1},\,L_{2}\to+\infty}\lambda_{p}(\mathcal{L}_{\\{(L_{1},\,L_{2}),\,d\\}}+a(x))\leq\sup\limits_{x\in\mathbb{R}}a(x).$

To prove $(iii)$, it suffices to prove that
$\liminf\limits_{-L_{1},\,L_{2}\to+\infty}\lambda_{p}(\mathcal{L}_{\\{(L_{1},\,L_{2}),\,d\\}}+a(x))\geq\sup\limits_{x\in\mathbb{R}}a(x)$
holds. In fact, by the continuity of $a(x)$ and the definition of sup, for
given $\epsilon>0$, there exists some $x_{0}\in\mathbb{R}$ such that

$\sup\limits_{x\in\mathbb{R}}a(x)-\epsilon\leq a(x_{0}).$

By $\mathbf{(J)}$, for given $\epsilon>0$, there exist $L_{1}<x_{0}-1$ and
$L_{2}>x_{0}+1$ such that

$\int_{L_{1}}^{L_{2}}J(z)dz>1-\epsilon.$

Now take

$\delta_{n}(x-x_{0})=\left\\{\begin{array}[]{ll}k_{1}e^{-1/(1-n^{2}(x-x_{0})^{2})}>0,&x\in(x_{0}-1/n,x_{0}+1/n),\\\\[5.69054pt]
=0,&x\notin(x_{0}-1/n,x_{0}+1/n),\end{array}\right.$

where $k_{1}$ is positive and satisfies
$\int_{\mathbb{R}}\delta_{n}(x-x_{0})dx=1$. It is easy to check that the
sequence $\\{\delta_{n}(x-x_{0})\\}$ weakly converges to some
$\delta(x-x_{0})$ in $L^{1}((L_{1},L_{2}))$. By the definition of
$\lambda_{p}(\mathcal{L}_{\\{(L_{1},\,L_{2}),\,d\\}}+a(x))$, one can easily
obtain

$\begin{array}[]{llllll}&&d\int_{L_{1}}^{L_{2}}J(x-y)\delta_{n}(y-x_{0})dy-d\delta_{n}(x-x_{0})+a(x)\delta_{n}(x-x_{0})\\\\[5.69054pt]
&\leq&\lambda_{p}(\mathcal{L}_{\\{(L_{1},\,L_{2}),\,d\\}}+a(x))\delta_{n}(x-x_{0}),\,\,x\in(L_{1},L_{2}).\end{array}$
(3.3)

Integrating the equation of (3.3) over $(L_{1},L_{2})$ yields

$\begin{array}[]{llllll}\lambda_{p}(\mathcal{L}_{\\{(L_{1},\,L_{2}),\,d\\}}+a(x))&\geq&\frac{d\int_{L_{1}}^{L_{2}}\int_{L_{1}}^{L_{2}}J(x-y)\delta_{n}(y-x_{0})dydx-d\int_{L_{1}}^{L_{2}}\delta_{n}(x-x_{0})dx+\int_{L_{1}}^{L_{2}}a(x)\delta_{n}(x-x_{0})dx}{\int_{L_{1}}^{L_{2}}\delta_{n}(x-x_{0})dx}\\\\[5.69054pt]
&\geq&\frac{d\int_{L_{1}}^{L_{2}}\int_{x_{0}-1}^{x_{0}+1}J(x-y)\delta_{n}(y-x_{0})dydx-d\int_{L_{1}}^{L_{2}}\delta_{n}(x-x_{0})dx+\int_{L_{1}}^{L_{2}}a(x)\delta_{n}(x-x_{0})dx}{\int_{L_{1}}^{L_{2}}\delta_{n}(x-x_{0})dx}\\\\[5.69054pt]
&\geq&\frac{d\int_{x_{0}-1}^{x_{0}+1}\delta_{n}(y-x_{0})[\int_{L_{1}-x_{0}+1}^{L_{2}-x_{0}-1}J(z)dz]dy-d\int_{L_{1}}^{L_{2}}\delta_{n}(x-x_{0})dx+\int_{L_{1}}^{L_{2}}a(x)\delta_{n}(x-x_{0})dx}{\int_{L_{1}}^{L_{2}}\delta_{n}(x-x_{0})dx}\\\\[5.69054pt]
&\geq&\frac{(d(1-\epsilon)-d)\int_{L_{1}}^{L_{2}}\delta_{n}(x-x_{0})dx+\int_{L_{1}}^{L_{2}}a(x)\delta_{n}(x-x_{0})dx}{\int_{L_{1}}^{L_{2}}\delta_{n}(x-x_{0})dx}\\\\[5.69054pt]
&=&-d\epsilon+\frac{\int_{L_{1}}^{L_{2}}a(x)\delta_{n}(x-x_{0})dx}{\int_{L_{1}}^{L_{2}}\delta_{n}(x-x_{0})dx},\end{array}$

where we have used that
$\int_{x_{0}-1}^{x_{0}+1}\delta_{n}(x-x_{0})dx=\int_{L_{1}}^{L_{2}}\delta_{n}(x-x_{0})dx$
for sufficiently large $n$. Therefore, by taking $n\rightarrow+\infty$,

$\begin{array}[]{llllll}\liminf\limits_{-L_{1},\,L_{2}\to+\infty}\lambda_{p}(\mathcal{L}_{\\{(L_{1},L_{2}),\,d\\}}+a(x))&\geq&-d\epsilon+\int_{-\infty}^{+\infty}a(x)\delta(x-x_{0})dx\\\\[5.69054pt]
&=&-d\epsilon+a(x_{0})\\\\[5.69054pt]
&\geq&-d\epsilon+\sup\limits_{x\in\mathbb{R}}a(x)-\epsilon.\end{array}$

It follows from the arbitrarily of $\epsilon$ that

$\liminf\limits_{-L_{1},L_{2}\to+\infty}\lambda_{p}(\mathcal{L}_{\\{(L_{1},\,L_{2}),\,d\\}}+a(x))\geq\sup\limits_{x\in\mathbb{R}}a(x).$

$\Box$

In the following, we discuss the monotonicity of the principal eigenvalue with
respect to $d$ and its limiting behaviors as $d\rightarrow 0$ or
$d\rightarrow+\infty$.

###### Theorem 3.4

Suppose that $\mathbf{(J)}$ and $\mathbf{(H)}$ hold. Then, the following
statements hold:

$(i)$ $\lambda_{p}(\mathcal{L}_{\\{(L_{1},\,L_{2}),\,d\\}}+a(x))$ is a
strictly monotone decreasing function of $d$.

$(ii)$ $\lim\limits_{d\to
0}\lambda_{p}(\mathcal{L}_{\\{(L_{1},\,L_{2}),\,d\\}}+a(x))=\max\limits_{x\in[L_{1},\,L_{2}]}a(x)$.

$(iii)$ If $\int_{-\infty}^{L_{1}-L_{2}}J(z)dz>0$(symmetrically,
$\int_{L_{2}-L_{1}}^{+\infty}J(z)dz>0$) holds, then
$\lim\limits_{d\to+\infty}\lambda_{p}(\mathcal{L}_{\\{(L_{1},\,L_{2}),\,d\\}}+a(x))=-\infty$.

Proof. $(i)$ Assume that
$\lambda_{p}(d_{1}):=\lambda_{p}(\mathcal{L}_{\\{(L_{1},L_{2}),\,d\\}}+a(x))$
is the principal eigenvalue and $\phi(x)$ is the corresponding positive
eigenfunction with $\|\phi\|_{L^{2}}=1$, we have

$\lambda_{p}(d_{1})\phi(x)=d_{1}\int_{L_{1}}^{L_{2}}J(x-y)\phi(y)dy-
d_{1}\phi(x)+a(x)\phi(x),\,\,\,x\in(L_{1},L_{2}).$

Suppose that $d<d_{1}$, then

$\begin{array}[]{llllll}\lambda_{p}(d_{1})&=&d_{1}\int_{L_{1}}^{L_{2}}\int_{L_{1}}^{L_{2}}J(x-y)\phi(y)\phi(x)dydx-
d_{1}+\int_{L_{1}}^{L_{2}}a(x)\phi^{2}(x)dx\\\\[5.69054pt]
&=&d\int_{L_{1}}^{L_{2}}\int_{L_{1}}^{L_{2}}J(x-y)\phi(y)\phi(x)dydx-
d_{1}+\int_{L_{1}}^{L_{2}}a(x)\phi^{2}(x)dx\\\\[5.69054pt]
&&+(d_{1}-d)\int_{L_{1}}^{L_{2}}\int_{L_{1}}^{L_{2}}J(x-y)\phi(y)\phi(x)dydx\\\\[5.69054pt]
&<&d\int_{L_{1}}^{L_{2}}\int_{L_{1}}^{L_{2}}J(x-y)\phi(y)\phi(x)dydx-
d_{1}+\int_{L_{1}}^{L_{2}}a(x)\phi^{2}(x)dx+d_{1}-d\\\\[5.69054pt]
&=&d\int_{L_{1}}^{L_{2}}\int_{L_{1}}^{L_{2}}J(x-y)\phi(y)\phi(x)dydx-d+\int_{L_{1}}^{L_{2}}a(x)\phi^{2}(x)dx\\\\[5.69054pt]
&\leq&\lambda_{p}(d).\end{array}$

Therefore $\lambda_{p}(d_{1})<\lambda_{p}(d)$.

$(ii)$ The idea of this proof is from Theorem 2.8 in [44]. For the eigenvalue
problem

$d\int_{L_{1}}^{L_{2}}J(x-y)\varphi(y)dy-d\varphi(x)+(\max\limits_{x\in[L_{1},\,L_{2}]}a(x))\,\varphi(x)=\lambda^{*}_{p}\varphi(x),\,\,x\in(L_{1},L_{2}).$
(3.4)

It follows from [19] that
$\lambda^{*}_{p}\leq\max\limits_{x\in[L_{1},\,L_{2}]}a(x)$. Therefore, we have
$\lambda_{p}(\mathcal{L}_{\\{(L_{1},\,L_{2}),\,d\\}}+a(x))\leq\lambda^{*}_{p}\leq\max\limits_{x\in[L_{1},\,L_{2}]}a(x)$
by $(ii)$ of Proposition 3.1.

Next, we prove that $\liminf\limits_{d\to
0}\lambda_{p}(\mathcal{L}_{\\{(L_{1},\,L_{2}),\,d\\}}+a(x))\geq\max\limits_{x\in[L_{1},\,L_{2}]}a(x)$.
Assume for the contrary that $\liminf\limits_{d\to
0}\lambda_{p}(\mathcal{L}_{\\{(L_{1},\,L_{2}),\,d\\}}+a(x))\leq\max\limits_{x\in[L_{1},\,L_{2}]}a(x)-\epsilon$
for some $\epsilon>0$. By the definition of $\liminf$, there exists some
$\widehat{d}>0$ such that if $d\leq\widehat{d}$, then

$\lambda_{p}(\mathcal{L}_{\\{(L_{1},\,L_{2}),\,d\\}}+a(x))\leq\max\limits_{x\in[L_{1},\,L_{2}]}a(x)-\frac{\epsilon}{2}.$

On the other hand, by the continuity of $a(x)$, there exist
$x_{0}\in(L_{1},L_{2})$ and $r>0$ such that

$\max\limits_{x\in[L_{1},\,L_{2}]}a(x)\leq a(x)+\frac{\epsilon}{4},\,\,x\in
U_{r}(x_{0})\subset(L_{1},L_{2}).$

Therefore,

$\lambda_{p}(\mathcal{L}_{\\{(L_{1},\,L_{2}),\,d\\}}+a(x))\leq
a(x)-\frac{\epsilon}{4}$

for $0<d<\widehat{d}$ and $x\in U_{r}(x_{0})$. Let
$(\lambda_{p}(\mathcal{L}_{\\{(L_{1},\,L_{2}),\,d\\}}+a(x)),\psi(x))$ be the
eigenpair of the following eigenvalue problem:

$d\int_{L_{1}}^{L_{2}}J(x-y)\psi(y)dy-d\psi(x)+a(x)\psi(x)=\lambda_{p}(\mathcal{L}_{\\{(L_{1},\,L_{2}),\,d\\}}+a(x))\psi(x),\
x\in(L_{1},L_{2}).$

Then,

$\int_{L_{1}}^{L_{2}}J(x-y)\psi(y)dy-\psi(x)=\frac{\lambda_{p}(\mathcal{L}_{\\{(L_{1},\,L_{2}),\,d\\}}+a(x))-a(x)}{d}\psi(x)\leq-\frac{\epsilon}{4d}\psi(x)\,\,\,\mbox{in}\,\,\,U_{r}(x_{0}).$

Let $\widetilde{\lambda}$ be the principal eigenvalue of the linear problem

$\left\\{\begin{array}[]{lll}\int_{\mathbb{R}}J(x-y)u(y)dy-u(x)=\lambda
u(x)&\,\mbox{in}\,\,\,U_{r}(x_{0}),\\\\[5.69054pt]
u(x)=0&\,\mbox{in}\,\,\,\mathbb{R}\backslash U_{r}(x_{0}).\end{array}\right.$
(3.5)

It is well known that $-1<\widetilde{\lambda}<0$ by Theorem 2.1 in [20]. Let
$\Psi(x)$ be the eigenfunction corresponding to $\widetilde{\lambda}$ and
$\|\Psi(x)\|_{L^{\infty}}=1$. Take

$\overline{\psi}(x)=\frac{\psi(x)}{\inf_{U_{r}(x_{0})}\psi(x)},\,\,\,\underline{\psi}(x)=\Psi(x).$

We consider the following problem

$\left\\{\begin{array}[]{lll}\int_{\mathbb{R}}J(x-y)u(y)dy-u(x)=-\frac{\epsilon}{4d}u(x)&\,\mbox{in}\,\,\,U_{r}(x_{0}),\\\\[5.69054pt]
u(x)=0&\,\mbox{in}\,\,\,\mathbb{R}\backslash U_{r}(x_{0}).\end{array}\right.$
(3.6)

Direct calculation yields

$\begin{array}[]{llllll}&&\int_{U_{r}(x_{0})}J(x-y)\overline{\psi}(y)dy-\overline{\psi}(x)+\frac{\epsilon}{4d}\overline{\psi}(x)\\\\[5.69054pt]
&=&\frac{1}{\inf\psi}[\int_{U_{r}(x_{0})}J(x-y)\psi(y)dy-\psi(x)]+\frac{\epsilon}{4d}\overline{\psi}(x)\\\\[5.69054pt]
&\leq&\frac{1}{\inf\psi}(-\frac{\epsilon}{4d}\psi(x)+\frac{\epsilon}{4d}\psi(x))\\\\[5.69054pt]
&=&0,\end{array}$

and

$\begin{array}[]{llllll}&&\int_{U_{r}(x_{0})}J(x-y)\underline{\psi}(y)dy-\underline{\psi}(x)+\frac{\epsilon}{4d}\underline{\psi}(x)\\\\[5.69054pt]
&=&\int_{U_{r}(x_{0})}J(x-y)\Psi(x)dy-\Psi(x)]+\frac{\epsilon}{4d}\Psi(x)\\\\[5.69054pt]
&=&\widetilde{\lambda}\Psi(x)+\frac{\epsilon}{4d}\Psi(x)\\\\[5.69054pt]
&\geq&0\end{array}$

provided $d<\min\\{\widehat{d},-\frac{\epsilon}{4\widetilde{\lambda}}\\}$.
Hence, by the super-sub solution method in [21], one can yield (3.6) has a
positive solution between $\overline{\psi}(x)$ and $\underline{\psi}$, which
implies that $\widetilde{\lambda}=-\frac{\epsilon}{4d}$. This contradicts to
the independence of $\widetilde{\lambda}$ from $d$. Therefore,
$\lim\limits_{d\to
0}\lambda_{p}(\mathcal{L}_{\\{(L_{1},\,L_{2}),\,d\\}}+a(x))=\max\limits_{x\in[L_{1},\,L_{2}]}a(x)$.

$(iii)$ We claim that if $\int_{-\infty}^{L_{1}-L_{2}}J(z)dz>0$ (or
$\int_{L_{2}-L_{1}}^{+\infty}J(z)dz>0$), then
$\lim\limits_{d\to+\infty}\lambda_{p}(\mathcal{L}_{\\{(L_{1},\,L_{2}),\,d\\}}+a(x)):=\lambda_{\infty}=-\infty$.
We argue by contradiction and suppose that $\lambda_{\infty}>-\infty$. Now let
$\varphi(x)=C_{1}$ (positive constant), without loss of generality, taking
$C_{1}=1$, then for $x\in(L_{1},L_{2})$,

$\begin{array}[]{llllll}&&d\int_{L_{1}}^{L_{2}}J(x-y)\varphi(y)dy-d\varphi(x)+a(x)\varphi(x)\\\\[5.69054pt]
&<&d\int_{L_{1}}^{L_{2}}J(x-y)dy-d+\max\limits_{x\in[L_{1},L_{2}]}a(x)\\\\[5.69054pt]
&=&-d\int_{\mathbb{R}\backslash[L_{1},L_{2}]}J(x-y)dy+\max\limits_{x\in[L_{1},L_{2}]}a(x)\\\\[5.69054pt]
&=&-d(\int_{-\infty}^{x-L_{2}}+\int_{x-L_{1}}^{+\infty})J(z)dz+\max\limits_{x\in[L_{1},L_{2}]}a(x)\\\
&\leq&-d(\int_{-\infty}^{L_{1}-L_{2}}+\int_{L_{2}-L_{1}}^{+\infty})J(z)dz+\max\limits_{x\in[L_{1},L_{2}]}a(x)\end{array}$

as $\int_{\mathbb{R}}J(x)dx=1$. Owing to the assumption that
$\int_{-\infty}^{L_{1}-L_{2}}J(z)dz>0$ (or
$\int_{L_{2}-L_{1}}^{+\infty}J(z)dz>0$), there exists a $d$ adequately large
such that

$d\int_{L_{1}}^{L_{2}}J(x-y)\varphi(y)dy-d\varphi(x)+a(x)\varphi(x)\leq(\lambda_{\infty}-1)\varphi(x).$

Thus, by definition of
$\lambda_{p}(\mathcal{L}_{\\{(L_{1},\,L_{2}),\,d\\}}+a(x))$, we have

$\lambda_{p}(\mathcal{L}_{\\{(L_{1},\,L_{2}),\,d\\}}+a(x))\leq\lambda_{\infty}-1,$

further,

$\lim\limits_{d\to+\infty}\lambda_{p}(\mathcal{L}_{\\{(L_{1},\,L_{2}),\,d\\}}+a(x))=\lambda_{\infty}\leq\lambda_{\infty}-1.$

We obtain the desired contradiction. Hence,
$\lim\limits_{d\to+\infty}\lambda_{p}(\mathcal{L}_{\\{(L_{1},\,L_{2}),\,d\\}}+a(x))=-\infty$.

$\Box$

###### Remark 3.5

Compared with [44], where the nonlocal operator is
$d\int_{\Omega}J(x-y)(\psi(y)-\psi(x))dy$, we consider the nonlocal operator
$d\int_{\Omega}J(x-y)\varphi(y)dy-d\varphi(x)$ and prove that when
$d\rightarrow+\infty$, the limit of the principal eigenvalue is $-\infty$ for
some cases, which is different from the result in [44]; its limit is the
average of $a(x)$ over $\Omega$.

##  4 Spreading-vanishing

Since $h(t)$ and $-g(t)$ are monotonically increasing with $t>0$, there exist
$h_{\infty}$ and $g_{\infty}$ such that
$\lim\limits_{t\rightarrow+\infty}g(t)=g_{\infty}\in[-\infty,-h_{0})$ and
$\lim\limits_{t\rightarrow+\infty}h(t)=h_{\infty}\in(h_{0},+\infty]$. Here we
define that $\mathbf{vanishing}$ occurs if $h_{\infty}-g_{\infty}<+\infty$ and
$\lim\limits_{t\rightarrow+\infty}\max\limits_{x\in[g(t),\,h(t)]}I(t,x)=0$;
and $\mathbf{spreading}$ happens provided that $h_{\infty}-g_{\infty}=+\infty$
and $\limsup\limits_{t\rightarrow+\infty}\|I(\cdot,t)\|_{C([g(t),\,h(t)])}>0$.
In this section, we always assume that $\mathbf{(J)}$ and $\mathbf{(H)}$ hold.
The following proposition directly holds from Theorem 3.5 in [6].

###### Proposition 4.1

Let $(S,I;g,h)$ be the unique solution of (1.3). If
$h_{\infty}-g_{\infty}<+\infty$, then
$\lim\limits_{t\rightarrow+\infty}g^{\prime}(t)=\lim\limits_{t\rightarrow+\infty}h^{\prime}(t)=0$.

Next, we discuss the asymptotic behavior of the solution of problem (1.3) when
$h_{\infty}-g_{\infty}<+\infty$.

###### Theorem 4.2

Let $(S,I;g,h)$ be the unique solution of problem (1.3) with
$h_{\infty}-g_{\infty}<+\infty$, then
$\lim\limits_{t\rightarrow+\infty}\max\limits_{x\in[g(t),\,h(t)]}I(t,x)=0$,
$\lim\limits_{t\rightarrow+\infty}S(t,x)=\frac{\sigma}{\mu_{1}}$ and
$\lambda_{p}(\mathcal{L}_{\\{(g_{\infty},\,h_{\infty}),\,d\\}}+a(x))\leq 0$.

Proof. Assume by contradiction that
$\lim\limits_{t\rightarrow+\infty}\max\limits_{x\in[g(t),\,h(t)]}I(t,x)>0$,
there exists $\epsilon_{1}>0$ and sequence
$\\{(t_{i},x_{i})\\}_{i=1}^{\infty}$ with $x_{i}\in[g(t),h(t)]$ and
$t_{i}\rightarrow+\infty$ as $i\rightarrow+\infty$ such that
$I(t_{i},x_{i})\geq\frac{\epsilon_{1}}{2}$ for $i\in\mathbb{N}$. Since
$g_{\infty}<g(t)<x_{i}<h(t)<h_{\infty}$, there exists a subsequence
$\\{x_{i_{j}}\\}_{j=1}^{\infty}$ such that $x_{i_{j}}\rightarrow
x_{0}\in(g_{\infty},h_{\infty})$ as $j\rightarrow+\infty$. For
$t\in(-t_{i},+\infty)$ and $x\in(g(t+t_{i}),h(t+t_{i}))$, define

$\overline{I_{i}}(t,x)=I(t+t_{i},x).$

Applying Theorem 2.2 gives that $I$ and $S$ are positive and bounded, and then
$\overline{I_{i}}(t,x)$ satisfies

$\overline{I_{it}}\geq
d\int_{g_{i}(t)}^{h_{i}(t)}J(x-y)\overline{I_{i}}(t,y)dy-d\overline{I_{i}}(t,x)-\mu_{2}\overline{I_{i}}-\gamma(b,0,x)\overline{I_{i}},\,t>-t_{i},\,x\in(g_{i}(t),h_{i}(t)).$

We next consider the following auxiliary problem

$\left\\{\begin{array}[]{lll}u_{t}=d\int_{g_{i}(t)}^{h_{i}(t)}J(x-y)u(t,y)dy-
du(t,x)-\mu_{2}u-\gamma(b,0,x)u,&\,t>-t_{i},\,x\in(g_{i}(t),h_{i}(t)),\\\\[5.69054pt]
u(0,x)=\overline{I}_{i}(0,x),&\,x\in(g_{i}(t),h_{i}(t)),\end{array}\right.$

it follows that $u(t,x)\rightarrow U(t,x)$ as $i\rightarrow+\infty$, and
$U(t,x)$ satisfies

$\left\\{\begin{array}[]{lll}U_{t}(t,x)=d\int_{g_{\infty}}^{h_{\infty}}J(x-y)U(t,y)dy-
dU(t,x)-\mu_{2}U-\gamma(b,0,x)U,&\,t\in\mathbb{R},\,\,\,x\in(g_{\infty},h_{\infty}),\\\\[5.69054pt]
U(0,x_{0})=\lim\limits_{i\rightarrow+\infty}\overline{I}_{i}(0,x_{i})=\lim\limits_{i\rightarrow+\infty}I(t_{i},x_{i})\geq\frac{\epsilon_{1}}{2}>0,\end{array}\right.$

and then $U(t,x)>0$ in $\mathbb{R}\times(g_{\infty},h_{\infty})$ by the
maximum principle [15] for the nonlocal problem.

On the other hand, considering $h_{\infty}-g_{\infty}<+\infty$ and Proposition
4.1, we have
$\lim\limits_{t\rightarrow+\infty}g^{\prime}(t)=\lim\limits_{t\rightarrow+\infty}h^{\prime}(t)=0$
as $t\rightarrow+\infty$, which means

$\begin{array}[]{llllll}0=\lim\limits_{i\rightarrow+\infty}h^{\prime}(t+t_{i})&=&k\lim\limits_{i\rightarrow+\infty}\int_{g(t+t_{i})}^{h(t+t_{i})}\int_{h(t+t_{i})}^{+\infty}J(x-y)\overline{I}_{i}(t,x)dydx\\\\[5.69054pt]
&\geq&k\int_{g(\infty)}^{h(\infty)}\int_{h(\infty)}^{+\infty}J(x-y)U(t,x)dydx\\\\[5.69054pt]
&>&0\end{array}$

and

$\begin{array}[]{llllll}0=\lim\limits_{i\rightarrow+\infty}g^{\prime}(t+t_{i})&=&-k\lim\limits_{i\rightarrow+\infty}\int_{g(t+t_{i})}^{h(t+t_{i})}\int_{-\infty}^{g(t+t_{i})}J(x-y)\overline{I}_{i}(t,x)dydx\\\\[5.69054pt]
&\leq&-k\int_{g(\infty)}^{h(\infty)}\int_{-\infty}^{g(\infty)}J(x-y)U(t,x)dydx\\\\[5.69054pt]
&<&0.\end{array}$

It is a contradiction. Hence,
$\lim\limits_{t\rightarrow+\infty}\max\limits_{x\in[g(t),\,h(t)]}I(t,x)=0$.

Next, we will prove that
$\lim\limits_{t\rightarrow+\infty}S(t,x)=\frac{\sigma}{\mu_{1}}$. Since
$\lim\limits_{t\rightarrow+\infty}\max\limits_{x\in[g(t),\,h(t)]}I(t,x)=0$,
for any $\epsilon>0$, one can choose a $T>0$ large such that

$0<I(t,x)<\epsilon$

for $t>T$, $x\in(g(t),h(t))$.

Obviously, $S(t,x)$ satisfies

$S_{t}\geq d\mathcal{L}_{1}[S]+\sigma-\mu_{1}S-\epsilon\beta(m,0,x)S,\,\,t>T,\
x\in\mathbb{R},$

and then $S(t,x)\geq\underline{S}(t)$, where $\underline{S}(t)$ is the
solution to problem

$\left\\{\begin{array}[]{ll}\underline{S}_{t}=\sigma-(\mu_{1}+\epsilon\sup\limits_{x\in\mathbb{R}}\beta(m,0,x))\underline{S},&t>T,\\\\[5.69054pt]
\underline{S}(T)=\inf\limits_{x\in\mathbb{R}}S(T,x).\end{array}\right.$

It follows from Lemma 2.4 in [26] that
$\lim\limits_{t\rightarrow+\infty}\underline{S}(t)=\frac{\sigma}{\mu_{1}+\epsilon\sup\limits_{x\in\mathbb{R}}\beta(m,0,x)}$.
Therefore,
$\liminf\limits_{t\rightarrow+\infty}S(t,x)\geq\frac{\sigma}{\mu_{1}+\epsilon\sup\limits_{x\in\mathbb{R}}\beta(m,0,x)}$.
Letting $\epsilon\rightarrow 0$ yields

$\liminf\limits_{t\rightarrow+\infty}S(t,x)\geq\frac{\sigma}{\mu_{1}}.$ (4.1)

On the other hand, $S(t,x)$ satisfies

$S_{t}\leq
d\mathcal{L}_{1}[S]+\sigma-\mu_{1}S+\gamma(b,\epsilon,x)\epsilon,\,\,t>T,\
x\in\mathbb{R}.$

Let $\overline{S}(t)$ be the solution of

$\left\\{\begin{array}[]{ll}\overline{S}_{t}=\sigma-\mu_{1}\overline{S}(t)+\sup\limits_{x\in\mathbb{R}}\gamma(b,\epsilon,x)\epsilon,&t>T,\\\\[5.69054pt]
\overline{S}(T)=\sup\limits_{x\in\mathbb{R}}S(T,x).\end{array}\right.$

Apparently, $S(t,x)\leq\overline{S}(t)$ by Lemma 2.4 in [26] and
$\lim\limits_{t\rightarrow+\infty}\overline{S}(t)=\frac{1}{\mu_{1}}[\sigma+\sup\limits_{x\in\mathbb{R}}\gamma(b,\epsilon,x)\epsilon]$.
Therefore,

$\limsup\limits_{t\rightarrow+\infty}S(t,x)\leq\lim\limits_{t\rightarrow+\infty}\overline{S}(t)=\frac{1}{\mu_{1}}[\sigma+\sup\limits_{x\in\mathbb{R}}\gamma(b,\epsilon,x)\epsilon].$

Letting $\epsilon\rightarrow 0$ gives

$\limsup\limits_{t\rightarrow+\infty}S(t,x)\leq\frac{\sigma}{\mu_{1}},$ (4.2)

which together with (4.1) yields
$\lim\limits_{t\rightarrow+\infty}S(t,x)=\frac{\sigma}{\mu_{1}}$ uniformly for
$x\in\mathbb{R}$.

In what follows, we prove that
$\lambda_{p}(\mathcal{L}_{\\{(g_{\infty},\,h_{\infty}),\,d\\}}+a(x))\leq 0$.
We argue by contradiction and suppose that
$\lambda_{p}(\mathcal{L}_{\\{(g_{\infty},\,h_{\infty}),\,d\\}}+a(x))>0$. Owing
to the continuous dependence of
$\lambda_{p}(\mathcal{L}_{\\{(g_{\infty},\,h_{\infty}),\,d\\}}+a(x))$ on
$a(x)$ and $(g_{\infty},h_{\infty})$, there exists a small $\epsilon$ such
that
$\lambda_{p}(\mathcal{L}_{\\{(g_{\infty}+\epsilon,\,h_{\infty}-\epsilon),\,d\\}}+a(x)-\beta(m,0,x)\epsilon)>0$.
Furthermore, in view of $S(t,x)\rightarrow\frac{\sigma}{\mu_{1}}$ as
$t\rightarrow+\infty$ and $h_{\infty}-g_{\infty}<+\infty$, there exists a
$T^{*}>0$ such that

$g(t)<g_{\infty}+\epsilon,\,\,h(t)>h_{\infty}-\epsilon,\,\,t>T^{*},$
$S(t,x)>\frac{\sigma}{\mu_{1}}-\epsilon,\,\,t>T^{*},\,\,x\in\mathbb{R}.$

Then, for $t>T^{*}$, $x\in(g(t),h(t))$,

$\begin{array}[]{llllll}I_{t}(t,x)&=&d\int_{g(t)}^{h(t)}J(x-y)I(t,y)dy-
dI(t,x)-\mu_{2}I+\beta(m,I,x)SI-\gamma(b,I,x)I\\\\[5.69054pt]
&\geq&d\int_{g_{\infty}+\epsilon}^{h_{\infty}-\epsilon}J(x-y)I(t,y)dy-
dI(t,x)-\mu_{2}I+\beta(m,I,x)(\frac{\sigma}{\mu_{1}}-\epsilon)I-\gamma(b,I,x)I\\\\[5.69054pt]
&\geq&d\int_{g_{\infty}+\epsilon}^{h_{\infty}-\epsilon}J(x-y)I(t,y)dy-
dI(t,x)-\mu_{2}I+\beta(m,0,x)(\frac{\sigma}{\mu_{1}}-\epsilon)I-\gamma(b,0,x)I.\end{array}$

Let $\phi(x)$ be the eigenfunction corresponding to
$\lambda_{p}(\mathcal{L}_{\\{(g_{\infty}+\epsilon,\,h_{\infty}-\epsilon),\,d\\}}+a(x)-\beta(m,0,x)\epsilon)$
and $\|\phi(x)\|_{L^{\infty}}=1$, and then $\phi(x)$ satisfies

$\begin{array}[]{llllll}&&d\int_{g_{\infty}+\epsilon}^{h_{\infty}-\epsilon}J(x-y)\phi(y)dy-d\phi-\mu_{2}\phi+\beta(m,0,x)(\frac{\sigma}{\mu_{1}}-\epsilon)\phi-\gamma(b,0,x)\phi\\\\[5.69054pt]
&=&d\int_{g_{\infty}+\epsilon}^{h_{\infty}-\epsilon}J(x-y)\phi(y)dy-d\phi+a(x)\phi-\beta(m,0,x)\epsilon\phi\\\\[5.69054pt]
&=&\lambda_{p}(\mathcal{L}_{\\{(g_{\infty}+\epsilon,\,h_{\infty}-\epsilon),\,d\\}}+a(x)-\beta(m,0,x)\epsilon)\phi.\end{array}$

If we choose $\delta$ sufficiently small such that $\delta\phi(x)\leq
I(T^{*},x)$ for $x\in[g_{\infty}+\epsilon,h_{\infty}-\epsilon]$, then

$I(t,x)\geq\delta\phi(x)>0\,\,\,\mbox{for}\,\,\,t>T^{*},\,\,\,x\in[g_{\infty}+\epsilon,h_{\infty}-\epsilon]$

by the comparison principle in [15], which leads to a contradiction to the
fact
$\lim\limits_{t\rightarrow+\infty}\max\limits_{x\in[g(t),\,h(t)]}I(t,x)=0$.

$\Box$

###### Theorem 4.3

Suppose $\lambda_{p}(\mathcal{L}_{\\{(-h_{0},\,h_{0}),\,d\\}}+a(x))<0$. Then
$h_{\infty}-g_{\infty}<+\infty$,
$\lim\limits_{t\rightarrow+\infty}\max\limits_{x\in[g(t),\,h(t)]}I(t,x)=0$ and
$\lim\limits_{t\rightarrow+\infty}S(t,x)=\frac{\sigma}{\mu_{1}}$ if
$\|S_{0}(x)\|_{L^{\infty}(\mathbb{R})}+\|I_{0}(x)\|_{C([-h_{0},\,h_{0}])}$ is
sufficiently small.

Proof. Applying Lemma 2.1 yields $S(t,x)\leq\frac{\sigma}{\mu_{1}}$ for
$t>0,\,x\in\mathbb{R}$ if
$\|S_{0}(x)\|_{L^{\infty}(\mathbb{R})}+\|I_{0}(x)\|_{C([-h_{0},\,h_{0}])}\leq\frac{\sigma}{\mu_{1}}$.
In view of $\lambda_{p}(\mathcal{L}_{\\{(-h_{0},\,h_{0}),\,d\\}}+a(x))<0$,
there exists a $\epsilon>0$ small such that
$\lambda_{p}(\mathcal{L}_{\\{(-h_{\epsilon},\,h_{\epsilon}),\,d\\}}+a(x))<0$
with $h_{\epsilon}=h_{0}+\epsilon$ by Theorem 3.3 $(i)$. Let $\phi(x)$ be the
eigenfunction of the principal eigenvalue
$\lambda_{p}(\mathcal{L}_{\\{(-h_{\epsilon},\,h_{\epsilon}),\,d\\}}+a(x))$,
which satisfies

$d\int_{-h_{\epsilon}}^{h_{\epsilon}}J(x-y)\phi(y)dy-d\phi(x)+a(x)\phi(x)=\lambda_{p}(\mathcal{L}_{\\{(-h_{\epsilon},\,h_{\epsilon}),\,d\\}}+a(x))\phi(x),\,\,x\in(-h_{\epsilon},h_{\epsilon}).$

Denote

$M=\delta
C(\int_{-h_{\epsilon}}^{h_{\epsilon}}\phi(x)dx)^{-1},\,\,\,C=\frac{h_{\epsilon}-h_{0}}{k},$
$\overline{h}(t)=h_{0}+kC[1-e^{-\delta
t}],\,\,\,\overline{g}(t)=-\overline{h}(t),$
$\overline{h}^{\prime}(t)=kC\delta e^{-\delta
t},\,\,\,\overline{I}(t,x)=Me^{-\delta t}\phi(x).$

By direct calculations, we have

$\begin{array}[]{llllll}&&k\int_{\overline{g}}^{\overline{h}}\int_{\overline{h}}^{\infty}J(x-y)\overline{I}(t,x)dydx\\\\[5.69054pt]
&\leq&k\int_{\overline{g}}^{\overline{h}}\overline{I}(t,x)dx\\\\[5.69054pt]
&=&k\delta Ce^{-\delta t}=\overline{h}^{\prime}(t),\,\,\,\,t>0.\end{array}$

In a similar way, one can deduce
$-k\int_{\overline{g}}^{\overline{h}}\int_{-\infty}^{\overline{g}}J(x-y)\overline{I}(t,x)dydx\geq\overline{g}^{\prime}(t)$
for $t>0$. Clearly, $\overline{I}(t,\overline{g}(t))>0$ and
$\overline{I}(t,\overline{h}(t))>0$ for $t>0$. For $t>0$ and
$x\in(\overline{g},\overline{h})$, we obtain

$\begin{array}[]{llllll}&&\overline{I}_{t}(t,x)-d\int_{\overline{g}(t)}^{\overline{h}(t)}J(x-y)\overline{I}(t,y)dy+d\overline{I}(t,x)+\mu_{2}\overline{I}-\beta(m,\overline{I},x)\overline{S}\overline{I}+\gamma(b,\overline{I},x)\overline{I}\\\\[5.69054pt]
&\geq&-[d\int_{\overline{g}(t)}^{\overline{h}(t)}J(x-y)Me^{-\delta
t}\phi(y)dy-dMe^{-\delta t}\phi(x)-\mu_{2}Me^{-\delta
t}\phi(x)]\\\\[5.69054pt]
&&-\delta\overline{I}+\gamma(b,\overline{I},x)\overline{I}-\beta(m,\overline{I},x)\overline{S}\overline{I}\\\\[5.69054pt]
&\geq&-\delta\overline{I}-(\lambda_{p}(\mathcal{L}_{(-h_{\epsilon},\,h_{\epsilon})}+a(x))+\gamma(b,0,x)-\beta(m,0,x)\frac{\sigma}{\mu_{1}})\overline{I}+\gamma(b,\overline{I},x)\overline{I}-\beta(m,\overline{I},x)\frac{\sigma}{\mu_{1}}\overline{I}\\\\[5.69054pt]
&=&\overline{I}(-\delta-\lambda_{p}(\mathcal{L}_{(-h_{\epsilon},\,h_{\epsilon})}+a(x))-\gamma(b,0,x)+\gamma(b,\overline{I},x)+\frac{\sigma}{\mu_{1}}(\beta(m,0,x)-\beta(m,\overline{I},x))).\end{array}$

Recalling that $\gamma(b,\overline{I},x)\rightarrow\gamma(b,0,x)$ and
$\beta(m,\overline{I},x)\rightarrow\beta(m,0,x)$ as $\delta\rightarrow 0$, we
can choose $\delta$ small enough so that

$\overline{I}_{t}(t,x)-d\int_{\overline{g}(t)}^{\overline{h}(t)}J(x-y)\overline{I}(t,y)dy+d\overline{I}(t,x)+\mu_{2}\overline{I}-\beta(x)\overline{S}\overline{I}+\gamma(x,b,\overline{I})\overline{I}\geq
0.$

Moreover, if
$\|S_{0}(x)\|_{L^{\infty}(\mathbb{R})}+\|I_{0}(x)\|_{C([-h_{0},\,h_{0}])}$
sufficiently small, then $I_{0}\leq M\phi(x)$, $x\in[-h_{0},h_{0}]$.

Applying the comparison principle gives

$\overline{g}(t)\leq g(t),\,\,\,h(t)\leq\overline{h}(t)\,\rm{and}\,\,\
I(t,x)\leq\overline{I}(t,x)$

for $t>0$ and $g(t)<x<h(t)$. Therefore,

$\lim\limits_{t\rightarrow+\infty}I(t,x)\leq\lim\limits_{t\rightarrow+\infty}\overline{I}(t,x)=0$
(4.3)

and

$h_{\infty}-g_{\infty}\leq 2h_{\epsilon}<+\infty,$

which gives that
$\lim\limits_{t\rightarrow+\infty}S(t,x)=\frac{\sigma}{\mu_{1}}$ uniformly for
$x\in\mathbb{R}$ by Theorem 4.2.
$\Box$

###### Remark 4.4

It follows from the proof of Theorem 4.3 that $M\rightarrow+\infty$ as
$k\rightarrow 0$. Therefore, there exists a $k_{*}>0$ such that (4.3) holds
for all $k\in(0,k_{*})$ for any given initial function pair
$(S_{0}(x),I_{0}(x))$.

###### Theorem 4.5

If $-g_{\infty}=h_{\infty}=+\infty$ and $\sup\limits_{x\in\mathbb{R}}a(x)>0$,
then
$\limsup\limits_{t\rightarrow+\infty}\|I(\cdot,t)\|_{C([g(t),\,h(t)])}>0$.

Proof. Conversely, suppose that
$\lim\limits_{t\rightarrow+\infty}\|I(t,\cdot)\|_{C([g(t),\,h(t)])}=0$. It
follows from Theorem 4.2 that

$\lim\limits_{t\rightarrow+\infty}S(t,x)=\frac{\sigma}{\mu_{1}}\,\,\
\textrm{uniformly for}\,\,\ x\in\mathbb{R}.$ (4.4)

Since $-g_{\infty}=h_{\infty}=+\infty$, from $(iii)$ of Theorem 3.3, there
exists a $T^{*}>0$ large enough such that

$\lambda_{p}(\mathcal{L}_{\\{(g(T^{*}),\,h(T^{*})),\,d\\}}+a(x))>0\,\,\,\
\textrm{for}\,\,\,\ t\geq T^{*}.$

Now, we consider the eigenvalue problem

$d\int_{g(T^{*})}^{h(T^{*})}J(x-y)\psi(y)dy-d\psi(x)+a(x)\psi(x)=\lambda_{p}(\mathcal{L}_{\\{(g(T^{*}),\,h(T^{*})),\,d\\}}+a(x))\psi(x),\,\,x\in(g(T^{*}),h(T^{*})),$

and positive function $\psi(x)$ with
$\|\psi\|_{L^{\infty}((g(T^{*}),\,h(T^{*})))}=1$ is its eigenfunction to the
principal eigenvalue
$\lambda_{p}(\mathcal{L}_{\\{(g(T^{*}),\,h(T^{*})),\,d\\}}+a(x))$.

In view of (4.4), for any given
$0<\epsilon<\min\\{\frac{\sigma}{\mu_{1}},\,\lambda_{p}(\mathcal{L}_{\\{(g(T^{*}),\,h(T^{*})),\,d\\}}+a(x))(\sup\limits_{x\in\mathbb{R}}\beta)^{-1}\\}$,
there exists a $T^{**}>T^{*}$ such that

$S(t,x)>\frac{\sigma}{\mu_{1}}-\epsilon,\,\,\,\ t\geq T^{**},\,\,\,\
x\in[g(T^{*}),h(T^{*})],$

and then $I(t,x)$ satisfies

$\left\\{\begin{array}[]{lll}I_{t}\geq
d\int_{g(T^{*})}^{h(T^{*})}J(x-y)I(t,y)dy-dI(t,x)-\mu_{2}I\\\\[5.69054pt]
\qquad+\beta(m,I,x)(\frac{\sigma}{\mu_{1}}-\epsilon)I-\gamma(b,I,x)I,&\,t>T^{**},\,\,x\in(g(T^{*}),h(T^{*})),\\\\[5.69054pt]
I(T^{**},x)>0,&\,g(T^{*})\leq x\leq h(T^{*}).\end{array}\right.$

We now construct a suitable lower solution for the following auxiliary problem

$\left\\{\begin{array}[]{lll}W_{t}=d\int_{g(T^{*})}^{h(T^{*})}J(x-y)W(t,y)dy-
dW(t,x)\\\\[5.69054pt]
\qquad-\mu_{2}W+\beta(m,W,x)(\frac{\sigma}{\mu_{1}}-\epsilon)W-\gamma(b,W,x)W,&\,t>T^{**},\
x\in(g(T^{*}),h(T^{*})),\\\\[5.69054pt]
W(T^{**},x)=I(T^{**},x),&\,g(T^{*})\leq x\leq h(T^{*}).\end{array}\right.$
(4.5)

Choose

$\underline{W}(t,x)=\delta\psi(x),\,\,\,\,\ t>T^{**},\,\,\ g(T^{*})\leq x\leq
h(T^{*}),$

where $\delta>0$ is small enough such that $\delta\psi(x)\leq I(T^{**},x)$ for
$x\in[g(T^{*}),h(T^{*})]$.

For $t>T^{**}$ and $g(T^{*})<x<h(T^{*})$, direct computation yields

$\begin{array}[]{llllll}&&\underline{W}_{t}-d\int_{g(T^{*})}^{h(T^{*})}J(x-y)\underline{W}(y)dy+d\underline{W}+\mu_{2}\underline{W}-\beta(m,\underline{W},x)(\frac{\sigma}{\mu_{1}}-\epsilon)\underline{W}+\gamma(b,\underline{W},x)\underline{W}\\\\[5.69054pt]
&=&-\delta(d\int_{g(T^{*})}^{h(T^{*})}J(x-y)\psi(t,y)dy-d\psi(x)-\mu_{2}\psi+\beta(m,\delta\psi,x)(\frac{\sigma}{\mu_{1}}-\epsilon)\psi-\gamma(b,\delta\psi,x)\psi)\\\\[5.69054pt]
&=&-\delta\psi(\lambda_{p}+\gamma(b,0,x)-\beta(m,0,x)\frac{\sigma}{\mu_{1}}+\beta(m,\delta\psi,x)(\frac{\sigma}{\mu_{1}}-\epsilon)-\gamma(b,\delta\psi,x))\\\\[5.69054pt]
&\leq&\delta\psi(-\lambda_{p}(\mathcal{L}_{\\{(g(T^{*}),h(T^{*})),\,d\\}}+a(x))+\beta(m,\delta\psi,x)\epsilon)\\\\[11.38109pt]
&<&0.\end{array}$

Note that the boundaries $g(T^{*})$ and $h(T^{*})$ are fixed, so there is no
need to compare the boundary values of $W(t,x)$ and $I(t,x)$ by Lemma 3.1 in
[15]. Applying the comparison principle in [15] gives

$I(t,x)\geq W(t,x)\geq\underline{W}(t,x)=\delta\psi(x)\,\,\ \textrm{in}\,\,\
[T^{**},+\infty)\times(g(T^{*}),h(T^{*})).$

Therefore,
$\liminf\limits_{t\rightarrow+\infty}I(t,x)\geq\liminf\limits_{t\rightarrow+\infty}W(t,x)\geq\delta\psi(0)>0$,
which is a contradiction.  $\Box$

###### Remark 4.6

Suppose that $a(x)$ is a positive constant. If
$h_{\infty}-g_{\infty}=+\infty$, then
$\limsup\limits_{t\rightarrow+\infty}\|I(\cdot,t)\|_{C([g(t),\,h(t)])}>0$. In
the special case, similar to Theorem 3.10 in [5], a spreading-vanishing
dichotomy holds.

Suppose that

$a(0)\geq d,$

we know that $\lambda_{p}(\mathcal{L}_{\\{(L_{1},\,L_{2}),\,d\\}}+a(x))>0$ for
any interval $(L_{1},L_{2})$ by Theorem 3.3 and $(i)$ of Proposition 3.1,
which yields the following conclusion by Theorem 4.2.

###### Theorem 4.7

If $a(0)\geq d$ holds, then spreading always occurs for (1.3).

Next we consider the case $0<a(0)<d$. According to Theorem 3.3 and $(i)$ of
Proposition 3.1, there exists $L^{*}>0$ such that

$\displaystyle\lambda_{p}(\mathcal{L}_{\\{(L_{1},\,L_{2}),\,d\\}}+a(x))\left\\{\begin{array}[]{ll}<0,&(L_{1},L_{2})\subset(-L^{*},L^{*}),\\\\[5.69054pt]
=0,&-L_{1}=L_{2}=L^{*},\\\\[5.69054pt]
>0,&(-L^{*},L^{*})\subset(L_{1},L_{2}).\end{array}\right.$

###### Theorem 4.8

Assume $0<a(0)<d$ holds, then

$(i)$ if $h_{0}\geq L^{*}$, then spreading always occurs for (1.3).

$(ii)$ if $h_{0}<L^{*}$, then there exists a positive constant $k^{*}$ such
that $h_{\infty}-g_{\infty}=\infty$ when $k>k^{*}$.

Proof. $(i)$ holds from Theorem 4.2 since

$\lambda_{p}(\mathcal{L}_{\\{(g_{\infty},\,h_{\infty}),\,d\\}}+a(x))>\lambda_{p}(\mathcal{L}_{\\{(-L^{*},\,L^{*}),\,d\\}}+a(x))=0.$

In what follows, we prove $(ii)$. Notice that

$-\mu_{2}+\beta(m(x),I,x)S-\gamma(b(x),I,x)>-\mu_{2}-\gamma(b(x),I,x)>-C$

for some $C>0$. Clearly $I(t,x)$ satisfies

$\left\\{\begin{array}[]{ll}I_{t}\geq d\int_{g(t)}^{h(t)}J(x-y)I(t,y)dy-
dI(t,x)-CI(t,x),&t>0,\,x\in(g(t),h(t)),\\\\[5.69054pt] I(t,x)=0,&t\geq 0,\
x\in\mathbb{R}\backslash(g(t),h(t)),\\\\[5.69054pt]
h^{\prime}(t)=k\int_{g(t)}^{h(t)}\int_{h(t)}^{+\infty}J(x-y)I(t,x)dydx,&t>0,\\\\[5.69054pt]
g^{\prime}(t)=-k\int_{g(t)}^{h(t)}\int_{-\infty}^{g(t)}J(x-y)I(t,x)dydx,&t>0,\\\\[5.69054pt]
g(0)=-h_{0},\ h(0)=h_{0},&x\in\mathbb{R},\\\\[5.69054pt]
I(0,x)=I_{0}(x),&x\in(-h_{0},h_{0}),\end{array}\right.$

thereby, for any given constant $M$, there exists $k^{*}>0$ such that
$h_{\infty}-g_{\infty}>M$ provided that $k>k^{*}$ by Lemma 3.9 in [16]. Then,
$h_{\infty}-g_{\infty}=+\infty$ by the arbitrariness of $M$.

$\Box$

Noting that the comparison principle for problem (1.3) is not valid, we cannot
obtain the monotonicity of the solution for (1.3) with $k$ and thus cannot
take $k$ as a sharp criterion for the spreading-vanishing dichotomy as in [5].
However, recalling Remark 4.4 and $(ii)$ in Theorem 4.8, we have the following
result:

###### Theorem 4.9

Suppose that $0<a(0)<d$ and $h_{0}<L^{*}$. For problem (1.3), there exists
$0<k_{*}\leq k^{*}$ such that vanishing occurs if $0<k<k_{*}$ and spreading
happens provided that $k>k^{*}$.

Finally, we will discuss the impact of the diffusion coefficient on the
vanishing and spreading of infectious disease.

Assume that $\int_{-\infty}^{-2h_{0}}J(z)dz>0$ (or
$\int_{2h_{0}}^{+\infty}J(z)dz>0$) holds. Using Theorem 3.4 with
$L_{1}=-h_{0}$ and $L_{2}=h_{0}$, if
$\max\limits_{x\in[-h_{0},\,h_{0}]}a(x)<0$, for any $d>0$, we have
$\lambda_{p}(\mathcal{L}_{\\{(-h_{0},\,h_{0}),\,d\\}}+a(x))<0$, while if
$\max\limits_{x\in[-h_{0},\,h_{0}]}a(x)>0$, there exists a $d^{*}>0$ such that

$\displaystyle\lambda_{p}(\mathcal{L}_{\\{(-h_{0},\,h_{0}),\,d\\}}+a(x))\left\\{\begin{array}[]{ll}<0,&d>d^{*},\\\\[5.69054pt]
=0,&d=d^{*},\\\\[5.69054pt] >0,&d<d^{*}.\end{array}\right.$

Inspired by the above analysis, combined with Theorem 4.2, we can obtain the
following result.

###### Theorem 4.10

Suppose $\int_{-\infty}^{-2h_{0}}J(z)dz>0$ (or
$\int_{2h_{0}}^{+\infty}J(z)dz>0$) holds. Then,

$(i)$ if $\max\limits_{x\in[-h_{0},\,h_{0}]}a(x)>0$, there exists a $d^{*}>0$
such that for $d<d^{*}$, then spreading occurs; conversely, if $d>d^{*}$, then
vanishing occurs provided that
$\|S_{0}(x)\|_{L^{\infty}(\mathbb{R})}+\|I_{0}(x)\|_{C([-h_{0},\,h_{0}])}$ is
small enough;

$(ii)$ if $\max\limits_{x\in[-h_{0},\,h_{0}]}a(x)\leq 0$, for any $d>0$, then
vanishing always occurs as long as
$\|S_{0}(x)\|_{L^{\infty}(\mathbb{R})}+\|I_{0}(x)\|_{C([-h_{0},\,h_{0}])}$ is
adequately small.

##  5 Discussion

In this paper, we study a free boundary problem (1.3) with media coverage and
hospital bed numbers, which describes a nonlocal diffusive SIS epidemic model.
The free boundary describes the moving front of the infected individuals, and
the nonlocal diffusion operator characterizes the long-distance spatial
movement of individuals.

For the SIS model with nonlocal diffusion and free boundaries (1.3), the
existence and uniqueness of the global solution are given by using two fixed
point theorems (see Theorem 2.2). Then, we define the principal eigenvalue of
the integral operator, and analyze the impacts of media coverage and hospital
bed number (Theorem 3.2), interval length (Theorem 3.3), and diffusion
coefficient (Theorem 3.4) on the principal eigenvalue. In addition, sufficient
conditions for disease spreading and vanishing (see Theorems 4.2, 4.3 and 4.5)
are given. Finally, we discuss the impact of the principal eigenvalue on the
spreading or vanishing of infectious diseases. If $a(0)\geq d$, then
$\lambda_{p}(\mathcal{L}_{\\{(L_{1},\,L_{2}),\,d\\}}+a(x))>0$ for any
$L_{1}<L_{2}$, and the disease is always spreading (see Theorem 4.7). If
$0<a(0)<d$, there exists an $L^{*}>0$, then spreading always appears for
$h_{0}\geq L^{*}$ (see Theorem 4.8); and when $h_{0}<L^{*}$, the impact of
expanding capability $k$ on the spreading or vanishing of disease is
discussed. That is, there exists $0<k_{*}\leq k^{*}$ such that vanishing
occurs if $0<k<k_{*}$, and spreading happens provided that $k>k^{*}$ (see
Theorem 4.9). If $\max_{x\in[-h_{0},\,h_{0}]}a(x)>0$, there exists a $d^{*}>0$
such that for $d<d^{*}$, then the disease spreads; if $d>d^{*}$ and
$\|S_{0}(x)\|_{L^{\infty}(\mathbb{R})}+\|I_{0}(x)\|_{C([-h_{0},\,h_{0}])}$ is
small enough, then the disease vanishes; if
$\max_{x\in[-h_{0},\,h_{0}]}a(x)\leq 0$, then $d^{*}=0$, that is, then
vanishing always appears provided that
$\|S_{0}(x)\|_{L^{\infty}(\mathbb{R})}+\|I_{0}(x)\|_{C([-h_{0},\,h_{0}])}$ is
adequately small (see Theorem 4.10).

Finally, we may conclude that the differences between nonlocal diffusion in
(1.3) and local diffusion in (1.2) in our mathematical analysis are as
follows: first, the existence and uniqueness of the global solution for (1.2)
are obtained by straightening the boundary and the first-order fixed point
theorem. However, owing to lack of compactness, the existence and uniqueness
of global solutions for (1.3) are given by using two fixed point theorems.
Second, for (1.2), the corresponding principal eigenvalue always exists.
However, for the nonlocal diffusion problem, the principal eigenvalue may not
exist. In this paper, for the nonlocal diffusion model (1.3), we first define
the generalized principal eigenvalue
$\lambda_{p}(\mathcal{L}_{\\{(L_{1},\,L_{2}),\,d\\}}+a(x))$ of the integral
operator, and then show that the generalized principal eigenvalue is the
principal eigenvalue under condition $(\mathbf{H})$. Third, unlike local
diffusion, whose principal eigenvalue is clear, the nonlocal operator leads to
more possibilities because of the choice of the kernel function.

It is worth mentioning that model (1.3) incorporates media coverage and
hospital bed numbers. Based on the monotonicity of the generalized principal
eigenvalue on media coverage and hospital bed numbers, we study the influence
of the principal eigenvalue on infectious diseases, which implies that large
media coverage and hospital bed numbers are beneficial to the prevention and
control of disease.

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